MetadataPastPaper.sampleTitle

Let \(H_n\) be the proposition that \(\mathbf{A}^n = \begin{pmatrix} 2^n & 3^n - 2^n \\ 0 & 3^n \end{pmatrix}\), where \(\mathbf{A} = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}\). For \(n = 1\), \(\mathbf{A}^1 = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}\) and the formula gives \(\begin{pmatrix} 2^1 & 3^1 - 2^1 \\ 0 & 3^1 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}\). Thus, \(H_1\) is true. Assume that \(H_k\) is true for some positive integer \(k\), so \(\mathbf{A}^k = \begin{pmatrix} 2^k & 3^k - 2^k \\ 0 & 3^k \end{pmatrix}\). We now consider the case \(n = k + 1\): \(\mathbf{A}^{k+1} = \mathbf{A}^k \mathbf{A} = \begin{pmatrix} 2^k & 3^k - 2^k \\ 0 & 3^k \end{pmatrix} \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}\). Multiplying these matrices gives: Row 1, Column 1: \(2^k \times 2 + (3^k - 2^k) \times 0 = 2^{k+1}\). Row 1, Column 2: \(2^k \times 1 + (3^k - 2^k) \times 3 = 2^k + 3^{k+1} - 3 \cdot 2^k = 3^{k+1} - 2 \cdot 2^k = 3^{k+1} - 2^{k+1}\). Row 2, Column 1: \(0 \times 2 + 3^k \times 0 = 0\). Row 2, Column 2: \(0 \times 1 + 3^k \times 3 = 3^{k+1}\). Thus, \(\mathbf{A}^{k+1} = \begin{pmatrix} 2^{k+1} & 3^{k+1} - 2^{k+1} \\ 0 & 3^{k+1} \end{pmatrix}\). Therefore, \(H_k \implies H_{k+1}\). Since the base case \(H_1\) is true and \(H_k \implies H_{k+1}\), by the principle of mathematical induction, the statement is true for all positive integers \(n\).

B1: Verifies the base case \(n = 1\) showing LHS = RHS.
M1: Assumes the formula holds for \(n = k\).
M1: Attempts to find \(\mathbf{A}^{k+1}\) by multiplying \(\mathbf{A}^k \mathbf{A}\) (or \(\mathbf{A} \mathbf{A}^k\)).
A1: Obtains the correct product entries before simplification, particularly showing the calculation of the top-right entry: \(2^k + 3(3^k - 2^k)\).
A1: Correctly simplifies all entries to show \(\mathbf{A}^{k+1}\) is in the required form.
A1: Provides a clear, logical concluding statement referencing the principle of mathematical induction.

(a) Let \( \frac{3r+2}{r(r+1)(r+2)} = \frac{A}{r} + \frac{B}{r+1} + \frac{C}{r+2} \). Multiply by the denominator: \( 3r+2 = A(r+1)(r+2) + Br(r+2) + Cr(r+1) \). Setting \( r = 0 \) gives \( 2 = 2A \implies A = 1 \). Setting \( r = -1 \) gives \( -1 = -B \implies B = 1 \). Setting \( r = -2 \) gives \( -4 = 2C \implies C = -2 \). Hence, \( u_r = \frac{1}{r} + \frac{1}{r+1} - \frac{2}{r+2} \).

(b) We can rewrite the term \( u_r \) as: \( u_r = \left(\frac{1}{r} - \frac{1}{r+1}\right) + 2\left(\frac{1}{r+1} - \frac{1}{r+2}\right) \). Summing from \( r=1 \) to \( n \): \( \sum_{r=1}^n u_r = \sum_{r=1}^n \left(\frac{1}{r} - \frac{1}{r+1}\right) + 2 \sum_{r=1}^n \left(\frac{1}{r+1} - \frac{1}{r+2}\right) \). Both sums telescope: \( \sum_{r=1}^n \left(\frac{1}{r} - \frac{1}{r+1}\right) = 1 - \frac{1}{n+1} \) and \( \sum_{r=1}^n \left(\frac{1}{r+1} - \frac{1}{r+2}\right) = \frac{1}{2} - \frac{1}{n+2} \). Thus, \( \sum_{r=1}^n u_r = 1 - \frac{1}{n+1} + 2\left(\frac{1}{2} - \frac{1}{n+2}\right) = 2 - \frac{1}{n+1} - \frac{2}{n+2} \). Combining these terms over a common denominator: \( \frac{2(n+1)(n+2) - (n+2) - 2(n+1)}{(n+1)(n+2)} = \frac{2n^2 + 6n + 4 - n - 2 - 2n - 2}{(n+1)(n+2)} = \frac{2n^2+3n}{(n+1)(n+2)} = \frac{n(2n+3)}{(n+1)(n+2)} \).

(c) As \( n \to \infty \), \( \frac{1}{n+1} \to 0 \) and \( \frac{2}{n+2} \to 0 \). Therefore, \( \sum_{r=1}^{\infty} u_r = 2 - 0 - 0 = 2 \).

(a)
M1: Attempting to express \( u_r \) in the form \( \frac{A}{r} + \frac{B}{r+1} + \frac{C}{r+2} \).
A1: Correct partial fractions: \( A = 1, B = 1, C = -2 \).

(b)
M1: Rewriting \( u_r \) in a telescoping form, e.g., \( \left(\frac{1}{r} - \frac{1}{r+1}\right) + 2\left(\frac{1}{r+1} - \frac{1}{r+2}\right) \).
M1: Attempting to sum and cancel terms.
A1: Obtaining the correct unsimplified sum: \( 2 - \frac{1}{n+1} - \frac{2}{n+2} \).
M1: Combining terms over a common denominator.
A0.5: Correct simplification to show \( \frac{n(2n+3)}{(n+1)(n+2)} \).

(c)
M1: Taking the limit as \( n \to \infty \) of their sum expression.
A1: Correct value of 2.

(a) From the given equation, we have \( \sum \alpha = \frac{3}{2} \) and \( \sum \alpha\beta = \frac{4}{2} = 2 \). Using the identity \( \alpha^2 + \beta^2 + \gamma^2 = (\sum \alpha)^2 - 2\sum \alpha\beta \), we get \( \alpha^2 + \beta^2 + \gamma^2 = \left(\frac{3}{2}\right)^2 - 2(2) = \frac{9}{4} - 4 = -\frac{7}{4} \).

