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12th Grade > Mathematics

INTEGRALS MCQs

Total Questions : 30 | Page 1 of 3 pages
Question 1. cos4xdxsin3x(sin5x+cos5x)35=12(1+cotAx)B+C then AB=
  1.    1
  2.    2
  3.    12
  4.    None of these
 Discuss Question
Answer: Option B. -> 2
:
B
I=cosx4dxsin3x(sin5x+cos5x)35=cosec2xcot4xdx(1+cot5x)3/51+cot5x=t.5cot4x(cosec2x)dx=dt=15dtt3/5=t2/52=12(1+cot5x)2/5+C.
Question 2. If 2cosxsinx+λcosx+sinx2dx=An|cosx+sinx2|+Bx+C. Then the ordered triplet A,B,λ  is
  1.    (12,32,−1)
  2.    (32,12,−1)
  3.    (12,−1,−32)
  4.    (32,−1,−12)
 Discuss Question
Answer: Option B. -> (32,12,−1)
:
B
ddx(A|cosx+sinx2|+Bx+C)=Acosxsinxcosx+sinx2+B=Acosx+BcosxAsinx+Bsinx2Bcosx+sinx2A+B=2,A+B=1,2B=λA=32,B=12,λ=1
Question 3. If f(x)=x,g(x)=ex1,andfog(x)dx=Afog(x)+Btan1(fog(x))+C, then A + B is equal to
  1.    1
  2.    2
  3.    3
  4.    0
 Discuss Question
Answer: Option D. -> 0
:
D
fog(x)=ex1I=ex1dx=2t2t2+1dt{wheret=ex1}=2t2tan1t+C=2ex12tan1(ex1)+C=2fog(x)2tan1(fog(x))+CA+B=2+(2)=0
Question 4. (2+sec x)sec x(1+2sec x)2dx
  1.    12cosec x+cot x+C
  2.    2cosec x+cot x+C
  3.    12cosec x−cot x+C
  4.    2cosec x−cot x+C  
 Discuss Question
Answer: Option A. -> 12cosec x+cot x+C
:
A
(2cosx+1)(2+cosx)2dx=(2+cosx)cosx+sin2x(2+cosx)2dx=cosx2+cosxdxsin2x(2+cosx)2dx=sinx2+cosx+c
Question 5. If Φ(x)=limnxnxnxn+xn,0<x<1,ϵN, then (sin1x)(Φ(x))dx is equal to
  1.    x sin−1(x)+√1−x2+C
  2.    −(x sin−1(x)+√1−x2)
  3.    x sin−1 x+√1−x2+C
  4.    None of these
 Discuss Question
Answer: Option A. -> x sin−1(x)+√1−x2+C
:
A
We have ϕ(x)=limnx2n1x2n+1=1,0<x<1sin1x.ϕ(x)dx=sin1xdx=[xsin1x+1x2]+c
Question 6. etan1x(1+x2)[(sec11+x2)2+cos1(1x21+x2)]dx(x>0)
  1.    etan−1x.tan−1x+C  
  2.    etan−1x.(tan−1x)22+C  
  3.    etan−1x.(sec−1(√1+x2))2+C  
  4.    etan−1x.(cos−1(√1+x2))2+C
 Discuss Question
Answer: Option C. -> etan−1x.(sec−1(√1+x2))2+C  
:
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
Question 7. 1+3x3x2dx is equal to
  1.    (1+x1/3)3/2+C
  2.    −(1+x1/3)3/2+C
  3.    2(1+x1/3)3/2+C
  4.    None of these
 Discuss Question
Answer: Option C. -> 2(1+x1/3)3/2+C
:
C
1+3x3x2dx=x23(1+x13)1/2dx
=x2/3(1+x1/3)1/2dx
1+x1/3=t2
x2/3dx=6tdt
=6t2dt
=2t3+c
=2(1+x3)3/2+c
Question 8. Let g(x) be an antiderivative for f(x). Then  In (1+(g(x))2) is an antiderivate for
  1.    2f(x)g(x)1+(f(x))2
  2.    2f(x)g(x)1+(g(x))2
  3.    2f(x)1+(f(x))2
  4.    None
 Discuss Question
Answer: Option B. -> 2f(x)g(x)1+(g(x))2
:
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)
Question 9. If dxx1x3=a log1x311x3+1+C then a =
  1.    13
  2.    23
  3.    −13
  4.    −23
 Discuss Question
Answer: Option A. -> 13
:
A
Put 1x3=t23x2dx=2tdtdxx1x3=x2x31x3dx=23dtt21=13logt1t+1+C=13log1x311x3+1+Ca=13
Question 10. If dxx1x3=aln(1x3+b1x3+1)+k, then
  1.    b=1,a=1
  2.    b=1,a=−13
  3.    b=1,a=−23
  4.    None of these
 Discuss Question
Answer: Option B. -> b=1,a=−13
:
B
I=dxx1x3Put1x3=t23x2dx=2tdtI=23dt(1t2)=13(11t+11+t)dt.=13log1+t1t+c=13log1+1x311x3+c.

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