| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2018 |
| Session | June |
| Marks | 12 |
| 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 FP2 polar coordinates question requiring the tangent condition dy/dx = 0, followed by routine area integration. Part (a) involves algebraic manipulation but follows a well-known method; part (b) is a direct application of the polar area formula with standard trigonometric integration. While more challenging than typical A-level questions due to the Further Maths content, it's a textbook exercise without novel insight required. |
| Spec | 4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(y = r\sin\theta = 2\sin\theta + \sqrt{3}\cos\theta\sin\theta\) | B1 | Multiplies \(r\) by \(\sin\theta\) |
| \(\frac{dy}{d\theta} = 2\cos\theta + \sqrt{3}\cos^2\theta - \sqrt{3}\sin^2\theta\) | M1A1 | M1 differentiates using product rule; A1 correct derivative |
| \(2\cos\theta + \sqrt{3}\cos^2\theta - \sqrt{3}(1-\cos^2\theta) = 0\) leading to \(2\sqrt{3}\cos^2\theta + 2\cos\theta - \sqrt{3} = 0\) | M1 | Use \(\sin^2\theta + \cos^2\theta = 1\) to form 3TQ in \(\cos\theta\) and attempt to solve |
| \(\cos\theta = \frac{-2 \pm \sqrt{28}}{4\sqrt{3}}\) or \(\frac{\sqrt{21}-\sqrt{3}}{6}\) oe | A1 | Accept \(\pm\) or \(+\); any exact equivalent |
| \(OP = r = 2 + \frac{-2+\sqrt{28}}{4} = \frac{1}{2}(3+\sqrt{7})\) | A1cso | Must show substitution of correct exact \(\cos\theta\) in \(r = 2+\sqrt{3}\cos\theta\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(y = r\sin\theta = (2+\sqrt{3}\cos\theta)\sin\theta\) | B1 | Leaves \(y\) as a product |
| \(\frac{dy}{d\theta} = (2+\sqrt{3}\cos\theta)\cos\theta - \sqrt{3}\sin\theta\sin\theta\) | M1A1 | M1 product rule; A1 correct derivative |
| \(2\sqrt{3}\cos^2\theta + 2\cos\theta - \sqrt{3} = 0\) | M1 | Use identity to form 3TQ in \(\cos\theta\) |
| \(\cos\theta = \frac{-2\pm\sqrt{28}}{4\sqrt{3}}\) or \(\frac{\sqrt{21}-\sqrt{3}}{6}\) oe | A1 | |
| \(OP = \frac{1}{2}(3+\sqrt{7})\) | A1cso |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(y = r\sin\theta = 2\sin\theta + \frac{\sqrt{3}}{2}\sin 2\theta\) | B1 | Uses double angle formula |
| \(\frac{dy}{d\theta} = 2\cos\theta + \sqrt{3}\cos 2\theta\) | M1A1 | M1 differentiates; A1 correct derivative |
| \(2\cos\theta + \sqrt{3}(2\cos^2\theta - 1) = 0 \Rightarrow 2\sqrt{3}\cos^2\theta + 2\cos\theta - \sqrt{3} = 0\) | M1 | Use double angle identity \(\cos 2\theta = 2\cos^2\theta -1\); attempt 3TQ |
| \(\cos\theta = \frac{-2\pm\sqrt{28}}{4\sqrt{3}}\) or \(\frac{\sqrt{21}-\sqrt{3}}{6}\) oe | A1 | |
| \(OP = \frac{1}{2}(3+\sqrt{7})\) | A1cso |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(\frac{dr}{d\theta} = -\sqrt{3}\sin\theta\) | B1 | Correct derivative |
| \(\frac{dy}{d\theta} = \frac{dr}{d\theta}\sin\theta + r\cos\theta\) | M1 | Differentiate using product rule |
| \(\frac{dy}{d\theta} = -\sqrt{3}\sin^2\theta + (2+\sqrt{3}\cos\theta)\cos\theta\) | A1 | Correct derivative as function of \(\theta\) |
| \(2\sqrt{3}\cos^2\theta + 2\cos\theta - \sqrt{3} = 0\) | M1 | Use identity; form 3TQ |
