| Exam Board | OCR MEI |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2021 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Exponential growth/decay model setup |
| Difficulty | Moderate -0.8 This question involves routine sketching of dy/dx for e^x (which is just e^x again), basic interpretation of exponential models, and straightforward substitution to find constants A and k. The conceptual demand is low—recognizing that dy/dx = e^x and substituting two data points into y = Ae^(kt) are standard textbook exercises. The reasoning parts require only simple biological common sense (exponential growth assumptions, carrying capacity limitations). No novel problem-solving or multi-step integration of techniques is required. |
| Spec | 1.06i Exponential growth/decay: in modelling context1.07b Gradient as rate of change: dy/dx notation |
| Answer | Marks |
|---|---|
| [Graph: \(\frac{dy}{dx}\) vs \(x\), curve in first quadrant approaching from bottom left, increasing, asymptotic shape] | B1 (1.2) |
| Answer | Marks | Guidance |
|---|---|---|
| [Graph: \(\frac{dy}{dx}\) vs \(y\), straight line with positive gradient through origin, first quadrant only] | M1 (2.2a) | Straight line (+ve gradient) through origin (may stop short of origin) |
| A1 (2.2a) | First quadrant only |
| Answer | Marks | Guidance |
|---|---|---|
| Suitable reason e.g.: reasonable to assume population growth is proportional to population; populations are often modelled by exponential growth | E1 (3.3) | Allow e.g. wolves give birth to more wolves than they started with; Do not allow e.g. population is proportional to time |
| Answer | Marks | Guidance |
|---|---|---|
| \(A = 21\) | B1 (3.4) | |
| \(51 = 21e^k\) | M1 (3.4) | |
| \(k = \ln\left(\frac{51}{21}\right) = \ln\left(\frac{17}{7}\right) \approx 0.887\) or better | A1 (1.1) | Allow 0.89 |
| Answer | Marks | Guidance |
|---|---|---|
| Suitable reason e.g.: population cannot keep growing; the wolves will run out of food if the population gets too big | E1 (3.5b) | Allow e.g. lack of resources or deforestation |
## Question 5:
**Part (a)(i):**
[Graph: $\frac{dy}{dx}$ vs $x$, curve in first quadrant approaching from bottom left, increasing, asymptotic shape] | B1 (1.2) |
[1 mark]
**Part (a)(ii):**
[Graph: $\frac{dy}{dx}$ vs $y$, straight line with positive gradient through origin, first quadrant only] | M1 (2.2a) | Straight line (+ve gradient) through origin (may stop short of origin)
| A1 (2.2a) | First quadrant only
[3 marks]
**Part (b)(i):**
Suitable reason e.g.: reasonable to assume population growth is proportional to population; populations are often modelled by exponential growth | E1 (3.3) | Allow e.g. wolves give birth to more wolves than they started with; Do not allow e.g. population is proportional to time
[1 mark]
**Part (b)(ii):**
$A = 21$ | B1 (3.4) |
$51 = 21e^k$ | M1 (3.4) |
$k = \ln\left(\frac{51}{21}\right) = \ln\left(\frac{17}{7}\right) \approx 0.887$ or better | A1 (1.1) | Allow 0.89
[3 marks]
**Part (b)(iii):**
Suitable reason e.g.: population cannot keep growing; the wolves will run out of food if the population gets too big | E1 (3.5b) | Allow e.g. lack of resources or deforestation
[1 mark]
---5
\begin{enumerate}[label=(\alph*)]
\item The diagram shows the curve $\mathrm { y } = \mathrm { e } ^ { \mathrm { x } }$.\\
\includegraphics[max width=\textwidth, alt={}, center]{a0d9573f-8273-4562-a2d3-07f15d9da1af-5_574_682_315_328}
On the axes in the Printed Answer Booklet, sketch graphs of
\begin{enumerate}[label=(\roman*)]
\item $\frac { \mathrm { dy } } { \mathrm { dx } }$ against $x$,
\item $\frac { \mathrm { dy } } { \mathrm { dx } }$ against $y$.
\end{enumerate}\item Wolves were introduced to Yellowstone National Park in 1995.
The population of wolves, $y$, is modelled by the equation\\
$y = A e ^ { k t }$,\\
where $A$ and $k$ are constants and $t$ is the number of years after 1995.
\begin{enumerate}[label=(\roman*)]
\item Give a reason why this model might be suitable for the population of wolves.
\item When $t = 0 , y = 21$ and when $t = 1 , y = 51$.
Find values of $A$ and $k$ consistent with the data.
\item Give a reason why the model will not be a good predictor of wolf populations many years after 1995.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 3 2021 Q5 [8]}}