(a) The curve is sketched with the correct shape, and the section \(\frac{1}{2} \pi \leq \theta \leq \pi\) is correct. The polar coordinates of the point furthest from the pole are \((\sqrt{\pi \arctan \pi}, 0) = (1.99, 0)\).

(b) Using the substitution \(u = \pi - \theta\), the area \(A\) is given by:

\(A = \frac{1}{2} \int_0^\pi (\pi - \theta) \arctan(\pi - \theta) \, d\theta\)

or

\(A = \frac{1}{2} \int_0^\pi u \arctan(u) \, du\)

Integrating by parts:

\(\frac{1}{2} \left[ \frac{1}{2} (u^2) \arctan(u) \right]_0^\pi - \frac{1}{2} \int_0^\pi \frac{u^2}{1 + u^2} \, du\)

\(\int_0^\pi \frac{u^2}{1 + u^2} \, du = \left[ u - \arctan(u) \right]_0^\pi\)

\(A = \frac{1}{4} \pi^2 \arctan(\pi) - \frac{1}{4} \left[ u - \arctan(u) \right]_0^\pi\)

\(A = \frac{1}{4} \pi^2 - \frac{1}{4} \pi = 2.65\)

(c) Let \(y = \sqrt{\pi - \theta} \arctan(\pi - \theta) \sin \theta\).

Find the derivative:

\(\frac{dy}{d\theta} = \sqrt{\pi - \theta} \arctan(\pi - \theta) \cos \theta - \frac{(\pi - \theta)(1 + (\pi - \theta)^2)^{-1} + \arctan(\pi - \theta)}{2\sqrt{\pi - \theta} \arctan(\pi - \theta)} \sin \theta\)

Set \(\frac{dy}{d\theta} = 0\) and form the equation:

\(2(\pi - \theta) \arctan(\pi - \theta) \cot \theta - \frac{\pi - \theta}{1 + (\pi - \theta)^2} - \arctan(\pi - \theta) = 0\)

Check sign change between \(\theta = 1.2\) and \(\theta = 1.3\):

\(2(\pi - 1.2) \arctan(\pi - 1.2) \cot 1.2 - \frac{\pi - 1.2}{1 + (\pi - 1.2)^2} - \arctan(\pi - 1.2) = 0.151\)

and

\(2(\pi - 1.3) \arctan(\pi - 1.3) \cot 1.3 - \frac{\pi - 1.3}{1 + (\pi - 1.3)^2} - \arctan(\pi - 1.3) = -0.395\)

Thus, there is a sign change, confirming a root exists between \(\theta = 1.2\) and \(\theta = 1.3\).

(a) By Vieta's formulas, the sum of the products of the roots taken two at a time is given by \(-\frac{b}{a}\) for the equation \(ax^3 + bx^2 + cx + d = 0\). Here, \(a = 2\), \(b = 1\), so \(\alpha\beta + \beta\gamma + \gamma\alpha = -\frac{1}{2}p\).

(b) The expression \(\alpha^2\beta\gamma + \alpha\beta^2\gamma + \alpha\beta\gamma^2\) can be factored as \(\alpha\beta\gamma(\alpha + \beta + \gamma)\). From Vieta's formulas, \(\alpha\beta\gamma = \frac{5}{2}\) and \(\alpha + \beta + \gamma = -\frac{1}{2}\). Therefore, \(\alpha^2\beta\gamma + \alpha\beta^2\gamma + \alpha\beta\gamma^2 = \frac{5}{2} \times -\frac{1}{2} = -\frac{5}{4}\).

(c) The roots of the new cubic equation are \(\alpha\beta, \beta\gamma, \alpha\gamma\). The sum of these roots is \(\alpha\beta + \beta\gamma + \alpha\gamma = -\frac{1}{2}p\), the sum of the products of the roots taken two at a time is \(\alpha^2\beta\gamma + \alpha\beta^2\gamma + \alpha\beta\gamma^2 = -\frac{5}{4}\), and the product of the roots is \((\alpha\beta)(\beta\gamma)(\alpha\gamma) = (\alpha\beta\gamma)^2 = \left(\frac{5}{2}\right)^2 = \frac{25}{4}\). The cubic equation is \(4z^3 + 2pz^2 - 5z - 25 = 0\).

