JEE Advanced 2024 Paper 2 Q04 Mathematics Differentiation & Applications Limits & Continuity Hard

JEE Advanced 2024 Paper 2 · Q04 · Limits & Continuity

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

  1. A. $f(x) = 0$ has infinitely many solutions in the interval $\left[\dfrac{1}{10^{10}},\ \infty\right)$.
  2. B. $f(x) = 0$ has no solutions in the interval $\left[\dfrac{1}{\pi},\ \infty\right)$.
  3. C. The set of solutions of $f(x) = 0$ in the interval $\left(0,\ \dfrac{1}{10^{10}}\right)$ is finite.
  4. D. $f(x) = 0$ has more than $25$ solutions in the interval $\left(\dfrac{1}{\pi^{2}},\ \dfrac{1}{\pi}\right)$.
Reveal answer + step-by-step solution

Correct answer:D

Solution

$f(x) = 0$ for $x \ne 0$ requires $\sin(\pi/x^{2}) = 0$, i.e. $\pi/x^{2} = n\pi$ for $n \in \mathbb{N}$, giving $x = 1/\sqrt{n}$. (A) Solutions of the form $1/\sqrt{n}$ accumulate near $0$, so only finitely many lie in $[1/10^{10}, \infty)$ — FALSE. (B) $x = 1/\sqrt{1} = 1 \in [1/\pi, \infty)$, so solutions exist — FALSE. (C) Infinitely many $1/\sqrt{n}$ lie in $(0, 1/10^{10})$ (take $n > 10^{20}$), so the set is infinite — FALSE. (D) Need $1/\pi^{2} < 1/\sqrt{n} < 1/\pi$, i.e. $\pi^{2} < n < \pi^{4}$, i.e. roughly $9.87 < n < 97.4$, giving $n \in \{10, 11, \ldots, 97\}$, which is $88$ values — well over $25$. TRUE.

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