Applications Of Derivatives Question 112
Question: The function $ \sin x(1+\cos x) $ at $ x=\frac{\pi }{3} $ , is
Options:
A) Maximum
B) Minimum
C) Neither maximum nor minimum
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ f(x)=\sin x(1+\cos x) $
therefore $ {f}’(x)=\cos 2x+\cos x $ and $ {f}’’(x)=-2\sin 2x-\sin x=-(2\sin 2x+\sin x) $
For maximum or minimum value of $ f(x) $ , $ f’(x)=0 $
$ \cos 2x+\cos x=0 $
therefore $ \cos x=-\cos 2x $
therefore $ \cos x=\cos (\pi \pm 2x) $
$ \therefore x=\pi \pm 2x $ or $ x=\frac{\pi }{3},-\pi $
Now $ {f}’’,( \frac{\pi }{3} )=-2\sin \frac{2\pi }{3}-\sin \frac{\pi }{3}=-2\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}=-\frac{3\sqrt{3}}{2}=-ve $
Hence $ f(x) $ is maximum at $ x=\frac{\pi }{3} $ .
BETA
