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

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Answer:  \frac{1}{e}

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


        \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}


        \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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