AS June 2019 Paper 1 Q4
4 In this question you must show detailed reasoning.
You are given that \(\mathrm{f}(z) = 4z^4 - 12z^3 + 41z^2 - 128z + 185\) and that \(2 + \mathrm{i}\) is a root of the equation \(\mathrm{f}(z) = 0\).
Two loci on an Argand diagram are defined by \(C_1 = \{z : |z| = r_1\}\) and \(C_2 = \{z : |z| = r_2\}\) where \(r_1 \gt r_2\). You are given that two of the points representing the roots of \(\mathrm{f}(z) = 0\) are on \(C_1\) and two are on \(C_2\). \(R\) is the region on the Argand diagram between \(C_1\) and \(C_2\).
Determine whether or not the point on the Argand diagram which represents \(\omega\) lies in \(R\). [2]
| Scheme | Marks | AO |
|---|---|---|
| \(2 - \mathrm{i}\) (is also a root) | B1 | 2.2a |
| \((z - (2 + \mathrm{i}))(z - (2 - \mathrm{i}))\) \((z - 2)^2 + 1\) | M1 | 1.1 |
| \(z^2 - 4z + 5\) | A1 | 2.1 |
| \((z^2 - 4z + 5)(4z^2 + \ldots + 37)\) | M1 | 1.1 |
| \((z^2 - 4z + 5)(4z^2 + 4z + 37)\) | A1 | 1.1 |
| [5] |
Notes
DR: Detailed reasoning required for this question
B1: Can be implied by \((z - (2 - \mathrm{i}))\)
M1: (1st) Correctly determining the form of the quadratic in factor form
A1: (1st) Working needs to be shown here (Show detailed reasoning). At least 1 step between factor form and final answer
M1: (2nd) With non-zero \(z\) term. Or attempt at symbolic division resulting in at least \(4z^2\). Or genuine attempt at method comparing coefficients leading to at least one useful equation (eg \(z^4\): \(A = 4\))
Must be a product of two quadratics
Multiplication table might be used (backwards)
Inspection is fine here, so candidates might just write down the correct two quadratic factors.
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{-4 \pm \sqrt{4^2 - 4 \times 4 \times 37}}{2 \times 4}\) soi | M1* | 1.1a |
| \(\dfrac{-4 \pm 24\mathrm{i}}{8}\) or \(\dfrac{-1 \pm 6\mathrm{i}}{2}\) etc. | A1dep(*) | 2.1 |
| \(\dfrac{-1 \pm 6\mathrm{i}}{2}\), \(2 \pm \mathrm{i}\) | A1(ft) dep(*) | 1.1 |
| [3] |
Notes
M1*: Correct method for finding a root of a quadratic (\(\pm\) not necessary)
or \(4\left(\left(z + \tfrac{1}{2}\right)^2 - \tfrac{1}{4}\right) + 37\) oe
Need to see some method
A1dep(*): Simplification of the square root to include i (\(\pm\) is now necessary)
Must be of the form \(z = a + b\mathrm{i}\) (or \(z = \frac{a + b\mathrm{i}}{c}\)) No square roots left in.
A1(ft) dep(*): All four - aef
Follow through on incorrect roots from previous mark as long as method mark awarded i.e. this mark is for realising that there are four roots and including the two from part (a). A0 if more than 4 roots given.
SC: If M0 awarded but all four correct roots are given with no extra then award SC B1
| Scheme | Marks | AO |
|---|---|---|
| \(r_2 = |2 \pm \mathrm{i}| = \sqrt{5}\) | B1 | 2.2a |
| \(r_1 = \left|\dfrac{-1 \pm 6\mathrm{i}}{2}\right| = \dfrac{\sqrt{37}}{2}\) | B1ft | 2.2a |
| \(\pi\left(\dfrac{\sqrt{37}}{2}\right)^2 - \pi\left(\sqrt{5}\right)^2\) | M1 | 3.1a |
| The area between the two circles is \(\dfrac{17}{4}\pi\) | A1ft | 3.2a |
| [4] |
Notes
B1: Need to see \(|2 + \mathrm{i}|\) or \(|2 - \mathrm{i}|\)
Or \(\sqrt{2^2 + 1^2} = \sqrt{5}\) etc.
B1ft: ft the magnitude of the other of their conjugate pairs
Same comment as above.
M1: Either way round. Their \(r_1\) and/or \(r_2\).
A1ft: \(\pi(r_1^2 - 5)\) simplified
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Positive area needed here. Must follow B1B1(ft)M1
| Scheme | Marks | AO |
|---|---|---|
| \(\omega = -\dfrac{-12}{4} = 3\ldots\) | B1 | 2.2a |
| \(\ldots\)and \(\sqrt{5} \lt 3 \lt \dfrac{\sqrt{37}}{2}\) so \(\omega\) is in \(R\) | E1 | 2.3 |
| [2] |
Notes
B1: or \(\dfrac{-1 + 6\mathrm{i}}{2} + \dfrac{-1 - 6\mathrm{i}}{2} + 2 + \mathrm{i} + 2 - \mathrm{i} = 3\)
Some justification needed, not just \(\omega = 3\). Allow \(3 + 0\mathrm{i}\)
E1: Or \(5 \lt 9 \lt 37/4\)
Or \(2.23\ldots \lt 3 \lt 3.04\ldots\)
Both end comparisons needed for E1
No follow through given in this part
Conclusion needed
Can be smaller radius is 2.24, larger radius is 3.04 so \(\omega\) is in \(R\)
Diagram is ok for E1 if it implies both end comparisons