已知函數(shù)f(x)=alnx+x2(a為實(shí)常數(shù)),
(1)若a=-2,求函數(shù)f(x)的單調(diào)遞增區(qū)間;
(2)當(dāng)a<-2時(shí),求函數(shù)f(x)在[1,e]上的最小值及相應(yīng)的x值;
(3)若存在x∈[1,e],使得f(x)≤(a+2)x成立,求a的取值范圍.
【答案】
分析:(1)由題設(shè)條件知
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181230578579252/SYS201310241812305785792020_DA/0.png)
,令f'(x)>0,可得到f(x)的單調(diào)遞增區(qū)間.
(2)由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181230578579252/SYS201310241812305785792020_DA/1.png)
=0得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181230578579252/SYS201310241812305785792020_DA/2.png)
.由此入手可推出當(dāng)x=e時(shí),f(x)
min=a+e
2.
(3)由f(x)≤(a+2)x知alnx+x
2-(a+2)x≤0,設(shè)g(x)=alnx+x
2-(a+2)x,據(jù)題意,當(dāng)x∈[1,e]時(shí),g(x)
min≤0,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181230578579252/SYS201310241812305785792020_DA/3.png)
.再通過分類討論可知a的取值范圍是[-1,+∞).
解答:解:(1)a=-2,f(x)=-2lnx+x
2,
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181230578579252/SYS201310241812305785792020_DA/4.png)
,
令f'(x)>0,由x>0得x>1,
∴f(x)的單調(diào)遞增區(qū)間是(1,+∞).(2分)
(2)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181230578579252/SYS201310241812305785792020_DA/5.png)
,
令f'(x)=0,由a<-2,x>0得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181230578579252/SYS201310241812305785792020_DA/6.png)
(3分)
①當(dāng)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181230578579252/SYS201310241812305785792020_DA/7.png)
,即-2e
2<a<-2時(shí),f(x)在
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181230578579252/SYS201310241812305785792020_DA/8.png)
遞減,在
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181230578579252/SYS201310241812305785792020_DA/9.png)
遞增,
∴當(dāng)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181230578579252/SYS201310241812305785792020_DA/10.png)
時(shí),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181230578579252/SYS201310241812305785792020_DA/11.png)
.(5分)
②當(dāng)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181230578579252/SYS201310241812305785792020_DA/12.png)
,即a≤-2e
2時(shí),f(x)在[1,e]遞減,
∴當(dāng)x=e時(shí),f(x)
min=a+e
2.(7分)
(3)f(x)≤(a+2)x化為:alnx+x
2-(a+2)x≤0,
設(shè)g(x)=alnx+x
2-(a+2)x,據(jù)題意,
當(dāng)x∈[1,e]時(shí),g(x)
min≤0,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181230578579252/SYS201310241812305785792020_DA/13.png)
,(9分)
(�。┊�(dāng)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181230578579252/SYS201310241812305785792020_DA/14.png)
即a≤2時(shí),當(dāng)x∈[1,e]時(shí),g'(x)≥0,∴g(x)遞增,
∴g(x)
min=g(1)=-1-a≤0,∴a≥-1,
∴-1≤a≤2;(11分)
(ⅱ)當(dāng)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181230578579252/SYS201310241812305785792020_DA/15.png)
即2<a<2e時(shí),g(x)在
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181230578579252/SYS201310241812305785792020_DA/16.png)
遞減,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181230578579252/SYS201310241812305785792020_DA/17.png)
遞增,
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181230578579252/SYS201310241812305785792020_DA/18.png)
,
∵
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181230578579252/SYS201310241812305785792020_DA/19.png)
,∴g(x)
min<0,
∴2<a<2e符合題意;(13分)
(ⅲ)當(dāng)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181230578579252/SYS201310241812305785792020_DA/20.png)
即a≥2e時(shí),g(x)在[1,e]遞減,
∴g(x)
min=g(e)=a+e
2-(a+2)e=(1-e)a+e
2-2e≤2e(1-e)+e
2-2e=-e
2<0,符合題意,(15分)
綜上可得,a的取值范圍是[-1,+∞).(16分)
點(diǎn)評(píng):本題考查數(shù)列知識(shí)的綜合運(yùn)用,解題時(shí)要認(rèn)真審題,注意分類討論思想的合理運(yùn)用.