For `x epsilon(0,(5pi)/2)`, definite `f(x)=int_(0)^(x)sqrt(t) sin t dt`. Then `f` has

Options

A. local minimum at `pi` and `2pi`
B. local minimum at `pi` and local maximum at `2pi`
C. local maximum at `pi` and local minimum at `2pi`
D. local maximum at `pi` and `2pi`

Answer

C

Explanation

Given `f(x)=int_(0)^(x)sqrt(t) sin t dt,` where `x epsilon (0,(5pi)/2)`
`impliesf'(x)={sqrt(x)sin x-0}`…………….i
[Using Newton-Leibnits formula]
Put`f'(x)=sqrt(x) sin x=0 impliessin x=0`
`impliesx=pi , 2pi`
Now `f''(x)=sqrt(x) cos x+1/(2sqrt(x) sin x`
At `x=pi,f''(pi)=-sqrt(pi) lt 0`
`:.` local maximum at `x=pi`
At `x=2pi, f''(2pi)=sqrt(2pi)gt0`
So local minimum at `x=2pi`