CALCULASCALCULAS

Vikasana - CET 2012

CALCULASCALCULASPART-A

MULTIPLE CHOICE QNSMULTIPLE CHOICE QNS

Vikasana - CET 2012

1 . . s i n ,I f y x t h e nd y

=

)

d yd x

c o s xa

=

)2 s i n

s i n)2 s i n

ax

xbx2 s i n

)s i n

xc o s xc

x2)

s i nc o s xd

x

Vikasana - CET 2012

SOLUTION: ANSWER ©

Vikasana - CET 2012

2. . ( ) ( 1)If f x x x x= − +( ) ( ). '(0)

f fthen f =

)0)1

ab)1) 1

bc −)

)d InfinityVikasana - CET 2012

( ) (0): '(0) lim f x fSolution f −= 0: (0) lim

0( 1)

xSolution fx

x x x

→=−

− +0

( 1)lim 1

( )

xx x x

xA

→

+= = −

: ( )Ans c

Vikasana - CET 2012

2 1 13 sin cotd x−⎡ ⎤+ =⎢ ⎥3. sin cot

1)0dx x

=⎢ ⎥−⎣ ⎦)0)1/ 2

ab)1/ 2) 1/ 2

bc −

) 1d −

Vikasana - CET 2012

1 1: cos cot xSolution put x θθ − +⇒

2

: . cos cot1 2

1

Solution put xx

θ

θ

= ⇒ =−

2 1sin (1 cos )2 2

1 1

y

d

θ θ⇒ = = −

1 1(1 )2 2

dyy xdx

−⇒ = − ⇒ =

:( )Ans c

Vikasana - CET 2012

...

4yey dyIf x e then

∞+

+

= =4. . .

1)

If x e thendx

xa

= =

−)

)1

axxb)

11)

xxc

x

−+

)1

xxd

x+Vikasana - CET 2012

1 x+

: logy xSolution x e x y x+= ⇒ = +

'

g1 1

y

y⇒ = +

1

yxdy x−

( )

ydx x

A

⇒ =

:( )Ans a

Vikasana - CET 2012

5. . cos sin & cos sin ,If x y thendy

θ θ θ θ θ θπ

= + = −

.2

)

dy atdx

a

πθ

π

=

)2

2)

a

b

−

)

)4

c

ππ44)dπ

Vikasana - CET 2012

: cos sin ; cos sinSolution x yθ θ θ θ θ θ= + = −: cos sin ; cos sin/ sin cos/ i

Solution x ydy dy dd d d

θ θ θ θ θ θθ θ θ θθ θ θ θ θ

+− −

= =/ sin cos cos

1 1 4dx dx d

dyθ θ θ θ θ− + +

− −⎡ ⎤⇒ = =⎢ ⎥⎣ ⎦ /2 /2: )dx

Ans dθ π π π=

⇒⎢ ⎥ −⎣ ⎦)

Vikasana - CET 2012

2 26. . . .sin . . . .(log )The derivative of x w r t x =.sin .cos)

logx x xa

x

2

2sin .cos)(log )

x xbx

2

( g )sin)2 log

xcx2log

) .logx

d x x

Vikasana - CET 2012

2 2: sin , (log )Solution u x v x= =: sin , (log )2sin .cos .sin .cos

1 l

Solution u x v xdu x x x x xd= =

⎛ ⎞1 log2.log .dv xxx⎛ ⎞⎜ ⎟⎝ ⎠

: )Ans a

Vikasana - CET 2012

( )7 xx dyIf y then⎛ ⎞= =⎜ ⎟( )1

7. . ,

)1 log2

x

x

If y thenx dxa

=

= =⎜ ⎟⎝ ⎠

+)1 log2)1

ab

+

) 1)1 log2

cd−−) g

Vikasana - CET 2012

( ): log log( )x xxSolution y y x xx= ⇒ =2

2

log .log1 1 2 log

y x xdy x x x

⇒ =

⇒ = +. 2 .log

(1 2log )

x x xy dx x

dy xy x

⇒ = +

+(1 2log )

1

xy xdx

dy

= +

⎛ ⎞⇒ =⎜ ⎟1

1

: )xdx

Ans b=

⇒ =⎜ ⎟⎝ ⎠

Vikasana - CET 2012

8. . ( ) . . . .nIf f x x then the valueof=( )'(1) ''(1) (1)(1) ............. ( 1)

