| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2021 |
| Session | October |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Exponential model with shifted asymptote |
| Difficulty | Moderate -0.3 This is a straightforward exponential model question requiring routine manipulation: finding when G=0 (solving a simple exponential equation), substituting t=70, and identifying the horizontal asymptote. All techniques are standard P3 content with no problem-solving insight needed, making it slightly easier than average. |
| Spec | 1.06i Exponential growth/decay: in modelling context1.08d Evaluate definite integrals: between limits |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0 = 40 - 30e^{1-0.05t} \Rightarrow 1 - 0.05t = \ln\frac{40}{30}\) | M1 | Sets \(G=0\), proceeds to linear equation in \(t\) or \(k\) using logs of both sides; condone slips in rearrangement |
| \((k=)\) awrt 14.2 | A1 | Allow exact value \(20\left(1-\ln\frac{4}{3}\right)\) or equivalents such as \(\frac{\ln\frac{4}{3}-1}{-0.05}\) or \(20\ln\left(\frac{3}{4}e\right)\); isw after correct answer |
| March 1814 | A1 | Allow April 1814 following "correct" \(k\); condone "third month of 1814" or "fourth month of 1814"; withhold if two answers given of which one is incorrect |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(G = 40 - 30e^{1-0.05\times70} = 37.5\) tonnes | M1 A1 | M1: attempts \(G\) with \(t=70\) (condone \(t=69\)); A1: awrt 37.5 tonnes (requires units) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| 40 (tonnes) | B1 | Condone \(G < 40\) or \(G \approx 40\); do not allow \(G > 40\) |
## Question 3:
### Part (a)(i) and (ii) - marked together:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0 = 40 - 30e^{1-0.05t} \Rightarrow 1 - 0.05t = \ln\frac{40}{30}$ | M1 | Sets $G=0$, proceeds to linear equation in $t$ or $k$ using logs of both sides; condone slips in rearrangement |
| $(k=)$ awrt 14.2 | A1 | Allow exact value $20\left(1-\ln\frac{4}{3}\right)$ or equivalents such as $\frac{\ln\frac{4}{3}-1}{-0.05}$ or $20\ln\left(\frac{3}{4}e\right)$; isw after correct answer |
| March 1814 | A1 | Allow April 1814 following "correct" $k$; condone "third month of 1814" or "fourth month of 1814"; withhold if two answers given of which one is incorrect |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $G = 40 - 30e^{1-0.05\times70} = 37.5$ tonnes | M1 A1 | M1: attempts $G$ with $t=70$ (condone $t=69$); A1: awrt 37.5 tonnes (requires units) |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| 40 (tonnes) | B1 | Condone $G < 40$ or $G \approx 40$; do not allow $G > 40$ |
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9b0b8db0-79fd-4ad5-88c9-737447d9f894-10_541_618_248_671}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
The total mass of gold, $G$ tonnes, extracted from a mine is modelled by the equation
$$G = 40 - 30 \mathrm { e } ^ { 1 - 0.05 t } \quad t \geqslant k \quad G \geqslant 0$$
where $t$ is the number of years after 1st January 1800.\\
Figure 2 shows a sketch of $G$ against $t$.
Use the equation of the model to answer parts (a), (b) and (c).
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the value of $k$.
\item Hence find the year and month in which gold started being extracted from the mine.
\end{enumerate}\item Find the total mass of gold extracted from the mine up to 1st January 1870.
There is a limit to the mass of gold that can be extracted from the mine.
\item State the value of this limit.\\
M
\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2021 Q3 [6]}}