Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • 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)

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

Posted by

infoexpert26

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE extends single girl child scholarship registration deadline till November 20
anam-khan

CBSE reminds schools to submit registration data of Class 9, 11 students by October 16

Latest Question