39th Putnam 1978

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Problem A3

Let p(x) = 2(x6 + 1) + 4(x5 + x) + 3(x4 + x2) + 5x3. Let a = ∫0 x/p(x) dx, b = ∫0 x2/p(x) dx, c = ∫0 x3/p(x) dx, d = ∫0 x4/p(x) dx. Which of a, b, c, d is the smallest?

 

Solution

Answer: b.

Let a1 = ∫01 x/p(x) dx, a2 = ∫1 x/p(x) dx. Similarly, b1, b2, c1, c2, d1, d2. Also define e1 = ∫01 1/p(x) dx.

Using the subtitution y = 1/x we find that a1 = c2, a2 = c1, b1 = b2, and e1 = d2. Hence, in particular, a = a1 + a2 = c2 + c1 = c.

But x - 2x2 + x3 = x(x - 1)2 > 0 on (0, 1), so a1 - 2b1 + c1 > 0. Hence a = a1 + a2 > b1 + b2 = b. Hence also c > b.

Similarly, 1 - 2x2 + x4 = (x2 - 1)2 > 0 on (0, 1). So e1 - 2b1 + d1 > 0. Hence d = d1 + d2 > b1 + b2 = b. So b is the smallest.

 


 

39th Putnam 1978

© John Scholes
jscholes@kalva.demon.co.uk
30 Nov 1999