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JEE Main 2021 · MathsMultiple choiceSingle correctHardMulti-step

JEE Main 26 August 2021, Shift 1, Maths Q68

Question 68 of 90 in this shift, Maths question 8 of 30, Section A.

Let

f(x) = \cos\left(2\tan^{-1}\sin\left(\cot^{-1}\sqrt{\frac{1-x}{x}}\right)\right)

,

0 < x < 1

. Then :

(1) $(1-x)^2 f'(x) - 2(f(x))^2 = 0$
(2) $(1+x)^2 f'(x) + 2(f(x))^2 = 0$
(3) $(1-x)^2 f'(x) + 2(f(x))^2 = 0$ Official answer
(4) $(1+x)^2 f'(x) - 2(f(x))^2 = 0$

Official answer

Option 3

NTA final key.

Chapter: Inverse Trigonometric Functions
Topic: Inverse Trig Identities
Idea tested: Multiple Angle Inverse ID

Same idea in other shifts

Asked 2× in all

9 Apr 2024, Shift 2 · Q11The integral $\int_{1/4}^{3/4}\cos\left(2\cot^{-1}\sqrt{\frac{1-x}{1+x}}\right)dx$ is equal toMediumSingle 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.