Question

Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be defined as $f(x) = |x^3 - 3x|[x]$, where $[x]$ denotes the greatest integer less than or equal to $x$. Which one of the following statements is TRUE?

• A. Every non-zero integer is a point of discontinuity of $f$
• B. $f$ is continuous at every real number
• C. Every integer is a point of discontinuity of $f$
• D. $f$ is continuous at every real number except for $0, \pm\sqrt{3}$

Hint:

A product of a continuous function $g(x)$ and a step function $h(x)$ (discontinuous at $x=k$) is continuous at $x=k$ if and only if $g(k) = 0$.
Here, $x^3 - 3x = 0$ at integer $x = 0$, making it the only integer where the discontinuity of $[x]$ is resolved.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The greatest integer function \([x]\) is discontinuous at every integer \(n\). For the product \(g(x)[x]\) to be continuous at \(x=n\), we generally require the continuous part \(g(x)\) to be zero at that integer.
Step 2: Checking Continuity at \(x=n\):
Let \(g(x) = |x^3 - 3x|\). The function \(f(x) = g(x)[x]\) is continuous at \(x=n\) if:
\(\lim_{x \to n^-} g(x)[x] = \lim_{x \to n^+} g(x)[x] = f(n)\)
\(g(n)(n-1) = g(n)(n) = g(n)n\)
This equality holds only if \(g(n) = 0\).
Step 3: Solving \(g(n) = 0\):
\(|x^3 - 3x| = 0 \implies x(x^2 - 3) = 0 \implies x = 0, \pm\sqrt{3}\).
Since \(\pm\sqrt{3}\) are not integers, the only integer where \(g(x)\) is zero is \(x = 0\).
Step 4: Final Answer:
At all other integers (\(n \neq 0\)), the function is discontinuous. Thus, every non-zero integer is a point of discontinuity.