Let $\mathrm{S}=\left\{\mathrm{t} \in \mathrm{R} / \mathrm{f}(x)=|x-\pi|\left(\mathrm{e}^{|x|}-1\right) \sin…
Let $\mathrm{S}=\left\{\mathrm{t} \in \mathrm{R} / \mathrm{f}(x)=|x-\pi|\left(\mathrm{e}^{|x|}-1\right) \sin |x|\right.$ is not differentiable at $\mathrm{t}\}$, then $\mathrm{S}$ is
$\phi$ (an empty set)
$\{0\}$
$\{\pi\}$
$\{0, \pi\}$
Solution
Differentiability at $x=\pi$ :
L.h.lim
$=\lim _{h \rightarrow 0} \frac{|\pi-h-\pi|\left(e^{|\pi-h|}-1\right) \sin |\pi-h|-0}{-h}=0$
R.h.lim
$=\lim _{h \rightarrow 0} \frac{|\pi+h-\pi|\left(e^{|\pi+h|}-1\right) \sin |\pi+h|-0}{h}=0$
Differentiability at $\boldsymbol{x}=\mathbf{0}$ :
L.h.lim $=\lim _{\mathrm{h} \rightarrow 0} \frac{|-\mathrm{h}-\pi|\left(\mathrm{e}^{|-\mathrm{h}|}-1\right) \sin |-\mathrm{h}|-0}{-\mathrm{h}}=0$
R.h.lim $=\lim _{h \rightarrow 0} \frac{|h-\pi|\left(e^{|h|}-1\right) \sin |h|-0}{h}=0$
The function $\mathrm{f}(x)$ is differentiable at $x=0, \pi$.
$\Rightarrow$ Set $\mathrm{S}$ is an empty set.