II Sem Mathematics All Solutions

CSVTU II SEMESTER SOLUTION ALL UNITS

Prepared By Mrityunjoy Dutta (Assist proff. Pragati Enginefering Mathematics)

Unit I – Complex Numbers

(May-Jun-2006)

1. Prove that

nninn

ii n

2sin

2cos

cossin1cossin1

.

Sol: We have

1 sin cos1 sin cos

nii

2

2

1 cos sin2 2

1 cos sin2 2

2cos 2sin cos4 2 4 2 4 2

2cos 2sin cos4 2 4 2 4 2

2cos co4 2

n

n

i

i

i

i

s sin4 2 4 2

2cos cos sin4 2 4 2 4 2

cos sin4 2 4 2

cos sin4 2 4 2

cos4 2

n

n

i

i

i

i

sin cos sin4 2 4 2 4 2

cos sin cos sin4 2 4 2 4 2 4 2

cos sin4 2 4 2

n n

i i

n n n n n n n ni i

n n n ni

2

cos sin2 2

n nn i n

2. If , prove that . )sin()tan( ivuiyx v

uy

xtanhtan

2sinh2sin



Ans: We have

sin 2 sinh 2tancos 2 cosh 2 cos 2 cosh 2

sin sin cosh cos sin

x i yx iyx y x y

u iv u v i u hv

Equating real and imaginary part, we get

sin 2 sin coshcos 2 cosh 2

sinh 2 cos sincos 2 cosh 2

x u vx y

y u hvx y

Dividing both the equation we get sin 2 sin cosh

sinh 2 cos sinx u vy u hv

sin 2 cot sinh 2 cothsin 2 tan

sinh 2 tan

x u y vx uy hv

Find sum of the series Ans: Let

2 3

1 1cos cos 2 cos32 31 1sin sin 2 sin 32 3

1 1 1 1cos cos 2 cos3 sin sin 2 sin 32 3 2 3

1 12 3

log 1

log 1 cos

i i i

i

c

s

c is

c is e e e

c is e

c is

2 2 1

sin1 sinlog 1 cos sin tan2 1 cos

i

c is i

...................3cos312cos

21cos



2 2 1

2

1

2 1

2sin cos1 2 2log 1 cos 2cos sin tan2 2cos

2

sin1 2log 2 1 2cos tan2 cos

21 log 2.2cos tan tan2 2 2

log cos2 2

log cos2

2

c is i

c is i

c is i

c is i

c

s

Equating real and imaginary parts we get

log cos2

c and

2s

(Nov-Dec-2006)

3. If ........iii A iB

prove that 2 2tan ,2

BA B and A B eA

.

Ans: ........iii A iB

A iBi A iB

log log( )A iB i A iB

1 2 2 1log(1) tan log tan BA iB i A B iA

2 2 1log tan2

BA iB i A B iA

2 2 1log tan2 2A B Bi A B i

A

1 2 2tan , log2 2A B B A B

A

2 21tan , log2 2 2A B B A B

A



2 2tan , log2A B B A B

A

2 2tan ,2

BA B A B eA

(Proved)

6. Show that

2 2 3 4 1 21 1 1sin sin 2 sin sin 3 sin sin 4 sin ....... tan (sin / (1 cos sin ))2 3 4

Ans: Let 2 2 3 41 1 1sin sin 2 sin sin 3 sin sin 4 sin .......2 3 4

S

2 3 41 1 1cos sin cos 2 sin cos3 sin cos 4 sin .......2 3 4

C

2 3sin sinsin (cos sin ) (cos 2 sin 2 ) (cos3 sin 3 ) ....

2 3C iS i i i

2 3

2 3sin sinsin ....2 3

i i iC iS e e e

2 3(sin ) (sin )sin ....2 3

i ii e eC iS e

log(1 sin )iC iS e

2log(1 sin cos sin )C iS i

log (cos sin )C iS r i -------------(1)

Where 2 4(1 sin cos ) sinr and 2

1 sintan1 sin cos

So, equation (1) becomes log iC iS re

logC iS r i

2

2 4 1 sinlog (1 sin cos ) sin tan1 sin cos

C iS i

2

1 sintan1 sin cos

S

2

2 2 3 4 11 1 1 sinsin sin 2 sin sin 3 sin sin 4 sin ....... tan2 3 4 1 sin cos



7. If cos( ) ii re prove that: 1 sin( )log2 sin( )

.

Ans: cos( ) ii re

cos cosh sin sinh cos sini r ir

cos cosh cos , sin sinh sinr r

cos sincosh , sinhcos sin

r r

coscosh cos

sinsinhsin

r

r

cosh cos sinsinh sin cos

cos sinsin cos

e ee e

cos sin sin coscos sin sin cos

e e e ee e e e

2 sin( )2 sin( )

ee

2 sin( )

sin( )e

sin( )2 logsin( )

1 sin( )log2 sin( )

(Proved)

(May-Jun 2007)

8. If 12cos xx

and 12cos yy

, show that one of the value of 1m nm nx y

x y is

2 cos( )m n .

Ans: 12cos xx

22 cos 1x x 2 2 cos 1 0x x



22cos 4cos 4

2x

2cos 2 sin cos sin

2ix i

----------------(1)

12cos yy

22 cos 1y y

2 2 cos 1 0y y

22cos 4cos 4

2y

2cos 2 sin cos sin

2iy i

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

Taking +ve sign we get

cos sin , cos sinm mx m i m x m i m and

cos sin , cos sinn ny n i n y n i n

cos sin cos sinm nx y m i m n i n

cos sinm nx y m n i m n and

1 cos sinm nm nx y m n i m n

x y

Now, 1m n m n m n

m nx y x y x yx y

1 cos sin cos sinm nm nx y m n i m n m n i m n

x y

1 2cosm nm nx y m n

x y (Proved)

9. If tan( ) ii e prove that:

i. 12 2

n

.

ii. 1 log tan2 4 2

Ans: tan( ) ii e



tan( ) cos sini i ------------(1)

So, tan( ) cos sini i ----------------(2)

(i) Now, tan 2 tan i i

tan tantan 2

1 tan tani i

i i

cos sin cos sintan 2

1 cos sin cos sini i

i i

2 costan 2 tan

1 1 2

22

n

1

2 2n

(ii) tan 2 tani i i

tan tantan 2

1 tan tani i

ii i


1 cos sin cos sini ii

i i

tanh 2 sini i tanh 2 sin

2 2

2 2

sin1

e ee e

2 2 2 2

2 2 2 2

1 sin1 sin

e e e ee e e e

24

2

cos( / 2) sin( / 2)cos( / 2) sin( / 2)

e

2 cos( / 2) sin( / 2)cos( / 2) sin( / 2)

e

2 1 tan( / 2)

1 tan( / 2)e

22

n 1

2 2n

(Proved)

10. Find the sum of the series2 3

sin sin 2 sin 3 .......2 3x xx .



Ans: Let 2 3

sin sin 2 sin 3 .......2 3x xS x

2 3

cos cos 2 cos3 .......2 3x xC x

2 3

(cos sin ) (cos 2 sin 2 ) (cos3 sin 3 ) .......2 3x xC iS x i i i

2 3

3 .......2 3

i i ix xC iS xe e e

2 3( ) ( ) .......

2 3

i ii xe xeC iS xe

log(1 )iC iS xe log(1 cos sin )C iS x ix

log (cos sin )C iS r i -------------(1)

Where 2 2 2(1 cos ) sinr x x 1 sintan1 cos

xx

21 2 cosr x x

So, equation (1) becomes log iC iS re logC iS r i

2 2 2 1 sin(1 cos ) sin tan1 cos

xC iS x x ix

1 sintan1 cos

xSx

2 3

1 sinsin sin 2 sin 3 ....... tan2 3 1 cosx x xx

x

(Ans)

(Nov-Dec 2007)

11. State De’Moivre’s Theorem. Ans: De’ Moivre’s Theorem: - For any real value of n, (cos sin ) cos sinni n i n .

12. Find sum of the series


S


C

........sin4sin41sin3sin

31sin2sin

21sin 4322




2 3C iS i i i

2 3


i i iC iS e e e

2 3(sin ) (sin )sin ....2 3

i ii e eC iS e

log(1 sin )iC iS e


log (cos sin )C iS r i -------------(1)


1 sintan1 sin cos


logC iS r i

2


C iS i

2

1 sintan1 sin cos

S

2


13. If 0coscoscossinsinsin , prove that

2 2 2 2 2 2cos cos cos sin sin sin . and cos2 cos2 cos2 sin2 sin2 sin2 .

Ans: - Let cos sincos sincos sin

x iy iz i

From the given information we will have 0sincossincossincos0 iiizyx …………1



0)sinsin(sin)coscos(cos i

0sinsinsin,0coscoscos

Now, sincos

1sincos

1sincos

1111iiizyx

sincossincossincos111 iiizyx

00)sinsin(sin)coscos(cos111 iizyx

0111

zyx ………………………………………………2

22 2 2 2( )x y z x y z xy yz zx

20 2( )xy yz zx ………….from …1

1 1 12 0xyzx y z

(from 2)………………………….3

2 2 22 2 2 cos sin cos sin cos sinx y z i i i

By applying the De’ Moivre’s Theorem we will have


= (cos 2 sin 2 ) cos 2 sin 2 cos 2 sin 2i i i

From 3 we will have

(cos 2 sin 2 ) cos 2 sin 2 cos 2 sin 2i i i =0

Therefore cos 2 cos 2 cos 2 sin 2 sin 2 sin 2 =0

cos 2 cos 2 cos 2 = 2 2 2 2 2 2cos sin cos sin cos sin =0

Therefore 2 2 2 2 2 2cos cos cos sin sin sin =0….Hence Proved

14. If , show that and .

Ans: tan( ) ii e

tan( ) cos sini i ------------(1)

So, tan( ) cos sini i ----------------(2)


iei )tan(22

1

n

24tanlog

21



tan tantan 2

1 tan tani i

i i



i i

2 costan 2 tan

1 1 2

22

n

1

2 2n

(ii) tan 2 tani i i

tan tantan 2

1 tan tani i

ii i



i i


2 2

2 2

sin1

e ee e

2 2 2 2

2 2 2 2

1 sin1 sin

e e e ee e e e

24

2

cos( / 2) sin( / 2)cos( / 2) sin( / 2)

e

2 cos( / 2) sin( / 2)cos( / 2) sin( / 2)

e

2 1 tan( / 2)

1 tan( / 2)e

22

n 1

2 2n

(Proved)

(May-Jun 2008)


16. Find sum of the series ........sin4sin41sin3sin

31sin2sin

21sin 4322




S


C


2 3C iS i i i

2 3


i i iC iS e e e

2 3(sin ) (sin )sin ....2 3

i ii e eC iS e

log(1 sin )iC iS e


log (cos sin )C iS r i -------------(1)


1 sintan1 sin cos


logC iS r i

2


C iS i

2

1 sintan1 sin cos

S

2


17. If 0coscoscossinsinsin , prove that

2 2 2 2 2 2cos cos cos sin sin sin . and cos2 cos2 cos2 sin2 sin2 sin2 .

Ans: - Let



cos sincos sincos sin

x iy iz i




Now, sincos

1sincos

1sincos

1111iiizyx



0111

zyx ………………………………………………2

22 2 2 2( )x y z x y z xy yz zx

20 2( )xy yz zx ………….from …1

1 1 12 0xyzx y z

(from 2)………………………….3


By applying the De’ Moivre’s Theorem we will have


= (cos 2 sin 2 ) cos 2 sin 2 cos 2 sin 2i i i

From 3 we will have

(cos 2 sin 2 ) cos 2 sin 2 cos 2 sin 2i i i =0

Therefore cos 2 cos 2 cos 2 sin 2 sin 2 sin 2 =0

cos 2 cos 2 cos 2 = 2 2 2 2 2 2cos sin cos sin cos sin =0

Therefore 2 2 2 2 2 2cos cos cos sin sin sin =0….Hence Proved

18. Find the sum of the series . 2 3

sin sin 2 sin 3 .......2 3x xx



Ans: Let 2 3

2 3

2 3 2 3

2 2 3 3



1 12 3

i i i

c x x x

s x x x

c is x x x x x

c is xe x e x e

c

2 2 2 1

2 1

log 1

log 1 cos sin1 sinlog 1 cos sin tan2 1 cos1 sinlog 1 2 cos tan2 1 cos

iis xe

c is x ixxc is x x i

xxc is x x i

x


2

1

1 log 1 2 cos2

sintan1 cos

c x x

xsx

(Nov - Dec 2008)

19. If 12cos xx

, find the value of 푥 .

Ans:

2

2

12cos

2 cos 1 0

2cos 4 4

cos sin

xx

x x

csox

x i

By De-moivers theorem

cos sin

cos sin

pp

p

x i

x p i p

20. If 2r rx Cis

, show that 1 2lim . ...... 1.

nn

x x x .

Ans: - Given that 2r rx Cis



Now, 1 2 1 2lim . ...... lim . ........2 2 2n nn n

x x x Cis Cis Cis

1 2 1 2lim . ...... lim .......2 2 2n nn n

x x x Cis

1 2 1 2 3lim . ...... .......2 2 2nn

x x x Cis

1 21 1lim . ...... ( ) cos sin 1 012 1

2

nnx x x Cis Cis i i

1 2lim . ...... 1nnx x x

(Ans).

21. Separate 푡푎푛 (푥 + 푖푦) into real and imaginary parts.

Ans: Let 1

1

tan ( ) (1)tan ( ) (2)

i x iyi x iy

Adding (1) and (2) , we get

1 1

1

12 2

12 2

12 2

2 tan ( ) tan ( )( ) ( )2 tan1 ( )( )

22 tan1

22 tanh1

1 2tan2 1

x iy x iyx iy x iy

x iy x iyx

x y

xx y

xx y

Subtracting equation (2) from equation (1), we get 1 1

1

12 2

12 2

12 2

2 tan ( ) tan ( )( ) ( )2 tan

1 ( )( )22 tan

122 tanh

11 2tanh2 1

i x iy x iyx iy x iyi

x iy x iyyi i

x yyi i

x yy

x y



Thus real part of 1tan ( )x iy = 12 2

1 2tan2 1

xx y

and imaginary part of 1tan ( )x iy =

12 2

1 2tanh2 1

yx y

.


Ans: - Let 퐶 = 1 − + ..푐표푠2휃 − . .

. .푐표푠3휃 + … … … … … … . .∞

Then 푆 = 0− + .

.푠푖푛2휃 − . .

. .푠푖푛3휃 + … … … … … … . .∞

Now 퐶 + 푖푠 = 1− (푐표푠휃 + 푖푠푖푛휃) + .

.(푐표푠2휃 + 푖푠푖푛2휃) − . .

. .(푐표푠3휃 + 푖푠푖푛3휃)+. .∞

⇒ 퐶 + 푖푠 = 1 − 푒 + ..푒 − . .

. .푒 +. .∞

⇒ 퐶 + 푖푠 = 1 + 푒 +−1

2 . − 12− 1

1.2푒 −

−12 . −1

2 − 1 . −12 − 2

1.2.3푒 +. .∞

⇒ 퐶 + 푖푠 = 1 + 푒 ⇒ 퐶 + 푖푠 = (1 + 푐표푠휃 + 푖푠푖푛휃)

⇒ 퐶 + 푖푠 = 2푐표푠 + 푖2푠푖푛 푐표푠 = 2푐표푠 푐표푠 + 푖푠푖푛

⇒ 퐶 + 푖푠 = 2푐표푠 푐표푠 − 푖푠푖푛 = 푐표푠 2푐표푠 − 푖푠푖푛 2푐표푠

So, ⇒ 퐶 = 푐표푠 2푐표푠

⇒ 1 − + ..푐표푠2휃 − . .

. .푐표푠3휃 + … … . .∞ = 푐표푠 2푐표푠 (Ans).

(May-Jun 2009)


24. Find all the roots of the equation: (i) cos 2z . Ans: cos 2z

22 4 4 1 02

iz iziz iz iz ize e e e e e

4 16 4 4 2 3 2 3

2 2ize

2 3 2 log 2 3iz Log n i

2 log 2 3z n i

...................6cos6.4.25.3.14cos

4.23.12cos

211



2 log 2 3z n i (Ans)

(ii) tanh 2z . Ans: tanh 2z

2 2 2z z

z z z zz z

e e e e e ee e

3 0z ze e 2 3ze 2 ( 3) (3) ( 1) 2 log 3z Log Log Log n i i 2 log 3 (2 1)z n i

1 1log32 2

z n i

(Ans)

25. Separate 1sin (cos sin )i into real and imaginary parts, where is a positive acute angle.

Ans: Let 1sin (cos sin )i x iy

(cos sin ) sin( )i x iy (cos sin ) sin cosh cos sinhi x y i x y

cos sin cosh ..........(1) sin cos sinh ..........(2)x y x y By squaring and adding (1) and (2) we get 2 2 2 2sin cosh cos sinh 1x y x y

2 2 2 2sin (1 sinh ) cos sinh 1x y x y

2 2 2 2 2sin sinh cos sinh 1 sinx y x y x

2 2 2 2sinh (sin cos ) cosy x x x

2 2cos sinhx y Now, from (2) we get

2 2 2 2 2 4sin cos sinh cos cos cosx y x x x

2cos sinx as is positive acute angle.

cos sinx

1cos sinx ………….(3)

By using (2) and (3) we get

sin sin sinh y

sinh sin sin2

y ye ey

2 2 sin 1y ye e



2 sin 4sin 4 2 sin 2 1 sin sin 1 sin

2 2ye

log sin 1 siny ………….(4)

So, 1 1sin (cos sin ) cos sin log sin 1 sini i (Ans)

26. Find the sum of the series ( 1) ( 1)( 2)sin sin 2 sin 3 .......1.2 1.2.3

n n n n nn .

Ans: Let ( 1) ( 1)( 2)sin sin 2 sin 3 .......1.2 1.2.3

n n n n nS n and

( 1) ( 1)( 2)cos cos 2 cos3 .......1.2 1.2.3

n n n n nC n

Then,

( 1) ( 1)( 2)(cos sin ) (cos 2 sin 2 ) (cos3 sin 3 ) ...1.2 1.2.3

n n n n nC iS n i i i

2 3( 1) ( 1)( 2) ...1.2 1.2.3

i in n n n nC iS ne e e

2 3( 1) ( 1)( 2)1 ... 11.2 1.2.3

i in n n n nC iS ne e e

1 1niC iS e

1 11

niC iS

e

1 11 cos sin nC iS

i

2

2 2 2

1 12sin 2sin cos

nC iSi

2 2 2

1 12 sin sin cos nn n

C iSi

2 2 2 2 2

1 12 sin cos sin

nn nC iS

i

2 2 2 2 2

1 1 12 sin cos sinn nC iS

n i n

2 2 2 2

2

cos sin1

2 sinn n

n i nC iS



2 2 2 2

2 2

cos sin1

2 sin 2 sinn n n n

n nC iS i

2 2

2

sin2 sinn n

nS

2 2

2

sin( 1) ( 1)( 2)sin sin 2 sin 3 .......1.2 1.2.3 2 sinn n

nn n n n nn

(Ans)

(Nov-Dec 2009)

27. If 2r rx Cis

. Find the value of 1 2lim . ...... nn

x x x

.

Ans: - Given that 2r rx Cis

Now, 1 2 1 2lim . ...... lim . ........2 2 2n nn n

x x x Cis Cis Cis

1 2 1 2lim . ...... lim .......2 2 2n nn n

x x x Cis

1 2 1 2 3lim . ...... .......2 2 2nn

x x x Cis

1 21 1lim . ...... ( ) cos sin 1 012 1

2

nnx x x Cis Cis i i

1 2lim . ...... 1nnx x x

(Ans).

28. If ........iii A iB


BA B and A B eA

.

Ans: -........iii A iB

A iBi A iB log log( )A iB i A iB


2 2 1log tan2

BA iB i A B iA


A


A




A


A

2 2tan ,2

BA B A B eA

(Proved)

OR

29. Find the sum to infinite of the series: 1 1.3 1.3.51 cos cos 2 cos3 .......2 2.4 2.4.6

Ans: - Let 퐶 = 1 − + ..푐표푠2휃 − . .

. .푐표푠3휃 + … … … … … … . .∞

Then 푆 = 0− + .

.푠푖푛2휃 − . .

. .푠푖푛3휃 + … … … … … … . .∞


.(푐표푠2휃 + 푖푠푖푛2휃) − . .

. .(푐표푠3휃 + 푖푠푖푛3휃)+. .∞

⇒ 퐶 + 푖푠 = 1 − 푒 + ..푒 − . .

. .푒 +. .∞

⇒ 퐶 + 푖푠 = 1 + 푒 +−1

2 . − 12− 1

1.2푒 −

−12 . −1

2 − 1 . −12 − 2

1.2.3푒 +. .∞





⇒ 1 − + ..푐표푠2휃 − . .

. .푐표푠3휃 + … … . .∞ = 푐표푠 2푐표푠 (Ans).

30. If log tan4 2

u

, proved that:

i. tanh tan2 2u

.

Ans: - 푢 = 푙표푔 푡푎푛 + ⇒ 푢 = 푙표푔 푡푎푛 + ⇒ 푒 = 푡푎푛 +

⇒/

/ = ⇒/ /

/ / = ⇒ 푡푎푛ℎ = 푡푎푛 (Proved)

ii. log tan4 2

iui

Ans: - From Ans (i)

⇒ 푡푎푛ℎ = 푡푎푛 ⇒ 푡푎푛 = 푡푎푛ℎ ⇒ = 푡푎푛ℎ 푡푎푛



⇒ = 푙표푔 ⇒ 휃 = 푙표푔.

⇒ 휃 = −푖푙표푔 푡푎푛 + 푖 (Ans)

May –June (2010)

31. What is the imaginary part of )34( iLog .

Ans: - 2 2 1( ) log 2 tan yLog x iy x y i nx

Here, Imaginary part of

xyniyxLog 1tan2)(

So, Imaginary part of 1 3(4 3) 2 tan4

Log i n

32. If ...........)1( 2210 xPxPPx n show that,

(i)

4cos2........... 2/

420nPPP n .

Ans: - ...........)1( 2210 xPxPPx n

...........)1( 2210 xPxPPx n

...........2)1()1( 44

220 xPxPPxx nn

Putting ix , we get ...........2)1()1( 420 PPPii nn

...........24

sin4

cos24

sin4

cos2 4202/12/1

PPPii

nn

...........24

sin4

cos24

sin4

cos2 4202/2/

PPPninnin nn

...........24

cos22 4202/ PPPnn

4

cos2........... 2/420

nPPP n (Proved).

(ii)

4sin2........... 2/

531nPPP n

Ans: - ...........)1( 2210 xPxPPx n

...........)1( 2210 xPxPPx n

...........2)1()1( 55

331 xPxPxPxx nn

Putting ix , we get



...........2)1()1( 531 iPiPiPii nn

...........24

sin4

cos24

sin4

cos2 5312/12/1

PPPiii

nn

...........24

sin4

cos24

sin4

cos2 5312/2/

PPPininnin nn

...........24

sin22 5312/ PPPinin

4

sin2........... 2/531

nPPP n (Proved).

33. If iivu )(sin 1 , prove that 2sin and 2cosh are the roots of the equation 0)1( 2222 uvuxx

Ans: - To prove 2sin and 2cosh are the roots of the equation

0)1( 2222 uvuxx

is same as to prove 2 2 2 2 2 2 2sin cosh (1 ), sin .coshu v u

1sin ( ) sin( )u iv i u iv i sin .cosh cos .sinhu iv i -------------------(1) From(1) modulus square is

2 2 2 2 2 2sin .cosh cos .sinhu v

2 2 2 2 2 2sin .cosh (1 sin ).(cosh 1)u v

2 2 2 2 2 2 2 2sin .cosh cosh 1 sin .cosh sinu v

2 2 2 2cosh 1 sinu v

2 2 2 2sin cosh (1 )u v (Proved -1) Form (1) sin .cosh , cos .sinhu v

2 2 2sin .coshu (proved – 2)

Hence 2sin and 2cosh are the roots of the equation 0)1( 2222 uvuxx .

34. Find sum of the series: ...............5sin5

3sin3

sin53

ccc .

Ans: - Let 3 5

sin sin 3 sin 5 ...............3 5c cS c

So, 3 5

cos cos3 cos5 ...............3 5c cC c

3 5

cos sin cos3 sin 3 cos5 cos5 ...............3 5c cC is c



3 5

3 5 .......3 3

i i ic cC iS ce e e

3 5

.......3 3

i ii ce ce

C iS ce

As

xxxxxx

11log

21tanh.................

531

53

1 1 1log log 1 log 12 1 2

ii i

i

ceC iS ce cece

2 2 2 1

2 2 2 1

1 sin(1 cos ) sin tan2 1 cos

1 sin(1 cos ) sin tan2 1 cos

cC iS c c ic

cc c ic

2 2 2 2 2 2

1 1

1 (1 cos ) sin (1 cos ) sin21 sin sintan tan2 1 cos 1 cos

C iS c c c c

c cic c

1 11 sin sintan tan2 1 cos 1 cos

c cSc c

1 11 sin sintan tan2 1 cos 1 cos

c cSc c

1 12

sin sin1 1 2 sin1 cos 1 costan tansin sin2 2 11 .

1 cos 1 cos

c ccc cS c c c

c c

3 5

12

1 2 sinsin sin 3 sin 5 ............... tan3 5 2 1c c cc

c

(Ans)

Nov Dec (2010)

35. Find :

4

5

(cos sin )(sin cos )

ii

Sol 4

5

(cos sin )(sin cos )

ii

We know from De’ Moivre’s Theorem: - For any real value of n,(cos sin ) cos sinni n i n .



4

55 5 5

(cos sin ) (cos 4 sin 4 ) (cos 4 sin 4 )(sin cos ) (sin cos ) cos sin

i i ii i i i

4

55 5

(cos sin ) (cos 4 sin 4 ) (cos 4 sin 4 )(sin cos ) (sin cos ) cos sin

i i ii i i i

54

5 5

(cos 4 sin 4 ) cos sin(cos sin ) (cos 4 sin 4 )(sin cos ) (sin cos )

i ii ii i i

54

5 5

(cos 4 sin 4 ) cos5 sin 5(cos sin ) (cos 4 sin 4 )(sin cos ) (sin cos )

i ii ii i i

4

5 5

(cos 4 5 sin 4 5(cos sin ) (cos 4 sin 4 )(sin cos ) (sin cos )

ii ii i i

4

5 5

(cos sin ) (cos 4 sin 4 ) (cos 4 5 sin 4 5(sin cos ) (sin cos )

i i i ii i

4

5 5

(cos sin ) (cos 4 sin 4 ) sin 4 5 (cos 4 5(sin cos ) (sin cos )

i i i ii i

36. If x = cos sini , y= cos sini , z= cos sini and x+ y +z =0, then prove that

1 1 1 0.x y z

Ans: - Let cos sincos sincos sin

x iy iz i




Now, sincos

1sincos

1sincos

1111iiizyx



0111

zyx ………………………………………………2 Hence Proved



37. If ( ) p x iya ib m then prove that one of the value of y/x is 2 2

2 tan ( / )log ( )e

b aa b

.

Ans:

2 2 1

log( ) loglog( ) log

1 log tan log log2

p x iy

p x iy

a ib m

a ib mp a ib x iy m

bp a b i x m iy ma

Equating real and imaginary parts

2 2

1

1log log ...........................(1)2

log tan ............................(2)

x m p a b

by m pa

Dividing (2 )by (1)

1 1

2 22 2

tan 2 tan

1 loglog2

b bpy a ax a bp a b

38. Sum the series 2 3

sin sin( ) sin( 2 ) sin( 3 ).......2 3

x xx

Ans: let

2

2

2

cos cos cos 2 ........................2!

sin sin sin 2 ........................2!

cos sin cos sin cos 2 sin 2 .........................2!

xc x

xs x

xc is i x i i

i xic is e xe

2

2

cos sin

sincos

cos

2........................................

