| Exam Board | OCR MEI |
|---|---|
| Module | Further Pure Core (Further Pure Core) |
| Year | 2023 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Argand & Loci |
| Type | Intersection of two loci |
| Difficulty | Challenging +1.2 This question requires understanding of loci in the complex plane and finding their intersection. Part (a) involves standard loci: a circle and a perpendicular bisector (half-plane). Part (b) requires recognizing that the unique solution occurs where the circle boundary meets the line boundary, then solving algebraically. While it involves multiple steps and geometric insight, the techniques are standard for Further Maths complex numbers topics, making it moderately above average difficulty. |
| Spec | 4.02k Argand diagrams: geometric interpretation4.02o Loci in Argand diagram: circles, half-lines |
| Answer | Marks | Guidance |
|---|---|---|
| 13 | (a) | (i) |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | circle, centre O |
| Answer | Marks | Guidance |
|---|---|---|
| 13 | (a) | (ii) |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | (–2, 4) and (2, 6) identified |
| Answer | Marks | Guidance |
|---|---|---|
| 13 | (b) | gradient = –2 |
| Answer | Marks |
|---|---|
| [unique solution is] z = 2 + i | M1 |
| Answer | Marks |
|---|---|
| A1dep | 1.1 |
| Answer | Marks |
|---|---|
| 3.2a | oe. Allow inequality. Could be obtained from diagram in (a). |
| Answer | Marks | Guidance |
|---|---|---|
| (𝑥+2)(cid:2870)+(𝑦−4)(cid:2870) = (𝑥−2)(cid:2870)+(𝑦−6)(cid:2870) | M1 | squaring both sides of equations or inequality |
| M1 | expanding all four sets of brackets | |
| y = –2x + 5 | A1 | oe. Allow inequality. |
| circle is x2 + y2 = 5 | B1 | allow inequality |
| x2 + (5 – 2x)2 = 5 | M1 | (cid:2870) |
| Answer | Marks | Guidance |
|---|---|---|
| 5x2 – 20x + 20 = 0 | M1 | simplifying to a three term quadratic equation |
| x = 2 [only] | A1* | or 𝑦 = 1 [only] |
| [unique solution is] z = 2 + i | A1dep |
Question 13:
13 | (a) | (i) | M1
A1
A1
[3] | 1.1
1.1
1.1 | circle, centre O
radius √5
shaded inside (oe). Candidates may shade the region that is
not required, but should clearly indicate that what they have
shaded is not required.
13 | (a) | (ii) | M1
A1
A1
[3] | 1.1
1.1
1.1 | (–2, 4) and (2, 6) identified
perpendicular bisector of (–2, 4) and (2, 6)
shaded on RHS of line (oe). Candidates may shade the region
that is not required, but should clearly indicate that what they
have shaded is not required.
13 | (b) | gradient = –2
passing through (0, 5)
equation y = –2x + 5
circle is x2 + y2 = 5
x2 + (5 – 2x)2 = 5
5x2 – 20x + 20 = 0
x = 2 [only]
[unique solution is] z = 2 + i | M1
B1
A1
B1
M1
M1
A1*
A1dep | 1.1
3.1a
1.1
3.1a
2.1
1.1
2.2a
3.2a | oe. Allow inequality. Could be obtained from diagram in (a).
allow inequality
(cid:2870)
or (cid:4672) (cid:2873) − (cid:3052) (cid:4673) +𝑦(cid:2870) = 5. Must be an equation.
(cid:2870) (cid:2870)
simplifying to a three-term quadratic equation
or 𝑦 = 1 [only]
Alternative method
(𝑥+2)(cid:2870)+(𝑦−4)(cid:2870) = (𝑥−2)(cid:2870)+(𝑦−6)(cid:2870) | M1 | squaring both sides of equations or inequality
M1 | expanding all four sets of brackets
y = –2x + 5 | A1 | oe. Allow inequality.
circle is x2 + y2 = 5 | B1 | allow inequality
x2 + (5 – 2x)2 = 5 | M1 | (cid:2870)
or (cid:4672) (cid:2873) − (cid:3052) (cid:4673) +𝑦(cid:2870) = 5. Must be an equation.
(cid:2870) (cid:2870)
5x2 – 20x + 20 = 0 | M1 | simplifying to a three term quadratic equation
x = 2 [only] | A1* | or 𝑦 = 1 [only]
[unique solution is] z = 2 + i | A1dep
[8]13
\begin{enumerate}[label=(\alph*)]
\item On separate Argand diagrams, show the set of points representing each of the following inequalities.
\begin{enumerate}[label=(\roman*)]
\item $| z | \leqslant \sqrt { 5 }$
\item $\quad | z + 2 - 4 i | \geqslant | z - 2 - 6 i |$
\end{enumerate}\item Show that there is a unique value of $z$, which should be determined, for which both $| z | \leqslant \sqrt { 5 }$ and $| z + 2 - 4 i | \geqslant | z - 2 - 6 i |$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Pure Core 2023 Q13 [14]}}