| Exam Board | Edexcel |
|---|---|
| Module | CP2 (Core Pure 2) |
| Session | Specimen |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Argand & Loci |
| Type | Area calculations in complex plane |
| Difficulty | Challenging +1.8 This question requires multiple sophisticated techniques: converting between Cartesian and polar forms of a circle, finding intersection points of a circle with a ray, and calculating an area using polar integration. The polar conversion proof and exact area calculation (likely requiring integration of r²/2 with trigonometric substitution) demand strong technical facility beyond routine A-level, though the individual components are within Core Pure 2 scope. |
| Spec | 4.02k Argand diagrams: geometric interpretation4.09a Polar coordinates: convert to/from cartesian4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Circle drawn passing through the origin | M1 | Draws a circle which passes through the origin |
| Fully correct diagram (circle centred in first quadrant) | A1 | Fully correct diagram |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \( | z-4-3i | =5 \Rightarrow |
| \((x-4)^2+(y-3)^2=25\) or any correct form | A1 | Correct equation in terms of \(x\) and \(y\) |
| \((r\cos\theta-4)^2+(r\sin\theta-3)^2=25 \Rightarrow r^2\cos^2\theta-8r\cos\theta+16+r^2\sin^2\theta-6r\sin\theta+9=25 \Rightarrow r^2-8r\cos\theta-6r\sin\theta=0\) | M1 | Introduces polar form, expands and uses \(\cos^2\theta+\sin^2\theta=1\) |
| \(\therefore r=8\cos\theta+6\sin\theta\) | A1* | Deduces the given equation (ignore reference to \(r=0\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct pair of rays added to diagram | B1 | Correct pair of rays added to their diagram |
| Area between pair of rays and inside circle shaded | B1ft | Shaded region correct, as long as there is an intersection |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(A=\frac{1}{2}\int r^2\,d\theta=\frac{1}{2}\int(8\cos\theta+6\sin\theta)^2\,d\theta = \frac{1}{2}\int(64\cos^2\theta+96\sin\theta\cos\theta+36\sin^2\theta)\,d\theta\) | M1 | Selects appropriate method using polar area formula |
| \(=\frac{1}{2}\int\left(32(\cos2\theta+1)+96\sin\theta\cos\theta+18(1-\cos2\theta)\right)d\theta\) | M1 | Uses double angle identities |
| \(=\frac{1}{2}\int(14\cos2\theta+50+48\sin2\theta)\,d\theta\) | A1 | Correct integral |
| \(=\frac{1}{2}\left[7\sin2\theta+50\theta-24\cos2\theta\right]_0^{\frac{\pi}{3}}=\frac{1}{2}\left\{\left(\frac{7\sqrt{3}}{2}+\frac{50\pi}{3}+12\right)-(-24)\right\}\) | M1 | Integrates and applies limits |
| \(=\frac{7\sqrt{3}}{4}+\frac{25\pi}{3}+18\) | A1 | Correct area |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Angle \(ACB=\frac{2\pi}{3}\), area sector \(ACB=\frac{1}{2}(5)^2\cdot\frac{2\pi}{3}\); area triangle \(OCB=\frac{1}{2}\times8\times3\) | M1 | Finds angle \(ACB\), area of sector \(ACB\) and area of triangle \(OCB\) |
| Sector area \(ACB\) + triangle area \(OCB = \frac{25\pi}{3}+12\) | A1 | Correct combined area |
| Angle \(ACO=2\pi-\frac{2\pi}{3}-\cos^{-1}\!\left(\frac{5^2+5^2-8^2}{2\times5\times5}\right)\); area \(OAC=\frac{1}{2}(5)^2\sin\!\left(\frac{4\pi}{3}-\cos^{-1}\!\left(\frac{-7}{25}\right)\right)\) | M1 | Starts finding area of triangle \(OAC\) by calculating angle \(ACO\) |
| \(=\frac{25}{2}\left(\sin\frac{4\pi}{3}\cos\!\left(\cos^{-1}\!\frac{-7}{25}\right)-\cos\frac{4\pi}{3}\sin\!\left(\cos^{-1}\!\frac{-7}{25}\right)\right) =\frac{25}{2}\left(\frac{7\sqrt{3}}{50}+\frac{1}{2}\sqrt{1-\left(\frac{7}{25}\right)^2}\right)=\frac{7\sqrt{3}}{4}+6\) | M1 | Uses addition formula; full correct method for area of shaded region |
| Total area \(=\frac{25\pi}{3}+\frac{1}{2}\times8\times3+6+\frac{7\sqrt{3}}{4}=\frac{7\sqrt{3}}{4}+\frac{25\pi}{3}+18\) | A1 | Correct area |
## Question 6:
### Part (a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Circle drawn passing through the origin | M1 | Draws a circle which passes through the origin |
| Fully correct diagram (circle centred in first quadrant) | A1 | Fully correct diagram |
