APEX DISTANCE LEARNING PROGRAM FOR IIT-JEE

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APEX INSTITUTE FOR IIT-JEE /AIEEE / PMT, 0120-4106067, +919990495952, +919910817866 ww.apexiit.co.in/

PREFACE

The Aim for writing this set of booklets is to teach Mathematics in the easiest and most practical

way. In order to increase the effectiveness and efficiency for both teaching and learning, the sequence

of skills was established from simple to complex by taking pedagogy into consideration.

TO THE STUDENT

In this section, as a teacher I will answer the most common question directed by student “How we learn

mathematics?” I hope that by expressing my ideas and giving appropriate advice I will enlighten the

learner. In my opinion, there is no exact answer to this question, but bear in mind that Mathematics is not a

skill that can be learned by observation. Learning mathematics is like learning swimming, in that; it is not

acquired by observation alone. Approach the way of learning mathematics with a positive attitude.

Avoid the common habit of trying to prepare yourself for the exam just two days before its date. Otherwise,

this will make you want to withdraw from mathematics.

Your competence in mathematics will mainly depend upon the number of problems you solve, the more the

better. So, try to solve as many possible. Being unsatisfied with one method for solving a problem and

looking for other techniques to solve the same problem are the characteristics of an able mathematician.

Strategies for solving Problems:

1. Read the question until you have understood it completely.

2. Write down what is asked in the question.

3. Determine the given theorems and identities that will guide you to the solution.

4. Use the data given in identities and theorems (appropriate to the operational sequence).

5. Find the solution and afterwards, show the proof.

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Definition of Sets

A set is well-defined collection of distinct objects.Sets are usually denoted by capital letters A, B, C, X, Y,Z.

Elements of sets:-

The elements of the sets are denoted by small letters in the alphabet of English language i.e., a, b, c, x,

y, z…..if x is an element of a set A, we write x A∈ (read as ‘x belongs to A’). If x is not an element of

A, write x A∉ (read as ‘x does not belongs to A’)

For example: if A= 1, 2, 3, 4, 5 then3 A∈ , 6 A∉

Represents of sets: -

There are two methods for representing a set.

(i) Tabulation method or Roster:-under this method, the elements are enclosed in curly bracket or

bracks after separating them by commas. a, b, c, b, a, c, c, a, b all denote the same set.

An element of set is not written more than once i.e., the set 1, 2, 3, 4, 3, 3, 2, 1, 2, 1, 4 is identical

with the set 1, 2, 3, 4.

ü If A is the set of prime numbers less than 10, then A= 2, 3, 5, 7

ü If A is the set of all even numbers lying b/w 3 and 20, then A=4, 6, 8, 10, 12, 14, 16, 18

(ii) Set builder method or property method:- under this method the stating properties which its

elements are to satisfy,

ü If A=1, 2, 3, 4, 5, 6, 7, 8, then we can write A= : 8x N x∈ ≤

ü A is the set of all odd integers lying b/w 2 and 51, then A= : 2 51,x x x< < is odd

Illustration1: Write the following in set -builder form:

(a) A= -3,-2,-1, 0, 1, 2, 3 (b) B = 3, 6, 9, 12

Solution: (a) : 3 3A x x Z and x= Î - £ £

(b) ( ) : 3 and , 4b B x x n n N n= = Î £

Illustration 2: Write the following in Roster form.

(a) C = x :x∈N and 50≤x≤60 (b) D = x :x∈R andx2−5x+6=0

Solution: (a) C =50, 51, 52,53,54,55,56,57,58,59,60

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x−5x+6=0 ⇒ (x−3) (x−2)=0

⇒ x= 3, 2. ∴ D= 2,3

Different types of sets:-

(i) Null set (or empty or void set): A set is having no element is called a null set or empty set it is

denoted by φ or .φ is called the null set .

ü φ is unique

ü φ is a subset of every set

ü φ is never written within braces. i.e. φ is not the null set.

(ii) Singleton set or unit set:

a set having one and only one element is called singleton or unit set.

ü :x x 3 4− = is a singleton set

Since x 3 4 x 7− = ⇒ = :x x 3 4 7− = =

(iii) Subset:

If every element of a set A is also an element of a set B., we write A B⊆ (read as A is subset of B

or A is contained in B).

[ ]A B x A x B⊆ ⇔ ∈ ⇒ ∈

ü Every set is subset of itself i.e., A A⊆

ü If A B⊆ , ,B C⊆ then A C⊆

ü If , ,A 2 3 4= and , , , , ,B 5 4 3 2 1= then A B⊆

(iv) Total number of subsets:

if a set A has n elements, then the number of subsets of nA=2

(v) Equal set:

two sets A and B are said to be equal if every element of A is an element of B, and every

element of B is an element of A, if A and B are equal then we write A=B ,it is clear that

A B⊆ and B A⊆ A=B⇔

(vi) Power set:

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the set of all the sub sets of a given set A is said to be the power set of A and is denoted by

P(A) or A2 Symbolically, : P(A) x x A= ⊆ Thus ( )x P A x A∈ ⇔ ⊆ .

