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P4 2024 Jan Q2

A Level / Edexcel / P4

IAL 2024 Jan Paper · Question 2

题目

Problem

Given that

 3x+4(x−2)(2x+1)2≡ Ax−2 +B2x+1 +C(2x+1)2.\begin{align*} &\,\frac{3x+4}{(x-2)(2x+1)^2}\\ \equiv&\,\frac{A}{x-2}\\ &\,\hspace{2pt}+\frac{B}{2x+1}\\ &\,\hspace{4pt}+\frac{C}{(2x+1)^2}. \end{align*}

(a) find the values of the constants AA, BB and CC.

(4)

(b) Hence find the exact value of

∫7123x+4(x−2)(2x+1)2 dx\int_7^{12}\frac{3x+4}{(x-2)(2x+1)^2}\,\mathrm{d}x

giving your answer in the form pln⁡q+rp\ln q+r where p,qp,q and rr are rational numbers.

(6)
题目中文翻译

已知

 3x+4(x−2)(2x+1)2≡ Ax−2 +B2x+1 +C(2x+1)2.\begin{align*} &\,\frac{3x+4}{(x-2)(2x+1)^2}\\ \equiv&\,\frac{A}{x-2}\\ &\,\hspace{2pt}+\frac{B}{2x+1}\\ &\,\hspace{4pt}+\frac{C}{(2x+1)^2}. \end{align*}

(a) 求常数 A,B,CA,B,C 的值。

(4)

(b) 进而求

∫7123x+4(x−2)(2x+1)2 dx\int_7^{12}\frac{3x+4}{(x-2)(2x+1)^2}\,\mathrm{d}x

的精确值,并将答案写成 pln⁡q+rp\ln q+r 的形式,其中 p,q,rp,q,r 都是有理数。

(6)

解答

(a)

解法一

思路

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先把恒等式两边同乘公共分母 (x−2)(2x+1)2(x-2)(2x+1)^2。然后代入能让某些因子变成 00 的 xx 值,先求出 AA 和 CC,最后再代入一个方便的值求 BB。

答题过程

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Given

 3x+4(x−2)(2x+1)2≡ Ax−2 +B2x+1 +C(2x+1)2.\begin{align*} &\,\frac{3x+4}{(x-2)(2x+1)^2}\\ \equiv&\,\frac{A}{x-2}\\ &\,\hspace{2pt}+\frac{B}{2x+1}\\ &\,\hspace{4pt}+\frac{C}{(2x+1)^2}. \end{align*}

multiply both sides by (x−2)(2x+1)2(x-2)(2x+1)^2:

3x+4≡ A(2x+1)2 +B(x−2)(2x+1) +C(x−2).\begin{align*} 3x+4\equiv&\,A(2x+1)^2\\ &\,\hspace{2pt}+B(x-2)(2x+1)\\ &\,\hspace{4pt}+C(x-2). \end{align*}

Put x=2x=2:

10= A(5)2A= 25.\begin{align*} 10=&\,A(5)^2\\ A=&\,\frac25. \end{align*}

Put x=−12x=-\frac12:

3(−12)+4= C(−12−2)52= −52CC= −1.\begin{align*} 3\big(-\frac12\big)+4 =&\,C\big(-\frac12-2\big)\\ \frac52=&\,-\frac52C\\ C=&\,-1. \end{align*}

Now put x=0x=0 in the identity:

4= A−2B−2C4= A−2B−2C.\begin{align*} 4=&\,A-2B-2C\\ 4=&\,A-2B-2C. \end{align*}

Using A=25A=\frac25 and C=−1C=-1,

4= 25−2B+2−2B= 85B= −45.\begin{align*} 4=&\,\frac25-2B+2\\ -2B=&\,\frac85\\ B=&\,-\frac45. \end{align*}

Therefore

A= 25,B= −45,C= −1.\begin{align*} A=&\,\boxed{\frac25},\\ B=&\,\boxed{-\frac45},\\ C=&\,\boxed{-1}. \end{align*}

解法二

思路

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也可以比较系数。这个方法计算稍多,但它适合检查部分分式答案是否完整一致。

答题过程

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From

3x+4≡ A(2x+1)2 +B(x−2)(2x+1) +C(x−2).\begin{align*} 3x+4\equiv&\,A(2x+1)^2\\ &\,\hspace{2pt}+B(x-2)(2x+1)\\ &\,\hspace{4pt}+C(x-2). \end{align*}

expand each product:

(2x+1)2=4x2+4x+1,(2x+1)^2=4x^2+4x+1, (x−2)(2x+1)=2x2−3x−2.(x-2)(2x+1)=2x^2-3x-2.

Thus the coefficients on the right are

x2: 4A+2B,x: 4A−3B+C,1: A−2B−2C.\begin{align*} x^2:\quad&\,4A+2B,\\ x:\quad&\,4A-3B+C,\\ 1:\quad&\,A-2B-2C. \end{align*}

Compare coefficients:

4A+2B= 0,4A−3B+C= 3,A−2B−2C= 4.\begin{align*} 4A+2B=&\,0,\\ 4A-3B+C=&\,3,\\ A-2B-2C=&\,4. \end{align*}

From 4A+2B=04A+2B=0,

B=−2A.B=-2A.

