Question

Let $\left[t\right]$ denote the greatest integer $\le t.$ The number of points where the function $f\left(x\right)=\left[x\right]|x^2-1|+\sin \left(\frac{\pi }{\left[x\right]+3}\right)-[x+1],x\in (-2,2)$ is not continuous is

Answer 1

$f\left(x\right)=\{\begin{array}{cc}-2|x^2-1|+1& x\in (-2,-1)\\ -|x^2-1|+1& x\in [-1,0)\\ \sin \left(\frac{\pi }{3}\right)-1& x\in [0,1)\\ |x^2-1|+\frac{1}{\sqrt{2}}-2& x\in [1,2)\end{array}$

$\underset{x\to -1^{-}}{lim}f\left(x\right)=1$ and $\underset{x\to -1^{+}}{lim}f\left(x\right)=1$
Hence continuous at $x=-1$

$\underset{x\to 0^{-}}{lim}f\left(x\right)=0$ and $\underset{x\to 0^{+}}{lim}f\left(x\right)=\frac{\sqrt{3}}{2}-1$
$\Rightarrow$ discontinuous

$\underset{x\to 1^{-}}{lim}f\left(x\right)=\frac{\sqrt{3}}{2}-1$ and $\underset{x\to 1^{+}}{lim}f\left(x\right)=\frac{1}{\sqrt{2}}-2$
$\Rightarrow$ discontinuous

Hence, $2$ points of discontinuity.