2024 Edexcel IAL Further Mathematics (YFM01) 模擬試題連答案詳解

We are given \(z = a + \mathrm{i}b\), which implies \(z^* = a - \mathrm{i}b\). Substituting these into the equation \(z + 3\mathrm{i}z^* = 12 + 4\mathrm{i}\) gives: \((a + \mathrm{i}b) + 3\mathrm{i}(a - \mathrm{i}b) = 12 + 4\mathrm{i}\). Expanding and simplifying: \(a + \mathrm{i}b + 3a\mathrm{i} - 3\mathrm{i}^2b = 12 + 4\mathrm{i}\). Since \(\mathrm{i}^2 = -1\), this becomes: \((a + 3b) + \mathrm{i}(3a + b) = 12 + 4\mathrm{i}\). Equating real and imaginary parts: Real parts: \(a + 3b = 12\) (Equation 1). Imaginary parts: \(3a + b = 4\) (Equation 2). From Equation 2, we have \(b = 4 - 3a\). Substituting this into Equation 1: \(a + 3(4 - 3a) = 12 \implies a + 12 - 9a = 12 \implies -8a = 0 \implies a = 0\). Substituting \(a = 0\) into \(b = 4 - 3a\) gives \(b = 4\). Therefore, \(a^2 + b^2 = 0^2 + 4^2 = 16\).

M1: Substitute \(z = a + \mathrm{i}b\) and \(z^* = a - \mathrm{i}b\) into the equation and equate real or imaginary parts to obtain at least one linear relation. A1: Correctly solve to find \(a = 0\) and \(b = 4\). A1 (0.5 marks): Correctly calculate \(a^2 + b^2 = 16\).

A matrix is singular if and only if its determinant is equal to zero. Thus, we require \(\det(\mathbf{M}) = 0\). Calculating the determinant: \((k-1)k - (2)(3) = 0 \implies k^2 - k - 6 = 0\). Factoring the quadratic equation: \((k - 3)(k + 2) = 0\). This gives the solutions \(k = 3\) and \(k = -2\).

M1: Set the determinant of \(\mathbf{M}\) equal to zero to form a quadratic equation in \(k\). A1: Correct quadratic equation, e.g., \(k^2 - k - 6 = 0\). A1 (0.5 marks): Correctly identify both values \(k = 3\) and \(k = -2\).

The area scale factor of a 2D transformation represented by a matrix \(\mathbf{A}\) is given by \(|\det(\mathbf{A})|\). First, calculate the determinant of \(\mathbf{A}\): \(\det(\mathbf{A}) = (3)(4) - (-1)(2) = 12 + 2 = 14\). Thus, the area of the image \(S'\) is given by: \(\text{Area}(S') = |\det(\mathbf{A})| \times \text{Area}(S) = 14 \times 5 = 70\text{ cm}^2\).

M1: Attempt to find the determinant of \(\mathbf{A}\). A1: Obtain \(\det(\mathbf{A}) = 14\). A1 (0.5 marks): Multiply the determinant by 5 to find the correct area of 70.

For the parabola \(y^2 = 4ax = 16x\), we have \(a = 4\). The focus is \(F(4, 0)\) and the equation of the directrix is \(x = -4\). By the focus-directrix property of a parabola, the distance from \(P\) to the focus \(F\) is equal to the perpendicular distance from \(P\) to the directrix. The distance from \(P(4t^2, 8t)\) to the directrix \(x = -4\) is \(4t^2 - (-4) = 4t^2 + 4\). Therefore, we have \(4t^2 + 4 = 13 \implies 4t^2 = 9 \implies t^2 = \frac{9}{4}\). Since \(t > 0\), we have \(t = \frac{3}{2} = 1.5\).

M1: State or use the relationship \(PF = 4t^2 + a\) or use the distance formula between \(P\) and \(F\) set equal to 13. A1: Form the equation \(4t^2 + 4 = 13\) or equivalent. A1 (0.5 marks): Deduce \(t = 1.5\), rejecting the negative root.

From the quadratic equation \(2x^2 - 5x + 6 = 0\), we can identify the sum and product of the roots: \(\alpha + \beta = \frac{5}{2}\) and \(\alpha\beta = \frac{6}{2} = 3\). Using the identity \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\), we substitute the known values: \(\alpha^2 + \beta^2 = \left(\frac{5}{2}\right)^2 - 2(3) = \frac{25}{4} - 6 = \frac{25}{4} - \frac{24}{4} = \frac{1}{4} = 0.25\).

M1: State or use the sum \(\alpha + \beta = 2.5\) and product \(\alpha\beta = 3\). A1: Correctly write the algebraic identity \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\) and substitute values. A1 (0.5 marks): Obtain the final correct value of 0.25.

We first find the derivative of \(f(x)\): \(f'(x) = 3x^2 - 3\). We then evaluate \(f(x)\) and \(f'(x)\) at \(x_0 = 2\): \(f(2) = 2^3 - 3(2) - 5 = 8 - 6 - 5 = -3\), and \(f'(2) = 3(2^2) - 3 = 12 - 3 = 9\). Applying the Newton-Raphson formula: \(x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 2 - \frac{-3}{9} = 2 + \frac{1}{3} = \frac{7}{3} \approx 2.333\).

M1: Differentiate \(f(x)\) to obtain \(f'(x) = 3x^2 - 3\) and evaluate both \(f(2)\) and \(f'(2)\). A1: Correct substitution into the Newton-Raphson formula: \(2 - \frac{-3}{9}\). A1 (0.5 marks): Correct final answer of 2.333.

We can expand the term inside the summation: \(\sum_{r=1}^{20} r(r + 1) = \sum_{r=1}^{20} (r^2 + r) = \sum_{r=1}^{20} r^2 + \sum_{r=1}^{20} r\). Using the standard formulae \(\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}\) and \(\sum_{r=1}^{n} r = \frac{n(n+1)}{2}\) with \(n = 20\): \(\sum_{r=1}^{20} r^2 = \frac{20 \times 21 \times 41}{6} = 2870\), and \(\sum_{r=1}^{20} r = \frac{20 \times 21}{2} = 210\). Adding these values together gives: \(2870 + 210 = 3080\).

M1: Split the sum into \(\sum r^2 + \sum r\) and attempt to use standard formulae. A1: Correct evaluation of at least one of the component sums (either 2870 or 210). A1 (0.5 marks): Correct final sum of 3080.

We have \(f(k) = 5^{2k} - 1\) and \(f(k+1) = 5^{2(k+1)} - 1 = 5^{2k+2} - 1\). We can rewrite this as: \(f(k+1) = 5^2 \cdot 5^{2k} - 1 = 25 \cdot 5^{2k} - 1\). Now we find the difference: \(f(k+1) - f(k) = (25 \cdot 5^{2k} - 1) - (5^{2k} - 1) = 25 \cdot 5^{2k} - 5^{2k} = 24 \cdot 5^{2k}\). Thus, the constant \(A\) is 24.

M1: Write an expression for \(f(k+1)\) in terms of \(5^{2k}\). A1: Attempt to find \(f(k+1) - f(k)\) and factor out \(5^{2k}\). A1 (0.5 marks): Conclude that \(A = 24\).

From the quadratic equation \(3x^2 - 7x + 12 = 0\), we have the sum and product of the roots:
\[\alpha + \beta = \frac{7}{3}\]
\[\alpha\beta = \frac{12}{3} = 4\]

We want to find the value of:
\[\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2}\]

First, we find \(\alpha^2 + \beta^2\):
\[\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(\frac{7}{3}\right)^2 - 2(4)\]
\[\alpha^2 + \beta^2 = \frac{49}{9} - 8 = -\frac{23}{9}\]

Now substitute this and the product \(\alpha\beta\) into the fraction:
\[\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{-\frac{23}{9}}{4^2} = \frac{-\frac{23}{9}}{16} = -\frac{23}{144}\]

M1: Identifies the correct sum and product of roots, \(\alpha + \beta = \frac{7}{3}\) and \(\alpha\beta = 4\).
M1: Correctly expresses \(\frac{1}{\alpha^2} + \frac{1}{\beta^2}\) in terms of \(\alpha + \beta\) and \(\alpha\beta\), i.e., \(\frac{(\alpha+\beta)^2 - 2\alpha\beta}{(\alpha\beta)^2}\).
A0.5: Correctly evaluates the expression to obtain \(-\frac{23}{144}\) (or equivalent exact fraction).

Let \(z = a + \mathrm{i}b\), where \(a, b \in \mathbb{R}\). Then \(z^* = a - \mathrm{i}b\).

Substitute \(z\) and \(z^*\) into the given equation:
\[(2 + \mathrm{i})(a + \mathrm{i}b) + 3\mathrm{i}(a - \mathrm{i}b) = 4 - 8\mathrm{i}\]

Expand each term:
\[(2a - b) + \mathrm{i}(a + 2b) + 3a\mathrm{i} + 3b = 4 - 8\mathrm{i}\]

Combine real and imaginary parts:
\[(2a + 2b) + \mathrm{i}(4a + 2b) = 4 - 8\mathrm{i}\]

Equating real and imaginary parts gives two simultaneous equations:
1) \(2a + 2b = 4 \implies a + b = 2\)
2) \(4a + 2b = -8 \implies 2a + b = -4\)

Subtracting the first equation from the second:
\[(2a + b) - (a + b) = -4 - 2 \implies a = -6\]

Substitute \(a = -6\) back into \(a + b = 2\):
\[-6 + b = 2 \implies b = 8\]

Thus, \(z = -6 + 8\mathrm{i}\).

M1: Substitutes \(z = a + \mathrm{i}b\) and \(z^* = a - \mathrm{i}b\) and attempts to expand and equate real and imaginary parts to form a pair of simultaneous equations in \(a\) and \(b\).
M1: Solves the system of equations to find values for both \(a\) and \(b\).
A0.5: Obtains the correct complex number \(z = -6 + 8\mathrm{i}\).

A matrix is singular if and only if its determinant is equal to zero.

Calculate the determinant of \(\mathbf{M}\):
\[\det(\mathbf{M}) = (k - 1)(k + 4) - (2)(-3)\]
\[\det(\mathbf{M}) = (k^2 + 3k - 4) + 6\]
\[\det(\mathbf{M}) = k^2 + 3k + 2\]

Setting the determinant to zero for a singular matrix:
\[k^2 + 3k + 2 = 0\]

Factorising the quadratic equation:
\[(k + 1)(k + 2) = 0\]

Therefore, the possible values of \(k\) are:
\[k = -1 \quad \text{and} \quad k = -2\]

M1: Attempts to find the determinant of \(\mathbf{M}\) and sets it equal to zero.
M1: Standard algebraic expansion and simplification to form a 3-term quadratic equation in \(k\), i.e., \(k^2 + 3k + 2 = 0\).
A0.5: Solves the quadratic to find both correct values, \(k = -1\) and \(k = -2\).

For \( y^2 = 20x \), we have \( 4a = 20 \implies a = 5 \). So the focus is \( S(5, 0) \). Differentiating \( y^2 = 20x \) with respect to \( x \): \( 2y \frac{\mathrm{d}y}{\mathrm{d}x} = 20 \implies \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{10}{y} \). At \( P(5t^2, 10t) \), the gradient of the tangent is: \( m = \frac{10}{10t} = \frac{1}{t} \). The equation of the tangent at P is: \( y - 10t = \frac{1}{t}(x - 5t^2) \implies ty - 10t^2 = x - 5t^2 \implies ty = x + 5t^2 \). The tangent cuts the \( y \)-axis at Q, so setting \( x = 0 \): \( ty = 5t^2 \implies y = 5t \) (since \( t \neq 0 \)). Thus, Q has coordinates \( (0, 5t) \). To find the area of triangle SPQ with vertices \( S(5, 0) \), \( P(5t^2, 10t) \), and \( Q(0, 5t) \), we can use the shoelace formula: \( \text{Area} = \frac{1}{2} |x_S(y_P - y_Q) + x_P(y_Q - y_S) + x_Q(y_S - y_P)| \implies \text{Area} = \frac{1}{2} |5(10t - 5t) + 5t^2(5t - 0) + 0| \implies \text{Area} = \frac{1}{2} |25t + 25t^3| = \frac{25}{2}|t(1 + t^2)| \). Since \( 1 + t^2 > 0 \) for all real \( t \), the area is indeed \( \frac{25}{2}|t|(1 + t^2) \).

