(a) The matrix \(\mathbf{M} = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 7 & 0 \\ 0 & 1 \end{pmatrix}\) represents a stretch followed by a shear. The first matrix \(\begin{pmatrix} 7 & 0 \\ 0 & 1 \end{pmatrix}\) is a stretch parallel to the \(x\)-axis with a scale factor of 7. The second matrix \(\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\) is a shear with the \(x\)-axis fixed, mapping \((0,1)\) to \((2,1)\).

(b) To find the invariant lines, consider \(\mathbf{M} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 7x + 2y \\ y \end{pmatrix}\). For a line \(y = mx\) to be invariant, \(mx = m(7x + 2mx)\). This simplifies to \(2m^2 + 6m = 0\), giving solutions \(m = 0\) and \(m = -3\). Thus, the invariant lines are \(y = 0\) and \(y = -3x\).

(c) The area of \(PQR\) is \(35 \text{ cm}^2\). The area of \(DEF\) is related by the determinant of the transformation matrix \(\mathbf{M}\), which is \(|7| = 7\). Therefore, \(\text{Area of } DEF = \frac{35}{7} = 5 \text{ cm}^2\).

(a) Base case: For \(n = 1\), \(\sum_{r=1}^{1} r^2 = 1^2 = 1\) and \(\frac{1}{6} \times 1 \times 2 \times 3 = 1\). So \(H_1\) is true.

Inductive step: Assume \(\sum_{r=1}^{k} r^2 = \frac{1}{6}k(k+1)(2k+1)\) is true for \(n = k\).

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

\(= \frac{1}{6}k(k+1)(2k+1) + (k+1)^2\)

\(= \frac{1}{6}(k+1)(2k^2 + k + 6k + 6)\)

\(= \frac{1}{6}(k+1)(2k^2 + 7k + 6)\)

\(= \frac{1}{6}(k+1)(k+2)(2k+3)\)

So \(H_{k+1}\) is true. By induction, \(H_n\) is true for all positive integers \(n\).

(b) Using the identity, expand and simplify:

\((2n+1)^5 - 1\)

\(= 160S_n + \frac{40}{3}n(n+1)(2n+1) + 2n\)

\(160S_n = (2n+1)^5 - \frac{40}{3}n(n+1)(2n+1) - (2n+1)\)

\(160S_n = (2n+1)((2n+1)^4 - \frac{40}{3}n(n+1) - 1)\)

\(\Rightarrow S_n = \frac{1}{30}n(n+1)(2n+1)(3n^2 + 3n - 1)\)

(c) \(n^{-5}S_n = \frac{1}{30}(1 + \frac{1}{n})(2 + \frac{1}{n})(3 + \frac{3}{n} - \frac{1}{n^2})\)

As \(n \to \infty\), \(n^{-5}S_n \to \frac{1}{5}\).

(a) The direction vectors of \(l_1\) and \(l_2\) are \(\begin{pmatrix} 0 \\ 1 \\ -2 \end{pmatrix}\) and \(\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}\) respectively. The vector between points on \(l_1\) and \(l_2\) is \(\begin{pmatrix} -4 \\ 0 \\ 1 \end{pmatrix}\). The cross product of the direction vectors is \(\begin{pmatrix} -2 \\ -2 \\ -1 \end{pmatrix}\). The shortest distance is given by \(\frac{1}{\sqrt{30}} \begin{vmatrix} -4 & 0 & 1 \\ -2 & -2 & -1 \end{vmatrix} = \frac{21}{\sqrt{30}}\).

(b) The normal to \(\Pi_1\) is the cross product of the direction vector of \(l_1\) and the direction vector of \(l_2\), which is \(\begin{pmatrix} -2 \\ -2 \\ -1 \end{pmatrix}\). Using the point \((1, 4, -1)\) on \(l_1\), the equation of \(\Pi_1\) is \(5x - 2y - z = -2\).

(c) The direction vector of the line of intersection is \(\begin{pmatrix} 1 \\ 4 \\ -3 \end{pmatrix}\). The normal to \(\Pi_2\) is \(\begin{pmatrix} -1 \\ -1 \\ 1 \end{pmatrix}\). Using the point \((1, 1, 1)\), the equation of \(\Pi_2\) is \(x + 2y + 3z = 6\).

(a) To find vertical asymptotes, set the denominator equal to zero: \(x^2 + 3 = 0\). This equation has no real roots, so there are no vertical asymptotes. The horizontal asymptote is found by considering the limit as \(x\) approaches infinity: \(y = \frac{x+1}{x^2+3} \to 0\). Thus, the horizontal asymptote is \(y = 0\).

(b) To find stationary points, differentiate \(y\) with respect to \(x\): \(\frac{dy}{dx} = \frac{(x^2+3) - (x+1)(2x)}{(x^2+3)^2}\). Set \(\frac{dy}{dx} = 0\) to find \(x^2 + 2x - 3 = 0\). Solving this quadratic equation gives \(x = -3\) and \(x = 1\). Substituting back into the original equation gives the stationary points \((-3, -\frac{1}{6})\) and \((1, \frac{1}{2})\).

(c) The intersections with the axes occur when \(y = 0\) and \(x = 0\). For \(y = 0\), solve \(x+1 = 0\) to get \(x = -1\), so the intersection is \((-1, 0)\). For \(x = 0\), \(y = \frac{1}{3}\), so the intersection is \((0, \frac{1}{3})\).

(d) For \(y^2 = \frac{x+1}{x^2+3}\), the intersections with the axes are the same as in part (c), but \(y\) can be positive or negative. Thus, the intersections are \((-1, 0)\) and \((0, \pm \frac{1}{3})\). The stationary points are \((1, \pm \frac{1}{2})\).

(a) The sketch of the curve shows two loops symmetric about the line \(\theta = \frac{1}{2} \pi\).

(b) Start with \(r^2 = \sin 2\theta \cos \theta\). Use \(\sin 2\theta = 2 \sin \theta \cos \theta\) and polar to Cartesian conversions: \(x = r \cos \theta\), \(y = r \sin \theta\). Thus, \(r^5 = 2x^2\). Therefore, \(\left( x^2 + y^2 \right)^{\frac{3}{2}} = 2x^2 y\).

(c) The total area enclosed by the curve is given by \(\frac{1}{2} \int_0^\pi \sin(2\theta) \cos \theta \, d\theta\). Simplifying, \(\int \sin \theta \cos^2 \theta \, d\theta = -\frac{1}{3} \cos^3 \theta\). Evaluating from 0 to \(\pi\), the area is \(\frac{2}{3}\).

(d) Differentiate \(r^2 = \sin 2\theta \cos \theta\) with respect to \(\theta\) to find the maximum distance. Solve \(6 \cos^3 \theta - 4 \cos \theta = 0\) to find \(\cos^2 \theta = \frac{2}{3}\). The greatest distance is \(r = \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{9} \approx 0.877\).