# then which of the following is true?

Question:

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?

1. (1) $f^{\prime}$ is increasing in $\left(-\frac{\pi}{2}, 0\right)$ and decreasing in $\left(0, \frac{\pi}{2}\right)$

2. (2) $f^{\prime}(0)=-\frac{\pi}{2}$

3. (3) $f^{\prime}$ is not differentiable at $x=0$

4. (4) $f^{\prime}$ is decreasing in $\left(-\frac{\pi}{2}, 0\right)$ and increasing in $\left(0, \frac{\pi}{2}\right)$

Correct Option: , 4

Solution:

$f^{\prime}(x)=x\left(\pi-\cos ^{-1}(\sin |x|)\right)$

$=x\left(\pi-\left(\frac{\pi}{2}-\sin ^{-1}(\sin |x|)\right)\right)=x\left(\frac{\pi}{2}+|x|\right)$

$f(x)= \begin{cases}x\left(\frac{\pi}{2}+x\right), & x \geq 0 \\ x\left(\frac{\pi}{2}-x\right), & x<0\end{cases}$

$f^{\prime}(x)= \begin{cases}\frac{\pi}{2}+2 x, & x \geq 0 \\ \frac{\pi}{2}-2 x, & x<0\end{cases}$

Hence, $f^{\prime}(x)$ is increasing in $\left(0, \frac{\pi}{2}\right)$ and decreasing

in $\left(\frac{-\pi}{2}, 0\right)$