Question 4 - A-Level Maths

Edexcel FP2 2002 June — Question 4 18 marks

Exam Board	Edexcel
Module	FP2 (Further Pure Mathematics 2)
Year	2002
Session	June
Marks	18
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Polar coordinates
Type	Area between two polar curves
Difficulty	Challenging +1.8 This is a substantial Further Maths polar coordinates question requiring sketching two curves, finding intersections, computing areas via integration, and combining given information to find a composite region. While the techniques are standard for FP2 (polar area formula, solving trigonometric equations), the multi-part structure, need to visualize the region correctly, and final synthesis in part (d) elevate it above routine exercises. It's challenging for Further Maths but follows established patterns.
Spec	4.09a Polar coordinates: convert to/from cartesian 4.09c Area enclosed: by polar curve

4. The curve $C$ has polar equation $r = 3 a \cos \theta , - \frac { \pi } { 2 } \leq \frac { \pi } { 2 }$. The curve $D$ has polar equation $r = a ( 1 + \cos \theta ) , - \pi \leq \theta < \pi$. Given that $a$ is a positive constant, (a) sketch, on the same diagram, the graphs of $C$ and $D$, indicating where each curve cuts the initial line. The graphs of $C$ intersect at the pole $O$ and at the points $P$ and $Q$.
(b) Find the polar coordinates of $P$ and $Q$.
(c) Use integration to find the exact area enclosed by the curve $D$ and the lines $\theta = 0$ and $\theta = \frac { \pi } { 3 }$ The region $R$ contains all points which lie outside $D$ and inside $C$.
Given that the value of the smaller area enclosed by the curve $C$ and the line $\theta = \frac { \pi } { 3 }$ is $$\frac { 3 a ^ { 2 } } { 16 } ( 2 \pi - 3 \sqrt { } 3 )$$ (d) show that the area of $R$ is $\pi a ^ { 2 }$.

Show mark scheme Show mark scheme source

Question 4:

Part (a)

Answer	Marks	Guidance
Answer	Marks	Guidance
Circle B1, Diameter $3a$	B1
Cardioid cusp at O	B1
Symmetry and $2a$	B1

Part (b)

Answer	Marks	Guidance
Answer	Marks	Guidance
$3a\cos\theta = a(1+\cos\theta)$, $\cos\theta = \frac{1}{2}$	M1
$\theta = \pm\frac{\pi}{3}$, $r = \frac{3a}{2}$	A1, A1

Part (c)

Answer	Marks	Guidance
Answer	Marks	Guidance
$A_1 = \frac{1}{2}\int a^2(1+\cos\theta)^2\,d\theta$	M1
$= \frac{1}{2}a^2\int(1+2\cos\theta+\frac{1}{2}(1+\cos 2\theta))\,d\theta$	M1, A1
$= \frac{1}{2}a^2\left[\frac{3\theta}{2}+2\sin\theta+\frac{1}{4}\sin 2\theta\right]$	A1, A1
Evaluate $A_1$ using 0 and $\frac{\pi}{3}$	M1
$A_1 = \frac{\pi a^2}{4}+\frac{9\sqrt{3}a^2}{16}$	A1

Part (d)

Answer	Marks	Guidance
Answer	Marks	Guidance
Area required $= \frac{9}{4}\pi a^2 - 2A_1 - 2\times\text{given}$	M1	$\frac{9}{4}\pi a^2$ B1
$= \frac{9\pi a^2}{4}-\frac{\pi a^2}{2}-\frac{9\sqrt{3}a^2}{8}-\frac{3\pi a^2}{4}+\frac{9\sqrt{3}a^2}{8}$	M1
$= \pi a^2$	A1

# Question 4:

## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Circle B1, Diameter $3a$ | B1 | |
| Cardioid cusp at O | B1 | |
| Symmetry and $2a$ | B1 | |

## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $3a\cos\theta = a(1+\cos\theta)$, $\cos\theta = \frac{1}{2}$ | M1 | |
| $\theta = \pm\frac{\pi}{3}$, $r = \frac{3a}{2}$ | A1, A1 | |

## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $A_1 = \frac{1}{2}\int a^2(1+\cos\theta)^2\,d\theta$ | M1 | |
| $= \frac{1}{2}a^2\int(1+2\cos\theta+\frac{1}{2}(1+\cos 2\theta))\,d\theta$ | M1, A1 | |
| $= \frac{1}{2}a^2\left[\frac{3\theta}{2}+2\sin\theta+\frac{1}{4}\sin 2\theta\right]$ | A1, A1 | |
| Evaluate $A_1$ using 0 and $\frac{\pi}{3}$ | M1 | |
| $A_1 = \frac{\pi a^2}{4}+\frac{9\sqrt{3}a^2}{16}$ | A1 | |

## Part (d)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Area required $= \frac{9}{4}\pi a^2 - 2A_1 - 2\times\text{given}$ | M1 | $\frac{9}{4}\pi a^2$ B1 |
| $= \frac{9\pi a^2}{4}-\frac{\pi a^2}{2}-\frac{9\sqrt{3}a^2}{8}-\frac{3\pi a^2}{4}+\frac{9\sqrt{3}a^2}{8}$ | M1 | |
| $= \pi a^2$ | A1 | |

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Show LaTeX source

4. The curve $C$ has polar equation $r = 3 a \cos \theta , - \frac { \pi } { 2 } \leq \frac { \pi } { 2 }$.

The curve $D$ has polar equation $r = a ( 1 + \cos \theta ) , - \pi \leq \theta < \pi$. Given that $a$ is a positive constant, (a) sketch, on the same diagram, the graphs of $C$ and $D$, indicating where each curve cuts the initial line.

The graphs of $C$ intersect at the pole $O$ and at the points $P$ and $Q$.\\
(b) Find the polar coordinates of $P$ and $Q$.\\
(c) Use integration to find the exact area enclosed by the curve $D$ and the lines $\theta = 0$ and $\theta = \frac { \pi } { 3 }$

The region $R$ contains all points which lie outside $D$ and inside $C$.\\
Given that the value of the smaller area enclosed by the curve $C$ and the line $\theta = \frac { \pi } { 3 }$ is

$$\frac { 3 a ^ { 2 } } { 16 } ( 2 \pi - 3 \sqrt { } 3 )$$

(d) show that the area of $R$ is $\pi a ^ { 2 }$.\\

\hfill \mbox{\textit{Edexcel FP2 2002 Q4 [18]}}

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Q1 5 Q2 10 Q3 13 Q4 18 Q5 7 Q6 11 Q7 14 Q8 15 Q9 7 Q10 12

Answer	Marks	Guidance
Answer	Marks	Guidance
\(3a\cos\theta = a(1+\cos\theta)\), \(\cos\theta = \frac{1}{2}\)	M1
\(\theta = \pm\frac{\pi}{3}\), \(r = \frac{3a}{2}\)	A1, A1