$f (x) = (cos x)^{cos x}$ is having local minimum value at $x = k$ , then the value of $k$ is equal to $(w h e r e - \frac{π}{2} < x < \frac{π}{2})$

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{cos}^{- 1} (\begin{matrix} \frac{1}{e} \end{matrix})

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{cos}^{- 1} (\begin{matrix} \frac{1}{e^{2}} \end{matrix})

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Solution

The correct option is D ${cos}^{- 1} (\begin{matrix} \frac{1}{e} \end{matrix})$
$ln f (x) = cos x ln cos x$
$f^{'} (x) = (cos x)^{cos x} (- sin x ln cos x - sin x) = 0$ for max. or min.
$\Rightarrow - sin x (ln cos x + 1) = 0$
$x = 0 & x = {cos}^{- 1} (\begin{matrix} \frac{1}{e} \end{matrix})$
Using number line, local minimum at $x = {cos}^{- 1} (\begin{matrix} \frac{1}{e} \end{matrix})$

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