ü If , , A a b c= then P(A) or A2 , , , , , , , , , , , , a b c a b b c c a a b cφ=

(vii) Superset:

the statement A B⊆ can be written as ,B A⊇ then B is called the superset of A and is

written as B A⊃ .

(viii) Proper subset:

a set A is said to be 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 B⊂ ( read as “A is a proper

subset of B”)

ü If 1,2,4A = and 5,1,2,4,3B = Then A B⊂ since 3,5 A∉

(ix) Finite and infinite sets:

a set in which the process of counting of elements comes to an end is called a finite set;

otherwise, it is called an infinite set.

ü Set of universities in India

ü Set of gold medalists’ student in C.S. in IIT Kanpur.

(x) Cardinal number of a finite set:

the number of distinct elements in a finite set A is called cardinal number of a set A is denoted

by n (A).

(xi) Comparability of sets:

two sets A and B are said to be comparable if either A B⊂ or B A⊂ or A B= otherwise

A and B is said to be said incomparable.

(xii) Universal sets:

All the set under consideration are likely to be subsets of a set is called the universal set and is

denoted by Ωor S or U.

(xiii) Operation on sets:

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the union of two set 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 B∪ or A B+ is (read as ‘A union B’ or ‘A cup B’or ‘A join

B’).Symbolically, :A B x x A∪ = ∈ or x B∈

ü If 1,2,3,4A = and 4,5,6B =

Then A B∪ 1,2,3,4,4,5,6 1,2,3,4,5,6= =

(xiv) Intersection of sets:

The intersection of two sets A and B is the set of all element that are common in A and B,

this set is denoted by A B∩ .

Symbolically, :A B x x A∩ = ∈ and x B∈ or : ^ A B x x A x B∩ = ∈ ∈

ü If 1,2,3,4A = and 4,5,6B = then 4A B∩ =

(xv) Disjoint set:

If two set A and B has no common element i.e., A B φ∩ = then the two sets called

disjoint set or mutually exclusive event.

(xvi) Difference of sets:

if A and B is two given sets then the set of all those elements of A which don’t belong to B is

called of difference of sets A and B it is written as A B− .it is denoted by A B: or A/B or

AC B (compliment of B in A).Symbolically, : & A B x x A x B− = ∈ ∉

(xvii) Symmetric difference of two sets:

Let A and B are two sets the symmetric difference of two sets A and B is the set

( ) ( )A B B A− ∪ − or ( ) ( )A B A B∪ − ∩ and is denoted by A B∆ or A B⊕ .i.e.,

A B⊕ or ( ) ( )A B A B B A∆ = − ∪ −

(xviii) Complement set:

Let U be the universal set and A be a set such that A U⊂ then the compliment of A with

respect to U is denoted by A’ or cA or ( )C A or U A− .

Symbolically, 'A orcA ( ) C A x x U= = ∈ and x A∉

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Demorgan’s law:

for any three sets A, B and C we have ( )' ' 'A B A B∪ = ∩

ü ( ) ' ' 'A B A B∩ = ∪

ü

( ) ( ) ( )A B C A B A C− ∪ = − ∩ −

ü

( ) ( ) ( )A B C A B A C− ∩ = − ∪ −

CARDINAL NUMBER OF SUM SETS:

If A, B and C are finite sets. And U be the finite universal set then

ü ( ') ( ) ( )n A n U n A= −

ü ( ) ( ) ( ) ( )n A B n A n B n A B∪ = + − ∩

ü ( ) ( ) ( )n A B n A n B∪ = + , if A and B are disjoint non-void sets.

ü ( ') ( ) ( )n A B n A n A B∩ = − ∩

ü ( ' ') ( )' ( ) ( )n A B n A B n U n A B∩ = ∪ = − ∪

ü ( ' ') ( )' ( ) ( )n A B n A B n U n A B∪ = ∩ = − ∩

ü ( ) ( ) ( )n A B n A n A B− = − ∩

ü ( ) ( ) ( ') ( ' )n A B n A B n A B n A B∩ = ∪ − ∩ − ∩

ü ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )n A B C n A n B n C n A B n B C n C A n A B C∪ ∪ = + + − ∩ − ∩ − ∩ + ∩ ∩

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1. If A and B two sets containing 3 and 6 elements respectively, what can be the minimum number of

elements A B∪ ? Find also the max. no. of elements in A B∪ .

2. In a group of 1000 people there are 750 who can speak Hindi and 400 can speak Bengali how any

can speak Hindi only? How many can speak Bengali only? How many can speak Bengali & Hindi

both?

3. A survey of 500 television watchers produced the following information285 watch football,195

watch hockey,115 watch basketball,45 watch football and basketball both,70 watch football and

hockey,50 watch hockey and basketball,50 don’t watch any of three games. How many watch all

three games? How many watch exactly one of three games?