1† 2† †

nnf f ff

n− + − + −

)1isa)1) 1)0

abc−

)0)

cd ∞

Vikasana - CET 2012

: (1) 1, '(1) , ''(1) ( 1)Solution f f n f nn= = = −: ( ) , ( ) , ( ) ( )'''(1) ( 1)( 2)

( 1) ( 2)( 3) †

S f f ff nn n= − −

( 1) ( 2)( 3) †. . 1 ........ ( 1)1† 2† 3† †

nn nn nn n nGEn

− − −= − + − + +−

1 2 3 .......... ( 1) (1 1) 0:( )

n no nnc nc nc nc nc

Ans c= − + − + +− = − =

:( )Ans c

Vikasana - CET 2012

21

2cos sin9. . tan , .

ydx xIf y then− −⎛ ⎞= =⎜ ⎟⎝ ⎠ 2,cos sin

)0

f ydxx x

a

⎜ ⎟+⎝ ⎠

)sin2)cos

b xc x)) cos2d x−

Vikasana - CET 2012

1 11 tan: tan tan tan1 tan 4 4

xSolution y x xx

π π− −−⎡ ⎤ ⎛ ⎞ ⎛ ⎞= = =− −⎜ ⎟ ⎜ ⎟⎢ ⎥+ ⎝ ⎠ ⎝ ⎠⎣ ⎦' ''1 0: )

y yAns a⇒ =− ⇒ =

: )Ans a

Vikasana - CET 2012

10 A point on the curve at2610.A point on the curve at which the tangent to the curve is

f

26y x x= −

0included at an angle of to theline x+y=0

045

a) (-3,9) b)(-3,-27) c) (3,9) d)(0,0)c) (3,9) d)(0,0)

Vikasana - CET 2012

: . .tan 6 2 ;Solution Slopeof gent xl f h l

= −

0

. . . . . . 1(6 2 ) 1

tan45

Slopeof thegivenlineisx

−

− +∴tan45

1 (6 2 )( 1)7 2

xx

∴ =+ − −

−7 2 1 4 12; 3, 92 5:( )

x x x yx

Ans c

⇒ = ⇒ = = =−

:( )Ans c

Vikasana - CET 2012

11 In the curve if SN varies2 2 2 2 2( )x y a x a= −11.In the curve if SN varies inversely as the nth power of the abscissa then n is equal to

( )x y a x a

then n is equal toa)n=2 b)n=3 c) n=4d) 1d)n=1

Vikasana - CET 2012

2 2 22 2 2( ): 1

x a aSolutiion y a a− ⎛ ⎞= = ⎜ ⎟2 2

2 2

: 1

2 12 ' ( )

Solutiion y a ax x

SN

= = −⎜ ⎟⎝ ⎠

2 2. 3 32 ' ( )

3

yy a a SNx x

n

α= ⇒

⇒ = 3: ( )

nAns b⇒

Vikasana - CET 2012

12 The surface area of a sphere when its12. The surface area of a sphere, when itsvolume if increasing at the same rate as its radius isa)1 b) c) 2 d) 41

2 π2 π

Vikasana - CET 2012

3 24 4dV drS l V 3 2

2

4: 43

dV drSolution V r rdt dt

dV dr

π π= ⇒ =

2

2

1 4

4 1

dV drrdt dt

S it

π⇒ = ∴ =

24 1 .: ( )

S r squnitsAns a

π= =

Vikasana - CET 2012

13.Tangent is drawn to the ellipse 2 2

127 1x y

+ =g pat Find the value of such that the sum of the intercepts on

27 1(3 3 cos ,sin )θ θ (0, /2)θ π∈ θ

such that the sum of the intercepts on the coordinate axes by the tangent is minimumminimuma) pi/3 b) pi/6 c) pi/8 d) pi/4

/ 3π

Vikasana - CET 2012

: . . . . .

