$y_1 = x^n w_1 + ax^{n-1}w, y_2 = x^n w_2 + 2ax^{n-1}w_1 + a(a-1)x^{n-2}w$ | B1B1 |
(Must see general results with $a$) |
Obtain any DE of the form $P(x,a)w_2 + Q(x,a)w_1 + R(x,a)w = f(x)$ | M1 |
Sets $a = -2$ and obtains required $x - w$ DE (AG) | M1A1 oew |
Complementary function $= Ae^{-x} + Be^{-x^2}$ | M1A1 |
Puts $P\sin 2x + Q\cos 2x$ to obtain 2 linear equations in $P$ and $Q$ | M1A1 |
Solves to obtain $P = 0, Q = -1$ | A1 |
$y = x^{-2}[Ae^{-x} + Be^{-x^2} - \cos 2x]$ | A1 |
Allow $x^2 y = [Ae^{-x} + Be^{-x^2} - \cos 2x]$ |

# Question 12 (EITHER)

**(i)** $\vec{AB} \times \vec{CD} = (\lambda - 2)\mathbf{i} + (4x - 12)\mathbf{j} - 4k$ | M1A1 |
$(-5\mathbf{i} + 2\mathbf{j} + 4k) \cdot [(\lambda - 2)\mathbf{i} + (4x - 12)\mathbf{j} - 4k] = 3\lambda - 30$ | M1A1 |
$|3\lambda - 30|/\sqrt{(\lambda - 2)^2 + (4x - 12)^2 + 16} = 3$ | M1A1 |
$\lambda^2 - 5\lambda + 4 = 0$ (AG) | A1 |

**(ii)** $\lambda = 1, 4$ | B1 |
$\lambda = 1$: $\mathbf{n}_1 = 5\mathbf{i} + 13\mathbf{j} - 7k$ | M1A1 cao |
$\lambda = 4$: $\mathbf{n}_2 = 8\mathbf{i} + 25\mathbf{j} - 7k$ | M1A1 cao |
Acute angle between planes $= \cos^{-1}|\mathbf{n}_1 \cdot \mathbf{n}_2/|\mathbf{n}_1||\mathbf{n}_2|| = 12.1°$ | M1A1 cao |
[For lines 2 and 3: $\frac{10 - 5\lambda - 24 + 8x - 16}{\sqrt{(2-\lambda)^2 + (12-4x)^2 + 4^2}} = 3$ | M1A1 |
$(\lambda - 10)^2 = 164 - 100\lambda + 17\lambda^2$ | M1A1] |

# Question 12 (OR)

**(i)** Transform given matrix to an echelon form, e.g., $\begin{pmatrix} 1 & 2 & -1 & -1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}$ | M1A1 |
$\dim(V) = 3$ **OR BY EQUIVALENT METHOD** If written down with no working 1/3 | B1 |

**(ii)** $a_1 \begin{pmatrix} 1 \\ 1 \\ 1 \\ 0 \end{pmatrix} + a_2 \begin{pmatrix} 2 \\ 3 \\ 0 \\ 3 \end{pmatrix} + a_3 \begin{pmatrix} -1 \\ -1 \\ 3 \\ -4 \end{pmatrix} = \mathbf{0} \Rightarrow a_1 = a_2 = a_3 = 0$, shown $\Rightarrow$ linear independence | M2A2 |

**(iii)** $\begin{Bmatrix} \begin{pmatrix} 1 \\ 1 \\ 1 \\ 0 \end{pmatrix} \begin{pmatrix} 2 \\ 3 \\ 0 \\ 3 \end{pmatrix} \begin{pmatrix} -1 \\ -1 \\ 3 \\ -4 \end{pmatrix} \end{Bmatrix}$ | B1 |

**(iv)** W not a vector space since W does not contain the zero vector, or equivalent or not closed wrt addition | B1 |

**(v)** Reduces $\begin{pmatrix} 1 & 2 & -1 & x \\ 1 & 3 & -1 & y \\ 1 & 0 & 3 & z \\ 0 & 3 & -4 & t \end{pmatrix} \to \begin{pmatrix} 1 & 2 & -1 & x \\ 0 & 1 & 0 & y-x \\ 0 & 0 & 4 & -3x+2y+z \\ 0 & 0 & 0 & -y+z+t \end{pmatrix}$ | M1A1 |
$\Rightarrow \mathbf{b} = \begin{pmatrix} x \\ y \\ z \\ t \end{pmatrix} \in V$ iff $y - z - t = 0$ or equivalent method | M1A1 |
so that $b \in W \Rightarrow y - z - t \neq 0$ | A1 |
**Alternative:** suppose $\begin{pmatrix} x \\ y \\ z \\ t \end{pmatrix} = \alpha \begin{pmatrix} 1 \\ 1 \\ 1 \\ 0 \end{pmatrix} + \beta \begin{pmatrix} 2 \\ 3 \\ 0 \\ 3 \end{pmatrix} + \gamma \begin{pmatrix} -1 \\ -1 \\ 3 \\ -4 \end{pmatrix}$ (i.e. in $V$) |
$x = ...$, $y = ...$, $z = ...$, $\Rightarrow y - z - t = ... = 0$ |
Hence $\begin{pmatrix} x \\ y \\ z \\ t \end{pmatrix} \in W \Rightarrow y - z - t \neq 0$ |