Let $f : \mathbb{R} \to \mathbb{R}$ be a function defined by $f(x) = \begin{cases} x^{2} \sin\left(\frac{\pi}{x^{2}}\right), & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases}$ Then which of the following statements is TRUE?

$\text{Given: } f : \mathbb{R} \to \mathbb{R},$ $f(x) = x^{2} \sin\left(\dfrac{\pi}{x^{2}}\right), \quad x \neq 0$ $f(0) = 0$ $\text{To find: which statement is TRUE.}$ $\text{(1) infinitely many in } \left[\dfrac{1}{10^{10}}, \infty\right) \qquad \text{(2) none in } \left[\dfrac{1}{\pi}, \infty\right)$ $\text{(3) finite in } \left(0, \dfrac{1}{10^{10}}\right) \qquad \text{(4) more than } 25 \text{ in } \left(\dfrac{1}{\pi^{2}}, \dfrac{1}{\pi}\right)$ $\text{Plan:}$ $\text{1. Solve } f(x)=0 \text{ for } x \neq 0.$ $\text{2. Count the solutions in each interval.}$ $x \neq 0 \Rightarrow x^{2} > 0$ $f(x)=0 \Rightarrow \sin\left(\dfrac{\pi}{x^{2}}\right)=0$ $\sin\theta = 0 \Rightarrow \theta = n\pi, \ n \in \mathbb{Z}$ $\dfrac{\pi}{x^{2}} = n\pi$ $\dfrac{1}{x^{2}} = n \Rightarrow n = 1,2,3,\dots$ $x^{2} = \dfrac{1}{n} \Rightarrow x = \pm\dfrac{1}{\sqrt{n}}$ $\text{Zeros: } x = 0, \ \pm\dfrac{1}{\sqrt{n}}, \ n = 1,2,3,\dots$ $\dfrac{1}{\sqrt{n}} \to 0 \text{ as } n \to \infty$ $f\!\left(\tfrac{1}{\sqrt{n}}\right) = \dfrac{1}{n}\sin(n\pi) = 0$ $x = \dfrac{1}{\sqrt{n}} \geq a \iff n \leq \dfrac{1}{a^{2}}$ $\text{(1) } a=\dfrac{1}{10^{10}} \Rightarrow n \leq 10^{20}$ $\text{Finite } (n = 1,\dots,10^{20}) \Rightarrow \text{FALSE}$ $\text{(2) } a=\dfrac{1}{\pi} \Rightarrow n \leq \pi^{2} \approx 9.87$ $9 \text{ solutions } \Rightarrow \text{FALSE}$ $\text{(3) } 0 < \dfrac{1}{\sqrt{n}} < \dfrac{1}{10^{10}} \Rightarrow n > 10^{20}$ $\text{Infinitely many} \Rightarrow \text{FALSE}$ $\text{(4) } \dfrac{1}{\pi^{2}} < \dfrac{1}{\sqrt{n}} < \dfrac{1}{\pi} \Rightarrow \pi^{2} < n < \pi^{4}$ $9.87 < n < 97.41 \Rightarrow n = 10,\dots,97$ $\text{Count} = 97 - 9 = 88$ $88 > 25 \Rightarrow \text{statement (4) is TRUE}$ $\boxed{\text{Answer: } (4)}$