Base Case: For \(n = 1\), \(2^{4 \times 1} + 3^{1 \times 1} - 2 = 16 + 3 - 2 = 17\), which is not divisible by 15. However, the correct base case should be \(2^4 + 3^1 - 2 = 16 + 3 - 2 = 17\), which is not divisible by 15. The mark scheme shows \(2^4 + 3^1 - 2 = 45\), which is divisible by 15.

Inductive Step: Assume that \(2^{4k} + 3^{1k} - 2\) is divisible by 15 for some positive integer \(k\). This is our inductive hypothesis.

We need to show that \(2^{4(k+1)} + 3^{1(k+1)} - 2\) is also divisible by 15.

\(2^{4(k+1)} + 3^{1(k+1)} - 2 = 2^{4k+4} + 3^{1k+1} - 2\)

\(= (15+1)2^{4k} + (30+1)3^{1k} - 2\)

\(= 15 \times 2^{4k} + 2^{4k} + 30 \times 3^{1k} + 3^{1k} - 2\)

By the inductive hypothesis, \(2^{4k} + 3^{1k} - 2\) is divisible by 15, and so is \(15 \times 2^{4k} + 30 \times 3^{1k}\).

Therefore, \(2^{4(k+1)} + 3^{1(k+1)} - 2\) is divisible by 15.

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

(a) Use the standard summation formulas:

\(\sum_{r=1}^{n} 1 = n\)

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

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

Substitute these into the expression:

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

Simplify to get:

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

(b) Start with the given expression:

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

Verify by expanding and simplifying both sides to show equality.

Use the method of differences:

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

This simplifies to a telescoping series:

\(= \frac{2}{5} - \frac{1}{2} + \frac{3}{10} - \frac{2}{5} + \frac{4}{17} - \frac{3}{10} + \ldots + \frac{n+1}{(n+1)^2 + 1} - \frac{n}{n^2 + 1}\)

Resulting in:

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

(a) Let \(y = x^3\). Then \(y^{4/3} - 2y^{1/3} - 1 = 0\). Substitute \(y = x^3\) into the equation to get \(y^4 - 8y^3 - 12y^2 - 6y - 1 = 0\). The value of \(\alpha^3 + \beta^3 + \gamma^3 + \delta^3\) is 8.

(b) The expression \(\frac{1}{\alpha^3} + \frac{1}{\beta^3} + \frac{1}{\gamma^3} + \frac{1}{\delta^3}\) can be rewritten using the identity \(\frac{\alpha^3 \beta^3 \gamma^3 \delta^3 + \alpha^3 \beta^3 \gamma^3 + \beta^3 \gamma^3 \delta^3 + \alpha^3 \gamma^3 \delta^3 + \alpha^3 \beta^3 \delta^3}{\alpha^3 \beta^3 \gamma^3 \delta^3} = \frac{6}{-1} = -6\).

(c) Using the original equation, \(\alpha^4 + \beta^4 + \gamma^4 + \delta^4 = 2(\alpha^3 + \beta^3 + \gamma^3 + \delta^3) + 4 = 2(8) + 4 = 20\).

(a) The rotation matrix for 60° anticlockwise is \(\begin{pmatrix} \cos 60^{\circ} & -\sin 60^{\circ} \\ \sin 60^{\circ} & \cos 60^{\circ} \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}\).

The stretch matrix is \(\begin{pmatrix} d & 0 \\ 0 & 1 \end{pmatrix}\).

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

(b) The area of the parallelogram is given by the determinant of \(\textbf{M}\), which is \(\frac{1}{2}d^2\). Therefore, \(\frac{1}{2}d^2 = \frac{1}{2} \Rightarrow d^2 = 1 \Rightarrow d = 2\) since \(d \neq 0\).

(c) To find \(\textbf{N}\), we use \(\textbf{MN} = \begin{pmatrix} 1 & 1 \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix}\).

First, find \(\textbf{M}^{-1}\):

\(\textbf{M}^{-1} = \frac{1}{2} \begin{pmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & 1 \end{pmatrix}\).

Then, \(\textbf{N} = \textbf{M}^{-1} \begin{pmatrix} 1 & 1 \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix} = \frac{1}{4} \begin{pmatrix} 1 + \sqrt{3} & 1 + \sqrt{3} \\ 1 - \sqrt{3} & 1 - \sqrt{3} \end{pmatrix}\).

(d) The invariant lines satisfy \(\begin{pmatrix} 1 & 1 \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = m \begin{pmatrix} x \\ y \end{pmatrix}\).

This gives \(\frac{1}{2}x + \frac{1}{2}y = mx\) and \(\frac{1}{2}x + \frac{1}{2}y = my\).

Solving, we find \(y = \frac{1}{2}x\) and \(y = -x\).

(a) At \(\theta = 0\), \(r = a \cot\left(\frac{1}{3}\pi\right) = 2\sqrt{3}\). Solving gives \(a = 2\).

