HMMT 二月 2011 · 冲刺赛 · 第 27 题
HMMT February 2011 — Guts Round — Problem 27
题目详情
- [ 16 ] Find the number of polynomials p ( x ) with integer coefficients satisfying 4 2 2 4 2 2 p ( x ) ≥ min { 2 x − 6 x + 1 , 4 − 5 x } and p ( x ) ≤ max { 2 x − 6 x + 1 , 4 − 5 x } for all x ∈ R .
解析
- [ 16 ] Find the number of polynomials p ( x ) with integer coefficients satisfying 4 2 2 4 2 2 p ( x ) ≥ min { 2 x − 6 x + 1 , 4 − 5 x } and p ( x ) ≤ max { 2 x − 6 x + 1 , 4 − 5 x } for all x ∈ R . Answer: 4 4 2 2 4 2 2 We first find the intersection points of f ( x ) = 2 x − 6 x +1 and g ( x ) = 4 − 5 x . If 2 x − 6 x +1 = 4 − 5 x , √ 4 2 2 2 3 then 2 x − x − 3 = 0, so (2 x − 3)( x + 1) = 0, and x = ± . Note that this also demonstrates 2 √ √ 3 3 that g ( x ) ≥ f ( x ) if and only if | x | ≤ and that p ( x ) must satisfy p ( x ) ≤ g ( x ) iff | x | ≤ . We 2 2 √ 3 7 7 2 must have p ( ) = − , so p ( x ) = − + (2 x − 3) q ( x ) for a polnomial q of degree 0 , 1 or 2. We now 2 2 2 examine cases. n 3 5 Case 1: q is constant. We have q = for an integer n since 1 ≤ p (0) ≤ 4, n = − or − . Clearly 2 2 2 5 7 3 2 2 q = − is an appropriate choice because then p = g . Let p ( x ) = − + − (2 x − 3) = 1 − 3 x . We 1 2 2 2 have p ( x ) is ≤ g ( x ) and ≤ f ( x ) in the right places, so this function works. Thus, there are 2 solutions. 1 Case 2: q is linear. We have p ( x ) is a cubic, so p ( x ) − g ( x ) is also a cubic, which means it can’t be positive for both arbitrarily large positive values of x and arbitrarily large negative values of x . Thus, there are no solutions. 2 Case 3: q is quadratic. We have q ( x ) = ax + bx + c . Apply the same argument for the case when q 3 5 is a constant, we have c = − or − . Since p must have integer coefficients, we have b must be an 2 2 integer. Since f ( x ) ≥ p ( x ) ≥ g ( x ) for large x , the leading coefficient of f must be greater than or equal 1 1 to the leading coefficient of p , which must be greater than 0. Thus a = 1 or a = . However, if a = , 2 2 then the quadratic term of p is not an integer, so a = 1. 5 Now if c = − , then p (0) = 4 = g (0). But this is the maximum value of g ( x ), so it must be a local 2 maximum of p . Thus p must not have a linear term odd (otherwise the function behaves like − 3 bx + 4 4 2 around x = 0). So p ( x ) must be 2 x − 8 x + 4. This is indeed bounded between f ( x ) and g ( x ) at all points. 3 Now suppose c = − . We have a must equal to 1. If b 6 = 0, then p will have a cubic term, which means 2 f ( x ) − p ( x ) can’t be positive for both arbitrarily large positive x and arbitrarily large negative x , so 7 3 2 2 b = 0. Therefore p ( x ) = − + (2 x − 3)( x − ) = f ( x ). It is easy to check that this choice of p is 2 2 indeed bounded between f ( x ) and g ( x ). Therefore, there are 4 solutions in total.