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CIE 9709 2025 November Paper 32 Q9

A Level / CIE / P3

CIE 9709 2025 Nov Paper 32 Paper · Question 9

题目

Problem

Let f(x)=x2+4ax+6a2(x+2a)(x+3a)\mathrm{f}(x)=\frac{x^2+4ax+6a^2}{(x+2a)(x+3a)}, where aa is a positive constant.

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

(5)

(b) Hence find the exact value of ∫−aaf(x) dx\displaystyle\int_{-a}^a\mathrm{f}(x)\,\mathrm{d}x. Give your answer in the form a(p+ln⁡q)a(p+\ln q), where pp and qq are rational.

(4)
题目中文翻译

设 f(x)=x2+4ax+6a2(x+2a)(x+3a)\mathrm{f}(x)=\frac{x^2+4ax+6a^2}{(x+2a)(x+3a)},其中 aa 为正常数。

(a) 将 f(x)\mathrm{f}(x) 表示为部分分式。

(5)

(b) 由此求 ∫−aaf(x) dx\displaystyle\int_{-a}^a\mathrm{f}(x)\,\mathrm{d}x 的精确值。答案写成 a(p+ln⁡q)a(p+\ln q) 的形式,其中 pp 和 qq 为有理数。

(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,先做除法取得常数项,再对两个一次因式拆分。

答题过程

展开

Since

(x+2a)(x+3a)=x2+5ax+6a2,(x+2a)(x+3a)=x^2+5ax+6a^2,

we have

f(x)=1−ax(x+2a)(x+3a).f(x)=1-\frac{ax}{(x+2a)(x+3a)}.

Write

−ax(x+2a)(x+3a)=Bx+2a+Cx+3a.-\frac{ax}{(x+2a)(x+3a)} =\frac{B}{x+2a}+\frac{C}{x+3a}.

Then

−ax=B(x+3a)+C(x+2a).-ax=B(x+3a)+C(x+2a).

Putting x=−2ax=-2a gives 2a2=aB2a^2=aB, so B=2aB=2a. Putting x=−3ax=-3a gives 3a2=−aC3a^2=-aC, so C=−3aC=-3a.

Thus

f(x)=1+2ax+2a−3ax+3a.\boxed{f(x)=1+\frac{2a}{x+2a}-\frac{3a}{x+3a}}.

(b)

解法一

思路

展开

在给定区间两个一次因式都为正,可以直接写对数。代入上下限后先合并同一系数的对数,再合成题目要求的一个对数。

答题过程

展开

From (a), since a>0a>0 and −a⩽x⩽a-a\leqslant x\leqslant a, both x+2ax+2a and x+3ax+3a are positive. Therefore

∫−aaf(x) dx= 2a+2a[ln⁡(3a)−ln⁡a] −3a[ln⁡(4a)−ln⁡(2a)]= 2a+2aln⁡3aa−3aln⁡4a2a= a(2+2ln⁡3−3ln⁡2)= a(2+ln⁡98).\begin{align*} \int_{-a}^a f(x)\,\mathrm{d}x =&\,2a+2a\big[\ln(3a)-\ln a\big]\\[4mm] &\,\hspace{2pt}-3a\big[\ln(4a)-\ln(2a)\big]\\[4mm] =&\,2a+2a\ln\frac{3a}{a}-3a\ln\frac{4a}{2a}\\ =&\,a(2+2\ln3-3\ln2)\\ =&\,\boxed{a\bigg(2+\ln\frac98\bigg)}. \end{align*}