JEE Main 2021MathematicsDifferentiationHardMCQ

JEE Main 2021 — Differentiation Question with Solution

JEE Main 2021 (26 Aug Shift 1)

Question

Let f(x)=cos2tan-1sincot-11-xx,0<x<1. Then:

Choose an option

Show full solutionCorrect option: A
Correct answer
A(1-x)2f'(x)+2(f(x))2=0

Step-by-step explanation

Put x=sin2θ,0<x<1

sinθ=x

⇒fx=cos2tan-1sincot-11-sin2θsin2θ

⇒fx=cos2tan-1(sinθ)

⇒fx=cos2tan-1x

=1-tan2tan-1x1+tan2tan-1x

⇒fx=1-x1+x

⇒f'x=1+x-1-1-x·11+x2

⇒f'x=-21+x2

Multiply, 1-x2 on both sides

⇒1-x2f'x=-21-x21+x2

Now, 2fx2=21-x21+x2

∴1-x2f'x+2fx2=0 option 1 satisfied

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About this question

This is a previous-year question from JEE Main 2021, covering the Differentiation chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.