3 3 i

Solution The parametric eqns of givenellipse

θ θ3 3cos , sincos. .tan . . sin 13 3

x yxEquationof gent at y

θ θθθ θ

= =

⇒ + =3 3

1cos3 3sec

x yecθθ

⇒ + =cos3 3sec

( ) 0 . .int 3 3sec cos

'( ) 0 3 3 t t 0

ec

f sumof ercepts ec

f

θθθ θ θ

θ θ θ θ θ

= ⇒ = +

⇒3

'( ) 0 3 3sec tan cos cot 0

cot 3 3

f ecθ θ θ θ θ

θ θ π

= ⇒ − =

⇒ = ⇒ = / 6( )A b Vikasana - CET 2012: ( )Ans b

4 41 4 . ( s i n c o s )x x d x− =∫c o s 2)

2s i n 2

xa c

x

+

s i n 2)2

s i n 2)

xb c

x

−+

s)2

c o s 2)

xc c

xd c

+

+)2

d c+

Vikasana - CET 2012

2 2 2 2: (sin cos )(sin cos )

sin2

Solution x x x x dx

x

− +∫

∫sin2cos2

2:( )

xI xdx c

Ans b

=− =− +∫:( )Ans b

Vikasana - CET 2012

2

2 2

2 315. x dx+∫ 2 2

1

15.( 1)( 4)

1. log tan . . &1 2

dxx x

x xIf I A B then A Bisx

−

− +−

= ++

∫

1 2) 1,1)1, 1

xab

+−−) ,

1 1) ,2 2

c

1 1) ,2 2

d −

Vikasana - CET 2012

1 1 1 1 1⎛ ⎞ 12 2

1 1 1 1 1: log tan2 1 2 21 4

x xSolution dxxx x

−−⎛ ⎞+ = +⎜ ⎟ +− +⎝ ⎠∫: )Ans c

⎝ ⎠

Vikasana - CET 2012

/ 2

1 6 .xa d x =∫

11) s in ( )

x x

x

a a

a a c

−

−

−

+

∫

1

) s in ( )lo g

1) ta n ( )x

a a ca

b a c−

+

+) ta n ( )lo g

) 2

x

x x

b a ca

c a a c−

+

+) 2) lo g ( )x x

c a a cd a a c−

− +

− +

Vikasana - CET 2012

/2 1 .logx xa a adxl i d∫ ∫ 22/2

g:1 log 11

xxx

solution dxa aa

a

=−−

∫ ∫

.1

x

aput t a=

11 sin ( )log

xa ca

−⇒ +

: )Ans a

Vikasana - CET 2012

117. dx =∫17.1 sin cos

) log 1 tan / 2

dxx x

a x c+ ++ +

∫) g

) log 1 sin cosb x x c+ + +

)2 log 1 tan / 2

1

c x c+ +

1) log 1 tan / 22

d x c+ +

Vikasana - CET 2012

: dx dxSolution I= =∫ ∫ 2

2

:(1 cos ) sin 2cos /2 2sin /2.cos /2

sec /2. log1 tan /2

Solution Ix x x x x

x dx x c

+ + +

⇒ = + +

∫ ∫

∫ log1 tan /22(1 tan /2): )

x cx

Ans a

⇒ = + ++∫

Vikasana - CET 2012

3 218 32 (log )x x dx =∫4 2

18. 32 .(log )

)8 (log )

x x dx

a x x c

=

+∫

4 2) . 8(log ) 4log 1b x x x c⎡ ⎤− + +⎣ ⎦⎡ ⎤4 2

3 2

) . 8(log ) 4log

) (log ) 2log

c x x x c

d x x x c

⎡ ⎤− +⎣ ⎦⎡ ⎤+ +⎣ ⎦) . (log ) 2logd x x x c⎡ ⎤+ +⎣ ⎦

Vikasana - CET 2012

4 4

: . . .32 32 2(log )

Solution Integrationbypartsx x x∫ ∫2 2 4 3

4 4

32 32 2(log )(log ) . . (log ) .8 16 .log4 4

16 16 1

x x xx dx x x x xdxx

x x

⇒ − = −

⎡ ⎤

∫ ∫

∫2 4

2 4 4 4

16 16 1(log ) .8 (log ). . .4 4

(log ) 8 4 (log ) 4

x xx x x dxx

x x x x x c

⎡ ⎤⇒ − −⎢ ⎥

⎣ ⎦⇒ + +

∫

4 2

(log ) .8 4 .(log ) 4

8(log ) 4log 1

x x x x x c

x x x c

⇒ − + +

⎡ ⎤⇒ − + +⎣ ⎦: )Ans b

Vikasana - CET 2012

31 9 2 d x∫ 01 9 . 2

) 5 / 2

d xx

a

=−∫))1 / 2) / 2

b) 7 / 2) 9

cd )

