HMMT 二月 2005 · CALC 赛 · 第 4 题

英文解析

4. Let f : R → R be a smooth function such that f ( x ) = f ( x ) f ( x ) for all x . Suppose
   (4) ′
   f (0) = 1 and f (0) = 9. Find all possible values of f (0).
   √
   Solution: ± 3
   ′ ′′ 2
   Let f (0) = a . Then the equation gives f (0) = a . Differentiating the given equationgives
   ′ ′′ ′′′ ′ ′′
   2 f ( x ) f ( x ) = f ( x ) f ( x ) + f ( x ) f ( x ) ,
   ′ ′′ ′′′
   or f ( x ) f ( x ) = f ( x ) f ( x ). Differentiating once more gives
   ′ ′′′ ′′ 2 (4) ′ ′′′
   f ( x ) f ( x ) + f ( x ) = f ( x ) f ( x ) + f ( x ) f ( x )
   √1
   ′′ 2 (4) (4) 4
   or f ( x ) = f ( x ) f ( x ), giving 9 = f (0) = a . Thus a = ± 3. These are indeed
   √
   ± x 3
   both attainable by f ( x ) = e .
   ′′ ′
   f ( x ) f ( x )
   Alternative Solution: Rewrite the given equation as = . Integrating both
   ′
   f ( x ) f ( x )
   ′ ′
   sides gives ln f ( x ) = ln f ( x ) + C , and exponentiating gives f ( x ) = Cf ( x ). This has
   Cx 1
   solution f ( x ) = Ae for constants A and C . Since f (0) = 1, A = 1, and differentiating
   √
   4 (4) ′
   we find that C = f (0) = 9, yielding f (0) = C = ± 3.