| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2017 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Logistic growth model |
| Difficulty | Standard +0.3 This is a standard logistic growth model question requiring straightforward substitution (parts a,b), algebraic manipulation to solve for t (part c), and quotient rule differentiation (part d). While it involves multiple techniques, each step follows routine procedures with no novel insight required, making it slightly easier than average for a C3/C4 question. |
| 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 = b1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| When \(t = 0\), \(N = 15\) | B1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Puts \(t = 10\), so \(N = 56.6\) (accept 56 or 57) | M1A1 | Substitutes \(t=10\) into correct formula; accept 56, 57 or awrt 56.6 |
| Answer | Marks | Guidance |
|---|---|---|
| \(82 = \frac{300}{3+17e^{-0.2t}} \Rightarrow e^{-0.2t} = \frac{54}{1394}\) = awrt 0.039 | M1 A1 | M1: substitutes 82, proceeds to obtain \(e^{\pm 0.2t} = C\); A1: \(e^{-0.2t} = \frac{27}{697}\) oe |
| \(-0.2t = \ln\left(\frac{54}{1394}\right) \Rightarrow t =\) | dM1 | Dependent on previous M; takes ln of positive value and proceeds to \(t=\) |
| \(t =\) awrt 16.3 | A1 | Accept 16 weeks, 16.25 weeks; accept \(t = 5\ln\left(\frac{1394}{54}\right)\) oe |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dN}{dt} = (-0.2)\times300\times(-1)\times17e^{-0.2t}(3+17e^{-0.2t})^{-2}\) | M1 A1 | M1: differentiates to give form \(ke^{-0.2t}(3+17e^{-0.2t})^{-2}\); A1: correct unsimplified derivative \(1020e^{-0.2t}(3+17e^{-0.2t})^{-2}\) |
| \(= 4.38\) so 4 insects per week | A1 cso | Obtains awrt 4 following a correct derivative |
# Question 10:
## Part (a)
| When $t = 0$, $N = 15$ | B1 | cao |
## Part (b)
| Puts $t = 10$, so $N = 56.6$ (accept 56 or 57) | M1A1 | Substitutes $t=10$ into correct formula; accept 56, 57 or awrt 56.6 |
## Part (c)
| $82 = \frac{300}{3+17e^{-0.2t}} \Rightarrow e^{-0.2t} = \frac{54}{1394}$ = awrt 0.039 | M1 A1 | M1: substitutes 82, proceeds to obtain $e^{\pm 0.2t} = C$; A1: $e^{-0.2t} = \frac{27}{697}$ oe |
| $-0.2t = \ln\left(\frac{54}{1394}\right) \Rightarrow t =$ | dM1 | Dependent on previous M; takes ln of positive value and proceeds to $t=$ |
| $t =$ awrt 16.3 | A1 | Accept 16 weeks, 16.25 weeks; accept $t = 5\ln\left(\frac{1394}{54}\right)$ oe |
## Part (d)
| $\frac{dN}{dt} = (-0.2)\times300\times(-1)\times17e^{-0.2t}(3+17e^{-0.2t})^{-2}$ | M1 A1 | M1: differentiates to give form $ke^{-0.2t}(3+17e^{-0.2t})^{-2}$; A1: correct unsimplified derivative $1020e^{-0.2t}(3+17e^{-0.2t})^{-2}$ |
| $= 4.38$ so 4 insects per week | A1 cso | Obtains awrt 4 **following a correct derivative** |
---10. A population of insects is being studied. The number of insects, $N$, in the population, is modelled by the equation
$$N = \frac { 300 } { 3 + 17 \mathrm { e } ^ { - 0.2 t } } \quad t \in \mathbb { R } , t \geqslant 0$$
where $t$ is the time, in weeks, from the start of the study.\\
Using the model,
\begin{enumerate}[label=(\alph*)]
\item find the number of insects at the start of the study,
\item find the number of insects when $t = 10$,
\item find the time from the start of the study when there are 82 insects. (Solutions based entirely on graphical or numerical methods are not acceptable.)
\item Find, by differentiating, the rate, measured in insects per week, at which the number of insects is increasing when $t = 5$. Give your answer to the nearest whole number.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2017 Q10 [10]}}