| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Repeated linear factor with distinct linear factor – decompose and integrate |
| Difficulty | Standard +0.3 This is a standard C4 partial fractions question with a repeated linear factor. Part (a) requires routine algebraic manipulation to decompose the fraction, and part (b) involves straightforward integration of logarithmic and rational terms followed by simplification. While it requires careful algebra across multiple steps, it follows a well-practiced template with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08d Evaluate definite integrals: between limits1.08j Integration using partial fractions |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\frac{7+3x+2x^2}{(1-2x)(1+x)^2} = \frac{A}{1-2x} + \frac{B}{1+x} + \frac{C}{(1+x)^2}\) | ||
| \(7 + 3x + 2x^2 = A(1+x)^2 + B(1-2x)(1+x) + C(1-2x)\) | ||
| \(x = \frac{1}{2} \Rightarrow 9 = \frac{9}{4}A \Rightarrow A = 4\) | B1 | |
| \(x = -1 \Rightarrow 6 = 3C \Rightarrow C = 2\) | B1 | |
| coeffs \(x^2 \Rightarrow 2 = A - 2B \Rightarrow B = 1\) | M1 | |
| \(\therefore f(x) = \frac{4}{1-2x} + \frac{1}{1+x} + \frac{2}{(1+x)^2}\) | A1 | |
| (b) \(= \int_1^2 (\frac{4}{1-2x} + \frac{1}{1+x} + \frac{2}{(1+x)^2}) dx\) | ||
| \(= [-2\ln | 1-2x | + \ln |
| \(= (-2\ln 3 + \ln 3 - \frac{2}{3}) - (0 + \ln 2 - 1)\) | M1 | |
| \(= -\ln 3 - \ln 2 + \frac{1}{3} = \frac{1}{3} - \ln 6\) | M1 A1 | \([p = \frac{1}{3}, q = 6]\) |
**(a)** $\frac{7+3x+2x^2}{(1-2x)(1+x)^2} = \frac{A}{1-2x} + \frac{B}{1+x} + \frac{C}{(1+x)^2}$ | |
$7 + 3x + 2x^2 = A(1+x)^2 + B(1-2x)(1+x) + C(1-2x)$ | |
$x = \frac{1}{2} \Rightarrow 9 = \frac{9}{4}A \Rightarrow A = 4$ | B1 |
$x = -1 \Rightarrow 6 = 3C \Rightarrow C = 2$ | B1 |
coeffs $x^2 \Rightarrow 2 = A - 2B \Rightarrow B = 1$ | M1 |
$\therefore f(x) = \frac{4}{1-2x} + \frac{1}{1+x} + \frac{2}{(1+x)^2}$ | A1 |
**(b)** $= \int_1^2 (\frac{4}{1-2x} + \frac{1}{1+x} + \frac{2}{(1+x)^2}) dx$ | |
$= [-2\ln|1-2x| + \ln|1+x| - 2(1+x)^{-1}]_1^2$ | M1 A3 |
$= (-2\ln 3 + \ln 3 - \frac{2}{3}) - (0 + \ln 2 - 1)$ | M1 |
$= -\ln 3 - \ln 2 + \frac{1}{3} = \frac{1}{3} - \ln 6$ | M1 A1 | $[p = \frac{1}{3}, q = 6]$ | (11)
---3.
$$f ( x ) = \frac { 7 + 3 x + 2 x ^ { 2 } } { ( 1 - 2 x ) ( 1 + x ) ^ { 2 } } , \quad | x | > \frac { 1 } { 2 }$$
\begin{enumerate}[label=(\alph*)]
\item Express $\mathrm { f } ( x )$ in partial fractions.
\item Show that
$$\int _ { 1 } ^ { 2 } \mathrm { f } ( x ) \mathrm { d } x = p - \ln q$$
where $p$ is rational and $q$ is an integer.\\
3. continued
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q3 [11]}}