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mdav, June 21, 2009

11 PM

SUPPLEMENTARY TEXTUAL MATERIALIN

MATHEMATICS

(CLASS XI)

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CONTENT

C H A P T E R 3; Trigonometric Functions

Law of sines, and law of cosines

C H A P T E R 5; Complex numbers and quadratic equations

Square - root of a complex number

C H A P T E R 9 : Sequences and series

Sum to infinity of a GP.

C HA PT ER 10; Straight line

Shifting of Origin

C H A PTE R 13; Limits and Derivative

Some important limits

ACKNOWLEDGEMENTS

CBSE Advisors

Sh, Vineet Joshi, Chairman, CBSE.

S h, S hash i B hu sh an, Director (Acad), CBSE

Convenor & EditorProf. K.P. Chinda, Retd; Delhi University

Development Team

Sh. GO. Dhall, Retd , NCERT

Sh. I.C. Nijhawall, Retd, GNCT of Delhi.

Member Coordinator

Dr. Srijata Das, Education officer, CBSE.

2

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C H A P T E R - 3

T R IG O N O M E T R IC F U N CT IO N S

I. Law of Sines (or Sine Rule)

The sine rule states that the lengths of the sides of a triangle are proportional to the sines of angles opposite to

ab cthem, i.e. in ~ABC, sin A = sin B = sin C .

Case (0 When L ' > A B C is acute angled triangle

~AIIIIII

CIIIII

~L- ~L- ~

Proof: Three cases arise

Let a = BC, b =AC and c =AB

From vertex A , d raw AD l-BC

ADIn t.ABD, AS = sin B ~AD=csin B ..... (i)

B D C....-------a ------- ....

ADIn AACD , . - = sin C =*AD =b sin C ..... (ii)

AC

From (i) an d (ii) w e get, c sin B '" b sin C

b cor~=~sin B sin C

....... (A)

Similarly, by draw ing BEl-A C , we can prove that

a c...........(B)

sin A sin C

From (A) an d (B), we see that

a b c

sin A sin B sin C A

Case (ii) When t.A BC is an obtuse angled triangle

From vertex A , draw A Dl-BC produced

AD .InMBD, A B =Sill (J80°·B)

AD . - - . B" (.)or ---- =Slfl B " '" A D =c sm .. ., ... I

AB

3

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Similarly .. in MCD, ~ =in C

or AD =b sin C.... . .. (ii)

From 0 ) an d (ii), w e get

. . b cc sin B = b Sill C or __ = - -

sin B sinC

Similarly, by drawing BEJ.A C, we can show that

a c

sinA sine

Hence, a b c

sin A sinB sinC

Case (iii) When LlABC is a right ang led t ri ang le

1 . 1 1 L 'o .A BC,right angle d at B

(i) AB = sin C or 5 _ = sin C ~ b = _c_AC b sill C

(ii) BC = sin A or ~ = sin A ~ b = ~

AC b sin A

(iii) sin B=sin~ = 1 ~ _b_ = b2 sin B

From (i). (ii) an d (iii). w e get

b= ~ = _c_ = _b_

sin A sin c sin B

a b c~--=---=-_

sin A sin B Sill C

From all the three cases, we see that

a b c

sin A sin B sin C

Note: (i) _a_ = _ _ I J _ _ =_c_ =ksin A sin B sin C

~ a= k sin A, b = k sin B, c = k sin C

4

A

c b

B a C

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[ii) sin A ~ sin B ~ sin C ~ '( )II -- ~-- ~-- ~l. say

abc -'

: ; ; : : > sin A = at... ,sin B = bt..., sin C = c A .

These can be used in solving problems

?

