AS October 2020 Paper 1 Q8
8 Two loci, \(C_1\) and \(C_2\), are defined by
\[\begin{aligned} C_1 &= \left\{z : |z| = |z - 4d^2 - 36|\right\} \\ C_2 &= \left\{z : \arg(z - 12d - 3\mathrm{i}) = \frac{1}{4}\pi\right\} \end{aligned}\]where \(d\) is a real number.
[You may assume that \(C_1 \cap C_2 \ne \varnothing\).] [6]
| Scheme | Marks | AO |
|---|---|---|
| \(C_1\) is (represented by) the line \(x = 2d^2 + 18\) | B1 | 3.1a |
| \(C_2\) is (represented by) the (half-)line \(y = x + c\) | M1 | 3.1a |
| \(3 = 12d + c\) | M1 | 3.1a |
| \(y = x + 3 - 12d\) | A1 | 1.1 |
| When \(x = 2d^2 + 18\), \(y = 2d^2 - 12d + 21\) | M1 | 1.1 |
| \(2d^2 + 18 + (2d^2 - 12d + 21)\mathrm{i}\) | A1 | 3.2a |
| [6] |
Notes
B1: Seen or implied in solution
M1: (1st) For understanding the \(C_2\) is a line or half-line whose gradient is 1.
M1: (2nd) Complete line would pass through the point \((12d, 3)\) or \(12d + 3\mathrm{i}\).
Half line starting at \((12d, c)\) and with angle \(\frac{\pi}{4}\)
M1: (3rd) Attempt at \(y\) coordinate
A1: (2nd) Must be in complex number form
or eg \(2(d^2 + 9) + (2(d - 3)^2 + 3)\mathrm{i}\)
Alternative Method
| Scheme | Marks |
|---|---|
| \(C_1\) is (represented by) the line \(x = 2d^2 + 18\) | B1 |
| \(C_2\) is (represented by) the (half-)line starting at the point \(12d + 3\mathrm{i}\) | M1 |
| \(C_2\) half line has gradient 1 | M1 |
| Right-angled triangle indicated with base length \((2d^2 + 18) - 12d\) | M1 |
| \(y\) coordinate at \(3 + ((2d^2 + 18) - 12d)\) | M1 |
| POI at \(2d^2 + 18 + (2d^2 - 12d + 21)\mathrm{i}\) | A1 |
B1: SOI
M1: (2nd) Correct half line needed here
Could be shown by making angle \(\frac{\pi}{4}\) with positive \(x\) direction
M1: (3rd) Follow through \(C_1\) line and start of \(C_2\) half line
M1: (4th) Attempt at \(y\) coordinate using base of triangle = height of triangle and adding on 3i
A1: Must be in complex number form
or eg \(2(d^2 + 9) + (2(d - 3)^2 + 3)\mathrm{i}\)
| Scheme | Marks | AO |
|---|---|---|
| When \(d = 3\), the PoI would be \(36 + 3\mathrm{i}\) and \(C_2 = \left\{z : \arg\left(z - (36 + 3\mathrm{i})\right) = \dfrac{1}{4}\pi\right\}\) | M1 | 3.1a |
| But \(36 + 3\mathrm{i}\) is not in \(C_2\) since \(\arg 0\) is not defined | A1 | 3.2b |
| [2] |
Notes
M1: Both
NB \(C_1 \cap C_2 = \varnothing\) without justification is M0A0.
A1: AG. Or \(\arg 0\) is not \(\pi/4\)
Alternative Method
| Scheme | Marks |
|---|---|
| For the intersection to exist we need \(2d^2 + 18 \gt 12d\) | M1 |
| \(d^2 - 6d + 9 \gt 0\) \((d - 3)^2 \gt 0\) \([d \ne 3]\) Hence \(d\) cannot be equal to 3 | A1 |
M1: Set up inequality using \(x\) coordinates of vertical line and starting point of half line
A1: Or “\(d = 3\)” is not valid etc.