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Please solve RD Sharma class 12 Chapter Maxima and Minima exercise Multiple choice question, question1 maths textbook solution.

Answers (1)

Answer: option (b) e^{\frac{1}{e}}

Hint: For local maxima or minima, we must have.\Rightarrow \frac{dy}{dx}=0

Given: y = x^{\frac{1}{x}}

Solution:

y = x^{\frac{1}{x}}

\log y =\frac{1}{x} \log _e x

\frac{1}{y} \frac{d y}{d x}=\frac{1}{x} \cdot \frac{1}{x}+\log _{e} x\left(\frac{-1}{x^{2}}\right)

For maximum or minimum \Rightarrow \frac{dy}{dx}=0

\begin{aligned} &\frac{d y}{d x}=y\left(\frac{1}{x^{2}}-\frac{1}{x^{2}}\left(\log _{e} x\right)\right)=0 \\ &\frac{1}{x^{2}}=\frac{1}{x^{2}} \log _{e} x \\ &\frac{1}{x^{2}}\left(\log _{e} x-1\right)=0 \end{aligned}

\log _ex-1

x = e is maximum value

Hence,

x^{\frac{1}{x}}=e^{\frac{1}{e}}

Maximum Value =f(e)=e^{\frac{1}{e}}.

Posted by

infoexpert24

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