【答案】
分析:(I)求導函數(shù),可得f'(x)=a
xlna+2x-lna=2x+(a
x-1)lna,確定f'(x)>0,即可得函數(shù)f(x)在(0,+∞)上單調(diào)遞增.
(II)先判斷函數(shù)F(x)的極小值,再由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124733321896235/SYS201310251247333218962021_DA/0.png)
有四個零點,進行等價轉化方程有解問題,去掉絕對值,變成兩個方程,根據(jù)F(x)≥1,解出b的范圍;
(Ⅲ)分析可得,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124733321896235/SYS201310251247333218962021_DA/1.png)
可以轉化為F(x)的最大值減去F(x)的最小值小于或等于e
2-2,由單調(diào)性知,F(xiàn)(x)的最大值是F(1)或F(-1),最小值F(0)=1,由F(1)-F(-1)的單調(diào)性,判斷F(1)與F(-1)的大小關系,再由F(x)的最大值減去最小值F(0)小于或等于e
2-2,構造方程即可求出a的取值范圍.
解答:(I)證明:∵函數(shù)f(x)=a
x+x
2,g(x)=xlna,
F(x)=a
x+x
2-xlna
求導函數(shù),可得F′(x)=a
xlna+2x-lna=2x+(a
x-1)lna,
由于a>1,
∴l(xiāng)na>0,當x>0時,a
x-1>0,
∴F′(x)>0,故函數(shù)F(x)在(0,+∞)上單調(diào)遞增.
(Ⅱ)解:令F′(x)=2x+(a
x-1)lna=0,得到x=0,
F′′(x)=a
x(lna)
2+2>0,F(xiàn)′(x)為單調(diào)增函數(shù),說明x=0是唯一的極值點,也是最小值點;F(0)=1,
∵F′(0)=0,∴當x<0時,F(xiàn)′(x)<0,為減函數(shù);
F(x),F(xiàn)′(x)的變化情況如下表:
x | (-∞,0) | | (0,+∞) |
F′(x) | - | | + |
F(x) | 遞減 | 極小值1 | 遞增 |
∵函數(shù)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124733321896235/SYS201310251247333218962021_DA/2.png)
=0,也即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124733321896235/SYS201310251247333218962021_DA/3.png)
,有四個零點;
∴等價于方程
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124733321896235/SYS201310251247333218962021_DA/4.png)
有解,∵F(x)≥F(0)=1,
由①得,F(xiàn)(x)=3+b-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124733321896235/SYS201310251247333218962021_DA/5.png)
≥1,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124733321896235/SYS201310251247333218962021_DA/6.png)
,解得b>
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124733321896235/SYS201310251247333218962021_DA/7.png)
-1或-1-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124733321896235/SYS201310251247333218962021_DA/8.png)
<b<0;
由②得,F(xiàn)(x)=-3+b-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124733321896235/SYS201310251247333218962021_DA/9.png)
≥1,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124733321896235/SYS201310251247333218962021_DA/10.png)
,解得,b>2+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124733321896235/SYS201310251247333218962021_DA/11.png)
或2-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124733321896235/SYS201310251247333218962021_DA/12.png)
<b<0;
綜上得:b>2+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124733321896235/SYS201310251247333218962021_DA/13.png)
或2-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124733321896235/SYS201310251247333218962021_DA/14.png)
<b<0;
(Ⅲ)解:問題等價于F(x)在[-1,1]的最大值與最小值之差小于等于e
2-2.
由(Ⅱ)可知F(x)在[-1,0]上遞減,在[0,1]上遞增,
∴F(x)的最小值為F(0)=1,最大值等于F(-1),F(xiàn)(1)中較大的一個,
F(-1)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124733321896235/SYS201310251247333218962021_DA/15.png)
+1+lna,F(xiàn)(1)=a+1-lna,F(xiàn)(1)-F(-1)=a-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124733321896235/SYS201310251247333218962021_DA/16.png)
-2lna,
記g(t)=t-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124733321896235/SYS201310251247333218962021_DA/17.png)
-2lnt(t>0),
∵g′(t)=1+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124733321896235/SYS201310251247333218962021_DA/18.png)
-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124733321896235/SYS201310251247333218962021_DA/19.png)
=(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124733321896235/SYS201310251247333218962021_DA/20.png)
)
2≥0(當t=1等號成立)
∴g(t)在t∈(0,+∞)上單調(diào)遞增,而g(1)=0,
所以當t>1時,g(t)>0;當0<t<1時,g(t)<0,
也就是當t>1時,F(xiàn)(1)-F(-1)>0,即F(1)>F(-1);
又由a>1時,則F(x)的最小值為F(0)=1,最大值為F(1)=a+1-lna,
則
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124733321896235/SYS201310251247333218962021_DA/21.png)
⇒F(1)-F(0)=a-lna≤e
2-2,
解得a≤e
2;
則a的取值范圍為a∈(1,e
2].
點評:本題考查導數(shù)知識的運用,考查函數(shù)的單調(diào)性,考查函數(shù)的零點,考查恒成立問題,考查學生分析解決問題的能力,解題的關鍵是利用導數(shù)確定函數(shù)的最值.