解:(Ⅰ)f
′(x)=1-
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-
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,
∵函數(shù)f(x)=x-alnx+
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在x=1處取得極值,∴f
′(1)=0,∴1-a-b=0,即b=1-a.
(Ⅱ)函數(shù)f(x)的定義域為(0,+∞),
由(Ⅰ)可得f
′(x)=
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=
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=
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.
令f
′(x)=0,則x
1=1,x
2=a-1.
①當a>2時,x
2>x
1,當x∈(0,1)∪(a-1,+∞)時,f
′(x)>0;當x∈(1,a-1)時,f
′(x)<0.
∴f(x)的單調遞增區(qū)間為(0,1),(a-1,+∞);單調遞減區(qū)間為(1,a-1).
②當a=2時,f
′(x)≥0,且只有x=1時為0,故f(x)在(0,+∞)上單調遞增.
③當a<2時,x
2<x
1,當x∈(0,1-a)∪(1,+∞)時,f
′(x)>0;當x∈(1-a,1)時,f
′(x)<0.
∴f(x)的單調遞增區(qū)間為(0,1-a),(1,+∞);單調遞減區(qū)間為(a-1,1).
分析:(Ⅰ)利用f
′(1)=0即可求得a與b的關系.
(Ⅱ)先求導得f
′(x)=
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,然后對參數(shù)a分a>2,a=2,a<2討論即可.
點評:本題考查了含有參數(shù)的函數(shù)的單調性,對參數(shù)恰當分類討論是解決問題的關鍵.