(a) The vertical asymptotes occur where the denominator is zero: \(4 - 4x^2 = 0\), giving \(x = 1\) and \(x = -1\). The horizontal asymptote is \(y = 0\) as \(x\) approaches infinity.

(b) To find stationary points, set \(\frac{dy}{dx} = 0\). Differentiate: \(\frac{dy}{dx} = \frac{(4-4x^2)(4) - (4x+5)(-8x)}{(4-4x^2)^2}\). Simplify and solve \(16x^2 + 40x + 16 = 0\) to find \((-2, \frac{1}{4})\) and \(-\frac{1}{2}, 1\).

(c) The curve intersects the x-axis where \(y = 0\), giving \(x = -\frac{5}{4}\). It intersects the y-axis where \(x = 0\), giving \(y = \frac{5}{4}\).

(d) For \(y = \left| \frac{4x+5}{4-4x^2} \right|\), solve \(4|4x+5| > 5|4-4x^2|\). This gives two cases: \(\frac{4x+5}{4-4x^2} > \frac{5}{4}\) and \(\frac{4x+5}{4-4x^2} < -\frac{5}{4}\). Solving these inequalities gives the ranges \(\frac{2}{3} - \frac{1}{3}\sqrt{6} < x < -\frac{4}{3}\) and \(0 < x < \frac{2}{3} + \frac{1}{3}\sqrt{6}\), excluding \(x = \pm 1\).

(a) The matrix \(\begin{pmatrix} 6 & 0 \\ 0 & 6 \end{pmatrix}\) represents an enlargement with a scale factor of 6.

(b) The determinant of \(\mathbf{A}\) is \(\det(\mathbf{A}) = 3 \times 2 - 4 \times 2 = 6 - 8 = -2\). The area of triangle PQR is \(2 \times 13 = 26 \text{ cm}^2\).

(c) To find \(\mathbf{B}\), we need \(\mathbf{AB} = \begin{pmatrix} 6 & 0 \\ 0 & 6 \end{pmatrix}\). First, find \(\mathbf{A}^{-1} = -\frac{1}{2} \begin{pmatrix} 2 & -4 \\ -2 & 3 \end{pmatrix}\). Then, \(\mathbf{B} = -3 \begin{pmatrix} 2 & -4 \\ -2 & 3 \end{pmatrix} = \begin{pmatrix} -6 & 12 \\ 6 & -9 \end{pmatrix}\).

(d) For \(\mathbf{A} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}\), we have \(\begin{pmatrix} 3x + 4y \\ 2x + 2y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}\). This leads to the equations \(3x + 4y = x\) and \(2x + 2y = y\). Solving these gives \(y = -2x\) and \(-6x = 0\), leading to \(x = 0, y = 0\). Thus, the origin is the only invariant point.

First, differentiate \(y = xe^{ax}\) once using the product rule:

\(\frac{dy}{dx} = axe^{ax} + e^{ax} = (ax + 1)e^{ax}\), so true when \(n = 1\).

Assume that \(\frac{d^k y}{dx^k} = \left( a^k x + ka^{k-1} \right) e^{ax}\).

Differentiate the \(k\)-th derivative:

\(\frac{d^{k+1} y}{dx^{k+1}} = a \left( a^k x + ka^{k-1} \right) e^{ax} + e^{ax} \left( a^k \right) = \left( a^{k+1} + (k+1)a^k \right) e^{ax}\).

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

(a) Consider the sum \(\sum_{r=1}^{n} \left( \ln r - 2 \ln (r+1) + \ln (r+2) \right)\).

This expands to:

• \(\ln 1 - 2 \ln 2 + \ln 3\)
• \(\ln 2 - 2 \ln 3 + \ln 4\)
• \(\ln 3 - 2 \ln 4 + \ln 5\)
• \(\ldots\)
• \(\ln (n-1) - 2 \ln n + \ln (n+1)\)
• \(\ln n - 2 \ln (n+1) + \ln (n+2)\)

Most terms cancel, leaving:

\([\ln 1] - \ln 2 - \ln (n+1) + \ln (n+2) = \ln \frac{n+2}{2(n+1)}\).

(b) The sum to infinity is \(S = -\ln 2\).

We need \(S_n - S = \ln \left( \frac{n+2}{n+1} \right) < 0.01\).

This leads to \(n+2 < e^{0.01}(n+1)\).

Solving gives the least value of \(n\) as 99.

(a) Using the formula 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, \(\alpha + \beta + \gamma = -2\) and \(\alpha\beta + \beta\gamma + \gamma\alpha = 3\).

Thus, \((-2)^2 - 2(3) = 4 - 6 = -2\).

(b) Using the formula for the sum of cubes of roots, \(\alpha^3 + \beta^3 + \gamma^3 = 3(\alpha\beta\gamma) + (\alpha + \beta + \gamma)(\alpha^2 + \beta^2 + \gamma^2 - \alpha\beta - \beta\gamma - \gamma\alpha)\).

\(\alpha\beta\gamma = -3\), so \(\alpha^3 + \beta^3 + \gamma^3 = 3(-3) + (-2)(-2 - 3) = -9 + 10 = 1\).

(c) Expanding \((\alpha + r)^3 = \alpha^3 + 3\alpha^2 r + 3\alpha r^2 + r^3\).

Thus, \(\sum_{r=1}^{n} ((\alpha + r)^3 + (\beta + r)^3 + (\gamma + r)^3) = \sum_{r=1}^{n} (1 + 3(-2)r + 3(-2)r^2 + 3r^3)\).

Using standard results, \(n - 6(\frac{1}{2}n(n+1)) - 6(\frac{1}{6}n(n+1)(2n+1)) + \frac{3}{4}n^2(n+1)^2\).

Simplifying gives \(n + \frac{1}{4}n(n+1)(-12 - 4(2n+1) + 3n(n+1))\).

Finally, \(n + \frac{1}{4}n(n+1)(3n^2 - 5n - 16)\).