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 Question:
The question asks us to identify the points of discontinuity of the function $f(x) = |x^3 - 3x|[x]$.

Step 2 : Key Formulas and Approach:
The function is a product of $g(x) = |x^3 - 3x|$ (which is continuous everywhere on $\mathbb{R}$) and the step function $[x]$ (which is continuous everywhere except at integer values $x = k \in \mathbb{Z}$).
We will evaluate the left-hand limit (LHL) and right-hand limit (RHL) at any integer point $x = k$ to determine whether the function is continuous at that point.

Step 3 : Detailed Explanation:
Let us test the continuity of $f(x)$ at an arbitrary integer $x = k \in \mathbb{Z}$:

• Left-Hand Limit (LHL):
As $x \to k^-$, the value of $[x] \to k - 1$. Since $g(x)$ is continuous:
\[ \lim_{x \to k^-} f(x) = |k^3 - 3k|(k - 1) \]

• Right-Hand Limit (RHL):
As $x \to k^+$, the value of $[x] \to k$. Thus:
\[ \lim_{x \to k^+} f(x) = |k^3 - 3k|k \]

• Function Value:
The value of the function at $x = k$ is:
\[ f(k) = |k^3 - 3k|k \]
For $f(x)$ to be continuous at $x = k$, we must have $\text{LHL} = \text{RHL}$:
\[ |k^3 - 3k|(k - 1) = |k^3 - 3k|k \]
\[ |k^3 - 3k| \left[ k - (k - 1) \right] = 0 \]
\[ |k^3 - 3k| = 0 \]
This equation is satisfied only when:
\[ k(k^2 - 3) = 0 \implies k = 0 \quad \text{or} \quad k = \pm\sqrt{3} \]
Since $k$ must be an integer, the only valid integer solution is $k = 0$ (as $\pm\sqrt{3}$ are irrational).
Let us verify continuity at $x = 0$:
\[ \text{LHL} = |0|(0 - 1) = 0 \]
\[ \text{RHL} = |0|(0) = 0 \]
Since $\text{LHL} = \text{RHL} = f(0) = 0$, the function is continuous at $x = 0$.
For any other integer $k \neq 0$, $|k^3 - 3k| \neq 0$, meaning $\text{LHL} \neq \text{RHL}$.
Thus, $f(x)$ is discontinuous at all non-zero integers.

Step 4 : Final Answer:
Every non-zero integer is a point of discontinuity of $f$.
This corresponds to Option (A).