25th Putnam 1964

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Problem A2

α is a real number. Find all continuous real-valued functions f :[0, 1] → (0, ∞) such that ∫01 f(x) dx = 1, ∫01 x f(x) dx = α, ∫01 x2 f(x) dx = α2.

 

Solution

We have ∫01 (α - x)2f(x) dx = α2 - 2α2 + α2 = 0. But the integrand is positive, except possibly for one point of the range, so the integral must also be positive. Contradiction. So there are no functions with this property.

 


 

25th Putnam 1964

© John Scholes
jscholes@kalva.demon.co.uk
5 Feb 2002