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,+∞].