JEE Main 2020MathematicsDifferentiationHardMCQ

JEE Main 2020 — Differentiation Question with Solution

JEE Main 2020 (08 Jan Shift 1)

Question

Let fx=sintan-1x+sincot-1x2-1, x>1. If dydx=12ddxsin-1fx and y3=π6, then y-3 is equal to:

Choose an option

Show full solutionCorrect option: C
Correct answer
C5π6

Step-by-step explanation

Let tan-1x=θ

⇒x=tanθ ⇒sinθ=x1+x2

So, y=x1+x2+11+x22-1

⇒y=x+121+x2-1

⇒y=2x1+x2=fx

Now, dydx=1211-f2×f'x

=1211-4x21+x22fx

=1+x22x2-1f'x

=1+x22x2-1×21+x2-2x21+x22

⇒dydx=1-x2x2-11+x2

⇒dy=1-x2x2-11+x2dx

Integrating both sides with respect to x

∫dy=∫1-x2x2-11+x2dx

⇒y=-tan-1x+c

Given, y3=π6⇒π6=-π3+c⇒c=π2

y=-tan-1x+π2=cot-1x

Now, y-3=cot-1x-3=5π6

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Differentiation chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

This is a previous-year question from JEE Main 2020, 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.