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

\(\frac{1}{(ar+1)(ar+a+1)} = \frac{1}{a} \left( \frac{1}{ar+1} - \frac{1}{ar+a+1} \right)\)

Now, sum from \(r=1\) to \(n\):

\(\sum_{r=1}^{n} \frac{1}{(ar+1)(ar+a+1)} = \frac{1}{a} \left( \frac{1}{a+1} + \frac{1}{2a+1} + \frac{1}{3a+1} + \ldots + \frac{1}{an+a+1} \right)\)

Using the method of differences, this simplifies to:

\(\frac{1}{a} \left( \frac{1}{a+1} - \frac{1}{an+a+1} \right)\)

(b) For the infinite series:

\(\sum_{r=1}^{\infty} \frac{1}{(ar+1)(ar+a+1)} = \frac{1}{a} \left( \frac{1}{a+1} \right) = \frac{1}{6}\)

This leads to:

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

Solving \(a^2 + a - 6 = 0\) gives \(a = 2\).

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

\(\overrightarrow{AB} = (-4 - 4)\mathbf{i} + (3 + 4)\mathbf{j} + (-4 - 1)\mathbf{k} = -8\mathbf{i} + 7\mathbf{j} - 5\mathbf{k}\)

\(\overrightarrow{AC} = (4 - 4)\mathbf{i} + (-1 + 4)\mathbf{j} + (-2 - 1)\mathbf{k} = 0\mathbf{i} + 3\mathbf{j} - 3\mathbf{k}\)

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

\(\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -8 & 7 & -5 \\ 0 & 3 & -3 \end{vmatrix} = \mathbf{i}(7(-3) - (-5)(3)) - \mathbf{j}(-8(-3) - (-5)(0)) + \mathbf{k}(-8(3) - 7(0))\)

\(= \mathbf{i}(-21 + 15) - \mathbf{j}(24) + \mathbf{k}(-24) = -6\mathbf{i} - 24\mathbf{j} - 24\mathbf{k}\)

Normalizing gives \(\mathbf{n} = \mathbf{i} + 4\mathbf{j} + 4\mathbf{k}\).

Using point \(A(4, -4, 1)\), the equation is:

\(x + 4y + 4z = -8\).

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

\(\frac{|0 + 4(0) + 4(0) + 8|}{\sqrt{1^2 + 4^2 + 4^2}} = \frac{8}{\sqrt{33}} \approx 1.39\).

(c) The line \(OD\) has the equation \(\mathbf{r} = t(2\mathbf{i} + 3\mathbf{j} - 3\mathbf{k})\).

Substitute into the plane equation:

\(2t + 12t - 12t = -8\)

\(2t = -8\)

\(t = -4\)

Coordinates of intersection: \((-8, -12, 12)\).

(a) Base case: \(u_1 > 4\) (given).

Inductive step: Assume \(u_k > 4\) for some positive integer \(k\).

Then \(u_{k+1} - 4 = \frac{u_k^2 + u_k + 12}{2u_k} - 4 = \frac{u_k^2 - 7u_k + 12}{2u_k}\).

\(u_{k+1} - 4 = \frac{(u_k - 3)(u_k - 4)}{2u_k} > 0\).

Hence, by induction, \(u_n > 4\) for all positive integers \(n\).

(b) \(u_{n+1} - u_n = \frac{-u_n^2 + u_n + 12}{2u_n} = \frac{-(u_n - 4)(u_n + 3)}{2u_n}\).

So \(u_n > 4 \Rightarrow u_{n+1} - u_n < 0\).

