JEE Main 2020 · MathsNumerical answerNumerical valueMediumCalculation

JEE Main 3 September 2020, Shift 1, Maths Q74

Question 74 of 75 in this shift, Maths question 24 of 25.

\lim_{x\to0}\left\{\frac{1}{x^8}\left(1-\cos\frac{x^2}{2}-\cos\frac{x^2}{4}\right.\right.

\left.\left.+\cos\frac{x^2}{2}\cos\frac{x^2}{4}\right)\right\}=2^{-k}

, then the value of k is __________.

Official answer

8.00

NTA final key (Sep 2020).

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

Same idea in other shifts

Asked 10× 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
31 Jan 2024, Shift 2 · Q25If $\lim_{x\to0}\frac{ax^2e^x-b\log_e(1+x)+cxe^{-x}}{x^2\sin x}=1$ , then $16(a^2+b^2+c^2)$ is equal to ________.HardNumerical 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
26 Jul 2022, Shift 2 · Q7Let $\beta=\lim_{x\to 0}\frac{\alpha x-(e^{3x}-1)}{\alpha x(e^{3x}-1)}$ for some $\alpha\in\mathbb{R}$ . Then the value of $\alpha+\beta$ …MediumSingle correct
26 Jun 2022, Shift 2 · Q4 $\lim_{x \to 0} \frac{\cos(\sin x) - \cos x}{x^4}$ is equal to :MediumSingle correct
27 Aug 2021, Shift 1 · Q65If $\alpha$ , $\beta$ are the distinct roots of $x^2+bx+c=0$ , then $\lim_{x\to\beta}\frac{e^{2(x^2+bx+c)}-1-2(x^2+bx+c)}{(x-\beta)^2}$ 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 108 of the 174 JEE Main shifts held from 2017 to 2026.