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CIE 9709 2023 March Paper 12 Q6

A Level / CIE / P1

CIE 9709 2023 March Paper 12 Paper · Question 6

题目

Problem

In the expansion of (xa+ax2)7\bigg(\dfrac{x}{a}+\dfrac{a}{x^2}\bigg)^7, it is given that

the coefficient of x4the coefficient of x=3.\frac{\text{the coefficient of }x^4}{\text{the coefficient of }x}=3.

Find the possible values of the constant aa.

(6)
题目中文翻译

在 (xa+ax2)7\bigg(\dfrac{x}{a}+\dfrac{a}{x^2}\bigg)^7 的展开式中,已知

x4 的系数x 的系数=3.\frac{x^4\text{ 的系数}}{x\text{ 的系数}}=3.

求常数 aa 的可能值。

(6)
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

每选一次 a/x2a/x^2 来替代 x/ax/a,总指数便减少 33。先找对应的选取次数,再计算系数之比。

答题过程

展开

If a/x2a/x^2 is selected jj times, the term is

(7j)a2j−7x7−3j.\binom7j a^{2j-7}x^{7-3j}.

For x4x^4 we have j=1j=1 and coefficient 7/a57/a^5. For xx we have j=2j=2 and coefficient 21/a321/a^3. Since a≠0a\ne0, their ratio is 1/(3a2)=31/(3a^2)=3, giving a=±1/3a=\pm1/3.