Answer:
Option A
Explanation:
We have
$f(x)=\mid x-\pi\mid.(e^{\mid x\mid}-1)\sin \mid x\mid$
$\begin{cases}\left(x-\pi\right)\left(e^{-x}-1\right)\sin x, & x < 0\\ -\left(x-\pi\right)\left(e^{x}-1\right)\sin x, & 0 \leq x <\pi \\ \left(x-\pi\right)\left(e^{x}-1\right)\sin x, & x\geq\pi\end{cases}$
We check the differentiability at x=0 and $\pi$
We have,
$\begin{cases}\left(x-\pi\right)\left(e^{-x}-1\right)\cos x +\left(e^{-x}-1\right)\sin x + \left(x-\pi\right) \sin xe^{-x}\left(-1\right), & x < 0\\ -\left[\left(x-\pi\right)\left(e^{-x}-1\right)\cos x +\left(e^{-x}-1\right)\sin x + \left(x-\pi\right) \sin xe^{-x}\right],& 0 < x <\pi \\ \left(x-\pi\right)\left(e^{-x}-1\right)\cos x +\left(e^{-x}-1\right)\sin x + \left(x-\pi\right) \sin xe^{-x}, & x>\pi\end{cases}$
Clearly, $\lim_{x \rightarrow 0^{-}}f'(x)=0=\lim_{x \rightarrow 0^{+}}f'(x)$
and $\lim_{x \rightarrow \pi^{-}}f'(x)=0=\lim_{x \rightarrow \pi^{+}}f'(x)$
$\therefore$ f is differentiable at x=0 and x= $\pi$
Hence, f is differentiable for all x
