(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 \(n \geq 1\).