Question

Assertion: The number of values of \[x\in \left[ 0,4\pi \right] \] satisfying \[\left| \sqrt{3}\cos x-\sin x \right|\ge 2\] is \[2\] .
Reason: \[\left| \cos \left( x+\dfrac{\pi }{6} \right) \right|=1\Rightarrow \] number of solution of \[\left| \sqrt{3}\cos x-\sin x \right|\ge 2\] is \[4\] .
A.Both assertion and reason are correct and the reason is the correct explanation for assertion.
B.Both assertion and the reason are correct but the 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

Hint: In the given question, we have given the inequality i.e. a type of equation that does not have an equals to sign. In order to solve the question, first we need to eliminate the modulus sign and write the absolute value by taking the positive and the negative value. Then we put \[\sqrt{3}=r\cos a,\ 1=r\sin a\] and we will get \[r=2\ and\ a=\dfrac{\pi }{6}\] . Later we get the new required equation i.e. \[2\cos \left( x+\dfrac{\pi }{6} \right)=\pm 2\] and solve for the value of ‘x’. in this way we will get the correct required answer.

Complete step-by-step answer:
We have given that,
 \[\Rightarrow \left| \sqrt{3}\cos x-\sin x \right|\ge 2\]
Therefore,
 \[\Rightarrow \sqrt{3}\cos x-\sin x=\pm 2\]
Now,
Putting \[\sqrt{3}=r\cos a,\ 1=r\sin a\]
Thus,
 \[\Rightarrow r=2\ and\ a=\dfrac{\pi }{6}\]
So,
The equation becomes;
 \[\Rightarrow 2\cos \left( x+\dfrac{\pi }{6} \right)=\pm 2\]
Dividing both the sides of the equation by 2, we will get
 \[\Rightarrow \cos \left( x+\dfrac{\pi }{6} \right)=\pm 1\]
Therefore,
 \[\Rightarrow x=\dfrac{5\pi }{6},\dfrac{11\pi }{6},\dfrac{17\pi }{6},\dfrac{23\pi }{6}\ for\,x\in \left[ 0,4\pi \right] \]
Thus,
There are four solutions.
Hence, the option (b) is the correct answer i.e. both assertion and the reason are correct but the reason is not the correct explanation for assertion.
So, the correct answer is “Option B”.

Note: Inequality basically defines the region of the quantity whosoever for which the inequality is used for, that is it gives you a range of possible values for the quantity you are finding for. In this range real as well as complex range also occurs. Here you have to be sure of the positive values of cos function because if you take negative values also then signs of the inequality will change accordingly.