Let f(x) = [2 x ^ 2 + 1] and g(x) = \ cases 2x - 3, & x < 0 2x + 3,...

Given,

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

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

= \left\{$\begin{array}{ll}\left[2{\left(2x - 3\right)}^{2}\right] + 1; x < 0 \\ \left[2{\left(2x + 3\right)}^{2}\right] + 1; x \ge 0\end{array}$\right)</p><p>Ifx < 0$then$- 1 < x < 0 ⇒ - 2 < x < 0$</p><p>$⇒ - 5 < 2x - 3 < - 3 ⇒ ⇒ 9 < {\left(2x - 3\right)}^{2} < 25$</p><p>$⇒ 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