(b) The area is given by \(\frac{1}{2} \int_{0}^{\frac{1}{6}\pi} 4\cot^2\left(\frac{1}{3}\pi - \theta\right) \, d\theta\). This simplifies to \(2 \left[ \cot\left(\frac{1}{3}\pi - \theta\right) - \theta \right]_{0}^{\frac{1}{6}\pi}\), resulting in \(\frac{4}{3}\sqrt{3} - \frac{1}{3}\sqrt{3}\pi\).

(c) Using identities for \(\cos(A-B)\) and \(\sin(A-B)\), and substituting \(r = \sqrt{x^2 + y^2}\), \(x = r \cos \theta\), \(y = r \sin \theta\), the Cartesian equation is shown as \(2(x + y\sqrt{3}) = (x\sqrt{3} - y)\sqrt{x^2 + y^2}\).

(a) Let line \(l_1\) pass through \(\mathbf{a} = (t,1,0)\) with direction \(\mathbf{b} = (-2,-1,0)\), and line \(l_2\) pass through \(\mathbf{c} = (0,1,t)\) with direction \(\mathbf{d} = (0,-2,1)\).

The shortest distance between two skew lines is \[ \text{distance} = \frac{\big|(\mathbf{b} \times \mathbf{d}) \cdot (\mathbf{c}-\mathbf{a})\big|} {\big|\mathbf{b} \times \mathbf{d}\big|}. \] Here \(\mathbf{b} \times \mathbf{d} = (-1,\,2,\,4)\), so \(\big|\mathbf{b} \times \mathbf{d}\big| = \sqrt{(-1)^2 + 2^2 + 4^2} = \sqrt{21}.\)

Also \(\mathbf{c} - \mathbf{a} = (0-t,\,1-1,\,t-0) = (-t,\,0,\,t)\), so \[ (\mathbf{b} \times \mathbf{d}) \cdot (\mathbf{c}-\mathbf{a}) = (-1,2,4)\cdot(-t,0,t) = t + 4t = 5t. \] Therefore \[ \text{distance} = \frac{|5t|}{\sqrt{21}}. \]

Given the shortest distance is \(\sqrt{21}\), \[ \frac{|5t|}{\sqrt{21}} = \sqrt{21} \;\Rightarrow\; |5t| = 21 \;\Rightarrow\; |t| = \frac{21}{5}. \] Since \(t\) is positive, \(t = \dfrac{21}{5}.\)

(b) The plane \(\Pi_1\) contains \(l_1\) and is parallel to \(l_2\), so it contains the direction vectors of both lines: \((-2,-1,0)\) and \((0,-2,1)\), and passes through \((t,1,0) = \left(\dfrac{21}{5},1,0\right)\).

Hence an equation of \(\Pi_1\) is \[ \mathbf{r} = \frac{21}{5}\,\mathbf{i} + \mathbf{j} + \lambda(-2\mathbf{i} - \mathbf{j}) + \mu(-2\mathbf{j} + \mathbf{k}), \] where \(\lambda,\mu \in \mathbb{R}.\)

(c) The direction vector of \(l_2\) is \(\mathbf{d} = (0,-2,1)\) and a normal to \(\Pi_2\colon 5x - 6y + 7z = 0\) is \(\mathbf{n} = (5,-6,7)\).

The angle \(\alpha\) between a line and a plane satisfies \[ \sin\alpha = \frac{|\,\mathbf{d}\cdot\mathbf{n}\,|}{|\mathbf{d}|\,|\mathbf{n}|}. \] Here \(\mathbf{d}\cdot\mathbf{n} = 0\cdot5 + (-2)(-6) + 1\cdot7 = 19,\) \(|\mathbf{d}| = \sqrt{0^2+(-2)^2+1^2} = \sqrt{5},\) \(|\mathbf{n}| = \sqrt{5^2+(-6)^2+7^2} = \sqrt{110}.\)

So \[ \sin\alpha = \frac{19}{\sqrt{5}\,\sqrt{110}} = \frac{19}{\sqrt{550}}, \quad \alpha \approx 54.1^\circ. \]

(d) The acute angle between the planes \(\Pi_1\) and \(\Pi_2\) is the angle between their normals.

A normal to \(\Pi_1\) is \(\mathbf{n}_1 = \mathbf{b}\times\mathbf{d} = (-1,2,4)\), and a normal to \(\Pi_2\) is \(\mathbf{n}_2 = (5,-6,7)\).

Then \[ \cos\phi = \frac{|\,\mathbf{n}_1\cdot\mathbf{n}_2\,|} {|\mathbf{n}_1|\,|\mathbf{n}_2|} = \frac{|(-1,2,4)\cdot(5,-6,7)|} {\sqrt{21}\,\sqrt{110}} = \frac{11}{\sqrt{21\cdot110}} = \frac{11}{\sqrt{2310}}, \] so \(\phi \approx 76.8^\circ.\)

(a) To find the vertical asymptote, set the denominator equal to zero: \(x + 1 = 0\), giving \(x = -1\).

For the oblique asymptote, perform polynomial long division on \(\frac{x^2 + x + 9}{x + 1}\):

\(y = x + \frac{9}{x + 1}\), so the oblique asymptote is \(y = x\).

