| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2021 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Tangent parallel/perpendicular to initial line |
| Difficulty | Challenging +1.2 This is a standard Further Maths polar coordinates question requiring the tangent perpendicularity condition (dy/dx = 0), solving a trigonometric equation, and computing a polar area integral. While it involves multiple steps and Further Maths content, the techniques are routine applications of the polar tangent formula and area formula with straightforward integration—no novel insight required beyond textbook methods. |
| 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 |
| \(x = r\cos\theta = 3\sin 2\theta\cos\theta\) | B1 | May be given by implication |
| \(\frac{dx}{d\theta} = 6\cos 2\theta\cos\theta - 3\sin 2\theta\sin\theta = 0\) | M1 | Attempt differentiation of \(x = r\cos\theta\) or \(x = r\sin\theta\); product rule must be used |
| \(2\cos\theta(\cos^2\theta - 2\sin^2\theta) = 0\) | M1 | Use correct double angle formula and equate derivative to 0 |
| ALT: \(x = 6\sin\theta\cos^2\theta \Rightarrow \frac{dx}{d\theta} = 6\cos^3\theta - 12\sin^2\theta\cos\theta = 0\) | M1 | Use correct double angle formula and equate derivative of \(r\cos\theta\) to 0 |
| \(\tan\phi = \frac{1}{\sqrt{2}}\) | A1* | (4) Complete to given answer; no extras; all values exact; accept \(\theta\) or \(\phi\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\tan\phi = \frac{1}{\sqrt{2}} \Rightarrow \sin\phi = \frac{1}{\sqrt{3}},\ \cos\phi = \frac{\sqrt{2}}{\sqrt{3}}\) | M1 | Attempt exact values for \(\sin\theta\) and \(\cos\theta\); values may have been seen in (a) |
| \(R = 3 \times 2 \times \frac{1}{\sqrt{3}} \times \frac{\sqrt{2}}{\sqrt{3}} = 2\sqrt{2}\) | A1 | (2) Correct exact value for \(R\); award M1A1 for correct exact answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Area of sector \(= \frac{1}{2}\int r^2\, d\theta = \frac{9}{2}\int \sin^2 2\theta\, d\theta\) | M1 | Use of Area \(= \frac{1}{2}\int r^2\, d\theta\); limits not needed |
| \(= \frac{9}{2}\int_0^{\arctan\!\left(\frac{1}{\sqrt{2}}\right)} \frac{1}{2}(1 - \cos 4\theta)\, d\theta\) | M1 | Use double angle formula to obtain \(k\int\frac{1}{2}(1\pm\cos 4\theta)\,d\theta\); ignore limits |
| \(= \frac{9}{2}\left[\frac{1}{2}\left(\theta - \frac{1}{4}\sin 4\theta\right)\right]_0^{\arctan\frac{1}{\sqrt{2}}}\) | M1A1 | Attempt integration \(\cos 4\theta \to \pm\frac{1}{4}\sin 4\theta\); correct integration of \(1-\cos 4\theta\) |
| \(= \frac{9}{4}\left[\arctan\frac{1}{\sqrt{2}} - \frac{1}{4}\sin 4\!\left(\arctan\frac{1}{\sqrt{2}}\right) - 0\right]\) | dM1 | Correct use of correct limits; depends on 2nd and 3rd M marks; 0 at lower limit need not be shown |
| \(\sin 4\phi = 2\sin 2\phi\cos 2\phi = 4\sin\phi\cos\phi(2\cos^2\phi - 1)\) | M1 | Attempt exact numerical value for \(\sin 4\!\left(\arctan\frac{1}{\sqrt{2}}\right)\) |
| \(= 4 \times \frac{1}{\sqrt{3}} \times \frac{\sqrt{2}}{\sqrt{3}}\!\left(2\times\frac{2}{3}-1\right) = \frac{4\sqrt{2}}{9}\) | ||
| Area \(= \frac{9}{4}\!\left(\arctan\frac{1}{\sqrt{2}} - \frac{1}{4}\times\frac{4\sqrt{2}}{9}\right) = \frac{9}{4}\arctan\frac{1}{\sqrt{2}} - \frac{\sqrt{2}}{4}\) | A1 | (7) Correct final answer |
