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CIE 9709 2024 November Paper 31 Q1

A Level / CIE / P3

CIE 9709 2024 Nov Paper 31 Paper · Question 1

题目

Problem

The polynomial 4x3+ax2+5x+b4x^3+ax^2+5x+b, where aa and bb are constants, is denoted by p(x)\mathrm{p}(x). It is given that (2x+1)(2x+1) is a factor of p(x)\mathrm{p}(x). When p(x)\mathrm{p}(x) is divided by (x−4)(x-4) the remainder is equal to 33 times the remainder when p(x)\mathrm{p}(x) is divided by (x−2)(x-2).

Find the values of aa and bb.

(5)
题目中文翻译

多项式 4x3+ax2+5x+b4x^3+ax^2+5x+b 记为 p(x)\mathrm{p}(x),其中 aa 和 bb 为常数。已知 (2x+1)(2x+1) 是 p(x)\mathrm{p}(x) 的一个因式。p(x)\mathrm{p}(x) 除以 (x−4)(x-4) 所得的余数等于 p(x)\mathrm{p}(x) 除以 (x−2)(x-2) 所得余数的 33 倍。

求 aa 和 bb 的值。

(5)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

解答

解法一

思路

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因式条件先给一个方程,余式的三倍关系给另一个方程。代入时要注意二倍一次因式的根为负二分之一。

答题过程

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Since 2x+12x+1 is a factor, p(−1/2)=0p(-1/2)=0. Hence

−12+a4−52+b=0,-\frac12+\frac a4-\frac52+b=0,

so

a4+b=3.(1)\frac a4+b=3. \tag{1}

By the remainder theorem, the other condition is p(4)=3p(2)p(4)=3p(2):

256+16a+20+b=3(32+4a+10+b).256+16a+20+b=3(32+4a+10+b).

Simplifying,

−2a+b=75.(2)-2a+b=75. \tag{2}

Using b=3−a/4b=3-a/4 from (1) in (2),

−2a+3−a4= 75,−9a= 288.\begin{align*} -2a+3-\frac a4=&\,75,\\[4mm] -9a=&\,288. \end{align*}

Therefore

a=−32,b=11.\boxed{a=-32,\qquad b=11}.