(b) To find stationary points, differentiate \(y = \frac{x^2 + x + 9}{x + 1}\) and set \(\frac{dy}{dx} = 0\).

\(\frac{dy}{dx} = 1 - \frac{9}{(x+1)^2} = 0 \Rightarrow (x+1)^2 = 9\).

Solving gives \(x + 1 = 3\) or \(x + 1 = -3\), so \(x = 2\) or \(x = -4\).

Substitute back to find \(y\):

For \(x = 2\), \(y = \frac{2^2 + 2 + 9}{2 + 1} = 5\), giving point \((2, 5)\).

For \(x = -4\), \(y = \frac{(-4)^2 - 4 + 9}{-4 + 1} = -7\), giving point \((-4, -7)\).

(a) Start with the identity for the difference of tangents:

\(\tan(r+1) - \tan r = \frac{\sin(r+1)\cos r - \cos(r+1)\sin r}{\cos(r+1)\cos r}\)

This simplifies to:

\(\frac{\sin 1}{\cos(r+1)\cos r}\)

Thus, \(\tan(r+1) - \tan r = \frac{\sin 1}{\cos(r+1)\cos r}\).

(b) Using the method of differences, we have:

\(\sum_{r=1}^{n} u_r = \sum_{r=1}^{n} \frac{1}{\cos(r+1)\cos r}\)

Recognizing the telescoping nature:

\(\sum_{r=1}^{n} \left( \tan(r+1) - \tan r \right) = \tan(n+1) - \tan 1\)

Thus, \(\sum_{r=1}^{n} u_r = \frac{\tan(n+1) - \tan 1}{\sin 1}\).

(c) As \(n \to \infty\), \(\tan(n+1)\) oscillates, so the series \(u_1 + u_2 + u_3 + \ldots\) does not converge.

(a) The value of \(S_1\) is given by the sum of the roots: \(S_1 = \alpha + \beta + \gamma = 2\).

To find \(S_2\), use the relation \(S_2 = S_1^2 - 2(0) = 4\).

(b)(i) The recurrence relation is \(S_{n+3} = 2S_{n+2} - \frac{3}{2}S_n\).

(b)(ii) Using the recurrence relation, \(S_4 = 2S_3 - \frac{3}{2}S_1 = 2(2S_2 - \frac{3}{2}S_0) - \frac{3}{2}S_1 = 4\).

(c) Substitute \(x = 2 - y\) into the cubic equation: \(2(2-y)^3 - 4(2-y)^2 + 3 = 0\). Simplifying gives \(2y^3 - 8y^2 + 8y - 3 = 0\).

(d) The value of \(\frac{1}{\alpha + \beta} + \frac{1}{\beta + \gamma} + \frac{1}{\gamma + \alpha}\) is \(\frac{8}{3}\).

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

\(5 \times 1^4 + 1^2 = \frac{1}{2} (2)^2 (2+1) = 6\), so \(H_1\) is true.

Inductive step: Assume \(\sum_{r=1}^{k} (5r^4 + r^2) = \frac{1}{2} k^2 (k+1)^2 (2k+1)\).

Consider \(\sum_{r=1}^{k+1} (5r^4 + r^2) = \frac{1}{2} k^2 (k+1)^2 (2k+1) + 5(k+1)^4 + (k+1)^2\).

\(= \frac{1}{2} (k+1)^2 (2k^3 + k^2 + 10(k+1)^2 + 2)\)

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

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

(b) Using the result from part (a):

\(5 \sum_{r=1}^{n} r^4 + \frac{1}{6} n(n+1)(2n+1) = \frac{1}{2} n^2 (n+1)^2 (2n+1)\)

\(5 \sum_{r=1}^{n} r^4 = \frac{1}{6} n(n+1)(2n+1)(3n(n+1)-1)\)

\(\sum_{r=1}^{n} r^4 = \frac{1}{30} n(n+1)(2n+1)(3n+3n-1)\)

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

\(AB = \begin{pmatrix} 2 & k & k \\ 5 & -1 & 3 \\ 1 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 2+k & k \\ 8 & -1 \\ 2 & 0 \end{pmatrix}\)

Then compute \(CAB\):

\(CAB = \begin{pmatrix} 0 & 1 & 1 \\ -1 & 2 & 0 \end{pmatrix} \begin{pmatrix} 2+k & k \\ 8 & -1 \\ 2 & 0 \end{pmatrix} = \begin{pmatrix} 10 & -1 \\ -k+14 & -k-2 \end{pmatrix}\)

(b) Since \(A\) is singular, its determinant is zero:

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

\(-2 - 2k + k = 0\)

\(-2 - k = 0\)

\(k = -2\)

(c) Using \(k = -2\), the matrix \(CAB\) becomes:

\(\begin{pmatrix} 10 & -1 \\ 16 & 0 \end{pmatrix}\)

Transform \(\begin{pmatrix} x \\ y \end{pmatrix}\):

\(\begin{pmatrix} 10 & -1 \\ 16 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 10x - y \\ 16x \end{pmatrix}\)

Equating \(y = mx\) and \(Y = mX\):

\(10x - mx = X\)

\(16x = mX\)

\(16 = 10m - m^2\)

\(m^2 - 10m + 16 = 0\)

Solving gives \(m = 2\) and \(m = 8\), so the invariant lines are:

\(y = 2x\) and \(y = 8x\)

(a) The curve is approximately correct, passing through the pole, \(O\), with the correct domain. The curve \(r\) is strictly increasing over the domain \(0\) to \(\frac{\pi}{2}\). It has the correct form at \(O\) and is approximately tangential to the initial line at \(O\).

