HMMT 二月 2003 · CALC 赛 · 第 9 题
HMMT February 2003 — CALC Round — Problem 9
题目详情
- Two differentiable real functions f ( x ) and g ( x ) satisfy ′ f ( x ) f ( x ) − g ( x ) = e ′ g ( x ) for all x , and f (0) = g (2003) = 1. Find the largest constant c such that f (2003) > c for all such functions f, g .
解析
- Two differentiable real functions f ( x ) and g ( x ) satisfy ′ f ( x ) f ( x ) − g ( x ) = e ′ g ( x ) for all x , and f (0) = g (2003) = 1. Find the largest constant c such that f (2003) > c for all such functions f, g . Solution: 1 − ln 2 ′ − f ( x ) ′ − g ( x ) d − f ( x ) Rearranging the given equation gives f ( x ) e = g ( x ) e for all x , so ( e − dx − g ( x ) ′ − f ( x ) ′ − g ( x ) − f ( x ) − g ( x ) e ) = − f ( x ) e + g ( x ) e = 0. Thus, e − e is a constant, and it must − f (0) − 1 − f (2003) − g (2003) − 1 − 1 ln 2 − 1 be less than e = e . Thus, e < e + e = 2 e = e ⇒ f (2003) > − f ( x ) − g ( x ) 1 − ln 2. On the other hand, we can find positive-valued functions e , e that take on the required values at 0 and 2003 and have constant difference arbitrarily close − 1 − f ( x ) − ( t (2003 − x )+1) − 1 to e . For example, for arbitrarily large t , we can set e = e + e − − (2003 t +1) − g ( x ) − ( t (2003 − x )+1) e and e = e , and we can check that the resulting functions f, g satisfy the required conditions. Thus, we can make f (2003) arbitrarily close to 1 − ln 2, so this is the answer.