| Exam Board | OCR MEI |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2020 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Modelling with logistic growth |
| Difficulty | Standard +0.8 This is a logistic growth differential equation requiring separation of variables with partial fractions (part (a) scaffolds this), integration involving logarithms, and solving for x explicitly using exponentials. While the partial fractions are guided and the model is standard, the algebraic manipulation to isolate x from ln terms and the multi-step solution process elevate this above routine questions, though it remains a recognizable exam pattern for Further Maths. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{A}{x(A-x)}\) | B1 | Denominator may be multiplied out |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\displaystyle\int \frac{400}{x(400-x)}\,dx = \int dt\) | M1 | Separates variables (inc \(dx\) and \(dt\)). 400 either side |
| \(\displaystyle\int\left(\frac{1}{x} + \frac{1}{400-x}\right)dx = \int dt\) | M1 | Use of result from (a) or method for partial fractions |
| \(\ln x - \ln(400-x) = t + c\) | M1 A1 | Attempt to integrate (at least one term correct); all integration correct. Or \(\ln\left |
| \(\ln\!\left(\dfrac{x}{400-x}\right) = t + c\) | ||
| When \(t=0\), \(x=100\) so \(c = \ln\!\left(\tfrac{1}{3}\right)\) | M1 | Finding constant (or \(A=3\)) |
| \(t = \ln\!\left(\dfrac{3x}{400-x}\right)\); when \(t=10\): \(10 = \ln\!\left(\dfrac{3x}{400-x}\right)\) | M1 | Using log laws to combine the \(c\) value |
| \(\dfrac{3x}{400-x} = e^{10} \Rightarrow 3x = e^{10}(400-x)\) | M1 | Attempt to rearrange to find \(x\) |
| \(400\) | A1 | Must be rounded to nearest whole number |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_{100}^{X} \frac{400}{x(400-x)}dx = \int_0^{10} dt\) | M1 | Separates variables |
| \(\int_{100}^{X} \left(\frac{1}{x} + \frac{1}{400-x}\right)dx = \int_0^{10} dt\) | M1 | Use result from (a) or method for partial fractions |
| M1 | Attempt to integrate (at least one term correct) | |
| \([\ln x - \ln(400-x)]_{100}^{X} = [t]_0^{10}\) | A1 | All integration correct |
| \(\left[\ln\frac{x}{(400-x)}\right]_{100}^{X} = [t]_0^{10}\) | ||
| \(\ln\frac{X}{(400-X)} - \ln\frac{100}{300} = 10-0\) | M1 | Limits applied (condone one error) |
| \(\ln\frac{X}{(400-X)} - \ln\frac{1}{3} = 10\) | Combining log terms together | |
| \(\ln\frac{3X}{(400-X)} = 10\) | M1 | Attempt to rearrange to find \(X\) |
| \(\frac{3X}{(400-X)} = e^{10}\) | M1 | |
| \(3X = e^{10}(400-X)\) | ||
| \(X = 400\) | A1 | Must be rounded to nearest whole number |
| [8] |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{A}{x(A-x)}$ | B1 | Denominator may be multiplied out |
---
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\displaystyle\int \frac{400}{x(400-x)}\,dx = \int dt$ | M1 | Separates variables (inc $dx$ and $dt$). 400 either side |
| $\displaystyle\int\left(\frac{1}{x} + \frac{1}{400-x}\right)dx = \int dt$ | M1 | Use of result from (a) or method for partial fractions |
| $\ln x - \ln(400-x) = t + c$ | M1 A1 | Attempt to integrate (at least one term correct); all integration correct. Or $\ln\left|\frac{Ax}{400-x}\right| = t$ |
| $\ln\!\left(\dfrac{x}{400-x}\right) = t + c$ | | |
| When $t=0$, $x=100$ so $c = \ln\!\left(\tfrac{1}{3}\right)$ | M1 | Finding constant (or $A=3$) |
| $t = \ln\!\left(\dfrac{3x}{400-x}\right)$; when $t=10$: $10 = \ln\!\left(\dfrac{3x}{400-x}\right)$ | M1 | Using log laws to combine the $c$ value |
| $\dfrac{3x}{400-x} = e^{10} \Rightarrow 3x = e^{10}(400-x)$ | M1 | Attempt to rearrange to find $x$ |
| $400$ | A1 | Must be rounded to nearest whole number |
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_{100}^{X} \frac{400}{x(400-x)}dx = \int_0^{10} dt$ | M1 | Separates variables |
| $\int_{100}^{X} \left(\frac{1}{x} + \frac{1}{400-x}\right)dx = \int_0^{10} dt$ | M1 | Use result from (a) or method for partial fractions |
| | M1 | Attempt to integrate (at least one term correct) |
| $[\ln x - \ln(400-x)]_{100}^{X} = [t]_0^{10}$ | A1 | All integration correct |
| $\left[\ln\frac{x}{(400-x)}\right]_{100}^{X} = [t]_0^{10}$ | | |
| $\ln\frac{X}{(400-X)} - \ln\frac{100}{300} = 10-0$ | M1 | Limits applied (condone one error) |
| $\ln\frac{X}{(400-X)} - \ln\frac{1}{3} = 10$ | | Combining log terms together |
| $\ln\frac{3X}{(400-X)} = 10$ | M1 | Attempt to rearrange to find $X$ |
| $\frac{3X}{(400-X)} = e^{10}$ | M1 | |
| $3X = e^{10}(400-X)$ | | |
| $X = 400$ | A1 | Must be rounded to nearest whole number |
| **[8]** | | |
---7
\begin{enumerate}[label=(\alph*)]
\item Express $\frac { 1 } { x } + \frac { 1 } { A - x }$ as a single fraction.
The population of fish in a lake is modelled by the differential equation\\
$\frac { d x } { d t } = \frac { x ( 400 - x ) } { 400 }$\\
where $x$ is the number of fish and $t$ is the time in years.\\
When $t = 0 , x = 100$.
\item In this question you must show detailed reasoning.
Find the number of fish in the lake when $t = 10$, as predicted by the model.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 3 2020 Q7 [9]}}