@`JEET

when $f'(x)= 0$ i.e. $2xcosx^2 = 0$ then either $x=0$ or $cos\;x^2 = 0 $

Now, $cos\;x^2 = 0$ means $x^2 = \frac{ \pi }{2} , \frac{3 \pi}{2} , \frac{5 \pi}{2} ,.......$

(or) $x = \pm \sqrt \frac{ \pi }{2} ,\pm \sqrt \frac{3 \pi}{2} , \pm \sqrt \frac{5 \pi}{2} ,.......$

We can also write it as :

$x = \pm \sqrt {0 \pi + \frac{ \pi }{2}} ,\pm \sqrt{ 1 \pi + \frac{ \pi}{2}} , \pm \sqrt {2\pi + \frac{ \pi}{2}} ,.......$

Now, $f''(x) = -4x^2sin\;x^2 + 2cos\;x^2$

1) at $x=0,$ $f''(0) = 2$ . So, $f''(0) > 0$ , So, at $x=0$ , $f(x)$ has local minima.

2) at $x= \pm \sqrt{ 1 \pi + \frac{ \pi}{2}}$, $f''(\pm \sqrt{ 1 \pi + \frac{ \pi}{2}}) = 6 \pi $ . So, $f''(0) > 0$ , So, at $x=\pm \sqrt{ 1 \pi + \frac{ \pi}{2}}$ , $f(x)$ has local minima.

3) at $x= \pm \sqrt {2\pi + \frac{ \pi}{2}}$, $f''(\pm \sqrt {2\pi + \frac{ \pi}{2}}) = -10 \pi $ . So, $f''(0) < 0$ , So, at $x=\pm \sqrt {2\pi + \frac{ \pi}{2}}$ , $f(x)$ has local maxima.

So, observe the pattern or we can prove it for even $k$ and odd $k$, answer should be $C$