Example 1 : InLlABC. if a =2, b =3 and sin A =~, find LB,, 3

Solution:abc

We know that ~ = = ~ = ~sin A sin B sin C

2Here a =2_ b = = 3 and sin A = = 'z:

:: >

2

2

3""" sin B = = 1 " '* B = = ~ or 90°

2in B

3

Example 2 : Inany triangle ABC, if the angles are in the ratio of I : 2 : 3 , prove that the corresponding sides are in

th e ratio of 1 :'; 3 :2 ,

Solution: Let the angles be 8, 28 and 38

As 8 + 28 + 38 = = 1800 = > 8 = = 30°

:_ The angles are 30°, 60°, 90°

Let the corresponding sides be a, b, C

abc abc__--= -- =-- or---= ---= ---

sin A sin B sin C sin 30° sin 60° sin 90°

abcor - =_. =' - =k (say)

l _ - . f 3 1 -'

22k .J3

"" a: b : c = - : - k : k or I : - . f 3 : 22 2

Example 3 : Inany triangle. prove that

sin (A-C)

sin (A+C)(ii) b cos B + c cos C =a cos (B-C)

Solution: (i) LHS = a2

- c2

= k2

sin2

A - Fsin'C [b y sine fOI1TIula]

b' k'sin'B

5

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sin 'A - sin' C sin (A+C) . sill (A - C)

sin' B sin' [1800 -(A+C)]

= sin (A+C) . sin (A-C) = sin (A-C) =RES

sin (A+C) . sin (A+C) sin (A+C)

(ii) LHS =b cos B + c cos C

=k [sin B cos B + Sill C cos C]

k [ . . 1- sin 2 B + srn 2 C2

= ~ [2 sin (B+C) cos (B-C)]2

=k [sin (l80"-A) cos (B-C)]

'" k sin A cos (B-C)

=a cos (B-C) = RES

Example 4 : Prove that

a (sin B- sin C) + b (sin C - Sill A ) + c (sin A - sin B) = 0

Solution:

abcWe know that --:---- = -.~ - = -.- . = K (say)

SlJJ A sin B SlIJ C -

--;> a =k sin A , b =k sin B, C =k sin C

:. LHS '" k sin A (sin B - sin C) + k sin B (sin C - sin A ) + k sin C (sin A - sin B)

=k [sin j I . . - s t i 1 B - sin j !ys'it l C + sin. _ B . - s i I I C - sin _ B . - s i I I A + sin ystil A - sin ysti1 BJ

= k.O = 0= RHS

Example 5 : In any MBC. prove that

1 + cos (A-B) cos C~-

1 + cos (A-C) cos B

a' + b'

Solution; We k now that in .6 .A BC , a =k sin A , b =k sin B, c = k sin C

.k ' (sin' A+ sin' B) 1- cos' A + sin' B.', LHS = . = --- _______

Y - (sin' A + sin' C) 1- cos-A + sin' C

1- (cos' A - sin' B)

I - (cos-A - sin' C)

6

= ~ [ I + cos 2A - I + cOslBJ

= ~ [.i cos (A+B) cos (A-B)]

1- cos (A-B) cos (A+B)

1 - cos (A-C) cos (A+C)

cos'A - sin' B

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1 + cos (A-B) cos C ~ RHS

1 + cos (A-C) cos C

EXERCISEl

I. In L I. ABC, if a = 18, b = 24 and c ~ 30 and LC = 90°, find sin A, sin B and sin C.

2. In any L I. ABC prove that

(A-B)cos --

2+b

c Csm

2

3.

b' -c' Sill (B - C)

a' sin (8 + C)

4. If a cos A =b cos B, then the triangle is either isosceles or right angled.

5 .

(A-B)tan --'_

')-b

a+ c tan (MB )2

6 . (B - C) b - c A

sm -- - -- cos -2 a 2

7.1 . + cos (A-B) cos C a2 + b2

1 . + cos (A-C) cos B a' - c'

8 .ABC

(b - c ) cot - + (c - a) cot- + (a - b) cot - ~ 02 2 2

9 . ( B - C ) Acos -- = (b + c) sin-2 2

10 .

c

a-b

A Btan - +tan -

2 2A B

tan - - tall -2 2

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11 . a (cos C - cos B) =2 (b - c) cos" ! 2 _2

12 . a sin A - b sin B =c sin (A - B)

Cosine RuleI.

In any triangle A BC, we have

(i)b' + c' -a '

a" '" b' + c ' - 2bc cos A a r cos A '" <

2bc

(ii)a' + c' -b2

b' =a'+ c' - lac cos B or cos B =<---~2ac

(ii) a' + b" - c'c' =a' + b ' - 2 ab cos C or cos C =---~2ab

Proof;

Case I :

Three cases arise :

W hen the L 'lA BC is an acute angled triangle.

