for Class XI & XII, Engineering Entrance and other

for Class XI amp XII Engineering Entranceand other Competitive Exams

Mathematics at a Glance

Sanjay MishraB Tech (IIT-Varanasi)

ISBN 978-93-325-2206-0

Copyright copy 2015 Pearson India Education Services Pvt Ltd Published by Pearson India Education Services Pvt Ltd CIN U72200TN2005PTC057128 formerly known as TutorVista Global Pvt Ltd licensee of Pearson Education in South Asia No part of this eBook may be used or reproduced in any manner whatsoever without the publisherrsquos prior written consent This eBook may or may not include all assets that were part of the print version The publisher reserves the right to remove any material in this eBook at any time

Head Office A-8 (A) 7th Floor Knowledge Boulevard Sector 62 Noida 201 309 URegistered Office Module G4 Ground Floor Elnet Software City TS-140 Block 2 Salai Taramani Chennai 600 113 Tamil Nadu India Fax 080-30461003 Phone 080-30461060 wwwpearsoncoin Email companysecretaryindiapearsoncom

eISBN 978-93-325-3736-1

Contents

Preface ivAcknowledgements v

1 Foundation of Mathematics 11-128 2 Exponential Logarithm 229-236 3 Sequence and Progression 337-347 4 Inequality 448-454 5 Theory of Equation 555-563 6 Permutation and Combination 664-678 7 Binomial Theorem 779-783 8 Infinite Series 884-886 9 Trigonometric Ratios and Identities 987-997 10 Trigonometric Equation 1098-10109 11 Properties of Triangle 11110-11120 12 Inverse Trigonometric Function 12121-12131 13 Properties of Triangle 13132-13139 14 Straight Line and Pair of Straight Line 14140-14151 15 Circle and Family of Circle 15152-15161 16 Parabola 16162-16172 17 Ellipse 17173-17179 18 Hyperbola 18180-18188 19 Complex Number 19189-19211 20 Sets and Relation 20212-20225 21 Functions 21226-21254 22 Limit Continuity and Differentiability 22255-22272 23 Method of Differentiation 23273-23277 24 Application of Derivatives 24278-24304 25 Indefinite Integration 25305-25321 26 Definite Integration and Area Under the Curve 26322-26336 27 Differential Equation 27337-27350 28 Vectors 28351-28365 29 Three Dimensional Geometry 29366-29381 30 Probability 30382-30391 31 Matrices and Determinants 31392-31411 32 Statistics 32412-32419

Any presentation or work on Mathematics must be conceived as an art rather than a text This is where this work holds it differently During my school days and throughout my long teaching career I realized that most of the JEE aspirants feel the need of a book that may provide them with rapid revision of all the concepts they learned and their important applications throughout their two years long time of preparation I prefer to call it Mathematics at a Glance The present book is written with sole objective of that The entire syllabus of Mathematics for AIEEE JEE Mains and JEE Advanced has been presented in an unprecedented format The reader ought to have the following pre requisites before going through it

(i) HeShe must have ample knowledge of high school Mathematics (ii) Must have conceptualtheoretical knowledge behind the various mathematical thoughts presented (iii) Must be confident enough that heshe is not the father of Mathematics and if not comfortable with

any concept or text we shall be thankful to have your valuable advice

As the name of this work suggests that it has been designed to help during revision It must be kept in mind that the motive of the text is to provide a recapitulation of the entire mathematics that you have studied in your mainstream syllabus While going through the book if you want detailed analysis of any thought or idea you must go for

ldquoFundamentals of Mathematics---By Sanjay MishrardquoAll the suggestions for improvement are welcome and shall be greatfully acknowledged

mdashSanjay Mishra

Preface

I am really grateful to ldquoPearson Educationrdquo for showing their faith in me and for providing me an opportunity to transform my yearning my years-long teaching experience and knowledge into the present rapid revision book ldquoMathematics at a Glancerdquo I would like to thank all teachers and my friends for their valuable criticism support and advice that was really helpful to carve out this work I wish to thank my parents and all my family members for their patience and support in bringing out this book and contributing their valuable share of time for this cause I extend my special thanks to my team including my assistant teachers Rakesh Gupta Parinika Mishra managers and computer operators for their hard work and dedication in completing this task

mdashSanjay Mishra

Acknowledgements

Chapter 1Foundation oF MatheMatiCs

MatheMatical Reasoning

11 INTRODUCTION

Mathematics is a pure application of brains To crack mathematical problems an analytical approach is required

12 PRE-REQUISITES

Flush out your thoughts of maintaining algorithms for mathematical problemsTry to connect the text and work in this chapter from high-school mathematics and make conclusive

analysis of applying basic principles of mathematics

121 Greek Words (Symbols)

Symbol Meaning Symbol Meaning Symbol Meaning

α Alpha β Beta g Gammaδ D Delta isin ε Epsilon ξ Zeta

η Eta θ Theta i Iotaκ Kappa λ Lambda micro Muv Nu ξ Xi o Omicronπ Pi ρ Rho σ sum Sigmaτ Tau υ Upsilon f Phiχ Chi ψ Psi ω Omega

13 UNDERSTANDING THE LANGUAGE OF MATHEMATICS

Well Obviously mathematics is no language by itself but as remarked by Albert Einstein ldquoMathematics is the language in which god has written the universerdquo

12 Mathematics at a Glance

131 Mathematical Symbols

Therefore int Single Integration D Triangle

∵ Because Since int int Double Integration rArr Implies

Such that Σ Sigma N The set of natural numbers So as a Proportionate to hArr Implies and is implied by Ratio f Function Z or I The set of integers

Proportion infin Infinity Q The set of rational numbers= Equal to _ Line bracket ℝ The set of real numbersne Not equal to () Small bracket |x| Absolute value of xgt Greater than Mid bracket ie ie (that is)lt Less than [] Large bracket eg example gratia (for example)

ge Greater than or equal to

isin Belongs to QED Quod erat demonstrandum

le Less than or equal to

notin Does not belong to nsub Is not a subset of

∢ Not less than sub Is a subset of cup Universal setnth root cup Union of sets ~ Similar toCube root cap Intersection of sets iff If and only if

ang Angle A times B Cartesian product of A and B

|| Parallel

^ Perpendicular A ndash B Difference of two sets A and B

f Null Set (phi)

Congruent to forall For all cap Arc$ There exists

14 STATEMENTS AND MATHEMATICAL STATEMENTS

141 Statement

It is a sentence which is complete in itself and explains its meanings completely eg Delhi is the capital of India

142 Mathematical StatementsA given statement is mathematical if either it is true or it is false but not both

143 Scientific StatementA given sentence will qualify as a scientific statement even if it may be true conditionally eg mass can be neither created nor destroyed

Foundation of Mathematics 13

15 CLASSIFICATION OF MATHEMATICAL STATEMENTS

1 axiom Mathematical statements which are accepted as truth without any formal proof given for it eg Equals added to equals are equals

2 Definition Mathematical statement which is used to explain the meaning of certain words used in the subject

Eg ldquoThe integers other than plusmn1 and 0 which are divisible by either one or by themselves are called prime integersrdquo

3 Theorems A mathematical statement which is accepted as lsquotruthrsquo only when a formal proof is given for it like summation of interior angles of a triangle is 180 degree is a theorem

151 ConjecturesIn mathematics a conjecture is an unproven proposition that appears correct For example every even integer greater than two can be expressed as a sum of two primes

152 Mathematical Reasoning

Reasoning is a process of logical steps that enables us to arrive at a conclusion In mathematics there are two types of reasoning These are as follows 1 inductive Reasoning Like that in mathematical induction 2 Deductive Reasoning Series of steps to deduct one mathematical statement from the other and

their proof which will be discussed in the text

16 WORKING ON MATHEMATICAL STATEMENTS

161 Negation of a Statement

The denial of a statement is called its negation To negate a statement we can use phrases like ldquoIt is falserdquo ldquois notrdquo Rita is at home rArr Rita is not at home

162 Compounding of Statements

Compounding of statements is defined as combining two or more component statements using the connecting words like lsquoandrsquo and lsquoorrsquo etc The new statement formed is called a compound statement

Compounding with OR

p x is odd prime numberq x is perfect square of integer

x is a odd prime or a perfect square integer

Compounding with AND

p 2 is a prime numberq 2 is an even number

2 is a prime and even natural number

NoteOR be inclusive or exclusive depending both conditions are simultaneously possible or not respectively

17 IMPLICATION OF A STATEMENT

If two statement p and q are connected by the group of words lsquoifhellip thenhelliprsquo the resulting compound statement

lsquoif p then qrsquo is called lsquoconditional implicationsrsquo of p and q is written in symbolic form as lsquop rarr qrsquo (read as lsquop implies qrsquo)

eg p the pressure increases q the volume increasesThen implication of the statements p and q is given by p rarr q if the pressure increases then the

volume decreases

171 Converse of a Statement

it is given by p rArr q means q rArr pIf a integer n is even then n2 is divisible by 4 Converse is ldquoIf n2 is divisible by 4 then n must be evenrdquo

172 Contra Positive of a Statement p rArr q is ~q rArr ~p

If a triangle has two equal sides then it is isosceles triangle Its contrapositive is lsquoif a triangle is not isosceles then it has no two sides equalrsquo

18 TRUTH VALUE

The truth (T) or falsity (F) of any statement is called its truth value Eg every mathematical statement is either true or false Truth value of a true statement is (T) and that of a false statement is (F)

Given below in the table are Venn Diagrams and truth tables of various mathematical and logical operations

Operation Venn Diagram Truth Table And p q p and qp ^ q

T T TT F FF T FF F F

Or p q p or qp or qT T TT F TF T TF F F

Operation Venn Diagram Truth Table Negation p ~p

T FF T

Implies and is Implied by

p q p rarr q q rarr p (prarrq) ^ (qrarrp)

T T T T TT F F T FF T T F FF F T T T

Implication p q p rarr q

T T TT F FF T TF F T

19 QUANTIFIERS

These are phrases like ldquothere exists $rdquo ldquofor all forallrdquo less than greater than etc For example there exist a polygon having its all sides equal is known as a regular polygon

191 Proofs in MathematicsWe can prove a mathematical statement in various ways which are categorized as straightforward Mathod of exhaustion Mathematical induction Using counter example Contradiction and Contrapositive statements

192 What is a Mathematical AssumptionA mathematical statement which is assumed to be true until a contradiction is achieved An assumed statement may prove to be false at a later stage of mathematical analysis

nuMbeR systeMWell Life without numbers is unpredictable Numbers have been used since ages to facilitate our transac-tions regarding trade exchange or other mathematical purposes Number system has successfully replaced the Bartar system of exchange In this text we will discuss the number system followed by mathematical analysis of real world problems Our present number system is known as Indo-Arabic number system

110 SET OF NATURAL NUMBERS

ℕ = x x is counting number) Counting numbers are called lsquonatural numbersrsquo and their set is denoted as = 1 2 3 4 5

If 0 is not included in the set of natural numbers then we obtain whole numbers (W) W = 0 1 2 3

1101 Algebraic Properties of Natural Numbers

They are associative and commutative ie for all a b c in the set of natural numbersassociative law a + (b + c) = (a + b) + c a(bc) = (ab)ccommutative law a + b = b + a ab = ba

The cancellation law holds for natural numbers If a b c are natural numbers a + c = b + crArr a = b ac = bc rArr a = b (c is not equal to zero)

Distribution of multiplication over addition a(b + c) = ac + bc Order properties (i) law of trichomy Given any two natural numbers a and b exactly one of the following

holds a gt b or a lt b or a = b (ii) transitivity For each triplet of natural numbers a b c a gt b and b gt c implies that a gt c (iii) Monotone Property for addition and Multiplication For each triplet of natural

numbers a gt b rArr a + c gt b + c and ac gt bc existence of additive and multiplicative identity Zero is an additive identity element and 1 is

a multiplicative identity element existence of additive and multiplicative inverse For every integer x there always exists its

negative ndashx which when added to x makes additive identity Multiplicative inverse of x is an element which when multiplied to x makes multiplicative identity 1

111 SET OF INTEGERS

When negatives of natural numbers are included in a set of whole numbers then a set of integers is formed ℤ = ndash4 ndash3 ndash2 ndash1 0 1 2 3 4

112 GEOMETRICAL REPRESENTATION OF INTEGERS

Greek Mathematicians invented Geometrical method of representing numbers on a line known as lsquonumber linersquo In this method a point is marked as zero (0) and with respect to zero the numbers are located in order of their magnitude The distance of number (x) from zero represents its magnitude (|x|)

1121 Properties of Integers (a) It is closed commutative associative and distributive for addition subtraction and multiplication (b) Zero is the identity element for addition and 1 is the identity for multiplication

(c) Additive inverse of x is equal to ndashx Q x+ (ndashx) = 0 (d) Multiplicative inverse of x is 1x provided x ne 0 as x 1x = 1 (e) Cancellation law holds for addition as well as multiplication (f) Property of order forall x y isinℤ either x gt y or x = y or x lt y Also known as law of trichomy

113 DIVISION ALGORITHM

Given are two integers a and b such that a gt b and b gt 0 then there exist two integers q and r such that a = bq + r where a dividend b divisor q quotient r remainder

Properties The remainder r is a non-negative integer which is less than the divisor b 0 le r lt b where r = 0 1 2 3 4 b ndash 1 If the remainder r = 0 then a = bq Then a is called completely divisible by b (ie multiple of b) and b and q are called factors of a

1131 Even and Odd Integers (a) Set of even integer = x x = 2k where k isin ℤ (b) Set of odd integers = x x = 2k + 1 where k isin ℤ

1132 Prime Integer

An integer x (other than 0 ndash1 and 1) is called prime iff it has only positive divisors as 1 and itself eg 2 3 5 7 etc

11321 Properties

An integer other than 0 ndash1 and 1 which are non-primes are called composite numbers A composite integer has atleast three factors

1 ndash1 0 are neither prime nor composite Twin Primes A pair of primes is said to be twin primes if they differ by 2 ie 3 5 and 11 13 etc Co-Primes A pair of integers is said to be co-primes if they have no common positive divisor except

1 eg 8 5 and 12 35 If p is prime and greater than or equal to 5 then p is either 6k + 1 or 6k ndash 1 but converse is not

necessarily true If p is prime and greater than 5 then p2 ndash 1 is always divisible by 24

114 FACTORIAL NOTATION

Factorial of r is denoted as r and is defined as product of first r natural numbers ie r = 1 2 3 4hellip (r ndash 1)reg 1 = 1 2 = 2 3 = 6 4 = 24 5 = 120 6 = 720 7 = 5040

Product of any r consecutive integer is always divisible by r

1141 Related TheoremsTheorem 1 xn ndash yn is divisible by (x ndash y) forall x isin ℕ since putting x = y makes expression xn ndash yn = yn ndash yn = 0 Therefore x ndash y must be factor in the above expression

Theorem 2 xn ndash yn is divisible by (x + y) forall odd natural numbers n Since putting x = ndashy makes expression xn + yn = yn + (ndashy)n = yn + (ndash1)n yn = yn ndash yn = 0 Thus x + y must be factor in the above expression (xn + yn) = (x + y)(xnndash1 ndash xnndash2y + xnndash3y2 ndashhellip+ (ndash1)nndash1 ynndash1)

Theorem 3 Given n isin ℕ and p and p is prime such that ldquon is co-prime to prdquo then np ndashn is always divisible by pFermatrsquos Theorem n = 2 and p = 5 rArr 5|25 ndash 2 rArr 5|24 ndash 1

corollary 1 np ndash n is also divisible by n and (n ndash 1)corollary 2 np ndash n is divisible by n(n ndash 1) Since n and (n ndash 1) are always co-primecorollary 3 npndash1 ndash 1 is always divisible by p

Theorem 4 (fundamental theorem of arithmetic) A natural number N can be expressed as product of non-negative exponent of primes N = pa qb rc sd hellip where p q r s are primes and a b c d are whole numbers eg 1800 = 23325270

Theorem 5 (Wilsonrsquos theorem) if p is a prime number then 1 + (p ndash 1) is divisible by p ie 16 + 1 is divisible b

1142 Divisors and Their PropertyA natural number x = pa qb rg is called divisor of N = pa qb rc rArr N is completely divisible by x

hArr all the prime factors of x are present in NhArr 0 le α le a 0 le b le b 0 le g le c where a b g are whole numbers Set of all divisors of N is given as x x = pα qb rg where 0 le α le a 0 le b le b 0 le g le c

1143 Number of Divisorsn (αbg) 0 le α le a 0 le b le b 0 le g le c = number of ways the integers a b g can take values applying the above restrictions = (a + 1)(b + 1)(c + 1)

sum of Divisor of n = pa qb rc (1 + p + p2 ++ pa) (1 + q + q2 ++ qb) (1 + r +r2 ++ rc)

improper Divisors of N = pa qb rc when a = b = g = 0 rArr x = 1 this is divisor of every integer and a = a b = b and g = c then x becomes number N itself These two are called lsquoimproper divisorrsquo The number of proper divisors of N = (a + 1)(b + 1)(c + 1) ndash 2

If p = 2 then number of even divisors = a(b + 1)(c + 1) Number of odd divisors = (b + 1)(c + 1)Number and sum of divisors of N divisible by a natural number 1 1 1a b cy p q r=

Let x = pa qb rg be such divisors ∵ 1 1 1a b cy | x p q r | p q rα β γrArrrArr a1 le α le a and b2 le b le b and c1 le g le c rArr Number of such divisors = (a ndash a1 + 1) (b ndash b1 + 1) (c ndash c1 + 1)

Sum of such divisors Sy = 1 1 1 1 1 1a a 1 b b 1 c c 1a b cyS (p p p )(q q q )(r r r )+ + += + + + + + + + + +

= 1 1 1a a b b c c2 2y(1 p p p )(1 q q q )(1 r r )minus minus minus+ + + + + + + + + + +

= 1 1 1a a 1 b b 1 c c 1p 1 q 1 r 1y

p 1 q 1 r 1

minus + minus + minus + minus minus minus

minus minus minus

1 The number of ways of resolving n into two factors is + + +1

( a 1)( b 1)( c 1)2

when n is not a perfect

square and + + + +1

( a 1)( b 1)( c 1) 12

when n is a perfect square

2 Every number n has two improper divisors 1 and n itself and the remaining divisors are called proper divisors Eg number of proper divisors of 108 is 10

1144 Least Common Multiple (LCM)

LCM of set of numbers is the smallest number (integerrational) which is completely divisible by each of them ie x is said to be LCM of y and z iff y divides x z also divides x and x is least positive of all such numbers Eg LCM of 6 4 9 is 36

Let x and y be two given integer x = paqβrgsd and 1 1 1 1y p q r sα β γ δ= where p q r are primes

If z is LCM of x and y then 1 1 1 1max( ) max( ) max( ) max( )z p q r sα α β β γ γ δ δ=

LCM LCM (a and c)a cand

b d HCF (b and d)=

1145 Greatest Common Divisor (GCD)Highest Common Factor (HCF)

HCF of a given set of numbers is the largest number which divides each of the given numbers HCF of y and z is also denoted as (y z) Therefore x is said to be GCD of y and z if x divides both y and z and x is largest of such numbers So clearly every common divisor of y and z also divides x and x ne 0

Eg HCF of 12 and 64 is 4 GCD of 6 and 35 is 1 (co-prime)

HFC HCF (a and c)a cand

b d LCM (b and d)=

Method to find hcF For two given integers x and y

Method 1 Consider their prime factors 1 1 1 2 2 2x p q r and y p q r α β γ α β γ= =If z is HCF of x and y rArr zx and zy z contains the least power for each corresponding prime factor rArr 1 2 1 2 1 2min min min z (x y) p q r α α β β γ γ= =

1146 Decimal Representation of Number

given a natural number x abcde= where e d c b a are respectively digits occupying unit tenrsquos hundredth thousandth ten thousandth places So the numerical value of x is defined as lsquosum of products of digits multiplied by their corresponding place valuesrsquo

th th th

4 3 2 1 0

tens place unit placetenthousand thousand hundredvalue valueplace value place value place value

x = a 10 + b 10 + c 10 + d 10 + e 10

minus minus minus

times times times times times

Theorem If an integer x is divided by 10 the reminder is a digit at the unit place of x

Proof = = + + + + = +4 3 2x abcde a(10 ) b(10 ) c(10 ) d(10) e 10m e rArr Remainder is e

Theorem The remainder if an integer x is divided by 5 is e 0 e 4e 5 5 e 9

le le minus le le

where e is are unit place

digit of the number 4 3 2x abcde a(10 ) b(10 ) c(10 ) d(10) e= = + + + +

= a(104) + b(103) + c(102) + d(10) + e = 5m + e 0 le e le 9

5m e 0 e 4 5m e 0 e 45m 5 e 5 5 e 9 5m (e 5) 5 e 9

+ le le + le le = = + + minus le le + minus le le

1147 Periodic Properties of Integers

Theorem 1 Unit digit of nth power of an integer having zero at its unit place is zero

rArr n1 1 1(abc0) (a b c 0)=

Theorem 2 Unit digit of nth power of an integer having one at its unit place is one

rArr n1 1 1(abc1) (a b c 1)=

Theorem 3 Unit digit of nth power of an integer having two at its unit place is described as follows

rArr cn1 1 1(abc2) (a b c 2)= if n = 4k + 1 ie n

1 1 1(abc2) (a b c 4)= if n = 4k + 2

ie n1 1 1(abc2) (a b c 8)= if n = 4k + 3 ie n

1 1 1(abc2) (a b c 6)= if n = 4k

Theorem 4 Unit digit of nth power of an integer having three at its unit place is described as follows

rArr n1 1 1(abc3) (a b c 3)= if n = 4k + 1 ie n

1 1 1(abc3) (a b c 9)= if n = 4k + 2

rArr ie n1 1 1(abc3) (a b c 7)= if n = 4k + 3 ie n

1 1 1(abc3) (a b c 1)= if n = 4k

Theorem 5 Unit digit of nth power of an integer having four at its unit place is described as follows

1 1 1(abc4) (a b c 6)= if n = 2k

Theorem 6 Unit digit of nth power of an integer having five at its unit place has five at unit place

rArr n1 1 1(abc5) (a b c 5)= if n isin ℕ

Theorem 7 Unit digit of nth power of an integer having six at its unit place has six at unit place

Theorem 8 Unit digit of nth power of an integer having seven at its unit place is described as follows

1 1 1(abc7) (a b c 9)= if n = 4k + 2

1 1 1(abc7) (a b c 1)= if n = 4k

Theorem 9 Unit digit of nth power of an integer having eight at its unit place is described as follows

1 1 1(abc8) (a b c 4)= if n = 4k + 2

ie n1 1 1(abc8) (a b c 2)= if 4k + 3 n

1 1 1ie (abc8) (a b c 6) if n 4k= =

Theorem 10 Unit digit of nth power of an integer having nine at its unit place is described as followsn

1 1 1ie (abc9) (a b c 9) if n 2k 1= = + n1 1 1ie (abc9) (a b c 1) if n 2k= =

115 TESTS OF DIVISIBILITy

1 Divisibility by 2 A number N is divisible by 2 if and only if its last digit is divisible by 2 (ie even) 2 Divisibility by 3 A number N is divisible by 3 if and only if the sums of all digits are divisible by 3 3 Divisibility by 4 A number N is divisible by 4 if its units digit plus twice its tenrsquos digit is divisible by 4 4 Divisibility by 5 A number N is divisible by 5 if and only if its last digit is divisible by 5 (ie if it

ends in 0 or 5) 5 Divisibility by 6 A number N is divisible by 6 if and only if its unitsrsquos digit is even and the sum of

its digits are divisible by 3 6 Divisibility by 7 A number N is divisible by 7 if and only if 3 times unitrsquos digit + 2 times tenrsquos digit ndash 1

times hundredrsquos digit ndash 3 times thousandrsquos digit -2 times ten thousandrsquos digit + 1 times hundred thousandrsquos digit is divisible by 7 ie 3(a0) + 2(a1) ndash 1(a2) ndash 3(a3) ndash 2(a4) + 1(a5) + 3(a6) + is divisible by 7

ie If there are more digits present in the sequence of multipliers 3 2 ndash 1 ndash 3 ndash 2 1 is repeated as often necessary

7 Divisibility by 8 A number N is divisible by 8 if and only if its unitrsquos digit + 2times tenrsquos digit + 4 times hundredrsquos digit is divisible

8 Divisibility by 9 A number N is divisible by 9 if and only if the sum of its digits is divisible by 9 9 Divisibility by 10 A number N is divisible by 10 if and only if the last digit is 0 10 Divisibility by 11 N is divisible by 11 if and only if the difference between the sum of the digits in

the odd places (starting from the right) and the sum of the digits in the even places (starting from the right) is a multiple of 11 eg 1221 123321 2783 etc

12 Divisibility by 13 A number N is divisible by 13 if and only if 10 times unitsrsquos digit ndash 4 times tenrsquos digit ndash 1 times hundredrsquos digit + 3 times thousandrsquos digit + 4 times ten thousandrsquos digit + 1 times hundred thousandrsquos digit is divisible by 13 (If there are more digits present the sequence of multipliers 10 ndash4 ndash1 3 4 1 is repeated as often as necessary)

116 RATIONAL (ℚ) AND IRRATIONAL NUMBERS (ℚprime)

A number x in the form pq where p and q are integers and q is not equal to 0 is called rational and

otherwise it is called irrational numbers ( or ) eg 1 3 5 0 25 1016 107 are rational while radic2 radic3 radic5hellip radicx x is not a perfect square of rational are irrationals Pie (p) is ratio of circumference of any circle to the diameter of the same circle It is an irrational number approximately equal to rational numbers 227 or 314

euler number (e) 1 1 1e 1 27 e 81 2 3

= + + + + infinrArr lt lt

1161 Properties of Rational and Irrational Numbers

If a number x in decimal form is written as x cdepqr= then

2 1 0 1 2 3

tens place unit place first decimal Second decimal third decimalhundredvalue value place value place value place valueplace value

x c 10 d 10 e 10 p 10 q 10 r 10minus minus minus

minus minus minusminus

= times + times + times + times + times + times

All terminating decimals are rational eg 4

abcdeabcde10

= = = 1 2 n1 2 3 n n

ax x xx ax x x x10

If a rational pq (in lowest term) is terminating decimal then q = 2m5n ie q must not contain any prime factor other than 2 or 5

Non-terminating but repeating decimals are also rationals eg y = xab ab ab helliphellip y xabrArr = helliphellip(i)

If number of repeating digits be n then multiply both side by 10n ie 210 y xabab= helliphellip(ii)

Subtracting (i) from (ii) we get xab xy99minus

= (which is a rational number)

Non-terminating and non-repeating decimals are irrationals 271354921275718 hellip (no periodic re-occurrence up to micro)

Set of rational numbers is countable while set of irrational numbers is uncountable

1162 nth Root of a NumberA real number y is called nth root of real number x where n is a natural number (n ge 2) Iff yn = x When n = 2 then it is called as square root and for n = 3 known as cube root All the numbers other than zero have more than one nth roots eg both 2 and ndash2 are square root of 4

1163 Principal nth RootThe principal nth root of a real number x (having atleast one n-th root) is that nth root which has its sign same as that of x It is denoted by a radical symbol n x

The positive integer n is known as the index of the radical symbol Usually we omit the index from the radical sign if index n = 2 and write as x

eg 27 336 68 2

= = and 5 4243 ( 3) 16 2minus = minus = whereas 4 16minus is a non-real number since fourth

power of no real number can be ndash16 which is negative

1164 Properties of nth Root

(i) Every positive real number x has exactly two real nth roots when n is a positive even natural number

(n = 2m) denoted by 2m 2mx and xminus are two real fourth roots of 256 eg 4 4256 4 256 4= minus = minus

(ii) Every real number x has only one real nth roots when n is a positive odd natural number

(n= 2m + 1) denoted by 2m 1 x+ eg 3 3125 5 125 5= = minus

(iii) nth root of a negative real number x is non-real when n is an even integer Eg 424 16minus minus has

no real values 1minus is a non-real number symbolized as i (iota)

(iv) Zero is only real number which has only one nth root and n 0 0= (v) Integers such as 1 4 9 16 25 and 49 are called perfect squares because they have integer

square roots (vi) Integers such as 1 8 27 64 are called perfect cubes as they have integer cube roots

square roots If b is the square root of a where a is the non-negative real number then b when squared must become equal to a

rArr b2 = a rArr b2 ndash a = 0 rArr (b ndash radica) (b + radica) = 0rArr b ndash radica = 0 or b + radica = 0 rArr b = radica (positive) or b = ndashradica (negative)

11641 Properties of Square Roots

(i) Zero has only one square root ie zero (ii) Every positive real number (except zero) has two square roots One of them is positive (called as

principle square root denoted as radica) and the other is negative denoted as (ndashradica) (iii) Magnitude of real number x denoted as |x| and defined as the quantity of x is

x if x 0| x | x 0 if x 0

x if x 0

gt= = =minus lt

1165 Algebraic Structure of and

closure law For addition and subtraction multiplication commutative law For addition and multiplication associative law For both addition and multiplication Distributive law For addition and subtraction operation

(i) Zero is the identity element for addition and 1 is the identity for multiplication

Q x + 0 = x and 1

times =

forall x isin ℚ x ne 0

(ii) Additive inverse of x = p | q is equal to ndashx Q x + (ndashx) = 0

(iii) Multiplicative inverse of = =p 1

provided x ne 0 as 1x 1

cancellation law holds for addition as well as multiplication 1 2 1 3 2 3

1 2 1 3 2 3

x x x x x xx x x x x x+ = + rArr =

= rArr =

provided x1 ne 0 Property of order forall x y isin ℚ either x gt y or x = y or x lt y Also known as law of trichotomy Union of set of rationals and set of irrationals is called set of real numbers ℝ

117 SURDS AND THEIR CONJUGATES

Sum of a rational and an irrational number is always irrational and called as surd denoted by s

rational irrationalpart of s part of s

s a b= + where b is not a perfect square of the rational number

For every surd s there exist element s s a b= minus where s a b= + called as conjugate of s

Rationalization of denominator of an irrational number2

s a b (a b)(a b) a b 2a bs a b a b(a b) (a b)(a b)

+ + + += = = +

minus minusminus minus +

118 REAL NUMBERS SySTEM

Union of set of Rationales and set of Irrationals is called set of Real numbers (ℝ) = cup

Properties

(i) Square of real numbers is always non-negative If x isin ℝ rArr x2 ge 0 (ii) Between any two real numbers there are infinitely many real numbers (iii) Magnitude of real number x is denoted as |x| and defined as the quantity of x

x if x 0| x | x 0 if x 0

x if x 0

gt= = =minus lt

(iv) They are represented on a straight line called as real number line in order of their magnitude such that distance of the number of x from zero is equal to magnitude of x (|x|)

(v) A real number line is infinitely dense and continuous line ie between two any two number (how so ever closed they are) there lies infinitely real number

1181 Concept of IntervalAs the set of all real numbers lying between two unequal real numbers a and b can never be expressed in roster form therefore these are expressed in set builder form using the concept of intervals

open interval Denoted as (a b) x isin (a b) = x a lt x lt b x isin ℝ ie end points are not included

closed interval [a b] x isin [a b] = x a le x le b xisin ℝ the end points are included

semi-open interval x isin (a b] rArr a lt x le b and x isin [a b) rArr a lt x le b

1182 Intersection and Union of Two or More Intervals To find the intersection or union of two or more intervals locate each interval over the same real number line and for intersection take the interval which is common to both and for union locate the interval which includes the numbers of all the interval considered

119 MATHEMATICAL INDUCTION

Mathematical induction is a mathematical tool by which we can prove the correctness of any mathematical statement or proposition It works on the principle that results for higher integers are induced from the results for lower integers

Statement Working RuleFirst principle of mathematical induction

The set of statements P(n) n isin N is true for each natural number n ge m is provided thatP(m) is trueP(k) is true for n = k (where k ge m)rArr P(n) is true for

n = k + 1

Let there be a proposition or a mathematical statement namely P(n) involving a natural num-ber n In order to prove that P(n) is true for all natural numbers n ge m we proceed as followsVerify that P(m) is trueAssume that P(k) is true (where k ge m)Prove that P(k + 1) is trueOnce step ndash (c) is completed after (a) and (b) we are through ie P(n) is true for all natural numbers n ge m

Second principle of mathematical induction

The set of statements P(n) n isin N is true for each natural number n ge m provided thatP(m) and P(m + 1) are true P(n) is true for n le k (where k ge m)rArr P(n) is true for

n = k + 1This is also called extended principle of Mathematical Induction

Verify that P(n) is true for n = m n = m + 1Assume that P(n) is true for n le k (where k ge m)Prove that P(n) is true for n = k + 1Once rule (c) is completed after (a) and (b) we are through That is P(n) is true for all natural numbers n ge m This method is to be used when P(n) can be expressed as a combination of P(n -1) and P(n - 2) In case P(n) turns out to be a combination of P(n -1) P(n - 2) and P(n -3) we can verify for n = m + 2 also in Rule(a)

1191 Ratio and ProportionRatio and proportions are algebraic operations which are operated on one or more variables as

Ratio It is a rational between two quantities that tells us what multiplepart one quantity is of the other Therefore if x and y are two quantities of the same kind then their ratio is x y which may be denoted by xy (This may be an integer or fraction)

1 A ratio may be represented in a number of ways eg x mx nxy my ny= = = where m nare

non-zero numbers 2 To compare two or more ratios always reduce them to a common denominator

3 Ratio of two fractions may be represented as the ratio of two integers eg x z xy xu y u zu yz

= or xu yz

4 Ratios are compounded by taking their product ie x z v xzv y u w yuw

5 Duplicatetriplicate ratio If x y is any ratio then its duplicate ratio is x2 y2 triplicate ratio is x3 y3 etc If xy is any ratio then its sub-duplicate ratio is x12 y12 sub triplicate ratio is x13 y13 etc

ProportionWhen two ratios ab and cd are equal then the four quantities composing them are said to be propor-tional If abcd are proportional then ab = cd and it is written as ab = cd or ab c d 1 lsquoarsquo and lsquodrsquo are known as extremes whereas lsquob and crsquo are called as means 2 Product of extremes = product of means

1192 Some Important Applications of Proportion

If four a b c d are proportional then many other useful proportions can be derived using various laws of fraction which are extremely useful in mathematical calculations and simplifications

invertando If a b = c d then b a = d calternando If a b = c d then a c = b d

componendo If ab = cd then a b c d

b d+ +

∵ a cb d= adding 1 from both sides a c1 1

b d+ = + rArr

a b c db d+ +

Dividendo If a b = c d then a b c db dminus minus

∵ a cb d= subtracting one to both sides a c1 1

b dminus = minus rArr

a b c db dminus minus

componendo and dividendo If a b = c d then applying both componendo and dividendo operations

together we get a b c da b c d+ +

=minus minus

If a c eb d f= = (say = l) then

1nn n n

xa yc zexb yd zf

+ + + +

1193 Linear EqualitiesAn expression of the form y = ax + b where a and b isin ℝ is called a linear polynomial function of x y and set of points (x y) satisfying the above relations if plotted on the xy plane a straight line is obtained An equation of the form ax + by + c = 0 is termed as linear equation in x and y

solving simultaneous linear equations in two unknowns

To solve a pair of linear equation a1x + b1y = c1 (i)a2x + b2y = c2 (ii)

The following three approaches are adopted

1194 Method of ComparisonFrom both equations find the value of any one variable say y in terms of other ie x

1 1 2 2

c a x c a xyb bminus minus

= = rArr 1 2 1 2

1 2 1 2

c c a a xb b b b

minus = minus

rArr 2 1 1 2

1 2 2 1

b c b ca b a b

minusminus

and similarly we get 2 1 1 2

1 2 1 2

a c a cyb a a b

minus=

1195 Method of SubstitutionTo solve equations (i) and (ii) substitute the value of y from equation (i) to (ii) get x and y then can also be

obtained 1 12 2 2

c a xa x b cb

minus+ =

a2b1x + b2c1 ndash b2a1x = b1c2

rArr (a2b1 ndash a1b2)x = b1c2 ndash b2c1 rArr 1 2 2 1

2 1 1 2

b c b cxa b a b

minus=

minus and so we get 1 2 2 1

1 2 2 1

a c a cya b a b

minus=

1196 Method of Eliminationa1x + b1y = c1 (i)a2x + b2y = c2 (ii)

Multiplying equation (i) by a2 and equation (ii) by a1 and subtracting x gets eliminated

a1a2x + b1a2y = a1c1 (iii)a1a2x + a1b2y = a1c2 (iv)

Subtracting equation (iii) and (iv) 2 1 1 2

2 1 1 2

a c a cya b a b

minus=

minus and thus 1 2 2 1

2 1 1 2

b c b cxa b a b

minus=

11961 Method of cross-multiplication

It is a very useful method for solving pair of linear equations in two or three variables Given two equations a1x + b1y + c1 helliphellip (i)a2x + b2y + c2z helliphellip (ii)

Dividing both equations by z and replacing 0x xz= and 0

y yz= we get

a1x0 + b1y0 + c1 helliphellip (iii)

a2x0 + b2y0 + c2z helliphellip (iv) Solving by any of the above mentioned three elementary methods we get

2 1 1 2 2 1 1 20 0

2 1 1 2 2 1 1 2

b c b c b c b cx xx xa b a b z a b a b z

minus minus= = = =

minus minus

that can be symmetrically expressed as 1 2 2 1 1 2 2 1 1 2 2 1

x y zb c b c c a c a a b a b

= =minus minus minus

Thus we can conclude that the set of solution of above pair of equation can always be expressed by the ratio x y z in terms of coefficients of the equations

step (1) Express the coefficients of x y z beginning with y in cyclic order as shown in the figure and take the product of the coefficients indicated by arrows

step (2) The product formed by descending arrows is considered positive and those by ascending arrows is taken negative

step (3) So we get x y z (b1c2 ndash b2c1) (c1a2 ndash c2a1) (a1b2 ndash a2b1)

FunDaMentals oF inequality

120 INTRODUCTION

The concept of inequality finds its origin from the property of order of real numbers An inequation is marked by the use of logical operations such as lt gt le ge ne etc An inequation can have one or more than one variables ax + by + c ge 0

inequation An inequation is a statement involving sign of inequality ie lt gt le ge ≮ ≯ ne

1201 Classification of InequalityInequalities are of four types

If a ndash b gt 0 rArr a gt b (read a greater than b)If a ndash b ge 0 rArr a ge b (read a greater than or equal to b)If a ndash b lt 0 rArr a lt b (read a is less than b)If a ndash b le 0 rArr a le b (read a is less than or equal to b)

linear inequality Inequality having variables in one degree eg 2x + 3y gt 5 x ndash 2y + 3z = 5 etc

solution of inequality The values of unknown variable which satisfies the given inequation are called solutions of inequality eg x = 2 y = 4 is a particular solution of inequality 2x + 3y gt 5

12011 Basic properties of inequality and laws

(i) transition property If a gt b and b gt c rArr a gt c (ii) law of trichotomy If x and y are two real numbers then exactly one of the three statements

hold ie x gt y or x lt y or x = y (iii) If a gt b then a + c gt b + c and a ndash c gt b ndash c forall c isin ℝ (iv) If x lt y lt 0 rArr |x| gt |y| (Larger the number smaller the magnitude) (v) If x gt y gt 0 rArr |x| gt |y| (Larger the number larger the magnitude) (vi) If a gt b then ac gt bc forall c gt 0 (sign of inequality does not change on multiplying by positive

real number) (vii) If a gt b then ac lt bc forall c lt 0 (sign of inequality gets reverse when multiplied both sides by negative

real number)

(viii) If a gt b then a b for c 0c cgt gt and a b for c 0

c clt lt

(ix) If a cb dge then ad ge bc if b and d same sign

(x) If a cb dge then ad le bc if b and d are opposite signs

(xi) law of addition If a1 gt b1 and a2 gt b2hellip and an gt bn rArr (a1+a2+hellip+ an) gt (b1+ b2 +hellip+ bn) (xii) law of Multiplication If a1 gt b1 gt 0 and a2 gt b2 gt 0hellip and an gt bn gt 0 rArr (a1a2a3hellipan) gt (b1b2b3hellipbn) (xiii) laws of reciprocal

(a) If 0 lt a lt b then 1 1a bgt (b) If b lt a lt 0 then 1 1

(c) If x isin [a b] then

gt lt minusinfin cup infin lt gt = isin infin = gt = minusinfin = lt =

1 1 for a b 0 or a b 0b a

1 1 for a 0 b 0 not defined at x 0a b1

x 1 for a 0 b 0 not defined at x 0b

1 for b 0 a 0not defined at x 0a

(xiv) laws of squares or positive even powers

a b if both ab 0agtb a b If |a| = |b|

a b If ab lt 0

rArr = lt

If a and b have opposite sign and a gt b ie a gt 0 and b lt 0 then

a b iff |a | | b |a b a b iff |a | | b |

a b iff |a | | b |

gt rArr = = lt lt

This law can be extended for any even natural power (2n)

If x isin [a b] then

2 22 2

[a b ] for a b 0[b a ] for a b 0

x x[0a ] for a b and ab 0

[0b ] for b a andab 0

ltisin isingt lt

similar is the case for x2n n isinℕ

(xv) law of square root If a and b both are non-negative and 2n 2n

a ba b

gtgt rArr gt

(xvi) laws of cubes or positive odd powers If x isin [a b] then x3 isin [a3 b3] similarly x2n+1 isin [a2n+1 b2n+1] for n isin ℕ

(xvii) law of cube root a gt b rArr a3 gt b3 and a13 gt b13 forall a b isinℝ a lt b rArr a3 lt b3 and a13 lt b13 forall a b isin ℝ this law can be extended for any odd natural power (2n+ 1) and odd root

(xviii) laws of exponential inequality (a) If 0 lt a lt 1 and r isin ℝ+ then 0 lt ar lt 1 lt andashr (b) If a gt 1 and r isin ℝ+ then ar gt 1 gt andashr gt 0(c) For a gt 1 ax gt ay for x gt y and x y isin ℝ(d) For 0 lt a lt 1 ax lt ay for x gt y and x y isin ℝ(e) For a bisin (0 1) or a b isin (1infin) if a gt b then ax lt bx for x lt 0 and ax gt bx for x gt 0(f) For a isin (0 1) and b isin (1 infin) ax gt bx for x lt 0 and ax lt bx for x gt 0

(xix) laws of logarithmic inequality (a) x ge y hArr logax ge logay for a gt 1 (b) x ge y hArr logax le logay for 0 lt a lt 1(c) ax ge y rArr x ge logay for a gt 1 (d) ax ge y rArr x le logay for 0 lt a lt 1

RemarkAbove two results follow from the fact that logarithmic and exponential function to the base a gt 1 are increasing function and when base lies between 0 and 1 then they become decreasing function

(xx) inequalities containing modulus functions(a) |x| lt a hArr ndasha lt x lt a where a gt 0 ie x isin (-a a)(b) |x| le a hArr ndasha le x le a where a gt 0 ie x isin [-a a](c) |x| gt a hArr x lt ndasha or x gt a ie x isin (ndashinfin ndasha] cup (a infin) (d) |x| ge a hArr x le ndasha or x ge a ie x isin (ndashinfinndasha] cup [a infin)(e) a lt |x| lt b hArr x isin (ndashb b) for a le 0(f) a lt |x| lt b hArr x isin (ndashb ndasha] cup [a b) for a gt 0

(xxi) triangle inequality | |x| ndash |y| | le |x plusmn y| le |x| + |y| forall x y isin ℝ Further(a) |x + y| = |x| + |y| for xy ge 0 (b) |x + y| lt |x| + |y| for xy lt 0(c) |x ndash y| = |x| + |y| for xy le 0 (d) |x ndash y| lt |x| + |y| for xy gt 0(e) | |x| ndash |y| | = |x + y| for xy le 0 (f) | |x| ndash |y| | lt |x + y| for xy gt 0(g) | |x| ndash |y| | lt |x ndashy| for xy lt 0 (h) | |x| ndash |y| | = |x ndashy| for xy ge 0

12012 Solutions of linear in equations in two variables

1 by graphical method Let L equiv ax + by + c = 0 be a line then by = ndashax ndash c Since the P point satisfies the equation of the line aa + bb + c = 0 From the given diagram we interpret that g gt b bg gt bb for b gt 0 rArr aa + bg gt aa + bb rArr aa + bg + c gt aa + bb + c rArr aa + bg + c gt 0 Thus all the points lying in the half plane II above the line ax + by + c = 0 satisfies the

inequality ax + by + c gt 0 Similarly in case b lt 0 we can prove that the point satisfying ax + by + c gt 0 lies in the

half plane I Hence we infer that all points satisfying ax + by + c gt 0 lies in one of the half plane II or I

according as b gt 0 or b lt 0 and conversely Thus the straight line ax + by + c = 0 divides the whole x ndashy plane into three regions (a) For b lt 0 (i) R1 = (a b) aa + bb + c = 0 (ii) R2 = (a b) aa + bb + c lt 0 (iii) R3 = (a b) aa + bb + c gt 0 (b) For b gt 0

2 short-cut method step i Consider the equation from the Inequality step ii Draw the straight line representing the Equation step iii Consider a Point P (a b) (not on the line) and find the sign of

linear expression for P (ab) step iV Check whether it satisfies the inequality or not If it satisfies

then the inequality represents the half plane which contains the point and shade the region

step V Otherwise the inequality represents that half plane which does not contain the point within it

For convenience the point (0 0) is preferred step Vi The set R1 is a straight line while the sets R2 and R3 are called open half planes The set

R1 cup R3 represent the points whose co-ordinates satisfy ax + by + c ge 0 (b gt 0) and R1 cup R2 represent the points whose co-ordinates satisfying ax + by + c le 0 (b gt 0) Here R2 is the solution region of inequality ax + by + c lt 0 b gt 0 and R3 is the solution region of inequality ax + by + c gt 0 b gt 0

+ndashndashndashndash ndash ndashndash ndash ndashndash ndash ndash ndash ndash ndashndash ndash

++++++++++++++++

ndash ndash ndashndash ndash ndashndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash

+++++++++++++++++

ndashndashndashndash ndashndashndash ndash ndashndash ndash ndash ndash ndash ndashndash ndash

+++++++++++++++++

1202 Rational Algebraic Inequalities

type 1 P(x) P(x) P(x) P(x)0 0 0 0Q(x) Q(x) Q(x) Q(x)

gt lt ge le P(x) Q(x) are polynomials

step 1 Factor P(x) and Q(x) into linear factors

step 2

(i) P(x) 0 P(x)Q(x) 0 P(x) 0Q(x) 0 or P(x) 0Q(x) 0Q(x)

gt rArr gt rArr gt gt lt lt

(ii) P(x) 0 P(x)Q(x) 0 P(x) 0 Q(x) 0 or P(x) 0Q(x) 0Q(x)

lt rArr lt rArr gt lt lt gt

(iii) P(x)Q(x) 0P(x) 0 P(x) 0 Q(x) 0 or P(x) 0 Q(x) 0

Q(x) 0Q(x)gege rArr rArr ge gt le lt ne

(iv) P(x)Q(x) 0P(x) 0 P(x) 0 Q(x) 0or P(x) 0 Q(x) 0

Q(x) 0Q(x)lele rArr rArr ge lt le gt ne

step 3 For solving the above inequalities formed eg P(x) Q(x) gt 0 use wavy curve method or solution set is given by x P(x) gt 0 Q(x) gt 0 cup x P(x) lt 0 Q(x) lt 0

type ii For solving inequality of the form P(x) R(x)Q(x) S(x)

step 1 P(x) R(x) R(x) P(x)0or 0Q(x) S(x) S(x) Q(x)

minus lt minus gt

rArr P(x)S(x) R(x)Q(x) R(x)Q(x) P(x)S(x)0 or 0Q(x)S(x) S(x)Q(x)

minus minus lt gt

Now solve as in Type 1

type iii For solving inequality of the form P(x) R(x) T(x)Q(x) S(x) M(x)

step 1 Solve the inequalities P(x) R(x) 0Q(x) S(x)

minus lt and R(x) T(x) 0S(x) M(x)

minus lt

rArr P(x)S(x) R(x)Q(x) 0

Q(x)S(x)minus lt

helliphellip(i) and R(x)M(x) T(x)S(x) 0

S(x)M(x)minus lt

helliphellip(ii)

Intersection of solution set of equations (i) and (ii) gives the solution set of the given inequality

Remarks

(i) If we have inequality of form gtP( x )

0Q( x )

and Q(x) gt 0 forall x isin ℝ then P(x) gt 0Q(x) rArr P(x) gt 0

(ii) If P( x )0

Q( x )gt and Q(x) lt 0 forall x isin ℝ then P(x)Q(x) lt 0 is multiplying by +ve real number does not

change the sign of inequality where as multiplying by ndashve real number reverses the sign of inequality

(iii) For all positive a b x

a x a if a bb x ba x a if a bb x b

+ gt lt + + lt gt +

121 POLyNOMIALS

An algebraic expression involving one or more variable that contains two mathematical operations multiplication and raising to a natural exponent (power) with respect to the variablevariables involved is called lsquomono-nomialrsquo

1211 Leading TermsLeading Coefficient The term containing highest power of variable x is called lsquoleading termrsquo and its coefficient is called leading coefficient Because it governs the value of f(x) where x rarr infin)

∵ n n 1 n 2 nn 2 n

a a af(x) x a x x xminus minus = + + + +

1212 Degree of PolynomialsHighest power of x in the polynomial expression is called lsquodegree of polynomialrsquo (ie power of x in leading term) Based on degree polynomials can be classfied as 0 (Constant) ax0 1 (linear) ax + b 2 (quadratic) ax2 + bx + c 3 (cubic) ax3 + ax2 + cx + d 4 (bi-quadratic) ax4 + bx3 + cx2 + dx + e

12121 Rational function and rational equation

An equation of the form f(x)g(x) where f(x) and g(x) are polynomials in x is known as rational function of x and when equated to zero it generates a rational equation

solving rational inequality While solving rational inequality the following facts must be always bear in mind

gt rArr gt lt

f (x) and g(x) have f (x) and g(x) havesame sign opposite sign

f (x) f(x)0 f(x)g(x) 0 0g(x) g(x)

f (x) and g(x) ofsamesign or f (x) 0

f (x)f(x)g(x) 0 0g(x)

gt le = ne

f (x) g(x) ofandopposite sign or

f (x) 0

f (x)g(x) 0f(x)or 0g(x)

f(x) 0 andg(x) 0 rArr

lt = ne

f(x)g(x) 0orf(x) 0 and g(x) 0

1213 Wavy-curve Method

To find the set of solution for inequality f(x) gt 0 (f(x) is polynomial)Factorize the polynomial and find all the roots eg f(x) = (x ndash a)3 (x ndash b)2 (x ndash d) (x ndash g)5 say a gt b gt d gt gLocate the roots (with their multiplicity) on the real number line Keep the sign expression in the

right-most interval same as that of the leading coefficient

Moving towards left change the sign of expression across the root with multiplicity odd and retain the same sign across the root with multiplicity even

there4 f(x) gt 0 rArr (a b) cup (b g) cup (d infin) Also f(x) ge 0 rArr (a b) cup (b g) cup (d infin) cup (a bgd)

rArr x isin [a g] cup [d infin) Similarly f(x) lt 0 rArr (ndashinfin a) cup (g d) and f(x) le 0 rArr (ndashinfin a) cup (g d) cup a b g d f(x) le 0 (ndashinfin a]

cup [g d] cup b

12131 Concept of continued sums and products

continued sum (sum) Sigma (Σ) stands for sum of indexed terms eg n

a=sum = a1 + a2 + a3 ++ an

In the above symbol ak is called lsquogeneral termrsquo and k is known as index

Properties

a=sum = a + a + a ++ a (n terms) = na

2 Sigma distributes on addition and subtraction n

k kk 1

(a b )=

plusmnsum = (a1 plusmn b1) + (a2 plusmn b2) ++ (an plusmn bn)

3 Sigma does not distribute on product and ratio of terms ie n

k kk 1

(a b )=

timessum = (a1 times b1) + (a2 times b2)

++ (an times bn) ne n n

k kk 1 k 1

a b= =

sum sum and

k kk 1

(a b )=sum = (a1b1) + (a2b2) ++ (anbn) ne

4 A constant factor can be taken out of sigma notation ien n

k kk 1 k 1

ma m a= =

=sum sum = m (a1 + a2 + a3

+ + an) cyclic and symmetric expressions An expression is called symmetric in variable x and y iff interchanging x and y does not changes the

expression x2 + y2 x2 + y2 ndash xy x3 + y3 + x2y + y2x x3 ndash y3 is not symmetric An expression is called cyclic in x y z iff cyclic replacement of variables does not change the

expression eg x + y + z xy + yz + zx etc Such expression can be abbreviated by cyclic sigma notation as follows Σx2 = x2 + y2 + z2 Σxy = xy + yz + zx

Σ(x ndash y) = 0 rArr x + y + z + x2 + y2 + z2 = Σx + Σx2

5 If sigma is defined for three variables say a b c occurring cyclically then it is evaluated as follows Σa = a + b + c = a + b + c Σ a b = ab + bc + ca Σa2 = a2 + b2 + c2

continued Products (π) Continued product of indexed termsn

a=prod is defined as product

of n number of indexed terms as n

k 1 2 3 nk 1

a a a a a=

Properties

=prodn

a aaaa (n times) = an

λ = λ λ λ = λ = λprod prodn n

nn 1 2 n 1 2 n k

k 1 k 1

a ( a )( a )( a ) (a a a ) a

3 π distributes over product and ratio of indexed terms but not over sum and difference of terms

ie = = =

= =prod prod prodn n n

k k k k 1 2 n 1 2 nk 1 k 1 k 1

a b a b (a a a )(b b b )

prodprod

kn1 2 3 nk k 1

nk 1 k 1 2 3 n

aa a a aa

b b b b bb

k k k kk 1 k 1 k 1

(a b ) a b= = =

plusmn ne plusmnprod prod prod

122 PARTIAL FRACTIONS

12231 Linear and non-repeating

Let D(x) = (x - a1) (x - a2) (x - a3) Then = + + + +minus minus minus minus

31 2 n

1 2 3 n

AA A AN(x) Q(x) D(x) x a x a x a x a

12232 Linear and repeated roots

Let D(x) = (x - a)K (x - a1) (x - a2)(x - an)

Then = + + + + + + + +minus minus minus minus minus minus

1 2 k 1 22 k

A A A B BN(x) BnQ(x) D(x) x a (x a) (x a) x a x a x a

12233 Quadratic and non-repeated roots

Let D(x) = (x2 + ax + b) (x ndash a1) (x ndash a2)(x ndash an) then+

= + + + + ++ + minus minus minus1 2 1 2 n

2 1 2 n

A x A B B BN(x) Q(x) D(x) (x ax b) x a x a x a

12234 Quadratic and repeated

Let D(x) = (x2 + a1 x + b1) (x2 + a2x + b2)(x2 + anx + bn)type V When both N(x) and D(x) contain only the even powers of x To solve these types of integrals follow the steps given belowstep 1 Put x2 = t in both N(x) and D(x) step 2 Make partial fractions of N(t)D(t)step 3 Put back t = x2 and solve the simplified integral now

123 THEOREMS RELATED TO TRIANGLES

Theorem 1 If two straight lines cut each other the vertically opposite angles are equalTheorem 2 If two triangles have two sides of the one equal to two sides of the other each to each and the angles included by those sides are equal then the triangles are equal in all respectsTheorem 3 If two angles of a triangle are equal to one another then the sides which are opposite to the equal angles are equal to one another

Theorem 4 If two triangles have the three sides of which one side is equal to three sides of another then they are equal in all respectsTheorem 5 If one side of a triangle is greater than other then the angle opposite to the greater side is greater then the angle opposite to the smaller sideTheorem 6 If one angle of a triangle is a greater than another then the side opposite to greater angle is greater than the side opposite to lessTheorem 7 Any two sides of a triangle are together greater they third sideTheorem 8 If all straight lines drawn from a given point to a given point on a given straight line then the perpendicular is the leastTheorem 9 If a straight line cuts two straight lines to make (i) The alternate angles equal or (ii) Exterior angles equal to the interior opposite angles on the same side of the cutting line or (iii) The interior angles on the same is side equal to two right angles then in each case the two straight

lines are parallelTheorem 10 If a straight line cuts two parallel lines it makes (i) The alternate angles equal to one another (ii) The exterior angle equal to the interior opposite angle on the same side of the cutting line (iii) The two interior angles on the same side together equal to two right anglesTheorem 11 The three angles of a triangle are together equal to two right anglesTheorem 12 If two triangles have two angles of one equal to two angles of the other each to each and any side of the first equal to the corresponding side of the other the triangles are equal in all respects called lsquoconjugatersquoTheorem 13 Two right angled triangles which have their hypotenuses equal and one side of one equal to one side of the other are equal in all respectsTheorem 14 If two triangles have two sides of the one equal to two sides of the other each to each but the angles included by the two sides of one greater than the angle included by the corresponding sides of the other then the base of that which has the greater angle is greater than the base of the other

12331 Theorems related to parallelograms

Theorem 15 The straight lines which join the extremities of two equal and parallel straight lines towards the same parts are themselves equal and parallelTheorem 16 The opposite sides and angles of a parallelogram are equal to one another and each diagonal bisects the parallelogramTheorem 17 If there are three or more parallel straight lines and the intersepts made by them on any transversal are equal then the corresponding intercept on any other transversal are also equalTheorem 18 Parallelograms on the same base and between the same parallels are equal in terms of area

12332 Theorems related to intersection of loci

The concurrence of straight lines in a triangle (i) The perpendiculars drawn to the sides of a triangle from their middle points are concurrent (ii) The bisectors of the angles of a triangles are concurrent (iii) The medians of a triangle are concurrentTheorem 19 Triangles on the same base and between the same parallel line are equal in area

Theorem 20 If two triangles are equal in area and stand on the same base and on the same side of it they are between the same parallel lineTheorem 21 Pythagorasrsquos theorem In any right-angled triangle the area of the square on the hypotenuse equals to the sum of the area of the squares on the other two sides

1231 Theorems Related to the Circle Definitions and First Principles

12311 Chords

Theorem 22 If a straight line drawn from the centre of a circle bisects a chord which does not pass through the centre it cuts the chord at right angles Conversely if it cuts the chords at right angles the straight line bisects itTheorem 23 One circle and only one can pass through any three points not in the same straight lineTheorem 24 If from a point within a circle more than two equal straight lines can be drawn to the circumference that point is the centre of the circleTheorem 25 Equal chords of a circle are equidistant from the centre Conversely chords which are equidistant from the centre than the equalTheorem 26 Of any two chords of a circle which is nearer to the centre is greater than one more remote Conversely the greater of two chords is nearer to the centre than the lessTheorem 27 If from any external point straight lines are drawn to the circumference of a circle the great-est is that which passes through the centre and the least is that which when produced passes through the centre And of any other two such lines the greater is that which subtends the greater angle at the centre

12312 Angles in a circleTheorem 28 The angle at the centre of a circle is double of an angle at the circumference standing on the same arcTheorem 29 Angles in the same segment of a circle are equal Coverse of this theorem states ldquoequal angles standing on the same base and on the same side of it have their vertices on an arc of a circle of which the given base is the chordrdquoTheorem 30 The opposite angles of quadrilateral inscribed in a circle are together equal to two right angles coverse of this theorem is also trueTheorem 31 The angle in a semi-circle is a right angleTheorem 32 In equal circles arcs which subtend equal angles either at the centres or at the circumferences are equalTheorem 33 In equal circles arcs which are cut-off by equal chords are equal the major arc equal to the major arc and the minor to the minorTheorem 34 In equal circles chords which cut-off equal arcs are equal

1232 TangencyTheorem 35 The tangent at any point of a circle is perpendicular to the radius drawn to the point of contactTheorem 36 Two tangent can be drawn to a circle from an external pointTheorem 37 If two circles touch one another the centres and the point of contact are in one straight line

Theorem 38 The angles made by a tangent to a circle with a chord drawn from the point of contact are respectively equal to the angles in the alternate segments of the circle Theorem 39 If two of straight lines one is divided into any number of parts the rectangle contained between the two lines is equal to the sum of the rectangles contained by the undivided line and the several parts of the divided lineTheorem 40 If a straight line is divided internally at any point the square on the given line is equal to the sum of the squares on the squares on the two segments together with twice the rectangle contained by the segmentsTheorem 41 If a straight line is divided externally at any point the square on the given line is equal to the sum of the squares on the two segments diminished by twice the rectangle contained by the segmentsTheorem 42 The difference of the squares on the two straight lines is equal to the rectangle contained by their sum and differenceTheorem 43 In an obtuse-angled triangle the square on the side subtending the obtuse angle is equal to the sum of the squares on the sides containing the obtuse angle together with twice the rectangle contained by one of those sides and the projection of the other side upon itTheorem 44 In every triangle the square on the side subtending an acute angle is equal to the sum of the squares on the sides containing that angle diminished by twice the rectangle contained by one of these sides and the projection of the other side upon itTheorem 45 stewardrsquos theorem If D is any point on the side BC of a then AB2DC + AC2 BD = AC (AD2 + BD DC)Theorem 46 In any triangle the sum of the squares on two sides is equal to twice the square on half the third side together with twice the square on the median which bisects the third side (Appolonius theorem which is a special case of Stewardrsquos theorem)

1233 Rectangles in Connection with CirclesTheorem 47 If two chords of a circle cut a point within it the rectangle contained by their segments are equalTheorem 48 If two chords of a circle when produced cut at a point outside it the rectangles contained by their segments are equal And each rectangle is equal to the square on the tangent from the point of intersectionTheorem 49 If from a point outside a circle two straight lines are drawn one of which cuts the circle and the other meets it and if the rectangle contained by the whole line which cuts the circle and the part of it outside the circle is equal to the square on the line which meets the circle then the line which meets the circle is a tangent to it

1234 Proportional Division of Straight LinesTheorem 50 A straight-line drawn parallel to one side of a triangle cuts the other two sides or those sides produced proportionallyTheorem 51 If the vertical angle of a triangle is bisected internally into segments which have the same ratio as the other sides of the triangle Conversely if the base is divided internally or externally into segments proportional to the other sides of the triangle the line joining the point of section to the vertex bisects the vertical angle internally or externally AD and ADrsquo are internal and external angle bisectors of the triangle

1235 Equiangular TrianglesTheorem 52 I f two triangles are equiangular to each other their corresponding sides are proportionalTheorem 53 If two triangles have their sides proportional when taken in order the triangles are equiangular to one another and those angles are equal which are opposite to the corresponding sides

Theorem 54 If two triangles have one angle of which one is equal to one angle of the other and the sides about the equal angles are proportionals then the triangles are similarTheorem 55 If two triangles have one angles of which one is equal to one angle of the other and the sides about another angle in one proportional to the corresponding sides of the other then the third angles are either equal or supplementary and in the former case the triangles are similarTheorem 56 In a right-angled triangle if a perpendicular is drawn from the right angle to the hypotenuse the triangles on each side of it are similar to the whole triangles and to each other

12351 Similar Figures

Theorem 57 Similar polygons can be divided into the same number of similar triangles and the lines joining the corresponding vertices in each figure are proportionalTheorem 58 Any two similar rectilinear figures may be placed in a way that the lines joining corre-sponding the vertices are concurrentTheorem 59 In equal circles angles whether at the centres or circumferences have the same ratio as the arcs on which they stand

12352 Proportion applied to area

Theorem 60 The areas of similar triangles are proportional to the squares on there corresponding sidesTheorem 61 The area of similar polygons are proportional to the squares on there corresponding sides

1236 Some Important Formulae 1 (a + b)2 = z2 + 2ab + b2 = (a ndash b)2 + 4ab 2 (a + b)2 = a2 ndash 2ab + b2 = (a ndash b)2 + 4ab 3 a2 ndash b2 = (a + b) (a ndash b) 4 (a + b)3 = a3 + b3 + 3ab (a + b) 5 (a ndash b)3 = a3 + b3 ndash 3ab(a ndash b) 6 a3 + b3 = (a + b)3 ndash 3ab(a + b) = (a + b) (a2 + b2 ndash ab) 7 a3 ndash b3 = (a ndash b)3 + 3ab (a ndash b) = (a ndash b) (a2 + b2 + ab)

8 2 2 2 2 2 2 2 1 1 1(a b c) a b c 1ab 2bc 2ca a b c 2abc

a b c + + = + + + + + = + + + + +

9 3 3 3 2 2 21a b c ab bc ca (a b) (b c) (c a)2 + + minus minus minus = minus + minus + minus

10 ( )( )3 3 2 2 2 2a b c 3abc a b c a b c ab bc ca+ + minus = + + + + minus minus minus = ( ) ( )2 2 21 a b c (a b) (b c) (c a)2

+ + minus + minus + minus 11 a4 ndash b4 = (a + b) (a ndash b) (a2 ndash b2)

12 a4 + a2 + 1 = (a2 + 1)2 ndash a2 = (1 + a + a2) (1 ndash a + a2)

13 2 2a b a b

ab2 2+ minus

= minus

14 a b (a b)(a b)minus = minus +

15 a2 + b2 + c2 ndash ab ndash bc ndash ca = (a ndash b2) + (b ndash c)2 + (c ndash a)2 16 (x + a) (x + b) = x2 + (a + b)x + ab 17 (a + b + c)3 = a3 +| b3 |+c3 + 3 (a + b) (b + c) (c + a) 18 a3 + b3 + c3 ndash3abc = (a + b + c) (a2 + b2 + c2 ndash ab ndash bc ndash ca) 19 (a + b)4 = (a +| b)2 |(a +| b)2 = a4 + b4 + 4a3b + 6a2b2 + 4ab3

20 (a ndash b)4 = (a ndash| b)2 |(a ndash| b)2 = a4 + b4 ndash 4a3b + 6a2b2 ndash 4ab3

21 (a + b)5 = (a +| b)3 |(a +| b)2 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5

Chapter 2eXpONeNtIaL LOGarIthM

21 ExponEntial Function

If a is a positive real number then ax (a ne 1) is always positive and it is called lsquoexponential function of xrsquo Here a is called lsquobasersquo and x is called index

211 Properties of Exponential Functions

(i) As we know that = times times times forall isin

n times

a a a a n where a is called lsquobasersquo and n is index or exponent

Exponential function f(x) = ax is generalisation of this law to facilitate some useful applications with some imposed functional restrictions ie a gt 0 and a ne 1

(ii) Domain of f(x) is set of real number and range of f(x) is (0 infin) ie forall x isinℝ f(x) = ax associates x to some positive real number uniquely ie exponential function f(x) is defined such that it is invertible

(iii) For a lt 0 and a = 0 the function f(x) = ax loses its meaning for some values of x isin ℝ For instance for a = ndash1 ndash12 ndash3 etc

f(x) = ax becomes non-real forall =pxq

where p and q are co-prime and q is an even integer

eg (ndash3)32 (ndash1)14 etcSimilarly when base a = 0 then f(x) = 0x does not remain an one-to-one function which is required for invariability same restriction also holds for a = 1 Since then f(x) = 1x again becomes many one function as all inputs x get associated to single output 1Therefore we conclude that for f(x) = ax the base a gt 0 a ne 1 and x isin ℝ thus y isin (0 infin)

(iv) If the base a is Euler number lsquoersquo then the exponential function ex is known as natural exponential function

212 Laws of Indices

(i) ax is defined and ax gt 0 forall x isin ℝ (ii) a0 = 1 We can observe that rarrn a 1 as n assumes very large value (n rarr infin) and it is true for both

cases ie a gt 1 or a isin (0 1) therefore when n rarr infin = = =1n 0n a a a 1

(iii) axtimesay = ax+y

(iv) minus=x

(v) (ax)y = axy = (ay)x

(vi) = qpq pa a where isinq and q ne 1 (vii) ax = ay rArr x = y or a = 1 (viii) ax = bx rArr either x = 0 or a = b

(ix) axbx = (ab)x and =

a ab b

(x) ax ge ay rArr ge gt

le isin

x y if a 1x y if a (01)

213 Graphical Representation of an Exponential Function

1 ax where a gt 1 behaves as an increasing nature function For example when a = 2 the value of function 2x increases as the

input x increases It can be understood from the table given below

x ndash3 ndash2 ndash1 0 1 2 3 4 2ax 18 14 12 1 2 4 8 16 32

2 If 0 lt a lt 1 behaves like a decreasing nature function For example when a = 12 the value of function 2ndashx decreases as the

input x increases which can be observed in the following table

x ndash5 ndash4 ndash3 ndash2 ndash1 0 1 2 3ax 32 16 8 4 2 1 12 14 18

3 If the base a gt 1 then ax ge 1 for all x ge 0 and ax lt 1 when x lt 0 if 0 lt a lt 1 then 0 lt ax lt 1 for x gt 0 and ax gt 1 for x lt 0 The above fact as well as the relative position of graphs of exponenital functions with different bases can be understood with the help of following figure

If the base a gt 1 then ax ge 1 for all x ge 0

Exponential Logarithm 231

214 Composite Exponential FunctionsA composite function is a function in which both the base and the exponent are the functions of x Generally any function of this form is a composite exponential function This function is also called an exponential power function or a power exponential function ie y = [u(x)]v(x) = uv In calculus the domain consists of such values of x for which u(x) and v(x) are defined and u(x) gt 0

215 Methods of Solving Exponential Equation

To solve an exponential equation we make use of the following facts

(i) If the equation is of the form ax = ay(a gt 0) rArr x = y or a = 1 (ii) If the equation is of the form ax = bx (a b gt 0) rArr either x = 0 or a = b (iii) If the equation is of the form ax = k (a gt 0) then

Case I If b ge 0 rArr x isin Case II If b gt 0 k ne 1 rArr x = logak Case III If a = 1 k ne 1 rArr x isin Case IV If a = 1 k = 1 rArr x isin ℝ(Since 1x = 1 rArr 1 = 1 x isin ℝ)

(iii) If the equation is of the form af(x) = ag(x) where a gt 0 and a ne 1 then the equation will be equivalent to the equation f(x) = g(x)

Remarksax = 1 rArr x = 0 is an incomplete conclusion it is only true if the base a ne 0 plusmn 1if a = 0 so equality does not holds as 00 is meaningless

Where as when a = 1 then an = 1 rArr 1x = 1 Thus x isin ℝ

In case a = ndash1 then (ndash1)x = 1 is true for x = pq when p is even and GCD of p and q = 1

22 Solving ExponEntial inEquality

(i) The value of ax increases as the value of x increases when base a isin (1 infin) but the value of ax de-

creases as the value of x increases when base a isin (0 1) ge gtge rArr le isin

x y x y if a 1a a

x y if a (01)

(ii) The elementary exponential inequalities are inequalities of form ax gt k ax lt k where a and k are certain numbers (a gt 0 a ne 1) Depending on the values of the parameters a and k the set of solutions of the inequality ax gt k can be in the following forms1 x isin (logak infin) for a gt 1 k gt 02 x isin (ndashinfin logak) for 0 lt a lt 1 k gt 03 x isin ℝ for a gt 0 k lt 0Depending on the values of a and k the set of solutions of the inequality ax lt k can be in the fol-lowing forms1 x isin (ndashinfin logak) for a gt 1 k gt 02 x isin (logak infin ) for 0 lt a lt 1 k gt 03 x isin for a gt 0 k lt 0 (ie the inequality has no solutions)

(iii) + = forall isinminus =

f(x y) f(x) f(y)x y

f(x y) f(x) f(y)

23 logaRitHMic Function

The logarithm of any number N to the given base a is the exponent or index or the power to which the base must be raised to obtain the number N Thus if ax = N x is called the logarithm of N to the base a It is denoted as logaN

loga N = x hArr ax = N a gt 0 a ne 1 and N gt 0

(a) The logarithm of a number is unique ie no number can have two different logarithms to a given base

(b) The base lsquoarsquo is a positive real number but excluding 1 ie a gt 0 a ne 1 As a consequence of the definition of exponential function we exclude a = 1

Since for a = 1 logax = y rArr x = ay = 1y which has no relevance to the cases of logax when x ne 1 ie for all values of exponent the value of x remains 1

(c) The number lsquoxrsquo represents result of exponentiation ie ay therefore it is also a positive real number ie x = ay gt 0

(d) The exponent lsquoyrsquo ie logarithm of lsquoxrsquo is a real number and neither a nor x equals to zero

(d) Domain of function y = logax is (0 infin) and the range (-infin infin)

when x rarr 0 then logax rarr-infin (for a gt 1) and logax rarr infin (for 0 lt a lt 1) because y = logax rArr x = ay which approaches to zero iff y rarr-infin as a-infin = 0 forall a gt 1 and when a isin (0 1) x = ay approaches to zero iff y rarr infin ∵ ainfin = 0 if 0 lt a lt 1

(e) Common Logarithms and Natural Logarithms The base of logarithm can be any positive number other than 1 but basically two bases are mostly used They are 10 and e (=2718 approximately) Logarithm of numbers to the base 10 are named as Common Logarithms whereas the logarithms of the numbers to the base e are called as Natural or Napierian logarithms

If a = 10 then we write log b instead of log10b

If a = e then we write ℓnb instead of logeb

We find logea = log10a loge10 or e10

log alog a 0434

log 10= = logea (this transformation is used to convert

natural logarithm to common logarithm)

231 Properties of Logarithm

P 1 loga 1 = 0 because 0 is the power to which a must be raised to obtain 1 P 2 logaa = 1 since 1 is the power to which a must be raised to obtain a P 3 alogaN = N and logaa

N = N as N is the power to which a must be raised to obtain aN P 4 logm(ab) = logm a + logmb (a gt 0 b gt 0) Logarithm of the product of two numbers to a certain base

is equal to the sum of the logarithms of the numbers to the same base

P 5 logm (ab) = logma-logmb logarithm of the quotient of two numbers is equal to the difference of their logarithms base remaining the same throughout

P 6 loga Nk = k logaN (k is any real number) Logarithm of the power of a number is equal to the product

of the power and logarithm of the number (base remaining the same) P 7 logak N = (1k)loga N

(1) The property 4 5 6 7 are not applicable conditionally because logaM + logaN is defined only when M and N are both positive whereas logaMN is defined even if M and N are both negative Therefore logaMN cannot be always replaced by logaM + logaN Therefore such replacement can lead to loss of root while solving an equation

(2) Thus to avoid the loss of root we consider the following transformations

(a) logaMN = loga |M| + loga |N| (when MN gt 0)

(b) a a a

Mlog log |M | log |N |

N= minus (when MN gt 0 N ne 0)

(c) logaN2k = 2k loga |N| (when N ne 0 k an integer)

(d) k |a|a

1log N log N

2k2 = (when N gt 0 k is an integer ne 0 a ne 0 |a| ne 1)

(3) The transformation 2(a) 2(b) or also valid conditionally as LHS is defined when M and N have same sign whereas the RHS is defined for any arbitrary values of M and N other than zero So such replacement while solving an equation can generate extraneous roots but since extraneous roots can be counter checked (and those not satisfying the parent equation can always be discarded) on the other hand the loss of root is difficult be traced therefore it is suggested to use the results of 2(a) 2(b) 2(c) 2(d) in place of property number 4 5 6 7 While simplifying and solving equationinequations

P 8 logba = logca logbc

P 9 = =cb

log a log alog alog b log b

(base remaining the same in numerator and denominator)

P 10 logbalogab = 1

P 11 alogmb = blogma

24 logaRitHMic EquationS

If we have an equation of the form loga f (x) = b (a gt 0) a ne 1 is equivalent to the equation f(x) = ab (f (x) gt 0)

241 Some Standard Forms to Solve Logarithmic Equations

Type 1 An equation of the form logxa = b a gt 0 has (a) Only root x = a1b if a ne 1 and b ne 0 (b) Any positive root different from unity if a = 1 and b = 0 (c) No roots if a = 1 b ne 0 (d) No roots if a ne 1 b = 0

Type 2 Equations of the form

(i) f (logax) = 0 a gt 0 a ne 1 (ii) g (logxa) = 0 a gt 0

Then equation (i) is equivalent to f(t) = 0 where t = loga xIf t1 t2 t3 tk are the roots of f(t) = 0 then logax = t1 logax = t2 logax = tk

rArr = 1 2 kt t tx a a a and equation (ii) is equivalent to g(y) = 0 where y = logxa If y1 y2 y3 yk are the roots of f(y) = 0 then logx a = y1 logxa = y2 logxa = yk = x

rArr = 1 2 k1y 1y 1yx a a a

Type 3 Equation of the form (i) loga f(x) = loga g(x) a gt 0 a ne 1 is equivalent to systems of equations and inequations as

given below

System 1 gt

g(x) 0f(x) g(x) System 2

f(x) 0f(x) g(x)

(Any one of the two systems can be used) (ii) logf(x)A = logg(x) A A gt 0 is equivalent to the systems of equations and inequations as given below

System 1 gt ne =

g(x) 0g(x) 1f(x) g(x)

System 2 gt ne =

f(x) 0f(x) 1f(x) g(x)

(Any one of the two systems can be used)

Type 4 Equation of the form

(i) logf(x) g(x) = logf(x) h(x) is equivalent to two systems of equations and inequations

System 1 gt gt =

g(x) 0f(x) 0g(x) h(x)

System 2 gt gt =

h(x) 0f(x) 0g(x) h(x)

(Any one of the two systems can be used) (ii) logg(x)f(x) = logh(x) f(x) is equivalent to two systems of equations and inequations

System 1

gt gt ne =

f(x) 0g(x) 0g(x) 1g(x) h(x)

System 2

gt gt ne =

f(x) 0h(x) 0h(x) 1g(x) h(x)

Type 5 An equation of the form logh(x) (logg(x) f(x)) = 0 is equivalent to the system

gt ne gt ne =

h(x) 0h(x) 1g(x) 0g(x) 1f(x) g(x)

Type 6 An equation of the form 2m loga f (x) = logag(x) a gt 0 a ne 1 m isin N is equivalent to the

system gt

f(x) 0[f(x)] g(x)

Type 7 An equation of the form (2m + 1) loga f (x) = logag (x) a gt 0 a ne 1 m isin N is equivalent to the

system +

f(x) 0[f(x)] g(x)

Type 8 An equation of the form loga f(x) + logag(x) = logam(x) a gt 0 a ne 1 is equivalent to the

system gt gt =

f(x) 0g(x) 0f(x)g(x) m(x)

Type 9 An equation of the form loga f(x)ndashlogag(x) = logah(x)ndashlogat(x) a gt 0 a ne 1 is equivalent to the

equation loga f(x) + logat(x) = logag(x) + logah(x) which is equivalent to the system

gt gt gt gt =

f(x) 0t(x) 0g(x) 0h(x) 0f(x)t(x) g(x)h(x)

25 logaRitHMic inEqualitiES

When base a gt 1 then logax is an increasing function where as when 0 lt a lt 1 then logax is a decreasing function

We can observe this by simple taking log2x and log12x and evaluating their value for various positive inputs and thus plotting the approximate graph of both the functions

x 116 18 14 12 1 2 4 8log2x ndash4 ndash3 ndash2 ndash1 0 1 2 3

log12x 4 3 2 1 0 ndash1 ndash2 ndash3

To solve a logarithmic inequality following facts must be kept in mindGive any positive real number then

(a) For a gt 1 the inequality logax gt logay rArr x gt y (Since for a gt 1 logax is an increasing function)

rArr If a gt 1 then logax lt a rArr 0 lt x lt aa

rArr If a gt 1 then logax gt a rArr x gt aa

(b) For 0 lt a lt 1 then inequality 0 lt x lt y rArr logax gt logay (∵ for 0 lt a lt 1 logax is a decreasing function) rArr If 0 lt a lt 1 then logax lt a rArr x gt aa

rArr If 0 lt a lt 1 then logax gt a rArr 0 lt x lt aa

251 Characteristic and Mantissa

Generally the logarithm of a number is partially integral and partially fractional The integral part to the logarithm of a number is called lsquocharacteristicrsquo and the decimal part is known as mantissa

(a) Given a number N Logarithms can be expressed as log10 N = Integer + fraction (+ ve) darr darr Characteristic Mantissa (i) The mantissa part of the log of a number is always kept positive the characteristic may be

positive or negative eg if loge x = ndash14 = ndash2 + 06 written as 26 (ii) If the characteristics of log10 N be n then the number of digits in N is (n + 1) (iii) If the characteristics of log10 N be (-n) then there exists (n ndash 1) number of zeros after the

decimal point of N (b) The number of positive integer having base lsquoarsquo and characteristic n = an+1-an (c) If the number and base are on the same side of the unity then the logarithm is positive If the number

and the base are on the opposite side of the unity then the logarithm is negative (d) Characteristic of the common logarithm of (i) any number greater than 1 is positive (ii) any positive number less than 1 is negative

Chapter 3SequenCe and progreSSion

31 Definition

ldquoSequence is a definite pattern of the numbers (defined by a function Tn ℕ rarr ℂ where ℕ is natural numbers ℂ is complex numbers) each of which is derived according to a definite law and whose general term (Tn) is expressible in terms of nrdquo It denoted by lt Tn gt where Tn is the general term eg ltngt 1 2 3 4ltn2gt 12 22 32

311 Types of Sequence rArr Increasing Sequence lttngt is called increasing sequence iff tn + 1 gt tn forall n isinℕ ie t1 lt t2 lt t3 lt t4hellip

eg lt 2n ndash 1gt 1 3 7 9 11 rArr Decreasing Sequence lt tngt is called decreasing iff tn+1 lt tn ie t1 gt t2 gt t3 gt t4hellip

1 1 1 1 1 1 2 2 4 8 16 32

rArr Converging Sequence A sequence is called converging sequence iff its term at infin is a finite real number ie Tn = finite when nrarrinfin

eg 1 1 1 1 2 4 8 16

is converging as 1T = 02infin infin = Also 1 1 1 1

2 3 4 5 is converging as 1T 0infin = =

infin rArr Diverging Sequence A sequence is called diverging sequence iff Tn rarrinfin when n rarrinfin

eg lt 2n gt 2 4 8 16 32 Tinfin = 2infin = infin rArr Oscillating Sequence A sequence is called oscillating sequence iff its value oscillates between two

numbers eg lt (ndash1)n gt ndash1 1 ndash1 1 ndash1 1 rArr Periodic Sequence If the term of sequence repeats after a fixed interval then the sequence is

called a periodic sequence

eg Periodic repeating

nsin 10 10 1 0 1 02π minus minus

312 Progression and SeriesProgression is a sequence in which each succeeding term bears a fixed relation with its proceeding one (ie a sequence following a certaindefinite pattern)

Types of Progression Progressions are generally of the following types

(i) Arithmetic Progression (AP) (ii) Geometric Progression (GP) (iii) Harmonic Progression (HP) (iv) Arithmetico Geometric Progression (AGP) (v) Arithmetico Arithmetic Progression (AAP) (vi) Arithmetico Arithmetic Geometric Progression(AAAP) (vi) Arithmetico Arithmetico Geometric Progression (AAGP) etc

32 SerieS

The term of a sequence are separated by positive sign is called series Hence a series is the summation of

terms of sequence denoted as Sn n

n kk 1

=sum ie the sum of the first lsquonrsquo terms of a sequence

(i) Arithmetic Progression It is the progression in which the difference of successive terms remain constant and this constant is known as common difference (eg sequence of odd natural numbers 1 3 5 7 )

321 Properties of Arithmetic Progression P1 If a is the first term and d is the common difference of the AP then AP can be written as a a + d

a + 2d a + ( n -1)d P2 General Terms nth term from beginning Tn = a + (n - 1)d = l (last term) where d = Tn - Tnndash1 n

th term from last Tn = l + (n -1) (-d)

bull If d gt 0 rArrincreasing Arithmetic Progression (AP) bull If d lt 0 rArrdecreasing Arithmetic Progression (AP) bull If d = 0 rArrall the terms remain same P3 Hence the nth term can also be written as Tn = Sn - Sn-1 P4 Sum of first n terms Sn = n2 [2a + (n -1)d] = n2 [a + l )

bull Sum of the first n natural number is n(n 1)2+ bull Sum of the first n odd natural number is n2

bull Sum of the first n even natural number is n(n + 1) P5 Summation of equidistant terms from beginning and end of an AP is always constant and is equal

to sum of the first and last term rArr n 1 n 2 n 1n nS (T T ) (T T ) 2 2 minus= + = + +

P6 If the nth term tn = an + b then the series so formed is an AP P7 If the sum of first n terms of a series is Sn = an2 + bn + c then the series so formed is an AP (provided

c = 0) If c ne 0 then series formed will be AP from the 2nd term onward P8 If every term of an AP is increased or decreased by the same quantity the resulting terms will also

be in AP with no change in common difference P9 If every term of an AP (CD = d) is multiplied or divided by the same non-zero quantity K then the

resulting terms will be in AP with new common difference equal to dK or dK P10 If the corresponding terms of two APrsquos are added or subtracted the resulting is also an AP

lt tn gt is AP with CD = d1 lt an gt in AP with CD = d2 lt tn + an gt is AP with CD = d1 + d2

Caution lttn angt n

is not necessarily an AP

Sequence and Progression 1339

P11 If equal number of terms (say k terms of an AP) are dropped alternately the resulting terms lie in AP with CD = (k + 1)d

P12 If equal number of terms say lsquokrsquo terms of an AP are grouped together and sum of terms in each group is obtained then the sum is in AP with common difference k2d

P13 If terms a1 a2an an+1a2n+1 are in AP then the sum of these terms will be equal to (2n + 1)an+1 Here total number of terms in the series is (2n + 1) and the middle term is an+1

P14 If terms a1 a2hellip a2nndash1 a2n are in AP The sum of these terms will be equal to n n 1a a(2n)2

where = n n 1a a2

AM of the middle terms

P15 If the ratio of the sum of Ist n term of 2 different APrsquos is given as a f(n) n

Sie f (n)S

= prime

then the ratio

of their kth terms is given by ( )kk

t f 2k 1t

= minus

P16 If the ratio of nth terms of two APrsquos is given as f(n) n

Tie f (n)T

= prime

then the ratio of their sum

of k terms is given by k

S k 1fS 2

Points to Remember

bull Selection of terms in AP When sum of given number of terms in AP is known then terms must be selected as bellow

bull Odd Number of Terms in AP Let the middle term be lsquoarsquo and CD = d 3 terms in AP a ndash d a a + d 5 terms in AP a ndash 2d a ndash d a a + d a + 2d

bull Even number of terms in AP The two middle terms (a ndash d) and (a + d) Consider the cd as 2d 4 terms in AP a ndash 3d a ndash d a + d a + 3d 6 terms in AP a ndash 5d a ndash 3d a ndash d a + d a + 3d a + 5d

33 Arithmetic meAn

331 Arithmetic Means of Numbers

Arithmetic mean for any n positive numbers a1 a2 a3 an is + + + +

= 1 2 3 na a a aAM

332 Insertion of n AMrsquos between Two Numbers

Arithmetic Mean between Two Numbers n arithmetic means between x and y is defined as a set of n numbers A1 A2 A3 An such that x A1 A2 A3 An y in AP

rArr kb aA a kn 1minus = + +

Remarks

1 Sum of n AMrsquos between a and b is equal to n times single AM between a and b =

= +sumn

nA ( a b)

2 If Tk and Tp of any AP are given then formula for Tn is minusminus

= =minus minus

p Kn Kn

T TT TT

n k p k

3 If p Tp = q Tq of an AP then Tp + q = 0

4 If pth term of an AP is q and qth terms is p then Tp + q = 0 and Tn = p + q ndash n

5 If pth terms of an AP is 1q and qth term is 1p then its pqth term is 1

6 If number of terms in any series is odd then only one middle term exists which is +

34 Geometric ProGreSSion

Geometric progression is a progression in which the ratio of the successive term remains the constant Such ratio is known as common ratio eg 3 6 12 24 48hellip

Therefore a1 a2 a3 an is in GP iff 32 4 n

1 2 3 n 1

aa a aa a a a minus

= = = = = constant (r) is called as common ratio

341 Properties of Geometric Progression

P1 If a is the first term and r is the common ratio then GP can be written as a ar ar2 ar3 ar4 arn-1

P2 nth term from the beginning Tn = arnndash1 = l (last term) where n

TrT minus

P3 nth term from the last with last term n n 1T r minus=

P4 The product of equidistant term from both ends of GP is constant T1Tn = T2Tnndash1= T3Tnndash2 = = a2rnndash1

P5 Sum of first n term n n

na(r 1) a(1 r )S(r 1) (1 r)

minus minus= =

minus minus If arnndash1 = l then n

a rS1 rminus

=minus where l is the last term in

the series

P6 Sum of infinite GP n

a(1 r )S (S )1 rinfin rarrinfin

rarrinfin

minus= = minus

plusmn infinplusmn infin minus

minus ltminus

if |r| gt 1 if r = 1

a if r = 1 and n odd0 if r = 1 and n evena if |r | 1

Remark

Sum of infinite GP infin =minusa

when | r | lt 1 ie ndash1 lt r lt 1 not finite when | r | gt 1

ie r gt 1 or r lt ndash1

P7 If every term of a GP is increased or decreased by the same non-zero quantity the resulting series may not be in GP

P8 If every term of a GP is multiplied or divided by the same non-zero quantity the resulting series is in GP with the same common ratio

P9 If a1 a2 a3 and b1 b2 b3 two GPrsquos of common ratio r1 r2 respectively then a1b1

a2 b2 31 2

aa a b b b

and will also form GP and the common ratio will be r1r2 and r1r2 respectively

P10 If each term of a GP be raised to the same power then the resulting series is also a GPie lt tn

kgt is also a the GP with CR = rk

P11 If a b c are in GP then logk a logk b logk c are in AP ie in general if a1 a2 a3 be a GP of positive terms then log a1 log a2 log a3 will be in AP and conversely

P12 If F a 1 a 2 a3 are in GP then 1 2 3

1 1 1 a a a

are in GP

P13 Value of a recurring decimal Let R denote the decimal representation of a number as given

x numbers y numbers

R 0XXXX YYYY=

where X0 denotes the figure consist of non-recurring digit appearing

x times whereas Y0 denote the recurring period consisting of y digits x0 010 R X Y=

where 0x times

X XXXX=

and 0y times

Y YYYY=

and x y0 0 010 R X Y Y+ times = Therefore by subtraction

0 0x y x

(10 10 )+

minus=

P14 Selection of Terms in GP When product of given number of terms in GP is known then the terms must be selected as followsrArr Odd Number of Terms in GP Let the middle term be lsquoarsquo and CR = r 3 terms in AP a r a ar 5 terms in AP a r2 a r a ar ar2rArr Even number of terms in GP The two middle terms are ar ar and CR = r2

4 terms in GP 33

a a ararr r

6 terms in GP 3 5a a a ar ar arr r r

35 Geometric meAn

351 Geometric Means of Numbers

Geometric Mean If three or more than three terms are in GP then all the numbers lying between first and last term are called geometrical means between them Geometric mean (G) of lsquonrsquo numbers x1 x2

x3xn is defined as nth root of their product rArr 1n

1 2 3 nG (x x x x )=

352 Geometric Mean between Two Numbers

If a b c are three positive numbers in a GP then b is called the geometrical mean between a c and b2 = ac

If a and b are two positive real and G is the single GM between them then G2 = ab

To insert lsquonrsquo GMrsquos between a and b Let a and b are two positive numbers and G1 G2hellip Gn are lsquonrsquo GMrsquos between them then a G1 G2 Gn b is a GP with lsquobrsquo as its (n + 2)th term

rArr b = arn+1

1n 1br

rArr G1 = ar G2 = ar2 hellip Gn = arn

Notes 1 Product of n GMrsquos inserted between a and b is equal to the nth power of a single GM between them

2 If a is positive and r gt 1 then GP is increasing but if 0lt r lt 1 then it is a decreasing GP

3 If a is negative and r is positive (r gt 1) then it is a decreasing GP but if 0 lt r lt 1 it is an increasing GP

36 hArmonic ProGreSSion

A sequence is said to be a harmonic progression if and only if the reciprocal of its terms form an arithmetic progression (eg 12 14 16 form a HP because 2 4 6 are in AP)

361 Properties of Harmonic Progression

P1 General form of a harmonic progression + + ++ +

1 1 1 a a d a 2d

P2 General Term Tn of HP = reciprocal of Tn of its corresponding AP (eg in the above series

a (n 1)d=

+ minus)

P3 If a b are the first two terms of an HP then n1t

1 1 1(n 1)a b a

= + minus minus

P4 If all the terms of an HP are multiplie or divided by a constant non-zero quantity the resulting series remains in HP

P5 If the term of an HP is infin this means that the corresponding term of the AP is zero P6 There is no general formula for finding the sum to n terms of HP

P7 If a b c are in HP 1 1 1 a b c

are in AP 2 1 1b a c= + rArr 2acb

+or a a b

c b cminus

=minus

P8 If a b c are in GP then logak logbk logck in HP

If terms are given in HP then the terms could be picked up in the following ways

(i) For three terms minus +1 1 1

a d a a d

(ii) For four terms minus minus + +1 1 1 1

a 3d a d a d a 3d

37 hArmonic meAn

If three or more than three terms are in HP then all the numbers lying between the first and last term are called harmonic means between them

371 Harmonic Mean of Numbers

(a) H of any two numbers a and b is given by 2abHa b

where a b are two non-zero numbers

(b) Also the HM of n non-zero numbers a1 a2 a3 an n

j 11 2 n j

n nH1 1 1 1a a a a=

= =+ + + sum

(c) Insertion of n harmonic mean between two numbers Let a and b be two given numbers and H1 H2

H3 Hn are HMrsquos between them Then a H1 H2 H3 Hn b ie r

1 1 rdH a

= + where (a b)d(n 1)ab

minus=

NoteSum of the reciprocals of all the n HMrsquos between a and b is equal to n times the reciprocal of single HM (H) between a and b

38 inequAlity of meAnS

rArr If A and B are positive numbers then A ge G ge H rArr If A G H are respectively AM GM HM between a and b both being unequal and positive then rArr G2 = AH ie A G H are in GP rArr For any set of positive real numbers x1 x2 xn

1 2 n 1n1 2 n

x x x n(x x x )1 1 1n x x x

+ + +ge ge

+ + + ie AM ge GM ge HM

Condition of Application rArr Equality holds (ie A = G = H) iff x1 = x2 = hellip = xn rArr If sum of the variable x1 x2 xn be x1 + x2 + hellip + xn = S then product x1x2hellipxn = P can be

maximized A ge G rArr 1nS (P)

nge rArr

rArr n

maxSPn

and it is obtained when x1= x2=hellip= xn= Sn

rArr Similarly if x1x2x3hellipxn = P is constant then minimum value of sum lsquoSrsquo can be obtained as

rArr 1

1 2 n n1 2 n

x x x(x x x )

n+ + +

ge rArr 1nS (P)

nge rArr S ge n(P)1n

rArr Smin ge n(P)1n and it is obtained when x1 = x2 =hellip = xn = (P)1n

Remarks

1 If A and G are two AM and GM between two positive questions a and b then the quadratic equation having a b as its roots is x2 ndash 2Ax + G2 = 0

2 If AG H are AM GH and HM respectively then the equation having three roots is

minus + minus =3 2 33Gx 3Ax x G 0

39 Arithmetic-Geometric ProGreSSion

A series formed by multiplying the corresponding terms of an AP and GP is called an Arithmetic Geometric progression eg 1 + 3 + 5 + 7 + is an AP 1 + x + x2 + x5 + is a GP Multiplying together the terms of these series we get 1+ 3x + 5x2 +7x3 + which is an Arithmetic Geometric Series

391 Standard Form

ab + (a + d)br + (a + 2d) br2 + + [a + (n + 1)d]brnndash1 n 1

ab dbr(1 r ) [a (n 1)d]S br1 r (1 r) 1 r

minusminus + minus= + minus

minus minus minus

392 Sum to Infinity Terms

When | r | lt 1 2

ab dbrS1 r (1 r)infin = +minus minus

310 (S) SiGmA notAtion

3101 Concept of Continued Sum [Sigma (S) Notation]

Continued Sum Sigma (S) stands for continued sum of indexed terms It is denoted as

general term

where k is called lsquoindex of termrsquo and it varies from 1 to n (where maximum value of k is n and minimum value of k is 1) thus indicating n number of terms in the series

rArr n

k 1 2 3 nk 1

a a a a a=

= + + + +sum rArr n

=sum as the general term is independent of k

a a a a a n times na=

= + + + + =sum

rArr A constant factor from the general term can be factored out of sigma notation ie n n

k kk 1 k 1

a a= =

λ = λsum sum

LHS = la1 + la2 + la3 + hellip + lan = l(a1 + a2 + a3 + hellip + an) = n

rArr Sigma is distributive over addition and subtraction of terms ie n n n

k k k kk 1 k 1 k 1

(a b ) a b= = =

plusmn = plusmnsum sum sum

LHS = (a1 plusmn b1) + (a2 plusmn b2) + hellip + (an plusmn bn) = (a1 + a2 + hellip +an) plusmn (b1 + b2 +hellip +bn) = n n

k kk 1 k 1

a b s= =

plusmnsum sum rArr Sigma does not distributes on multiplication and division of terms

k k k kk 1 k 1 k 1

a b a b= = =

nesum sum sum Similarly

knk k 1

nk 1 k

sumsum

Application of Sigma The concept of sigma is used to find sum of series whose general term is given or known For example let general term of a series be Tn = an2 + bn + c

Sn = T1 + T2 + T3 + +Tn = n

T=sum = 2

kak bk c

+ +sum = n n n

k 1 k 1 k 1

a k b k c 1= = =

+ +sum sum sum

nn(n 1)(2n 1) n(n 1)S a b cn

6 2+ + + = + +

Usually sum of n terms of any series is represented by placing S before the nth term of the series But

if we have to find the sum of k terms of a series whose nth term is un then this will be represented by k

Shortly S is written in place of Σn

311 ProPertieS

r 1 2 3 n=

= + + + +sum = n(n 1)2+ P2

n2 2 2 2 2

r 1 2 3 n=

= + +sum = n(n 1)(2n 1)6

n(n 1)r2=

+ = sum P4

nr (n 1)(2n 1)(3n 3n 1)30=

= + + + minussum

312 Double SiGmA notAtion

iji 1 j 1

T= =sumsum stands for summation of elements of a two-dimensional array (arrangement) of terms

It can also be regarded as summation of summation of a series rArr General Element The general term is represented by Tij where i denotes the row index

(row position) and j denotes the column index (column position) of the term

row columnindex index

i jT is the element placed in the ith Row and jth column

3121 Representation

iji 1 j 1

sumsum can be represented as a two dimensional array of

numbers on a rectangular matrix with m rows and n columns

For example T14 is element placed in the 1st row and the 4th column T41 is element placed in the 4th row and the 1st column rArr Now consider square matrix of size n times n Elements (terms)

on the principal diagonal are addressed with i = j bull Tij

i lt j is the term that lies above the principal diagonal

Tij i gt j is the term that lies below the principal diagonalTij i = j is the term that lies on the principal diagonal

31211 Conclusion

rArr Total number of squares abovebelow the principal diagonal line

rArr Total number of squares on or below the diagonal = 2 2n n n nn2 2minus +

+ = = n 12

n(n 1) C 2

3122 Properties of Double Sigma

P1 n n n n

j 1 i 1 j 1 i 1

a a= = = =

sumsum sum sum =

i 1 i 1

na na 1= =

=sum sum = nan = n2a ie summation of a a in n2 places on matrix

P2 n n n n n

1 i j n j 2 j 3 j ni 1 i 2 i n 1

a a a ale lt le = = =

= = = minus

= + + +sumsum sum sum sum = (n ndash 1)a + (n ndash 2)a ++ a = a(1 + 2 + 3++ (n ndash 1)) =an(n 1)

2minus

(ie Number of terms above the Principle Diagonal)

P3 2n n

1 i j n

(n n)aa C a2

le le le

+= =sumsum

2n n n(n 1)a na a2 2

minus ++ =

(ie Look at the sum of all the terms on or above the principle diagonal = (number of terms)a = n+1C2a)

P4 n n n n

i j i ji 1 j 1 i 1 j 1

a a a a= = = =

sumsum sum sum let

k 1 2 nk 1

a a a a S=

= + + + =sum = 2n n n

2i i k

i 1 i 1 k 1

Sa S a SS S a= = =

= = = =

sum sum sum

| P5 Sum of Product taken two at a time of any set given n numbers a1 a2 a3 an

= i j 01 i j n

a a S (say)le lt le

=sumsum rArr 2n n

20 i k

i 1 k 1

2S a a= =

sum sum

k kk 1 k 1

=sum sum

k kn nk 1 k 1

i j1 j j n

a aa a

le le le

=sum sum

sumsum = Sum of terms on or above the diagonal

P7 n n n

i j ki 1 j 1 k 1

(a a ) 2n a= = =

sumsum sum Q Each term is written 2n times in the matrix = nS + nS = 2nS

P8 A constant factor can always be factored out of double sigma n n n n

i j i ji i j n i i j n

(a a ) (a a )le lt le le lt le

+ λ = λ +sumsum sumsum P9 Double sigma distributes on sum and difference of element provided the elements are

symmetric in the variable i and j i j i j i j i j0 i j n 0 i j n 0 i j n

(a a a a ) (a a ) (a a )le lt le le lt le le lt le

+ + = + +sumsum sumsum sumsum P10 i j

0 i j n

(a a )le lt lesumsum = Sum of product of n numbers a1 a2an taking two at a time

k kk 1 k 1

i j1 i j n

a aa a

le lt le

=sum sum

sumsum

313 methoDS of Difference

Given a series with nth term unknown eg

1 2 3 4 5 n 1

1 2 3 4 5 6 n 1 n

d d d d d d

t t t t t t t t minus

If the differences of the successive terms (dkrsquos) of a series are in AP or GP then we can find nth term of the series by the following procedureStep 1 Denote the nth term and the sum of the series upto n terms of the series by Tn and Sn respectivelyStep 2 Rewrite the given series with each term shifted by one place to the rightStep 3 Then substract the second expression of Sn from the first expression to obtain Tn

Remarks

(a) Difference of successive terms is constant then nth term is given by Tn = a + bn (where a and b is constant)Sn = S Tn

(b) If difference of difference is constant then Tn = an2 + bn + c (where a b c are constant)

(c) If difference of difference is constant then Tn=an3 + bn2 + cn + d (where a b c d are constant)

314 Vn methoD

A method to find sum of an unknown series whose general term tn is known

eg to compute n1 1 1 1S

1middot2 2middot3 3middot4 4middot5= + + + +

Step 1 Write the general term n1t

n(n 1)=

Step 2 Express tn as difference of two consecutive terms of another series lt vn gt

eg n1 (n 1) n 1 1t

n(n 1) n(n 1) n n 1+ minus

= = = minus+ + +

rArr n n n 11 1t V Vn n 1 += minus = minus

Step 3 rArr 1 1 21 1t V V1 2

= minus = minus rArr 2 2 31 1t V V2 3

= minus = minus

rArr n 1 n 1 n1 1t V V

n 1 nminus minus= minus = minusminus

+rArr n 1 n 1

1 nS V V 1n 1 n 1+= minus = minus =+ +

Chapter 4InequalIty

41 InequalIty contaInIng modulus functIon

Type 1 The inequality of the type f(|x|) lt g(x) is equivalent to the collection (union) of

system lt ge

minus lt lt

f(x) g(x) if x 0f( x) g(x) if x 0

Type 2 The inequality of the form |f(x)| lt g(x) is equivalent to collection (union) of the

systems lt ge

minus lt lt

f(x) g(x) if f(x) 0f(x) g(x) if f(x) 0

Aliter ndashg(x) lt f(x) lt g(x) for g(x) gt 0 and no solution for g(x) le 0

In particular |f(x)| lt a has no solution for a le 0 and for a gt 0 it is equivalent to the

system lt ge

minus lt lt

f(x) a for f(x) 0f(x) a for f(x) 0 or ndasha lt f(x) lt a for a gt 0 and no solution for a le 0

Type 3 The inequation of the form |f(x)| gt g (x) is equivalent to the systems

Aliter f(x) lt ndashg(x) or f(x) gt g(x) for g(x) ge 0 and solution will be the domain set Df of f(x) for g(x) lt 0

In particular |f(x)| gt a has solution x isin domain of f(x) if a lt 0 and for a ge 0 equation is equivalent to

collection (union) of the system gt ge

lt minus lt

f(x) a for f(x) 0f(x) a for f(x) 0

Type 4 The inequation of the form | f (| x |)| gt g (x) or | f (| x |)| lt g(x) is equivalent to the collection

(union) of systems gt ge

minus gt lt

| f (x)| g(x) if x 0| f( x)| g(x) if x 0 or

lt ge minus lt lt

| f (x)| g(x) if x 0| f( x)| g(x) if x 0 respectively

Aliter minus lt lt gt le

g(x) f(| x |) g(x) for g(x) 0Nosolution for g(x) 0 or

minus lt lt gt geminus lt minus lt gt lt le

g(x) f(x) g(x) for g(x) 0 x 0g(x) f( x) g(x) for g(x) 0 x 0

Nosolution for g(x) 0

Inequality 449

Type 5 The inequation of the form | f (x) | ge | g (x) | is equivalent to the collection of system f 2 (x) ge g 2 (x)

Aliter f(x) le ndash|g(x)| or f(x) ge |g(x)| or

lt minus ge lt lt lt lt gt ge gt gt minus lt gt

f(x) g(x) for g(x) 0 f(x) 0f(x) g(x) for g(x) 0 f(x) 0f(x) g(x) for g(x) 0 f(x) 0f(x) g(x) for g(x) 0 f(x) 0

Type 6 The inequation of the form h (x | f (x) |) lt g (x) or h (x | f (x) |) gt g(x) is equivalent to the

collection of systems

lt gt ge ge minus lt minus gt lt lt

h(x f(x)) g(x) h(x f(x)) g(x)if f (x) 0 if f(x) 0

orh(x f(x)) g(x) h(x f(x)) g(x)if f (x) 0 if f(x) 0

42 IrratIonal InequalItIes

The inequalities which contain the unknown under the radical sign There are some standard forms to solve these irrational inequalities

Type 1 The equation of the type lt isin2n 2nf(x) g(x) n is equivalent to the system ge

f (x) 0g(x) f(x) and

inequation of the type + +lt isin2n 1 2n 1f (x) g(x) n is equivalent to the f (x) lt g (x)

Type 2 An inequation of the type lt2n f(x) g(x) n isin ℕ is equivalent to the system

f(x) 0g(x) 0f(x) g (x)

inequation of the type + lt2n 1 f (x) g(x) n isin ℕ is equivalent to the equation f(x) lt g2n + 1(x)

Type 3 An inequation of the form gt isin2n f(x) g(x)n is equivalent to the collection of two systems

of inequations ie ge

g(x) 0f(x) g (x)

g(x) 0f(x) 0 and inequation of the form + gt isin2n 1 f (x) g(x)n is

equivalent to the inequation f (x) gt g 2n + 1 (x)

421 Exponential Inequalities

Type 1 To solve exponential inequation af(x) gt b (a gt 0) we have

(i) x isin Df if b le 0

(ii) if b gt 0 then we have

lt lt lt gt gt = ge isin = lt lt

f (x) log b if 0 a 1f(x) log b if a 1no solution if a 1and b 1x D if a 1and 0 b 1

Type 2 af(x) lt b (a gt 0)

(i) No solution for b le 0 (ii) x isin Df for a = 1 b gt 1 (iii) No solution for a = 1 (iv) f(x) lt logab for b gt 0 a gt 1 (v) f(x) gt logab for b gt 0 0 lt a lt 1

Type 3 An equation of the form f(ax) ge 0 or f(ax) le 0 is equivalent to the system of

collection gt = ge le

xt 0 where t af(t) 0 or f(t) 0

Type 4 An inequation of the form aaf(x) + bbf(x) + gcf(x) ge 0 or aaf(x) + bbf(x) + gc f(x) le 0 when a b g isin R a b g ne 0 and the bases satisfy the condition b2 = ac is equivalent to the inequation at2 + bt + g ge 0 or at2 + bt + g le 0 when t = (ab)f(x)

Type 5 An equation of the form aaf(x) + bbf(x) + g ge 0 or aaf(x) + bbf(x) + g le 0 where a b g isin R and a b g ne 0 and ab = 1

(a b are inverse + ve numbers) is equivalent to the inequation at 2 + gt + b ge 0 or at 2 + gt + b le 0 where t = af(x)

Type 4 If an inequation of the exponential form reduces to the solution of homogeneous algebraic inequation ie aof

n(x) + a1fnndash1(x) g(x) + a2f

nndash2(x) g2(x) + + anndash1 f(x) gnndash1(x) + angn (x) ge 0 when a0 a1 a2an

are constants (a0 ne 0) and f (x) and g (x) are functions of x

rArr minus minus

minus minus+ + + + gen n 1 n 2

0 1 2 nn n 1 n 2

f (x) f (x) f (x)a a a a 0g (x) g (x) g (x)

rArr a0tn + a1t

nndash1 + a2tnndash2 + hellip + an ge 0 where =

f(x)tg(x)

and hence gives n values of t = t1 t2 t3helliptn (say)

rArr = if (x) tg(x)

i = 1 2 3 helliphelliphellip n

rArr f(x) ndash tig(x) = 0 solve for x corresponding to each i

Type (iii) Logh(x) f(x) gt logh(x)g(x)

rArr lt lt lt gt

gt gt gt

0 h(x) 1 f(x) g(x) f(x) 0h(x) 1 f(x) g(x)g(x) 0

Type (iv) Logh(x) f(x) lt a

rArr α

lt lt lt lt

0 h(x) 10 f(x) (h(x))h(x) 1 f(x) (h(x))

422 Canonical Forms of Logarithmic Inequality

(a) gt gt rArr gt gt

alog x 0 x 1a 1 a 1 (b)

gt lt lt rArr lt lt lt lt

alog x 0 0 x 10 a 1 0 a 1

(c) lt lt lt rArr gt gt

alog x 0 0 x 1a 1 a 1 (d)

lt gt rArr lt lt lt lt

alog x 0 x 10 a 1 0 a 1

Inequality 451

423 Some Standard Forms to Solve Logarithmic Inequality

Type 1 Equation of the type

Type Collection of Systems

(a) logg(x) f(x) gt 0 hArr gt lt lt

gt lt lt

f(x) 1 0 f(x) 1

g(x) 1 0 g(x) 1

(b) logg(x) f(x) ge 0 hArr ge lt le

gt lt lt

f(x) 1 0 f(x) 1

g(x) 1 0 g(x) 1

(c) logg(x) f(x) lt 0 hArr gt lt lt

lt lt gt

f(x) 1 0 f(x) 1

0 g(x) 1 g(x) 1

(d) logg(x) f(x) le 0 hArr ge lt le

lt lt gt

f(x) 1 0 f(x) 1

0 g(x) 1 g(x) 1

Type Collection of systems

(a) logf(x) f(x) gt logf(x) g(x) hArr

gt lt gt gt φ gt lt φ lt

f(x) g(x) f(x) g(x)g(x) 0 f(x) 0

(x) 1 0 (x) 1

(b) logf(x) f(x) ge logf(x) g(x) hArr

ge le gt gt φ gt lt φ lt

f(x) g(x) f(x) g(x)g(x) 0 f(x) 0

(x) 1 0 (x) 1

(c) logf(x) f(x) lt logf(x) g(x) hArr

lt gt gt gt φ gt lt φ lt

f(x) g(x) f(x) g(x)f(x) 0 g(x) 0

(x) 1 0 (x) 1

(d) logf(x) f(x) le logf(x) g(x) hArr

le ge gt gt φ gt lt φ lt

f(x) g(x) f(x) g(x)f(x) 0 g(x) 0

(x) 1 0 (x) 1

424 Inequalities of Mean of Two Positive Real Numbers

If a and b are two positive real numbers then AM ge GM ge HM ie +

ge ge+

a b 2abab2 a b

Remarks

(i) AM gt GM gt HM if a ne b (ii) AM = GM = HM if a = b

425 Inequality of Means of n Positive Real Number

If = sum ixA

n = AM of x1 x2 x3 hellipxn

= prod

G x = GM of x1 x2 x3 hellipxn

= HM of x1 x2 x3 hellipxn then A ge G ge H

Remark

(i) A gt G gt H iff x1 x2 x3 hellipxn are not all equal

(ii) A = G = H iff x1 = x2 = x3 =hellip= xn

43 theorem of weIghted mean

Theorem of weighted mean implies + ++ + +ge

+ + +31 2 n 1 2 n

1mm m m m m m1 1 2 2 n n

1 2 3 n1 2 n

m a m a m a (a a a a )m m m

forall ai gt 0

where i = 1 2 3n and mi be +ve real numbers ( )sumgesum prodsum1

i i mi mii

m ge equality holds where airsquos are

equal Here a1 a2 a3 an are positive real numbers and m1m2mn are positive real numbers

431 Theorem

(a) (Inequality of the mean of mth power and mth power of mean) If a and b are two positive real numbers Then

(i) + + gt

mm ma b a b2 2

if m lt 0 or m gt 1 (ii) + + lt

mm ma b a b2 2

if 0 lt m lt 1

(iii) + + =

mm ma b a b2 2

if m = 0 or 1 or a = b

(b) If a1 a2 a3 hellipan are n positive real numbers then

sum summm

i ia an n

if m isin (ndashinfin 0) cup (1 infin) (ii)

sum summm

i ia an n

if m isin (0 1)

sum summm

i ia an n

if m = 0 or 1 or all airsquos are equal

Inequality 453

432 Weighted Power Mean Inequality

If a1 a2an b1 b2bn are two sets of n rationals airsquos are not all equal m isin Q (rational)

Then + + + + + +

gt + + + + + +

mm m m1 1 2 2 n n 1 1 2 2 n n

1 2 n 1 2 n

b a b a b a b a b a b ab b b b b b

when m notin (01) and

+ + + + +lt

+ + + + + +

mm m1 1 n n 1 1 2 2 n n

1 2 n 1 2 n

b a b a b a b a b ab b b b b b

when m isin (0 1) Equality occurs when either a1 = a2 = = an or m isin 0 1

433 Cauchy-Schwarz Inequality

If a1 a2an and b1b2bn are two sets of n real numbers then (a1 b1 + a2b2 + + an bn)2 le

(a12 + a2

2 + + an2) (b2

1 + b22 ++ b2

n) with the equality holding if and only if = = =1 2 n

a a ab b b

434 Tchebysheffrsquos Inequality

If x1 x2 xn and y1 y2 yn are real numbers such that x 1 le x2 lele xn and y1 le y2 le leyn then n(x1y1 + + xn yn) ge (x1 + + xn) (y1 ++ yn) For i ne j xi ndash xj and yi ndash yi are both non-positive or non-negative

For the equality to hold at least one in every pair of xi - xj and yi - yj must be zero This certainly hap-pens if x1 = x2 = = xn or if y1 = y2 = = yn and these are the only possibilities

Corollary If x1xn and y1yn are any real numbers such that x1 le x2 lele xn and y1 le y2 lele yn

then + + + + + ge

1 1 n n 1 n 1 nx y x y x x y y

44 weIerstrass InequalIty

For all ai isin IR + and n gt 1 and ai lt 1 If Sn = a1 + a2 + a3 +an then (1 + Sn) lt (1 + a1) (1 + a2) (1 + a3)

(1 + an) lt minus n

if Sn lt 1 otherwise (1 - Sn) lt (1 - a1) (1 - a2) (1 - a3)(1 - an) lt + n

441 Application to Problems of Maxima and Minima

Suppose that a1 a2 a3 an are n positive variables and k is a constant then

(a) If a1 + a2 + a3 + + an = k (constant) the value of a1 a2 a3an is greatest when a1 = a2 = a3 = = an so that the greatest value of a1 a2 a3 an is (kn)n

(b) If a1 a2 a3an = k (constant) the value of a1 + a2 + a3 + + an is least when a1 = a2 = a3 = = an So the least value of a1 + a2 + a3 + + an is n (k)1n

(c) If a1 + a2 + a3 + + an = k (constant) then as m does not or does lie between 0 and 1 the least or the greatest value of + + + +

1 2 3 n

m m m ma a a a occurs when a1 = a2 = a3 = = an the value in question being n1ndashmkm

(d) If + + + +1 2 3 n

m m m ma a a a = k then according as m does not or does lie between 0 and 1 the greatest or the least value of a1 + a2 + a3 + + an occurs when a1 = a2 = a3 = = an the value in question being n1ndash1mk1m

Theorem 4 If a b are two angles in the 1st quadrant with a given constant sum f then maximum value

of minus φ

α β =1 cossin sin

φ2sin2

and that of φ

α + β =sin sin 2sin2

and it occurs when φ

α =β =2

similar result also holds good for cosine

Theorem 5 If a1a2a3 are n angles each lying between (0p2) whose sum is constant A To find maxi-

mum value of ==

α αsumprodn n

k kk 1K 1

sin and sin Suppose that any two of the angles (say) α1 and α2 are unequal then

if we replace two unequal factors sin α1 and sin α2 in the given product by two equal factors α +α1 2sin

α +α1 2sin2

the value of product is increasing but the sum of angles remains unaltered as long as any

two of the angles are unequal the product is not maxm this indicalies that the product is maxn when all the

angles are equal so each angle is An Therefore=

kk 1 max

Asin sinn

kk 1 max

Asin nsinn

45 use of calculus In ProVIng InequalItIes

451 Monotonicity

A function f is defined on an interval [a b] said to be (a) Monotonically increasing function If x2 ge x1 rArr f(x2) ge f(x1) for all x1 x2 isin [a b] (b) Strictly increasing function If x2 gt x1 rArr f(x2) gt f(x1) for all x1 x2 isin [a b] (c) Monotonically decreasing function If x2 ge x1 rArr f(x2) le f(x1) for all x1 x2 isin [a b] (d) Strictly decreasing function If x2 gt x1 rArr f(x2) lt f(x1) for all x1 x2 isin [a b]

452 Test of Monotonicity

(a) The function f (x) is monotonically increasing in the interval [a b] if fprime(x) ge 0 in[a b] (b) The function f (x) is strictly increasing in the interval [a b] if fprime(x) gt 0 in [a b] (c) The function f (x) is monotonically decreasing in the interval [a b] if fprime(x) le 0 in [a b] (d) The function f (x) is strictly decreasing in the interval [a b] if fprime(x) lt 0 in [a b]

Chapter 5theory of equation

51 Polynomial ExPrEssion

An algebraic expression involving one or more variable that contains two mathematical operations multiplication and raising to a natural exponent (power) with respect to the variablevariables involved

is called lsquomono-nomialrsquo For example 2 22ax bx 3xy x yz3

An expression that involves many such mono-nomials separated by positive sign is known as multinomial

For example 3 2 2 3 3 3ax bx yz cxy z dy z+ + + + etc A multinomial having single unknown variable is called lsquopolynomialrsquo An algebraic expression of

type f(x) = a0 + a1x + a2x2 + a3x

3 +hellip+ anxn is called lsquopolynomialrsquo in variable x provided that the powers of x are whole numbers The numerical constants a0 a1 a2hellip an are known as coefficients

511 Leading TermsLeading Coefficient

The term containing highest power of variable x is called leading term and its coefficient is called lsquoleading coefficientrsquo Because it governs the value of f(x) where x rarr infin

n n 1 n 2 nn 2 n

5111 Degree of polynomials

Highest power of x in the polynomial expression is called degree of polynomial (ie power of x in leading term)

5112 Root of polynomial

Roots are the value of the variable x for which the polynomial expression vanishesGeometrically roots are the x-coordinate of the points where the graph of the polynomial

meets axis of x

52 ClassifiCation of Polynomials

521 Polynomial Equation

When a polynomials expression is equated to zero then it generates corresponding equation Roots of polynomial expression are the solution of its corresponding equation A Polynomial equation of nth degree has exactly n roots not necessarily all real (Because it can be

factorized into exactly n linear factors) Two polynomials are equal if they have same degree and same coefficients corresponding to same

power of x If sum of coefficients of a polynomial equation vanishes then x = 1 is one of its roots If sum of coefficients of odd power term of x is equal to the sum of coefficients of even power term

of x then x = ndash1 is one of its roots

522 Polynomials Identity

If an equation is true for all values of variable for which it is defined then it is called identity 2

Rational identity

x 3x 2 (x 1)(x 2)x 1 x 1minus + minus minus

ax3 + bx2 + cx + d = 0 is identity rArr a = b = c = d = 0 If has more number of roots than its degree

Theory of Equation 557

5221 Conclusion

Therefore to prove a nth degree polynomial equation to be an identity there are two ways Either show that number of roots ge n + 1 Show that all the coefficients are zero

NotesIn an identity in x coefficients of similar powers of x on the two sides are equal

Thus if ax3 + bx3 + cx + d = 7x3 ndash 5x2 + 8x ndash 6 be an identity in x then a =7 b = ndash5 c = 8 d = ndash 6

53 Equation stanDarD Equation anD quaDratiC

ax2 + bx + c = 0 is known as quadratic equation if a is non-zero a b c isin R The roots of this equation can be obtained by ax2 + bx + c = 0 (i)

rArr 2b b 4acx

2a 2aminus minus

= plusmn (b2 ndash 4ac = D is known as Discriminant of quadratic)

531 Quadratic Equation

Consider the quadratic expression y = ax2 + bx + c (a ne 0) and a b c are real numbers Thus y = ax2 + bx + c

= 2 b ca x 2 x2a a

b b c ba x 2 x2a 4a a 4a

+ + + minus

b 4ac ba x2a 4a

minus + +

rArr 2D by a x

4a 2a + = +

Where D = b2 ndash 4ac is the discriminant of the quadratic equation shifting the origin at