4. A class has 175 students following table shows the no. of students studying one or more of the

following subjects in this case.

Subjects No. of students

Mathematics 100

Physics 70

Chemistry 46

Mathematics and physics 30

Mathematics and chemistry 28

Physics and chemistry 23

Mathematics, physics and chemistry 18

How many students are enrolled in mathematics alone? Physics alone? Chemistry alone? Are there

students who have not offered any of these subjects?

5. in a pollution study of 1500 rivers the following data were reported ,520 were polluted by sulphar

compound, 335 were polluted by phosphate, 425 were polluted by crude oil , 100 were polluted by

both crude oil and sulphar compound, 180 were polluted by both phosphate and sulphar compound,

100 were polluted by both crude oil and phosphate, and 28 were polluted by phosphate, sulphar

compound, crude oil, how many rivers were polluted by at least one of three impurities? How many

rivers were polluted by exactly one of three impurities?

6. An analysis of 100 people’s injury claims made upon a mator insurance company revealed that loss

of injury in respect of an eye, an arm, an leg occurred in 30, 50, and 70 respectively. Claims of this

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injury to two of these members numbered 44. How many claimed involve in injury to all the three?

We must assume that one or another of three was mentioned in each of the following 100 claims.

7. An investigator interviewed 100 students to determine the preferences of the drink (three).milk (M)

coffee(C), tea (T) he reported the following 10 student had all the three drink M, C, T.20 had M and

C, 30 had C and T, 25 had M and T, 12 had M only. 5 had C only, 8 had T only using a Venn

diagram find how many did not the any of the drink?

8. In a group of 45 persons 22 can speak Hindi only, 12 can speak English only, how many can speak

both language?

9. At a certain conference of 100 people there are 29 Indian women, and 23 were Indian men, of these

Indians 4 are doctors, 24 are either men or doctors, there are no foreign doctors . How many

foreigners were attending the conference how many women doctors were attending the conference?

10. The report on a survey of 100 students stated that the no. of studying the language were Sanskrit,

Hindi and Tamil 5; Hindi and Sanskrit 10;tamil and Sanskrit 8; Hindi and Tamil 20; Sanskrit 30;

Hindi 23; Tamil 50; the surveyor who prepared the report was fired why ?

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Natural numbers:

The numbers 1, 2, 3, 4... are called natural numbers, their set is denoted by N. Thus N 1,2,3,4,5...)

Integers:

The numbers . . -3, -2, -1, 0, 1, 2, 3... are called integers and the set is denoted by I or Z. Thus I (or Z)

=... - 3, -2, -1,0,1,2, 3.... Including among set of integers are

v Set of positive integers denoted by I + and consists of 1, 2, 3, . . . ( N)

v Set of negative integers, denoted by I - and consists of ..., -3, -2, -1

v Set of non-negative integers 0, 1, 2 ), called as set of whole numbers.

v Set of non-positive integers ...., -3, -2, -1, 0)

Rational Numbers:

All numbers of the form p/q where p and q are integers and q ≠ 0, are called rational numbers their set is

denoted by Q. Thus,

It may be noted that every integer is a rational number since it can be written as p/i. It may also be noted

that all recurring decimals are rational numbers. e.g., = 0.33333...

then 10p -p = 3 ⇒ 9p = 3

⇒ p = 3/9 ⇒ p – 1/3, which is a rational number.

Irrational Numbers:

There are numbers which cannot be expressed in p/q form. These numbers are called irrational numbers

and their set i& denoted by QC (i.e. complementary set of Q) e.g. 2 ,1+ 3 ,π etc.

Irrational numbers cannot be expressed as recurring decimals.

Real Numbers:

The complete set of rational and irrational numbers is the set of real numbers and is denoted by R.

Thus cR Q Q= ∪

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It may be noted that N I Q R⊂ ⊂ ⊂ . The real numbers can also be expressed in terms of position of a

point on the real line. The real line is the number line wherein the position of a point relative to the origin

(i.e. 0) represents a unique real number and vice versa.

Real Line

All the numbers defined so far follow the order property i.e. if there are two numbers a and b then either

a < b or a= b or a> b.

Intervals:

Intervals are basically subsets of R and are of very much importance in calculus as you will get to know

shortly. If there are two numbers a, b e R such that a < b, we can define four types of intervals as follows:

v Open Interval: (a, b) = x: a< x< b) i.e. end points are not included.

v Closed Interval: [a, b] = x: a ≤ x ≤ b i.e. end points are also included. This is possible only

when both a and b are finite.

v Open-closed Interval: (a, b] = x: a <x ≤ b

v Closed —open Interval: [a, b) = x: a ≤ x < b

The infinite intervals are defined as follows

v (a, ∞ ) x:x>a

v [a, ∞ )=x:x ≥ a

v (- ∞ , b) = x: x< b)

v ( ∞ ,b] =x:x ≤ b

Note: Intervals are particularly important in solving inequalities or in finding domains etc.

