# Free Objective Test 01 Practice Test - 11th and 12th

If dxx1x3=aln(1x3+b1x3+1)+k, then

A. b=1,a=1
B. b=1,a=13
C. b=1,a=23
D. None of these

#### SOLUTION

Solution : B

I=dxx1x3Put 1x3=t23x2dx=2tdtI=23dt(1t2)=13(11t+11+t)dt.=13 log 1+t1t+c=13 log 1+1x311x3+c.

sin3x(cos4x+3cos2x+1)tan1(secx+cosx)dx=

A. tan1(secx+cosx)+c
B. log|tan1(secx+cosx)|+c
C. 1(secx+cos2x)2+c
D. log|secx+cosx|+c

#### SOLUTION

Solution : B

Put tan1(secx+cosx)=f(x)f(x)=sin3xcos4x+3cos2x+1f(x)f(x)dx=log|f(x)|+cIntegral=log|tan1(secx+cosx)|+c

If 2cosxsinx+λcosx+sinx2dx=An|cosx+sinx2|+Bx+C. Then the ordered triplet A,B,λ  is

A. (12,32,1)
B. (32,12,1)
C. (12,1,32)
D. (32,1,12)

#### SOLUTION

Solution : B

ddx(A|cosx+sinx2|+Bx+C)=Acos xsin xcos x+sin x2+B=Acosx+BcosxAsinx+Bsinx2Bcosx+sinx2A+B=2,A+B=1,2B=λA=32,B=12,λ=1

cos4xdxsin3x(sin5x+cos5x)35=12(1+cotAx)B+C then AB=

A. 1
B. 2
C. 12
D. None of these

#### SOLUTION

Solution : B

I=cosx4dxsin3x(sin5x+cos5x)35=cosec2xcot4xdx(1+cot5x)3/51+cot5x=t.5 cot4x(cosec2x)dx=dt=15dtt3/5=t2/52=12(1+cot5x)2/5+C.

If f(x)=x,g(x)=ex1,andfog(x)dx=Afog(x)+Btan1(fog(x))+C, then A + B is equal to

A. 1
B. 2
C. 3
D. 0

#### SOLUTION

Solution : D

fog(x)=ex1I=ex1dx=2t2t2+1dt{where t=ex1}=2t2tan1t+C=2ex12tan1(ex1)+C=2fog(x)2tan1(fog(x))+CA+B=2+(2)=0

(2+sec x)sec x(1+2sec x)2dx

A. 12cosec x+cot x+C
B. 2cosec x+cot x+C
C. 12cosec xcot x+C
D. 2cosec xcot x+C

#### SOLUTION

Solution : A

(2cos x+1)(2+cos x)2dx=(2+cos x)cos x+sin2x(2+cos x)2dx=cos x2+cos xdxsin2x(2+cos x)2dx=sin x2+cos x+c

Let g(x) be an antiderivative for f(x). Then  In (1+(g(x))2) is an antiderivate for

A. 2f(x)g(x)1+(f(x))2
B. 2f(x)g(x)1+(g(x))2
C. 2f(x)1+(f(x))2
D. None

#### SOLUTION

Solution : B

Given f(x)dx=g(x)g(x)=f(x)
Now ddx(In(1+g2(x))=2g(x)g(x)1+g2(x)2f(x)g(x)1+g2(x)

If (x1x+1)dxx3+x2+x=2tan1f(x)+C, find f(x).

A. x+1x+1
B. x+1x+2
C. x1x+1
D. x1x1

#### SOLUTION

Solution : A

I=(x1)dx(x+1)xx+1+1x=(x1)(x+1)dx(x+1)2xx+1+1x=(11x2)dx(x+1x+2)x+1x+1Put x+1+1x=t2(11x2)dx=2t dt=2t dt(t2+1)t=2tan1t+c=2tan1(x+1x+1)+c
f(x) = x+1x+1

etan1x(1+x2)[(sec11+x2)2+cos1(1x21+x2)]dx(x>0)

A. etan1x.tan1x+C

B. etan1x.(tan1x)22+C

C. etan1x.(sec1(1+x2))2+C

D. etan1x.(cos1(1+x2))2+C

#### SOLUTION

Solution : C

note that sec11+x2=tan1x;cos1(1x21+x2)=2tan1x for x > 0
I=etan1x1+x2((tan1x)2+2tan1x)dx
Put tan1x=t
=et(t2+2t)dt=et.t2=etan1x((tan1x)2)+C

Let f(x)=2sin2x1cosx+cosx(2sinx+1)1+sinx then ex(f(x)+f(x))dx equals
(where c is the constant of integeration)

A. extanx+c
B. excotx+c
C. excosec2x+c
D. None of these

#### SOLUTION

Solution : A

cosx(1+2sinx)1+sinxcos2xsin2xcosx=cos2x(1+2sinx)(1+sinx)(cos2xsin2x)cosx(1+sinx)
=sinxcos2x+sin2x(1+sinx)cosx(1+sinx)=sinx(1sinx)+sin2xcosx=tanx

(1+tanx)(1+tan2x)2tanxdx equal to

A. logtan2x+tanx+c

B. logtan2x+12tanx+c

C. log|tanx|+2tanx+c

D. log|tanx|+tanx+c

#### SOLUTION

Solution : A

12sinxcosxdx+12tanxsinxcosxdx=12sin2x+cos2xsinxcosxdx+12sec2xtanxdx=log(tan2x)+tanx+c

x22x3x21dx is equal to

A. x2x21
B. x2x21
C. x21x2
D. x21x2

#### SOLUTION

Solution : D

x22x3x21dx=dxxx212dxx3x21=sec1x2sec θ tan θsec3 θ tan θdθ=[put x=sec θdx=sec θ tan θ dθ]=sec1x2cos2 θ dθ=sec1x2(1+cos 2θ)dθ=sec1x(θ+sin 2 θ2)+C=sec1xsec1xx21x2+C=x21x2+C

1+3x3x2dx is equal to

A. (1+x1/3)3/2+C
B. (1+x1/3)3/2+C
C. 2(1+x1/3)3/2+C
D. None of these

#### SOLUTION

Solution : C

1+3x3x2dx=x23(1+x13)1/2dx
=x2/3(1+x1/3)1/2dx
1+x1/3=t2
x2/3dx=6 t dt
=6 t2dt
=2t3+c
=2(1+x3)3/2+c

If Φ(x)=limnxnxnxn+xn,0<x<1,ϵN, then (sin1x)(Φ(x))dx is equal to

A. x sin1(x)+1x2+C
B. (x sin1(x)+1x2)
C. x sin1 x+1x2+C
D. None of these

#### SOLUTION

Solution : A

We have ϕ(x)=limnx2n1x2n+1=1,0<x<1sin1x.ϕ(x) dx=sin1x dx=[x sin1 x+1x2]+c

The value of ax2bxc2x2(ax2+b)2dx is equal to

A. sin1[ax+bxc]+k
B. sin1[ax2+bx2c]+k
C. cos1[ax+bxc]+k
D. cos1[ax2+bx2c]+k

#### SOLUTION

Solution : A

I=abx2c2(ax2+bx)2dx=(abx2)c2(ax+bx)2dx
Put ax+bx=t(abx2)dx=dtI=dtc2t2=sin1[ax+bxc]+k