(- b2a - D4a) ie substituting bX x2a

and DY y4a

The parabola opens upwards or downwards as a gt 0 or a lt 0

54 naturE of roots

1 If a b c isin R and a ne 0 then (a) If D lt 0 then roots of equation (i) will be non-real complex conjugate

(b) If D gt 0 then the roots of equation (i) are real and distinct namely b D2a

minus +α = b D

2aminus minus

and then ax2 + bx + c = a(x - a) (x - b) (ii) (c) If D1 and D2 are discriminants of equation a1x2 + b1x + c1 = 0 (i) a2x2 + b2x + c2 = 0 (ii) Case I D1 + D2 ge 0 then (i) At least one of D1 or D2 ge 0 (must be greater than zero) (ii) If D1 lt 0 then D2 gt 0 and if D1 gt 0 then D2 lt 0 ie at least one of equation has both

roots real and distinct (d) If D1 + D2 lt 0 then (i) at least one of D1 and D2 lt 0 (ii) If D1 lt 0 then D2 gt 0 and if D1 gt 0 then D2 lt 0 (must be less than zero) ie at least one of equations has both roots imaginary (ie complex conjugates) (e) If D = 0 then equation (i) has real and equal rootsa + b = -b2a and then

ax2 + bx + c = a (x - a)2 (iii) 2 If a b c isin Q and D is a perfect square of a rational number then the roots are rational and in case

it is not a perfect square then the roots are irrational Conjugate Roots 3 If a b c isin R and p + iq is one root of equation (i) (q ne 0) then the other must be the conjugate p ndash iq

and vice versa (p q isin R and i = radicndash1) Irrational Roots

4 If a b c isin Q and p + q is one root of equation (i) then the other must be the conjugate p qminus

and vice versa (where p is a rational and q is irrational) 5 ax2 + bx + c = 0 equiv a(x - a) (x - b) (if a and b are roots of the equation) Q a ne 0 dividing both sides of the equation by a and comparing the coefficient a + b = - ba and ab = ca

rArr a - b (difference of roots) = radicDa 6 If the equation ax2 + bx + c = 0 has more than two roots then its degree then it will becomes an

identity and this implies a = b = c = 0 7 If a = 1 and b c are integers and the root of equation (i) are rational numbers ie D gt 0 and perfect

square then these roots must be integers Q a + b = - ba isin I and ab = ca isin I a and b must be integers 8 If a + b + c = 0 and a b c are rational then 1 is a root of the equation (i) and roots of the

equation (i) are rational

541 Formation of Quadratic EquationA quadratic equation whose summation of roots is S and product of roots is P can be written as x2 - Sx + P = 0 Hence a b be the roots of equation ax2 + bx + c = 0 then to obtain the equation whose roots are (i) 1a 1b (ii) -a -b (iii) ka kb (iv) - 1a -1b (v) pa + q pb + q

We proceed as below

Since a + b = -ba and ab = ca and the equation whose summation of roots is S and product of roots is P can be written as x2 ndash Sx + P = 0 Therefore

(i) S β+α=

αβ P = 1ab rArr 2x β+α

minusαβ

x + 1ab = 0

rArr abx2 - (b + a) x + 1 = 0 rArr cx2 + bx + a = 0 (The reciprocal equation of ax2 + bx + c = 0 can be obtained by replacing x with 1x in the

later equation ie i interchanging the coefficients of equidistant terms from beginning and end)

(ii) S = -(a + b) P = ab rArr x2 + (a + b)x +ab = 0 rArr ax2 - bx + c = 0 (The equation whose roots are negative of the roots of equation ax2 + bx + c = 0 can be obtained by

replacing x with ndashx is the ax2 + bx + c = 0) (iii) If a b g are roots the roots of cubic equation then the equation is x3 ndash (a + b + g) x2 +

(ab + bg + ag) x ndash abg = 0

542 Sum and Product of the Roots

Since a + b = - ba and ab = ca are the sum and product of the equation x2 ndash Sx + P = 0 where a and b are the roots of this equation

55 ConDition for Common roots

(i) One roots to be common Consider two quadratic equations ax2 + bx + c = 0 and aprimex2 + bprimex + cprime = 0 (where aaprime ne 0 and abprime ndash aprimeb ne 0) Let a be a common root then aa2 + ba + c = 0 (i) and aprimea2 + bprimea + cprime = 0 (ii)

Solving the above equations we get 2 1

bc b c ca c a ab a bα α

= =prime prime prime prime prime primeminus minus minus

From first two relations we get bc b cca c aprime primeminus

α =prime primeminus

and from last two relations we get ca c aab a bprime primeminus

eliminating a we get bc b cca c aprime primeminusprime primeminus

= ca c aab a bprime primeminusprime primeminus

rArr 2(bc b c)(ab a b) (ca c a)prime prime prime prime prime primeminus minus = minus or

rArr 2a b b c c a

a b b c c atimes =

prime prime prime prime prime prime (Remember) this is the required condition for one root of two

quadratic equation to be common (ii) Both roots to be common

If a + b = ndashba = ndashbprimeaprime and ab = ca = cprimeaprime ie a b ca b c= =prime prime prime

this is the required condition for both

roots of two quadratic equations to be identical

NoteTo find the common root between the two equations make the same coefficient of x2 in both equations and then subtract the two equations

Detail Analysis of Quadratic Equation If b2 ndash 4ac gt 0 then

Coefficients Graphs Analysis of Nature of Roots

a gt 0 b gt 0 c gt 0a gt 0 b gt 0 c = 0a gt 0 b gt 0 c lt 0

a + b lt 0 ab gt 0a + b lt 0 ab = 0a + b lt 0 ab lt 0

Both roots are negativeOne root is ndashve and the other is zeroRoots are opposite in sign and magnitude of negative root is more than the magnitude of positive root

a gt 0 b lt 0 c gt 0a gt 0 b lt 0 c =0a gt 0 b lt 0 c lt 0

a + b gt 0 ab gt 0a + b gt 0 ab = 0a + b gt 0 ab lt 0

Both roots are positiveOne root is +ve and the other is zeroRoots are opposite in sign and magnitude of positive root is more than the magnitude of negative root

56 symmEtriC funCtion of thE roots

A function of a and b is said to be a symmetric function if it remains unchanged when a and b are interchanged

In order to find the value of a symmetric function of a and b express the given function in terms of a + b and ab The following results might be useful 1 a2 + b2 = (a + b)2 ndash 2ab 2 a3 + b3 = (a + b)3 ndash 3 ab (a + b)

3 a4 + b4 = (a3 + b3) ndash (a2 + b2) -2a2b2 (a2 +b2) 4 2( ) 4αminusβ = α+β minus αβ

5 (a3 - b3) = (a + b) [(a - b)2 - ab] 6 (a4 ndash b4) = (a + b) (a - b) (a2 + b2) 7 a5 + b5 = (a2 + b2) (a2 + b2) ndasha2b2(a + b)

561 MaximumMinimum Value and Sign of Quadratic Equation

Extreme value of any quadratic expression y = ax2 + bx + c is given by y-coordinate of vertex of corresponding parabola and it occurs at x-coordinate of vertex

(i) For a gt 0 The curve y = ax2 + bx + c is a parabola opening upwards

such that minD by at x

4a 2aminus minus

= = and ymax rarr infin

(ii) For a lt 0 The curve y = ax2 + bx + c is a parabola opening downward such that

maxD by at x

4a 2aminus minus

= = and ymin rarr ndash infin

57 loCation of roots

Let f(x) = ax2 + bx + c where a b c isin R be a quadratic expression and k k1 k2 be real numbers such that k1 lt k2 and if a b be the roots of equation

f(x) = 0 Then b D2a

minus minusα = and b D

2aminus +

β = where D is the discriminant

of the equation

(a) Conditions for a number k to lie between the roots of a quadratic equation OR under what condition do the roots of akx2 + bx + c = 0 lie on either side of number k

If a number k lies between the roots of a quadratic equation f(x) = ax2 + bx + c = 0 then the equation must have real roots and the sign of f(k) must be opposite to the sign of lsquoarsquo as is evident from the

(i) D gt 0 and (ii) a f(k) lt 0 (b) Condition for both the roots of a quadratic equation to lie

between numbers k1 and k2 or in the interval k1lt x lt k2

If both the roots aand b of a quadratic equation lie between number k1 and k2

(i) D gt 0 (ii) a f(k1) gt 0 a f(k2) gt 0 and (iii) k1 lt ndashb2a lt k2 (c) Conditions for a number k to be less than the roots of a

quadratic equation or under what condition will both roots of ax2 + bx + c = 0 be greater than a certain specified number k

Thus a number k is smaller than the roots of a quadratic equation ax2 + bx + c = 0 iff (i) D gt 0 (ii) a f(k) gt 0 (iii) k lt ndashb2a

(d) Condition for exactly one root of a quadratic equation to lie in the interval (k1 k2) where k1 lt k2

If exactly one root of the equation ax2 + bx + c = 0 lies in the interval (k1 k2) then the equation ax2 + bx + c = 0 must have real roots and f(k1) and f(k2) must be of opposite signs Thus exactly one root of the equation ax2 + bx + c = 0 lies in the interval (k1 k2) if

(i) D gt 0 (ii) f(k1) f(k2) lt 0

(e) Condition for a number lsquokrsquo to be more than the roots of a quadratic equation

If a number k is more than the roots of a quadratic equation ax2 + bx + c then (i) D gt 0 (ii) a f(k) gt 0 (iii) k gt ndashb2a

58 DEsCartEs rulE

Step 1 To check at most positive roots in f(x) = 0 Check change in sign = most positive roots eg f(x) = x9 + 5x8 ndash x3 + 7x + 2 = 0 There are 2 changes in sign at most 2 positive roots

Step 2 Check at most negative roots in f(x) = 0 The numbers of changes in sign = most negative roots eg f(x) = x9 + 5x8 - x3 + 7x + 2rArr f(- x) = - x9 + 5x8 + x3 - 7x + 2 There are 3 changes in sign at most 3 negative roots

581 Some Important Forms of Quadratic Equations

An equation f(x) = 0 cannot have more positive roots then there are changes of sign in f(x) and cannot have more negative roots than there are changes of sign in f(ndashx)

1 An equation of the form (x ndash a) (x ndash b) (x ndash c) (x ndash d) = A where a lt b lt c lt d a + b = c + d can be solved by a change of variable

ie (x a) (x b) (x c) (x d)y4

minus + minus + minus + minus= or (a b c d)y x

4+ + +

= minus

2 Equation of type (x ndash a) (x ndash b) (x ndash c)(x ndash d) = Ax2 where ab = cd can be reduced to a collection

of two quadratic equations by a change of variable aby x4

3 An equation of the form (x - a)4 + (x - b)4 = A can also be solved by a change of variable

ie making a substitution (x a) (x b)y2

minus + minus=

4 A reciprocal equation of the standard form can be reduced to an equation of half of its dimensions

5 An equation of the form af(x) + bf(x) = c where a b c isin R and a b c satisfies the condition a2 + b2 = c then solution of the equation is f(x) = 2 and no other solution of the equation is possible

582 Position of Roots of a Polynomial Eqution

(a) If f(x) = 0 is an equation and a b are two real numbers such that f(a) f(b) lt 0 Then the equation f(x) = 0 has at least one real root or an odd number of real roots between a and b (b) If f(a) f(b) gt 0 then either no real root or an even number of real roots of f(x) = 0 lies

between a and b

59 Equation of highEr DEgrEE

The equation a0 + a1x + a2x2 + + an xn (an 0) when a0a1a2an are constant but an ne 0 is a polynomial of digree n a1a2an an be n roots then

a1 + a2 + a3 + + an = 1

aaminus a1a2 + a2a3 + a3a4 + + an- 1an = 2

a1a2 a3 an = 1

aaminus

rArr Cubic and Biquadratic

Tips and TricksThe truth of the following statements will be readily admitted

1 If all the coefficients are real then the imaginary roots occurs in pairs (ie number of complex roots is always even)

2 If the degree of a polynomial equation is odd then the number of real roots will also be odd It follows than at least one of the roots will be real

3 Every equation of an odd degree has at least one real root whose sign is opposite to that of its last term

4 Every equation which is of even degree and has its last term negative has at least two real roots one positive and one negative

6 If a b c k are roots of the equation f(x) = 0 then = + + + +minus minus minus minus

f ( x ) f ( x ) f ( x ) f ( x )f ( x )

x a x b x c x k

8 If the coefficients are all positive the equation has no positive root Thus the equation x5 + x3 + 2x +1 = 0 cannot have a positive root

9 If the coefficients of the even powers of x are all of one sign and the coefficients of the odd powers are all of the contrary sign the equation has no negative root Thus thee quation x7 + x5

ndash 2x4 + x3 ndash 3x2 + 7x ndash 5 = 0 cannot have a negative root

10 If the equation contains only even powers of x and the coefficients are all of the same sign the equation has no real root Thus the equation 2x6 + 3x4 + x2 + 7 = 0 cannot have a real root

11 If the equation contains only odd powers of x and the coefficients are all of the same sign the equation has no real root except x = 0 Thus the equation x9 + 2x5 + 3x3 + x = 0 has no real root except x = 0

12 If there is no change in sign then all the roots are imaginary

13 If in the polynomial of degree n the maximum number of possible positive real roots is k1 and maximum number of possible negative real roots is k2 and zero is not the root of polynomial then the minimum number of complex roots will be equal to n ndash (k1 + k2)

Chapter 6permutation and

Combination

61 introduction

Permutations and combinations is the art of counting without counting ie we study various principles and techniques of counting to obtain the total number of ways an event can occur without counting each and every way individually

62 Fundamental PrinciPles oF counting

621 Addition Rule

If an event (operation) E1 can occur in n1 ways E2 can occur in n2 ways hellip and Ek can occur in nk ways (where k ge 1) And these ways for the above events to occur are pair-wise disjoint then the number of

ways for at least one of the events (E1 E2 E3 hellip or Ek) to occurs is (n1 + n2 + n3 + hellip + nk) = i k

bull An equivalent form of above rule using set-theoretic terminology is given belowLet A1 A2 hellip Ak be any k finite sets where k ge 1 If the given sets are pairs wise disjoint

ie Ai cap Aj = f for i j = 1 2 hellip k i ne j then k k

i 1 2 k ii 1i 1

A | A A A | | A |==

= cup cup cup =sum

where |Ai|

denotes the number of elements in the set Ai

622 Multiplication RuleIf an event E can be decomposed into n ordered event E1 E2 hellip Er and that there are n1 ways for the event E1 to occurs n2 ways for the event E2 to occur hellip nr ways for the event Er to occur Then the total number

of ways for the event E to occur is given by n(E1 and E2 and hellip and Er) = r

1 2 r ii 1

n n n n=

times times times =prod

bull An equivalent form of (MP) using set-theoretic terminology is stated belowr

i 1 2 ri 1

A A A A=

= times times timesprod = (a1 a2 hellip an) | ai isin Ai i = 1 2 hellip r denote the cartesian product of the

finite sets A1 A2 Ar Then r r

i 1 2 r ii 1 i 1

A | A | | A | | A | A= =

= times times times =prod prod

Permutation and Combination 1665

bull And stands for intersection (cap) or multiplication

bull Or stands for union (cup) or addition

bull Both addition and multiplication rules can be extended to any finite number of mutually exclusive operations

623 Complementation Rule

If A and A are two complementary sets and S be universal set thenQ ( )n(A) + n A = n(S) rArr = minusn(A) n(S) n(A)

So we count n(A) or n(A) whichever is easier to count then subtract from n(S) to get the other

624 Principles of Inclusion-Exclusion

Let X be a finite set of m elements and x1 x2 x3 xr be some properties which the elements of X may or may not have if the subset of X having the property xi (where i = 1 2 3 r) is Xi and those having both

properties xi and xj is denoted by i jX Xcap and so on

Then the number of elements of X which have at least one of the properties x1 x2 x3 xr is given

= S1 - S2 + S3 - S4 + + (-1)rndash1Sr and the number of elements of U which have none of the

properties x1 x2 x3 xr is given byr

n X m=

- S1 - S2 + S3 - S4 + + (-1)rndash1Sr where

1 ii 1

S n(X )=

=sum r r

2 i j1 i 1 r

S n(X X )le lt le

= capsumsumeg For r = 2 n(X1 cup X2) = n(X1) + n(X2) ndash n(X1 cap X2)For r = 3 n(X1 cup X2 cup X3) = n(X1) + n(X2) + n(X3) ndash n(X1 cap X2) ndash n(X1 cap X3) ndash n(X2 cap X3) +

n(X1 cap X2 cap X3)

625 Injection and Bijection Principles

Suppose that a group of n students attend a lecture in a lecture theater which has 100 seats assuming that no student occupies more than one seat and no two students share a seat if it is known that every student has a seat then we must have n le 100 If it known furthermore that no seat is vacant then we are sure that n = 100 without actually counting the number of students

6251 Injection principle (IP)

Let A and B be two finite sets if there is an injection from A to B then |A| le |B|

6252 Bijection principle (BP)

Let A and B be two finite sets if there is a bijection from A to B then |A| = |B|

63 combinations and Permutations

Each of the groups or selections which can be made by taking some or all of a number of things without considering the order in which the objects are taken is called a combination Whereas a selection of objects where the order in which the objects are taken is also taken into account is called as an arrangementpermutation

To understand the concept of combination and permutation let us consider the combinations which can be made by taking the letters from a b c d two at a time namely

Combinations (total no 6)

Permutations (totalnumber 12)

ab ac ad da ca babc bd bd cb

Number of combinations of lsquonrsquo distinct objects taken r at a time denoted as nr

nCr(n r)

=minus

NoteFrom the above illustration it is simply clear that in combinations we are only concerned with the number of things each selection contains without taking into account the order in which the objects are being selected (ie ab and ba are regarded as same selection) Whereas in permutation the order of objects is taken into account

64 Permutation oF diFFerent objects

Case I When repetition of objects is not allowedNumber of permutation of n distinct things taken r at a time (0 le r le n) is denoted by nPr and it is equivalent to filling up of r vacancies by n different person clearly first place can be filled in n ways and after which 2nd place can be filled in (n -1) ways and 3rd place can be filled in (n - 2) ways and similarly rth place can be filled in (n - r + 1) ways

rArr nPr = n(n - 1) (n - 2) (n - r + 1) = n(n r)minus

nr C rr(n r)

= timesminus

Case II When repetition of objects is allowedNumber of permutation =

n n n ntimes times times times = nr because now each of the vacancies can be filled

up in n ways

bull The word indicating permutation are arrangement standing in a line seated in a row problems on digits word formation rank of word number of vectors joining given points and number of greetings sent among a group etc

bull The number of permutations of n distinct objects taken all at a time = n

bull The number of all permutations of n different object taken r at a time when a particular set of k objects is to be always included in each arrangement is r nndashkCrndashk

bull Number of permutations of n different things taken all at a time when r specified things always remain together is r(n ndash r +1)

bull Number of permutations of n different things taken all at a time when r specified things never occur together is n ndash r(n ndash r +1)

bull The number of permutations of n different things taken all at a time when no two of the r particular things come together is nndashr+1Cr (n ndash r) r

65 Permutation oF identical objects (taking all oF them at a time)

Number of permutations (N) of lsquonrsquo things taken all at a time when lsquoprsquo are of one kind lsquoqrsquo of a second kind

lsquorrsquo of a third kind and so on is given by nNpqr

Explanation let N be the required number of permutations From any of these if the p like things were different we could make p new permutations Thus if the p like things were all different we would have got N(p) new permutations Similarly if the q like things were different we would get N (q) new permutations from each of the second set of permutations

Thus if the p like things and the q like things were all different we would have got Npq permutations in all The process is continued untill all the sets of like things are different and we then get the number of permutations of n things taken all at a time when they are all different (which is n)

Npqr n= rArr nNpqr

66 rank oF words

When all the letters of a word are arranged in all possible ways to form different words and the words formed are further arranged as per the order of ordinary dictionary then the position occupied by that word is called as its rank eg rank of the word MAT is 3 because it occupied third position in the alpha-betical list (AMT ATM MAT MTA TAM TMA) of words formed using letters A M T

Shortcut to Find Rank of a Word Example Banana Example Large

1 Write the letters of the word in alphabetical order

AAABNN A E G L R

2 Pick the letters one-by-one in the order in which they are heard while speaking

B A N A N A L A R G E

3 For each of the letters in this order using representation in Step (1) find

number of letter in left on nx

p q are number of identical letters

Cross the letters as done with it

13x AAABNN

32= rarr

x2 = 0 rarr AANN

32x AAN

22= rarr

x4 = 0x5 = 1 rarr A

1x 3 AEGER= rarr

2x 0 AEGER= rarr

3x 2 EGR= rarr

x4 = 1

Rank = (x1)5 + (x2)4 + (x3)3 + (x4)2 + x5 + 0 5 3 3 2 1 1 0 3432 22times times

+ + times + =Rank 3 times 4 + 0 times 3 + 22 + 11 + 0 = 78

67 circular Permutation

The arrangement of objects around a circle is called lsquocircular permutationrsquo Two circular permutations are called identical iff one of them can be super imposed on the other by a suitable rotation without overturn-ing and without changing the relative position of object eg following 5 circular permutations are identical

671 Circular Permutation of n ObjectsWhen lsquonrsquo distinct objects (A1 A2 A3 An) are to be arranged around a circle then each circular arrangement generates lsquonrsquo number of distinct linear arrangements by rotating the objects around the

circle by 0360

at a time (keeping their relative position fixed)

rArr Each circular array generates lsquonrsquo linear permutation Let the total number of circular array be x

rArr Number of linear arrays = nx rArr nx = n rArr nx = n 1n

= minus

Remark bull As in circular permutation (unlike linear permutations) there is no initial and terminal position

therefore fixing the position of one object around the circle its position acts as a terminal consequently the remaining (n ndash 1) positions become as distinct as in linear permutations Therefore rest of (n ndash 1) object can be arranged in these position in (n ndash 1) ways

ExplanationIn a circular permutation the relative position among the things is important whereas the place of a thing has no significance Thus in a circular permutation the first thing can be placed anywhere This operation can be done only in one way then relative order begins Thus the ways for performing remaining parts of the operation can be calculated just like the calculation of linear permutation for an example to place 8 different things round a circle first we place any one thing at any place there will be only one numbers of ways = 7 Thus required number of circular permutations if 7

bull Since each circular arrangement has its unique counter-clockwise arrangement therefore the number

of clockwise array = number of counter-clockwise arrays = minus( n 1)2

bull In a garland of flowers or a necklace of beads (since the overturning of permutations is possible) It is difficult to distinguish clockwise and anti-clockwise orders of things so a circular permutation under both these orders (the clockwise and anti-clockwise) is considered to be the same

Therefore the number of ways of arranging n beads along a circular wire is minus( n 1)2

bull The total number of circular arrangements of n distinct objects taken r at a time is

(i) minus =n

PC ( r 1)

r when clockwise and anticlockwise orders are treated as different

(ii) minus =n

P1C ( r 1)

2 2r when the above two orders are treated as same

68 number oF numbers and their sum

Case I Number of r digit numbers formed using n digits D1 D2 Dn when repetition allowed bull Number of numbers = nr

bull Sum of all numbers = r r

10 1 D n9

Proof When all the numbers formed are arranged vertically for summation Any digit gets repeated nrndash1 times in each column keeping a particular digit say Dk

fixed at some place out of r then remaining (r ndash 1)

places can be arranged using n digits in nrndash1 ways

Summation of digits in any column = sum of all digits times repetition of digit ( )n

D n minus

rArr Sum of all numbers ( )n

D n minus

sum (1 + 10 + 102 + hellip + 10rndash1)

Case II Number of r digit numbers formed using n digits D1 D2 Dn when repetition not allowed

bull Number of numbers = n

rP if r n0 if r n

n 1k r 1

10 1 D P9

minusminus

Proof When all the numbers formed are arranged vertically for summation Any digit gets repeated nndash1Prndash1

times in each column keeping a particular digit say Dk

fixed at some place out of r

then remaining (r ndash 1) places can be arranged using n ndash 1 digits in nndash1Prndash1 ways

Summation of digits in any column = sum of all digits times repetition of digit = minusminus

k r 1k 1

rArr Sum of all numbers = ( )n

n 1k r 1

D Pminusminus

sum (1 + 10 + 102 + hellip + 10rndash1) =

r rn 1

k r 1k 1

10 1 D P9

minusminus

681 Divisor of Composite NumberA natural number x = pa qb rg is called divisor of N = pa qb rc iff N is completely divisible by x For Example when all the prime factors of x are present in N which is possible only if 0 le a le a 0 le b le b and 0 le g le c where a b g isin ℕ cup 0

bull Set of all divisors of N is given as x x = pa qb rg where 0 le a le a 0 le b le b 0 le g le c

bull Number of divisor number of divisors = n(a b g) 0 le a le a 0 le b le b 0 le g le c= na 0 le a le a times nb 0 le b le b times ng 0 le g le c = (a + 1) (b + 1) (c + 1)

bull Number of divisors are given by number of distinct terms in the product= (1+ p + p2 + + pa) (1+ q + q2 + + qb) (1+ r + r2 + + rc)= (a + 1) (b + 1) (c + 1) (which includes 1 and the N it self)

682 Sum of Divisor

Since each individual divisor is given as terms of the expansion (p0 + p1 + p2 + + pa) (1+ q + q2 + + qb) (1+ r + r2 + + rc) therefore the sum of all divisors is = 1 + p + q + r + p2 + q2 + r2 + pq + pr ++ pa qb rc

= a 1 b 1 c 1p 1 q 1 r 1p 1 q 1 r 1

+ + + minus minus minus minus minus minus

Notes bull ImproperProper divisors of N = pa qb rc When a = b = g = 0

rArr x = 1 which is divisor of every integer and a = a b = b and g = c then x becomes number N itself These two are called lsquoimproper divisorrsquo

rArr The number of proper divisors of N = (a + 1)(b + 1)(c + 1) ndash 2

bull If p = 2 then number of even divisors = a(b + 1)(c + 1) number of odd divisors = (b + 1)(c + 1)

683 NumberSum of Divisors Divisible by a Given NumberIf x = pa qb rg is divisor of N = pa qb rc and completely divisible by 1 1 1y = p q r α β γ

bull Set of all divisors of N is given as x x = pa qb rg where a1 le a le a b1 le b le b g1 le g le c rArr Number of divisors = n(a b g) a1 le a le a b1 le b le b g1 le g le c = (a ndash a1 + 1) (b ndash b1 + 1)

(c ndash g1 + 1)

684 Factorizing a Number into Two Integer FactorsIf x and y be two factors of the Natural Number N = pa qb rc N = xy

rArr x and y are divisors of N

Case I If number N is not a perfect square

Number of two factor products (number of total divisors)= 2

Case II If number N is a perfect square

Number of two factor products (number of total divisors) + 1= 2

Case III Number of integer solution of equation xy = pa qb rc sd = 2 times total number of divisor

Since number of natural number solution of the equation

xy = pa qb rc sd = Number of divisors = (a + 1) (b + 1) (c + 1) (d + 1)

rArr Number of integer solution of the equation = 2(a + 1) (b + 1) (c + 1) (d + 1)

69 combination

Combination of n objects taken r at a time is denoted as nCr and defined as nr

nCr(n r)

=minus

691 Properties of Combinations

1 The number of combination of n different things taken r at a time is denoted by nCr or C(n r)

and it is empirically calculated as =minus

r(n r) (0 le r le n) where n isin N and r isin W

whole numbers = 0 (if r gt n) 2 nCr is always an integer The following important conclusions can be made out of the above statement (a) Product of r consecutive integers is always divisible by r

∵ nr

n(n 1)(n 2)(n 3)(n r 1)C Ir

minus minus minus minus += isin

Clearly the numerator is completely divisible by r

(b) 0 = 1 n n0 n

nC C 1

n 0 = = =

and nC1 = n

(c) k = infin if k lt 0 (Think why) 3 nCr = nCnndashr this is simply selection of r things means rejection of n ndash r at the same time 4 nCx + nCy rArr x = y or x + y = n 5 nCr + nCrndash1 = n+1Cr (1 lt r lt n) this is also known as Pascal Rule

6 rnCr = nnndash1Crndash1 rArr n n 1 n 2

r r 1 r 2n n n 1C ( C ) C r r r 1

minus minusminus minus

minus = = = minus Thus we can work out as

Choosing r MPrsquos from n citizens (nCr ways)Choosing 1 PM from r Choosen MPrsquos (r ways)r times rCr waysMP Member of Parliament

equivChoosing 1 PM from n citizens (n ways) and Choosing remaining (rndash1) MPrsquos fromremaining (n ndash 1) citizens (nndash1Crndash1) waysMP Prime Minister

7 n n n 2r r 1 r 2

r 1 (r 1(r 2)C C Cn 1 (n 1)(n 2)

+ + + = = + + +

8 nCr rCs = nCs

nndashsCrndashs (n ge r ge s) This we can work out as

Choosing r MPrsquos (nCr ways) andChoosing s ministers out of rMPrsquos (rCs ways) nCr times rCs

equivChoosing s ministers (rCs ways) andChoosing remaining (r ndash s) MPrsquos out ofremaining (n ndash s) citizens nCs times nndashsCrndashs

C n r 1C rminus

minus +=

10 nC0 + nC1 + nC2 + + nCn = 2n this is selection of any number of objects out of given n objects For each object we have only two possibilities selection or rejection which is 2n

11 nC0 + nC2 + nC4 + = nC1 + nC3 + nC5 + hellip = 2nndash1 12 nCm + nndash1Cm + nndash2Cm + + mCm = n+1Cm+1

692 Restricted Combinations

The number of combinations of n different things taking r at a time (a) When p particular things are always to be excluded = nndashpCr (b) When p particular things are always to be included = nndashpCrndashp (c) When p particular things are always included and q particular things are always excluded = nndashpndashqCrndashz

693 Combination of Objects Taking any Number of Them at a Time

bull Number of selections of objects when any number of them can be selected is given as nC0 + nC1 + + nCn = 2n

Where nCr corresponds to the case when r objects are selected out of n different objects In above case r varies from 0 to n The right hand side value 2n can be explained as number of ways of dealing with all n objects each in exactly two ways either selected or rejected

bull Number of selection of objects (at least one) out of n different objects n

n n n n nr 1 2 n

C C C C 2 1=

= + + + = minussum

bull Number of selection of atleast two object out of n = 2nndashnC0 ndash nC1

694 Combination when Some Objects are Identical (Taking any Number of Them at a Time)

1 Combination when some objects are identical The total number of ways in which it is possible to make a selection taking some or all out of (p + q + r) things where p are alike of the first kind q are alike of the second kind and r alike of the third kind and s are different = (p + 1) (q + 1) (r + 1) 2s ways

Explanation Out of p alike things we may select none or one or two or three or all p Hence they may be disposed off in (p + 1) ways Similarly q alike things may be disposed of in (q + 1) ways similarly for r And s different things may be disposed of in 2s ways (This includes the case in which all of them are rejected)

bull Number of ways (if at least one object to be selected) = (p + 1) (q + 1) (r + 1) 2s ndash1 bull Number of ways (if at least one from s different object to be selected) = (p + 1) (q + 1)(r + 1) (2s ndash1) bull Number of ways (if at least one object of each identical type lot is to be selected) = (p q r)2s

695 Combination when Some Objects are Identical (Taking specific number of them at a time)

Case 1 If a group has n things in which p are identical then the number of ways of selecting r things

from a group is r

Cminus

=sum or

kk r p

Cminus

= minussum according as r le p or r gt p

Explanation It can be obtained by assuming the selection of k distinct object and rest r ndash k objects identical and taking the values of variable k from 0 to r (or p) whichever is less

For an instance when no object is selected from identical objects (k = 0) then the number of selection = nndashpCr

And when one object is selected from identical objects (k = 2) then the number of selection = nndashpCrndash1Similarly for k = 3 the number of selection = nndashpCrndash2 and so on

(i) The number of ways of selecting r objects out of n identical objects is 1

(ii) The number of ways of selecting any number of objects out of n identical objects is n + 1

Case 2 If there are p1 objects of one kind p2 objects of second kind pn objects of nth kind then the number of ways of choosing r objects out of these (p1 + p2 + + pn) objects

= coefficients of xr in 1 2 np p p2(1 x x )(1 x x )(1 x x )+ + + + + + + + +If one object of each kind is to be included in such a collection then the number of ways of choosing r objects

= coefficients of xr in the product 1 2 np p p2(x x )(x x )(x x )+ + + + + +This problem can also be stated as Let there be n distinct objects x1 xn x1 can be used at the most p1 times x2 at the

most p2 times xn at the most pn times then the number of ways to have r things

Renarks bull Given n distinct points in a plane no three of which are collinear then the number of line segments

they determine is nC2

bull The number of diagonals in n-polygon (n sides closed polygon) is nC2 ndash n

If in which m points are collinear (m ge 3) then the number of line segments is (nC2 ndash mC2) + 1

bull Given n distinct points in a plane no three of which are collinear then the number of triangles formed = nC3 If in which m points are collinear (m ge 3) then the number of triangles is nC3 ndash mC3

bull Given n distinct points of which no three points are collinear

(i) Number of straight lines = nC2

(ii) Number of triangles = nC3

(iii) Number of quadrilaterals = nC4

(iv) Number of pentagon = nC5

bull Given n points in a plane out of which r of them are collinear Except these r points no other three points are collinear Then number of different geometric figures constructed by joining these points are expressed as below

(i) number of line segments (LS) = nC2

(ii) number of directed line segments vectors (DLS) = nP2

(iii) number of lines formed = nC2 ndash rC2 + 1 or nndashrC2 + (n ndash r)r + 1

(iv) number of triangles formed = nC3 ndash rC3 or nndashrC3 + (n ndash rC2)r + (nndashr) rC2

(v) number of quadrilateral = nC4 ndash (rC4 + (n ndash r) rC3 )) or nndashrC4 + nndashrC3

rC1 + nndashrC2rC2

(vi) number of rectanglessquares formed put of m horizontal lines and n vertical lines such that distance between conjugative line both set of parallel lines is unity

bull Given A1 A2 A3 An are horizontal lines B1 B2 B3 Bm are vertical lines as shown in figure

(i) Number of rectangles = number of ways of

choosing two lines from each set = ( )i j k ln A A and B B

= nC2 times mC2 (ii) Number of square of size k times k = number of

ways of choosing two lines i j jA A + horizontal

line = ( ) ( )i j k j j k

1 i n k 1 j m k

n A A n B B+ +

le le minus le le minus

= (n ndash k) (m ndash k)

(iii) Total number of squares = ( ) ( )k r

n ndash k m ndash k=

=sum where r = min m ndash 1 n ndash 1

610 distribution

6101 Distribution Among Unequal Groups

To find the number of ways in which m + n things can be divide into two groups containing m and n things respectively This is clearly equivalent to finding the number of combinations of m + n things taking m at a time for every time we select a group of m things we leave a group of n things behind

Thus the required number = (m n)mn+

6102 To Find the Number of Ways in Which the m + n + p Things Can be Divided into Three Groups Containing m n p Things Separately

First divide the m + n + p things into two groups containing m and n + p things respectively the number

of ways in which this can be done is m+n+pCm = (m n p)m(n p)+ +

+ And the number of ways in which the group of

n + p things can be divided into two groups containing n and p things respectively is n pn

(n p)C np

+ += Hence

the number of ways in which the subdivision into three groups containing m n p things can be made follows

6103 Distribution Among Equal Groups

When name of groups is not specified If 2m objects are to be distributed among two equal groups

containing m objects each Then it can be done in (m m)m m2

2m(m) 2

because each division it is possible

to arrange the groups into 2 ways without obtaining new distributionExplanation Then we divide the total number of arrangements obtained normally by k where k is |

number of groups among which the objects are distributed If we put n = p = m we obtain3m

but since this include 3 times the actual number of divisions because of the arrangement of groups among them selves therefore the number of different ways in which subdivision into three equal groups can

be made is =3m

mm m 3

6104 When Name of Groups Specified If the name of groups among which the objects are distributed are specified (eg distributing books to students dividing soldiers into regiment distributing students into sections etc) If we put n = p = m

we obtain 3mm m m

bull The number of ways of dividing pq objects among p groups of same size each group containing q

objects = p

(pq)(q) p

bull The number of ways of distributing pq objects among n people each person getting q

objects = p

(pq)(q)

611 multinomial theorem

The expansion of [x1 + x2 + x3 + + xn]r where n and r are integers (0 lt r le n) is a homogenous

expression in x1 x2 x3 xn and given as [x1+ x2 + x3 + + xn]r = 31 2 n1 2 3 n

1 2 3 n

r x x x x

λλ λ λ λ λ λ λ

(where n and r are integers 0 lt r le n and l1 l2 ln are non-negative integers) Such that l1 + l2 + + ln = r (valid only if x1 x2 x3 xn are independent of each other) coefficient of 31 2

1 2 3x x x λλ λ = total number of arrangements of r objects out of which l1 number of x1rsquos are identical l2 number of x2rsquos are identical and

so on = 1 2 3 n

1 2 3 n 1 2 3 n

( ) (r)

λ +λ +λ + λ=

λ λ λ λ λ λ λ λ

6111 Number of Distinct TermsSince (x1 + x2 + x3 + + xn)r is multiplication of (x1 + x2 + x3 + + xn)r times and will be a homogeneous expansion of rth degree in x1 x2 xn So in each term sum of powers of variables must be r

So number of distinct terms will be total number of non-negative integral solution of equation is l1 + l2 + l3 + + ln = r = Number of ways of distributing r identi-cal objects among n persons = number of ways of distributing r balls among n people

= number of arrangements of r balls and n ndash 1 identical separators = (n 1 r)(n 1)rminus +minus

= n+rndash1Cr = n+rndash1Cnndash1

612 dearrangements and distribution in Parcels

Any change in the order of the things in a group is called a derangement If n things are arranged in a row the number of ways in which they can be dearranged so that none of them occupies its original position

is n1 1 1 1 1n 1 ( 1)1 2 3 4 n

minus + minus + minus + minus

bull If r objects go to wrong places out of n thing then (n ndash r) objects go to their original place If Dn rarr number of group and if all objects go to the wrong places and Dr rarr number of ways if r objects go to wrong places out of n then (n ndash r) objects go to correct places

Then Dn = nCnndashr Dt where Dr = r1 1 1 1 1r 1 ( 1)1 2 3 4 r

bull Derangement of a given n-permutations minus

1 2 3 n 1 n

n permutation

P P P P P is an arrangement in which at least one

object does not occupy its assigned position rArr Total number of dearrangements = n ndash 1 bull Let Ai denotes set of arrays when ith objects occupies ith place n(Ai) = (n ndash 1)

rArr n(A1 cup Aj) = (n ndash 2)rArr Number of arrays in which atleast one object occupies its correct place = n(A1 cap A2 cap A3 hellip cap

An) = Σn(Ai) ndash Σn(Ai cup Aj) + Σn(Ai cup Aj cup Ak) ndash hellip + (ndash1)nndash1 n (A1 cup A2 cup A3 hellip cup An)= nC1 (n ndash 1) ndash nC2(n ndash 2) + nC3 (n ndash 3) ndash hellip + (ndash1)nndash1 nCnO

= minusminus + minus +

nn n n ( 1) n1 2 3 n

= minus minus

minus + minus +

n 11 1 1 ( 1)n 1 2 3 n

the total number of dearrangement in which no object occupies its correct place = n ndash n (A1 cap A2 cap A3 hellip cap An)

= minus minus

minus minus + +

n 11 1 1 ( 1)n n 1 2 3 n

= minusminus + minus + +

n1 1 1 1 ( 1)n 1 1 2 3 4 n

= minus

minus + +

n1 1 1 ( 1)n 2 3 4 n

(n 1)separators

bull Number of dearrangement in which exactly r objects occupy their assigned places

minustimes minus minus + minus + minus

choo singr objectsand placing them Arrangingg n r objects so that noneat their correct places oft hem occupies their assigned positions

1 1 1 ( 1)C (n r) 2 3 4 (n r)

613 distribution in Parcels

6131 Distribution in Parcels When Empty Parcels are Allowed The number of ways in which n different objects can be distributed in r different groups (here distributed means order of objects inside a group is not important) under the condition that empty groups are allowed = rn Take any one of the objects which can be put in any one of the groups in r ways Similarly all the objects can be put in any one of groups in r number of ways So number of ways = r r rn times = rn

= coefficient of xn in n (ex)r = r 1

k r nk

( 1) C (r k)minus

minus minussum

6132 When at Least One Parcel is EmptyNumber of distribution when at least one parcel is empty

= n (A1 cup A2 cup A3 hellip cup Ar) Ai is the set of distribution when ith parcel is emptyn(Ai) = (r ndash n)n and n (Ai cap Aj) = (r ndash 2)n = Sn (Ai) ndash S n (Ai cap Aj) + Sn (Ai cap Aj cap Ak) + hellip + (ndash1)rndash1 n (A1 cap A2 cap cap Ar)

= nC1 (r ndash1)n ndash rC2(r ndash2)n + rC3 (r ndash 3)n + helliphellip +(ndash1)rndash1 rCrndash1 = r 1

k 1 r nk

( 1) C (r k)minus

minus minussumThe number of ways in which n different objects can be arranged in r different groups= n r 1

r 1n C+ minusminustimes if empty groups are allowed = n 1

r 1n Cminusminustimes if empty groups are not allowed

The number of ways in which n different things can be distributed into r different places blank roots being admissible is rn

RemarksGiven two sets A = a1 a2 an and B = b1 b2 b3 br then following holds good

(i) n(A times B) = n(A) n(B) = n times r (ii) Number of relation R A rarr B = number of subsets of A times B = 2nr (iii) Number of functions f A rarr B = number of ways of distributing n elements

(objects) of A in to elements (boxes) of B = rn (iv) Number of injective functions f A rarr B = number of permutations of n elements

of A (objects) over r elements of B (places) =r

nP if r n

0 if r n

(v) Number of into (non surjective) functions f A rarr B = number of ways of distributing n elements

(objects) of A into elements (boxes) of B such that atleast one box is empty = r 1

k 1 r nk

( 1) C ( r k )minus

minus minussum

(vi) Number of on-to (surjective) functions f A rarr B = number of ways of distributing n elements (objects)

of A in to elements (boxes) of B such that no box is empty= r 1

k r nk

( 1) C ( r k )minus

minus minussum

614 exPonent oF a Prime in n

Exponent of prime p in n is denoted by Ep (n) where n is natural number so the last integer amongest 1 2(n - 1)n which is divisible by p is [np] p when [n] le x

rArr s

n n nE (n) p p p

= + + +

where s is the largest number such that ps le n lt ps+1

6141 Exponent of Prime lsquoPrsquo in n

Exponent of prime number lsquoprsquo in n is defined as power of p when n is factorized into prime factor using unique factorization theorem and it is denoted as Ep (n)

Theorem The largest natural number divisible by p is less than or equal to lsquonrsquo is given as n pp

Proof Division algorithm as n le p thus there existTwo natural number q and r such that n = pq + r

where 0 le r le p rArr n rqp p= + where r0 1

ple lt

q is called integer part of number np denoted as n randp p

is known as fractional part of

number np denoted as n p

Observe the situation on ℝ number lies

Conclusion ie np

is the quotient in the division of n by p

Theorem The number of natural numbers divisible by p less than or equal to lsquonrsquo is equal to np

rArr The number of natural numbers divisible by p2 less than or equal to lsquonrsquo is equal to 2

Exponent of prime p in n p 2 3

n n nE (n) p p p

= + + +

Chapter 7binomial theorem

71 IntroductIon

We have dealt with expansions of (x + a)2 while dealing with quadratic equations Herein we will study expansions of the form (x + a)n Any power of binomial expression (a + x)2 can be expanded in the form of a series which is obtained the by process of continuous multiplication as shown here (a + x)2 = (a + x) (a + x) = a2 + ax + ax + x2 = a2 + 2ax + x2 which can be explained as the terms of expansion are obtained when any one of two terms a or x are selected from each factor and finally they are multiplied together

72 BInomIal

Any algebraic expression containing two terms is called lsquobinomial expressionrsquo [Bi (two ) + Nomial (terms)] is an expression containing sum of two different terms

721 Binomial Expansion (Natural Index)

Binomial expansion is a polynomial equivalent of powers of a given binomial expression The expressions for (a + x)n has been obtained as (a + x)n = nC0 a

n x0 + nC1 anndash1 x1 + nC2 a

n ndash2 x2 + + nCr an - r xr + +nCn a

bull Where n is a positive integer which is given by n

n n n r rr

(a x) C a xminus

+ =sum and

nn r n n r r

(a x) ( 1) C a xminus

minus = minussum

bull n

n r nr

(1 x) x C=

+ =sum n

n r n rr

(1 x) ( 1) C x=

minus = minussum where n isin I+ is known as index of binomial

and nCr is binomial coefficient) bull nCr are known as binomial coefficients bull n is called index of binomial bull The binomial expansion is homogenous in a and x ie sum of powers of a and x in each term

remains constant and this constant is equal to index of binomial bull Number of distinct terms in the expansion is equal to (n + 1)

bull The equidistant binomial coefficients from beginning and end are equal

bull The number of terms in the expansion (a + x)n + (a ndash x)n will be n2+1 when n is even n 12+ and

when n is odd bull The number of terms in the above expansion (a + x)n ndash (a ndash x)n will be n2 when n is even

and n 12+ when n is odd

73 General term

A general term is known as representative term of binomial and it is (r + 1)th term of the expansion and is given by Tr +1 = nCr a

n - r xr in expansion of (a + x )n

731 rth Term from Beginning

The term nCr xnndashr y r is the ( r + 1)th term from beginning in the expansion of (x + y)n It is usually called

the general term and it is denoted by Tr+1 ie Tr +1 = nCr xnndashr yr

732 kth Term from End

kth term from end in the expansion of (x + y)n = (n - k + 2)th term from beginning

74 mIddle term

The middle term depends upon the value of n

Case I If n is even Then total number of terms in the expansion of (x + y)n is n+1 (odd) So there is only one middle term ie (n2 +1)th term is the middle term ie Tn2 +1= nCn2 x

n2 yn2

Case II If n is odd Then total number of terms in the expansion of (x + y)n is n+1 (even) So there are

two middle terms ie n 12+

th and n 32+

th are two middle terms They are given by n 1 n 1

2 2n 1C x y

minus +

and n 1 n 1

n 2 2n 1

C x y+ minus

75 numBer of terms In expansIons

bull (a + x)n = nC0an + nC1a

nndash1 x + nC2anndash2x2 + hellip + nCnndash1a

1xnndash1 + nCna0xn = n

n n r rr

C a xminus

bull (a ndash x)n = nC0anx0 ndash nC1a

nndash1x + nC2anndash2 x2 + hellip + nCnndash1a(ndashx)nndash1 + nCna0 (ndashx)n =

nr n n r r

( 1) C a xminus

minussum

bull (a + x)n + (a ndash x)n = m

n n 2r 2r2r

2 C a xminus

=sum where

n 2m if n is evenn 1 2m if n is odd

= minus =

Binomial Theorem 1781

rArr Number of terms

n 2 if n is even2m 1

n 1 if n is odd2

++ = +

bull (a + x)n ndash (a ndash x)n = m

n n 2r 1 2r 12r 1

2 C a xminus minus ++

=sum where

n 2m 1 if n is oddn 1 2m 1 if n is even

= + minus = +

n if n is even2m 1

n 1 if n is odd2

76 Greatest term

If Tr and Tr+1 be the rth and (r + 1)th terms in the expansion of (1 + x)n then n r

r 1 rn r 1

T C x n r 1 xT C x r+

minusminus

minus += =

Let numerically Tr+1 be the greatest term in the above expansion Then Tr+1 ge Tr or r 1

T 1T+ ge

n r 1 | x| 1

rminus +

ge to find the value of r ie (n 1)r | x|(1 | x |)

Now substituting values of n and x in (i) we get r le m + f or r le m where m is a positive integer and f is a fraction such that 0 lt f lt 1 In the first case Tm+1 is the greatest term while in the second case Tm and Tm+1 are the greatest terms and both are equal

761 To Find the Greatest Term in the Expansion of (1 + x)n

bull Calculate m = bull If m is integer then Tm and Tm+1 are equal and both are greatest term bull If m is not integer then T[m]+1 is the greatest term where [] denotes the greatest integral part

NoteTo find the greatest term in the expansion of (x +y)n since (x +y)n = xn(1+ yx)n and then find the greatest term in (1+yx)n

77 Greatest coeffIcIent

To determine the greatest coefficient in the binomial expansion of (1 + x)n consider the following

T C n r 1 n 1 1T C r r+

minus + += = = minus

Now the (r + 1)th binomial coefficient will be greater than the rth binomial

coefficient when n 1 1 1r+

minus gt

rArr n 1 r2+

gt (i)

But r must be an integer and therefore when n is even the greatest binomial coefficient is given by the greatest value of r consistent with (i) ie r = n2 and hence the greatest binomial coefficient is nCn2

bull If n is even then greatest coefficient = nCn2 bull If n is odd then greatest coefficients are nC(n ndash 1)2 and nC(n + 1)2

78 propertIes of BInomIal coeffIcIent

The binomial coefficient for general term of the expansion (a + x)n is given as nCr which states the number of ways the term an - r xr occurs in the expansion

781 Properties of nCr

It is defined as number of selections of r objects out of n different objects and is given by

r(n r)=

minus when n gt r (= 0 if n lt r)

bull nCr is always an integer Product of r consecutive integers is always divisible by r

n(n 1)(n 2)(n 3)(n r 1)C Ir

minus minus minus minus += isin (Clearly the numerator is completely divisible by r)

bull nCr = nCnndashr

bull nCx = nCy rArr x = y or x + y = n bull nCr + nCr-1 = n+1Cr

bull ( )n n 1 n 2r r 1 r 2

n n n 1C C Cr r r 1

minus = = minus = helliphelliphelliphellip

bull n n 1 n 2r r 1 r 2

(r 1)(r 2)r 1C C Cn 1 (n 1)(n 2)

+ ++ +

+ ++ = = + + +

bull r nCr = n nndash1Crndash1 and n n 1

r r 1C Cr 1 n 1

79 propertIes of coeffIcIents

Properties of binomial expression are derived from

bull n

n n rr

(1 x) C x=

+ =sum = nC0 + nC1x + nC2x2 + +nCr x

r + + nCnxn (i)

bull n

n r n rr

(1 x) ( 1) C x=

minus = minussum = nC0 - nC1x + nC2x2 -+ (ndash1)n nCn x

n (ii)

bull n

n r n n n n nr 0 1 2 n

C x C C C C 2=

= + + + + =sum

bull n

r n n n n n nr 0 1 2 n

( 1) C C C C ( 1) C 0=

minus = minus + minus + minus =sum bull The sum of the binomial coefficients of the odd terms in the expansion of(1 + x)n is equal to the sum

of the coefficients of the even terms and each is equal to 2nndash1 bull C0 + C2 + C4 + hellip = C1 + C3 + C5 + hellip = 2nndash1

bull n

r C=sum =1C1 + 2C2 + 3C3 + + nCn = n 2n ndash 1

bull n

2 n 2 2 2 2r 1 2 3 n

r C 1 C 2 C 3 C n C=

= + + + +sum

bull 0r 1 2 nCC C C Cr 1 1 2 3 n 1

= + + + ++ +sum

bull nn n

r 1 k 0r 1

C n(n 1)r kC 2= =minus

+= =sum sum

710 multInomIal theorem

bull The general term in the multinomial expansion is 1 2 kr r r1 2 k

n x x xr r r

bull The total number of terms in the multinomial expansion = number of non -negative integral solutions of the equation r1 + r2 + + rk = n = n + k ndash 1Cn or n + k ndash 1Ckndash 1

bull Coefficient of x1r1 x2

r2 x5r5 in the expansion of a1x1 + a2x2 + + akxk =

bull Greatest coefficient in the expansion of (x1 + x2 + + xk)n where q is the quotient and r the remainder when n is divided by k

bull The number of terms in the expansion of (x + y + z)n where n is a positive integer is 12 (n + 1) (n + 2) bull Sum of all the coefficients is obtained by putting all the variables xi equal to 1 and it is equal to nm

711 tIps and trIcks

1 (x + y)n = sum of odd terms + sum of even terms

2 In the expansion of (x + y)n r 1

T n r 1 yn NT r x+ minus + isin =

3 The coefficient of xn ndash 1 in the expansion of (x + 1) (x + 2) (x + n) = n(n 1)2+

4 The coefficient of xn ndash 1 in the expansion of (x + 1) (x ndash 2) (x ndash n) = n(n 1)2

minus +

5 Greatest term in (x +y)n = xn Greatest terms in ny1

6 The number of terms in the expansion of (x1 + x2 + + xn) n = n+rndash1Crndash1

7 If the coefficients of the rth (r + 1) and (r + 2) th terms in the expansion of (1 + x)n are in HP then n + (n ndash 2r)2 = 0

8 If the coefficients of the rth (r + 1) th and (r + 2) th terms in the expansion of (1 + x)n are in AP then n2 ndash n(4r + 1) + 4r2 ndash 2 = 0

Chapter 8InfInIte SerIeS

81 Binomial theorem for any index (n)

|x| lt 1 ie ndash1 lt x lt 1minus minus minus minus minus minus +

+ = + + + + + + infin2 3 r

n n(n 1)x n(n 1)(n 2)x n(n 1)(n 2)(n r 1)x(1 x) 1 nx to2 3 r

General term of (1 + x)nrn(n 1)(n 2)(n r 1)x

rminus minus minus +

Expansion of (x + a)n for any index

Case I When x gt a ie ax lt 1

In this case (x + a)n = x (1 + ax)n = x n (1 + ax)n = xn

a n(n 1)1 n (ax)x 2

n(n 1)(n 2) (ax) 3

minus + + + minus minus +

Case II When x lt a ie xa lt 1

In this case (x + a)n = a (1 + xa)n = a n (1 + xa)n = an

x n(n 1) x1 na 2 a

n(n 1)(n 2) x 3 a

minus + + +

minus minus +

Remarks

q nCr cannot be used because it is defined only for natural number

q If x be so small then its square and higher powers may be neglected then the approximate value of (1 + x)n = 1 + nx

82 Greatest term

To find the greatest term numerically in the expansion of (1 + x)n |x| lt 1 If Tr + 1 is the required term

then | Tr + 1| ge |Tr| or r 1

T 1T+ ge gives

| x |(x 1)r m| x | 1

+(say)

Infinite Series 885

(a) Calculate | x |(n 1)m

| x | 1+

(b) If m is integer then Tm and Tm+1 are equal and both are greatest terms (c) If m is not integer then T[m]+1 is the greatest term where [] denotes the greatest integer

Remarks

1 (1 ndash x)ndash1 = 1 + x + x2 + x3 + + xr +

2 (1 ndash x)ndash2 = 1 + 2x + 3x2 + + (r + 1)xr + and

3 (1 ndash x)ndash3 = 1 + 3x + 6x2 + + + +( r 1)( r 2)

2 xr + helliphellip

83 taylor expansion

For any function f(x) we have

(i) 2 3hf (a) h f (a) h f (a)f(a h) f(a)

1 2 3+ = + + + +

(ii) replacing (a + h) by x 2 3(x a)f (a) (x a) f (a) (x a) f (a)f(x) f(a)

1 2 3minus minus minus

= + + + +

That is function f(x) expressed as a polynomial of infinite degree in (x ndash a)

831 Maclaurins Expansions

In taylors expansions replace a by 0 and h hy x we have 2 3xf (0) x f (0) x f (0)f(x) f(0)

1 2 3= + + + +

That is 2 3x cos0 x ( sin(0)) x ( cos0)sin x sin0

1 2 3minus minus

= + + + +

(i) 3 5x xsin x x

3 5= minus + +

(ii) 2 4 6x x xcos x 1

2 4 6= minus + minus +

(iii) 3 5 7x 2x 17xtan x x

3 15 315= + + + +

(iv) 3 5 7

1 x x xtan x x 3 5 7

minus = minus + minus +

(v) + = minus + minus + minus lt le

2 3 4x x xn(1 x) x ( 1 x 1)2 3 4

832 Eulerrsquos Number

The summation of the infinite series + + + + + infin1 1 1 11 1 2 3 4

is denoted by e which is equal to the limiting

value of (1 + 1n)n as n tends to infinity

833 Properties of e

(a) e lies between 2 and 3 ie 2 lt e lt 3 n 1

1 1since for n 2n 2 minus

(b) The value of e correct to 10 places of decimals is 27182818284 (c) e is an irrational (incommensurable) number (d) e is the base of natural logarithm (Napier logarithm) ie ln x = loge x

834 Expansion of ex

For x isin R 2 3 r

x x x x xe 1 1 2 3 r

= + + + + + + infin or n

The above series is known as exponential series and ex is called exponential function Exponential function is also denoted by exp ie exp A = eA exp x = ex

835 Important Deduction from Exponential Series

(i) 2 3 r

x x x xe 1 1 2 3 r

= + + + + +infin =sum (ii) 2 3 r r

x x x ( 1) xe 1 1 2 3 r

infinminus

minus= minus + minus + +infin =sum

(iii) x x 2 4 6 2r

e e x x x x1 2 2 4 6 (2r)

minus infin

+= + + + + =sum (iv)

x x 3 5 2r 1

e e x x x x2 1 3 5 (2r 1)

minus +infin

minus= + + + =

(v) r 0

1 1 1e 1 1 2 r

= + + + +infin =sum (vi) r

1 1 1 ( 1)e 1 1 2 3 r

infinminus

(vii) 1

e e 1 1 1 11 2 2 4 6 (2r)

minus infin

+= + + + + +infin =sum (viii)

e e 1 1 1 12 1 3 5 (2r 1)

minus infin

minus= + + + +infin =

(ix) n 2 3n(n 1) n(n 1)(n 2)(1 x) 1 nx x x 2 3minus minus minus

+ = + + + + If given that x is so smalll as compared to 1

that x2 and higher powers of x can be neglected then it is called as binomial approximation of Binomial expression

84 loGarithmic series

For ndash1 lt x le 1 loge (1 + x) = ln (1 + x) = 2 3 4 r 1 r

x x x ( 1) xx 2 3 4 r

minusinfin

minusminus + minus + infin =sum

841 Important Deduction from Logarithmic Series

(i) 2 3 4

ex x xlog (1 x) x ( 1 x 1)2 3 4

minus = minus minus minus minus minus le lt

(ii) 2 4 6

ex x xlog (1 x)(1 x) 2 ( 1 x 1)2 4 6

+ minus = minus + + + minus lt lt

(iii) 3 5

e(1 x) x xlog 2 x ( 1 x 1)(1 x) 3 5

+ = + + + minus lt lt minus

Chapter 9trigonometriC ratios and identities

91 INTRODUCTION

The word lsquotrigonometryrsquo is derived from two Greek words (i) trigon (means a triangle) and (ii) metron (means a measure) Therefore trigonometry means science of measuring the sides of angles and study of the relations between side and angles of triangle

92 ANgle

Angle is defined as the measure of rotation undergone by a given revolving ray OX in a plane about its initial point O The original ray OX is called the initial side and the final position (OP) of the ray after rotation is called the terminal side of the angle angXOP The point of rotation (O) is called the vertex

921 Rules for Signs of Angles

bull If initial ray OA rotates to terminal ray OA then angle = q (rotation anti clockwise)

bull If initial ray OA rotates to terminal ray OB then angle = ndashq (rotation clockwise) where q is the measure of rotation

922 Measurement of AngleThe measurement of angle is done under the following three systems of measurement of angles

9221 Sexagesimal or english system

1 right angle = 900 (degrees) 10 = 60 (minutes)1 = 60 (seconds)

9222 Centesimal or french system (Grade)

1 right angle = 100g (grades) 1g = 100 (minutes)1 = 100 (seconds)

RemarkThe minutes and seconds in the sexagesimal system are different them the respective minutes and seconds in the centesimal system Symbols in both there systems are also different

9223 Radian measure or circular measurement

One radian corresponds to the angle subtended by arc of length r (radius) at the centre of the circle with radius r Since the ratio is independent of the size of a circle it follows that a radian is a dimensionless quantity

Length of an arc of a circle θ = =arc lengthlAngle (in radians)

r radiusRelation between radian and degree πc = 180deg

In hand working tips

bull The unit radian is denoted by c (circular measure) and it is customary to omit this symbol c Thus

when an angle is denoted as 2π

it means that the angle is 2π

radians where p is the number with approximate value 314159

bull D G R

180 200= =

deg π where D G and R denotes degree grades and radians respectively

bull The angle between two consecutive digits in a clock is 30deg (p6 rad) The hour hand rotates through an angle of 30deg in one hour

bull The minute hand rotate through an angle of 6deg in one minute

93 POlygON AND ITs PROPeRTIes

A closed figure surrounded by n straight lines is called a polygon It is classified in two ways bull A closed figure surrounded by n straight lines bull If all sides of a polygon are equal then it is regular polygon bull Convex Polygon A polygon in which all the internal angles are smaller than 180deg bull Concave Polygon A polygon in which at least one internal angles is larger

than 180deg

Properties bull An angle is called reflexive angle if it is greater than or equal to 180deg or p radians bull Sum of all internal angles of a convex polygon = (n ndash 2) pc = (n ndash 2) 180deg

bull Each internal angle of regular polygon of n sides = (n 2)

nminus π

Nomenclature of Polygons

Name of Polygon

Number of Sides

Name of Polygon

Number of Sides

Name of Polygon

Number of Sides

1 Triangle 3 7 Nonagon 9 13 Penta-decagon 152 Quadrilateral 4 8 Decagon 10 14 Hexa-decagon 163 Pentagon 5 9 Hendecagon 11 15 Hepta-decagon 17

Trigonometric Ratios and Identities 1989

Name of Polygon

Number of Sides

Name of Polygon

Number of Sides

Name of Polygon

Number of Sides

4 Hexagon 6 10 Duodecagon 12 16 Octa-decagon 185 Heptagon 7 11 Tri-decagon 13 17 Nona-decagon 196 Octagon 8 12 Tetra-decagon 14 18 Ico-sagon 20

Circular Sector bull Perimeter of a circular sector of sectoral angle qc = r(2 + q)

bull Area of a circular sector of sectoral angle c 21q r q2

94 TRIgONOmeTRIC RATIOs

Consider an angle q = angXOA as shown in figure P be any point other than O on its terminal side OA and let PM be perpendicular from P on x-axis Let length OP = r OM = x and MP = y We take the length OP = r always positive while x and y can be positive or negative depending upon the position of the terminal side OA of angXOA

In the right-angled triangle OMP we have Base = OM = x perpendicular = PM = y and Hypotenuse = OP = r

We define the following trigonometric ratios which are also known as trigonometric functions

Perpendicular ysinHypotenuse r

θ = = Base xcos

Hypotenuse rθ = =

Perpendicular ytanBase x

θ = = Hypotaneuse rcosecPerpendicular y

θ = =

Hypotaneuse rsecBase x

θ = = Base xcot

Perpendicular yθ = =

941 Signs of Trigonometric Ratios

Consider a unit circle with centre at origin of the coordinate axes Let P(a b) be any point on the circle with angle AOP = x radian ie length of arc AP = x as shown in the following figure

We defined cos x = a and sin x = b Since DOMP is a right triangle we have OM2 + MP2 = OP2 or

a2 + b2 = 1Thus for every point on the unit circle we have a2 + b2 = 1 or cos2x + sin2x = 1 Accordingly we can

judge the sign of a trigonometric function by comparing it with the sign of respective coordinates in that particular quadrant

Remark

The sign conventions can be kept in mind by the sentence ldquoAfter School To Collegerdquo where A stands for All S stands for Sine T stands for Tangent C stands for Cosine

942 Range of Trigonometric Ratios

943 Trigonometric Ratios of Allied Angles

9431 Trigonometric ratios of ndashq

Sin(ndashq) = ndashsinq

cos(ndashq) = cosq

tan(ndashq) = ndashtanq

cot(ndashq) = ndashcotq

sec(ndashq) = secq

cosec(ndashq) = ndashcosecq

9432 Trigonometric ratios of p ndash q

Sin(p ndash q) = sinq

cos(p ndashq) = ndashcosq

tan(p ndash q) = ndashtanq

cot(p ndash q) = ndashcotq

sec(pndash q) = ndashsecq

cosec(pndashq) =cosecq

Similarly

Sin(p + q)= ndashsinq cos(p+ q) = ndashcosq tan(p+ q) = tanq cot(p+ q) = cotq

sin cos2π minusθ = θ

cos sin

2π minusθ = θ

tan cot

2π minusθ = θ

cot tan

2π minusθ = θ

sin cos2π + θ = θ

cos sin

2π + θ = minus θ

tan cot

2π +θ =minus θ

cot tan

2π +θ =minus θ

sec(p+ q) = ndashsecq cosec(p+ q)= ndashcosec q

sec cosec2π minusθ = θ

cosec sec

2π minusθ = θ

cosec sec

2π + θ = θ

Think yourself Try to evaluate the conversions for f(270 ndash q) f(270 + q) f(360 ndash q) f(360 + q) where f is a trigonometric function

Generalized Results The values of trigonometric functions of any angle can be represented in terms of

an angle in the first quadrant as follows Let A n2π

= plusmnθ where n isin Z 02π

le θ lt Then

(i) Sin p = 0 cosn p= (ndash1)n (ii)

(n 1)2

( 1) cos if n is oddsin n2 ( 1) sin if n is even

minusπ minus θ+ θ = minus θ

(iii) ( )(n 1)

1 sin if n is oddcos n2 ( 1) cos if n is even

+π minus θ+ θ = minus θ

(iv) tan if n is even

tan ncot if n is odd2

plusmn θπ plusmn θ = plusmn θ

(v) cot if n is even

cot ntan if n is odd2

plusmn θπ plusmn θ = plusmn θ (vi)

sec if n is evensec n

cosec if n is odd2plusmn θπ plusmn θ = plusmn θ

Think and fill up the blank blocks in the following table

Angles Functions

p 76π 4

53π 11

sinq 0 12 1radic2 radic32 1cosq 1 radic32 1radic2 12 0tanq 0 1radic3 1 radic3 ND (infin)cotq infin radic3 1 1radic3 0secq 1 2radic3 radic2 2 ND (infin)tanq infin 2 radic2 2radic3 1

95 gRAPhs Of DIffeReNT TRIgONOmeTRIC RATIOs

951 y = sin x

x 0 p6 p4 p3 p2 2p3 3p4 5p6 psin x 0 12 1radic2 radic32 1 radic32 1radic2 12 0

Properties

P1 Domain of sinx is R and range is [ndash1 1] P2 sinx is periodic function with period 2p

P3 Principle domain 2 2π π minus

P4 It is an odd function P5 It is a continuous function and increases in first and fourth quadrants while decreases in second and

third quadrants

952 y = cos x

X 0 p6 p4 p3 p2 2p3 3p4 5p6 pcos x 1 radic32 1radic2 12 0 ndash12 ndash1radic2 ndashradic32 ndash1

Properties P1 The domain of cosx is R and the range is [ndash1 1] P2 Principle domain is [0 p] P3 cosx is periodic with period 2p P4 It is an even function so symmetric about the

y-axis

9521 y = tan x

X 0 p6 p4 p3 p2 2p3 3p4 5p6 ptan x 0 1radic3 1 radic3 infin ndashradic3 ndash1 ndash1radic3 0

Properties P1 The domain of tanx is R ndash (2n + 1) p2 and range

R or (ndashinfin infin) Principal domain is (ndashp2 p2) P2 It is periodic with period p P3 It is discontinuous x = R ndash (2n + 1) p2 and it is

strictly increasing function in its domain

953 y = cot x

X 0 p6 p4 p3 p2 2p3 3p4 5p6 pcot x infin radic3 1 1radic3 0 ndash1radic3 ndash1 ndashradic3 ndashinfin

Properties

P1 The domain of f(x) = cotx is domain isin R ~ np Range isin ℝ P2 It is periodic with period p and has x = np n isin z as its

asymptotes P3 Principal domain is (0 p) P4 It is discontinuous at x = np P5 It is strictly decreasing function in its domain

954 y = cosec x

x 0 p6 p4 p3 p2 2p3 3p4 5p6 pcosec x infin 2 radic2 2radic3 1 2radic3 radic2 2 infin

Properties P1 The domain is R ~ np | n isin z P2 Range of cosecx is R ndash (ndash1 1)

P3 Principal domain is 02 2π π minus minus

P4 The cosecx is periodic with period 2p

955 y = sec x

X 0 p6 p4 p3 p2 2p3 3p4 5p6 psec x 1 2radic3 radic2 2 infin ndash2 ndashradic2 ndash2radic3 ndash1

Properties

P1 The domain of sec x is R (2n 1) n z2π minus + isin

range is R ndash (ndash1 1) P2 The sec x is periodic with period 2p P3 Principal domain is [0 p] ndash p2 P4 It is discontinuous at x = (2n + 1) p2

956 Trigonometric Identities

9561 Pythagorean identities

The following three trigonometric identities are directly derived from the pythagoras theorem

1 sin2x + cos2 x = 1 x isin ℝ rArr cos2 A = 1 ndash sin2 x or sin2 x = 1 ndash cos2 x or cos x 1 sin x

1 sin x cos x+

=minus

2 1+ tan2 x = sec2 x x ~ (2n 1) n2π isin + isin

rArr sec2x ndash tan2x = 1 or 1sec x tan x

sec x tan xminus =

3 cot2 x + 1 = cosec2 x x isin ℝ ~ np n isin ℤ rArr cosec2 x ndash cot2 x = 1 or1cosec x cot x

cosec x cot xminus =

NoteIt is possible to express trigonometrical ratios in terms of any one of them as

θ =+ θ2

cotcos

θθ =

+ θ 1

tancot

θ =θ

2cosec 1 cotθ = + θ

Remember sign of the dependent function will depend upon the location of angle in one or the other quadrant

957 Trigonometric Ratios of Compound AnglesAn angle made up of the sum of the algebraic sum of the two or more angles is called a lsquocompound anglersquo Some of the formulae on various trigonometric functions are given below

1 sin (A + B) = sin A cos B + cos A sin B 2 sin (A ndash B) = sin A cos B - cos A sin B 3 cos (A + B) = cos A cos B ndash sin A sin B 4 cos (A ndash B) = cos A cos B + sin A sin B

5 tan A tanBtan(A B)

1 tan A tanB+

+ =minus

1 tan A tanBminus

minus =+

7 cot A cot B 1cot(A B)cot B cot A

minus+ =

cot A cot B 1cot(A B)cot B cot A

+minus =

minus 9 sin(A + B) sin (A ndash B) = sin2 A ndash sin2 B = cos2 B ndash cos2 A

10 cos (A + B) cos (A ndash B) = cos2 A ndash sin2 B = cos2 B ndash sin2 A

958 Trigonometric Ratios of Multiples of Angles

2 tan Asin A 2sin A cos A1 tan A

2 tan Atan2A1 tan A

=minus

where A (2n 1)4π

3 1 cos A Atan

sin A 2minus =

where A ne (2n + 1)p 4

1 cos A Acotsin A 2+ =

where A ne (2np)

5 21 cos A Atan1 cos A 2minus = +

where A ne (2n + 1)p 6 21 cos A Acot1 cos A 2+ = minus

where A ne2np

7 A Asin cos 1 sin A2 2+ = plusmn + 8

A Asin cos 1 sin A2 2minus = plusmn minus

9 cos2A = cos2 A ndash sin2 A = 1 ndash 2 sin2 A = 2

1 tan A2cos A 11 tan Aminus

minus =+

10 1 + cos 2A = 2 cos2 A 1 ndash cos2A = sin2A or 21 cos2A cos A2

+= 21 cos2A sin A

2minus

11 sin 3A = 3 sin A ndash 4 sin3 A = 4 sin (60deg ndash A)sin Asin( 60deg + A) 12 cos 3A = 4cos3 A ndash 3cosA = 4 cos (60deg ndash A) cos Acos (60deg + A)

3tan A tan Atan3A1 3tan A

minus=

minus= tan (60deg ndash A)tan Atan (60deg + A)

14 sin A cos A 2 sin A 2 cos A2 4π π plusmn = plusmn =

959 Transformation Formulae

9591 Expressing the product of trigonometric ratio sum or difference

(i) 2 sin A cos B = sin (A + B) + sin (A ndash B) (ii) 2 cos A sin B = sin (A + B) ndash sin (A ndash B) (iii) 2 cos A cos B = cos (A + B) + cos (A ndash B) (iii) 2 sin A sin B = cos (A ndash B) ndash cos (A + B)

9592 Expressing the sum or difference of trigonometric ratios into product

1 C D C DsinC sinD 2sin cos

2 2+ + + =

C D C DsinC sinD 2cos sin2 2+ minus minus =

3 C D C DcosC cosD 2cos cos

2 2+ minus + =

C D C DcosC cosD 2sin sin2 2+ minus minus =

5 sin(A B)tan A tanBcos AcosB

++ = where AB n

ne π+

minusminus = where A B ne np+ AB n

ne π+

7 sin(A B)cot A cot Bsin AsinB

++ = where A B nen n isinz

minusminus = where A B ne npn isinz

9510 Conditional Identities

If A + B + C = p then

(i) sin2A + sin2b + sin2C = 4sinA sinB sinC (ii) A B Csin A sinB sinC 4cos cos cos22 2 2

(iii) cos2A + cos2B + cos2C = ndash1 ndash 4 cosA cosB cosC (iv) A B Ccos A cosB cosC 1 4sin sin sin2 2 2

+ + = +

(v) tanA + tanB + tanC = tanA tanB tanC (vi) A B B C C Atan tan tan tan tan 12 2 2 2 2 2

(vii) A B C A B Ccot cot cot cot cot cot2 2 2 2 2 2+ + = (viii) cotA cotB + cotB cotC + cotC cotA = 1

(ix) A B C2π

+ + = then tanA tanB + tanB tanC + tanC tanA = 1

96 sOme OTheR UsefUl ResUlTs

(i) sin a + sin (a + b) + sin (a + 2b) + hellip + hellip to n terms =

( )n 1 nsin sin2 2

minus β β α +

(ii) cos a + cos (a + b) + cos (a + 2b) + hellip + hellip to n term =

( )n 1 ncos sin2 2

minus β β α +

(iii) cos A cos 2A cos23 A hellip n

sin2 Acos2 A2 sin A

minus = when n rarr infin minus

θ θ θ θinfin =

θ2 n 1

sincos cos cos 2 2 2

(iv) If A B C = π then bull cosA + cosB + cosC le 32 bull sinA2 sinB2 sinC2 le 18 equality holds good if A = B = C = 60deg bull tan2A2 + tan2B2 + tan2C2 ge 1

97 sOme OTheR ImPORTANT VAlUes

SNo Angle Value SNo Angle Value

1 sin 15deg 3 12 2minus 2 cos 15deg 3 1

3 tan 15deg 2 3minus = cot 75deg 4 cot 15deg 2 3+ = tan 75deg

5 sin 2212

deg ( )1 2 22

minus 6 cos 2212

deg ( )1 2 22

7 tan 2212

deg 2 1minus 7 cot 2212

deg 2 1+

9 sin 18deg5 14minus

= cos 72deg 10 cos 18deg 10 2 54+

= sin 72deg

11 sin 36deg10 2 5

4minus = cos 54deg 12 cos 36deg 5 1

4+ = sin 54deg

13 sin 9deg 3 5 5 54

+ minus minus

or cos 81deg

14 cos 9deg3 5 5 5

4+ + minus

or sin 81deg 15 cos 36deg ndash cos 72deg 12 16 cos 36degcos 72deg 14

98 mAxImUm AND mINImUm VAlUes Of A COs q + b sIN q

Consider a point (a b) on the cartesian plane Let its distance from origin be r and the line joining the point and the origin make an angle a with the positive direction of x axis Then a = r cos a and b = r sin a

Squaring and adding 2 2r a b= + So a cos q + b sin q = r [cos a cos q + sin a sin q] = r cos (a ndash q)

but ndash 1 le cos (a ndash q)le 1rArr ndash r le a cos q + b sin q le r

So maximum value is 2 2a b+ and minimum value is 2 2a bminus +

99 TIPs AND TRICs

1 if x = secq + tanq Then 1x = secq ndash tanq 2 if x = cosecq + cotq Then 1x = cosecq ndash cotq

3 cos A cos2Acos22A n

sin2 Acos2 A2 sin A

minus = if A ne n p = 1 if A = 2n p = (ndash1)n if A = (2n + 1) p

4 sinA2 plusmn cosA2 = radic2 = sin[p 4 plusmn A] = radic2cos [A p4]

5 cos a+ cos b + cos g + cos (a+b+g) = ( ) ( ) ( )4cos cos cos

2 2 2α+β β+ γ γ +α

6 sin a+ sin b + sin g ndash sin (a+b+g) = ( ) ( ) ( )4sin sin sin

2 2 2α+β β+ γ γ +α

Chapter 10trigonometriC

equation

101 IntroductIon

The equations involving trigonometric functions of one or more unknown variables are known as lsquotrigonometric equationsrsquo For example cosq = 0 cos2q - 4 cosq = 1 sin2q + sinq = 2 cos2q - 4sinq = 1 etc

102 SoLutIon oF trIGonoMEtrIc EQuAtIon

A solution of a trigonometric equation is the value of the unknown variable (angle) that satisfies the

equation For example 1sin2

θ = rArr 4π

θ = or 3 9 11

4 4 4 4π π π π

Thus the trigonometric equation may have infinite number of solutions

1021 Classification of Solutions of Trigonometric Equations

(i) Particular solution (ii) Principal solution (iii) General solution

103 PArtIcuLAr SoLutIon

Any specific solution that satisfies a given trigonometric equation is called a particular solution

For example sin x = has a particular solution π

104 PrIncIPAL SoLutIon

The solutions of a trigonometric equation having least magnitude that is belonging to principal domain of

trigonometric function are called principal solution For example sin 1x2

= has principal solution 6π

Parallely cos 1x2

= minus has principal solutions 23π

The following figures represent principal domains of trigonometric functions

Trigonometric Equation 1099

Principal Domain 2 2π π minus

Principal Domain [0 π]

Principal Domain (0 π)

Principal Domain [0 ]~2π π

Principal Domain ~ 02 2π π minus

105 GEnErAL SoLutIon

Since trigonometric functions are periodic a solution can be generalized by means of periodicity of the trigonometric functions An expression which is a function of integer n and a particular solution a representing all possible particular solutions of a trigonometric equation is called its lsquogeneral solutionrsquo We use the following results for solving the trigonometric equations

Result 1 sinq = 0 hArr q = n p n isin ℤ

General Solutions for Some Standard Equations

Sin q = 0 rArr q = n π sin 1 (4n 1)2π

θ = rArrθ = + π

θ = minus rArrθ = minussin 1 (4n 1)2

Result 2 cos q = 0 hArr (2n 1) n2π

θ = + isin

cos 0 (2n 1)2π

θ = rArrθ = + cos q = 1 rArr q = 2nπ cos q = ndash1 rArr q =(2n + 1)π

Result 3 tan q = 0 hArr q = n p n isin ℤ

tan q = 0

rArr q = nπ tan 1 (4n 1)4π

θ = rArrθ = + tan 1 (4n 1)4π

θ = minus rArrθ = minus

Result 4 sin q = sin a hArr q = n p +(-1)n a where n isin ℤ and a is a particular solution preferably taken least non-negative or

that having least magnitude

Result 5 cos q = cos a hArr q = 2n p plusmn a n isin ℤ

ndash +ndash

Result 6 tan q = tan a hArr q = n p + a n isin ℤ

Result 7 sin2 q = sin2 a cos2 q = cos2 a tan2 q = tan2 a hArr q = n p plusmn a n isinℤ

RemarkIn formulae if we take any of a the set of all possible solutions represented by general solution remains unique

Theorem 1 sin q = k where k isin [ndash1 1] has general solution q = nπ +(ndash1)na

Where 2 2π π αisin minus

st sin a = k

Theorem 2 cos q = k where k isin [ndash1 1] has general solution q = 2nπ plusmn α where α isin [0 π] st cos α = k

+ndash

Theorem 3 tan q = k where k isin ℝ has general solution q = nπ + α where 2 2π π αisin minus

st tan α = k

Theorem 4 sin2q = k where k isin [0 1] has general solution q = nπ plusmn α where 02π αisin

st sin2α = k

Theorem 5 cos2 q = k where k isin [0 1] has general solution q = nπ plusmn α where 02π αisin

cos2 α = k

ndash +

Theorem 6 tan2 q = k where k isin [0 infin) has general solution q = nπ plusmn α where 02π αisin

tan2 α = k

ndash+ndash

106 SuMMAry oF thE AbovE rESuLtS

1 sin q = 0 hArr q = np n isin ℤ

2 cos q = 0 hArr (2n 1) n2π

θ = + isin

3 tan q = 0 hArr q = np n isin ℤ 4 sin q = sin a hArr q = n p +(-1)n a n isin ℤ 5 cos q = cos a hArr q = 2n p plusmn a n isin ℤ 6 tan q = tan a hArr q = n p + a n isin ℤ 7 sin2 q = sin2 a cos2 q = cos2 a tan2q = tan2a n isin ℤ

8 sin q = 1 hArr (4n 1) n2π

θ = + isin

9 sin q = ndash1 hArr q =(4n + 3)2π

n isin ℤ

10 cos q = 1 hArr q = 2n p 11 cos q = -1 hArr q =(2n + 1) p n isin ℤ 12 sin q = sin a and cos q = cos a hArr q = 2n p + a n isin ℤ

1 The general solution should be given unless the solution is required in a specified interval or range

2 a is a particular solution preferably taken least positive or that having least magnitude

107 tyPE oF trIGonoMEtrIc EQuAtIonS

Type 1 Trigonometric equations which can be solved by use of factorization eg (2 cos x ndash sin x)(1 ndash sin x) = cos2 x rArr (2 cos x ndash sin x)(1 + sin x) = 1 ndash sin2 x

rArr (1 + sin x)(2 cos x ndash 1) = 0 rArr sin x = ndash1 or 1cos x2

rArr x (4n 3)2π

= + or 2n n3π

πplusmn isin are the general solutions

Type 2 Trigonometric equations which can be solved by reducing them to quadratic equations eg 2 sin2 x + 2 sin x = 5 cos2 x rArr 2 sin2 x + 2 sin x = 5(1 ndash sin2 x)

rArr 7 sin2 x + 2 sin x ndash 5 = 0 rArr sin x = ndash1 or 5sin x7

rArr x (4n 3) n2π

= + isin or x = nπ +(ndash1)n α n isin ℤ

And 5sin7

α = are the required general solutions

Type 3 Trigonometric equation which can be solved by transforming a sum or difference of trigonomet-ric ratios into their product

eg cosx ndash sin3x = cos2x rArr cosx ndash cos2x = sin3x

rArr 3x x 3x 3x2sin sin sin3x 2sin cos2 2 2 2

rArr 3x x x2sin sin cos3 02 2 2

minus = rArr

3xsin 02=

rArr 3x n n2= π isin rArr

2nx n3π

= isin helliphellip(i)

or x 3xsin cos 02 2minus = rArr

x 3xcos cos 02 2 2π minus minus =

rArr x2sin sin x 0

4 2 4π π + minus =

rArr x m4π

= π+ hellip(ii)

Combining equation (i) and (ii) general solutions are given by 2nx 2n n n

3 2 4π π π

= πminus π+ isin

Type 4 Trigonometric equations which can be solved by transforming a product of trigonometric ratios into their sum or difference For example sin x cos 5x = sin4x cos2 x

rArr sin6x + sin(ndash4x) = sin6x + sin2x rArr sin2x + sin4x = 0 rArr 2sin(3x) cos x = 0

rArr nx3π

= or x (2n 1) n Z2π

= + isin

Type 5 Trigonometric equations of the form a sinx + b cosx = c where a b c isin ℝ can be solved by

dividing both sides of the equation by 2 2a b+

To solve the equation a cosq + b sinq = c put a = r cos f b = r sin f such that 2 2r a b= + 1 btana

minusφ =

ie take 2 2π π φisin minus

such that

Substituting these values in the equation we have r cos f cos q + r sin f sin q = c

rArr ccos( )r

θminusφ = rArr 2 2

ccos( )a b

θminusφ =+

1 If gt +2 2c a b then the equation a cos q + b sin q = c has no solution

2 If 2 2c a ble + then put 2 2

a b+ = cos a so that cos(q ndash f) = cos a

rArr (q - f) = 2n p plusmn a rArr q = 2n p plusmn a + f where n isin ℤ eg sin x cos x 2+ =

rArr a = b = 1 Let a = r cosq b = r sinq

rArr 2 2r a b 2= + = 1 2 cos 1 2 sinθ θ= =

rArr tanq = 1 rArr q = tanndash11 rArr 4πθ =

2 cos x 24π minus =

rArr cos x 1

4π minus =

rArr x 2n n

4ππ= + isin

3 Trigonometric equation of the form a sinx + cosx = c can also be solved by changing sinx and cosx into their corresponding tangent of half the angle and solving for tan x2 ie we substitute

x1 tan

2cos xx

1 tan2

minus=

2sin xx

1 tan2

Type 6 Equation of the form R(sin x plusmn cos x sin x cos x) = 0 Where R is a rational function of the arguments in the brackets Put sin x + cos x = t (i) and use the following identity (sin x + cos x)2 = sin2 x + cos2 x + 2 sin x cos x = 1 + 2 sin x cos x

rArr 2t 1sin x cos x2minus

= (ii)

Taking equation (i) and (ii) into account we can reduce given equation into R(t(t2 ndash 1)2) = 0 Similarly by the substitution(sin x - cos x) = t we can reduce the equation of the form R(sin x - cos x sin x cos x) = 0 to an equation R(t(1 ndash t2)2) = 0

Type 7 Trigonometric equations which can be solved by the use boundedness of the trigonometric

ratios sinx and cosx eg 5xsin cos x 24+ = Now the above equation is true if

5xsin 14= and cos x = 1

rArr 5x 2n n4 2

π= π+ isin and x = 2mp m isin z

rArr (8n 2)x n

= isin helliphelliphelliphellip(iii)

and x= 2mp m isin ℤ helliphelliphellip(iv)

Now to find general solution of equation (i) (8n 2) 2m

rArr 8n + 2 = 10 m rArr 5m 1n

4minus

If m = 1 then n = 1 m = 5 then n = 6 hellip helliphellip helliphellip hellip helliphellip helliphellip If m = 4p ndash 3 p isin ℤ then n = 5p ndash 4 p isin ℤ General solution of a given equation can be obtained by

(8n 2)x 2m m n ~ 2m m 4p 3p

5+ = π isin cup π isin π = minus isin

or (8n 2) (8n 2)x 2m m n ~ n 5p 4p

5 5+ + = π isin cup π isin π = minus isin

Type 8 A trigonometric equation of the form R(sin kx cos nx tan mx cot l(x) = 0 l m n then use the following formulae

2 tan x 2sin x

1 tan x 2 2

1 tan x 2cos x

1 tan x 2minus

2 tan x 2tan x

1 tan x 2=

21 tan x 2cot x

2tan x 2minus

108 hoMoGEnEouS EQuAtIon In SInx And coSx

The equation of the form a0 sinn x + a1 sinn-1 x cos x + a2 sinn-2 x cos2 x + + an cosn x = 0 where a0 a1 an are real numbers and the sum of the exponents in sin x and cos x in each term is equal to n are said to be homogeneous with respect to sin x and cos x For cos x ne 0 the above equation can be written as a0 tann x + a1 tann-1 x + + an = 0

109 SoLvInG SIMuLtAnEouS EQuAtIonS

Here we discuss problems related to the solution of two equations satisfied simultaneously We may divide the problems into two categories as shown by the following diagram

When number of equations is more than or equal to number of variables

∎ Single variable problems with intermediate values

Step 1 Find the values of variable x satisfying both equations

Step 2 Find common period of function used in both the equation say T and obtain x = α isin(0 T] sat-isfying both the equations

Step 3 Generalizing the value of α we get x = nT + α

∎ Single variable problem with extreme values

Step 1 When LHS and RHS of a equation have their ranges say R1 and R2 in common domain and R1 cap R2 = f then the equations have no solution

Step 2 If R1 cap R2 have finitely many elements and the number of elements are few then individual cases can be analyzed and solved

1091 More Than One Variable Problems

bull Substitute one variable (say y) in terms of other variable x ie eliminate y and solve as the trigonometric equations in one variables

bull Extract the linearalgebraic simultaneous equations from the given trigonometric equations and solve as simultaneous algebric equations

bull Many times you may need to make appropriate substitutions bull When number of variables is more than number of equations To solve an equation involving more than one variable definite solutions can be obtained if extreme

values (range) of the functions are used

10911 Some important results

1 While solving a trigonometric equation squaring the equations at any step should be avoided as far as possible If squaring is necessary check the solution for extraneous values

2 Never cancel terms containing unknown terms on the two sides which are in product It may cause loss of the genuine solution

3 The answer should not contain any such values of angles which make any of the terms undefined or infinite

4 Domain should not change If it changes necessary corrections must be made 5 Check that denominator is not zero at any stage while solving equations

1010 trAnScEdEntAL EQuAtIonS

To solve the equation when the terms on the two side (LHS and RHS) of the equation are of different nature eg trigonometric and algebraic we use inequality method Which is used to verify whether the given equation has any real solution or not In this method we follow the steps given below

Step I If given equation is f(x) = g(x) then let y = f(x) and y = g(x) ie break the equation in two parts

Step II Find the extreme values of both sides of equation giving range of values of y for both side If there is any value of y satisfying both the inequalities then there will be a real solution otherwise there will be no real solution

1011 GrAPhIcAL SoLutIonS oF EQuAtIonS

For solution of equation f(x) ndash g(x) = 0

Let a is root rArr α = α =f( ) g( ) k(say)

rArr y f(x) and y g(x)= =

have same output for input x = α

rArr ( k) satisfying both the curves y f(x) and y g(x)α = =

Solutions of equation f(x) ndash g(x) = 0 are abscissa (x-co-ordinate) of the point of intersection of the graph y = f(x) and y = g(x)

Algorithm To solve the equation f(x) ndash g(x) = 0 eg 10sinx ndash x = 0

Step 1 Write the equation f(x) = g(x) ie sinx = x10

Step 2 Draw the graph of y = f(x) and y = g(x) on same x ndash y plane

Let f(x) = sinx and g(x) = x

also we know that -1 le sinx le 1

-1 le x

10 le 1

rArr -10 le x le 10

Thus sketching both the curves when x isin [minus10 10]

Step 3 Count the number of the points of intersection If graphs of y = f(x) and y = g(x) cuts at one two three no points then number of solutions are one two three zero respectively

From the given graph we can conclude that f(x) = sinx and g(x) = x

10 intersect at 7 points So number

of solutions are 7

1012 SoLvInG InEQuALItIES

To solve trigonometric inequalities including trigonometric functions it is good to practice periodicity and monotonicity of functions Thus first solve the inequality for one period and then get the set of all solutions by adding numbers of the form 2np n isin ℤ to each of the solutions obtained on that interval

-3π -2π2π

3π-π

f(x) = sinxg(x) = x10

(-frac1234)

frac12 10

(101)(3 3 10)ππ(2 2 10)ππ

( 10)ππ

For example Find the solution set of inequality sinx gt 12

Solution When sinx = 12 the two values of x between 0 and 2p are p6 and 5p6 from the grpah of y = sinx it is obvious that between 0 and 2p

sinx gt 12

for p6 lt x lt 5p6

Hence sinx gt 12 rArr 52n x 2n

6 6π π

π+ lt lt π+

The required solution set is n Z

52n 2n6 6isin

π π π+ π+

10121 Review of Some Important Trigonometric Values

1 3 1sin15

2 2minus

deg = 2 3 1cos 15

3 tan 15deg = 2 - radic3 = cot 75deg 4 cot 15deg = 2 + radic3 = tan 75deg

5 ( )1 1sin 22 2 22 2

= minus

6 ( )1 1cos22 22 2 22 2

deg = = +

7 1tan 22 2 12

= minus

8 1cot 22 2 12

9 5 1sin18 cos724minus

deg = = deg 10 10 2 5cos18 sin72

deg = = deg

11 10 2 5sin36 cos54

4minus

deg = = deg 12 5 1cos36 sin544+

deg = = deg

13 3 5 5 5sin9 cos81

4+ minus minus

deg = = deg 14 3 5 5 5cos9 sin81

4+ + minus

deg = = deg

15 cos 36deg - cos 72deg = 12 16 cos 36deg cos 72deg = 14

Chapter 11properties of

triangles

111 IntroductIon

Here we shall discuss the various properties of tringels

1111 Sine Formula

In any triangle ABC the sides are proportional to the sines of the opposite angles

ie a b c 2Rsin A sinB sinC

= = = R = circumradius of DABC

1112 Cosine Formula

In any triangle ABC to find the cosine of an angle in terms of the sides

2 2 2b c acos A

2bc+ minus

= 2 2 2a c bcosB

2ac+ minus

= 2 2 2a b ccosC

2ab+ minus

1113 Projection FormulaIn any triangle ABC a = c cos B + b cos C b = a cos C + c cos A c = a cos B + b cos A the sine cosine and Tangent of the half-anlges in terms of the sides

(i) (s b)(s c)Asin2 bc

minus minus= (s a)(s c)Bsin

2 acminus minus

= (s a)(s b)sin

2 abminus minus

(ii) s(s a)Acos

2 bcminus

= s(s b)Bcos

2 acminus

= s(s c)Ccos

2 abminus

(iii) (s b)(s c) s(s a)A sin A 2tan

2 cos A 2 bc bcminus minus minus

= = divide (s b)(s c)Atan

2 s(s a)minus minus

=minus

(s a)(s c)Btan

2 s(s b)minus minus

=minus

and (s a)(s b)Ctan

2 s(s c)minus minus

=minus

Properties of Triangles 11111

11131 sin A in terms of the sides of the triangle

(s b)(s c) s(s a)A Asin A 2sin cos 22 2 bc bc

minus minus minus= = times

rArr 2 2sin A s(s a)(s b)(s c)bc bc

∆= minus minus minus = Similarly 2 2sinB s(s a)(s b)(s c)

ca ca∆

= minus minus minus = minus

2 2sinC s(s a)(s b)(s c)ab ab

∆= minus minus minus = D = area of D ABC

112 nAPIErrsquoS AnALoGY

In any triangle ABC (A B) a b Ctan cot2 a b 2minus minus

(B C) b c Atan cot2 b c 2minus minus

(C A) c a Btan cot2 c a 2minus minus

1121 Solution of Triangle

Case 1 When three sides of a triangle are givenIn this case the following formulae are generally used

(i) minus minus

=(s b)(s c)Asin

2 bc (ii)

s(s a)Acos2 bc

minus=

(iii) (s b) (s c)Atan

2 s(s a)minus minus

=minus

(iv) 2 2 2b c acos A

2bc+ minus

Case 2 When two sides and the included angle of the triangle are given Let b c and A be given then lsquoarsquo can be found from the formula a2 = b2 + c2 ndash 2bc cos A

Now angle B can be found from the formulae 2 2 2c a bcosB

2ac+ minus

= or bsin AsinBa

= and C from

C = 180deg ndash A ndash B

Another way to solve such triangle is first to find B C2minus by using the formulae

B C b c Atan cot2 b c 2minus minus = +

and therefore by addition and subtraction B and C and the third side lsquoarsquo by

cosine formula a2 = b2 + c2 ndash 2bc cos A or bsin A

= or a = b cos C + c cos B

Case 3 When two angles and the included side of a triangle are givenLet angle B C and side a be given The angle A can be found fromA = 180deg - B - C and the sides b and c from sine rule

a b csin A sinB sinC

= = ie a sinBbsin A

= and a sinCcsin A

Case 4 Ambiguous CaseWhen two sides (say) a and b and the angle (say) A opposite to one side a are given There are following three possibilities

(i) Either there is no such triangle (ii) One triangle (iii) Two triangles which have the same given elements

We have b asinB sin A

= rArr bsin AsinB

a= hellip (1)

Also c2 ndash 2 (b cos A) c + b2 ndash a2 = 0 (2)

gives 2 2 2c bcos A a b sin A= plusmn minus (3)Now the following cases may raise

(a) When a lt b sin A rArr sin B gt 1 form equation (1) or from equation (3) c is imaginary which is impossible Hence no triangle is possible

(b) When b sin A = a rArr from equation (1) sin B = 1 rArr B = 90deg and from equation (3) c = b cos A This value of c is admissible only when b cos A is positive ie when the angle A is acute In such a

case a lt b (b sin A = a) or A lt B Hence only one definite triangle is possible

In this case a = b is not possible since A = B = 90deg which is not possible Since no triangle can have two right angles

(c) When b sin A lt a and sin B lt 1 from (4) In this case there are three possibilities (i) If a = b then A = B and from equation (3) we get c = 2b cos A or 0 Hence in this case we get

only one triangle (since in this case it is must that A and B are acute angles) (ii) If a lt b then A lt B Therefore A must be an acute angle b cos A gt 0 Further a2 lt b2 rArr a2 lt b2 (cos2 A + sin2 A)

rArr 2 2 2a b sin A bcos Aminus lt From equation (3) it is clear that both values of c are positive so we get two triangles such that

and 2 22c bcos A a b sin A= minus minus

It is also clear from equation (1) that there are two values of B which are supplementary

(iii) If a gt b then A gt B also a2 - b2 sin2 A gt b2 cos2 A or 2 2 2a b sin A bcos Aminus gt

Hence one value of c is positive and other is negative for any value of angle A Therefore we get only one solution Since for given values of a b and A there is a doubt or ambiguity in the determination of the triangle Hence this case is called ambiguous case of the solution of triangles

113 GEomEtrIc dIScuSSIon

Let a b and the angle A be given Draw a line AX At A construct angle angXAY = A Cut a segment AC = b from AY Now describe a circular arc with its centre C and radius a Also draw CD perpendicular to AX

CD = b sin A The following cases may arise

(a) If a lt b sin A ie a lt CD then the circle will not meet AX and hence there is no triangle satisfying the given condition

(b) If a = b sin A the circle will touch AX at D (or B) and only one right angled triangle is possible In this case B = 90deg and A lt 90deg

(c) If a = b (angA ne 90deg) then the circle will cut AX at B and passes through A Hence here we get only one solution of given data (as shown in the figure)

(d) If a gt b sin A then the circle will cut AX at two distinct points (other than A) Let the point be B1 and B2Sub-case 1 If b sin A lt a lt b then both B1 and B2 are on the same side of A as shown in the figure and we get two distinct triangles ACB1 and ACB2

Sub-case 2 If a gt b then the two points B1 and B2 are on the opposite sides of A and only one of the triangle ACB1 or ACB2 will satisfy the given data If A is an acute angle then DCAB2 is the required triangle and if A is obtuse angle then DAB1C is the required triangle

114 ArEA of trIAnGLE ABc

If D represents the area of a triangle ABC then D = 12 (BCAD) 1 AD 1a(csinB) as sinB acsinB2 c 2

Also ADsinCb

= rArr AD = b sin C

1 a bsinC2

∆ = Similarly 1 bcsin A2

1 1 1absinC bcsin A ca sinB2 2 2

∆ = = =

(i) Area of a triangle in terms of sides (Herorsquos formula)

1 1 A Abcsin A bc2sin cos2 2 2 2

∆ = = = (s b)(s c) s(s a)

bcbc bc

minus minus minus

rArr s(s a)(s b)(s c)∆ = minus minus minus

(ii) Area of triangle in terms of one side and sine of three angles

1 1bcsin A (k sinB)(k sinC)sin A2 2

∆ = = = 21 k sin AsinBsinC2

= 21 a sin AsinBsinC

2 sin A

= 2a sinBsinC

2 sin A

Thus 2a sinBsinC

2 sin A∆ = =

2 2b sin AsinC c sin AsinB2 sinB 2 sinC

115 mndashn thEorEm

In any triangle ABC if D is any point on the base BC such that BD DC m n angBAD = α angCAD = b angCDA = q then (m + n) cot q = m cot α ndash n cot b = n cot B ndash m cot C

1151 Some Definitions

11511 Circumcircle

The circle which passes through the angular points of a triangle is called its circumscribing circle or more briefly circumcircle The centre of this circle is called circumcentre Generally it is denoted by O and its radius always denoted by R Another property of circum centre is that it is the point of concurrency of perpendicular bisectors of sides of a triangle

11512 Radius of circum circle lsquoRrsquo of any triangle

In DABC we have 2sin sin sin

a b c RA B C= = =

The circumradius may be expressed in terms of sides of the trianglea abc abcR

2sin A 2bcsin A 4= = =

1 sin2

bc A ∆ = Thus abcR

11513 Incircle

The circle which can be inscribed within the triangle so as to touch each of the sides is called its in-scribed circle or more briefly its incircle The centre of this circle is called incentre It is denoted by I and its radius always denoted by r In-centre is the point of concurrency of internal angles bisectors of the triangle

Radius r of the incircle of triangle ABCSince D = Area DIBC + ar(D ICA) + ar(D IAB)rArr D = (12) ar + (12) br + (12) cr = 12 (a + b + c)r

rArr D = sr rArr r = Ds a b cs2

+ += = semi-perimeter

The radius of incircle in terms of sides and tangent of the half angleAr (s a)tan2

= minus = B(s b)tan2

minus = C(s c)tan2

The radius of incircle in terms of one side and the functions of the half anglesa sin(B 2)sin(A 2)r

cos(A 2)= = bsin(B 2)sin(C 2)

cos(B 2) = Csin(A 2)sin(B 2)

cos(C 2)since a = 2R sinA = 4R sinA2 cosA2 r = 4R sinA2 sin B2 sin C2

11514 Escribed circles

The circle which touches the sides BC and two sides AB and AC (produced) of triangle ABC is called escribed circle opposite the angle A The centre of escribed circle is called ex-centre and is denoted by I1 or IA and radius by r1 or rA

Radii of escribed circles of a triangle 1r s a∆

=minus

2r s b∆

=minus

3r s c∆

=minus

Radii of the Escribed circles in terms of sides and the tangents of half angler1

= s tan A2 r2 = s tan B2 r3 = s tan C2

Radii of the escribed circles in terms of one side and function of half angles

1a cos(B 2)cos(C 2)r

cos(A 2)= 2

bcos(C 2)cos(A 2)rcos(B 2)

= 3ccos(A 2)cos(B 2)r

cos(C 2)=

Now Since a = 2R sin A = 4R sin A2 cosA2rArr r1 = 4R sin A2 cosB2 cosC2 r2 = 4R cosA2 sinB2 cos C2 and r3 = 4RcosA2 cosB2 sinC2

116 orthocEntrE And PEdAL trIAnGLE

Let ABC be any triangle and let D E F be the feet of the perpendiculars from the angular points on the opposite sides of the triangle ABC DEF is known as Pedal Triangle of ABC

The three perpendiculars AD BE and CF always meet in a single point H which is called the ortho-centre of triangle

1161 Sides and Angles of the Pedal Triangle

angFDE = 180deg ndash 2A angDEF = 180deg ndash 2B angDFE = 180deg ndash 2CFD = b cos B DE = c cos C FE = a cos Aor FD = R sin 2B DE = R sin 2C FE = R sin 2A

11611 Perimeter of pedal triangle

R(sin 2A + sin 2B + sin 2C) = 4R sinA sinB sinC

NoteIf the angle ACB of the given triangle is obtuse the expressions 180deg ndash 2C and c cosC are both negative and the values we have obtained require some modification In this case the angles are 2A 2B 2C ndash 180deg and the sides are a cosA b cos B ndash c cos C

Distance of the orthocentre from the angular points of the triangleAH = 2R cos ABH = 2R cos B CH = 2R cosC

11612 Distances of the orthocentre from the sides of the triangle

HD = 2R cosB cosC HE = 2R cosA cosC HF = 2R cosA cosB

sin AAH 2R cos A cosBcosCHD 2R cosBcosC sin A cos A

sin(B C)tanB tanCcosBcosC

tan A tan A

Area and Circum-radius of the Pedal Triangle (a) Area of triangle = 12 (product of two sides)times (sin of included angle) = 12 (Rsin 2B) (R Sin 2C)

sin(180deg - 2A) 21 R sin2Asin2Bsin2C2=

(b) Circumradius = EF R sin2A R2sinFDE 2sin(180 2A) 2

= =degminus

(c) The in-radius of the Pedal Triangle Area of ( DEF)

DEFSemi Perimeter of DEF

= 21 R sin2Asin2Bsin2C

2 2R cos A cosB cosC2R sin AsinBsinC

117 In-cEntrE of PEdAL trIAnGLE

Since HD HE and HF bisect the angles FDE DEF and EFD respectively So that H is the in-centre of the triangle DEF Thus the orthocentre of a triangle is the in-centre of the pedal triangle

118 cIrcumcIrcLE of PEdAL trIAnGLE (nInE-PoInt cIrcLE)

The circumcircle of pedal triangle for any DABC is called a nine-point circle

1181 Properties of Nine-point Circle

1 If passes through nine points of triangle L M N (feet of altitudes) D E F (mid points of sides) and midpoints of HA HB HC where H is orthocentre of triangle ABC

2 Its centre is called nine-points centre (N) It is the circumcentre of a pedal triangle

3 Its radius is 91 2

4 O (orthocentre) N G C (circumcentre) are collinear bull N divides OC in ratio 11 bull G divides OC in ratio 21 5 If circumcentre of triangle be origin and centroid has coordinate (x y) then coordinate of

orthocentre = (3x 3y) coordinate of nine point centre 3 32 2x y =

119 thE Ex-cEntrAL trIAnGLE

Let ABC be a triangle and I be the centre of incircle Let IA IB IC be the centres of the escribed circles which are opposite to A B and C respectively then IA IB IC is called the ex-central triangle of D ABC By geometry IC bisects the angle ACB and IBC bisects the angle ACM

ang ICIB = angACI + angACIB = 12

ang ACB + 12

ang ACM = 12

ang (180deg) = 90deg

Similarly ang ICIA = 90deg

Hence IA IB is a straight line perpendicular to IC Similarly AI is perpendicular to the straight line IBIC and BI is perpendicular to the straight line IA IC

Also since IA and IAA both bisect the angle BAC hence A I and IA are collinear Similarly BIIB and CIIC are straight lines

Hence IA IB IC is a triangle thus the triangle ABC is the pedal triangle of its ex-central triangle IA IB IC The angles IBIA and ICIA are right angles hence the points B I C IA are concyclic Similarly C I A IB and the points A I B IC are concyclic

The lines AIA BIB CIC meet at the incentre I which is therefore the orthocentre of the ex-central triangle IA IB IC

Remarks

1 Each of the four points I IA IB IC is the orthocente of the triangle formed by joining the other three points

2 The circumcentre the centroid the centre of the nine point circle and the orthocentre all lie on a straight line

1110 cEntroId And mEdIAnS of AnY trIAnGLE

In triangle ABC the midpoint of sides BC CA and AB are D E and F respectively The lines AD BE and CF are called medians of the triangle ABC the point of concurrency of three medians is called centroid Generally it is represented by G

By geometry 2 2AG AD BG BE3 3

= = and 2CG CF3

1111 LEnGth of mEdIAnS

= + minus2 2 21AD 2b 2c a2

2 2 21BE 2c 2a b2

= + minus and 2 2 21CF 2a 2b c2

= + minus

The angles that the median makes with sides

Let angBAD = b and angCAD = g we have sin DC asinC AD 2x

γ= = (Let AD = x)

a sinC a sinCsin2x 2b 2c a

γ = =+ minus

a sinBsin2b 2c a

β =+ minus

Again sin AC bsinC AD x

bsinC 2bsinCsinx 2b 2c a

θ = =+ minus

11111 The Centroid Lies on the Line Joining the Circumcentre to the Orthocentre

Let O and H represent the circum-centre and orthocenter respectively OM is perpendicular to BC Let AM meets HO at G The two triangles AHG and GMO are equiangular

AH = 2R cosA and in DOMC OM = RcosA

rArr AH 2R cos A 2OM R cos A

Hence by similar triangles AG HG AH 2GM GO OM

rArr G divides AM in the ratio 2 1 Clearly G is the centroid of DABC and G divides HA in the ratio 2 1 Thus centroid lies on the

line joining the orthocentre to the circum-centre and divides it in the ratio 2 1

The distance of the orthocentre from the circum-centre

OH R 1 8cos A cosBcosC= minus

The distance between the incentre and circumcentre

OI R 1 8sinB 2sinC 2sin A 2= minus

The distance of an ex-centre from the circum-centre

OI1 = A B CR 1 8sin cos cos2 2 2

+ OI2 = A B CR 1 8cos sin cos2 2 2

OI2 = R 1 8cos(A 2)cos(B 2)sin(C 2)+

111111 The length of angle bisector and the angle that the bisector makes with the sides

Let AD be the bisector of angle A and x and y be the portions of base BC From geometry BD ABDC AC

or x y x y ac b b c b c

+= = =

acxb c

and abyb c

Further DABC = DABD + DADC

rArr 1 1 A 1 Abcsin A czsin bzsin2 2 2 2 2

bc sin A 2bcz cos A2b c sin A 2 b c

= = = + + (ii)

Also q = angBAD + B = A2 + B

The Perimeter and Area of a Regular Polygon of n-sides Inscribed in a circle of radius r

Perimeter of polygon = nAB = 2nR sin pn

Area of polygon = n(Area of triangle AOB) = 2nR 2sin

The Perimeter amp Area of Regular Polygon of n-sides Circumscribed about a given circle of radius lsquorrsquo

Perimeter of Polygon = n AB = 2n AL = 2nOL tannπ = 2n tan

Area of Polygon = n(Area of triangle AOB) = 2(OLAB)n nr tan2 n

The Radii of the inscribed and circumscribing circles of a regular polygon having n sides each of length lsquoarsquo

a aR cosec2sin n 2 n

π a ar cot

2tan n 2 nπ

1112 rESuLt rELAtEd to cYcLIc quAdrILAtrAL

(a) Ptolemyrsquos Theorem In a cyclic quadrilateral ABCD ACBD = ABCD + BCDA ie the product of diagonals is equal to the sum of product of opposite sides

(b) D = area of cyclic quadrilateral

= 1 (ab cd)sinB2

+ = (s a)(s b)(s c)(s d)minus minus minus minus where a b c d2

(c) (ac bd)(ad bc)AC(ab cd)+ +

(d) Circum-radius (R) of cyclic-quadrilateral ACABCD

2sinB= = AC (ab cd)AC

2 4A2ab cd

1 (ac bd)(ad bc)(ab cd)R4 (s a)(s b)(s c)(s d)

+ + +=

minus minus minus minus

(e) 2 2 2 2a b c dcosB

2(ab cd)+ minus minus

Chapter 12Inverse trIgonometrIC FunCtIon

121 INVerse FuNctIoN

If a function is one-to-one and onto from A to B then function g which associates each element y isin B to one and only one element x isin A such that y = f(x) hArr x = g(y) then g is called the inverse function of f denoted by g = fndash1 [Read as f inverse] Thus if f A rarr B then g B rarr A

1211 Inverse Trigonometric Functions

The equation sin x = y and x = sinndash1 y are not identical because the former associates many values of x of a single value of y while the latter associates a single x to a particular value of y To assign a unique angle to a particular value of trigonometric ratio we introduce a term called principle range

We list below the domain (values of x) and principle ranges (values of y) of all the inverse trigonometric functions and their graph

Remarks

1 sin 5π6 = 12 But 5π6 ne sinminus1(12) there4 sinndash1x cosndash1x tanndash1x denotes angles or real number lsquowhose sine is xrsquo lsquowhose cosine is xrsquo and lsquowhose tangent is xrsquo provided that the answers given are numerically smallest available

2 If there are two angles one positive and the other negative having same numerical value Then

we shall take the positive value For example cos 1

π= and cos 1

π minus =

But we write cosndash1

and cosndash1 1

ne minus4π

3 I quadrant is common to all the inverse functions

4 III quadrant is not used in inverse function

5 IV quadrant is used in the clockwise direction ie minusπ2 le y le 0

122 DomaIN aND raNge oF INVerse FuNctIoNs

Function Domain Range Principal Value Branch

y = sinndash1x [ndash1 1] [ndashπ2π2] ndashπ2 le y le π2

y = cosndash1x [ndash1 1] [0π] 0 le y le π

y = tanndash1x ℝ (ndashπ2π2) minusπ2 lt y lt π2

y = cotndash1x ℝ (0 π) 0 lt y lt π

y = secndash1x (ndashinfinndash1]cup[1infin) [0 π] ndash π2 0 le y le π y ne π2

y = cosecndash1x (ndashinfinndash1] cup[1infin) [ndashπ2 π2]ndash0 ndashπ2 le y le π2 y ne 0

RemarkIf no branch of an inverse trigonometric function is mentioned then it means the principal value branch of the function

123 graphs oF INVerse cIrcular FuNctIoNs aND theIr DomaIN aND raNge

1 Graph of function y = sin x y = sinndash1x

ndash1

ndashπ2 π2

y=sinx

ndashinfin infinO

π2 (1π2)

y = sinndash1 x(ndash1ndashπ2) ndashπ2

ndash1 O 1x

y = sinx and y = sinndash1 x(shown in single graph)

ndash110

(1 π2)

(ndashπ2ndash1)

(ndash1ndashπ2)

(π21)

π2ndashπ2 X

2 Graph of function y = cos x y = cosndash1x

ndash10

y = cos x

Y(ndash1π) π

(1 0)X0

y = cosndash1x

1ndash1

y=cosx and y=cosndash1x(shown in single graph)

(ndash1π)

(01) 1

ndash1ndash1

(0π2) π2

(π20) (πndash1)(1 0)

Inverse Trigonometric Function 112123

3 Graph of function y = tan x y = tanndash1x

Y infin

Xndashπ2

ndashinfin

ndash1

ndashπ2

ndashinfin infinO x

ndashπ2

ndashπ2ndashinfin

ndashinfin

+infin

infin xx

y = tanx y = tanndash1x y = tanx and y = tanndash1x

4 Graph of function y = cot x y = cotndash1x

xndashπ2 π

ndashinfin

+infin

πndashinfin

ndashπ2π2

ndashinfin

+infin

y = cot x y = cotndash1 x y = cotx and y = cotndash1x

4 Graph of function y = sec x y = secndash1x

0ndash1

π2 xπ

+infin

ndashinfin

O 1ndash1

infinndashinfin

Y+infin

+infin

ndashinfin

0 1ndash1

ndash1

y = sec x y = secndash1x y = secx and y =secndash1x

5 Graph of function y = cosec x y = cosecndash1x

ndashπ2ndash10 π2

+infin

ndashinfin

π2(1π2)

(π21)

(ndashπ2ndash1)

(ndash1 ndashπ2)

π2ndashπ2

ndashπ2

ndash1

+infin

ndashinfin y = cosec x y = cosecndash1x y = cosecx and y = cosecndash1x

124 composItIoNs oF trIgoNometrIc FuNctIoNs aND theIr INVerse FuNctIoNs

1241 Trigonometric Functions of Their Corresponding Circular Functions

(i) sin (sinminus1 x) = x for all x isin [minus1 1]

(ii) cos (cosminus1 x) = x for all x isin [minus1 1]

(iii) tan (tanminus1 x) = x for all x isin ℝ

(iv) cot (cotminus1 x) = x for all x isin ℝ

(v) cosec (cosecminus1 x) = x for all x isin (minusinfin minus1] cup [1 infin)

(vi) sec (secminus1 x) = x for all x isin (minusinfin minus1] cup [1 infin)

125 INVerse cIrcular FuNctIoNs oF theIr correspoNDINg trIgoNometrIc FuNctIoNs oN prINcIpal DomaIN

(i) sinndash1 (sin x) = x for all x isin [minusπ2 π2] (ii) cosndash1 (cos x) = x for all x isin [0 π] (iii) tanndash1 (tanx) = x for all x isin (minusπ2 π2) (iv) cotndash1 (cot x) = x for all x isin (0 π) (v) secndash1 (sec x) = x for all x isin [0 π] ~ π2 (vi) cosecndash1 (cosec x) = x for all x isin [minusπ2 π2] ~ 0

126 INVerse cIrcular FuNctIoNs oF theIr correspoNDINg trIgoNometrIc FuNctIoNs oN DomaIN

1 sinminus1 (sin x) =

minusπminus isin minus π minusπ isin minusπ ππminus isin π πminus π+ isin π π

x if x [ 3 2 2]x if x [ 2 2]

x if x [ 2 3 2]2 x if x [3 2 5 2]

and so on as shown below

Ondash1

ndash1

y=sin(sinndash1x)=cos(cosndash1x)=x

O45deg

ndash1

ndash11

y=tan(tanndash1x)=cot(cotndash1x)=x

minusπ 2 y=x

πminus

y=-(x)

πminusy=x-2

minus3π

y=ndash(3π+x)

minus5π2π2

3π25π2πminusπ

minus2πminus3π2π 3π0

y=sinndash1(sinx)

Domain ℝ Range 2 2π π minus

Period 2π

Remarky = sinndash1 (sinx) can be formed by tangents of y = sinx at x = nπ as shown below

ndash3π

ndash3π2

ndash5π2 ndashπ2π2 y=πndashx y=

xndash2π y=3πndashx

π23π2

5π22π 3πx

ndashπ2ndashπ πO

y y=sinndash1(sinx)

y=2π+

y=(3π+x

y=ndash(π+x)ndash2π

2 cosminus1 (cos x) =

x if x [ 0]x if x [0 ]2 x if x [ 2 ]

2 x if x [2 3 ]

minus isin minusπ isin π πminus isin π πminus π+ isin π π

and so on as shown

π 2πndashπndash2πndash3π 3πx

y=2πndashx

y=xndash

2πy=ndash(x+2π)

y=ndashx

y = cosndash1(cosx)

Domain ℝ Range [0 π] Period 2π

3 tanminus1 (tan x) =

( )( )( )( )

x if x 3 2 2

x if x 2 2

x if x 2 3 2

x 2 if x 3 2 5 2

π+ isin minus π minusπ

isin minusπ π

minusπ isin π π minus π isin π π

and so on as shown

minusπ 2minusπ 2minus3π2

π2 3π2π

minusπminus2π2πO

y=tanndash1(tanx)

Domain ~ (2n 1)2π +

Range 2 2π π minus

Period π

x 2 for x ( 2 )x for x ( 0)

y cot (cot x) x for x (0 )x for x ( 2 )x 2 for x (2 3 )

+ π isin minus π minusπ + π isin minusπ= = isin π minusπ isin π π minus π isin π π

The graph of cotndash1 (cotx) is as shown Domain x isin R minus n π n isin ℤ Range y isin (0 π) Period periodic with period π and and cotminus1 (cot x ) = x forall x isin (0π)

5 y = secndash1 (sec x) =

x for x [ 0]

x for x [0 ]~232 x for x [ 2 ]~2

minus isin minusπ π isin π

π πminus isin π π

minusπ 2minus3π 2 3π2

y=2πndashx

2π y=ndashx

ndashπminus2π 2πO

y=secndash1(secx)

The graph of y = secndash1 (secx) is as shown

Domain x isin ℝ minus (2n 1) n2π + isin

Range y isin [0 π2) cup (π2 π]

Period Periodic with period 2π and secminus1(sec x) = x forall x isin [0 π2) cup (π2 π]

6 y = cosecndash1 (cosec x)

3( x) for x ~ 2 2

x for x ~ 02 2

3x for x ~ 2 2

minus π minusπ minus π+ isin minusπ minusπ π = isin π π πminus isin π

minusπ 2

πminus

y= ndash (x)

π+y=2x

y=x ndash2ππ 2

πminusπminus2π 2π0

y=cosecndash1(cosecx)

Domain x isin ℝ sim nπn isin ℤ Range y isin [ndash π2 π2] sim 0 Period Periodic with period 2π and cosecminus1(cosec x) = x for x isin [ndash π2 π2] sim 0

127 INVerse trIgoNometrIc FuNctIoNs oF NegatIVe INputs

(i) sinndash1 (ndashx) = ndashsinndash1 (x) for all x isin [ndash1 1] (ii) cosndash1 (ndashx) = π ndash cosndash1 (x) for all x isin [ndash1 1] (iii) tanndash1(ndashx) = ndashtanndash1 x for all x isin R (iv) cosecndash1(ndashx) = ndashcosecndash1 x for all x isin (minusinfin minus1] cup[1 infin) (v) secndash1 (ndashx) = π ndash secndash1x for all x isin (minusinfin minus1] cup [1 infin) (vi) cotndash1(ndashx) = π minus cotndash1 x for all x isin R

y=xndash

y=cotndash1(cotx)

128 INVerse trIgoNometrIc FuNctIoNs oF recIprocal INputs

(i) sinminus1 (1x) = cosecminus1 x for all x isin (minusinfin ndash1] cup [1 infin) (ii) cosminus1 (1x) = secndash1 x for all x isin (minusinfin ndash1] cup [1 infin)

(iii) tanminus1(1x) = 1

cot x for x 0cot x for x 0

gtminusπ+ lt

129 INter coNVersIoN oF INVerse trIgoNometrIc FuNctIoNs

(a) sinndash1x = 1 2

cos 1 x if 0 x 1

cos 1 x if 1 x 0

minus le le minus minus minus le le

xtan1 x

minus if forall x isin (ndash1 1)

1 xcot if 0 x 1x

1 xcot if 1 x 0x

minuslt le

minus minusπ+ minus le lt

1sec if x 0 11 x

1sec if x 1 01 x

minus isin minus minus

= 1 1cosecx

if x isin [ndash1 1] ~ 0

(b) minus

minus isin = πminus minus isin minus

sin 1 x for x 0 1cos x

sin 1 x for x 1 0 =

1 xtan for x (0 1]x

1 xtan for x [ 1 0)x

minus isin

minusπ+ isin minus

xcot for x ( 1 1)1 x

minus isin minus

minus = 1 1sec for x 1 1 ~ 0

xminus isin minus

1cosec for x [0 1)1 x

1cosec for x ( 1 0]1 x

πminus isin minus minus

(c) 1 1

xtan x sin for x1 x

minus minus = isin

1cos for x [0 1]1 x

1cos for x [ 1 0]1 x

minus isin minus +

1cot for x 0x

minusπ+ lt

= ( )( )

sec 1 x for x 0

+ gtminus + lt

1 1 xcosec for x ~ 0x

minus + isin

1sin for x 01 xcot x

1sin for x 01 x

+ = πminus le +

xcos x1 x

minus forall isin

1tan for x 0x

π+ lt

1 1 xsec x ~ 0x

minus + forall isin

( )( )

cosec 1 x for x 0

+ gtπminus + lt

x 1sin for x 0x

sec xx 1sin for x 0

minus gt =

minus π+ lt

= 1 1cos x ~ 0x

minus forall isin

= ( )1 2

tan x 1 for x 0

minus gtπminus minus lt

1cot for x 0 x 1x 1

πminus lt ne minus minus

xcosec for x 0x 1

π+ lt minus

(f) 1 1 1cosec x sin for x ~ 0x

minus minus= isin =

x 1cos for x 0x

minus gt

minus minusπ+ lt

1tan for x 0 1x 1

minus lt ne minus minus

= ( )( )

cot x 1 for x 0

minus gtminus minus lt

xsec for x 0 1x 1

minusπ+ lt ne minus minus

1210 three ImportaNt IDeNtItIes oF INVerse trIgoNometrIc FuNctIoNs

(i) sinndash1x + cosndash1 x = π2 for all x isin[ndash1 1] (ii) tanndash1x + cotndash1 x = π2 for all x isin R (iii) secndash1x + cosecndash1 x = π2 for all x isin(ndashinfin ndash1] cup [1 infin)

