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HMMT 二月 2005 · 代数 · 第 1 题

HMMT February 2005 — Algebra — Problem 1

专题
Discrete Math / 离散数学
难度
L3
来源
HMMT

题目详情

  1. How many real numbers x are solutions to the following equation? | x − 1 | = | x − 2 | + | x − 3 |
解析
  1. How many real numbers x are solutions to the following equation? | x − 1 | = | x − 2 | + | x − 3 | Solution: 2 If x < 1, the equation becomes (1 − x ) = (2 − x ) + (3 − x ) which simplifies to x = 4, contradicting the assumption x < 1. If 1 ≤ x ≤ 2, we get ( x − 1) = (2 − x ) + (3 − x ), which gives x = 2. If 2 ≤ x ≤ 3, we get ( x − 1) = ( x − 2) + (3 − x ), which again gives x = 2. If x ≥ 3, we get ( x − 1) = ( x − 2) + ( x − 3), or x = 4. So 2 and 4 are the only solutions, and the answer is 2.