#### Please solve R.D Sharma class 12 Chapter 17 Maxima and Minima  Excercise 17.3  question 1 Sub Question 12 maths textbook solution.

x=$\frac{3}{4}$ is a point of local maxima and its maximum value is 5/4.

Hint:

Using chain rule of Differentiation

Given:

\begin{aligned} &\mathrm{f}(\mathrm{x})=\mathrm{x}+\sqrt{1-\mathrm{x}} \\ &\mathrm{x} \leq 1 \end{aligned}

Explanation:

Differentiating f(x) with respect to x

\begin{aligned} &\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}+\sqrt{1-\mathrm{x}}) \\ &=\frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}(\mathrm{x})+\frac{\mathrm{d}}{\mathrm{dx}}(1-\mathrm{x})^{1 / 2} \\ &=1+\frac{1}{2}(1-\mathrm{x})^{\frac{1}{2}-1}(-1) \end{aligned}                                                      [  Using Chain Rule Diffrentiation    ]

\begin{aligned} &=1-\frac{1}{2 \sqrt{1-x}} \\ &=\frac{2 \sqrt{1-x}-1}{2 \sqrt{1-x}} \\ &\text { Put } f^{\prime}(x)=0 \\ &\frac{2 \sqrt{1-x}-1}{2 \sqrt{1-x}}=0 \\ &\Rightarrow 2 \sqrt{1-x}-\mid 1=0 \text { if } x \neq 1 \\ &\Rightarrow \sqrt{1-x}=\frac{1}{2} \end{aligned}                   [Squaring Both Sides]

\begin{aligned} &\text { i. } e 1-x=\frac{1}{4} \\ &x=\frac{3}{4} \end{aligned}

Thus ,  $x=\frac{3}{4}$   is the point of local maxima and minima

Differentiating f’ with respect to x

\begin{aligned} &\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left(1-\frac{1}{2 \sqrt{1-\mathrm{x}}}\right)\\ &=\frac{\mathrm{d}}{\mathrm{dx}}(1)-\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{2 \sqrt{1-\mathrm{x}}}\right)\\ &=0-\frac{1}{2} \frac{\mathrm{d}}{\mathrm{dx}}(1-\mathrm{x})^{-1 / 2}\\ &=\frac{1}{2} \times-\frac{1}{2}(1-x)^{-3 / 2}(-1)\\ &\text { (Using chain rule) }\\ &=-\frac{1}{4(1-x)^{\frac{3}{2}}} \end{aligned}

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

So $x = \frac{3}{4}$is a point of local maxima

The local maximum value at f(x)
\begin{aligned} &\mathrm{f}\left(\frac{1}{2}\right)=\frac{3}{4}+\sqrt{1-\frac{3}{4}} \\ &=\frac{3}{4}+\frac{1}{2}=\frac{3+2}{4}=\frac{5}{4} \end{aligned}