Mathematics - Continuity - Differentiability Question with Solution | TestHub

As $[.]$ is discontinuous at integers so we will check at $x=-1, 0, 1$ only 

at $x=-1$ 

$[-1^{-}]=-2$ and $[-1^{+}]=-1$

we have 

$\mathrm{LHL}=(-2)·\left(0\right)+\sin \left(\mathrm{\pi }\right)+1=1$

$\mathrm{RHL}=(-1)·\left(0\right)+\sin \left(\frac{\pi }{2}\right)-0=1$

$f(-1)=1\Rightarrow$ Continuous at $x=-1$

Again at $x=0$

$\left[0^{-}\right]=-1 \& \left[0^{+}\right]=0$

$\mathrm{LHL}=(-1)·\left(1\right)+\sin \left(\frac{\pi }{2}\right)-0=0$

$\mathrm{RHL}=0+\sin \left(\frac{\pi }{3}\right)-1\ne \mathrm{LHL}$

Hence, discontinuous at $x = 0$

Again at $x=1$, 

$\left[1^{-}\right]=0 \& \left[1^{+}\right]=1$

We have

$\mathrm{LHL}=0+\sin \left(\frac{\pi }{3}\right)-1$

$\mathrm{RHL}=0+\sin \left(\frac{\pi }{4}\right)-2\ne \mathrm{LHL}$ 

Hence, discontinuous at $x=1$

Hence, discontinuous at exactly $2$ points