| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Chemical reaction kinetics |
| Difficulty | Challenging +1.2 This is a separable differential equation requiring partial fractions and integration, followed by solving for t when x=2. Part (b) requires qualitative analysis of the DE. While it involves multiple steps and partial fractions (beyond basic C4), the setup is standard and the question guides students through specific values. It's moderately harder than average but not exceptionally challenging for C4 level. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks |
|---|---|
| (a) \(\int \frac{1}{(x-6)(x-3)} dx = \int 2 dt\) | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = 6 \Rightarrow A = \frac{1}{3}\), \(x = 3 \Rightarrow B = -\frac{1}{3}\) | M1 A2 | |
| \(\frac{1}{3}\int \left(\frac{1}{x-6} - \frac{1}{x-3}\right) dx = \int 2 dt\) | M1 A1 | |
| \(\ln | x-6 | - \ln |
| \(t = 0\), \(x = 0\): \(\ln 6 - \ln 3 = c\), \(c = \ln 2\) | M1 A1 | |
| \(x = 2 \Rightarrow \ln 4 - 0 = 6t + \ln 2\) | M1 | |
| \(t = \frac{1}{6}\ln 2 = 0.1155\) hrs \(= 0.1155 \times 60\) mins \(= 6.93\) mins \(\approx 7\) mins | A1 | |
| (b) \(\ln\left | \frac{x-6}{2(x-3)}\right | = 6t\), \(t = \frac{1}{6}\ln\left |
| As \(x \to 3\), \(t \to \infty\) \(\therefore\) cannot make 3 g | B2 | (12 marks) |
**(a)** $\int \frac{1}{(x-6)(x-3)} dx = \int 2 dt$ | M1 |
$\frac{1}{(x-6)(x-3)} = \frac{A}{x-6} + \frac{B}{x-3}$
$1 = A(x-3) + B(x-6)$
$x = 6 \Rightarrow A = \frac{1}{3}$, $x = 3 \Rightarrow B = -\frac{1}{3}$ | M1 A2 |
$\frac{1}{3}\int \left(\frac{1}{x-6} - \frac{1}{x-3}\right) dx = \int 2 dt$ | M1 A1 |
$\ln|x-6| - \ln|x-3| = 6t + c$ | M1 A1 |
$t = 0$, $x = 0$: $\ln 6 - \ln 3 = c$, $c = \ln 2$ | M1 A1 |
$x = 2 \Rightarrow \ln 4 - 0 = 6t + \ln 2$ | M1 |
$t = \frac{1}{6}\ln 2 = 0.1155$ hrs $= 0.1155 \times 60$ mins $= 6.93$ mins $\approx 7$ mins | A1 |
**(b)** $\ln\left|\frac{x-6}{2(x-3)}\right| = 6t$, $t = \frac{1}{6}\ln\left|\frac{x-6}{2(x-3)}\right|$ | M1 |
As $x \to 3$, $t \to \infty$ $\therefore$ cannot make 3 g | B2 | (12 marks)
---4. During a chemical reaction, a compound is being made from two other substances. 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
$$\frac { \mathrm { d } x } { \mathrm {~d} t } = 2 ( x - 6 ) ( x - 3 )$$
\begin{enumerate}[label=(\alph*)]
\item show that it takes approximately 7 minutes to produce 2 g of the compound.
\item Explain why it is not possible to produce 3 g of the compound.\\
4. continued
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q4 [12]}}