HMMT 二月 2008 · 代数 · 第 3 题

HMMT February 2008 — Algebra — Problem 3

[ 4 ] Determine all real numbers a such that the inequality | x + 2 ax + 3 a | ≤ 2 has exactly one solution in x .

解析

[ 4 ] Determine all real numbers a such that the inequality | x + 2 ax + 3 a | ≤ 2 has exactly one solution in x . 2 Answer: 1 , 2 Let f ( x ) = x + 2 ax + 3 a . Note that f ( − 3 / 2) = 9 / 4, so the graph of f is a parabola 2 that goes through ( − 3 / 2 , 9 / 4). Then, the condition that | x + 2 ax + 3 a | ≤ 2 has exactly one solution means that the parabola has exactly one point in the strip − 1 ≤ y ≤ 1, which is possible if and only 2 if the parabola is tangent to y = 1. That is, x + 2 ax + 3 a = 2 has exactly one solution. Then, the 2 2 discriminant ∆ = 4 a − 4(3 a − 2) = 4 a − 12 a + 8 must be zero. Solving the equation yields a = 1 , 2.