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UPSC NDA II 2015 Mathematics Solved Paper

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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]

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Question : 91 of 120

Marks: +1, -0

At what value of x does the function attain maximum value ?

e
$\sqrt{e}$
$\frac{1}{\sqrt{e}}$
$\frac{1}{e}$

Solution:

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

(say)

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

\frac{1}{y}\frac{dy}{dx}=2\left[2x\log\left(\frac{1}{x}\right)+x^{2}\cdot\frac{1}{\frac{1}{x}}\cdot\left(\frac{-1}{x^{2}}\right)\right]

\frac{dy}{dx}=2y\left[2x\log\left(\frac{1}{x}\right)-x\right]

For max. or min. value

\frac{dy}{dx}=0

2x\log\left(\frac{1}{x}\right)-x=0

x\left[2\log\left(\frac{1}{x}\right)-1\right]=0

\because x=0

\Rightarrow 2\log\left(\frac{1}{x}\right)=1

\Rightarrow \frac{1}{x}=e^{\frac{1}{2}}

\Rightarrow x=e^{-\frac{1}{2}}

Again

\frac{d^{2}y}{dx^{2}}=\left\{2xy\left[\log\left(\frac{1}{x}\right)-1\right]\right\}

=(2xy)'\left(\log\left(\frac{1}{x}\right)-1\right)+2xy\left[x\cdot\left(-\frac{1}{x^{2}}\right)\right]

=(2xy)'\left(\log\left(\frac{1}{x}\right)-1\right)-2y

=2\left[(y+xy^{1})\left(\log\left(\frac{1}{x}\right)-1\right)-y\right]<0

So at

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

function is maximum

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