(a) The vertical asymptotes occur where the denominator is zero: \(x^2 - x - 2 = 0\). Solving gives \(x = -1\) and \(x = 2\). The horizontal asymptote is found by considering the limits as \(x\) approaches infinity, giving \(y = 1\).

(b) To find stationary points, set \(\frac{dy}{dx} = 0\). The derivative is \(\frac{dy}{dx} = \frac{(x^2 - x - 2)(2x) - (x^2 + 2)(2x - 1)}{(x^2 - x - 2)^2}\). Solving \(x^2 + 8x - 2 = 0\) gives the stationary points \((-8.2, 0.9)\) and \((0.2, -0.9)\).

(c) The sketch shows the curve with asymptotes and intersection at \((0, -1)\).

(d) The curve \(y = \frac{1}{f(x)}\) is sketched by inverting the values of \(f(x)\) from part (c).

(e) Solve \(\frac{x^2 + 2}{x^2 - x - 2} = 1\) and \(\frac{x^2 + 2}{x^2 - x - 2} = -1\) to find critical points. The solution is \(-4 < x < -1, \ 0 < x < \frac{1}{2}, \ x > 2\).

(a) Base case: For \(n = 1\),

\(2\mathbf{A} = \begin{pmatrix} 6 & 0 \\ 2 & 2 \end{pmatrix} = \begin{pmatrix} 2 \times 3^1 & 0 \\ 3^1 - 1 & 2 \end{pmatrix}.\)

Assume true for \(n = k\), so

\(2\mathbf{A}^k = \begin{pmatrix} 2 \times 3^k & 0 \\ 3^k - 1 & 2 \end{pmatrix}.\)

Then for \(n = k + 1\),

\(2\mathbf{A}^{k+1} = \begin{pmatrix} 2 \times 3^{k+1} & 0 \\ 3^{k+1} - 3 + 2 & 2 \end{pmatrix}.\)

Thus, true for \(n = k + 1\). By induction, true for all positive integers.

(b) \(\det \mathbf{A}^n = \det \begin{pmatrix} 3^n & 0 \\ \frac{1}{2}(3^n - 1) & 1 \end{pmatrix} = 3^n\).

\(\mathbf{A}^{-n} = 3^{-n} \begin{pmatrix} 1 & 0 \\ \frac{1}{2}(1 - 3^n) & 3^n \end{pmatrix}\).

(a) Using the identity for the sum of squares of roots: \(\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\).

From the cubic equation, \(\alpha + \beta + \gamma = -4\) and \(\alpha\beta + \beta\gamma + \gamma\alpha = 6\).

Thus, \(\alpha^2 + \beta^2 + \gamma^2 = (-4)^2 - 2(6) = 16 - 12 = 4\).

(b) Expand \((\alpha + r)^2 = \alpha^2 + 2\alpha r + r^2\).

Therefore, \(\sum_{r=1}^{n} ((\alpha + r)^2 + (\beta + r)^2 + (\gamma + r)^2) = \sum_{r=1}^{n} (\alpha^2 + 2\alpha r + r^2 + \beta^2 + 2\beta r + r^2 + \gamma^2 + 2\gamma r + r^2)\).

Using \(\alpha + \beta + \gamma = -4\) and \(\alpha^2 + \beta^2 + \gamma^2 = 4\), we have:

\(\sum_{r=1}^{n} (3r^2 + 2(-4)r + 4) = \sum_{r=1}^{n} (3r^2 - 8r + 4)\).

Using standard summation formulas:

\(\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}\) and \(\sum_{r=1}^{n} r = \frac{n(n+1)}{2}\).

Thus, \(3 \sum_{r=1}^{n} r^2 - 8 \sum_{r=1}^{n} r + 4n = 3 \cdot \frac{n(n+1)(2n+1)}{6} - 8 \cdot \frac{n(n+1)}{2} + 4n\).

Simplifying gives:

\(\frac{n(n+1)(2n+1)}{2} - 4n(n+1) + 4n\).

\(= n(n^2 - \frac{5}{2}n + \frac{1}{2})\).

Thus, \(a = -\frac{5}{2}, b = \frac{1}{2}\).

(a) Start by expressing \(\frac{1}{(kr+1)(kr-k+1)}\) using partial fractions:

\(\frac{1}{(kr+1)(kr-k+1)} = \frac{1}{k} \left( \frac{1}{kr-k+1} - \frac{1}{kr+1} \right)\)

Now, apply the method of differences:

\(\sum_{r=1}^{n} \frac{1}{(kr+1)(kr-k+1)} = \frac{1}{k} \left( 1 - \frac{1}{k+1} + \frac{1}{k+1} - \frac{1}{2k+1} + \ldots + \frac{1}{k(n-1)+1} - \frac{1}{kn+1} \right)\)

This simplifies to:

\(\frac{1}{k} \left( 1 - \frac{1}{kn+1} \right)\)

(b) For the infinite series, the terms after the first cancel out, leaving:

\(\frac{1}{k}\)

(c) Use the same method of differences for the range \(r=n\) to \(r=n^2\):

\(\sum_{r=n}^{n^2} \frac{1}{(kr+1)(kr-k+1)} = \sum_{r=1}^{n^2} \frac{1}{(kr+1)(kr-k+1)} - \sum_{r=1}^{n-1} \frac{1}{(kr+1)(kr-k+1)}\)

This results in:

\(\frac{1}{k} \left( \frac{1}{k(n-1)+1} - \frac{1}{kn^2+1} \right)\)

(a) For a matrix to represent a rotation, it must be of the form \(\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}\). Comparing with \(\mathbf{M} = \begin{pmatrix} a & b^2 \\ c^2 & a \end{pmatrix}\), we equate \(b^2 = -c^2\), which is impossible for real \(b\) and \(c\) with \(b \neq 0\).

(b) Consider the transformation \(\begin{pmatrix} a & b^2 \\ c^2 & a \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} ax + b^2y \\ c^2x + ay \end{pmatrix}\). For invariant lines, \(y = mx\) and \(Y = mX\). Solving \(c^2x + amx = m(ax + b^2mx)\) gives \(c^2 + am = ma + b^2m^2 \Rightarrow c^2 = b^2m^2\). Thus, \(y = \frac{c}{b}x\) and \(y = -\frac{c}{b}x\).

(c) The enlargement matrix is \(\begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix}\) and the shear matrix is \(\begin{pmatrix} 1 & 5 \\ 0 & 1 \end{pmatrix}\). The combined transformation is \(\begin{pmatrix} 1 & 5 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} = \begin{pmatrix} 5 & 25 \\ 0 & 5 \end{pmatrix}\).

(d) The area of triangle PQR is \(12 \times \det(\mathbf{M}) = 12 \times 25 = 300 \text{ cm}^2\).