Edexcel FP2 2023 June — Question 2 6 marks

Exam BoardEdexcel
ModuleFP2 (Further Pure Mathematics 2)
Year2023
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicReduction Formulae
TypeEstablish bounds or inequalities
DifficultyStandard +0.3 This is a standard loci question requiring algebraic manipulation to find a circle equation, verification of specific points, and sketching. While it involves multiple parts, each step follows routine A-level Further Maths techniques with no novel problem-solving required. The inequality shading is straightforward once the locus is identified.
Spec4.02k Argand diagrams: geometric interpretation4.02o Loci in Argand diagram: circles, half-lines

  1. A complex number \(z\) is represented by the point \(P\) in the complex plane.
Given that \(z\) satisfies $$| z - 6 | = 2 | z + 3 i |$$
  1. show that the locus of \(P\) passes through the origin and the points - 4 and - 8 i
  2. Sketch on an Argand diagram the locus of \(P\) as \(z\) varies.
  3. On your sketch, shade the region which satisfies both $$| z - 6 | \geqslant 2 | z + 3 i | \text { and } | z | \leqslant 4$$

Question 2:
Part (a)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\-4-6\ =10\) and \(2\
Both \(-4\) and \(-8i\) verified and \(\-6\ =6\), \(\
Alternative: \((x-6)^2+y^2=4x^2+4(y+3)^2 \Rightarrow x^2+y^2+8y+4x=0 \Rightarrow (x+2)^2+(y+4)^2=20\); substitutes at least one of \((-4,0)\) or \((0,-8)\)M1 Cartesian equation approach
Substitutes \((-4,0)\) and \((0,-8)\) and \((0,0)\) to show all holdA1
Parts (b)+(c)
AnswerMarks Guidance
Answer/WorkingMark Guidance
Circle passing through the originM1
Correct circle drawn with correct axis interceptsA1 Centre in third quadrant, passing through \(-4\), \(0\), \(-8i\)
Circle with centre \((0,0)\) and radius \(4\)B1 Can sketch part of circle sufficient to show overlap
Shades area inside both circlesB1ft Follow through; allow if overlapping circles present
# Question 2:

## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\|-4-6\|=10$ and $2\|-4+3i\|=\left\{2\sqrt{4^2+3^2}\right\}=2\times5=10$, so $-4$ is on locus; OR $\|-8i-6\|=\left\{\sqrt{64+36}\right\}=10$ and $2\|-8i+3i\|=2\|-5i\|=2\times5=10$, so $-8i$ is on locus | M1 | Verifies at least one non-zero point with correct modulus method |
| Both $-4$ and $-8i$ verified **and** $\|-6\|=6$, $\|3i\|=6$ so origin also on locus | A1 | All three points correctly checked |
| **Alternative:** $(x-6)^2+y^2=4x^2+4(y+3)^2 \Rightarrow x^2+y^2+8y+4x=0 \Rightarrow (x+2)^2+(y+4)^2=20$; substitutes at least one of $(-4,0)$ or $(0,-8)$ | M1 | Cartesian equation approach |
| Substitutes $(-4,0)$ **and** $(0,-8)$ **and** $(0,0)$ to show all hold | A1 | — |

## Parts (b)+(c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Circle passing through the origin | M1 | — |
| Correct circle drawn with correct axis intercepts | A1 | Centre in third quadrant, passing through $-4$, $0$, $-8i$ |
| Circle with centre $(0,0)$ and radius $4$ | B1 | Can sketch part of circle sufficient to show overlap |
| Shades area inside both circles | B1ft | Follow through; allow if overlapping circles present |

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\begin{enumerate}
  \item A complex number $z$ is represented by the point $P$ in the complex plane.
\end{enumerate}

Given that $z$ satisfies

$$| z - 6 | = 2 | z + 3 i |$$

(a) show that the locus of $P$ passes through the origin and the points - 4 and - 8 i\\
(b) Sketch on an Argand diagram the locus of $P$ as $z$ varies.\\
(c) On your sketch, shade the region which satisfies both

$$| z - 6 | \geqslant 2 | z + 3 i | \text { and } | z | \leqslant 4$$

\hfill \mbox{\textit{Edexcel FP2 2023 Q2 [6]}}