| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2019 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration with Partial Fractions |
| Type | Basic partial fractions then integrate |
| Difficulty | Standard +0.3 This is a standard C3/C4 partial fractions question with straightforward decomposition (repeated linear factor), routine integration of logarithmic and rational terms, and a parametric volume of revolution that directly uses part (b). All steps are textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08j Integration using partial fractions4.08e Mean value of function: using integral |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{9}{t^2(t-3)} = \frac{A}{t} + \frac{B}{t^2} + \frac{C}{t-3}\) | Setup of partial fractions | |
| \(9 = At(t-3) + B(t-3) + Ct^2\) | B1 | A correct equation (may be implied) |
| \(t=3 \Rightarrow C=...\) or \(t=0 \Rightarrow B=...\) or expanding and comparing | M1 | Finds one constant by either substitution or use of simultaneous equations |
| \(A=-1, B=-3, C=1\) | A1 | Correct values or correct fractions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int -\frac{1}{t} + \frac{1}{t-3}\,dt = -\ln t + \ln(t-3)\) | M1 | Allow for \(\int \frac{A}{t} + \frac{C}{t-3}\,dt = \alpha\ln t + \beta\ln(t-3)\) |
| \(\int -\frac{3}{t^2}\,dt = \frac{3}{t}\) | M1 | Allow for \(\int \frac{B}{t^2}\,dt = \pm\frac{\alpha}{t}\) |
| \(\int \frac{9}{t^2(t-3)}\,dt = -\ln t + \frac{3}{t} + \ln(t-3) + (c)\) | A1ft | Correct integration (possibly unsimplified) or correct follow through for non-zero \(A\), \(B\), \(C\) |
| \(I = \left[-\ln t + \frac{3}{t} + \ln(t-3)\right]_4^{12}\) substituting 12 and 4 | M1 | For substituting in 12 and 4 into a "changed" function and subtracting either way round |
| \(= \ln\!\left(\frac{9\times4}{12}\right) - \frac{1}{2}\) | dddM1 | Dependent on all previous method marks; must be fully correct log work combining ln's into single logarithm |
| \(= \ln(3) - \frac{1}{2}\) | A1 | Condone lack of brackets; allow equivalents for \(\frac{1}{2}\) e.g. \(0.5\) or \(\frac{2}{4}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = 2\ln(t-3) \Rightarrow \frac{dx}{dt} = \frac{2}{t-3}\) | B1, M1 on ePEN | Correct expression for \(\frac{dx}{dt}\) (may be implied) |
| \(V = \int \pi y^2 \frac{dx}{dt}\,dt = \int \pi \times \frac{36}{t^2} \times \frac{2}{(t-3)}\,dt\) | M1 | Uses \((\pi\times)\int y^2\frac{dx}{dt}\,dt = \int\left(\frac{6}{t}\right)^2 \times their\,\frac{2}{(t-3)}\,dt\); condone missing brackets, missing \(\pi\) and missing \(dt\) |
| \(= 8\pi \times I\) | A1 | Correct volume in terms of \(\pi\); allow \(k=8\pi\); must reference limits (\(x=0\Rightarrow t=4\) and \(x=2\ln 9\Rightarrow t=12\)) |
| \(V = 8\pi\!\left(\ln 3 - \frac{1}{2}\right)\) | A1 can be awarded for this provided above conditions met |
## Question 11:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{9}{t^2(t-3)} = \frac{A}{t} + \frac{B}{t^2} + \frac{C}{t-3}$ | | Setup of partial fractions |
| $9 = At(t-3) + B(t-3) + Ct^2$ | B1 | A correct equation (may be implied) |
| $t=3 \Rightarrow C=...$ or $t=0 \Rightarrow B=...$ or expanding and comparing | M1 | Finds one constant by either substitution or use of simultaneous equations |
| $A=-1, B=-3, C=1$ | A1 | Correct values or correct fractions |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int -\frac{1}{t} + \frac{1}{t-3}\,dt = -\ln t + \ln(t-3)$ | M1 | Allow for $\int \frac{A}{t} + \frac{C}{t-3}\,dt = \alpha\ln t + \beta\ln(t-3)$ |
| $\int -\frac{3}{t^2}\,dt = \frac{3}{t}$ | M1 | Allow for $\int \frac{B}{t^2}\,dt = \pm\frac{\alpha}{t}$ |
| $\int \frac{9}{t^2(t-3)}\,dt = -\ln t + \frac{3}{t} + \ln(t-3) + (c)$ | A1ft | Correct integration (possibly unsimplified) or correct follow through for non-zero $A$, $B$, $C$ |
| $I = \left[-\ln t + \frac{3}{t} + \ln(t-3)\right]_4^{12}$ substituting 12 and 4 | M1 | For substituting in 12 and 4 into a "changed" function and subtracting either way round |
| $= \ln\!\left(\frac{9\times4}{12}\right) - \frac{1}{2}$ | dddM1 | Dependent on all previous method marks; must be fully correct log work combining ln's into single logarithm |
| $= \ln(3) - \frac{1}{2}$ | A1 | Condone lack of brackets; allow equivalents for $\frac{1}{2}$ e.g. $0.5$ or $\frac{2}{4}$ |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = 2\ln(t-3) \Rightarrow \frac{dx}{dt} = \frac{2}{t-3}$ | B1, M1 on ePEN | Correct expression for $\frac{dx}{dt}$ (may be implied) |
| $V = \int \pi y^2 \frac{dx}{dt}\,dt = \int \pi \times \frac{36}{t^2} \times \frac{2}{(t-3)}\,dt$ | M1 | Uses $(\pi\times)\int y^2\frac{dx}{dt}\,dt = \int\left(\frac{6}{t}\right)^2 \times their\,\frac{2}{(t-3)}\,dt$; condone missing brackets, missing $\pi$ and missing $dt$ |
| $= 8\pi \times I$ | A1 | Correct volume in terms of $\pi$; allow $k=8\pi$; must reference limits ($x=0\Rightarrow t=4$ and $x=2\ln 9\Rightarrow t=12$) |
| $V = 8\pi\!\left(\ln 3 - \frac{1}{2}\right)$ | | A1 can be awarded for this provided above conditions met |
---11. (a) Given
$$\frac { 9 } { t ^ { 2 } ( t - 3 ) } \equiv \frac { A } { t } + \frac { B } { t ^ { 2 } } + \frac { C } { ( t - 3 ) }$$
find the value of the constants $A , B$ and $C$.\\
(b)
$$I = \int _ { 4 } ^ { 12 } \frac { 9 } { t ^ { 2 } ( t - 3 ) } \mathrm { d } t$$
Find the exact value of $I$, giving your answer in the form $\ln ( a ) - b$, where $a$ and $b$ are positive constants.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a9870c94-0910-46ec-a54a-44a431cb324e-34_535_880_959_525}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a sketch of part of the curve $C$ with parametric equations
$$x = 2 \ln ( t - 3 ) , \quad y = \frac { 6 } { t } \quad t > 3$$
The finite region $R$, shown shaded in Figure 3, is bounded by the curve $C$, the $y$-axis, the $x$-axis and the line with equation $x = 2 \ln 9$
The region $R$ is rotated $360 ^ { \circ }$ about the $x$-axis to form a solid of revolution.\\
(c) Show that the exact volume of the solid generated is
$$k \times I$$
where $k$ is a constant to be found.\\
\hfill \mbox{\textit{Edexcel C34 2019 Q11 [12]}}