【答案】
分析:首先算出|x-a|<1的解,即a-1<x<a+1.由題意說(shuō)明,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213412400339719/SYS201310232134124003397002_DA/0.png)
是a-1<x<a+1的真子集,求解即可.
解答:解:由|x-a|<1,可得a-1<x<a+1.
它的充分非必要條件是
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213412400339719/SYS201310232134124003397002_DA/1.png)
<x<
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213412400339719/SYS201310232134124003397002_DA/2.png)
,
也就是說(shuō)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213412400339719/SYS201310232134124003397002_DA/3.png)
<x<
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213412400339719/SYS201310232134124003397002_DA/4.png)
是a-1<x<a+1的真子集,則a須滿足屬于{a|a-1≤
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213412400339719/SYS201310232134124003397002_DA/5.png)
且a+1>
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213412400339719/SYS201310232134124003397002_DA/6.png)
}或{a|a-1<
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213412400339719/SYS201310232134124003397002_DA/7.png)
且a+1≥
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213412400339719/SYS201310232134124003397002_DA/8.png)
};
解得a∈(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213412400339719/SYS201310232134124003397002_DA/9.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213412400339719/SYS201310232134124003397002_DA/10.png)
]∪[
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213412400339719/SYS201310232134124003397002_DA/11.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213412400339719/SYS201310232134124003397002_DA/12.png)
),
即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213412400339719/SYS201310232134124003397002_DA/13.png)
≤a≤
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213412400339719/SYS201310232134124003397002_DA/14.png)
故選B.
點(diǎn)評(píng):本題考查絕對(duì)值不等式的解法,必要條件、充分條件與充要條件的判斷,考查計(jì)算能力,是中檔題.