(a) The matrix \(\mathbf{M} = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}\) represents a rotation followed by a shear. The rotation is anticlockwise, centered at the origin, through an angle \(\theta\), with the x-axis fixed and \((0,1)\) mapped to \((2,1)\).

(b) The combined matrix is \(\mathbf{M} = \begin{pmatrix} \cos \theta + 2\sin \theta & 2\cos \theta - \sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}\).

\(To find the line of invariant points, solve:\)

\(\begin{pmatrix} \cos \theta + 2\sin \theta & 2\cos \theta - \sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}\)

This gives the equations:

\(x\sin \theta + y\cos \theta = y\)

\(x\cos \theta + 2x\sin \theta - y\sin \theta + 2y\cos \theta = x\)

Rearranging and simplifying, we find:

\(x\cos \theta + 2x\sin \theta + \frac{x\sin \theta}{1 - \cos \theta}(2\cos \theta - \sin \theta) = x\)

\((\cos \theta + 2\sin \theta)(1 - \cos \theta) + 2\sin \theta\cos \theta - \sin^2 \theta = 1 - \cos \theta\)

\(\cos^2 \theta - \cos^2 \theta + 2\sin \theta - \sin^2 \theta = 1 - \cos \theta\)

\(\Rightarrow \sin \theta + \cos \theta = 1\)

Solving gives \(\theta = \frac{1}{2} \pi\).

(a) The curve is a spiral starting at the pole \(\theta = 0\) and spiraling outward as \(\theta\) increases to \(2\pi\). The radius \(r\) is strictly increasing, and the gradient is correct at points of intersection with the initial line.

(b) The area \(A\) is given by:

\(A = \frac{1}{2} \int_{0}^{2\pi} \theta^2 e^{\frac{1}{4} \theta} d\theta\)

Integrate by parts:

\(= \frac{1}{2} \left[ 4\theta^2 e^{\frac{1}{4} \theta} \right]_{0}^{2\pi} - \int_{0}^{2\pi} 8\theta e^{\frac{1}{4} \theta} d\theta\)

Integrate by parts again:

\(= \frac{1}{2} \left[ 4\theta^2 e^{\frac{1}{4} \theta} - 4(4\theta e^{\frac{1}{4} \theta} + 16)e^{\frac{1}{4} \theta} \right]_{0}^{2\pi}\)

\(= \left[ 2\theta^2 e^{\frac{1}{4} \theta} - 16\theta e^{\frac{1}{4} \theta} + 64e^{\frac{1}{4} \theta} \right]_{0}^{2\pi}\)

\(= (8\pi^2 - 32\pi + 64)e^{\frac{1}{2}\pi} - 64\)

(c) Let \(y = \theta e^{\frac{1}{8} \theta} \sin \theta\). Differentiate:

\(\frac{dy}{d\theta} = e^{\frac{1}{8} \theta} \cos \theta + \sin \theta \left( \frac{1}{8} \theta e^{\frac{1}{8} \theta} + e^{\frac{1}{8} \theta} \right)\)

Set \(\frac{dy}{d\theta} = 0\):

\(e^{\frac{1}{8} \theta} (\theta \cos \theta + (\frac{1}{8} \theta + 1) \sin \theta) = 0\)

Check sign change:

\(5 \cos 5 + (\frac{5}{8} + 1) \sin 5 = -0.1399\)

\(5.05 \cos 5.05 + (\frac{5.05}{8} + 1) \sin 5.05 = 0.1336\)

There is a sign change between 5 and 5.05, confirming a root exists.

(a) To find the equation of the plane ABC, we first determine the direction vectors \(\overrightarrow{AB} = 2\mathbf{j} + 4\mathbf{k}\) and \(\overrightarrow{AC} = \mathbf{i} - \mathbf{j} + \mathbf{k}\).

The normal vector \(\mathbf{n}\) to the plane is given by the cross product \(\overrightarrow{AB} \times \overrightarrow{AC}\):

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 2 & 4 \\ 1 & -1 & 1 \end{vmatrix} = \begin{pmatrix} 6 \\ 4 \\ -2 \end{pmatrix}\)

Substituting point \(A(1, 0, -2)\) into the plane equation \(3x + 2y - z = d\), we find \(d = 5\).

