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Solution for Write the Maximum Value of F(X) = X1/X. - CBSE (Science) Class 12 - Mathematics

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Question

Write the maximum value of f(x) = x1/x.

Solution

\[\text { Given }: \hspace{0.167em} f\left( x \right) = x^\frac{1}{x} \]

\[\text { Taking log on both sides, we get }\]

\[\log f\left( x \right) = \frac{1}{x}\log x\]

\[\text { Differentiating w . r . t . x, we get }\]

\[\frac{1}{f\left( x \right)}f'\left( x \right) = \frac{- 1}{x^2}\log x + \frac{1}{x^2}\]

\[ \Rightarrow f'\left( x \right) = f\left( x \right)\frac{1}{x^2}\left( 1 - \log x \right)\]

\[ \Rightarrow f'\left( x \right) = x^\frac{1}{x} \left( \frac{1}{x^2} - \frac{1}{x^2}\log x \right) . . . \left( 1 \right)\]

\[ \Rightarrow f'\left( x \right) = x^\frac{1}{x} - 2 \left( 1 - \log x \right) \]

\[\text { For a local maxima or a local minima, we must have }\]

\[f'\left( x \right) = 0\]

\[ \Rightarrow x^\frac{1}{x} - 2 \left( 1 - \log x \right) = 0\]

\[ \Rightarrow \log x = 1\]

\[ \Rightarrow x = e\]

\[\text { Now }, \]

\[f''\left( x \right) = x^\frac{1}{x} \left( \frac{1}{x^2} - \frac{1}{x^2}\log x \right)^2 + x^\frac{1}{x} \left( \frac{- 2}{x^3} + \frac{2}{x^3}\log x - \frac{1}{x^3} \right) = x^\frac{1}{x} \left( \frac{1}{x^2} - \frac{1}{x^2}\log x \right)^2 + x^\frac{1}{x} \left( - \frac{3}{x^3} + \frac{2}{x^3}\log x \right)\]

\[\text { At }x = e\]

\[f''\left( e \right) = e^\frac{1}{e} \left( \frac{1}{e^2} - \frac{1}{e^2}\log e \right)^2 + e^\frac{1}{e} \left( - \frac{3}{e^3} + \frac{2}{e^3}\log e \right) = - e^\frac{1}{e} \left( \frac{1}{e^3} \right) < 0\]

\[\text { So, x = e is a point of local maximum }. \]

\[\text { Thus, the maximum value is given by }\]

\[f\left( e \right) = e^\frac{1}{e} \]

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Solution for question: Write the Maximum Value of F(X) = X1/X. concept: Graph of Maxima and Minima. For the courses CBSE (Science), CBSE (Commerce), CBSE (Arts), PUC Karnataka Science
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