| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2019 |
| Session | June |
| 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 differential equation with standard integration. Part (i) is a hint for part (ii). The separation is immediate, and the integration involves recognizing that ∫cosec²θ dθ = -cot θ and ∫(1/x)dx = ln x. Applying the initial condition to find the constant is routine. Slightly above average difficulty due to the trigonometric functions and rearrangement needed, but still a standard textbook exercise. |
| Spec | 1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use chain rule | M1 | \(k\cos\theta\sin^{-3}\theta\left(= -k\text{cosec}^2\theta\cot\theta\right)\). Allow M1 for \(-2\cos\theta\sin^{-1}\theta\) |
| Obtain correct answer in any form | A1 | e.g. \(-2\text{cosec}^2\theta\cot\theta\), \(\frac{-2\cos\theta}{\sin^3\theta}\). Accept \(\frac{-2\sin\theta\cos\theta}{\sin^4\theta}\) |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Separate variables correctly and integrate at least one side | B1 | \(\int x\,dx = \int -\text{cosec}^2\theta\cot\theta\,d\theta\) |
| Obtain term \(\frac{1}{2}x^2\) | B1 | |
| Obtain term of the form \(\frac{k}{\sin^2\theta}\) | M1* | or equivalent |
| Obtain term \(\frac{1}{2\sin^2\theta}\) | A1 | or equivalent |
| Use \(x=4\), \(\theta=\frac{1}{6}\pi\) to evaluate a constant, or as limits, in a solution with terms \(ax^2\) and \(\frac{b}{\sin^2\theta}\), where \(ab\neq 0\) | DM1 | Dependent on the preceding M1 |
| Obtain solution \(x = \sqrt{(\text{cosec}^2\theta + 12)}\) | A1 | or equivalent |
| Total | 6 |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use chain rule | M1 | $k\cos\theta\sin^{-3}\theta\left(= -k\text{cosec}^2\theta\cot\theta\right)$. Allow M1 for $-2\cos\theta\sin^{-1}\theta$ |
| Obtain correct answer in any form | A1 | e.g. $-2\text{cosec}^2\theta\cot\theta$, $\frac{-2\cos\theta}{\sin^3\theta}$. Accept $\frac{-2\sin\theta\cos\theta}{\sin^4\theta}$ |
| **Total** | **2** | |
## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Separate variables correctly and integrate at least one side | B1 | $\int x\,dx = \int -\text{cosec}^2\theta\cot\theta\,d\theta$ |
| Obtain term $\frac{1}{2}x^2$ | B1 | |
| Obtain term of the form $\frac{k}{\sin^2\theta}$ | M1* | or equivalent |
| Obtain term $\frac{1}{2\sin^2\theta}$ | A1 | or equivalent |
| Use $x=4$, $\theta=\frac{1}{6}\pi$ to evaluate a constant, or as limits, in a solution with terms $ax^2$ and $\frac{b}{\sin^2\theta}$, where $ab\neq 0$ | DM1 | Dependent on the preceding M1 |
| Obtain solution $x = \sqrt{(\text{cosec}^2\theta + 12)}$ | A1 | or equivalent |
| **Total** | **6** | |5 (i) Differentiate $\frac { 1 } { \sin ^ { 2 } \theta }$ with respect to $\theta$.\\
(ii) The variables $x$ and $\theta$ satisfy the differential equation
$$x \tan \theta \frac { d x } { d \theta } + \operatorname { cosec } ^ { 2 } \theta = 0$$
for $0 < \theta < \frac { 1 } { 2 } \pi$ and $x > 0$. It is given that $x = 4$ when $\theta = \frac { 1 } { 6 } \pi$. Solve the differential equation, obtaining an expression for $x$ in terms of $\theta$.\\
\hfill \mbox{\textit{CAIE P3 2019 Q5 [8]}}