(a) Expand \((2-3r)(5-3r) = 9r^2 - 21r + 10\).

Use standard summation formulas: \(\sum r^2 = \frac{1}{6}n(n+1)(2n+1)\) and \(\sum r = \frac{1}{2}n(n+1)\).

Substitute: \(9\left(\frac{1}{6}n(n+1)(2n+1)\right) - 21\left(\frac{1}{2}n(n+1)\right) + 10n\).

Simplify to get \(3n^3 - 6n^2 + n\).

(b) Express \(\frac{1}{(2-3r)(5-3r)}\) as partial fractions: \(\frac{1}{3} \left( \frac{1}{2-3r} - \frac{1}{5-3r} \right)\).

Write the sum: \(\sum_{r=1}^{n} \frac{1}{3} \left( \frac{1}{2-3r} - \frac{1}{5-3r} \right)\).

Observe cancellation: \(-\frac{1}{6} + \frac{1}{3(2-3n)}\).

(c) As \(n \to \infty\), the term \(\frac{1}{3(2-3n)} \to 0\).

Thus, the sum converges to \(-\frac{1}{6}\).

(a) Let \(y = x^3 - 1\). Then \(x^3 = y + 1\).

Substitute into the original equation: \((y+1) + 2(y+1)^{1/3} + 1 = 0\).

This simplifies to \(2(y+1)^{1/3} = -y - 2\).

Cube both sides: \(8(y+1) = -(y+2)^3\).

Expand and simplify: \(8y + 8 = -y^3 - 6y^2 - 12y - 8\).

Rearrange to get the cubic equation: \(y^3 + 6y^2 + 20y + 16 = 0\).

(b) Use the identity \(\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\).

Given \(\alpha + \beta + \gamma = 0\), \(\alpha\beta + \beta\gamma + \gamma\alpha = 2\), and \(\alpha\beta\gamma = -1\), calculate \((\alpha^3 - 1)^2 + (\beta^3 - 1)^2 + (\gamma^3 - 1)^2 = 6^2 - 2(20) = -4\).

(c) Use the result from (a) or a valid formula for the sum of cubes: \((\alpha^3 - 1)^3 + (\beta^3 - 1)^3 + (\gamma^3 - 1)^3 = -6(-6) - 20(-6) - 16(3) = 96\).

(a) Base case: For \(n = 1\), \(u_1 = 5 = 6^1 - 1\). Assume it is true for \(n = k\), so \(u_k = 6^k - 1\).

Then \(u_{k+1} = 6(6^k - 1) + 5 = 6^{k+1} - 1\).

So, it is also true for \(n = k+1\). Hence, by induction, \(u_n = 6^n - 1\) is true for all positive integers.

(b) \(\frac{u_{2n}}{u_n} = \frac{6^{2n} - 1}{6^n - 1} = \frac{(6^n - 1)(6^n + 1)}{6^n - 1} = 6^n + 1\).

Thus, \(u_{2n}\) is divisible by \(u_n\).

(a) The matrix \(\mathbf{M}\) can be decomposed into two transformations: a rotation and a shear. The rotation is represented by \(\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}\), which is an anticlockwise rotation about the origin through an angle \(\theta\). The shear is represented by \(\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\), which fixes the x-axis and maps \((0,1)\) to \((2,1)\).

(b) The matrix \(\mathbf{M}\) is given by:

\(\mathbf{M} = \begin{pmatrix} \cos \theta + 2 \sin \theta & 2 \cos \theta - \sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}\)

Applying \(\mathbf{M}\) to a point \((x, y)\) gives:

\(\begin{pmatrix} \cos \theta + 2 \sin \theta & 2 \cos \theta - \sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \cos \theta + 2x \sin \theta - y \sin \theta + 2y \cos \theta \\ x \sin \theta + y \cos \theta \end{pmatrix}\)

\(For a line of invariant points, we require:\)

\(x \sin \theta + y \cos \theta = y\) and \(x \cos \theta + 2x \sin \theta - y \sin \theta + 2y \cos \theta = x\).

Solving these equations gives:

\(x \cos \theta + 2x \sin \theta + \frac{x \sin \theta}{1 - \cos \theta} (2 \cos \theta - \sin \theta) = x\)

Simplifying, we find:

\((\cos \theta + 2 \sin \theta)(1 - \cos \theta) + 2 \sin \theta - \sin^2 \theta = 1 - \cos \theta\)

\(\cos \theta - \cos^2 \theta + 2 \sin^2 \theta - \sin^2 \theta = 1 - \cos \theta \Rightarrow \sin \theta + \cos \theta = 1\)

Thus, \(\theta = \frac{1}{2} \pi\).

(a) The curve is a spiral on \([0, 2\pi]\), starting at the pole with \(\theta = 0\) and increasing as \(\theta\) increases, intersecting the initial line at one other point.

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

\(A = \frac{1}{2} \int_0^{2\pi} \theta^2 e^{\frac{1}{4}\theta} \, d\theta\)

Integrate by parts twice:

\(= \frac{1}{2} \left[ 4\theta^2 e^{\frac{1}{4}\theta} \right]_0^{2\pi} - \frac{1}{2} \int_0^{2\pi} 8\theta e^{\frac{1}{4}\theta} \, d\theta\)

\(= \frac{1}{2} \left[ 4\theta^2 e^{\frac{1}{4}\theta} \right]_0^{2\pi} - 4 \left[ 4\theta e^{\frac{1}{4}\theta} \right]_0^{2\pi} + 16 \int_0^{2\pi} e^{\frac{1}{4}\theta} \, d\theta\)

\(= \left[ 2\theta^2 e^{\frac{1}{4}\theta} - 16\theta e^{\frac{1}{4}\theta} + 64e^{\frac{1}{4}\theta} \right]_0^{2\pi}\)

\(= (8\pi^2 - 32\pi + 64)e^{\frac{1}{2}\pi} - 64\)

(c) Use \(y = r\sin \theta\) and differentiate:

\(\frac{dy}{d\theta} = \theta e^{\frac{1}{8}\theta} \cos \theta + \sin \theta \left( \frac{1}{8}\theta e^{\frac{1}{8}\theta} + e^{\frac{1}{8}\theta} \right)\)

\(e^{\frac{1}{8}\theta} \neq 0 \Rightarrow \theta \cos \theta + \left( \frac{1}{8}\theta + 1 \right) \sin \theta = 0\)

Check sign change:

\(5\cos 5 + \left( \frac{5}{8} + 1 \right)\sin 5 = -0.1399\)

\(5.05\cos 5.05 + \left( \frac{5.05}{8} + 1 \right)\sin 5.05 = 0.1336\)

(a) Find direction vectors \(\overrightarrow{AB} = 2\mathbf{j} + 4\mathbf{k}\) and \(\overrightarrow{AC} = \mathbf{i} - \mathbf{j} + \mathbf{k}\).

Calculate the normal to the plane ABC using the cross product:

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 2 & 4 \\ 1 & -1 & 1 \end{vmatrix} = 3\mathbf{i} + 2\mathbf{j} - \mathbf{k}\)

Substitute point \(A(1, 0, -2)\) into \(3x + 2y - z = d\) to find \(d = 5\).

(b) Find normal to plane ABD:

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 2 & 4 \\ 0 & 0 & t+2 \end{vmatrix} = (2t+4)\mathbf{i}\)

Use dot product to find angle:

\(\begin{pmatrix} 3 \\ 2 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} 2t+4 \\ 0 \\ 0 \end{pmatrix} = \sqrt{14} \cos \theta\)

\(\theta = 36.7^\circ\) when \(t \neq -2\).

(c) Find common perpendicular:

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 2 & 4 \\ -1 & 1 & t+1 \end{vmatrix} = \begin{pmatrix} -4 \\ 2 \\ 1 \end{pmatrix}\)

Use formula for perpendicular distance:

\(\frac{l + 2}{\sqrt{t^2 - 2t + 6}} = \sqrt{2}\)

Solve \(t^2 - 8t + 8 = 0\) to find \(t = 6.83, \; t = 1.17\).

(a) Vertical asymptotes occur where the denominator is zero: \(2x^2 - 7x - 4 = 0\). Solving gives \(x = -\frac{1}{2}\) and \(x = 4\). The horizontal asymptote is found by comparing the degrees of the numerator and denominator, giving \(y = 1\).

(b) To find stationary points, set \(\frac{dy}{dx} = 0\). Differentiate using the quotient rule: \(\frac{dy}{dx} = \frac{(2x^2 - 7x - 4)(4x - 5) - (2x^2 - 5x)(4x - 7)}{(2x^2 - 7x - 4)^2}\). Simplify and solve \(4x^2 + 16x - 20 = 0\) to find \(x = -5\) and \(x = 1\). Substitute back to find \(y\)-coordinates: \((-5, \frac{25}{7})\) and \((1, \frac{1}{3})\).

(c) The curve intersects the axes at \((0,0)\) and \((\frac{5}{2}, 0)\).

(d) Sketch the curve \(y = \left| \frac{2x^2 - 5x}{2x^2 - 7x - 4} \right|\) with correct shape and position.

(e) Solve \(\left| \frac{2x^2 - 5x}{2x^2 - 7x - 4} \right| < \frac{1}{9}\). This leads to solving \(16x^2 - 38x + 4 = 0\) and \(20x^2 - 52x - 4 = 0\). The critical points are \(x = \frac{19}{16} - \frac{3}{16} \sqrt{33}, \ x = \frac{19}{16} + \frac{3}{16} \sqrt{33}, \ x = \frac{13}{10} - \frac{3}{10} \sqrt{21}, \ x = \frac{13}{10} + \frac{3}{10} \sqrt{21}\). The solution is \(\frac{13}{10} - \frac{3}{10} \sqrt{21} < x < \frac{19}{16} - \frac{3}{16} \sqrt{33}\) or \(\frac{19}{16} + \frac{3}{16} \sqrt{33} < x < \frac{13}{10} + \frac{3}{10} \sqrt{21}\).

(a) Expand \((2-3r)(5-3r) = 9r^2 - 21r + 10\).

Use standard summation formulas: \(\sum r^2 = \frac{1}{6}n(n+1)(2n+1)\) and \(\sum r = \frac{1}{2}n(n+1)\).

Substitute: \(9\left(\frac{1}{6}n(n+1)(2n+1)\right) - 21\left(\frac{1}{2}n(n+1)\right) + 10n\).

Simplify to get \(3n^3 - 6n^2 + n\).

(b) Partial fraction decomposition: \(\frac{1}{(2-3r)(5-3r)} = \frac{1}{3}\left(\frac{1}{2-3r} - \frac{1}{5-3r}\right)\).

Sum: \(\sum_{r=1}^{n} \frac{1}{3}\left(\frac{1}{2-3r} - \frac{1}{5-3r}\right)\).

Telescoping series: \(-\frac{1}{6} + \frac{1}{3(2-3n)}\).

(c) As \(n \to \infty\), the term \(\frac{1}{3(2-3n)} \to 0\).

Thus, the sum is \(\frac{1}{6}\).

(a) Let \(y = x^3 - 1\). Then \(x^3 = y + 1\). Substitute into the original equation:

\((y+1)^3 + 2(y+1) + 1 = 0\)

\(\Rightarrow 2(y+1) = -y - 2\)

\(8(y+1) = -(y+2)^3\)

\(\Rightarrow 8y + 8 = -y^3 - 6y^2 - 12y - 8\)

\(y^3 + 6y^2 + 20y + 16 = 0\)

(b) \((\alpha^3 - 1)^2 + (\beta^3 - 1)^2 + (\gamma^3 - 1)^2 = 6^2 - 2(20) = -4\)

(c) \((\alpha^3 - 1)^3 + (\beta^3 - 1)^3 + (\gamma^3 - 1)^3 = -6(-6) - 20(-6) - 16(3) = 96\)

(a) Base case: For \(n = 1\), \(u_1 = 5 = 6^1 - 1\). Thus, the base case holds.

Inductive step: Assume \(u_k = 6^k - 1\) is true for \(n = k\). Then,

\(u_{k+1} = 6u_k + 5 = 6(6^k - 1) + 5 = 6^{k+1} - 6 + 5 = 6^{k+1} - 1.\)

Thus, \(u_{k+1} = 6^{k+1} - 1\) is true. By induction, \(u_n = 6^n - 1\) for all positive integers \(n\).

(b) We have \(u_{2n} = 6^{2n} - 1\) and \(u_n = 6^n - 1\). Consider:

\(\frac{u_{2n}}{u_n} = \frac{6^{2n} - 1}{6^n - 1} = \frac{(6^n - 1)(6^n + 1)}{6^n - 1} = 6^n + 1.\)

Since \(6^n + 1\) is an integer, \(u_{2n}\) is divisible by \(u_n\) for \(n \geq 1\).

(a) The matrix \(\mathbf{M} = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}\) represents a rotation followed by a shear. The rotation is anticlockwise, centered at the origin, through an angle \(\theta\), with the x-axis fixed and \((0,1)\) mapped to \((2,1)\).

(b) The combined matrix is \(\mathbf{M} = \begin{pmatrix} \cos \theta + 2\sin \theta & 2\cos \theta - \sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}\).

\(To find the line of invariant points, solve:\)

\(\begin{pmatrix} \cos \theta + 2\sin \theta & 2\cos \theta - \sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}\)

This gives the equations:

\(x\sin \theta + y\cos \theta = y\)

\(x\cos \theta + 2x\sin \theta - y\sin \theta + 2y\cos \theta = x\)

Rearranging and simplifying, we find:

\(x\cos \theta + 2x\sin \theta + \frac{x\sin \theta}{1 - \cos \theta}(2\cos \theta - \sin \theta) = x\)

\((\cos \theta + 2\sin \theta)(1 - \cos \theta) + 2\sin \theta\cos \theta - \sin^2 \theta = 1 - \cos \theta\)

\(\cos^2 \theta - \cos^2 \theta + 2\sin \theta - \sin^2 \theta = 1 - \cos \theta\)

\(\Rightarrow \sin \theta + \cos \theta = 1\)

Solving gives \(\theta = \frac{1}{2} \pi\).

(a) The curve is a spiral starting at the pole \(\theta = 0\) and spiraling outward as \(\theta\) increases to \(2\pi\). The radius \(r\) is strictly increasing, and the gradient is correct at points of intersection with the initial line.

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

\(A = \frac{1}{2} \int_{0}^{2\pi} \theta^2 e^{\frac{1}{4} \theta} d\theta\)

Integrate by parts:

\(= \frac{1}{2} \left[ 4\theta^2 e^{\frac{1}{4} \theta} \right]_{0}^{2\pi} - \int_{0}^{2\pi} 8\theta e^{\frac{1}{4} \theta} d\theta\)

Integrate by parts again:

\(= \frac{1}{2} \left[ 4\theta^2 e^{\frac{1}{4} \theta} - 4(4\theta e^{\frac{1}{4} \theta} + 16)e^{\frac{1}{4} \theta} \right]_{0}^{2\pi}\)

\(= \left[ 2\theta^2 e^{\frac{1}{4} \theta} - 16\theta e^{\frac{1}{4} \theta} + 64e^{\frac{1}{4} \theta} \right]_{0}^{2\pi}\)

\(= (8\pi^2 - 32\pi + 64)e^{\frac{1}{2}\pi} - 64\)

(c) Let \(y = \theta e^{\frac{1}{8} \theta} \sin \theta\). Differentiate:

\(\frac{dy}{d\theta} = e^{\frac{1}{8} \theta} \cos \theta + \sin \theta \left( \frac{1}{8} \theta e^{\frac{1}{8} \theta} + e^{\frac{1}{8} \theta} \right)\)

Set \(\frac{dy}{d\theta} = 0\):

\(e^{\frac{1}{8} \theta} (\theta \cos \theta + (\frac{1}{8} \theta + 1) \sin \theta) = 0\)

Check sign change:

\(5 \cos 5 + (\frac{5}{8} + 1) \sin 5 = -0.1399\)

\(5.05 \cos 5.05 + (\frac{5.05}{8} + 1) \sin 5.05 = 0.1336\)

There is a sign change between 5 and 5.05, confirming a root exists.

(a) To find the equation of the plane ABC, we first determine the direction vectors \(\overrightarrow{AB} = 2\mathbf{j} + 4\mathbf{k}\) and \(\overrightarrow{AC} = \mathbf{i} - \mathbf{j} + \mathbf{k}\).

The normal vector \(\mathbf{n}\) to the plane is given by the cross product \(\overrightarrow{AB} \times \overrightarrow{AC}\):

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 2 & 4 \\ 1 & -1 & 1 \end{vmatrix} = \begin{pmatrix} 6 \\ 4 \\ -2 \end{pmatrix}\)

Substituting point \(A(1, 0, -2)\) into the plane equation \(3x + 2y - z = d\), we find \(d = 5\).

(b) The normal to the plane ABD is found using the cross product:

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 2 & 4 \\ 0 & 0 & t+2 \end{vmatrix} = \begin{pmatrix} 2t + 4 \\ 0 \\ 0 \end{pmatrix}\)

The angle \(\theta\) between the planes is given by:

\(\cos \theta = \frac{\left( \begin{pmatrix} 3 \\ 2 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \right)}{\sqrt{14} \cdot 1}\)

\(\theta = 36.7^\circ\).

