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Prove that f(x)=\sin x+\sqrt{3} \cos x has maximum value at x=\frac{\pi}{6}.

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Given: f(x)=\sin x+\sqrt{3} \cos x

To prove: the given function has maximum value at x=\frac{\pi}{6}

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

While calculating the first derivative we get,

\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}(\sin \mathrm{x})}{\mathrm{dx}}+\frac{\mathrm{d}(\sqrt{3} \cos \mathrm{x})}{\mathrm{d} \mathrm{x}}

Then putting the derivative, we get

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

Critical point ca be calculated by equating the derivative with 0,

f'(x) = 0

\Rightarrow cos x- \sqrt 3sin x =0 \Rightarrow \sqrt 3sin x = cos x
\\ \Rightarrow \frac{\sin x}{\cos x}=\frac{1}{\sqrt{3}}$ \\$\Rightarrow \tan x=\frac{1}{\sqrt{3}}$ \\$x=\frac{\pi}{6}
This is possible only when
The second derivative if the c=function can be calculated by,
\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{\mathrm{d}(\cos \mathrm{x})}{\mathrm{dx}}-\sqrt{3} \frac{\mathrm{d}(\sin \mathrm{x})}{\mathrm{dx}}
Putting the value of derivative, we get
\\ f^{\prime \prime}(x)=-\sin x-\sqrt{3} \cos x
Then, we will substitute \\x=\frac{\pi}{6} in the above equation, we get
\\ f^{\prime \prime}(x)=-\sin x-\sqrt{3} \cos x$ \\$f^{\prime}\left(\frac{\pi}{6}\right)=-\sin \frac{\pi}{6}-\sqrt{3} \cos \frac{\pi}{6}

Putting the corresponding value, we get

\\ \mathrm{f}^{\prime}\left(\frac{\pi}{6}\right)=-\frac{1}{2}-\sqrt{3} \times \frac{\sqrt{3}}{2} \\ \mathrm{f}^{\prime}\left(\frac{\pi}{6}\right)=-\frac{1}{2}-\frac{3}{2} \\ \mathrm{f}^{\prime}\left(\frac{\pi}{6}\right)=-2<0

Hence f(x) has a maximum value at x=\frac{\pi}{6}.

Hence proved.

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