| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2024 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration with Partial Fractions |
| Type | Partial fractions with repeated linear factor |
| Difficulty | Standard +0.3 This is a standard partial fractions question with a repeated linear factor followed by routine integration. Part (a) requires algebraic manipulation to find constants (cover-up method or equating coefficients), and part (b) involves integrating standard forms (ln and power functions) then substituting limits. While it requires multiple steps and careful algebra, it follows a well-practiced procedure with no novel insight required, making it slightly easier than average. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08d Evaluate definite integrals: between limits1.08j Integration using partial fractions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{3x+4}{(x-2)(2x+1)^2} = \frac{A}{x-2}+\frac{B}{2x+1}+\frac{C}{(2x+1)^2}\); multiplies through and substitutes \(x=2\) or \(x=-\frac{1}{2}\) or compares coefficients | M1 | Correct method leading to one constant; implied by correct value of \(A\) by cover-up rule |
| One of \(A=\frac{2}{5},\ B=-\frac{4}{5},\ C=-1\) | A1 | Any one of the three constants correct |
| Obtains all three constants \(A=...,\ B=...,\ C=...\) | M1 | Full correct method to obtain all three constants |
| \(\frac{3x+4}{(x-2)(2x+1)^2}=\frac{2}{5(x-2)}-\frac{4}{5(2x+1)}-\frac{1}{(2x+1)^2}\) or \(A=\frac{2}{5},\ B=-\frac{4}{5},\ C=-1\) | A1 | Correct full partial fraction expression or correct values stated |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int\frac{p}{x-2}dx = \alpha\ln(x-2)\) or \(\int\frac{q}{2x+1}dx=\beta\ln(2x+1)\) | M1 | Integrating to obtain at least one ln term; \(\alpha\ln(5x-10)\) and \(\beta\ln(10x+5)\) equally fine |
| \(\int\frac{k}{(2x+1)^2}dx = \frac{K}{2x+1}\) | M1 | For \(\frac{k}{(2x+1)^2}\to\frac{K}{2x+1}\) (any \(K\)) |
| \(\int_7^{12}\frac{3x+4}{(x-2)(2x+1)^2}dx = \left[\frac{"2"}{5}\ln(x-2)-\frac{"4"}{5}\times\frac{1}{2}\ln(2x+1)+\frac{"1"}{4x+2}\right]_7^{12}\) | A1ft | Fully correct integration following through on their non-zero constants |
| \(= \left(\frac{2}{5}\ln(10)-\frac{2}{5}\ln(25)+\frac{1}{50}\right)-\left(\frac{2}{5}\ln(5)-\frac{2}{5}\ln(15)+\frac{1}{30}\right)=...\) | DM1 | Depends on at least one of first two M's; substitutes limits and subtracts either way round |
| \(= \frac{2}{5}\ln\frac{10}{25}+\frac{1}{50}-\frac{2}{5}\ln\frac{5}{15}-\frac{1}{30}=\frac{2}{5}\ln\frac{10\times15}{25\times5}+...\) | M1 | Achieves single log term with evidence of correct use of at least one log law |
| \(\frac{2}{5}\ln\frac{6}{5}-\frac{1}{75}\) (oe) | A1 | Correct answer; accept equivalents in correct form |
# Question 2(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{3x+4}{(x-2)(2x+1)^2} = \frac{A}{x-2}+\frac{B}{2x+1}+\frac{C}{(2x+1)^2}$; multiplies through and substitutes $x=2$ or $x=-\frac{1}{2}$ or compares coefficients | M1 | Correct method leading to one constant; implied by correct value of $A$ by cover-up rule |
| One of $A=\frac{2}{5},\ B=-\frac{4}{5},\ C=-1$ | A1 | Any one of the three constants correct |
| Obtains all three constants $A=...,\ B=...,\ C=...$ | M1 | Full correct method to obtain all three constants |
| $\frac{3x+4}{(x-2)(2x+1)^2}=\frac{2}{5(x-2)}-\frac{4}{5(2x+1)}-\frac{1}{(2x+1)^2}$ or $A=\frac{2}{5},\ B=-\frac{4}{5},\ C=-1$ | A1 | Correct full partial fraction expression or correct values stated |
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# Question 2(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int\frac{p}{x-2}dx = \alpha\ln(x-2)$ or $\int\frac{q}{2x+1}dx=\beta\ln(2x+1)$ | M1 | Integrating to obtain at least one ln term; $\alpha\ln(5x-10)$ and $\beta\ln(10x+5)$ equally fine |
| $\int\frac{k}{(2x+1)^2}dx = \frac{K}{2x+1}$ | M1 | For $\frac{k}{(2x+1)^2}\to\frac{K}{2x+1}$ (any $K$) |
| $\int_7^{12}\frac{3x+4}{(x-2)(2x+1)^2}dx = \left[\frac{"2"}{5}\ln(x-2)-\frac{"4"}{5}\times\frac{1}{2}\ln(2x+1)+\frac{"1"}{4x+2}\right]_7^{12}$ | A1ft | Fully correct integration following through on their non-zero constants |
| $= \left(\frac{2}{5}\ln(10)-\frac{2}{5}\ln(25)+\frac{1}{50}\right)-\left(\frac{2}{5}\ln(5)-\frac{2}{5}\ln(15)+\frac{1}{30}\right)=...$ | DM1 | Depends on at least one of first two M's; substitutes limits and subtracts either way round |
| $= \frac{2}{5}\ln\frac{10}{25}+\frac{1}{50}-\frac{2}{5}\ln\frac{5}{15}-\frac{1}{30}=\frac{2}{5}\ln\frac{10\times15}{25\times5}+...$ | M1 | Achieves single log term with evidence of correct use of at least one log law |
| $\frac{2}{5}\ln\frac{6}{5}-\frac{1}{75}$ (oe) | A1 | Correct answer; accept equivalents in correct form |
---\begin{enumerate}
\item Given that
\end{enumerate}
$$\frac { 3 x + 4 } { ( x - 2 ) ( 2 x + 1 ) ^ { 2 } } \equiv \frac { A } { x - 2 } + \frac { B } { 2 x + 1 } + \frac { C } { ( 2 x + 1 ) ^ { 2 } }$$
(a) find the values of the constants $A , B$ and $C$.\\
(b) Hence find the exact value of
$$\int _ { 7 } ^ { 12 } \frac { 3 x + 4 } { ( x - 2 ) ( 2 x + 1 ) ^ { 2 } } \mathrm {~d} x$$
giving your answer in the form $p \ln q + r$ where $p$, $q$ and $r$ are rational numbers.
\hfill \mbox{\textit{Edexcel P4 2024 Q2 [10]}}