(a) Let \(y = x^{-3}\), then \(x = y^{-\frac{1}{3}}\). Substitute into the original equation:

\(2y^{-1} + 5y^{-\frac{2}{3}} - 6 = 0\)

Multiply through by \(y\) to eliminate the fractions:

\(5y^{-\frac{2}{3}} = 6 - 2y^{-1}\)

\(5y^{-\frac{2}{3}} = 6 - 2y^{-1}\)

\(125y = (6y - 2)^3\)

Expanding and simplifying gives:

\(216y^3 - 216y^2 - 53y - 8 = 0\)

(b) Using the identity \(\alpha^{-3} + \beta^{-3} + \gamma^{-3} = 1\) and \(\alpha^{-3}\beta^{-3} + \beta^{-3}\gamma^{-3} + \gamma^{-3}\alpha^{-3} = -\frac{53}{216}\), we find:

\(\alpha^{-6} + \beta^{-6} + \gamma^{-6} = (\alpha^{-3} + \beta^{-3} + \gamma^{-3})^2 - 2(\alpha^{-3}\beta^{-3} + \beta^{-3}\gamma^{-3} + \gamma^{-3}\alpha^{-3})\)

\(\alpha^{-6} + \beta^{-6} + \gamma^{-6} = 1^2 - 2\left(-\frac{53}{216}\right)\)

\(\alpha^{-6} + \beta^{-6} + \gamma^{-6} = \frac{161}{108}\)

(c) Using the equation \(216S_9 = 216S_6 + 53S_3 + 24\) and substituting the values from parts (a) and (b):

\(S_9 = \frac{399}{216} = \frac{133}{72}\)

(a) To find vertical asymptotes, set the denominator equal to zero: \(x^2 + x + 1 = 0\). The discriminant is \(b^2 - 4ac = 1^2 - 4 \times 1 \times 1 = -3\), which is less than zero, so no real roots exist. Therefore, there are no vertical asymptotes. The horizontal asymptote is found by considering the degrees of the polynomials. As \(x \to \infty\), \(y \to \frac{2x^2}{x^2} = 2\). Thus, the horizontal asymptote is \(y = 2\).

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

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

Set \(\frac{dy}{dx} = 0\) to find critical points:

\(3x^2 + 6x = 0\).

Solving gives \(x = 0\) or \(x = -2\). Substituting back into the original equation gives points \((0, -1)\) and \((-2, 3)\).

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

\(2x^2 - x - 1 = 0\).

Solving gives \(x = 1\) and \(x = -\frac{1}{2}\). The curve intersects the y-axis where \(x = 0\), giving \((0, -1)\).

(d) For \(\left| \frac{2x^2 - x - 1}{x^2 + x + 1} \right| = k\) to have 4 distinct real solutions, the graph must intersect the line \(y = k\) at 4 points. This occurs when \(0 < k < 1\).

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

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

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

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

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

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

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

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

Check sign change between 0.6 and 0.7:

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

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

Thus, a root exists between 0.6 and 0.7.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

We use the identity:

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

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

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

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

Substituting the known values:

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

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

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

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

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

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

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

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

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

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

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

(a) Consider \(\sum_{r=1}^{n} e^{rx}(e^{2x} - 2e^x + 1) = e^x - e^{2x} - e^{(n+1)x} + e^{(n+2)x}\).

Using the method of differences, the terms cancel out in a pattern, leading to the final result.

(b) For convergence, \(e^{nx} \to 0\) as \(n \to \infty\), which occurs when \(x < 0\). The sum to infinity is \(e^x - e^{2x}\).

(c) Using the laws of logarithms, \(\sum_{r=1}^{n} \ln u_r = \sum_{r=1}^{n} (rx + \ln(e^x - 1)^2)\).

Applying the formula \(\sum_{r=1}^{n} r = \frac{1}{2} n(n+1)\), we get \(\frac{1}{2} xn(n+1) + n \ln(e^x - 1)^2\).

(a) The matrix \(\mathbf{A} = \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}\) represents a shear in the \(x\)-direction.

(b) Base case: For \(n = 1\), \(\mathbf{A}^1 = \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}\), which is true.

Inductive step: Assume true for \(n = k\), so \(\mathbf{A}^k = \begin{pmatrix} 1 & ka \\ 0 & 1 \end{pmatrix}\).