2!21 ........................

2!

cos sin sin sin

x i

i xx

x

ie

xi iic is e xe e

ic is e e

c is e ec is e x i x

equating real and imaginary parts



cos

cos

cos sin

sin sin

x

x

c e x

s e x

May June (2011)

39. Find real and imagiant parts of exp(z2).

Ans: We have

22

2 2 2

2 2 2

2 2 2

2 2 2 2 2

exp exp

exp exp 2

exp exp .exp 2

exp exp . cos 2 sin 2

exp cos 2 exp sin 2 exp

z x iy

z x y ixy

z x y ixy

z x y xy i xy

z xy x y i xy x y

Hence real part is 2 2cos 2 expxy x y and imaginary part is 2 2sin 2 expxy x y .

40. If (1 + 푥) = 푃 + 푃 푥 + 푃 푥 + 푃 푥 + ⋯… … ..

Show that

(i) p0 - p2 + p4 - ---------------------- = 2n/2 cos n/4 , (iii) p1 – p3 + p5 – ---------------------- = 2n/2 sin n/4 .

(i)

4cos2........... 2/

420nPPP n .

Ans: - ...........)1( 2210 xPxPPx n

...........)1( 2210 xPxPPx n

...........2)1()1( 44

220 xPxPPxx nn

Putting ix , we get ...........2)1()1( 420 PPPii nn

...........24

sin4

cos24

sin4

cos2 4202/12/1

PPPii

nn

...........24

sin4

cos24

sin4

cos2 4202/2/

PPPninnin nn



...........24

cos22 4202/ PPPnn

4

cos2........... 2/420

nPPP n (Proved).

(ii)

4sin2........... 2/

531nPPP n

Ans: - ...........)1( 2210 xPxPPx n

...........)1( 2210 xPxPPx n

...........2)1()1( 55

331 xPxPxPxx nn

Putting ix , we get ...........2)1()1( 531 iPiPiPii nn

...........24

sin4

cos24

sin4

cos2 5312/12/1

PPPiii

nn

...........24

sin4

cos24

sin4

cos2 5312/2/

PPPininnin nn

...........24

sin22 5312/ PPPinin

4

sin2........... 2/531

nPPP n (Proved).

41. If ........iii A iB


BA B and A B eA

.

Ans: -........iii A iB

A iBi A iB log log( )A iB i A iB


2 2 1log tan2

BA iB i A B iA


A


A


A


A



2 2tan ,2

BA B A B eA

(Proved)

42. Find the sum to infinite of the series:

Cos + Cos ( ) + Cos ( 2 ) +….. to n terms.

Cos + Cos ( ) + Cos ( 2 ) +….. to n terms.

\

cos cos cos 2 ........................

sin sin sin 2 ........................

cos sin cos sin cos 2 sin 2 ............................

c nterms

s nterms

c is i i i nterms

iic is e e

2

........................................2

. . ................................

1

1

ie nterms

i ii i ic is e e e e e nterms

inie ec is ie

We know that

12

2 2 2 21 2 sin2

2 2 2 21 2 sin2

sin cos2 2

in

i n

i i i iie e e e e i

n n nin i i i ne e e e e i

nc is e ec


1cos sin cos

2 2 2

1sin sin cos

2 2 2

n nc ec

n ns ec

Nov Dec (2011)

43. State the De Moiver’s theorem.



Ans: De’ Moivre’s Theorem: - For any real value of n, (cos sin ) cos sinni n i n

44. Prove that 1(1 cos sin ) (1 cos sin ) 2 cos cos2 2

n n n n ni i .

Ans:

2 2

(1 cos sin ) (1 cos sin )

2cos .2sin cos 2cos .2sin cos2 2 2 2 2 2

2 cos cos sin 2 cos cos sin2 2 2 2 2 2

2 cos cos sin 2 cos cos sin2 2 2 2 2 2

n n

n n

n nn n n n

n n n n

i i

i i

i i

n n n ni i

1

2 cos 2cos2 2

2 cos cos2 2

n n

n n

n

n

45. If , prove that .

Ans: We have

sin 2 sinh 2tancos 2 cosh 2 cos 2 cosh 2

sin sin cosh cos sin

x i yx iyx y x y

u iv u v i u hv

Equating real and imaginary part, we get

sin 2 sin coshcos 2 cosh 2

sinh 2 cos sincos 2 cosh 2

x u vx y

y u hvx y

Dividing both the equation we get sin 2 sin cosh

sinh 2 cos sinx u vy u hv

sin 2 cot sinh 2 cothsin 2 tan

sinh 2 tan

x u y vx uy hv

46. Find the sum to infinite of the series: 1 1.3 1.3.51 cos cos 2 cos3 .......2 2.4 2.4.6

Ans: - Let 퐶 = 1 − + ..푐표푠2휃 − . .

. .푐표푠3휃 + … … … … … … . .∞

)sin()tan( ivuiyx v

uy

xtanhtan

2sinh2sin



Then 푆 = 0− + ..푠푖푛2휃 − . .

. .푠푖푛3휃 + … … … … … … . .∞


.(푐표푠2휃 + 푖푠푖푛2휃) − . .

. .(푐표푠3휃 + 푖푠푖푛3휃)+. .∞

⇒ 퐶 + 푖푠 = 1 − 푒 + ..푒 − . .

. .푒 +. .∞

⇒ 퐶 + 푖푠 = 1 + 푒 +−1

2 . − 12− 1

1.2푒 −

−12 . −1

2 − 1 . −12 − 2

1.2.3푒 +. .∞





⇒ 1 − + ..푐표푠2휃 − . .

. .푐표푠3휃 + … … . .∞ = 푐표푠 2푐표푠 (Ans).

May –June (2012)

47. Prove that : 4cos sin cos8 sin8

sin cosi ii

Sol. L.H.S.

=

4cos sinsin cos

ii

4

4

4

cos sin

s sin2 2

cos sin

s 4 sin 42 2

i

co i

i

co i

4

4

4

cos sins 2 4 sin 2 4

cos sin

s sin

ico i

ico i



4 4

8

cos sin cos sin

cos sin

cos8 sin8

i i

i

i

48. If 12cos xx

and 12cos yy

show that : 1 2 cos( )m nm nx y m n

x y

Ans: 12cos xx

22 cos 1x x 2 2 cos 1 0x x

22cos 4cos 4

2x

2cos 2 sin cos sin

2ix i

----------------(1)

12cos yy

22 cos 1y y

2 2 cos 1 0y y

22cos 4cos 4

2y

2cos 2 sin cos sin

2iy i

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

Taking +ve sign we get

cos sin , cos sinm mx m i m x m i m and

cos sin , cos sinn ny n i n y n i n

cos sin cos sinm nx y m i m n i n

cos sinm nx y m n i m n and

1 cos sinm nm nx y m n i m n

x y

Now, 1m n m n m n

m nx y x y x yx y

1 cos sin cos sinm nm nx y m n i m n m n i m n

x y



1 2cosm nm nx y m n

x y (Proved)

49. If tan( ) ii e , show that : 12 2

n

and 1 log tan2 4 2

Ans: tan( ) ii e

tan( ) cos sini i ------------(1)

So, tan( ) cos sini i ----------------(2)


tan tantan 2

1 tan tani i

i i



i i

2 costan 2 tan

1 1 2

22

n

1

2 2n

(ii) tan 2 tani i i

tan tantan 2

1 tan tani i

ii i



i i


2 2

2 2

sin1

e ee e

2 2 2 2

2 2 2 2

1 sin1 sin

e e e ee e e e

24

2

cos( / 2) sin( / 2)cos( / 2) sin( / 2)

e

2 cos( / 2) sin( / 2)cos( / 2) sin( / 2)

e

2 1 tan( / 2)

1 tan( / 2)e



22

n 1

2 2n

(Proved)

50. Sum of the following series: 1 1cos cos 2 cos3 ..........2 3

Ans: Let

2 3



1 12 3

log 1

log 1 cos

i i i

i

c

s

c is

c is e e e

c is e

c is

2 2 1

sin1 sinlog 1 cos sin tan2 1 cos

i

c is i

2 2 1

2

1

2 1

2sin cos1 2 2log 1 cos 2cos sin tan2 2cos

2

sin1 2log 2 1 2cos tan2 cos

21 log 2.2cos tan tan2 2 2

log cos2 2

log cos2

2

c is i

c is i

c is i

c is i

c

s

Dec –Jan (2012)

51. Define De Moiver’s theorem Ans: De’ Moivre’s Theorem: - For any real value of n, (cos sin ) cos sinni n i n



52. If and are the roots of the equation 2 2sin sin 1 0.z z then prove that i. 2cos cosn n nn ec

ii. 2cosn n nec Ans: It is given those roots of the equation 2 2sin sin 1 0.z z Are and therefor by applying the property of relations between roos and

coeffecfint of the equation we will have

cosb eca

And 2cos ec ………………………..1

As 2cos ec therefore 2cosn n n nec ……………………2

cosb eca

; 2cos ec

2cos ec

From above 2cos cosec ec

2 2cos cos 0ec ec 2 2cos cos 4cos

2ec ec ec

1 3cos2iec

1 3 1 3cos & cos2 2i iec ec

cos cos sin & cos cos sinec r i ec r i

221

31 3 21 & tan 12 2 3

2

where r

1 3 1 3cos & cos2 2

cos cos sin & cos cos sinn nn n n n

i iec ec

ec i ec i

2cos cosn n nec

In the given problem instead of 2cos cosn n nn ec it should be

2cos cosn n nec

53. f ( ) p x iya ib m then prove that one of the value of y/x is 2 2

2 tan ( / )log ( )e

b aa b

Ans:



log( ) log

p x iy

p x iy

a ib m

a ib m

2 2 1

log( ) log

1 log tan log log2

p a ib x iy m

bp a b i x m iy ma

Equating real and imaginary parts

2 2

1

1log log ...........................(1)2

log tan ............................(2)

x m p a b

by m pa

Dividing (2 )by (1)

1 1

2 22 2

tan 2 tan

1 loglog2

b bpy a ax a bp a b



S


C


2 3C iS i i i

2 3


i i iC iS e e e

2 3(sin ) (sin )sin ....2 3

i ii e eC iS e

log(1 sin )iC iS e


log (cos sin )C iS r i -------------(1)


1 sintan1 sin cos


........sin4sin41sin3sin

31sin2sin

21sin 4322



logC iS r i

2


C iS i

2

1 sintan1 sin cos

S

2


Unit II – Differential Equations of higher order

April -May, 2006

1. Solve: 2

2 32 2 cos 2x xd y y x e e x

dx

Ans: 2 2 32 cos 2x xD y x e e x -----------(1)

Here Auxiliary equation is 2 2 0 2D D i

So, cos 2 sin 2hy A x B x

2 32

1. . cos 22

x xpP I y x e e x

D

3 22 2

1 1 cos 2( 3) 2 ( 1) 2

x xpy e x e x

D D

3

222

1 1 cos 211 2 36 1

11

xx

pey x e x

D DD D

3

22 2

1 1 cos 2611 4 2 31

11

xx

pey x e x

D D D



2 32 23 2

22

6 66 2 131 ... cos 211 11 121 1331 4 169

xx

p

D D D De D D Dy x e xD

3

2 22

6 1 36 2 131 ... cos 211 11 11 121 4 169

xx

pe D Dy D x e x

D

3

2 22

6 47 2 131 ... cos 211 11 121 4 4 169

xx

pe D Dy D x e x

3

2 6 47 2 132 2 0 0 ... cos 211 11 121 233

xx

pe Dy x x e x

3

2 12 94 4sin 2 13cos 211 11 121 233

x x

pe x ey x x x

So, Solution h py y y

3

2 12 94cos 2 sin 2 4sin 2 13cos 211 11 121 233

x xe x ey A x B x x x x

2. Solve: 3 2

3 23 2

12 2 10( )d y d yx x y xdx dx x

Ans: 3 2

3 23 2

12 2 10d y d yx x y xdx dx x

-------------(1)

Let logzx e z x

So, 2 3

2 32 3', '( ' 1) , '( ' 1)( ' 2)dy d y d yx D x D D y x D D D y

dx dx dx where

' dDdz

Putting all these values in (1) we get

'( ' 1)( ' 2) 2 '( ' 1) 2 10( )z zD D D D D y e e

3 2' ' 2 10( )z zD D y e e -------(2)

Its auxiliary equation is 3 2 2 0m m

2( 1)( 2 2) 0m m m



2 41,

2m

1, 1m i

So, . . cos sin )z zhC F y Ae e B z C z

. . cos(log ) sin(log )hAC F y x B x C xx

-------------(3)

Now, 3 2

1. . 10( )' ' 2

z zpP I y e e

D D

3 2 3 2

1 110 10' ' 2 ' ' 2

z zpy e e

D D D D

3 2 3 2

1 110 10 (1)1 1 2 ( ' 1) ( ' 1) 2

z zpy e e

D D

2

15 10 (1)'( ' 4 ' 5)

z zpy e e

D D D

1

21 4 15 10 1 ' ' (1)5 ' 5 5

z zpy e e D D

D

1 45 2 1 ' ..... (1)

' 5z z

py e e DD

15 2 (1)

'z z

py e eD

5 2z zpy e ze

2log5p

xy xx


2 logcos(log ) sin(log ) 5A xy x B x C x xx x

(Ans)

3. Solve the following simultaneous equation: 2 , 2 , 2dx dy dzy z xdt dt dt

(Ans.) The given equation is:

2dx ydt

………..(1), 2 ..........(2)dy zdt

,

2 ..........(3)dz xdt

Differentiating (1)w . r. “t.”, we get 2

2 2 2(2 )d x dy zdt dt

using(2)



Differentiating again w. r. t. “t”,, we get 3

3 4 4(2 )d x dz xdt dt

3

3 2

21 2 3

21 2 3 2 3

21 2 3 3

21 2

( 8) 0,

8 0 ( 2)() 2 4) 0

2, 1 3

cos( 3 )12

1 2 cos( 3 ) 3 sin( 3 )2

2 2cos cos( 3 ) sin sin( 3 )3 3

cos 3

t t

t t t

t t

t t

dD x whereDdt

D or D D D

D i

x c e c e t cdxydt

c e c e t c c e t c

c e c e t c t c

c e c e t

323

c

From(2)….. 21 2 3 2 3

21 2 3

12

1 2 22 cos( 3 ) 3 sin( 3 )2 3 3

4cos( 3 )3

t t t

t t

dyzdt

c e c e t c c e t c

c e c e t c

Nov-Dec 2006

4. Solve the following : sin , cosdx dxy t x tdt dt

, Given that , 2, 0, , 0x y when t

Ans: Given simultaneous differential equation are

sindx y tdt

-------------(1)

cosdy x tdt

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

From equation (2) we get cosdyx tdt

---------(3)

Differentiating (3) with respect to t we get

2

2 sindx d y tdt dt

---------(4)

From (1) and (4) we get

2

2sin sind yt y tdt



2

2 2sind y y tdt

2( 1) 2sinD y t -----------(5)

Its Auxiliary equation is 2 1 0 1,1m m

So, . . t thC F y Ae Be

And 2

1. . ( 2sin )1pP I y t

D

2

1( 2sin ) sin1 1py t t

So, h py y y

sint ty Ae Be t -------(6)

Putting the value of y in (2) we get

cos cost tAe Be t x t

t tx Ae Be ---------(7)

Given that 2, 0, 0x y when t

So, equation (6) and (7) becomes 0 A B and 2 A B

1, 1A B

So, solution t tx e e and sint ty e e t (Ans)

5. Solve the differential equation: 3 2

3 23 2

12 2 10( )d y d yx x y xdx dx x

Ans: 3 2

3 23 2

12 2 10d y d yx x y xdx dx x

-------------(1)

Let logzx e z x

So, 2 3

2 32 3', '( ' 1) , '( ' 1)( ' 2)dy d y d yx D x D D y x D D D y

dx dx dx where ' dD

dz

Putting all these values in (1) we get

'( ' 1)( ' 2) 2 '( ' 1) 2 10( )z zD D D D D y e e



3 2' ' 2 10( )z zD D y e e -------(2)

Its auxiliary equation is 3 2 2 0m m

2( 1)( 2 2) 0m m m

2 41,

2m

1, 1m i

So, . . cos sin )z zhC F y Ae e B z C z

. . cos(log ) sin(log )hAC F y x B x C xx

-------------(3)

Now, 3 2

1. . 10( )' ' 2

z zpP I y e e

D D

3 2 3 2

1 110 10' ' 2 ' ' 2

z zpy e e

D D D D

3 2 3 2

1 110 10 (1)1 1 2 ( ' 1) ( ' 1) 2

z zpy e e

D D

2

15 10 (1)'( ' 4 ' 5)

z zpy e e

D D D

1

21 4 15 10 1 ' ' (1)5 ' 5 5

z zpy e e D D

D

1 45 2 1 ' ..... (1)

' 5z z

py e e DD

15 2 (1)

'z z

py e eD

5 2z zpy e ze

2log5p

xy xx


2 logcos(log ) sin(log ) 5A xy x B x C x xx x

(Ans)

6. Using method of variation of parameters: 2

2 4 tan 2d y y xdx

Ans: Homogeneous equation is 4 0y y



Its characteristics equation is 2 4 0 2i Hence, Homogeneous solution is cos 2 sin 2hy A x B x

1 2cos 2 , sin 2y x y x

1 1 2 2

2 2

cos 2 sin 22cos 2 2sin 2 2

2sin 2 2cos 2y y x x

W x xx xy y

Wrdxyy

Wrdxyyy p

.. 12

21

sin 2 .tan 2 cos 2 tan 2cos 2 sin 2

2 2px xdx x xdxy x x

cos 2 sin 2(sec 2 cos 2 ) sin 2 .

2 2px xy x x dx x dx

cos 2 log(sec 2 tan 2 ) sin 2 sin 2 cos 2

2 2 2 2 2px x x x x xy

cos 2 .log(sec 2 tan 2 )

4px x xy

So, Solution cos 2 .log(sec 2 tan 2 )cos 2 sin 2

4h px x xy y y A x B x

(Ans)

April -May, 2007

7. Solve the differential equation2

2 2 sinxd y dy y xe xdx dx

.

Ans: 2


Its auxiliary equation is 2 22 1 0 ( 1) 0 1,1m m m m

So, . . x xhC F y Ae Bxe

Now, 2

1. . sin( 1)

xpP I y xe x

D

2 2

1 1sin sin( 1 1)

x xpy e x x e x x

D D

1 sinx

py e x xdxD

1 cos sinxpy e x x x

D

cos sinxpy e x x x dx

sin cos cosxpy e x x x x

sin 2cosxpy e x x x



So, general solution h py y y

sin 2cosx x xy Ae Bxe e x x x (Ans)

8. Solve by method of variation of parameters of : - 2

2 sind y y x xdx

.

Ans: Homogeneous equation is 0y y

Its characteristics equation is 2 1 0 2i Hence, Homogeneous solution is cos sinhy A x B x


1 1 2 2

2 2

cos sincos sin 1

sin cosy y x x

W x xx xy y

Wrdxyy

Wrdxyyy p

.. 12

21

sin . sin cos . sincos sin

1 1px x xdx x x xdxy x x

2cos sin . sin sin cospy x x x dx x x x xdx

cos sin(1 cos 2 ) sin 2 .

2 2px xy x x dx x x dx

2cos sin 2 cos 2 sin cos 2 sin 2

2 2 2 4 2 2 4px x x x x x x x xy

2 cos cos sin 2 cos cos 2 sin cos 2 sin sin 2

4 4 8 4 8px x x x x x x x x x x xy

2 cos sin 2 sin cos 2cos cos cos 2 sin sin 2

4 4 8p

x x x x xx x x x x xy

2 cos sin cos

4 4 8px x x x xy

So, Solution 2 cos sin coscos sin

4 4 8h px x x x xy y y A x B x (Ans)


2 22 3 4 (1 )d y dyx x y x

dx dx .

Ans: 2

2 22 3 4 (1 )d y dyx x y x

dx dx ---------(1)

Let logzx e z x and 2

22' , '( ' 1)dy d yx D y x D D y

dx dx where ' dD

dz



Putting all the values in (1) we get

2'( ' 1) 3 ' 4 (1 )zD D D y e 2 2' 4 ' 4 1 2 z zD D y e e -----------(2)

Its auxiliary equation is 2 24 4 0 ( 2) 0 2, 2m m m m

So, 2 2 2 2. . logz zhC F y Ae BZe Ax Bx x

Now, 22

1. . (1 2 )( ' 2)

z zpP I y e e

D

0 22 2 2

1 1 12( ' 2) ( ' 2) ( ' 2)

z z zpy e e e

D D D

22 2 2

1 1 1.1 2 1(0 2) (1 2) ( ' 2 2)

z zpy e e

D

22

1 12 14 '

z zpy e e

D 21 2 1

4z z

py e e dzdz

2 2 2 21 1 (log )2 2

4 2 4 2

zz

pz e x xy e x

So, general solution is 2 2

2 2 1 (log )log 24 2h p

x xy y y Ax Bx x x (Ans).

Nov-Dec 2007

10. Solve: 2

2 3 4 0d y dy ydx dx

Sol. It’s symbolic form is: 2

2

3 4 0( 3 4) 0D y Dy yD D y

It’s auxiliary equation is :

2 3 4 0D D 4, 1D

Hence C.F.= 4

1 2x xy c e c e (Ans.)

11. Solve: 2

2 5 6 sin 3d y dy y xdx dx

Sol. It’ s symbolic form is 2 5 6 tan 2D D y x

It’s auxiliary equation is: 2 5 6 0D D

( 2)( 3) 02,3

D DD

Hence C.F.= y=2 3

1 2x xc e c e

And P.I.=



2 2

2 2

1 1sin 3 sin 3( 5 6) 3 5 6

1 1 5 3sin 3 [ ][ ]sin 35 3 5 3 5 35 3 5 3sin 3 sin 3

25 9 25( 3 ) 91 15cos3 3sin 3(5 3)sin 3

234 2345cos3 sin 3

78

x xD D D

Dx xD D D

D Dx xD

x xD x

x x

Hence complete solution is:y=C.F.+P.I.

2 31 2

1 (5cos3 sin 3 )78

x xy c e c e x x

12. Solve by method variation of parameters:

2

2 4 tan 2d y y xdx




1 1 2 2

2 2



W x xx xy y

Wrdxyy

Wrdxyyy p

.. 12

21






2 2 2 2 2px x x x x xy


4px x xy



(Ans)

13. Solve the simultaneously equation:



2 5 , 4 3tdx dxx y e x y tdt dt

Sol. The given equation can be expressed as:

( 5) 3 ............(1)2 ( 5) ......(2)t

D x y tx D y e

To eliminate y , operating equation (1) by (D+5) and equation (2) by 3 then subtracting , we get

2

( 5)( 4) 6 ( 5) 3( 9 14) 1 5 3

t

t

D D x x D t eD D x t e

…………………..(3)

The root of auxiliary equation of the equation corresponding homogeneous equation 2( 9 14) 0D D x

of the equation (3) is given by 2( 9 14) 0, 2, 7

D Dor D

Hence the complementary function of equation (3) is: 2 7

1 2. . t tC F c e c e

The particular integral of equation (3) is

2

2 2

2

1. . (1 5 3 )( 9 14)

1 1(1 5 ) 3( 9 14) ( 9 14)

1 9 31 ............. (1 5 )14 14 14 1 9 14

1 91 5 .514 14 81 315

14 14 8

t

t

t

t

t

P I t eD D

t eD D D D

D eD t

et

et

Hence the general solution of equation (3) is:

2 71 2

2 71 2

. . . .1 315

14 14 85, 2 7

14 8

tt t

tt t

x C F P Iex c e c e t

dx eNow c e c edt

…………………….(4)

Substituting the above values of x and dx/dt in equation (1), we get;



2 7 2 71 2 1 2

2 71 2

5 20 1243 2 7 4 414 8 14 196 2

1 3 5 272 3 .......................(5)3 7 8 98

t tt t t t

t t t

e ey c e c e c e c e t t

y c e c e t e

Since the degree of D in the determinant 4 3

, 22 5

Dis

D

it follows that the number of

independent constant in general solution must be two. Hence (4) and (5) together constitute the general solution of the given system.

April -May, 2008

14. State Cauchy’s Linear equation.

Ans: - A Differential equation of the form

Xyadx

ydxadx

ydxadx

ydx nn

nn

n

nn

n

nn

........2

22

21

11

1 is known as Cauchy’s

Linear differential equation.

15. Solve2

22 4 sin2xd y dy

y e xdx dx

.

(ANS). It’s symbolic form is : 2 2( 4 1) sin 2xD D y e x

It’s auxiliary equation is: 2( 4 1) 0D D

4 16 4( ) 2 2 32

D

21 2. cosh 2 3 sinh 2 3xC F e C C

And P.I.=

2 2

2 2

1 1sin 2 sin 2( 4 1) 1 4 1 1

x xe x e xD D D D

2 2

2 2sin 2 sin 21 2 4 4 1 2 2

x xe ex xD D D D D

2 2

2 sin 2 sin 24 2 2 16 2 2

x xe ex xD D

2 2

sin 2 sin 214 2 14 2

x xe ex xD D



22

2

2 14sin 2 sin 2

14 2 4 196

xx e De x xD D

2

2

2 14sin 2

4 2 196

xe Dx

2 2 14sin 2

212

xe Dx

2 4cos 2 14sin 2212

xe x x

Hence complete solution is:y=C.F.+P.I.

22

1 2

4cos 2 14sin 2cosh 2 3 sinh 2 3

212

xx e x x

y e C C

16. Solve by method of variation of parameters of : -2

2 4 4tan2 .d y

y xdx




1 1 2 2

2 2



W x xx xy y

Wrdxyy

Wrdxyyy p

.. 12

21






2 2 2 2 2px x x x x xy


4px x xy



(Ans)

17. Solve 2 0dydx

x ydt dt

and 5 3 0dydx

x ydt dt



(Ans). The given equation are:

2 0.....................(1)dydx

x ydt dt

5 3 0.....................(2)dydx

x ydt dt

Subtracting equation (2) from equation (1), we get

3 2 0...............(3)dx x ydt

Writing equation (2)and (3) symbolically (i.e;D=d/dt),we have 5 ( 3) 0..........(4)( 3) 2 0..........(5)

x D yD x y

Operating equation (5)by D+3 ,we get 2( 9) 2( 3) 0.........(6)D x D y

Now, multiplying equation(4) by 2 and adding in equation (6),we get 2( 1) 0D x ………………………….(7)

This is the linear equation in x with constant coefficient s for which the auxiliary equation is 2( 1) 0D

D i

Hence the general solution of linear equation is: 0

1 2

1 2

( cos sin )( cos sin )..........................(8)

tx e c t c tx c t c t

From equation (8)

1 2

1 2

sin cos

sin cos

dx c t c tdtDx c t c t

Substituting these values in equation(5) we get

1 2 1 2

2 1 1 2

1 1( 3 ) ( sin cos ) 3( cos sin )2 21 1( 3 )cos ( 3 )sin2 2

y Dx x c t c t c t c t

y c c t c c t

Nov-Dec 2008 18. Define the linear differential equation. Ans. A differential equation in which the dependent variable & its derivatives occur only in the first

degree & are not multiplied together is known as Linear Differential equation.

General form of Linear Differential Equation of nth order is:

XyPdx

ydPdx

ydPdx

ydnn

n

n

n

n

n

........2

2

21

1

1 , where XPPP n ,,........,, 21 are functions of x.



19. Solve the differential equation:

2

2 3 2xd y dy ey e

dx dx

Ans: - Its symbolic form is 2 3 2xeD D y e .

