Question

The number of values of $x\in [0,4\pi ]$ satisfying $|\sqrt{3}\cos x-\sin x|\ge 2$ is

• A. 2
• B. 0
• C. 4
• D. 8

Answer 1

The correct option is C 4

given, $\mid \sqrt{3}cosx-sinx\mid \ge 2$

$\Rightarrow \mid cos\frac{\pi }{6}cosx-cos\frac{\pi }{6}sinx\mid \ge 1$

$\Rightarrow \mid cos(x+\frac{\pi }{6})\mid \ge 1$

but we know $\mid cos(x+\frac{\pi }{6})\mid \le 1$

Hence,
$\Rightarrow \mid cos(x+\frac{\pi }{6})\mid =1\Rightarrow cos(x+\frac{\pi }{6})=\pm 1$

Hence solutions in the given interval are,
$x=\frac{3\pi }{4},\frac{7\pi }{4},\frac{11\pi }{4},\frac{15\pi }{4}$
Hence, option 'C' is correct.