(a) The curve is sketched with \(r\) strictly increasing from \(\theta = 0\) to \(\theta = 2\). The maximum distance is found by evaluating \(r\) at \(\theta = 2\), giving \(r = \sqrt{\frac{1}{2}\arctan(1)} = \frac{1}{2}\sqrt{\pi}\).

(b) The area is calculated using the integral \(\frac{1}{2} \int_0^2 \arctan\left(\frac{1}{2}\theta\right) d\theta\). Integration by parts is applied:

\(\frac{1}{2}\left[ \frac{1}{2}\theta\arctan\left(\frac{1}{2}\theta\right) - \int \frac{\theta}{\theta^2 + 4} d\theta \right]_0^2\)

\(= \frac{1}{2}\left[ \frac{1}{2}\theta\arctan\left(\frac{1}{2}\theta\right) - \ln(\theta^2 + 4) \right]_0^2\)

\(= \frac{1}{4}\pi - \frac{1}{2}\ln 2\)

(c) To find the point furthest from \(\theta = \frac{1}{2}\pi\), set \(x = (\arctan\left(\frac{1}{2}\theta\right))^{\frac{1}{4}} \cos\theta\) and differentiate:

\(\frac{dx}{d\theta} = (\arctan\left(\frac{1}{2}\theta\right))^{\frac{1}{4}} \sin\theta + \cos\theta (\arctan\left(\frac{1}{2}\theta\right))^{\frac{1}{4}} (\theta^2 + 4)^{-1} = 0\)

\((\theta^2 + 4)\arctan\left(\frac{1}{2}\theta\right)\sin\theta - \cos\theta = 0\)

Check sign change between 0.6 and 0.7:

\((0.6^2 + 4)\arctan(0.3)\sin 0.6 - \cos 0.6 = -0.108\)

\((0.7^2 + 4)\arctan(0.35)\sin 0.7 - \cos 0.7 = 0.209\)

Thus, a root exists between 0.6 and 0.7.

(a) To find when \(A\) is non-singular, set \(\det(A) \neq 0\). Calculate \(\det(A) = 1(k \cdot 9 - 6 \cdot 8) - 2(4 \cdot 9 - 6 \cdot 7) + 3(4 \cdot 8 - k \cdot 7)\).

Simplifying, \(\det(A) = -12k + 60\). Set \(-12k + 60 \neq 0\), giving \(k \neq 5\).

(b) The inverse of \(A\) is \(A^{-1} = \frac{1}{\det(A)} \text{adj}(A)\). The top row of \(\text{adj}(A)\) is \(\begin{pmatrix} k \cdot 9 - 6 \cdot 8 & -(4 \cdot 9 - 6 \cdot 7) & 4 \cdot 8 - k \cdot 7 \end{pmatrix}\).

Thus, the top row of \(A^{-1}\) is \(\begin{pmatrix} \frac{1}{60 - 12k} & \frac{6}{60 - 12k} & \frac{3k - 16}{4(5-k)} \end{pmatrix}\).

(c) Calculate \(BA = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 \\ 4 & k & 6 \\ 7 & 8 & 9 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 \\ 4 & k & 6 \end{pmatrix}\).

Set \(C = \begin{pmatrix} a & b & c \\ d & e & f \end{pmatrix}\) such that \(BAC = \begin{pmatrix} 2 & 1 \\ k & 4 \end{pmatrix}\).

Choose \(C = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}\).

(d) The transformation matrix is \(\begin{pmatrix} 2 & 1 \\ k & 4 \end{pmatrix}\). The characteristic equation is \(\lambda^2 - 6\lambda + (8-k) = 0\).

For two distinct invariant lines, the discriminant must be positive: \(6^2 - 4(8-k) > 0\).

Simplifying, \(36 - 32 + 4k > 0\) gives \(4k > -4\), so \(k > 1\). Also, \(k \neq 8\) to avoid repeated roots.

(a) The curve \(y = \frac{x+1}{x-1}\) has a vertical asymptote at \(x = 1\) and a horizontal asymptote at \(y = 1\). The branches are in the first and third quadrants.

(b) The curve \(y = \frac{|x|+1}{|x|-1}\) is symmetrical about \(x = 0\). The critical points are found by solving \(\frac{|x|+1}{|x|-1} = -2\), leading to \(|x| = \frac{1}{3}\). Thus, the set of values of \(x\) for which \(\frac{|x|+1}{|x|-1} < -2\) is \(-1 < x < -\frac{1}{3}, \frac{1}{3} < x < 1\).

(a) By Vieta's formulas, we have:

\(\alpha + \beta + \gamma = -5\)

\(\alpha\beta + \beta\gamma + \gamma\alpha = 10\)

\(\alpha\beta\gamma = 2\)

We use the identity:

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

\(\alpha^2 + \beta^2 + \gamma^2 = (-5)^2 - 2(10) = 25 - 20 = 5\)

(b) To show the matrix is singular, we find its determinant:

\(\begin{vmatrix} 1 & \alpha & \beta \\ \alpha & 1 & \gamma \\ \beta & \gamma & 1 \end{vmatrix} = 1 - \alpha^2 - \beta^2 - \gamma^2 + 2\alpha\beta\gamma\)

Substituting the known values:

\(1 - 5 + 2(2) = 0\)

Thus, the determinant is zero, and the matrix is singular.

(a) The vertical asymptote occurs where the denominator is zero, so \(x = 1\).

For the oblique asymptote, perform polynomial long division on \(\frac{ax^2 + x - 1}{x - 1}\):

\(y = (x - 1)(ax + a + 1) + a\) gives \(y = ax + a + 1 + \frac{a}{x - 1}\).

Thus, the oblique asymptote is \(y = ax + a + 1\).

(b) Rewrite the equation as \(ax^2 + x - 1 = y(x - 1)\), leading to \(ax^2 + x - 1 = yx - y\).

This simplifies to \(ax^2 - (y - 1)x + (y - 1) = 0\).

The discriminant \((y - 1)^2 - 4a(y - 1)\) must be less than zero for no real roots, giving \(1 < y < 1 + 4a\).

Thus, there is no point on \(C\) for which \(1 < y < 1 + 4a\).

(c) The sketch shows the curve with the vertical asymptote at \(x = 1\) and the oblique asymptote \(y = ax + a + 1\). The curve has two branches, one in each of the regions determined by the asymptotes.