JEE Main 2024 · MathsMultiple choiceSingle correctMediumCalculation

JEE Main 6 April 2024, Shift 1, Maths Q8

Question 8 of 90 in this shift, Maths question 8 of 30, Section A.

f(x)=\begin{cases} x^3\sin\left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x = 0 \end{cases}

, then

(1) $f''(0)=0$
(2) $f''\left(\frac{2}{\pi}\right)=\frac{24-\pi^2}{2\pi}$ Official answer
(3) $f''(0)=1$
(4) $f''\left(\frac{2}{\pi}\right)=\frac{12-\pi^2}{2\pi}$

Official answer

Option 2

NTA final key.

Chapter: Continuity and Differentiability
Topic: Continuity and Differentiability
Idea tested: Existence of Higher Order Derivatives

Same idea in other shifts

Asked 3× in all

24 Jun 2022, Shift 2 · Q17If $y = \tan^{-1}\left(\sec x^3 - \tan x^3\right)$ , $\frac{\pi}{2} < x^3 < \frac{3\pi}{2}$ , thenHardSingle correct
5 Sep 2020, Shift 1 · Q59If the function $f(x)=\begin{cases}k_1(x-\pi)^2-1, & x\le\pi\\ k_2\cos x, & x>\pi\end{cases}$ is twice differentiable, then the ordered…MediumSingle correct

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.