AS June 2023 Paper 1 Q7
7.
| Scheme | Marks | AO |
|---|---|---|
 | M1 A1 B1 | 3.1a 1.1b 1.1b |
| (3) |
Notes
M1: Draws a single straight line through both axes with a negative gradient. Ignore any line joining (3, 0) and (0, −6)
A1: Draws a single straight line through both axes with a negative gradient which has a negative \(y\) intercept. Ignore any intercept marked on the axes. Ignore any line joining (3, 0) and (0, −6)
B1: Shades the area above their straight line (not a bounded region such as a triangle bounded by the axes and the line)
| Scheme | Marks | AO |
|---|---|---|
| \(m = \tan\left(\dfrac{\pi}{3}\right)\left\{= \sqrt{3}\right\}\) and \(y - 0 = m(x - 2)\) leads to \(y - 0 = \sqrt{3}(x - 2)\) or \(y = \sqrt{3}x - 2\sqrt{3}\) \(m = \tan\left(\dfrac{\pi}{6}\right)\left\{= \dfrac{\sqrt{3}}{3}\right\}\) and \(y - 0 = m(x - (-1))\) leads to \(y - 0 = \dfrac{\sqrt{3}}{3}(x - (-1))\) or \(y = \dfrac{\sqrt{3}}{3}x + \dfrac{\sqrt{3}}{3}\) | M1 A1 A1 | 3.1a 1.1b 1.1b |
| \(\sqrt{3}x - 2\sqrt{3} = \dfrac{\sqrt{3}}{3}x + \dfrac{\sqrt{3}}{3} \Rightarrow x = \ldots\) | M1 | 1.1b |
| \(y = \sqrt{3}\left(\dfrac{7}{2}\right) - 2\sqrt{3} = \ldots\) | M1 | 1.1b |
| \(\{w =\}\ \dfrac{7}{2} + \dfrac{3\sqrt{3}}{2}\mathrm{i}\) | A1 | 2.1 |
| (6) | ||
| (9 marks) |
Notes
M1: Finds the Cartesian equations for both loci by using the gradient as tan(argument) and correct coordinate. Must be an attempt at both equations but one correct equation scores this mark
A1: One equation correct, need not be simplified
A1: Both equations correct, need not be simplified
M1: Solve simultaneously to find either the real or imaginary component.
M1: Finds the other component to complete the process of finding \(w\).
A1: Correct exact answer
Note: If leaves the answer as a coordinate this is A0. If defines \(w = a + b\mathrm{i}\) and then states \(a = \dfrac{7}{2}\) and \(b = \dfrac{3\sqrt{3}}{2}\) this is A1
Note: If candidates use decimal instead of exact values throughout allow the method marks \(y = 1.73x - 3.46\) and \(y = 0.58x + 0.58\)
Alternative
| Scheme | Marks | AO |
|---|---|---|
| \(\tan\left(\dfrac{\pi}{6}\right) = \dfrac{\sqrt{3}}{3} = \dfrac{y}{x_{-1}}\) and \(\tan\left(\dfrac{\pi}{3}\right) = \sqrt{3} = \dfrac{y}{x_2}\) | M1 | 1.1b |
 | A1 A1 | 1.1b 1.1b |
| \(y\sqrt{3} = y\dfrac{\sqrt{3}}{3} + 3 \Rightarrow y = \ldots\) | M1 | 3.1a |
| Uses \(x = y\sqrt{3} - 1\) or \(x = \dfrac{\sqrt{3}}{3}y + 2\) with their value of \(y\) leading to a value for \(x\) | M1 | 1.1b |
| \((w =)\ \dfrac{7}{2} + \dfrac{3\sqrt{3}}{2}\mathrm{i}\) | A1 | 2.1 |
| (6) |
M1: Use both arguments to form equations involving \(x\) and \(y\)
A1: (One correct triangle) value for \(x\) in terms of \(y\)
A1: (Two correct triangles), values for \(x\) in terms of \(y\)
M1: Forms and solves an equation \(y\sqrt{3} = y\dfrac{\sqrt{3}}{3} + 3 \Rightarrow y = \ldots\) must be come from \(x_2 = x_{-1} + 3\)
M1: Uses their \(y\) value and \(x = y\sqrt{3} - 1\) or \(x = \dfrac{\sqrt{3}}{3}y + 2\) to find a value for \(x\)
A1: Correct exact answer
Q7(ii) Two alternatives seen
Alternative 3
| Scheme | Marks | AO |
|---|---|---|
| \(b = 3\sin\left(\dfrac{\pi}{3}\right)\) and \(c = 3\cos\left(\dfrac{\pi}{3}\right)\) | M1 A1 A1 | 3.1a 1.1b 1.1b |
| \(b = 3\sin\left(\dfrac{\pi}{3}\right) = \ldots\) | M1 | 1.1b |
| \(a = 2 + 3\cos\left(\dfrac{\pi}{3}\right) = \ldots\) | M1 | 1.1b |
| \((w =)\ \dfrac{7}{2} + \dfrac{3\sqrt{3}}{2}\mathrm{i}\) | A1 | 2.1 |
| (6) |
M1: Uses correct geometry to form equations involving \(a\) and \(c\)
A1: One correct equation
A1: Two correct equations
M1: Finds the imaginary component
M1: Uses 2 + their \(c\) to find the real component
A1: Correct exact answer
Alternative 4
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{3}{\sin 30} = \dfrac{AB}{\sin 120}\) \(AB = 3\sqrt{3}\) | M1 A1 | 3.1a 1.1b |
| \(\sin 30 = \dfrac{BC}{3\sqrt{3}}\), \(BC = \dfrac{3}{2}\sqrt{3}\) \(\sin 60 = \dfrac{AC}{3\sqrt{3}}\), \(AC = \dfrac{9}{2}\) | M1 A1 | 1.1b 1.1b |
| Uses trigonometry to find the other component | M1 | 1.1b |
| \((w =)\ \dfrac{7}{2} + \dfrac{3\sqrt{3}}{2}\mathrm{i}\) | A1 | 2.1 |
| (6) |
(Corrected from the printed mark scheme: \(AC = \dfrac{7}{2}\). \(AC = 3\sqrt{3}\sin 60 = \dfrac{9}{2}\); the real part of \(w\) is then \(\dfrac{9}{2} - 1 = \dfrac{7}{2}\).)
M1: Uses the sine rule to find the length \(AB\)
A1: Correct length \(AB\)
M1: Uses trigonometry to find either the real or imaginary component
A1: Correct real or imaginary component
M1: Uses trigonometry to find the other component
A1: Correct exact answer