Test Index

Army Group X Practice Set 4

© examsnet.com

Question : 70 of 70

Marks: +1, -0

The maximum value of

\sin x(1+\cos x)

will be at

$x=\frac{\pi}{2}$
$x=\frac{\pi}{6}$
$x=\frac{\pi}{3}$
$x=\pi$

Solution:

y=\sin x(1+\cos x)=\sin x+\frac{1}{2}\sin 2x

[\because \sin 2\theta =2\sin\theta \cos\theta ]

\therefore \frac{dy}{dx}=\cos x+\cos 2x

and

\frac{d^2y}{dx^2}=-\sin x-2\sin 2x

For maximum and minimum ,

\frac{dy}{dx}=0

\cos x+\cos 2x=0

\Rightarrow \cos x=-\cos 2x=\cos(\pi -2x)

\Rightarrow x=\pi-2x

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

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

=-\frac{\sqrt{3}}{2}-2\cdot\frac{\sqrt{3}}{2}

=-\frac{3\sqrt{3}}{2},

which is negative

\therefore \text{ At } x=\frac{\pi}{3},

the function is maximum.

© examsnet.com

Go to Question:

Prev Question