JEE Main 2019MathematicsDifferentiationMediumMCQ

JEE Main 2019 — Differentiation Question with Solution

JEE Main 2019 (12 Jan Shift 1)

Question

For x>1, if 2x2y=4e2x-2y, then 1+loge⁡2x2 dydx is equal to

Choose an option

Show full solutionCorrect option: B
Correct answer
Bxloge⁡2x-loge⁡2x

Step-by-step explanation

Given, 2x2y=4.e2x-2y

Taking natural logarithm on both sides, we get

2y loge2x= loge4+2x-2y

⇒2y=loge4+2x1+loge2x

Differentiating both sides with respect to x, we get

2dydx=1+loge2x.2-loge4+2x1x1+loge2x2 (Using quotient rule)

⇒1+loge2x2dydx=x.loge2x-loge2x.

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About this question

This is a previous-year question from JEE Main 2019, 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.