(1)函數(shù)的定義域?yàn)椋?,+∞）．
當(dāng)a=1時(shí),f(x)＝x−lnx,f′(x)＝1−1/x=(x-1)/x
令f′（x）=0,得x=1．
當(dāng)0＜x＜1時(shí),f′（x）＜0；當(dāng)x＞1時(shí),f′（x）＞0．
∴f（x）極小值=f（1）=1,無(wú)極大值
（2）f′(x)＝(1−a)x+a−1/x =[(1−a)x2+ax−1]/x=[(1−a)x+1](x−1)/ x
=(1−a)[x−1/ (a−1)](x−1)/ x
當(dāng)1 / (a−1)＝1,即a=2時(shí),f′(x)＝−(x−1)2 / x ≤0,f（x）在（0,+∞）上是減函數(shù)；
當(dāng)1 / (a−1) ＜1,即a＞2時(shí),令f′（x）＜0,得0＜x＜1 / (a−1) 或x＞1；令f′（x）＞0,得
1 / (a−1) ＜x＜1．
當(dāng)1 / (a−1) ＞1,即1＜a＜2時(shí),令f′（x）＜0,得0＜x＜1或x＞1 / (a−1) ;令f′（x）＞0,得
1＜x＜1 / (a−1)
綜上,當(dāng)a=2時(shí),f（x）在定義域上是減函數(shù)；
當(dāng)a＞2時(shí),f（x）在(0,1 / (a−1))和（1,+∞）單調(diào)遞減,在(1 / (a−1),1)上單調(diào)遞增；
當(dāng)1＜a＜2時(shí),f（x）在（0,1）和(1 / (a−1),+∞)單調(diào)遞減,在(1,1 / (a−1))上單調(diào)遞增　
（Ⅲ）由（Ⅱ）知,當(dāng)a∈（2,3）時(shí),f（x）在[1,2]上單調(diào)遞減,
∴當(dāng)x=1時(shí),f（x）有最大值,當(dāng)x=2時(shí),f（x）有最小值．
∴|f(x1)−f(x2)|≤f(1)−f(2)＝a/2−3/2+ln2
∴ma+ln2＞a/2−3/2+ln2
而a＞0經(jīng)整理得m＞1/2−3/2a
由2＜a＜3得−1/4＜1/2−3/2a＜0,所以m≥0.