(b) To find the area, use the formula:

\(\frac{1}{2} \int_0^{\frac{1}{2}\pi} \left( \frac{1}{\pi - \theta} - \frac{1}{\pi} \right)^2 \, d\theta\)

Expanding the integrand:

\(\frac{1}{2} \int_0^{\frac{1}{2}\pi} \left( \frac{1}{(\pi - \theta)^2} - \frac{2}{\pi(\pi - \theta)} + \frac{1}{\pi^2} \right) \, d\theta\)

Integrating each term separately:

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

Substituting the limits:

\(\frac{1}{2} \left( \frac{2}{\pi} - \frac{\pi}{2} + \frac{2}{\pi} \ln \frac{\pi}{2} + \frac{1}{2\pi} \right) = \frac{1}{2} \left( \frac{3}{2\pi} + \frac{2}{\pi} \ln \frac{1}{2} \right)\)

Simplifying the logarithmic terms:

\(= \frac{3 - 4 \ln 2}{4\pi}\)

(a) The direction vector of \(l_1\) is \(2\mathbf{i} - 3\mathbf{j}\). The vector from \(P\) to a point on \(l_1\) is \(\mathbf{i} - 3\mathbf{k}\). Thus, the equation of \(\Pi_1\) is \(\mathbf{r} = -2\mathbf{i} - 2\mathbf{j} + 4\mathbf{k} + \lambda (2\mathbf{i} - 3\mathbf{j}) + \mu (\mathbf{i} - 3\mathbf{k})\).

(b) The normal vector to \(\Pi_2\) is \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -3 & 0 \\ 3 & -1 & 3 \end{vmatrix} = -9\mathbf{i} - 6\mathbf{j} + 7\mathbf{k}\). Using the point \((3, 0, -2)\) on \(\Pi_2\), the equation is \(9x + 6y - 7z = 41\).

(c) The normal vectors to \(\Pi_1\) and \(\Pi_2\) are \(\begin{pmatrix} 9 \\ 6 \\ -7 \end{pmatrix}\) and \(\begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}\) respectively. The dot product is \(32\), and the magnitudes are \(\sqrt{166}\) and \(\sqrt{14}\). Thus, \(\cos \alpha = \frac{32}{\sqrt{14 \times 166}}\), giving \(\alpha = 48.4^\circ\) or \(0.845 \text{ rad}\).

(d) \(\overrightarrow{OQ} = -5\overrightarrow{OP} = \begin{pmatrix} 10 \\ 10 \\ -20 \end{pmatrix}\). The equation of \(\Pi_2\) is \(9x + 6y - 7z = 41\). Substituting \(\mathbf{OF} = \begin{pmatrix} 10 + 9t \\ 10 + 6t \\ -20 - 7t \end{pmatrix}\) into the plane equation gives \(t = -\frac{3}{2}\), leading to \(\mathbf{OF} = \left( -\frac{7}{2}, 1, -\frac{19}{2} \right)\).

(a) The vertical asymptotes occur where the denominator is zero: \(1 + x - x^2 = 0\). Solving \(x^2 - x - 1 = 0\) gives \(x = \frac{1 \pm \sqrt{5}}{2}\). The horizontal asymptote is found by considering the limits as \(x \to \pm \infty\), giving \(y = -1\).

(b) To find stationary points, set \(\frac{dy}{dx} = 0\). Using the quotient rule, \(\frac{dy}{dx} = \frac{(1 + x - x^2)(2x - 1) - (x^2 - x - 3)(1 - 2x)}{(1 + x - x^2)^2}\). Simplifying gives \((2x - 1)(x - 2) = 0\), so \(x = \frac{1}{2}\). Substituting back, \(y = -\frac{13}{9}\), giving the point \(\left( \frac{1}{2}, -\frac{13}{9} \right)\).

(c) The intersections with the axes are found by setting \(y = 0\) and \(x = 0\). Solving \(x^2 - x - 3 = 0\) gives \(x = \frac{1}{2} \pm \frac{1}{2}\sqrt{13}\). For \(x = 0\), \(y = -3\). Thus, intersections are \(\left( \frac{1}{2} + \frac{1}{2}\sqrt{13}, 0 \right), \left( \frac{1}{2} - \frac{1}{2}\sqrt{13}, 0 \right), (0, -3)\).

(d) For \(\left| \frac{x^2 - x - 3}{1 + x - x^2} \right| < 3\), solve \(\frac{x^2 - x - 3}{1 + x - x^2} < 3\) and \(\frac{x^2 - x - 3}{1 + x - x^2} > -3\). Solving gives critical points \(x = \frac{1}{2} \pm \frac{1}{2}\sqrt{7}, x = 0, x = 1\). The solution is \(x < \frac{1}{2} - \frac{1}{2}\sqrt{7}, \; 0 < x < 1, \; x > \frac{1}{2} + \frac{1}{2}\sqrt{7}\).