From vertex A , draw A D..lBC

BDIn L l.A BD , cos B =~ => BD =c cos B

c

CDIn i'l.A CD . cos C = b" => CD =b cos C

Also, Ae" =CD" + AD'

=AD ' + (BC - BD)'

=BC ' + (AD ' + BD ') - 2BC .BD

AC" =Be" + AB' - 2BC.BD

a?'+ c2 _ b2or. cos B =---~

. 2 ac

Case II : When i'l.A BC is an obtuse angled triangle.

From vertex A , draw A D..lCB produced

I II i 'l .ABD,

BD- =cos (1800 - B) =- cos Bc

~ BD = -c cos B

Also, AC' =AD" + CD'

8

1fAIIIIII

C,,III, . .

B D C+-------a -------.

A

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'" AD' + (Be + BD),

'" A D' + BD ' + BCZ + 2BC.BD

AU= ABo + BU + 2BC.BD

or b' = c' + a' + 2a (·c cos B)

c' + a"·b'or cos B = ----

2ac

Case Ill: When L'lABCis a right triangle.

b' '" c' + a'

A

As B '" 1 r _ ~ cos B '" 02

b

:. b' =c' + a' . 2ac cos B [.: cos B =0 ]

c'+ a'· b'~ cos B> ----

2ac

B a c

c'+ a'· b 'Thus, in all the three cases cos B = ----

2ac

By follow ing the same procedure, w e can prove that

b 2 + c2 _ a"2 a2 + b 2 _ c2

cos A = . and cos C= ----2bc 2ab

Let us now take some examples:

Example 6; In a aABC, prove that a (b cos C . c cos B) = b' . c'

Solution: LHS : a (b cos C • c cos B)

b+c c+a a+b . casA cosB caseExample 7; If in an)' a ABC, -- = --= --, then prove that -- = -- =

12 13 15 2 7 11

9

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Solution:b+c

12

c+a

13

a+b-"k15

~ b+c '" 12k, c+a= 13k, a+b '" 15k

(b+c) + (c+a) + (a+b) =40k

'* a+b+c '"20k

'* a= Sk , b " . 7k , c= 5k

Cos A'"7

CosB= ----S O k ' 2ac

CasC= ----2ab

64k'+49k'- 25k]

112k2

88

112

11

14

1Cos A: Cos B : Cos C .., -::;

11- : - "'2: 7: 112 14

CosA CosB CosC--=--=--.

2 7 11

Examples 8: Ina D . ABC, prove thatb- c ccs A cos C

c-b cosA cos B

Solution: LHS=b-c cosA

c-b cosA

(b'+c'-a')c-b -'-------'-

)bc

(b ' +c' -a')b-e -'-------'-

2bc

=LHS

Example 9 : In a L \. ABC, if a = 18, b =24 and c =30, find cosA, cosls and casC

Solution: Here a '" 18. b '" 24 and c '" 30

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cos A'"576+900-324

2bc 1440

IIS2 4

1440 5

a'+c'-b'cos B= ----

2ac

324+900+576 6483--=-

1080 1080 5

324 + 576 - 900-----=0osC= ----

2ac 864

4 3.. cos A = - cos B = - cos C = 05 ~ 5 ~

Example 10 : In any 6. ABC, prove that

2 (be cosA + ea cos B + ab cos C) '" a~ + V + c'

{

(b2+c1_ a1) a"+c'_b" al+b1_cl)

Solution . LHS '" 2 be + ca + ab. 2bc 2ac 2ab

= a' + b'+ c' = RHS

Example 11 : In any flABC, prove that

b' -c' c' _a' a' -b'--, -. sin 2A + --, -.sin 2B + --, -. sin 2C =0a- b ' c-

b '_C1

[b1

+c1

_a'] c'-a' [a1+c'-b'] ft.' -,b'.2 (kc) [a2

+.b'-C'.]LHS=-,-·2(ka) + ---:-o--·2(kb) . . + .'

a- 2bc b- 2ac c 2ab

= ~[J/-~-#+/cJ+~-/-y / +~ +/-y-/c]+~J=O~RHSabc

cosA cosB cosC a2 + b2 + c'Example 12. In anv LlABC, prove tbat--+---+-- = ----

. 0 abc 2abc

11

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casA casB casCSolution; LHS =~~ +b' ~ +-

a c

2abc 2abe 2abe

1 [ ;I .r' , j ;I ,;I j ] a' +b" +c"= - b'+c'- +a'+,% -,0" + , +)3 -,% = --- =RHS

2~c 2~c

EXERCISE 2

In a 6ABC, if a=J, b=5 and c=I, find cosA, cosE and cosc.

