OCR FP2 2006 June — Question 7 11 marks

Exam BoardOCR
ModuleFP2 (Further Pure Mathematics 2)
Year2006
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPolar coordinates
TypeArea of region with line boundary
DifficultyStandard +0.8 This is a multi-part Further Maths polar coordinates question requiring: finding tangent at pole (solving r=0 with tan θ), identifying maximum r, sketching an unfamiliar curve (r = √3 + tan θ), and computing area with the polar integral formula. While the individual techniques are standard for FP2, the curve itself is non-standard and the multi-step nature with 4 parts requiring different skills makes this moderately challenging for Further Maths students.
Spec4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve

7 The equation of a curve, in polar coordinates, is $$r = \sqrt { 3 } + \tan \theta , \quad \text { for } - \frac { 1 } { 3 } \pi \leqslant \theta \leqslant \frac { 1 } { 4 } \pi$$
  1. Find the equation of the tangent at the pole.
  2. State the greatest value of \(r\) and the corresponding value of \(\theta\).
  3. Sketch the curve.
  4. Find the exact area of the region enclosed by the curve and the lines \(\theta = 0\) and \(\theta = \frac { 1 } { 4 } \pi\).

AnswerMarks Guidance
(i) Attempt to solve \(r=0\), \(\tan \theta = - \sqrt{3}\)M1 Allow \(\pm \sqrt{3}\)
Get \(\theta = -\frac{1}{3}\pi\) onlyA1 Allow \(-60°\)
(ii) \(r = \sqrt{3} + 1\) when \(\theta = \frac{1}{4}\pi\)B1, B1 AEF for \(r\), \(45°\) for \(\theta\)
(iii)B1 Correct \(r\) at correct end-values of \(\theta\); Ignore extra \(\theta\) used
[Graph showing correct shape]B1 Correct shape with \(r\) not decreasing
(iv) Formula with correct \(r\) usedM1 \(l^2\) may be implied
Replace \(\tan^2 \theta = \sec^2 \theta - 1\)B1
Attempt to integrate their expressionM1 Must be 3 different terms leading to any 2 of \(a\theta + b\ln(\sec\theta/\cos\theta) + c\tan\theta\)
Get \(\theta + \sqrt{3}\ln\sec\theta + \frac{1}{2}\tan\theta\)A1 Condone answer x2 if \(\frac{1}{2}\) seen elsewhere
Correct limits to \(\frac{1}{4}\pi + \sqrt{3}\ln 2 + \frac{1}{2}\)A1 cao; AEF
**(i)** Attempt to solve $r=0$, $\tan \theta = - \sqrt{3}$ | M1 | Allow $\pm \sqrt{3}$
Get $\theta = -\frac{1}{3}\pi$ only | A1 | Allow $-60°$

**(ii)** $r = \sqrt{3} + 1$ when $\theta = \frac{1}{4}\pi$ | B1, B1 | AEF for $r$, $45°$ for $\theta$

**(iii)** | B1 | Correct $r$ at correct end-values of $\theta$; Ignore extra $\theta$ used
[Graph showing correct shape] | B1 | Correct shape with $r$ not decreasing

**(iv)** Formula with correct $r$ used | M1 | $l^2$ may be implied
Replace $\tan^2 \theta = \sec^2 \theta - 1$ | B1 |
Attempt to integrate their expression | M1 | Must be 3 different terms leading to any 2 of $a\theta + b\ln(\sec\theta/\cos\theta) + c\tan\theta$
Get $\theta + \sqrt{3}\ln\sec\theta + \frac{1}{2}\tan\theta$ | A1 | Condone answer x2 if $\frac{1}{2}$ seen elsewhere
Correct limits to $\frac{1}{4}\pi + \sqrt{3}\ln 2 + \frac{1}{2}$ | A1 | cao; AEF
7 The equation of a curve, in polar coordinates, is

$$r = \sqrt { 3 } + \tan \theta , \quad \text { for } - \frac { 1 } { 3 } \pi \leqslant \theta \leqslant \frac { 1 } { 4 } \pi$$

(i) Find the equation of the tangent at the pole.\\
(ii) State the greatest value of $r$ and the corresponding value of $\theta$.\\
(iii) Sketch the curve.\\
(iv) Find the exact area of the region enclosed by the curve and the lines $\theta = 0$ and $\theta = \frac { 1 } { 4 } \pi$.

\hfill \mbox{\textit{OCR FP2 2006 Q7 [11]}}