Then \(\mathbf{A}^{k+1} = \mathbf{A}^k \mathbf{A} = \begin{pmatrix} 1 & ka \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & a + ak \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & (k+1)a \\ 0 & 1 \end{pmatrix}\).

Thus, by induction, \(\mathbf{A}^n = \begin{pmatrix} 1 & na \\ 0 & 1 \end{pmatrix}\) for all positive integers \(n\).

(c) \(\mathbf{A}^n \mathbf{B} = \begin{pmatrix} 1 & na \\ 0 & 1 \end{pmatrix} \begin{pmatrix} b & b \\ a^{-1} & a^{-1} \end{pmatrix} = \begin{pmatrix} b+n & b+n \\ \frac{1}{a} & \frac{1}{a} \end{pmatrix}\).

Transform \(\begin{pmatrix} x \\ y \end{pmatrix}\) by multiplying matrices to find \(\begin{pmatrix} X \\ Y \end{pmatrix}\).

\(\begin{pmatrix} b+n & b+n \\ \frac{1}{a} & \frac{1}{a} \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} (b+n)(x+y) \\ \frac{1}{a}(x+y) \end{pmatrix}\).

\(\frac{1}{a}(1+m) = m(b+n)(1+m)\)

\(y = -x\)

\(y = \frac{1}{a(b+n)}x\)

(a) Start with the polar coordinates: \(x = r \cos \theta\) and \(y = r \sin \theta\). Substitute into the Cartesian equation:

\(x^2 + xy + y^2 = r^2 \cos^2 \theta + r^2 \cos \theta \sin \theta + r^2 \sin^2 \theta = a\).

This simplifies to \(r^2 (\cos^2 \theta + \sin^2 \theta + \cos \theta \sin \theta) = a\).

Using \(\cos^2 \theta + \sin^2 \theta = 1\) and \(\sin 2\theta = 2 \sin \theta \cos \theta\), we have:

\(r^2 (1 + \frac{1}{2} \sin 2\theta) = a\).

Thus, \(r^2 = \frac{2a}{2 + \sin 2\theta}\).

(b) The sketch should show a polar graph with the curve in the correct domain \(0 \leq \theta \leq \frac{1}{4}\pi\), with \(r\) strictly decreasing.

(c) Given \(\sin 2\theta = \frac{2 \tan \theta}{1 + \tan^2 \theta}\), substitute \(t = \tan \theta\), then \(\frac{d\theta}{dt} = \frac{1}{1 + t^2}\).

The integral becomes:

\(R = \frac{1}{2} a \int_{0}^{1} \frac{1}{t^2 + t + 1} \, dt\).

Complete the square: \(t^2 + t + 1 = (t + \frac{1}{2})^2 + \frac{3}{4}\).

Integrate: \(R = \frac{1}{\sqrt{3}} [\arctan(\frac{2t + 1}{\sqrt{3}})]_{0}^{1}\).

Evaluate: \(R = \frac{\pi a}{6\sqrt{3}}\).

(a) The direction vector of line AB is \(\mathbf{AB} = 4\mathbf{i} - \mathbf{j} + \mathbf{k}\).

The direction vector of line CD is \(\mathbf{CD} = \mathbf{j} + (\lambda - 3)\mathbf{k}\).

Find the common perpendicular using the cross product and the formula for perpendicular distance:

\(\frac{1}{\sqrt{(\lambda - 2)^2 + (4\lambda - 12)^2 + 16}} \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -1 & 1 \\ 0 & 1 & \lambda - 3 \end{vmatrix} = 3\)

Solving gives \(\lambda^2 - 5\lambda + 4 = 0\).

(b)(i) The equation of \(\Pi_1\) is \(\mathbf{r} = 7\mathbf{i} + 4\mathbf{j} - \mathbf{k} + s(4\mathbf{i} - \mathbf{j} + \mathbf{k}) + t(-5\mathbf{i} + 3\mathbf{j} + 2\mathbf{k})\).

