JEE Main 2021 · MathsMultiple choiceSingle correctMediumCalculation

JEE Main 4 August 2021, Shift 2, Maths Q66

Question 66 of 90 in this shift, Maths question 6 of 30, Section A.

The value of

\lim_{x\to2}\left(2-\cos\left(2x^2-5x+2\right)\right)^{\frac{1}{\left(x^2-4x+4\right)}}

is equal to

(1) $e^{\frac{9}{4}}$
(2) $e^{\frac{9}{2}}$ Official answer
(3) $e^{\frac{3}{4}}$
(4) $e^{\frac{3}{2}}$

Official answer

Option 2

NTA final key.

Chapter: Limits and Derivatives
Topic: L'Hopital's Rule
Idea tested: e-Power Shortcut for 1^infinity Forms

Same idea in other shifts

Asked 6× in all

28 Jul 2022, Shift 1 · Q29 $\lim_{x\to0}\left(\frac{(x+2\cos x)^3+2(x+2\cos x)^2+3\sin(x+2\cos x)}{(x+2)^3+2(x+2)^2+3\sin(x+2)}\right)^{\frac{100}{x}}$ is equal to…HardNumerical value
4 Aug 2021, Shift 1 · Q67 $\lim_{x\to 2}\left(x^3-x^2-8x+13\right)^{\frac{1}{x^3-3x^2+4}}$ MediumSingle correct
25 Feb 2021, Shift 1 · Q66 $\lim_{n \to \infty}\left(1 + \frac{1 + \frac{1}{2} + ....... + \frac{1}{n}}{n^2}\right)^n$ is equal to :MediumSingle correct
8 Jan 2020, Shift 1 · Q57 $\lim_{x\to 0}\left(\dfrac{3x^2+2}{7x^2+2}\right)^{1/x^2}$ is equal to :EasySingle correct
2 Sep 2020, Shift 2 · Q59 $\lim_{x \to 0}\left(\tan\left(\frac{\pi}{4}+x\right)\right)^{1/x}$ is equal to :EasySingle 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.