f(x)=lnx-(1/2)ax^2-2x ,(x>0)
求导f'(x)=1/x-ax-2=(-ax^2-2x+1)/x,
若函数f(x)在定义域内单调递增,则有f'(x)>=0,且f'(x)不恒为0得
-ax^2-2x+1>=0,即ax^2+2x-1