Substitute this into the other two equations:

10A+C= 3,5A−2C= 4.\begin{align*} 10A+C=&\,3,\\ 5A-2C=&\,4. \end{align*}

From C=3−10AC=3-10A, the second equation gives

5A−2(3−10A)= 425A= 10A= 25.\begin{align*} 5A-2(3-10A)=&\,4\\ 25A=&\,10\\ A=&\,\frac25. \end{align*}

Thus

B=−45,C=3−10(25)=−1.B=-\frac45, \qquad C=3-10\big(\frac25\big)=-1.

So

A= 25,B= −45,C= −1.\begin{align*} A=&\,\boxed{\frac25},\\ B=&\,\boxed{-\frac45},\\ C=&\,\boxed{-1}. \end{align*}

(b)

解法一

思路

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“Hence” 表示要使用 (a) 的部分分式结果。把 integrand 分成三个较简单的项后分别积分:前两个给出对数项,最后一个是 (2x+1)−2(2x+1)^{-2} 的积分。最后按题目要求把对数合并成单个 ln⁡q\ln q。

答题过程

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Using part (a), let

I=∫7123x+4(x−2)(2x+1)2 dx.I=\int_7^{12} \frac{3x+4}{(x-2)(2x+1)^2}\,\mathrm{d}x.

The partial fractions give

 3x+4(x−2)(2x+1)2= 25(x−2) −45(2x+1) −1(2x+1)2.\begin{align*} &\,\frac{3x+4}{(x-2)(2x+1)^2}\\ =&\,\frac{2}{5(x-2)}\\ &\,\hspace{2pt}-\frac{4}{5(2x+1)}\\ &\,\hspace{4pt}-\frac{1}{(2x+1)^2}. \end{align*}

The required antiderivatives are

∫1x−2 dx=ln⁡∣x−2∣+C,\int\frac{1}{x-2}\,\mathrm{d}x =\ln|x-2|+C, ∫12x+1 dx=12ln⁡∣2x+1∣+C,\int\frac{1}{2x+1}\,\mathrm{d}x =\frac12\ln|2x+1|+C,

and

∫(2x+1)−2 dx=−12(2x+1)+C.\int(2x+1)^{-2}\,\mathrm{d}x =-\frac{1}{2(2x+1)}+C.

Since 7≤x≤127\leq x\leq12, both logarithm arguments are positive. An antiderivative is

F(x)= 25ln⁡(x−2) −25ln⁡(2x+1) +12(2x+1).\begin{align*} F(x)=&\,\frac25\ln(x-2)\\ &\,\hspace{2pt}-\frac25\ln(2x+1)\\ &\,\hspace{4pt}+\frac{1}{2(2x+1)}. \end{align*}

At the two limits,

F(12)= 25ln⁡10−25ln⁡25 +150= 25ln⁡25+150,\begin{align*} F(12)=&\,\frac25\ln10-\frac25\ln25\\ &\,\hspace{2pt}+\frac1{50}\\ =&\,\frac25\ln\frac25+\frac1{50}, \end{align*}

and

F(7)= 25ln⁡5−25ln⁡15 +130= 25ln⁡13+130.\begin{align*} F(7)=&\,\frac25\ln5-\frac25\ln15\\ &\,\hspace{2pt}+\frac1{30}\\ =&\,\frac25\ln\frac13+\frac1{30}. \end{align*}

The log difference is

ln⁡25−ln⁡13=ln⁡(25×3)=ln⁡65,\ln\frac25-\ln\frac13 =\ln\bigg(\frac25\times3\bigg) =\ln\frac65,

and

150−130=−175.\frac1{50}-\frac1{30}=-\frac1{75}.

Therefore

I= F(12)−F(7)= 25ln⁡65−175.\begin{align*} I=&\,F(12)-F(7)\\ =&\,\boxed{\frac25\ln\frac65-\frac1{75}}. \end{align*}

也可以先合并后两项,再拆分分子。令

J=∫8x+9(2x+1)2 dx.J=\int\frac{8x+9}{(2x+1)^2}\,\mathrm{d}x.

Using 8x+9=4(2x+1)+58x+9=4(2x+1)+5,

J= 4∫12x+1 dx +5∫(2x+1)−2 dx= 2ln⁡∣2x+1∣−52(2x+1)+C.\begin{align*} J=&\,4\int\frac{1}{2x+1}\,\mathrm{d}x\\ &\,\hspace{2pt}+5\int(2x+1)^{-2}\,\mathrm{d}x\\ =&\,2\ln|2x+1|-\frac5{2(2x+1)}+C. \end{align*}

The last two partial fractions contribute −15J-\frac15J, which gives the same F(x)F(x) and hence the same value of II.