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  • Explain Solution R.D. Sharma Class 12 Chapter 17 Maxima and Minima Exercise 17.3 Question 2 Sub Question 2 Maths Textbook Solution.

Explain Solution R.D. Sharma Class 12 Chapter 17  Maxima and Minima  Exercise 17.3 Question 2 Sub Question 2 Maths Textbook Solution.

Answers (1)

Answer:

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

Hint:

Put f^{\prime}(x)=0

Given:

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

\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,\mathrm{f}^{\prime}(\mathrm{x})=0

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

\begin{aligned} &\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-\mathrm{x}} \frac{\mathrm{d}}{\mathrm{dx}}(2-3 \mathrm{x})-(2-3 \mathrm{x}) \frac{\mathrm{d}}{\mathrm{dx}} \sqrt{1-\mathrm{x}}}{(1-\mathrm{x})}\right) \cdots \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{\frac{v d u}{d x}+\frac{u d v}{d x}}{v^{2}} \end{aligned}

=\frac{1}{2}\left[\frac{(-3) \sqrt{1-x}-(2-3 x) \cdot 1 / 2(1-x)^{\frac{1}{2}}(-1)}{(1-x)}\right]

\begin{aligned} &=\frac{1}{2}\left[\frac{-3}{\sqrt{1-x}}+\frac{2-3 x}{2(1-x)^{8 / 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} \text { in } f^{\prime \prime}(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}

=-\frac{3}{2} \sqrt{3}<0

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

Posted by

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