| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2017 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration with Partial Fractions |
| Type | Partial fractions in differential equations |
| Difficulty | Standard +0.3 Part (i) is a routine partial fractions decomposition with linear factors. Part (ii) is a standard separable differential equation requiring integration using the partial fractions from (i), followed by applying initial conditions and substitution. This is a textbook application of partial fractions to differential equations with no novel insight required, making it slightly easier than average. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Carry out a relevant method to obtain \(A\) and \(B\) such that \(\frac{1}{x(2x+3)} \equiv \frac{A}{x} + \frac{B}{2x+3}\), or equivalent | M1 | |
| Obtain \(A = \frac{1}{3}\) and \(B = -\frac{2}{3}\), or equivalent | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Separate variables and integrate one side | B1 | |
| Obtain term \(\ln y\) | B1 | |
| Integrate and obtain terms \(\frac{1}{3}\ln x - \frac{1}{3}\ln(2x+3)\), or equivalent | B2 FT | |
| Use \(x = 1\) and \(y = 1\) to evaluate a constant, or as limits, in a solution containing \(a\ln y\), \(b\ln x\), \(c\ln(2x+3)\) | M1 | |
| Obtain correct solution in any form, e.g. \(\ln y = \frac{1}{3}\ln x - \frac{1}{3}\ln(2x+3) + \frac{1}{3}\ln 5\) | A1 | |
| Obtain answer \(y = 1.29\) (3 s.f. only) | A1 |
## Question 9(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Carry out a relevant method to obtain $A$ and $B$ such that $\frac{1}{x(2x+3)} \equiv \frac{A}{x} + \frac{B}{2x+3}$, or equivalent | M1 | |
| Obtain $A = \frac{1}{3}$ and $B = -\frac{2}{3}$, or equivalent | A1 | |
## Question 9(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Separate variables and integrate one side | B1 | |
| Obtain term $\ln y$ | B1 | |
| Integrate and obtain terms $\frac{1}{3}\ln x - \frac{1}{3}\ln(2x+3)$, or equivalent | B2 FT | |
| Use $x = 1$ and $y = 1$ to evaluate a constant, or as limits, in a solution containing $a\ln y$, $b\ln x$, $c\ln(2x+3)$ | M1 | |
| Obtain correct solution in any form, e.g. $\ln y = \frac{1}{3}\ln x - \frac{1}{3}\ln(2x+3) + \frac{1}{3}\ln 5$ | A1 | |
| Obtain answer $y = 1.29$ (3 s.f. only) | A1 | |9 (i) Express $\frac { 1 } { x ( 2 x + 3 ) }$ in partial fractions.\\
(ii) The variables $x$ and $y$ satisfy the differential equation
$$x ( 2 x + 3 ) \frac { \mathrm { d } y } { \mathrm {~d} x } = y$$
and it is given that $y = 1$ when $x = 1$. Solve the differential equation and calculate the value of $y$ when $x = 9$, giving your answer correct to 3 significant figures.\\
\hfill \mbox{\textit{CAIE P3 2017 Q9 [9]}}