| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2021 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Exponential model with shifted asymptote |
| Difficulty | Moderate -0.3 This is a straightforward exponential model question requiring substitution (part a), solving for a constant using logarithms (part b), and identifying the horizontal asymptote (part c). All techniques are standard AS-level procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.06i Exponential growth/decay: in modelling context |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(35\ (\text{km}^2)\) | B1 | Uses equation of model to find area on 1st January 2005; do not accept \(35\ \text{m}^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Sets \(60 = 80 - 45e^{14c} \Rightarrow 45e^{14c} = 20\) | M1, A1 | Sets \(Ae^{14c} = B\) |
| \(\Rightarrow c = \frac{1}{14}\ln\!\left(\frac{20}{45}\right) = \cdots \approx -0.0579\) | dM1 | Full method using correct log laws; \(e^x\) and \(\ln x\) are inverse functions |
| \(A = 80 - 45e^{-0.0579t}\) | A1 | Complete equation for model |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Suitable answer, e.g. maximum area covered by trees is only \(80\ \text{km}^2\); the "80" would need to be "100"; substituting 100 shows formula fails (cannot take log of negative number) | B1 | Gives suitable interpretation |
## Question 11:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $35\ (\text{km}^2)$ | B1 | Uses equation of model to find area on 1st January 2005; do not accept $35\ \text{m}^2$ |
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Sets $60 = 80 - 45e^{14c} \Rightarrow 45e^{14c} = 20$ | M1, A1 | Sets $Ae^{14c} = B$ |
| $\Rightarrow c = \frac{1}{14}\ln\!\left(\frac{20}{45}\right) = \cdots \approx -0.0579$ | dM1 | Full method using correct log laws; $e^x$ and $\ln x$ are inverse functions |
| $A = 80 - 45e^{-0.0579t}$ | A1 | Complete equation for model |
### Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Suitable answer, e.g. maximum area covered by trees is only $80\ \text{km}^2$; the "80" would need to be "100"; substituting 100 shows formula fails (cannot take log of negative number) | B1 | Gives suitable interpretation |
---\begin{enumerate}
\item The owners of a nature reserve decided to increase the area of the reserve covered by trees.
\end{enumerate}
Tree planting started on 1st January 2005.\\
The area of the nature reserve covered by trees, $A \mathrm {~km} ^ { 2 }$, is modelled by the equation
$$A = 80 - 45 \mathrm { e } ^ { c t }$$
where $c$ is a constant and $t$ is the number of years after 1st January 2005.\\
Using the model,\\
(a) find the area of the nature reserve that was covered by trees just before tree planting started.
On 1st January 2019 an area of $60 \mathrm {~km} ^ { 2 }$ of the nature reserve was covered by trees.\\
(b) Use this information to find a complete equation for the model, giving your value of $c$ to 3 significant figures.
On 1st January 2020, the owners of the nature reserve announced a long-term plan to have $100 \mathrm {~km} ^ { 2 }$ of the nature reserve covered by trees.\\
(c) State a reason why the model is not appropriate for this plan.
\hfill \mbox{\textit{Edexcel AS Paper 1 2021 Q11 [6]}}