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CIE 9709 2026 June Paper 31 Q10

A Level / CIE / P3

CIE 9709 2026 June Paper 31 Paper · Question 10

题目

Problem

The polynomial ax3−11x2+bx+4ax^3-11x^2+bx+4, where aa and bb are constants, is denoted by p(x)\mathrm{p}(x). It is given that (x−2)(x-2) is a factor of p(x)\mathrm{p}(x), and that when p′(x)\mathrm{p}'(x) is divided by (x−2)(x-2) the remainder is 2424.

(a) Find the value of aa and the value of bb.

(5)

(b) (i) When aa and bb have these values, factorise p(x)\mathrm{p}(x) completely.

(3)

(ii) Hence, solve the inequality p(x)≥0\mathrm{p}(x)\ge0.

(2)
题目中文翻译

多项式 ax3−11x2+bx+4ax^3-11x^2+bx+4 记为 p(x)\mathrm{p}(x),其中 aa 和 bb 为常数。已知 (x−2)(x-2) 是 p(x)\mathrm{p}(x) 的一个因式,且 p′(x)\mathrm{p}'(x) 除以 (x−2)(x-2) 的余数为 2424。

(a) 求 aa 和 bb 的值。

(5)

(b) (i) 当 aa 和 bb 取上述值时,将 p(x)\mathrm{p}(x) 完全因式分解。

(3)

(ii) 由此解不等式 p(x)≥0\mathrm{p}(x)\ge0。

(2)
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

(a)

解法一

思路

展开

由因式定理得到 p(2)=0\mathrm p(2)=0。对导函数应用余式定理,得到 p′(2)=24\mathrm p'(2)=24,联立求参数。

答题过程

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The factor condition gives

p(2)= 8a−44+2b+4=0,4a+b= 20.\begin{align*} \mathrm p(2)=&\,8a-44+2b+4=0,\\[4mm] 4a+b=&\,20. \end{align*}

Also p′(x)=3ax2−22x+b\mathrm p'(x)=3ax^2-22x+b, so the remainder condition gives

p′(2)= 12a−44+b=24,12a+b= 68.\begin{align*} \mathrm p'(2)=&\,12a-44+b=24,\\[4mm] 12a+b=&\,68. \end{align*}

Subtracting the equations gives 8a=488a=48, hence a=6a=6 and b=−4b=-4.

(b)(i)

解法一

思路

展开

除去已知因式 x−2x-2 后,继续分解得到的二次式。

答题过程

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With the values found in part (a), p(x)=6x3−11x2−4x+4\mathrm p(x)=6x^3-11x^2-4x+4. Since x−2x-2 is a factor, write p(x)=(x−2)(6x2+Bx+C)\mathrm p(x)=(x-2)(6x^2+Bx+C). Expanding gives

p(x)=6x3+(B−12)x2+(C−2B)x−2C.\mathrm p(x)=6x^3+(B-12)x^2+(C-2B)x-2C.

Comparing coefficients gives B−12=−11B-12=-11, so B=1B=1, and −2C=4-2C=4, so C=−2C=-2. The xx coefficient is then C−2B=−4C-2B=-4, as required. Hence

p(x)= (x−2)(6x2+x−2)= (x−2)(6x2+4x−3x−2)= (x−2)(3x+2)(2x−1).\begin{align*} \mathrm p(x)=&\,(x-2)(6x^2+x-2)\\[4mm] =&\,(x-2)(6x^2+4x-3x-2)\\[4mm] =&\,(x-2)(3x+2)(2x-1). \end{align*}

(b)(ii)

解法一

思路

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三个根都是单根,首项系数为正。按根从小到大分区间判断三个因子的乘积符号,并包含等号。

答题过程

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The roots are −2/3-2/3, 1/21/2 and 22. The sign of p(x)\mathrm p(x) on the four successive intervals is negative, positive, negative, positive. Including the zeros, the solution is

−23≤x≤12orx≥2.-\frac23\le x\le\frac12 \quad\text{or}\quad x\ge2.