Explain Solution R.D. Sharma Class 12 Chapter 17  Maxima and Minima  Exercise 17.3 Question 1 Sub Question 11 Maths Textbook Solution.

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 $x=1,-1$ in $\mathrm{f}^{\prime \prime}(\mathrm{x})$

Given:

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

Explanation:

Differentiating f(x) with respect to x

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

using

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

Using

\begin{aligned} &\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{d} x}[\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})) \end{aligned}

\begin{aligned} &\frac{d}{d x} \sqrt{2-x^{2}}=\frac{d}{d x}\left(2-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) \end{aligned}

\begin{aligned} &=\frac{1}{2}\left(2-\mathrm{x}^{2}\right)^{1 / 2}(-2 \mathrm{x}) \\ &=-\frac{\mathrm{x}}{\sqrt{2-\mathrm{x}^{2}}} \end{aligned}

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

\begin{aligned} &=\frac{-x^{2}+2-x^{2}}{\sqrt{2-x^{2}}} \\ &=\frac{2-2 x^{2}}{\sqrt{2-x^{2}}} \end{aligned}

Put,

\begin{aligned} &\mathrm{f}^{\prime}(\mathrm{x})=0 \\ &\frac{2-2 \mathrm{x}^{2}}{\sqrt{2-\mathrm{x}^{2}}}=0 \\ &\Rightarrow 2-2 \mathrm{x}^{2}=0 \\ &\Rightarrow 2 \mathrm{x}^{2}=2 \\ &\Rightarrow \mathrm{x}^{2}=1 \\ &\Rightarrow \mathrm{x}=\pm 1 \end{aligned}

$\text { if } x \neq \sqrt{2} \text { and } x \neq-\sqrt{2}$

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{d} x}\left(\frac{\mathrm{u}}{\mathrm{v}}\right)=\frac{\mathrm{v} \frac{\mathrm{d}}{\mathrm{dx}} \mathrm{u}-\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}} \end{aligned}                (Applying chain rule)

$=\frac{-4 \mathrm{x}}{\sqrt{2-\mathrm{x}^{2}}}+\frac{\mathrm{x}\left(2-2 \mathrm{x}^{2}\right)}{\left(2-\mathrm{x}^{2}\right)^{3 / 2}}$

\begin{aligned} &=\frac{-4 x\left(2-x^{2}\right)+2 x-2 x^{3}}{\left(2-x^{2}\right)^{3 / 2}} \\ &=\frac{-8 x+4 x^{3}+2 x-2 x^{3}}{\left(2-x^{2}\right)^{3 / 2}} \end{aligned}

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

$=\frac{-6+2}{(1)^{3 / 2}}=-4<0$

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

\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}{(\sqrt{1})^{3}}=4>0 \end{aligned}

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