APPLICATION OF DERIVATIVES — Maths STD 12 Science — Question
Gujarat BoardEnglish MediumSTD 12 ScienceMathsAPPLICATION OF DERIVATIVES5 Marks
Question
Prove that $\text{f}(\text{x})=\sin\text{x}+\sqrt{3}\cos\text{x}$ has maximum value at $\text{x}=\frac{\pi}{6}$.
✓
Answer
We have, $\text{f}(\text{x})=\sin\text{x}+\sqrt{3}\cos\text{x}$
$\Rightarrow\text{f}'(\text{x})=\cos\text{x}+\sqrt{3}(-\sin\text{x})$
$\Rightarrow\text{f}'(\text{x})=\cos\text{x}-\sqrt{3}-\sin\text{x}$
For f(x) to have maximum to minimum value, We must have f'(x) = 0
$\Rightarrow\cos\text{x}-\sqrt{3}\sin\text{x}=0$
$\Rightarrow\cot\text{x}=\sqrt{3}$
$\Rightarrow\text{x}=\frac{\pi}{6}$
Also, $\text{f}''(\text{x})=-\sin\text{x}-\sqrt{3}\cos\text{x}$
$\text{f}''\Big(\frac{\pi}{6}\Big)=-\sin\frac{-\pi}{6}-\sqrt{3}\cos\frac{\pi}{6}$
$=-\frac{1}{2}-\sqrt{3}\Big(\frac{\sqrt{3}}{2}\Big)=-\frac{1}{2}-\frac{3}{2}=-2<0$
So, $\text{x}=\frac{\pi}{6}$ is point of maxima.
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