问题 解答题
对定义在区间D上的函数f(x),若存在闭区间[a,b]⊆D和常数C,使得对任意的x∈[a,b]都有f(x)=C,且对任意的x∉[a,b]都有f(x)>C恒成立,则称函数f(x)为区间D上的“U型”函数.
(1)求证函数f(x)=|x-1|+|x-3|是R上的“U型”函数;
(2)设函数f(x)是(1)中的“U型”函数,若不等式|t-1|+|t-2|≤f(x)对一切t∈R恒成立,求实数t的取值范围.
(3)若函数g(x)=mx+
x2+2x+n
是区间[-2,+∞)上的“U型”函数,求实数m和n的值.
答案

(1)当x∈[1,3]时,f(x)=x-1+3-x=2,

当x∉[1,3]时,f(x)=|x-1|+|x-3|>|x-1+3-x|=2,

故存在闭区间[a,b]=[1,3]⊆R和常数C=2符合条件,

所以函数f(x)=|x-1|+|x-3|是R上的“U型”函数;

(2)因为不等式|t-1|+|t-2|≤f(x)对一切x∈R恒成立,

所以|t-1|+|t-2|≤f(x)min

由(1)可知f(x)min=(|x-1|+|x-3|)min=2,

所以|t-1|+|t-2|≤2,

解得:

1
2
≤t≤
5
2

(3)由“U型”函数定义知,存在闭区间[a,b]⊆[-2,+∞)和常数c,使得对任意的x∈[a,b],

都有g(x)=mx+

x2+2x+n
=c,即
x2+2x+n
=c-mx,

所以x2+2x+n=(c-mx)2恒成立,即x2+2x+n=m2x2-2cmx+c2对任意的x∈[a,b]成立,

所以

m2=1
-2cm=2
c2=n
,所以
m=1
c=-1
n=1
m=-1
c=1
n=1

①当

m=1
c=-1
n=1
时,g(x)=x+|x+1|.

当x∈[-2,-1]时,g(x)=-1,当x∈(-1,+∞)时,g(x)=2x+1>-1恒成立.

此时,g(x)是区间[-2,+∞)上的“U型”函数;

②当

m=-1
c=1
n=1
时,g(x)=-x+|x+1|.

当x∈[-2,-1]时,g(x)=-2x-1≥1,当x∈(-1,+∞)时,g(x)=1.

此时,g(x)不是区间[-2,+∞)上的“U型”函数.

综上分析,m=1,n=1为所求;

阅读理解

阅读理解.

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