HMMT 二月 2011 · ALGCALC 赛 · 第 17 题
HMMT February 2011 — ALGCALC Round — Problem 17
题目详情
- Let f : (0 , 1) → (0 , 1) be a differentiable function with a continuous derivative such that for every n n positive 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 2 π 2 π
解析
- Let f : (0 , 1) → (0 , 1) be a differentiable function with a continuous derivative such that for every n n positive 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 all 2 a a +1 x ∈ ( c, d ). Now there exists some n, a such that , are both in the interval ( c, d ). From the n n 2 2 a +1 a ′ f ( ) − f ( ) b b n n 2 2 n ′ ′ definition, = 2 ( − ) = b − b where b, b are integers; one is odd, and one is a +1 a n 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 2 ′ 1 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 a and we can show by induction that indeed f ( ) = for all integers a, n . Since numbers of this form n 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. 2 Query: if the condition that the derivative is continuous were omitted, would the same result still hold? 2 π 2 π