(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 \(p = \frac{1}{12}\).