HMMT 二月 2006 · TEAM2 赛 · 第 15 题

HMMT February 2006 — TEAM2 Round — Problem 15

[40] Let a, b, c, d be real numbers so that c, d are not both 0. Define the function ax + b m ( x ) = cx + d on all real numbers x except possibly − d/c , in the event that c 6 = 0. Suppose that the equation x = m ( m ( x )) has at least one solution that is not a solution of x = m ( x ). Find all possible values of a + d . Prove that your answer is correct. 2

解析

[40] Let a, b, c, d be real numbers so that c, d are not both 0. Define the function ax + b m ( x ) = cx + d 4 on all real numbers x except possibly − d/c , in the event that c 6 = 0. Suppose that the equation x = m ( m ( x )) has at least one solution that is not a solution of x = m ( x ). Find all possible values of a + d . Prove that your answer is correct. Answer: 0 Solution: That 0 is a possible value of a + d can be seen by taking m ( x ) = − x , i.e., a = − d = 1, b = c = 0. We will now show that 0 is the only possible value of a + d . 2 ( a + bc ) x + ( a + d ) b The equation x = m ( m ( x )) implies x = , which in turn implies 2 ( a + d ) cx + ( bc + d ) 2 ( a + d )[ cx + ( − a + d ) x − b ] = 0 . Suppose for the sake of contradiction that a + d 6 = 0. Then the above equation would further imply 2 cx + ( − a + d ) x − b = 0 , x ( cx + d ) = ax + b, which would imply x = m ( x ) for any x except possibly − d/c . But of course − d/c is not a root of x = m ( m ( x )) anyway, so in this case, all solutions of x = m ( m ( x )) are also solutions of x = m ( x ), a contradiction. So our assumption was wrong, and in fact a + d = 0, as claimed. 5