34th Putnam 1973

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Problem A4

How many real roots does 2x = 1 + x2 have?

 

Solution

Answer: 3. x = 0, x = 1 and a value just over 4.

Clearly there are no roots for negative x, since for such x, 2x < 1, whereas 1 + x2 > 1. There are certainly roots at x = 0 and 1. Also 24 < 42 + 1, whereas 25 > 52 + 1, so there is a root between 4 and 5. We have to show that there are no other roots. Put f(x) = 2x - x2 - 1. Then f ''(x) = (ln 2)2 2x - 2. This is strictly increasing with a single zero. f '(0) > 0, so f '(x) starts positive, decreases through zero to a minimum, then increases through zero. So it has just two zeros. Hence f(x) has at most three zeros, which we have already found.

 


 

34th Putnam 1973

© John Scholes
jscholes@kalva.demon.co.uk
22 Aug 2001