| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Session | Specimen |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Rate of change in exponential model |
| Difficulty | Standard +0.3 This is a straightforward exponential modeling question requiring standard techniques: sketching exponential functions, identifying asymptotes, interpreting limits, basic differentiation to compare rates of change, and solving a simple exponential equation. Part (d) requires differentiation and algebraic manipulation but follows directly from the given equations. All parts are routine applications of A-level techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02z Models in context: use functions in modelling1.06a Exponential function: a^x and e^x graphs and properties1.06i Exponential growth/decay: in modelling context1.07b Gradient as rate of change: dy/dx notation1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07l Derivative of ln(x): and related functions |
| Answer | Marks | Guidance |
|---|---|---|
| [Graph: \(P_G\) shape through O; \(P_R\) shape through \((0, 20000)\)] | M1, M1 | \(P_G\) shape through O; \(P_R\) shape through \((0, 20000)\); condone graphs for \(-ve\ t\) |
| Answer | Marks | Guidance |
|---|---|---|
| Asymptote for \(P_G = 10000\) | A1 | Or \(p = 10000\) |
| Asymptote for \(P_R = 0\) | A1 [4] | Or \(p = 0\) |
| Answer | Marks |
|---|---|
| Red squirrels zero | B1 |
| Grey \(10000\) | B1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| One relevant comment evaluating the validity of the model | B1 [1] | E.g. Grey population increases as would be expected; Red population decreases as would be expected; Number of squirrels tends to a limit as would be expected; Would expect grey population to grow slower at first; Would expect red population to fall slower at first |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dP_G}{dt} = 10000ke^{-kt}\) | A1 | Attempts to differentiate either or both |
| \(\frac{dP_R}{dt} = -20000ke^{-kt}\) | A1 | |
| so \(\frac{dP_R}{dt} = -2\frac{dP_G}{dt}\) | E1 [4] | Or in words |
| Answer | Marks | Guidance |
|---|---|---|
| \(10000(1 - e^{-3k}) = 20000e^{-3k}\) | M1 | |
| \(\Rightarrow 1 - e^{-3k} = 2e^{-3k}\) | A1 | |
| \(\Rightarrow e^{-3k} = \frac{1}{3}\) | ||
| \(\Rightarrow -3k = \ln\left(\frac{1}{3}\right)\) | M1 | Taking natural logs of both sides |
| \(\Rightarrow k = -\frac{1}{3}\ln\left(\frac{1}{3}\right) = 0.366\) or \(\frac{1}{3}\ln 3\) | A1 [4] | cao |
## Question 10:
### Part (a)(i):
[Graph: $P_G$ shape through O; $P_R$ shape through $(0, 20000)$] | **M1, M1** | $P_G$ shape through O; $P_R$ shape through $(0, 20000)$; condone graphs for $-ve\ t$
### Part (a)(ii):
Asymptote for $P_G = 10000$ | **A1** | Or $p = 10000$
Asymptote for $P_R = 0$ | **A1** [4] | Or $p = 0$
### Part (b):
Red squirrels zero | **B1** |
Grey $10000$ | **B1** [2] |
### Part (c):
One relevant comment evaluating the validity of the model | **B1** [1] | E.g. Grey population increases as would be expected; Red population decreases as would be expected; Number of squirrels tends to a limit as would be expected; Would expect grey population to grow slower at first; Would expect red population to fall slower at first
### Part (d):
$\frac{dP_G}{dt} = 10000ke^{-kt}$ | **A1** | Attempts to differentiate either or both
$\frac{dP_R}{dt} = -20000ke^{-kt}$ | **A1** |
so $\frac{dP_R}{dt} = -2\frac{dP_G}{dt}$ | **E1** [4] | Or in words
### Part (e):
$10000(1 - e^{-3k}) = 20000e^{-3k}$ | **M1** |
$\Rightarrow 1 - e^{-3k} = 2e^{-3k}$ | **A1** |
$\Rightarrow e^{-3k} = \frac{1}{3}$ | |
$\Rightarrow -3k = \ln\left(\frac{1}{3}\right)$ | **M1** | Taking natural logs of both sides
$\Rightarrow k = -\frac{1}{3}\ln\left(\frac{1}{3}\right) = 0.366$ or $\frac{1}{3}\ln 3$ | **A1** [4] | cao10 In a certain region, the populations of grey squirrels, $P _ { \mathrm { G } }$ and red squirrels $P _ { \mathrm { R } }$, at time $t$ years are modelled by the equations:\\
$P _ { \mathrm { G } } = 10000 \left( 1 - \mathrm { e } ^ { - k t } \right)$\\
$P _ { \mathrm { R } } = 20000 \mathrm { e } ^ { - k t }$\\
where $t \geq 0$ and $k$ is a positive constant.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item On the axes in your Printed Answer Book, sketch the graphs of $P _ { \mathrm { G } }$ and $P _ { \mathrm { R } }$ on the same axes.
\item Give the equations of any asymptotes.
\end{enumerate}\item What does the model predict about the long term population of
\begin{itemize}
\item grey squirrels
\item red squirrels?
\end{itemize}
Grey squirrels and red squirrels compete for food and space. Grey squirrels are larger and more successful than red squirrels.
\item Comment on the validity of the model given by the equations, giving a reason for your answer.
\item Show that, according to the model, the rate of decrease of the population of red squirrels is always double the rate of increase of the population of grey squirrels.
\item When $t = 3$, the numbers of grey and red squirrels are equal. Find the value of $k$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 Q10 [15]}}