Vikasana - CET 2012

1∫

3

1: . . .2

1

Solution Weknowthat dx xx x=∫3

3

00

1 ( 2)2 22

dx xx x∴ = −− −∫1 5(1 4)2 2

= + =

: ( )Ans a

Vikasana - CET 2012

/ 4

020. . tan ,n

nIf I xdx thenπ

= ∫02lim

) /

n

n n n

f

I I→∞ + =⎡ ⎤+⎣ ⎦

∫

)1/ 2)1

ab))0

cd∞

Vikasana - CET 2012

/ 4

0: tan ,n

nSolution I xdxπ

= ∫/ 4 2

2 0/ 4

2

tan ,

(1 )

nn

n

I xdx

I I d

π

π

++ = ∫

∫ 22

0/ 4/ 4 1

2

tan (1 tan )

1tantn

nn n

n

I I x x dx

xdππ

+

+

+ = +

⎡ ⎤⎢ ⎥

∫

∫[ ]

2

00

2

tan sec11

lim 1n n n

x xdxnn

I I→∞ +

⎡ ⎤= = =⎢ ⎥ ++⎣ ⎦+ =

∫

: ( )Ans b

Vikasana - CET 2012

21 dxTh l fπ

∫0

21. . .5 3cos

) /8

Thevalueofx=

+∫) /8) /4

abππ

)0) /2

cd) /2d π

Vikasana - CET 2012

: . .Solution FromDirectObservation

2 2

2 2( . . )

cosdx a greaterthanb

a b x b

π π=

+∫ 2 20 cosa b x a b

dxπ π π

+ −

= =

∫

∫0 5 3cos 425 4

: )x

Ans b

= =+ −∫: )Ans b

Vikasana - CET 2012

222. . . . . . 2 . . . .Theareabetweenthecurves y x x and thex axisis= − −)8/5)4/3

ab)4/3)5/3

bc)7/3d

Vikasana - CET 2012

2: . 0 2 0Solution Now y x x= ⇒ − =

2 22

. . . 0, 2

(2

simiplifing we get x x

A d d

= =

∫ ∫ 2

0 023

(2A ydx x x dx= = −

⎡ ⎤

∫ ∫3

2

0

4 8 / 3 4 / 33

)

xx

A b

⎡ ⎤= = − =−⎢ ⎥⎣ ⎦: )Ans b

Vikasana - CET 2012

23. . . . . . .The area enclosed between the curvelog ( ) & . . .

)2ey x e the coordinate axes is

a= +

)2)1)4

ab

)4)3

cd

Vikasana - CET 2012

: . log ( ). . . .eSolution clearly y x e cuts the x axis at= + −

0 0

(1 , 0) & . (0,1)

log( )

e y axis at

A ydx x e dx

− −

= = +∫ ∫

( )

1 10

0

g( )

1log( )

e e

y

x dxx e x

− −

= −+

∫ ∫

∫( )

[ ]

11

00

.log( ).

0 log( )1

ee

x dxx e xx e

e dx x e x e

−−

++

⎛ ⎞= − = − − +⎜ ⎟

∫

∫ [ ]

[ ]

11

0 log( )1

1 : : )(1 )

ee

dx x e x ex e

Ans be e

−−

= − = − − +−⎜ ⎟−⎝ ⎠

= =− − −

∫

Vikasana - CET 2012

sin sin 2 sin3 sin 2 sin323 ( ) 3 4sin 3 4sin

x x x x xIf f x x x

+ ++

/2

23. . ( ) 3 4sin 3 4sin1 4sin sin 1

If f x x xx x

π

= ++

/2

0

. . . . ( )

)3

then the value of f x dxπ

=∫)3)2 / 3)1/ 3

abc)1/ 3

))0cd

Vikasana - CET 2012

s in s in 2 s in 3 s in 2 s in 3: ( ) 3 4 s in 3 4 s in

x x x x xSo lu tio n f x x x

+ += +

1 1 2 3

: ( ) 3 4 s in 3 4 s in1 4 sin s in 1

. ( )

