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CIE 9709 2024 June Paper 33 Q5

A Level / CIE / P3

CIE 9709 2024 June Paper 33 Paper · Question 5

题目

Problem

Express 6x2−2x+2(x−1)(2x+1)\dfrac{6x^2-2x+2}{(x-1)(2x+1)} in partial fractions.

(5)
题目中文翻译

将 6x2−2x+2(x−1)(2x+1)\dfrac{6x^2-2x+2}{(x-1)(2x+1)} 表示为部分分式。

(5)
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

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分子、分母同次,先设常数项加两个简单分式。代入使一个分母因子为零的值求系数,再比较最高次项。

答题过程

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Write

6x2−2x+2(x−1)(2x+1)=A+Bx−1+C2x+1.\frac{6x^2-2x+2}{(x-1)(2x+1)} =A+\frac{B}{x-1}+\frac{C}{2x+1}.

Multiplying by (x−1)(2x+1)(x-1)(2x+1),

6x2−2x+2= A(x−1)(2x+1) +B(2x+1)+C(x−1).\begin{align*} 6x^2-2x+2=&\,A(x-1)(2x+1)\\[4mm] &\,\hspace{2pt}+B(2x+1)+C(x-1). \end{align*}

Comparing coefficients of x2x^2 gives 2A=62A=6, so A=3A=3.

Putting x=1x=1 gives 6=3B6=3B, so B=2B=2. Putting x=−1/2x=-1/2 gives 9/2=−3C/29/2=-3C/2, so C=−3C=-3.

Therefore

6x2−2x+2(x−1)(2x+1)=3+2x−1−32x+1.\boxed{\frac{6x^2-2x+2}{(x-1)(2x+1)} =3+\frac{2}{x-1}-\frac{3}{2x+1}}.