JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Continuity and Differentiability

Writing $f$ in full, $f(x)=x+1$ for $x<0$, $f(x)=1-x$ for $0\le x<1$ and $f(x)=x-1$ for $x\ge 1$. Since $g(t)=t+1$ when $t<0$ and $g(t)=1$ when $t\ge 0$, the value of $g(f(x))$ depends only on the sign of $f(x)$. For $x<-1$, $f(x)=x+1<0$, so $g(f(x))=f(x)+1=x+2$; for every $x\ge -1$, $f(x)\ge 0$, so $g(f(x))=1$. Thus $g\circ f$ equals $x+2$ for $x<-1$ and $1$ for $x\ge -1$. At $x=-1$ both pieces take the value $1$, so the composite is continuous there, while the left slope is $1$ and the right slope is $0$, so differentiability fails only at $x=-1$. Hence $g\circ f$ is continuous everywhere but not differentiable at exactly one point.