what is the range of y = sec⁻¹(x)? o -π/2,0 and 0,π/2 o -π/2,0 and 0,π/2 o 0,π/2 and (π/2,π o 0,π/2 and π/2,π

Answer:

The range of (y = \sec^{-1}(x)) is (\left[0,\frac{\pi}{2}\right)\cup\left(\frac{\pi}{2},\pi\right]). So the correct option is the one that represents (\left[0,\frac{\pi}{2}\right)) and (\left(\frac{\pi}{2},\pi\right]) (assuming one of the options is written in this correct interval - notation form).

Explanation:

Step1: Recall sec - inverse definition

The function (y=\sec^{-1}(x)) is the inverse of the secant function (y = \sec(x)) restricted to certain intervals.

Step2: Consider secant function

The secant function (y=\sec(x)=\frac{1}{\cos(x)}). The cosine function (y = \cos(x)) has a range of ([- 1,1]). The secant function is undefined when (\cos(x)=0).

Step3: Determine range of sec - inverse

The range of (y = \sec^{-1}(x)) is chosen such that for every (x\in(-\infty,-1]\cup[1,\infty)), there is a unique (y) value. We choose (y\in\left[0,\frac{\pi}{2}\right)\cup\left(\frac{\pi}{2},\pi\right]) because on this set of values, the secant function is one - to - one and covers all possible values of the secant function's range ((-\infty,-1]\cup[1,\infty)).