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The following are some very useful points to remember:

1. We may add the same number to both sides of an inequality, and the sense will not change.

If a > b, then a + c > b + c.

2. We may multiply both sides of an inequality by the same positive number, and the sense will not

change.

If a > b, and c > 0, then ca > cb.

3. If we multiply both sides of an inequality by the same negative number, the sense of the

inequality changes. If a > b, and c < 0, then ca < cb.

4. If we change the signs on both sides of an inequality, then the sense of the inequaltiy will

change.

If −a < −b, then a > b.

5. If a, b are both positive or both negative, then on taking reciprocals, the sense of the inequality

changes.

6. a<b and b<c ⇒ a<c

7. a<b ⇒ a+c < b+c ∀ c∈R

8. 0<a<b ⇒ ar <br if r >0 and ar > br if r<0

9. (a+1

a) ≥ 2 ∀ a>0 and equality holds for a= 1.

10. (a+1

a) ≤ 2 ∀ a<0 and equality holds for a = -1.

Illustration1: Solve 5x + 7 < 3(x + 1).

Solution: First I'll multiply through on the right-hand side, and then solve as usual:

5x + 7 < 3(x + 1) ⇒ 5x + 7 < 3x + 3 2x + 7 < 3 ⇒ 2x < –4 x < –2

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Illustration 2: Solve 3(x – 2) + 4 > 2(2x – 3).

First I'll multiply through and simplify; then I'll solve:

3(x – 2) + 4 > 2(2x – 3) 3x – 6 + 4 > 4x – 6 3x – 2 > 4x – 6 –2 > x – 6 (*) 4 > x x < 4

Illustration 3: Solve 14(x-2) 132 - 281x Solution:

( )14 2 132 281 1 4 28 132 281

1 4 160 281 295 160

160 / 295 32 / 59

x x x x

x x x

x x

- £ - Þ - £ -

£ - Þ £

£ Þ £

The solution set is x : x 32/59.

Illustration4: Find x such that 3x — 3 ≥ x + 5 and 1 1

1 8x≥

−

Solution: 3x —3 ≥ x +5 ⇒ 2x ≥ 8 ⇒ x ≥ 4 ……….. (1)

1 1

0 1 8 1 91 8

x xx

≥ ⇒ < − ≤ ⇒ < ≤−

………….(2)

from (1)and (2), we have 4 ≤ x ≤ 9.

Illustration 5: Solve the inequality 2x - 3 < x2 - 5x

Solution: 2x - 3 > x2 - 5x 0 > x2 - 7x + 3

We use the quadratic formula to find the roots:

So x2 - 7x + 3 =. . Thus, we want to solve

0.>.

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_________________ _____________________ _________________

Sign of x< <x< x>

x- - - +

x- - + +

+ - +

Thus, 0 > if .<.x.<. .

The solution set to 2x - 3 > x2 - 5x is .

Illustration 6: Solve the inequality 1

2

x

x x£

+

Solution:21 1 – 2

0 02 2 ( 2)

x x x x

x x x x x x

-£ Û - £ Û £

+ + +

( 1)( – 2)

0( 2)

x x

x x

+Þ £

+

From wavy curve method

Also at 2x = - & 0x = given inequality becomes undefined

Therefore final solution is ( ] ( ]2, 1 0,1x Î - - È

Illustration 7: Solve the inequality1

3

x

x x£

-.

Solution:21 1 3

0 03 3 ( 3)

x x x x

x x x x x x

- +£ Û - £ Û £

- - -

The function f(x) = x2 – x + 3 / x (x-3) changes sign with variable x passes through the points

x1 = 0 and x2 = 3 (we have x2 – x + 3 > 0 for all x ∈ R).

( 3) 0x xÞ - £ From wavy curve method

Therefore final solution is ( )0,3 .x Î

Illustration 8: Solve the inequality ( ) ( ) ( )2 3

1 1 4 0.x x x- + - <

Solution: The function f(x) = (x-1)2 (x+1)3 (x-4) changes sign only when x passes through

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the points x1 = -1, x2 = 4. We have f(x) > 0 on the interval (4, +∞),

f(x) <0 on the next interval (-1, 4), excluding the point x = 1

at which f(x) = 0 and f(x) > 0 on the last interval (-∞, -1).

Answer is (-1, -1) ∪ (1, 4).

Illustration 9: Solve the inequality ( ) ( )

( )

2 3

34

1 10.

2

x x

x x

- +£

-

Solution: The function f(x) = (x-1)2 (x+1)3 / x4 (x-2)2 changes sign only when the x passes

through the points x1 = -1, x2 = 2. When x passes through points x3 = 0 and x4 = 1,

the function f(x) does not change sign.

We have f(x) >0 on the interval (2, + ∞), f(x) < 0 on the next intervals (1, 2), (0, 1),

(-1, 0) and f(x) > 0 on the interval (-∞, -1). At the point x4 = 1 the inequality is

satisfied and at the point x3 = 0 the function f(x) is not defined.

