APPLICATION OF DERIVATIVES — Maths STD 12 Science — Question
Gujarat BoardEnglish MediumSTD 12 ScienceMathsAPPLICATION OF DERIVATIVES3 Marks
Question
Prove that $\text{f(x)}=\sin\text{x}+\sqrt{3}\cos\text{x}$ has maximum value at $\text{x}=\frac{\pi}{6}.$
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Answer
We have, $\text{f(x)}=\sin\text{x}+\sqrt{3}\cos\text{x}$ $\therefore\ \text{f}'(\text{x})=\cos\text{x}-\sqrt{3}\sin\text{x}$ For $\text{f}'(\text{x)}=0,\cos\text{x}=\sqrt{3}\sin\text{x}$ $\Rightarrow\ \tan\text{x}=\frac{1}{\sqrt{3}}$ $\Rightarrow\ \text{x}=\frac{\pi}{6}$ Differentiating f'(x), we get $\text{f}''(\text{x})=-\sin\text{x}-\sqrt{3}\cos\text{x}$ $\text{f}''\Big(\frac{\pi}{6}\Big)=-\sin\frac{\pi}{6}-\sqrt{3}\cos\frac{\pi}{6}<0$ $=-\frac{1}{2}-\sqrt{3}.\frac{\sqrt{3}}{2}$ $=-\frac{1}{2}-\frac{3}{2}=-2<0$ Hence, at $\text{x}=\frac{\pi}{6},\text{f(x)}$ has maxima value and $\frac{\pi}{6}$ is the point of local maxima.
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