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CIE 9709 2025 June Paper 35 Q9

A Level / CIE / P3

CIE 9709 2025 June Paper 35 Paper · Question 9

题目

Problem

(a) Express 12x2+55x−2(3x−2)(x+6)\frac{12x^2+55x-2}{(3x-2)(x+6)} in partial fractions.

(5)

(b) Hence obtain the expansion of 12x2+55x−2(3x−2)(x+6)\frac{12x^2+55x-2}{(3x-2)(x+6)} in ascending powers of xx, up to and including the term in x2x^2.

(4)
题目中文翻译

(a) 将 12x2+55x−2(3x−2)(x+6)\frac{12x^2+55x-2}{(3x-2)(x+6)} 表示为部分分式。

(5)

(b) 由此求 12x2+55x−2(3x−2)(x+6)\frac{12x^2+55x-2}{(3x-2)(x+6)} 按 xx 的升幂展开的展开式,保留至 x2x^2 项(包括该项)。

(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

(a)

解法一

思路

展开

分子分母同次,先做除法,再对余下真分式拆分。

答题过程

展开

The denominator expands to 3x2+16x−123x^2+16x-12, so division gives

12x2+55x−2(3x−2)(x+6)=4+−9x+46(3x−2)(x+6).\frac{12x^2+55x-2}{(3x-2)(x+6)} =4+\frac{-9x+46}{(3x-2)(x+6)}.

Write

−9x+46(3x−2)(x+6)=B3x−2+Cx+6.\frac{-9x+46}{(3x-2)(x+6)} =\frac B{3x-2}+\frac C{x+6}.

Then

−9x+46=B(x+6)+C(3x−2).-9x+46=B(x+6)+C(3x-2).

Putting x=2/3x=2/3 gives 40=(20/3)B40=(20/3)B, so B=6B=6. Putting x=−6x=-6 gives 100=−20C100=-20C, so C=−5C=-5.

Therefore

12x2+55x−2(3x−2)(x+6)=4+63x−2−5x+6.\boxed{\frac{12x^2+55x-2}{(3x-2)(x+6)} =4+\frac6{3x-2}-\frac5{x+6}}.

(b)

解法一

思路

展开

承接(a),分别写成等比形式并展开,最后把同次项合并。

答题过程

展开

Using (a),

63x−2= −3(1−32x)−1= −3−92x−274x2+⋯ ,\begin{align*} \frac6{3x-2} =&\,-3\bigg(1-\frac32x\bigg)^{-1}\\[4mm] =&\,-3-\frac92x-\frac{27}{4}x^2+\cdots, \end{align*}

and

−5x+6= −56(1+x6)−1= −56+536x−5216x2+⋯ .\begin{align*} -\frac5{x+6} =&\,-\frac56\bigg(1+\frac x6\bigg)^{-1}\\[4mm] =&\,-\frac56+\frac5{36}x-\frac5{216}x^2+\cdots. \end{align*}

Adding the constant term 44,

f(x)= 4−3−56+(−92+536)x +(−274−5216)x2+⋯= 16−15736x−1463216x2+⋯.\begin{align*} f(x)=&\,4-3-\frac56 +\bigg(-\frac92+\frac5{36}\bigg)x\\ &\,\hspace{2pt}+\bigg(-\frac{27}{4}-\frac5{216}\bigg)x^2+\cdots\\ =&\,\boxed{\frac16-\frac{157}{36}x-\frac{1463}{216}x^2+\cdots}. \end{align*}

The common validity interval is ∣x∣<2/3|x|<2/3.