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CIE 9709 2025 June Paper 33 Q7

A Level / CIE / P3

CIE 9709 2025 June Paper 33 Paper · Question 7

题目

Problem

Let f(x)=3a−5x(3a+2x)(2a−x)\mathrm{f}(x)=\frac{3a-5x}{(3a+2x)(2a-x)}, where aa is a positive constant.

(a) Express f(x)\mathrm{f}(x) in partial fractions.

(3)

(b) Hence obtain the expansion of f(x)\mathrm{f}(x) in ascending powers of xx, up to and including the term in x2x^2.

(4)

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

(1)
题目中文翻译

设 f(x)=3a−5x(3a+2x)(2a−x)\mathrm{f}(x)=\frac{3a-5x}{(3a+2x)(2a-x)},其中 aa 为正常数。

(a) 将 f(x)\mathrm{f}(x) 表示为部分分式。

(3)

(b) 由此求 f(x)\mathrm{f}(x) 按 xx 的升幂展开的展开式,保留至 x2x^2 项(包括该项)。

(4)

(c) 写出使 (b) 问中的展开式成立的 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

对两个一次因式分别设常数分子,再代入使其中一项消失的值。

答题过程

展开

Write

f(x)=A3a+2x+B2a−x.f(x)=\frac A{3a+2x}+\frac B{2a-x}.

Then

3a−5x=A(2a−x)+B(3a+2x).3a-5x=A(2a-x)+B(3a+2x).

Putting x=2ax=2a gives −7a=7aB-7a=7aB, so B=−1B=-1.

Putting x=−3a/2x=-3a/2 gives 21a/2=7aA/221a/2=7aA/2, so A=3A=3. Hence

f(x)=33a+2x−12a−x.\boxed{f(x)=\frac3{3a+2x}-\frac1{2a-x}}.

(b)

解法一

思路

展开

承接(a),分别提出常数使分母成为一加或一减一个量,再用等比展开。

答题过程

展开

Using (a),

33a+2x= 1a(1+2x3a)−1= 1a−2x3a2+4x29a3+⋯ ,\begin{align*} \frac3{3a+2x} =&\,\frac1a\bigg(1+\frac{2x}{3a}\bigg)^{-1}\\[4mm] =&\,\frac1a-\frac{2x}{3a^2}+\frac{4x^2}{9a^3}+\cdots, \end{align*}

and

−12a−x= −12a(1−x2a)−1= −12a−x4a2−x28a3+⋯ .\begin{align*} -\frac1{2a-x} =&\,-\frac1{2a}\bigg(1-\frac{x}{2a}\bigg)^{-1}\\[4mm] =&\,-\frac1{2a}-\frac{x}{4a^2}-\frac{x^2}{8a^3}+\cdots. \end{align*}

Adding,

f(x)= 12a−(23+14)xa2 +(49−18)x2a3+⋯= 12a−11x12a2+23x272a3+⋯.\begin{align*} f(x)=&\,\frac1{2a} -\bigg(\frac23+\frac14\bigg)\frac{x}{a^2}\\[4mm] &\,\hspace{2pt}+\bigg(\frac49-\frac18\bigg)\frac{x^2}{a^3} +\cdots\\[4mm] =&\,\boxed{\frac1{2a}-\frac{11x}{12a^2} +\frac{23x^2}{72a^3}+\cdots}. \end{align*}

(c)

解法一

思路

展开

两个展开条件都要满足,取较小的半径。

答题过程

展开

The two expansions require

∣2x3a∣<1,∣x2a∣<1.\bigg|\frac{2x}{3a}\bigg|<1,\qquad \bigg|\frac{x}{2a}\bigg|<1.

Since a>0a>0, their intersection is

∣x∣<3a2.\boxed{|x|<\frac{3a}{2}}.