The number of points where the function  $f\left(x\right)=\{\begin{array}{l}|2x^2-3x-7|,x\le -1\\ [4x^2-1],-1<x<1\\ |x+1|+|x-2|,x\ge 1\end{array}$
(Where  $\left[t\right]$  denotes the greatest integer $\le t)$ , is discontinuous is___

 $f(-1^{-})=2,f(-1^{+})=2,f(-1)=2$
So continous at  $x=-1$
$[4x^2-1]$  is discontinuous at  $x=\pm \frac{\sqrt{3}}{2},\pm \frac{1}{\sqrt{2}},\pm \frac{1}{2}$
Also  $f\left(1^{-}\right)=2,f\left(1^{+}\right)=3$
So discontinuous  at x=1
So points of discontinuity are
 $x=\pm \frac{\sqrt{3}}{2},\pm \frac{1}{\sqrt{3}},\pm \frac{1}{2},1$