| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2024 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Area of region with line boundary |
| Difficulty | Challenging +1.2 This is a standard Further Maths polar coordinates question requiring finding an intersection point, then computing area using the polar area formula. Part (a) involves solving 1+cos θ = k sec θ with given r-value (straightforward algebra). Part (b) requires integrating ½r² which involves standard techniques (∫cos²θ, ∫cos θ) but with careful bookkeeping across two regions. More routine than average FM questions but requires competent execution of multiple steps. |
| Spec | 4.09a Polar coordinates: convert to/from cartesian4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(1 + \cos\theta = 1 + \frac{\sqrt{3}}{2} \Rightarrow \cos\theta = \frac{\sqrt{3}}{2}\); \(1 + \frac{\sqrt{3}}{2} = k \times \frac{2}{\sqrt{3}} \Rightarrow k = \ldots\) | M1 | Correct method to find value of \(k\) |
| \(k = \frac{\sqrt{3}}{2} + \frac{3}{4}\) oe | A1 | Correct value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| At \(P\), \(\theta = \frac{\pi}{6}\) | B1 | Deduces correct value of \(\theta\) at \(P\); may be seen anywhere — in (a), on figure, or as limit on integral |
| \(\int(1+\cos\theta)^2\,d\theta = \int\left(1 + 2\cos\theta + \frac{1}{2} + \frac{1}{2}\cos 2\theta\right)d\theta\) | M1 | Attempts \(\frac{1}{2}\int r^2\,d\theta\), achieves at least three terms from squaring and applies \(\cos^2\theta = \pm\frac{1}{2} \pm \frac{1}{2}\cos 2\theta\); the \(\frac{1}{2}\) in area formula may be omitted for this mark |
| \(\int(1+\cos\theta)^2\,d\theta = \frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta\) (oe) | A1 | Correct integration of \(r^2\) |
| \(\frac{1}{2}\left[\frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta\right]_{\frac{\pi}{6}}^{\pi}\) \(= \frac{1}{2}\left[\left(\frac{3\pi}{2}+0+0\right) - \left(\frac{\pi}{4}+1+\frac{\sqrt{3}}{8}\right)\right]\) | M1 | Applies limits \(\frac{\pi}{6}\) and \(\pi\) to integral (evidence of subtraction required); alternatively use limits 0 to \(\pi\) and 0 to \(\frac{\pi}{6}\); not dependent, allow if incorrect double angle identity used or only two terms from squaring |
| Triangle: \(\frac{1}{2}\times\left(1+\frac{\sqrt{3}}{2}\right)\sin\frac{\pi}{6}\times\left(1+\frac{\sqrt{3}}{2}\right)\cos\frac{\pi}{6} = \frac{1}{2}\left(\frac{1}{2}+\frac{\sqrt{3}}{4}\right)\left(\frac{\sqrt{3}}{2}+\frac{3}{4}\right)\) \(\left(=\frac{7\sqrt{3}}{32}+\frac{3}{8}\right)\) | M1 | Uses correct strategy for exact area of triangle, via \(\frac{1}{2}ab\) or \(\frac{1}{2}ab\sin C\) or integration \(\frac{1}{2}\int_0^{\pi/6} k^2\sec^2\theta\,d\theta\) |
| Area of \(R = \frac{5\pi}{8} - \frac{\sqrt{3}}{16} - \frac{1}{2} + \frac{7\sqrt{3}}{32} + \frac{3}{8}\) | dM1 | Fully correct method for required area; depends on all previous method marks; \(\frac{1}{2}\) in area formula must have been included |
| \(\frac{5\pi}{8} + \frac{5\sqrt{3}}{32} - \frac{1}{8}\) | A1 | cao; terms may be in different order; accept if put over common denominator |
