| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2014 |
| Session | November |
| Marks | 9 |
| 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 requiring standard separation of variables, integration (including ln rules), and finding a particular solution using initial conditions. Part (ii) requires recognizing that maximum R occurs when dR/dx = 0. All techniques are routine for P3 level with no novel insight required, making it slightly easier than average. |
| Spec | 1.02z Models in context: use functions in modelling1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Separate variables correctly and attempt to integrate at least one side | B1 | |
| Obtain term \(\ln R\) | B1 | |
| Obtain \(\ln x - 0.57x\) | B1 | |
| Evaluate a constant or use limits \(x = 0.5, R = 16.8\), in a solution containing terms of the form \(a\ln R\) and \(b\ln x\) | M1 | |
| Obtain correct solution in any form | A1 | |
| Obtain a correct expression for \(R\), e.g. \(R = xe^{(3.80 - 0.57x)}, R = 44.7xe^{-0.57x}\) or \(R = 33.6xe^{(0.285 - 0.57x)}\) | A1 | [6] |
| (ii) Equate \(\frac{dR}{dx}\) to zero and solve for \(x\) | M1 | |
| State or imply \(x = 0.57^{-1}\), or equivalent, e.g. 1.75 | A1 | |
| Obtain \(R = 28.8\) (allow 28.9) | A1 | [3] |
**(i)** Separate variables correctly and attempt to integrate at least one side | B1 |
Obtain term $\ln R$ | B1 |
Obtain $\ln x - 0.57x$ | B1 |
Evaluate a constant or use limits $x = 0.5, R = 16.8$, in a solution containing terms of the form $a\ln R$ and $b\ln x$ | M1 |
Obtain correct solution in any form | A1 |
Obtain a correct expression for $R$, e.g. $R = xe^{(3.80 - 0.57x)}, R = 44.7xe^{-0.57x}$ or $R = 33.6xe^{(0.285 - 0.57x)}$ | A1 | [6]
**(ii)** Equate $\frac{dR}{dx}$ to zero and solve for $x$ | M1 |
State or imply $x = 0.57^{-1}$, or equivalent, e.g. 1.75 | A1 |
Obtain $R = 28.8$ (allow 28.9) | A1 | [3]7 In a certain country the government charges tax on each litre of petrol sold to motorists. The revenue per year is $R$ million dollars when the rate of tax is $x$ dollars per litre. The variation of $R$ with $x$ is modelled by the differential equation
$$\frac { \mathrm { d } R } { \mathrm {~d} x } = R \left( \frac { 1 } { x } - 0.57 \right)$$
where $R$ and $x$ are taken to be continuous variables. When $x = 0.5 , R = 16.8$.\\
(i) Solve the differential equation and obtain an expression for $R$ in terms of $x$.\\
(ii) This model predicts that $R$ cannot exceed a certain amount. Find this maximum value of $R$.
\hfill \mbox{\textit{CAIE P3 2014 Q7 [9]}}