Question

The domain of the function $f(x)=\sqrt{\cos x-1}$ is
A) $R-\\{n\pi :n\in Z\\}$
B) $\\{n\pi :n\in Z\\}$
C) $\\{2n\pi :n\in Z\\}$
D) $\left( -\infty ,\infty \right)$

Answer 1

Hint: The domain of a function is the set from which all of the values of input of the function are taken. We know that a defined function never has an overall negative term inside the square root. So, $\cos x-1\ge 0$ and solve further. We know that if $\cos x=1$ then x=0, $2\pi ,4\pi ,6\pi ,.............$
So, the general solution of $\cos x=1$ is $x=2n\pi$ .
Complete step-by-step answer:
According to the question, we have the function, $f(x)=\sqrt{\cos x-1}$ ……………….(1)
We know that a defined function never has an overall negative term inside the square root. So, the term inside the square root must be greater than zero.
From equation (1) we can get the term inside the square root and make it positive.
So, $\cos x-1\ge 0$ ……………….(2)
Solving equation (2), we get
$\cos x-1\ge 0$
$\Rightarrow \cos x\ge 1$
We know that the range of cosine function is [-1,1]. The maximum value of cosine function can be equal to 1 but not more than 1.
So, $\cos x$ cannot be greater than 1.
It can only be equal to 1.
So, $\cos x=1$………………….(3)
We know that $\cos 0=1$ . Now using this in equation (3), we get

& \cos x=1 \\\ & \Rightarrow \cos x=\cos 0 \\\ & \Rightarrow x=0 \\\ \end{aligned}$$ We also know that $$\cos 2\pi =1$$ . Now using this in equation (3), we get $$\begin{aligned} & \cos x=1 \\\ & \Rightarrow \cos x=\cos 2\pi \\\ & \Rightarrow x=2\pi \\\ \end{aligned}$$ We also know that, $$\cos 4\pi =1$$ . Now using this in equation (3), we get $$\begin{aligned} & \cos x=1 \\\ & \Rightarrow \cos x=\cos 4\pi \\\ & \Rightarrow x=4\pi \\\ \end{aligned}$$ We can see that x can be 0, $$2\pi $$ , $$4\pi $$ , $$6\pi $$ etc. We can observe that x is the multiple of $$2\pi $$ ……………….(4) Now, from equation (3) and equation (4), we get $$\cos x=1$$ $$\begin{aligned} & \Rightarrow \cos x=\cos 2n\pi \\\ & \Rightarrow x=2n\pi \\\ \end{aligned}$$ Therefore, option (C) is the correct option. Note: In this question, one may write the general solution of $$\cos x=1$$ as $$x=n\pi $$ . When we put x=0 in the general solution of x then it satisfies the given equation. But if we put x=1, then it doesn’t satisfy because $$\cos \pi =-1$$ . So, we cannot take $$x=n\pi $$ as the general solution. We have to take only those values of x which is a multiple of $$2\pi $$ . Hence, the general solution of $$\cos x=1$$ is $$x=2n\pi $$ .