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# The Maximum Value of X1/X, X > 0 is - Mathematics

#### Question

The maximum value of x1/x, x > 0 is __________ .

##### Options
• e^(1/e)

• (1/e)^e

• 1

• none of these

#### Solution

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

$\text { Given }: f\left( x \right) = x^\frac{1}{x}$
$\text { Taking log on both sides, we get }$
$\log f\left( x \right) = \frac{1}{x}\log x$
$\text { Differentiating w . r . t . x, we get }$
$\frac{1}{f\left( x \right)}f'\left( x \right) = \frac{- 1}{x^2}\log x + \frac{1}{x^2}$
$\Rightarrow f'\left( x \right) = f\left( x \right)\frac{1}{x^2}\left( 1 - \log x \right)$
$\Rightarrow f'\left( x \right) = x^\frac{1}{x} \left( \frac{1}{x^2} - \frac{1}{x^2}\log x \right) . . . \left( 1 \right)$
$\Rightarrow f'\left( x \right) = x^\frac{1}{x} - 2 \left( 1 - \log x \right)$
$\text { For a local maxima or a local minima, we must have }$
$f'\left( x \right) = 0$
$\Rightarrow x^\frac{1}{x} - 2 \left( 1 - \log x \right) = 0$
$\Rightarrow \log x = 1$
$\Rightarrow x = e$
$\text { Now,}$
$f''\left( x \right) = x^\frac{1}{x} \left( \frac{1}{x^2} - \frac{1}{x^2}\log x \right)^2 + x^\frac{1}{x} \left( \frac{- 2}{x^3} + \frac{2}{x^3}\log x - \frac{1}{x^3} \right) = x^\frac{1}{x} \left( \frac{1}{x^2} - \frac{1}{x^2}\log x \right)^2 + x^\frac{1}{x} \left( - \frac{3}{x^3} + \frac{2}{x^3}\log x \right)$
$\text { At }x = e:$
$f''\left( e \right) = e^\frac{1}{e} \left( \frac{1}{e^2} - \frac{1}{e^2}\log e \right)^2 + e^\frac{1}{e} \left( - \frac{3}{e^3} + \frac{2}{e^3}\log e \right) = - e^\frac{1}{e} \left( \frac{1}{e^3} \right) < 0$
$\text { So, x = e is a point of local maxima }.$
$\therefore \text { Maximum value } = f\left( e \right) = e^\frac{1}{e}$
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