| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Partial fractions for differential equations |
| Difficulty | Standard +0.3 This is a structured multi-part question with clear scaffolding. Part (a) is routine partial fractions with linear factors. Part (b) is a standard separable differential equation where the partial fractions result is directly applied, followed by straightforward integration and using initial conditions. Parts (c)(i) and (c)(ii) require basic interpretation of the model (finding when V=0 and the horizontal asymptote). While it spans multiple techniques, each step is procedural with no novel insight required, making it slightly easier than average. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08k Separable differential equations: dy/dx = f(x)g(y)1.08l Interpret differential equation solutions: in context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{3}{(2x-1)(x+1)} = \frac{A}{2x-1} + \frac{B}{x+1} \Rightarrow A = ..., B = ...\) | M1 | Correct method of partial fractions leading to values for \(A\) and \(B\) |
| Either \(A = 2\) or \(B = -1\) | A1 | Correct value for \(A\) or \(B\) |
| \(\frac{3}{(2x-1)(x+1)} = \frac{2}{2x-1} - \frac{1}{x+1}\) | A1 | Correct partial fractions; must be seen as fractions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int \frac{1}{V}\,dV = \int \frac{3}{(2t-1)(t+1)}\,dt\) | B1 | Separates variables; may be implied by later work |
| \(\int \frac{2}{2t-1} - \frac{1}{t+1}\,dt = ...\ln(2t-1) - ...\ln(t+1)(+c)\) | M1 | Correct attempt at integration of partial fractions |
| \(\ln V = \ln(2t-1) - \ln(t+1)(+c)\) | A1ft | Fully correct equation following through their \(A\) and \(B\) only; no requirement for \(+c\) |
| Substitutes \(t=2, V=3 \Rightarrow c = (\ln 3)\) | M1 | Attempts to find \(c\); condone poor algebra as long as \(t=2, V=3\) is used |
| \(\ln V = \ln(2t-1) - \ln(t+1) + \ln 3\), \(V = \dfrac{3(2t-1)}{(t+1)}\) | A1* | Correct processing leading to given answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int \frac{1}{3V}\,dV = \int \frac{1}{(2t-1)(t+1)}\,dt\) | B1 | Separates variables; may be implied by later work |
| \(\frac{1}{3}\int \frac{2}{2t-1} - \frac{1}{t+1}\,dt = ...\ln(2t-1) - ...\ln(t+1)(+c)\) | M1 | Correct attempt at integration |
| \(\frac{1}{3}\ln 3V = \frac{1}{3}\ln(2t-1) - \frac{1}{3}\ln(t+1)(+c)\) | A1ft | Fully correct equation following through their \(A\) and \(B\) only |
| Substitutes \(t=2, V=3 \Rightarrow c = \left(\frac{1}{3}\ln 3\right)\) | M1 | Attempts to find \(c\) |
| \(V = \dfrac{3(2t-1)}{(t+1)}\) | A1* | Correct processing leading to given answer; note B0M1A1M1A1 is not possible |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (i) 30 (minutes) | B1 | Units not required; allow equivalents e.g. \(\frac{1}{2}\) an hour; if units given must be correct |
| (ii) \(6\ (\text{m}^3)\) | B1 | Units not required; condone \(V < 6\) or \(V \leq 6\); if units given must be correct |
## Question 14:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{3}{(2x-1)(x+1)} = \frac{A}{2x-1} + \frac{B}{x+1} \Rightarrow A = ..., B = ...$ | M1 | Correct method of partial fractions leading to values for $A$ and $B$ |
| Either $A = 2$ or $B = -1$ | A1 | Correct value for $A$ or $B$ |
| $\frac{3}{(2x-1)(x+1)} = \frac{2}{2x-1} - \frac{1}{x+1}$ | A1 | Correct partial fractions; must be seen as fractions |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \frac{1}{V}\,dV = \int \frac{3}{(2t-1)(t+1)}\,dt$ | B1 | Separates variables; may be implied by later work |
| $\int \frac{2}{2t-1} - \frac{1}{t+1}\,dt = ...\ln(2t-1) - ...\ln(t+1)(+c)$ | M1 | Correct attempt at integration of partial fractions |
| $\ln V = \ln(2t-1) - \ln(t+1)(+c)$ | A1ft | Fully correct equation following through their $A$ and $B$ only; no requirement for $+c$ |
| Substitutes $t=2, V=3 \Rightarrow c = (\ln 3)$ | M1 | Attempts to find $c$; condone poor algebra as long as $t=2, V=3$ is used |
| $\ln V = \ln(2t-1) - \ln(t+1) + \ln 3$, $V = \dfrac{3(2t-1)}{(t+1)}$ | A1* | Correct processing leading to given answer |
### Part (b) Alternative:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \frac{1}{3V}\,dV = \int \frac{1}{(2t-1)(t+1)}\,dt$ | B1 | Separates variables; may be implied by later work |
| $\frac{1}{3}\int \frac{2}{2t-1} - \frac{1}{t+1}\,dt = ...\ln(2t-1) - ...\ln(t+1)(+c)$ | M1 | Correct attempt at integration |
| $\frac{1}{3}\ln 3V = \frac{1}{3}\ln(2t-1) - \frac{1}{3}\ln(t+1)(+c)$ | A1ft | Fully correct equation following through their $A$ and $B$ only |
| Substitutes $t=2, V=3 \Rightarrow c = \left(\frac{1}{3}\ln 3\right)$ | M1 | Attempts to find $c$ |
| $V = \dfrac{3(2t-1)}{(t+1)}$ | A1* | Correct processing leading to given answer; note B0M1A1M1A1 is not possible |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| (i) 30 (minutes) | B1 | Units not required; allow equivalents e.g. $\frac{1}{2}$ an hour; if units given must be correct |
| (ii) $6\ (\text{m}^3)$ | B1 | Units not required; condone $V < 6$ or $V \leq 6$; if units given must be correct |
---\begin{enumerate}
\item (a) Express $\frac { 3 } { ( 2 x - 1 ) ( x + 1 ) }$ in partial fractions.
\end{enumerate}
When chemical $A$ and chemical $B$ are mixed, oxygen is produced.\\
A scientist mixed these two chemicals and measured the total volume of oxygen produced over a period of time.
The total volume of oxygen produced, $V \mathrm {~m} ^ { 3 } , t$ hours after the chemicals were mixed, is modelled by the differential equation
$$\frac { \mathrm { d } V } { \mathrm {~d} t } = \frac { 3 V } { ( 2 t - 1 ) ( t + 1 ) } \quad V \geqslant 0 \quad t \geqslant k$$
where $k$ is a constant.\\
Given that exactly 2 hours after the chemicals were mixed, a total volume of $3 \mathrm {~m} ^ { 3 }$ of oxygen had been produced,\\
(b) solve the differential equation to show that
$$V = \frac { 3 ( 2 t - 1 ) } { ( t + 1 ) }$$
The scientist noticed that
\begin{itemize}
\item there was a time delay between the chemicals being mixed and oxygen being produced
\item there was a limit to the total volume of oxygen produced
\end{itemize}
Deduce from the model\\
(c) (i) the time delay giving your answer in minutes,\\
(ii) the limit giving your answer in $\mathrm { m } ^ { 3 }$
\hfill \mbox{\textit{Edexcel Paper 2 2022 Q14 [10]}}