(a) To find vertical asymptotes, set the denominator equal to zero: \(x^2 + 1 = 0\). This has no real roots, so there are no vertical asymptotes. The horizontal asymptote is found by considering the limits as \(x \to \pm \infty\), giving \(y = 1\).

(b) Rewrite \(y = 1 + \frac{2}{x^2 + 1}\). Since \(\frac{2}{x^2 + 1} > 0\), it follows that \(y > 1\). The maximum value of \(y\) occurs when \(\frac{2}{x^2 + 1}\) is maximized, which is 2, giving \(y = 3\). Thus, \(1 < y \leq 3\).

(c) Differentiate \(y = \frac{x^2 + 3}{x^2 + 1}\) to find stationary points: \(\frac{d}{dx} \left( \frac{x^2 + 3}{x^2 + 1} \right) = \frac{(x^2 + 1)(2x) - (x^2 + 3)(2x)}{(x^2 + 1)^2} = 0\). Solving gives \(x = 0\), and substituting back gives \(y = 3\). Thus, the stationary point is \((0, 3)\).

(d) The sketch shows the curve with a horizontal asymptote at \(y = 1\) and a stationary point at \((0, 3)\). The curve intersects the y-axis at \((0, 3)\).

(e) For \(y = \frac{x^2 + 1}{x^2 + 3} < \frac{1}{2}\), solve \(2(x^2 + 1) < x^2 + 3\), giving \(x^2 < 1\). Thus, \(-1 < x < 1\).

(a) The polar equation is \(r = a(\cos \theta + \sin \theta)\). Using \(x = r\cos \theta\) and \(y = r\sin \theta\), we have:

\(x = a\cos \theta (\cos \theta + \sin \theta) = a(\cos^2 \theta + \cos \theta \sin \theta)\)

\(y = a\sin \theta (\cos \theta + \sin \theta) = a(\sin \theta \cos \theta + \sin^2 \theta)\)

Adding these, \(x + y = a(\cos^2 \theta + 2\cos \theta \sin \theta + \sin^2 \theta) = a\).

Thus, \(x + y = a\). The Cartesian equation is \(x^2 + y^2 = ax + ay\), which simplifies to \((x - \frac{a}{2})^2 + (y - \frac{a}{2})^2 = (\frac{a}{\sqrt{2}})^2\).

(b) The greatest distance from the pole is when \(\theta = \frac{\pi}{4}\), giving \(r = a\sqrt{2}\).

(c) To verify \(1.25 < \phi < 1.26\), solve \(\cos \phi + \sin \phi - \phi = 0\). Check values: \(\cos 1.25 + \sin 1.25 - 1.25 = 0.01\) and \(\cos 1.26 + \sin 1.26 - 1.26 = -0.002\), showing a sign change.

(d) The area of the smaller region is:

\(\frac{1}{2} \int_0^\phi \theta^2 \, d\theta + \frac{a^2}{2} \int_\phi^{\frac{3}{4}\pi} (\cos \theta + \sin \theta)^2 \, d\theta\)

Calculate each integral and simplify to get:

\(\frac{1}{2}a^2 \left( \frac{3}{4}\pi + \frac{1}{3}\phi^3 - \phi + \frac{1}{2}\cos 2\phi \right)\)

The area of the larger region is:

\(\frac{a^2}{2} \left( \frac{\pi}{3} + \frac{1}{3}\phi - \frac{1}{2}\cos 2\phi \right)\)

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

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

(b) The line of invariant points satisfies \(\begin{pmatrix} k & k \\ 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}\). This gives:

\(kx + ky = x\) and \(y = y\)

Solving \(kx + ky = x\) gives \(y = \frac{1-k}{k}x\).

(c) The inverse of M is:

\(\text{If } \text{det}(\textbf{M}) = k, \text{ then } \textbf{M}^{-1} = \frac{1}{k} \begin{pmatrix} 1 & -k \\ 0 & k \end{pmatrix}\)

(d) Given \(|k| = 3k^2\), solving gives \(k = \pm \frac{1}{3}\).

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

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

which is true.

Inductive step: Assume

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

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

Then,

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

This can be written as

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

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

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

(a) Let \(y = x^4\). Then the equation becomes \(y + 2y^{3/4} - 1 = 0\). Substitute \(y^{3/4} = (1-y)^4\) to get \(16y^3 = 1 - 4y + 6y^2 - 4y^3 + y^4\). Rearrange to form \(y^4 - 20y^3 + 6y^2 - 4y + 1 = 0\). The value of \(\alpha^4 + \beta^4 + \gamma^4 + \delta^4 = 20\).

(b) Using \(\alpha + \beta + \gamma + \delta = -2\), multiply the original equation by \(x\) to get \(x^5 + 2x^4 - x = 0\). Thus, \(\alpha^5 + \beta^5 + \gamma^5 + \delta^5 = -2(20) + (-2) = -42\).

(c) Use the formula for the sum of squares: \(\alpha^8 + \beta^8 + \gamma^8 + \delta^8 = 20^2 - 2(6) = 388\).