(b)(ii) The normal to \(\Pi_2\) is found using the cross product:

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -1 & 1 \\ -5 & 3 & 7 \end{vmatrix} = -8\mathbf{i} - 25\mathbf{j} + 7\mathbf{k}\)

Substitute a point to find \(-8x - 25y + 7z = -163\).

(c) The angle between \(\Pi_1\) and \(\Pi_2\) is found using the dot product of normals:

\(\cos \theta = \frac{414}{\sqrt{243 \times 738}}\)

\(\theta = 12.1^\circ\).

(a) From the roots \(\alpha, \alpha, \gamma\), we have:

\(2\alpha + \gamma = -b\)

\(\alpha^2 + 2\alpha\gamma = 0\)

Solving, \(\alpha = -2\gamma\) leading to \(-4\gamma + \gamma = -b\), so \(\gamma = \frac{1}{3}b\) and \(\alpha = -\frac{2}{3}b\).

\(\alpha^2\gamma = -d\) leading to \(\frac{4}{27}b^3 = -d\) leading to \(4b^3 + 27d = 0\).

(b) Given \(2\alpha^2 + \gamma^2 = 3b\), substitute for \(\alpha\) and \(\gamma\):

\(3b = b^2\) leading to \(b = 3\).

Substitute back to find \(d\):

\(d = -4\).

1. Base Case: For \(n = 1\), \(7^{2 \times 1} + 97^1 - 50 = 49 + 97 - 50 = 96\), which is divisible by 48.

2. Inductive Hypothesis: Assume \(7^{2k} + 97^k - 50\) is divisible by 48 for some positive integer \(k\).

3. Inductive Step: Consider \(7^{2(k+1)} + 97^{k+1} - 50\).

4. \(7^{2(k+1)} + 97^{k+1} - 50 = 7^{2k+2} + 97^{k+1} - 50 = (48+1)7^{2k} + (96+1)97^k - 50\).

5. This expression is divisible by 48 because \(48 \times 7^{2k} + 96 \times 97^k\) is divisible by 48.

6. Therefore, by induction, \(7^{2n} + 97^n - 50\) is divisible by 48 for all positive integers \(n\).

(a) Expand \((2r+1)^3 - (2r-1)^3\):

\(8r^3 + 12r^2 + 6r + 1 - (8r^3 - 12r^2 + 6r - 1) = 24r^2 + 2\).

Sum both sides and use the method of differences:

\(24 \sum_{r=1}^{n} r^2 + 2n = (2n+1)^3 - 1\).

\(24 \sum_{r=1}^{n} r^2 = 8n^3 + 12n^2 + 4n = 4n(n+1)(2n+1)\).

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

(b) Express \(S_{2n}\) in terms of sums:

\(S_{2n} = \sum_{r=1}^{2n} r^2 + 2 \sum_{r=1}^{n} (2r)^2 = \sum_{r=1}^{2n} r^2 + 8 \sum_{r=1}^{n} r^2\).

Apply \(\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n+1)(2n+1)\):

\(S_{2n} = \frac{1}{6}(2n)(2n+1)(4n+1) + \frac{8}{6}n(n+1)(2n+1)\).

\(S_{2n} = \frac{1}{3}n(2n+1)(4n+4+4) = \frac{1}{3}n(2n+1)(8n+5)\).

(c) Evaluate the limit:

\(\lim_{n \to \infty} \frac{S_{2n}}{n^3} = \frac{16}{3}\).

(a) To find the Cartesian equation of the plane \(\Pi\), we first find a normal vector to the plane by taking the cross product of the direction vectors of the lines:

Direction vectors: \(\begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix}\) and \(\begin{pmatrix} 3 \\ 2 \\ -1 \end{pmatrix}\).

Cross product: \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 2 & 1 \\ 3 & 2 & -1 \end{vmatrix} = \begin{pmatrix} -4 \\ 2 \\ -8 \end{pmatrix}\).

Normal vector: \(\begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix}\).

