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CIE 9709 2022 June Paper 12 Q1

A Level / CIE / P1

CIE 9709 2022 June Paper 12 Paper · Question 1

题目

Problem

The coefficient of x4x^4 in the expansion of (3+x)5(3+x)^5 is equal to the coefficient of x2x^2 in the expansion of (2x+ax)6\big(2x+\frac{a}{x}\big)^6.

Find the value of the positive constant aa.

(4)
题目中文翻译

(3+x)5(3+x)^5 的展开式中 x4x^4 的系数,等于 (2x+ax)6\big(2x+\frac{a}{x}\big)^6 的展开式中 x2x^2 的系数。

求正常数 aa 的值。

(4)
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

展开

先找第一式的四次项。第二式中每选一次 a/xa/x 代替 2x2x,总次数减少 22。

答题过程

展开

The coefficient of x4x^4 in the first expansion is (54)(3)=15\binom54(3)=15. In the second, choosing a/xa/x twice gives the power 6−2(2)=26-2(2)=2, with coefficient (62)24a2=240a2\binom62 2^4a^2=240a^2. Thus 240a2=15240a^2=15, so a2=1/16a^2=1/16. The positive value is a=1/4a=1/4.