2. Ifthe sides of a 6ABC are a=l, b=6 and c=S. shaw that 6 cosC =4+3 casB

.. ,3.

f :::- "J AIn any 6ABC, prove that a- =(b+c): -4bceas--

2

4. In a 6ABC, if LB = 60° , prove that (a+b+c) (a-b+c) =3ac

5. In any ll.ABC, prove that

(b'-c') cotA+ (c'-a') eotB+ (a'-b') eotC=O

sinA6. Ina AABC if casC = = 2 '. D prove that th e triangle is isosceles.

SltLD

7. In any 6ABC, prove that

(' b)' ,C ( 'b)' . ,C 1a- cas--+a+ -Slll--=C-

2 2

8. In any triangle ABC, prove that 2 ( b cos' ~+ e cos' ~) =a+b+e

9. In any A ABC, prove that

10. In any 6ABC, if LC =60", prove that

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1 1 1-+-=--a+c b+c a+b+c

ANSWERS:

1. sin A = Y s , sinB = ~ , sin C = 1

13 11 -IAns.l. cosA= 14,cosB=14,cosC="2

EXERCISE: 1

EXERCISE: 2

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C H A P T E R - S

C O M P L E X NUMBERS A ND Q U A D R AT IC E Q U AT IO N S .

1 . SQUARE- ROOT OF A COMPLEX NUMBER

We know that every negative real number has exactly two square - roots. For example-

.j:i =±J2 i, +6 = ± . J 6 ietc.

Let us now tr y to find the square - root of a complex.number suppose, we have to find .Ja + bi

Let =. J a + b i = = x + yi -

Squaring we get

a+bi = = (x'"i) + Zixy

Equating real and imaginary parts, we get

(i ) x'-y' = = J, and (ii) 2xy =b ;=} xy= ~

Now Cx'+i)' = (Cx'-y')' +( 2xy)') =a 2 + b'

==? (iii) x'+y' = = . J a' + b 2 [Positive sign as LHS is always positive

From (i) and (iii), we can find x and y [using (ii)]

Let us consider some examples

Examplel : find the square - root of 3+4i

Solution: Let .h+ 4i= x+yi

:. (x+yi)? '" 3+4i or x'-y'=3 and 2xy '" 4 (i)

Now, (x'+y')' = = (x'-y'P + (2xy)' = = 3'+4' =25

:. x'+y' =5 (ii)

, , '+y ') + (x'-yO ) = = 8 ==? X = = ± 2 and y' = = 1 ==? Y = = ± 1

A s xy is positive ==? when x = = 2 , y'" I and when x '" -2 , Y '" -1

==? The two square roots of 3 + 4i are

2+i and -2-i

Example 2: Find the square - root of -15 + 8i

Solution: Let .J-lS + 8i = = x+yi (i)

Squaring (i), we get -15+8i =(x'-y') + 2xyi

Equating real and imaginary parts, we get

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XC _y' '" -15 and Zxy >8

Now (x'+y')' '" (x'-y')' + (2xy)' '" (-15)' + (8)'

= 225+64 = 289 = (± 17 P

:. x '+ i = 17 (Rejecting negative sign)

We have found x '- y O = -1 5

:. 2x' = 2 ::::}x = ± 1

and 2y' = 32 ::::} Y = ± 4

As x.y is positive ~ x+yi has values

1+4iand-1-4i:

Example 3:

Find~

Solution: Let ~ = x + yi

: : ::} .5-12i = (x? - y') + Zxy i

Equating real and imaginary parts, we get

(i) x '-y ' =.5 and (n ) 2xy= -12

Now (x'+y'), = (x'f), + (2xy)' = 5'+ 12' = 169

:. x'+y'= 13 (iii)

From (i), (ii) and (iii), we get

2x ' = 18 ::::} x =± 3

andy=±2

As xy is negativc ee when x '" 3, Y '" -2

andwhen x '" -3, y = 2

The required square - roots are 3-2 i and -3 + 2i

or ± (3- 2i)