(b) Rearranging \( 2x^3 - 3x^2 + 4x - 1 = 0 \) to isolate terms with odd powers of \( x \): \( 2x^3 + 4x = 3x^2 + 1 \implies x(2x^2 + 4) = 3x^2 + 1 \). Substitute \( u = x^2 \): \( x(2u + 4) = 3u + 1 \). Squaring both sides: \( x^2(2u + 4)^2 = (3u + 1)^2 \). Substituting \( x^2 = u \) again: \( u(4u^2 + 16u + 16) = 9u^2 + 6u + 1 \implies 4u^3 + 16u^2 + 16u = 9u^2 + 6u + 1 \implies 4u^3 + 7u^2 + 10u - 1 = 0 \).

(c) The roots of the equation \( 4u^3 + 7u^2 + 10u - 1 = 0 \) are \( \alpha^2, \beta^2, \gamma^2 \). From the coefficients: \( \sum \alpha^2 = -\frac{7}{4} \) and \( \sum \alpha^2\beta^2 = \frac{10}{4} = \frac{5}{2} \). Using the identity \( \alpha^4 + \beta^4 + \gamma^4 = (\sum \alpha^2)^2 - 2\sum \alpha^2\beta^2 \), we obtain: \( \alpha^4 + \beta^4 + \gamma^4 = \left(-\frac{7}{4}\right)^2 - 2\left(\frac{5}{2}\right) = \frac{49}{16} - 5 = -\frac{31}{16} \).

(a)
B1: Stating \( \sum \alpha = 1.5 \) and \( \sum \alpha\beta = 2 \) (or equivalent).
M1: Using \( \alpha^2 + \beta^2 + \gamma^2 = (\sum \alpha)^2 - 2\sum \alpha\beta \).
A0.5: Obtaining the correct value \( -\frac{7}{4} \) (or \(-1.75\)).

(b)
M1: Correctly rearranging the equation to isolate odd powers of \( x \).
M1: Substituting \( u = x^2 \) and squaring both sides.
A1: Correctly expanding to obtain \( u(4u^2 + 16u + 16) = 9u^2 + 6u + 1 \).
A0.5: Completing the algebraic reduction to the given answer.

(c)
M1: Finding the sum and product-sum of the roots of the \( u \)-equation: \( \sum \alpha^2 = -\frac{7}{4} \) and \( \sum \alpha^2\beta^2 = \frac{5}{2} \).
M1: Applying the identity for the sum of squares of \( \alpha^2, \beta^2, \gamma^2 \).
A0.5: Correctly calculating \( -\frac{31}{16} \) (or \(-1.9375\)).

(a) (i) For a reflection in \(y = x\tan\theta\), the matrix is \(\begin{pmatrix} \cos 2\theta &
\sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}\). Here, \(\tan\theta = \sqrt{3}\), so \(\theta = \frac{\pi}{3}\). Thus, \(2\theta = \frac{2\pi}{3}\).
\[\mathbf{M}_1 = \begin{pmatrix} \cos(2\pi/3) & \sin(2\pi/3) \\ \sin(2\pi/3) & -\cos(2\pi/3) \end{pmatrix} = \begin{pmatrix} -1/2 & \sqrt{3}/2 \\ \sqrt{3}/2 & 1/2 \end{pmatrix}\]
A shear parallel to the \(x\)-axis has the matrix form \(\begin{pmatrix} 1 & s \\ 0 & 1 \end{pmatrix}\). Since \((0,1)\) maps to \((2,1)\):
\[\begin{pmatrix} 1 & s \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} s \\ 1
\end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix} \implies s = 2\]
So, \(\mathbf{M}_2 = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\).

(ii) The combined transformation is represented by \(\mathbf{M} = \mathbf{M}_2 \mathbf{M}_1\):
\[\mathbf{M} = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} -1/2 & \sqrt{3}/2 \\ \sqrt{3}/2 & 1/2 \end{pmatrix} = \begin{pmatrix} \sqrt{3} - 1/2 & \sqrt{3}/2 + 1 \\ \sqrt{3}/2 & 1/2 \end{pmatrix}\]

(b) The system does not have a unique solution when the determinant of the coefficient matrix is zero.
\[\det \begin{pmatrix} 1 & 3 & 1 \\ 2 & k & -1 \\ 1 & 1 & k \end{pmatrix} = 0\]
\[1(k^2 + 1) - 3(2k + 1) + 1(2 - k) = 0\]
\[k^2 + 1 - 6k - 3 + 2 - k = 0\]
\[k^2 - 7k = 0 \implies k(k-7) = 0\]
Thus, \(k = 0\) or \(k = 7\).

(c) Case 1: \(k = 0\)
The equations are:
(1) \(x + 3y + z = 2\)
(2) \(2x - z = 5 \implies z = 2x - 5\)
(3) \(x + y = 7 \implies y = 7 - x\)
Substitute into (1):
\[x + 3(7 - x) + (2x - 5) = 2 \implies 16 = 2\]
This is a contradiction, so the system is inconsistent. Since no two planes are parallel, the planes form a triangular prism.

Case 2: \(k = 7\)
The equations are:
(1) \(x + 3y + z = 2\)
(2) \(2x + 7y - z = 5\)
(3) \(x + y + 7z = 0\)
Add (1) and (2):
\[3x + 10y = 7\]
Calculate \(7 \times (1) - (3)\):
\[7(x + 3y + z) - (x + y + 7z) = 7(2) - 0 \implies 6x + 20y = 14 \implies 3x + 10y = 7\]
Since these equations are identical, the system is consistent with infinitely many solutions. Geometrically, the three planes meet in a single straight line (they form a sheaf of planes).

(a)(i) M1: 1 mark for correct angle/formulas, 1 mark for correct matrices M1 and M2. (ii) M1: 1 mark for matrix multiplication order, A1 for correct combined matrix.
(b) M1: 1 mark for attempting to find the determinant of the 3x3 matrix, A1 for correct expansion k^2 - 7k, A1 for setting determinant to 0 and solving, A1 for k = 0 and k = 7.
(c) M1: 1 mark for substituting k=0 and showing inconsistency, A1 for geometric description (triangular prism); M1: 1 mark for substituting k=7 and showing consistency, A1 for geometric description (line of intersection / sheaf of planes).

(a) The curve is a limaçon with an inner loop. It crosses the initial line at \(\theta = 0\) with \(r = 3\) (point \((3,0)\)) and at \(\theta = \pi\) with \(r = -1\) (point \((1,0)\) in Cartesian coordinates). It passes through the pole when \(r = 0 \implies \cos\theta = -1/2 \implies \theta = \pm 2\pi/3\).

