Question

Let $A=\{x\in ,R/|\sqrt{3}\cos x-\sin x|\ge 2,0\le x\le 2\pi \}$ If $x_1\in A,x_2\in A$, then $\frac{x_1}{x_2}=$

• A. $\frac{5}{23}$
• B. $\frac{11}{17}$
• C. $\frac{5}{11}$
• D. $\frac{11}{23}$

Answer 1

Answer: (C)

We have,
$|\sqrt{3}\cos x-\sin x|\ge 2$
It is possible only when
$\sqrt{3}\cos x-\sin x=2$ or $\sqrt{3}\cos x-\sin x=-2$
$\frac{\sqrt{3}}{2}\cos x-\frac{1}{2}\sin x=1$
or $\frac{\sqrt{3}\cos x}{2}-\frac{\sin x}{2}=-1$
$\cos \left(x+\frac{\pi }{6}\right)=1$ or $\cos \left(x+\frac{\pi }{6}\right)=-1$
$x+\frac{\pi }{6}=2\pi$ or $x+\frac{\pi }{6}=\pi$
$\Rightarrow x=\frac{11\pi }{6}$ or $x=\frac{5\pi }{6}$
$\therefore$ Let $x_1=\frac{5\pi }{6}$ and $x_2=\frac{11\pi }{6}$
$\Rightarrow \frac{x_1}{x_2}=\frac{5}{11}$