1211 multIples oF INVerse trIgoNometrIc FuNctIoNs

Property (1)

1 1sin (2x 1 x ) if x2 2

12sin x sin (2x 1 x ) if x 12

1sin (2x 1 x ) if 1 x2

minus minus

minus minus le le

= πminus minus le leminusπminus minus minus le le minus

Property (2)

1 1sin (3x 4x ) if x2 2

13sin x sin (3x 4x ) if x 12

1sin (3x 4x ) if 1 x2

minus minus

minus minus le le= πminus minus le leminusπminus minus minus le le minus

Property (3) 2cosndash1 x = 1 2

cos (2x 1) if 0 x 12 cos (2x 1) if 1 x 0

minus le leπminus minus minus le le

Property (4) 3 cosndash1 x =

1cos (4x 3x) if x 121 12 cos (4x 3x) if x2 2

12 cos (4x 3x) if 1 x2

minus le le πminus minus minus le le π+ minus minus le le minus

Property (5)

2xtan if 1 x 11 x

2xtan if x 11 x2tan x

2xtan if x 11 x

for x 12

minus lt lt minus π+ gt minus =

minusπ+ lt minus minus π =

Property (6) 3 tanndash1 x =

3x x 1 1tan if x1 3x 3 3

3x x 1tan if x1 3x 33x x 1tan if x1 3x 3

1for x2 3

minusminus lt lt minus

minus π+ gt minus minusminusπ+ lt minus minus

2xsin if 1 x 11 x

2xsin if x 11 x

minus le le + πminus gt + minusπminus lt minus +

1 xcos if 0 x1 x

1 xcos if x 01 x

minusle ltinfin +

minusminus minusinfin lt le +

1212 sum aND DIFFereNce oF INVerse trIgoNometrIc FuNctIoNs

Property (1)

2 21 2 2

1 1 1 2 2 2 2

1 2 2 2 2

if x y 1sin x 1 y y 1 x

or if xy 0 and x y 1 where x y 11

sin x sin y sin x 1 y y 1 x if 0 x y 1 and x y 1

sin x 1 y y 1 x if 1 x y 0 and x y 1

minus minus minus

+ leminus + minus

lt + gt isin minus + = πminus minus + minus lt le + gtminusπminus minus + minus minus le lt + gt

Property (2)

2 21 2 2

1 1 1 2 2 2 2

1 2 2 2 2

or xy 0 and x y 1 where x y 1 1

sin x sin y sin x 1 y y 1 x if 0 x 1 1 y 0 and x y 1

sin x 1 y y 1 x if 1 x 0 0 y 1 and x y 1

minus minus minus

+ leminus minus minus

gt + gt isin minus minus = πminus minus minus minus lt le minus le le + gtminusπminus minus minus minus minus le lt lt le + gt

Property (3)

cosndash1 x + cosndash1y =

cos xy 1 x 1 y if 1 x y 1 and x y 0

2 cos xy 1 x 1 y if 1 x y 1 and x y 0

minus minus minus

minus le le + geπminus minus minus minus

minus le le + le

Property (4)

cosndash1x ndash cosndash1y =

cos xy 1 x 1 y if 1 x y 1 and x y

+ minus minus

minus le le leminus + minus minus minus le le ge

Property (5)

tanndash1x + tanndash1 y =

x ytan if xy 11 xy

x ytan if x 0 y 0 and xy 11 xy

for x 0 y 0 and xy 12

+lt minus

+π+ gt gt gt minus +minusπ+ lt lt gt minus π gt gt =

πminus lt lt =

Property (6)

tanndash1x ndash tanndash1 y =

x ytan if xy 11 xy

minusgt minus +

minusπ+ gt lt gt minus + minusminusπ+ lt gt gt minus + π gt gt = minus

πminus lt lt = minus

Chapter 13point and

Cartesian system

131 IntroductIon

The study of co-ordinate geometry begins with the study of ldquoconcept of pointrdquo which is defined as a geometrical construction having no dimensions Several methods have been developed by mathematicians to uniquely locate the position of a point in the space

132 FrAME oF rEFrEncE

It is a set of fixed pointslinesurfaces with respect to which the following observations are made ∎ Rectangular co-ordinate System ∎ Oblique co-ordinate System ∎ Polar co-ordinate System

1321 Rectangular Co-ordinate SystemAny point P in (x y) plane can be represented by unique ordered pair of two real numbers (x y) Here x is abscissa of point (OM or PN) Y is ordinate of point (ON or PM)

Sign ConventionTherefore the x-y plane (Cartesian plane) is algebraically represented as Cartesian product of two set of real numbers

So called as ℝ times ℝ (ℝ2) plane ℝ times ℝ = (x y) x isin ℝ and y isin ℝ

ℝ+ times ℝ+ = 1 quadrant ℝndash times ℝ+ = 2nd quadrant ℝndash times ℝndash = 3rd quadrant ℝ+ times ℝndash = 4th quadrant

1322 Polar Co-ordinate SystemIt consist of a fixed point O which is known as pole and semi-infinite ray OX which is called initial line ∎ The polar coordinate of any point P is given as (r q) where r is the distance

of point P from pole O is lsquorrsquo and the angle angXOP = q

Point and Cartesian System 13133

133 dIstAncE ForMulA

The distance between any two points P and Q when coordinate of two points is given in Cartesian form Let P(x1 y1) and Q(x2 y2) be two given points then

2 22 1 2 1PQ d (x x ) (y y )= = minus + minus

1331 Applications of Distance Formula

Position of three points Let A B C are points lying in a plane then two condition arises either they are collinear or they form a triangle

13311 Collinearity of three given points

The three given points A B C are collinear ie lie on the same straight line if ∎ any of the three points (say B) lie on the straight line joining the other two points

∎ area of DABC is zero It means 1 1

x y 1x y 1 0x y 1

∆ = =

rArr [x1(y2 ndash y3) + x2(y3 ndash y1) + x3(y1 ndash y2)] = 0 ∎ slope of line AB(mAB) = slope of line BC(mBC) = slope of line AC(mAC) ∎ coordinates of any of the points x1 and y1 can be written as linear combination of other two x2 x3 and

y2 y3 as x1=lx2+mx3 and y1 =ly2+my3 such that l + m = 1

134 sEctIon ForMulA IntErnAl dIvIsIon

Co-ordinates of a point which divides the line seg-ment joining two points P(x1 y1) and Q(x2 y2) in the

ratio m n internally are 2 1 2 1mx nx my nym n m n+ +

∎ If P is the mid-point of AB then it divides AB in the ratio 11 so its coordinates

are 1 2 1 2x x y y

2 2+ +

∎ The given diagram helps in remembering the section formula

Coordinates of a point which divides the line segment joining two points P(x1 y1) and Q(x2 y2) in the

ratio m n externally are 2 1 2 1mx nx my ny

m n m nminus minus

minus minus

∎ To get the point of the external division only replace the n of internal division by -n

∎ Co-ordinates of any point on the line segment joining two points P(x1 y1) and Q(x2 y2) and dividing it

in the ratio l1 is given by 1 2 1 2x x y y ( 1)

1 1λ λ λ

λ λ+ + ne minus + +

∎ Lines formed by joining (x1 y1) and (x2 y2) is divided by

(a) x-axis in the ratio hArr -y1y2 (b) y-axis in the ratio hArr -x1x2

If the ratio is positive the axis divide it internally and if negative then divides externally

∎ Line ax + by + c = 0 divides the line joining the points P(x1 y1) and Q(x2 y2) in the ratio l 1

then 1 1

ax by c

ax by cλ

+ += minus + +

If l is positive it divides internally if l is negative then externally

135 slopE oF lInE sEgMEnt

Slope of a line segment is a physical quantity that measures the amount of inclination of the line with respect to the x axis and defined as rate of change of ordinate with respect to the abscissa

Denoted as ym x

∆ = ∆ bull Slope can be obtained as tangent of angle that line

segment makes with positive direction of x axis in anticlockwise sense

rArr 2 1

y ym tanx xminus

= θ =minus

bull If Line is horizontal rArr q = 0 rArr m = 0 bull If line vertical rArr q = 90deg rArr m rarr infin bull If the points A and B coincide rArr Slope is indeterminate

1351 Area of Triangle

Area of triangle when the coordinates of vertices A B C of triangle are A(x1 y1) B(x2 y2) and C(x3 y3) is given as

1 2 3 2 3 1 3 1 21[ x (y y ) x (y y ) x (y y )]2

∆ = minus + minus + minus

This expression for the area can also be written in the

determinant form 1 1

x y 11 x y 12

x y 1∆ =

∎ If area of D is zero then the point are collinear Hence for three points to be collinear the essential

condition is area of D = 0 rArr 1 1

x y 1 0

∎ If the coordinate of vertices of D are given in polar form (r1 q1) (r2 q2) (r3 q3) then the area of D will

be given by [ ]2 3 1 1 3 1 1 3 1 2 2 1

1r r sin( ) r r sin( ) r r sin( )

2θ θ θ θ θ θ∆ = minus + minus + minus

1352 Area of General Quadrilateral

If A(x1 y1) B(x2 y3) C(x3 y3) and D(x4 y4) are vertices of a quadrilateral then its area will be given

x y 1x y 11x y 12x y 1

= 1 2 2 1 2 3 3 2 3 4 4 3 4 1 1 41[(x y x y ) (x y x y ) (x y x y ) (x y x y )]2

minus + minus + minus + minus =

x y 1x y 11x y 12x y 1

NoteIf area of a quadrilateral joining four points is zero then four points are collinear

1353 Area of Polygon

The area of polygon whose vertices are (x1 y1) (x2 y2) (x3 y3)(xn yn) is |(x1y2 ndash x2y1) + (xny3 ndash x3y2) ++(xny1 ndash x1yn)|

Stair method Repeat first co-ordinate one time in last for down arrow use +ve sign and for up arrow use -ve sign

Area of polygon =

x yx yx y

x yx y

= 1 2 2 3 n 1 1 2 2 3 n 11 |(x y x y x y ) (y x y x y x )|2

+ + + minus + + +

bull Area of a triangle can also be found by easy method ie stair method

x yx y1x y2x y

∆ = = 1 2 2 3 3 1 1 2 2 3 3 11 |(x y x y x y ) (y x y x y x )|2

+ + minus + +

bull If one vertex (x3 y3) is at (0 0) then D = 1 2 2 11 |(x y x y )|2

∆ = minus

136 locus oF poInt And EquAtIon oF locus

The path traced by a moving point P(x y) is called locus of P The equation of locus is a relation in the variable x and y which is satisfied by the coordinates of the moving point P(which moves under given geometrical restriction) at any position on its path

That is if f(x y) = 0 is satisfied by (a b) forall (a b) lying on the path then its called equation of locus

Method to Find Equation of Locus

Step I Let coordinate of point P be (h k) and apply the condition given to express h and k as a function of some parameter (q a b t l etc )

Step II Eliminate the parameters to relate h and k

Step III In the equation between h and k therefore obtained replace h by x and k by y to get equation of locus

1361 Union of LociLocus is a set of points that follow a given relation in x and y

Given two loci S1 and S2 defined as belowS1 (x y) S = f (x y) = 0 and S2 (x y) S = g(x y) =0Union of loci S = 0 and S = 0 is set of those points which lie ether

on S = 0 or S = 0 rArr S1 cup S2 = (x y) f(x y) = 0 or g(x y) = 0 And its equation is given

as S S = 0 ie f(x y) g(x y) = 0

1362 Intersection of LociIntersection of loci S = 0 and S = 0 is defined as set of those points which lie on both the curves S = 0 and S = 0 That is set of common points

rArr S1 cap S2 = (x y) f(x y) = 0 or g(x y) = 0 and its equation is given as

rArr |f(x y)| + |g(x y)| = 0 or |S| + |S| = 0 or radicS + radicS = 0 or S2 + S2 = 0

1363 Locus Passing Through Intersection of Two Locus

Given two loci S = 0 and S = 0 defined as ( )( )x y S f(x y) 0x y S g(x y) 0

= = The equation S + lS = 0 represents a family of curve

passing through A and B that is intersection of S = 0 and S = 0 where l is a real parameterDiscussion S + lS = 0 rArr f(x y) + lg(x y) = 0Represent infinitely many curve due to parameter l and since both point A and B satisfy the above equation because f(a b) = g(a b) = f(g d) = g(g d) = 0

rArr f(a b) + lg(a b) = 0 + l0 = 0

Ellipse Ellipse is a locus of a point which moves so that the summation of its distances from two fixed points A and B remains constant l

Hyperbola Hyperbola is locus of a point which moves so that the difference of its distances from two fixed points A and B remains constant l

Parabola It is the locus of all points such that the distance from a fixed point and perpendicular distance from a fixed line is always equal

Circle Locus of all points which are equidistant from a given point in a plane

137 cHoIcE oF orIgIn And sElEctIon oF coordInAtE AXEs

In order to solve any general geometric problem conveniently a suitable choice of origin and proper selection of coordinate axes can be considered but care must be taken that during such selection the generality of the problem is not lost So any assumption is regarded as perfectly general iff by shifting the origin to a suitable point and rotating the coordinate axes by some angle the most general case can be transformed to assumed case

138 gEoMEtrIcAl trAnsForMAtIons

Any geometric operation undergoing through which the coordinate of the point changes It is of two types (i) Linear Transformation A transformation in which the origin of reference frame does not

change and the new coordinate obtained are linear function of old coordinate ie xrsquo = ax + by and yrsquo = cx + dy is called linear transformation

(ii) Non-linear Transformation In such a transformation the straight line remains straight The remaining transformations are called non-linear transformation

1381 Transformations in Cartesian Plane

T1 Reflection of point in x-axis

1T(x y) (x y)rarr minus

T2 Reflection of point in y-axis

2T(x y) ( x y)rarr minus

T3 Reflection of point in origin

3T(x y) ( x y)rarr minus minus

T4 Reflection of point in the line y = x

4T(x y) (x y)rarr

T5 Rotation of point about origin

5T(x y) (x y )rarr

T6 Reflection of point in the line y = xtan q

6T(x y) (x y )rarr

1382 Transformation of Coordinates Axis

Shifting of origin without rotating axes If origin of coordinate frame is shifted to O to O (h k) keeping the coordinate axis respectively parallel regional axes

Conclusion ∎ New coordinate of point P in terms of old x = x ndash h and

y = y ndash k ∎ Old coordinate of point P in terms of new x = x + h and

y = y + k ∎ The transformation equation of a locus f(x y) = 0 is

obtained by replacing x by x + h and y by y + k x x h

y y kf(x y) 0 f(x h y k) 0rarr +rarr += rarr + + =

1383 Rotation of the Axes (Without Changing Origin)

To change the direction of the axis of coordinates without changing the origin let OX and OY be the old axes and OX and OY be the new axes obtained by rotating the old axes through an angle q in anti-clock wise sense about origin

The old coordinate of P(x y) with respect to new coordi-nate axes will be given by

x = ON ndash NL y = PQ + QL

x x cos y siny y cos x sin

prime prime= θminus θprime prime= θ+ θ

helliphellip (i)

139 gEoMEtrIcAl tIps And trIcks

Method to Find Circum Centre

Step I Consider (OA)2 = (OB)2 = (OC)2 rArr (x ndash x1)2 + (y ndash y1)

2 = (x ndash x2)

2 + (y ndash y2)2 = (x ndash x3)

2 + (y ndash y3)2

Step II Solving two linear equations obtained we can get coordinates of circum-centre

Step III The obtained value of x and y always satisfy third equation that indicates the concurrency of ^ bisectors

1391 The Coordinates of Centroid

In a DABC the coordinates of centroid are given by a b c a b cx x x y y y

3 3+ + + +

bull If mid-points of the sides of a triangle ABC are D E F respectively of BC CA AB as shown in the figure then A(xE + xF ndash xD yE + yF ndash yD) B(xD + xF ndash xE yD + yF ndash yE) and C(xD + xE ndash xE yD + yE ndash yF)

Area of DABC = 4 times Area of DDEF ie area of a D is four times the area of the D formed by joining the midpoints of its sides

bull If two vertices of a D are (x1 y1) and (x2 y2) and the coordinates of Centroid are (a b) then co-ordinates of the third vertices are (3a ndash x1 ndash x2 3b ndash y1 ndash y2)

1392 Coordinates of Incentre

If A(x1 y1) B(x2 y2) and C(x3 y3) are the vertices of the DABC with sides BC CA AB of lengths a b c

respectively then the coordinates of the incentre 1 2 3 1 2 3ax bx cx ay by cyI

a b c a b c+ + + + = + + + +

1393 Coordinates of Ex-centre

The coordinates of ex-centres of the triangle are given by

rArr A

(b a)y cyy

b a cminus +

=minus +

rArr A

B A CI

bx ax cxx

b a cminus +

=minus +

and minus +=

minus +A

B A CI

by ay cyy

Chapter 14Straight line and

pair of Straight line

141 Definition

A straight line is a curve such that every point on the line segment joining any two points lie on it or in other words straight line is the locus of a point which moves such that the slope of line segment joining any two of its position remains constant

1411 Equation of Straight Line

A relation between x and y which is satisfied by coordinates of every point lying on a line is called the equation of straight line Every first degree equation in x y ie ax + by + c = 0 represents a line Thus a line which is also defined as the locus of a point satisfying the condition ax + by + c = 0 where a b c are constant

∎ Equation of straight line parallel to axes (i) Equation of a straight line which is parallel to x-axis and at a distance b units from it is given by

y = b b gt or lt 0 according as it is in positive or negative side

equation of x-axis is y = 0 (ii) Similarly for any line parallel to the y axis and at a distance a unit from it is given by x = a

a gt or a lt 0 according as the line lies on positive or negative sides of the x-axis ∎ The combined equation of the coordinate axis is xy = 0

Straight Line and Pair of Straight Line 14141

1412 Different Forms of the Equation of Straight Line

Two Point From Straight line passing through A(x1 y1) and B(x2

2 11 1

y yy y (x x )x xminus

minus = minusminus

or 1 1

x y 1x y 1 0x y 1

determinant form

Slope Point From Straight line passing through A(x1 y1) and having slope m

y ndash y1 = m(x ndash x1)

Slope Intercept Form Equation of line having slope lsquomrsquo and making an intercept c on y-axis

y = mx + c where q is the angle made by line with +ve direction of x-axis in counter-clockwise sense

Two Intercept From Equation of line making intercepts a and b respectively on x and y axis

x y 1a b+ = or

x y 1a 0 1 00 b 1

in determinant form

PerpendicularNormal Form Equation of line upon which the length of perpendicular form origin is p and perpendicular makes a angle with +ve direction of x-axis

x cos a + y sin a = prArr If equation of line be x cos a + y sin a = ndashp (p gt 0) the equation will not be in normal form to convert it to normal form multiply both sides by ndash1rArr x(ndashcosa) + y(ndashsina)

= prArr x cos(p + a) +

ysin(p+a) = p

Symmetric (Parametric) From Straight line passing through A(x1 y1) and making angle q with positive x-axis

1 1x x y y rcos sinminus minus

= =θ θ

where r is distance of the point P(x y) from the fixed point A(x1 y1)

rArr Using symmetric form bull To find coordinate of any point P(x y) from the fixed point A(x1y1) on the line if AP is given as r rArr x = x1+ r cosq and y = y1 + r sinq bull To find distance of a point from a fixed point on the line along the line bull To find distance r if qis known q if r is given

1413 Angle Between Two Lines

Given two lines

11 1 1 1 1

aL a x b y c 0 m tanb

+ + = = minus = α

22 2 2 2 2

+ + = = minus = β

The angle between L1 = 0 and L2 = 0 q = b ndash a

rArr 2 1

m mtan1 m m

minusθ =

rArr 1 2 2 1

1 2 1 2

a b a btana a b b

minusθ =

14131 Conclusion

∎ There are two angles formed between any pair of line q and p ndash q (say) then tangent of acute angle q

m mtan1 m m

minusθ =

+ and 2 1

m mtan( )1 m m

minusπminusθ = minus

∎ Lines are parallel if rArr tanq = 0 rArr m1 = m2

∎ Lines are perpendicular rArr tanq rarr infin rArr m1 m2 = ndash1

∎ Lines are coincident if they have same slope and intercept

rArr 1 1 1

a b ca b c

∎ Lines L1 = 0 and L2 = 0 are perpendicular when q = 90 ∎ If m1m2 = 1 then angle of L1 with x-axis is same as angle of L2 with y-axis Hence both lines make same angle with y = x + k and y = ndashx + k ∎ If m1 + m2 = 0 Lines L1 and L2 make supplementary angles with x and y-axis when extended to

intersect they form an isosceles triangle with the coordinates axis (x or y)

1414 Equation of a Line Perpendicular and Parallel to Given Line

rArr Let m be the slope of the line ax + by + c = 0 Then m = -ab Since the required line is parallel to the given line The slope of the required line is also m Let C1 be the intercept by the line on y-axis Then its equation is y = mx + c1

rArr 1ay x c

bminus

rArr ax + by - bc1 = 0 rArr ax + by + l = 0 where l = - bc1 = constant The equation of line parallel to a given line is ax + by + l = 0

To find the equation of a line parallel to a given line keep the expression containing x and y same and simply replace the given constant by a new arbitrary constant l The value of a l can be determined by same given condition

rArr The equation of line perpendicular to given line ax+ by + cz = 0 is bx ndash ay + l = 0

ie interchange the coefficient of x and y by reversing the sign of exactly of them one and replace the constant term by parameter l

1415 Straight Line Through (x1 y1) Making an Angle α with y = mx + c

Equation of line passing through a point A(x1 y1) and making a given angle q with the line y = mx + c

Let slope of the line be mrsquo

m mtan1 mm

minusθ =

+ rArr m m tan

1 mmminus

= plusmn θ+

rArr m ndash m = plusmn tan q plusmn m m tan q rArr m tan m(1 m tan )θ = plusmn θ

rArr m tanm1 m tan

plusmn θ

So equation of lines are 1 1m tany y (x x )

1 m tanθ

minus = minusplusmn θ

1416 Position of Two Points wrt a Straight Line

Two points P(x1 y1) and Q(x2 y2) lie on the same side or on the opposite side of the line ax + by + c = 0 according as ax1 + by1 + c and ax2 + by2 + c are of the same sign or opposite signs respectively The coordinates

of the point R which divides the line joining P and Q sides in the ratio mn are 2 1 2 1mx nx my nym n m n+ +

If this point lie on (i) then 2 1 2 1mx nx my nya b c 0m n m n+ + + + = + +

rArr m(ax2 + by2 +c) + n(ax1 + b1 + c) = 0

rArr 1 1

ax by cmn ax by c

+ += minus

If the point R is between the points P and Q

Then the ratio m n is positive So from the above equation we get 1 1

ax by c 0ax by c

+ + rArr ax1 + by1 + c and ax2 + by2 + c are of opposite sign

If point R is not between P and Q then the ratio m n is negative

rArr 1 1

ax by c 0ax by c

rArr ax1 + by1 + c and ax2 + by2 + c are of same sign

NoteIf the location of a single point is to be defined then the other point is taken as the origin and wrt the origin The location of the point wrt the line is defined

Two points P(x1 y1) and Q(x2 y2) will be located at the same side of the line If they give the same sign of the expression when they are used in the line otherwise they will lie on the opposite side of the line

1417 Distance of a Point From a LineLet the given line be ax + by + c = 0 then the distance of any point P(x1 y1) from the given line be

rArr 1 12 2

|ax by c |PNa b

The length of the perpendicular from the origin to the line ax + by + c = 0 is 2 2

1418 Distance Between Two Parallel Straight LinesLet ax + by + c = 0 and ax + by + c = 0 be the parallel straight lines then the distance between them is

given by 2 2

a b+ rArr Oblique distance of a point from a line Distance of a point

P(x1 y1) from a line L1 = ax + by + c = 0 along L2 = y = mx + cMethod I Let line parallel to y = mx + c through P cuts ax + by + c = 0 at Q(x0 y0) rArr equation of PQ y ndash y1 = m(x ndash x1) hellip(i)

Solving (i) and (ii) get coordinates of Q and applying distance

formula ( ) ( ) ( )2 21 0 1 0pqd x x y y= minus + minus

Method II Let m = tan qEquation of PQ is 1 1x x y y rcos sinminus minus

= =θ θ

For Q (x1 + r cosq y1 + rsin q) Must satisfy L1 ax + by + c = 0rArr a(x1 + r cosq) + b(y1 + r sinq) + c = 0

rArr 1 1(ax by c)ra cos bsin

+ + = minus θ+ θ The sign of r indicates the position of point wrt Line and |r| is

required distance

1419 Intersection of Two LinesThe point of intersection of two lines a1x + b1y + c1 = 0 and a2x + b2y +c2 = 0 is 1 2 2 1 1 2 2 1

1 2 2 1 1 2 2 1

b c b c c a c aa b a b a b a b

minus minus

minus minus ∎ Condition for concurrency of Lines Three lines are said to be concurrent if they pass through a

common point Thus if three lines are concurrent the point of intersection of two lines lies on the third line Let a1x + b1y + c1 = 0 a2x + b2y + c2 = 0 and a3x + b3y + c3 = 0

rArr 1 1 1

a b ca b c 0a b c

= which is the required condition for concurrency of lines

NoteAnother condition of concurrency of three lines L1 a1x + b1y + c1 = 0 L2 a2x + b2y + c2 = 0 and

L3 a3x + b3y + c3 = 0 are concurrent iff there exists constants l1 l2 l3 not all zero such that

l1L1 + l2 L2 + l3 L3 = 0 l1(a1x + b1y + c1) + l2 (a2x + b2y + c2) + l3(a3x + b3y + c3) = 0

14110 Equation of the Bisectors of the Angles Between LinesMethod 1 Let L1 equiv a1x + b1y + c1 = 0 and L2 equiv a2x + b2y + c2 = 0 be two intersecting lines then the

equations of the lines bisecting the angles between L1 and L2 are given by 1 1 1 2 2 22 2 2 21 1 2 2

a x + b y c a x + b y c

a + b a + b

+ += plusmn

If a1a2 + b1b2 = 0 then the given lines are perpendicular to each other else they will contain acute and obtuse angle

ie a1a2 + b1b2 ne 0 Let q be the angle between L1 and L2 which is bisected by one of the bisectors say L3 Then angle between L1 and L3 is q2 Now find tan q2

Two Cases Arise

(i) If tan 1 then 2 2θ πlt θ lt Thus L3 will be bisecting the acute angles between L1 and L2

(ii) If tan 12θgt then

θ gt Thus L3 will be bisecting the obtuse angle between L1 and L2

Method 2 If c1 ne 0 c2 ne 0 then origin must lie in one of the angles between L1 and L2 Let us assume

c1 c2 gt 0 Then 1 1 1 2 2 22 2 2 21 1 2 2

a x b y c a x b y c

a b a b

+ + + += +

+ + is one of the bisectors of L1 and L2 If a1a2 + b1b2 gt 0 the given

equation represents obtuse angle bisector otherwise it represents acute angle bisector (if a1 a2 + b1b2 lt 0)

141101 Bisector of angle containing the origin

Let the equations of the two lines be a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 To find the bisectors of the angle containing the origin the following steps are taken

Step 1 See whether the constant terms c1 and c2 in the equations of two lines are positive or not If not then multiply both the sides of the equations by ndash1 to make the constant term positive

Step 2 Now obtain the bisector corresponding to the positive sign 1 1 1 2 2 22 2 2 21 1 2 2

a x b y c a x b y c a b a b

+ + + += +

This is the required bisector of the angle containing the origin and negative sign bisector of that angle which does not contain origin

141102 Bisector of acute and obtuse angle

Let the equations of the two lines be a1x + b1y +c1 = 0 and a1x + b2y + c2 = 0 To separate the bisectors of the obtuse and acute angles between the lines we proceed as follows

Step 1 See whether the constant terms c1 and c2 in the equations of two lines are positive or not If not then multiply both the sides of the equations by -1 to make the constant term positive

Step 2 Determine the sign of the expression a1a2 + b1b2

Step 3 If a1a2 + b1b2 gt 0 then the bisector corresponding to + sign gives the obtuse angle bisector and the bisector corresponding to ndash sign is the bisector of acute angle between the lines

ie 1 1 1 2 2 22 2 2 21 1 2 2

a x b y c a x b y c

a b a b

+ + + +=

+ + and 1 1 1 2 2 2

2 2 2 21 1 2 2

a x b y c a x b y c

a b a b

+ + + += minus

+ + are the bisectors of

obtuse and acute angles respectively

Step 4 If a1a2 + b1b2 lt 0 then the bisector corresponding to + sign gives the acute and obtuse angle

bisectors respectively 1 1 1 2 2 22 2 2 21 1 2 2

a x b y c a x b y c

a b a b

+ + + +=+

and 1 1 1 2 2 22 2 2 21 1 2 2

a x b y c a x b y c

a b a b

+ + + += minus

the bisectors of acute and obtuse angles respectively

141103 Whether the origin lies in the obtuse angle or acute angle

Let the equations of the two lines be a1x + b1y +c1 = 0 and a2x + b2y + c2 = 0 To determine whether the origin lies in the acute angle or obtuse angle between the lines we proceed as follows

Step 1 See whether the constant terms c1 and c2 in the equations of two lines are positive if not then multiply both the sides of the equations by ndash1 to make the constant term positive

Step 3 If a1a2 + b1b2 gt 0 then the origin lies in the obtuse angle and the lsquo+rsquo sign gives the bisector of obtuse angle If a1a2 + b1b2 lt 0 then the origin lies in the acute angle and the lsquo+rsquo sign gives the bisector of acute angle

Tips and Tricks

Equation of a Reflected Ray in a Mirror Given a line mirror LM = ax + by + c = 0 and a ray is incident along the line L1 = a1x + b1y + c1 = 0

The equation of the reflected ray is LR = (y ndash b) ndash m0 (x ndash α) = 0

In general if a point (x2 y2) lies at a distance k times the distance of P(x1 y1) from M (xm ym) then

2 1 2 1 1 12 2

y y x x (ax by c)(k 1)b a a bminus minus + +

= = minus ++

Foot of perpendicular and image of a point in a line If point P is reflected with respect to line Lm then the coordinates of its reflection are given by Q (xQ = 2xm ndash xp yQ = 2ym ndash yp) bull Equation of a Reflected Ray in a Mirror Choose a point P(p q) on the incident ray (preferably

any one of p or q taken zero) and get the image in line mirror Q(r s) In the line mirror

rArr ( )

2 ap bq cr p s qa b a b

minus + +minus minus= =

Equation of reflected ray is sy (x )rminusβ

minusβ = minusαminusα

rArr Yet another way the equation of the reflected ray is given as LI + lLM = 0 ie (a1x + b1y + c1) + l (ax + by + c) = 0

rArr minus +λ = λ = +1 1

2(aa bb )0 (incidentray) or (reflectedray)a b

Equation of reflected ray is minus ++ =

I M2 2

2(aa bb )L L 0a b

14111 Family of Straight Lines

The general equation of line has two effective parameters Therefore two conditions are needed to repre-sent a line uniquely But if only one condition is given then the resulting equation consist of a parameter and termed as lsquofamily of straight linesrsquo ∎ If L1 equiv a1x + b1y + c1 = 0 and L2 equiv a2x + b2y + c2 = 0 are two straight lines (not parallel) then

L1 + lL2 equiv a1x + b1y + c1 + l (a2x + b2y + c2) = 0 represents family of lines passing through the point of intersection of L1 = 0 and L2 = 0 (Here l is a parameter)

∎ Family of straight lines parallel to the line ax + by + c = 0 is given by ax + by + k = 0 where k is a parameter

∎ Family of straight lines perpendicular to the line ax + by + c = 0 is given by bx ndash y + k = 0 where k is a parameter

∎ If a1x + b1y + c1 = 0 a2x + b2y + c2 = 0 a3x + b3y + c3 = 0 are concurrent then p (a1x + b1y + c1)

+ q(a2x + b2y + c2) + r(a3x + b3y + c3) = 0 rArr p + q + r = 0 ie 1 1 1

a b ca b b 0a b c

142 General equation of SeconD DeGree anD Pair of StraiGht lineS

The general equation of pair of a straight lines is represented by the most general equation of second degree in x and y but any equation in x and y in degree two does not always represent pair of straight lines

Considering the following equation as a quadratic equation in y

rArr by2 + 2(hx + f) y + ax2 + 2gx + c = 0

rArr = minus + plusmn + minus + +2 2by (hx f ) (hx f ) b(ax 2gx c)

+ + = plusmn minus minusα minusβ2hx by f (h ab) (x )(x ) (1)

where a and b are roots of quadratic (h2 ndash ab)x2 + 2(hf ndash bg)x + f2 ndash bc

This equation (1) represents pair of straight lines if a = b ie D = 0

rArr D = 4 (hf ndash bg)2 ndash 4(h2 ndash ab) (f2 ndash bc) = 0

rArr b2g2 ndash 2hfgb + h2bc + abf2 ndash ab2c = 0

D= abc + 2fgh ndash af2 ndash bg2 ndash ch2 = 0rArr

a h gh b f 0g f c

∆ = =

rArr The lines represented are given as + + = plusmn minus minusα2hx by f h ab(x )

ConclusionsIf h2 ndash ab gt 0 rArr two real and distinct linesIf h2 ndash ab lt 0 rArrtwo imaginary linesIf h2 ndash ab = 0 rArrtwo parallel lines if atleast one of bg ndash hf ne 0 af ndash gh ne 0If h2 ndash ab = 0 and bg ndash hf = 0 af ndash gh = 0 rArrtwo coincident linesa + b = 0 rArrboth lines are perpendicular

1421 Pair of Straight Lines Through the Origin

The homogenous equation of second degree ax2 + 2hxy + by2 = 0 always represent a pair of straight lines through the origin ax2 + 2hxy + by2 = 0

rArr b(yx) 2 + 2h(yx) + a = 0 rArr 2y 2h 4h 4ab

x 2bminus plusmn minus

rArr y = m1x or y = m2x where2

1h h abm

bminus + minus

= and 2

2h h abm

bminus minus minus

Since h2 le ab therefore values of m1 and m2 are real Clearly y = m1x and y = m2x are straight lines passing through the origin Hence ax2 + 2hxy + by2 = 0 represents a pair of straight lines passing through the origin

rArr According to the value of m1 and m2 then line are Real and distinct if h2 gt ab = 0 and h2 gt ab If ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents two straight lines they can be found by considering

the lines as (lx + my + n) (lprimex + mprimey + nprime) = 0 After multiplying and comparing the coefficients of like power we can find l lprime m mprime n nprime to find the required equations

1422 Angle Between the Pair of Straight Lines ax2 + 2hxy + by2 = 0 (i)

1 2 (h ab)tan|a b |

minus minus

θ = +

(i) Condition for the lines to be parallel If the two lines are parallel then q = 0 ie tanq = 0 Hence the two lines are parallel if h2 = ab (ii) Condition for the lines to be perpendicular If the two lines are perpendicular then q = 90deg ie tanq=infina + b = 0 ie coefficient of x2 + coefficient of y2 = 0

NoteThe above conditions are also valid for general equation of second degree

1 Equation of angle bisector of the pair of straight lines ax2 + 2hxy + by2 = 0 is given by minus=

2 2x y xya b h

rArr Condition for coincidence of lines The lines will be coincident if 1 1 1

l m nl m n= = Taking the

above ratios in pairs the conditions are h2 - ab = 0 g2 - ac = 0 and f 2 - bc = 0 rArr Point of intersection of the lines The point of intersection of ax2 + 2hxy + by2 + 2gx + 2fy

+ c = 0 is 2 2

bg hf af ghh ab h ab

or 2 2

f bc g cah ab h ab

2 Bisectors of the angles between the lines given by ax2+ 2hxy + by2 + 2gx + 2fy + c = 0 If (xprime yprime) be the point of intersection of the lines then we shift the origin to the point (xprime yprime) The

transformed equation will be ax2 + 2hxy + by2 = 0 of the bisectors which are given by 2 2x y xya b hminus

=minus

The above bisectors are referred to (xprime yprime) as origin Now we have to write x - xprime from x and y -yprime for y Hence the equation of the bisectors of the angle between the lines is

2 2(x x ) (y y ) (x x )(y y )a b hprime prime prime primeminus minus minus minus minus

=minus

2 2 2 2

2 2 2 2shifting origin to

ax 2hxy by 2gx 2fy c 0 ax 2hxy by 0(x ) (y ) (x )(y ) x y xy

a b h a b hα β

+ + + + + = + + = rarr minusα minus minusβ minusα minusβ minus

= = minus minus

Tips and Tricks

rArr Point of Intersection Given a pair of straight lines S = ax2 + 2hxy + by2

+ 2gx + 2fy + c = 0

Let (ab) be the point of intersection of both lines represented by S = 0

Shifting origin to (a b) the equation S = 0 must transform to homogenous form

ie a(x + a)2 + b (y + b)2 + 2h (x + a)(y + b) + 2g(x + a) +2 f(y + b) + c = 0

coefficient of x 0 a h g 0coefficient of y 0 h b f 0

= rArr α+ β+ = = rArr α+ β+ =

rArr ( )

S 0x α β

part = part and

part= part

The point of intersection of POSL if D = 0

rArr 2

hf bgab hminus

α =minus

af gh h abminus

β =minus

Homogeneous equation of degree 2 in x and yax2 + 2hxy + by2

= 0 always represents POSL

(real or imaginary) passing through origin

A homogeneous equation of degree n represents n straight lines through origin If two POSL have same homogeneous part of degree two in their equation then they always

construct a parallelogram If two POSL S = 0 (L1L2 = 0) and Srsquo = 0 (L1L2 = 0) have common angle

bisectors (B1B2 = 0) then their lines are iso-inclined to each other respectively ie angle between L1 and L1 is equal to angle between L2 and L2 also angle between L1 and L2 is equal to angle between L2 and L1 angle between L1 = 0 and L1 = 0 = angle between L2 = 0 and L2 = 0 = f ndash q also angle between L1 = 0 and L2 = 0 = angle between L2 = 0 and L1 = 0 = f + q

Equation of POSL joining origin to the point of intersection of a curve and a straight line

( ) + +

Homogeneous LinearHomogeneous

Homogeneous

lx my lx myS = ax + 2hxy + by +2 gx + fy +c = 0n n

Equation of POSL joining origin to the point of intersection of a curve and a straight line Given a straight line lx + my = n hellip (i) and a conic S = ax2 + 2hxy + by2 +2gx + 2fy +c = 0 hellip (ii)

Required a homogeneous equation of degree two that satisfies the coordinates of A(a b) and B(g d)

Since la + mb = n and S(a b) = aa2 + 2hab + bb2 + 2ga + 2fb +c = 0 ∎ If A (x1 y1) B( x2 y2) and C(x3 y3) are the vertices of a DABC

rArr equation of median through A is given by 1 1 1 1

2 2 3 3

x y 1 x y 1x y 1 x y 1 0x y 1 x y 1

rArr equation of the internal bisector of angle A is (where b = AC and c = AB)

1 1 1 1

2 2 3 3

x y 1 x y 1b x y 1 c x y 1 0

x y 1 x y 1+ =

Chapter 15CirCle and

Family oF CirCle

151 introduction

A circle is the most regular object we know Each point on a circlersquos circumference is equidistant from its centre The shape and symmetry of circle has been fascinating mathematicians since ages

152 definiton of circle

A circle is the locus of a point moving in a plane so that its distance from a fixed point remains constant The fixed point is called centre of the circle and the constant distance is called the radius of the circle

1521 Equation of a Circle in Various Forms

Centrendashradius form Equation of a circle with Centre at (h k) and radius lsquorrsquo is (x ndash h)2 + (y ndash k)2 = r2 Standard Form When centre is (0 0) and radius is lsquoarsquo then the standard

form becomes x2 + y2 = a2

1522 General EquationThe equation x2 + y2 + 2gx + 2fy + c = 0 is called general equation of circle in canonical form Comparing with equation x2 + y2 ndash 2αx ndash 2βy + α2+ β2 ndash r2 = 0 The equation x2 + y2 + 2gx + 2fy + c = 0

can also be written as ( ) ( )22 2 2 2x g (y f ) g f c+ + + = + minus

Hence centre equiv (ndashg ndashf) ie 1 1 coefficient of x coefficient of y2 2

minus minus

and radius equiv + minus2 2g f c

g2 + f2 ndash c gt 0 rArrreal circle with positive radiusg2 + f2 ndash c = 0 rArrrepresent a point circleg2 + f2 ndash c lt 0 rArrrepresent an imaginary

Circle and Family of Circle 15153

NoteA general equation of second degree non-homogenous is ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 in x y represents a circle if

Coefficients of x2 = coefficients of y2 ie a = b ne 0

Coefficient of xy is zero ie h = 0

g2 + f2 ndash c le 0

The general equation may be of the form Ax2 + Ay2 + 2Gx + 2Fy + c = 0 represent a equation of circle

Centre = G F

minus minus

and radius = 2 21G F AC

A+ minus

1523 Diametric FormIf (x1 y1) and (x2 y2 ) are the extremities of one of the diameter of a circle then its equation is (x ndash x1) (x ndash x2) + (y ndashy1) (y ndashy2 )= 0

1524 Equation of Circle Thorugh Three Points The equation of circle through three non-collinear points

A(x1 y1) B(x2 y2) C(x3 y3) is

2 21 1 1 12 22 2 2 22 33 3 3 3

x y x y 1x y x y 1

0x y x y 1x y x y 1

1525 The Carametric Coordinates of any Point on the CircleParametric Equation of Circle When both x and y coordinates of the point on the circle are expressed as a function of single parameter eg t or θ etc then the equation is called parametric equation of circle

Case 1 Standard Equation x2+ y2 = r2 parametric equation x = r cosθ and y = r sinθBy restricting the values of parameter we can express the part of curve (the arc of circle

line segment etc) very conveniently which is not as easy in case of Cartesian equation of curveqisin[0 2p) full circle qisin(0p) upper semicircleqisin(p2p) lower semicircle qisin(a b) circular arc

Case 2 General equation (x ndash a)2 + (y ndash b)2 = r2 parametric equation x = a + r cosq and y = b + r sinq

x y rcos sinminusα minusβ

= =θ θ

where q is parameter and constant represents circle

= =θ θ

where r is parameter and q is constant represents

straight line

Parametric coordinates of any point on the circle x2 + y2 + 2gx + 2fy + c = 0 are 2 2x g g f c cos= minus + + minus θ2 2y f g f c sin= minus + + minus θ (ndashg ndash f) is the centre and 2 2g f c+ minus is the radius of the circle

(x1y1) (x3y3)

(x2y2)

1526 Position of a Point with Respect to a CirclePoint P( x1 y1) lies inside on or outside the circle

S = x2 + y2 + 2gx + 2fy +c = 0 accordingly as S1 = x12 + y1

2 + 2gx1 + 2fy1 + c is lt 0 = 0 or gt 0 respectively

rArr 2 2 2 21 1(x g) (y f ) g f c+ + + hArr + minus

rArr (x1 + g)2 + (y1 + f)2 hArr g2 + f 2 ndash crArr x1

2 + y12 + 2gx1 + 2fy1 + c hArr 0 or S1 hArr 0 where

S1 = x12 + y1

2 + 2gx1 + 2fy1 + c So S1 gt 0 rArr (x1 y1) is outside the circle S1 = 0 rArr (x1 y1) is on the circle S1 lt 0 rArr (x1 y1) is inside the circle

Length of tangent from point P to the circleS = x2 + y2 + 2gx + 2fy + c = 0

2 2 2 2 2 2T 1 1L PT PC r (x g) (y f ) (g f c)= = minus = + + + minus + minus

= 2 21 1 1 1 1(x y 2gx 2fy c S+ + + + =

If S1 is called power of point P wrt circle S = 0 radic S1= length of tangent drawn from P to circlebull If P lies outside S1 then is + ve rArrtwo tangents drawnbull If P lies on circle S1 = 0 rArr only one tangent bull If P lies inside circle S1 lt 0 rArrno (imaginary) tangent

1527 Position of a Line with Respect to a CircleLet L = 0 be a line and S = 0 be a circle if lsquorrsquo be the radius of a circle and p be the length of perpendicular from the centre of circle on the line then if

p gt r rArrLine is outside the circle p = r rArrLine touches circlep lt r rArrLine is the chord of circle p = 0 rArrLine is diameter of circle

(i) Length of the intercept made by the circle on the line is 2 22 r pminus

(ii) The length of the intercept made by the line y = mx +c with the circle x2 + y2 = a2 is 2 2 2

a (1 m ) c21 m+ minus+

15271 Condition for Tangency

(i) The line y = mx + c is tangent to the circle x2 + y2 = a2 if and only if c2 = a2(1 + m2) If it is tangent

then the point of contact is given by 2 2ma a

c c minus

(ii) The line lx + my + n = 0 is tangent to the circle x2 + y2 = a2 if and only if n2 = a2 (l2 + m2) If it is

tangent then point of contact is given by 2 2la ma

n n minus minus

Note2y mx a 1 m m= plusmn + forall isin is called family of tangents or tangent in term of slope In case the slope of

tangent is given or tangents passing from a given point are to be obtained this formula can be applied

153 equation of tangent and normal

1531 TangentsTangent line to a circle at a point P(x1 y1) is defined as a limiting case of a chord PQ where Q is (x2 y2) such that Q rarr P As Q rarr P ie x2 rarr x1 and y2 rarr y1

Then chord PQ rarrtangent at P rArr Slope of chord PQ rarrslope of tangent at P

rArr 2 12 1

2 1t x x

2 1y y

y ym limx xrarr

minus=

minus =

2 12 1

x x1 2 1y y

x x xlimy y yrarr

+minus = minus

( )11 1

xy y x xy

minus =minus minus rArr T = xx1 + yy1 ndash a2 = 0

Q 2 2 21 1x y a+ = (1)

2 2 22 2x y a+ = (2)

rArr 2 2 2 22 1 2 1(x x ) (y y )minus = minus minus rArr 2 1 1 2

2 1 1 2

y y x xx x y y

minus += minus

minus + If the equations of the circle are given in general form then the equation of tangent to S = x2 + y2 + 2gx + 2fy + c = 0 at a point (x1 y1) is T = xx1 + yy1 + g(x + x1) + f(y + y1) + c = 0

1532 Parametric FormEquation of tangent to circle x2 + y2 = a2 at (a cos a a sin a) is x cos a + y sin a = a

Point of intersection of the tangent drawn to the circle x2 + y2= a2 at the point P(a) and Q(b) is

a cos2x

=αminusβ

a sin2y

=αminusβ

1533 Pair of Tangents

Combined equation of the pair of tangents drawn from an external point lsquoPrsquo to a given circle is SS1 = T2 2 2 2

1 1 1S x y a= + minus and T equiv xx1 + yy1 ndash a2 = 0

Q(h k)

(x1 y1)

1534 Normals

Normal is defined as a line perpendicular to the tangent line to the circle at the point of tangency P(x1 y1)

If the equation of the circle be x2 + y2 + 2gx + 2fy + c = 0

rArr slope of the normal 1

y fmx g+

rArr Equation of normal +minus = minus

y f(y y ) (x x )x g

Equation of normal in determinant form is given by 1 1

x y 1x y 1 0g f 1

=minus minus

Director Circle The locus of point of intersection of two perpendicular tangents is called the director circle The director circle of the circle x2 + y2 = a2 is x2 + y2 = 2a2

Diameter of a circle The locus of middle points of a system of parallel chords of a circle is called the diameter of a circle The diameter of the circle x2 + y2 = r2 corresponding to the system of parallel chords y = mx + c is x + my = 0

(i) Every diameter passes through the centre of the circle

(ii) A diameter is perpendicular to the system of parallel chords

1535 Equation of Chord with Mid-point as (x1y1)

Slope of chord = 1

minus rArr equation of chord minus = minus minus11 1

x(y y ) (x x )y

rArr 2 21 1 1 1yy y xx xminus = minus + rArr 2 2

1 1 1 1yy xx x y+ = +

rArr 2 2 2 21 1 1 1xx yy a x y a+ minus = + minus ie T = S1

For any conic section the equation of chord whose mid point is (x1 y1) is given by T = S1

154 chord of contact

From a point P(x1 y1) exterior to a circle two tangents can be drawn to the circle Let these tangents be PA and PB Then the line segment AB is a chord of the circle and is called chord of contact of P(x1 y1) with respect to the circle

If S = 0 is the circle then equation of the chord of contact of P(x1 y1) wrt the circle S = 0 is T = 0

Equation of locus through intersection of S = 0 and Sprime = 0 is S + lSprime = 0 ie (x2 + y2 ndash a2) + l(x2 + y2 ndash xx1 ndash yy1) = 0

For l = ndash1 the curve becomes x x1 + y y1 = a2

1541 Relative Position of Two CirclesS1 = x2 + y2 + 2g1x + 2f1y +c1 = 0 and S2 = x2 + y2 + 2g2x + 2f2y + c2 = 0

Case 1 Two circle lies outside each other Distance between centres d gt r1 + r2 Four common tangent (two direct two transverse) PQ divides C1C2 in ratio r1 r2 externallyinternally

Equation of direct common tangent

minusβ = minusα α β

Two values of m can be obtained from condition thatthis line touches both the circles S 0 and S 0

y m(x ) where P is( )

Similarly we get equation of TCT

bullDirect Common Tangent Length of direct common tangent is defined as distance between

point of contacts ie ( )22D 1 2L MN d r r = = minus minus

Angle between DCT = 2q = 1 1 2| r r |2sind

minus minus

bullTransverse Common Tangent Length of transverse common tangent is defined as distance between point of

contacts ie S and T ( )22T 1 2L ST d r r = = minus + Angle

between TCT = 2a = 1 1 2r r2sind

minus +

Case 2 Two circles touch each other externally C1 C2 = d = r1 + r2

Three common tangent (two DCT and one TCT)

Equation of DCT (obtained as in case I) Equation of TCT is S1 ndash S2 = 0

S direct commontangentstransverse

common tangents

Direct Common Tangent ( )22D 1 2 1 2 1 2L (r r ) r r 2 r r = + minus minus = Angle between DCT = 2q

= 1 1 2

r r2sinr r

minus minus

Transverse Common Tangent ( )22T 1 2 1 2L (r r ) r r 0= + minus + = Angle between TCT = 2a

= 1 1 2

r r2sinr r

minus minus= π

Case 3 Two Circles intersect each other |r1 ndash r2 | lt C1C2 lt r1 + r2 Two common tangent (two DCT and no TCT) Equation of common chord is S1 ndash S2 = 0

2 1 1 2 2 1 1 2

1 2 1 2

r g r g r f r fP r r r r

minus minus

minus minus Equation of DCT

1 2Two values of m can be obtained fromcondition that this

line touches both the circles S 0 and S 0

y m(x ) where P is ( )

Direct Common Tangent ( )22D 1 2L MN d r r= = minus minus

Angle between DCT 1 1 2| r r |2 2sind

minus minus θ =

Case IV Two Circles touch each other internally C1 C2 = |r1 ndash r2| Two direct common tangents Equation of DCT S1 ndash S2 = 0

1542 Direct Common Tangent

( ) ( )2 22 2D 1 2 1 2 1 2L d r r (r r ) r r 0= minus minus = minus minus minus =

Angle between DCT = 2q = 1 1 2

r r2sinr r

minus minus= π

Case V If 0 lt C1C2 = d lt |r1 ndash r2| then the circle lies completely inside other bullAngle of Intersection Angle of intersection (q) between two curve is defined as angle between

their tangents at their point of intersection which is same as angle between their normals at the point of intersection

2 2 21 2

r r dcos2r r+ minus

2 2 21 1 2

r r dcos2r r

minus + minus

rArr θ =

bullOrthogonal Intersection If the angle of intersection is p2 then it is called as orthogonal intersection Condition of orthogonality of the above two circles is

2 2 21 2 r r d+ = 2 2 2 2 2 2

1 2 1 2 2 2 2 1 2 1g f c g f c (g g ) (f f )rArr + minus + + minus = minus + minus

1 2 1 2 1 22(g g f f ) c crArr + = +

155 intercept made on coordinate axes by the circle

The intercept made by the circle x2 + y2 + 2gx + 2fy + c = 0Let circle intersect x-axis at two points (x1 0) and ( x2 0) then x1 x2 are roots

x2 + 2gx + c = 0

Length of x-intercept = |x1 ndash x2| = 2 2g cminus

Similarly length of y-intercept = | y1 ndash y2| = 2 2f cminus Conditions that given circle touches

(i) x-axis is g2 = c(ii) y-axis is f2 = c

NotesCircle x2 + y2 + 2gx + 2fy + c = 0 cuts

(i) x-axis in two real coincident or imaginary points according as g2 gt = lt c

(ii) y-axis in two real coincident or imaginary points according as f2 gt = lt c

156 family of circles

General Equation of Circle x2 + y2 + 2gx + 2fy + c = 0 contains three unknown parameters (effective) Therefore three conditions are necessary in order to determine a circle uniquely and if only two conditions are given then the obtained equation contains a parameter and it is described as family of circle

Following are the ways of expressing some known family of circles

1 Equation of circle through intersection of a circle S = 0 and a line L = 0 S + lL = 0

2 Equation of family of circle passing through intersection of two circles S1 = 0 and S2 = 0 is given as 1 1 2S (S S ) 0+λ minus =

3 Family of concentric circles The family of circles with the same centre and different radii is called a family of concentric circles (xndasha)2 + (y ndash b)2 = r2 where (a b) is the fixed point and r is a parameter

4 Equation of any circle passes through two points (x1y1) and

(x2y2) 1 2 1 2 1 1

x y 1(x x )(x x ) (y y )(y y ) x y 1 0

x y 1minus minus + minus minus +λ =

5 Equation of family of circle touching the line with slope m at the point (x1y1) is

2 21 1 1 1(x x ) (y y ) (y y ) m(x x ) 0minus + minus +λ minus minus minus = and if m is

infinite the family of circle is 2 21 1 1(x x ) (y y ) (x x ) 0minus + minus +λ minus =

where lis a parameter

6 Equation of circle circumscribing a triangle with sides L1= 0 L2 = 0 and L3 = 0 is 1 2 2 3 3 1L L L L L L 0+λ +micro = where l m is obtained by applying the condition that coefficient x2 = coefficient y2 and coefficient of xy = 0

7 Family of conic circumscribing a quadrilateral with sides L1 = 0 L2 = 0 L3 = 0 and L4 = 0 taken in order is 1 3 2 4L L L L 0+λ = and condition of concyclic ness and equation of possible circumcircle can be obtained by applying the condition that coefficient of x2 = coefficient of y2 and coefficient xy = 0 and analyzing the outcome mathematically

157 radical axes and radical centre

Radical axis of S = 0 and Sprime = 0 is the locus of the point from which the tangents drawn to the two circles are of equal lengths Its equation is given by S ndash Sprime = 0 ( only if coefficients of x2 y2 in both circles are same)

Remarks

(i) If the circles S = 0 and Sprime= 0 intersect each other then their common chord and their radical axis coincide Thus they have the same eqn S ndash Sprime = 0

(ii) If two circles touch each other then their radical axis coincides with their common tangent at their point of contact The equation is again S ndashSprime= 0

bullRadical Centre The common point of intersection of the radical axes of three circles taken two at a time is called the radical centre of the three circles

Tips and Tricks

1 If two circles do not intersect (c1c2 gt r1 + r2) then they have two transverse and two direct common tangents

2 If two circles intersect (c1c2 lt r1 + r2) then they have two direct tangents only

3 If two circles touch externally (c1c2 = r1 + r2) then they have one transverse and two direct common tangents

4 If two circles touch internally (c1c2 = r1 ndash r2) then they have only one common tangent

5 If the point P lies outside the circle then the polar and the chord of contact of this point P are same straight line

6 If the point P lies on the circle then the polar and the tangent to the circle at P are same straight line

7 The coordinates of the pole of the line lx + my + n = 0 with respect to the circel x2 + y2 = a2 are

8 If (x1 y1) is the pole of the line lx + my + n = 0 wrt the circle x2 + y2 + 2gx + 2fy +c = then where r is the radius of the circle

Chapter 16parabola

161 IntroductIon to conIc SectIonS

A conic section or conic is the locus of a point which moves in a plane so that its distance from a fixed point is in a constant ratio to its perpendicular distance from a fixed straight line Conic sections are section obtained when a pair of two vertical cones with same vertex are intersected by a plane in various orientation The point V is called vertex and the line L1 is Axis

The rotating line L2 is called as generator of the cone the vertex separates the cone into two parts known as nappes

Nature of conic sections depends on the position of the intersecting plane with respect to the cone and the angle f made by it with the vertical axis of the cone

Circle When f = 90deg the section is a circle

Ellipse When q ltflt 90deg the section is an ellipse

Parabola If plane is parallel to a generator of the cone (ie when f = q) then section is a parabola

Parabola 16163

Hyperbola When 0 le f lt q the plane cuts through both the nappes and the curves of intersection is hyperbola

Degenerated Conics

When the plane cuts at the vertex of the cone we have the different cases

When q lt f le 90deg then the section is a point

When 0 le f lt q then the section is a pair of intersecting straight lines It is the degenerated case of a hyperbola

Whenf= q then the plane contains a generator of the cone and the section is a coincident straight line

1611 Definition of Various Terms Related to Conics

Focus The fixed point is called the focus of the conic section

Eccentricity The constant ratio (e) is called the eccentricity of the conic section

Directrix The fixed straight line is called the directrix

Axis The straight line passing through the focus and perpendicular to the directrix is called the axis of the conic section

Vertex The point of intersection of conic and the axis are called vertices of conic section

Centre The point which bisects every chord of the conic passing through it is called centre of the conic

Double Ordinate A chord perpendicular to the axis is called double ordinate (normal chord) of the conic section The double ordinate passing through the focus is called the latus rectum

1612 General Equation of a ConicIf the focus is (a b) and the directrix is ax + by + c = 0 then the equation of the conic section whose eccentricity is e is given by

According to the definition of conic SP costant ePM

= = or

SP = e PM 2 2

|ax by c |(x ) (y ) e(a b )

+ +minusα + minusβ =

+ where P(x y) is a

point lying on the conic or 2

2 2 22 2

(ax by c)(x ) (y ) e(a b )+ +

minusα + minusβ =+

The equation of conics is represented by the general equation of second degree ax2 +2hxy + by2 + 2gx + 2fy + c = 0

We know that the discriminant of the above equation is represented by D where

2 2 2abc 2fgh af bg ch∆ = + minus minus minus or a h gh b fg f c

Case I When D = 0 then the equation represents degenerate conic

Condition Conic

D = 0 and h2 ndash ab = 0 A pair of coincident lines or parallel linesD = 0 and h2 ndash ab gt 0 A pair of intersecting straight linesD = 0 and h2 ndash ab lt 0 Imaginary pair of straight lines with real point of intersection also

known as point locus

Case II When D ne 0 the equation represents non-degenerate conic

Condition ConicD ne 0 and h = 0 a = b A circleD ne 0 and h2 ndash ab = 0 A parabolaD ne 0 and h2 ndash ab lt 0 An ellipse or empty setD ne 0 and h2 ndash ab gt 0 A hyperbola D ne 0 and h2 ndash ab gt 0 and a + b = 0 A rectangular hyperbola

162 PArABoLA

A parabola is the locus of a point which moves in a plane so that its distance from a fixed point (called focus) is equal to its distance from a fixed straight line (called directrix) It is the conic with e = 1

1621 Standard EquationGiven S(a 0) as focus and the line x + a = 0 as directrix Standard Equation Given S(a 0) as focus and the line x + a = 0

as directrix

Parabola 16165

Focal distance SP PM a h= = + ( )2 2h a k a hrArr minus + = +

2 2 2 2 2a h 2ah k a h 2ahrArr + minus + = + + 2k 4ahrArr = 2y 4axrArr =

Equation of parabola y2 = 4ax a gt0 Opening rightwards passing through origin Parametric equation x = at2y = 2 at where t isin ℝ Focus S(a o) vertex (0 0) Axis y = 0 Directrix x + a = 0 TV x = 0 Focal distance=a + h Latus rectum Equation x ndash a = 0 and length 4a extremities (a plusmn2a)

Equation (a gt 0) Axis Focus Directrix Latus rectum Graph

y2 = 4axx = at2

y = 2at

y = 0 (a 0) x + a = 0 x = a 4a (a plusmn2a)

y2 = ndash4axx = ndashat2

y = 2at

y = 0 (ndasha 0) x ndash a = 0 x = ndasha 4a(ndasha plusmn2a)

x2 = 4ayy = at2

x = 2at

x = 0 (0 a) y + a = 0 y = a 4a(plusmn2a a)

x2 = ndash4ayy = ndashat2

x = 2at

x = 0 (0 ndasha) y ndash a = 0 y = ndasha 4a(plusmn2a ndasha)

Equation of parabola with length of LR (latus rectum) = 4a vertex at (a b) and axis is given as (y ndash β)2 = plusmn 4a(x ndash α)

Focus (α plusmn a β) Axis y ndash β = 0 TV (transverse axis) x ndash α = 0 Parametric equation 2 2( at 2at)( at 2at)α+ β+ αminus β+ Directrix x a= α Extremetric ( a 2a)αminus βplusmn ( a 2a)αminus βplusmn

Focus lies at 14th of the latus rectum away from vertex along axis towards parabola

Equation of parabola with length of LR = 4a vertex at (a b) and axis parallel to y-axis is given as (x ndash α)2 = plusmn4a(y ndash β) Focus (α β plusmn a) Axis x ndash α = 0 TV y ndash β = 0 Directrix y a=β

Ends of LR ( 2a a)( 2a a)αplusmn β+ αplusmn βminus

Parametric equation 2 2( 2at y at )( 2at y at )α+ =β+ α+ =βminus

NoteEquation of general parabola with axis lx + my + n = 0 and TV is mx ndash ly + k = 0 and LR is of length 4a

is given as 2 2 2( lx my n) LR l m ( mx ly k )+ + = plusmn + minus +

1622 Position of Point wrt ParabolaThe region towards focus is defined as inside region of parabola and towards directrix is outside region of parabola

Given a parabola y2 = 4ax and a point P(x1 y1) Point P lies inside hArr S1 lt 0 Point P lies on parabola hArr S1 = 0 Point P lies outside parabola hArr S1 gt 0

1623 Position of Line wrt ParabolaWhether the straight line y = mx + c cutstoucheshas no contact with the parabola y2 = 4ax can be determined by solving the parabola and straight line together

2 y cy 4a 0mminus minus =

(mx + c)2 ndash 4ax = 0 which is m2x2 + (2cm ndash 4a) x + c2 = 0

2 4a 4acy y 0m m

rArr minus + = (i) D gt 0 rArr line cuts at two distinct point

1 2 1 24a 4acy y and y ym m

rArr + = =

(ii) D = 0 rArr line touches the parabolaCondition of tangency D =0

16a a cm 0m

minus =

rArr = (iii) D lt 0 rArr line has no contact

Parabola 16167

rArr ay mx m ~ 0m

= + forall isin known as family of tangent with slope m is tangent to the parabola

y2 = 4ax

rArr Point of contact 2

a 2am m

a 2a 1 (at 2at) mm m t

hArr rArr =

rArr Parametric equation of tangent at point lsquotrsquo is given as 2xy at yt x att

= + rArr = +

163 chordS of PArABoLA And ItS ProPertIeS

Given a parabola y2 = 4ax let AB be the chord joining A(x1 y1) and B(x2 y2)

Q 2 21 1 2 2y 4ax and y 4ax= = rArr y2

2 ndash y12 = 4a(x2 ndash x1) rArr 2 1

2 1 1 2

y y 4ax x y yminus

=minus +

rArr Slope of chord 2 1

y yABx xminus

=minus

= 1 21 2

4a 2ay yy y

Equation of chord ( )1 11 2

4ay y x xy y

minus = minus+

Condition to be a focal chord rArr y1y2 = ndash4a2 and x1x2 = a2 ie t1t2 = ndash1

1631 Chord of Parabola in Parametric Form

2Slope of chordt t

Equation of chord ( )21 1

2y 2at x att t

minus = minus+

For focal chord Put y = 0 x = a rArr 0 = 2a(1 + t1t2) rArr t1t2 = ndash1

1632 Properties of Focal Chord

A focal chord is basically a chord passing through the focus of the parabola

Extremeties of focal chord P(at2 2at) and minus 2

a 2aQ t t

Segments of focal chord SP = l1 = a + at2 2 2

aSQ l at

HM of segments of focal chord is semi latus-rectum 2a

Length of focal chord 21L a t

t rArr = +

Slope of focal chord 2tt 1

Equation of focal chord ( )2

2ty x at 1

= minusminus

(i) Equation of chord with mid-point M (x1 y1)

rArr 21 1 1 1yy 2a( x x ) y 4axminus + = minus ie T = S1

(ii) Equation of a chord of contact formed by joining the points of contacts of the tangents drawn form point A to the parabola

Chord of contact is yy1 ndash 2a(x + x1) = 0 ie T = 0

164 tAnGent of PArABoLA And ItS ProPertIeS

Tangent to a parabola at P(x1 y1) rArr T yy1 ndash 2a(x + x1) = 0

Tangent to the parabola at P(at2 2at)

rArr yt = x + at2

1641 Properties of Tangents of a Parabola

If the point of intersection of tangents at t1 and t2 on the parabola be T then T (at1 t2 a (t1 + t2))

If T be the point of intersection of tangent at P and Q then SP ST SQ are in GP

ie ST = SPSQ

Consider the parabola shown in the diagram below

Coordinate of T (ndashat2 0) coordinate of Y (0 at) SP = ST = PM = SG = a + at2

angMPT = angSTP = angSPT = q

Parabola 16169

Reflection Property of Parabola Light rays emerging from focus after reflection become parallel to the axis of parabolic mirror and all light rays coming parallel to axis of parabola converge at focus

Foot of perpendicular from focus upon any tangent lies at Y(0 at) on the tangent at vertex (TV)

SY is median and DSPT is isosceles SY is altitude ie SY is perpendicular to PT angTSY = angYSP = p2 ndash q and SY = MY rArr SPMT is rhombus

Points A B and C lie on the parabola y2 = 4ax The tangents

to the parabola at A B and C taken in pairs intersect at points P Q and R respectively then the ratio of the areas of the DABC and DPQR is 2 1

Tangent at any point on parabola bisects the internal angle between focal distances SP and PM

Normal at P bisects the external angle between SP and PM The portion of the tangent intercepted between axis and point

of contact is bisected by tangent at vertex Y is the mid-point of PT SY is median and DSPT is isosceles

SY is altitude angTSY = ang YSP = p2 ndash q and SY = MY rArr SPMT is rhombus

Equation of a pair of tangents to the parabola form P(x1y1) SS1 =T2

2 2 21 1 1 1(y 4ax)(y 4ax ) [yy 2a(x x )]minus minus = minus +

165 norMALS And theIr ProPertIeS

Given a parabola y2 = 4ax at point lsquotrsquoSlope of normal m = ndashtEquation of normal y ndash 2at = ndasht(x ndash at2)rArr y + xt = 2at + at3

1651 Properties

Coordinate of G = (2a + at2 0)

If the normal at P(t) meets the parabola at Q(t1) then = minus minust t

If the normal to the parabola y2 = 4ax at point P(t1) and Q(t2) cuts the parabola at some point R (t3) then

(i) t1t2 = 2 (ii) t3 = ndash(t1 + t2)

1652 Normals in Terms of SlopeSince Equation of normal y + xt = 2at + at3 at (at2 2at)Put t = ndashm rArr y = mx ndash 2am ndash am3 where foot of normal is (am2 ndash2am) From any point P(h k) in the plane of the parabola three normals can be drawn to the parabola The

foot of these normals are called co-normal points of the parabola rArr Sum of ordinate of foot of conformal points yP + yQ + yR = ndash2a (m1 + m2 + m3) = 0 where m1 m2 m3

are the slopes of the three normals Sum of the slopes of the concurrent normals to a parabola is zero Centroid of the triangle joining the

co-normal point P Q R lies on the axis of the parabola Necessary condition for existence of three real normal through the point (h k) is h gt 2a if a gt 0

and h lt 2a if a lt 0But the converse of statement is not true ie if h gt 2a if a gt 0 and h lt 2a if a lt 0 does not necessarily implies that the three normals are real

Sufficient condition for 3 real normals from (h k) f(m) = am3 + (2a ndash h)m + k it has 3 real and distinct rootsIf f rsquo(m) = 3am2 + 2a ndash h = 0 has 2 real and distinct roots ie

h 2am say 3aminus

= plusmn α βsufficient

condition for 3 real slopes is f (a) f(b) lt 0 rArr f(a) f(b) lt 0 rArr f(a)(ndasha) lt 0 rArr 27ak2 lt 4(h ndash 2a)3

Atmost there are four concylic point on the parabola and sum of ordinates of these points vanishes

rArr Sum of ordinates of four concyclic points on parabola Since 2a(t1 + t2 + t3 + t4) = 0

Parabola 16171

Pair of chord obtained by joining any four concyclic points are equally inclined to the axis of the parabola

Circle passing through any three co-normal points on the parabola also passes through the vertex of the parabola

Table representing the equations of tangents in different forms and related terms

Equation y2 = 4ax y2 = ndash4ax x2 = 4ay x2 = ndash4ay

Tangent in point form yy1 = 2a(x + x1) yy1 = ndash2a(x + x1) xx1 = 2a(y + y1) xx1 = ndash2a(y + y1)

Parametric co-ordinate (at2 2at) (ndashat2 2at) (2at at2) (2at ndashat2)

Tangent in parametric form

ty = x + at2 ty = ndashx + at2 tx = y + at2 tx = ndashx + at2

Point of contact in terms of slope (m) 2

a 2am m

minus minus

(2am am2) (ndash2am ndasham2)

Condition of tangency acm

= minusc = ndasham2 c = am2

Tangent in slope form ay mxm

= +ay mxm

= minusy = mx ndash am2 y = mx + am2

Table representing the equations of tangents in different forms and related terms to parabolas having vertex at (h k) and axes parallel to co-ordinate axes

Equation (y ndash k)2 = 4a(x ndash h) (y ndash k)2 = ndash4a(x ndash h) (x ndash h)2 = 4a(y ndash k) (x ndash h)2 = ndash4a(y ndash k)

Tangent in point form

(y ndash y1)(y ndash k) = 2a(x ndash x1)

(y ndash y1)(y ndash k) = ndash2a (x ndash x1)

(x ndash x1)(x1 ndash h) = 2a (y ndash y1)

(x ndash x1)(x1 ndash h) = ndash2a(y ndash y1)

Parametric co-ordinate

(h + at2 k + 2at) (h ndash at2 k + 2at) (h + 2at k + at2) (h + 2at k ndash at2)

t(y ndash k) = (x ndash h) + at2

t(y ndash k) = ndash(x ndash h) + at2

t(x ndash h) = (y ndash k) + at2

t(x ndash h) = ndash(y ndash k) + at2

Point of con-tact in terms of slope (m)

a 2ah km m

minus minus

(h + 2am k + am2) (h ndash 2am k ndash am2)

Condition of tangency

ac mh km

+ = +ac mh km

+ = minusc + mh = k ndash am2 c + mh = k + am2

Tangent in slope form

ay mx mh km

= minus + +ay mx mh km

= minus + minusy = mx ndash mh + k ndash am2

y = mx ndash mh + k + am2

y2 = 4

axy2 =

ndash4a

x2 = ndash

)2aminus

minus=

minus1

2aminus

=minus

xminusminus

=minus

xminus

=minus

ric co

t2 2at

)(ndash

ndashat

ric fo

y ndash

x ndash

m2 ndash

)(ndash

ityc =

ndash2a

m ndash

m=minus

ndash 2a

m ndash

minusminus

(y ndash

x ndash

ndash k

)2 = ndash

ndash h

)2 = 4

ndash k)

(x ndash

ndash4a

(y ndash

2aminus

minusminus

=minus

2aminusminus

=minus

minusminus

=minus

minus1

hminus

=minus

ndash k

) + t(

x ndash

t + at

ndash k

) ndash t(

x ndash

t + at

ndash h

) + t(

y ndash

t + at

ndash h

) ndash t(

y ndash

t + at

t of c

ndash 2

(h ndash

ityc =

k ndash

ndash 2a

m ndash

k ndash

minus+

=minus

minusminus

(y ndash

ndash h

)ndash 2

ndash am

ndash k

(x ndash

minus=

minus+

minus=

minusminus

Chapter 17ellipse

171 Definition

Ellipse is the locus of a point which moves in a plane such that the ratio of its distance from a fixed point (Focus) to its distance from the fixed line (Directrix) is always constant and equal to a quantity which is less than 1

172 StAnDARD eQUAtion of eLLiPSe

Given focus S(ae 0) and the x ndash (ae) = 0 as directrix

1721 Focal DistanceFocal distance (SP) of a point P is given as

aSP ePM e h a ehe

= = minus = minus

rArr 2 2(h ae) k a ehminus + = minus

rArr 2 2 2 2 2 2 2a e h 2aeh k a e h 2aeh+ minus + = + minus

rArr 2 2 2 2 2 2 2a e h k a e h+ + = + rArr 2 2 2 2 2 2 2h e h k a a eminus + = minus

rArr 2 2 2 2 2h (1 e ) k a (1 e )minus + = minus rArr 2 2

h k 1a a (1 e )

+ =minus

Let 2 2 2a (1 e ) bminus = rArr 2 2

x y 1a b

173 tRAcing of eLLiPSe

Equation of Ellipse 2 2

x y 1a b

Eccentricity 2

= minus

Symmetry Since curve is even wrt variable x and y the graph is symmetric about both the co-ordinate axes There are two foci and two directrices

Foci S1 (ae 0) S2 (ndashae 0) Directrices D1 x = ae D2 x = ndashae

Focal distances S1P = ePM = a ndash eh 2aS P = ePM = e he

= a + eh

AAprimeis called major axis length = 2a equation y = 0 BBprimeis called minor axis length= 2b equation x = 0 The point of intersection of major and minor is called

centre All the chords passing through the centre get bisected at the centre

Normal chord Chord normal to the major axis is called normal chord or double ordinate If it passes through the focus it is called latus rectum

Length of 22bLR =

a equation of LR x = ae

Ellipse 17175

Ellipse is a locus of the point that moves in such a manner so that the summation of its distances from two fixed points S1 and S2 (foci) remains constant (2a)

S1P + S2P = 2a where 2a is length of major axis Case I If 2a gt S1S2 = 2ae locus ellipse Case II S1P + S2P = S1S2 locus segment SSprime Case III S1P + S2P lt S1S1 no locus

If equation of ellipse is 2 2

x y 1a b

+ = where b gt a

Eccentricity e = radic1ndash(a2b2) Major axis x = 0

Length of Major axis 2b Minor axis y = 0Length of Minor axis 2a foci (0 plusmn be)LR y = plusmn be length of LR = 2a2b Extremities (plusmna2b be)

Equation of ellipse where centre lies at (a b) and major axis is parallel to the x-axis of length 2a and

minor axis of 2b (a gt b) 2 2

(x ) (y ) 1a bminusα minusβ

Major axis y = b Length of Major axis 2aMinor axis x = a Length of Minor axis 2bFoci S1 = (a + ae b)

S2 = (a ndash ae b)Directrix x = a + ae

x = a ndashae

Auxiliary Circle of an Ellipse A circle drawn on major axis of the ellipse as diameter is called

Auxiliary circle of ellipse Given the equation of ellipse 2 2

x yS 1a b

The equation of auxiliary circle 2 2 2x y a + =

Eccentric Angle Of any point P on the ellipse is angle (q) made by CPprimewith positive direction of major axis in anti-clockwise sense (where C is centre and Pprimeis corresponding point of P on Auxiliary circle)

Q P Px x a cos= = θ rArr 2 2 2

a cos y 1a b

θ+ = rArr y2 = b2 sin2q

Parametric equation xp = a cosq and yp = b sinq isin [0 2p) (a cosq b sinq) is called point q an the ellipse

174 PRoPeRtieS ReLAteD to eLLiPSe AnD AUxiLiARy ciRcLe

The ratio of ordinate of point P on ellipse and its corresponding point on AC is constant PM bsin bPM a cos a

The ratio of area of triangle inscribed in ellipse (DPQR) to the area of triangle (DPprimeQprimeRprime) formed by its corresponding point an AC is constant = ba

The above property is valid even for an n-sided polygon inscribed in the ellipse As n rarrinfin is the polygon that coincides with the ellipse and its corresponding polygon coincides with auxiliary circle

Ellipse 17177

1741 Position of a Point with Respect to Ellipse + minusyxS 1 = 0a b

A point P(x1y1) lies insideonoutside of ellipse as S1 lt 0S1 = 0S1 gt 0

1742 Position of a Line with Respect to EllipseThe Straight line y = mx + c cutstoucheshas no contact with ellipse

x yS 1 0a b

+ minus = as the equation b2x2 + a2 (mx + c) 2 ndash a2b2 = 0 has D gt 0D = 0D lt 0

Condition of tangency 2 2 2a m bplusmn + Thus all lines of the form 2 2 2y mx a m b= plusmn + will always be tangent to the ellipse where m is real

Equation of tangent in terms of slope also known as ever tangent 2 2 2y mx a m b= plusmn + and point

of contact is 2 2a m b c c

Chord of ellipse joining point q and f

Slope of chord of joining point q and f b cota 2

θ+ φ = minus

Equation of chord x ycos sin cosa 2 b 2 2

θ+ φ θ+ φ θminusφ+ =

Condition of focal chord If Passes through (ae 0)

or (ndashae 0) rArr e 1 e 1tan tan or2 2 e 1 e 1θ φ minus +

=+ minus

Equation of tangent at q (a cosq b sinq) x ycos sin 1a b

θ+ θ =

Equation of Normal at q Slope am tanb

= θ rArr Equation a siny bsin (x a cos )b cos

θminus θ = minus θ

θ rArr ax sec q ndash by cosec q = a2 ndash b2 = a2e2

Equation of tangent 1 12 2

xx yyT 1 0a b

+ minus = and equation of Normal 2 2

2 2 2 2

a x b y a b a e x y

minus = minus =

175 PRoPeRtieS of tAngentS AnD noRmALS

The slopes and equations of various tangents and normals are given by

Construction Slope Equation

Tangent at (x1y1)2

b xa y

minus 1 12 2

xx yy 1 0a b

+ minus =

Tangent at qb cota

minus θx ycos sin 1a b

θ+ θ =

Normal at (x1y1)2

a yb x

2 22 2

1 1 2 2a e

a x b y a bx y

minus = minus

Normal at qa tanb

2 2a eax sec bycosec a bθminus θ = minus

Point of Intersection of Tangent Point of intersection of tangent at

lsquofrsquoand lsquoqrsquoon the ellipse 2 2

x y 1a b

+ = is

a cos bsin2 2

cos cos2 2

θ+ φ θ+ φ

θminusφ θminusφ

Locus of foot of perpendicular from either foci upon any tangent is auxiliary circle of ellipse

Locus of point of intersection of a perpendicular tangents is the director circle of ellipse in fact the locus of point of intersection of perpendicular tangents (in case of conic sections other than parabola) is called lsquodirector circle of conic sectionrsquo

Product of length of perpendiculars from both foci upon any tangent is constant (b2) where b is length of semi-major axis of ellipseproduct of the lengthrsquos of the perpendiculars from either foci on a variable tangent to an EllipseHyperbola = (semi minor axis)2(semi conjugate axis)2 = b2

Tangent at any point (P) bisects the external angle and nor-mal at same point bisects the internal angle between fo-cal distances of P This refers to the well-known reflection property of the ellipse which states that rays from one are reflected through other focus and vice-versa

Ellipse 17179

In general four normals can be drawn to an ellipse from any point and if a b d g are the eccentric angles of these four co-normal points then a + b + d + g is an odd multiple of p

In general there are four concyclic points on an ellipse and if a b d g are the eccentric angles of these four points then a + b + d + g is an even multiple of p

The circle on any focal distance as diameter touches the auxiliary circle The straight lines joining each focus to the foot of the perpendicular from the other focus upon the

tangent at any point P meet on the normal PG and bisects it where G is the point where normal at P meets the major axis

Chord of contact 1 12 2

xx yyT 1 0a b

= + minus =

Pair of tangents SS1 = T2 2 2

x y 1a b

+ minus

xx yy 1a b

+ minus

Chord with a given middle point T = S1 2 2

1 1 1 12 2 2 2

xx yy x y1 1a b a b

+ minus = + minus

rArr 2 2

1 1 1 12 2 2 2

xx yy x ya b a b

Diameter The locus of the mid points of a system of parallel chords of an ellipse is called the diameter and the point where the diameter intersects the ellipse is called the vertex of the diameter

If y = mx + c is the system of parallel chords to 2 2

x y 1a b

then the locus of the midpoint is given

b xya m

= minus

Conjugate diameter Two diameters are said to be conjugate if each bisects all chords parallel to the other

If conjugate diameters are perpendicular to each other then ellipse becomes a circle The eccentric angles of the ends of a pair of conjugate diameters of an ellipse differ by a right angle The sum of squares of any two conjugate semi-diameters of an ellipse is constant and is equal to sum

of the squares of the semi-axes of the ellipse The product of the focal distances of a point on an ellipse is equal to the square of the semi-diameter

which is conjugate to the diameter through the point The tangents at the extremities of a pair of conjugate diameters form a parallelogram whose area is

constant and is equal to the area of rectangle formed by major and minor axis lengths

Chapter 18hyperbola

181 Definition

It is the locus of a point P whose ratio of distance from a fixed point (S) to a fixed line (Directrix) remains constant (e) is known as the eccentricity of hyperbola (e gt 1)

1811 Standard EquationGiven S(ae 0) as focus and the line x ndash (ae) = 0 as directrix

Focal Distance Focal distance of a point P is given as Q SP = ePM = eh ndash a rArr a2e2 + h2 ndash 2aeh + k2 = a2 + e2h2 ndash 2aeh

rArr h2(1 ndash e2) + k2 = a2(1 ndash e2) rArr minus =minus

h k 1a a (e 1)

rArr minus =2 2

x y 1a b

where a2(e2 ndash 1) = b2

1812 Tracing of Hyperbola

Equation of hyperbola minus =2 2

x y 1a b

Eccentricity = +2

Symmetry Since equation is even wrt variable x and y so graph is symmetric about both co-ordinate axes Hence there should be two foci and two directrix

Hyperbola 18181

Foci S1(ae 0) S2 (ndashae 0) Directrices D1 x = ae D2 x = ndashae Intersection with x-axis y = 0 rArr x = plusmn a rArr A(a 0) Aprime(ndasha 0)

AAprime is called transverse axis of hyperbola length = 2a equation y = 0 Intersection with y-axis x = 0 rArr y = plusmnbi rArr B(0 b) Bprime(0 ndashb)

BBprime is called conjugate axis length = 2b equation x = 0The point of intersection of transverse and conjugate is called centre

Normal chord Chord normal to transverse axis is called normal chord or double ordinate If it passes through focus it is called latus rectum

Extremities of Latus rectum

1bL aea

= minus

bL aea

Length of =22bRR

aequation x = +ae ndashae

Focal distances S1P = ePM = eh ndash a S2P = ePMprime = eh + a|S2P ndash S1P| = 2a where 2a is length of transverse axisCase I If 2a lt S1S2 = 2ae rArr hyperbolaCase II If S1P + S2P = S1S2 rArr union of two raysCase III If S1P + S2P gt S1S2

rArr No locus

Conjugate hyperbola of a hyperbola H = 0 is a hyperbola C = 0 whose transverse axis is conjugate axis of H = 0 and conjugate axis is transverse axis of H = 0 both in the sense of length and equation

Equation hyperbola minus =2 2

x y 1a b

Conjugate hyperbola minus = minus2 2

x y 1a b

Eccentricity = + 2 22e 1 (a b ) Foci (0 plusmnbe2)

Transverse axis x = 0 Length = 2b Conjugate axis y = 0 Length = 2a

Latus Rectum y = plusmn be2 LprimeL plusmn

2a beb

and length = 22a

+ = + =+ +

2 2 2 2 2 22 1

1 1 b a 1e e a b a b

The foci of a hyperbola and its conjugate are con-cylic and form the vertices of a square

If a = b hyperbola is said to be equilateral or rectangular and has the equation x2 ndash y2 = a2 Eccentricity for such a hyperbola is radic2Equation of hyperbola whose centre lies at (a b) and trans-verse axis is parallel to x-axis of length 2a and conjugate axis of

length 2b equation minusα minusβminus =

(x ) (y ) 1a b

Transverse axis y = b Length = 2a Conjugate axis x = a Length = 2b Foci S1 = (a + ae b) S2 = (a ndash ae b) Directrix D1 x = a + ae x = a ndashae

Equation of Hyperbola Referred to two perpendicular straight lines as their axes but not parallel to coordinate axes

( ) + +minus + + + minus =

1 1 11 1 22 2 2 21 1 1 1

l x m y nm x l y nm l l m

Centre C is the point of intersection of line l1x + m1y + n1 = 0 and m1x ndash l1y + n2 = 0

Equations of Directrices If (x y) is any point on a directrix then its ^r distance from conjugate axis ie m1x ndash l1y + n2 = 0 is ae

Equation of directrices are given by minus +

= plusmn+

1 1 22 21 1

m x l y n aem l

Hyperbola 18183

Foci Foci can be obtained by solving the equation l1x + m1y + n1 = 0 and the pair of normal chords

(Latera Recta) minus +

= plusmn+

1 1 22 21 1

m x l y n aem l

Length of each Latera Recta =22b

a Equations of Latera Recta are given by

minus += plusmn

+1 1 2

2 21 1

m x l y n aem l

1813 Auxiliary Circle of HyperbolaA circle drawn on transverse axis of the hyperbola as diameter is called auxiliary circle of hyperbola for

hyperbola minus =2 2

x y 1a b

auxiliary circle is given by x2 + y2 = a2

Eccentric Angle Of any point P on the hyperbola is angle (q) made by CPprime with positive direction of transverse axis in anticlockwise (where C is centre and Pprime is point of contact of tangent drawn from foot of ordinate of P to the Auxiliary circle)

Parametric Equation x = a secq and y = btanq π π θisin π

3[02 ) 2 2

and (a secq b tanq) is called

point q an the hyperbola The ratio of ordinate of point P on hyperbola and length of tangent from the foot of ordinate (M) to

the Auxiliary circle is constant (ba) θ= =

θPM bsin bPM a cos a

182 Director circle

The locus of the point of intersection of the tangents to the

x y 1a b

which are perpendicular to each other

is called director circleThe equation of director circle is P(h k) is x2 + y2 = a2 ndash b2

(a gt b)

1821 Position of a Point with Respect to Hyperbola

Given hyperbola minus minus =2 2

x y 1 0a b

rArr = minus2

2 2 22

by (x a )a

A point P(x1 y1) lies inside (towards centre)on

outside (towards focus) of hyperbola as S1 lt 0S1 = 0S1 gt 0

1822 Position of a Line with Respect to Hyperbola minus minus22

yxS 1 = 0a b

The straight line y = mx + c cutstoucheshas no contact with hyperbola minus =2 2

x y 1a b

as the equation

b2x2 ndash a2 (mx + c)2 ndash a2b2 = 0 has D gt 0D = 0D lt 0

Condition of tangency = plusmn minus2 2 2c a m b

Equation of tangent in terms of slope = plusmn minus2 2 2y mx a m b and

point of contact is minus minus

2 2a m b c c

Chord of Hyperbola Joining Point q and fEquation of chord of hyperbola Joining Point q and f

is θ θ =φ φ

x y 1a sec btan 1 0a sec btan 1

which can also be written as

θminusφ θ+ φ θ+ φminus =

x ycos sin cos a 2 b 2 2

Condition for Focal Chord Chord becomes focal chord if it passes through (ae0) or (ndashae 0) Sup-

pose it passes through (ae0) then rArr θ φ minus = +

1 etan tan 2 2 1 e

or +minus

1 e1 e

if it passes through (ndashae 0)

1823 Properties of Tangents and Normals

Tangent at (x1y1)2

b xa y

minus minus =1 12 2

xx yy 1 0a b

Tangent at q θb coseca

θminus θ =x ysec tan 1a b

Normal at (x1y1) minus2

a yb x

2 22 2

1 1 2 2a e

a x b y a bx y

Normal at q minus θa sinb

θ+ θ = +

2 2a eax cos by cot a b

Point of intersection of tangent at q and f on

the hyperbola minus =2 2

x y 1a b

θminusφ

=θ+φ1

cos2x a

θ+ φ

=θ+φ1

sin2y b

Tangent at any point (P) bisects the internal angle and normal at same point bisects the external angle between focal distances of P This refers to reflection property of the hyperbola which states that rays from one Focus are reflected such that they appear to be coming from other focus

Hyperbola 18185

An ellipse and hyperbola if con-focal always intersect orthogonally

Chord of contact = minus minus =1 12 2

xx yyT 1 0a b

Pair of tangents = minus minus minus minus = minus minus

22 22 21 1 1 1

1 2 2 2 2 2 2 2

x y xx yyx ySS T 1 1 1a b a b a b

Chord with a given middle point = minus minus = minus minus

2 21 1 1 1

1 2 2 2 2

xx yy x yT S 1 1a b a b

rArr minus = minus2 2

1 1 1 12 2 2 2

xx yy x ya b a b

1824 Asymptote Hyperbola

Asymptote to any curve is straight line at finite distance that touches the curve at infinity (micro)

Let y = mx + c be asymptote to hyperbola then both roots of the equation (b2 ndash a2m2)x2 ndash 2a2cmx ndash a2 (c2 + b2) = 0 approach to micro

rArr minus =

sum of root infinity

b a m 0 and 2 2 2 2

condition of tangency

c a m b= minus

rArr = plusmnbma

and c = 0

= minus =

are pair of asymptotes

18241 Properties of asymptote hyperbola

Both the asymptotes are pair of tangents to a hyperbola from its centre Axis of Hyperbola bisects the angle between asymptotes

If lines be drawn through A Aprime parallel to C axis and through B Bprime parallel to T axis then asymptotes lie along the diagonal of rectangle thus formed

Combined equation of asymptotes (A = 0) differs from equation of hyperbola (H = 0) and conjugate hyperbola (C = 0) by same constant ie A = H + l and A = C ndash l

As minus =2 2

x yH 1a b

and minus =2 2

x yA 0 2a b

minus = minus2 2

x yC 1a b

Relation between A C H+

2 Angle between Asymptote Included angle between two asymptotes is

minus minus = minus 1 1

2ab btan 2tana b a

or 2 Secndash1(e)

If the angle asymptotes is 90deg then b = a and hyperbola is called rectangular hyperbola The product of the perpendicular drawn from any point on a hyperbola to its asymptotes is constant The foot of the perpendicular from a focus to an asymptote is a point of intersection of the auxiliary

circle and the corresponding directix

The portion of any tangent to hyperbola intercepted between asymptote is bisected at the point of contact

Any tangent to the hyperbola makes with asymptote a triangle of constant area

183 rectangular hyperbola

A hyperbola whose asymptotes include a right angle is called rectangular hyperbola or if the lengths of transverse and conjugate axes of a hyperbola be equal it is called rectangular or equilateral hyperbola Equation x2 ndash y2 = a2 TA y = 0 Length 2a CA x = 0 Length 2a Eccentricity (e) = radic2 Foci (plusmnaradic2 0) Directrix x = plusmnaradic2 Asymptote y = x and y = x

1831 Rectangular Hyperbola where Asymptote are Coordinate Axis

Given rectangular hyperbola x2 ndash y2 = a2 If axes rotating by p4 about the origin+

rarrx yx

2 and minus +

rarrx yy

2 the equation transforms to + minus

minus =2 2

2(x y) (x y) a2 2

Eccentricity = radic2 Transverse axis Equation

y = x Length 2radic2c

Conjugate axis Equation y + x = 0 Length 2radic2c

Foci S(cradic2 cradic2) and Sprime(-cradic2 ndash cradic2)

Directrix x + y = plusmncradic2

Parametric equation x = ct y = ct t isin Rndash0 Centre (0 0)

Vertex (c c) and (ndash

c ndashc)

Conjugate Hyperbola of Rectangular Hyperbola xy = c2

It is given by xy = ndashc2 Centre (0 0)

Hyperbola 18187

Vertex (ndashc c) and (c ndashc) Eccentricity = radic2

TA Equation y + x = 0 Length 2radic2c CA Equation y = x Length 2radic2c

Foci S(ndashcradic2 cradic2) and Sprime(cradic2 ndashcradic2) Directrix x ndash y = plusmncradic2 Parametric equation x = ct y = ndashct t isin R ndash 0

1832 Parametric Equations of Chord Tangents and Normal

Slope of chord joining the points P(t1) and Q(t2) = minus1 2

1m t t

Equation of chord x + t1t2

(t1 + t2)

Condition for focal chord += plusmn

t t 21 t t

Equation of the tangent at P(x1 y1) + =

x y 2x y

Equation of tangent at P(t) x + yt2 = 2ct

Equation of normal at P (t) minus = minus2cy t (x ct)

rArr xt3 ndash yt = c(t4 ndash 1) If normal of hyperbola xy = c2 at the point P(T) meet the hyperbola again at Tprime the T3Tprime = ndash1

Chord with a given middlepoint as (h k) is kx + hy = 2hk

1833 Co-normal Points

In general four normals can be drawn on a hyperbola each passing through a common point The feets of perpendicular of these four normals lying on the hyperbola are called co-normal points

18331 Properties of co-normal points

1 In general four normals can be drawn to a hyperbola from any point and if a b g d be the eccentric angles of these four co-nomal points then a + b + g + d is an odd multiple of p

2 If a b g are the eccentric angles of three points on the hyperbola minus =2 2

x y 1a b

the normals at which

are concurrent then sin (a + b) + sin (b + g) + sin (a + g) = 0

18332 Diameter of a hyperbola

The locus of the middle points of a system of parallel chords of a hyperbola is called a diameterThe equation of a diameter bisecting a system of parallel chords of slope m of the hyperbola

minus =2 2

x y 1a b

18333 Conjugate diameters

Two diameters are said to be conjugate when each bisects all chords parallel to the othersTwo diameters y = mx and y = kx are said to be conjugate if their gradients are related as

km = b2a2

1834 Properties of Conjugate Diameters

If a pair of diameters are conjugate with respect to a hyperbola then they are also conjugate with respect to its conjugate hyperbola

If a pair of diameters be conjugate with respect to a hyperbola then one of those diametsrs will meet the hyperbola in real points while the other diameter will meet the conjugate hyperbola in real points

If a pair of conjugate diameters meet the hyperbola

minus =

x y 1a b

and its conjugate hyperbola

minus + =

x y 1 0a b

in PPprime and D Dprime respective then

(i) CP2 ndash CD2 = a2 ndash b2

(ii) The parallelogram formed by the tangents at the extremities of conjugate diameters has its vertices lying on the asymptotes and its of constant area

(iii) Show that the asymptotes to the hyperbola bisect PD PDprime PprimeD and PprimeDprime

Chapter 19Complex Number

191 IntroductIon

While working with real numbers (ℝ) we would not find relations to equations such as x2 + 9 = 0 () So to look forward we have to difine another set of number systems

1911 Imaginary Numbers (Non-real Numbers)

A number whose square is non-positive is termed as an imaginary number eg 2 or (1 2)minus + minus

Iota Euler introduced the symbol i for the number 1minus It is known as iota (a Greek word for

lsquoimaginaryrsquo) Thus 2 2iminus = and + minus = +1 2 1 2i are imaginary numbers

Remark (i) Imaginary numbers do not follow the property of order ie for z1 and z2 imaginary numbers we

cannot say which one is greater Since i is neither positive nor negative nor zero

(ii) Here non-possible does not imply negative eg 1 2+ minus is also non-positive

1912 Purely Imaginary Numbers (I)

The number z whose square is non positive real number (negative or zero) is termed as purely imaginary

number For example 5minus ie I = z z = ai where a isin ℝ and i =

19121 Geometrical representation of purely imaginary numbers

Single multiplication by i is equivalent to geometrical rotation of number by p2 radians anti-clockwise

Therefore purely imaginary numbers are represented as points lying on y axis of argand plane For example z = ai is represented by point (0 a) on y axis as shown here

Remarks 1 The plane formed by real and imaginary axes is called ArgandGaussianComplex Plane

2 It should be kept in mind that any equation not having real roots does not necessarily posses imaginary roots For example the equation x + 5 = x + 7 is neither satisfied by real numbers nor is satisfied by imaginary numbers

1913 Properties of Iota

1 i0 = 1 i2 = ndash1 i3 = ndashi i4 = 1 2 Periodic properties of i i4n = 1 i4n + 1 = i i4n + 2 = ndash 1 i4n + 3 = ndashi forall n isin ℤ 3 i ndash 1 = ndash i 4 Sum of four consecutive power terms of i is zero that is in + in + 1 + in + 2 + in + 3 = 0 forall n isin ℤ 5 For any two real numbers a and b times =a b ab is true only when atleast one of a and b is

non-negative real number ie both a and b are non-negative

192 complex number

A number z resulting as a sum of a purely real number x and a purely imaginary number iy is called a

complex number ie a number of the form z = x + iy where x y isin ℝ and = minusi 1 is called a complex number Here x is called real part and y is called imaginary part of the complex number and they are expressed as Re(z) = x Im (z) = y A complex z = x + iy number may also be defined as an ordered pair of real numbers and may be denoted by the symbol (x y)

The set of complex numbers is denoted by ℂ and is given by = z z = x + iy where x y isin ℝ and = minusi 1

193 ArgAnd plAne

Any complex number z = a + ib can be written as an ordered pair (a b) which can be represented on a plane by the point P(a b) (known as affix of point P) as shown in the figure This plane is called Argand plane complex plane or the Gaussian plane

1931 Representation of Complex Numbers

Complex numbers can be represented by following forms 1 Cartesian form (rectangular form) A complex number z = x + iy can be represented by the

point P having coordinate (x y) 2 Vector form (Algebraic form) Every complex number z is regarded as a position vec-

(OP) which is sum of two position vectors Purely real vector x

(OA) and purely imaginary

vector iy

= + = +

OP OA AP OA OB rArr z = x + iy

Modulus of z Distance of point P from the origin is called modulus of complex number z and is denoted by |z| It is length of vector

(OP) It is distance of P(z) from origin

Complex Number 19191

( )( ) ( )( )there4 = = + = + 2 22 2z OP x y Re z Im z

Argument of z Argument of z is the angle made by

OP with the positive direction of real axis Also known as amplitude z and is denoted by amp (z)

Arg(z) = q where θ =ytanx

q lies in the quadrant in

which complex number z lies

NoteThe principal arguments q isin (ndashp p]

3 Polar form (amplitude modulus form) In DOAP OP = |z| = r rArr OA = x = r cosq and AP = y = r sinq rArr z = x + iy = r (cosq + i sinq) = r cisq

Remarkcis q is unimodular complex number and acts as unit vector in the direction of q where q is arg z

4 Euler form (Exponential form) Euler represented complex number z as an exponential function of its argument q (radians) and described here As we know that using Taylorrsquos series expansion cos q and sinq can be expanded in terms of polynomial in q as given below

θ θ θ

θ = minus + minus +2 4 6

cos 1 2 4 6

and θ θ θθ = θminus + minus +

sin 3 5 7

(cosq + isinq) = ( ) ( ) ( ) θθ θ θ

+ θ+ + + + infin =2 3 4

ii i i1 i to e

2 3 4 rArr z = x + iy = r (cosq + i sinq) = reiq

Advantages of using Euler form Convenient for division and multiplication of complex numbers Suitable for exponential logarithmic and irrational functions involving complex numbers

19311 Inter-conversion from polartrigonometric to algebraic form

(i) Algebraic form to polar form Given z = x + iy then

= +2 2r x y θ = θ =x ycos sinr r

gives q = f (say)

In polar form = + φ+ φ2 2z x y (cos isin )

(ii) Polar form to algebraic form Given z = r(cosq + isinq) = rcosq + i(rsinq)

rArr z = x + iy where x = rcosq and y = rsinq

1932 Properties of Complex Numbers

(i) Equality Two complex numbers z1 and z2 are equal only when their real and imaginary parts are respectively equal ie Re(z1) = Re(z2) and I(z1) = I(z2) or |z1| = |z2| and arg (z1) = arg (z2)

RemarksStudents must note that x y isin ℝ and x y ne 0 If x + y = 0 rArr x = ndash y is correct but x + iy = 0 rArr x = ndash iy is incorrect (unless both x and y are zero)

Hence a real number cannot be equal to the imaginary number unless both are zero

(ii) Inequality Inequality in complex number is not defined because lsquoirsquo is neither positive zero nor negative So 4 + 3i gt 1 + 2i or i lt 0 or i gt 0 is meaningless

(iii) If Re(z) = 0 then z is purely imaginary and if Im (z) = 0 then z is purely real (iv) z = 0 rArr Re(z) = Im (z) = 0 therefore the complex number 0 is purely real and purely imaginary or both (v) If z = x + iy then iz = ndashy + ix rArr Re(iz) = ndash Im(z) and Im(iz) = Re(z) (vi) Conjugate of complex number z = x + i y is denoted

as z = (x ndash iy) ie a complex number with same real part as of z and negative imaginary part as that of z

(vii) If z is purely real positive rArr Arg(z) = 0 (viii) If z is purely real negative rArr Arg(z) = p (ix) If z is purely imaginary with positive imaginary part

rArr Arg(z) = p2 (x) If z is purely imaginary with negative imaginary part

rArr Arg(z) = ndashp2 (xi) Arg(0) is not defined

19321 Binary operations defined on set of complex numbers

Binary operation on set of complex number is a function from set of complex numbers to itself That is if z1 z2 isin C and is a binary operation on the set of complex numbers then z1 z2 isin C Following binary operations are defined on set of complex numbers

Addition of two complex numbers Let z1 = x1 + iy1 and z2 = x2 + iy2 rArr z1 + z2 = (x1 + iy1) + (x2 + iy2)= (x1 + x2) + i (y1 + y2) ie z1 + z2 = [R(z1) + R(z2)] + i[I(z1) + I(z2)] isin C

19322 Geometric representation

Consider two complex numbers z1 = (x1 + iy1) and z2 = (x2 + iy2) represented by

vector =

z OB as shown in figure

Then by parallelogram law of vector addition + = + =

1 2z z OA OB OC Hence C represents the affix of z1 + z2

NotesIn DOAC [Since sum of two sides of a D is always greater than the third side] OA + AC ge OC

rArr |OA| |OB| |OC|+ ge

rArr | z1 | + | z2 | ge | z1 + z2| This is called triangle inequality Also considering OAB OA + OB ge AB

rArr + ge rArr + ge minus

1 2 1 2|OA| |OB| |BA| |z | |z | |z z |

Subtraction of two complex numbers Let z1 = x1 + iy1 and z2 = x2 + iy2 then z1ndash z2 =(x1 + iy1) ndash (x2 + iy2) = (x1 ndash x2) + i (y1 ndash y2) ie z1 ndash z2 = [R(z1) ndash R(z2)] + i[I(z1) ndash I(z2)] isin C

Using vector representation again we have = minus = minus =

1 2BA OA OB z z OC Hence the other diagonal of the parallelogram represents the difference

vector of z1 and z2

1 While BA

represents the free vector corresponding to z1 ndash z2 OC

represents the position vector of z1 ndash z2

rArr C is affix of complex number z1 ndash z2

2 In a triangle the difference of two sides is always less than the third side

rArr OB OA ABminus le

rArr ||z2| ndash |z1|| le |z2 + z1|

3 Triangle Inequality ||z1| ndash |z2|| le |z1 plusmn z2| le |z1| + |z2 |

Multiplication of two complex numbers Let z1 = x1 + iy1 and z2 = x2 + iy2 then z1z2 = (x1 + iy1) (x2 + iy2) = [R(z1)R(z2) ndash I(z1)I(z2)] + i[R(z2)I(z1) + R(z1) I(z2)] isin C

Geometric representation Let A and B are two points in the complex plane respectively affixes of z1 and z2 where z1 = r1(cos q1 + i sinq1) and z2 = r2(cos q2 + i sinq2) z1z2 = r1r2(cosq1 + isin q1) (cosq2 + i sinq2)

1933 ResultThe product rule can be generalized to n complex numbers Let zn = rn(cosqn + i sinqn) where n = 1 2

Now |z1 z2zn| = r1r2rn = |z1| | z2 ||zn| and arg (z1 z2zn) = q1 + q2 + + qn = arg z1 + arg z2 + + arg zn

As special case arg zn = n arg z

Division of two complex numbers Let z1 = x1 + iy1 and z2 = x2 + iy2 rArr z1z2 = (x1 + iy1)(x2 + iy2)

= + minus+ isin

+ +1 2 1 2 2 1 1 2

2 2 2 22 2 2 2

(x x y y ) i(x y x y ) C(x y ) (x y )

Geometric representation Let A and B are two points in the complex plane which are affixes of z1 and z2 respectively where z1 = r1(cos q1 + i sinq1) and z2 = r2(cos q2 + i sinq2)

Then we get =2 2

z rz r

[cos(q2 ndash q1) + i sin(q2 ndash q1)]

1 If q1 and q2 are principal values of argument of z1 and z2 then q1 + q2 may not necessarily be the principal value of argument of z1 z2 and q1 ndash q2 may not necessarily be principal value of argument of z1z2 To make this argument as principal value add or subtract 2np where n is such an integer which makes the argument as principal value

2 Note that angle a between two vectors OA

and OB

is a = q2 ndash q1 a = arg z2 ndash arg z1

194 AlgebrAIc Structure of Set of complex numberS

(i) Complex numbers obey closure law (for addition subtraction and multiplication) commutative law (for addition and multiplication) associative law (for addition and multiplication) existence of additive and multiplicative identitiy and inverse

(ii) Existence of conjugate element Every complex number z = x + iy has unique conjugate denoted as x ndash iy

1941 Conjugate of a Complex NumberConjugate of a complex number z = x + iy is defined as = minusz x iy It is mirror image of z in real axis as mirror shown in the figure given here

Let z = r (cosq + isinq) rArr = θminus θz r(cos isin ) = r [cos(ndashq) + isin(ndashq)]rArr z has its affix point having magnitude r and argument (ndashq)

1942 Properties of Conjugate of a Complex Number 1 = = minusR(z) R(z) I(z) I(z)

2 = = = +2 2 2 2zz | z | | z | (R(z)) (I(z))

3 = =(z) z (z) z and so on

4 = minus =|z| |z|and Agr z Arg z

5 If =z z ie arg z = arg z rArr z is purely real 6 If z = ndash z ie arg (ndashz) = arg( z ) rArr z is purely imaginary

7 += = =

z zR(z) x R(z)2

= = = minusz zIm(z) y Im(z)

8 θ minus θ +

i ie ecos2

θ minus θ minus

i ie esin2i

9 plusmn plusmn plusmn plusmn = plusmn plusmn plusmn plusmn1 2 3 n 1 2 3 n(z z z z ) z z z z

10 =1 2 3 n 1 2 3 n(z z z z ) (z )(z )(z )(z )

11 = 11 2

(z )(z z )(z )

12 =n n(z ) (z) 13 If w = f(z) then ω= f(z) where f(z) is algebraic polynomial

14 + =1 2 2 1 2 1z z z z 2R(z z )

15 + = + +2 21 2 1 2 1 2| z z | | z | | z | 2Re(z z )

16 |z1 + z2|2 + |z1 ndash z2|

2 = 2(|z1|2 + |z2|

1943 Modulus of a Complex NumberModulus of a complex number z = x + iy is denoted by |z| If point p(x y) represents the complex number

z on Argandrsquos plane then = = +2 2z OP x y = distance between origin and point = +2 2P [R(z)] [I(z)]

19431 Properties of modulus of complex numbers

1 Modulus of a complex numbers is distance of complex number from the origin and hence is non-negative and |z| ge 0 rArr | z | = 0 iff z = 0 and | z | gt 0 iff z ne 0

2 ndash| z | le Re(z) le | z |and ndash | z | le Im(z) le | z | 3 = = minus = minus| z | | z | | z | | z |

4 = 2zz | z | 5 | z1z2 | = | z1 || z2 | In general |z1z2z3 zn| = |z1||z2|| z3 ||zn| 6 (z2 ne 0) 7 Triangle inequality | z1 plusmn z2 | le | z1 | + | z2 | In general | z1 plusmn z2 plusmn z3plusmn zn| le | z1 |plusmn| z2 |

plusmn | z3 | plusmnplusmn | zn | 8 Similarly | z1 plusmn z2 | ge | z1 | ndash | z2 | 9 | zn | = | z |n

10 || z1 | ndash | z2 || le | z1 plusmn z2| le | z1 | + | z2 | Thus | z1 | + | z2 | is the greatest possible value of | z1 plusmn z2 | and || z1 | ndash | z2 || is the least possible value of | z1 plusmn z2|

11 plusmn = + plusmn +2 2 21 2 1 2 1 2 1 2| z z | | z | | z | (z z z z ) or + plusmn2 2

1 2 1 2| z | | z | 2Re(z z ) or | z1 |2 + | z2 |

2 plusmn 2 | z1 || z2 |

cos (q1 ndash q2)

12 + = θ minusθ21 2 1 2 1 2 1 2| z z z z | 2 |z | | z | cos( ) where q1 = arg (z1) and q 2 = arg (z2)

13 | z1 + z2 |2 = | z1 |

2 + | z2 |2 hArr 1

is purely imaginary

14 | z1 + z2 |2 + | z1 ndash z2 |

2 = 2| z1 |2 + | z2 |

2 15 | az1 + bz2 |

2 + | bz1 ndash az2 |2 = (a2 + b2)(| z1 |

2 + | z2 |2) where a b isin R

16 Unimodular If z is unimodular then | z | = 1 Now if f (z) is a unimodular then it can always be expressed as f (z) = cosq + isinq qisinℝ

19432 Argument and principal argument of complex number

Argument of z (arg z) is also known as amp(z) is angle which the radius vector

OP makes with positive direction of real axis

Principle Argument In general argument of a complex number is not unique if q is the argument then 2nπ + q is also the argument of the complex number where n = 0 plusmn 1 plusmn 2 Hence we define principle value of argument q which satisfies the condition ndashp lt q le p Hence Principle value of arg(z) is taken as an angle lying in (ndashp p] It is denoted by Arg(z) Thus arg(z) = Arg(z) plusmn 2kp k isin ℤ

A complex number z given as (x + iy) lies in different quadrant depending upon the sign of x and y Based on the quadrantal location of the complex number its principle argument are given as follows

Sign of x and y Location of z Principal Argument

x gt 0 y gt 0 Ist quadrant minusθ = α = 1 ytanx

x lt 0 y gt 0 IInd quadrant minusθ = πminusα = πminus 1 y( ) tanx

x lt 0 y lt 0 IIIrd quadrant 1 ytanx

minusθ = minusπ+

x gt 0 y lt 0 IVth quadrant minusθ = minusα = minus 1 ytanx

19433 Caution

An a usual mistake is to take the argument of z = x + iy as tanndash1 (yx) is irrespective of the value of x and y

Remember that tanndash1 (yx) lies in the interval π π minus

Whereas the principal value of argument of z (Arg(z)) lies in the interval (ndashπ π]

Thus if z = x + iy then

tan (yx) if x 0 y 0tan (yx) if x 0 y 0tan (yx) if x 0 y 0Arg(z)

2 if x 0 y 02 if x 0 y 0

Not defined for x 0 y 0

+ π lt ge minusπ lt lt= π = gtminusπ = lt

19434 Properties of argument of complex number

1 arg (z1z2) = arg z1 + arg z2

2 arg(zn) = n (argz)

= minus

zarg arg z arg zz

4 arg(z) = 0 hArr complex number z is purely real and positive 5 arg(z) = p hArr complex number z is purely real and negative 6 arg(z) = p2 hArr complex number z is purely imaginary with positive Im(z) 7 arg(z) = ndash p2 hArr complex number z is purely imaginary with negative Im(z) 8 arg(z) = not defined hArr z = 0 9 arg(z) = p4 hArr z = (1 + i) or (x + xi) etc for (x gt 0)

Properties of Principal Arguments (Principal argument of complex number is denoted by arg (z))

1 If θ= θ + θ = kik k k k kz r (cos isin ) r e are number of complex numbers then

= plusmn π

sumprod

k kk 1k 1

Arg z Arg z 2k

where k isin ℤ choose k suitably such that principal Arg of the resultant number lies in principal range

zArg 2Arg (z)z

3 Arg (zn) = n Arg z plusmn 2kp 4 Arg (ndashz) = ndashp + Arg z or p + Arg z respectively when Arg z gt 0 or lt 0 5 Arg (1z) = ndashArg z

Method of Solving Complecs EquationsLet the given equation be f(z) = g(z) To solve this equation we have the following four methods

Method 1 Put z = x + iy in the given equation and equate the real and imaginary parts of both sides and solve to find x and y hence z = x + iy

Method 2 Put z = r(cosq + isinq) and equate the real and imaginary parts of both sides solve to get r and q hence z

Method 3 Take conjugate of both sides of given equations Thus we get two equations f(z) = g(z) (1) and =f(z) g(z) (2)

Adding and Subtracting the above two equations we get two new equations solving then we get z

Method 4 Geometrical Solution From the given equation we follow the geometry of complex number z and find its locus

1944 Square Roots of a Complex Number

Square roots of z = a + ib are given by + minus

plusmn +

| z | a | z | ai

2 2 b gt 0 and

+ minusplusmn minus

| z | a | z | ai

2 2 b lt 0

19441 Shortcut method

Step 1 Consider =0Im(z ) b2 2

Step 2 Factorize b2 into factors x yx2 ndash y2 = Re(z0) = a

Step 3 Therefore a + ib = (x + iy)2

rArr + = plusmn +a ib (x iy) eg minus8 15i a = 8 b = ndash15 lt 0

rArr = minusb 152 2

= xy such that x2 ndash y2 = 8 rArr = = minus5 3x y2 2

minus = plusmn minus = plusmn minus

5 3i 18 15i (5 3i)2 2 2

19442 Cube root of unity

Let 3 1 = cube root of unity

rArr x3 = 1 where minus +ω=

and minus minusω =2 1 3i

2 Cube roots of unity are 1 w w2 and w w2 are called the

imaginary cube roots of unity

19443 Properties of cube root of unity

P(1) |w| = |w2| = 1 P(2) ω=ω2

P(3) w3 = 1 P(4) w3n = 1 w3n + 1 = w and w3n + 2 = w2 forallnisinℤP(5) Sum of cube roots of unity is 0 That is 1 + w + w2 = 0

Remarks

∵ 2ω ω= ∵ 1 0ω ω+ + =

∵ 2ω ω= and 3 4 2 2 21 ( ) ( )ω ω ωω ω ω ω= = = = =

∵ 2 21 1 ( )ω ω ω ω+ + = + + ∵ 21 ( ) 0ω ω+ + =

P(6) +ω +ω =

n 2n 3 when n is multiple of 31

0 when n is not a multiple of 3

P(7) 1 w w2 are the vertices of an equilateral D having each side = radic3

P(8) Circumcentre of D ABC with vertices as cube roots of unity lies at origin and the radius of circumcircle is 1 unit Clearly OA = OB = OC = 1

RemarkFrom the above properties clearly cube roots of unity are the vertices of an equilateral D having each side = radic3 and circumscribed by circle of unit radius and having its centre at origin

P(9) π

ω = minus + =

1 3iarg( ) arg2 2 3

ω = minus minus =

2 1 3 4arg( ) i2 2 3

P(10) Any complex number a + ib for which =1(a b)3

or 3 1 can always be expressed in

terms of i w w2

eg + = minus ω21 i 3 2 + minus + ω+ = + = = =

i 1 i 3 2 1 i 3 23 i (1 i 3) 2i2 2i i 2 i

19444 Important relation involving complex cubic roots of unity

(a) x2 + x + 1 = (x ndash w) (x ndash w2) (b) x2 ndash x + 1 = (x + w) (x + w2) (c) x2 + xy + y2 = (x ndash yw) (x ndash yw2) (d) x2 ndash xy + y2 = (x + yw) (x + yw2) (e) x2 + y2 = (x + iy) (x ndash iy) (f) x3 + y3 = (x + y) (x + yw) (x + yw2) (g) x3 ndash y3 = (x ndash y) (x ndash yw) (x ndash yw2) (h) x2 + y2 + z2 ndash xy ndash yz ndash zx = (x + yw + zw2) (x + yw2 + zw) (i) x3 + y3 + z3 ndash 3xyz = (x + y + z)(x + yw + zw2)(x + w2y + wz)

195 de moIVerrsquoS tHeorem

This theorem states that (i) (cosq + isinq)n = cosnq + isinnq if n is an rational number (ii) (cosq + isinq)1n = [cos(q + 2kp) + isin (q + 2kp)]1n

(∵ period of sinq and cosq is 2p) = π θ π θ+

(2k + ) (2k + )cos i sin n n

k = 0 1 2 n ndash 1

1951 nth Root of Unity

Let x be an nth root of unity then ( ) ( )= = +1 1n nx 1 cos0 isin0 = π+ π+ + =

2r 0 2r 0cos isin r 0n n

1 2 n ndash 1

= ππ+ π+ + = =

i2rn2r 0 2r 0cos isin r 0 e r 0

n n 12n ndash 1=

π π minus π2 4 2(n 1)i i in n n1 e e e = 1 a a2

an ndash 1 where π

α =2ine

1952 Properties of nth Root of Unity

P(1) nth roots of unity form a GP

P(2) 1 + a + a2 + + an ndash 1 = 0

P(3) 1 aa2an ndash 1 = (ndash1)n ndash 1

P(4) nth roots of unity are vertices of n-sided regular polygon circumscribed by a unit circle having its centre at the origin

Case (i) When n is oddLet n = 2m + 1 m is some positive integers then only one root is real that is 1 and remaining 2m roots are non real complex conjugates

The 2m non-real roots are (a a2m) (a2 a2mndash1) (a3 a2mndash2) (am am+1) where the ordered pairs are (z z) ie non-real roots and their

conjugates and π

α =2ine

NoteThe nth roots given as ordered pairs are conjugate and reciprocal of each other

m2m 1 2m 11 2m 2m 1 m m 1

1 1 11

α αα α α α αα α α α α

+ +minus + +

= = = = = = = = =

Case (ii) When n is even

Let n = 2m π π α = =

2cis cisn m

except 1 and ndash1 other roots are non-real

complex conjugate pairs

NoteThe nth roots arranged vertically below are conjugate and reciprocal of each other and diagonally (passing through origin) are negative of each other

19521 nth root of a complex number n z

Let z = r cis q z1n = (r1n) (cis(2kπ + q))1n = (r1n) π θ +

2kcisn n

where r1n is positive nth root of r

= π θ

1n 2k(r ) cis cisn n

where π2kcisn

is the nth root of unity k = 0 1 2 n ndash 1

19522 To find logarithm of a complex number

Consider z = x + iy converting lsquox + iyrsquo into Eulerrsquos form such that q = principal value of argument of z then we get loge (x + iy) = loge (|z|eiq)

rArr loge(x + iy) = loge |z| + logeeiq rArr loge (x + iy) = loge |z| + iq

In general loge(x + iy) = loge|z| + i(q + 2np) nisinℤ To find (x + iy)(a+ib) ie 2z

Let u + iv = (x + iy)(a+ib)

rArr ln (u + iv) = (a + ib) ln (x + iy) rArr (u + iv) = e (a + ib) ln (x + iy) now solve for u and v by expressing (x + iy) in polar form

For example x = (i)i rArr lnx = ilni = ππ π π + = =

i 2 2i n cos isin i n(e ) i ne2 2 2

rArr π= minusnx

2 rArr

πminus

= 2x e Thus (i)i = endashp2

Alternatively ππ π π π+ minus

= = = = = =

i n cos isin iii n(i) i ni i n(e )2 2 2 2(i) e e e e e e

196 geometry of complex number

1961 Line Segment in Argandrsquos Plane

Any line segment joining the complex numbers z1 and z2 (directed towards z2) represents a complex number given by z2 ndash z1 Since every complex number has magnitude and direction therefore z2 ndash z1 also

|z2 ndash z1| represents the length of line segment BA ie the distance between z1 and z2 and arg(z2 ndash z1) represents the angle which line segment AB (on producing) makes with positive direction of real axis

19611 Angle between to lines segments (Roation theorm or conirsquos theorem)

Consider three complex numbers z1 z2 and z3 such that the angle between line segments joining z1 to z2 and z3 to z1 is equal to q

Then q = a ndash b = Arg(z3 ndash z1) ndash Arg(z2 ndash z1) = 3 1

z z Post-rotation vectorArg Argz z Pre-rotation vector

minus = minus

rArr i3 1

z zArg Arg( e )

z zθ minus

= θ = ρ minus

rArr (z3 ndash z1) = (z2 ndash z1) r eiq where minus

ρ =minus

z zz z

If z1 = 0

rArr z3 = rz2 eiq arg(z3z2) is an angle through which z2 is to be rotated to

coincide it with z3If a complex number (z2 ndash z1) is multiplied by another complex number reiq then the complex

number (z2 ndash z1) gets rotated by the argument (q) of multiplying complex number in anti-clockwise direc-tion (It is called Rotation Theorem or Conirsquos Theorem)

