题干

已知f(x)=xlnx,g(x)=﹣x2+ax﹣3.
(1)求函数f(x)在[t,t+2](t>0)上的最小值;
(2)对一切x∈(0,+∞),2f(x)≥g(x)恒成立,求实数a的取值范围;
(3)证明:对一切x∈(0,+∞),都有lnx>
1
e
x
2
e
x
成立.
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答案(点此获取答案解析)

解:(1)f(x)=xlnx,
∴f'(x)=lnx+1
当x∈(0,
1
e
),f′(x)<0,f(x)单调递减,
当x∈(

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完形填空

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