Question

Let $f\left(x\right)=x{\cos }^{-1}\left(\sin \right(-|x|\left)\right),x\in (-\frac{\pi }{2},\frac{\pi }{2})$, then which of the following is true ?

• A. $f^{\prime }\left(0\right)=-\frac{\pi }{2}$
• B. $f^{\prime }$ is decreasing in $(-\frac{\pi }{2},0)$ and increasing in $(0,\frac{\pi }{2})$
• C. $f$ is not differentiable at $x=0$
• D. $f^{\prime }$ is increasing in $(-\frac{\pi }{2},0)$ and decreasing in $(0,\frac{\pi }{2})$

Answer 1

The correct option is B $f^{\prime }$ is decreasing in $(-\frac{\pi }{2},0)$ and increasing in $(0,\frac{\pi }{2})$

$f\left(x\right)=x{\cos }^{-1}\left(\sin \right(-|x|\left)\right)$
$\Rightarrow f\left(x\right)=x{\cos }^{-1}(-\sin |x|)$
$\Rightarrow f\left(x\right)=x[\pi -{\cos }^{-1}(\sin |x|\left)\right]$
$\Rightarrow f\left(x\right)=x[\pi -(\frac{\pi }{2}-{\sin }^{-1}(\sin |x|\left)\right)]$
$\Rightarrow f\left(x\right)=x(\frac{\pi }{2}+|x|)$

$\Rightarrow f\left(x\right)=\{\begin{array}{cc}x(\frac{\pi }{2}+x),& x\ge 0\\ x(\frac{\pi }{2}-x),& x<0\end{array}$

$\Rightarrow f^{\prime }\left(x\right)=\{\begin{array}{cc}\frac{\pi }{2}+2x,& x\ge 0\\ \frac{\pi }{2}-2x,& x<0\end{array}$

$\Rightarrow f^{\prime \prime }\left(x\right)=\{\begin{array}{cc}2,& x\ge 0\\ -2,& x<0\end{array}$

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