AS June 2023 Paper 1 Q3
3 In this question you must show detailed reasoning.
The function \(\mathrm{f}(z)\) is given by \(\mathrm{f}(z) = 2z^3 - 7z^2 + 16z - 15\).
By first evaluating \(\mathrm{f}\left(\frac{3}{2}\right)\), find the roots of \(\mathrm{f}(z) = 0\). [6]
| Scheme | Marks | AO |
|---|---|---|
| DR \(\mathrm{f}\left(\dfrac{3}{2}\right) = \dfrac{27}{4} - \dfrac{63}{4} + 24 - 15 = -9 + 9 = 0\) [so 3/2 is a root] | B1 | 1.1 |
| \(\Rightarrow 2z - 3\) is a factor | M1 | 2.2a |
| \(\mathrm{f}(z) = (2z - 3)(z^2 - 2z + 5)\) | M1 A1 | 1.1 1.1 |
| \(z = \dfrac{2 \pm \sqrt{-16}}{2}\) | M1 | 1.1 |
| \(z = 1 + 2\mathrm{i},\ 1 - 2\mathrm{i},\ \dfrac{3}{2}\) | B1 | 2.2a |
| [6] |
Notes
B1: (1st) must see some substitution, [so \(\mathrm{f}\left(\frac{3}{2}\right) = 0\) alone is B0]
M1: (1st) or \(z - 3/2\)
M1 A1: attempt to factorise (oe, e.g. long division)
or \((z - 3/2)(2z^2 - 4z + 10)\)
M1: (3rd) or by completing the square or using sum and prod of roots
[\(z = a \pm \mathrm{i}b\), \(2a = 2\), \(a^2 + b^2 = 5 \Rightarrow a = 1\), \(b = \pm 2\)]
B1: (2nd) [3/2 may be stated as a root earlier]
If no working shown then award no marks
Alternative solution
| Scheme | Marks |
|---|---|
| Other roots are \(\alpha\) and \(\beta\) where \(\alpha + \beta + \dfrac{3}{2} = \dfrac{7}{2},\ \dfrac{3}{2}\alpha + \dfrac{3}{2}\beta + \alpha\beta = 8,\ \dfrac{3}{2}\alpha\beta = \dfrac{15}{2}\) | M1 |
| so \(\alpha\beta = 5,\ \alpha + \beta = 2\) | A1 |
| \(\Rightarrow \alpha^2 - 2\alpha + 5 = 0\) | A1 |
| \(\Rightarrow \alpha = \dfrac{2 \pm \sqrt{-16}}{2}\) | M1 |
| \(z = 1 + 2\mathrm{i},\ 1 - 2\mathrm{i},\ \dfrac{3}{2}\) | B1 |
M1: (1st) symmetric property of roots used (condone 7, 15, 16 or sign errors) – must have at least 2 of the 3
M1: (2nd) or by completing the square or using sum and prod of roots
[\(z = a \pm \mathrm{i}b\), \(2a = 2\), \(a^2 + b^2 = 5 \Rightarrow a = 1\), \(b = \pm 2\)]