(a) The transformation matrix for a stretch in the x-direction by a factor of 3 is \(\begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix}\). The reflection in the line y = x is represented by \(\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\). Therefore, \(\mathbf{M} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 3 & 0 \end{pmatrix}\).

(b) The inverse of \(\mathbf{M}\) is \(\mathbf{M}^{-1} = \begin{pmatrix} 0 & \frac{1}{3} \\ 1 & 0 \end{pmatrix}\). This represents a reflection in the line y = x followed by a stretch in the x-direction by a factor of \(\frac{1}{3}\).

(c) Given \(\mathbf{MN} = \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix}\), we find \(\mathbf{N}\) by multiplying by \(\mathbf{M}^{-1}\) on the left: \(\mathbf{N} = \mathbf{M}^{-1} \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix} = \begin{pmatrix} 0 & \frac{1}{3} \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix} = \begin{pmatrix} 1 & \frac{1}{2} \\ 1 & 2 \end{pmatrix}\).

(d) The invariant lines satisfy \(\begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x + 2y \\ 3x + 2y \end{pmatrix}\). Setting \(y = mx\), we have \(3x + 2mx = m(x + 2y)\). This leads to the quadratic \(2m^2 - m - 3 = 0\), which factors to \((2m + 3)(m - 1) = 0\). Thus, \(m = \frac{3}{2}\) or \(m = -1\), giving the lines \(y = \frac{3}{2}x\) and \(y = -x\).

(a) The curve is a spiral shape on \([0, 2\pi]\), starting at the pole with one other point of intersection with the initial line. The maximum distance from the pole is \(a\).

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

\(A = \frac{1}{2} \int_0^{2\pi} r^2 \, d\theta = \frac{1}{2} a^2 \int_0^{2\pi} \tan^2\left(\frac{1}{8}\theta\right) \, d\theta\)

Using \(\tan^2 A = \sec^2 A - 1\), we have:

\(= \frac{1}{2} a^2 \left[ 8 \tan\left(\frac{1}{8}\theta\right) - \theta \right]_0^{2\pi}\)

\(= \frac{1}{2} a^2 (8 - 2\pi) = a^2 (4 - \pi)\)

(c) Let \(y = a \tan\left(\frac{1}{8}\theta\right) \sin\theta\). The derivative is:

\(\frac{dy}{d\theta} = a \left( \tan\left(\frac{1}{8}\theta\right) \cos\theta + \frac{1}{8} \sin\theta \sec^2\left(\frac{1}{8}\theta\right) \right) = 0\)

\(\Rightarrow \cos\left(\frac{1}{8}\theta\right) \sin\left(\frac{1}{8}\theta\right) \cos\theta + \frac{1}{8} \sin\theta = 0\)

\(\Rightarrow \sin\left(\frac{1}{4}\theta\right) \cos\theta + \frac{1}{4} \sin\theta = 0\)

\(\Rightarrow 4 \sin\left(\frac{1}{4}\theta\right) \cos\theta + \sin\theta = 0\)

Checking values:

\(4 \sin\left(\frac{1}{4} \times 4.95\right) \cos 4.95 + \sin 4.95 \approx -0.0822\)

\(4 \sin\left(\frac{1}{4} \times 5\right) \cos 5 + \sin 5 \approx 0.118\)

There is a sign change, confirming a root between 4.95 and 5.

(a) The horizontal asymptote is found by considering the limits as \(x \to \pm \infty\). The leading coefficients of the numerator and denominator are equal, so \(y = 1\) is the horizontal asymptote.

(b) Consider \(y = \frac{x^2 + x - 4}{x^2 + x + 2}\). Rearrange to form a quadratic in \(x\): \((y-1)x^2 + (y-1)x + (2y + 4) = 0\). The discriminant \(\Delta = (y-1)^2 - 4(y-1)(2y+4) \geq 0\) gives \((y-1)(7y + 17) \leq 0\). Solving this inequality gives \(-\frac{17}{7} \leq y < 1\).

(c) Find stationary points by setting \(\frac{dy}{dx} = 0\). Differentiate: \(\frac{dy}{dx} = \frac{12x + 6}{(x^2 + x + 2)^2} = 0\). Solving gives \(x = -\frac{1}{2}\). Substitute back to find \(y = -\frac{17}{7}\). Stationary point is \(\left(-\frac{1}{2}, -\frac{17}{7}\right)\).

(d) Sketch the curve, noting the asymptote \(y = 1\) and intersections at \((0, -2), \left(\frac{1}{2}(1-\sqrt{7}), 0\right), \left(\frac{1}{2}(1+\sqrt{7}), 0\right)\).

(e) For \(y = \frac{|x|^2 + |x| - 4}{|x|^2 + |x| + 2} < -\frac{1}{2}\), solve \(3|x|^2 + 3|x| - 6 = 0\). Critical points are \(|x| = 1\). Thus, \(-1 < x < 1\).

(a) The transformation matrix for a stretch parallel to the x-axis with scale factor k is \(\begin{pmatrix} k & 0 \\ 0 & 1 \end{pmatrix}\). The shear transformation with the x-axis fixed and (0, 1) mapped to (k, 1) is \(\begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}\). The combined transformation matrix M is the product:

\(\mathbf{M} = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix} \begin{pmatrix} k & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} k & k \\ 0 & 1 \end{pmatrix}\)

(b) The transformation \(\mathbf{M} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} kx + ky \\ y \end{pmatrix}\) implies invariant points satisfy \(kx + ky = x\). Solving gives \(y = \frac{1-k}{k}x\).

(c) The inverse matrix \(\mathbf{M}^{-1}\) is found by inverting \(\mathbf{M}\):

\(\mathbf{M}^{-1} = \frac{1}{k} \begin{pmatrix} 1 & -k \\ 0 & k \end{pmatrix}\)

(d) The area of P is given by the determinant of M, \(|k| = 3k^2\). Solving \(|k| = 3k^2\) gives \(k = \pm \frac{1}{3}\).

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

\(\frac{d}{dx}(\arctan x) = \frac{1}{1+x^2},\)

which is true.

Inductive step: Assume

\(\frac{d^k}{dx^k}(\arctan x) = P_k(x)(1+x^2)^{-k},\)

where \(\deg P_k(x) = k-1\).

Then,

\(\frac{d^{k+1}}{dx^{k+1}}(\arctan x) = P_k'(x)(1+x^2)^{-k} - 2kxP_k(x)(1+x^2)^{-k-1}.\)

This can be written as

\(\left(P_k'(x)(1+x^2) - 2kxP_k(x)\right)(1+x^2)^{-k-1},\)

which implies \(\deg P_{k+1}(x) = k\).

Thus, the statement is true for \(n = k+1\). By induction, it is true for all positive integers \(n\).