HMMT 十一月 2021 · GEN 赛 · 第 7 题
HMMT November 2021 — GEN Round — Problem 7
题目详情
- Define the function f : R → R by { √ 3 1 √ if x 6 ∈ ( − 2 , 0] 2 4 x + x +2 x f ( x ) = . 0 otherwise √ a + b c 10 The sum of all real numbers x for which f ( x ) = 1 can be written as , where a, b, c, d are integers, d d is positive, c is square-free, and gcd( a, b, d ) = 1. Find 1000 a + 100 b + 10 c + d . n 3 (Here, f ( x ) is the function f ( x ) iterated n times. For example, f ( x ) = f ( f ( f ( x ))).)
解析
- Define the function f : R → R by { √ 3 1 √ if x 6 ∈ ( − 2 , 0] 2 4 x + x +2 x f ( x ) = . 0 otherwise √ a + b c 10 The sum of all real numbers x for which f ( x ) = 1 can be written as , where a, b, c, d are integers, d d is positive, c is square-free, and gcd( a, b, d ) = 1. Find 1000 a + 100 b + 10 c + d . n 3 (Here, f ( x ) is the function f ( x ) iterated n times. For example, f ( x ) = f ( f ( f ( x ))).) Proposed by: Sean Li Answer: 932 √ 3 Solution: If x ∈ ( − 2 , 0], it is evidently not a solution, so let us assume otherwise. Then, we find √ 2 4 x + 2 x − x f ( x ) = , 2 x 2 2 which implies that xf ( x ) + x f ( x ) − 1 / 2 = 0, by reverse engineering the quadratic formula. Therefore, 2 2 if x > 0, f ( x ) is the unique positive real t so that xt + x t = 1 / 2. However, then x is the unique 2 2 positive real so that xt + x t = 1 / 2, so f ( t ) = x . This implies that if x > 0, then f ( f ( x )) = x . 10 10 11 Suppose that f ( x ) = 1. Then, since f ( x ) > 0, we find that f ( x ) = f ( f ( x )) = f ( x ) = f (1). 10 9 9 Conversely, if f ( x ) = f (1), then f ( x ) = f ( f ( x )) = f ( f (1)) = 1, so we only need to solve f ( x ) = f (1). This is equivalent to √ √ √ √ √ √ 2 2 4 4 2 2 4 4 x + x + 2 x = 1 + 3 ⇐⇒ x + 2 x = 1 + 3 − x = ⇒ x + 2 x = x − 2(1 + 3) x + (1 + 3) , which is equivalent to √ √ 2 2 2(1 + 3) x + 2 x − (1 + 3) = 0 . Obviously, if x = 1 then f ( x ) = f (1), so we already know 1 is a root. This allows us to easily factor √ 1+ 3 the quadratic and find that the other root is − . This ends up not being extraneous—perhaps the 2 √ 1+ 3 shortest way to see this is to observe that if x = − , 2 ( ) √ √ √ 1 + 3 2 1 + 3 − x = (1 + 3) 1 − > 0 , 4 so since we already know √ 4 2 2 x + 2 x = (1 + 3 − x ) , we have √ √ 2 4 x + 2 x = 1 + 3 − x . √ 1 − 3 Therefore, the sum of solutions is . 2