HMMT 二月 2006 · CALC 赛 · 第 10 题

HMMT February 2006 — CALC Round — Problem 10

Suppose f and g are differentiable functions such that ′ ′ ′ ′ xg ( f ( x )) f ( g ( x )) g ( x ) = f ( g ( x )) g ( f ( x )) f ( x ) for all real x . Moreover, f is nonnegative and g is positive. Furthermore, ∫ a − 2 a e f ( g ( x )) dx = 1 − 2 0 for all reals a . Given that g ( f (0)) = 1, compute the value of g ( f (4)).

解析

Suppose f and g are differentiable functions such that ′ ′ ′ ′ xg ( f ( x )) f ( g ( x )) g ( x ) = f ( g ( x )) g ( f ( x )) f ( x ) for all real x . Moreover, f is nonnegative and g is positive. Furthermore, ∫ a − 2 a e f ( g ( x )) dx = 1 − 2 0 for all reals a . Given that g ( f (0)) = 1, compute the value of g ( f (4)). 1 − 16 Answer: e or 16 e − 2 a Solution: Differentiating the given integral with respect to a gives f ( g ( a )) = e . Now ′ ′ ′ ′ d [ln ( f ( g ( x )))] f ( g ( x )) g ( x ) g ( f ( x )) f ( x ) d [ln ( g ( f ( x )))] x = x = = dx f ( g ( x )) g ( f ( x )) dx where the second equals sign follows from the given. Since ln ( f ( g ( x ))) = − 2 x , we have 2 2 − x − 16 − x + C = ln ( g ( f ( x ))), so g ( f ( x )) = Ke . It follows that K = 1 and g ( f (4)) = e . 4