| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2009 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Repeated linear factor with distinct linear factor – decompose and integrate |
| Difficulty | Moderate -0.3 This is a standard partial fractions question with a repeated linear factor, followed by routine integration. Part (i) requires systematic algebraic manipulation to find constants A, B, and C using the cover-up method or equating coefficients. Part (ii) applies standard integration formulas for logarithms and powers. While it requires careful algebra and knowledge of multiple techniques, it's a textbook exercise with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08j Integration using partial fractions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(4x \equiv A(x-3)^2+B(x-3)(x-5)+C(x-5)\) | M1 | |
| \(A=5\) | A1 | 'cover-up' rule, award B1 |
| \(B=-5\) | A1 | |
| \(C=-6\) | A1 4 | 'cover-up' rule, award B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int\frac{A}{x-5}dx = A\ln(5-x)\) or \(A\ln | 5-x | \) or \(A\ln |
| \(\int\frac{B}{x-3}dx = B\ln(3-x)\) or \(B\ln | 3-x | \) or \(B\ln |
| \(\int\frac{C}{(x-3)^2}dx = -\frac{C}{x-3}\) | \(\sqrt{}\)B1 | |
| \(5\ln\frac{3}{4}+5\ln 2\) aef, isw; \(\sqrt{}A\ln\frac{3}{4}-B\ln 2\) | \(\sqrt{}\)B1 | Allow if SR B1 awarded |
| \(-3\); \(\sqrt{}\frac{1}{2}C\) | \(\sqrt{}\)B1 5 |
# Question 6:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4x \equiv A(x-3)^2+B(x-3)(x-5)+C(x-5)$ | M1 | |
| $A=5$ | A1 | 'cover-up' rule, award B1 |
| $B=-5$ | A1 | |
| $C=-6$ | A1 **4** | 'cover-up' rule, award B1 |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int\frac{A}{x-5}dx = A\ln(5-x)$ or $A\ln|5-x|$ or $A\ln|x-5|$ | $\sqrt{}$B1 | but not $A\ln(x-5)$ |
| $\int\frac{B}{x-3}dx = B\ln(3-x)$ or $B\ln|3-x|$ or $B\ln|x-3|$ | $\sqrt{}$B1 | but not $B\ln(x-3)$ |
| $\int\frac{C}{(x-3)^2}dx = -\frac{C}{x-3}$ | $\sqrt{}$B1 | |
| $5\ln\frac{3}{4}+5\ln 2$ aef, isw; $\sqrt{}A\ln\frac{3}{4}-B\ln 2$ | $\sqrt{}$B1 | Allow if SR B1 awarded |
| $-3$; $\sqrt{}\frac{1}{2}C$ | $\sqrt{}$B1 **5** | |
---6 The expression $\frac { 4 x } { ( x - 5 ) ( x - 3 ) ^ { 2 } }$ is denoted by $\mathrm { f } ( x )$.\\
(i) Express f $( x )$ in the form $\frac { A } { x - 5 } + \frac { B } { x - 3 } + \frac { C } { ( x - 3 ) ^ { 2 } }$, where $A , B$ and $C$ are constants.\\
(ii) Hence find the exact value of $\int _ { 1 } ^ { 2 } \mathrm { f } ( x ) \mathrm { d } x$.\\
\hfill \mbox{\textit{OCR C4 2009 Q6 [9]}}