Let $f\left(x\right)=\left[2x^2+1\right]$ and $g\left(x\right)=\left\{\begin{array}{ll}2x-3,& x<0\\ 2x+3,& x\ge 0\end{array}\right.$, where $\left[t\right]$ is the greatest integer $\le t$. Then, in the open interval $\left(-1,1\right)$, the number of points where fog is discontinuous is equal to ______.

Given,

$f\left(x\right)=\left[2x^2+1\right]$ and $g\left(x\right)=\left\{\begin{array}{ll}2x-3,& x<0\\ 2x+3,& x\ge 0\end{array}\right.$

Now, $f\left(g\left(x\right)\right)=\left[2g^2\left(x\right)\right]+1$

$=\left\{\begin{array}{l}\left[2{\left(2x-3\right)}^2\right]+1;x<0\\ \left[2{\left(2x+3\right)}^2\right]+1;x\ge 0\end{array}\right.$

If $x<0$ then $-1<x<0\Rightarrow -2<x<0$

$\Rightarrow -5<2x-3<-3\Rightarrow \Rightarrow 9<{\left(2x-3\right)}^2<25$

$\Rightarrow 18<2{\left(2x-3\right)}^2<50$, So there will be $31$ values where $2{\left(2x-3\right)}^2$ is integer, Similarly for $2{\left(2x+3\right)}^2$ there will be $31$, so total $62$ points of discontinuity