| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2007 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration with Partial Fractions |
| Type | Improper algebraic form then partial fractions |
| Difficulty | Standard +0.3 This is a standard partial fractions question with an improper fraction requiring polynomial division first, followed by routine integration. While it requires multiple steps (division, finding constants, integration, simplification), each step follows a well-practiced algorithm with no novel insight needed. Slightly easier than average due to the straightforward nature of the algebraic manipulation and the helpful 'show that' format in part (b). |
| Spec | 1.02y Partial fractions: decompose rational functions1.08j Integration using partial fractions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{2(4x^2+1)}{(2x+1)(2x-1)} \equiv 2 + \frac{4}{(2x+1)(2x-1)}\), so \(A=2\) | B1 | |
| \(4 \equiv B(2x-1) + C(2x+1)\) or \(Dx + E \equiv B(2x-1) + C(2x+1)\) | M1 | Forming any one of these two identities. Can be implied. |
| Let \(x = -\frac{1}{2}\): \(4 = -2B \Rightarrow B = -2\) | A1 | Either one of \(B=-2\) or \(C=2\) correct |
| Let \(x = \frac{1}{2}\): \(4 = 2C \Rightarrow C = 2\) | A1 | Both \(B\) and \(C\) correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{2(4x^2+1)}{(2x+1)(2x-1)} \equiv A + \frac{B}{(2x+1)} + \frac{C}{(2x-1)}\), \(A=2\) | B1 | Decide to award B1 here, for \(A=2\) |
| \(2(4x^2+1) \equiv A(2x+1)(2x-1) + B(2x-1) + C(2x+1)\) | M1 | Forming this identity. Can be implied. |
| Equate \(x^2\): \(8=4A \Rightarrow A=2\) | ||
| \(B=-2\), \(C=2\) | A1, A1 | Either one / both correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int \frac{2(4x^2+1)}{(2x+1)(2x-1)}\,dx = \int 2 - \frac{2}{(2x+1)} + \frac{2}{(2x-1)}\,dx\) | ||
| \(= 2x - \frac{2}{2}\ln(2x+1) + \frac{2}{2}\ln(2x-1) \ (+c)\) | M1* | Either \(p\ln(2x+1)\) or \(q\ln(2x-1)\) present |
| \(A \rightarrow Ax\) | B1\(\checkmark\) | |
| \(-\frac{2}{2}\ln(2x+1) + \frac{2}{2}\ln(2x-1)\) or \(-\ln(2x+1)+\ln(2x-1)\) | A1 | cso & aef |
| \(\int_1^2 = \left[2x - \ln(2x+1) + \ln(2x-1)\right]_1^2\) | ||
| \(= (4 - \ln 5 + \ln 3)-(2 - \ln 3 + \ln 1)\) | depM1* | Substitutes limits 2 and 1, subtracts correct way round |
| \(= 2 + \ln 3 + \ln 3 - \ln 5\) | ||
| \(= 2 + \ln\!\left(\frac{3(3)}{5}\right)\) | M1 | Use of correct product/power and/or quotient laws for logarithms to obtain a single logarithmic term for their numerical expression |
| \(= 2 + \ln\!\left(\frac{9}{5}\right)\) | A1 | Or \(2 - \ln\!\left(\frac{5}{9}\right)\) with \(k\) stated as \(\frac{9}{5}\) |
## Question 4(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{2(4x^2+1)}{(2x+1)(2x-1)} \equiv 2 + \frac{4}{(2x+1)(2x-1)}$, so $A=2$ | B1 | |
| $4 \equiv B(2x-1) + C(2x+1)$ or $Dx + E \equiv B(2x-1) + C(2x+1)$ | M1 | Forming any one of these two identities. Can be implied. |
| Let $x = -\frac{1}{2}$: $4 = -2B \Rightarrow B = -2$ | A1 | Either one of $B=-2$ or $C=2$ correct |
| Let $x = \frac{1}{2}$: $4 = 2C \Rightarrow C = 2$ | A1 | Both $B$ and $C$ correct |
**Way 2 (Aliter):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{2(4x^2+1)}{(2x+1)(2x-1)} \equiv A + \frac{B}{(2x+1)} + \frac{C}{(2x-1)}$, $A=2$ | B1 | Decide to award B1 here, for $A=2$ |
| $2(4x^2+1) \equiv A(2x+1)(2x-1) + B(2x-1) + C(2x+1)$ | M1 | Forming this identity. Can be implied. |
| Equate $x^2$: $8=4A \Rightarrow A=2$ | | |
| $B=-2$, $C=2$ | A1, A1 | Either one / both correct |
---
## Question 4(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int \frac{2(4x^2+1)}{(2x+1)(2x-1)}\,dx = \int 2 - \frac{2}{(2x+1)} + \frac{2}{(2x-1)}\,dx$ | | |
| $= 2x - \frac{2}{2}\ln(2x+1) + \frac{2}{2}\ln(2x-1) \ (+c)$ | M1* | Either $p\ln(2x+1)$ or $q\ln(2x-1)$ present |
| $A \rightarrow Ax$ | B1$\checkmark$ | |
| $-\frac{2}{2}\ln(2x+1) + \frac{2}{2}\ln(2x-1)$ or $-\ln(2x+1)+\ln(2x-1)$ | A1 | cso & aef |
| $\int_1^2 = \left[2x - \ln(2x+1) + \ln(2x-1)\right]_1^2$ | | |
| $= (4 - \ln 5 + \ln 3)-(2 - \ln 3 + \ln 1)$ | depM1* | Substitutes limits 2 and 1, subtracts correct way round |
| $= 2 + \ln 3 + \ln 3 - \ln 5$ | | |
| $= 2 + \ln\!\left(\frac{3(3)}{5}\right)$ | M1 | Use of correct product/power and/or quotient laws for logarithms to obtain a single logarithmic term for their numerical expression |
| $= 2 + \ln\!\left(\frac{9}{5}\right)$ | A1 | Or $2 - \ln\!\left(\frac{5}{9}\right)$ with $k$ stated as $\frac{9}{5}$ |
---4.
$$\frac { 2 \left( 4 x ^ { 2 } + 1 \right) } { ( 2 x + 1 ) ( 2 x - 1 ) } \equiv A + \frac { B } { ( 2 x + 1 ) } + \frac { C } { ( 2 x - 1 ) } .$$
\begin{enumerate}[label=(\alph*)]
\item Find the values of the constants $A , B$ and $C$.
\item Hence show that the exact value of $\int _ { 1 } ^ { 2 } \frac { 2 \left( 4 x ^ { 2 } + 1 \right) } { ( 2 x + 1 ) ( 2 x - 1 ) } \mathrm { d } x$ is $2 + \ln k$, giving the value of the constant $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2007 Q4 [10]}}