JEE Main 2024 · MathsMultiple choiceSingle correctMediumCalculation

JEE Main 27 January 2024, Shift 1, Maths Q9

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

a=\lim_{x\to0}\dfrac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2}}{x^4}

and

b=\lim_{x\to0}\dfrac{\sin^2 x}{\sqrt{2}-\sqrt{1+\cos x}}

, then the value of

ab^3

is :

(1) $25$
(2) $30$
(3) $36$
(4) $32$ Official answer

Official answer

Option 4

NTA final key.

Chapter: Limits and Derivatives
Topic: L'Hopital's Rule
Idea tested: Conjugation vs L'Hopital for Radicals

Same idea in other shifts

Asked 4× in all

4 Apr 2024, Shift 1 · Q24If $\lim_{x\to1}\frac{(5x+1)^{1/3}-(x+5)^{1/3}}{(2x+3)^{1/2}-(x+4)^{1/2}}=\frac{m\sqrt{5}}{n(2n)^{2/3}}$ , where $\gcd(m, n)=1$ , then…HardNumerical value
25 Jun 2022, Shift 2 · Q5…HardSingle correct
27 Aug 2021, Shift 2 · Q66If $\lim_{x \to \infty} \left(\sqrt{x^2 - x + 1} - ax\right) = b$ , then the ordered pair (a, b) 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.