A2 October 2021 Paper 1 Q1
1. The transformation \(P\) is an enlargement, centre the origin, with scale factor \(k\), where \(k \gt 0\)
The transformation \(Q\) is a rotation through angle \(\theta\) degrees anticlockwise about the origin.
The transformation \(P\) followed by the transformation \(Q\) is represented by the matrix
\[\mathbf{M} = \begin{pmatrix}-4 & -4\sqrt{3}\\ 4\sqrt{3} & -4\end{pmatrix}\]A square \(S\) has vertices at the points with coordinates \((0, 0)\), \((a, -a)\), \((2a, 0)\) and \((a, a)\) where \(a\) is a constant.
The square \(S\) is transformed to the square \(S^{\prime}\) by the transformation represented by \(\mathbf{M}\).
| Scheme | Marks | AO |
|---|---|---|
| Way 1 \(\det\mathbf{M} = -4 \times -4 - 4\sqrt{3} \times -4\sqrt{3} = \ldots \Rightarrow k = \sqrt{\det\mathbf{M}} = \ldots\) | M1 | 3.1a |
| \(k = 8\) | A1 | 1.1b |
| \(\Rightarrow \mathbf{Q} = \begin{pmatrix}-\tfrac{1}{2} & -\tfrac{\sqrt{3}}{2}\\ \tfrac{\sqrt{3}}{2} & -\tfrac{1}{2}\end{pmatrix} = \begin{pmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{pmatrix} \Rightarrow \cos\theta = -\dfrac{1}{2} \Rightarrow \theta = \ldots\) | M1 | 1.1b |
| \(\left(\cos\theta \lt 0, \sin\theta \gt 0 \Rightarrow \text{ Quadrant 2 so}\right)\ \theta = 120^\circ\) | A1 | 1.1b |
| (4) |
Notes
Way 1
M1: A full method to find \(k\) such as attempting the square root of the determinant of \(\mathbf{M}\). It is immediately deducible so the method may be implied by \(k = 8\).
A1: \(k = 8\)
M1: A full method to find a value of \(\theta\) using their \(k\), no need to justify quadrant. Only one equation needed for this mark. Allow if a radians answer is given. May be implied by a correct angle.
A1: Correct angle in degrees.
Way 2
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{pmatrix}\begin{pmatrix}k & 0\\ 0 & k\end{pmatrix} = k\begin{pmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{pmatrix} = \begin{pmatrix}-4 & -4\sqrt{3}\\ 4\sqrt{3} & -4\end{pmatrix}\) | M1 | 3.1a |
| Achieves both the equations \(k\cos\theta = -4\) and \(k\sin\theta = 4\sqrt{3}\) | A1 | 1.1b |
| \(\dfrac{k\sin\theta}{k\cos\theta} = \dfrac{4\sqrt{3}}{-4} \Rightarrow \tan\theta = -\sqrt{3} \Rightarrow \theta = \ldots\) | M1 | 1.1b |
| \(\theta = 120^\circ\) and \(k = 8\) | A1 | 1.1b |
| (4) |
M1: Multiplies the correct matrix representing transformation \(Q\) by the matrix representing transformation \(P\) and sets equal to matrix \(\mathbf{M}\). Allow for the matrices either way round as the transformations commute. No need to see the identity matrix, just multiplying through by \(k\) is sufficient.
A1: Both correct equations. Note that if a correct value of \(k\) is found, this A is scored under Way 1.
M1: Solves their simultaneous equations to find a value for \(\theta\) (or \(k\))
A1: \(\theta = 120^\circ\) and \(k = 8\)
| Scheme | Marks | AO |
|---|---|---|
| Area of \(S^{\prime}\) = area of \(S \times k^2\) (The area of the square \(S = 2a^2\)) | M1 | 1.1b |
| Area of \(S^{\prime} = 128a^2\) | A1ft | 2.2a |
| (2) | ||
| (6 marks) |
Notes
M1: Complete method to find the area of \(S^{\prime}\): \(\text{``}\text{their } k^2\text{''} \times \text{``}\text{their } 2a^2\text{''}\). Must be an attempt at the area of \(S\) but it need not be correct.
A1ft: Deduces the correct area for \(S^{\prime}\), follow through their value of \(k\)