rule

δ=(a- 1)^2-4a^2<； 0

Solution: a

or

a & gt 1/3

The function y = (2a 2-a) x is increasing function.

rule

2a^2-a>； 1

Solution: a

or

a & gt 1

A. At least one true proposition of B is to find the union of these two solution sets.

or

a & gt 1/3

A.b has one and only one true proposition.

- 1≤a & lt； - 1/2

Or 1/3

For reference only:

First of all, we won't discuss it. Find out the range where A satisfies A and B before discussing.

For a:

The solution set of x 2+(A- 1) x+A 2 ≤ 0 is an empty set.

Explain that the discriminant on the left is less than 0.

So this time: (a- 1) 2-4 * a 2

After simplification, we get:

3a^2+2a- 1>； 0

(a+ 1)(3a- 1)>0

So: a> 1/3

or

a & lt- 1.

For b:

The function y = (2a 2-a) x is increasing function.

This function is exponential, so when 2a 2-a >1,it is increasing function.

2a^2-a〉 1

2a^2-a- 1〉0

(a- 1)*(2a+ 1)>0

So: a> 1

or

a & lt- 1/2.

Now let's discuss that (1) A.B. has at least one true proposition.

It means that one or both of A and B can be established. By plotting on the number axis,

Easy to obtain:

a & lt- 1/2

Or a> 1/3.

Discussion (2) A.B. has one and only one true proposition.

It means that when A is established, B is not; Or b is true and a is not.

Look at it first: A holds, but B does not.

Draw through several axes: when drawing, A draws according to the above interval, and because B is not established, it draws its complementary set range.

If it intersects with A, it is easy to get: 1/3.

Similarly, when B holds, A does not.

Draw through several axes: when drawing, B draws according to the above interval, because A is not established, so draw its complement range.

When it intersects with B, it is easy to get:-1