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  • Prove that the function f given by f (x) = log sin x is increasing on 0, 2 pi and decreasing on pi by 2 ,pie

16) Prove that the function f given by f (x) = \log \sin x  is increasing on 

\left ( 0 , \pi /2 \right )\: \: and \: \: decreasing \: \: on \: \: \left ( \pi/2 , \pi \right )
 

Answers (1)

best_answer

Given function is,
f (x) = \log \sin x
f^{'}(x) = \frac{1}{\sin x}\cos x = \cot x
Now, we know that  cot x is+ve in the interval  \left ( 0 , \pi /2 \right )   and -ve  in the interval \left ( \pi/2 , \pi \right )
f^{'}(x) > 0 \ in \ \left ( 0,\frac{\pi}{2} \right ) \ and \ f^{'}(x) < 0 \ in \ \left ( \frac{\pi}{2} , \pi \right )
Hence, f (x) = \log \sin x  is increasing in the interval \left ( 0 , \pi /2 \right ) and   decreasing in interval  \left ( \pi/2 , \pi \right )

Posted by

Gautam harsolia

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