Question

Assertion :The number of values of $xϵ[0,4\pi ]$ satisfying $\mid \sqrt{3}\cos x-\sin x\mid \ge 2$ is 2. Reason: $\mid cos(x+\pi /6)\mid =1\Rightarrow$ number of solutions of $\mid \sqrt{3}\cos x-\sin x\mid \ge 2$ is 4.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is D Assertion is incorrect but Reason is correct

Given,$\mid \sqrt{3}\cos x-\sin x\mid \ge 2$
$\Rightarrow \sqrt{3}\cos x-\sin x=\pm 2$

Put $\sqrt{3}=r\cos \alpha ,1=r\sin \alpha$

$\Rightarrow r=2,\alpha =\frac{\pi }{6}$

So, the equation becomes $2\cos (x+\frac{\pi }{6})=\pm 2$
$\Rightarrow \cos (x+\frac{\pi }{6})=\pm 1$
$\Rightarrow x=\frac{5\pi }{6},\frac{11\pi }{6},\frac{17\pi }{6},\frac{23\pi }{6}$ for $x\in [0,4\pi ]$
Hence, there are four solutions.