JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Continuity and Differentiability

On each unit interval the greatest-integer value is constant, so $f(x)=-2x$ on $(-2,-1)$, $f(x)=-x$ on $[-1,0)$, $f(x)=0$ on $[0,1)$ and $f(x)=x-1$ on $[1,2)$. Each of these expressions is already non-negative, so $|f(x)|$ equals the same four expressions. As $x\to -1^{-}$, $|f|\to -2(-1)=2$, whereas $|f(-1)|=1$, so $|f|$ has a jump and is discontinuous at $x=-1$; at $x=0$ and $x=1$ the adjacent pieces meet at the value $0$, so $|f|$ is continuous there. Thus it is discontinuous at exactly one point and $m=1$. For differentiability, $|f|$ fails at the discontinuity $x=-1$, and it has corners at $x=0$ (left slope $-1$, right slope $0$) and at $x=1$ (left slope $0$, right slope $1$). Hence $n=3$, and $m+n=4$.