| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2011 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Find stationary/turning points |
| Difficulty | Challenging +1.2 This is a multi-part parametric curves question requiring sketching, symmetry analysis, differentiation, and self-intersection investigation. While it covers several techniques (parametric differentiation, finding turning points, angle between tangents), each part follows standard FP2 methods without requiring novel insights. The sketching and pattern recognition in part (i) adds some challenge, but overall this is a typical Further Pure question that's moderately harder than average A-level due to the Further Maths content and multi-step nature. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks |
|---|---|
| G1 | |
| Total | 7 |
| Answer | Marks |
|---|---|
| M1 | Considering effect on \(t\) |
| A1 (ag) | Effect on \(y\) |
| B1 | |
| Total | 3 |
| Answer | Marks |
|---|---|
| M1 | Using Chain Rule |
| A1 | |
| A1 | Values of \(t\) |
| A1 | |
| A1 | Both, in any form |
| Total | 5 |
| Answer | Marks |
|---|---|
| B1 (ag) | Verification |
| M1 | Complete method for angle |
| A1 | Accept \(115°\) (or \(65°\)) |
| Total | 3 |
## (i)
[Graphs for $a = 1$, $a = 2$, $a = 0.5$ shown]
| | G1 | |
| **Total** | **7** | |
## (ii)
If $x \to -x, t \to -t$
but $y(-t) = y(t)$
Curve is symmetrical in the $y$-axis
| | M1 | Considering effect on $t$ |
| | A1 (ag) | Effect on $y$ |
| | B1 | |
| **Total** | **3** | |
## (iii)
$\frac{dy}{dx} = \frac{a\sin t}{1 + a\cos t}$
$\frac{dy}{dx} = 0 \Rightarrow a\sin t = 0 \Rightarrow t = 0$ and $\pm\pi$
$t = 0 \Rightarrow$ T.P. is $(0, 1-a)$
$t = \pm\pi \Rightarrow$ T.P. are $(\pm\pi, 1+a)$
| | M1 | Using Chain Rule |
| | A1 | |
| | A1 | Values of $t$ |
| | A1 | |
| | A1 | Both, in any form |
| **Total** | **5** | |
## (iv)
$a = \frac{\pi}{2}$: both $t = \frac{\pi}{2}$ and $\frac{3\pi}{2}$ give the point $(\pi, 1)$
Gradients are $a$ and $-a$ (or $\frac{\pi}{2}$ and $-\frac{\pi}{2}$)
Hence angle is $2\arctan(\frac{\pi}{2}) \approx 2.01$ radians
| | B1 (ag) | Verification |
| | M1 | Complete method for angle |
| | A1 | Accept $115°$ (or $65°$) |
| **Total** | **3** | |
**TOTAL FOR QUESTION 5: 18**A curve has parametric equations
$$x = t + a \sin t, \quad y = 1 - a \cos t,$$
where $a$ is a positive constant.
\begin{enumerate}[label=(\roman*)]
\item Draw, on separate diagrams, sketches of the curve for $-2\pi < t < 2\pi$ in the cases $a = 1$, $a = 2$ and $a = 0.5$.
By investigating other cases, state the value(s) of $a$ for which the curve has
\begin{enumerate}[label=(\Alph*)]
\item loops,
\item cusps. [7]
\end{enumerate}
\item Suppose that the point P$(x, y)$ lies on the curve. Show that the point P$'(-x, y)$ also lies on the curve. What does this indicate about the symmetry of the curve? [3]
\item Find an expression in terms of $a$ and $t$ for the gradient of the curve. Hence find, in terms of $a$, the coordinates of the turning points on the curve for $-2\pi < t < 2\pi$ and $a \neq 1$. [5]
\item In the case $a = \frac{1}{2}\pi$, show that $t = \frac{1}{3}\pi$ and $t = \frac{5}{3}\pi$ give the same point. Find the angle at which the curve crosses itself at this point. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI FP2 2011 Q5 [18]}}