Question

f(x) = |x - $\pi$| (e^|x| - 1) sin |x|
we check differentiability at x = $\pi$ & x = 0 
$\underline{at x =\pi }$
R.H.D = $\underset{h\to 0^{+}}{\lim }\frac{|\mathrm{\pi }+\mathrm{h}-\mathrm{\pi }|(e^{|\mathrm{\pi }+\mathrm{h}|}-1)\sin |\mathrm{\pi }+\mathrm{h}|-0}{h}$= 0
L.H.D = $\underset{h\to 0^{+}}{\lim }\frac{|\mathrm{\pi }-\mathrm{h}-\mathrm{\pi }|(e^{|\mathrm{\pi }-\mathrm{h}|}-1)\sin |\mathrm{\pi }-\mathrm{h}|-0}{-h}$ = 0
$\because$RHD = LHD so function is differentiable at x = $\mathrm{\pi }$
$\underline{at x = 0}$
R.H.D =  $\underset{h\to 0^{+}}{\lim }\frac{|h-\mathrm{\pi }|(e^{\left|h\right|}-1)\sin \left|h\right|-0}{h}=0$
L.H.D =   $\underset{h\to 0^{+}}{\lim }\frac{|-h-\mathrm{\pi }|(e^{|-h|}-1)\sin |-h|-0}{-h}=0$
$\because$ RHD = LHD so function is differentiable at x= 0
set S is empty set, $\varphi$