f(x) = $\left[x^2\right]-{\left[x\right]}^2$ is discontinuous on 

$\begin{array}{l}f\left(0\right)=0,=\underset{x\to 0}{Lt}\left[x^2\right]-{\left[x\right]}^2\\ \underset{x\to 0^{+}}{Lt}f\left(x\right)=\underset{x\to 0^{+}}{Lt}\left[x^2\right]-{\left[x\right]}^2\\ \underset{x\to 0^{+}}{Lt}f\left(x\right)=\underset{x\to 0^{+}}{Lt}\left[x^2\right]-{\left[x\right]}^2=\underset{x\to 0^{+}}{Lt}(0-0)=0\\ \underset{x\to 0^{-}}{Lt}f\left(x\right)=\underset{x\to 0^{-}}{Lt}(\left[x^2\right]-{\left[x\right]}^2)=(0-1)=-1\\ =0-0=0\end{array}$

$\underset{x\to 1^{+}}{Lt}f\left(x\right)=\underset{x\to 1^{+}}{Lt}\left[x^2\right]-\left[x^2\right]$

$=1-1=0$

$\therefore f\left(1\right)=\left[1^2\right]-{\left[1\right]}^2=1-1=0$

$f\left(x\right)$is discontinuous at $x=0$and also discontinuous on z

$\therefore f\left(x\right)$ is continuous at $x=1$

$\therefore f\left(x\right)$  is discontinuous at all integer except 1