The function f(x)=[x]n−xn,n≥2 (where[x] is the greatest integer less than or equal to x), is discontinuousat all points of

$f (x) = 0 for x \in I$

If $0 < x < 1,$ then $0 < x^{n} < 1, so$

$\begin{array}{l} = 0 and [x^{n}] = 0 \\ \Rightarrow f (x) = 0 for 0 < x < 1 \end{array}$

If $1 < x < 2^{1 / n}$ then $1 < x^{n} < 2 \Rightarrow [x] = 1$

Thus $f (x) = [x]^{n} - [x^{n}] = 0 if 1 < x < 2^{1 / n}$

So $f (x) = 0 if 0 \leq x < 2^{1 / n}$

Therefore, f is continuous at $x = 1$
To the left of any integral value $m \neq 1$ but close to $m, f (x)$

$\neq 0$ but to the right of m and close to $m, f (x) = 0 .$ Hence $f$ is

discontinuous for all $m \in I ~ {1}$

The function $f (x) = [x]^{n} - [x^{n}], n \geq 2$ (where
$[x]$ is the greatest integer less than or equal to $x$ ), is discontinuous
at all points of