Question

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

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

Answer 1

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

Given
$y=x-2\sin x,0\le x\le 2\pi$
On differentiating both sides w.r.t $x$ we get
$\frac{dy}{dx}=1-\cos x$
$\frac{dy}{dx}=0$ (Given)
$1-2\cos x=0$
$\cos x=\frac{1}{2}$
$\because 0\le x\le 2\pi$
For $x=\frac{\pi }{3},\frac{5\pi }{3},\cos x=\frac{1}{2}$
$\therefore \frac{dy}{dx}>0$ for $x\in (\frac{\pi }{3},\frac{5\pi }{3})$
$\frac{dy}{dx}<0$ for $x\in (0,\frac{\pi }{3})\cup (\frac{5\pi }{3},2\pi )$
$y=x-2\sin x$ increases in interval $(\frac{\pi }{3},\frac{5\pi }{3})$