【答案】
分析:(I)根據(jù)對數(shù)函數(shù)的定義求得函數(shù)的定義域,再根據(jù)f(x)的解析式求出f(x)的導(dǎo)函數(shù),然后分別令導(dǎo)函數(shù)大于0和小于0得到關(guān)于x的不等式,求出不等式的解集即可得到相應(yīng)的x的范圍即分別為函數(shù)的遞增和遞減區(qū)間;
(II)假設(shè)函數(shù)f(x)的圖象上存在兩點(diǎn)A(x
1,y
1),B(x
2,y
2),使得AB存在“中值相依切線”,根據(jù)斜率公式求出直線AB的斜率,利用導(dǎo)數(shù)的幾何意義求出直線AB的斜率,它們相等,再通過構(gòu)造函數(shù),利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性和最值即可證明結(jié)論.
解答:解:(Ⅰ)函數(shù)f(x)的定義域是(0,+∞).…(1分)
由已知得,
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112019866269030/SYS201312021120198662690019_DA/0.png)
.…(2分)
(1)當(dāng)a>0時(shí),令f'(x)>0,解得0<x<1; 令f'(x)<0,解得x>1.
所以函數(shù)f(x)在(0,1)上單調(diào)遞增,在(1,+∞)上單調(diào)遞減.…(3分)
(2)當(dāng)a<0時(shí),
①當(dāng)
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112019866269030/SYS201312021120198662690019_DA/1.png)
時(shí),即a<-1時(shí),令f'(x)>0,解得
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112019866269030/SYS201312021120198662690019_DA/2.png)
或x>1;
令f'(x)<0,解得
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112019866269030/SYS201312021120198662690019_DA/3.png)
.
所以,函數(shù)f(x)在
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112019866269030/SYS201312021120198662690019_DA/4.png)
和(1,+∞)上單調(diào)遞增,在
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112019866269030/SYS201312021120198662690019_DA/5.png)
上單調(diào)遞減;…(4分)
②當(dāng)
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112019866269030/SYS201312021120198662690019_DA/6.png)
時(shí),即a=-1時(shí),顯然,函數(shù)f(x)在(0,+∞)上單調(diào)遞增; …(5分)
③當(dāng)
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112019866269030/SYS201312021120198662690019_DA/7.png)
時(shí),即-1<a<0時(shí),令f'(x)>0,解得0<x<1或
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112019866269030/SYS201312021120198662690019_DA/8.png)
;
令f'(x)<0,解得
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112019866269030/SYS201312021120198662690019_DA/9.png)
.
所以,函數(shù)f(x)在(0,1)和
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112019866269030/SYS201312021120198662690019_DA/10.png)
上單調(diào)遞增,在
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112019866269030/SYS201312021120198662690019_DA/11.png)
上單調(diào)遞減.…(6分)
綜上所述,(1)當(dāng)a>0時(shí),函數(shù)f(x)在(0,1)上單調(diào)遞增,在(1,+∞)上單調(diào)遞減;
(2)當(dāng)a<-1時(shí),函數(shù)f(x)在
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112019866269030/SYS201312021120198662690019_DA/12.png)
和(1,+∞)上單調(diào)遞增,在
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112019866269030/SYS201312021120198662690019_DA/13.png)
上單調(diào)遞減;
(3)當(dāng)a=-1時(shí),函數(shù)f(x)在(0,+∞)上單調(diào)遞增;
(4)當(dāng)-1<a<0時(shí),函數(shù)f(x)在(0,1)和
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112019866269030/SYS201312021120198662690019_DA/14.png)
上單調(diào)遞增,在
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112019866269030/SYS201312021120198662690019_DA/15.png)
上單調(diào)遞減.…(7分)
(Ⅱ)假設(shè)函數(shù)f(x)存在“中值相依切線”.
設(shè)A(x
1,y
1),B(x
2,y
2)是曲線y=f(x)上的不同兩點(diǎn),且0<x
1<x
2,
則
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112019866269030/SYS201312021120198662690019_DA/16.png)
,
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112019866269030/SYS201312021120198662690019_DA/17.png)
.
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112019866269030/SYS201312021120198662690019_DA/18.png)
=
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112019866269030/SYS201312021120198662690019_DA/19.png)
=
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112019866269030/SYS201312021120198662690019_DA/20.png)
…(8分)
曲線在點(diǎn)M(x
,y
)處的切線斜率k=f'(x
)=
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112019866269030/SYS201312021120198662690019_DA/21.png)
=
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112019866269030/SYS201312021120198662690019_DA/22.png)
,…(9分)
依題意得:
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112019866269030/SYS201312021120198662690019_DA/23.png)
=
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112019866269030/SYS201312021120198662690019_DA/24.png)
.
化簡可得:
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112019866269030/SYS201312021120198662690019_DA/25.png)
=
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112019866269030/SYS201312021120198662690019_DA/26.png)
,
即
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112019866269030/SYS201312021120198662690019_DA/27.png)
=
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112019866269030/SYS201312021120198662690019_DA/28.png)
=
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112019866269030/SYS201312021120198662690019_DA/29.png)
.…(11分)
設(shè)
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112019866269030/SYS201312021120198662690019_DA/30.png)
(t>1),上式化為:
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112019866269030/SYS201312021120198662690019_DA/31.png)
,
即
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112019866269030/SYS201312021120198662690019_DA/32.png)
.…(12分)
令
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112019866269030/SYS201312021120198662690019_DA/33.png)
,
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112019866269030/SYS201312021120198662690019_DA/34.png)
=
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112019866269030/SYS201312021120198662690019_DA/35.png)
.
因?yàn)閠>1,顯然g'(t)>0,所以g(t)在(1,+∞)上遞增,
顯然有g(shù)(t)>2恒成立.
所以在(1,+∞)內(nèi)不存在t,使得
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202112019866269030/SYS201312021120198662690019_DA/36.png)
成立.
綜上所述,假設(shè)不成立.所以,函數(shù)f(x)不存在“中值相依切線”.…(14分)
點(diǎn)評:此題考查學(xué)生會(huì)利用導(dǎo)函數(shù)的正負(fù)求出函數(shù)的單調(diào)區(qū)間,靈活運(yùn)用中點(diǎn)坐標(biāo)公式化簡求值,掌握反證法進(jìn)行命題證明的方法,是一道綜合題,屬難題.