Question
$\text{f}(\text{x})=\sin +\sqrt{3}\cos\text{x}$ is maximum when x =
- $\frac{\pi}{3}$
- $\frac{\pi}{4}$
- $\frac{\pi}{6}$
- $0$
✓
Answer
-
$\frac{\pi}{6}$
Solution:
$\text{f}(\text{x})=\sin +\sqrt{3}\cos\text{x}$
$\Rightarrow\text{f}'(\text{x})=\cos\text{x}-\sqrt{3}\sin\text{x}$
For maxima or maxima,
f'(x) = 0
$\cos\text{x}-\sqrt{3}\sin\text{x}=0$
$\Rightarrow\ \tan\text{x}=\frac{1}{\sqrt{3}}\Rightarrow\text{x}=\frac{\pi}{6}$
$\text{f}''\Big(\frac{\pi}{6}\Big)=-\sin\frac{\pi}{6}-\sqrt{3}\cos\frac{\pi}{6}=\frac{-1-\sqrt{3}}{2}<0$
function has local maima at $\text{x}=\frac{\pi}{6}$
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