JEE Main 2024 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 6 April 2024, Shift 2, Maths Q12

Question 12 of 90 in this shift, Maths question 12 of 30, Section A.

\int\frac{1}{a^2\sin^2x+b^2\cos^2x}dx=\frac{1}{12}\tan^{-1}(3\tan x)+\text{constant}

, then the maximum value of

a\sin x+b\cos x

, is :

(1) $\sqrt{42}$
(2) $\sqrt{39}$
(3) $\sqrt{40}$ Official answer
(4) $\sqrt{41}$

Official answer

Option 3

NTA final key.

Chapter: Integrals
Topic: Integration by Substitution
Idea tested: Dividing by Cosine Squared Substitution

Same topic in other shifts

All Integration by Substitution questions

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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 42 of the 174 JEE Main shifts held from 2017 to 2026.