Answer :
To determine which function has the range [tex]\((-\infty, -2] \cup [0, \infty)\)[/tex], we need to analyze the range of each function one by one.
Option A: [tex]\( y = \csc(x) - 1 \)[/tex]
First, recall that [tex]\( \csc(x) = \frac{1}{\sin(x)} \)[/tex], which makes the range of [tex]\( \csc(x) \)[/tex] be [tex]\((-\infty, -1] \cup [1, \infty)\)[/tex]. Now, if we subtract 1 from [tex]\( \csc(x) \)[/tex]:
- For [tex]\( \csc(x) \)[/tex] values in [tex]\((-\infty, -1]\)[/tex]: [tex]\( \csc(x) - 1 \leq -2 \)[/tex]
- For [tex]\( \csc(x) \)[/tex] values in [tex]\([1, \infty)\)[/tex]: [tex]\( \csc(x) - 1 \geq 0 \)[/tex]
Thus, the range of [tex]\( y = \csc(x) - 1 \)[/tex] is [tex]\((-\infty, -2] \cup [0, \infty)\)[/tex].
Option B: [tex]\( y = \sec(x) + 1 \)[/tex]
Next, note that [tex]\( \sec(x) = \frac{1}{\cos(x)} \)[/tex], which makes the range of [tex]\( \sec(x) \)[/tex] be [tex]\((-\infty, -1] \cup [1, \infty)\)[/tex]. Adding 1 to [tex]\( \sec(x) \)[/tex]:
- For [tex]\( \sec(x) \)[/tex] values in [tex]\((-\infty, -1]\)[/tex]: [tex]\( \sec(x) + 1 \leq 0 \)[/tex]
- For [tex]\( \sec(x) \)[/tex] values in [tex]\([1, \infty)\)[/tex]: [tex]\( \sec(x) + 1 \geq 2 \)[/tex]
So, the range of [tex]\( y = \sec(x) + 1 \)[/tex] is [tex]\((-\infty, 0] \cup [2, \infty)\)[/tex], which does not match the required range.
Option C: [tex]\( y = \cos(x + 1) \)[/tex]
For [tex]\( y = \cos(x + 1) \)[/tex], the function essentially has the same range as [tex]\( \cos(x) \)[/tex] since horizontal shifts do not affect the range of cosine. The range of [tex]\( \cos(x) \)[/tex] is [tex]\([-1, 1]\)[/tex], which does not include values outside of this interval.
Therefore, the range of [tex]\( y = \cos(x + 1) \)[/tex] is [tex]\([-1, 1]\)[/tex].
Option D: [tex]\( y = \cot(2x) - 1 \)[/tex]
Finally, recall that [tex]\( \cot(x) = \frac{\cos(x)}{\sin(x)} \)[/tex] and its range is [tex]\((-\infty, \infty)\)[/tex] because [tex]\( \cot(x) \)[/tex] can take any real value. Subtracting 1 does not change the fact that the range will remain all real numbers:
The range of [tex]\( y = \cot(2x) - 1 \)[/tex] is still [tex]\((-\infty, \infty)\)[/tex].
Conclusion:
From the analysis above, the function that has the range [tex]\( (-\infty, -2] \cup [0, \infty) \)[/tex] is:
A. [tex]\( y = \csc(x) - 1 \)[/tex]
Option A: [tex]\( y = \csc(x) - 1 \)[/tex]
First, recall that [tex]\( \csc(x) = \frac{1}{\sin(x)} \)[/tex], which makes the range of [tex]\( \csc(x) \)[/tex] be [tex]\((-\infty, -1] \cup [1, \infty)\)[/tex]. Now, if we subtract 1 from [tex]\( \csc(x) \)[/tex]:
- For [tex]\( \csc(x) \)[/tex] values in [tex]\((-\infty, -1]\)[/tex]: [tex]\( \csc(x) - 1 \leq -2 \)[/tex]
- For [tex]\( \csc(x) \)[/tex] values in [tex]\([1, \infty)\)[/tex]: [tex]\( \csc(x) - 1 \geq 0 \)[/tex]
Thus, the range of [tex]\( y = \csc(x) - 1 \)[/tex] is [tex]\((-\infty, -2] \cup [0, \infty)\)[/tex].
Option B: [tex]\( y = \sec(x) + 1 \)[/tex]
Next, note that [tex]\( \sec(x) = \frac{1}{\cos(x)} \)[/tex], which makes the range of [tex]\( \sec(x) \)[/tex] be [tex]\((-\infty, -1] \cup [1, \infty)\)[/tex]. Adding 1 to [tex]\( \sec(x) \)[/tex]:
- For [tex]\( \sec(x) \)[/tex] values in [tex]\((-\infty, -1]\)[/tex]: [tex]\( \sec(x) + 1 \leq 0 \)[/tex]
- For [tex]\( \sec(x) \)[/tex] values in [tex]\([1, \infty)\)[/tex]: [tex]\( \sec(x) + 1 \geq 2 \)[/tex]
So, the range of [tex]\( y = \sec(x) + 1 \)[/tex] is [tex]\((-\infty, 0] \cup [2, \infty)\)[/tex], which does not match the required range.
Option C: [tex]\( y = \cos(x + 1) \)[/tex]
For [tex]\( y = \cos(x + 1) \)[/tex], the function essentially has the same range as [tex]\( \cos(x) \)[/tex] since horizontal shifts do not affect the range of cosine. The range of [tex]\( \cos(x) \)[/tex] is [tex]\([-1, 1]\)[/tex], which does not include values outside of this interval.
Therefore, the range of [tex]\( y = \cos(x + 1) \)[/tex] is [tex]\([-1, 1]\)[/tex].
Option D: [tex]\( y = \cot(2x) - 1 \)[/tex]
Finally, recall that [tex]\( \cot(x) = \frac{\cos(x)}{\sin(x)} \)[/tex] and its range is [tex]\((-\infty, \infty)\)[/tex] because [tex]\( \cot(x) \)[/tex] can take any real value. Subtracting 1 does not change the fact that the range will remain all real numbers:
The range of [tex]\( y = \cot(2x) - 1 \)[/tex] is still [tex]\((-\infty, \infty)\)[/tex].
Conclusion:
From the analysis above, the function that has the range [tex]\( (-\infty, -2] \cup [0, \infty) \)[/tex] is:
A. [tex]\( y = \csc(x) - 1 \)[/tex]