x??????212121))(1(2xxxxnxkx????222222122214)1(21222knkknnknk????????22)1(4)22(222?????nnknnk22)1(4)22(222?????nnknnk.1)1(22???nnk.若1?n,则直线nkxy??过定点)1,0(,不合题意.若1?n,则由21221????nkkk,得1??kn.于是直线l的方程nkxy??可化简为1)1(???xky.所以直线l过定点)1,1(??.…………………………………………………………12分高二数学(文科)参考答案及评分细则第6页(共6页)22.(本小题满分12分)已知函数12ln)(???xaxxf,R?a.(Ⅰ)若函数)(xf在),0(??上为单调增函数,求a的取值范围;(Ⅱ)设0??nm,求证:nmnmnm????)(2lnln.解析:(Ⅰ)?12ln)(???xaxxf,?222)1(1)22()1(21)(?????????xxxaxxaxxf.要使函数函数)(xf在),0(??上为单调增函数,只需0)1(1)22()(22???????xxxaxxf在),0(???x时恒成立.因为0)1(2??xx,所以只需01)22(2????xax在),0(???x时恒成立.由01)22(2????xax,得)1(22xxa????.因为当0?x时,21??xx,所以222???a,即2?a.所以,所求a的取值范围为]2,(??.…………………………………………………6分(Ⅱ)由(Ⅰ)知,当2?a时,14ln)(???xxxf在),0(??上单调递增.因为0??nm,所以1?nm,所以)1()(fnmf?.于是214ln???nmnm,即nmnnm????42lnln.所以nmnmnm????)(2lnln.…………………………………………………………12分
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