HMMT 二月 2011 · ALGCALC 赛 · 第 15 题

英文解析

15. Let f : [0 , 1] → [0 , 1] be a continuous function such that f ( f ( x )) = 1 for all x ∈ [0 , 1]. Determine the
    ∫
    set of possible values of f ( x ) dx .1
    ∫0
    Answer: ( , 1] Since the maximum value of f is 1, f ( x ) dx ≤ 1.31
    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 off , so f (1) = 1. Now f ( x ) is continuous on a closed interval so it attains a minimum value c . Since cis in the range of f , f ( c ) = 1.
    ∫
    If c = 1, f ( x ) = 1 for all x and f ( x ) dx = 1.1
    Now assume c < 1. By the intermediate value theorem, since f is continuous it attains all values 0
    between c and 1. So for all x ≥ c , f ( x ) = 1. Therefore,
    ∫ ∫
    1 cf ( x ) dx = f ( x ) dx + (1 − c ) .
    0 0
    ∫
    2 c
    Since f ( x ) ≥ c , f ( x ) dx > c , and the equality is strict because f is continuous and thus cannot bec for all x < c and 1 at c . So 0
    ∫
    1 3 31
    2 2
    f ( x ) dx > c + (1 − c ) = ( c − ) + ≥ .
    2 4 4
    ∫0
    Therefore < f ( x ) dx ≤ 1, and it is easy to show that every value in this interval can be reached.31
    4 0
    Algebra & Calculus Individual Test 2