June 2025 Paper 2 Q9
9.
In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.
A new type of car is released for sale.
The total number of this type of car sold, \(N\), in a particular region, \(t\) months after the cars were released for sale, is modelled by the equation\[N = 5000 - 5000\mathrm{e}^{-0.075t} \qquad t \geqslant 0\]
Use the equation of the model to answer parts (a), (b), (c) and (d).
Given that \(N = 3000\) when \(t = T\)
After a marketing campaign, the total number of cars sold is expected to rise and have an upper limit of 6500
| Scheme | Marks | AO |
|---|---|---|
| \(5000 - 5000\mathrm{e}^{-0.075 \times 3} = \ldots\) | M1 | 3.4 |
| 1007 | A1 | 1.1b |
| (2) |
Notes
M1: Substitutes \(t = 3\) into the given model and proceeds to a value.
Implied by awrt 1007, 1008 or 1010
Condone copying slips e.g. \(-0.75\) or \(-0.0075\) in place of \(-0.075\) or 500 in place of 5000 – in such cases you may need to check their calculation if the substitution isn’t shown.
A1: 1007 but allow 1008 or 1010 (3sf). Must be a whole number.
Correct answer only scores both marks. A0 for e.g. 1007.4
Do not ISW if they go on to sum their values of \(N\) from \(t = 1\) to 3
| Scheme | Marks | AO |
|---|---|---|
| \(5000 - 5000\mathrm{e}^{-0.075T} = 3000 \rightarrow 5000\mathrm{e}^{-0.075T} = 2000\) | M1 | 1.1b |
| \(\Rightarrow T = \dfrac{1}{-0.075}\ln\dfrac{2000}{5000}\) | dM1 | 1.1b |
| \(\Rightarrow T = 12.22\) | A1 | 1.1b |
| (3) |
Notes
M1: Sets \(5000 - 5000\mathrm{e}^{-0.075T}\) equal to 3000 and proceeds to either \(A\mathrm{e}^{-0.075T} = B\) or \(\mathrm{e}^{-0.075T} = k\) where \(A\), \(B\), \(k\) are constants with no restrictions for this mark.
May use \(t\) instead of \(T\). Condone slips.
Condone copying slips e.g. \(-0.75\) or \(-0.0075\) in place of \(-0.075\) or 500 in place of 5000.
dM1: Uses the correct order of operations and correct log work from an equation of the form \(A\mathrm{e}^{-0.075T} = B\) with \(AB \gt 0\) or \(\mathrm{e}^{-0.075T} = k\) with \(k \gt 0\) to proceed to a value for \(T\) or \(t\) which may be a numerical expression.
e.g. \(A\mathrm{e}^{-0.075T} = B \Rightarrow \ln A - 0.075T = \ln B \Rightarrow T = \dfrac{\ln A - \ln B}{0.075}\) or \(= \dfrac{\ln\left(A/B\right)}{0.075}\)
Condone log being used in place of ln unless there is clear evidence that they have used an inconsistent base.
May not be scored if “recovered” from e.g. \(\mathrm{e}^{-0.075T} = -\dfrac{2}{5}\)
Condone copying slips e.g. \(-0.75\) or \(-0.0075\) in place of \(-0.075\) or 500 in place of 5000.
A1: 12.22 cao but must see a correct equation following use of logs, e.g., \(-0.075T = \ln\dfrac{2}{5}\) or \(T = \dfrac{1}{-0.075} \times -0.916\) which may have intermediate rounding. Ignore any units given.
This mark is available following the occasional slip in writing \(-0.075\) as \(-0.75\) or \(-0.0075\) provided it is not consistently used.
| Scheme | Marks | AO |
|---|---|---|
| \(\left(\dfrac{\mathrm{d}N}{\mathrm{d}t} =\right)\ -5000 \times -0.075\mathrm{e}^{-0.075 \times 3} = \ldots\) | M1 | 3.4 |
| 299 (car sales per month) | A1 | 1.1b |
| (2) |
Notes
M1: Differentiates once to the form \(\lambda\mathrm{e}^{-0.075t}\), \(\lambda\) a constant (even \(\pm 5000\)), and substitutes in \(t = 3\)
Award for an expression of the form \(k\mathrm{e}^{-0.075 \times 3}\) with constant \(k\) if no incorrect working is seen.
The substitution of \(t = 3\) may be implied by a correct value for their derivative of the required form, provided the derivative is seen. Do not be concerned with what they call their derivative.
A1: awrt 299 (car sales per month). Answer only scores no marks. Condone 300 following 299.4
Requires a correct derivative to be seen, e.g., \(375\mathrm{e}^{-0.075t}\) o.e., with or without the substitution of \(t = 3\) present.
Units may be omitted, but score A0 if an incorrect time frame is given e.g. cars per week
| Scheme | Marks | AO |
|---|---|---|
| Change the constant 5000 to 6500. | B1 | 3.5c |
| (1) | ||
| (8 marks) |
Notes
B1: Acceptable refinement. Some examples:
- Change the 5000 to 6500 (condone ambiguity about which 5000)
- Change the first 5000 to 6500
- Change both 5000s to 6500s
- Multiply the model by 1.3 (or e.g. \(\dfrac{65}{50}\) or \(\dfrac{13}{10}\))
- Add 1500 to the model
- Change the constant to 6500
The statement of an acceptable refined model e.g. \((N =)\ 6500 - 6500\mathrm{e}^{-0.075t}\) or e.g. \((N =)\ 6500 - 5000\mathrm{e}^{-0.075t}\) scores B1.
Ignore any extra unnecessary refinements such as increase/decrease the 0.075.
The following score B0:
- Change 5000 in the model (not specific enough – they haven’t said to 6500)
- Change the second 5000 to 6500 (incorrect)