Using point \((3, -2, 1)\) on the plane, substitute into \(2x - y + 4z = d\):

\(2(3) - (-2) + 4(1) = 12\).

Thus, the equation is \(2x - y + 4z = 12\).

(b) The line \(l\) is parallel to \(\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\). The normal to the plane is \(\begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix}\).

Dot product: \(\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} = \sqrt{2} \sqrt{21} \cos \alpha\).

\(\cos \alpha = \frac{3}{\sqrt{42}}\).

Acute angle: \(90^\circ - \alpha = 27.6^\circ\).

(c) The position vector of the foot of the perpendicular \(\overrightarrow{OF}\) is given by:

\(\overrightarrow{OF} = \overrightarrow{OP} + t \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} = \begin{pmatrix} 2 + 2t \\ 3 - t \\ 1 + 4t \end{pmatrix}\).

Substitute into the plane equation: \(2(2 + 2t) - (3 - t) + 4(1 + 4t) = 12\).

Solve for \(t\): \(2t + 5 = 12\), \(t = \frac{7}{2}\).

\(\overrightarrow{OF} = \frac{1}{3} \begin{pmatrix} 8 \\ 8 \\ 7 \end{pmatrix}\).

(a) The matrix \(M\) is a product of two matrices: \(\begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}\) represents a shear in the x-direction, and \(\begin{pmatrix} \frac{1}{2}\sqrt{2} & -\frac{1}{2}\sqrt{2} \\ \frac{1}{2} & \frac{1}{2}\sqrt{2} \end{pmatrix}\) represents a rotation anticlockwise about the origin through 45°.

(b) To find \(M^{-1}\), we calculate the inverse of the product of the two matrices: \(M^{-1} = \begin{pmatrix} \frac{k+1}{\sqrt{2}} & \frac{k-1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} \end{pmatrix}\).

(c) For no invariant lines through the origin, the characteristic equation of \(M\) must have no real roots. The characteristic equation is derived from \(\det(M - \lambda I) = 0\). Solving gives \(k^2 + 4k - 4 < 0\), leading to the solution \(2(-1-\sqrt{2}) < k < 2(-1+\sqrt{2})\).

(a) Start with the Cartesian equation \((x^2 + y^2)^2 = 36(x^2 - y^2)\).

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

Substitute: \(r^4 = 36(r^2 \cos^2 \theta - r^2 \sin^2 \theta)\).

Factor out \(r^2\): \(r^4 = 36r^2(\cos^2 \theta - \sin^2 \theta)\).

Use the identity \(\cos 2\theta = \cos^2 \theta - \sin^2 \theta\): \(r^2 = 36 \cos 2\theta\).

(b) The sketch is a closed curve symmetrical about the initial line. The maximum distance from the pole is 6.

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

\(\frac{1}{2} \int_{-\frac{1}{4}\pi}^{\frac{1}{4}\pi} r^2 \, d\theta = \frac{1}{2} \int_{-\frac{1}{4}\pi}^{\frac{1}{4}\pi} 36 \cos 2\theta \, d\theta\).

\(= 18 \int_{-\frac{1}{4}\pi}^{\frac{1}{4}\pi} \cos 2\theta \, d\theta\).

\(= 18 [\frac{1}{2} \sin 2\theta]_{-\frac{1}{4}\pi}^{\frac{1}{4}\pi}\).

\(= 18 [\sin \frac{1}{2}\pi - \sin(-\frac{1}{2}\pi)]\).

\(= 18 [1 - (-1)] = 18 \times 2 = 18\).

(d) To find the maximum distance from the initial line, set \(y = 6 \cos^{\frac{1}{2}} 2\theta \sin \theta\).

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

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

Use trigonometric identities to solve for \(\theta\):

\(\theta = \pm \frac{1}{4}\pi\).

The maximum distance is \(3\sqrt{2}\).

(a) The vertical asymptote occurs where the denominator is zero: \(5x - 2 = 0 \Rightarrow x = \frac{2}{5}\). For the oblique asymptote, divide \(5x^2\) by \(5x - 2\) to get \(y = x + \frac{2}{5}\).

