Differentiation under Integral Sign
Trending Questions
Q. limx→0x(e(√1+x2+x4−1)/x−1)√1+x2+x4−1
- is equal to √e
- is equal to 1
- is equal to 0
- does not exist
Q. If limx→∞xln(e(1+1x)1−x)=mn where m and n are relatively prime positive integers, then the value of m+n is
Q.
The period of is
Q. If ∫3ex+5e−x4ex−5e−xdx=Ax+Bln∣∣4e2x−5∣∣+K for constant of integration K, then
- A=−1 and B=−78
- A=1 and B=78
- A=−18 and B=78
- A=−1 and B=78
Q.
In is equal to
None of these
Q.
The negative of the compound proposition is
None of these
Q. For 3A ---> xB, d[B]/dt is found to be 2/3rd of d[A]/dt. Then, the value of x is A)1.5 B)3 C)1/2 D)2
Q. Differentiate with respect to x from first principles f(x)=logsinx
Q. Let y=y(x) be the solution of the differential equation xtan(yx)dy=(ytan(yx)−x)dx, −1≤x≤1, y(12)=π6. Then the area of region bounded by the curves x=0, x=1√2 and y=y(x) in the upper half plane is
- 16(π−1)
- 112(π−3)
- 18(π−1)
- 14(π−2)
Q. If x∫0f(t) dt=x2+1∫xt2f(t) dt, then f′(12) is:
- 45
- 2425
- 1825
- 625
Q.
ddx[sinnx.sin.nx]=
n sinn-1 x.sin(n+1)x
-n sinn-1 x. sin(n+1)x
nsinn-1 x.sin(n-1)x
nSinx
Q. If f(x)=∫x2+1x2 e−t2 dt, then f(x) increases in
- (0, 2)
- no value of x
- (0, ∞)
- (−∞, 0)