解.(1)按題意,得

.
∴

即 α>2. (3分)
又

∴關(guān)于x的方程

.
在(2,+∞)內(nèi)有二不等實根x=α、β.
?關(guān)于x的二次方程ax
2+(a-1)x+2(1-a)=0在(2,+∞)內(nèi)有二異根α、β.

.
故

. (6分)
(2)令Φ(x)=ax
2+(a-1)x+2(1-a),
則Φ(2)•Φ(4)=4a•(18a-2)=8a(9a-1)<0.
∴2<α<4<β. (10分)
(3)∵

,

=

.
∵lna<0,
∴當x∈(α,4)時,g'(x)>0;
當x∈(4,β)是g'(x)>0.
又g(x)在[α,β]上連接,
∴g(x)在[α,4]上遞增,在[4,β]上遞減.
故 M=g(4)=log
a9+1=log
a9a. (12分)
∵

,
∴0<9a<1.
故M>0.
若M≥1,則9a=a
M.
∴9=a
M-1≤1,矛盾.
故0<M<1. (15分)
分析:(1)由已知中f(x)在[α,β]上為減函數(shù)函數(shù)f(x)=

的定義域為[α,β],值域為[log
aa(β-1),log
aa(α-1)],我們可得

,根據(jù)對數(shù)式中底數(shù)及真數(shù)的限制條件,可得α>2,同理β>2,故關(guān)于x的方程

在(2,+∞)內(nèi)有二不等實根α、β.由此構(gòu)造關(guān)于a的不等式組,解不等式組即可求出a的取值范圍;
(2)令Φ(x)=ax
2+(a-1)x+2(1-a),我們易得Φ(2)•Φ(4)<0,進而根據(jù)零點存在定理,結(jié)合(1)中的結(jié)論,得到答案;
(3)由已知中函數(shù)g(x)=log
aa(x-1)-

,x∈[α,β]的解析式,我們利用導(dǎo)數(shù)法,可以判斷出函數(shù)的單調(diào)性,進而得到M=g(4)=log
a9+1,結(jié)合(1)中a的取值范圍,即可得到答案.
點評:本題考查的知識點是利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性,導(dǎo)數(shù)的運算,利用導(dǎo)數(shù)求閉區(qū)間上函數(shù)的最值,其中(1)的關(guān)鍵是根據(jù)函數(shù)的單調(diào)性將問題轉(zhuǎn)化為關(guān)于x的方程

在(2,+∞)內(nèi)有二不等實根α、β.并由此構(gòu)造關(guān)于a的不等式組,(2)的關(guān)鍵是構(gòu)造函數(shù)Φ(x)=ax
2+(a-1)x+2(1-a),將問題轉(zhuǎn)化為函數(shù)零點判斷問題,(3)的關(guān)鍵是利用導(dǎo)數(shù)法,判斷出M=g(4).