OCR FP2 2009 January — Question 7 8 marks

Exam BoardOCR
ModuleFP2 (Further Pure Mathematics 2)
Year2009
SessionJanuary
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPolar coordinates
TypeArea of region with line boundary
DifficultyChallenging +1.3 This is a Further Pure 2 polar coordinates question requiring understanding of symmetry properties and polar area integration. Part (i) involves recognizing that perpendicular lines correspond to angles differing by π/2 and using the cosine addition formula—straightforward for FP2 students. Part (ii) is a standard polar area integral with substitution. While this requires FP2-specific knowledge (making it harder than typical A-level), it's a routine application of polar techniques without requiring novel insight or complex multi-step reasoning.
Spec4.09a Polar coordinates: convert to/from cartesian4.09c Area enclosed: by polar curve

\includegraphics{figure_7} The diagram shows the curve with equation, in polar coordinates, $$r = 3 + 2\cos \theta, \quad \text{for } 0 \leq \theta < 2\pi.$$ The points \(P\), \(Q\), \(R\) and \(S\) on the curve are such that the straight lines \(POR\) and \(QOS\) are perpendicular, where \(O\) is the pole. The point \(P\) has polar coordinates \((r, \alpha)\).
  1. Show that \(OP + OQ + OR + OS = k\), where \(k\) is a constant to be found. [3]
  2. Given that \(\alpha = \frac{1}{4}\pi\), find the exact area bounded by the curve and the lines \(OP\) and \(OQ\) (shaded in the diagram). [5]

AnswerMarks Guidance
(i) \(OP = 3 + 2\cos \alpha\). \(OQ = 3 + 2\cos(\frac{1}{2}\pi+\alpha) = 3 - 2\sin \alpha\). Similarly \(OR = 3 - 2\cos \alpha\). \(OS = 3 + 2\sin \alpha\). Sum = 12M1, M1, M1, A1 Any other unsimplified value; Attempt at simplification of at least two correct expressions; cao
(ii) Correct formula with attempt at \(r^2\). Square \(r\) correctly. Attempt to replace \(\cos^2\theta\) with \(a(\cos 2\theta \pm 1)\). Integrate their expression. Get \(\frac{1}{3}r^2\theta - 1\)M1, A1, M1, A1V, A1 Need not be expanded, but three terms if it is; Need three terms; cao
(i) $OP = 3 + 2\cos \alpha$. $OQ = 3 + 2\cos(\frac{1}{2}\pi+\alpha) = 3 - 2\sin \alpha$. Similarly $OR = 3 - 2\cos \alpha$. $OS = 3 + 2\sin \alpha$. Sum = 12 | M1, M1, M1, A1 | Any other unsimplified value; Attempt at simplification of at least two correct expressions; cao

(ii) Correct formula with attempt at $r^2$. Square $r$ correctly. Attempt to replace $\cos^2\theta$ with $a(\cos 2\theta \pm 1)$. Integrate their expression. Get $\frac{1}{3}r^2\theta - 1$ | M1, A1, M1, A1V, A1 | Need not be expanded, but three terms if it is; Need three terms; cao
\includegraphics{figure_7}

The diagram shows the curve with equation, in polar coordinates,
$$r = 3 + 2\cos \theta, \quad \text{for } 0 \leq \theta < 2\pi.$$

The points $P$, $Q$, $R$ and $S$ on the curve are such that the straight lines $POR$ and $QOS$ are perpendicular, where $O$ is the pole. The point $P$ has polar coordinates $(r, \alpha)$.

\begin{enumerate}[label=(\roman*)]
\item Show that $OP + OQ + OR + OS = k$, where $k$ is a constant to be found. [3]

\item Given that $\alpha = \frac{1}{4}\pi$, find the exact area bounded by the curve and the lines $OP$ and $OQ$ (shaded in the diagram). [5]
\end{enumerate}

\hfill \mbox{\textit{OCR FP2 2009 Q7 [8]}}