So lu tio n f x x xx x

C ons id er C C C C

++

→ − +

s in s in 2 s in 3( ) 0 3 4 sin

0 s in 1

x x xf x x

x=

2

/ 2/ 2

(3 4 s in ) s in 3

co s 3s in 3 1 / 3(0 1) 1 / 3

S in x x x

xxd xππ

= − =

−⎡ ⎤∴ = = − − =⎢ ⎥∫0 0

s in 3 1 / 3(0 1) 1 / 33

: ( )

xd x

A ns c

∴ ⎢ ⎥⎣ ⎦∫

Vikasana - CET 2012

24. . . . .deg . . . .Theorderand the reeof thedifferential equation1/3 3

34 .1 3d ydy aredxd

⎛ ⎞ =+⎜ ⎟⎝ ⎠)(1,2/3)

dxdxa⎝ ⎠

)(3,4))(3,3)

bc)( ))(1,2)d

Vikasana - CET 2012

333

3:1 3 4dy d ySolution

d d⎛ ⎞+ = ⎜ ⎟⎝ ⎠3

3, 3dx dx

Order Degree

⎜ ⎟⎝ ⎠

= =,: )

gAns c

Vikasana - CET 2012

2

25. . . . . . . . . . .Thedifferential equationof all non vertical linesina planeisd

−2

2) 0d yadxd

=

) 0dxbdyd

=

2

) 0dycdxd

=

2

2) 0d xddy

=

Vikasana - CET 2012

: . . . . . . . . . .Solution Thedifferential equationof all non vertical linesina planeis−.

, 0givenbyax by c b+ = ≠

2

20 . 0dy d ya b bdx dx

+ = ⇒ =

2

2 0d ydx

⇒ =

: )Ans a

Vikasana - CET 2012

2

2 6 . . . . . .T h e s o l u t i o n o f t h e e q u a t i o nd y 2

2

2

.

)

x

x

d y e i s yd x

e

−

−

= =

2

)4

)x

a

eb c x d−

+ +

22

)4

)4

x

b c x d

ec c x d−

+ +

+ +

2

)4

)4

xed c d−

+ +

Vikasana - CET 20124

22: xd ySo lu tion e −=2

2

:

1 x

S o lu tion edx

dy e c−

=

−⇒ = +

2

21 1 x

e cd x

y e cx d−

⇒ +

⎛ ⎞⇒ = − + +−⎜ ⎟

2

2 21 x

y e cx d

y e cx d−

⇒ + +⎜ ⎟⎝ ⎠

⇒ = + +4

: )

y e cx d

A ns b

⇒ = + +

Vikasana - CET 2012

CALCULASCALCULASPART-B

MULTIPLE CHOICE QNSMULTIPLE CHOICE QNS

Vikasana - CET 2012

101. . ( ) , ( ) sin & ( ) ( ( ))xIf f x e g x x h x f g x−= = ='( ).( )

h xthenh x

=

1) sin1)

a x

b

−

2)

11)

bx

c

−

1

2

sin

)1

) x

cx

d e−

−

Vikasana - CET 2012

1: ( ) , ( ) sinxSolution f x e g x x−= =

1

( ) ( ( ))( ) (sin )

h x f g xh x f x−

⇒ =

⇒ =1 1sin sin

2

1( ) '( ) .1

x xh x e h x ex

− −

⇒ = ⇒ =−

2

'( ) 1( ) 1

h xh x x

⇒ =−

: )Ans b

Vikasana - CET 2012

02 . . s in h ,xIf y e th en=dydx

=

) co sh) co sh

x

x

a eb e−)

) co sh1

x xc e e

2

1)1

dy+

Vikasana - CET 2012

Answer:c)

Vikasana - CET 2012

1 1/203. . ( ) cot (cos 2 )If f x thenx−= ⎡ ⎤⎣ ⎦

'( )62

f π=

2)33

a

3)2

)2 / 3

b

c)2 / 3)3 / 2

cd

Vikasana - CET 2012

1 1/2: ( ) cot (cos 2 )Solution f x x−= ⎡ ⎤⎣ ⎦1 1'( ) . .( sin 2 ).2

1 cos 2 2 cos 2f x x

x x= − −

+

1 1 3' . .( ).21 26 2 1/ 21f π −⎛ ⎞⇒ = −⎜ ⎟⎝ ⎠ +1

22 1 2.3 3

+

⇒ =3 32: )Ans a

Vikasana - CET 2012

204 . . log sin , .2x dyIf y then

d= =

2

)2 co t2

dxxa2

) tan2xb2

) co t2xc

2) tan2xd

Vikasana - CET 2012

2: log sin xSolution y = g2

1 1.2 sin .cos .2 2 2

y

dy x xd

=2 2 2 2sin

2xdx

xcot2

: )

x

Ans c

=

: )Ans c

Vikasana - CET 2012

1 105. . . .sin . .1

xThe derivative of w r t x− −⎡ ⎤⎢ ⎥⎣ ⎦

2

11)

