| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2018 |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration with Partial Fractions |
| Type | Partial fractions with repeated linear factor |
| Difficulty | Standard +0.3 This is a standard partial fractions question with a repeated linear factor, following a routine template: decompose, integrate, then evaluate definite integral. The techniques are well-practiced at A-level Further Maths, requiring only methodical application of the cover-up rule and standard integration formulas. Slightly above average difficulty due to the repeated factor and logarithm simplification, but still a textbook exercise. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08d Evaluate definite integrals: between limits1.08j Integration using partial fractions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(1 = A(3x-1)^2 + B(3x-1) + Cx\) | B1 | Obtaining this identity at any stage |
| \(x \to 0\): \(1 = A\) | M1 | Complete method for finding any one constant |
| \(x \to \frac{1}{3}\): \(1 = \frac{1}{3}C \Rightarrow C = 3\) | A1 | Any two constants correct |
| \(0 = 9A + 3B \Rightarrow B = -3\) | A1 | All three constants correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int\left(\frac{1}{x} - \frac{3}{3x-1} + \frac{3}{(3x-1)^2}\right)dx\) | ||
| \(= \ln x - \frac{3}{3}\ln(3x-1) + \frac{3}{(-1)3}(3x-1)^{-1}\) \((+C)\) | M1 A1ft A1ft | |
| \(= \ln x - \ln(3x-1) - \frac{1}{3x-1}\) \((+C)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int_1^2 f(x)dx = \left[\ln x - \ln(3x-1) - \frac{1}{3x-1}\right]_1^2\) | ||
| \(= \left(\ln 2 - \ln 5 - \frac{1}{5}\right) - \left(\ln 1 - \ln 2 - \frac{1}{2}\right)\) | M1 | Substituting correct limits and subtracting; evidence both 1 and 2 used |
| \(= \ln\frac{2\times 2}{5} + \ldots\) | M1 | Applies log addition/subtraction rules correctly at least once |
| \(= \frac{3}{10} + \ln\left(\frac{4}{5}\right)\) | A1 | Accept \(\ln\frac{4}{5} + \frac{3}{10}\). Note: \(\frac{3}{10} - \ln\frac{5}{4}\) is not acceptable |
## Question 3(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $1 = A(3x-1)^2 + B(3x-1) + Cx$ | B1 | Obtaining this identity at any stage |
| $x \to 0$: $1 = A$ | M1 | Complete method for finding any one constant |
| $x \to \frac{1}{3}$: $1 = \frac{1}{3}C \Rightarrow C = 3$ | A1 | Any two constants correct |
| $0 = 9A + 3B \Rightarrow B = -3$ | A1 | All three constants correct |
## Question 3(b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int\left(\frac{1}{x} - \frac{3}{3x-1} + \frac{3}{(3x-1)^2}\right)dx$ | | |
| $= \ln x - \frac{3}{3}\ln(3x-1) + \frac{3}{(-1)3}(3x-1)^{-1}$ $(+C)$ | M1 A1ft A1ft | |
| $= \ln x - \ln(3x-1) - \frac{1}{3x-1}$ $(+C)$ | | |
## Question 3(b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_1^2 f(x)dx = \left[\ln x - \ln(3x-1) - \frac{1}{3x-1}\right]_1^2$ | | |
| $= \left(\ln 2 - \ln 5 - \frac{1}{5}\right) - \left(\ln 1 - \ln 2 - \frac{1}{2}\right)$ | M1 | Substituting correct limits and subtracting; evidence both 1 and 2 used |
| $= \ln\frac{2\times 2}{5} + \ldots$ | M1 | Applies log addition/subtraction rules correctly at least once |
| $= \frac{3}{10} + \ln\left(\frac{4}{5}\right)$ | A1 | Accept $\ln\frac{4}{5} + \frac{3}{10}$. Note: $\frac{3}{10} - \ln\frac{5}{4}$ is not acceptable |
---3.
$$\mathrm { f } ( x ) = \frac { 1 } { x ( 3 x - 1 ) ^ { 2 } } = \frac { A } { x } + \frac { B } { ( 3 x - 1 ) } + \frac { C } { ( 3 x - 1 ) ^ { 2 } }$$
\begin{enumerate}[label=(\alph*)]
\item Find the values of the constants $A , B$ and $C$
\item \begin{enumerate}[label=(\roman*)]
\item Hence find $\int \mathrm { f } ( x ) \mathrm { d } x$
\item Find $\int _ { 1 } ^ { 2 } \mathrm { f } ( x ) \mathrm { d } x$, giving your answer in the form $a + \ln b$, where $a$ and $b$ are constants.\\
(6)
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel P4 2018 Q3 [10]}}