AS June 2024 Paper 1 Q9
9 In this question you must show detailed reasoning.
You are given that \(a\) is a real root of the equation \(x^4 + x^3 + 3x^2 - 5x = 0\).
You are also given that \(a + 2 + 3\mathrm{i}\) is one root of the equation
\(z^4 - 2(1 + a)z^3 + (21a - 10)z^2 + (86 - 80a)z + (285a - 195) = 0\).
Determine all possible values of \(z\). [8]
| Scheme | Marks | AO |
|---|---|---|
| DR \(a = 0\) (is one possibility) | B1 | 1.1 |
| \(1^4 + 1^3 + 3 \times 1^2 - 5 \times 1 = 0\) so \(a = 1\) (is another possibility) | B1 | 3.1a |
| \(x^4 + x^3 + 3x^2 - 5x = x(x^2(x - 1) + 2x(x - 1) + 5(x - 1)) = x(x - 1)(x^2 + 2x + 5)\) and discriminant of quadratic \(= 2^2 - 4 \times 1 \times 5 = -16 \lt 0\) so no further real roots. | B1 | 2.3 |
| \(a = 0 \Rightarrow 2 + 3\mathrm{i}\) is a root of \(z^4 - 2z^3 - 10z^2 + 86z - 195\) [\(= 0\)] so \(2 - 3\mathrm{i}\) is also a root OR \(a = 1 \Rightarrow 3 + 3\mathrm{i}\) is a root of \(z^4 - 4z^3 + 11z^2 + 6z + 90\) [\(= 0\)] so \(3 - 3\mathrm{i}\) is also a root | M1 | 3.1a |
| \(2 + 3\mathrm{i} + 2 - 3\mathrm{i} = 4\) and \((2 + 3\mathrm{i})(2 - 3\mathrm{i}) = 13\) so \(z^2 - 4z + 13\) is a factor OR \(3 + 3\mathrm{i} + 3 - 3\mathrm{i} = 6\) and \((3 + 3\mathrm{i})(3 - 3\mathrm{i}) = 18\) so \(z^2 - 6z + 18\) is a factor | M1 | 1.1 |
| \(z^4 - 2z^3 - 10z^2 + 86z - 195 = (z^2 - 4z + 13)(z^2 + 2z - 15)\) OR \(z^4 - 4z^3 + 11z^2 + 6z + 90 = (z^2 - 6z + 18)(z^2 + 2z + 5)\) | M1 | 1.1 |
| \(z^2 + 2z - 15 = 0 \Rightarrow z = -5\), \(z = 3\) and \(2 \pm 3\mathrm{i}\) stated as roots (possibly earlier). | A1 | 2.2a |
| \(z^2 + 2z + 5 = 0 \Rightarrow z = -1 \pm 2\mathrm{i}\) and \(3 \pm 3\mathrm{i}\) stated as roots (possibly earlier). | A1 | 2.2a |
| [8] |
Notes
B1: (2nd) or eg \(\mathrm{f}(1) = 0\) if intent is clear but must be some justification.
B1: (3rd) Some justification must be given that there are no more real roots.
Allow for finding the two complex roots \((-1 \pm 2\mathrm{i})\) - must have seen the correct \(x^2 + 2x + 5\)
If B1B0B0 or B0B0B0 then SC1 for “\(a = 1\) and no others” or “\(a = 1\), \((-1 \pm 2\mathrm{i})\)” without justification.
M1: Condone small errors in calculation of coefficients in equation
Need both the pair of complex roots SOI and the quartic shown (allow sign slips). Only need one case for the M1.
M1: oe eg expanding \((z - (2 + 3\mathrm{i}))(z - (2 - 3\mathrm{i}))\) or \((z - (3 + 3\mathrm{i}))(z - (3 - 3\mathrm{i}))\)
Attempt to find quadratic factor from the complex roots. Only one case needed for M1
Allow with no working
M1: Attempt to factorise their quartic with their quadratic factor (at least \(z^3\) and constant terms consistent).
Only one case needed.
MUST see some evidence of factorisation here
A1: All four roots SOI for the \(a = 0\) case
DR so need to see evidence of where the roots came from i.e. factorisation into two quadratics
A1: All four roots SOI for the \(a = 1\) case
If extra values of \(z\) found (from complex \(a\) or incorrect \(a\) values) then A0
Alternate for 1st and 2nd M mark
| Scheme | Marks |
|---|---|
| \((a + 2 + 3\mathrm{i}) + (a + 2 - 3\mathrm{i}) = 2a + 4\) and \((a + 2 + 3\mathrm{i})(a + 2 - 3\mathrm{i}) = (a + 2)^2 + 9\) so \(z^2 - (2a + 4)z + [a^2 + 4a + 13]\) is a factor | M1 |
| \(a = 0 \Rightarrow z^2 - 4z + 13\) is a factor OR \(a = 1 \Rightarrow z^2 - 6z + 18\) is a factor | M1 |
M1: Quadratic factor found in general case. Can also be found by expanding \([z - (a + 2 + 3\mathrm{i})][z - (a + 2 - 3\mathrm{i})]\)
Constant term can be either \((a + 2)^2 + 9\) or \([a^2 + 4a + 13]\)
Alternate 2 if only the quartic in \(z\) with \(a = 0\) considered via linear factors
| Scheme | Marks |
|---|---|
| \(a = 0\) (is one possibility) | B1 |
| \(1^4 + 1^3 + 3 \times 1^2 - 5 \times 1 = 0\) so \(a = 1\) (is another possibility) | B1 |
| \(x^4 + x^3 + 3x^2 - 5x = x(x^2(x - 1) + 2x(x - 1) + 5(x - 1)) = x(x - 1)(x^2 + 2x + 5)\) and discriminant of quadratic \(= 2^2 - 4 \times 1 \times 5 = -16 \lt 0\) so no further real roots. | B1 |
| If \(a = 0\) quartic is \(z^4 - 2z^3 - 10z^2 + 86z - 195\) \(\mathrm{f}(3) = 0\) so \((z - 3)\) is a factor \(\mathrm{f}(z) = (z - 3)(z^3 + z^2 - 7z + 65)\) | M1 |
| \(\mathrm{f}(-5) = 0\) so \((z + 5)\) is a factor \(\mathrm{f}(z) = (z - 3)(z + 5)(z^2 - 4z + 13)\) | M1 |
| So the roots are \(3\), \(-5\), \(2 + 3\mathrm{i}\), \(2 - 3\mathrm{i}\) | A1 |
B1: (2nd) or eg \(\mathrm{f}(1) = 0\) if intent is clear but must be some justification.
B1: (3rd) Some justification must be given that there are no more real roots. Allow for finding the two complex roots \((-1 \pm 2\mathrm{i})\) (must have seen correct quadratic)
If B1B0B0 or B0B0B0 then SC1 for “\(a = 1\) and no others” or “\(a = 1\), \((-1 \pm 2\mathrm{i})\)” without justification.
M1: (1st) Identifying linear factor and factorising
A1: Identifying 4 roots
Final 2 marks unavailable