| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 1 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Polar curve with substitution integral |
| Difficulty | Challenging +1.8 This is a multi-part Further Maths polar coordinates question requiring area integration with trigonometric substitution, finding curve intersections, and geometric reasoning about angles in triangles. While it involves several techniques (polar area formula, solving equations, angle calculations), each step follows standard FP3 methods without requiring exceptional insight. The integration and algebraic manipulation are moderately challenging but routine for Further Maths students. |
| Spec | 4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(r = (1 - \tan^2\theta)\sec\theta = \cos^{-1}\theta - \sin^2\theta\cos^{-1}\theta\) | ||
| Area \(= \frac{1}{2}\int_{-\pi/4}^{\pi/4} r^2 \, d\theta = \frac{1}{2}\int_{-\pi/4}^{\pi/4}(1-\tan^2\theta)^2\sec^2\theta \, d\theta\) | M1 | Correct area formula with limits |
| \(= \int_0^{\pi/4}(1-\tan^2\theta)^2\sec^2\theta \, d\theta\) | M1 | Using symmetry |
| Let \(u = \tan\theta\), \(du = \sec^2\theta \, d\theta\), limits \(0\) to \(1\) | M1 | Substitution \(u = \tan\theta\) |
| \(= \int_0^1 (1-u^2)^2 \, du = \int_0^1 (1 - 2u^2 + u^4) \, du\) | A1 | Correct integral |
| \(= \left[u - \frac{2u^3}{3} + \frac{u^5}{5}\right]_0^1 = 1 - \frac{2}{3} + \frac{1}{5} = \frac{8}{15}\) | A1 | Correct completion to \(\frac{8}{15}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Set \((1-\tan^2\theta)\sec\theta = \frac{1}{2}\sec^3\theta\) | M1 | Equating the two expressions |
| \((1-\tan^2\theta) = \frac{1}{2}\sec^2\theta = \frac{1}{2}(1+\tan^2\theta)\) | M1 | Rearranging |
| \(1 - \tan^2\theta = \frac{1}{2} + \frac{1}{2}\tan^2\theta \Rightarrow \frac{3}{2}\tan^2\theta = \frac{1}{2} \Rightarrow \tan^2\theta = \frac{1}{3}\) | ||
| \(\theta = \pm\frac{\pi}{6}\), \(r = \frac{1}{2}\sec^3(\pi/6) = \frac{1}{2}\cdot\left(\frac{2}{\sqrt{3}}\right)^3 = \frac{4}{3\sqrt{3}}\) | A1 | Correct \(r\) value |
| \(A = \left(\frac{4}{3\sqrt{3}}, \frac{\pi}{6}\right)\), \(B = \left(\frac{4}{3\sqrt{3}}, -\frac{\pi}{6}\right)\) | A1 | Both coordinates correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Point \(P\) is on initial line: \(r = \sec^3(0) \cdot \frac{1}{2} = \frac{1}{2}\), so \(P = \left(\frac{1}{2}, 0\right)\) or check \(C\): \(r=(1-0)(1)=1\), so \(P=(1,0)\) | M1 | Finding \(P\) |
| \(\tan\alpha = \frac{r\sin\theta}{r\cos\theta - r_P}\) using appropriate coordinate geometry | M1 | Method for finding \(\tan\alpha\) |
| Cartesian coords of \(A\): \(x = r\cos\theta = \frac{4}{3\sqrt{3}}\cdot\frac{\sqrt{3}}{2} = \frac{2}{3}\), \(y = r\sin\theta = \frac{4}{3\sqrt{3}}\cdot\frac{1}{2} = \frac{2}{3\sqrt{3}}\) | A1 | Correct coordinates |
| \(\tan\angle OAP\): gradient \(OP\) along x-axis, use \(\tan\alpha = \frac{y}{x_P - x_A}\)... leading to \(\tan\alpha = k\sqrt{3}\) with \(k\) integer | M1 A1 | Correct derivation showing integer \(k\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(k = -1\) (or appropriate integer value from (b)(ii)), \(\tan\alpha = -\sqrt{3}\)... | ||
| Point \(A\) lies outside (or inside) the circle with diameter \(OP\) because angle \(OAP\) is acute/obtuse (not \(90°\)) | B1 | Correct conclusion with valid reason referencing angle \(OAP \neq 90°\) |
## Question 8:
### Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $r = (1 - \tan^2\theta)\sec\theta = \cos^{-1}\theta - \sin^2\theta\cos^{-1}\theta$ | | |
| Area $= \frac{1}{2}\int_{-\pi/4}^{\pi/4} r^2 \, d\theta = \frac{1}{2}\int_{-\pi/4}^{\pi/4}(1-\tan^2\theta)^2\sec^2\theta \, d\theta$ | M1 | Correct area formula with limits |
| $= \int_0^{\pi/4}(1-\tan^2\theta)^2\sec^2\theta \, d\theta$ | M1 | Using symmetry |
| Let $u = \tan\theta$, $du = \sec^2\theta \, d\theta$, limits $0$ to $1$ | M1 | Substitution $u = \tan\theta$ |
