Post on 08-May-2019
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
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: . . . . . . . .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
++
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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− +
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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
−⎛ ⎞⇒ + =⎜ ⎟⎝ ⎠
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Vikasana - CET 2012