(b) The normal to the plane ABD is found using the cross product:

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 2 & 4 \\ 0 & 0 & t+2 \end{vmatrix} = \begin{pmatrix} 2t + 4 \\ 0 \\ 0 \end{pmatrix}\)

The angle \(\theta\) between the planes is given by:

\(\cos \theta = \frac{\left( \begin{pmatrix} 3 \\ 2 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \right)}{\sqrt{14} \cdot 1}\)

\(\theta = 36.7^\circ\).

(c) To find \(t\) such that the shortest distance between lines AB and CD is \(\sqrt{2}\), we use the formula for the perpendicular distance between skew lines:

\(\frac{1}{\sqrt{t^2 - 2t + 6}} \left| \begin{pmatrix} 1 \\ -1 \\ -2 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix} \right| = \frac{t + 2}{\sqrt{t^2 - 2t + 6}}\)

Setting this equal to \(\sqrt{2}\) and solving gives:

\(t^2 - 8t + 8 = 0 \Rightarrow t = 2(2 \pm \sqrt{2})\)

Thus, \(t = 6.83\) and \(t = 1.17\).

(a) The vertical asymptotes occur where the denominator is zero: \(2x^2 - 7x - 4 = 0\). Solving gives \(x = -\frac{1}{2}\) and \(x = 4\). The horizontal asymptote is found by considering the degrees of the polynomials: \(y = 1\).

(b) To find stationary points, set \(\frac{dy}{dx} = 0\). Differentiate using the quotient rule: \(\frac{dy}{dx} = \frac{(2x^2 - 7x - 4)(4x - 5) - (2x^2 - 5x)(4x - 7)}{(2x^2 - 7x - 4)^2}\). Simplifying gives \(4x^2 + 16x - 20 = 0\), leading to points \((-5, \frac{25}{3})\) and \((1, \frac{1}{3})\).

(c) The curve intersects the axes at \((0,0)\) and \((\frac{5}{2},0)\).

(d) The sketch of \(y = \left| \frac{2x^2 - 5x}{2x^2 - 7x - 4} \right|\) reflects the negative parts of the curve above the x-axis.

(e) Solve \(\left| \frac{2x^2 - 5x}{2x^2 - 7x - 4} \right| < \frac{1}{9}\) by considering \(\frac{2x^2 - 5x}{2x^2 - 7x - 4} < \frac{1}{9}\) and \(\frac{2x^2 - 5x}{2x^2 - 7x - 4} > -\frac{1}{9}\). Solving these inequalities gives the intervals \(\frac{13}{10} < x < \frac{19}{16}\) and \(\frac{\sqrt{33}}{16} < x < \frac{13}{10} + \frac{3}{10} \sqrt{21}\).

(a) The stretch parallel to the x-axis with scale factor 14 is represented by the matrix \(\begin{pmatrix} 14 & 0 \\ 0 & 1 \end{pmatrix}\).

The rotation anticlockwise about the origin through angle \(\frac{1}{3} \pi\) is represented by \(\begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}\).

Thus, \(\mathbf{M} = \begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix} \begin{pmatrix} 14 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 7 & -\frac{\sqrt{3}}{2} \\ 7\sqrt{3} & \frac{1}{2} \end{pmatrix}\).

Therefore, \(2\mathbf{M} = \begin{pmatrix} 14 & -\sqrt{3} \\ 14\sqrt{3} & 1 \end{pmatrix}\).

(b) The transformation \(\mathbf{M}\) transforms \(\begin{pmatrix} x \\ y \end{pmatrix}\) to \(\begin{pmatrix} 7x - \frac{\sqrt{3}}{2}y \\ 7\sqrt{3}x + \frac{1}{2}y \end{pmatrix}\).

For invariant lines, \(y = mx\) and \(Y = mX\).

\(7\sqrt{3}x + \frac{1}{2}mx = m(7x - \frac{\sqrt{3}}{2}mx)\).

\(7\sqrt{3} + \frac{1}{2}m = 7m - \frac{\sqrt{3}}{2}m^2\).

\(\sqrt{3}m^2 - 13m + 14\sqrt{3} = 0\).

Solving gives \(m = 2\sqrt{3}\) and \(m = \frac{7}{\sqrt{3}}\).

Thus, the invariant lines are \(y = 2\sqrt{3}x\) and \(y = \frac{1}{\sqrt{3}}x\).

(c) The inverse of \(2\mathbf{M}\) is \(\mathbf{M}^{-1} = \frac{1}{14} \begin{pmatrix} 1 & \sqrt{3} \\ -14\sqrt{3} & 7 \end{pmatrix}\).