【答案】
分析:(1)y=f(x)在[-1,1]上單調(diào)遞減函數(shù),要存在零點只需f(1)≤0,f(-1)≥0即可
(2)存在性問題,只需函數(shù)y=f(x)的值域為函數(shù)y=g(x)的值域的子集即可
(3)研究函數(shù)y=f(x)(x∈[t,4])的值域,需要對t進行討論,研究單調(diào)性
解答:解:(Ⅰ):因為函數(shù)f(x)=x
2-4x+a+3的對稱軸是x=2,
所以f(x)在區(qū)間[-1,1]上是減函數(shù),
因為函數(shù)在區(qū)間[-1,1]上存在零點,
則必有:
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103173839477705406/SYS201311031738394777054019_DA/0.png)
即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103173839477705406/SYS201311031738394777054019_DA/1.png)
,解得-8≤a≤0,
故所求實數(shù)a的取值范圍為[-8,0].
(Ⅱ)若對任意的x
1∈[1,4],總存在x
2∈[1,4],
使f(x
1)=g(x
2)成立,只需函數(shù)y=f(x)的值域為函數(shù)y=g(x)的值域的子集.
f(x)=x
2-4x+3,x∈[1,4]的值域為[-1,3],下求g(x)=mx+5-2m的值域.
①當m=0時,g(x)=5-2m為常數(shù),不符合題意舍去;
②當m>0時,g(x)的值域為[5-m,5+2m],要使[-1,3]⊆[5-m,5+2m],
需
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103173839477705406/SYS201311031738394777054019_DA/2.png)
,解得m≥6;
③當m<0時,g(x)的值域為[5+2m,5-m],要使[-1,3]⊆[5+2m,5-m],
需
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103173839477705406/SYS201311031738394777054019_DA/3.png)
,解得m≤-3;
綜上,m的取值范圍為(-∞,-3]∪[6,+∞)
(Ⅲ)由題意知
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103173839477705406/SYS201311031738394777054019_DA/4.png)
,可得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103173839477705406/SYS201311031738394777054019_DA/5.png)
.
①當t≤0時,在區(qū)間[t,4]上,f(t)最大,f(2)最小,
所以f(t)-f(2)=7-2t即t
2-2t-3=0,解得t=-1或t=3(舍去);
②當0<t≤2時,在區(qū)間[t,4]上,f(4)最大,f(2)最小,
所以f(4)-f(2)=7-2t即4=7-2t,解得t=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103173839477705406/SYS201311031738394777054019_DA/6.png)
;
③當2<t<
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103173839477705406/SYS201311031738394777054019_DA/7.png)
時,在區(qū)間[t,4]上,f(4)最大,f(t)最小,
所以f(4)-f(t)=7-2t即t
2-6t+7=0,解得t=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103173839477705406/SYS201311031738394777054019_DA/8.png)
(舍去)
綜上所述,存在常數(shù)t滿足題意,t=-1或
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103173839477705406/SYS201311031738394777054019_DA/9.png)
.
點評:本題綜合考查了函數(shù)的零點,值域與恒成立問題.