Answer is [-1, 0) ∪ (0, 2).

Illustration 10: Solve the inequality ( ) ( ) ( )

( )

2012 103 101

2000205

2 1 1 3 20.

2

x x x

x x

- - -³

-

Solution: The function ( ) ( ) ( )

( )

2012 103 101

2000205

2 1 1 3 2( ) 0.

2

x x xf x

x x

- - -= ³

-changes sign only when

the x passes through the points x1 = 0, x2 = 2/3 & x3 =1. When x passes through

points x4 = 1/2 and x5 =2, the function f(x) does not change sign.

We have f(x) >0 on the interval [ ) ( )2

0, 1, 2 2,3

xæ ùç úÎ È È +¥çç úè û

.

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1. Solve the inequality ( ) ( )2 12 1 .

3 5 3 6

x x+ - -- < -

23. ,

20Ans

æ ö- ÷ç-¥ - ÷ç ÷çè ø

2. Solve the inequality ( )

( )

11 0

4 ( 3)

x

x x

- -³

+ - ( ] ( ). , 11 4,3Ans -¥ - È -


( )

1 0

4 (2 3 )

x

x x

-³

- -

( ) [ ) , 2 / 3 1, 4Ans x Î -¥ È

4. Solve the inequality ( )( ) ( )( )

1 10

1 2 2 3x x x x- ³

- - - - ( ) ( ). ,1 2,3Ans -¥ È

5. Solve the inequality 2 3 2 0.x x- + > ( ) ( ). ,1 2,Ans -¥ È ¥

6. Solve the inequality 2 4 5.x x− <

7. Solve the inequality 2 2 2 9 9.x x x− ≤ − +

8. Solve the inequality 4 2 4

.2 2 3 9 3

x xxx x x

+− <

+− −

9. Solve the inequality 4 19 4 17

. 5 3

x xx x

+ −<

+ −

10. Solve the inequality 21

0.2 5 6

x

x x

+<

− +

11. Solve the inequality 2 1

1.2 1

x

x x

−<

+ +


( )

32 2 3 0.

2 1 3 4

x x

x x x

− −≥

+ − −

13. Solve the inequality ( ) ( )

20 10 1 0.

4 3 4 xx x+ + >

−− −

14. Solve the inequality 23 7 8

1 2.2 1

x x

x

− +< ≤

+

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Absolute Value:

The absolute value of a number x, denoted by x is defined by the formula

0

0

x xx

x x

ì- £ïï= íï+ ³ïî

Basic Properties of |x|:

v x x=

v x a a x a< Þ - < < if a R+Î and no solution if 0a R-Î U .

v x a x a> Þ > or x a< - if a R+Î and x RÎ if a R-Î

v x y x y+ £ +

v x y x y- ³ :

v The last two properties can be put in one compact form namely, |x| — |y| ≤ |x ± y| ≤ |x| + |y|

v xy x y=

v xx

y y=

v 0x x x> " <

v | | ( , ) ( , )x xα β β α α β< < ⇒ ∈ − − ∪

v | | | | | |x y x y+ = + if both are same sign.

Illustration 6: Solve 2 3x - <

Solution: 2 3x - < may be rewritten as -3 < x-2 < 3. Now solve this

3 2 – 2 2 3 2x- + < + < + Addition Property of Inequalities

1 5x- < <

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Illustration 7: Solve 2 1 4x + ³

Solution: Rewrite as 2 1 4x + > AND 2 1 4x + <

Now, solve both of these inequalities.

2 1 4x + ³ Given

2 3x ³ Add –3 to both sides using Addition Property of Inequalities

3 / 2x ³ Divide both sides by 2 using Division Property of Inequalities

Now 2 1 4x + £ - ⇒ 2 5x £ - ⇒ 5 / 2x £ -

The solution is the Union of these two solutions, which we write as

3 / 2x ³ Or 5 / 2x £ -

Illustration 8: Find the set of real values of x satisfying 1 3x − ≤ and 1 1x − ≥

Solution: 1 3 3 1 3

1 1 1 11 1

x xx xx

− ≤ − ≤ − ≤⇒− ≤− ∪ − ≥− ≥

2 4 2,0 2,4 .0 2

x xx x

− ≤ ≤⇒ ⇒ ∈ − ∪

≤ ∪ ≥

Illustration 9: Find the set of real values of x satisfying 1 1 1x − − ≤

Solution: 1 1 1 1 1 1 1x x− − ≤ ⇒ − ≤ − − ≤

0 1 2x⇒ ≤ − ≤

1 3x⇒ − ≤ ≤ [ ]1,3x ∈ −

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1. Find the solution sets of the following equations.

(i) 2 x x 2 0− − = (ii) x + 1 = 5 - x (iii) x - 1 2 = 1+

(iv) - 2 = 3x x− (v) 2 =8x x −

2. Find the solution sets of the following inequalities

(i) 2 2 x − ≤ (ii) 2 1x + ≥ (iii) 1 x + 1 3≤ ≤

(iv) 3 1 4 x− ≤ − ≤ (v) x-4 < x-1 (vi) 2 -5 6 0x x + <

3. Find the value of a if the sum of the roots of the equation - 2 3 - x a x a= is equal to

4.

