| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Logistic growth model |
| Difficulty | Standard +0.8 This logistic growth question requires multiple techniques beyond standard C3: part (a) is routine substitution, but (b) requires algebraic manipulation to isolate exponential terms and use logarithms non-trivially, (c) demands quotient rule differentiation with exponentials (algebraically intensive), and (d) requires understanding of horizontal asymptotes and limits—testing conceptual depth rather than just procedural fluency. The multi-step nature and need to work with the logistic model structure elevates this above typical C3 fare. |
| Spec | 1.02z Models in context: use functions in modelling1.06a Exponential function: a^x and e^x graphs and properties1.06g Equations with exponentials: solve a^x = b1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P = \frac{800e^0}{1+3e^0} = \frac{800}{1+3} = 200\) | M1, A1 | Sub \(t=0\) into \(P\) and use \(e^0=1\) in at least one case. Accept \(P=\frac{800}{1+3}\) as evidence. A1: 200. (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(250 = \frac{800e^{0.1t}}{1+3e^{0.1t}}\) | Sub \(P=250\), cross multiply, collect terms in \(e^{0.1t}\) and proceed to \(Ae^{0.1t}=B\) | |
| \(250(1+3e^{0.1t}) = 800e^{0.1t} \Rightarrow 50e^{0.1t}=250 \Rightarrow e^{0.1t}=5\) | M1, A1 | A1: \(e^{0.1t}=5\) or \(e^{-0.1t}=0.2\) |
| \(t = \frac{1}{0.1}\ln(5)\) | M1 | Dependent on \(e^{0.1t}=E\); take ln of both sides |
| \(t = 10\ln(5)\) | A1 | Accept exact equivalents e.g. \(t=5\ln(25)\). (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dP}{dt} = \frac{(1+3e^{0.1t})\times 800\times 0.1e^{0.1t} - 800e^{0.1t}\times 3\times 0.1e^{0.1t}}{(1+3e^{0.1t})^2}\) | M1, A1 | Full application of quotient rule; \(\frac{d}{dt}e^{0.1t}=ke^{0.1t}\) not \(kte^{0.1t}\) |
| At \(t=10\): \(\frac{dP}{dt} = \frac{(1+3e)\times 80e - 240e^2}{(1+3e)^2} = \frac{80e}{(1+3e)^2}\) | M1, A1 | Sub \(t=10\) into \(\frac{dP}{dt}\), NOT \(P\). Accept \(\frac{dP}{dt}=80e(1+3e)^{-2}\), \(\frac{80e}{1+6e+9e^2}\). Numerical value 2.59. (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P = \frac{800e^{0.1t}}{1+3e^{0.1t}} = \frac{800}{e^{-0.1t}+3} \Rightarrow P_{\max} = \frac{800}{3} = 266\). Hence P cannot be 270 | B1 | Accept: substituting \(P=270\) giving unsolvable equation; sight of \(\frac{800}{3}\) with comment it cannot reach 270; large \(t\) giving 266.6 or 267; graph with asymptote at 266.6 or 267. Do not accept "exp's cannot be negative" without numerical evidence. (1) (11 marks) |
## Question 8(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P = \frac{800e^0}{1+3e^0} = \frac{800}{1+3} = 200$ | M1, A1 | Sub $t=0$ into $P$ and use $e^0=1$ in at least one case. Accept $P=\frac{800}{1+3}$ as evidence. A1: 200. (2) |
## Question 8(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $250 = \frac{800e^{0.1t}}{1+3e^{0.1t}}$ | | Sub $P=250$, cross multiply, collect terms in $e^{0.1t}$ and proceed to $Ae^{0.1t}=B$ |
| $250(1+3e^{0.1t}) = 800e^{0.1t} \Rightarrow 50e^{0.1t}=250 \Rightarrow e^{0.1t}=5$ | M1, A1 | A1: $e^{0.1t}=5$ or $e^{-0.1t}=0.2$ |
| $t = \frac{1}{0.1}\ln(5)$ | M1 | Dependent on $e^{0.1t}=E$; take ln of both sides |
| $t = 10\ln(5)$ | A1 | Accept exact equivalents e.g. $t=5\ln(25)$. (4) |
## Question 8(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dP}{dt} = \frac{(1+3e^{0.1t})\times 800\times 0.1e^{0.1t} - 800e^{0.1t}\times 3\times 0.1e^{0.1t}}{(1+3e^{0.1t})^2}$ | M1, A1 | Full application of quotient rule; $\frac{d}{dt}e^{0.1t}=ke^{0.1t}$ not $kte^{0.1t}$ |
| At $t=10$: $\frac{dP}{dt} = \frac{(1+3e)\times 80e - 240e^2}{(1+3e)^2} = \frac{80e}{(1+3e)^2}$ | M1, A1 | Sub $t=10$ into $\frac{dP}{dt}$, NOT $P$. Accept $\frac{dP}{dt}=80e(1+3e)^{-2}$, $\frac{80e}{1+6e+9e^2}$. Numerical value 2.59. (4) |
## Question 8(d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P = \frac{800e^{0.1t}}{1+3e^{0.1t}} = \frac{800}{e^{-0.1t}+3} \Rightarrow P_{\max} = \frac{800}{3} = 266$. Hence P cannot be 270 | B1 | Accept: substituting $P=270$ giving unsolvable equation; sight of $\frac{800}{3}$ with comment it cannot reach 270; large $t$ giving 266.6 or 267; graph with asymptote at 266.6 or 267. Do not accept "exp's cannot be negative" without numerical evidence. (1) **(11 marks)** |8. A rare species of primrose is being studied. The population, $P$, of primroses at time $t$ years after the study started is modelled by the equation
$$P = \frac { 800 \mathrm { e } ^ { 0.1 t } } { 1 + 3 \mathrm { e } ^ { 0.1 t } } , \quad t \geqslant 0 , \quad t \in \mathbb { R }$$
\begin{enumerate}[label=(\alph*)]
\item Calculate the number of primroses at the start of the study.
\item Find the exact value of $t$ when $P = 250$, giving your answer in the form $a \ln ( b )$ where $a$ and $b$ are integers.
\item Find the exact value of $\frac { \mathrm { d } P } { \mathrm {~d} t }$ when $t = 10$. Give your answer in its simplest form.
\item Explain why the population of primroses can never be 270
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2014 Q8 [11]}}