已知函數(shù)f(x)=(lnx-12)x2-6a(lnx-1)x,a為常數(shù),a∈R.
(Ⅰ)當a=13時,求f(x)在x=e處的切線方程;
(Ⅱ)(?。┯懻摵瘮?shù)f(x)的單調(diào)性;
(ⅱ)?x∈(e,+∞),不等式f(x)>2a2恒成立,求a的取值范圍.
1
2
1
3
【答案】(Ⅰ)y=2(e-1)x+2e-.
(Ⅱ)(i)a≤0時f(x)在(0,1)上單調(diào)遞減,在(1,+∞)上單調(diào)遞增;0<a時f(x)在(0,3a)上單調(diào)遞增,在(3a,1)上單調(diào)遞減,在(1,+∞)上單調(diào)遞增;a=時f(x)在(0,+∞)上單調(diào)遞增;a>時f(x)在(0,1)上單調(diào)遞增,在(1,3a)上單調(diào)遞減,在(3a,+∞)上單調(diào)遞增.
(ii)[-,).
3
e
2
2
(Ⅱ)(i)a≤0時f(x)在(0,1)上單調(diào)遞減,在(1,+∞)上單調(diào)遞增;0<a
≤
1
3
1
3
1
3
(ii)[-
e
2
e
23
18
3
【解答】
【點評】
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發(fā)布:2024/9/1 3:0:8組卷:140引用:3難度:0.5
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