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Exponential Functions:

By an exponential function we mean a function of the form

( ) xf x a=

where a is a positive constant called the base of the function.

Properties of Exponents:

( )

0

( )

( ) /

( )

( )

( ) 1

( ) 0

x y x y

x y x y

yx xy

x y

x y

i a a a

ii a a a

iii a a

iv a a x y

v a

vi a b x y

+

-

=

=

=

= Û =

=

= Û = =

Graph of xa v If a > 1, the graph of f(x) = xa will rise to the right.

v If 0< a < 1, then the graph of f(x) = xa will fall to the right.

v The graph of g(x) = ( )1 /x

a is the reflection of the graph of y = xa across the y-axis.

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Exponential Inequalities:

( ) ( )

( ) 1 ( ) ( )

( ) 0< 1 ( ) ( )

f x g xa a

i if a then f x g x

ii if a then f x g x

>

> >

< <

Illustration 10: Solve the equation 4 2 12 0x x- - = .

Solution:

( )2

2

4 2 12 0 2 2 12 0

12 0 ( 4)( 3) 0

4 3 2 4 or 2 3

2.

x x x x

x x

t t t t

t or t

x

- - = Û - - =

- - = Û - + =

= = - Û = = -

=

Illustration 11: Find the number of solutions of the equation 2 2 2sinx x x-+ =

Solution:

2 2 2sin

1/ 2 0

2 1 / 2 2 max 2 sin 2

0.

0 . . 2 & . . 0

.

x x

x x

x

we know that x x x

therefore but imum value of x is

so given equatin has solution only for x

for x L H S R H S

therefore given equation has no solutions

-+ =

+ ³ " >

+ ³

=

= = =

1. Solve the equation 3x2-5x+6 = 1.

2. Solve the equation (3/7)3x-7 = (7/3)7x-3.

3. Solve the equation 0.125.42x-8 = (0.25/√2)-x.

4. Solve the equation 52x-1 + 5x+1 = 250.

5. Solve the equation 9x + 6x = 2.4x.

6. Solve the equation 5x+1 – 5x-1 = 24.

7. Solve the equation 6x + 6x+1 = 2x +2x+1 + 2x+2.

8. Solve the equation 2x+2 > (1/4)1/x.

9. Solve the equation (1.25)1-x < (0.64)2(1+√x).

21

Logarithmic Functions:

v The expression logb a is meaningful for a>0 and for either 0<b<1 or b>1.

v logb aa b=

v 1

loglog

b

a

ab

=

v 1

loglog

b

a

ab

= provided both a and b are non-unity

Graph of logarithmic function:

Properties:

1. log 1 0, 0, 1a a a= > ≠

2. log 0, 0, 1a a a a= > ≠

3. loglog 0

xx ee e x x R and e x x= ∀ ∈ = ∀ > i.e. loge x is the inverse function of exponential function

xe .

4. log ( ) log loga a axy x y= + for all x, y >0 and a > 0, 1a ≠

5. log (1 / ) log 0, 0, 1a ax x x a a= − ∀ > > ≠

6. log ( / ) log log , 0, 0, 1a a ax y x yx y a a= − > > ≠

7. If 0a R and x∈ > then we define log

0

( log )ea xa ne

n

x e a x∞

=

= = ∑

8. log log 0,yx y x x y R= ∀ > ∈

9. (Change of base of formula) if 0, 1a a> ≠ then loga x (if a = e than we write log loge x x= )by

log

loge

e

x

ain fact if a, b, c > 0 and , 1a c ≠ then

loglog

loge

a

e

bb

a= .

v ( ) ( ) 0 1

log ( ) log ( )0 ( ) ( ) 0 1

b b

f x g x if bf x g x

f x g x if b

≥ > >≥ ⇒

< ≤ < <

22


Illustration 12: Find the solution of the equation ( )( )3 3log 5 4log 1 2.x+ - =

Solution: The logarithm of the number b to the base a (a > 0, a ≠ 1) is the power to which

the number a must be raised in order to obtain the number b. Consequently,

( ) 235 4log 1 3 ,x+ - =

Or ( ) ( )3 34 log 1 9 – 5 log 1 1.x x- = Þ - =

And again by the definition of a logarithm, we have

1 1 3 , 4.x x- = =

The verification ( 1 0)x- > , which is a part of the solution, confirms the truth of the result

obtained. Answer is x = 4.

Illustration 13: Find the solution of the equation ( ) ( )2 2 log 3 log 1 3.x x- + - =

Solution: To find the domain of the function appearing on the left hand side, we form a

system of inequalities

3 0

11 0

xx

x

ì - >ïï Þ <íï - >ïî

…………………………..(1)

we can write ( )( ) 2 3 – 1 – 3,log x x = and by the definition of a logarithm

we have ( )( ) 3 23 – 1– 2 , – 4 – 5 0.x x or x x= =

Then 1 25, 1.x x= = - Since the first value of x does not belong to the

domain of the definition, the final result is x = -1. Answer is -1.

