题目内容
已知
为实数,
,
为
的导函数.
(Ⅰ)若
,求
在
上的最大值和最小值;
(Ⅱ)若
在
和
上均单调递增,求
的取值范围
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231624837264.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231624853795.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231624868425.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231624884402.png)
(Ⅰ)若
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231624915454.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231624884402.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231624978381.png)
(Ⅱ)若
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231624884402.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231625149455.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231625165466.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231624837264.png)
(Ⅰ)
,
(Ⅱ)![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231625695401.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231625664413.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231625680515.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231625695401.png)
本试题主要是考查了导数在研究函数中的运用。
(1)根据导数的符号与函数单调性的关系得到函数的极值,进而得到最值。
(2)因为函数给定区间是单调的,则必有导数恒大于等于零或者恒小于等于零,得到参数的范围。
解:(1)
.
(2)
,
.
由
,得
,此时
,
,
由
,得
或
.
又
,
,
,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231626148195.png)
在
上的最大值为
,最小值为
.
(3)解法一![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231626257235.png)
,
依题意:
对
恒成立,即
,所以![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231626428436.png)
对
恒成立,即
,所以![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231626506422.png)
综上:
.
解法二![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231626257235.png)
,![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231626148195.png)
的图像是开口向上且过点
的抛物线,由条件得
,
,
,
.解得
. ![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231626148195.png)
的取值范围为
.
(1)根据导数的符号与函数单调性的关系得到函数的极值,进而得到最值。
(2)因为函数给定区间是单调的,则必有导数恒大于等于零或者恒小于等于零,得到参数的范围。
解:(1)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408232316257421258.png)
(2)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231625773865.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231625789839.png)
由
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231625820529.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231625851454.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231625867918.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231625882735.png)
由
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231625914551.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231625929479.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231625960333.png)
又
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231625976818.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231626007636.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231626023656.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231626148195.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231626163447.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231625695401.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231625664413.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231625680515.png)
(3)解法一
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231626257235.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231625789839.png)
依题意:
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231626319931.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231626366533.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408232316263971543.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231626428436.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231626319931.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231626460526.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408232316264911535.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231626506422.png)
综上:
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231626538520.png)
解法二
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231626257235.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231625789839.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231626148195.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231626647479.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231626662470.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231626725602.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231626756580.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231626772594.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231626787553.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231626803483.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231626148195.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231626834283.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823231625695401.png)
![](http://thumb.zyjl.cn/images/loading.gif)
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