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HMMT 二月 2008 · 代数 · 第 3 题

HMMT February 2008 — Algebra — Problem 3

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. [ 4 ] Determine all real numbers a such that the inequality | x + 2 ax + 3 a | ≤ 2 has exactly one solutionin x .
解析

英文解析

  1. [ 4 ] Determine all real numbers a such that the inequality | x + 2 ax + 3 a | ≤ 2 has exactly one solutionin x .
    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 2
    means that the parabola has exactly one point in the strip − 1 ≤ y ≤ 1, which is possible if and onlyif the parabola is tangent to y = 1. That is, x + 2 ax + 3 a = 2 has exactly one solution. Then, the 2
    2 2
    discriminant ∆ = 4 a − 4(3 a − 2) = 4 a − 12 a + 8 must be zero. Solving the equation yields a = 1 , 2.