| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Separable variables - partial fractions |
| Difficulty | Standard +0.8 This C4 separable differential equation requires partial fractions decomposition, integration, and manipulation of logarithms to find an explicit solution. The context requires careful interpretation of equilibrium behavior. While the techniques are standard for C4, the multi-step algebraic manipulation and conceptual understanding of why x=3 is unattainable elevates this above routine exercises. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int \frac{1}{(x-6)(x-3)} \, dx = \int 2 \, dt\) | M1 | |
| \(\frac{1}{(x-6)(x-3)} \equiv \frac{A}{x-6} + \frac{B}{x-3}\), \(\quad 1 \equiv A(x-3) + B(x-6)\) | M1 | |
| \(x = 6 \Rightarrow A = \frac{1}{3}\), \(x = 3 \Rightarrow B = -\frac{1}{3}\) | A2 | |
| \(\frac{1}{3}\int\left(\frac{1}{x-6} - \frac{1}{x-3}\right) dx = \int 2 \, dt\) | ||
| \(\ln\ | x-6\ | - \ln\ |
| \(t = 0\), \(x = 0\) \(\therefore \ln 6 - \ln 3 = c\), \(c = \ln 2\) | M1 A1 | |
| \(x = 2 \Rightarrow \ln 4 - 0 = 6t + \ln 2\) | M1 | |
| \(t = \frac{1}{6}\ln 2 = 0.1155 \text{ hrs} = 0.1155 \times 60 \text{ mins} = 6.93 \text{ mins} \approx 7 \text{ mins}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\ln\left\ | \frac{x-6}{2(x-3)}\right\ | = 6t\), \(\quad t = \frac{1}{6}\ln\left\ |
| as \(x \to 3\), \(t \to \infty\) \(\therefore\) cannot make 3 g | B2 | (12) |
# Question 7:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \frac{1}{(x-6)(x-3)} \, dx = \int 2 \, dt$ | M1 | |
| $\frac{1}{(x-6)(x-3)} \equiv \frac{A}{x-6} + \frac{B}{x-3}$, $\quad 1 \equiv A(x-3) + B(x-6)$ | M1 | |
| $x = 6 \Rightarrow A = \frac{1}{3}$, $x = 3 \Rightarrow B = -\frac{1}{3}$ | A2 | |
| $\frac{1}{3}\int\left(\frac{1}{x-6} - \frac{1}{x-3}\right) dx = \int 2 \, dt$ | | |
| $\ln\|x-6\| - \ln\|x-3\| = 6t + c$ | M1 A1 | |
| $t = 0$, $x = 0$ $\therefore \ln 6 - \ln 3 = c$, $c = \ln 2$ | M1 A1 | |
| $x = 2 \Rightarrow \ln 4 - 0 = 6t + \ln 2$ | M1 | |
| $t = \frac{1}{6}\ln 2 = 0.1155 \text{ hrs} = 0.1155 \times 60 \text{ mins} = 6.93 \text{ mins} \approx 7 \text{ mins}$ | A1 | |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\ln\left\|\frac{x-6}{2(x-3)}\right\| = 6t$, $\quad t = \frac{1}{6}\ln\left\|\frac{x-6}{2(x-3)}\right\|$ | | |
| as $x \to 3$, $t \to \infty$ $\therefore$ cannot make 3 g | B2 | **(12)** |
---\begin{enumerate}
\item During a chemical reaction, a compound is being made from two other substances.
\end{enumerate}
At time $t$ hours after the start of the reaction, $x \mathrm {~g}$ of the compound has been produced. Assuming that $x = 0$ initially, and that
$$\frac { \mathrm { d } x } { \mathrm {~d} t } = 2 ( x - 6 ) ( x - 3 )$$
(i) show that it takes approximately 7 minutes to produce 2 g of the compound.\\
(ii) Explain why it is not possible to produce 3 g of the compound.\\
\hfill \mbox{\textit{OCR C4 Q7 [12]}}