| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2016 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Area of region with line boundary |
| Difficulty | Challenging +1.2 This is a multi-part Further Maths polar coordinates question requiring standard techniques: evaluating r at a point, using perpendicular geometry to find coordinates, calculating triangle area, and finding area between curve and line using integration. While it has multiple steps and requires careful algebraic manipulation (especially the trigonometric identity in part b), all techniques are standard for FP3 polar coordinates with no novel insights required. The 7-mark allocation and structured parts make it moderately above average difficulty but well within typical Further Maths scope. |
| 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 | Mark | Guidance |
| \((OA=)\; 1+\tan\frac{\pi}{3}\) | M1 | \(1+\tan\frac{\pi}{3}\) PI by \(1+\sqrt{3}\) as final answer |
| \(OA = 1+\sqrt{3}\) | A1 | Or \(1+\sqrt{3}\) as final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Polar eqn of \(AN\): \(r\cos\theta = OA\cos\frac{\pi}{3}\) | M1 | Or \(ON=OA\cos\frac{\pi}{3}\) OE |
| \(r\cos\theta = \frac{1+\sqrt{3}}{2}\) | A1F | Ft from (a). OE \(ON=\frac{1+\sqrt{3}}{2}\) |
| At \(B\): \((1+\tan\alpha)\cos\alpha = \frac{1+\sqrt{3}}{2}\) | m1 | \((1+\tan\alpha)\cos\alpha = \frac{1}{2}c\)'s \(OA\). OE |
| \((\cos\alpha+\sin\alpha)^2 = \left(\frac{1+\sqrt{3}}{2}\right)^2 = 1+\frac{\sqrt{3}}{2}\) | A1 | AG. Printed result convincingly obtained. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(1+\sin 2\alpha = 1+\frac{\sqrt{3}}{2} \Rightarrow \sin 2\alpha = \frac{\sqrt{3}}{2}\) | M1 | \(\sin 2\alpha = \frac{\sqrt{3}}{2}\) OE eg \(\sin\left(\alpha+\frac{\pi}{4}\right)=\frac{1+\sqrt{3}}{2\sqrt{2}}\) |
| Since \(0<\alpha(=\theta_B)<\theta_A\left(=\frac{\pi}{3}\right)\), so \(\alpha=\frac{\pi}{6}\) | A1 | Condone \(\theta_B=\frac{\pi}{6}\); NMS scores 0/2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Area of \(\triangle OAB = \frac{1}{2}OA(1+\tan\theta_B)\sin(\theta_A-\theta_B)\) | M1 | Valid method; correct expression in terms of known lengths/angles OE eg \(\frac{1}{2}\left(OA\cos\frac{\pi}{3}\right)^2\left(\tan\frac{\pi}{3}-\tan\alpha\right)\) |
| \(= \frac{1}{2}(1+\sqrt{3})\left(1+\frac{1}{\sqrt{3}}\right)\sin\frac{\pi}{6} = \frac{3+2\sqrt{3}}{6}\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Area \(= \frac{1}{2}\int_{(\theta_B)}^{(\frac{\pi}{3})}(1+\tan\theta)^2\,(d\theta)\) | M1 | Use of \(\frac{1}{2}\int r^2\,(d\theta)\) |
| \(= \frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\left(1 + 2\tan\theta + \tan^2\theta\right)(d\theta)\) | B1F | Correct expansion of \([1+\tan\theta]^2\) and limits \(\frac{\pi}{3}\) and c's \(\alpha\), in terms of \(\pi\) such that \(0 < \alpha < \frac{\pi}{3}\) used with \(k\int r^2\,(d\theta)\) |
| \(\int\left(\sec^2\theta + 2\tan\theta\right)(d\theta)\) | M1 | \(1 \pm \tan^2\theta = \pm\sec^2\theta\) with \(k\int r^2\,(d\theta)\) |
| \(= \left[\tan\theta + 2\ln\sec\theta\right]\) | A1 | Correct integration of \(\sec^2\theta + 2\tan\theta\) following use of correct identity |
| Shaded area \(= \frac{1}{2}\int_{\theta_B}^{\frac{\pi}{3}}(1+\tan\theta)^2\,(d\theta) - \dfrac{3+2\sqrt{3}}{6}\) | M1 | Condone difference taken in wrong order for this M mark |
| \(= \frac{1}{2}\left[\left(\sqrt{3}+2\ln 2\right)-\left(\frac{1}{\sqrt{3}}+2\ln\frac{2}{\sqrt{3}}\right)\right] - \dfrac{3+2\sqrt{3}}{6}\) | A1 | Any correct numerical expression for the shaded area, can be unsimplified but must be exact |
| \(= \dfrac{\sqrt{3}}{3}+\ln\sqrt{3}-\dfrac{1}{2}-\dfrac{\sqrt{3}}{3} = \ln\sqrt{3}-\dfrac{1}{2}\) | A1 | Simplified to the difference of two correct exact terms condoning one of them to be unsimplified; OE factorised form |
| Total | 7 | |
| TOTAL | 75 |
# Question 8:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $(OA=)\; 1+\tan\frac{\pi}{3}$ | M1 | $1+\tan\frac{\pi}{3}$ PI by $1+\sqrt{3}$ as final answer |
| $OA = 1+\sqrt{3}$ | A1 | Or $1+\sqrt{3}$ as final answer |