Its Auxiliary equation is 2 3 2 0D D 1, 2D

Hence, Complementary function is 2x xcy Ae Be

Now, 2

1 1 1 13 2 ( 2)( 1) ( 2) ( 1)

x x xe e ePI e e eD D D D D D

2 2 21 12 2

x x x x x x x xx x x xe e e ePI e e e dx e e e e e e dx e e e dxD D

2 x xePI e e

So, Solution 2 2x x xc p

xey y y Ae Be e e (Ans)

20. Solve by method variation of parameters:

2 3

2 26 9xd y dy ey

dx dx x

Ans: - 2

3

2

2

96xey

dxdy

dxyd x

Its Homogeneous equation is 0962

2

ydxdy

dxyd

Its symbolic form is 0)96( 2 yDD

Its characteristics equation is 0)96( 2 DD 3,3 D

Hence, Homogeneous solution is 213)( ByAyeBxAy x

h

xx xeyey 32

31 ,

xxxxxxx

xx

exexeexeee

xeeyyyy

W 6666333

33

21

21 3333''

Wrdxyy

Wrdxyyy p

.. 12

21

dxxe

eexedx

xe

exeey

x

x

xx

x

x

xx

p 2

3

6

33

2

3

6

33 ..

dxx

xedxx

ey xxp 2

33 11

xxexey x

ex

p1log 33 x

ex

p exey 33 log



So, Solution xe

xxxph exeBxeAeyyy 3333 log (Ans).


5 2 , 2 0dx dyx y t x ydt dt

being x=y=0 when t=0.

Ans: - + 5푥 − 2푦 = 푡, + 2푥 + 푦 = 0 Its symbolic form is (퐷 + 5)푥 − 2푦 = 푡------------------(1) 2푥 + (퐷 + 1)푦 = 0-------------------(2) Now, eq(1) X (D+1) + eq(2) X 2 we get ⇒ [(퐷 + 1)(퐷 + 5) + 4]푥 = (퐷 + 1)푡 + 0 ⇒ (퐷 + 6퐷 + 9)푥 = 1 + 푡, which is linear differential equation. Its auxiliary equation is (퐷 + 6퐷 + 9) = 0 ⇒ 퐷 = −3,−3 So, 퐶.퐹. = (퐶 + 퐶 푡)푒

Now, 푃. 퐼. = ( ) (1 + 푡) = 1 + (1 + 푡) = 1− 2 + 3 + … … … … . (1 + 푡)

⇒ 푃. 퐼. = 1 + 푡 − = 푡 + = +

So, 푥 = (퐶 + 퐶 푡)푒 + + ---------------------------------(3) Now, from (1) 2푦 = (퐷 + 5)푥 − 푡 = 퐷푥 + 5푥 − 푡 ⇒ 2푦 = 5(퐶 + 퐶 푡)푒 + + − 3(퐶 + 퐶 푡)푒 + 퐶 푒 + − 푡

⇒ 푦 = (2퐶 + 퐶 ) + 푡퐶 푒 − + ---------------(4) Given that at 푡 = 0, 푥 = 푦 = 0, so equation (3) and (4) becomes 퐶 + = 0 ⇒ 퐶 = − and

(2퐶 + 퐶 ) + = 0 ⇒ 퐶 = − − 2퐶 == − + = − = − So, Solution is

푥 = (− − 푡)푒 + + and 푦 = − + + 푡푒 − +

⇒ 푥 = − {(1 + 6푡)푒 − (3푡 + 1)} and 푦 = − {(4 + 6푡)푒 + (6푡 − 4)}

April -May, 2009

22. Explain briefly the method of variation of parameter.

Ans: Let us solve 2

2

d y dyP Qy Xdx dx

by variation of parameter method.

Let its complementary function is 1 1 2 2cy c y c y

Then find out wronskian 1 21 2

1 2

( , )' '

y yW y y

y y

Now, particular integral 2 11 2

1 2 1 2

. .( , ) ( , )py X y Xy y dx y dx

W y y W y y



Then general solution is c py y y (Ans).

23. Solve the differential equation 2 2 32 cos 2x xD y x e e x .

Ans: 2 2 32 cos 2x xD y x e e x -----------(1)

Here Auxiliary equation is 2 2 0 2D D i

So, cos 2 sin 2hy A x B x

2 32

1. . cos 22

x xpP I y x e e x

D

3 22 2

1 1 cos 2( 3) 2 ( 1) 2

x xpy e x e x

D D

3

22 2

1 1 cos 2611 2 31

11

xx

pey x e x

D D D D

3

22 2

1 1 cos 2611 4 2 31

11

xx

pey x e x

D D D

2 32 23 2

22

6 66 2 131 ... cos 211 11 121 1331 4 169

xx

p

D D D De D D Dy x e xD

3

2 22

6 1 36 2 131 ... cos 211 11 11 121 4 169

xx

pe D Dy D x e x

D

3

2 22

6 47 2 131 ... cos 211 11 121 4 4 169

xx

pe D Dy D x e x

3

2 6 47 2 132 2 0 0 ... cos 211 11 121 233

xx

pe Dy x x e x

3

2 12 94 4sin 2 13cos 211 11 121 233

x x

pe x ey x x x


3

2 12 94cos 2 sin 2 4sin 2 13cos 211 11 121 233

x xe x ey A x B x x x x

24. Solve the equation : - 2

22(1 ) (1 ) sin 2 log(1 )d y dyx x y x

dx dx .

Ans: 2

22(1 ) (1 ) sin 2 log(1 )d y dyx x y x

dx dx ……(1) is a Legendre’s equation.

Now let us put 1 tx e we get



2

22(1 ) ( 1) , (1 )d y dyx D D y x Dy

dx dx where

dDdt

So, equation (1) becomes ( 1) sin 2D D y Dy y t

2( 1) sin 2D y t which is linear differential equation with constant coefficients.

Its auxiliary equation is 2( 1) 0D D i

So, 1 2. . cos sinC F c t c t ……………..(2)

Now, 2 2

1 sin 2 sin 2. . sin 21 2 1 3

t tP I tD

So, solution 1 2sin 2. . . . cos sin

3ty C F P I c t c t (Ans)

25. Solve the simultaneous equation: -

2 , 2t tdx dyy e x edt dt

.

Ans: 2 , 2t tdx dyy e x edt dt

2 ...............(1) 2 ..............(2)t tDx y e Dy x e where dDdt

(1) 2 (2) D we get

22 4 2 2 t tDx y D y Dx e e

2( 4) 2 t tD y e e …………..(3)

Here Its auxiliary equation is 2( 4) 0 2D D i

So, 1 2. . cos 2 sin 2C F c t c t ……………..(4)

Now, 2 2 2

1 1 1. . 2 21 1 1

t t t tP I e e e eD D D

2 2

1 1. . 21 1 ( 1) 1 2

tt t t eP I e e e

So, solution 1 2. . . . cos 2 sin 22

tt ey C F P I c t c t e

……………..(5)

Now putting the value of y in (2) we get 2 tDy x e

1 22 sin 2 2 cos 2 22

tt tec t c t e x e

1 22 2 sin 2 2 cos 22

tt ex c t c t e

1 2sin 2 cos 22 4

t te ex c t c t

………………(6)



So, solution 1 2 1 2sin 2 cos 2 , cos 2 sin 22 4 2

t t tte e ex c t c t y c t c t e

(Ans)

Nov-Dec 2009

26. Solve the equation: 2( 2) 0D D y . Ans: - 2( 2) 0D D y

Its auxiliary equation is 2( 2) 0 1, 2D D D

So, solution is 2x xy Ae Be (Ans). 27. Solve the following differential equation:

2

22 logd y dyx x y x

dx dx .

Ans: - 푥 − 푥 + 푦 = 푙표푔푥

--------------------(1)

Given equation is Cauchy’s Homogeneous Linear differential equation.

Put 푥 = 푒 ⇒ 푡 = 푙표푔푥, if 퐷 = , 푥 = 퐷푦, 푥 = 퐷(퐷 − 1)푦

So, equation (1) is 퐷(퐷 − 1)푦 − 퐷푦 + 푦 = 푡 ⇒ (퐷 − 2퐷 + 1)푦 = 푡 Which is linear differential equation with constant coefficients. So, its Auxiliary equation is 퐷 − 2퐷 + 1 = 0 ⇒ 퐷 = 1,1 So, 퐶.퐹. = 퐶 푒 + 퐶 푡푒 ------------------------(2) Now, 푃. 퐼. =

( )푡 = (1− 퐷) 푡 = (1 + 2퐷 + 3퐷 + … … … … . )푡

⇒ 푃. 퐼. = 푡 + 2 So, solution is 푦 = 퐶.퐹. +푃. 퐼. = 퐶 푒 + 퐶 푡푒 + 푡 + 2 ⇒ 푦 = 퐶 푥 + 퐶 푥푙표푔푥 + 푙표푔푥 + 2 (Ans).

OR 28. Solve the simultaneous differential equation:

5 2 , 2 0dx dyx y t x ydt dt

, given that 0x y when 0t .


Now, 푃. 퐼. = ( ) (1 + 푡) = 1 + (1 + 푡) = 1− 2 + 3 + … … … … . (1 + 푡)

⇒ 푃. 퐼. = 1 + 푡 − = 푡 + = +






푥 = (− − 푡)푒 + + and 푦 = − + + 푡푒 − +

⇒ 푥 = − {(1 + 6푡)푒 − (3푡 + 1)} and 푦 = − {(4 + 6푡)푒 + (6푡 − 4)}

29. Solve the following differential equation:2

2 4 sinhd y y x xdx

.

Ans: - − 4푦 =

⇒ − 4푦 =

Its Symbolic form is ⇒ (퐷 − 4)푦 =

Its Auxiliary equation is ⇒ (퐷 − 4) = 0 ⇒ 퐷 = ±2

So, 퐶.퐹. = 퐶 푒 + 퐶 푒

Now, 푃. 퐼. = = 푥푒 − 푥푒

⇒ 푃. 퐼. = 푒( )

푥 − 푒( )

푥 = 푒 푥 − 푒 푥

⇒ 푃. 퐼. = 푥 + 푥

⇒ 푃. 퐼. = 1 + + … … 푥 + 1 + + … … 푥

⇒ 푃. 퐼. = 푥 + + 푥 − = ( ) − ( )

⇒ 푃. 퐼. = [−3푥푠푖푛ℎ푥 − 2푐표푠ℎ푥]

So, solution is 푦 = 퐶.퐹. +푃. 퐼. = 퐶 푒 + 퐶 푒 + [−3푥푠푖푛ℎ푥 − 2푐표푠ℎ푥] (Ans).

April -May, 2010 30. Define Cauchy and Legendre Linear differential equation.




Xyadx

ydxadx

ydxadx

ydx nn

nn

n

nn

n

nn

........2

22

21

11


Linear differential equation. A Differential equation of the form

Xyadx

ydbaxadx

ydbaxadx

ydbax nn

nn

n

nn

n

nn

........)()()( 2

22

21

11

1 is

known as Legendre’s Linear differential equation.

31. Solve 3

3 3 2xd y dy ey e

dx dx .




Now, 2

1 1 1 13 2 ( 2)( 1) ( 2) ( 1)


2 2 21 12 2

x x x xx x e x e x x x e x x ePI e e e dx e e e e e e dx e e e dxD D

2 xx ePI e e

So, Solution 2 2 xx x x ec py y y Ae Be e e (Ans)

32. Solve, by the method of variation of parameters: xeDD x log)12( 2 . Ans: - Its Auxiliary equation is 2 2 1 0D D 1, 1D

Hence, Complementary function is xcy A Bx e

1 2,x xy e y xe

1 2 2 2 2 2

1 2

x xx x x x

x x x

y y e xeW e xe xe e

e e xey y

WXdxyy

WXdxyyy p

.. 12

21

2 2

. log . logx x x xx x

p x x

xe e xdx e e xdxy e xee e

log logx xpy e x xdx xe xdx

2 2

log log2 4

x xp

x xy e x xe x x x

2 2 23log 2 log 3

2 4 4x x x

px x xy e x e e x



So, Solution 2

( ) 2log 34

x xc p

xy y y A Bx e e x (Ans)

33. Solve the simultaneous equation: -

0 ydtdxt , 0 x

dtdyt , given that 0)1(,1)1( yx .

Ans: - 0 ydtdxt -------------------(1), 0 x

dtdyt -------------(2)

Differentiating (1) w.e.t. t we get 2

2 0dx d x dytdt dt dt

, by multiplying t both side we get

2

22 0dx d x dyt t t

dt dt dt --------------------(3)

By putting (2) in (3) we get2 2

2 22 2 0d x dx d xt t t x

dt dt dt -----(4) which is Cauchy

Linear Differential Equation.

Put logzt e z t and 2

22 ( 1) ,d x dxt D D x t Dx

dt dt where

dDdz

So, eqn (4) becomes 2( 1) 0 ( 1) 0D D x Dx x D x

Its Auxiliary equation is 2( 1) 0 1,1D D

So, z z Bx Ae Be Att

-------(5)

Putting in (1) we get 2

dx B By t t A Atdt t t

-----------(6)

So, Solution is Bx Att

and By Att

(Ans).

Given x(1) = 1 and y(-1) = 0, so 21,

210,1 BAABBA

Hence solution is

t

tyBA

ttx 1

21,1

21

(Ans).

Nov-Dec 2010 34. Write the formula for P.I. for the method of variation of parameters.

Ans: - Let equation is rybaDD )( 2 , then P.I is given by

Wrdxyy

Wrdxyyy p

.. 12

21 where

'' 21

21

yyyy

W .

35. Solve )2sin(8)2( 222 xxeyD x . Ans: - )2sin(8)2( 222 xxeyD x ---------------------(1)

Its characteristics equation is 2,20)2( 2 DD



So, xxh BxeAeyFC 22.. -----------------------(2)

Now, )2sin(8)2(

1.. 222 xxe

DyIP x

p

222

22 )2(

12sin)2(

1)2(

18 xD

xD

eD

y xp

222

2

21

1412sin

441

)2(218 x

Dx

DDe

Dxy x

p

2

2

222 ..........

43

221

412sin

4421

218 xDDx

Dexy x

p

..........00

462

41

22cos

41

28 22

2

xxxexy xp

3422cos4 222 xxxexy xp

So, Solution 3422cos4 22222 xxxexBxeAeyyy xxxph (Ans).

36. Solve, by the method of variation of parameters: 2

3

2

2

96xey

dxdy

dxyd x

.

Ans: - 2

3

2

2

96xey

dxdy

dxyd x

Its Homogeneous equation is 0962

2

ydxdy

dxyd

Its symbolic form is 0)96( 2 yDD

Its characteristics equation is 0)96( 2 DD 3,3 D

Hence, Homogeneous solution is 213)( ByAyeBxAy x

h

xx xeyey 32

31 ,

xxxxxxx

xx

exexeexeee

xeeyyyy

W 6666333

33

21

21 3333''

Wrdxyy

Wrdxyyy p

.. 12

21

dxxe

eexedx

xe

exeey

x

x

xx

x

x

xx

p 2

3

6

33

2

3

6

33 ..

dxx

xedxx

ey xxp 2

33 11



xxexey x

ex

p1log 33 x

ex

p exey 33 log

So, Solution xe

xxxph exeBxeAeyyy 3333 log (Ans).

37. Solve: xydxdyx

dxydx elog2

22 .

Ans: - xydxdyx

dxydx elog2

22 ---------(1)

Let logzx e z x and 2

22' , '( ' 1)dy d yx D y x D D y

dx dx where ' dD

dz

Putting all the values in (1) we get zyDDzyDDD ]12[1)1( 2 -----------(2)

Its auxiliary equation is 1,10122 mmm

So, xBxAxBzeAeyFC ezz

h log..

Now, zD

yIP p 2)1(1..

zDDzDy p .........)321()1( 22

xzzy ep log22........0002

So, general solution is xxBxAxyyy eeph log2log (Ans).

April -May, 2011

38. Find the particular integral of Solve 3

3. 4 sin 2d y dy xdx dx

Ans. 3


P.I will be 3

1 sin 24

PI xD D

3 2

1 1sin 2 sin 24 4

x xD D D D

3 2

1 1sin 2 sin 24 2 4

x xD D D

3

1 1sin 2 sin 24 0

x xD D D

By differentiating 3 4D D as 1, 0in f D can not be

f D



3 2

1 1sin 2 sin 24 3 4

x x xD D D

3 2

1 1sin 2 sin 24 3( 2 ) 4

x x xD D

3

1 sin 2 sin 24 8

xx xD D

3

1 sin 2sin 24 8

x xxD D

P.I of3


will be sin 28

x x

39. Solve 2

2 3 2xd y dy ex y e

dx dx




Now, 2

1 1 1 13 2 ( 2)( 1) ( 2) ( 1)


2 2 21 12 2

x x x xx x e x e x x x e x x ePI e e e dx e e e e e e dx e e e dxD D

2 xx ePI e e

So, Solution 2 2x xc p

xx ey y y Ae Be e e (Ans)

40. Solve the differential equation: 2

22 5 4 . log .

d y dyx x y x x

dx dx

(Ans). Let

zx e and d Ddz

. Then 2

22, ( 1)dy d yx Dy x D D y

dx dx

Thus the given differential equation reduces to the following fotm:

2

( 1) 5 4

( 4 4)

z

z

D D D y ze

D D y ze

Which is the linear differential equation with constant coefficient s, for which the auxiliary equation is:

2 4 4 0D D 2, 2D

C.F= 2 21 2 1 2 logzc c z e c c x x

P.I Will be



1( )

zzef D

i.e

21.

( 4 4)zP I ze

D D

2 2 2

1.( 4 4) ( 1) 4( 1) 4 ( 6 9)

z zz e eP I ze z z

D D D D D D

26. (1 )9 9 9

ze D DP I z

6.9 9

zeP I z

2. log9 3xP I x

Complete Ans is C.F +P.I

21 2 logc c x x +

2log9 3x x

41. Solve sin , cos ,dx dyy t x tdt dt

Given that x=2 and y=0 when t=0.

Ans: Given simultaneous differential equation are

sindx y t

dt

-------------(1)

cosdy x t

dt

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

From equation (2) we get cosdyx tdt

---------(3)

Differentiating (3) with respect to t we get

2

2 sindx d y tdt dt

---------(4)

From (1) and (4) we get

2

2sin sind yt y tdt

2

2 2sind y y tdt

2( 1) 2sinD y t -----------(5)

Its Auxiliary equation is 2 1 0 1,1m m

So, . . t thC F y Ae Be



And 2

1. . ( 2sin )1pP I y t

D

2

1( 2sin ) sin1 1py t t

So, h py y y

sint ty Ae Be t -------(6) Putting the value of y in (2) we get

cos cost tAe Be t x t

t tx Ae Be ---------(7)

Given that 2, 0, 0x y when t

So, equation (6) and (7) becomes 0 A B and 2 A B

1, 1A B So, solution t tx e e and sint ty e e t (Ans)

Nov-Dec 2011

42. Explain Cauchy’s homogeneous linear differential equation.


Xyadx

ydxadx

ydxadx

ydx nn

nn

n

nn

n

nn

........2

22

21

11


Linear differential equation.



.

Ans: 2


Its auxiliary equation is 2 22 1 0 ( 1) 0 1,1m m m m

So, . . x xhC F y Ae Bxe

Now, 2

1. . sin( 1)

xpP I y xe x

D

2 2

1 1sin sin( 1 1)

x xpy e x x e x x

D D

1 sinx

py e x xdxD

1 cos sinxpy e x x x

D

cos sinxpy e x x x dx



sin cos cosxpy e x x x x

sin 2cosxpy e x x x

So, general solution h py y y

sin 2cosx x xy Ae Bxe e x x x (Ans)

44. Solve the differential equation 2

22 log .sin(log )

d y dyx y x xdx dx

zx e and d Ddz

. Then 2


dx dx


2

( 1) 1 (sin )

( 1) sin

D D D y z z

D y z z


2 1 0D ,D i i

C.F= 1 2 1 2cos sin cos log sin logc z c z c x c x

P.I Will be 1 sin( )

z zf D

i.e

2 2

1 1. sin .( 1) ( 1)

izP I z z I P zeD D

221. sin .

( 1) 1)

izeP I z z I P zD D i

2

1. sin .( 1) 2

izeP I z z I P zD D i

21 1 1. sin .

( 1) 2 12

izP I z z I Pe zDD ii

1

2

1 1. sin . 1( 1) 2 2

DizP I z z I Pe zD i i

2

1 1. sin . 1( 1) 2 2

DizP I z z I Pe zD i i



2

1 1 1. sin .( 1) 2 2

izP I z z I Pe zD i i

2

1. sin .( 1) 2 2

i iizP I z z I Pe zD

2

1 1. sin .( 1) 2 4

iizP I z z I Pe zD

2

1 1. sin . cos sin( 1) 2 4

iP I z z I P z i z zD

2

1 cos sin sin. sin . cos( 1) 2 4 2 4

i z z z i zP I z z I P z zD

2

1 cos sin sin 1. sin . cos( 1) 4 2 4 2

z z z zP I z z I P i z zD

2

1 sin 1. sin cos( 1) 4 2

zP I z z z zD

2

1 sin log 1. sin log cos log( 1) 4 2

xP I z z x xD

Ans.= C.F+P.I

1 2cos log sin logc x c x +sin log 1 log cos log

4 2x x x

45. Solve the following simultaneous equation 2dx ydt

, 2dy zdt ,

2 .dz xdt

(Ans.) The given equation is:

2dx ydt

………..(1), 2 ..........(2)dy zdt

,

2 ..........(3)dz xdt

Differentiating (1)w . r. “t.”, we get 2

2 2 2(2 )d x dy zdt dt

using(2)

Differentiating again w. r. t. “t”,, we get 3

3 4 4(2 )d x dz xdt dt

3

3 2

21 2 3

( 8) 0,

8 0 ( 2)() 2 4) 0

2, 1 3

cos( 3 )12

t t

dD x whereDdt

D or D D D

D i

x c e c e t cdxydt



21 2 3 2 3

21 2 3 3

21 2 3

1 2 cos( 3 ) 3 sin( 3 )2

2 2cos cos( 3 ) sin sin( 3 )3 3

2cos 33

t t t

t t

t t

c e c e t c c e t c

c e c e t c t c

c e c e t c

From(2)….. 21 2 3 2 3

21 2 3

12

1 2 22 cos( 3 ) 3 sin( 3 )2 3 3

4cos( 3 )3

t t t

t t

dyzdt

c e c e t c c e t c

c e c e t c

April -May, 2012

46. Solve 3

3 0d y ydx

Ans: - Its symbolic form is Solve 013 yD .

Its Auxiliary equation is 0)1)(1(01 23 DDDD

231,

231,1 iiD

So, solution 23

23221 sincos..x

xxx eCCeCyFC (Ans).

47. Solve 2

22 5 4 logd y dyx x y x x

dx dx

(Ans). Let

zx e and d Ddz

. Then 2


dx dx


2

( 1) 5 4

( 4 4)

z

z

D D D y ze

D D y ze


2 4 4 0D D 2, 2D

C.F= 2 21 2 1 2 logzc c z e c c x x

P.I Will be 1( )

zzef D

i.e



21.

( 4 4)zP I ze

D D

2 2 2

1.( 4 4) ( 1) 4( 1) 4 ( 6 9)

z zz e eP I ze z z

D D D D D D

26. (1 )9 9 9

ze D DP I z

6.9 9

zeP I z

2. log9 3xP I x

Complete Ans is C.F +P.I

21 2 logc c x x +

2log9 3x x

48. Solve :

3 2

3. 22 sin 2xd y d y dy e xdx dx dx

Sol. It’s symbolic form is xeyDDD x 2sin2 23

Its Auxiliary equation is 0)12(02 223 DDDDDD

1,1,0 D

So, xxc exCCeCyFC 32

01..

xc exCCCy 321 ---------------------(1)

Now, xeDDD

yIP xp 2sin

21.. 23

xDDD

eDDD

y xp 2sin

21

21

2323

xDDDD

eDD

xy xp 2sin

2.1

1431

222

xDD

eD

xy xp 2sin

)4(2)4.(1

4612

xD

exyx

p 2sin83

14)1(6

2

xD

Dexyx

p 2sin64983

2 2

2

xDexyx

p 2sin64)4(983

2

2



10

2sin82cos62

2 xxexyx

p

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

So, solution 10

2sin82cos62

2

321xxexexCCCyyy

xx

pc

(Ans).

49. Solve the equations simultaneously

5 2 , 2 0, 0dx dyx t t x y x y

dt dt when t=0


Now, 푃. 퐼. = ( ) (1 + 푡) = 1 + (1 + 푡) = 1− 2 + 3 + … … … … . (1 + 푡)

⇒ 푃. 퐼. = 1 + 푡 − = 푡 + = +




푥 = (− − 푡)푒 + + and 푦 = − + + 푡푒 − +

⇒ 푥 = − {(1 + 6푡)푒 − (3푡 + 1)} and 푦 = − {(4 + 6푡)푒 + (6푡 − 4)}

Nov -Dec, 2012 50. Write the formula for P.I. for the method of variation of parameters.

Ans: - Let equation is rybaDD )( 2 , then P.I is given by

Wrdxyy

Wrdxyyy p

.. 12

21 where

'' 21

21

yyyy

W .

51. Solve 2 1 sin cosD y x x

Ans: 2 1 sin cosD y x x x



2( 1) 0D therefore 1D

1 2. . x xcC F y C e C e

2

1. . sin cos1

P I x x xD

2 2

1 1. . sin cos1 1

P I x x xD D

2 2

1 1. . . cos1 1

ixP I I P xe xD D

2 2

1. . . cos11

ixeP I I P x xDD i

22

1. . . cos1 11 2 1

ixeP I I P x xD Di

22

1. . . cos1 11 2 1

ixeP I I P x xD Di

2

1. . . cos22 2

ixeP I I P x xD Di

2

1. . . cos2

2 12

ixeP I I P x xDDi

12 1. . . 1 cos2 2 2

ixe DP I I P Di x x

1. . . 1 cos2 2

ixeP I I P Di x x

1. . . cos2 2

ixeP I I P x i x

cos sin 1. . . cos2 2

x i xP I I P x i x

cos cos sin sin 1. . . cos2 2

x x i x ix x xP I I P x

By selecting only imaginary part from above we will have



cos sin 1. . cos2 2

x x xP I x

Y = C.F + P.I

Y= 1 2. . x xcC F y C e C e +

cos sin 1 cos2 2

x x xx

52. Solve the equation : - 2

22(1 ) (1 ) sin 2 log(1 )d y dyx x y x

dx dx .

Ans: 2

22(1 ) (1 ) sin 2 log(1 )d y dyx x y x

dx dx ……(1) is a Legendre’s equation.

Now let us put 1 tx e we get

2

22(1 ) ( 1) , (1 )d y dyx D D y x Dy

dx dx where

dDdt

So, equation (1) becomes ( 1) sin 2D D y Dy y t

2( 1) sin 2D y t which is linear differential equation with constant coefficients.

Its auxiliary equation is 2( 1) 0D D i

So, 1 2. . cos sinC F c t c t ……………..(2)

Now, 2 2

1 sin 2 sin 2. . sin 21 2 1 3

t tP I tD

So, solution 1 2sin 2. . . . cos sin

3ty C F P I c t c t (Ans)


22 3 , 3 2 tdx dxx y t x y edt dt

Sol. The given equation can be expressed as:

2

( 2) 3 ............(1)3 ( 2) ......(2)t

D x y tx D e

To eliminate y , operating equation (1) by (D+2) and equation (2) by 3 then adding , we get 2 2

2 2

( 2) 9 ( 2) 3( 4 5) 1 2 3

t

t

D x x D t eD D x t e

…………………..(3)

The root of auxiliary equation of the equation corresponding homogeneous equation 2( 4 5) 0D D x

of the equation (3) is given by 2( 4 5) 0, 5,1

D Dor D

Hence the complementary function of equation (3) is:



51 2. . t tC F c e c e

The particular integral of equation (3) is

22

22 2

12

2

2

1. . 1 2 3( 4 5)

1 1(1 2 ) 3( 4 5) ( 4 5)

1 4 31 (1 2 )5 5 5 4 8 5

1 4 31 (1 2 )5 5 71 8 3(1 2 )

5 5 7

t

t

t

t

t

P I t eD D

t eD D D D

D eD t

eD t

et

Hence the general solution of equation (3) is:

25

1 2

. . . .1 8 3(1 2 )

5 5 7

tt t

x C F P Iex c e c e t

…………………….(4)

25

1 22 6, 5

5 8

tt tdx eNow c e c e

dt

2 25 5

1 2 1 21 1 2 6 1 8 3, 2 5 2 (1 2 )3 3 5 8 5 5 7

t tt t t tdx e eNow y x t c e c e c e c e t t

dt

2 25

1 21 3 2 8 6 33 23 5 5 5 8 7

t tt t t e ey c e c e t

25

1 21 2 153 2 13 5 8

tt t t ey c e c e t

Unit III – Multiple Integral

(May-Jun-2006)

1. Evaluate by changing the order of integration.

Ans 0 0

2xx yxe dydx

0 0

/ .2

xyx dydxxe



0 0

2xx yxe dydx

by changing the order of the equation we will have 0

2

y

xyxe dydx

0 0

2 2

y y

x xy yxe dydx xe dx dy

0 0

2

2 2

2

2

2

22

2 2 2

y

y

t

y y

x xyy yxe dydx e dy

y

xxyxe dx taking ty

xdx ydt

xx yy y yyxdx dt xe dx e dt e

y

0 0 0 0

2 20

2 2 2y

x xy y yy y y yxe dydx e dy e dy e dy

yy

0 0

21 1. 1/ 22 1y

x yey yxe dydx y e

(Ans).

