Question

含参数、绝对值的不等式|x+3|-|x-1| ≤ a^-3a 对任意实数x恒成立 求a的取值范围

Answer 1

|x+3|-|x-1|≤a^2-3a

→|x+3|-|x-1|≤|(x+3)+(1-x)|≤a^2-3a

→a^2-3a-4≥0

→(a+1)(a-4)≥0

→a≤-1,a≥4

故实数a取值范围为{a|a≤-1}∪{a|a≥4}