| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Exponential growth/decay model setup |
| Difficulty | Moderate -0.3 This is a straightforward exponential model question requiring standard techniques: substituting initial conditions to find constants, differentiating to find rate of change, and solving an exponential equation using logarithms. All steps are routine A-level procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.06i Exponential growth/decay: in modelling context1.07k Differentiate trig: sin(kx), cos(kx), tan(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| States or implies \(A = 2500\) | B1 | e.g. award for \(N = 2500e^{kt}\) |
| \(10000 = 2500e^{k\times8} \Rightarrow 8k = \ln 4 \Rightarrow k = ...\) | M1 | Use \(N = Ae^{kt}\) with \(t=8\), \(N=10000\) and their \(A\); must use correct ln work |
| \(k = \frac{1}{8}\ln 4\) or awrt \(0.1733\) | A1 | Accept exact value \(\frac{1}{8}\ln 4\); isw after seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dN}{dt} = 60000 \times -0.6e^{-0.6\times5} = -1792\), so decrease is \(1790\) | M1, A1 | M1: \(\frac{dN}{dt} = Ce^{-0.6\times5}\) where \(C\) is constant; must be correct index (not \(kt\)). A1: awrt 1790; condone awrt \(-1790\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(60000e^{-0.6t} = 2500e^{0.1733t}\) | M1 | Sets equal; allow slip e.g. 60000 as 6000; allow \(k\) in place of 0.1733 if \(k\) found in (a) |
| \(24 = e^{0.1733t+0.6t} \Rightarrow 0.1733t + 0.6t = \ln 24 \Rightarrow t = ...\) | dM1 | Rearranges to \(e^{mt} = D\) (\(D>0\)) and applies ln; ln work must be correct |
| \(T = 4.11\) | A1 | awrt 4.11; must be a value not expression in ln terms; answer only scores no marks |
# Question 7:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| States or implies $A = 2500$ | B1 | e.g. award for $N = 2500e^{kt}$ |
| $10000 = 2500e^{k\times8} \Rightarrow 8k = \ln 4 \Rightarrow k = ...$ | M1 | Use $N = Ae^{kt}$ with $t=8$, $N=10000$ and their $A$; must use correct ln work |
| $k = \frac{1}{8}\ln 4$ or awrt $0.1733$ | A1 | Accept exact value $\frac{1}{8}\ln 4$; isw after seen |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dN}{dt} = 60000 \times -0.6e^{-0.6\times5} = -1792$, so decrease is $1790$ | M1, A1 | M1: $\frac{dN}{dt} = Ce^{-0.6\times5}$ where $C$ is constant; must be correct index (not $kt$). A1: awrt 1790; condone awrt $-1790$ |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $60000e^{-0.6t} = 2500e^{0.1733t}$ | M1 | Sets equal; allow slip e.g. 60000 as 6000; allow $k$ in place of 0.1733 if $k$ found in (a) |
| $24 = e^{0.1733t+0.6t} \Rightarrow 0.1733t + 0.6t = \ln 24 \Rightarrow t = ...$ | dM1 | Rearranges to $e^{mt} = D$ ($D>0$) and applies ln; ln work must be correct |
| $T = 4.11$ | A1 | awrt 4.11; must be a value not expression in ln terms; answer only scores no marks |\begin{enumerate}
\item A scientist is studying two different populations of bacteria.
\end{enumerate}
The number of bacteria $N$ in the first population is modelled by the equation
$$N = A \mathrm { e } ^ { k t } \quad t \geqslant 0$$
where $A$ and $k$ are positive constants and $t$ is the time in hours from the start of the study.
Given that
\begin{itemize}
\item there were 2500 bacteria in this population at the start of the study
\item there were 10000 bacteria 8 hours later\\
(a) find the exact value of $A$ and the value of $k$ to 4 significant figures.
\end{itemize}
The number of bacteria $N$ in the second population is modelled by the equation
$$N = 60000 \mathrm { e } ^ { - 0.6 t } \quad t \geqslant 0$$
where $t$ is the time in hours from the start of the study.\\
(b) Find the rate of decrease of bacteria in this population exactly 5 hours from the start of the study. Give your answer to 3 significant figures.
When $t = T$, the number of bacteria in the two different populations was the same.\\
(c) Find the value of $T$, giving your answer to 3 significant figures.\\
(Solutions relying entirely on calculator technology are not acceptable.)
\hfill \mbox{\textit{Edexcel P3 2023 Q7 [8]}}