分析:(I)求f(x)的導(dǎo)數(shù)f′(x),討論
0≤a≤時,f′(x)的正負(fù),從而判定f(x)的單調(diào)性;
(II)由題意,要使f(x
1)≥g(x
2)成立,只需f
min(x)≥g
min(x)即可,求出f
min(x),
方法一:g(x)是二次函數(shù),求出g(x)在[2,3]上的最小值,得出b的取值范圍;
方法二:參變量分離得
b≥x+,從而求出b的取值范圍.
解答:解:(I)∵f(x)=lnx-a(x+
)+
+1(x>0,a∈R),
∴
f′(x)=-a+-=
=-(x>0)
①當(dāng)a=0時,f′(x)=
(x>0),∴0<x<1時,f′(x)<0,f(x)是減函數(shù),x>1時,f′(x)>0,f(x)是增函數(shù);
②當(dāng)
a=時,f′(x)=-
≤0,∴f(x)在(0,+∞)單調(diào)遞減;
③當(dāng)
0<a<時,
>1,x∈(0,1]時,f′(x)<0,函數(shù)f(x)在(0,1]上單調(diào)遞減;
x∈(1,]時,f′(x)>0,函數(shù)f(x)在
(1,]上單調(diào)遞增;
x∈(,+∞)時,f′(x)<0,函數(shù)f(x)在
(,+∞)上單調(diào)遞減.
(II)若對任意x
1∈(0,2],存在x
2∈[2,3],使f(x
1)≥g(x
2)成立,
只需f
min(x)≥g
min(x);
由(I)知,當(dāng)
a=時,f(x)在(0,1]單調(diào)遞減,在(1,2]單調(diào)遞增.
∴
fmin(x)=f(1)=,
方法一:g(x)=x
2-bx+2,對稱軸
x=,①當(dāng)
≤2,即b≤4時,
gmin(x)=g(2)≤,得:
≤b≤4;
②當(dāng)
≥3,即b≥6時,
gmin(x)=g(3)≤,得:b≥6;
③當(dāng)
2<<3,即4<b<6時,
gmin(x)=g()≤,得:4<b<6.
綜上:
b≥.
方法二:參變量分離:
b≥x+,
令
h(x)=x+,只需b≥h
min(x),可知h(x)在[2,3]上單調(diào)遞增,
∴
hmin(x)=h(2)=,
b≥.
點(diǎn)評:本題考查了利用導(dǎo)數(shù)判定函數(shù)的單調(diào)性以及利用函數(shù)的單調(diào)性解含參數(shù)的不等式的問題,是較難的題目.