JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Continuity and Differentiability

For $x\le 0$: $f(x)=-x+1$ and $g(x)=x^2+1$. We have $f(x)=g(x)$ when $-x+1=x^2+1$, giving $x^2+x=0$, so $x=0$ or $x=-1$. For $x<-1$, $f(x)>g(x)$; for $-1<x<0$, $g(x)>f(x)$. So $h(x)=f(x)$ for $x\le -1$ and $h(x)=g(x)$ for $-1\le x\le 0$. For $x>0$: $f(x)=x+1$ and $g(x)=x^2+1$. We have $f(x)=g(x)$ when $x+1=x^2+1$, giving $x=x^2$, so $x=0$ or $x=1$. For $0<x<1$, $f(x)<g(x)$; for $x>1$, $g(x)<f(x)$. So $h(x)=f(x)$ for $0<x\le 1$ and $h(x)=g(x)$ for $x\ge 1$. The function $h$ has potential non-differentiability at $x=-1,0,1$. At each of these points, the left and right derivatives differ, so $h$ is not differentiable at 3 points.