【答案】
分析:(Ⅰ)首先函數(shù)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/0.png)
是二次函數(shù),再利用二次函數(shù)的性質(zhì)解決對一切x∈R,都有f(x)≥0;根據(jù)f(1)=0得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/1.png)
,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/2.png)
,從而可得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/3.png)
,進而可得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/4.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/5.png)
,
另解:首先函數(shù)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/6.png)
是二次函數(shù),再利用二次函數(shù)的性質(zhì)解決對一切x∈R,都有f(x)≥0;由f(1)=0,得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/7.png)
,代入上式得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/8.png)
,根據(jù)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/9.png)
,可得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/10.png)
,從而有
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/11.png)
,故可求a、c的值;
(Ⅱ)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/12.png)
.該函數(shù)圖象開口向上,且對稱軸為x=2m+1.假設(shè)存在實數(shù)m使函數(shù)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/13.png)
在區(qū)間[m,m+2]上有最小值-5.根據(jù)函數(shù)的對稱軸與區(qū)間的關(guān)系進行分類討論,從而可求m的值
解答:解:(Ⅰ)當(dāng)a=0時,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/14.png)
.
由f(1)=0得:
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/15.png)
,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/16.png)
,∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/17.png)
.
顯然x>1時,f(x)<0,這與條件②相矛盾,不合題意.
∴a≠0,函數(shù)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/18.png)
是二次函數(shù). …(2分)
由于對一切x∈R,都有f(x)≥0,于是由二次函數(shù)的性質(zhì)可得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/19.png)
即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/20.png)
(*)…(4分)
由f(1)=0得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/21.png)
,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/22.png)
,代入(*)得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/23.png)
.
整理得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/24.png)
,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/25.png)
.
而
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/26.png)
,∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/27.png)
.
將
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/28.png)
代入(*)得,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/29.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/30.png)
. …(7分)
另解:(Ⅰ)當(dāng)a=0時,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/31.png)
.
由f(1)=0得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/32.png)
,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/33.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/34.png)
.
顯然x>1時,f(x)<0,這與條件②相矛盾,
∴a≠0,因而函數(shù)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/35.png)
是二次函數(shù). …(2分)
由于對一切x∈R,都有f(x)≥0,于是由二次函數(shù)的性質(zhì)可得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/36.png)
即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/37.png)
…(4分)
由此可知 a>0,c>0,
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/38.png)
.
由f(1)=0,得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/39.png)
,代入上式得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/40.png)
.
但前面已推得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/41.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/42.png)
.
由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/43.png)
解得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/44.png)
. …(7分)
(Ⅱ)∵
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/45.png)
,∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/46.png)
.
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/47.png)
.
該函數(shù)圖象開口向上,且對稱軸為x=2m+1. …(8分)
假設(shè)存在實數(shù)m使函數(shù)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/48.png)
在區(qū)間[m,m+2]上有最小值-5.
①當(dāng)m<-1時,2m+1<m,函數(shù)g(x)在區(qū)間[m,m+2]上是遞增的,
∴g(m)=-5,
即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/49.png)
,
解得 m=-3或m=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/50.png)
.
∵
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/51.png)
>-1,∴m=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/52.png)
舍去. …(10分)
②當(dāng)-1≤m<1時,m≤2m+1<m+1,函數(shù)g(x)在區(qū)間[m,2m+1]上是遞減的,而在區(qū)間[2m+1,m+2]上是遞增的,
∴g(2m+1)=-5,
即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/53.png)
.
解得 m=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/54.png)
或m=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/55.png)
,均應(yīng)舍去. …(12分)
③當(dāng)m≥1時,2m+1≥m+2,函數(shù)g(x)在區(qū)間[m,m+2]上是遞減的,
∴g(m+2)=-5,
即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/56.png)
.
解得 m=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/57.png)
或m=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/58.png)
,其中m=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/59.png)
應(yīng)舍去.
綜上可得,當(dāng)m=-3或m=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101230816345277123/SYS201311012308163452771019_DA/60.png)
時,函數(shù)g(x)=f(x)-mx在區(qū)間[m,m+2]上有最小值-5.
…(14分)
點評:本小題主要考查函數(shù)、方程、不等式等基本知識,考查綜合運用數(shù)學(xué)知識分析和解決問題的能力,本題考查的重點是函數(shù)的解析式的求解與函數(shù)最值的研究,解題的關(guān)鍵是合理運用函數(shù)的性質(zhì),正確分類,同時考查學(xué)生分析解決問題的能力,有一定的綜合性.