| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2017 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Region bounded by curve and tangent lines |
| Difficulty | Hard +2.3 This is a challenging Further Maths polar coordinates question requiring: (a) finding tangent points by solving dr/dθ = r tan φ where φ is the tangent angle, (b) computing polar area integrals for a four-petaled rose curve, and (c) combining rectangular and polar area calculations. The multi-step nature, need for tangent condition in polar form, and integration of cos²(2θ) across multiple intervals places this well above average difficulty, even for Further Maths students. |
| Spec | 4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = r\sin\theta = 4\cos 2\theta \sin\theta\) | M1 | Attempts to use \(r\sin\theta\) |
| \(\frac{dy}{d\theta} = 4\cos 2\theta\cos\theta - 8\sin 2\theta\sin\theta\) | B1 | Correct expression for \(\frac{dy}{d\theta}\) or any multiple |
| \(\frac{dy}{d\theta} = 0 \Rightarrow \theta = ...\) | M1 | Set \(\frac{dy}{d\theta} = 0\) and attempt to solve for \(\theta\) |
| \(r = \frac{8}{3}\), \(\theta = 0.421\), \(\theta = 2.72\) | A1 | Any one of: \(r = \frac{8}{3}\) (or awrt 2.7) or \(\theta = 0.421...\) or \(\theta = 2.72...\) |
| \(r = \frac{8}{3}\), \(\theta = 0.421, 2.72\) | A1 | Correct \(r\) and both angles; coordinates need not be paired; awrt 0.421, 2.72; awrt 2.7 for \(\frac{8}{3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(A = ...\int(4\cos 2\theta)^2 \, d\theta\) | M1 | Indication that integration of \((4\cos 2\theta)^2\) is required |
| \(\cos^2 2\theta = \frac{1}{2}(1+\cos 4\theta)\) | A1 | Correct identity seen or implied |
| \(A = ...[\alpha\theta + \beta\sin 4\theta]\) | dM1 | Integrates to form \(\alpha\theta + \beta\sin 4\theta\); dependent on first method mark |
| \(= 16\left[\theta + \frac{1}{4}\sin 4\theta\right]_0^{\frac{\pi}{4}}\) | ddM1 | Fully correct method; if evaluated correctly gives \(4\pi\); dependent on all previous method marks |
| \(= 4\pi\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(PQ = 2r\sin\theta = \frac{16}{3\sqrt{6}}\) | B1 | Correct expression or value for \(PQ\) or \(PQ/2\) |
| \(SP = 8\) or \(\frac{SP}{2} = 4\) | B1 | Correct value for \(SP\) or \(SP/2\) |
| Area \(PQRS = \frac{16}{3\sqrt{6}} \times 8 \left(= \frac{64\sqrt{6}}{9}\right)\) | M1 | Their \(PQ \times SP\); must be complete rectangle |
| Required area \(= \frac{128}{3\sqrt{6}} - 4\pi\) | M1A1 | M1: rectangle area minus part (b); A1: correct exact answer e.g. \(\frac{64\sqrt{6}}{9} - 4\pi\); allow awrt 4.8 or 4.9 |
## Question 7:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = r\sin\theta = 4\cos 2\theta \sin\theta$ | M1 | Attempts to use $r\sin\theta$ |
| $\frac{dy}{d\theta} = 4\cos 2\theta\cos\theta - 8\sin 2\theta\sin\theta$ | B1 | Correct expression for $\frac{dy}{d\theta}$ or any multiple |
| $\frac{dy}{d\theta} = 0 \Rightarrow \theta = ...$ | M1 | Set $\frac{dy}{d\theta} = 0$ and attempt to solve for $\theta$ |
| $r = \frac{8}{3}$, $\theta = 0.421$, $\theta = 2.72$ | A1 | Any one of: $r = \frac{8}{3}$ (or awrt 2.7) or $\theta = 0.421...$ or $\theta = 2.72...$ |
| $r = \frac{8}{3}$, $\theta = 0.421, 2.72$ | A1 | Correct $r$ and both angles; coordinates need not be paired; awrt 0.421, 2.72; awrt 2.7 for $\frac{8}{3}$ |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $A = ...\int(4\cos 2\theta)^2 \, d\theta$ | M1 | Indication that integration of $(4\cos 2\theta)^2$ is required |
| $\cos^2 2\theta = \frac{1}{2}(1+\cos 4\theta)$ | A1 | Correct identity seen or implied |
| $A = ...[\alpha\theta + \beta\sin 4\theta]$ | dM1 | Integrates to form $\alpha\theta + \beta\sin 4\theta$; dependent on first method mark |
| $= 16\left[\theta + \frac{1}{4}\sin 4\theta\right]_0^{\frac{\pi}{4}}$ | ddM1 | Fully correct method; if evaluated correctly gives $4\pi$; dependent on all previous method marks |
| $= 4\pi$ | A1 | cao |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $PQ = 2r\sin\theta = \frac{16}{3\sqrt{6}}$ | B1 | Correct expression or value for $PQ$ or $PQ/2$ |
| $SP = 8$ or $\frac{SP}{2} = 4$ | B1 | Correct value for $SP$ or $SP/2$ |
| Area $PQRS = \frac{16}{3\sqrt{6}} \times 8 \left(= \frac{64\sqrt{6}}{9}\right)$ | M1 | Their $PQ \times SP$; must be complete rectangle |
| Required area $= \frac{128}{3\sqrt{6}} - 4\pi$ | M1A1 | M1: rectangle area minus part (b); A1: correct exact answer e.g. $\frac{64\sqrt{6}}{9} - 4\pi$; allow awrt 4.8 or 4.9 |7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{2026c49f-243b-497a-b702-e40d012ad308-20_465_1070_255_507}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of the curve $C$ with polar equation
$$r = 4 \cos 2 \theta , \quad - \frac { \pi } { 4 } \leqslant \theta \leqslant \frac { \pi } { 4 } \text { and } \frac { 3 \pi } { 4 } \leqslant \theta \leqslant \frac { 5 \pi } { 4 }$$
The lines $P Q , Q R , R S$ and $S P$ are tangents to $C$, where $Q R$ and $S P$ are parallel to the initial line and $P Q$ and $R S$ are perpendicular to the initial line.
\begin{enumerate}[label=(\alph*)]
\item Find the polar coordinates of the points where the tangent SP touches the curve. Give the values of $\theta$ to 3 significant figures.
\item Find the exact area of the finite region bounded by the curve $C$, shown unshaded in Figure 1.
\item Find the area enclosed by the rectangle $P Q R S$ but outside the curve $C$, shown shaded in Figure 1.
\end{enumerate}
\hfill \mbox{\textit{Edexcel F2 2017 Q7 [15]}}