E X ER C ISE 1

Find th e square - roots of following complex numbers

ei) -15 - 8i

(ii) -3 -Ai

(iii) 2-2.J3i

15

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(iv) 8 + 6 i

(v) 7 - 2 4 i

Examples4: Solve the following quadratic equation:

,.'+.JISi x-i=O

Solution:

Here b' - 4ac = (.Jl5 i r +4.2 i

=-15+8i

-,flS i± . J -15+81

:::::>x=-------4

Let .J -15+8i =a+bi

____ 0 ) and2ab =8 (ii)

(a'+b')"= (a'-b")'+ (Zab)? = (-15)' + (8)' = 289

____ (iii)

.',a' = 1 :::::>'"± 1, b = ± 4

When a = I, b =4

When a = -1. b =-4

.. a + ib = 1+ 4i or -1-4i

•• X =-.Jl5 i± (1+4i)

4

.'. x =-1 - (.Ji5+4) i

or ----'--__!_-

4

Example 5 : Solve the quadratic equation

x' - x + (l+i) =0

Solution: Here a = 1, b = -1, c= l+l

Discriminant= b' - 4ac = (-1)' - 4 (1+i)

=1- 4 - 4i

'" (1)' + (2i)' - 2.(1) (2i)

= (2i - 1)'

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~ J b ' -4ac "" ± (2i-I)

:. x = 1 ± (2i-l) = - i + 1,i

2

EXERCISE 2

Solve the following quadratic equations:

(i) j x' - x+ 12i =0

(ii) x' - (3.J2 - 2i)x - . J 2 i = 0

(iii) x ~ - ( . J 2 + i)x + ..fi i=0

(iv) 2x' - (3+7i) x + (91- 3) =0

(v) x' - (3.fi -2i)x + 6.fi i= 0

ANSWERS:

EXERCISE 1:

(j) ± (l-4i)

(li) ± (1-2i)

(iii) ±(.fi - i)

(iv) ± (3+.i)

(v) ±.(4 - 3i)

EXERCISE 2

(i) -4i,3i

(iii) .fi.,I

3+ i .(iv) - 31

2 '

(v) 3.fi ?, _1

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CHAPTER-9

SEQUENCES AND SERIES

I. Sum to infinity of a GP.

Let us consider the G Po

3 9

1 , 5 ' 2 5 ' - - - -

Here a =1, r =Y s

~ r 1- ( Y s r ] - - - (i)

Let us study the behaviour of ( X r as n becomes larger and larger,

n 5 10 20

Y s ( Y s T (X r

= 006 =0007776, =0006047 0,00003656

( 3 ) "e observe that as 11 becomes larger and larger. 5 becomes closer and closer to zero 0

(3 ) "n other words, we can say that as n~ =, 5 " ~ 0

Thus, from 0 ) we find that the sum to infinitely many terms (S~) of the above geometric progression is

5aiven bv S =~'= ' _ , - 2

Now, for a geometric progression a, ar, ar", if 1 1 ' 1 < 1, then

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sn

a (1 - r" )

1 - r l-r l-r

As n ---7 00 r" ---7 0, as I r 1 < 1

.. S~ ---7a.

1 - [

F or ex am ple

1 1 1 1 3 11+- + - + - + =- =- asr=-

(i) 3 3' 3' ---- l-.!. 2 3

3

(li)

I ]1--+-3 3

2

1 I 3 1--+ =--=- as[=--3 ' - - - - 1 - ( ~ ) 4' 3

2 4Example 1 ; Find the sum of the infinite GP 1':3' " 9 ' _

Solution;2

Here a = 1 r = - ie r < 1

, 3'·

a 1. S =--=--=3.. ~ l-r 1_2/

/3

Example 2 ; Find the sum to infinity of the GP

10, -9,8.1, -7.29,

Solution: Here a = 10, r = -0.9

Since I f I < 1

.. S = _a_- l-r

10-------1-(-0.9)

10 100

1.9 19

=5.263

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Example 3 : Find the sum to infinity of the series

1 1 I 1 1 I-+- +- +-+- +- +3 5' 3' 54 3' 56

Solution: We here

5 2 _ 1 9 . 1 25+--. = - x -+- X -

l-_!_ 3" 8 25 ~ 24

5'

3 1 10 5-+-""-""-8 24 24 12

Example 4 : Prove that 3 /5 . 3J',; . 3Ys _ - - _ - =3

Solution: We have 3 1 5 . 3;;; . ){

_ [ , - + i,i, 1- 3 2 4 $ ---

'-

=3 t-!; = 31 =3

[I. Recurrmg decimal numbers as geometric series.

