| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2017 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Find constant from given area |
| Difficulty | Standard +0.8 This is a Further Maths FP2 polar coordinates question requiring knowledge of the area formula (½∫r²dθ), expansion of (6+a sin θ)², integration of sin²θ using double angle formula, and solving a resulting quadratic equation. While the individual techniques are standard, the multi-step process and algebraic manipulation make it moderately challenging for Further Maths students. |
| Spec | 4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(A=\frac{1}{2}\int(6+a\sin\theta)^2\,d\theta\) | B1 | Use of \(\frac{1}{2}\int r^2\,d\theta\); limits not needed; can be gained if \(\frac{1}{2}\) appears later |
| \((6+a\sin\theta)^2=36+12a\sin\theta+a^2\sin^2\theta\) | — | Expansion |
| \((6+a\sin\theta)^2=36+12a\sin\theta+a^2\left(\frac{1-\cos 2\theta}{2}\right)\) | M1A1 | M1: Squares (\(36+k\sin^2\theta\) where \(k=a^2\) or \(a\) as minimum) and attempts to change \(\sin^2\theta\) to expression in \(\cos 2\theta\); A1: Correct expression |
| \(\left(\frac{1}{2}\right)\left[36\theta-12a\cos\theta+\frac{a^2}{2}\theta-\frac{a^2}{4}\sin 2\theta\right]\) | dM1A1 | dM1: Attempt to integrate; \(\cos 2\theta\to\pm\frac{1}{2}\sin 2\theta\); limits not needed; A1: Correct integration, limits not needed |
| \(=36\pi+\frac{\pi a^2}{2}\) | A1 | Correct area from correct integration and correct limits; trig functions must be evaluated |
| \(36\pi+\frac{\pi a^2}{2}=\frac{97\pi}{2}\Rightarrow a=...\) | ddM1 | Set area \(=\frac{97\pi}{2}\) and attempt to solve for \(a\); depends on both M marks; if \(\frac{1}{2}\) omitted initially and area set \(=97\pi\), give B1 by implication |
| \(a=5\) | A1cso | cao and cso; \(a=\pm5\) or \(a=-5\) scores A0 |
# Question 6:
$$r=6+a\sin\theta$$
| Answer/Working | Mark | Guidance |
|---|---|---|
| $A=\frac{1}{2}\int(6+a\sin\theta)^2\,d\theta$ | B1 | Use of $\frac{1}{2}\int r^2\,d\theta$; limits not needed; can be gained if $\frac{1}{2}$ appears later |
| $(6+a\sin\theta)^2=36+12a\sin\theta+a^2\sin^2\theta$ | — | Expansion |
| $(6+a\sin\theta)^2=36+12a\sin\theta+a^2\left(\frac{1-\cos 2\theta}{2}\right)$ | M1A1 | M1: Squares ($36+k\sin^2\theta$ where $k=a^2$ or $a$ as minimum) and attempts to change $\sin^2\theta$ to expression in $\cos 2\theta$; A1: Correct expression |
| $\left(\frac{1}{2}\right)\left[36\theta-12a\cos\theta+\frac{a^2}{2}\theta-\frac{a^2}{4}\sin 2\theta\right]$ | dM1A1 | dM1: Attempt to integrate; $\cos 2\theta\to\pm\frac{1}{2}\sin 2\theta$; limits not needed; A1: Correct integration, limits not needed |
| $=36\pi+\frac{\pi a^2}{2}$ | A1 | Correct area from correct integration and correct limits; trig functions must be evaluated |
| $36\pi+\frac{\pi a^2}{2}=\frac{97\pi}{2}\Rightarrow a=...$ | ddM1 | Set area $=\frac{97\pi}{2}$ and attempt to solve for $a$; depends on both M marks; if $\frac{1}{2}$ omitted initially and area set $=97\pi$, give B1 by implication |
| $a=5$ | A1cso | cao and cso; $a=\pm5$ or $a=-5$ scores A0 |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9019397a-a9c2-4b69-97fd-ea9eb9132244-18_364_695_260_756}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of a curve with polar equation
$$r = 6 + a \sin \theta$$
where $0 < a < 6$ and $0 \leqslant \theta < 2 \pi$\\
The area enclosed by the curve is $\frac { 97 \pi } { 2 }$\\
Find the value of the constant $a$.\\
\hfill \mbox{\textit{Edexcel FP2 2017 Q6 [8]}}