| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2006 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Exponential growth/decay - approach to limit (dN/dt = k(N - Nâ‚€)) |
| Difficulty | Standard +0.3 This is a straightforward first-order differential equation problem with clear setup. Part (i) requires simple algebraic manipulation to form the DE from given conditions. Part (ii) is a standard separable variables problem with initial conditions—routine for P3 level with no novel insight required, though slightly above average due to the 8-mark allocation and multi-step solution. |
| Spec | 4.10b Model with differential equations: kinematics and other contexts4.10c Integrating factor: first order equations |
| Answer | Marks |
|---|---|
| State or imply that \(\frac{dx}{dt} = kx - 25\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Show that \(k = 0.1\) and justify the given statement | B1 | 2 |
| Separate variables and attempt integration | M1 | |
| Obtain \(\ln(x - 250)\), or equivalent | A1 | |
| Obtain \(0.1t\), or equivalent | A1 | |
| Evaluate a constant or use limits \(t = 0, x = 1000\) with a solution containing terms \(a\ln(x - 250)\) and \(bt\) | M1 | |
| Obtain any correct form of solution, e.g. \(\ln(x - 250) = 0.1t + \ln 750\) | A1 | |
| Rearrange and obtain \(x = 250(3e^{0.1t} + 1)\), or equivalent | A1 | 6 |
**(i)**
| State or imply that $\frac{dx}{dt} = kx - 25$ | B1 |
**(ii)**
| Show that $k = 0.1$ and justify the given statement | B1 | 2 |
| Separate variables and attempt integration | M1 |
| Obtain $\ln(x - 250)$, or equivalent | A1 |
| Obtain $0.1t$, or equivalent | A1 |
| Evaluate a constant or use limits $t = 0, x = 1000$ with a solution containing terms $a\ln(x - 250)$ and $bt$ | M1 |
| Obtain any correct form of solution, e.g. $\ln(x - 250) = 0.1t + \ln 750$ | A1 |
| Rearrange and obtain $x = 250(3e^{0.1t} + 1)$, or equivalent | A1 | 6 |In a certain industrial process, a substance is being produced in a container. The mass of the substance in the container $t$ minutes after the start of the process is $x$ grams. At any time, the rate of formation of the substance is proportional to its mass. Also, throughout the process, the substance is removed from the container at a constant rate of 25 grams per minute. When $t = 0$, $x = 1000$ and $\frac{dx}{dt} = 75$.
\begin{enumerate}[label=(\roman*)]
\item Show that $x$ and $t$ satisfy the differential equation
$$\frac{dx}{dt} = 0.1(x - 250).$$ [2]
\item Solve this differential equation, obtaining an expression for $x$ in terms of $t$. [6]
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2006 Q5 [8]}}