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)

$\left(0,\frac{\pi}{2}\right)$

Answer 2

We have, $f(x) = cos^2\, x$
On differentiating both sides w.r.t. ?$x$?, we get
$f'(x) = -2 \,cosx \, sinx = -sin2x$
$\because -1 \le sin\,2x \le 1$
$\therefore f'(x) > 0$ 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 $( 0, \frac{\pi}{2})$.