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  • Please solve rd  sharma class 12 Chapter 17  Maxima and Minima excercise 17.3 question 2 subquestion (ii) maths textbook solution

Please solve rd  sharma class 12 Chapter 17  Maxima and Minima excercise 17.3 question 2 subquestion (ii)  maths textbook solution

Answers (1)

Answer:

x=\frac{2}{3}  is a point of local maxima

Hint:

Put f'(x) = 0

Given:

f(x)=x \sqrt{1-x}, x \leq 1

Explanation:

Differentiating with respect to x

\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{x} \frac{\mathrm{d}}{\mathrm{dx}} \sqrt{1-\mathrm{x}}+(\sqrt{1-\mathrm{x}}) \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}) \mid

\begin{aligned} &=\frac{-\mathrm{x}}{2 \sqrt{1-\mathrm{x}}}+\sqrt{1-\mathrm{x}} \\ &=\frac{-\mathrm{x}+2-2 \mathrm{x}}{2 \sqrt{1-\mathrm{x}}} \\ &=\frac{2-3 \mathrm{x}}{2 \sqrt{1-\mathrm{x}}} \end{aligned}

Put f'(x) = 0

\begin{aligned} &\frac{2-3 \mathrm{x}}{2 \sqrt{1-\mathrm{x}}}=0 \\ &\Rightarrow 2-3 \mathrm{x}=0 \quad \text { if } \mathrm{x} \neq 1 \\ &\Rightarrow 3 \mathrm{x}=2 \\ &\Rightarrow \mathrm{x}=\frac{2}{3} \end{aligned}

Thus x=\frac{2}{3}is the possible point of local maxima and minima

Differentiating with respect to x

\begin{aligned} &\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{2-3 \mathrm{x}}{2 \sqrt{1-\mathrm{x}}}\right)\\ &=\frac{1}{2}\left(\frac{\sqrt{1-x} \frac{d}{d x}(2-3 x)-(2-3 x) \frac{d}{d x} \sqrt{1-x}}{(1-x)}\right) \cdot \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{1}{2}\left[\frac{(-3) \sqrt{1-\mathrm{x}}-(2-3 \mathrm{x}) \cdot 1 / 2(1-\mathrm{x})^{-\frac{1}{2}}(-1)}{(1-\mathrm{x})}\right]\\ &\text { [Using chain rule] }\\ &=\frac{1}{2}\left[\frac{-3}{\sqrt{1-x}}+\frac{2-3 x}{2(1-x)^{3 / 2}}\right]\\ &=\frac{1}{2}\left[\frac{-6+6 x+2-3 x}{2(1-x)^{\frac{3}{2}}}\right]\\ &=\frac{1}{2}\left[\frac{-4+3 x}{2(1-x)^{3 / 2}}\right] \end{aligned}

Put x=\frac{2}{3}   in f’’ (x)

\begin{aligned} &\mathrm{f}^{\prime \prime}\left(\frac{2}{3}\right)=\frac{1}{4}\left(\frac{-4+3\left(\frac{2}{3}\right)}{\left(1-\frac{2}{3}\right)^{3 / 2}}\right) \\ &=\frac{1}{4} \times \frac{-4+2}{\left(\sqrt{\frac{1}{3}}\right)^{3}} \end{aligned}

So x=\frac{2}{3} is a point of local maxima

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