AS June 2022 Paper 1 Q8
8 A transformation T of the plane has matrix \(\mathbf{M}\), where \(\mathbf{M} = \begin{pmatrix} \cos\theta & 2\cos\theta - \sin\theta \\ \sin\theta & 2\sin\theta + \cos\theta \end{pmatrix}\).
The matrix \(\mathbf{N}\) is \(\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\).
| Scheme | Marks | AO |
|---|---|---|
| \(\det\mathbf{M} = \cos\theta(2\sin\theta + \cos\theta) - \sin\theta(2\cos\theta - \sin\theta)\) | M1 | 2.1 |
| \(= \cos^2\theta + \sin^2\theta = 1\) so T preserves area | A1 | 2.2a |
| [2] |
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix} \cos\theta & 2\cos\theta - \sin\theta \\ \sin\theta & 2\sin\theta + \cos\theta \end{pmatrix}\begin{pmatrix} 0 \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ y^{\prime} \end{pmatrix}\) | M1 | 3.1a |
| \(\Rightarrow (2\cos\theta - \sin\theta)y = 0\) [for all \(y\)] \(\Rightarrow 2\cos\theta = \sin\theta\) | A1 | 1.1 |
| \(\Rightarrow \tan\theta = 2\) | M1 | 3.1a |
| \(\theta = 1.11\) rads | A1 | 3.2a |
| [4] |
Notes
M1: (1st) condone \(\begin{pmatrix} \cos\theta & 2\cos\theta - \sin\theta \\ \sin\theta & 2\sin\theta + \cos\theta \end{pmatrix}\begin{pmatrix} 0 \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ y \end{pmatrix}\) for this M1…
A1: (1st) … but not this A1
M1: (2nd) solved using tan = sin/cos
A1: (2nd) cao 1.11 or better condone \(63.4^\circ\)
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\mathbf{N}^{-1} = \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}\) | B1 | 1.1 |
| \(\mathbf{M}\mathbf{N}^{-1} = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}\) | B1 | 1.1 |
| [2] | ||
| (ii) T is a shear | M1 | 2.2a |
| with \(x\)-axis fixed mapping \((0, 1)\) to \((2, 1)\) | A1 | 1.1 |
| followed by a rotation | M1 | 2.2a |
| [about O] anticlockwise through \(\theta\) | A1 | 1.1 |
| [4] |
Notes
(c)(ii)
A1: (1st) not ‘scale factor 2’ or ‘shear factor 2’
A1: (2nd) Maximum 3 out of 4 if order is wrong