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The Maximum Value of X1/X, X > 0 is - CBSE (Science) Class 12 - Mathematics

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Question

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

  • `e^(1/e)`

  • `(1/e)^e`

  • 1

  • none of these

Solution

\[e^\frac{1}{e}\]

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

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Solution The Maximum Value of X1/X, X > 0 is Concept: Graph of Maxima and Minima.
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