(a) The curve is sketched as a semicircle with \(\theta = 0\) at the rightmost point. The polar coordinates of the point furthest from the pole are \((1, 0)\).

(b) The area \(A\) is given by:

\(A = \frac{1}{2} \int_{0}^{\pi} \frac{1}{\theta^2 + 1} \, d\theta\)

Integrating, we have:

\(\frac{1}{2} \left[ \arctan \theta \right]_{0}^{\pi} = \frac{1}{2} \arctan \pi = 0.631\)

(c) Let \(y = \frac{\sin \theta}{\sqrt{\theta^2 + 1}}\). The derivative is:

\(\frac{dy}{d\theta} = \left( \theta^2 + 1 \right)^{-\frac{1}{2}} \cos \theta - \theta \left( \theta^2 + 1 \right)^{-\frac{3}{2}} \sin \theta = 0\)

Solving gives:

\(\theta \neq 0 \Rightarrow \left( \theta + \frac{1}{\theta} \right) \cot \theta - 1 = 0\)

Checking for a root between 1.1 and 1.2:

\(\left( 1.1 + \frac{1}{1.1} \right) \cot 1.1 - 1 = 0.02 \ldots\)

\(\left( 1.2 + \frac{1}{1.2} \right) \cot 1.2 - 1 = -0.209 \ldots\)

There is a sign change, confirming a root exists between 1.1 and 1.2.

(a) To find the vertical asymptote, set the denominator equal to zero: \(x - 2 = 0 \Rightarrow x = 2\). For the oblique asymptote, divide the numerator by the denominator: \(x^2 + 2x - 15 = (x + 4)(x - 2) + 7 \Rightarrow y = x + 4\).

(b) Differentiate \(y = \frac{x^2 + 2x - 15}{x - 2}\) using the quotient rule: \(\frac{dy}{dx} = \frac{(x-2)(2x+2) - (x^2+2x-15)}{(x-2)^2}\). Simplify to get \(\frac{dy}{dx} = \frac{x^2 - 4x + 11}{(x-2)^2}\). The quadratic \(x^2 - 4x + 11\) has no real roots (discriminant \(16 - 44 < 0\)), so no stationary points.

(c) Intersections: Set \(y = 0\) to find \(x\)-intercepts: \(x^2 + 2x - 15 = 0 \Rightarrow (x+5)(x-3) = 0 \Rightarrow x = -5, 3\). For \(y\)-intercept, set \(x = 0\): \(y = \frac{-15}{-2} = 7.5\).

(d) Sketch the curve \(y = \left| \frac{x^2 - 2x - 15}{x - 2} \right|\) using the shape from part (c) and reflecting any negative parts above the \(x\)-axis.

(e) Solve \(\left| \frac{2x^2 + 4x - 30}{x - 2} \right| < 15\). Consider \(\frac{2x^2 + 4x - 30}{x - 2} = 15\) and \(\frac{2x^2 + 4x - 30}{x - 2} = -15\). Solve \(2x^2 + 4x - 30 = 15(x-2)\) and \(2x^2 + 4x - 30 = -15(x-2)\) to find critical points \(x = 0, \frac{11}{2}, -12, \frac{5}{2}\). Test intervals to find \(-12 < x < 0\) or \(\frac{5}{2} < x < \frac{11}{2}\).

(a) The equation of the plane \(\Pi_1\) can be found by setting up the system of equations from the given vectors:

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix}\)

Solving gives \(y + z = -7\).

(b) To find the line of intersection, find a point common to both planes, e.g., \(\begin{pmatrix} -7 \\ -2 \\ -5 \end{pmatrix}\), and the direction vector by solving:

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 1 & 1 \\ -5 & 3 & 5 \end{vmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 2 \\ -5 \\ 5 \end{pmatrix}\)

The vector equation is \(\mathbf{r} = \begin{pmatrix} -7 \\ -2 \\ -5 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -5 \\ 5 \end{pmatrix}\).

(c) The perpendicular distance from \(A\) to \(\Pi_1\) is given by:

\(\frac{|a + a - 7 + 7|}{\sqrt{2}} = \sqrt{2}\)

Solving gives \(a = 1\).

(d) The obtuse angle condition gives:

\(\frac{|b + b|}{\sqrt{2((1-b)^2 + 2b^2)}} = \frac{1}{2}\sqrt{2}\)

Solving \(2b = \sqrt{(1-b)^2 + 2b^2}\) gives \(b = -1 + \sqrt{2}\).

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

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

Assume true for \(n = k\), so \(2A^k = \begin{pmatrix} 2 \times 3^k & 0 \\ 3^k - 1 & 2 \end{pmatrix}\).

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

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

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

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

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

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

Given \(\alpha + \beta + \gamma = -4\) and \(\alpha\beta + \beta\gamma + \gamma\alpha = 6\), we have:

\((-4)^2 - 2(6) = 16 - 12 = 4\).

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

Thus, \(\sum_{r=1}^{n} ((\alpha + r)^2 + (\beta + r)^2 + (\gamma + r)^2) = \sum_{r=1}^{n} (4 + 2(-4)r + 3r^2)\).

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

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

\(= n(-4n + \frac{1}{2}(2n^2 + 3n + 1))\).

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