【答案】
分析:(1)函數(shù)f(x)在區(qū)間(a,a+1)上有極值⇒f′(x)=0在(a,a+1)上有根,結(jié)合條件由函數(shù)的單調(diào)性可得函數(shù)有唯一極值點x=1,1∈(a,a+1).
(2)構(gòu)造函數(shù)g(x)=x
2-2x+k,若關(guān)于x的方程f(x)=x
2-2x+k有實數(shù)解⇒f(x)=g(x)有實數(shù)解⇒g(x)
min=g(1)≤f(x)
max(法二)由f(x)=x
2-2x+k分離系數(shù)k=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103100412497668279/SYS201311031004124976682019_DA/0.png)
,構(gòu)造函數(shù)h(x)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103100412497668279/SYS201311031004124976682019_DA/1.png)
,由題意可得,k≤h(x)
max.
(3)結(jié)合函數(shù)f(x)在(1,+∞)上的單調(diào)性可得,f
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103100412497668279/SYS201311031004124976682019_DA/2.png)
⇒1+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103100412497668279/SYS201311031004124976682019_DA/3.png)
⇒
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103100412497668279/SYS201311031004124976682019_DA/4.png)
,利用該結(jié)論分別把n=1,2,3,…代入疊加可證.
解答:解:(1)∵
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103100412497668279/SYS201311031004124976682019_DA/5.png)
,∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103100412497668279/SYS201311031004124976682019_DA/6.png)
∴當(dāng)x∈(0,1)時,f'(x)>0;當(dāng)x∈(1,+∞)時,f'(x)<0;
∴函數(shù)f(x)在區(qū)間(0,1)上為增函數(shù);在區(qū)間(1,+∞)為減函數(shù)(3分)
∴當(dāng)x=1時,函數(shù)f(x)取得極大值,而函數(shù)f(x)在區(qū)間(a,a+1)有極值.
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103100412497668279/SYS201311031004124976682019_DA/7.png)
,解得0<a<1
(2)由(1)得f(x)的極大值為f(1)=1,令g(x)=x
2-2x+k,
所以當(dāng)x=1時,函數(shù)g(x)取得最小值g(1)=k-1,
又因為方程f(x)=x
2-2x+k有實數(shù)解,那么k-1≤1,即k≤2,所以實數(shù)k的取值范圍是:k≤2
解法二:∵f(x)=x
2-2x+k,∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103100412497668279/SYS201311031004124976682019_DA/8.png)
,
令h(x)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103100412497668279/SYS201311031004124976682019_DA/9.png)
,所以h'(x)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103100412497668279/SYS201311031004124976682019_DA/10.png)
+2-2x,當(dāng)x=1時,h'(x)=0
當(dāng)x∈(0,1)時,h'(x)>0;
當(dāng)x∈(1,+∞)時,h'(x)<0
∴當(dāng)x=1時,函數(shù)h(x)取得極大值為h(1)=2
∴當(dāng)方程f(x)=x
2-2x+k有實數(shù)解時,k≤2.)
(3)∵函數(shù)f(x)在區(qū)間(1,+∞)為減函數(shù),而
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103100412497668279/SYS201311031004124976682019_DA/11.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103100412497668279/SYS201311031004124976682019_DA/12.png)
,∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103100412497668279/SYS201311031004124976682019_DA/13.png)
,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103100412497668279/SYS201311031004124976682019_DA/14.png)
∴l(xiāng)nn=ln2-ln1+ln3-ln2+…+lnn-ln(n-1)<
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103100412497668279/SYS201311031004124976682019_DA/15.png)
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103100412497668279/SYS201311031004124976682019_DA/16.png)
而n•f(n)=1+lnn,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103100412497668279/SYS201311031004124976682019_DA/17.png)
,結(jié)論成立
點評:本題考查函數(shù)存在極值的性質(zhì),函數(shù)與方程的轉(zhuǎn)化,及利用函數(shù)的單調(diào)性證明不等式,要注意疊加法及放縮法在證明不等式中的應(yīng)用.