Question

The function $f(x)=[x{]}^2-[x^2]$ (where $[y]$ is the greatest integer 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

Answer: (D)

Note that $f(x)=0$ for each integral value of $x$.
Also, if $0\le x<1$, then $0\le x^2<1$
$\therefore [x]=0$ and $\left[x^2\right]=0\Rightarrow f(x)=0$ for $0\le x<1$
Next, if $1\le x<\sqrt{2}$, then
$1\le x^2<2\Rightarrow [x]=1$ and $\left[x^2\right]=1$
Thus, $f(x)=[x{]}^2-\left[x^2\right]=0$ if $1\le x<\sqrt{2}$
It follows that $f(x)=0$ if $0\le x<\sqrt{2}$
This shows that $f(x)$ must be continuous at $x=1$.
However, at points $x$ other than integers and not lying between 0 and $\sqrt{2},f(x)\equiv 0$
Thus, $f$ is discontinuous at all integers except 1 .