2. Find the volume common to the cylinder 2 2 2 2 2 2,x y a x z a .



Ans:

2 2 2 2 2 2 2 2

2 2 2 2 0 0 0

Volume 8a a x a x a a x a x

a a x a x

dxdydz dxdydz

2 2 2 2

2 22 2

00 0 0 0

Volume 8 8a a x a a x

a xz dxdy a x dxdy

2 2

2 2 2 2 2 2 2 20

0 0 0

Volume 8 . 8 . 8 ( )a a a

a xa x y dx a x a x dx a x dx

3 3 3 3

2 2

0

2 16Volume 8 8 83 3 3 3

ax a a aa x a

(Ans).

3. Prove that .

Sol.

1

10

1 !log

1

nn

n

nmx x dxm

1

0

log nmx x dx ……………………………………………….1

Let log tx t x e tdx e dt

1 becomes 0

0

1 1n nm t m te t dt e t dt

……………………………….2

1,0,)1(

!.)1().(log 1

1

0

mn

mndxxx n

nnm



1

1dpm t p dt

m

2 becomes 0

11 1

nppe dpm m

0

11 11

nn

p ne p dpm

=

0

1 1 11 11

nn npe p dp

m

0

1 11 11 11 1 11 1

n nn nnpe p dp n

m m

1 1 !1 1 1 11 1

nn

n nn nm m

Hence Proved

(Nov –Dec 2006)

4. Evaluate the following integral by changing of order of integration 2

4 2 20

a a

ax

y dxdyy a x

.

Ans:

Here region of integration is x varies from 0 to a, y varies from ax to a.

After changing order y varies from 0 to a, x varies from 0 to2y

a.

2 /2 2

4 2 2 4 2 20 0 0

y aa a a

ax

y dxdy y dydxy a x y a x

putting 2a as common we get



2

2

4 2 2 220 0 0 2

2a a a

ax

y y dydxay dxdy ay a x y x

a

2

2 21

24 2 2 220 0 0 02

0

22

sina a a a

ax

yay y dxay dxdy y xa dy dy

a yy a x y x aa

2

2 2 21 1 1

2 24 2 20 0 0

0

2

sin sin sin 0a a a a

ax

ya y

y dxdy y x y ady dya ay yy a x

a a

2 2 2

4 2 20 0 0

02 2

a a a a

ax

y dxdy y ydy dya ay a x

2 2 2

3

04 2 20 02 6 6

a a aa

ax

y dxdy y ady ya ay a x

(Ans)

5. Find the volume common to gas cylinder 2 2 2 2 2 2,x y a x z a .

Ans:



2 2 2 2 2 2 2 2

2 2 2 2 0 0 0


a a x a x

dxdydz dxdydz

2 2 2 2

2 22 2

00 0 0 0


a xz dxdy a x dxdy

2 2

2 2 2 2 2 2 2 20

0 0 0

Volume 8 . 8 . 8 ( )a a a


3 3 3 3

2 2

0

2 16Volume 8 8 83 3 3 3

ax a a aa x a

(Ans).

6. Define Beta function and show that ( ) ( )( , )( )m nm nm n

.

Ans: The Beta function is defined as 1 1

1 1

0 0

( , ) (1 ) , , 0(1 )

nm n

m nxB m n x x dx dx m nx

.

We know that 1

0

( ) .x nn e x dx

By putting x az dx adz we get

1

0

( ) .( )az nn e az adz

1

0

( ) .n az nn a e z dz

Putting z = x we get

1

0

( ) .n ax nn a e x dx

Now, taking a = z we get

1

0

( ) .zx n nn e x z dx

By multiplying 1z me z both side we get

1 (1 ) 1 1

0

( ) .z m z x n m nn e z e x z dx



Now taking integration with respect to z both side from 0z we get

1 1 (1 ) 1

0 0 0

( ) .z m n z x m nn e z dz x e z dzdx

Now, let (1 )1 1

y dyz x y z dzx x

we get

1

1

0 0

.( ) ( )(1 )

y m nn

m n

e y dyn m x dxx

1

1

0 0

( ) ( ) .(1 )

ny m n

m n

xn m e y dy dxx

1

0

( ) ( ) ( )(1 )

n

m nxm n m n dxx

1

0

( ) ( ) ( ) ( ). ( , )(1 )

n

m nxm n m n dx m n B m nx

( ) ( )( , ) .( )m nB m nm n

(proved)

(May-Jun-2007)

7. Change the order of integration in 2

1 2

0

x

x

I xydxdy

and hence evaluate the same.

Ans: From 2

1 2

0

x

x

I xydxdy

, region of integration is 2 , 2 , 0, 1y x y x x x



By changing of order of integration we get two regions

In one region limit of x is 0,x x y and limit of y is 0, 1y y

In another region limit of x is 0, 2x x y and limit of y is 1, 2y y

So, 2

21 2 1 2

0 0 0 1 0

y yx

x

I xydxdy xydydx xydydx

21 2

0 0 1 0

y y

I xydx dy xydx dy

21 22 2

0 10 02 2

y yx y x yI dy dy

1 22 2

0 1

0 (2 ) 02 2

y y yI dy dy

1 22 3

2

0 1

2 22 2y yI dy y y dy

1 23 3 4

2

0 1

26 3 8y y yI y

1 16 16 2 10 4 16 3 8 3 8

I

4 96 128 48 24 16 3

24I

164 153 11

24 24I

(Ans)

8. Find the volume bounded by the cylinder 2 2 4x y and the planes 4, 0y z z

Ans:

From the figure 4z y is to be integrated over the circle 2 2 4x y in the xy plane.



To cover the shaded half of the circle, x varies from 0 to 24 y

So, 2 24 42 2

2 0 2 0

Volume 2 2 (4 )y y

zdxdy y dxdy

2

24

02

Volume 2 (4 ) yy x dy

2

2

2

Volume 2 (4 ) 4y y dy

2 2

2 2

2 2

Volume 8 4 2 4y dy y y dy

22

1

2

4 4Volume 8 sin 02 2 2

y y y

{second term is zero because of odd function}

1 1Volume 8 0 0 2sin 1 2sin ( 1)

1Volume 8 4sin 1 8 4 162 (Ans).

9. Solve 2 2 /2( )nxy x y dxdy over the positive quadrant of 2 2 4x y supposing n+3>0. Ans:



2 2

2 2 /2 2 2 /2

0 0

( ) ( )a a x

n nxy x y dxdy xy x y dxdy

2 2

32 2 2

2 2 /2

0

0

( )( ) 322

a xn

an

y

x x yxy x y dxdy dxn

3

2 2 /2 2 2 2 2

0

1( ) ( 0)3

a nnxy x y dxdy x x a x dx

n

3 322 2

2 2 /2

0 0

( )3 3 2

n n aan a a xxy x y dxdy xdx

n n

73

222 2 /2( )

3 2 2( 3)

nn

naa axy x y dxdy

n n

(Ans)

(Nov –Dec 2007)

10. Define Beta Function. Ans. The Beta function ,m n is defined as

1

0

1 1, 1m nm n x x dx for m>0 and n>0 It is also known as first Eularian Integral

11. Evaluate .

Ans:

2 2 2c b a

x y z dxdydzc ab

3 32 2 2 2 22 2

3 3

ac b c bz ax z y z dxdy ax ay dxdyc cab b

3 3 3 3

2 2 2 2 22 2 4 42 43 3 3 3

bc b a c cay a y ab a bx y z dxdydz ax y dx abx dxc a c cbb

3 3 3 3 3 3

2 2 2 4 4 4 8 8 83 3 3 3 3 3

cc b a abx ab x a bx abc ab c a bcx y z dxdydzc a cb

2 2 2 2 2 283

c b a abcx y z dxdydz a b cc ab

(Ans)

c

c

b

b

a

a

dxdydzzyx )( 222



12. Prove that .

Sol.

1

10

1 !log

1

nn

n

nmx x dxm

1

0

log nmx x dx ……………………………………………….1


1 becomes 0

0


……………………………….2

1

1dpm t p dt

m

2 becomes 0

11 1

nppe dpm m

0

11 11

nn

p ne p dpm

=

0

1 1 11 11

nn npe p dp

m

0

1 11 11 11 1 11 1

n nn nnpe p dp n

m m

1 1 !1 1 1 11 1

nn

n nn nm m

Hence Proved

13. Find the area enclosed by the parabolas 푦 = 4푥 − 푥 , 푦 = 푥 .

Ans:

1,0,)1(

!.)1().(log 1

1

0

mnm

ndxxx n

nnm



23 4

0

Areax x

x

dxdy

2

3 3 34 2 2

0 0 0

Area (4 ) (3 )x x

xy dx x x x dx x x dx

32 3

0

27 27 27 9Area 3 0 02 3 2 3 6 2x x

(Ans)

(May-Jun-2008)

14. Define Gamma Function.

Ans. Gamma function is defined as 1

0

, 0x nn e x dx n

15. Evaluate the integral: 1

loglog

1 1

xye ezdxdydz

Sol. 1

loglog

1 1

xye ezdxdydz

1 1 1

log log loglog log 0 11

1 1 1 1

xx

x xy y ye e e eezdz dxdy z z z dxdy e x e dx dy

1 1 1

log log loglog 1 1

1 1 1

xx x x x x

y ye e e e yzdxdydz e x e dx dy e x e e x dy

1

1 1

1 1

loglog log 2 log 2 1

1 1log

log log 2 log 11 1

x

x

ye e ezdxdydz y y y y e e dy

ye e ezdxdydz y y y y e dy

2

2 2

1

loglog log log 1

4 11 1

xy

eye e yzdxdydz y y y y y e

2 2

1

log 5log 1 0 5 / 4 0 12 41 1

xye e e ezdxdydz e e e e

2

1

loglog 2 13 / 4

41 1

xye e ezdxdydz e

Ans.



16. Given 1

0 1 sin( )

nx dxx n

Show that 1

sin( )n n

n

Hence evaluate 40 1

dyy

Sol. Given that

1

0 1 sin( )

nx dxx n

We know that ( , ) m nm nm n

1

0 1

n

m nx m ndx

m nx

1 1let m n m n

1

0

1 111

n

m nx n ndx n nx

1

0

1sin( )1

nx dx n nnx

Therefore 1sin( )

n nn

Now for

40 1

dyy

……………………………………….1

Let 4y =x

Y= 1/4x

Y= 3/414

dy x dx

3/4 1/4 1

40 0 0

1 1 1 1 111 4 1 4 1 4 sin 4 4 2 2sin

4 2

dy x dx x dxy x x

(Ans)

17. Find the area enclosed by the parabolas 푦 = 4푎푥,푥 = 4푎푦 .



Area is given by 4

0

2

2/4

a axdxdy

x a

4 4

0 0

2 22/42/4

a aax axdxdy y dxx a

x a

4 42

0 0

22 / 4

2/4

a aaxdxdy ax x a dx

x a

44 3 2 2 2

0 0

2 3/2 1 32 16 1623 / 2 4 3 3 3 32/4

aa ax x x a a adxdy a

ax a

(Ans)

(Nov –Dec 2008)

18. Write only the value of .

Ans By using Legendre duplication formula

1 22 12 2

m m mm

14

1 1 1 1 12 214 4 2 4 22 122 2

19. Evaluate .

Ans:

43.

41

c

c

b

b

a

a

dxdydzzyx )( 222



2 2 2c b a

x y z dxdydzc ab

3 32 2 2 2 22 2

3 3

ac b c bz ax z y z dxdy ax ay dxdyc cab b

3 3 3 3

2 2 2 2 22 2 4 42 43 3 3 3

bc b a c cay a y ab a bx y z dxdydz ax y dx abx dxc a c cbb

3 3 3 3 3 3

2 2 2 4 4 4 8 8 83 3 3 3 3 3

cc b a abx ab x a bx abc ab c a bcx y z dxdydzc a cb

2 2 2 2 2 283

c b a abcx y z dxdydz a b cc ab

(Ans)

20. Change the order of integration 0

y

x

e dxdyy

and hence evaluate it.

Ans 0

y

x

e dxdyy

0

y

x

e dxdyy

by changing the order of the equation we will have 0 0

y ye dxdyy

0 0 0 0

y yy ye edxdy dx dyy y

0 0 0 0

yy y ye edxdy x dyy y

0 0 0 0

y y yye edxdy y dy e dy

y y



0

0 0 0 0

1y y y

y ye edxdy y dy e dy ey y

(Ans).

21. Find the area of the loop of the curve 푎푦 = 푥 (푎 − 푥) . Ans.

Area of the loop of the curve 푎푦 = 푥 (푎 − 푥) .will be given by 0 0

a xx aadxdy

0 0 0 0

a x a xx xa aa adxdy dy dx

……………………………..1

0 0 0 0 0 0

a x a x a xx x xa aa a a adxdy dy dx y dx

0 0 0 0 0 00 0

a x a x a xx x xa aa a a aa a xdxdy dy dx y dx x dxa

0 0 0 0

a xx aa a a xdxdy x dxa

…………………………..2

Take 2sinx a

Becomes /2 /2

2 2 2 4 2

0 0 0 0

sin cos 2 sin cos 2 sin cos

a xx aadxdy a a d a a d



/2 22 4 2 2

0 0 0

3 1 12 sin cos 2

6 4 2 2 16

a xx aa adxdy a a d a

(Ans).

(May-Jun-2009)

22. Write the relation between Beta and Gamma function.

Ans: ( ). ( )( , )( )m nB m nm n

23. Evaluate the following integral by changing of order of integration 2

4 2 20

a a

ax

y dxdyy a x

.

Ans:



a.

2 /2 2

4 2 2 4 2 20 0 0

y aa a a

ax



2

2

4 2 2 220 0 0 2

2a a a

ax


a



2

2 21

24 2 2 220 0 0 02

0

22

sina a a a

ax


a yy a x y x aa

2

2 2 21 1 1

2 24 2 20 0 0

0

2


ax

ya y


a a

2 2 2

4 2 20 0 0

02 2

a a a a

ax


2 2 2

3

04 2 20 02 6 6

a a aa

ax


(Ans)

24. Evaluate 2 2( )

0 0

x ye dxdy

by changing to polar coordinates. Hence show that 2

0 2xe dx

.

Ans: 2 2 2

/2( )

0 0 0 0

x y r

r

e dxdy e rdrd

2 2 2

/2( )

0 0 0 0

1 22

x y r

r

e dxdy re dr d

2 2 2

/2( )

00 0 0

12

x y re dxdy e d

2 2

/2( )

0 0 0

1 (0 1)2

x ye dxdy d

2 2/2

/2( )0

0 0 0

1 1 12 2 2 2 4

x ye dxdy d

(Ans)

Now, 2 2( )

0 0 4x ye dxdy

2 2( ) ( )

0 0 4x ye dx e dy



2 2( ) ( )

0 0 4x xe dx e dx

2

2

( )

0 4xe dx

2

0 2xe dx

(Proved)

25. Find by double integration, the area lying between the parabolas 24y x x and the line y x .

Ans:

23 4

0

Areax x

x

dxdy

2

3 3 34 2 2

0 0 0

Area (4 ) (3 )x x


32 3

0

27 27 27 9Area 3 0 02 3 2 3 6 2x x

(Ans)

(Nov –Dec 2009)

26. Find the value of 2

0

axe x dx

.

Ans: - ∫ 푒 푥 푑푥∞ Let 푎푥 = 푡 ⇒ 푎푑푥 = 푑푡 ⇒ 푑푥 = 푑푡

= ∫ 푒 푡 푑푥∞ = Γ(3) = × 2 = (Ans).

27. Change the order of integration and evaluate 43

0 1

( )y

x y dxdy



Ans: - ∫ ∫ (푥 + 푦)푑푥 푑푦 = ∫ ∫ (푥 + 푦)푑푦 푑푥 = ∫ 푥푦 + 푑푥

= ∫ 푥(4 − 푥 ) + ( ) 푑푥

= ∫ 4푥 − 푥 + 8− 4푥 + 푑푥

= 2푥 − + 8푥 − +

= 2(4 − 1) − (16 − 1) + 8(2 − 1) − (8 − 1) + (32 − 1)

= 6− + 8− + = = (Ans).

28. Find the volume bounded by the plane xy, the cylinder 2 2 1x y and plane 3x y z .

Ans: - Volume = ∭푑푥푑푦푑푧

= ∫ ∫ ∫ 푑푥푑푦푑푧√√ = ∫ ∫ 푧] 푑푥푑푦√

√

= ∫ ∫ (3 − 푥 − 푦)푑푥푑푦 = ∫ 3푦 − 푥푦 −√

√푑푥√

√

= ∫ (3− 푥)(√1− 푥 + √1− 푥 ) − ( )푑푥

= ∫ 2(3− 푥)√1− 푥 푑푥 = ∫ 6√1 − 푥 − 푥√1− 푥 푑푥

= 6 √ + 6 푠푖푛 푥 + ( ) /

= 6(0 − 0) + 3 + + (0 − 0) = 3휋 (Ans).

29. Prove that: 1 12

4 40 0 4 21 1

x dx dxx x

.

Ans: - ∫ √ Let 푥 = 푠푖푛푡 ⇒ 2푥푑푥 = 푐표푠푡푑푡 ⇒ 푑푥 = =

√푑푡

⇒ ∫ √= ∫ /

√푑푡 = ∫ √푠푖푛푡푑푡/ =

Γ Γ

Γ=

Γ

Γ√

⇒ ∫ √=

Γ

Γ√ ----------------------------------(1)

∫ √ Let 푥 = 푡푎푛푡 ⇒ 2푥푑푥 = 푠푒푐 푡푑푡 ⇒ 푑푥 = =

√푑푡

⇒ ∫ √= ∫ /

√푑푡 = ∫

√푑푡/ = ∫ √

푑푡/

⇒ ∫ √=

√∫ √

푑푡/ Let 2푡 = 휃 ⇒ 2푑푡 = 푑휃 ⇒ 푑푡 =



⇒ ∫ √=

√∫ √

/ =√

Γ Γ

Γ=

√

Γ

Γ√ --------------------(2)

Now, ∫ √× ∫ √

=Γ

Γ√ ×

√

Γ

Γ√

⇒ ∫ √× ∫ √

=√

Γ

Γ휋 =

√

Γ

Γ휋 =

√ (Ans).

(May-Jun-2010) 30. Compute, 훽 , .

Ans: -

9 7 7 5 3 1 5 3 1. . . . . . . .9 7 52 2 2 2 2 2 2 2 2,9 72 2 7.6.5.4.3.2.1 20482 2

(Ans).

31. Show that : ∫ √= √ ∫ √

=.

.

Ans: -

1

061 x

dx putting ddxxx cossin

31sinsin 3/23/13

2/

0

1)2/1(21)6/1(22/

0

3/22/

0

3/21

06

cossin31sin

31

cos

cossin31

1

dd

d

xdx

3/2

.6/161

3/222/1.6/1

31

1

1

06

xdx

--------------------(1)

By Legendre’s Duplication method nnn n 222

112

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

Now, putting 61

n in eqn (2) we get 3/12

3/26/1 3/2

3/2

3/126/13/2

So, equation (1) becomes

3/23/23/12

61

3/23/12

3/261

1

3/23/21

06

xdx



233/2

2

23/21

06 )3/sin(/

3/1.261

3/23/13/13/12

61

1

x

dx as

nrn

rnr

sin1.

3/7

333/21

06 2

3/1.4.

3.3/1.261

1

xdx

(Proved)--------(I)

Again

1

0312

3

xdx

putting ddxx cossin32sin 3/13/2

2/

0

1)2/1(21)3/1(22/

0

3/12/

0

3/11

03

cossin3

1cossin3

1cos

cossin32

23

123

dd

d

xdx

6/5

.3/132

16/52

2/1.3/13

1

123 1

03

xdx

-----------------(3)

By Legendre’s Duplication method nnn n 222

112

---------(4)

Now, putting 31

n in eqn (4) we get 3/22

6/53/1 3/2

3/1

3/226/53/2

So, equation (3) becomes

3/2

3/132

1

3/13/22

3/132

112

3 2

3/53/2

1

03

xdx

)3/sin(/

3/132

13/23/13/13/1

321

123 3

3/5

2

3/5

1

03

xdx

as

nrn

rnr

sin1.

3/7

33

3/5

1

03 2

3/1233/1

321

123

xdx

(Proved)--------------(II)

From (I) and (II)



3/7

3

1

03

1

06 2

31

123

1

xdx

xdx

(Proved)

32. Change the order of following integration: ∫ ∫ 푥푦 푑푥 푑푦 and evaluate.

Ans: From 2

1 2

0

x

x

I xydxdy

, region of integration is 2 , 2 , 0, 1y x y x x x

By changing of order of integration we get two regions

In one region limit of x is 0,x x y and limit of y is 0, 1y y

In another region limit of x is 0, 2x x y and limit of y is 1, 2y y

So, 2

21 2 1 2

0 0 0 1 0

y yx

x

I xydxdy xydydx xydydx

21 2

0 0 1 0

y y

I xydx dy xydx dy

21 22 2

0 10 02 2

y yx y x yI dy dy

1 22 2

0 1

0 (2 ) 02 2

y y yI dy dy

1 22 3

2

0 1

2 22 2y yI dy y y dy

1 23 3 4

2

0 1

26 3 8y y yI y



1 16 16 2 10 4 16 3 8 3 8

I

83

249

24155164

243162448128964

I (Ans).

33. Find the volume enclosed by the cylinders

x2 + y2 = 2ax and z2 = 2ax.

Ans: - 2 2 2 2 2 2 22 2 0 ( )x y ax x y ax x a y a

At 2 20, ( ) 0,2y x a a x a

Volume = 2

2

2 2 2

0 22

a ax x ax

axax x

dxdydz

Volume = 2

2

2 22

20 2

a ax xax

axax x

z dxdy

Volume = 2

2

2

2

2 2 22

20 02

2 2 2 2a ax x a

ax x

ax xax x

axdxdy a x y dx

Volume = 2

2

0

2 2 .2 2a

a x ax x dx

Volume = 2

2

0

4 2 . 2a

a x ax x dx Let 22 sin 4 sin cosx a dx a d

Volume = /2

0

4 2 2 sin .2 sin cos .4 sin cosa a a a d

Volume = /2 3

3 3 2 3

0

2 12864 sin .cos 64 .5 3 15

aa d a

(Ans).

(Nov –Dec 2010)

34. Write the relation between beta and gamma function. ( ) ( )( , ) .( )m nB m nm n

35. Change the order of integration and evaluate : 2

4 2

0 /4

a ax

x a

I dydx



2

4 2

0 /4

a ax

x a

I dydx By changing the order of the integration we will have figure below and the

solution will be 2 2

24 2 4

0 0/4 /4

aya ax a

x a y a

dydx dydx

=

24

0 2/4

ayadydx

y a =

4

0

24 22/40 2/4

aaya aydydx x dyy ay a

4 42

0 0

22 / 4

2/4

a aaydxdy ay y a dy

y a

44 3 2 2 2

0 0

2 3/2 1 32 16 1623 / 2 4 3 3 3 32/4

aa ay

y y a a adxdy aa

y a

(Ans)

36. Evaluate :

2 22 11 1

0 0 0

x yx

I xyzdydxdz

Sol.

2 22 22 2 111 1 1 1 2

0 0 0 0 0 02

x yx yx x xyzI xyzdydxdz dydx



2 22 2 2 211 1 1 1

0 0 0 0 0

12

x yx x xy x yI xyzdydxdz dydx

2 22 2 3 311 1 1 1

0 0 0 0 0 2

x yx x xy x y y xI xyzdydxdz dydx

2

2 22

12 3 2 4

11 1 1

0 0 0 0

0

2 2 42

x

x yxxy x y y x

I xyzdydxdz dx

2 222 2 4

2 3 2 211 1 1

0 0 0 0

1 1 1

4 4 8

x yx x x x x x xI xyzdydxdz dx

2 22 22 3 2 211 1 1

0 0 0 0

1 1 14 4 8

x yx x x x x x xI xyzdydxdz dx

2 22 3 3 5 5 311 1 1

0 0 0 0

24 4 8

x yx x x x x x x xI xyzdydxdz dx

2 22

12 4 4 6 2 6 4

11 1

0 0 0

0

22 4 4 6 2 6 4

4 4 8

x yxx x x x x x x

I xyzdydxdz

2 22 11 1

0 0 0

1 1 1 1 1 1 18 16 16 24 16 48 16

x yx

I xyzdydxdz

2 22 11 1

0 0 0

1 1 1 1 1 1 1 18 16 16 24 16 48 16 48

x yx

I xyzdydxdz

(Ans)

37. Given 1

0 1 sin( )

nx dxx n

Show that 1

sin( )n n

n

Hence evaluate 40 1

dyy

Sol. Given that

1

0 1 sin( )

nx dxx n




1

0 1

n

m nx m ndx

m nx

1 1let m n m n

1

0

1 111

n

m nx n ndx n nx

1

0

1sin( )1

nx dx n nnx

Therefore 1sin( )

n nn

Now for

40 1

dyy

……………………………………….1

Let 4y =x

Y= 1/4x

Y= 3/414

dy x dx

3/4 1/4 1

40 0 0

1 1 1 1 111 4 1 4 1 4 sin 4 4 2 2sin

4 2

dy x dx x dxy x x

(Ans)

(May-Jun-2011)

38. Evaluate 3

2

Ans. We know that

( 1) ( )n n n

( 1)( ) nnn

3 13 2

322



13 2

322

………………………..1

As we know that

( 1)( ) nnn

Therefore we will have

1 111 2 2

1 122 2

…………………..2

From 1 and 2 we will have

121 1

3 2 23 3 12

2 2 2

143 2

2 3

Ans

39. Evaluate the following integral by changing of order of integration

2

4 2 20

a a

ax

y dxdyy a x

.

Ans:





a.

2 /2 2

4 2 2 4 2 20 0 0

y aa a a

ax



2

2

4 2 2 220 0 0 2

2a a a

ax


a

2

2 21

24 2 2 220 0 0 02

0

22

sina a a a

ax


a yy a x y x aa

2

2 2 21 1 1

2 24 2 20 0 0

0

2


ax

ya y


a a

2 2 2

4 2 20 0 0

02 2

a a a a

ax


2 2 2

3

04 2 20 02 6 6

a a aa

ax


(Ans)

40. State and prove relation between beta and gamma function. Ans: The Beta function is defined as

1 11 1

0 0

( , ) (1 ) , , 0(1 )

nm n


.