(c) To find \(t\) such that the shortest distance between lines AB and CD is \(\sqrt{2}\), we use the formula for the perpendicular distance between skew lines:

\(\frac{1}{\sqrt{t^2 - 2t + 6}} \left| \begin{pmatrix} 1 \\ -1 \\ -2 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix} \right| = \frac{t + 2}{\sqrt{t^2 - 2t + 6}}\)

Setting this equal to \(\sqrt{2}\) and solving gives:

\(t^2 - 8t + 8 = 0 \Rightarrow t = 2(2 \pm \sqrt{2})\)

Thus, \(t = 6.83\) and \(t = 1.17\).

(a) The vertical asymptotes occur where the denominator is zero: \(2x^2 - 7x - 4 = 0\). Solving gives \(x = -\frac{1}{2}\) and \(x = 4\). The horizontal asymptote is found by considering the degrees of the polynomials: \(y = 1\).

(b) To find stationary points, set \(\frac{dy}{dx} = 0\). Differentiate using the quotient rule: \(\frac{dy}{dx} = \frac{(2x^2 - 7x - 4)(4x - 5) - (2x^2 - 5x)(4x - 7)}{(2x^2 - 7x - 4)^2}\). Simplifying gives \(4x^2 + 16x - 20 = 0\), leading to points \((-5, \frac{25}{3})\) and \((1, \frac{1}{3})\).

(c) The curve intersects the axes at \((0,0)\) and \((\frac{5}{2},0)\).

(d) The sketch of \(y = \left| \frac{2x^2 - 5x}{2x^2 - 7x - 4} \right|\) reflects the negative parts of the curve above the x-axis.

(e) Solve \(\left| \frac{2x^2 - 5x}{2x^2 - 7x - 4} \right| < \frac{1}{9}\) by considering \(\frac{2x^2 - 5x}{2x^2 - 7x - 4} < \frac{1}{9}\) and \(\frac{2x^2 - 5x}{2x^2 - 7x - 4} > -\frac{1}{9}\). Solving these inequalities gives the intervals \(\frac{13}{10} < x < \frac{19}{16}\) and \(\frac{\sqrt{33}}{16} < x < \frac{13}{10} + \frac{3}{10} \sqrt{21}\).

(a) The stretch parallel to the x-axis with scale factor 14 is represented by the matrix \(\begin{pmatrix} 14 & 0 \\ 0 & 1 \end{pmatrix}\).

The rotation anticlockwise about the origin through angle \(\frac{1}{3} \pi\) is represented by \(\begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}\).

Thus, \(\mathbf{M} = \begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix} \begin{pmatrix} 14 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 7 & -\frac{\sqrt{3}}{2} \\ 7\sqrt{3} & \frac{1}{2} \end{pmatrix}\).

Therefore, \(2\mathbf{M} = \begin{pmatrix} 14 & -\sqrt{3} \\ 14\sqrt{3} & 1 \end{pmatrix}\).

(b) The transformation \(\mathbf{M}\) transforms \(\begin{pmatrix} x \\ y \end{pmatrix}\) to \(\begin{pmatrix} 7x - \frac{\sqrt{3}}{2}y \\ 7\sqrt{3}x + \frac{1}{2}y \end{pmatrix}\).

For invariant lines, \(y = mx\) and \(Y = mX\).

\(7\sqrt{3}x + \frac{1}{2}mx = m(7x - \frac{\sqrt{3}}{2}mx)\).

\(7\sqrt{3} + \frac{1}{2}m = 7m - \frac{\sqrt{3}}{2}m^2\).

\(\sqrt{3}m^2 - 13m + 14\sqrt{3} = 0\).

Solving gives \(m = 2\sqrt{3}\) and \(m = \frac{7}{\sqrt{3}}\).

Thus, the invariant lines are \(y = 2\sqrt{3}x\) and \(y = \frac{1}{\sqrt{3}}x\).

(c) The inverse of \(2\mathbf{M}\) is \(\mathbf{M}^{-1} = \frac{1}{14} \begin{pmatrix} 1 & \sqrt{3} \\ -14\sqrt{3} & 7 \end{pmatrix}\).

Base case: For \(n = 1\), \(2025^1 + 47^1 - 2 = 2070 = 45 \times 46\), which is divisible by 46.

Inductive step: Assume \(2025^k + 47^k - 2\) is divisible by 46 for some positive integer \(k\).

Then, consider \(2025^{k+1} + 47^{k+1} - 2\).

\(2025^{k+1} + 47^{k+1} - 2 = (2024 + 1)2025^k + (46 + 1)47^k - 2\)

This can be rewritten as:

\(2024 \times 2025^k + 46 \times 47^k + 2025^k + 47^k - 2\)

\(= 46(44 \times 2025^k + 47^k) + (2025^k + 47^k - 2)\)

By the inductive hypothesis, \(2025^k + 47^k - 2\) is divisible by 46.

Hence, \(2025^{k+1} + 47^{k+1} - 2\) is divisible by 46.

Therefore, by induction, the statement is true for every positive integer \(n\).

(a) Using the identity \((\alpha + \beta + \gamma + \delta)^2 = \alpha^2 + \beta^2 + \gamma^2 + \delta^2 + 2(\alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta)\), we have:

\(0 = \alpha^2 + \beta^2 + \gamma^2 + \delta^2 + 2(7)\)

Thus, \(\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = -14\).

(b) Using the given equation, \(S_4 + 7S_2 + 3S_1 + 88 = 0\), where \(S_4 = \alpha^4 + \beta^4 + \gamma^4 + \delta^4\), we find:

\(S_4 = 10\).

(c) Expanding \((r + \alpha^2)^2 = r^2 + 2\alpha^2 r + \alpha^4\), we have:

\(\sum_{r=1}^{10} ((r + \alpha^2)^2 + (r + \beta^2)^2 + (r + \gamma^2)^2 + (r + \delta^2)^2) = \sum_{r=1}^{10} (4r^2 - 28r + 10)\)

\(= 4\left(\frac{10}{6}(10+1)(2(10)+1)\right) - 14(10)(10+1) + 10(10)\)

\(= 100\).

(a) Consider \(w_{r+1} = (r+1)(r+2)\ldots(r+10)\) and \(w_r = r(r+1)(r+2)\ldots(r+9)\).

Then, \(w_{r+1} - w_r = (r+1)(r+2)\ldots(r+10) - r(r+1)(r+2)\ldots(r+9)\).

Factor out \((r+1)(r+2)\ldots(r+9)\):

\(= (r+1)(r+2)\ldots(r+9)((r+10) - r) = 10(r+1)(r+2)\ldots(r+9)\).

(b) The sum \(\sum_{r=1}^{n} u_r = \frac{1}{10}((w_2 - w_1) + (w_3 - w_2) + \ldots + (w_{n+1} - w_n))\).

This telescopes to \(\frac{1}{10}(w_{n+1} - w_1)\).

\(w_{n+1} = (n+1)(n+2)\ldots(n+10)\) and \(w_1 = 1 \cdot 2 \cdot \ldots \cdot 10 = 10!\).

Thus, \(\sum_{r=1}^{n} u_r = \frac{1}{10}((n+1)(n+2)\ldots(n+10) - 10!)\).

(c) The series \(v_1 + v_2 + \ldots + v_n = x^{w_2} - x^{w_1} + x^{w_3} - x^{w_2} + \ldots + x^{w_{n+1}} - x^{w_n}\).

This telescopes to \(x^{w_{n+1}} - x^{w_1}\).

For convergence as \(n \to \infty\), \(|x| < 1\) is required.

Thus, for \(-1 < x < 1\), the sum to infinity is \(-x^{w_1} = -x^{10!}\).

For \(x = \pm 1\), the sum to infinity is 0.

(a) To find the Cartesian equation of the plane \(\Pi\), we need a normal vector. The direction vectors are \(\mathbf{i} - 2\mathbf{j} - \mathbf{k}\) and \(3\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}\). The normal vector is the cross product:

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -2 & -1 \\ 3 & 2 & -2 \end{vmatrix} = \begin{pmatrix} 6 \\ -1 \\ 8 \end{pmatrix}\)

Using the point \((2, 3, -2)\) on the plane, substitute into \(6x - y + 8z = d\):

\(6(2) - 3 + 8(-2) = -7\)

Thus, the equation is \(6x - y + 8z = -7\).

(b) The position vector of the foot of the perpendicular \(\mathbf{F}\) is given by:

\(\mathbf{OF} = \mathbf{OP} + t \begin{pmatrix} 6 \\ -1 \\ 8 \end{pmatrix}\)

\(\mathbf{OF} = \begin{pmatrix} 4 \\ 2 \\ 9 \end{pmatrix} + t \begin{pmatrix} 6 \\ -1 \\ 8 \end{pmatrix} = \begin{pmatrix} 4 + 6t \\ 2 - t \\ 9 + 8t \end{pmatrix}\)

Substitute into the plane equation:

\(6(4 + 6t) - (2 - t) + 8(9 + 8t) = -7\)

\(101t + 94 = -7\)

\(t = -1\)

\(\mathbf{OF} = \begin{pmatrix} -2 \\ 3 \\ 1 \end{pmatrix}\)

(c) The acute angle \(\alpha\) between \(l\) and \(\Pi\) is found using:

\(\cos \alpha = \frac{\mathbf{d} \cdot \mathbf{n}}{\|\mathbf{d}\| \|\mathbf{n}\|}\)

\(\mathbf{d} = \begin{pmatrix} 3 \\ 5 \\ -1 \end{pmatrix}, \mathbf{n} = \begin{pmatrix} 6 \\ -1 \\ 8 \end{pmatrix}\)

\(\cos \alpha = \frac{5}{\sqrt{35} \sqrt{101}}\)

\(\alpha = 4.8^\circ\)

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

(b) Differentiate \(y = \frac{x^2 + a}{x + a}\) to find stationary points: \(\frac{dy}{dx} = \frac{(x+a)^2 - (x^2+a)}{(x+a)^2} = \frac{-a^2}{(x+a)^2}\). Set \(\frac{dy}{dx} = 0\) to find \(x = -a \pm \sqrt{a^2 + a}\).

(c) Sketch the curve showing the asymptotes \(x = -a\) and \(y = x - a\). The curve intersects the y-axis at \((0, 1)\).

(d) Sketch the absolute value of the curve from (c), reflecting any negative parts above the x-axis.

(e) Solve \(\left| \frac{x^2 + a}{x + a} \right| = a\). This leads to \(x^2 - ax + a - a^2 = 0\). The discriminant \(a^2 - 4(a-a^2) = 5a^2 - 4a > 0\) gives \(a > \frac{3}{4}\).

(a) Differentiate \(r^2 = e^{\sin \theta} \cos \theta\) with respect to \(\theta\) and set \(\frac{dr}{d\theta} = 0\). This gives:

\(-e^{\sin \theta} \sin \theta + e^{\sin \theta} \cos^2 \theta = 0\)

\(\cos^2 \theta = \sin \theta\)

\(1 - \sin^2 \theta = \sin \theta\)

\(\sin \theta = \frac{1}{2}(-1 + \sqrt{5})\)

\(\theta = 0.666\)

\(r = \sqrt{e^{\sin(0.666)} \cos(0.666)} = 1.208\)

(b) Use \(x = e^{\sin \theta} \cos \theta\) and differentiate with respect to \(\theta\):

\(\frac{dx}{d\theta} = -\frac{1}{2}e^{\sin \theta} \cos \theta \sin \theta + e^{\sin \theta} \cos \theta \cos \theta - \frac{1}{2}e^{\sin \theta} \sin \theta \cos \theta = 0\)

\(-3\sin \theta + \cos^2 \theta = 0\)

\(-3\sin \theta + 1 - \sin^2 \theta = 0\)

\(\sin \theta = \frac{1}{2}(-3 + \sqrt{13})\)

\(\theta = 0.308\)

\(r = \sqrt{e^{\sin(0.308)} \cos(0.308)} = 1.136\)

(c) The sketch shows a closed loop passing through O and one other point on the initial line. It does not go into the 2nd and 3rd quadrants and is more in the 1st than the 4th quadrant.

(d) The area is given by:

\(\frac{1}{2} \int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi} e^{\sin \theta} \cos \theta \, d\theta\)

\(= \frac{1}{2} \left[ e^{\sin \theta} \right]_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi}\)

\(= \frac{1}{3}(e - e^{-1})\)

(a) We start with \(\sum_{r=1}^{n} (2r+1) = 2\sum_{r=1}^{n} r + n\). Using the formula for the sum of the first \(n\) natural numbers, \(\sum_{r=1}^{n} r = \frac{n(n+1)}{2}\), we have:

\(2\sum_{r=1}^{n} r + n = 2 \times \frac{n(n+1)}{2} + n = n(n+1) + n = n(n+2)\).

(b) To show \(\frac{2r+1}{(r^2+1)(r^2+2r+2)} = \frac{1}{r^2+1} - \frac{1}{r^2+2r+2}\), we find a common denominator:

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

(c) Using the result from (b), the series becomes a telescoping series:

\(\sum_{r=1}^{n} \left( \frac{1}{r^2+1} - \frac{1}{r^2+2r+2} \right) = \frac{1}{2} - \frac{1}{n^2+2n+2}\).

(d) As \(n \to \infty\), the term \(\frac{1}{n^2+2n+2} \to 0\), so the sum converges to \(\frac{1}{2}\).

First, find the first derivative:

\(\frac{d}{dx}(x \ln x) = 1 + \ln x\).

Check the base case for \(n = 2\):

\(\frac{d^2}{dx^2}(x \ln x) = x^{-1} = (-1)^2 (2-2)! x^{1-2}\).

Assume that \(\frac{d^k}{dx^k}(x \ln x) = (-1)^k (k-2)! x^{1-k}\) holds for some integer \(k \geq 2\).

Differentiate the \(k\)-th derivative:

\(\frac{d^{k+1}}{dx^{k+1}}(x \ln x) = (-1)^k (k-2)!(1-k)x^{1-k-1}\).

Simplify:

\(= (-1)^{k+1} ((k+1)-2)! x^{1-(k+1)}\).

Thus, \(\frac{d^n}{dx^n}(x \ln x) = (-1)^n (n-2)! x^{1-n}\) is also true for \(n = k+1\).

By induction, the statement is true for every integer \(n \geq 2\).

(a) To find the equation of the plane, we first determine the direction vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\):

\(\overrightarrow{AB} = (-5\mathbf{i} + 3\mathbf{j} + \mathbf{k}) - (2\mathbf{j} + 3\mathbf{k}) = -5\mathbf{i} + \mathbf{j} - 2\mathbf{k}\)

\(\overrightarrow{AC} = (\mathbf{i} + 2\mathbf{j} + 5\mathbf{k}) - (2\mathbf{j} + 3\mathbf{k}) = \mathbf{i} + 2\mathbf{k}\)

The normal vector \(\mathbf{n}\) to the plane is given by the cross product \(\overrightarrow{AB} \times \overrightarrow{AC}\):

\(\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -5 & 1 & -2 \\ 1 & 0 & 2 \end{vmatrix} = 2\mathbf{i} + 8\mathbf{j} - \mathbf{k}\)

Using point \(A(0, 2, 3)\), the equation of the plane is:

\(2(0) + 8(2) - 1(3) = 13 \Rightarrow 2x + 8y - z = 13\)

(b) The perpendicular distance from the origin \(O(0, 0, 0)\) to the plane is:

\(\frac{|2(0) + 8(0) - 1(0) - 13|}{\sqrt{2^2 + 8^2 + (-1)^2}} = \frac{13}{\sqrt{69}} = 1.57\)

(c) The acute angle \(\theta\) between the line \(OA\) and the plane is found using the dot product:

\(\cos \theta = \frac{\mathbf{n} \cdot \overrightarrow{OA}}{\|\mathbf{n}\| \|\overrightarrow{OA}\|}\)

\(\overrightarrow{OA} = 2\mathbf{j} + 3\mathbf{k}\)

\(\mathbf{n} \cdot \overrightarrow{OA} = 2(0) + 8(2) - 1(3) = 13\)

\(\|\mathbf{n}\| = \sqrt{69}, \quad \|\overrightarrow{OA}\| = \sqrt{13}\)

\(\cos \theta = \frac{13}{\sqrt{69} \sqrt{13}}\)

\(\theta = \cos^{-1}\left(\frac{13}{\sqrt{69} \sqrt{13}}\right) = 0.449\) radians

(a) The determinant of the matrix \(\begin{pmatrix} 1 & \alpha & \beta \\ \alpha & 1 & \gamma \\ \beta & \gamma & 1 \end{pmatrix}\) is calculated as:

\(\left| \begin{array}{ccc} 1 & \alpha & \beta \\ \alpha & 1 & \gamma \\ \beta & \gamma & 1 \end{array} \right| = 1 \cdot \left| \begin{array}{cc} 1 & \gamma \\ \gamma & 1 \end{array} \right| - \alpha \cdot \left| \begin{array}{cc} \alpha & \gamma \\ \beta & 1 \end{array} \right| + \beta \cdot \left| \begin{array}{cc} \alpha & 1 \\ \beta & \gamma \end{array} \right|\)

\(= 1 - \alpha^2 - \beta^2 - \gamma^2 + 2\alpha\beta\gamma\)

Since the matrix is singular, the determinant is zero:

\(1 - \alpha^2 - \beta^2 - \gamma^2 + 2\alpha\beta\gamma = 0\)

Given \(\alpha\beta\gamma = 1\), we have:

\(\alpha^2 + \beta^2 + \gamma^2 = 1 + 2(1) = 3\)

(b) We use the identity:

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

Given \(\alpha^2 + \beta^2 + \gamma^2 = 3\), we have:

\(3 = b^2 - 2c\)

Also, given \(\alpha^3 + \beta^3 + \gamma^3 = 3\), we use:

\(\alpha^3 + \beta^3 + \gamma^3 = -b(3) - c(-b) + 3\)

\(3 = -3b + cb + 3\)

Solving these simultaneous equations, we find:

\(c = 3, \ b = 3\)

(a) The transformation matrix for a stretch in the x-direction by a factor of 3 is \(\begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix}\). The reflection in the line y = x is represented by \(\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\). Therefore, \(\mathbf{M} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 3 & 0 \end{pmatrix}\).

