问题 解答题

定义:已知函数f(x)在[m,n](m<n)上的最小值为t,若t≤m恒成立,则称函数f(x)在[m,n](m<n)上具有“DK”性质,

(1)判断函数f(x)=x2-2x+2在[1,2]上是否具有“DK”性质,说明理由;

(2)若f(x)=x2-ax+2在[a,a+1]上具有“DK”性质,求a的取值范围.

答案

解:(1)∵f(x)=x2-x2x+2,x∈[1,2],

∴f(x)min=1≤1,

∴函数f(x)在[1,2]上具有“DK”性质。

(2)f(x)=x2-ax+2,x∈[a,a+1],其图象的对称轴方程为

①当,即a≥0时,f(x)min=f(a)=a2-a2+2=2,

若函数f(x)具有“DK”性质,则有2≤a总成立,即a≥2;

②当,即-2<a<0时,

若函数f(x)具有“DK”性质,则有总成立,解得a不存在;

③当,即a≤-2时,f(x)min=f(a+1)=a+3,

若函数f(x)具有“DK”性质,则有a+3≤a,解得a不存在;

综上所述,若f(x)=x2-ax+2在[a,a+1]上具有“DK”性质,则a≥2。

阅读理解
Reading comprehension.
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Choose the best topic. (选择最佳标题)
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