JEE Main 2024 · MathsNumerical answerNumerical valueHardMulti-step

JEE Main 31 January 2024, Shift 2, Maths Q25

Question 25 of 90 in this shift, Maths question 25 of 30, Section B.

\lim_{x\to0}\frac{ax^2e^x-b\log_e(1+x)+cxe^{-x}}{x^2\sin x}=1

, then

16(a^2+b^2+c^2)

is equal to ________.

Official answer

NTA final key.

Chapter: Limits and Derivatives
Topic: L'Hopital's Rule
Idea tested: Choosing Series Expansion over L'Hopital

Same idea in other shifts

Asked 6× in all

6 Apr 2026, Shift 1 · Q17The value of $\lim_{x\to 0}\left(\frac{x^2\sin^2 x}{x^2-\sin^2 x}\right)$ is:MediumSingle correct
8 Apr 2025, Shift 2 · Q16Given below are two statements: Statement I :…MediumTwo statements
8 Apr 2024, Shift 1 · Q26The value of $\lim_{x\to0}2\left(\frac{1-\cos x\sqrt{\cos2x}\sqrt[3]{\cos3x}\cdots\sqrt[10]{\cos10x}}{x^2}\right)$ is ____________.MediumNumerical value
13 Apr 2023, Shift 2 · Q9If $\lim_{x\to0}\dfrac{e^{ax}-\cos(bx)-\dfrac{cxe^{-cx}}{2}}{1-\cos(2x)}=17$ , then $5a^2+b^2$ is equal toHardSingle correct
18 Mar 2021, Shift 1 · Q72If $\lim_{x\to0}\dfrac{\sin^{-1}x-\tan^{-1}x}{3x^3}$ is equal to L, then the value of $(6L+1)$ is :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 42 of the 174 JEE Main shifts held from 2017 to 2026.