M1: Identifies the focus \( S(5, 0) \) or uses \( a = 5 \).
M1: Differentiates or uses the general tangent equation to find the equation of the tangent \( ty = x + 5t^2 \).
A1: Correct coordinates of \( Q(0, 5t) \).
M1: Applies a valid method to find the area of the triangle with vertices \( (5, 0) \), \( (5t^2, 10t) \), and \( (0, 5t) \).
A1: Obtains the correct given area \( \frac{25}{2}|t|(1+t^2) \) with clear working.

(a) The determinant of \( \mathbf{M} \) is given by: \( \det(\mathbf{M}) = 4(-3) - 2p = -12 - 2p \). Given that \( \det(\mathbf{M}) = -26 \), we have: \( -12 - 2p = -26 \implies 2p = 14 \implies p = 7 \). (b) With \( p = 7 \), \( \mathbf{M} = \begin{pmatrix} 4 & 7 \\ 2 & -3 \end{pmatrix} \). We compute: \( \mathbf{M}^2 = \begin{pmatrix} 4 & 7 \\ 2 & -3 \end{pmatrix} \begin{pmatrix} 4 & 7 \\ 2 & -3 \end{pmatrix} = \begin{pmatrix} 16 + 14 & 28 - 21 \\ 8 - 6 & 14 + 9 \end{pmatrix} = \begin{pmatrix} 30 & 7 \\ 2 & 23 \end{pmatrix} \). Subtracting \( \mathbf{M} \): \( \mathbf{M}^2 - \mathbf{M} = \begin{pmatrix} 30 & 7 \\ 2 & 23 \end{pmatrix} - \begin{pmatrix} 4 & 7 \\ 2 & -3 \end{pmatrix} = \begin{pmatrix} 26 & 0 \\ 0 & 26 \end{pmatrix} = 26\mathbf{I} \). Thus, \( k = 26 \).

Part (a):
M1: Sets up the equation for the determinant: \( 4(-3) - 2p = -26 \).
A1: Correctly finds \( p = 7 \).

Part (b):
M1: Multiplies \( \mathbf{M} \) by itself to find \( \mathbf{M}^2 \) (at least 2 elements correct with their value of \( p \)).
M1: Subtracts \( \mathbf{M} \) from their \( \mathbf{M}^2 \) and sets equal to \( k\mathbf{I} \).
A1: Correctly identifies \( k = 26 \).

(a) Since the coefficients of the cubic equation are real, complex roots must occur in conjugate pairs. Therefore, the other complex root is: \( z_2 = 3 + 2\mathrm{i} \). (b) Let the third root be \( z_3 \). Using the sum of roots: \( z_1 + z_2 + z_3 = 8 \implies (3 - 2\mathrm{i}) + (3 + 2\mathrm{i}) + z_3 = 8 \implies 6 + z_3 = 8 \implies z_3 = 2 \). (c) The roots are \( 2 \), \( 3 - 2\mathrm{i} \), and \( 3 + 2\mathrm{i} \). We can find \( c \) and \( d \) by expanding the product of the factors: \( (z - 2)(z - (3 - 2\mathrm{i}))(z - (3 + 2\mathrm{i})) = (z - 2)((z - 3)^2 + 4) = (z - 2)(z^2 - 6z + 13) = z^3 - 6z^2 + 13z - 2z^2 + 12z - 26 = z^3 - 8z^2 + 25z - 26 = 0 \). Comparing coefficients gives \( c = 25 \) and \( d = -26 \).

Part (a):
B1: States \( z_2 = 3 + 2\mathrm{i} \).

Part (b):
M1: Uses the sum of roots formula \( z_1 + z_2 + z_3 = 8 \) (or alternative valid method such as polynomial division).
A1: Obtains \( z_3 = 2 \).

Part (c):
M1: Expands the three factors or uses formulas for \( \sum \alpha\beta \) and \( \alpha\beta\gamma \) to find \( c \) and/or \( d \).
A1: Correct values: \( c = 25 \) and \( d = -26 \).

(a) A shear parallel to the \( x \)-axis is represented by a matrix of the form \( \mathbf{Q} = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix} \). Since the point \( (0, 1) \) is mapped to \( (4, 1) \): \( \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} k \\ 1 \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \end{pmatrix} \implies k = 4 \). Thus, \( \mathbf{Q} = \begin{pmatrix} 1 & 4 \\ 0 & 1 \end{pmatrix} \). (b) The matrix representing the transformation \( V \) followed by \( U \) is: \( \mathbf{R} = \mathbf{P}\mathbf{Q} = \begin{pmatrix} -3 & 0 \\ 0 & -3 \end{pmatrix} \begin{pmatrix} 1 & 4 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} -3 & -12 \\ 0 & -3 \end{pmatrix} \). (c) The area of the image triangle \( T' \) is: \( \text{Area of } T' = |\det(\mathbf{R})| \times \text{Area of } T \). We calculate \( \det(\mathbf{R}) = (-3)(-3) - (-12)(0) = 9 \). Thus, \( \text{Area of } T' = |9| \times 5 = 45 \text{ cm}^2 \).

Part (a):
B1: Correctly writes down \( \mathbf{Q} = \begin{pmatrix} 1 & 4 \\ 0 & 1 \end{pmatrix} \).

Part (b):
M1: Forms the product \( \mathbf{P}\mathbf{Q} \) in the correct order.
A1: Correct matrix \( \mathbf{R} = \begin{pmatrix} -3 & -12 \\ 0 & -3 \end{pmatrix} \).

Part (c):
M1: Finds \( \det(\mathbf{R}) \) and multiplies its absolute value by the area of \( T \).
A1: Correct area of \( 45 \text{ cm}^2 \) (units not required).

From the equation, we have: \( \alpha + \beta = \frac{5}{2} \) and \( \alpha\beta = 2 \). We want to find a quadratic equation with roots \( u = \frac{\alpha}{\beta} + 1 \) and \( v = \frac{\beta}{\alpha} + 1 \). Sum of the new roots: \( u + v = \frac{\alpha}{\beta} + \frac{\beta}{\alpha} + 2 = \frac{\alpha^2 + \beta^2}{\alpha\beta} + 2 = \frac{(\alpha+\beta)^2 - 2\alpha\beta}{\alpha\beta} + 2 \). Substituting the values: \( \alpha^2 + \beta^2 = \left(\frac{5}{2}\right)^2 - 2(2) = \frac{25}{4} - 4 = \frac{9}{4} \). Thus, \( u + v = \frac{9/4}{2} + 2 = \frac{9}{8} + 2 = \frac{25}{8} \). Product of the new roots: \( uv = \left(\frac{\alpha}{\beta} + 1\right)\left(\frac{\beta}{\alpha} + 1\right) = 1 + \frac{\alpha}{\beta} + \frac{\beta}{\alpha} + 1 = \frac{\alpha^2 + \beta^2}{\alpha\beta} + 2 = \frac{25}{8} \). The new quadratic equation is: \( x^2 - (u+v)x + uv = 0 \implies x^2 - \frac{25}{8}x + \frac{25}{8} = 0 \). Multiplying by 8 to obtain integer coefficients: \( 8x^2 - 25x + 25 = 0 \).

M1: States \( \alpha + \beta = 2.5 \) and \( \alpha\beta = 2 \).
M1: Expresses the sum of the new roots in terms of \( \alpha+\beta \) and \( \alpha\beta \), and attempts to evaluate it.
A1: Correct sum of \( \frac{25}{8} \).
M1: Expresses the product of the new roots in terms of \( \alpha+\beta \) and \( \alpha\beta \), and evaluates it.
A1: Correct quadratic equation with integer coefficients: \( 8x^2 - 25x + 25 = 0 \) (or any non-zero integer multiple).

(a) Let \( f(x) = x^3 - 5x^2 + 5x + 2 \). Then \( f'(x) = 3x^2 - 10x + 5 \). With \( x_0 = 3 \): \( f(3) = 3^3 - 5(3)^2 + 5(3) + 2 = 27 - 45 + 15 + 2 = -1 \) and \( f'(3) = 3(3)^2 - 10(3) + 5 = 27 - 30 + 5 = 2 \). Applying the Newton-Raphson formula: \( x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 3 - \frac{-1}{2} = 3.5 \). (b) For linear interpolation on the interval \( [3, 4] \), we have: \( f(3) = -1 \) and \( f(4) = 4^3 - 5(4)^2 + 5(4) + 2 = 64 - 80 + 20 + 2 = 6 \). Setting up the linear interpolation ratio: \( \frac{x_1 - 3}{4 - x_1} = \t\frac{0 - (-1)}{6 - 0} = \frac{1}{6} \implies 6(x_1 - 3) = 4 - x_1 \implies 7x_1 = 22 \implies x_1 = \frac{22}{7} \approx 3.142857... \). To 3 decimal places, the approximation is \( 3.143 \).

Part (a):
M1: Differentiates to find \( f'(x) = 3x^2 - 10x + 5 \).
M1: Correctly substitutes \( x_0 = 3 \) into both \( f(x) \) and \( f'(x) \) and applies the Newton-Raphson formula.
A1: Correct answer of \( 3.5 \).

Part (b):
M1: Evaluates \( f(4) = 6 \) and sets up a linear interpolation ratio equation.
A1: Correctly evaluates the approximation to 3 decimal places as \( 3.143 \) (accept \( \frac{22}{7} \)).

We expand the summation as: \( \sum_{r=1}^n (3r^2 - 5r + 1) = 3 \sum_{r=1}^n r^2 - 5 \sum_{r=1}^n r + \sum_{r=1}^n 1 \). Using the standard formulae: \( = 3 \frac{n(n+1)(2n+1)}{6} - 5 \frac{n(n+1)}{2} + n = \frac{n(n+1)(2n+1)}{2} - \frac{5n(n+1)}{2} + \frac{2n}{2} \). Factor out \( \frac{n}{2} \): \( = \frac{n}{2} [ (n+1)(2n+1) - 5(n+1) + 2 ] = \frac{n}{2} [ 2n^2 + 3n + 1 - 5n - 5 + 2 ] = \frac{n}{2} [ 2n^2 - 2n - 2 ] \). This is in the required form with \( a = 2, b = -2, c = -2 \).

M1: Applies standard formulae for \( \sum r^2 \), \( \sum r \), and \( \sum 1 \).
M1: Factors out \( \frac{n}{2} \) from the expression.
A1: Correctly expands the quadratic terms inside the bracket: \( (2n^2 + 3n + 1) - 5(n+1) + 2 \).
M1: Simplifies the terms inside the bracket to a single quadratic expression.
A1: Identifies \( a = 2, b = -2, c = -2 \) clearly.

Let \( f(n) = 3^{2n} + 7 \). Step 1: For \( n = 1 \), \( f(1) = 3^2 + 7 = 16 \). Since 16 is divisible by 8 (\( 16 = 8 \times 2 \)), the statement is true for \( n = 1 \). Step 2: Assume the statement is true for \( n = k \), where \( k \) is a positive integer. That is, \( f(k) = 3^{2k} + 7 = 8m \) for some integer \( m \). Step 3: We want to show the statement is true for \( n = k + 1 \). We have: \( f(k+1) = 3^{2(k+1)} + 7 = 3^{2k+2} + 7 = 9 \times 3^{2k} + 7 \). Substituting \( 3^{2k} = 8m - 7 \): \( f(k+1) = 9(8m - 7) + 7 = 72m - 63 + 7 = 72m - 56 = 8(9m - 7) \). Since \( 9m - 7 \) is an integer, \( f(k+1) \) is divisible by 8. (Alternative method: \( f(k+1) - f(k) = (3^{2k+2} + 7) - (3^{2k} + 7) = 3^{2k}(9 - 1) = 8 \times 3^{2k} \), which is divisible by 8.) Step 4: Since the statement is true for \( n = 1 \), and if true for \( n = k \) then it is true for \( n = k + 1 \), the statement is true for all positive integers \( n \) by mathematical induction.

B1: Shows the base case \( n=1 \) is true with clear working.
M1: States a clear inductive hypothesis (assumes statement is true for \( n=k \)).
M1: Attempts to express \( f(k+1) \) in terms of \( f(k) \), or evaluates \( f(k+1) - f(k) \).
A1: Obtains a correct algebraic expression showing divisibility by 8: e.g., \( 8(9m - 7) \) or \( f(k+1) - f(k) = 8 \times 3^{2k} \).
A1: Gives a complete and correct conclusion stating that the result holds for all positive integers \( n \) by induction.

