| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2014 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Repeated factor with x² in denominator |
| Difficulty | Standard +0.3 Part (a) is a standard partial fractions decomposition with repeated linear factor in denominator—routine C4 technique. Part (b) requires setting up a volume of revolution integral, but after squaring the function and using the partial fractions result, it becomes straightforward integration. This is a typical textbook-style C4 question slightly easier than average due to its predictable structure. |
| Spec | 1.02y Partial fractions: decompose rational functions4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| At least one of "\(B\)" or "\(C\)" correct | B1 | |
| \(B = 25\), \(C = 100\) | B1 cso | Breaks up partial fraction correctly into three terms and both \(B = 25\) and \(C = 100\) |
| Writes correct identity \(25 \equiv Ax(2x+1) + B(2x+1) + Cx^2\) and attempts to find \(A\), \(B\) or \(C\) | M1 | Can be achieved by substitution or comparing coefficients |
| \(x=0\): \(25 = B\); \(x = -\frac{1}{2}\): \(25 = \frac{1}{4}C \Rightarrow C = 100\); \(x^2\) terms: \(0 = 2A + C \Rightarrow A = -50\) | A1 | Correct value for \(A\) from correct identity and correct partial fraction decomposition |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(V = \pi\int_1^4\left(\dfrac{5}{x\sqrt{(2x+1)}}\right)^2 dx\) | B1 | For \(\pi\int\left(\dfrac{5}{x\sqrt{(2x+1)}}\right)^2\). Ignore limits and \(dx\) |
| \(\int \dfrac{25}{x^2(2x+1)}\,dx = \int\left(-\dfrac{50}{x} + \dfrac{25}{x^2} + \dfrac{100}{(2x+1)}\right)dx\) | M1 | Either \(\pm\dfrac{A}{x} \to \pm a\ln x\) or \(\pm a\ln kx\), or \(\pm\dfrac{B}{x^2} \to \pm bx^{-1}\), or \(\dfrac{C}{(2x+1)} \to \pm c\ln(2x+1)\) |
| \(= -50\ln x - \dfrac{25x^{-1}}{(-1)} + \dfrac{100}{2}\ln(2x+1) + c\) | A1ft | At least two terms correctly integrated |
| A1ft | All three terms correctly integrated | |
| Applies limits of 4 and 1, subtracts correct way | dM1 | Dependent on previous M mark |
| \(= \left(-50\ln 4 - \dfrac{25}{4} + 50\ln 9\right) - (0 - 25 + 50\ln 3)\) | ||
| \(= 50\ln\left(\dfrac{3}{4}\right) + \dfrac{75}{4}\) | ||
| \(V = \dfrac{75}{4}\pi + 50\pi\ln\left(\dfrac{3}{4}\right)\) | A1 oe | Final exact answer in form \(a + b\ln c\), e.g. \(\dfrac{75}{4}\pi + 50\pi\ln\left(\dfrac{3}{4}\right)\) or equivalent forms |
# Question 4:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| At least one of "$B$" or "$C$" correct | B1 | |
| $B = 25$, $C = 100$ | B1 cso | Breaks up partial fraction correctly into **three terms** and both $B = 25$ and $C = 100$ |
| Writes correct identity $25 \equiv Ax(2x+1) + B(2x+1) + Cx^2$ and attempts to find $A$, $B$ or $C$ | M1 | Can be achieved by substitution or comparing coefficients |
| $x=0$: $25 = B$; $x = -\frac{1}{2}$: $25 = \frac{1}{4}C \Rightarrow C = 100$; $x^2$ terms: $0 = 2A + C \Rightarrow A = -50$ | A1 | Correct value for $A$ from correct identity and correct partial fraction decomposition |
$$\frac{25}{x^2(2x+1)} \equiv -\frac{50}{x} + \frac{25}{x^2} + \frac{100}{(2x+1)}$$
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $V = \pi\int_1^4\left(\dfrac{5}{x\sqrt{(2x+1)}}\right)^2 dx$ | B1 | For $\pi\int\left(\dfrac{5}{x\sqrt{(2x+1)}}\right)^2$. Ignore limits and $dx$ |
| $\int \dfrac{25}{x^2(2x+1)}\,dx = \int\left(-\dfrac{50}{x} + \dfrac{25}{x^2} + \dfrac{100}{(2x+1)}\right)dx$ | M1 | Either $\pm\dfrac{A}{x} \to \pm a\ln x$ or $\pm a\ln kx$, or $\pm\dfrac{B}{x^2} \to \pm bx^{-1}$, or $\dfrac{C}{(2x+1)} \to \pm c\ln(2x+1)$ |
| $= -50\ln x - \dfrac{25x^{-1}}{(-1)} + \dfrac{100}{2}\ln(2x+1) + c$ | A1ft | At least two terms correctly integrated |
| | A1ft | All three terms correctly integrated |
| Applies limits of 4 and 1, subtracts correct way | dM1 | Dependent on previous M mark |
| $= \left(-50\ln 4 - \dfrac{25}{4} + 50\ln 9\right) - (0 - 25 + 50\ln 3)$ | | |
| $= 50\ln\left(\dfrac{3}{4}\right) + \dfrac{75}{4}$ | | |
| $V = \dfrac{75}{4}\pi + 50\pi\ln\left(\dfrac{3}{4}\right)$ | A1 oe | Final exact answer in form $a + b\ln c$, e.g. $\dfrac{75}{4}\pi + 50\pi\ln\left(\dfrac{3}{4}\right)$ or equivalent forms |
---4. (a) Express $\frac { 25 } { x ^ { 2 } ( 2 x + 1 ) }$ in partial fractions.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e14881c1-5ba5-4868-92ee-8bc58d4884dc-06_623_849_408_561}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of part of the curve $C$ with equation $y = \frac { 5 } { x \sqrt { } ( 2 x + 1 ) } , x > 0$\\
The finite region $R$ is bounded by the curve $C$, the $x$-axis, the line with equation $x = 1$ and the line with equation $x = 4$
This region is shown shaded in Figure 2
The region $R$ is rotated through $360 ^ { \circ }$ about the $x$-axis.\\
(b) Use calculus to find the exact volume of the solid of revolution generated, giving your answer in the form $a + b \ln c$, where $a , b$ and $c$ are constants.\\
\hfill \mbox{\textit{Edexcel C4 2014 Q4 [10]}}