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CIE 9709 2022 June Paper 31 Q2

A Level / CIE / P3

CIE 9709 2022 June Paper 31 Paper · Question 2

题目

Problem

(a) Expand (2−x2)−2(2-x^2)^{-2} in ascending powers of xx, up to and including the term in x4x^4, simplifying the coefficients.

(4)

(b) State the set of values of xx for which the expansion is valid.

(1)
题目中文翻译

(a) 将 (2−x2)−2(2-x^2)^{-2} 按 xx 的升幂展开,写到并包括 x4x^4 项,并化简各项系数。

(4)

(b) 写出该展开式成立时 xx 的取值集合。

(1)
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

先提出常数 22,把剩余部分写成标准二项式。由于其中变量是 x2x^2,展开到它的二次项就是 x4x^4。

答题过程

展开

The binomial expansion is (1+u)−2=1−2u+3u2+⋯(1+u)^{-2}=1-2u+3u^2+\cdots for ∣u∣<1|u|<1. Taking u=−x2/2u=-x^2/2 gives

(2−x2)−2= 14(1−x22)−2= 14[1+x2+34x4+⋯]= 14+14x2+316x4+⋯.\begin{align*} (2-x^2)^{-2} =&\,\frac14\bigg(1-\frac{x^2}{2}\bigg)^{-2}\\[4mm] =&\,\frac14\bigg[ 1+x^2+\frac34x^4+\cdots\bigg]\\[4mm] =&\,\boxed{\frac14+\frac14x^2+\frac3{16}x^4+\cdots}. \end{align*}

(b)

解法一

思路

展开

非整数或负整数幂的无穷二项展开要求附加量绝对值小于 11。

答题过程

展开

The expansion requires

∣−x22∣< 1,x2< 2.\begin{align*} \bigg|-\frac{x^2}{2}\bigg|<&\,1,\\[4mm] x^2<&\,2. \end{align*}

Therefore −2<x<2\boxed{-\sqrt2<x<\sqrt2}.