191问答库 > 已知函数f(x)是偶函数,且f(x)在[0,+∞)上是增函数,如果x∈[1⼀2,1]时,不等式f(ax+1)≤f(x-2)恒成立，则实

已知函数f(x)是偶函数,且f(x)在[0,+∞)上是增函数,如果x∈[1⼀2,1]时,不等式f(ax+1)≤f(x-2)恒成立，则实

2025-05-16 15:05:46

推荐回答（1个）

回答1：

|ax+1|<=|x-2|
(ax+1)^2<=(x-2)^2
[(a+1)x-1][(a-1)x+3]<=0
a=1时，x<=1/2(舍去)
a=-1时，x<=3/2，符合条件
a>1时，-3/(a-1)<=x<=1/(a+1)<1/2(舍去)
a<-1时，1/(a+1)<=a<=-3/(a-1)
-3/(a-1)>=1
(-2-a)(a-1)>=0
-2<=a<=1
-2<=a<-1
-1 1/(a+1)<=1/2且-3/(a-1)<=1/2 或 1/(a+1)>=1且-3/(a-1)>=1
(1-a)(a+1)<=0且(-5-a)(a-1)<=0(舍去) 或 -a(a+1)>=0且(-2-a)(a-1)>=0
-1<=a<=0
-1故-2<=a<=0