PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
Share

# Prove that F(X) = Sinx + √ 3 Cosx Has Maximum Value at X = π 6 ? - PUC Karnataka Science Class 12 - Mathematics

#### Question

Prove that f(x) = sinx + $\sqrt{3}$ cosx has maximum value at x = $\frac{\pi}{6}$ ?

#### Solution

$\text{We have },$

$f\left( x \right) = \sin x + \sqrt{3}\cos x$

$\Rightarrow f'\left( x \right) = \cos x + \sqrt{3}\left( - \sin x \right)$

$\Rightarrow f'\left( x \right) = \cos x - \sqrt{3}\sin x$

$\text { For } f\left( x \right) \text { to have maximum or minimum value, we must have } f'\left( x \right) = 0$

$\Rightarrow cos x - \sqrt{3}sin x = 0$

$\Rightarrow cos x = \sqrt{3}sin x$

$\Rightarrow \cot x = \sqrt{3}$

$\Rightarrow x = \frac{\pi}{6}$

$\text { Also }, f''\left( x \right) = -\text { sin } x - \sqrt{3}\cos x$

$\Rightarrow f''\left( \frac{\pi}{6} \right) = - \sin\frac{\pi}{6} - \sqrt{3}\cos\frac{\pi}{6} = - \frac{1}{2} - \sqrt{3}\left( \frac{\sqrt{3}}{2} \right) = - \frac{1}{2} - \frac{3}{2} = - 2 < 0$

$\text { So, x } = \frac{\pi}{6} \text { is point of maxima } .$

Is there an error in this question or solution?

#### Video TutorialsVIEW ALL [1]

Solution Prove that F(X) = Sinx + √ 3 Cosx Has Maximum Value at X = π 6 ? Concept: Graph of Maxima and Minima.
S