【答案】
分析:(Ⅰ)依題意可得f(x)=lnx+x
2-bx,由f(x)在定義域(0,+∞)上遞增,可得
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202111915909708041/SYS201312021119159097080020_DA/0.png)
≥0對x∈(0,+∞)恒成立,即
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202111915909708041/SYS201312021119159097080020_DA/1.png)
對x∈(0,+∞)恒成立,只需
(Ⅱ)當(dāng)a=1,b=-1時,f(x)=lnx-x
2+x,其定義域是(0,+∞),
對函數(shù)求導(dǎo),利用導(dǎo)數(shù)的知識判斷函數(shù)f(x)在區(qū)間(0,1),(1,+∞)上單調(diào)性可知
當(dāng)x=1時,函數(shù)f(x)取得最大值,其值為f(1)=ln1-1+1=0,當(dāng)x≠1時,f(x)<f(1)=0即函數(shù)f(x)只有一個零點
(Ⅲ)由已知得
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202111915909708041/SYS201312021119159097080020_DA/3.png)
兩式相減,得
由
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202111915909708041/SYS201312021119159097080020_DA/5.png)
及2x
=x
1+x
2,得
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202111915909708041/SYS201312021119159097080020_DA/6.png)
=
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202111915909708041/SYS201312021119159097080020_DA/7.png)
,結(jié)合導(dǎo)數(shù)的知識可證明
解答:解:(Ⅰ)依題意:f(x)=lnx+x
2-bx
f(x)在(0,+∞)上遞增,∴
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202111915909708041/SYS201312021119159097080020_DA/8.png)
≥0對x∈(0,+∞)恒成立
即
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202111915909708041/SYS201312021119159097080020_DA/9.png)
對x∈(0,+∞)恒成立,只需
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202111915909708041/SYS201312021119159097080020_DA/10.png)
…(2分)
∵x>0,
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202111915909708041/SYS201312021119159097080020_DA/11.png)
當(dāng)且僅當(dāng)
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202111915909708041/SYS201312021119159097080020_DA/12.png)
時取=
∴
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202111915909708041/SYS201312021119159097080020_DA/13.png)
∴b的取值范圍為
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202111915909708041/SYS201312021119159097080020_DA/14.png)
…(4分)
(Ⅱ)當(dāng)a=1,b=-1時,f(x)=lnx-x
2+x,其定義域是(0,+∞)
∴
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202111915909708041/SYS201312021119159097080020_DA/15.png)
=
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202111915909708041/SYS201312021119159097080020_DA/16.png)
…(6分)
∴0<x<1時,f′(x)>0當(dāng)x>1時,f′(x)<0
∴函數(shù)f(x)在區(qū)間(0,1)上單調(diào)遞增,在區(qū)間(1,+∞)上單調(diào)遞減
∴當(dāng)x=1時,函數(shù)f(x)取得最大值,其值為f(1)=ln1-1+1=0
當(dāng)x≠1時,f(x)<f(1)=0即
∴函數(shù)f(x)只有一個零點 …(8分)
(Ⅲ)由已知得
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202111915909708041/SYS201312021119159097080020_DA/17.png)
兩式相減,得
由
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202111915909708041/SYS201312021119159097080020_DA/19.png)
及2x
=x
1+x
2,得
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202111915909708041/SYS201312021119159097080020_DA/20.png)
=
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202111915909708041/SYS201312021119159097080020_DA/21.png)
=
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202111915909708041/SYS201312021119159097080020_DA/22.png)
=
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202111915909708041/SYS201312021119159097080020_DA/23.png)
=
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202111915909708041/SYS201312021119159097080020_DA/24.png)
…(10分)
令
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202111915909708041/SYS201312021119159097080020_DA/25.png)
∈(0,1)且
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202111915909708041/SYS201312021119159097080020_DA/26.png)
(0<t<1)
∴
![](http://thumb.zyjl.cn//pic6/res/gzsx/web/STSource/20131202111915909708041/SYS201312021119159097080020_DA/27.png)
∴∅(t)在(0,1)上遞減,∴∅(t)>∅(1)=0
x
1<x
2,f′(x
)<0(12分)
點評:導(dǎo)數(shù)與函數(shù)的單調(diào)性的結(jié)合是導(dǎo)數(shù)最為基本的考查,而函數(shù)的恒成立問題常轉(zhuǎn)化為利用相關(guān)知識求解函數(shù)的最值問題,體現(xiàn)了轉(zhuǎn)化思想在解題中的應(yīng)用,還考查了運用基本知識進行推理論證的能力