(b) The inverse of \(\mathbf{M}\) is \(\mathbf{M}^{-1} = \begin{pmatrix} 0 & \frac{1}{3} \\ 1 & 0 \end{pmatrix}\). This represents a reflection in the line y = x followed by a stretch in the x-direction by a factor of \(\frac{1}{3}\).

(c) Given \(\mathbf{MN} = \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix}\), we find \(\mathbf{N}\) by multiplying by \(\mathbf{M}^{-1}\) on the left: \(\mathbf{N} = \mathbf{M}^{-1} \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix} = \begin{pmatrix} 0 & \frac{1}{3} \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix} = \begin{pmatrix} 1 & \frac{1}{2} \\ 1 & 2 \end{pmatrix}\).

(d) The invariant lines satisfy \(\begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x + 2y \\ 3x + 2y \end{pmatrix}\). Setting \(y = mx\), we have \(3x + 2mx = m(x + 2y)\). This leads to the quadratic \(2m^2 - m - 3 = 0\), which factors to \((2m + 3)(m - 1) = 0\). Thus, \(m = \frac{3}{2}\) or \(m = -1\), giving the lines \(y = \frac{3}{2}x\) and \(y = -x\).

(a) The curve is a spiral shape on \([0, 2\pi]\), starting at the pole with one other point of intersection with the initial line. The maximum distance from the pole is \(a\).

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

\(A = \frac{1}{2} \int_0^{2\pi} r^2 \, d\theta = \frac{1}{2} a^2 \int_0^{2\pi} \tan^2\left(\frac{1}{8}\theta\right) \, d\theta\)

Using \(\tan^2 A = \sec^2 A - 1\), we have:

\(= \frac{1}{2} a^2 \left[ 8 \tan\left(\frac{1}{8}\theta\right) - \theta \right]_0^{2\pi}\)

\(= \frac{1}{2} a^2 (8 - 2\pi) = a^2 (4 - \pi)\)

(c) Let \(y = a \tan\left(\frac{1}{8}\theta\right) \sin\theta\). The derivative is:

\(\frac{dy}{d\theta} = a \left( \tan\left(\frac{1}{8}\theta\right) \cos\theta + \frac{1}{8} \sin\theta \sec^2\left(\frac{1}{8}\theta\right) \right) = 0\)

\(\Rightarrow \cos\left(\frac{1}{8}\theta\right) \sin\left(\frac{1}{8}\theta\right) \cos\theta + \frac{1}{8} \sin\theta = 0\)

\(\Rightarrow \sin\left(\frac{1}{4}\theta\right) \cos\theta + \frac{1}{4} \sin\theta = 0\)

\(\Rightarrow 4 \sin\left(\frac{1}{4}\theta\right) \cos\theta + \sin\theta = 0\)

Checking values:

\(4 \sin\left(\frac{1}{4} \times 4.95\right) \cos 4.95 + \sin 4.95 \approx -0.0822\)

\(4 \sin\left(\frac{1}{4} \times 5\right) \cos 5 + \sin 5 \approx 0.118\)

There is a sign change, confirming a root between 4.95 and 5.

(a) The horizontal asymptote is found by considering the limits as \(x \to \pm \infty\). The leading coefficients of the numerator and denominator are equal, so \(y = 1\) is the horizontal asymptote.

(b) Consider \(y = \frac{x^2 + x - 4}{x^2 + x + 2}\). Rearrange to form a quadratic in \(x\): \((y-1)x^2 + (y-1)x + (2y + 4) = 0\). The discriminant \(\Delta = (y-1)^2 - 4(y-1)(2y+4) \geq 0\) gives \((y-1)(7y + 17) \leq 0\). Solving this inequality gives \(-\frac{17}{7} \leq y < 1\).

(c) Find stationary points by setting \(\frac{dy}{dx} = 0\). Differentiate: \(\frac{dy}{dx} = \frac{12x + 6}{(x^2 + x + 2)^2} = 0\). Solving gives \(x = -\frac{1}{2}\). Substitute back to find \(y = -\frac{17}{7}\). Stationary point is \(\left(-\frac{1}{2}, -\frac{17}{7}\right)\).

(d) Sketch the curve, noting the asymptote \(y = 1\) and intersections at \((0, -2), \left(\frac{1}{2}(1-\sqrt{7}), 0\right), \left(\frac{1}{2}(1+\sqrt{7}), 0\right)\).

(e) For \(y = \frac{|x|^2 + |x| - 4}{|x|^2 + |x| + 2} < -\frac{1}{2}\), solve \(3|x|^2 + 3|x| - 6 = 0\). Critical points are \(|x| = 1\). Thus, \(-1 < x < 1\).

(a) The transformation matrix for a stretch parallel to the x-axis with scale factor k is \(\begin{pmatrix} k & 0 \\ 0 & 1 \end{pmatrix}\). The shear transformation with the x-axis fixed and (0, 1) mapped to (k, 1) is \(\begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}\). The combined transformation matrix M is the product:

\(\mathbf{M} = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix} \begin{pmatrix} k & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} k & k \\ 0 & 1 \end{pmatrix}\)

(b) The transformation \(\mathbf{M} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} kx + ky \\ y \end{pmatrix}\) implies invariant points satisfy \(kx + ky = x\). Solving gives \(y = \frac{1-k}{k}x\).

(c) The inverse matrix \(\mathbf{M}^{-1}\) is found by inverting \(\mathbf{M}\):

\(\mathbf{M}^{-1} = \frac{1}{k} \begin{pmatrix} 1 & -k \\ 0 & k \end{pmatrix}\)

(d) The area of P is given by the determinant of M, \(|k| = 3k^2\). Solving \(|k| = 3k^2\) gives \(k = \pm \frac{1}{3}\).

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

\(\frac{d}{dx}(\arctan x) = \frac{1}{1+x^2},\)

which is true.

Inductive step: Assume

\(\frac{d^k}{dx^k}(\arctan x) = P_k(x)(1+x^2)^{-k},\)

where \(\deg P_k(x) = k-1\).

Then,

\(\frac{d^{k+1}}{dx^{k+1}}(\arctan x) = P_k'(x)(1+x^2)^{-k} - 2kxP_k(x)(1+x^2)^{-k-1}.\)

This can be written as

\(\left(P_k'(x)(1+x^2) - 2kxP_k(x)\right)(1+x^2)^{-k-1},\)

which implies \(\deg P_{k+1}(x) = k\).

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

(a) Let \(y = x^4\). Then the equation becomes \(y + 2y^{3/4} - 1 = 0\). Substituting \(y = (1-y)^4\), we expand using the binomial theorem: \(16y^3 = 1 - 4y + 6y^2 - 4y^3 + y^4\). Rearranging gives \(y^4 - 20y^3 + 6y^2 - 4y + 1 = 0\). The sum \(\alpha^4 + \beta^4 + \gamma^4 + \delta^4 = 20\).

(b) Using \(\alpha + \beta + \gamma + \delta = -2\), multiply the original equation by \(x\) to get \(x^5 + 2x^4 - x = 0\). Thus, \(\alpha^5 + \beta^5 + \gamma^5 + \delta^5 = -2(20) + (-2) = -42\).

(c) Using the formula for the sum of squares, \(\alpha^8 + \beta^8 + \gamma^8 + \delta^8 = 20^2 - 2(6) = 388\).

(a) Start by expressing \(\frac{5k}{(5r+k)(5r+5+k)}\) using partial fractions:

\(\frac{5k}{(5r+k)(5r+5+k)} = k \left( \frac{1}{5r+k} - \frac{1}{5r+5+k} \right)\)

Then, the sum becomes:

\(\sum_{r=1}^{n} \frac{5k}{(5r+k)(5r+5+k)} = k \left( \frac{1}{5+k} + \frac{1}{10+k} + \frac{1}{15+k} + \ldots + \frac{1}{5n+k} - \frac{1}{5n+5+k} \right)\)

This simplifies to:

\(= k \left( \frac{1}{5+k} - \frac{1}{5n+5+k} \right)\)

(b) Given \(\sum_{r=1}^{\infty} \frac{5k}{(5r+k)(5r+5+k)} = \frac{1}{3}\), we have:

\(\frac{k}{5+k} = \frac{1}{3}\)

Solving for \(k\):

\(3k = 5 + k \Rightarrow 2k = 5 \Rightarrow k = \frac{5}{2}\)

(c) Using the result from (a), find:

\(\sum_{r=n}^{n^2} \frac{5k}{(5r+k)(5r+5+k)} = \sum_{r=1}^{n^2} \frac{5k}{(5r+k)(5r+5+k)} - \sum_{r=1}^{n-1} \frac{5k}{(5r+k)(5r+5+k)}\)

\(= k \left( \frac{1}{5+k} - \frac{1}{5n^2+5+k} \right) - k \left( \frac{1}{5+k} - \frac{1}{5n+5+k} \right)\)

\(= k \left( \frac{1}{5n+5+k} - \frac{1}{5n^2+5+k} \right)\)

Substitute \(k = \frac{5}{2}\):

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

(a) Substitute \(x = r \cos \theta\) and \(y = r \sin \theta\) into the Cartesian equation \((x^2 + y^2)^2 = 6xy\).

This gives \((r^2)^2 = 6r^2 \cos \theta \sin \theta\).

Thus, \(r^4 = 6r^2 \cos \theta \sin \theta\).

Divide both sides by \(r^2\) (assuming \(r \neq 0\)) to get \(r^2 = 6 \cos \theta \sin \theta\).

Using the identity \(\sin 2\theta = 2 \sin \theta \cos \theta\), we have \(r^2 = 3 \sin 2\theta\).

(b) The sketch shows a single loop symmetrical about \(\theta = \frac{1}{4}\pi\). The maximum distance from the pole is \(\sqrt{3}\).

(c) The area enclosed by \(C\) is given by \(\frac{1}{2} \int_0^{\frac{1}{2}\pi} r^2 \, d\theta\).

Substitute \(r^2 = 3 \sin 2\theta\) to get \(\frac{3}{2} \int_0^{\frac{1}{2}\pi} \sin 2\theta \, d\theta\).

Integrate to find \(\frac{3}{2} \left[ -\frac{1}{2} \cos 2\theta \right]_0^{\frac{1}{2}\pi} = \frac{3}{2}\).

(d) To find the maximum distance from the initial line, set \(y = r \sin \theta = 3^{\frac{3}{4}} \sin^{\frac{1}{2}} 2\theta \sin \theta\).

Set \(\frac{dy}{d\theta} = 0\) and solve \(\sin^2 2\theta \cos \theta + \sin^{-\frac{1}{2}} 2\theta \cos 2\theta \sin \theta = 0\).

Using trigonometric identities, solve for \(\theta = \frac{1}{3}\pi\).

The maximum distance is \(\frac{3^{\frac{3}{4}}}{2^{\frac{1}{2}}} = 1.40\).

(a) To find the vertical asymptotes, set the denominator equal to zero: \(2x^2 - 7x + 3 = 0\). Solving gives \(x = \frac{1}{2}\) and \(x = 3\). The horizontal asymptote is found by considering the degrees of the polynomials: \(y = \frac{4}{2} = 2\).

(b) To find stationary points, differentiate \(y\) with respect to \(x\):

\(\frac{dy}{dx} = \frac{(2x^2 - 7x + 3)(8x + 1) - (4x^2 + x + 1)(4x - 7)}{(2x^2 - 7x + 3)^2}\).

Set \(\frac{dy}{dx} = 0\) and solve \(-3x^2 + 2x + 1 = 0\) to find \(x = -\frac{1}{3}\) and \(x = 1\). Substitute back to find \(y\) values: \(\left(-\frac{1}{3}, \frac{1}{3}\right)\) and \((1, -3)\).

(c) Sketch the curve using the asymptotes and stationary points. The curve intersects the y-axis at \((0, \frac{1}{3})\).

(d) For \(\left| \frac{4x^2 + x + 1}{2x^2 - 7x + 3} \right| = k\) to have 4 distinct real solutions, \(k > 3\).

(a) To find the equation of \(\Pi_1\), we need a normal vector to the plane. Since \(\Pi_1\) contains \(l_1\) and is parallel to \(l_2\), the normal vector can be found using the cross product of the direction vectors of \(l_1\) and \(l_2\):

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & 1 \\ 1 & -4 & 2 \end{vmatrix} = \begin{pmatrix} 6 \\ -3 \\ -9 \end{pmatrix}\)

Substitute the point \((1, 3, -2)\) from \(l_1\) into the plane equation \(6x - 3y - 9z = d\) to find \(d\):

\(6(1) - 3(3) - 9(-2) = d\)

\(6 - 9 + 18 = d\)

\(d = 15\)

Thus, the equation of \(\Pi_1\) is \(2x - y - 3z = 5\).

(b) The normal to \(\Pi_2\) is the direction vector of \(l_2\), \(\begin{pmatrix} 1 \\ -4 \\ 2 \end{pmatrix}\). The angle \(\theta\) between \(\Pi_1\) and \(\Pi_2\) is given by:

\(\cos \theta = \frac{\mathbf{n_1} \cdot \mathbf{n_2}}{\|\mathbf{n_1}\| \|\mathbf{n_2}\|}\)

\(\cos \theta = \frac{\begin{pmatrix} 6 \\ -3 \\ -9 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -4 \\ 2 \end{pmatrix}}{\sqrt{6^2 + (-3)^2 + (-9)^2} \sqrt{1^2 + (-4)^2 + 2^2}}\)

\(\cos \theta = \frac{-7}{\sqrt{14 \times 77}}\)

\(\theta = 77.7^\circ\)

(c) To find \(\mathbf{PQ}\), use the points \(\mathbf{P}\) and \(\mathbf{Q}\) on \(l_1\) and \(l_2\) respectively:

\(\mathbf{OP} = \begin{pmatrix} 1 + 2\lambda \\ 3 + \lambda \\ -2 + \lambda \end{pmatrix}\)

\(\mathbf{OQ} = \begin{pmatrix} 1 + \mu \\ -2 - 4\mu \\ 9 + 2\mu \end{pmatrix}\)

\(\mathbf{PQ} = \mathbf{OQ} - \mathbf{OP} = \begin{pmatrix} \mu - 2\lambda \\ -5 - 4\mu - \lambda \\ 11 + 2\mu - \lambda \end{pmatrix}\)

Set \(\mathbf{PQ}\) perpendicular to both direction vectors:

\(\begin{pmatrix} \mu - 2\lambda \\ -5 - 4\mu - \lambda \\ 11 + 2\mu - \lambda \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} = 0\)

\(\begin{pmatrix} \mu - 2\lambda \\ -5 - 4\mu - \lambda \\ 11 + 2\mu - \lambda \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -4 \\ 2 \end{pmatrix} = 0\)

Solve these equations to find \(\lambda\) and \(\mu\), then substitute back to find \(\mathbf{PQ}\):

\(\mathbf{r} = \begin{pmatrix} -1 \\ 6 \\ 5 \end{pmatrix} + \lambda \begin{pmatrix} 6 \\ -3 \\ -9 \end{pmatrix}\)

Base case: For \(n = 1\), \(u_1 = 4 = 3^1 + 1\). This holds true.

Inductive step: Assume it is true for \(n = k\), so \(u_k = 3^k + 1\).

Then \(u_{k+1} = 3u_k - 2 = 3(3^k + 1) - 2 = 3^{k+1} + 3 - 2 = 3^{k+1} + 1\).

Thus, it is true for \(n = k+1\). Hence, by induction, \(u_n = 3^n + 1\) for all positive integers \(n\).

(a) To find the equation of the plane \(\Pi\), we need a normal vector to the plane. The direction vector of \(l_1\) is \(\mathbf{i} - \mathbf{j} - 4\mathbf{k}\), and the plane is parallel to \(2\mathbf{i} + 5\mathbf{j} - 4\mathbf{k}\). The normal vector is the cross product of these two vectors:

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & -4 \\ 2 & 5 & -4 \end{vmatrix} = \mathbf{i}(16 - 20) - \mathbf{j}(-4 - (-8)) + \mathbf{k}(5 + 2) = -4\mathbf{i} + 4\mathbf{j} + 7\mathbf{k}\)

Using the point \((1, 3, -1)\) on \(l_1\), the equation of the plane is:

\(24(1) - 4(3) + 7(-1) = d \Rightarrow 24x - 4y + 7z = 5\).

