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

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

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