已知函数f(x)=[ax2+(a-1)2x-a2+3a-1]ex(a∈R).(Ⅰ)若函数f(x)在(2,3)上单调递增,求实数a
已知函数f(x)=[ax2+(a-1)2x-a2+3a-1]ex(a∈R).(Ⅰ)若函数f(x)在(2,3)上单调递增,求实数a的取值范围;(Ⅱ)若a=0,设g(x)=f...
已知函数f(x)=[ax2+(a-1)2x-a2+3a-1]ex(a∈R).(Ⅰ)若函数f(x)在(2,3)上单调递增,求实数a的取值范围;(Ⅱ)若a=0,设g(x)=f(x)ex+lnx-x,斜率为k的直线与曲线y=g(x)交于A(x1,y1),B(x2,y2)(其中x1<x2)两点,证明:(x1+x2)k>2.
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(Ⅰ)函数f(x)的导数f'(x)=[2ax+(a-1)2]?ex+[ax2+(a-1)2x+a-(a-1)2]?ex
=[ax2+(a2+1)x+a]?ex,
当a≥0时,∵x∈(2,3),∴f'(x)>0,∴f(x)在(2,3)上单调递增,
当a<0时,∵f(x)在(2,3)上单调递增,∴f'(x)=a(x+a)(x+
)?ex≥0,
①当-1<a<0时,解得-a≤x≤-
,由题意知(2,3)?[-a,-
],得?
≤a<0,
②当a=-1时,f'(x)=-(x-1)2?ex≤0,不合题意,舍去,
③当a<-1时,解得?
≤x≤-a,则由题意知(2,3)?[-?
,-a],得a≤-3,
综上可得,实数a的取值范围是(-∞,-3]∪[-
,+∞);
(Ⅱ)a=0时,g(x)=
+lnx-x=lnx-1,k=
,
∵x2-x1>0,要证(x2+x1)k>2,即证(x1+x2)
>2,
即证ln
-
>0(
>1),
设h(x)=lnx-
(x>1),h'(x)=
-
=
>0,
∴h(x)在(1,+∞)上单调递增,h(x)>h(1)=0,
∴ln
-
>0(
>1)成立,
即(x1+x2)k>2成立.
=[ax2+(a2+1)x+a]?ex,
当a≥0时,∵x∈(2,3),∴f'(x)>0,∴f(x)在(2,3)上单调递增,
当a<0时,∵f(x)在(2,3)上单调递增,∴f'(x)=a(x+a)(x+
1 |
a |
①当-1<a<0时,解得-a≤x≤-
1 |
a |
1 |
a |
1 |
3 |
②当a=-1时,f'(x)=-(x-1)2?ex≤0,不合题意,舍去,
③当a<-1时,解得?
1 |
a |
1 |
a |
综上可得,实数a的取值范围是(-∞,-3]∪[-
1 |
3 |
(Ⅱ)a=0时,g(x)=
f(x) |
ex |
lnx2?lnx1 |
x2?x1 |
∵x2-x1>0,要证(x2+x1)k>2,即证(x1+x2)
lnx2?lnx1 |
x2?x1 |
即证ln
x2 |
x1 |
2(
| ||
|
x2 |
x1 |
设h(x)=lnx-
2(x?1) |
x+1 |
1 |
x |
4 |
(x+1)2 |
(x?1)2 |
x(x+1)2 |
∴h(x)在(1,+∞)上单调递增,h(x)>h(1)=0,
∴ln
x2 |
x1 |
2(
| ||
|
x2 |
x1 |
即(x1+x2)k>2成立.
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