The sum to infinity of a geometric progression, with 1 1 " 1 < 1, can be applied in the infinite recurring

non:terminating decimal expansion of some real numbers. Let us take the simple case of

0.3 = 0.3333

We can write 0.3333 =0.3 +003 +0.003 + (i)

The RHS of (i), is the SLUn of infinite OP with a = 0.3 and r =0.1 ( I r l <1) .

.. S = _22_ = 0.3 = _ l_

- 1-0.1 0.9 3

1 1Thus. 0.3 = ::;-or we can say that the rationalnumber ::;-, when expressed as a decimal will have O . '3

~ ~

as its expansion.

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Example 5 : Find a rational number, which when expressed as a decimal, will have 0.68 as its expansion.

Solution: We write

0.68 =.68888- - - -

=0.6 + [0.08 + 0.008 + 0.0008 + }

=0.6 + 0.08 =0.6 + 0.081-0.1 0.9

6 8 54+8 62 31=-+-=--=-=-- 10 90 90 90 45

3J

Hence, the required rational number is 45

Example 6: The first term of a GP. is 2 and sum to infinity in 6. Find the conunon ratio.

Solution: Here a=Z S N = = 6

?: .. 6 = = - - = - arl-r = = 2~ = = I~ .

l-r /6 73

Exercise 1.

Find the sum to infinity in each of the following geometric progressions:

20 805 .:__ -, 7' 49'

2. 6, 1.2,0.24, _ ~ ~

3 .

-1

1, 3' 3" ",+'---

"

4.-5 5 5

4' 16' 64'---

-3 3 -35.

4' 16' 64' - --

6 . 0.3.0.18,0.108,

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7 . ( J 2 + 1),1, ( J 2 - 1 ) , ( J 2 - If, _

8. The COlIU110natio of a GP is

7sand the sum to infinity is 89° . Find the first term.

9. Find an infinite GP whose first term is 1 and each term is the sum of all the terms which follow it.

10. The sum of first two terms of an infinite GP is 5 and each term is three times the sum of the succeeding

terms. Find the GP.

11. Find the rational number having th e follo win g decim al expansions:

(i ) 0.15 (ii) 0.712 (iii) 3.52 (iv) 0.231 (v) 0.356

12 . Let x = 1+a+a~ + and y = I +b+b? , where lal <1 and Ibl < 1. Prove that

l+ab+a'b'+

xy

x+y-L:

13. If the sum of an infinite geometric series is 15 and the sum of the squares of three terms is 45. Find the

series.

14. The sum of an infinite GP. is 57 and the sum of their cubes is 9747, find the Gp.

I I I

15. Prove that 62, 6~' 68,--- = 6

ANSWERS:

EXERCISE: 1

35 3 -31.

32 . 7 .5 3 .

44. -1 5 .

5

4 + 3 J 2 I 16. 0.75 7. 8 . 16 9 . 1 ~ ~ ---

2 ' 2' 4'

4 I ~ ~ 1[4 712 317 231 35310.

, '4' 16'11 . (i) 99 (ii) 999 (iii) 90 (iv)

999(v)

990

10 20 19 38 7612 . 5+-+-+--- 14. - ,

3 9 ' , ' 9)

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CHAPTER~10

ST RAIG HT L IN E

L SHIFTING OF ORIGIN

The position of origin and the direction of axes plays a major role in describing a curve in terms of

equations. An equation corresponding to a set of points with reference to a system of coordinate axes

may be simplified by taking the set of points in some other suitable coordinate system. One such transfor-

mation is when origin is shifted to a new point and new axes are transformed parallel to th e original axes.

To see how the coordinates of a point of the

plane changed under shifting of origin (or trans-

lation of axes).

y.

