(Ⅰ)
f'(
x)=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231406267861035.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140626801723.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140626832330.gif)
…………2分
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140626832936.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231406271601286.gif)
—1<x<1时,f'(x)>0;x>1时,f'(x)<0,
∴
f(x)在(—1,1)上为增函数,在(1,+∞)上为减函数
所以
f(
1)为函数
f(
x)的极大值 …………4分
(Ⅱ)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231406273631150.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140627394617.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231406273941222.gif)
…………5分
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231406274251171.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231406274411205.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140627456848.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140627472969.gif)
+∞)是为减函数
因此
f(
x)在
x=—1+
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140627768331.gif)
处取得区间(—1,+∞)上的最大值 ——6分
由
f(—1+
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140627768331.gif)
)=0得
a=—
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140627800263.gif)
…………7分
(1)当
a<
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140627815272.gif)
时,
f(—1+
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140627831854.gif)
所以方程
f(
x)=0在区间(0,3)内无实数根 …………8分
(2)当
a=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140627815272.gif)
时,
f(—1+
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140627862435.gif)
所以方程
f(
x)=0在区间(0,3)内有且仅有1个实数根
—1+
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140627878467.gif)
…………9分
(3)当
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140627893295.gif)
a≤
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140625865252.gif)
时,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140627924507.gif)
≤1,
又
f(0)
a<0,f(—1+
f(3)=ln4+16
a≤ln4-2<0,
所以方程
f(
x)=0在区间(0,3)内有2个实数根. …………11分
综上所述,
当
a<
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140627815272.gif)
时,方程
f(
x)=0在区间(0,3)内无实数根;
当
a![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140627971283.gif)
时,方程
f(
x)=0在区间(0,3)内有1个实数根;
当
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140627987332.gif)
≤
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140625865252.gif)
时,方程
f(
x)=0 在区间(0,3),内有2个实数根.
…………12分