Let A(m, f(m)), B(n, f(n)) (m≠n≠0).

According to the meaning of the question: f'(m)=f'(n)

That's 3: 00 a. m. -b=3an？ -B.

∴m？ =n？

∵m≠n∴n=-m

∵L 1, L2 is perpendicular to AB.

∴AB slope k*f'(m)=- 1

That is, [f(n)-f(m)]/(n-m)*(3am? -b)=- 1

∴-2(am？ -bm)/(-2m)*(3am？ -b)=- 1

∴(am？ B) (3 am? -b)=- 1

∴3(am？ )? -4b(am？ )+b？ + 1=0

∵a & gt； 0，m≠0，am？ & gt0

Let t=am? & gt0

∴ equation about t 3t? -4bt+b？ +1=0 has a positive solution.

∵b？ + 1 & gt； 0

∴{δ= 16b？ - 12(b？ + 1)≥0

{ b & gt0

Solution: b? ≥3，b≥√3

The value range of ∴b is [√3,+∞].