(a) The vertical asymptote occurs where the denominator is zero: \(5x - 2 = 0 \Rightarrow x = \frac{2}{5}\). For the oblique asymptote, divide \(5x^2\) by \(5x - 2\) to get \(y = x + \frac{2}{5}\).

(b) Differentiate \(y = \frac{5x^2}{5x-2}\) using the quotient rule: \(\frac{dy}{dx} = \frac{(5x-2)(10x) - (5x^2)(5)}{(5x-2)^2}\). Set \(\frac{dy}{dx} = 0\) to find stationary points: \(5x^2 - 4x = 0 \Rightarrow x(5x - 4) = 0\). Thus, \(x = 0\) or \(x = \frac{4}{5}\). The coordinates are \((0,0)\) and \(\left( \frac{4}{5}, \frac{8}{5} \right)\).

(c) Sketch the curve showing the asymptotes and stationary points.

(d) Solve \(\left| \frac{5x^2}{5x-2} \right| < 2\). This gives two cases: \(\frac{5x^2}{5x-2} < 2\) and \(\frac{5x^2}{5x-2} > -2\). Solving these inequalities gives the intervals \(-1 - \frac{1}{3}\sqrt{5} < x < -1 + \frac{1}{3}\sqrt{5}\) and \(1 - \frac{1}{3}\sqrt{5} < x < 1 + \frac{1}{3}\sqrt{5}\).

(a) We start with \(\sum_{r=1}^{n} r(r+2) = \sum_{r=1}^{n} (r^2 + 2r)\).

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

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

Simplifying gives \(\frac{1}{6}n(n+1)(2n+7)\).

(b) Express \(\frac{1}{r(r+2)}\) in partial fractions: \(\frac{1}{r(r+2)} = \frac{1}{2} \left( \frac{1}{r} - \frac{1}{r+2} \right)\).

Then, \(\sum_{r=1}^{n} \frac{1}{r(r+2)} = \frac{1}{2} \left( 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots + \frac{1}{n} - \frac{1}{n+2} \right)\).

This simplifies to \(\frac{1}{2} \left( \frac{3}{2} - \frac{1}{n+1} - \frac{1}{n+2} \right)\).

(c) As \(n \to \infty\), the terms \(\frac{1}{n+1}\) and \(\frac{1}{n+2}\) approach zero.

Thus, the sum converges to \(\frac{1}{2} \times \frac{3}{2} = \frac{3}{4}\).

(a) Let \(y = x^{-2}\), then \(x = y^{-\frac{1}{2}}\). Substitute into the original equation:

\(y^{-2} + 3y^{-1} + 2y^{-\frac{1}{2}} + 6 = 0\)

Multiply through by \(y^2\) to clear fractions:

\(1 + 3y + 2y^{\frac{3}{2}} + 6y^2 = 0\)

Rearrange to form a quartic equation:

\(36y^4 + 32y^3 + 21y^2 + 6y + 1 = 0\)

The value of \(\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} + \frac{1}{\delta^2}\) is \(-\frac{8}{9}\).

(b) Using the relation \(\alpha \beta \gamma \delta = 6\), we find:

\(\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} + \frac{1}{\delta^2} = \frac{\alpha^2 \beta^2 \gamma^2 \delta^2 + \alpha^2 \beta^2 \gamma^2 \delta^2 + \alpha^2 \beta^2 \gamma^2 \delta^2 + \alpha^2 \beta^2 \gamma^2 \delta^2}{\alpha^2 \beta^2 \gamma^2 \delta^2}\)

\(\beta^2 \gamma^2 \delta^2 + \alpha^2 \gamma^2 \delta^2 + \alpha^2 \beta^2 \delta^2 + \alpha^2 \beta^2 \gamma^2 = -32\)

(c) Using the formula for the sum of squares:

\(\frac{1}{\alpha^4} + \frac{1}{\beta^4} + \frac{1}{\gamma^4} + \frac{1}{\delta^4} = \left(\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} + \frac{1}{\delta^2}\right)^2 - 2(\alpha^{-2}\beta^{-2} + \alpha^{-2}\gamma^{-2} + \alpha^{-2}\delta^{-2} + \beta^{-2}\gamma^{-2} + \beta^{-2}\delta^{-2} + \gamma^{-2}\delta^{-2})\)

\(= \left(-\frac{8}{9}\right)^2 - 2\left(\frac{21}{36}\right)\)

= \(-\frac{61}{162}\)

(a) The matrix \(\begin{pmatrix} 1 & 0 \\ k & 1 \end{pmatrix}\) represents a shear in the y-direction. The matrix \(\begin{pmatrix} 1 & 0 \\ 0 & k \end{pmatrix}\) represents a stretch parallel to the y-axis with scale factor \(k\). The transformations are applied in the order of shear followed by stretch.

(b) To find \(\mathbf{M}^{-1}\), we first find the inverse of each matrix: \(\begin{pmatrix} 1 & 0 \\ k & 1 \end{pmatrix}^{-1} = \begin{pmatrix} 1 & 0 \\ -k & 1 \end{pmatrix}\) and \(\begin{pmatrix} 1 & 0 \\ 0 & k \end{pmatrix}^{-1} = \begin{pmatrix} 1 & 0 \\ 0 & \frac{1}{k} \end{pmatrix}\). Thus, \(\mathbf{M}^{-1} = \begin{pmatrix} 1 & 0 \\ -k & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & \frac{1}{k} \end{pmatrix}\).

(c) The transformation \(\mathbf{M} = \begin{pmatrix} 1 & 0 \\ k^2 & k \end{pmatrix}\) transforms \(\begin{pmatrix} x \\ y \end{pmatrix}\) to \(\begin{pmatrix} x \\ k^2x + ky \end{pmatrix}\). Setting \(\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ k^2x + ky \end{pmatrix}\), we get \(k^2x + ky = y\), which simplifies to \(y = \frac{k^2}{1-k}x\).

(d) The area of triangle DEF is equal to the area of triangle ABC when \(|\det(\mathbf{M})| = 1\). The determinant \(\det(\mathbf{M}) = 1 \times k - 0 \times k^2 = k\). Thus, \(|k| = 1\), giving \(k = -1\).

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

\(\frac{d}{dx}(xf(x)) = xf'(x) + f(x) = xf'(x) + (2(1)-1)f(x).\)

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

\(\frac{d^{2k-1}}{dx^{2k-1}}(xf(x)) = xf'(x) + (2k-1)f(x).\)

Then,

\(\frac{d^{2k}}{dx^{2k}}(xf(x)) = xf(x) + 2kf'(x).\)

\(\frac{d^{2k+1}}{dx^{2k+1}}(xf(x)) = xf'(x) + f(x) + 2kf(x)\)

\(= xf'(x) + (2k+1)f(x).\)

So, it is also true for \(n = k+1\). Hence, by induction, true for all positive integers.