HMMT 二月 2011 · ALGCALC 赛 · 第 17 题

英文解析

17. Let f : (0 , 1) → (0 , 1) be a differentiable function with a continuous derivative such that for everyn npositive integer n and odd positive integer a < 2 , there exists an odd positive integer b < 2 such
    ( ) ( )
    a b 1
    ′
    that f = . Determine the set of possible values of f .
    n n
    2 2 2
    ′
    Answer: {− 1 , 1 } The key step is to notice that for such a function f , f ( x ) 6 = 0 for any x .
    ′ ′
    Assume, for sake of contradiction that there exists 0 < y < 1 such that f ( y ) = 0. Since f is
    ′1
    a continuous function, there is some small interval ( c, d ) containing y such that | f ( x ) | ≤ for alla a +12
    x ∈ ( c, d ). Now there exists some n, a such that , are both in the interval ( c, d ). From then n
    2 2
    a +1 a
    ′
    f ( ) − f ( )
    b bn n
    2 2 n ′
    ′
    definition, = 2 ( − ) = b − b where b, b are integers; one is odd, and one isa +1 an n
    2 2
    n n−
    2 2
    ′
    even. So b − b is an odd integer. Since f is differentiable, by the mean value theorem there exists a
    ′ ′ ′ 1
    point where f = b − b . But this point is in the interval ( c, d ), and | b − b | > . This contradicts the
    ′ 12
    assumption that | f ( x ) | ≤ for all x ∈ ( c, d ).
    ′ ′ ′2
    Since f ( x ) 6 = 0, and f is a continuous function, f is either always positive or always negative. So f is
    1 1 1 1 3 3
    either increasing or decreasing. f ( ) = always. If f is increasing, it follows that f ( ) = , f ( ) = ,
    2 2 4 4 4 4
    a aand we can show by induction that indeed f ( ) = for all integers a, n . Since numbers of this formn n
    2 2
    are dense in the interval (0 , 1), and f is a continuous function, f ( x ) = x for all x .
    It can be similarly shown that if f is decreasing f ( x ) = 1 − x for all x . So the only possible values of
    ′ 1
    f ( ) are − 1 , 1.
    Query: if the condition that the derivative is continuous were omitted, would the same result still 2
    hold?
    2 π 2 π