1

x

ax

⎢ ⎥+⎣ ⎦

−1

2)1

x

bx

−−+

1)1

x

cx

−−

1)1

dx−

Vikasana - CET 2012

1 1: sin &1

xSolution u v xx

− −⎡ ⎤= =⎢ ⎥+⎣ ⎦

1 2

11sin . . tan1

xxu Put xx

θ−

+⎣ ⎦−⎡ ⎤= =⎢ ⎥+⎣ ⎦

1 1sin 2 2 tansin( 2 )22

u xππ θθ− −⎛ ⎞⇒ = = − = −−⎜ ⎟⎝ ⎠2 1 1 1. &

1 2 (1 ) 22

du dvdx x dxx x x xdu

⇒ = − = − =+ +2

1: )

dudv x

Ans b

⇒ = −+

Vikasana - CET 2012

21 1 3 2loglog( ) ye x d⎡ ⎤ +⎡ ⎤⎢ ⎥1 12

22

3 2loglog( )06. . tan tan ,1 6loglog

x dIf y thenx x dxex

− − +⎡ ⎤⎢ ⎥= + =⎢ ⎥−⎢ ⎥ ⎣ ⎦⎣ ⎦)1) 1

ab −))0) 1/ 2

cd) 1/ 2d −

Vikasana - CET 2012

1 12 3 2loglog( )t te x

S l ti − −⎡ ⎤ +⎡ ⎤⎢ ⎥+ ⎢ ⎥

1 12

2

2

gg( ): tan tan1 6loglog

1 l

Solution y x xex⎢ ⎥= + ⎢ ⎥−⎢ ⎥ ⎣ ⎦⎣ ⎦

⎛ ⎞21 1 1 2

2

1 1 2 1 1 2

1 logtan tan 3 tan (log )1 log

1 (l ) 3 (l )

xy xx

− − −⎛ ⎞−= + +⎜ ⎟+⎝ ⎠

1 1 2 1 1 2tan 1 tan (log ) tan 3 tan (log )tan

' 0 '' 0

y x xy cons t

− − − −= − + +=' 0 '' 0

: )y yAns c= ⇒ =

Vikasana - CET 2012

( )207 tan 2 11/ 2The gent tothecurvex yat makesan= ( )07. .tan . . . . 2 . . .1,1/ 2. . .

The gent tothecurvex yat makesananglewiththex axis

=

−0

0

)30)90

ab

0

)90)45

bc

0)60d

Vikasana - CET 2012

2: 2 2 2dy dySolution x y x x= ⇒ = ⇒ =

1 tan tan

ydx dx

dy slopeof gent θ⎛ ⎞⇒ = = =⎜ ⎟(1,1/2)

0

1 . .tan tan

45

slopeof gentdx

θ

θ

⇒ = = =⎜ ⎟⎝ ⎠

⇒ =45: )Ans c

θ⇒ =

Vikasana - CET 2012

2 2 2 2

08 1& 1x y x yTheCurves cutorthogonally+ = + =08. . . 1& 1. . ,16 25 16

TheCurves cutorthogonallya

then a

+ = + =

=,)6

then aa)4)7

bc)7)9

cd

Vikasana - CET 2012

2 2 2 2

: . . 1& 1x y x ySolution The ellipses + = + =: . . 1& 1

. , .

Solution The ellipsesA B a b

cut orthogonally then A B a b

+ +

− = −16 25 16

32 25a

a⇒ − = −⇒ = −

7: )

aAns c⇒ =

)

Vikasana - CET 2012

09. . . . . . int . . .If the subnormal at any po on the curve. . . . .

