HMMT 二月 2016 · 团队赛 · 第 7 题

HMMT February 2016 — Team Round — Problem 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?

解析

[ 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 2 Define g ( x ) = 2 x − 1, so that q ( x ) = − + g x + . Thus 2 2 ( ) 1 1 N N q ( 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 equation N q ( x ) = 0 is equivalent to ( ( )) 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