Multiply numerator and denominator by \( 1 + 2\mathrm{i} \): \( z = \frac{(3 + a\mathrm{i})(1 + 2\mathrm{i})}{(1 - 2\mathrm{i})(1 + 2\mathrm{i})} = \frac{3 - 2a + \mathrm{i}(6 + a)}{5} \). Since \( \operatorname{Re}(z) = \operatorname{Im}(z) \), we have \( \frac{3 - 2a}{5} = \frac{6 + a}{5} \implies 3 - 2a = 6 + a \implies 3a = -3 \implies a = -1 \). For \( a = -1 \), we have \( z = \frac{3 - 2(-1) + \mathrm{i}(6 - 1)}{5} = \frac{5 + 5\mathrm{i}}{5} = 1 + \mathrm{i} \). Thus, the modulus is \( |z| = \sqrt{1^2 + 1^2} = \sqrt{2} \), and the argument is \( \arg(z) = \arctan(1) = \frac{\pi}{4} \).

M1: Multiplies numerator and denominator of \( z \) by \( 1 + 2\mathrm{i} \). A1: Correct expression for \( z \) in terms of \( a \) with real and imaginary parts separated. A1: Sets real part equal to imaginary part and solves to find \( a = -1 \). B1: Finds the correct modulus \( |z| = \sqrt{2} \). B1: Finds the correct argument \( \arg(z) = \frac{\pi}{4} \).

Differentiating \( y^2 = 12x \) with respect to \( x \) gives \( 2y\frac{\mathrm{d}y}{\mathrm{d}x} = 12 \), so \( \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{6}{y} \). At the point \( P(3t^2, 6t) \), the gradient of the tangent is \( \frac{6}{6t} = \frac{1}{t} \). Since the gradient is \( \frac{1}{2} \), we have \( \frac{1}{t} = \frac{1}{2} \), which gives \( t = 2 \). Thus, the coordinates of \( P \) are \( (3(2)^2, 6(2)) = (12, 12) \). The equation of the tangent at \( P \) is \( y - 12 = \frac{1}{2}(x - 12) \), which simplifies to \( y = \frac{1}{2}x + 6 \). At the \( y \)-axis, \( x = 0 \), so \( Q \) has coordinates \( (0, 6) \). Triangle \( OPQ \) has vertices at \( O(0,0) \), \( Q(0,6) \), and \( P(12,12) \). Taking \( OQ \) as the base (length 6) and the \( x \)-coordinate of \( P \) as the perpendicular height (height 12), the area is \( \frac{1}{2} \times 6 \times 12 = 36 \).

M1: Differentiates the equation of the parabola or uses parametric differentiation to find \( \frac{\mathrm{d}y}{\mathrm{d}x} \) in terms of \( t \). A1: Solves \( \frac{1}{t} = \frac{1}{2} \) to find \( t = 2 \) and correctly finds the coordinates of \( P \) as \( (12, 12) \). M1: Finds the equation of the tangent at \( P \) and substitutes \( x = 0 \) to find the \( y \)-intercept. A1: Correctly identifies the coordinates of \( Q \) as \( (0, 6) \). B1: Calculates the correct area of triangle \( OPQ \) as \( 36 \).

The determinant of \( \mathbf{M} \) is \( \det(\mathbf{M}) = k(k - 3) - 4 = k^2 - 3k - 4 \). The area scale factor of the transformation is given by \( |\det(\mathbf{M})| \). Since the area of \( T \) is \( 8\text{ cm}^2 \) and the area of \( T' \) is \( 48\text{ cm}^2 \), the area scale factor is \( \frac{48}{8} = 6 \). This gives \( |k^2 - 3k - 4| = 6 \). Case 1: \( k^2 - 3k - 4 = 6 \implies k^2 - 3k - 10 = 0 \implies (k - 5)(k + 2) = 0 \), which gives \( k = 5 \) or \( k = -2 \). Case 2: \( k^2 - 3k - 4 = -6 \implies k^2 - 3k + 2 = 0 \implies (k - 1)(k - 2) = 0 \), which gives \( k = 1 \) or \( k = 2 \). Thus, the possible values of \( k \) are \( -2, 1, 2, \) and \( 5 \).

M1: Finds the determinant of \( \mathbf{M} \) in terms of \( k \). A1: Sets up the relation \( |\det(\mathbf{M})| = 6 \). M1: Solves the quadratic equation \( k^2 - 3k - 10 = 0 \) to find \( k = 5 \) and \( k = -2 \). A1: Solves the quadratic equation \( k^2 - 3k + 2 = 0 \) to find \( k = 1 \) and \( k = 2 \). A1: Clearly lists all four correct values: \( -2, 1, 2, 5 \).

From the given quadratic equation, we have \( \alpha + \beta = \frac{5}{2} \) and \( \alpha\beta = 2 \). Let the new roots be \( \gamma = \frac{\alpha}{\beta} + 1 = \frac{\alpha + \beta}{\beta} \) and \( \delta = \frac{\beta}{\alpha} + 1 = \frac{\alpha + \beta}{\alpha} \). The sum of the new roots is \( \gamma + \delta = \frac{\alpha + \beta}{\beta} + \frac{\alpha + \beta}{\alpha} = (\alpha + \beta)\left(\frac{1}{\alpha} + \frac{1}{\beta}\right) = \frac{(\alpha + \beta)^2}{\alpha\beta} = \frac{(\frac{5}{2})^2}{2} = \frac{25}{8} \). The product of the new roots is \( \gamma\delta = \left(\frac{\alpha + \beta}{\beta}\right)\left(\frac{\alpha + \beta}{\alpha}\right) = \frac{(\alpha + \beta)^2}{\alpha\beta} = \frac{25}{8} \). A quadratic equation with these roots is \( x^2 - (\text{sum})x + (\text{product}) = 0 \), which is \( x^2 - \frac{25}{8}x + \frac{25}{8} = 0 \). Multiplying by 8 to obtain integer coefficients yields \( 8x^2 - 25x + 25 = 0 \).

B1: Identifies \( \alpha + \beta = \frac{5}{2} \) and \( \alpha\beta = 2 \). M1: Expresses the sum of the new roots in terms of \( (\alpha + \beta) \) and \( \alpha\beta \). A1: Evaluates the sum of the new roots to be \( \frac{25}{8} \). M1: Expresses the product of the new roots in terms of \( (\alpha + \beta) \) and \( \alpha\beta \) and evaluates it to be \( \frac{25}{8} \). A1: Formulates the final quadratic equation with integer coefficients: \( 8x^2 - 25x + 25 = 0 \).

For \( n = 1 \): LHS \( = 1(2(1) - 1) = 1 \). RHS \( = \frac{1(1+1)(4(1)-1)}{6} = \frac{1(2)(3)}{6} = 1 \). Since LHS \( = \) RHS, the statement is true for \( n = 1 \). Assume the statement is true for \( n = k \), so \( \sum_{r=1}^k r(2r-1) = \frac{k(k+1)(4k-1)}{6} \). For \( n = k + 1 \): \( \sum_{r=1}^{k+1} r(2r-1) = \sum_{r=1}^k r(2r-1) + (k+1)(2(k+1)-1) = \frac{k(k+1)(4k-1)}{6} + (k+1)(2k+1) \). Factorising out \( \frac{k+1}{6} \) gives \( \frac{k+1}{6}[k(4k-1) + 6(2k+1)] = \frac{k+1}{6}[4k^2 - k + 12k + 6] = \frac{k+1}{6}[4k^2 + 11k + 6] \). Factorising the quadratic term gives \( 4k^2 + 11k + 6 = (k+2)(4k+3) \). Thus, the sum is \( \frac{(k+1)(k+2)(4(k+1)-1)}{6} \). This is the same formula with \( n \) replaced by \( k + 1 \). If the statement is true for \( n = k \), it is also true for \( n = k + 1 \). Since it is true for \( n = 1 \), it is true for all positive integers \( n \) by mathematical induction.

B1: Shows the statement is true for \( n = 1 \). M1: Assumes statement is true for \( n = k \) and adds the \( (k+1) \)-th term to the assumed sum. M1: Factors out \( \frac{k+1}{6} \) or attempts algebraic simplification to form a single fraction. A1: Correctly simplifies and factorises the numerator to obtain the final form for \( n = k + 1 \). B1: Concludes with a complete and correct inductive statement.

From \(3x^2 - 4x + 6 = 0\), we have:
\(\alpha + \beta = \frac{4}{3}\)
\(\alpha\beta = 2\)

Let the roots of the new equation be \(p = \alpha^2 + \beta\) and \(q = \beta^2 + \alpha\).

The sum of the new roots is:
\(p + q = (\alpha^2 + \beta) + (\beta^2 + \alpha) = (\alpha^2 + \beta^2) + (\alpha + \beta)\)

Since \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\):
\(\alpha^2 + \beta^2 = \left(\frac{4}{3}\right)^2 - 2(2) = \frac{16}{9} - 4 = -\frac{20}{9}\)

Therefore, the sum is:
\(p + q = -\frac{20}{9} + \frac{4}{3} = -\frac{8}{9}\)

The product of the new roots is:
\(pq = (\alpha^2 + \beta)(\beta^2 + \alpha) = \alpha^2\beta^2 + \alpha^3 + \beta^3 + \alpha\beta\)

Since \(\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)\):
\(\alpha^3 + \beta^3 = \left(\frac{4}{3}\right)^3 - 3(2)\left(\frac{4}{3}\right) = \frac{64}{27} - 8 = -\frac{152}{27}\)

We also know \(\alpha^2\beta^2 = (\alpha\beta)^2 = 4\).

Therefore, the product is:
\(pq = 4 - \frac{152}{27} + 2 = 6 - \frac{152}{27} = \frac{162 - 152}{27} = \frac{10}{27}\)

The new quadratic equation is given by:
\(x^2 - (p + q)x + pq = 0\)
\(x^2 - \left(-\frac{8}{9}\right)x + \frac{10}{27} = 0\)
\(x^2 + \frac{8}{9}x + \frac{10}{27} = 0\)

Multiplying by 27 to obtain integer coefficients:
\(27x^2 + 24x + 10 = 0\)

M1: For writing down the sum \(\alpha+\beta = 4/3\) and product \(\alpha\beta = 2\) (implied by later correct work).
M1: For attempting to find the sum of the new roots, expanding \(\alpha^2 + \beta^2\) correctly as \((\alpha+\beta)^2 - 2\alpha\beta\) and substituting correct values.
A1: For obtaining a sum of \(-\frac{8}{9}\).
M1: For attempting to find the product of the new roots, expanding correctly, expressing \(\alpha^3 + \beta^3\) in terms of sum and product, and substituting.
A1: For obtaining a product of \(\frac{10}{27}\).
A1: For the final equation \(27x^2 + 24x + 10 = 0\) (must include \(= 0\), accept any integer multiple).

(a) Evaluating the function at the boundaries:
\(f(2.4) = (2.4)^3 - 5(2.4) - 3 = 13.824 - 12 - 3 = -1.176\)
\(f(2.5) = (2.5)^3 - 5(2.5) - 3 = 15.625 - 12.5 - 3 = 0.125\)

Since \(f(2.4) < 0\) and \(f(2.5) > 0\), there is a sign change in the interval \([2.4, 2.5]\). Since \(f(x)\) is a polynomial, it is continuous. Therefore, by the Intermediate Value Theorem, a root \(\alpha\) must lie in the interval \([2.4, 2.5]\).

(b) To apply the Newton-Raphson method, we first find the derivative:
\(f'(x) = 3x^2 - 5\)

Using \(x_0 = 2.5\):
\(f(2.5) = 0.125\)
\(f'(2.5) = 3(2.5)^2 - 5 = 18.75 - 5 = 13.75\)

The Newton-Raphson formula is:
\(x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}\)

Substituting the values:
\(x_1 = 2.5 - \frac{0.125}{13.75} = 2.5 - \frac{1}{110} \approx 2.490909...\)

Rounding to 3 decimal places:
\(x_1 \approx 2.491\)

(a)
M1: Attempt to evaluate both \(f(2.4)\) and \(f(2.5)\).
A1: Both values correct (\(-1.176\) and \(0.125\)) and a valid conclusion mentioning change of sign and continuity.
(b)
M1: Differentiates \(f(x)\) to obtain \(f'(x) = kx^2 - 5\).
A1: Correct derivative \(f'(x) = 3x^2 - 5\).
M1: Correct use of the Newton-Raphson formula with their values of \(f(2.5)\) and \(f'(2.5)\).
A1: Correct value of \(2.491\) to 3 decimal places.

