Question

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?

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

Hint:

To solve trigonometric equations, express the argument in terms of known solutions, like \( n\pi \) for \(\sin(x) = 0\), and analyze the constraints on \( n \) based on the interval.

Answer 1

The Correct Option is D

We are given: \[ x^2 \sin\left(\frac{\pi}{x^2}\right) = 0. \] This implies: \[ \sin\left(\frac{\pi}{x^2}\right) = 0 \quad \text{(for \( x \neq 0 \))}. \] This is satisfied when: \[ \frac{\pi}{x^2} = n\pi, \quad \text{so } x = \frac{1}{\sqrt{n}}, \quad (n>0). \] If \( x \in \left(\frac{1}{\pi^2}, \frac{1}{\pi}\right] \), then: \[ \frac{1}{\pi^2}<\frac{1}{\sqrt{n}} \leq \frac{1}{\pi}. \] Squaring all sides: \[ \frac{1}{\pi^4}<\frac{1}{n} \leq \frac{1}{\pi^2}. \] Taking reciprocals: \[ \pi^4>n \geq \pi^2. \] Approximating \( \pi^4 \approx 97 \) and \( \pi^2 \approx 9 \), we find: \[ n = 9, 10, 11, \ldots, 97. \] Thus, there are more than 25 solutions for \( f(x) = 0 \) in the interval \( \left(\frac{1}{\pi^2}, \frac{1}{\pi}\right]. \)