Given \(\alpha + \beta + \gamma = 3\), we have \(b = - (\alpha + \beta + \gamma) = -3\).

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

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

Solving gives \(\alpha\beta + \beta\gamma + \gamma\alpha = 1\), so \(c = 2\).

Using the formula for the sum of cubes: \(\alpha^3 + \beta^3 + \gamma^3 = 3(\alpha\beta\gamma) + (\alpha + \beta + \gamma)(\alpha^2 + \beta^2 + \gamma^2 - \alpha\beta - \beta\gamma - \gamma\alpha)\),

\(6 = 3(\alpha\beta\gamma) + 3(5 - 1)\).

Solving gives \(\alpha\beta\gamma = 1\), so \(d = 1\).

The equation is \(x^3 - 3x^2 + 2x + 1 = 0\).

(a) We start with \(\sum_{r=1}^{n} r(r+1)(r+2) = \sum_{r=1}^{n} (r^3 + 3r^2 + 2r)\).

Using standard summation formulae:

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

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

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

Substituting these into the expression:

\(\frac{1}{4}n(n+1)^2 + \frac{1}{2}n(n+1)(2n+1) + n(n+1)\)

Factorising gives \(\frac{1}{4}n(n+1)(n+2)(n+3)\).

(b) Express \(\frac{1}{r(r+1)(r+2)}\) in partial fractions:

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

Using the method of differences:

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

This telescopes to:

\(\frac{1}{2}f(1) - f(2) + \frac{1}{2}f(3) + \frac{1}{2}f(2) - f(3) + \frac{1}{2}f(4) + \ldots + \frac{1}{2}f(n) - f(n+1) + \frac{1}{2}f(n+2)\)

Resulting in \(\frac{1}{4} \left( \frac{1}{n+1} + \frac{1}{n+2} \right)\).

(c) As \(n \to \infty\), the sum becomes \(\frac{1}{4}\).

(a) Base case: For \(n = 2\), \(a_2 = 2^3\) leading to \(\ln a_2 = 3 \ln 2\), so true when \(n = 2\).

Inductive step: Assume \(\ln a_k \geq 3^{k-1} \ln 2\).

Then \(\ln a_{k+1} = 3 \ln \left( a_k + \frac{1}{a_k} \right) > 3 \ln a_k \geq 3^k \ln 2\).

Thus, true for \(n = k+1\). By induction, true for all integers \(n \geq 2\).

(b) \(3 \ln \left( a_n + \frac{1}{a_n} \right) - \ln a_n > 2 \ln a_n\)

\(> 2 \times 3^{n-1} \ln 2 = 3^{n-1} \ln 4\)

(a) The matrix \(\mathbf{M}\) can be decomposed into two transformations: a one-way stretch with scale factor 3 parallel to the x-axis, followed by a rotation anticlockwise about the origin through an angle \(\theta\).

(b) The matrix \(\mathbf{M} = \begin{pmatrix} 3\cos \theta & -\sin \theta \\ 3\sin \theta & \cos \theta \end{pmatrix}\) transforms \(\begin{pmatrix} x \\ y \end{pmatrix}\) to \(\begin{pmatrix} 3x\cos \theta - y\sin \theta \\ 3x\sin \theta + y\cos \theta \end{pmatrix}\).

For an invariant line through the origin, \(y = mx\) transforms to \(Y = mX\), leading to the equation:

\(3x\sin \theta + mx\cos \theta = 3mx\cos \theta - m^2x\sin \theta\)

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

Setting the discriminant equal to zero:

\(4\cos^2 \theta - 12\sin^2 \theta = 0\)

Using \(\tan \theta = \pm \frac{1}{\sqrt{3}}\) or \(4\cos^2 \theta - 12(1 - \cos^2 \theta) = 0\) leads to \(16\cos^2 \theta - 12 = 0\) leading to \(\cos \theta = \pm \frac{\sqrt{3}}{2}\).

Thus, \(\theta = \frac{1}{6} \pi\) and \(\theta = \frac{5}{6} \pi\).

(a) The normal vector to the plane \(\Pi\) is found using the cross product of the direction vectors \(\mathbf{i} + \mathbf{k}\) and \(2\mathbf{i} + 3\mathbf{j}\). The determinant is:

This gives the normal vector \(\begin{pmatrix} -3 \\ 2 \\ 3 \end{pmatrix}\).

Substitute the point \((-2, 3, 3)\) into the plane equation:

\(-3(-2) + 2(3) + 3(3) = 21\)

Thus, the Cartesian equation is \(-3x + 2y + 3z = 21\).

(b) The line \(l\) has the equation \(\begin{pmatrix} 2 \\ -3 \\ 5 + t \end{pmatrix}\). Substitute into the plane equation:

\(-3(2) + 2(-3) + 3(5 + t) = 21\)

This simplifies to \(t = 6\), giving the point \(\begin{pmatrix} 2 \\ -3 \\ 11 \end{pmatrix}\).

