HMMT 二月 2021 · ALGNT 赛 · 第 9 题
HMMT February 2021 — ALGNT Round — Problem 9
题目详情
- Let f be a monic cubic polynomial satisfying f ( x ) + f ( − x ) = 0 for all real numbers x . For all real numbers y , define g ( y ) to be the number of distinct real solutions x to the equation f ( f ( x )) = y. Suppose that the set of possible values of g ( y ) over all real numbers y is exactly { 1 , 5 , 9 } . Compute the sum of all possible values of f (10).
解析
- Let f be a monic cubic polynomial satisfying f ( x ) + f ( − x ) = 0 for all real numbers x . For all real numbers y , define g ( y ) to be the number of distinct real solutions x to the equation f ( f ( x )) = y. Suppose that the set of possible values of g ( y ) over all real numbers y is exactly { 1 , 5 , 9 } . Compute the sum of all possible values of f (10). Proposed by: Sujay Kazi Answer: 970 3 Solution: We claim that we must have f ( x ) = x − 3 x. First, note that the condition f ( x )+ f ( − x ) = 0 3 implies that f is odd. Combined with f being monic, we know that f ( x ) = x + ax for some real number a. Note that a must be negative; otherwise f ( x ) and f ( f ( x )) would both be increasing and 1 would be the only possible value of g ( y ). Now, consider the condition that the set of possible values of g ( y ) is { 1 , 5 , 9 } . The fact that we can have g ( y ) = 9 means that some horizontal line crosses the graph of f ( f ( x )) 9 times. Since f ( f ( x )) has degree 9 , this means that its graph will have 4 local maxima and 4 local minima. Now, suppose we start at some value of y such that g ( y ) = 9 , and slowly increase y. At some point, the value of g ( y ) will decrease. This happens when y is equal to a local maximum of f. Since g ( y ) must jump from 9 down to 5 , all four local maxima must have the same value. Similarly, all four local minima must also have the same value. Since f is odd, it suffices to just consider the four local maxima. 2 2 3 2 The local maximum of f ( x ) occurs when 3 x + a = 0 . For convenience, let a = − 3 b , so f ( x ) = x − 3 b x. 3 Then, the local maximum is at x = − b, and has a value of f ( − b ) = 2 b . We consider the local maxima of f ( f ( x )) next. They occur either when x = − b (meaning f ( x ) is at 3 a local maximum) or f ( x ) = − b. If f ( x ) = − b, then f ( f ( x )) = f ( − b ) = 2 b . Thus, we must have 3 3 f ( f ( − b )) = f (2 b ) = 2 b . This yields the equation 3 9 2 3 3 f (2 b ) = 8 b − 3 b · 2 b = 2 b ( ) ( ) 2 3 2 2 2 3 which factors as 2 b b − 1 2 b + 1 . The only possible value of b is 1 . Thus, f ( x ) = x − 3 x, and 3 our answer is 10 − 3 · 10 = 970 .