(d) Given \(\alpha^2 + \beta^2 + \gamma^2 = \frac{1}{3}\), we use the identity \(\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\). Substituting the known values, \(\frac{1}{3} = \left(-\frac{1}{2}\right)^2 - 2\left(-\frac{1}{2}p\right)\). Solving for \(p\), \(\frac{1}{3} = \frac{1}{4} + p\), so \(p = \frac{1}{12}\).

Base Case: For \(n = 1\), \(6^{4 \times 1} + 38^1 - 2 = 1296 + 38 - 2 = 1332\). Since \(1332 \div 74 = 18\), it is divisible by 74.

Inductive Step: Assume \(6^{4k} + 38^k - 2\) is divisible by 74 for some positive integer \(k\). This is the inductive hypothesis.

Consider \(6^{4(k+1)} + 38^{k+1} - 2\):

\(6^{4(k+1)} + 38^{k+1} - 2 = 6^{4k+4} + 38^{k+1} - 2\)

\(= (1295 + 1) \times 6^{4k} + (37 + 1) \times 38^k - 2\)

\(= 1295 \times 6^{4k} + 6^{4k} + 37 \times 38^k + 38^k - 2\)

By the inductive hypothesis, \(6^{4k} + 38^k - 2\) is divisible by 74, and since \(1295 \times 6^{4k} + 37 \times 38^k\) is also divisible by 74, the entire expression is divisible by 74.

Therefore, by induction, \(6^{4n} + 38^n - 2\) is divisible by 74 for every positive integer \(n\).

(a) Expand the expression \(\sum_{r=1}^{N} r(r+1)(3r+4)\) as follows:

\(3 \sum_{r=1}^{N} r^3 + 7 \sum_{r=1}^{N} r^2 + 4 \sum_{r=1}^{N} r.\)

Using standard formulae:

\(\frac{3}{4} N^2 (N+1)^2 + \frac{7}{6} N(N+1)(2N+1) + 2N(N+1).\)

Simplify to:

\(\frac{1}{12} N (9N^3 + 46N^2 + 75N + 38) = \frac{1}{12} N(N+1)(N+2)(9N+19).\)

(b) Express \(\frac{3r+4}{r(r+1)}\) in partial fractions:

\(\frac{3r+4}{r(r+1)} = \frac{4}{r} - \frac{1}{r+1}.\)

Then,

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

This simplifies to:

\(\frac{1}{4} - \frac{1}{4^{N+1}(N+1)}.\)

(c) As \(N \to \infty\), the term \(\frac{1}{4^{N+1}(N+1)} \to 0\), so the sum converges to:

\(\frac{1}{4}\).

(a) The matrix \(\mathbf{M}\) can be decomposed into two transformations: a stretch and a rotation. The first matrix \(\begin{pmatrix} 14 & 0 \\ 0 & 1 \end{pmatrix}\) represents a stretch parallel to the x-axis with a scale factor of 14. The second matrix \(\begin{pmatrix} \frac{1}{2} & -\frac{1}{2}\sqrt{3} \\ \frac{1}{2}\sqrt{3} & \frac{1}{2} \end{pmatrix}\) represents a rotation by \(\frac{\pi}{3}\) anticlockwise about the origin.

(b) To find \(\mathbf{M}^{-1}\), we take the inverse of each matrix in the product: \(\begin{pmatrix} 14 & 0 \\ 0 & 1 \end{pmatrix}^{-1} = \begin{pmatrix} \frac{1}{14} & 0 \\ 0 & 1 \end{pmatrix}\) and \(\begin{pmatrix} \frac{1}{2} & -\frac{1}{2}\sqrt{3} \\ \frac{1}{2}\sqrt{3} & \frac{1}{2} \end{pmatrix}^{-1} = \begin{pmatrix} \frac{1}{2} & \frac{1}{2}\sqrt{3} \\ -\frac{1}{2}\sqrt{3} & \frac{1}{2} \end{pmatrix}\).

(c) The invariant lines are found by solving \(\mathbf{M} \begin{pmatrix} x \\ y \end{pmatrix} = \lambda \begin{pmatrix} x \\ y \end{pmatrix}\). Solving gives the equations \(y = 2\sqrt{3}x\) and \(y = \frac{2}{3}\sqrt{3}x\).

(d) The area of triangle \(ABC\) is found using the determinant of \(\mathbf{M}\). Given \(|\mathbf{DEF}| = |\det \mathbf{M}| \times |\mathbf{ABC}|\), we have \(28 = 14 \times |\mathbf{ABC}|\), so \(|\mathbf{ABC}| = 2\) cm².