(b) The inner loop is traced for \(\frac{2\pi}{3} \le \theta \le \frac{4\pi}{3}\) (or by symmetry, we can double the area from \(\frac{2\pi}{3}\) to \(\pi\)):
\[A = \frac{1}{2} \int_{2\pi/3}^{4\frac{\pi}{3}} r^2 \, d\theta = \int_{2\pi/3}^{\pi} (1 + 2\cos\theta)^2 \, d\theta\]
\[(1 + 2\cos\theta)^2 = 1 + 4\cos\theta + 4\cos^2\theta = 1 + 4\cos\theta + 2(1 + \cos 2\theta) = 3 + 4\cos\theta + 2\cos 2\theta\]
\[A = \int_{2\pi/3}^{\pi} (3 + 4\cos\theta + 2\cos 2\theta) \, d\theta = \left[ 3\theta + 4\sin\theta + \sin 2\theta \right]_{2\pi/3}^{\pi}\]
\[= (3\pi + 0 + 0) - \left( 3\left(\frac{2\pi}{3}\right) + 4\sin\left(\frac{2\pi}{3}\right) + \sin\left(\frac{4\pi}{3}\right) \right)\]
\[= 3\pi - \left( 2\pi + 4\left(\frac{\sqrt{3}}{2}\right) - \frac{\sqrt{3}}{2} \right) = \pi - \frac{3\sqrt{3}}{2}\]

(c) Tangents perpendicular to the initial line occur when \(\frac{dx}{d\theta} = 0\) (provided \(\frac{dy}{d\theta} \neq 0\)).
\[x = r\cos\theta = (1 + 2\cos\theta)\cos\theta = \cos\theta + 2\cos^2\theta\]
\[\frac{dx}{d\theta} = -\sin\theta - 4\cos\theta\sin\theta = -\sin\theta(1 + 4\cos\theta) = 0\]
This gives \(\sin\theta = 0 \implies \theta = 0, \pi\) or \(\cos\theta = -1/4\).
- At \(\theta = 0\): \(r = 3 \implies x = 3, y = 0\).
- At \(\theta = \pi\): \(r = -1 \implies x = 1, y = 0\).
- At \(\cos\theta = -1/4\): \(r = 1 + 2(-1/4) = 1/2\).
\(x = r\cos\theta = \frac{1}{2}\left(-\frac{1}{4}\right) = -\frac{1}{8}\).
\(\sin\theta = \pm\sqrt{1 - (-1/4)^2} = \pm\frac{\sqrt{15}}{4}\).
\(y = r\sin\theta = \frac{1}{2}\left(\pm\frac{\sqrt{15}}{4}\right) = \pm\frac{\sqrt{15}}{8}\).
Thus, the points are \((3,0)\), \((1,0)\), \((-\frac{1}{8}, \frac{\sqrt{15}}{8})\), and \((-\frac{1}{8}, -\frac{\sqrt{15}}{8})\).

(a) B1 for correct limaçon shape with inner loop, B1 for correct intercepts on initial line.
(b) M1 for setting up integral with limits from 2pi/3 to 4pi/3 (or 2pi/3 to pi), M1 for expanding r^2 correctly, A1 for correct integration, M1 for substituting limits, A1 for final exact area.
(c) M1 for expressing x in terms of theta, M1 for differentiating and setting to 0, A1 for theta values, A1 for Cartesian coordinates of all 4 points (accept decimals if 3sf, but exact expected).

(a) Let \(\mathbf{d}_1 = \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}\) and \(\mathbf{d}_2 = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}\) be the direction vectors. Since \(\mathbf{d}_1\) is not a scalar multiple of \(\mathbf{d}_2\), the lines are not parallel.
Find a vector perpendicular to both lines:
\[\mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2 = \begin{pmatrix} 2 - (-1) \\ 1 - 4 \\ -2 - 1 \end{pmatrix} = \begin{pmatrix} 3 \\ -3 \\ -3 \end{pmatrix} \sim \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}\]
Let \(A(1,0,-1)\) be a point on \(l_1\) and \(B(2,3,1)\) be a point on \(l_2\). The vector \(\overrightarrow{AB} = \begin{pmatrix} 1 \\ 3 \\ 2 \end{pmatrix}\).
The shortest distance between the lines is:
\[d = \frac{|\overrightarrow{AB} \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{|1(1) + 3(-1) + 2(-1)|}{\sqrt{1^2 + (-1)^2 + (-1)^2}} = \frac{|1 - 3 - 2|}{\sqrt{3}} = \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3}\]
Since this distance is non-zero and the lines are not parallel, they are skew.

(b) The plane \(\Pi\) has normal vector \(\mathbf{n} = \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}\) and contains \(A(1, 0, -1)\).
\[1(x - 1) - 1(y - 0) - 1(z + 1) = 0 \implies x - y - z = 2\]

(c) The perpendicular distance from \(P(5, 1, 5)\) to \(\Pi\) is:
\[D = \frac{|5 - 1 - 5 - 2|}{\sqrt{1^2 + (-1)^2 + (-1)^2}} = \frac{|-3|}{\sqrt{3}} = \sqrt{3}\]
The equation of the line passing through \(P\) perpendicular to \(\Pi\) is:
\[\mathbf{r} = \begin{pmatrix} 5 \\ 1 \\ 5 \end{pmatrix} + t \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}\]
Substitute this into the equation of \(\Pi\):
\[(5 + t) - (1 - t) - (5 - t) = 2 \implies 3t - 1 = 2 \implies t = 1\]
Thus, the foot of the perpendicular \(F\) has coordinates:
\[\mathbf{r}_F = \begin{pmatrix} 5 + 1 \\ 1 - 1 \\ 5 - 1 \end{pmatrix} = \begin{pmatrix} 6 \\ 0 \\ 4 \end{pmatrix}\]
So, the foot of the perpendicular is \((6, 0, 4)\).

(a) M1 for finding vector product of directions, A1 for correct normal vector. M1 for calculating projection of AB onto normal, A1 for correct shortest distance, A1 for stating conclusion (skew because d is non-zero and non-parallel).
(b) M1 for using the normal vector in plane equation, A1 for substituting point on l1, A1 for correct equation x - y - z = 2.
(c) M1 for perpendicular distance formula, A1 for distance = sqrt(3). M1 for set-up of line perpendicular to plane, M1 for substituting into plane equation to find t, A1 for foot of perpendicular (6,0,4).

(a) By algebraic division:
\[y = \frac{x^2 - 4x + 8}{x - 2} = \frac{x(x - 2) - 2(x - 2) + 4}{x - 2} = x - 2 + \frac{4}{x - 2}\]
As \(x \to 2\), \(y \to \pm \infty\), so the vertical asymptote is \(x = 2\).
As \(x \to \pm \infty\), \(y \to x - 2\), so the oblique asymptote is \(y = x - 2\).

