Question

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

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

Answer 1

The Correct Option is D

Step 1: Given Equation

We are given the equation: \[ x^2 \sin \left( \frac{\pi}{x^2} \right) = 0. \] 

Step 2: Solving for \( x \)

Since \( x^2 \neq 0 \), we set: \[ \sin \left( \frac{\pi}{x^2} \right) = 0. \] This implies: \[ \frac{\pi}{x^2} = n\pi, \quad \text{for some integer } n > 0. \] Solving for \( x \): \[ x = \frac{1}{\sqrt{n}}, \quad (n > 0). \]

Step 3: Applying Interval Constraints

Given \( x \in \left( \frac{1}{\pi^2}, \frac{1}{\pi} \right) \), we write: \[ \frac{1}{\pi^2} < \frac{1}{\sqrt{n}} \leq \frac{1}{\pi}. \] Squaring all sides: \[ \frac{1}{\pi^4} < \frac{1}{n} \leq \frac{1}{\pi^2}. \] Taking reciprocals: \[ \pi^4 > n \geq \pi^2. \]

Step 4: Approximating Values

Approximating \( \pi^4 \approx 97 \) and \( \pi^2 \approx 9 \), we find: \[ n = 9, 10, 11, \dots, 97. \] The total number of valid \( n \) values is: \[ 97 - 9 + 1 = 89. \]

Final Conclusion

Since there are **more than 25 solutions**, we confirm that the function has at least **89** solutions in the given interval.