| Exam Board | OCR MEI |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2020 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Linear transformation to find constants |
| Difficulty | Moderate -0.3 This is a standard exponential modeling question requiring routine application of logarithmic linearization. Part (a)(i) is trivial differentiation, part (b) is showing ln(y) = ln(A) + kx (bookwork), and parts (c)-(d) involve reading a graph and substituting values. Only part (e) requires brief critical thinking about extrapolation. Slightly easier than average due to the guided structure and minimal calculation complexity. |
| Spec | 1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form1.06i Exponential growth/decay: in modelling context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(ke^{kx}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Reason referring to growth proportional to profit | E1 | E.g. As more people hear about the business, they will sell more so it is reasonable that the rate of growth is proportional to the profits |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\ln y = \ln A + kx\) | M1 | |
| Equation is of form "\(y = mx + c\)" so a straight line — hence model is consistent with graph | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\ln A = 1.9\) | M1 | Intercept \(= 1.9\) not enough for M1 |
| \(A = 6.686\) | A1 | One or more d.p. Allow \(e^{1.9}\) |
| \(k = \frac{0.4}{1.6}\) | M1 | Attempt to find gradient. \(\frac{11}{45}\) gets M1A1 |
| \(k = 0.24\) to \(0.25\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = 11\) | B1 | |
| \(y = 6.686 \times e^{0.25x}\), so \(y = 6.686 \times e^{0.25\times11}\) | M1 | Use of model with *their* \(A\) and \(k\) |
| \(104.586...\), £104 587 | A1 | Translation into pounds (may be to nearest thousand) (FT their \(k\) and \(a\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Not reliable. Extrapolation may not be valid. OR: Fit of model is good so far and it's only two more years o.e. | E1 | E.g. relationship between sales and time may change; there is a limit to the market for revision resources; changes in the economy may affect the business. 'Reliable' plus 'extrapolation' does not score |
## Question 6(a)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $ke^{kx}$ | B1 | |
---
## Question 6(a)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Reason referring to growth proportional to profit | E1 | E.g. As more people hear about the business, they will sell more so it is reasonable that the rate of growth is proportional to the profits |
---
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\ln y = \ln A + kx$ | M1 | |
| Equation is of form "$y = mx + c$" so a straight line — hence model is consistent with graph | E1 | |
---
## Question 6(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\ln A = 1.9$ | M1 | Intercept $= 1.9$ not enough for M1 |
| $A = 6.686$ | A1 | One or more d.p. Allow $e^{1.9}$ |
| $k = \frac{0.4}{1.6}$ | M1 | Attempt to find gradient. $\frac{11}{45}$ gets M1A1 |
| $k = 0.24$ to $0.25$ | A1 | |
---
## Question 6(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = 11$ | B1 | |
| $y = 6.686 \times e^{0.25x}$, so $y = 6.686 \times e^{0.25\times11}$ | M1 | Use of model with *their* $A$ and $k$ |
| $104.586...$, £104 587 | A1 | Translation into pounds (may be to nearest thousand) (FT their $k$ and $a$) |
---
## Question 6(e):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Not reliable. Extrapolation may not be valid. OR: Fit of model is good so far and it's only two more years o.e. | E1 | E.g. relationship between sales and time may change; there is a limit to the market for revision resources; changes in the economy may affect the business. 'Reliable' plus 'extrapolation' does not score |
---6
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Write down the derivative of $\mathrm { e } ^ { \mathrm { kx } }$, where $k$ is a constant.
\item A business has been running since 2009. They sell maths revision resources online.
Give a reason why an exponential growth model might be suitable for the annual profits for the business.
Fig. 6 shows the relationship between the annual profits of the business in thousands of pounds ( $y$ ) and the time in years after $2009 ( x )$. The graph of lny plotted against $x$ is approximately a straight line.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a13f7a05-e2d3-4354-a0c7-ef7283eff514-07_1052_1157_751_242}
\captionsetup{labelformat=empty}
\caption{Fig. 6}
\end{center}
\end{figure}
\end{enumerate}\item Show that the straight line is consistent with a model of the form $\mathbf { y } = \mathrm { Ae } ^ { \mathrm { kx } }$, where $A$ and $k$ are constants.
\item Estimate the values of $A$ and $k$.
\item Use the model to predict the profit in the year 2020.
\item How reliable do you expect the prediction in part (d) to be? Justify your answer.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 3 2020 Q6 [12]}}