(b) Differentiating \(y\):
\[\frac{dy}{dx} = 1 - \frac{4}{(x-2)^2}\]
Setting \(\frac{dy}{dx} = 0\):
\[1 - \frac{4}{(x-2)^2} = 0 \implies (x-2)^2 = 4 \implies x - 2 = \pm 2\]
So \(x = 4\) or \(x = 0\).
- At \(x = 4\): \(y = 4 - 2 + \frac{4}{4-2} = 4\).
- At \(x = 0\): \(y = 0 - 2 + \frac{4}{0-2} = -4\).
To determine the nature, find \(\frac{d^2y}{dx^2}\):
\[\frac{d^2y}{dx^2} = \frac{8}{(x-2)^3}\]
- At \(x = 4\): \(\frac{d^2y}{dx^2} = 1 > 0 \implies (4, 4)\) is a local minimum.
- At \(x = 0\): \(\frac{d^2y}{dx^2} = -1 < 0 \implies (0, -4)\) is a local maximum.

(c) The sketch should show:
- Vertical asymptote \(x = 2\) and oblique asymptote \(y = x - 2\).
- Turning points at \((4, 4)\) (local minimum) and \((0, -4)\) (local maximum).
- Intersects \(y\)-axis at \((0, -4)\). No \(x\)-intercepts as \(x^2 - 4x + 8 = 0\) has no real roots.
- Branches in the top-right and bottom-left regions formed by the asymptotes.

(d) Setting the line equal to the curve:
\[mx - 4 = \frac{x^2 - 4x + 8}{x - 2}\]
\[(mx - 4)(x - 2) = x^2 - 4x + 8\]
\[mx^2 - 2mx - 4x + 8 = x^2 - 4x + 8\]
\[mx^2 - 2mx = x^2 \implies (m - 1)x^2 - 2mx = 0\]
\[x((m - 1)x - 2m) = 0\]
One intersection point is always at \(x = 0\). For there to be a second distinct intersection point:
1. The coefficient of \(x^2\) must not be zero, so \(m - 1 \neq 0 \implies m \neq 1\).
2. The second root \(x = \frac{2m}{m - 1}\) must be distinct from the first root \(x = 0\), which requires \(m \neq 0\).
Thus, the set of values is \(m \in \mathbb{R}\) where \(m \neq 0\) and \(m \neq 1\).

(a) M1 for algebraic division or identifying vertical asymptote, A1 for x = 2, A1 for y = x - 2.
(b) M1 for finding dy/dx and setting to 0, A1 for correct x values, A1 for correct y values and classifying both points using the second derivative.
(c) B1 for correct asymptotes drawn and labeled, B1 for correct turning points positioned, B1 for overall correct shape of two branches.
(d) M1 for equating line and curve and simplifying, A1 for factoring to find roots, A1 for stating the correct conditions m != 0 and m != 1.

Let \(y = \arctan(1+x)\).

**Step 1: Find the value of \(y\) at \(x = 0\)**
\[y(0) = \arctan(1) = \frac{\pi}{4}\]

**Step 2: Find the first derivative**
Using direct differentiation:
\[y' = \frac{1}{1 + (1+x)^2} = \frac{1}{x^2 + 2x + 2}\]
At \(x = 0\):
\[y'(0) = \frac{1}{2}\]

**Step 3: Find the second derivative**
Using the chain rule or quotient rule on \(y' = (x^2 + 2x + 2)^{-1}\):
\[y'' = -(x^2 + 2x + 2)^{-2}(2x + 2) = \frac{-2(x+1)}{(x^2+2x+2)^2}\]
At \(x = 0\):
\[y''(0) = \frac{-2(1)}{2^2} = -\frac{1}{2}\]

**Step 4: Find the third derivative**
Using the quotient rule on \(y''\):
\[y''' = \frac{-2(x^2+2x+2)^2 - (-2(x+1)) \cdot 2(x^2+2x+2)(2x+2)}{(x^2+2x+2)^4}\]
Simplifying by dividing the numerator and denominator by \((x^2+2x+2)\):
\[y''' = \frac{-2(x^2+2x+2) + 8(x+1)^2}{(x^2+2x+2)^3}\]
At \(x = 0\):
\[y'''(0) = \frac{-2(2) + 8(1)^2}{2^3} = \frac{-4 + 8}{8} = \frac{1}{2}\]

*(Alternative method for derivatives using implicit differentiation:)*
\(\tan y = 1+x\)
Differentiating with respect to \(x\):
\(\sec^2 y \cdot y' = 1 \implies y' = \cos^2 y\)
At \(x=0, y=\frac{\pi}{4} \implies y'(0) = \cos^2\left(\frac{\pi}{4}\right) = \frac{1}{2}\).

Differentiating again:
\(y'' = -2\cos y \sin y \cdot y' = -y' \sin(2y)\)
At \(x=0 \implies y''(0) = -\frac{1}{2} \sin\left(\frac{\pi}{2}\right) = -\frac{1}{2}\).

Differentiating once more:
\(y''' = -y'' \sin(2y) - 2y' \cos(2y) \cdot y'\)
At \(x=0 \implies y'''(0) = -\left(-\frac{1}{2}\right)\sin\left(\frac{\pi}{2}\right) - 2\left(\frac{1}{2}\right)^2 \cos\left(\frac{\pi}{2}\right) = \frac{1}{2} - 0 = \frac{1}{2}\).

**Step 5: Apply the Maclaurin's series formula**
The Maclaurin's series is given by:
\[y(x) = y(0) + x y'(0) + \frac{x^2}{2!} y''(0) + \frac{x^3}{3!} y'''(0) + \dots\]
Substituting our values in:
\[y(x) = \frac{\pi}{4} + \frac{1}{2}x + \frac{-1/2}{2}x^2 + \frac{1/2}{6}x^3 + \dots\]
\[y(x) = \frac{\pi}{4} + \frac{1}{2}x - \frac{1}{4}x^2 + \frac{1}{12}x^3\]

M1: For evaluating \(y(0) = \frac{\pi}{4}\) and attempting first derivative.
A1: For correct first derivative and evaluating \(y'(0) = \frac{1}{2}\).
M1: For attempting to find the second derivative \(y''\) and evaluating \(y''(0)\).
A1: For obtaining \(y''(0) = -\frac{1}{2}\) correctly.
M1: For attempting to find the third derivative \(y'''\) and evaluating \(y'''(0)\).
A1: For obtaining \(y'''(0) = \frac{1}{2}\) and writing down the correct series: \(\frac{\pi}{4} + \frac{1}{2}x - \frac{1}{4}x^2 + \frac{1}{12}x^3\).

