Question

Prove that the function $f$ given by $f\left(x\right)=\log \cos x$ is strictly decreasing on $(0,\frac{\pi }{2})$and strictly increasing on $(\frac{\pi }{2},\pi )$.

Answer 1

We have, $f\left(x\right)=\log \cos x$

$\therefore f^{\prime }\left(x\right)=\frac{1}{\cos x}(-\sin x)=-\tan x$

In the interval $(0,\frac{\pi }{2})$, $\tan x>0\Rightarrow -\tan x<0.$

$\therefore f^{\prime }\left(x\right)<0$ on $(0,\frac{\pi }{2})$

$\therefore$ $f$ is strictly decreasing on $(0,\frac{\pi }{2})$.

In interval $(\frac{\pi }{2},\pi )$, $\tan x<0\Rightarrow -\tan x>0.$

$\therefore f^{\prime }\left(x\right)>0$ on $(\frac{\pi }{2},\pi )$

$\therefore$ $f$ is strictly increasing on $(\frac{\pi }{2},\pi )$