(b) To find the acute angle between \(l_2\) and \(\Pi\), we use the dot product of the direction vector of \(l_2\), \(5\mathbf{i} - 5\mathbf{j} - 2\mathbf{k}\), and the normal vector of \(\Pi\), \(24\mathbf{i} - 4\mathbf{j} + 7\mathbf{k}\):

\((24, -4, 7) \cdot (5, -5, -2) = 24 \times 5 + (-4) \times (-5) + 7 \times (-2) = 120 + 20 - 14 = 126\).

The magnitudes are \(\sqrt{24^2 + (-4)^2 + 7^2} = \sqrt{641}\) and \(\sqrt{5^2 + (-5)^2 + (-2)^2} = \sqrt{54}\).

\(\cos \alpha = \frac{126}{\sqrt{641} \times \sqrt{54}}\).

The angle \(\alpha\) is the angle between the line and the normal, so the acute angle between \(l_2\) and \(\Pi\) is \(90^\circ - \alpha = 42.6^\circ\).

(a) We use the identity \(\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = (\alpha + \beta + \gamma + \delta)^2 - 2(\alpha \beta + \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta + \gamma \delta)\).

Substituting the given values, \(3 = 2^2 - 2(\alpha \beta + \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta + \gamma \delta)\).

Solving, \(\alpha \beta + \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta + \gamma \delta = \frac{1}{2}\).

(b) We use the identity \(\alpha^2(\beta + \gamma + \delta) + \beta^2(\alpha + \gamma + \delta) + \gamma^2(\alpha + \beta + \delta) + \delta^2(\alpha + \beta + \gamma)\).

This simplifies to \(\alpha^2(2-\alpha) + \beta^2(2-\beta) + \gamma^2(2-\gamma) + \delta^2(2-\delta)\).

Substituting the given values, \(2(3) - 4 = 2\).

(c)(i) Using the polynomial equation, \(6S_4 - 12S_3 + 3S_2 + 2S_1 + 24 = 0\).

Substitute \(S_3 = 4, S_2 = 3, S_1 = 2\) to find \(S_4 = \frac{11}{6}\).

(c)(ii) Using the polynomial equation, \(6S_5 - 12S_4 + 3S_3 + 2S_2 + S_1 = 0\).

Substitute \(S_4 = \frac{11}{6}, S_3 = 4, S_2 = 3, S_1 = 2\) to find \(S_5 = -\frac{4}{3}\).

(a) To find \(CAB\), first calculate \(AB\):

\(AB = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \\ 3 & 2 & 5 \end{pmatrix} \begin{pmatrix} 0 & -2 \\ -1 & 3 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} -2 & 4 \\ -1 & 3 \\ -4 & 7 \end{pmatrix}\)

Then calculate \(CAB\):

\(CAB = \begin{pmatrix} -2 & -1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} -2 & 4 \\ -1 & 3 \\ -4 & 7 \end{pmatrix} = \begin{pmatrix} 3 & -7 \\ -9 & 3 \end{pmatrix}\)

(b) For invariant lines, solve \(\begin{pmatrix} 3 & -7 \\ -9 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = m \begin{pmatrix} x \\ y \end{pmatrix}\):

\(-9x + 3mx = m(3x - 7y)\)

\(-9 + 3m = 3m - 7m^2 \Rightarrow 7m^2 = 9\)

\(m = \pm \frac{3}{7}\)

Thus, \(y = \frac{3}{7}x\) and \(y = -\frac{3}{7}x\).

(c) The matrix \(M\) represents a stretch parallel to the x-axis with scale factor 3.

(d) Find \(N\) such that \(NM = CAB\):

\(M^{-1} = \begin{pmatrix} \frac{1}{3} & 0 \\ 0 & 1 \end{pmatrix}\)

\(N = CAB \cdot M^{-1} = \begin{pmatrix} 3 & -7 \\ -9 & 3 \end{pmatrix} \begin{pmatrix} \frac{1}{3} & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -7 \\ -3 & 3 \end{pmatrix}\)

(a) The series \(S_n = \sum_{r=1}^{n} (x^{f(r)} - x^{f(r+1)})\) telescopes to \(x^{f(1)} - x^{f(n+1)}\).

(b) With \(f(r) = \ln r\), the series becomes \(\sum_{r=1}^{\infty} (x^{\ln r} - x^{\ln(r+1)})\). For convergence, \(0 < x \leq 1\). If \(x < 1\), the series converges to \(1\). If \(x = 1\), the series converges to \(0\).

(c) With \(f(r) = 2 \log_x r\), \(x^{2 \log_x r} = r^2\). Thus, \(S_n = 1 - (n+1)^2 = -n^2 - 2n\). Using standard results, \(\sum_{n=1}^{N} (-n^2 - 2n) = -\frac{1}{6} N(N+1)(2N+7)\).

(a) To find vertical asymptotes, set the denominator equal to zero: \(x^2 + 1 = 0\). This has no real roots, so there are no vertical asymptotes. The horizontal asymptote is found by considering the limits as \(x \to \pm \infty\), giving \(y = 1\).

(b) Rewrite \(y = 1 + \frac{2}{x^2 + 1}\). Since \(\frac{2}{x^2 + 1} > 0\), it follows that \(y > 1\). The maximum value of \(y\) occurs when \(\frac{2}{x^2 + 1}\) is maximized, which is 2, giving \(y = 3\). Thus, \(1 < y \leq 3\).

(c) Differentiate \(y = \frac{x^2 + 3}{x^2 + 1}\) to find stationary points: \(\frac{d}{dx} \left( \frac{x^2 + 3}{x^2 + 1} \right) = \frac{(x^2 + 1)(2x) - (x^2 + 3)(2x)}{(x^2 + 1)^2} = 0\). Solving gives \(x = 0\), and substituting back gives \(y = 3\). Thus, the stationary point is \((0, 3)\).

(d) The sketch shows the curve with a horizontal asymptote at \(y = 1\) and a stationary point at \((0, 3)\). The curve intersects the y-axis at \((0, 3)\).

(e) For \(y = \frac{x^2 + 1}{x^2 + 3} < \frac{1}{2}\), solve \(2(x^2 + 1) < x^2 + 3\), giving \(x^2 < 1\). Thus, \(-1 < x < 1\).

(a) The polar equation is \(r = a(\cos \theta + \sin \theta)\). Using \(x = r\cos \theta\) and \(y = r\sin \theta\), we have:

\(x = a\cos \theta (\cos \theta + \sin \theta) = a(\cos^2 \theta + \cos \theta \sin \theta)\)

\(y = a\sin \theta (\cos \theta + \sin \theta) = a(\sin \theta \cos \theta + \sin^2 \theta)\)

Adding these, \(x + y = a(\cos^2 \theta + 2\cos \theta \sin \theta + \sin^2 \theta) = a\).

Thus, \(x + y = a\). The Cartesian equation is \(x^2 + y^2 = ax + ay\), which simplifies to \((x - \frac{a}{2})^2 + (y - \frac{a}{2})^2 = (\frac{a}{\sqrt{2}})^2\).

(b) The greatest distance from the pole is when \(\theta = \frac{\pi}{4}\), giving \(r = a\sqrt{2}\).

(c) To verify \(1.25 < \phi < 1.26\), solve \(\cos \phi + \sin \phi - \phi = 0\). Check values: \(\cos 1.25 + \sin 1.25 - 1.25 = 0.01\) and \(\cos 1.26 + \sin 1.26 - 1.26 = -0.002\), showing a sign change.

(d) The area of the smaller region is:

\(\frac{1}{2} \int_0^\phi \theta^2 \, d\theta + \frac{a^2}{2} \int_\phi^{\frac{3}{4}\pi} (\cos \theta + \sin \theta)^2 \, d\theta\)

Calculate each integral and simplify to get:

\(\frac{1}{2}a^2 \left( \frac{3}{4}\pi + \frac{1}{3}\phi^3 - \phi + \frac{1}{2}\cos 2\phi \right)\)

The area of the larger region is:

\(\frac{a^2}{2} \left( \frac{\pi}{3} + \frac{1}{3}\phi - \frac{1}{2}\cos 2\phi \right)\)

(a) The transformation is a stretch parallel to the x-axis with scale factor k, represented by \(\begin{pmatrix} k & 0 \\ 0 & 1 \end{pmatrix}\), followed by a shear with (0, 1) mapped to (k, 1), represented by \(\begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}\). The combined transformation matrix M is:

\(\begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix} \begin{pmatrix} k & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} k & k \\ 0 & 1 \end{pmatrix}\)

(b) The line of invariant points satisfies \(\begin{pmatrix} k & k \\ 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}\). This gives:

\(kx + ky = x\) and \(y = y\)

Solving \(kx + ky = x\) gives \(y = \frac{1-k}{k}x\).

(c) The inverse of M is:

\(\text{If } \text{det}(\textbf{M}) = k, \text{ then } \textbf{M}^{-1} = \frac{1}{k} \begin{pmatrix} 1 & -k \\ 0 & k \end{pmatrix}\)

(d) Given \(|k| = 3k^2\), solving gives \(k = \pm \frac{1}{3}\).

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

\(\frac{d}{dx} \left( \arctan x \right) = \frac{1}{1 + x^2},\)

which is true.

Inductive step: Assume

\(\frac{d^k}{dx^k} \left( \arctan x \right) = P_k(x) (1 + x^2)^{-k},\)

where \(\deg P_k(x) = k - 1\).

Then,

\(\frac{d^{k+1}}{dx^{k+1}} \left( \arctan x \right) = P_k'(x) (1 + x^2)^{-k} - 2kx P_k(x) (1 + x^2)^{-k-1}.\)

This can be written as

\(\left( P_k'(x) (1 + x^2) - 2kx P_k(x) \right) (1 + x^2)^{-k-1},\)

so \(\deg P_{k+1}(x) = k\).

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

(a) Let \(y = x^4\). Then the equation becomes \(y + 2y^{3/4} - 1 = 0\). Substitute \(y^{3/4} = (1-y)^4\) to get \(16y^3 = 1 - 4y + 6y^2 - 4y^3 + y^4\). Rearrange to form \(y^4 - 20y^3 + 6y^2 - 4y + 1 = 0\). The value of \(\alpha^4 + \beta^4 + \gamma^4 + \delta^4 = 20\).

(b) Using \(\alpha + \beta + \gamma + \delta = -2\), multiply the original equation by \(x\) to get \(x^5 + 2x^4 - x = 0\). Thus, \(\alpha^5 + \beta^5 + \gamma^5 + \delta^5 = -2(20) + (-2) = -42\).

(c) Use the formula for the sum of squares: \(\alpha^8 + \beta^8 + \gamma^8 + \delta^8 = 20^2 - 2(6) = 388\).

(a) Start by expressing \(\frac{5k}{(5r+k)(5r+5+k)}\) using partial fractions:

\(\frac{5k}{(5r+k)(5r+5+k)} = k \left( \frac{1}{5r+k} - \frac{1}{5r+5+k} \right)\)

Then, the sum becomes:

\(\sum_{r=1}^{n} \frac{5k}{(5r+k)(5r+5+k)} = k \left( \frac{1}{5+k} + \frac{1}{10+k} + \frac{1}{15+k} + \ldots + \frac{1}{5n+k} - \frac{1}{5+k} - \frac{1}{10+k} - \ldots - \frac{1}{5n+5+k} \right)\)

By the method of differences, this simplifies to:

\(k \left( \frac{1}{5+k} - \frac{1}{5n+5+k} \right)\)

(b) Given \(\sum_{r=1}^{\infty} \frac{5k}{(5r+k)(5r+5+k)} = \frac{1}{3}\), we have:

\(k \left( \frac{1}{5+k} \right) = \frac{1}{3}\)

Solving for \(k\):

\(\frac{k}{5+k} = \frac{1}{3}\)

\(3k = 5 + k\)

\(2k = 5\)

\(k = \frac{5}{2}\)

(c) Using the result from (a), find:

\(\sum_{r=n}^{n^2} \frac{5k}{(5r+k)(5r+5+k)} = \sum_{r=1}^{n^2} \frac{5k}{(5r+k)(5r+5+k)} - \sum_{r=1}^{n-1} \frac{5k}{(5r+k)(5r+5+k)}\)

\(= k \left( \frac{1}{5+k} - \frac{1}{5n^2+5+k} \right) - k \left( \frac{1}{5+k} - \frac{1}{5n+5+k} \right)\)

\(= k \left( \frac{1}{5n+5+k} - \frac{1}{5n^2+5+k} \right)\)

Substitute \(k = \frac{5}{2}\):

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

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

(a) Substitute \(x = r \cos \theta\) and \(y = r \sin \theta\) into the Cartesian equation:

\((r^2)^2 = 6r^2 \sin \theta \cos \theta\)

\(r^4 = 6r^2 \sin \theta \cos \theta\)

\(r^2 = 3 \sin 2\theta\)

(b) The sketch shows a single loop symmetrical about \(\theta = \frac{1}{4}\pi\). The maximum distance from the pole is \(\sqrt{3}\).

(c) The area enclosed by \(C\) is given by:

\(\frac{1}{2} \int_0^{\frac{1}{2}\pi} r^2 \, d\theta = \frac{1}{2} \int_0^{\frac{1}{2}\pi} 3 \sin 2\theta \, d\theta\)

\(= \frac{3}{2} \left[ -\frac{1}{2} \cos 2\theta \right]_0^{\frac{1}{2}\pi}\)

= \(\frac{3}{2}\)

(d) To find the maximum distance from the initial line, set \(y = 3^{\frac{1}{2}} \sin^{\frac{1}{2}} 2\theta \sin \theta\) and solve \(\frac{dy}{d\theta} = 0\).

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

\(\sin 2\theta \cos \theta + \cos 2\theta \sin \theta = 0\)

\(\tan 2\theta = -\tan \theta\)

\(\frac{2 \tan \theta}{1 - \tan^2 \theta} = -\tan \theta\)

\(\theta = \frac{1}{3}\pi\)

Maximum distance is \(\frac{3^{\frac{3}{2}}}{2^{\frac{3}{2}}} = 1.40\).

(a) The vertical asymptotes occur where the denominator is zero: \(2x^2 - 7x + 3 = 0\). Solving gives \(x = \frac{1}{2}\) and \(x = 3\). The horizontal asymptote is found by considering the limits as \(x \to \pm \infty\), giving \(y = 2\).

(b) To find stationary points, set \(\frac{dy}{dx} = 0\). Differentiate using the quotient rule: \(\frac{dy}{dx} = \frac{(2x^2 - 7x + 3)(8x + 1) - (4x^2 + x + 1)(4x - 7)}{(2x^2 - 7x + 3)^2}\). Simplifying the numerator gives \(-3x^2 + 2x + 1 = 0\). Solving this quadratic gives the stationary points \(\left( -\frac{1}{3}, \frac{1}{3} \right)\) and \((1, -3)\).

(c) The sketch should show the asymptotes at \(x = \frac{1}{2}, x = 3\), and \(y = 2\). The curve intersects the y-axis at \((0, \frac{1}{3})\).

(d) For \(y = \left| \frac{4x^2 + x + 1}{2x^2 - 7x + 3} \right| = k\) to have 4 distinct real solutions, the graph must intersect the line \(y = k\) at four points. This occurs when \(k > 3\).

(a) To find the equation of \(\Pi_1\), we need a normal vector to the plane. Since \(\Pi_1\) contains \(l_1\) and is parallel to \(l_2\), the direction vector of \(l_2\), \(\begin{pmatrix} 1 \\ -4 \\ 2 \end{pmatrix}\), is parallel to \(\Pi_1\). The direction vector of \(l_1\) is \(\begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}\). The normal to \(\Pi_1\) is the cross product of these two vectors:

\(\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & 1 \\ 1 & -4 & 2 \end{vmatrix} = \begin{pmatrix} 6 \\ -3 \\ 3 \end{pmatrix}\)

Substitute the point on \(l_1\), \((1, 3, -2)\), into the plane equation \(6x - 3y + 3z = d\) to find \(d\):

\(6(1) - 3(3) + 3(-2) = -5\)

Thus, the equation of \(\Pi_1\) is \(2x - y - 3z = 5\).

(b) The normal to \(\Pi_2\) is the direction vector of \(l_2\), \(\begin{pmatrix} 1 \\ -4 \\ 2 \end{pmatrix}\). The angle \(\theta\) between \(\Pi_1\) and \(\Pi_2\) is given by:

\(\cos \theta = \frac{\mathbf{n}_1 \cdot \mathbf{n}_2}{\|\mathbf{n}_1\| \|\mathbf{n}_2\|} = \frac{\begin{pmatrix} 6 \\ -3 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -4 \\ 2 \end{pmatrix}}{\sqrt{14} \sqrt{77}} = \frac{-7}{\sqrt{14 \times 77}}\)

\(\theta = 77.7^\circ\).

(c) To find \(\mathbf{PQ}\), we need points \(P\) and \(Q\) such that \(\mathbf{PQ}\) is perpendicular to both lines. Let \(\mathbf{OP} = \begin{pmatrix} 1 + 2\lambda \\ 3 + \lambda \\ -2 + \lambda \end{pmatrix}\) and \(\mathbf{OQ} = \begin{pmatrix} 1 + \mu \\ -2 - 4\mu \\ 9 + 2\mu \end{pmatrix}\).

