| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2008 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Partial fractions for differential equations |
| Difficulty | Moderate -0.3 This is a standard multi-part question that guides students through partial fractions, integration, and a separable differential equation. The partial fractions decomposition is straightforward (linear factors), the integration follows directly, and the differential equation is separable with clear steps. While it requires multiple techniques and careful algebraic manipulation to reach the given form, each individual step is routine for C4 level, making it slightly easier than average. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08j Integration using partial fractions1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{2}{(x^2-1)} = \frac{A}{x-1}+\frac{B}{x+1}\) | ||
| \(2 = A(x+1)+B(x-1)\) | M1 | |
| \(x=1\quad x=-1\) | m1 | Use two values of \(x\) or equate coefficients and solve; \(A+B=0\) and \(A-B=2\) |
| \(A=1\quad B=-1\) | A1 | Both \(A\) and \(B\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int\frac{2}{x^2-1}\,dx = p\ln(x-1)+q\ln(x+1)\) | M1 | ln integrals |
| \(= \ln(x-1)-\ln(x+1)\) | A1F | F on \(A\) and \(B\); condone missing brackets |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int\frac{dy}{y} = \int\frac{2}{3(x^2-1)}\,dx\) | M1 | Separate and attempt to integrate on one side |
| \(\ln y = \frac{1}{3}(\ln(x-1)-\ln(x+1))\,(+C)\) | A1, A1F | Left hand side; F from part (b) on right hand side |
| \((3,1)\quad \ln 1 = \frac{1}{3}(\ln 2-\ln 4)+C\) | m1 | Use \((3,1)\) to attempt to find a constant |
| Answer | Marks | Guidance |
|---|---|---|
| \(y^3 = \frac{2(x-1)}{x+1}\) | A1 | CSO AG |
# Question 6:
## Part 6(a)
$\frac{2}{(x^2-1)} = \frac{A}{x-1}+\frac{B}{x+1}$ | |
$2 = A(x+1)+B(x-1)$ | M1 |
$x=1\quad x=-1$ | m1 | Use two values of $x$ or equate coefficients and solve; $A+B=0$ and $A-B=2$
$A=1\quad B=-1$ | A1 | Both $A$ and $B$
## Part 6(b)
$\int\frac{2}{x^2-1}\,dx = p\ln(x-1)+q\ln(x+1)$ | M1 | ln integrals
$= \ln(x-1)-\ln(x+1)$ | A1F | F on $A$ and $B$; condone missing brackets
## Part 6(c)
$\int\frac{dy}{y} = \int\frac{2}{3(x^2-1)}\,dx$ | M1 | Separate and attempt to integrate on one side
$\ln y = \frac{1}{3}(\ln(x-1)-\ln(x+1))\,(+C)$ | A1, A1F | Left hand side; F from part (b) on right hand side
$(3,1)\quad \ln 1 = \frac{1}{3}(\ln 2-\ln 4)+C$ | m1 | Use $(3,1)$ to attempt to find a constant
$3\ln y = \ln(x-1)-\ln(x+1)-(\ln 2-\ln 4)$
$3\ln y = \left(\ln\left(\frac{x-1}{x+1}\right)+\ln 2\right)$
$\ln y^3 = \ln\left(\frac{2(x-1)}{x+1}\right)$
$y^3 = \frac{2(x-1)}{x+1}$ | A1 | CSO AG
---6
\begin{enumerate}[label=(\alph*)]
\item Express $\frac { 2 } { x ^ { 2 } - 1 }$ in the form $\frac { A } { x - 1 } + \frac { B } { x + 1 }$.
\item Hence find $\int \frac { 2 } { x ^ { 2 } - 1 } \mathrm {~d} x$.
\item Solve the differential equation $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 y } { 3 \left( x ^ { 2 } - 1 \right) }$, given that $y = 1$ when $x = 3$. Show that the solution can be written as $y ^ { 3 } = \frac { 2 ( x - 1 ) } { x + 1 }$.
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2008 Q6 [10]}}