| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2007 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration with Partial Fractions |
| Type | Partial fractions with repeated linear factor |
| Difficulty | Moderate -0.3 This is a standard partial fractions question with a repeated linear factor, followed by routine integration. Part (i) is straightforward algebraic manipulation, and part (ii) requires integrating ln and a power function—both standard C4 techniques with no conceptual challenges. Slightly easier than average due to the mechanical nature of the process. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08j Integration using partial fractions |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(2x + 1 = l = A(x - 3) + B\) | M1 | |
| \(A = 2\) | A1 | |
| \(B = 7\) | A/B 1 | 3 marks |
| (ii) \(\int \frac{1}{x-3}(dx) = \ln(x - 3)\) or \(\ln | x - 3 | \) |
| \(\int \frac{1}{(x-3)}(dx) = -\frac{1}{x-3}\) | B1 | Accept B or \(\frac{1}{B}\) as a multiplier |
| \(6 + 2 \ln 7\) | √B2 | 4 marks |
(i) $2x + 1 = l = A(x - 3) + B$ | M1 |
$A = 2$ | A1 |
$B = 7$ | A/B 1 | 3 marks | Cover-up rule acceptable for B1
(ii) $\int \frac{1}{x-3}(dx) = \ln(x - 3)$ or $\ln|x - 3|$ | B1 | Accept A or $\frac{1}{4}$ as a multiplier
$\int \frac{1}{(x-3)}(dx) = -\frac{1}{x-3}$ | B1 | Accept B or $\frac{1}{B}$ as a multiplier
$6 + 2 \ln 7$ | √B2 | 4 marks | Follow-through $\frac{6}{7}B + A \ln 7$
---\begin{enumerate}[label=(\roman*)]
\item Express $\frac{2x + 1}{(x - 3)^2}$ in the form $\frac{A}{x - 3} + \frac{B}{(x - 3)^2}$, where $A$ and $B$ are constants. [3]
\item Hence find the exact value of $\int_4^{10} \frac{2x + 1}{(x - 3)^2} \, dx$, giving your answer in the form $a + b \ln c$, where $a$, $b$ and $c$ are integers. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR C4 2007 Q6 [7]}}