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?

The given function is

$$f(x)=\begin{cases}x^{2}\sin\!\left(\dfrac{\pi}{x^{2}}\right), & x\ne 0,\\[4pt] 0, & x=0.\end{cases}$$

For $$x\ne 0$$ the factor $$x^{2}$$ is never zero, so the zeros of $$f(x)$$ (apart from $$x=0$$) come only from the sine term:

$$\sin\!\left(\dfrac{\pi}{x^{2}}\right)=0 \;\Longrightarrow\; \dfrac{\pi}{x^{2}}=n\pi,\; n\in\mathbb{Z}.$$ Cancelling $$\pi$$ (and discarding $$n=0$$ because it would give the undefined value $$x=\infty$$) we get

$$\dfrac{1}{x^{2}}=n,\; n=1,2,3,\dots$$ $$\Longrightarrow\; x=\pm\dfrac{1}{\sqrt{n}}.$$

Therefore, for positive $$x$$ the zeros are exactly

$$x_{n}=\dfrac{1}{\sqrt{n}},\quad n=1,2,3,\dots$$

We now test each option.

Option A: $$f(x)=0$$ has infinitely many solutions in $$\Bigl[\dfrac{1}{10^{10}},\infty\Bigr).$$
The condition $$x_{n}\ge\dfrac{1}{10^{10}}$$ gives $$\dfrac{1}{\sqrt{n}}\ge\dfrac{1}{10^{10}} \;\Longrightarrow\; \sqrt{n}\le 10^{10} \;\Longrightarrow\; n\le 10^{20}.$$ There are only $$10^{20}$$ such integers, a (very) large but finite number. Hence Option A is false.

Option B: $$f(x)=0$$ has no solutions in $$\Bigl[\dfrac{1}{\pi},\infty\Bigr).$$
Require $$x_{n}\ge\dfrac{1}{\pi} \;\Longrightarrow\; \dfrac{1}{\sqrt{n}}\ge\dfrac{1}{\pi} \;\Longrightarrow\; \sqrt{n}\le\pi \;\Longrightarrow\; n\le\pi^{2}\approx 9.87.$$ The integers $$n=1,2,\dots,9$$ all satisfy this, so there are nine zeros in the stated interval. Hence Option B is false.

Option C: The set of solutions of $$f(x)=0$$ in $$\Bigl(0,\dfrac{1}{10^{10}}\Bigr)$$ is finite.
Here we need $$x_{n}\lt\dfrac{1}{10^{10}} \;\Longrightarrow\; \dfrac{1}{\sqrt{n}}\lt\dfrac{1}{10^{10}} \;\Longrightarrow\; \sqrt{n}\gt 10^{10} \;\Longrightarrow\; n\gt 10^{20}.$$ All integers $$n=10^{20}+1,10^{20}+2,\dots$$ work, and there are infinitely many of them. So Option C is false.

Option D: $$f(x)=0$$ has more than 25 solutions in $$\Bigl(\dfrac{1}{\pi^{2}},\dfrac{1}{\pi}\Bigr).$$
The inequalities $$\dfrac{1}{\pi^{2}}\lt x_{n}\lt\dfrac{1}{\pi}$$ give $$\dfrac{1}{\pi^{2}}\lt\dfrac{1}{\sqrt{n}}\lt\dfrac{1}{\pi} \;\Longrightarrow\; \pi\lt\sqrt{n}\lt\pi^{2} \;\Longrightarrow\; \pi^{2}\lt n\lt\pi^{4}.$$ Numerically, $$\pi^{2}\approx 9.87,\qquad \pi^{4}\approx 97.41.$$ Thus $$n=10,11,\dots,97$$ are all admissible—there are $$97-10+1=88$$ such integers, far exceeding 25. Therefore Option D is true.

Hence the only correct statement is:
Option D which is: $$f(x)=0$$ has more than 25 solutions in the interval $$\Bigl(\dfrac{1}{\pi^{2}},\dfrac{1}{\pi}\Bigr).$$