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 )

Answers (1)

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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