First, we find the integrating factor, \(I(x)\):
\[I(x) = \mathrm{e}^{\int 2\tanh x \,\mathrm{d}x} = \mathrm{e}^{2\ln(\cosh x)} = \cosh^2 x.\]

Multiply both sides of the differential equation by the integrating factor:
\[\cosh^2 x \frac{\mathrm{d}y}{\mathrm{d}x} + 2y \cosh x \sinh x = \cosh^3 x\]
\[\frac{\mathrm{d}}{\mathrm{d}x} (y \cosh^2 x) = \cosh^3 x.\]

Integrate both sides with respect to \(x\):
\[y \cosh^2 x = \int \cosh^3 x \,\mathrm{d}x.\]

To evaluate the integral, write \(\cosh^3 x = \cosh^2 x \cosh x = (1 + \sinh^2 x)\cosh x\):
\[\int (1 + \sinh^2 x)\cosh x \,\mathrm{d}x = \sinh x + \frac{1}{3}\sinh^3 x + C.\]

Thus, the general solution is:
\[y \cosh^2 x = \sinh x + \frac{1}{3}\sinh^3 x + C.\]

We apply the initial condition \(y = 1\) when \(x = 0\):
\[(1)(\cosh 0)^2 = \sinh 0 + \frac{1}{3}\sinh^3 0 + C\]
\[1(1) = 0 + 0 + C \implies C = 1.\]

Hence, the particular solution is:
\[y \cosh^2 x = \sinh x + \frac{1}{3}\sinh^3 x + 1\]
\[y = \left(\sinh x + \frac{1}{3}\sinh^3 x + 1\right)\operatorname{sech}^2 x.\]

M1: Find the integrating factor by calculating \(\mathrm{e}^{\int 2\tanh x \,\mathrm{d}x}\).
A1: Obtain the correct integrating factor of \(\cosh^2 x\).
M1: Write the differential equation in the form \(\frac{\mathrm{d}}{\mathrm{d}x}(y \cdot \text{I.F.}) = \text{I.F.} \cdot \cosh x\) and attempt integration of \(\cosh^3 x\).
A1: Correct integration of \(\cosh^3 x\) to get \(\sinh x + \frac{1}{3}\sinh^3 x\).
M1: Substitute \(x = 0, y = 1\) to determine the constant of integration \(C\).
A1: Obtain \(C = 1\).
A1: Express the final particular solution in a correct explicit form for \(y\).

(i) Using de Moivre's theorem and the binomial expansion:
\[\cos(4\theta) + \mathrm{i}\sin(4\theta) = (\cos \theta + \mathrm{i}\sin \theta)^4\]
\[= \cos^4 \theta + 4\mathrm{i}\cos^3 \theta \sin \theta - 6\cos^2 \theta \sin^2 \theta - 4\mathrm{i}\cos \theta \sin^3 \theta + \sin^4 \theta.\]

Equating real and imaginary parts:
\[\cos(4\theta) = \cos^4 \theta - 6\cos^2 \theta \sin^2 \theta + \sin^4 \theta,\]
\[\sin(4\theta) = 4\cos^3 \theta \sin \theta - 4\cos \theta \sin^3 \theta.\]

We know that:
\[\tan(4\theta) = \frac{\sin(4\theta)}{\cos(4\theta)} = \frac{4\cos^3 \theta \sin \theta - 4\cos \theta \sin^3 \theta}{\cos^4 \theta - 6\cos^2 \theta \sin^2 \theta + \sin^4 \theta}.\]

Dividing both the numerator and the denominator by \(\cos^4 \theta\):
\[\tan(4\theta) = \frac{4\tan \theta - 4\tan^3 \theta}{1 - 6\tan^2 \theta + \tan^4 \theta}.\]

(ii) Let \(t = \tan \theta\). The given equation is:
\[t^4 - 4t^3 - 6t^2 + 4t + 1 = 0 \implies 1 - 6t^2 + t^4 = 4t^3 - 4t.\]

Dividing both sides by \(1 - 6t^2 + t^4\) (assuming it is non-zero):
\[\frac{4t^3 - 4t}{1 - 6t^2 + t^4} = 1 \implies \frac{4t - 4t^3}{1 - 6t^2 + t^4} = -1.\]

Comparing this with the expression for \(\tan(4\theta)\), we have:
\[\tan(4\theta) = -1.\]

Solving this for \(4\theta\):
\[4\theta = -\frac{\pi}{4} + k\pi \implies \theta = -\frac{\pi}{16} + \frac{k\pi}{4} \quad \text{for } k \in \mathbb{Z}.\]

For the range \(-\frac{\pi}{2} < \theta < \frac{\pi}{2}\), the four distinct values of \(\theta\) are:
\[\theta = -\frac{\pi}{16}, \quad \theta = -\frac{\pi}{16} + \frac{\pi}{4} = \frac{3\pi}{16}, \quad \theta = -\frac{\pi}{16} + \frac{\pi}{2} = \frac{7\pi}{16}, \quad \theta = -\frac{\pi}{16} - \frac{\pi}{4} = -\frac{5\pi}{16}.\]

Thus, the four roots of the equation are:
\[\tan\left(-\frac{5\pi}{16}\right), \quad \tan\left(-\frac{\pi}{16}\right), \quad \tan\left(\frac{3\pi}{16}\right), \quad \tan\left(\frac{7\pi}{16}\right).\]

Part (i):
M1: Apply de Moivre's theorem and expand \((\cos\theta + \mathrm{i}\sin\theta)^4\) binomially.
A1: Correct expressions for \(\cos(4\theta)\) and \(\sin(4\theta)\) by equating real and imaginary parts.
M1: Write \(\tan(4\theta)\) as \(\sin(4\theta)/\cos(4\theta)\) and divide through by \(\cos^4\theta\).
A1: Obtain the given identity for \(\tan(4\theta)\).

Part (ii):
M1: Rearrange the given equation \(t^4 - 4t^3 - 6t^2 + 4t + 1 = 0\) to relate it to \(\tan(4\theta)\).
M1: Deduce that \(\tan(4\theta) = -1\) and solve for \(\theta\).
A1: Correctly state all four values of \(\alpha\) in the interval \((-\pi/2, \pi/2)\).

