Given,
f(x)=[2x2+1] and g(x)={\begin{matrix}2x-3, & x<0 \\ 2x+3, & x\geq 0\end{matrix}
Now, f(g(x))=[2g2(x)]+1
={\begin{matrix}[2{(2x-3)}^{2}]+1;x<0 \\ [2{(2x+3)}^{2}]+1;x\geq 0\end{matrix}
If x<0 then −1<x<0⇒−2<x<0
⇒−5<2x−3<−3⇒⇒9<(2x−3)2<25
⇒18<2(2x−3)2<50, So there will be 31 values where 2(2x−3)2 is integer, Similarly for 2(2x+3)2 there will be 31, so total 62 points of discontinuity