1962 Application of the Rotation Theorem (i) Section Formula Let us rotate the line BC about the point C so that it becomes parallel to

the line CA The corresponding equation of rotation will be ( )πminus minus= = minus

minus minusi1 1

z z | z z | m e 1z z |z z | n

rArr nz1 ndash nz = ndash mz2 + mz rArr +=

+1 2nz mzzm n

Similarly if C(z) divides the segment AB externally in the ratio of m n

then minus=

minus1 2nz mzzm n

In the specific case if C(z) is the mid point of AB then += 1 2z zz

(ii) Condition for Collinearity If there are three real numbers (other than 0) l m and n such that lz1 + mz2 + nz3 = 0 and l + m + n = 0 then complex numbers z1 z2 and z3 will be collinear

(iii) To find the conditions for perpendicularity of two straight lines Condition that angA of DABC where A(z1) B(z2) C(z3) is right angle and can be obtained by applying Rotation Theorem at A

minus π π= minus

minus 3 1

z zArg

z z 2 2 (i)

rArr π

plusmn minus minus= ρ = plusmnρ ρ = minus minus

i3 1 3 12

2 1 2 1

z z z ze iz z z z

rArr minus

= minus

z zR 0

rArr minus minus

+ =minus minus

3 1 3 1

2 1 2 1

z z z z0

z z z z rArr |z2 ndash z3|

2 = |z3 ndash z1|2 + |z2 ndash z1|

If ABC is right-angled isosceles triangle with AB = AC

rArr r = 1 rArr minus

= plusmnminus

(iv) Conditions for ∆ABC to be an equilateral triangle Let the DABC where A(z1) B(z2) C(z3) be an equilateral triangle

The following conditions hold

(i) |z1 ndash z2| = |z2 ndash z3| = |z3 ndash z1|

(ii) minus π= plusmn minus = minus

minus 3 1

3 1 2 12 1

z zArg and |z z | |z z |

(Applying the rotation theorem at A and knowing CA = BA)

(iii) minus minus π

= = minus minus

3 1 1 2

2 1 3 2

z z z zArg Argz z z z 3 (Applying rotation theorem at A and B)

(iv) + + = + +2 2 21 2 3 1 2 2 3 3 1z z z z z z z z z

(v) πminus

= = +minus

i1 2 3

z z 1 3e iz z 2 2

(vi) + + =minus minus +2 3 3 1 1 2

1 1 1 0z z z z z z

(vii) Conditions for four points to be concyclic or condition for points z1 z2 z3 z4 to represent a cyclic quadrilateralIf points A(z1) B(z2) C(z3) D(z4) are con-cyclic then the following two cases may occur

Case I If z3 and z4 lies on same segment with respect to the chord joining z1 and z2

minusminusminus =

minus minus 2 32 4

1 4 1 3

z zz zArg Arg 0z z z z

rArr minusminus

= minus minus

1 32 4

1 4 2 3

z zz zArg 0z z z z

rArr w is real and positive or Im(w) = 0 and Re(w) gt 0

Case II If z3 and z4 lie on opposite segment of circle with respect to chord joining z1 and z2

minus minus

+ = π minus minus

2 3 1 4

1 3 2 4

z z z zArg Argz z z z

rArr Arg (1w) = π rArr Arg (w) = ndashπ So the principal argument of w = πrArr w is negative real number or Im(w) = 0 and Re(w) lt 0

Conclusion Four complex numbers z1 z2 z3 z4 to be concyclic

minus minus

= π minus minus

1 3 2 4

2 3 1 4

(z z )(z z )Arg 0 or

(z z )(z z ) rArr w is purely real I(w) = 0 rArr =w w

1963 Loci in Argand Plane

A(1) Straight line in Argand plane A line through z0 making angle a with the positive real axisArg(z ndash z0) = α or ndash π + α

The given equation excludes the point z0 Arg (z ndash z0) = a represents the right-ward ray Arg (z ndash z0) = ndashp + a represents the left-ward ray

A(2) Line through points A(z1) and B(z2) Consider a straight line passing through A(z1) and B(z2) taking a variable point P(z) on it

∵ for each position of P

AP is collinear with

AB rArr = λ

AP AB rArr = λ minus

2 1AP (z z )

∵ = +

OP OA AP z = z1 + l(z2 ndash z1) z = z1(1 ndash l) + lz2

19631 Conclusion

1 if z = xz1 + yz2 x + y = 1 and x and yisinℝ then z z1 z2 are collinear

2 Equation represents line segment AB if l isin [0 1] 3 Right-ward ray through B if lisin (1 infin) 4 Left-ward ray through A if lisin (ndashinfin 0)

(i) Equation of straight line with the help of coordinate geometry

Writing + minus= =

z z z zx y2 2i

etc in minus minus

=minus minus

2 1 2 1

y y x xy y x x

and re-arranging the terms we find that the

equation of the line through z1 and z2 is given by minus minus

=minus minus

2 1 2 1

z z z zz z z z

or =1 1

z z 1z z 1 0z z 1

(ii)Equation of a straight line with the help of rotation formulaLet A(z1) and B(z2) be any two points lying on any line and we have to obtain the equation of this line For this purpose let us take any point C(z) lying on

this line Since Arg minus

= minus

z z 0z z

minus minus=

minus minus1 1

2 1 2 1

z z z zz z z z

hellip (i)

This is the equation of the line that passes through A(z1) and B(z2) After rearranging the terms

it can also be put in the following form =1 1

z z 1z z 1 0z z 1

(iii) Line segment AB The equation of the line segment AB is given as minus

= π minus

z zArgz z

(iv) Equation of two rays excluding the line segment AB minus

= π minus

z zArgz z

(v) Complete line except z1 and z2 (general equation of line)

The equation is given as 1 1

z z z zArg 0 ie I 0z z z z

minus minus= π =

minus minus

rArr minus minus

=minus minus

z z z zz z z z

rArr minus minus +2 1 1 2zz z z z z z z = minus minus +1 2 2 1zz z z z z z z

rArr minus + minus + minus =1 2 2 1 1 2 2 1(z z )z (z z )z z z z z 0 rArr minus minus

+ + =1 2 2 11 2

(z z ) (z z )z z I(z z ) 02i 2i

rArr + + =az az b 0 where rArr where minus= 2 1z za

2i and minus minus

= =minus2 1 1 2z z z za

RemarkTwo points P(z1) and Q(z2) lie on the same side or opposite side of the line + +az az b accordingly as

+ +1 1az az b and + +2 2az az b have the same sign or opposite sign

197 tHeorem

Perpendicular distance of P(c) (where c = c1 + ic2) from the straight line is

given by+ +

=|ac ac b |p

2 |a | Slope of a given line Let the given line be + + =za za b 0

Replacing z by x + iy we get ( ) ( )+ + minus + =x iy a x iy a b 0

rArr ( ) ( )+ + minus + =a a x iy a a b 0

Itrsquos slope is = ( ) ( )+

= = minusminus 2

a a 2Re(a) Re(a)i a a 2i lm a lm(a)

Equation of a line parallel to a given line Equation of a line parallel to the line + + =za za b 0 is + +λ =za za 0 (where l is a real number)

Equation of a line perpendicular to a given line Equation of a line perpendicular to the line + + =za za b 0 is minus + λ =za za i 0 (where l is a real number)

Equation of perpendicular bisectorConsider a line segment joining A(z1) and B(z2) Let the line lsquoLrsquo be itrsquos perpendicular bisectorIf P(z) be any point on the lsquoLrsquo then we havePA = PB rArr | z ndash z1 | = | z ndash z2 |rArr ( ) ( )minus + minus + minus =2 1 2 1 1 1 2 2z z z z z z z z z z 0

198 complex Slope of tHe lIne

If z1 and z2 are two unequal complex numbers represented by points P and Q then minusminus

z zz z

is called the

complex slope of the line joining z1 and z2 (ie line PQ) It is denoted by w Thus minus

=minus

z zwz z

1 The equation of line PQ is 1 1z z w( z z )minus = minus Clearly 1 2 1 2

1 2 1 2

z z z zw 1

z z z z

minus minus= = =

minus minus

2 The two lines having complex slopes w1 and w2 are parallel if and only if w1 = w2

3 Two lines with complex slopes w1 and w2 are perpendicular if w1 + w2 = 0

1981 Circle in Argand PlaneA(1) Centre radius form

The equation of circule with z0 as centre and a positive real number k as radius as given as |z ndash z0| = k

rArr |z ndash z0|2 = k2

rArr minus minus = 20 0(z z )(z z ) k rArr minus minus + minus =2 2

0 0 0zz z z z z |z | k 0 (1)

If z0 = 0 then |z| = K

A(2) General Equation of CircleReferring to equation (1) thus we can say

+ + + =zz az az b 0 (2)where a is a complex constant and bisinℝ represents a general circle

Comparing (2) with (1) we note that centre = ndasha and radius = minus2a b

A(3) Diametric Form of CircleAs we know that diameter of any circle subtends right angle at any point on the circumference Equation of circle with A(z1) and B(z2) as end points of diameter

π minus = πminus minus

Case Iz z 2Argz z Case II

rArr minus minus

= plusmn =minus minus

z z z zki where kz z z z rArr

minus minus= minus

minus minus2 2

z z z zz z z z

rArr minus minus + minus minus =1 2 2 1(z z )(z z ) (z z )(z z ) 0 further minus minus

+ =minus minus

z z z z 0z z z z

is diametric form

rArr |z ndash z1|2 + |z ndash z2|

2 = |z1 ndash z2|2

199 AppoloneouS cIrcle

If minus=

minus1

z z kz z

ie |z ndash z1| = k |z ndash z2| Then equation represents apploloneous

circle of A (z1) B(z2) with respect to ratio k when k = 1 this gives |z ndash z1| = |z ndash z2| which is straight line ie perpendicular bisector of line segment joining z1 to z2

1910 eQuAtIon of cIrculAr Arc

As per the figure equation of circular arc at which chord AB (where A(z1) and B(z2)) subtends angle a is

given as minus

= α minus

z zArgz z

Case I If 0 lt a lt p2 or ndash p2 lt a lt 0 (Major arc of circle)

Case II πα = plusmn

2 (Semicircular arc)

Case III π π αisin minusπ cup π

(Minor arc of circle)

Case IV a = 0 (Major arc of infin radius)

Case V a = p (Minor arc of infin radius)

19101 Equation of Tangent to a Given Circle

Let | z ndash z0 | = r be the given circle and we have to obtain the tangent at A(z1) Let us take any point P(z) on the tangent line at A(z1)

Clearly angPAB = p2 arg minus π

= plusmn minus

z zz z 2

rArr minusminus

z zz z

is purely imaginary

rArr ( ) ( )minus + minus + minus minus =20 1 0 1 1 1 0 1 0z z z z z z 2 |z | z z z z 0

In particular if given circle is | z | = r equation of the tangent at z = z1 would be + = =2 21 1 1zz zz 2 |z | 2r

If minus= λ

minus1

z zz z

(l isin R+ l ne 1) where z1 and z2 are given complex numbers and z is a arbitrary

complex number then z would lie on a circle

19102 ExplanationLet A(z1) and B(z2) be two given complex numbers and P(z) be any arbitrary complex number Let PA1 and PA2 be internal and external bisectors of angle angAPB respectively Clearly angA2PA1 = p2

Now minus minus= = = λ

minus minus1 1

| z z | z zAPBP |z z | z z

Thus points A1 and A2 would divide AB in the ratio of l 1 internally and externally respectively Hence P(z) would be lying on a circle with A1A2 being itrsquos diameter Note If we take lsquoCrsquo to be the mid-point of A2A1 it can be easily prove that CA CB = (CA1)

2 ie | z1 ndash z0 || z2 ndash z0 | = r2 where the point C is denoted by z0 and r is the radius of the circle

Notes (i) If we take lsquoCrsquo to be the mid-point of A2A1 it can be easily proved that CA CB = (CA1)

(ii) If l = 1 rArr | z ndash z1 | = | z ndash z2 | hence P(z) would lie on the right bisector of the line A(z1) and B(z2) Note that in this case z1 and z2 are the mirror images of each other with respect to the right bisector

19103 Equation of Parabola

Equation of parabola with directrix + + =az az b 0 and focus z0 is given as SP = PM

+ +minus =0

|az az b || z z |2 |a |

0 04aa(z z )(z z ) (az az b)

rArr minus minus + = + + 20 0 0 04aa(zz zz z z z z ) (az az b)

19104 Equation of EllipseEllipse is locus of point P(z) such that sum of its distances from two fixed points A(z1) and B(z2) (ie foci of ellipse) remains constant (2a)

rArr PA + PB = 2a rArr |z ndash z1| + |z ndash z2| = 2a where 2a is length of major axis

Case I If 2a gt |z1 ndash z2| = AB (Locus is ellipse)

Case II 2a = |z1 ndash z2| (Locus is segment AB)

Case III 2a lt |z1 ndash z2| (No locus)

1911 eQuAtIon of HyperbolA

Hyperbola is locus of point P(z) such that difference of its distances from two fixed point A(z1) and B(z2) (foci of hyperbola) remains constant (2a)

rArr PA ndash PB = 2a rArr ||z ndash z1| ndash |z ndash z2|| = 2a where 2a is length of major axis

Case I If 2a lt |z1 ndash z2| = AB (locus is branch of hyperbola)

Case II 2a = |z1 ndash z2| (Locus is union of two rays)

Case III 2a gt |z1 ndash z2| (No locus)

Case IV If ||z ndash z1| ndash |z ndash z2|| gt 2a 2a lt |z1 ndash z2| (Exterior of hyperbola)

1912 Some ImpotAnt fActS

A (1) If A B C are the vertices of a triangle represented by complex numbers z1 z2

z3 respectively in anti-clockwise sense and DBAC = a then αminus minus

=minus minus

i3 1 2 1

3 1 2 1

z z z z ez z z z

A(2) If z1 and z2 are two complex numbers representing the points A and B then

the point on AB which divides line segment AB in ratio m n is given by ++

1 2nz mzm n

A(3) If a b c are three real numbers not all simultaneously zero such that az1 + bz2 + cz3 = 0 and a + b + c = 0 then z1 z2 z3 will be collinear

A(4) If z1 z2 z3 represent the vertices ABC of DABC then

(i) Centroid of + +∆ = 1 2 3z z z

(ii) In centre of + +∆ =

+ +1 2 3az bz cz

ABCa b c

(iii) Orthocentre of + +

∆ =+ +

1 2 3(a secA)z (bsecB)z (csecC)zABC

(a secA) (bsecB) (csecC)= + +

+ +1 2 3(z tan A z tanB z tanC)

tan A tanB tanC

(iv) Circumcentre of + +

∆ =+ +

1 2 3z sin2A z sin2B z sin2CABC

sin2A sin2B sinC (v) If z1z2z3 are the vertices of an equilateral triangle then the circumcentre z0 may be given

as z21 + z2

2 + z23 = 3z0

2 (vi) If z1z2z3 are the vertices of an isosceles triangle right angled at z2 then z2

1 + z22 + z2

3 = 2z2( z1 + z3) (vii) If z1z2z3 are the vertices of right-angled isosceles triangle then (z1 ndash z2)

2 = 2 (z1 ndash z3)(z3 ndash z2)

(viii) Area of triangle formed by the points z1 z2 and z3 is 1 1

z z 11 z z 14i

19121 Dot and Cross ProductLet z1 = x1+ iy1 and z2 = x2 + iy2 be two complex numbers ie (vectors) The dot product (also called the

scalar product) of z1 and z2 is defined by z1 z2 = |z1| |z2| cosq = x1x2 + y1y2 = Re = +1 2 1 2 1 21z z z z z z 2

Where q is the angle between z1 and z2 which lies between 0 and p

If vectors z1 z2 are perpendicular then z1z2 = 0 rArr + =1 2

z z 0z z

ie Sum of complex slopes = 0

The cross product of z1 and z2 is defined by z1z2 = |z1| |z2| sinq = x1y2ndashy1x2 = = minus1 2 1 2 1 2Imz z z z z z 2i

If vectors z1 z2 are parallel then z1 z2 = 0 rArr =1 2

z zz z

ie complex slopes are equal

A(5) amp(z) = q represents a ray emanating from the origin and inclined at an angle q with the positive direction of x-axis

Also arg(z ndash z1) = q represents the ray originating from A(z1) inclined at an angle q with positive direction of x-axis as shown in the above diagram

A(6) |z ndash z1| = |z ndash z2| represents perpendicular bisector of line segment joining the points A(z1) and B(z2) as shown here

A(7) The equation of a line passing through the points A(z1) and B(z2) can be expressed in determinant

form as =1 1

z z 1z z 1 0z z 1

it is also the condition for three points z1 z2 z3 (when z is replaced by z3) to be

collinear

A(8) Reflection Points for a Straight LinesTwo given points P and Q are the reflection points of a given straight line if the given line is the right bisector of the segment PQ Note that the two points denoted by the complex number z1 and z2 will be the reflection points for the straight line α +α + =z z r 0 if and only if α +α + =1 2z z r 0 where r is real and a is non-zero constant

19122 Inverse Points wrt a CircleTwo points P and Q are said to be inverse wrt a circle with centre O and radius r if

(i) The point O P Q are collinear and P Q are on the same side of O (ii) OP OQ = r2

NoteThat the two points z1 and z2 will be the inverse point wrt the circle zz z z r 0α α+ + + = if and only if

1 2 1 2z z z z r 0α α+ + + =

19123 Ptolemys Theoremrsquos It states that the product of the length of the diagonal of a convex quadrilateral in scribed in a circle is equal to the sum of the products of lengths of the two pairs of its opposite sides ie |z1ndashz3||z2ndashz4| = |z1ndashz2| |z3ndashz4| + |z1ndashz4| |z2ndashz3|

A(10) a lt |z| lt b represents points lying inside the circular annulus bounded by circles having radii a and b and having their centres at origin as shown below

A(11) |z + z1| = |z| + |z1| represents the ray originating from origin and passing through the point A(z1) as shown below |z + z1| = PPprime = PO + OPprime = |z| + OA = |z| + |z1| (∵ OPprime = OA)

A(12) |z ndash z1| = |z| ndash |z1| represents a ray originating from A(z1) but not passing through the origin as shown below |z ndash z1| = OP ndash OA = |z| ndash |z1|

A(13) Re(z) ge a represents the half-plane to the right of straight line x = a including the line itself as shown below

Re(z) le a represents the half-plane to the left of straight line x = a including the line itself as shown here

Im(z) le a represents the half-plane below the straight line y = a including the line itself as shown below

Im(z) ge a represents the half-plane above the straight line y = a including the line itself as shown below

A(13) Inverse points wrt a circleTwo points A and B are said to be inverse wrt a circle with its centre lsquoOrsquo and radius a if

(i) The points O A B are collinear and on the same side of O and (ii) OAOB = a2

RemarkTwo points z1 and z2 will be the inverse points wrt the circle zz z z r 0β β+ + + = if and only

if 1 2 1 2z z z z r 0β β+ + + =

A(14) If l is a positive real constant and z satisfies minus= λ

minus1

z zz z

then the point z describes a circle of

which A B are inverse points unless l = 1 in which case z describes the perpendicular bisector of AB

A(15) To convert an equation in cartesian to complex form put +=

and minus=

z zy2i

and to convert

an equation complex form to Cartesian form put z = x + iy and = minusz x iy

Chapter 20SetS and

relationS

201 SetS

lsquolsquoA set is any collection of distinct and distinguishable objects of our intuition or thoughtrsquorsquo By the term lsquodistinctrsquo we mean that no object is repeated By the term lsquodistinguishablersquo we mean that given an object we can decide whether that object is in our collection or not

202 RePReSeNtAtION OF SetS

A set is represented by listing all its elements between braces and by separating them from each other by commas (if there are more than one element)

203 NOtAtION OF SetS

Sets are usually denoted by capital letters of the English alphabet while the elements are denoted in gen-eral by small letters eg set of vowels = A = a e i o u

204 NOtAtION FOR SOMe SPeCIAL SetS

W Whole Number ℤ Integer ℚ Rational Numbers ℝ Real Numbers

ℕ Nutural Numbers I Integer Number ℚc Irrational Number C Complex Numbers

If x is an element of a set A we write x isin A (read as lsquox belongs to Arsquo) If x is not an element of A we write x notin A (read as lsquox does not belong to Arsquo) The symbol lsquoisinrsquo is called the membership relation a isin A but d notin A

206 MetHOD RePReSeNtAtION OF SetS

(i) Tabular Form or Roster Form Under this method elements are enclosed in curly brackets after separating them by commas For example if A is a set of naturals number which is less than 5 then A = 1 2 3 4

Sets and Relation 20213

(ii) Set Builder Method Under this method set may be represented with the help of certain property or properties possessed by all the elements of that set

A = x | P(x) or A = x P(x) This signifies A is the set of element x such that x has the property P For example the set

A = 1 2 3 4 5 can be written as A = x | x isin N and x le 5

207 CARDINAL NuMbeR OF A SetS

Cardinal number of a set X is the number of distinct elements in a set and it is denoted by n(X) For example for X = x1 x2 x3 n (X) = 3

208 tyPeS OF SetS

Finite Set A set lsquoXrsquo is called lsquofinitersquo if it haslimited number of elements in it That is ifits all elements are labeled with the helpof natural numbers the processterminates at certain finite naturalnumber eg set of living people on earth

Null Set A set lsquoXrsquo iscalled nullvoidemptyif it has no element init It is denoted By φ or For example A = x x isin amp x2 + 2 = 0B = xx isin amp x2 lt0

Singletion Set A set Xis called singleton set if ithas only one element init For example A = xx isin and x2 + 4 = 0B = xx isin and x2 le0

Infinite Set A set lsquoXrsquo is calledinfnite if it has unlimited numberof elements in it For exampleset of rational numbers or set

of points in a plane

Classification of Set

Countably infiniteSet A set lsquoXrsquo is called countableif its elements can belabeled with the helpof natural numbersThat is its elementsare function ofnatural numbers Forexample a set of oddnatural numbers

Uncountable A set lsquoXrsquo is calleduncountable if itselements cannot belabeled with the helpof Natural numbersie Its elements cannot be written asfunction of naturalnumbers eg set ofreal numbers set ofirrational numbers

Some Important Remarks

Equivalent Sets Two finite sets A and B are equivalent if their cardinal numbers are same That is n (A) = n (B)

Equal Sets Two sets A and B are said to be equal if every element of A is a member of B and every element of B is a member of A That is A = B if A and B are equal and A ne B if they are not equal

Every finite set is countable but every countable set is not necessarily finite

Infinite sets may or may not be countable

Uncountable sets are always infinite

Every subset of a countable set is countable

Every superset of an uncountable set is also uncountable

Intersection of countable sets is always countable

Countable union of countable sets is always countable

209 SubSetS

A set A is said to be a subset of B if all the elements of A are present in B and is denoted by A sube B (read as A is subset of B) and symbolically written as x isin A rArr x isin B hArr A sube B

2010 NuMbeR OF SubSetS

Consider a set X containing n elements as x1 x2 xn then the total number of subsets of X = 2n

Proof Number of subsets of the above set is equal to the number of selections of elements taking any number of them at a time out of the total n elements and it is equal to 2n Q

nC0 + nC1 + nC2++ nCn = 2n

2011 tyPeS OF SubSetS

(i) Proper Subset A non-empty set A is said to be a proper subset of a set B if every element of A is an element

of B and B has at least one element which is not an element of A and is denoted by A sub B (ii) Improper Subset The set A itself and the empty set is known as improper subset For example if X = x1 x2 xn

then total number of proper sub-sets = 2n - 2 (excluding itself and the null set) The statement A sube B can be written as B supe A then B is called the super set of A

2012 POweR SetS

The collection of all subsets of set A is called the power set of A and is denoted by P(A) ie P(A) =

X X is a subset of A If A = x1 x2 x3 xn then n(P(A)) = 2n n(P(P(A)) = 22n

Thus X isin P(A) hArr X sube A ie the elements of P(A) are the subset of A

2013 DISjOINt SetS

Sets A and B are said to be disjoint iff A and B have no common element or A cap B = f If A cap B ne f then A and B are said to be intersecting or overlapping sets Eg

(i) If A = 1 2 3 B = 4 5 6 and C= 4 7 9 then A and B are disjoint set where B and C are intersecting sets

(ii) Set of even natural numbers and odd natural numbers are disjoint sets

2014 uNIVeRSAL SetS

It is a set which includes all the sets under considerations To explain this it is a super set of each of the given set Thus a set that contains all sets in a given context is called the universal set It is denoted by U For example if A = 1 2 3 B = 2 4 5 6 and C = 1 3 5 7 then U = 1 2 3 4 5 6 7 can be taken as the universal set

2015 COMPLeMeNt Set OF A gIVeN Set

Complement set of a set A is a set containing all those elements of universal set which are not in A It is denoted by cA A or Aprime So Ac = x x isin U but x notin A For example if set A = 1 2 3 4 5 and universal set

U = 1 2 3 4 50 then A = 6 7 50

2016 COMPLeMeNtRy Set OF A gIVeN SetS

Two sets A and B are said to be complementry sets if A B and B A= = To explain this if elements of A are removed from universal set U we get the elements of set B and if elements of A are removed from U we get elements of set B

Remarks (i) Two disjoint sets need not be complementry eg if U = 1 2 3 4 5 A = 1 3 B = 2 4 then A

and B are disjoint but Ac = 2 4 5 ne B and Bc = 1 3 5 ne A

(ii) Two complementary sets are always disjoint

2017 COMPARAbLe SetS

Two set A and B are said to be comparable if either A sub B or B sub A or A = B If neither (A sub B or B sub A) nor A = B then A and B are said to be incomparable

2018 VeNN (euLeR) DIAgRAMS

Here we represent the universal set U as the set of all points within rectangle and the subset A of the set U is represented by the interior of a circle If a set A is a subset of a set B then the circle representing A is drawn inside the circle representing B If A and B are not equal but they have some common elements then we represent A and B by two intersecting circles

2019 OPeRAtIONS ON SetS

20191 Union of Two SetsThe union of two sets A and B is the set of all those elements which are either in A or in B or in both This set is denoted by A cup B (read as lsquoArsquo union Brsquo)

Symbolically A cup B = x x isin A or x isin Bor A cup B = x x isin A (cup v denotes OR) x isin BClearly x isin A hArr x isin A or x isin B and x notin A cup B hArr x notin A and x notin BThe union of two sets can be represented by a Venn diagram as shown in the following figures

The shaded region represents A cup B

20192 Intersection of Two SetsThe intersection of two sets A and B is the set of all those elements which are common in A and B This set is denoted by A cap B (read as lsquoA intersection Brsquo)

Symbolically A cap B = x x isin A and x isin Bor A cap B = x x isin A cap x isin B [cap denotes lsquoandrsquo]Clearly x isin A cap B hArr x isin A and x isin BBut x notin A cap B hArr x notin A or x notin B ie x is not found in atleast one of A and BThe intersection of two sets can be represented by a Venn diagram as shown in above figure The

shaded region represents A cap B

20193 Difference of Two SetsThe difference of two sets A and B in this order (also called lsquorelative complementrsquo of B in A) is the set of all those elements of A which are not elements of B It is denoted by A ndash B and is read as lsquoA minus Brsquo

Symbolically A ndash B = x x isin A and x notin BThus x isin A ndash B hArr x isin A and x notin BSimilarly B ndash A = x x isin B and x notin A Thus x isin B ndash A hArr x isin B and x notin AA ndash B can be represented by Venn diagram as shown in the given figure The shaded region represents A ndash B

20194 Symmetric Difference of Two SetsSet of those elements which are obtained by taking the union of the difference of A and B ie (A - B) and the difference of B and A ie (B - A) is known as the symmetric differerence of two sets A and B and it is denoted by (A D B) Thus A D B = (A - B) cup (B - A) = x x isin (A cup B) ndash (A cap B)

Representation through the Venn diagram is given in the figure here

20195 Complement of a Set

The complement of a set A (also called lsquoabsolute complementrsquo of A) is the set of all those elements of the universal set S which are not elements of A It is denoted by Aprime or Ac

Clearly Aprime or Ac = S ndash ASymbolically Aprime or Ac = x x isin S and x notin AThus x isin Aprime hArr x notin AComplement of a set can be represented by a Venn diagram as shown in the figure here The shaded

region represents Aprime

2020 LAwS FOLLOweD by Set OPeRAtIONS cup cap AND D

(i) Idempotent Operation For any set A we have(a) A cup A = A and (b) A cap A = A

(ii) Existence of identity element wrt set operationFor any set A we have(a) A cup f = A and (b) A cap U = A(c) A ndash f = A (d) A D f = AThat is f and U are identity elements for (union difference symmetric difference) and intersection respectively

(iii) Commutativity For any set A and B we have(a) A cup B = B cup A (b) A cap B = B cap A(c) A D B = B D AThat is union and intersection and symmetric difference are commutative Note that A ndash B ne B ndash A

(iv) AssociativityIf A B and C are any three sets then(a) (A cup B) cup C = A cup (B cup C) (b) (A cap B) cap C = A cap (B cap C)(c) (A D B) D C = A D (B D C)ie union and intersection are associativeNote that (A ndash B) ndash C ne A ndash (B ndash C) eg for A = 2 3 4 5 6 7 8 B = 6 7 8 9 10 C = 4 5 6 7 10 12 (A ndash B) ndash C = 2 3 A ndash (B ndash C) = 2 3 4 5 6 7

(v) Divisibility If A B and C are any three sets then(a) A cup (B cap C) = (A cup B) cap (A cup C) (b) A cap (B cup C) = (A cap B) cup (A cap C)(c) A cup (B cup C) = (A cup B) cup (A cup C) (d) A cap (B cap C) = (A cap B) cap (A cap C)ie union and intersection are distributive over intersection and union and on themselves

(vi) Complement law(a) A cup Aprime = cup (Universal set) (b) A cap Aprime = f(c) (Aprime)prime = A (d) fprime = cup and cupprime = f

2021 De-MORgANrsquoS PRINCIPLe

If A and B are any two sets then

(i) (A cup B)prime = Aprime cap Bprime (ii) (A cap B)prime = Aprime cup Bprime

2022 INCLuSIVe-exCLuSIVe PRINCIPLe

(i) For set A and B n(A cup B) = n(A) + n(B) ndash n(A cap B) (ii) For sets A B and C n(A cup B cup C) = n(A) + n(B) + n(C) ndash n(A cap B) ndash n(B cap C) ndash n(C cap A) +

n(A cap B cap C)

2023 SOMe ReSuLtS ON CARDINAL NuMbeRS

(i) max n(A) + n(B) ndash n(S) 0 le n (A cap B) le min n(A) n(B) (ii) max n(A) n(B) le n (A cup B) le min n(A) + n(B) n(S) (iii) n(Ac) = n(U) ndash n(A)

20231 Cartesian Product of Two SetsCartesian product of two sets A and B is a set containing the ordered pairs (a b) such that a isin A and b isin B It is denoted by

A times B ie A times B = (a b) a isin A and b isin B If set A = a1 a2 a3 and B = b1 b2 thenA times B = (a1 b1) (a1 b2) (a2 b1) (a2 b2) (a3 b1) (a3 b2) andB times A = (b1 a1) (b1 a2) (b1 a3) (b2 a1) (b2 a2) (b2 a3)Clearly A times B ne B times A until A and B are equal

Remarks

1 Since A times B has elements as ordered pairs therefore it can be geometrically located on X ndash Y plane by considering set A on X-axis and set B on Y-axis

2 Cartesian product of n sets A1 A2 A3An is denoted by A1 x A2 x A3 x x An and is the set of n ordered tuples ie A1 x A2 x A3 x x An = (a1 a2 a3 an) ai isin Ai i = 1 2 3 n Cartesian product of n sets represents n-dimensional space

3 A times B times C and (A times B) times C are not same

A times B times C = (a b c) a isin A b isin B c isin C whereas

(A times B) times C = (a b c) a isin A b isin B c isin C

20232 Number of Elements in Cartesian Product A times B

If number of elements in A denoted by n(A) = m and number of elements in B denoted by n(B) = n then number of elements in (A times B) = m times n ie n(A times B) = n(A) times n(B)

Since A times B contains all such ordered pairs of the type (a b) such that a isin A and b isin B that means it includes all possibilities in which the elements of set A can be related with the elements of set B Therefore A times B contains n(A) times n(B) number of elements

20233 Properties and Laws of Cartesian Product

202331 Distributive laws

1 (a) Cartesian product distributes over union and intersection of sets That is A times (B cup C) = (A times B) cup (A times C) and A times (B cap C) = (A times B) cap (A times C) for every group

of sets A B and C

(b) Cartesian product distributes over subtraction of sets That is A times (B ndash C) = (A times B) ndash (A times C) 2 Cartesian Product is not Associative Cartesian product of sets is not associative in nature

That is A times (B x C) ne (A times B) times C As the elements of A times (B times C) are of the type (a (b c)) whereas the elements of (A times B) times C are of

the type ((a b) c) a isin A b isin B c isin C 3 Cartesian Product is not Commutative Cartesian product of sets is not commutative in nature That is A times B ne B times A until A = B 4 Cardinality of Cartesian Product (a) If A and B are two sets then n(A times B) = n(A) times n(B) (b) If A and B are sets having k number of common elements ie n(A cap B) = k then the number

of elements common to A times B and B times A = k2 5 Intersection of cross product is equal to cross product of intersection That is for sets A B S and T (A times B) cap (S times T) = (A cap S) times (B cap T) 6 For subset A of B and C of D We have (a) (A times C) sube (B cap C) for every set C (b) (A times C) sube (B cap D) (c) (A times A) sube (A times B) cap (B times A) 7 For complementary sets B and C of sets B and C (a) A times (B cup C) = (A times B) cap (A times C) (b) A times (B cap C) = (A times B) cup (A times C) 8 A times (B D C) = (A times B) D (A times C)

2024 ReLAtIONS

A relation R from set X to Y (R X rarr Y) is a correspondence between set X to set Y by which none one or more elements of X are associated with none one or more elements of Y Therefore a relation (or binary relation) R from a non-empty set X to another non-empty set Y is a subset of X times Y That is R X rarr Y is nothing but subset of A times B For example consider set X and Y as set of all males and females members of a royal family of the Ayodhya kingdom

X = Dashrath Ram Bharat Laxman Shatrughan and Y = Koshaliya Kaikai Sumitra Sita Mandavi Urmila Shrutkirti and a relation R is defined as was husband of from set X to set Y

Then R = (Dashrath Koshaliya) (Ram Sita) (Bharat Mandavi) (Laxman Urmila) (Shatrughan Shrutkirti) (Dashrath Kaikai ) (Dashrath Sumitra)

2025 DOMAIN CO-DOMAIN AND RANge OF ReLAtION

Domain Domain of a relation R from set A to set B is the collection of elements of set A which are participating in the correspondence ie it is set of all pre-images under the relation R For example domain of R = (1 5) (2 10) (3 6) is

DR = 1 2 3 where R is a relation from set A = 1 2 3 4 to set B = 5 6 7 8 9 10

Co-domain Co-domain of a relation R from set A to set B is set B itself irrespective of the fact whether an element of set B is related with any element of A or not For example B =5 6 7 8 9 10 is co-domain of above relation R

Range Range of a relation R from set A to set B is the set of those elements of set B which are participating in the correspondence ie set of all images under the relation R For the above relation range is given by the set RR = 5610

2026 uNIVeRSAL ReLAtION FROM Set A tO Set b

Since A times B contains all possible ordered pairs which relate each element of A to every element of B therefore (A times B) is largest possible relation defined from set A to set B and hence also known as Universal relation from A to B

2027 NuMbeR OF ReLAtIONS FROM Set A tO Set b

Since each relation from A to B is a subset of Cartesian product A times B therefore number of relations that can be defined from set A to set B is equal to the number of subsets of A times B Thus the number of relations from A to B = 2n(A times B) = 2n(A) x n(B)

2028 ReLAtION ON A Set

A relation R from set A to itself is called relation on set AFor example let A = 1234916 Define a relation from set A to itself as a R b if b is square of a

but a ne b thenR = (2 4) (3 9)(4 16) Here domain = 2 3 4 co-domain = A range = 4 9 16

2029 RePReSeNtAtION OF ReLAtION IN DIFFeReNt FORMS

(i) By representing the relation as a set of ordered pairs (Roster form)In this method we represent the relation by a set containing ordered pairs (a b) where a isin A and b isin B such that aRb as shown for the relation R from A = 1 2 3 4 to set B = 2 3 4 5 6 7 when b isin B is to be related to a isin A here such that b = 2a + 1 R = (13) (2 5) (3 7)

(ii) Analytical method or set builder from In this method we represent the relation as R = (a b) a isin A b isin B ahellipb where the dots are replaced by an equation connecting image b with its pre-image a For example let R be a relation from set A = 1 2 3 4 to set B = 2 3 4 5 6 7 given by R = (13)(25)(37) then it can be represented by R = (x y) x isin A y isin B x R y iff y = 2x + 1

(iii) Graphical representation or representation by lattice In this method we take set X along x-axis and set B along y-axis then plot the points (a b) isin R in x y plane For example in the above illustration the relation can be represented as shown in the diagram given below

(iv) By arrow diagram In this method we represent set A and set B by two circles or by two ellipses and join the images and their pre-images by using arrows as shown below for above illustration

(v) Tabular form In this form of representation of a relation R from set A to set B elements of A and B are written in the first column and first row respectivelyIf (a b) isin ℝ then we write lsquo1rsquo in the row containing a and the column containing b and if (a b) notin ℝ then we write lsquo0rsquo in the row containing a and the column containing bFor example for the relation R = (1 3) (2 5) (3 7) from set A = 1 2 3 4 to set B = 2 3 4 5 6 7 we have the following tabular representation

R 2 3 4 5 6 7

1 0 1 0 0 0 02 0 0 0 1 0 03 0 0 0 0 0 14 0 0 0 0 0 0

2030 CLASSIFICAtION OF ReLAtIONS

One-one or Injective Relation

If different elements of set X are related with different elements of set Y ie no two different elements of domain are related to same element of set Y then R is said to be one-one relation or injective relation from set X to set Y

Many-one Relation

When there exists at least one group having more than one element of set X which are related with same element of set Y then R is said to be many one relation from set X to set Y

One-many Relation

Relation R from set X to set Y is said to be one-many if there exists an element in set X which is related with more than one element of set Y

Many-many Relation

Relation R from set X to set Y is said to be many-many if it is many-one as well as one-many

Onto Relation (Surjective Relation)

A relation R X rarr Y is said to be onto or surjective relation if there is no such element y isin Y which is not related with any x isin X ie for each y isin Y there exist at least one element x in X which is related with y In such a relation

Range (RR) = co-domain ie range of onto relation is nothing but the co-domain of the relation

RemarkIn onto relation all elements of set X may or may not participate in relation but all elements of co-domain set Y participate in relation

2031 INtO ReLAtION

A relation R X rarr Y is said to be into iff there exist at least one y isin Y which is not related with any x isin X

That is if range (RR) sub co-domain that is range of relation is a proper subset of co-domain

That is R6 (x1 y1) (x1 y2) (x2 y3)Clearly under relation R6 y4 has no pre-image in X

20311 One-One-Onto Relation (Bijective Relation)

A relation R X rarr Y is said to be bijective relation iff it is both onendashone as well as onto relation

For example R7 (x1 y2) (x2 y1) (x3 y3) where X = x1 x2 x3 x4 and Y = y1 y2 y3)

2032 tyPeS OF ReLAtIONS

20321 Reflexive Relation

R X rarr Y is said to be reflexive iff x R x x isin X That is every element of X must be related to itselfTherefore if for each x isin X (x x) isin R then relation R is called reflexive relation

RemarkIf R X rarr Y is a reflexive relation then its domain is X For example if R is a relation on set of integers (ℤ) defined by ldquoxRy iff x divides yrdquo then it is reflexive and hence its domain set is ℤ

20322 Identity RelationA relation R X rarr Y is said to be an identity relation if each element of X is related to itself only For example if X = x1 x2 x3 and Y = x1 x2 x3 x4 then the relation R = (x1 x1) (x2 x2) (x3 x3) is an identity relation from set X to set Y

Remarks 1 Every identity relation from set X to set Y is reflexive relation from set X to set Y but converse is

not true That is every reflexive relation need not be identity For example R X rarr Y where X = x1 x2 x3 and Y = x1 x2 x3 x4 then the relation R = (x1 x1) (x2 x2) (x3 x3) (x1 x2) is reflexive but not identity relation from set X to set Y because x1 R x1 as well as x1Rx2

2 If R is a relation from set X to itself then the relation is called relation on set X

(a) R is said to be reflexive on set X if xRx x isin X

(b) R is said to be identity relation on set X if x R x x isin X and x is not related to any other element and it is denoted by Ix

3 Symmetric Relation R X rarr Y is said to be symmetric iff (x y) isin R rArr (y x) isin R

That is x R y rArr y R x For example perpendicularity of lines in a plane is symmetric relation

20323 Transitive Relation

R X rarr Y is said to be transitive iff (x y) isin R and (y z) isin R rArr (x z) isin RThat is x R y and yR zrArr x R z For example the relation ldquobeing sister ofrdquo among the members of a family is always transitive

Notes (i) Every null relation is a transitive relation

(ii) Every singleton relation is a transitive relation

(iii) Universal and identity relations are reflexive as well as transitive

20324 Anti-symmetric RelationA relation R from set X to set Y is said to be an anti-symmetric relation iff (a b) isin R and (b a) isin R rArr a = b

That is for two different elements x isin X and y isin Y the relation R does not contain the ordered pairs (x y) and (y x) simultaneously

For example relations ldquobeing subset of rdquo ldquois greater than or equal tordquo and ldquoidentity relationrdquo are anti-symmetric relations

RemarkA relation R from set X to set Y may be both symmetric as well as anti-symmetric any one or not bothFor example let X = 1 2 3 4 and Y = 1 2 3 4 5 6

203241 Consider the relations

(i) R1 = (11) (22) (ii) R2 = (1 2) (2 1) (2 3) (3 2) (iii) R3 = (1 1) (2 2) (3 4) (iv) R4 = (1 2) (2 1) (3 4)

1 R1 is symmetric as whenever ordered pair (x y) isin R1 rArr (y x) isin R1Also R1 is anti-symmetric as for no two different elements x y the ordered pairs (x y) and (y x) occur in R1

2 R2 is symmetric but not anti-symmetric as (1 2) (2 3) isin R2 rArr (2 1) (3 2) isin R2 but 1 ne 2 and 2 ne 3 3 R3 is anti-symmetric but not symmetric as (3 4) isin R3 but (4 3) notin R3 4 R4 is neither symmetric nor anti-symmetric as (3 4) isin R3 but (4 3) notin R3 and (1 2) (2 1) both are

in R3 but 1 ne 2

20325 Equivalence RelationA relation R from a set X to set Y (R X rarr Y) is said to be an equivalence relation iff it is reflexive symmetric as well as transitive The equivalence relation is denoted by ~ For example relation ldquois equal tordquo Equality Similarity and congruency of triangles parallelism of lines are equivalence relations

2033 COMPOSItION OF ReLAtIONS

Let R and S be two relations from set A to B and B to C respectively Then we can define a relation SoR from A to C such that (a c) isin SoR hArr exist b isin B such that (a b) isin R and (b c) isin S

This relation is called the composition of R and S Diagrammatically it is shown in the following figure

2034 INVeRSe OF A ReLAtION

Let A B be two sets and let R be a relation from a set A to B Then the inverse of R denoted by R-1 is a relation from B to A and is defined by R-1 = (b a) (a b)isinR Clearly (a b) isin R hArr (b a) isin R-1

Also Dom (R) = Range (R-1) and range (R) = Dom (R-1)For example let A = 1 2 3 4 and B = 2 3 4 5Define a relation R from A to B as xRy iff y = x + 1 then R = (1 2)

(2 3) (3 4) (4 5)rArr Rndash1 = (2 1) (3 2) (4 3) (5 4)Thus we can define Rndash1 a relation from B to A as xRy iff y = x ndash 1

The arrow diagram represents the relations R and Rndash1

Remark(SoR)ndash1 = Rndash1oSndash1 where R is a relation from A to B and S is a relation from B to C

Tips and Tricks

If number of elements in A n(A) = m and n(B) = n then number of elements in (A times B) = m times n A times B is termed as the largest possible relation defined from set A to set B it is also known as the

universal relation from A to B If A sube B then (A times B) cap (B times A) = A2 = A times A If A has m elements and B has n elements then number of relations that can be defined

from A to B = 2m times n If A is a set containing n elements then the number of relations that can be defined

on set ( )2nA 2=

If A and B are two non-empty sets having n elements in common then A times B and B times A have n2 elements in common

If A is related to B then symbolically it is written as (aRb) where a is pre-image and b is image If A is not related to B then symbolically it is written as a R b All identity relations are reflexive but all reflexive relations are not identity Every null relation is a transitive relation Every singleton relation is a transitive relation Universal and identity relations are reflexive as well as transitive Identity relation is symmetric as well as anti-symmetric or both Union of two reflexive (or symmetric) relations on a set A also reflexive (or symmetric) on set A Union of two transitive relations need not be transitive on set A Union of two equivalence relations need not be equivalence 1 If R and S are two equivalence relations on a set A then R cap S is also an equivalence

relation on A 2 The inverse of an equivalence relation is an equivalence relation 3 The set (a a)a isin A = D is called the diagonal line of A times A Then lsquothe relation R in a is

antisymmetric iff R cap Rndash1 sube D

Chapter 21FunCtions

211 Definition of function

Let X and Y be two non-empty sets Then a function lsquof rsquo from set X to set Y is denoted as f X rarr Y or y = f(x) x isin X and y isin Y A function f(x) from X (domain) to Y (co-domain) is defined as a relation f from set X to set Y such that each and every element of X is related with exactly one element of set Y

Image and Pre-image Let f be a function from set X to set Y ie f X rarr Y and let an element x of set X be associated to the element y of set Y through the rule lsquof rsquo then (x y) isin f ie f(x) = y then y is called lsquoimage of x under f rsquo and x is called lsquopre-image of y under f rsquo

Natural Domain The natural domain of a function is the largest set of real number inputs that give real number outputs of the function

Co-domain Set Y is called co-domain of function f

Range of Function If f Df (sube X) rarr Y is a function with domain Df then the set of images y (output isinY) generated corresponding to input x isin Df is called range of function and it is denoted by Rf

ie Rf = f(x) xisinDf sube Y

Remarks

(i) Every function is a relation but every relation read not be a function

(ii) A relation R A rarr B is a function if its domain = A and it is not one-many ie either one-one or many-many

Functions 21227

(iii) To find domain of function we need to know when does a function become undefined and when it is defined

ie we need to find those values of x where f(x) is finite and real and those values of x where f(x) is either infinite or imaginary

(iv) When its value tends to infinity (infin)

eg =minus2

x 1 at x = plusmn1 f(x) is not defined at x = plusmn1 and defined forall x isin ℝ except for plusmn1 therefore

domain of f(x) = ℝ ~ 1 ndash1

(v) When it takes imaginary value eg = minusy x 1 at x isin (ndashinfin 1) f(x) is not defined on (ndashinfin 1) and

defined on [1 infin) therefore domain of f(x) = [1 infin)

(vi) When it takes indeterminate form ie becomes of the form infininfininfin infin minus infin

infin0 00

1 0 etc0

212 RepResentation of a function

A fanction can be represented analytcally as orduced pass parametrically wita arrow diagram praphibly

Remarks All function cannot be represented by all the above methods

(i) The Drichlet-Function which is defined as f(x) =

0 when x is rational

1 when x is irrational cannot be graphed since there exist

infinite number of rationals as well as irrationals between any two real numbers

(ii) Consider the Eulerrsquos totient or Eulerrsquos phi function f(n) = Number of positive integers less than or equal to n and co-prime to n where n is a natural number

The domain of f is the set of positive integers Its range is the set of positive integers 1 2 3 hellip

We cannot represent this function analytically A portion of the graph of f(n) as shown here for understanding of the function

(iii) Consider another function called prime number function defined by f(x) = number of prime numbers less than or equals to x where x is non-negative real number

Then domain of f(x) is (0 infin) and range is the set of non-negative integers ie 0 1 2 3 hellip

The graph of function is shown here

As x increases the function f(x) remains constant until x reaches a prime at which the graph of function jumps by 1Therefore the graph of f consists of horizontal line segments This is an example of a class of function called step functions

(iv) Another function which is so complicated that it is impossible to draw its graph

h(x) = minus

x if x is rational

x is irrationalif x2

As we know that between any two real numbers there lie infinitely many relations and irrational number so it is impossible to draw its graph

isin ℝ

b2 ndash

d isin

Functions 21229

x3 + cx

k) infin

r a gt

(ndashinfin

a 0xn + a

1xnndash1 +

a2xnndash

2 + hellip

nndash1 x

a n ai isin

0 ne 0

n isin

k) infin

a 0 gt

(ndashinfin

r a 0 gt

= minus

infin)

minuslt

Rndash

t of a

Functions 21231

isinisin

f(x) =

ries o

(0 infin

f(x) =

log ax

(0 infin

ℝ ndash

Functions 21233

= xk k

isin ℝ

n isin

[0 infin

ndash(2

isin ℕ

ℝ ndash

ndash(2

isin ℕ

ℝ ndash

(0 infin

infin)

[0 infin

Functions 21235

1)=minus

ℝ ndash

=minus

(0 infin

infin)

isinminus

infin)

1minus

ltisin

1minus

2mminus

=minus

(0 infin

infin)

1)minus

=minus

isinminus

ℝ ndash

Functions 21237

=minus

isinminus

R ndash

(0 infin

infin)

(0 infin

infin)

) = si

s xf(x

ndash co

sec2 x

ndash times

divide)

xminus

infin)

minus=

ℝ ndash

s x in

f y fo

3)minus

=minus

minus is

+minus

minus=

=minus

Functions 21239

214 equal oR iDentical functions

Two functions f and g are said to be equal if

1 The domain of f = the domain of g 2 The range of f = the range of g 3 f(x) = g(x) for every x belonging to their common domain eg f (x) =1x and g(x) = xx2 are identical

functions f(x) =log(x2) and g(x) =2log(x) are not-identical functions as domain of f(x) = (ndashinfin infin) ~ 0 whereas

that of g(x) = (0 infin)

RemarkGraphs of trigonometric function and inverse trigonometric functions with their domain and range are givenin the same book under corresponding topics

215 pRopeRties of GReatest inteGeR function (BRacket function)

(i) Domain of [x] ℝ Range of [x] ℤ (ii) [[x]] = [x] (iii) [x + m] = [x] + m provided m isin ℤ (iv) [x + [y + [z]]] = [x] + [y] +[z] (v) [x] gt n n isin ℤ rArr [x] isinn + 1 n + 2 n + 3 rArr x isin [n + 1 infin) (vi) [x] ge n rArr x isin [n infin) (vii) [x] lt n rArr x isin (ndashinfin n) (viii) [x] le n rArr n isin (ndashinfin n + 1)

(ix) [x] x if x

[ x]1 [x] if x

minus = minus isinminus = minus minus notin

(x) x ndash 1 lt [x] le x equality holds iff x isin ℤ (xi) [x] le x lt [x] + 1

(xii) xc

for c isin ℕ and x isin ℝ

(xiii) [x] + [y] le [x + y] le [x] + [y] + 1

(xiv) [x] = x x 12 2

+ + forall x isin ℝ

(xv) The number of positive integers less than or equal to n and divisible by m is given by nm

and n are positive integers

(xvi) If p is a prime number and e is the largest exponent of p such that pe divides n then k

infin =

2151 Properties of Least Integer Function

1 The domain of the function is (-infin + infin) 2 The range is the set of all integers 3 [x] converts x = (I + f) into I while x converts it into I + 1 Eg If x = 24 then 2lt x lt3 rArr x = 3 = I + 1 4 When x is an integer [x] = x = x

5 x + n = x + n where n is an integer

2152 Properties of Fractional Part Function

(i) Domain of fractional part function = Df = ℝ Range of fractional part function = Rf = [0 1) (ii) x is periodic function with period 1 (iii) [x] = 0 (iv) [x] = 0 (v) x = x this result is true when fractional part function is applied on x on left hand side more

than or equal to twice

(vi) 0 x

x1 x x

isinminus = minus notin

(vii) [x] [y] 0 x y 1

[x y][x] [y] 1 1 x y 2

+ le + lt+ = + + le + lt

2153 Properties of Nearest Integer Function

(i) (x) =

1[x] if 0 x2

1[x] 1 if x 12

le lt + le lt

(ii) (x + n) = (x) + n if n isin ℤ

2n 1(x) x ~ x n2( x)

2n 1(x) 1 for x n2

+ minus forall isin = isin minus = + minus + = isin

1[x] n if n x n2(x)

1[x] 1 n 1 if n x n 12

= le lt += + = + + le lt +

Properties of Modulus of a real number 1 |x1 x2 x3 xn| = | x1 | | x2 | | x3 | | xn| forall xi isin ℝ

2 x | x |y | y |= forall x y isin ℝ and y ne 0

3 | xn | = | x |n forall n isin ℤ 4 | ndashx | = | x | forall x isin ℝ 5 | x | = d rArr x = d or x = -d

6 | x | lt d rArr x isin (- d d) and | x | gt d 7 | x - a | lt d rArr x isin (a ndash d a + d)

Functions 21241

8 | x ndash a | = d rArr x = a + d or a ndash d 9 | x ndash a | gt d rArr x gt a + d or x lt a ndash d

10 2x | x |= forall x isin ℝ 11 |x| = maxndashx x forall x isin ℝ 12 |x| = |y| hArr x2 = y2

13 |x + y| is not always equal to | x | + | y | 14 (Triangle inequality) | x + y | le | x | + | y | for all real x and y inequality holds if xy lt 0 ie x and y are

of opposite signs equality holds if xy ge 0 ie x and y are of same sign or at least one of x and y is zero 15 |x ndash y| le |x| + |y| for real x and y inequality holds if xy gt 0 ie x and y are of same sign equality holds

if xy le 0 ie x and y are of opposite sign or at least one of x and y is zero 16 ||x| ndash |y|| le |x + y| for real x and y Equality holds if x and y are of opposite signs and for same sign

inequality holds 17 ||x| ndash |y|| le |x ndash y| for real x and y Equality holds if x and y are of same sign and for opposite signs

inequality holds

21531 Methods of testing a relation to be a function

Method 1 When the relation to be tested is represented analytically A relation f X rarr Y defined as y = f(x) will be function iff x1 = x2 rArr f(x1) = f(x2) since otherwise an element of X would have two different image

Method 2 When the relation to be tested is represented as a set of ordered pairs

A relation f X rarr Y represented as a set of ordered pairs will be function from X to Y iffSet of abscissa of all ordered pairs is equal to XNo two ordered pairs should have same abscissa

RemarkBecause f is a relation from X rarr Y therefore abscissa of ordered pairs must belong to X where as ordinates of ordered pairs must belong to Y

Method 3 When the relation to be tested is represented graphically relation f X rarr Y y = f(x) is function iff all the straight line x = a forall a isin X intersect the graph of function exactly once as shown below

A relation f X rarr Y will not be a function in following two conditions 1 If for some a isin X line x = a does not cut the curve y = f(x) eg in the graph of function shown below

the line x = a does not cut the graph of function and a isin X (Df) = [a b] ie no output for input x = a

rArr f(x) is not a function from X to Y 2 If for atleast one a isin X line x = a intersects y = f(x) more than once ie there exists an input having

more than one output say at (a y1) (a y2) and (a y3) rArr For input x = a f(x) has three outputs y1 y2 as well as y3 Hence f(x) is not function

Method 4 When the relation to be tested is represented diagrammatically A relation f X rarr Y is a function if no input has two or more outputs in Y and no x isin X is un-related

216 classification of functions

2161 One-one (Injective) Function

f X rarr Y is called injective when different elements in set X are related with different elements of set Y ie no two elements of domain have same image in co-domain In other words we can also say that no element of co-domain is related with two or more elements of domain

217 Many-one functions

f X rarr Y is many-one when there exist at least two elements in the domain set X which are related with same element of co-domain Y

2171 Onto (Surjective) FunctionA function f X rarr Y is called surjective only when each element in the co-domain is f-image of at least one element in the domain ie f X rarr Y is onto iff y isin Y there exists x isin X such that f(x) = y ie iff Rf = co-domain (Y)

Surjective f X rarr Y reduces the co-domain set to range of function

218 MethoD of testinG foR injectivity

(a) Analytical Method A function f X rarr Y is injective (one-one) iff whenever two images are equal then it means that they are outputs of same pre-image ie f(x1) = f(x2) hArr x1 = x2 forall x1 x2 isin X Or by using contra-positive of the above condition ie x1 ne x2 hArr f(x1) ne f(x2) forall x1 x2 isin X

1 If f (x) is not one-one then it is many-one function If we go according to definition consider f(x1) = f(x2) rArr x1 is not necessarily equal to x2

ie If two f-images are equal then their pre-images may or may not be equal

2 To test injectivity of f(x) consider f(x1) = f(x2) and solve the equation and get x2 in terms of x1 If x2 = x1 is only solution then function f is injective but if other real solutions also exist then f is many-one function

Functions 21243

(b) Graphical Method For one-one every line parallel to x-axis y = k isin Rf cuts the graph of function exactly once then the function is one-one or injective

For many-one If there exists a line parallel to x-axis which cuts the graph of function at least twice then the function is many one

(c) Method of Monotonicity for one-one If a function f(x) is continuous and monotonic

(f (x) ge 0 f (x) = 0 occures at isolated points) on an interval I then it is always one-one on interval I because any straight line parallel to x-axis y = k isin I intersects the graph of such functions exactly once

For many-one

(i) If a function is continuous and non-monotonic on interval I then it must be many-one on interval I

(ii) If a function is discontinuous and monotonic on interval I then it can be one-one or many-one on I as is clear from the figures given below

(iii) Even functions and periodic functions are always many-one in their natural domains whereas they are one-one in their principal domain They can be made one-one by restricting the domain

eg cosx is many one on ℝ but is one-one on [0 p] or 02π

Similarly fraction part function

x is periodic function with period 1 It is many one on ℝ but one-one on [n n + 1) for each integer n

(iv) If a function is discontinuous and non-monotonic on an interval I then it can be one-one or many one on I It can be understood well by the graph shown as follows

(v) All polynomials of even degree defined in ℝ have at least one local maxima or minima and hence

are many-one in the domain ℝ Polynomials of odd degree can be one-one or many-one in ℝ (d) Hit and trial method to test many-one functions It is possible to find an element in the range of function which is f image of two or more than two

elements in the domain of function

219 into (non-suRjective) function

While defining function we have mentioned that there may exist some element in the co-domain which are not related to any element in the co-domain

f X rarr Y is into iff there exists at least one y isin Y which is not related with any x isin X

Thus the range of the into function is proper subset of the co-domain ie range sub co-domain (properly)

2110 one-one onto function (Bijective function)

If a function is both one-one as well as onto then f(x) is set to be bijective function or simply bijection

2111 testinG of a function foR suRjective

Method 1 The equality of range of function to co-domain forms the condition to test surjectivity of function For instance to test surjectivity of f [0 infin) rarr [2 infin) such that f(x) = x2 + 2

Using the analytic formula we obtain the rule of function for argument x in terms of y as shown below

∵ y = x2 + 2 x2 = y ndash 2 ie |x| = y 2minus

rArr x y 2= minus ∵ x ge 0

Functions 21245

Now we check whether the expression of x in forms of y is valid for all elementary co-domain If it is so then f is surjective otherwise it is non-surjective

Thus x to be real and positive RHS ie y 2minus must be real and positive thus y isin [2 infin) Hence the given function f is onto

Method 2 Hit and Trial Method Sometimes we choose an element of co-domain which may not be an image of any element in domain and we test it for same If it comes out to be true then f is into function

RemarkIn order to convert a function from many-one to an injective function its domain must be transformed to principal domain In order to convert a function from into to onto the co-domain of function must be replaced by its range

2112 nuMBeR of Relations anD functions

Number of Relations No of relations = Number of subsets of A times B = 2n(AtimesB) = 2nm

Number of Functions Since each element of set A can be mapped in m waysrArr Number of ways of mapping all n elements of A

n times

m m m m ways m ways= times times times times =

Conclusion 2nm ge mn forall m n isin ℕ

211211 Number of one-one function (injective)

rArr Number of injective functions

= m(m ndash 1) (m ndash 2) hellip (m ndash n + 1) = m

nP m n0 m n

Conclusion m nnP mle total number of functions

211212 Number of non-surjective functions (into functions)

Number of into function (N) = Number of ways of distributing n different objects into m distinct boxes

so that at least one box is empty N = minus

minus minussumm

m r 1 nr

C ( 1) (m r)

211213 Number of surjective functions

Number of surjective functions = Total number of functions ndash Number of into functionsm

n m r 1 nr

m C ( 1) (m r)minus

= minus minus minussumm

m r nr

C ( 1) (m r)=

= minus minussum

Conclusion In case when n(A) = n(B) the onto function will be bijectionNumber of onto function = Number of one-one function

rArr n

n r nr

C ( 1) (n r) n=

minus minus =sum

Remarks 1 If n(X) lt n(Y) then after mapping different elements of X to different elements of Y we are left with

at least one element of Y which is not related with any element of X and hence there will be no onto function from X to Y ie all the functions from X to Y will be into

2 If f from X to Y is a bijective function then n(X) = n(Y)

21121 Composite of Uniformly Defined FunctionsGiven two functions f X rarr Y and g Y rarr Z then there exists a function h = gof X rarr Z such that h(x) = (gof) (x) = g(f(x)) forall x isin X It is also called as lsquofunction of functionrsquo or lsquocomposite function of g and f rsquo or lsquog composed with f lsquoand diagrammatically shown as

2113 coMposition of non-unifoRMly DefineD functions

2x 1 0 x 2f(x)

x 1 2 x 4minus le lt= + le le

and x 1 1 x 1

g(x)2x 1 x 3+ minus le lt= le le

then 2

2x 1 1 x 1fog(x)

4x 1 1 x 2+ minus le lt= + le le

2114 pRopeRties of coMposition of function

(a) fog(x) is not necessarily equal to gof(x) ie generally not commutative (b) The composition of functions is associative in nature ie fo(goh) = (fog) oh (c) The composite of two bijections is a bijection (d) If gof is one-one then f is one-one and g need not be one-one (e) If gof is onto then g is onto but f need not be onto (f) If f(x) and g(x) are both continuous functions then g(f(x)) is also continuous (g) Monotonicity of composite function Composition of two functions having same monotonicity is a

monotonically increasing function (h) Composition of two functions having opposite monotonicity is a decreasing function

21141 Definition of Inverse of a FunctionA function f X rarr Y is said to be invertible iff there exists another functiong Y rarr X such that f(x) = y hArr g(y) = x forall x isin X and y isin YThen g Y rarr X is called inverse of f X rarr Y and is denoted by fndash1rArr g = fndash1 = (f(x) x) (x f(x)) isin f

Functions 21247

2115 conDition foR invisiBility of a function

For a function to be invertible it should be one-one and onto ie bijective function

21151 Method to Find Inverse of a Given FunctionStep 1 Check the injectivity (one-one) Take f(x1) = f(x2) and show that x1 = x2 or show that f is continuous and monotonic on its domain

Step 2 Surjectivity (onto) Find the Range of the function (Rf) and compare it with co-domain

If Rf = Co-domain then f is onto

Step 3 Using equation y = f(x) express x in terms of y hellip (1)

Step 4 Replace x by y and y by x in the obtained relation (1) to get y = fndash1(x)

RemarkSince to each (x y) isin f there exists (y x) isin fndash1 and (y x) and (x y) are mirror images of each other in the line y = x therefore the graph of f-1(x) is obtained by reflecting the graph of f(x) in the line y = x as shown below

2116 pRopeRties of inveRse of a function

(i) The inverse of a bijection is unique (ii) The inverse of a bijection is also a bijection (iii) If f and g are two bijections f A rarr B g B rarr C then the inverse of gof exists

and (gof)ndash1 = fndash1ogndash1 (iv) Inverse of inverse of a given function is the given function itself ie (fndash1)ndash1 = f (v) f(x) and fndash1(x) if intersect then the point of intersection should be on the line

y = x or y = ndashx + k for some real value of k (vi) f(x) and fndash1(x) have same monotonic nature ie either both increasing or

both decreasing (vii) If f(x) is increasing function then fndash1(x) is also an increasing function but

f(x) and fndash1(x) have opposite curvatures(viii) If f(x) is a decreasing function then fndash1(x) is also a decreasing function but

f(x) and fndash1(x) have same curvatures (ix) If the graph of a function f(x) is symmetric about the line y = x then f(x)

and fndash1(x) are equal functions ie f(x) will be self invertible function or (involution) (x) If f A rarr B is a bijection then fndash1 B rarr A is an inverse function of f then fndash1of = IA and fofndash1 = IB

Here IA is an identity function on set A and IB is an identity function on set B

2117 even function

A function f X rarr Y is said to be an even function iff f(ndashx) = f(x) forall x ndashx isin X (Domain)ie f(x) ndash f(ndashx) = 0eg x2n sin2x cosx secx 2x + 2ndashx

21171 Properties of even functions (i) Graph of even function is symmetric about y-axis (ii) For any function f(x) if g(x) = f(x) + f(ndashx) then g(x) is always an even function (iii) The domain of even function must be symmetric about zero (iv) Even functions are non invertible as they can not be strictly monotonic when taken in their natural

domain however even functions can be made invertible by restricting their domains (v) If f(x) is even function then f (x) is odd function (vi) f(x) = c where lsquocrsquo is a constant defined on symmetrical domain is an even function

21172 Odd FunctionA function f X rarr Y is said to be an odd function iff f(ndashx) = ndash f(x) forall x ndashx isin Xie f(x) + f(ndashx) = 0 forall x ndashx isin Xeg x3 sin x tan x 2x ndash 2ndashx are odd functions

21173 Properties of Odd Functions (i) Graph of odd function is symmetric about origin Also known as symmetric in opposite quadrants (ii) For any function f(x) if g(x) = f(ndashx) ndash f(x) then g(x) is always an odd function (iii) The domain of odd function must be symmetric about zero (iv) f(x) is odd then f rsquo(x) is an even function (v) If x = 0 lies in the domain of an odd function then f(0) = 0

2118 alGeBRa of even-oDD functions

1 f(x) = 0 (identically zero function) is the only function which is both an odd and an even function provided it is defined in a symmetric domain

2 A linear combination of two or more even functions is an even function ie in particular for two even functions f(x) and g(x) the function (af + bg) is an even function where a b isin ℝ

3 A linear combination of two or more odd functions is an odd function ie in particular for two odd functions f(x) and g(x) the function (af + bg) is an odd function where a b isin ℝ

4 The product of two or more even functions is an even function 5 The product of an odd and an even function is an odd function 6 The quotient of two even functions (or two odd functions) is an even function 7 The nature (odd or even) of product of odd functions depends upon the number of functions taken

in the product ie product of odd number of odd functions is an odd function and that of even number of odd functions is an even function

8 Composition of several functions f(g(h(p(x))))) is odd iff all are odd functions

Functions 21249

9 Composition of several functions is even iff at least one function is even provided the function composed of either even or odd functions after that even function

10 Any function f(x) can always be written as sum of an even function and an odd function

Remarks

(i) The functions having no symmetry like oddeven functions are called as lsquoneither even nor odd functionsrsquo

(ii) Before testing the evenodd symmetry of the function it is essential to observe whether the domain of function is symmetric about y-axis ie if the domain is of the type [ndashx0 x0] or [ndashx2 ndashx1] cup [x1 x2] etc

2119 even extension of function

Extending the domain of function f(x) and defining such that the function obtained is even

ie f(x) if x

h(x)f( x) if x

α le leβ= minus minusβ le le minusα

2120 oDD extension of function

Extending the domain of function and redefining it such that the new function obtained becomes odd

ie h(x) = f(x) if xf( x) if x

α le leβminus minus minusβ le le minusα

21201 Definition of Periodic FunctionA function f(x) is said to be a periodic function if there exists a real positive and finite constant T inde-pendent of x such that f(x + T) = f(x) forall x isin Df provided (x + T) isin Df (domain)

The least positive value of such T (if exists) is called the periodprincipal period or fundamental period of f(x)

eg f(x) = tan x f(x) = sin x are periodic functions with period p and 2p respectively

2121 facts anD pRopeRties ReGaRDinG peRioDicity

(a) Trigonometric functions The function sin x cos x sec x cosec x are periodic with period 2p Whereas tan x cot x are periodic functions with period p

(b) There may be periodic functions which have no fundamental period eg

(i) Dirichlet function 1 when x isrational

f(x)0 when x is irrational=

(ii) Constant function Consider a function f(x) = c

(c) No rational function (except constant function) can be a periodic function (d) Algebraic function (Except Constant Function) cannot be a periodic function (i) If f (x) is periodic with period T then a f (x + k) + b is also periodic with same period T where

a b are real constants and a gt 0

(ii) If f(x) is periodic with period T then f(kx + b) is periodic with period T

| k |provided lsquokrsquo is

non-zero real number and b isin ℝ

2122 peRioD of coMposite functions

Theorem If f(x) is periodic function with fundamental period T and g(x) is monotonic function over the range of f(x) then g(f(x)) is also periodic with fundamental period T

If f(x) is periodic with period T then

(i) 1f(x)

is also periodic with same period T

(ii) f(x) is also periodic with same period T

1 Composition of a non-monotonic function g(x) over a periodic function f(x) having period T is always a periodic function with period T (But fundamental period may be less than T)

eg if g(x) = x2 and f(x)= cosx then gof(x) = cos2x is periodic with period 2p But its fundamental period is p

2 Composition of a non-periodic function g(x) with a periodic function f(x) may be a periodic function eg if g(x) = [x] and f(x) = cospx then fog(x) = cos[x] is periodic with period 2

3 Composition of two non-periodic functions may be a periodic function

eg consider g(x) = 3[x] ndash 2 and ( )

minusnotin minus=

+ isin

x 2f ( x )3 sin x cos x x

we have fog(x) = 3 forall xisinℝ

which is a periodic function indeed

2123 peRioDicity of MoDuluspoweR of a function

(i) Period if f(x)2n+1 If the fundamental period of f(x) is T then the fundamental period of f(x)2n+1 n isin ℤ will also be T ie the fundamental period of function remains same on raising it to an odd integer power

(ii) Period of f(x)2n If the fundamental period of f(x) is T then the fundamental period of f(x)2n n isin ℤ may not be T

ie the fundamental period of function may change on raising it to an even integer power For example we know that the period of the functions sin x cos x sec x cosec x is 2p and that of

(iii) If f (x) be periodic with period T1 and g(x) with period T2 such that LCM of T1 and T2 exist and is equal to T then af(x) + bg(x) is a periodic function with period T (a and b are non-zeros)

Functions 21251

Remarks

(i) LCM of two or more fractional numbers = LCM of a c e LCM of ( ace )

b d f HCF of ( bd f )

eg the LCM of 7 3and

30 20 is

LCM of 7 and3 21HCF of 30 and 20 10

(ii) LCM of rational and irrational number does not exist

eg The function x + cos x is non-periodic because the period of x is 1 and the period of cosx is 2p And the LCM (1 2p) does not exist

Also the function = sin x + tan px + sin x3 is not periodic because LCM of (2p 1 6p) does not exists

(iii) The LCM of two irrational quantities may or may not exist

1 The sumdifference of a periodic and an non-periodic function can be periodic

2 The sumdifference of two non-periodic functions can be periodic function

3 The productquotient of a periodic and an non-periodic function can be periodic

eg consider f(x) = cot x and 1 x 0

g( x )3 x 0

= minus= =

then the function f(x) g(x) and f ( x )g( x )

periodic

Clearly f(x) is periodic with period p but g(x) is non-periodic function

The domain of f(x) g(x) and f ( x )g( x )

is ℝ ~ np n isin ℤ hence f(x)f ( x )

g( x )g( x )

= = cotx which is

periodic function in its natural domain with fundamental period p

4 The productquotient of two non-periodic functions can be periodic function

eg consider 1 x 0

f ( x )1 x 0

lt= minus ge

and 1 x 0

g( x )1 x 0

minus lt= ge

then the function f(x)g(x) and

f ( x )1

g( x )= minus which being a constant function is a periodic function with no fundamental period

2124 exception to lcM Rule

Case I If f (x) be periodic with period T1 and g(x) with period T2 such that LCM of T1 and T2 exist and is equal to T and f (x) and g (x) can be interchanged by adding a least positive constant K (lt T)

ie f(x + K) = g(x) and g(x + K) = f(x) then K is period of f(x) + g(x) otherwise period will be T

Case II If f (x) be periodic with period T1 and g(x) with period T2 such that LCM of T1 and T2 exist and

is equal to T then the period of F(x) = f (x) plusmn g (x) or f(x)g(x) or f(x)g(x)

is necessarily T but the

fundamental period can be given by a positive constant K (lt T) if F(x) gets simplified to an equivalent function F(x + K) = F(x)

2125 peRioDicity of functions expResseD By functional equations

(i) If a function f(x) is defined such that f(x + T) = ndashf(x) where T is a positive constant then f is periodic with period 2T (Converse is not true)

(ii) If a function f(x) is defined such that f(x + T) = 1f(x)

or f(x + T) = 1

f(x)minus

where T is a positive

constant then f is periodic with period 2T (Converse is not true)

(iii) If f(x + l) = g(f(x)) such that ( )( )( )( )composed n times

g g g g(x) x=

then prove that f(x) is periodic with period

nl (where l is fixed positive real constant)

2126 tips foR finDinG DoMain anD RanGe of a function

If f(x) and g(x) be two functions such that f(x) has domain Df and g(x) has domain Dg then the following results always hold good

Rule 1 Domain (k f(x)) = Df for all k isin set of non-zero real numbers

Rule 2 Domain f1 D ~x f(x) 0

f(x) = =

Rule 3 Domain (f(x) plusmn g(x)) = Df cap Dg

Rule 4 Domain (f(x) g(x)) = Df cap Dg

Rule 5 Domain f(g(x)) = x x isin Dg and g(x) isin Df = Dg ~x g(x) notin Df

Rule 6 Domain of even root of f(x) = 2mff (x) D ~ x f(x) 0= lt

Rule 7 Domain 2m 1ff (x) D+ =

Rule 8 Domain (log f(x)) = Df ~ x f(x) le 0

Rule 9 Domain of composite exponential function y = [f(x)]g(x) = x isin ℝ x isin Df cap Dg and f(x) gt 0

Remarks

= xy f ( x ) is defined for x isin 2 3 4 hellip and f(x) gt 0 where as y = (f(x))1x is defined for x ne 0 and f(x) gt 0

Rule 10 Methods to find Range of Functions Given a function f X rarr Y where y = f(x)

Method I

Step 1 Find domain of f(x) say a le x le b

Step 2 Express x in terms of y using equation of function ie x = fndash1 (y)

Step 3 Apply the domain restriction ie a le x le b rArr a le fndash1 (y) le b

Step 4 Find the set of all possible y satisfying above inequality

Method II

For composition of continuous functions

Step 1 Identify the function as composite function of constituent functions f g and h say f(x) = h(f(g(x)))

Step 2 Test the monotonicity of f and g and h say g(ndash(increasing)) f(darr (decreasing)) h(darr (decreasing))

Functions 21253

Step 3 Find domain of h(f(g(x))) say a le x le b

Step 4 ∵ a le x le b rArr Rf = [h(f(g(a))) h(f(g(b)))]

Rule 5 If domain is a set having only finite number of points then the range will be the set of corre-sponding values of f (x)

Rule 6 If domain of y = f (x) is R or Rndashsome finite points or an infinite interval then with the help of given relation express x in terms of y and from there find the values of y for which x is defined and belongs to the domain of the function f (x) The set of corresponding values of y constitute the range of function

Rule 7 If domain is not an infinite interval find the least and the greatest values of f (x) using monoto-nicity (This method is applicable only for continuous functions and is the most general method)

Rule 8 For the quadratic function f(x) = ax2 + bx + c domain is ℝ and range is given

minus infin gt = minus minusinfin lt

D for a 04aR

D for a 04a

Rule 9 For the quadratic function 2f (x) ax bx c= + + domain is given by f

for a 0D 0D

for a 0 D 0gt lt = φ lt lt

and range is given by

0 for D 0 a 0

D for D 0 a 04a

RD0 for D 0 a 0

4afor D 0 a 0

infin ge gt minus infin lt gt = minus ge lt

φ lt lt

Rule 10 For odd degree polynomial domain and range both are ℝ

Rule 11 For even degree polynomial domain is ℝ and range is given by [k infin) if the leading coefficient is positive where k is the minimum value of polynomial occurring at one of the points of local minima whereas range is (ndashinfin k] if the leading coefficient is negative where k is maximum value of polynomial occurring at one of the points of local maxima

Rule 12 For QuadraticQuadratic

or Linear

Quadratic or Quadratic

Linear expression put Qy

Q= cross-multiply convert

into a quadratic and use the knowledge of quadratic equations

Rule 13 For discontinuous functions only method is to draw the graph and find the range known as graphical method of finding out range

Rule 14 Range of function f(x) = asinx + bcosx is 2 2 2 2a b a b minus + +

Rule 15 (i) If f(x) and g(x) are increasing functions in their respective domain then gof(x) is also an increasing function in its domain Further if both f(x) and g(x) are continuous in

their respective domain then gof is also continuous in its domain Now if common domain of f(x) and gof(x) is [a b] or (a b) then range of f(x) is [f(a) f(b)] or (f(a) f(b)) which in turn is domain of g(x) Then range of fog(x) will be [g(f(a) g(f(b))] or (g(f(a) g(f(b)))

(ii) If f(x) and g(x) both are decreasing functions in their respective domain then gof is also a decreasing function Further if both f(x) and g(x) are continuous in their respective domain then gof is continuous and increasing function in its domain If common domain of f(x) and gof(x) is [a b] or (a b) then range of f(x) is [f(b) f(a)] or (f(b) f(a)) which in turn in domain of g(x) which is decreasing and continuous function Thus range of gof will be [g(f(a)) g(f(b))] or (g(f(a)) g(f(a)))

(iii) If f(x) and g(x) are functions of opposite monotonicity in their respective domain then gof is a decreasing function on its domain Further if f(x) and g(x) are continuous functions then gof is continuous and decreasing function If [a b] or (a b) is common domain of gof(x) and decreasing function f(x) (say) then range of f(x) is [f(b) f(a)] or (f(b) f(a)) which in turn in domain of g(x) g(x) being continuous and increasing (say) range of gof(x) will be [g(f(b)) g(f(a))] or (g(f(b)) g(f(a))) Same will be the range of gof(x) if f(x) is increasing and g(x) is decreasing

(iv) If f(x) is an increasing and continuous function in its domain and g(x) is non-monotonic having range [a b] or (a b) then the range of fog(x)) will be [f(a) f(b)] or (f(a) f(b)) Similarly if f(x) is decreasing and continuous function in its domain and g(x) is non-monotonic having range [a b] or (a b) then the range of fog(x) will be [f(b) f(a)] or (f(b) f(a))

(v) If f(x) is non-monotonic function and continuous in its domain and g(x) is any function (monotonic or non-monotonic) for which the composition function fog is defined then range of fog can be obtained by analyzing the behaviour of function f(x) on the range set of function g(x) ie by determining the intervals of monotonicity lub gub of f(x) in range set of g(x)

(vi) If f(x) is monotonic and continuous in its domain and g(x) is non-monotonic for which fog(x) is defined and range of g(x) is [a b] or (a b) then the range of fog(x) will be [f(a) f(b)] or (f(a) f(b)) if f(x) is increasing and it will be [f(b) f(a)] or (f(b) f(a)) if f(x) is decreasing

(vii) If f(x) and g(x) both are non-monotonic and continuous for which fog(x) is defined then the range of f(x) can be obtained by analyzing the behaviour of f(x) on the range set of g(x) ie by determining the intervals of monotonicity lub and glb of f(x) in the range set of g(x)

Chapter 22Limits Continuity and differentiabiLity

221 Limit of a function

Limit at x = a means value of function at x = a if the function is without any break and if it is not defined at x = ay then limit of function means to find the real number to which function tends when x tends to a independent of the way whether x approaches to a by taking smalles values then lsquoarsquo or greater then lsquoarsquo 1 meaning of a x is approaching nearer and nearer to lsquoarsquo (fixed real number) but x ne a 2 meaning of a+ x is approaching to lsquoarsquo by taking values greater than lsquoarsquo 3 meaning of andash x is approaching to lsquoarsquo by taking values less than lsquoarsquo

222 Limit of function f(x) at x = a

Is denoted by x 0lim f(x) l(say)rarr

= means ldquoas x tends to a f(x) tends to lrdquo ie the number lsquolrsquo is said to be limit

of f (x) at x = a if for any arbitrary chosen positive numberisin however small but not zero there exists cor-responding number d greater than zero such that |f (x) ndash l |lt isin for all values of x for which 0 lt |x ndash a| lt d

223 ExistEncE of Limit of a function

Limit of a function f(x) is said to exist at x = a if x a x alim f(x) lim f(x) l(lt )

minus +rarr rarr= = infin

ie LHL (Left Hand Limit) = RHL (Right Hand Limit) = l(lt infin) Symbolically f(andash) + f(a+) = l(ltinfin)

224 non-ExistEncE of Limit of a function

x 0lim f(x)rarr

does not exist due to the following reason(s)

(i) If one of the two sided limits is finite and other is infinite (ii) If both sided limits are finite but unequal (iii) If both limits are infinite but of opposite signs (iv) If at least one of the two limits does not exist uniquely

Remarks (i) For the existence of limit function must be defined in at least one of the two deleted

neighbourhood of lsquoarsquo ie (a ndash d a) (left deleted neighbourhood of a) or (a a + d) (right deleted neighbourhood of a) In such cases limit is taken equal to one-sided limit for which x tends to a taking those values of x in deleted neighbourhood of a in which function is defined eg

f ( x )x

then minus

rarr rarr= = =

π π1

x 1 x 1

sin x 2lim f ( x ) lim

x 1 2 as f(x) is defined for x isin [ndash1 1] Similarly

rarrminus rarrminus

minus= = =

minusπ π1

x 1 x 1

x 1 2 (ii) If both one-sided limits of function are infinite of same sign then we say that limit exists infinitely

eg if =minus 2

1f ( x )

( x 1) then

x 1 x 1lim f ( x ) lim f ( x )

(iii) If f (x) is defined at x = a it does not imply that rarrx a

lim f ( x ) exist

eg = minusf ( x ) sin x 1 rarr

minusπx 2

lim sin x 1 f(x) is defined at p2 ie = minus =ππf ( 2) sin 1 02

butisnotdefinedinsurroundingofx=π2thereforerarr πx 2lim f ( x ) does not exist

(iv) Even if f (x) is not defined at x = a but rarrx a

lim f ( x ) may exist If + minusrarr rarr

=x a x alim f ( x ) lim f ( x )

eg rarr

minus= =

x 9f ( x ) lim f ( x ) 6

x 3 although f(3) is not defined

(v) Limit may be finite as well as infinite eg rarr

= infinx 0

225 aLGEbra of Limits

If x alim f(x)rarr

= l and lim g (x) = m (l and m are finite real numbers) then the following statements hold good

(i) Sum Rule x alimrarr

(f (x) + g (x)) = l + m eg x 2lim (x sin x)rarr

+ = rarr

+x 2 x 2lim x lim sin x = 2 + sin2

(ii) Difference Rule rarrx a

lim (f (x) - g(x)) = l - m eg rarr

minus x

x 3lim (tan x 2 ) =

rarr rarrminus x

x 3 x 3lim tan x lim 2 = tan3 - 23

(iii) Constant multiple Rule rarrx a

lim kf (x) = kl eg rarr

x 5lim 5e = 5 times x

x 5lim erarr

(iv) Product Rule x alimrarr

(f (x) g(x)) = lm eg x 34lim x cos xrarr

= x 34 x 34lim (x) lim cos xrarr

= 34 cos34

(v) Quotient Rule x a

f (x)lim m 0g(x) mrarr

= ne eg

x 1x 1

lim ee elimsin x lim sin x sin1

rarrrarr

(vi) Power Rule If p and q are integers then p q p q

x alim(f(x)) rarr

= provided lpq is a real number

(vii) Rule for composite functions x alim f(g(x))rarr

= ( )x a

f ( lim g(x)) f mrarr

= provided lsquof rsquo is a function

continuous at g(x) = m eg x alim ln[f(x)] ln( )rarr

= only if l gt 0

Notesin

x alim

sin(f(x)) The above said is also valid for cos (f(x)) tan (f(x)) cot (f(x)) sec (f(x)) cosec (f(x))

Pn (f(x)) provided these functions are defined at that point

Limit Continuity and Differentiability 22257

226 indEtErminatE forms

Sometimes we come across the functions which do not have definite value corresponding to some particular value of the independent variable (If by substituting x = a in any function f(x) it takes up any one of form 00 infin infin 0 times infin infin - infin 1infin 00 infin0 then the limit of function f(x) as x rarr a is

called indeterminate form) There are two basic indeterminate forms 0 0infin

infin and all the other forms

can be converted to these two basic forms In such cases value of function at x = a does not exist while

x alimrarr

f(x) may exist

(a) 2( x 9)

f(x)x 3minus

=minus

Here 2

x 3lim x 9 0rarr

minus = and x 3lim x 3 0rarr

minus = So rarrx 3

lim f (x) is called an indeterminate form

of the type 00

(b) rarrinfinx

ln xlim

x is an indeterminate form of type infininfin

(c) rarr

x 0lim (1 x) is an indeterminate form of the type 1infin

Forms which are sometimes mistaken as indeterminate but are well defined

(i) +infinminusinfin

if c is positive cons tan tif c is negative cons tan t

(ii) =infinc 0

(iii) infin + infin = infin (iv) infin times infin = infin (v) 0infin = 0 (vi) 00 = 0 (vii) infin + 0 = infin (viii) (infin)infin = infin

NoteInfinity is a symbol and not a number It does not obey laws of elementary algebra

227 somE standard Limits

(a) rarr rarr rarr

= =x 0 x 0 x 0

tan xsin x xlim lim limx sin x x

= 1 -1

1 -1x 0 x 0 x 0 x 0

tan x x sin x xlim lim lim lim 1x tan x x sin x

minusrarr rarr rarr rarr= = =

e 1lim 1

minus= (c) ( )

b 1lim ln b b 0

minus= gt

(d) x 0

ln (1 x)lim 1xrarr

(e) aax 0

log (1 x)lim log e

(f) 1x

x 0lim(1 x) erarr

+ = (g) ( )rarrinfin

xlim 1 1 x e

(h) rarr

minus= isin

m mm-1

x alim ma m Q(set of relations)x a

(i) rarr

minus= isin ne

m mm-n

n nx a

x a mlim a mn Qn 0x a n

(j) rarr

minus= isin

x 1lim mm Qx 1

0 if 0 a 11 if a 1

lim a if a 1

does not exist if a 0

rarrinfin

lt lt == infin gt lt

(l) x a

sin f(x)lim 1f(x)rarr

= (m) x a

Limcos f(x) 1rarr

(n) rarr

tan f(x)lim 1f(x)

(o) rarr

minus=

e 1lim 1f(x)

(p) rarr

minus= gt

b 1lim nb(b 0)f(x)

(q) rarr

n(1 f(x))lim 1f(x)

(r) rarr

+ =1f (x)

x alim(1 f(x) e

(s) rarr

= gtx alim f(x) A 0 and

x alim (x) Brarrφ = (a finite quantity) then (x) B

x alim f(x) Aφ

228 Limits of somE standard compositE functions

If x alim f(x)rarr

exists and is equal to L then the following will always hold good

(a) x alimsin f(x) sinLrarr

= (b) x alimcos f(x) cosLrarr

(c) x alim tan f(x) tanLrarr

= (d) x alim f (x)f (x) L

x alima a a (a 0)rarr

rarr= = gt

(e) n nx alimP f(x) P (L)rarr

= (f) rarr

=x alim log f(x) log L (provided f(x) gt 0)

229 somE usEfuL transformations

transformation 1 Cancellation of a term approaching towards zero from numerator and denominator

eg rarr rarr

minus + minus minus= = minus

+ minus minus +

2x 2 x 2

x 5x 6 (x 2)(x 3) 1lim limx 3x 10 (x 2)(x 5) 7

[Q (x ndash 2) ne 0 ]

transformation 2 Dividing and multiplying by the input (of sin or tan) if it is approaching

towards zero eg

rarr rarr= =

33x 0 x 03

sin x xsin x xlim lim 0(tan x) tan x x

transformation 3 Changing the variable of limit eg minusrarr

minus minusminus +1

2x sin 3

sin x 2sin x 3limsin x 4sin x 3

Here we take sin x = t As x = sinndash1 3 t = 3 rarr rarr

minus minus minus += = = =

minus + minus minus

2t 3 t 3

t 2t 3 (t 3)(t 1) 4L lim lim 2t 4t 3 (t 3)(t 1) 2

2210 somE important Expansions

Limits of various functions can be evaluated by expanding the functions using the binomial exponential and Logarithmic expansion and expansion of functions like sinx cosx and tanx etc The following results are to be remembered and can be used directly to evaluate limits unless otherwise mentioned

(a) = + + + + gt2 2 3 3

x x lna x ln a x ln aa 1 a 01 2 3

(b) = + + + +2 3

x x x xe 1 1 2 3

(c) + = minus + minus + minus lt le2 3 4x x xln(1 x) x for 1 x 1

2 3 4 (d) = minus + minus +

3 5 7x x xsin x x 3 5 7

(e) = minus + minus +2 4 6x x xcos x 1

2 4 6 (f) = + + +

3 5x 2xtan x x 3 15

(g) minus = minus + minus +3 5 7

(h) minus = + + + +2 2 2 2 2 2

1 3 5 71 1 3 1 3 5sin x x x x x 3 5 7

(i) minus = + + + +2 4 6

1 x 5x 61xsec x 1 2 4 6

(j) + = minus + +

1x 2x 11(1 x) e 1 x 2 24

(k) for |x| lt 1 n isin ℝ (1 + x)n = minus minus minus+ + + + infin2 3n(n 1) n(n 1)(n 2)1 nx x x

12 123

2211 somE standard approachEs to find Limit of a function

(a) By Directly Substituting x = a eg (i)

rarr+ +4 3

x 2lim(x 3x 2) can be obtained by substituting x = 2 in x4 + 3x3 + 2 So the limit is 42

(ii) x a

P(x) P(a)limQ(x) Q(a)rarr

= and 3 2

x 4x 3 0lim 0x 5 6rarrminus

+ minus= =

(b) By Factorization eg ( )2

x 2 x 2 x 2

x 4 (x 2) (x 2)lim lim lim x 2 4x 2 x 2rarr rarr rarr

minus minus += = + =

minus minus

(c) By substituting x = a + h h rarr 0

(i) For the limit 00

form we can substitute x = a + h provided f(x) is continuous in the deleted neigh-

bourhood of a eg 2

x 5x 6lim(x 2)rarr

minus +minus

+ minus + ++ minus

(2 h) 5(2 h) 6lim(2 h 2)

=rarr rarr

minus= minus = minus

h 0 h 0

h hlim lim(h 1) 1h

(ii) If there is a possibility of f(x) to be discontinuous across a then evaluate LHL and RHL

separately by substituting x = a ndash h h rarr 0+ and x = a + h h rarr 0+ respectively eg minus=

minus(x 2)f(x)x 2

Then minus + + +

rarr rarr rarr rarr

minus minus minus minus minus= = = = = minus

minus minus minus minusx 2 h 0 h 0 h 0

(x 2) (2 h 2) ( h) ( h)f(2 ) lim lim lim lim 1x 2 2 h 2 h h

f(2ndash) ne f(2+) rArr limit of f(x) does not exist at x = 2

RemarksIn the following cases both sided limits should be calculated separately to find the existencenon-existence of limit

(i) In case the function is defined piecewise and we are to find limit at extreme point

(ii) In case the function involves modulus function

(iii) In case the function involves greatest integer function

(d) By Rationalization eg minusminus =

+x ax ax a

minusminus =

+ +13 13

13 13 13 13

x ax ax a x a

(e) By application of Standard Limits

eg rarr rarr rarr

minus = =

2 2x 0 x 0 x 0

1 cos 4x 2sin 2x sin2xlim lim lim2x x x

times = times times =

sin2x2 lim 2 2 (1 2) 82x

(Hence we used the standard limit rarr

sin xlim 1x

(f) to solve limit at infin eg 3 2

6x 5x 2x 1lim form2x 3x 9rarrinfin

+ + minus infin + + infin

5 2 16x x xlim

3 92x x

rarrinfin

+ + minus

= 6 0 0 0 6 32 0 0 2+ + minus

= =+ +

(Dividing numerator and denominator by x3)

RemarksIf m n are positive integers and a0 b0 ne 0 and non-zero real numbers then

minusminus

minusrarrinfinminus

+ + + + = =+ + + + infin gt

m m 10 1 m 1 m 0

n n 1x0 1 n 1 n 0

a x a x a x a alim m n

b x b x b x b b

(g) By using expansion Sometimes it is easy to use expansions to evaluate limits

eg rarr rarr

+ + + + minus minus + minus

minus =

2 3 2 4

x 0 x 0

x x x x1 x 1 2 3 2 4e cos xlim lim

= rarr rarr

+ + + + +

minus =

2 3 5 6

x 0 x 0

x x x xx 2 2 2 3 5 6e cos xlim lim

(h) Using Lrsquo Hospitals Rule

If f (x) and g (x) are functions of x such that f(a)g(a)

is either infin infin

( )( )

( )( )rarr rarr rarr

= =x a x a x a

f x f x f xlim lim lim

g x g x g x

till a determinate is obtained

Note 1 Any indeterminate form may be converted into 00 form and then one can apply L - Hospitals Rule to

find out Limits

eg rarr

+ minus + x 0

sin x cos x 1 0lim form

tan x x 0 =

minus=

cos x sin x 1lim

sec x 1 2

(i) method to evaluate limits of the form (1)infin

(a) rarr rarr rarr

+ = =1

g(x)x a x a x alim 1 f(x) where lim f(x) limg(x) 0 Then times

rarr rarr+ = +

1 1 f (x)g(x) f (x) g(x)

x a x alim 1 f(x) lim 1 f(x)

= rarr

f (x)limg(x)e eg ( ) x 0

xlim1x x

x 0lim 1 x e erarr

rarr+ λ = =

(b) ( )1

g(x)x 0 x a x alim f(x) where lim f(x) 1 limg(x) 0rarr rarr rarr

= = then ( ) ( )11

g(x)g(x)x 0 x alim f(x) lim 1 f(x) 1rarr rarr

= + minus

= ( ) ( ) x a

f (x) 11 f (x) 1 lim g(x)g(x)f (x) 1x alim 1 f(x) 1 e rarr

minus minus times minusrarr + minus = eg ( )

x 0lim cos x = rarr rarr

minus minus

= =x 0 x 0

cos x 1 xlim lim tansinx 2e e 1

(j) method to evaluate limits of the form (0)0

To evaluate rarr rarr

= = g(x)

x a x 0L lim f(x) where lim f(x) 0 but f(x) gt 0 in neighbourhood of x = a and

x 0limg(x) 0 then we write rarr

rarr rarr= = = =

g ( x ) n f ( x )g ( x )x 0limn f (x) g(x) nf (x)

x a x 0L lime lime 0 e

RemarkIf f(x) is not positive throughout the neighbourhood of a then limit does not exist because in this case function will not be defined in the neighbourhood of x = a

(k) method to evaluate limit of the form (infin)0

Let rarr rarr

= =infin = h(x)

x a x af (x) g(x) where limg(x) and limh(x) 0 then

h(x)lim1

lng(x)h(x)lng(x)

0L lime e form0

(i) rarrinfin

n xlim 0

x (ii)

rarrinfin=xx

xlim 0

e (iii)

rarrinfin=

xlim 0

e (iv)

( )rarrinfin

n xlim 0

(v) ( )+rarr

x 0lim x n x 0 As x rarr infin ln x increases much slower than any (+ve) power of x where ex increases

much faster than (+ve) power of x

(vi) rarrinfin

minus =n

nlim(1 h) 0 and

rarrinfin+ rarr infinn

nlim(1 h) where h rarr 0+

22111 Sandwitch Theorem or Squeeze Play Theorem

If f(x) le g(x) le h(x) forall x and rarr rarr rarr

x a x a x alim f(x) limh(x) then limg(x)

ContinUity

2212 continuity of f(x) at x = a

f(x) is said to be continuous at x = a if (i) f(a) exists (ii)

x a x alim f(x) lim f(x) f(a)

minus +rarr rarr= =

22121 Reasons of Discontinuity of f(x) at x = a (i) Removable discontinuity f(andash) = f(a+) ne f(a) bull If f(a) is not defined then f(x) is said to have missing point removable discontinuity bull If f(a) exist and is different from f(andash) and f(a+) then f(x) is said to have isolated point removable

discontinuity (ii) Jump Discontinuity f(andash) and f(a+) exists finitely but f(andash) ne f(a+) jump | f(andash) ndash f(a+)| (iii) infinite Discontinuity If at least one of f(andash) and f(a+) is infinite ie either f(andash) = plusmninfin

or f(a) = plusmninfin (iv) oscillatory discontinuity

If atleast one of f(andash) and f(a+) oscillates and is not unique eg 1f(x) sinx

= has oscillatory

discontinuity at x = 0 as for x rarr 0 both f(andash) and f(0+) oscillates in between ndash1 and 1

2213 discontinuity of first Kind

A function f(x) is said to have discontinuity of first kind if f(andash) and f(a+) exist finitely but condition of continuity is not satisfied

ie (i) Either f(andash) ne f(a+) but f(andash) and f(a+) are finite (ii) f(andash) = f(a+) = l ne f(a) where l lt infin

22131 Discontinuity of Second KindA function f(x) is said to have discontinuity of second kind if atleast one of f(andash) and f(a+) does not exist or is infinite

Pictorial Diagram Representing Classification of Discontinuity

2214 poLE discontinuity

If a function f(x) becomes infinite (+infin or ndashinfin) as x tends to a then f(x) is said to have pole discontinuity

at x = a ie x a x alim f(x) lim f(x)

minus +rarr rarr= =infin

2215 sinGLE point continuity

Functions which are continuous only at a single point are called single point continuous function

2216 onE sidEd continuity

(a) Function left continuous (or right discontinuous) at x = a If f(andash) = f(a) ne f(a+) ie LHL f(a) (ltinfin) and f(a) ne f(a+) (b) Function right continuous (or left discontinuous) at x = a If f(andash) ne f(a) = f(a+) ie RHL = f(a) (ltinfin) and f(a) ne f(andash)

22161 Continuity of an Even and Odd Function

If a function f(x) is even or odd then it is simultaneously continuous (or discontinuous) at x = a and x = ndasha (i) If f(x) is an even function then f(ndasha+) = f(andash) f(ndashandash) = f(a+) (ii) If f(x) is an odd function then f(ndasha+) = ndashf(andash) f(ndashandash) = ndashf(a+)

2217 aLGEbra of continuity

If f(x) and g(x) are two continuous functions ie f(a+) = f(andash) = f(a) g(a+) = g(andash) = g(a) then the following results always hold good 1 kf(x) is continuous at x = a (k is finite real constant) 2 f(x) plusmn g(x) is continuous at x = a 3 f(x) g(x) is also continuous at x = a

4 f(x)g(x)

is also continuous at x = a iff g(a) ne 0

5 The sum of a finite number of functions continuous at a point is a continuous function at that point

RemarkFrom the above theorem it is obvious that difference of finite number of continuous functions at x = a is also continuous at x = a

6 The product of a finite number of functions continuous at a point is a continuous function at that point

7 If f1f2f3hellipfn and g1 g2 g3hellipgn are two sets of continuous functions at x = a such that gi(a) ne 0

for any i then 31 2 n

1 2 3 n

k kk k

m m m m

f ff f

g g g g is continuous at x = a where ki miisin 1 2 3hellip n

8 (a) (Chain rule for continuity or continuity of composite functions) If f(x) is continuous at x = a and g(y) is continuous at y = f(a) then the composite function (gof)(x) is continuous at x = a

(b) Let a function f(x) be continuous at all points in the interval [ab] and let its range be the interval [m M] and further the function g(x) be contininuous in the interval [m M] then the composite function (gof) (x) is continuous in the interval [a b]

(c) If the function f is continuous everywhere and the function g is continuous everywhere then the composition gof is continuous everywhere

(d) All polynomials trigonometric functions inverse trigonometric functions exponential and logarithmic functions are continuous at all points in their respective domains

(e) If f(x) is continuous at x = a then |f(x)| is also continuous at x = a

9 Sum of the two discontinuous functions may be continuous Example (i) f(andash) = l1 f(a+) = l2 l1 ne l2 and g(andash) = l2 g(a+) = l1 then (f + g)(andash) = (f + g)(a+) = (f + g)(a) = l1 + l2

(ii) f(x) = [x] + x = x is a continuous function where as [x] and x are discontinuous functions at integer points

10 Summation of a continuous and a discontinuous function is always discontinuous 11 Product of a continuous function with a discontinuous function may be continuous and this is

possible only when the continuous function attains zero at that point 12 Quotient of a continuous and discontinuous function may be continuous may be discontinuous

2218 continuity of a function on a sEt

A function f(x) is said to be continuous on a set A if f(x) is continuous at every point of set A If a function has discontinuity even at single point of set A then f(x) is said to be discontinuous on set A

Domain of continuity of function The set of all those points where the function f(x) is continuous is called Domain of continuity of function f(x) Every function is continuous on its domain of continuity

22181 Domain of Continuity of Some Standard Function

f(x) Domain of Continuity f(x) Domain of Continuity

Polynomial P(x) ℝ sec x ℝ ndash (2n + 1) π2

n isin ℤ

P(x)Q(x) ℝ ndash x Q(x) = 0 cosec x ℝ ndash np n isin ℤ

ax a gt 0 ℝ sinndash1 x [ndash1 1]

log x (0 infin) cosndash1 x [ndash1 1]sin x ℝ tanndash1 x ℝcos x ℝ cotndash1 x ℝ

tan x ℝ ndash (2n + 1) π2

n isin ℤ secndash1 x (ndashinfin ndash1] cup [1 infin)

cot x ℝ ndash np n isin ℤ cosecndash1 x (ndashinfin ndash1] cup [1 infin)

Continuity of a function on its domain A function f(x) is said to be continuous on its domain if it is continuous at every point of its domain

22182 Continuity in an Open Interval

A function f(x) is said to be continuous in (a b) when f(x) is continuous at each point c isin (a b) ie f(cndash) = f(c+) = f(c) forallc isin (a b)

221821 Method of testing of continuity in open interval

1 First of all make sure that every point of open interval is in the domain of given function ie each constituent function is defined at each point of open interval (a b) eg f(x) = x2 + sinx ndash tanx then x2 sinx tan x each is defined in open interval (0 1) but same function is not defined in open interval

(1 2) as tan x is not defined at π157

2 Thus f(x) cannot be continuous in open interval (1 2) due

to discontinuity at π

2 Use the knowledge of domain of continuity of standard constituent functions involved and

algebra of continuity eg if lt le

+ lt lt

x sin x 0 x 1f(x)

x 21 x 4

Now x and sin x have their domain of continuity ℝ and the product of two continuous functions at a point is also continuous at that point Thus x sin x is continuous in (0 1)

Also (x2 + 2) being a polynomial function is also continuous at each real number x2 + 2 is also continuous on (1 4)

3 Test the continuity of f(x) at suspicious points (ie points splitting the function into two different definitions) For example in above step (2) x = 1 is the suspesious point

2219 continuity of a function on a cLosEd intErvaL

A function f(x) is said to be continuous on closed interval [a b] if

(i) f(x) is continuous in (a b) (ii) f(x) is right continuous at x = a (iii) f(x) is left continuous at x = b

Thus f(x) is continuous on [a b] if

(i) f(cndash) = f(c+) = f(c) forall c isin (a b) (ii) f(a) = f(a+) (iii) f(bndash) = f(b)

2220 propErtiEs of continuous function

P1 (Fermatrsquos Theorem) Every function f(x) which is continuous in [a b] is always bounded

Remark

If a function is continuous in open interval then it is not necessarily bounded eg tan x forall x isin π π minus

is continuous but not bounded as its range is (ndashinfin infin) 4

f ( x ) x (02)x 2

= forall isinminus

is continuous but not bound-

ed as its range is (2infin)

P 2 intermediate Value TheoremIf f is continuous on [a b] and f(a) ne f(b) then for any value c lying in between f(a) and f(b) there exist at least one number x0 in [a b] for which f(x0) = c

P 3 Weierstrass Theorem (Extreme Value Theorem)If f is continuous on [a b] then f takes on a least value m and a greatest value M on this interval

P 4 Bolzanos TheoremIf f(a) and f(b) possesses opposite signs then $ at least one solution of the equation f(x) = 0 in the open interval (a b) provided that f is continuous in [a b]

P 5 A continuous functions whose domain is some closed interval must have its range also a closed interval

RemarkIf a function f(x) is continuous on an open interval (a b) or on real number line R and m and M are respectively the greatest lower bound and least upper bounds of f(x) then Range of f(x) = [m M] if f(x) attains m and M and it is (m M) if f(x) does not attain its bounds m or M are included in range if m or M are attained by the function

P 6 Continuity of inverse FunctionIf the function y = f(x) is defined continuous and strictly monotonic on the domain of function f(x) then there exists a single-valued inverse function x = f(y) defined continuous and also strictly monotonic in the range of the function y = f(x)

P 7 If a function f(x) is integrable on [a b] then intx

af (t)dt x isin [a b] is continuous function

DiFFEREntiABiLity

2221 diffErEntiabiLity at a point

A function f(x) is said to be differentiable at a point x = a iff +

minus minus = minus

slopeof lefthand tangentat(a f (a))

f (a h) f(a)lim f (a )h

ins tantaneous rateof changeinleftneighbourhoodof a

left hand derivative LHD +

+ minus = =

slopeof righthand tangentat(a f (a))

ins tantaneous rateof changeinrightneighbourhoodof a

right hand derivative RHD = a finite real number

22211 Physical Significance

Since minusminus

f(x) f(a)x a

is an average rate of change of f(x) wrt lsquox in [a x] therefore x rarr a the interval [a x]

converts to an instant and rarr

minusminusx a

f (x) f(a)limx a

becomes instantaneous rate of change of f(x) wrt x at x = a

So differentiability physically signifies that no sudden change in the instantaneous rate of change at x = a

22212 Geometrical SignificanceDifferentiability of f(x) at x = a implies LHD = RHD This geometrically means that a unique tangent with finite slope can be drawn at x = a Therefore graph of f(x) must be smooth without any sharp edgecorner at x = a and tangent line at x = a is not vertical

2222 concEpt of tanGEnt and its association with dErivabiLity

Slope of the line joining P(a f(a)) and Q(a + h f(a + h)) = + minusf(a h) f(a)

Slope of tangent at P = f (a) rarr

+ minus=

f(a h) f(a)limh

The tangent to the graph of a continuous function f at the point P (a f(a)) is (i) the line through P with slope f (a) if f (a) exists

(ii) the line x = a if rarr

+ minus= =infin

f(a h) f(a)limh

If neither (i) nor (ii) holds then the graph of f does not have a tangent at the point P In case (i) the equation of tangent is (y ndash f(a)) = f (a) (x ndash a) In case (ii) it is x = a

Remarks (i) Tangent is also defined as the line joining two infinitely small close points on a curve

(ii) A function is said to be derivable at x = a if there exists a tangent of finite slope at that point ie f(a+) = f(a-) = finite real number

22221 Theorem Relating to Continuity and Differentiability Differentiability rArr Continuity ie if a function is differentiable at x = a then it is continuous at x = a Thus if a function is discontinuous at x = a then it must be non-differentiable at x = a

22222 Reasons of Non-differentiability of a Function at x = a

1 Discontinuity of function at x = a

2 Sharp points on graph The point P on graph is called sharp corner or kink At such

points the graph changes its direction abruptly In such case LHD ne RHD

3 Vertical tangent

4 oscillation Point If a function f(x) is continuous but left and right

derivative do not exist at x = a due to high frequency oscillations in neighbourhood of x = a then the function f(x) is non-differentiable at x = a and such a point is called oscillation point

2223 aLGEbra of diffErEntiabiLity

If f(x) and g(x) are differentiable functions at x = a then the following statements holds good

1 K f(x) is always differentiable (k is finite) at x = a 2 f(x) plusmn g(x) is always differentiable at x = a 3 f(x) g(x) is always differentiable at x = a

4 f(x)g(x)

is differentiable at x = a provided g(a) ne 0

5 f(g(x)) is differentiable at x = a if f is differentiable at x = g(a) and g(x) is differenable at x = a 6 Sum of two non-differentiable functions can be differentiable 7 Sum of differentiable function and non-differentiable function is always non-differentiable 8 Product of two non-diff functions may be differentiable eg f(x) = |x| and g(x) = |x| f(x)g(x) = (|x|)2 = |x2| = x2 which is always differentiable 9 Product of a diff and non-differentiable function may be differentiable f(x) = |x| and g(x) = x

eg f(x) = x|x| = gtminus lt

x x 0x x 0

22231 Domain of DifferentiabilityThe set containing all the points at which the function is differentiable is called domain of differentiability of a given function for example if f(x) = ||x| ndash 1| then its graph is given below

The graph of f(x) has corner points at x = ndash1 0 and 1 Except for all these points f(x) has smooth and continuous graph at all real points Thus domain of differentiability of f(x) is ℝ ~ ndash1 0 1

2224 domain of diffErEntiabiLity of somE standard functions

f(x) Domain of Differentiability f(x) Domain of Differentiability

Polynomial P(x) ℝ sec x ℝ ndash (2n + 1) p2 nisinℤP(x)Q(x) ℝ ndashx Q(x) = 0 cosec x ℝ ndash np nisinℤ

ax ℝ sinndash1 x (ndash1 1)

log x (0infin) cos-1 x (ndash1 1)sin x ℝ tan-1 x ℝcos x ℝ cot-1 x ℝtan x ℝ ndash (2n+1)p2 nisinℤ secndash1 x (ndashinfinndash1)cup (1infin)

cot x ℝ ndash np nisinℤ cosecndash1 x (ndashinfinndash1)cup (1infin)

22241 Differentiability in Open and Closed Interval

A function is differentiable in open interval (a b) if f (cndash) = f (c+) real and finite c isin (a b) A function is differentiable in closed interval [a b] if f is differentiable in (a b) and RHD at x = a and LHD at x = b should be real and finite

method to check the differentiability of a given function on a set or to find domain of differentiability

1 From the graph theory or using standard functionrsquos domain of continuity find all those points where the function is discontinuous (say) x = x1 x2 x3 xn Then f(x) will be non-differentiable at these points

2 Find all those points where the function f(x) takes a sharp turn ie have kink points At these points function will be non-differentiable

3 Also find all those points where the function f(x) has vertical tangent At such points f(x) will be non-differentiable

4 Find all points where f(x) oscillates with infinite frequency At such points f(x) will be non-differentiable 5 The set ℝ except for the points of non-differentiability will be the domain of differentiability

of given function 6 If f(x) is a multi-formula function then remove the sign of equality at the points where the definition

of function changes Find the corresponding derivative functions The continuity of function at the point of separation of two different branches and continuity of derivative function implies the dif-ferentiability of function at that point

2225 miscELLanEous rEsuLts on diffErEntiabiLity

1 Differentiability of a function does not imply the continuity of derivative function 2 Continuity of derivative function does not imply differentiability of function Thus derivative function f (x) is continuous at p4 Thus continuity of derivative function does not imply differentiability of function however continuity

of derivative of continuous function which are non-oscillating implies differentiability of function

22251 Alternative Limit Form of Derivatives

We know that the derivative of a function f(x) at x = a is given by f (a) = rarr

+ minush 0

f(a h) f(a)limh

On substituting a + h = x x rarr a we get f (a) = rarr

minusminusx a

f (x) f(a)limx a

and we have

LHD = f (andash) = minusrarr

minusminusx a

f (x) f(a)limx a

and RHD = f (a+) = +rarr

minusminusx a

f (x) f(a)limx a

Another alternative form of derivative by using centered difference quotient Let (a ndash h a + h) be neighbourhood of lsquoa of radius lsquoh and centre lsquoa then the quotient

+ minus minus + minus minus=

+ minus minusf(a h) f(a h) f(a h) f(a h)(a h) (a h) 2h

is called centered difference quotient

Consider the limit rarr

+ minus minush 0

f(a h) f(a h)lim2h

= rarr

+ minus + minus minush 0

f(a h) f(a) f(a) f(a h)lim2h

= rarr

+ minus minus minus minus h 0

f(a h) f(a) f(a h) f(a)lim

rarr rarr

+ minus minus minus + minus h 0 h 0

f (a h) f(a)1 f(a h) f(a)lim lim2 h h

If f(x) is differentiable x = a then rarr rarr

+ minus minus minus= =

minush 0 h 0

f (a h) f(a) f(a h) f(a)lim lim f (a)h h

+ minus minus= + =

f(a h) f(a h) 1lim f (a) f (a) f (a)2h 2

Thus rarr

+ minus minus=

f(a h) f(a h)lim f (a)2h

Remarks

f ( a g( h)) f ( a)lim f ( a)

g( h)rarr

+ minus= provided g(h) rarr 0 as h rarr 0

f ( a g( h)) f ( a ( h))lim f ( a)

g( h) ( h)φ

φrarr

+ minus +=

minus provided g(h) f (h) rarr as h rarr 0

2226 diffErEntiabiLity of paramEtric functions

Let the function y = f(x) be defined parametrically as x = f (t) and y = y(t) Then y = f(x) is differentiable at x = f(t) on at t

if + +rarr rarr

ψ minus minusψ ψ + minusψ =φ minus minusφ φ + minusφ h 0 h 0

(t h) (t) (t h) (t)lim lim

(t h) (t) (t h) (t) = a finite real number

Remarks (i) If x = f(t) is an increasing function of (t) then the above terms are LHD and RHD however

if x = f (t) is a decreasing function of t then the above term are RHD and LHD respectively

(ii) Alternatively we can eliminate the parameter lsquot and get y = f(x) and then we can investigate the differentiability at x

2227 rEpEatEdLy diffErEntiabLE functions

A function f(x) is said to be twice differentiable at x = a if f (x) is also differentiable at x = a ie

+ minus

f (a h) f (a)limh

exists finitely or x a

f (x) f (a)limx ararr

minusminus

exists finitely Similarly a functions f(x) is said to be

thrice differentiable at x = a if f (x) is differentiable at x = a ie rarr rarr

+ minus minusminush 0 x a

f (a h) f (a) f (x) f (a)lim or limh x a

exist finitely In general f(x) is said to be differentiable n-times at x = a if minus minus

+ minus(n 1) (n 1)

f (a h) f (a)lim or

minus minus

minusminus

(n 1) (n 1)

f (x) f (a)limx a

exists finitely

Remarks

1 Note that fn(x) stands for function f applied n-times whereas f(n)(x) stands for nth derivative of f(x)

2 If a function f(x) is such that derivative function f(x) is not defined at x = a then it is possible that f(x) is differentiable at x = a

eg If f(x) = (x)15 tan x then f(x) = x15 (sec2x) + (tan x) 4 51( x )

5minus

= x15 sec2 x + 4 5

1tan x

5( x )

Clearly f(x) is not defined at x = 0 but h 0 h 0

f (0 h) f (0 ) f ( h) (0 )lim lim

h hrarr rarr

+ minus minus=

= 15 15

h 0 h 0

h tan h 0 h tan hlim lim 01 0

h hrarr rarr

minus= = = f(0) = 0 ie f(a) is differentiable at x = 0

3 If limit of a derivative function exists and is equal to the value of derivative then the function is called continuously differentiable or f(x) is continuous ie continuity of derivative function

4 It may happen that a function f(x) is differentiable but not continuously differential

2228 functionaL Equation

An equation involving unknown functions is called a functional equation For example

(i) f(x) = f(ndashx) holds for every even function f(x) eg f(x) = x2 f(x) = |x| f(x) = cos x f(x) = sin2x etc (ii) f(ndashx) = ndashf(x) holds for every odd function eg f(x) = x3 f(x) = x|x| f(x) = sin x f(x) = tan3x etc (iii) fof(x) = x holds for every self invertible function eg f(x) = ndashx + k k isin ℝ fof(x) = f(f(x)) = ndash f(x) + k = ndash(ndashx + k) + k = x (iv) fog(x) = x holds when g(x) = fndash1(x) Q fog(x) = f(fmdash1(x)) [let f(y) = x rArr fndash1(x) = y] = f(y) = x eg f(x) = sin(sinndash1 x) = x forall x isin [ndash1 1] f(x) = exp (ln x) = x forall x gt 0

22281 Solution of a Functional Equation

By solution of a functional equation we mean to find a function satisfying the given functional equation Usually a given functional equation has more than one solution as is clear from illustrations Unique solution can exist when some additional conditions are given like continuity differentiability at a point values of functions at some particular points For example let the given functional equation be f(x + y) = f(x) + f(y) forall x y isin ℝ and f(x) is a differentiable function forall x isin ℝ and f(2) = 8

Now f (x) = rarr

+ minush 0

f(x h) f(x)h

lim h 0 h 0 h 0

f(x) f(h) f(x) f(h) f(h) f(0) f (0) k(say)h h h

lim lim limrarr rarr rarr

+ minus minus= = = = =

+ = + rArr = = rArr = rArr =

f(x y) f(x) f(y)for x y 0f(0) 2f(0)f(0) 0

f (x) = k rArr = +int intf (x)dx k dx C

rArr f(x) = kx + C now f(0) = 0 rArr C = 0rArr f(x) = kx (family of straight lines through origin) f(2) = 8 rArr f(2) = 2k = 8 rArr k = 4 f(x) = 4x Solution is f(x) = 4x

Some famous functional equations in two variable and their corresponding solutions (a) f(x + y) = f(x) + f(y) rArr f(x) = kx k isin ℝ (b) f(x + y) = f(x) f(y) rArr f(x) = 0 f(x) = akx a gt 0 ne 1 (c) f(xy) = f(x) + f(y) forall x y isin ℝ ~ 0 then f(x) = k loga |x| a gt 0 ne or f(x) = 0 (d) f(xy) = f(x) f(y) x gt 0 y gt 0 rArr f(x) = xn n isin ℝ (e) Jensenrsquos functional equation

x y f(x) f(y)f2 2

rArr f(x) = ax + b

(f) D Alambertrsquos functional equation f(x + y) + f(x ndash y) = 2f(x) f(y) rArr f(x) = 0 forall x or f(x) = cos kx or f(x) = cos h kx (cos hyperbolic kx) (g) (i) g(x + y) = g(x) f(y) + f(x) g(y) (ii) f(x + y) = f(x) f(y) ndash g(x) g(y) (iii) g(x ndash y) = g(x) f(y) ndash g(y) f(x) (iv) f(x ndash y) = f(x) f(y) + g(x) g(y)

These four functional equations represent the addition and subtraction theorem for the trigonometric functions f(x) = cos kx and g(x) = sin kx

Chapter 23Method of differentiation

231 Method of differentiation

2311 Derivatives Using First Principle (Ab-initio) Method

h 0 h 0

dy f(x h) f(x) f(x h) f(x)f (x) lim limdx h hrarr rarr

+ minus minus minus = = =

2312 Method of Using First Principle

Step I Let y = f(x) (i)

Step II Giving increment of δy and δx y + δy = f(x + δx) (ii)

Step III (ii) ndash (i) gives δy = f(x + δx) ndash f(x)

rArr dy f(x x) f(x)dx x

+ δ minus=

Step IV Taking limit as δx rarr 0 on both sides x 0 x 0

y f(x x) f(x)lim limx xδ rarr δ rarr

δ + δ minus=

rArr h 0

dy f(x h) f(x)lim f (x)dx hrarr

+ minus= =

232 algebra of differentiation

(i) d d d(cons tan t) 0 (kf(x)) k (f(x))k cons tan tdx dx dx

= = = =

(ii) d d d(u(x) v(x)) (u(x)) (v(x))

dx dx dx= plusmn = plusmn

(iii) Product rule d dv du(uv) u vdx dx dx

= = + where u and v are function of x

(iv) (Quotient Rule) 2

du dvv ud u dx dxdx v v

minus = =

where u and v are function of x

233 Chain rule

If lsquoyrsquo is a function of lsquoursquo and lsquoursquo is a function of lsquoxrsquo ie lets say y = f(u) and u = g(x) ie y = f(g(x)) then dy dy du f (u) g (x)dx du dx

= = times = f (g(x)) times g(x)

Remarks 1 It is important to realize that the cancellation is valid because the chain rule is incomplete in the sense

that it does not say clearly at what points to evaluate the derivatives We can add this information by writing

( )( ) ( )= = ==

= x a h h a x ag g h a

dy dy dg dhdx dg dh dx

While applying chain rule we work from the outside to inside

( )( )

( )( ) ( )( ) =

outter function derivative ofevaluated at evaluated at derivative ofoutter functioninner function inner function inner function

df g x f g x g x

2 ( )= = + +ωω ω ωd du dv d

uv ( v ) (u ) (uv )dx dx dx dx

same can be generalised to product of finite number

of function

3 Chain rule can be generalized to any finite number of function

eg ( )= = times timesω ω ω ωdu( v( ( x ) )) u( v( ( x )) v ( ( x )) ( x )

