17) Prove that the function f given by f (x) = log |cos x| is decreasing on $( 0 , \pi /2 )$ and increasing on $( 3 \pi/2 , 2\pi )$

Given function is,
f(x) =  log|cos x|
value of cos x is always +ve in both these  cases
So, we can write   log|cos x| = log(cos x)
Now,
$f^{'}(x) = \frac{1}{\cos x}(-\sin x) = -\tan x$
We know that  in interval $\left ( 0,\frac{\pi}{2} \right )$  ,  $\tan x > 0 \Rightarrow -\tan x< 0$
$f^{'}(x) < 0$
Hence, f(x) =  log|cos x| is decreasing in interval   $\left ( 0,\frac{\pi}{2} \right )$

We know that  in interval  $\left ( \frac{3\pi}{2},2\pi \right )$  , $\tan x < 0 \Rightarrow -\tan x> 0$
$f^{'}(x) > 0$
Hence, f(x) =  log|cos x| is increasing in interval  $\left ( \frac{3\pi}{2},2\pi \right )$

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