HMMT 二月 2003 · CALC 赛 · 第 7 题

HMMT February 2003 — CALC Round — Problem 7

解析

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