| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 14 |
| 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 straightforward linear factors, followed by routine integration and binomial expansion. All parts follow predictable textbook patterns with no novel problem-solving 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 |
|---|---|---|
| \((a)\) \(\frac{8 - x}{(1+x)(2-x)} = \frac{A}{1+x} + \frac{B}{2-x}\) | ||
| \(8 - x \equiv A(2 - x) + B(1 + x)\) | M1 | |
| \(x = -1 \Rightarrow 9 = 3A \Rightarrow A = 3\) | A1 | |
| \(x = 2 \Rightarrow 6 = 3B \Rightarrow B = 2 \therefore f(x) = \frac{3}{1+x} + \frac{2}{2-x}\) | A1 | |
| \((b)\) \(= \int_0^1 \left(\frac{3}{1+x} + \frac{2}{2-x}\right) dx = [3\ln | 1+x | - 2\ln |
| \(= (3\ln\frac{3}{2} - 2\ln 1) - (0 - 2\ln 2)\) | M1 | |
| \(= \ln\frac{3}{2} + \ln 4 = \ln 6\) | M1 A1 | |
| \((c)\) \(f(x) = 3(1 + x)^{-1} + 2(2 - x)^{-1}\) | B1 | |
| \((1 + x)^{-1} = 1 - x + x^2 - x^3 + \ldots\) | M1 | |
| \((2 - x)^{-1} = 2^{-1}(1 - \frac{x}{2})^{-1}\) | M1 | |
| \(= \frac{1}{2}[1 + (-1)(-\frac{x}{2}) + \frac{(-1)(-2)(-\frac{x}{2})^2}{2!} + \ldots] = \frac{1}{2}(1 + \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{8} + \ldots)\) | M1 | |
| \(= \frac{1}{2}(1 + x + \frac{1}{4}x^2 + \frac{1}{8}x^3 + \ldots)\) | A1 | |
| \(\therefore f(x) = 3(1 - x + x^2 - x^3 + \ldots) + (1 + \frac{1}{2}x + \frac{1}{4}x^2 + \frac{1}{8}x^3 + \ldots)\) | M1 | |
| \(= 4 - \frac{5}{2}x + \frac{13}{4}x^2 - \frac{23}{8}x^3 + \ldots\) | A1 | (14 marks) |
$(a)$ $\frac{8 - x}{(1+x)(2-x)} = \frac{A}{1+x} + \frac{B}{2-x}$ | |
$8 - x \equiv A(2 - x) + B(1 + x)$ | M1 |
$x = -1 \Rightarrow 9 = 3A \Rightarrow A = 3$ | A1 |
$x = 2 \Rightarrow 6 = 3B \Rightarrow B = 2 \therefore f(x) = \frac{3}{1+x} + \frac{2}{2-x}$ | A1 |
$(b)$ $= \int_0^1 \left(\frac{3}{1+x} + \frac{2}{2-x}\right) dx = [3\ln|1+x| - 2\ln|2-x|]_0^1$ | M1 A1 |
$= (3\ln\frac{3}{2} - 2\ln 1) - (0 - 2\ln 2)$ | M1 |
$= \ln\frac{3}{2} + \ln 4 = \ln 6$ | M1 A1 |
$(c)$ $f(x) = 3(1 + x)^{-1} + 2(2 - x)^{-1}$ | B1 |
$(1 + x)^{-1} = 1 - x + x^2 - x^3 + \ldots$ | M1 |
$(2 - x)^{-1} = 2^{-1}(1 - \frac{x}{2})^{-1}$ | M1 |
$= \frac{1}{2}[1 + (-1)(-\frac{x}{2}) + \frac{(-1)(-2)(-\frac{x}{2})^2}{2!} + \ldots] = \frac{1}{2}(1 + \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{8} + \ldots)$ | M1 |
$= \frac{1}{2}(1 + x + \frac{1}{4}x^2 + \frac{1}{8}x^3 + \ldots)$ | A1 |
$\therefore f(x) = 3(1 - x + x^2 - x^3 + \ldots) + (1 + \frac{1}{2}x + \frac{1}{4}x^2 + \frac{1}{8}x^3 + \ldots)$ | M1 |
$= 4 - \frac{5}{2}x + \frac{13}{4}x^2 - \frac{23}{8}x^3 + \ldots$ | A1 | (14 marks)
**Total: (75 marks)**7.
$$f ( x ) = \frac { 8 - x } { ( 1 + x ) ( 2 - x ) } , \quad | x | < 1$$
\begin{enumerate}[label=(\alph*)]
\item Express $\mathrm { f } ( x )$ in partial fractions.
\item Show that
$$\int _ { 0 } ^ { \frac { 1 } { 2 } } \mathrm { f } ( x ) \mathrm { d } x = \ln k$$
where $k$ is an integer to be found.
\item Find the series expansion of $\mathrm { f } ( x )$ in ascending powers of $x$ up to and including the term in $x ^ { 3 }$, simplifying each coefficient.\\
7. continued\\
7. continued
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q7 [14]}}