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 $0 \leq x<1$, then $0 \leq x^{2}<1$
$\therefore[x]=0$ and $\left[x^{2}\right]=0 \Rightarrow f(x)=0$ for $0 \leq x<1$
Next, if $1 \leq x <\sqrt{2}$, then
$1 \leq x^{2}<2 \Rightarrow[x]=1$ and $\left[x^{2}\right]=1$
Thus, $f(x)=[x]^{2}-\left[x^{2}\right]=0$ if $1 \leq x<\sqrt{2}$
It follows that $f ( x )=0$ if $0 \leq x <\sqrt{2}$
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 .