JEE Advanced · 2016 · 22 May 2016 · Paper 2 · Continuity and Differentiability

The function $f(x)=[x^2-3]$ is discontinuous where $x^2-3$ is an integer. In $[-1/2,2]$, $x^2\in [0,4]$, so $x^2-3\in [-3,1]$. The function jumps at $x^2=1,2,3$, i.e., at $x=1,\sqrt{2},\sqrt{3}$, and also at the boundary $x=2$ where $x^2-3=1$. So $f$ is discontinuous at exactly four points: $1,\sqrt{2},\sqrt{3},2$, confirming (B). For $g(x)=f(x)[|x|+|4x-7|]$, note that $f(x)=0$ for $x\in [\sqrt{3},2)$. The function $g$ is non-differentiable where $f$ is discontinuous (at $x=1,\sqrt{2},\sqrt{3}$) and where $|x|$ or $|4x-7|$ are non-differentiable (at $x=0,7/4$). However, since $f(x)=0$ for $x\in [\sqrt{3},2)$, the point $x=7/4$ does not contribute. Thus $g$ is non-differentiable at exactly four points: $0,1,\sqrt{2},\sqrt{3}$, confirming (C).