| Exam Board | OCR |
|---|---|
| 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 three-part C4 question covering routine partial fractions, straightforward integration using logarithms, and binomial expansion of each fraction. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08d Evaluate definite integrals: between limits |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{8-x}{(1+x)(2-x)} \equiv \frac{A}{1+x} + \frac{B}{2-x}\) | ||
| \(8 - x \equiv A(2-x) + B(1+x)\) | 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 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(= \int_0^{\frac{1}{2}} \left(\frac{3}{1+x} + \frac{2}{2-x}\right)dx = \left[3\ln | 1+x | - 2\ln |
| \(= (3\ln\frac{3}{2} - 2\ln\frac{3}{2}) - (0 - 2\ln 2)\) | M1 | |
| \(= \ln\frac{3}{2} + \ln 4 = \ln 6\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(x) = 3(1+x)^{-1} + 2(2-x)^{-1}\) | ||
| \((1+x)^{-1} = 1 - x + x^2 - x^3 + \ldots\) | B1 | |
| \((2-x)^{-1} = 2^{-1}(1 - \frac{1}{2}x)^{-1}\) | M1 | |
| \(= \frac{1}{2}\left[1 + (-1)(-\frac{1}{2}x) + \frac{(-1)(-2)}{2}(-\frac{1}{2}x)^2 + \frac{(-1)(-2)(-3)}{3\times2}(-\frac{1}{2}x)^3 + \ldots\right]\) | M1 | |
| \(= \frac{1}{2}\left(1 + \frac{1}{2}x + \frac{1}{4}x^2 + \frac{1}{8}x^3 + \ldots\right)\) | 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 | (13) |
# Question 9:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{8-x}{(1+x)(2-x)} \equiv \frac{A}{1+x} + \frac{B}{2-x}$ | | |
| $8 - x \equiv A(2-x) + B(1+x)$ | 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 | |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $= \int_0^{\frac{1}{2}} \left(\frac{3}{1+x} + \frac{2}{2-x}\right)dx = \left[3\ln|1+x| - 2\ln|2-x|\right]_0^{\frac{1}{2}}$ | M1 A1 | |
| $= (3\ln\frac{3}{2} - 2\ln\frac{3}{2}) - (0 - 2\ln 2)$ | M1 | |
| $= \ln\frac{3}{2} + \ln 4 = \ln 6$ | A1 | |
## Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) = 3(1+x)^{-1} + 2(2-x)^{-1}$ | | |
| $(1+x)^{-1} = 1 - x + x^2 - x^3 + \ldots$ | B1 | |
| $(2-x)^{-1} = 2^{-1}(1 - \frac{1}{2}x)^{-1}$ | M1 | |
| $= \frac{1}{2}\left[1 + (-1)(-\frac{1}{2}x) + \frac{(-1)(-2)}{2}(-\frac{1}{2}x)^2 + \frac{(-1)(-2)(-3)}{3\times2}(-\frac{1}{2}x)^3 + \ldots\right]$ | M1 | |
| $= \frac{1}{2}\left(1 + \frac{1}{2}x + \frac{1}{4}x^2 + \frac{1}{8}x^3 + \ldots\right)$ | 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 | **(13)** |
**Total: (72)**9.
$$f ( x ) = \frac { 8 - x } { ( 1 + x ) ( 2 - x ) } , \quad | x | < 1$$
(i) Express $\mathrm { f } ( x )$ in partial fractions.\\
(ii) Show that
$$\int _ { 0 } ^ { \frac { 1 } { 2 } } \mathrm { f } ( x ) \mathrm { d } x = \ln k$$
where $k$ is an integer to be found.\\
(iii) 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.
\hfill \mbox{\textit{OCR C4 Q9 [13]}}