(1)
∵f(x)=㏑x–ax
∴f'(x)=1/x-a
∴f'(2)=1/2-a=3/2
∴a=-1
(2)
由(1)可求出直线l方程为y=(3/2)x+ln2-1
构建函数h(x)=(3/2)x+ln2-1-f(x)=-lnx+(1/2)x+ln2-1
h'(x)=-(1/x)+1/2 (由题设知x>0)
∵x>2时h'(x)>0
x=2时h'(x)=0
x