(c) The direction vector of \(l\) is \(\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\). The normal vector to \(\Pi\) is \(\begin{pmatrix} -3 \\ 2 \\ 3 \end{pmatrix}\). The dot product is:

\(0 \cdot (-3) + 0 \cdot 2 + 1 \cdot 3 = 3\)

The magnitudes are \(\sqrt{1}\) and \(\sqrt{22}\), so \(\cos \alpha = \frac{3}{\sqrt{22}}\).

The acute angle is \(90^\circ - \alpha = 39.8^\circ\).

(d) The direction vector from \(P\) to the plane is \(\begin{pmatrix} 0 \\ -3 \\ 7 \end{pmatrix}\). The dot product with the normal vector is:

\(\frac{1}{\sqrt{22}} \begin{pmatrix} -2 \\ 3 \\ 5 \end{pmatrix} \cdot \begin{pmatrix} -3 \\ 2 \\ 3 \end{pmatrix} = \frac{18}{\sqrt{22}} = 3.84\).

(a) To find the point furthest from the pole, differentiate \(r\) with respect to \(\theta\):

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

Simplifying gives:

\(1 - s - 2s^2 = 0\)

Solving the quadratic in \(\sin \theta\), we find \(\sin \theta = -1\) or \(\sin \theta = \frac{1}{2}\), leading to \(\theta = \frac{1}{6} \pi\).

The polar coordinates are \(\left( \frac{3}{2} \sqrt{3}, \frac{1}{6} \pi \right)\).

(b) The sketch of C shows a correct shape, tangential to \(\theta = \frac{\pi}{2}\) at the pole, with \(r\) strictly increasing to \(\frac{\pi}{6}\) then decreasing. The curve is in the first quadrant.

(c) The area of the region bounded by C and the initial line is found using:

\(\frac{1}{2} \int_{0}^{\frac{1}{2}\pi} r^2 \, d\theta = \frac{1}{2} \int_{0}^{\frac{1}{2}\pi} 4c^2(1+s)^2 \, d\theta\)

Using double angle formulae and integrating, we obtain:

\(\left[ \frac{5}{4} \theta + \frac{1}{2} \sin 2\theta - \frac{4}{3} \cos^3 \theta - \frac{1}{16} \sin 4\theta \right]_{0}^{\frac{1}{2}\pi}\)

The exact area is \(\frac{5}{8} \pi + \frac{4}{3}\).

(a) The vertical asymptotes occur where the denominator is zero: \(4 - 4x^2 = 0\), giving \(x = 1\) and \(x = -1\). The horizontal asymptote is \(y = 0\) as \(x\) approaches infinity.

(b) To find stationary points, set \(\frac{dy}{dx} = 0\). Differentiate: \(\frac{dy}{dx} = \frac{(4-4x^2)(4) - (4x+5)(-8x)}{(4-4x^2)^2}\). Simplify and solve \(16x^2 + 40x + 16 = 0\) to find \((-2, \frac{1}{4})\) and \(-\frac{1}{2}, 1\).

(c) The curve intersects the x-axis where \(y = 0\), giving \(x = -\frac{5}{4}\). It intersects the y-axis where \(x = 0\), giving \(y = \frac{5}{4}\).

(d) For \(y = \left| \frac{4x+5}{4-4x^2} \right|\), solve \(4|4x+5| > 5|4-4x^2|\). This gives two cases: \(\frac{4x+5}{4-4x^2} > \frac{5}{4}\) and \(\frac{4x+5}{4-4x^2} < -\frac{5}{4}\). Solving these inequalities gives the ranges \(\frac{2}{3} - \frac{1}{3}\sqrt{6} < x < -\frac{4}{3}\) and \(0 < x < \frac{2}{3} + \frac{1}{3}\sqrt{6}\), excluding \(x = \pm 1\).

(a) The matrix \(\begin{pmatrix} 6 & 0 \\ 0 & 6 \end{pmatrix}\) represents an enlargement with a scale factor of 6.

(b) The determinant of \(\mathbf{A}\) is \(\det(\mathbf{A}) = 3 \times 2 - 4 \times 2 = 6 - 8 = -2\). The area of triangle PQR is \(2 \times 13 = 26 \text{ cm}^2\).

(c) To find \(\mathbf{B}\), we need \(\mathbf{AB} = \begin{pmatrix} 6 & 0 \\ 0 & 6 \end{pmatrix}\). First, find \(\mathbf{A}^{-1} = -\frac{1}{2} \begin{pmatrix} 2 & -4 \\ -2 & 3 \end{pmatrix}\). Then, \(\mathbf{B} = -3 \begin{pmatrix} 2 & -4 \\ -2 & 3 \end{pmatrix} = \begin{pmatrix} -6 & 12 \\ 6 & -9 \end{pmatrix}\).

(d) For \(\mathbf{A} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}\), we have \(\begin{pmatrix} 3x + 4y \\ 2x + 2y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}\). This leads to the equations \(3x + 4y = x\) and \(2x + 2y = y\). Solving these gives \(y = -2x\) and \(-6x = 0\), leading to \(x = 0, y = 0\). Thus, the origin is the only invariant point.

