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NDA Math Application of Derivatives

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Consider the function

f(x)=\left(\frac{1}{x}\right)^{2x^2}

,where x>0

[NDA II 2015]

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Question : 56 of 73

Marks: +1, -0

The maximum value of the function is

e
$e^{\frac{2}{e}}$
$e^{\frac{1}{e}}$
$\frac{1}{e}$

Solution:

f(x)=\left(\frac{1}{x}\right)^{2x^{2}}

f\left(e^{-\frac{1}{2}}\right)=\left(e^{\frac{1}{2}}\right)^{2\times e^{-1}}

=\left(e^{\frac{1}{2}}\right)^{\frac{2}{e}}

=e^{\frac{1}{2}\times\frac{2}{e}}

=e^{\frac{1}{e}}

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