(a) Expanding the term inside the summation:
\(\sum_{r=1}^n r(3r-1) = \sum_{r=1}^n (3r^2 - r) = 3\sum_{r=1}^n r^2 - \sum_{r=1}^n r\)

Using the standard formulae:
\(\sum_{r=1}^n r^2 = \frac{1}{6}n(n+1)(2n+1)\)

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

Substitute these into the expression:
\(3\left[\frac{1}{6}n(n+1)(2n+1)\right] - \frac{1}{2}n(n+1) = \frac{1}{2}n(n+1)(2n+1) - \frac{1}{2}n(n+1)\)

Factor out \(\frac{1}{2}n(n+1)\):
\(= \frac{1}{2}n(n+1)\left[(2n+1) - 1\right]\)

\(= \frac{1}{2}n(n+1)(2n)\)

\(= n^2(n+1)\) (as required)

(b) We want to evaluate:
\(\sum_{r=10}^{20} r(3r-1)\)

We can write this as:
\(\sum_{r=1}^{20} r(3r-1) - \sum_{r=1}^{9} r(3r-1)\)

Using the result from part (a):
\(\sum_{r=1}^{20} r(3r-1) = 20^2(20+1) = 400(21) = 8400\)

\(\sum_{r=1}^{9} r(3r-1) = 9^2(9+1) = 81(10) = 810\)

Therefore:
\(\sum_{r=10}^{20} r(3r-1) = 8400 - 810 = 7590\)

(a)
M1: Expands and splits the sum into \(3\sum r^2 - \sum r\).
M1: Substitutes the standard summation formulae for \(\sum r^2\) and \(\sum r\).
A1: Correct algebraic expression: \(\frac{1}{2}n(n+1)(2n+1) - \frac{1}{2}n(n+1)\) (or equivalent).
M1: Factors out a common factor, e.g., \(\frac{1}{2}n(n+1)\).
A1*: Fully correct simplification leading to \(n^2(n+1)\) with no errors seen.
(b)
M1: Expresses the required sum as the difference of two sums from \(r=1\): \(S_{20} - S_9\).
A1: Correct value of \(7590\).

To solve \(\frac{3x}{x-2} - (x+1) > 0\), we first express it as a single fraction:

\(\frac{3x - (x+1)(x-2)}{x-2} > 0\)

\(\frac{3x - (x^2 - x - 2)}{x-2} > 0\)

\(\frac{-x^2 + 4x + 2}{x-2} > 0\)

Multiply by \(-1\) and reverse the inequality:

\(\frac{x^2 - 4x - 2}{x-2} < 0\)

Find the critical values by setting the numerator and denominator to zero:

Denominator: \(x = 2\)

Numerator: \(x^2 - 4x - 2 = 0 \implies x = \frac{4 \pm \sqrt{16 - 4(1)(-2)}}{2} = 2 \pm \sqrt{6}\)

We analyze the sign of the quotient \(\frac{x^2 - 4x - 2}{x-2}\) across the intervals formed by the critical values \(2-\sqrt{6}\), \(2\), and \(2+\sqrt{6}\):

- For \(x < 2-\sqrt{6}\), the numerator is positive and the denominator is negative, so the expression is negative. (Satisfies the inequality)
- For \(2-\sqrt{6} < x < 2\), both are negative, so the expression is positive.
- For \(2 < x < 2+\sqrt{6}\), the numerator is negative and the denominator is positive, so the expression is negative. (Satisfies the inequality)
- For \(x > 2+\sqrt{6}\), both are positive, so the expression is positive.

Thus, the solution set is:

\(x < 2 - \sqrt{6} \text{ or } 2 < x < 2 + \sqrt{6}\)

M1: Multiplies by \((x-2)^2\) (or equivalent method) and attempts to find the critical values by equating the terms.

A1: Obtains the correct critical values \(x = 2\) and \(x = 2 \pm \sqrt{6}\).

A0.5: Correct final inequalities with correct inequality signs.

Let \(z = x + \mathrm{i}y\). Substituting this into the transformation formula:

\(w = \frac{x + \mathrm{i}(y-1)}{(x+2) + \mathrm{i}y}\)

Multiply the numerator and denominator by the complex conjugate of the denominator, \((x+2) - \mathrm{i}y\):

\(w = \frac{[x + \mathrm{i}(y-1)][(x+2) - \mathrm{i}y]}{(x+2)^2 + y^2}\)

Since \(w\) is purely imaginary, the real part of \(w\) must be equal to zero:

\(\operatorname{Re}(w) = \frac{x(x+2) + y(y-1)}{(x+2)^2 + y^2} = 0\)

This gives:

\(x(x+2) + y(y-1) = 0 \implies x^2 + 2x + y^2 - y = 0\)

Completing the square for both \(x\) and \(y\):

\((x+1)^2 - 1 + \left(y - \frac{1}{2}\right)^2 - \frac{1}{4} = 0\)

\((x+1)^2 + \left(y - \frac{1}{2}\right)^2 = \frac{5}{4}\)

M1: Substitutes \(z = x + \mathrm{i}y\), rationalises the denominator, and sets the real part of \(w\) to zero.

A1: Correct Cartesian equation in expanded form, \(x^2 + 2x + y^2 - y = 0\).

A0.5: Correctly completes the square to find the final equation in the required form.

Using partial fractions, we write:

\(\frac{1}{r(r+2)} = \frac{1}{2}\left(\frac{1}{r} - \frac{1}{r+2}\right)\)

Now, sum from \(r=1\) to \(n\):

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

Listing the terms for cancellation:

\(r=1: \quad \frac{1}{2} \left(1 - \frac{1}{3}\right)\)

\(r=2: \quad \frac{1}{2} \left(\frac{1}{2} - \frac{1}{4}\right)\)

\(r=3: \quad \frac{1}{2} \left(\frac{1}{3} - \frac{1}{5}\right)\)

\(\dots\)

\(r=n-1: \quad \frac{1}{2} \left(\frac{1}{n-1} - \frac{1}{n+1}\right)\)

\(r=n: \quad \frac{1}{2} \left(\frac{1}{n} - \frac{1}{n+2}\right)\)

Summing these terms, we see that intermediate terms cancel out, leaving:

\(\frac{1}{2} \left( 1 + \frac{1}{2} - \frac{1}{n+1} - \frac{1}{n+2} \right) = \frac{1}{2} \left( \frac{3}{2} - \frac{2n+3}{(n+1)(n+2)} \right)\)

\(= \frac{3(n+1)(n+2) - 2(2n+3)}{4(n+1)(n+2)}
= \frac{3(n^2+3n+2) - 4n-6}{4(n+1)(n+2)}
= \frac{3n^2 + 5n}{4(n+1)(n+2)}
= \frac{n(3n+5)}{4(n+1)(n+2)}\)

Thus, \(a = 3\) and \(b = 5\).

M1: Splits the term into partial fractions and writes down enough terms to show the cancellation structure.

A1: Obtains the correct unsimplified sum of remaining terms, \(\frac{1}{2}\left(\frac{3}{2} - \frac{1}{n+1} - \frac{1}{n+2}\right)\).

A0.5: Combines into a single fraction and identifies the correct values for \(a\) and \(b\).

Let \(f(x) = \ln(\sin x)\).

Evaluate \(f(x)\) at \(x = \frac{\pi}{4}\):

\(f\left(\frac{\pi}{4}\right) = \ln\left(\sin\frac{\pi}{4}\right) = \ln\left(\frac{1}{\sqrt{2}}\right) = -\frac{1}{2}\ln 2\)

First derivative:

\(f'(x) = \frac{\cos x}{\sin x} = \cot x\)

Evaluate at \(x = \frac{\pi}{4}\):

\(f'\left(\frac{\pi}{4}\right) = \cot\left(\frac{\pi}{4}\right) = 1\)

Second derivative:

\(f''(x) = -\csc^2 x\)

Evaluate at \(x = \frac{\pi}{4}\):

\(f''\left(\frac{\pi}{4}\right) = -\csc^2\left(\frac{\pi}{4}\right) = -2\)

Using the Taylor series formula:

\(f(x) \approx f\left(\frac{\pi}{4}\right) + f'\left(\frac{\pi}{4}\right)\left(x - \frac{\pi}{4}\right) + \frac{f''\left(\frac{\pi}{4}\right)}{2!}\left(x - \frac{\pi}{4}\right)^2\)

\(f(x) \approx -\frac{1}{2}\ln 2 + 1\left(x - \frac{\pi}{4}\right) + \frac{-2}{2}\left(x - \frac{\pi}{4}\right)^2\)

\(= -\frac{1}{2}\ln 2 + \left(x - \frac{\pi}{4}\right) - \left(x - \frac{\pi}{4}\right)^2\)

M1: Differentiates \(f(x)\) to find expressions for \(f'(x)\) and \(f''(x)\).

A1: Evaluates \(f\left(\frac{\pi}{4}\right)\), \(f'\left(\frac{\pi}{4}\right)\), and \(f''\left(\frac{\pi}{4}\right)\) correctly.

A0.5: Uses the Taylor series formula correctly to state the final expansion.

The differential equation is first-order linear in the form \(\frac{\mathrm{d}y}{\mathrm{d}x} + P(x)y = Q(x)\), where \(P(x) = \cot x\).

Find the integrating factor, \(I(x)\):

\(I(x) = \mathrm{e}^{\int \cot x \mathrm{d}x} = \mathrm{e}^{\ln(\sin x)} = \sin x\) (since \(0 < x < \pi\), \(\sin x > 0\))

Multiply the differential equation by \(I(x)\):

\(\sin x \frac{\mathrm{d}y}{\mathrm{d}x} + y \cos x = 2 \sin x \cos x\)

\(\frac{\mathrm{d}}{\mathrm{d}x}(y \sin x) = 2 \sin x \cos x\)

Integrate both sides with respect to \(x\):

\(y \sin x = \int 2 \sin x \cos x \mathrm{d}x = \sin^2 x + C\)

Using the boundary condition \(y\left(\frac{\pi}{2}\right) = 1\):

\((1)\sin\left(\frac{\pi}{2}\right) = \sin^2\left(\frac{\pi}{2}\right) + C\)

\(1 = 1 + C \implies C = 0\)

So:

\(y \sin x = \sin^2 x \implies y = \sin x\)

M1: Finds the correct integrating factor \(\sin x\) and multiplies the equation to set up the product rule derivative.

A1: Integrates correctly to find the general solution equation (e.g., \(y \sin x = \sin^2 x + C\) or \(y \sin x = -\frac{1}{2}\cos 2x + C'\)).

A0.5: Uses the boundary condition to find \(C\) and correctly gives the final particular solution as \(y = \sin x\).

To solve \(\left|\frac{2x+3}{x-1}\right| < 3\), we find the critical values where \(\frac{2x+3}{x-1} = 3\) or \(\frac{2x+3}{x-1} = -3\).

Case 1:
\[\frac{2x+3}{x-1} = 3 \implies 2x+3 = 3x-3 \implies x = 6\]

Case 2:
\[\frac{2x+3}{x-1} = -3 \implies 2x+3 = -3x+3 \implies 5x = 0 \implies x = 0\]

The expression is also undefined at \(x = 1\), which is a vertical asymptote.

Now, test the intervals defined by these critical points (\(x=0, 1, 6\)):
- For \(x < 0\): Let \(x = -1\). \(\left|\frac{1}{-2}\right| = \frac{1}{2} < 3\) (True)
- For \(0 < x < 1\): Let \(x = 0.5\). \(\left|\frac{4}{-0.5}\right| = 8 < 3\) (False)
- For \(1 < x < 6\): Let \(x = 2\). \(\left|\frac{7}{1}\right| = 7 < 3\) (False)
- For \(x > 6\): Let \(x = 7\). \(\left|\frac{17}{6}\right| < 3\) (True)

Thus, the solution set is \(x < 0\) or \(x > 6\).

M1: Attempt to find critical values by equating the fraction to \(\pm 3\) or by squaring both sides.
A1: Correct critical values \(x = 0\) and \(x = 6\).
M1: Identifies \(x = 1\) as a critical boundary (asymptote) and considers the four resulting intervals.
A1.5: Fully correct solution: \(x < 0\) or \(x > 6\) (or equivalent set notation).

