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Provide solution RD Sharma maths class 12 chapter maxima and minima exercise 17.2 question 12 maths textbook solution

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$$ x=2 / 3 \text { is the point of local maxima and the value of local maxima is } \frac{2 \sqrt{3}}{9}


Use first derivative test to find the value and point of local maxima and local minima.


$$ f(x)=x \sqrt{1-x} \quad, x>0

Differentiating $$ f(x) w \cdot r . t^{\prime} x^{\prime} \text { then }

$$ \begin{aligned} \frac{d}{d x}\{f(x)\} &=\frac{d}{d x}(x \sqrt{1-x}) \\ &=x \frac{d}{d x}(\sqrt{1-x})+\sqrt{1-x} \frac{d}{d x}(x) \quad\left[\because \frac{d}{d x}(y x)=y \frac{d}{d x} x+x \frac{d}{d x} y\right] \\ &=x \frac{d}{d x}(\sqrt{1-x})+(\sqrt{1-x}) \frac{d}{d x}(x) \\ &=x \frac{1}{2}(1-x)^{\frac{1}{2}-1} \frac{d}{d x}(1-x)+\sqrt{1-x} \cdot 1\left[\because \frac{d}{d x}(x+a)^{n}=n(x+a)^{n-1} \frac{d}{d x}(x+a)\right] \\ \end{aligned}

                        =x \cdot \frac{1}{2}(1-x)^{\frac{1-2}{1}}\left\{\frac{d}{d x}(1)-\frac{d}{d x}(x)\right\}+\sqrt{1-x} \quad\left[\because \frac{d}{d x}(x+a)=\frac{d}{d x}(x)+\frac{d}{d x}(a)\right] \\

                        =\frac{x}{2}(1-x)^{-1 / 2}\{0-1\}+\sqrt{1-x} \quad\left[\because \frac{d}{d x} \text { Constant }=0, \frac{d}{d x}(x)=1\right] \\

                        =\frac{x}{2 \sqrt{1-x}}(0-1)+\sqrt{1-x} \\

                        =\frac{-x}{2 \sqrt{1-x}}+\sqrt{1-x} \\

                        =\frac{-x+2(1-x)}{2 \sqrt{1-x}} \\

                        =\frac{-x+2-2 x}{2 \sqrt{1-x}} \\

                        =\frac{2-3 x}{2 \sqrt{1-x}}

           {f}'\left ( x \right )=\frac{-\left ( 3x-2 \right )}{2 \sqrt{1-x}}     

By first derivative test, for local maxima and local minima, we have

\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow \frac{-(3 x-2)}{2 \sqrt{1-x}}=0 \quad \Rightarrow \frac{3 x-2}{2 \sqrt{1-x}}=0 \Rightarrow \frac{3 x-2}{\sqrt{1-x}}=0 \quad\left[\because \frac{1}{2} \neq 0\right] \\ &\Rightarrow 3 x-2=0 \quad \Rightarrow \quad 3 x=2 \quad x=2 / 3 \end{aligned}

\text { Since, } f^{\prime}(x) \text { changes } f \text { rom }+\text { ve to }-\text { ve when } x \text { increases through } 2 / 3 \text { is the point of local minima }

\begin{aligned} &\therefore \text { the value of the local maxima of } f(x) \text { at } x=2 / 3 \text { is }\\ &f\left(\frac{2}{3}\right)=\frac{2}{3} \sqrt{1-\frac{2}{3}}=\frac{2}{3} \sqrt{\frac{3-2}{3}}=\frac{2}{3} \sqrt{\frac{1}{3}}\\ &=\frac{2}{3} \cdot \frac{1}{\sqrt{3}}=\frac{2}{3 \sqrt{3}}\\ &=\frac{2}{3 \sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{2 \sqrt{3}}{3 \times 3}=\frac{2 \sqrt{3}}{9} \end{aligned}


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