A2 June 2025 Paper 1 Q6
6.
\[\mathrm{f}(z) = z^3 + az^2 + bz + c \qquad \text{where } a,\ b \text{ and } c \text{ are real constants}\]The roots of the equation \(\mathrm{f}(z) = 0\) are \(z_1\), \(z_2\) and \(z_3\)
When plotted on an Argand diagram, the points representing these roots form the vertices of a triangle.
Given that
- \(z_1 = 2 + 4\mathrm{i}\)
- the area of the triangle is 12
Determine the two possible functions \(\mathrm{f}(z)\) (5)
| Scheme | Marks | AO |
|---|---|---|
| Method 1: using roots | ||
| \(z_2 = 2 - 4\mathrm{i}\) | B1 | 1.1b |
| \(12 = \dfrac{1}{2}(4 - {-4}) \times h \Rightarrow h = \ldots \Rightarrow z_3 = 2 + h\) or \(2 - h\) | M1 | 3.1a |
| \(\big(z - (2 - 4\mathrm{i})\big)\big(z - (2 + 4\mathrm{i})\big)(z - 5)\) or \(\big(z - (2 - 4\mathrm{i})\big)\big(z - (2 + 4\mathrm{i})\big)\big(z - (-1)\big)\) | dM1 | 2.1 |
| \((\mathrm{f}(z)) = z^3 - 9z^2 + 40z - 100\) \((\mathrm{f}(z)) = z^3 - 3z^2 + 16z + 20\) | ddM1 A1 | 1.1b 2.2a |
| (5) | ||
| (5 marks) |
Notes
Method 2: using sum, pair sum and product of roots
| Scheme | Marks | AO |
|---|---|---|
| \(z_2 = 2 - 4\mathrm{i}\) | B1 | 1.1b |
| \(12 = \dfrac{1}{2}(4 - {-4}) \times h \Rightarrow h = \ldots \Rightarrow z_3 = 2 + h\) or \(2 - h\) | M1 | 3.1a |
| Sum \(= 2 + 4\mathrm{i} + 2 - 4\mathrm{i} + \text{``}5\text{''} = 9 = -a\) Pair sum \(= (2 + 4\mathrm{i})(2 - 4\mathrm{i}) + \text{``}5\text{''}(2 + 4\mathrm{i}) + \text{``}5\text{''}(2 - 4\mathrm{i}) = 40 = b\) Product \(= \text{``}5\text{''}(2 + 4\mathrm{i})(2 - 4\mathrm{i}) = 100 = -c\) OR Sum \(= 2 + 4\mathrm{i} + 2 - 4\mathrm{i} + \text{``}(-1)\text{''} = 3 = -a\) Pair sum \(= (2 + 4\mathrm{i})(2 - 4\mathrm{i}) + \text{``}(-1)\text{''}(2 + 4\mathrm{i}) + \text{``}(-1)\text{''}(2 - 4\mathrm{i}) = 16 = b\) Product \(= \text{``}(-1)\text{''}(2 + 4\mathrm{i})(2 - 4\mathrm{i}) = -20 = -c\) | dM1 | 2.1 |
| \((\mathrm{f}(z)) = z^3 - 9z^2 + 40z - 100\) \((\mathrm{f}(z)) = z^3 - 3z^2 + 16z + 20\) | ddM1 A1 | 1.1b 2.2a |
| (5) |
Method 1:
B1: States the other complex root. May be implied by later working.
M1: Uses the area of the triangle to determine at least one value for the real root. You can award for sight of one correct value of \(z_3\)
(Note: \(h = 3 \Rightarrow z_3 = 5\) or \(-1\))
dM1: Dependent on the previous method mark. Deduces a correct expression for one of their real roots. Award for \(\big(z - (2 - 4\mathrm{i})\big)\big(z - (2 + 4\mathrm{i})\big)(z - z_3)\) or equivalent e.g. \(\left(z^2 - 4z + 20\right)(z - z_3)\)
\(z_3\) must be real.
ddM1: Dependent on previous method marks. Multiplies out to find values for the constants \(a\), \(b\) and \(c\). Award for the sight of at least 2 correct values of the constants \(a\), \(b\) and \(c\) in any one of their attempts at the function.
A1: Deduces the correct two functions.
Method 2:
B1: States the other complex root. May be implied by later working.
M1: Uses the area of the triangle to determine at least one value for the real root, You can award for sight of one correct value of \(z_3\) (Note: \(h = 3 \Rightarrow z_3 = 5\) or \(-1\))
dM1: Dependent on the previous method mark. Attempts sum, pair sum and product for at least one of their real roots.
ddM1: Dependent on previous method marks. Multiplies out to find values for the constants \(a\), \(b\) and \(c\)
Award for the sight of at least 2 correct values of the constants \(a\), \(b\) and \(c\) in any one of their attempts at the function.
A1: Deduces the correct two functions.
Note: There are several methods that can be used to obtain the first method mark:
- Uses vertices \((2,\ 4)\), \((2,\ -4)\) and \((x_3,\ 0)\) to determine at least one value for the real root. Alternatively uses \((x_3,\ y_3)\). This may come from a diagram. Do not be concerned about the variables they use.
e.g. \(12 = \dfrac{1}{2} \times 8 \times \lvert x_3 - 2\rvert \Rightarrow z_3 =\) or \(x_3 =\) - Uses the formula for the area of a triangle with vertices \((x_1,\ y_1),\ (x_2,\ y_2),\ (x_3,\ y_3)\) to determine at least one value for the real root. Do not be concerned about the variables they use.
\(\left[\dfrac{1}{2}\left\lvert x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\right\rvert\right]\)
\(12 = \dfrac{1}{2}\left\lvert 2(-4 - y_3) + 2(y_3 - 4) + x_3(4 - {-4})\right\rvert\)
\(24 = \lvert -16 + 8x_3\rvert\)
\(\Rightarrow z_3 =\) or \(x_3 =\) - Uses the shoelace method with no sign errors.
e.g. Area \(= \dfrac{1}{2}\begin{vmatrix}2 & 2 & x_3 & 2\\ 4 & -4 & 0 & 4\end{vmatrix}\)
\(12 = \dfrac{1}{2}\left\lvert (2 \times {-4} + 4x_3) - (2 \times 4 - 4x_3)\right\rvert\)
\(24 = \lvert -16 + 8x_3\rvert\)
\(\Rightarrow z_3 =\) or \(x_3 =\)
If you are uncertain in how to apply a method, then please send to review.