Number of Solutions for Piecewise Trigonometric Function

To determine which of the given statements is true, we first analyze the solutions to the equation $f(x) = 0$.

The function is given by:
$f(x) = \begin{cases} x^2 \sin\left(\frac{\pi}{x^2}\right), & \text{if } x \neq 0, \\ 0, & \text{if } x = 0. \end{cases}$

For $x \neq 0$, $f(x) = 0$ implies:
$x^2 \sin\left(\frac{\pi}{x^2}\right) = 0 \implies \sin\left(\frac{\pi}{x^2}\right) = 0$

Since $\frac{\pi}{x^2} > 0$, the solutions satisfy:
$\frac{\pi}{x^2} = k\pi \quad \text{for } k \in \{1, 2, 3, \dots\}$ $x^2 = \frac{1}{k} \implies x = \pm \frac{1}{\sqrt{k}}, \quad k \in \mathbb{N}$

For positive $x$, the solutions are given by $x_k = \frac{1}{\sqrt{k}}$ where $k \in \{1, 2, 3, \dots\}$.

Now, let's evaluate each option:

1. Option (A):
   We consider $x_k \in \left[\frac{1}{10^{10}}, \infty\right)$: $\frac{1}{\sqrt{k}} \ge \frac{1}{10^{10}} \implies \sqrt{k} \le 10^{10} \implies k \le 10^{20}$ Thus, $k \in \{1, 2, 3, \dots, 10^{20}\}$. This gives exactly $10^{20}$ solutions, which is a finite number of solutions. Therefore, statement (A) is FALSE.

2. Option (B):
   We consider $x_k \in \left[\frac{1}{\pi}, \infty\right)$: $\frac{1}{\sqrt{k}} \ge \frac{1}{\pi} \implies \sqrt{k} \le \pi \implies k \le \pi^2 \approx 9.87$ Thus, $k \in \{1, 2, 3, \dots, 9\}$. This interval contains $9$ solutions (for example, $x_1 = 1 \ge \frac{1}{\pi}$). Therefore, statement (B) is FALSE.

3. Option (C):
   We consider $x_k \in \left(0, \frac{1}{10^{10}}\right)$: $0 < \frac{1}{\sqrt{k}} < \frac{1}{10^{10}} \implies \sqrt{k} > 10^{10} \implies k > 10^{20}$ The integers satisfying $k > 10^{20}$ form an infinite set, so there are infinitely many solutions in this interval. Therefore, statement (C) is FALSE.

4. Option (D):
   We consider $x_k \in \left(\frac{1}{\pi^2}, \frac{1}{\pi}\right)$: $\frac{1}{\pi^2} < \frac{1}{\sqrt{k}} < \frac{1}{\pi} \implies \pi < \sqrt{k} < \pi^2 \implies \pi^2 < k < \pi^4$ Using the approximation $\pi \approx 3.14159$: $\pi^2 \approx 9.87 \quad \text{and} \quad \pi^4 \approx 97.41$ Since $k$ must be an integer, we have: $k \in \{10, 11, 12, \dots, 97\}$ The number of solutions is: $97 - 10 + 1 = 88$ Since $88 > 25$, $f(x) = 0$ has more than $25$ solutions in the interval $\left(\frac{1}{\pi^2}, \frac{1}{\pi}\right)$. Therefore, statement (D) is TRUE.

Thus, the correct option is D.