(i) We differentiate \(x^3 + y^3 - 3xy = 3\) implicitly with respect to \(x\):
\[3x^2 + 3y^2 \frac{\mathrm{d}y}{\mathrm{d}x} - 3\left(y + x\frac{\mathrm{d}y}{\mathrm{d}x}\right) = 0.\]

Dividing through by 3 gives:
\[x^2 + y^2 \frac{\mathrm{d}y}{\mathrm{d}x} - y - x\frac{\mathrm{d}y}{\mathrm{d}x} = 0.\]

Substituting the coordinates of \(P(2, 1)\):
\[2^2 + 1^2 \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right) - 1 - 2\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right) = 0\]
\[4 + \frac{\mathrm{d}y}{\mathrm{d}x} - 1 - 2\frac{\mathrm{d}y}{\mathrm{d}x} = 0\]
\[3 - \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \implies \frac{\mathrm{d}y}{\mathrm{d}x} = 3.\]

(ii) We differentiate \(x^2 + y^2 y' - y - x y' = 0\) implicitly with respect to \(x\) again:
\[2x + \left(2y (y')^2 + y^2 y''\right) - y' - \left(y' + x y''\right) = 0,\]
where \(y' = \frac{\mathrm{d}y}{\mathrm{d}x}\) and \(y'' = \frac{\mathrm{d}^2 y}{\mathrm{d}x^2}\).

Simplifying the expression:
\[2x + 2y (y')^2 + y^2 y'' - 2y' - x y'' = 0.\]

Substituting \(x = 2\), \(y = 1\), and \(y' = 3\):
\[2(2) + 2(1)(3)^2 + (1)^2 y'' - 2(3) - 2 y'' = 0\]
\[4 + 18 + y'' - 6 - 2 y'' = 0\]
\[16 - y'' = 0 \implies y'' = 16.\]

Thus, \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 16\) at \(P\).

Part (i):
M1: Differentiate the implicit equation with respect to \(x\), using the product rule correctly on the \(3xy\) term.
A1: Correct first derivative equation: \(3x^2 + 3y^2 y' - 3y - 3xy' = 0\) (or simplified form).
A1: Substitute \(x = 2, y = 1\) and correctly evaluate \(\frac{\mathrm{d}y}{\mathrm{d}x} = 3\).

Part (ii):
M1: Differentiate the first derivative relation implicitly with respect to \(x\) again, using product/chain rules.
A1: Correct second derivative expression: \(2x + 2y(y')^2 + y^2 y'' - 2y' - x y'' = 0\) (or equivalent).
M1: Substitute the values \(x = 2\), \(y = 1\), and \(y' = 3\) into their second derivative equation.
A1: Obtain \(\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = 16\).

(i) Use the addition formula \(\sinh(A+B) = \sinh A \cosh B + \cosh A \sinh B\):
\(\sinh(3x) = \sinh(2x)\cosh x + \cosh(2x)\sinh x\)

Substitute the double angle formulas \(\sinh(2x) = 2\sinh x\cosh x\) and \(\cosh(2x) = 2\sinh^2 x + 1\):
\(\sinh(3x) = (2\sinh x\cosh x)\cosh x + (2\sinh^2 x + 1)\sinh x\)
\(= 2\sinh x\cosh^2 x + 2\sinh^3 x + \sinh x\)

Substitute \(\cosh^2 x = 1 + \sinh^2 x\):
\(\sinh(3x) = 2\sinh x(1 + \sinh^2 x) + 2\sinh^3 x + \sinh x\)
\(= 4\sinh^3 x + 3\sinh x\).

(ii) The equation \(\sinh(3x) = 13\sinh x\) becomes:
\(4\sinh^3 x + 3\sinh x = 13\sinh x \implies 4\sinh^3 x - 10\sinh x = 0\).

Factorising gives \(\sinh x (4\sinh^2 x - 10) = 0\).
This gives \(\sinh x = 0\) or \(\sinh^2 x = \frac{5}{2}\).

For \(\sinh x = 0\), the only real solution is \(x = 0\).

For \(\sinh^2 x = \frac{5}{2}\), since we require real solutions, \(\sinh x = \pm \sqrt{\frac{5}{2}}\).

Using the logarithmic definition \(\sinh^{-1} u = \ln(u + \sqrt{u^2 + 1})\):
If \(\sinh x = \sqrt{\frac{5}{2}}\), then \(x = \ln\left(\sqrt{\frac{5}{2}} + \sqrt{\frac{5}{2} + 1}\right) = \ln\left(\frac{\sqrt{5} + \sqrt{7}}{\sqrt{2}}\right)\).

By symmetry, if \(\sinh x = -\sqrt{\frac{5}{2}}\), \(x = -\ln\left(\frac{\sqrt{5} + \sqrt{7}}{\sqrt{2}}\right) = \ln\left(\frac{\sqrt{7} - \sqrt{5}}{\sqrt{2}}\right)\).

So the real solutions are \(x = 0, \pm \ln\left(\frac{\sqrt{5} + \sqrt{7}}{\sqrt{2}}\right)\).

(iii) For \(y = \sinh(3x) - 13\sinh x\), we have:
\(\frac{dy}{dx} = 3\cosh(3x) - 13\cosh x\).

Using \(\cosh(3x) = 4\cosh^3 x - 3\cosh x\):
\(\frac{dy}{dx} = 3(4\cosh^3 x - 3\cosh x) - 13\cosh x = 12\cosh^3 x - 22\cosh x\).

Setting \(\frac{dy}{dx} = 0 \implies \cosh x (12\cosh^2 x - 22) = 0\).

Since \(\cosh x \ge 1\) for all real \(x\), \(\cosh x \neq 0\).

Thus \(\cosh^2 x = \frac{11}{6}\).
Since \(\frac{11}{6} > 1\), this has real solutions: \(\cosh x = \sqrt{\frac{11}{6}}\).

The \(x\)-coordinates of the stationary points are \(x = \pm \cosh^{-1}\left(\sqrt{\frac{11}{6}}\right) = \pm \ln\left(\frac{\sqrt{11} + \sqrt{5}}{\sqrt{6}}\right)\).

To find the \(y\)-coordinates:
\(y = 4\sinh^3 x - 10\sinh x\).

Since \(\sinh^2 x = \cosh^2 x - 1 = \frac{5}{6}\), we have \(\sinh x = \pm \sqrt{\frac{5}{6}}\).

For \(x > 0\), \(\sinh x = \sqrt{\frac{5}{6}}\), so:
\(y = 4\left(\frac{5}{6}\right)^{3/2} - 10\sqrt{\frac{5}{6}} = \sqrt{\frac{5}{6}}\left(\frac{10}{3} - 10\right) = -\frac{20}{3}\sqrt{\frac{5}{6}}\).

By symmetry, for \(x < 0\), \(\sinh x = -\sqrt{\frac{5}{6}}\), so \(y = \frac{20}{3}\sqrt{\frac{5}{6}}\).