234 derivatives of soMe standard funCtions

2341 Algebraic Functions

1 n n 1d (x ) n x ndx

minus= isin

2 n n 1d (ax b) n(a)(ax b) ndx

minus= + = + isin

2342 Logarithmic and Exponential Functions

1 x xd (e ) edx

2 x xd (a ) a log adx

3 ed 1(log | x |)

dx x= a gt 0 4 a

d 1(log | x |)dx x log a

= a gt 0 a ne 1

2343 Trigonometric Functions

1 d (sin x) cos xdx

2 d (cos x) sin xdx

= minus

3 2d (tan x) sec xdx

4 (cot x) cosec xdx

= minus

5 d (sec x) sec x tan xdx

6 d (cosec x) cosec xcot xdx

= minus

Method of Differentiation 23275

2344 Inverse Circular Functions

d 1(sin x) 1 x 1 or |x| 1dx 1 x

minus = minus lt lt ltminus

d 1(cos x) 1 x 1 or |x| 1dx 1 x

minus minus= minus lt lt lt

d 1(tan x) x or x Rdx 1 x

minus = minusinfinlt ltinfin isin+

d 1(cot x) xdx 1 x

minus minus= minusinfinlt ltinfin

d 1(sec x) | x | 1 or x R [ 11]dx | x | x 1

minus += gt isin minus minus

d 1(cosec x) | x | 1 or x R [ 11]dx | x | x 1

minus minus= gt isin minus minus

235 differentiation of a funCtion With resPeCt to another funCtion

To find the derivative of f(x) wrt g(x) we first differentiate both wrt x and then divide the derivative of f(x)

wrt x by the derivative of g(x) wrt x ie if y = f(x) and u = g(x) there4 To find

d (f(x))d(f(x)) dxdd(g(x)) (g(x))

236 logarithMiC and eXPonential differentiation

Differentiation of function which are either product of a number of functions or are in the form (f(x))g(x) is usually done by application of logarithmsCase I y = u1u2u3u4un then ℓy = ℓnu1 + ℓnu2 + ℓnu3 + + ℓnun

rArr 1 2 n

du du du1 dy 1 1 1 y dx u dx u dx u dx

= + + + rArr 1 2 n

du du dudy 1 1 1y dx u dx u dx u dx

= + + +

Case II y = [f(x)]g(x) rArr ℓny = g(x) ℓn f(x)

rArr 1 dy g(x) f (x) g (x) n f(x)y dx f(x)

= + rArr dy g(x)y f (x) g (x) n f(x)dx f(x)

237 differentiation of inverse funCtion

If g(x) = fndash1(x) then 1d 1(f (x)) g (x)dx f (g(x))

minus= = =

2371 Rules of Higher Order Derivative

1 If k is a constant then ( )( )( )2

d k f xdx

= k ( )( )2

d f xdx

2 ( ) ( )( ) ( )( ) ( )( )2 2 2

d d df x g x f x g xdx dx dx

plusmn = plusmn

3 ( )2 2 2

d d v du dv d uuv u 2 vdx dx dx dx dx

= + times times + where u v are function of lsquoxrsquo

4 ( )3 3 2 2 3

3 3 2 2 3

d uv d v du d v dv d u d uu 3 3 vdx dx dx dx dx dx dx

= + times times + times times + times where u v are functions of lsquoxrsquo

5 If y = f(u) and u = g(x) then 22 2 2

d y d y du dy d udx du dx du dx

= + times

238 iMPliCit differentiation

Implicit functions are those in which y cannot be expressed exclusively in terms of x ie if the relation between the variables x and y are given by an equation containing both and this equation is not immediately solvable for y then y is called an implicit function of x For example y2 + x2 + 2xy - 3x2y = 0 or x2 y = sinxy etc

2381 Procedure to Find dydx for Implicit Function

(a) To get dydx

differentiate entire function with respect to x treating y as a function of x

(b) Collect the coefficient of dydx

at one place and transfer the remaining terms to the right hand side

(c) Find dydx

in terms of x and y

2382 Short cut Method to Find dydx for Implicit Functions

For implicit function put d f xf(x y)dx f y

minuspart part=part part

where fxpartpart

is partial differential of a given function with

respect to x (ie differentiating f with respect to x keeping y constant) and fypartpart

means partial differential

of a given function with respect to y (ie differentiating f with respect to y keeping x constant)

239 ParaMetriC differentiation

If y = f(t) x = g(t) then dy dy dt f (t)dx dx dt g (t)

= = and 2

d y d dy d dy dt d f (t) dtdx dx dx dt dx dx dt g (t) dx

= = times = times

there4 2

d y g (t)f (t) f (t)g (t)dx (g (t))

minus=

2310 deterMinant forMs of differentiation

(i) If f(x) g(x)

yu(x) v(x)

= rArr f (x) g (x) f(x) g(x)dyu(x) v(x) u (x) v (x)dx

(ii) If y is a function of x given in determinant form as f(x) g(x)

y f(x)v(x) u(x)g(x)u(x) v(x)

= = minus

rArr dy f(x) u (x) v(x)f (x) u(x)g (x) g(x)u (x)dx

prime prime prime prime= + minus minus = f (x) v(x) ndash u(x) g(x) + v(x) f(x0 ndash u(x) g(x)

=f (x) g (x) f(x) g(x)u(x) v(x) u (x) v (x)prime prime

+prime prime

Similarly y = u(x) v(x) w(x)p(x) q(x) r(x)

(x) (x) (x)λ micro γ then

u (x) v (x) w (x) u(x) v(x) w(x) u(x) v(x) w(x)

dy p(x) q(x) r(x) p (x) q (x) r (x) p(x) q(x) r(x)dx

(x) (x) (x) (x) (x) (x) (x) (x) (x)

prime prime primeprime prime prime= + +

prime prime primeλ micro γ λ micro γ λ micro γ

The differentiation can also be done column-wise

2311 leibnitzrsquos theoreM for the nth derivative of the ProduCt of tWo funCtions of X

Theorem If y = fg where f and g are functions of x having derivatives of nth order then n

nn r n r r

y C f gminus=

=sum ie yn = fn g + nC1 fn-1 g1 + nCn-2 fnndash2 g2 +helliphellip+ nCrfn-r gr +helliphellip+ nCn fgn where suffixes

denote order of derivatives with respect to x

2312 suCCessive differentiation

(a) If y = xm rArr y1 = mxm ndash1 rArr y2 = m(m ndash 1)xm ndash 2

rArr yn = m(m ndash 1)(m ndash 2) (m ndash n + 1)xm ndash n for n lt m rArr ym = m rArr ym + 1 = ym + 2 = = 0 (b) (af + bg)n = afn + bgn

where a b are constant and fn gn denotes nth derivatives of f and g respectively

2313 soMe standard substitution

In many functions direct differentiation becomes very tedious whereas some suitable substitution may reduce the calculation considerably Following are some substitutions which are useful in finding the derivatives

Expression substitution

a2 + x2 or 2 2a x+ x = a tanq where ndash p 2 lt q lt p2 or x = a cot q where 0 lt q lt p

a2 ndash x2 or 2 2a xminus x = a sinq where ndash p 2 le q le p2x = a cos q where 0 le q le p

x2 ndash a2 or 2 2x aminus x = a secq where q isin [0 p] ~ p2 x = a cosec q where 0 isin [- p2p2]

( )22ax x x a 1 cosminus = minus θ where 0 le q le p a x a xor a x a x+ minusminus +

x = a cos2q where 0 lt q le p2

2 2 2 2

a x a xor a x a xminus ++ minus

x2 = a2 cosq where 0 lt q le p2

1 Take care of the fact that substitution may sometimes violate the domain restrictions Therefore one need to be careful while applying these substitution

Chapter 24appliCation of

Derivatives

Rate of Change

241 Instantaneous rate of change of quantItIes

If y = f (x) is a differentiable function of x then dydx

is called the instantaneous rate of change of y with

respect to x

242 applIcatIon of rate of change of quantItIes

We will be given y = f(x) and dxdt and asked to find dydt at x = x0 We can find 0x

by the following

procedure Differentiating both sides wrt t we get dy dxf (x)dt dt

= ie 0

dy dxf (x )dt dt=

Example Rate of change of area of circle when rate of change of radius is known at the instant r = r0

2 dA drA r 2 rdT dt

= π rArr = π rArr 0

dA dr2 rdT dt=

Velocity of a Moving Body Given by dsvdt

= where s = displacement ie velocity is the time rate of

change of displacement of body

acceration Given by 2

dv d s dvf vdt dt ds

= = = ie acceralation is the time rate of change of velocity

243 errors and approXIMatIons

Let a function y = f(x) be defined and if Dx be the error occurred while calculating x then we may also get an error in calculation of y ie f(x) The correct value of y should have been y = (x + Dx) But the value that we have obtained because of the error in calculation of x will be y = f(x) Therefore f(x + Dx) ndash f(x) will be the error in calculation of y and is denoted Dy

Application of Derivatives 24279

2431 Types of Errors 1 absolute errors It is the deviation of measured value of a physical quantity from its actual

value ie error = Dy = f(x + Dx) ndash f(x)

2 Relative errors It is the ratio of error to the total quantity measured eg yyδ where dy is absolute

error and y is actual value

3 Percentage errors It is given by relative error times 100 ie y 100yδ

4 Maximum probable error It is the error encountered in the final measured quantity assuming that all the errors occurring in the measurement of component quantities have same sign ie cumulative in nature eg if z = f(x) + f(y) then maximum probable error in z = |error in f(x)| + |error in f(y)|

244 calculatIon of dy correspondIng to dX

y dylimx dxδ rarr

δ rArr for small values of δx and δy dyy xdx

δ = timesδ

tangent and noRMal

PT rarr Tangent to curve y = f(x) at point P(x1 y1)PN rarr Normal to curve y = f(x) at point P(x1 y1)TM rarr Sub-tangent to curve y = f(x) at point P(x1 y1)

(projection of tangent on x-axis)MN rarr Sub-normal to curve y = f(x) at point P(x1 y1)

(Projection of normal on x-axis)

Length of tangent at point P(x1 y1) =

1(x y )

dxy 1dy

Length of normal at point P(x1 y1) = 2

1(x y )

dyy 1dx

Length of sub-tangent at point P(x1 y1) = 1 1

1(x y )

dxy dy

Length of sub-normal at point P(x1 y1) = 1 1

1(x y )

dyy dx

Slope of tangent at point P(x1 y1) = 1 1(x y )

Slope of normal at point P(x1 y1) =

1 1(x y )

A P(x1y1)

Normal

Sub-normalSub-tangent

Tangent

NMT 0 X90degndashθθ

Equation of tangent at point P(x1 y1) is given by 1 1

1 1(x y )

dy(y y ) (x x )dx

minus = minus

Equation of normal at point P(x1 y1) is given by

(x y )

1(y y ) (x x )dydx

minus = minus minus

Remarks

(i) Tangent parallel to x-axis rArr =dy

(ie horizontal tangent)

(ii) Tangent parallel to y-axis rArr dy dxor 0

dx dyrarrinfin = (ie vertical tangent)

(iii) Two curves y = f1(x) and y = f2(x) touch each other at point (x1 y1) iff 1 1( x y )

for f1(x) and that for

f2(x) are equal

Method (a) Find point of intersection P(x1 y1)

(b) Find 1

at (x1 y1) for curve y = f1(x) and y = f2(x) and show that m1 = m2

(iv) The basic property of a tangent line is that it indicates the direction of a curve at a point

(v) If the tangent at any point on the curve is equally inclined to both the axes then dydx

= plusmn1

(vi) For finding the intercepts on the axes by a tangent the write equation of tangent in intercept form

ie x y1

a b+ = Example intercept on x-axis = a and intercept on y-axis = b

(vii) line ax + by + c = 0 will be tangent to a curve y = f(x) at (x1 y1) if ax + by + c = 0 and

1 1( x y )

dy( y y ) ( x x )

dx minus = minus

are identical ie

1 11 1

1 1( x y ) ( x y )

a b cdy 1 dy

y xdx dx

= =minus minus

245 tangents froM an eXternal poInt

Given a point P(a b) which does not lie on the curve y = f(x) then the equa-tion of possible tangents to the curve y = f(x) passing through (a b) can be found by first finding the point of contact Q of the tangent with the curve

Let point Q be (x1 y1)Since Q lies on the curve we have y1 = f(x1) hellip(1)

Also the slope of PQ = the slop of the tangent at the point Q on the

curve y = f(x) = 1 1(x y )

Slope of PQ 1 1

(x y )1

y b dyx a dxminus

equiv =minus

hellip(2)

Solving (1) (2) we can get the point of contact (x1 y1)

246 tangentsnorMals to second degree

1 To find the equation of tangent at (x1 y1) substitute xx1 for x2 yy1 for y2 1x x2+

for x 1y y2+ for

y and 1 1xy x y2+

for xy and keep the constant as such This method is applicable only for second

degree conics ie ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 2 Easy method to find normal at (x1 y1) of second degree conics ax2 + 2hxy + by2 + 2gx +

2fy + c = 0 (i)

then according to determinant a h gh b fg f c

Write first two rows ax + hy + g and hx + by + f then

equation of normal at (x1 y1) of (i) is 1 1

1 1 1 1

x x y yax hy g hx by f

minus minus=

+ + + +

2461 Tangent to Parametric Functions

Given the equation of the curve x = f(t) and y = g(t) then ( )( )

dyg tdy dt

dxdx f tdt

The equation of tangent at any point lsquotrsquo on the curve is given by ( ) ( )( ) ( )( )g t

y g t x f tf t

minus = minus

The equation of normal at point lsquotrsquo is given by y ndash g(t) = ( )( ) ( )( )f t

x f tg t

Some common parametric coordinate on a curve are as follows

(a) For x2 + y2 = a2 x = a cos q y = a sinq

(b) For x2 ndash y2 = a2 x = a sec q y = a tanq

(c) For 2 2

x y 1a b

minus = x = a cosq y = b sinq

(d) For y2 = 4ax x = at2 y = 2at

(e) For 2 2

x y 1a b

minus = x = a secq y = btanq

(f) For x23 + y23 = a23 x = a cos3q y = asin3q

(g) For x y a+ = x = a cos4q y = asin4q

(h) For n n

x y 1a b

+ = x = a (cosq)2n and y = b(sinq)

(i) For n n

x y 1a b

minus = x = a (secq)2n and y = b(tanq)2n

(j) For c2 (x2+y2) = x2y2 rArr x = c sec q and y = c cosec q (k) For ay2 = x3 rArr x = at2 and y = at3

247 tangent at orIgIn

If a rational integral algebraic equation of a curve is passing through the origin then the equations of the tangent at the origin is obtained by equating the lowest terms in the equation of curve to be equal to zero

248 angles of IntersectIon of two curves

The angle of intersection of two curves is defined as the angle between the tangents to the two curves at thier common point of intersection

2481 Algorithm to Find Angle of Intersection

Step 1 Solve both the curves to get pointpoints of intersection P

Step 2 Find slope of tangents at P ie m1 =1dy

dx tana and m2 = 2dy tan

dx= β

Step 3 b+ q = a rArr q = a ndash b

( ) 1 2

m mtan tantan tan tan1 tan tan 1 m m

minusα minus βθ = α minusβ = rArr θ =

+ α β +

(i) Two curve are said to be orthogonal (q = 90deg) at a point P if m1m2 = ndash1

(ii) Two curves touch each other (q = 0deg) at P if m1 = m2

249 orthogonal curves

If the angle of intersection of two curves is right angle then the two curves are said to be intersecting orthogonally and such curves are called orthogonal curves For example y = mx and x2 + y2 = r2 are two orthogonal curves for any value of m and r If the curves are orthogonal then angle of intersection q = p2

rArr 1 2

dy dy1 0dx dx

rArr 1 2C C

dy dy 1dx dx

= minus

2410 coMMon tangent

Common tangent to two curves are of two types (i) Tangent common to two curves with same

point of contact (ii) Tangent common to two curves with different

point of contacts

Here 1 2

1 1 2 2

c c2 1

(x y ) (x y ) 2 1

y ydy dydx dx x x

minus = = minus

2411 shortest dIstance between two non-IntersectIng curves

The shortest distance between two non-intersecting curves is found along the common normal to the two curves In fact if the two curves also have the largest distance between them then it is also found along the common normal to the two curves This can be established with the help the concept of maximum minima

In the figure we notice that the shortest distance between the curves is AB and the largest distance between them is PQ both of which are found along a common normal Note that the common normal may be different in two cases

MonotoniCity

lsquoIt is study of increasing decreasingconstant behavior of function over an interval as we travel form left to right along its graphrsquo

For example the function shown in the figure is decreasing forall x isin (ndashinfin a) and increasing forall x isin (a b) Again decreasing forall x isin (b c) and remains con-stant over the interval (c infin)

24111 Strictly Increasing Function at a Point x = a

If f(a ndash h) lt f(a) lt f(a + h) h rarr 0+ Strictly decreasing function at a point x = aIf f(a ndash h) gt f(a) gt f(a + h) h rarr 0+ Non decreasing functionIf f(a ndash h) le f(a) le f(a + h) h rarr 0+ Non increasing functionIf f(a ndash h) ge f(a) ge f(a + h) h rarr 0+

Conditions for monotonicity of differentiable function at a point x = a

(i) If f (a) gt 0 then f(x) is stictly increasing at x = a (ii) If f (a) lt 0 then f(x) is strictly decreasing at x = a (iii) If f (a) = 0 then

Case 1 If f (a ndash h) gt 0 and f (a + h) gt 0 then f(x) is strictly increasing at x = a

Case 2 If f (a ndash h) lt 0 and f (a + h) lt 0 then f(x) is strictly decreasing at x = a

Case 3 If f (a ndash h) and f (a + h) are of opposite signs then f(x) is neither increasing nor decreasing at x = a ie f(x) is a critical point

Case 4 If f (a ndash h) and f (a + h) = 0 then f(x) is constant function

2412 MonotonIcIty of a functIon on an Interval

(i) f(x) is said to be strictly increasing on open interval (a b) if f(x) is strictly increasing at each x isin (a b)

(ii) f(x) is said to be strictly decreasing on open interval (a b) if f(x) is strictly decreasing at each x isin (a b)

(iii) f(x) is said to be strictly increasing on close interval [a b] if (a) f(x) is strictly increasing on (a b) (b) f(a) lt f(a + h) h rarr 0+

(c) f(b ndash h) lt f(b) h rarr 0+

(iv) f(x) is said to be strictly decreasing on close interval [a b] if (a) f(x) is strictly decreasing on (a b) (b) f(a) gt f(a + h) h rarr 0+

(c) f(b ndash h) gt f(b) h rarr 0+

2413 condItIon for MonotonIcIty of dIfferentIable functIons on an Interval

(i) If f (x) ge 0 forall x isin (a b) and f (x) = 0 at isolated point then f(x) is strictly increasing on (a b) (ii) If f (x) le 0 forall x isin (a b) and f (x) = 0 at isolated point then f(x) is strictly decreasing on (a b) (iii) If f (x) ge 0 forall x isin (a b) f (a+) ge 0 f (bndash) ge 0 and f (x) = 0 at isolated point then f(x) is strictly

increasing on [a b] (iv) If f (x) le 0 forall x isin (a b) f (a) le 0 f (b) le 0 and f (x) = 0 at isolated point then f(x) is strictly

decreasing on [a b] (v) If f (x) ge 0 and f (x) = 0 in any subinterval of (a b) or [a b] then f(x) is non-decreasing function (vi) If f (x) le 0 and f (x) = 0 in any subinterval of (a b) or [a b] then f(x) is non-increasing function

2414 MonotonIcIty of functIon on Its doMaIn

(i) f(x) is said to be strictly increasing in its domain Df if f(x2) gt f(x1) forall x1 x2 isin Df and x2 gt x1 (ii) f(x) is said to be strictly decreasing in its domain Df if f(x2) lt f(x1) forall x1 x2 isin Df and x2 gt x1 (iii) f(x) is said to be strictly non-decreasing in its domain Df if f(x2) ge f(x1) forall x1 x2 isin Df and x2 gt x1 (iv) f(x) is said to be strictly non-increasing in its domain Df if f(x2) le f(x1) forall x1 x2 isin Df and x2 gt x1

2415 doMaIn of MonotonIcIty of a functIon

(i) Set of points of domain of function in which the function is monotonically increasing is called domain of monotonic increasing (Interval of monotonic increasing if it is an interval)

(ii) Set of points of domain of function in which the function is monotonically decreasing is called domain of monotonic decreasing (Interval of monotonic decreasing if it is an interval)

2416 crItIcal poInt

The points at which f (x) = 0 or f (x) does not exist are called critical point The point where f (x) = 0 are called stationary points

Remarks

(i) At critical points function may change its monotonicity but it is not always the case

(ii) If x = a is a critical point of f(x) then it is also for the function g(x) = f(x) + k k = constant and x = a + k is critical point for g(x) = f(x + k)

(iii) For a function discontinuous at x = a derivative test does not work for such function monotonicity is tested by using the basic definition of monotonic function at a point

(iv) If a function is strictly monotonic then it may be discontinuous

(v) For a discontinuous function if f(x) gt 0 forall x isin Df then it is not necessary that function is increasing

(vi) If f(x) is discontinuous then f(x) lt 0 forall x isin Df ⇏ f(x) is decreasing

(vii) For a continuous function on ℝ if f(a) = 0 f(a) ne 0 or f(a) does not exist due to sharp point then x = a is a critical point

Case 1 If f (x) lt 0 forall x isin (ndashinfin a) and f(x) gt 0 forall x isin (a infin) then (ndashinfin a] is the interval of monotonic decreasing of f(x) and [a infin) is the interval of monotonic increasing of f(x)

Case 2 If f lsquo(x) gt 0 forall x isin (ndashinfin a) and f(x) lt 0 forall x isin (a infin) then (ndash infin a] is the interval of monotonic increasing of f(x) and [a infin) is the interval of monotonic decreasing of f(x)

(viii) If f (a) = 0 and f (a) = 0 then x = a is called point of inflexion and is not a critical point In such case monotonicity of f(x) in left neighbourhood (a ndash h a) and right neighbourhood (a + h a) remain same

(ix) If f(x) is a continuous function then its intervals of monotonicity can be obtained by first finding the critical point ie point where f (x) = 0 or f (x) does not exist (f (x) ne infin) and then analysing the behaviour of f (x) to be positive or negative in the neighbourhood of critical point

(x) If f(x) is continuous function and f(a) = plusmninfin then f(x) has same monotonicity in left and right neighbourhoods of x = a

If f(a ndashh) gt 0 and frsquo(a + h) gt 0 h rarr 0+ then f(x) is monotonically increasing at x = a and f(x) is strictly increasing in (a ndash h a + h) h rarr 0+

Simillary If frsquo(a ndashh) lt 0 and frsquo(a + h) lt 0 h rarr 0+ then f(x) is monotonically decreasing at x = a and f(x) is strictly decreasing in (a ndash h a + h) h rarr 0+

(xi) While presenting the answer for intervals of monotonic increasedecrease do not use union symbols without taking adequate care

∵ It may happen that f(x) decreases in two intervals but fail to behave so in their union

Consider f(x) = x2 endashx Here f(x) darr for (ndashinfin 0) and also for (2 infin)

But as is evident from the graph of the function x1 lt x2 rArr f(x) gt f(x2) but x2 lt x3 ⇏ f(x2) gt f(x3)

Rather x2 lt x3 rArr f(x2) lt f(x3)

f(x) is not decreasing on (ndashinfin 0) cup (2 infin)

(xii) (Although when f (x) is discontinuous then this may happen that if f(x) increases in [a b] and [c d] both so it is also increases in [a b] cup [c d]

For instance see the graph function y = f(x)

Here the function increases in the intervals (a b) (c d) and we may proceed to write that it in (a b) cup (c d)

∵ Here we have f(b) ge f(c)

2417 Intervals of MonotonIcIty for dIscontInuous functIon

1 If f(x) is uarr in [a b] decreases in [c d] again uarr in [d e] Then f(x) increases in [a b] cup [d e] is true iff maximum value forall x isin [a b] should be less than min value of f(x) forall x isin [d e] ie f(b) lt f(d)

For example consider f(x) as shown in the figure 2 If f(x) is uarr in [a b] in [c d] and again uarr in [d e] then f(x) is decreasing

in [a b] cup [d e] iff minimum value of f(x) forall x isin [a b] is less then the maximum value of f(x) forall x isin [d e] ie f(b) gt f(d)

3 Conventionally interval of monotonicity is expressed using open interval but ideally use of closed interval is more informative particularly for discontinuous functions

4 For continuous functions (defined over closed interval) the open intervals of monotonicity can be replaced by closed interval

2418 propertIes of MonotonIc functIon

P1 By application of increasing (uarr) function the sign of inequality does not change But the sign of inequality reverse on the application of a decreasing (darr) function To explain this if a le x le b

f(a) f(x) f(b) if f isf(a) f(x) f(b) if f is

le le uarr

ge ge darr

P2 If f(x) is continuous and increasing function for all x isin[a b] then Rf [f(a) f(b)] P3 If f(x) is continuous and decreasing forall x isin Df[a b] then Rf [f(b) f(a)] P4 If f is increasing x isin [a α] and f darr x isin (a b] and f(x) is continuous then Rf [min f(a) f(b) f(a)] P5 If f(x) is monotonically decreasing function forall x isin [a α) and increasing function forall x isin (α b] and

is continuous in [a b] then Rf [f(α) max f(a) f(b)]

P6 If f(x) is monotonically increasing then kf(x) is when k 0kf(x) is when k 0

uarr gt

darr lt

P7 If f and g are both increasing function then (f(x) + g(x)) is increasing Converse is not true P8 If f and g are both decreasing function then (f(x) + g(x)) is decreasing Converse is not true P9 If f is increasing and g is decreasing function then (f(x) ndash g(x)) is increasing P10 If f is decreasing and g is increasing function then (f(x) ndash g(x)) is decreasing

P11 f(x) and g(x) 0 and both

Iff(x) and g(x) 0 and both

gt uarr

lt darr rArr y = f(x)g(x) uarr Both converse is not true

P12 If f is rArr 1f is decreasing function wherever defined

P13 If f and g gt 0 and f is increasing and g is decreasing f(x) isg(x)

rArr uarr

P14 Composition of two monotonically increasing function is always an uarr function P15 Composition of two monotonically decreasing function is always an uarr function P16 When f and g have opposite monotonicity then f(g(x)) is a decreasing function P17 (a) If f(x) is strictly uarr in [a b] and g(x) is strictly uarr in [f(a) f(b)] then gof is strictly uarr in [a b] (b) If f is strictly decreasing in [a b] and g is strictly decreasing in [f(b) f(a)] then gof is strictly

increasing uarr for all x (c) If f is strictly uarr in [a b] and g is strictly decreasing in [f(a) f(b)] then gof is strictly decreasing in

[a b] (d) If f is strictly decreasing in [a b] and g is strictly increasing in [f(x) f(b)] then gof is strictly de-

creasing in [a b] (e) f and fndash1 have same monotonic nature ie either both are increasing or both are decreasing

table representing monotonicity of f(x) g(x) and functions obtained from f(x) and g(x)

Increasing (uarr) Decreasing (darr) Neither Increasing Nor Decreasing (X)f(x) uarr darr uarr darrg(x) uarr uarr darr darrndashf(x) darr uarr darr uarrndashg(x) darr darr uarr uarrf(x) + g(x) uarr uarr or darr or X uarr or darr or X darrf(x) ndash g(x) uarr or darr or X darr uarr uarr or darr or Xf(x) g(x) uarr uarr or darr or X uarr or darr or X darrf(x)g(x) uarr or darr or X darr uarr uarr or darr or X1f(x) darr uarr darr uarr1g(x) darr darr uarr uarr(fog)x uarr darr darr uarr

2419 applIcatIon of MonotonIcIty

1 In order to prove that a function f(x) ge k for all x ge x0 it is sufficient to prove that f(x0) ge k and fprime(x) ge 0 forall x ge x0

2 In order to prove some inequalities any of the following two methods can be conveniently adopted

Method 1 Rearrange the terms so that LHS and RHS become the value of a function f(x) at two different inputs a b

ie inequality becomes the type f(α) gt f(b) (say) then it is sufficient to prove that f (x) 0 iff (x) 0 ifprime gt α gtβ

prime lt α ltβ

Method ii To prove f (x) ge g (x) for all x ge a Consider the function h (x) = f (x) ndash g (x) hprime(x) = f prime(x) ndash g prime(x) Test the monotonicity of h(x) If hrsquo(x) gt 0 forall x ge a and h(a) ge 0 then h(x) ge h(a) ge 0 rArr h(x) ge 0 forall x ge a rArr f(x) ndash g(x) ge 0 forall x ge a rArr f(x) ge g(x) forall x ge a

CuRVatuRe of a funCtion

The rate of bending of curves at a point is known as curvature of the curve at that point

Curvature of f(x) at P = ddsφ

Curvature of a circle of radius lsquorrsquo = 1r

Radius of curvature (P) =

dy1dx1 dsd ycurvature ddx

δφδs

φ+δφφ

Clearly P gt 0 if 2

d y 10 0dx

gt rArr gtρ

and P lt 0 if 2

d y 10 0dx

lt rArr ltρ

That is curvature of a curve at a point is positive if the point is situated on concave upwards parts and is negative if the point is situated on concave downwards parts of a curve or in other wards if the curve bends upon its tangents then curvature is positive and if it bends below its tangnet then the curvature is negative

Positive curvature at P

Negative curvature at P

Remarks (i) f and fndash1 have same monotonic nature but is not same for thier curvature

(ii) f is uarr then f and fndash1 have same sign of cauvature

(iii) if f is darr then f and fndash1 have opposite sign to curvature

2420 hyper crItIcal poInt

A hyper critical point or cirtical point of second kind or second order critical point are those values x for which f (x) = 0 or f (x) does not exist

2421 poInts of InfleXIon

The point of inflexion is a point which separates the convex portion of the curve from its concave portion

Remarks 1 At the point of inflexion tangent (if exist) cuts the curve

2 Continuous function lsquof rsquo need not have an inflection point at all the points satisfying f(x) = 0 If f(x) = x4 we have f(0) = 0 but the graph of f is always concave up and hence there is no point of inflection

Let us take the function y = x5 ndash 5x4 Here y = 20x2 (x ndash 3)

Now y= 0 for x = ndash3 the second derivative changes sign and thus x = 3 is a point of inflection But when x passes through the point x = 0 the second derivative retains constant sign and therefore the origin is not a point of inflection (since the graph of the given function is concave up on both sides of the origin)

3 If x = c is a point of inflection of a curve y = f(x) and at this point there exists the second derivative f(c) then f(c) is necessarily equal to zero (f(c) = 0)

4 The point (1 0) in y = (x ndash 1)3 being both a critical point and a point of inflection is a point of horizontal inflection (Q the tangent at (1 0) on y is parallel to x-axis)

5 If a function f is such that the derivative f is continuous at x = c and f(c) = 0 while f(c) ne 0 then the curve y = f(x) has a point of inflection for x = c

6 It should be noted that a point separating a concave up arc of a curve from a concave down arc may be such that the tangent at that point is perpendicular to the x-axis ie vertical tangent or such that the tangent does not exist

This can be demonstrated easily by the behavior of the graph of the

function = 3 x in the vicinity of the origin In such a case we speak of a point of inflection with vertical tangent

7 A number c such that f(c) is not defined and the concavity of f changes at c will correspond to an inflection point if and only if f(c) is defined In other words for a point lsquocrsquo to be a point of inflection f(x) must be defined at x = c even if f(x) is not defined at x = c

2422 Method to fInd the poInts of InfleXIon of the curve y = f(X)

Step 1 Find 2

and find all possible x where 2

d y 0dx

= (say a b hellip) or where 2

does not exist

(Say a b hellip)

Step 2 Locate them on real number line and find the sign scheme for 2

Step 3 The point x = a is a point of inflexion if 2

changes it sign at x = a

2423 type of MonotonIc functIon

(i) Monotonically increasing function with increasing rate of increase

(a) dy 0dx

gt (b) 2

d y 0dx

(ii) Monotonically increasing function with decreasing rate of increase

(a) dy 0dx

gt (b) 2

d y 0dx

(iii) Monotonically decreasing function with decreasing rate of decrease

(a) dy 0dx

lt (b) 2

d y 0dx

(iv) Monotonically decreasing function with increasing rate of decrease

(a) dy 0dx

lt (b) 2

d y 0dx

(v) Monotonically increasing function with constant rate of increase

(a) dy 0dx

gt (b) 2

d y 0dx

(vi) Monotonically decreasing function with constant rate of decrease

(a) dy 0dx

lt (b) 2

d y 0dx

RolleS and Mean Value theoReM

2424 rollersquos theoreM

Let a function f(x) defined on [a b] such that It is continuous in the interval [a b] It is differentiable in the interval (a b) and satisfies f(a) = f(b) then there exist at least one c isin (a b)

where f(c) = 0ConclusionThere is atleast one point lying between A and B the tangent at which is parallel to x-axis

Remarks

Rollersquos theorem fails for the function which does not satisfy at least one of the three conditions

The converse of Rollersquos theorem may not be true ie f(c) may be zero at a point in (a b) without

satisfying all the three conditions

Case I Case II Case III

a0 c b

f(a)nef(b)

f(c)=0

Rollersquos theorem algerbraically states that between any two consecutive roots of a polynomial there exist at least one root of f(x) = 0

2425 applIcatIon of rollersquos theoreM

If f(x) is a polynomial function then as we already know that all polynomial functions are continuous and differentiable in their domain thereby the following deduction can be made

1 If all the roots of f(x) = 0 are real then all the roots of f (x) = 0 are also real and the roots of f (x) = 0 separate the roots of f(x) = 0

Here a b c d e are the 5 roots of f(x) = 0 and a b g d are the 4 roots of f (x) = 0 2 If f(x) is of degree lsquonrsquo then f (x) is of degree lsquon ndash 1rsquo and a root of f (x) = 0 exists in each of them n ndash 1

interval between the n roots of f(x) = 0 and in such a case the root of f (x) = 0 f (x) = 0hellip are also real and the roots of any one of these equations separate those of the preceding equation

3 Not more than one root of f(x) = 0 can lie between two consecutive roots of f (x) = 0 4 If f (x) = 0 has n real roots then f(x) = 0 cannot have more than (n + 1) real roots 5 If f(n) (x) is the nth derivative of f(x) and the equation f(n) (x) = 0 has some imaginary roots then

f(x) = 0 has atleast as many imaginary roots 6 If all the real roots a b g d of f (x) = 0 are known we can find the number of real roots of

f(x) = 0 by considering the signs of f(a) f(b)hellip A single root of f(x) = 0 or no root lies between a and b according as f(a) and f(b) have opposite

signs or the same sign

2426 lagrangersquos Mean value theoreM

If a function f(x) defined on [a b] such that it is

bull Continuous over the interval [a b]

bull Differentiable in the interval (a b) then $ at least one cisin (a b) where f(b) f(a)f (c)b aminusprime =minus

ie where slope of tangent becomes equal to slope of the chord AB

Remarks

Rollersquos theorem is a special case of LMVT since f(a) = f(b) rArr minusprime =minus

f ( b) f ( a)f ( c )

b a rArr 0

f ( c ) 0b a

prime = =minus

Lagrangersquos mean value theorem fails for the function which does not satisfy atleast one of the two conditions

The function is discontinuous at x = x1

The function is non-differentiable at x = x1

The converse of LMVT may not be true ie f(x) may be equal to f ( b) f ( a)b aminusminus

at a point c in (a b)

without satisfying both the conditions of LMVT

2427 alternatIve forM of lMvt

If a function f(x) is continuous in a closed interval [a a + h] and derivable in the open interval (a a + h) then there exists at least one number lsquoqrsquoisin (0 1) such that f (a + h) = f(a) + h f (a + q h)

MaxiMa and MiniMa

2428 local MaXIMa

A function f(x) is said to have a local maxima at x = a if f(a) is greater than every other value assumed by f(x) in the immediate neighbourhood of x = a

f(a) f(a h)f(a) f(a h)

ge +ge minus

for a sufficiently small positive h

24281 Local Minima

A function f(x) is said to have a minima at x = b if f(b) f(b h)f(b) f(b h)

le + le minus

Remarks

The term lsquoextremumrsquo or lsquoextremalrsquo or lsquoturning valuersquo is used both for maximumminimum value

The above definition is applicable to all functions continuous or discontinuous differentiable or non-differentiable at x = a

If the graph of a function f attains a local maximum at the point (a f(a)) then x = a is called the point of local maximum and f(a) is called the local maximum value A similar terminology is used for local minimum

A function can have several local maximum and minimum values

If a function is strictly increasing or strictly decreasing at an interior point x = a it cannot have an extremum at x = a and vice versa

A local maximum (local minimum) value of a function may not be the greatest (least) value in a finite interval A local minimum value may be greater than a local maximum value

For a continuous function there must exist one local minima between any two local maxima and vice-versa

However this may or may not be the case for discontinuous functions

necessary and sufficient condition for local maxima and local minima (for differentiable function) (a) for local Maxima f (x0) = 0 and f (x) changes its sign from positive to negative as we go from left to right

crossing x0 ie f (x0 ndash h) gt 0 f (x0 ) = 0 and f (x0 + h) lt 0 (b) for local Minima f (x0) = 0 and f (x) changes its sign from negative to positive as we go from left to right crossing x0

ie f (x0 ndash h) lt 0 f (x0 ) = 0 and f (x0 + h) lt 0

Remarks

If =dy

at x = x0 for a differential function y = f(x) and sign of f (x) does not changes as we move from left

to right crossing x = x0 then x = x0 is a point of inflection

ie neither point of maxima nor point of minima

necessary and sufficient condition for local maxima and local minima (for continuous and non-differentiable function)

(a) for local Maxima Let f(x) be non-differentiable at x = x0 then x = x0 will be the point of local

maxima iff (i) f (x) ge 0 forall x isin (x0ndashhx0) (ii) f (x) le 0 forall x isin (x0 x0+ h)

(b) for local Minima Let f(x) be non-differentiable at x = x0 then x = x0 will be the point of local

minima iff (i) f (x) le 0 forall x isin (x0 ndash h x0) h gt 0 and h rarr 0 (ii) f (x) ge 0 forall x isin (x0 x0 + h) h gt 0 and h rarr 0

2429 fIrst derIvatIve test (for contInous functIons)

Step 1 Find dydx

and find critical points ie points where dy dy0 ordx dx

= does not exist (sharp turn)

Step 2 If x = x0 is a critical points then it will be a point of local minima if f (x0 ndash h) lt 0 and f (x0 + h) gt 0 h gt 0 h rarr 0 and it will be a point of local maxima if f (x0 ndash h) gt 0 and f (x0 + h) lt 0 h lt 0 h rarr 0

f(x0) does not exist

x = x0 a point of local minima

x0+h x0+h

f(x0+h)gt0f(x0+h)gt0f(x0ndashh)lt0 f(x0ndashh)lt0

x0ndashh x0ndashh X0

Continous functiondifferentiable at x = x0

Continous functionnon-differentiable at x = x0

f(x0) =0 f(x0) does not exist

x=x0 a point of local maxima

f(x0+h) lt0f(x0+h) lt0

f(x0ndashh) gt0

x0ndashh X0

RemarkIf f(x) does not change its sign while crossing x0 ie f(x0 ndash h) f(x0 + h) gt 0 then x = x0 while a point of neither maxima nor minima

2430 poInt of InflectIon and saddle poInt

Point seperating the concave and convex part of function is called a point of inflection

Point of inflection where f (x0) = 0 is called saddle point represented in figure

2431 global or absolute MaXIMa and MInIMa

Let y = f(x) be a given function in an interval [a b] and a1 a2 a3 a4 be the critical points and f(a1) f(a2) f(a3)hellip f(an) be the values of the function at critical points The greatestlargestglobal maximumabsolute maximum values of a function in a closed interval [a b] is given by M = maxf(a) f(a1) f(a2) f(a3)hellip f(an) f(b) and the leastsmallestGlobal minimumabsolute minimum of the function f(x) in [a b] is given by m = min f(a) f(a1) f(a2) f(a3)hellip f(an) f(b) Let y = f(x) be a given function in an interval (a b)

The greatestlargestglobal maximumabsolute maximum values of a function in a closed interval [a b] is given by M = max f(a+) f(a1) f(a2) f(a3) f(an) f(bndash) the leastsmallestglobal minimumabsolute minimum of the function f(x) in [a b] is given by m = max f(a+) f(a1) f(a2) f(a3) f(an) f(bndash)

2432 algebra of global eXtreMa

i If y = f(x) has a local maximum at x = a then y = ndashf(x) has a local minimum at x = a and vice-versa

ii If f and g are non-negative function which attain their greatest (least) values at x = a then y = f(x) g(x) also attains its greatest (least) values at x = a

iii If f is such that f(x) is maximum (minimum) at x = a provided f(a) ne 0 then k

f(x)is minimum

(maximum) at x = a (where k is a positive constant) and if k is a negative constant then k

f(x)is

maximumminimum at the point x = a where f(x) is maximumminimum (provide f(a) ne 0) ie

k f(x) at x = a kf(x) at x = aPositive maximum minimum

minimum maximumNegative maximum maximum

minimum minimum

iV If f is non-negative and g is positive so that f attains its greatest (least) value at x = a and g attains its

least (greatest) value at x = a then f(x)yg(x)

= attains its greatest (least) value at x = a

V If f(x) is continuous on [a b] and g(x) is continuous on [m M] where m and M are the absolute minimum and the absolute maximum of f on [a b] then max gof = max g(x)

x isin [a b] x isin [m M] and min gof = min g(x) x isin [a b] x isin [m M]

2433 evenodd functIons

(i) An even function has an extremum at x = 0 provided it is defined in the immediate neighbourhood of x = 0

(ii) If an even function f has a local maximum (minimum) at x = a then it also has a local maximum

(minimum) at x = ndasha (iii) If an odd function f has a local maximum (minimum) at x = a then it has a local minimum

(maximum) at x = ndasha

2434 MIscellaneous Method

Many problems of maximaminimarange can be solved using elementary methods and without using calculus It is essential for students to know these methods as it may reduce the calculations and hence speedup your solution

For an example it is obvious that if f(x) = 2

| x |1 x+

then 1 1f(x) f(1)1 2xx

is the only maximum

value of f which is achieved when x = plusmn1It is to be noticed that some important problems of maxima and minima can be solved by elementary

algebraical methods without recourse to calculus

2435 secondhIgher order derIvatIve test

Step i Find the derivative of the function and find the root of fprime(x) = 0 (Say x = x0 x1 x2 hellip)

Step ii Now find fprimeprime(x) at x = x0 then the following cases may arise If f primeprime(x0)lt 0 then f (x) is maximum at x = x0 If f primeprime(x0) gt 0 then f (x) is minimum at x = x0 If f primeprime(x0) = 0 then the second derivative test fails to conclude

Step iii Now find f primeprimeprime(x) at x = x0 and the following two cases may arise If f primeprimeprime(x0) ne 0 then f (x) has neither maximum nor minimum (inflexion point) at x = x0 But if f primeprimeprime(x0) = 0 then go for the next higher derivative test

Step iV Find fiv (x0) and analyzing the following cases If fiv (x0) = 0 then similar analysis of higher derivative continues If fiv (x0) = positive then f (x) is minimum at x = x0 If fiv (x0) = negative then f (x) is maximum at x = x0

In general let f prime(x0) = fprimeprime(x0) = = f(nndash1)(x0) = 0 f(n)(x0) ne 0 If n is odd then there is neither maximum nor minimum at x = x0 and if n is even then f (n)(x0) gt 0rArr min at x0 and f (n) (x0) lt 0 rArr max at x0

2436 fIrst derIvatIve test for paraMetrIc functIons

Assume that the function is continuous the following steps should be followed

1 Find the critical points dy dy dtdx dx dt

2 Find values of t where dydx is zero or does not exist 3 Find the sign scheme of dydx on the number line of t 4 Now convert the sign scheme of dydx on the number line of x 5 If x =x(t) is a strictly increasing function t then the sign scheme in x is the same as the sign scheme in t 6 If x = x (t) is a strictly decreasing function of t then the sign scheme in x is obtained by reversing

the number line in t

2437 second derIvatIve test for paraMetrIc functIon

Assume that the function is differentiable dy dy dtdx dx dt

= First we get the stationary points we find the

values of t = tc where dy 0dt

= but dx 0dt

ne If dx 0dt

= then this test is not applicable

d y yx xydx (x)

minus=

rArr c cc

2 3 2t t t tt t

d y yx xy ydx (x) (x)= ==

minus= =

Now c c

2 2t t t t

d y d y0 if 0dx dt

gt gt then x = x(tc) is a point of local minimum

Further c

d y 0dx

lt if c

d y 0dx

lt then x = x(tc) is a point of local maximum

Consider x = tanndash1 t + 1 y = ln (4 ndash t2) dy 0 t 0dt

= rArr =

We confirm that at t = 0 dx 0dt

ne The sign of 2

is same as sign of 2

2 2 2t 0

d y 2(t 4) 0dt (4 t )

+= minus lt

minus Hence x = x(0) = 1 is a point of local maximum

2438 darbouX theoreM

If f(x) is differentiable for a le x le b f (a) = a f (b) = b and g lies between a and b then there is a x between a and b for which f (x) = g

2439 forK eXtreMuM theoreM

If f is a continuous function defined on a finite or infinite interval I such that f has a unique local extremum in I then that local extremum is also an absolute extremum on I

ie if f(x) has a unique critical point in interval I (infinite or finite) then f (x) gt 0 forall x isin IrArr a is a point of local minima and f (x) lt 0 forall x isin IrArr a is a point of local and maxima

2440 eXtreMa of dIscontInuous functIons

Minimum at x = a f(a) lt f(a ndash h) and f(a) lt f(a + h)

Maximum at x = a f(a) gt f(a ndash h) and f(a) gt f(a + h)

neither maximum nor minimum at x = af(a ndash h) lt f(a) lt f(a + h) or f(a ndash h) gt f(a) gt f(a + h)

2441 MaXIMuM and MInIMuM for dIscrete valued functIons

discrete values function A real valued function whose domain is a finite or countable set is called discrete valued function Since the function can give exactly one image of every point of domain the range of discrete-valued functions is also finite or countable

For example f(x) = n2 n isin ℕ is a discrete function with domain ℕ set of natural numbers and range = n2 n isin ℕ = 1 4 9 16hellip

For such function f (x) is evaluated and we find the intervals of monotonic increasing and monotonic decreasing ie intervals for which f (n) gt 0 and f (n) lt 0 Let they be [n1 n2] and [n3 n4] Then we observed f(n1) f(n2) f(n3) f(n4) and max f(n2) f(n3) gives us maximum value of function and minf(n1) f(n4) gives us the minimum value of function If n ( = 1 2 3 4) is not in the domain of function then the real number in the domain nearest to n serves the purpose but it should be in the same interval of monotonicity in which n lies

eg let f(x) = 2

n (n )n 90

then f (x) gt 0 for x (0 90)isin and f (x) lt 0 for x ( 90 )isin infin

rArr f(n) has greatest value at n 90 948=

but x isin IrArr f(x) has its greatest value either at x = 9 or at n = 10 Now f(9) = 119 and f(10) = 119 Thus f(9) = f(10) = 119 is the greatest term

1 absin

th lsquoa

f a ci

lsquoarsquo

a2 sinq

times h

us lsquor

rsquopr

ndashsid

h lsquoarsquo

sinnπ

us lsquor

rsquo4p

r with

lsquorrsquo

t tria

de lsquoa

minusminus

ndashsid

l slan

(i) If

a tria

1 (na)

indashpe

1 3 (A

Some important Points

(i) Among rectangles of given perimeter (costant) square has maximum area (ii) Among rectangles of given area (constant) square has minimum perimeter (iii) Area of triangle with given base lsquoarsquo (diameter) inscribed in a circle is maximum when it is an

equilateral triangle (iv) Area of triangle circumscribing a given circle is minimum when it is an equilateral triangle (v) Generally we are to deal with following type of questions (a) To find the cone with maximum volume inscribed in a cylinder (b) To find the cone with maximum area inscribed in a sphere (c) To find the cylinder with maximum area inscribed in a cone (d) To find the cylinder with maximum area inscribed in a sphere (e) Triangle circumscribing a given a circle (f) Triangle inscribed in a circle (g) Cone around sphere (h) Cone around cylinder (i) Rectangle inscribed in a triangle with one side coincident (j) Rectangle inscribed in a semi-circle (k) Triangle inscribed in a ellipse

2443 general concept (shortest dIstance of a poInt froM a curve)

Given a fixed point A(a b) and a moving point P(x f(x)) on the curve y = f(x) Then AP will be maximum or minimum if it is normal to the curve at P

Proof F (x) = (x ndash a)2 + (f (x) ndash b))2

rArr F(x) = 2(x ndash a) + 2(f (x) ndash b) middot f (x)

there4 F(x) = 0 rArr (x a)f (x)f (x) b

minus= minus

Also APf(x) bm

x aminus

=minus

Hence f (x) middot mAP = ndash 1rArr AP perp tan gent to f(x) at P

Chapter 25IndefInIte IntegratIon

251 INTRODUCTION

Integration is the inverse process of differentiation Instead of differentiating a function we are given the derivative of a function and asked to find its primitive ie the original function Such a process is called integration or anti-differentiation

2511 Anti-derivative of a Function

A function F(x) is called an anti-derivative of the function f(x) on the interval I say [x1 x2] iff at all the points of the interval I Fprime(x) = f(x) if x is terminal point of the interval ie x1 or x2 then Fprime(x) shall be only one-sided derivative eg Fprime(x1

+) at x1 and Fprime(x2ndash) at x = x2 Thus if f(x) is derivative

of F(x) then F(x) is an aniderivative of f(x) In fact F(x) + C C = arbitraly constant is the set of all antiderivatives of f(x)

2512 Notation of Anti-derivatives or Indefinite Integral

If Fprime(x) = f(x) forall x isin I (closed interval) then = +int f(x)dx F(x) C (anti-derivative of f(x)) where C is arbitrary constant called constant of integration

Remarks

(i) Anti-derivatives of a function f(x) if exist then they are infinitely many If F(x) is one of the anti-derivative then F(x) + C C = arbitrary constant is the set of all antiderivates

(ii) Any two anti-derivaties of a fanction differ by a constant

(iii) For anti-derivative of a function f(x) to exist over an interval [x1 x2] it is necessary that the function is continuous in that interval

(iv) Geometrically an indefinite integral int f ( x )dx is a family of curves y = F(x) + C each of whose

members is obtained by translating any one of the curves parallel to itself vertically upwards or downwards (that is along the y-axis)

2513 Algebra of Integration

Constant Rule = = +int int 0cdx c x dx cx b

Constant Multiple Rule kf(x)dx k f(x)dx=int intAddition Rule ( ) f x g(x) dx f(x)dx g(x)dxplusmn = plusmnint int int

The above rule can be extended to any finite number of functions For instance

( ) u(x) v(x) w(x) dx u(x)dx v(x)dx w(x) dxplusmn plusmn = plusmn plusmnint int int int

Product Rule (integration by parts) ( )f(x)g(x)dx uv dx u vdx u vdx dx= = minusint int int int int u = f(x) v = g(x)

ie (Ist function) (IInd function)dxtimesint

= d(Ist function) (IInd function)dx (Ist function) (IInd function)dx dxdx minus int int int

we choose first function and second function according to order of preferance given by word lsquoILATErsquo where I = inverse function eg sinndash1x tanndash1x etcL = logorithmic function eg logx log (x + 1) etc

A = algerabic function eg 2

x2 x3 etc

T = Trigonometric function eg tanx sinx (1 + cosx) etcE = Exponetial function eg ax ex a(x2 + 1) etc

Power Rule If r is any rational number except ndash1 then r 1

r xx dx Cr 1

= ++int

Generalized Power Rule Case 1 Let f be a differentiable function and n a rational number different from ndash1

Then n 1

n [f(x)][f(x)] f (x)dx Cn 1

= ++int

Case 2 n = ndash1 ie f (x)dx n f(x) Cf(x)

= +int

Theorem If f(x) and g(x) are two continuous functions such that f(x)dx g(x) C= +int then

integral of f(ax + b) with respect to x is given by ( )1 g(ax b) Ca

Remarks

(a) ( ) =intd

f ( x )dx f ( x )dx

or ( )d f ( x )dx f ( x )=int (b) f ( x )dx df ( x ) f ( x ) C= = +int int

Indefinite Integration 25307

252 INTEGRAL OF SOME STANDARD FUNCTIONS

(a) Algebraic and Exponential Functions

f(x) int f(x)dx f(ax + b) int f(ax + b)dx

xn n ne ndash1 n 1x C

++ (ax + b)n n ne ndash1 ( )n 1ax b

C(n 1)a

1x ln|x| + C

1ax b+

ln(ax b)C

ax a gt 0 a ne 1xa C

lna+ abx+c a gt 0 a ne 1 b ne 0

bx c1 a Cb lna

ex ex + C eax+b a ne 0 ax b1 e Ca

(b) Trigonometric Functions

sin x ndash cos x + C sin(ax + b)1 cos(ax b) Ca

minus + +

cos x sin x + C cos(ax + b)1 sin(ax b) Ca

tan x ln |sec x| + C tan(ax + b)1 ln | sec(ax b) | Ca

cot x ln |sin x| + C cot(ax + b)1 ln |sin(ax b)| Ca

sec x ln |sec x + tan x| + C or

ndashln xtan C4 2π minus +

sec(ax + b) 1 ln |sec(ax b) tan(ax b)| C ora

1 (ax b)ln tan Ca 4 2

+ + + +

π + minus minus +

cosec x ln |cosec x ndash cot x| + C

or xln tan C2

cosec(ax + b)1a

ln |cosec (ax + b) ndash cot (ax + b)| +

C or 1 ax bln tan Ca 2

sec2 x tan x + C sec2(ax + b)1 tan(ax b) Ca

cosec2 x ndashcot x + C cosec2(ax + b)1 cot(ax b) Ca

minus + +

sec x tan x sec x + C sec(ax + b) tan (ax + b)

1 sec(ax b) Ca

cosec x cot x ndashcosec x + C cosec(ax + b) cot(ax + b)

1 cosec(ax b) Ca

minus + +

(c) Rational and Irrational Functions

1 xminus sinndash1x + C or ndashcosndash1x + C 2 2

a xminus1 xsin C

aminus +

x 1minus2ln | x x 1 | C+ minus + 2 2

x aminus2 2ln | x x a | C+ minus +

x 1+2ln|x x 1 | C+ + + 2 2

x a+2 2ln|x x a | C+ + +

11 xminus

1 1 xln C2 1 x

minus 2 2

1a xminus

1 a xln C2 a x

1x 1minus

1 x 1ln C2 x 1

minus+

1x aminus

1 x aln C2a x a

minus+

1x 1+ tanndash1x + C or ndashcotndash1x + C 2 2

11 xtan Ca a

minus +

or 11 xcot Ca a

minus minus +

21 xminus2

1x 1 x 1 sin x C2 2

minusminus+ + 2 2a xminus

2 2 21x a x a xsin C

2 2 aminusminus + +

2x 1minus2

2x x 1 1 ln x x 1 C2 2minus

minus + minus + 2 2x aminus2 2 2

2 2x x a a ln x x a C2 2minus

minus + minus +

21 x+2

2x 1 x 1 ln x x 1 C2 2+

+ + + + 2 2a x+2 2 2

2 2x a x a ln x a x C2 2minus

+ + + +

x x 1minus1 1sec x C or cosec x Cminus minus+ minus + 2 2

x x aminus11 xsec C

a aminus + or 11 xcosec C

a aminus minus +

253 ThE METhOD OF SUbSTITUTION

If the integrand is of the type f(g(x)) gprime(x) where g be a differentiable function with range set Rg and f(x) is continuous over interval Rg Suppose F is an anti-derivative of f over the interval Rg then to find the integral we substitute t = g(x) and proceed as follows

g(x) t= dtg (x)

dxrArr = g (x)dx dtrArr =

( )f g(x) g (x)dx f(t)dt=int int = F(t) + C = F(g(x)) + C

2531 List of Some Standard Substitutions

Integrand Standard Substitutions Mutation of Differential

f(ex) ex = t rArremsp dx = tndash1 dtf(log x) log x = t rArremsp dx = et dt

f(tan x) tan x = t rArremsp 2

dtdx1 t

f(cot x) cot x = t rArremsp2

dtdx1 t

= minus+

f(xx) (1 + ln x) xx = t rArremsp (1 + ln x) dx = tndash1 dt

Integrand

Standard Substitutions

Mutation of Differential

( )2f x x 1+ + 2x x 1 t+ + = rArremsp2 2

(1 t ) 1 1 t 1dx dt x t and x 12t 2 t 2t+ + = = minus + =

1 1f x 1x x

+ minus

+ = rArremsp 2

11 dx dtx

minus =

1 1f x 1x x

minus +

minus = rArremsp 2

11 dx dtx

1 1f x xx x

+ minus

+ = rArremsp 3

12 x dx dtx

minus =

1 1f x xx x

minus +

minus = rArremsp 3

12 x dx dtx

1 2n n

1f(x a) (x b)

(n1 n2 isinemspℕemsp(and gt 1)

Put (x + a) = t(x + b)

rArremsp2

1 dtdx(x b) (b a)

=+ minus

1nrsax bR x x

where R is a rational function of its arguments

Put nax b tcx d+

rArremspn 1

1 nt dtdx(cx d) (ad bc)

=+ minus

Integrand

( )xe f(x) f (x)+ ex f(x) = t rArremsp ex (f(x) + fprime(x))dx = dt

( )f(x) xf (x)+ xf(x) = t rArremsp (f(x) + xfprime (x)) dx = dt

2533 List of Some Standard Substitutions for Integrand Function

Involving minus2 2a x 2 2x aminus 2 2a x+ a xa xminus+

( )minus2 2f a x a sinx

a cosθ= θ

rArremsp dx = a cos q dq or ndasha sinq dq

( )2 2f a x+ a tanx

a cotθ= θ

rArremsp emspdx = a sec2q dq or ndasha cosec2q dq or ndash a cosecq cotq dq

( )2 2f x aminus a secx

a cosecθ= θ

rArremsp dx = a secq tanqemspdqor ndasha cosecqemsp cot q

a xfa x

minus +

a cosx

a sinθ= θ

rArremsp dx = ndasha sinq dq or a cosq dq

2534 List of Some Standard Substitutions for Integrand Having

Function of (x a) or (b x)minus divide times plusmn 1 1nx a

minus+ +

( ) minusminus minus minus

x af (x a)(b x) or fb x

x = a cos2q + b sin2 qemsp rArremspdx = 2(b ndash a) sinq cosq dq

( ) x af (x a)(b x) or fb x

minusminus + +

x = a sec2q + b tan2 q rArremspdx = 2(a + b) sec2q tanq dq

( ) ( )1 11 1n nx a x bminus minus minus + +

x a tx b+

(b a)dt dx(x b)

minus=

(a b)or dt dx(x a)

minus=

2535 Substitution after Taking xn Common

Many integrals can be evaluated by taking xn common from some bracketed expression and then using substitution Some of the suggested forms are given as follows

dx nx(x 1)

isin+int

Take xn common and put 1 + xndashn = t

(ii) 2 n (n 1)n

dx nx (x 1) minus isin

+int Take xn common and put 1 + xndashn = tn

(iii) n n 1n

dxx (1 x )+int Take xn common and put 1 + xndashn = t

254 INTREGRATION OF

m nsin x cos x dxmnisinintIf one of m and n is odd positive integer (say) (m) and (n) is any integer (+ve or -ve) then

Case (i) minus= int m 1 nI sin x cos xsin x dx = m 1

2 n2(1 cos x) cos x sin x dxminus

minusintPut cosx = t rArr sinxdx = dt

2 n2I (1 t ) t dtminus

= minusint

Expained m 1

2 2(1 t )minus

minus binomially and integrate term by term Similar is the case when m is even and n is odd

Case (ii) If both m and n are odd positive integer then take out one power from any of sinmx or cosnx and put respectively cosx or sinx equal to t For quick solution put the function sinx or cosx with greater power equal to t

Case (iii) If both m and n are even then convert the integraand as trinogometic raios of multiple angles by using the formula 2sin2x = 1 ndash cos2x 2cos2x = 1+ cos2x and sin2x cos2x = 14 sin2x etc

Case (iv) If (m + n) is a negative even integerm

m n m nm

sin xI sin x cos x dx cos x dxcos x

+= =int int = ( )m m n 2 2tan x (cos x) sec x dx+ minusint

= ( )m n 2

m 2 22tan x (sec x) sec x dx+ minus minus

int = ( )m n 2

m 2 2t (1 t ) dt+ minus minus

+int t = tanx

Now expand binomially and integrate term by term

2541 To Slove Integral of the Form m n

1I tan x sec x dx= int m n2I cot x cos ec x dx= int

(i) When n is even positive integer put tanx = sec2xdx = dz and solve as follows

( ) ( )minus= intm n 2 2

1I tan x sec x sec x dx = ( ) ( )n 2

m 2 22tan x tan x 1 sec x dxminus

= ( ) ( )n 2

m 2 2z z 1 dzminus

+int expand binomially and integrate term by term

(ii) When m is odd positive integer put secx = z secx tanx dx = dz and slove as followsm 1 n 1

1I (tan x) (sec x) (sec x tan x)dxminus minusint

2 n 12(tan x) (sec x) (sec x tan x)dxminus

minusint = m 1

2 n 12(sec x 1) (sec x) (sec x tan x)dxminus

minusminusint

2 n 12(z 1) (z) dzminus

minusminusint expand binomially and integrate term by term

Simiarlly to solve I2 when n is evenput cot x zandwhen m is odd put cose x z

255 INTEGRATION by pARTIAL FRACTION

An integral can be reduced into simple one by using partial fraction

2551 Integration of Rational Functions by Using Ostrogradsky Method

Let P(x) dxQ(x)int be the integral to be found where Q(x) has repeated roots then we set

G (x) H (x)P(x) dx dxQ(x) G (x) Q(x) G (x)

= +int int (1)

where G1(x) = greatest common divisor of Q(x) and Qprime(x) (derivative of Q(x))G0(x) = polynomial of degree 1 less than that of G1(x)H0(x) = polynomial of degree 1 less than that of Q(x)G1(x)The unknown coefficients of G0(x) and H0(x) are obtained by differentiating the identity (1) wrt (x)

256 INTEGRATION OF RATIONAL AND IRRATIONAL ExpRESSIONS

Type (i) Integration of rational expressions of the type 2

1 dxax bx c+ +

where b2 ndash 4ac lt 0

ie quadratic in denominator is irreducible

dxIb Dx

= minus + +

int = 12a 2ax btan CD D

minus + +

minus minus

Type (ii) Integration of rational expression of the type 2

1 dxax bx c+ +int where b2 ndash 4ac = 0

ie denominator has repeated roots Let ax2 + bx + c = a(x ndash α)2 then 2 2

1 1dx dxax bx c a(x )

=+ + minusαint int

2 121 1 (x )(x ) dx C

a a ( 2 1)

minus +minus minusα

minusα = +minus +int = 1 1 C

a (x )minus +

minusα

Type (iii) Integration of rational expression of type 2

1 dxax bx c+ +int where b2 ndash 4ac gt 0

ie denominator contains a reducible quadratic factors

Method 1 Let ax2 + bx + c = 4(x ndash α) (x ndash β) then 2

dx 1 dxI(x )a(x )(x ) a (x )(x )

= =minusαminusα minusβ minusβminusβ

int int hellip (i)

Let x txminusα

=minusβ

rArr 2

(x ) (x ) dx dt(x )minusβ minus minusα

=minusβ

rArr 2

(x ) dx dt(x )

minusβ =minusβ

rArr 2

dx dt(x ) (x )minusβ minusβ

hellip (ii)

Using (ii) in (i) we get 1 dt 1I ln | t | Ca (t)( ) a( )

= = +α minusβ αminusβint

1 xI ln Ca( ) x

minusα= +

αminusβ minusβ

Method 2 (By using partial fractions)

Type (iv) Integration of irrational expression of type 2ax bx c dx+ +int 2I ax bx c dx= + +int

Completing square inside the root we have 2

b Da x dx2a (2a)

+ minus

Case (i) When a D gt 0 then I = 2

2 Da t dt2a

int where bt x

2a = +

Case (ii) When a gt 0 D lt 0 then 2

2 DI a t dt2a

minus= +

int where

bt x2a

Case (iii) When a lt 0 D gt 022b DI ( a) x dx

= minus minus + minus int

2 2D b( a) x dx2a 2a

= minus minus + int

2Da t dt2a

minus minus

int where bt x

2a = +

Case (iv) When a lt 0 D lt 0 integration is impossible as the integrand becomes imaginary

Type (v) Integration of irrational expression of type 2

ax bx c+ +int

Case (i) When a gt 0 D gt 0

b Da x2a 4a

= + minus

int 22

11 dxa b Dx

= + minus

21 Dln t t C2aa

= + minus +

where bt x

2a = +

Case (ii) When a gt 0 D lt 0 minus

21 Dln t t C2aa

where bt x2a

Case (iii) When a lt 0 D gt 0 2

1I dxb D( a) x

= minus minus + minus

1 1 dta D t

=minus

int bt x2a

Type (vi) Integration of rational and irrational expressions of the type

px q dxax bx c

++ +int (b)

px q dxax bx c

+ +int (c) 2(px q) ax bx c dx+ + +int

Algorithm Let px + q = 2dA (ax bx c) Bdx

+ + + px = A(2ax + b) + B

rArr pA2a

= and pcB q2a

= minus

solve as discussedearlierput ax bx c t

px q 2ax b dxdx A dx B ax bx c ax bx c ax bx c

+ += +

+ + + + + +int int int

px q 2ax b dxdx A dx Bax bx c ax bx c ax bx c

+ += +

solve as disccused earlierput ax bx c t

(px q) ax bx c dx A (2ax b) ax bx c dx B ax bx c

+ + + = + + + + + +int int int

Type (vii) Integration of rational expression of type 4 2

1 dxax bx c+ +int 4 2

1I dxax bx c

=+ +int

Here ax4 + bx2 + c is a quadratic in x2 with discriminant D = b2 ndash 4ac

Case I D gt 0 For D gt 0 bi-quadratic ax4 + bx2 + c can be factorized as a(x2 ndash α) (x2 ndash β)

1 dxIa (x )(x )

=minusα minusβint

1 1 1 dxa( ) x x

= minus α minusβ minusα minusβ

which can be further solved by using the standard integral 2 2

1 1 x adx ln Cx a 2a x a

minus= +

minus +int

Case II D lt 0The bi-quadratic ax4 + bx2 + C cannot be further factorized as the above So we proceed as follows

dxIcx ax bx

int hellip (1)

Now depending upon the nature of lsquoarsquo three arise two sub cases

Sub case (i) a gt 0 (∵ b2 ndash 4ac lt 0 a lt 0 rArr c gt 0)

dxIcx ax 2 a c 2 a c b

= + + minus +

int = 2

cx ax b 2 a cx

plusmn +

hellip (2)

Put cax tx

plusmn = hellip (3)

I = 2 2

1 dt 1 dt2 c t b 2 a c 2 c t b 2 a c

minus+ + + minusint int hellip(5)

When can be solved by using the standard integral 12 2

dx 1 xtan Cx a a a

minus = + + int

Sub case (ii) a lt 0 (∵ b2 ndash 4ac lt 0 a lt 0 rArr c lt 0)

I = 2 2

dx( c)x ( a)x bxminus minus minus + +

int = 2 2

dxCx Ax Bx

minus + +

where ndasha = A gt 0 and ndashc = C gt 0 and b = B

From sub-case (i) we have I = 2 2

1 dt 1 dt 2 C t B 2 AC 2 C t B 2 AC

minus ++ + + minusint int

where Ct Axx

= minus in 1st integral and Ct Axx

= + in 2nd integral and A = ndasha C = ndashc B = b

Type (viii) Integration of the of type 2

ax bx c dxpx qx r

+ ++ +int

ax bx c dxpx qx r

+ +int ( )2 2ax bx c px qx r dx+ + + +int

(a) For integrals of the form 2

ax bx c dxpx qx r

+ ++ +int and

ax bx c dxpx qx r

+ +int

Put (ax2 + bx + c) = A(px2 + qx + r) + B ddx

(px2 + qx + r) + C

By comparing the coefficients of like terms on both sides we obtain the values of constant A B and C

Type (ix) Integration of the type 4 3 2

f (x) dxax bx cx bx a+ + plusmn +

where f(x) is a rational function of x and is expressible in the form 1 1x g xx x

minus +

or 1 1x g xx x

+ minus

Algorithm Given integral is

I = 2 3 2

int = 2

f (x) dx

1 1x a x b x cx x

+ + plusmn +

These a arise two cases

Case (i) If I = 2

1 1x g x dxx x

1 1x a x b x cx x

minus + + + + +

int = 2

1 11 g x dxx x

1 1a x b x cx x

minus + + + + +

Putting 1x tx

+ = and 2

11 dx dtx

minus =

we have I = ( ) 22

g(t) dt g(t) dt

at bt (c 2a)a t 2 bt c=

+ + minusminus + +int int

which can be solved further by one of the methods discussed earlier

Case (ii) If I = 2

1 1x g x dxx x

1 1x a x b x cx x

+ minus + + minus +

int putting 1x tx

minus =

11 dx dtx

we have

I = ( ) 22

g(t) dt g(t) dt

at bt (2a c)a t 2 bt c=

+ + ++ + +int int

Type (x) Integration of the type m2

P (x) dxpx qx r+ +

int where Pm(x) is a polynomial of degree (m)

Algorithm Consider m 2m 12 2

P (x) dx dxP (x) px qx r Kpx qx r px qx r

minus= + + ++ + + +

int int hellip(1)

where Pmndash1(x) is a polynomial of degree (m ndash 1)Differentiating both sides of (1) wrt x we get

( )2m m 1m 12 2 2

P (x) P (x)(2px q) Kpx qx r P (x)px qx r 2 px qx r px qx r

minusminus

+= + + + +

+ + + + + +

rArr 2Pm(x) = Pmndash1 (x) (2px + q) + 2(px2 + qx + r) Pprimemndash1(x) + 2K hellip (2)Comparing the coefficients of like terms on both sides of (2) we get the coefficients of polynomial Pmndash1(x) and K and further the integral can be solved

257 TO SOLvE INTEGRAL OF ThE FORM

2 2 2 2

dx dx dx a sin x bcos x a bcos x a sin x b+ + +int int int 2

dx (a sin x bcos x)+int 2 2

dxa sin x bcos x csin x cos x d+ + +int

Algorithm Divide numerator and denominator by cos2x and put tanx = z and sec2xdx = dz To solve integral of the form

dx a bsin x+int dx

a bcos x+int dx a sin x bcos x+int

cos x dx

a bcos x+int sin x dx

a bsin x+int

( )f tan x 2dx

a sin x bcos x c+ +int

Algorithm Step I Put tan x2

= t and 21 xsec dx dt2 2

= and 2

2dtdx1 t

Step II Take 2 2

2 tan(x 2) 2tsin x1 tan (x 2) 1 t

= =+ +

and 2 2

1 tan (x 2) 1 tcos x1 tan (x 2) 1 tminus minus

= =+ +

Step III The integral is reduced to the form 2

f (t)dtAt Bt C+ +int where f(t) is a polynomial in t

Integral of the form

a sin x bcos xI dxcsin x d cos x

Algorithm

Step I Express the numerator ( )r r rdN A D B Ddx

Step II Obtain the constants A and B which reduces the integral to the sum of two integrals

Step III Solve the above integrals as I = A(ln | Dr | + Bx + C

Integral of type IV a sin x bcos x c dxpsin x q cos x r

+ ++ +int

Algorithm

Step I Express the numerator in the form ( )d (Dr) m Dr ndx

Step II a sinx + b cosx + c = l(p cos x ndash q sin x) + m (p sin x + q cosx + r) + n where l m and n are con-stants Comparing the coefficients of sin x cos x and constant terms on both sides and determine l m n

Step III We have I = l ln (denominator) + mx + n dx Cpsin x q cos x r

++ +int

Integral of type (VI) R(sin xcos x sin x cos x)dxplusmnint

Algorithm Substitute sin x plusmn cos x = t and (cos xemspplusmnemspsin x)dx = dt

rArremsp 21 2sin xcos x t plusmn = 2

2 2t 1R t t 2(t 1)2

minusplusmn minus plusmn

2571 Integral of Type

(i) sinaxsin bx dxint (ii) sinaxcos bx dxint (iii) cosaxcos bx dxint

Use sinax sinbx = 1 (cos(a b)x cos(a b)x)2

minus minus +

Use cosax cosbx = 1 (cos(a b)x cos(a b)x)2

minus + + and

Use sinax cosbx = 1 (sin(a b)x sin(a b)x)2

minus + +

258 INTEGRAL OF IRRATIONAL FUNCTIONS

Integral of type dx linear linearint

dx(linear) linearint

dxeg(ax b) cx d+ +int put cx + d = t2

which can be solved further by using the standard integrals 2 2

1 dtt a+int 2 2

1 dtt aminusint 2 2

dta tminusint

Integral of type dxlinear quadraticint and

dx(linear) quadraticint eg

(ax b) cx d+ +int

Algorithm Put 1ax bt

+ = which can be integrated by using the method of solving integrals of the

type 2

at bt c+ +int

Integral of type dxquadratic linearint Let I = 2

a x b x c p x q( )+ + +int put px + q = t2

Integral of type dx

quadratic quadraticint ie 2 2

(ax bx c) px qx r+ + + +int

Case I When (ax2 + bx + c) breaks up into two linear factors say a(x ndash α) (x ndash β)

rArr 2 2

1 1put (x ) put (x )t t

A dx B dxIa a(x ) px qx r (x ) px qx r

minusα = minusβ =

= +minusα + + minusβ + +

int int

which can be further solved by using the method of solving integral of the form dx

linear quadraticint

Case II If ax2 + bx + c is a perfect square say (lx + m)2

dxI(lx m ) px qx r

=+ + +

int put 1lx mt

+ = ( ) ( )2 2 2

p 1 tm qlt 1 tm rl t

minus=

minus + minus +int

which can be solved further by using the integral of type linear dxquadraticint

Case III If b = 0 q = 0 eg 2 2

dxI(ax c) px r

int then put 1xt

rArr 2

1dx dtt

= minus put rt2 + p = z2 rArremspzdztdt

zdz dzIcz ar cpz pc a z

= minus = minus + minus minus

int int which is a standard form

Integral of type mn pq rsR(x x x )dxint where p q r s m nisinℤ and R(x) is a rational algebraic

function of x Put x = tk such that k = LCM (n q shellip) and integrand reduced to rational function of t

Integral of type ( )pm nx a bx dx+int where m n p are rationals

Case I If p isin ℤ expand binomially and integrate term-by-term

m p p r n rr

I x C a (bx ) dxminus

= sumintp nr m 1

p p r rr

xC a b Cnr m 1

+ +minus

= + + + sum

Case II If p minusisin ie I = ( )pm nx a bx dx+int then put x = tk where k = LCM of denominator of m and n

Case (III) If pisin ( )pm nI x a bx dx= +int where rps

Case (a) m 1

isin The substitution a + bxn = ts reduces the integral into integrable form

1nst ax

b minus

mnsm t ax

b minus

rArr =

rArr s 1

st dtdxnbx

minus=

m 1 1s nr s 1s t a t dt

+minus

+ minus minus=

Case (b) If ( )pm nI x a bx dx= +int where rps

= and m 1n+

notin and m 1 pn+

+ isin

The substitution a + bxn = ts xn reduces the integral into integrable form

rArr sn

a t bx

= minus

m np n 1n r s 1

s a t dtna t b

+ minus = minus minus int

2581 Eulers Substitution

If the integrand is of the form ( )2R x ax bx c+ + ie to evaluate ( )2R x ax bx c dx+ +int

Case (i) If D = b2 ndash 4ac gt 0 then put2

Where is a root of ax + bx + c = 0

ax bx c t(x )α

+ + = minusα

Case (ii) If D = b2 ndash 4ac lt 0 rArr ax2 + bx + c gt 0 forallemspxisinemspℝ iff a c gt 0 put 2 t x aax bx c

plusmn+ + = plusmn

or try

to carry out the trigonometric substitutions as given below

1 Substitute

D sin tb 2ax2a D cos t

if a lt 0 D gt 0 2 Substitute

D tan tb 2ax2a D cot t

minus+ =

if a gt 0 D lt 0

3 Substitute

D sectb 2ax2a D cosec t

if a gt 0 D gt 0

259 INTEGRATING INvERSES OF FUNCTIONS

Integration by parts leads to a rule for integrating inverse that usually gives good results

Let 1I f (x)dxminus= int to be evaluated

Let y = fndash1 (x) rArr x = f(y) rArr dx = fprime(y) dy

rArr I y f (y)dy= int = 1yf(y) f(y)dy xf (x) f(y)dyminusminus = minusint int Let y = ln x rArr x = ey dx = ey dy rArr yln x dx ye dy=int int = yey ndash ey + C = x ln x ndash x + C

2510 INTEGRATION OF A COMpLEx FUNCTION OF A REAL vARIAbLE

The definite integral of a complex function of a real variable f(x) = u(x) + iv(x) is defined as followsb b b

f (x)dx u(x)dx i v(x)dx= +int int intNow this provides an alternative method to find the integrals axe cos bx dxint and axe sin bx dxint

as given below

Let P = axe cos bx dxint and Q = axe sin bx dxint P + iQ = ( )axe cos bx isin bx dx+int = ax ibx (a ib)xe e dx e dx+=int int

( ) ( )ax ax ax ax

ae cos bx be sin bx i ae sin bx be cos bxC iD

+ + minus= + +

Equating real and imaginary parts we get the values of P and Q as before ( )ax

e a cos bx bsin bxP C

e a sin bx bcos bxQ D

a bminus

where C and D are constants of integration

2511 MULTIpLE INTEGRATION by pARTS

While calculating several integrals we are to use integration by parts number of times successively The same calculation work can be done more rapidly and conveniently by using the so called generalized formula for multiple integration by parts which is given by consider u v as two differential function of x

n 1 n 1 n n1 2 3 4 n nuv dx uv u v u v u v ( 1) u v ( 1) u v dxminus minus= minus + minus + + minus + minusint int hellip (6)

where u(n) denotes nth order derivative of u

Reduction formula for int nsin x dx and int ncos x dx and Wallirsquos Formulae

(a) nnI sin x dx= int

emsp rArr n 1

n n 2cos x sin x (n 1)I I

minus= minus +

(b) Let nnI cos x dx= int

n n 2sin x cos x (n 1)I I

minus= +

Reduction Formula for tannx

rArr n 1

n n 2tan xI I

minus= minusminus

Reduction Formula for secnx

rArr n 2

n n 2sec x tan x (n 2)I I

n 1 n 1

minus= +

minus minus

Wallirsquos Formula is

rArr rarrinfin

π = minus minus minus n

2224466(2n)(2n)lim13355(2n 1)(2n 1)(2n 1) 2

This can be proved by using integration and reusing in = π

cos (x)

π π π

minus= minus minus minusint int int2 2 2

n n 2 n

cos (x)dx (n 1) cos (x)dx n 1 cos (x)dx

Chapter 26Definite integration

anD area unDer the Curve

261 AreA Function

If f(x) is continuous on [a b] then the function = isinintx

f (x)dx A(x) x [ab] is called area function and it

represents the algebraic sum of areas bounded by function f(x) ordinates x = a and x = x such that the area bounded by function above the x-axis is positive and that is bounded by the function below the x-axis is negative

262 First FundAmentAl theorem

If f(x) is continuous function on [a b] and = geintx

A(x) f(x)dx x a is the area function then

A(x) = f(x) forall x isin [a b]

263 second FundAmentAl theorem

If f(x) is continuous function on [a b] then = minusintb

f (x)dx F(b) F(a) where = +int f(x)dx F(x) C

Definite Integral as limit of sum (Integrating by first principle or ab-initio)

(a) By using subinterval of equal length

rarrrarrinfin

= + + + + + + + minus intb

h 0a n

f(x)dx limh f(a) f(a h) f(a 2h) f (a n 1h) where minus

= rarr rarrinfinb ah h 0asn

rarrrarrinfin

= + + + + + + intb

h 0a n

f(x)dx limh f(a h) f(a 2h) f (a nh) where or

rarrinfin rarrinfin

= =rarr

minus minus = + minus = + minus

sum sumint

n nr 1 r 1a h 0

b a b af(x)dx limh f(a (r 1)h lim f a ( 1)n n

= left and estimation of intb

f (x)dx

and ( )rarrinfin rarrinfin

= =rarr

minus = + = +

sum sumintb n n

n nr 1 r 1a h 0

b af(x)dx limh f(a rh) lim f a rhn

Definite Integration and Area Under the Curve 26323

(b) By using subintervals of unequal length such that their end point are forming a GP Let [a b] be divided into n-subintervals with partition a0 a1 a2 a3 an such that a0 = a and

ai = aRi and an = b

rArr aRn = b rArr =

= common ratio then

Length of rth subintervals = Dr = ar ndash arndash1 = aRr ndash aRrndash1 = aRrndash1(R ndash 1)

= minus rarr rarrinfin

r 1nba (R 1) 0asn

a then

rarrinfin= ∆ + ∆ + + ∆ int

1 1 2 2 n nna

f (x)dx lim f(a ) f(a ) f (a )

rarrinfin=

∆ ∆ = minus

r r rn 1

blim f(a ) where a (R )a

For Example if =1f(x)x

then int3

can be evaluated by above GP method

Remark

1f ( x )

x= then

dx( a b)

xltint can be evaluated by using the inequality

h h h[a ( r 1)h][a rh] [a ( r 1)h] [a ( r 2)h][a ( r 1)h]

lt lt+ minus + + minus + minus + minus

Substituting l = 1 2 3n and adding we get n

1 1 h 1 1a b [a ( 1)h] a h b h=

minus lt lt minus+ minus minus minussum

2 2h 01a n

1 h 1 1dx lim

x [a ( 1)h] a brarr=rarrinfin

= = minus+ minussumint

264 lineArity oF deFinite integrAl

Suppose f and g are integrable on [a b] and that k is a constant then kf and f + g are integrable and

(i) =int intb b

kf(x)dx k f(x)dx (ii) +int intb b

f (x)dx g(x)dx and consequently

(iii) minus = minusint int intb b b

[f(x) g(x)]dx f(x)dx g(x)dx

265 ProPerties oF deFinite integrAl

Property 1 Mere change of variable does not change the value of integral ie =int intb b

a af (x) dx f(t)dt

Property 2 By interchanging the limits of integration the value of integral becomes negative ie

=minusint intb a

a bf (x) dx f(x)dx

Property 3 = +int int intb c b

f (x) dx f(x) dx f(x) dx provided that lsquocrsquo lies in the domain of continuity of f(x)

2651 GeneralizationThe property can be generalized into the following form

b c c b

a a c cf (x)dx f(x)dx f(x)dx f (x)dx= + + +int int int int where c1 c2 c3cn lies in the domain of continuity of f(x)

Conclusion Although we can break limit of integration at any point but it is necessary to break limit at following points1 where f(x) is discontinuous 2 where f(x) is not defined3 where f(x) changes its definition

Property 4 =intb

f (x) dx 0 and f(x) is continuous then f(x) has at least one root isin (a b)

Remarks

Converse of above property is not true ie if f(x) has a root in (a b) then b

f ( x ) dxint need not be zero

Example if f(x) = x2 ndash 2x has a root x = 2 isin (1 3) but 3

1 26 2( x 2x ) dx ( 27 1) (9 1) 8 0

3 3 3minus = minus minus minus = minus = neint

Property 5 Substitution Property To evaluate intb

f (x) dx if we decide to substitute g(x) = t

then x = gndash1(t) then the following conditions must be kept in mind

2652 Condition of Substitutionq g(x) must be continuous and defined forall x isin [a b]q g(x) must be monotonic forall x isin [a b] (to ensure invertibility) If the above two conditions are fulfilled then we may take the following steps Step 1 Change integrand g(x) = t grsquo(x) dx = dt

Step 2 Change the limits of integration minusminusint

dtf(g (t))g (g (t))

Property 6 intb

f (x)dx is called improper integral if

q f(x) is discontinuous at at least one point c isin (a b) whether the discontinuity is of first kind or infinite discontinuity

q If intb

f (x)dx is such that f(x) is unbounded as x rarr a+ then we take +rarr

=int intb b

t 0a a t

f (x)dx lim f(x)dx

q If intb

f (x) dx is such that f(x) is unbounded infin as x rarr bndash then we take +

rarr=int int

t 0a a

f (x) dx lim f(x) dx

266 convergent And divergent imProPer integrAls

A definite integral having either or both limits infinite (improper integral) is said to be convergent if its value is finite ie if the area bounded by the continuous function f(x) x-axis and between its limits is finite otherwise it is said to be divergent Thus

(i) infin

f (x)dx is said to be divergent if rarrinfin

lim f(x)dx L (finite)

(ii) minusinfinintb

f(x)dx is said to be divergent if rarrminusinfin int

lim f(x)dx = L (finite)

(iii) infin

minusinfinint f(x)dx is said to be convergent if

minusinfin rarrminusinfin rarrinfin= +int int int

a ba 0

f (x)dx lim f(x)dx lim f(x)dx and each of the two

integrals on right hand side is convergent Note that if at least one of the two improper integrals on

right side is divergent then infin

minusinfinint f(x)dx is said to be divergent

Property 7 Reflection Property minus

= minusint intb a

f (x)dx f( x)dx

Property 8 Shifting Property +

= minusint intb b c

f (x)dx f(x c)dx ie area under a part of function and above

x-axis remains same when graph of function is shifted horizontally without having any change in the shape of curve

Property 9 = minusint inta a

f (x)dx f(a x)dx ie area under a part of function above x-axis and that under its

reversed part above x-axis are same

Property 10 = + minusint intb b

f (x)dx f(a b x)dx

Remark

If a = 0 and we take b = a then b b

f ( x )dx f ( a b x )dx= + minusint int rArr a a

f ( x )dx f ( a x )dx= minusint int ie property 9

267 APPlicAtions

Application I If f(a + b ndash x) = f(x) Then to evaluate = intb

I xf(x)dx helliphellip(i)

By above property

rArr = + minus + minusintb

I (a b x)f(a b x)dx = + minusintb

(a b x)f(x)dx helliphellip(ii)

as f(a + b ndashx) = f(x) rArr +

= intb

(a b)I f(x)dx2

Application II If f(x) + f(a + b ndash x) = λ then evaluate = intb

I f(x)dx helliphelliphellip(i)

By above property = + minusintb

I f(a b x)dx hellip(ii)

Adding (i) and (ii) we have λ minus

=(b a)I

Property 11 = = gt int int int

bkb bk

a akak

(Streching) (contraction)

1 xf(x)dx f dx k f(kx)dx k 1k k

ie when we stretch graph k times area

increases lsquokrsquo times Therefore we divide by lsquokrsquo to keep the value of integral unchanged

Property 12 Transformation of a definite integral into other with new limits 0 to 1

Let = intb

I f(x)dx be the given definite integral

Let x = lt + m (l m constants) ie we can always choose a linear substitution such that t = 0 at x = a and t = 1 at x = b

a = l(0) + m and b = l(1) + m rArr m = a and l = b ndash m = b ndash a x = (b ndash a)t + a rArr dx = (b ndash a)dt

Thus ( ) = minus minus + int intb 1

f x dx (b a) f (b a)x a dx

Property 13 minus

minus = minus= minus =

int inta

0 if f ( x) f(x) ie f is odd functionf(x)dx

2 f(x)dx if f( x) f(x) ie f is even function

Property 14 (a) = + minusint int int2a a a

f (x)dx f(x)dx f(2a x)dx

(b) = minus + +int int int2a a a

f (x)dx f(a x)dx f(a x)dx

Property 15 minus = minus= minus =

int int

0 if f (2a x) f(x)f(x)dx

2 f(x)dx if f(2a x) f(x)

or + = minus minus= + = minus

int int

0 if f (a x) f(a x)f(x)dx

2 f(x)dx if f(a x) f(a x)

Equivalently int2a

f (x)dx =

= inta

0 if graph of f(x) is symmetric about point (a 0)

2 f(x)dx if graph of f(x) is symmetric about line x a

Property 16 Integral of an Inverse Function If f is an invertible function and f is continuous then definite integral of fndash1 can be expressed in terms of definite integral of function f(x)

ie minus = minus minusint intf (b) b

f (a) a

f (y)dy bf(b) af(a) f(x)dx

minus= int

cf (x)dx = int

af (x)dx

2671 Evaluation of Limit Under Integral SignThe limit of a function expressed in the form of definite integral can also be evaluated by first finding the limit of the integrand function wrt a quantity of which the limit of integration are independent and

subsequently integrating the result thus obtained eg ( )β β

rarr rarrα α

=int intx k x klim f(x t)dt lim f(x t) dt

2672 Leibnitzrsquos Rule for the Differentiation Under the Integral Sign

(a) If f is continuous on [a b] and f(x) and y(x) are differentiable functions of x whose values lie in

[a b] then ψ

ψ φ= ψ minus φint

d d df(t)dt f (x) f (x)dx dx dx

(b) If the function f(x) and y(x) are defined on [a b] and differentiable at each point x isin(a b) and f(x t)

is continuous then ( )ψ

φint(x)

d f(x t)dtdx

part ψ φ+ ψ minus φ

partint(x)

d (x) d (x)f(x t)dt f(x (x)) f(x (x))x dx dx

(c) If f(x a) be a continuous function of x for x isin [a b] and a isin [c d] let α = αintb

I( ) f(x )dx is a function

of a then part

α = αpartαint

I ( ) f(x )dx

Property 17 If f(x) is an odd function of x then intx

af (t)dt is an even function of x

Property 18 If f (x) is an even function of x then intx

af (t)dt is an odd function of x iff =int

0f (t)dt 0

Property 19 If f(x) is a periodic function with period T ie f(x) = f(x + T) then the following properties hold good

q =int intnT T

f(x)dx n f(x)dx where n is a positive integer

Property 20 If f (x) is a periodic function with

period T then +

inta T

af (x)dx is independent of a

Hence prove that +

=int inta T T

f (x)dx f(x)dx

Corollary +

=int inta nT T

a 0f (x)dx n f(x)dx Where n isin ℤ+

Property 21 If f(x) is a function such that f(x) ge 0 forall x isin [a b] then geintb

f (x)dx 0

Property 22 If f(x) gt g(x) forall x isin [a b] then gtint intb b

f (x)dx g(x)dx

Property 23 If f(x) g(x) h(x) are continuous functions such that

g(x) le f(x) le h(x) in [a b] then le leint int intb b b

g(x)dx f(x)dx h(x)dx

Application To prove that lt ltintb

k f(x)dx k where k1 k2 isin ℝ It is

suggested to find two functions g(x) and h(x) Such that

geintb

g(x)dx k and leintb

h(x)dx k then prove that g(x) le f(x) le h(x)

rArr lt ltint int intb b b

g(x)dx f(x)dx h(x)dx rArr le lt lt leint int intb b b

1 2a a a

k g(x)dx f(x)dx h(x)dx k rArr lt ltintb

k f(x)dx k

Property 24 leint intb b

f (x)dx | f(x)|dx where f(x) is continuous and bounded on (a b)

Discussion This is derived from generalized form of polygonal inequality and can be understood as below

= minus + le = + +int intb b

1 2 3 1 2 3a a

f (x)dx | A A A | | f(x)|dx A A A

where A1 A2 A3 are magnitudes of areas as shown above

Here leint intb b

f (x)dx | f(x)|dx = =

∆ le ∆sum sumn n

k 1 k 1

f (x ) x | f(x ) x |

Property 25 (Max-Min inequality) If m and M are respectively the global minmax values of f(x) in [a b] then

minus le le minusintb

m(b a) f(x)dx M(b a)

Property 26 If the function f(x) increases and has a concave graph in the interval [a b] that is f (x) and f (x) both positive

(+ve) then + minus lt lt minus

f (a) f(b)(b a)f(a) f(x)dx (b a)2

Property 27 If the function f(x) increases and has a convex upwards (or concave downwards) graph in the interval [a b] that is f (x) is positive (+ve) and f (x) is negative (ndashve) then

+ minus lt lt minus int

f (a) f(b)(b a) f(x)dx (b a)f(b)2

Property 28 SchwarzndashBunyakovsky Inequality If f(x) and g(x) are two functions such that f2(x) and

g2(x) are integrable then ( )( )leint int intb b b2 2

a a af (x)g(x)dx f (x)dx g (x)dx

Property 29 If f(x) is continuous in [a b] then there exists a

point c isin (a b) such that = minusintb

f (x)dx f(c)(b a) and the number

=minus int

1f(c) f(x)dxb a

is called mean value of the function f(x) on the

interval [a b]

Evaluating Integrals Dependent on a Parameter

Property 30 Suppose f (x a) and fprime(x a) are continuous functions when c le a le d and a le x le b

then primeα = αintb

aI ( ) f (x )dx (where Iprime(a)) is the derivative of I(a) wrt a and fprime(x a) is the derivative of

f(x a) wrt a keeping x constant α = αintb

I( ) f(x )dx then ( )part= α = α

α partαintb

dI I ( ) f(x ) dxd

2673 Evaluate of Limit of Infinite Sum Using Integration

To evaluate rarrinfinn

limg(n) (when g(n) can be expressed as infinite sum) using definite integral follow the steps given here

Step I Express the function g(n) in terms of infinite summation using sigma notation

minus minus = +

(b a) b ag(n) f a rn n

Step II Replace minus + rarr

b aa r xn

and minus rarr

b a dxn

Step III β

rarrinfin= αsum int

lim converts to where rarrinfin

minus β = + =

b alim a r bn

rarrinfin

minus α = + =

b alim a r an

Step IV rarrinfin

minus minus + =

sum intbn

n r 1 a

b a b alim a r f(x)dxn n

q When domain of f(x) is divided into unit length sub-intervals each of which further divided into n subintervals Interval [a b] contains p(n)th to q(n)th stripes Then algorithm becomes

Step II Replace rarrr xn

and rarr

Step III β=

r q(n)

n r p(n)

lim converts to where rarrinfin rarrinfin rarrinfin rarrinfin

β = = α = =

max min

n n n n

r rq(n) p(n)lim lim lim limn n n n

Step IV β

rarrinfin= α

sum intq(n)

n r p(n)

r 1lim f f(x)dxn n

268 WAllirsquos FormulAe

1 For π π

isin =int int

2 2n n

n sin x dx cos x dx =

minus times minus times minus π times minus times minus

minus times minus times minus times minus times minus

(n 1) (n 3) (n 5) If n is evenn (n 2) (n 4) 2

(n 1) (n 3) (n 5) If n is oddn (n 2) (n 4)

2 For π

isin int

m n sin x cos x dx = minus times minus minus times minus+ + minus

(m 1) (m 3)(n 1) (n 3) p(m n)(m n 2)

Where p = p2 if both m and n are even otherwise p = 1

2681 Wallirsquos Product

We can express p2 in the form of infinite product given by rarrinfin

π = minus +

2462n 1lim 2 135(2n 1) (2n 1)

2682 Some Important Expansion

+ + + + infin =2

2 2 2 2

1 1 1 1 1 2 3 4 6

+ + + + infin =2

2 2 2 2

1 1 1 1 1 3 5 7 8

minus + minus + infin =2

2 2 2 2

1 1 1 1 1 2 3 4 12

+ + + + infin =2

2 2 2 2

1 1 1 1 2 4 6 8 24

5 = minus + minus + infin1 1 1ln2 1 2 3 4

26821 Root mean square value (RMSV)

RMSV of a function y = f(x) in the range (a b) is given by minus

intb 2

a[f(x)] dx

269 BetA Function

It is denoted by B (m n) and is given by minus minus= minusint1 m 1 n 1

0B(m n) x (1 x) dx where m n gt 0 It can be proved that

for m n isin (0 1) the above improper integral is convergent however the proof is beyond the scope of this book Clearly B (m n) is proper for m n ge 1

If (2mndash1) and (2nndash1) are positive integers then minus minus= minusint1 m 1 n 1

0B(m n) x (1 x) dx

Let us substitute x = sin2q

rArr π minus minus= θ θint

2 2m 2 2n 2

0B(m n) sin (cos ) 2sinq cos qdq

rArr π minus minus= θ θ θint

2 2m 1 2n 1

0B(m n) sin cos d

By Wallirsquos formula

minus minus minus minus=

+ minus + minus

2 (2m 2)(2m 4) (2n 2)(2n 4B(mn) p

(2m 2n 2)(2m 2n 4)

Where p = p2 if both (2mndash1) and (2nndash1) are even integers otherwise p = 1

2610 gAmmA Function

The improper integralinfin

minus minusint x n 1

e x dx where n is a positive rational number is called gamma function and is

denoted by n

Thus = minus minus = minus minus minusn (n 1) (n 1) (n 1)(n 2) (n 2) and so on (By previous illustration)

26101 Properties of Gamma Function

(i) = minus(n) (n 1) if n is a positive integer

(ii) = minus = =1 (1 1) 0 1

(iii) = minus minus(n) (n 1) n 1 eg = =5 4 4 4(3) 3 = 4(3)(2) 2

(iv) =infin0

(v) = π12

26102 Relation Between Beta and Gamma Functions

For gt =+

m nm n 0 B(m n)

Remark

If m n are positive integers then m 1 n 1

( m n)m n 1

β minus minus=

+ minusas n n 1= minus for n isinℕ

2611 Weighted meAn vAlue theorem

If f(x) and g(x) are two continuous functions on [a b] such that g(x) does not change its sign in [a b] then

there exists c isin [a b] such that =int intb b

a af (x)g(x)dx f(c) g(x)dx

26111 Generalized Mean Value Theorem

If g(x) is continuous [a b] and f(x) has derivative function which is continuous and never changes its sign

in [a b] Then there exists some c isin [a b] such that = +int int intb c b

a a cf (x)g(x)dx f(a) g(x)dx f(b) g(x)dx

2612 determinAtion oF Function By using integrAtion

Let f(x) be a given continuous and differentiable function Sometimes we are given a functional equation connecting the functional values at different points or function with some definite integral having integrand as f(x) or f (x) or any other algebraic or trigonometric or exponential function Then by differentiating and integrating we can find the function f(x)

AREA unDER thE CuRVE

2613 AreA Bounded By single curve With x-Axis

(a) If f(x) is a continuous function in [a b] then area bounded by

f(x) with x-axis in between the ordinates x = a and x = b is given

by = intb

A f(x) dx

(b) If f(x) is discontinuous function in [a b] say at x = c isin (a b) then

= +int intc b

A f(x)dx f(x)dx

26131 Area Bounded by Single Curve with y-axis

(a) If f(x) is a continuous function in [a b] such that f(a) = c and

f(b) = d then the area bounded by the function f(x) with y-axis and

abscissa y = c and y = d is given by minus= =int intd d

A (x) dy f (y) dy

(b) If f(x) is discontinuous function in [a b] at x = c then fndash1(y) is also

discontinuous at y = f(c) = e(say) then minus minus= +int inte d

A f (y)dy f (y)dy

26132 Sign Conversion for Finding the Area Using Integration

For the intervals where f(x) ge 0 take integrand f(x) and for the intervals where f(x) le 0 take integrand ndashf(x) eg as given in the figure given below

= = + minus + + minus +int int int int int intb c d e f b

a a c d e f

A f(x)dx f(x)dx f(x)dx f(x)dx f(x)dx f(x)dx

26133 Area Bounded Between Two Curves

(a) If f(x) and g(x) are two continuous function functions on [a b] then the area bounded between two

curves and the ordinates x= a and x = b is given by = minusintb

A f(x) g(x)dx

(b) Area bounded between the curves f(x) g(x) and the abscissa y = c and y = d are given by

minus minus= minusintd

A | f (y) g (y)|dy

26134 Area Enclosed by Inverse Function

Area enclosed by y = fndash1 (x) and x-axis between ordinate x = a and x = b is same as area enclosed y = f(x) and y-axis from y = a to y = b Clearly from above figure the area bounded by y = f(x) with y-axis from y = a to y = b and fndash1(x) with x-axis from x = a to x = b are same as y = f(x) and fndash1(x) are reflection of each other on line y = x

26135 Variable Area its Optimization and Determination of ParametersIf the region bounded by curve is continuously changing due to some variable ordinate or abscissae or any other parameter present in the boundary curve then we obtain a variable area function that can be optimized with respect to involved parameters eg

Area = =3

1tOAB A4

and Area of parabolic region =3

Thus A1 and A2 can be optimized for parameter t

261351 Least value of variable area

Let f(x) be a monotonic function with f rsquo(x) ne 0 in (a b) then the area

bounded by function y = f(x) y = f(c) (a lt c lt b) And ordinates

x = a x = b is minimum for +

261352 Method of tracing the region represented by inequality

Each curve f(x y) = 0 divides the entire x ndash y plane into three set of points as given in figure

R1 = (x y) f(x y) = 0 x y isinℝ ie the points lying on the curve f(x y) = 0

R2 = (x y) f(x y) gt 0 x y isinℝ ie the points lying on one side of the curve f(x y) = 0 (outside the curve if closed)

R3 = (x y) f(x y) lt 0 x y isinℝ ie the points lying on other side of f(x y) = 0 (inside the curve if closed)

Steps to Identify the Region Represented by a Given Inequality (say) f(x y) gt 0

Step I Consider the equality and draw the curve using the symmetry and other concepts of curve sketching and transformation of graphs

Step II Consider any points (a b) not lying on the curve preferably (0 0) or point on coordinate axis and determine the sign of f(a b)

Step III If f(a b) gt 0 then f(x y) gt 0 represents the region containing (a b) If f(a b) lt 0 then the region which does not contains point (a b) will be represented by inequality f(x y) gt 0

Note that the region represented by inequality f(x y) gt 0 or f(x y) lt 0 does not contain the points on the curve whereas the region represented by inequality f(x y) ge 0 and f(x y) le 0 contains the points on the curve

26136 Determination of Curve When Area Function is Given

If the area bounded by some function and x-axis between x = a and x = b is given g(a b) forall a gt b where a is a given real number and b is a real parameter then the function can be obtained as described below Let the unknown function be y = f(x)

Q intb

f (x)dx represents area enclosed between f(x) and x-axis between the semi-variable boundaries

x = a and x = b as b is a real parameter and it is given as g(a b) Of course area changes by variation in lsquobrsquo but always the value of area shall be represented by a function g(a b)

Thus =intb

f (x)dx g(a b) Now differentiating both sides wrt b we get ( )=df(b) g(ab) =g (ab)

rArr f(b) = plusmn g(a b) consequently determining two curves f(x) = g(a x) or f(x) = ndashg(a x)

2614 AreA enclosed in curved looP

Any curve forming loop is multi-valued function so first of all solve the equation of curve for y to find its functional branches and obtain the domain of function say [a b] eg ax2 + by2 + 2hxy + 2gx + 2fy + c = 0

(say) solving for y we get+

f (x) g(x)y

minus=2

f (x) g(x)y

Clearly there two functions are forming the loop Area of loop = β β

minus =int int2 1| y y |dx g(x) dx

Area enclosed by curve between two radius vectors when its equation is given in polar form

If r = f(q) is the equation of curve in polar form where f(q) is a continuous function of q then the area enclosed by curve r = f(q) and the radius vectors r = f(a) and r = f(b) (a lt b) is given by

= θ = θ θint int2 21 1A r d [f( )] d2 2

NoteIn order to transform the Cartesian equation of a curve to polar form we replace x by r cosq and y by rsinq

( ) ( ) ===

Catesianequation

f r cos r sin 0x r cosf x y

y r sin polar equation

θ θθθ

26141 Graphical Solution of the Intersection of Polar Curves

The following steps are taken to find the points of intersection of polar curves

Step 1 Find all simultaneous solutions of the given system of equations

Step 2 Determine whether the pole lies on the two graphs

Step 3 Graph the curves to look for other points of intersection

Area enclosed by curve having their equations in parametric formLet y = f(x) be a continuous function on closed interval [a b] and let x = g(t) and y = h(t) be its parametric equations with domain t isin [t1 t2] such that g(t1) = a and g(t2) = b Let the traced curve be simple Its derivative function g(t) is continuous on [t1 t2] then the area under the curve is given by

= = = ge isinint int int2

1 2a a t

A y dx h(t)d(g(t)) h(t)g (t)dt y 0 for t [t t ]

Area bounded by a closed curve defined in parametric form

Consider a closed curve represented by the parametric equations x = f (t) y = f (t) lsquotrsquo being the parameter We suppose that the curve does not intersect itself Also suppose that as the parameterlsquotrsquo increases from value t1 to the value t2 the point P(x y) describes the curve completely in the counter clockwise sense The curve being closed the point on it corresponding to the value t2 of the parameter is the same as the point corresponding to the value t1 of the parameter Let this point be C

It will now be shown that the area of the region bounded by such a curve is minus int

1 dy dxx y dt2 dt dt

Chapter 27Differential equation

271 IntroductIon

ldquoDifferential Equationrdquo as the term signifies is an equation involving derivatives of dependent variables (y) wrt dependent variables and other trigonometric and algebraic functions of independence variables (x)

272 dIfferentIal equatIon

An equation involving independent variable x dependent variable y and differential coefficients of one or more dependent variables with respect to one or more than one independent variables is called a differential equation

Example 32 2

dy d y xdx dx

2 2dy dz dyy x z y xdx dx dx

+ = + = or 2 2 2

z z z4xx x ypart part part minus = part part part

2721 Types of Differential Equation

27211 Ordinary differential equations

An equation involving only one independent variable and ordinary derivatives with respect to that is

known as ordinary differential equation (ODE) For instance 2

dy d yF x y 0dx dx

is a standard form

of an ordinary differential equation

Example sin2x dx + e2y dy = 0 or

32 22 3

dy d y1dx dx

27212 Partial differential equations

Differential equation involving two or more independent variables and partial derivatives with respect to

these is known as partial differential equation (PDE) For example 2 2z xy xy x zx ypart part

+ =part part

2722 Order and Degree of Differential Equation

27221 Order

It is the order of the highest order derivative appearing in the differential equation The order of differential equation states about the number of times the family of curve has been differentiated in order to eliminate all its parameters to form the given differential equation Therefore the order of differential equation is same as the number of effective arbitrary constants present in the family of curves from which the differential equation is derived

Example 2dyx x y sin x log xdx

+ = is of 1st order where as 2

d y dy3x x edx dx

minus = is of second

27222 Degree

Degree of differential equation is the degree ie power of the highest order derivative present in the differential equation after the equation is made free from all radicals and fractions as far as derivatives are concerned and is written in terms of a polynomial in all differential coefficients (ie derivative involved in differential equation

To get the degree of the differential equation we first try to convert it into the following form

d yf x ydx

+ ( ) ( ) 1nm 1

d yg x y

+ ( ) ( ) 2nm 2

d yh x y 0

hellip(1)

is of order m and degree p Where m n1 n2 hellip nk are positive integers

273 lInear dIfferentIal equatIon

If the dependent variables and all its derivatives present occur in the first degree only that is neither the product of dependent variables nor product of derivative nor product of derivative and dependent vari-ables are present in the differential equation

274 non-lInear dIfferentIal equatIons

A differential equation which is not linear is termed as a non linear differential equation For instance the

differential equation 2

d y dy2x x x y 0dx dx

minus + = is linear while 3 2

d y d y dyx 3y 0dx dx dx

+ minus = is non linear

2741 Formation of Family of Curves

We know that differential equations are used to represent a family of curves Family of curves is defined as a set of infinite curves having some common characteristics and atleast one variable characteristic which is represented by unknown parameter involved in the equation of family of curves (ie atleast one parameter) By assigning different values to these parameters (arbitrary constants) different members of the family of curves can be obtained

Differential Equation 27339

27411 Single parameter family

The family of curves having only one arbitrary constant eg y = mx + 2 family of lines passing through y = 2 x2 + y2 = k2 family of concentric circles y2 = kx family of parabolas y = x + k set of parallel lines

27412 Double parameter familyThese are the equations having two parameters so called family of family of curves

eg y = mx + c denoting family of family of lines in xy planeeg two families of the above family of family of lines are shown in the figure y = x + k when m = 1 y = mx + 2 when c = 2Similarly other two parameter family of curves are (x ndash h)2 + y2 = r2 (circles) y = A sin x + B cos x

y = Ax2 + Bx etc

2742 Formation of Differential EquationThe differential equation of the family of curves f(x y C1 C2 C3Cn = 0) with parameters C1 C2 C3Cn can be found by differentiating it n-times and eliminating the n-parameters by using above (n + 1) equations (One given equation and n other equation obtained by differentiating it n-times)

For instance y = Asinx + Bcosx helliphelliphellip(i)

rArr dy A cos x Bsin xdx

= minus helliphelliphellip(ii)

rArr 2

d y Asin x Bcos xdx

= minus minus helliphelliphellip(iii)

From (iii) 2

d y (Asin x Bcos x) ydx

= minus minus = minus

rArr 2

d y y 0dx

+ = is the differential equation of gives family of curves (i)

RemarksIt is evident that a differential equation of the nth order cannot have more than n arbitrary constants in its solution for if it had say (n + 1) on eliminating them there would appear not an equation of the nth order but one of the (n + 1)th order Finally the differential equation corresponding to a family of curves is obtained by using the following steps

(a) Identify the number of essential arbitrary constants (say n) in the equation of the curve

(b) Differentiate the equation n times

(c) Eliminate the arbitrary constants from the equation of curve and n additional equations obtained in step (b)

275 solutIon of dIfferentIal equatIon

Solving a differential equation is an inverse process of forming differential equation of a family of curves by integrating the given differential equation to obtain a general relation between the independent and de-pendent variables Therefore the solutionintegralprimitive of a differential equation is a family of curves is satisfying the differential equation in the domain of differential equation

Example y = sin x + c is solution of differential equation dy = cos x dx

276 classIfIcatIon of solutIon

2761 General SolutionA family of curves (ie relation in variables x and y) satisfying the given differential equation which contains exactly as many effective arbitrary constants as the order of differential equation is known as general solution or complete integral or complete primitives Let the equation involving the variables x y and n independent arbitrary constants be f(x y C1 C2 Cn) = 0 (i)

and the differential equation obtained from (i) be 2 n

dy d y d yF x y 0dx dx dx

Then (i) is called the general solution of (ii)

(i) The general solution of an ordinary differential equation of nth order contains n independent parameters (essential arbitrary constant) which means the general solution of an ordinary differential equation of order one contains one arbitrary parameter and of second order contains two independent parameters and so on

(ii) The general solution of a differential equation contains exactly as many essential parameters as the degree of equation While counting the parameters in the general solution it must be seen that they are essential and are not equivalent to a lesser number of parameters The parameters in the solution of a differential equation are said to be essential if it is impossible to deduce from the solution an equivalent relation containing lesser number of parameters

eg y = (A + B)ex + CeDndashx = has apparently four parameters but number of essential parameter is two this can be observed as below y = (A + B)ex + CeDendashx rArr y = C1e

x + C2endashx where C1 = A + B and C2 = CeD

27611 Particular solution

A solution obtained from the general solution by giving particular values to the arbitrary constants eg y = sinx + 1 is one particular solution of equation dy ndash cos x dx = 0

27612 Cauchyrsquos initial value problem

The problem of determining the particular solution of equation dy F(x y)dx

= (1) satisfying the

condition y(x0) = y0 (2) where x0 y0 are given numbers is referred to as Cauchyrsquos initial value problem Condition (2) is called the initial condition (boundary condition) The particular integral of equation (1) satisfying the initial condition (2) is called the solution of Cauchy problem

27613 Singular solutions

General solution of differential equations may not include all possible solutions consequently the differential equation may also possess other solutions than the general solution The solution which cannot be obtained by giving any value to arbitrary constants present in the general solution are termed as singular solution

In order to realize the existence of singular solution consider an example of family of circles (x ndash h)2 + y2 = a2 where h is parameter and a is fixed constant

It is the general solution of differential equation 2 2

dy a 1dx y

= minus

Clearly y = plusmna also satisfies the above

differential equation but it can never be obtained by providing any real value for the parameter h from the general solution Therefore y = plusmna is indeed the singular solution of the above differential equation

Method of solving a differential equations of order and degree one

27614 When f(x y) is function of only x or only y

If the equation is of the form dy f(x)dx

= hellip(i)

and dy f(y)dx

= hellip(ii)

The equation of type (i) will reduce to y f(x)dx C= +int whereas the equation of type (ii) will reduce to

dy dxf(y)

= that can be solved as dy x Cf(y)

= +int where C is parameter

277 VarIable separable form

The differential equation of the form dy f(x)g(y)dx

= or dy f(x)dx g(y)

= is called variable separable form

So the general form of such equation is N(y)dy = M(x) dx which can be solved by integrating

both sides ie M(x)dx N(y)dy=int int as described as here for dy f(x)g(y)dx

Step I Rearrange the expression to express it in the form f(x)dx = g(y)dy

Step II Integrating both sides we get f(x)dx g(y)dy=int int Say F(x) G(y) be some anti-derivatives of

f(x) and g(y) respectively we get G(y) = F(x) + CStep III Solving the equation G(y) = F(x) + C for y we express the general solution as y = H(x C)

2771 Equations Reducible to Variable Separable Form

Type A dy f(ax by c)dx

= + + where b ne 0

Algorithm The differential equations are of the form dy f(ax by c)dx

= + + are reducible to variable sepa-

rable form by substituting ax + by + c = t The substitution reduces the differential equation to dy dta bdx dx

rArr 1 dt a f(t)b dx minus =

rArr dt bf(t) adx

rArr dt dx

bf(t) a=

+int int which can be solved as it is clearly in variable separable form

Type B Equation of type 1 1 1

a x b y cdydx a x b y c

+ + where b1 + a2 = 0

Consider 1 1 1

+ + where b1 + a2 = 0 cross multiply and observe the perfect differential of

xy and then integrate term by term rArr a2(xdy + ydx) + (b2y + c2)dy = (a1x + c1) dx Now integrate a2d(xy) + (b2y + x2)dy = (a1x + c1)dx

on integration we get 2 2

2 12 2 1

b y a xa (xy) c y c x C2 2

+ + = + +

Type C Equation of type Any equation of the form R(x2 + y2 x dx + ydy x dy ndash ydx) = 0Some times transformation to the polar co-ordinates facilitates separation of variablesSubstitute x = r cos q andy = r sin q

rArr x2 + y2 = r2 (1)

and y tanx= θ hellip (2)

Differentiating (1) wrt any variable we get xdx + ydy = rdr (3)

Differenting (ii) wrt x rArr 22

xdy y ddx secx dx

minus θ= θ

rArr xdy ndash ydx = x2sec2qdq = r2dq (4)Now the equation reduces to R (r2 rdr r2dq) = 0

Type D Equation of type Any equation of the form R(x2 ndash y2 x dx ndash ydy xdy ndash ydx) = 0Substitute x = r secq and y = r tan q

rArr x2 ndash y2 = r2 (1)

And y sinx= θ hellip(2)

differentiate equation (1) wrt any variable we get xdx ndash ydy = rdr hellip (3)

Differentiate equation (2) wrt x we get 2

xdy ydx cos dxminus

= θ θ

rArr xdy ndash ydx = r2 sec qdq hellip(4)Now the equation gets reduced to R(r2 rdr r2 sec qdq) = 0

RemarksMemorising the following differentials of course helps If x = r cosq y = r sinq then (a) x dx + y dy = r dr (b) (dx)2 + (dy)2 = (dr)2 + r2 (dq)2

(c) x dy ndash y dx = r2 dq If x = r secq and y = r tanq then x dx - y dy = r dr (b) x dy ndash y dx = r2 secq dq

Type E Equation of type In an equation of the form yf1 (xy) dx + xf2 (xy) dy = 0 the variable can be separated by the substitution xy = v and finding an equation in x and v (variable separated form)

2772 Homogeneous Differential Equation

27721 Homogeneous function

f(x y) is said to be homogenous expression of its variable of degree n iff it can expressed as

n yf(x y) xx

or n xyy

it satisfies the identity f(tx ty) = tn f(x y)

27722 Homogeneous differential equation

A differential equation of the form dy f(x y)dx (x y)

where f(x y) and f(x y) are homogenous functions of x

y and of the same degree or dy F(x y)dx

= iff F is homogenous function of zero degree in variable x and y

Since the above equation can be reduced to the form dy yGdx x

Therefore it can be solved by reducing

them to variable separable form using the substitution y vx= ie y = vx

278 solutIon of Homogeneous dIfferentIal equatIon

By using proper substitution each homogeneous differential equation can be converted to variables sepa-rable type differential equation

Type A Differential equation of type dy yfdx x

To solve this kind of equation substitute y vx=

Type B Differential equation of type y f(x y) dx + xg(xy)dy = 0 (i)Substitute xy = u

279 equatIons reducIble to tHe Homogeneous form

To solve the equation of form 1 1 1

dy ax by cdx a x b y c

Case I If 1 1 1

a b ca b c= ne then ax + by + c = 0 and a1x + b1y + c1 = 0 represent parallel

lines let a 1 1

a b ka b= =

rArr a = ka1 b = kb1 rArr ax + by = k(a1x + b1y) helliphelliphellip(i)

Now substitut a1x + b1y = v rArr 11

b (kv c)dv adx v c

+ rArr 1

1 1 1 1 1

(v c )dv dx[(kb a )v (b c a c )

+ + + int

which can be solved further to get the general solution of given differential equation

Case II If 1 1 1

a b c ka b c= = = then ax + by + c = 0 and a1x + b1y + c1 = 0 are coincident lines

rArr ax + by + c = k(a1x + b1y + c1) on substituting a1x + b1y + c1 = v helliphelliphellip(i)

rArr 1dv k adx

= + which gives us the required solutions of given differential equation

Case III If a1 + b = 0 then on cross multiplication we have a1(xdy + ydx) = (ax + c)dx ndash (b1y + c1)dyor a1d(xy) = (ax + c)dx ndash (b1y + c1)dy

rArr 22

b yaxa xy cx c y C 2 2

= + minus + +

Which gives us the general solution of given differential equation

Case IV Equation of the form 1 1 1

+ + where

nea ba b

can be reduced to

a homogeneous form by substituting x = X + h and y = Y + k where h and k are constants which are to be determined

Such that the given differential equation has no constant terms in numerator and denominator

Now 1 1

a ba bne

rArr ax + by + c = 0 and a1x + b1y + c1 = 0 have unique solution say (h k) ie unique point of intersection

Now x = X + h y = Y + k rArr dx = dX dy = dY

gives dy dYdx dX

= now given differential equation becomes 1 1

dY a(X h) b(Y k) cdX a (X h) b(Y k) c

+ + + +=

+ + + +

rArr 1 1

dY aX bYdX a X b Y

+ choosing h k so that ah + bk + c = 0 and a1h + bk + c1 = 0(say) (h = h1 k = k1)

Now put Y = VX rArr dY dVV XdX dX

rArr 1 1

dV a bVV XdX a b V

+ which is in variable separable form and can be solved further for giving a rela-

tion in X Y and V Resubstituting X = x ndash h1 and Y = y ndash k1 and V = YX we get required general solution of given differential equation

2710 exact and non-exact dIfferentIal equatIon

A differential equation is called exact iff it can be written as complete differential of some function of x y without any rearrangement ie without any further operation of elimination or reduction by multiplying with any function of x and y

Theorem Mdx + Ndy = 0 where M and N are function of x and y is exact iff M N y x

part part=

part partSolution of exact differential equation is given by

y constt

Mdx (terms of N notcontaining x)dy cminus

+ =int int

27101 Method of Solving an Exact Differential Equation

(a) General method Step I Integrate M with respect to x regarding y as a constant Step II Then integrate with respect to y those terms in N which do not involve x Step III The sum of the two expressions thus obtained equated to a constant is the required

solution (b) By method of inspection To solve the exact differential equations we use the knowledge of standard complete differential

expressionq dx plusmn dy = d(x plusmn y) q xdy + ydx = d(x y)

q 2 21xdx ydy d(x y )2

plusmn = plusmn q 2

ydx xdy xdy y

minus=

xdy ydx ydx xminus =

x dy y dx1dxy x y

+ minus =

2xy dx x dyxdy y

minus =

2xy dy y dxydx x

minus =

q 2 2 2

x 2xy dx 2x ydydy y

minus=

y 2yx dy 2y xdxdx x

minus=

q 12 2

x ydx xdyd tany x y

minus minus= +

q 12 2

xdy ydx yd tanx y x

minusminus = +

q x dy y dx

d(ln(xy))xy+

= q 2 22 2

1 xdx ydyd ln(x y )2 x y

+ + = +

q y dx x dyxd ln

y xy minus

x dy y dxyd lnx xy

minus =

q x x x

e ye dx e dydy y

minus=

e xe dy e dxdx x

minus=

1 x y xdy ydxd log2 x y x y

+ minus= minus minus

q ( )2 2

x dx y dyd x y

d[f(x y)] f (x y)1 n (f(x y))

minus prime=

2711 non-exact dIfferentIal equatIon

If Mdx + Ndy = 0 is a non-exact equation if Mdx + Ndy is not a complete differential of a function f (x y)

and that is only when M Ny x

part partne

part part eg ydx ndash x dy = 0 is non-exact equation

27111 Integrating FactorIf a non-exact equation is multiplied by a function of x y and the equation becomes exact (ie differential) then such function is known as integrating factor

27112 Leibnitz Linear Differential EquationA differential equation in which the dependent variable and its differential coefficients occur only in first degree and are not multiplied together is called a linear differential equation Linear equation of nth order

is given as n n 1

0 1 n 1 nn n 1

d y d y dya (x) a (x) a (x) a (x) (x) 0dx dx dx

minusminus+ + + + + +ϕ = where a0 a1an are functions of

only x is nth order linear differential equation (ie no term contains product of powers of y and derivatives or higher power of derivatives) Those which are not linear are termed as non linear differential equations

Remark

A linear differential equation is always of the first degree but every differential equation of the first degree need

not be linear eg the differential equation

d y d y2 5 y 0

d x d x is not linear though its degree is 1

27113 First Order Linear Differential Equation

The differential equation of the form ( ) ( )dy P x y Q xdx

+ = where P(x) and Q(x) are functions of only x is

called a first order linear differential equation It is non-exact equation The integrating factor (IF) for such

equation is ( )P x dx

eint For example the GS of the above equation is ( )y(IF) Q(x) IF dx c= +int

RemarkSome times a given differential equation becomes linear if we take y as the independent variable and x as the dependent variable

271131 Equation reducible to linear form (bernoullirsquos differential equation)

An equation of the form ndy P(x)y Q(x)ydx

+ = (where P and Q are either functions of x or constants

and n is a constant other than 0 or 1) is called Bernoullis differential equation

Given differential equation can be written as n n 1

1 dy 1 P(x) Q(x)y dx y minus+ = helliphelliphellip(i)

