June 2025 Paper 3 Q7
7 A group of scientists is studying how a population of insects grows in the laboratory.
At the start of the study, there are 48 insects.
At the end of the first month, there are 72 insects.
At the end of the second month, there are 108 insects.
The group develops two models, \(P\) and \(Q\), to model the population.
The first model, \(P\), is of the form \(P_{t+1} = kP_t\), where \(k\) is a constant.
\(P_0\) denotes the number of insects at the start of the study.
\(P_t\) denotes the number of insects at the end of month \(t\), \(t \geqslant 1\).
Model \(P\) fits the data for \(t = 0\), 1 and 2 exactly.
The second model, \(Q\), is of the form \(Q = 48\mathrm{e}^{0.4t}\), where \(t\) is the time, in months, after the start of the study and \(Q\) denotes the number of insects at time \(t\).
| Scheme | Marks | AO |
|---|---|---|
| \((k =)\ 1.5\) | B1 | 3.3 |
| [1] |
Notes
B1: Allow embedded in equation e.g. \(P_{t+1} = \frac{3}{2}P_t\)
| Scheme | Marks | AO |
|---|---|---|
| 162 | B1 | 3.4 |
| [1] |
Notes
B1: cao
| Scheme | Marks | AO |
|---|---|---|
| (i) \(48\mathrm{e}^{0.4} = 71.6\ (\approx 72)\) \(48\mathrm{e}^{0.8} = 106.8\ (\approx 107)\) | M1 | 3.4 |
| Comment comparing their values with the data e.g. • Reasonably close • Not exactly right | A1 | 3.5a |
| [2] | ||
| (ii) \(\dfrac{\mathrm{d}Q}{\mathrm{d}t} = 19.2\mathrm{e}^{0.4t}\) | M1 | 1.1 |
| \(\dfrac{\mathrm{d}Q}{\mathrm{d}t} = 0.4Q\) so the rate of change is proportional to the population | A1 | 2.2a |
| [2] | ||
| (iii) Method to find multiplier for Q At time \(t + 1\), \(Q = 48\mathrm{e}^{0.4(t+1)}\) \(\quad = 48\mathrm{e}^{0.4t} \times \mathrm{e}^{0.4}\) | M1 | 3.1b |
| \(\mathrm{e}^{0.4}\) or 1.49 ... (is multiplier for Q) | B1 | 1.1 |
| Multiplier for sequence P is 1.5. Both sequences are equal at \(t = 0\) or 1 So \(\times 1.49\) each month will give smaller answers than \(\times 1.5\) | A1 | 2.4 |
| [3] |
Notes
(i) M1: Both values – might not round answers
Allow \(106.8 \approx 108\)
Condone incorrect values if \(48\mathrm{e}^{0.4}\) and \(48\mathrm{e}^{0.8}\) seen
(i) A1: Must have correct values
e.g. 71.6 is close to 72 and 106.8 is close to 108
Ignore attempts to compare the two models
Condone e.g. ‘close for \(t = 1\)’ and ‘not close for \(t = 2\)’
(ii) M1: Differentiation – might not simplify e.g. \(\frac{\mathrm{d}Q}{\mathrm{d}t} = 19.2\mathrm{e}^{0.4t}\) or \(\frac{\mathrm{d}Q}{\mathrm{d}t} = \frac{96}{5}\mathrm{e}^{0.4t}\)
Implied by \(\frac{\mathrm{d}Q}{\mathrm{d}t} = ke^{0.4t}\)
Condone poor/no notation
(ii) A1: Showing the rate of change is proportional to population e.g. \(\frac{\mathrm{d}Q}{\mathrm{d}t} = 0.4 \times 48\mathrm{e}^{0.4t}\) and \(Q = 48\mathrm{e}^{0.4t}\) soi or \(\dfrac{\frac{\mathrm{d}Q}{\mathrm{d}t}}{48\mathrm{e}^{0.4t}} = 0.4\)
0.4 must be exact and not rounded e.g. from numerical trials
Must see \(\frac{\mathrm{d}Q}{\mathrm{d}t}\) or e.g. \(Q^\prime\)
(iii) M1: Use of exponential equation one month later – may award for correct use of equation for Q to find populations for at least 2 consecutive months (for \(t = 2\) onwards)
For info: \(t = 2, Q = 106.8\); \(t = 3, Q = 159.4\); \(t = 4, Q = 237.7\) to 3 s.f. or better
This M1 may be implied by finding multiplier of \(\mathrm{e}^{0.4}\)
(iii) B1: Finding multiplier for Q
If ‘multiplier’ omitted, then multiplication must be implied by later work
(iii) A1: Completion with convincing general explanation including consideration of an early value (could be \(P > Q\) at \(t = 1\) or both sequences start at 48 or at \(t = 2\) the values are 107 for \(Q\) and 108 for \(P\))
Alternative 1 for A1 (iii)
| Scheme | Marks |
|---|---|
| Model P is \(48(1.5)^t\) oe \(1.49\ldots < 1.5\) So, \(48(1.49\ldots)^t < 48(1.5)^t\) | A1 |
A1: Convincing completion, must see ‘48’ or both equal populations at \(t = 0\) (or \(t = 1\))
oe e.g. showing \(\dfrac{1.5^t}{\mathrm{e}^{0.4t}} > 1\) and starting values of 48 soi or writing model for P as \(48\mathrm{e}^{(\ln 1.5)t} = 48\,\mathrm{e}^{0.405\ldots t}\)
Alternative 2 for A1 (iii)
| Scheme | Marks |
|---|---|
| Model P is \(48(1.5)^t\) \(\dfrac{\mathrm{dP}}{\mathrm{d}t} = 48 \times \ln(1.5) \times 1.5^t = \ln(1.5) \times P\) Rate of population growth for P is greater than rate of population growth for Q as \(\ln 1.5 > 0.4\) and both equal populations at \(t = 0\) (or \(t = 1\)) | A1 |
A1: Differentiates may use ln 1.5 or 0.405…
Convincing completion
Alternative method using inequalities (iii)
| Scheme | Marks |
|---|---|
| Model for P is \(48 \times 1.5^t\) | B1 |
| \((48)\mathrm{e}^{0.4t} < (48\times)1.5^t\) oe | M1 |
| \(0.4t < \ln 1.5^t\) \(0.4t < t\ln 1.5\) True for \(t > 1\) since \(\ln 1.5 = 0.405\ldots > 0.4\) | A1 |
B1: B1 for finding correct model for P
M1: Forming inequality for P and Q
A1: Convincing completion