# Free Objective Test 02 Practice Test - 11th and 12th

If In=tannxdx,then I0+I1+2(I2+...I8)+I9+I10, is equal to

A. 9n=1tannxn
B. 1+8n=1tannxn
C. 9n=1tannxn+1
D. 10n=2tannxn+1

#### SOLUTION

Solution : A

We have        In=tannxdx=tann2x(sec2x1)dx=tann1xn1In2i.e.In2+In=tann1xn1(n2)
Thus, we have
I0+I1+2(I2+....+I8)+I9+I10=(I0+I2)+(I1+I3)+(I2+I4)+(I3+I5)+(I4+I6)+(I5+I7)+(I6+I8)+(I7+I9)+(I8+I10).=10n=2tann1xn1=9n=1tannxn

Primitive of f(x)=x.2In(x2+1) with respect to x  is

A. 2In(x2+1)(x2+1)+C
B. (x2+1)2In(x2+1)In2+1+C
C. (x2+1)In2+12(In2+1)+C
D. (x2+1)In22(In2+1)+C

#### SOLUTION

Solution : C

I=x2In(x2+1)dx let x2+1=t;xdx=dt2
Hence I=122Intdt=12tIn2dt=12.tIn2+1In2+1+C=12.(x2+1)In2+1In2+1+C

cos3x+cos5xsin2x+sin4xdx equals

A. sinx6tan1(sinx)+c

B. sinx2sin1x+c

C. sinx2(sinx)16tan1(sinx)+c

D. sinx2(sinx)1+5tan1(sinx)+c

#### SOLUTION

Solution : C

sinx=t;I=(1t2)(2t2)t2(1+t2)dt=(1+2t261+t2)dt
=sinx2(sinx)16tan1(sinx)+c

dx(x3)(4/5)(x+1)6/5=

A. ((x3)(x+1)(1/5))+c
B. 54(x3x+1)1/5+c
C. (x3x+1)1/5+c
D. (x3)6/5(x+1)4/5+c

#### SOLUTION

Solution : B

I=dx(x3)4/5(x+1)6/5Put t=x3x+1dt=4(x+1)2dx.I=14dtt4/5=14(t4/5+14/5+1)=54 t1/5=54(x3x+1)1/5

Let x2nπ1,nϵN. Then, the value of x2sin(x2+1)sin2(x2+1)2sin(x2+1)+sin2(x2+1)dx is equal to

A. log|12sec(x2+1)|+C
B. log|sec(x2+12)|+C
C. 12log|sec(x2+1)|+C
D. None of these

#### SOLUTION

Solution : B

We have,        x2sin(x2+1)sin2(x2+1)2sin(x2+1)+sin2(x2+1)dx    =x2sin(x2+1)2sin(x2+1)cos(x2+1)2sin(x2+1)+2sin(x2+1)cos(x2+1)dx    =x1cos(x2+1)1+cos(x2+1)dx    =xtan(x2+12)dx    =tan(x2+12)d(x2+12)    =log|sec(x2+12)|+C

dxcos(2x)cos(4x)is equal to

A. 122log|1+2 sin 2x12 sin 2x|12(log|sec 2x+tan 2x|)+C
B. 122log|1+2 sin 2x1+2 sin 2x|12(log|sec 2x+tan 2x|)+C
C. 12log|1+2 sin 2x1+2 sin 2x|12(log|sec 2xtan 2x|)+C
D. None of these

#### SOLUTION

Solution : A

sin(4x2x)dxsin(2x)cos(2x)cos(4x)=sin(4x)dxsin(2x)cos(4x)sec 2x dx=2cos2x dxcos4x12(log|sec 2x+tan 2x|)

17 cos2xsin7xcos2xdx=f(x)(sinx)7+C, then f(x) is equal to

A. sin x
B. cos x
C. tan x
D. cot x

#### SOLUTION

Solution : C

17 cos2xsin7xcos2xdx=(sec2xsin7x7sin7x)dx=sec2xsin7xdx7sin7dx=I1+I2Now,I1=sec2xsin7dx=tan xsin7x+7tanx cosxsin8x=tan xsin7xI2I1+I2=tan xsin7x+Cf(x)=tanx

ex[x3+x+1(1+x2)3/2]dx is equal to

A.