To reduce (i) to linear form substitute n 1

1 uy minus = (ii)

rArr du (1 n)P(x)u (1 n)Q(x)dx

+ minus = minus

Which is a linear differential equation in u and hence can be solved for u Resubstituting n 1

1uy minus=

we get solution to original differential equation

271132 Differential equation reducible to linear differential equation by substitution

dyf (y) Pf(y) Qdx

+ = where P and Q are functions of x or constants can be reduce to linear differential

equation if we put f(y) = v so that dy dvf (y)dx dx

The equation becomesdv Pv Qdx

+ = which is a linear equation in v and x

NoteIn each of these equations single out Q (function on the right) and then make suitable substitution to reduce the equation in linear form

27114 Differential Equation of First Order and Higher Degree

Type (A) Equation solvable for p

If (p ndash f1(x y)) (p ndash f2(x y)) (f ndash fn(x y)) = 0 then pk = fk(x y) each of these is of first order Let gk(x y

Ck) = 0 where k = 1 2n are solution then general solution is n

k kk 1

g (x yC ) 0=

=prod It contains n arbitrary

constants but being an equation of first order it must contain only one arbitrary constant so without loss

of generality we can take C1 = C2 = hellip = Cn = C Hence general solution is n

k kk 1

g (x yC ) 0=

Type (B) Equation solvable for yEquation can be expressed y = f (x p) (i)

Differentiating both sides with respect to x dy dph x pdx dx

Let the solution of (ii) be g (x p c) = 0 (iii)Eliminating p between (i) and (iii) we get relation between x y and c and is general solution

Type (C) Clairauts equationIt is equation of first degree in x and y of the form y = px + f (p) (i)

where dypdx

= Differenting both sides wrt x we get dy dpp (x f (p))dx dx

prime= + +

rArr dpp p (x f (p))dx

prime= + + rArr dp(x f (p)) 0dx

prime+ =

rArr either (x + f (p)) = 0 or dp 0dx

= if dp 0dx

rArr p = c (constant) helliphelliphellip(ii)Eliminating p from (i) and (ii) we have y = Cx + f(c) as a solutionIf x + f (p) = 0 then by eliminating p again we obtain another solution

Remarks 1 Some given differential equation can be reduced to clairauts form by suitable substitution

2 To obtain general solution of differential equation in clairauts form simply replace p by (c)) in the given equation y = px + f(p)

Type (D) Equation solvable for xLet the given differential equation be of the form x = f (y p) (i)

Differentiating with respect to y we get dx 1 dph ypdy p dy

Linear is in y and p so solve to get solution g (y p c) = 0 (iii)Eliminate p from (i) and (iii) to get relation set x y and cIf it is difficult to eliminate then eliminate of (i) and (iii) will the required general solution

2712 HIgHer order dIfferentIal equatIon

q Equation of Type 2

d y f(x)dx

= This requires merely ordinary integrations wrt x

q Equation of Type2

d y f(y)dx

= helliphellip(1)

Method 1 Multiply both sides by dydx

rArr 2

dy d y dyf(y)dx dx dx

= rArr 2

dy d y dx f(y)dydx dx

= and then integrating both sides

rArr dy dyd f(y)dydx dx

= int int rArr

21 dy f(y)dy C2 dx = + int (2)

Method 2 Given 2

d y f(y)dx

Let symbol p denote dydx

rArr 2

d y dp dp dy dpp dx dx dy dx dy

= = = therefore equation (1) becomes dpp f(y)dy

Consider it as a first order differential equation with p as dependent and y as independent variable

rArr pdp f(y)dy=int int rArr 21 p f(y)dy A2

= + int which is equivalent to equation (2)

rArr dy dx

2 f(y)dy 2A= plusmn

+int Now solve as the variable are separated

2713 Integral equatIons and tHeIr solVIng metHod

Some equations involve the unknown function f(x) under integral sign and are called integral equations To solve such equations differentiate the equation and form a differential equation and thereafter solve the obtained equation

27131 Orthogonal Trajectory of a Given Curve

271311 Trajectory

A curve of family or curves f(x y c) = 0 which cuts every member of a given family of curves f (x y c) = 0 according to a fixed rule is called a trajector of the family of curves

If we consider only the trajectories cutting each member of fam-ily of curves f (x y c) = 0 at a constant angle then the curve which cuts every member of a given family of curves at right angle is called an orthogonal trajectory of the family

In order to find out the orthogonal trajectories the following steps are taken

Step 1 Let f(x y c) = 0 be the equation where c is an arbitrary parameter

Step 2 Differentiate the given equation wrt x and eliminate c

Step 3 Substitute dxdy

minus for dydx

in the equation obtained in Step 2

Step 4 Solve the differential equation obtained from Step 3

2714 applIcatIon of dIfferentIal equatIon

1 Radioactive Decay If f(t) = 0 is the amount present at a time t then f (t) represents the rate of

change of amount at time t then law of decay states that df(t) Kf(t)

dt= minus

2 Falling body in a resisting medium If a body of weights m falling in a resisting medium then according to Newtonrsquos law we get the following equation ma = mg ndash kv

rArr dv k v gdt m

+ = It represents a linear different equation

3 Cooling problem If f(t) is the temperature of the body at time t and if M(t) denotes the (known) temperature of the surrounding medium then according to Newtons law of cooling df(t) k[ f(t) M(t)]

dt= minus minus

4 Dilution problem If f(t) denote the number of points of salt in the tank at time t minutes after mixing begins Then according to dilution problem two factors which cause f(t) to change the incoming mixture per minute (k) and outgoing mixture which removes salt R f(t)

Then df(t) k R f(t)

dt= minus

5 If voltage is denoted by v(t) and current by I(t) then according to Kirchhoff s law (here L and R are

constants) LdI(t) RI(t) v(t)dt

+ = It represents a linear differential equation

Chapter 28VeCtors

281 Physical Quantities

A property of phenomenon body or substance which has magnitude that can be expressed as a number and a reference

Type of Physical Quantities

Directed Line Segment A line segment drawn in a given direction is called a directed line segment

A directed line segment has the following three properties

Length OA ie length of line segment OA

Supportline of supportline of action The line of which OA is a line segment

Sense The sense of directed line segment is from O to A

Representation of a vector A vector is represented by a directed line segment OA where O is called initial point and A is called terminal point of vector Length of the line segment OA is called magnitude of vector and an arrow gives the direction of a vectorThe above vector is expressed as

Notation of a vector A vector is denoted by small letters of the English alphabet under an arrow For example above

OA can be denoted by a ie =

OA a a or simply lsquoarsquo represents the magnitude of

vector called modulus of vectors

2811 Equality of Two VectorsTwo vectors are said to be equal if and only if they have (a) equal magnitudes (ie same length) (b) same direction (ie same or parallel support their lines of action may be different)

(c) same sense

Triangle law of vector addition If two vectors are represented by two adjacent sides of a triangle taken in the same order then the closing side of the triangle taken in the opposite order represents the sum of the first two vectors

28111 Parallelogram law of vector addition

If two vectors are represented by the two adjacent sides of a parallelogram both in magnitude and direction then their resultant will be given by the diagonal through the intersection of these sides (in both senses ie magnitude and direction)

Remarks

(i) Number of line segments obtained by joining two of n points (no three lying on a line) = nc2

(ii) Maximum number of vectors obtained by joining two of the n-points (no three lying on a line) = 2 times nc2

(iii) Number of diagonal obtained by joining two of n-vertices of an n-sided convex polygon = (nc2 ndash n)

(iv) Maximum number of diagonal vectors obtained by joining two on n-vertices of n-sided convex polygon = 2(nc2 ndash n)

282 classification of Vectors

Opposite Vectors (Negative Vectors)

The negative of a vector a is defined as a vector having same magnitude that of a and the direction opposite to a It is denoted as - a

Zero Vector (Null Vector)

A vector whose initial and terminal points are same is called a null vector eg

AA Such vector has zero magnitude and arbitrary (indefinite) direction It is denoted by

O + + =

AB BC CA AA or + + =

AB BC CA O Unit Vector A unit vector is a vector whose magnitude is

unity We write a unit vector in the direction

of a as a which is given by

vector along x-axis y-axis and z-axis are

denoted by i j and z respectively

Vectors 28353

CollinearParallel Vectors

Vectors having same or parallel line of action irrespective of their magnitude

Like Parallel Vectors

Two vectors having parallel line of action drawn in the same sense irrespective of their magnitude are called like parallel vectors

Unlike Parallel Vectors

Two vectors having parallel line of action drawn in the opposite sense irrespective of their magnitude are called Unlike parallel vectorsOpposite vectors are unlike parallel vectors

Free Vectors A vector a which can be represented by

any one of the two directed line segments

AB and

PQ whose lengths are equal and are in the same direction is known as a free vector Such vectors have freedom to have their initial point any where

Localized Vector If a vector is restricted to pass through a specified point (ie a fixed point) then it is called localized vector An example of a localized vector is a force as its effect depends on the point of its application Co-terminus vectors position vectors etc are examples of localized vectors

Co-initial Vectors Vectors having same initial point (say origin) are called co-initial vectors If vectors in plane (or shape) are free vectors then they can be shifted parallely and can be converted to co-initial vectors having their initial points at origin

Position Vector If P is a point having co-ordinates (x y) or (x y z) (accordingly P is in plane or space) then position vectors of point P is denoted

by r and is given by = = + +

OP r xi y j zk

Length of position vector

= = = = + +

2 2 2OP OP r r x y z

P(xyz)

A(0y0)

M(xy0)(x00)B

(00z) zk

2821 Representation of a Free Vector in Component Form

PQ is a vector with initial point P(x1 y1 z1) and terminal point Q(x2 y2 z2) then = minus + minus + minus

2 1 2 1 2 1PQ (x x )i (y y )j (z z )k

2822 Direction cosine and Direction Ratios of Vectors

Direction of a vector

OP is defined as the smallest angles which the vector

OP makes with the positive direction of co-ordinates axes

Direction cosines of

OP along x-axis = cos a = l (denotes)

OP along y-axis = cos b = m (denotes)

OP along z-axis = cos g = n (denotes)

Thus direction cosine are lt α β γ gt equivx y zcos cos cos r r r

where P(x y z) and

= + + =

2 2 2r x y z OP

Properties of Direction cosines of

1 Direction cosines have values in [ndash1 1] 2 l2 + m2 + n2 = 1 where ltl m ngt are direction cosines 3 If x = lr y = mr z = nr where ltl m ngt are direction cosines

4 If r = unit vector along r then = + +

r i mj nk where ltl m ngt are direction of r

5 Direction cosine of like parallel vectors are same eg for a and 3a 6 Direction cosine of unlike parallel vectors are numerically same but opposite sign

eg for a and minus 3a

283 addition of Vectors

If and = + +

2 2 2 2r x i y j z k then + = + + + + +

1 2 1 2 1 2 1 2r r (x x )i (y y )j (z z )k

Geometrically +

a b is the vector given by triangle law and parallelogram law of vector addition

Vectors 28355

28321 Properties of vector addition

(i) Commutative + = +

a b b a

(ii) Associative + + = + +

(a b) c a (b c) can be generalized for any number of vector

(iii) Additive Identity

0 (Null vector) is additive identity ie + = = + forall

a 0 a 0 a a

(iv) Additive Inverse minusa is additive inverse of a ie + minus = = minus +

a ( a) 0 ( a) a (v) Triangle inequality

(a) a b a b+ le +

(b) + ge minus

a b a b (c) minus le + le +

a b a b a b

(vi) The negative of a vector sum and difference of two vectors ie plusmn plusmn plusmn +

a b (a b) all lie in same plane or parallel plane

284 subtraction of Vectors

If = + +

1 1 1r x i y j z k and = + +

2 2 2 2r x i y j z k then minus = minus + minus + minus

1 2 1 2 1 2 1 2r r (x x )i (y y )j (z z )k

Geometrically Subtraction of

1 2r from r is nothing but addition of minus

1 2r and r

2841 Properties of Vector Subtraction

(i) Not commutative minus ne minus

a b b a but minus = minus minus

(a b) (b a)

(ii) Not associative minus minus ne minus minus

a (b c) (a b) c)

(iii) = minus minus = minus

a a a b b a

(iv) Triangle inequality (a) minus le +

a b a b (b) minus ge minus

a b a b (c) minus le minus le +

a b a b a b

Multiplication of a vector by a scalar l (real number)It is the product of scalar l with a

λ = λ + + = λ +λ +λ

1 2 3 1 2 3a (a i a j a k) a i a j a k rArr λ = λ a a ie length of λ

a is l times that of a

Remarks

(i) aa

is a vector along a

having unit length

= plusmn

according as b

is along or in opposite direction to that of a

(ii) Division of a

by non-zero scalar l is multiplication of a

by 1λ

ie a 1

( a)λ λ=

(iii) ( a b) a bλ λ λ+ = +

(ie scalar multiplication distributes over vector addition)

Unit vector along diagonal of a parallelogram

ie unit vector along +

a bACa b

Unit vector along angle bisector of parallelogram (a) Unit vector along internal angle bisector of angO

= unit vector along the diagonal of rhombus OLMN of unit

length = +

a ba b

(along the internal angle bisector of angO

(b) Unit vector along the internal angle bisector of angO

outwards = + = minus +

a bONa b

(c) Unit vector along the external angle bisector at O along minus +=

minus +

ˆˆ( a b)OT ˆa b

285 collinear Vectors

Vectors which are parallel to the same line are called collinear vectors irrespective of their magnitude and sense of direction

a b c d are representing collinear vectors and for collinear vectors the line of action is either same or parallel

2851 Conditions for Vectors to be CollinearTwo vectors are said to be collinear if any one of the following conditions is satisfied

(a) There exists a relation =

a mb where m is a non-zero scalar

(b) If

a and b are non-zero collinear vectors then there exists a set of x and y other than (0 0) such that

xa yb 0 Here converse is also true ie if + =

xa yb 0 and x y are non-zero scalars then

a and b are collinear vectors

(c) For two vectors

a and b to be collinear times =

a b 0 ie =

ˆˆ ˆi j ka a a 0b b b

Vectors 28357

1 If a and b

are non-zero and non-collinear then xa yb 0+ =

rArr x = 0 y = 0 as proved in the theorem

as given below

2 If three points A( a) B( b ) C( c )

are collinear then ( b a) ( c b )λminus = minus

or equivalently ( b a) ( c b )λminus = minus

ie ( b a) and ( c b )minus minus

are collinear vectors

Theorem If a and

b are two non-collinear non-zero vectors m and n are scalars such that

ma nb 0 then m = 0 and n = 0

286 section formula

Let P and Q points have their position vectors a and

b respectively then the position vector of point R dividing the line segment PQ internally in the

ratios m n is given by +

na mbcm n

If R divides PQ externally in the ratio m n (or internally in the

ratio-mn)) thenminus

=minus

na mbcn m

Remarks

0ngt then division is internal

(ii) m

0nlt then division is external

(iii) If m

( 10 )nisin minus then R lies outside PQ near P

(iv) If m

( 1)nisin minusinfin minus then R lies outside PQ near Q

1n= then

ie R is mid-point of PQ

(vi) m

1n= minus then PR RQ= minus

rArr no such point R exist

(vii) If positions vectors of vertices A B C of DABC are respectively a b and c

then position vector of

centroid of DABC is given by a b c

+ +equiv

(viii) OP ( a) OQ( b ) and OR( c )=

lie on same plane

(ix) na nb

rArr nc mc na mb+ = +

rArr na mb ( n m)c 0+ minus + =

helliphellip(i)

Clearly section formula is applicable iff points P Q R lie on a straight line Thus from this fact we

have necessary and sufficient condition for three different point P Q R with position vector a b and c

to be collinear (ie lying on a straight line) there exist non-zero scalers l + m + n = 0

Hence a mb nc 0+ + =

ensures coplanrity of a b and c

where as along with above the additional condition l + m + n = 0 ensures collinearality of point P Q R

ie existence of non-zero l m n such that a mb nc 0+ + =

coplanarity of a b and c

a mb nc 0 and

rArr collinearity of P Q R rArr coplanarity of a b and c

(x) If R( c )

divides the line joining P( a)

and Q( b )

in the ratio mn n

isin minus

then a b c

on same plane confining the line passing through points PQR and the origin Thus if any three co-terminus (Co-initial vector) or free vectors are non-coplanar(ie do not lie on same or parallel plane) then terminal point of none of three vectors can divide the line segment joining the terminal point of other two vectors Also it three co-terminus vectors having non parallel line or action are coplanar but there terminal points are non-collinear even then none of the terminal point of three vectors can divide the line segment joining the terminal points of other two vectors

Thus four section formula to be valid four point P Q and R with position vectors a b and c

the position

vector a b c

must be coplanar and PQR must be collinear However if PQR are collinear then

will be coplanar Thus for section formula to be applied for three different points PQR

collinearity of points PQR is necessary and sufficient condition However coplanarity of a b c

necessary condition but not sufficient a b c

are coplanar and point PQR are collinear

2861 Collinearity of the Points Point lying on same line are called collinear Two points are always collinear Thus necessary and sufficient condition for three different points A B and C to be collinear is that there exist three non-zero scalars x y z such that + + =

xa yb zc o and x + y + z = 0

However in above condition any one scalar is zero say x then + =

yb zc o and y = ndashz rArr =

b c rArr we have points A and B C coincidentrArr equivAB( C) are collinear If any two scalars are zero (say x and y) then the third are one z = 0rArr which holds for every three vectors

ab and c

Conclusion The necessary and sufficient condition for three point

A(a) B(b) C(c) to be collinear is

that there exist three scalars x y z not all zeros (at most one scalar can be zero) such that + + + =

Vectors 28359

Notes 1 If the points A( a) B( b ) C( c )

are collinear then AB BCλ=

where l is a scalar

or equivalently area of triangle

ABC is zero ie ( b a) ( c b ) 0minus times minus =

2862 Linear Combination of VectorsLinear combination of vectors

1 2 3 na a a a is a vector written as = λ +λ +λ + λ

1 1 2 2 3 3 n nr a a a a where l1 l2 ln are scalars

2863 Linearly Dependent Vectors

A system of vectors 1 2 3 na a a a is said to be linearly dependent if there exist n scalars 1 2 n λ λ λ (not

all zero) such that 1 1 2 2 3 3 n na a a a 0λ +λ +λ + +λ =

(ie above system is linearly dependent if one or some of them can be written as linear combination of the remaining)

Two collinear vectors are always linearly dependent Three co-planar vectors are always linearly dependent

2864 Linearly Independent Vectors

A system of n vectors 1 2 3 na a a a

is said to be linearly independent if none of them can be written as the linear combination of the remaining Therefore mathematically it means

If 1 1 2 2 3 3 n na a a a 0λ +λ +λ + +λ =

rArr λ = λ = = λ =1 2 n 0 where 1 2 n λ λ λ are n scalars

For example two non-collinear vectors are always linearly independent three non-coplanar vectors are always linearly independent

2865 Product of Two VectorsThese are of two types (a) Scalar Product (dot product) of two vectors

Quantity definition = θ

ab a b cos q is the angle between

a and b 0 le q le p

Geometrical interpretation

ab is the product of length of one vector and the projection of other

vector in the direction of the former vector ie ( ) ( )ab a b cos or b a cos= θ θ

Remarks (i) If q lt 90deg rArr ab 0gt

(ii) If q = 90deg rArr ab 0=

(iii) If q gt 90deg rArr ab 0lt

Properties of dot product of two vectors

(i) Dot product is commutative =

(ii) ne

(ab)c a(bc) in general Q λ ne micro c a

(iii) (Distributive law) Dot product distributes our vectors addition and subtraction ie plusmn = plusmn

a(b c) (ab) (ac)

(iv) 22 2a aa a a= = =

but no other powers of a vector are defined = = =ˆ ˆˆˆ ˆˆii jj kk 1

(v) If = + +

1 2 3ˆˆ ˆa a i a j a k and = + +

1 2 3ˆˆ ˆb b i b j b k

( ) ( )= + + + + = + +

1 2 3 1 2 3 1 1 2 2 3 3ˆ ˆˆ ˆ ˆ ˆab a i a j a k b i b j b k a b a b a b

(vi) + +

θ = =+ + + +

1 1 2 2 3 32 2 2 2 2 21 2 3 1 2 3

a b a b a babcosa b a a a b b b

ie minusθ = 1 ˆˆcos (ab)

(vii) =

ab 0 therefore = = =ˆ ˆˆˆ ˆ ˆij jk ki 0 (vector

a and b are perpendicular to each other provided that

a and b are non-zero vectors

(viii) = + + = + +

x y zˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆa a i a j a k (ai)i (aj)j (ak)k

(ix) plusmn = plusmn plusmn = + plusmn

2 2 2(a b) (a b)(a b) a b 2ab

Scalar projection of

a on b = θ = θ = θ =

ˆa cos a cos a b cos ab

Similarly scalar projection =

ˆbon a ba

Vector projection of

a on b θ = ˆ ˆ ˆ( a cos )b (ab)b is a vector along or

opposite to

b accordingly q is acute or optus

Similarly vector projection of =

ˆ ˆb on a (ba)a

Scalar projection of a perpendicular to = = θ = times

ˆb MA a sin a b

Vector projection of a perpendicular to ndash (vector projection

a on b )

= minus ˆ ˆa (ab)b

Work done

work done = θ = =

( F cos ) s F s

(b) Vector product (or cross product) of two vectors

Skew product outer product is denoted by times θ

a b( a b sin ) (unit vector n ) 0 le q le p where direction of

n is perpendicular to plane containing

a and b and is directed as given by right handed thumb rule as shown in figure given below

Magnitude of times = times = θ = θ θisin π

ˆa b a b a b sin n a b sin as [0 ]

Vectors 28361

Remarks (i) If q gt p then a b ( b a)times = minus times

Now while evaluating b a [0 ]θ πtimes isin

(ii) Unit vectors along a b

( a b )a b

plusmn timestimes =

Where 1 2 3 1 2 3 1 2 3

ˆˆ ˆi j kˆ ˆˆ ˆ ˆ ˆa b a a a a a i a j a k b b i b j b k

times = = + + = + +

28651 Properties of vector product

1 Anticommutative times = minus times

a b (b a)

2 times = times = times

(ma) b m(a b) a (mb) (where m is a scalar)

3 If two vectors

a and b are parallel we have times =

4 times =

a b 0 rArr

a and b are parallel vectors (provided

a and b are both non-zero vectors)

5 times = times = times =ˆ ˆˆ ˆ ˆ ˆi i j j k k 0 times = = minus timesˆˆ ˆ ˆ ˆi j k (j i) times = = minus timesˆ ˆˆ ˆ ˆj k i (k j) times = = minus timesˆ ˆˆ ˆ ˆ(k i) j (i k)

6 Cross product is distributive over addition or substraction times plusmn = times plusmn times

a (b c) a b a c Cross product

of three vectors is not associative

7 Let = + +

rArr times = = minus + minus + minus

1 2 3 2 3 3 2 3 1 1 3 1 2 2 1

ˆˆ ˆi j kˆˆ ˆa b a a a i(a b a b ) j(a b a b ) k(a b a b )

8 times

a bsin

Remarks

Since a b

sin [0 ]a b

θ θ πtimes

= isin

rArr 1a b

sina b

minus times =

or 1a b

sina b

π minus times minus =

show that it is suggested to use

dot product instead of cross product while finding the angle between two vectors

Geometrical interpretation times

a b represents the area of

parallelogram with two adjacent sides represented by

a and b

Area of D with two sides represented by

a and b

times = minus times +

1 1a b (a b) (a b)2 4

rArr times = minus times +

1a b (a b) (a b)2

= times

1 21 d d 2

1 2d and d are diagonal vector

Scalar triple product times = times = times =

a(b c) b(c a) c(a b) [a b c] (notation)

If = + +

1 2 3ˆˆ ˆa a i a j a k = + +

1 2 3ˆˆ ˆb b i b j b k = + +

1 2 3ˆˆ ˆc c i c j c k then =

a a aa b c b b b

Geometrical interpretation scalar triple product

Geometrically

[a b c] represents the volume of above parallopiped with

co-terminus edges represented by

a b and c

Properties of scalar triple product (a) Dot and cross can be interchanged without changing the value of

scalar triple proudct times = times

a(b c) (a b)c (b) Scalar triple proudct remains same if cyclic order of three vectors

do not changed = =

a b c b c a c a b

(c) + + = +

a b c d a c d b c d

(d) Scalar triple product vanishes when two of its vector are equal we have =

a a b 0

(e) The value of a scalar triple product if two of its vectors are parallel is zero ie =

a b c 0 if = λ

(f) For three co-planar vectors =

a b c 0 (even if

a b c are non-zero vectors)

(g) If = + +

a b c d a b d b c d c a rArr

ab c and d are co-planar

(h) If l is a scalar then λ = λ

a b c a b c

(i) Volume of tetrahedron =

1 a b c6

(j) The volume of the triangular prism (diagonally half of parallopiped) whose adjacent sides are

represented by the vectors

ab c is

1 a b c2

It is composed of two similar triangles of sides a and

b two rectangles of sides a c and bc and rectangle having sides |a ndash b| and c)

Vector triple product times times times times

a (b c) or (a b) c however times times

a b c is meaningless

Properties of vector tipple product

(i) times times = minus

a (b c) b(ac) c(ab)

(ii) times times = minus

(a b) c b(ac) a(bc)

(iii) times times ne times times

a (b c) (a b) c equality holds when a and c are collinear

(iv) times times

a (b c) represents vector normal to plane containing

b and c and also perpendicular to a

(v) If a perpendicular (plane containing and c )

Vectors 28363

ie times

a ||(b c) then times times =

a (b c) o

(vi) times times = times times = times times =ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆi (j k) j (i k) k (i j) o

(vii) times times

a (b c) is a linear combination of those two vectors which are with in brackets

(viii) If = times times

r a (b c) then r perpendicular to

a and lie in the parallel to that of

b and c

2866 Scalar Product of Four Vectors

times times

(a b)(c d) let times =

(a b) n therefore ( )times = times = times times

n(c d) (n c)d (a b) c d

= minus times times = minus minus

(c (a b))d ((c b)a (ca) b)d = minus =

ac bc(c a)(bd) (b c)(a d)

It is also called as Lagrangersquos identity

2867 Vector Product of Four Vectors

a b c d are four vectors the products times times times

(a b) (c d) is called vector product of four vectors

ie times times times = minus

(a b) (c d) [abd]c [ab c]d also times times times = minus

(a b) (c d) [a cd]b [b cd]a

NotesWe can look upon the above product as vector product in two ways one shown as above and other as shown below

Let c d p product ( a b ) ptimes = = times times

= ( ap)b ( pb )a [ac d ]b [c d b ]aminus = minus

So it can be defined either as linear combination of aand b

or as linear combination of

c and d

Reciprocal system of vectors

ab c be a system of three non-coplanar vectors Then the system of vectors

a b c which satisfy =

aa bb c c 1 and = = = = =

ab bc ba ca cb 0 is called the reciprocal system to the vector in term

ab c the vector

ab c are given bytimes times times

b c c a a ba b c [a b c] [a b c] [a b c]

Properties of reciprocal system of vectors

(i) = = =

aa bb c c 1 (ii) = = =

ab bc ca 0

(iii) =

1[a b c][a b c ]

(iv) times

b c a[a b c ]

(v) = = = = = =

ab ac ba bc ca cb 0 (vi) times =

[a b c] [a b c ] 1

(vii) System of unit vector ˆˆ ˆi j k is its own reciprocal = = =ˆ ˆˆ ˆ ˆ ˆi i j j k k

(viii) The orthogonal triad of vectors ˆˆ ˆi j k is self reciprocal

a b c are non-coplanar iff

a b c are non coplanar

Geometrical Application (i) Vector equation of straight line A line passing through a point A with position

vector a and parallel to another vector

b is given by the equation = +λ

r a (b)

If co-ordinates of point A (x1 y1z1) and direction cosine of b

is (l m n) respectively then the Cartesian

equation of the above line can also be derived as 1 1 1ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ( xi yj zk ) ( x i y j z k ) ( li mj nk )λ+ + = + + + + +

since i j k are linearly independent

Therefore (x ndash x1) ndash ll = 0 (y ndash y1) ndash lm = 0 and (z ndash z1) ndash ln = 0

rArr 1 1 1( x x ) ( y y ) ( z z )m n

λminus minus minus= = =

(ii) A line passing through two points A with position vector a and B with position vec-tor

b is given by the equation = +λ minus

r a (b a) where l is any real scalar parameter

If co-ordinates of point A (x1 y1 z1) and A (x2 y2 z2) Therefore direction ratio of line will be (x2 ndashx1) (y2 ndash y1) (z2 ndashz1) respectively then the Cartesian equation of the above line can also be derived as

1 1 1 2 1 2 1 2 1ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ( xi yj zk ) ( x i y j z k ) (( x x )i ( y y )j ( z z )k )λ+ + = + + + minus + minus + minus

Since ijk are linearly independent

Therefore (x ndash x1) ndash l (x2 ndash x1) = 0 (y ndash y1) ndashl (y2 ndash y1) = 0 and (z ndashz1) ndash l (z2 ndashz1) = 0

rArr 1 1 1

2 1 2 1 2 1

( x x ) ( y y ) ( z z )( x x ) ( y y ) ( z z )

minus minus minus

Internal and external angle bisectors at a line

The internal bisector of angle between unit vectors ˆa and b is along the vector + ˆa b The external bisector

is along minus ˆa b Equation of internal and external bisectors of the line = +λ

1r a b and = +micro

2r a b

internally at A(a) are given by

= + plusmn

b br a tb b

Vector equation of a plane

(i) The vector equation of plane passing through origin and containing

a and b is = λ +λ

1 2r a b

rArr times =

r(a b) 0

(ii) Vector equation of the plane passing through some other point C(c) and co-planar with two vector

a and b is = +λ +λ

1 2r c a b Taking dot product with times

a b minus times = rArr times =

(r c)(a b) 0 r(a b) [a b c]

(iii) Vector equation of a plane passing through three points A B C having position vector

a b and crespectively

= minus = minus

AB b a AC c a Therefore = λ minus +micro minus

r (b a) (c a)

Vectors 28365

287 Vector eQuation and method of solVing

A vector equation is a relation between some unknown vector(s) and some known quantities and the values of the unknown vectors satisfying the equation is called the solution of equation Solving a vector equation means determining an unknown vector (or a number of vectors satisfying the given conditions)

Type I times = times

r b a b rArr = +

r a tb t is any scalar

Type II times = perp

r b a a b rArr = minus times +

1r (a b) ybbb

Type III times = times times = perp

r b c b r a 0 a

b rArr = minus

car c bba

Type IV + times = ne

k a b k 0 (scalar) rArr

= + + times +

a b1r a kb a bk a k

Cevarsquos Theorem

If D E F are three points on the sides BC CA AB respectively of a triangle ABC

such that the lines AD BE and CF are

concurrent then = minusBD CE AF 1CD AE BF

conversely

Menelaursquos Theorem

If D E F are three points on the sides BC CA AB respectively of a triangle ABC such that the points D E F are collinear

then =BD CE AF 1CD AE BF

and conversely

Deasargue Theorem

If ABC A1 B1 C1 are two triangles such that the three lines AA1 BB1 and CC1 are concurrent then the points of intersection of the three pairs of sides BC B1C1 CA C1 A1 AB A1B1 are collinear and conversely

Chapter 29three-Dimensional

Geometry

291 IntroductIon

Since all points in a 3D space do not lie in a plane therefore to locate these points two co-ordinates are not sufficient Therefore to locate a point in a three-dimensional space we need three co-ordinates corresponding to three mutually perpendicular co-ordinate axes

These three co-ordinate axes divide the entire space into 8 parts each known as octant as shown here in the figure

Octant Sign Convention

OXYZ (I) (+ + +)OXprimeYZ (II) (ndash + +)OXYprimeZ (III) (+ ndash +)OXYZprime (IV) (+ + ndash)OXprimeYprimeZ (V) (ndash ndash +)OXprimeYZprime (VI) (ndash + ndash)OXYprimeZprime (VII) (+ ndash ndash)OXprimeYprimeZprime (VIII) (ndash ndash ndash)

Distance of point P(xyz) from origin = = + +2 2 2OP x y z

(i) Shifting of origin keeping axes parallel to origin (translation of axes) If origin is shifted to point O(a b g) keeping the axes parallel to then the co-ordinates of any point P wrt new co-ordinate system are given by (X = x ndash a Y = y ndash b Z = z ndash g) where (x y z) are co-ordinates of point P wrt original co-ordinates system

Three-Dimensional Geometry 29367

(ii) Rotation of axes (keeping the origin fixed)If the axes are rotated by an angle q keeping the origin fixed then the co-ordinates of point P wrt new co-ordinates system are given by X = xcosq + y sinq Y = ndashxsinq + ycosq It can be remember by the following box

Also x = Xcosq ndash Ysinq y = Xsinq + Ycosq

2911 Section Formula

If P(x1 y1 z1) and Q(x2 y2 z2) are two points in space and point R(x y z) divides PQ in ratio m n (m n gt 0)

(i) Internally Then 2 1 2 1 2 1mx nx my ny mz nzx y zm n m n m n+ + +

= = =+ + +

(ii) externally Then 2 1 2 1 2 1mx nx my ny mz nzx y zm n m n m nminus minus minus

= = =minus minus minus

2912 Corollary

(a) If R(x y z) divides the join of P(x1 y1 z1) and Q(x2 y2 z2) in ratio of l 1 then

2 1 2 1 2 1x x y y z zx y z1 1 1

λ plusmn λ plusmn λ plusmn= = =

λ plusmn λ plusmn λ plusmn

positive sign is taken for internal division and negative sign is taken for external division

(b) The mid-point of PQ is 1 2 1 2 1 2x x y y z z 2 2 2+ + +

2913 Centroid of a Triangle

The centroid of a triangle ABC whose vertices are A (x1 y1 z1) B(x2 y2 z2) and C(x3 y3 z3) are

1 2 3 1 2 3 1 2 3x x x y y y z z z

3 3 3+ + + + + +

2914 Centroid of a Tetrahedron

The centroid of a tetrahedron ABCD whose vertices are A (x1 y1 z1) B(x2 y2 z2) C(x3 y3 z3) and

D(x4 y4 z4) are 1 2 3 4 1 2 3 4 1 2 3 4x x x x y y y y z z z z

4 4 4+ + + + + + + + +

Direction cosine of a line Direction cosines are the cosines of the angles subtended by the line with the positive direction of axes If line subtends angle abg respectively with positive directions of x-axis y-axis and z-axis then its direction cosines are l = cosa m = cosb n = cosg

Remarks

(a) The direction cosines of the x-axis are cos0 cos cos2 2π π

ie 1 0 0 Similarly the dcrsquos of y and z

axis are (0 1 0) and (0 0 1) respectively

(b) If l m n be the dcrsquos of a line OP and OP = r then the co-ordinates of the point P are (lr mr nr)

(c) l2 + m2 + n2 = 1 or cos2 a + cos2 b + cos2 g = 1

2915 Direction Ratios (DRrsquos)Direction ratios of a line are numbers which are proportional to the dcrsquos of a line Direction ratios of a line PQ (where P and Q are (x1 y1 z1) and (x2 y2 z2) respectively are ( x2 ndash x1) (y2 ndash y1) (z2 ndash z1)

2916 Relation Between the DCrsquos and DRrsquos

If a b c are the drrsquos and l m n are the dcrsquos then2 2 2 2 2 2 2 2 2

a b c m n a b c a b c a b c

= plusmn = plusmn = plusmn+ + + + + +

Remarks 1 If a b c are the DRrsquos of AB then DCrsquos of AB are given by the +ve sign and those of the line BA

by ndashve sign

2 The unit vector along the line can be written as + +

ˆ ˆ ˆi m j n j

3 If DCrsquos of line AB is (l m n) then direction cosinersquos of line BA will be (ndashl ndashm ndashn) 4 The direction ratios of the line segment joining points (x1 y1 z1) and (x2 y2 z2) are proportional

to x2 ndash x1 y2 ndash y1 z2 ndash z1

5 Two parallel vectors have proportional direction ratios

6 If a vector

r has direction ratios proportional to a b c then = + ++ +

| r| ˆˆ ˆr ( ai bj ck )a b c

2917 The Angle Between Two LinesAngle between two lines is defined as angle between their direction vectors If (l1 m1 n1) and (l2 m2 n2) be the direction cosines of any two lines and q be the angle between then them cosq = l1l2 + m1m2 + n1n2

Deductions (i) If lines are perpendicular then l1l2 + m1m2 + n1n2 = 0

(ii) If lines are parallel then 1 1 1

m nm n

(iii) If the direction ratios of two lines are a1 b1c1 and a2b2c2 then 1 2 1 2 1 22 2 2 2 2 21 1 1 2 2 2

a a b b c ccosa b c a b c

+ +θ =

+ + + +

bull If cosq gt 0 gives acute angle q between the lines bull If cosq lt 0 gives obtuse angle q between the lines

bull If cosq = 0 lines are perpendicular to each other Further 2

1 2 2 1

2 2 2 2 2 21 1 1 2 2 2

(b c b c )sin

a b c a b c

Σ minusθ =

+ + + +

(iv) Lines are perpendicular to each other if a1a2 + b1b2 + c1c2 = 0

(v) Lines are parallel to each other if 1 1 1

a b ca b c

= = and Σ(b1c2 ndash b2c2)2 = 0

(vi) If (l1 m1 n1) and (l2 m2 n2) are the dcrsquos of two lines then drrsquos of the line which are perpendicular to both of them are m1n2 ndash m2n1 n1l2 ndash n2l1 l1m2 ndash l2m1It can be kept in memory by using the following method

1 2 2 1 1 2 2 1 1 2 2 1

m n(m n m n ) n n m m

Here denominator are obtained by subtracting the product of terms on off diagonal from the

product of tems on principal diagonal of the matrices 1 1 1 1 1 1

2 2 2 2 2 2

m n n m and

m n n m

respectively

2918 Projection of a Line Joining Two Points

P (x1 y1 z1) and Q(x2 y2 z2) on other line with direction cosines ltl m n gt is given by |(x2 ndash x1) l + (y2 ndash y1) m + (z2 ndash z1)n|

Corollary

(a) If P is a point (x1 y1 z1) then the projection of OP on a line whose direction cosines are (l1 m1 n1) is | l1 x1 + m1 y1 + n1 z1 | where O is origin

(b) The projections of PQ when P is (x1 y1 z1) and Q is (x2 y2 z2) on the co-ordinates axes are = (x2 ndash x1) (y2 ndash y1) (z2 ndash z1)

(c) If Projections of PQ on AB is zero then PQ is perpendicular to AB

2919 Vector Equation of a Curve

29191 Cartesian equation of a curve

Replacing ˆˆ ˆr xi yj zk= + + in the obtained vector equation and comparing scalar coefficient of ˆˆ ˆi j k from

both side of the equation we get an equation in x y z as F(x y z) = 0 called as Cartesian equation of curve

The straight line in 3-dimensional geometry A straight line is generated by the intersection of any two planes (non-parallel) A straight line in space is uniquely determined if

(i) It passes through a fixed point and is parallel to a fixed line (ii) It passes through two fixed points (iii) It is the intersection of two given non-parallel planes

(i) Equation of straight line passing through a point A(x1 y1 z1)Vector equation r a b= +λ

Cartesian equation 1 1 1x x y y z zm n

minus minus minus= = = λ

(ii) Equation of line passing through two points P (x1 y1 z1) and Q (x2 y2 z2)

Vector equation r a (b a)= +λ minus

or r b (b a)= +λ minus

Cartesian equation 1 1 1

2 1 2 1 2 1

x x y y z zx x y y z zminus minus minus

= = = λminus minus minus

Remarks

(i) For each scalars l isin ℝ +

a bλ is the position vector a point lying on straight line = +

r a bλ by varying the values of l we can obtain different points on the above line

(ii) For each scalars l isin ℝ + minus

a ( b a)λ is the position vector a point lying on straight line joining

P( a) andQ( b )

(iii) The points A(x1 y1 z1) B(x2 y2 z2) and C(x3 y3 z3) are said to be collinear if the third point satisfies

the equation of line formed by 2 points ie minus minus minus= =

minus minus minus3 1 3 1 3 1

2 1 2 1 2 1

x x y y z zx x y y z z

If 1L r a b= +λ

or 1 1 1

x x y y z zm n

minus minus minus= =

and 2L r c d= +micro

or 2 2 2

x x y y z zm n

Are two straight line in space then

1 2 1 2 1 22 2 2 2 2 21 1 1 2 2 2

m m n nbdcosb d m n m n

+ +θ = =

+ + + +

rArr 1 1 1 2 1 2 1 22 2 2 2 2 21 1 1 2 2 2

m m n nbdcos cosb d m n m n

minus minus + + θ = = + + + +

29111 Condition of Parallelism

r a b and r c d= +λ = +micro

are parallel iff b ||d

ie b kd=

for some scalar k ne 0 or 1 1 1

m nm n

29112 Condition of Perpendicularity

r a band r c d= +λ = +micro

are perpendicular for each other iff b dperp

ie bd 0=

Or l1l2 +m1m2 + n1n2 = 0

θrarr

A(x1y1z1)(a)

C(x2y2z2)(c)

rarr d= 2im 2jn 2k

rarrb=1im1jn1k

29113 Condition of Coincidence

will be coincidence iff b d and (c a)minus

are parallel ie 1 2b k d and d k (c a)= = minus

for k1k2 isin ℝ ~0 ie two lines if parallel and have a common point are coincident

29114 Foot of Perpendicular Drawn From a Point P(x1 y1 z1)

Having position vector c on line L r a b= +λ

ie x y z

m nminusα minusβ minus γ

= = = λ

where ˆ ˆˆ ˆ ˆ ˆa i j k b i mj nk= α +β + γ = + +

Vector method Position vector of foot M of perpendicular

mr a AM= +

(projection of AP on b

) Or mˆ ˆr a ((c a)b)b= + minus

Cartesian Method Consider the foot of perpendicularM ( m n )equiv λ +α λ +β λ + γ hellip(i)

Then PM bperp

rArr (PM)b 0=

rArr 1 1 12 2 2

(x ) m(y ) n(z )m n

minusα + minusβ + minus γλ =

hellip(ii)

Substituting value of l form equation (ii) in (i) we get co-ordinates of foot M

Image of point 1 1 1P(x y z ) P(c)equiv on line L r a b= +λ

where ˆ ˆˆ ˆ ˆ ˆa i j k and b i mj nk= α +β + γ = + +

Vector method Since M is mid-point of PP

rArr Position vector of mP ( q) 2 r c= = minus

where mˆ ˆr a ((c a)b)b= + minus

Cartesian Method Suppose M(xm ym zm) be the co-ordinates of foot of perpendicular drawn from P on line L then M being mid-point of PP co-ordinates P are given by (2xm ndash x1 2ym ndash ym 2zm ndash z1)

Remark

Thus in order to find co-ordinates of image of a point first find the co-ordinates of foot of perpendicular and then image of point by using mid-point formula

29115 Distance of a Point P (x1y1z1) From the Line L

r a b= +λ

Method 1 2 2d AP AM= minus AM = scalar projection of AP on b

(unit vector) = AP cos APbθ =

22 2 21 1 1 1 1 1d (x ) (y ) (z ) (x ) m(y ) n(z )= minusα + minusβ + minus γ minus minusα + minusβ + minus γ

Method 2 After finding the coordinates foot of perpendicular drawn of point P on line L say

M(xm ym zm) then distance of point P from line L is given by 2 2 21 m 1 m 1 mPL (x x ) (y y ) (z z )= minus + minus minus

29116 Co-ordinates of Point of Intersection

Of two lines in space can be obtained they are non-parallel non-coincident but are intersecting

Method 1

Step I Compare the position vector of both lines ie let position vector of point of intersection be r

Step II Compare the scalar coefficient of linearly independent vectors to get three linear equations in l and m

Step III Solving any two to get l and m and if the values obtained satisfy 3rd equation then lines are intersecting and for the obtained value of l get the position vector of the point

Method 2

Step I Take a general point of L1 = 0 (ll1+x1 lm1+y1 ln1+z1)

Step II Substituted in equation L2 = 0 to get two equations in l

Step III If the values of l obtained from both equation are same then the lines Intersect otherwise they are parallel or skew

Step IV If the lines intersect then the values of l obtained generate point of intersection

Special Cases If [c a)b d] 0 and b kdminus = ne

lines intersect

Case I c a bminus = λ

point of intersection is C

Case II c a dminus = λ

point of intersection is A

Distance between to parallel lines

Vector form CL = scalar projection of (a c) bminus perp

= ˆ ˆ(a c) ((a c)b)b a c cosminus minus minus = minus θ

= a c b sin (a c) b

minus θ minus times=

Cartesian form If 1 2 3ˆˆ ˆa a i a j a k= + +

1 2 3ˆˆ ˆb b i b j b k= + +

1 2 3ˆˆ ˆc c i c j c k= + +

Then distance between parallel lines = 1 1 2 2 3 3

ˆˆ ˆi j k1 a c a c a cb b b b

minus minus minus or 1 1 2 2 3 3

ˆˆ ˆi j k1 a c a c a cd d d d

minus minus minus

292 Skew lIneS

Skew lines are defined as pair of lines in the space which are neither parallel nor intersecting

Two straight lines in space are called skew lines when they are non-coplanar

Shortest distance between two skew lines 1 2L r a b L r c d= +λ = +micro

Vector form Scalar projection of (a c)on(b d)minus times

= (a c)(b d)

minus times

Cartesian form

1 1 2 2 3 3

a c a c a cb b bd d d

ˆˆ ˆi j kb b bd d d

minus minus minus

Condition for intersecting line in space

Two non-parallel lines in space are intersecting iff their shortest distance is zero ie

b kd and[(a c)bd] 0ne minus =

Co-planarity of lines in space

Two lines L1 and L2 in space are coplanar (lies on same place) if

(i) Line are parallel (ii) Lines are intersecting (shortest distance = 0)

are coplanar

Iff either b kd or (a c)(b d) 0= minus times =

ie b kd=

or [a c b d] 0minus =

(scalar triple product)

Non-co planarity of line in space

Lines 1 2L a b and L c d= +λ +micro

are non-coplanar (do not lie on same plane)

Iff b kdne

(ie non-parallel)

and [a c bd] 0minus ne

(ie non-intersecting)

2921 Equation of Line of Shortest Distance

(Line of intersection of planes APL and CPQ)

Line of intersection of plane [r abb d] 0minus times =

(plane APL) and

[r cdb d] 0minus times =

(plane CPQ)

Gives the equation of line shortest distance

293 Plane

Plane is a locus of a point which moves so that any point on the line segment joining two position of moving point always lie on the same locus

Properties of plane bull It has a unique normal vector defining its orientation in the space bull The normal vector of the plane remains perpendicular to all the line lying in that plane bull Unique plane passes through three points bull Unique plane contains two intersecting lines bull Unique plane passes through a line and normal to a given vector n bull Unique plane passes through a line and parallel to a given vector bull Unique plane passes through a point and parallel to two lines bull Unique plane passes through a point and normal to a given vector

Equation of plane

General form Ax + By + Cz + D = 0General form can be obtained when any one of

the condition discussed in properties of plane is given The equation of plane under the following cases are given here

Case (i) Equation of a plane passing through a point a 1 1 1A (x y z ) and normal to vector ˆˆ ˆn ai bj ck= + +

Vector equation (r a)n 0minus = or rn an d= =

(constant)

Cartesian equation Taking ˆˆ ˆr xi yj zk= + + and 1 1 1

ˆˆ ˆa x i b j z k= + + and ˆˆ ˆn ai bj ck= + +

we have ax + bz + cz = ax1 + by1 + cz1 Here lta b cgt are direction ratio of vector normal to plane

Case (ii) Nomralperpendicular formEquation of plane upon which the length of perpendicular from origin is p and normal vectors has direction cosine ltl m ngt

Vector form ˆ ˆ(r pu)u 0minus =

or ˆru p=

where ˆ ˆˆ ˆ ˆ ˆˆr xi yj zk and u i mj nk= + + = + +

Cartesian form lx + mj + nz = pTo convert general equation of plane rn d=

to normal form

Step 1 ax + by + cz = d make d gt 0 if not so by multiplying with ndash1

Step 2 2 2 2 2 2 2 2 2 2 2 2 2

ax by cz d

a b c a b c a b c a b c+ + =

+ + + + + + + +

Step 3 lx + my + nz = p where2 2 2

a b c=

+ + and

dpa b c

= distance of plane form origin

Case (iii) Intercept form of the plane the equation of a plane which cuts on intercepts a on x-axis b on y-axis and c on z-axis

Vector form Let A(a)B(b)C(c)

be the point of intersection of

required plane with coordinates axes and P(r) be arbitrary point on

plane then [PA AB AC] 0=

rArr[(r a) (a b) (a c)] 0minus minus minus =

Cartesian form x a y 0 z 0

a 0 b 0 0a 0 c

minus minus minusminus =

Remarks

If Ax + By + Cz = d is the general equation of plane then + + =x y z

1dA dB dC

is the intercept form with

intercept d d d

on coordinates axes

Case (iv) Equation of plane passing through three points

Let 1 1 1A(a) (x y z )equiv

2 2 2B(b) (x y z )equiv

3 3 3C(c) (x y z )equiv

be three fixed (given points) on plane and P(r) be

an arbitrary point on plane

Vector form [AP AB AC] 0=

rArr [(r a) (b a) (c a)] 0minus minus minus =

Cartesian form 1 1 1

2 1 2 1 2 1

3 1 3 1 3 1

x x y y z zx x y y z z 0x x y y z z

minus minus minusminus minus minus =minus minus minus

RemarkCondition for four points A(x1y1z1) B(x2y2z2) C(x3y3z3) and D(x4 y4 z4)

with position vector

a b c and d to be coplanar is =

[ AB AC AD] 0

Or alternatively prove that the plane passing through any of the three points through the four points

1 2 1 2 1 2

1 3 1 3 1 3

x x y y z z

x x y y z z 0

x x y y z z

is satisfied by (x y z) equiv (x4 y4 z4)

B b P(xyz)

Case (v) Equation of plane passing through two points and normal to a given plane

Let A(x1 y1 z1) and B(x2 y2 z2) be two points lying on the

plane with position vectors a and b

and P(x y z)(r) be

arbitrary point on plane ˆˆ ˆi mj nkα = + +

be vector normal

to given plane

Vector form Normal vector torequired plane

(AP) (AB ) 0timesα =

or Scalar trippleproduct

[(r a) (b a) ] 0minus minus α =

2 1 2 1 2 1

x x y y z zx x y y z z 0

minus minus minusminus minus minus =

RemarksThis case is similar to the case of finding the Equation of a plane passing through two points A(x1 y1 z1) and

B(x2 y2 z2) and parallel to a vector + + =ˆˆ ˆli mj nk 0

Case (vi) Equation of a plane passing through a point and parallel to two lines (or vectors)

Let the plane passes through a fixed point R(x1y1z1) having its position vector a Let the plane be parallel to vector band c

Vector form PR(b c) 0times =

rArr (r a)(b c) 0minus times =

or [r a b c] 0minus =

x x y y z zb b bc c c

minus minus minus

294 area of trIangle

If Ayz Azx Axy be the projections of an area A on the co-ordinate

planes yz zx and xy respectively then 2 2 2yz zx xyA (A A A )= + +

If vertices of a triangle are (x1 y1 z1) (x2 y2 z2) and (x3 y3 z3) then

1 1 1 1

yz 2 2 zx 2 2

3 3 3 3

y z 1 z x 11 1A y z 1 A z x 12 2

y z 1 z x 1= = and

xy 2 2

x y 11A x y 12

x y 1=

2941 Angle Between Two Planes (Angle Between the Normal Vector)

Let 1 1 2 2rn d and r n d= =

be two planes

Then 1 21 1 1 2 1 2 1 22 2

1 2 1 2

n n a a b b c ccos cosn n a a

minus minus + + α = =

a is acute obtuse accordingly 1 2 1 2n n 0 or n n 0gt lt

Corollary

1 If planes are perpendicular then 1 2n n 0=

or a1a2 + b1b2 + c1c2 = 0

2 If planes are parallel then 1 2n k n=

3 If planes are coincident then 1 1 1 1

2 2 2 2

a b c da b c d

2942 Angle Between Line and Plane

Let x y zL m n

minusα minusβ minus γ= =

and p ax + by + cz + d = 0

2 2 2 2 2 2

a mb ncsin sin cos

2 m n a b c

+ +π θ = minusα = α + + + +

rArr 1

2 2 2 2 2 2

a mb ncsin

m n a b cminus + +

θ = + + + +

where a and q are

acute angles

Remarks

Line is perpendicular to plane = =a b cl m n

hellip(i)

Line is paralel to plane al + bm + cn = 0 helliphellip(ii)

Line is coincident in the plane + + = + + + =

al bm cn 0 and

a b c d 0α β γ hellip(iii)

Case (vi) Equation of a plane parallel to a plane Equation of family of planes parallel to plane ax + by + cz + d = 0 or rn d= are given by ax + by + cz + d = 0 and rn = λ

2943 Distance Between Two Parallel Planes

1 22 2 2

d dda b c

minus=

2944 Distance of a Point From a Given Plane

Let rn d 0+ =

pn dPMn+

Cartesian form

1 1 12 2 2

ax by cz dPMa b c

+ + +=

+ + where 1 1 1

ˆ ˆˆ ˆ ˆ ˆp x i y j z k n ai bj ck= + + = + +

and equation of plane is ax + by + cz + d = 0

2945 Foot of Perpendicular Drawn From a Point on Plane

From above m 2

pn dr p nn

+ = minus

where equation of plane is rn d 0+ =

and position vector of p is p

Cartesian form

Equation of line PAMQ is 1 1 1x x y y z za b cminus minus minus

= = = λ

rArr x = al + x1 y = bl + y1 z = cl + z1 if it is M then lies

on planerArr a (al + x1) + b(bl + y1) + c(cl + z1) + d = 0

rArr 1 1 12 2 2

(ax by cz d)a b c

minus + + +λ =

+ + Foot of perpendicular M is given by

( )1 1 11 1 1

ax by cz dx x y y z za b c a b c

+ + +minus minus minus= = = minus

Image of point P (x1 y1 z1) (p) on plane rn d 0+ =

From mid-point formula (vector form)Position vector of image Q is given by mq 2r p= minus

rArr 2

p n dq p 2 nn

minus + = minus

Cartesian formFor A (xyz) to be the image of point P mid-point of AP must lie on plane

rArr 1 1 1a b cx y z2 2 2λ λ λ + + +

lie on ax + by + cz + d = 0

rArr 1 1 1ay by ca x b y c z d 02 2 2

λ + + + + + + =

rArr 1 1 12 2 2

2(ax by cz d)a b c

minus + + +λ =

Image of point M is given by 1 1 1 1 1 12 2 2

x x y y z z 2(ax by cz d)a b c a b cminus minus minus minus + + +

= = =+ +

Image of a line in a plane and projection of a line in a plane

Step I Given a plane ax + by + cz + d = 0 and a line 1 1 1x x y y z za b cminus minus minus

Step II Find the point of intersection of line and plane ie point A (say (x0 y0 z0))

Setp III Take a point P(x1 y1 z1) on line and find the image and foot of perpendicular of P(x1 y1 z1) in plane Q and M respectively

Step IV Write the equation of line AQ (image of the line) and AM (line of projection)

Equation of plane passing through the intersection of two given planeIf P1 = 0 and P2 = 0 are two planes then the equation of plane passing

through the line of intersection P1 = 0 and P2 = 0 is given by P1 + l P2 = 0 l isin ℝ ie 1 1 2 2(rn d ) (r n d ) 0+λ + =

Equation of line of intersection of two planes 1 1 2 2rn d 0 rn d 0+ = + =

Step 1 Find any point on the line of intersection let it be on x ndash y plane ie put z = 0 and solve a1x + b1y + d1 = 0 and a2x + b2y + d2 = 0 Let it be (a b 0)

Step 2 Find the direction ratios of line of intersection given by 1 2 1 1 1

ˆˆ ˆi j kˆˆ ˆn n a b c i mj nk

a b ctimes = = + +

Step 3 Line of intersection is given by x y z 0m n

minusα minusβ minus= =

Remarks (i) If n = 0 and l ne 0 then point P may be taken as the point of intersection of line AB and yz-plane and

if m ne 0 then P may be taken as the point where AB cuts zx-plane

(ii) If one line in symmetrical form and other in general form Let lines are minus minus minus

= =1 1 1x x y y z zl m n

and a1x + b1y + c1z + d1 = 0 = a2x + b2y + c2z + d2 The condition for co-planarity is

+ + + + +=

+ + + + +1 1 1 1 1 1 1 1 1 1

2 1 2 1 2 1 2 2 2 2

a x b y c z d a l b m c na x b y c z d a l b m c n

(iii) If both lines in general form Let lines are a1x + b1y + c1z + d1 = 0 = a2x + b2y + c2z + d2 and a3x + b3y + c3z + d3 = 0 = a4x + b4y + c4z +d4

The condition that this pair of lines is co-planar is =

1 1 1 1

2 2 2 2

3 3 3 3

4 4 4 4

a b c d

a b c d0

a b c d

29451 Condition of intersection of three planes

Given three planesP1 = 0 ie a1x + b1y + c1z = d1 hellip(i)P2 = 0 ie a2x + b2y + c2 z = d2 hellip(ii)P3 = 0 ie a3x + b3y + c3z = d3 hellip(iii)

Solving equation (iii) by Crammerrsquos rule we get Dx = D1 Dy = D2 Dz = D3

Case I The given three planes cut at one point iff D ne 0 That is a unique solution and the point of

intersection is given by (a b g) where 31 2 ∆∆ ∆

α = β = γ =∆ ∆ ∆

Case II The given three planes does not have a common point iff a set of equations have no solutionThat is D = 0 and atleast one of D1 D2 D3 is non-zero

Case III The given three planes have then infinitely many solutions D = 0 = D1 = D2 = D3All three rowrsquos of D are identical or two rowrsquos of D are identical

2946 Equation of Bisectors of the Angle Between Two Planes

Equations of the bisectors of the planesP1 ax + by + cz + d = 0P2 a1x + b1y + c1z + d1 = 0where d gt 0 and d1 gt 0 are

1 1 12 2 2 2 2 2

|a x b y c z d ||ax by cz d |

(a b c ) (a b c )

+ + ++ + +=

+ + + +

Conditions Acute Angle Bisectors Obtuse Angle Bisectors

aa1 + bb1 + cc1 gt 0 ndash +aa1 + bb1 + cc1 lt 0 + ndash

Remarks

(i) Equation of bisector of the angle between the planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z

+ d2 = 0 is + + + + + += plusmn

+ + + +1 1 1 1 2 2 2 2

2 2 2 2 2 21 1 1 2 2 2

a x b y c z d a x b y c z d

a b c a b c

(ii) Bisector of the acute and obtuse angles between two planes Let the two planes be a1x + b1y + c1z = d1 = 0 and a2x + b2y + c2z + d2 = 0 where d1 d2 gt 0

+ + + + + +

=+ + + +

1 1 1 1 2 2 2 2

2 2 2 2 2 21 1 1 2 2 2

a b c a b c is the equation of acute and obtuse angle between the two

planes according as a1a2 + b1b2 + c1c2 lt 0 or gt 0

Other bisector will be the bisector of the other angle between the two planes

(iii) To test whether origin lies in the acute or obtuse angle between two planes Let the equation of two planes be a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 where the origin lies in the acute or obtuse angle between two planes accordingly as a1a2 + b1b2 + c1c2lt 0 or gt 0

Chapter 30probability

301 ExpErimEnts

An experiment is a set of processes which are carried out under stipulated conditions to study the phenomenon associated with it It is defined as below

ldquoA mathematical operation which results in some well-defined outcomes is known as experimentrdquo Broadly there can be two types of experiments These are as follows (a) Random experiments Prediction of any specific outcome is impossible before completion of

experiments For example tossing of a coin throwing of a die etc (b) Non-random experiments Prediction of some of the outcomes can be made before performing

the experiments For example ratio of hydrogen and oxygen in a molecule of H2O throwing of a two-dimentioal die etc

Sample Space

Set(S) of all possible outcomes of random experiments For example S = 1 2 3 4 5 6 for die and S = HT for tossing a coin

Infinite Sample Space

Sample space having infinite number of elements These are of two types (a) Discrete Sample Space Having elements which can be put into a set with onendashone

correspondence with the set of natural numbers (ie sample space is countability infinite) For example tossing of a coin till a head appears S = HTHTTHTTTHhellip

(b) Continuous Sample Space Sample space is an interval or union of interval for example lifetime of a computer hard-disk manufactured by HCL has sample space S =[0 infin)

NoteThe problems related with continuous sample space are generally solved using geometrical equivalent of sample space and event space and they will be dealt in our coming article under the heading Geometrical Probability

3011 EventA subset of sample space For example E1 = 2 4 6 is the event of getting even numbers in the experi-ments of throwing of a die where S = 1 2 3 4 5 6 Clearly E1 sube S

Probability 30383

Single Event An event having single point

Compound Event An event having more than one point

Impossible Event An event having elements outside the sample space or which is emptyFor example event of getting 7 while throwing a die

Possible Event An event having elements from the sample space

Sure or Certain Event An event which is equal to the sample space

Mutually Exclusive EventsA set of events is said to be mutually exclusive if occurrence of one of them precludes the occurrence of any of the remaining events If a set of events E1 E2 En are mutually exclusive events ie Ei cap Ej = f for all i j = 1 2 n and i ne j then P(E1 cup E2 cup cup En) = P(E1) + P(E2) + P(E3) + + P(En)

3012 Exhaustive EventsA set of events is said to be exhaustive if the performance of the experiment results in the occurrence of at least one of them Therefore if a set of events E1 E2 En are exhaustive events then

P(E1 cup E2 cup cup En) = 1 =

Two events A and B are said to be mutually exhaustive if P(A cup B) = 1 ie A cup B = S

3013 Equally Likely EventsThe given events say E1 E2 En are said to be equally likely if all the factors are taken into consideration we have no reason to believe that anyone of these factors has better chances of occurrence than the other That is P (E1) = P(E2) = = P(En)

3014 Disjoint Events

Events E1 and E2 are said to be disjoint when they have no common elements That is E1 cap E 2 = f

Complementary Event Of E is Ec = (S ndash E)

3015 Independent and Dependent Events

Two events are said to be dependent if the occurence or non-occurrence of one does decides and disturbs the occurrence or non-occurrence of the other For example in the withdrawl of cards from a deck of cards without replacement the outcomes will be dependent events but if the withdrawl is done with replacement the outcomes will be independent If a set of events E1 E2 En are independent Events then P(E1 cap E2 cap E3 cap cap En) = P(E1) P(E2) P (En)

3016 Mutually Exclusive and Exhaustive EventsA set of events is said to be mutually exclusive as well as exhaustive if the conditions as given below are satisfied

Ei cap Ej = f V i j such that i ne j and E1 cup E2 cupcup En = Sfor mutually exclusive and exhaustive events E1 E2 EnProbability of occurrence of atleast one of them

= P(E1 cup E2 cup cup En) = P(E1) + P(E2) + + P(En) = 1

Trials Experiments performed repeatedly are known as trialsProbability of occurrence of an eventIf an event can happen in x ways and fail to happen in y ways and each of these ways is equally likely

then the probability or the chance of its occurrence will be +x

x y and that of its non-occurrence

will be +y

Mathematical and Classical Definition of Probability

For an experiment with continuous finite sample space S the probability of occurrence of an event E is

denoted by P(E) and it is defined as n(E) number of elements in EP(E)n(S) number of elementsinspace S

= number of outcomes favourable to E in sample space S

total number of outcomes (elements) in S

For example in the experiment of throwing a dice the probability of getting 2 as an outcome is 16

30161 Properties of probability of event

1 The probability P(E) of occurrence of any event E lies between 0 and 1

2 Complementary event of E is denoted as Ec or Eprime or E which literally means non-occurrence of E Thus E occurs only when E does not occurs Therefore + =P(E) P(E) 1

3 If E is an impossible event then P(E) = 0 4 If E is a possible event then 0 lt P(E) lt 1 5 If E is a certain event then P(E) = 1

Remarks

1 A die is a solid cube with six faces and numbers 1 2 3 4 5 and 6 marked on the faces respectively In throwing or rolling a die any of the above numbers would shown on the uppermost face

2 A pack of cards consists of 52 cards in 4 suits ie (a) spades spades (b) Clubs clubs (c) Hearts hearts (d) diamonds diams Each suit consists of 13 cards Out of these spades and clubs are black faced cards while hearts and diamonds are red-faced cards The aces kings queens jack (or knave) are called face cards or honour cards king queen and jack are known as court cards

3 Game of Bridge It is played by 4 players each player is given 13 cards

4 Game of Whist It is played by two pairs of persons

30162 Statistical definition of probability

When a random experiment is repeated n times under similar conditions ie n trials are made and n is very large and an event E occurs r times out of the n trials then the probability of occurrence of the

event E is defined as rarrinfin

rP(E) limn

Probability 30385

30163 Odds in favour and odds against an event

If in an experiment the number of outcomes favourable to an event E is x and number of outcomes not favourable to event E is y then

(a) Odds in favour of = = =Enumber of outcomes favourable(n(E)) P(E) x

ynumber of outcomes unfavourable(n(E)) P(E)

(b) Odds against = = =Enumber of unfavourable outcomes (n(E)) yP(E)

number of favourable outcomes (n(E)) P(E) x For example Odds in favour of getting a spade when a card is drawn from a well-shuffled pack

of 52 cards are = =13

C 13 1C 39 3

Remarks

If odds in favour of an event are m n then the probability of the occurence of that event ism

m n+ and the

probability of non-occurence of that event is n

30164 Property of compound events

(i) capcap = 1 2

1 2n(E E )P(E E )

(ii) cup + minus cap

cup = =1 2 1 2 1 21 2

n(E E ) n(E ) n(E ) n(E E )P(E E )n(S) n(S)

= P(E1) + P(E2) = P(E1 cap E2)

30165 Set theoretic principle

Let E1 E2 E3 E4 be four events then (a) E1 cup E2 stands for occurrence of atleast one of E1 E2 (b) E1 cap E2 stands for simultaneous occurrence of E1 and E2

(c) Eprime or E or EC stands for non occurrence of event E

(d) cap = cup1 2 1 2(E E ) E E = stands for non-occurrence of both E1 and E2 ie

the occurence of neither E1 nor E2

(e) E1 - E2 denotes the occurrence of event E1 but not of E2

(f) cap1 2E E denotes the occurence of event E1 but not of E2

(g) E1 cup E2 cup E3 denotes the occurence of at least one of the events E1 or E2 or E3

(h) cap cup cap1 2 1 2(E E ) (E E ) denotes the occurence of exactly one of E1 and E2

(j) E1 cap E2 cap E3 denotes the occurence of all three E1 E2 and E3 (k) cap cap cup cap cap cup cap cap1 2 3 1 2 3 1 2 3(E E E ) (E E E ) (E E E ) denotes the occurence of exactly two of E1 E2 and E3

30166 Probability of events based on theoritic principle

(i) = minusP(E) 1 P(E)

(ii) cap = minus cap1 2 1 1 2P(E E ) P(E ) P(E E ) (iii) Probability of simultaneous non-occurrence of events E1 and E2

= ( )cap = cup = minus cup1 2 1 2 1 2p E E P(E E ) 1 P(E E )

(iv) Probability of occurrence of exactly of the events E1 and = cap + cap2 1 2 1 2E P(E E ) P(E E )

= + minus cap1 2 1 2P(E ) P(E ) 2P(E E )

30167 Probability of independent events (i) P(A cap B) = P(A) P(B) for independent events A and B (ii) A and B are independent events iff (A and B) or (Aand B) or(A and B) are independent events

Thus P(A cap B) = P(A) P(B)hArr cap =P(A B) P(A)P(B) hArr cap =P(A B) P(A)P(B) hArr cap =P(A B) P(A)P(B)

30168 Mutually independent eventsThree events E1 E2 E3 are said to be mutualy independent iff

P(E1 cap E2) = P(E1) P(E2) P(E1 cap E3) = P(E1) P(E3)P(E2 cap E3) = P(E2) P(E3) and P(E1 cap E2 cap E3) = P(E1) P(E2) P(E3)

Probability 30387

30169 Pairwise independent events

These events would said to be pairwise independent whenP(E1 cap E2) = P(E1) P(E2) P(E2 cap E3) = P(E2) P(E3) and P(E1 cap E3) = P(E1) P(E3)Thus mutually independent events are always pairwise independent but the converse may not be true

301610 Probability dependent events

If the events are not independent they are dependent and for such events A and B P(A cap B) ne P(A) P(B)

3017 Conditional Probability

Probability of occurrence of E2 when E1 has already occurred is denoted by

= cap cap

= ne2 1 1 21

n(E E ) P(E E ) P(E ) 0n(E ) P(E )

21 2 1

EP(E E ) P(E )PE

30171 Properties of conditional probability

1 If E1 and E2 are independent events then

EP P(E )E

2 If E1 E2En are independent events then P(E1 cup E2 cup cup En) = 1 - P(E1 cup E2 cup cup En)c = c c c

1 2 n1 P(E )P(E )P(E )

3 If E1 and E2 are two events such that E2 ne f then

E EP P 1E E

4 If E1 and E2 are two events such that E1 ne f then P(E2) = P(E1)

E EP P(E )PE E

5 If E1 and E2 and E3 are three events such that E1 ne f E1 cap E2 ne f then P(E1 cap E2 cap E3) =

cap 32

EEP PE E E

3018 Generalized Form

If E1 E2 En are n events such that E1 ne f E1 cap E2 ne f E1 cap E2 cap E3 ne f E1 cap E2 cap E3 cap cap Enndash1 ne

f then P(E1 cap E2 cap E3 cap En) = P(E1) minus

cap cap cap cap 32 n

1 1 2 1 2 n 1

EE EP P PE E E E E E

Total Probability Theorem (for dependent events)Let A be any events of S and A1 A2 A3helliphellip An be n mutually exclusive as well as exhaustive event

and A depends upon them individually then

= + + +

1 2 n1 2 n

A A AP(A) P(A )P P(A )P P(A )PA A A

ii 1 i

AP(A) P(A )PA

Remarks

1 We have already discussed that mutually exclusive set of events are strongly dependent because occurrence of one precludes the occurrence of the other

2 The concept of mutual exclusive is set theoretic in nature while the concept of dependenceindependence is probablistic in nature

Some important probabilities

(a) If A and B are any events in S then P(A cup B) = P(A) + P(B) ndash P (A cap B) If A and B are mutually exclusive then P(A cap B) = 0 and P(A cup B) = P(A) + P(B) (b) If A B C are any three events of the sample space then P(A cup B cup C) = P(A) + P(B) + P(C) - P(A cap B) - P(A cap C) - P(B cap C) + P(A cap B cap C) (c) If A1 A2An are n events then P(A1 cup A2cup An)

= = le lt le le lt lt le

minus cap + cap cap minussum sum sum sumsumsum1 2 1 2 31 2 1 2 3

i r r r r ri 1 1 r r n 1 r r r n

P(A ) (P(A A )) (P(A A A ))

(d) The probability that one of several mutually exclusive events A1 A2An will occur is sum of the probabilities of the occurrence of separate events P(A1 cup A2 cup An) = P(A1) + P(A2) ++P(An)

(e) Probability that exactly one of A B C occurs = P(A) + P(B) + P(C) - 2P(A cap B) - 2 P(B cap C) - 2 P(A cap C) + 3 P(A cap B cap C) (f) Probability that exactly two of A B C occurs P(A cap B) + P(B cap C) + P(A cap C) - 3P(A cap B cap C) (g) Probability that atleast two of A B C occurs P(A cap B) + P(B cap C) + P(A cap C) - 2 P(A cap B cap C) (h) If A1 A2An are n events then (i) P(A1 cup A2 cup cup An) le P(A1) + P(A2) ++ P(An) (ii) P(A1 cap A2 cap cap An) ge 1 - P(Aprime1) - P(Aprime2) -- P(Aprimen) (i) If A1 A2 An are n events then P(A1 cap A2 cap cap An) ge P(A1) + P(A2) ++ P(An) - (n -1) (j) If A and B are two events such that A sube B then P(A) le P(B) (k) Max [(P(A) + P(B) - 1 P(A) P(B)] le P(A cup B) le P(A) + P(B)

30181 Partition of sample space

A family of non-empty event sets E1 E2En is said to form a partition of set S (sample space) if they are mutually exclusive as well as exhaustive rArr Ei cap Ej = f for all i ne j and 1 le i j le n and E1 cup E2 cup E3cup En = S

3019 Bayersquos TheoremBayersquos theorem revises (reassigns) the probabilities of the events A1 A2An related to a sample space when there is an information about the outcome beforehand The earlier probabilities of the events P(Ai) i = 1 2 n are called a priori probabilites and the probabilities of events calculated after the information is received ie (AiA) is called posteriori probabilites

If E1 E2 En be n mutually exclusive and exhaustive events and E is an event which occurs together (in conjugation with) either of Ei ie if events E1 E2En form a partition of S and E be any event

then ( )( )

i ii 1

P(E )P EEEPE P(E )P EE

Probability 30389

RemarkIf in a problem some event has already happened and then the probability of another event is to be found it is an application of Bayersquos theorem

Random Variable and Their Probability Distribution

It is a real-valued function having its domain as the sample space of a random experiment eg while tossing two coins if X denotes the number of heads then S = HH HT TH TT then X (HH) = 2 X(HT) = 1 X(TH) = 1 X(TT) = 0

RemarkMore than one random variables can be defined on same sample space For example in the above sample space if Y denotes the number of head minus number of tails then Y(HH) = 2 Y(HT) = 0 Y (TH) = 0 = Y(TT)= ndash2

Probability Distribution of a Random VariableIt is a table representing the possible values of random variable X along the first row and their corresponding probabilities in the second row as shown below

X x1 x2 x3 helliphelliphelliphelliphelliphellip xn

P(x = xi) p1 p2 p3 helliphelliphelliphelliphelliphellip pn

Hence Pi gt 0 for each I and =

p 1 and pi represents the probability where X = xi

Remarks

(i) ProbabilitywhenXgexk (ie X has values atleast xk) is given by

k k 1 k 2 n ii k

p p p p P+ +=

+ + + + = sum

(ii) Probability when X le xk (ie X has values atmost xk) is given by k

1 2 3 k ii 1

p p p p p=

+ + + + = sum

(iii) Probability when xm le X le xk (ie X has values atleast xm but atmost xk) is given by k

m m 1 m 2 k ii m

p p p p p+ +=

+ + + + = sum

Mean (m) or Expectation E(X) of a Random Variable X It is the sum of products of all possible values

of X by their corresponding probabilities ie mean = =

micro = =sumn

i ii 1

E(X) p x

Variance of a Random Variable X It is the weighted mean of the squares of derivations of possible values of X from mean (m) each being weighted by its probabilities with which it occurs

= σ = minusmicro = minusmicrosumn

2 2 2i i

Var(X) x (x ) p E((X ) ) = = =

sum sum2n n

2i i i i

i 1 i 1

p x p x

= E(X2) ndash [E(X)]2 where =

2 2i i

E(X ) p x

Standard deviation of a random variable X It is given by σ = = minusmicro 2x variance(X) E((X ) )

= = = =

minusmicro = minus

sum sum sum

2n n n2 2

i i i i i ii 1 i 1 i 1

(x ) p p x p x = minus2 2E(X ) [E(X)]

Bernoullirsquos Trails The trails of a random experiments are called bernoullirsquos trails if

(i) There are finite number of trails (ii) The trails are independent (iii) Each trails has two outcomes ie success or failure (iv) The probability of success remains the same in each trails

For example while throwing a die 50 times there are 50 Bernoullirsquos trails when success is the event of getting and odd number and failure is the event of getting and even number Probability of suc-cess in each trails is 12 Moreover successive trails are independent experiments

Binomial Distribution If there are n ndash Bernoullirsquos trails and P = Probability of success in each trails q = probability of failure of each trails = 1 ndash p then are successes and (n ndash r) failures can result in the

number of page in which r times S and (n ndash r) times F can be arrange ie

= = minus

r (n r)

nSSSSS FFFF Cr(n r)

Thus probability of r successes = minus

minus minus

n n r n rr r

r times (n r) times

C pppp qqqq C (p) (q) = Tr+1 of (q + p)n

Thus minus minus minus+ = + + + + + +n n n 0 n n 1 1 n n 2 2 n n r r n 0 n0 1 2 r n(q p) C (q) (p) C (q) (p) C (q) (p) C (q) (p) C (q) (p)

gives us the probabilities of 0 successes 1 successes 2 successes r successeshellip n successes as the 1st 2nd 3rd 4th hellip (r + 1)th hellip (n + 1)th terms of expansion in the right hand side P(r ndash successes) = Tr+1 of (q + p)n = nCr (q)nndashrpr Clearly as (q + p)n = 1 sum of 0 1 2 3 hellip n

successes equals 1

Probability distribution of the number of successes in an experiments consisting of n-Bernoullirsquos trails shown below (binomial distribution)

X 0 1 2 3 hellip r hellip nP(X) nC0q

n nC1qnndash1 p nC2q

nndash2 p2 nC3qnndash3 p3 helliphellip nCrq

nndashr pr helliphellip nCnpn

Remarks

(i) Here n and r is called parameter of binomial distributions

(ii) Probability of r success = P(r) = nCr(q)nndashr (p)r rArr P(x) = nCx(q)nndashx (p)x is called probability function of binomial distributions

(iii) A binomial distributions with nndashBernoullirsquos trails and with probability of success in each trail as p is denoted by B (np)

(iv) Probability of atleast m-successes = n

n r n rr

C p q 1minus

=sum ndash probability of atmost (m ndash 1) successes

n r n rr

1 C p qminus

minus sum

Probability 30391

(v) Probability of atmost m-successes = m

n r n rr

C p q minus

=sum = 1 ndash probability of atleast (m + 1) successes

n r n rr

r ( m 1)

1 C p q minus

minus sum

(vi) Probability of atleast m-success and atmost k successes is given by k

n r n rr

C p q minus

(vii) P (atmost m successes) + P (atleast (m + 1) success) = 1

n r n r n r n rr r

r 0 r m 1

C p q C p q 1minus minus

+ =sum sum

Standard Deviation of Binomial Distribution

σ = minussumn

2 n r n r 2x r

r C p q [E(X)] = minus

minus + minussumn

2 n r n r 2 2r

(r r r) C p q n p

= minus minus minus minus minus minusminus minus

= minus

minusminus + minus

minussum sumn n

n 2 2 r 2 n r n 1 r 1 n r 2 2r 2 r 1

r 2 r 1

n(n 1) nr(r 1) C p p q r C pp q n pr(r 1) r

= minus minusminus + + + minusn 2 2 n 1 2 2n(n 1)(q p) p n(q p) p n p

= minus + + = minus =2 2 2 2 2n p np np n p np(1 p) npq

=SD npq

Variance of Binomial Distribution (SD)2 = npq

Mode of Binomial Distribution Mode of binomial distribution is the value of r when P(X = r) is maximum (n + 1) p -1 le r le (n + 1) p

302 GEomEtrical probability

When the number of points in the sample space is infinite it becomes difficult to apply classical definition of probability For instance if we are interested in finding the probability that a point selected at random from the interval [1 5] lies either in the interval [1 2] or [4 5] we cannot apply the classical definition of probability In this case we define the probability as follows

isin =Measure of region A

Px AMeasure of the sample space S

where measure stands for length area or volume depending upon whether S is an one-dimensional two-dimensional or three dimensional region

Here the required probability = + +

= =lengthof interval [12] lengthof interval [45] 1 1 1

lengthof interval [15] 4 2

Chapter 31MatriCes and

deterMinants

311 Matrix

A rectangular array of (m times n) objects arranged along m-horizontal lines (called rows) and along n-vertical lines (called columns) as shown below

11 12 ln

1 2 3 mn

am am am a

Here aij = elements in ith row and jth column The matrix as shown here is denoted by [aij]mtimesnOrder of Matrix Matrix having m-rows and nndashcolumns is said to have order m times nReal Matrix A matrix having all real elementsComplex Matrix A matrix having atleast one imaginary elementComplex Conjugate of a Matrix A matrix obtained by replacing the elements of a complex matrix

A = [aij]mtimes n by their conjugate is called complex conjugate of matrix A and it is denoted by times

= ij m nA a

Rectangular Matrix A matrix of order m times n where m n isin ℕ and m ne n These are of two types (a) Horizontal Matrix A matrix of order m times n where n gt m ie number of columns is greater than

number of rows (b) Vertical Matrix A matrix of order m times n where m gt n ie number of rows is greater than number

of columnsRow Matrix A matrix of order 1 times n that is a matrix having one row onlyColumn Matrix A matrix of order n times 1 that is a matrix having one column only

Remark

Clearly row matrix is horizontal whereas column matrix is vertical

Square Matrix Matrix of order m times n that is a matrix having equal number of rows and columns Such a matrix is called mndashrowed square matrix

Matrices and Determinants 31393

Principal (Leading) Diagonal and Off-diagonal of Square MatrixDiagonal along which the elements a11 a22 a33hellip ann lie is called principal diagonal or simply diagonal when there is number chance of confusion The other diagonal is called off-diagonal

The elements lying diagonal are called diagonal elements

Trace of a Square Matrix The sum of diagonal elements

ii 11 22 33 nni 1

(notation)

a (a a a a ) Tr(A)=

= + + + + =sum

Diagonal Matrix A square matrix having all non-diagonal elements zeros ie

a 0 0 00 a 0 00 0 a 0 0 0 0 a

= diagonal [a11 a22 a33 hellip ann] or 11 22 33 nn

(Notation)

diagonal (a a a a )

Scalar Matrix A diagonal matrix having all diagonal elements equal ie

k 0 0 00 k 0 00 0 k 0

diagonal(k k k k) 0 0 0 k

Unit Matrix (Identity Matrix) A scalar matrix having each diagonal element unit ie 1

1 0 0 00 1 0 00 0 1 0 0 0 0 1

Remark

I1 = [1]

= = 2 3

1 0 01 0

I I 0 1 00 1

are called unit matrix of order 1 order 2

and order 3 and so on respectively

Null Matrix (Zero Matrix) A matrix having its all elements zero

Triangular Matrix A square matrix in which all the elements above the principal diagonal or below the principal diagonal are zero is called triangular matrix

Lower Triangular Matrix A square matrix having its all elements above diagonal zeros ie having non-zero elements (if not only null) on principal diagonal or below it ie aij = 0 for a11 i lt j

Upper Triangular Matrix Square matrix having its all elements below diagonal zero ie having non-zero elements (if non-null) on or above the principal diagonal ie aij = 0 for all i gt j

ExampleUpper triangle contains non-zero elements if non-null matrix

Remarks

(i) Null square matrix is simultaneously both upper as well as lower triangular matrix

(ii) Minimum number of zeros in a triangular matrix of order minus

=n( n 1)

(iii) Maximum number of non-zero entries in a triangular matrix of order +

=n( n 1)

(iv) Diagonal matrix is simultaneously both upper as well as lower triangular matrix

(v) Minimum number of zero entries in a diagonal matrix = (n2 ndash n) = n (nndash1)

(vi) Maximum number of non-zero entries in a diagonal matrix of order n = n

(vii) Maximum number of zero entries in a diagonal matrix of order n = n2 (when its is null)

(viii) Maximum number of different elements in a triangular matrix of order + +

=2n n 2

(ix) Minimum number of different elements in a non-null diagonal matrix of order n = 2 (x) Minimum number of different elements in a non-null triangular matrix = 2 (xi) Minimum number of zeros in a scalar matrix = (n2 ndash n) (xii) Number of zeros in a non-null scalar matrix = (n2 ndash n) (xiii) Number of different entries in a non null scalar matrix = 2 (xiv) A triangle matrix is called strictly triangular iff aii = 0 for all i 1 le i le n

312 Sub Matrix

Matrix obtained by leaving some rows or columns or both of a matrix A is called a sub-matrix of matrix A

For example 2 57 9

is a sub-matrix of matrix

2 5 87 9 41 3 5

3121 Equal Matrices

Two matrices are said to be equal iff they are of same order and the elements on their corresponding positions are same ie A = [aij]m times n = B [bij]r times p hArr m = r n = p and aij = bij

31211 Addition of matrices

Two matrices A = [aij] and B =[Bij] are said to be conformable for addition iff they are of same order Further A + B = [aij + bij]m times n where A = [aij]mtimesn and B = [bij]m timesn

Properties of Matrix Addition 1 Matrix addition is commutative A + B = B + A Matrix addition is associative A + (B + C) = (A + B) + C 2 Null matrix of order m times n additive identity in the set of matrices of order m times n If [aij]mtimesn = 0 and B = [bij]mtimesn then [aij]mtimesn + [bij]mtimesn = [bij]mtimesn = [bij]mtimesn + [aij]mtimesn where aij = O for all i and j 3 ndashA = [ndashaij]mtimesn is additive inverse of A = [aij]mtimesn 4 Left cancellation law A + B = A + C rArr B = C Right cancellation law A + B = C + B rArr A = C 5 A + X = O has a unique solution X = ndashA of order m times n and X = [ndashaij]m timesn if A = [aij]m timesn

Subtraction of Matrices If A = [aij]m timesn and B = [bij]m timesn ie A and B are of same order (conformable for subtraction) then A ndash B = [aij ndash bij]m timesn

Properties of Subtraction of Matrices 1 Neither commutative nor associative 2 Follows left concellation and right concllation 3 Left cancellation law A ndash B = A ndash C rArr B = C 4 Right cancellation law A ndash B = C ndash B rArr A = C 5 Equation A ndash X = O where O is a null matrix of order m times n and A and X are matrices of order

(m times n) has a unique solution X = A

Multiplication of Matrix by a ScalarlA = l[aij]mtimesn = [laij]mtimesn ie scalar multiplication of a matrix A gives a new matrix of same order whose elements are scalar (l) times the corresponding elements of matrix A

Scalar Multiplication is Commutative and Distributive

(i) Matrix addition is commutative and associative (ii) Follows cancellation and right cancellation law

313 Multiplication of Matrix

Two matrices A and B are said to be conformable for the product AB if A = (aij) is of the order m times n and

B = (bij) is of the order n times p the resulting matrix is of the order m times p and AB = (Cij) where n

ij ik kjk 1

(C ) a b=

= ai1 b1j + ai2 b2j + ainbnj for i = 1 2 3m and j = 1 2 3pAs an aid to memory denote the rows of matrix A by R1 R2

R3 and columns of B by C1 C2 and C3

Also 1 1 1 1 2 1 3

2 1 2 3 2 1 2 2 2 3

3 3 1 3 2 3 3

R R C R C R CA B R (C C C ) R C R C R C

R R C R C R C

times = times =

where Ri Cj is the scalar product of Ri and CjThe diagrammatical working of product of two matrices is

shown as in the figure

Remarks (i) In the product AB A is called post-multiplied by B and B is called P multiplied by A

(II) A = [a1 a2an] and

rArr AB = [a1 b1 + a2 b2 + + an bn]

3131 Properties of Multiplication of Matrices 1 AB and BA both may be defined yet AB ne BA 2 AB and BA both may be defined and AB = BA 3 One of the products AB or BA may not be defined 4 If A be a square matrix of the same order as I then IA = A I = A and OA = AO = O where O is a null

matrix ie multiplication by identity and null matrix is commutative 5 AB may be a zero matrix and BA may be a non-zero matrix or vice versa when A ne O

B ne O 6 AB and BA both may be a zero matrix when A ne 0 B ne 0 7 Multiplication of matrices is associative and distributive over addition 8 The matrix AB is the matrix B pre-multiplied by A and the matrix BA is the B post multiplied

by A 9 If A B are suitable matrices and λ is a scalar then λ (AB) = (λA)B = A(λB) 10 Existence of multiplicative Identity If A = [aij ] is an m times n matrix then ImA = A = AIn 11 The product of any matrix and null matrix of a suitable order is a null matrix If A = [aij] is an m times n matrix then Op times m A = Op times n and AOn times q = Om times q 12 Powers of a square matrix Let A be a square matrix of order n then AA makes sense and it is also

a square matrix of order n We define A1 = A A2 = AAAm = Am ndash 1A = AAm ndash 1 for all positive integers m

3132 Transpose of a Matrix

A matrix obtained by interchanging rows and columns of a matrix A is called the transpose of a matrix

If A is a matrix then its transpose must be denoted as Aprime or AT eg if 2 3 5

A5 6 8

then T

2 5A 3 6

Properties of Transpose of a Matrix

(i) (AT)T = A ie the transpose of the transpose of a matrix is the matrix itself (ii) (A + B) T = AT + BT ie the transpose of the sum of two matrices is the sum of their transpose (iii) (kA)T = kAT (where k is a scalar) (iv) (AB)T = BTAT ie the transpose of the product of two matrice is the product in reverse order of

their transpose (v) (ndashA)T = ((ndash1) A)T = (ndash1)AT = ndashAT (vi) (A ndash B)T = (A + (ndashB))T = AT + (ndashB)T = AT + (ndashBT) = AT ndash BT) (vii) If A is m times n matrix then AT is n times m matrix

3133 Symmetric MatrixA square matrix will be called symmetric if the elements across principal diagonal are symmetrically equal

Skew Symmetric Matrix A square matrix A = [aij]mtimesn is said to be skew symmetric iff aij = ndashaij forall i and jrArr aii= 0 forall i ie the diagonal elements are zeros

31331 Properties of symmetricskew-symmetric matrix

1 A symmetricskew-symmetric matrix is necessarily a square matrix 2 Symmetric matrix does not change by interchanging the rows and columns ie symmetric matrices are transpose of themselves 3 A is symmetric if AT = A and A is skew-symmetric if AT = ndash A 4 A + AT is a symmetric matrix and A ndash AT is a skew-symmetric matrix Consider (A + AT) = AT + (AT) T = AT + A = A + AT = A + AT is symmetric Similarly we can prove that A ndash AT is skew-symmetric 5 The sum of two symmetric matrix is a symmetric matrix and the sum of two skew-symmetric matrix

is a skew symmetric matrix 6 If A and B are symmetric matrices then AB + BA is a symmetric matrix and AB ndash BA is a skew

symmetric matrix 7 Every square matrix can be uniquely expressed as the sum of symmetric and skew-symmetric matrix

8 Maximum number of distinct entries in a symmetric matrix of order n is n(n 1)

9 Maximum number of distinct elements in a skew symmetric matrix of order n = n2 ndash n +1 10 Maximum number of distinct nonndashzero elements in a skew-symmetric matrix of order

n = (n2 ndash n) = n (n ndash 1)

11 Maximum number of elements with distinct magnitude in a skew-symmetric matrix = 2n n 12

minus+

12 The matrix (B) AB is symmetric or skew-symmetric according as A is symmetric or non-symmetric respectively

13 The determinant of a skew-symmetric matrix with real entries and odd order always vanishes 14 The determinant of a skew-symmetric matrix with even real entries order is always a perfect square

31332 Properties of trace of a matrices

(i) tr(lA) = ltr(A) (ii) tr (A plusmn B) = tr (A) plusmn tr(B) (iii) tr(AB) = tr(BA) (iv) tr (skew-symmetric matrix) = 0 (v) tr(A) = na where A is a scalar matrix of order n and with diagonal elements a (vi) tr [diagonal (a b c) diagonal (d e f)] = tr [diagonal (ad be cf)] = (ad + be + cf)

(vii) tr(A) tr(A) A= = conjugate matrix of A

(viii) tr(Aprime) = tr(A) Aprime = transpose of matrix A

314 HerMitian Matrix

If A = [aij]mtimesn is such that ija aji= ie ( ) =A A ie Aq = A where Aq = ( ) ( )A A =

eg ( )2 3 2i 2 3 2i 2 3 2iA A A A

3 2i 7 3 2i 7 3 2i 7+ minus +

rArr rArr = = minus + minus

3141 Properties of Hermitian Matrices 1 Diagonal elements are purely real ii ii ii iia a a a 0= rArr minus = rArr 2Im (aii) = 0

2 Every symmetric matrix with real number as elements is hermitian eg ij ij jia a a A= = rarr is hermitian

3142 Skew-Hermitian Matrix

If A = [aij]mtimesn is such that ij ija aminus = ie (A ) A= minus ie Aq = ndashA eg 3i 1 3i 2

A 1 3i 0 4 i2 4 i 2i

minus = minus minus + minus minus +

31421 Properties of hermitianskewndashhermitian matrix

1 Elements on principal diagonal are either purely imaginary or zero eg for i = j

ii iia a= minus rArr ℝ(aii) = 0 rArr aii is purly imaginary 2 Every skew-symmetric matrix with real numbers as elements is skew-Hermitian 3 Every square matrix can be uniquely represented as the sum of a hermitian and skew-Hermitian

matrices

4 If A is any matrix then 1 1A A A A A 2 2

θ θ= + + minus = Hermitian + skew-Hermitian

3143 Orthogonal MatrixA square matrix A is called an orthogonal matrix if the product of the matrix A and its transpose A is an identity matrix ie AA = AA = I

31431 Properties of Orthogonal Matrix

(i) If AAprime = I then Andash1 = Aprime Q AAprime = I rArr Andash1 (AAprime) = Andash1I = Andash1 rArr Aprime = Andash1 (ii) If A and B are orthogonal then AB is also orthogonal Q (AB) (AB) = (AB) (BprimeAprime) = A(BBprime)Aprime = AIAprime = AAprime = I similarly (ABprime) (AB) = I (iii) Value of corresponding determinant of orthogonal matrix is plusmn1

3144 Idempotent MatrixA square matrix A is called idempotent provided that it satisfies the relation A2 = A

Properties

(i) If A and B are idempotent matrices then AB is as idempotent matrix if AB = BA (ii) If A and B are idempotent matrices then A + B is an idempotent if AB + BA = O (iii) A is idempotent and A + B = I then B is also idempotent and AB = BA = O

3145 Periodic MatrixA square matrix A is called periodic if Ak+1 = A where k is a positive integer If k is the least positive integer for which Ak+1 = A then k is said to be period of A For k = 1 we get A2 = A and we called it to be an idempotent matrix

3146 Nilpotent MatrixA square matrix A is called Nilpotent matrix of order k provided that it satisfies the relation Ak = O and Akndash1 ne A where k is positive integer and O is null matrix and k is the order of the nilpotent matrix A

3147 Involutory MatrixA square matrix A is called involutory matrix provided that it satisfies the relation A2 = I where I is

identity matrix eg 1 0

= minus and 2 1 0

A I0 1

Properties (i) A is involutory iff (A + I) (A ndash I) = O (ii) Identity matrix is a trivial example of involutory matrix

3148 Unitary Matrix

A square matrix A is called a unitary matrix if A Aq = I where I is an identity matrix and Aq is the trans-pose conjugate of A

31481 Properties of Unitary Matrix

(i) If A is unitary matrix then Aprime is also unitary (ii) If A is unitary matrix then Andash1 is also unitary (iii) If A and B are unitary matrices then AB is also unitary

31482 Determinant of a square matrix

A number associate with every square matrix A is called its determinant and denoted by |A| or det (A)

Let a b

then a b

A (ad bc)c d

= = minus

Evaluation of Determinant of Order 3

Let 11 12 13 11 12 13

21 22 23 21 22 23

31 32 33 31 32 33

a a a a a aA a a a then A a a a

a a a a a a

= 22 23 21 23 21 2211 12 13

32 33 31 33 31 32

a a a a a aa a a

a a a a a aminus +

Singular Matrix Square matrix having its determinant = 0

3149 Non-singular Matrix

Square matrix A for which |A| ne 0

31491 Minor of elements of a square matrix

The determinant obtained by deleting the ith row and jth column passing through the aij element is called

minor of element aij and is denoted by Mij eg 11 1223

a a= = (a11a32 ndash a31a12) = minor element a23

where 11 12

Co-factors of Element of Square Matrix The determinant obtained by deleting ith row and jth column when multiplied by (ndash1)i+j gives us the co-factors of element aij and is denoted by Aij or Cij In other words

Cij = (ndash1)i + j Mij ie (ndash1)i+j times the minor of element aij

eg 11 122 3 523 23 11 32 31 12

a aC ( 1) M ( 1) (a a a a )

a a+= minus = minus = minus minus = co-factor of element a23

Remarks

(i) |A| = a11C11 + a12 C12 + a13 C13 in general =

= sum3

ik ikk 1

A a C i = 1 or 2 or 3 (expansion along rows) or

= sum3

kj kjk 1

A a C j = 1 or 2 or 3 (expansion along columns)

(ii) = =

= = nesum sum3 3

ik jk ki kjk 1 k 1

a C a C 0 for i j

315 adjoint of a Square Matrix

The transpose of the matrix containing co-factors of elements of square matrix A It is denoted by Adj(A)

ie 11 12 13

21 22 23

31 32 33

C C CAdj(A) C C C

where Cij = co-factors of aij rArr 11 12 13

21 22 23

31 32 33

C C CAdj(A) C C C

3151 Properties of Adjoint of Square Matrix A

(i) A(adj A) = (adj A) (A) = |A| In where A is a square matrix of order n (ii) If A is a singular matrix then A(Adj A) = (Adj A) A = 0 (Q |A| = 0) (iii) |Adj A| = |A|nndash1 (iv) Adj (AB) = (Adj B) (Adj A) provided that A B are non-singular square matrices of order n (v) Adj (AT) = (Adj A)T (vi) Adj( Adj A) = |A|nndash2 A

(vii) 2(n 1)Adj (Adj A) A minus

= (viii) Adjoint of a diagonal matrix is a diagonal matrix (ix) adj(lA) = lnndash1 (Adj A) where l is a scalar and (A)ntimesn

3152 Inverse of Non-singular Square Matrix

A square matrix B of order n is called inverse of non-singular square matrixA of order n iff AB = BA = In

Let AdjA

B A 0A

= ne rArr nn

A IA(AdjA)AB I

A A= = = similarly n

A I(AdjA)BA A I

A A= = =

Thus AB = BA = In rArr B = Andash1 Thus 1 AdjAA

Aminus = provided that |A| ne 0

Invertible Matrix A square matrix iff it is non-singular ie |A| ne 0

31521 Properties of inverse of square matrix

1 Every invertible matrix possesses a unique inverse 2 A square matrix is invertible if and only if it is non-singular 3 If A B be two non-singular matrices of the same order then AB is also non-singular and

(AB)ndash1 = Bndash1 Andash1 (reversal law of inverse) 4 (i) AB = AC rArr B = C (ii) BA = CA rArr B = C 5 Since we already know that (AB)ndash1 = Bndash1 Andash1 therefore in general we can say that

(ABCZ)ndash1 = Zndash1 Yndash1 Bndash1 Andash1 6 If A is an invertible square matrix then adj (Aprime) = (adj A)prime 7 (AT)ndash1 = (Andash1)T

8 T 1 1 T(A ) (A )minus minus= 9 AAndash1 = Andash1A = I 10 (Andash1)ndash1 = A

316 Matrix polynoMial

Let f(x) = a0 xm + a1 x

mndash1 + + amndash1 x + am be a polynomial in x and A be a square matrix of order n then f(A) = a0 A

m + a1 Amndash1 + + amndash1 A + am In is called a matrix polynomial in A Thus to obtain f (A) replace

x by A in f(x) and the constant term is multiplied by the identity matrix of the order equal to that of AThe polynomial equation f (x) = 0 is said to be satisfied by the matrix A iff f(A) = Oeg if f(x) = 2x2 ndash 3x + 7 and A is a square matrix of order 3 then f(A) = 2A2 ndash 3A + 7I3The polynomial | A ndash x In| is called characteristic polynomial of square matrix AThe equation | A ndash x In| = O is called characteristic equation of matrix A

So a0 An + a1 A

n ndash 1 + + a2 An ndash 2 + + an I = O rArr 1 n 1 n 20 1

a aA A A a a

minus minus minus = minus + +

3162 Elementry Transformation bull Interchange of any two rows or columns Denotion by i jR Rharr or

i jC Charr

bull Multiplication by non-zero scalar Denotion i iR kRharr or

j jC kCharr bull Replacing the ith row (or column) by the sum of its elements and scalar multiplication of

corresponding elements of any other row (or column)Denotion Ri rarr Ri + kRj or Ci rarr Ci + k Cj

bull Transformed matrix using sequence of elementary transformations (one or more) is known as equivalent matrix of A

3163 Elementary MatrixElementary matrix obtained from identities matrix by single elementary transformation

eg 1 3

1 0 0 0 0 10 1 0 R R 0 1 00 0 1 1 0 0

3164 Equivalent MatricesTwo matrices A and B are equivalent if one can be obtained from the other by a sequence of elementary transformations denoted by A ~ B

31641 Inverse of a matrix A by using elementary row operations

Step 1 Write A = In A ie

11 12 1n

21 22 2n

n1 n2 nn

a a a 1 0 0 0a a a 0 1 0 0

a a a 0 0 0 1

Step 2 Now applying the sequence of elementary row operation on matrix A and matrix In simultaneously till matrix A on LHS of the above equation get converted to identity matrix InStep 3 After (Step 2) reaching at In = BA rArr B = Andash1

31642 Inverse of matrix A by using elementary column operations

Step 1 Write A = A InStep 2 Now apply as above sequence of elementary column operations on matrix A on the left hand side and same sequence of elementary column operations on identity matrix In on the right hand side of the above equation till matrix A on the left hand side gets converted to InStep 3 After (Step 2) reaching at In = AB rArr B = Andash1

31643 System of simultaneous equations

The system of n equations in n-unknown given by a11x1 + a12x2 + a13x3 + hellip + a1nxn = b1a21x1 + a22x2 + a23x3 + hellip + a2nxn = b2an1x1 + an2x2 + an3x3 + hellip + annxn = bn where b1 b2 b3 helliphellip bn are not all zeros is called non-homogenous system of equations

This system of equation can be written in matrix form as

12 1n11 1 1

22 2n21 2 2

n2 nnn1 n n

a aa x ba aa x b

a aa x b

AX = B Here A is a square matrix A system is said to be consistent if it has atleast one set of solution otherwise known as inconsistent equation

31644 Solutions of non-homogenous systems of equation

There are three methods of solving non-homogenous equations in three variables

(i) Matrix method (ii) Determinant method (Cramerrsquos rule) (iii) By using elementary row and column operations

31645 Matrix method of solving non-homogeneous system of equations

Let the given system of equation be AX = B rArr X = Andash1B gives us 1 Unique solution of system of non-homogenous equations provided |A| ne 0 2 No solution if |A| = 0 and (adj A) B ne 0 (null matrix) 3 Infinitely many solutions if |A| = 0 and (adj A) B = 0 For getting infinitely many solutions take

any (n ndash1) equations Take any one variable say xn = k and solve these (n ndash1) equations for x1 x2 x3hellip xnndash1 in terms of k

The infinitely many solutions are given by x1 = f1(k) x2 = f2(k) hellip xnndash1 = fnndash1(k) xn = k and k isin ℝ

317 deterMinant MetHod (craMerprimeS rule) for Solving non-HoMogenouS equationS

3171 For Two Variables

Let a1x + b1y = C1 and a2x + b2y = C2 then take 1 1 1 1 1 11 2

2 2 2 2 2 2

a b c b a c

a b c b a c∆ = ∆ = ∆ =

ie D is determinant formed by coefficient of x and yD1 is determinant formed by replacing elements of first column of D by C1 and C2 and D2 is

determinant formed by replacing elements of second column of D by C1 and C2

Case (i) If D ne 0 then system of equation has a unique solution given by 1 2x y∆ ∆= =∆ ∆

Case (ii) If D = 0 (a) If D1 D2 both are not zeros ie atleast one of D1 and D2 is non zero then there is no solution (b) If D1 = D2 = 0 then the system of equation has infinitely many solution Take x or y say y = k

rArr 1 1

C b kxaminus

= Thus 1 1

C b kxaminus

= y = k k isin ℝ gives infinitely many solutions

3172 For Three Variables

a1x + b1y + c1z = d1 a2x + b2y + c2z = d2 a3x + b3y + c3z = d3

1 1 1 1 1 1 1 1 1 1 1 1

2 2 2 1 2 2 2 2 2 2 2 3 2 2 2

3 3 3 3 3 3 3 3 3 3 3 3

a b c d b c a d c a b da b c d b c a d c a b da b c d b c a d c a b d

∆ = ∆ = ∆ = ∆ =

Case (i) For D ne 0 there will be unique solutions 31 2x y z∆∆ ∆

= = =∆ ∆ ∆

Case (ii) For D = 0 (a) If atleast one of D1 D2 D3 is non-zero there is no solution ie system of equations is consistent (b) If D1 = D2 = D3 = 0 then there will be infinitely many solutions For these infinitely many solu-

tions take any two equations say (i) and (ii) and put z = k to obtain a1x + b1y = d1 ndash c1k and

a2x + b2y = d2 ndash c2k Solving we get x and y in term of k (say) x = f1(k) and y = f2(k) Thus x = f1(k) y = f2(k) z = k k isin ℝ gives us infinitely many solutions

318 Solution of non-HoMogeneouS linear equationS by eleMentary row or coluMn operationS

Let 11 12 13 1 1

21 22 23 2 2

31 32 33 3 3

a a a x bA a a a X x B b

a a a x b

be such that AX = B ie 11 12 13 1 1

21 22 23 2 2

31 32 33 3 3

a a a x ba a a x ba a a x b

(by using elementary row operations)Apply elementary row operations on matrix A and same operations simultaneously on B to reduce

it into a b c x0 d e y0 0 f z

λ = micro α

rArr ax + by + cz = l (i) dy + ez = m hellip(ii) fz + a hellip(iii)

from equation (iii) we get z = af from equation (ii) we get e

α micro minus =

And from equation (i) we get

α micro minus α λ minus minus = (By using elementary column

operations)Now applying elementary column operations to Aprime and simultaneously same elementary column

operationrsquos to Bprime to get

rArr 1 2 3

a 0 0[x x x ] b d 0 [ ]

= λ micro α

rArr ax1 + bx2 + cx3 = l helliphellip(i) dx2 + ex3 = m helliphellip(ii) fx3 = a hellip(iii)

From (iii) 3xfα

= from (ii) 2

α micro minus = from (iii) 1

α micro minus α λ minus minus =

3181 Solutions of Homogenous System of EquationConsider the following system of homogenous linear equation in n unknowns x1 x2 xn

11 1 22 2 1n m

22 1 22 2 2n m

m1 1 m2 2 mn n

a x a x a x 0a x a x a x 0a x a x a x 0

+ + + =+ + + =

+ + + =

This system of equation can be written in matrix form as follows

12 1n11 1

22 2n21 2

n2 nnn1 n

a aa x 0a aa x 0

a aa x 0

rArr AX = O

(i) If | A | ne 0 the system of equations has only trivial solution and that will be the only solution (ii) If | A | = 0 the system of equations has non-trivial solution and it has infinite solutions (iii) If number of equations lt Number of unknowns then it has non-trivial solution

RemarkIf numbers of equations lt number of unknown variables then either the system of equations have no solutions or infinitely many solutions

319 eliMinant

Eliminant of a given number of equation in some variables is an expression which is obtained by eliminating the variables out of these equations

3191 Linear TransformationThe transformation in which the straight line remains straight and origin does not change its position

We represent point (x y) by column matrix

and transformation mapping is denoted by a matrix

operation which transform x X

Definition Any transformation of xy

that can be expressed by the linear equation

a1x + b1y = X and a2x + b2y = Y is called linear transformation

a b x Xa b y Y

operator 1 1

is matrix of transformation

Origin remains invariant of such transformation Some common linear transformations are 1 Drag by a factor k along x-axis 4 Rotation through any angle about origin 2 Enlargment or reduction 5 Shearing parallel to x-axisy-axis 3 Reflection in any line through origin

3192 Compound TransformationWhen a transformation (2) is carried out after (1) the compound transformation is denoted by a matrix operator M2 o M1 = M2 M1 where M2 and M1 are respective matrix operators for (i) and (ii) operation M2 o M1 is known as composition of M2 with M1 (order of performance of operations must be mentioned)

Matrix representing reflection in x-axis If P(xy) be any point and Pprime (XY) is its reflection on x ndashaxis then X = 1(x) + 0(y) and Y = 0(x) + (ndash1)y

rArr X 1 0 xY 0 1 y

= minus Thus

1 00 1 minus

described reflection of point P(xy) on x-axis

Matrix representing reflection in y-axisHere X = (ndash1)x + 0(y) and Y = (0) x + 1(y)

Matrix representing reflection through the origin If P(xy) is any point then Pprime (XY) ie reflection of P(xy) on origin is given by X = ndash1(x) + 0(y) and

Y = 0(x) + (ndash1)y rArr X 1 0 xY 0 1 y

minus = minus

Matrix representing reflection in the line y = x

Let P(xy) be any point and (XY) be its reflection on line y = xHere X = y and Y = x

rArr X = 0(x) + 1(y) and Y =1(x) + 0(y) rArr X 0 1 xY 1 0 y

Matrix representing reflection in the line y = x tanq

rArr X cos2 sin2 xY sin2 cos2 y

θ θ = θ minus θ

Matrix representing rotation through an angle q

rArr X cos sin xY sin cos y

θ minus θ = θ θ

Expansion of determinant using co-factor (Laplace method)

Let 11 12 13

21 22 23

31 32 33

a a aa a aa a a

∆ = be determinant or order 3 times 3 then

i1 i1 i2 i2 i3 i3 ik ikk 1

a C a C a C a C=

∆ = + + + =sum = expansion of D along ith

rows and 3

1j 1j 2 j 2 j 3j 3j kj kjk 1

a C a C a C a C=

∆ = + + + =sum = expansion of D along

jth column

Sarrus rule of expanding a determinant of third order

Sarrus gave a rule for evaluating a determinant of the order three mentioned as follows

P(x y)

Rule Write down the three rows of the determinant and rewrite the first two rows just below them The three diagonals sloping down to the right give the three positive terms and the three diagonals

sloping down to the left give the three negative terms If 11 12 13

21 22 23

31 32 33

a a aa a aa a a

∆ = then

3193 Application of Determinant

Out of wide applications of determinants a few are given belowbull Area of D with vertices A(x1 y1) B(x2 y2) C(c3 y3)

rArr 1 1

x y 11 x y 12

x y 1∆ = where |x| denotes absolute value of x

bull Cross product of vectors x y z x y zˆ ˆˆ ˆ ˆ ˆa a i a j a kb b i b j b k= + + = + +

rArr x y z

ˆˆ ˆi j ka b a a a

b b btimes =

It is also used to find the scalar triple product of three vector a(b c)times

is STP of x y z

a a a[abc] b b b

c c c=

3194 Properties of DeterminantsProperty 1 The value of determinant remains unaltered if the rows are changed into columns and

columns into rows For example if 1 1

a ba b

∆ = (a1 b2 ndash b1 a2) and 1 21 2 2 1

a a(a b a b )

b bprime∆ = = minus rArr D = Dprime

Property 2 If all the elements of a rowcolumn are zero then the value of determinant will be zero

Property 3 Reduction and increase of order of determinant (a) If all the elements in a row (or a column) except one element are zeros the determinant reduces

to a determinant of an order less by one (b) A determinant can be replaced by a determinant of a higher order by one as per the requirment

Property 4 If any two rows or two columns of a determinant are interchanged the determinant retains its absolute value but changes its sign and symbolically the interchange of ith and jth rows or ith and jth columns is written as

i jR R∆ = minus∆ i jC C(or )minus∆

Property 5 The value of a determinant is zero if any two rows or columns are identical Symbolically it is written as DRi equiv Rj

= 0 or DCi equiv Cj = 0

Property 6 (a) If every element of a given row of matrix A is multiplied by a number l the matrix thus obtained has determinant equal to l (det A) As a consequence if every element in a row of a determinant has the same factor this can be factored out of the determinant Symbolically it is written as

i i1R Rm

∆ = ∆

(b) If all the elements of a row (column) of a determinant are multiplied by a constant (k) then the determinant gets multiplied by that constant

Property 7 The value of the determinant corresponding to a triangular determinant is equal to product of its principal diagonal elements

Property 8 If any row or column of a determinant be passed over n rows or columns the resulting determinant will be (ndash1)n times the original determinantProperty 9 (a) If every element of a column or (row) is the sum (difference) of two terms then the determinant is equal to the sum (difference) of two determinants of same order one containing only the first term in place of each sum the other only the second term The remaining elements of both determinants are the same as in the given determinant (b) A determinant having two or more terms in the elements of a row (or column) can be written as the

sum of two or more determinantsProperty 10 The value D of a determinant A remains unchanged if all the elements of one row (column) are multiplied by a scalar and added or subtracted to the corresponding elements of another row (column) Symbolically it is written as

i i jR R mRrarr +∆ = ∆ (or j j iC C mCrarr +∆ ) and operation is also symbolically written as

Ri rarr Ri + mRj or Cj = Cj + mCi

Property 11 (a) The sum of the products of elements of a row (or column) with their corresponding co-factors is equal to the value of the determinant For example a11C11 + a12 C12 + a13 C13 = a21 C21 + a22 C22 + a23 C23 = D (b) Sum of the products of elements of any row (or column) with the co-factors of the corres sponding

elements of a parallel row (or column) is always zero For example a11 C21 + a12 C22 + a13C23 = 0Property 12 If the elements of a determinant D involve x ie the determinant is a polynomial in x and if it vanishes for x = a then (x ndash a) must be a factor of D In other words if two rows (or two column) become identical for x = a then (x ndash a) is a factor of D Generalizing this result we can say if r rows (or r columns) become identical when a is substituted for x then (x ndash a)r ndash 1 should be a factor of D

For example if 2

x 5 2x 9 4x 16 8

∆ = at x = 2 D = 0 (∵ C1 and C2 become identical at x = 2)

3195 CautionWhile applying all the above properties from property 1 to property 10 atleast one row (or column) must remain unchanged

3110 Special deterMinant

31101 Symmetric DeterminantSymmetric determinant is a determinant in which the elements situated at equal distance (symmetrically) from the principle diagonal are equal both in magnitude and sign ie (i j)th element

(aij) = (j i)th element (aji) eg 2 2 2

a h gh b f abc 2fgh af bg chg f c

+ minus minus minus

31102 Skew-Symmetric DeterminantAll the diagonal elements are zero and the elements situated at equal distance from the diagonal are equal in magnitude but opposite in sign ie (i j)th element = ndash(j i)th element ie aij = ndashaji The value of a

skew-symmetric determinant of odd order is zero eg 0 b cb 0 a 0c a 0

minus∆ = minus =

31103 Cyclic DeterminantsDeterminants in which if a is replaced by b b by c and c by a then value of determinants remains unchanged are called cyclic determinants

(i) 2 2 2

1 1 1a b c (a b)(b c)(c a)a b c

= minus minus minus (Already proved in previous article)

(ii) 3 3 3

1 1 1a b c (a b)(b c)(c a)(a b c)a b c

= minus minus minus + + (can be proved using factorization)

(iii) 2 2 2

1 1 1a b c (a b)(b c)(c a)(ab bc ca)a b c

31104 CirculantsCirculants are those determinants in which the elements of rows (or columns) are cyclic arrangements of letters

x a x b x cx b x c x ax c x a x b

+ + ++ + ++ + +

a b c db c d ac d a bd a b c

ega b cb c a (a b c 3abc)c a b

= minus + + minus

(iii) 3 3 3

a b cb c a (a b c 3abc)c a b

= minus + + minus (iv)

a b c x a y b z cb c a y b z c x a

z c x a y ba b c

+ + ++ + ++ + +

Remarks 1 An expression is called cyclic in x y z iff cyclic replacement of variables does not change the expression

eg x + y + z xy + yz + zx etc Such expression can be abbreviated by cyclic sigma notation as below

= + + = + + minus =sum sum sum2 2 2 2x x y z xy xy yz zx ( x y ) 0 = + + + + + = +sum sum2 2 2 2x y z x y z x x

2 An expression is called symmetric in variable x and y iff interchanging x and y does not change the expression x2 + y2 x2 + y2 ndash xy x3 + y3 + x2y + y2x x3 ndash y3 is not symmetric

31105 Product of Two DeterminantTwo determinants are conformable to multiply iff they are of same size Since |A| |B| = |AB| = |ATBT | = |AT

B| = |A| BT| There are four method of taking product of two determinant

Let 1 1 1

1 2 2 2

a b ca b ca b c

∆ = and 1 2 3

2 1 2 3

m m mn n n

∆ =l l l

and D = [Pij]3times3

Method 1 Method of Multiplication (Row by column) D = |AB|

1 1 1 1 1 1 1 2 1 2 1 2 1 3 1 3 1 3

1 2 2 1 2 1 2 1 2 2 2 2 2 2 2 3 2 3 2 3

3 1 3 1 3 1 3 2 3 2 3 2 3 3 3 3 3 3

a b m c n a b m c n a b m c n a b m c n a b m c n a b m c n

a b m c n a b m c n a b m c n

+ + + + + +∆ = ∆ ∆ = + + + + + +

+ + + + + +

l l ll l ll l l

pij = scalar product of ith row vector and jth column vectors of D1 and D2 respectively

RemarkSince |AB| = |A||B| = |B||A| = |BA| = |BTA| = |BAT| = |BTAT| thus |AB| can also be obtained by row-column row-row column-row or column-column multiplication of B and A Thus there are eight ways of obtaining (D1 D2)

31106 Adjoint or Adjugate of Determinant If D = |aij|ntimesn is a determinant of order n times n then Dprime = |Cij|3times3 where Cij is co-factor of element aij is called Adjoint or Adjugate of determinant

311061 Jacobiprimes theorm

Its states that Dprime = Dn-1 D ne 0 where Dprime = adjoint of D = determinant |Cij| Cij = co-factor of aij

311062 Reciprocal determinant

If D = |aij| ne 0 then ijC∆ =

∆ where Cij is the cofactor of aij is called the reciprocal determinant of D

ijn n n

C 1 1 |C |minus∆ ∆

∆ = = = = =∆ ∆ ∆ ∆ ∆

311063 Method to break a determinant as the product of two determinants

(a) Observe the diagonal symmetry of the elements and apply the following facts q The determinant of skew symmetric determinant with odd order always vanishes Therefore any

odd order skew symmetric determinant can be broken into product of two matrices of which atleast one is singular

q The determinant of skew symmetric determinant with even order is a perfect square Therefore an even ordered skew symmetric determinant can be written as a square of a determinant having symmetrical elements

(b) Observe the symmetry of the elements and make sure whether (i j)th element of the given determinant can be written as Ri Cj where Ri is the i th row of the first factor (determinant) and Cj is the jth column of the second factor (determinant)

(c) While applying the approach (b) it is advised to choose the (i j)th element to be diagonal elements

3111 differentiation of deterMinantS

The differentiation of a determinant can be obtained as the sum of as many determinants as the order The process can be carried out along the rowcolumn by differentiating one rowcolumn at a time and retaining the others as they are

If 1 2

f (x) f (x)g (x) g (x)

∆ = of order 2 which is a function of x then

1 21 2 1 2

f (x) f (x)d d d (f (x)g (x) g (x)f (x))g (x) g (x)dx dx dx

∆= = minus

= 1 2 2 1 1 2 2 1(f (x)g (x) g (x)f (x) g (x)f (x) f (x)g (x)prime prime prime primeminus minus minus = 1 2 1 2

1 2 1 2

f (x) f (x) f (x) f (x)g (x) g (x) g (x) g (x)prime prime

+prime prime

NoteIn order to find out the coefficient of xr in any polynomial f(x) differentiate the given polynomial f(x) r times successively and then substitute x = 0

ie the coefficient of

rr f (0 )

d f ( x )f (0 )

dx at x = 0

31111 Integration of a DeterminantIntegration of a determinant As determinant is a numerical value so it can always be integrated by expanding but the integration of the determinant can be done without expansion if it has only one variable rowcolumn

Given a determinant D (x) = (where a b c l m and n are constants) as a function of x

a a ab

f (x)dx g(x)dx h(x)dx

(x)dx a b c

int int intint

Chapter 32StatiStiCS

321 Measures of Central tendenCy

For a given date a single value of the variable which describes its characteristics is identified This single value is known as the average An average value generally lies in the central part of the distribution and therefore such values are called the measures of central tendency The commonly used measures of central tendency are (a) Arithmetic Mean (b) Geometric Mean (c) Harmonic Mean (d) Median (e) Mode

322 types of distribution

(i) IndividualDiscrete Distribution (Ungrouped Data) Here we are given x1 x2 x3 helliphellip xn different values

(ii) Discrete Series with Frequency Distribution (Ungrouped Data with Frequency Distribution) Here we are given

xi x1 x2 x3 hellip xn

fi f1 f2 f3 hellip fn

where fi is frequency of xi (iii) Continuous series with frequency distribution (grouped data)

Here we are given class intervals with corresponding frequencies

Class interval 0 ndash10 10 ndash 20 20 ndash 30 helliphellipFrequency f1 f2 f3 helliphellip

Range Range = Largest observation ndash smallest observation

Class sizelength of clan-interval (a ndash b) is defined as (b ndash a) eg class size of (40 ndash50) is (50 ndash 40) = 10

Classndashmark of class interval Midndashpoint of class interval eg class mark of class interval (40 ndash 50) is minus

+ =(50 40)40 45

2 In general classndashmark of class interval (a ndashb) is minus +

+ =(b a) a ba

Statistics 32413

3221 Arithmetic Mean (i) For discrete series

(a) Direct method = + + + += =sum

i1 2 3 ni 1

xx x x x

(b) Short-cut method =

= + = + = minus

ux a u a u (x a)

Here a is a suitable number which makes the greater values of xirsquos to smaller values For example if given data is 202 219 238 258 279 299 It will be convenient to take a = 250 This method helps to find means by reducing calculations when given values of xi are larger

(c) Step deviation method =

minus= + = + =sum

u (x a)x a hu a h u n h

where a and h are suitable

real numbers eg in data 210 220 230 260 280 290 take a = 250 and h = 10 (ii) For discrete series with frequency distribution

(a) Direct method ==sumsum

i ii 1

(b) Short-cut method = + = + = minussumsum

i ii i

f xx a u a u (x a)

f a = suitability chosen real number

(c) Step deviation method = + = + sumsum

f ux a hu a h

f minus

x au h

where a and h are suitably

chosen real number (iii) For continuous series (grouped data)

