试题分析: ⑴由 a=2,b=1得,f (x)=(2+
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055452940314.png)
)e
x, 定义域为(-∞,0)∪(0,+∞);从而可求得 f ′(x)=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055452987807.png)
e
x, 令f ′(x)=0,得x
1=-1,x
2=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055452909338.png)
,列表可求得f (x)的极值.
⑵①当a=1时,g (x)=(x-
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055453034396.png)
-2)e
x,由已知得不等式g (x)≥1在x∈(0,+∞)上恒成立,即b≤x
2-2x-
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055453049419.png)
在x∈(0,+∞)上恒成立,从而b≤(x
2-2x-
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055453049419.png)
)
min x∈(0,+∞),令h(x)=x
2-2x-
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055453049419.png)
(x>0)利用函数导数求出h(x)的最小值即可.
②由于g (x)=(ax-
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055453034396.png)
-2a)e
x,所以g ′(x)=(
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055453127452.png)
+ax-
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055453034396.png)
-a)e
x; 由g (x)+g ′(x)=0,得(ax-
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055453034396.png)
-2a)e
x+(
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055453127452.png)
+ax-
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055453034396.png)
-a)e
x=0,整理得2ax
3-3ax
2-2bx+b=0.
存在x>1,使g (x)+g ′(x)=0成立,等价于存在x>1,2ax
3-3ax
2-2bx+b=0成立.
注意到a>0,所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055452893396.png)
=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055453252668.png)
(x>1);设u(x)=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055453252668.png)
(x>1),则问题等价于
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055453299678.png)
的最小值(或下确界),利用函数导数可判断u(x)在
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055453314509.png)
上的单调性可求得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055453330668.png)
从而可得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055452893396.png)
的取值范围为(-1,+∞).
试题解析:⑴当a=2,b=1时,f (x)=(2+
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055452940314.png)
)e
x,定义域为(-∞,0)∪(0,+∞).
所以f ′(x)=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055452987807.png)
e
x.令f ′(x)=0,得x
1=-1,x
2=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055452909338.png)
,列表
x
| (-∞,-1)
| -1
| (-1,0)
| (0, )
| ![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055452909338.png)
| ( ,+∞)
|
f ′(x)
| ![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055453502180.png)
| ![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055453517260.png)
| -
| -
| ![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055453517260.png)
| ![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055453502180.png)
|
f (x)
| ↗
| 极大值
| ↘
| ↘
| 极小值
| ↗
|
由表知f (x)的极大值是f (-1)=e
-1,f (x)的极小值是f (
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055452909338.png)
)=4
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055452924329.png)
.
⑵① 因为g (x)=(ax-a)e
x-f (x)=(ax-
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055453034396.png)
-2a)e
x,当a=1时,g (x)=(x-
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055453034396.png)
-2)e
x.
因为g (x)≥1在x∈(0,+∞)上恒成立,所以b≤x
2-2x-
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055453049419.png)
在x∈(0,+∞)上恒成立.
记h(x)=x
2-2x-
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055453049419.png)
(x>0),则h′(x)=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055453720915.png)
.
当0<x<1时,h′(x)<0,h(x)在(0,1)上是减函数;
当x>1时,h′(x)>0,h(x)在(1,+∞)上是增函数.
所以h(x)
min=h(1)=-1-e
-1.所以b的最大值为-1-e
-1.
②因为g (x)=(ax-
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055453034396.png)
-2a)e
x,所以g ′(x)=(
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055453127452.png)
+ax-
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055453034396.png)
-a)e
x.
由g (x)+g ′(x)=0,得(ax-
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055453034396.png)
-2a)e
x+(
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055453127452.png)
+ax-
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055453034396.png)
-a)e
x=0,整理得2ax
3-3ax
2-2bx+b=0.
存在x>1,使g (x)+g ′(x)=0成立,等价于存在x>1,2ax
3-3ax
2-2bx+b=0成立.
因为a>0,所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055452893396.png)
=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055453252668.png)
.设u(x)=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055453252668.png)
(x>1),则u′(x)=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240554539231523.png)
.
因为x>1,u′(x)>0恒成立,所以u(x)在(1,+∞)是增函数,所以u(x)>u(1)=-1,
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055452893396.png)
>-1,即
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824055452893396.png)
的取值范围为(-1,+∞).