F (x) < b× g (x), where: x? < b(x- 1)→x？ -bx+b 0 → b < 0 or b > 4

That is, the range of b is: (-infinity, 0)∩(4,+infinity).

2、F(x)=f(x)-m×g(x)+ 1-m-m？ =x？ -mx+m+ 1-m-m？ =x？ -mx+ 1-m？ =(x-m/2)？ + 1-(5/4)m？

The absolute value F(x) monotonically increases at 0, 1:

△=(-m)？ -4( 1-m？ )=3m？ -4=3(m-2 root 3/3)(m+2 root 3/3)

(1) When △≤0 → -2 root 3/3≤m≤2 root 3/3

When m/2≤0, that is, m≤0, the absolute value F(x) monotonically increases at 0, 1

The prerequisite is: -2 radical 3/3≤m≤0, and the absolute value F(x) is monotonically increasing at 0, 1

(2) When △ > 0 has 3/3 m 2 roots.

1) When m/2≤ 1 and the root number of m/2-2 is 3/3≤ 1→m≤2, the absolute value of F(x) increases monotonically at 0, 1.

The prerequisite is: when m

2) When the root number is 3/3≤0, the absolute value F(x) monotonically increases at 0, 1, but this is impossible!

Therefore, the absolute value F(x) monotonically increases at 0, 1, and the value range of the real number m is (-infinity, 0).