HMMT 二月 2005 · CALC 赛 · 第 10 题

HMMT February 2005 — CALC Round — Problem 10

Let f : R → R be a smooth function such that f ( x ) = f (1 − x ) for all x and f (0) = 1. Find f (1). 1

解析

Let f : R → R be a smooth function such that f ( x ) = f (1 − x ) for all x and f (0) = 1. Find f (1). Solution: sec 1 + tan 1 ′′ Differentiating the given equation gives f ( x ) = − f ( x ). This has solutions of the form ′ A cos( x ) + B sin( x ). Since f (0) = 1, A = 1. Then f ( x ) = B cos( x ) − sin( x ) and f (1 − x ) = cos(1 − x ) + B sin(1 − x ) = cos 1 cos x + sin 1 sin x + B sin 1 cos x − B cos 1 sin x = (cos 1 + B sin 1) cos x + (sin 1 − B cos 1) sin x. Therefore, B = cos 1+ B sin 1 and − 1 = sin 1 − B cos 1, both of which yield as solutions cos 1 1 + sin 1 B = = = sec 1 + tan 1 . 1 − sin 1 cos 1 4