f'(x)=1/x-ax=(1-ax^2)/x
(1)a=<0時(shí),f'(x)恒>0,則有單調(diào)增區(qū)間是(0,+無窮)
(2)a>0時(shí),f'(x)>0時(shí)有:1-ax^2>0,ax^2<1,x^2<1/a,得到0
2.
由(1)得到,當(dāng)a>0時(shí),f(x)有極大值,是f(根號(hào)1/a)
即有f(根號(hào)1/a)=ln根號(hào)(1/a)-1/2a*1/a>0
1/2ln(1/a)>1/2
-lna>1
lna<-1
0