HMMT 二月 2010 · 代数 · 第 10 题
HMMT February 2010 — Algebra — Problem 10
题目详情
- [ 8 ] Let p ( x ) and q ( x ) be two cubic polynomials such that p (0) = − 24, q (0) = 30, and p ( q ( x )) = q ( p ( x )) for all real numbers x . Find the ordered pair ( p (3) , q (6)).
解析
- [ 8 ] Let p ( x ) and q ( x ) be two cubic polynomials such that p (0) = − 24, q (0) = 30, and p ( 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 composition is 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. The polynomials 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 entire family of commuting cubic polynomials with p (0) = − 24 and q (0) = 30. They are of the form p ( 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 in this 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