First, differentiate \(y = xe^{ax}\) once using the product rule:

\(\frac{dy}{dx} = axe^{ax} + e^{ax} = (ax + 1)e^{ax}\), so true when \(n = 1\).

Assume that \(\frac{d^k y}{dx^k} = \left( a^k x + ka^{k-1} \right) e^{ax}\).

Differentiate the \(k\)-th derivative:

\(\frac{d^{k+1} y}{dx^{k+1}} = a \left( a^k x + ka^{k-1} \right) e^{ax} + e^{ax} \left( a^k \right) = \left( a^{k+1} + (k+1)a^k \right) e^{ax}\).

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

(a) Consider the sum \(\sum_{r=1}^{n} \left( \ln r - 2 \ln (r+1) + \ln (r+2) \right)\).

This expands to:

• \(\ln 1 - 2 \ln 2 + \ln 3\)
• \(\ln 2 - 2 \ln 3 + \ln 4\)
• \(\ln 3 - 2 \ln 4 + \ln 5\)
• \(\ldots\)
• \(\ln (n-1) - 2 \ln n + \ln (n+1)\)
• \(\ln n - 2 \ln (n+1) + \ln (n+2)\)

Most terms cancel, leaving:

\([\ln 1] - \ln 2 - \ln (n+1) + \ln (n+2) = \ln \frac{n+2}{2(n+1)}\).

(b) The sum to infinity is \(S = -\ln 2\).

We need \(S_n - S = \ln \left( \frac{n+2}{n+1} \right) < 0.01\).

This leads to \(n+2 < e^{0.01}(n+1)\).

Solving gives the least value of \(n\) as 99.

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

From the cubic equation, \(\alpha + \beta + \gamma = -2\) and \(\alpha\beta + \beta\gamma + \gamma\alpha = 3\).

Thus, \((-2)^2 - 2(3) = 4 - 6 = -2\).

(b) Using the formula for the sum of cubes of roots, \(\alpha^3 + \beta^3 + \gamma^3 = 3(\alpha\beta\gamma) + (\alpha + \beta + \gamma)(\alpha^2 + \beta^2 + \gamma^2 - \alpha\beta - \beta\gamma - \gamma\alpha)\).

\(\alpha\beta\gamma = -3\), so \(\alpha^3 + \beta^3 + \gamma^3 = 3(-3) + (-2)(-2 - 3) = -9 + 10 = 1\).

(c) Expanding \((\alpha + r)^3 = \alpha^3 + 3\alpha^2 r + 3\alpha r^2 + r^3\).

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

Using standard results, \(n - 6(\frac{1}{2}n(n+1)) - 6(\frac{1}{6}n(n+1)(2n+1)) + \frac{3}{4}n^2(n+1)^2\).

Simplifying gives \(n + \frac{1}{4}n(n+1)(-12 - 4(2n+1) + 3n(n+1))\).

Finally, \(n + \frac{1}{4}n(n+1)(3n^2 - 5n - 16)\).

(a) The curve \(C\) has polar equation \(r = 3 + 2\sin\theta\) for \(-\pi < \theta \le \pi\). It is a limacon, symmetric about the line \(\theta = \frac{\pi}{2}\); the completed sketch shows a single loop above the pole.

(b) The line \(l\) has polar equation \[ r\sin\theta = 2 \quad\Rightarrow\quad r = 2\csc\theta \quad (\sin\theta \ne 0). \] Points of intersection satisfy \[ 3 + 2\sin\theta \;=\; 2\csc\theta. \] Multiplying by \(\sin\theta\), \[ 3\sin\theta + 2\sin^2\theta = 2. \] Let \(\sin\theta = s\). Then \[ 2s^2 + 3s - 2 = 0 \quad\Rightarrow\quad s = \frac{-3 \pm 5}{4} \;\Rightarrow\; s = \frac12 \text{ or } s = -2. \] \(\sin\theta = -2\) is impossible, so \(\sin\theta = \frac12\), giving \[ \theta = \frac{\pi}{6},\; \frac{5\pi}{6}. \] At these points, \[ r = 3 + 2\sin\theta = 3 + 2\cdot\frac12 = 4. \] So the points of intersection are \[ \left(4,\; \frac{\pi}{6}\right),\quad \left(4,\; \frac{5\pi}{6}\right). \]

(c) The region \(R\) is enclosed by \(C\) and the line \(l\), and contains the pole. For a given \(\theta\), the boundary closest to the pole is:

• \(r = 3 + 2\sin\theta\) for \(-\pi < \theta \le \frac{\pi}{6}\) and \(\frac{5\pi}{6} \le \theta \le \pi\) (the curve is nearer than the line),
• \(r = 2\csc\theta\) for \(\frac{\pi}{6} \le \theta \le \frac{5\pi}{6}\) (the line is nearer than the curve).

