问题 解答题

选修4-5:不等式选讲

已知函数f(x)=|x-a|.

(1)若f(x)≤m的解集为{x|-1≤x≤5},求实数a,m的值;

(2)当a=2且t≥0时,解关于x的不等式f(x)+t≥f(x+2).

答案

(1)由于函数f(x)=|x-a|,由f(x)≤m可得-m≤x-a≤x+a,即a-m≤x≤a+m.

再由f(x)≤m的解集为{x|-1≤x≤5},可得

a-m=-1
a+m=5
,解得 
a=2
m=3

(2)当a=2时,f(x)=|x-2|,关于x的不等式f(x)+t≥f(x+2),即|x|-|x-2|≤t.

令h(t)=|x|-|x-2|=

2 , x≥2
2x-2 ,0<x<2
-2 , x≤0
,故函数h(x)的最大值为2,最小值为-2,不等式即 h(x)≤t.

①当t≥2时,不等式 h(x)≤t恒成立,故原不等式的解集为R.

②当 0≤t<2时,(1)若x≤0,则h(x)=-2,h(x)≤t 恒成立,不等式的解集为{x|x≤0}.

                (2)若 0<x<2,此时,h(x)=2x-2,不等式即 2x-2≤t,解得 x≤

t
2
+1,即此时不等式的解集为 {x|0<x≤
t
2
+1 }.

综上可得,当t≥2时,不等式的解集为R; ②当 0≤t<2时,不等式的解集为 {x|x≤

t
2
+1 }.

完形填空
完形填空
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