题目

F(X)是偶函数,且f(x)在(0,+∞)是增函数
如果X∈[0.5,1]f(ax+1)≤ f(X-2)恒成立,则实数a的取值范围是.(注,答案为[-2,0])非此答案者也必乱写

提问时间：2020-10-17

答案

F(X)是偶函数,则f(x)=f(|x|)
f(ax+1)≤ f(X-2),即f(|ax+1|)<=f(|x-2|)
f(x)在(0,+∞)是增函数,则|ax+1|=<|x-2|对于X∈[0.5,1]恒成立.
0.5<=x<=1时,|x-2|=2-x
即|ax+1|<=2-x
(1)ax+1<=2-x
a<=(1-x)/x=1/x-1,又1<=1/x<=2
故0<=1/x-1<=1
所以有:a<=0
(2)ax+1>=-(2-x)=x-2
a>=(x-3)/x=1-3/x
又:3<=3/x<=6,-6<=-3/x<=-3
故:-5<=1-3/x<=-2
所以有:a>=-2
综合(1)(2)得:-2<=a<=0
即a的范围是[-2,0]