Let \(z = x + \mathrm{i}y\). Substituting this into the given equation:
\[\left|\frac{x + \mathrm{i}(y-4)}{(x-2) + \mathrm{i}y}\right| = 2\]
Squaring both sides:
\[\frac{x^2 + (y-4)^2}{(x-2)^2 + y^2} = 4\]
\[x^2 + (y-4)^2 = 4\left((x-2)^2 + y^2\right)\]
Expanding both sides:
\[x^2 + y^2 - 8y + 16 = 4(x^2 - 4x + 4 + y^2)\]
\[x^2 + y^2 - 8y + 16 = 4x^2 - 16x + 16 + 4y^2\]
Rearranging terms to one side:
\[3x^2 + 3y^2 - 16x + 8y = 0\]
Dividing by 3:
\[x^2 - \frac{16}{3}x + y^2 + \frac{8}{3}y = 0\]
Completing the square for both \(x\) and \(y\):
\[\left(x - \frac{8}{3}\right)^2 - \frac{64}{9} + \left(y + \frac{4}{3}\right)^2 - \frac{16}{9} = 0\]
\[\left(x - \frac{8}{3}\right)^2 + \left(y + \frac{4}{3}\right)^2 = \frac{80}{9}\]
This is the equation of a circle.
- The centre is \(\left(\frac{8}{3}, -\frac{4}{3}\right)\), which corresponds to the complex number \(\frac{8}{3} - \frac{4}{3}\mathrm{i}\).
- The radius is \(\sqrt{\frac{80}{9}} = \frac{\sqrt{80}}{3} = \frac{4\sqrt{5}}{3}\).

M1: Substitute \(z = x + \mathrm{i}y\) and square both sides to eliminate the modulus.
A1: Obtain a correct Cartesian equation in expanded form before grouping terms.
M1: Rearrange and form a quadratic equation in \(x\) and \(y\) with coefficients of \(x^2\) and \(y^2\) equal.
A1: Complete the square correctly to find the standard form of the circle.
A1.5: Identify the centre as \(\frac{8}{3} - \frac{4}{3}\mathrm{i}\) and the radius as \(\frac{4\sqrt{5}}{3}\) (both must be exact and in simplified form).

(a) Equating numerators:
\[A(2r+1)^2 + B(2r-1)^2 = 4r\]
\[A(4r^2 + 4r + 1) + B(4r^2 - 4r + 1) = 4r\]
Comparing coefficients:
- For \(r^2\): \(4A + 4B = 0 \implies B = -A\)
- For \(r\): \(4A - 4B = 4 \implies 8A = 4 \implies A = \frac{1}{2}\)
- Therefore, \(B = -\frac{1}{2}\).
This yields:
\[\frac{4r}{(2r-1)^2(2r+1)^2} = \frac{1}{2}\left( \frac{1}{(2r-1)^2} - \frac{1}{(2r+1)^2} \right)\]

(b) Using the result from (a):
\[\sum_{r=1}^n u_r = \frac{1}{2} \sum_{r=1}^n \left[ \frac{1}{(2r-1)^2} - \frac{1}{(2r+1)^2} \right]\]
Listing terms:
- For \(r=1\): \(\frac{1}{2}\left( 1 - \frac{1}{9} \right)\)
- For \(r=2\): \(\frac{1}{2}\left( \frac{1}{9} - \frac{1}{25} \right)\)
...
- For \(r=n\): \(\frac{1}{2}\left( \frac{1}{(2n-1)^2} - \frac{1}{(2n+1)^2} \right)\)
Summing these terms cancels out intermediate fractions, leaving:
\[\sum_{r=1}^n u_r = \frac{1}{2} \left( 1 - \frac{1}{(2n+1)^2} \right) = \frac{1}{2} \left( \frac{(2n+1)^2 - 1}{(2n+1)^2} \right)\]
\[= \frac{1}{2} \left( \frac{4n^2 + 4n}{(2n+1)^2} \right) = \frac{2n(n+1)}{(2n+1)^2}\]

Part (a):
M1: Set up the algebraic system to solve for \(A\) and \(B\) by equating numerators.
A1: Correct values: \(A = \frac{1}{2}\) and \(B = -\frac{1}{2}\).

Part (b):
M1: Expresses the sum using the partial fractions and writes out terms to demonstrate cancellation.
A1: Obtains the correct simplified expression before factoring: \(\frac{1}{2} \left( 1 - \frac{1}{(2n+1)^2} \right)\).
M1: Places over a common denominator and expands the numerator.
A0.5: Reaches the required form \(\frac{2n(n+1)}{(2n+1)^2}\) with clear, rigorous algebra.

We are given \(y(0) = 1\) and \(y'(0) = -2\).
Substitute these into the differential equation at \(x = 0\):
\[y''(0) + (1+0)y'(0) + (y(0))^2 = 0 \implies y''(0) + 1(-2) + 1^2 = 0 \implies y''(0) = 1\]

Differentiate the original differential equation with respect to \(x\):
\[\frac{\mathrm{d}^3y}{\mathrm{d}x^3} + (1+x)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + (1)\frac{\mathrm{d}y}{\mathrm{d}x} + 2y\frac{\mathrm{d}y}{\mathrm{d}x} = 0\]

Substitute \(x = 0\), \(y(0) = 1\), \(y'(0) = -2\), and \(y''(0) = 1\) into this equation:
\[y'''(0) + (1)(1) + (-2) + 2(1)(-2) = 0\]
\[y'''(0) + 1 - 2 - 4 = 0 \implies y'''(0) = 5\]

The Taylor series expansion of \(y\) around \(x = 0\) is:
\[y(x) = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \dots\]
Substituting our values:
\[y(x) = 1 - 2x + \frac{1}{2}x^2 + \frac{5}{6}x^3\]

B1: Obtains \(y''(0) = 1\) using the given values in the differential equation.
M1: Correctly differentiates the given DE with respect to \(x\) (requiring both product and chain rules).
A1: Correct differentiated expression: \(\frac{\mathrm{d}^3y}{\mathrm{d}x^3} + (1+x)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + \frac{\mathrm{d}y}{\mathrm{d}x} + 2y\frac{\mathrm{d}y}{\mathrm{d}x} = 0\).
A1: Substitutes values to obtain \(y'''(0) = 5\).
M1: Applies the Taylor series formula correctly.
A0.5: Obtains the correct simplified expansion: \(y = 1 - 2x + \frac{1}{2}x^2 + \frac{5}{6}x^3\).

First, find the boundaries of integration by finding the points of intersection:
\[2(1 + \cos\theta) = 3 \implies \cos\theta = \frac{1}{2} \implies \theta = \pm\frac{\pi}{3}\]

For the region inside \(C_1\) but outside \(C_2\), we integrate over the interval \(-\frac{\pi}{3} \le \theta \le \frac{\pi}{3}\). Due to symmetry, the area \(A\) is:
\[A = \int_{0}^{\pi/3} \left( r_1^2 - r_2^2 \right) \mathrm{d}\theta\]
\[A = \int_{0}^{\pi/3} \left( 4(1+\cos\theta)^2 - 9 \right) \mathrm{d}\theta\]
\[A = \int_{0}^{\pi/3} \left( 4 + 8\cos\theta + 4\cos^2\theta - 9 \right) \mathrm{d}\theta\]
\[A = \int_{0}^{\pi/3} \left( 8\cos\theta + 4\cos^2\theta - 5 \right) \mathrm{d}\theta\]
Using the identity \(\cos^2\theta = \frac{1+\cos 2\theta}{2}\):
\[A = \int_{0}^{\pi/3} \left( 8\cos\theta + 2(1+\cos 2\theta) - 5 \right) \mathrm{d}\theta\]
\[A = \int_{0}^{\pi/3} \left( 8\cos\theta + 2\cos 2\theta - 3 \right) \mathrm{d}\theta\]
Integrating:
\[A = \left[ 8\sin\theta + \sin 2\theta - 3\theta \right]_{0}^{\pi/3}\]
Evaluating at the limits:
\[A = \left( 8\sin\frac{\pi}{3} + \sin\frac{2\pi}{3} - 3\left(\frac{\pi}{3}\right) \right) - (0)\]
\[A = 8\left(\frac{\sqrt{3}}{2}\right) + \frac{\sqrt{3}}{2} - \pi = \frac{9\sqrt{3}}{2} - \pi\]

M1: Set \(r_1 = r_2\) and solve to find the intersection points \(\theta = \pm\frac{\pi}{3}\).
M1: Apply the polar area difference formula with correct limits of integration.
A1: Correct unsimplified integrand: \(8\cos\theta + 4\cos^2\theta - 5\).
M1: Apply the double angle identity \(\cos^2\theta = \frac{1+\cos 2\theta}{2}\) to make the terms integrable.
A1: Integrate correctly to obtain \(8\sin\theta + \sin 2\theta - 3\theta\).
A0.5: Correctly substitute limits to get the exact answer \(\frac{9\sqrt{3}}{2} - \pi\).

This is a first-order linear differential equation with integrating factor \(I(x)\):
\[I(x) = \mathrm{e}^{\int \frac{2}{x} \mathrm{d}x} = \mathrm{e}^{2\ln x} = x^2\]

Multiplying both sides of the differential equation by \(x^2\):
\[x^2\frac{\mathrm{d}y}{\mathrm{d}x} + 2xy = \sin x\]
\[\frac{\mathrm{d}}{\mathrm{d}x}\left(x^2 y\right) = \sin x\]

Integrating both sides with respect to \(x\):
\[x^2 y = \int \sin x \mathrm{d}x = -\cos x + C\]

Apply the initial condition \(y = 1\) when \(x = \pi\):
\[\pi^2 (1) = -\cos\pi + C\]
\[\pi^2 = -(-1) + C \implies C = \pi^2 - 1\]

Thus, the particular solution is:
\[x^2 y = -\cos x + \pi^2 - 1 \implies y = \frac{\pi^2 - 1 - \cos x}{x^2}\]

M1: Finds the correct integrating factor \(I(x) = x^2\).
A1: Converts the differential equation to the correct form \(\frac{\mathrm{d}}{\mathrm{d}x}(x^2 y) = \sin x\).
M1: Integrates both sides with respect to \(x\) and includes an integration constant.
A1: Obtains the general solution \(x^2 y = -\cos x + C\).
M1: Evaluates the constant \(C\) using the given initial condition.
A0.5: Expresses the final answer in terms of \(y\) correctly: \(y = \frac{\pi^2 - 1 - \cos x}{x^2}\).

First, find the complementary function (CF) by solving the auxiliary equation:
\[m^2 - 4m + 4 = 0 \implies (m-2)^2 = 0 \implies m = 2 \quad (\text{repeated root})\]
So, the complementary function is:
\[y_c = (A + Bx)\mathrm{e}^{2x}\]

Next, find a particular integral (PI). Since \(2\) is a double root of the auxiliary equation, we try:
\[y_p = C x^2 \mathrm{e}^{2x}\]

Differentiating \(y_p\):
\[y_p' = 2Cx\mathrm{e}^{2x} + 2Cx^2\mathrm{e}^{2x} = 2C(x + x^2)\mathrm{e}^{2x}\]
\[y_p'' = 2C(1 + 2x)\mathrm{e}^{2x} + 4C(x + x^2)\mathrm{e}^{2x} = 2C(1 + 4x + 2x^2)\mathrm{e}^{2x}\]

Substituting these into the original differential equation:
\[2C(1 + 4x + 2x^2)\mathrm{e}^{2x} - 4\left[2C(x + x^2)\mathrm{e}^{2x}\] + 4Cx^2\mathrm{e}^{2x} = 8\mathrm{e}^{2x}\]
Divide through by \(\mathrm{e}^{2x}\):
\[2C(1 + 4x + 2x^2) - 8C(x + x^2) + 4Cx^2 = 8\]
\[C(2 + 8x + 4x^2 - 8x - 8x^2 + 4x^2) = 8 \implies 2C = 8 \implies C = 4\]

So the particular integral is:
\[y_p = 4x^2 \mathrm{e}^{2x}\]

The general solution is:
\[y = (A + Bx)\mathrm{e}^{2x} + 4x^2 \mathrm{e}^{2x}\]

M1: Solves auxiliary equation to find the repeated root \(m = 2\).
A1: Formulates the correct complementary function \(y_c = (A + Bx)\mathrm{e}^{2x}\).
M1: Proposes a correct Particular Integral form \(y_p = C x^2 \mathrm{e}^{2x}\) and differentiates to find \(y_p'\) and \(y_p''\).
A1: Correctly substitutes into the DE and solves to find \(C = 4\).
A1.5: Combines CF and PI to give the correct general solution.