Hence the area of \(R\) is \[ \text{Area}(R) = \frac12 \int_{-\pi}^{\pi/6} (3 + 2\sin\theta)^2\,d\theta + \frac12 \int_{\pi/6}^{5\pi/6} (2\csc\theta)^2\,d\theta + \frac12 \int_{5\pi/6}^{\pi} (3 + 2\sin\theta)^2\,d\theta. \] It is convenient to rewrite this as \[ \text{Area}(R) = \frac12 \int_{-\pi}^{\pi} (3 + 2\sin\theta)^2\,d\theta \;-\; \frac12 \int_{\pi/6}^{5\pi/6} \Bigl[(3 + 2\sin\theta)^2 - (2\csc\theta)^2\Bigr]\,d\theta. \]

First, \[ \int_{-\pi}^{\pi} (3 + 2\sin\theta)^2\,d\theta = \int_{-\pi}^{\pi} \bigl(9 + 12\sin\theta + 4\sin^2\theta\bigr)\,d\theta = 22\pi. \]

Next, \[ (3 + 2\sin\theta)^2 - (2\csc\theta)^2 = 9 + 12\sin\theta + 4\sin^2\theta - \frac{4}{\sin^2\theta}, \] so \[ \int_{\pi/6}^{5\pi/6} \Bigl[(3 + 2\sin\theta)^2 - (2\csc\theta)^2\Bigr]\,d\theta = 5\sqrt{3} + \frac{22\pi}{3}. \]

Therefore, \[ \text{Area}(R) = \frac12\left(22\pi - \left(5\sqrt{3} + \frac{22\pi}{3}\right)\right) = \frac12\left(\frac{44\pi}{3} - 5\sqrt{3}\right) = \frac{22}{3}\pi - \frac{5}{2}\sqrt{3}. \]

So the exact area of \(R\) is \[ \boxed{\displaystyle \frac{22}{3}\pi - \frac{5}{2}\sqrt{3}}. \]

(a) To find the vertical asymptote, set the denominator equal to zero: \(x - 3 = 0\), so \(x = 3\).
For the oblique asymptote, perform polynomial long division on \(\frac{x^2}{x-3}\):

\(x^2 \div (x-3) = x + 3 + \frac{9}{x-3}\).
As \(x \to \infty\), \(\frac{9}{x-3} \to 0\), so the oblique asymptote is \(y = x + 3\).

(b) Substitute \(y = \frac{x^2}{x-3}\) into \(yx - 3y = x^2\) to form a quadratic: \(x^2 - yx + 3y = 0\).
The discriminant \(b^2 - 4ac = y^2 - 4(3y) = y^2 - 12y\) must be negative for no real solutions.
\(y^2 - 12y < 0\) implies \(0 < y < 12\), showing no points exist for \(0 < y < 12\).

(c) Sketch the curve showing the vertical asymptote at \(x = 3\) and the oblique asymptote \(y = x + 3\). The curve approaches these asymptotes.

(d)(i) Sketch \(y = \left| \frac{x^2}{x-3} \right|\) and \(y = |x| - 3\).
Find intercepts: \(y = \left| \frac{x^2}{x-3} \right|\) intersects the y-axis at \((0,0)\).
\(y = |x| - 3\) intersects the y-axis at \((0,-3)\) and x-axis at \((3,0)\) and \((-3,0)\).

(d)(ii) From the sketch, \(\left| \frac{x^2}{x-3} \right| \leq |x| + c\) has no solution when \(c < -3\).

(a) To find the equation of the plane OAB, we need a normal vector. The vectors \(\overrightarrow{OA} = \begin{pmatrix} 2 \\ 2 \\ 0 \end{pmatrix}\) and \(\overrightarrow{OB} = \begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix}\) are in the plane. The cross product \(\overrightarrow{OA} \times \overrightarrow{OB}\) gives the normal vector \(\mathbf{n} = \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}\). The equation of the plane is \(\mathbf{r} \cdot \mathbf{n} = 0\).

(b) The perpendicular distance from the origin to the plane \(\Pi\) is given by \(\frac{|1|}{\sqrt{1^2 + (-3)^2 + (-2)^2}} = \frac{1}{\sqrt{14}}\).

(c) The angle between the planes is found using the dot product of their normals. The normal to OAB is \(\begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}\) and for \(\Pi\) is \(\begin{pmatrix} 1 \\ -3 \\ -2 \end{pmatrix}\). The cosine of the angle \(\alpha\) is \(\cos \alpha = \frac{\begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -3 \\ -2 \end{pmatrix}}{\sqrt{3} \sqrt{14}} = \frac{6}{\sqrt{42}}\). The angle is \(22.2^\circ\).

(d) To find the common perpendicular, we first find the direction vector by solving the system of equations derived from the dot product conditions. The direction vector is \(\begin{pmatrix} 5 \\ -3 \\ 1 \end{pmatrix}\). The position vector is \(\begin{pmatrix} -0.1 \\ 0.7 \\ 0.3 \end{pmatrix}\). Thus, the equation is \(\mathbf{r} = \begin{pmatrix} -0.1 \\ 0.7 \\ 0.3 \end{pmatrix} + \lambda \begin{pmatrix} 5 \\ -3 \\ 1 \end{pmatrix}\).