| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2007 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Partial fractions with algebraic division first |
| Difficulty | Moderate -0.3 This is a standard C4 partial fractions question with routine algebraic division and integration. Part (a) requires simple polynomial division to get A + B/(x-3) form, then direct integration. Part (b) involves factorising a difference of squares and standard partial fractions decomposition. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average but still requiring multiple steps and careful algebra. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08j Integration using partial fractions |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{3x-5}{x-3} = 3 + \frac{4}{x-3}\) | B1, B1 (2 marks) | B1 for 3, B1 for \(\frac{4}{x-3}\) or \(B=4\). By partial fractions: M1 multiply by \(x-3\) and use 2 values of \(x\), A1 both correct |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int 3 + \frac{4}{x-3}\,dx = 3x + 4\ln(x-3)\ (+c)\) | M1A1F (2 marks) | M1 \(\int 3 + \frac{4}{x-3}\,dx\) and attempt at integrals; ft on \(A\) and \(B\); condone omission of brackets around \(x-3\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(6x - 5 = P(2x-5) + Q(2x+5)\) | M1 | Clear evidence of use of cover-up rule M2 |
| \(x = \frac{5}{2}\), \(x = -\frac{5}{2}\); \(10 = 10Q\), \(-20 = -10P\) | m1 | |
| \(Q = 1\), \(P = 2\) | A1 (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int \frac{2}{2x+5} + \frac{1}{2x-5}\,dx\) | M1 | Attempt at ln integral: \(\left(a\ln(2x+5) + b\ln(2x-5)\right)\) |
| \(\ln(2x+5) + \frac{1}{2}\ln(2x-5)\ (+c)\) | M1, A1F (3 marks) | ft on \(P\) and \(Q\); must have brackets |
## Question 4:
**Part (a)(i)**
$\frac{3x-5}{x-3} = 3 + \frac{4}{x-3}$ | B1, B1 (2 marks) | B1 for 3, B1 for $\frac{4}{x-3}$ or $B=4$. By partial fractions: M1 multiply by $x-3$ and use 2 values of $x$, A1 both correct
**Part (a)(ii)**
$\int 3 + \frac{4}{x-3}\,dx = 3x + 4\ln(x-3)\ (+c)$ | M1A1F (2 marks) | M1 $\int 3 + \frac{4}{x-3}\,dx$ and attempt at integrals; ft on $A$ and $B$; condone omission of brackets around $x-3$
Alternative: $u = x-3$: $\int\frac{3x-5}{x-3}\,dx = \int\frac{3u+4}{u}\,du$ (M1); $= 3(x-3)+4\ln(x-3)$ (A1) correct in $x$
**Part (b)(i)**
$6x - 5 = P(2x-5) + Q(2x+5)$ | M1 | Clear evidence of use of cover-up rule M2
$x = \frac{5}{2}$, $x = -\frac{5}{2}$; $10 = 10Q$, $-20 = -10P$ | m1 |
$Q = 1$, $P = 2$ | A1 (3 marks) |
**Part (b)(ii)**
$\int \frac{2}{2x+5} + \frac{1}{2x-5}\,dx$ | M1 | Attempt at ln integral: $\left(a\ln(2x+5) + b\ln(2x-5)\right)$
$\ln(2x+5) + \frac{1}{2}\ln(2x-5)\ (+c)$ | M1, A1F (3 marks) | ft on $P$ and $Q$; must have brackets
**Total: 10 marks**
---4
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Express $\frac { 3 x - 5 } { x - 3 }$ in the form $A + \frac { B } { x - 3 }$, where $A$ and $B$ are integers. (2 marks)
\item Hence find $\int \frac { 3 x - 5 } { x - 3 } \mathrm {~d} x$.\\
(2 marks)
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Express $\frac { 6 x - 5 } { 4 x ^ { 2 } - 25 }$ in the form $\frac { P } { 2 x + 5 } + \frac { Q } { 2 x - 5 }$, where $P$ and $Q$ are integers.\\
(3 marks)
\item Hence find $\int \frac { 6 x - 5 } { 4 x ^ { 2 } - 25 } \mathrm {~d} x$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C4 2007 Q4 [10]}}