(b) Differentiate \(y = \frac{5x^2}{5x-2}\) using the quotient rule: \(\frac{dy}{dx} = \frac{(5x-2)(10x) - (5x^2)(5)}{(5x-2)^2}\). Set \(\frac{dy}{dx} = 0\) to find stationary points: \(5x^2 - 4x = 0 \Rightarrow x(5x - 4) = 0\). Thus, \(x = 0\) or \(x = \frac{4}{5}\). The coordinates are \((0,0)\) and \(\left( \frac{4}{5}, \frac{8}{5} \right)\).

(c) Sketch the curve showing the asymptotes and stationary points.

(d) Solve \(\left| \frac{5x^2}{5x-2} \right| < 2\). This gives two cases: \(\frac{5x^2}{5x-2} < 2\) and \(\frac{5x^2}{5x-2} > -2\). Solving these inequalities gives the intervals \(-1 - \frac{1}{3}\sqrt{5} < x < -1 + \frac{1}{3}\sqrt{5}\) and \(1 - \frac{1}{3}\sqrt{5} < x < 1 + \frac{1}{3}\sqrt{5}\).

(a) We start with \(\sum_{r=1}^{n} r(r+2) = \sum_{r=1}^{n} (r^2 + 2r)\).

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

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

Simplifying gives \(\frac{1}{6}n(n+1)(2n+7)\).

(b) Express \(\frac{1}{r(r+2)}\) in partial fractions: \(\frac{1}{r(r+2)} = \frac{1}{2} \left( \frac{1}{r} - \frac{1}{r+2} \right)\).

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

This simplifies to \(\frac{1}{2} \left( \frac{3}{2} - \frac{1}{n+1} - \frac{1}{n+2} \right)\).

(c) As \(n \to \infty\), the terms \(\frac{1}{n+1}\) and \(\frac{1}{n+2}\) approach zero.

Thus, the sum converges to \(\frac{1}{2} \times \frac{3}{2} = \frac{3}{4}\).

(a) Let \(y = x^{-2}\), then \(x = y^{-\frac{1}{2}}\). Substitute into the original equation:

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

Multiply through by \(y^2\) to clear fractions:

\(1 + 3y + 2y^{\frac{3}{2}} + 6y^2 = 0\)

Rearrange to form a quartic equation:

\(36y^4 + 32y^3 + 21y^2 + 6y + 1 = 0\)

The value of \(\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} + \frac{1}{\delta^2}\) is \(-\frac{8}{9}\).

(b) Using the relation \(\alpha \beta \gamma \delta = 6\), we find:

\(\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} + \frac{1}{\delta^2} = \frac{\alpha^2 \beta^2 \gamma^2 \delta^2 + \alpha^2 \beta^2 \gamma^2 \delta^2 + \alpha^2 \beta^2 \gamma^2 \delta^2 + \alpha^2 \beta^2 \gamma^2 \delta^2}{\alpha^2 \beta^2 \gamma^2 \delta^2}\)

\(\beta^2 \gamma^2 \delta^2 + \alpha^2 \gamma^2 \delta^2 + \alpha^2 \beta^2 \delta^2 + \alpha^2 \beta^2 \gamma^2 = -32\)

(c) Using the formula for the sum of squares:

\(\frac{1}{\alpha^4} + \frac{1}{\beta^4} + \frac{1}{\gamma^4} + \frac{1}{\delta^4} = \left(\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} + \frac{1}{\delta^2}\right)^2 - 2(\alpha^{-2}\beta^{-2} + \alpha^{-2}\gamma^{-2} + \alpha^{-2}\delta^{-2} + \beta^{-2}\gamma^{-2} + \beta^{-2}\delta^{-2} + \gamma^{-2}\delta^{-2})\)

\(= \left(-\frac{8}{9}\right)^2 - 2\left(\frac{21}{36}\right)\)

= \(-\frac{61}{162}\)

(a) The matrix \(\begin{pmatrix} 1 & 0 \\ k & 1 \end{pmatrix}\) represents a shear in the y-direction. The matrix \(\begin{pmatrix} 1 & 0 \\ 0 & k \end{pmatrix}\) represents a stretch parallel to the y-axis with scale factor \(k\). The transformations are applied in the order of shear followed by stretch.

