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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Prove that $f(x)=\sin x+\sqrt 3\cos x$ has maximum value at $x=\Large\frac{1}{6}$.

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  • Let $f(x)$ be a function with domain $D \subset R$. Then $f(x)$ is said to attain the maximum value at a point, $ a \in D,$ if $ f'(x) \leq f(a)$ for all $ x \in D$.
Step 1
$ f(x)= \sin x+ \sqrt 3 \cos x$
differentiating w.r.t $x$ we get,
$ f'(x)= \cos x+ \sqrt 3(-\sin x)$
$ = \cos x-\sqrt 3 \sin x$
at $ x = \large\frac{\pi}{6}$
$ f'(x)= \cos \large\frac{\pi}{6}-\sqrt 3 \times \sin \large\frac{\pi}{6}$
But $ \cos \large\frac{\pi}{6}=\large\frac{\sqrt 3}{2} \: and \: \sin \large\frac{\pi}{6}=\large\frac{1}{2}$
$ \therefore f'(x)=\large\frac{\sqrt 3}{2}-\large\frac{\sqrt 3}{2}=0$
$ f(x)=\sin\large\frac{\pi}{6}+\sqrt 3 \times \cos \large\frac{\pi}{6}$
$ = \large\frac{1}{2}+\large\frac{\sqrt 3}{2}$
$ = \large\frac{\sqrt 3+1}{2}$
Step 2
Again differentiating $f(x)$ w.r.t $x$ we get,
$ f''(x)-\sin x - \sqrt 3 \cos x$
$ f''(x)=-( \sin x+ \sqrt 3 \cos x)$
$ f''(x)=-f(x)$
Since $f''(x)<0,$ the given function is maximum.
Hence the maximum value is $ \large\frac{\sqrt 3+1}{2}$


answered Aug 6, 2013 by thanvigandhi_1
edited Aug 6, 2013 by thanvigandhi_1

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