| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Long-term behaviour analysis |
| Difficulty | Standard +0.3 This is a structured differential equations question with clear steps: (a) requires simple substitution at t=0, (b) involves algebraic manipulation to find k (answer given), and (c) requires differentiating the logistic function and solving. While it involves multiple techniques (implicit differentiation, logarithms, solving equations), each part is guided and uses standard A-level methods without requiring novel insight. Slightly easier than average due to the scaffolding and given answer in part (b). |
| Spec | 1.02z Models in context: use functions in modelling1.06i Exponential growth/decay: in modelling context1.07k Differentiate trig: sin(kx), cos(kx), tan(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((F=)\ 35\) | B1 | Correct value, 35 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(200 = \frac{350e^{15k}}{9+e^{15k}} \Rightarrow 1800+200e^{15k}=350e^{15k} \Rightarrow 150e^{15k}=1800\) | M1 | Uses \(F=200\) and \(t=15\) and reaches \(Ae^{15k}=B\) where \(A\times B>0\) |
| \(e^{15k} = \frac{1800}{150} \Rightarrow 15k = \ln 12 \Rightarrow k = \frac{1}{15}\ln 12\) * | dM1, A1* | dM1: proceeds using correct order of operations to obtain a value for \(k\); A1: correct proof with all necessary steps shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F = \frac{350e^{kt}}{9+e^{kt}} \Rightarrow \frac{dF}{dt} = \frac{350ke^{kt}\left(9+e^{kt}\right)-350e^{kt}\left(ke^{kt}\right)}{\left(9+e^{kt}\right)^2}\) | M1, A1 | M1: correct attempt at quotient (product or chain) rule; for quotient rule look for \(\frac{dF}{dt}=\frac{Pe^{kt}(9+e^{kt})-Qe^{kt}(e^{kt})}{(9+e^{kt})^2}\), \(P,Q>0\); for product rule look for \(Pe^{kt}(9+e^{kt})^{-1}\pm Qe^{2kt}(9+e^{kt})^{-2}\); A1: correct differentiation, may be unsimplified |
| \(\frac{3150ke^{kt}}{\left(9+e^{kt}\right)^2}=10 \Rightarrow 315ke^{kt}=81+18e^{kt}+e^{2kt} \Rightarrow e^{2kt}+(18-315k)e^{kt}+81=0\) | M1 | Sets derivative \(=10\) and obtains a 3TQ in \(e^{kt}\); dependent on reasonable attempt to differentiate; allow whole of part (c) with exact \(k\), or \(k=\) awrt 0.166 |
| \(e^{2kt}+(18-315k)e^{kt}+81=0 \Rightarrow e^{kt}=\frac{315k-18\pm\sqrt{(18-315k)^2-4\times81}}{2} \Rightarrow kt=\ldots\) | M1 | Solving 3TQ in \(e^{kt}\) by any method including calculator (check accuracy to 2sf rounded or truncated); then taking ln's to obtain at least one value for \(kt\) |
| \(T=\) awrt \(5.7,\ 20.8\) | A1 | Both values awrt 5.7, 20.8 |
# Question 10:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(F=)\ 35$ | B1 | Correct value, 35 |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $200 = \frac{350e^{15k}}{9+e^{15k}} \Rightarrow 1800+200e^{15k}=350e^{15k} \Rightarrow 150e^{15k}=1800$ | M1 | Uses $F=200$ and $t=15$ and reaches $Ae^{15k}=B$ where $A\times B>0$ |
| $e^{15k} = \frac{1800}{150} \Rightarrow 15k = \ln 12 \Rightarrow k = \frac{1}{15}\ln 12$ * | dM1, A1* | dM1: proceeds using correct order of operations to obtain a value for $k$; A1: correct proof with all necessary steps shown |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F = \frac{350e^{kt}}{9+e^{kt}} \Rightarrow \frac{dF}{dt} = \frac{350ke^{kt}\left(9+e^{kt}\right)-350e^{kt}\left(ke^{kt}\right)}{\left(9+e^{kt}\right)^2}$ | M1, A1 | M1: correct attempt at quotient (product or chain) rule; for quotient rule look for $\frac{dF}{dt}=\frac{Pe^{kt}(9+e^{kt})-Qe^{kt}(e^{kt})}{(9+e^{kt})^2}$, $P,Q>0$; for product rule look for $Pe^{kt}(9+e^{kt})^{-1}\pm Qe^{2kt}(9+e^{kt})^{-2}$; A1: correct differentiation, may be unsimplified |
| $\frac{3150ke^{kt}}{\left(9+e^{kt}\right)^2}=10 \Rightarrow 315ke^{kt}=81+18e^{kt}+e^{2kt} \Rightarrow e^{2kt}+(18-315k)e^{kt}+81=0$ | M1 | Sets derivative $=10$ and obtains a 3TQ in $e^{kt}$; dependent on reasonable attempt to differentiate; allow whole of part (c) with exact $k$, or $k=$ awrt 0.166 |
| $e^{2kt}+(18-315k)e^{kt}+81=0 \Rightarrow e^{kt}=\frac{315k-18\pm\sqrt{(18-315k)^2-4\times81}}{2} \Rightarrow kt=\ldots$ | M1 | Solving 3TQ in $e^{kt}$ by any method including calculator (check accuracy to 2sf rounded or truncated); then taking ln's to obtain at least one value for $kt$ |
| $T=$ awrt $5.7,\ 20.8$ | A1 | Both values awrt 5.7, 20.8 |\begin{enumerate}
\item In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
\end{enumerate}
A population of fruit flies is being studied.\\
The number of fruit flies, $F$, in the population, $t$ days after the start of the study, is modelled by the equation
$$F = \frac { 350 \mathrm { e } ^ { k t } } { 9 + \mathrm { e } ^ { k t } }$$
where $k$ is a constant.\\
Use the equation of the model to answer parts (a), (b) and (c).\\
(a) Find the number of fruit flies in the population at the start of the study.
Given that there are 200 fruit flies in the population 15 days after the start of the study,\\
(b) show that $k = \frac { 1 } { 15 } \ln 12$
Given also that, when $t = T$, the number of fruit flies in the population is increasing at a rate of 10 per day,\\
(c) find the possible values of $T$, giving your answers to one decimal place.
\hfill \mbox{\textit{Edexcel P3 2023 Q10 [9]}}