题目内容
已知函数![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231302443558608/SYS201311012313024435586020_ST/0.png)
(1)若函数f(x)在[1,+∞)上为递增函数,求正实数a的取值范围;
(2)当a=1时,求函数f(x)在
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231302443558608/SYS201311012313024435586020_ST/1.png)
(3)试比较
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231302443558608/SYS201311012313024435586020_ST/2.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231302443558608/SYS201311012313024435586020_ST/3.png)
【答案】分析:(1)求导函数,确定函数的单调区间,利用函数f(x)在[1,+∞)上为递增函数,可得[1,+∞)是单调增区间的子集,由此可确定正实数a的取值范围;
(2)确定函数在
上的单调性,进而可求函数f(x)在
上的最大值和最小值;
(3)令a=1,由(1)得
(当且仅当t=1时等号成立),
两边同除t(t>0)得
,再令t=n2,进而利用累加法,即可得到结论.
解答:解:(1)由
,可得
,∴函数f(x)在
递增,
∵函数f(x)在[1,+∞)上为递增函数
∴[1,+∞)是
的子集,
∴![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231302443558608/SYS201311012313024435586020_DA/8.png)
∵a>0,∴a≥1.
(2)当a=1,由(1)的函数f(x)在
上递减,在[1,2]上递增.
则ymin=f(1)=2;
又因为
,
∴![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231302443558608/SYS201311012313024435586020_DA/11.png)
∴
.
(3)令a=1,
由(1)得
(当且仅当t=1时等号成立),
两边同除t(t>0)得
,
令t=n2,
可得![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231302443558608/SYS201311012313024435586020_DA/15.png)
由累加法得![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231302443558608/SYS201311012313024435586020_DA/16.png)
=![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231302443558608/SYS201311012313024435586020_DA/18.png)
点评:本题重点考查导数知识的运用,考查函数的单调性,考查函数的最值,考查大小比较,解题的关键是正确求出导函数,合理构建不等式,属于中档题.
(2)确定函数在
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231302443558608/SYS201311012313024435586020_DA/0.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231302443558608/SYS201311012313024435586020_DA/1.png)
(3)令a=1,由(1)得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231302443558608/SYS201311012313024435586020_DA/2.png)
两边同除t(t>0)得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231302443558608/SYS201311012313024435586020_DA/3.png)
解答:解:(1)由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231302443558608/SYS201311012313024435586020_DA/4.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231302443558608/SYS201311012313024435586020_DA/5.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231302443558608/SYS201311012313024435586020_DA/6.png)
∵函数f(x)在[1,+∞)上为递增函数
∴[1,+∞)是
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231302443558608/SYS201311012313024435586020_DA/7.png)
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231302443558608/SYS201311012313024435586020_DA/8.png)
∵a>0,∴a≥1.
(2)当a=1,由(1)的函数f(x)在
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231302443558608/SYS201311012313024435586020_DA/9.png)
则ymin=f(1)=2;
又因为
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231302443558608/SYS201311012313024435586020_DA/10.png)
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231302443558608/SYS201311012313024435586020_DA/11.png)
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231302443558608/SYS201311012313024435586020_DA/12.png)
(3)令a=1,
由(1)得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231302443558608/SYS201311012313024435586020_DA/13.png)
两边同除t(t>0)得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231302443558608/SYS201311012313024435586020_DA/14.png)
令t=n2,
可得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231302443558608/SYS201311012313024435586020_DA/15.png)
由累加法得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231302443558608/SYS201311012313024435586020_DA/16.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231302443558608/SYS201311012313024435586020_DA/17.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231302443558608/SYS201311012313024435586020_DA/18.png)
点评:本题重点考查导数知识的运用,考查函数的单调性,考查函数的最值,考查大小比较,解题的关键是正确求出导函数,合理构建不等式,属于中档题.
![](http://thumb.zyjl.cn/images/loading.gif)
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