【答案】
分析:(1)由f(1)=g(1),f′(1)=g′(1)得到a與b的值,因?yàn)镕(x)=f(x)-g(x)求出導(dǎo)函數(shù)討論在區(qū)間上的增減性得到函數(shù)的極值即可;
(2)因f(x)與g(x)有一個公共點(diǎn)(1,1),而函數(shù)f(x)=x
2在點(diǎn)(1,1)的切線方程為y=2x-1,
下面驗(yàn)證
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125022012638290/SYS201310251250220126382020_DA/0.png)
都成立即可.由x
2-2x+1≥0,得x
2≥2x-1,知f(x)≥2x-1恒成立.設(shè)h(x)=lnx+x-(2x-1),即h(x)=lnx-x+1,易知其在(0,1)上遞增,在(1,+∞)上遞減,所以h(x)=lnx+x-(2x-1)的最大值為h(1)=0,所以lnx+x≤2x-1恒成立.故存在;
(3)因?yàn)镚(x)=f(x)+2-g(x)有兩個零點(diǎn)x
1,x
2,把兩個零點(diǎn)代入到G(x)中,得一式子,然后求出導(dǎo)函數(shù)討論兩個零點(diǎn)的大小得到G'(x
)值的符號為正.
解答:解:(1)由f(1)=g(1),f′(1)=g′(1)得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125022012638290/SYS201310251250220126382020_DA/1.png)
,解得a=b=1則F(x)=f(x)-g(x)=x
2-lnx-x,F(xiàn)′(x)=2x-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125022012638290/SYS201310251250220126382020_DA/2.png)
-1
x=1或x=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125022012638290/SYS201310251250220126382020_DA/3.png)
,當(dāng)x<-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125022012638290/SYS201310251250220126382020_DA/4.png)
或x>1時,f′(x)>0,函數(shù)為增函數(shù);當(dāng)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125022012638290/SYS201310251250220126382020_DA/5.png)
<x<1時,f′(x)<0,函數(shù)為減函數(shù).
得到F(x)
極小值=F(1)=0;
(2)因f(x)與g(x)有一個公共點(diǎn)(1,1),而函數(shù)f(x)=x
2在點(diǎn)(1,1)的切線方程為y=2x-1,
下面驗(yàn)證
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125022012638290/SYS201310251250220126382020_DA/6.png)
都成立即可.由x
2-2x+1≥0,得x
2≥2x-1,知f(x)≥2x-1恒成立.設(shè)h(x)=lnx+x-(2x-1),即h(x)=lnx-x+1,易知其在(0,1)上遞增,在(1,+∞)上遞減,所以h(x)=lnx+x-(2x-1)的最大值為h(1)=0,所以lnx+x≤2x-1恒成立.故存在這樣的k和m,且k=2,m=-1.
(3)G′(x
)的符號為正,理由為:因?yàn)镚(x)=x
2+2-alnx-bx有兩個零點(diǎn)x
1,x
2,則有
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125022012638290/SYS201310251250220126382020_DA/7.png)
,兩式相減得x
22-x
12-a(lnx
2-lnx
1)-b(x
2-x
1)=0,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125022012638290/SYS201310251250220126382020_DA/8.png)
,
于是G′(x
)=2x
-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125022012638290/SYS201310251250220126382020_DA/9.png)
-b=(x
1+x
2-b)-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125022012638290/SYS201310251250220126382020_DA/10.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125022012638290/SYS201310251250220126382020_DA/11.png)
-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125022012638290/SYS201310251250220126382020_DA/12.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125022012638290/SYS201310251250220126382020_DA/13.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125022012638290/SYS201310251250220126382020_DA/14.png)
①當(dāng)0<x
1<x
2時,令
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125022012638290/SYS201310251250220126382020_DA/15.png)
=t,則t>1,且u′(t)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125022012638290/SYS201310251250220126382020_DA/16.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125022012638290/SYS201310251250220126382020_DA/17.png)
>0,則u(t)=lnt-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125022012638290/SYS201310251250220126382020_DA/18.png)
在(1,+∞)上為增函數(shù),
而u(1)=0,所以u(t)>0,即lnt-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125022012638290/SYS201310251250220126382020_DA/19.png)
>0,又因?yàn)閍>0,x
2-x
1>0
所以G′(x
)>0;
②當(dāng)0<x
2<x
1時,同理可得:G′(x
)>0
綜上所述:G′(x
)的符號為正.
點(diǎn)評:考查學(xué)生利用導(dǎo)數(shù)研究函數(shù)極值的能力,利用導(dǎo)數(shù)求閉區(qū)間上函數(shù)極值的能力.