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CIE 9709 2022 June Paper 32 Q8

A Level / CIE / P3

CIE 9709 2022 June Paper 32 Paper · Question 8

题目

Problem

Let

f(x)=x2+9x(3x−1)(x2+3).\mathrm{f}(x)=\frac{x^2+9x}{(3x-1)(x^2+3)}.

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

(5)

(b) Hence find ∫13f(x) dx\displaystyle\int_1^3\mathrm{f}(x)\,\mathrm{d}x, giving your answer in a simplified exact form.

(5)
题目中文翻译

设

f(x)=x2+9x(3x−1)(x2+3).\mathrm{f}(x)=\frac{x^2+9x}{(3x-1)(x^2+3)}.

(a) 将 f(x)\mathrm{f}(x) 分解为部分分式。

(5)

(b) 由此求 ∫13f(x) dx\displaystyle\int_1^3\mathrm{f}(x)\,\mathrm{d}x,答案以化简后的精确形式表示。

(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

(a)

解法一

思路

展开

二次因子的分子先设一次式。清分母后代入 x=1/3x=1/3,再比较常数项与二次项。

答题过程

展开

Write

f(x)=A3x−1+Bx+Cx2+3.\mathrm f(x)=\frac{A}{3x-1}+\frac{Bx+C}{x^2+3}.

Then

x2+9x= A(x2+3) +(Bx+C)(3x−1).\begin{align*} x^2+9x=&\,A(x^2+3)\\[4mm] &\,\hspace{2pt}+(Bx+C)(3x-1). \end{align*}

At x=1/3x=1/3, 28/9=A(28/9)28/9=A(28/9), so A=1A=1.

Comparing constant terms gives 0=3A−C0=3A-C, so C=3C=3. Comparing coefficients of x2x^2 gives 1=A+3B1=A+3B, so B=0B=0.

Therefore

f(x)=13x−1+3x2+3.\boxed{\mathrm f(x)=\frac1{3x-1}+\frac3{x^2+3}}.

(b)

解法一

思路

展开

承接(a)分别积分得到对数与反正切;上下限给出两个特殊角,最后合并对数。

答题过程

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Using part (a), the derivatives of ln⁡(3x−1)\ln(3x-1) and tan⁡−1(x/3)\tan^{-1}(x/\sqrt3) are 3/(3x−1)3/(3x-1) and 3/(x2+3)\sqrt3/(x^2+3) respectively. Their integral coefficients are therefore 1/31/3 and 3\sqrt3. Also 3x−1>03x-1>0 on 1≤x≤31\le x\le3. Hence

I= [13ln⁡(3x−1)+3tan⁡−1x3]13= 13(ln⁡8−ln⁡2)+3(π3−π6)= 23ln⁡2+3π6.\begin{align*} I =&\,\bigg[\frac13\ln(3x-1) +\sqrt3\tan^{-1}\frac{x}{\sqrt3}\bigg]_1^3\\[4mm] =&\,\frac13(\ln8-\ln2) +\sqrt3\bigg(\frac\pi3-\frac\pi6\bigg)\\[4mm] =&\,\boxed{\frac23\ln2+\frac{\sqrt3\pi}{6}}. \end{align*}