HMMT 二月 2011 · CALCGEO 赛 · 第 14 题

HMMT February 2011 — CALCGEO Round — Problem 14

Let f : [0 , 1] → [0 , 1] be a continuous function such that f ( f ( x )) = 1 for all x ∈ [0 , 1]. Determine the ∫ 1 set of possible values of f ( x ) dx . 0 2

解析

Let f : [0 , 1] → [0 , 1] be a continuous function such that f ( f ( x )) = 1 for all x ∈ [0 , 1]. Determine the ∫ 1 set of possible values of f ( x ) dx . 0 ∫ 1 3 Answer: ( , 1] Since the maximum value of f is 1, f ( x ) dx ≤ 1. 4 0 By our condition f ( f ( x )) = 1, f is 1 at any point within the range of f . Clearly, 1 is in the range of f , so f (1) = 1. Now f ( x ) is continuous on a closed interval so it attains a minimum value c . Since c is in the range of f , f ( c ) = 1. ∫ 1 If c = 1, f ( x ) = 1 for all x and f ( x ) dx = 1. 0 Now assume c < 1. By the intermediate value theorem, since f is continuous it attains all values between c and 1. So for all x ≥ c , f ( x ) = 1. Therefore, ∫ ∫ 1 c f ( x ) dx = f ( x ) dx + (1 − c ) . 0 0 ∫ c 2 Since f ( x ) ≥ c , f ( x ) dx > c , and the equality is strict because f is continuous and thus cannot be 0 c for all x < c and 1 at c . So ∫ 1 1 3 3 2 2 f ( x ) dx > c + (1 − c ) = ( c − ) + ≥ . 2 4 4 0 ∫ 1 3 Therefore < f ( x ) dx ≤ 1, and it is easy to show that every value in this interval can be reached. 4 0 2