\(\mathbf{PQ} = \mathbf{OQ} - \mathbf{OP} = \begin{pmatrix} \mu - 2\lambda \\ -5 - 4\mu - \lambda \\ 11 + 2\mu - \lambda \end{pmatrix}\).

Set \(\mathbf{PQ} \cdot \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} = 0\) and \(\mathbf{PQ} \cdot \begin{pmatrix} 1 \\ -4 \\ 2 \end{pmatrix} = 0\) to find \(\lambda\) and \(\mu\):

\(-6\lambda + 6 = 0 \rightarrow \lambda = 1\)

\(21\mu + 42 = 0 \rightarrow \mu = -2\)

Substitute back to find \(\mathbf{PQ}\):

\(\mathbf{r} = \begin{pmatrix} -1 \\ 6 \\ 5 \end{pmatrix} + \mu \begin{pmatrix} 6 \\ -3 \\ -9 \end{pmatrix}\).

(a) By Vieta's formulas, the sum \(\alpha\beta + \beta\gamma + \gamma\alpha = -\frac{b}{a} = -\frac{1}{2}p\).

(b) The expression \(\alpha^2\beta\gamma + \alpha\beta^2\gamma + \alpha\beta\gamma^2\) can be factorized as \(\alpha\beta\gamma(\alpha + \beta + \gamma)\). Using Vieta's formulas, \(\alpha + \beta + \gamma = -\frac{1}{2}\) and \(\alpha\beta\gamma = -\frac{5}{2}\). Thus, the value is \(-\frac{5}{2} \times -\frac{1}{2} = \frac{5}{4}\).

(c) The roots \(\alpha\beta, \beta\gamma, \alpha\gamma\) imply a new cubic equation. Using Vieta's formulas, the sum of the 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 \(p\), we get \(\frac{1}{3} = \frac{1}{4} + p\), leading to \(p = \frac{1}{12}\).

Base Case: For \(n = 1\), \(6^4 + 38^1 - 2 = 1332\), which is divisible by 74.

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

Then, consider \(6^{4(k+1)} + 38^{k+1} - 2\):

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

This expression is divisible by 74 because \(1295 \times 6^{4k} + 37 \times 38^k\) is divisible by 74.

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

(a) Expand \(r(r+1)(3r+4)\) to get \(3r^3 + 7r^2 + 4r\).

Use standard summation formulae:

\(\sum_{r=1}^{N} r^3 = \left( \frac{N(N+1)}{2} \right)^2, \quad \sum_{r=1}^{N} r^2 = \frac{N(N+1)(2N+1)}{6}, \quad \sum_{r=1}^{N} r = \frac{N(N+1)}{2}.\)

Substitute these into the expanded expression:

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

Simplify to get:

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

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

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

Substitute into the series:

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

Write out terms and simplify using the method of differences:

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

The series telescopes to:

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

(c) As \(N \to \infty\), the series becomes:

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

(a) The matrix \(\mathbf{M}\) represents a stretch followed by a rotation. The stretch is parallel to the x-axis with a scale factor of 14. The rotation is \(\frac{\pi}{3}\) anticlockwise about the origin.

(b) \(\mathbf{M}^{-1} = \begin{pmatrix} \frac{1}{14} & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} \frac{1}{2} & \frac{1}{2}\sqrt{3} \\ -\frac{1}{2}\sqrt{3} & \frac{1}{2} \end{pmatrix}\)

(c) The equations of the invariant lines are \(y = 2\sqrt{3}x\) and \(y = \frac{\sqrt{3}}{3}x\).

(d) Given \(28 = 14 \times |\triangle ABC|\), the area of triangle \(ABC\) is \(2 \text{ cm}^2\).

(a) To find the equation of the plane ABC, we first find the direction vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\):

\(\overrightarrow{AB} = (2\mathbf{i} + 4\mathbf{j} - \mathbf{k}) - (2\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}) = 2\mathbf{j} - 5\mathbf{k}\)

\(\overrightarrow{AC} = (-3\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}) - (2\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}) = -5\mathbf{i} - 5\mathbf{j}\)

The normal vector to the plane is given by the cross product \(\overrightarrow{AB} \times \overrightarrow{AC}\):

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 2 & -5 \\ -5 & -5 & 0 \end{vmatrix} = -5\mathbf{i} + 5\mathbf{j} + 2\mathbf{k}\)

Using point \(A(2, 2, 4)\), the equation of the plane is:

\(-5(x - 2) + 5(y - 2) + 2(z - 4) = 0\)

Simplifying gives \(-5x + 5y + 2z = 8\).

(b) The perpendicular distance from point D to the plane is given by:

\(\frac{|8 - (2\mathbf{i} + \mathbf{j} + 3\mathbf{k}) \cdot (-5\mathbf{i} + 5\mathbf{j} + 2\mathbf{k})|}{\sqrt{(-5)^2 + 5^2 + 2^2}} = \frac{7}{\sqrt{54}} = 0.953\)

(c) The direction vector \(\overrightarrow{CD}\) is:

\(\overrightarrow{CD} = (2\mathbf{i} + \mathbf{j} + 3\mathbf{k}) - (-3\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}) = 5\mathbf{i} + 4\mathbf{j} - \mathbf{k}\)

Using the formula for the shortest distance between two skew lines:

\(\frac{|\overrightarrow{AB} \cdot (\overrightarrow{CD} \times \overrightarrow{AC})|}{|\overrightarrow{CD} \times \overrightarrow{AC}|} = \frac{35}{\sqrt{1049}} = 1.08\)

(a) The vertical asymptote is at \(x = -2\). The oblique asymptote is found by dividing \(x^2 + ax + 1\) by \(x + 2\), giving \(y = x + a - 2\).

(b) Differentiate \(y = \frac{x^2 + ax + 1}{x + 2}\) using the quotient rule: \(\frac{dy}{dx} = \frac{(x+2)(2x+a) - (x^2+ax+1)}{(x+2)^2}\). Simplifying gives \(\frac{dy}{dx} = \frac{x^2 + 4x + 2a - 1}{(x+2)^2}\). Setting the numerator to zero gives \(x^2 + 4x + 2a - 1 = 0\). The discriminant is \(16 - 4(2a-1) = 20 - 8a\), which is less than zero for \(a > \frac{5}{2}\), so no stationary points exist.

(c) The curve intersects the y-axis at \((0, \frac{1}{2})\). The asymptotes are \(x = -2\) and \(y = x + a - 2\).

(d)(i) Sketch the curve \(y = \left| \frac{x^2 + ax + 1}{x + 2} \right|\) with the same asymptotes and a reflection in the x-axis where the original curve is negative.

(d)(ii) Draw the line \(y = a\) on the sketch.

(d)(iii) Solve \(\frac{x^2 + ax + 1}{x + 2} = a\) and \(\frac{x^2 + ax + 1}{x + 2} = -a\). This gives the quadratic \(x^2 + 1 - 2a = 0\) or \(x^2 + 2ax + 1 = 0\). The roots are \(-a - \sqrt{a^2 - 2a - 1} < x < -a + \sqrt{a^2 - 2a - 1}\). Solving for \(a\) gives \(a = 5\).

(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.

(a) To find the equation of the plane ABC, we first find the direction vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\):

\(\overrightarrow{AB} = (2\mathbf{i} + 4\mathbf{j} - \mathbf{k}) - (2\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}) = 2\mathbf{j} - 5\mathbf{k}\)

\(\overrightarrow{AC} = (-3\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}) - (2\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}) = -5\mathbf{i} - 5\mathbf{j}\)

The normal vector \(\mathbf{n}\) to the plane is given by the cross product \(\overrightarrow{AB} \times \overrightarrow{AC}\):

\(\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 2 & -5 \\ -5 & -5 & 0 \end{vmatrix} = (-5(2) - 5(-5))\mathbf{i} + (0(-5) - (-5)(-5))\mathbf{j} + (0(-5) - (-5)(2))\mathbf{k}\)

\(= -10\mathbf{i} + 25\mathbf{j} + 10\mathbf{k}\)

\(= -5\mathbf{i} + 5\mathbf{j} + 2\mathbf{k}\)

Using point \(A(2, 2, 4)\), the equation of the plane is:

\(-5(x - 2) + 5(y - 2) + 2(z - 4) = 0\)

\(-5x + 5y + 2z = 8\)

(b) The perpendicular distance from point \(D(2, 1, 3)\) to the plane is given by:

\(\frac{|(-5)(2) + 5(1) + 2(3) - 8|}{\sqrt{(-5)^2 + 5^2 + 2^2}} = \frac{7}{\sqrt{54}} \approx 0.953\)

(c) The shortest distance between the lines AB and CD is found using the vector \(\overrightarrow{CD}\) and the cross product:

\(\overrightarrow{CD} = (2\mathbf{i} + \mathbf{j} + 3\mathbf{k}) - (2\mathbf{i} + 4\mathbf{j} - \mathbf{k}) = -3\mathbf{j} + 4\mathbf{k}\)

\(\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 2 & -5 \\ 5 & 4 & -1 \end{vmatrix} = (18)\mathbf{i} + (-25)\mathbf{j} + (-10)\mathbf{k}\)

The shortest distance is:

\(\frac{|\overrightarrow{AB} \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{35}{\sqrt{1049}} \approx 1.08\)

(a) The vertical asymptote occurs where the denominator is zero: \(x + 2 = 0 \Rightarrow x = -2\).

The horizontal asymptote is found by dividing the leading coefficients: \(y = \frac{x^2 + ax + 1}{x + 2} = x + a - 2\).

(b) Differentiate \(y = \frac{x^2 + ax + 1}{x + 2}\) using the quotient rule:

\(\frac{dy}{dx} = \frac{(x+2)(2x+a) - (x^2+ax+1)}{(x+2)^2} = \frac{x^2 + 4x + 2a - 1}{(x+2)^2}\).

Set \(\frac{dy}{dx} = 0\):

\(x^2 + 4x + 2a - 1 = 0\).

The discriminant \(16 - 4(2a-1) = 20 - 8a < 0\) implies no real roots, hence no stationary points.

(c) The curve intersects the \(y\)-axis at \(x = 0\):

\(y = \frac{0^2 + a(0) + 1}{0 + 2} = \frac{1}{2}\).

Asymptotes: vertical at \(x = -2\), oblique at \(y = x + a - 2\).

(d)(i) Sketch \(y = \left| \frac{x^2 + ax + 1}{x + 2} \right|\) with the same asymptotes.

(ii) Draw the line \(y = a\) on the sketch.

(iii) Solve \(\left| \frac{x^2 + ax + 1}{x + 2} \right| < a\) for given intervals:

\(\frac{x^2 + ax + 1}{x + 2} = a\) or \(\frac{x^2 + ax + 1}{x + 2} = -a\).

\(x^2 + 1 - 2a = 0\) or \(x^2 + 2ax + 1 = 0\).

\(-a - \sqrt{a^2 - 2a - 1} < x < -\sqrt{2a - 1}\) and \(-a + \sqrt{a^2 - 2a - 1} < x < \sqrt{2a - 1}\).

Given intervals imply \(a = 5\).

(a) The polar coordinates of the point of C 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:

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

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

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

(c) Let \(y = r \sin \theta\), then:

\(\frac{dy}{d\theta} = \frac{d}{d\theta} \left( \sqrt{(\pi - \theta) \arctan(\pi - \theta)} \sin \theta \right)\)

\(= \frac{(\pi - \theta) \arctan(\pi - \theta) \cos \theta}{\sqrt{(\pi - \theta) \arctan(\pi - \theta)}} + \arctan(\pi - \theta) \sin \theta\)

Setting \(\frac{dy}{d\theta} = 0\) gives:

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

For \(\theta = 1.2\),

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

For \(\theta = 1.3\),

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

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

(a) To show that A is non-singular, we need to show that its determinant is non-zero for all real \(k\).

The determinant of \(A\) is calculated as follows:

\(\det(A) = k \begin{vmatrix} 5 & 2 \\ 3 & -k \end{vmatrix} - 1 \begin{vmatrix} 6 & 2 \\ -1 & -k \end{vmatrix} + 0 \begin{vmatrix} 6 & 5 \\ -1 & 3 \end{vmatrix}\)

\(= k(5(-k) - 6) - (6(-k) + 2)\)

\(= k(-5k - 6) - (-6k + 2)\)

\(= -5k^2 - 6k + 6k - 2\)

\(= -5k^2 - 2\)

Since \(-5k^2 - 2\) is always negative for all real \(k\), the determinant is never zero. Therefore, A is non-singular.

(b) Given \(A^{-1} = \begin{pmatrix} 3 & 0 & -1 \\ 1 & 0 & 0 \\ -\frac{23}{2} & \frac{1}{2} & 3 \end{pmatrix}\), we use the property \(AA^{-1} = I\).

From the first row of \(A\) and the first column of \(A^{-1}\), we have:

\(k \cdot 3 + 1 \cdot 1 + 0 \cdot -\frac{23}{2} = 1\)

\(3k + 1 = 1\)

\(3k = 0\)

\(k = 0\)

(a) Let \(y = \alpha^2 + 1\). Then \(\alpha^2 = y - 1\). Substitute into the original equation:

\((y - 1)^3 + 2(y - 1)^2 + 3(y - 1) + 1 = 0\)

Expanding and simplifying gives:

\(y^3 - y^2 + 4y - 5 = 0\)

(b) Using the result from part (a), we have:

\((\alpha^2 + 1)^2 + (\beta^2 + 1)^2 + (\gamma^2 + 1)^2 = 1 - 2(4) = -7\)

(c) Again using the result from part (a):

\((\alpha^2 + 1)^3 + (\beta^2 + 1)^3 + (\gamma^2 + 1)^3 = -7 - 4 + 5(3) = 4\)

(a) The matrix \(\mathbf{M} = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 7 & 0 \\ 0 & 1 \end{pmatrix}\) represents a stretch followed by a shear. The first matrix \(\begin{pmatrix} 7 & 0 \\ 0 & 1 \end{pmatrix}\) is a stretch parallel to the \(x\)-axis with a scale factor of 7. The second matrix \(\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\) is a shear with the \(x\)-axis fixed, mapping \((0,1)\) to \((2,1)\).

(b) To find the invariant lines, consider \(\mathbf{M} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 7x + 2y \\ y \end{pmatrix}\). For a line \(y = mx\) to be invariant, \(mx = m(7x + 2mx)\). This simplifies to \(2m^2 + 6m = 0\), giving solutions \(m = 0\) and \(m = -3\). Thus, the invariant lines are \(y = 0\) and \(y = -3x\).

(c) The area of \(PQR\) is \(35 \text{ cm}^2\). The area of \(DEF\) is related by the determinant of the transformation matrix \(\mathbf{M}\), which is \(|7| = 7\). Therefore, \(\text{Area of } DEF = \frac{35}{7} = 5 \text{ cm}^2\).

(a) Base case: For \(n = 1\), \(\sum_{r=1}^{1} r^2 = 1^2 = 1\) and \(\frac{1}{6} \times 1 \times 2 \times 3 = 1\). So \(H_1\) is true.

Inductive step: Assume \(\sum_{r=1}^{k} r^2 = \frac{1}{6}k(k+1)(2k+1)\) is true for \(n = k\).

Consider \(\sum_{r=1}^{k+1} r^2 = \sum_{r=1}^{k} r^2 + (k+1)^2\).

\(= \frac{1}{6}k(k+1)(2k+1) + (k+1)^2\)

\(= \frac{1}{6}(k+1)(2k^2 + k + 6k + 6)\)

\(= \frac{1}{6}(k+1)(2k^2 + 7k + 6)\)

\(= \frac{1}{6}(k+1)(k+2)(2k+3)\)

So \(H_{k+1}\) is true. By induction, \(H_n\) is true for all positive integers \(n\).

(b) Using the identity, expand and simplify:

\((2n+1)^5 - 1\)

\(= 160S_n + \frac{40}{3}n(n+1)(2n+1) + 2n\)

\(160S_n = (2n+1)^5 - \frac{40}{3}n(n+1)(2n+1) - (2n+1)\)

\(160S_n = (2n+1)((2n+1)^4 - \frac{40}{3}n(n+1) - 1)\)

\(\Rightarrow S_n = \frac{1}{30}n(n+1)(2n+1)(3n^2 + 3n - 1)\)

(c) \(n^{-5}S_n = \frac{1}{30}(1 + \frac{1}{n})(2 + \frac{1}{n})(3 + \frac{3}{n} - \frac{1}{n^2})\)

As \(n \to \infty\), \(n^{-5}S_n \to \frac{1}{5}\).

(a) The direction vectors of \(l_1\) and \(l_2\) are \(\begin{pmatrix} 0 \\ 1 \\ -2 \end{pmatrix}\) and \(\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}\) respectively. The vector between points on \(l_1\) and \(l_2\) is \(\begin{pmatrix} -4 \\ 0 \\ 1 \end{pmatrix}\). The cross product of the direction vectors is \(\begin{pmatrix} -2 \\ -2 \\ -1 \end{pmatrix}\). The shortest distance is given by \(\frac{1}{\sqrt{30}} \begin{vmatrix} -4 & 0 & 1 \\ -2 & -2 & -1 \end{vmatrix} = \frac{21}{\sqrt{30}}\).

