Question

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

• 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 $1$

Note that $f\left(x\right)=0$ for each integral value of $x.$

Also, if $0\le x<1,$ then $0\le x^2<1$

$\therefore \left[x\right]=0$ and $\left[x^2\right]=0\Rightarrow f\left(x\right)=0$ for $0\le x<1$

Next, if $1\le x<\sqrt{2},$ then

$1\le x^2<2\Rightarrow \left[x\right]=1$ and $\left[x^2\right]=1$

Thus, $f\left(x\right)={\left[x\right]}^2-\left[x^2\right]=0$ if $1\le x<\sqrt{2}$

It follows that $f\left(x\right)=0$ if $0\le x<\sqrt{2}$

This shows that $f\left(x\right)$ must be continuous at $x=1.$

However, at points $x$ other than integers and not lying between $0$ and $\sqrt{2},f\left(x\right)\ne 0$

Thus, $f$ is discontinuous at all integers except $1.$