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  • Provide solution for RD Sharma maths class 12 chapter 17 Maxima and Minima exercise Very short answer type question 6

Provide solution for RD Sharma maths class 12 chapter 17 Maxima and Minima exercise Very short answer type question  6

Answers (1)

Answer: \left(\frac{1}{e},-\frac{1}{e}\right)

Given: f(x)=x \log _{\mathrm{e}} x

Solution:

        \begin{gathered} f(x)=x \log _{\mathrm{e}} x \\\\ f^{\prime}(x)=\log _{\mathrm{e}} x+1 \end{gathered}

For a local maxima or a local minima we must have

        \begin{aligned} &f^{\prime}(x)=0 \\\\ &\log _{\mathrm{e}} x+1=0 \\\\ &\log _{\mathrm{e}} x=-1 \end{aligned}

        \begin{aligned} &x=\frac{1}{e} \\\\ &f\left(\frac{1}{e}\right)=\frac{1}{e} \log _{e}\left(\frac{1}{\theta}\right)=-\frac{1}{e} \end{aligned}

Now

        \begin{aligned} &f^{\prime \prime}\left(\frac{1}{e}\right)=\frac{1}{x} \\\\ &\text { At } x=\frac{1}{e} \\\\ &f^{\prime \prime}\left(\frac{1}{e}\right)=\frac{1}{\frac{1}{e}}=e>0 \end{aligned}

So, \left(\frac{1}{e},-\frac{1}{e}\right) Is a point of local maximum

 

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