# Question 7:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = r\cos\theta = 3\sin 2\theta\cos\theta$ | B1 | May be given by implication |
| $\frac{dx}{d\theta} = 6\cos 2\theta\cos\theta - 3\sin 2\theta\sin\theta = 0$ | M1 | Attempt differentiation of $x = r\cos\theta$ or $x = r\sin\theta$; product rule must be used |
| $2\cos\theta(\cos^2\theta - 2\sin^2\theta) = 0$ | M1 | Use correct double angle formula **and** equate derivative to 0 |
| **ALT:** $x = 6\sin\theta\cos^2\theta \Rightarrow \frac{dx}{d\theta} = 6\cos^3\theta - 12\sin^2\theta\cos\theta = 0$ | M1 | Use correct double angle formula **and** equate derivative of $r\cos\theta$ to 0 |
| $\tan\phi = \frac{1}{\sqrt{2}}$ | A1* | (4) Complete to given answer; no extras; all values exact; accept $\theta$ or $\phi$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\tan\phi = \frac{1}{\sqrt{2}} \Rightarrow \sin\phi = \frac{1}{\sqrt{3}},\ \cos\phi = \frac{\sqrt{2}}{\sqrt{3}}$ | M1 | Attempt exact values for $\sin\theta$ and $\cos\theta$; values may have been seen in (a) |
| $R = 3 \times 2 \times \frac{1}{\sqrt{3}} \times \frac{\sqrt{2}}{\sqrt{3}} = 2\sqrt{2}$ | A1 | (2) Correct exact value for $R$; award M1A1 for correct exact answer |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Area of sector $= \frac{1}{2}\int r^2\, d\theta = \frac{9}{2}\int \sin^2 2\theta\, d\theta$ | M1 | Use of Area $= \frac{1}{2}\int r^2\, d\theta$; limits not needed |
| $= \frac{9}{2}\int_0^{\arctan\!\left(\frac{1}{\sqrt{2}}\right)} \frac{1}{2}(1 - \cos 4\theta)\, d\theta$ | M1 | Use double angle formula to obtain $k\int\frac{1}{2}(1\pm\cos 4\theta)\,d\theta$; ignore limits |
| $= \frac{9}{2}\left[\frac{1}{2}\left(\theta - \frac{1}{4}\sin 4\theta\right)\right]_0^{\arctan\frac{1}{\sqrt{2}}}$ | M1A1 | Attempt integration $\cos 4\theta \to \pm\frac{1}{4}\sin 4\theta$; correct integration of $1-\cos 4\theta$ |
| $= \frac{9}{4}\left[\arctan\frac{1}{\sqrt{2}} - \frac{1}{4}\sin 4\!\left(\arctan\frac{1}{\sqrt{2}}\right) - 0\right]$ | dM1 | Correct use of correct limits; depends on 2nd and 3rd M marks; 0 at lower limit need not be shown |
| $\sin 4\phi = 2\sin 2\phi\cos 2\phi = 4\sin\phi\cos\phi(2\cos^2\phi - 1)$ | M1 | Attempt exact numerical value for $\sin 4\!\left(\arctan\frac{1}{\sqrt{2}}\right)$ |
| $= 4 \times \frac{1}{\sqrt{3}} \times \frac{\sqrt{2}}{\sqrt{3}}\!\left(2\times\frac{2}{3}-1\right) = \frac{4\sqrt{2}}{9}$ | | |
| Area $= \frac{9}{4}\!\left(\arctan\frac{1}{\sqrt{2}} - \frac{1}{4}\times\frac{4\sqrt{2}}{9}\right) = \frac{9}{4}\arctan\frac{1}{\sqrt{2}} - \frac{\sqrt{2}}{4}$ | A1 | (7) Correct final answer |
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4d3e1c8e-c659-4cfe-82ac-5bfce0f58ba3-24_445_597_248_676}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of curve $C$ with polar equation
$$r = 3 \sin 2 \theta \quad 0 \leqslant \theta \leqslant \frac { \pi } { 2 }$$
The point $P$ on $C$ has polar coordinates $( R , \phi )$. The tangent to $C$ at $P$ is perpendicular to the initial line.
\begin{enumerate}[label=(\alph*)]
\item Show that $\tan \phi = \frac { 1 } { \sqrt { 2 } }$
\item Determine the exact value of $R$.
The region $S$, shown shaded in Figure 1, is bounded by $C$ and the line $O P$, where $O$ is the pole.
\item Use calculus to show that the exact area of $S$ is
$$p \arctan \frac { 1 } { \sqrt { 2 } } + q \sqrt { 2 }$$
where $p$ and $q$ are constants to be determined.
Solutions relying entirely on calculator technology are not acceptable.
\end{enumerate}
\hfill \mbox{\textit{Edexcel F2 2021 Q7 [13]}}