(b) To find \(\mathbf{M}^{-1}\), we first find the inverse of each matrix: \(\begin{pmatrix} 1 & 0 \\ k & 1 \end{pmatrix}^{-1} = \begin{pmatrix} 1 & 0 \\ -k & 1 \end{pmatrix}\) and \(\begin{pmatrix} 1 & 0 \\ 0 & k \end{pmatrix}^{-1} = \begin{pmatrix} 1 & 0 \\ 0 & \frac{1}{k} \end{pmatrix}\). Thus, \(\mathbf{M}^{-1} = \begin{pmatrix} 1 & 0 \\ -k & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & \frac{1}{k} \end{pmatrix}\).

(c) The transformation \(\mathbf{M} = \begin{pmatrix} 1 & 0 \\ k^2 & k \end{pmatrix}\) transforms \(\begin{pmatrix} x \\ y \end{pmatrix}\) to \(\begin{pmatrix} x \\ k^2x + ky \end{pmatrix}\). Setting \(\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ k^2x + ky \end{pmatrix}\), we get \(k^2x + ky = y\), which simplifies to \(y = \frac{k^2}{1-k}x\).

(d) The area of triangle DEF is equal to the area of triangle ABC when \(|\det(\mathbf{M})| = 1\). The determinant \(\det(\mathbf{M}) = 1 \times k - 0 \times k^2 = k\). Thus, \(|k| = 1\), giving \(k = -1\).

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

\(\frac{d}{dx}(xf(x)) = xf'(x) + f(x) = xf'(x) + (2(1)-1)f(x).\)

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

\(\frac{d^{2k-1}}{dx^{2k-1}}(xf(x)) = xf'(x) + (2k-1)f(x).\)

Then,

\(\frac{d^{2k}}{dx^{2k}}(xf(x)) = xf(x) + 2kf'(x).\)

\(\frac{d^{2k+1}}{dx^{2k+1}}(xf(x)) = xf'(x) + f(x) + 2kf(x)\)

\(= xf'(x) + (2k+1)f(x).\)

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

(a) The polar equation is \(r = a \sec^2 \theta\). At \(\theta = 0\), \(r = a \sec^2(0) = a\), giving the point \((a, 0)\). At \(\theta = \frac{1}{4} \pi\), \(r = a \sec^2(\frac{1}{4} \pi) = 2a\), giving the point \((2a, \frac{1}{4} \pi)\).

(b) The maximum distance from the initial line is when \(\theta = \frac{1}{4} \pi\), where \(r = 2a\). The distance is \(y = r \sin \theta = 2a \sin(\frac{1}{4} \pi) = a\sqrt{2}\).

(c) The area \(A\) is given by \(\frac{1}{2} \int_0^{\frac{1}{4} \pi} r^2 \, d\theta\). Substituting \(r = a \sec^2 \theta\), we have:

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

\(= \frac{1}{2} a^2 \int_0^{\frac{1}{4} \pi} (\sec^2 \theta)(\tan^2 \theta + 1) \, d\theta\)

\(= \frac{1}{2} a^2 \left[ \tan \theta + \frac{1}{3} \tan^3 \theta \right]_0^{\frac{1}{4} \pi}\)

\(= \frac{3}{8} a^2\)

(d) The Cartesian form is derived by substituting \(x = r \cos \theta\) and \(y = r \sin \theta\). From \(r = a \sec^2 \theta\), we have:

\(x^2 + y^2 = r^2\)

\(r^2 \cos^2 \theta = a^2 \Rightarrow x^2 = a^2 - y^2\)

\(x^4 = a^2 (x^2 + y^2)\)

\(y = \sqrt{a^2 x^4 - x^2} = x \sqrt{a^2 x^2 - 1}\)

(a) To find the length \(PQ\), we first find a vector joining any point on \(l_1\) to any point on \(l_2\). The vector is \(\begin{pmatrix} 0 \\ 2 \\ 6 \end{pmatrix} - \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -2 \\ 2 \\ 5 \end{pmatrix}\).

