Question:
The maximum value of $\left( \frac{1}{x} \right)^x$ is
The maximum value of $\left( \frac{1}{x} \right)^x$ is
Updated On: Apr 8, 2024
- $e $
- $e^e$
- $e^{1/e}$
- $\left( \frac{1}{e}\right)^e$
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The Correct Option is C
Solution and Explanation
Let $y=\left(\frac{1}{x}\right)^{x}$
$ \Rightarrow y=x^{-x}$
$\therefore \frac{d y}{d x}=x^{-x}(-1-\log x)$
$\Rightarrow \frac{d y}{d x}=-x^{-x}(1+\log x)$
$\left[\because \frac{d}{d x} f(x)^{d(x)}=f(x)^{g(x)}\right.$
$\left.\left\{g(x) \cdot \frac{1}{f(x)} \cdot f'(x)+g'(x) \log f(x)\right\}\right]$
For maxima,
$\frac{dy}{dx} = 0$
$\Rightarrow 1 + \log\,x = 0\,\,[\because x^{-x} \ne0 ]$
$\Rightarrow \log\,x = -1$
$\Rightarrow x = e^{-1}$
Hence, the maximum value of $\left(\frac{1}{x}\right)^x$ is $\left(\frac{1}{e^{-1}}\right)^{e^{-1}}$
i.e., $(e)^{1/e}$ at $x = e^{-1}$
$ \Rightarrow y=x^{-x}$
$\therefore \frac{d y}{d x}=x^{-x}(-1-\log x)$
$\Rightarrow \frac{d y}{d x}=-x^{-x}(1+\log x)$
$\left[\because \frac{d}{d x} f(x)^{d(x)}=f(x)^{g(x)}\right.$
$\left.\left\{g(x) \cdot \frac{1}{f(x)} \cdot f'(x)+g'(x) \log f(x)\right\}\right]$
For maxima,
$\frac{dy}{dx} = 0$
$\Rightarrow 1 + \log\,x = 0\,\,[\because x^{-x} \ne0 ]$
$\Rightarrow \log\,x = -1$
$\Rightarrow x = e^{-1}$
Hence, the maximum value of $\left(\frac{1}{x}\right)^x$ is $\left(\frac{1}{e^{-1}}\right)^{e^{-1}}$
i.e., $(e)^{1/e}$ at $x = e^{-1}$
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