JEE Advanced 2024 · Paper 2 · Mathematics Q4

MathematicsSingle correct+3 / -1Question 4 of 51

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)

Answer and worked solution

Official answer: D

For $x>0$ , $f(x)=0$ exactly when $\sin\frac{\pi}{x^2}=0$ , i.e. $\frac{\pi}{x^2}=n\pi$ , so $x=\frac{1}{\sqrt n}$ for $n\in\mathbb N$ .

(A) $\frac1{\sqrt n}\ge10^{-10}$ means $n\le10^{20}$ : finitely many solutions. False.
(B) $\frac1{\sqrt n}\ge\frac1\pi$ means $n\le\pi^2\approx9.87$ , giving $n=1,\dots,9$ . There are solutions, so false.
(C) $\frac1{\sqrt n}<10^{-10}$ for every $n>10^{20}$ : infinitely many solutions. False.
(D) $\frac1{\pi^2}<\frac1{\sqrt n}<\frac1\pi$ means $\pi^2<n<\pi^4\approx97.4$ , giving $n=10,\dots,97$ . That is $88$ solutions, more than $25$ . True.

JEE Advanced 2024 · Paper 2, Mathematics question 4