We know that 1

0

( ) .x nn e x dx

By putting x az dx adz we get



1

0

( ) .( )az nn e az adz

1

0

( ) .n az nn a e z dz

Putting z = x we get

1

0

( ) .n ax nn a e x dx

Now, taking a = z we get

1

0

( ) .zx n nn e x z dx

By multiplying 1z me z both side we get

1 (1 ) 1 1

0

( ) .z m z x n m nn e z e x z dx

Now taking integration with respect to z both side from 0z we get

1 1 (1 ) 1

0 0 0

( ) .z m n z x m nn e z dz x e z dzdx

Now, let (1 )1 1

y dyz x y z dzx x

we get

1

1

0 0

.( ) ( )(1 )

y m nn

m n

e y dyn m x dxx

1

1

0 0

( ) ( ) .(1 )

ny m n

m n

xn m e y dy dxx

1

0

( ) ( ) ( )(1 )

n

m nxm n m n dxx

1

0

( ) ( ) ( ) ( ). ( , )(1 )

n

m nxm n m n dx m n B m nx

( ) ( )( , ) .( )m nB m nm n

(proved)



41. Find the volume common to gas cylinder 2 2 2 2 2 2,x y a x z a .

Ans:

2 2 2 2 2 2 2 2

2 2 2 2 0 0 0


a a x a x

dxdydz dxdydz

2 2 2 2

2 22 2

00 0 0 0


a xz dxdy a x dxdy

2 2

2 2 2 2 2 2 2 20

0 0 0

Volume 8 . 8 . 8 ( )a a a


3 3 3 3

2 2

0

2 16Volume 8 8 83 3 3 3

ax a a aa x a

(Ans).

(Nov –Dec 2011)

42. Write the relation between Beta and Gamma function

Ans: - The relation between beta & gamma function is ( ) ( )( , )( )m nm nm n

.

43. Change the order of integration 0

y

x

e dxdyy

and hence evaluate it.

Ans 0

y

x

e dxdyy



0

y

x

e dxdyy


y ye dxdyy

0 0 0 0


0 0 0 0


0 0 0 0


y y

0

0 0 0 0

1y y y


(Ans).

44. Find the volume common to the cylinder 2 2 2 2 2 2,x y a x z a .

Ans:



2 2 2 2 2 2 2 2

2 2 2 2 0 0 0


a a x a x

dxdydz dxdydz

2 2 2 2

2 22 2

00 0 0 0


a xz dxdy a x dxdy

2 2

2 2 2 2 2 2 2 20

0 0 0

Volume 8 . 8 . 8 ( )a a a


3 3 3 3

2 2

0

2 16Volume 8 8 83 3 3 3

ax a a aa x a

(Ans).

45. Prove that 2

1 2 12 2 1n

nnn

Ans.

2 11 2 1 12 2 2

nnn

2 1 2 1 2 1 2 1 2 3 2 31 1 ..........2 2 2 2 2 2 2

n n n n n nn

2 1 2 1 2 3 2 5 2 71 1 11 ............2 2 2 2 2 2 2 2

n n n n nn

2 1 . 2 3 . 2 5 . 2 7 ................112 2

n n n nn n

2 . 2 1 . 2 2 . 2 3 . 2 4 . 2 5 . 2 6 . 2 7 ................112 2 .2 . 2 2 . 2 4 . 2 6 ......................2

n n n n n n n nn n n n n n

1 2 !2 2 .2 . 1 . 2 . 3 ......................1

nn n n n n n

1 2 !22 2 . !

nn n n

1 2 122 2 . 1

nn n n

hence Proved

(May-Jun-2012)

46. Evaluate the following Integral:



2 3 2

0 1xy dxdy

3 23 22 3 2 22

0 1 0 01 0

26 26 52 / 33 3 6

xy x xxy dy dx dx dx

47. Evaluate the following integral by changing the order of integration:

0

y

x

e dydxy

Ans 0

y

x

e dxdyy

0

y

x

e dxdyy


y ye dxdyy

0 0 0 0


0 0 0 0


0 0 0 0


y y

0

0 0 0 0

1y y y


(Ans).

48. Given 1

0 1 sin

nx dxx n

, show that ( ) (1 )sin

n nn



Sol. Given that

1

0 1 sin( )

nx dxx n


1

0 1

n

m nx m ndx

m nx

1 1let m n m n

1

0

1 111

n

m nx n ndx n nx

1

0

1sin( )1

nx dx n nnx

Therefore 1sin( )

n nn

Now for

40 1

dyy

……………………………………….1

Let 4y =x

Y= 1/4x

Y= 3/414

dy x dx

3/4 1/4 1

40 0 0

1 1 1 1 111 4 1 4 1 4 sin 4 4 2 2sin

4 2

dy x dx x dxy x x

(Ans)

49. Find the area included between the parabola: 24y x x and the line y x Ans:



23 4

0

Areax x

x

dxdy

2

3 3 34 2 2

0 0 0

Area (4 ) (3 )x x


32 3

0

27 27 27 9Area 3 0 02 3 2 3 6 2x x

(Ans)

(Nov-Dec2012)

50. Define Beta and Gamma Function.

Ans:

Beta function

The Beta function is defined as 1 1

1 1

0 0

( , ) (1 ) , , 0(1 )

nm n


.

Gamma Function: - The Gamma function )(n is defined as dxxen nx

0

1)( for 0n .

It is also know as Second Eulerian Integral

51. Evaluate0

y

x

e dydxy

Ans 0

y

x

e dxdyy

0

y

x

e dxdyy


y ye dxdyy



0 0 0 0


0 0 0 0


0 0 0 0


y y

0

0 0 0 0

1y y y


(Ans).

52. Prove that

Hence evaluate 1 3

0(log )x x dx

Sol.

1

10

1 !log

1

nn

n

nmx x dxm

1

0

log nmx x dx ……………………………………………….1


1 becomes 0

0


……………………………….2

1

1dpm t p dt

m

2 becomes 0

11 1

nppe dpm m

0

11 11

nn

p ne p dpm

=

0

1 1 11 11

nn npe p dp

m

0

1 11 11 11 1 11 1

n nn nnpe p dp n

m m

1 1 !1 1 1 11 1

nn

n nn nm m

Hence Proved

1,0,)1(

!.)1().(log 1

1

0

mnm

ndxxx n

nnm



As we know that

1

10

1 !log

1

nn

n

nmx x dxm

there fore

31 3

3 10

1 3! 6 3(log )16 81 1

x x dx

=

53. Find the volume of the cylinder 2 2 4,x y and the planes 4y z and z=0

Ans:

From the figure 4z y is to be integrated over the circle 2 2 4x y in the xy plane.

To cover the shaded half of the circle, x varies from 0 to 24 y

So, 2 24 42 2

2 0 2 0

Volume 2 2 (4 )y y

zdxdy y dxdy

2

24

02

Volume 2 (4 ) yy x dy

2

2

2

Volume 2 (4 ) 4y y dy

2 2

2 2

2 2

Volume 8 4 2 4y dy y y dy

22

1

2

4 4Volume 8 sin 02 2 2

y y y

{second term is zero because of odd function}

1 1Volume 8 0 0 2sin 1 2sin ( 1)



1Volume 8 4sin 1 8 4 162 (Ans).

APPLIED MATHS – II

300214 B.E.(Second Semester)

Unit IV – Vector Calculus

(May-Jun-2006)

1. Show that 2 2( 1)n nr n n r where 푟 = 푥푖 + 푦푗 + 푧푘.

Ans: - 222ˆˆˆ zyxkzjyixRr

2222222nn

n zyxzyxr

2 2 2 2 2 2 2 2 22 2 2ˆ ˆˆ ˆ ˆ ˆ

n n nn n n

nx y z x y z x y zr r rr i j k i j k

x y z x y z

1 1 12 2 2 2 2 2 2 2 22 2 2ˆˆ ˆ.2 .2 .22

n n nn nr i x y z x j x y z y k x y z z

2

2 2 2 2 ˆˆ ˆn

nr n x y z ix jy kz

2n nr nr R

…………………………………1

2 2n nr nr R

2 2ˆ ˆ ˆn nr i j k nr Rx y z

2 2 2 2n n n nr nr R i nr R j nr R kx y z

2 2 2 2n n n nr nr x nr y nr z

x y z because ˆˆ ˆR xi yj zk

2 2 2 2n n n nr n r x r y r z

x y z

2 2 2 2 2 2 2n n n n n n nr n r x x r r y y r r z z r

x x y y z z



2 2 3 2 3 2 3.1 2 .1 2 .1 2

r r rn n n n n n nr n r x n r r y n r r z n rx y z

2 2 3 2 3 2 3.1 2 .1 2 .1 2yx zn n n n n n nr n r x n r r y n r r z n r

r r r

2 2 2 4 2 2 4 2 2 4.1 2 .1 2 .1 2n n n n n n nr n r x n r r y n r r z n r

2 2 4 2 2 23 2n n nr n r n r x y z

2 2 4 23 2n n nr n r x n r r

2 2 23 2n n nr n r n r

2 2 3 2n nr nr n

2 2 1n nr nr n

2 21n nr n n r hence proved

2. Verify Stoke’s theorem for the vector field on the upper half surface of , bounded by its projection on the xy-plane.

Ans: Stoke’s theorem states that ˆ. ( ).C S

F dr curl F nds

1. Here C is unit circle 2 2 1, 0x y z

2 2 ˆ ˆˆ ˆ ˆ ˆ. (2 ) .F dr x y i yz j y zk idx jdy kdz

2 2. (2 )F dr x y dx yz dy y zdz

Here C is z = 0, dz = 0 and cos , sinx y

. (2 )F dr x y dx

So, 2

0

. (2 ) (2cos sin )( sin )C C

F dr x y dx d

2

0

1 cos 2. sin 22C

F dr d

2

0

cos 2 sin 2.2 2 4C

F dr

1 1 2 0 0 0.2 2 4C

F dr

-------(1)

2. Now, 2 2 ˆˆ ˆ(2 )F x y i yz j y zk

2 2 ˆˆ ˆ(2 )F x y i yz j y zk

2 2 2 1x y z



2 2

ˆˆ ˆ

( )

2

i j k

curl Fx y z

x y yz y z

ˆ ˆˆ ˆ( ) ( 2 2 ) (0 0) (0 1)curl F i yz yz j k k

By using spherical co-ordinate we have

ˆˆ ˆˆ sin cos sin sin cosn i j k

So, ˆ( ). coscurl F n

Hence /2 2

0 0

ˆ( ). cos .sinS

curl F nds d d

/2 2

0 0

ˆ( ). cos .sinS

curl F nds d d

/22

2

00

sinˆ( ).2S

curl F nds

1 0ˆ( ). (2 0)2S

curl F nds

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

From (1) and (2) it verifies Stoke’s theorem. (Ans)

3. Use divergence theorem to evaluate where and S is the

surface of the sphere .

Ans: By divergence theorem .S V

F ds divFdv

3 3 3. ( ) ( ) ( )S V

F ds x y z dvx y z

2 2 2. 3S V

F ds x y z dxdydz

2 2 2 2 2. 3S V

F ds a dxdydz x y z a given

3 3

2 4 4. 33 3S V

a aF ds a volume of sphere dxdydz

5. 4S

F ds a Ans

S

dsF

kzjyixF ˆˆˆ 333

2222 azyx



Nov-Dec, 2006

4. Verify Green’s theorem for: 2 2( )C

xy y dx x dy where C is bounded by 2,y x y x .

Ans:

By Green’s theorem 2 11 2

C S

F FF dx F dy dxdyx y

(1) Here 2 2ˆ ˆ( )F xy y i x j

So, 2 12 , 2F Fx x yx y

2

12 1

0

2 2x

S x

F F dxdy x x y dxdyx y

2

122 1

0

x

xS

F F dxdy xy y dxx y

1

2 2 3 42 1

0S

F F dxdy x x x x dxx y

11 5 4

4 32 1

0 05 4S

F F x xdxdy x x dxx y

2 1 1 1 15 4 20S

F F dxdyx y

----------------(1)

(2) Now we have to find line integral along two curves. Along 2 , 2y x dy xdx .

So, 1 1

2 21 2 ( )

C C

F dx F dy xy y dx x dy



1

12 4 2

1 20

( . ) .2C

F dx F dy x x x dx x xdx

1

11 4 53 4

1 20 0

3(3 )4 5C

x xF dx F dy x x dx

1

1 23 1 194 5 20C

F dx F dy

Along ,y x dy dx

So, 2 2

2 21 2 ( )

C C


2

02 2

1 21

( . ) .C

F dx F dy x x x dx x dx

2

002 3

1 2 11

3 0 1 1C

F dx F dy x dx x

Now, 1 2

1 2 1 2 1 219 1120 20C C C

F dx F dy F dx F dy F dx F dy ----------(2)

From (1) and (2) it verifies Green’s theorem.

5. Evaluate .S

F dS

where 2 2 ˆˆ ˆ4 2F xi y j z k

and S is the surface bounding the region

2 2 4x y , 0, 3z z . Ans:



By divergence theorem .S V

F ds divFdv

2 2. (4 ) ( 2 ) ( )S V

F ds x y z dvx y z

. 4 4 2S V

F ds y z dxdydz

2

2

2 4 3

2 04

. 4 4 2x

S x

F ds y z dxdydz

2

2

2 432

02 4

. 4 4x

zS x

F ds z yz z dxdy

2 2

2 2

2 4 2 4

2 24 4

. 12 12 9 21 12x x

S x x

F ds y dxdy y dxdy

2

2

242

42

. 21 6x

xS

F ds y y dx

2

2

2

. 42 4 0S

F ds x dx

2

2

2

. 42 4S

F ds x dx

22

1

2

4 4. 42 sin2 2 2S

x x xF ds

1 1. 42 0 0 2sin 1 2sin ( 1)S

F ds

1. 42 4sin 1 42 4 842S

F ds (Ans).



6. Evaluate C

F dR

, where ˆˆ ˆ2F yi zj xk

and the curve cos , sin , 2cosx t y t z t

from 0 / 2t to .

Ans: Here ˆˆ ˆ2F yi zj xk

So,

ˆˆ ˆˆˆ ˆ2 ( ) (2 ) (2 )

i j kF dR y z x i zdz xdy j ydz xdx k ydy zdx

dx dy dz

Now, ˆˆ ˆ( ) (2 ) (2 )C C

F dR i zdz xdy j ydz xdx k ydy zdx

2 2/2

0

ˆ ˆ(4sin cos cos ) ( 4sin sin cos )ˆ(2sin cos 2sin cos )C

i t tdt tdt j tdt t tdtF dR

k t tdt t tdt

/2

2 2

0

ˆ ˆ(4sin cos cos ) (4sin sin cos )C

F dR i t t t j t t t dt

/2

0

1 cos 2 sin 2ˆ ˆ2sin 2 2(1 cos 2 )2 2C

t tF dR i t j t dt

/2

0

sin 2cos 22ˆ ˆcos 2 2 sin 2

2 4C

tt tF dR i t j t t

0 0 0 1 12ˆ ˆ1 1 2 0 0 02 2 4C

F dR i j

1ˆ ˆ24 2C

F dR i j

(Ans)

May-June, 2007

7. Find the work done in moving a particle in the force field: 2 ˆˆ ˆ3 (2 )F x i xz y j zk

, along:

(a) The straight line from (0, 0, 0) to (2, 1, 3). Ans: The equation of straight line joining (0, 0, 0) and (2, 1, 3) is

0 0 0 2 , , 3

2 0 1 0 3 0x y z t x t y t z t

where 0t to 1t

So, Work done .C

F dr



1

2

0

Work done 3(2 ) .2 (4 (3 )) 3 .3t dt t t t dt t dt

1 1

2 2 2

0 0

Work done 24 12 8 36 8t t t dt t t dt

13 2

0Work done 12 4 12 4 0 0 16t t

(b) The curve defined by 2 34 ,3 8x y x z from 0, 2x x . Ans: 2 34 ,3 8x y x z

Let 2 33,4 8t tx t y z where 0, 2t t

So, Work done .C

F dr

1 4 2 3 2

2

0

3 3 9Work done 3 .4 4 2 8 8t t t t tt dt dt dt

1 15 3 5 3 5

2 2

0 0

3 27 51Work done 3 38 8 64 8 64t t t t tt dt t dt

24 6

3

0

17 16 17 64 1 17Work done 8 8 1632 128 32 128 2 2t tt

(Ans)

8. Evaluate divergence and curl of 2 2 3 ˆˆ ˆ3 5F x i xy j xyz k

at the point (1, 2, 3).

Ans: 2 2 3 ˆˆ ˆ3 5F x i xy j xyz k

2( ) 6 10 3div F x xy xyz

(1,2,3)( ) 6 20 54 80div F

(Ans).

Now,

2 2 3

ˆˆ ˆ

( )

3 5

i j k

curl Fx y zx xy xyz

3 3 2ˆˆ ˆ( ) ( 0) ( 0) (5 0)curl F i xz j yz k y

3 3 2 ˆˆ ˆ( ) 5curl F xz i yz j y k

(1,2,3)ˆˆ ˆ( ) 9 54 20curl F i j k

(Ans).

9. Use divergence theorem to evaluate .S

F dS


and S is the

surface bounding the region 2 2 4x y , 0, 3z z . Ans:




F ds divFdv

2 2. (4 ) ( 2 ) ( )S V

F ds x y z dvx y z

. 4 4 2S V

F ds y z dxdydz

2

2

2 4 3

2 04

. 4 4 2x

S x

F ds y z dxdydz

2

2

2 432

02 4

. 4 4x

zS x

F ds z yz z dxdy

2 2

2 2

2 4 2 4

2 24 4

. 12 12 9 21 12x x

S x x

F ds y dxdy y dxdy

2

2

242

42

. 21 6x

xS

F ds y y dx

2

2

2

. 42 4 0S

F ds x dx

2

2

2

. 42 4S

F ds x dx

22

1

2

4 4. 42 sin2 2 2S

x x xF ds



1 1. 42 0 0 2sin 1 2sin ( 1)S

F ds

1. 42 4sin 1 42 4 842S

F ds (Ans).

Nov - Dec, 2007 10. Define Gradient.

Ans. Gradient of a scalar point function : -Let f be a scalar point function then vector point function f is known as gradient of f. Its geometrical interpretation is grad(f) is the normal to the

surface f = const.So, ˆˆ ˆ( ) f f fgrad f f i j kx y z

11. Find the values of a and b such that the surfaces xabyzax )2(2 and 44 32 zyx cut orthogonally at (1, -1, 2).

Ans: - Here 0)2(2 xabyzaxF and 044 32 zyxG

So, kbyjbziaaxF ˆˆˆ)22( and kzjxixyG ˆ3ˆ4ˆ8 22

kbjbiaF ˆˆ2ˆ)2()2,1,1( kjiG ˆ12ˆ4ˆ8)2,1,1(

As surfaces are orthogonal, so 01281680 bbaGF

04201684 baab ---------------(1)

As point (1, -1, 2) lies on the surface, so it satisfies surface F.

10220)2(2 bbaba

Putting in (1) we get 2/50412 aa

So, 1,2/5 ba (Ans).

12. If 2 2 2 ˆˆ ˆ( ) 2F x y i xyj y xy k

then find dif F and Curl F . Ans: If 2 2 2 ˆˆ ˆ( ) 2F x y i xyj y xy k

then dif F will be defined as

2 2 2ˆ ˆˆ ˆ ˆ ˆ. . ( ) 2f f fF i j k x y i xyj y xy kx y z

. (2 2F x x



If 2 2 2 ˆˆ ˆ( ) 2F x y i xyj y xy k

then Curl F will be defined as

2 2 2ˆ ˆˆ ˆ ˆ ˆ( ) 2f f fF i j k x y i xyj y xy kx y z

2 2 2

ˆˆ ˆ

( ) (2 )

i j k

Fx y z

x y xy y xy

ˆˆ ˆ2 2 2F y x i y j y y k

ˆ ˆ2F y x i y j

Ans.

13. Verify Stoke’s theorem for the vector field 2 2 ˆ ˆ( ) 2F x y i xyj

taken around the rectangle bounded by , 0x a y to y b


F dr curl F nds

……………………………1

Here C is Rectangle bounded by ,x a y b

2 2 ˆˆ ˆ ˆ ˆ. ( ) 2 .F dr x y i xyj idx jdy kdz

2 2. ( ) 2F dr x y dx xydy

2 2. ( ) 2F dr x y dx xydy

So, 2 2. ( ) 2C C DA AB BC CD

F dr x y dx xydy

0

2 2 2

0

2 2a b a

DA AB BC CD a a b

x dx aydy x b dx aydy



0

2 2 2

0 0

4 2 2a b a b

DA AB BC CD a a b

x dx aydy x b dx aydy aydy

3 2 3

2

0

43 2 3

a b a

DA AB BC CD a a

x ay x b x

3 3

2 22 22 23 3DA AB BC CD

a aab ab

24DA AB BC CD

ab …………………………………………2

ˆ( ).S

curl F nds

( )curl F F

=

2 2

ˆˆ ˆ

( ) ( 2 ) 0

i j k

Fx y z

x y xy

2 2 4F y y k yk

0

ˆ( ). ( )b a

S a

curl F nds curl F Nds

0 0 0

( ) 4 4b a b a b a

a a a

curl F Nds yk kds y dxdy

0 0 0

( ) 4 4b a b a b

a

aa a

curl F Nds y dxdy yx dy

0 0 0

( ) 4 4 2b a b a b

a a

curl F Nds y dxdy ay dy

2 20

0 0 0

( ) 4 4 2 4 4b a b a b

b

a a

curl F Nds y dxdy ay dy a y ab

………………………………..3

From 1 , 2, and 3rd we can Verify the Stokes’s theorem

(May-Jun-2008)

14. Define gradient.



Ans. Gradient of a scalar point function : -Let f be a scalar point function then vector point function f is known as gradient of f. Its geometrical interpretation is grad(f) is the normal to the

surface f = const.So, ˆˆ ˆ( ) f f fgrad f f i j kx y z

15. Find the values of a and b such that the surfaces and cut orthogonally at (1, -1, 2).




As surfaces are orthogonal, so 01281680 bbaGF 04201684 baab ---------------(1) As point (1, -1, 2) lies on the surface, so it satisfies surface F. 10220)2(2 bbaba

Putting in (1) we get 2/50412 aa So, 1,2/5 ba (Ans).

16. Prove that : 2 2( ) ''( ) '( )f r f r f r

r


2 ( ) . ( )f r f r

( )f r =

ˆ ˆ ˆ ( )i j k f rx y z

ˆ ˆ ˆ( ) ( ) ( ) ( )f r i f r j f r k f rx y z

/ / /ˆ( ) ( )r r r

f r f r i f j f kx y z

/ ˆ( ) ( )yx z

f r f r i j kr r r

/( )

( )f r R

f rr

2

/( )( ) . ( ) .

f r Rf r f r

r

xabyzax )2(2 44 32 zyx



2 / /( ) ( ) . . ( )

R Rf r f r f r

r r

2 / /ˆ ˆ ˆ ˆ ˆ ˆ( ) ( ) . ( )

R Rf r f r i j k i j k f r

x y z r r x y z

2 / /ˆ ˆ ˆ ˆ ˆ ˆ ˆ( ) ( ) . ( )

2 2 2 2 2 2 2 2 2x x x R

f r f r i j k i j k i j k f rx y z r x y zx y z x y z x y z

2 / // ˆ ˆ ˆ( ) ( ) ( )

2 2 2 2 2 2 2 2 2yx z R r r r

f r f r f r i j kx y z r x y zx y z x y z x y z

2 / // ˆ ˆ ˆ( ) ( ) ( )

2 2 2x R r r r

f r f r f r i j kx r x y zx y z

2

2 2 2 2 2 2/ // ˆ ˆ ˆ( ) ( ) ( )22 2 2

x y z x x x y z yR x zx xf r f r f r i j kr r r r

x y z

2

22 2 22 2 22/ //( ) ( ) ( )22 2 2

xx y z x

x y z R Rf r f r f r

r rx y z

2

22 2 2 2/ //( ) ( ) ( )

2 2 2 2 2 2

x y z x Rf r f r f r

rx y z x y z

2

221/ //( ) ( ) ( )3/22 2 2 2 2 2

x Rf r f r f r

rx y z x y z

2

22 2 23/ //( ) ( ) ( )3/22 2 2 2 2 2

x y z Rf r f r f r

rx y z x y z



2

223/ //( ) ( ) ( )3/22 2 2 2

r Rf r f r f r

rx y z r

2

23 1/ //( ) ( ) ( )

2 2 2

Rf r f r f r

r rx y z

2

23 1/ //( ) ( ) ( )

2 2 2 2 2 2

Rf r f r f r

rx y z x y z

2

22/ //( ) ( ) ( )

2 2 2

Rf r f r f r

rx y z

222

2/ // 2 2 2 2ˆ ˆ ˆ ˆ ˆ ˆ( ) ( ) ( ) ( ).( )2 2 2

rf r f r f r R xi yj zk xi yj zk x y z r

rx y z

2 2/ //( ) ( ) ( ) Prf r f r f r Hence oved

r

17. Evaluate the line integral 2 2 2( ) ( )

C

x xy dx x y dy where C is square formed by the

lines 푦 = ±1, 푥 = ± 1. Ans. Here C is Square bounded by 1, 1x y



So, 2 2 2. ( ) ( )C C DA AB BC CD

F dr x xy dx x y dy

02 2 2

0

2 2a b a

DA AB BC CD a a b

x dx aydy x b dx aydy

1 1 1 1

2 2 2 2

1 1 1 1

1 1DA AB BC CD

x x dx y dy x x dx y dy

1 1 1 1 1 1

2 2 2 2 2 2

1 1 1 1 1 1

1 1 1 1DA AB BC CD

x x dx y dy x x dx y dy y dy y dy

1 1

2 2

1 1DA AB BC CD

x x dx x x dx

1 1 1 13 2 3 2

1 1 1 13 2 3 2DA AB BC CD

x x x x

2 2 03 3DA AB BC CD

2 2 2. ( ) ( )C C DA AB BC CD

F dr x xy dx x y dy

= 0

(Nov –Dec 2008)

18. Evaluate 푑푖푣 (3푥 횤̂ + 5푥푦 횥̂ + 5푥푦푧 푘) at (1, 2, 3).



Ans: - 2 2 3 2 2 3ˆˆ ˆ3 5 5 3 5 5div x i xy j xyz k x xy xyzx y z

2 2 3 2ˆˆ ˆ3 5 5 6 10 15div x i xy j xyz k x xy xyz

2 2 3

(1,2,3)ˆˆ ˆ3 5 5 6 20 270 296div x i xy j xyz k (Ans).

19. What is the directional derivative of 2 3xy yz at the point (2, -1, 1) in the direction of

normal to the surface at (-1, 2, 1)? Ans: Let 2 3xy yz

The gradient of above is and its directional derivative is in the direction of unit vector normal to the surface

(2, 1,1)ˆˆ ˆi j k

x y z

2 3(2, 1,1)

ˆˆ ˆi j k xy yzx y z

2 3 2(2, 1,1)

ˆˆ ˆ2 3y i yx z j yz k

(2, 1,1)ˆˆ ˆ3 3i j k

Normal vector to the surface 2log 4x z y is given by 2log 4x z y

2 2ˆˆ ˆlog 4 log 4x z y i j k x z yx y z

2 ˆˆlog 4 4x z y j k

Now directional derivative of 2 3xy yz in the direction of ˆˆ4 j k will be

ˆˆ4ˆˆ ˆ3 3 .17

j ki j k

Which is equal to 1517

20. If 푓 = (푥 + 푦 + 푧 ) , find 푑푖푣 푔푟푎푑(푓) and determine 푛 if 푑푖푣 푔푟푎푑(푓) =

0.


