设f(x)在[0,π/2]上连续,在(0,π/2)内可导,证明:存在ξ,η∈(0,π/2),使得π/2f’(ξ)=f’(η)/sinη.

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问题 设f(x)在[0,π/2]上连续,在(0,π/2)内可导,证明:存在ξ,η∈(0,π/2),使得π/2f’(ξ)=f’(η)/sinη.

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答案令g(x)=cosx,g’(x)=-sinx≠0(0<x<π/2),由柯西中值定理,存在η∈(0,π/2),使得[f(π/2)-f(0)]/[g(π/2)-g(0)],即f(π/2)-f(0)=f’(η)/sinη;由拉格朗日中值定理,存在ξ∈(0,π/2),使得f(π/2)-f(0)=f’(ξ)(π/2-0),故π/2f’(ξ)=f’(η)/sinη.

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