JEE Main 2020 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 5 September 2020, Shift 1, Maths Q58

Question 58 of 75 in this shift, Maths question 8 of 25.

\alpha

is the positive root of the equation,

p(x)=x^2-x-2=0

, then

\lim_{x\to\alpha^+}\frac{\sqrt{1-\cos(p(x))}}{x+\alpha-4}

is equal to :

(1) $\frac{1}{2}$
(2) $\frac{1}{\sqrt{2}}$
(3) $\frac{3}{\sqrt{2}}$ Official answer
(4) $\frac{3}{2}$

Official answer

Option 3

NTA final key (Sep 2020).

Chapter: Limits and Derivatives
Topic: L'Hopital's Rule
Idea tested: Trigonometric Simplification Before Differentiation

Same topic in other shifts

All L'Hopital's Rule questions

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5 Apr 2026, Shift 1 · Q17The product of all possible values of $\alpha$ , for which…HardSingle correct
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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
2 Apr 2025, Shift 1 · Q7For $\alpha, \beta, \gamma\in\mathbf{R}$ , if $\lim_{x\to0}\frac{x^2\sin\alpha x+(\gamma-1)e^{x^2}}{\sin2x-\beta x}=3$ , then…MediumSingle correct
2 Apr 2025, Shift 2 · Q17If $\lim_{x\to0} \frac{\cos(2x) + a\cos(4x) - b}{x^4}$ is finite, then $(a+b)$ is equal to :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.