Thus, the stationary points are \(\left(\ln\left(\frac{\sqrt{11}+\sqrt{5}}{\sqrt{6}}\right), -\frac{20}{3}\sqrt{\frac{5}{6}}\right)\) and \(\left(-\ln\left(\frac{\sqrt{11}+\sqrt{5}}{\sqrt{6}}\right), \frac{20}{3}\sqrt{\frac{5}{6}}\right)\).

(i)
M1: For applying addition formula for \(\sinh(3x)\).
M1: For using double angle identities for \(\sinh(2x)\) and \(\cosh(2x)\).
M1: For using \(\cosh^2 x - \sinh^2 x = 1\) to express entirely in \(\sinh x\).
A1: Correctly obtains the given result.

(ii)
M1: For substituting the identity from (i) and getting a cubic in \(\sinh x\).
A1: For obtaining \(\sinh x = 0\) and \(\sinh^2 x = 2.5\).
M1: For applying the inverse hyperbolic logarithmic formula to \(\sinh x = \pm \sqrt{2.5}\).
A1: For obtaining the non-zero solutions in exact form.
A1: For obtaining \(x=0\).

(iii)
M1: For differentiating and setting derivative to 0.
M1: For obtaining \(\cosh^2 x = \frac{11}{6}\).
A1: For correct \(x\)-coordinates in exact logarithmic form.
A1: For correct \(y\)-coordinates in exact simplified form.

(i) We can write the sum as:
\(\sum_{r=1}^{n} \frac{n}{n^2 + 3r^2} = \sum_{r=1}^{n} \frac{1}{n} \frac{1}{1 + 3(r/n)^2}\).

As \(n \to \infty\), this sum converges to the definite integral:
\(\int_0^1 \frac{1}{1 + 3x^2} \, dx\).

Using the standard integral \(\int \frac{1}{1+a^2x^2} \, dx = \frac{1}{a}\arctan(ax)\):
\(\int_0^1 \frac{1}{1 + 3x^2} \, dx = \left[ \frac{1}{\sqrt{3}} \arctan(\sqrt{3}x) \right]_0^1 = \frac{1}{\sqrt{3}}\left(\arctan(\sqrt{3}) - 0\right) = \frac{1}{\sqrt{3}} \cdot \frac{\pi}{3} = \frac{\pi}{3\sqrt{3}}\).

(ii) Similarly, the sum can be written as:
\(\sum_{r=1}^{n} \frac{r^2}{n(n^2 + 3r^2)} = \sum_{r=1}^{n} \frac{1}{n} \frac{(r/n)^2}{1 + 3(r/n)^2}\).

As \(n \to \infty\), this sum converges to:
\(\int_0^1 \frac{x^2}{1 + 3x^2} \, dx\).

To integrate this, write:
\(\frac{x^2}{1 + 3x^2} = \frac{1}{3} \frac{3x^2}{1 + 3x^2} = \frac{1}{3} \left( 1 - \frac{1}{1 + 3x^2} \right)\).

So the integral is:
\(\frac{1}{3} \int_0^1 \left( 1 - \frac{1}{1 + 3x^2} \right) \, dx = \frac{1}{3} \left[ x \right]_0^1 - \frac{1}{3} \int_0^1 \frac{1}{1 + 3x^2} \, dx\)
\(= \frac{1}{3}(1) - \frac{1}{3} \left( \frac{\pi}{3\sqrt{3}} \right) = \frac{1}{3} - \frac{\pi}{9\sqrt{3}}\).

(iii) The function \(f(x) = \frac{1}{1+3x^2}\) is strictly decreasing on \([0, 1]\), since \(f'(x) = \frac{-6x}{(1+3x^2)^2} < 0\) for \(0 < x \le 1\).

For any subinterval \(\left[\frac{r-1}{n}, \frac{r}{n}\right]\), because \(f(x)\) is decreasing, the minimum value of \(f(x)\) on this subinterval is at the right endpoint, i.e., \(f(x) \ge f\left(\frac{r}{n}\right)\) for \(x \in \left[\frac{r-1}{n}, \frac{r}{n}\right]\), with equality only at \(x = \frac{r}{n}\).

Therefore, the area under the curve is strictly greater than the right-hand Riemann sum:
\(\int_{\frac{r-1}{n}}^{\frac{r}{n}} f(x) \, dx > \frac{1}{n} f\left(\frac{r}{n}\right)\).

Summing from \(r = 1\) to \(n\):
\(\int_0^1 f(x) \, dx > \sum_{r=1}^n \frac{1}{n} f\left(\frac{r}{n}\right) = \sum_{r=1}^n \frac{n}{n^2 + 3r^2}\).

Since \(\int_0^1 f(x) \, dx = \frac{\pi}{3\sqrt{3}}\), we must have:
\(\sum_{r=1}^n \frac{n}{n^2 + 3r^2} < \frac{\pi}{3\sqrt{3}}\).

(i)
M1: For rewriting the sum with factor \(1/n\) and terms in \(r/n\).
A1: Identifies the limit as the definite integral of \(\frac{1}{1+3x^2}\) from 0 to 1.
M1: Integrates correctly using the arctan substitution.
A1: Shows the final limit value clearly.

(ii)
M1: For rewriting the second sum with factor \(1/n\) and terms in \(r/n\).
A1: Identifies the limit as the definite integral of \(\frac{x^2}{1+3x^2}\).
M1: Performs algebraic division or equivalent manipulation to integrate.
A1: Obtains \(\frac{1}{3} - \frac{\pi}{9\sqrt{3}}\).

(iii)
M1: For showing or stating that \(f(x)\) is strictly decreasing on \([0, 1]\).
M1: For explaining that the right-hand Riemann sum is an underestimate for a decreasing function.
A1: Completes the proof to show \(S_n < \frac{\pi}{3\sqrt{3}}\).

(i) The characteristic equation is \(\det(\mathbf{A} - \lambda\mathbf{I}) = 0\):
\(\det \begin{pmatrix} 2-\lambda & 1 & -1 \\ 0 & 3-\lambda & 0 \\ 0 & 1 & 1-\lambda \end{pmatrix} = (2-\lambda)(3-\lambda)(1-\lambda) = 0\).
So the eigenvalues are \(\lambda = 1, 2, 3\).

(ii) For \(\lambda = 1\):
\(\begin{pmatrix} 1 & 1 & -1 \\ 0 & 2 & 0 \\ 0 & 1 & 0 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies y = 0, x = z\).
An eigenvector is \(\mathbf{e}_1 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}\).

