June 2022 Paper 2 Q14
14.
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
\[\dfrac{\mathrm{d}V}{\mathrm{d}t} = \dfrac{3V}{(2t - 1)(t + 1)} \qquad V \geqslant 0 \qquad t \geqslant k\]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,
The scientist noticed that
- there was a time delay between the chemicals being mixed and oxygen being produced
- there was a limit to the total volume of oxygen produced
Deduce from the model
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{3}{(2x - 1)(x + 1)} = \dfrac{A}{2x - 1} + \dfrac{B}{x + 1} \Rightarrow A = \ldots,\ B = \ldots\) | M1 | 1.1b |
| Either \(A = 2\) or \(B = -1\) | A1 | 1.1b |
| \(\dfrac{3}{(2x - 1)(x + 1)} = \dfrac{2}{2x - 1} - \dfrac{1}{x + 1}\) | A1 | 1.1b |
| (3) |
Notes
M1: Correct method of partial fractions leading to values for their \(A\) and \(B\)
E.g. substitution: \(\dfrac{3}{(2x - 1)(x + 1)} = \dfrac{A}{2x - 1} + \dfrac{B}{x + 1} \Rightarrow 3 = A(x + 1) + B(2x - 1) \Rightarrow A = \ldots,\ B = \ldots\)
Or compare coefficients \(\dfrac{3}{(2x - 1)(x + 1)} = \dfrac{A}{2x - 1} + \dfrac{B}{x + 1} \Rightarrow 3 = x(A + 2B) + A - B \Rightarrow A = \ldots,\ B = \ldots\)
Note that \(\dfrac{3}{(2x - 1)(x + 1)} = \dfrac{A}{2x - 1} + \dfrac{B}{x + 1} \Rightarrow 3 = A(2x - 1) + B(x + 1) \Rightarrow A = \ldots,\ B = \ldots\) scores M0
A1: Correct value for “\(A\)” or “\(B\)”
A1: Correct partial fractions not just values for “\(A\)” and “\(B\)”. \(\dfrac{2}{2x - 1} - \dfrac{1}{x + 1}\) or e.g. \(\dfrac{2}{2x - 1} + \dfrac{-1}{x + 1}\)
Must be seen as fractions but if not stated here, allow if the correct fractions appear later.
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle\int \dfrac{1}{V}\,\mathrm{d}V = \int \dfrac{3}{(2t - 1)(t + 1)}\,\mathrm{d}t\) | B1 | 1.1a |
| \(\displaystyle\int \dfrac{2}{2t - 1} - \dfrac{1}{t + 1}\,\mathrm{d}t = \ldots\ln(2t - 1) - \ldots\ln(t + 1)\ (+c)\) | M1 | 3.1a |
| \(\ln V = \ln(2t - 1) - \ln(t + 1)\ (+c)\) | A1ft | 1.1b |
| Substitutes \(t = 2,\ V = 3 \Rightarrow c = (\ln 3)\) | M1 | 3.4 |
| \(\begin{aligned}&\ln V = \ln(2t - 1) - \ln(t + 1) + \ln 3\\[6pt]&V = \dfrac{3(2t - 1)}{(t + 1)}\ *\end{aligned}\) | A1* | 2.1 |
| (5) |
(b) Alternative separation of variables:
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle\int \dfrac{1}{3V}\,\mathrm{d}V = \int \dfrac{1}{(2t - 1)(t + 1)}\,\mathrm{d}t\) | B1 | 1.1a |
| \(\displaystyle\dfrac{1}{3}\int \dfrac{2}{2t - 1} - \dfrac{1}{t + 1}\,\mathrm{d}t = \ldots\ln(2t - 1) - \ldots\ln(t + 1)\ (+c)\) | M1 | 3.1a |
| \(\dfrac{1}{3}\ln 3V = \dfrac{1}{3}\ln(2t - 1) - \dfrac{1}{3}\ln(t + 1)\ (+c)\) | A1ft | 1.1b |
| Substitutes \(t = 2,\ V = 3 \Rightarrow c = \left(\dfrac{1}{3}\ln 3\right)\) | M1 | 3.4 |
| \(\begin{aligned}&\dfrac{1}{3}\ln V = \dfrac{1}{3}\ln(2t - 1) - \dfrac{1}{3}\ln(t + 1) + \dfrac{1}{3}\ln 3\\[6pt]&V = \dfrac{3(2t - 1)}{(t + 1)}\ *\end{aligned}\) | A1* | 2.1 |
| (5) |
Notes
B1: Separates variables \(\displaystyle\int \dfrac{1}{V}\,\mathrm{d}V = \int \dfrac{3}{(2t - 1)(t + 1)}\,\mathrm{d}t\). May be implied by later work.
