【答案】
分析:(1)由∵f(0)=0可得c=0而函數(shù)對于任意x∈R都有
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182503078885886/SYS201310241825030788858019_DA/0.png)
,可得函數(shù)f(x)的對稱軸從而可得a=b
結(jié)合f(x)≥x,即ax
2+(b-1)x≥0對于任意x∈R都成立,可轉(zhuǎn)化為二次函數(shù)的圖象可得a>0,且△=(b-1)
2≤0.
(2)由(1)可得g(x)=f(x)-|λx-1|=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182503078885886/SYS201310241825030788858019_DA/1.png)
根據(jù)函數(shù)g(x)需討論:
①當
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182503078885886/SYS201310241825030788858019_DA/2.png)
時,函數(shù)g(x)=x
2+(1-λ)x+1的對稱軸為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182503078885886/SYS201310241825030788858019_DA/3.png)
,
則要比較對稱軸與區(qū)間端點的大小,為此產(chǎn)生討論:
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182503078885886/SYS201310241825030788858019_DA/4.png)
,與
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182503078885886/SYS201310241825030788858019_DA/5.png)
分別求單調(diào)區(qū)間
②當
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182503078885886/SYS201310241825030788858019_DA/6.png)
時,函數(shù)g(x)=x
2+(1+λ)x-1的對稱軸為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182503078885886/SYS201310241825030788858019_DA/7.png)
,
同①的討論思路
(3)結(jié)合(2)中的單調(diào)區(qū)間及零點存在定理進行判斷函數(shù)g(x)的零點
解答:(1)解:∵f(0)=0,∴c=0.(1分)
∵對于任意x∈R都有
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182503078885886/SYS201310241825030788858019_DA/8.png)
,
∴函數(shù)f(x)的對稱軸為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182503078885886/SYS201310241825030788858019_DA/9.png)
,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182503078885886/SYS201310241825030788858019_DA/10.png)
,得a=b.(2分)
又f(x)≥x,即ax
2+(b-1)x≥0對于任意x∈R都成立,
∴a>0,且△=(b-1)
2≤0.
∵(b-1)
2≥0,∴b=1,a=1.
∴f(x)=x
2+x.(4分)
(2)解:g(x)=f(x)-|λx-1|=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182503078885886/SYS201310241825030788858019_DA/11.png)
(5分)
①當
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182503078885886/SYS201310241825030788858019_DA/12.png)
時,函數(shù)g(x)=x
2+(1-λ)x+1的對稱軸為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182503078885886/SYS201310241825030788858019_DA/13.png)
,
若
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182503078885886/SYS201310241825030788858019_DA/14.png)
,即0<λ≤2,函數(shù)g(x)在
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182503078885886/SYS201310241825030788858019_DA/15.png)
上單調(diào)遞增;(6分)
若
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182503078885886/SYS201310241825030788858019_DA/16.png)
,即λ>2,函數(shù)g(x)在
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182503078885886/SYS201310241825030788858019_DA/17.png)
上單調(diào)遞增,在
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182503078885886/SYS201310241825030788858019_DA/18.png)
上單調(diào)遞減.
(7分)
②當
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182503078885886/SYS201310241825030788858019_DA/19.png)
時,函數(shù)g(x)=x
2+(1+λ)x-1的對稱軸為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182503078885886/SYS201310241825030788858019_DA/20.png)
,
則函數(shù)g(x)在
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182503078885886/SYS201310241825030788858019_DA/21.png)
上單調(diào)遞增,在
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182503078885886/SYS201310241825030788858019_DA/22.png)
上單調(diào)遞減.(8分)
綜上所述,當0<λ≤2時,函數(shù)g(x)單調(diào)遞增區(qū)間為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182503078885886/SYS201310241825030788858019_DA/23.png)
,單調(diào)遞減區(qū)間為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182503078885886/SYS201310241825030788858019_DA/24.png)
;(9分)
當λ>2時,函數(shù)g(x)單調(diào)遞增區(qū)間為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182503078885886/SYS201310241825030788858019_DA/25.png)
和
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182503078885886/SYS201310241825030788858019_DA/26.png)
,單調(diào)遞減區(qū)間為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182503078885886/SYS201310241825030788858019_DA/27.png)
和
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182503078885886/SYS201310241825030788858019_DA/28.png)
.(10分)
(3)解:①當0<λ≤2時,由(2)知函數(shù)g(x)在區(qū)間(0,1)上單調(diào)遞增,
又g(0)=-1<0,g(1)=2-|λ-1|>0,
故函數(shù)g(x)在區(qū)間(0,1)上只有一個零點.(11分)
②當λ>2時,則
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182503078885886/SYS201310241825030788858019_DA/29.png)
,而g(0)=-1<0,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182503078885886/SYS201310241825030788858019_DA/30.png)
,g(1)=2-|λ-1|,
(�。┤�2<λ≤3,由于
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182503078885886/SYS201310241825030788858019_DA/31.png)
,
且
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182503078885886/SYS201310241825030788858019_DA/32.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182503078885886/SYS201310241825030788858019_DA/33.png)
,
此時,函數(shù)g(x)在區(qū)間(0,1)上只有一個零點;(12分)
(ⅱ)若λ>3,由于
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182503078885886/SYS201310241825030788858019_DA/34.png)
且g(1)=2-|λ-1|<0,此時,函數(shù)g(x)在區(qū)間(0,1)
上有兩個不同的零點.(13分)
綜上所述,當0<λ≤3時,函數(shù)g(x)在區(qū)間(0,1)上只有一個零點;
當λ>3時,函數(shù)g(x)在區(qū)間(0,1)上有兩個不同的零點.(14分)
點評:本題主要考查了函數(shù)的解析式的求解,函數(shù)的單調(diào)區(qū)間,零點存在的判定定理,考查了分類討論思想的在解題中的應用.屬于綜合性較強的試題.