JEE Main 2020MathematicsDifferentiationMediumMCQ

JEE Main 2020 — Differentiation Question with Solution

JEE Main 2020 (07 Jan Shift 1)

Question

Let xk+yk=ak,a,k>0 and dydx+yx13=0, then k is

Choose an option

Show full solutionCorrect option: C
Correct answer
C23

Step-by-step explanation

Differentiating the given equation with respect to x

k.xk-1+k.yk-1dydx=0

⇒dydx=-xyk-1

⇒dydx+xyk-1=0

⇒k-1=-13

⇒k=1-13=23

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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.