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  • Provide solution for rd sharma class 12 chapter 17 Maxima and Minima excercise 17.3 question 4

Provide solution for rd sharma class 12 chapter 17 Maxima and Minima excercise 17.3 question 4

Answers (1)

best_answer

Answer:

We need to prove that x = e is local maxima

Given:

The given function \frac{\log x}{x}

Explanation:

\text { Let } y=\frac{\log x}{x}

Then

\begin{aligned} \frac{d y}{d x} &=\frac{x \cdot \frac{d}{d x}(\log x)-\log (x) \frac{d}{d x}(x)}{x^{2}}, \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{\frac{v d u}{d x}+\frac{u d v}{d x}}{v^{2}} \\ &=\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{d y}{d x}  with respect to x

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

At x=e,

\begin{aligned} \frac{d^{2} y}{d x^{2}} &=\frac{-3 e+2 e \log 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

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