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  • Need solution for RD Sharma maths class 12 chapter 17 Maxima and minima exercise 17 point 3 question 1 subquestion xi

 Need solution for RD Sharma maths class 12 chapter 17 Maxima and minima  exercise 17 point 3 question 1 subquestion (xi)

Answers (1)

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Answer:

x=1 is a point of local maxima and its maximum value is 1 and x=-1 is a point of local minima and its minimum value is -1.

Hint:

Putting x=1,-1in f''(x)

Given:

\begin{aligned} &f(x)=x \sqrt{2-x^{2}} \\ &-\sqrt{2} \leq x \leq \sqrt{2} \end{aligned}

Explanation:

Differentiating f(x) with respect to x

f^{\prime}(x)=\frac{d}{d x}\left(x \sqrt{2-x^{2}}\right)

\begin{aligned} &\text { Using, }\\ &\frac{\mathrm{d}}{\mathrm{dx}}[\mathrm{f}(\mathrm{x}) \cdot \mathrm{g}(\mathrm{x})]=\mathrm{f}(\mathrm{x}) \frac{\mathrm{d}}{\mathrm{dx}}\{\mathrm{g}(\mathrm{x})\}+\mathrm{g}(\mathrm{x}) \frac{\mathrm{d}}{\mathrm{dx}}\{\mathrm{f}(\mathrm{x})\}\\ &\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{x} \frac{\mathrm{d}}{\mathrm{dx}} \sqrt{2-\mathrm{x}^{2}}+\sqrt{2-\mathrm{x}^{2}} \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})-\mathrm{-}(1) \end{aligned}

\begin{aligned} &\text { Using, }\\ &\frac{\mathrm{df}(\mathrm{x})^{\mathrm{n}}}{\mathrm{dx}}=\mathrm{nf}(\mathrm{x})^{\mathrm{n}-1} \frac{\mathrm{d}}{\mathrm{dx}} \mathrm{f}(\mathrm{x}) \text { and }\\ &\frac{\mathrm{d}}{\mathrm{dx}}[\mathrm{f}(\mathrm{x}) \pm \mathrm{g}(\mathrm{x})]=\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{f}(\mathrm{x})) \pm \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{g}(\mathrm{x}))\\ &\frac{\mathrm{d}}{\mathrm{dx}} \sqrt{2-\mathrm{x}^{2}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(2-\mathrm{x}^{2}\right)^{1 / 2}\\ &=\frac{1}{2}\left(2-x^{2}\right)^{1 / 2-1} \frac{d}{d x}\left(2-x^{2}\right)\\ &=\frac{1}{2}\left(2-x^{2}\right)^{1 / 2}(-2 x)\\ &=-\frac{x}{\sqrt{2-x^{2}}} \end{aligned}

\begin{aligned} &\therefore \operatorname{From}(1), \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{x}\left(-\frac{\mathrm{x}}{\sqrt{2-\mathrm{x}^{2}}}\right)+\sqrt{2-\mathrm{x}^{2} \times 1} \\ &=\frac{-\mathrm{x}^{2}+2-\mathrm{x}^{2}}{\sqrt{2-\mathrm{x}^{2}}} \\ &=\frac{2-2 \mathrm{x}^{2}}{\sqrt{2-\mathrm{x}^{2}}} \end{aligned}

\begin{aligned} &\text { Put, }\\ &\mathrm{f}^{\prime}(\mathrm{x})=0\\ &\frac{2-2 x^{2}}{\sqrt{2-x^{2}}}=0\\ &\Rightarrow 2-2 x^{2}=0\\ &\Rightarrow 2 \mathrm{x}^{2}=2\\ &\Rightarrow \mathrm{x}^{2}=1\\ &\Rightarrow x=\pm 1\\ &\text { if } x \neq \sqrt{2} \text { and } x \neq-\sqrt{2} \end{aligned}

These x=1 and x=-1 we the possible point of local maxima and minima

Differentiating f’(x) with respect to x

\begin{aligned} \mathrm{f}^{\prime \prime}(\mathrm{x}) &=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{2-2 \mathrm{x}^{2}}{\sqrt{2-\mathrm{x}^{2}}}\right) \\ &=\frac{\sqrt{2-\mathrm{x}^{2}} \frac{\mathrm{d}}{\mathrm{dx}}\left(2-2 \mathrm{x}^{2}\right)-\left(2-2 \mathrm{x}^{2}\right) \frac{\mathrm{d}}{\mathrm{dx}} \sqrt{2-\mathrm{x}^{2}}}{\left(2-\mathrm{x}^{2}\right)} \end{aligned}

Using the formula,

\begin{aligned} &\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{u}}{\mathrm{v}}\right)=\frac{\mathrm{v} \frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}-\mathrm{u} \frac{\mathrm{d}}{\mathrm{ax}} \mathrm{v}}{\mathrm{v}^{2}} \\ &=\frac{\sqrt{2-\mathrm{x}^{2}} \times(-4 \mathrm{x})-\frac{\left(2-2 \mathrm{x}^{2}\right) 1}{2}\left(2-\mathrm{x}^{2}\right)^{-1 / 2}(-2 \mathrm{x})}{\sqrt{2-\mathrm{x}^{2}} 2} \quad \text { (Applying chain rule) } \\ &=\frac{-4 \mathrm{x}}{\sqrt{2-\mathrm{x}^{2}}}+\frac{\mathrm{x}\left(2-2 \mathrm{x}^{2}\right)}{\left(2-\mathrm{x}^{2}\right)^{3 / 2}} \\ &=\frac{-4 \mathrm{x}\left(2-\mathrm{x}^{2}\right)+2 \mathrm{x}-2 \mathrm{x}^{3}}{\left(2-\mathrm{x}^{2}\right)^{3 / 2}} \\ &=\frac{-8 \mathrm{x}+4 \mathrm{x}^{3}+2 \mathrm{x}-2 \mathrm{x}^{3}}{\left(2-\mathrm{x}^{2}\right)^{3 / 2}} \end{aligned}

    \begin{aligned} &\mathrm{f}^{\prime \prime}(\mathrm{x}) \text { at } \mathrm{x}=1 \\ &\mathrm{f}^{\prime \prime}(1)=\frac{-6(1)+2(1)^{3}}{\left(2-(1)^{2}\right)^{3 / 2}} \\ &=\frac{-6+2}{(1)^{3 / 2}}=-4<0 \end{aligned}

So x=1 is local maxima and its maximum value f(1)=1

    \begin{aligned} &f^{\prime \prime}(x) \text { at } x=-1 \\ &f^{\prime \prime}(-1)=\frac{-6(-1)+(2)(-1)^{3}}{\left(2-(-1)^{2}\right)^{3 / 2}} \\ &=\frac{6-2}{(\sqrt{1})^{3}}=4>0 \end{aligned}

So , x = 1 is a point of local maxima and its minimum value f(-1)=-1

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