HMMT 二月 2011 · ALGCALC 赛 · 第 10 题
HMMT February 2011 — ALGCALC Round — Problem 10
题目详情
- For all real numbers x , let 1 f ( x ) = √ . 2011 2011 1 − x 2011 Evaluate ( f ( f ( . . . ( f (2011)) . . . ))) , where f is applied 2010 times. ( ) ∫ 2011 ∞ ln x
解析
- For all real numbers x , let 1 f ( x ) = √ . 2011 2011 1 − x 2011 Evaluate ( f ( f ( . . . ( f (2011)) . . . ))) , where f is applied 2010 times. √ 2011 2011 1 − x 2011 Answer: 2011 Direct calculation shows that f ( f ( x )) = and f ( f ( f ( x ))) = x . − x 2011 Hence ( f ( f ( . . . ( f ( x )) . . . ))) = x , where f is applied 2010 times. So ( f ( f ( . . . ( f (2011)) . . . ))) = 2011 2011 . ( ) ∫ 2011 ∞ ln x