# Question 10:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $1 + \cos\theta = 1 + \frac{\sqrt{3}}{2} \Rightarrow \cos\theta = \frac{\sqrt{3}}{2}$; $1 + \frac{\sqrt{3}}{2} = k \times \frac{2}{\sqrt{3}} \Rightarrow k = \ldots$ | M1 | Correct method to find value of $k$ |
| $k = \frac{\sqrt{3}}{2} + \frac{3}{4}$ oe | A1 | Correct value |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| At $P$, $\theta = \frac{\pi}{6}$ | B1 | Deduces correct value of $\theta$ at $P$; may be seen anywhere — in (a), on figure, or as limit on integral |
| $\int(1+\cos\theta)^2\,d\theta = \int\left(1 + 2\cos\theta + \frac{1}{2} + \frac{1}{2}\cos 2\theta\right)d\theta$ | M1 | Attempts $\frac{1}{2}\int r^2\,d\theta$, achieves at least three terms from squaring and applies $\cos^2\theta = \pm\frac{1}{2} \pm \frac{1}{2}\cos 2\theta$; the $\frac{1}{2}$ in area formula may be omitted for this mark |
| $\int(1+\cos\theta)^2\,d\theta = \frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta$ (oe) | A1 | Correct integration of $r^2$ |
| $\frac{1}{2}\left[\frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta\right]_{\frac{\pi}{6}}^{\pi}$ $= \frac{1}{2}\left[\left(\frac{3\pi}{2}+0+0\right) - \left(\frac{\pi}{4}+1+\frac{\sqrt{3}}{8}\right)\right]$ | M1 | Applies limits $\frac{\pi}{6}$ and $\pi$ to integral (evidence of subtraction required); alternatively use limits 0 to $\pi$ **and** 0 to $\frac{\pi}{6}$; not dependent, allow if incorrect double angle identity used or only two terms from squaring |
| Triangle: $\frac{1}{2}\times\left(1+\frac{\sqrt{3}}{2}\right)\sin\frac{\pi}{6}\times\left(1+\frac{\sqrt{3}}{2}\right)\cos\frac{\pi}{6} = \frac{1}{2}\left(\frac{1}{2}+\frac{\sqrt{3}}{4}\right)\left(\frac{\sqrt{3}}{2}+\frac{3}{4}\right)$ $\left(=\frac{7\sqrt{3}}{32}+\frac{3}{8}\right)$ | M1 | Uses correct strategy for exact area of triangle, via $\frac{1}{2}ab$ or $\frac{1}{2}ab\sin C$ or integration $\frac{1}{2}\int_0^{\pi/6} k^2\sec^2\theta\,d\theta$ |
| Area of $R = \frac{5\pi}{8} - \frac{\sqrt{3}}{16} - \frac{1}{2} + \frac{7\sqrt{3}}{32} + \frac{3}{8}$ | dM1 | Fully correct method for required area; depends on all previous method marks; $\frac{1}{2}$ in area formula must have been included |
| $\frac{5\pi}{8} + \frac{5\sqrt{3}}{32} - \frac{1}{8}$ | A1 | cao; terms may be in different order; accept if put over common denominator |
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There is no mark scheme content to extract from this page.10.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{09582a82-cd57-4c2f-aefa-8412d4f4cb64-32_497_919_292_573}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of the curve $C$ with polar equation
$$r = 1 + \cos \theta \quad 0 \leqslant \theta \leqslant \pi$$
and the line $l$ with polar equation
$$r = k \sec \theta \quad 0 \leqslant \theta < \frac { \pi } { 2 }$$
where $k$ is a positive constant.\\
Given that
\begin{itemize}
\item $\quad C$ and $l$ intersect at the point $P$
\item $O P = 1 + \frac { \sqrt { 3 } } { 2 }$
\begin{enumerate}[label=(\alph*)]
\item determine the exact value of $k$.
\end{itemize}
The finite region $R$, shown shaded in Figure 1, is bounded by $C$, the initial line and $l$.
\item Use algebraic integration to show that the area of $R$ is
$$p \pi + q \sqrt { 3 } + r$$
where $p , q$ and $r$ are simplified rational numbers to be determined.
\end{enumerate}
\hfill \mbox{\textit{Edexcel F2 2024 Q10 [9]}}