| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2010 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Argand & Loci |
| Type | Complex arithmetic operations |
| Difficulty | Moderate -0.8 This is a routine FP1 question testing standard complex number operations: division by multiplying by conjugate, finding modulus, and finding argument. All three parts follow textbook procedures with no problem-solving required, making it easier than average even for Further Maths students. |
| Spec | 4.02c Complex notation: z, z*, Re(z), Im(z), |z|, arg(z)4.02e Arithmetic of complex numbers: add, subtract, multiply, divide |
| Answer | Marks |
|---|---|
| (a) \(y = \frac{c^2}{x} = -c^2x^{-2}\) | B1 |
| \(\frac{dy}{dx} = -\frac{c^2}{(ct)^2} = -\frac{1}{t^2}\) without \(x\) or \(y\) | M1 |
| \(y - \frac{c}{t} = -\frac{1}{t^2}(x - ct) \Rightarrow t^2y + x = 2ct\) (*) | M1 A1cso |
| Answer | Marks |
|---|---|
| (b) Substitute \((15c, -c)\): \(-ct^2 + 15c = 2ct\) | M1 |
| Answer | Marks |
|---|---|
| \((t + 5)(t - 3) = 0 \Rightarrow t = -5\) or \(t = 3\) | A1 |
| Points are \(\left(-5c, -\frac{c}{5}\right)\) and \(\left(3c, \frac{c}{3}\right)\) both | M1 A1 A1 |
| Answer | Marks |
|---|---|
| (a) \(\frac{dx}{dt} = c\) and \(\frac{dy}{dt} = -ct^{-2}\) | B1 |
| \(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt} = -\frac{1}{t^2}\) | M1, then as in main scheme. |
| (a) \(y + x\frac{dy}{dx} = 0\) | B1 |
| \(\frac{dy}{dx} = -\frac{y}{x} = -\frac{1}{t^2}\) | M1, then as in main scheme. |
(a) $y = \frac{c^2}{x} = -c^2x^{-2}$ | B1
$\frac{dy}{dx} = -\frac{c^2}{(ct)^2} = -\frac{1}{t^2}$ without $x$ or $y$ | M1
$y - \frac{c}{t} = -\frac{1}{t^2}(x - ct) \Rightarrow t^2y + x = 2ct$ (*) | M1 A1cso
(4)
(b) Substitute $(15c, -c)$: $-ct^2 + 15c = 2ct$ | M1
$t^2 + 2t - 15 = 0$
$(t + 5)(t - 3) = 0 \Rightarrow t = -5$ or $t = 3$ | A1
Points are $\left(-5c, -\frac{c}{5}\right)$ and $\left(3c, \frac{c}{3}\right)$ both | M1 A1 A1
(5)
[9]
**Notes**
(a) Use of $y - y_1 = m(x - x_1)$ where $m$ is their gradient expression in terms of $c$ and/or $t$ only for second M1. Accept $y = mx + k$ and attempt to find $k$ for second M1.
(b) Correct absolute factors for their constant for second M1. Accept correct use of quadratic formula for second M1.
**Alternatives:**
(a) $\frac{dx}{dt} = c$ and $\frac{dy}{dt} = -ct^{-2}$ | B1
$\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt} = -\frac{1}{t^2}$ | M1, then as in main scheme.
(a) $y + x\frac{dy}{dx} = 0$ | B1
$\frac{dy}{dx} = -\frac{y}{x} = -\frac{1}{t^2}$ | M1, then as in main scheme.
---\begin{enumerate}
\item The complex numbers $z _ { 1 }$ and $z _ { 2 }$ are given by
\end{enumerate}
$$z _ { 1 } = 2 + 8 i \quad \text { and } \quad z _ { 2 } = 1 - i$$
Find, showing your working,\\
(a) $\frac { Z _ { 1 } } { Z _ { 2 } }$ in the form $a + b \mathrm { i }$, where $a$ and $b$ are real,\\
(b) the value of $\left| \frac { z _ { 1 } } { z _ { 2 } } \right|$,\\
(c) the value of $\arg \frac { Z _ { 1 } } { Z _ { 2 } }$, giving your answer in radians to 2 decimal places.\\
\hfill \mbox{\textit{Edexcel FP1 2010 Q1 [7]}}