(a) Direct method = sumsum

f where xirsquos are class-makes of intervals

i i1 1

f ux a u a u (x a)

f a is suitably chosen real numbers

f ux a hu a h

f minus

x au h

a and h are suitably choosen

real numbers Generally h = width of classndashintervals Here minus= i

h defines mean

deviation of variate xi form assumed mean lsquoarsquo

3222 Weighted Arithmetic Mean

If w1 w2 w3 wn are the weights assigned to the values x1 x2 x3xn respectively then the weighted

average is defined as Weighted+ + + +

=+ + + +

1 1 2 2 3 3 n n

1 2 3 n

w x w x w x w xAM

w w w w

323 CoMbined Mean

If we are given the AM of two data sets and their sizes then the combined AM of two data sets can be

obtained by the formula +=

+1 1 2 2

n x n xxn n

where 12x = combined mean of the two date sets 1 and 2

1x = Mean of the first data 2x = mean of the second data

1n = Size of the first data 2n = Size of the second data

3231 Properties of Arithmetic Mean (i) In a statistical data the sum of the deviations of individual values from AM is always zero

That is =

minus =sumn

i ii 1

f (x x) 0 where fi is the frequency of xi (1 le i le n)

(ii) In a statistical data the sum of square of the deviations of individual values from real number lsquoarsquo is the least when a is mean (x) That is minus ge minussum sum2 2

i i i if (x a) f (x x)

(iii) If each observation xi is increased (decreased) by lsquodrsquo then AM also increases (decreases) by lsquodrsquo

Q =sumsum

plusmn= = plusmn = plusmnsum sum sum

sum sum sumi i i i i

f (x d) f x fA d A d

(iv) If each observation xi is multiplied (or divided) by d (ne 0 for division) then the new AM

1d ord

times of original AM

sumsum

then = = =sum sumsum sum

i i i i

f (x d) d f xA dA

324 GeoMetriC Mean

(a) For ungrouped data GM of x1 x2 x3helliphellip xn x ne 0 is given by GM = (x1 x2 x3helliphellip xn)1n

(i) If (x1 x2 x3helliphellip xn) lt 0 and n is even then GM is not defined (ii) If (x1 x2 x3helliphellip xn) lt 0 and n is odd then GM is defined given by GM = ndash(|x1||x2|

|x3|helliphellip|xn|)1n

rArr GM = ndashAntilog + + +

1 2 nlog x log x log xn

(iii) If each xi ge 0 then GM = Antilog + + +

(iv) If each xi is non-zero and x1 x2 x3hellipxn gt 0 then GM = Antilog + + +

(b) For ungrouped data with frequency distribution or grouped data (continuous series) It is given by =sum1 2 nf f f 1N

1 2 n iGM((x ) (x ) (x ) ) N f when defined In case of continuous series

xi = class-mark (mid-value of interval)

Statistics 32415

rArr GM = Antilog

sum i if log x

N for ( ) gt1 2 n

f f f1 2 nx (x ) (x ) 0

and GM = ndashAntilog

sum i if log x

N for ( ) lt1 2 n

f f f1 2 nx (x ) (x ) 0 N = odd

325 HarMoniC Mean

The harmonic mean of n observation x1 x2hellip xn is defined as HM + + +1 2 n

n1 x 1 x 1 x

If x1 x2hellip xn are n observations which occur with frequencies f1 f2 fn respectively their HM

is given by =

i ii 1

(f x )

326 order of aM GM and HM

The arithmetic mean (AM) geometric mean (GM) and harmonic mean (HM) for a given set of obser-vations are related as under AM ge GM ge HM

Equality sign holds only when all the observations are equalRelation between GM HM of two numbers a and b GM of two numbers a and b is also the GM of AM and HM of a and b

+ = + 2 a b 2ab( ab)

2 a b ie (GM)2 = (AM) (HM)

rArr = timesGM AM HM

327 Median

Median is the middle most or the central value of the variate in a set of observations when the observations are arranged either in ascending or in descending order of their magnitudes It divides the arranged series in two equal parts (a) For individualdiscrete series

Step I Arrange the variables in ascending or descending order

Step II Median =

+ = + +

n 1 term for n odd2

n nterm 1 term2 2 for n even

(b) For discrete series with frequency distribution Step I Arrange the variables xirsquos in ascending or descending order keeping frequencies along

with them Step II Prepare a cumulative frequency table and find Sfi = N

Step III Median =

ththth

N 1 observation if N odd

N N2 2 term if N even

For thN

2terms see the value of xi corresponding to

cumulative frequency similar for

thN 12

(c) For continuous series (Grouped data) Step I Prepare the cumulative frequency table

Step II Find median class ie class corresponding tothN

2observation

Step III Median = + minus times

N hC2 f

where l = lower limit of median class

=sum iN f h = width of class-intervals

f = frequency of median class C = cumulative frequency of class preceding the median class

Remarks

1 Median is also known as 2nd quartile (Q2) ie median = N h2 C

4 f + minus times

2 1st quartile = N h1 C

4 f + minus times

3 3rd quartile = N h3 C

4 f + minus times

4 Similarly we have deciles D1 D2 D3D9 where i

N hD i C

10 f = + minus times

rArr D5 = 5th decile = median

5 In the same way we have percentile P1 P2 P3P99 where i

N hP i C

rArr P50 = 50th percentile = median Thus median Q2 = D5 = P50

328 Mode

Mode is that value in a series which occurs most frequently In a frequency distribution mode is that vari-ate which has the maximum frequency

Statistics 32417

3281 Computation of Mode

(a) Mode of Individual Series In the case of individual series the value which is repeated maximum number of times is the mode of the series

(b) Mode of Discrete Series In the case of discrete frequency distribution mode is the value of the variate corresponding to the maximum frequency

Case (i) If a group has two or more scores with the same frequency and that frequency is the maximum positive distribution is bimodal or multimodal that is to say it has several modes eg 1 1 1 4 4 5 5 5 7 8 9 9 9 has modes 1 5 and 9

Case (ii) When the scores of a group all have the same frequency there is no mode eg 2 2 3 3 6 6 9 9 has no mode

Case (iii) If two adjacent values are the maximum frequency the average of two adjacent scores is

the mode 0 1 3 3 5 5 7 8 mode = +=

3 5 42

(c) Mode of Continuous Series Case 1 When classes have the same width Step 1 Find the modal class ie the class which has maximum frequency The modal class can be

determined either by inspection or with the help of grouping table Step 1 The mode is given by the formula

Mode = minus

minus +

minus+ times

minus minusm m 1

m m 1 m 1

f fl h2f f f

where l = the lower limit of the modal class

h = the width of the modal class minusm 1f = the frequency of the class preceding modal class

mf = the frequency of the modal class

+m 1f = the frequency of the class succeeding modal class In case the modal value lies in a class other than the one containing maximum frequency we take

the help of the following formula Mode = +

minus +

+ times+m 1

m 1 m 1

fl hf f

where symbols have usual meaning

Case (ii) When classes have different width Let ai be the width of ith interval

Step I First find the heights = ii

The nodal class is the one with the greatest height and mode = ( ) ( )

minus +

minus+

minus + minus

m m 1i

m m 1 m m 1

h h ah h h h

329 Measures of dispersion

The degree to which numerical values in the set of values tend to spread about an average value is called the dispersion of variation The commonly used measures of dispersion are (a) Range (b) Quartile Deviation or Semi-inter-quartile Range (c) Mean Deviation (d) Standard Deviation

Range It is the difference between the greatest and the smallest observations of the distributionIf L is the largest and s is the smallest observation in a distribution then its Range = L ndash S Also

Coefficient of range = minus+

L SL S

Quartile Deviation Quartile Deviation or semi-inter-quartile range is given by = minus3 11QD (Q Q )2

coefficient of minus=

(Q Q )QD

(Q Q )

Mean Deviation For a frequency distribution the mean deviation from an average (median or arithmetic mean) is given by

(i) For individual series

MP from mean = =

minussumn

x mean

n MD from median = =

minussumn

x median

n (ii) For discrete series with frequency distribution and continuous series

MD from mean = =

minussum

i ii 1

f x median

f MD from median =

minussumsum

f x medianf

(iii) For continuous series xi = classndashmark

Coefficient of MD from mean = MD(Mean)mean

Coefficient of MD from median = MD(medain)mean

3210 standard deviation

The standard deviation of a statistical data is defined as the positive square root of the squared deviations of observations from the AM of the series under consideration (a) For ungrouped dataindividualdiscrete series

(i) Direct Method =

minus + minusσ = =

sum sumn

22 2ii ii 1

(x x) (x x 2x x )n n

= ( )minus

+ minus = + minussum sumsum2 22

2i i 2i

x xnx 2x x x 2(x)n n n n

minus = minus

sum sum sum22 2

2i i ix x xx

Thus minus

σ = = minus

sum sum sum22 2

i i i(x x) (x ) xn n n

(ii) Short-cut Method If observations are larger select a = any suitable number and take

ui = (xi ndash a) then

+ minus + σ =

sumsum ii

u(a u ) a

rArr minus

σ = = minus = =

sum sum sum22

i i ii i

(u u) u u u (x a)

Statistics 32419

(iii) Step Deviation Method Take minus= i

h a and h are suitably choosen real numbers then

minus σ = = minus

sum sum sumn

2 22ii ii 1

(u u) u uh h

n n n (b) For discrete series with frequency distribution or continuous series

(i) Direct Method minus

σ = = minus

sum sum sumsum sum sum

22 2i i i i i i

f (x x) f x f xf f f

(ii) Short-cut Method Take ui = (xi ndasha)

σ = minus

sum sumsum sum

22i i i i

f u f uf f

σ = minus

sum sumsum sum

22i i i i

f u f uh

In case of continuous series xi = class-mark of ith class-interval

Remark

SD of first n-natural numbers = 2n 112minus

3211 varianCe

That is variance of a statistical data is square of standard deviation ie variance = (SD)2 = (s)2 or

σ = variance Coefficient of variance (CV) σtimes = times

SD 100 100Mean x

NoteCV is expressed as per centage

3212 CoMbined standard deviation

Let A1 and A2 be two series having n1 and n2 observations respectively Let their AM be 1x and 2x and standard deviations be s1 and s2 Then the combined standard deviation s or s12 of A1 and A2 is given by

s or σ + σ + + σ σ + + σ +σ = =

2 2 2 2 2 2 2 21 1 2 2 1 1 2 2 1 1 1 2 2 2

121 2 1 2

n n n d n n ( d ) n ( d )n n n n

where = minus = minus1 1 12 2 2 12d x x d x x and +=

+1 1 2 2

n x n xxn n

is the combined mean

Remarks (i) Coefficient of variation and consistency are reciprocal of each other Higher is the CV lower will be

the consistency (stability) again lower is the CV higher will be the stability

(ii) If we are given scores of two players and the number of matches in which the given scores were attained and we are asked to find better run getter the player with best average (mean) Also we are asked to find most stable player or most consistent player the player with lower CV (Coefficient of variation)

Contents

Preface

Acknowledgements

Chapter 1 Foundation of Mathematics

Mathematical Reasoning

11 Introduction

12 Pre-Requisites


13 Understanding the Language of Mathematics


14 Statements and Mathematical Statemens

141 Statement

142 Mathematical Statements

143 Scientific Statement

15 Classification of Mathematical Statements

151 Conjectures


16 Working on Mathematical Statements



17 Implication of a Statement



18 Truth Value

19 Quantifiers

191 Proofs in Mathematics

192 What is a Mathematical Assumption

Number System

110 Set of Natural Numbers


111 Set of Integers

112 Geometrical Representation of Integers

1121 Properties of Integers

113 Division Algorithm

1131 Even and Odd Integers

1132 Prime Integer

114 Factorial Notation

1141 Related Theorems

1142 Divisors and Their Property

1143 Number of Divisors





115 Tests of Divisibility

116 Rrational (ℚ) and Irrational Numbers (ℚprime)


1162 nth Root of a Number

1163 Principal nth Root


1165 Algebraic Structure of ℚ and ℚ

117 Surds and Their Conjugates

118 Real Numbers System

1181 Concept of Interval

1182 Intersection and Union of Two or More Intervals

119 Mathematical Induction

1191 Ratio and Proportion


1193 Linear Equalities

1194 Method of Comparison

1195 Method of Substitution

1196 Method of Elimination

Fundamentals of Inequality

120 Introduction

1201 Classification of Inequality

121 Polynomials


1212 Degree of Polynomials


122 Partial Fractions

123 Theorems Related to Triangles


1232 Tangency

1233 Rectangles in Connection with Circles

1234 Proportional Division of Straight Lines

1235 Equiangular Triangles

1236 Some Important Formulae

Chapter 2 Exponential13Logarithm

21 Exponential Function


212 Laws of Indices


214 Composite Exponential Functions


22 Solving Exponential Inequality

23 Logarithmic Function


24 Logarithmic Equations


25 Logarithmic Inequalities



Chapter 3 Sequence and Progression

31 Definition

311 Types of Sequence

312 Progression and Series

32 Series

321 Properties of Arithmetic Progression

33 Arithmetic Mean



34 Geometric Progression


35 Geometric Mean



36 Harmonic Progression


37 Harmonic Mean

38 Inequality of Means

39 Arithmetic-Geometric Progression

391 Standard Form


310 (Σ) Sigma Notation

3101 Concept of Continued Sum [Sigma (Σ) Notation]

311 Properties

312 Double Sigma Notation

3121 Representation

313 Methods of Difference

314 Vn Method

Chapter 4 Inequality

41 Inequality Containing Modulus Function

42 Irrational Inequalities





43 Theorem of Weighted Mean

431 Theorem




44 Weierstrass Inequality


45 Use of Calculus In Proving Inequalities

451 Monotonicity


Chapter 5 Theory of Equation

51 Polynomial Expression


52 Classification of Polynomials



53 Equation Standard Equation and Quadratic


54 Nature of Roots

541 Formation of Quadratic Equation


55 Condition for Common Roots

56 Symmetric Function of the Roots


57 Location of Roots

58 Descartes Rule



59 Equation of Higher Degree

Chapter 6 Permutation and Combination

61 introduction

62 Fundamental Principles of Counting

621 Addition Rule

622 Multiplication Rule




63 Combinations and PermutationS

64 Permutation of Different Objects

65 Permutation of Identical Objects (Taking all of them at a Time)

66 Rank of Words

67 Circular Permutation

671 Circular Permutation of n Objects

68 Number of Numbers and their Sum

681 Divisor of Composite Number

682 Sum of Divisor

683 NumberSum of Divisors Divisible by a Given Number

684 Factorizing a Number into Two Integer Factors

69 Combination




694 Combination When Some Objects are Identical(Taking any Number of Them at a Time)

695 Combination When Some Objects are Identical(Taking specific number of them at a time)

610 Distribution




6104 When Name of Groups Specified

611 Multinomial Theorem

6111 Number of Distinct Terms

612 Dearrangements and Distribution in Parcels

613 Distribution in Parcels

6131 Distribution in Parcels When Empty Parcels are Allowed

6132 When at Least One Parcel is Empty

614 Exponent of a Prime in N


Chapter 7 Binomial Theorem

71 Introduction

72 Binomial


73 General Term



74 Middle Term

75 Number of Terms in Expansions

76 Greatest Term

77 Greatest Coefficient

78 Properties of Binomial Coefficient


79 Properties of Coefficients


711 Tips and Tricks

Chapter 8 Infinite Series

81 Binomial theorem for any index (N)

82 Greatest Term

83 Taylor Expansion



833 Properties of e

834 Expansion of ex


84 Logarithmic Series


Chapter 9 Trigonometric Ratios and Identities

91 Introduction

92 Angle


922 Measurement of Angle

93 Polygon and its Properties

94 Trigonometric Ratios




95 Graphs of Different Trigonometric Ratios

951 y = sin x

952 y = cos x

953 y = cot x

954 y = cosec x

955 y = sec x


957 Trigonometric Ratios of Compound Angles




96 Some Other Useful Results

97 Some Other Important Values

98 Maximum and Minimum Values of a Cos θ + B Sin θ

99 Tips and Trics

Chapter 10 Trigonometric Equation

101 Introduction

102 Solution of Trigonometric Equation

103 Particular Solution

104 Principal Solution

105 General Solution

106 Summary of the above Results

107 Type of Trigonometric Equations

108 Homogeneous Equation in Sinx and Cosx

109 Solving Simultaneous Equations


1010 Transcedental Equations

1011 Graphical Solutions of Equations

1012 Solving Inequalities


Chapter 11 Properties of Triangles

111 Introduction

112 Napierrsquos Analogy


113 Geometric Discussion

114 Area of Triangle ABC

115 MndashN Theorem


116 Orthocentre and Pedal Triangle


117 In-Centre of Pedal Triangle

118 Circumcircle of Pedal Triangle (Nine-Point Circle)


119 The Ex-Central Triangle

1110 Centroid and Medians of Any Triangle

1111 Length of Medians

1112 Result Related To Cyclic Quadrilatral

Chapter 12 Inverse Trigonometric Function

121 Inverse Function


122 Domain and Range of Inverse Functions

123 Graphs of Inverse Circular Functions and their Domain and Range

124 Compositions of Trigonometric Functions and their Inverse Functions

1241 Trigonometric Functions of their Corresponding Circular Functions

125 Inverse Circular Functions of their Corresponding Trigonometric Functions on Principal Domain

126 Inverse Circular Functions of their Corresponding Trigonometric Functions on Domain

127 Inverse Trigonometric Functions of Negative Inputs

128 Inverse Trigonometric Functions of Reciprocal Inputs

129 Inter Conversion of Inverse Trigonometric Functions

1210 Three Important Identities of Inverse Trigonometric Functions

1211 Multiples of Inverse Trigonometric Functions

1212 Sum and Difference of Inverse Trigonometric Functions

Chapter 13 Point and Cartesian System

131 Introduction

132 Frame of Refrence

1321 Rectangular Co-ordinate System

1322 Polar Co-ordinate System

133 Distance Formula


134 Section Formula Internal Division

135 Slope of Line Segment




136 Locus of Point and Equation of Locus

1361 Union of Loci

1362 Intersection of Loci


137 Choice of Origin and Selection of Coordinate Axes

138 Geometrical Transformations




139 Geometrical Tips and Tricks




Chapter 14 Straight Line and Pair of Straight Line

141 Definition







1417 Distance of a Point From a Line

1418 Distance Between Two Parallel Straight Lines

1419 Intersection of Two Lines

14110 Equation of the Bisectors of the Angles Between Lines


142 General Equation of Second Degreeand Pair of Straight Lines


1422 Angle Between the Pair of Straight Lines

Chapter 15 Circle and Family of Circle

151 Introduction

152 Definiton of Circle


1522 General Equation

1523 Diametric Form

1524 Equation of Circle Thorugh Three Points

1525 The Carametric Coordinates of any Point on the Circle

1526 Position of a Point with Respect to a Circle

1527 Position of a Line with Respect to a Circle

153 Equation of Tangent and Normal

1531 Tangents

1532 Parametric Form


1534 Normals

154 Chord of Contact

1541 Relative Position of Two Circles


155 Intercept Made on Coordinate Axes by the Circle

156 Family of Circles

157 Radical Axes and Radical Centre

Chapter 16 Parabola

161 Introduction to Conic Sections


162 Parabola

1621 Standard Equation

1622 Position of Point wrt Parabola

1623 Position of Line wrt Parabola

163 Chords of Parabola and Its Properties



164 Tangent of Parabola and Its ProPerties


165 Normals and their Properties

1651 Properties

1652 Normals in Terms of Slope

Chapter 17 Ellipse

171 Definition

172 Standard Equation of Ellipse

1721 Focal Distance

173 Tracing of Ellipse

174 Properties Related to Ellipse and Auxiliary Circle

1741 Position of a Point with Respect to Ellipse S X2a2 + y2b2 minus1 =0

1742 Position of a Line with Respect to Ellipse


Chapter 18 Hyperbola

181 Definition



1813 Auxiliary Circle of Hyperbola

182 Director Circle


1822 Position of a Line with Respect to Hyperbola S x2a2minusy2b2-1=0



183 Rectangular Hyperbola





Chapter 19 Complex Number

191 Introduction




192 Complex Number

193 Argand Plane



1933 Result

194 Algebraic Structure of Set of Complex Numbers

1941 Conjugate of a Complex Number

1942 Properties of Conjugate of a Complex Number

1943 Modulus of a Complex Number

195 De Moiverrsquos Theorem



196 Geometry of Complex Number


1962 Application of the Rotation Theorem


197 Theorem

198 Complex Slope of the Line

1981 Circle in Argand Plane

199 Appoloneous Circle

1910 Equation of Circular Arc


19102 Explanation


19104 Equation of Ellipse

1911 Equation of Hyperbola

1912 Some Impotant Facts

19121 Dot and Cross Product

19122 Inverse Points wrt a Circle

19123 Ptolemys Theoremrsquos

Chapter 20 Sets and Relations

201 Sets

202 Representation of Sets

203 Notation of Sets

204 Notation for Some Special Sets

205 Notation For Some Special Sets

206 Method Representation of Sets

207 Cardinal Number of a Sets

208 Types of Sets

209 Subsets

2010 Number of Subsets

2011 Types of Subsets

2012 Power Sets

2013 Disjoint Sets

2014 Universal Sets

2015 Complement Set of a Given Set

2016 Complementry Set of a Given Sets

2017 Comparable Sets

2018 Venn (Euler) Diagrams

2019 Operations on Sets

20191 Union of Two Sets

20192 Intersection of Two Sets

20193 Difference of Two Sets

20194 Symmetric Difference of Two Sets


2020 LAWS Followed by Set Operations cup cap and Δ

2021 De-Morganrsquos Principle

2022 Inclusive-Exclusive Principle

2023 Some Results on Cardinal Numbers

20231 Cartesian Product of Two Sets



2024 Relations

2025 Domain Co-Domain and Range of Relation

2026 Universal Relation from Set A to Set B

2027 Number of Relations from Set A to Set B

2028 Relation on a Set

2029 Representation of Relation in Different Forms

2030 Classification of Relations

2031 Into Relation


2032 Types of Relations


20322 Identity Relation


20324 Anti-symmetric Relation

20325 Equivalence Relation

2033 Composition of Relations

2034 Inverse of a Relation

Chapter 21 Functions

211 Definition of Function

212 Representation of a Function

213 Some Standard Function

214 Equal or Identical Functions

215 Properties of Greatest Integer Function(Bracket Function)




216 Classification of Functions


217 Many-One Functions

2171 Onto (Surjective) Function

218 Method of Testing for Injectivity

219 Into (Non-Surjective) Function

2110 One-One Onto Function (Bijective Function)

2111 Testing of a Function for Surjective

2112 Number of Relations and Functions

2113 Composition of Non-Uniformly Defined Functions

2114 Properties of Composition of Function

21141 Definition of Inverse of a Function

2115 Condition for Invisibility of a Function

21151 Method to Find Inverse of a Given Function

2116 Properties of Inverse of a Function

2117 Even Function

21171 Properties of Even Functions

21172 Odd Function

21173 Properties of Odd Functions

2118 Algebra of Even-Odd Functions

2119 Even Extension of Function

2120 Odd Extension of Function

21201 Definition of Periodic Function

2121 Facts and Properties Regarding Periodicity

2122 Period of Composite Functions

2123 Periodicity of ModulusPower of a Function

2124 Exception to LCM Rule

2125 Periodicity of Functions Expressed by Functional Equations

2126 Tips for Finding Domain and Range of a Function

Chapter 22 Limits Continuity and Differentiability

Limit

221 Limit of a Function

222 Limit of Function F(X) At X = A

223 Existence of Limit of a Function

224 Non-Existence of Limit of a Function

225 Algebra of Limits

226 Indeterminate Forms

227 Some Standard Limits

228 Limits of Some Standard Composite Functions

229 Some Useful Transformations

2210 Some Important Expansions

2211 Some Standard Approaches to Find Limit of a Function


Continuity

2212 Continuity of F(X) At X = A

22121 Reasons of Discontinuity of f(x) at x = a

2213 Discontinuity of First Kind

22131 Discontinuity of Second Kind

2214 Pole Discontinuity

2215 Single Point Continuity

2216 One Sided Continuity


2217 Algebra of Continuity

2218 Continuity of a Function on a Set



2219 Continuity of a Function on a Closed Interval

2220 Properties of Continuous Function

Differentiability

2221 Differentiability at a Point


22212 Geometrical Significance

2222 Concept of Tangent and Its Association with Derivability

22221 Theorem Relating to Continuity and Differentiability

22222 Reasons of Non-Differentiability of a Function at x = a

2223 Algebra of Differentiability

22231 Domain of Differentiability

2224 Domain of Differentiability of Some Standard Functions


2225 Miscellaneous Results on Differentiability


2226 Differentiability of Parametric Functions

2227 Repeatedly Differentiable Functions

2228 Functional Equation


Chapter 23 Method of Differentiation

231 Method of Differentiation



232 Algebra of Differentiation

233 Chain Rule

234 Derivatives of Some Standard Functions





235 Differentiation of a Function with Respect13to Another Function

236 Logarithmic and Exponential Differentiation

237 Differentiation of Inverse Function


238 Implicit Differentiation


2382 Shortcut Method to Find dydx for Implicit Functions

239 Parametric Differentiation

2310 Determinant Forms of Differentiation

2311 Leibnitzrsquos Theorem for the Nth Derivative of the Product of Two Functions of X

2312 Successive Differentiation

2313 Some Standard Substitution

Chapter 24 Application of Derivatives

Rate of Change

241 Instantaneous Rate of Change of Quantities

242 Application of Rate of Change of Quantities

243 Errors and Approximations

2431 Types of Errors

244 Calculation of δY Corresponding to δX

Tangent and Normal

245 Tangents from an External Point

246 TangentsNormals to Second Degree

247 Tangent at Origin

248 Angles of Intersection of two Curves


249 Orthogonal Curves

2410 Common Tangent

2411 Shortest Distance Between Two Non-Intersecting Curves

Monotonicity

2412 Monotonicity of a Function on an Interval

2413 Condition for Monotonicity of Differentiable Functions on an Interval

2414 Monotonicity of Function on its Domain

2415 Domain of Monotonicity of a Function

2416 Critical Point

2417 Intervals of Monotonicity for Discontinuous Function

2418 Properties of Monotonic Function

2419 Application of Monotonicity

Curvature of a Function

2420 Hyper Critical Point

2421 Points of Inflexion

2422 Method to Find the Points of Inflexion of the Curve Y = F(X)

2423 Type of Monotonic Function

Rolles and Mean Value Theorem

2424 Rollersquos Theorem

2425 Application of Rolle rsquos Theorem

2426 Lagrangersquos Mean Value Theorem

2427 Alternative form of LMVT

Maxima and Minima

2428 Local Maxima

24281 Local Minima

2429 First Derivative Test (For Continous Functions)

2430 Point of Inflection and Saddle Point

2431 Global or Absolute Maxima and Minima

2432 Algebra of Global Extrema

2433 EvenOdd Functions

2434 Miscellaneous Method

2435 SecondHigher Order Derivative Test

2436 First Derivative Test for Parametric Functions

2437 Second Derivative Test for Parametric Function

2438 Darboux Theorem

2439 Fork Extremum Theorem

2440 Extrema of Discontinuous Functions

2441 Maximum and Minimum for Discrete Valued Functions

2442 Surface Area and Volume of Solids and Area Perimeters of Plane Figures

2443 General Concept (Shortest Distance of a Pointfrom a Curve)

Chapter 25 Indefinite Integration

251 Introduction




252 Integral of Some Standard Functions

253 The Method of Substitution






254 Intregration of sinm x cosn x dx mn Є Z

2541 To Slove Integral of the Form

255 Integration by Partial Fraction

256 Integration of Rational and Irrational Expressions

257 To Solve Integral of the Form


258 Integral of Irrational Functions


259 Integrating Inverses of Functions

2510 Integration of a Complex Function of a Real Variable

2511 Multiple Integration by Parts

Chapter 26 Definite Integration and Area Under the Curve

261 Area Function

262 First Fundamental Theorem

263 Second Fundamental Theorem

264 Linearity of Definite Integral

265 Properties of Definite Integral

2651 Generalization

2652 Condition of Substitution

266 Convergent and Divergent Improper Integrals

267 Applications

2671 Evaluation of Limit Under Integral Sign



268 Wallirsquos Formulae



269 Beta Function

2610 Gamma Function



2611 Weighted Mean Value Theorem


2612 Determination of Function by Using Integration

Area Under the Curve

2613 Area Bounded by Single Curve with X-Axis





26135 Variable Area its Optimization and Determination of Parameters


2614 Area Enclosed in Curved Loop


Chapter 27 Differential Equation

271 Introduction

272 Differential Equation



273 Linear Differential Equation

274 Non-Linear Differential Equations


2742 Formation of Differential Equation

275 Solution of Differential Equation

276 Classification of Solution


277 Variable Separable Form



278 Solution of Homogeneous Differential Equation

279 Equations Reducible to the Homogeneous Form

2710 Exact and Non-Exact Differential Equation


2711 Non-Exact Differential Equation

27111 Integrating Factor

27112 Leibnitz Linear Differential Equation



2712 Higher Order Differential Equation

2713 Integral Equations and their Solving Method


2714 Application of Differential Equation

Chapter 28 Vectors


2811 Equality of Two Vectors

282 Classification of Vectors


2822 Direction Cosine and Direction Ratios of Vectors

283 Addition of Vectors

284 Subtraction of Vectors


285 Collinear Vectors

2851 Conditions for Vectors to be Collinear

286 Section Formula

2861 Collinearity of the Points

2862 Linear Combination of Vectors



2865 Product of Two Vectors



287 Vector Equation and Method of Solving

Chapter 29 Three-Dimensional Geometry

291 Introduction


2912 Corollary



2915 Direction Ratios (DRrsquos)


2917 The Angle Between Two Lines



29110 Angle Between two Lines







292 Skew lines


293 Plane





2944 Distance of a Point from a Given Plane



Chapter 30 Probability

301 Experiments

3011 Event

3012 Exhaustive Events

3013 Equally Likely Events



3016 Mutually Exclusive and Exhaustive Events



3019 Bayersquos Theorem

302 Geometrical Probability

Chapter 31 Matrices and Determinants

311 Matrix

312 Sub Matrix

3121 Equal Matrices


3131 Properties of Multiplication of Matrices


3133 Symmetric Matrix

314 Hermitian Matrix

3141 Properties of Hermitian Matrices


3143 Orthogonal Matrix

3144 Idempotent Matrix

3145 Periodic Matrix

3146 Nilpotent Matrix

3147 Involutory Matrix

3148 Unitary Matrix


315 Adjoint of a Square Matrix



316 Matrix Polynomial

3161 Cayley Hamilton Theorem

3162 Elementry Transformation

3163 Elementary Matrix

3164 Equivalent Matrices

317 Determinant Method (Cramers Rule) for Solving Non-Homogenous Equations



318 Solution of Non-Homogeneous Linear Equations by Elementary Row or Column Operations

3181 Solutions of Homogenous System of Equation

319 Eliminant

3191 Linear Transformation

3192 Compound Transformation


3194 Properties of Determinants

3195 Caution

3110 Special Determinant

31101 Symmetric Determinant

31102 Skew-Symmetric Determinant

31103 Cyclic Determinants

31104 Circulants

31105 Product of two Determinant

31106 Adjoint or Adjugate of Determinant

3111 Differentiation of Determinants

31111 Integration of a Determinant

Chapter 32 Statistics

321 Measures of Central Tendency

322 Types of Distribution

3221 Arithmetic Mean


323 Combined Mean

3231 Properties of Arithmetic Mean

324 Geometric Mean

325 Harmonic Mean

326 Order of AM GM and HM

327 Median

328 Mode


329 Measures of Dispersion

3210 Standard Deviation

3211 Variance

3212 Combined Standard Deviation

Mathematics at a Glance

Sanjay MishraB Tech (IIT-Varanasi)

ISBN 978-93-325-2206-0

eISBN 978-93-325-3736-1

Contents

mdashSanjay Mishra

Preface

mdashSanjay Mishra

Acknowledgements

11 INTRODUCTION

12 PRE-REQUISITES

|| Parallel

f Null Set (phi)

141 Statement

Compounding with OR

volume decreases

18 TRUTH VALUE

T FF T

19 QUANTIFIERS

111 SET OF INTEGERS

1132 Prime Integer

11321 Properties

= 1 1 1a a 1 b b 1 c c 1p 1 q 1 r 1y

p 1 q 1 r 1

minus minus minus

( a 1)( b 1)( c 1)2

square and + + + +1

( a 1)( b 1)( c 1) 12

b d HCF (b and d)=

b d LCM (b and d)=

th th th

4 3 2 1 0

x = a 10 + b 10 + c 10 + d 10 + e 10

minus minus minus

le le minus le le

= a(104) + b(103) + c(102) + d(10) + e = 5m + e 0 le e le 9

5m e 0 e 4 5m e 0 e 45m 5 e 5 5 e 9 5m (e 5) 5 e 9

rArr n1 1 1(abc0) (a b c 0)=

rArr n1 1 1(abc1) (a b c 1)=

1 1 1(abc2) (a b c 4)= if n = 4k + 2

1 1 1(abc2) (a b c 6)= if n = 4k

1 1 1(abc3) (a b c 9)= if n = 4k + 2

1 1 1(abc3) (a b c 1)= if n = 4k

1 1 1(abc4) (a b c 6)= if n = 2k

1 1 1(abc7) (a b c 9)= if n = 4k + 2

1 1 1(abc7) (a b c 1)= if n = 4k

1 1 1(abc8) (a b c 4)= if n = 4k + 2

ie n1 1 1(abc8) (a b c 2)= if 4k + 3 n

1 1 1ie (abc8) (a b c 6) if n 4k= =

euler number (e) 1 1 1e 1 27 e 81 2 3

2 1 0 1 2 3

abcdeabcde10

= = = 1 2 n1 2 3 n n

ax x xx ax x x x10

eg 27 336 68 2

x if x 0| x | x 0 if x 0

x if x 0

gt= = =minus lt

Q x + 0 = x and 1

times =

1 2 1 3 2 3

= rArr =

+ + + += = = +

Properties

x if x 0| x | x 0 if x 0

x if x 0

gt= = =minus lt

n = k + 1

= or xu yz

b d+ +

b d+ = + rArr

a b c db d+ +

=minus minus

1nn n n

xa yc zexb yd zf

+ + + +

1 1 2 2

= = rArr 1 2 1 2

1 2 1 2

c c a a xb b b b

minus = minus

rArr 2 1 1 2

1 2 2 1

b c b ca b a b

minusminus

1 2 1 2

a c a cyb a a b

minus=

obtained 1 12 2 2

c a xa x b cb

minus+ =

2 1 1 2

b c b cxa b a b

minus=

1 2 2 1

a c a cya b a b

minus=

2 1 1 2

a c a cya b a b

minus=

2 1 1 2

b c b cxa b a b

minus=

y yz= we get

2 1 1 2 2 1 1 20 0

2 1 1 2 2 1 1 2

minus minus= = = =

minus minus

120 INTRODUCTION

real number)

c clt lt

a b If ab lt 0

rArr = lt

a b iff |a | | b |

gt rArr = = lt lt

2 22 2

ltisin isingt lt

a ba b

gtgt rArr gt

++++++++++++++++

+++++++++++++++++

step 2

minus lt minus gt

minus minus lt gt

minus lt

Q(x)S(x)minus lt

S(x)M(x)minus lt

helliphellip(ii)

Remarks

0Q( x )

(ii) If P( x )0

+ gt lt + + lt gt +

121 POLyNOMIALS

∵ n n 1 n 2 nn 2 n

gt rArr gt lt

f (x) f(x)0 f(x)g(x) 0 0g(x) g(x)

f (x)f(x)g(x) 0 0g(x)

gt le = ne

f (x) 0

lt = ne

cup [g d] cup b

a=sum = a1 + a2 + a3 ++ an

Properties

k kk 1

(a b )=

k kk 1

(a b )=

k kk 1 k 1

a b= =

sum sum and

k kk 1

k kk 1 k 1

ma m a= =

=sum sum = m (a1 + a2 + a3

k 1 2 3 nk 1

a a a a a=

Properties

=prodn

nn 1 2 n 1 2 n k

k 1 k 1

a ( a )( a )( a ) (a a a ) a

ie = = =

k k k k 1 2 n 1 2 nk 1 k 1 k 1

prodprod

kn1 2 3 nk k 1

nk 1 k 1 2 3 n

aa a a aa

b b b b bb

k k k kk 1 k 1 k 1

(a b ) a b= = =

31 2 n

1 2 3 n

1 2 k 1 22 k

2 1 2 n

12311 Chords

a b c + + = + + + + + = + + + + +

13 2 2a b a b

ab2 2+ minus

= minus

n times

212 Laws of Indices

(iv) minus=x

a ab b

le isin

x y x y if a 1a a

x y if a (01)

f(x y) f(x) f(y)x y

f(x y) f(x) f(y)

log alog a 0434

(b) a a a

(d) k |a|a

1log N log N

P 9 = =cb

P 10 logbalogab = 1

given below

System 1 gt

f(x) 0f(x) g(x)

System 1 gt ne =

g(x) 0g(x) 1f(x) g(x)

System 2 gt ne =

f(x) 0f(x) 1f(x) g(x)

System 1 gt gt =

g(x) 0f(x) 0g(x) h(x)

System 2 gt gt =

h(x) 0f(x) 0g(x) h(x)

System 1

gt gt ne =

f(x) 0g(x) 0g(x) 1g(x) h(x)

System 2

gt gt ne =

f(x) 0h(x) 0h(x) 1g(x) h(x)

gt ne gt ne =

h(x) 0h(x) 1g(x) 0g(x) 1f(x) g(x)

system gt

f(x) 0[f(x)] g(x)

system +

f(x) 0[f(x)] g(x)

system gt gt =

gt gt gt gt =

31 Definition

1 1 1 1 1 1 2 2 4 8 16 32

eg 1 1 1 1 2 4 8 16

32 SerieS

n kk 1

Caution lttn angt n

where = n n 1a a2

Sie f (n)S

= prime

then the ratio

t f 2k 1t

= minus

Tie f (n)T

= prime

S k 1fS 2

Points to Remember

33 Arithmetic meAn

= 1 2 3 na a a aAM

Remarks

= +sumn

nA ( a b)

= =minus minus

p Kn Kn

T TT TT

n k p k

1 2 3 n 1

aa a aa a a a minus

TrT minus

na(r 1) a(1 r )S(r 1) (1 r)

minus minus= =

a rS1 rminus

the series

rarrinfin

minus= = minus

minus ltminus

Remark

a2 b2 31 2

aa a b b b

1 1 1 a a a

are in GP

x numbers y numbers

R 0XXXX YYYY=

where 0x times

X XXXX=

and 0y times

Y YYYY=

0 0x y x

(10 10 )+

minus=

4 terms in GP 33

a a ararr r

35 Geometric meAn

1 2 3 nG (x x x x )=

rArr b = arn+1

1n 1br

1 1 1 a a d a 2d

a (n 1)d=

+ minus)

1 1 1(n 1)a b a

= + minus minus

+or a a b

c b cminus

=minus

a d a a d

a 3d a d a d a 3d

37 hArmonic meAn

j 11 2 n j

n nH1 1 1 1a a a a=

= =+ + + sum

1 1 rdH a

minus=

1 2 n 1n1 2 n

+ + +ge ge

nge rArr

rArr n

maxSPn

rArr 1

1 2 n n1 2 n

x x x(x x x )

n+ + +

ge rArr 1nS (P)

Remarks

391 Standard Form

minus minus minus

When | r | lt 1 2

general term

rArr n

k 1 2 3 nk 1

a a a a a=

= + + + +sum rArr n

= + + + + =sum

k kk 1 k 1

a a= =

λ = λsum sum

k k k kk 1 k 1 k 1

(a b ) a b= = =

k kk 1 k 1

a b s= =

k k k kk 1 k 1 k 1

a b a b= = =

knk k 1

nk 1 k

sumsum

Sn = T1 + T2 + T3 + +Tn = n

T=sum = 2

kak bk c

+ +sum = n n n

k 1 k 1 k 1

a k b k c 1= = =

+ +sum sum sum

nn(n 1)(2n 1) n(n 1)S a b cn

6 2+ + + = + +

311 ProPertieS

r 1 2 3 n=

= + + + +sum = n(n 1)2+ P2

n2 2 2 2 2

r 1 2 3 n=

= + +sum = n(n 1)(2n 1)6

n(n 1)r2=

+ = sum P4

nr (n 1)(2n 1)(3n 3n 1)30=

= + + + minussum

iji 1 j 1

3121 Representation

iji 1 j 1

31211 Conclusion

+ = = n 12

n(n 1) C 2

P1 n n n n

j 1 i 1 j 1 i 1

a a= = = =

sumsum sum sum =

i 1 i 1

na na 1= =

P2 n n n n n

1 i j n j 2 j 3 j ni 1 i 2 i n 1

= = = minus

2minus

P3 2n n

1 i j n

(n n)aa C a2

le le le

+= =sumsum

minus ++ =

P4 n n n n

i j i ji 1 j 1 i 1 j 1

a a a a= = = =

sumsum sum sum let

k 1 2 nk 1

a a a a S=

= + + + =sum = 2n n n

2i i k

i 1 i 1 k 1

Sa S a SS S a= = =

= = = =

sum sum sum

= i j 01 i j n

a a S (say)le lt le

=sumsum rArr 2n n

20 i k

i 1 k 1

2S a a= =

sum sum

k kk 1 k 1

=sum sum

k kn nk 1 k 1

i j1 j j n

a aa a

le le le

=sum sum

P7 n n n

i j ki 1 j 1 k 1

(a a ) 2n a= = =

0 i j n

k kk 1 k 1

i j1 i j n

a aa a

le lt le

=sum sum

sumsum

1 2 3 4 5 n 1

1 2 3 4 5 6 n 1 n

d d d d d d

Remarks

314 Vn methoD

n(n 1)=

eg n1 (n 1) n 1 1t

= = = minus+ + +

Step 3 rArr 1 1 21 1t V V1 2

= minus = minus

+rArr n 1 n 1

Chapter 4InequalIty

system lt ge

minus lt lt

systems lt ge

minus lt lt

system lt ge

minus lt lt

lt minus lt

minus gt lt

lt ge minus lt lt

Inequality 449

f(x) 0g(x) 0f(x) g (x)

g(x) 0f(x) g (x)

rArr minus minus

0 1 2 nn n 1 n 2

rArr a0tn + a1t

f(x)tg(x)

rArr = if (x) tg(x)

rArr lt lt lt gt

gt gt gt

rArr α

lt lt lt lt

0 h(x) 10 f(x) (h(x))h(x) 1 f(x) (h(x))

alog x 0 x 1a 1 a 1 (b)

alog x 0 0 x 10 a 1 0 a 1

alog x 0 0 x 1a 1 a 1 (d)

alog x 0 x 10 a 1 0 a 1

Inequality 451

gt lt lt

f(x) 1 0 f(x) 1

g(x) 1 0 g(x) 1

gt lt lt

f(x) 1 0 f(x) 1

g(x) 1 0 g(x) 1

lt lt gt

f(x) 1 0 f(x) 1

0 g(x) 1 g(x) 1

lt lt gt

f(x) 1 0 f(x) 1

0 g(x) 1 g(x) 1

f(x) g(x) f(x) g(x)g(x) 0 f(x) 0

(x) 1 0 (x) 1

f(x) g(x) f(x) g(x)g(x) 0 f(x) 0

(x) 1 0 (x) 1

f(x) g(x) f(x) g(x)f(x) 0 g(x) 0

(x) 1 0 (x) 1

f(x) g(x) f(x) g(x)f(x) 0 g(x) 0

(x) 1 0 (x) 1

ge ge+

a b 2abab2 a b

Remarks

If = sum ixA

= prod

Remark

+ + +31 2 n 1 2 n

1mm m m m m m1 1 2 2 n n

1 2 3 n1 2 n

forall ai gt 0

i i mi mii

431 Theorem

(i) + + gt

mm ma b a b2 2

if 0 lt m lt 1

(iii) + + =

mm ma b a b2 2

sum summm

i ia an n

sum summm

i ia an n

if m isin (0 1)

sum summm

i ia an n

Inequality 453

Then + + + + + +

gt + + + + + +

mm m m1 1 2 2 n n 1 1 2 2 n n

1 2 n 1 2 n

+ + + + +lt

+ + + + + +

mm m1 1 n n 1 1 2 2 n n

1 2 n 1 2 n

(a12 + a2

2 + + an2) (b2

1 + b22 ++ b2

a a ab b b

then + + + + + ge

(1 + an) lt minus n

1 2 3 n

(d) If + + + +1 2 3 n

of minus φ

α β =1 cossin sin

φ2sin2

and that of φ

α =β =2

mum value of ==

α αsumprodn n

k kk 1K 1

α +α1 2sin2

kk 1 max

Asin sinn

kk 1 max

Asin nsinn

451 Monotonicity

type f(x) = a0 + a1x + a2x2 + a3x

n n 1 n 2 nn 2 n

meets axis of x

Rational identity

5221 Conclusion

rArr 2b b 4acx

2a 2aminus minus

= 2 b ca x 2 x2a a

+ + + minus

b 4ac ba x2a 4a

minus + +

rArr 2D by a x

4a 2a + = +

and DY y4a

54 naturE of roots

minus +α = b D

2aminus minus

We proceed as below

(i) S β+α=

minusαβ

x + 1ab = 0

rArr 2a b b c c a

a b b c c atimes =

4a 2aminus minus

maxD by at x

4a 2aminus minus

f(x) = 0 Then b D2a

2aminus +

of the equation

(i) D gt 0 (ii) f(k1) f(k2) lt 0

58 DEsCartEs rulE

ie (x a) (x b) (x c) (x d)y4

4+ + +

= minus

minus + minus=

between a and b

a1 + a2 + a3 + + an = 1

a1a2 a3 an = 1

aaminus

f ( x ) f ( x ) f ( x ) f ( x )f ( x )

x a x b x c x k

Combination

61 introduction

621 Addition Rule

i 1 2 k ii 1i 1

A | A A A | | A |==

= cup cup cup =sum

where |Ai|

1 2 r ii 1

n n n n=

i 1 2 ri 1

A A A A=

i 1 2 r ii 1 i 1

A | A | | A | | A | A= =

n X m=

1 ii 1

S n(X )=

=sum r r

2 i j1 i 1 r

S n(X X )le lt le

n(X1 cap X2 cap X3)

nCr(n r)

=minus

nr C rr(n r)

= timesminus

up in n ways

Npqr n= rArr nNpqr

66 rank oF words

AAABNN A E G L R

13x AAABNN

32= rarr

x2 = 0 rarr AANN

32x AAN

22= rarr

x4 = 0x5 = 1 rarr A

1x 3 AEGER= rarr

2x 0 AEGER= rarr

3x 2 EGR= rarr

x4 = 1

circle by 0360

= minus

(i) minus =n

PC ( r 1)

(ii) minus =n

P1C ( r 1)

10 1 D n9

D n minus

rP if r n0 if r n

n 1k r 1

10 1 D P9

minusminus

k r 1k 1

n 1k r 1

D Pminusminus

r rn 1

k r 1k 1

10 1 D P9

minusminus

682 Sum of Divisor

= a 1 b 1 c 1p 1 q 1 r 1p 1 q 1 r 1

(c ndash g1 + 1)

69 combination

nCr(n r)

=minus

∵ nr

n(n 1)(n 2)(n 3)(n r 1)C Ir

(b) 0 = 1 n n0 n

nC C 1

n 0 = = =

and nC1 = n

r r 1 r 2n n n 1C ( C ) C r r r 1

7 n n n 2r r 1 r 2

r 1 (r 1(r 2)C C Cn 1 (n 1)(n 2)

+ + + = = + + +

8 nCr rCs = nCs

C n r 1C rminus

minus +=

n n n n nr 1 2 n

C C C C 2 1=

= + + + = minussum

from a group is r

Cminus

=sum or

kk r p

Cminus

rC1 + nndashrC2rC2

1 i n k 1 j m k

n A A n B B+ +

610 distribution

(n p)C np

+ += Hence

2m(m) 2

be made is =3m

mm m 3

we obtain 3mm m m

objects = p

(pq)(q) p

objects = p

(pq)(q)

1 2 3 n

r x x x x

so on = 1 2 3 n

1 2 3 n 1 2 3 n

( ) (r)

λ +λ +λ + λ=

is n1 1 1 1 1n 1 ( 1)1 2 3 4 n

1 2 3 n 1 n

n permutation

nn n n ( 1) n1 2 3 n

= minus minus

minus + minus +

n 11 1 1 ( 1)n 1 2 3 n

= minus minus

minus minus + +

n 11 1 1 ( 1)n n 1 2 3 n

n1 1 1 1 ( 1)n 1 1 2 3 4 n

= minus

minus + +

n1 1 1 ( 1)n 2 3 4 n

(n 1)separators

1 1 1 ( 1)C (n r) 2 3 4 (n r)

k r nk

( 1) C (r k)minus

minus minussum

k 1 r nk

( 1) C (r k)minus

nP if r n

0 if r n

k 1 r nk

( 1) C ( r k )minus

minus minussum

k r nk

( 1) C ( r k )minus

minus minussum

rArr s

n n nE (n) p p p

= + + +

ple lt

Conclusion ie np

n n nE (n) p p p

= + + +

71 IntroductIon

72 BInomIal

n n n r rr

(a x) C a xminus

+ =sum and

nn r n n r r

minus = minussum

bull n

n r nr

(1 x) x C=

+ =sum n

n r n rr

(1 x) ( 1) C x=

73 General term

74 mIddle term

n2 yn2

th and n 32+

2 2n 1C x y

minus +

and n 1 n 1

n 2 2n 1

C x y+ minus

n n r rr

C a xminus

nr n n r r

( 1) C a xminus

minussum

n n 2r 2r2r

2 C a xminus

=sum where

= minus =

n 1 if n is odd2

++ = +

n n 2r 1 2r 12r 1

=sum where

= + minus = +

n if n is even2m 1

n 1 if n is odd2

76 Greatest term

r 1 rn r 1

minusminus

minus += =

T 1T+ ge

n r 1 | x| 1

rminus +

T C n r 1 n 1 1T C r r+

minus gt

rArr n 1 r2+

gt (i)

r(n r)=

n(n 1)(n 2)(n 3)(n r 1)C Ir

bull ( )n n 1 n 2r r 1 r 2

n n n 1C C Cr r r 1

(r 1)(r 2)r 1C C Cn 1 (n 1)(n 2)

+ ++ +

+ ++ = = + + +

r r 1C Cr 1 n 1

bull n

n n rr

(1 x) C x=

r + + nCnxn (i)

bull n

n r n rr

(1 x) ( 1) C x=

n (ii)

bull n

C x C C C C 2=

= + + + + =sum

bull n

( 1) C C C C ( 1) C 0=

bull n

2 n 2 2 2 2r 1 2 3 n

r C 1 C 2 C 3 C n C=

= + + + +sum

bull 0r 1 2 nCC C C Cr 1 1 2 3 n 1

= + + + ++ +sum

bull nn n

r 1 k 0r 1

+= =sum sum

n x x xr r r

711 tIps and trIcks

minus +

+ = + + + + + + infin2 3 r

a n(n 1)1 n (ax)x 2

n(n 1)(n 2) (ax) 3

x n(n 1) x1 na 2 a

n(n 1)(n 2) x 3 a

minus + + +

minus minus +

Remarks

82 Greatest term

T 1T+ ge gives

| x |(x 1)r m| x | 1

+(say)

Infinite Series 885

| x | 1+

Remarks

2 xr + helliphellip

83 taylor expansion

1 2 3+ = + + + +

= + + + +

1 2 3= + + + +

1 2 3minus minus

= + + + +

(i) 3 5x xsin x x

3 5= minus + +

3 15 315= + + + +

(iv) 3 5 7

2 3 4x x xn(1 x) x ( 1 x 1)2 3 4

833 Properties of e

834 Expansion of ex

For x isin R 2 3 r

x x x x xe 1 1 2 3 r

= + + + + + + infin or n

(i) 2 3 r

x x x xe 1 1 2 3 r

= + + + + +infin =sum (ii) 2 3 r r

x x x ( 1) xe 1 1 2 3 r

infinminus

(iii) x x 2 4 6 2r

e e x x x x1 2 2 4 6 (2r)

minus infin

+= + + + + =sum (iv)

x x 3 5 2r 1

e e x x x x2 1 3 5 (2r 1)

minus +infin

minus= + + + =

(v) r 0

1 1 1e 1 1 2 r

1 1 1 ( 1)e 1 1 2 3 r

infinminus

(vii) 1

e e 1 1 1 11 2 2 4 6 (2r)

minus infin

e e 1 1 1 12 1 3 5 (2r 1)

minus infin

x x x ( 1) xx 2 3 4 r

minusinfin

(i) 2 3 4

ex x xlog (1 x) x ( 1 x 1)2 3 4

(ii) 2 4 6

ex x xlog (1 x)(1 x) 2 ( 1 x 1)2 4 6

(iii) 3 5

e(1 x) x xlog 2 x ( 1 x 1)(1 x) 3 5

91 INTRODUCTION

92 ANgle

bull D G R

180 200= =

than 180deg

nminus π

Name of Polygon

Number of Sides

Name of Polygon

Number of Sides

Name of Polygon

Number of Sides

Name of Polygon

Number of Sides

Name of Polygon

Number of Sides

Name of Polygon

Number of Sides

θ = = Base xcos

Hypotenuse rθ = =

θ = =

θ = = Base xcot

Remark

cos(ndashq) = cosq

sec(ndashq) = secq

Similarly

cos sin

2π minusθ = θ

tan cot

2π minusθ = θ

cot tan

2π minusθ = θ

cos sin

2π + θ = minus θ

tan cot

2π +θ =minus θ

cot tan

2π +θ =minus θ

cosec sec

2π minusθ = θ

cosec sec

2π + θ = θ

le θ lt Then

(n 1)2

(iii) ( )(n 1)

Angles Functions

p 76π 4

53π 11

951 y = sin x

Properties

third quadrants

952 y = cos x

y-axis

9521 y = tan x

953 y = cot x

Properties

954 y = cosec x

955 y = sec x

Properties

1 sin x cos x+

=minus

sec x tan xminus =

θ =+ θ2

cotcos

θθ =

+ θ 1

tancot

θ =θ

1 tan A tanB+

+ =minus

1 tan A tanBminus

minus =+

minus+ =

+minus =

2 tan Atan2A1 tan A

=minus

where A (2n 1)4π

3 1 cos A Atan

sin A 2minus =

where A ne (2np)

where A ne2np

minus =+

+= 21 cos2A sin A

2minus

minus=

2 2+ + + =

2 2+ minus + =

++ = where AB n

ne π+

+ + = +

(ix) A B C2π

( )n 1 nsin sin2 2

minus β β α +

( )n 1 ncos sin2 2

minus β β α +

sin2 Acos2 A2 sin A

θ θ θ θinfin =

θ2 n 1

5 sin 2212

deg ( )1 2 22

minus 6 cos 2212

deg ( )1 2 22

7 tan 2212

deg 2 1+

= cos 72deg 10 cos 18deg 10 2 54+

= sin 72deg

11 sin 36deg10 2 5

4+ = sin 54deg

13 sin 9deg 3 5 5 54

+ minus minus

or cos 81deg

14 cos 9deg3 5 5 5

4+ + minus

99 TIPs AND TRICs

sin2 Acos2 A2 sin A

2 2 2α+β β+ γ γ +α

equation

101 IntroductIon

θ = rArr 4π

θ = or 3 9 11

4 4 4 4π π π π

Parallely cos 1x2

θ = rArrθ = + π

θ = + isin

cos 0 (2n 1)2π

tan q = 0

θ = rArrθ = + tan 1 (4n 1)4π

ndash +ndash

st sin a = k

+ndash

st tan α = k

st sin2α = k

cos2 α = k

ndash +

tan2 α = k

ndash+ndash

θ = + isin

n isin ℤ

rArr x (4n 3)2π

= + or 2n n3π

rArr x (4n 3) n2π

And 5sin7

minus = rArr

3xsin 02=

2nx n3π

rArr x2sin sin x 0

4 2 4π π + minus =

rArr x m4π

= π+ hellip(ii)

3 2 4π π π

= πminus π+ isin

rArr nx3π

= or x (2n 1) n Z2π

= + isin

minusφ =

such that

rArr ccos( )r

ccos( )a b

θminusφ =+

rArr cos x 1

4π minus =

rArr x 2n n

4ππ= + isin

x1 tan

2cos xx

1 tan2

minus=

2sin xx

1 tan2

= (ii)

rArr 5x 2n n4 2

rArr (8n 2)x n

4minus

(8n 2)x 2m m n ~ 2m m 4p 3p

or (8n 2) (8n 2)x 2m m n ~ n 5p 4p

2 tan x 2sin x

1 tan x 2 2

1 tan x 2cos x

1 tan x 2minus

2 tan x 2tan x

1 tan x 2=

21 tan x 2cot x

2tan x 2minus

-1 le x

10 le 1

rArr -10 le x le 10

of solutions are 7

-3π -2π2π

3π-π

(-frac1234)

frac12 10

(101)(3 3 10)ππ(2 2 10)ππ

( 10)ππ

sinx gt 12

for p6 lt x lt 5p6

6 6π π

π+ lt lt π+

52n 2n6 6isin

π π π+ π+

1 3 1sin15

2 2minus

deg = 2 3 1cos 15

5 ( )1 1sin 22 2 22 2

= minus

6 ( )1 1cos22 22 2 22 2

deg = = +

7 1tan 22 2 12

= minus

8 1cot 22 2 12

deg = = deg 10 10 2 5cos18 sin72

deg = = deg

11 10 2 5sin36 cos54

4minus

deg = = deg 12 5 1cos36 sin544+

deg = = deg

13 3 5 5 5sin9 cos81

4+ minus minus

deg = = deg 14 3 5 5 5cos9 sin81

4+ + minus

deg = = deg

triangles

111 IntroductIon

1111 Sine Formula

1112 Cosine Formula

2 2 2b c acos A

2bc+ minus

= 2 2 2a c bcosB

2ac+ minus

= 2 2 2a b ccosC

2ab+ minus

2 acminus minus

= (s a)(s b)sin

2 abminus minus

(ii) s(s a)Acos

2 bcminus

= s(s b)Bcos

2 acminus

= s(s c)Ccos

2 abminus

2 s(s a)minus minus

=minus

(s a)(s c)Btan

2 s(s b)minus minus

=minus

and (s a)(s b)Ctan

2 s(s c)minus minus

=minus

ca ca∆

(i) minus minus

=(s b)(s c)Asin

2 bc (ii)

s(s a)Acos2 bc

minus=

2 s(s a)minus minus

=minus

2bc+ minus

2ac+ minus

= or bsin AsinBa

= and C from

= = ie a sinBbsin A

= and a sinCcsin A

= rArr bsin AsinB

a= hellip (1)

Also ADsinCb

= rArr AD = b sin C

1 a bsinC2

∆ = = =

∆ = = = (s b)(s c) s(s a)

bcbc bc

minus minus minus

2 sin A

= 2a sinBsinC

2 sin A

Thus 2a sinBsinC

2 sin A∆ = =

115 mndashn thEorEm

11511 Circumcircle

a b c RA B C= = =

1 sin2

11513 Incircle

minus = C(s c)tan2

=minus

2r s b∆

=minus

3r s c∆

=minus

cos(A 2)= 2

cos(C 2)=

tan A tan A

= =degminus

ang ACB + 12

ang ACM = 12

Remarks

= = and 2CG CF3

= + minus2 2 21AD 2b 2c a2

2 2 21BE 2c 2a b2

= + minus

γ= = (Let AD = x)

γ = =+ minus

a sinBsin2b 2c a

β =+ minus

θ = =+ minus

+= = =

acxb c

and abyb c

= = = + + (ii)

π a ar cot

2tan n 2 nπ

= 1 (ab cd)sinB2

2 4A2ab cd

+ + +=

Remarks

π= and cos 1

π minus =

and cosndash1 1

ne minus4π

ndash1

ndashπ2 π2

y=sinx

ndashinfin infinO

π2 (1π2)

ndash1 O 1x

ndash110

(1 π2)

(ndashπ2ndash1)

(ndash1ndashπ2)

(π21)

π2ndashπ2 X

ndash10

y = cos x

Y(ndash1π) π

(1 0)X0

y = cosndash1x

1ndash1

(ndash1π)

(01) 1

ndash1ndash1

(0π2) π2

(π20) (πndash1)(1 0)

Y infin

Xndashπ2

ndashinfin

ndash1

ndashπ2

ndashinfin infinO x

ndashπ2

ndashπ2ndashinfin

ndashinfin

+infin

infin xx

xndashπ2 π

ndashinfin

+infin

πndashinfin

ndashπ2π2

ndashinfin

+infin

0ndash1

π2 xπ

+infin

ndashinfin

O 1ndash1

infinndashinfin

Y+infin

+infin

ndashinfin

0 1ndash1

ndash1

ndashπ2ndash10 π2

+infin

ndashinfin

π2(1π2)

(π21)

(ndashπ2ndash1)

(ndash1 ndashπ2)

π2ndashπ2

ndashπ2

ndash1

+infin

x if x [ 3 2 2]x if x [ 2 2]

x if x [ 2 3 2]2 x if x [3 2 5 2]

Ondash1

ndash1

O45deg

ndash1

ndash11

minusπ 2 y=x

πminus

y=-(x)

πminusy=x-2

minus3π

y=ndash(3π+x)

minus5π2π2

3π25π2πminusπ

y=sinndash1(sinx)

Period 2π

ndash3π

ndash3π2

π23π2

5π22π 3πx

ndashπ2ndashπ πO

y y=sinndash1(sinx)

y=2π+

y=(3π+x

x if x [ 0]x if x [0 ]2 x if x [ 2 ]

2 x if x [2 3 ]

and so on as shown

y=2πndashx

y=xndash

2πy=ndash(x+2π)

y=ndashx

y = cosndash1(cosx)

( )( )( )( )

x if x 3 2 2

x if x 2 2

x if x 2 3 2

x 2 if x 3 2 5 2

isin minusπ π

and so on as shown

π2 3π2π

minusπminus2π2πO

y=tanndash1(tanx)

Period π

x 2 for x ( 2 )x for x ( 0)

x for x [ 0]

x for x [0 ]~232 x for x [ 2 ]~2

y=2πndashx

2π y=ndashx

y=secndash1(secx)

3( x) for x ~ 2 2

x for x ~ 02 2

3x for x ~ 2 2

minusπ 2

πminus

y= ndash (x)

π+y=2x

y=x ndash2ππ 2

y=xndash

y=cotndash1(cotx)

gtminusπ+ lt

cos 1 x if 0 x 1

cos 1 x if 1 x 0

xtan1 x

1 xcot if 0 x 1x

1 xcot if 1 x 0x

minuslt le

1sec if x 0 11 x

1sec if x 1 01 x

= 1 1cosecx

(b) minus

sin 1 x for x 1 0 =

1 xtan for x (0 1]x

minus isin

minusπ+ isin minus

minus isin minus

xminus isin minus

(c) 1 1

xtan x sin for x1 x

minus minus = isin

1cos for x [0 1]1 x

minus isin minus +

1cot for x 0x

minusπ+ lt

= ( )( )

sec 1 x for x 0

+ gtminus + lt

minus + isin

1sin for x 01 x

+ = πminus le +

xcos x1 x

minus forall isin

1tan for x 0x

π+ lt

1 1 xsec x ~ 0x

minus + forall isin

( )( )

cosec 1 x for x 0

+ gtπminus + lt

x 1sin for x 0x

sec xx 1sin for x 0

minus gt =

minus π+ lt

= 1 1cos x ~ 0x

minus forall isin

= ( )1 2

tan x 1 for x 0

1cot for x 0 x 1x 1

xcosec for x 0x 1

π+ lt minus

minus minus= isin =

x 1cos for x 0x

minus gt

minus minusπ+ lt

1tan for x 0 1x 1

= ( )( )

cot x 1 for x 0

xsec for x 0 1x 1

Property (1)

1 1sin (2x 1 x ) if x2 2

1sin (2x 1 x ) if 1 x2

minus minus

minus minus le le

Property (2)

1 1sin (3x 4x ) if x2 2

1sin (3x 4x ) if 1 x2

minus minus

cos (2x 1) if 0 x 12 cos (2x 1) if 1 x 0

12 cos (4x 3x) if 1 x2

Property (5)

2xtan if 1 x 11 x

2xtan if x 11 x

for x 12

3x x 1 1tan if x1 3x 3 3

1for x2 3

2xsin if 1 x 11 x

2xsin if x 11 x

1 xcos if 0 x1 x

1 xcos if x 01 x

minusle ltinfin +

Property (1)

2 21 2 2

1 1 1 2 2 2 2

1 2 2 2 2

minus minus minus

+ leminus + minus

Property (2)

2 21 2 2

1 1 1 2 2 2 2

1 2 2 2 2

minus minus minus

Property (3)

minus minus minus

minus le le + le

Property (4)

+ minus minus

Property (5)

x ytan if xy 11 xy

+lt minus

πminus lt lt =

Property (6)

x ytan if xy 11 xy

minusgt minus +

Chapter 13point and

Cartesian system

131 IntroductIon

x y 1x y 1 0x y 1

∆ = =

are 1 2 1 2x x y y

2 2+ +

m n m nminus minus

minus minus

1 1λ λ λ

then 1 1

ax by c

ax by cλ

+ += minus + +

Denoted as ym x

rArr 2 1

y ym tanx xminus

= θ =minus

1 2 3 2 3 1 3 1 21[ x (y y ) x (y y ) x (y y )]2

x y 11 x y 12

x y 1∆ =

x y 1 0

be given by [ ]2 3 1 1 3 1 1 3 1 2 2 1

x y 1x y 11x y 12x y 1

Area of polygon =

x yx yx y

x yx y

= 1 2 2 3 n 1 1 2 2 3 n 11 |(x y x y x y ) (y x y x y x )|2

+ + + minus + + +

x yx y1x y2x y

∆ = = 1 2 2 3 3 1 1 2 2 3 3 11 |(x y x y x y ) (y x y x y x )|2

+ + minus + +

∆ = minus

as S S = 0 ie f(x y) g(x y) = 0

rArr f(a b) + lg(a b) = 0 + l0 = 0

4T(x y) (x y)rarr

5T(x y) (x y )rarr

6T(x y) (x y )rarr

helliphellip (i)

2 = (x ndash x2)

2 + (y ndash y3)2

3 3+ + + +

a b c a b c+ + + + = + + + +

rArr A

(b a)y cyy

b a cminus +

=minus +

rArr A

B A CI

bx ax cxx

b a cminus +

=minus +

and minus +=

minus +A

B A CI

by ay cyy

141 Definition

2 11 1

minus = minusminus

or 1 1

x y 1x y 1 0x y 1

determinant form

x y 1a b+ = or

x y 1a 0 1 00 b 1

in determinant form

ysin(p+a) = p

= =θ θ

Given two lines

11 1 1 1 1

+ + = = minus = α

22 2 2 2 2

+ + = = minus = β

rArr 2 1

m mtan1 m m

minusθ =

rArr 1 2 2 1

1 2 1 2

a b a btana a b b

minusθ =

14131 Conclusion

m mtan1 m m

minusθ =

+ and 2 1

m mtan( )1 m m

rArr 1 1 1

a b ca b c

rArr 1ay x c

bminus

m mtan1 mm

minusθ =

+ rArr m m tan

1 mmminus

= plusmn θ+

rArr m tanm1 m tan

plusmn θ

1 m tanθ

rArr 1 1

ax by cmn ax by c

+ += minus

ax by c 0ax by c

rArr 1 1

ax by c 0ax by c

rArr 1 12 2

|ax by c |PNa b

given by 2 2

= =θ θ

required distance

1 2 2 1 1 2 2 1

minus minus

rArr 1 1 1

a b ca b c 0a b c

a + b a + b

+ += plusmn

Two Cases Arise

c1 c2 gt 0 Then 1 1 1 2 2 22 2 2 21 1 2 2

a x b y c a x b y c

a b a b

+ + + += +

ie 1 1 1 2 2 22 2 2 21 1 2 2

a x b y c a x b y c

a b a b

+ + + +=

+ + and 1 1 1 2 2 2

2 2 2 21 1 2 2

a x b y c a x b y c

a b a b

+ + + += minus

a x b y c a x b y c

a b a b

+ + + +=+

and 1 1 1 2 2 22 2 2 21 1 2 2

a x b y c a x b y c

a b a b

+ + + += minus

Tips and Tricks

2 1 2 1 1 12 2

= = minus ++

rArr ( )

I M2 2

2(aa bb )L L 0a b

a b ca b b 0a b c

a h gh b f 0g f c

∆ = =

1h h abm

bminus + minus

= and 2

2h h abm

bminus minus minus

1 2 (h ab)tan|a b |

minus minus

θ = +

2 2x y xya b h

+ c = 0 is 2 2

or 2 2

f bc g cah ab h ab

=minus

2 2 2 2

a b h a b hα β

= = minus minus

Tips and Tricks

+ 2gx + 2fy + c = 0

rArr ( )

S 0x α β

part = part and

part= part

rArr 2

hf bgab hminus

α =minus

af gh h abminus

β =minus

( ) + +

Homogeneous

2 2 3 3

x y 1 x y 1x y 1 x y 1 0x y 1 x y 1

1 1 1 1

2 2 3 3

x y 1 x y 1b x y 1 c x y 1 0

x y 1 x y 1+ =

Family oF CirCle

151 introduction

minus minus

Centre = G F

minus minus

A+ minus

A(x1 y1) B(x2 y2) C(x3 y3) is

2 21 1 1 12 22 2 2 22 33 3 3 3

x y x y 1x y x y 1

0x y x y 1x y x y 1

= =θ θ

straight line

(x1y1) (x3y3)

(x2y2)

S1 = x12 + y1

= 2 21 1 1 1 1(x y 2gx 2fy c S+ + + + =

c c minus

n n minus minus

rArr 2 12 1

2 1t x x

2 1y y

y ym limx xrarr

minus=

minus =

2 12 1

x x1 2 1y y

x x xlimy y yrarr

+minus = minus

( )11 1

xy y x xy

Q 2 2 21 1x y a+ = (1)

2 2 22 2x y a+ = (2)

2 1 1 2

y y x xx x y y

minus += minus

a cos2x

=αminusβ

a sin2y

=αminusβ

Q(h k)

(x1 y1)

1534 Normals

y fmx g+

y f(y y ) (x x )x g

x y 1x y 1 0g f 1

=minus minus

Slope of chord = 1

x(y y ) (x x )y

1 1 1 1yy xx x y+ = +

minus minus

minus +

common tangents

= 1 1 2

r r2sinr r

minus minus

= 1 1 2

r r2sinr r

minus minus= π

2 1 1 2 2 1 1 2

1 2 1 2

minus minus

minus minus θ =

r r2sinr r

minus minus= π

2 2 21 2

r r dcos2r r+ minus

2 2 21 1 2

r r dcos2r r

minus + minus

rArr θ =

2 2 21 2 r r d+ = 2 2 2 2 2 2

1 2 1 2 1 22(g g f f ) c crArr + = +

x2 + 2gx + c = 0

(x2y2) 1 2 1 2 1 1

x y 1(x x )(x x ) (y y )(y y ) x y 1 0

Remarks

Tips and Tricks

Chapter 16parabola

Parabola 16163

Degenerated Conics

= = or

SP = e PM 2 2

|ax by c |(x ) (y ) e(a b )

+ where P(x y) is a

2 2 22 2

(ax by c)(x ) (y ) e(a b )+ +

Condition Conic

162 PArABoLA

as directrix

Parabola 16165

y2 = 4axx = at2

y = 2at

x2 = 4ayy = at2

x = 2at

2 4a 4acy y 0m m

rArr + = =

16a a cm 0m

minus =

Parabola 16167

rArr ay mx m ~ 0m

y2 = 4ax

a 2am m

hArr rArr =

= + rArr = +

2 1 1 2

y y 4ax x y yminus

=minus +

y yABx xminus

=minus

= 1 21 2

4a 2ay yy y

4ay y x xy y

minus = minus+

2Slope of chordt t

2y 2at x att t

minus = minus+

a 2aQ t t

aSQ l at

t rArr = +

2ty x at 1

= minusminus

rArr yt = x + at2

ie ST = SPSQ

Parabola 16169

1651 Properties

(i) t1t2 = 2 (ii) t3 = ndash(t1 + t2)

h 2am say 3aminus

Parabola 16171

a 2am m

minus minus

= +ay mxm

a 2ah km m

minus minus

ac mh km

+ = +ac mh km

ay mx mh km

y2 = 4

axy2 =

ndash4a

x2 = ndash

)2aminus

minus=

minus1

2aminus

=minus

xminusminus

=minus

xminus

=minus

ric co

t2 2at

)(ndash

ndashat

ric fo

y ndash

x ndash

m2 ndash

)(ndash

ityc =

ndash2a

m ndash

m=minus

ndash 2a

m ndash

minusminus

(y ndash

x ndash

ndash k

)2 = ndash

ndash h

)2 = 4

ndash k)

(x ndash

ndash4a

(y ndash

2aminus

minusminus

=minus

2aminusminus

=minus

minusminus

=minus

minus1

hminus

=minus

ndash k

) + t(

x ndash

t + at

ndash k

) ndash t(

x ndash

t + at

ndash h

) + t(

y ndash

t + at

ndash h

) ndash t(

y ndash

t + at

t of c

ndash 2

(h ndash

ityc =

k ndash

ndash 2a

m ndash

k ndash

minus+

=minus

minusminus

(y ndash

ndash h

)ndash 2

ndash am

ndash k

(x ndash

minus=

minus+

minus=

minusminus

Chapter 17ellipse

171 Definition

aSP ePM e h a ehe

= = minus = minus

h k 1a a (1 e )

+ =minus

x y 1a b

Eccentricity 2

= minus

= a + eh

Length of 22bLR =

Ellipse 17175

x y 1a b

+ = where b gt a

x = a ndashae

x yS 1a b

a cos y 1a b

Ellipse 17177

x yS 1 0a b

θ+ φ = minus

=+ minus

θ+ θ =

xx yyT 1 0a b

2 2 2 2

a x b y a b a e x y

minus = minus =

Tangent at (x1y1)2

b xa y

minus 1 12 2

xx yy 1 0a b

+ minus =

Tangent at qb cota

θ+ θ =

Normal at (x1y1)2

a yb x

2 22 2

1 1 2 2a e

a x b y a bx y

minus = minus

Normal at qa tanb

x y 1a b

+ = is

a cos bsin2 2

cos cos2 2

θ+ φ θ+ φ

θminusφ θminusφ

Ellipse 17179

xx yyT 1 0a b

= + minus =

x y 1a b

+ minus

xx yy 1a b

+ minus

1 1 1 12 2 2 2

xx yy x y1 1a b a b

+ minus = + minus

rArr 2 2

1 1 1 12 2 2 2

xx yy x ya b a b

x y 1a b

b xya m

= minus

Chapter 18hyperbola

181 Definition

h k 1a a (e 1)

rArr minus =2 2

x y 1a b

Eccentricity = +2

Hyperbola 18181

1bL aea

= minus

bL aea

Length of =22bRR

rArr No locus

x y 1a b

2a beb

and length = 22a

+ = + =+ +

2 2 2 2 2 22 1

(x ) (y ) 1a b

( ) + +minus + + + minus =

1 1 11 1 22 2 2 21 1 1 1

= plusmn+

1 1 22 21 1

m x l y n aem l

Hyperbola 18183

= plusmn+

1 1 22 21 1

m x l y n aem l

minus += plusmn

+1 1 2

2 21 1

m x l y n aem l

x y 1a b

3[02 ) 2 2

182 Director circle

x y 1a b

(a gt b)

x y 1 0a b

rArr = minus2

2 2 22

by (x a )a

yxS 1 = 0a b

x y 1a b

as the equation

2 2a m b c c

is θ θ =φ φ

1 etan tan 2 2 1 e

or +minus

1 e1 e

Tangent at (x1y1)2

b xa y

minus minus =1 12 2

xx yy 1 0a b

a yb x

2 22 2

1 1 2 2a e

a x b y a bx y

θ+ θ = +

x y 1a b

θminusφ

=θ+φ1

cos2x a

θ+ φ

=θ+φ1

sin2y b

Hyperbola 18185

xx yyT 1 0a b

22 22 21 1 1 1

1 2 2 2 2 2 2 2

2 21 1 1 1

1 2 2 2 2

1 1 1 12 2 2 2

xx yy x ya b a b

rArr minus =

b a m 0 and 2 2 2 2

c a m b= minus

rArr = plusmnbma

and c = 0

= minus =

As minus =2 2

x yH 1a b

and minus =2 2

x yA 0 2a b

minus = minus2 2

x yC 1a b

2ab btan 2tana b a

or 2 Secndash1(e)

rarrx yx

2 and minus +

rarrx yy

minus =2 2

2(x y) (x y) a2 2

c ndashc)

Hyperbola 18187

1m t t

(t1 + t2)

t t 21 t t

x y 2x y

x y 1a b

minus =2 2

x y 1a b

km = b2a2

minus =

x y 1a b

minus + =

x y 1 0a b

191 IntroductIon

192 complex number

193 ArgAnd plAne

vector iy

= + = +

θ θ θ

cos 1 2 4 6

sin 3 5 7

(cosq + isinq) = ( ) ( ) ( ) θθ θ θ

+ θ+ + + + infin =2 3 4

ii i i1 i to e

gives q = f (say)

vector =

1 2 1 2|OA| |OB| |BA| |z | |z | |z z |

vector of z1 and z2

1 While BA

= + minus+ isin

+ +1 2 1 2 2 1 1 2

2 2 2 22 2 2 2

(x x y y ) i(x y x y ) C(x y ) (x y )

Then we get =2 2

z rz r

and OB

2 = = = +2 2 2 2zz | z | | z | (R(z)) (I(z))

7 += = =

z zR(z) x R(z)2

8 θ minus θ +

i ie ecos2

θ minus θ minus

i ie esin2i

10 =1 2 3 n 1 2 3 n(z z z z ) (z )(z )(z )(z )

11 = 11 2

(z )(z z )(z )

14 + =1 2 2 1 2 1z z z z 2R(z z )

15 + = + +2 21 2 1 2 1 2| z z | | z | | z | 2Re(z z )

16 |z1 + z2|2 + |z1 ndash z2|

2 = 2(|z1|2 + |z2|

1 2 1 2| z | | z | 2Re(z z ) or | z1 |2 + | z2 |

2 plusmn 2 | z1 || z2 |

cos (q1 ndash q2)

13 | z1 + z2 |2 = | z1 |

2 + | z2 |2 hArr 1

is purely imaginary

14 | z1 + z2 |2 + | z1 ndash z2 |

2 = 2| z1 |2 + | z2 |

2 15 | az1 + bz2 |

2 + | bz1 ndash az2 |2 = (a2 + b2)(| z1 |

minusθ = minusπ+

19433 Caution

2 if x 0 y 02 if x 0 y 0

= minus

zarg arg z arg zz

= plusmn π

sumprod

k kk 1k 1

Arg z Arg z 2k

zArg 2Arg (z)z

plusmn +

| z | a | z | ai

2 2 b gt 0 and

+ minusplusmn minus

| z | a | z | ai

2 2 b lt 0

rArr = minusb 152 2

5 3i 18 15i (5 3i)2 2 2

P(1) |w| = |w2| = 1 P(2) ω=ω2

Remarks

∵ 2ω ω= ∵ 1 0ω ω+ + =

∵ 2ω ω= and 3 4 2 2 21 ( ) ( )ω ω ωω ω ω ω= = = = =

∵ 2 21 1 ( )ω ω ω ω+ + = + + ∵ 21 ( ) 0ω ω+ + =

P(6) +ω +ω =

P(9) π

ω = minus + =

1 3iarg( ) arg2 2 3

ω = minus minus =

2 1 3 4arg( ) i2 2 3

terms of i w w2

i 1 i 3 2 1 i 3 23 i (1 i 3) 2i2 2i i 2 i

(2k + ) (2k + )cos i sin n n

k = 0 1 2 n ndash 1

1 2 n ndash 1

= ππ+ π+ + = =

n n 12n ndash 1=

an ndash 1 where π

α =2ine

P(2) 1 + a + a2 + + an ndash 1 = 0

conjugates and π

α =2ine

m2m 1 2m 11 2m 2m 1 m m 1

1 1 11

+ +minus + +

= = = = = = = = =

2cis cisn m

2kcisn n

= π θ

where π2kcisn

rArr π= minusnx

2 rArr

πminus

= = = = = =

minus = minus

rArr i3 1

z zArg Arg( e )

z zθ minus

= θ = ρ minus

ρ =minus

z zz z

If z1 = 0

minus minusi1 1

z z | z z | m e 1z z |z z | n

+1 2nz mzzm n

then minus=

minus1 2nz mzzm n

minus π π= minus

minus 3 1

z zArg

z z 2 2 (i)

rArr π

i3 1 3 12

2 1 2 1

z z z ze iz z z z

rArr minus

= minus

z zR 0

rArr minus minus

+ =minus minus

3 1 3 1

2 1 2 1

z z z z0

= plusmnminus

minus 3 1

3 1 2 12 1

= = minus minus

3 1 1 2

2 1 3 2

(iv) + + = + +2 2 21 2 3 1 2 2 3 3 1z z z z z z z z z

(v) πminus

= = +minus

i1 2 3

z z 1 3e iz z 2 2

(vi) + + =minus minus +2 3 3 1 1 2

1 1 1 0z z z z z z

minusminusminus =

minus minus 2 32 4

1 4 1 3

rArr minusminus

= minus minus

1 32 4

1 4 2 3

z zz zArg 0z z z z

minus minus

+ = π minus minus

2 3 1 4

1 3 2 4

minus minus

= π minus minus

1 3 2 4

2 3 1 4

AB rArr = λ

2 1AP (z z )

∵ = +

19631 Conclusion

Writing + minus= =

z z z zx y2 2i

etc in minus minus

=minus minus

2 1 2 1

y y x xy y x x

=minus minus

2 1 2 1

z z z zz z z z

or =1 1

z z 1z z 1 0z z 1

= minus

z z 0z z

minus minus=

minus minus1 1

2 1 2 1

z z z zz z z z

hellip (i)

z z 1z z 1 0z z 1

= π minus

z zArgz z

= π minus

z zArgz z

minus minus= π =

minus minus

rArr minus minus

=minus minus

z z z zz z z z

+ + =1 2 2 11 2

(z z ) (z z )z z I(z z ) 02i 2i

2i and minus minus

197 tHeorem

given by+ +

=|ac ac b |p

= = minusminus 2

z zz z

is called the

=minus

z zwz z

1 2 1 2

z z z zw 1

z z z z

minus minus= = =

minus minus

0 0 0zz z z z z |z | k 0 (1)

rArr minus minus

minus minus= minus

minus minus2 2

z z z zz z z z

+ =minus minus

z z z z 0z z z z

is diametric form

2 = |z1 ndash z2|2

If minus=

minus1

z z kz z

given as minus

= α minus

z zArgz z

= plusmn minus

z zz z 2

rArr minusminus

z zz z

is purely imaginary

If minus= λ

minus1

z zz z

minus minus1 1

+ +minus =0

|az az b || z z |2 |a |

0 04aa(z z )(z z ) (az az b)

=minus minus

i3 1 2 1

3 1 2 1

z z z z ez z z z

1 2nz mzm n

+ +1 2 3az bz cz

ABCa b c

∆ =+ +

tan A tanB tanC

∆ =+ +

as z21 + z2

2 + z23 = 3z0

1 + z22 + z2

z z 11 z z 14i

z z 0z z

z zz z

form as =1 1

z z 1z z 1 0z z 1

collinear

1 2 1 2z z z z r 0α α+ + + =

if 1 2 1 2z z z z r 0β β+ + + =

minus1

z zz z

and minus=

z zy2i

and to convert

Chapter 20SetS and

relationS

201 SetS

208 tyPeS OF SetS

209 SubSetS

nC0 + nC1 + nC2++ nCn = 2n

2012 POweR SetS

2013 DISjOINt SetS

2014 uNIVeRSAL SetS

U = 1 2 3 4 50 then A = 6 7 50

n(A cap B cap C)

Remarks

of sets A B and C

2024 ReLAtIONS

R 2 3 4 5 6 7

1 0 1 0 0 0 02 0 0 0 1 0 03 0 0 0 0 0 14 0 0 0 0 0 0

Many-one Relation

One-many Relation

Many-many Relation

2031 INtO ReLAtION

(i) R1 = (11) (22) (ii) R2 = (1 2) (2 1) (2 3) (3 2) (iii) R3 = (1 1) (2 2) (3 4) (iv) R4 = (1 2) (2 1) (3 4)

in R3 but 1 ne 2

Tips and Tricks

on set ( )2nA 2=

Chapter 21FunCtions

Remarks

Functions 21227

eg =minus2

infin0 00

1 0 etc0

h(x) = minus

x if x is rational

isin ℝ

b2 ndash

d isin

Functions 21229

x3 + cx

k) infin

r a gt

(ndashinfin

a 0xn + a

1xnndash1 +

a2xnndash

2 + hellip

nndash1 x

a n ai isin

0 ne 0

n isin

k) infin

a 0 gt

(ndashinfin

r a 0 gt

= minus

infin)

minuslt

Rndash

t of a

Functions 21231

isinisin

f(x) =

ries o

(0 infin

f(x) =

log ax

(0 infin

ℝ ndash

Functions 21233

= xk k

isin ℝ

n isin

[0 infin

ndash(2

isin ℕ

ℝ ndash

ndash(2

isin ℕ

ℝ ndash

(0 infin

infin)

[0 infin

Functions 21235

1)=minus

ℝ ndash

=minus

(0 infin

infin)

isinminus

infin)

1minus

ltisin

1minus

2mminus

=minus

(0 infin

infin)

1)minus

=minus

isinminus

ℝ ndash

Functions 21237

=minus

isinminus

R ndash

(0 infin

infin)

(0 infin

infin)

) = si

s xf(x

ndash co

sec2 x

ndash times

divide)

xminus

infin)

minus=

ℝ ndash

s x in

f y fo

3)minus

=minus

minus is

+minus

minus=

=minus

Functions 21239

(ix) [x] x if x

[ x]1 [x] if x

(xii) xc

(xiii) [x] + [y] le [x + y] le [x] + [y] + 1

(xiv) [x] = x x 12 2

infin =

(vi) 0 x

x1 x x

(vii) [x] [y] 0 x y 1

[x y][x] [y] 1 1 x y 2

+ le + lt+ = + + le + lt

(i) (x) =

1[x] if 0 x2

1[x] 1 if x 12

le lt + le lt

2n 1(x) x ~ x n2( x)

2n 1(x) 1 for x n2

1[x] n if n x n2(x)

1[x] 1 n 1 if n x n 12

= le lt += + = + + le lt +

Functions 21241

inequality holds

Functions 21243

For many-one

Functions 21245

n times

nP m n0 m n

minus minussumm

m r 1 nr

C ( 1) (m r)

n m r 1 nr

m C ( 1) (m r)minus

m r nr

C ( 1) (m r)=

= minus minussum

rArr n

n r nr

C ( 1) (n r) n=

minus minus =sum

2x 1 0 x 2f(x)

and x 1 1 x 1

then 2

2x 1 1 x 1fog(x)

Functions 21247

2117 even function

Functions 21249

Remarks

ie f(x) if x

h(x)f( x) if x

(i) 1f(x)

minusnotin minus=

+ isin

Functions 21251

Remarks

30 20 is

g( x )3 x 0

= minus= =

periodic

g( x )g( x )

= = cotx which is

eg consider 1 x 0

f ( x )1 x 0

lt= minus ge

and 1 x 0

g( x )1 x 0

minus lt= ge

f ( x )1

or f(x + T) = 1

f(x)minus

g g g g(x) x=

f(x) = =

Remarks

Method I

Method II

Functions 21253

D for a 04aR

D for a 04a

for a 0D 0D

0 for D 0 a 0

D for D 0 a 04a

RD0 for D 0 a 0

4afor D 0 a 0

φ lt lt

or Linear

x 0lim f(x)rarr

f ( x )x

then minus

rarr rarr= = =

π π1

x 1 x 1

rarrminus rarrminus

minus= = =

minusπ π1

x 1 x 1

eg if =minus 2

1f ( x )

( x 1) then

lim f ( x ) exist

minusπx 2

eg rarr

minus= =

x 9f ( x ) lim f ( x ) 6

= infinx 0

If x alim f(x)rarr

+ = rarr

minus x

x 3lim (tan x 2 ) =

rarr rarrminus x

x 5lim erarr

= 34 cos34

= ne eg

x 1x 1

rarrrarr

x alim(f(x)) rarr

= ( )x a

= only if l gt 0

Notesin

x alim

x alimrarr

f(x) may exist

(a) 2( x 9)

f(x)x 3minus

=minus

Here 2

x 3lim x 9 0rarr

minus = So rarrx 3

of the type 00

(b) rarrinfinx

ln xlim

(c) rarr

(ii) =infinc 0

(a) rarr rarr rarr

= =x 0 x 0 x 0

= 1 -1

1 -1x 0 x 0 x 0 x 0

e 1lim 1

minus= (c) ( )

b 1lim ln b b 0

minus= gt

(d) x 0

ln (1 x)lim 1xrarr

(e) aax 0

log (1 x)lim log e

(f) 1x

x 0lim(1 x) erarr

xlim 1 1 x e

(h) rarr

minus= isin

m mm-1

(i) rarr

minus= isin ne

m mm-n

n nx a

(j) rarr

minus= isin

x 1lim mm Qx 1

0 if 0 a 11 if a 1

lim a if a 1

rarrinfin

(l) x a

= (m) x a

Limcos f(x) 1rarr

(n) rarr

tan f(x)lim 1f(x)

(o) rarr

minus=

e 1lim 1f(x)

(p) rarr

minus= gt

b 1lim nb(b 0)f(x)

(q) rarr

n(1 f(x))lim 1f(x)

(r) rarr

+ =1f (x)

x alim(1 f(x) e

(s) rarr

x alim f(x) Aφ

If x alim f(x)rarr

rarr= = gt

= (f) rarr

eg rarr rarr

+ minus minus +

2x 2 x 2

x 5x 6 (x 2)(x 3) 1lim limx 3x 10 (x 2)(x 5) 7

towards zero eg

rarr rarr= =

33x 0 x 03

minus minusminus +1

2x sin 3

minus + minus minus

2t 3 t 3

(a) = + + + + gt2 2 3 3

(b) = + + + +2 3

x x x xe 1 1 2 3

3 5 7x x xsin x x 3 5 7

2 4 6 (f) = + + +

3 5x 2xtan x x 3 15

(h) minus = + + + +2 2 2 2 2 2

1 3 5 71 1 3 1 3 5sin x x x x x 3 5 7

(i) minus = + + + +2 4 6

1 x 5x 61xsec x 1 2 4 6

(j) + = minus + +

1x 2x 11(1 x) e 1 x 2 24

12 123

rarr+ +4 3

(ii) x a

= and 3 2

+ minus= =

x 2 x 2 x 2

minus minus

bourhood of a eg 2

x 5x 6lim(x 2)rarr

minus +minus

+ minus + ++ minus

(2 h) 5(2 h) 6lim(2 h 2)

=rarr rarr

h 0 h 0

h hlim lim(h 1) 1h

minus(x 2)f(x)x 2

Then minus + + +

rarr rarr rarr rarr

+x ax ax a

minusminus =

+ +13 13

13 13 13 13

x ax ax a x a

eg rarr rarr rarr

minus = =

2 2x 0 x 0 x 0

sin2x2 lim 2 2 (1 2) 82x

sin xlim 1x

5 2 16x x xlim

3 92x x

rarrinfin

+ + minus

= 6 0 0 0 6 32 0 0 2+ + minus

= =+ +

minusminus

minusrarrinfinminus

+ + + + = =+ + + + infin gt

m m 10 1 m 1 m 0

n n 1x0 1 n 1 n 0

b x b x b x b b

eg rarr rarr

minus =

2 3 2 4

x 0 x 0

= rarr rarr

+ + + + +

minus =

2 3 5 6

x 0 x 0

( )( )

= =x a x a x a

g x g x g x

find out Limits

eg rarr

+ minus + x 0

tan x x 0 =

minus=

cos x sin x 1lim

sec x 1 2

(a) rarr rarr rarr

+ = =1

rarr rarr+ = +

1 1 f (x)g(x) f (x) g(x)

= rarr

xlim1x x

x 0lim 1 x e erarr

rarr+ λ = =

(b) ( )1

= = then ( ) ( )11

= + minus

= ( ) ( ) x a

minus minus

= =x 0 x 0

= = g(x)

rarr rarr= = = =

Let rarr rarr

= =infin = h(x)

h(x)lim1

lng(x)h(x)lng(x)

0L lime e form0

(i) rarrinfin

n xlim 0

x (ii)

rarrinfin=xx

xlim 0

e (iii)

rarrinfin=

xlim 0

e (iv)

( )rarrinfin

n xlim 0

(v) ( )+rarr

(vi) rarrinfin

minus =n

nlim(1 h) 0 and

ContinUity

minus +rarr rarr= =

= has oscillatory

4 f(x)g(x)

1 2 3 n

k kk k

m m m m

f ff f

n isin ℤ

+ lt lt

x sin x 0 x 1f(x)

x 21 x 4

Remark

f ( x ) x (02)x 2

= forall isinminus

DiFFEREntiABiLity

minus minus = minus

+ minus = =

Since minusminus

f(x) f(a)x a

minusminusx a

f (x) f(a)limx a

+ minus=

f(a h) f(a)limh

+ minus= =infin

f(a h) f(a)limh

3 Vertical tangent

4 f(x)g(x)

x x 0x x 0

+ minush 0

f(a h) f(a)limh

minusminusx a

f (x) f(a)limx a

and we have

minusminusx a

f (x) f(a)limx a

minusminusx a

f (x) f(a)limx a

+ minus minush 0

f(a h) f(a h)lim2h

= rarr

rarr rarr

minush 0 h 0

+ minus minus= + =

Thus rarr

+ minus minus=

Remarks

f ( a g( h)) f ( a)lim f ( a)

g( h)rarr

f ( a g( h)) f ( a ( h))lim f ( a)

g( h) ( h)φ

φrarr

+ minus +=

if + +rarr rarr

+ minus

f (a h) f (a)limh

minusminus

+ minus(n 1) (n 1)

f (a h) f (a)lim or

minus minus

minusminus

(n 1) (n 1)

f (x) f (a)limx a

exists finitely

Remarks

5minus

= x15 sec2 x + 4 5

1tan x

5( x )

f (0 h) f (0 ) f ( h) (0 )lim lim

h hrarr rarr

+ minus minus=

= 15 15

h 0 h 0

h hrarr rarr

Now f (x) = rarr

+ minush 0

f(x h) f(x)h

lim h 0 h 0 h 0

x y f(x) f(y)f2 2

rArr f(x) = ax + b

h 0 h 0

+ δ minus=

δ + δ minus=

rArr h 0

+ minus= =

= = = =

(ii) d d d(u(x) v(x)) (u(x)) (v(x))

minus = =

233 Chain rule

( )( ) ( )= = ==

( )( )

( )( ) ( )( ) =

df g x f g x g x

2 ( )= = + +ωω ω ωd du dv d

of function

minus= isin

minus= + = + isin

1 x xd (e ) edx

3 ed 1(log | x |)

dx x= a gt 0 4 a

= a gt 0 a ne 1

1 d (sin x) cos xdx

2 d (cos x) sin xdx

= minus

4 (cot x) cosec xdx

= minus

d 1(sin x) 1 x 1 or |x| 1dx 1 x

d 1(cos x) 1 x 1 or |x| 1dx 1 x

d 1(cot x) xdx 1 x

d 1(sec x) | x | 1 or x R [ 11]dx | x | x 1

rArr 1 2 n

= + + + rArr 1 2 n

= + + +

minus= = =

d k f xdx

= k ( )( )2

d f xdx

2 ( ) ( )( ) ( )( ) ( )( )2 2 2

plusmn = plusmn

3 ( )2 2 2

4 ( )3 3 2 2 3

3 3 2 2 3

= + times

(a) To get dydx

(c) Find dydx

in terms of x and y

where fxpartpart

= = and 2

= = times = times

there4 2

minus=

(i) If f(x) g(x)

yu(x) v(x)

= = minus

+prime prime

(x) (x) (x) (x) (x) (x) (x) (x) (x)

nn r n r r

y C f gminus=

2 2 2 2

Derivatives

Rate of Change

respect to x

by the following

= ie 0

dy dxf (x )dt dt=

2 dA drA r 2 rdT dt

dA dr2 rdT dt=

y dylimx dxδ rarr

δ = timesδ

tangent and noRMal

1(x y )

dxy 1dy

1(x y )

dyy 1dx

1(x y )

dxy dy

1(x y )

dyy dx

1 1(x y )

A P(x1y1)

Normal

Tangent

1 1(x y )

dy(y y ) (x x )dx

minus = minus

(x y )

1(y y ) (x x )dydx

minus = minus minus

Remarks

f2(x) are equal

(b) Find 1

= plusmn1

ie x y1

1 1( x y )

dy( y y ) ( x x )

dx minus = minus

are identical ie

1 11 1

1 1( x y ) ( x y )

a b cdy 1 dy

y xdx dx

= =minus minus

curve y = f(x) = 1 1(x y )

Slope of PQ 1 1

(x y )1

y b dyx a dxminus

equiv =minus

hellip(2)

for x 1y y2+ for

y and 1 1xy x y2+

2fy + c = 0 (i)

1 1 1 1

minus minus=

+ + + +

dyg tdy dt

dxdx f tdt

y g t x f tf t

minus = minus

x f tg t

(c) For 2 2

x y 1a b

(e) For 2 2

x y 1a b

(h) For n n

x y 1a b

(i) For n n

x y 1a b

dx= β

( ) 1 2

+ α β +

rArr 1 2

dy dy1 0dx dx

rArr 1 2C C

dy dy 1dx dx

= minus

2410 coMMon tangent

point of contacts

Here 1 2

1 1 2 2

c c2 1

(x y ) (x y ) 2 1

y ydy dydx dx x x

minus = = minus

MonotoniCity

2416 crItIcal poInt

Remarks

le le uarr

ge ge darr

uarr gt

darr lt

gt uarr

rArr uarr

prime lt α ltβ

δφδs

φ+δφφ

Clearly P gt 0 if 2

d y 10 0dx

gt rArr gtρ

and P lt 0 if 2

d y 10 0dx

lt rArr ltρ

Step 1 Find 2

d y 0dx

does not exist

(Say a b hellip)

(a) dy 0dx

gt (b) 2

d y 0dx

(a) dy 0dx

gt (b) 2

d y 0dx

(a) dy 0dx

lt (b) 2

d y 0dx

(a) dy 0dx

lt (b) 2

d y 0dx

(a) dy 0dx

gt (b) 2

d y 0dx

(a) dy 0dx

lt (b) 2

d y 0dx

Remarks

a0 c b

f(a)nef(b)

f(c)=0

Remarks

f ( b) f ( a)f ( c )

b a rArr 0

f ( c ) 0b a

prime = =minus

MaxiMa and MiniMa

2428 local MaXIMa

ge +ge minus

24281 Local Minima

le + le minus

Remarks

If =dy

Step 1 Find dydx

x0+h x0+h

f(x0ndashh) gt0

x0ndashh X0

f(x)is minimum

f(x)is

minimum minimum

| x |1 x+

is the only maximum

= but dx 0dt

ne If dx 0dt

d y yx xydx (x)

minus=

rArr c cc

2 3 2t t t tt t

minus= =

Now c c

2 2t t t t

d y d y0 if 0dx dt

Further c

d y 0dx

lt if c

d y 0dx

= rArr =

ne The sign of 2

2 2 2t 0

d y 2(t 4) 0dt (4 t )

+= minus lt

eg let f(x) = 2

n (n )n 90

1 absin

th lsquoa

f a ci

lsquoarsquo

a2 sinq

times h

us lsquor

rsquopr

ndashsid

h lsquoarsquo

sinnπ

us lsquor

rsquo4p

r with

lsquorrsquo

t tria

de lsquoa

minusminus

ndashsid

l slan

(i) If

a tria

1 (na)

indashpe

1 3 (A

minus= minus

Also APf(x) bm

x aminus

=minus

251 INTRODUCTION

Remarks

x2 x3 etc

r xx dx Cr 1

= ++int

Then n 1

= ++int

= +int

Remarks

(a) ( ) =intd

f ( x )dx f ( x )dx

C(n 1)a

1x ln|x| + C

1ax b+

ln(ax b)C

bx c1 a Cb lna

minus + +

+ + + +

π + minus minus +

or xln tan C2

cosec(ax + b)1a

minus + +

1 sec(ax b) Ca

1 cosec(ax b) Ca

minus + +

a xminus1 xsin C

aminus +

x 1+2ln|x x 1 | C+ + + 2 2

x a+2 2ln|x x a | C+ + +

11 xminus

1 1 xln C2 1 x

minus 2 2

1a xminus

1 a xln C2 a x

1x 1minus

1 x 1ln C2 x 1

minus+

1x aminus

1 x aln C2a x a

minus+

11 xtan Ca a

minus +

or 11 xcot Ca a

minus minus +

21 xminus2

1x 1 x 1 sin x C2 2

2 2 aminusminus + +

2x 1minus2

minus + minus +

21 x+2

2x 1 x 1 ln x x 1 C2 2+

+ + + + 2 2a x+2 2 2

+ + + +

x x aminus11 xsec C

a aminus minus +

g(x) t= dtg (x)

dtdx1 t

= minus+

Integrand

( )2f x x 1+ + 2x x 1 t+ + = rArremsp2 2

1 1f x 1x x

+ minus

+ = rArremsp 2

11 dx dtx

minus =

1 1f x 1x x

minus +

minus = rArremsp 2

11 dx dtx

1 1f x xx x

+ minus

+ = rArremsp 3

12 x dx dtx

minus =

1 1f x xx x

minus +

minus = rArremsp 3

12 x dx dtx

1 2n n

1f(x a) (x b)

Put (x + a) = t(x + b)

rArremsp2

1 dtdx(x b) (b a)

=+ minus

1nrsax bR x x

Put nax b tcx d+

rArremspn 1

=+ minus

Integrand

a cosθ= θ

( )2 2f a x+ a tanx

a cotθ= θ

a cosecθ= θ

a xfa x

minus +

a cosx

a sinθ= θ

minus+ +

minusminus + +

x a tx b+

(b a)dt dx(x b)

minus=

(a b)or dt dx(x a)

minus=

dx nx(x 1)

isin+int

(ii) 2 n (n 1)n

(iii) n n 1n

254 INTREGRATION OF

= minusint

Expained m 1

2 2(1 t )minus

m n m nm

= ( )m n 2

int = ( )m n 2

+int t = tanx

= ( ) ( )n 2

m 2 2z z 1 dzminus

minusint = m 1

minusminusint

= +int int (1)

1 dxax bx c+ +

dxIb Dx

= minus + +

minus + +

minus minus

=+ + minusαint int

2 121 1 (x )(x ) dx C

a a ( 2 1)

a (x )minus +

minusα

dx 1 dxI(x )a(x )(x ) a (x )(x )

int int hellip (i)

Let x txminusα

=minusβ

rArr 2

=minusβ

rArr 2

(x ) dx dt(x )

minusβ =minusβ

rArr 2

hellip (ii)

1 xI ln Ca( ) x

minusα= +

αminusβ minusβ

b Da x dx2a (2a)

+ minus

2 Da t dt2a

int where bt x

2a = +

2 DI a t dt2a

minus= +

int where

bt x2a

= minus minus + int

2Da t dt2a

minus minus

int where bt x

2a = +

ax bx c+ +int

b Da x2a 4a

= + minus

int 22

11 dxa b Dx

= + minus

21 Dln t t C2aa

= + minus +

where bt x

2a = +

21 Dln t t C2aa

where bt x2a

1I dxb D( a) x

1 1 dta D t

=minus

int bt x2a

px q dxax bx c

++ +int (b)

px q dxax bx c

+ + + px = A(2ax + b) + B

rArr pA2a

= and pcB q2a

= minus

+ += +

+ + + = + + + + + +int int int

1I dxax bx c

=+ +int

1 dxIa (x )(x )

=minusα minusβint

1 1 1 dxa( ) x x

minus= +

minus +int

dxIcx ax bx

int hellip (1)

= + + minus +

int = 2

cx ax b 2 a cx

plusmn +

hellip (2)

Put cax tx

plusmn = hellip (3)

I = 2 2

1 dt 1 dt2 c t b 2 a c 2 c t b 2 a c

dx 1 xtan Cx a a a

minus = + + int

I = 2 2

int = 2 2

dxCx Ax Bx

minus + +

where Ct Axx

ax bx c dxpx qx r

+ ++ +int

ax bx c dxpx qx r

+ ++ +int and

ax bx c dxpx qx r

+ +int

(px2 + qx + r) + C

minus +

or 1 1x g xx x

+ minus

I = 2 3 2

int = 2

f (x) dx

1 1x a x b x cx x

+ + plusmn +

Case (i) If I = 2

1 1x g x dxx x

1 1x a x b x cx x

minus + + + + +

int = 2

1 11 g x dxx x

1 1a x b x cx x

minus + + + + +

Putting 1x tx

+ = and 2

11 dx dtx

minus =

we have I = ( ) 22

g(t) dt g(t) dt

Case (ii) If I = 2

1 1x g x dxx x

1 1x a x b x cx x

+ minus + + minus +

int putting 1x tx

minus =

11 dx dtx

we have

I = ( ) 22

g(t) dt g(t) dt

+ + ++ + +int int

P (x) dxpx qx r+ +

minus= + + ++ + + +

int int hellip(1)

( )2m m 1m 12 2 2

minusminus

+= + + + +

+ + + + + +

2 2 2 2

dx a bsin x+int dx

cos x dx

a bsin x+int

( )f tan x 2dx

= and 2

2dtdx1 t

Step II Take 2 2

= =+ +

and 2 2

= =+ +

Algorithm

+ ++ +int

Algorithm

++ +int

2 2t 1R t t 2(t 1)2

minus minus +

minus + + and

minus + +

1 dtt a+int 2 2

1 dtt aminusint 2 2

dta tminusint

(ax b) cx d+ +int

type 2

at bt c+ +int

Integral of type dx

rArr 2 2

minusα = minusβ =

int int

linear quadraticint

dxI(lx m ) px qx r

=+ + +

int put 1lx mt

+ = ( ) ( )2 2 2

minus=

minus + minus +int

dxI(ax c) px r

int then put 1xt

rArr 2

1dx dtt

m p p r n rr

= sumintp nr m 1

p p r rr

xC a b Cnr m 1

+ +minus

= + + + sum

Case (a) m 1

1nst ax

b minus

mnsm t ax

b minus

rArr =

rArr s 1

st dtdxnbx

minus=

+minus

+ minus minus=

= and m 1n+

notin and m 1 pn+

+ isin

rArr sn

a t bx

= minus

m np n 1n r s 1

s a t dtna t b

ax bx c t(x )α

+ + = minusα

plusmn+ + = plusmn

or try

1 Substitute

minus+ =

if a gt 0 D lt 0

3 Substitute

if a gt 0 D gt 0

as given below

( ) ( )ax ax ax ax

+ + minus= + +

a bminus

emsp rArr n 1

minus= minus +

minus= +

rArr n 1

n n 2tan xI I

minus= minusminus

rArr n 2

n 1 n 1

minus= +

minus minus

rArr rarrinfin

2224466(2n)(2n)lim13355(2n 1)(2n 1)(2n 1) 2

cos (x)

π π π

n n 2 n

261 AreA Function

rarrrarrinfin

= + + + + + + + minus intb

h 0a n

rarrrarrinfin

= + + + + + + intb

h 0a n

rarrinfin rarrinfin

= =rarr

sum sumint

n nr 1 r 1a h 0

f (x)dx

= =rarr

minus = + = +

sum sumintb n n

n nr 1 r 1a h 0

ai = aRi and an = b

rArr aRn = b rArr =

= common ratio then

r 1nba (R 1) 0asn

a then

1 1 2 2 n nna

rarrinfin=

∆ ∆ = minus

r r rn 1

then int3

Remark

1f ( x )

x= then

dx( a b)

1 1 h 1 1a b [a ( 1)h] a h b h=

2 2h 01a n

1 h 1 1dx lim

(i) =int intb b

a af (x) dx f(t)dt

=minusint intb a

a bf (x) dx f(x)dx

b c c b

Property 4 =intb

Remarks

1 26 2( x 2x ) dx ( 27 1) (9 1) 8 0

dtf(g (t))g (g (t))

Property 6 intb

q If intb

=int intb b

t 0a a t

f (x)dx lim f(x)dx

q If intb

rarr=int int

t 0a a

(i) infin

(ii) minusinfinintb

(iii) infin

a ba 0

= minusint intb a

f (x)dx f( x)dx

= minusint intb b c

f (x)dx f(a b x)dx

Remark

267 APPlicAtions

By above property

= intb

(a b)I f(x)dx2

=(b a)I

bkb bk

a akak

Let = intb

Property 13 minus

int inta

int int

Equivalently int2a

f (x)dx =

= inta

f (a) a

minus= int

cf (x)dx = int

af (x)dx

rarr rarrα α

[a b] then ψ

φint(x)

d f(x t)dtdx

partint(x)

of a then part

α = αpartαint

I ( ) f(x )dx

0f (t)dt 0

q =int intnT T

period T then +

inta T

Hence prove that +

=int inta T T

f (x)dx f(x)dx

Corollary +

=int inta nT T

f (x)dx 0

f (x)dx g(x)dx

geintb

g(x)dx k and leintb

1 2a a a

k f(x)dx k

1 2 3 1 2 3a a

Here leint intb b

f (x)dx | f(x)|dx = =

k 1 k 1

f (x ) x | f(x ) x |

=minus int

1f(c) f(x)dxb a

interval [a b]

α partαintb

dI I ( ) f(x ) dxd

minus minus = +

b aa r xn

and minus rarr

b a dxn

Step III β

minus β = + =

b alim a r bn

rarrinfin

minus α = + =

b alim a r an

Step IV rarrinfin

minus minus + =

sum intbn

n r 1 a

and rarr

Step III β=

r q(n)

n r p(n)

β = = α = =

max min

n n n n

Step IV β

rarrinfin= α

sum intq(n)

n r p(n)

r 1lim f f(x)dxn n

1 For π π

isin =int int

2 2n n

(n 1) (n 3) (n 5) If n is oddn (n 2) (n 4)

2 For π

isin int

(m 1) (m 3)(n 1) (n 3) p(m n)(m n 2)

π = minus +

2462n 1lim 2 135(2n 1) (2n 1)

+ + + + infin =2

2 2 2 2

1 1 1 1 1 2 3 4 6

+ + + + infin =2

2 2 2 2

1 1 1 1 1 3 5 7 8

2 2 2 2

1 1 1 1 1 2 3 4 12

+ + + + infin =2

2 2 2 2

1 1 1 1 2 4 6 8 24

intb 2

a[f(x)] dx

269 BetA Function

0B(m n) x (1 x) dx

2 2m 2 2n 2

2 2m 1 2n 1

0B(m n) sin cos d

+ minus + minus

2 (2m 2)(2m 4) (2n 2)(2n 4B(mn) p

(2m 2n 2)(2m 2n 4)

2610 gAmmA Function

denoted by n

(ii) = minus = =1 (1 1) 0 1

(iv) =infin0

(v) = π12

For gt =+

m nm n 0 B(m n)

Remark

( m n)m n 1

β minus minus=

by = intb

A f(x) dx

= +int intc b

A f(x)dx f(x)dx

A (x) dy f (y) dy

A f (y)dy f (y)dy

a a c d e f

A f(x) g(x)dx

A | f (y) g (y)|dy

Area = =3

1tOAB A4

Q intb

Thus =intb

f (x) g(x)y

minus=2

f (x) g(x)y

= θ = θ θint int2 21 1A r d [f( )] d2 2

( ) ( ) ===

Catesianequation

θ θθθ

1 2a a t

271 IntroductIon

Example 32 2

dy d y xdx dx

+ = + = or 2 2 2

dy d yF x y 0dx dx

is a standard form

32 22 3

dy d y1dx dx

+ =part part

27221 Order

d y dy3x x edx dx

27222 Degree

d yf x ydx

+ ( ) ( ) 1nm 1

d yg x y

+ ( ) ( ) 2nm 2

d yh x y 0

hellip(1)

y = Ax2 + Bx etc

rArr 2

d y Asin x Bcos xdx

From (iii) 2

rArr 2

d y y 0dx

dy a 1dx y

= minus

= hellip(i)

and dy f(y)dx

= hellip(ii)

dy dxf(y)

= or dy f(x)dx g(y)

= + + where b ne 0

rArr dt bf(t) adx

rArr dt dx

bf(t) a=

+ + where b1 + a2 = 0

Consider 1 1 1

2 12 2 1

+ + = + +

rArr x2 + y2 = r2 (1)

xdy y ddx secx dx

minus θ= θ

xdy ydx cos dxminus

= θ θ

n yf(x y) xx

or n xyy

Case I If 1 1 1

lines let a 1 1

a b ka b= =

b (kv c)dv adx v c

+ rArr 1

1 1 1 1 1

+ + + int

Case II If 1 1 1

rArr 1dv k adx

rArr 22

= + minus + +

+ + where

nea ba b

can be reduced to

Now 1 1

a ba bne

gives dy dYdx dX

+ + + +=

+ + + +

rArr 1 1

dY aX bYdX a X b Y

rArr 1 1

dV a bVV XdX a b V

part part=

y constt

+ =int int

plusmn = plusmn q 2

ydx xdy xdy y

minus=

x dy y dx1dxy x y

+ minus =

2xy dx x dyxdy y

minus =

2xy dy y dxydx x

minus =

q 2 2 2

x 2xy dx 2x ydydy y

minus=

y 2yx dy 2y xdxdx x

minus=

q 12 2

x ydx xdyd tany x y

minus minus= +

q 12 2

xdy ydx yd tanx y x

minusminus = +

q x dy y dx

d(ln(xy))xy+

= q 2 22 2

+ + = +

q y dx x dyxd ln

y xy minus

x dy y dxyd lnx xy

minus =

q x x x

e ye dx e dydy y

minus=

e xe dy e dxdx x

minus=

q ( )2 2

x dx y dyd x y

d[f(x y)] f (x y)1 n (f(x y))

minus prime=

part partne

is given as n n 1

0 1 n 1 nn n 1

Remark

d y d y2 5 y 0

1 uy minus = (ii)

+ minus = minus

1uy minus=

dyf (y) Pf(y) Qdx

k kk 1

g (x yC ) 0=

k kk 1

g (x yC ) 0=

where dypdx

prime= + +

prime+ =

= if dp 0dx

d y f(x)dx

q Equation of Type2

d y f(y)dx

= helliphellip(1)

rArr 2

= rArr 2

= int int rArr

Method 2 Given 2

d y f(y)dx

rArr 2

rArr dy dx

2 f(y)dy 2A= plusmn

271311 Trajectory

minus for dydx

dt= minus

rArr dv k v gdt m

dt= minus minus

Then df(t) k R f(t)

dt= minus

Chapter 28VeCtors

(c) same sense

Remarks

O + + =

Vectors 28353

AB and

OP r xi y j zk

= = = = + +

P(xyz)

A(0y0)

M(xy0)(x00)B

(00z) zk

2 1 2 1 2 1PQ (x x )i (y y )j (z z )k

where P(x y z) and

= + + =

2 2 2r x y z OP

If and = + +

2 2 2 2r x i y j z k then + = + + + + +

1 2 1 2 1 2 1 2r r (x x )i (y y )j (z z )k

Geometrically +

Vectors 28355

a b b a

a 0 a 0 a a

(a) a b a b+ le +

(b) + ge minus

a b a b a b

If = + +

1 1 1r x i y j z k and = + +

1 2 1 2 1 2 1 2r r (x x )i (y y )j (z z )k

1 2r and r

(a b) (b a)

a (b c) (a b) c)

a a a b b a

a b a b a b

λ = λ + + = λ +λ +λ

Remarks

(i) aa

is a vector along a

having unit length

= plusmn

according as b

(ii) Division of a

by 1λ

ie a 1

( a)λ λ=

(iii) ( a b) a bλ λ λ+ = +

a bACa b

length = +

a ba b

a bONa b

minus +

ˆˆ( a b)OT ˆa b

(b) If

(c) For two vectors

a b 0 ie =

Vectors 28357

1 If a and b

as given below

Theorem If a and

286 section formula

na mbcm n

=minus

na mbcn m

Remarks

(ii) m

(iii) If m

(iv) If m

1n= then

(vi) m

+ +equiv

lie on same plane

(ix) na nb

helliphellip(i)

a mb nc 0 and

(x) If R( c )

and Q( b )

in the ratio mn n

isin minus

then a b c

the position

vector a b c

ab and c

Vectors 28359

where l is a scalar

If 1 1 2 2 3 3 n na a a a 0λ +λ +λ + +λ =

a and b 0 le q le p

(ii) ne

a(b c) (ab) (ac)

(iv) 22 2a aa a a= = =

(v) If = + +

( ) ( )= + + + + = + +

(vi) + +

θ = =+ + + +

1 1 2 2 3 32 2 2 2 2 21 2 3 1 2 3

(vii) =

(viii) = + + = + +

2 2 2(a b) (a b)(a b) a b 2ab

a on b = θ = θ = θ =

ˆbon a ba

opposite to

ˆ ˆb on a (ba)a

ˆb MA a sin a b

a on b )

Work done

work done = θ = =

( F cos ) s F s

Vectors 28361

( a b )a b

plusmn timestimes =

Where 1 2 3 1 2 3 1 2 3

times = = + + = + +

a b (b a)

3 If two vectors

4 times =

a b 0 rArr

7 Let = + +

1 2 3 2 3 3 2 3 1 1 3 1 2 2 1

8 times

a bsin

Remarks

Since a b

sin [0 ]a b

θ θ πtimes

= isin

rArr 1a b

sina b

minus times =

or 1a b

sina b

a and b

1 1a b (a b) (a b)2 4

1a b (a b) (a b)2

= times

1 21 d d 2

If = + +

1 2 3ˆˆ ˆa a i a j a k = + +

1 2 3ˆˆ ˆb b i b j b k = + +

a a aa b c b b b

Geometrically

a b and c

do not changed = =

a b c b c a c a b

(c) + + = +

a b c d a c d b c d

a a b 0

a b c 0 if = λ

a b c 0 (even if

(g) If = + +

a b c a b c

1 a b c6

ab c is

1 a b c2

a (b c) b(ac) c(ab)

(a b) c b(ac) a(bc)

(iv) times times

Vectors 28363

ie times

a (b c) o

(vii) times times

b and c

times times

c and d

aa bb c c 1 and = = = = =

ab c the vector

(i) = = =

aa bb c c 1 (ii) = = =

ab bc ca 0

(iii) =

1[a b c][a b c ]

(iv) times

b c a[a b c ]

(v) = = = = = =

[a b c] [a b c ] 1

r a (b)

rArr 1 1 1( x x ) ( y y ) ( z z )m n

rArr 1 1 1

2 1 2 1 2 1

( x x ) ( y y ) ( z z )( x x ) ( y y ) ( z z )

minus minus minus

1r a b and = +micro

2r a b

= + plusmn

b br a tb b

a and b is = λ +λ

1 2r a b

rArr times =

r(a b) 0

(r c)(a b) 0 r(a b) [a b c]

= minus = minus

r (b a) (c a)

Vectors 28365

r b a b rArr = +

1r (a b) ybbb

r b c b r a 0 a

b rArr = minus

car c bba

= + + times +

a b1r a kb a bk a k

Cevarsquos Theorem

conversely

and conversely

Deasargue Theorem

Geometry

291 IntroductIon

= = =+ + +

2912 Corollary

2 1 2 1 2 1x x y y z zx y z1 1 1

1 2 3 1 2 3 1 2 3x x x y y y z z z

3 3 3+ + + + + +

4 4 4+ + + + + + + + +

Remarks

by ndashve sign

ˆ ˆ ˆi m j n j

6 If a vector

m nm n

+ +θ =

+ + + +

1 2 2 1

2 2 2 2 2 21 1 1 2 2 2

(b c b c )sin

a b c a b c

Σ minusθ =

+ + + +

a b ca b c

1 2 2 1 1 2 2 1 1 2 2 1

2 2 2 2 2 2

m n n m and

m n n m

respectively

Corollary

2 1 2 1 2 1

Remarks

P( a) andQ( b )

2 1 2 1 2 1

If 1L r a b= +λ

or 1 1 1

x x y y z zm n

or 2 2 2

x x y y z zm n

1 2 1 2 1 22 2 2 2 2 21 1 1 2 2 2

+ +θ = =

+ + + +

rArr 1 1 1 2 1 2 1 22 2 2 2 2 21 1 1 2 2 2

minus minus + + θ = = + + + +

ie b kd=

m nm n

ie bd 0=

Or l1l2 +m1m2 + n1n2 = 0

θrarr

A(x1y1z1)(a)

C(x2y2z2)(c)

rarr d= 2im 2jn 2k

rarrb=1im1jn1k

ie x y z

= = = λ

mr a AM= +

Then PM bperp

rArr (PM)b 0=

rArr 1 1 12 2 2

(x ) m(y ) n(z )m n

hellip(ii)

Remark

r a b= +λ

Method 1

Method 2

lines intersect

= a c b sin (a c) b

1 2 3ˆˆ ˆb b i b j b k= + +

1 2 3ˆˆ ˆc c i c j c k= + +

minus minus minus

292 Skew lIneS

= (a c)(b d)

minus times

Cartesian form

1 1 2 2 3 3

minus minus minus

are coplanar

ie b kd=

Iff b kdne

(ie non-parallel)

(plane APL) and

(plane CPQ)

