| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2005 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Logistic growth model |
| Difficulty | Standard +0.3 This is a structured logistic growth question with clear scaffolding through parts (a)-(d). Part (a) requires simple substitution at t=0 and solving a linear equation. Part (b) involves substituting p=1850 and solving for t using logarithms—standard C3 technique. Parts (c) and (d) require algebraic manipulation and limit analysis, but the 'show that' format provides the target answer. While logistic models are conceptually richer than basic exponentials, the question guides students through each step with no novel problem-solving required, making it slightly easier than a typical C3 question. |
| Spec | 1.06g Equations with exponentials: solve a^x = b1.06i Exponential growth/decay: in modelling context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Setting \(p = 300\) at \(t = 0 \Rightarrow 300 = \frac{2800a}{1+a}\) | M1 | |
| \(300 = 2500a\); \(\quad a = 0.12\) | dM1 A1 | c.s.o. (3 marks total) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(1850 = \frac{2800(0.12)e^{0.2t}}{1 + 0.12e^{0.2t}}\); \(\quad e^{0.2t} = 16.2...\) | M1 A1 | |
| Correctly taking logs to \(0.2\,t = \ln k\) | M1 | |
| \(t = 14 \quad (13.9...)\) | A1 | (4 marks total) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Correct derivation showing division of numerator and denominator by \(e^{0.2t}\); using \(a\) | B1 | (1 mark total) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Using \(t \to \infty\), \(e^{-0.2t} \to 0\) | M1 | |
| \(p \to \frac{336}{0.12} = 2800\) | A1 | (2 marks total) [Total: 10] |
## Question 7:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Setting $p = 300$ at $t = 0 \Rightarrow 300 = \frac{2800a}{1+a}$ | M1 | |
| $300 = 2500a$; $\quad a = 0.12$ | dM1 A1 | c.s.o. (3 marks total) |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $1850 = \frac{2800(0.12)e^{0.2t}}{1 + 0.12e^{0.2t}}$; $\quad e^{0.2t} = 16.2...$ | M1 A1 | |
| Correctly taking logs to $0.2\,t = \ln k$ | M1 | |
| $t = 14 \quad (13.9...)$ | A1 | (4 marks total) |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Correct derivation showing division of numerator and denominator by $e^{0.2t}$; using $a$ | B1 | (1 mark total) |
### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Using $t \to \infty$, $e^{-0.2t} \to 0$ | M1 | |
| $p \to \frac{336}{0.12} = 2800$ | A1 | (2 marks total) **[Total: 10]** |\begin{enumerate}
\item A particular species of orchid is being studied. The population $p$ at time $t$ years after the study started is assumed to be
\end{enumerate}
$$p = \frac { 2800 a \mathrm { e } ^ { 0.2 t } } { 1 + a \mathrm { e } ^ { 0.2 t } } , \text { where } a \text { is a constant. }$$
Given that there were 300 orchids when the study started,\\
(a) show that $a = 0.12$,\\
(b) use the equation with $a = 0.12$ to predict the number of years before the population of orchids reaches 1850.\\
(c) Show that $p = \frac { 336 } { 0.12 + \mathrm { e } ^ { - 0.2 t } }$.\\
(d) Hence show that the population cannot exceed 2800.\\
\hfill \mbox{\textit{Edexcel C3 2005 Q7 [10]}}