| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2022 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Region bounded by curve and tangent lines |
| Difficulty | Challenging +1.8 This is a challenging Further Maths polar coordinates question requiring: (a) finding where tangent is horizontal using dy/dθ = 0 (involving product rule and trig identities), and (b) computing area using polar integration with non-trivial limits and algebraic manipulation. The multi-step nature, integration of (1+cos θ)² terms, and exact form answer requirement make this substantially harder than typical A-level questions, though it follows standard Further Pure techniques. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{d(r\sin\theta)}{d\theta} = 4a\cos\theta + 4a\cos^2\theta - 4a\sin^2\theta\) or \(4a\cos\theta + 4a\cos 2\theta\) | M1 | Attempt differentiation of \(r\sin\theta\) using product rule or \(\sin 2\theta = 2\sin\theta\cos\theta\); allow errors in coefficients as long as form is correct |
| \(4a\cos\theta + 4a\cos^2\theta - 4a\sin^2\theta = 0 \Rightarrow \cos\theta + \cos^2\theta - (1-\cos^2\theta) = 0\) | M1 | Sets derivative of \(r\sin\theta\) equal to zero and achieves a quadratic in \(\cos\theta\) |
| \(2\cos^2\theta + \cos\theta - 1 = 0\) (terms in any order) | A1 | Correct 3-term quadratic in \(\cos\theta\) (any multiple, including \(a\)) |
| \((2\cos\theta - 1)(\cos\theta + 1) = 0 \Rightarrow \cos\theta = \ldots\) | ddM1 | Dep on both M marks; solve quadratic (usual rules) |
| \(\cos\theta = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{3}\) (\(\theta = \pi\) need not be seen) | A1 | Correct quadratic solved to give \(\theta = \frac{\pi}{3}\) |
| \(r = 4a \times \frac{3}{2} = 6a\) | A1 (6) | Correct \(r\) from intermediate step; accept \(r = 4a\left(1+\cos\frac{\pi}{3}\right) = 6a\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Area \(= \frac{1}{2}\int r^2 d\theta = \frac{1}{2}\int_{\pi/6}^{\pi/3} 16a^2(1+\cos\theta)^2\, d\theta\) | — | — |
| \(= \frac{16a^2}{2}\int_{\pi/6}^{\pi/3}(1 + 2\cos\theta + \cos^2\theta)\,d\theta\) | M1 | Use of correct area formula \(\frac{1}{2}\); squaring bracket to obtain 3 terms; limits need not be shown |
| \(= 8a^2\int_{\pi/6}^{\pi/3}\left(1 + 2\cos\theta + \frac{1}{2}(\cos 2\theta + 1)\right)d\theta\) | M1 | Use double angle formula of form \(\cos^2\theta = \pm\frac{1}{2}(\cos 2\theta \pm 1)\) to obtain integrable function |
| \(= 8a^2\left[\theta + 2\sin\theta + \frac{1}{2}\left(\frac{1}{2}\sin 2\theta + \theta\right)\right]_{\pi/6}^{\pi/3}\) | dM1A1 | Attempt integration \(\cos\theta \to \pm k\sin\theta\) and \(\cos 2\theta \to \pm m\sin 2\theta\); correct integration |
| \(8a^2\left[\frac{\pi}{3}+\sqrt{3}+\frac{1}{4}\times\frac{\sqrt{3}}{2}+\frac{\pi}{6} - \left(\frac{\pi}{6}+1+\frac{1}{4}\times\frac{\sqrt{3}}{2}+\frac{\pi}{12}\right)\right]\) | A1 | Include the \(\frac{1}{2}\) and substitute correct limits in a correct integral |
| \(8a^2\left[\frac{\pi}{4}+\sqrt{3}-1\right]\) | — | — |
| Area \(R = 8a^2\left[\frac{\pi}{4}+\sqrt{3}-1\right] - 6a^2\left(1+\frac{\sqrt{3}}{2}\right) = a^2\left(2\pi + 5\sqrt{3} - 14\right)\) | M1A1 (7) | M1: attempt area of triangle (accept valid attempt even if not subtracted); A1: correct final answer in required form |