(b) The normal to \(\Pi_1\) is the cross product of the direction vector of \(l_1\) and the direction vector of \(l_2\), which is \(\begin{pmatrix} -2 \\ -2 \\ -1 \end{pmatrix}\). Using the point \((1, 4, -1)\) on \(l_1\), the equation of \(\Pi_1\) is \(5x - 2y - z = -2\).

(c) The direction vector of the line of intersection is \(\begin{pmatrix} 1 \\ 4 \\ -3 \end{pmatrix}\). The normal to \(\Pi_2\) is \(\begin{pmatrix} -1 \\ -1 \\ 1 \end{pmatrix}\). Using the point \((1, 1, 1)\), the equation of \(\Pi_2\) is \(x + 2y + 3z = 6\).

(a) To find vertical asymptotes, set the denominator equal to zero: \(x^2 + 3 = 0\). This equation has no real roots, so there are no vertical asymptotes. The horizontal asymptote is found by considering the limit as \(x\) approaches infinity: \(y = \frac{x+1}{x^2+3} \to 0\). Thus, the horizontal asymptote is \(y = 0\).

(b) To find stationary points, differentiate \(y\) with respect to \(x\): \(\frac{dy}{dx} = \frac{(x^2+3) - (x+1)(2x)}{(x^2+3)^2}\). Set \(\frac{dy}{dx} = 0\) to find \(x^2 + 2x - 3 = 0\). Solving this quadratic equation gives \(x = -3\) and \(x = 1\). Substituting back into the original equation gives the stationary points \((-3, -\frac{1}{6})\) and \((1, \frac{1}{2})\).

(c) The intersections with the axes occur when \(y = 0\) and \(x = 0\). For \(y = 0\), solve \(x+1 = 0\) to get \(x = -1\), so the intersection is \((-1, 0)\). For \(x = 0\), \(y = \frac{1}{3}\), so the intersection is \((0, \frac{1}{3})\).

(d) For \(y^2 = \frac{x+1}{x^2+3}\), the intersections with the axes are the same as in part (c), but \(y\) can be positive or negative. Thus, the intersections are \((-1, 0)\) and \((0, \pm \frac{1}{3})\). The stationary points are \((1, \pm \frac{1}{2})\).

(a) The sketch of the curve shows two loops symmetric about the line \(\theta = \frac{1}{2} \pi\).

(b) Start with \(r^2 = \sin 2\theta \cos \theta\). Use \(\sin 2\theta = 2 \sin \theta \cos \theta\) and polar to Cartesian conversions: \(x = r \cos \theta\), \(y = r \sin \theta\). Thus, \(r^5 = 2x^2\). Therefore, \(\left( x^2 + y^2 \right)^{\frac{3}{2}} = 2x^2 y\).

(c) The total area enclosed by the curve is given by \(\frac{1}{2} \int_0^\pi \sin(2\theta) \cos \theta \, d\theta\). Simplifying, \(\int \sin \theta \cos^2 \theta \, d\theta = -\frac{1}{3} \cos^3 \theta\). Evaluating from 0 to \(\pi\), the area is \(\frac{2}{3}\).

(d) Differentiate \(r^2 = \sin 2\theta \cos \theta\) with respect to \(\theta\) to find the maximum distance. Solve \(6 \cos^3 \theta - 4 \cos \theta = 0\) to find \(\cos^2 \theta = \frac{2}{3}\). The greatest distance is \(r = \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{9} \approx 0.877\).

(a) Consider \((r+1)^2 - r^2 = r^2 + 2r + 1 - r^2 = 2r + 1\).

Using the method of differences, sum both sides:

\(2 \sum_{r=1}^{n} r + n = (n+1)^2 - 1^2\)

\(\Rightarrow 2 \sum_{r=1}^{n} r = n^2 + n(n+1)\)

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

(b) \(\sum_{r=1}^{n} (r+a) = \sum_{r=1}^{n} r + an\)

\(\frac{1}{2} n(n+1) + an = n\)

Solving for \(a\):

\(a = \frac{1}{2}(1-n)\)

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

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

So \(H_1\) is true.

Inductive step: Assume \(\sum_{r=1}^{k} rx^{r-1} = \frac{1 - (k+1)x^k + kx^{k+1}}{(1-x)^2}\).

Consider \(\sum_{r=1}^{k+1} rx^{r-1} = \frac{1 - (k+1)x^k + kx^{k+1}}{(1-x)^2} + (k+1)x^k\).

\(= \frac{1 - (k+1)x^k + kx^{k+1} + (k+1)x^k(1-2x+x^2)}{(1-x)^2}\)

\(= \frac{1 + kx^{k+1} + (k+1)x^k(-2x+x^2)}{(1-x)^2}\)

\(= \frac{1 - (k+2)x^{k+1} + (k+1)x^{k+2}}{(1-x)^2}.\)

So \(H_{k+1}\) is true. By induction, \(H_n\) is true for all positive integers \(n\).

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

\(b = - (\alpha + \beta + \gamma + \delta) = -3\).

For \(c\), use the formula for the sum of squares:

\(5 = (-3)^2 - 2(\alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta)\).

Solving gives \(c = 2\).

For \(d\), use the given inverse sum:

\(6 = \frac{\alpha\beta\gamma + \beta\gamma\delta + \gamma\delta\alpha + \delta\alpha\beta}{\alpha\beta\gamma\delta} = \frac{-d}{-2}\).

Solving gives \(d = 12\).

The equation is \(x^4 - 3x^3 + 2x^2 + 12x - 2 = 0\).

(b) Using the derived equation and given \(\alpha^3 + \beta^3 + \gamma^3 + \delta^3 = -27\):

\(\alpha^4 + \beta^4 + \gamma^4 + \delta^4 = 3(-27) - 2(5) - 12(3) + 2(4)\).

Calculating gives \(-119\).

(a) To find the shortest distance between \(l_1\) and \(l_2\), we first find the direction vector from one line to another: \(\begin{pmatrix} 2 \\ -2 \\ 3 \end{pmatrix} - \begin{pmatrix} -2 \\ -3 \\ -5 \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \\ 8 \end{pmatrix}\).

Next, find the common perpendicular by taking the cross product of the direction vectors of \(l_1\) and \(l_2\):

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -4 & 3 & 5 \\ 2 & -3 & 1 \end{vmatrix} = \begin{pmatrix} 18 \\ 14 \\ 6 \end{pmatrix} \sim \begin{pmatrix} 9 \\ 7 \\ 3 \end{pmatrix}\).

Use the formula for shortest distance:

\(\frac{1}{\sqrt{139}} \left| \begin{pmatrix} 4 \\ 1 \\ 8 \end{pmatrix} \cdot \begin{pmatrix} 9 \\ 7 \\ 3 \end{pmatrix} \right| = \frac{67}{\sqrt{139}} \approx 5.68\).

(b) To find the equation of the plane \(\Pi\), find a vector perpendicular to the plane using the cross product of the direction vector of \(l_1\) and the vector from the point \(-\mathbf{i} - 3\mathbf{j} - 4\mathbf{k}\) to a point on \(l_1\):

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 1 \\ -4 & 3 & 5 \end{vmatrix} = \begin{pmatrix} 3 \\ -3 \\ -1 \end{pmatrix} \sim \begin{pmatrix} 9 \\ -3 \\ -1 \end{pmatrix}\).

Use the point \(-\mathbf{i} - 3\mathbf{j} - 4\mathbf{k}\) in the plane equation:

\(1(-1) + 3(-3) - 1(-4) = -6 \Rightarrow x + 3y - z = -6\).

(a) Since \(\mathbf{A}\) is singular, its determinant is zero. Calculate the determinant:

\(\det(\mathbf{A}) = 1(1 \cdot 5 - 3 \cdot 2) - k(2 \cdot 5 - 3 \cdot 3) + 3(2 \cdot 2 - 1 \cdot 3) = 0\)

Solving gives \(k = 2\).

Calculate \(\mathbf{CAB}\):

\(\mathbf{C} \cdot \mathbf{A} = \begin{pmatrix} -2 & -1 & 1 \\ 1 & 1 & 3 \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \\ 3 & 2 & 5 \end{pmatrix} = \begin{pmatrix} -2 & -4 & -4 \\ 12 & 9 & 18 \end{pmatrix}\)

\(\mathbf{CAB} = \begin{pmatrix} -2 & -4 & -4 \\ 12 & 9 & 18 \end{pmatrix} \begin{pmatrix} 0 & -2 \\ -1 & 3 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 3 & -7 \\ -9 & 3 \end{pmatrix}\)

(b) For invariant lines, solve \(\begin{pmatrix} 3 & -7 \\ -9 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = m \begin{pmatrix} x \\ y \end{pmatrix}\).

\(-9x + 3mx = m(3x - 7mx)\)

\(-9 + 3m = 3m - 7m^2 \Rightarrow 7m^2 = 9\)

\(m = \pm \frac{3}{\sqrt{7}}\)

Invariant lines: \(y = \frac{3}{\sqrt{7}}x\) and \(y = -\frac{3}{\sqrt{7}}x\).

(c) Given \(\mathbf{CAB} = \mathbf{D} - 9\mathbf{EF}\), set \(\mathbf{D} = \begin{pmatrix} \alpha & 0 \\ 0 & \alpha \end{pmatrix}\), \(\mathbf{E} = \begin{pmatrix} \beta & 0 \\ 0 & 1 \end{pmatrix}\), \(\mathbf{F} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\).

\(\begin{pmatrix} 3 & -7 \\ -9 & 3 \end{pmatrix} = \begin{pmatrix} \alpha & 0 \\ 0 & \alpha \end{pmatrix} - 9 \begin{pmatrix} 0 & \beta \\ 1 & 0 \end{pmatrix}\)

\(\alpha = 3, \beta = \frac{7}{9}\)

\(\mathbf{D} = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}, \mathbf{E} = \begin{pmatrix} \frac{7}{9} & 0 \\ 0 & 1 \end{pmatrix}, \mathbf{F} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\)

(a) Start with the Cartesian equation \(\left( x - \frac{1}{2} \right)^2 + y^2 = \frac{1}{4}\). Expand and rearrange to get \(x^2 - x + \frac{1}{4} + y^2 = \frac{1}{4}\), which simplifies to \(x^2 + y^2 = x\). In polar coordinates, \(x = r \cos \theta\) and \(x^2 + y^2 = r^2\). Substitute to get \(r^2 = r \cos \theta\), which simplifies to \(r(r - \cos \theta) = 0\). Since \(r \neq 0\), we have \(r = \cos \theta\).

(b) Set \(\sin 2\theta = \cos \theta\). Using the identity \(\sin 2\theta = 2\sin \theta \cos \theta\), we have \(2\sin \theta \cos \theta = \cos \theta\). Assuming \(\cos \theta \neq 0\), divide by \(\cos \theta\) to get \(2\sin \theta = 1\), so \(\sin \theta = \frac{1}{2}\). Therefore, \(\theta = \frac{\pi}{6}\). The polar coordinates of \(P\) are \(\left( \frac{1}{2}, \frac{\sqrt{3}}{6} \pi \right)\).

(c) Sketch the curves \(C_1: r = \cos \theta\) and \(C_2: r = \sin 2\theta\), marking the intersection point \(P\).

(d) The area \(R\) is given by \(\frac{1}{2} \int_{0}^{\frac{\pi}{6}} \sin^2 2\theta \, d\theta + \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \cos^2 \theta \, d\theta\). Use the identities \(\sin^2 2\theta = \frac{1}{2}(1 - \cos 4\theta)\) and \(\cos^2 \theta = \frac{1}{2}(1 + \cos 2\theta)\) to integrate. The result is \(\frac{1}{4} (\pi - 3\sqrt{3})\).

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

(b) To find stationary points, set \(\frac{dy}{dx} = 0\). The derivative is \(\frac{dy}{dx} = \frac{(x^2 - x - 2)(2x) - (x^2 + 2)(2x - 1)}{(x^2 - x - 2)^2}\). Solving \(x^2 + 8x - 2 = 0\) gives the stationary points \((-8.2, 0.9)\) and \((0.2, -0.9)\).

(c) The sketch shows the curve with asymptotes at \(x = -1\), \(x = 2\), and \(y = 1\). The curve intersects the y-axis at \((0, -1)\).

(d) The sketch of \(y = \frac{1}{f(x)}\) is derived from the sketch in (c), showing the reciprocal relationship.

(e) Solve \(\frac{x^2 + 2}{x^2 - x - 2} = 1\) and \(\frac{x^2 + 2}{x^2 - x - 2} = -1\) to find critical points. The solution gives the intervals \(-4 < x < -1\), \(0 < x < \frac{1}{2}\), and \(x > 2\).

(a) Use the standard results for summation: \(\sum_{r=1}^{n} r = \frac{n(n+1)}{2}\) and \(\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}\).

Calculate \(\sum_{r=1}^{n} (3r^2 + 3r + 1) = 3 \sum_{r=1}^{n} r^2 + 3 \sum_{r=1}^{n} r + \sum_{r=1}^{n} 1\).

Substitute the standard results: \(3 \left( \frac{n(n+1)(2n+1)}{6} \right) + 3 \left( \frac{n(n+1)}{2} \right) + n\).

Simplify to get \(\frac{1}{2} n(n+1)(2n+1) + \frac{3}{2} n(n+1) + n = n^3 + 3n^2 + 3n\).

(b) Start with \(\frac{1}{r^3} - \frac{1}{(r+1)^3} = \frac{(r+1)^3 - r^3}{r^3 (r+1)^3}\).

Expand \((r+1)^3 - r^3 = r^3 + 3r^2 + 3r + 1 - r^3 = 3r^2 + 3r + 1\).

Thus, \(\frac{1}{r^3} - \frac{1}{(r+1)^3} = \frac{3r^2 + 3r + 1}{r^3 (r+1)^3}\).

Use the method of differences: \(\sum_{r=1}^{n} \left( \frac{1}{r^3} - \frac{1}{(r+1)^3} \right) = 1 - \frac{1}{(n+1)^3}\).

(c) As \(n \to \infty\), \(\frac{1}{(n+1)^3} \to 0\), so the sum converges to 1.

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

\(\frac{d}{dx} \left( x^2 e^x \right) = x^2 e^x + 2x e^x = (x^2 + 2x) e^x.\)

So true when \(n = 1\).

Inductive step: Assume that

\(\frac{d^k}{dx^k} \left( x^2 e^x \right) = \left( x^2 + 2kx + k(k-1) \right) e^x\)

for some value of \(k\).

Then,

\(\frac{d^{k+1}}{dx^{k+1}} \left( x^2 e^x \right) = \left( x^2 + 2kx + k(k-1) \right) e^x + e^x (2x + 2k).\)

Simplifying gives:

\(\left( x^2 + 2(k+1)x + k(k+1) \right) e^x.\)

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

(a) The matrix \(\mathbf{M}\) is the product of two matrices: \(\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}\) represents a shear parallel to the x-axis, and \(\begin{pmatrix} k & 0 \\ 0 & 1 \end{pmatrix}\) represents a stretch parallel to the x-axis by a factor of \(k\). The shear is applied first, followed by the stretch.

(b) The area of the parallelogram \(OPQR\) is given by the absolute value of the determinant of \(\mathbf{M}\). Calculating the determinant: \(\det(\mathbf{M}) = \det\left(\begin{pmatrix} k & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}\right) = k\). Thus, \(|OPQR| = |k|\).

The inverse matrix \(\mathbf{M}^{-1}\) is calculated as follows: \(\mathbf{M}^{-1} = \frac{1}{k} \begin{pmatrix} 1 & 0 \\ -1 & k \end{pmatrix}\).

(c) To show the line is invariant, consider the transformation of a point \(\begin{pmatrix} x \\ \frac{1}{k-1}x \end{pmatrix}\) under \(\mathbf{M}\):

\(\begin{pmatrix} k & 0 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ \frac{1}{k-1}x \end{pmatrix} = \begin{pmatrix} kx \\ x + \frac{1}{k-1}x \end{pmatrix} = \begin{pmatrix} kx \\ \frac{k}{k-1}x \end{pmatrix} = k \begin{pmatrix} x \\ \frac{1}{k-1}x \end{pmatrix}\).

This shows that the line \(y = \frac{1}{k-1}x\) is invariant under the transformation.

(a) Let \(y = 3x + 1\), then \(x = \frac{1}{3}(y - 1)\).

