【答案】
分析:(1)求出F(x)的解析式,根據(jù)對數(shù)函數(shù)的性質(zhì),求出其定義域,把n=2代入F(x),利用導(dǎo)數(shù)研究函數(shù)的極值點;
(2)已知對于任意的正整數(shù)n,當(dāng)s≥2,x≥2時,f(s)+g(x)≤x-1,將其轉(zhuǎn)化為1≤x-1-aln(x-1),即只需x-2-aln(x-1)≥0對x≥2成立,再對a進行討論,求出a的范圍;
解答:解:(1)由已知得函數(shù)F(x)的定義域為{x|x>1},
當(dāng)n=2時,F(xiàn)(x)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103104422614928992/SYS201311031044226149289020_DA/0.png)
+aln(x-1),所以F′(x)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103104422614928992/SYS201311031044226149289020_DA/1.png)
,
①當(dāng)a>0時,由F′(x)=0得x
1=1+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103104422614928992/SYS201311031044226149289020_DA/2.png)
>1,x
2=1-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103104422614928992/SYS201311031044226149289020_DA/3.png)
<1,
此時F′(x)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103104422614928992/SYS201311031044226149289020_DA/4.png)
,
當(dāng)x∈(1,x
1)時,F(xiàn)′(x)<0,F(xiàn)(x)單調(diào)遞減;
當(dāng)x∈(x
1,+∞)時,F(xiàn)′(x)>0,F(xiàn)(x)單調(diào)遞增;
從而F(x)在x
1=1+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103104422614928992/SYS201311031044226149289020_DA/5.png)
處取得極小值,極小值為:F(1+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103104422614928992/SYS201311031044226149289020_DA/6.png)
)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103104422614928992/SYS201311031044226149289020_DA/7.png)
(1+ln
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103104422614928992/SYS201311031044226149289020_DA/8.png)
),
②當(dāng)a≤0時,F(xiàn)′(x)<0恒成立,所以F(x)無極值.
綜上所述,n=2時;
當(dāng)a>0時,F(xiàn)(x)在x=1+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103104422614928992/SYS201311031044226149289020_DA/9.png)
處取得極小值,極小值為F(1+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103104422614928992/SYS201311031044226149289020_DA/10.png)
)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103104422614928992/SYS201311031044226149289020_DA/11.png)
(1+ln
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103104422614928992/SYS201311031044226149289020_DA/12.png)
)
當(dāng)a≤0時,函數(shù)為減函數(shù),F(xiàn)(x)無極值;
(2)當(dāng)x≥2時,對任意的正整數(shù)n,恒有f(s)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103104422614928992/SYS201311031044226149289020_DA/13.png)
≤1,故對任意的正整數(shù)n,當(dāng)s≥2,x≥2時,
有f(s)+g(x)≤x-1,只需1≤x-1-aln(x-1),即只需x-2-aln(x-1)≥0對x≥2成立,
令h(x)=x-2-aln(x-1),因為h′(x)=1-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103104422614928992/SYS201311031044226149289020_DA/14.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103104422614928992/SYS201311031044226149289020_DA/15.png)
(x≥2),又h(2)=0,
所以當(dāng)x∈[2,+∞)時,h(x)≥h(2),即h(x)當(dāng)x∈[2,+∞)時最小值為h(2)=0,
①當(dāng)a≤1,h′(x)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103104422614928992/SYS201311031044226149289020_DA/16.png)
≥0,h(x)當(dāng)x∈[2,+∞)單調(diào)遞增,結(jié)論成立;
②當(dāng)a>1時,當(dāng)x∈[2,1+a),h′(x)<0,x∈[1+a,+∞),h′(x)≥0,又h(2)=0,
故結(jié)論不成立,
綜合得a≤1;
點評:此題主要考查利用導(dǎo)數(shù)研究函數(shù)的極值,解題過程中也用到了分類討論和轉(zhuǎn)化的思想,考查的知識點比較多,這類綜合題,也是高考的熱點問題;