JEE Main 2023 · MathsMultiple choiceSingle correctHardMulti-step

JEE Main 8 April 2023, Shift 2, Maths Q9

Question 9 of 90 in this shift, Maths question 9 of 30, Section A.

\alpha > \beta > 0

are the roots of the equation

ax^2 + bx + 1 = 0

, and

\lim_{x \to \frac{1}{\alpha}} \left(\frac{1 - \cos(x^2 + bx + a)}{2(1 - \alpha x)^2}\right)^{\frac{1}{2}} = \frac{1}{k}\left(\frac{1}{\beta} - \frac{1}{\alpha}\right)

, then

k

is equal to

(1) $\alpha$
(2) $\beta$
(3) $2\alpha$ Official answer
(4) $2\beta$

Official answer

Option 3

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 6× in all

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7 Apr 2025, Shift 1 · Q16…MediumSingle correct
7 Apr 2025, Shift 2 · Q23For $t>-1$ , let $\alpha_t$ and $\beta_t$ be the roots of the equation…HardNumerical value
31 Jan 2024, Shift 1 · Q6 $\lim_{x\to0}\frac{e^{2|\sin x|}-2|\sin x|-1}{x^2}$ 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.