【答案】
分析:(1)先求函數(shù)f(x)的值域,然后對函數(shù)f(x)進(jìn)行求導(dǎo),根據(jù)導(dǎo)函數(shù)的正反判斷函數(shù)的單調(diào)性,進(jìn)而可得到最小值;
(2)先由(1)可判斷函數(shù)在不同區(qū)間的不同取值,然后對m的范圍進(jìn)行分析可確定方程f(x)-m=0(m∈R)的解的個數(shù).
(3)先將不等式f(a)+f(b)≥f(a+b)-(a+b)ln2轉(zhuǎn)化為f(a)+f[(a+b)-a]≥f(a+b)-(a+b)ln2,然后令函數(shù)g(x)=f(x)+f(k-x)并將函數(shù)f(x)的解析式代入后求導(dǎo)數(shù),根據(jù)導(dǎo)數(shù)的正負(fù)判斷函數(shù)的單調(diào)性從而求出函數(shù)g(x)的最小值,并且任意x有g(shù)(x)大于等于g(x)的最小值,得證.
解答:解:(I)f(x)的定義域為(0,+∞)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181059671885608/SYS201310241810596718856019_DA/0.png)
,
當(dāng)x∈(0,+∞)時,f′(x),f(x)的變化的情況如下:
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181059671885608/SYS201310241810596718856019_DA/images1.png)
所以,f(x)在(0,+∞)最小值是
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181059671885608/SYS201310241810596718856019_DA/1.png)
.
(Ⅱ)當(dāng)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181059671885608/SYS201310241810596718856019_DA/2.png)
,f(x)單調(diào)遞減且f(x)的取值范圍是
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181059671885608/SYS201310241810596718856019_DA/3.png)
;
當(dāng)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181059671885608/SYS201310241810596718856019_DA/4.png)
時,f(x)單調(diào)遞增且f(x)的取值范圍是
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181059671885608/SYS201310241810596718856019_DA/5.png)
下面討論f(x)-m=0的解;
所以,當(dāng)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181059671885608/SYS201310241810596718856019_DA/6.png)
時,原方程無解;
當(dāng)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181059671885608/SYS201310241810596718856019_DA/7.png)
時,原方程有唯一解;
當(dāng)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181059671885608/SYS201310241810596718856019_DA/8.png)
時,原方程有兩解
(Ⅲ)原不等式可化為:f(a)+f[(a+b)-a]≥f(a+b)-(a+b)ln2
設(shè)函數(shù)g(x)=f(x)+f(k-x)(k>0)
則g(x)=xlnx+(k-x)ln(k-x)(0<x<k)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181059671885608/SYS201310241810596718856019_DA/9.png)
令g'(x)>0,則
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181059671885608/SYS201310241810596718856019_DA/10.png)
,∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181059671885608/SYS201310241810596718856019_DA/11.png)
,∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181059671885608/SYS201310241810596718856019_DA/12.png)
,
解得:
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181059671885608/SYS201310241810596718856019_DA/13.png)
,
令g'(x)<0,解得:0<x<
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181059671885608/SYS201310241810596718856019_DA/14.png)
∴函數(shù)g(x)在
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181059671885608/SYS201310241810596718856019_DA/15.png)
上單調(diào)遞減,在
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181059671885608/SYS201310241810596718856019_DA/16.png)
上單調(diào)遞增,
∴g(x)在(0,k)上的最小值為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181059671885608/SYS201310241810596718856019_DA/17.png)
∴當(dāng)x∈(0,k)時,總有g(shù)(x)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181059671885608/SYS201310241810596718856019_DA/18.png)
,
即:
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181059671885608/SYS201310241810596718856019_DA/19.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181059671885608/SYS201310241810596718856019_DA/20.png)
令x=a,k-x=b,則有:f(a)+f(b)≥f(a+b)-(a+b)ln2.
點評:本題主要考查函數(shù)的單調(diào)性與其導(dǎo)函數(shù)的正負(fù)之間的關(guān)系.導(dǎo)數(shù)是高考的熱點題,每年必考,要給予充分重視.