Explanation

If a range of values for a can be found, then the range of values for $a^2$ can be found.

Start by testing the end values of b, –3 and 1.

Plug in –3 for b in the first given inequality then solve for a.

You find that –4 < a < 7. If b = 1, 0 < a <11; b could be any integer in the range –3 ≤ b ≤ 1, this means –4 < a < 11 overall.
Remember to take the last step, though!
The question is looking for the range of $a^2$, not a; $a^2$ is always positive (i.e., $0 < a^2$).
Because a < 11, $a^2 < 121$.


This means $0 < a^2 < 121$; the answer is choice (D).