CAIE FP1 2012 June — Question 4 9 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2012
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPolar coordinates
TypeArea of region with line boundary
DifficultyStandard +0.8 This is a multi-part polar coordinates question requiring sketching a cardioid, computing a standard polar area integral, then finding areas of two regions divided by a half-line. While the integration itself is routine (using standard formulas for ∫cos²θ), the question requires understanding of polar area concepts, careful setup of integral limits, and numerical evaluation. It's moderately challenging for Further Maths students but follows standard techniques without requiring novel insight.
Spec4.09a Polar coordinates: convert to/from cartesian4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve
4 The curve \(C\) has polar equation \(r = 2 + 2 \cos \theta\), for \(0 \leqslant \theta \leqslant \pi\). Sketch the graph of \(C\). Find the area of the region \(R\) enclosed by \(C\) and the initial line. The half-line \(\theta = \frac { 1 } { 5 } \pi\) divides \(R\) into two parts. Find the area of each part, correct to 3 decimal places.
Question 4:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Shows \((4,0)\) and \((0,\pi)\) lie on \(C\); correct shapeB1, B1 Full cardioid is B1 unless clear evidence of plotting up to \(2\pi\) or \(-\pi\) to \(\pi\)
\(\frac{1}{2}\displaystyle\int_0^{\pi}(4+8\cos\theta+4\cos^2\theta)\,d\theta\)M1 Uses \(\frac{1}{2}\int_\alpha^\beta r^2\,d\theta\)
\(= \displaystyle\int_0^{\pi}(2+4\cos\theta+2\cos^2\theta)\,d\theta\)
\(= \displaystyle\int_0^{\pi}(3+4\cos\theta+\cos2\theta)\,d\theta\)M1 Uses double angle formula
\(= \left[3\theta + 4\sin\theta + \frac{\sin2\theta}{2}\right]_0^{\pi} = 3\pi\)A1A1 (CWO) A1 for correct integral; integrates and obtains area
\(\frac{3\pi}{5} + 4\sin\frac{\pi}{5} + \sin\frac{\pi}{5}\cos\frac{\pi}{5} = 4.712\)M1A1 Finds areas
\(3\pi - 4.712 = 4.713\)A1
Total: [9]
## Question 4:

| Answer/Working | Marks | Guidance |
|---|---|---|
| Shows $(4,0)$ and $(0,\pi)$ lie on $C$; correct shape | B1, B1 | Full cardioid is B1 unless clear evidence of plotting up to $2\pi$ or $-\pi$ to $\pi$ |
| $\frac{1}{2}\displaystyle\int_0^{\pi}(4+8\cos\theta+4\cos^2\theta)\,d\theta$ | M1 | Uses $\frac{1}{2}\int_\alpha^\beta r^2\,d\theta$ |
| $= \displaystyle\int_0^{\pi}(2+4\cos\theta+2\cos^2\theta)\,d\theta$ | | |
| $= \displaystyle\int_0^{\pi}(3+4\cos\theta+\cos2\theta)\,d\theta$ | M1 | Uses double angle formula |
| $= \left[3\theta + 4\sin\theta + \frac{\sin2\theta}{2}\right]_0^{\pi} = 3\pi$ | A1A1 (CWO) | A1 for correct integral; integrates and obtains area |
| $\frac{3\pi}{5} + 4\sin\frac{\pi}{5} + \sin\frac{\pi}{5}\cos\frac{\pi}{5} = 4.712$ | M1A1 | Finds areas |
| $3\pi - 4.712 = 4.713$ | A1 | |

**Total: [9]**

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4 The curve $C$ has polar equation $r = 2 + 2 \cos \theta$, for $0 \leqslant \theta \leqslant \pi$. Sketch the graph of $C$.

Find the area of the region $R$ enclosed by $C$ and the initial line.

The half-line $\theta = \frac { 1 } { 5 } \pi$ divides $R$ into two parts. Find the area of each part, correct to 3 decimal places.

\hfill \mbox{\textit{CAIE FP1 2012 Q4 [9]}}
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