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JEE Advance - Mathematics (2024 - Paper 2 Online - No. 4)

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function defined by

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

Then which of the following statements is TRUE?

$f(x)=0$ has infinitely many solutions in the interval $\left[\frac{1}{10^{10}}, \infty\right)$.
$f(x)=0$ has no solutions in the interval $\left[\frac{1}{\pi}, \infty\right)$.
The set of solutions of $f(x)=0$ in the interval $\left(0, \frac{1}{10^{10}}\right)$ is finite.
$f(x)=0$ has more than 25 solutions in the interval $\left(\frac{1}{\pi^2}, \frac{1}{\pi}\right)$.

Explanation

Option-A : $$\mathrm{f}(\mathrm{x})=\mathrm{x}^2 \sin \frac{\pi}{\mathrm{x}^2}$$

$$\mathrm{f}(\mathrm{x})=0 \Rightarrow \sin \frac{\pi}{\mathrm{x}^2}=0 \Rightarrow \frac{\pi}{\mathrm{x}^2}=\mathrm{n} \pi, \mathrm{n} \in \mathrm{N}$$

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

$$\frac{1}{\sqrt{\mathrm{n}}} \geq \frac{1}{10^{10}} \Rightarrow 10^{10} \geq \sqrt{\mathrm{n}} \Rightarrow \mathrm{n} \leq 10^{20}$$, finite number of solutions

Option-B : $$\mathrm{x}=\frac{1}{\sqrt{\mathrm{n}}} \Rightarrow \frac{1}{\sqrt{\mathrm{n}}}>\frac{1}{\pi} \Rightarrow \pi>\sqrt{\mathrm{n}} \Rightarrow \mathrm{n}<\pi^2$$, Number of solutions is 9

Option-C : $$\mathrm{x}=\frac{1}{\sqrt{\mathrm{n}}}, \frac{1}{\sqrt{\mathrm{n}}}<\frac{1}{10^{10}} \Rightarrow \sqrt{\mathrm{n}}>10^{10} \Rightarrow \mathrm{n}>10^{20}$$, Infinite number of solutions

Option-D : $$\frac{1}{\pi^2}<\frac{1}{\sqrt{\mathrm{n}}}<\frac{1}{\pi} \Rightarrow \sqrt{\mathrm{n}} \in\left(\pi, \pi^2\right) \Rightarrow \mathrm{n} \in\left(\pi^2, \pi^4\right)$$, Definitely more than 25 solutions

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