Question 4

To find the solutions of $f(x) = 0$ for $x \neq 0$, we have: $x^2 \sin\left(\frac{\pi}{x^2}\right) = 0$ $\sin\left(\frac{\pi}{x^2}\right) = 0 \implies \frac{\pi}{x^2} = n\pi \text{ for } n \in \mathbb{Z}^+$ $x^2 = \frac{1}{n} \implies x = \pm \frac{1}{\sqrt{n}}$ For $x > 0$, the solutions are $x_n = \frac{1}{\sqrt{n}}$.

(A) In $[\frac{1}{10^{10}}, \infty)$, we need $\frac{1}{\sqrt{n}} \ge \frac{1}{10^{10}} \implies n \le 10^{20}$. This is a finite number of solutions. Statement (A) is false.

(B) In $[\frac{1}{\pi}, \infty)$, we need $\frac{1}{\sqrt{n}} \ge \frac{1}{\pi} \implies \sqrt{n} \le \pi \implies n \le \pi^2 \approx 9.87$. So $n \in \{1, 2, \dots, 9\}$. There are 9 solutions, so (B) is false.

(C) In $(0, \frac{1}{10^{10}})$, we need $0 < \frac{1}{\sqrt{n}} < \frac{1}{10^{10}} \implies n > 10^{20}$. There are infinitely many such integers $n$. Statement (C) is false.

(D) In $(\frac{1}{\pi^2}, \frac{1}{\pi})$, we need: $\frac{1}{\pi^2} < \frac{1}{\sqrt{n}} < \frac{1}{\pi} \implies \pi < \sqrt{n} < \pi^2 \implies \pi^2 < n < \pi^4$ Since $\pi^2 \approx 9.87$ and $\pi^4 \approx 97.41$, the possible integer values for $n$ are $\{10, 11, \dots, 97\}$. Number of solutions = $97 - 10 + 1 = 88$. Since $88 > 25$, statement (D) is true.