| \(\cos\theta = \frac{-2\pm\sqrt{28}}{4\sqrt{3}}\) or \(\frac{\sqrt{21}-\sqrt{3}}{6}\) oe | A1 | |
| \(OP = \frac{1}{2}(3+\sqrt{7})\) | A1cso |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(y = r\cos\theta\) (NOT \(x = r\cos\theta\)) | B0 | |
| \(r\cos\theta = 2\cos\theta + \sqrt{3}\cos^2\theta\) | ||
| \(\frac{dy}{d\theta} = -2\sin\theta - 2\sqrt{3}\sin\theta\cos\theta\) | M1, A0 | Differentiates but cannot obtain correct derivative |
| No further marks available |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \((2+\sqrt{3}\cos\theta)^2 = 4 + 4\sqrt{3}\cos\theta + 3\cos^2\theta\) | M1 | Attempt \(r^2\) as 3-term quadratic and use double angle formula \(\cos^2\theta = \frac{1}{2}(\cos 2\theta \pm 1)\) |
| \(= 4 + 4\sqrt{3}\cos\theta + \frac{3}{2}(\cos 2\theta + 1)\) | A1 | Correct result |
| \(\int r^2\,d\theta = 4\theta + 4\sqrt{3}\sin\theta + 3\left(\frac{1}{4}\sin 2\theta + \frac{1}{2}\theta\right)\) oe | dM1A1 | dM1 attempts to integrate \(r^2\); depends on first M1; \(\cos\theta \to \pm\sin\theta\); \(\cos 2\theta \to \pm k\sin 2\theta\), \(k=1\) or \(\frac{1}{2}\); A1 correct integral |
| \(\frac{1}{2}\int_0^{2\pi} r^2\,d\theta = \frac{1}{2}(8\pi + 3\pi - 0)\) | ddM1 | Substitutes correct limits; \(\frac{1}{2}\int_0^{2\pi}\) or \(2\times\frac{1}{2}\int_0^{\pi}\) or \(\frac{1}{2}\int_{-\pi}^{\pi}\) |
| \(= \frac{11\pi}{2}\) | A1cso | Correct exact answer; accept \(5.5\pi\); no errors in working |
# Question 7:
## Part (a) — Way 1:
| Working | Mark | Notes |
|---------|------|-------|
| $y = r\sin\theta = 2\sin\theta + \sqrt{3}\cos\theta\sin\theta$ | B1 | Multiplies $r$ by $\sin\theta$ |
| $\frac{dy}{d\theta} = 2\cos\theta + \sqrt{3}\cos^2\theta - \sqrt{3}\sin^2\theta$ | M1A1 | M1 differentiates using product rule; A1 correct derivative |
| $2\cos\theta + \sqrt{3}\cos^2\theta - \sqrt{3}(1-\cos^2\theta) = 0$ leading to $2\sqrt{3}\cos^2\theta + 2\cos\theta - \sqrt{3} = 0$ | M1 | Use $\sin^2\theta + \cos^2\theta = 1$ to form 3TQ in $\cos\theta$ and attempt to solve |
| $\cos\theta = \frac{-2 \pm \sqrt{28}}{4\sqrt{3}}$ or $\frac{\sqrt{21}-\sqrt{3}}{6}$ oe | A1 | Accept $\pm$ or $+$; any exact equivalent |
| $OP = r = 2 + \frac{-2+\sqrt{28}}{4} = \frac{1}{2}(3+\sqrt{7})$ | A1cso | Must show substitution of correct exact $\cos\theta$ in $r = 2+\sqrt{3}\cos\theta$ |
**(6 marks)**
## Part (a) — Way 2:
| Working | Mark | Notes |
|---------|------|-------|
| $y = r\sin\theta = (2+\sqrt{3}\cos\theta)\sin\theta$ | B1 | Leaves $y$ as a product |
| $\frac{dy}{d\theta} = (2+\sqrt{3}\cos\theta)\cos\theta - \sqrt{3}\sin\theta\sin\theta$ | M1A1 | M1 product rule; A1 correct derivative |
| $2\sqrt{3}\cos^2\theta + 2\cos\theta - \sqrt{3} = 0$ | M1 | Use identity to form 3TQ in $\cos\theta$ |
| $\cos\theta = \frac{-2\pm\sqrt{28}}{4\sqrt{3}}$ or $\frac{\sqrt{21}-\sqrt{3}}{6}$ oe | A1 | |
| $OP = \frac{1}{2}(3+\sqrt{7})$ | A1cso | |
**(6 marks)**
## Part (a) — Way 3:
| Working | Mark | Notes |
|---------|------|-------|
| $y = r\sin\theta = 2\sin\theta + \frac{\sqrt{3}}{2}\sin 2\theta$ | B1 | Uses double angle formula |
