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Convexity

The maximum v...

Question

The maximum value of ${(\frac{1}{x})}^{x}$ is $e^{1 / e}$ .

A

True

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B

False

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Solution

The correct option is A True
$y = {((\frac{1}{x}))}^{x} l o g y = x l o g (\frac{1}{x}) (\frac{1}{y}) (\frac{d y}{d x}) = (\frac{d (- x l o g x)}{d x}) (\frac{1}{y}) (\frac{d y}{d x}) = - [l o g x + x \times (\frac{1}{x})] (\frac{d y}{d x}) = - y [l o g x + 1] = 0 f o r p o i n t o f i n c l i n a t i o n l o g x + 1 = 0 l o g x = - 1 ∴ x = (\frac{1}{e}) ∴ m a x i m u m v a l u e = {((\frac{1}{(\frac{1}{e})}))}^{(\frac{1}{e})} = e^{(\frac{1}{e})}$

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