The maximum value of f x - x 1-x isA- eB- 1 C- e 1-eD- 1-p

The correct option is C e^1/ef(x) = x^1/x ⇒ln f(x) = 1/xln x ⇒f'/f = 1/x^2 1/x^2ln x For maxima or minima nbsp; f' = 0 ⇒ 1 ln x = 0 ⇒ x = e Checking for ...

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Standard XII

Mathematics

Logarithmic Differentiation

The maximum v...

Question

The maximum value of $f (x) = x^{\frac{1}{x}}$ is

e

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

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\frac{1}{e}

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Solution

The correct option is C $e^{\frac{1}{e}}$
$f (x) = x^{\frac{1}{x}} \Rightarrow ln f (x) = \frac{1}{x} ln x \Rightarrow \frac{f^{'}}{f} = \frac{1}{x^{2}} - \frac{1}{x^{2}} ln x$
For maxima or minima
$f^{'} = 0 \Rightarrow 1 - ln x = 0 \Rightarrow x = e$
Checking for maxima,
$\frac{f^{''} f - (f^{'})^{2}}{f^{2}} = - \frac{2}{x^{3}} - \frac{1}{x^{3}} + \frac{2}{x^{3}} ln x$
Putting $f^{'} = 0$
$\Rightarrow \frac{f^{''}}{f} = - \frac{3}{e^{3}} + \frac{2}{e^{3}} \Rightarrow f^{''} = e^{\frac{1}{e}} (- \frac{3}{e^{3}} + \frac{2}{e^{3}}) < 0$
So at $x = e$ the function has maxima
The maxima is $= e^{\frac{1}{e}}$

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The maximum value of f(x)=x1x is

The correct option is C e1ef(x)=x1x⇒lnf(x)=1xlnx⇒f′f=1x2−1x2lnx For maxima or minima f′=0⇒1−lnx=0⇒x=e Checking for maxima, f′′f−(f′)2f2=−2x3−1x3+2x3lnx Putting f′=0 ⇒f′′f=−3e3+2e3⇒f′′=e1e(−3e3+2e3)<0 So at x=e the function has maxima The maxima is =e1e

The maximum value of $f (x) = x^{\frac{1}{x}}$ is