(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 \(\mathbf{PQ}\):

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