已知函数f(x)=ax²+bx+lnx,若b=-2a,函数f(x)在(0,+∞)上是单调函数,求实数a的取值范围

f(x)=ax^2-2ax+lnx在(0,+∞)上是单调函数,则有x>0,f'(x)>=0f'(x)=2ax-2a+1/x>=0---> 2ax^2-2ax+1>=0---> g(x)=2a(x-1/2)^2+1-a/2>=0因此必有a>0,且g(x)的极小值g(1/2)=1-a/2>=0----> a