Question

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

Answer 1

We have,

f(x)=log cos x

$fx=\frac {1}{cosx} (-sinx)=-tan x$

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

f'(x)<0 on $(0,\frac{\pi}{2})$

∴ f is strictly decreasing in $(0,\frac{\pi}{2})$

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

f'(x)>0 on $(\frac{\pi}{2},\pi)$

∴f is strictly increasing in $(\frac{\pi}{2},\pi)$.