【答案】
分析:(1)設(shè)x∈[-e,0),則-x∈(0,e],從而可得f(-x)=-ax+ln(-x),結(jié)合f(x)為奇函數(shù),可求f(x),x∈[-e,0)
(2)由a=-1時(shí),可得f(x)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231631037382684/SYS201311012316310373826019_DA/0.png)
,g(x)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231631037382684/SYS201311012316310373826019_DA/1.png)
,而x∈(0,e]時(shí),f(x)=-x+lnx
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231631037382684/SYS201311012316310373826019_DA/2.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231631037382684/SYS201311012316310373826019_DA/3.png)
,結(jié)合導(dǎo)數(shù)可得f(x)
max=f(1)=-1,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231631037382684/SYS201311012316310373826019_DA/4.png)
,結(jié)合導(dǎo)數(shù)可得g(x)
min=g(e)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231631037382684/SYS201311012316310373826019_DA/5.png)
,要證明當(dāng)x∈(0,e]時(shí),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231631037382684/SYS201311012316310373826019_DA/6.png)
恒成立,即證f(x)
max![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231631037382684/SYS201311012316310373826019_DA/7.png)
即可
(3)假設(shè)存在負(fù)數(shù)a滿足條件,由(1)可得,x∈(0,e],f(x)=ax+lnx,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231631037382684/SYS201311012316310373826019_DA/8.png)
,令f′(x)>0可得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231631037382684/SYS201311012316310373826019_DA/9.png)
,f′(x)<0可得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231631037382684/SYS201311012316310373826019_DA/10.png)
,要判斷函數(shù)的單調(diào)區(qū)間,需要比較e與
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231631037382684/SYS201311012316310373826019_DA/11.png)
的大小,故需要討論:①
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231631037382684/SYS201311012316310373826019_DA/12.png)
,②
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231631037382684/SYS201311012316310373826019_DA/13.png)
兩種情況分別求解函數(shù)的最大值,進(jìn)而可求a
解答:解:(1)當(dāng)x∈[-e,0)時(shí)可得,-x∈(0,e]
∵x∈(0,e]時(shí),f(x)=ax+lnx
f(-x)=-ax+ln(-x)
∵函數(shù)f(x)為奇函數(shù)可得f(-x)=-f(x)
-f(x)=-ax+ln(-x)
f(x)=ax-ln(-x)
f(x)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231631037382684/SYS201311012316310373826019_DA/14.png)
證明:(2)a=-1時(shí),f(x)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231631037382684/SYS201311012316310373826019_DA/15.png)
,g(x)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231631037382684/SYS201311012316310373826019_DA/16.png)
,
x∈(0,e]時(shí),f(x)=-x+lnx
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231631037382684/SYS201311012316310373826019_DA/17.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231631037382684/SYS201311012316310373826019_DA/18.png)
令f′(x)>0可得0<x<1,f′(x)<0可得1<x≤e
函數(shù)f(x)在(0,1]單調(diào)遞增,在(1,e]單調(diào)遞減
f(x)
max=f(1)=-1
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231631037382684/SYS201311012316310373826019_DA/19.png)
,由x∈(0,e]可得g′(x)≤0
g(x)在(0,e]上單調(diào)遞減
g(x)
min=g(e)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231631037382684/SYS201311012316310373826019_DA/20.png)
-1<
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231631037382684/SYS201311012316310373826019_DA/21.png)
即f(x)
max![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231631037382684/SYS201311012316310373826019_DA/22.png)
當(dāng)x∈(0,e]時(shí),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231631037382684/SYS201311012316310373826019_DA/23.png)
恒成立;
解:(3)假設(shè)存在負(fù)數(shù)a滿足條件
由(1)可得,x∈(0,e],f(x)=ax+lnx,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231631037382684/SYS201311012316310373826019_DA/24.png)
令f′(x)>0可得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231631037382684/SYS201311012316310373826019_DA/25.png)
,f′(x)<0可得
①若
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231631037382684/SYS201311012316310373826019_DA/27.png)
,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231631037382684/SYS201311012316310373826019_DA/28.png)
,則函數(shù)在(0,-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231631037382684/SYS201311012316310373826019_DA/29.png)
]上單調(diào)遞增,在(-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231631037382684/SYS201311012316310373826019_DA/30.png)
,e]上單調(diào)遞減
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231631037382684/SYS201311012316310373826019_DA/31.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231631037382684/SYS201311012316310373826019_DA/32.png)
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231631037382684/SYS201311012316310373826019_DA/33.png)
②若
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231631037382684/SYS201311012316310373826019_DA/34.png)
即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231631037382684/SYS201311012316310373826019_DA/35.png)
,則函數(shù)在(0,e]單調(diào)遞增,則f(x)
max=f(e)=ae+1=-3
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131101231631037382684/SYS201311012316310373826019_DA/36.png)
(舍)
故
點(diǎn)評(píng):本題主要考查了利用函數(shù)的奇偶性求解函數(shù)的解析式,及利用函數(shù)的導(dǎo)數(shù)判斷函數(shù)的單調(diào)性,求解函數(shù)的最值,利用單調(diào)性證明不等式,解題的關(guān)鍵是熟練應(yīng)用函數(shù)的性質(zhì).是綜合性較強(qiáng)的試題.