Let 0 be the origin and P (x.y) be a point re- N N' P (X, Y)

ferred to the axes OX and OY. I- -- - - - _. - - -- - - - - .,

M' ~x'

0' h,k)I

I

IX

L M

Let 0' X' and 0' Y' be the new axes parallel

to OX and OY respectively, where 0 is the neworigin --+------1--------...,.---+

Let O L =hand 0' L =k

Let the coordinates of P referred to new axes--+------+------- ......--+

be (X,Y)tben O'M'= X and P'M' =Y. OM

= x and PM = Y

o

:. x = OM = OL +LM= h + OM = h+X

y =PM = :MM'+-PM'=K+PM' =k + Y

Thus, x =X + h, y =Y+k give the relation between the old and new coordinates.

Thus. if the equation of the set of points P with respect to OX and OY be f (x, y) =0, the equation to the

same set of points when. origin is shifted to 0 becomes f (X+h, Y+k) =0, where X, Yare coordinates

with reference to new axes O' X' and 0' Y'.

If, therefore, the origin is shifted at a point (h.k), we should substitude X + hand Y + k for x and yr espec ti v e ly .

The transformation formula from new axes to old axes is X =x-h, Y =y-k. The coordinates of old origin,

referred to new axes are (-h, -k).

Example l: Find the new coordinates of the point (3,-5) if origin is shifted to the point (2,3) by a translation of

axes.

Solution: Coordinates of new origin are (h,k) = (2,3) and the original coordinates ofpoint are (3,-5) = (x.y)

:. new coordinates (X, Y) are given by

x = X + h, y = Y + k i.e, 3 = X + 2. -5 =Y + 3:. X = 3-2 = I, Y = ·5·3 =-8

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Hence, the coordinates of the point (3,-5) in new system are (L-8)

Example 2 : Find what the equation x' + xy - 3y' - Y + 2 =0 becomes when the origin is shifted to the point

(l,m

Solution: Let the coordinates of a point P changes from (x.y) to (X,Y) when origin is shifted to (1,1)

Substiluting in the given equation, we get

(X+1), + (X+l) (Y+l) -3(Y+lr' - (y+l) + 2 = 0

~ x' + 2x + 1 + XY + X + Y + 1 - 3(y' + 2y + 1) - (Y + l) + 2 =0

~ X ' - 3Y' + XY + 3X -6Y -1 = 0

:. Equation in new system is X' - 3Y' + XY + 3X - 6Y = 0

Example 3: Find the point to which the origin should be shifted after shifting of origin so that tile equation

X C - 12;;:+ 4 = 0 w ill have no first degree term.

Solution: Let origin be shifted to (h, k) and P ex, y) becomes

P (X + h, Y +k). Substituting in the given equation we get

(X+h), -12 (X+h) + 4 =0

~ X' + 2hX + h"-12X -12h + 4 =0

Since there is no first degree term: 2h - 12 = 0

or h = 6

Hence origin shouLd be shifted to (ri.k) for any real value k.

Example 4: Verify that the area of the triangle wi th ver ti ce s (4,6), (7,10) an d (1,-2) remains invariant under the

translation of axes when origin is shifted to tile point (-2, 1)

Solution: Let P (4. 6), Q (7,10) and R (- L,2) be the given points

1.'; Area of t. PQR = '2 [4 (10-2) + 7(2-6) -1 (6-10)]

2 [32 - 28 + 4] =4 sq.U.

Now shifting (x.y) to (X-2, Y+l)

New coordinates are X =x+2, Y =y-J

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:. P (4, 6) ---7 (6, 5)

Q (7, 10) ---7 (9, 9)

R (-1,2) ---7 (1, 1)

1:. Area of Ll "" " 2 [6 (9-1) + 9 (1-5) + 1 (5-9)]

1" " " 2 [48 - 36-4] ""4 sq. units

Hence the area remains invarient.

Exercise 1.

