| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2006 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Sketch polar curve |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring: (i) sketching a three-petaled rose curve with sign conventions, (ii) polar area integration with cos²3θ requiring double angle formulas, (iii) trigonometric substitutions for non-standard integrals. While these are standard FP2 techniques, the combination of polar curves (less familiar than Cartesian), careful handling of negative r values, and multiple integration techniques requiring algebraic manipulation places this above average difficulty for A-level, though still within expected FP2 scope. |
| Spec | 1.08h Integration by substitution4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Three loops, roughly correct shape | B1 | |
| Continuous line for \(r > 0\) (loops at \(\theta = 0, \pm\frac{2\pi}{3}\)) | B1 | |
| Broken line for \(r < 0\) (loops at \(\theta = \pm\frac{\pi}{3}\)) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(A = \frac{1}{2}\int r^2 \, d\theta\) | M1 | Correct formula stated or implied |
| \(= \frac{1}{2}\int_{-\pi/6}^{\pi/6} a^2\cos^2 3\theta \, d\theta\) | A1 | Correct limits and integrand |
| \(= \frac{a^2}{2}\int_{-\pi/6}^{\pi/6} \frac{1}{2}(1 + \cos 6\theta) \, d\theta\) | M1 | Using double angle formula |
| \(= \frac{a^2}{4}\left[\theta + \frac{\sin 6\theta}{6}\right]_{-\pi/6}^{\pi/6}\) | A1 | Correct integration |
| \(= \frac{\pi a^2}{12}\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{\sqrt{3-4x^2}} = \frac{1}{\sqrt{3}\sqrt{1-\frac{4x^2}{3}}}\), let \(x = \frac{\sqrt{3}}{2}\sin\theta\) | M1 | Suitable substitution |
| \(dx = \frac{\sqrt{3}}{2}\cos\theta \, d\theta\), denominator \(= \sqrt{3}\cos\theta\) | A1 | Correct derivative and simplification |
| Integral \(= \frac{1}{2}\int d\theta\) | A1 | Simplification to \(\frac{1}{2}\) |
| Limits: \(x=0 \Rightarrow \theta=0\); \(x=\frac{3}{4} \Rightarrow \theta = \frac{\pi}{3}\) | M1 | Changing limits correctly |
| \(= \frac{\pi}{6}\) | A1 | Correct exact answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Let \(x = \frac{1}{\sqrt{3}}\tan\theta\) | M1 | Correct substitution stated |
| \(1 + 3x^2 = \sec^2\theta\), \(dx = \frac{1}{\sqrt{3}}\sec^2\theta \, d\theta\) | A1 | Correct expressions |
| Integral \(= \int \frac{\frac{1}{\sqrt{3}}\sec^2\theta}{\sec^3\theta} d\theta = \frac{1}{\sqrt{3}}\int\cos\theta \, d\theta\) | M1 | Correct simplification |
| \(= \frac{1}{\sqrt{3}}[\sin\theta]\), limits \(0\) to \(\frac{\pi}{3}\) | A1 | Correct integration with limits |
| \(= \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{2} = \frac{1}{2}\) | A1 | Correct exact answer |
# Question 1:
## Part (a)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Three loops, roughly correct shape | B1 | |
| Continuous line for $r > 0$ (loops at $\theta = 0, \pm\frac{2\pi}{3}$) | B1 | |
| Broken line for $r < 0$ (loops at $\theta = \pm\frac{\pi}{3}$) | B1 | |
## Part (a)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $A = \frac{1}{2}\int r^2 \, d\theta$ | M1 | Correct formula stated or implied |
| $= \frac{1}{2}\int_{-\pi/6}^{\pi/6} a^2\cos^2 3\theta \, d\theta$ | A1 | Correct limits and integrand |
| $= \frac{a^2}{2}\int_{-\pi/6}^{\pi/6} \frac{1}{2}(1 + \cos 6\theta) \, d\theta$ | M1 | Using double angle formula |
| $= \frac{a^2}{4}\left[\theta + \frac{\sin 6\theta}{6}\right]_{-\pi/6}^{\pi/6}$ | A1 | Correct integration |
| $= \frac{\pi a^2}{12}$ | A1 | Correct answer |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{\sqrt{3-4x^2}} = \frac{1}{\sqrt{3}\sqrt{1-\frac{4x^2}{3}}}$, let $x = \frac{\sqrt{3}}{2}\sin\theta$ | M1 | Suitable substitution |
| $dx = \frac{\sqrt{3}}{2}\cos\theta \, d\theta$, denominator $= \sqrt{3}\cos\theta$ | A1 | Correct derivative and simplification |
| Integral $= \frac{1}{2}\int d\theta$ | A1 | Simplification to $\frac{1}{2}$ |
| Limits: $x=0 \Rightarrow \theta=0$; $x=\frac{3}{4} \Rightarrow \theta = \frac{\pi}{3}$ | M1 | Changing limits correctly |
| $= \frac{\pi}{6}$ | A1 | Correct exact answer |
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Let $x = \frac{1}{\sqrt{3}}\tan\theta$ | M1 | Correct substitution stated |
| $1 + 3x^2 = \sec^2\theta$, $dx = \frac{1}{\sqrt{3}}\sec^2\theta \, d\theta$ | A1 | Correct expressions |
| Integral $= \int \frac{\frac{1}{\sqrt{3}}\sec^2\theta}{\sec^3\theta} d\theta = \frac{1}{\sqrt{3}}\int\cos\theta \, d\theta$ | M1 | Correct simplification |
| $= \frac{1}{\sqrt{3}}[\sin\theta]$, limits $0$ to $\frac{\pi}{3}$ | A1 | Correct integration with limits |
| $= \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{2} = \frac{1}{2}$ | A1 | Correct exact answer |
---1
\begin{enumerate}[label=(\alph*)]
\item A curve has polar equation $r = a \cos 3 \theta$ for $- \frac { 1 } { 2 } \pi \leqslant \theta \leqslant \frac { 1 } { 2 } \pi$, where $a$ is a positive constant.
\begin{enumerate}[label=(\roman*)]
\item Sketch the curve, using a continuous line for sections where $r > 0$ and a broken line for sections where $r < 0$.
\item Find the area enclosed by one of the loops.
\end{enumerate}\item Find the exact value of $\int _ { 0 } ^ { \frac { 3 } { 4 } } \frac { 1 } { \sqrt { 3 - 4 x ^ { 2 } } } \mathrm {~d} x$.
\item Use a trigonometric substitution to find $\int _ { 0 } ^ { 1 } \frac { 1 } { \left( 1 + 3 x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } } \mathrm {~d} x$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI FP2 2006 Q1 [18]}}