JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Continuity and Differentiability

Build $g$ piece by piece. For $-2<x<0$, $f(x)=-1$ so $|f(x)|=1$, while $f(|x|)=x^2-1$, giving $g=x^2$. For $0\le x<1$, $f(x)=x^2-1$ is negative, so $|f(x)|=1-x^2$ and $f(|x|)=x^2-1$ cancel to $g=0$. For $1\le x<2$, both equal $x^2-1$, so $g=2(x^2-1)$. At $x=0$ the two one-sided derivatives are both $0$, so $g$ is smooth there. At $x=1$ the left derivative is $0$ but the right derivative is $4$, so $g$ fails to be differentiable there. Hence $g$ is not differentiable at exactly one point.