Substitute into the original equation:

\(27\left(\frac{y-1}{3}\right)^3 + 18\left(\frac{y-1}{3}\right)^2 + 6\left(\frac{y-1}{3}\right) - 1 = 0\)

\(\Rightarrow (y-1)^3 + 2(y-1)^2 + 2(y-1) - 1 = 0\)

\(\Rightarrow y^3 - 3y^2 + 3y - 1 + 2y^2 - 4y + 2 + 2y - 2 - 1 = 0\)

\(\Rightarrow y^3 - y^2 + y - 2 = 0\)

(b) \(S_2 = 1^2 - 2(1) = -1\)

\(S_3 = (3\alpha + 1)^3 + (3\beta + 1)^3 + (3\gamma + 1)^3 = -1 - (1) + 6 = 4\)

(c) \(S_{-1} = \frac{(3\alpha + 1)(3\beta + 1) + (3\beta + 1)(3\gamma + 1) + (3\gamma + 1)(3\alpha + 1)}{(3\alpha + 1)(3\beta + 1)(3\gamma + 1)} = \frac{1}{2}\)

\(2S_{-2} = S_1 + S_{-1} = 1 - 3 + \frac{1}{2}\)

\(S_{-2} = -\frac{3}{4}\)

(a) The normal vector to \(\Pi_1\) can be found using the cross product of the direction vectors \(\mathbf{i} - 2\mathbf{j} - 3\mathbf{k}\) and \(3\mathbf{i} - \mathbf{k}\). This gives \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -2 & -3 \\ 3 & 0 & -1 \end{vmatrix} = \mathbf{i}(2 - 0) - \mathbf{j}(-1 - (-9)) + \mathbf{k}(0 - (-6)) = 2\mathbf{i} - 8\mathbf{j} + 6\mathbf{k}\).

Using the point \((1, -1, -2)\), the equation is \(1(x - 1) - 4(y + 1) + 3(z + 2) = 0\), simplifying to \(x - 4y + 3z = -1\).

(b) The direction vector of \(l\) is \(\mathbf{i} + \mathbf{j} + \mathbf{k}\). The dot product with the normal vector \(1\mathbf{i} - 4\mathbf{j} + 3\mathbf{k}\) is \(1(1) - 4(1) + 3(1) = 0\), showing \(l\) is parallel to \(\Pi_1\).

(c) The distance from a point on \(l\) to \(\Pi_1\) is \(\frac{|(-3)(1) + 0(-4) + 1(3) + 1|}{\sqrt{1^2 + (-4)^2 + 3^2}} = \frac{1}{\sqrt{26}}\).

(d) A point common to both planes is found by solving \(x - 4y + 3z = -1\) and \(3x + 3y + 2z = 1\). Solving these gives a point \(\begin{pmatrix} \frac{5}{7} \\ 0 \\ -\frac{4}{7} \end{pmatrix}\). The direction vector of the line of intersection is the cross product of the normals \(\begin{pmatrix} 1 \\ -4 \\ 3 \end{pmatrix}\) and \(\begin{pmatrix} 3 \\ 3 \\ 2 \end{pmatrix}\), which is \(\begin{pmatrix} -17 \\ 7 \\ 15 \end{pmatrix}\). Thus, the line is \(\mathbf{r} = \begin{pmatrix} \frac{5}{7} \\ 0 \\ -\frac{4}{7} \end{pmatrix} + \lambda \begin{pmatrix} -17 \\ 7 \\ 15 \end{pmatrix}\).

(a) The curve is sketched with the initial line drawn. The shape is correct with \(r\) strictly decreasing. The greatest distance from the pole is \(1 - e^{-\frac{1}{2}\pi}\).

(b) The area is calculated using the integral:

\(\frac{1}{2} \int_{0}^{\frac{1}{2}\pi} \left( e^{-\theta} - e^{-\frac{1}{2}\pi} \right)^2 \, d\theta\)

Expanding and integrating:

\(\frac{1}{2} \int_{0}^{\frac{1}{2}\pi} \left( e^{-2\theta} - 2e^{-\theta - \frac{1}{2}\pi} + e^{-\pi} \right) \, d\theta\)

\(\frac{1}{2} \left[ -\frac{1}{2}e^{-2\theta} + 2e^{-\theta - \frac{1}{2}\pi} + e^{-\pi}\theta \right]_{0}^{\frac{1}{2}\pi}\)

\(\frac{1}{2} \left( -\frac{1}{2}e^{-\pi} + 2e^{-\pi} + \frac{1}{2}\pi e^{-\pi} - 2e^{-\frac{1}{2}\pi} \right) = \frac{3}{4}e^{-\pi} + \frac{1}{4}\pi e^{-\pi} - e^{-\frac{1}{2}\pi} + \frac{1}{4}\)

(c) Using \(y = (e^{-\theta} - e^{-\frac{1}{2}\pi}) \sin \theta\), differentiate:

\(\frac{dy}{d\theta} = (e^{-\theta} - e^{-\frac{1}{2}\pi}) \cos \theta + \sin \theta (-e^{-\theta}) = 0\)

\(\left[ \theta = \frac{1}{2}\pi \Rightarrow 1 + \frac{-e^{-\theta}}{e^{-\theta} - e^{-\frac{1}{2}\pi}} \right] \tan \theta = 0 \Rightarrow 1 - e^{\theta - \frac{1}{2}\pi} - \tan \theta = 0\)

Checking values:

\(1 - e^{0.56 - \frac{1}{2}\pi} - \tan 0.56 = 0.00912\)

\(1 - e^{0.57 - \frac{1}{2}\pi} - \tan 0.57 = -0.00856\)

There is a sign change, confirming a root between 0.56 and 0.57.

(a) The vertical asymptote occurs where the denominator is zero, so \(x = -1\). For the oblique asymptote, perform polynomial long division on \(\frac{x^2}{x+1}\) to get \(y = x - 1\).

(b) To find stationary points, set \(\frac{dy}{dx} = 0\). Differentiate \(y = \frac{x^2}{x+1}\) to get \(\frac{dy}{dx} = \frac{x^2 + 2x}{(x+1)^2}\). Solve \(x^2 + 2x = 0\) to find \(x = 0\) and \(x = -2\). Substitute back to find \(y\) values: \((0, 0)\) and \((-2, -4)\).

(c) Sketch the curve using the asymptotes and stationary points. The curve approaches the asymptotes and passes through the stationary points.

(d) For \(y = \frac{1}{f(x)}\), differentiate and set \(\frac{dy}{dx} = 0\) to find stationary points. The stationary point is \((-2, -\frac{1}{4})\).

(e) Sketch \(y = \frac{1}{f(x)}\) using the asymptotes and stationary points. Solve \(\frac{1}{f(x)} > f(x)\) by setting \(\frac{x^2}{x+1} = 1\) and \(\frac{x^2}{x+1} = -1\). Solve \(x^2 - x - 1 = 0\) to find critical points \(x = \frac{1}{2} \pm \frac{1}{2} \sqrt{5}\). The solution set is \(x < -1, -\frac{1}{2} \sqrt{5} < x < \frac{1}{2} \sqrt{5}, x \neq 0\).

(a) Consider \((r+1)^2 - r^2 = r^2 + 2r + 1 - r^2 = 2r + 1\).

Using the method of differences, sum both sides from \(r=1\) to \(n\):

\(2 \sum_{r=1}^{n} r + n = (n+1)^2 - 1^2.\)

This simplifies to:

\(2 \sum_{r=1}^{n} r = n^2 + n = n(n+1).\)

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

(b) Given \(\sum_{r=1}^{n} (r+a) = \sum_{r=1}^{n} r + an = n\).

Substitute \(\sum_{r=1}^{n} r = \frac{1}{2} n(n+1)\):

\(\frac{1}{2} n(n+1) + an = n.\)

Solving for \(a\):

\(an = n - \frac{1}{2} n(n+1)\)

\(a = \frac{1}{2} (1-n).\)

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

\(1 = \frac{1 - 2x + x^2}{(1-x)^2} = \frac{(1-x)^2}{(1-x)^2},\)

so \(H_1\) is true.

Inductive step: Assume that

\(\sum_{r=1}^{k} r x^{r-1} = \frac{1 - (k+1)x^k + kx^{k+1}}{(1-x)^2}.\)

Consider \(\sum_{r=1}^{k+1} r x^{r-1}\):

\(\sum_{r=1}^{k+1} r x^{r-1} = \frac{1 - (k+1)x^k + kx^{k+1}}{(1-x)^2} + (k+1)x^k.\)

Combine over a common denominator:

\(\frac{1 - (k+1)x^k + kx^{k+1} + (k+1)x^k(1-2x+x^2)}{(1-x)^2}.\)

Simplify:

\(\frac{1 + kx^{k+1} + (k+1)x^k(-2x + x^2)}{(1-x)^2} = \frac{1 - (k+2)x^{k+1} + (k+1)x^{k+2}}{(1-x)^2}.\)

Thus, \(H_{k+1}\) is true. By induction, \(H_n\) is true for all positive integers \(n\).

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

\(b = - (\alpha + \beta + \gamma + \delta) = -3\).

For \(c\), use the formula for the sum of squares:

\(5 = (-3)^2 - 2(\alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta)\).

Solving gives \(c = 2\).

For \(d\), use the given inverse sum:

\(6 = \frac{\alpha\beta\gamma + \beta\gamma\delta + \gamma\delta\alpha + \delta\alpha\beta}{\alpha\beta\gamma\delta} = \frac{-d}{-2}\).

Solving gives \(d = 12\).

The equation is \(x^4 - 3x^3 + 2x^2 + 12x - 2 = 0\).

(b) Using the derived quartic equation:

\(\alpha^4 + \beta^4 + \gamma^4 + \delta^4 = 3(-27) - 2(5) - 12(3) + 2(4)\).

Calculating gives \(-119\).

(a) To find the shortest distance between \(l_1\) and \(l_2\), we first find the direction vectors of the lines: \(\mathbf{d_1} = \begin{pmatrix} -4 \\ 3 \\ 5 \end{pmatrix}\) and \(\mathbf{d_2} = \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix}\).

The vector between a point on \(l_1\) and a point on \(l_2\) is \(\mathbf{b} = \begin{pmatrix} 4 \\ 1 \\ 8 \end{pmatrix}\).

The cross product \(\mathbf{d_1} \times \mathbf{d_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -4 & 3 & 5 \\ 2 & -3 & 1 \end{vmatrix} = \begin{pmatrix} 9 \\ 7 \\ 3 \end{pmatrix}\).

The shortest distance is given by \(\frac{|\mathbf{b} \cdot (\mathbf{d_1} \times \mathbf{d_2})|}{|\mathbf{d_1} \times \mathbf{d_2}|} = \frac{67}{\sqrt{139}} \approx 5.68\).

(b) The plane \(\Pi\) contains \(l_1\) and the point \(-\mathbf{i} - 3\mathbf{j} - 4\mathbf{k}\). The normal vector to the plane is \(\mathbf{n} = \mathbf{d_1} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 1 \\ -4 & 3 & 5 \end{vmatrix} = \begin{pmatrix} 3 \\ -1 \\ -1 \end{pmatrix}\).

The equation of the plane is \(1(-1) + 3(-3) - 1(-4) = -6 \Rightarrow x + 3y - z = -6\).

(a) Since \(\mathbf{A}\) is singular, its determinant is zero. Calculate the determinant:

\(\det(\mathbf{A}) = 1(1 \cdot 5 - 3 \cdot 2) - k(2 \cdot 5 - 3 \cdot 3) + 3(2 \cdot 2 - 1 \cdot 3) = 0.\)

This simplifies to \(-1 - k + 3 = 0 \Rightarrow k = 2.\)

Substitute \(k = 2\) into \(\mathbf{A}\) and compute \(\mathbf{CAB}\):

\(\mathbf{C} \cdot \mathbf{A} = \begin{pmatrix} -2 & -1 \\ 1 & 1 \\ 1 & 3 \end{pmatrix} \cdot \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \\ 3 & 2 & 5 \end{pmatrix} = \begin{pmatrix} -2 & -4 & -11 \\ 1 & 3 & 6 \\ 1 & 5 & 12 \end{pmatrix}.\)

Then, \(\mathbf{CAB} = \begin{pmatrix} -2 & -4 & -11 \\ 1 & 3 & 6 \end{pmatrix} \cdot \begin{pmatrix} 0 & -2 \\ -1 & 3 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 3 & -7 \\ -9 & 3 \end{pmatrix}.\)

(b) For invariant lines, solve \(\begin{pmatrix} 3 & -7 \\ -9 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = m \begin{pmatrix} x \\ y \end{pmatrix}.\)

This gives \(-9x + 3mx = m(3x - 7mx)\) and \(-9 + 3m = 3m - 7m^2 \Rightarrow 7m^2 = 9.\)

Thus, \(m = \pm \frac{3}{\sqrt{7}}\), giving lines \(y = \frac{3}{\sqrt{7}}x\) and \(y = -\frac{3}{\sqrt{7}}x\).

(c) Assume \(\mathbf{D} = \begin{pmatrix} \alpha & 0 \\ 0 & \alpha \end{pmatrix}, \mathbf{E} = \begin{pmatrix} \beta & 0 \\ 0 & 1 \end{pmatrix}, \mathbf{F} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.\)

Given \(\mathbf{CAB} = \mathbf{D} - 9\mathbf{EF},\) solve:

\(\begin{pmatrix} 3 & -7 \\ -9 & 3 \end{pmatrix} = \begin{pmatrix} \alpha & 0 \\ 0 & \alpha \end{pmatrix} - 9 \begin{pmatrix} 0 & \beta \\ 1 & 0 \end{pmatrix}.\)

Equating gives \(\alpha = 3\) and \(\beta = \frac{7}{9}.\)

Thus, \(\mathbf{D} = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}, \mathbf{E} = \begin{pmatrix} \frac{7}{9} & 0 \\ 0 & 1 \end{pmatrix}, \mathbf{F} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.\)

(a) Start with the Cartesian equation \(\left( x - \frac{1}{2} \right)^2 + y^2 = \frac{1}{4}\).

Expand and rearrange: \(x^2 - x + \frac{1}{4} + y^2 = \frac{1}{4}\) which simplifies to \(x^2 + y^2 = x\).

In polar coordinates, \(x = r \cos \theta\) and \(x^2 + y^2 = r^2\), so \(r^2 = r \cos \theta\).

Factor to get \(r(r - \cos \theta) = 0\). Since \(r \neq 0\), \(r = \cos \theta\).

(b) Set \(r = \cos \theta = \sin 2\theta\).

Use the identity \(\sin 2\theta = 2 \sin \theta \cos \theta\), so \(\cos \theta = 2 \sin \theta \cos \theta\).

Thus, \(\cos \theta (1 - 2 \sin \theta) = 0\). Since \(\cos \theta \neq 0\), \(\sin \theta = \frac{1}{2}\).

Therefore, \(\theta = \frac{\pi}{6}\) and \(r = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}\).

The polar coordinates of \(P\) are \(\left( \frac{\sqrt{3}}{2}, \frac{\pi}{6} \right)\).

(c) Sketch the curves \(C_1: r = \cos \theta\) and \(C_2: r = \sin 2\theta\) on the same diagram, marking the point \(P\).

(d) The area \(R\) is given by:

\(\frac{1}{2} \int_0^{\frac{\pi}{6}} \sin^2 2\theta \, d\theta + \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \cos^2 \theta \, d\theta\).

Use identities: \(\sin^2 2\theta = \frac{1}{2}(1 - \cos 4\theta)\) and \(\cos^2 \theta = \frac{1}{2}(1 + \cos 2\theta)\).

Integrate to find:

\(\frac{1}{4} \left[ \theta - \frac{1}{4} \sin 4\theta \right]_0^{\frac{\pi}{6}} + \frac{1}{4} \left[ \theta + \frac{1}{2} \sin 2\theta \right]_{\frac{\pi}{6}}^{\frac{\pi}{2}}\).

Calculate to get \(\frac{1}{4} \left( \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right) + \frac{1}{4} \left( \frac{\pi}{2} - \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right)\).

Simplify to \(\frac{1}{4} \left( \pi - \sqrt{3} \right)\).

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

(b) To find stationary points, set \(\frac{dy}{dx} = 0\). The derivative is \(\frac{dy}{dx} = \frac{(x^2 - x - 2)(2x) - (x^2 + 2)(2x - 1)}{(x^2 - x - 2)^2}\). Solving \(x^2 + 8x - 2 = 0\) gives the stationary points \((-8.2, 0.9)\) and \((0.2, -0.9)\).

(c) The sketch shows the curve with asymptotes and intersection at \((0, -1)\).

(d) The curve \(y = \frac{1}{f(x)}\) is sketched by inverting the values of \(f(x)\) from part (c).

(e) Solve \(\frac{x^2 + 2}{x^2 - x - 2} = 1\) and \(\frac{x^2 + 2}{x^2 - x - 2} = -1\) to find critical points. The solution is \(-4 < x < -1, \ 0 < x < \frac{1}{2}, \ x > 2\).

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

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

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

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

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

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

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

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

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

(a) Using the identity 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, \(\alpha + \beta + \gamma = -4\) and \(\alpha\beta + \beta\gamma + \gamma\alpha = 6\).

Thus, \(\alpha^2 + \beta^2 + \gamma^2 = (-4)^2 - 2(6) = 16 - 12 = 4\).

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

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

Using \(\alpha + \beta + \gamma = -4\) and \(\alpha^2 + \beta^2 + \gamma^2 = 4\), we have:

\(\sum_{r=1}^{n} (3r^2 + 2(-4)r + 4) = \sum_{r=1}^{n} (3r^2 - 8r + 4)\).

Using standard summation formulas:

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

Thus, \(3 \sum_{r=1}^{n} r^2 - 8 \sum_{r=1}^{n} r + 4n = 3 \cdot \frac{n(n+1)(2n+1)}{6} - 8 \cdot \frac{n(n+1)}{2} + 4n\).