Part (a):
Divide the equation by \(x^2 - 4\) to obtain the standard form:
\[\frac{\mathrm{d}y}{\mathrm{d}x} + \frac{4x}{x^2 - 4}y = \frac{2x}{x^2 - 4}\]

Find the integrating factor \(I(x)\):
\[I(x) = \mathrm{e}^{\int \frac{4x}{x^2-4} \mathrm{d}x} = \mathrm{e}^{2 \ln|x^2 - 4|} = (x^2 - 4)^2\]

Multiply both sides of the differential equation by \(I(x)\):
\[(x^2 - 4)^2 \frac{\mathrm{d}y}{\mathrm{d}x} + 4x(x^2 - 4)y = 2x(x^2 - 4)\]
\[\frac{\mathrm{d}}{\mathrm{d}x}\left[ y(x^2 - 4)^2 \right] = 2x^3 - 8x\]

Integrate both sides with respect to \(x\):
\[y(x^2 - 4)^2 = \int (2x^3 - 8x) \mathrm{d}x = \frac{1}{2}x^4 - 4x^2 + C\]

Solve for \(y\):
\[y = \frac{\frac{1}{2}x^4 - 4x^2 + C}{(x^2 - 4)^2}\]
Alternatively, we can write the integrated right-hand side as \(\frac{1}{2}(x^2 - 4)^2 + A\), yielding:
\[y = \frac{1}{2} + \frac{A}{(x^2 - 4)^2}\]

Part (b):
Using the boundary condition \(y = 1\) when \(x = 3\):
\[1 = \frac{1}{2} + \frac{A}{(3^2 - 4)^2}\]
\[\frac{1}{2} = \frac{A}{25} \implies A = \frac{25}{2}\]

Substitute \(A\) back into the equation:
\[y = \frac{1}{2} + \frac{25}{2(x^2 - 4)^2} = \frac{x^4 - 8x^2 + 41}{2(x^2 - 4)^2}\]

Part (a):
M1: Attempt to write the differential equation in standard form and find the integrating factor \(I(x)\).
A1: Correct integrating factor \((x^2 - 4)^2\).
M1: Multiply by the integrating factor and attempt to integrate the RHS.
A1: Correct integration: \(y(x^2 - 4)^2 = \frac{1}{2}x^4 - 4x^2 + C\) (or equivalent).
A1: Correct general solution for \(y\).

Part (b):
M1: Use the boundary condition \(y = 1\) when \(x = 3\) to set up an equation for the constant of integration.
A1: Correct value of the constant (e.g., \(A = 12.5\) or \(C = 20.5\)).
A1: Correct particular solution in the required form, e.g., \(y = \frac{1}{2} + \frac{25}{2(x^2 - 4)^2}\) or \(y = \frac{x^4 - 8x^2 + 41}{2(x^2 - 4)^2}\).

Part (a):
To find the point of intersection, equate \(r\) for both curves:
\[2(1 + \cos\theta) = 3\]
\[1 + \cos\theta = 1.5\n\implies \cos\theta = 0.5\]
Since \(0 \le \theta \le \pi\):
\[\theta = \frac{\pi}{3}\]
Thus, the polar coordinates of the point of intersection are \(\left(3, \frac{\pi}{3}\right)\).

Part (b):
The region inside \(C_1\) but outside \(C_2\) lies between \(\theta = 0\) and \(\theta = \frac{\pi}{3}\), where \(r_{C_1} \ge r_{C_2}\).
The area \(A\) of this region is:
\[A = \frac{1}{2} \int_{0}^{\frac{\pi}{3}} \left( r_{C_1}^2 - r_{C_2}^2 \right) \mathrm{d}\theta\]
\[A = \frac{1}{2} \int_{0}^{\frac{\pi}{3}} \left( 4(1 + \cos\theta)^2 - 9 \right) \mathrm{d}\theta\]

Expand the integrand:
\[4(1 + 2\cos\theta + \cos^2\theta) - 9 = 8\cos\theta + 4\cos^2\theta - 5\]
Use the identity \(\cos^2\theta = \frac{1 + \cos 2\theta}{2}\):
\[8\cos\theta + 2(1 + \cos 2\theta) - 5 = 8\cos\theta + 2\cos 2\theta - 3\]

Now integrate:
\[\int_{0}^{\frac{\pi}{3}} (8\cos\theta + 2\cos 2\theta - 3) \mathrm{d}\theta = \left[ 8\sin\theta + \sin 2\theta - 3\theta \right]_{0}^{\frac{\pi}{3}}\]
Evaluate at the upper limit \(\theta = \frac{\pi}{3}\):
\[8\sin\left(\frac{\pi}{3}\right) + \sin\left(\frac{2\pi}{3}\right) - 3\left(\frac{\pi}{3}\right) = 8\left(\frac{\sqrt{3}}{2}\right) + \frac{\sqrt{3}}{2} - \pi = \frac{9\sqrt{3}}{2} - \pi\]
Evaluate at the lower limit \(\theta = 0\):
\[0\]

Thus, the area is:
\[A = \frac{1}{2} \left( \frac{9\sqrt{3}}{2} - \pi \right) = \frac{9\sqrt{3}}{4} - \frac{\pi}{2}\]

Part (a):
M1: Equate the two polar equations and solve for \(\cos\theta\).
A1: Correct coordinates \(\left(3, \frac{\pi}{3}\right)\).

Part (b):
M1: Formulate the area integral with correct limits and integrand structure \(\frac{1}{2} \int (r_1^2 - r_2^2) \mathrm{d}\theta\).
M1: Expand \(r_1^2\) and use the double-angle identity \(\cos^2\theta = \frac{1}{2}(1 + \cos 2\theta)\) to rewrite the integrand.
A1: Correct simplified integrand: \(8\cos\theta + 2\cos 2\theta - 3\).
M1: Integrate the terms correctly to obtain \(8\sin\theta + \sin 2\theta - 3\theta\).
M1: Substitute the limits \(0\) and \(\frac{\pi}{3}\) into their integrated expression.
A1: Correct final exact area: \\frac{9\\sqrt{3}}{4} - \\frac{\\pi}{2} (or equivalent exact expression).

Part (a):
Combine the fractions on the right-hand side over a common denominator:
\[\frac{1}{r^2} - \frac{1}{(r+1)^2} = \frac{(r+1)^2 - r^2}{r^2(r+1)^2} = \frac{r^2 + 2r + 1 - r^2}{r^2(r+1)^2} = \frac{2r+1}{r^2(r+1)^2}\]
This completes the proof.

Part (b):
Using the result from (a), write the sum as:
\[\sum_{r=1}^{n} \frac{2r+1}{r^2(r+1)^2} = \sum_{r=1}^{n} \left( \frac{1}{r^2} - \frac{1}{(r+1)^2} \right)\]
Write out the terms explicitly to observe the cancellation:
\[\left( 1 - \frac{1}{4} \right) + \left( \frac{1}{4} - \frac{1}{9} \right) + \dots + \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} \right)\]
All intermediate terms cancel, leaving:
\[1 - \frac{1}{(n+1)^2} = \frac{(n+1)^2 - 1}{(n+1)^2} = \frac{n(n+2)}{(n+1)^2}\]

Part (c):
As \(n \to \infty\), \(\frac{1}{(n+1)^2} \to 0\).
Therefore, the sum to infinity is:
\[\lim_{n \to \infty} \left( 1 - \frac{1}{(n+1)^2} \right) = 1\]

Part (d):
The sum from \(r=5\) to \(r=20\) is:
\[\sum_{r=5}^{20} \frac{2r+1}{r^2(r+1)^2} = \sum_{r=1}^{20} \frac{2r+1}{r^2(r+1)^2} - \sum_{r=1}^{4} \frac{2r+1}{r^2(r+1)^2}\]
\[= \left( 1 - \frac{1}{21^2} \right) - \left( 1 - \frac{1}{5^2} \right) = \frac{1}{25} - \frac{1}{441}\]
\[= \frac{441 - 25}{11025} = \frac{416}{11025} \approx 0.0377324...\]
To 5 decimal places, this is \(0.03773\).

Part (a):
M1: Combine the RHS fractions over a common denominator.
A1: Fully correct simplification to obtain the LHS (show intermediate steps clearly).

Part (b):
M1: Apply the method of differences by writing out several terms of the sum to show the cancellation.
A1: Correctly identify the remaining terms: \(1 - \frac{1}{(n+1)^2}\).
A1: Combine into a single algebraic fraction: \\frac{n(n+2)}{(n+1)^2}\.

Part (c):
B1: State the sum to infinity is 1.

Part (d):
M1: Express the sum as a difference of two sums from 1, or use \\left( \\frac{1}{5^2} - \\frac{1}{21^2} \\right) directly.
A1: Correct numerical value to 5 decimal places: 0.03773.

First, express \(z\) in terms of \(w\):
\[w(z - 3\mathrm{i}) = z + \mathrm{i}\]
\[wz - 3\mathrm{i}w = z + \mathrm{i}\]
\[z(w - 1) = \mathrm{i}(3w + 1)\]
\[z = \frac{\mathrm{i}(3w + 1)}{w - 1}\]

Since \(|z| = 1\), we have:
\[\left| \frac{\mathrm{i}(3w + 1)}{w - 1} \right| = 1\]
Since \(|\mathrm{i}| = 1\), this simplifies to:
\[|3w + 1| = |w - 1|\]

Let \(w = u + \mathrm{i}v\) where \(u\) and \(v\) are real:
\[|3(u + \mathrm{i}v) + 1| = |(u + \mathrm{i}v) - 1|\]
\[|(3u + 1) + 3\mathrm{i}v| = |(u - 1) + \mathrm{i}v|\]

Square both sides:
\[(3u + 1)^2 + 9v^2 = (u - 1)^2 + v^2\]
\[9u^2 + 6u + 1 + 9v^2 = u^2 - 2u + 1 + v^2\]
\[8u^2 + 8v^2 + 8u = 0\]

Divide the entire equation by 8:
\[u^2 + v^2 + u = 0\]

Complete the square for \(u\):
\[\left(u + \frac{1}{2}\right)^2 + v^2 = \frac{1}{4}\]

This is the equation of a circle. Therefore, the curve \(C\) is a circle with:
Center: \(w = -0.5\) (or \(-0.5 + 0\mathrm{i}\))
Radius: \(0.5\) (or \(\frac{1}{2}\))

M1: Rearrange the transformation equation to make \(z\) the subject.
A1: Obtain the correct expression for \(z = \frac{\mathrm{i}(3w + 1)}{w - 1}\) (or equivalent form).
M1: Apply the condition \(|z| = 1\) to get an equation in terms of \(w\), e.g., \(|3w + 1| = |w - 1|\).
M1: Let \(w = u + \mathrm{i}v\) (or \(x + \mathrm{i}y\)) and square both sides.
A1: Correctly expand and simplify to get \(8u^2 + 8v^2 + 8u = 0\) (or equivalent).
M1: Complete the square for their quadratic expression.
A1: Correct standard form of the circle equation, e.g., \((u + 0.5)^2 + v^2 = 0.25\).
A1: Correctly identify the center as \(-0.5\) and the radius as \(0.5\).

Using the hyperbolic identity \(\cosh^2 x = \frac{1}{2}(\cosh 2x + 1)\), we can rewrite the integral: \(\int_0^{\ln 3} \cosh^2 x \, \mathrm{d}x = \int_0^{\ln 3} \left(\frac{1}{2}\cosh 2x + \frac{1}{2}\right) \mathrm{d}x = \left[ \frac{1}{4}\sinh 2x + \frac{1}{2}x \right]_0^{\ln 3}\). Evaluating at the upper limit: \(\frac{1}{4}\sinh(2\ln 3) + \frac{1}{2}\ln 3 = \frac{1}{4}\sinh(\ln 9) + \frac{1}{2}\ln 3\). Since \(\sinh(\ln 9) = \frac{e^{\ln 9} - e^{-\ln 9}}{2} = \frac{9 - 1/9}{2} = \frac{40}{9}\), the term becomes \(\frac{1}{4} \left(\frac{40}{9}\right) = \frac{10}{9}\). Evaluating at the lower limit gives 0. Thus, the exact value is \(\frac{10}{9} + \frac{1}{2}\ln 3\).

