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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) ? - Mathematics

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प्रश्न

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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उत्तर

\[f\left( x \right) = \log \cos x\]

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

\[ = - \tan x\]

\[\text { Now,} \]

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

\[ \Rightarrow \tan x < 0\]

\[ \Rightarrow - \tan x > 0 \]

\[ \Rightarrow f'(x) > 0\]

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

\[\text { Now,} \]

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

\[ \Rightarrow \tan x > 0\]

\[ \Rightarrow - \tan x < 0 \]

\[ \Rightarrow f'(x) < 0\]

\[\text { So, f(x)is  strictly decreasing on }\left( 0, \frac{\pi}{2} \right).\]

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अध्याय 17: Increasing and Decreasing Functions - Exercise 17.2 [पृष्ठ ३५]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 17 Increasing and Decreasing Functions
Exercise 17.2 | Q 31 | पृष्ठ ३५

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