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CIE 9709 2024 June Paper 32 Q2

A Level / CIE / P3

CIE 9709 2024 June Paper 32 Paper · Question 2

题目

Problem

Express 6x2−9x−162x2−5x−12\dfrac{6x^2-9x-16}{2x^2-5x-12} in partial fractions.

(5)
题目中文翻译

将 6x2−9x−162x2−5x−12\dfrac{6x^2-9x-16}{2x^2-5x-12} 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,先除得常数,再分解二次分母并拆成两个简单分式。

答题过程

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Factoring the denominator,

2x2−5x−12= 2x2+3x−8x−12= x(2x+3)−4(2x+3)= (2x+3)(x−4).\begin{align*} 2x^2-5x-12 =&\,2x^2+3x-8x-12\\[4mm] =&\,x(2x+3)-4(2x+3)\\[4mm] =&\,(2x+3)(x-4). \end{align*}

Polynomial division gives

6x2−9x−162x2−5x−12=3+6x+20(2x+3)(x−4).\frac{6x^2-9x-16}{2x^2-5x-12} =3+\frac{6x+20}{(2x+3)(x-4)}.

Write the remainder fraction as B/(2x+3)+C/(x−4)B/(2x+3)+C/(x-4). Then

6x+20=B(x−4)+C(2x+3).6x+20=B(x-4)+C(2x+3).

Putting x=−3/2x=-3/2 gives 11=−11B/211=-11B/2, so B=−2B=-2. Putting x=4x=4 gives 44=11C44=11C, so C=4C=4.

Hence

6x2−9x−162x2−5x−12=3−22x+3+4x−4.\boxed{\frac{6x^2-9x-16}{2x^2-5x-12} =3-\frac2{2x+3}+\frac4{x-4}}.