M1: Applies the identity \(\cosh^2 x = \frac{1}{2}(\cosh 2x + 1)\) and integrates successfully. M1: Substitutes the limits and uses the exponential form of \(\sinh(2\ln 3)\) to simplify the first term. A1: Obtains the correct final exact answer \(\frac{10}{9} + \frac{1}{2}\ln 3\).

Let \(y = \operatorname{artanh}(\sin x)\). Using the chain rule, we have: \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{1 - \sin^2 x} \cdot \cos x\). Since \(1 - \sin^2 x = \cos^2 x\), the derivative simplifies to: \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\cos x}{\cos^2 x} = \sec x\). Evaluating at \(x = \frac{\pi}{6}\) gives: \(\sec\left(\frac{\pi}{6}\right) = \frac{1}{\cos(\pi/6)} = \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}\).

M1: Applies the chain rule to differentiate \(\operatorname{artanh}(\sin x)\). M1: Simplifies the resulting expression to \(\sec x\) or equivalent. A1: Obtains the correct exact value of \(\frac{2\sqrt{3}}{3}\) or equivalent.

The parametric coordinates on the ellipse are \(x = 4\cos\theta\) and \(y = 3\sin\theta\). At \(\theta = \frac{\pi}{4}\), the point \(P\) is \((2\sqrt{2}, \frac{3\sqrt{2}}{2})\). We find the gradient of the tangent using: \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y/\mathrm{d}\theta}{\mathrm{d}x/\mathrm{d}\theta} = \frac{3\cos\theta}{-4\sin\theta} = -\frac{3}{4}\cot\theta\). At \(\theta = \frac{\pi}{4}\), the gradient of the tangent is \(-\frac{3}{4}\), so the gradient of the normal is \(m = \frac{4}{3}\). Using the equation of a straight line: \(y - \frac{3\sqrt{2}}{2} = \frac{4}{3}(x - 2\sqrt{2}) \implies 3y - \frac{9\sqrt{2}}{2} = 4x - 8\sqrt{2} \implies 4x - 3y = \frac{7\sqrt{2}}{2}\). Multiplying by 2 to obtain integer coefficients gives: \(8x - 6y = 7\sqrt{2}\).

M1: Finds the coordinates of the point and the gradient of the tangent or normal. M1: Sets up the equation of the normal with their point and gradient. A1: Obtains the correct equation in the specified form with integer coefficients.

Let \(u = 2x\), which means \(\mathrm{d}u = 2\,\mathrm{d}x\). Substituting these into the integral gives: \(\int \frac{1}{\sqrt{u^2 - 9}} \frac{1}{2} \, \mathrm{d}u = \frac{1}{2} \int \frac{1}{\sqrt{u^2 - 3^2}} \, \mathrm{d}u\). Using the standard formula \(\int \frac{1}{\sqrt{y^2 - a^2}} \, \mathrm{d}y = \operatorname{arcosh}\left(\frac{y}{a}\right) + C\), we get: \(\frac{1}{2} \operatorname{arcosh}\left(\frac{u}{3}\right) + C = \frac{1}{2} \operatorname{arcosh}\left(\frac{2x}{3}\right) + C\).

M1: Employs a substitution or recognizes the standard integral form. M1: Integrates correctly, taking care of the scaling factor of 2. A1: Provides the correct final expression including the constant of integration.

The parallel planes have a common normal vector \(\mathbf{n} = 2\mathbf{i} - \mathbf{j} + 2\mathbf{k}\). The magnitude of this normal vector is \(|\mathbf{n}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{9} = 3\). The equations of the planes in normalized form \(\mathbf{r} \cdot \hat{\mathbf{n}} = d\) are \(\mathbf{r} \cdot \hat{\mathbf{n}} = \frac{5}{3}\) and \(\mathbf{r} \cdot \hat{\mathbf{n}} = -\frac{4}{3}\). The shortest distance between them is the absolute difference between these values: \(d = \left|\frac{5}{3} - \left(-\frac{4}{3}\right)\right| = \frac{9}{3} = 3\).

M1: Calculates the magnitude of the normal vector to be 3. M1: Utilizes a correct method to find the perpendicular distance between the planes. A1: Obtains the correct distance of 3.

To find the eigenvalues, we solve the characteristic equation \(\det(\mathbf{M} - \lambda \mathbf{I}) = 0\): \(\det\begin{pmatrix} 1-\lambda & 0 & 2 \\ 0 & 3-\lambda & 0 \\ 2 & 0 & 1-\lambda \end{pmatrix} = 0\). Expanding along the second row: \((3-\lambda) \begin{vmatrix} 1-\lambda & 2 \\ 2 & 1-\lambda \end{vmatrix} = 0 \implies (3-\lambda)((1-\lambda)^2 - 4) = 0\). This gives \(3-\lambda = 0\) or \((1-\lambda)^2 = 4 \implies 1-\lambda = \pm 2\). Therefore, \(\lambda = 3\) or \(\lambda = -1\). The eigenvalues are \(-1\) and \(3\) (where 3 is a repeated eigenvalue).

M1: Sets up the determinant equation for the characteristic polynomial. M1: Expands the determinant and factorizes the resulting cubic equation. A1: Identifies the correct eigenvalues of \(-1\) and \(3\).

Using the definitions of hyperbolic functions in terms of exponentials: \(3\left(\frac{e^x + e^{-x}}{2}\right) - \left(\frac{e^x - e^{-x}}{2}\right) = 3 \implies 3(e^x + e^{-x}) - (e^x - e^{-x}) = 6\). Simplifying this yields: \(2e^x + 4e^{-x} = 6 \implies e^x + 2e^{-x} = 3\). Multiplying through by \(e^x\) gives a quadratic in \(e^x\): \(e^{2x} - 3e^x + 2 = 0\). Factoring the quadratic: \((e^x - 1)(e^x - 2) = 0\), which leads to \(e^x = 1\) or \(e^x = 2\). Taking natural logarithms, we obtain the real solutions: \(x = 0\) or \(x = \ln 2\).

M1: Substitutes the exponential definitions of \(\cosh x\) and \(\sinh x\) and simplifies to form a quadratic equation in \(e^x\). M1: Solves the quadratic equation to find the values of \(e^x\). A1: Correctly identifies the solutions \(x = 0\) and \(x = \ln 2\).

The formula for the arc length of a curve is \(s = \int_{a}^{b} \sqrt{1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2} \, \mathrm{d}x\). Here, \(\frac{\mathrm{d}y}{\mathrm{d}x} = \sinh x\), so \(1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = 1 + \sinh^2 x = \cosh^2 x\). Since \(\cosh x > 0\) for all real \(x\), the integrand simplifies to \(\cosh x\). The arc length is: \(s = \int_0^{\ln 3} \cosh x \, \mathrm{d}x = \left[ \sinh x \right]_0^{\ln 3} = \sinh(\ln 3) - \sinh(0)\). Using \ \(\sinh(\ln 3) = \frac{e^{\ln 3} - e^{-\ln 3}}{2} = \frac{3 - 1/3}{2} = \frac{4}{3}\) and \(\sinh(0) = 0\), we get \(s = \frac{4}{3}\).

M1: Uses the arc length formula and applies the identity \(1 + \sinh^2 x = \cosh^2 x\) to simplify the integrand. M1: Integrates to find \(\sinh x\) and substitutes the limits 0 and \(\ln 3\). A1: Obtains the correct exact value of \(\frac{4}{3}\).

Using the logarithmic definition of the inverse hyperbolic tangent function: \(\operatorname{artanh} x = \frac{1}{2} \ln \left(\frac{1+x}{1-x}\right)\), for \(|x| < 1\). For \(x = \frac{1}{3}\): \(\operatorname{artanh}\left(\frac{1}{3}\right) = \frac{1}{2} \ln \left(\frac{1 + \frac{1}{3}}{1 - \frac{1}{3}}\right) = \frac{1}{2} \ln \left(\frac{\frac{4}{3}}{\frac{2}{3}}\right) = \frac{1}{2} \ln 2\). For \(x = \frac{1}{2}\): \(\operatorname{artanh}\left(\frac{1}{2}\right) = \frac{1}{2} \ln \left(\frac{1 + \frac{1}{2}}{1 - \frac{1}{2}}\right) = \frac{1}{2} \ln \left(\frac{\frac{3}{2}}{\frac{1}{2}}\right) = \frac{1}{2} \ln 3\). Adding the two values: \(\operatorname{artanh}\left(\frac{1}{3}\right) + \operatorname{artanh}\left(\frac{1}{2}\right) = \frac{1}{2} \ln 2 + \frac{1}{2} \ln 3 = \frac{1}{2}(\ln 2 + \ln 3) = \frac{1}{2} \ln 6 = \ln \sqrt{6}\).

M1: Applies the logarithmic identity \(\operatorname{artanh} x = \frac{1}{2} \ln \left(\frac{1+x}{1-x}\right)\) to at least one term. A1: Obtains both correct simplified logarithmic terms, namely \(\frac{1}{2} \ln 2\) (or \(\ln \sqrt{2}\)) and \(\frac{1}{2} \ln 3\) (or \(\ln \sqrt{3}\)). A1: Combines the terms correctly to obtain the final answer of \(\ln \sqrt{6}\).

Using the identity \(\cosh^2 x = 1 + \sinh^2 x\), we can rewrite the equation as \(2(1 + \sinh^2 x) - 5\sinh x = 5\). Rearranging gives the quadratic equation \(2\sinh^2 x - 5\sinh x - 3 = 0\). Factoring the quadratic, we obtain \((2\sinh x + 1)(\sinh x - 3) = 0\), which yields \(\sinh x = -\frac{1}{2}\) or \(\sinh x = 3\). Since \(x > 0\), we must have \(\sinh x > 0\), so we reject \(\sinh x = -\frac{1}{2}\) and choose \(\sinh x = 3\). Using the logarithmic form of the inverse hyperbolic sine, we get \(x = \text{arsinh}(3) = \ln(3 + \sqrt{3^2 + 1}) = \ln(3 + \sqrt{10})\). Thus, \(a = 3\) and \(b = 10\).

M1: Substitutes the identity \(\cosh^2 x = 1 + \sinh^2 x\) into the given equation.
A1: Obtains a correct quadratic equation in terms of \(\sinh x\), such as \(2\sinh^2 x - 5\sinh x - 3 = 0\).
M1: Factorises or uses the quadratic formula to find the values of \(\sinh x\).
A1: Correctly identifies \(\sinh x = 3\) and justifies rejecting \(\sinh x = -1/2\) because \(x > 0\).
A1: Applies the logarithmic form of \(\text{arsinh}\) to obtain the correct final exact answer \(\ln(3 + \sqrt{10})\).

Differentiating \(y\) with respect to \(x\) using the chain rule gives \(\frac{dy}{dx} = 2(\text{arcosh } x) \frac{1}{\sqrt{x^2 - 1}}\). Squaring both sides and multiplying by \((x^2 - 1)\), we obtain \((x^2 - 1)\left(\frac{dy}{dx}\right)^2 = 4(\text{arcosh } x)^2 = 4y\). Differentiating implicitly with respect to \(x\) gives \(2x\left(\frac{dy}{dx}\right)^2 + 2(x^2 - 1)\frac{dy}{dx}\frac{d^2 y}{dx^2} = 4\frac{dy}{dx}\). Dividing both sides by \(2\frac{dy}{dx}\) (since \(\frac{dy}{dx} \neq 0\) for \(x > 1\)), we obtain \((x^2 - 1)\frac{d^2 y}{dx^2} + x\frac{dy}{dx} = 2\). Thus, \(k = 2\).

M1: Differentiates \(y\) using the chain rule to obtain the first derivative.
A1: Correct expression for the first derivative \(\frac{dy}{dx} = \frac{2\text{arcosh } x}{\sqrt{x^2 - 1}}\).
M1: Rearranges the derivative expression, squares both sides to obtain \((x^2 - 1)\left(\frac{dy}{dx}\right)^2 = 4y\), or correctly attempts second differentiation with quotient/product rule.
M1: Differentiates implicitly with respect to \(x\) to find the relationship involving \(\frac{d^2y}{dx^2}\).
A1: Simplifies the resulting expression to show \((x^2 - 1)\frac{d^2 y}{dx^2} + x\frac{dy}{dx} = 2\), concluding that \(k = 2\).

