JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Continuity and Differentiability

The composition is $f(g(x))=[2g(x{)}^2+1]$. For $x<0$, $g(x)=2x-3$, so $f(g(x))=[2(2x-3{)}^2+1]$. For $x\ge 0$, $g(x)=2x+3$, so $f(g(x))=[2(2x+3{)}^2+1]$. The greatest integer function $[t]$ is discontinuous at integer values of $t$. So $f(g(x))$ is discontinuous whenever $2(2x-3{)}^2+1$ or $2(2x+3{)}^2+1$ is an integer, except at $x=0$ where we must check separately. For $x\in (-1,0)$, $2x-3\in (-5,-3)$, so $(2x-3{)}^2\in (9,25)$, and $2(2x-3{)}^2\in (18,50)$. This gives 31 integer values. For $x\in (0,1)$, $2x+3\in (3,5)$, so $(2x+3{)}^2\in (9,25)$, and $2(2x+3{)}^2\in (18,50)$. This also gives 31 integer values. Total: $31+31=62$ points of discontinuity.