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A topic of mathematics study plan for senior two in Nanjing.
f'(x)=3ax? -B.

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,+∞].