Question

Prove that the function $f$ given by $f\left(x\right)=\log |\cos x|$ is decreasing on $(0,\frac{\pi }{2})$ and increasing on $(\frac{3\pi }{2},2\pi )$

Answer 1

Given: $f\left(x\right)=\log |\cos x|$

we need to know the sign of the modulus, so

When $xϵ(0,\frac{\pi }{2})$ or $x\in (\frac{3\pi }{2},2\pi )$

$\cos x>0\Rightarrow |\cos x|=\cos x$

$\Rightarrow f\left(x\right)=\log \cos x$

Differentiating w.r.t $x,$

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

When $xϵ(0,\frac{\pi }{2})$

$\Rightarrow \tan x>0$

$\Rightarrow f^{\prime }\left(x\right)<0$

Thus, $f\left(x\right)$ is decreasing in $(0,\frac{\pi }{2})$

When $xϵ(\frac{3\pi }{2},2\pi )$

$\Rightarrow \tan x<0$

$\Rightarrow f^{\prime }\left(x\right)>0$

Thus, $f\left(x\right)$ is increasing in $(\frac{3\pi }{2},2\pi )$

Hence proved.