题目内容
设a为实数,函数f(x)=2x2+(x-a)·|x-a|.
(1)若f(0)≥1,求a的取值范围;
(2)求f(x)的最小值;
(3)设函数h
(
x)=f(x),x∈(a,+∞),直接写出(不需给出步骤)不等式h(x)≥1的解集.
(1)若f(0)≥1,求a的取值范围;
(2)求f(x)的最小值;
(3)设函数h
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/2014082317135665265.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/2014082317135665265.gif)
略
解:(1)因为f(0)=-a|-a|≥1,所以-a>0,即a<0.由a2≥1知a≤-1.因此,a的取值范围为(-∞,-1].
(2)记f(x)的最小值为g(a).则有f(x)=2x2+(x-a)|x-a|
=
(ⅰ)当a≥0时,f(-a)=-2a2
,由①②知f(x)≥-2a2,此时g(a)=-2a2.
(ⅱ)当a<0时,f()=a2.若x>a,则由①知f(x)≥a2;
若x≤a,则x+a≤2a<0,由②知f(x)≥2a2>a2.此时g(a)=a2.
综上,得g(a)=
(3)(ⅰ)当a∈(-∞,-]∪[,+∞)时,解集为(a,+∞);
(ⅱ)当a∈[-,)时,解集为[,+∞);
(ⅲ)当a∈(-,-)时,解集
为(a,]∪[,+∞).
(2)记f(x)的最小值为g(a).则有f(x)=2x2+(x-a)|x-a|
=
(ⅰ)当a≥0时,f(-a)=-2a2
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/2014082317135668372.gif)
(ⅱ)当a<0时,f()=a2.若x>a,则由①知f(x)≥a2;
若x≤a,则x+a≤2a<0,由②知f(x)≥2a2>a2.此时g(a)=a2.
综上,得g(a)=
(3)(ⅰ)当a∈(-∞,-]∪[,+∞)时,解集为(a,+∞);
(ⅱ)当a∈[-,)时,解集为[,+∞);
(ⅲ)当a∈(-,-)时,解集
为(a,]∪[,+∞).
![](http://thumb2018.1010pic.com/images/loading.gif)
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