Next, find the common perpendicular using the cross product: \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 2 \\ 1 & 2 & -2 \end{vmatrix} = \begin{pmatrix} 2 \\ -4 \\ -3 \end{pmatrix}\).

The length \(PQ\) is given by \(\frac{1}{\sqrt{29}} \begin{pmatrix} -2 \\ 2 \\ 5 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -4 \\ -3 \end{pmatrix} = \frac{27}{\sqrt{29}} = 5.01\).

(b)(i) The equation of \(\Pi_1\) is \(\mathbf{r} = 2\mathbf{i} + \mathbf{k} + s(\mathbf{i} - \mathbf{j} + 2\mathbf{k}) + t(2\mathbf{i} - 4\mathbf{j} - 3\mathbf{k})\).

(b)(ii) The normal to \(\Pi_2\) is found using the cross product: \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -2 \\ 2 & -4 & -3 \end{vmatrix} = \begin{pmatrix} -14 \\ -1 \\ -8 \end{pmatrix}\).

Substitute a point to find the equation: \(14(0) + (2) + 8(6) = 50 \Rightarrow 14x + y + 8z = 50\).

(c) The normals to \(\Pi_1\) and \(\Pi_2\) are \(\begin{pmatrix} 11 \\ 7 \\ -2 \end{pmatrix}\) and \(\begin{pmatrix} 14 \\ 1 \\ 8 \end{pmatrix}\) respectively.

Use the dot product to find the angle: \(\cos \theta = \frac{145}{\sqrt{174} \sqrt{261}}\).

The acute angle is \(47.1^\circ\).

(a) The vertical asymptote occurs where the denominator is zero: \(x + 1 = 0 \Rightarrow x = -1\).
The oblique asymptote is found by dividing \(x^2 - x\) by \(x + 1\), giving \(y = x - 2\).

(b) Differentiate \(y = \frac{x^2 - x}{x + 1}\) using the quotient rule:
\(\frac{dy}{dx} = \frac{(2x - 1)(x + 1) - (x^2 - x)}{(x + 1)^2} = \frac{1 - 2(x + 1)^2}{(x + 1)^2}\).
Set \(\frac{dy}{dx} = 0\) to find stationary points:
\(1 - 2(x + 1)^2 = 0 \Rightarrow (x + 1)^2 = \frac{1}{2}\).
Solving gives \(x = -1 \pm \sqrt{2}\).
Substitute back to find \(y\):
\(y = \frac{(-1 + \sqrt{2})^2 - (-1 + \sqrt{2})}{-1 + \sqrt{2} + 1} = -3 + 2\sqrt{2}\),
\(y = \frac{(-1 - \sqrt{2})^2 - (-1 - \sqrt{2})}{-1 - \sqrt{2} + 1} = -3 - 2\sqrt{2}\).

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

(d) Solve \(\left| \frac{x^2 - x}{x + 1} \right| < 6\):
\(\frac{x^2 - x}{x + 1} < 6\) or \(\frac{x^2 - x}{x + 1} > -6\).
For \(\frac{x^2 - x}{x + 1} < 6\):
\(x^2 - 7x - 6 = 0 \Rightarrow x = -3, -2\).
For \(\frac{x^2 - x}{x + 1} > -6\):
\(x^2 + 5x + 6 = 0 \Rightarrow x = -2, -3\).
Thus, \(-3 < x < -2 \text{ and } \frac{1}{2} - \frac{1}{2}\sqrt{73} < x < \frac{1}{2} + \frac{1}{2}\sqrt{73}\).