What is the range of [tex]\( y = \sec^{-1}(x) \)[/tex]?

A. [tex]\( \left[-\frac{\pi}{2}, 0\right) \)[/tex] and [tex]\( \left(0, \frac{\pi}{2}\right] \)[/tex]

B. [tex]\( \left[-\frac{\pi}{2}, 0\right] \)[/tex] and [tex]\( \left[0, \frac{\pi}{2}\right] \)[/tex]

C. [tex]\( \left[0, \frac{\pi}{2}\right) \)[/tex] and [tex]\( \left(\frac{\pi}{2}, \pi\right] \)[/tex]

D. [tex]\( \left[0, \frac{\pi}{2}\right] \)[/tex] and [tex]\( \left[\frac{\pi}{2}, \pi\right] \)[/tex]



Answer :

To determine the range of the function [tex]\( y = \sec^{-1}(x) \)[/tex], we need to understand the behavior and properties of the inverse secant function. The secant function, [tex]\( \sec(\theta) \)[/tex], is defined as [tex]\( \sec(\theta) = \frac{1}{\cos(\theta)} \)[/tex].

Firstly, for [tex]\( y = \sec^{-1}(x) \)[/tex] to be defined, [tex]\( \sec(y) = x \)[/tex].

Now, we need to consider where the secant function is invertible, taking into account its periodicity and that [tex]\( \sec(\theta) \)[/tex] is undefined for [tex]\( \theta \)[/tex] where [tex]\( \cos(\theta) = 0 \)[/tex], i.e., [tex]\( \theta = \frac{\pi}{2} \)[/tex] and [tex]\( \theta = \frac{3\pi}{2} \)[/tex], etc.

Since [tex]\( \sec(\theta) = \frac{1}{\cos(\theta)} \)[/tex], the values of [tex]\( \theta \)[/tex] where [tex]\( \sec(\theta) \)[/tex] is defined and the function is monotonic are in the intervals:
- [tex]\( [0,\frac{\pi}{2}) \)[/tex]
- [tex]\( (\frac{\pi}{2}, \pi] \)[/tex]

These intervals avoid the asymptote at [tex]\( \frac{\pi}{2} \)[/tex] and are sufficient to cover all the range positively and negatively without overlapping. This ensures each [tex]\( x \)[/tex] value maps to a unique [tex]\( \theta \)[/tex].

From these considerations, the range of [tex]\( y = \sec^{-1}(x) \)[/tex] is therefore:
[tex]\[ \left[ 0, \frac{\pi}{2} \right) \text{ and } \left( \frac{\pi}{2}, \pi \right] \][/tex]

Thus, the correct answer is:
[tex]\[ \left[0, \frac{\pi}{2}\right) \text{ and } \left(\frac{\pi}{2}, \pi\right] \][/tex]