✨ 最佳解答 ✨
|ax + 1| ≦ b
-b ≦ ax + 1 ≦ b
(-b - 1) ≦ ax ≦ (b - 1)
1. 若 a > 0
(-b - 1)/a ≦ x ≦ (b - 1)/a
等同 -2 ≦ x ≦ 6
可得
-2a + b = -1 ...(1)
6a - b = -1 ...(2)
--> a = -1/2, b = -2
a < 0 與前面的假設矛盾。
2. 若 a < 0
(b - 1)/a ≦ x ≦ (-b - 1)/a (除以負數,不等號要轉向)
等同 -2 ≦ x ≦ 6
可得
-2a - b = -1 ...(1)
6a + b = -1 ...(2)
--> a = -1/2, b = 2 #
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