16th Australian 1995 Problem B4 Find all real-valued functions f on the positive reals such that f(1) = 1/2 and f(xy) = f(x) f(3/y) + f(y) f(3/x).
f(x) = 1/2 for all x Solution Put x = 1, y = 3, then f(3) = 1/4 + f(3) 2 , or (f(3) - 1/2) 2 = 0, so f(3) = 1/2. Put x = 1, then f(y) = ½f(3/y) + ½f(y), so f(3/y) = f(y). Substituting into the relation gives f(xy) = 2f(x)f(y). Putting y = 3/x, we get 1/2 = 2f(x) 2 , so f(x) = 1/2 or -1/2 for all x. Hence f(x 2 ) = 2f(x) 2 = 1/2 for all x. So if x is any positive real, put y = √x, then f(x) = f(y 2 ) = 1/2. 16th Aus 1995 © John Scholes jscholes@kalva.demon.co.uk 7 February 2004 Last updated/corrected 7 Feb 04