| $\frac{dy}{d\theta} = 2\cos\theta + \sqrt{3}\cos 2\theta$ | M1A1 | M1 differentiates; A1 correct derivative |
| $2\cos\theta + \sqrt{3}(2\cos^2\theta - 1) = 0 \Rightarrow 2\sqrt{3}\cos^2\theta + 2\cos\theta - \sqrt{3} = 0$ | M1 | Use double angle identity $\cos 2\theta = 2\cos^2\theta -1$; attempt 3TQ |
| $\cos\theta = \frac{-2\pm\sqrt{28}}{4\sqrt{3}}$ or $\frac{\sqrt{21}-\sqrt{3}}{6}$ oe | A1 | |
| $OP = \frac{1}{2}(3+\sqrt{7})$ | A1cso | |
**(6 marks)**
## Part (a) — Way 4:
| Working | Mark | Notes |
|---------|------|-------|
| $\frac{dr}{d\theta} = -\sqrt{3}\sin\theta$ | B1 | Correct derivative |
| $\frac{dy}{d\theta} = \frac{dr}{d\theta}\sin\theta + r\cos\theta$ | M1 | Differentiate using product rule |
| $\frac{dy}{d\theta} = -\sqrt{3}\sin^2\theta + (2+\sqrt{3}\cos\theta)\cos\theta$ | A1 | Correct derivative as function of $\theta$ |
| $2\sqrt{3}\cos^2\theta + 2\cos\theta - \sqrt{3} = 0$ | M1 | Use identity; form 3TQ |
| $\cos\theta = \frac{-2\pm\sqrt{28}}{4\sqrt{3}}$ or $\frac{\sqrt{21}-\sqrt{3}}{6}$ oe | A1 | |
| $OP = \frac{1}{2}(3+\sqrt{7})$ | A1cso | |
**(6 marks)**
## Part (a) — Special Case ($y = r\cos\theta$):
| Working | Mark | Notes |
|---------|------|-------|
| $y = r\cos\theta$ (NOT $x = r\cos\theta$) | B0 | |
| $r\cos\theta = 2\cos\theta + \sqrt{3}\cos^2\theta$ | | |
| $\frac{dy}{d\theta} = -2\sin\theta - 2\sqrt{3}\sin\theta\cos\theta$ | M1, A0 | Differentiates but cannot obtain correct derivative |
| No further marks available | | |
## Part (b):
| Working | Mark | Notes |
|---------|------|-------|
| $(2+\sqrt{3}\cos\theta)^2 = 4 + 4\sqrt{3}\cos\theta + 3\cos^2\theta$ | M1 | Attempt $r^2$ as 3-term quadratic and use double angle formula $\cos^2\theta = \frac{1}{2}(\cos 2\theta \pm 1)$ |
| $= 4 + 4\sqrt{3}\cos\theta + \frac{3}{2}(\cos 2\theta + 1)$ | A1 | Correct result |
| $\int r^2\,d\theta = 4\theta + 4\sqrt{3}\sin\theta + 3\left(\frac{1}{4}\sin 2\theta + \frac{1}{2}\theta\right)$ oe | dM1A1 | dM1 attempts to integrate $r^2$; depends on first M1; $\cos\theta \to \pm\sin\theta$; $\cos 2\theta \to \pm k\sin 2\theta$, $k=1$ or $\frac{1}{2}$; A1 correct integral |
| $\frac{1}{2}\int_0^{2\pi} r^2\,d\theta = \frac{1}{2}(8\pi + 3\pi - 0)$ | ddM1 | Substitutes correct limits; $\frac{1}{2}\int_0^{2\pi}$ or $2\times\frac{1}{2}\int_0^{\pi}$ or $\frac{1}{2}\int_{-\pi}^{\pi}$ |
| $= \frac{11\pi}{2}$ | A1cso | Correct exact answer; accept $5.5\pi$; no errors in working |
**(6 marks) — Total: 12**
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5aa7f449-215b-4a21-9fdc-df55d26abc9d-24_508_896_212_525}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The curve $C$ shown in Figure 1 has polar equation
$$r = 2 + \sqrt { 3 } \cos \theta , \quad 0 \leqslant \theta < 2 \pi$$
The tangent to $C$ at the point $P$ is parallel to the initial line.
\begin{enumerate}[label=(\alph*)]
\item Show that $O P = \frac { 1 } { 2 } ( 3 + \sqrt { 7 } )$
\item Find the exact area enclosed by the curve $C$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2018 Q7 [12]}}