HMMT 二月 2016 · 团队赛 · 第 7 题

英文解析

7. [ 40 ] Let q ( x ) = q ( x ) = 2 x + 2 x − 1, and let q ( x ) = q ( q ( x )) for n > 1. How many negative real
   2016
   roots does q ( x ) have?
   Proposed by: Ernest Chiu
   2017
   2 +1
   Answer:
   ( )3
   1 1
   Define g ( x ) = 2 x − 1, so that q ( x ) = − + g x + . Thus 2
   2 2
   ( )
   1 1
   N Nq ( x ) = 0 ⇐⇒ = g x +
   2 2
   where N = 2016.
   But, viewed as function g : [ − 1 , 1] → [ − 1 , 1] we have that g ( x ) = cos(2 arccos( x )). Thus, the equationq ( x ) = 0 is equivalent to n
   ( ( ))
   1 1
   2016
   cos 2 arccos x + = .
   2 2
   Thus, the solutions for x are
   ( )
   1 π/ 3 + 2 πn
   2016
   x = − + cos n = 0 , 1 , . . . , 2 − 1 .
   2016
   2 2
   So, the roots are negative for the values of n such that
   1 π/ 3 + 2 πn 5
   π < < π
   2016
   3 2 3
   which is to say
   1 1
   2016 2016
   (2 − 1) < n < (5 · 2 − 1) .
   6 6
   1 2016 1 2016 1 2016
   The number of values of n that fall in this range is (5 · 2 − 2) − (2 + 2) + 1 = (4 · 2 + 2) =
   6 6 6
   1 2017
   (2 + 1). 3