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Question
Prove that the function f given by f(x) = log cos x is strictly increasing on (−π/2, 0) and strictly decreasing on (0, π/2) ?
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Solution
\[f\left( x \right) = \log \cos x\]
\[f'\left( x \right) = \frac{1}{\cos x}\left( - \sin x \right)\]
\[ = - \tan x\]
\[\text { Now,} \]
\[x \in \left( - \frac{\pi}{2}, 0 \right)\]
\[ \Rightarrow \tan x < 0\]
\[ \Rightarrow - \tan x > 0 \]
\[ \Rightarrow f'(x) > 0\]
\[\text { So, f(x) is strictly increasing on } \left( - \frac{\pi}{2}, 0 \right) . \]
\[\text { Now,} \]
\[x \in \left( 0, \frac{\pi}{2} \right)\]
\[ \Rightarrow \tan x > 0\]
\[ \Rightarrow - \tan x < 0 \]
\[ \Rightarrow f'(x) < 0\]
\[\text { So, f(x)is strictly decreasing on }\left( 0, \frac{\pi}{2} \right).\]
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