AS June 2024 Paper 1 Q2
2 In this question you must show detailed reasoning.
| Scheme | Marks | AO |
|---|---|---|
| DR \(\dfrac{8 + \mathrm{i}}{2 - \mathrm{i}} \times \dfrac{2 + \mathrm{i}}{2 + \mathrm{i}} = \dfrac{16 + 8\mathrm{i} + 2\mathrm{i} - 1}{4 + 1}\) | M1 | 1.1 |
| \(= \dfrac{15 + 10\mathrm{i}}{5} = 3 + 2\mathrm{i}\) cao | A1 | 1.1 |
| [2] |
Notes
M1: Multiplying top and bottom by the conjugate of the bottom and attempting expansion involving \(\mathrm{i}^2 = -1\).
Allow appearance of 5 in denominator with no working shown as long as clear what numerator multiplied by.
A1: Answer must be in the requested form.
Must have at least one line of working before answer (DR)
Alternate method
| Scheme | Marks |
|---|---|
| DR \((8 + \mathrm{i}) = (2 - \mathrm{i})(a + b\mathrm{i})\) \(8 = 2a + b\) \(1 = -a + 2b\) | M1 |
| \(\Rightarrow a = 3, b = 2\) \(3 + 2\mathrm{i}\) cao | A1 |
M1: Equating to \(a + b\mathrm{i}\) and rearranging to real and imaginary parts
A1: Answer must be in the requested form.
Must have at least one line of working before answer (DR)
| Scheme | Marks | AO |
|---|---|---|
| DR \(x = \dfrac{-(-8) \pm \sqrt{(-8)^2 - 4 \times 4 \times 5}}{2 \times 4} = \dfrac{8 \pm \sqrt{-16}}{8} = \dfrac{8 \pm 4\mathrm{i}}{8}\) | M1 | 1.1 |
| \(= 1 + \dfrac{1}{2}\mathrm{i}\) or \(1 - \dfrac{1}{2}\mathrm{i}\) | A1 | 1.1 |
| [2] |
Notes
M1: Use of formula and finding the square root of a negative number in terms of i (can be awarded even if error in calculation under square root).
Condone one sign error (eg \(-8\) rather than \(-(-8)\) or \((-8)^2 = -64\)).
Could also use completing the square. Must get as far as attempting “\(x = \ldots\)” but might make some slips.
A1: or \(1 \pm \dfrac{1}{2}\mathrm{i}\) or \(1 \pm 0.5\mathrm{i}\) but must be in correct form so eg \(\dfrac{2 \pm \mathrm{i}}{2}\) is A0
Condone \(1 \pm \dfrac{\mathrm{i}}{2}\)