| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration with Partial Fractions |
| Type | Partial fractions to find specific parameter |
| Difficulty | Challenging +1.2 This is a two-part question requiring partial fraction decomposition followed by integration to find a parameter. While it involves multiple steps (decomposition, integration of logarithms, and solving for k), each step follows standard A-level procedures without requiring novel insight. The integration bounds avoid the discontinuity at x=2, making it straightforward. This is moderately harder than average due to the reverse-engineering aspect of finding k from a given integral value, but remains a standard Further Maths Pure question type. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08j Integration using partial fractions |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{3kx-18}{(x+4)(x-2)} \equiv \frac{A}{x+4}+\frac{B}{x-2} \Rightarrow 3kx-18 \equiv A(x-2)+B(x+4)\) | B1 | Correct form and correct corresponding identity |
| \(6k-18=6B \Rightarrow B=\ldots\) or \(-12k-18=-6A \Rightarrow A=\ldots\) or comparing coefficients to find \(A\) or \(B\) | M1 | Substitute \(x=2\) or \(x=-4\); or expand, collect and compare coefficients |
| \(\frac{2k+3}{x+4}+\frac{k-3}{x-2}\) | A1 | Correct partial fractions; correct answer implies B1M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\int\left(\frac{\text{"2k+3"}}{x+4}+\frac{\text{"k-3"}}{x-2}\right)dx = \ldots\ln(x+4)+\ldots\ln(x-2)\) | M1 | Score for either term integrating to appropriate \(\ln\) form |
| \((\text{"2k+3"})\ln(x+4)+(\text{"k-3"})\ln(x-2)\) | A1ft | Follow through numerators in terms of \(k\); condone missing brackets |
| \((\text{"2k+3"})\ln 5 - (\text{"k-3"})\ln 5 \Rightarrow (\text{"k+6"})\ln 5 = 21 \Rightarrow k=\ldots\) | dM1 | Must have integrated, substituted correct limits, subtracted, set \(=21\), solve for \(k\) |
| \(k = \dfrac{21}{\ln 5}-6\) | A1 | Exact equivalents e.g. \(\frac{21-6\ln 5}{\ln 5}\), \(\frac{21-3\ln 25}{\ln 5}\) acceptable |
# Question 10(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{3kx-18}{(x+4)(x-2)} \equiv \frac{A}{x+4}+\frac{B}{x-2} \Rightarrow 3kx-18 \equiv A(x-2)+B(x+4)$ | **B1** | Correct form and correct corresponding identity |
| $6k-18=6B \Rightarrow B=\ldots$ or $-12k-18=-6A \Rightarrow A=\ldots$ **or** comparing coefficients to find $A$ or $B$ | **M1** | Substitute $x=2$ or $x=-4$; or expand, collect and compare coefficients |
| $\frac{2k+3}{x+4}+\frac{k-3}{x-2}$ | **A1** | Correct partial fractions; correct answer implies B1M1A1 |
**(3 marks)**
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# Question 10(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\int\left(\frac{\text{"2k+3"}}{x+4}+\frac{\text{"k-3"}}{x-2}\right)dx = \ldots\ln(x+4)+\ldots\ln(x-2)$ | **M1** | Score for either term integrating to appropriate $\ln$ form |
| $(\text{"2k+3"})\ln(x+4)+(\text{"k-3"})\ln(x-2)$ | **A1ft** | Follow through numerators in terms of $k$; condone missing brackets |
| $(\text{"2k+3"})\ln 5 - (\text{"k-3"})\ln 5 \Rightarrow (\text{"k+6"})\ln 5 = 21 \Rightarrow k=\ldots$ | **dM1** | Must have integrated, substituted correct limits, subtracted, set $=21$, solve for $k$ |
| $k = \dfrac{21}{\ln 5}-6$ | **A1** | Exact equivalents e.g. $\frac{21-6\ln 5}{\ln 5}$, $\frac{21-3\ln 25}{\ln 5}$ acceptable |
**(4 marks)**
---\begin{enumerate}
\item $\mathrm { f } ( x ) = \frac { 3 k x - 18 } { ( x + 4 ) ( x - 2 ) } \quad$ where $k$ is a positive constant\\
(a) Express $\mathrm { f } ( x )$ in partial fractions in terms of $k$.\\
(b) Hence find the exact value of $k$ for which
\end{enumerate}
$$\int _ { - 3 } ^ { 1 } f ( x ) d x = 21$$
\hfill \mbox{\textit{Edexcel Paper 2 2023 Q10 [7]}}