| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem and Partial Fractions |
| Type | Improper fraction partial fractions |
| Difficulty | Standard +0.3 This is a standard C4 question testing partial fractions, integration using logarithms, and binomial series expansion. Part (a) is routine partial fraction decomposition, part (b) applies standard integration of the decomposed form, and part (c) requires binomial expansion of two terms and collecting coefficients. While multi-step with 14 marks total, each component uses well-practiced techniques 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 |
|---|---|
| \(x(3x-7) = A(1-x)(1-3x) + B(1-3x) + C(1-x)\) | M1 |
| \(x=1 \Rightarrow -4=-2B \Rightarrow B=2\) | A1 |
| \(x=\frac{1}{3} \Rightarrow -2=\frac{2}{3}C \Rightarrow C=-3\) | A1 |
| coeffs \(x^2 \Rightarrow 3=3A \Rightarrow A=1\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \int_0^1 \left(1+\frac{2}{1-x}-\frac{3}{1-3x}\right) dx = \left[x-2\ln | 1-x | +\ln |
| \(= \left(\frac{1}{4}-2\ln\frac{1}{4}+\ln\frac{1}{4}\right)-(0)\) | M1 | |
| \(= \frac{1}{4}+\ln\frac{16}{9}+\ln\frac{1}{4} = \frac{1}{4}+\ln\frac{9}{4}\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(x) = 1+2(1-x)^{-1}-3(1-3x)^{-1}\) | M1 | |
| \((1-x)^{-1} = 1+x+x^2+x^3+\ldots\) | B1 | |
| \((1-3x)^{-1} = 1+3x+(3x)^2+(3x)^3+\ldots = 1+3x+9x^2+27x^3+\ldots\) | M1 A1 | |
| \(\therefore f(x) = 1+2(1+x+x^2+x^3+\ldots)-3(1+3x+9x^2+27x^3+\ldots)\) | M1 | |
| \(= -7x-25x^2-79x^3+\ldots\) | A1 | (14 marks) |
## (a)
$x(3x-7) = A(1-x)(1-3x) + B(1-3x) + C(1-x)$ | M1 |
$x=1 \Rightarrow -4=-2B \Rightarrow B=2$ | A1 |
$x=\frac{1}{3} \Rightarrow -2=\frac{2}{3}C \Rightarrow C=-3$ | A1 |
coeffs $x^2 \Rightarrow 3=3A \Rightarrow A=1$ | A1 |
## (b)
$= \int_0^1 \left(1+\frac{2}{1-x}-\frac{3}{1-3x}\right) dx = \left[x-2\ln|1-x|+\ln|1-3x|\right]_0^1$ | M1 A1 |
$= \left(\frac{1}{4}-2\ln\frac{1}{4}+\ln\frac{1}{4}\right)-(0)$ | M1 |
$= \frac{1}{4}+\ln\frac{16}{9}+\ln\frac{1}{4} = \frac{1}{4}+\ln\frac{9}{4}$ | M1 A1 |
## (c)
$f(x) = 1+2(1-x)^{-1}-3(1-3x)^{-1}$ | M1 |
$(1-x)^{-1} = 1+x+x^2+x^3+\ldots$ | B1 |
$(1-3x)^{-1} = 1+3x+(3x)^2+(3x)^3+\ldots = 1+3x+9x^2+27x^3+\ldots$ | M1 A1 |
$\therefore f(x) = 1+2(1+x+x^2+x^3+\ldots)-3(1+3x+9x^2+27x^3+\ldots)$ | M1 |
$= -7x-25x^2-79x^3+\ldots$ | A1 | **(14 marks)**
---
**Total: 75 marks**$$\text{f}(x) = \frac{x(3x-7)}{(1-x)(1-3x)}, \quad |x| < \frac{1}{3}.$$
\begin{enumerate}[label=(\alph*)]
\item Find the values of the constants $A$, $B$ and $C$ such that
$$\text{f}(x) = A + \frac{B}{1-x} + \frac{C}{1-3x}.$$ [4]
\item Evaluate
$$\int_0^{\frac{1}{4}} \text{f}(x) \, dx,$$
giving your answer in the form $p + \ln q$, where $p$ and $q$ are rational. [5]
\item Find the series expansion of f(x) in ascending powers of $x$ up to and including the term in $x^3$, simplifying each coefficient. [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q8 [14]}}