Idea 1:
? If you don't have a clue when you encounter this problem, you might as well use the reverse method.
? According to the conditions in the problem and λ f' (zeta)+μ f' (η) = 0:
f'(ζ)=-μ/λf'(η)
Make a change on this basis? f'(ζ)=-μ(λ+μ)/λ(λ+μ)f'(η)
λ/(λ+μ)f '(ζ)+μ/(λ+μ)f '(η)= 0。
Order? K=λ/(λ+μ) Known μ/(λ+μ) =1-K.
Namely. The above problem is equivalent to proving that kf'(ζ)+( 1-k)f'(η)=0.
You can construct a function. F(x)=kf(x)，G(x)=( 1-k)f(x)
According to the conditions in the problem, these two functions satisfy Lagrange theorem, so we can know from the theorem that:
The existence of zeta point in (0, 1) makes
? F' (zeta) = 0, that is, KF' (zeta) = 0.
The existence of (0, 1) midpoint η makes
? G' (η) = 0, that is, (1-k)f'(η)=0.
Therefore, it is available. . .
Idea 2:
? After seeing the topic, directly construct the function (in fact, it is similar to the previous idea).
? A builder? F(x)=？ λ/(λ+μ)f(x)，
? G(x)=μ/(λ+μ)f(x)
According to the conditions in the problem, these two functions satisfy Lagrange theorem, so we can know from the theorem that:
The existence of zeta point in (0, 1) makes
? F' (zeta) = 0, that is, λ/(λ+μ) f' (zeta) = 0.
The existence of (0, 1) midpoint η makes
G' (η) = 0, that is μ/(λ+μ)f'(η)=0.
So that there are two points zeta and eta in (0, 1), so that
λ/(λ+μ)f'(ζ)+μ/(λ+μ)f'(η)=0？ Eliminate 1/(λ+μ)
Understand? λf'(ζ)+μf'(η)=0
Well, you can look at the picture:
That is to say, there are two points ζ and η in (0, 1). If the slopes of the curves corresponding to these two points are K 1 and K2 respectively, there are:
K 1+nk2=0, that is? K 1=-nk2 is what it means.
This is from a geometric point of view.
? The second kind, if skilled, should be able to get it out quickly without much effort. The key is to proceed from the problem and make full use of the conditions given by the topic.
I don't think the geometric meaning I proved before is obvious, and I don't know the answer.
? I haven't touched textbooks for more than a year, and I haven't paid attention to these things anymore. I don't know if what I said is useful. Here, I wish the postgraduate entrance examination a success! ~~