Q126 / 200 · Manipal (MET) 2011 · Mathematics
If $f(x)=|x|^{|\sin x|}$, then $f^{\prime}\left(-\frac{\pi}{4}\right)$ is equal to
- Correct answerA$\left(\frac{\pi}{4}\right)^{1 / \sqrt{2}}\left(\frac{\sqrt{2}}{2} \log \frac{4}{\pi}-\frac{2 \sqrt{2}}{\pi}\right)$
- B$\left(\frac{\pi}{4}\right)^{1 / \sqrt{2}}\left(\frac{\sqrt{2}}{2} \log \frac{4}{\pi}+\frac{2 \sqrt{2}}{\pi}\right)$
- C$\left(\frac{\pi}{4}\right)^{1 / \sqrt{2}}\left(\frac{\sqrt{2}}{2} \log \frac{\pi}{4}-\frac{2 \sqrt{2}}{\pi}\right)$
- D$\left(\frac{\sqrt{2}}{2} \log \frac{\pi}{4}+\frac{2 \sqrt{2}}{\pi}\right)$
Solution
Since, $f(x)=|x|^{\sin x}$
when $x \lt 0$
$\begin{aligned} f(x) & =(-x)^{(-\sin x)} \\ & =e^{-\sin x \log (-x)} \\ \Rightarrow \quad f^{\prime}(x) & =e^{-\sin x \log (-x)}\end{aligned}$
$\left(-\cos x \log (-x)-\frac{\sin x}{x}\right)$
$\Rightarrow \quad f^{\prime}(x)=(-x)^{-\sin x}(-\cos x \cdot \log (-x)-\frac{\sin x}{x})$
$\Rightarrow \quad f^{\prime}\left(-\frac{\pi}{4}\right)=\left(\frac{\pi}{4}\right)^{1 / \sqrt{2}}\left(-\frac{1}{\sqrt{2}} \log \frac{\pi}{4}+\frac{4}{\pi}\left(-\frac{1}{\sqrt{2}}\right)\right)$ $\begin{aligned} & \quad \\ & =\left(\frac{\pi}{4}\right)^{1 / \sqrt{2}}\left(\frac{\sqrt{2}}{2} \log \frac{4}{\pi}-\frac{2 \sqrt{2}}{\pi}\right)\end{aligned}$
$\Rightarrow \quad f^{\prime}\left(-\frac{\pi}{4}\right)=\left(\frac{\pi}{4}\right)^{1 / \sqrt{2}}\left(-\frac{1}{\sqrt{2}} \log \frac{\pi}{4}+\frac{4}{\pi}\left(-\frac{1}{\sqrt{2}}\right)\right)$ $\begin{aligned} & \quad \\ & =\left(\frac{\pi}{4}\right)^{1 / \sqrt{2}}\left(\frac{\sqrt{2}}{2} \log \frac{4}{\pi}-\frac{2 \sqrt{2}}{\pi}\right)\end{aligned}$
Hence, the correct answer is option: 'A'.