【答案】
分析:(Ⅰ)解方程|f(x)|=g(x),根據(jù)積商符號法則轉(zhuǎn)化為兩個絕對值不等式的根的問題;
(Ⅱ)不等式f(x)≥g(x)恒成立即(x
2-1)≥a|x-1|對x∈R恒成立,對x進行討論,分離參數(shù),轉(zhuǎn)化為函數(shù)的最值問題求解;(Ⅲ)去絕對值,分段求函數(shù)的最值.
解答:解:(Ⅰ)方程|f(x)|=g(x),
即|x
2-1|=a|x-1|,變形得|x-1|(|x+1|-a)=0,
顯然,x=1已是該方程的根,
從而欲原方程有兩個不同的解,即要求方程|x+1|=a
“有且僅有一個不等于1的解”或
“有兩解,一解為1,另一解不等于1”
得a=0或a=2
(Ⅱ)不等式f(x)≥g(x)對x∈R恒成立,
即(x
2-1)≥a|x-1|(*)對x∈R恒成立,
①當x=1時,(*)顯然成立,此時a∈R
②當x≠1時,(*)可變形為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181151188323697/SYS201310241811511883236019_DA/0.png)
,
令
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181151188323697/SYS201310241811511883236019_DA/1.png)
,
因為當x>1時,φ(x)>2;而當x<1時,φ(x)>-2.
所以g(x)>-2,故此時a≤-2
綜合①②,得所求a的取值范圍是a≤-2
(Ⅲ)因為h(x)=|f(x)|+g(x)=|x
2-1|+a|x-1|
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181151188323697/SYS201310241811511883236019_DA/2.png)
,
1)當
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181151188323697/SYS201310241811511883236019_DA/3.png)
,即a>2時,
h(x)在[-2,1]上遞減,在[1,2]上遞增,
且h(-2)=3a+3,h(2)=a+3,
經(jīng)比較,此時h(x)在[-2,2]上的最大值為3a+3
2)當
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181151188323697/SYS201310241811511883236019_DA/4.png)
,即0≤a≤2時,
h(x)在[-2,-1],
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181151188323697/SYS201310241811511883236019_DA/5.png)
上遞減,
在
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181151188323697/SYS201310241811511883236019_DA/6.png)
上[1,2]上遞增,
且h(-2)=3a+3,h(2)=a+3,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181151188323697/SYS201310241811511883236019_DA/7.png)
,
經(jīng)比較,知此時h(x)在[-2,2]上的最大值為3a+3
3)當
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181151188323697/SYS201310241811511883236019_DA/8.png)
,即-2≤a<0時,
h(x)在[-2,-1],
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181151188323697/SYS201310241811511883236019_DA/9.png)
上遞減,
在
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181151188323697/SYS201310241811511883236019_DA/10.png)
,[1,2]上遞增,
且h(-2)=3a+3,h(2)=a+3,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181151188323697/SYS201310241811511883236019_DA/11.png)
,
經(jīng)比較知此時h(x)在[-2,2]上的最大值為a+3
4)當
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181151188323697/SYS201310241811511883236019_DA/12.png)
,即-3≤a<-2時,
h(x)在
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181151188323697/SYS201310241811511883236019_DA/13.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181151188323697/SYS201310241811511883236019_DA/14.png)
上遞減,
在
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181151188323697/SYS201310241811511883236019_DA/15.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181151188323697/SYS201310241811511883236019_DA/16.png)
上遞增,且h(-2)=3a+3<0,h(2)=a+3≥0,
經(jīng)比較知此時h(x)在[-2,2]上的最大值為a+3
5)當
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181151188323697/SYS201310241811511883236019_DA/17.png)
,即a<-3時,
h(x)在[-2,1]上遞減,在[1,2]上遞增,
故此時h(x)在[-2,2]上的最大值為h(1)=0
綜上所述,當a≥0時,h(x)在[-2,2]上的最大值為3a+3;
當-3≤a<0時,h(x)在[-2,2]上的最大值為a+3;
當a<-3時,h(x)在[-2,2]上的最大值為0.
點評:考查絕對值方程、不等式和最值問題的求法,體現(xiàn)了分類討論、等價轉(zhuǎn)化的數(shù)學思想方法,特別是(Ⅲ)難度較大,很好的考查分析問題、解決問題的能力.屬難題.