Test Index

UPSC NDA II 2015 Mathematics Solved Paper

Show Para Hide Para

For the next two (02) items that follow:

Consider the function

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

,where x>0

[NDA II 2015]

© examsnet.com

Question : 92 of 120

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}}

© examsnet.com

Go to Question:

Prev Question

Next Question