For $$x \in \left( {0,{{5\pi } \over 2}} \right),$$ define $$f\left( x \right) = \int\limits_0^x {\sqrt t \sin t\,dt.} $$ Then $$f$$ has
A
local minimum at $$\pi $$ and $$2\pi $$
B
local minimum at $$\pi $$ and local maximum at $$2\pi $$
C
local maximum at $$\pi $$ and local minimum at $$2\pi $$
D
local maximum at $$\pi $$ and $$2\pi $$
Explanation
$$f'\left( x \right) = \sqrt x \sin x$$
At local maxima or minima, $$f'\left( x \right) = 0$$
$$ \Rightarrow x = 0$$ or $$sin$$ $$x=0$$
$$ \Rightarrow x = 2\pi ,\,\,\pi \in \left( {0,{{5\pi } \over 2}} \right)$$
$$f''\left( x \right) = \sqrt x \cos \,x + {1 \over {2\sqrt x }}\sin \,x$$
$$ = {1 \over {2\sqrt x }}\left( {2x\,\cos \,x + \sin \,x} \right)$$
At $$x = \pi ,$$ $$f''\left( x \right) < 0$$
Hence, local maxima at $$x = \pi $$
At $$x = 2\pi ,\,\,\,f''\left( x \right) > 0$$
Hence local minima at $$x = 2\pi $$