分析:(1)借助于導(dǎo)數(shù),討論參數(shù),得到函數(shù)的單調(diào)區(qū)間和極值;
(2)借助于(1)的單調(diào)區(qū)間可知函數(shù)在(0,
)的單調(diào)性,構(gòu)建新函數(shù),再借助其導(dǎo)數(shù),判斷新函數(shù)的單調(diào)性,即得證.
解答:解:(1)由
f′(x)=-+=①當(dāng)a≤0時,f′(x)>0,故f(x)在(0,+∞)上單調(diào)遞增,無極值;
②當(dāng)a>0時,若0<x<a,f
′(x)<0,故函數(shù)f(x)在(0,a)上單調(diào)遞減,
若x>a,f′(x)>0,故f(x)在(a,+∞)上單調(diào)遞增,
所以極小值f(a)=1+lna,無極大值.
(2)證明:不妨設(shè)x
1≥x
2,而0<a
≤,由(1)知f(x)在(0,
)的單調(diào)遞減,
故對任意
x1,x2∈(0,),|f(x
1)-f(x
2)|≥a|x
1-x
2|等價于:
對任意x1:x2∈(0,),f(x2)-f(x1)≥a(x1-x2)即f(x
1)+ax
1≤f(x
2)+ax
2令g(x)=f(x)+ax,則
g′(x)=,
令h(x)=ax
2+x-a,∵0<a
≤,
∴h(0)=-a<0,
h()=+-a=≤0,
則
g′(x)<0(0<x<),故g(x)在(0,
)上單調(diào)遞減,
又由x
1≥x
2,∴g(x
2)≥g(x
1),即f(x
2)+ax
2≥f(x
1)+ax
1
∴
對任意x1:x2∈(0,),|f(x1)-f(x2)|≥a|x1-x2|.
點評:本題主要考查導(dǎo)函數(shù)的正負(fù)與原函數(shù)的單調(diào)性之間的關(guān)系,即當(dāng)導(dǎo)函數(shù)大于0時原函數(shù)單調(diào)遞增,當(dāng)導(dǎo)函數(shù)小于0時原函數(shù)單調(diào)遞減.