| $= \int_0^1 (1-u^2)^2 \, du = \int_0^1 (1 - 2u^2 + u^4) \, du$ | A1 | Correct integral |
| $= \left[u - \frac{2u^3}{3} + \frac{u^5}{5}\right]_0^1 = 1 - \frac{2}{3} + \frac{1}{5} = \frac{8}{15}$ | A1 | Correct completion to $\frac{8}{15}$ |
### Part (b)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Set $(1-\tan^2\theta)\sec\theta = \frac{1}{2}\sec^3\theta$ | M1 | Equating the two expressions |
| $(1-\tan^2\theta) = \frac{1}{2}\sec^2\theta = \frac{1}{2}(1+\tan^2\theta)$ | M1 | Rearranging |
| $1 - \tan^2\theta = \frac{1}{2} + \frac{1}{2}\tan^2\theta \Rightarrow \frac{3}{2}\tan^2\theta = \frac{1}{2} \Rightarrow \tan^2\theta = \frac{1}{3}$ | | |
| $\theta = \pm\frac{\pi}{6}$, $r = \frac{1}{2}\sec^3(\pi/6) = \frac{1}{2}\cdot\left(\frac{2}{\sqrt{3}}\right)^3 = \frac{4}{3\sqrt{3}}$ | A1 | Correct $r$ value |
| $A = \left(\frac{4}{3\sqrt{3}}, \frac{\pi}{6}\right)$, $B = \left(\frac{4}{3\sqrt{3}}, -\frac{\pi}{6}\right)$ | A1 | Both coordinates correct |
### Part (b)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Point $P$ is on initial line: $r = \sec^3(0) \cdot \frac{1}{2} = \frac{1}{2}$, so $P = \left(\frac{1}{2}, 0\right)$ or check $C$: $r=(1-0)(1)=1$, so $P=(1,0)$ | M1 | Finding $P$ |
| $\tan\alpha = \frac{r\sin\theta}{r\cos\theta - r_P}$ using appropriate coordinate geometry | M1 | Method for finding $\tan\alpha$ |
| Cartesian coords of $A$: $x = r\cos\theta = \frac{4}{3\sqrt{3}}\cdot\frac{\sqrt{3}}{2} = \frac{2}{3}$, $y = r\sin\theta = \frac{4}{3\sqrt{3}}\cdot\frac{1}{2} = \frac{2}{3\sqrt{3}}$ | A1 | Correct coordinates |
| $\tan\angle OAP$: gradient $OP$ along x-axis, use $\tan\alpha = \frac{y}{x_P - x_A}$... leading to $\tan\alpha = k\sqrt{3}$ with $k$ integer | M1 A1 | Correct derivation showing integer $k$ |
### Part (b)(iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $k = -1$ (or appropriate integer value from (b)(ii)), $\tan\alpha = -\sqrt{3}$... | | |
| Point $A$ lies **outside** (or inside) the circle with diameter $OP$ because angle $OAP$ is acute/obtuse (not $90°$) | B1 | Correct conclusion with valid reason referencing angle $OAP \neq 90°$ |
These images show blank answer spaces (pages 22-24) from an AQA exam paper (P71700/Jun14/MFP3) — they contain no mark scheme content, only ruled lines for student responses and the "END OF QUESTIONS" notice on the final page.
To extract mark scheme content, you would need to provide images of the **actual mark scheme document**, not the question paper's answer pages.8 The diagram shows a sketch of a curve $C$, the pole $O$ and the initial line. The curve $C$ intersects the initial line at the point $P$.\\
\includegraphics[max width=\textwidth, alt={}, center]{0eb3e96e-528c-4a99-b164-31cc865f0d68-20_432_949_402_525}
The polar equation of $C$ is $r = \left( 1 - \tan ^ { 2 } \theta \right) \sec \theta , - \frac { \pi } { 4 } \leqslant \theta \leqslant \frac { \pi } { 4 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that the area of the region bounded by the curve $C$ is $\frac { 8 } { 15 }$.
\item The curve whose polar equation is
$$r = \frac { 1 } { 2 } \sec ^ { 3 } \theta , \quad - \frac { \pi } { 4 } \leqslant \theta \leqslant \frac { \pi } { 4 }$$
intersects $C$ at the points $A$ and $B$.
\begin{enumerate}[label=(\roman*)]
\item Find the polar coordinates of $A$ and $B$.
\item Given that angle $O A P =$ angle $O B P = \alpha$, show that $\tan \alpha = k \sqrt { 3 }$, where $k$ is an integer.
\item Using your value of $k$ from part (b)(ii), state whether the point $A$ lies inside or lies outside the circle whose diameter is $O P$. Give a reason for your answer.\\[0pt]
[1 mark]
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{0eb3e96e-528c-4a99-b164-31cc865f0d68-21_2484_1707_221_153}
\end{center}
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{0eb3e96e-528c-4a99-b164-31cc865f0d68-23_2484_1707_221_153}
\end{center}
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2014 Q8 [1]}}