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# Solution for 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) ? - CBSE (Commerce) Class 12 - Mathematics

ConceptIncreasing and Decreasing Functions

#### 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) ?

#### Solution

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

$f'\left( x \right) = \frac{1}{\cos x}\left( - \sin x \right)$

$= - \tan x$

$\text { Now,}$

$x \in \left( - \frac{\pi}{2}, 0 \right)$

$\Rightarrow \tan x < 0$

$\Rightarrow - \tan x > 0$

$\Rightarrow f'(x) > 0$

$\text { So,f(x)is strictly increasing on } \left( - \frac{\pi}{2}, 0 \right) .$

$\text { Now,}$

$x \in \left( 0, \frac{\pi}{2} \right)$

$\Rightarrow \tan x > 0$

$\Rightarrow - \tan x < 0$

$\Rightarrow f'(x) < 0$

$\text { So,f(x)isstrictly decreasing on }\left( 0, \frac{\pi}{2} \right).$

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Solution for 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) ? concept: Increasing and Decreasing Functions. For the courses CBSE (Commerce), CBSE (Arts), PUC Karnataka Science, CBSE (Science)
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