【答案】
分析:(I)根據(jù)奇函數(shù)的定義g(x)=-g(-x)列出關(guān)于b的等式,由函數(shù)的奇偶性定義求出b的值;
(II)分當(dāng)a>1和當(dāng)0<a<1兩種情況討論,利用分離參數(shù)法,結(jié)合導(dǎo)數(shù)在最大值、最小值問題中的應(yīng)用來解m的取值范圍.
(Ⅲ)先得出:
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172855676418930/SYS201311031728556764189021_DA/0.png)
,再分情況討論:當(dāng)n=2時,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172855676418930/SYS201311031728556764189021_DA/1.png)
,2
n-2=2,∴a
f(2)+f(3)++f(n)>2
n-2;當(dāng)n=3時,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172855676418930/SYS201311031728556764189021_DA/2.png)
,2
n-2=6,∴a
f(2)+f(3)++f(n)=2
n-2;當(dāng)n≥4時,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172855676418930/SYS201311031728556764189021_DA/3.png)
2
n-2進(jìn)行證明即可.
解答:解:(Ⅰ)由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172855676418930/SYS201311031728556764189021_DA/4.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172855676418930/SYS201311031728556764189021_DA/5.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172855676418930/SYS201311031728556764189021_DA/6.png)
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172855676418930/SYS201311031728556764189021_DA/7.png)
恒成立,b
2=1,b=±1經(jīng)檢驗b=1
(Ⅱ)由x∈[2,4]時,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172855676418930/SYS201311031728556764189021_DA/8.png)
恒成立,
①當(dāng)a>1時
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172855676418930/SYS201311031728556764189021_DA/9.png)
對x∈[2,4]恒成立
∴0<m<(x+1)(x-1)(7-x)在x∈[2,4]恒成立
設(shè)g(x)=(x+1)(x-1)(7-x),x∈[2,4]
則g(x)=-x
3+7x
2+x-7
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172855676418930/SYS201311031728556764189021_DA/10.png)
∴當(dāng)x∈[2,4]時,g'(x)>0
∴y=g(x)在區(qū)間[2,4]上是增函數(shù),g(x)
min=g(2)=15
∴0<m<15
②當(dāng)0<a<1時
由x∈[2,4]時,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172855676418930/SYS201311031728556764189021_DA/11.png)
恒成立,
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172855676418930/SYS201311031728556764189021_DA/12.png)
對x∈[2,4]恒成立
∴m>(x+1)(x-1)(7-x)在x∈[2,4]恒成立
設(shè)g(x)=(x+1)(x-1)(7-x),x∈[2,4]
由①可知y=g(x)在區(qū)間[2,4]上是增函數(shù),g(x)
max=g(4)=45
∴m>45
綜上,當(dāng)a>1時,0<m<15;
當(dāng)0<a<1時,m>45
(Ⅲ)∵
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172855676418930/SYS201311031728556764189021_DA/13.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172855676418930/SYS201311031728556764189021_DA/14.png)
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172855676418930/SYS201311031728556764189021_DA/15.png)
當(dāng)n=2時,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172855676418930/SYS201311031728556764189021_DA/16.png)
,2
n-2=2,∴a
f(2)+f(3)++f(n)>2
n-2
當(dāng)n=3時,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172855676418930/SYS201311031728556764189021_DA/17.png)
,2
n-2=6,∴a
f(2)+f(3)++f(n)=2
n-2
當(dāng)n≥4時,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172855676418930/SYS201311031728556764189021_DA/18.png)
2
n-2
下面證明:當(dāng)n≥4時,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172855676418930/SYS201311031728556764189021_DA/19.png)
2
n-2
當(dāng)n≥4時,2
n-2=C
n+C
n1+C
n2++C
nn-1+C
nn-2=C
n1+C
n2++C
nn-1![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172855676418930/SYS201311031728556764189021_DA/20.png)
∴當(dāng)n≥4時,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172855676418930/SYS201311031728556764189021_DA/21.png)
2
n-2
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172855676418930/SYS201311031728556764189021_DA/22.png)
n≥4時,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172855676418930/SYS201311031728556764189021_DA/23.png)
,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172855676418930/SYS201311031728556764189021_DA/24.png)
2
n-2
∴當(dāng)n≥4時,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172855676418930/SYS201311031728556764189021_DA/25.png)
2
n-2.
點評:本題是函數(shù)性質(zhì)的綜合題,本小題主要考查函數(shù)奇偶性的性質(zhì)、函數(shù)奇偶性的應(yīng)用、不等式的解法、導(dǎo)數(shù)在最大值、最小值問題中的應(yīng)用等基礎(chǔ)知識,考查運算求解能力,考查化歸與轉(zhuǎn)化思想.屬于中檔題.