Let $f(x)=x \cos ^{-1}(-\sin |x|), x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, then which of the following is true?
Correct Option: , 4
$f^{\prime}(x)=x\left(\pi-\cos ^{-1}(\sin |x|)\right)$
$=x\left(\pi-\left(\frac{\pi}{2}-\sin ^{-1}(\sin |x|)\right)\right)=x\left(\frac{\pi}{2}+|x|\right)$
$f(x)= \begin{cases}x\left(\frac{\pi}{2}+x\right), & x \geq 0 \\ x\left(\frac{\pi}{2}-x\right), & x<0\end{cases}$
$f^{\prime}(x)= \begin{cases}\frac{\pi}{2}+2 x, & x \geq 0 \\ \frac{\pi}{2}-2 x, & x<0\end{cases}$
Hence, $f^{\prime}(x)$ is increasing in $\left(0, \frac{\pi}{2}\right)$ and decreasing
in $\left(\frac{-\pi}{2}, 0\right)$
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