Explain Solution R.D.Sharma Class 12 Chapter 17  Maxima and Minima  Exercise 17.3 Question 8 Maths Textbook Solution.

Hence proved.

Hint:

Put, $f^{\prime}(x)=0$

Given:

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

Explanation:

Differentiating $f\left ( x \right )$ with respect to x

$\mathrm{f}^{\prime}(\mathrm{x})=\cos \mathrm{x}-\sqrt{3} \sin \mathrm{x}$

Put,$f^{\prime}(x)=0$

\begin{aligned} &\cos x-\sqrt{3} \sin x=0 \\ &\Rightarrow \sqrt{3} \sin x=\cos x \\ &\Rightarrow \tan x=\frac{1}{\sqrt{3}} \\ &\Rightarrow x=\frac{\pi}{6} \end{aligned}

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} a t f^{\prime \prime}(x) \end{aligned}

\begin{aligned} &\mathrm{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} \end{aligned}

\begin{aligned} &=-\frac{1}{2}-\frac{3}{2} \\ &=-\frac{4}{2} \\ &=-2<0 \end{aligned}

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