| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2011 |
| 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 Further Maths question combining polar curve sketching and area calculation with non-trivial integration. Part (a) requires understanding of cardioid curves and applying the polar area formula ∫½r²dθ, yielding 3πa²/2. Part (b) involves recognizing arctangent and trigonometric substitution patterns. While these are standard FP2 techniques, the combination of topics and exact form requirements elevate this above average A-level difficulty but remain within expected Further Maths scope. |
| Spec | 4.08h Integration: inverse trig/hyperbolic substitutions4.09a Polar coordinates: convert to/from cartesian4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Correct general shape including symmetry in vertical axis | G1 | |
| Correct form at O and no extra sections | G1 | Dependent on first G1. For an otherwise correct curve with a sharp point at the bottom, award G1G0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\text{Area} = \frac{1}{2}a^2\int_0^{2\pi}(1-\sin\theta)^2\,d\theta\) | M1 | Integral expression involving \((1-\sin\theta)^2\) |
| \(= \frac{1}{2}a^2\int_0^{2\pi}(1-2\sin\theta+\sin^2\theta)\,d\theta\) | M1, A1 | Expanding; correct integral expression, incl. limits |
| \(= \frac{1}{2}a^2\int_0^{2\pi}\left(\frac{3}{2}-2\sin\theta-\frac{1}{2}\cos 2\theta\right)d\theta\) | M1 | Using \(\sin^2\theta = \frac{1}{2}-\frac{1}{2}\cos 2\theta\) |
| \(= \frac{1}{2}a^2\left[\frac{3}{2}\theta+2\cos\theta-\frac{1}{4}\sin 2\theta\right]_0^{2\pi}\) | A2 | Correct result of integration; give A1 for one error |
| \(= \frac{3}{2}\pi a^2\) | A1 | Dependent on previous A2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\int_{-\frac{1}{2}}^{\frac{1}{2}}\frac{1}{1+4x^2}\,dx = \frac{1}{4}\int_{-\frac{1}{2}}^{\frac{1}{2}}\frac{1}{\frac{1}{4}+x^2}\,dx = \frac{1}{4}\left[2\arctan 2x\right]_{-\frac{1}{2}}^{\frac{1}{2}}\) | M1, A1 | arctan alone, or any tan substitution; \(\frac{1}{4}\times 2\) and \(2x\) |
| \(= \frac{1}{2}\left(\frac{\pi}{4}-\left(-\frac{\pi}{4}\right)\right)\) | ||
| \(= \frac{\pi}{4}\) | A1 | Evaluated in terms of \(\pi\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x = \frac{1}{2}\tan\theta \Rightarrow dx = \frac{1}{2}\sec^2\theta\,d\theta\) | M1 | Any tan substitution |
| \(\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{1}{(\sec^2\theta)^{\frac{3}{2}}}\times\frac{\sec^2\theta}{2}\,d\theta\) | A1A1 | \(\frac{1}{(\sec^2\theta)^{\frac{3}{2}}}\), \(\frac{\sec^2\theta}{2}\) |
| \(= \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{1}{2}\cos\theta\,d\theta\) | ||
| \(= \left[\frac{1}{2}\sin\theta\right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\) | M1, A1ft | Integrating \(a\cos b\theta\) using consistent limits; \(\frac{a}{b}\sin b\theta\) |
| \(= \frac{1}{2}\left(\frac{1}{\sqrt{2}}-\left(-\frac{1}{\sqrt{2}}\right)\right)\) | ||
| \(= \frac{1}{\sqrt{2}}\) | A1 |
# Question 1:
## Part (a)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| Correct general shape including symmetry in vertical axis | G1 | |
| Correct form at O and no extra sections | G1 | Dependent on first G1. For an otherwise correct curve with a sharp point at the bottom, award G1G0 |
## Part (a)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{Area} = \frac{1}{2}a^2\int_0^{2\pi}(1-\sin\theta)^2\,d\theta$ | M1 | Integral expression involving $(1-\sin\theta)^2$ |
| $= \frac{1}{2}a^2\int_0^{2\pi}(1-2\sin\theta+\sin^2\theta)\,d\theta$ | M1, A1 | Expanding; correct integral expression, incl. limits |
| $= \frac{1}{2}a^2\int_0^{2\pi}\left(\frac{3}{2}-2\sin\theta-\frac{1}{2}\cos 2\theta\right)d\theta$ | M1 | Using $\sin^2\theta = \frac{1}{2}-\frac{1}{2}\cos 2\theta$ |
| $= \frac{1}{2}a^2\left[\frac{3}{2}\theta+2\cos\theta-\frac{1}{4}\sin 2\theta\right]_0^{2\pi}$ | A2 | Correct result of integration; give A1 for one error |
| $= \frac{3}{2}\pi a^2$ | A1 | Dependent on previous A2 |
## Part (b)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\int_{-\frac{1}{2}}^{\frac{1}{2}}\frac{1}{1+4x^2}\,dx = \frac{1}{4}\int_{-\frac{1}{2}}^{\frac{1}{2}}\frac{1}{\frac{1}{4}+x^2}\,dx = \frac{1}{4}\left[2\arctan 2x\right]_{-\frac{1}{2}}^{\frac{1}{2}}$ | M1, A1 | arctan alone, or any tan substitution; $\frac{1}{4}\times 2$ and $2x$ |
| $= \frac{1}{2}\left(\frac{\pi}{4}-\left(-\frac{\pi}{4}\right)\right)$ | | |
| $= \frac{\pi}{4}$ | A1 | Evaluated in terms of $\pi$ |
## Part (b)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $x = \frac{1}{2}\tan\theta \Rightarrow dx = \frac{1}{2}\sec^2\theta\,d\theta$ | M1 | Any tan substitution |
| $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{1}{(\sec^2\theta)^{\frac{3}{2}}}\times\frac{\sec^2\theta}{2}\,d\theta$ | A1A1 | $\frac{1}{(\sec^2\theta)^{\frac{3}{2}}}$, $\frac{\sec^2\theta}{2}$ |
| $= \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{1}{2}\cos\theta\,d\theta$ | | |
| $= \left[\frac{1}{2}\sin\theta\right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}}$ | M1, A1ft | Integrating $a\cos b\theta$ using consistent limits; $\frac{a}{b}\sin b\theta$ |
| $= \frac{1}{2}\left(\frac{1}{\sqrt{2}}-\left(-\frac{1}{\sqrt{2}}\right)\right)$ | | |
| $= \frac{1}{\sqrt{2}}$ | A1 | |
---1
\begin{enumerate}[label=(\alph*)]
\item A curve has polar equation $r = a ( 1 - \sin \theta )$, where $a > 0$ and $0 \leqslant \theta < 2 \pi$.
\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, in an exact form, the value of the integral $\int _ { - \frac { 1 } { 2 } } ^ { \frac { 1 } { 2 } } \frac { 1 } { 1 + 4 x ^ { 2 } } \mathrm {~d} x$.
\item Find, in an exact form, the value of the integral $\int _ { - \frac { 1 } { 2 } } ^ { \frac { 1 } { 2 } } \frac { 1 } { \left( 1 + 4 x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } } \mathrm {~d} x$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI FP2 2011 Q1 [18]}}