JEE Main 2020MathematicsDifferentiationMediumMCQ

JEE Main 2020 — Differentiation Question with Solution

JEE Main 2020 (07 Jan Shift 2)

Question

Let y=yx be a function of x satisfying y1-x2=k-x1-y2 where k is a constant and y12=-14.Then dydx at x=12 , is equal to

Choose an option

Show full solutionCorrect option: B
Correct answer
B-52

Step-by-step explanation

x=12,y=-14⇒xy=-18

y.-2x21-x2+y'.1-x2=0-1.1-y2+x.-2y21-y2y'

-xy1-x2+y'1-x2=-1-y2+xy.y'1-y2

y'1-x2-xy1-y2=xy1-x2-1-y2

y'32+18.154=-18.34-154

y'45+1215=-1+4543

y'=-1523=-52

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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.