(a) Start with the Cartesian equation \(\left( x - \frac{1}{2} \right)^2 + y^2 = \frac{1}{4}\). Expand and rearrange to get \(x^2 - x + \frac{1}{4} + y^2 = \frac{1}{4}\), which simplifies to \(x^2 + y^2 = x\). In polar coordinates, \(x = r \cos \theta\) and \(x^2 + y^2 = r^2\). Substitute to get \(r^2 = r \cos \theta\), which simplifies to \(r(r - \cos \theta) = 0\). Since \(r \neq 0\), we have \(r = \cos \theta\).

(b) Set \(\sin 2\theta = \cos \theta\). Using the identity \(\sin 2\theta = 2\sin \theta \cos \theta\), we have \(2\sin \theta \cos \theta = \cos \theta\). Assuming \(\cos \theta \neq 0\), divide by \(\cos \theta\) to get \(2\sin \theta = 1\), so \(\sin \theta = \frac{1}{2}\). Therefore, \(\theta = \frac{\pi}{6}\). The polar coordinates of \(P\) are \(\left( \frac{1}{2}, \frac{\sqrt{3}}{6} \pi \right)\).

(c) Sketch the curves \(C_1: r = \cos \theta\) and \(C_2: r = \sin 2\theta\), marking the intersection point \(P\).

(d) The area \(R\) is given by \(\frac{1}{2} \int_{0}^{\frac{\pi}{6}} \sin^2 2\theta \, d\theta + \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \cos^2 \theta \, d\theta\). Use the identities \(\sin^2 2\theta = \frac{1}{2}(1 - \cos 4\theta)\) and \(\cos^2 \theta = \frac{1}{2}(1 + \cos 2\theta)\) to integrate. The result is \(\frac{1}{4} (\pi - 3\sqrt{3})\).

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

(b) To find stationary points, set \(\frac{dy}{dx} = 0\). The derivative is \(\frac{dy}{dx} = \frac{(x^2 - x - 2)(2x) - (x^2 + 2)(2x - 1)}{(x^2 - x - 2)^2}\). Solving \(x^2 + 8x - 2 = 0\) gives the stationary points \((-8.2, 0.9)\) and \((0.2, -0.9)\).

(c) The sketch shows the curve with asymptotes at \(x = -1\), \(x = 2\), and \(y = 1\). The curve intersects the y-axis at \((0, -1)\).

(d) The sketch of \(y = \frac{1}{f(x)}\) is derived from the sketch in (c), showing the reciprocal relationship.

(e) Solve \(\frac{x^2 + 2}{x^2 - x - 2} = 1\) and \(\frac{x^2 + 2}{x^2 - x - 2} = -1\) to find critical points. The solution gives the intervals \(-4 < x < -1\), \(0 < x < \frac{1}{2}\), and \(x > 2\).

(a) Use the standard results for summation: \(\sum_{r=1}^{n} r = \frac{n(n+1)}{2}\) and \(\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}\).

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

Substitute the standard results: \(3 \left( \frac{n(n+1)(2n+1)}{6} \right) + 3 \left( \frac{n(n+1)}{2} \right) + n\).

Simplify to get \(\frac{1}{2} n(n+1)(2n+1) + \frac{3}{2} n(n+1) + n = n^3 + 3n^2 + 3n\).

(b) Start with \(\frac{1}{r^3} - \frac{1}{(r+1)^3} = \frac{(r+1)^3 - r^3}{r^3 (r+1)^3}\).

Expand \((r+1)^3 - r^3 = r^3 + 3r^2 + 3r + 1 - r^3 = 3r^2 + 3r + 1\).

Thus, \(\frac{1}{r^3} - \frac{1}{(r+1)^3} = \frac{3r^2 + 3r + 1}{r^3 (r+1)^3}\).

Use the method of differences: \(\sum_{r=1}^{n} \left( \frac{1}{r^3} - \frac{1}{(r+1)^3} \right) = 1 - \frac{1}{(n+1)^3}\).

(c) As \(n \to \infty\), \(\frac{1}{(n+1)^3} \to 0\), so the sum converges to 1.

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

\(\frac{d}{dx} \left( x^2 e^x \right) = x^2 e^x + 2x e^x = (x^2 + 2x) e^x.\)

So true when \(n = 1\).

Inductive step: Assume that

\(\frac{d^k}{dx^k} \left( x^2 e^x \right) = \left( x^2 + 2kx + k(k-1) \right) e^x\)

for some value of \(k\).

Then,

\(\frac{d^{k+1}}{dx^{k+1}} \left( x^2 e^x \right) = \left( x^2 + 2kx + k(k-1) \right) e^x + e^x (2x + 2k).\)

Simplifying gives:

\(\left( x^2 + 2(k+1)x + k(k+1) \right) e^x.\)

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

(a) The matrix \(\mathbf{M}\) is the product of two matrices: \(\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}\) represents a shear parallel to the x-axis, and \(\begin{pmatrix} k & 0 \\ 0 & 1 \end{pmatrix}\) represents a stretch parallel to the x-axis by a factor of \(k\). The shear is applied first, followed by the stretch.

(b) The area of the parallelogram \(OPQR\) is given by the absolute value of the determinant of \(\mathbf{M}\). Calculating the determinant: \(\det(\mathbf{M}) = \det\left(\begin{pmatrix} k & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}\right) = k\). Thus, \(|OPQR| = |k|\).

The inverse matrix \(\mathbf{M}^{-1}\) is calculated as follows: \(\mathbf{M}^{-1} = \frac{1}{k} \begin{pmatrix} 1 & 0 \\ -1 & k \end{pmatrix}\).

(c) To show the line is invariant, consider the transformation of a point \(\begin{pmatrix} x \\ \frac{1}{k-1}x \end{pmatrix}\) under \(\mathbf{M}\):

\(\begin{pmatrix} k & 0 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ \frac{1}{k-1}x \end{pmatrix} = \begin{pmatrix} kx \\ x + \frac{1}{k-1}x \end{pmatrix} = \begin{pmatrix} kx \\ \frac{k}{k-1}x \end{pmatrix} = k \begin{pmatrix} x \\ \frac{1}{k-1}x \end{pmatrix}\).

This shows that the line \(y = \frac{1}{k-1}x\) is invariant under the transformation.