Verified Solution • 2025–26 Session moderate

QJEEM21Maf205: JEE Main 2021 Mathematics - Class 12 JEEM21Maf205 Solution: The local maximum value of the function , x > 0, is

The local maximum value of the function f(x)=(2x)x2f(x) = {\left( {{2 \over x}} \right)^{{x^2}}} , x > 0, is

Multiple Choice Options

  • A (2e)1e{\left( {2\sqrt e } \right)^{{1 \over e}}}
  • B (4e)e4{\left( {{4 \over {\sqrt e }}} \right)^{{e \over 4}}}
  • C (e)2e{(e)^{{2 \over e}}}
  • D 1

Step-by-Step Verified Solution

f(x)=(2x)x2f(x) = {\left( {{2 \over x}} \right)^{{x^2}}} ; x > 0

ln⁡f(x)=x2(ln⁡2−ln⁡x)\ln f(x) = {x^2}(\ln 2 - \ln x)

f′(x)=f(x){−x+(ln⁡2−ln⁡x)2x}f'(x) = f(x)\{ - x + (\ln 2 - \ln x)2x\}

f′(x)=f(x)⏟+ . x⏟+(2ln⁡2−2ln⁡x−1)⏟g(x)f'(x) = \underbrace {f(x)}_ + \,.\,\underbrace x_ + \underbrace {(2\ln 2 - 2\ln x - 1)}_{g(x)}

g(x)=2ln⁡2−2ln⁡x−1g(x) = 2{\ln ^2} - 2\ln x - 1

=ln⁡4x2−1=0⇒x=2e= \ln {4 \over {{x^2}}} - 1 = 0 \Rightarrow x = {2 \over {\sqrt e }}

JEE Main 2021 (Online) 26th August Evening Shift Mathematics - Application of Derivatives Question 92 English Explanation

LM=2eLM = {2 \over {\sqrt e }}

Local maximum value = (22/e)4e=e2e{\left( {{2 \over {2/\sqrt e }}} \right)^{{4 \over e}}} = {e^{{2 \over e}}}

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