#### Need Solution for R.D.Sharma Maths Class 12 Chapter 17 Maxima and Minima  Exercise 17.3 Question 4 Maths Textbook Solution.

We need to prove that x = e is local maxima

Given:

The given function $\frac{\log \mathrm{x}}{\mathrm{x}}$

Explanation:

Let $y=\frac{\log \mathrm{x}}{\mathrm{x}}$

Then

$\frac{\mathrm{dy}}{\mathrm{d} \mathrm{x}}=\frac{\mathrm{x} \cdot \frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}(\log \mathrm{x})-\log (\mathrm{x}) \frac{\mathrm{d}}{\mathrm{d} \mathrm{s}}(\mathrm{x})}{\mathrm{x}^{2}} \ldots \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{\frac{v d u}{d x}+\frac{u d v}{d x}}{v^{2}}$

\begin{aligned} &=\frac{x \times \frac{1}{x}-\log x}{x^{2}} \\ &=\frac{1-\log x}{x^{2}} \end{aligned}

Put,

\begin{aligned} &\frac{d y}{d x}=0 \\ &\frac{1-\log x}{x^{2}}=0 \\ &\Rightarrow 1-\log x=0, x \neq 0 \\ &\Rightarrow \log x=1 \\ &x=e^{1}=e \end{aligned}

Differentiating $\frac{dy}{dx}$ with respect to x

$\frac{\mathrm{d}^{2} y}{\mathrm{dx}^{2}}=\frac{\mathrm{x}^{2} \frac{\mathrm{d}}{\mathrm{dx}}(1-\log \mathrm{x})-(1-\log \mathrm{x}) \frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}\left(\mathrm{x}^{2}\right)}{\mathrm{x}^{4}} . \cdot \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{\frac{v d u}{d x}+\frac{u d v}{d x}}{v^{2}}$

\begin{aligned} &=\frac{x^{2}\left(-\frac{1}{x}\right)-2 x(1-\log x)}{x^{4}} \\ &=\frac{-x-2 x+2 x \log x}{x^{4}} \\ &=\frac{-3 x-2 x \log x}{x^{4}} \end{aligned}

At $x=e$

\begin{aligned} \frac{d^{2} y}{d x^{2}} &=\frac{-3 e+2 e l o g e}{e^{4}} \\ &=\frac{-3 e+2 e(1)}{e^{4}} \ldots \log e=1 \\ &=-\frac{e}{e^{4}}=-\frac{1}{e^{3}}<0 \end{aligned}

So  $x=e$ is a point of local maxima.