# Question 6:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{d(r\sin\theta)}{d\theta} = 4a\cos\theta + 4a\cos^2\theta - 4a\sin^2\theta$ or $4a\cos\theta + 4a\cos 2\theta$ | M1 | Attempt differentiation of $r\sin\theta$ using product rule or $\sin 2\theta = 2\sin\theta\cos\theta$; allow errors in coefficients as long as form is correct |
| $4a\cos\theta + 4a\cos^2\theta - 4a\sin^2\theta = 0 \Rightarrow \cos\theta + \cos^2\theta - (1-\cos^2\theta) = 0$ | M1 | Sets derivative of $r\sin\theta$ equal to zero and achieves a quadratic in $\cos\theta$ |
| $2\cos^2\theta + \cos\theta - 1 = 0$ (terms in any order) | A1 | Correct 3-term quadratic in $\cos\theta$ (any multiple, including $a$) |
| $(2\cos\theta - 1)(\cos\theta + 1) = 0 \Rightarrow \cos\theta = \ldots$ | ddM1 | Dep on both M marks; solve quadratic (usual rules) |
| $\cos\theta = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{3}$ ($\theta = \pi$ need not be seen) | A1 | Correct quadratic solved to give $\theta = \frac{\pi}{3}$ |
| $r = 4a \times \frac{3}{2} = 6a$ | A1 (6) | Correct $r$ from intermediate step; accept $r = 4a\left(1+\cos\frac{\pi}{3}\right) = 6a$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Area $= \frac{1}{2}\int r^2 d\theta = \frac{1}{2}\int_{\pi/6}^{\pi/3} 16a^2(1+\cos\theta)^2\, d\theta$ | — | — |
| $= \frac{16a^2}{2}\int_{\pi/6}^{\pi/3}(1 + 2\cos\theta + \cos^2\theta)\,d\theta$ | M1 | Use of correct area formula $\frac{1}{2}$; squaring bracket to obtain 3 terms; limits need not be shown |
| $= 8a^2\int_{\pi/6}^{\pi/3}\left(1 + 2\cos\theta + \frac{1}{2}(\cos 2\theta + 1)\right)d\theta$ | M1 | Use double angle formula of form $\cos^2\theta = \pm\frac{1}{2}(\cos 2\theta \pm 1)$ to obtain integrable function |
| $= 8a^2\left[\theta + 2\sin\theta + \frac{1}{2}\left(\frac{1}{2}\sin 2\theta + \theta\right)\right]_{\pi/6}^{\pi/3}$ | dM1A1 | Attempt integration $\cos\theta \to \pm k\sin\theta$ and $\cos 2\theta \to \pm m\sin 2\theta$; correct integration |
| $8a^2\left[\frac{\pi}{3}+\sqrt{3}+\frac{1}{4}\times\frac{\sqrt{3}}{2}+\frac{\pi}{6} - \left(\frac{\pi}{6}+1+\frac{1}{4}\times\frac{\sqrt{3}}{2}+\frac{\pi}{12}\right)\right]$ | A1 | Include the $\frac{1}{2}$ and substitute correct limits in a correct integral |
| $8a^2\left[\frac{\pi}{4}+\sqrt{3}-1\right]$ | — | — |
| Area $R = 8a^2\left[\frac{\pi}{4}+\sqrt{3}-1\right] - 6a^2\left(1+\frac{\sqrt{3}}{2}\right) = a^2\left(2\pi + 5\sqrt{3} - 14\right)$ | M1A1 (7) | M1: attempt area of triangle (accept valid attempt even if not subtracted); A1: correct final answer in required form |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ff9ff379-78d8-41c0-a177-ec346e359249-20_497_1196_260_520}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The curve shown in Figure 1 has polar equation
$$r = 4 a ( 1 + \cos \theta ) \quad 0 \leqslant \theta < \pi$$
where $a$ is a positive constant.\\
The tangent to the curve at the point $A$ is parallel to the initial line.
\begin{enumerate}[label=(\alph*)]
\item Show that the polar coordinates of $A$ are $\left( 6 a , \frac { \pi } { 3 } \right)$
The point $B$ lies on the curve such that angle $A O B = \frac { \pi } { 6 }$\\
The finite region $R$, shown shaded in Figure 1, is bounded by the line $A B$ and the curve.
\item Use calculus to determine the area of the shaded region $R$, giving your answer in the form $a ^ { 2 } ( n \pi + p \sqrt { 3 } + q )$, where $n , p$ and $q$ are integers.
\end{enumerate}
\hfill \mbox{\textit{Edexcel F2 2022 Q6 [13]}}