If $f(x)=\cos 3 x, 0 ≤ x ≤ \frac{\pi}{2}$, then :

(A) f is strictly increasing on $\left[0, \frac{\pi}{3}\right)$
(B) f is strictly decreasing on $\left(\frac{\pi}{3}, \frac{\pi}{2}\right]$
(C) f is strictly increasing on $\left(\frac{\pi}{3}, \frac{\pi}{2}\right]$
(D) f is strictly decreasing on $\left[0, \frac{\pi}{3}\right)$
(E) f is strictly increasing on $\left[0, \frac{\pi}{2}\right]$

Choose the correct answer from the options given below:

(C) and (D) only

The correct answer is Option (1) → (C) and (D) only

$f(x)=\cos 3 x$

$f'(x)=-3\sin 3x=0$

$⇒\sin 3x=0$

$3x=0, \pi$

$x=0,\frac{\pi}{3}$

$f(x)$ increasing on $\left(\frac{\pi}{3}, \frac{\pi}{2}\right]$ and decreasing on $\left[0, \frac{\pi}{3}\right)$