Consider the function $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ defined by $f(x)=\left\{\begin{array}{cc}\left(2-\sin \left(\frac{1}{x}\right)\right)|x|, x \neq 0 \\ 0 & , x=0\end{array}\right.$. Then $f$ is :
Correct Option: , 2
$f(x)=\left\{\begin{array}{cc}-x\left(2-\sin \left(\frac{1}{x}\right)\right) & x<0 \\ 0 & x=0 \\ x\left(2-\sin \left(\frac{1}{x}\right)\right) & \end{array}\right.$
$f^{\prime}(x)= \begin{cases}-\left(2-\sin \frac{1}{x}\right)-x\left(-\cos \frac{1}{x} \cdot\left(-\frac{1}{x^{2}}\right)\right) & x<0 \\ \left(2-\sin \frac{1}{x}\right)+x\left(-\cos \frac{1}{x}\left(-\frac{1}{x^{2}}\right)\right) & x>0\end{cases}$
$f^{\prime}(x)= \begin{cases}-2+\sin \frac{1}{x}-\frac{1}{x} \cos \frac{1}{x} & x<0 \\ 2-\sin \frac{1}{x}+\frac{1}{x} \cos \frac{1}{x} & x>0\end{cases}$
$f^{\prime}(x)$ is an oscillating function which is non-monotonic in $(-\infty, 0) \cup(0, \infty)$.