HMMT 二月 2010 · 代数 · 第 10 题

英文解析

10. [ 8 ] Let p ( x ) and q ( x ) be two cubic polynomials such that p (0) = − 24, q (0) = 30, andp ( q ( x )) = q ( p ( x ))
    for all real numbers x . Find the ordered pair ( p (3) , q (6)).
    3 3
    Answer: (3 , − 24) Note that the polynomials f ( x ) = ax and g ( x ) = − ax commute under com-
    − 1
    position. Let h ( x ) = x + b be a linear polynomial, and note that its inverse h ( x ) = x − b is also a − 1 − 1
    linear polynomial. The composite polynomials h f h and h gh commute, since function compositionis associative, and these polynomials are also cubic.
    − 1 − 1
    We solve for the a and b such that ( h f h )(0) = − 24 and ( h gh )(0) = 30. We must have:
    3 3
    ab − b = − 24 , − ab − b = 30 ⇒ a = 1 , b = − 3
    3 3
    These values of a and b yield the polynomials p ( x ) = ( x − 3) + 3 and q ( x ) = − ( x − 3) + 3. Thepolynomials take on the values p (3) = 3 and q (6) = − 24.
    Remark: The pair of polynomials found in the solution is not unique. There is, in fact, an entirefamily of commuting cubic polynomials with p (0) = − 24 and q (0) = 30. They are of the formp ( x ) = tx ( x − 3)( x − 6) − 24 , q ( x ) = − tx ( x − 3)( x − 6) + 30
    where t is any real number. However, the values of p (3) and q (6) are the same for all polynomials inthis family. In fact, if we give the initial conditions p (0) = k and q (0) = k , then we get a general
    1 2
    solution of
    ( )
    3 1 k − k
    2 2 1
    3 2
    p ( x ) = t x − ( k + k ) x + ( k + k ) x + x + k
    1 2 1 2 1
    2 2 k + k
    2 1
    ( )
    3 1 k − k
    2 2 1
    3 2
    q ( x ) = − t x − ( k + k ) x + ( k + k ) x − x + k .
    1 2 1 2 2
    2 2 k + k
    2 1
    Algebra Subject Test