### Part (a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $|z-4-3i|=5 \Rightarrow |x+iy-4-3i|=5 \Rightarrow (x-4)^2+(y-3)^2=\ldots$ | M1 | Uses $z=x+iy$ and uses modulus to find equation in $x$ and $y$ only |
| $(x-4)^2+(y-3)^2=25$ or any correct form | A1 | Correct equation in terms of $x$ and $y$ |
| $(r\cos\theta-4)^2+(r\sin\theta-3)^2=25 \Rightarrow r^2\cos^2\theta-8r\cos\theta+16+r^2\sin^2\theta-6r\sin\theta+9=25 \Rightarrow r^2-8r\cos\theta-6r\sin\theta=0$ | M1 | Introduces polar form, expands and uses $\cos^2\theta+\sin^2\theta=1$ |
| $\therefore r=8\cos\theta+6\sin\theta$ | A1* | Deduces the given equation (ignore reference to $r=0$) |
**(6 marks total)**
---
### Part (b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct pair of rays added to diagram | B1 | Correct pair of rays added to their diagram |
| Area between pair of rays and inside circle shaded | B1ft | Shaded region correct, as long as there is an intersection |
---
### Part (b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $A=\frac{1}{2}\int r^2\,d\theta=\frac{1}{2}\int(8\cos\theta+6\sin\theta)^2\,d\theta = \frac{1}{2}\int(64\cos^2\theta+96\sin\theta\cos\theta+36\sin^2\theta)\,d\theta$ | M1 | Selects appropriate method using polar area formula |
| $=\frac{1}{2}\int\left(32(\cos2\theta+1)+96\sin\theta\cos\theta+18(1-\cos2\theta)\right)d\theta$ | M1 | Uses double angle identities |
| $=\frac{1}{2}\int(14\cos2\theta+50+48\sin2\theta)\,d\theta$ | A1 | Correct integral |
| $=\frac{1}{2}\left[7\sin2\theta+50\theta-24\cos2\theta\right]_0^{\frac{\pi}{3}}=\frac{1}{2}\left\{\left(\frac{7\sqrt{3}}{2}+\frac{50\pi}{3}+12\right)-(-24)\right\}$ | M1 | Integrates and applies limits |
| $=\frac{7\sqrt{3}}{4}+\frac{25\pi}{3}+18$ | A1 | Correct area |
**(7 marks total)**
---
### Part (b)(ii) Alternative:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Angle $ACB=\frac{2\pi}{3}$, area sector $ACB=\frac{1}{2}(5)^2\cdot\frac{2\pi}{3}$; area triangle $OCB=\frac{1}{2}\times8\times3$ | M1 | Finds angle $ACB$, area of sector $ACB$ and area of triangle $OCB$ |
| Sector area $ACB$ + triangle area $OCB = \frac{25\pi}{3}+12$ | A1 | Correct combined area |
| Angle $ACO=2\pi-\frac{2\pi}{3}-\cos^{-1}\!\left(\frac{5^2+5^2-8^2}{2\times5\times5}\right)$; area $OAC=\frac{1}{2}(5)^2\sin\!\left(\frac{4\pi}{3}-\cos^{-1}\!\left(\frac{-7}{25}\right)\right)$ | M1 | Starts finding area of triangle $OAC$ by calculating angle $ACO$ |
| $=\frac{25}{2}\left(\sin\frac{4\pi}{3}\cos\!\left(\cos^{-1}\!\frac{-7}{25}\right)-\cos\frac{4\pi}{3}\sin\!\left(\cos^{-1}\!\frac{-7}{25}\right)\right) =\frac{25}{2}\left(\frac{7\sqrt{3}}{50}+\frac{1}{2}\sqrt{1-\left(\frac{7}{25}\right)^2}\right)=\frac{7\sqrt{3}}{4}+6$ | M1 | Uses addition formula; full correct method for area of shaded region |
| Total area $=\frac{25\pi}{3}+\frac{1}{2}\times8\times3+6+\frac{7\sqrt{3}}{4}=\frac{7\sqrt{3}}{4}+\frac{25\pi}{3}+18$ | A1 | Correct area |
**(13 marks total for Question 6)**
---\begin{enumerate}
\item (a) (i) Show on an Argand diagram the locus of points given by the values of $z$ satisfying
\end{enumerate}
$$| z - 4 - 3 \mathbf { i } | = 5$$
Taking the initial line as the positive real axis with the pole at the origin and given that $\theta \in [ \alpha , \alpha + \pi ]$, where $\alpha = - \arctan \left( \frac { 4 } { 3 } \right)$,\\
(ii) show that this locus of points can be represented by the polar curve with equation
$$r = 8 \cos \theta + 6 \sin \theta$$
The set of points $A$ is defined by
$$A = \left\{ z : 0 \leqslant \arg z \leqslant \frac { \pi } { 3 } \right\} \cap \{ z : | z - 4 - 3 \mathbf { i } | \leqslant 5 \}$$
(b) (i) Show, by shading on your Argand diagram, the set of points $A$.\\
(ii) Find the exact area of the region defined by $A$, giving your answer in simplest form.
\hfill \mbox{\textit{Edexcel CP2 Q6 [13]}}