| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Partial fractions with linear factors – decompose, integrate, and expand as series |
| Difficulty | Standard +0.3 This is a standard C4 partial fractions question with routine integration and binomial expansion. Part (a) is straightforward decomposition, part (b) requires integrating logarithms and simplifying (standard but requires care), and part (c) applies binomial expansion to each fraction separately. All techniques are textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 1.02y Partial fractions: decompose rational functions1.04c Extend binomial expansion: rational n, |x|<11.08j Integration using partial fractions |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1+14x}{(1-x)(1+2x)} \equiv \frac{A}{1-x} + \frac{B}{1+2x}\) and attempt \(A\) and or \(B\) | M1 | |
| \(A = 5, B = -4\) | A1, A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int \left(\frac{5}{1-x} - \frac{4}{1+2x}\right) dx = [-5\ln | 1-x | - 2\ln |
| \(= \left(-5\ln\frac{2}{3} - 2\ln\frac{5}{3}\right) - \left(-5\ln\frac{5}{6} - 2\ln\frac{4}{3}\right)\) | M1 | |
| \(= 5\ln\frac{5}{4} + 2\ln\frac{4}{5}\) | ||
| \(= 3\ln\frac{5}{4} = \ln\frac{125}{64}\) | M1 A1 | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| \(5(1-x)^{-1} - 4(1+2x)^{-1}\) | B1 | |
| \(= 5(1 + x + x^2 + x^3) - 4\left(1 - 2x + \frac{(-1)(-2)(2x)^2}{2} + \frac{(-1)(-2)(-3)(2x)^3}{6} + \ldots\right)\) | M1 A1 | |
| \(= 1 + 13x - 11x^2 + 37x^3 + \ldots\) | M1 A1 | (5) |
## (a)
$\frac{1+14x}{(1-x)(1+2x)} \equiv \frac{A}{1-x} + \frac{B}{1+2x}$ and attempt $A$ and or $B$ | M1 |
$A = 5, B = -4$ | A1, A1 | (3)
## (b)
$\int \left(\frac{5}{1-x} - \frac{4}{1+2x}\right) dx = [-5\ln|1-x| - 2\ln|1+2x|]$ | M1 A1 |
$= \left(-5\ln\frac{2}{3} - 2\ln\frac{5}{3}\right) - \left(-5\ln\frac{5}{6} - 2\ln\frac{4}{3}\right)$ | M1 |
$= 5\ln\frac{5}{4} + 2\ln\frac{4}{5}$ | |
$= 3\ln\frac{5}{4} = \ln\frac{125}{64}$ | M1 A1 | (5)
## (c)
$5(1-x)^{-1} - 4(1+2x)^{-1}$ | B1 |
$= 5(1 + x + x^2 + x^3) - 4\left(1 - 2x + \frac{(-1)(-2)(2x)^2}{2} + \frac{(-1)(-2)(-3)(2x)^3}{6} + \ldots\right)$ | M1 A1 |
$= 1 + 13x - 11x^2 + 37x^3 + \ldots$ | M1 A1 | (5)
**Total: (13 marks)**
---$$f(x) = \frac{1 + 14x}{(1 - x)(1 + 2x)}, \quad |x| < \frac{1}{2}.$$
\begin{enumerate}[label=(\alph*)]
\item Express $f(x)$ in partial fractions.
[3]
\item Hence find the exact value of $\int_{-\frac{1}{6}}^{\frac{1}{4}} f(x) \, dx$, giving your answer in the form $\ln p$, where $p$ is rational.
[5]
\item Use the binomial theorem to expand $f(x)$ in ascending powers of $x$, up to and including the term in $x^5$, simplifying each term.
[5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q3 [13]}}