$\text{Let}f:R\to R\text{be a function defined by}$

$\text{<br>}$

$f\left(x\right)=\left\{\begin{array}{ll}{x}^2\sin \left(\frac{\pi }{x^2}\right)& \text{; if}x\ne 0\\ 0& \text{; if}x=0\end{array}\right.$

$\text{Then which of the following statements is TRUE?}$

The correct statement is that $f\left(x\right)=0$ has more than  25 solutions in the interval $(\frac{1}{{\pi }^2},\frac{1}{\pi })$.

Step-by-Step Explanation:

Step 1: Find the general condition for $f\left(x\right)=0$
For non-zero values of $x$, the function is defined as:

$f\left(x\right)=x^2\sin \left(\frac{\pi }{x^2}\right)$

Setting $f\left(x\right)=0$ for $x\ne 0$:

$x^2\sin \left(\frac{\pi }{x^2}\right)=0\Rightarrow \sin \left(\frac{\pi }{x^2}\right)=0$

Since $\sin ()=0$ when $=k\pi$ for any integer $k$:

Canceling $\pi$ from both sides:

$\frac{1}{x^2}=k\Rightarrow x^2=\frac{1}{k}\Rightarrow x=\frac{1}{\sqrt{k}}$

Step 2: Determine the number of solutions in the interval $(\frac{1}{{\pi }^2},\frac{1}{\pi })$
We need to find how many positive integer values of $k$ satisfy:

$\frac{1}{{\pi }^2}<\frac{1}{\sqrt{k}}<\frac{1}{\pi }$

Taking reciprocals reverses the inequality signs:

$\pi <\sqrt{k}<{\pi }^2$

Squaring all parts:

${\pi }^2<k<{\pi }^4$

Step 3: Calculate the numerical range for $k$
Using the approximation $\pi \approx 3.14159$:

${\pi }^2\approx 9.87$

${\pi }^4={\left({\pi }^2\right)}^2\approx {\left(9.87\right)}^2\approx 97.41$

Therefore, the inequality becomes:

$9.87<k<97.41$

The integers $k$ in this range are $10,11,12,\dots ,97$.

The total count of these integers is:

$97-10+1=88$

Since $88>25$, there are indeed more than 25 solutions in the interval $(\frac{1}{{\pi }^2},\frac{1}{\pi })$.