The function $\mathrm{f}\left(\mathrm{x}\right)=[\mathrm{x}{]}^2-\left[{\mathrm{x}}^2\right]$
(Where [.] is the greatest integer function) is discontinuous at

Clearly, f (x) = 0 for each integral value of x .
Also, if $0<\mathrm{x}<1\text{, then }0<{\mathrm{x}}^2<1$
$\begin{array}{l}\Rightarrow \left[\mathrm{x}\right]=0\text{ and }\left[{\mathrm{x}}^2\right]=0\\ \therefore \mathrm{f}\left(\mathrm{x}\right)=0\text{ for }0<\mathrm{x}<1\end{array}$
Again, if $1\le \mathrm{x}<\sqrt{2}\text{ then }1\le {\mathrm{x}}^2<2$
$\begin{array}{l}\Rightarrow \left[\mathrm{x}\right]=1,\left[{\mathrm{x}}^2\right]=1\\ \therefore \mathrm{f}\left(\mathrm{x}\right)=0\text{ for }1\le \mathrm{x}<\sqrt{2}\\ \therefore \mathrm{f}\left(\mathrm{x}\right)=0\text{ is continuous at }\mathrm{x}=1\end{array}$
However, at points x other than integers and not lying between
0 and $\sqrt{2},\mathrm{f}\left(\mathrm{x}\right)\ne 0$