Let $\left[t\right]$denote the greatest integer $\le t$ and $\underset{x\to 0}{\lim }x\left[\frac{4}{x}\right]=A.$Then the function, $f\left(x\right)=\left[x^2\right]\sin \left(\pi x\right)\text{ is discontinuous, when }x\text{ is equal to : }$

$\begin{array}{l}\frac{4}{x}-1<\left[\frac{4}{x}\right]\le \frac{4}{x}\\ x>0 \Rightarrow x\left(\frac{4}{x}-1\right)<x\left[\frac{4}{x}\right]\le x·\frac{4}{x}\\ \underset{x\to 4}{\lim }x\le \underset{x\to 4}{\lim }x\left[\frac{4}{x}\right]\le \underset{x\to 4}{\lim }4\\ \underset{x\to 4}{\lim }x\left[\frac{4}{x}\right]=4\end{array}$

$\begin{array}{l}\\ \underset{x\to 4}{\lim }x\left[\frac{4}{x}\right]=4=A\\ f\left(x\right)=\left[x^2\right]\sin \left(\pi x\right)\text{ check the }\end{array}$

$\begin{array}{l}\text{ At i) }\sqrt{A+5}=\sqrt{9}=3\\ \text{ ii) }\sqrt{A+21}=5\end{array}$

$\text{ iii) }\sqrt{A+1}=\sqrt{5}$

$\text{ iv) }\sqrt{A}=2$

when

$\begin{array}{l}\mathrm{x}=3,5,2\\ {lit}_{x+\sqrt{5}}\left[x^2\right]\sin \pi x\\ {\text{ lt }}_{x+\sqrt{5}}5\sin \pi x=5·\sin \pi \sqrt{5}\\ l_{x+\sqrt{5}}4\sin \pi x=4·\sin \pi \sqrt{5}\\ \text{ L.H. }L\ne R.H.L\\ x=\sqrt{5}\end{array}$