| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2022 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Logistic growth model |
| Difficulty | Standard +0.3 This is a standard logistic growth model question requiring substitution to find a parameter, solving an exponential equation using logarithms, and finding a limit as tââ. All techniques are routine for P3 level with clear signposting, though it requires careful algebraic manipulation across multiple steps. |
| Spec | 1.06g Equations with exponentials: solve a^x = b1.06i Exponential growth/decay: in modelling context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Sets \(t=0\), \(e^0=1\): \(30=\frac{80p}{p+4}\) | M1 | Sets \(A=30\) and \(e^{0.15\times0}=1\) to form equation in \(p\) |
| \(30p+120=80p \Rightarrow p=\frac{120}{50}=2.4\) | A1* | No significant errors; one correct linear equation in \(p\); condone minor slips recovered before answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(50=\frac{80\times2.4\,e^{0.15T}}{2.4\,e^{0.15T}+4}\Rightarrow 72e^{0.15T}=200\) | M1 A1 | M1: Sets \(A=50\), \(p=2.4\), proceeds to form \(ce^{0.15t}=d\), \(c\times d>0\); A1: \(72e^{0.15T}=200\) |
| \(0.15T=\ln\!\left(\frac{200}{72}\right)\Rightarrow T=6.8\) | dM1 A1 | dM1: Correct order of operations to find \(T\); A1: AWRT 6.8 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(80\text{ m}^2\) | B1 | Units required |
## Question 4:
$$A = \frac{80p\,e^{0.15t}}{p\,e^{0.15t}+4}$$
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Sets $t=0$, $e^0=1$: $30=\frac{80p}{p+4}$ | M1 | Sets $A=30$ and $e^{0.15\times0}=1$ to form equation in $p$ |
| $30p+120=80p \Rightarrow p=\frac{120}{50}=2.4$ | A1* | No significant errors; one correct linear equation in $p$; condone minor slips recovered before answer |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $50=\frac{80\times2.4\,e^{0.15T}}{2.4\,e^{0.15T}+4}\Rightarrow 72e^{0.15T}=200$ | M1 A1 | M1: Sets $A=50$, $p=2.4$, proceeds to form $ce^{0.15t}=d$, $c\times d>0$; A1: $72e^{0.15T}=200$ |
| $0.15T=\ln\!\left(\frac{200}{72}\right)\Rightarrow T=6.8$ | dM1 A1 | dM1: Correct order of operations to find $T$; A1: AWRT 6.8 |
### Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $80\text{ m}^2$ | B1 | Units required |
---4. The growth of a weed on the surface of a pond is being studied.
The surface area of the pond covered by the weed, $A \mathrm {~m} ^ { 2 }$, is modelled by the equation
$$A = \frac { 80 p \mathrm { e } ^ { 0.15 t } } { p \mathrm { e } ^ { 0.15 t } + 4 }$$
where $p$ is a positive constant and $t$ is the number of days after the start of the study.\\
Given that
\begin{itemize}
\item $30 \mathrm {~m} ^ { 2 }$ of the surface of the pond was covered by the weed at the start of the study
\item $50 \mathrm {~m} ^ { 2 }$ of the surface of the pond was covered by the weed $T$ days after the start of the study
\begin{enumerate}[label=(\alph*)]
\item show that $p = 2.4$
\item find the value of $T$, giving your answer to one decimal place.\\
(Solutions relying entirely on graphical or numerical methods are not acceptable.)
\end{itemize}
The weed grows until it covers the surface of the pond.
\item Find, according to the model, the maximum possible surface area of the pond.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2022 Q4 [7]}}