| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2012 |
| Session | June |
| 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 C4 partial fractions question with a repeated linear factor. While it requires multiple techniques (finding constants, integration, definite integral evaluation), each step follows a well-practiced procedure with no novel insight needed. The repeated factor adds slight complexity over basic partial fractions, placing it slightly above average difficulty. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08j Integration using partial fractions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(1 = A(3x-1)^2 + Bx(3x-1) + Cx\) | B1 | |
| \(x \to 0\): \((1 = A)\) | M1 | |
| \(x \to \frac{1}{3}\): \(1 = \frac{1}{3}C \Rightarrow C = 3\) | A1 | any two constants correct |
| Coefficients of \(x^2\): \(0 = 9A + 3B \Rightarrow B = -3\) | A1 | all three constants correct (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int \left(\frac{1}{x} - \frac{3}{3x-1} + \frac{3}{(3x-1)^2}\right)dx\) | ||
| \(= \ln x - \frac{3}{3}\ln(3x-1) + \frac{3}{(-1)3}(3x-1)^{-1} \quad (+C)\) | M1 A1ft A1ft | |
| \(\left(= \ln x - \ln(3x-1) - \frac{1}{3x-1} \quad (+C)\right)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | 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 | |
| \(= \ln\frac{2\times2}{5} + \ldots\) | M1 | |
| \(= \frac{3}{10} + \ln\left(\frac{4}{5}\right)\) | A1 | (6) [10] |
**Question 1:**
**Part (a):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $1 = A(3x-1)^2 + Bx(3x-1) + Cx$ | B1 | |
| $x \to 0$: $(1 = A)$ | M1 | |
| $x \to \frac{1}{3}$: $1 = \frac{1}{3}C \Rightarrow C = 3$ | A1 | any two constants correct |
| Coefficients of $x^2$: $0 = 9A + 3B \Rightarrow B = -3$ | A1 | all three constants correct **(4)** |
**Part (b)(i):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int \left(\frac{1}{x} - \frac{3}{3x-1} + \frac{3}{(3x-1)^2}\right)dx$ | | |
| $= \ln x - \frac{3}{3}\ln(3x-1) + \frac{3}{(-1)3}(3x-1)^{-1} \quad (+C)$ | M1 A1ft A1ft | |
| $\left(= \ln x - \ln(3x-1) - \frac{1}{3x-1} \quad (+C)\right)$ | | |
**Part (b)(ii):**
| Answer/Working | Marks | 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 | |
| $= \ln\frac{2\times2}{5} + \ldots$ | M1 | |
| $= \frac{3}{10} + \ln\left(\frac{4}{5}\right)$ | A1 | **(6)** **[10]** |1.
$$\mathrm { f } ( x ) = \frac { 1 } { x ( 3 x - 1 ) ^ { 2 } } = \frac { A } { x } + \frac { B } { ( 3 x - 1 ) } + \frac { C } { ( 3 x - 1 ) ^ { 2 } }$$
\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 $\int _ { 1 } ^ { 2 } \mathrm { f } ( x ) \mathrm { d } x$, leaving your answer in the form $a + \ln b$, where $a$ and $b$ are constants.
1\\
$f ( x ) = \frac { 1 } { x ( 3 x - 1 ) ^ { 2 } } = \frac { A } { x } + \frac { } { ( 3 x }$\\
(a) Find the values of the constants $A , B$ and $C$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2012 Q1 [10]}}