39th Putnam 1978 Problem A3 Let p(x) = 2(x 6 + 1) + 4(x 5 + x) + 3(x 4 + x 2 ) + 5x 3 . Let a = ∫ 0 ∞ x/p(x) dx, b = ∫ 0 ∞ x 2 /p(x) dx, c = ∫ 0 ∞ x 3 /p(x) dx, d = ∫ 0 ∞ x 4 /p(x) dx. Which of a, b, c, d is the smallest? Solution
: b. Let a 1 = ∫ 0 1 x/p(x) dx, a 2 = ∫ 1 ∞ x/p(x) dx. Similarly, b 1 , b 2 , c 1 , c 2 , d 1 , d 2 . Also define e 1 = ∫ 0 1 1/p(x) dx. Using the subtitution y = 1/x we find that a 1 = c 2 , a 2 = c 1 , b 1 = b 2 , and e 1 = d 2 . Hence, in particular, a = a 1 + a 2 = c 2 + c 1 = c. But x - 2x 2 + x 3 = x(x - 1) 2 > 0 on (0, 1), so a 1 - 2b 1 + c 1 > 0. Hence a = a 1 + a 2 > b 1 + b 2 = b. Hence also c > b. Similarly, 1 - 2x 2 + x 4 = (x 2 - 1) 2 > 0 on (0, 1). So e 1 - 2b 1 + d 1 > 0. Hence d = d 1 + d 2 > b 1 + b 2 = b. So b is the smallest. 39th Putnam 1978 © John Scholes jscholes@kalva.demon.co.uk 30 Nov 1999