f( 1)=ln 1- 1？ +1=0, f' (x) < 0 when x > 1, and f' (x) > 0 when 0 < x < 1.
∴f(x) monotonically increases at (0, 1) and monotonically decreases at (1, +∞), so f(x)≤f( 1)=0.
That is, the maximum value of f(x) in the definition domain is zero, and there is only one maximum point of x= 1.
Therefore, when a= 1, f(x) has only one zero.
(2) Solution: f'(x)= 1/x-2a? x+a =-(2ax+ 1)(ax- 1)/x
F(x) only decreases at (1, +∞), which means that when x > 1, f' (x) < 0, that is-(2ax+1) (ax-1)/x < 0.
When x > 1, this inequality is equivalent to (2ax+ 1) (AX- 1) > 0. Obviously, the inequality does not hold when a=0, so a≠0, divide both sides of the inequality by 2a? , an image with (x+ 1/2a) (x- 1/a) > 0, and the quadratic function y=(x+ 1/2a)(x- 1/a) is an image with an upward opening and intersecting the x axis =-
1>- 1/2a
1> 1/a
Solve this set of inequalities and get a > 1 or a.
The range of ∴a is (-∞,-1/2)∩( 1,+∞).