(a) Let \(y = 3x + 1\), then \(x = \frac{1}{3}(y - 1)\).

Substitute into the original equation:

\(27\left(\frac{y-1}{3}\right)^3 + 18\left(\frac{y-1}{3}\right)^2 + 6\left(\frac{y-1}{3}\right) - 1 = 0\)

\(\Rightarrow (y-1)^3 + 2(y-1)^2 + 2(y-1) - 1 = 0\)

\(\Rightarrow y^3 - 3y^2 + 3y - 1 + 2y^2 - 4y + 2 + 2y - 2 - 1 = 0\)

\(\Rightarrow y^3 - y^2 + y - 2 = 0\)

(b) \(S_2 = 1^2 - 2(1) = -1\)

\(S_3 = (3\alpha + 1)^3 + (3\beta + 1)^3 + (3\gamma + 1)^3 = -1 - (1) + 6 = 4\)

(c) \(S_{-1} = \frac{(3\alpha + 1)(3\beta + 1) + (3\beta + 1)(3\gamma + 1) + (3\gamma + 1)(3\alpha + 1)}{(3\alpha + 1)(3\beta + 1)(3\gamma + 1)} = \frac{1}{2}\)

\(2S_{-2} = S_1 + S_{-1} = 1 - 3 + \frac{1}{2}\)

\(S_{-2} = -\frac{3}{4}\)

(a) The normal vector to \(\Pi_1\) can be found using the cross product of the direction vectors \(\mathbf{i} - 2\mathbf{j} - 3\mathbf{k}\) and \(3\mathbf{i} - \mathbf{k}\). This gives \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -2 & -3 \\ 3 & 0 & -1 \end{vmatrix} = \mathbf{i}(2 - 0) - \mathbf{j}(-1 - (-9)) + \mathbf{k}(0 - (-6)) = 2\mathbf{i} - 8\mathbf{j} + 6\mathbf{k}\).

Using the point \((1, -1, -2)\), the equation is \(1(x - 1) - 4(y + 1) + 3(z + 2) = 0\), simplifying to \(x - 4y + 3z = -1\).

(b) The direction vector of \(l\) is \(\mathbf{i} + \mathbf{j} + \mathbf{k}\). The dot product with the normal vector \(1\mathbf{i} - 4\mathbf{j} + 3\mathbf{k}\) is \(1(1) - 4(1) + 3(1) = 0\), showing \(l\) is parallel to \(\Pi_1\).

(c) The distance from a point on \(l\) to \(\Pi_1\) is \(\frac{|(-3)(1) + 0(-4) + 1(3) + 1|}{\sqrt{1^2 + (-4)^2 + 3^2}} = \frac{1}{\sqrt{26}}\).

(d) A point common to both planes is found by solving \(x - 4y + 3z = -1\) and \(3x + 3y + 2z = 1\). Solving these gives a point \(\begin{pmatrix} \frac{5}{7} \\ 0 \\ -\frac{4}{7} \end{pmatrix}\). The direction vector of the line of intersection is the cross product of the normals \(\begin{pmatrix} 1 \\ -4 \\ 3 \end{pmatrix}\) and \(\begin{pmatrix} 3 \\ 3 \\ 2 \end{pmatrix}\), which is \(\begin{pmatrix} -17 \\ 7 \\ 15 \end{pmatrix}\). Thus, the line is \(\mathbf{r} = \begin{pmatrix} \frac{5}{7} \\ 0 \\ -\frac{4}{7} \end{pmatrix} + \lambda \begin{pmatrix} -17 \\ 7 \\ 15 \end{pmatrix}\).

(a) The curve is sketched with the initial line drawn. The shape is correct with \(r\) strictly decreasing. The greatest distance from the pole is \(1 - e^{-\frac{1}{2}\pi}\).

