| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2012 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Logistic/bounded growth |
| Difficulty | Standard +0.3 This is a standard logistic differential equation question with clear scaffolding: part (a) gives the partial fractions decomposition needed for part (b), which follows a routine separation of variables method. The final part (c) requires only inspection of the solution form as tââ. While it involves multiple steps, each is algorithmic with no novel insight required, making it slightly easier than average. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| \(1 = A(5-P)+BP\) | M1 | Forming correct identity; can be implied |
| \(A=\frac{1}{5},\ B=\frac{1}{5}\) | A1 | Either one |
| \(\frac{\frac{1}{5}}{P}+\frac{\frac{1}{5}}{(5-P)}\) | A1 cao, aef | Must be stated in part (a) only |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int\frac{1}{P(5-P)}\,dP = \int\frac{1}{15}\,dt\) | B1 | Separates variables; \(dP\) and \(dt\) in correct positions |
| \(\frac{1}{5}\ln P - \frac{1}{5}\ln(5-P) = \frac{1}{15}t\ (+c)\) | M1*, A1ft | Both \(\pm\lambda\ln P\) and \(\pm\mu\ln(\pm5\pm P)\) |
| Use \(t=0,\ P=1\) to find \(c=-\frac{1}{5}\ln 4\) | dM1* | |
| \(\frac{1}{5}\ln\!\left(\frac{P}{5-P}\right)=\frac{1}{15}t-\frac{1}{5}\ln 4\) | dM1* | Using subtraction/addition log laws correctly |
| \(\ln\!\left(\frac{4P}{5-P}\right)=\frac{1}{3}t\) | ||
| \(\frac{4P}{5-P}=e^{\frac{1}{3}t}\) | dM1* | Eliminate ln's correctly |
| Make \(P\) the subject | dM1* | Full acceptable rearrangement |
| \(P=\dfrac{5}{1+4e^{-\frac{1}{3}t}}\) or \(P=\dfrac{25}{5+20e^{-\frac{1}{3}t}}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(1+4e^{-\frac{1}{3}t}>1 \Rightarrow P<5\), so population cannot exceed 5000 | B1 | Must have correct \(a,b,c\) values in \(P=\frac{a}{b+ce^{-\frac{1}{3}t}}\) and conclusion relating to 5000 |
# Question 8:
## Part (a):
| $1 = A(5-P)+BP$ | M1 | Forming correct identity; can be implied |
| $A=\frac{1}{5},\ B=\frac{1}{5}$ | A1 | Either one |
| $\frac{\frac{1}{5}}{P}+\frac{\frac{1}{5}}{(5-P)}$ | A1 cao, aef | Must be stated in part (a) only |
## Part (b):
| $\int\frac{1}{P(5-P)}\,dP = \int\frac{1}{15}\,dt$ | B1 | Separates variables; $dP$ and $dt$ in correct positions |
| $\frac{1}{5}\ln P - \frac{1}{5}\ln(5-P) = \frac{1}{15}t\ (+c)$ | M1*, A1ft | Both $\pm\lambda\ln P$ and $\pm\mu\ln(\pm5\pm P)$ |
| Use $t=0,\ P=1$ to find $c=-\frac{1}{5}\ln 4$ | dM1* | |
| $\frac{1}{5}\ln\!\left(\frac{P}{5-P}\right)=\frac{1}{15}t-\frac{1}{5}\ln 4$ | dM1* | Using subtraction/addition log laws correctly |
| $\ln\!\left(\frac{4P}{5-P}\right)=\frac{1}{3}t$ | | |
| $\frac{4P}{5-P}=e^{\frac{1}{3}t}$ | dM1* | Eliminate ln's correctly |
| Make $P$ the subject | dM1* | Full acceptable rearrangement |
| $P=\dfrac{5}{1+4e^{-\frac{1}{3}t}}$ or $P=\dfrac{25}{5+20e^{-\frac{1}{3}t}}$ | A1 | |
## Part (c):
| $1+4e^{-\frac{1}{3}t}>1 \Rightarrow P<5$, so population cannot exceed 5000 | B1 | Must have correct $a,b,c$ values in $P=\frac{a}{b+ce^{-\frac{1}{3}t}}$ and conclusion relating to 5000 |
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There is **no mark scheme content** (no questions, answers, mark allocations, or guidance notes) visible on this page to extract.
Could you please share the **actual mark scheme pages** containing the questions and answers? I'd be happy to format those for you once you upload them.\begin{enumerate}
\item (a) Express $\frac { 1 } { P ( 5 - P ) }$ in partial fractions.
\end{enumerate}
A team of conservationists is studying the population of meerkats on a nature reserve. The population is modelled by the differential equation
$$\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 15 } P ( 5 - P ) , \quad t \geqslant 0$$
where $P$, in thousands, is the population of meerkats and $t$ is the time measured in years since the study began.
Given that when $t = 0 , P = 1$,\\
(b) solve the differential equation, giving your answer in the form,
$$P = \frac { a } { b + c \mathrm { e } ^ { - \frac { 1 } { 3 } t } }$$
where $a$, $b$ and $c$ are integers.\\
(c) Hence show that the population cannot exceed 5000\\
\hfill \mbox{\textit{Edexcel C4 2012 Q8 [12]}}