Question

The function $ f(x) = \cos^2x $ is strictly decreasing on

• A. $ \left[0,\frac{\pi}{2}\right] $
• B. $ \left[0,\frac{\pi}{2}\right) $
• C. $ \left(0,\frac{\pi}{2}\right) $
• D. $ \left(0,\frac{\pi}{2}\right] $

Answer 1

Answer: (C)

We have, $f(x) = \cos^2\, x$
On differentiating both sides w.r.t. ‘$x$’, we get
$f'(x) = -2 \,\cos x \, \sin x = -\sin2x$
$\because -1 \le \sin\,2x \le 1$
$\therefore f'(x) > 0 $ for $x \in (\frac{-\pi}{2} , 0)$
and $f'(x) < 0 $ for $ x\in (0, \frac{\pi}{2})$
$\therefore f(x) = \cos^2\,x$ is strictly decreasing on $( 0, \frac{\pi}{2})$.