Question

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

• A. All integers
• B. All integers except $0$ and $1$
• C. All integers except $0$
• D. All integers except $1$

Answer 1

The correct option is D All integers except $1$

We have $x\in I$

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

$f\left(x^{+}\right)=\underset{x\to 0}{lim}{[x+h]}^2-\left[{x+h}^2\right]=\underset{x\to 0}{lim}{x}^2-x^2-2hx-h^2=\underset{x\to 0}{lim}-2hx-h^2=0$

$f\left(x^{-}\right)=\underset{x\to 0}{lim}{[x-h]}^2-\left[{x-h}^2\right]$

$=\underset{x\to 0}{lim}{(x-1)}^2-x^2+2hx-h^2$ since $-2hx+h^2<0$

$={(x-1)}^2-x^2+1$

$=x^2-2x+1-x^2+1$

$=-2x+2$

$\therefore f\left(x^{-}\right)=0$ only when $x=1$

$\therefore$ At $x=0,f\left(x^{+}\right)=f\left(x^{-}\right)=f\left(x\right)=0$

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

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