JEE Main 2019 — Differentiation Question with Solution
From: JEE Main 2019 (Online) 12th January Morning Slot
Question
For x > 1, if (2x)2y = 4e2x2y,
then (1 + loge 2x)2 is equal to :
then (1 + loge 2x)2 is equal to :
Choose an option
Show full solutionCorrect option: A
Correct answer
A
Step-by-step explanation
(2x)2y = 4e2x-2y
2yn2x = n4 + 2x 2y
y =
y ' =
y '
2yn2x = n4 + 2x 2y
y =
y ' =
y '
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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.