CAIE FP1 2011 November — Question 8 10 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2011
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPolar coordinates
TypeArea of region with line boundary
DifficultyChallenging +1.2 This is a Further Maths polar coordinates question requiring a sketch and area calculation using the standard formula ½∫r²dθ. While it involves integration of trigonometric expressions and comparing two regions, the method is direct and the integration is routine (involving sin²θ). The conceptual demand is moderate for FM students who have learned polar areas, making it slightly above average difficulty overall but not requiring novel insight.
Spec4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve

8 The curve \(C\) has polar equation \(r = 1 + \sin \theta\) for \(- \frac { 1 } { 2 } \pi \leqslant \theta \leqslant \frac { 1 } { 2 } \pi\). Draw a sketch of \(C\). The area of the region enclosed by the initial line, the half-line \(\theta = \frac { 1 } { 2 } \pi\), and the part of \(C\) for which \(\theta\) is positive, is denoted by \(A _ { 1 }\). The area of the region enclosed by the initial line, and the part of \(C\) for which \(\theta\) is negative, is denoted by \(A _ { 2 }\). Find the ratio \(A _ { 1 } : A _ { 2 }\), giving your answer correct to 1 decimal place.

Question 8:
Polar Curve Area
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Arc above initial line; Arc below initial lineB1, B1 2 marks for sketching graph
\(\frac{1}{2}\int(1+\sin\theta)^2\,d\theta = \frac{1}{2}\int(1+2\sin\theta+\sin^2\theta)\,d\theta\)M1 Uses \(\frac{1}{2}\int r^2\,d\theta\)
\(= \frac{1}{2}\int\left(\frac{3}{2}+2\sin\theta - \frac{1}{2}\cos 2\theta\right)d\theta\)M1 Uses double angle formula
\(= \frac{1}{2}\left[\frac{3\theta}{2}-2\cos\theta-\frac{1}{4}\sin 2\theta\right] + c\)M1A1 Integrates
\(A_1 = \left[\frac{1}{2}\left(\frac{3\theta}{2}-2\cos\theta-\frac{1}{4}\sin 2\theta\right)\right]_0^{\frac{\pi}{2}} = \frac{3\pi}{8}+1\)M1A1 Inserts limits
\(A_2 = \left[\frac{1}{2}\left(\frac{3\theta}{2}-2\cos\theta-\frac{1}{4}\sin 2\theta\right)\right]_{-\frac{\pi}{2}}^{0} = \frac{3\pi}{8}-1\)A1
\(n = \left(\frac{3\pi}{8}+1\right)\div\left(\frac{3\pi}{8}-1\right) = 12.2\ \text{(1 d.p.)}\)A1 8 marks, total [10]
# Question 8:

## Polar Curve Area

| Answer/Working | Marks | Guidance |
|---|---|---|
| Arc above initial line; Arc below initial line | B1, B1 | 2 marks for sketching graph |
| $\frac{1}{2}\int(1+\sin\theta)^2\,d\theta = \frac{1}{2}\int(1+2\sin\theta+\sin^2\theta)\,d\theta$ | M1 | Uses $\frac{1}{2}\int r^2\,d\theta$ |
| $= \frac{1}{2}\int\left(\frac{3}{2}+2\sin\theta - \frac{1}{2}\cos 2\theta\right)d\theta$ | M1 | Uses double angle formula |
| $= \frac{1}{2}\left[\frac{3\theta}{2}-2\cos\theta-\frac{1}{4}\sin 2\theta\right] + c$ | M1A1 | Integrates |
| $A_1 = \left[\frac{1}{2}\left(\frac{3\theta}{2}-2\cos\theta-\frac{1}{4}\sin 2\theta\right)\right]_0^{\frac{\pi}{2}} = \frac{3\pi}{8}+1$ | M1A1 | Inserts limits |
| $A_2 = \left[\frac{1}{2}\left(\frac{3\theta}{2}-2\cos\theta-\frac{1}{4}\sin 2\theta\right)\right]_{-\frac{\pi}{2}}^{0} = \frac{3\pi}{8}-1$ | A1 | |
| $n = \left(\frac{3\pi}{8}+1\right)\div\left(\frac{3\pi}{8}-1\right) = 12.2\ \text{(1 d.p.)}$ | A1 | 8 marks, total [10] |

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8 The curve $C$ has polar equation $r = 1 + \sin \theta$ for $- \frac { 1 } { 2 } \pi \leqslant \theta \leqslant \frac { 1 } { 2 } \pi$. Draw a sketch of $C$.

The area of the region enclosed by the initial line, the half-line $\theta = \frac { 1 } { 2 } \pi$, and the part of $C$ for which $\theta$ is positive, is denoted by $A _ { 1 }$. The area of the region enclosed by the initial line, and the part of $C$ for which $\theta$ is negative, is denoted by $A _ { 2 }$. Find the ratio $A _ { 1 } : A _ { 2 }$, giving your answer correct to 1 decimal place.

\hfill \mbox{\textit{CAIE FP1 2011 Q8 [10]}}