A2 October 2020 Paper 2 Q1
1 In this question you must show detailed reasoning.
Solve the equation \(4z^2 - 20z + 169 = 0\). Give your answers in modulus-argument form. [5]
| Scheme | Marks | AO |
|---|---|---|
| DR \(z = \dfrac{--20 \pm \sqrt{(-20)^2 - 4 \times 4 \times 169}}{2 \times 4}\) | M1 | 1.1 |
| \(z = \dfrac{5 \pm 12\mathrm{i}}{2}\) | A1 | 1.1 |
| \(r = \sqrt{\left(\dfrac{5}{2}\right)^2 + \left(\dfrac{12}{2}\right)^2} = \dfrac{13}{2}\) oe | B1ft | 1.1 |
| \(\theta = \tan^{-1}\dfrac{6}{2.5}\) oe | M1 | 1.1 |
| \(\dfrac{13}{2}(\cos 1.18 + \mathrm{i}\sin 1.18)\) | A1 | 2.5 |
| \(\dfrac{13}{2}(\cos(-1.18) + \mathrm{i}\sin(-1.18))\) | ||
| [5] |
Notes
M1: Term by term substituting into formula.
If formula quoted, allow one slip …
Or correctly completes the square
Condone anything correct of the form \(\dfrac{p \pm \sqrt{q}}{r}\)
eg \(4\left(\left(z - \dfrac{5}{2}\right)^2 - \dfrac{25}{4}\right) + 169 = 0\)
B1ft: Ft workings from complex conjugate distinct pair (with real component)
M1: Attempting to find argument using trigonometry
\(\theta = \cos^{-1}\dfrac{2.5}{6.5},\quad \theta = \sin^{-1}\dfrac{6}{6.5}\)
A1: Angle must be in radians.
oe eg \(\dfrac{13}{2}\operatorname{cis}1.18\) or \(\left[\dfrac{13}{2}, 1.18\right]\)
Argument could be 5.11 but both angles must be the same.
Not 5.10 (rounding error)
Not e.g. \(\cos(-1.18) + \mathrm{i}\sin(5.11)\)