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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 : 55 of 73

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 \cdot \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^{-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^{-1/2}

function is maximum

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