### 37th Putnam 1976

**Problem A6**

Let R be the real line. f **:** R → [-1, 1] is twice differentiable and f(0)^{2} + f '(0)^{2} = 4. Show that f(x_{0}) + f ''(x_{0}) = 0 for some x_{0}.

**Solution**

Let k(x) = f(x)^{2} + f '(x)^{2}. By the mean value theorem for some a in the interval (0, 2) we have f '(a) = 1/2 (f(2) - f(0) ). But |f(x)| ≤ 1 for all x, so |f '(a)| ≤ 1. Hence k(a) ≤ 1 + 1 = 2. Similarly, we can find b in the interval (-2, 0) with k(b) ≤ 2. We are given that k(0) = 4. Hence k(x) has a maximum at some interior point of (-2, 2). Let this point be c. Then certainly k(c) ≥ k(0) = 4, f(c)^{2} ≤ 1, so |f '(c)| > 0. We have k '(c) = 0. But k '(c) = 2 f '(c) ( f(c) + f ''(c) ). We have just shown that f '(c) is non-zero, so f(c) + f ''(c) = 0.

37th Putnam 1976

© John Scholes

jscholes@kalva.demon.co.uk

23 Jan 2001