(b) The area is calculated using the integral:

\(\frac{1}{2} \int_{0}^{\frac{1}{2}\pi} \left( e^{-\theta} - e^{-\frac{1}{2}\pi} \right)^2 \, d\theta\)

Expanding and integrating:

\(\frac{1}{2} \int_{0}^{\frac{1}{2}\pi} \left( e^{-2\theta} - 2e^{-\theta - \frac{1}{2}\pi} + e^{-\pi} \right) \, d\theta\)

\(\frac{1}{2} \left[ -\frac{1}{2}e^{-2\theta} + 2e^{-\theta - \frac{1}{2}\pi} + e^{-\pi}\theta \right]_{0}^{\frac{1}{2}\pi}\)

\(\frac{1}{2} \left( -\frac{1}{2}e^{-\pi} + 2e^{-\pi} + \frac{1}{2}\pi e^{-\pi} - 2e^{-\frac{1}{2}\pi} \right) = \frac{3}{4}e^{-\pi} + \frac{1}{4}\pi e^{-\pi} - e^{-\frac{1}{2}\pi} + \frac{1}{4}\)

(c) Using \(y = (e^{-\theta} - e^{-\frac{1}{2}\pi}) \sin \theta\), differentiate:

\(\frac{dy}{d\theta} = (e^{-\theta} - e^{-\frac{1}{2}\pi}) \cos \theta + \sin \theta (-e^{-\theta}) = 0\)

\(\left[ \theta = \frac{1}{2}\pi \Rightarrow 1 + \frac{-e^{-\theta}}{e^{-\theta} - e^{-\frac{1}{2}\pi}} \right] \tan \theta = 0 \Rightarrow 1 - e^{\theta - \frac{1}{2}\pi} - \tan \theta = 0\)

Checking values:

\(1 - e^{0.56 - \frac{1}{2}\pi} - \tan 0.56 = 0.00912\)

\(1 - e^{0.57 - \frac{1}{2}\pi} - \tan 0.57 = -0.00856\)

There is a sign change, confirming a root between 0.56 and 0.57.

(a) The vertical asymptote occurs where the denominator is zero, so \(x = -1\). For the oblique asymptote, perform polynomial long division on \(\frac{x^2}{x+1}\) to get \(y = x - 1\).

(b) To find stationary points, set \(\frac{dy}{dx} = 0\). Differentiate \(y = \frac{x^2}{x+1}\) to get \(\frac{dy}{dx} = \frac{x^2 + 2x}{(x+1)^2}\). Solve \(x^2 + 2x = 0\) to find \(x = 0\) and \(x = -2\). Substitute back to find \(y\) values: \((0, 0)\) and \((-2, -4)\).

(c) Sketch the curve using the asymptotes and stationary points. The curve approaches the asymptotes and passes through the stationary points.

(d) For \(y = \frac{1}{f(x)}\), differentiate and set \(\frac{dy}{dx} = 0\) to find stationary points. The stationary point is \((-2, -\frac{1}{4})\).

(e) Sketch \(y = \frac{1}{f(x)}\) using the asymptotes and stationary points. Solve \(\frac{1}{f(x)} > f(x)\) by setting \(\frac{x^2}{x+1} = 1\) and \(\frac{x^2}{x+1} = -1\). Solve \(x^2 - x - 1 = 0\) to find critical points \(x = \frac{1}{2} \pm \frac{1}{2} \sqrt{5}\). The solution set is \(x < -1, -\frac{1}{2} \sqrt{5} < x < \frac{1}{2} \sqrt{5}, x \neq 0\).

(a) Consider \((r+1)^2 - r^2 = r^2 + 2r + 1 - r^2 = 2r + 1\).

Using the method of differences, sum both sides from \(r=1\) to \(n\):

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

This simplifies to:

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

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

(b) Given \(\sum_{r=1}^{n} (r+a) = \sum_{r=1}^{n} r + an = n\).

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

\(\frac{1}{2} n(n+1) + an = n.\)

Solving for \(a\):

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

\(a = \frac{1}{2} (1-n).\)