| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2006 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Area enclosed by polar curve |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring polar curve sketching, area integration with trigonometric identities, Maclaurin series derivation using chain rule, and approximation of a definite integral. While each technique is standard for FP2, the combination of topics and the need for exact algebraic manipulation (especially the polar area integral requiring cos²θ expansion) places it moderately above average difficulty. |
| Spec | 4.08a Maclaurin series: find series for function4.08b Standard Maclaurin series: e^x, sin, cos, ln(1+x), (1+x)^n4.08d Volumes of revolution: about x and y axes4.09a Polar coordinates: convert to/from cartesian4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Correct sketch of curve (limaçon shape, passing through origin at \(\theta = \pm\frac{3}{4}\pi\), symmetric about initial line) | B2 | B1 for basic shape; B1 for correct intercepts/symmetry |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Area \(= \frac{1}{2}\int_{-3\pi/4}^{3\pi/4} r^2 \, d\theta\) | M1 | Correct integral form |
| \(r^2 = a^2(\sqrt{2}+2\cos\theta)^2 = a^2(2 + 4\sqrt{2}\cos\theta + 4\cos^2\theta)\) | A1 | Expansion |
| \(= a^2(2 + 4\sqrt{2}\cos\theta + 2 + 2\cos 2\theta)\) | M1 | Use of \(\cos^2\theta = \frac{1}{2}(1+\cos 2\theta)\) |
| Integrating: \(a^2\left[4\theta + 4\sqrt{2}\sin\theta + \sin 2\theta\right]_{0}^{3\pi/4}\) | M1 A1 | Correct integration |
| \(= a^2(3\pi + 4\sqrt{2}\cdot\frac{\sqrt{2}}{2} + \sin\frac{3\pi}{2}) \times 2\) ... | DM1 | Substituting limits |
| \(= a^2(3\pi + 4)\) | A1 | Exact final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f(0) = \tan\frac{\pi}{4} = 1\) | B1 | |
| \(f'(x) = \sec^2(\frac{\pi}{4}+x)\), \(f'(0) = 2\) | M1 A1 | |
| \(f''(x) = 2\sec^2(\frac{\pi}{4}+x)\tan(\frac{\pi}{4}+x)\), \(f''(0) = 4\) | M1 A1 | |
| \(f(x) \approx 1 + 2x + 2x^2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_{-h}^{h} x^2(1+2x+2x^2+\ldots)\,dx\) | M1 | Substituting series |
| Odd terms vanish on \([-h,h]\) | M1 | Symmetry argument |
| \(= \int_{-h}^{h}(x^2 + 2x^4)\,dx = \left[\frac{x^3}{3}+\frac{2x^5}{5}\right]_{-h}^{h}\) | A1 | |
| \(= \frac{2h^3}{3} + \frac{4h^5}{5}\) | A1 | Completion |
# Question 1:
## Part (a)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Correct sketch of curve (limaçon shape, passing through origin at $\theta = \pm\frac{3}{4}\pi$, symmetric about initial line) | B2 | B1 for basic shape; B1 for correct intercepts/symmetry |
## Part (a)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Area $= \frac{1}{2}\int_{-3\pi/4}^{3\pi/4} r^2 \, d\theta$ | M1 | Correct integral form |
| $r^2 = a^2(\sqrt{2}+2\cos\theta)^2 = a^2(2 + 4\sqrt{2}\cos\theta + 4\cos^2\theta)$ | A1 | Expansion |
| $= a^2(2 + 4\sqrt{2}\cos\theta + 2 + 2\cos 2\theta)$ | M1 | Use of $\cos^2\theta = \frac{1}{2}(1+\cos 2\theta)$ |
| Integrating: $a^2\left[4\theta + 4\sqrt{2}\sin\theta + \sin 2\theta\right]_{0}^{3\pi/4}$ | M1 A1 | Correct integration |
| $= a^2(3\pi + 4\sqrt{2}\cdot\frac{\sqrt{2}}{2} + \sin\frac{3\pi}{2}) \times 2$ ... | DM1 | Substituting limits |
| $= a^2(3\pi + 4)$ | A1 | Exact final answer |
## Part (b)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(0) = \tan\frac{\pi}{4} = 1$ | B1 | |
| $f'(x) = \sec^2(\frac{\pi}{4}+x)$, $f'(0) = 2$ | M1 A1 | |
| $f''(x) = 2\sec^2(\frac{\pi}{4}+x)\tan(\frac{\pi}{4}+x)$, $f''(0) = 4$ | M1 A1 | |
| $f(x) \approx 1 + 2x + 2x^2$ | A1 | |
## Part (b)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_{-h}^{h} x^2(1+2x+2x^2+\ldots)\,dx$ | M1 | Substituting series |
| Odd terms vanish on $[-h,h]$ | M1 | Symmetry argument |
| $= \int_{-h}^{h}(x^2 + 2x^4)\,dx = \left[\frac{x^3}{3}+\frac{2x^5}{5}\right]_{-h}^{h}$ | A1 | |
| $= \frac{2h^3}{3} + \frac{4h^5}{5}$ | A1 | Completion |
---1
\begin{enumerate}[label=(\alph*)]
\item A curve has polar equation $r = a ( \sqrt { 2 } + 2 \cos \theta )$ for $- \frac { 3 } { 4 } \pi \leqslant \theta \leqslant \frac { 3 } { 4 } \pi$, where $a$ is a positive constant.
\begin{enumerate}[label=(\roman*)]
\item Sketch the curve.
\item Find, in an exact form, the area of the region enclosed by the curve.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Find the Maclaurin series for the function $\mathrm { f } ( x ) = \tan \left( \frac { 1 } { 4 } \pi + x \right)$, up to the term in $x ^ { 2 }$.
\item Use the Maclaurin series to show that, when $h$ is small,
$$\int _ { - h } ^ { h } x ^ { 2 } \tan \left( \frac { 1 } { 4 } \pi + x \right) \mathrm { d } x \approx \frac { 2 } { 3 } h ^ { 3 } + \frac { 4 } { 5 } h ^ { 5 }$$
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI FP2 2006 Q1 [18]}}