HMMT 二月 2007 · 代数 · 第 6 题

英文解析

6. [ 5 ] Consider the polynomial P ( x ) = x + x − x + 2. Determine all real numbers r for which thereexists a complex number z not in the reals such that P ( z ) = r.
   Answer: r > 3 , r < . Because such roots to polynomial equations come in conjugate pairs, we 49
   seek the values r such that P ( x ) = r has just one real root x . Considering the shape of a cubic, we are 27
   interested in the boundary values r such that P ( x ) − r has a repeated zero. Thus, we write
   3 2 2 3 2 2
   P ( x ) − r = x + x − x + (2 − r ) = ( x − p ) ( x − q ) = x − (2 p + q ) x + p ( p + 2 q ) x − p q.
   Then q = − 2 p − 1 and 1 = p ( p + 2 q ) = p ( − 3 p − 2) so that p = 1 / 3 or p = − 1. It follows that the graphof P ( x ) is horizontal at x = 1 / 3 (a maximum) and x = − 1 (a minimum), so the desired values r arer > P ( − 1) = 3 and r < P (1 / 3) = 1 / 27 + 1 / 9 − 1 / 3 + 2 = 49 / 27 . 1