Continuity and Differentiability
PYP_JEE_ADV_2024_P2
Grade None
Question:
Let $f : \mathbb{R} \to \mathbb{R}$ be a function defined by
$$f(x) = \begin{cases} x^2 \sin\left(\dfrac{\pi}{x^2}\right), & \text{if } x \neq 0, \\ 0, & \text{if } x = 0. \end{cases}$$
Then which of the following statements is TRUE?
$f(x) = 0$ has infinitely many solutions in the interval $\left[ \dfrac{1}{10^{10}}, \infty \right)$.
$f(x) = 0$ has no solutions in the interval $\left[ \dfrac{1}{\pi}, \infty \right)$.
The set of solutions of $f(x) = 0$ in the interval $\left( 0, \dfrac{1}{10^{10}} \right)$ is finite.
$f(x) = 0$ has more than 25 solutions in the interval $\left( \dfrac{1}{\pi^2}, \dfrac{1}{\pi} \right)$.
Step-by-Step Solution
Key Concept: Solving $f(x)=0$ by finding the roots of $\sin(\pi/x^2)$ and using inequalities to count the number of integer solutions $m$ in the given intervals.
For $x \neq 0$, the roots of $f(x) = 0$ are given by:
$$\sin\left(\dfrac{\pi}{x^2}\right) = 0 \implies \dfrac{\pi}{x^2} = m\pi \implies x^2 = \dfrac{1}{m} \implies x_m = \dfrac{1}{\sqrt{m}} \quad \text{for } m \in \mathbb{Z}^+$$
Let's analyze the options:
- Option A: $x_m \ge 1/10^{10} \implies m \le 10^{20}$. This represents a finite number of solutions ($10^{20}$ solutions). Thus, A is false.
- Option B: $x_m \ge 1/\\pi \implies m \le \\pi^2 \approx 9.87$. The integers $m \in \{1, 2, \dots, 9\}$ satisfy this, so solutions exist. Thus, B is false.
- Option C: $x_m < 1/10^{10} \implies m > 10^{20}$. There are infinitely many integers $m > 10^{20}$, so there are infinitely many solutions. Thus, C is false.
- Option D: $\dfrac{1}{\pi^2} < x_m < \dfrac{1}{\pi} \implies \pi < \sqrt{m} < \pi^2 \implies \pi^2 < m < \pi^4$.
Since $\pi^2 \approx 9.87$ and $\pi^4 \approx 97.4$, the integer $m$ must satisfy $10 \le m \le 97$.
The number of such integers is $97 - 10 + 1 = 88$.
Since $88 > 25$, there are indeed more than 25 solutions in this interval.
Thus, the correct option is D.
Correct Answer: D