| Exam Board | Edexcel |
|---|---|
| Module | FP1 AS (Further Pure 1 AS) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Logistic/bounded growth |
| Difficulty | Standard +0.3 Part (a) is a straightforward single-step Euler method application with given values. Part (b) requires recognizing that as t→∞, dx/dt→0, leading to x=0.8, which is a standard long-term behavior question. Both parts are routine FP1 techniques with no novel insight required, making this slightly easier than average. |
| Spec | 1.09g Numerical methods in context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t_0=3\), step is half a year so \(h=\frac{1}{2}\) and \(x_0=\frac{1}{2}\) | B1 | Correct parameters seen or implied |
| \(\left(\frac{dx}{dt}\right)_0 = \frac{0.5\times3\times(0.8-0.5)}{0.5^2+5\times3} = \frac{9}{305} = 0.0295\ldots\) | M1 A1 | M1: uses their \(x\) and \(t\) in equation; condone one slip. A1: \(\frac{9}{305}\) or awrt 0.03 |
| So when \(t=3.5\), \(x \approx \frac{1}{2}+\frac{1}{2}\times\frac{9}{305} = \ldots\) | M1 | Applies approximation formula with their \(x\), \(h\), and \(\left(\frac{dx}{dt}\right)_0\) |
| \(= \frac{157}{305} \approx 0.51475\ldots\) | A1 | Accept as fraction or awrt 0.515 or 51.5% |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Long term proportion is \(\frac{4}{5}\) | B1 | Allow equivalents e.g. 0.8 or 80% |
# Question 2(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t_0=3$, step is half a year so $h=\frac{1}{2}$ and $x_0=\frac{1}{2}$ | B1 | Correct parameters seen or implied |
| $\left(\frac{dx}{dt}\right)_0 = \frac{0.5\times3\times(0.8-0.5)}{0.5^2+5\times3} = \frac{9}{305} = 0.0295\ldots$ | M1 A1 | M1: uses their $x$ and $t$ in equation; condone one slip. A1: $\frac{9}{305}$ or awrt 0.03 |
| So when $t=3.5$, $x \approx \frac{1}{2}+\frac{1}{2}\times\frac{9}{305} = \ldots$ | M1 | Applies approximation formula with their $x$, $h$, and $\left(\frac{dx}{dt}\right)_0$ |
| $= \frac{157}{305} \approx 0.51475\ldots$ | A1 | Accept as fraction or awrt 0.515 or 51.5% |
# Question 2(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Long term proportion is $\frac{4}{5}$ | B1 | Allow equivalents e.g. 0.8 or 80% |
---\begin{enumerate}
\item An area of woodland contains a mixture of blue and yellow flowers.
\end{enumerate}
A study found that the proportion, $x$, of blue flowers in the woodland area satisfies the differential equation
$$\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { x t ( 0.8 - x ) } { x ^ { 2 } + 5 t } \quad t > 0$$
where $t$ is the number of years since the start of the study.\\
Given that exactly 3 years after the start of the study half of the flowers in the woodland area were blue,\\
(a) use one application of the approximation formula $\left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) _ { n } \approx \frac { y _ { n + 1 } - y _ { n } } { h }$ to estimate the proportion of blue flowers in the woodland area half a year later.\\
(b) Deduce from the differential equation the proportion of flowers that will be blue in the long term.
\hfill \mbox{\textit{Edexcel FP1 AS 2024 Q2 [6]}}