For \(\lambda = 2\):
\(\begin{pmatrix} 0 & 1 & -1 \\ 0 & 1 & 0 \\ 0 & 1 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies y = 0, z = 0\).
An eigenvector is \(\mathbf{e}_2 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\).

For \(\lambda = 3\):
\(\begin{pmatrix} -1 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 1 & -2 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies y = 2z, x = y - z = z\).
An eigenvector is \(\mathbf{e}_3 = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}\).

(iii) The matrix \(\mathbf{P}\) is formed by the eigenvectors as columns, and \(\mathbf{D}\) is the diagonal matrix of the corresponding eigenvalues:
\(\mathbf{P} = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 0 & 2 \\ 1 & 0 & 1 \end{pmatrix}\) and \(\mathbf{D} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}\).

To find \(\mathbf{P}^{-1\), we calculate \(\det(\mathbf{P}) = 2\) and find the inverse:
\(\mathbf{P}^{-1} = \begin{pmatrix} 0 & -1/2 & 1 \\ 1 & 0 & -1 \\ 0 & 1/2 & 0 \end{pmatrix}\).

(iv) Since \(\mathbf{A}^n = \mathbf{P} \mathbf{D}^n \mathbf{P}^{-1}\):
\(\mathbf{A}^n = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 0 & 2 \\ 1 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1^n & 0 & 0 \\ 0 & 2^n & 0 \\ 0 & 0 & 3^n \end{pmatrix} \begin{pmatrix} 0 & -1/2 & 1 \\ 1 & 0 & -1 \\ 0 & 1/2 & 0 \end{pmatrix}\)
\(= \begin{pmatrix} 1^n & 2^n & 3^n \\ 0 & 0 & 2 \cdot 3^n \\ 1^n & 0 & 3^n \end{pmatrix} \begin{pmatrix} 0 & -1/2 & 1 \\ 1 & 0 & -1 \\ 0 & 1/2 & 0 \end{pmatrix}\)
\(= \begin{pmatrix} 2^n & -\frac{1}{2} + \frac{1}{2} 3^n & 1 - 2^n \\ 0 & 3^n & 0 \\ 0 & -\frac{1}{2} + \frac{1}{2} 3^n & 1 \end{pmatrix}\) (noting that \(1^n = 1\)).

(i)
M1: For setting up characteristic equation \(\det(\mathbf{A}-\lambda\mathbf{I}) = 0\).
A1: For correct expansion of determinant.
A1: For finding eigenvalues \(1, 2, 3\).

(ii)
M1: For attempting to solve \((\mathbf{A}-\lambda\mathbf{I})\mathbf{x} = \mathbf{0}\) for any eigenvalue.
A1: Correct eigenvector for \(\lambda = 1\).
A1: Correct eigenvector for \(\lambda = 2\).
A1: Correct eigenvector for \(\lambda = 3\).

(iii)
B1: For writing down correct \(\mathbf{P}\) and \(\mathbf{D}\) (consistent with order).
M1: For attempting to find the inverse of \(\mathbf{P}\).
A1: Correct \(\mathbf{P}^{-1}\).

(iv)
M1: For performing matrix multiplication \(\mathbf{P}\mathbf{D}^n\mathbf{P}^{-1}\).
A1: Correct intermediate matrix product.
A1: Correct final matrix \(\mathbf{A}^n\).

(i) Using integration by parts on \(I_n = \int_0^1 (1-x^2)^n \cdot 1 \, dx\):
Let \(u = (1-x^2)^n \implies du = n(1-x^2)^{n-1}(-2x) \, dx = -2nx(1-x^2)^{n-1} \, dx\)
Let \(dv = dx \implies v = x\).

Applying integration by parts:
\(I_n = \left[ x (1-x^2)^n \right]_0^1 - \int_0^1 x \left( -2nx(1-x^2)^{n-1} \right) \, dx\)

The boundary term vanishes: \(\left[ x (1-x^2)^n \right]_0^1 = 1(0) - 0(1) = 0\).
So:
\(I_n = 2n \int_0^1 x^2 (1-x^2)^{n-1} \, dx\).

Write \(x^2 = 1 - (1-x^2)\):
\(I_n = 2n \int_0^1 \left( 1 - (1-x^2) \right) (1-x^2)^{n-1} \, dx\)
\(= 2n \int_0^1 (1-x^2)^{n-1} \, dx - 2n \int_0^1 (1-x^2)^n \, dx\)
\(= 2n I_{n-1} - 2n I_n\).

Rearranging gives:
\((2n+1) I_n = 2n I_{n-1} \implies I_n = \frac{2n}{2n+1} I_{n-1}\).

(ii) Find \(I_0 = \int_0^1 1 \, dx = 1\).
Using the reduction formula:
\(I_1 = \frac{2}{3} I_0 = \frac{2}{3}\)
\(I_2 = \frac{4}{5} I_1 = \frac{8}{15}\)
\(I_3 = \frac{6}{7} I_2 = \frac{48}{105} = \frac{16}{35}\)
\(I_4 = \frac{8}{9} I_3 = \frac{128}{315}\).

(iii) The volume of revolution \(V\) generated by rotating \(R\) about the \(x\)-axis is:
\(V = \pi \int_0^a y^2 \, dx\).

Since \(y = 0\) at \(x = 1\) (for \(x \ge 0\)), the limits are from \(x=0\) to \(x=1\).
\(V = \pi \int_0^1 \left( (1-x^2)^2 \right)^2 \, dx = \pi \int_0^1 (1-x^2)^4 \, dx = \pi I_4\).

From part (ii), \(I_4 = \frac{128}{315}\).
Therefore, \(V = \frac{128\pi}{315}\).

(i)
M1: For attempting integration by parts with \(u = (1-x^2)^n\) and \(dv = dx\).
A1: Correct derivative \(du\) and correct boundary term evaluation.
M1: For rewriting \(x^2\) as \(1 - (1-x^2)\).
M1: For expressing the resulting integrals in terms of \(I_{n-1}\) and \(I_n\).
A1: Correctly completes the algebraic steps to obtain the given formula.

(ii)
B1: For correctly evaluating \(I_0 = 1\) or \(I_1 = \frac{2}{3}\).
M1: For applying the reduction formula step-by-step to find \(I_4\).
A1: Correct exact value \(\frac{128}{315}\).

(iii)
M1: For using the formula \(V = \pi \int y^2 \, dx\) with correct limits.
A1: For identifying that the integral is exactly \(I_4\).
M1: For calculating the volume by multiplying \(I_4\) by \(\pi\).
A1: Correct final volume \(\frac{128\pi}{315}\).