Question

The maximum value of $\sin x(1 + \cos x)$ will be at

• A. $x = \frac{\pi}{2}$
• B. $x = \frac{\pi}{6}$
• C. $x = \frac{\pi}{3}$
• D. ) $x = \pi$

Answer 1

Answer: (C)

$x = \frac{\pi}{3}$

Answer 2

$y = \sin x(1 + \cos x) = \sin x + \frac{1}{2}\sin 2x\therefore$ $\frac{dy}{dx} = \cos x + \cos 2x$ and $\frac{d^{2}y}{dx^{2}} = - \sin x - 2\sin 2x$

On putting $\frac{dy}{dx} = 0,\cos x + \cos 2x = 0$

⇒ $\cos x = - \cos 2x = \cos(\pi - 2x)$ ⇒ $x = \pi - 2x$

$\therefore$ $x = \frac{\pi}{3}$, $\therefore$ $\left( \frac{d^{2}y}{dx^{2}} \right)_{x = \pi/3} = - \sin\left( \frac{1}{3}\pi \right) - 2\sin\left( \frac{2}{3}\pi \right)$

= $\frac{- \sqrt{3}}{2} - 2.\frac{\sqrt{3}}{2} = \frac{- 3\sqrt{3}}{2}$ which is negative.

$\therefore$ at $x = \frac{\pi}{3}$ the function is maximum.