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题目(2011年“北约”自主招生压轴题)求f(x)=|x-1|+|2x-1|+…+|2011-x|的最小值.此题短小精悍,可难度不小.文[1]立足绝对值的定义,去掉绝对值符号,给出了解法一.文[2]根据此题命题背景:绝对值不等式,也漂亮地给出了解法二.下面笔者将原题进行推广、变式及溯源:推广一求f(x)=|x-a_1|+|x-a_2+…+|x-a_n|(其中a_n≥a_(n-1)≥…≥a_2≥a_1)的最小值.解由绝对值的几何意义:|xa|表示数轴上任意一点x与a的距离.当n=2时,如图1:
Title (2011 “NATO ” self-enrollment finale) Find the minimum f (x) = | x-1 | + | 2x-1 | + ... + | 2011-x | Small. [1] based on the definition of absolute value, remove the absolute value of the symbol, gives the solution of a. [2] According to the title proposition background: absolute inequality, but also gives a beautiful solution II. To promote, variant and traceability: to promote a seeking f (x) = | x-a_1 | + | x-a_2 + ... + | x-a_n | (where a_n ≥ a_ (n-1) ≥ ... ≥ a_2 ≥ a_1 ) The solution to the geometric meanings of absolute value: | xa | represents the distance between any point x and a on the axis If n = 2, as shown in Figure 1: