JEE Main 2021MathematicsApplication Of DerivativesMaxima And MinimamediumMCQ
JEE Main 2021 — Application Of Derivatives Question with Solution
From: JEE Main 2021 (Online) 26th August Evening Shift
Question
The local maximum value of the function f(x)=(x2)x2, x > 0, is
Choose an option
▸Show full solutionCorrect option: C
Correct answer
C(e)e2
Step-by-step explanation
f(x)=(x2)x2 ; x > 0
lnf(x)=x2(ln2−lnx)
f′(x)=f(x){−x+(ln2−lnx)2x}
f′(x)=+f(x).+xg(x)(2ln2−2lnx−1)
g(x)=2ln2−2lnx−1
=lnx24−1=0⇒x=e2
LM=e2
Local maximum value = (2/e2)e4=ee2
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