HMMT 二月 2003 · CALC 赛 · 第 9 题

英文解析

9. 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) > cfor 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 −
   − g ( x ) ′ − f ( x ) ′ − g ( x ) − f ( x ) − g ( x )dxe ) = − 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 thattake 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 functionsf, g satisfy the required conditions. Thus, we can make f (2003) arbitrarily close to
   1 − ln 2, so this is the answer.