The set of real solutions of the inequality |x – 1| + |x + 2| < 5 is:
- -3 < x < 2
- -1 < x < 2
- -2 < x < 1
- -3/2 < x < 7/2
- empty (no solutions)
Given: |x – 1| + |x + 2| < 5.
There are three cases to consider: x ≥ 1, -2 < x < 1, and x ≤ -2.
x ≥ 1: x – 1 + x + 2 < 5 → 2x + 1 < 5 → 2x < 4 → x < 2, so 1 ≤ x < 2.
-2 < x < 1: 1 – x + x + 2 < 5 → 3 < 5, so -2 < x < 1.
x ≤ -2: 1 – x – x – 2 < 5 → -2x – 1 < 5 → -2x < 6 → x > -3, so -3 < x ≤ -2.
Taking the union of all of these solutions, we get -3 < x < 2. The answer is (a).