Question

If the point $(2\cos \theta ,2\sin \theta )$, for $\theta ϵ(0,2\pi )$ lies in the region between the lines $x+y=2$ and $x-y=2$ containing the origin, then $\theta$ lies in

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

Answer 1

The correct option is C $(\frac{\pi }{2},\frac{3\pi }{2})$

Given point $P(2\cos \theta ,2\sin \theta )$ lies in between $x+y-2=0$ and $x-y-2=0$

Hence $(2\cos \theta +2\sin \theta -2)(2\cos \theta -2\sin \theta -2)<0$

${\cos }^2\theta -{\sin }^2\theta -1<0$

${\cos }^2\theta -1+{\cos }^2\theta -1<0$

${\cos }^2\theta <1$

$\cos \theta <1$ and $\cos \theta <-1$

$\theta <\frac{\pi }{2}$ and $-\cos \theta >1$

$\theta <\frac{\pi }{2}$ and $\cos \theta >\cos \frac{3\pi }{2}$ Since $-\cos \theta =\cos \theta$

$\theta <\frac{\pi }{2}$ and $\theta >\frac{3\pi }{2}$

$\frac{\pi }{2}<\theta <\frac{3\pi }{2}$

$\theta ϵ(\frac{\pi }{2},\frac{3\pi }{2})$