Illustration 14: Find the solution of the equation ( ) log (3 )2log 9 2 10 .x x-- =

Solution: The domain of definition is

9 2 0,3 0,

⇒ 2 9, 3,

⇒ 9,

3, ⇒ x < 3

log2 (9 – 2x) = 3 – x.

23

By the definition of a logarithm,

3 3 2 9 – 2 , 2 / 2 9 – 2 ,

22 – 9. 2 8 0,

x x x x

x x

- = =

Þ + =

2x = 1, x = 0; 2x = 8, x = 3

x = 3 is an extraneous root. Answer is x = 0.

Illustration 15: Find the solution of the equation ( ) ( )log ( 1)

1 100 1 .x

x x+

+ = +

Solution: The domain of the definition is x + 1 > 0 ⇒ x > -1.

We take logarithms of both sides of the equation to the base 10:

( ) ( ) ( )1 .log 1 log100 log 1 .log x x x+ + = + +

We designate log (x + 1) = t and then the equation assumes the form t2 – t – 2 = 0.

Its solutions are t1 = -1, t2 = 2, i.e.

( )

( )

log 1 1 1 1 /10 0.9;

log 1 2 1 100 99. 0.9, 99.

x x x

x x x Answer is

+ = - Þ + = Þ = -

+ = Þ + = Þ = -

Illustration 16: Solve the equation ( )2 3log – log 2 0.x x + =

Solution: We introduce the variable t = log x, x > 0, and the initial equation assumes the

form t2 – 3t + 2 = 0. Its solutions are t1 = 1, t2 = 2.

Whence we have log x = 1, x = 10;

log x = 2, x = 100. Answer is 10, 100.

Illustration 17: Solve the equation ( )2 ( 1)1 1 4.xlog x log -+ - =

Solution: The domain of definition is x > 1, x ≠ 2.

( ) ( )

2( 1)

2 2

log 4 2 4 .

log –1 log –1xlog

x x- = =

We designate log2 (x – 1) = y and then the initial equation assumes the form

1 + y = 2 / y, or y2 + y – 2 =0, whence y1 = -2, y2 = 1.

log2 (x – 1) = -2, x – 1 = 1 / 4, x1 = 5 / 4,

log2 (x – 1) = 1, x – 1 = 2, x2 = 3. Answer is 5 / 4, 3.

Illustration 18: Find the solution of the inequality: ( ) ( ) ( ) ( )2 2log 2 3 log 24 6

x xx x

− −− > −

24

Solution: Case I if 0 < x - 2 < 1 ⇒ 2 < x < 3

2x-3 < 24-6x

⇒ 8 x < 27

⇒ x <27

8

But 2 < x < 3

⇒ 2 < x < 3 ------------ (1 )

Case II if x – 2 > 1 ⇒ x > 3

Then 2x-3 > 24-6x ⇒ 8x >27

⇒ x > 27

8 But, x > 3

⇒ x > 27

8 ----------- (2)

Therefore x ∈ (2,3) ∪ (27

8, ∞ ).

Illustration 19: 1/5

4 6 log 0

xSolve for x

x

+≥

Solution. Since 1/ 5log 1 0= , the given inequality can be written as

1/ 5 1/ 5

4 6log log 1

x

x

+≥

When the domain of the function is taken into account, the inequality equivalent to the

system of inequalities

4 6 4 60, 0,

4 6 4 6

1 0

x x

x xor

x x

x x

+ + > >

+ + ≤ ≤

We solve the inequalities by the method of intervals

3 34 . 0, 0 3

2,2 22

3 ( 2) 0, ( 2) 0,

x x x xx

x x x x

+ > + + > ⇒ ⇒ ∈ − − + ≤ + ≤

25

Answer 3

2,2

x

∈ − −

Illustration 20: Solve 2

2 3 2 3log log (2 3)x xx x+ +< +

Solution. 22 3 2 3log log (2 3)x xx x+ +< +

,The given inequality is equivalent to the collection of the

systems

2

2

0 2 3 1, (1)

2 3,

2 3 1, (2)

0 2 3

x

x x

x

x x

< + <

> + + > < < +

Solving system (1), we obtain

31

(3)2

( 3)( 1) 0

x

x x

− < < −

− + >

System (3) is equivalent to the collection of two systems

31, (4)

2

3;

31, (5)

2

1

x

x

x

x

− < < −

>

− < < −

< −

System (4) has no solutions. The solution of system (5) is3

, 12

x

∈ − −


1 1,. . ( 1,3).

