| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem and Partial Fractions |
| Type | Partial fractions with validity range |
| Difficulty | Standard +0.3 This is a standard C4 partial fractions question with binomial expansion. Part (a) is routine decomposition with a repeated linear factor. Part (b) requires expanding two binomial terms and collecting coefficients—methodical but straightforward. Part (c) is direct recall of convergence conditions. Slightly above average due to the repeated factor and algebraic manipulation required, but follows standard textbook patterns with no novel insight needed. |
| Spec | 1.02y Partial fractions: decompose rational functions1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\frac{5-8x}{(1+2x)(1-x)^2} \equiv \frac{A}{1+2x} + \frac{B}{1-x} + \frac{C}{(1-x)^2}\) | M1 | |
| \(5 - 8x = A(1-x)^2 + B(1+2x)(1-x) + C(1+2x)\) | M1 | |
| \(x = -\frac{1}{2} \Rightarrow 9 = \frac{9}{4}A \Rightarrow A = 4\) | A1 | |
| \(x = 1 \Rightarrow -3 = 3C \Rightarrow C = -1\) | A1 | |
| coeffs \(x^2 \Rightarrow 0 = A - 2B \Rightarrow B = 2\) | M1 A1 | |
| \(f(x) = \frac{4}{1+2x} + \frac{2}{1-x} - \frac{1}{(1-x)^2}\) | ||
| (b) \(f(x) = 4(1+2x)^{-1} + 2(1-x)^{-1} - (1-x)^{-2}\) | M1 | |
| \((1+2x)^{-1} = 1 + (-1)(2x) + \frac{(-1)(-2)}{2}(2x)^2 + \frac{(-1)(-2)(-3)}{3\times 2}(2x)^3 + \ldots\) | ||
| \(= 1 - 2x + 4x^2 - 8x^3 + \ldots\) | A1 B1 | |
| \((1-x)^{-1} = 1 + x + x^2 + x^3 + \ldots\) | ||
| \((1-x)^{-2} = 1 + (-2)(-x) + \frac{(-2)(-3)}{2}(-x)^2 + \frac{(-2)(-3)(-4)}{3\times 2}(-x)^3 + \ldots\) | ||
| \(= 1 + 2x + 3x^2 + 4x^3 + \ldots\) | A1 M1 | |
| \(f(x) = 4(1 - 2x + 4x^2 - 8x^3) + 2(1 + x + x^2 + x^3) - (1 + 2x + 3x^2 + 4x^3)\) | ||
| \(= 5 - 8x + 15x^2 - 34x^3 + \ldots\) | A1 | |
| (c) \( | x | < \frac{1}{2}\) |
**(a)** $\frac{5-8x}{(1+2x)(1-x)^2} \equiv \frac{A}{1+2x} + \frac{B}{1-x} + \frac{C}{(1-x)^2}$ | M1 |
$5 - 8x = A(1-x)^2 + B(1+2x)(1-x) + C(1+2x)$ | M1 |
$x = -\frac{1}{2} \Rightarrow 9 = \frac{9}{4}A \Rightarrow A = 4$ | A1 |
$x = 1 \Rightarrow -3 = 3C \Rightarrow C = -1$ | A1 |
coeffs $x^2 \Rightarrow 0 = A - 2B \Rightarrow B = 2$ | M1 A1 |
$f(x) = \frac{4}{1+2x} + \frac{2}{1-x} - \frac{1}{(1-x)^2}$ |
**(b)** $f(x) = 4(1+2x)^{-1} + 2(1-x)^{-1} - (1-x)^{-2}$ | M1 |
$(1+2x)^{-1} = 1 + (-1)(2x) + \frac{(-1)(-2)}{2}(2x)^2 + \frac{(-1)(-2)(-3)}{3\times 2}(2x)^3 + \ldots$ |
$= 1 - 2x + 4x^2 - 8x^3 + \ldots$ | A1 B1 |
$(1-x)^{-1} = 1 + x + x^2 + x^3 + \ldots$ |
$(1-x)^{-2} = 1 + (-2)(-x) + \frac{(-2)(-3)}{2}(-x)^2 + \frac{(-2)(-3)(-4)}{3\times 2}(-x)^3 + \ldots$ |
$= 1 + 2x + 3x^2 + 4x^3 + \ldots$ | A1 M1 |
$f(x) = 4(1 - 2x + 4x^2 - 8x^3) + 2(1 + x + x^2 + x^3) - (1 + 2x + 3x^2 + 4x^3)$ |
$= 5 - 8x + 15x^2 - 34x^3 + \ldots$ | A1 |
**(c)** $|x| < \frac{1}{2}$ | A1 | (12) |
---$$f(x) = \frac{5 - 8x}{(1 + 2x)(1 - x)^2}.$$
\begin{enumerate}[label=(\alph*)]
\item Express $f(x)$ in partial fractions. [5]
\item Find the series expansion of $f(x)$ in ascending powers of $x$ up to and including the term in $x^2$, simplifying each coefficient. [6]
\item State the set of values of $x$ for which your expansion is valid. [1]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q5 [12]}}