| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Session | Specimen |
| Marks | 11 |
| 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 C4 partial fractions question with a repeated linear factor. Part (a) requires routine algebraic manipulation (cover-up method or substitution), and part (b) involves straightforward integration of the partial fractions form followed by logarithm evaluation. While it requires multiple steps and careful arithmetic, it follows a well-practiced procedure with no novel insight needed, making it slightly easier than average. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08j Integration using partial fractions |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(11x - 1 \equiv A(2+3x) + B(1-x)(2+3x) + C(1-x)^2\) | ||
| Putting \(x=1 \Rightarrow A = 2\) | B1 | |
| Putting \(x = -\frac{2}{3} \Rightarrow -\frac{25}{3} = \frac{25}{9}C \Rightarrow C = -3\) | B1 | |
| cf \(x^2\): \(0 = -3B + C \Rightarrow B = -1\) | M1 A1 | (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(\int_0^{\frac{1}{2}} \frac{2}{(1-x)^2} - \frac{1}{(1-x)} - \frac{3}{(2+3x)}\,dx\) | ||
| \(= \left[\frac{2}{1-x} + \ln | 1-x | - \ln |
| \(= \left[4 + \ln\frac{1}{2} - \ln 3\frac{1}{2}\right] - (2 - \ln 2)\) | M1 | |
| \(= 2 + \ln\frac{\frac{1}{2}\times 2}{3\frac{1}{2}}\) | M1 | |
| \(= 2 + \ln\frac{2}{7}\) | A1 | (7 marks) |
# Question 6:
## Part (a):
| Working | Marks | Notes |
|---------|-------|-------|
| $11x - 1 \equiv A(2+3x) + B(1-x)(2+3x) + C(1-x)^2$ | | |
| Putting $x=1 \Rightarrow A = 2$ | B1 | |
| Putting $x = -\frac{2}{3} \Rightarrow -\frac{25}{3} = \frac{25}{9}C \Rightarrow C = -3$ | B1 | |
| cf $x^2$: $0 = -3B + C \Rightarrow B = -1$ | M1 A1 | (4 marks) |
## Part (b):
| Working | Marks | Notes |
|---------|-------|-------|
| $\int_0^{\frac{1}{2}} \frac{2}{(1-x)^2} - \frac{1}{(1-x)} - \frac{3}{(2+3x)}\,dx$ | | |
| $= \left[\frac{2}{1-x} + \ln|1-x| - \ln|2+3x|\right]$ | M1 A1ft A1ft A1ft | |
| $= \left[4 + \ln\frac{1}{2} - \ln 3\frac{1}{2}\right] - (2 - \ln 2)$ | M1 | |
| $= 2 + \ln\frac{\frac{1}{2}\times 2}{3\frac{1}{2}}$ | M1 | |
| $= 2 + \ln\frac{2}{7}$ | A1 | (7 marks) |
---6. Given that
$$\frac { 11 x - 1 } { ( 1 - x ) ^ { 2 } ( 2 + 3 x ) } \equiv \frac { A } { ( 1 - x ) ^ { 2 } } + \frac { B } { ( 1 - x ) } + \frac { C } { ( 2 + 3 x ) }$$
\begin{enumerate}[label=(\alph*)]
\item find the values of $A , B$ and $C$.
\item Find the exact value of $\int _ { 0 } ^ { \frac { 1 } { 2 } } \frac { 11 x - 1 } { ( 1 - x ) ^ { 2 } ( 2 + 3 x ) } \mathrm { d } x$, giving your answer in the form $k + \ln a$, where $k$ is an integer and $a$ is a simplified fraction.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q6 [11]}}