2 2 2 2n nx y z r

( ) .div gradf f

4log 2 yzx

4log 2 yzx



2 2 2 2n nf x y z r

2 2 2ˆˆ ˆ

n n nr r rf i j k

x y z

2 1 2 1 2 1 ˆˆ ˆ2 2 2r r rn n nf n r i n r j n r kx y z

2 1 ˆˆ ˆ2 r r rnf n r i j kx y z

2 1 ˆˆ ˆ2 x y znf n r i j kr r r

2 12 Rnf n rr

2 12 Rnf n rr

2 1ˆˆ ˆ 2 Rnf i j k n rx y z r

2 1ˆˆ ˆ2 Rnf n i j k rx y z r

2 1 2 1ˆ ˆˆ ˆ ˆ ˆ2 x y z Rn nf n r i j k i j k rx r y r z r r x y z

2 1 2 1ˆ2 2 2 2

r x x r r y y r r z z r Rn nx x x x x xf n r i rr xr r r

2 1 2 2 ˆ2 2 12 2 2

x y zr x r y r z R rn nr r rf n r n r ir xr r r

2 1 2 2 ˆ2 2 12 2 2

x y zr x r y r z R xn nr r rf n r n r ir rr r r

2 2 2 2 2 22 1 2 22 2 13 3 3

r x r y r z R Rn nf n r n rr rr r r



22 2 2 2 2 22 1 2 22 2 13 3 3

y z x z y x Rn nf n r n rrr r r

2 2 2 222 1 2 22 2 13

x y z Rn nf n r n rrr

2

2 2 2 22 .2 2 1 .1 1Rn nf n r n rr

2 22 .2 2 1nf n r n

2 22 . 2 1nf n r n

0f

2 22 . 2 1 0 . 2 1 0 1/ 2nn r n n n

21. Verify Green’s theorem for: where C is bounded by

. Ans:


C S


Here 2 2ˆ ˆ( )F xy y i x j

So, 2 12 , 2F Fx x yx y

2

12 1

0

2 2x

S x


2

122 1

0

x

xS

F F dxdy xy y dxx y

2 2( )C

xy y dx x dy 2,y x y x



1

2 2 3 42 1

0S


11 5 4

4 32 1

0 05 4S


2 1 1 1 15 4 20S

F F dxdyx y

----------------(1)

Now we have to find line integral along two curves. Along 2 , 2y x dy xdx .

So, 1 1

2 21 2 ( )

C C


1

12 4 2

1 20

( . ) .2C


1

11 4 53 4

1 20 0

3(3 )4 5C

x xF dx F dy x x dx

1

1 23 1 194 5 20C

F dx F dy

Along ,y x dy dx

So, 2 2

2 21 2 ( )

C C


2

02 2

1 21

( . ) .C


2

002 3

1 2 11

3 0 1 1C

F dx F dy x dx x

Now, 1 2

1 2 1 2 1 219 1120 20C C C

F dx F dy F dx F dy F dx F dy ---------- (2)

From (1) and (2) it verifies Green’s theorem.

May-June 2009

22. State the Green’s theorem in the plane. Ans: If C be the regular closed curve in xy plane bounding any region R and 1( , )F x y and

2 ( , )F x y be continuous functions on C and its interior, having continuous partial derivatives

1Fy

and 2Fx

in R, then 2 11 2

C R


, the line integral being



taken along the boundary C of R such that R is on the left as one advances in the direction of integration.

23. Prove that 5 3

1 3( )( )A R B R A BA Br r r

where A and B are constants vectors

ˆˆ ˆR xi yj zk and 2 2 2r x y z .

Ans: 2 2 2r x y z

3/22 2 2 2 2 2

1 1 1 1 ˆˆ ˆ2 2 22

xi yj kr x y z x y z

3 32 2 2

1 1 1 RRr r rx y z

3

1 B RBr r

3

1 B RBr r

3 3

1 1 1B B R B Rr r r

-----------(1)

Where ˆ ˆˆ ˆ ˆ ˆB R ai cj ck xi yj zk ax by cz

ˆˆ ˆB R ax by cz ai cj ck B

So equation (1) becomes

3 4

1 3BB B R rr r r

2 2 2

3 4

1 3B RBB x y z

r r r

3 4 2 2 2

1 1 ˆˆ ˆ3 2 2 22

B RBB xi yj kr r r x y z

3 5

1 3B RBB R

r r r

3 5

1 3A R B RA BA B

r r r



5 3

1 3A R B R A BA B

r r r

(Proved)

24. A vector is given by 2 2 ˆ ˆ( ) (2 )F x y x i xy y j

. Show that the field is irrotational and find its scalar potential. Hence evaluate the line integral from (1, 2) to (2, 1).

Ans: Here 2 2 ˆ ˆ( ) (2 )F x y x i xy y j

2 2

ˆˆ ˆ

( ) (2 ) 0

i j k

Fx y z

x y x xy y

ˆˆ ˆ(0 0) (0 0) ( 2 2 ) 0F i j k y y

So, 2 2 ˆ ˆ( ) (2 )F x y x i xy y j

is irrotational and can be expressed as the gradient of a scalar potential.

Hence, F

2 2 ˆ ˆ ˆ ˆ( ) (2 )x y x i xy y j i jx y

2 2( )..........(1) (2 )..........(2)x y x xy yx y

Now, integrating (1) we get

3 2

2 ( )............(3)3 2x xy x f y

Now, differentiating (3) with respect to y we get

2 '( )yx f yy

(2 ) 2 '( )xy y yx f y

2

'( ) ( )2yf y y f y

So, 3 2 2

2

3 2 2x x yy x (Ans)

Again .F dR

from (1, 2) to (2, 1) (1,2) (2,1)

1 1 4 8 4 1. 4 23 2 2 3 2 2

F dR

2 24 3 12 16 12 12 3 31 13 44 22.6 6 6 6 6 3

F dR

(Ans)



25. Verify Stoke’s theorem for the vector field 2 2 ˆˆ ˆ(2 )F x y i yz j y zk

on the upper half surface of 2 2 2 1x y z , bounded by its projection on the xy-plane.


F dr curl F nds

Here C is unit circle 2 2 1, 0x y z

2 2 ˆ ˆˆ ˆ ˆ ˆ. (2 ) .F dr x y i yz j y zk idx jdy kdz

2 2. (2 )F dr x y dx yz dy y zdz

Here C is z = 0, dz = 0 and cos , sinx y

. (2 )F dr x y dx

So, 2

0

. (2 ) (2cos sin )( sin )C C

F dr x y dx d

2

0

1 cos 2. sin 22C

F dr d

2

0

cos 2 sin 2.2 2 4C

F dr

1 1 2 0 0 0.2 2 4C

F dr

-------(1)

Now, 2 2 ˆˆ ˆ(2 )F x y i yz j y zk

2 2

ˆˆ ˆ

( )

2

i j k

curl Fx y z

x y yz y z

ˆ ˆˆ ˆ( ) ( 2 2 ) (0 0) (0 1)curl F i yz yz j k k

By using spherical co-ordinate we have

ˆˆ ˆˆ sin cos sin sin cosn i j k

So, ˆ( ). coscurl F n



Hence /2 2

0 0

ˆ( ). cos .sinS

curl F nds d d

/2 2

0 0

ˆ( ). cos .sinS

curl F nds d d

/22

2

00

sinˆ( ).2S

curl F nds

1 0ˆ( ). (2 0)2S

curl F nds

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

From (1) and (2) it verifies Stoke’s theorem. (Ans)

Nov-Dec, 2009

26. Find 2 2 3 ˆˆ ˆ3 5 3div x i xy j xyz k at the point (1,2,3) .

Ans: - 2 2 3 2 2 3ˆˆ ˆ3 5 3 3 5 3div x i xy j xyz k x xy xyzx y z

2 2 3 2ˆˆ ˆ3 5 3 6 10 9div x i xy j xyz k x xy xyz

2 2 3

(1,2,3)ˆˆ ˆ3 5 3 6 20 162 188div x i xy j xyz k (Ans).

27. Verify Green’s theorem for: 2 2( )C

xy y dx x dy where C is bounded by 2,y x y x .

Ans: -


C S



So, 2 12 , 2F Fx x yx y



2

12 1

0

2 2x

S x


2

122 1

0

x

xS

F F dxdy xy y dxx y

1

2 2 3 42 1

0S


11 5 4

4 32 1

0 05 4S


2 1 1 1 15 4 20S

F F dxdyx y

----------------(1)

(4) Now we have to find line integral along two curves. Along 2 , 2y x dy xdx .

So, 1 1

2 21 2 ( )

C C


1

12 4 2

1 20

( . ) .2C


1

11 4 53 4

1 20 0

3(3 )4 5C

x xF dx F dy x x dx

1

1 23 1 194 5 20C

F dx F dy

Along ,y x dy dx

So, 2 2

2 21 2 ( )

C C


2

02 2

1 21

( . ) .C


2

002 3

1 2 11

3 0 1 1C

F dx F dy x dx x

Now, 1 2

1 2 1 2 1 219 1120 20C C C


From (1) and (2) it verifies Green’s theorem. OR

28. Evaluate ˆ.S

F ndS



2 2 4x y , 0, 3z z . Ans: -




F ds divFdv

2 2. (4 ) ( 2 ) ( )S V

F ds x y z dvx y z

. 4 4 2S V

F ds y z dxdydz

2

2

2 4 3

2 04

. 4 4 2x

S x

F ds y z dxdydz

2

2

2 432

02 4

. 4 4x

zS x

F ds z yz z dxdy

2 2

2 2

2 4 2 4

2 24 4

. 12 12 9 21 12x x

S x x

F ds y dxdy y dxdy

2

2

242

42

. 21 6x

xS

F ds y y dx

2

2

2

. 42 4 0S

F ds x dx

2

2

2

. 42 4S

F ds x dx

2

21

2

4 4. 42 sin2 2 2S

x x xF ds

1 1. 42 0 0 2sin 1 2sin ( 1)S

F ds 1. 42 4sin 1 42 4 842S

F ds

(Ans). 29. Find the directional derivative of 2 3( , , )f x y z xy yz at (2, 1,1) in the direction of

vector ˆˆ ˆ2 2i j k .

Ans: - 2 3( , , )f x y z xy yz

2 3[ ( , , )] [ ]grad f x y z grad xy yz

2 3 2 ˆˆ ˆ(2 ) 3f y i xy z j yz k (2, 1,1)ˆˆ ˆ3 3f i j k

Given that ˆˆ ˆ2 2a i j k



So, directional derivative of 2 3( , , )f x y z xy yz at (2,-1,1) in the direction of

ˆˆ ˆ2 2a i j k

is (2, 1,1)

ˆˆ ˆ2 2 1 6 6 11ˆˆ ˆ3 3ˆˆ ˆ 3 32 2

a i j kf i j ka i j k

(Ans).

May-June 2010

30. Explain in brief the directional derivatives.

It is the maximum rate of change of a scalar point function in the direction of a vector.

So, directional derivative of f at a point P in the direction of vector a is given by a fa

31. Evaluate C

dszxy )( 2 , where C is the arc of the helix tztytx ,sin,cos , which joins

the points )0,0,1( and ),0,1( .

Ans: - Here z t , so 0 0z t

2 2 2

2 2sin cos 1 2dx dy dzds dt t t dt dtdt dt dt

Now, 2 2

0

( ) (sin .cos ) 2C

xy z ds t t t dt

3

2 2

0 0

sin 2 cos 2( ) 2 22 4 3C

t t txy z ds t dt

3 3

2 1 1 2( ) 2 04 3 4 3C

xy z ds

(Ans).

32. Evaluate S

dSF

, where kzjyixF ˆˆ2ˆ4 22


3,0,422 zzyx . Ans:




F ds divFdv

2 2. (4 ) ( 2 ) ( )S V

F ds x y z dvx y z

. 4 4 2S V

F ds y z dxdydz 2

2

2 4 3

2 04

. 4 4 2x

S x

F ds y z dxdydz

2

2

2 432

02 4

. 4 4x

zS x

F ds z yz z dxdy

2 2

2 2

2 4 2 4

2 24 4

. 12 12 9 21 12x x

S x x

F ds y dxdy y dxdy

2

2

242

42

. 21 6x

xS

F ds y y dx

2

2

2

. 42 4 0S

F ds x dx

2

2

2

. 42 4S

F ds x dx

2

21

2

4 4. 42 sin2 2 2S

x x xF ds

1 1. 42 0 0 2sin 1 2sin ( 1)S

F ds

1. 42 4sin 1 42 4 842S

F ds (Ans).

33. If kxjiyF ˆˆ3ˆ2 2

and S is the surface of the parabolic cylinder xy 82 in the first

octant bounded by the planes 6,4 zy , show that 132S

dSNF

.

Ans: -



Here Let 28 yx , jyigrad ˆ2ˆ8)(

16

ˆˆ4464

ˆ2ˆ8ˆ2ˆ8

ˆ2ˆ8)()(ˆ

22

yjyi

yjyi

jyijyi

gradgradn

164ˆ

16

ˆˆ4ˆˆ22

yi

yjyiin

16

1116

3816

ˆˆ4ˆˆ3ˆ2ˆ222

2

yy

yyy

yjyikxjiynF

dydzy

y

dydzy

yin

dydzNFdSNFS

4

0

6

02

4

0

6

02

4

0

6

0 411

16416

11ˆˆ

.132684

1124

11 60

4

0

2

zydSNF

S

(Proved).

(Nov –Dec 2010)

34. State the Green’s theorem in the plane. Ans: If C be the regular closed curve in xy plane bounding any region R and 1( , )F x y and

2 ( , )F x y be continuous functions on C and its interior, having continuous partial derivatives 1Fy

and 2Fx

in R, then 2 11 2

C R


, the line integral being taken along

the boundary C of R such that R is on the left as one advances in the direction of integration Or



Ans: - Green’s theorem in the plane: If xyyxyx ,),,(),,( be continuous in a region

E of the xy plane bounded by a closed curve C, then dxdyyx

dydxEC

)..( .

35. Find the values of a and b such that the surfaces xabyzax )2(2 and 44 32 zyx cut orthogonally at (1, -1, 2).




As surfaces are orthogonal, so 01281680 bbaGF 04201684 baab ---------------(1) As point (1, -1, 2) lies on the surface, so it satisfies surface F. 10220)2(2 bbaba

Putting in (1) we get 2/50412 aa So, 1,2/5 ba (Ans).

36. Using Green’s theorem, evaluate C

xdydxxy cos)sin( , where C is the plane triangle

enclosed by the lines

xyxy 2,2

,0

Ans: -

For

C

xdydxxy cos)sin(

Here xxy

xxy sin,1cos,sin



So, by Green’s theorem dxdyyx

xdydxxyEC

cos)sin(

2/

0

/2

0

)1sin(cos)sin( x

x

xy

yC

dxdyxxdydxxy

2/

0

/20)1sin(cos)sin(

x

x

x

C

dxyxxdydxxy

2/

0

2/

0

)1(sin22)1sin(cos)sin(

x

x

x

xC

dxxxdxxxxdydxxy

2/

0

2

2sin)cos(2cos)sin(

xxxxxxdydxxyC

000

81

20

22cos)sin(

2C

xdydxxy

24

18

2cos)sin(2

C

xdydxxy (Ans).

37. Find the work done in moving a particle in the force field kzjyxzixF ˆˆ)2(ˆ3 2

along:

i. The straight line from (0, 0, 0) to (2, 1, 3). ii. The curved defined by zxyx 83,4 32 from 0x to 2x .

Ans: - The equation of straight line joining (0, 0, 0) and (2, 1, 3) is

0 0 0 2 , , 3

2 0 1 0 3 0x y z t x t y t z t

where 0t to 1t

So, Work done .C

F dr

1

2

0


1 1

2 2 2

0 0


13 2

0Work done 12 4 12 4 0 0 16t t

i. The curved defined by zxyx 83,4 32 from 0x to 2x . Ans: - 2 34 ,3 8x y x z

Let 2 33,4 8t tx t y z where 0, 2t t



So, Work done .C

F dr

1 4 2 3 2

2

0

3 3 9Work done 3 .4 4 2 8 8t t t t tt dt dt dt

1 15 3 5 3 5

2 2

0 0

3 27 51Work done 3 38 8 64 8 64t t t t tt dt t dt

24 6

3

0

17 16 17 64 1 17Work done 8 8 1632 128 32 128 2 2t tt

(Ans)

(May-Jun-2011)

38. What are irrational and solenoidal vectors?

Irrotational Field (Conservative): -If F is irrotational field, then

1. 0

F and RdF

is path independent.

2. 0C

RdF

i.e. circulation along every closed surface is zero.

3. F , where is a scalar function.

Solenoidal Field (Incompressible): - If F is solenoidal field, then

1. 0 F .

2. Flux dsNF

across every closed surface is zero.

3. VF

, where V

is a Vector function.

39. The temperature of points in space is given by 푇(푥, 푦, 푧) = 푥 + 푦 − 푧. A mosquito located at (1,1,2) desires of fly in such a direction that it will get warm as soon as possible. In what direction should it move?

Ans:The temperature of points in space is given by 푇(푥, 푦, 푧) = 푥 + 푦 − 푧. A mosquito located at (1,1,2) desires of fly in such a direction that it will get warm as soon as possible. In direction of directional derivative of surface 푇(푥, 푦, 푧) = 푥 + 푦 − 푧.

The direction will be given by T

(1,1,2)ˆˆ ˆT i j k T

x y z

(1,1,2)2 2ˆˆ ˆT i j k

x y zx y z



(1,1,22 2

)2 2 2 2ˆˆ ˆT i j k

x yx y z x y z x y

zz

(1,1,2)ˆˆ ˆ2 2 2T xi yj zk

(1,1,2)ˆˆ ˆ2 2 4T i j k

A mosquito located at (1,1,2) desires of fly in direction along the vector ˆˆ ˆ2 2 4i j k

40. Verify Green’s theorem for ∫ [(3푥 − 8푦 )푑푥 + (4푦 − 6푥푦)푑푦] where C is the boundary of the region bounded by x=0, y=0 and x+y=1.

Ans: -


C S


(5) Here 2(3 8 ) (4 6 )ˆ ˆF i x jx y y y

So, 2 16 , 16F Fx yx y

1

2 1

0 0

16 16

S

xF F dxdy y y dxdyx y

1

2 1

0 0

110

S

xF F dxdy y dxdyx y

2 1

00

1 125S

xF F dxdy y dxx y

2 1

0

1 25 1S

F F dxdy x dxx y

2 1

0

125 1 2

S

F F dxdy x x dxx y



1

2 1

0

3 25 5 53S

F F xdxdy x xx y

2 1 5 55 53 3S

F F dxdyx y

----------------(1)

Verification of Green’s theorem by line integral 2(3 8 ) ( 6 ). 4x y dx y xy dyF dr

So, 2(3. 8 ) (4 6 )C C OA AB BO

x y dx yF r xyd dy

1

0

20

1

(3 8 ) (4 6 )3 4OA AB BO AB

x yxd dx y xy dyx ydy

220

1

(3 8 ) (4 6 ) (3 8 1 ) (4 1 6 1 )( )AB

x y dx y xy dy x x dx x x x dx

1 0 0

0 1

2

1

(3 8 1 ) (4 1 6 1 ) 4(3 )OA AB BO

x x dxdx x x x ydyx dx

1 002 2

2

10 1

3 8 1 2 4 1 12

63 42OA AB BO

x x dx x x x dx yx x

0

2 2

1

19 8 83 22

4 4 6 6OA AB BO

xx dx x x dxx

0

2 2

1

19 8 8 4 4 6 2632OA AB BO

x x x dxx x

0 02 3

0

11 1

29 13 22 2

4 123OA AB BO

x xx

3 29 14 22 2

1 13

2OA AB BO

3 29 14 22 2

1 13

2OA AB BO

=

53

…………………………….2

From 1 and 2 Greens theorem verified

41. Apply divergence theorem to evaluate ˆ.S

F ndS


and S is the

surface bounding the region 2 2 4x y , 0, 3z z .

Ans: -




F ds divFdv

2 2. (4 ) ( 2 ) ( )S V

F ds x y z dvx y z

. 4 4 2S V

F ds y z dxdydz

2

2

2 4 3

2 04

. 4 4 2x

S x

F ds y z dxdydz

2

2

2 432

02 4

. 4 4x

zS x

F ds z yz z dxdy

2 2

2 2

2 4 2 4

2 24 4

. 12 12 9 21 12x x

S x x

F ds y dxdy y dxdy

2

2

242

42

. 21 6x

xS

F ds y y dx

2

2

2

. 42 4 0S

F ds x dx

2

2

2

. 42 4S

F ds x dx

2

21

2

4 4. 42 sin2 2 2S

x x xF ds

1 1. 42 0 0 2sin 1 2sin ( 1)S

F ds 1. 42 4sin 1 42 4 84

2S

F ds (Ans).

(Nov –Dec 2011)

42. State the Gauss Divergent theorem



Ans: Gauss Divergence Theorem: - If F is a continuously differentiable vector function in

region E bounded by closed surface S, then ( )S E

F Nds div F dv

Where N

is the unit

normal vector at any point of S and kfjfifF ˆˆˆ321

31 21 2 3( )

S E

ff ff dydz f dzdx f dxdy dxdydzx y z

43. Prove that : 2 2( ) ''( ) '( )f r f r f r

r


2 ( ) . ( )f r f r

( )f r =

ˆ ˆ ˆ ( )i j k f rx y z

ˆ ˆ ˆ( ) ( ) ( ) ( )f r i f r j f r k f rx y z

/ / /ˆ( ) ( )r r r

f r f r i f j f kx y z

/ ˆ( ) ( )yx z

f r f r i j kr r r

/( )

( )f r R

f rr

2

/( )( ) . ( ) .

f r Rf r f r

r

2 / /( ) ( ) . . ( )

R Rf r f r f r

r r

2 / /ˆ ˆ ˆ ˆ ˆ ˆ( ) ( ) . ( )

R Rf r f r i j k i j k f r

x y z r r x y z

2 / /ˆ ˆ ˆ ˆ ˆ ˆ ˆ( ) ( ) . ( )

2 2 2 2 2 2 2 2 2x x x R

f r f r i j k i j k i j k f rx y z r x y zx y z x y z x y z



2 / // ˆ ˆ ˆ( ) ( ) ( )

2 2 2 2 2 2 2 2 2yx z R r r r

f r f r f r i j kx y z r x y zx y z x y z x y z

2 / // ˆ ˆ ˆ( ) ( ) ( )

2 2 2x R r r r

f r f r f r i j kx r x y zx y z

2

2 2 2 2 2 2/ // ˆ ˆ ˆ( ) ( ) ( )22 2 2

x y z x x x y z yR x zx xf r f r f r i j kr r r r

x y z

2

22 2 22 2 22/ //( ) ( ) ( )22 2 2

xx y z x

x y z R Rf r f r f r

r rx y z

2

22 2 2 2/ //( ) ( ) ( )

2 2 2 2 2 2

x y z x Rf r f r f r

rx y z x y z

2

221/ //( ) ( ) ( )3/22 2 2 2 2 2

x Rf r f r f r

rx y z x y z

2

22 2 23/ //( ) ( ) ( )3/22 2 2 2 2 2

x y z Rf r f r f r

rx y z x y z

2

223/ //( ) ( ) ( )3/22 2 2 2

r Rf r f r f r

rx y z r

2

23 1/ //( ) ( ) ( )

2 2 2

Rf r f r f r

r rx y z



2

23 1/ //( ) ( ) ( )

2 2 2 2 2 2

Rf r f r f r

rx y z x y z

2

22/ //( ) ( ) ( )

2 2 2

Rf r f r f r

rx y z

222

2/ // 2 2 2 2ˆ ˆ ˆ ˆ ˆ ˆ( ) ( ) ( ) ( ).( )2 2 2

rf r f r f r R xi yj zk xi yj zk x y z r

rx y z

2 2/ //( ) ( ) ( ) Prf r f r f r Hence oved

r

44. Evaluate ∮ 퐹⃗ 푑푟⃗ by Stoke’s theorem, where 퐹⃗ = 푦 횤⃗ + 푥 횥⃗ − (푥 + 푧)푘⃗ and c is the boundary of the triangle with vertices at (0,0,0), (1,0,0) and (1,1,0).

Ans: We know that from Stoke’s theorem ˆ. ( ).C S

F dr curl F nds

As ˆ. ( ).

C S

F dr curl F nds

Therefore ( ) ( )curl F F



2 2

ˆˆ ˆ

ˆˆ 2 2

i j k

F j x y kx y z

y x x z

ˆˆ 2 2F j x y k

ˆ ˆˆˆ( ). 2 2S S

curl F nds j x y k kdxdy

ˆ( ). 2 2S S

curl F nds x y dxdy

1 1

0 0

ˆ( ). 2 2x

S

curl F nds x y dxdy

11 2

0 0

ˆ( ). 22

x

S

ycurl F nds xy dx

21

0

1ˆ( ). 2 1

2S

xcurl F nds x x dx

212

0

1 2ˆ( ). 2 2

2S

x xcurl F nds x x dx

1

2 2

0

1ˆ( ). 4 4 1 22S

curl F nds x x x x dx

1

2

0

1ˆ( ). 6 5 12S

curl F nds x x dx

12 3

0

1 3 5ˆ( ).2 2 3S

x xcurl F nds x

1

0

1 3 5 1 9 10 6 13ˆ( ). 12 2 3 2 2 4S

curl F nds

Therefore ˆ. ( ).C S

F dr curl F nds

= 134

Ans.

45. Evaluate ∬ 푦 푧 횤⃗ + 푧 푥 횥⃗ + 푧 푦 푘⃗ . 푛 푑푠 where S is the point if the sphere 푥 + 푦 + 푧 = 1 above XY-plane and bounded by this plane. Ans: By divergence theorem .

S V

F ds divFdv



2 2 2 2 2 2. ( )S V

y z i z xF ds i j k dvx

zy

kz

j y

2 2 2 2 2 2

V

dvx y z

y z z x z y

22

V

zy dv

2

2 22 112

1

1 021

x yxdxdzy ydz

x

2 2

211 2 210

1 21

x x ydxdz y y

x

22 2

2111

1 21

xx y dxdy

x

y

2 42 2

211

1 21

xx y dxdyy y

x

2 4 2 42 2 2 2

0

2 21 11 12

1 121

x xx y dxdy x y dxdyy y

x

y y

2 4 2 4

0

2 2 2

2 21 11 12

1 121

1

x xx y dxdy x y dxdyy y

x

y

3 5

2

0

2112

3 511

xyyx dx

22 2 2 22

1 1 1 1 12

3 511

x x x xx dx

2 22 2 22 21 11 1 11 1 423 3 5 15 31 1

x x xx x dx dx

22 2

0

18 1 115

x x dx

sin cos ,x dx d then



22 2 6

0 0

1 /28 81 1 cos15 15

x x dx d

By using reduction formula we will have

2 2 2 2 2 2( ).12S V

y z i z x j z yF ds i j k dvy z

kx

April -May, 2012

46. State Gauss divergence theorem.

Ans: Gauss Divergence Theorem: - If F is a continuously differentiable vector function in

region E bounded by closed surface S, then ( )S E

F Nds div F dv

Where N

is the unit

normal vector at any point of S and kfjfifF ˆˆˆ321

31 21 2 3( )

S E

ff ff dydz f dzdx f dxdy dxdydzx y z

47. Find the directional derivative of 2 24x yz xz at the point (1, 2,1) in the direction of the

vector 2 2I J k . Ans: The directional derivative of 2 24x yz xz at the point (1, 2,1) in the direction of

the vector 2 2I J k is given by

2 24A i j k x yz xz Ax y z

where A is 2 2I J k ………………..1

2 24i j k x yz xzx y z

2 2 22 4 8xyz z i x z j x y xz k

4 4 2 8 6(1, 2,1) i j k j k …………………..2

2 2 2 2 2 22 2 39

I J k I J k I J kAI J k

…………………………3

From 1,2 and 3 we will have directional derivative of 2 24x yz xz at the point (1, 2,1) in

the direction of the vector 2 2I J k


= 2 26 .

3I J kj k



A = 6 .j k2 2 13 / 3

3I J k

48. Verify Green theorem for 2 2 ,c

xy y dx x dy where C is bounded by 2,y x y x

Ans: -


C S



So, 2 12 , 2F Fx x yx y

2

12 1

0

2 2x

S x


2

122 1

0

x

xS

F F dxdy xy y dxx y

1

2 2 3 42 1

0S


11 5 4

4 32 1

0 05 4S


2 1 1 1 15 4 20S

F F dxdyx y

----------------(1)

Now we have to find line integral along two curves. Along 2 , 2y x dy xdx .

