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CIE 9709 2026 June Paper 11 Q2

A Level / CIE / P1

CIE 9709 2026 June Paper 11 Paper · Question 2

题目

Problem

(a) Find the expansion of each of the following expressions, in ascending powers of xx, up to and including the term in x2x^2.

(i) (1+2x)5(1+2x)^5

(2)

(ii) (3+x2)6(3+x^2)^6

(1)

(b) Hence, find the coefficient of x2x^2 in the expansion of (1+2x)5(3+x2)6(1+2x)^5(3+x^2)^6.

(2)
题目中文翻译

(a) 将下列各式按 xx 的升幂展开,保留至 x2x^2 项(包括该项)。

(i) (1+2x)5(1+2x)^5

(2)

(ii) (3+x2)6(3+x^2)^6

(1)

(b) 由此,求 (1+2x)5(3+x2)6(1+2x)^5(3+x^2)^6 的展开式中 x2x^2 的系数。

(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)(i)

解法一

思路

展开

按二项式定理取常数项、一次项和二次项。

答题过程

展开

The first three terms are

1+(51)(2x)+(52)(2x)2=1+10x+40x2.1+\binom51(2x)+\binom52(2x)^2=1+10x+40x^2.

(a)(ii)

解法一

思路

展开

因子中只有常数和 x2x^2。取一个 x2x^2 得到二次项,取两个就已经超过所需次数。

答题过程

展开

Up to and including degree two, the expansion is

36+(61)35x2=729+1458x2.3^6+\binom61 3^5x^2=729+1458x^2.

(b)

解法一

思路

展开

利用 (a) 两个展开式中常数与二次项的配对。第二个展开式没有一次项。

答题过程

展开

The coefficient of x2x^2 is

1(1458)+40(729)=30618.1(1458)+40(729)=30618.