AS June 2022 Paper 1 Q5
5 In this question you must show detailed reasoning.
Give your answer in the form \(a + b\mathrm{i}\). [2]
| Scheme | Marks | AO |
|---|---|---|
| DR \((a + b\mathrm{i})^2 = a^2 - b^2 + 2ab\mathrm{i}\) | B1 | 1.1 |
| \(a^2 - b^2 = -16\) and \(2ab = 30\) (where \(a\) and \(b\) are real) | M1 | 1.1 |
| \(b = \dfrac{15}{a} \Rightarrow a^2 - \left(\dfrac{15}{a}\right)^2 = -16\) \(\Rightarrow a^4 + 16a^2 - 225 = 0\) | M1 | 1.1 |
| \(a\) (\(b\)) real so \(a^2 = 9\) (or \(b^2 = 25\)) only | A1 | 1.1 |
| \(3 + 5\mathrm{i}\) and \(-3 - 5\mathrm{i}\) | A1 | 2.2a |
| [5] |
Notes
B1: Seen or implied in solution
M1: (1st) Comparing real and imaginary parts (no i unless later recovered) from a 3 (or 4) term expansion. Allow sign slips
M1: (2nd) Eliminating \(b\) or \(a\) to obtain 3 term quadratic in \(a^2\) or \(b^2\). Unknowns must not be in denominator. Must be an equation.
\((b^4 - 16b^2 - 225 = 0)\)
Factorised forms:
\((a^2 - 9)(a^2 + 25)\)
\((b^2 - 25)(b^2 + 9)\)
A1: (1st) Rogue solutions; \(a^2 = -25\), \(b^2 = -9\)
A1: (2nd) Both roots. Can be \(\pm(3 + 5\mathrm{i})\) but not \(\pm 3 \pm 5\mathrm{i}\) or \(\pm 3 + 5\mathrm{i}\).
Note: 4/5 possible following B0
| Scheme | Marks | AO |
|---|---|---|
| DR \((3 + 5\mathrm{i})^3 = (-16 + 30\mathrm{i})(3 + 5\mathrm{i}) = -198 + 10\mathrm{i}\) | M1 | 1.1 |
| \(= \dfrac{-99 + 5\mathrm{i}}{4} \times 2^3\) so \(\dfrac{3}{2} + \dfrac{5\mathrm{i}}{2}\) | A1 | 3.1a |
| [2] |
Notes
M1: Can awarded this for cubing an incorrect answer to 5(a)
Or \((-3 - 5\mathrm{i})^3 = 198 - 10\mathrm{i}\ldots\)
Could be done by expansion either in two steps or binomial:
\(3^3 + 3 \times 3^2 \times 5\mathrm{i} + 3 \times 3 \times (5\mathrm{i})^2 + (5\mathrm{i})^3\)
For binomial want to see 4 terms and either 2nd or 3rd term correct (up to sign error)
Do not need to see intermediate step before correct answer. If incorrect need to see proof of expanding three brackets
A1: Correct answer must follow a correct root
\(= \dfrac{-99 + 5\mathrm{i}}{4} \times (-2)^3\)