So, 1 1

2 21 2 ( )

C C


1

12 4 2

1 20

( . ) .2C


1

11 4 53 4

1 20 0

3(3 )4 5C

x xF dx F dy x x dx

1

1 23 1 194 5 20C

F dx F dy

Along ,y x dy dx



So, 2 2

2 21 2 ( )

C C


2

02 2

1 21

( . ) .C


2

002 3

1 2 11

3 0 1 1C

F dx F dy x dx x

Now, 1 2

1 2 1 2 1 219 1120 20C C C


From (1) and (2) it verifies Green’s theorem. 49. Find the work done in moving a particle in the force field 23 (2 )F x I xz y J zK , along

the straight line from (0,0,0) to (2,1,3). Ans: The equation of straight line joining (0, 0, 0) and (2, 1, 3) is

0 0 0 2 , , 3

2 0 1 0 3 0x y z t x t y t z t

where 0t to 1t

So, Work done .C

F dr

1

2

0


1 1

2 2 2

0 0


13 2

0Work done 12 4 12 4 0 0 16t t

Nov-Dec, 2012

50. Evaluate divF and curl of F at point (1,2,3) where F is 2 2 2 ˆˆ ˆF x yzi xy zj xyz k

Ans: 2 2 2 ˆˆ ˆF x yzi xy zj xyz k

( ) 6 2 2 8 2div F xyz xy xyz xyz xy

(1,2,3)( ) 8 2 52div F xyz xy

(Ans).

Now,

2 2 2

ˆˆ ˆ

( )

i j k

curl Fx y z

x yz xy z xyz



2 2 2 2 2 2ˆˆ ˆ( ) ( ) ( ) ( )curl F i xz xy j yz yx k yz zx

(1,2,3)ˆˆ ˆ( ) 16 15curl F i j k

(Ans).

51. Find the directional derivative of 2 24x yz xz at (1, 2,1, ) in the direction of vectorˆˆ ˆ2 2i j k .

Ans: The directional derivative of 2 24x yz xz at the point (1, 2,1) in the direction of

the vector 2 2I J k is given by


where A is 2 2I J k ………………..1

2 24i j k x yz xzx y z

2 2 22 4 8xyz z i x z j x y xz k

4 4 2 8 6(1, 2,1) i j k j k …………………..2

2 2 2 2 2 22 2 39

I J k I J k I J kAI J k

…………………………3

From 1,2 and 3 we will have directional derivative of 2 24x yz xz at the point (1, 2,1) in

the direction of the vector 2 2I J k


= 2 26 .

3I J kj k

A = 6 .j k2 2 13 / 3

3I J k

52. Find the value of a if the vector 2 2 2 2 22 2ax z yz i xy z x j xyz x y k has zero

divergence find the curl of the above vector which has zero divergence.

Ans: 2 2 2 2 22 2F ax z yz i xy z x j xyz x y k

it is given that divF is Zero

2 2 2 2 2( ) 2 2 0div F ax z yz i xy z x j xyz x y k

( ) 2 2 2 0div F ax xy xy

( ) 2 4 0div F ax xy

2a y Curl of above vector whose divergence is equal to zero is given by



2 2 2 2 2

ˆˆ ˆ

( )

2 2 2

i j k

curl Fx y z

yx z yz xy z x xyz x y

2 2 2 2 2 2 ˆˆ ˆ( ) 2 4 2 2 4 2 2curl F xz x y xz i yz xy yx y j y z x z z k

2 2 2 2 2 2 ˆˆ ˆ( ) 4 4 2 4 2 2curl F xz x y i yz y xy yx j y z x z z k

53. Evaluate ∫ 퐹⃗.푑푠 where 퐹⃗ = 4x횤̂ - 2y2횥̂ + z2푘 and s is the surface bounding the region x2 + y2 = 4, z = 0 and z = 3.

Ans: -


F ds divFdv

2 2. (4 ) ( 2 ) ( )S V

F ds x y z dvx y z

. 4 4 2S V

F ds y z dxdydz

2

2

2 4 3

2 04

. 4 4 2x

S x

F ds y z dxdydz

2

2

2 432

02 4

. 4 4x

zS x

F ds z yz z dxdy

2 2

2 2

2 4 2 4

2 24 4

. 12 12 9 21 12x x

S x x

F ds y dxdy y dxdy

2

2

242

42

. 21 6x

xS

F ds y y dx

2

2

2

. 42 4 0S

F ds x dx



2

2

2

. 42 4S

F ds x dx

2

21

2

4 4. 42 sin2 2 2S

x x xF ds

1 1. 42 0 0 2sin 1 2sin ( 1)S

F ds 1. 42 4sin 1 42 4 842S

F ds

(Ans).

Unit V – Theory Of Equations April -May, 2006

1. If O, A, B, C are the four point on a straight line such that the distance of A, B and C from

O are the roots of the equation 3 23 3 0ax bx cx d . If B is the middle point of AC show that 3 33 2 0a d abc b Ans . Let us assume , , are the roots of the equation 3 23 3 0ax bx cx d and

, ,OA OB OC

According to the given condition 2 --------------(1) According to the property of roots

Now, 3ba

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

3ca

------------(3)

da

-------------(4)

From 2 and 3 we will have 33 b ba a

-------------(5)

3 3c ca a



From 1 2

2 3 32 2c b ca a a

2

2

3 2c ba a

-------------(6)

again from 4 we have

d

a

da

From 5 and 6 we have 2

2

3 2b c b da a a a

By solving above we will have

2

2

3 2ca bb da

Hence we will have 3 33 2 0a d abc b

2. Solve by Cardan’s method of equation 3 23 3 0x x . Ans: 3 23 3 0x x --------------(1)

Here we can remove second term of equation (1) by diminishing its roots by

3 1

3 1bhna

We can diminished each root by 1 by synthetic division method



Hence the transformed equation is 3 3 1 0y y ---------(2)where 1y x

Let 1/3 1/3y p q be the solution of equation (2).

3 1/3 1/3 1/3 1/33y p q p q p q

3 1/3 1/33y p q p q y

3 1/3 1/33 ( ) 0y p q y p q -------------(3)

By comparing equation (2) and (3) we get

1/3 1/3 1, ( ) 1p q p q

1, ( ) 1pq p q

So, p and q are the roots of the equation 2 1 0t t

1 1 4 1 3

2 2it

.

So, let1 3 1 3,

2 2i ip q

.

So roots of given cubic equation (1) are

(i) 1/3 1/3

1/3 1/3 1 3 1 32 2i ip q

1/3 1/32 2 2 2cos sin cos sin

3 3 3 3i i

2 2 2 2cos sin cos sin9 9 9 9

i i

22cos9

(ii) 1/3 1/3

1/3 2 1/3 1 3 1 3 1 3 1 32 2 2 2i i i iwp w q

4/3 4/3

1 3 1 32 2i i




3 3 3 3i i


i i

82 cos9

(iii) 1/3 1/3

2 1/3 1/3 1 3 1 3 1 3 1 32 2 2 2i i i iw p wq

1/3 1/32 2 2 2 2 2 2 2cos sin cos sin cos sin cos sin

3 3 3 3 3 3 3 3i i i i

2 2 2 2 2 2 2 2cos sin cos sin cos sin cos sin3 3 9 9 3 3 9 9

i i i i

2 2 2 2 2 2 2 2cos sin cos sin9 3 9 3 9 3 9 3

i i


i i

4 4 142cos 2cos 2 2cos9 9 9

So, roots of equation (2) are 2 8 142cos , 2 cos , 2 cos9 9 9

Now, roots of given equation (1) are 2 8 141 2cos , 1 2 cos , 1 2cos9 9 9

(Ans)

3. Solve the equation by Ferrari’s method: 4 3 212 41 18 72 0x x x x . Ans: 4 3 212 41 18 72 0x x x x

Let 4 3 2 2 2 2( ) 12 41 18 72 ( 6 ) ( ) 0f x x x x x x x mx n ----------(1)

4 3 2 4 2 2 3 2 2 2 212 41 18 72 36 12 12 2 2x x x x x x x x x m x mnx n

4 3 2 4 3 2 2 2 212 41 18 72 12 (36 2 ) ( 12 2 )x x x x x x m x mn x n

Equating the coefficients we get 2 2 2(36 2 ) 41, ( 12 2 ) 18, 72m mn n

2 2 22 5, 9 6 , 72m mn n



222 5 72 9 6

3 2 22 144 5 360 81 36 108 0 3 22 41 252 441 0 2( 3)(2 35 147) 0

2123,7,

By taking 3 1, 9m n equation (1) becomes

4 3 2 2 2 212 41 18 72 ( 6 3) ( 9) 0x x x x x x x

2 2( 6 3 9)( 6 3 9) 0x x x x x x

2 2( 5 6)( 7 12) 0x x x x ( 6)( 1)( 3)( 4) 0x x x x

1,3, 4,6x (Ans)

Nov-Dec, 2006 4. Solve by Cardan’s method of equation 3 23 3 0x x .

Ans: 3 23 3 0x x --------------(1)


3 1

3 1bhna




3 1/3 1/3 1/3 1/33y p q p q p q



3 1/3 1/33y p q p q y

3 1/3 1/33 ( ) 0y p q y p q -------------(3)


1/3 1/3 1, ( ) 1p q p q

1, ( ) 1pq p q


1 1 4 1 3

2 2it

.

So, let1 3 1 3,

2 2i ip q

.


(i) 1/3 1/3

1/3 1/3 1 3 1 32 2i ip q


3 3 3 3i i


i i

22cos9

(ii) 1/3 1/3

1/3 2 1/3 1 3 1 3 1 3 1 32 2 2 2i i i iwp w q

4/3 4/3

1 3 1 32 2i i


3 3 3 3i i


i i

82 cos9



(iii) 1/3 1/3

2 1/3 1/3 1 3 1 3 1 3 1 32 2 2 2i i i iw p wq


3 3 3 3 3 3 3 3i i i i


i i i i

2 2 2 2 2 2 2 2cos sin cos sin9 3 9 3 9 3 9 3

i i


i i

4 4 142cos 2cos 2 2cos9 9 9



(Ans)

5. Solve the equation 5 4 3 26 43 43 6 0x x x x x . Ans: 5 4 3 26 43 43 6 0x x x x x

The given equation is a reciprocal equation of odd degree having coefficient of terms equidistance from beginning and end.

So, 1x is its root.

4 3 2( 1)(6 5 38 5 6) 0x x x x x 4 3 2(6 5 38 5 6) 0x x x x

Dividing by 2x we get

22

5 66 5 38 0x xx x

22

1 16 5 38 0x xx x

21 16 2 5 38 0x xx x



26 2 5 38 0t t where 1x tx

26 5 50 0t t 10 5(3 10)(2 5) 0 ,3 2

t t t

1 10 1 53 2

x and xx x

2 23 3 10 2 2 5x x and x x 2 23 10 3 0 2 5 2 0x x and x x

10 100 36 5 25 166 4

x and x

10 8 5 36 4

x and x

1 13, 2,2 2

x and x

So, roots are 1 12, 1, , , 32 2

x (Ans)

6. If , , are the roots of the equation 3 0x qx r . Find the equation whose roots are

, , .

Ans: Given that , , are the roots of the equation 3 0x qx r .

Now, 0 --------------(1)

q ------------(2)

r -------------(3)

Now 2 2 2 2( ) 2( ) 0 q q ----------(4)

Let , ,A B C

A B C



2 2 2 2 2 2

A B C

2 2 2 2 2 2

A B C

2 2 2 2 2 2

A B C

( ) ( ) ( )A B C

A B C

as 0

3 3A B C

------------------(5)

AB BC CA

2 2 2 2 2 2 2 2 2 2 2 2

AB BC CA

2 2 2 2 2 2 2 2 2 2 2 2

AB BC CA

2 2 2 2 2 2q q q q q qAB BC CA

r r r


r r r

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2( ) ( ) ( )q q q q q qAB BC CA

r r r

2 3 2 2 2 2 3 2 2 2 2 3 2 2 2q q q q q q q q qAB BC CA

r r r




r

2 3 3 3 3 3 3 3( )( ) ( )q q qAB BC CA

r

2 3 2( ) ( ) 0 ( ) ( ) 3 ( )q q q q rAB BC CA

r

3 4 2 3( )( )

q q q q r qAB BC CA

r r

3 4 2

2

2 3q q q rAB BC CAr

-----------(6)

ABC

2 2 2 2 2 2

ABC

2 2 2q q qABC

3 2

2 2 2

( ) ( )q q qABC

3 2 3 2

2 2

0q q r q q rABCr r

Equation whose roots are A, B, C is

3 2( ) ( ) 0x A B C x AB BC CA x ABC

3 4 2 3 2

3 22 2

2 33 0q q q r q q rx x xr r

2 3 2 2 3 4 2 3 23 (2 3 ) ( ) 0r x r x q q q r x q q r (Ans)

7. Solve by Cardan’s method of equation 3 23 3 0x x .



Ans: 3 23 3 0x x --------------(1)


3 1

3 1bhna




3 1/3 1/3 1/3 1/33y p q p q p q

3 1/3 1/33y p q p q y

3 1/3 1/33 ( ) 0y p q y p q -------------(3)


1/3 1/3 1, ( ) 1p q p q

1, ( ) 1pq p q


1 1 4 1 3

2 2it

.

So, let1 3 1 3,

2 2i ip q

.




(i) 1/3 1/3

1/3 1/3 1 3 1 32 2i ip q


3 3 3 3i i


i i

22cos9

(ii) 1/3 1/3

1/3 2 1/3 1 3 1 3 1 3 1 32 2 2 2i i i iwp w q

4/3 4/3

1 3 1 32 2i i


3 3 3 3i i


i i

82 cos9

(iii) 1/3 1/3

2 1/3 1/3 1 3 1 3 1 3 1 32 2 2 2i i i iw p wq


3 3 3 3 3 3 3 3i i i i


i i i i

2 2 2 2 2 2 2 2cos sin cos sin9 3 9 3 9 3 9 3

i i


i i



4 4 142cos 2cos 2 2cos9 9 9



(Ans)

8. Solve the equation 5 4 3 26 43 43 6 0x x x x x .

Ans: 5 4 3 26 43 43 6 0x x x x x


So, 1x is its root.

4 3 2( 1)(6 5 38 5 6) 0x x x x x 4 3 2(6 5 38 5 6) 0x x x x


22

5 66 5 38 0x xx x

22

1 16 5 38 0x xx x

21 16 2 5 38 0x xx x

26 2 5 38 0t t where 1x tx

26 5 50 0t t 10 5(3 10)(2 5) 0 ,3 2

t t t

1 10 1 53 2

x and xx x

2 23 3 10 2 2 5x x and x x 2 23 10 3 0 2 5 2 0x x and x x

10 100 36 5 25 166 4

x and x

10 8 5 36 4

x and x



1 13, 2,2 2

x and x

So, roots are 1 12, 1, , , 32 2

x (Ans)

9. If , , are the roots of the equation 3 0x qx r . Find the equation whose roots are

, , .

Ans: Given that , , are the roots of the equation 3 0x qx r .

Now, 0 --------------(1)

q ------------(2)

r -------------(3)

Now 2 2 2 2( ) 2( ) 0 q q ----------(4)

Let , ,A B C

A B C

2 2 2 2 2 2

A B C

2 2 2 2 2 2

A B C

2 2 2 2 2 2

A B C

( ) ( ) ( )A B C

A B C

as 0



3 3A B C

------------------(5)

AB BC CA

2 2 2 2 2 2 2 2 2 2 2 2

AB BC CA

2 2 2 2 2 2 2 2 2 2 2 2

AB BC CA


r r r


r r r

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2( ) ( ) ( )q q q q q qAB BC CA

r r r


r r r


r

2 3 3 3 3 3 3 3( )( ) ( )q q qAB BC CA

r

2 3 2( ) ( ) 0 ( ) ( ) 3 ( )q q q q rAB BC CA

r

3 4 2 3( )( )

q q q q r qAB BC CA

r r

3 4 2

2

2 3q q q rAB BC CAr

-----------(6)



ABC

2 2 2 2 2 2

ABC

2 2 2q q qABC

3 2

2 2 2

( ) ( )q q qABC

3 2 3 2

2 2

0q q r q q rABCr r

Equation whose roots are A, B, C is

3 2( ) ( ) 0x A B C x AB BC CA x ABC

3 4 2 3 2

3 22 2

2 33 0q q q r q q rx x xr r

2 3 2 2 3 4 2 3 23 (2 3 ) ( ) 0r x r x q q q r x q q r (Ans)

May-June, 2007

10. Solve the equation by Ferrari’s method: 4 3 212 41 18 72 0x x x x . Ans: 4 3 212 41 18 72 0x x x x

Let 4 3 2 2 2 2( ) 12 41 18 72 ( 6 ) ( ) 0f x x x x x x x mx n ----------(1)


4 3 2 4 3 2 2 2 212 41 18 72 12 (36 2 ) ( 12 2 )x x x x x x m x mn x n


2 2 22 5, 9 6 , 72m mn n

222 5 72 9 6

3 2 22 144 5 360 81 36 108 0



3 22 41 252 441 0 2( 3)(2 35 147) 0

2123,7,


4 3 2 2 2 212 41 18 72 ( 6 3) ( 9) 0x x x x x x x

2 2( 6 3 9)( 6 3 9) 0x x x x x x

2 2( 5 6)( 7 12) 0x x x x ( 6)( 1)( 3)( 4) 0x x x x

1,3, 4,6x (Ans)

11. Solve by Cardan’s method of equation 3 15 126 0x x . Ans: 3 15 126 0x x

Here the equation having terms involving 2x term missing. Let 1/3 1/3x p q be the solution of equation (1).

3 1/3 1/3 1/3 1/33x p q p q p q

3 1/3 1/33x p q p q x

3 1/3 1/33 ( ) 0x p q x p q -------------(2) By comparing equation(1) and (2) we get 1/3 1/3 5, ( ) 126p q p q 125, ( ) 126pq p q

So, p and q are the roots of the equation 2 126 125 0t t ( 1)( 125) 0 1,125t t t . So, let p = 1 and q = 125.

So roots of given cubic equation (1) are (i) 1/3 1/3 1 5 6p q

(ii) 1/3 2 1/3 1 3 1 3 6 4 35 3 2 32 2 2i i iwp w q i

(iii) 2 1/3 1/3 1 3 1 3 6 4 35 3 2 32 2 2i i iw p wq i

So, roots are 6, 3 2 3, 3 2 3i i (Ans).

12. Solve 3 24 20 48 0x x x given that the roots , are connected by the relation

2 0 . Ans: Let , , are the roots of 3 24 20 48 0x x x . Given that 2 0 2 Now, 4 --------------(1)



20 ------------(2) 48 -------------(3) From (1) 2 4 4 ----------(4) From (3) 2 . .(4 ) 48

3 24 24

3 24 24 0

2( 2)( 6 12) 0

6 36 482, 2, 3 3

2i

So, 2 4, 6 So, roots are -4, 2, 6.

Nov-Dec, 2007 13. Find the number of real roots the equation 3 2 x x 4x 4 0

Ans: number of real roots the equation 3 2 x x 4x 4 0 are 3 , two positive roots and 1

negative roots.

14. Find the condition that the equation 3 2x x 4x 4 0p had roots , , which satisfy

1 0 .

Ans: Let , , are the roots of 3 2 0x px qx r . Given that 1 0 1 --------------(1)

Now, p --------------(2) 4 ------------(3) 4 -------------(4) From (1) and (4) 4 ----------(4) Putting the value of 4 in the equation 3 2 0x px qx r we will have

64 16 16 4 0p

84 2116 4

p

15. If , , are roots of the equation 3 0x qx r Find the equation whose roots are

2 2 2, ,

Ans: Let , , are the roots of 3 0x qx r …………………(1) Now, 0 --------------(2)



q ------------(3) r -------------(4)

Let 2y

2 2 4

2 2 4

2 2 40 r

2 2 4rxx

2 4ry xx

3 4 0x yx r ……………………………..(5)

Subtracting 4 from 1 we will have

3 0q y x r

3rx

q y

………………..(6)

Put the value of x in the equation 1 we will have 3

3 3 0r rq rq y q y

2 3327 3 0r rq q y r q y

3 2 2 3 3 2 227 3 2 3 3 0r rq q r rq r q r r q rq

3 3 3 2 2 3 4 3 2 227 3 3 6 3 3 0r rq r q r q rq r r q r q 3 3 3 2 2 427 4 6 9 0r rq r q r q r

16. Solve by Cardan’s methods: 3 15 126 0x x

Ans: 3 15 126 0x x


3 1/3 1/3 1/3 1/33x p q p q p q

3 1/3 1/33x p q p q x

3 1/3 1/33 ( ) 0x p q x p q -------------(2) By comparing equation(1) and (2) we get



1/3 1/3 5, ( ) 126p q p q 125, ( ) 126pq p q



(ii) 1/3 2 1/3 1 3 1 3 6 4 35 3 2 32 2 2i i iwp w q i

(iii) 2 1/3 1/3 1 3 1 3 6 4 35 3 2 32 2 2i i iw p wq i


April -May 2008

17. Form the equation whose roots are 1, 2, 3.

Ans .The equation will be 1 2 3x x x 3 21 2 3 6 11 6x x x x x x

18. Solve the equation 2x2+x2-7x-6 = 0 when the difference of two roots is 3.

Ans: Let , , are the roots of 3 22x x 7x 6 0 . Given that 3 -------------- (1) Now, 1/ 2 -------------- (2) 7 / 2 ------------ (3) 3 ------------- (4) From (1) and (2) 2 7 / 2 ------------- (5) From (4) and (1) . .(3 ) 3 ------------- (6)

From (5) and (6) 7 2 .(3 ) 3

2

From

3 24 24

3 24 24 0

2( 2)( 6 12) 0

6 36 482, 2, 3 3

2i

So, 2 4, 6



So, roots are -4, 2, 6.

19. If 휶,휷,휸 are the roots of the cubic 3 2 0x px qx r find the equation whose roots are

1 1 1, ,

Ans. As , , are the roots of the cubic 3 2 0x px qx r .

Here , ,p q r

Let root of the new equation is y, and according to the given condition 1 1 1ry y

x

1 1r ry xx y

and x is the root of the equation 3 2 0x px qx r

3 21 1 1 0r r rp q r

y y y

3 2 2 31 1 1 0r p r y q r y ry

3 2 2 2 31 3 3 1 2 1 0r r r p r r y q r y ry

3 2 2 3 21 1 2 1 3 3 0ry q r y p r r y r r r

Required equation will be

3 2 2 3 21 1 2 1 3 3 0ry q r y p r r y r r r

20. Solve by Cardan’s methods: 3 29 +6x -1=0 x Ans: 3 29 +6x -1=0 x

Here the equation having terms involving 2x term. We can have another equation 1= yx

And by putting the value of y in the equation 3 29 +6x -1=0 x we will have 3 2

31 19 +6 -1=0 y 6 9 0yy y

Here in the 3y 6 9 0y ……………………………………………(2)

equation term involving 2y is missing .

Let y u v be the solution of equation (2).

3 3 3 3y u v uv u v

3 3 3 3y u v uvy

3 3 33 0y uvy u v -------------(2)



By comparing equation(1) and (2) we get 3 32, ( ) 9uv u v ------------(3)

3 3 3 3 3 32, ( ) 9 8, ( ) 9uv u v u v u v

So, 3u and 3v are the roots of the equation 2 9 8 0t t 2 9 8 0 ( 8) 1t t t t .

So, let 3u = 8 and 22, 2 , 2u w w therefore from 3 we will have 21, ,v w w

So roots of given cubic equation (1) are (i) 2 1 3u v

(ii) 2 1 3 1 3 3 32 22 2 2i i iu v w w

(iii) 2 1 3 1 3 3 32 22 2 2i i iu v w w

So, roots of the equation 3y 6 9 0y are 3 3 3 33, ,

2 2i i

And roots of the equation 3 29 +6x -1=0 x are

1 11 3 3 3 3, ,3 2 2

i i

as

1= yx

Nov -Dec 2008

21. From the equation whose roots are reciprocal of the roots of the equation5 3 22 4 3 7 6 0x x x x

Ans: The equation whose roots are reciprocal of the roots of the equation5 3 22 4 3 7 6 0x x x x is y

1yx

by putting the value of y in the above equation we will have

5 3 22 1/ 4 1/ 3 1/ 7 1/ 6 0x x x x

4 35 26 7 3 4 2 0x x x x 22. If 1 2 3, ,r r r are the root of the equation 3 22 3 1 0x x kx find constant k if sum of

two roots is 1.

Let 1 2 3, ,r r r are the roots of 3 22 3 1 0x x kx .

Given that 1 2 1r r -------------- (1)



Now, 1 2 3 3 / 2r r r -------------- (2)

1 2 2 3 1 3 / 2r r r r r r k ------------ (3)

1 2 3 1/ 2r r r ------------- (4)

From (1) and (2) 3 1/ 2r ------------- (5)

From (5) and (4) 1 2 1r r ------------- (6)

From (5) and (6)we can construct a quadratic equation that is 2 1 0x x

By solving equation 2 1 0x x we will have 2 4 1 3

2 2b b ac ix

a

1 2 31 3 1 3

2, , 1 / 2

2i ir r r

23. Find the equation whose roots are the roots of 풙ퟒ + 풙ퟑ − ퟑ풙ퟐ − 풙 + ퟐ = ퟎ.each diminished by 3. Ans. The above equation can be solved by using synthetic division then we will have

3y x Let , , , are the roots of 풙ퟒ + 풙ퟑ − ퟑ풙ퟐ − 풙 + ퟐ = ퟎ.

Now, 1 -------------- (1) As roots are diminished by 3 therefore new root will be 3, 3, 3, 3y

3 3 3 3 1 12

3 3 3 3 13

3 1 1 3 1 2

0 3 12 27 78

1 4 9 26 80

0 3 21 90

1 7 30 116

0 3 30

1 10 60

0 3

1 13

Hence the new equation is 4 3 213 60 116 80x x x x

24. Solve by Cardan’s method of equation 3 15 126 0x x .



Ans: 3 15 126 0x x


3 1/3 1/3 1/3 1/33x p q p q p q

3 1/3 1/33x p q p q x

3 1/3 1/33 ( ) 0x p q x p q -------------(2) By comparing equation(1) and (2) we get 1/3 1/3 5, ( ) 126p q p q 125, ( ) 126pq p q



(ii) 1/3 2 1/3 1 3 1 3 6 4 35 3 2 32 2 2i i iwp w q i

(iii) 2 1/3 1/3 1 3 1 3 6 4 35 3 2 32 2 2i i iw p wq i


May-June 2009

25. Write the relation between roots and coefficients of the equation.

1 20 1 2 1.......... 0n n n

n na x a x a x a x a

Ans: Let 1 2, ,........., n are roots of 1 20 1 2 1.......... 0n n n

n na x a x a x a x a .

Then 1

1 0i

i n

aa

, 2 2

0

( 1)i ji j

aa

3 3

0

( 1)i j ki j k

aa

Similarly, 1 20

......... ( 1)n nn

aa

26. If , , are the roots of the cubic 3 0x px q . Show that i. 5 5 5 5

ii. 2 5 3 43 5

Ans: As , , are the roots of the cubic 3 0x px q .--------(1)



So, 0, ,p q

Now, 22 2 0 2 2p p .

