$If f (x) = | x |^{| \sin x |} then f^{1} (- \frac{π}{4}) =$

Moderate

A

${(\frac{π}{5})}^{\frac{1}{\sqrt{2}}} (\frac{\sqrt{2}}{3} \log \frac{5}{π} - \frac{2 \sqrt{2}}{π})$
B

${(\frac{π}{3})}^{\frac{1}{\sqrt{2}}} (\frac{\sqrt{2}}{2} \log \frac{3}{π} - \frac{3 \sqrt{3}}{π})$
C

${(\frac{π}{4})}^{\frac{1}{\sqrt{2}}} (\frac{\sqrt{2}}{2} \log \frac{4}{π} - \frac{2 \sqrt{2}}{π})$
D

None of these

Solution

$f (x) = {(- x)}^{- \sin x} = e^{- \sin x \log (- x)}$
$\begin{array}{l} f^{'} (x) = e^{- \sin x \log (- x)} [- \cos x \log (- x) + \frac{\sin x}{x}] \\ f^{1} (- \frac{π}{4}) = {(\frac{π}{4})}^{\frac{1}{\sqrt{2}}} [\frac{- 1}{\sqrt{2}} \log \frac{π}{4} + \frac{4}{π} \times (\frac{- 1}{\sqrt{2}})] \\ = {(\frac{π}{4})}^{\frac{1}{\sqrt{2}}} [\frac{\sqrt{2}}{2} \log \frac{4}{π} - \frac{2 \sqrt{2}}{π}] \end{array}$

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