| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Logistic/bounded growth |
| Difficulty | Standard +0.8 This is a multi-step logistic differential equation problem requiring: (1) interpreting a verbal rate condition to find the constant of proportionality, (2) separating variables with partial fractions, (3) integrating and applying initial conditions, (4) solving for x at a specific time. While the DE type is standard for C4, the setup in part (i) requires careful interpretation of 'if the rate remained constant,' and part (ii) involves non-trivial algebra to extract the final answer. This is above-average difficulty for C4. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)1.08l Interpret differential equation solutions: in context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| when \(x = \frac{1}{4}\), \(\frac{dx}{dt} = \frac{3}{4} \div 6 = \frac{1}{8}\) | M1 | |
| \(\frac{dx}{dt} = kx(1-x) \quad \therefore \frac{1}{8} = k \times \frac{1}{4} \times \frac{3}{4},\ k = \frac{2}{3} \quad \therefore \frac{dx}{dt} = \frac{2}{3}x(1-x)\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int \frac{1}{x(1-x)}\ dx = \int \frac{2}{3}\ dt\) | M1 | |
| \(\frac{1}{x(1-x)} \equiv \frac{A}{x} + \frac{B}{1-x}, \quad 1 \equiv A(1-x) + Bx\) | M1 | |
| \(x = 0 \implies A = 1\) | A1 | |
| \(x = 1 \implies B = 1\) | A1 | |
| \(\therefore \int\left(\frac{1}{x} + \frac{1}{1-x}\right)\ dx = \int \frac{2}{3}\ dt\) | ||
| \(\ln | x | - \ln |
| \(t=0,\ x=\frac{1}{4} \implies \ln\frac{1}{4} - \ln\frac{3}{4} = c,\quad c = \ln\frac{1}{3}\) | M1 | |
| \(t=3 \implies \ln | x | - \ln |
| \(\ln\left | \frac{3x}{1-x}\right | = 2, \quad \frac{3x}{1-x} = e^2\) |
| \(3x = e^2(1-x), \quad x(e^2+3) = e^2\) | M1 | |
| \(x = \frac{e^2}{e^2+3} \quad \therefore \%\ \text{destroyed} = \frac{e^2}{e^2+3} \times 100\% = 71.1\%\ \text{(3sf)}\) | A1 | (13) |
# Question 8:
## Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| when $x = \frac{1}{4}$, $\frac{dx}{dt} = \frac{3}{4} \div 6 = \frac{1}{8}$ | M1 | |
| $\frac{dx}{dt} = kx(1-x) \quad \therefore \frac{1}{8} = k \times \frac{1}{4} \times \frac{3}{4},\ k = \frac{2}{3} \quad \therefore \frac{dx}{dt} = \frac{2}{3}x(1-x)$ | M1 A1 | |
## Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int \frac{1}{x(1-x)}\ dx = \int \frac{2}{3}\ dt$ | M1 | |
| $\frac{1}{x(1-x)} \equiv \frac{A}{x} + \frac{B}{1-x}, \quad 1 \equiv A(1-x) + Bx$ | M1 | |
| $x = 0 \implies A = 1$ | A1 | |
| $x = 1 \implies B = 1$ | A1 | |
| $\therefore \int\left(\frac{1}{x} + \frac{1}{1-x}\right)\ dx = \int \frac{2}{3}\ dt$ | | |
| $\ln|x| - \ln|1-x| = \frac{2}{3}t + c$ | M1 A1 | |
| $t=0,\ x=\frac{1}{4} \implies \ln\frac{1}{4} - \ln\frac{3}{4} = c,\quad c = \ln\frac{1}{3}$ | M1 | |
| $t=3 \implies \ln|x| - \ln|1-x| = 2 + \ln\frac{1}{3}$ | | |
| $\ln\left|\frac{3x}{1-x}\right| = 2, \quad \frac{3x}{1-x} = e^2$ | M1 | |
| $3x = e^2(1-x), \quad x(e^2+3) = e^2$ | M1 | |
| $x = \frac{e^2}{e^2+3} \quad \therefore \%\ \text{destroyed} = \frac{e^2}{e^2+3} \times 100\% = 71.1\%\ \text{(3sf)}$ | A1 | **(13)** |
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**Total: (72)**8. When a plague of locusts attacks a wheat crop, the proportion of the crop destroyed after $t$ hours is denoted by $x$. In a model, it is assumed that the rate at which the crop is destroyed is proportional to $x ( 1 - x )$.
A plague of locusts is discovered in a wheat crop when one quarter of the crop has been destroyed.
Given that the rate of destruction at this instant is such that if it remained constant, the crop would be completely destroyed in a further six hours,\\
(i) show that $\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { 2 } { 3 } x ( 1 - x )$,\\
(ii) find the percentage of the crop destroyed three hours after the plague of locusts is first discovered.
\hfill \mbox{\textit{OCR C4 Q8 [13]}}