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Mere Teacher · Previous Year Questions
JEE Main Mathematics · Indefinite Integration · Set 5 (Q121-150)
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121.
When x>0, then ∫cos−1(1+x21−x2)dx is
(1)2(xtan−1x−log(1+x2))+c
(2)2(xtan−1x+log(1+x2))+c
(3)2xtan−1x+log(1+x2)+c
(4)2xtan−1x−log(1+x2)+c
122.
∫sin(x−3π)cosxdx=
(1)2log(2tanx−3)+C
(2)2log(sin(x−3π)cosx)+C
(3)2log(tanx+3)+C
(4)2log(sinx+3cosx)+C
123.
∫59log3x2+log(588−84x+3x2)2dx is equal to
(1)2
(2)1
(3)21
(4)4
124.
The value of ∫ex(1+cosx1+sinx)dx is equal to
(1)extan2x+C
(2)extanx+C
(3)ex(1+cosx)+C
(4)ex(1+sinx)+C
125.
Find integration∫(x−3)6/7⋅(x+4)8/7dx
(1)(x+4x−3)71+c
(2)7(x+4x−3)71+c
(3)7(x+4x−3)76+c
(4)7(x−3x+4)76+c
126.
Consider the following statements (A) and (B)(A) ∫abdxd(f(x))dx=dxd∫ab(f(x))dx(B) dxd(∫f(x)dx)=f(x)+C Which one of the following is true?
(1)Only (A) is true
(2)Only (B) is true
(3)Both (A) and (B) are true
(4)Both (A) and (B) are false
127.
∫cosecx−sinxdx=
(1)2sinx+C
(2)sinx+C
(3)sinx2+C
(4)2sinx+C
128.
The integral ∫sec2/3xcosec4/3xdx is equal to:
(1)−3tan−1/3x+C
(2)−43tan−4/3x+C
(3)−3cot−1/3x+C
(4)3tan−1/3x+C
129.
∫(1+x)2022dx=
(1)(1+x)20212(20201+x−20211)+C
(2)(1+x)20222(20201+x−2021x)+C
(3)(1+x)2(2022(1+x)2022−2021(1+x)2021)+C
(4)(1+x)21((1+x)10101−(1+x)10111)+C
130.
If x2+x−642−13x=lx+mA+px+qB where lm>0 and pq<0 then BmqAlp=
(1)3227
(2)827
(3)2438
(4)32243
131.
∫−33cot−1xdx=
(1)3π
(2)6π
(3)0
(4)3
132.
∫1+x2x3dx is equal to
(1)1+x2−3x(1+x2)3/2+C
(2)x1+x2+32(1+x2)3/2+C
(3)x21+x2−32(1+x2)3/2+C
(4)x21+x2−31(1+x2)1/2+C
133.
If [x] is the greatest integer function not greater than x then ∫08[x]dx is equal to
(1)28
(2)29
(3)30
(4)20
134.
If ∫5+7sinθ−2cos2θcosθdθ=Aloge|f(θ)|+c (where c is a constant of integration), then Af(θ) can be
(1)sinθ+32sinθ+1
(2)5(sinθ+3)2sinθ+1
(3)2sinθ+15(sinθ+3)
(4)sinθ+35(2sinθ+1)
135.
limn→∞(1+n51+25+n524+35+n534+…+n5+n5n4)=
(1)51log3
(2)31log5
(3)21log5
(4)log52
136.
∫0πcos4x+sin4xcos4xdx=
(1)4π
(2)2π
(3)8π
(4)π
137.
If f(x)=∫xx+1e−t2dt, then the interval in which f(x) is decreasing is
(1)(−21,∞)
(2)(−∞,2)
(3)(−∞,0)
(4)(−2,2)
138.
If (2x−1)(x+2)(x−3)x3=A+2x−1B+x+2C+x−3D then A is equal to
(1)21
(2)50−1
(3)25−8
(4)2527
139.
If In=∫0π/4tannxdx, then I2+I41+I3+I51+I4+I61=
(1)I9+I111
(2)I10+I121
(3)I12+I141
(4)I11+I131
140.
For x2=nπ+1,n∈N(the set of natural numbers), the integral ∫x2sin(x2−1)+sin2(x2−1)2sin(x2−1)−sin2(x2−1)dx is equal to:
(1)loge(21sec2(x2−1))+c
(2)21loge(sec(x2−1))+c
(3)21loge(sec2(2x2−1))+c
(4)loge(sec(2x2−1))+c
141.
Evaluate: ∫x4−9x3+xdx
(1)41log(x4−9)+121log(x2−3x2+3)+C
(2)41log(x4−9)−121log(x2+3x2−3)+C
(3)41log(x4−9)+121log(x2+3x2−3)+C
(4)None of these
142.
If the equivalent partial fraction of (2x−1)(x+2)(x−3)x3 is of the form A+2x−1B+x+2C+x−3D then the value of A+B+C=
(1)−8/25
(2)4/25
(3)−1/50
(4)1/2
143.
If ∫e−xtan−1(ex)dx=f(x)−21log(1+e2x)+C, then f(x) equals
(1)ex−e−xtan−1(ex)
(2)x2+e−xtan−1(ex)
(3)−e−xtan−1(ex)
(4)x−e−xtan−1(ex)
144.
If x2(2x−3)x−2=xA+x2B+2x−3C then 2(A−C)=
(1)3B
(2)2B
(3)0
(4)B
145.
If ∫(x+2)(x2+1)dx=alog(1+x2)+btan−1x+51log|x+2|+c, then
(1)a=10−1,b=52
(2)a=10−1,b=5−2
(3)a=101,b=52
(4)a=101,b=5−2
146.
∫0πsecx⋅cosecxxtanxdx=
(1)2π
(2)4π
(3)2π2
(4)4π2
147.
If x>0 and x=(2n+1)2π then ∫(xx−elog(secxtanx)+x23x2−2x+1)dx=
(1)xx−secx+3x−2logx−x1+c
(2)52x2x−secx+3x+x22−x1+c
(3)xx−secx+3x+x22−x1+c
(4)52x2x−secx+3x−2logx−x1+c
148.
∫(1+cosx)9/7(1−cosx)2/7dx is equal to
(1)117(tan2x)711+C
(2)117(tan2x)117+C
(3)117(cot2x)711+C
(4)711(cot2x)117+C
149.
If (2x+3)(x2+2)3x2+ax+3=2x+33+x2+2Bx+C then a(B+C)=
(1)-2
(2)3
(3)-3
(4)2
150.
If ∫x2+x+12x2+5x+9dx=xx2+x+1+αx2+x+1+βloge(x+21+x2+x+1)+C, where C is the constant of integration, then α+2β is equal to .