HMMT 二月 2007 · CALC 赛 · 第 5 题

HMMT February 2007 — CALC Round — Problem 5

[ 5 ] The function f : R → R satisfies f ( x ) f ( x ) = f ( x ) f ( x ) for all real x . Given that f (1) = 1 and ′′′ ′ ′′ f (1) = 8, determine f (1) + f (1). 2 3

解析

[ 5 ] The function f : R → R satisfies f ( x ) f ( x ) = f ( x ) f ( x ) for all real x . Given that f (1) = 1 and ′′′ ′ ′′ f (1) = 8, determine f (1) + f (1). ′ ′′ 2 Answer: 6 . Let f (1) = a and f (1) = b . Then setting x = 1 in the given equation, b = a . Differentiating the given yields ′ 2 ′′ 2 ′′′ ′′ ′ 2 ′ ′′ 2 2 xf ( x ) f ( x ) + f ( x ) f ( x ) = f ( x ) f ( x ) + 2 xf ( x ) f ( x ) . Plugging x = 1 into this equation gives 2 ab + 8 = ab + 2 ab , or ab = 8. Then because a and b are real, we obtain the solution ( a, b ) = (2 , 4). 2 x − 2 Remarks. A priori, the function needn’t exist, but one possibility is f ( x ) = e . 1 2 3