x2ex(1+x2)1/2+c

B.

exx(1+x2)1/2+c

C.

ex(1+x2)1/2+c

D.

xex(1+x2)1/2+c

#### SOLUTION

Solution : D

I=ex[x1+x2+1(1+x2)3/2]dx

Let f(x)=x1+x2f(x)=1+x2x21+x2(1+x2)

=1(1+x2)3/2

I=exf(x)+c

=exx1+x2+c

dxsin4x+cos4 x is equal to

A. 12tan1(12tan 2x)+C
B. 2tan1(12tan 2x)+C
C. 12tan1(12cot 2x)+C
D. None of these

#### SOLUTION

Solution : A

dxsin4 x +cos4 x=dx(sin2 x+cos2 x)22 sin2 x cos2 x=2 dx2sin2 2x=2 sec2 2x2 sec2 2xtan2 2xdx=2sec2 2x2+tan2 2xdx=dt(2)2+t2[putting tan 2x=t2 sec2 2x dx=dt]=12tan1(t2)+C=12tan1(12tan 2x)+C

If dxx1x3=a log1x311x3+1+C then a =

A.

13

B.

23

C.

13

D.

23

#### SOLUTION

Solution : A

Put<br>        1x3=t23x2 dx=2t dt    dxx1x3=x2x31x3dx=23dtt21    =13 logt1t+1+C    =13 log1x311x3+1+C    a=13

If cos 8x+1tan 2xcot 2xdx=a cos 8x+C, then

A. a=116
B. a=18
C. a=116
D. a=18

#### SOLUTION

Solution : C

cos 8x+1tan 2xcot 2xdx=2 cos2 4xsin2 2xcos2 2x.sin 2x cos 2x dx=sin 4x cos 4x dx=12sin 8x dx=116cos 8x+Ca=116

The anti – derivative of the function (3x + 4) |sin x|, when 0<x<π, is given by

A. 3 sin x(3x+4) cos x+c
B. 3 sin x+(3x+4) cos x+c
C. 3 sin x(3x+4) cos x+c
D. None of these

#### SOLUTION

Solution : A

In the interval (0,π), sin x is positive, therefore, (3x+4)|sin x|=(3x+4) sin x.
The antiderivative of (3x+4)|sin x| is
=(3x+4) sin x dx=(3x+4) cos x+3 cos x dx=(3x+4) cos x+3 sin x+c

If Φ(x)=dxsin12x cos72x, then Φ(π4)Φ(0)=

A. 125
B. 95
C. 65
D. 0

#### SOLUTION

Solution : A

tan x=tsec2 x dx=dt  f(x)=dxsin12xcos12x.cos4x=(1+tan2 x)sec2 xtan xdx  =(1+t2)tdt=(t1/2+t3/2)dt  =2t1/2+25t5/2=2tan x+25(tan x)5/2  ϕ(π4)ϕ(0)=2+25=125

x{f(x2)g′′(x2)f′′(x2)g(x2)}dx

A. f(x2)g(x2)g(x2)f(x2)+c
B. 12{f(x2)g(x2)f(x2)}+c
C. 12{f(x2) g(x2)g(x2) f(x2)}+c
D. None of these

#### SOLUTION

Solution : C

Put x2=tI=12{f(t)g′′(t)g(t)f′′(t)}dt=12{f(t) g(t)f(t) g(t)g(t) f(t)+g(t) f(t)dt}+cI=12{f(t) g(t)g(t) f(t)}+c=12{f(x2) g(x2)g(x2) f(x2)}+c

x4+11+x6 dx=

A. tan1(x)tan1(x3)+c
B. tan1(x)13tan1(x3)+c
C. tan1(x)+tan1(x3)+c
D. tan1(x)+13tan1(x3)+c

#### SOLUTION

Solution : D

I=x4+11+x6 dx=(x4x2+1)+x2(1+x6) dx=x4x2+11+x6 dx+x21+x6 dx=11+x2 dx+133x21+x6 dx=tan1(x)+13tan1x3+c