Applications Of Derivatives Question 314
Question: The maximum value of $ \sin x(1+\cos x) $ will be at the
[UPSEAT 1999]
Options:
A) $ x=\frac{\pi }{2} $
B) $ x=\frac{\pi }{6} $
C) $ x=\frac{\pi }{3} $
D) $ x=\pi $
Show Answer
Answer:
Correct Answer: C
Solution:
$ y=\sin x(1+\cos x) $
$ =\sin x+\frac{1}{2}\sin 2x $ $ \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 $
therefore $ \cos x=-\cos 2x=\cos (\pi -2x) $
therefore $ x=\pi -2x $ $ x=\frac{\pi }{3} $ ; $ {{( \frac{d^{2}y}{dx^{2}} )}_{x=\pi /3}}=-\sin ( \frac{1}{3}\pi )-2\sin ( \frac{2}{3}\pi ) $ = $ \frac{-\sqrt{3}}{2}-2.\frac{\sqrt{3}}{2} $ = $ \frac{-3\sqrt{3}}{2} $ , which is negative.
At $ x=\frac{\pi }{3} $ the function is maximum.
BETA
