A2 June 2023 Paper 2 Q8
8. Given that a cubic equation has three distinct roots that all lie on the same straight line in the complex plane,
where \(b\), \(c\) and \(d\) are real constants.
The roots of \(\mathrm{f}(z)\) are distinct and lie on a straight line in the complex plane.
Given that one of the roots is \(\dfrac{3}{2} + \dfrac{3}{2}\mathrm{i}\)
where \(P\) and \(Q\) are real constants, has 3 distinct roots.
The roots of \(\mathrm{g}(z)\) lie on a different straight line in the complex plane than the roots of \(\mathrm{f}(z)\)
Given that
- \(\mathrm{f}(z)\) and \(\mathrm{g}(z)\) have one root in common
- one of the roots of \(\mathrm{g}(z)\) is \(-4\)
| Scheme | Marks | AO |
|---|---|---|
| The real axis. Horizontal line through (0, 0) Line \(y = 0\) Accept on a diagram The other possibility is that all three roots have the same real part so lie on a vertical line/perpendicular to the real axis/parallel to the imaginary axis Line \(x = k\) where \(k\) is a real number Accept on a diagram | B1 B1 | 3.1a 2.2a |
| (2) |
Notes
B1: One correct line described
B1: Two correct lines described
Special case: If candidate states that "any line is possible" score B1B1 (as they may be considering a cubic with complex coefficients).
| Scheme | Marks | AO |
|---|---|---|
| Other roots are \(\dfrac{3}{2}\) and \(\dfrac{3}{2} - \dfrac{3}{2}\mathrm{i}\) | B1 | 3.2a |
| (1) |
Notes
B1: Interprets the conclusion from (a) in context by identifying the correct two roots.
| Scheme | Marks | AO |
|---|---|---|
| (i) Common root must be \(\dfrac{3}{2}\) | B1 | 2.2a |
| (1) | ||
| (ii) Sets product of roots = – 12 using their \(\dfrac{3}{2} \times -4 \times \alpha = -12\) Or \(\mathrm{g}(z) = \left(z - \dfrac{3}{2}\right)(z + 4)(z - \alpha)\) | M1 | 1.1b |
| Solves to find a value of the third root their \(\dfrac{3}{2} \times -4 \times \alpha = -12 \Rightarrow \alpha = 2\) Or \(\mathrm{g}(z) = \left(z - \dfrac{3}{2}\right)(z \pm 4)(z - \alpha) \Rightarrow -\dfrac{3}{2} \times 4 \times -\alpha = 12 \Rightarrow \alpha = 2\) | M1 A1 | 3.1a 1.1b |
| (3) |
Notes
B1: Deduces the real root is the one in common.
M1: Sets product of roots = – 12 using their \(\dfrac{3}{2} \times -4 \times \alpha = -12\). Alternatively forms an equation for \(\mathrm{g}(z)\) using the roots
M1: Solves their equation to find the third root, condone use of 12 for this mark. Alternatively multiply their constant and sets = 12 to find the third root. Condone a sign slip for this mark
A1: Correct third root
| Scheme | Marks | AO |
|---|---|---|
| \(8\left\{z - \dfrac{3}{2}\right\}\left(z - \dfrac{3}{2} - \dfrac{3}{2}\mathrm{i}\right)\left(z - \dfrac{3}{2} + \dfrac{3}{2}\mathrm{i}\right) = 8\left\{z - \dfrac{3}{2}\right\}\left(z^2 - 3z + \dfrac{9}{2}\right)\) Or \(b = -8\left(\dfrac{3}{2} + \dfrac{3}{2} + \dfrac{3}{2}\mathrm{i} + \dfrac{3}{2} - \dfrac{3}{2}\mathrm{i}\right) = \ldots\{-36\}\) \(c = 8\left[\left(\dfrac{3}{2} \times \left(\dfrac{3}{2} + \dfrac{3}{2}\mathrm{i}\right)\right) + \left(\dfrac{3}{2} \times \left(\dfrac{3}{2} - \dfrac{3}{2}\mathrm{i}\right)\right) + \left(\left(\dfrac{3}{2} + \dfrac{3}{2}\mathrm{i}\right) \times \left(\dfrac{3}{2} - \dfrac{3}{2}\mathrm{i}\right)\right)\right] = \ldots\{72\}\) \(d = -8\left[\dfrac{3}{2} \times \left(\dfrac{3}{2} + \dfrac{3}{2}\mathrm{i}\right) \times \left(\dfrac{3}{2} - \dfrac{3}{2}\mathrm{i}\right)\right] = \ldots\{-54\}\) | M1 | 1.1b |
| \(\mathrm{f}(z) = \mathrm{g}(z) \Rightarrow 8\left(z - \dfrac{3}{2}\right)\left(z^2 - 3z + \dfrac{9}{2}\right) = \left(z - \dfrac{3}{2}\right)(z + 4)(z - 2)\) \(\Rightarrow 8z^2 - 24z + 36 = (z + 4)(z - 2)\) Or Either \(\mathrm{g}(z) = \left(z - \dfrac{3}{2}\right)(z + 4)(z - 2) = \ldots\) or \(P = -\left(\dfrac{3}{2} - 4 + 2\right) = \ldots\left\{\dfrac{1}{2}\right\}\) and \(Q = \left(\dfrac{3}{2} \times -4\right) + \left(\dfrac{3}{2} \times 2\right) + (2 \times -4) = \ldots\{-11\}\) to find \(\mathrm{g}(z)\) and sets their \(\mathrm{f}(z)\) = their \(\mathrm{g}(z)\) \(8z^3 - 36z^2 + 72z - 54 = z^3 + \dfrac{1}{2}z^2 - 11z + 12\) | M1 | 3.1a |
| \(7z^2 - 26z + 44 = 0 \Rightarrow z = \ldots\) or \(7z^3 - \dfrac{73}{2}z^2 + 83z - 66 = 0 \Rightarrow z = \ldots\) | M1 | 1.1b |
| So solutions are \(\dfrac{3}{2}\), \(\dfrac{13 \pm \mathrm{i}\sqrt{139}}{7}\) | A1 | 1.1b |
| (4) | ||
| (11 marks) |
Notes
M1: Uses their roots of \(\mathrm{f}(z)\) to form a cubic expression for \(\mathrm{f}(z)\), and expands to at least a linear term times a quadratic with real coefficients (which may be seen later). May just expand the complex brackets.
They must have the factor 8 for this mark
Alternative uses \(b = -8(\text{sum of their roots})\) \(c = 8(\text{pair sum of their roots})\) and \(d = -8(\text{product of their roots})\)
Note: \(\mathrm{f}(z) = 8z^3 - 36z^2 + 72z - 54\)
M1: Sets their expressions equal and factorises out or cancels the common term to achieve a quadratic expression in \(z\). Allow if \(\mathrm{f}(z)\) is not yet expanded or factor of 8 is missing.
Alternatively finds the expression for \(\mathrm{g}(z)\) by multiplying out bracket or uses P = - sum roots and Q = pair sum. Sets their \(\mathrm{f}(z)\) = their \(\mathrm{g}(z)\) both must be cubic
M1: Expands, gathers terms and solves the resulting quadratic. Allow this mark if the \(z = \dfrac{3}{2}\) solution is not given.
Alternatively simplifies for form a cubic = 0 and solves using calculator to find complex roots
A1: All three correct solutions given. Note decimals \(\dfrac{13}{7} \pm \mathrm{i}1.68\ldots\) is A0
Special case If the candidate just forgets the factor of 8 for \(\mathrm{f}(z)\) this scores M0M1M0A0