1. Find the new coordinates of the points in each of the following, ifthe origin is shifted to the point

(1,2) by translation of axes:

ei) (4,4) (ii) (4,5)

(iii) (9,4) (iv) (3,2)

(v) (7, -I) (vi) (2,5)

2. Find what the following equations become when origin is shifted to the point (2,3)

(I) x' + 2xy - Y' + Y + 3 "" 0 (ii) 3xy - x' - Y + x "" 0

(iii) 4x y + 2x - 3y + 2 "" 0 (iv) x" + Y' - 3x + 4y "" 0

3. If the origin is shifted to the point (1.-2), what do the following equations become?

(i) 2x' + y : - 4x + 4y "" 0 (ii) y' - 4x + 4y + 8 ""0

4. At what point the origin be shifted, if the coordinates of a point (4,5) becomes (-3,9) ?

5. Prove that the area of a triangle is invariant under the translation of the axes.

ANSWERS:

I. (i ) (3,2) (ii) (3) (iii) (8.2) (iv) (2,0) (v) (6,-3) (vi) (1,3)

2. (I) x"-y'+2xy+lOx-y-5=O

(ii) 3xy - x~ + 6x + 5y + 13=0

(iii) 4xy+14x+5y+21=0

(iv) x' + y'+ x+ 10y+ 19=0

3 . (i) 2x' + y' = 6 ,

(7, -4)

(ii) y' = 4x

4 .

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CHAPTER-I3

L IM I T A N D D E RIV A T IV E S

I. SOME IMPORTANT LIMITS

. 1I. CUTI -

,--') 0 x

1We can easily observe that as x ~ ° from the left hand side, -, gets smaller and smaller i.e. ~ - =,

x x

1and as x - - - + 0 from right band side, - gets greater and greater i.e. - - - + + ""

x x

1So, as x approaches to 0, either from left hand side or from right hand side, - never approaches to a

x

f in it e numbe r.

. I eim ]Hence, we say that e 1111_ - and both do not exist i.e, eim - does not exist.

. ~ _ _ _ ; ; O x : - : _ _ _ _ _ , o X X - - - - c l - 0 x

2 . eimX-Jo<l<l X

When x takes positive values only and successive values of x increase and become greater than any pre

- assigned positive real number, however large itmay be, then we say that x tends to infinity, i.e. x - - - + oo

1Clearly, as x ~ 00 - ~ O .

'x

. 1Thus we say that e 1111 - =0

: \ . - - - - 7 IlEJ X

3 . eim (I)1+ -x

1Let x > 1 then - < 1

x

:. Using Binornial Theorem, we have

( 1 ) ' _ . 1 x (x - 1 ) ( 1 ) 2 X (x- 1 ) (x - 2 ) ( 1 ) 31+- - 1 + x - + - + . - +

x x 2! x 3! x

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[

1 ] [ ( 1 ) ( 2 ) 1-- 1-- I--

I + I + 2 !x .. + x 3! x +~~~

1 2 3When x ---'t =«, each one of -,

x x xetc will tend to zero

. (. 1 ) X 1 1 I:. tun 1 + - = I + - + - + -. + =e.

x...;~ x I! 2 ! 31 ~--

. ( 1 )'hus, :. eJill I + - = e;< --'io " '" X

!

4. eim ( 1 + xFx --:) 0

ITaking - '" y, we have, as x ~ 0, y ~ CoO

X

I ( l J Y.eim(l + x ) ; ' : = . 6 1 1 1 1+- =e:-;~O '1-t..... Y

. ! .Hence tim (1+x)" = = e

.\·~o

5.log ( I + x )

e i m ----=--- ' -_ __f_

~ - - - - ' J . a x

e . log (1 + x) . 1 () ( ) . ! .im = tl111 - log 1 + x =eim log 1 + x .x

:-\-)-0 X X~OX :'-)00

= = log e = = 1 {log (1 + x) means log, (1 + x)}

e" - 16 . C i r n - -

; . > - - 7 0 X

Taking e-' - 1 = = y, we have, as x ~ 0, y ~ °ande'=l+y:::::> x=log(l+y)

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6 1 1 1 e' - 1 = C im y

,-->0 X )->0 log (i+v)

. 1 1Clln = - = 1y-> 0 log (1+y) J

Y

. e' - 1Hence elm -- = 1

x -)0 0 x

. a x _ I7. fun --, a> 0

,\~ 0 X

Taking a' - 1= y, we have. as x --7 0, y --7 0 and a' = 1+ Y =? x log a = log ( l + y)

. a X - ICll1-- = eim:-;----)0 X y - - - - ) 0

y loga = eim

log (1+y) y-> 0

log a .. ]log (1 + y) = og a

y

Hence,

a x _ ICim --- = log a; .;~ 0 x

2 8