| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2006 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Partial fractions with linear factors – decompose and integrate (definite) |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question testing standard C4 techniques: partial fractions decomposition (routine), definite integration using logarithms (standard), and interpretation of a zero integral (basic conceptual understanding). All parts follow textbook methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08d Evaluate definite integrals: between limits |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{A}{x} + \frac{B}{3-x}\) & c-u rule or \(A(3-x) + Bx = 3 - 2x\) | M1 | Correct format + suitable method |
| \(\frac{1}{x}\) | A1 | seen in (i) or (ii) |
| \(-\frac{1}{3-x}\) | A1 | 3 ditto; \(\frac{1}{x} - \frac{1}{3-x}\) scores 3 immediately |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int\frac{1}{x}(dx) = \ln x\) or \(\ln | x | \) |
| \(\int\frac{1}{3-x}(dx) = -\ln(3-x)\) or \(-\ln | 3-x | \) |
| Correct method idea of substitution of limits | M1 | Dep on an attempt at integrating |
| \(\ln 2 + (\ln 1 - \ln 1) - \ln 2 = 0\) | A1 | 4 Clearly seen; WWW AG |
| Answer | Marks | Guidance |
|---|---|---|
| If ignoring PFs, \(\ln x(3-x)\) immediately | B2 | \(\ln x(3-x) \to 0\) |
| As before | M1, A1 (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Suitable statement or clear implication e.g. | B1 | 1 |
### (i)
$\frac{A}{x} + \frac{B}{3-x}$ & c-u rule or $A(3-x) + Bx = 3 - 2x$ | M1 | Correct format + suitable method
$\frac{1}{x}$ | A1 | seen in (i) or (ii)
$-\frac{1}{3-x}$ | A1 | 3 ditto; $\frac{1}{x} - \frac{1}{3-x}$ scores 3 immediately
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### (ii)
$\int\frac{1}{x}(dx) = \ln x$ or $\ln|x|$ | B1 |
$\int\frac{1}{3-x}(dx) = -\ln(3-x)$ or $-\ln|3-x|$ | B1 | Check sign carefully; do not allow $\ln(x-3)$
Correct method idea of substitution of limits | M1 | Dep on an attempt at integrating
$\ln 2 + (\ln 1 - \ln 1) - \ln 2 = 0$ | A1 | 4 Clearly seen; WWW AG
**Alternative Method**
If ignoring PFs, $\ln x(3-x)$ immediately | B2 | $\ln x(3-x) \to 0$
As before | M1, A1 (4) |
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### (iii)
Suitable statement or clear implication e.g. | B1 | 1
Equal amounts (of area) above and below (axis)
or graph crosses axis or there's a root
---\begin{enumerate}[label=(\roman*)]
\item Express $\frac{3 - 2x}{x(3 - x)}$ in partial fractions. [3]
\item Show that $\int_1^2 \frac{3 - 2x}{x(3 - x)} dx = 0$. [4]
\item What does the result of part (ii) indicate about the graph of $y = \frac{3 - 2x}{x(3 - x)}$ between $x = 1$ and $x = 2$? [1]
\end{enumerate}
\hfill \mbox{\textit{OCR C4 2006 Q3 [8]}}