JEE Main · 2018 · 08 Apr 2018 · Offline · Continuity and Differentiability

The function is a product of continuous factors, so only the corners at $x=0$ and $x=\pi$ can threaten differentiability. Near $x=0$ the factors $(e^{|x|}-1)$ and $\sin |x|$ each vanish like $|x|$, so the product behaves like $x^2$ and the one-sided derivatives both equal $0$. Near $x=\pi$ the factor $\sin |x|$ has a simple zero that cancels the kink of $|x-\pi |$, again leaving matching one-sided derivatives of $0$. The function is differentiable at both points and everywhere else, so the set $S$ is empty.