293 Plane

Equation of plane

(constant)

or ˆru p=

to normal form

Step 2 2 2 2 2 2 2 2 2 2 2 2 2

ax by cz d

+ + + + + + + +

a b c=

+ + and

dpa b c

a 0 b 0 0a 0 c

Remarks

1dA dB dC

intercept d d d

on coordinates axes

2 1 2 1 2 1

3 1 3 1 3 1

[ AB AC AD] 0

1 2 1 2 1 2

1 3 1 3 1 3

x x y y z z

x x y y z z 0

x x y y z z

B b P(xyz)

and P(x y z)(r) be

be vector normal

to given plane

2 1 2 1 2 1

minus minus minus

1 1 1 1

yz 2 2 zx 2 2

3 3 3 3

y z 1 z x 11 1A y z 1 A z x 12 2

y z 1 z x 1= = and

xy 2 2

x y 11A x y 12

x y 1=

be two planes

Then 1 21 1 1 2 1 2 1 22 2

1 2 1 2

Corollary

or a1a2 + b1b2 + c1c2 = 0

2 2 2 2

a b c da b c d

Let x y zL m n

2 2 2 2 2 2

a mb ncsin sin cos

2 m n a b c

+ +π θ = minusα = α + + + +

rArr 1

2 2 2 2 2 2

a mb ncsin

m n a b cminus + +

θ = + + + +

where a and q are

acute angles

Remarks

hellip(i)

al bm cn 0 and

1 22 2 2

d dda b c

minus=

Let rn d 0+ =

pn dPMn+

Cartesian form

1 1 12 2 2

ax by cz dPMa b c

+ + +=

+ + where 1 1 1

From above m 2

pn dr p nn

+ = minus

Cartesian form

= = = λ

rArr 1 1 12 2 2

(ax by cz d)a b c

minus + + +λ =

( )1 1 11 1 1

rArr 2

p n dq p 2 nn

minus + = minus

rArr 1 1 1a b cx y z2 2 2λ λ λ + + +

λ + + + + + + =

rArr 1 1 12 2 2

2(ax by cz d)a b c

minus + + +λ =

= = =+ +

a b ctimes = = + +

= =1 1 1x x y y z zl m n

+ + + + +=

+ + + + +1 1 1 1 1 1 1 1 1 1

2 1 2 1 2 1 2 2 2 2

1 1 1 1

2 2 2 2

3 3 3 3

4 4 4 4

a b c d

a b c d0

a b c d

α = β = γ =∆ ∆ ∆

1 1 12 2 2 2 2 2

(a b c ) (a b c )

+ + ++ + +=

+ + + +

Remarks

+ d2 = 0 is + + + + + += plusmn

+ + + +1 1 1 1 2 2 2 2

2 2 2 2 2 21 1 1 2 2 2

a b c a b c

+ + + + + +

=+ + + +

1 1 1 1 2 2 2 2

2 2 2 2 2 21 1 1 2 2 2

301 ExpErimEnts

Sample Space

Probability 30383

will be +y

Remarks

rP(E) limn

Probability 30385

C 13 1C 39 3

Remarks

m n+ and the

(i) capcap = 1 2

1 2n(E E )P(E E )

cup = =1 2 1 2 1 21 2

= P(E1) + P(E2) = P(E1 cap E2)

Probability 30387

= cap cap

= ne2 1 1 21

n(E E ) P(E E ) P(E ) 0n(E ) P(E )

21 2 1

EP(E E ) P(E )PE

EP P(E )E

1 2 n1 P(E )P(E )P(E )

E EP P 1E E

E EP P(E )PE E

cap 32

EEP PE E E

1 1 2 1 2 n 1

= + + +

1 2 n1 2 n

ii 1 i

AP(A) P(A )PA

Remarks

P(A ) (P(A A )) (P(A A A ))

then ( )( )

i ii 1

Probability 30389

Remarks

k k 1 k 2 n ii k

p p p p P+ +=

+ + + + = sum

1 2 3 k ii 1

p p p p p=

+ + + + = sum

m m 1 m 2 k ii m

p p p p p+ +=

+ + + + = sum

micro = =sumn

i ii 1

E(X) p x

2 2 2i i

Var(X) x (x ) p E((X ) ) = = =

sum sum2n n

2i i i i

i 1 i 1

p x p x

2 2i i

E(X ) p x

= = = =

minusmicro = minus

sum sum sum

2n n n2 2

= = minus

r (n r)

nSSSSS FFFF Cr(n r)

minus minus

n n r n rr r

r times (n r) times

successes equals 1

Remarks

n r n rr

C p q 1minus

n r n rr

1 C p qminus

minus sum

Probability 30391

n r n rr

C p q minus

n r n rr

r ( m 1)

1 C p q minus

minus sum

n r n rr

C p q minus

n r n r n r n rr r

r 0 r m 1

+ =sum sum

σ = minussumn

2 n r n r 2x r

minus + minussumn

2 n r n r 2 2r

(r r r) C p q n p

= minus

minusminus + minus

minussum sumn n

n 2 2 r 2 n r n 1 r 1 n r 2 2r 2 r 1

r 2 r 1

=SD npq

deterMinants

311 Matrix

11 12 ln

1 2 3 mn

am am am a

= ij m nA a

Remark

ii 11 22 33 nni 1

(notation)

a (a a a a ) Tr(A)=

= + + + + =sum

a 0 0 00 a 0 00 0 a 0 0 0 0 a

(Notation)

diagonal (a a a a )

k 0 0 00 k 0 00 0 k 0

1 0 0 00 1 0 00 0 1 0 0 0 0 1

Remark

I1 = [1]

= = 2 3

1 0 01 0

I I 0 1 00 1

Remarks

=n( n 1)

=2n n 2

312 Sub Matrix

For example 2 57 9

2 5 87 9 41 3 5

3121 Equal Matrices

ij ik kjk 1

(C ) a b=

Also 1 1 1 1 2 1 3

2 1 2 3 2 1 2 2 2 3

3 3 1 3 2 3 3

R R C R C R C

times = times =

A5 6 8

then T

2 5A 3 6

minus+

eg ( )2 3 2i 2 3 2i 2 3 2iA A A A

3 2i 7 3 2i 7 3 2i 7+ minus +

A 1 3i 0 4 i2 4 i 2i

matrices

Properties

= minus and 2 1 0

A I0 1

3148 Unitary Matrix

Let a b

then a b

A (ad bc)c d

= = minus

Let 11 12 13 11 12 13

21 22 23 21 22 23

31 32 33 31 32 33

a a a a a a

= 22 23 21 23 21 2211 12 13

32 33 31 33 31 32

a a a a a aa a a

a a a a a aminus +

where 11 12

eg 11 122 3 523 23 11 32 31 12

a aC ( 1) M ( 1) (a a a a )

Remarks

= sum3

ik ikk 1

= sum3

kj kjk 1

(ii) = =

= = nesum sum3 3

ik jk ki kjk 1 k 1

a C a C 0 for i j

ie 11 12 13

21 22 23

31 32 33

C C CAdj(A) C C C

21 22 23

31 32 33

C C CAdj(A) C C C

Let AdjA

B A 0A

= ne rArr nn

A IA(AdjA)AB I

A I(AdjA)BA A I

A A= = =

So a0 An + a1 A

a aA A A a a

i jC Charr

eg 1 3

1 0 0 0 0 10 1 0 R R 0 1 00 0 1 1 0 0

11 12 1n

21 22 2n

n1 n2 nn

a a a 1 0 0 0a a a 0 1 0 0

a a a 0 0 0 1

12 1n11 1 1

22 2n21 2 2

n2 nnn1 n n

a aa x ba aa x b

a aa x b

2 2 2 2 2 2

a b c b a c

a b c b a c∆ = ∆ = ∆ =

rArr 1 1

C b kxaminus

= Thus 1 1

C b kxaminus

1 1 1 1 1 1 1 1 1 1 1 1

2 2 2 1 2 2 2 2 2 2 2 3 2 2 2

3 3 3 3 3 3 3 3 3 3 3 3

∆ = ∆ = ∆ = ∆ =

= = =∆ ∆ ∆

Let 11 12 13 1 1

21 22 23 2 2

31 32 33 3 3

a a a x b

21 22 23 2 2

31 32 33 3 3

λ = micro α

α micro minus =

rArr 1 2 3

a 0 0[x x x ] b d 0 [ ]

= λ micro α

From (iii) 3xfα

= from (ii) 2

11 1 22 2 1n m

22 1 22 2 2n m

m1 1 m2 2 mn n

+ + + =+ + + =

+ + + =

12 1n11 1

22 2n21 2

n2 nnn1 n

a aa x 0a aa x 0

a aa x 0

rArr AX = O

319 eliMinant

a b x Xa b y Y

operator 1 1

rArr X 1 0 xY 0 1 y

= minus Thus

1 00 1 minus

minus = minus

θ θ = θ minus θ

θ minus θ = θ θ

Let 11 12 13

21 22 23

31 32 33

a a aa a aa a a

a C a C a C a C=

rows and 3

1j 1j 2 j 2 j 3j 3j kj kjk 1

a C a C a C a C=

jth column

P(x y)

21 22 23

31 32 33

a a aa a aa a a

∆ = then

rArr 1 1

x y 11 x y 12

rArr x y z

b b btimes =

is STP of x y z

a a a[abc] b b b

c c c=

a ba b

∆ = (a1 b2 ndash b1 a2) and 1 21 2 2 1

a a(a b a b )

i i1R Rm

∆ = ∆

For example if 2

x 5 2x 9 4x 16 8

+ minus minus minus

minus∆ = minus =

(i) 2 2 2

1 1 1a b c (a b)(b c)(c a)a b c

(ii) 3 3 3

(iii) 2 2 2

+ + ++ + ++ + +

= minus + + minus

(iii) 3 3 3

z c x a y ba b c

+ + ++ + ++ + +

Let 1 1 1

1 2 2 2

a b ca b ca b c

∆ = and 1 2 3

2 1 2 3

m m mn n n

∆ =l l l

1 1 1 1 1 1 1 2 1 2 1 2 1 3 1 3 1 3

1 2 2 1 2 1 2 1 2 2 2 2 2 2 2 3 2 3 2 3

3 1 3 1 3 1 3 2 3 2 3 2 3 3 3 3 3 3

+ + + + + +∆ = ∆ ∆ = + + + + + +

+ + + + + +

l l ll l ll l l

ijn n n

C 1 1 |C |minus∆ ∆

∆ = = = = =∆ ∆ ∆ ∆ ∆

If 1 2

f (x) f (x)g (x) g (x)

1 21 2 1 2

∆= = minus

1 2 1 2

+prime prime

rr f (0 )

d f ( x )f (0 )

dx at x = 0

a a ab

(x)dx a b c

int int intint

+ =(50 40)40 45

+ =(b a) a ba

Statistics 32413

i1 2 3 ni 1

xx x x x

= + = + = minus

ux a u a u (x a)

minus= + = + =sum

i ii 1

i ii i

f xx a u a u (x a)

f ux a hu a h

f minus

x au h

i i1 1

f ux a u a u (x a)

f ux a hu a h

f minus

x au h

h defines mean

=+ + + +

1 1 2 2 3 3 n n

1 2 3 n

w x w x w x w xAM

w w w w

323 CoMbined Mean

+1 1 2 2

n x n xxn n

That is =

minus =sumn

i ii 1

Q =sumsum

1d ord

sumsum

i i i i

f (x d) d f xA dA

324 GeoMetriC Mean

Statistics 32415

rArr GM = Antilog

sum i if log x

N for ( ) gt1 2 n

f f f1 2 nx (x ) (x ) 0

sum i if log x

N for ( ) lt1 2 n

f f f1 2 nx (x ) (x ) 0 N = odd

325 HarMoniC Mean

n1 x 1 x 1 x

is given by =

i ii 1

(f x )

+ = + 2 a b 2ab( ab)

2 a b ie (GM)2 = (AM) (HM)

327 Median

Step II Median =

+ = + +

n 1 term for n odd2

Step III Median =

ththth

For thN

thN 12

2observation

N hC2 f

Remarks

4 f + minus times

N hD i C

N hP i C

328 Mode

Statistics 32417

the mode 0 1 3 3 5 5 7 8 mode = +=

3 5 42

Mode = minus

minus +

minus+ times

minus minusm m 1

m m 1 m 1

f fl h2f f f

minus +

+ times+m 1

m 1 m 1

fl hf f

minus +

minus+

minus + minus

m m 1i

m m 1 m m 1

h h ah h h h

L SL S

(Q Q )QD

(Q Q )

MP from mean = =

minussumn

x mean

minussumn

x median

MD from mean = =

minussum

i ii 1

f x median

f MD from median =

minussumsum

f x medianf

(i) Direct Method =

minus + minusσ = =

sum sumn

22 2ii ii 1

= ( )minus

2i i 2i

minus = minus

sum sum sum22 2

2i i ix x xx

Thus minus

σ = = minus

sum sum sum22 2

+ minus + σ =

sumsum ii

u(a u ) a

rArr minus

σ = = minus = =

sum sum sum22

i i ii i

(u u) u u u (x a)

Statistics 32419

minus σ = = minus

sum sum sumn

2 22ii ii 1

(u u) u uh h

σ = = minus

22 2i i i i i i

σ = minus

sum sumsum sum

22i i i i

f u f uf f

σ = minus

sum sumsum sum

22i i i i

f u f uh

Remark

3211 varianCe

SD 100 100Mean x

s or σ + σ + + σ σ + + σ +σ = =

2 2 2 2 2 2 2 21 1 2 2 1 1 2 2 1 1 1 2 2 2

121 2 1 2

+1 1 2 2

n x n xxn n

Contents

Preface

Acknowledgements


11 Introduction

12 Pre-Requisites





141 Statement




151 Conjectures








18 Truth Value

19 Quantifiers



Number System



111 Set of Integers





1132 Prime Integer




























120 Introduction


121 Polynomials







1232 Tangency








212 Laws of Indices













31 Definition



32 Series


33 Arithmetic Mean





35 Geometric Mean





37 Harmonic Mean



391 Standard Form




311 Properties


3121 Representation


314 Vn Method









431 Theorem







451 Monotonicity










54 Nature of Roots







58 Descartes Rule





61 introduction


621 Addition Rule








66 Rank of Words





682 Sum of Divisor



69 Combination






610 Distribution














71 Introduction

72 Binomial


73 General Term



74 Middle Term


76 Greatest Term






711 Tips and Tricks



82 Greatest Term

83 Taylor Expansion



833 Properties of e

834 Expansion of ex





91 Introduction

92 Angle









951 y = sin x

952 y = cos x

953 y = cot x

954 y = cosec x

955 y = sec x









99 Tips and Trics


101 Introduction















111 Introduction





115 MndashN Theorem



























131 Introduction












1361 Union of Loci













141 Definition
















151 Introduction




1523 Diametric Form






1531 Tangents



1534 Normals







Chapter 16 Parabola



162 Parabola










1651 Properties


Chapter 17 Ellipse

171 Definition


1721 Focal Distance







181 Definition




182 Director Circle











191 Introduction




192 Complex Number

193 Argand Plane



1933 Result












197 Theorem






19102 Explanation









201 Sets







208 Types of Sets

209 Subsets



2012 Power Sets

2013 Disjoint Sets

2014 Universal Sets


















2024 Relations







2031 Into Relation


































2117 Even Function


21172 Odd Function













Limit













Continuity















Differentiability






















233 Chain Rule



















Rate of Change






Tangent and Normal







2410 Common Tangent


Monotonicity





2416 Critical Point














Maxima and Minima

2428 Local Maxima

24281 Local Minima

















251 Introduction























261 Area Function





2651 Generalization



267 Applications







269 Beta Function

2610 Gamma Function

















271 Introduction



























Chapter 28 Vectors











286 Section Formula










291 Introduction


2912 Corollary















292 Skew lines


293 Plane









301 Experiments

3011 Event











311 Matrix

312 Sub Matrix

3121 Equal Matrices













3148 Unitary Matrix















319 Eliminant





3195 Caution





31104 Circulants










323 Combined Mean


324 Geometric Mean

325 Harmonic Mean


327 Median

328 Mode




3211 Variance


ISBN 978-93-325-2206-0

eISBN 978-93-325-3736-1

Contents

mdashSanjay Mishra

Preface

mdashSanjay Mishra

Acknowledgements

11 INTRODUCTION

12 PRE-REQUISITES

|| Parallel

f Null Set (phi)

141 Statement

Compounding with OR

volume decreases

18 TRUTH VALUE

T FF T

19 QUANTIFIERS

111 SET OF INTEGERS

1132 Prime Integer

11321 Properties

= 1 1 1a a 1 b b 1 c c 1p 1 q 1 r 1y

p 1 q 1 r 1

minus minus minus

( a 1)( b 1)( c 1)2

square and + + + +1

( a 1)( b 1)( c 1) 12

b d HCF (b and d)=

b d LCM (b and d)=

th th th

4 3 2 1 0

x = a 10 + b 10 + c 10 + d 10 + e 10

minus minus minus

le le minus le le

= a(104) + b(103) + c(102) + d(10) + e = 5m + e 0 le e le 9

5m e 0 e 4 5m e 0 e 45m 5 e 5 5 e 9 5m (e 5) 5 e 9

rArr n1 1 1(abc0) (a b c 0)=

rArr n1 1 1(abc1) (a b c 1)=

1 1 1(abc2) (a b c 4)= if n = 4k + 2

1 1 1(abc2) (a b c 6)= if n = 4k

1 1 1(abc3) (a b c 9)= if n = 4k + 2

1 1 1(abc3) (a b c 1)= if n = 4k

1 1 1(abc4) (a b c 6)= if n = 2k

1 1 1(abc7) (a b c 9)= if n = 4k + 2

1 1 1(abc7) (a b c 1)= if n = 4k

1 1 1(abc8) (a b c 4)= if n = 4k + 2

ie n1 1 1(abc8) (a b c 2)= if 4k + 3 n

1 1 1ie (abc8) (a b c 6) if n 4k= =

euler number (e) 1 1 1e 1 27 e 81 2 3

2 1 0 1 2 3

abcdeabcde10

= = = 1 2 n1 2 3 n n

ax x xx ax x x x10

eg 27 336 68 2

x if x 0| x | x 0 if x 0

x if x 0

gt= = =minus lt

Q x + 0 = x and 1

times =

1 2 1 3 2 3

= rArr =

+ + + += = = +

Properties

x if x 0| x | x 0 if x 0

x if x 0

gt= = =minus lt

n = k + 1

= or xu yz

b d+ +

b d+ = + rArr

a b c db d+ +

=minus minus

1nn n n

xa yc zexb yd zf

+ + + +

1 1 2 2

= = rArr 1 2 1 2

1 2 1 2

c c a a xb b b b

minus = minus

rArr 2 1 1 2

1 2 2 1

b c b ca b a b

minusminus

1 2 1 2

a c a cyb a a b

minus=

obtained 1 12 2 2

c a xa x b cb

minus+ =

2 1 1 2

b c b cxa b a b

minus=

1 2 2 1

a c a cya b a b

minus=

2 1 1 2

a c a cya b a b

minus=

2 1 1 2

b c b cxa b a b

minus=

y yz= we get

2 1 1 2 2 1 1 20 0

2 1 1 2 2 1 1 2

minus minus= = = =

minus minus

120 INTRODUCTION

real number)

c clt lt

a b If ab lt 0

rArr = lt

a b iff |a | | b |

gt rArr = = lt lt

2 22 2

ltisin isingt lt

a ba b

gtgt rArr gt

++++++++++++++++

+++++++++++++++++

step 2

minus lt minus gt

minus minus lt gt

minus lt

Q(x)S(x)minus lt

S(x)M(x)minus lt

helliphellip(ii)

Remarks

0Q( x )

(ii) If P( x )0

+ gt lt + + lt gt +

121 POLyNOMIALS

∵ n n 1 n 2 nn 2 n

gt rArr gt lt

f (x) f(x)0 f(x)g(x) 0 0g(x) g(x)

f (x)f(x)g(x) 0 0g(x)

gt le = ne

f (x) 0

lt = ne

cup [g d] cup b

a=sum = a1 + a2 + a3 ++ an

Properties

k kk 1

(a b )=

k kk 1

(a b )=

k kk 1 k 1

a b= =

sum sum and

k kk 1

k kk 1 k 1

ma m a= =

=sum sum = m (a1 + a2 + a3

k 1 2 3 nk 1

a a a a a=

Properties

=prodn

nn 1 2 n 1 2 n k

k 1 k 1

a ( a )( a )( a ) (a a a ) a

ie = = =

k k k k 1 2 n 1 2 nk 1 k 1 k 1

prodprod

kn1 2 3 nk k 1

nk 1 k 1 2 3 n

aa a a aa

b b b b bb

k k k kk 1 k 1 k 1

(a b ) a b= = =

31 2 n

1 2 3 n

1 2 k 1 22 k

2 1 2 n

12311 Chords

a b c + + = + + + + + = + + + + +

13 2 2a b a b

ab2 2+ minus

= minus

n times

212 Laws of Indices

(iv) minus=x

a ab b

le isin

x y x y if a 1a a

x y if a (01)

f(x y) f(x) f(y)x y

f(x y) f(x) f(y)

log alog a 0434

(b) a a a

(d) k |a|a

1log N log N

P 9 = =cb

P 10 logbalogab = 1

given below

System 1 gt

f(x) 0f(x) g(x)

System 1 gt ne =

g(x) 0g(x) 1f(x) g(x)

System 2 gt ne =

f(x) 0f(x) 1f(x) g(x)

System 1 gt gt =

g(x) 0f(x) 0g(x) h(x)

System 2 gt gt =

h(x) 0f(x) 0g(x) h(x)

System 1

gt gt ne =

f(x) 0g(x) 0g(x) 1g(x) h(x)

System 2

gt gt ne =

f(x) 0h(x) 0h(x) 1g(x) h(x)

gt ne gt ne =

h(x) 0h(x) 1g(x) 0g(x) 1f(x) g(x)

system gt

f(x) 0[f(x)] g(x)

system +

f(x) 0[f(x)] g(x)

system gt gt =

gt gt gt gt =

31 Definition

1 1 1 1 1 1 2 2 4 8 16 32

eg 1 1 1 1 2 4 8 16

32 SerieS

n kk 1

Caution lttn angt n

where = n n 1a a2

Sie f (n)S

= prime

then the ratio

t f 2k 1t

= minus

Tie f (n)T

= prime

S k 1fS 2

Points to Remember

33 Arithmetic meAn

= 1 2 3 na a a aAM

Remarks

= +sumn

nA ( a b)

= =minus minus

p Kn Kn

T TT TT

n k p k

1 2 3 n 1

aa a aa a a a minus

TrT minus

na(r 1) a(1 r )S(r 1) (1 r)

minus minus= =

a rS1 rminus

the series

rarrinfin

minus= = minus

minus ltminus

Remark

a2 b2 31 2

aa a b b b

1 1 1 a a a

are in GP

x numbers y numbers

R 0XXXX YYYY=

where 0x times

X XXXX=

and 0y times

Y YYYY=

0 0x y x

(10 10 )+

minus=

4 terms in GP 33

a a ararr r

35 Geometric meAn

1 2 3 nG (x x x x )=

rArr b = arn+1

1n 1br

1 1 1 a a d a 2d

a (n 1)d=

+ minus)

1 1 1(n 1)a b a

= + minus minus

+or a a b

c b cminus

=minus

a d a a d

a 3d a d a d a 3d

37 hArmonic meAn

j 11 2 n j

n nH1 1 1 1a a a a=

= =+ + + sum

1 1 rdH a

minus=

1 2 n 1n1 2 n

+ + +ge ge

nge rArr

rArr n

maxSPn

rArr 1

1 2 n n1 2 n

x x x(x x x )

n+ + +

ge rArr 1nS (P)

Remarks

391 Standard Form

minus minus minus

When | r | lt 1 2

general term

rArr n

k 1 2 3 nk 1

a a a a a=

= + + + +sum rArr n

= + + + + =sum

k kk 1 k 1

a a= =

λ = λsum sum

k k k kk 1 k 1 k 1

(a b ) a b= = =

k kk 1 k 1

a b s= =

k k k kk 1 k 1 k 1

a b a b= = =

knk k 1

nk 1 k

sumsum

Sn = T1 + T2 + T3 + +Tn = n

T=sum = 2

kak bk c

+ +sum = n n n

k 1 k 1 k 1

a k b k c 1= = =

+ +sum sum sum

nn(n 1)(2n 1) n(n 1)S a b cn

6 2+ + + = + +

311 ProPertieS

r 1 2 3 n=

= + + + +sum = n(n 1)2+ P2

n2 2 2 2 2

r 1 2 3 n=

= + +sum = n(n 1)(2n 1)6

n(n 1)r2=

+ = sum P4

nr (n 1)(2n 1)(3n 3n 1)30=

= + + + minussum

iji 1 j 1

3121 Representation

iji 1 j 1

31211 Conclusion

+ = = n 12

n(n 1) C 2

P1 n n n n

j 1 i 1 j 1 i 1

a a= = = =

sumsum sum sum =

i 1 i 1

na na 1= =

P2 n n n n n

1 i j n j 2 j 3 j ni 1 i 2 i n 1

= = = minus

2minus

P3 2n n

1 i j n

(n n)aa C a2

le le le

+= =sumsum

minus ++ =

P4 n n n n

i j i ji 1 j 1 i 1 j 1

a a a a= = = =

sumsum sum sum let

k 1 2 nk 1

a a a a S=

= + + + =sum = 2n n n

2i i k

i 1 i 1 k 1

Sa S a SS S a= = =

= = = =

sum sum sum

= i j 01 i j n

a a S (say)le lt le

=sumsum rArr 2n n

20 i k

i 1 k 1

2S a a= =

sum sum

k kk 1 k 1

=sum sum

k kn nk 1 k 1

i j1 j j n

a aa a

le le le

=sum sum

P7 n n n

i j ki 1 j 1 k 1

(a a ) 2n a= = =

0 i j n

k kk 1 k 1

i j1 i j n

a aa a

le lt le

=sum sum

sumsum

1 2 3 4 5 n 1

1 2 3 4 5 6 n 1 n

d d d d d d

Remarks

314 Vn methoD

n(n 1)=

eg n1 (n 1) n 1 1t

= = = minus+ + +

Step 3 rArr 1 1 21 1t V V1 2

= minus = minus

+rArr n 1 n 1

Chapter 4InequalIty

system lt ge

minus lt lt

systems lt ge

minus lt lt

system lt ge

minus lt lt

lt minus lt

minus gt lt

lt ge minus lt lt

Inequality 449

f(x) 0g(x) 0f(x) g (x)

g(x) 0f(x) g (x)

rArr minus minus

0 1 2 nn n 1 n 2

rArr a0tn + a1t

f(x)tg(x)

rArr = if (x) tg(x)

rArr lt lt lt gt

gt gt gt

rArr α

lt lt lt lt

0 h(x) 10 f(x) (h(x))h(x) 1 f(x) (h(x))

alog x 0 x 1a 1 a 1 (b)

alog x 0 0 x 10 a 1 0 a 1

alog x 0 0 x 1a 1 a 1 (d)

alog x 0 x 10 a 1 0 a 1

Inequality 451

gt lt lt

f(x) 1 0 f(x) 1

g(x) 1 0 g(x) 1

gt lt lt

f(x) 1 0 f(x) 1

g(x) 1 0 g(x) 1

lt lt gt

f(x) 1 0 f(x) 1

0 g(x) 1 g(x) 1

lt lt gt

f(x) 1 0 f(x) 1

0 g(x) 1 g(x) 1

f(x) g(x) f(x) g(x)g(x) 0 f(x) 0

(x) 1 0 (x) 1

f(x) g(x) f(x) g(x)g(x) 0 f(x) 0

(x) 1 0 (x) 1

f(x) g(x) f(x) g(x)f(x) 0 g(x) 0

(x) 1 0 (x) 1

f(x) g(x) f(x) g(x)f(x) 0 g(x) 0

(x) 1 0 (x) 1

ge ge+

a b 2abab2 a b

Remarks

If = sum ixA

= prod

Remark

+ + +31 2 n 1 2 n

1mm m m m m m1 1 2 2 n n

1 2 3 n1 2 n

forall ai gt 0

i i mi mii

431 Theorem

(i) + + gt

mm ma b a b2 2

if 0 lt m lt 1

(iii) + + =

mm ma b a b2 2

sum summm

i ia an n

sum summm

i ia an n

if m isin (0 1)

sum summm

i ia an n

Inequality 453

Then + + + + + +

gt + + + + + +

mm m m1 1 2 2 n n 1 1 2 2 n n

1 2 n 1 2 n

+ + + + +lt

+ + + + + +

mm m1 1 n n 1 1 2 2 n n

1 2 n 1 2 n

(a12 + a2

2 + + an2) (b2

1 + b22 ++ b2

a a ab b b

then + + + + + ge

(1 + an) lt minus n

1 2 3 n

(d) If + + + +1 2 3 n

of minus φ

α β =1 cossin sin

φ2sin2

and that of φ

α =β =2

mum value of ==

α αsumprodn n

k kk 1K 1

α +α1 2sin2

kk 1 max

Asin sinn

kk 1 max

Asin nsinn

451 Monotonicity

type f(x) = a0 + a1x + a2x2 + a3x

n n 1 n 2 nn 2 n

meets axis of x

Rational identity

5221 Conclusion

rArr 2b b 4acx

2a 2aminus minus

= 2 b ca x 2 x2a a

+ + + minus

b 4ac ba x2a 4a

minus + +

rArr 2D by a x

4a 2a + = +

and DY y4a

54 naturE of roots

minus +α = b D

2aminus minus

We proceed as below

(i) S β+α=

minusαβ

x + 1ab = 0

rArr 2a b b c c a

a b b c c atimes =

4a 2aminus minus

maxD by at x

4a 2aminus minus

f(x) = 0 Then b D2a

2aminus +

of the equation

(i) D gt 0 (ii) f(k1) f(k2) lt 0

58 DEsCartEs rulE

ie (x a) (x b) (x c) (x d)y4

4+ + +

= minus

minus + minus=

between a and b

a1 + a2 + a3 + + an = 1

a1a2 a3 an = 1

aaminus

f ( x ) f ( x ) f ( x ) f ( x )f ( x )

x a x b x c x k

Combination

61 introduction

621 Addition Rule

i 1 2 k ii 1i 1

A | A A A | | A |==

= cup cup cup =sum

where |Ai|

1 2 r ii 1

n n n n=

i 1 2 ri 1

A A A A=

i 1 2 r ii 1 i 1

A | A | | A | | A | A= =

n X m=

1 ii 1

S n(X )=

=sum r r

2 i j1 i 1 r

S n(X X )le lt le

n(X1 cap X2 cap X3)

nCr(n r)

=minus

nr C rr(n r)

= timesminus

up in n ways

Npqr n= rArr nNpqr

66 rank oF words

AAABNN A E G L R

13x AAABNN

32= rarr

x2 = 0 rarr AANN

32x AAN

22= rarr

x4 = 0x5 = 1 rarr A

1x 3 AEGER= rarr

2x 0 AEGER= rarr

3x 2 EGR= rarr

x4 = 1

circle by 0360

= minus

(i) minus =n

PC ( r 1)

(ii) minus =n

P1C ( r 1)

10 1 D n9

D n minus

rP if r n0 if r n

n 1k r 1

10 1 D P9

minusminus

k r 1k 1

n 1k r 1

D Pminusminus

r rn 1

k r 1k 1

10 1 D P9

minusminus

682 Sum of Divisor

= a 1 b 1 c 1p 1 q 1 r 1p 1 q 1 r 1

(c ndash g1 + 1)

69 combination

nCr(n r)

=minus

∵ nr

n(n 1)(n 2)(n 3)(n r 1)C Ir

(b) 0 = 1 n n0 n

nC C 1

n 0 = = =

and nC1 = n

r r 1 r 2n n n 1C ( C ) C r r r 1

7 n n n 2r r 1 r 2

r 1 (r 1(r 2)C C Cn 1 (n 1)(n 2)

+ + + = = + + +

8 nCr rCs = nCs

C n r 1C rminus

minus +=

n n n n nr 1 2 n

C C C C 2 1=

= + + + = minussum

from a group is r

Cminus

=sum or

kk r p

Cminus

rC1 + nndashrC2rC2

1 i n k 1 j m k

n A A n B B+ +

610 distribution

(n p)C np

+ += Hence

2m(m) 2

be made is =3m

mm m 3

we obtain 3mm m m

objects = p

(pq)(q) p

objects = p

(pq)(q)

1 2 3 n

r x x x x

so on = 1 2 3 n

1 2 3 n 1 2 3 n

( ) (r)

λ +λ +λ + λ=

is n1 1 1 1 1n 1 ( 1)1 2 3 4 n

1 2 3 n 1 n

n permutation

nn n n ( 1) n1 2 3 n

= minus minus

minus + minus +

n 11 1 1 ( 1)n 1 2 3 n

= minus minus

minus minus + +

n 11 1 1 ( 1)n n 1 2 3 n

n1 1 1 1 ( 1)n 1 1 2 3 4 n

= minus

minus + +

n1 1 1 ( 1)n 2 3 4 n

(n 1)separators

1 1 1 ( 1)C (n r) 2 3 4 (n r)

k r nk

( 1) C (r k)minus

minus minussum

k 1 r nk

( 1) C (r k)minus

nP if r n

0 if r n

k 1 r nk

( 1) C ( r k )minus

minus minussum

k r nk

( 1) C ( r k )minus

minus minussum

rArr s

n n nE (n) p p p

= + + +

ple lt

Conclusion ie np

n n nE (n) p p p

= + + +

71 IntroductIon

72 BInomIal

n n n r rr

(a x) C a xminus

+ =sum and

nn r n n r r

minus = minussum

bull n

n r nr

(1 x) x C=

+ =sum n

n r n rr

(1 x) ( 1) C x=

73 General term

74 mIddle term

n2 yn2

th and n 32+

2 2n 1C x y

minus +

and n 1 n 1

n 2 2n 1

C x y+ minus

n n r rr

C a xminus

nr n n r r

( 1) C a xminus

minussum

n n 2r 2r2r

2 C a xminus

=sum where

= minus =

n 1 if n is odd2

++ = +

n n 2r 1 2r 12r 1

=sum where

= + minus = +

n if n is even2m 1

n 1 if n is odd2

76 Greatest term

r 1 rn r 1

minusminus

minus += =

T 1T+ ge

n r 1 | x| 1

rminus +

T C n r 1 n 1 1T C r r+

minus gt

rArr n 1 r2+

gt (i)

r(n r)=

n(n 1)(n 2)(n 3)(n r 1)C Ir

bull ( )n n 1 n 2r r 1 r 2

n n n 1C C Cr r r 1

(r 1)(r 2)r 1C C Cn 1 (n 1)(n 2)

+ ++ +

+ ++ = = + + +

r r 1C Cr 1 n 1

bull n

n n rr

(1 x) C x=

r + + nCnxn (i)

bull n

n r n rr

(1 x) ( 1) C x=

n (ii)

bull n

C x C C C C 2=

= + + + + =sum

bull n

( 1) C C C C ( 1) C 0=

bull n

2 n 2 2 2 2r 1 2 3 n

r C 1 C 2 C 3 C n C=

= + + + +sum

bull 0r 1 2 nCC C C Cr 1 1 2 3 n 1

= + + + ++ +sum

bull nn n

r 1 k 0r 1

+= =sum sum

n x x xr r r

711 tIps and trIcks

minus +

+ = + + + + + + infin2 3 r

a n(n 1)1 n (ax)x 2

n(n 1)(n 2) (ax) 3

x n(n 1) x1 na 2 a

n(n 1)(n 2) x 3 a

minus + + +

minus minus +

Remarks

82 Greatest term

T 1T+ ge gives

| x |(x 1)r m| x | 1

+(say)

Infinite Series 885

| x | 1+

Remarks

2 xr + helliphellip

83 taylor expansion

1 2 3+ = + + + +

= + + + +

1 2 3= + + + +

1 2 3minus minus

= + + + +

(i) 3 5x xsin x x

3 5= minus + +

3 15 315= + + + +

(iv) 3 5 7

2 3 4x x xn(1 x) x ( 1 x 1)2 3 4

833 Properties of e

834 Expansion of ex

For x isin R 2 3 r

x x x x xe 1 1 2 3 r

= + + + + + + infin or n

(i) 2 3 r

x x x xe 1 1 2 3 r

= + + + + +infin =sum (ii) 2 3 r r

x x x ( 1) xe 1 1 2 3 r

infinminus

(iii) x x 2 4 6 2r

e e x x x x1 2 2 4 6 (2r)

minus infin

+= + + + + =sum (iv)

x x 3 5 2r 1

e e x x x x2 1 3 5 (2r 1)

minus +infin

minus= + + + =

(v) r 0

1 1 1e 1 1 2 r

1 1 1 ( 1)e 1 1 2 3 r

infinminus

(vii) 1

e e 1 1 1 11 2 2 4 6 (2r)

minus infin

e e 1 1 1 12 1 3 5 (2r 1)

minus infin

x x x ( 1) xx 2 3 4 r

minusinfin

(i) 2 3 4

ex x xlog (1 x) x ( 1 x 1)2 3 4

(ii) 2 4 6

ex x xlog (1 x)(1 x) 2 ( 1 x 1)2 4 6

(iii) 3 5

e(1 x) x xlog 2 x ( 1 x 1)(1 x) 3 5

91 INTRODUCTION

92 ANgle

bull D G R

180 200= =

than 180deg

nminus π

Name of Polygon

Number of Sides

Name of Polygon

Number of Sides

Name of Polygon

Number of Sides

Name of Polygon

Number of Sides

Name of Polygon

Number of Sides

Name of Polygon

Number of Sides

θ = = Base xcos

Hypotenuse rθ = =

θ = =

θ = = Base xcot

Remark

cos(ndashq) = cosq

sec(ndashq) = secq

Similarly

cos sin

2π minusθ = θ

tan cot

2π minusθ = θ

cot tan

2π minusθ = θ

cos sin

2π + θ = minus θ

tan cot

2π +θ =minus θ

cot tan

2π +θ =minus θ

cosec sec

2π minusθ = θ

cosec sec

2π + θ = θ

le θ lt Then

(n 1)2

(iii) ( )(n 1)

Angles Functions

p 76π 4

53π 11

951 y = sin x

Properties

third quadrants

952 y = cos x

y-axis

9521 y = tan x

953 y = cot x

Properties

954 y = cosec x

955 y = sec x

Properties

1 sin x cos x+

=minus

sec x tan xminus =

θ =+ θ2

cotcos

θθ =

+ θ 1

tancot

θ =θ

1 tan A tanB+

+ =minus

1 tan A tanBminus

minus =+

minus+ =

+minus =

2 tan Atan2A1 tan A

=minus

where A (2n 1)4π

3 1 cos A Atan

sin A 2minus =

where A ne (2np)

where A ne2np

minus =+

+= 21 cos2A sin A

2minus

minus=

2 2+ + + =

2 2+ minus + =

++ = where AB n

ne π+

+ + = +

(ix) A B C2π

( )n 1 nsin sin2 2

minus β β α +

( )n 1 ncos sin2 2

minus β β α +

sin2 Acos2 A2 sin A

θ θ θ θinfin =

θ2 n 1

5 sin 2212

deg ( )1 2 22

minus 6 cos 2212

deg ( )1 2 22

7 tan 2212

deg 2 1+

= cos 72deg 10 cos 18deg 10 2 54+

= sin 72deg

11 sin 36deg10 2 5

4+ = sin 54deg

13 sin 9deg 3 5 5 54

+ minus minus

or cos 81deg

14 cos 9deg3 5 5 5

4+ + minus

99 TIPs AND TRICs

sin2 Acos2 A2 sin A

2 2 2α+β β+ γ γ +α

equation

101 IntroductIon

θ = rArr 4π

θ = or 3 9 11

4 4 4 4π π π π

Parallely cos 1x2

θ = rArrθ = + π

θ = + isin

cos 0 (2n 1)2π

tan q = 0

θ = rArrθ = + tan 1 (4n 1)4π

ndash +ndash

st sin a = k

+ndash

st tan α = k

st sin2α = k

cos2 α = k

ndash +

tan2 α = k

ndash+ndash

θ = + isin

n isin ℤ

rArr x (4n 3)2π

= + or 2n n3π

rArr x (4n 3) n2π

And 5sin7

minus = rArr

3xsin 02=

2nx n3π

rArr x2sin sin x 0

4 2 4π π + minus =

rArr x m4π

= π+ hellip(ii)

3 2 4π π π

= πminus π+ isin

rArr nx3π

= or x (2n 1) n Z2π

= + isin

minusφ =

such that

rArr ccos( )r

ccos( )a b

θminusφ =+

rArr cos x 1

4π minus =

rArr x 2n n

4ππ= + isin

x1 tan

2cos xx

1 tan2

minus=

2sin xx

1 tan2

= (ii)

rArr 5x 2n n4 2

rArr (8n 2)x n

4minus

(8n 2)x 2m m n ~ 2m m 4p 3p

or (8n 2) (8n 2)x 2m m n ~ n 5p 4p

2 tan x 2sin x

1 tan x 2 2

1 tan x 2cos x

1 tan x 2minus

2 tan x 2tan x

1 tan x 2=

21 tan x 2cot x

2tan x 2minus

-1 le x

10 le 1

rArr -10 le x le 10

of solutions are 7

-3π -2π2π

3π-π

(-frac1234)

frac12 10

(101)(3 3 10)ππ(2 2 10)ππ

( 10)ππ

sinx gt 12

for p6 lt x lt 5p6

6 6π π

π+ lt lt π+

52n 2n6 6isin

π π π+ π+

1 3 1sin15

2 2minus

deg = 2 3 1cos 15

5 ( )1 1sin 22 2 22 2

= minus

6 ( )1 1cos22 22 2 22 2

deg = = +

7 1tan 22 2 12

= minus

8 1cot 22 2 12

deg = = deg 10 10 2 5cos18 sin72

deg = = deg

11 10 2 5sin36 cos54

4minus

deg = = deg 12 5 1cos36 sin544+

deg = = deg

13 3 5 5 5sin9 cos81

4+ minus minus

deg = = deg 14 3 5 5 5cos9 sin81

4+ + minus

deg = = deg

triangles

111 IntroductIon

1111 Sine Formula

1112 Cosine Formula

2 2 2b c acos A

2bc+ minus

= 2 2 2a c bcosB

2ac+ minus

= 2 2 2a b ccosC

2ab+ minus

2 acminus minus

= (s a)(s b)sin

2 abminus minus

(ii) s(s a)Acos

2 bcminus

= s(s b)Bcos

2 acminus

= s(s c)Ccos

2 abminus

2 s(s a)minus minus

=minus

(s a)(s c)Btan

2 s(s b)minus minus

=minus

and (s a)(s b)Ctan

2 s(s c)minus minus

=minus

ca ca∆

(i) minus minus

=(s b)(s c)Asin

2 bc (ii)

s(s a)Acos2 bc

minus=

2 s(s a)minus minus

=minus

2bc+ minus

2ac+ minus

= or bsin AsinBa

= and C from

= = ie a sinBbsin A

= and a sinCcsin A

= rArr bsin AsinB

a= hellip (1)

Also ADsinCb

= rArr AD = b sin C

1 a bsinC2

∆ = = =

∆ = = = (s b)(s c) s(s a)

bcbc bc

minus minus minus

2 sin A

= 2a sinBsinC

2 sin A

Thus 2a sinBsinC

2 sin A∆ = =

115 mndashn thEorEm

11511 Circumcircle

a b c RA B C= = =

1 sin2

11513 Incircle

minus = C(s c)tan2

=minus

2r s b∆

=minus

3r s c∆

=minus

cos(A 2)= 2

cos(C 2)=

tan A tan A

= =degminus

ang ACB + 12

ang ACM = 12

Remarks

= = and 2CG CF3

= + minus2 2 21AD 2b 2c a2

2 2 21BE 2c 2a b2

= + minus

γ= = (Let AD = x)

γ = =+ minus

a sinBsin2b 2c a

β =+ minus

θ = =+ minus

+= = =

acxb c

and abyb c

= = = + + (ii)

π a ar cot

2tan n 2 nπ

= 1 (ab cd)sinB2

2 4A2ab cd

+ + +=

Remarks

π= and cos 1

π minus =

and cosndash1 1

ne minus4π

ndash1

ndashπ2 π2

y=sinx

ndashinfin infinO

π2 (1π2)

ndash1 O 1x

ndash110

(1 π2)

(ndashπ2ndash1)

(ndash1ndashπ2)

(π21)

π2ndashπ2 X

ndash10

y = cos x

Y(ndash1π) π

(1 0)X0

y = cosndash1x

1ndash1

(ndash1π)

(01) 1

ndash1ndash1

(0π2) π2

(π20) (πndash1)(1 0)

Y infin

Xndashπ2

ndashinfin

ndash1

ndashπ2

ndashinfin infinO x

ndashπ2

ndashπ2ndashinfin

ndashinfin

+infin

infin xx

xndashπ2 π

ndashinfin

+infin

πndashinfin

ndashπ2π2

ndashinfin

+infin

0ndash1

π2 xπ

+infin

ndashinfin

O 1ndash1

infinndashinfin

Y+infin

+infin

ndashinfin

0 1ndash1

ndash1

ndashπ2ndash10 π2

+infin

ndashinfin

π2(1π2)

(π21)

(ndashπ2ndash1)

(ndash1 ndashπ2)

π2ndashπ2

ndashπ2

ndash1

+infin

x if x [ 3 2 2]x if x [ 2 2]

x if x [ 2 3 2]2 x if x [3 2 5 2]

Ondash1

ndash1

O45deg

ndash1

ndash11

minusπ 2 y=x

πminus

y=-(x)

πminusy=x-2

minus3π

y=ndash(3π+x)

minus5π2π2

3π25π2πminusπ

y=sinndash1(sinx)

Period 2π

ndash3π

ndash3π2

π23π2

5π22π 3πx

ndashπ2ndashπ πO

y y=sinndash1(sinx)

y=2π+

y=(3π+x

x if x [ 0]x if x [0 ]2 x if x [ 2 ]

2 x if x [2 3 ]

and so on as shown

y=2πndashx

y=xndash

2πy=ndash(x+2π)

y=ndashx

y = cosndash1(cosx)

( )( )( )( )

x if x 3 2 2

x if x 2 2

x if x 2 3 2

x 2 if x 3 2 5 2

isin minusπ π

and so on as shown

π2 3π2π

minusπminus2π2πO

y=tanndash1(tanx)

Period π

x 2 for x ( 2 )x for x ( 0)

x for x [ 0]

x for x [0 ]~232 x for x [ 2 ]~2

y=2πndashx

2π y=ndashx

y=secndash1(secx)

3( x) for x ~ 2 2

x for x ~ 02 2

3x for x ~ 2 2

minusπ 2

πminus

y= ndash (x)

π+y=2x

y=x ndash2ππ 2

y=xndash

y=cotndash1(cotx)

gtminusπ+ lt

cos 1 x if 0 x 1

cos 1 x if 1 x 0

xtan1 x

1 xcot if 0 x 1x

1 xcot if 1 x 0x

minuslt le

1sec if x 0 11 x

1sec if x 1 01 x

= 1 1cosecx

(b) minus

sin 1 x for x 1 0 =

1 xtan for x (0 1]x

minus isin

minusπ+ isin minus

minus isin minus

xminus isin minus

(c) 1 1

xtan x sin for x1 x

minus minus = isin

1cos for x [0 1]1 x

minus isin minus +

1cot for x 0x

minusπ+ lt

= ( )( )

sec 1 x for x 0

+ gtminus + lt

minus + isin

1sin for x 01 x

+ = πminus le +

xcos x1 x

minus forall isin

1tan for x 0x

π+ lt

1 1 xsec x ~ 0x

minus + forall isin

( )( )

cosec 1 x for x 0

+ gtπminus + lt

x 1sin for x 0x

sec xx 1sin for x 0

minus gt =

minus π+ lt

= 1 1cos x ~ 0x

minus forall isin

= ( )1 2

tan x 1 for x 0

1cot for x 0 x 1x 1

xcosec for x 0x 1

π+ lt minus

minus minus= isin =

x 1cos for x 0x

minus gt

minus minusπ+ lt

1tan for x 0 1x 1

= ( )( )

cot x 1 for x 0

xsec for x 0 1x 1

Property (1)

1 1sin (2x 1 x ) if x2 2

1sin (2x 1 x ) if 1 x2

minus minus

minus minus le le

Property (2)

1 1sin (3x 4x ) if x2 2

1sin (3x 4x ) if 1 x2

minus minus

cos (2x 1) if 0 x 12 cos (2x 1) if 1 x 0

12 cos (4x 3x) if 1 x2

Property (5)

2xtan if 1 x 11 x

2xtan if x 11 x

for x 12

3x x 1 1tan if x1 3x 3 3

1for x2 3

2xsin if 1 x 11 x

2xsin if x 11 x

1 xcos if 0 x1 x

1 xcos if x 01 x

minusle ltinfin +

Property (1)

2 21 2 2

1 1 1 2 2 2 2

1 2 2 2 2

minus minus minus

+ leminus + minus

Property (2)

2 21 2 2

1 1 1 2 2 2 2

1 2 2 2 2

minus minus minus

Property (3)

minus minus minus

minus le le + le

Property (4)

+ minus minus

Property (5)

x ytan if xy 11 xy

+lt minus

πminus lt lt =

Property (6)

x ytan if xy 11 xy

minusgt minus +

Chapter 13point and

Cartesian system

131 IntroductIon

x y 1x y 1 0x y 1

∆ = =

are 1 2 1 2x x y y

2 2+ +

m n m nminus minus

minus minus

1 1λ λ λ

then 1 1

ax by c

ax by cλ

+ += minus + +

Denoted as ym x

rArr 2 1

y ym tanx xminus

= θ =minus

1 2 3 2 3 1 3 1 21[ x (y y ) x (y y ) x (y y )]2

x y 11 x y 12

x y 1∆ =

x y 1 0

be given by [ ]2 3 1 1 3 1 1 3 1 2 2 1

x y 1x y 11x y 12x y 1

Area of polygon =

x yx yx y

x yx y

= 1 2 2 3 n 1 1 2 2 3 n 11 |(x y x y x y ) (y x y x y x )|2

+ + + minus + + +

x yx y1x y2x y

∆ = = 1 2 2 3 3 1 1 2 2 3 3 11 |(x y x y x y ) (y x y x y x )|2

+ + minus + +

∆ = minus

as S S = 0 ie f(x y) g(x y) = 0

rArr f(a b) + lg(a b) = 0 + l0 = 0

4T(x y) (x y)rarr

5T(x y) (x y )rarr

6T(x y) (x y )rarr

helliphellip (i)

2 = (x ndash x2)

2 + (y ndash y3)2

3 3+ + + +

a b c a b c+ + + + = + + + +

rArr A

(b a)y cyy

b a cminus +

=minus +

rArr A

B A CI

bx ax cxx

b a cminus +

=minus +

and minus +=

minus +A

B A CI

by ay cyy

141 Definition

2 11 1

minus = minusminus

or 1 1

x y 1x y 1 0x y 1

determinant form

x y 1a b+ = or

x y 1a 0 1 00 b 1

in determinant form

ysin(p+a) = p

= =θ θ

Given two lines

11 1 1 1 1

+ + = = minus = α

22 2 2 2 2

+ + = = minus = β

rArr 2 1

m mtan1 m m

minusθ =

rArr 1 2 2 1

1 2 1 2

a b a btana a b b

minusθ =

14131 Conclusion

m mtan1 m m

minusθ =

+ and 2 1

m mtan( )1 m m

rArr 1 1 1

a b ca b c

rArr 1ay x c

bminus

m mtan1 mm

minusθ =

+ rArr m m tan

1 mmminus

= plusmn θ+

rArr m tanm1 m tan

plusmn θ

1 m tanθ

rArr 1 1

ax by cmn ax by c

+ += minus

ax by c 0ax by c

rArr 1 1

ax by c 0ax by c

rArr 1 12 2

|ax by c |PNa b

given by 2 2

= =θ θ

required distance

1 2 2 1 1 2 2 1

minus minus

rArr 1 1 1

a b ca b c 0a b c

a + b a + b

+ += plusmn

Two Cases Arise

c1 c2 gt 0 Then 1 1 1 2 2 22 2 2 21 1 2 2

a x b y c a x b y c

a b a b

+ + + += +

ie 1 1 1 2 2 22 2 2 21 1 2 2

a x b y c a x b y c

a b a b

+ + + +=

+ + and 1 1 1 2 2 2

2 2 2 21 1 2 2

a x b y c a x b y c

a b a b

+ + + += minus

a x b y c a x b y c

a b a b

+ + + +=+

and 1 1 1 2 2 22 2 2 21 1 2 2

a x b y c a x b y c

a b a b

+ + + += minus

Tips and Tricks

2 1 2 1 1 12 2

= = minus ++

rArr ( )

I M2 2

2(aa bb )L L 0a b

a b ca b b 0a b c

a h gh b f 0g f c

∆ = =

1h h abm

bminus + minus

= and 2

2h h abm

bminus minus minus

1 2 (h ab)tan|a b |

minus minus

θ = +

2 2x y xya b h

+ c = 0 is 2 2

or 2 2

f bc g cah ab h ab

=minus

2 2 2 2

a b h a b hα β

= = minus minus

Tips and Tricks

+ 2gx + 2fy + c = 0

rArr ( )

S 0x α β

part = part and

part= part

rArr 2

hf bgab hminus

α =minus

af gh h abminus

β =minus

( ) + +

Homogeneous

2 2 3 3

x y 1 x y 1x y 1 x y 1 0x y 1 x y 1

1 1 1 1

2 2 3 3

x y 1 x y 1b x y 1 c x y 1 0

x y 1 x y 1+ =

Family oF CirCle

151 introduction

minus minus

Centre = G F

minus minus

A+ minus

A(x1 y1) B(x2 y2) C(x3 y3) is

2 21 1 1 12 22 2 2 22 33 3 3 3

x y x y 1x y x y 1

0x y x y 1x y x y 1

= =θ θ

straight line

(x1y1) (x3y3)

(x2y2)

S1 = x12 + y1

= 2 21 1 1 1 1(x y 2gx 2fy c S+ + + + =

c c minus

n n minus minus

rArr 2 12 1

2 1t x x

2 1y y

y ym limx xrarr

minus=

minus =

2 12 1

x x1 2 1y y

x x xlimy y yrarr

+minus = minus

( )11 1

xy y x xy

Q 2 2 21 1x y a+ = (1)

2 2 22 2x y a+ = (2)

2 1 1 2

y y x xx x y y

minus += minus

a cos2x

=αminusβ

a sin2y

=αminusβ

Q(h k)

(x1 y1)

1534 Normals

y fmx g+

y f(y y ) (x x )x g

x y 1x y 1 0g f 1

=minus minus

Slope of chord = 1

x(y y ) (x x )y

1 1 1 1yy xx x y+ = +

minus minus

minus +

common tangents

= 1 1 2

r r2sinr r

minus minus

= 1 1 2

r r2sinr r

minus minus= π

2 1 1 2 2 1 1 2

1 2 1 2

minus minus

minus minus θ =

r r2sinr r

minus minus= π

2 2 21 2

r r dcos2r r+ minus

2 2 21 1 2

r r dcos2r r

minus + minus

rArr θ =

2 2 21 2 r r d+ = 2 2 2 2 2 2

1 2 1 2 1 22(g g f f ) c crArr + = +

x2 + 2gx + c = 0

(x2y2) 1 2 1 2 1 1

x y 1(x x )(x x ) (y y )(y y ) x y 1 0

Remarks

Tips and Tricks

Chapter 16parabola

Parabola 16163

Degenerated Conics

= = or

SP = e PM 2 2

|ax by c |(x ) (y ) e(a b )

+ where P(x y) is a

2 2 22 2

(ax by c)(x ) (y ) e(a b )+ +

Condition Conic

162 PArABoLA

as directrix

Parabola 16165

y2 = 4axx = at2

y = 2at

x2 = 4ayy = at2

x = 2at

2 4a 4acy y 0m m

rArr + = =

16a a cm 0m

minus =

Parabola 16167

rArr ay mx m ~ 0m

y2 = 4ax

a 2am m

hArr rArr =

= + rArr = +

2 1 1 2

y y 4ax x y yminus

=minus +

y yABx xminus

=minus

= 1 21 2

4a 2ay yy y

4ay y x xy y

minus = minus+

2Slope of chordt t

2y 2at x att t

minus = minus+

a 2aQ t t

aSQ l at

t rArr = +

2ty x at 1

= minusminus

rArr yt = x + at2

ie ST = SPSQ

Parabola 16169

1651 Properties

(i) t1t2 = 2 (ii) t3 = ndash(t1 + t2)

h 2am say 3aminus

Parabola 16171

a 2am m

minus minus

= +ay mxm

a 2ah km m

minus minus

ac mh km

+ = +ac mh km

ay mx mh km

y2 = 4

axy2 =

ndash4a

x2 = ndash

)2aminus

minus=

minus1

2aminus

=minus

xminusminus

=minus

xminus

=minus

ric co

t2 2at

)(ndash

ndashat

ric fo

y ndash

x ndash

m2 ndash

)(ndash

ityc =

ndash2a

m ndash

m=minus

ndash 2a

m ndash

minusminus

(y ndash

x ndash

ndash k

)2 = ndash

ndash h

)2 = 4

ndash k)

(x ndash

ndash4a

(y ndash

2aminus

minusminus

=minus

2aminusminus

=minus

minusminus

=minus

minus1

hminus

=minus

ndash k

) + t(

x ndash

t + at

ndash k

) ndash t(

x ndash

t + at

ndash h

) + t(

y ndash

t + at

ndash h

) ndash t(

y ndash

t + at

t of c

ndash 2

(h ndash

ityc =

k ndash

ndash 2a

m ndash

k ndash

minus+

=minus

minusminus

(y ndash

ndash h

)ndash 2

ndash am

ndash k

(x ndash

minus=

minus+

minus=

minusminus

Chapter 17ellipse

171 Definition

aSP ePM e h a ehe

= = minus = minus

h k 1a a (1 e )

+ =minus

x y 1a b

Eccentricity 2

= minus

= a + eh

Length of 22bLR =

Ellipse 17175

x y 1a b

+ = where b gt a

x = a ndashae

x yS 1a b

a cos y 1a b

Ellipse 17177

x yS 1 0a b

θ+ φ = minus

=+ minus

θ+ θ =

xx yyT 1 0a b

2 2 2 2

a x b y a b a e x y

minus = minus =

Tangent at (x1y1)2

b xa y

minus 1 12 2

xx yy 1 0a b

+ minus =

Tangent at qb cota

θ+ θ =

Normal at (x1y1)2

a yb x

2 22 2

1 1 2 2a e

a x b y a bx y

minus = minus

Normal at qa tanb

x y 1a b

+ = is

a cos bsin2 2

cos cos2 2

θ+ φ θ+ φ

θminusφ θminusφ

Ellipse 17179

xx yyT 1 0a b

= + minus =

x y 1a b

+ minus

xx yy 1a b

+ minus

1 1 1 12 2 2 2

xx yy x y1 1a b a b

+ minus = + minus

rArr 2 2

1 1 1 12 2 2 2

xx yy x ya b a b

x y 1a b

b xya m

= minus

Chapter 18hyperbola

181 Definition

h k 1a a (e 1)

rArr minus =2 2

x y 1a b

Eccentricity = +2

Hyperbola 18181

1bL aea

= minus

bL aea

Length of =22bRR

rArr No locus

x y 1a b

2a beb

and length = 22a

+ = + =+ +

2 2 2 2 2 22 1

(x ) (y ) 1a b

( ) + +minus + + + minus =

1 1 11 1 22 2 2 21 1 1 1

= plusmn+

1 1 22 21 1

m x l y n aem l

Hyperbola 18183

= plusmn+

1 1 22 21 1

m x l y n aem l

minus += plusmn

+1 1 2

2 21 1

m x l y n aem l

x y 1a b

3[02 ) 2 2

182 Director circle

x y 1a b

(a gt b)

x y 1 0a b

rArr = minus2

2 2 22

by (x a )a

yxS 1 = 0a b

x y 1a b

as the equation

2 2a m b c c

is θ θ =φ φ

1 etan tan 2 2 1 e

or +minus

1 e1 e

Tangent at (x1y1)2

b xa y

minus minus =1 12 2

xx yy 1 0a b

a yb x

2 22 2

1 1 2 2a e

a x b y a bx y

θ+ θ = +

x y 1a b

θminusφ

=θ+φ1

cos2x a

θ+ φ

=θ+φ1

sin2y b

Hyperbola 18185

xx yyT 1 0a b

22 22 21 1 1 1

1 2 2 2 2 2 2 2

2 21 1 1 1

1 2 2 2 2

1 1 1 12 2 2 2

xx yy x ya b a b

rArr minus =

b a m 0 and 2 2 2 2

c a m b= minus

rArr = plusmnbma

and c = 0

= minus =

As minus =2 2

x yH 1a b

and minus =2 2

x yA 0 2a b

minus = minus2 2

x yC 1a b

2ab btan 2tana b a

or 2 Secndash1(e)

rarrx yx

2 and minus +

rarrx yy

minus =2 2

2(x y) (x y) a2 2

c ndashc)

Hyperbola 18187

1m t t

(t1 + t2)

t t 21 t t

x y 2x y

x y 1a b

minus =2 2

x y 1a b

km = b2a2

minus =

x y 1a b

minus + =

x y 1 0a b

191 IntroductIon

192 complex number

193 ArgAnd plAne

vector iy

= + = +

θ θ θ

cos 1 2 4 6

sin 3 5 7

(cosq + isinq) = ( ) ( ) ( ) θθ θ θ

+ θ+ + + + infin =2 3 4

ii i i1 i to e

gives q = f (say)

vector =

1 2 1 2|OA| |OB| |BA| |z | |z | |z z |

vector of z1 and z2

1 While BA

= + minus+ isin

+ +1 2 1 2 2 1 1 2

2 2 2 22 2 2 2

(x x y y ) i(x y x y ) C(x y ) (x y )

Then we get =2 2

z rz r

and OB

2 = = = +2 2 2 2zz | z | | z | (R(z)) (I(z))

7 += = =

z zR(z) x R(z)2

8 θ minus θ +

i ie ecos2

θ minus θ minus

i ie esin2i

10 =1 2 3 n 1 2 3 n(z z z z ) (z )(z )(z )(z )

11 = 11 2

(z )(z z )(z )

14 + =1 2 2 1 2 1z z z z 2R(z z )

15 + = + +2 21 2 1 2 1 2| z z | | z | | z | 2Re(z z )

16 |z1 + z2|2 + |z1 ndash z2|

2 = 2(|z1|2 + |z2|

1 2 1 2| z | | z | 2Re(z z ) or | z1 |2 + | z2 |

2 plusmn 2 | z1 || z2 |

cos (q1 ndash q2)

13 | z1 + z2 |2 = | z1 |

2 + | z2 |2 hArr 1

is purely imaginary

14 | z1 + z2 |2 + | z1 ndash z2 |

2 = 2| z1 |2 + | z2 |

2 15 | az1 + bz2 |

2 + | bz1 ndash az2 |2 = (a2 + b2)(| z1 |

minusθ = minusπ+

19433 Caution

2 if x 0 y 02 if x 0 y 0

= minus

zarg arg z arg zz

= plusmn π

sumprod

k kk 1k 1

Arg z Arg z 2k

zArg 2Arg (z)z

plusmn +

| z | a | z | ai

2 2 b gt 0 and

+ minusplusmn minus

| z | a | z | ai

2 2 b lt 0

rArr = minusb 152 2

5 3i 18 15i (5 3i)2 2 2

P(1) |w| = |w2| = 1 P(2) ω=ω2

Remarks

∵ 2ω ω= ∵ 1 0ω ω+ + =

∵ 2ω ω= and 3 4 2 2 21 ( ) ( )ω ω ωω ω ω ω= = = = =

∵ 2 21 1 ( )ω ω ω ω+ + = + + ∵ 21 ( ) 0ω ω+ + =

P(6) +ω +ω =

P(9) π

ω = minus + =

1 3iarg( ) arg2 2 3

ω = minus minus =

2 1 3 4arg( ) i2 2 3

terms of i w w2

i 1 i 3 2 1 i 3 23 i (1 i 3) 2i2 2i i 2 i

(2k + ) (2k + )cos i sin n n

k = 0 1 2 n ndash 1

1 2 n ndash 1

= ππ+ π+ + = =

n n 12n ndash 1=

an ndash 1 where π

α =2ine

P(2) 1 + a + a2 + + an ndash 1 = 0

conjugates and π

α =2ine

m2m 1 2m 11 2m 2m 1 m m 1

1 1 11

+ +minus + +

= = = = = = = = =

2cis cisn m

2kcisn n

= π θ

where π2kcisn

rArr π= minusnx

2 rArr

πminus

= = = = = =

minus = minus

rArr i3 1

z zArg Arg( e )

z zθ minus

= θ = ρ minus

ρ =minus

z zz z

If z1 = 0

minus minusi1 1

z z | z z | m e 1z z |z z | n

+1 2nz mzzm n

then minus=

minus1 2nz mzzm n

minus π π= minus

minus 3 1

z zArg

z z 2 2 (i)

rArr π

i3 1 3 12

2 1 2 1

z z z ze iz z z z

rArr minus

= minus

z zR 0

rArr minus minus

+ =minus minus

3 1 3 1

2 1 2 1

z z z z0

= plusmnminus

minus 3 1

3 1 2 12 1

= = minus minus

3 1 1 2

2 1 3 2

(iv) + + = + +2 2 21 2 3 1 2 2 3 3 1z z z z z z z z z

(v) πminus

= = +minus

i1 2 3

z z 1 3e iz z 2 2

(vi) + + =minus minus +2 3 3 1 1 2

1 1 1 0z z z z z z

minusminusminus =

minus minus 2 32 4

1 4 1 3

rArr minusminus

= minus minus

1 32 4

1 4 2 3

z zz zArg 0z z z z

minus minus

+ = π minus minus

2 3 1 4

1 3 2 4

minus minus

= π minus minus

1 3 2 4

2 3 1 4

AB rArr = λ

2 1AP (z z )

∵ = +

19631 Conclusion

Writing + minus= =

z z z zx y2 2i

etc in minus minus

=minus minus

2 1 2 1

y y x xy y x x

=minus minus

2 1 2 1

z z z zz z z z

or =1 1

z z 1z z 1 0z z 1

= minus

z z 0z z

minus minus=

minus minus1 1

2 1 2 1

z z z zz z z z

hellip (i)

z z 1z z 1 0z z 1

= π minus

z zArgz z

= π minus

z zArgz z

minus minus= π =

minus minus

rArr minus minus

=minus minus

z z z zz z z z

+ + =1 2 2 11 2

(z z ) (z z )z z I(z z ) 02i 2i

2i and minus minus

197 tHeorem

given by+ +

=|ac ac b |p

= = minusminus 2

z zz z

is called the

=minus

z zwz z

1 2 1 2

z z z zw 1

z z z z

minus minus= = =

minus minus

0 0 0zz z z z z |z | k 0 (1)

rArr minus minus

minus minus= minus

minus minus2 2

z z z zz z z z

+ =minus minus

z z z z 0z z z z

is diametric form

2 = |z1 ndash z2|2

If minus=

minus1

z z kz z

given as minus

= α minus

z zArgz z

= plusmn minus

z zz z 2

rArr minusminus

z zz z

is purely imaginary

If minus= λ

minus1

z zz z

minus minus1 1

+ +minus =0

|az az b || z z |2 |a |

0 04aa(z z )(z z ) (az az b)

=minus minus

i3 1 2 1

3 1 2 1

z z z z ez z z z

1 2nz mzm n

+ +1 2 3az bz cz

ABCa b c

∆ =+ +

tan A tanB tanC

∆ =+ +

as z21 + z2

2 + z23 = 3z0

1 + z22 + z2

z z 11 z z 14i

z z 0z z

z zz z

form as =1 1

z z 1z z 1 0z z 1

collinear

1 2 1 2z z z z r 0α α+ + + =

if 1 2 1 2z z z z r 0β β+ + + =

minus1

z zz z

and minus=

z zy2i

and to convert

Chapter 20SetS and

relationS

201 SetS

208 tyPeS OF SetS

209 SubSetS

nC0 + nC1 + nC2++ nCn = 2n

2012 POweR SetS

2013 DISjOINt SetS

2014 uNIVeRSAL SetS

U = 1 2 3 4 50 then A = 6 7 50

n(A cap B cap C)

Remarks

of sets A B and C

2024 ReLAtIONS

R 2 3 4 5 6 7

1 0 1 0 0 0 02 0 0 0 1 0 03 0 0 0 0 0 14 0 0 0 0 0 0

Many-one Relation

One-many Relation

Many-many Relation

2031 INtO ReLAtION

(i) R1 = (11) (22) (ii) R2 = (1 2) (2 1) (2 3) (3 2) (iii) R3 = (1 1) (2 2) (3 4) (iv) R4 = (1 2) (2 1) (3 4)

in R3 but 1 ne 2

Tips and Tricks

on set ( )2nA 2=

Chapter 21FunCtions

Remarks

Functions 21227

eg =minus2

infin0 00

1 0 etc0

h(x) = minus

x if x is rational

isin ℝ

b2 ndash

d isin

Functions 21229

x3 + cx

k) infin

r a gt

(ndashinfin

a 0xn + a

1xnndash1 +

a2xnndash

2 + hellip

nndash1 x

a n ai isin

0 ne 0

n isin

k) infin

a 0 gt

(ndashinfin

r a 0 gt

= minus

infin)

minuslt

Rndash

t of a

Functions 21231

isinisin

f(x) =

ries o

(0 infin

f(x) =

log ax

(0 infin

ℝ ndash

Functions 21233

= xk k

isin ℝ

n isin

[0 infin

ndash(2

isin ℕ

ℝ ndash

ndash(2

isin ℕ

ℝ ndash

(0 infin

infin)

[0 infin

Functions 21235

1)=minus

ℝ ndash

=minus

(0 infin

infin)

isinminus

infin)

1minus

ltisin

1minus

2mminus

=minus

(0 infin

infin)

1)minus

=minus

isinminus

ℝ ndash

Functions 21237

=minus

isinminus

R ndash

(0 infin

infin)

(0 infin

infin)

) = si

s xf(x

ndash co

sec2 x

ndash times

divide)

xminus

infin)

minus=

ℝ ndash

s x in

f y fo

3)minus

=minus

minus is

+minus

minus=

=minus

Functions 21239

(ix) [x] x if x

[ x]1 [x] if x

(xii) xc

(xiii) [x] + [y] le [x + y] le [x] + [y] + 1

(xiv) [x] = x x 12 2

infin =

(vi) 0 x

x1 x x

(vii) [x] [y] 0 x y 1

[x y][x] [y] 1 1 x y 2

+ le + lt+ = + + le + lt

(i) (x) =

1[x] if 0 x2

1[x] 1 if x 12

le lt + le lt

2n 1(x) x ~ x n2( x)

2n 1(x) 1 for x n2

1[x] n if n x n2(x)

1[x] 1 n 1 if n x n 12

= le lt += + = + + le lt +

Functions 21241

inequality holds

Functions 21243

For many-one

Functions 21245

n times

nP m n0 m n

minus minussumm

m r 1 nr

C ( 1) (m r)

n m r 1 nr

m C ( 1) (m r)minus

m r nr

C ( 1) (m r)=

= minus minussum

rArr n

n r nr

C ( 1) (n r) n=

minus minus =sum

2x 1 0 x 2f(x)

and x 1 1 x 1

then 2

2x 1 1 x 1fog(x)

Functions 21247

2117 even function

Functions 21249

Remarks

ie f(x) if x

h(x)f( x) if x

(i) 1f(x)

minusnotin minus=

+ isin

Functions 21251

Remarks

30 20 is

g( x )3 x 0

= minus= =

periodic

g( x )g( x )

= = cotx which is

eg consider 1 x 0

f ( x )1 x 0

lt= minus ge

and 1 x 0

g( x )1 x 0

minus lt= ge

f ( x )1

or f(x + T) = 1

f(x)minus

g g g g(x) x=

f(x) = =

Remarks

Method I

Method II

Functions 21253

D for a 04aR

D for a 04a

for a 0D 0D

0 for D 0 a 0

D for D 0 a 04a

RD0 for D 0 a 0

4afor D 0 a 0

φ lt lt

or Linear

x 0lim f(x)rarr

f ( x )x

then minus

rarr rarr= = =

π π1

x 1 x 1

rarrminus rarrminus

minus= = =

minusπ π1

x 1 x 1

eg if =minus 2

1f ( x )

( x 1) then

lim f ( x ) exist

minusπx 2

eg rarr

minus= =

x 9f ( x ) lim f ( x ) 6

= infinx 0

If x alim f(x)rarr

+ = rarr

minus x

x 3lim (tan x 2 ) =

rarr rarrminus x

x 5lim erarr

= 34 cos34

= ne eg

x 1x 1

rarrrarr

x alim(f(x)) rarr

= ( )x a

= only if l gt 0

Notesin

x alim

x alimrarr

f(x) may exist

(a) 2( x 9)

f(x)x 3minus

=minus

Here 2

x 3lim x 9 0rarr

minus = So rarrx 3

of the type 00

(b) rarrinfinx

ln xlim

(c) rarr

(ii) =infinc 0

(a) rarr rarr rarr

= =x 0 x 0 x 0

= 1 -1

1 -1x 0 x 0 x 0 x 0

e 1lim 1

minus= (c) ( )

b 1lim ln b b 0

minus= gt

(d) x 0

ln (1 x)lim 1xrarr

(e) aax 0

log (1 x)lim log e

(f) 1x

x 0lim(1 x) erarr

xlim 1 1 x e

(h) rarr

minus= isin

m mm-1

(i) rarr

minus= isin ne

m mm-n

n nx a

(j) rarr

minus= isin

x 1lim mm Qx 1

0 if 0 a 11 if a 1

lim a if a 1

rarrinfin

(l) x a

= (m) x a

Limcos f(x) 1rarr

(n) rarr

tan f(x)lim 1f(x)

(o) rarr

minus=

e 1lim 1f(x)

(p) rarr

minus= gt

b 1lim nb(b 0)f(x)

(q) rarr

n(1 f(x))lim 1f(x)

(r) rarr

+ =1f (x)

x alim(1 f(x) e

(s) rarr

x alim f(x) Aφ

If x alim f(x)rarr

rarr= = gt

= (f) rarr

eg rarr rarr

+ minus minus +

2x 2 x 2

x 5x 6 (x 2)(x 3) 1lim limx 3x 10 (x 2)(x 5) 7

towards zero eg

rarr rarr= =

33x 0 x 03

minus minusminus +1

2x sin 3

minus + minus minus

2t 3 t 3

(a) = + + + + gt2 2 3 3

(b) = + + + +2 3

x x x xe 1 1 2 3

3 5 7x x xsin x x 3 5 7

2 4 6 (f) = + + +

3 5x 2xtan x x 3 15

(h) minus = + + + +2 2 2 2 2 2

1 3 5 71 1 3 1 3 5sin x x x x x 3 5 7

(i) minus = + + + +2 4 6

1 x 5x 61xsec x 1 2 4 6

(j) + = minus + +

1x 2x 11(1 x) e 1 x 2 24

12 123

rarr+ +4 3

(ii) x a

= and 3 2

+ minus= =

x 2 x 2 x 2

minus minus

bourhood of a eg 2

x 5x 6lim(x 2)rarr

minus +minus

+ minus + ++ minus

(2 h) 5(2 h) 6lim(2 h 2)

=rarr rarr

h 0 h 0

h hlim lim(h 1) 1h

minus(x 2)f(x)x 2

Then minus + + +

rarr rarr rarr rarr

+x ax ax a

minusminus =

+ +13 13

13 13 13 13

x ax ax a x a

eg rarr rarr rarr

minus = =

2 2x 0 x 0 x 0

sin2x2 lim 2 2 (1 2) 82x

sin xlim 1x

5 2 16x x xlim

3 92x x

rarrinfin

+ + minus

= 6 0 0 0 6 32 0 0 2+ + minus

= =+ +

minusminus

minusrarrinfinminus

+ + + + = =+ + + + infin gt

m m 10 1 m 1 m 0

n n 1x0 1 n 1 n 0

b x b x b x b b

eg rarr rarr

minus =

2 3 2 4

x 0 x 0

= rarr rarr

+ + + + +

minus =

2 3 5 6

x 0 x 0

( )( )

= =x a x a x a

g x g x g x

find out Limits

eg rarr

+ minus + x 0

tan x x 0 =

minus=

cos x sin x 1lim

sec x 1 2

(a) rarr rarr rarr

+ = =1

rarr rarr+ = +

1 1 f (x)g(x) f (x) g(x)

= rarr

xlim1x x

x 0lim 1 x e erarr

rarr+ λ = =

(b) ( )1

= = then ( ) ( )11

= + minus

= ( ) ( ) x a

minus minus

= =x 0 x 0

= = g(x)

rarr rarr= = = =

Let rarr rarr

= =infin = h(x)

h(x)lim1

lng(x)h(x)lng(x)

0L lime e form0

(i) rarrinfin

n xlim 0

x (ii)

rarrinfin=xx

xlim 0

e (iii)

rarrinfin=

xlim 0

e (iv)

( )rarrinfin

n xlim 0

(v) ( )+rarr

(vi) rarrinfin

minus =n

nlim(1 h) 0 and

ContinUity

minus +rarr rarr= =

= has oscillatory

4 f(x)g(x)

1 2 3 n

k kk k

m m m m

f ff f

n isin ℤ

+ lt lt

x sin x 0 x 1f(x)

x 21 x 4

Remark

f ( x ) x (02)x 2

= forall isinminus

DiFFEREntiABiLity

minus minus = minus

+ minus = =

Since minusminus

f(x) f(a)x a

minusminusx a

f (x) f(a)limx a

+ minus=

f(a h) f(a)limh

+ minus= =infin

f(a h) f(a)limh

3 Vertical tangent

4 f(x)g(x)

x x 0x x 0

+ minush 0

f(a h) f(a)limh

minusminusx a

f (x) f(a)limx a

and we have

minusminusx a

f (x) f(a)limx a

minusminusx a

f (x) f(a)limx a

+ minus minush 0

f(a h) f(a h)lim2h

= rarr

rarr rarr

minush 0 h 0

+ minus minus= + =

Thus rarr

+ minus minus=

Remarks

f ( a g( h)) f ( a)lim f ( a)

g( h)rarr

f ( a g( h)) f ( a ( h))lim f ( a)

g( h) ( h)φ

φrarr

+ minus +=

if + +rarr rarr

+ minus

f (a h) f (a)limh

minusminus

+ minus(n 1) (n 1)

f (a h) f (a)lim or

minus minus

minusminus

(n 1) (n 1)

f (x) f (a)limx a

exists finitely

Remarks

5minus

= x15 sec2 x + 4 5

1tan x

5( x )

f (0 h) f (0 ) f ( h) (0 )lim lim

h hrarr rarr

+ minus minus=

= 15 15

h 0 h 0

h hrarr rarr

Now f (x) = rarr

+ minush 0

f(x h) f(x)h

lim h 0 h 0 h 0

x y f(x) f(y)f2 2

rArr f(x) = ax + b

h 0 h 0

+ δ minus=

δ + δ minus=

rArr h 0

+ minus= =

= = = =

(ii) d d d(u(x) v(x)) (u(x)) (v(x))

minus = =

233 Chain rule

( )( ) ( )= = ==

( )( )

( )( ) ( )( ) =

df g x f g x g x

2 ( )= = + +ωω ω ωd du dv d

of function

minus= isin

minus= + = + isin

1 x xd (e ) edx

3 ed 1(log | x |)

dx x= a gt 0 4 a

= a gt 0 a ne 1

1 d (sin x) cos xdx

2 d (cos x) sin xdx

= minus

4 (cot x) cosec xdx

= minus

d 1(sin x) 1 x 1 or |x| 1dx 1 x

d 1(cos x) 1 x 1 or |x| 1dx 1 x

d 1(cot x) xdx 1 x

d 1(sec x) | x | 1 or x R [ 11]dx | x | x 1

rArr 1 2 n

= + + + rArr 1 2 n

= + + +

minus= = =

d k f xdx

= k ( )( )2

d f xdx

2 ( ) ( )( ) ( )( ) ( )( )2 2 2

plusmn = plusmn

3 ( )2 2 2

4 ( )3 3 2 2 3

3 3 2 2 3

= + times

(a) To get dydx

(c) Find dydx

in terms of x and y

where fxpartpart

= = and 2

= = times = times

there4 2

minus=

(i) If f(x) g(x)

yu(x) v(x)

= = minus

+prime prime

(x) (x) (x) (x) (x) (x) (x) (x) (x)

nn r n r r

y C f gminus=

2 2 2 2

Derivatives

Rate of Change

respect to x

by the following

= ie 0

dy dxf (x )dt dt=

2 dA drA r 2 rdT dt

dA dr2 rdT dt=

y dylimx dxδ rarr

δ = timesδ

tangent and noRMal

1(x y )

dxy 1dy

1(x y )

dyy 1dx

1(x y )

dxy dy

1(x y )

dyy dx

1 1(x y )

A P(x1y1)

Normal

Tangent

1 1(x y )

dy(y y ) (x x )dx

minus = minus

(x y )

1(y y ) (x x )dydx

minus = minus minus

Remarks

f2(x) are equal

(b) Find 1

= plusmn1

ie x y1

1 1( x y )

dy( y y ) ( x x )

dx minus = minus

are identical ie

1 11 1

1 1( x y ) ( x y )

a b cdy 1 dy

y xdx dx

= =minus minus

curve y = f(x) = 1 1(x y )

Slope of PQ 1 1

(x y )1

y b dyx a dxminus

equiv =minus

hellip(2)

for x 1y y2+ for

y and 1 1xy x y2+

2fy + c = 0 (i)

1 1 1 1

minus minus=

+ + + +

dyg tdy dt

dxdx f tdt

y g t x f tf t

minus = minus

x f tg t

(c) For 2 2

x y 1a b

(e) For 2 2

x y 1a b

(h) For n n

x y 1a b

(i) For n n

x y 1a b

dx= β

( ) 1 2

+ α β +

rArr 1 2

dy dy1 0dx dx

rArr 1 2C C

dy dy 1dx dx

= minus

2410 coMMon tangent

point of contacts

Here 1 2

1 1 2 2

c c2 1

(x y ) (x y ) 2 1

y ydy dydx dx x x

minus = = minus

MonotoniCity

2416 crItIcal poInt

Remarks

le le uarr

ge ge darr

uarr gt

darr lt

gt uarr

rArr uarr

prime lt α ltβ

δφδs

φ+δφφ

Clearly P gt 0 if 2

d y 10 0dx

gt rArr gtρ

and P lt 0 if 2

d y 10 0dx

lt rArr ltρ

Step 1 Find 2

d y 0dx

does not exist

(Say a b hellip)

(a) dy 0dx

gt (b) 2

d y 0dx

(a) dy 0dx

gt (b) 2

d y 0dx

(a) dy 0dx

lt (b) 2

d y 0dx

(a) dy 0dx

lt (b) 2

d y 0dx

(a) dy 0dx

gt (b) 2

d y 0dx

(a) dy 0dx

lt (b) 2

d y 0dx

Remarks

a0 c b

f(a)nef(b)

f(c)=0

Remarks

f ( b) f ( a)f ( c )

b a rArr 0

f ( c ) 0b a

prime = =minus

MaxiMa and MiniMa

2428 local MaXIMa

ge +ge minus

24281 Local Minima

le + le minus

Remarks

If =dy

Step 1 Find dydx

x0+h x0+h

f(x0ndashh) gt0

x0ndashh X0

f(x)is minimum

f(x)is

minimum minimum

| x |1 x+

is the only maximum

= but dx 0dt

ne If dx 0dt

d y yx xydx (x)

minus=

rArr c cc

2 3 2t t t tt t

minus= =

Now c c

2 2t t t t

d y d y0 if 0dx dt

Further c

d y 0dx

lt if c

d y 0dx

= rArr =

ne The sign of 2

2 2 2t 0

d y 2(t 4) 0dt (4 t )

+= minus lt

eg let f(x) = 2

n (n )n 90

1 absin

th lsquoa

f a ci

lsquoarsquo

a2 sinq

times h

us lsquor

rsquopr

ndashsid

h lsquoarsquo

sinnπ

us lsquor

rsquo4p

r with

lsquorrsquo

t tria

de lsquoa

minusminus

ndashsid

l slan

(i) If

a tria

1 (na)

indashpe

1 3 (A

minus= minus

Also APf(x) bm

x aminus

=minus

251 INTRODUCTION

Remarks

x2 x3 etc

r xx dx Cr 1

= ++int

Then n 1

= ++int

= +int

Remarks

(a) ( ) =intd

f ( x )dx f ( x )dx

C(n 1)a

1x ln|x| + C

1ax b+

ln(ax b)C

bx c1 a Cb lna

minus + +

+ + + +

π + minus minus +

or xln tan C2

cosec(ax + b)1a

minus + +

1 sec(ax b) Ca

1 cosec(ax b) Ca

minus + +

a xminus1 xsin C

aminus +

x 1+2ln|x x 1 | C+ + + 2 2

x a+2 2ln|x x a | C+ + +

11 xminus

1 1 xln C2 1 x

minus 2 2

1a xminus

1 a xln C2 a x

1x 1minus

1 x 1ln C2 x 1

minus+

1x aminus

1 x aln C2a x a

minus+

11 xtan Ca a

minus +

or 11 xcot Ca a

minus minus +

21 xminus2

1x 1 x 1 sin x C2 2

2 2 aminusminus + +

2x 1minus2

minus + minus +

21 x+2

2x 1 x 1 ln x x 1 C2 2+

+ + + + 2 2a x+2 2 2

+ + + +

x x aminus11 xsec C

a aminus minus +

g(x) t= dtg (x)

dtdx1 t

= minus+

Integrand

( )2f x x 1+ + 2x x 1 t+ + = rArremsp2 2

1 1f x 1x x

+ minus

+ = rArremsp 2

11 dx dtx

minus =

1 1f x 1x x

minus +

minus = rArremsp 2

11 dx dtx

1 1f x xx x

+ minus

+ = rArremsp 3

12 x dx dtx

minus =

1 1f x xx x

minus +

minus = rArremsp 3

12 x dx dtx

1 2n n

1f(x a) (x b)

Put (x + a) = t(x + b)

rArremsp2

1 dtdx(x b) (b a)

=+ minus

1nrsax bR x x

Put nax b tcx d+

rArremspn 1

=+ minus

Integrand

a cosθ= θ

( )2 2f a x+ a tanx

a cotθ= θ

a cosecθ= θ

a xfa x

minus +

a cosx

a sinθ= θ

minus+ +

minusminus + +

x a tx b+

(b a)dt dx(x b)

minus=

(a b)or dt dx(x a)

minus=

dx nx(x 1)

isin+int

(ii) 2 n (n 1)n

(iii) n n 1n

254 INTREGRATION OF

= minusint

Expained m 1

2 2(1 t )minus

m n m nm

= ( )m n 2

int = ( )m n 2

+int t = tanx

= ( ) ( )n 2

m 2 2z z 1 dzminus

minusint = m 1

minusminusint

= +int int (1)

1 dxax bx c+ +

dxIb Dx

= minus + +

minus + +

minus minus

=+ + minusαint int

2 121 1 (x )(x ) dx C

a a ( 2 1)

a (x )minus +

minusα

dx 1 dxI(x )a(x )(x ) a (x )(x )

int int hellip (i)

Let x txminusα

=minusβ

rArr 2

=minusβ

rArr 2

(x ) dx dt(x )

minusβ =minusβ

rArr 2

hellip (ii)

1 xI ln Ca( ) x

minusα= +

αminusβ minusβ

b Da x dx2a (2a)

+ minus

2 Da t dt2a

int where bt x

2a = +

2 DI a t dt2a

minus= +

int where

bt x2a

= minus minus + int

2Da t dt2a

minus minus

int where bt x

2a = +

ax bx c+ +int

b Da x2a 4a

= + minus

int 22

11 dxa b Dx

= + minus

21 Dln t t C2aa

= + minus +

where bt x

2a = +

21 Dln t t C2aa

where bt x2a

1I dxb D( a) x

1 1 dta D t

=minus

int bt x2a

px q dxax bx c

++ +int (b)

px q dxax bx c

+ + + px = A(2ax + b) + B

rArr pA2a

= and pcB q2a

= minus

+ += +

+ + + = + + + + + +int int int

1I dxax bx c

=+ +int

1 dxIa (x )(x )

=minusα minusβint

1 1 1 dxa( ) x x

minus= +

minus +int

dxIcx ax bx

int hellip (1)

= + + minus +

int = 2

cx ax b 2 a cx

plusmn +

hellip (2)

Put cax tx

plusmn = hellip (3)

I = 2 2

1 dt 1 dt2 c t b 2 a c 2 c t b 2 a c

dx 1 xtan Cx a a a

minus = + + int

I = 2 2

int = 2 2

dxCx Ax Bx

minus + +

where Ct Axx

ax bx c dxpx qx r

+ ++ +int

ax bx c dxpx qx r

+ ++ +int and

ax bx c dxpx qx r

+ +int

(px2 + qx + r) + C

minus +

or 1 1x g xx x

+ minus

I = 2 3 2

int = 2

f (x) dx

1 1x a x b x cx x

+ + plusmn +

Case (i) If I = 2

1 1x g x dxx x

1 1x a x b x cx x

minus + + + + +

int = 2

1 11 g x dxx x

1 1a x b x cx x

minus + + + + +

Putting 1x tx

+ = and 2

11 dx dtx

minus =

we have I = ( ) 22

g(t) dt g(t) dt

Case (ii) If I = 2

1 1x g x dxx x

1 1x a x b x cx x

+ minus + + minus +

int putting 1x tx

minus =

11 dx dtx

we have

I = ( ) 22

g(t) dt g(t) dt

+ + ++ + +int int

P (x) dxpx qx r+ +

minus= + + ++ + + +

int int hellip(1)

( )2m m 1m 12 2 2

minusminus

+= + + + +

+ + + + + +

2 2 2 2

dx a bsin x+int dx

cos x dx

a bsin x+int

( )f tan x 2dx

= and 2

2dtdx1 t

Step II Take 2 2

= =+ +

and 2 2

= =+ +

Algorithm

+ ++ +int

Algorithm

++ +int

2 2t 1R t t 2(t 1)2

minus minus +

minus + + and

minus + +

1 dtt a+int 2 2

1 dtt aminusint 2 2

dta tminusint

(ax b) cx d+ +int

type 2

at bt c+ +int

Integral of type dx

rArr 2 2

minusα = minusβ =

int int

linear quadraticint

dxI(lx m ) px qx r

=+ + +

int put 1lx mt

+ = ( ) ( )2 2 2

minus=

minus + minus +int

dxI(ax c) px r

int then put 1xt

rArr 2

1dx dtt

m p p r n rr

= sumintp nr m 1

p p r rr

xC a b Cnr m 1

+ +minus

= + + + sum

Case (a) m 1

1nst ax

b minus

mnsm t ax

b minus

rArr =

rArr s 1

st dtdxnbx

minus=

+minus

+ minus minus=

= and m 1n+

notin and m 1 pn+

+ isin

rArr sn

a t bx

= minus

m np n 1n r s 1

s a t dtna t b

ax bx c t(x )α

+ + = minusα

plusmn+ + = plusmn

or try

1 Substitute

minus+ =

if a gt 0 D lt 0

3 Substitute

if a gt 0 D gt 0

as given below

( ) ( )ax ax ax ax

+ + minus= + +

a bminus

emsp rArr n 1

minus= minus +

minus= +

rArr n 1

n n 2tan xI I

minus= minusminus

rArr n 2

n 1 n 1

minus= +

minus minus

rArr rarrinfin

2224466(2n)(2n)lim13355(2n 1)(2n 1)(2n 1) 2

cos (x)

π π π

n n 2 n

261 AreA Function

rarrrarrinfin

= + + + + + + + minus intb

h 0a n

rarrrarrinfin

= + + + + + + intb

h 0a n

rarrinfin rarrinfin

= =rarr

sum sumint

n nr 1 r 1a h 0

f (x)dx

= =rarr

minus = + = +

sum sumintb n n

n nr 1 r 1a h 0

ai = aRi and an = b

rArr aRn = b rArr =

= common ratio then

r 1nba (R 1) 0asn

a then

1 1 2 2 n nna

rarrinfin=

∆ ∆ = minus

r r rn 1

then int3

Remark

1f ( x )

x= then

dx( a b)

1 1 h 1 1a b [a ( 1)h] a h b h=

2 2h 01a n

1 h 1 1dx lim

(i) =int intb b

a af (x) dx f(t)dt

=minusint intb a

a bf (x) dx f(x)dx

b c c b

Property 4 =intb

Remarks

1 26 2( x 2x ) dx ( 27 1) (9 1) 8 0

dtf(g (t))g (g (t))

Property 6 intb

q If intb

=int intb b

t 0a a t

f (x)dx lim f(x)dx

q If intb

rarr=int int

t 0a a

(i) infin

(ii) minusinfinintb

(iii) infin

a ba 0

= minusint intb a

f (x)dx f( x)dx

= minusint intb b c

f (x)dx f(a b x)dx

Remark

267 APPlicAtions

By above property

= intb

(a b)I f(x)dx2

=(b a)I

bkb bk

a akak

Let = intb

Property 13 minus

int inta

int int

Equivalently int2a

f (x)dx =

= inta

f (a) a

minus= int

cf (x)dx = int

af (x)dx

rarr rarrα α

[a b] then ψ

φint(x)

d f(x t)dtdx

partint(x)

of a then part

α = αpartαint

I ( ) f(x )dx

0f (t)dt 0

q =int intnT T

period T then +

inta T

Hence prove that +

=int inta T T

f (x)dx f(x)dx

Corollary +

=int inta nT T

f (x)dx 0

f (x)dx g(x)dx

geintb

g(x)dx k and leintb

1 2a a a

k f(x)dx k

1 2 3 1 2 3a a

Here leint intb b

f (x)dx | f(x)|dx = =

k 1 k 1

f (x ) x | f(x ) x |

=minus int

1f(c) f(x)dxb a

interval [a b]

α partαintb

dI I ( ) f(x ) dxd

minus minus = +

b aa r xn

and minus rarr

b a dxn

Step III β

minus β = + =

b alim a r bn

rarrinfin

minus α = + =

b alim a r an

Step IV rarrinfin

minus minus + =

sum intbn

n r 1 a

and rarr

Step III β=

r q(n)

n r p(n)

β = = α = =

max min

n n n n

Step IV β

rarrinfin= α

sum intq(n)

n r p(n)

r 1lim f f(x)dxn n

1 For π π

isin =int int

2 2n n

(n 1) (n 3) (n 5) If n is oddn (n 2) (n 4)

2 For π

isin int

(m 1) (m 3)(n 1) (n 3) p(m n)(m n 2)

π = minus +

2462n 1lim 2 135(2n 1) (2n 1)

+ + + + infin =2

2 2 2 2

1 1 1 1 1 2 3 4 6

+ + + + infin =2

2 2 2 2

1 1 1 1 1 3 5 7 8

2 2 2 2

1 1 1 1 1 2 3 4 12

+ + + + infin =2

2 2 2 2

1 1 1 1 2 4 6 8 24

intb 2

a[f(x)] dx

269 BetA Function

0B(m n) x (1 x) dx

2 2m 2 2n 2

2 2m 1 2n 1

0B(m n) sin cos d

+ minus + minus

2 (2m 2)(2m 4) (2n 2)(2n 4B(mn) p

(2m 2n 2)(2m 2n 4)

2610 gAmmA Function

denoted by n

(ii) = minus = =1 (1 1) 0 1

(iv) =infin0

(v) = π12

For gt =+

m nm n 0 B(m n)

Remark

( m n)m n 1

β minus minus=

by = intb

A f(x) dx

= +int intc b

A f(x)dx f(x)dx

A (x) dy f (y) dy

A f (y)dy f (y)dy

a a c d e f

A f(x) g(x)dx

A | f (y) g (y)|dy

Area = =3

1tOAB A4

Q intb

Thus =intb

f (x) g(x)y

minus=2

f (x) g(x)y

= θ = θ θint int2 21 1A r d [f( )] d2 2

( ) ( ) ===

Catesianequation

θ θθθ

1 2a a t

271 IntroductIon

Example 32 2

dy d y xdx dx

+ = + = or 2 2 2

dy d yF x y 0dx dx

is a standard form

32 22 3

dy d y1dx dx

+ =part part

27221 Order

d y dy3x x edx dx

27222 Degree

d yf x ydx

+ ( ) ( ) 1nm 1

d yg x y

+ ( ) ( ) 2nm 2

d yh x y 0

hellip(1)

y = Ax2 + Bx etc

rArr 2

d y Asin x Bcos xdx

From (iii) 2

rArr 2

d y y 0dx

dy a 1dx y

= minus

= hellip(i)

and dy f(y)dx

= hellip(ii)

dy dxf(y)

= or dy f(x)dx g(y)

= + + where b ne 0

rArr dt bf(t) adx

rArr dt dx

bf(t) a=

+ + where b1 + a2 = 0

Consider 1 1 1

2 12 2 1

+ + = + +

rArr x2 + y2 = r2 (1)

xdy y ddx secx dx

minus θ= θ

xdy ydx cos dxminus

= θ θ

n yf(x y) xx

or n xyy

Case I If 1 1 1

lines let a 1 1

a b ka b= =

b (kv c)dv adx v c

+ rArr 1

1 1 1 1 1

+ + + int

Case II If 1 1 1

rArr 1dv k adx

rArr 22

= + minus + +

+ + where

nea ba b

can be reduced to

Now 1 1

a ba bne

gives dy dYdx dX

+ + + +=

+ + + +

rArr 1 1

dY aX bYdX a X b Y

rArr 1 1

dV a bVV XdX a b V

part part=

y constt

+ =int int

plusmn = plusmn q 2

ydx xdy xdy y

minus=

x dy y dx1dxy x y

+ minus =

2xy dx x dyxdy y

minus =

2xy dy y dxydx x

minus =

q 2 2 2

x 2xy dx 2x ydydy y

minus=

y 2yx dy 2y xdxdx x

minus=

q 12 2

x ydx xdyd tany x y

minus minus= +

q 12 2

xdy ydx yd tanx y x

minusminus = +

q x dy y dx

d(ln(xy))xy+

= q 2 22 2

+ + = +

q y dx x dyxd ln

y xy minus

x dy y dxyd lnx xy

minus =

q x x x

e ye dx e dydy y

minus=

e xe dy e dxdx x

minus=

q ( )2 2

x dx y dyd x y

d[f(x y)] f (x y)1 n (f(x y))

minus prime=

part partne

is given as n n 1

0 1 n 1 nn n 1

Remark

d y d y2 5 y 0

1 uy minus = (ii)

+ minus = minus

1uy minus=

dyf (y) Pf(y) Qdx

k kk 1

g (x yC ) 0=

k kk 1

g (x yC ) 0=

where dypdx

prime= + +

prime+ =

= if dp 0dx

d y f(x)dx

q Equation of Type2

d y f(y)dx

= helliphellip(1)

rArr 2

= rArr 2

= int int rArr

Method 2 Given 2

d y f(y)dx

rArr 2

rArr dy dx

2 f(y)dy 2A= plusmn

271311 Trajectory

minus for dydx

dt= minus

rArr dv k v gdt m

dt= minus minus

Then df(t) k R f(t)

dt= minus

Chapter 28VeCtors

(c) same sense

Remarks

O + + =

Vectors 28353

AB and

OP r xi y j zk

= = = = + +

P(xyz)

A(0y0)

M(xy0)(x00)B

(00z) zk

2 1 2 1 2 1PQ (x x )i (y y )j (z z )k

where P(x y z) and

= + + =

2 2 2r x y z OP

If and = + +

2 2 2 2r x i y j z k then + = + + + + +

1 2 1 2 1 2 1 2r r (x x )i (y y )j (z z )k

Geometrically +

Vectors 28355

a b b a

a 0 a 0 a a

(a) a b a b+ le +

(b) + ge minus

a b a b a b

If = + +

1 1 1r x i y j z k and = + +

1 2 1 2 1 2 1 2r r (x x )i (y y )j (z z )k

1 2r and r

(a b) (b a)

a (b c) (a b) c)

a a a b b a

a b a b a b

λ = λ + + = λ +λ +λ

Remarks

(i) aa

is a vector along a

having unit length

= plusmn

according as b

(ii) Division of a

by 1λ

ie a 1

( a)λ λ=

(iii) ( a b) a bλ λ λ+ = +

a bACa b

length = +

a ba b

a bONa b

minus +

ˆˆ( a b)OT ˆa b

(b) If

(c) For two vectors

a b 0 ie =

Vectors 28357

1 If a and b

as given below

Theorem If a and

286 section formula

na mbcm n

=minus

na mbcn m

Remarks

(ii) m

(iii) If m

(iv) If m

1n= then

(vi) m

+ +equiv

lie on same plane

(ix) na nb

helliphellip(i)

a mb nc 0 and

(x) If R( c )

and Q( b )

in the ratio mn n

isin minus

then a b c

the position

vector a b c

ab and c

Vectors 28359

where l is a scalar

If 1 1 2 2 3 3 n na a a a 0λ +λ +λ + +λ =

a and b 0 le q le p

(ii) ne

a(b c) (ab) (ac)

(iv) 22 2a aa a a= = =

(v) If = + +

( ) ( )= + + + + = + +

(vi) + +

θ = =+ + + +

1 1 2 2 3 32 2 2 2 2 21 2 3 1 2 3

(vii) =

(viii) = + + = + +

2 2 2(a b) (a b)(a b) a b 2ab

a on b = θ = θ = θ =

ˆbon a ba

opposite to

ˆ ˆb on a (ba)a

ˆb MA a sin a b

a on b )

Work done

work done = θ = =

( F cos ) s F s

Vectors 28361

( a b )a b

plusmn timestimes =

Where 1 2 3 1 2 3 1 2 3

times = = + + = + +

a b (b a)

3 If two vectors

4 times =

a b 0 rArr

7 Let = + +

1 2 3 2 3 3 2 3 1 1 3 1 2 2 1

8 times

a bsin

Remarks

Since a b

sin [0 ]a b

θ θ πtimes

= isin

rArr 1a b

sina b

minus times =

or 1a b

sina b

a and b

1 1a b (a b) (a b)2 4

1a b (a b) (a b)2

= times

1 21 d d 2

If = + +

1 2 3ˆˆ ˆa a i a j a k = + +

1 2 3ˆˆ ˆb b i b j b k = + +

a a aa b c b b b

Geometrically

a b and c

do not changed = =

a b c b c a c a b

(c) + + = +

a b c d a c d b c d

a a b 0

a b c 0 if = λ

a b c 0 (even if

(g) If = + +

a b c a b c

1 a b c6

ab c is

1 a b c2

a (b c) b(ac) c(ab)

(a b) c b(ac) a(bc)

(iv) times times

Vectors 28363

ie times

a (b c) o

(vii) times times

b and c

times times

c and d

aa bb c c 1 and = = = = =

ab c the vector

(i) = = =

aa bb c c 1 (ii) = = =

ab bc ca 0

(iii) =

1[a b c][a b c ]

(iv) times

b c a[a b c ]

(v) = = = = = =

[a b c] [a b c ] 1

r a (b)

rArr 1 1 1( x x ) ( y y ) ( z z )m n

rArr 1 1 1

2 1 2 1 2 1

( x x ) ( y y ) ( z z )( x x ) ( y y ) ( z z )

minus minus minus

1r a b and = +micro

2r a b

= + plusmn

b br a tb b

a and b is = λ +λ

1 2r a b

rArr times =

r(a b) 0

(r c)(a b) 0 r(a b) [a b c]

= minus = minus

r (b a) (c a)

Vectors 28365

r b a b rArr = +

1r (a b) ybbb

r b c b r a 0 a

b rArr = minus

car c bba

= + + times +

a b1r a kb a bk a k

Cevarsquos Theorem

conversely

and conversely

Deasargue Theorem

Geometry

291 IntroductIon

= = =+ + +

2912 Corollary

2 1 2 1 2 1x x y y z zx y z1 1 1

1 2 3 1 2 3 1 2 3x x x y y y z z z

3 3 3+ + + + + +

4 4 4+ + + + + + + + +

Remarks

by ndashve sign

ˆ ˆ ˆi m j n j

6 If a vector

m nm n

+ +θ =

+ + + +

1 2 2 1

2 2 2 2 2 21 1 1 2 2 2

(b c b c )sin

a b c a b c

Σ minusθ =

+ + + +

a b ca b c

1 2 2 1 1 2 2 1 1 2 2 1

2 2 2 2 2 2

m n n m and

m n n m

respectively

Corollary

2 1 2 1 2 1

Remarks

P( a) andQ( b )

2 1 2 1 2 1

If 1L r a b= +λ

or 1 1 1

x x y y z zm n

or 2 2 2

x x y y z zm n

1 2 1 2 1 22 2 2 2 2 21 1 1 2 2 2

+ +θ = =

+ + + +

rArr 1 1 1 2 1 2 1 22 2 2 2 2 21 1 1 2 2 2

minus minus + + θ = = + + + +

ie b kd=

m nm n

ie bd 0=

Or l1l2 +m1m2 + n1n2 = 0

θrarr

A(x1y1z1)(a)

C(x2y2z2)(c)

rarr d= 2im 2jn 2k

rarrb=1im1jn1k

ie x y z

= = = λ

mr a AM= +

Then PM bperp

rArr (PM)b 0=

rArr 1 1 12 2 2

(x ) m(y ) n(z )m n

hellip(ii)

Remark

r a b= +λ

Method 1

Method 2

lines intersect

= a c b sin (a c) b

1 2 3ˆˆ ˆb b i b j b k= + +

1 2 3ˆˆ ˆc c i c j c k= + +

minus minus minus

292 Skew lIneS

= (a c)(b d)

minus times

Cartesian form

1 1 2 2 3 3

minus minus minus

are coplanar

ie b kd=

Iff b kdne

(ie non-parallel)

(plane APL) and

(plane CPQ)

293 Plane

Equation of plane

(constant)

or ˆru p=

to normal form

Step 2 2 2 2 2 2 2 2 2 2 2 2 2

ax by cz d

+ + + + + + + +

a b c=

+ + and

dpa b c

a 0 b 0 0a 0 c

Remarks

1dA dB dC

intercept d d d

on coordinates axes

2 1 2 1 2 1

3 1 3 1 3 1

[ AB AC AD] 0

1 2 1 2 1 2

1 3 1 3 1 3

x x y y z z

x x y y z z 0

x x y y z z

B b P(xyz)

and P(x y z)(r) be

be vector normal

to given plane

2 1 2 1 2 1

minus minus minus

1 1 1 1

yz 2 2 zx 2 2

3 3 3 3

y z 1 z x 11 1A y z 1 A z x 12 2

y z 1 z x 1= = and

xy 2 2

x y 11A x y 12

x y 1=

be two planes

Then 1 21 1 1 2 1 2 1 22 2

1 2 1 2

Corollary

or a1a2 + b1b2 + c1c2 = 0

2 2 2 2

a b c da b c d

Let x y zL m n

2 2 2 2 2 2

a mb ncsin sin cos

2 m n a b c

+ +π θ = minusα = α + + + +

rArr 1

2 2 2 2 2 2

a mb ncsin

m n a b cminus + +

θ = + + + +

where a and q are

acute angles

Remarks

hellip(i)

al bm cn 0 and

1 22 2 2

d dda b c

minus=

Let rn d 0+ =

pn dPMn+

Cartesian form

1 1 12 2 2

ax by cz dPMa b c

+ + +=

+ + where 1 1 1

From above m 2

pn dr p nn

+ = minus

Cartesian form

= = = λ

rArr 1 1 12 2 2

(ax by cz d)a b c

minus + + +λ =

( )1 1 11 1 1

rArr 2

p n dq p 2 nn

minus + = minus

rArr 1 1 1a b cx y z2 2 2λ λ λ + + +

λ + + + + + + =

rArr 1 1 12 2 2

2(ax by cz d)a b c

minus + + +λ =

= = =+ +

a b ctimes = = + +

= =1 1 1x x y y z zl m n

+ + + + +=

+ + + + +1 1 1 1 1 1 1 1 1 1

2 1 2 1 2 1 2 2 2 2

1 1 1 1

2 2 2 2

3 3 3 3

4 4 4 4

a b c d

a b c d0

a b c d

α = β = γ =∆ ∆ ∆

1 1 12 2 2 2 2 2

(a b c ) (a b c )

+ + ++ + +=

+ + + +

Remarks

+ d2 = 0 is + + + + + += plusmn

+ + + +1 1 1 1 2 2 2 2

2 2 2 2 2 21 1 1 2 2 2

a b c a b c

+ + + + + +

=+ + + +

1 1 1 1 2 2 2 2

2 2 2 2 2 21 1 1 2 2 2

301 ExpErimEnts

Sample Space

Probability 30383

will be +y

Remarks

rP(E) limn

Probability 30385

C 13 1C 39 3

Remarks

m n+ and the

(i) capcap = 1 2

1 2n(E E )P(E E )

cup = =1 2 1 2 1 21 2

= P(E1) + P(E2) = P(E1 cap E2)

Probability 30387

= cap cap

= ne2 1 1 21

n(E E ) P(E E ) P(E ) 0n(E ) P(E )

21 2 1

EP(E E ) P(E )PE

EP P(E )E

1 2 n1 P(E )P(E )P(E )

E EP P 1E E

E EP P(E )PE E

cap 32

EEP PE E E

1 1 2 1 2 n 1

= + + +

1 2 n1 2 n

ii 1 i

AP(A) P(A )PA

Remarks

P(A ) (P(A A )) (P(A A A ))

then ( )( )

i ii 1

Probability 30389

Remarks

k k 1 k 2 n ii k

p p p p P+ +=

+ + + + = sum

1 2 3 k ii 1

p p p p p=

+ + + + = sum

m m 1 m 2 k ii m

p p p p p+ +=

+ + + + = sum

micro = =sumn

i ii 1

E(X) p x

2 2 2i i

Var(X) x (x ) p E((X ) ) = = =

sum sum2n n

2i i i i

i 1 i 1

p x p x

2 2i i

E(X ) p x

= = = =

minusmicro = minus

sum sum sum

2n n n2 2

= = minus

r (n r)

nSSSSS FFFF Cr(n r)

minus minus

n n r n rr r

r times (n r) times

successes equals 1

Remarks

n r n rr

C p q 1minus

n r n rr

1 C p qminus

minus sum

Probability 30391

n r n rr

C p q minus

n r n rr

r ( m 1)

1 C p q minus

minus sum

n r n rr

C p q minus

n r n r n r n rr r

r 0 r m 1

+ =sum sum

σ = minussumn

2 n r n r 2x r

minus + minussumn

2 n r n r 2 2r

(r r r) C p q n p

= minus

minusminus + minus

minussum sumn n

n 2 2 r 2 n r n 1 r 1 n r 2 2r 2 r 1

r 2 r 1

=SD npq

deterMinants

311 Matrix

11 12 ln

1 2 3 mn

am am am a

= ij m nA a

Remark

ii 11 22 33 nni 1

(notation)

a (a a a a ) Tr(A)=

= + + + + =sum

a 0 0 00 a 0 00 0 a 0 0 0 0 a

(Notation)

diagonal (a a a a )

k 0 0 00 k 0 00 0 k 0

1 0 0 00 1 0 00 0 1 0 0 0 0 1

Remark

I1 = [1]

= = 2 3

1 0 01 0

I I 0 1 00 1

Remarks

=n( n 1)

=2n n 2

312 Sub Matrix

For example 2 57 9

2 5 87 9 41 3 5

3121 Equal Matrices

ij ik kjk 1

(C ) a b=

Also 1 1 1 1 2 1 3

2 1 2 3 2 1 2 2 2 3

3 3 1 3 2 3 3

R R C R C R C

times = times =

A5 6 8

then T

2 5A 3 6

minus+

eg ( )2 3 2i 2 3 2i 2 3 2iA A A A

3 2i 7 3 2i 7 3 2i 7+ minus +

A 1 3i 0 4 i2 4 i 2i

matrices

Properties

= minus and 2 1 0

A I0 1

3148 Unitary Matrix

Let a b

then a b

A (ad bc)c d

= = minus

Let 11 12 13 11 12 13

21 22 23 21 22 23

31 32 33 31 32 33

a a a a a a

= 22 23 21 23 21 2211 12 13

32 33 31 33 31 32

a a a a a aa a a

a a a a a aminus +

where 11 12

eg 11 122 3 523 23 11 32 31 12

a aC ( 1) M ( 1) (a a a a )

Remarks

= sum3

ik ikk 1

= sum3

kj kjk 1

(ii) = =

= = nesum sum3 3

ik jk ki kjk 1 k 1

a C a C 0 for i j

ie 11 12 13

21 22 23

31 32 33

C C CAdj(A) C C C

21 22 23

31 32 33

C C CAdj(A) C C C

Let AdjA

B A 0A

= ne rArr nn

A IA(AdjA)AB I

A I(AdjA)BA A I

A A= = =

So a0 An + a1 A

a aA A A a a

i jC Charr

eg 1 3

1 0 0 0 0 10 1 0 R R 0 1 00 0 1 1 0 0

11 12 1n

21 22 2n

n1 n2 nn

a a a 1 0 0 0a a a 0 1 0 0

a a a 0 0 0 1

12 1n11 1 1

22 2n21 2 2

n2 nnn1 n n

a aa x ba aa x b

a aa x b

2 2 2 2 2 2

a b c b a c

a b c b a c∆ = ∆ = ∆ =

rArr 1 1

C b kxaminus

= Thus 1 1

C b kxaminus

1 1 1 1 1 1 1 1 1 1 1 1

2 2 2 1 2 2 2 2 2 2 2 3 2 2 2

3 3 3 3 3 3 3 3 3 3 3 3

∆ = ∆ = ∆ = ∆ =

= = =∆ ∆ ∆

Let 11 12 13 1 1

21 22 23 2 2

31 32 33 3 3

a a a x b

21 22 23 2 2

31 32 33 3 3

λ = micro α

α micro minus =

rArr 1 2 3

a 0 0[x x x ] b d 0 [ ]

= λ micro α

From (iii) 3xfα

= from (ii) 2

11 1 22 2 1n m

22 1 22 2 2n m

m1 1 m2 2 mn n

+ + + =+ + + =

+ + + =

12 1n11 1

22 2n21 2

n2 nnn1 n

a aa x 0a aa x 0

a aa x 0

rArr AX = O

319 eliMinant

a b x Xa b y Y

operator 1 1

rArr X 1 0 xY 0 1 y

= minus Thus

1 00 1 minus

minus = minus

θ θ = θ minus θ

θ minus θ = θ θ

Let 11 12 13

21 22 23

31 32 33

a a aa a aa a a

a C a C a C a C=

rows and 3

1j 1j 2 j 2 j 3j 3j kj kjk 1

a C a C a C a C=

jth column

P(x y)

21 22 23

31 32 33

a a aa a aa a a

∆ = then

rArr 1 1

x y 11 x y 12

rArr x y z

b b btimes =

is STP of x y z

a a a[abc] b b b

c c c=

a ba b

∆ = (a1 b2 ndash b1 a2) and 1 21 2 2 1

a a(a b a b )

i i1R Rm

∆ = ∆

For example if 2

x 5 2x 9 4x 16 8

+ minus minus minus

minus∆ = minus =

(i) 2 2 2

1 1 1a b c (a b)(b c)(c a)a b c

(ii) 3 3 3

(iii) 2 2 2

+ + ++ + ++ + +

= minus + + minus

(iii) 3 3 3

z c x a y ba b c

+ + ++ + ++ + +

Let 1 1 1

1 2 2 2

a b ca b ca b c

∆ = and 1 2 3

2 1 2 3

m m mn n n

∆ =l l l

1 1 1 1 1 1 1 2 1 2 1 2 1 3 1 3 1 3

1 2 2 1 2 1 2 1 2 2 2 2 2 2 2 3 2 3 2 3

3 1 3 1 3 1 3 2 3 2 3 2 3 3 3 3 3 3

+ + + + + +∆ = ∆ ∆ = + + + + + +

+ + + + + +

l l ll l ll l l

ijn n n

C 1 1 |C |minus∆ ∆

∆ = = = = =∆ ∆ ∆ ∆ ∆

If 1 2

f (x) f (x)g (x) g (x)

1 21 2 1 2

∆= = minus

1 2 1 2

+prime prime

rr f (0 )

d f ( x )f (0 )

dx at x = 0

a a ab

(x)dx a b c

int int intint

+ =(50 40)40 45

+ =(b a) a ba

Statistics 32413

i1 2 3 ni 1

xx x x x

= + = + = minus

ux a u a u (x a)

minus= + = + =sum

i ii 1

i ii i

f xx a u a u (x a)

f ux a hu a h

f minus

x au h

i i1 1

f ux a u a u (x a)

f ux a hu a h

f minus

x au h

h defines mean

=+ + + +

1 1 2 2 3 3 n n

1 2 3 n

w x w x w x w xAM

w w w w

323 CoMbined Mean

+1 1 2 2

n x n xxn n

That is =

minus =sumn

i ii 1

Q =sumsum

1d ord

sumsum

i i i i

f (x d) d f xA dA

324 GeoMetriC Mean

Statistics 32415

rArr GM = Antilog

sum i if log x

N for ( ) gt1 2 n

f f f1 2 nx (x ) (x ) 0

sum i if log x

N for ( ) lt1 2 n

f f f1 2 nx (x ) (x ) 0 N = odd

325 HarMoniC Mean

n1 x 1 x 1 x

is given by =

i ii 1

(f x )

+ = + 2 a b 2ab( ab)

2 a b ie (GM)2 = (AM) (HM)

327 Median

Step II Median =

+ = + +

n 1 term for n odd2

Step III Median =

ththth

For thN

thN 12

2observation

N hC2 f

Remarks

4 f + minus times

N hD i C

N hP i C

328 Mode

Statistics 32417

the mode 0 1 3 3 5 5 7 8 mode = +=

3 5 42

Mode = minus

minus +

minus+ times

minus minusm m 1

m m 1 m 1

f fl h2f f f

minus +

+ times+m 1

m 1 m 1

fl hf f

minus +

minus+

minus + minus

m m 1i

m m 1 m m 1

h h ah h h h

L SL S

(Q Q )QD

(Q Q )

MP from mean = =

minussumn

x mean

minussumn

x median

MD from mean = =

minussum

i ii 1

f x median

f MD from median =

minussumsum

f x medianf

(i) Direct Method =

minus + minusσ = =

sum sumn

22 2ii ii 1

= ( )minus

2i i 2i

minus = minus

sum sum sum22 2

2i i ix x xx

Thus minus

σ = = minus

sum sum sum22 2

+ minus + σ =

sumsum ii

u(a u ) a

rArr minus

σ = = minus = =

sum sum sum22

i i ii i

(u u) u u u (x a)

Statistics 32419

minus σ = = minus

sum sum sumn

2 22ii ii 1

(u u) u uh h

σ = = minus

22 2i i i i i i

σ = minus

sum sumsum sum

22i i i i

f u f uf f

σ = minus

sum sumsum sum

22i i i i

f u f uh

Remark

3211 varianCe

SD 100 100Mean x

s or σ + σ + + σ σ + + σ +σ = =

2 2 2 2 2 2 2 21 1 2 2 1 1 2 2 1 1 1 2 2 2

121 2 1 2

+1 1 2 2

n x n xxn n

Contents

Preface

Acknowledgements


11 Introduction

12 Pre-Requisites





141 Statement




151 Conjectures








18 Truth Value

19 Quantifiers



Number System



111 Set of Integers





1132 Prime Integer




























120 Introduction


121 Polynomials







1232 Tangency








212 Laws of Indices













31 Definition



32 Series


33 Arithmetic Mean





35 Geometric Mean





37 Harmonic Mean



391 Standard Form




311 Properties


3121 Representation


314 Vn Method









431 Theorem







451 Monotonicity










54 Nature of Roots







58 Descartes Rule





61 introduction


621 Addition Rule








66 Rank of Words





682 Sum of Divisor



69 Combination






610 Distribution














71 Introduction

72 Binomial


73 General Term



74 Middle Term


76 Greatest Term






711 Tips and Tricks



82 Greatest Term

83 Taylor Expansion



833 Properties of e

834 Expansion of ex





91 Introduction

92 Angle









951 y = sin x

952 y = cos x

953 y = cot x

954 y = cosec x

955 y = sec x









99 Tips and Trics


101 Introduction















111 Introduction





115 MndashN Theorem



























131 Introduction












1361 Union of Loci













141 Definition
















151 Introduction




1523 Diametric Form






1531 Tangents



1534 Normals







Chapter 16 Parabola



162 Parabola










1651 Properties


Chapter 17 Ellipse

171 Definition


1721 Focal Distance







181 Definition




182 Director Circle











191 Introduction




192 Complex Number

193 Argand Plane



1933 Result












197 Theorem






19102 Explanation









201 Sets







208 Types of Sets

209 Subsets



2012 Power Sets

2013 Disjoint Sets

2014 Universal Sets


















2024 Relations







2031 Into Relation


































2117 Even Function


21172 Odd Function













Limit













Continuity















Differentiability






















233 Chain Rule



















Rate of Change






Tangent and Normal







2410 Common Tangent


Monotonicity





2416 Critical Point














Maxima and Minima

2428 Local Maxima

24281 Local Minima

















251 Introduction























261 Area Function





2651 Generalization



267 Applications







269 Beta Function

2610 Gamma Function

















271 Introduction



























Chapter 28 Vectors











286 Section Formula










291 Introduction


2912 Corollary















292 Skew lines


293 Plane









301 Experiments

3011 Event











311 Matrix

312 Sub Matrix

3121 Equal Matrices













3148 Unitary Matrix















319 Eliminant





3195 Caution





31104 Circulants










323 Combined Mean


324 Geometric Mean

325 Harmonic Mean


327 Median

328 Mode




3211 Variance


Documents

for Class XI & XII, Engineering Entrance and other