(2) The function y=2f2(x)+2bf(x)+ 1 about x has eight different zeros.

If △=0, then f(x) has a value such that y=0. Let this value be f(x)=m, which means that there are eight different questions according to the meaning.

The value of x makes f(x)=m, which is impossible from f(x) image;

If △ > 0, that is, f(x) has two values, so y=0. Let these two values be: f(x)=k and f(x)=h, according to the meaning of the question.

That is, there are four x values that make f(x)=k and four x values that make f(x)=h, so it is possible to combine f(x) images.

, that is, at 0

So: the original problem can be transformed into the following questions:

The function f(x) = y = 2x 2+2bx+ 1 There are two different intersections between x (x is equivalent to F (x) in the original formula) and X axis.

And the intersection points are all within (0, 1), so find the range of B.

As long as the following four conditions are met:

△& gt； 0 0 & lt-2b/(2 * 2)& lt； 1 (symmetry axis) F(0)>0 F( 1)>0

Simultaneous solution: -3/2