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CIE 9709 2023 Nov Paper 32 Q3

A Level / CIE / P3

CIE 9709 2023 Nov Paper 32 Paper · Question 3

题目

Problem

The polynomial 2x3+ax2−11x+b2x^3+ax^2-11x+b is denoted by p(x)\mathrm{p}(x). It is given that p(x)\mathrm{p}(x) is divisible by (2x−1)(2x-1) and that when p(x)\mathrm{p}(x) is divided by (x+1)(x+1) the remainder is 1212.

Find the values of aa and bb.

(5)
题目中文翻译

多项式 2x3+ax2−11x+b2x^3+ax^2-11x+b 记为 p(x)\mathrm{p}(x)。已知 p(x)\mathrm{p}(x) 能被 (2x−1)(2x-1) 整除,且 p(x)\mathrm{p}(x) 除以 (x+1)(x+1) 的余数为 1212。

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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因式定理给出 p(1/2)=0p(1/2)=0,余式定理给出 p(−1)=12p(-1)=12,由此解出两个常数。

答题过程

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Since 2x−12x-1 is a factor,

p(12)=0.\mathrm p\bigg(\frac12\bigg)=0.

Hence

14+a4−112+b= 0,a+4b= 21.\begin{align*} \frac14+\frac a4-\frac{11}{2}+b=&\,0,\\[4mm] a+4b=&\,21. \end{align*}

By the remainder theorem, p(−1)=12\mathrm p(-1)=12, so

−2+a+11+b= 12,a+b= 3.\begin{align*} -2+a+11+b=&\,12,\\[4mm] a+b=&\,3. \end{align*}

Subtracting the second equation from the first gives 3b=183b=18, so b=6b=6 and a=3−6=−3a=3-6=-3.

Thus a=−3,b=6\boxed{a=-3,\qquad b=6}.