题目内容
若
<
<0,则下列不等式:①
<
;②|a|+b>0;③a-
>b-
;④lna2>lnb2中,正确的是( )
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824040134776298.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824040134791313.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824040134807366.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824040134838351.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824040134776298.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824040134791313.png)
A.①④ | B.②③ | C.①③ | D.②④ |
D
先由
<
<0得到a与b的大小关系,再根据不等式的性质,对各个不等式进行逐一判断.
由
<
<0,可知b<a<0.
①中,a+b<0,ab>0,所以
<0,
>0.
故有
<
,即①正确.
②中,∵b<a<0,∴-b>-a>0,故-b>|a|,即|a|+b<0,故②错误.
③中,∵b<a<0,即0>a>b,
又∵
<
<0,∴-
>-
>0,
∴a-
>b-
,故③正确.
④中,∵b<a<0,根据y=x2在(-∞,0)上为单调递减函数,可得b2>a2>0,而y=lnx在定义域上为增函数.∴lnb2>lna2,故④错,综上分析,②④错误,①③正确.
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824040134776298.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824040134791313.png)
由
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824040134776298.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824040134791313.png)
①中,a+b<0,ab>0,所以
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824040134807366.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824040134838351.png)
故有
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824040134807366.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824040134838351.png)
②中,∵b<a<0,∴-b>-a>0,故-b>|a|,即|a|+b<0,故②错误.
③中,∵b<a<0,即0>a>b,
又∵
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824040134776298.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824040134791313.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824040134776298.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824040134791313.png)
∴a-
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824040134776298.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824040134791313.png)
④中,∵b<a<0,根据y=x2在(-∞,0)上为单调递减函数,可得b2>a2>0,而y=lnx在定义域上为增函数.∴lnb2>lna2,故④错,综上分析,②④错误,①③正确.
![](http://thumb.zyjl.cn/images/loading.gif)
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