The function $f(x)=[x]^2-\left[x^2\right]$ (where [y] is the greatest integer less than or equal to y), is discontinuous at

all integers except 1

We know that the function [x] is discontinuous at all integer points. Therefore, $[x]^2$ is discontinuous at all integer points.

Also, $\left[x^2\right]$ is discontinuous at $x= \pm \sqrt{|n|}$, where $n \in Z$.

We observe that

$\lim\limits_{x \rightarrow 1^{-}} f(x)=\lim\limits_{h \rightarrow 0} f(1-h)=\lim\limits_{h \rightarrow 0}[1-h]^2-\left[(1-h)^2\right]$

$\Rightarrow \lim\limits_{x \rightarrow 1^{-}} f(x)=0-0=0$

$\lim\limits_{x \rightarrow 1^{+}} f(x)=\lim\limits_{h \rightarrow 0} f(1+h)=\lim\limits_{h \rightarrow 0}[1+h]^2-\left[(1+h)^2\right]$

$\Rightarrow \lim\limits_{x \rightarrow 1^{+}} f(x)=1-1=0$

and, $f(1)=[1]^2-\left[1^2\right]=0$

∴  $\lim\limits_{x \rightarrow 1^{-}} f(x)=\lim\limits_{x \rightarrow 1^{+}} f(x)=f(1)$

So, f(x) is continuous at x = 1.

f(x) is discontinuous at all other integer points.