1问题的提出在选修4-5《不等式选讲》的模块测试中,有这样一道题:已知不等式|3x-a|>x-1对x∈[0,2]恒成立,求实数a的取值范围.学生的答卷中有下面两种解答:解答1由绝对值不等式的等价形式|f(x)|>g(x)f(x)>g(x)或f(x)<-g(x)可知:原不等式等价于3x-a>x-1或3x-a<1-x,即a<2x+1或a>4x-1.已知不等式|3x-a|>x-1对x∈[0,2]恒成立等价于a<2x+1或a>4x-1对x∈[0,2]恒成立,即a<2x+1对x∈[0,2]恒成立或a>4x-1对x∈[0,2]恒成立.则 1Problem raised In the module test of elective 4-5 “Inequalities”, there is such a problem: the known inequality |3x-a|>x-1 is fixed for x∈[0,2], and the real number a The range of values. There are two kinds of answers in the student’s answer sheet: Answer 1 The equivalent form of the absolute value inequality |f(x)|>g(x)f(x)>g(x) or f(x) <-g(x) shows that the original inequality is equivalent to 3x-a>x-1 or 3x-a<1-x, that is, a<2x+1 or a>4x-1. The known inequality |3x-a| >x-1 is equal to x∈[0,2] is equal to a<2x+1 or a>4x-1 is fixed for x∈[0,2], ie, a<2x+1 vs. x∈[0 , 2] is constant or a>4x-1 is fixed for x∈[0,2].