Question

Find the domain of definition of $f(x)={{\cos }^{-1}}\left( {{x}^{2}}-4 \right)$ i.e. the values of x for which $f(x)$ will be well defined.

Answer 1

Hint: In this question, we are given the definition of the function as an inverse trigonometric function. The inverse trigonometric functions are defined only on the domain which the corresponding trigonometric function can take, in this case. In this case, as $f(x)$ is of the form of inverse cosine function, the domain of f(x) will the values of x for which the value of ${{x}^{2}}-1$ lies within the range of the cosine function i.e. -1 and 1. Thus, using this range, we can solve the resulting quadratic equation to obtain the range of $f(x)$.

Complete step-by-step answer:
We know that the value of $\cos \theta $ always lies between -1 and 1 for any angle $\theta $. Therefore, ${{\cos }^{-1}}\theta $ will be defined only on the domain -1 to 1…………………(1.1)

In this question the function is given as$f(x)={{\cos }^{-1}}\left( {{x}^{2}}-4 \right)$. We see that it is in the same form as stated in equation (1.1) with ${{x}^{2}}-4$ in place of $\theta $. Therefore, using (1.1), the domain of f(x) will be

$\begin{align}

  & -1\le {{x}^{2}}-4\le 1 \\

 & \Rightarrow -1+4\le {{x}^{2}}\le 1+4 \\

 & \Rightarrow 3\le {{x}^{2}}\le 5.............(1.2) \\

\end{align}$

Therefore, the domain of f(x) will be all the values of x whose square lies between 3 to 5.

Now, we can rewrite the inequalities separately to obtain

$\begin{align}

  & {{x}^{2}}\le 5 \\

 & \Rightarrow -\sqrt{5}\le x\le \sqrt{5}..................(1.3) \\

\end{align}$

And

$\begin{align}

  & 3\le {{x}^{2}} \\

 & \Rightarrow x\ge \sqrt{3}\text{ or }x\le -\sqrt{3}..................(1.4) \\

\end{align}$

Now, as the value of $\sqrt{5}$ is greater than the value of $\sqrt{3}$ and from (1.2), we find that both equations (1.3) and (1.4) must be satisfied, we get that the possible values of x should lie in between

$-\sqrt{5}\le x\le -\sqrt{3}\text{ or }\sqrt{3}\le x\le \sqrt{5}\text{ }.........\text{(1}\text{.5)}$

Which can be written in interval form as

$x\in \left[ -\sqrt{5},-\sqrt{3} \right]\cup \left[ \sqrt{3},\sqrt{5} \right]$

Which is the answer to the given question.

Note: In this question, we should be careful to also consider the negative values of x for which equation (1.2) is satisfied because equation (1.2) is a relation of ${{x}^{2}}$, so negative values of x can also give positive values of ${{x}^{2}}$ lying in the interval as given in equation (1.2).