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CIE 9709 2024 June Paper 31 Q1

A Level / CIE / P3

CIE 9709 2024 June Paper 31 Paper · Question 1

题目

Problem

Expand (3+x)(1−2x)12(3+x)(1-2x)^{\frac12} in ascending powers of xx, up to and including the term in x2x^2, simplifying the coefficients.

(4)
题目中文翻译

将 (3+x)(1−2x)12(3+x)(1-2x)^{\frac12} 按 xx 的升幂展开，直至并包括 x2x^2 项，并化简各项系数。

(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

先展开平方根到二次项，再乘一次因子并按幂次合并。

答题过程

展开

The binomial expansion gives

(1−2x)1/2= 1+12(−2x)+(1/2)(−1/2)2!(−2x)2+⋯= 1−x−12x2+⋯ .\begin{align*} (1-2x)^{1/2} =&\,1+\frac12(-2x) +\frac{(1/2)(-1/2)}{2!}(-2x)^2+\cdots\\ =&\,1-x-\frac12x^2+\cdots. \end{align*}

Hence

(3+x)(1−2x)1/2= (3+x)(1−x−12x2+⋯)= 3+(−3+1)x+(−32−1)x2+⋯= 3−2x−52x2+⋯.\begin{align*} (3+x)(1-2x)^{1/2} =&\,(3+x)\bigg(1-x-\frac12x^2+\cdots\bigg)\\ =&\,3+(-3+1)x+\bigg(-\frac32-1\bigg)x^2+\cdots\\ =&\,\boxed{3-2x-\frac52x^2+\cdots}. \end{align*}