CAIE FP1 2014 June — Question 5 6 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2014
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPolar coordinates
TypeArea of region with line boundary
DifficultyStandard +0.8 This is a Further Maths polar coordinates question requiring sketching a cardioid and computing an area integral with trigonometric simplification. While the integral setup is standard (½∫r²dθ), evaluating ∫(1+sinθ)²dθ requires expanding, integrating sin²θ using double angle formulas, and careful algebraic manipulation to reach an exact answer. The cardioid sketch adds complexity. This is moderately challenging for FM students but follows established techniques.
Spec4.09a Polar coordinates: convert to/from cartesian4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve
5 The curve \(C\) has polar equation \(r = a ( 1 + \sin \theta )\), where \(a\) is a positive constant and \(0 \leqslant \theta < 2 \pi\). Draw a sketch of \(C\). Find the exact value of the area of the region enclosed by \(C\) and the half-lines \(\theta = \frac { 1 } { 3 } \pi\) and \(\theta = \frac { 2 } { 3 } \pi\).
Question 5:
AnswerMarks Guidance
Working/AnswerMark Guidance
Closed curve through \((a,0)\), \(\left(2a, \frac{1}{2}\pi\right)\), \((a,\pi)\), \(\left(0, \frac{3}{2}\pi\right)\)B1
Completely correct shape for cardioidB1
\(\text{Area} = \frac{a^2}{2}\int_{\frac{1}{3}\pi}^{\frac{2}{3}\pi}(1+\sin\theta)^2\,d\theta = \frac{a^2}{2}\int_{\frac{1}{3}\pi}^{\frac{2}{3}\pi}(1 + 2\sin\theta + \sin^2\theta)\,d\theta\)M1 LNR
\(= \frac{a^2}{2}\int_{\frac{1}{3}\pi}^{\frac{2}{3}\pi}\left(\frac{3}{2} + 2\sin\theta - \frac{1}{2}\cos 2\theta\right)d\theta\)M1 LR
\(= \frac{a^2}{2}\left[\frac{3}{2}\theta - 2\cos\theta - \frac{1}{4}\sin 2\theta\right]_{\frac{1}{3}\pi}^{\frac{2}{3}\pi}\)M1
\(= a^2\left(\frac{1}{4}\pi + 1 + \frac{1}{8}\sqrt{3}\right)\)A1 CAO
Total: [4]
## Question 5:

| Working/Answer | Mark | Guidance |
|---|---|---|
| Closed curve through $(a,0)$, $\left(2a, \frac{1}{2}\pi\right)$, $(a,\pi)$, $\left(0, \frac{3}{2}\pi\right)$ | B1 | |
| Completely correct shape for cardioid | B1 | |
| $\text{Area} = \frac{a^2}{2}\int_{\frac{1}{3}\pi}^{\frac{2}{3}\pi}(1+\sin\theta)^2\,d\theta = \frac{a^2}{2}\int_{\frac{1}{3}\pi}^{\frac{2}{3}\pi}(1 + 2\sin\theta + \sin^2\theta)\,d\theta$ | M1 | LNR |
| $= \frac{a^2}{2}\int_{\frac{1}{3}\pi}^{\frac{2}{3}\pi}\left(\frac{3}{2} + 2\sin\theta - \frac{1}{2}\cos 2\theta\right)d\theta$ | M1 | LR |
| $= \frac{a^2}{2}\left[\frac{3}{2}\theta - 2\cos\theta - \frac{1}{4}\sin 2\theta\right]_{\frac{1}{3}\pi}^{\frac{2}{3}\pi}$ | M1 | |
| $= a^2\left(\frac{1}{4}\pi + 1 + \frac{1}{8}\sqrt{3}\right)$ | A1 | CAO |

**Total: [4]**

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5 The curve $C$ has polar equation $r = a ( 1 + \sin \theta )$, where $a$ is a positive constant and $0 \leqslant \theta < 2 \pi$. Draw a sketch of $C$.

Find the exact value of the area of the region enclosed by $C$ and the half-lines $\theta = \frac { 1 } { 3 } \pi$ and $\theta = \frac { 2 } { 3 } \pi$.

\hfill \mbox{\textit{CAIE FP1 2014 Q5 [6]}}
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