【答案】
分析:(Ⅰ)把a和b的值代入解析式確定出f(x),求出f(x)的導(dǎo)函數(shù),把x=1代入f(x)中求出f(1)的值即為切點的縱坐標,得到切點坐標,把x=1代入導(dǎo)函數(shù)中求出的導(dǎo)函數(shù)值即為切線的斜率,由切點坐標和斜率寫出切線的方程即可;(Ⅱ)把a=-2-b代入解析式表示出f(x),求出f(x)的導(dǎo)函數(shù),又根據(jù)負數(shù)沒有對數(shù)求出f(x)的定義域,令導(dǎo)函數(shù)等于0求出x的值為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181817039465535/SYS201310241818170394655019_DA/0.png)
和1,分四種情況考慮:
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181817039465535/SYS201310241818170394655019_DA/1.png)
小于等于0;
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181817039465535/SYS201310241818170394655019_DA/2.png)
大于0小于1;
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181817039465535/SYS201310241818170394655019_DA/3.png)
等于1;
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181817039465535/SYS201310241818170394655019_DA/4.png)
大于1,分別討論導(dǎo)函數(shù)的正負即可得到函數(shù)的單調(diào)區(qū)間.
解答:解:(Ⅰ)因為a=1,b=-1,所以函數(shù)f(x)=x
2+x-lnx,f(1)=2
又
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181817039465535/SYS201310241818170394655019_DA/5.png)
,f′(1)=2(2分)
所以y-2=2(x-1)
即f(x)在x=1處的切線方程為2x-y=0(5分)
(Ⅱ)因為a=-2-b,所以f(x)=x
2-(2+b)x+blnx,
則
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181817039465535/SYS201310241818170394655019_DA/6.png)
(x>0)
令f'(x)=0,得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181817039465535/SYS201310241818170394655019_DA/7.png)
,x
2=1.(7分)
①當(dāng)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181817039465535/SYS201310241818170394655019_DA/8.png)
,即b≤0時,函數(shù)f(x)的單調(diào)遞減區(qū)間為(0,1),單調(diào)遞增區(qū)間為(1,+∞);(8分)
②當(dāng)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181817039465535/SYS201310241818170394655019_DA/9.png)
,即0<b<2時,f'(x),f(x)的變化情況如下表:
x | ![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181817039465535/SYS201310241818170394655019_DA/10.png) | ![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181817039465535/SYS201310241818170394655019_DA/11.png) | (1,+∞) |
f'(x) | + | - | + |
f(x) | ↗ | ↘ | ↗ |
所以,函數(shù)f(x)的單調(diào)遞增區(qū)間為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181817039465535/SYS201310241818170394655019_DA/12.png)
∪(1,+∞),單調(diào)遞減區(qū)間為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181817039465535/SYS201310241818170394655019_DA/13.png)
;(9分)
③當(dāng)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181817039465535/SYS201310241818170394655019_DA/14.png)
,即b=2時,函數(shù)f(x)的單調(diào)遞增區(qū)間為(0,+∞);(10分)
④當(dāng)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181817039465535/SYS201310241818170394655019_DA/15.png)
,即b>2時,f'(x),f(x)的變化情況如下表:
x | (0,1) | ![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181817039465535/SYS201310241818170394655019_DA/16.png) | ![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181817039465535/SYS201310241818170394655019_DA/17.png) |
f'(x) | + | - | + |
f(x) | ↗ | ↘ | ↗ |
所以函數(shù)f(x)的單調(diào)遞增區(qū)間為(0,1),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181817039465535/SYS201310241818170394655019_DA/18.png)
,單調(diào)遞減區(qū)間為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181817039465535/SYS201310241818170394655019_DA/19.png)
;(12分)
綜上,當(dāng)b≤0時,函數(shù)f(x)的單調(diào)遞減區(qū)間為(0,1),單調(diào)遞增區(qū)間為(1,+∞);
當(dāng)0<b<2時,函數(shù)f(x)的單調(diào)遞增區(qū)間為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181817039465535/SYS201310241818170394655019_DA/20.png)
∪(1,+∞),單調(diào)遞減區(qū)間為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181817039465535/SYS201310241818170394655019_DA/21.png)
;
當(dāng)b=2時,函數(shù)f(x)的單調(diào)遞增區(qū)間為(0,+∞);
當(dāng)b>2時,函數(shù)f(x)的單調(diào)遞增區(qū)間為(0,1),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181817039465535/SYS201310241818170394655019_DA/22.png)
,單調(diào)遞減區(qū)間為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181817039465535/SYS201310241818170394655019_DA/23.png)
.(13分)
點評:此題考查學(xué)生會利用導(dǎo)數(shù)求曲線上過某點切線的斜率,會利用導(dǎo)函數(shù)的正負確定函數(shù)的單調(diào)區(qū)間,考查了分類討論的數(shù)學(xué)思想,是一道中檔題.