| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2009 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration with Partial Fractions |
| Type | Basic partial fractions then integrate |
| Difficulty | Moderate -0.3 This is a standard C4 partial fractions question with straightforward cover-up method for finding constants, followed by routine integration of logarithmic terms. The algebra is mechanical and the integration technique is direct application of standard results, making it slightly easier than average but still requiring competent execution across multiple steps. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08j Integration using partial fractions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{4-2x}{(2x+1)(x+1)(x+3)} = \frac{A}{2x+1}+\frac{B}{x+1}+\frac{C}{x+3}\) | ||
| \(4-2x = A(x+1)(x+3)+B(2x+1)(x+3)+C(2x+1)(x+1)\) | M1 | |
| M1 | A method for evaluating one constant | |
| \(x\to -\frac{1}{2},\ 5=A\left(\frac{1}{2}\right)\left(\frac{5}{2}\right)\Rightarrow A=4\) | A1 | any one correct constant |
| \(x\to -1,\ 6=B(-1)(2)\Rightarrow B=-3\) | ||
| \(x\to -3,\ 10=C(-5)(-2)\Rightarrow C=1\) | A1 (4) | all three constants correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int\left(\frac{4}{2x+1}-\frac{3}{x+1}+\frac{1}{x+3}\right)dx\) | ||
| \(= \frac{4}{2}\ln(2x+1)-3\ln(x+1)+\ln(x+3)+C\) | M1 A1ft | A1 two ln terms correct |
| A1ft (3) | All three ln terms correct and "\(+C\)"; ft constants |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\left[2\ln(2x+1)-3\ln(x+1)+\ln(x+3)\right]_0^2\) | ||
| \(= (2\ln 5-3\ln 3+\ln 5)-(2\ln 1-3\ln 1+\ln 3)\) | M1 | |
| \(= 3\ln 5-4\ln 3\) | ||
| \(= \ln\left(\frac{5^3}{3^4}\right)\) | M1 | |
| \(= \ln\left(\frac{125}{81}\right)\) | A1 (3) |
# Question 3:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{4-2x}{(2x+1)(x+1)(x+3)} = \frac{A}{2x+1}+\frac{B}{x+1}+\frac{C}{x+3}$ | | |
| $4-2x = A(x+1)(x+3)+B(2x+1)(x+3)+C(2x+1)(x+1)$ | M1 | |
| | M1 | A method for evaluating one constant |
| $x\to -\frac{1}{2},\ 5=A\left(\frac{1}{2}\right)\left(\frac{5}{2}\right)\Rightarrow A=4$ | A1 | any one correct constant |
| $x\to -1,\ 6=B(-1)(2)\Rightarrow B=-3$ | | |
| $x\to -3,\ 10=C(-5)(-2)\Rightarrow C=1$ | A1 (4) | all three constants correct |
## Part (b)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int\left(\frac{4}{2x+1}-\frac{3}{x+1}+\frac{1}{x+3}\right)dx$ | | |
| $= \frac{4}{2}\ln(2x+1)-3\ln(x+1)+\ln(x+3)+C$ | M1 A1ft | A1 two ln terms correct |
| | A1ft (3) | All three ln terms correct and "$+C$"; ft constants |
## Part (b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left[2\ln(2x+1)-3\ln(x+1)+\ln(x+3)\right]_0^2$ | | |
| $= (2\ln 5-3\ln 3+\ln 5)-(2\ln 1-3\ln 1+\ln 3)$ | M1 | |
| $= 3\ln 5-4\ln 3$ | | |
| $= \ln\left(\frac{5^3}{3^4}\right)$ | M1 | |
| $= \ln\left(\frac{125}{81}\right)$ | A1 (3) | |
---3.
$$\mathrm { f } ( x ) = \frac { 4 - 2 x } { ( 2 x + 1 ) ( x + 1 ) ( x + 3 ) } = \frac { A } { 2 x + 1 } + \frac { B } { x + 1 } + \frac { C } { x + 3 }$$
\begin{enumerate}[label=(\alph*)]
\item Find the values of the constants $A , B$ and $C$.
\item \begin{enumerate}[label=(\roman*)]
\item Hence find $\int f ( x ) \mathrm { d } x$.
\item Find $\int _ { 0 } ^ { 2 } \mathrm { f } ( x ) \mathrm { d } x$ in the form $\ln k$, where $k$ is a constant.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2009 Q3 [10]}}