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

Options

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

Answer

A

Explanation

We have
`f(x)=int_(0)^(x)sqrt(t)sin t dt`
`impliesf'(x)=sqrt(x)sinx`
For maximum of minimum value of `f(x), f'(x)=0`
`impliesx=2npi,n epsilonZ`
Sign scheme of `f'(x)` for `xepsilon(0,(5pi)/2)` is

`f'(x)` changes its sign from +ve to -ve in the neighborhood of `pi` and from -to + in the neighborhood of `2pi`.
Hence `f(x)` has local maximum at `x=pi` and local minima at `x=2pi`.