Question

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

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

Answer 1

The correct option is B

$\mathrm{f}\text{'}$is decreasing in $\left(\frac{-\mathrm{\pi }}{2},0\right)$and increasing in $\left(0,\frac{\mathrm{\pi }}{2}\right)$

Explanation for the correct option.

Given, $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{xcos}}^{-1}\left(\sin \right(-\left|\mathrm{x}\right|\left)\right)$

Since,${\cos }^{-1}(-\mathrm{x})=\mathrm{\pi }-{\cos }^{-1}\left(\mathrm{x}\right)$

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

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

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

$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}\left[\frac{\mathrm{\pi }}{2}+\left|\mathrm{x}\right|\right]$

$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}\left[\frac{\mathrm{\pi }}{2}\pm \mathrm{x}\right]$

$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}\frac{\mathrm{\pi }}{2}\pm {\mathrm{x}}^{2}0\le 0<0$

$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{\pi }}{2}\pm 2x0\le 0<0$

$\therefore \mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}\mathrm{x}\left(\frac{\mathrm{\pi }}{2}+\mathrm{x}\right)\frac{\mathrm{\pi }}{2}>\mathrm{x}\ge 0\\ \mathrm{x}\left(\frac{\mathrm{\pi }}{2}-\mathrm{x}\right)\frac{\mathrm{\pi }}{2}<\mathrm{x}<0\end{array}\right.$

And

$$\begin{gathered} {\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)=\left(\frac{\mathrm{\pi }}{2}+2\mathrm{x}\right)\frac{\mathrm{\pi }}{2}>\mathrm{x}\ge 0 \\ =\left(\frac{\mathrm{\pi }}{2}-2\mathrm{x}\right)-\frac{\mathrm{\pi }}{2}\le \mathrm{x}<0 \end{gathered}$$

Therefore,

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

Hence, the correct option is (B).