HMMT 二月 2007 · CALC 赛 · 第 5 题
HMMT February 2007 — CALC Round — Problem 5
题目详情
英文原题
- [ 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
英文解析
- [ 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 .
2 31