解:(1)函數(shù)f(x)定義域為x>0,
=
.
由f'(x)>0且x>0
得
即
(i)當(dāng)a-1=1即a=2時,f(x)在(0,+∞)上為增函數(shù);
(ii)當(dāng)a>2時,x>a-1或0<x<1,∴f(x)在(a-1,+∞),(0,1)上為增函數(shù);
(iii)當(dāng)1<a<2時,0<x<a-1或x>1,∴f(x)在(0,a-1),(1,+∞)上為增函數(shù).
綜上可知:f(x)的單調(diào)區(qū)間為:當(dāng)a=2時,(0,+∞)
當(dāng)a>2時,(a-1,+∞),(0,1)
當(dāng)1<a<2時,(0,a-1),(1,+∞).
(2)x=2是f(x)極值點,∴f'(2)=0,即
,解得a=3.
∴
(x>0),
.
∵
,且當(dāng)2<x<e
2時,f
′(x)>0;當(dāng)1<x<2時,f
′(x)<0;當(dāng)
時,f′(x)>0.
∴函數(shù)f(x)在區(qū)間
及(2,e
2]上單調(diào)遞增,在區(qū)間(1,2)上單調(diào)遞減.
∴f(x)在
最大值應(yīng)在x=1和x=e
2處取得
又
,
,
∴
.
分析:(1)利用f
′(x)>0即可求出其單調(diào)遞增區(qū)間;
(2)利用函數(shù)取得極值點的條件先求出a的值,再利用導(dǎo)數(shù)得出其單調(diào)區(qū)間、極值,進(jìn)而即可求出其最值.
點評:熟練掌握分類討論的思想方法、利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性與極值是解題的關(guān)鍵.