JEE Main 2020MathematicsDifferential EquationsMediumMCQ

JEE Main 2020 — Differential Equations Question with Solution

JEE Main 2020 (08 Jan Shift 1)

Question

Let y=yx be a solution of the differential equation, 1-x2dydx+1-y2=0,x<1. If y12=32, then y-12 is equal to

Choose an option

Show full solutionCorrect option: C
Correct answer
C12

Step-by-step explanation

dy1-y2+dx1-x2=0

⇒sin-1y+sin-1x=c
At,x=12, y=32⇒ c=π2⇒ sin-1y=cos-1x
Hence, y-12=sincos-1-12=12

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About this question

This is a previous-year question from JEE Main 2020, covering the Differential Equations 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.