2 2 p -----------(2)

As 3 30 3 0 3( ) 0q

3 3q -------------(3)

Let us multiply equation (1) by x we get 4 2 0x px qx

4 2 0p q

4 2 2( 2 ) (0) 2p q p p q p

4 22 p ---------------(3)

(i) 5 5 5 5

Ans: Now by multiplying (1) by 2x we get 5 3 2 0x px qx

5 3 2 0p q

5 3 2 ( 3 ) ( 2 ) 5p q p q q p pq ----------(4)

5 5( )q p

5 5

5 5 5 5 (Proved)

(ii) 2 5 3 43 5

Ans: 2 5 23 3.( 2 ).(5 ) 30p pq p q -------------(5)

3 4 2 25 5( 3 )(2 ) 30q p p q ------------(6)

From (5) and (6) 2 5 3 43 5 (Proved)

27. Show that the equation 4 3 210 23 6 15 0x x x x can be transformed into reciprocal

equation by diminishing the roots by 2. Hence solve the equation. Ans: Let us diminished the roots of 4 3 210 23 6 15 0x x x x --------------(1) by 2. Putting 2 2y x x y By Using synthetic division method



The transformed equation is 4 3 22 13 2 1 0y y y y ------------(2) Equation (2) is a reciprocal equation of even degree having coefficients of terms equidistant from the beginning and end equal.

Dividing equation (2) by 2y we get 22

2 12 13 0y yy y

22

1 12 13 0y yy y

Now putting 2 22

1 1 2y z y zy y

2 2 2 13 0z z 2 2 15 0 ( 5)( 3) 0z z z z 3,5z

1 3yy

1 5yy

2 1 3y y 2 1 5y y

2 3 1 0y y 2 5 1 0y y

3 9 42

y

5 25 42

y

3 52

y

5 212

y



3 5 3 5 5 21 5 21, , ,2 2 2 2

y

So, solution for 2x y

3 5 3 5 5 21 5 212, 2, 2, 22 2 2 2

x

1 5 1 5 9 21 9 21, , ,2 2 2 2

x (Ans)

28. Solve by Cardan’s method, the equation 3 227 54 198 73 0x x x .

Ans: 3 227 54 198 73 0x x x

3 2 22 732 03 27

x x x ------------(1)

Let us compare this equation with 3 2 0ax bx cx d

Then 22 731, 2, ,3 27

a b c d

Putting 2 23 3

z ax b x x z in (1) we get

3 22 2 22 2 732 0

3 3 3 3 27z z z

3 2 24 8 8 8 22 44 732 2 03 27 3 9 3 9 27

z z z z z z

3 6 7 0z z ------------------------(2) Now comparing with 3 3 0z Hz G Where 2, 7H G

3 21 1 14 7 32 49 7 9 82 2 2

p G H G


(i) 1/31/3

22 12

Hpp

(ii) 1/3 2 2 21/3

22 2 2( ) 32

Hwp w w w w w wwp w

1/31/3

1 3 4 3 3 3 1 3 32 32 2 2

H i i iwpwp

(iii) 2 1/3 2 2 22 1/3 2

22 2 2( ) 32

Hw p w w w w w ww p w



2 1/32 1/3

1 3 4 3 3 3 1 3 32 32 2 2

H i i iw pw p

So, roots of equation (2) are 1 3 3 1 3 31, ,

2 2i i

.

Now, roots of given equation (1) are 2 1 3 3 2 1 3 3 21 , ,3 2 3 2 3

i i

1 3 9 3 4 3 9 3 4, ,3 6 6

i i

1 7 9 3 7 9 3, ,3 6 6

i i (Ans).

Nov-Dec, 2009

29. Write the relation between roots and coefficients of the equation. 1 2

0 1 2 1.......... 0n n nn na x a x a x a x a

Ans: Let 1 2, ,........., n are roots of 1 20 1 2 1.......... 0n n n

n na x a x a x a x a .

Then 1

1 0i

i n

aa

, 2 2

0

( 1)i ji j

aa

3 3

0

( 1)i j ki j k

aa

Similarly, 1 20

......... ( 1)n nn

aa

30. If ,, are the roots of the equation 3 0x px q , then show that i. 5 5 5 5

ii. 2 5 3 43 5 .


So, 0, ,p q

Now, 22 2 0 2 2p p .

2 2 p -----------(2)

As 3 30 3 0 3( ) 0q

3 3q -------------(3)




4 2 0p q

4 2 2( 2 ) (0) 2p q p p q p

4 22 p ---------------(3)

(i) 5 5 5 5


5 3 2 0p q

5 3 2 ( 3 ) ( 2 ) 5p q p q q p pq ----------(4)

5 5( )q p

5 5

5 5 5 5 (Proved)

(ii) 2 5 3 43 5

Ans: 2 5 23 3.( 2 ).(5 ) 30p pq p q -------------(5)

3 4 2 25 5( 3 )(2 ) 30q p p q ------------(6)

From (5) and (6) 2 5 3 43 5 (Proved)

31. Show that the equation 4 3 210 23 6 15 0x x x x can be transformed into reciprocal equation by diminishing the roots by 2. Hence solve the equation

Ans: Let us diminished the roots of 4 3 210 23 6 15 0x x x x --------------(1) by 2. Putting 2 2y x x y By Using synthetic division method



The transformed equation is 4 3 22 13 2 1 0y y y y ------------(2) Equation (2) is a reciprocal equation of even degree having coefficients of terms equidistant from the beginning and end equal.

Dividing equation (2) by 2y we get 22

2 12 13 0y yy y

22

1 12 13 0y yy y

Now putting 2 22

1 1 2y z y zy y

2 2 2 13 0z z 2 2 15 0 ( 5)( 3) 0z z z z 3,5z

1 3yy

1 5yy

2 1 3y y 2 1 5y y

2 3 1 0y y 2 5 1 0y y

3 9 42

y

5 25 42

y

3 52

y

5 212

y



3 5 3 5 5 21 5 21, , ,2 2 2 2

y

So, solution for 2x y

3 5 3 5 5 21 5 212, 2, 2, 22 2 2 2

x

1 5 1 5 9 21 9 21, , ,2 2 2 2

x (Ans)

32. Solve by Cardan’s method, the equation 3 227 54 198 73 0x x x .

Ans: 3 227 54 198 73 0x x x

3 2 22 732 03 27

x x x ------------(1)

Let us compare this equation with 3 2 0ax bx cx d

Then 22 731, 2, ,3 27

a b c d

Putting 2 23 3

z ax b x x z in (1) we get

3 22 2 22 2 732 0

3 3 3 3 27z z z

3 2 24 8 8 8 22 44 732 2 03 27 3 9 3 9 27

z z z z z z

3 6 7 0z z ------------------------(2) Now comparing with 3 3 0z Hz G Where 2, 7H G

3 21 1 14 7 32 49 7 9 82 2 2

p G H G


(i) 1/31/3

22 12

Hpp

(ii) 1/3 2 2 21/3

22 2 2( ) 32

Hwp w w w w w wwp w

1/31/3

1 3 4 3 3 3 1 3 32 32 2 2

H i i iwpwp

(iii) 2 1/3 2 2 22 1/3 2

22 2 2( ) 32

Hw p w w w w w ww p w



2 1/32 1/3

1 3 4 3 3 3 1 3 32 32 2 2

H i i iw pw p

So, roots of equation (2) are 1 3 3 1 3 31, ,

2 2i i

.

Now, roots of given equation (1) are 2 1 3 3 2 1 3 3 21 , ,3 2 3 2 3

i i

1 3 9 3 4 3 9 3 4, ,3 6 6

i i

1 7 9 3 7 9 3, ,3 6 6

i i (Ans).

May-June 2010

33. If ,, are the roots of the equation 08126 23 xxx , find an equation whose roots are 2,2,2 .

Ans: - It is same as to find an equation whose roots are diminished by 2 of 08126 23 xxx 0)2( 3 x -----------------(1) Let 22 yxxy ,

so eqn (1) becomes 03 y . Then the equation is 3 0x .

34. If the roots of the equation 023 rqxpxx are in H.P., then prove that

02927 32 qpqrr .

Ans: - Let ,, are the roots of the equation 023 rqxpxx . Here , ,p q r Given that roots are in HP.

So, 1 1 2

2 2 3

3 33 r

q

As is a root of the given eqn so it must satisfy the equation.

Hence 3 2 0p q r

3 2 3 2 3

3

3 3 3 27 9 20 0r r r r pqr rqp q rq q q q

3 2 327 9 2 0r pqr rq

2 327 9 2 0r pqr q (Proved).



35. If ,, are the roots of the equation 03 rqxx , form an equation whose roots are:

(i)

,, .

(ii) 2 2 2, ,

Ans: - Here 0, ,q r

i. , ,

Let 22 2 2 2 2 2 2, , , , , ,r r r r ry x

x y

Now, 03 rqxx 3 6 4 2 2 22x qx r x qx q x r

3 22 22r r rq q r

y y y

3 22 2

3 22r r rq q ry y y

2 2 2 32r qry q y ry

3 2 2 22 0ry q y qry r

New eqn is 3 2 2 22 0rx q x qrx r

(ii) 222 ,,

Ans: - Here 0, ,q r

Let 2 2 2 2 2 2

1 1 1, , , ,y

1,1,110,10,10 y as 0

yx

xy 11

So, eqn 03 rqxx becomes 01011 323

ryqyr

yq

y

0123 qyry

New eqn is 0123 qxrx



36. Solve by Cardan’s method: 01928 23 xx Ans: - 01928 23 xx ------------------(1)

Let 1 1y xx y

, then eqn (1) becomes

3 9 28 0y y ---------------(2)

Here the equation having terms involving 2y term missing. Let y u v be the solution of equation (2).

3 3 3 3 ( )y u v uv u v 3 3 3 3y u v uvy

3 3 33 0y uvy u v -------------(3)

By comparing equation(2) and (3) we get 3 33, 28uv u v 3 3 3 327, 28u v u v

So, 3 3,u v are the roots of the equation 2 28 27 0t t

( 1)( 27) 0 1, 27t t t . So, let 3 31, 27u v .

21, ,u w w , then 23 3, 3 , 3v w wu

Then 2 21 3, 3 , 3y u v w w w w 24,1 2 ,1 2y u v w w

1 3 1 34,1 2 ,1 22 2i iy u v

4, 2 3,2 3y i i

1 1 2 3 2 3, ,

4 7 7i ix

y

So, roots are 1 2 3 2 3, ,

4 7 7i ix

. (Ans).

Nov-Dec 2010

37. State the intermediate property.

Ans Intermediate property of roots says that for the equation ( ) 0f x for two numbers

let a and b such that, )(af and )(bf have different sign then equation 0)( xf has at least one root lies between a and b.

38. If ,, are the roots of the equation 3 0x px q , then show that i. 5 5 5 5



ii. 2 5 3 43 5 .


So, 0, ,p q

Now, 22 2 0 2 2p p .

2 2 p -----------(2)

As 3 30 3 0 3( ) 0q

3 3q -------------(3)


4 2 0p q

4 2 2( 2 ) (0) 2p q p p q p

4 22 p ---------------(3)

(i) 5 5 5 5


5 3 2 0p q

5 3 2 ( 3 ) ( 2 ) 5p q p q q p pq ----------(4)

5 5( )q p

5 5

5 5 5 5 (Proved)

(ii) 2 5 3 43 5

Ans: 2 5 23 3.( 2 ).(5 ) 30p pq p q -------------(5)

3 4 2 25 5( 3 )(2 ) 30q p p q ------------(6)

From (5) and (6) 2 5 3 43 5 (Proved)

39. Solve: 5 4 3 26 41 97 97 41 6 0x x x x x . Ans: 5 4 3 26 43 43 6 0x x x x x


So, 1x is its root.



4 3 2( 1)(6 5 38 5 6) 0x x x x x 4 3 2(6 5 38 5 6) 0x x x x


22

5 66 5 38 0x xx x

22

1 16 5 38 0x xx x

21 16 2 5 38 0x xx x

26 2 5 38 0t t where 1x tx

26 5 50 0t t 10 5(3 10)(2 5) 0 ,3 2

t t t

1 10 1 53 2

x and xx x

2 23 3 10 2 2 5x x and x x 2 23 10 3 0 2 5 2 0x x and x x

10 100 36 5 25 166 4

x and x

10 8 5 36 4

x and x

1 13, 2,2 2

x and x

So, roots are 1 12, 1, , , 32 2

x (Ans)

40. Solve by Ferrari’s method: 031052 234 xxxx .

Ans: - 031052 234 xxxx ---------------------------(1) Equation (1) can be written as by combining 푥 and 푥 into perefect square

0)()( 222 nmxxx ----------------------------(2)

0)22()12(2 222234 nmnxmxxx ----------(3) 퐹푟표푚 (1)푎푛푑 (3)



33

5102262512

2222

22

nnmnmnmm

--------------------(4)

From (4) 222 )(. mnnm 22 )5()3).(62(

07452 23 Then 1 is one of the root. So, 246)1(22 mm (one of the value)

245 nmn So, (2) can be written as 0)22()1( 222 xxx

0)221)(221( 22 xxxxxx

0)13)(3( 22 xxxx

0)13(,0)3( 22 xxxx

2493,

21211

xx

253,

2131

xx (Ans)

April -May-2011

41. What is geometrical meaning of root of the equation f(x)=0

Ans. For any value of x which satisfies 0)( xf are called as solution of 0)( xf or root of

0)( xf .

Example: - For 062 x

a. If 2x , it is not satisfies 062 x . So, 2x is not a solution of 062 x . If 3x , it is not satisfies 062 x . So, 3x is a solution of 062 x .

Geometrically for the curve f(x) =0 the roots are the curve of , , ,................x a x b x c which touches the curve at a, b, and c respectively.

42. Solve the following equations : 6x5-41x4+97x3-97x2+41x-6=0

Ans: 5 4 3 26 43 43 6 0x x x x x


So, 1x is its root.



4 3 2( 1)(6 5 38 5 6) 0x x x x x 4 3 2(6 5 38 5 6) 0x x x x


22

5 66 5 38 0x xx x

22

1 16 5 38 0x xx x

21 16 2 5 38 0x xx x

26 2 5 38 0t t where 1x tx

26 5 50 0t t 10 5(3 10)(2 5) 0 ,3 2

t t t

1 10 1 53 2

x and xx x

2 23 3 10 2 2 5x x and x x 2 23 10 3 0 2 5 2 0x x and x x

10 100 36 5 25 166 4

x and x

10 8 5 36 4

x and x

1 13, 2,2 2

x and x

So, roots are 1 12, 1, , , 32 2

x (Ans)

43. Find the equation of squared difference of the roots of the cubic 3 2x 6x 7x 2 0.

Ans: Let , , are the roots of 3 2x 6x 7x 2 0. …………………(1) Now, 6 --------------(2) 7 ------------(3) 2 -------------(4)



As sum of the roots is -6 therefore to remove the 2nd term we will make a new equation whose roots will be increased by 2

Here by removing the term containing 2x We will have

2 1 6 7 2

0 2 8 2

1 4 1 4

0 2 4

1 2 5

0 2

1 0

The transformed equation will be 3-5y 4 0.y and roots of it will be 2, 2, 2 …(5)

Let 2, 2, 2a b c

Now, 0a b c --------------(6) 5ab bc ca ------------(7) 4abc -------------(8)

2 2

2 2

2 2

b c

c a

a b

And we are here to find 2 2 2, , which is equivalent to

2 2 2, ,a b b c c a if z be the root of the equation squared differences we will have

2 2 4z b c b c bc

2 2 4z b c b c bc from equation 6 and 8 we will have

3

2 2 2 4 164 abc az b c b c bc aa a

3 16aza

3 16yzy

as y is the root of the equation 3-5y 4 0.y and equivalent to a, b, and c

3 16 0y zy ………………………(9)

Subtracting the equation 9 from equation 8 we will have 125 12 0

5z y y

z



Putting the value of y in the equation 8 we will have

312 12-5 4 0.

5 5z z

3 2-30z 225 68 0.z z and roots of the obtained equation are 2 2 2, ,a b b c c a

which is equivalent to 2 2 2, , .

Hence 3 2-30z 225 68 0.z z is the required equation.

44. Solve by Cardan’s methods: x3-3x+1=0.

Ans: 3 3 1 0x x --------------(1)

Hence the transformed equation is 3 3 1 0x x ---------(2)where

Let 1/3 1/3x p q be the solution of equation (2).

3 1/3 1/3 1/3 1/33x p q p q p q

3 1/3 1/33x p q p q x

3 1/3 1/33 ( ) 0y p q y p q -------------(3)


1/3 1/3 1, ( ) 1p q p q

1, ( ) 1pq p q


1 1 4 1 3

2 2it

.

So, let1 3 1 3,

2 2i ip q

.


(i) 1/3 1/3

1/3 1/3 1 3 1 32 2i ip q




3 3 3 3i i


i i

22cos9

(ii) 1/3 1/3

1/3 2 1/3 1 3 1 3 1 3 1 32 2 2 2i i i iwp w q

4/3 4/3

1 3 1 32 2i i


3 3 3 3i i


i i

82 cos9

(iii) 1/3 1/3

2 1/3 1/3 1 3 1 3 1 3 1 32 2 2 2i i i iw p wq


3 3 3 3 3 3 3 3i i i i


i i i i

2 2 2 2 2 2 2 2cos sin cos sin9 3 9 3 9 3 9 3

i i


i i

4 4 142cos 2cos 2 2cos9 9 9




Dec-Jan-2011

45. Explain Descartes’ rule of sign of equation. Ans Descarte’s Rule of Sign:- a. Maximum number of positive real root of 0)( xf is same as the number of sign changes

in )(xf . b. Maximum number of negative real root of 0)( xf is same as the number of sign changes

in )( xf . c. If an equation having degree n such that at most p number of positive roots, at most q

number of negative roots, then equation has atleast )( qpn number of imaginary roots.

46. If O, A, B, C are the four point on a straight line such that the distance of A, B and C from

O are the roots of the equation 3 3 0ax cx d . If B is the middle point of AC show that 3 33 2 0a d abc b

Ans . Let us assume , , are the roots of the equation 3 23 3 0ax bx cx d and , ,OA OB OC

According to the given condition 2 --------------(1) According to the property of roots

Now, 3ba

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

3ca

------------(3)

da

-------------(4)

From 2 and 3 we will have 33 b ba a

-------------(5)

3 3c ca a

From 1 2

2 3 32 2c b ca a a



2

2

3 2c ba a

-------------(6)

again from 4 we have

d

a

da

From 5 and 6 we have 2

2

3 2b c b da a a a

By solving above we will have

2

2

3 2ca bb da

Hence we will have 3 33 2 0a d abc b

47. Solve by Cardan’s methods : X3 – 3X2 + 3 =0.

Ans: 3 23 3 0x x --------------(1)


3 1

3 1bhna




3 1/3 1/3 1/3 1/33y p q p q p q



3 1/3 1/33y p q p q y

3 1/3 1/33 ( ) 0y p q y p q -------------(3)


1/3 1/3 1, ( ) 1p q p q

1, ( ) 1pq p q


1 1 4 1 3

2 2it

.

So, let1 3 1 3,

2 2i ip q

.


(i) 1/3 1/3

1/3 1/3 1 3 1 32 2i ip q


3 3 3 3i i


i i

22cos9

(ii) 1/3 1/3

1/3 2 1/3 1 3 1 3 1 3 1 32 2 2 2i i i iwp w q

4/3 4/3

1 3 1 32 2i i


3 3 3 3i i


i i

82 cos9



(iii) 1/3 1/3

2 1/3 1/3 1 3 1 3 1 3 1 32 2 2 2i i i iw p wq


3 3 3 3 3 3 3 3i i i i


i i i i

2 2 2 2 2 2 2 2cos sin cos sin9 3 9 3 9 3 9 3

i i


i i

4 4 142cos 2cos 2 2cos9 9 9



(Ans)

48. Solve by Ferrari’s method: 4 3 212 41 18 72 0x x x x .

Ans: 4 3 212 41 18 72 0x x x x

Let 4 3 2 2 2 2( ) 12 41 18 72 ( 6 ) ( ) 0f x x x x x x x mx n ----------(1)


4 3 2 4 3 2 2 2 212 41 18 72 12 (36 2 ) ( 12 2 )x x x x x x m x mn x n


2 2 22 5, 9 6 , 72m mn n

222 5 72 9 6

3 2 22 144 5 360 81 36 108 0 3 22 41 252 441 0 2( 3)(2 35 147) 0

2123,7,




4 3 2 2 2 212 41 18 72 ( 6 3) ( 9) 0x x x x x x x

2 2( 6 3 9)( 6 3 9) 0x x x x x x

2 2( 5 6)( 7 12) 0x x x x ( 6)( 1)( 3)( 4) 0x x x x

1,3, 4,6x (Ans)

April -May-2012

49. Solve the equation 3 6x 20 0 x , one root being 1 3 .i Ans . As we know that from the general properties of roots of an equation that In an equation with real coefficients, if i is a root then i is also one of the root, therefore two

roots of the equation 3( ) 6x 20 f x x are 1 3 .i and 1 3 .i

( ) 1-3i 1+3if x x x x a

( ) 1 3 1 3f x x i x i x a

( ) 1 3 1 3f x x i x i x a

2( ) 1 9f x x x a

2( ) 2 1 9f x x x x a

2( ) 2 8f x x x x a

2

( )2 8

f x x ax x

3

2

6x 20

2 8

xx a

x x

2 2x x a a

Roots of the equation 3 6x 20x are 1 3 .,1 3 2i i and and

50. Solve the equation Solve the equation 3 2x 4x 20x 48 0 , given that the roots 훼 푎푛푑 훽 are connected by the relation 0 Ans . Let us assume , , are the roots of the equation 3 2x 4x 20x 48 0 and According to the given condition 0 --------------(1) According to the property of roots Now, 4 --------------(2) From 1 we will have 0 4 4

20 ------------(3)



48 -------------(4)

From 2 and 4 we will have 4 48 12 -------------(5)

By using the equation 1 and 5 we 0 and 12

We can have quadratic equation from above i.e 2 -12 0y and 2 3y

2 3 2 3and and three roots of the equation 3 2x 4x 20x 48 0 are

2 3 , 2 3 4and

51. Solve the equation 3 26x 11x 3x 2 0 , given that its roots are in H.P. Ans. We know that H.P is reciprocal of A.P then by taking 1/y x the new equation will be

3 23 21 1 16 11 3 2 0 2 3y 11 6 0y y

y y y

now roots of the equation

3 22 3y 11 6 0y y are in A.P As , ,d d are the roots of the cubic equation 3 22 3y 11 6 0y y .

By properties of relation between roots and coefficient of the equation we will have

3 / 2d d

3 3 / 2 1/ 2

Again we will have 3d d

2 2 3d

221 1 3

2 2d

21 64

d

2 1 2564 4

d

52

d by putting the value of 5 / 2d in , ,d d we will have roots of the

equation 3 22 3y 11 6 0y y are 2,1/ 2,3 and roots of the equation 3 26x 11x 3x 2 0 are 1/ 2, 2 1/ 3and

52. Solve by Cardan’s methods: 3 x 18x 35 0.

Ans: 3 x 18x 35 0. Here the equation having terms involving 2x term missing.

Let x u v be the solution of equation (1).



3 3 3 3x u v uv u v

3 3 3 3x u v uvx

3 3 33 0x uvx u v -------------(2)

By comparing equation(1) and (2) we get 3 36, ( ) 35uv u v ------------(3)3 3 3 3 3 36, ( ) 35 216, ( ) 35uv u v u v u v

So, 3u and 3v are the roots of the equation 2 35 216 0t t 2 35 216 ( 27)( 8) 27, 8t t t t t .

So, let 3u = -27 and 23, 3 , 3u w w therefore from 3 we will have 22, 2 , 2v w w

So roots of given cubic equation (1) are (iv) 3 2 5u v

(v) 2 1 3 1 3 5 33 2 3 22 2 2i i iu v w w

(vi) 2 1 3 1 3 5 33 2 3 22 2 2i i iu v w w

So, roots are 5 3 5 35, ,

2 2i i

(Ans).

Dec –Jan -2012

53. From the equation of the fourth degree whose roots are 3 + i and √7 . Ans . As we know that from the general properties of roots of an equation that In an equation with real coefficients, if i is a root then i is also one of the root, and for the

irrational roots if an equation has ba is one of the root, then ba is also another root.

Therefore two roots of the equation we will have 3 i and 3-i , 7 7and and the

equation will be ( ) 3+i 3-i 7 7 f x x x x x

( ) 3 3 7 7 f x x i x i x x

22 2 2( ) 3 7f x x i x

22 2 2( ) 3 7f x x i x

2 2( ) 9 6 1 7f x x x x



4 2 2 3 2( ) 7 9 63 6 42 7f x x x x x x x 4 3 2( ) 6 3 42 70f x x x x x

54. Solve the equations: 5 4 3 2 6x x 43x – 43x 6 0 .x

Ans: 5 4 3 26 43 43 6 0x x x x x

The given equation is a reciprocal equation of odd degree having coefficient of terms equidistance from beginning and end.So, 1x is its root.

4 3 2( 1)(6 5 38 5 6) 0x x x x x 4 3 2(6 5 38 5 6) 0x x x x


22

5 66 5 38 0x xx x

22

1 16 5 38 0x xx x

21 16 2 5 38 0x xx x

26 2 5 38 0t t where 1x tx

26 5 50 0t t 10 5(3 10)(2 5) 0 ,3 2

t t t

1 10 1 53 2

x and xx x

2 23 3 10 2 2 5x x and x x 2 23 10 3 0 2 5 2 0x x and x x

10 100 36 5 25 166 4

x and x

10 8 5 36 4

x and x

1 13, 2,2 2

x and x

So, roots are 1 12, 1, , , 32 2

x (Ans)



55. Solve by Cardan’s methods : 3 – 27x 54=0 x

Ans: 3 – 27x 54=0 x Here the equation having terms involving 2x term missing.

Let x u v be the solution of equation (1). 3 3 3 3x u v uv u v

3 3 3 3x u v uvx

3 3 33 0x uvx u v -------------(2)

By comparing equation(1) and (2) we get 3 39, ( ) 54uv u v ------------(3)

3 3 3 3 3 39, ( ) 54 729, ( ) 54uv u v u v u v

So, 3u and 3v are the roots of the equation 2 54 729 0t t 2 2 254 27 ( 27) 27t t t t .

So, let 3u = -27 and 23, 3 , 3u w w therefore from 3 we will have 23, 3 , 3v w w

So roots of given cubic equation (1) are (i) 3 3 6u v

(ii) 2 1 3 1 3 63 3 3 3 32 2 2i iu v w w

(iii) 2 1 3 1 3 63 3 3 3 32 2 2i iu v w w

So, roots are 6, 3, 3 (Ans).

56. Solve by Ferrari;s method : X4 – 4X3 –X2 + 16X – 12=0.

Ans: 4 3 2 – 4x – 16x – 12 0.x x

Let 4 3 2 2 2 2( ) – 4x – 16x – 12 ( 2 ) ( ) 0f x x x x x mx n ----------(1)

4 3 2 4 2 2 3 2 2 2 2– 4x – 16x – 12 4 4 4 2 2x x x x x x x m x mnx n 4 3 2 4 3 2 2 2 2– 4x – 16x – 12 4 (4 2 ) ( 4 2 )x x x x m x mn x n


2 2 22 5, 8 2 , 12m mn n ……………………..(2)

222 5 12 8 2

3 2 22 24 5 60 64 4 32



3 22 8 4 0

By solving the above equation we will have 2, 2, 1 / 2

By taking 2, in the equation 2 we will have 2 29, 12, 16m mn n

23, 12, 4m mn n

3, 4 3, 4m n or m n

By putting the value of above in the equation 1 we will have 4 3 2 2 2 2( ) – 4x – 16x – 12 ( 2 2) (3 4) 0f x x x x x x 4 3 2 2 2( ) – 4x – 16x – 12 ( 2 2) (3 4) ( 2 2) (3 4)f x x x x x x x x x 4 3 2 2 2( ) – 4x – 16x – 12 2 2 3 4 2 2 3 4f x x x x x x x x x

4 3 2 2 2( ) – 4x – 16x – 12 5 6 2f x x x x x x x

2 25 6 3 2 , 2 2 1x x x x x x x x

4 3 2 ( ) – 4x – 16x – 12 3 2 2 1f x x x x x x x

Roots of the equation 4 3 2 – 4x – 16x – 12 0.x x 3,2,-2and 1

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II Sem Mathematics All Solutions