【答案】
分析:(1)先求函數(shù)f(x)的導(dǎo)數(shù),再根據(jù)導(dǎo)函數(shù)的正負(fù)和原函數(shù)的關(guān)系可得答案.
(2)(i)先求出b
n的值然后代入到an=ln(l+n)-bn放縮可得答案.
(ii)根據(jù)(i)知
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023211835040348516/SYS201310232118350403485021_DA/0.png)
.,然后用數(shù)學(xué)歸納法證明即可.
解答:解:(I)因為f(x)=ln(1+x)-x,所以函數(shù)定義域為(-1,+∞),且f′(x)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023211835040348516/SYS201310232118350403485021_DA/1.png)
-1=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023211835040348516/SYS201310232118350403485021_DA/2.png)
.
由f′(x)>0得-1<x<0,f(x)的單調(diào)遞增區(qū)間為(-1,0);
由f’(x)<0得x>0,f(x)的單調(diào)遞減區(qū)間為(0,+∞).
(II)因為f(x)在[0,n]上是減函數(shù),所以b
n=f(n)=ln(1+n)-n,
則a
n=ln(1+n)-b
n=ln(1+n)-ln(1+n)+n=n.
(i)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023211835040348516/SYS201310232118350403485021_DA/3.png)
>
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023211835040348516/SYS201310232118350403485021_DA/4.png)
.
又lim
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023211835040348516/SYS201310232118350403485021_DA/5.png)
,
因此c<1,即實數(shù)c的取值范圍是(-∞,1).
(Ⅱ)因為f(x)在[0,n]上是減函數(shù),所以b
n=f(n)=ln(1+n)-n,
則a
n=ln(1+n)-b
n=ln(1+n)-ln(1+n)+n=n.
(i)因為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023211835040348516/SYS201310232118350403485021_DA/6.png)
對n∈N*恒成立.所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023211835040348516/SYS201310232118350403485021_DA/7.png)
對n∈N*恒成立.
則
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023211835040348516/SYS201310232118350403485021_DA/8.png)
對n∈N*恒成立.
設(shè)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023211835040348516/SYS201310232118350403485021_DA/9.png)
,n∈N*,則c<g(n)對n∈N*恒成立.
考慮
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023211835040348516/SYS201310232118350403485021_DA/10.png)
.
因為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023211835040348516/SYS201310232118350403485021_DA/11.png)
=0,
所以g(x)在[1,+∞)內(nèi)是減函數(shù);則當(dāng)n∈N*時,g(n)隨n的增大而減小,
又因為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023211835040348516/SYS201310232118350403485021_DA/12.png)
=1.
所以對一切n∈N,g(n)>1因此c≤1,即實數(shù)c的取值范圍是(-∞,1].
(ⅱ)由(ⅰ)知
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023211835040348516/SYS201310232118350403485021_DA/13.png)
.
下面用數(shù)學(xué)歸納法證明不等式
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023211835040348516/SYS201310232118350403485021_DA/14.png)
(n∈N
+)
①當(dāng)n=1時,左邊=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023211835040348516/SYS201310232118350403485021_DA/15.png)
,右邊=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023211835040348516/SYS201310232118350403485021_DA/16.png)
,左邊<右邊.不等式成立.
②假設(shè)當(dāng)n=k時,不等式成立.即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023211835040348516/SYS201310232118350403485021_DA/17.png)
.
當(dāng)n=k+1時,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023211835040348516/SYS201310232118350403485021_DA/18.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023211835040348516/SYS201310232118350403485021_DA/19.png)
,
即n=k+1時,不等式成立
綜合①、②得,不等式
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023211835040348516/SYS201310232118350403485021_DA/20.png)
成立.
所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023211835040348516/SYS201310232118350403485021_DA/21.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023211835040348516/SYS201310232118350403485021_DA/22.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023211835040348516/SYS201310232118350403485021_DA/23.png)
.
即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023211835040348516/SYS201310232118350403485021_DA/24.png)
.
點評:本小題主要考查函數(shù)的單調(diào)性、最值、不等式、數(shù)列等基本知識,考查運(yùn)用導(dǎo)數(shù)研究函數(shù)性質(zhì)的方法,考查分析問題和解決問題的能力.此題為壓軸題,所以平時可以讓學(xué)生學(xué)會放棄一些自己能力范圍之外的題目,把多余的時間多花點在中低檔題目上,可是80%的分?jǐn)?shù)呀,多么可觀,可是縱觀歷年的高考成績來看又有多少人真正的做到了.