CAIE FP1 2011 November — Question 10 13 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2011
SessionNovember
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPolar coordinates
TypeArea of region with line boundary
DifficultyChallenging +1.2 This is a multi-part Further Maths polar coordinates question requiring: (1) sketching a limaçon and vertical line, (2) solving simultaneous equations to find intersections, and (3) computing area using the polar area formula with appropriate limits. While it involves several steps and FM content, the techniques are standard and methodical—substitute the line equation into the curve, solve the resulting trigonometric equation, then integrate ½r² with correct bounds. More routine than problems requiring geometric insight or novel approaches.
Spec4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve
10 The curve \(C\) has polar equation \(r = 3 + 2 \cos \theta\), for \(- \pi < \theta \leqslant \pi\). The straight line \(l\) has polar equation \(r \cos \theta = 2\). Sketch both \(C\) and \(l\) on a single diagram. Find the polar coordinates of the points of intersection of \(C\) and \(l\). The region \(R\) is enclosed by \(C\) and \(l\), and contains the pole. Find the area of \(R\).
Question 10:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Closed loop through \((5,0)\) and \((1,\pi)\); Correct shape near \((1,\pi)\); Perpendicular to initial line, through \((2,0)\)B1, B1, B1 Curve C and straight line
\((3 + 2\cos\theta)\cos\theta = 2 \Rightarrow 2\cos^2\theta + 3\cos\theta - 2 = 0\) (aef)M1 Forms quadratic equation in usual form
\((2\cos\theta - 1)(\cos\theta + 2) = 0\)
\(\Rightarrow \cos\theta = 0.5\) (since \(\cos\theta > 0\))A1 Solves quadratic equation
Intersections at \(\left(4, \frac{\pi}{3}\right)\) and \(\left(4, -\frac{\pi}{3}\right)\)A1A1 Writes down points of intersection
Two sectors \(= 2 \times \frac{1}{2}\int_{\frac{\pi}{3}}^{\pi}(9 + 12\cos\theta + 4\cos^2\theta)\,d\theta\)M1 Finds required area
\(= \int_{\frac{\pi}{3}}^{\pi}(11 + 12\cos\theta + 2\cos 2\theta)\,d\theta\)A1
\(= \left[11\theta + 12\sin\theta + \sin 2\theta\right]_{\frac{\pi}{3}}^{\pi}\)M1
\(= \frac{22\pi}{3} - \frac{13\sqrt{3}}{2}\) (= 11.78)A1
Triangle \(= 2\sqrt{3} \times 2 = 4\sqrt{3}\) (= 6.928)B1
Total Area \(= \frac{22\pi}{3} - \frac{5\sqrt{3}}{2}\) (= 18.708 = 18.7 (3sf))A1 
# Question 10:

| Answer/Working | Marks | Guidance |
|---|---|---|
| Closed loop through $(5,0)$ and $(1,\pi)$; Correct shape near $(1,\pi)$; Perpendicular to initial line, through $(2,0)$ | B1, B1, B1 | Curve C and straight line | **Part total: 3** |
| $(3 + 2\cos\theta)\cos\theta = 2 \Rightarrow 2\cos^2\theta + 3\cos\theta - 2 = 0$ (aef) | M1 | Forms quadratic equation in usual form |
| $(2\cos\theta - 1)(\cos\theta + 2) = 0$ | | |
| $\Rightarrow \cos\theta = 0.5$ (since $\cos\theta > 0$) | A1 | Solves quadratic equation |
| Intersections at $\left(4, \frac{\pi}{3}\right)$ and $\left(4, -\frac{\pi}{3}\right)$ | A1A1 | Writes down points of intersection | **Part total: 4** |
| Two sectors $= 2 \times \frac{1}{2}\int_{\frac{\pi}{3}}^{\pi}(9 + 12\cos\theta + 4\cos^2\theta)\,d\theta$ | M1 | Finds required area |
| $= \int_{\frac{\pi}{3}}^{\pi}(11 + 12\cos\theta + 2\cos 2\theta)\,d\theta$ | A1 | |
| $= \left[11\theta + 12\sin\theta + \sin 2\theta\right]_{\frac{\pi}{3}}^{\pi}$ | M1 | |
| $= \frac{22\pi}{3} - \frac{13\sqrt{3}}{2}$ (= 11.78) | A1 | |
| Triangle $= 2\sqrt{3} \times 2 = 4\sqrt{3}$ (= 6.928) | B1 | |
| Total Area $= \frac{22\pi}{3} - \frac{5\sqrt{3}}{2}$ (= 18.708 = 18.7 (3sf)) | A1 | | **Part total: 6** |
10 The curve $C$ has polar equation $r = 3 + 2 \cos \theta$, for $- \pi < \theta \leqslant \pi$. The straight line $l$ has polar equation $r \cos \theta = 2$. Sketch both $C$ and $l$ on a single diagram.

Find the polar coordinates of the points of intersection of $C$ and $l$.

The region $R$ is enclosed by $C$ and $l$, and contains the pole. Find the area of $R$.

\hfill \mbox{\textit{CAIE FP1 2011 Q10 [13]}}
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