【答案】
分析:(1)利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性,可得函數(shù)f(x)=lnx-ax在(0,1)上是增函數(shù),在(1,+∞)上是減函數(shù),故f
max(x)=f(1).
(2)由 y=f(x)=0 可得lnx=ax,故函數(shù)y=f(x)的零點(diǎn)個(gè)數(shù)即為 y=lnx與 y=ax 的交點(diǎn)的個(gè)數(shù).結(jié)合圖象可得,當(dāng)a≤0或a=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024191514088684919/SYS201310241915140886849019_DA/0.png)
時(shí),y=f(x)有1個(gè)零點(diǎn); 當(dāng) 0<a<
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024191514088684919/SYS201310241915140886849019_DA/1.png)
時(shí),y=f(x)有2個(gè)零點(diǎn); 當(dāng)a>
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024191514088684919/SYS201310241915140886849019_DA/2.png)
時(shí),y=f(x)沒有零點(diǎn).
(3)由(1)可得,當(dāng)x∈(0,+∞)時(shí),lnx≤x-1,可得 lna
k<a
k-1,故 b
k•lna
k<b
k(a
k-1)=b
k•a
k-b
k.可得 ln
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024191514088684919/SYS201310241915140886849019_DA/3.png)
+ln
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024191514088684919/SYS201310241915140886849019_DA/4.png)
+…+ln
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024191514088684919/SYS201310241915140886849019_DA/5.png)
<a
1b
1+a
2b
2+…+a
nb
n -( b
1+b
2+…+b
n),再由已知條件證得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024191514088684919/SYS201310241915140886849019_DA/6.png)
…
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024191514088684919/SYS201310241915140886849019_DA/7.png)
≤1成立.
解答:解:(1)當(dāng)a=1時(shí),f′(x)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024191514088684919/SYS201310241915140886849019_DA/8.png)
-1,當(dāng)x>1時(shí),f′(x)<0,當(dāng)0<x<1時(shí),f′(x)>0.
故函數(shù)f(x)=lnx-ax在(0,1)上是增函數(shù),在(1,+∞)上是減函數(shù).
故f
max(x)=f(1)=ln1-1=-1.
(2)由 y=f(x)=0 可得lnx=ax,故函數(shù)y=f(x)的零點(diǎn)個(gè)數(shù)即為 y=lnx與 y=ax 的交點(diǎn)的個(gè)數(shù).
結(jié)合圖象可得,當(dāng)a≤0時(shí),f(x)的零點(diǎn)個(gè)數(shù)僅有一個(gè).
當(dāng)a>0時(shí),令f′(x)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024191514088684919/SYS201310241915140886849019_DA/9.png)
-a=0,可得 x=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024191514088684919/SYS201310241915140886849019_DA/10.png)
.
由于 當(dāng)x>
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024191514088684919/SYS201310241915140886849019_DA/11.png)
時(shí),f′(x)<0,當(dāng)0<x<
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024191514088684919/SYS201310241915140886849019_DA/12.png)
時(shí),f′(x)>0. 故 f(x)在(0,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024191514088684919/SYS201310241915140886849019_DA/13.png)
)上是增函數(shù),在(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024191514088684919/SYS201310241915140886849019_DA/14.png)
,+∞)上是減函數(shù).
故f
max(x)=f(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024191514088684919/SYS201310241915140886849019_DA/15.png)
)=-lna-1.
故當(dāng)-lna-1>0時(shí),即 0<a<
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024191514088684919/SYS201310241915140886849019_DA/16.png)
時(shí),y=f(x)有2個(gè)零點(diǎn);當(dāng)a=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024191514088684919/SYS201310241915140886849019_DA/17.png)
時(shí),y=f(x)有1個(gè)零點(diǎn); 當(dāng)a>
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024191514088684919/SYS201310241915140886849019_DA/18.png)
時(shí),y=f(x)沒有零點(diǎn).
綜上可得,當(dāng)a≤0或a=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024191514088684919/SYS201310241915140886849019_DA/19.png)
時(shí),y=f(x)有1個(gè)零點(diǎn); 當(dāng) 0<a<
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024191514088684919/SYS201310241915140886849019_DA/20.png)
時(shí),y=f(x)有2個(gè)零點(diǎn); 當(dāng)a>
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024191514088684919/SYS201310241915140886849019_DA/21.png)
時(shí),y=f(x)沒有零點(diǎn).
(3)由(1)可得,當(dāng)x∈(0,+∞)時(shí),lnx≤x-1.
∵a
k,b
k 都是正數(shù),∴l(xiāng)na
k<a
k-1,
∴b
k•lna
k<b
k(a
k-1)=b
k•a
k-b
k.
∴l(xiāng)n
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024191514088684919/SYS201310241915140886849019_DA/22.png)
+ln
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024191514088684919/SYS201310241915140886849019_DA/23.png)
+…+ln
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024191514088684919/SYS201310241915140886849019_DA/24.png)
<a
1b
1+a
2b
2+…+a
nb
n -( b
1+b
2+…+b
n).
又因?yàn)?a
1b
1+a
2b
2+…+a
nb
n≤b
1+b
2+…+b
n,
∴l(xiāng)n
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024191514088684919/SYS201310241915140886849019_DA/25.png)
+ln
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024191514088684919/SYS201310241915140886849019_DA/26.png)
+…+ln
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024191514088684919/SYS201310241915140886849019_DA/27.png)
≤0,即ln(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024191514088684919/SYS201310241915140886849019_DA/28.png)
•
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024191514088684919/SYS201310241915140886849019_DA/29.png)
…
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024191514088684919/SYS201310241915140886849019_DA/30.png)
)≤0,
故
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024191514088684919/SYS201310241915140886849019_DA/31.png)
…
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024191514088684919/SYS201310241915140886849019_DA/32.png)
≤1.
點(diǎn)評(píng):本題主要考查方程的根的存在性及個(gè)數(shù)判斷,利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性,用放縮法證明不等式,屬于中檔題.