| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 12 |
| 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, routine integration of logarithms, and binomial expansion of two simple terms. All techniques are textbook exercises requiring no novel insight, though part (iii) requires careful algebraic manipulation. Slightly easier than average due to the simple denominators and clear structure. |
| Spec | 1.02y Partial fractions: decompose rational functions1.04c Extend binomial expansion: rational n, |x|<11.08j Integration using partial fractions |
| Answer | Marks |
|---|---|
| \(1 + 3x \equiv A(1-3x) + B(1-x)\) | M1 |
| \(x = 1 \Rightarrow 4 = -2A \Rightarrow A = -2\) | A1 |
| \(x = \frac{1}{3} \Rightarrow 2 = \frac{2}{3}B \Rightarrow B = 3\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_0^1 \left(\frac{3}{1-3x} - \frac{2}{1-x}\right) dx\) | ||
| \(= [-\ln | 1-3x | + 2\ln |
| \(= (-\ln\frac{1}{4} + 2\ln\frac{3}{4}) - (0)\) | M1 | |
| \(= \ln\frac{9}{16} - \ln\frac{1}{4} = \ln\frac{9}{4}\) | A1 |
| Answer | Marks |
|---|---|
| \(f(x) = 3(1-3x)^{-1} - 2(1-x)^{-1}\) | |
| \((1-x)^{-1} = 1 + x + x^2 + x^3 + \ldots\) | B1 |
| \((1-3x)^{-1} = 1 + 3x + (3x)^2 + (3x)^3 + \ldots = 1 + 3x + 9x^2 + 27x^3 + \ldots\) | M1 A1 |
| \(\therefore f(x) = 3(1 + 3x + 9x^2 + 27x^3 + \ldots) - 2(1 + x + x^2 + x^3 + \ldots)\) | M1 |
| \(= 1 + 7x + 25x^2 + 79x^3 + \ldots\) | A1 |
| (12) |
**(i)**
$1 + 3x \equiv A(1-3x) + B(1-x)$ | M1 |
$x = 1 \Rightarrow 4 = -2A \Rightarrow A = -2$ | A1 |
$x = \frac{1}{3} \Rightarrow 2 = \frac{2}{3}B \Rightarrow B = 3$ | A1 |
**(ii)**
$\int_0^1 \left(\frac{3}{1-3x} - \frac{2}{1-x}\right) dx$ | |
$= [-\ln|1-3x| + 2\ln|1-x|]_0^1$ | M1 A1 |
$= (-\ln\frac{1}{4} + 2\ln\frac{3}{4}) - (0)$ | M1 |
$= \ln\frac{9}{16} - \ln\frac{1}{4} = \ln\frac{9}{4}$ | A1 |
**(iii)**
$f(x) = 3(1-3x)^{-1} - 2(1-x)^{-1}$ | |
$(1-x)^{-1} = 1 + x + x^2 + x^3 + \ldots$ | B1 |
$(1-3x)^{-1} = 1 + 3x + (3x)^2 + (3x)^3 + \ldots = 1 + 3x + 9x^2 + 27x^3 + \ldots$ | M1 A1 |
$\therefore f(x) = 3(1 + 3x + 9x^2 + 27x^3 + \ldots) - 2(1 + x + x^2 + x^3 + \ldots)$ | M1 |
$= 1 + 7x + 25x^2 + 79x^3 + \ldots$ | A1 |
| | **(12)** |6.
$$f ( x ) = \frac { 1 + 3 x } { ( 1 - x ) ( 1 - 3 x ) } , \quad | x | < \frac { 1 } { 3 }$$
(i) Find the values of the constants $A$ and $B$ such that
$$\mathrm { f } ( x ) = \frac { A } { 1 - x } + \frac { B } { 1 - 3 x }$$
(ii) Evaluate
$$\int _ { 0 } ^ { \frac { 1 } { 4 } } f ( x ) d x$$
giving your answer as a single logarithm.\\
(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 Q6 [12]}}