JEE Main 2022 · MathsMultiple choiceSingle correctMediumCalculation

JEE Main 29 June 2022, Shift 2, Maths Q5

Question 5 of 90 in this shift, Maths question 5 of 30, Section A.

The value of

\lim_{x \to 1} \frac{(x^2 - 1)\sin^2(\pi x)}{x^4 - 2x^3 + 2x - 1}

is equal to:

(1) $\frac{\pi^2}{6}$
(2) $\frac{\pi^2}{3}$
(3) $\frac{\pi^2}{2}$
(4) $\pi^2$ Official answer

Official answer

Option 4

NTA final key.

Chapter: Limits and Derivatives
Topic: L'Hopital's Rule
Idea tested: Simplifying with Standard Limits Before L'Hopital

Same idea in other shifts

Asked 13× in all

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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.