CAIE FP1 2009 June — Question 5 7 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2009
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPolar coordinates
TypeArea of region with line boundary
DifficultyStandard +0.8 This question requires sketching the Archimedean spiral r=θ, setting up and evaluating a polar area integral (∫½r²dθ), then solving an equation involving π² to find where areas are equal. It combines visualization, integration technique, and algebraic manipulation beyond routine exercises, though the concepts are standard for Further Maths polar coordinates.
Spec4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve

5 Draw a sketch of the curve \(C\) whose polar equation is \(r = \theta\), for \(0 \leqslant \theta \leqslant \frac { 1 } { 2 } \pi\). On the same diagram draw the line \(\theta = \alpha\), where \(0 < \alpha < \frac { 1 } { 2 } \pi\). The region bounded by \(C\) and the line \(\theta = \frac { 1 } { 2 } \pi\) is denoted by \(R\). Find the exact value of \(\alpha\) for which the line \(\theta = \alpha\) divides \(R\) into two regions of equal area.

AnswerMarks
Sketch with correct shape, location and orientationB1
Shows tangency to the initial line at the poleB1
Ignore extra in diagram
Draws, in the same diagram, a straight line passing through the origin and with positive gradient (distinct half-line and not a construction line)B1
\((1/2)\int_0^{\pi/2} \theta^2 d\theta = \pi^3/48\)M1A1
\((1/2)\int_0^\alpha \theta^2 d\theta = \alpha^3/6\)A1
\(\alpha^3/6 = \pi^3/96 \Rightarrow \alpha = \pi 2^{-4/3}\) (acf)A1
or for previous 4 marks:
AnswerMarks
\((1/2)\int_0^\alpha \theta^2 d\theta = (1/2)\int_{\pi/2}^\pi \theta^2 d\theta\)M1
\(\alpha^3/6 = (1/6)[(\pi/2)^3 - \alpha^3]\)A1A1
\(\Rightarrow \alpha^3 = \pi^3/16 \Rightarrow \alpha = \pi(16)^{-1/3}\), or equivalentA1
Sketch with correct shape, location and orientation | B1 |

Shows tangency to the initial line at the pole | B1 |

Ignore extra in diagram |

Draws, in the same diagram, a straight line passing through the origin and with positive gradient (distinct half-line and not a construction line) | B1 |

$(1/2)\int_0^{\pi/2} \theta^2 d\theta = \pi^3/48$ | M1A1 |

$(1/2)\int_0^\alpha \theta^2 d\theta = \alpha^3/6$ | A1 |

$\alpha^3/6 = \pi^3/96 \Rightarrow \alpha = \pi 2^{-4/3}$ (acf) | A1 |

**or for previous 4 marks:**

$(1/2)\int_0^\alpha \theta^2 d\theta = (1/2)\int_{\pi/2}^\pi \theta^2 d\theta$ | M1 |

$\alpha^3/6 = (1/6)[(\pi/2)^3 - \alpha^3]$ | A1A1 |

$\Rightarrow \alpha^3 = \pi^3/16 \Rightarrow \alpha = \pi(16)^{-1/3}$, or equivalent | A1 |
5 Draw a sketch of the curve $C$ whose polar equation is $r = \theta$, for $0 \leqslant \theta \leqslant \frac { 1 } { 2 } \pi$.

On the same diagram draw the line $\theta = \alpha$, where $0 < \alpha < \frac { 1 } { 2 } \pi$.

The region bounded by $C$ and the line $\theta = \frac { 1 } { 2 } \pi$ is denoted by $R$. Find the exact value of $\alpha$ for which the line $\theta = \alpha$ divides $R$ into two regions of equal area.

\hfill \mbox{\textit{CAIE FP1 2009 Q5 [7]}}