Condone omission of the integral signs but the \(\mathrm{d}V\) and \(\mathrm{d}t\) must be in the correct positions if awarding this mark in isolation but they may be implied by subsequent work.
M1: Correct attempt at integration of the partial fractions.
Look for \(\ldots\ln(2t - 1) + \ldots\ln(t + 1)\) where … are constants.
Condone missing brackets around the \((2t - 1)\) and/or the \((t + 1)\) for this mark
A1ft: Fully correct equation following through their \(A\) and \(B\) only.
No requirement for \(+c\) here.
The brackets around the \((2t - 1)\) and/or the \((t + 1)\) must be seen or implied for this mark
M1: Attempts to find “\(c\)” or e.g. “\(\ln k\)” using \(t = 2,\ V = 3\) following an attempt at integration.
Condone poor algebra as long as \(t = 2,\ V = 3\) is used to find a value of their constant.
Note that the constant may be found immediately after integrating or e.g. after the ln’s have been combined.
A1*: Correct processing leading to the given answer \(V = \dfrac{3(2t - 1)}{(t + 1)}\)
Alternative:
B1: Separates variables \(\displaystyle\int \dfrac{1}{3V}\,\mathrm{d}V = \int \dfrac{1}{(2t - 1)(t + 1)}\,\mathrm{d}t\). May be implied by later work.
Condone omission of the integral signs but the \(\mathrm{d}V\) and \(\mathrm{d}t\) must be in the correct positions if awarding this mark in isolation but they may be implied by subsequent work.
M1: Correct attempt at integration of the partial fractions.
Look for \(\ldots\ln(2t - 1) + \ldots\ln(t + 1)\) where … are constants.
Condone missing brackets around the \((2t - 1)\) and/or the \((t + 1)\) for this mark
A1ft: Fully correct equation following through their \(A\) and \(B\) only.
No requirement for \(+c\) here.
The brackets around the \((2t - 1)\) and/or the \((t + 1)\) must be seen or implied for this mark
M1: Attempts to find “\(c\)” or e.g. “\(\ln k\)” using \(t = 2,\ V = 3\) following an attempt at integration.
Condone poor algebra as long as \(t = 2,\ V = 3\) is used to find a value of their constant.
Note that the constant may be found immediately after integrating or e.g. after the ln’s have been combined.
A1*: Correct processing leading to the given answer \(V = \dfrac{3(2t - 1)}{(t + 1)}\)
(Note the working may look like this:
\[\dfrac{1}{3}\ln 3V = \dfrac{1}{3}\ln(2t - 1) - \dfrac{1}{3}\ln(t + 1) + c,\quad \dfrac{1}{3}\ln 9 = \dfrac{1}{3}\ln(3) - \dfrac{1}{3}\ln 3 + c,\quad c = \dfrac{1}{3}\ln 9\]\[\ln 3V = \ln\dfrac{9(2t - 1)}{(t + 1)} \Rightarrow 3V = \dfrac{9(2t - 1)}{(t + 1)} \Rightarrow V = \dfrac{3(2t - 1)}{(t + 1)}\ *\text{)}\]Note that B0M1A1M1A1 is not possible in (b) as the B1 must be implied if all the other marks have been awarded.
Note also that some candidates may use different variables in (b) e.g. \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3y}{(2x - 1)(x + 1)} \Rightarrow \displaystyle\int \dfrac{1}{y}\,\mathrm{d}y = \int \dfrac{3}{(2x - 1)(x + 1)}\,\mathrm{d}x\) etc. In such cases you should award marks for equivalent work but they must revert to the given variables at the end to score the final mark.
Also if e.g. a “\(t\)” becomes an “\(x\)” within their working but is recovered allow full marks.
| Scheme | Marks | AO |
|---|---|---|
| (i) 30 (minutes) | B1 | 3.2a |
| (ii) 6 (\(\mathrm{m}^3\)) | B1 | 3.4 |
| (2) | ||
| (10 marks) |
Notes
B1: Deduces 30 minutes. Units not required so just look for 30 but allow equivalents e.g. ½ an hour.
If units are given they must be correct so do not allow e.g. 30 hours.
B1: Deduces \(6\,\mathrm{m}^3\). Units not required so just look for 6. Condone \(V \lt 6\) or \(V \leqslant 6\)
If units are given they must be correct so do not allow e.g. 6 m.