| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2017 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Exponential growth/decay - approach to limit (dN/dt = k(N - N₀)) |
| Difficulty | Standard +0.3 This is a straightforward applied differential equations question requiring students to (i) set up a simple DE from a word problem using given constraints, and (ii) solve a standard separable first-order DE with initial conditions. The setup is direct (x + y = 50, so y = 50 - x), and the solution involves routine separation of variables and exponential manipulation. Slightly easier than average due to the guided structure and standard technique. |
| Spec | 1.02z Models in context: use functions in modelling1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Justify the given differential equation | B1 | |
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Separate variables correctly and attempt to integrate one side | B1 | |
| Obtain term \(kt\), or equivalent | B1 | |
| Obtain term \(-\ln(50 - x)\), or equivalent | B1 | |
| Evaluate a constant, or use limits \(x = 0,\ t = 0\) in a solution containing terms \(a\ln(50-x)\) and \(bt\) | M1* | |
| Obtain solution \(-\ln(50 - x) = kt - \ln 50\), or equivalent | A1 | |
| Use \(x = 25,\ t = 10\) to determine \(k\) | DM1 | |
| Obtain correct solution in any form, e.g. \(\ln 50 - \ln(50-x) = \frac{1}{10}(\ln 2)t\) | A1 | |
| Obtain answer \(x = 50\!\left(1 - \exp(-0.0693t)\right)\), or equivalent | A1 | |
| Total: 8 |
## Question 8(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Justify the given differential equation | B1 | |
| **Total: 1** | | |
## Question 8(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Separate variables correctly and attempt to integrate one side | B1 | |
| Obtain term $kt$, or equivalent | B1 | |
| Obtain term $-\ln(50 - x)$, or equivalent | B1 | |
| Evaluate a constant, or use limits $x = 0,\ t = 0$ in a solution containing terms $a\ln(50-x)$ and $bt$ | M1* | |
| Obtain solution $-\ln(50 - x) = kt - \ln 50$, or equivalent | A1 | |
| Use $x = 25,\ t = 10$ to determine $k$ | DM1 | |
| Obtain correct solution in any form, e.g. $\ln 50 - \ln(50-x) = \frac{1}{10}(\ln 2)t$ | A1 | |
| Obtain answer $x = 50\!\left(1 - \exp(-0.0693t)\right)$, or equivalent | A1 | |
| **Total: 8** | | |8 In a certain chemical reaction, a compound $A$ is formed from a compound $B$. The masses of $A$ and $B$ at time $t$ after the start of the reaction are $x$ and $y$ respectively and the sum of the masses is equal to 50 throughout the reaction. At any time the rate of increase of the mass of $A$ is proportional to the mass of $B$ at that time.\\
(i) Explain why $\frac { \mathrm { d } x } { \mathrm {~d} t } = k ( 50 - x )$, where $k$ is a constant.\\
It is given that $x = 0$ when $t = 0$, and $x = 25$ when $t = 10$.\\
(ii) Solve the differential equation in part (i) and express $x$ in terms of $t$.\\
\hfill \mbox{\textit{CAIE P3 2017 Q8 [9]}}