已知函數(shù)f(x)=x3-3x+1,x∈R,,A={x|t≤x≤t+1},B={x||f(x)|≥1}集合A∩B只含有一個(gè)元素,則實(shí)數(shù)t的取值范圍是 .
【答案】
分析:由|f(x)|≥1 得|x
3-3x+1|≥1,故x
3-3x+1≥1①,或x
3-3x+1≤-1②.由①②求得-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122951725027060/SYS201310251229517250270016_DA/0.png)
≤x≤0 或x≥
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122951725027060/SYS201310251229517250270016_DA/1.png)
,或 x=1,或 x≤-2,再畫出數(shù)軸如圖,結(jié)合數(shù)軸即可得實(shí)數(shù)t的取值范圍.
解答:解:由|f(x)|≥1 得|x
3-3x+1|≥1,∴x
3-3x+1≥1①,或x
3-3x+1≤-1②,∴由①得:-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122951725027060/SYS201310251229517250270016_DA/2.png)
≤x≤0 或x≥
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122951725027060/SYS201310251229517250270016_DA/3.png)
.
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122951725027060/SYS201310251229517250270016_DA/images4.png)
由②得x=1,x≤-2,綜合可得:-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122951725027060/SYS201310251229517250270016_DA/4.png)
≤x≤0 或x≥
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122951725027060/SYS201310251229517250270016_DA/5.png)
,或 x=1,或 x≤-2.
畫出數(shù)軸如圖,又∵t≤x≤t+1,結(jié)合數(shù)軸得:實(shí)數(shù)t的取值范圍是(0,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122951725027060/SYS201310251229517250270016_DA/6.png)
-1),
故答案為(0,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025122951725027060/SYS201310251229517250270016_DA/7.png)
).
點(diǎn)評(píng):本題考查了集合關(guān)系中的參數(shù)取值問(wèn)題、一元不等式的解法,主要根據(jù)集合元素的特征進(jìn)行求解,對(duì)于集合關(guān)系中的參數(shù)取值問(wèn)題的問(wèn)題,需要數(shù)形結(jié)合幫助求解或說(shuō)明.