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  • Provide solution for RD Sharma maths class 12 chapter 17 Maxima and Minima exercise Very short answer type question 10

Provide solution for RD Sharma maths class 12 chapter 17 Maxima and Minima exercise Very short answer type question  10

Answers (1)

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Answer:  \frac{1}{e}

Given: f(x)=\frac{\log x}{x}

Solution:

        \begin{aligned} &f(x)=\frac{\log x}{x} \\\\ &f^{\prime}(x)=\frac{1-\log x}{x^{2}} \end{aligned}

For a local maxima or a local minima we must have     

        \begin{aligned} &f^{\prime}(x)=0 \\\\ &\frac{1-\log x}{x^{2}}=0 \\\\ &1-\log x=0 \end{aligned}

        \begin{aligned} &\log x=1 \\\\ &\log x=\log e \\\\ &x=e \end{aligned}

Now

        \begin{aligned} &f^{\prime \prime}(x)=\frac{-x-2 x(1-\log x)}{x^{4}} \\\\ &=\frac{-3 x-2 x \log x}{x^{4}} \\\\ &\text { At } x=e \quad f^{\prime \prime}(x)=\frac{-3 e-2 e \text { loge }}{e^{4}}=-\frac{5}{e^{3}}<0 \end{aligned}

So x= e is a point local maximum.

Thus the local maximum value is given by

        \begin{aligned} &f(e)=\frac{\operatorname{loge}}{e} \\\\ &=\frac{1}{e} \end{aligned}

        \therefore f(e)=1 / e

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