( 3)( 1) 0, 1 3

x xor i e x

x x x

> − > − ∈ −

− + < − < <

Answers: 3

, 1 ( 1,3)2

x

∈ − − ∪ −

Illustration 21:solve 4 2

2

2 1log log 0

3x

x

x+

−<

+

Solution. The solution is the union of the sets of solution of the systems

26

2 2

2 2

4 40 1, 1,

2 2

2 1 2 1log 0 log 0

3 3 (1) & (2)

2 1 2 10 0

3 3

2 1 2 1log 1; log 1;

3 3

x x

x x

x x

x x

x x

x x

x x

+ + < < >

− − > >

+ +

− − > > + + − − > <

+ +


0<x+4<2,

2 1 4 2,13

1 ( 3) 0, 1

( 3) 0, 22

2 1 2 602 1

2; 33

x xx

x xx x

x xx

xx

− − < < − >+

⇒ − + > − + >

− − −>−

> + +

( )4 2,

3 0, 4, 3 .......... ( )

10

2

x

x x a

x

− > < −

+ < ⇒ ∈ − − − <


2, 2

2 1 2 1 31 0,

3 3 1 1

2 ( 3) 0, ( 3) 02 2

2 1 2 1 2 62; 0;

3 3

x x

x x x

x x

x x x x

x x x

x x

> − > − − − − − > >

+ +

⇒ − + > − + >

− − − −

< < + +

( )2,

4, 4,

1

2

x

x x

x

> −

⇒ > ⇒ ∈ +∞ >

Answer: ( )( 4, 3) 4,− − ∪ +∞

Illustration 22: Solve 2 ( ) 1,

( ) 2

y x

x y

x y

x y

−

−

+ =

+ =

Solution. The domain of definition is x + y >0

27

1

2

( ) 2

y x

x y

x y

x y

−

−

+ =

+ =

From the first equation we find ( ) 2x yx y −+ = and substitute it into the second equation.

Then

( ) ( ) ( )

2 22 2 2 2 1

x yx y x y x y−− −= ⇒ = ⇒ − =

The solution of the given system is the solution of the collection of systems

1,1,

& 12, ,

2

x yx y

x y x y

− = −− =

+ = + =

Answers: 3 1 1 3

, , ,2 2 4 4

−

Illustration 23: Solve 3 3 3

27

log log 2 log 2,

2log ( )

3

x y

x y

+ = +

+ =

Solution. The domain of definition is x > 0, y> 0.

3 3 3 3

27

log log log 2 log 9,

2log ( )

3

x y

x y

+ = +

+ =

In the first equation we use identity (7) and in the second equation we use the definition of a

logarithm. We obtain

( )

( )

3 3

2 / 3

log log 18, 18,

9.( ) 3 ,

xy xy

x yx y

= =⇒

+ =+ =

Solving the last system, we also get a solution of the given system.

Answers: ( ) ( ) 6,3 , 3,6

Illustration 24: Find the no. of solutions of the equation :

( ) 2 2

2

( 6 8) (2 2 3)log log 2

x x x xx x

+ + + +− = 0

28


Solution: Here

2

2

2

6 8 0 & 1

2 2 3 0 & 1

2 0

x x

x x

x x

+ + > ≠

+ + > ≠

− >

Now, 2 2

2

( 6 8) (2 6 3)log log ( 2 ) 0

x x x xx x

+ + + +

− =

2

2

(2 2 3)log ( 2 ) 1

x xx x

+ +− =

2 2

2

2 2 2 3

4 3 0

( 3)( 1) 0

1, 3

x x x x

x x

x x

x

⇒ − = + +

⇒ + + =

⇒ + + =

⇒ = − −

But for 3x = − ,2 6 8 0x + + < so 3x = − is not a solution.

Therefore 1x = − is only solution of the given equation.

1. Simplify of the followings :

4 3( ) log 5.log 9.log 2bi ( )1/4 2 3( ) log log .3log 4ii

2. Find the solution of the equation : 3log (3 8) 2x x− = −

3. Find the solution set of the in equation : 1

3

2log ( 1) 1 0x x+ + + >

29

Functions; Domain and Range: The temperature at which water boils depends on the elevation above

sea level (the boiling point drops as you ascend). The interest paid on a cash investment depends on the

length of time the investment is held. The area of a circle depends on the radius of the circle. The distance

an object travels from an initial location along a straight line path depends on its speed. In each case, the

value of one variable quantity, which we might call y, depends on the value of another variable quantity,

which we might call x. Since the value of y is completely determined by the value of x, we say that y is a

function of x. Often the value of y is given by a rule or formula that says how to calculate it from the

variable x. The letter x, called the independent variable, represents the input value of ƒ, and y, the

dependent variable, represents the corresponding output value of ƒ at x.

Definition Function

A function from a set D to a set Y is a rule that assigns a unique (single) element ( )f x YÎ

to each element x DÎ .

The set D of all possible input values is called the domain of the function. The set of all values of ƒ(x) as

x varies throughout D is called the range of the function. The range may not include every element in

the set Y.

30


1. Constant function f (x) = k Domain = _ Range = k

31


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