HMMT 二月 2003 · CALC 赛 · 第 7 题

英文解析

7. For what value of a > 1 is
   ∫
   a 2
   1 x − 1
   log dxa x 32
   minimum?
   Solution: 3
   ∫ 2
   a df
   1 x − 1
   Let f ( a ) = log dx . Then we want = 0; by the Fundamental Theorem ofax 32 da
   Calculus and the chain rule, this implies that
   ( ) ( )
   ∫ ∫
   a a 22
   1 a − 1 1 a − 1 d 1 x − 1 1 x − 1
   2 a log − log = log dx − log dx = 0 ,
   a 32 a 32 da x 32 x 322
   c c
   2 2
   a − 1 a − 1 a − 1
   where c is any constant with 1 < c < a . Then 2 log = log , so that ( ) =2
   32 32 32
   a − 1
   . After canceling factors of ( a − 1) / 32 (since a > 1), this simplifies to ( a − 1)( a +1) =2
   3 2 232
   32 ⇒ a + a − a − 33 = 0, which in turn factors as ( a − 3)( a + 4 a + 11) = 0. Thequadratic factor has no real solutions, so this leaves only a = 3. However, we have thata > 1, and we can check that f (1) = 0, lim f ( a ) > 0, and f (3) < 0, so the globala →∞
   minimum does occur at a = 3.