| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 1 (Further AS Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Sketch polar curve |
| Difficulty | Standard +0.3 This is a straightforward polar coordinates question requiring only routine techniques: substitution to verify a point lies on the curve, finding maximum r by differentiating or using cosine properties, and converting polar to Cartesian coordinates using standard formulas. All parts are direct applications of standard methods with no problem-solving insight required, making it slightly easier than average. |
| Spec | 4.09a Polar coordinates: convert to/from cartesian4.09b Sketch polar curves: r = f(theta) |
| Answer | Marks | Guidance |
|---|---|---|
| \(4-2\cos\!\left(\frac{\pi}{3}\right) = 4-2\times\frac{1}{2} = 3\); \(\therefore \left(3,\frac{\pi}{3}\right)\) lies on \(C\) | B1 | Verifies \(r=3\) and \(\theta=\frac{\pi}{3}\) satisfies the polar equation; condone missing conclusion; accept \(4-2\times\frac{1}{2}=3\) as sufficient verification |
| Answer | Marks | Guidance |
|---|---|---|
| \(r = 4-2\times(-1) = 6\); \(\cos\theta = -1 \Rightarrow \theta = \pi\); furthest from \(O\) is \((6,\pi)\) | M1, A1, A1 | M1: Selects method, e.g. substitutes \(\cos\theta=-1\) to find \(r\) or solves \(\cos\theta=-1\) to find \(\theta\); PI by correct value of \(r\) or \(\theta\). A1: Correct value of \(r\) or \(\theta\). A1: Correct polar coordinates. |
| Answer | Marks | Guidance |
|---|---|---|
| \(r = 4-2\cos\!\left(\frac{\pi}{6}\right) = 4-2\times\frac{\sqrt{3}}{2} = 4-\sqrt{3}\) | M1 | Substitutes \(\theta=\frac{\pi}{6}\) to find \(r\); PI by 2.27 or 1.96 or 1.13 or better |
| \(x = (4-\sqrt{3})\cos\!\left(\frac{\pi}{6}\right) = 2\sqrt{3}-\frac{3}{2}\) | M1 | Obtains expression for \(x\) or \(y\) coordinate; PI by 1.96 or 1.13 or better; 2.2a |
| \(y = (4-\sqrt{3})\sin\!\left(\frac{\pi}{6}\right) = 2-\frac{\sqrt{3}}{2}\) | A1 | Obtains correct exact Cartesian coordinates; \(\left(2\sqrt{3}-\frac{3}{2},\ 2-\frac{\sqrt{3}}{2}\right)\); ACF |
## Question 8(a):
$4-2\cos\!\left(\frac{\pi}{3}\right) = 4-2\times\frac{1}{2} = 3$; $\therefore \left(3,\frac{\pi}{3}\right)$ lies on $C$ | B1 | Verifies $r=3$ and $\theta=\frac{\pi}{3}$ satisfies the polar equation; condone missing conclusion; accept $4-2\times\frac{1}{2}=3$ as sufficient verification
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## Question 8(b):
$r = 4-2\times(-1) = 6$; $\cos\theta = -1 \Rightarrow \theta = \pi$; furthest from $O$ is $(6,\pi)$ | M1, A1, A1 | M1: Selects method, e.g. substitutes $\cos\theta=-1$ to find $r$ or solves $\cos\theta=-1$ to find $\theta$; PI by correct value of $r$ or $\theta$. A1: Correct value of $r$ or $\theta$. A1: Correct polar coordinates.
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## Question 8(c):
$r = 4-2\cos\!\left(\frac{\pi}{6}\right) = 4-2\times\frac{\sqrt{3}}{2} = 4-\sqrt{3}$ | M1 | Substitutes $\theta=\frac{\pi}{6}$ to find $r$; PI by 2.27 or 1.96 or 1.13 or better
$x = (4-\sqrt{3})\cos\!\left(\frac{\pi}{6}\right) = 2\sqrt{3}-\frac{3}{2}$ | M1 | Obtains expression for $x$ or $y$ coordinate; PI by 1.96 or 1.13 or better; 2.2a
$y = (4-\sqrt{3})\sin\!\left(\frac{\pi}{6}\right) = 2-\frac{\sqrt{3}}{2}$ | A1 | Obtains correct exact Cartesian coordinates; $\left(2\sqrt{3}-\frac{3}{2},\ 2-\frac{\sqrt{3}}{2}\right)$; ACF8 The curve $C$ has the polar equation
$$r = 4 - 2 \cos \theta \quad - \pi < \theta \leq \pi$$
8
\begin{enumerate}[label=(\alph*)]
\item Verify that the point with polar coordinates $\left( 3 , \frac { \pi } { 3 } \right)$ lies on $C$\\
8
\item Find the exact polar coordinates of the point on $C$ which is furthest from the pole, $O$ [3 marks]\\
8
\item Find the exact Cartesian coordinates of the point on $C$ where $\theta$ is $\frac { \pi } { 6 }$
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 1 2022 Q8 [7]}}