)2

ny ax is a constant then na

= =)2)1)3 / 2

ab)3 / 2) 2

cd −

Vikasana - CET 2012

i tS l i Th l h f b l h2

: . . . . . . int. . .. . . tan , . . . . . . . 4

Solution Thelengthof subnormalatany po onthecurvewillbeacons t if onlythecurveisaparabolay ax=

2

,. . . . . . : 2n

f y p yTheCurvey axwillbeof form y ax n= = ⇒ =

: )Ans a

Vikasana - CET 2012

10. . . . . . . . sin . . .Thesidesof anequilateral triangleareincrea g at the. .2 / sec. . . . . . . . .

. . .10 .rateof cm Therateat whichtheareais increases whenthesideis cmis

) 3 . / sec)10 . / sec

a squnitsb squnits

)10 3 . / sec10

c squnits10) .3

d squni / sects

Vikasana - CET 2012

: . . . . . . . .Solution Let l lengthof the sidesof theequilateral triagnle=

23 3 . .4 2

dA dlArea A l ldt dt

= = ⇒ =

3 . .2 32

dA l ldt

⇒ = =

10

10 3 .l

dA squnitsdt =

⎛ ⎞⇒ =⎜ ⎟⎝ ⎠: )Ans c

Vikasana - CET 2012

log11 xThe Maximum value of =11. . . . .

1) l 2

The Maximum value ofx

=

) log 22

)0

a

b)0)1 /

bc e

)1d

Vikasana - CET 2012

log: xSolution y =

22

1 1 11log . . (log 1)

S yx

dy x xdx x x x

⎛ ⎞= + = − −−⎜ ⎟⎝ ⎠2

2

20 & 0

dx x x xxdy d yx edx dx

⎝ ⎠

= ⇒ = ∠

log. 1 /

dx dxeMax Value e

e= =

: )Ans c

Vikasana - CET 2012

cos4 112 cosx dx A ecx B then+= +∫12. cos ,

cot tan) 1 / 2

dx A ecx B thenx x

a A

+−

= −

∫

) 1 / 8) 1 / 4

b Ac A

= −= −) 1 / 4

) 1 / 4c Ad A

==

Vikasana - CET 2012

(cos4 1)cos sinx x x+∫ 2 2

2

(cos4 1)cos .sin:cos sini

x x xSolution dxx x+−∫

21 2cos 2 .sin2 1 1sin2 .cos2 . sin4 . cos42 cos2 2 8

x xdx x xdx xdx x cx

−= = = = +∫ ∫ ∫

: )Ans b

Vikasana - CET 2012

3 log 4 113. .( 1) .xe x dx−+ =∫4

13. .( 1) .

)1 / 4 log( 1)

e x dx

a x c

+

+ +∫

4

4

) log( 1)) l ( 1)

b x c− + +4) log( 1)

1)

c x c

d

+ +

4)1

d cx

++

Vikasana - CET 2012

3 341 4 1l ( )x xl d d∫ ∫ 4

4 4

1 4 1: log( 1)4 41 1

: )

x xSolution dx dx x cx x

Ans a

= = + ++ +∫ ∫

: )Ans a

Vikasana - CET 2012

414 sin 6 cos 2xe x xdx =∫4 4

14. sin 6 .cos 2

(sin8 2cos8 ) (sin 4 cos 4 ))40 16

x x

e x xdx

e x x e x xa c

=

− −+ +

∫

4 4

4 4

40 16(sin8 2cos8 ) (sin 4 cos 4 ))

40 16

x xe x x e x xb c− −− +

4 4

4 4

(sin 8 2cos8 ) (sin 4 cos 4 ))40 16

(sin8 2cos8 ) (sin 4 cos 4 )

x x

x x

e x x e x xc c

e x x e x x

− −− + +

− −(sin 8 2cos8 ) (sin 4 cos 4 ))40 16

e x x e x xd c− +

Vikasana - CET 2012

4 4 41 1 1: (sin8 sin4 ) sin8 sin4x x xSolution I e x x dx e xdx e xdx= + = +∫ ∫ ∫

[ ]

: (s 8 s ) s 8 s2 2 2

sin( ) sin( ) cos( )ax

ax

Solution e x x dx e xdx e xdx

ee bx c dx a bx c b bx c c+ = + − + +

∫ ∫ ∫

∫ [ ]2 2

4 4

sin( ) sin( ) cos( )

(sin8 2cos8 ) (sin4 cos4 )sin :x x

e bx c dx a bx c b bx c ca b

e x x e x xu gweget c

+ + + ++− −

+ +

∫

sin . . .:40 16

u gweget c+ +

Vikasana - CET 2012

(1 lo g )1 5xe x x d x+

∫( g )1 5 .

