23rd Australian 2002 Problem B1 Find all real-valued functions f(x) on the reals such that f(2002x - f(0) ) = 2002 x 2 for all x. Solution
: f(x) = x 2 /2002 or f(x) = x 2 /2002 + 2x + 2002. Let f(0) = k. Put x = k/2002, then k = f(0) = 2002 k 2 /2002 2 , so k = 0 or 2002. If k = 0, then f(2002x) = 2002x 2 , so f(x) = x 2 /2002, which is obviously a solution. If k = 2002, then put x = y/2002 + 1 and we get f(y) = 2002 (y/2002 + 1) 2 = y 2 /2002 + 2y + 2002, which is also a solution. 23rd Aus 2002 (C) John Scholes jscholes@kalva.demon.co.uk 26 July 2002