Question

The interval in which the function $y=x-2\sin x;0\le x\le 2\pi$ increases throughout is :

• A. $(0,\frac{\pi }{4})$
• B. $(\frac{5\pi }{3},2\pi )$
• C. $(0,\frac{\pi }{3})$
• D. $(\frac{\pi }{3},\frac{5\pi }{3})$

Answer 1

The correct option is D $(\frac{\pi }{3},\frac{5\pi }{3})$

$y=x-2\sin x\Rightarrow \frac{dy}{dx}=1-2\cos x$
For function to be increasing
$\frac{dy}{dx}>0\Rightarrow 1-2\cos x>0\Rightarrow \cos x<\frac{1}{2}\therefore x\in (\frac{\pi }{3},\frac{5\pi }{3})$