HMMT 二月 2011 · 冲刺赛 · 第 27 题

英文解析

27. [ 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
    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 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
    2 2 21
    have p ( x ) is ≤ g ( x ) and ≤ f ( x ) in the right places, so this function works. Thus, there are 2 solutions.
    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 1
    positive for both arbitrarily large positive values of x and arbitrarily large negative values of x . Thus,
    there are no solutions.
    Case 3: q is quadratic. We have q ( x ) = ax + bx + c . Apply the same argument for the case when q 2
    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.
    Now if c = − , then p (0) = 4 = g (0). But this is the maximum value of g ( x ), so it must be a local 5
    maximum of p . Thus p must not have a linear term odd (otherwise the function behaves like − 3 bx + 42
    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 allpoints.
    Now suppose c = − . We have a must equal to 1. If b 6 = 0, then p will have a cubic term, which means 3
    f ( x ) − p ( x ) can’t be positive for both arbitrarily large positive x and arbitrarily large negative x , so 2
    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.