JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Continuity and Differentiability

Differentiability need only be tested where the definition changes. At $x=\pi$ the two pieces join smoothly, so $f$ is differentiable there. At $x=0$ the right-hand derivative is $f^{\prime }(0^{+})=\cos 0-1+2\cos 0=2$ and the left-hand derivative is $f^{\prime }(0^{-})=-\cos 0+1+2\cos 0=2$; since they agree, $f$ is differentiable at $x=0$ as well. The set of points of non-differentiability is therefore empty.