#### Please solve rd  sharma class 12 Chapter 17  Maxima and Minima  exercise 17.3 question 8  maths textbook solution

Hence proved.

Hint: $\mathrm{f}^{\prime}(\mathrm{x})=0$

Put,

Given:

$\mathrm{f}(\mathrm{x})=\sin \mathrm{x}+\sqrt{3} \cos \mathrm{x}$

Explanation:

Differentiating fx with respect to x

\begin{aligned} &\mathrm{f}^{\prime}(\mathrm{x})=\cos x-\sqrt{3} \sin \mathrm{x} \\ &\text { Put } \mathrm{f}^{\prime}(\mathrm{x})=0 \\ &\cos \mathrm{x}-\sqrt{3} \sin \mathrm{x}=0 \\ &\Rightarrow \sqrt{3} \sin \mathrm{x}=\cos \mathrm{x} \\ &\Rightarrow \tan \mathrm{x}=\frac{1}{\sqrt{3}} \end{aligned}

$\Rightarrow x=\frac{\pi}{6}$

Differentiating f’(x) with respect to x ,

\begin{aligned} &f^{\prime \prime}(x)=-\sin x-\sqrt{3} \cos x \\ &\text { Putt } x=\frac{\pi}{6} \text { at } f^{\prime \prime}(x) \\ &f^{\prime \prime}\left(\frac{\pi}{6}\right)=-\sin \frac{\pi}{6}-\sqrt{3} \cos \frac{\pi}{6} \\ &=-\frac{1}{2}-\sqrt{3} \times \frac{\sqrt{3}}{2} \\ &=-\frac{1}{2}-\frac{3}{2} \\ &=-\frac{4}{2} \\ &=-2<0 \end{aligned}

So, $x = \frac{\pi}{6}$ is a point local maxima