已知函數(shù)f(x)=aln(1+ex)-(a+1)x,(其中a>0),點(diǎn)A(x1,f(x1)),B(x2,f(x2)),C(x3,f(x3))從左到右依次是函數(shù)y=f(x)圖象上三點(diǎn),且2x2=x1+x3.
(Ⅰ)證明:函數(shù)f(x)在(-∞,+∞)上是減函數(shù);
(Ⅱ)求證:△ABC是鈍角三角形;
(Ⅲ)試問△ABC能否是等腰三角形?若能,求△ABC面積的最大值;若不能,請(qǐng)說明理由.
【答案】
分析:(Ⅰ)∵f(x)=aln(1+e
x)-(a+1)x,欲證函數(shù)f(x)在(-∞,+∞)上是單調(diào)減函數(shù),只須證明其導(dǎo)數(shù)f′(x)<0即可;
(Ⅱ)先設(shè)A(x
1,f(x
1)),B(x
2,f(x
2)),C(x
3,f(x
3))且x
1<x
2<x
3,欲證:△ABC是鈍角三角形,只須證明其中一個(gè)內(nèi)角為鈍角即可,結(jié)合向量的坐標(biāo)運(yùn)算,只須證明:
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182207330693375/SYS201310241822073306933017_DA/0.png)
即得;
(Ⅲ)假設(shè)△ABC為等腰三角形,則只能是
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182207330693375/SYS201310241822073306933017_DA/1.png)
,再利用平面內(nèi)兩點(diǎn)的距離公式將點(diǎn)的坐標(biāo)代入計(jì)算,如出現(xiàn)矛盾,則△ABC不可能為等腰三角形,如不矛盾,則△ABC能是等腰三角形.
解答:解:(Ⅰ)∵f(x)=aln(1+e
x)-(a+1)x,∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182207330693375/SYS201310241822073306933017_DA/2.png)
恒成立,
所以函數(shù)f(x)在(-∞,+∞)上是單調(diào)減函數(shù).(3分)
(Ⅱ)證明:據(jù)題意A(x
1,f(x
1)),B(x
2,f(x
2)),C(x
3,f(x
3))且x
1<x
2<x
3,
由(Ⅰ)知f(x
1)>f(x
2)>f(x
3),x
2=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182207330693375/SYS201310241822073306933017_DA/3.png)
(4分)
可得A(x
1,f(x
1)),B(x
2,f(x
2)),C(x
3,f(x
3))三點(diǎn)不共線
(反證法:否則
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182207330693375/SYS201310241822073306933017_DA/4.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182207330693375/SYS201310241822073306933017_DA/5.png)
,得x
1=x
3)
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182207330693375/SYS201310241822073306933017_DA/6.png)
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182207330693375/SYS201310241822073306933017_DA/7.png)
(6分)
∵x
1-x
2<0,x
3-x
2>0,f(x
1)-f(x
2)>0,f(x
3)-f(x
2)<0,∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182207330693375/SYS201310241822073306933017_DA/8.png)
,∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182207330693375/SYS201310241822073306933017_DA/9.png)
即△ABC是鈍角三角形(8分)
(Ⅲ)假設(shè)△ABC為等腰三角形,則只能是
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182207330693375/SYS201310241822073306933017_DA/10.png)
即:(x
1-x
2)
2+[f(x
1)-f(x
2)]
2=(x
3-x
2)
2+[f(x
3)-f(x
2)]
2∵x
2-x
1=x
3-x
2∴[f(x
1)-f(x
2)]
2=[f(x
3)-f(x
2)]
2即2f(x
2)=f(x
1)+f(x
3)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182207330693375/SYS201310241822073306933017_DA/11.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182207330693375/SYS201310241822073306933017_DA/12.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182207330693375/SYS201310241822073306933017_DA/13.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182207330693375/SYS201310241822073306933017_DA/14.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182207330693375/SYS201310241822073306933017_DA/15.png)
①(11分)
而事實(shí)上,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182207330693375/SYS201310241822073306933017_DA/16.png)
②
由于
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182207330693375/SYS201310241822073306933017_DA/17.png)
,故(2)式等號(hào)不成立.這與(1)式矛盾.
所以△ABC不可能為等腰三角形.(13分)
點(diǎn)評(píng):本小題主要考查利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性、數(shù)量積表示兩個(gè)向量的夾角、兩點(diǎn)間距離公式的應(yīng)用等基礎(chǔ)知識(shí),考查運(yùn)算求解能力,考查數(shù)形結(jié)合思想、化歸與轉(zhuǎn)化思想.