| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2021 |
| Session | October |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Exponential growth/decay model setup |
| Difficulty | Moderate -0.3 This is a standard exponential modeling question requiring routine application of initial conditions to find constants, differentiation for rate of change, and solving a simple exponential equation. All steps are textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.02y Partial fractions: decompose rational functions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(A = 1000\) | B1 | Correct value of \(A\); award if model is of form \(N = 1000e^{...t}\) |
| \(2000 = 1000e^{5k}\) or \(e^{5k} = 2\) | M1 | Uses model to set up correct equation in \(k\); substituting \(N=2000, t=5\) following their \(A\) |
| \(e^{5k} = 2 \Rightarrow 5k = \ln 2 \Rightarrow k = ...\) | M1 | Uses correct ln work to solve \(ae^{5k} = b\) and obtain value for \(k\) |
| \(N = 1000e^{(\frac{1}{5}\ln 2)t}\) or \(N = 1000e^{0.139t}\) | A1 | Correct equation of model; also accept \(N = 1000 \times 2^{\frac{t}{5}}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dN}{dt} = 1000 \times \left(\frac{1}{5}\ln 2\right)e^{\left(\frac{1}{5}\ln 2\right)t}\) or \(\frac{dN}{dt} = 1000 \times 0.139e^{0.139t}\); substitutes \(t=8\) | M1 | Differentiates \(\alpha e^{kt}\) to \(\beta e^{kt}\) and substitutes \(t=8\); condone \(\alpha = \beta\) |
| \(= \text{awrt } 420\) | A1 | For awrt 420 (2sf) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(500e^{1.4\times(\frac{1}{5}\ln 2)T} = 1000e^{(\frac{1}{5}\ln 2)T}\) or \(500e^{1.4\times"0.139"t} = 1000e^{"0.139"t}\) | M1 | Uses both models to set up equation in \(T\) using their \(k\); also allow in terms of \(k\) |
| E.g. \(0.08T\ln 2 = \ln 2\) or \(1.4\times"0.339"T = \ln 2 + "0.339"t\) | M1 | Correct processing using lns to obtain a linear equation in \(T\) |
| \(T = 12.5\) hours | A1 | Awrt 12.5 |
## Question 8:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $A = 1000$ | B1 | Correct value of $A$; award if model is of form $N = 1000e^{...t}$ |
| $2000 = 1000e^{5k}$ or $e^{5k} = 2$ | M1 | Uses model to set up correct equation in $k$; substituting $N=2000, t=5$ following their $A$ |
| $e^{5k} = 2 \Rightarrow 5k = \ln 2 \Rightarrow k = ...$ | M1 | Uses correct ln work to solve $ae^{5k} = b$ and obtain value for $k$ |
| $N = 1000e^{(\frac{1}{5}\ln 2)t}$ or $N = 1000e^{0.139t}$ | A1 | Correct equation of model; also accept $N = 1000 \times 2^{\frac{t}{5}}$ |
**(4 marks)**
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dN}{dt} = 1000 \times \left(\frac{1}{5}\ln 2\right)e^{\left(\frac{1}{5}\ln 2\right)t}$ or $\frac{dN}{dt} = 1000 \times 0.139e^{0.139t}$; substitutes $t=8$ | M1 | Differentiates $\alpha e^{kt}$ to $\beta e^{kt}$ and substitutes $t=8$; condone $\alpha = \beta$ |
| $= \text{awrt } 420$ | A1 | For awrt 420 (2sf) |
**(2 marks)**
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $500e^{1.4\times(\frac{1}{5}\ln 2)T} = 1000e^{(\frac{1}{5}\ln 2)T}$ or $500e^{1.4\times"0.139"t} = 1000e^{"0.139"t}$ | M1 | Uses both models to set up equation in $T$ using their $k$; also allow in terms of $k$ |
| E.g. $0.08T\ln 2 = \ln 2$ or $1.4\times"0.339"T = \ln 2 + "0.339"t$ | M1 | Correct processing using lns to obtain a linear equation in $T$ |
| $T = 12.5$ hours | A1 | Awrt 12.5 |
**(3 marks)**
---\begin{enumerate}
\item A scientist is studying the growth of 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 1000 bacteria in this population at the start of the study
\item it took exactly 5 hours from the start of the study for this population to double\\
(a) find a complete equation for the model.\\
(b) Hence find the rate of increase in the number of bacteria in this population exactly 8 hours from the start of the study. Give your answer to 2 significant figures.
\end{itemize}
The number of bacteria, $M$, in the second population is modelled by the equation
$$M = 500 \mathrm { e } ^ { 1.4 k t } \quad t \geqslant 0$$
where $k$ has the value found in part (a) and $t$ is the time in hours from the start of the study.\\
Given that $T$ hours after the start of the study, the number of bacteria in the two different populations was the same,\\
(c) find the value of $T$.
\hfill \mbox{\textit{Edexcel Paper 1 2021 Q8 [9]}}