| Exam Board | OCR |
|---|---|
| Module | H240/03 (Pure Mathematics and Mechanics) |
| Year | 2018 |
| Session | December |
| Marks | 15 |
| Topic | Parametric integration |
| Type | Show integral then evaluate area |
| Difficulty | Challenging +1.2 This is a multi-part parametric integration question requiring standard techniques: converting area integral to parameter form, partial fractions decomposition, and eliminating the parameter. While it involves several steps and careful algebraic manipulation (especially the partial fractions in part b), all techniques are standard M1/C4 content with no novel insights required. The 'show detailed reasoning' instruction and 8 marks for part (b) indicate extended working, but the methods are textbook applications. Slightly above average difficulty due to length and the need for careful execution across multiple techniques. |
| Spec | 1.02y Partial fractions: decompose rational functions1.03g Parametric equations: of curves and conversion to cartesian1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = \ln(t^2 - 4) \Rightarrow \frac{dx}{dt} = \frac{2t}{t^2 - 4}\) | M1 | Attempt differentiation of \(x\) using chain rule – must be of the form \(\frac{kt}{t^2 - 4}\) |
| \(\text{Area} = \int \frac{4}{t^2}(\frac{2t}{t^2 - 4})dt\) | M1 | Use of \(\int y \frac{dx}{dt} dt\) with their \(\frac{dx}{dt}\) |
| \(= \int \frac{8}{t(t^2 - 4)}dt\) | A1 | AG |
| \(a = 3, b = 4\) | B1 | Correct limits |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{8}{t(t^2 - 4)} = \frac{A}{t} + \frac{B}{t - 2} + \frac{C}{t + 2}\) | B1 | Correct form of partial fractions |
| \(8 = A(t - 2)(t + 2) + B(t + 2) + C(t - 2)\) | M1 | Cover up, substituting or equating coefficients – must be a complete method for finding one of \(A, B\) or \(C\) |
| \(A = -2, B = 1, C = 1\) | A2 | A1 for one correct |
| \(\int(\frac{-2}{t} + \frac{1}{t - 2} + \frac{1}{t + 2})dt = -2\ln t + \ln(t - 2) + \ln(t + 2)\) | M1* | Attempt to integrate all three terms – must be of the form \(\alpha \ln t + \beta \ln(t - 2) + \gamma \ln(t + 2)\) |
| \((-2\ln 4 + \ln 2 + \ln 6) - (-2\ln 3 + \ln 1 + \ln 5)\) | M1dep* | Applying their limits correctly |
| \(\ln(\frac{27}{20})\) | A1 | \(k = \frac{27}{20}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(t^2 = \frac{4}{y} \Rightarrow x = \ln(\frac{4}{y} - 4) = y = K\) | M1* | Re-arrange and eliminate \(t\) |
| \(e^x = \frac{4}{y} - 4 \Rightarrow y = K\) | M1dep* | Remove logs and attempt to make \(y\) the subject |
| \(y = \frac{4}{e^x + 4}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(e^x = t^2 - 4\) | M1* | Remove logs |
| \(t^2 = e^x + 4 \Rightarrow y = K\) | M1dep* | Rearrange and eliminate \(t\) |
| \(y = \frac{4}{e^x + 4}\) | A1 |
## Part (a)
$x = \ln(t^2 - 4) \Rightarrow \frac{dx}{dt} = \frac{2t}{t^2 - 4}$ | M1 | Attempt differentiation of $x$ using chain rule – must be of the form $\frac{kt}{t^2 - 4}$
$\text{Area} = \int \frac{4}{t^2}(\frac{2t}{t^2 - 4})dt$ | M1 | Use of $\int y \frac{dx}{dt} dt$ with their $\frac{dx}{dt}$
$= \int \frac{8}{t(t^2 - 4)}dt$ | A1 | AG
$a = 3, b = 4$ | B1 | Correct limits
## Part (b)
**DR**
$\frac{8}{t(t^2 - 4)} = \frac{A}{t} + \frac{B}{t - 2} + \frac{C}{t + 2}$ | B1 | Correct form of partial fractions
$8 = A(t - 2)(t + 2) + B(t + 2) + C(t - 2)$ | M1 | Cover up, substituting or equating coefficients – must be a complete method for finding one of $A, B$ or $C$
$A = -2, B = 1, C = 1$ | A2 | A1 for one correct
$\int(\frac{-2}{t} + \frac{1}{t - 2} + \frac{1}{t + 2})dt = -2\ln t + \ln(t - 2) + \ln(t + 2)$ | M1* | Attempt to integrate all three terms – must be of the form $\alpha \ln t + \beta \ln(t - 2) + \gamma \ln(t + 2)$
$(-2\ln 4 + \ln 2 + \ln 6) - (-2\ln 3 + \ln 1 + \ln 5)$ | M1dep* | Applying their limits correctly
$\ln(\frac{27}{20})$ | A1 | $k = \frac{27}{20}$
# Question 6 (continued)
## Part (c)
$t^2 = \frac{4}{y} \Rightarrow x = \ln(\frac{4}{y} - 4) = y = K$ | M1* | Re-arrange and eliminate $t$
$e^x = \frac{4}{y} - 4 \Rightarrow y = K$ | M1dep* | Remove logs and attempt to make $y$ the subject
$y = \frac{4}{e^x + 4}$ | A1 |
**Alternative solution**
$e^x = t^2 - 4$ | M1* | Remove logs
$t^2 = e^x + 4 \Rightarrow y = K$ | M1dep* | Rearrange and eliminate $t$
$y = \frac{4}{e^x + 4}$ | A1 |\includegraphics{figure_6}
The diagram shows the curve with parametric equations $x = \ln(t^2 - 4)$, $y = \frac{4}{t}$, where $t > 2$.
The shaded region $R$ is enclosed by the curve, the $x$-axis and the lines $x = \ln 5$ and $x = \ln 12$.
\begin{enumerate}[label=(\alph*)]
\item Show that the area of $R$ is given by
$$\int_a^b \frac{8}{t(t^2 - 4)} dt,$$
where $a$ and $b$ are constants to be determined. [4]
\item In this question you must show detailed reasoning.
Hence find the exact area of $R$, giving your answer in the form $\ln k$, where $k$ is a constant to be determined. [8]
\item Find a cartesian equation of the curve in the form $y = \text{f}(x)$. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR H240/03 2018 Q6 [15]}}