The function $f(x) = x^2 \left[x^2 - \frac{1}{x^2}\right], x \neq 0$ is ($[x]$ represents the greatest integer $\leq x$)
continuous at $x = 1$
discontinuous at $x = -1$
discontinuous at infinitely many points
discontinuous at finite number of points
Step-by-Step Solution
Key Concept: The function is discontinuous at infinitely many isolated points determined by the floor function applied to the reciprocal of the square.
Given $f(1) = f(-1) = 1$, we establish that for $|x| > 1$, $0 1$ implies $\frac{1}{x^2} < 1$, also giving $f(x) = 0$ except at isolated points. Analyzing intervals like $\frac{1}{3} < x^2 < 1$ and $\frac{1}{2} < x^2 < \frac{1}{3}$ yields points where $[\frac{1}{x^2}] = k$ for various integers $k$, causing $f(x) = 0$. At points $x = \pm\frac{1}{\sqrt{n}}$ for $n \in \mathbb{N}$, we have $x^2 = \frac{1}{n}$, giving $f(x) = 1$. Therefore, $f(x)$ is discontinuous at the infinite set of points $\{\pm\frac{1}{\sqrt{n}} : n \in \mathbb{N}\}$.
Correct Answer: 2,3