| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2015 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric integration |
| Type | Partial fractions parametric area |
| Difficulty | Standard +0.3 This is a standard parametric equations question requiring routine application of the area formula with parameter substitution (finding limits from x-values), followed by partial fractions integration and a straightforward cartesian conversion. All techniques are textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 1.02y Partial fractions: decompose rational functions1.03g Parametric equations: of curves and conversion to cartesian1.08h Integration by substitution1.08j Integration using partial fractions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dx}{dt} = \frac{1}{t+2}\) | B1 | States or implies \(\frac{dx}{dt} = \frac{1}{t+2}\); accept embedded in integral before final answer |
| Area \(= \int y\,dx = \int \frac{4}{t^2} \times \frac{1}{(t+2)}\,dt\) | M1 | States and uses Area \(= \int y\,dx\) replacing both \(y\) and \(dx\) by functions of \(t\) |
| Correct proof with limits: Area \(= \int_1^3 \frac{4}{t^2(t+2)}\,dt\) | A1* | Correct proof with no errors on any line for integrand; must have \(dt\); limits need only be correct on final line |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{4}{t^2(t+2)} = \frac{A}{t} + \frac{B}{t^2} + \frac{C}{(t+2)}\) or \(\frac{A}{t^2} + \frac{B}{(t+2)}\) | B1 | Scored for use of partial fractions; accept either form |
| \(4 = At(t+2)+B(t+2)+Ct^2\); sub \(t=0 \Rightarrow B=2\); sub \(t=-2 \Rightarrow C=1\); compare \(t^2\): \(A+C=0 \Rightarrow A=-1\) | M1A1 | Substitute values of \(t\) and/or inspection to determine \(A,B,C\); partial fraction must be correct form. For \(\frac{4}{t^2(t+2)} = \frac{-1}{t}+\frac{2}{t^2}+\frac{1}{(t+2)}\) |
| \(\int_1^3 \frac{-1}{t}+\frac{2}{t^2}+\frac{1}{(t+2)}\,dt = \left[-\ln t - \frac{2}{t} + \ln(t+2)\right]_1^3\) | M1A1 | \(\int \frac{A}{t^2}+\frac{C}{(t+2)}\,dt = \ldots+\ln(t+2)\) |
| \(= \left(-\ln 3 - \frac{2}{3} + \ln 5\right) - \left(-\ln 1 - \frac{2}{1} + \ln 3\right)\) | ||
| \(= \ln\left(\frac{5}{9}\right) + \frac{4}{3}\) | dM1A1 | Dependent on previous M; sub limits, subtract, use correct log law to get form \(a + \ln b\). Correct solution only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sub \(t = e^x - 2\) into \(y = \frac{4}{t^2} \Rightarrow y = \frac{4}{(e^x-2)^2}\), \((x > \ln 2)\) | M1A1 | Rearranges \(x = \ln(t+2)\) to \(t = e^x \pm 2\) and subs into \(y = \frac{4}{t^2}\). Ignore domain reference |
# Question 9:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dx}{dt} = \frac{1}{t+2}$ | B1 | States or implies $\frac{dx}{dt} = \frac{1}{t+2}$; accept embedded in integral before final answer |
| Area $= \int y\,dx = \int \frac{4}{t^2} \times \frac{1}{(t+2)}\,dt$ | M1 | States and uses Area $= \int y\,dx$ replacing both $y$ and $dx$ by functions of $t$ |
| Correct proof with limits: Area $= \int_1^3 \frac{4}{t^2(t+2)}\,dt$ | A1* | Correct proof with no errors on any line for integrand; must have $dt$; limits need only be correct on final line |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{4}{t^2(t+2)} = \frac{A}{t} + \frac{B}{t^2} + \frac{C}{(t+2)}$ or $\frac{A}{t^2} + \frac{B}{(t+2)}$ | B1 | Scored for use of partial fractions; accept either form |
| $4 = At(t+2)+B(t+2)+Ct^2$; sub $t=0 \Rightarrow B=2$; sub $t=-2 \Rightarrow C=1$; compare $t^2$: $A+C=0 \Rightarrow A=-1$ | M1A1 | Substitute values of $t$ and/or inspection to determine $A,B,C$; partial fraction must be correct form. For $\frac{4}{t^2(t+2)} = \frac{-1}{t}+\frac{2}{t^2}+\frac{1}{(t+2)}$ |
| $\int_1^3 \frac{-1}{t}+\frac{2}{t^2}+\frac{1}{(t+2)}\,dt = \left[-\ln t - \frac{2}{t} + \ln(t+2)\right]_1^3$ | M1A1 | $\int \frac{A}{t^2}+\frac{C}{(t+2)}\,dt = \ldots+\ln(t+2)$ |
| $= \left(-\ln 3 - \frac{2}{3} + \ln 5\right) - \left(-\ln 1 - \frac{2}{1} + \ln 3\right)$ | | |
| $= \ln\left(\frac{5}{9}\right) + \frac{4}{3}$ | dM1A1 | Dependent on previous M; sub limits, subtract, use correct log law to get form $a + \ln b$. Correct solution only |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sub $t = e^x - 2$ into $y = \frac{4}{t^2} \Rightarrow y = \frac{4}{(e^x-2)^2}$, $(x > \ln 2)$ | M1A1 | Rearranges $x = \ln(t+2)$ to $t = e^x \pm 2$ and subs into $y = \frac{4}{t^2}$. Ignore domain reference |
---9.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{03548211-79cb-4629-b6ca-aa9dfcc77a33-15_618_899_262_566}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
The curve $C$ has parametric equations
$$x = \ln ( t + 2 ) , \quad y = \frac { 4 } { t ^ { 2 } } \quad t > 0$$
The finite region $R$, shown shaded in Figure 2, is bounded by the curve $C$, the $x$-axis and the lines with equations $x = \ln 3$ and $x = \ln 5$
\begin{enumerate}[label=(\alph*)]
\item Show that the area of $R$ is given by the integral
$$\int _ { 1 } ^ { 3 } \frac { 4 } { t ^ { 2 } ( t + 2 ) } \mathrm { d } t$$
\item Hence find an exact value for the area of $R$.
Write your answer in the form ( $a + \ln b$ ), where $a$ and $b$ are rational numbers.
\item Find a cartesian equation of the curve $C$ in the form $y = \mathrm { f } ( x )$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2015 Q9 [12]}}