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UPSC NDA II 2015 Mathematics Paper

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Question Numbers: 72-73

For the next two (02) items that follow:

Consider the function

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

, where x > 0

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Question : 73 of 120

Marks: +1, -0

The maximum value of the function is

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

Solution:

Concept:

Maximum value is found by differentiating

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

and evaluating at the critical point.

Explanation:

Take natural log:

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

.

Differentiate:

\frac{f'(x)}{f(x)} = -4x \ln x - 2x

.

Set

f'(x)=0

:

-2x(2\ln x + 1) = 0

gives

x = e^{-1/2} = \frac{1}{\sqrt{e}}

(since

x>0

).

Substitute into

f

:

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

.

The maximum value is

e^{1/e}

.

Answer:

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

(Option C)

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