JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Continuity and Differentiability

For $x\in (-1,1)$: $[|x|]=0$, so $f(x)=\max \{2x,0\}$.

This gives $f(x)=\{\begin{array}{ll}0& ,x\in [-1,0]\\ 2x& ,x\in (0,1)\end{array}$

At $x=-1$: $f(-1^{-})=1$, $f(-1)=1$, $f(-1^{+})=0$ → discontinuous

At $x=0$: $f(0^{-})=0$, $f(0)=0$, $f(0^{+})=0$ → continuous but not differentiable (corner)

At $x=1$: $f(1^{-})=2$, $f(1)=1$ → discontinuous

Therefore, $f$ is discontinuous at 2 points ($x=-1,1$) and not differentiable at 3 points ($x=-1,0,1$).

$(m,n)=(2,3)$