| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2022 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Modelling with rate proportional to amount |
| Difficulty | Standard +0.3 This is a separable first-order differential equation requiring straightforward separation of variables, integration (including exponential functions), and application of initial conditions. While it involves multiple steps and exponential manipulation, the techniques are standard for A-level Further Maths, making it slightly easier than average difficulty. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.06g Equations with exponentials: solve a^x = b1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Separate variables correctly | B1 | \(\int \frac{1}{x}\,dx = \int ke^{-0.1t}\,dt\) |
| Obtain term \(\ln x\) | B1 | |
| Obtain term \(-10ke^{-0.1t}\) | B1 | Not from \(\int xe^{-0.1t}\,dt\) |
| Use \(x=20\), \(t=0\) to evaluate a constant or as limits in a solution containing terms \(a\ln x,\, be^{-0.1t}\) where \(ab\neq 0\) | M1 | |
| Obtain \(\ln x = 10k\left(1 - e^{-0.1t}\right) + \ln 20\) | A1 | or equivalent ISW |
| Total | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use \(x=40\), \(t=10\) to find \(k\) or \(10k\) | M1 | Available for their function of the correct structure even if they found no constant in (a) |
| Obtain \(10k = 1.09654\) | A1 | or equivalent e.g. \(10k = \frac{\ln 2}{1-e^{-1}}\) |
| State that \(x\) tends to \(59.9\) | A1 | Need a number, not an expression for that value. 3 sf or better \(59.87595\ldots\) |
| Total | 3 |
## Question 8(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Separate variables correctly | B1 | $\int \frac{1}{x}\,dx = \int ke^{-0.1t}\,dt$ |
| Obtain term $\ln x$ | B1 | |
| Obtain term $-10ke^{-0.1t}$ | B1 | Not from $\int xe^{-0.1t}\,dt$ |
| Use $x=20$, $t=0$ to evaluate a constant or as limits in a solution containing terms $a\ln x,\, be^{-0.1t}$ where $ab\neq 0$ | M1 | |
| Obtain $\ln x = 10k\left(1 - e^{-0.1t}\right) + \ln 20$ | A1 | or equivalent ISW |
| **Total** | **5** | |
## Question 8(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use $x=40$, $t=10$ to find $k$ or $10k$ | M1 | Available for their function of the correct structure even if they found no constant in (a) |
| Obtain $10k = 1.09654$ | A1 | or equivalent e.g. $10k = \frac{\ln 2}{1-e^{-1}}$ |
| State that $x$ tends to $59.9$ | A1 | Need a number, not an expression for that value. 3 sf or better $59.87595\ldots$ |
| **Total** | **3** | |8 In a certain chemical reaction the amount, $x$ grams, of a substance is increasing. The differential equation satisfied by $x$ and $t$, the time in seconds since the reaction began, is
$$\frac { \mathrm { d } x } { \mathrm {~d} t } = k x \mathrm { e } ^ { - 0.1 t }$$
where $k$ is a positive constant. It is given that $x = 20$ at the start of the reaction.
\begin{enumerate}[label=(\alph*)]
\item Solve the differential equation, obtaining a relation between $x , t$ and $k$.
\item Given that $x = 40$ when $t = 10$, find the value of $k$ and find the value approached by $x$ as $t$ becomes large.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2022 Q8 [8]}}