(a) Start by expressing \(\frac{1}{(kr+1)(kr-k+1)}\) using partial fractions:

\(\frac{1}{(kr+1)(kr-k+1)} = \frac{1}{k} \left( \frac{1}{kr-k+1} - \frac{1}{kr+1} \right)\)

Thus, the sum becomes:

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

This simplifies to:

\(\frac{1}{k} \left( 1 - \frac{1}{kn+1} \right)\)

(b) For the infinite sum, as \(n \to \infty\), the term \(\frac{1}{kn+1} \to 0\), so:

\(\sum_{r=1}^{\infty} \frac{1}{(kr+1)(kr-k+1)} = \frac{1}{k}\)

(c) Using the result from part (a), the sum from \(n\) to \(n^2\) is:

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

Using the formula from part (a):

\(= \frac{1}{k} \left( 1 - \frac{1}{kn^2+1} \right) - \frac{1}{k} \left( 1 - \frac{1}{k(n-1)+1} \right)\)

This simplifies to:

\(\frac{1}{k} \left( \frac{1}{k(n-1)+1} - \frac{1}{kn^2+1} \right)\)

(a) For \(\mathbf{M}\) to represent a rotation, it must be of the form \(\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}\). Comparing with \(\mathbf{M} = \begin{pmatrix} a & b^2 \\ c^2 & a \end{pmatrix}\), we have \(b^2 = -c^2\), which is impossible since \(b\) and \(c\) are real and \(b \neq 0\).

(b) The transformation of \(\begin{pmatrix} x \\ y \end{pmatrix}\) by \(\mathbf{M}\) gives \(\begin{pmatrix} ax + b^2y \\ c^2x + ay \end{pmatrix}\). For invariant lines through the origin, \(y = mx\) gives \(c^2x + amx = m(ax + b^2y)\). Simplifying, \(c^2 + am = ma + b^2m^2\) leads to \(c^2 = b^2m^2\), giving \(y = \frac{c}{b} x\) and \(y = -\frac{c}{b} x\).

(c) The enlargement matrix is \(\begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix}\) and the shear matrix is \(\begin{pmatrix} 1 & 5 \\ 0 & 1 \end{pmatrix}\). Thus, \(\mathbf{M} = \begin{pmatrix} 1 & 5 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} = \begin{pmatrix} 5 & 5^2 \\ 0 & 5 \end{pmatrix}\).

(d) The area of \(PQR\) is \(12 \times \det(\mathbf{M})\). \(\det(\mathbf{M}) = 5 \times 5 - 0 \times 25 = 25\). Therefore, the area is \(12 \times 25 = 300 \text{ cm}^2\).

(a) The sketch shows a polar graph with \(\theta = 0\) at the pole and the curve extending to \(\theta = \pi\). The point furthest from the pole is \((1, 0)\).

(b) The area \(A\) is given by:

\(A = \frac{1}{2} \int_{0}^{\pi} \frac{1}{\theta^2 + 1} \, d\theta\)

\(= \frac{1}{2} \left[ \arctan \theta \right]_{0}^{\pi}\)

\(= \frac{1}{2} \arctan \pi = 0.631\)

(c) Let \(y = \frac{\sin \theta}{\sqrt{\theta^2 + 1}}\).

Differentiate and set \(\frac{dy}{d\theta} = 0\):

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

\(\theta \neq 0 \Rightarrow \left( \theta + \frac{1}{\theta} \right) \cot \theta - 1 = 0\)

Check for root between 1.1 and 1.2:

\(\left( 1.1 + \frac{1}{1.1} \right) \cot 1.1 - 1 = 0.02 \ldots\)

\(\left( 1.2 + \frac{1}{1.2} \right) \cot 1.2 - 1 = -0.209 \ldots\)

There is a sign change, confirming a root exists between 1.1 and 1.2.

(a) The vertical asymptote occurs where the denominator is zero: \(x = 2\). For the oblique asymptote, divide the numerator by the denominator: \(x^2 + 2x - 15 = (x - 2)(x + 4) - 7\), giving \(y = x + 4\).

(b) Differentiate \(y\) to find stationary points: \(\frac{dy}{dx} = \frac{(x - 2)(2x + 2) - (x^2 + 2x - 15)}{(x - 2)^2} = \frac{x^2 - 4x + 11}{(x - 2)^2}\). The quadratic \(x^2 - 4x + 11 = 0\) has no real roots (discriminant \(4^2 - 4 \times 1 \times 11 = -28 < 0\)), so no stationary points.

(c) The curve intersects the y-axis at \(x = 0\): \(y = \frac{0^2 + 2 \times 0 - 15}{0 - 2} = 7.5\). Intersects the x-axis where \(y = 0\): \(x^2 + 2x - 15 = 0\), giving roots \(x = -5\) and \(x = 3\).

(d) Sketch the absolute value function based on the sketch from (c), reflecting the negative parts above the x-axis.

(e) Solve \(\left| \frac{2x^2 + 4x - 30}{x - 2} \right| < 15\). Critical points from \(2x^2 + 4x - 30 = 15(x - 2)\) and \(2x^2 + 4x - 30 = -15(x - 2)\) give \(x = 0, \frac{11}{2}, -12, \frac{5}{2}\). Test intervals to find \(-12 < x < 0\) or \(\frac{5}{2} < x < \frac{11}{2}\).

(a) The direction vectors of \(\Pi_1\) are \(\begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}\) and \(\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}\). The normal vector is the cross product:

\(\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} = \begin{pmatrix} 0 \\ 2 \\ 2 \end{pmatrix}\)

Substitute \(\mathbf{r} = \begin{pmatrix} 0 \\ -4 \\ -3 \end{pmatrix}\) into \(\mathbf{n} \cdot \mathbf{r} = d\):

\(0 \cdot 0 + 2(-4) + 2(-3) = d\)

\(d = -8 - 6 = -14\)

Equation: \(0x + 2y + 2z = -14\) simplifies to \(y + z = -7\).

(b) A point on both planes is \(\begin{pmatrix} -7 \\ -2 \\ -5 \end{pmatrix}\). The direction vector is the cross product of normals:

\(\mathbf{n_1} = \begin{pmatrix} 0 \\ 2 \\ 2 \end{pmatrix}, \mathbf{n_2} = \begin{pmatrix} -5 \\ 3 \\ 5 \end{pmatrix}\)

\(\mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 2 & 2 \\ -5 & 3 & 5 \end{vmatrix} = \begin{pmatrix} 2 \\ -5 \\ 5 \end{pmatrix}\)

Vector equation: \(\mathbf{r} = \begin{pmatrix} -7 \\ -2 \\ -5 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -5 \\ 5 \end{pmatrix}\).

(c) Distance from \(A\) to \(\Pi_1\):

\(\frac{|a + a - 7 + 7|}{\sqrt{2}} = \sqrt{2}\)

\(|2a| = 2 \Rightarrow a = 1\)

(d) Angle between \(l\) and \(\Pi_1\):

\(\frac{|b + b|}{\sqrt{2((1-b)^2 + 2b^2)}} = \frac{1}{2} \sqrt{2}\)

\(2b = \sqrt{(1-b)^2 + 2b^2} \Rightarrow b^2 + 2b - 1 = 0\)

\(b = -1 + \sqrt{2}\)