题目内容
已知
,
(
且
).
(1)过
作曲线
的切线,求切线方程;
(2)设
在定义域上为减函数,且其导函数
存在零点,求实数
的值.
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823205053954771.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823205053970655.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823205053985392.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823205054001359.png)
(1)过
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823205054063534.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823205054079573.png)
(2)设
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823205054095783.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823205054126613.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823205054157283.png)
(1)∵f(0)=0,∴P(0,2)不在曲线y=f(x)上,
设切点为Q(x0,y0),∵f′(x)=2-x,
∴k=f′(x0)=2-x0,且y0=f(x0)=2x0-
,
∴切线方程为y-2x0+
=(2-x0)(x-x0),即y=(2-x0)x+
, ……3分
∵(0,2)在切线上,代入可得x0=±2,……………………………5分
∴切线方程为y=2或y=4x+2. …………………………………7分
(2)h(x)=2x-
x2-logax在(0,+∞)上递减,
∴h′(x)=2-x-
≤0在(0,+∞)上恒成立,
∵x>0,∴
≥2x-x2在(0,+∞)上恒成立.
又2x-x2∈(-∞,1],∴
≥1,∴0<lna≤1,①…………………10分
又∵h′(x)=2-x-
存在零点,
即方程lna·x2-2lna·x+1=0有正根,
∴Δ=4ln2a-4lna≥0,∴lna≥1或lna<0,②…………………12分
由①②知lna=1,∴a=e.
设切点为Q(x0,y0),∵f′(x)=2-x,
∴k=f′(x0)=2-x0,且y0=f(x0)=2x0-
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823205054157499.png)
∴切线方程为y-2x0+
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823205054157499.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823205054157499.png)
∵(0,2)在切线上,代入可得x0=±2,……………………………5分
∴切线方程为y=2或y=4x+2. …………………………………7分
(2)h(x)=2x-
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823205054219338.png)
∴h′(x)=2-x-
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823205054235495.png)
∵x>0,∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823205054251413.png)
又2x-x2∈(-∞,1],∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823205054251413.png)
又∵h′(x)=2-x-
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823205054235495.png)
即方程lna·x2-2lna·x+1=0有正根,
∴Δ=4ln2a-4lna≥0,∴lna≥1或lna<0,②…………………12分
由①②知lna=1,∴a=e.
略
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