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Need solution for RD Sharma maths class 12 chapter Maxima and Minima exercise Multiple Choice question, question 18.

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Answer: option(c) \frac{\pi}{6}

Hint: For local maxima or minima, we must have f'(x)=0.

Given: f(x)=\sin x+\sqrt{3}\cos x


We have,

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

f'(x)=\cos x-\sqrt{3}\sin x                                             

For maxima and minima  f'(x)=0

\Rightarrow \cos x-\sqrt{3}\sin x=0

\Rightarrow \cos x=\sqrt{3}\sin x

\Rightarrow \tan x=\frac{1}{\sqrt{3}}

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


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

So x=\frac{\pi}{6} is local maxima.

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