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

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

$\therefore$fog is discontinuous whenever $2(2x-3{)}^2$ or

$2(2x+3{)}^2$ belongs to integer except x=0.

$$\begin{gathered} if -1<x<0\Rightarrow f\left(g\left(x\right)\right)\in \left(19,51\right) , \\ \therefore number of integers=31 \\ if 0<x<1\Rightarrow f\left(g\left(x\right)\right)\in \left(19,51\right) , \\ \therefore number of integers=31 \end{gathered}$$

$\therefore$62 points of discontinuity.