| Exam Board | OCR MEI |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2019 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Partial fractions with linear factors – decompose and integrate (indefinite) |
| Difficulty | Moderate -0.8 This is a straightforward partial fractions question with simple linear factors and standard integration. Part (a) is routine algebraic manipulation, and part (b) requires integrating the partial fractions to get logarithms then combining them using log laws. Both parts are textbook exercises with no problem-solving required, making this easier than average. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08j Integration using partial fractions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{(x+2)(x+3)} = \frac{A}{(x+2)} + \frac{B}{(x+3)}\) | Method marks implied by correct answer | |
| \(1 = A(x+3) + B(x+2)\) | M1 | For clearing the fractions |
| \(x = -3 \Rightarrow 1 = -B \Rightarrow B = -1\); \(x = -2 \Rightarrow 1 = A\) | M1 | For one appropriate substitution |
| \(\frac{1}{(x+2)(x+3)} = \frac{1}{(x+2)} - \frac{1}{(x+3)}\) | A1 | For correct completion |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int\left(\frac{1}{(x+2)} - \frac{1}{(x+3)}\right)dx = \ln\ | x+2\ | - \ln\ |
| \(\ln\left\ | \frac{x+2}{x+3}\right\ | + c\) |
| [3] |
## Question 3:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{(x+2)(x+3)} = \frac{A}{(x+2)} + \frac{B}{(x+3)}$ | | Method marks implied by correct answer |
| $1 = A(x+3) + B(x+2)$ | M1 | For clearing the fractions |
| $x = -3 \Rightarrow 1 = -B \Rightarrow B = -1$; $x = -2 \Rightarrow 1 = A$ | M1 | For one appropriate substitution |
| $\frac{1}{(x+2)(x+3)} = \frac{1}{(x+2)} - \frac{1}{(x+3)}$ | A1 | For correct completion |
| **[3]** | | |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int\left(\frac{1}{(x+2)} - \frac{1}{(x+3)}\right)dx = \ln\|x+2\| - \ln\|x+3\| \;[+c]$ | M2 | For both terms correct FT their fractions; M1 for one of their terms correct |
| $\ln\left\|\frac{x+2}{x+3}\right\| + c$ | A1 | A0 if modulus sign missing or if further work e.g. to try and find $c$; Ignore '$f(x) =$'; Need brackets or better (e.g. single fraction) |
| **[3]** | | |
---3
\begin{enumerate}[label=(\alph*)]
\item Express $\frac { 1 } { ( x + 2 ) ( x + 3 ) }$ in partial fractions.
\item Find $\int \frac { 1 } { ( x + 2 ) ( x + 3 ) } \mathrm { d } x$ in the form $\ln ( \mathrm { f } ( x ) ) + c$, where $c$ is the constant of integration and $\mathrm { f } ( x )$ is a function to be determined.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 3 2019 Q3 [6]}}