Question

The domain of $\sqrt {{{\sec }^{ - 1}}(\dfrac{{1 - | \times |}}{2})}$ is.
(1) $( - \infty , - 3) \cup [3,\infty ]$
(2) $[3,\infty )$
(3) $( - \infty , - 3]$
(4) R

Answer 1

Hint : To solve this question, i.e., to find the domain of given expression, we will start with taking the range of ${\sec ^{ - 1}}t,$ where t $=$ $\dfrac{{1 - |x|}}{2}$, either $t \leqslant - 1$ or $t \geqslant 1$. So, we will get two cases here, and by solving both the cases separately, we will be able to find the domain of the given function.

Complete step-by-step answer :
We have been given a function $\sqrt {{{\sec }^{ - 1}}(\dfrac{{1 - | \times |}}{2})}$ , we need to find the domain of the function.
We know that the range of ${\sec ^{ - 1}}t \in [0,\pi ]$ except $\dfrac{\pi }{2}$.
The first rule of finding a domain is that the value under square root should be positive, which is true because, ${\sec ^{ - 1}}t \in [0,\pi ]$.
Now, ${\sec ^{ - 1}}t \in [0,\pi ]$ is defined when $t \leqslant - 1$ or $t \geqslant 1$.
Here, we have t $=$ $\dfrac{{1 - |x|}}{2}$
So, $\dfrac{{1 - |x|}}{2} \leqslant - 1$ or $\dfrac{{1 - |x|}}{2} \geqslant 1$
We got two cases, here, let's check one by one.
Case 1. $\dfrac{{1 - |x|}}{2} \leqslant - 1$
$\begin{gathered} 1 - |x| \leqslant - 2 \\\ \- |x| \leqslant - 2 - 1 \\\ \- |x| \leqslant - 3 \\\ |x| \geqslant 3 \\\ \end{gathered}$
$\Rightarrow x \in ( - \infty , - 3] \cup [3,\infty )$
Case 2. $\dfrac{{1 - |x|}}{2} \geqslant 1$
$\begin{gathered} 1 - |x| \geqslant 2 \\\ \- |x| \geqslant 2 - 1 \\\ \- |x| \geqslant 1 \\\ |x| \leqslant - 1 \\\ \end{gathered}$
Here in this case, $|x| \leqslant - 1$, but we know that $\left| x \right|$ is always positive. Hence, this case is not possible.
Thus, option (1) $( - \infty , - 3) \cup [3,\infty ]$, is the domain of $\sqrt {{{\sec }^{ - 1}}(\dfrac{{1 - | \times |}}{2})}$.
So, the correct answer is “Option A”.

Note : We have mentioned in the solution, that, $\left| x \right|$ is always positive, so let us understand that clearly. Assume x be any real number, then the absolute value of any real number x, notated as $\left| x \right|$, is defined to be x, if $x \geqslant 0$ (i.e., if x is non-negative), and it will be −x, if $x < 0$ (i.e., if x is negative). Thus, if x is positive, it is kept as is, that is positive. Then the only other choice for real x is to be negative, in that case we take the negative of the given negative value, which will always be positive.