lo g)x

d xx

e x

=∫lo g)

) (1 lo g )x

e xa cx

b e x c

+

+ +) (1 lo g )) lo gx

b e x cc e x c

+ +

+

) lo gxd x e x c+

Vikasana - CET 2012

1: log logx xSolution I x e dx e x c⎛ ⎞= + = +⎜ ⎟∫: log log

: )

Solution I x e dx e x cx

Ans c

= + = +⎜ ⎟⎝ ⎠∫

: )Ans c

Vikasana - CET 2012

1 1 21 216. . sin & sin 1 ,If I xdx I x dx then− −= = −∫ ∫

1 2)

)

a I I

b π=

∫ ∫

2 1)2

)

b I I

I I

π

π

=

+1 2)2

)

c I I x

d I I π

+ =

+ =1 2)2

d I I+ =

Vikasana - CET 2012

1 2 1 1 1 π∫ ∫1 2 1 1 1

1 2:sin 1 cos (sin cos )2

Solution x x I I x xdx dxπ− − − −− = ⇒ + = + =∫ ∫

1 2 2I I xπ+ =

: )Ans c

Vikasana - CET 2012

1 7 d x∞

=∫ 2 20

1 7 .( 4 ) ( 9 )

)

x x

a π+ +∫

)6 0

)

a

b π)2 0

)4 0

b

c π)4 0

)8 0

d π

Vikasana - CET 20128 0

1 1 1:Solution I dx∞⎛ ⎞= −⎜ ⎟∫ 2 2

0

:5 4 9

1 1

Solution I dxx x

xx ∞∞

⎜ ⎟+ +⎝ ⎠

⎡ ⎤⎡ ⎤

∫

11

0 0

1 1tantan

5.2 5.3 32xx −− ⎡ ⎤⎡ ⎤= − ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

⎛ ⎞ ⎛ ⎞1 110 15 602 2

ππ π⎛ ⎞ ⎛ ⎞= − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠: )Ans a

Vikasana - CET 2012

/ 2

1 8 .E v a lu a t eπ / 2

0

s i ns in c o s

x d xx x

π

=+∫

)4

a π

)2

)

b

Z

π

))1

c Z e r od

Vikasana - CET 2012

: . .Solution By direct observation

4I π=

: )Ans a

Vikasana - CET 2012

2 2 2 2 2

19 . . . . . . .( )

T he A rea o f the reg ion bounded byi2 2 2 2 2

2

( ). :

) .2

a y x a x isaa sq un it

= −

22

2) .3ab sq un it

2

34) .

3ac sq un it

2

3

) .4

ad sq un it

Vikasana - CET 2012

2 2: Re . . 4a xSolution quired Area A a x dx= −∫0

2 2

: Re . . 4

2 2a

Solution quired Area A a x dxa

x a x dx= −

∫

∫0

2 2

0

. 2

a

Put a x t xdx dt− = ⇒ = −

∫

[ ] 2

2

003 / 22 2 2.

34

aa

A t dt ta a= − = −∫

24 . .3: )

A a sq unit

Ans c

=

Vikasana - CET 2012

20. . . . . . cos( )dyThe Solution of the equation x ydx

= −

:

) cot

dxis

x ya y c−⎛ ⎞+ =⎜ ⎟) cot2

) cot2

a y c

x yb x c

+ =⎜ ⎟⎝ ⎠

−⎛ ⎞+ =⎜ ⎟⎝ ⎠2

) tan2

x yc y c

⎝ ⎠−⎛ ⎞+ =⎜ ⎟

⎝ ⎠

) tan2

x yd x c−⎛ ⎞+ =⎜ ⎟⎝ ⎠

Vikasana - CET 2012

: cos( )dySolution x ydx

= −

.

1 1 cos

Put x y tdt dy dt tdx dx dx

− =

⇒ − = ⇒ = −

21 1 cos1 cos 2 2

dx dx dxtdt dx ec dx

t⇒ = ⇒ =

−∫ ∫ ∫ ∫

cot2t x c

x y

⇒ − = +

⎛ ⎞cot2

: )

x yx c

Ans b

−⎛ ⎞⇒ + =⎜ ⎟⎝ ⎠

Vikasana - CET 2012

Vikasana - CET 2012