Simplifying gives:

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

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

Thus, \(a = -\frac{5}{2}, b = \frac{1}{2}\).

(a) Start by expressing \(\frac{1}{(kr+1)(kr-k+1)}\) using partial fractions:

\(\frac{1}{(kr+1)(kr-k+1)} = \frac{1}{k} \left( \frac{1}{kr-k+1} - \frac{1}{kr+1} \right)\)

Now, apply the method of differences:

\(\sum_{r=1}^{n} \frac{1}{(kr+1)(kr-k+1)} = \frac{1}{k} \left( 1 - \frac{1}{k+1} + \frac{1}{k+1} - \frac{1}{2k+1} + \ldots + \frac{1}{k(n-1)+1} - \frac{1}{kn+1} \right)\)

This simplifies to:

\(\frac{1}{k} \left( 1 - \frac{1}{kn+1} \right)\)

(b) For the infinite series, the terms after the first cancel out, leaving:

\(\frac{1}{k}\)

(c) Use the same method of differences for the range \(r=n\) to \(r=n^2\):

\(\sum_{r=n}^{n^2} \frac{1}{(kr+1)(kr-k+1)} = \sum_{r=1}^{n^2} \frac{1}{(kr+1)(kr-k+1)} - \sum_{r=1}^{n-1} \frac{1}{(kr+1)(kr-k+1)}\)

This results in:

\(\frac{1}{k} \left( \frac{1}{k(n-1)+1} - \frac{1}{kn^2+1} \right)\)

(a) For a matrix to represent a rotation, it must be of the form \(\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}\). Comparing with \(\mathbf{M} = \begin{pmatrix} a & b^2 \\ c^2 & a \end{pmatrix}\), we equate \(b^2 = -c^2\), which is impossible for real \(b\) and \(c\) with \(b \neq 0\).

(b) Consider the transformation \(\begin{pmatrix} a & b^2 \\ c^2 & a \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} ax + b^2y \\ c^2x + ay \end{pmatrix}\). For invariant lines, \(y = mx\) and \(Y = mX\). Solving \(c^2x + amx = m(ax + b^2mx)\) gives \(c^2 + am = ma + b^2m^2 \Rightarrow c^2 = b^2m^2\). Thus, \(y = \frac{c}{b}x\) and \(y = -\frac{c}{b}x\).

(c) The enlargement matrix is \(\begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix}\) and the shear matrix is \(\begin{pmatrix} 1 & 5 \\ 0 & 1 \end{pmatrix}\). The combined transformation is \(\begin{pmatrix} 1 & 5 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} = \begin{pmatrix} 5 & 25 \\ 0 & 5 \end{pmatrix}\).

(d) The area of triangle PQR is \(12 \times \det(\mathbf{M}) = 12 \times 25 = 300 \text{ cm}^2\).

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

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

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

(a) Start by expressing \(\frac{1}{(kr+1)(kr-k+1)}\) using partial fractions:

\(\frac{1}{(kr+1)(kr-k+1)} = \frac{1}{k} \left( \frac{1}{kr-k+1} - \frac{1}{kr+1} \right)\)

Thus, the sum becomes:

\(\sum_{r=1}^{n} \frac{1}{(kr+1)(kr-k+1)} = \frac{1}{k} \left( 1 - \frac{1}{k+1} + \frac{1}{k+1} - \frac{1}{2k+1} + \ldots + \frac{1}{k(n-1)+1} - \frac{1}{kn+1} \right)\)

This simplifies to:

\(\frac{1}{k} \left( 1 - \frac{1}{kn+1} \right)\)

(b) For the infinite sum, as \(n \to \infty\), the term \(\frac{1}{kn+1} \to 0\), so:

\(\sum_{r=1}^{\infty} \frac{1}{(kr+1)(kr-k+1)} = \frac{1}{k}\)

(c) Using the result from part (a), the sum from \(n\) to \(n^2\) is:

\(\sum_{r=n}^{n^2} \frac{1}{(kr+1)(kr-k+1)} = \sum_{r=1}^{n^2} \frac{1}{(kr+1)(kr-k+1)} - \sum_{r=1}^{n-1} \frac{1}{(kr+1)(kr-k+1)}\)

Using the formula from part (a):

\(= \frac{1}{k} \left( 1 - \frac{1}{kn^2+1} \right) - \frac{1}{k} \left( 1 - \frac{1}{k(n-1)+1} \right)\)

This simplifies to:

\(\frac{1}{k} \left( \frac{1}{k(n-1)+1} - \frac{1}{kn^2+1} \right)\)

(a) For \(\mathbf{M}\) to represent a rotation, it must be of the form \(\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}\). Comparing with \(\mathbf{M} = \begin{pmatrix} a & b^2 \\ c^2 & a \end{pmatrix}\), we have \(b^2 = -c^2\), which is impossible since \(b\) and \(c\) are real and \(b \neq 0\).

(b) The transformation of \(\begin{pmatrix} x \\ y \end{pmatrix}\) by \(\mathbf{M}\) gives \(\begin{pmatrix} ax + b^2y \\ c^2x + ay \end{pmatrix}\). For invariant lines through the origin, \(y = mx\) gives \(c^2x + amx = m(ax + b^2y)\). Simplifying, \(c^2 + am = ma + b^2m^2\) leads to \(c^2 = b^2m^2\), giving \(y = \frac{c}{b} x\) and \(y = -\frac{c}{b} x\).

(c) The enlargement matrix is \(\begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix}\) and the shear matrix is \(\begin{pmatrix} 1 & 5 \\ 0 & 1 \end{pmatrix}\). Thus, \(\mathbf{M} = \begin{pmatrix} 1 & 5 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} = \begin{pmatrix} 5 & 5^2 \\ 0 & 5 \end{pmatrix}\).

(d) The area of \(PQR\) is \(12 \times \det(\mathbf{M})\). \(\det(\mathbf{M}) = 5 \times 5 - 0 \times 25 = 25\). Therefore, the area is \(12 \times 25 = 300 \text{ cm}^2\).

(a) The sketch shows a polar graph with \(\theta = 0\) at the pole and the curve extending to \(\theta = \pi\). The point furthest from the pole is \((1, 0)\).

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

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

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

\(= \frac{1}{2} \arctan \pi = 0.631\)

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

Differentiate and set \(\frac{dy}{d\theta} = 0\):

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

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

Check for 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) The vertical asymptote occurs where the denominator is zero: \(x = 2\). For the oblique asymptote, divide the numerator by the denominator: \(x^2 + 2x - 15 = (x - 2)(x + 4) - 7\), giving \(y = x + 4\).

(b) Differentiate \(y\) to find stationary points: \(\frac{dy}{dx} = \frac{(x - 2)(2x + 2) - (x^2 + 2x - 15)}{(x - 2)^2} = \frac{x^2 - 4x + 11}{(x - 2)^2}\). The quadratic \(x^2 - 4x + 11 = 0\) has no real roots (discriminant \(4^2 - 4 \times 1 \times 11 = -28 < 0\)), so no stationary points.

(c) The curve intersects the y-axis at \(x = 0\): \(y = \frac{0^2 + 2 \times 0 - 15}{0 - 2} = 7.5\). Intersects the x-axis where \(y = 0\): \(x^2 + 2x - 15 = 0\), giving roots \(x = -5\) and \(x = 3\).

(d) Sketch the absolute value function based on the sketch from (c), reflecting the negative parts above the x-axis.

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

(a) The direction vectors of \(\Pi_1\) are \(\begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}\) and \(\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}\). The normal vector is the cross product:

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

Substitute \(\mathbf{r} = \begin{pmatrix} 0 \\ -4 \\ -3 \end{pmatrix}\) into \(\mathbf{n} \cdot \mathbf{r} = d\):

\(0 \cdot 0 + 2(-4) + 2(-3) = d\)

\(d = -8 - 6 = -14\)

Equation: \(0x + 2y + 2z = -14\) simplifies to \(y + z = -7\).

(b) A point on both planes is \(\begin{pmatrix} -7 \\ -2 \\ -5 \end{pmatrix}\). The direction vector is the cross product of normals:

\(\mathbf{n_1} = \begin{pmatrix} 0 \\ 2 \\ 2 \end{pmatrix}, \mathbf{n_2} = \begin{pmatrix} -5 \\ 3 \\ 5 \end{pmatrix}\)

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

Vector equation: \(\mathbf{r} = \begin{pmatrix} -7 \\ -2 \\ -5 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -5 \\ 5 \end{pmatrix}\).

(c) Distance from \(A\) to \(\Pi_1\):

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

\(|2a| = 2 \Rightarrow a = 1\)

(d) Angle between \(l\) and \(\Pi_1\):

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

\(2b = \sqrt{(1-b)^2 + 2b^2} \Rightarrow b^2 + 2b - 1 = 0\)

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

Base case: For \(n = 1\), \(5^{3 \times 1} + 32^1 - 33 = 5^3 + 32 - 33 = 124\), which is divisible by 31.

Inductive step: Assume \(5^{3k} + 32^k - 33\) is divisible by 31 for some positive integer \(k\).

Consider \(5^{3(k+1)} + 32^{k+1} - 33\):

\(5^{3(k+1)} + 32^{k+1} - 33 = 5^{3k+3} + 32^{k+1} - 33\)

\(= (124 + 1)5^{3k} + (31 + 1)32^k - 33\)

\(= 124 \times 5^{3k} + 31 \times 32^k + 5^{3k} + 32^k - 33\)

Since \(124 \times 5^{3k} + 31 \times 32^k\) is divisible by 31, \(5^{3(k+1)} + 32^{k+1} - 33\) is divisible by 31.

Thus, by induction, the statement is true for every positive integer \(n\).

(a) Use the standard summation formulas:

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

\(\sum_{r=1}^{n} r = \frac{1}{2}n(n+1)\)

Substitute into the expression:

\(6 \left( \frac{1}{6}n(n+1)(2n+1) \right) + 6 \left( \frac{1}{2}n(n+1) \right) - 5n\)

Simplify to get \(2n^3 + 6n^2 - n\).

(b) Divide by the denominator:

\(\frac{6r^2 + 6r - 5}{r^2 + r} = 6 - \frac{5}{r(r+1)}\)

Find partial fractions:

\(\frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1}\)

Sum the series:

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

Result: \(6n - 5 + \frac{5}{n+1}\).

(c) Use the result from (b):

\(\sum_{r=n+1}^{2n} \frac{6r^2 + 6r - 5}{r^2 + r} = \sum_{r=1}^{2n} \frac{6r^2 + 6r - 5}{r^2 + r} - \sum_{r=1}^{n} \frac{6r^2 + 6r - 5}{r^2 + r}\)

\(= 12n - 5 + \frac{5}{2n+1} - (6n - 5 + \frac{5}{n+1})\)

Simplify to get \(6n - \frac{5n}{(n+1)(2n+1)}\).

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

\(y^2 - y + 2y^{\frac{1}{2}} + 5 = 0\).

Rearrange and square both sides to eliminate the square root:

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

Expand and simplify to obtain:

\(y^4 - 2y^3 + 11y^2 - 14y + 25 = 0\).

The value of \(\alpha^2 + \beta^2 + \gamma^2 + \delta^2\) is 2.

(b) Using the relation \(\alpha^2 \beta^2 \gamma^2 \delta^2 = 25\), we have:

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

Substitute the known values to get \(\frac{14}{25}\).

(c) Using the identity:

\(\alpha^4 + \beta^4 + \gamma^4 + \delta^4 = (\alpha^2 + \beta^2 + \gamma^2 + \delta^2)^2 - 2(\alpha^2 \beta^2 + \alpha^2 \gamma^2 + \alpha^2 \delta^2 + \beta^2 \gamma^2 + \beta^2 \delta^2 + \gamma^2 \delta^2)\).

Substitute the known values to get \(2^2 - 2(11) = -18\).

(a) The matrix \(\begin{pmatrix} \cos 2\theta & -\sin 2\theta \\ \sin 2\theta & \cos 2\theta \end{pmatrix}\) represents a rotation about the origin through angle \(2\theta\). The transformation sequence is a shear in the x-direction followed by this rotation.

(b) The inverse of \(\mathbf{M}\) is given by \(\mathbf{M}^{-1} = \begin{pmatrix} 1 & -k \\ 0 & 1 \end{pmatrix} \begin{pmatrix} \cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{pmatrix}\).

(c) To find when \(\mathbf{M} - \mathbf{I}\) is singular, calculate the determinant: \(\det(\mathbf{M} - \mathbf{I}) = (\cos 2\theta - 1)(k \sin 2\theta + \cos 2\theta - 1) - k \sin 2\theta \cos 2\theta + \sin^2 2\theta = 0\). Simplifying gives \(2 - 2\cos 2\theta - k \sin 2\theta = 0\), leading to \(4\sin^2 \theta = 2k \sin \theta \cos \theta\). Using double angle formulas, \(\tan \theta = \frac{1}{2}k\).

(d) Given \(k = 2\sqrt{3}\) and \(\theta = \frac{1}{3}\pi\), the matrix \(\mathbf{M}\) becomes \(\begin{pmatrix} -\frac{1}{2} & -\frac{2\sqrt{3}}{3} \\ \frac{\sqrt{3}}{2} & -\frac{1}{2} \end{pmatrix}\). Transforming \(\begin{pmatrix} x \\ y \end{pmatrix}\) to \(\begin{pmatrix} x \\ y \end{pmatrix}\) gives \(\begin{pmatrix} -\frac{1}{2}x - \frac{2\sqrt{3}}{3}y \\ \frac{\sqrt{3}}{2}x - \frac{1}{2}y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}\). Solving gives \(\sqrt{3}x + 3y = 0\).

(a) Start with the Cartesian equation \(x^2 - y^2 = a\).

In polar coordinates, \(x = r \cos \theta\) and \(y = r \sin \theta\).

Substitute these into the equation: \((r \cos \theta)^2 - (r \sin \theta)^2 = a\).

This simplifies to \(r^2 (\cos^2 \theta - \sin^2 \theta) = a\).

Using the double angle identity, \(\cos 2\theta = \cos^2 \theta - \sin^2 \theta\), we have \(r^2 \cos 2\theta = a\).

Thus, \(r^2 = a \sec 2\theta\).

(b) The sketch shows the curve starting at \(\theta = 0\) and increasing, with the minimum distance from the pole being \(\sqrt{a}\).

(a) The direction vectors of the lines in the plane are:

\(\overrightarrow{AB} = \begin{pmatrix} -2 \\ 1 \\ 4 \end{pmatrix}, \ \overrightarrow{AC} = \begin{pmatrix} -3 \\ 0 \\ 2 \end{pmatrix}\)

The normal vector to the plane is found using the cross product:

\(\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 1 & 4 \\ -3 & 0 & 2 \end{vmatrix} = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}\)

The equation of the plane is \(x - 2y + z = d\). Substituting point \(A(1, 1, 0)\) gives \(d = -1\).

(b) The perpendicular distance from \(O\) to the plane is:

\(\frac{1}{\sqrt{1^2 + (-2)^2 + 1^2}} = \frac{1}{\sqrt{6}} \approx 0.408\)

(c) Let \(\overrightarrow{OP} = \begin{pmatrix} -2\lambda \\ \lambda \\ 3\lambda \end{pmatrix}\) and \(\overrightarrow{OQ} = \begin{pmatrix} 1 - 2\mu \\ 1 + \mu \\ 4\mu \end{pmatrix}\).

\(\overrightarrow{PQ} = \begin{pmatrix} 1 + \mu - \lambda \\ 4\mu - 3\lambda \end{pmatrix}\).

Using the dot product with the line direction \(\begin{pmatrix} -2 \\ 1 \\ 3 \end{pmatrix}\):

\((1 - 2\mu + 2\lambda) \cdot (-2) + (1 + \mu - \lambda) \cdot 1 + (4\mu - 3\lambda) \cdot 3 = 0\)

Solving gives \(\lambda = \frac{4}{5}\).

The vector equation is \(\mathbf{r} = \frac{-4}{5} \begin{pmatrix} -2 \\ 1 \\ 3 \end{pmatrix} + k \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix}\).

(a) The vertical asymptote occurs where the denominator is zero, so \(x = 3\). For the oblique asymptote, perform polynomial long division on \(\frac{x^2 + 2x + 1}{x - 3}\):

\(y = \frac{(x-3)(x+5) + 16}{x-3} = x + 5 + \frac{16}{x-3}\)

Thus, the oblique asymptote is \(y = x + 5\).

(b) Differentiate \(y = \frac{x^2 + 2x + 1}{x - 3}\) to find turning points:

\(\frac{dy}{dx} = 1 - \frac{16}{(x-3)^2}\)

Set \(\frac{dy}{dx} = 0\):

\(1 - \frac{16}{(x-3)^2} = 0 \Rightarrow (x-3)^2 = 16\)

Solving gives \(x = -1\) and \(x = 7\). Substitute back to find \(y\)-coordinates: \((-1, 0)\) and \((7, 16)\).

(c) Sketch the curve using the asymptotes and turning points found.

(d) For \(y = \left| \frac{x^2 + 2x + 1}{x - 3} \right|\), reflect the negative parts of the curve above the x-axis. For \(y^2 = \frac{x^2 + 2x + 1}{x - 3}\), consider both positive and negative square roots, sketching both branches.