Given the ranges for X and Y:

To determine the range of X² – Y²:

Determine the range of X²:

Given –3 < X < –1. Squaring these values yields positive results. The square of –3 is 9, and the square of –1 is 1. Therefore, the range for X² is 1 < X² < 9.
Determine the range of Y²:

Given –1 < Y < 1. Squaring Y results in values between 0 (inclusive, as Y can be 0) and 1 (exclusive, as Y cannot be –1 or 1). Therefore, the range for Y² is 0 ≤ Y² < 1.
Determine the range of X² – Y²:

To find the minimum value of X² – Y², we consider the smallest possible value of X² and subtract the largest possible value of Y²:
- Minimum of (X² – Y²) = (lower bound of X²) – (upper bound of Y²)
- Since X² > 1 and Y² < 1, the expression X² – Y² > 1 – 1 = 0.
To find the maximum value of X² – Y², we consider the largest possible value of X² and subtract the smallest possible value of Y²:
- Maximum of (X² – Y²) = (upper bound of X²) – (lower bound of Y²)
- Since X² < 9 and Y² ≥ 0, the expression X² – Y² < 9 – 0 = 9.
Combining these, the range for X² – Y² is 0 < X² – Y² < 9. This corresponds to the interval between 0 and 9.

Option 1 (–9 & 1) and Option 2 (–9 & –1): These options are incorrect because X² is always positive (1 < X² < 9) and Y² is always non-negative (0 ≤ Y² < 1). Consequently, X² – Y² must always be positive. The minimum value of X² – Y² is greater than 0, not –9.
Option 3 (0 & 8): This option incorrectly specifies the upper bound. While the lower bound of 0 is correct (as X² – Y² > 0), the maximum value of X² – Y² is less than 9, not 8. This maximum is achieved when X² approaches its upper bound (9) and Y² approaches its lower bound (0).