HMMT 十一月 2010 · 冲刺赛 · 第 27 题

HMMT November 2010 — Guts Round — Problem 27

[ 14 ] Let f ( x ) = − x + 10 x − 20. Find the sum of all 2 solutions to f ( f ( . . . ( x ) . . . )) = 2. ︸︷︷︸ 2010 f s

解析

[ 14 ] Let f ( x ) = − x + 10 x − 20. Find the sum of all 2 solutions to f ( f ( . . . ( x ) . . . )) = 2. ︸︷︷︸ 2010 f s 2010 Answer: 5 · 2 Define g ( x ) = f ( f ( . . . ( x ) . . . )). We calculate: 2 2 2 f (10 − x ) = − (10 − x ) + 10(10 − x ) − 20 = − 100 + 20 x − x + 100 − 10 x − 20 = − x + 10 x − 20 = f ( x ) . This implies that g (10 − x ) = g ( x ). So if g ( x ) = 2, then g (10 − x ) = 2. Moreover, we can calculate f (5) = − 25 + 50 − 20 = 5, so g (5) = 5 6 = 2. Thus the possible solutions to g ( x ) = 2 can be grouped into pairs, ( x , 10 − x ) , ( x , 10 − x ) , . . . . The sum of the members of each pair is 10, and there are 1 1 2 2 2009 2 pairs, so the sum is 2009 2010 10 · 2 = 5 · 2 .