Question 8 - A-Level Maths

CAIE FP1 2014 November — Question 8 11 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2014
Session	November
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Polar coordinates
Type	Area between two polar curves
Difficulty	Challenging +1.2 This is a standard Further Maths polar coordinates question requiring sketching, finding intersections by equating equations (giving cos θ = 0, so θ = π/2, 3π/2), and computing area using the polar area formula. While it involves multiple steps and integration, the techniques are routine for FM students and the symmetry simplifies the calculation. The 'show that' format removes algebraic uncertainty.
Spec	4.09a Polar coordinates: convert to/from cartesian 4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve

8 A circle has polar equation $r = a$, for $0 \leqslant \theta < 2 \pi$, and a cardioid has polar equation $r = a ( 1 - \cos \theta )$, for $0 \leqslant \theta < 2 \pi$, where $a$ is a positive constant. Draw sketches of the circle and the cardioid on the same diagram. Write down the polar coordinates of the points of intersection of the circle and the cardioid. Show that the area of the region that is both inside the circle and inside the cardioid is $$\left( \frac { 5 } { 4 } \pi - 2 \right) a ^ { 2 }$$

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Question 8:

Answer	Marks	Guidance
Working/Answer	Marks	Guidance
Circle sketched	B1
Cardioid — correct location and orientation — correct indentation near pole	B1B1 (3)
$\left(a,\frac{\pi}{2}\right)$ and $\left(a,\frac{3\pi}{2}\right)$	B1B1 (2)	B1 for reverse, or $a=a(1-\cos\theta)\Rightarrow\theta=\frac{\pi}{2},\frac{3\pi}{2}$ seen
Area $=\frac{1}{2}\pi a^2+2\times\frac{1}{2}\int_0^{\frac{\pi}{2}}a^2(1-\cos\theta)^2\,\mathrm{d}\theta$	B1M1	Half circle + Area of sector
$=\frac{1}{2}\pi a^2+a^2\int_0^{\frac{\pi}{2}}(1-2\cos\theta+\cos^2\theta)\,\mathrm{d}\theta$	A1
$=\frac{1}{2}\pi a^2+a^2\int_0^{\frac{\pi}{2}}\!\left(\frac{3}{2}-2\cos\theta+\frac{1}{2}\cos 2\theta\right)\mathrm{d}\theta$	M1	Use of double angle formula
$=\frac{1}{2}\pi a^2+a^2\!\left[\frac{3\theta}{2}-2\sin\theta+\frac{1}{4}\sin 2\theta\right]_0^{\frac{\pi}{2}}$	M1	Integration
$=\frac{1}{2}\pi a^2+a^2\!\left(\frac{3}{4}\pi-2\right)=\left(\frac{5}{4}\pi-2\right)a^2$	A1 (6)	AG

## Question 8:

| Working/Answer | Marks | Guidance |
|---|---|---|
| Circle sketched | B1 | |
| Cardioid — correct location and orientation — correct indentation near pole | B1B1 (3) | |
| $\left(a,\frac{\pi}{2}\right)$ and $\left(a,\frac{3\pi}{2}\right)$ | B1B1 (2) | B1 for reverse, or $a=a(1-\cos\theta)\Rightarrow\theta=\frac{\pi}{2},\frac{3\pi}{2}$ seen |
| Area $=\frac{1}{2}\pi a^2+2\times\frac{1}{2}\int_0^{\frac{\pi}{2}}a^2(1-\cos\theta)^2\,\mathrm{d}\theta$ | B1M1 | Half circle + Area of sector |
| $=\frac{1}{2}\pi a^2+a^2\int_0^{\frac{\pi}{2}}(1-2\cos\theta+\cos^2\theta)\,\mathrm{d}\theta$ | A1 | |
| $=\frac{1}{2}\pi a^2+a^2\int_0^{\frac{\pi}{2}}\!\left(\frac{3}{2}-2\cos\theta+\frac{1}{2}\cos 2\theta\right)\mathrm{d}\theta$ | M1 | Use of double angle formula |
| $=\frac{1}{2}\pi a^2+a^2\!\left[\frac{3\theta}{2}-2\sin\theta+\frac{1}{4}\sin 2\theta\right]_0^{\frac{\pi}{2}}$ | M1 | Integration |
| $=\frac{1}{2}\pi a^2+a^2\!\left(\frac{3}{4}\pi-2\right)=\left(\frac{5}{4}\pi-2\right)a^2$ | A1 (6) | AG |

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8 A circle has polar equation $r = a$, for $0 \leqslant \theta < 2 \pi$, and a cardioid has polar equation $r = a ( 1 - \cos \theta )$, for $0 \leqslant \theta < 2 \pi$, where $a$ is a positive constant. Draw sketches of the circle and the cardioid on the same diagram.

Write down the polar coordinates of the points of intersection of the circle and the cardioid.

Show that the area of the region that is both inside the circle and inside the cardioid is

$$\left( \frac { 5 } { 4 } \pi - 2 \right) a ^ { 2 }$$

\hfill \mbox{\textit{CAIE FP1 2014 Q8 [11]}}

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