## Part (b)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Polar eqn of $AN$: $r\cos\theta = OA\cos\frac{\pi}{3}$ | M1 | Or $ON=OA\cos\frac{\pi}{3}$ OE |
| $r\cos\theta = \frac{1+\sqrt{3}}{2}$ | A1F | Ft from (a). OE $ON=\frac{1+\sqrt{3}}{2}$ |
| At $B$: $(1+\tan\alpha)\cos\alpha = \frac{1+\sqrt{3}}{2}$ | m1 | $(1+\tan\alpha)\cos\alpha = \frac{1}{2}c$'s $OA$. OE |
| $(\cos\alpha+\sin\alpha)^2 = \left(\frac{1+\sqrt{3}}{2}\right)^2 = 1+\frac{\sqrt{3}}{2}$ | A1 | AG. Printed result convincingly obtained. |
## Part (b)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $1+\sin 2\alpha = 1+\frac{\sqrt{3}}{2} \Rightarrow \sin 2\alpha = \frac{\sqrt{3}}{2}$ | M1 | $\sin 2\alpha = \frac{\sqrt{3}}{2}$ OE eg $\sin\left(\alpha+\frac{\pi}{4}\right)=\frac{1+\sqrt{3}}{2\sqrt{2}}$ |
| Since $0<\alpha(=\theta_B)<\theta_A\left(=\frac{\pi}{3}\right)$, so $\alpha=\frac{\pi}{6}$ | A1 | Condone $\theta_B=\frac{\pi}{6}$; NMS scores 0/2 |
## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Area of $\triangle OAB = \frac{1}{2}OA(1+\tan\theta_B)\sin(\theta_A-\theta_B)$ | M1 | Valid method; correct expression in terms of known lengths/angles OE eg $\frac{1}{2}\left(OA\cos\frac{\pi}{3}\right)^2\left(\tan\frac{\pi}{3}-\tan\alpha\right)$ |
| $= \frac{1}{2}(1+\sqrt{3})\left(1+\frac{1}{\sqrt{3}}\right)\sin\frac{\pi}{6} = \frac{3+2\sqrt{3}}{6}$ | A1 | AG |
## Question (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Area $= \frac{1}{2}\int_{(\theta_B)}^{(\frac{\pi}{3})}(1+\tan\theta)^2\,(d\theta)$ | M1 | Use of $\frac{1}{2}\int r^2\,(d\theta)$ |
| $= \frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\left(1 + 2\tan\theta + \tan^2\theta\right)(d\theta)$ | B1F | Correct expansion of $[1+\tan\theta]^2$ and limits $\frac{\pi}{3}$ and c's $\alpha$, in terms of $\pi$ such that $0 < \alpha < \frac{\pi}{3}$ used with $k\int r^2\,(d\theta)$ |
| $\int\left(\sec^2\theta + 2\tan\theta\right)(d\theta)$ | M1 | $1 \pm \tan^2\theta = \pm\sec^2\theta$ with $k\int r^2\,(d\theta)$ |
| $= \left[\tan\theta + 2\ln\sec\theta\right]$ | A1 | Correct integration of $\sec^2\theta + 2\tan\theta$ following use of correct identity |
| Shaded area $= \frac{1}{2}\int_{\theta_B}^{\frac{\pi}{3}}(1+\tan\theta)^2\,(d\theta) - \dfrac{3+2\sqrt{3}}{6}$ | M1 | Condone difference taken in wrong order for this M mark |
| $= \frac{1}{2}\left[\left(\sqrt{3}+2\ln 2\right)-\left(\frac{1}{\sqrt{3}}+2\ln\frac{2}{\sqrt{3}}\right)\right] - \dfrac{3+2\sqrt{3}}{6}$ | A1 | Any correct numerical expression for the shaded area, can be unsimplified but must be exact |
| $= \dfrac{\sqrt{3}}{3}+\ln\sqrt{3}-\dfrac{1}{2}-\dfrac{\sqrt{3}}{3} = \ln\sqrt{3}-\dfrac{1}{2}$ | A1 | Simplified to the difference of two correct exact terms condoning one of them to be unsimplified; OE factorised form |
| **Total** | **7** | |
| **TOTAL** | **75** | |8 The diagram shows the sketch of part of a curve, the pole $O$ and the initial line.\\
\includegraphics[max width=\textwidth, alt={}, center]{0b9b947d-824b-4d3a-b66d-4bfd8d49be17-20_609_670_358_703}
The polar equation of the curve is $r = 1 + \tan \theta$.\\
The point $A$ is the point on the curve at which $\theta = \frac { \pi } { 3 }$.\\
The perpendicular, $A N$, from $A$ to the initial line intersects the curve at the point $B$.
\begin{enumerate}[label=(\alph*)]
\item Find the exact length of $O A$.
\item \begin{enumerate}[label=(\roman*)]
\item Given that, at the point $B , \theta = \alpha$, show that $( \cos \alpha + \sin \alpha ) ^ { 2 } = 1 + \frac { \sqrt { 3 } } { 2 }$.
\item Hence, or otherwise, find $\alpha$ in terms of $\pi$.
\end{enumerate}\item Show that the area of triangle $O A B$ is $\frac { 3 + 2 \sqrt { 3 } } { 6 }$.
\item Find, in an exact simplified form, the area of the shaded region bounded by the curve and the line segment $A B$.\\[0pt]
[7 marks]
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{0b9b947d-824b-4d3a-b66d-4bfd8d49be17-23_2488_1709_219_153}
\end{center}
\section*{DO NOT WRITE ON THIS PAGE ANSWER IN THE SPACES PROVIDED}
\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2016 Q8 [17]}}