| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2007 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Area of region with line boundary |
| Difficulty | Challenging +1.2 This is a multi-part Further Maths question covering standard FP2 techniques: polar area calculation with exponential curve, standard arctangent integration, and Maclaurin series application. While it requires competence across multiple topics and careful algebraic manipulation (especially the polar area involving the straight line boundary), each component is a textbook-style exercise without requiring novel insight. The polar area is the most challenging part, requiring subtraction of a triangular region, but this is a standard FP2 technique. Overall, moderately harder than average A-level due to being Further Maths content and multi-step reasoning. |
| Spec | 1.08d Evaluate definite integrals: between limits4.08b Standard Maclaurin series: e^x, sin, cos, ln(1+x), (1+x)^n4.08e Mean value of function: using integral4.09a Polar coordinates: convert to/from cartesian4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Spiral starting at \(r = a\) when \(\theta = 0\), decreasing to \(r = ae^{-k\pi}\) at \(\theta = \pi\), correct shape | B1, B1 | Correct endpoints A and B marked |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Area} = \frac{1}{2}\int_0^{\pi} r^2 \, d\theta - \text{triangle}\) | M1 | Setting up area integral |
| \(= \frac{1}{2}\int_0^{\pi} a^2e^{-2k\theta} \, d\theta\) | M1 | Correct integrand |
| \(= \frac{a^2}{4k}(1 - e^{-2k\pi})\) minus triangle area | A1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \frac{1}{3}\int_0^{\frac{1}{2}} \frac{1}{1+\frac{4x^2}{3}} \, dx\) | M1 | Factoring out \(\frac{1}{3}\) |
| Let \(x = \frac{\sqrt{3}}{2}\tan\theta\) or use standard form \(\frac{1}{a^2+b^2x^2}\) | M1 | Correct substitution |
| \(= \left[\frac{1}{2\sqrt{3}}\arctan\left(\frac{2x}{\sqrt{3}}\right)\right]_0^{\frac{1}{2}}\) | A1 | Correct integral form |
| \(= \frac{1}{2\sqrt{3}}\arctan\left(\frac{1}{\sqrt{3}}\right) = \frac{1}{2\sqrt{3}} \cdot \frac{\pi}{6} = \frac{\pi}{12\sqrt{3}}\) | A1, A1 | Exact value |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(0)=0,\ f'(x)=\sec^2 x,\ f'(0)=1\) | M1 | Differentiating |
| \(f''(x)=2\sec^2 x \tan x,\ f''(0)=0\) | A1 | |
| \(f'''(x)=2\sec^4 x+4\sec^2 x\tan^2 x,\ f'''(0)=2\) | A1 | |
| \(\tan x \approx x + \frac{x^3}{3}\) | A1 | Correct series |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{\tan x}{x} \approx 1 + \frac{x^2}{3}\) | M1 | Using series |
| \(\int_h^{4h}\left(1+\frac{x^2}{3}\right)dx = \left[x + \frac{x^3}{9}\right]_h^{4h}\) | M1 | Integrating |
| \(= 3h + \frac{64h^3 - h^3}{9} = 3h + 7h^3\) | A1 | Correct result |
## Question 1:
**Part (a)(i):** Sketch $r = ae^{-k\theta}$ for $0 \leq \theta \leq \pi$
| Spiral starting at $r = a$ when $\theta = 0$, decreasing to $r = ae^{-k\pi}$ at $\theta = \pi$, correct shape | B1, B1 | Correct endpoints A and B marked |
**Part (a)(ii):** Area enclosed by curve and line AB
| $\text{Area} = \frac{1}{2}\int_0^{\pi} r^2 \, d\theta - \text{triangle}$ | M1 | Setting up area integral |
| $= \frac{1}{2}\int_0^{\pi} a^2e^{-2k\theta} \, d\theta$ | M1 | Correct integrand |
| $= \frac{a^2}{4k}(1 - e^{-2k\pi})$ minus triangle area | A1, A1 | |
**Part (b):** $\int_0^{\frac{1}{2}} \frac{1}{3+4x^2} \, dx$
| $= \frac{1}{3}\int_0^{\frac{1}{2}} \frac{1}{1+\frac{4x^2}{3}} \, dx$ | M1 | Factoring out $\frac{1}{3}$ |
| Let $x = \frac{\sqrt{3}}{2}\tan\theta$ or use standard form $\frac{1}{a^2+b^2x^2}$ | M1 | Correct substitution |
| $= \left[\frac{1}{2\sqrt{3}}\arctan\left(\frac{2x}{\sqrt{3}}\right)\right]_0^{\frac{1}{2}}$ | A1 | Correct integral form |
| $= \frac{1}{2\sqrt{3}}\arctan\left(\frac{1}{\sqrt{3}}\right) = \frac{1}{2\sqrt{3}} \cdot \frac{\pi}{6} = \frac{\pi}{12\sqrt{3}}$ | A1, A1 | Exact value |
**Part (c)(i):** Maclaurin series for $\tan x$ up to $x^3$
| $f(0)=0,\ f'(x)=\sec^2 x,\ f'(0)=1$ | M1 | Differentiating |
| $f''(x)=2\sec^2 x \tan x,\ f''(0)=0$ | A1 | |
| $f'''(x)=2\sec^4 x+4\sec^2 x\tan^2 x,\ f'''(0)=2$ | A1 | |
| $\tan x \approx x + \frac{x^3}{3}$ | A1 | Correct series |
**Part (c)(ii):** Show $\int_h^{4h} \frac{\tan x}{x} \, dx \approx 3h + 7h^3$
| $\frac{\tan x}{x} \approx 1 + \frac{x^2}{3}$ | M1 | Using series |
| $\int_h^{4h}\left(1+\frac{x^2}{3}\right)dx = \left[x + \frac{x^3}{9}\right]_h^{4h}$ | M1 | Integrating |
| $= 3h + \frac{64h^3 - h^3}{9} = 3h + 7h^3$ | A1 | Correct result |
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\begin{enumerate}[label=(\alph*)]
\item A curve has polar equation $r = a \mathrm { e } ^ { - k \theta }$ for $0 \leqslant \theta \leqslant \pi$, where $a$ and $k$ are positive constants. The points A and B on the curve correspond to $\theta = 0$ and $\theta = \pi$ respectively.
\begin{enumerate}[label=(\roman*)]
\item Sketch the curve.
\item Find the area of the region enclosed by the curve and the line AB .
\end{enumerate}\item Find the exact value of $\int _ { 0 } ^ { \frac { 1 } { 2 } } \frac { 1 } { 3 + 4 x ^ { 2 } } \mathrm {~d} x$.
\item \begin{enumerate}[label=(\roman*)]
\item Find the Maclaurin series for $\tan x$, up to the term in $x ^ { 3 }$.
\item Use this Maclaurin series to show that, when $h$ is small, $\int _ { h } ^ { 4 h } \frac { \tan x } { x } \mathrm {~d} x \approx 3 h + 7 h ^ { 3 }$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI FP2 2007 Q1 [18]}}