A2 June 2024 Paper 2 Q2
2 In this question you must show detailed reasoning.
| Scheme | Marks | AO |
|---|---|---|
| \(x = \dfrac{--6 \pm \sqrt{(-6)^2 - 4 \times 1 \times 58}}{2 \times 1}\) | M1 | 1.1 |
| \(\sqrt{-196} = 14\mathrm{i}\) | B1FT | 1.1 |
| \(x = \dfrac{6 \pm 14\mathrm{i}}{2} = 3 \pm 7\mathrm{i}\) | A1 | 2.5 |
| [3] |
Notes
M1: DR. Correctly using formula or completing the square. Condone “\(-6 \pm \sqrt{\Delta}\)” for M1. Condone missing brackets under root if \(-6\) squares to 36 (ie \(\Delta = -196\) rather than \(-268\)).
B1FT: FT their negative discriminant. Writing square root of negative number as the correct multiple of i. “\(-196\)” (or “\(-49\)”) must be seen.
A1: Must be \(a + b\mathrm{i}\).
Alternative method
| Scheme | Marks |
|---|---|
| The roots are \(\alpha = a + b\mathrm{i}\) and \(\beta = a - b\mathrm{i}\) (where \(a\) and \(b\) are real) | B1 |
| \(\alpha + \beta = -(-6)/1 = 6\) and \(\alpha\beta = 58/1 = 58\) So \(2a = 6\) \((\Rightarrow a = 3)\) and \(a^2 + b^2 = 58\) | M1 |
| So \(b^2 = 58 - 9 = 49\) so roots are \(3 \pm 7\mathrm{i}\) | A1 |
| [3] |
B1: Using the fact that the roots of a real quadratic form a complex conjugate pair. May be embedded.
M1: Finding the numerical value of the sum and product of the roots.
Could also be found by expanding \((x - (a + b\mathrm{i}))(x - (a - b\mathrm{i}))\) and comparing with equation.
Could also substitute \((a + b\mathrm{i})\) into the equation to derive \((2ab - 6b) = 0\) and \(a^2 - b^2 - 6a + 58 = 0\).
A1: Must be \(a + b\mathrm{i}\). \(a\) real \(\Rightarrow b \neq 0 \Rightarrow a = 3 \Rightarrow b = \pm 7\)
| Scheme | Marks | AO |
|---|---|---|
| \(\arg\left(-10 + 5\sqrt{12}\mathrm{i}\right) = \tan^{-1}\dfrac{5\sqrt{12}}{-10}\ \left(= \dfrac{2}{3}\pi\right)\) | M1 | 1.1 |
| \(\left(\arg\left(-10 + 5\sqrt{12}\mathrm{i}\right)^5 = 5\arg\left(-10 + 5\sqrt{12}\mathrm{i}\right) =\right)\) \(5 \times \dfrac{2}{3}\pi\ \left(= \dfrac{10}{3}\pi\right)\) | M1 | 1.1 |
| so required angle is \(-\dfrac{2}{3}\pi\ \left(\text{or } \dfrac{4}{3}\pi\right)\) | A1 | 1.1 |
| [3] |
Notes
M1: DR. Using correct formula for argument of complex number with non-zero real and imaginary parts.
Condone \(\tan\alpha = \frac{5\sqrt{12}}{-10} \Rightarrow \alpha = -\frac{\pi}{3}\) or \(\tan^{-1}\frac{5\sqrt{12}}{10} \Rightarrow \alpha = \frac{2\pi}{3}\) or \(-\frac{\pi}{3}\) for M1.
M1: Using De Moivre’s Theorem for their angle. Condone error in modulus if shown.
or using a valid method for finding \(z^5\) explicitly (eg by expansion or by writing \(-10 + 5\sqrt{12}\mathrm{i} = 20\mathrm{e}^{\frac{2}{3}\pi\mathrm{i}}\))
\(z^5 = -1600000 - 1600000\sqrt{3}\,\mathrm{i}\) or \(\left(20^5\right)\mathrm{e}^{5 \times \frac{2}{3}\pi\mathrm{i}}\)
A1: cao