| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2020 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Separable variables |
| Difficulty | Standard +0.3 This is a straightforward separable variables question requiring standard techniques: separate variables, integrate both sides (using the identity cos²(2x) = (1+cos(4x))/2), apply initial conditions, and rearrange. Part (b) requires simple limit analysis. Slightly above average due to the trigonometric integration and exponential manipulation, but still a routine textbook exercise with no novel insight required. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Correct separation of variables | B1 | \(\int \sec^2 2x\, dx = \int e^{-3t}\, dt\); needs correct structure |
| Obtain term \(-\frac{1}{3}e^{-3t}\) | B1 | |
| Obtain term of the form \(k\tan 2x\) | M1 | From correct working |
| Obtain term \(\frac{1}{2}\tan 2x\) | A1 | |
| Use \(x=0,\, t=0\) to evaluate a constant, or as limits in a solution containing terms of the form \(a\tan 2x\) and \(be^{-3t}\), where \(ab \neq 0\) | M1 | |
| Obtain correct solution in any form | A1 | e.g. \(\frac{1}{2}\tan 2x = -\frac{1}{3}e^{-3t} + \frac{1}{3}\) |
| Obtain final answer \(x = \frac{1}{2}\tan^{-1}\!\left(\frac{2}{3}(1-e^{-3t})\right)\) | A1 | |
| Total: 7 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State that \(x\) approaches \(\frac{1}{2}\tan^{-1}\!\left(\frac{2}{3}\right)\) | B1 FT | Correct value. Accept \(x \to 0.294\). The FT is dependent on letting \(e^{-3t} \to 0\) in a solution containing \(e^{-3t}\) |
| Total: 1 |
## Question 7(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Correct separation of variables | B1 | $\int \sec^2 2x\, dx = \int e^{-3t}\, dt$; needs correct structure |
| Obtain term $-\frac{1}{3}e^{-3t}$ | B1 | |
| Obtain term of the form $k\tan 2x$ | M1 | From correct working |
| Obtain term $\frac{1}{2}\tan 2x$ | A1 | |
| Use $x=0,\, t=0$ to evaluate a constant, or as limits in a solution containing terms of the form $a\tan 2x$ and $be^{-3t}$, where $ab \neq 0$ | M1 | |
| Obtain correct solution in any form | A1 | e.g. $\frac{1}{2}\tan 2x = -\frac{1}{3}e^{-3t} + \frac{1}{3}$ |
| Obtain final answer $x = \frac{1}{2}\tan^{-1}\!\left(\frac{2}{3}(1-e^{-3t})\right)$ | A1 | |
| **Total: 7** | | |
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## Question 7(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| State that $x$ approaches $\frac{1}{2}\tan^{-1}\!\left(\frac{2}{3}\right)$ | B1 FT | Correct value. Accept $x \to 0.294$. The FT is dependent on letting $e^{-3t} \to 0$ in a solution containing $e^{-3t}$ |
| **Total: 1** | | |
---7 The variables $x$ and $t$ satisfy the differential equation
$$\mathrm { e } ^ { 3 t } \frac { \mathrm {~d} x } { \mathrm {~d} t } = \cos ^ { 2 } 2 x$$
for $t \geqslant 0$. It is given that $x = 0$ when $t = 0$.
\begin{enumerate}[label=(\alph*)]
\item Solve the differential equation and obtain an expression for $x$ in terms of $t$.
\item State what happens to the value of $x$ when $t$ tends to infinity.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2020 Q7 [8]}}