| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Session | Specimen |
| Marks | 14 |
| Topic | Polar coordinates |
| Type | Area of region with line boundary |
| Difficulty | Standard +0.3 This is a straightforward polar area problem requiring sketching r=θ (Archimedean spiral), setting up the standard polar area integral ½∫r²dθ, and solving a simple equation to find where areas are equal. The integration is routine (∫θ²dθ) and the algebra is elementary, making this easier than average despite being a Further Maths topic. |
| Spec | 4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
**(i)** Sketch with the correct shape, location and orientation B1
Shows tangency to the initial line at the pole (lenient) B1 **[2]**
**(ii)** Draws, in the same diagram, a straight line starting at the origin and with positive gradient (must be a half line) B1 **[1]**
**(iii)** EITHER: $\frac{1}{2}\int_0^{\frac{1}{2}\pi} \theta^2\, d\theta = \frac{1}{48}\pi^3$ M1A1
$\frac{1}{2}\int_0^\alpha \theta^2\, d\theta = \frac{1}{6}\alpha^3$ A1
$\frac{1}{6}\alpha^3 = \frac{1}{96}\pi^3 \Rightarrow \alpha = \frac{\pi}{\sqrt[3]{16}}$ (acf) M1A1 **[5]**
OR: $\frac{1}{2}\int_0^\alpha \theta^2\, d\theta = \frac{1}{2}\int_\alpha^{\frac{1}{2}\pi} \theta^2\, d\theta$ M2
$\frac{1}{6}\alpha^3 = \frac{1}{6}\left[(\frac{1}{2}\pi)^3 - \alpha^3\right]$ A1A1
$\Rightarrow \alpha^3 = \frac{1}{16}\pi^3 \Rightarrow \alpha = \frac{\pi}{\sqrt[3]{16}}$ (acf) A1 **(5)**4 (i) Draw a sketch of the curve $C$ whose polar equation is $r = \theta$, for $0 \leqslant \theta \leqslant \frac { 1 } { 2 } \pi$.\\
(ii) On the same diagram draw the line $\theta = \alpha$, where $0 < \alpha < \frac { 1 } { 2 } \pi$.
The region bounded by $C$ and the line $\theta = \frac { 1 } { 2 } \pi$ is denoted by $R$.\\
(iii) Find the exact value of $\alpha$ for which the line $\theta = \alpha$ divides $R$ into two regions of equal area.
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 Q4 [14]}}