First, we complete the square for the quadratic expression under the square root: \(x^2 - 2x + 5 = (x-1)^2 + 4\). Let \(u = x-1\), so \(du = dx\). The limits transform as follows: when \(x = 1\), \(u = 0\); when \(x = 2\), \(u = 1\). The integral becomes \(\int_{0}^{1} \frac{1}{\sqrt{u^2 + 4}} \, du = \left[ \text{arsinh}\left(\frac{u}{2}\right) \right]_0^1 = \text{arsinh}\left(\frac{1}{2}\right) - \text{arsinh}(0) = \text{arsinh}\left(\frac{1}{2}\right)\). Using the logarithmic identity \(\text{arsinh}(x) = \ln\left(x + \sqrt{x^2 + 1}\right)\), we obtain \(\text{arsinh}\left(\frac{1}{2}\right) = \ln\left(\frac{1}{2} + \sqrt{\frac{1}{4} + 1}\right) = \ln\left(\frac{1}{2} + \sqrt{\frac{5}{4}}\right)\). Thus, \(a = \frac{1}{2}\) and \(b = \frac{5}{4}\).

M1: Completes the square on the quadratic term under the radical to obtain \((x-1)^2 + 4\).
A1: Integrates correctly to obtain \(\text{arsinh}\left(\frac{x-1}{2}\right)\) (or equivalent log form).
M1: Substitutes the upper and lower integration limits of \(2\) and \(1\) into their integrated expression.
M1: Uses the logarithmic form of the inverse hyperbolic sine to convert their expression.
A1: Obtains the correct exact final answer in the form \(\ln\left(\frac{1}{2} + \sqrt{\frac{5}{4}}\right)\) (or equivalent).

(a) Since \(\lambda = 2\) is an eigenvalue of \(\mathbf{M}\), the determinant of \(\mathbf{M} - 2\mathbf{I}\) must be zero. \(\mathbf{M} - 2\mathbf{I} = \begin{pmatrix} -1 & -1 & 3 \\ 2 & k-2 & 1 \\ 0 & 1 & -1 \end{pmatrix}\). Expanding along the first row: \(-1((k-2)(-1) - 1) - (-1)(2(-1) - 0) + 3(2(1) - 0) = 0 \Rightarrow -1(-k + 2 - 1) + 1(-2) + 6 = 0 \Rightarrow k - 1 - 2 + 6 = 0 \Rightarrow k = -3\). (b) For \(k = -3\), we solve \((\mathbf{M} - 2\mathbf{I})\mathbf{x} = \mathbf{0}\): \
\(\begin{pmatrix} -1 & -1 & 3 \\ 2 & -5 & 1 \\ 0 & 1 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\). From the third row: \(y - z = 0 \Rightarrow y = z\). From the first row: \(-x - y + 3z = 0 \Rightarrow x = 2z\). Choosing \(z = 1\), we get \(y = 1\) and \(x = 2\). Thus, a corresponding eigenvector is \(\begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}\).

M1: Sets up the equation \\det(\mathbf{M} - 2\mathbf{I}) = 0\\ to find the unknown parameter \(k\).
A1: Solves the determinant equation to find \(k = -3\).
M1: Formulates the linear system of equations for the eigenvector using \(k = -3\).
M1: Solves the system of equations to find a relationship between the variables, e.g., \(x = 2z\) and \(y = z\).
A1: Gives a correct non-zero eigenvector, such as \(\begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}\) or any scalar multiple of it.

For the ellipse \(E\), we have \(a^2 = 25\) (so \(a = 5\)) and \(b^2 = 9\) (so \(b = 3\)).
(i) The eccentricity \(e\) satisfies the relation \(b^2 = a^2(1 - e^2)\). Hence, \(9 = 25(1 - e^2) \Rightarrow 1 - e^2 = \frac{9}{25} \Rightarrow e^2 = \frac{16}{25} \Rightarrow e = \frac{4}{5}\).
(ii) The coordinates of the foci are given by \((\pm ae, 0)\). Since \(ae = 5 \times \frac{4}{5} = 4\), the foci are at \((\pm 4, 0)\).
(iii) The equations of the directrices are given by \(x = \pm \frac{a}{e}\). Substituting our values, we obtain \(x = \pm \frac{5}{4/5} = \pm \frac{25}{4}\).

M1: Uses the correct relationship \(b^2 = a^2(1 - e^2)\) with their values of \(a^2\) and \(b^2\).
A1: Calculates the correct eccentricity \(e = \frac{4}{5}\) (or \(0.8\)).
A1: Identifies the coordinates of the foci as \((\pm 4, 0)\) (accept separate points).
M1: Uses the standard formula \(x = \pm \frac{a}{e}\) to find the equations of the directrices.
A1: Finds the correct equations of the directrices as \(x = \pm \frac{25}{4}\) (or \(x = \pm 6.25\)).

The direction vectors of the lines are \(\mathbf{d}_1 = 2\mathbf{i} + \mathbf{j} - \mathbf{k}\) and \(\mathbf{d}_2 = \mathbf{i} - \mathbf{j} + 2\mathbf{k}\). A vector \(\mathbf{n}\) perpendicular to both lines is their cross product: \(\mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2 = \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} \times \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} = \begin{pmatrix} (1)(2) - (-1)(-1) \\ -( (2)(2) - (1)(-1) ) \\ (2)(-1) - (1)(1) \end{pmatrix} = \begin{pmatrix} 1 \\ -5 \\ -3 \end{pmatrix}\). The magnitude of \(\mathbf{n}\) is \(|\mathbf{n}| = \sqrt{1^2 + (-5)^2 + (-3)^2} = \sqrt{35}\). Let \(\mathbf{a}_1 = \mathbf{i} + \mathbf{k}\) and \(\mathbf{a}_2 = 2\mathbf{i} + \mathbf{j} + 3\mathbf{k}\). The vector between starting points is \(\mathbf{a}_2 - \mathbf{a}_1 = \mathbf{i} + \mathbf{j} + 2\mathbf{k}\). The shortest distance \(d\) is the projection of this vector onto the common normal: \(d = \frac{|(\mathbf{a}_2 - \mathbf{a}_1) \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{|(1)(1) + (1)(-5) + (2)(-3)|}{\sqrt{35}} = \frac{|1 - 5 - 6|}{\sqrt{35}} = \frac{10}{\sqrt{35}} = \frac{10\sqrt{35}}{35} = \frac{2\sqrt{35}}{7}\).

M1: Attempts to find the cross product of the direction vectors of the two lines.
A1: Obtains the correct perpendicular normal vector \(\mathbf{i} - 5\mathbf{j} - 3\mathbf{k}\) (or any non-zero multiple).
M1: Correctly calculates a position difference vector between the two lines, e.g., \(\mathbf{a}_2 - \mathbf{a}_1 = \mathbf{i} + \mathbf{j} + 2\mathbf{k}\).
M1: Applies the shortest distance formula involving the scalar product of the difference vector and the unit normal.
A1: Simplifies to find the exact shortest distance \(\frac{2\sqrt{35}}{7}\).

We use integration by parts with \(u = x^n\) and \(\frac{dv}{dx} = (1-x)^{1/2}\). This gives \(\frac{du}{dx} = n x^{n-1}\) and \(v = -\frac{2}{3}(1-x)^{3/2}\). Applying the integration by parts formula: \(I_n = \left[ -\frac{2}{3} x^n (1-x)^{3/2} \right]_0^1 + \frac{2n}{3} \int_{0}^{1} x^{n-1} (1-x)^{3/2} \, dx\). Evaluating the boundary term at both limits gives \(0\). Thus, we have: \(I_n = \frac{2n}{3} \int_{0}^{1} x^{n-1} (1-x)\sqrt{1-x} \, dx = \frac{2n}{3} \left( \int_{0}^{1} x^{n-1}\sqrt{1-x} \, dx - \int_{0}^{1} x^n\sqrt{1-x} \, dx \right) = \frac{2n}{3} (I_{n-1} - I_n)\). Rearranging to solve for \(I_n\): \(I_n = \frac{2n}{3} I_{n-1} - \frac{2n}{3} I_n \Rightarrow I_n \left(1 + \frac{2n}{3}\right) = \frac{2n}{3} I_{n-1} \Rightarrow I_n \left(\frac{2n+3}{3}\right) = \frac{2n}{3} I_{n-1} \Rightarrow I_n = \frac{2n}{2n+3} I_{n-1}\).

M1: Applies integration by parts to \(I_n\) with a correct selection of \(u\) and \(dv/dx\).
A1: Obtains the correct integrated part \(v = -\frac{2}{3}(1-x)^{3/2}\).
M1: Evaluates the boundary term to show it is zero at both limits, and sets up the remaining integral term.
M1: Splits the remaining integral term by writing \((1-x)^{3/2} = (1-x)\sqrt{1-x}\) and relates it to \(I_{n-1}\) and \(I_n\).
A1: Rearranges the algebraic expression correctly to show that \(I_n = \frac{2n}{2n+3} I_{n-1}\).

First, we find the characteristic equation by setting \(\det(\mathbf{M} - \lambda\mathbf{I}) = 0\):
\(\det \begin{pmatrix} 1-\lambda & 0 & 2 \\ 0 & 3-\lambda & 0 \\ 2 & 0 & 1-\lambda \end{pmatrix} = (3-\lambda)\left[ (1-\lambda)^2 - 4 \right] = (3-\lambda)(\lambda^2 - 2\lambda - 3) = -\lambda^3 + 5\lambda^2 - 3\lambda - 9 = 0\).
Thus, the characteristic equation of \(\mathbf{M}\) is \(\lambda^3 - 5\lambda^2 + 3\lambda + 9 = 0\).
By the Cayley-Hamilton theorem, \(\mathbf{M}\) satisfies its own characteristic equation: \(\mathbf{M}^3 - 5\mathbf{M}^2 + 3\mathbf{M} + 9\mathbf{I} = \mathbf{0}\).
We multiply this matrix equation by \(\mathbf{M}^{-1}\):
\(\mathbf{M}^2 - 5\mathbf{M} + 3\mathbf{I} + 9\mathbf{M}^{-1} = \mathbf{0}\).
Rearranging to make \(\mathbf{M}^{-1}\) the subject:
\(9\mathbf{M}^{-1} = -\mathbf{M}^2 + 5\mathbf{M} - 3\mathbf{I}\)
\(\mathbf{M}^{-1} = -\frac{1}{9}\mathbf{M}^2 + \frac{5}{9}\mathbf{M} - \frac{1}{3}\mathbf{I}\).
So, \(a = -\frac{1}{9}\), \(b = \frac{5}{9}\), and \(c = -\frac{1}{3}\).

M1: Set up the characteristic determinant equation \(\det(\mathbf{M} - \lambda\mathbf{I}) = 0\).
A1: Expand and solve to find the correct characteristic equation \(\lambda^3 - 5\lambda^2 + 3\lambda + 9 = 0\) (or any non-zero multiple).
M1: Apply the Cayley-Hamilton theorem to obtain the matrix equation \(\mathbf{M}^3 - 5\mathbf{M}^2 + 3\mathbf{M} + 9\mathbf{I} = \mathbf{0}\).
M1: Multiplies the matrix equation through by \(\mathbf{M}^{-1}\) and rearranges to solve for \(\mathbf{M}^{-1}\).
A1: Obtains the correct coefficients: \(a = -\frac{1}{9}\), \(b = \frac{5}{9}\), and \(c = -\frac{1}{3}\).

We use the identity \( \cosh(2x) = 1 + 2\sinh^2 x \) to rewrite the equation in terms of \( \sinh x \): \\ \( 2(1 + 2\sinh^2 x) - 7\sinh x = 4 \) \\ \( 4\sinh^2 x - 7\sinh x - 2 = 0 \) \\ Factoring this quadratic equation: \\ \( (4\sinh x + 1)(\sinh x - 2) = 0 \) \\ This gives two possible values: \\ \( \sinh x = 2 \) or \( \sinh x = -\frac{1}{4} \) \\ Using the logarithmic definition \( \text{arsinh } y = \ln\left(y + \sqrt{y^2 + 1}\right) \): \\ For \( \sinh x = 2 \): \\ \( x = \ln\left(2 + \sqrt{2^2 + 1}\right) = \ln(2 + \sqrt{5}) \) \\ For \( \sinh x = -\frac{1}{4} \): \\ \( x = \ln\left(-\frac{1}{4} + \sqrt{\left(-\frac{1}{4}\right)^2 + 1}\right) = \ln\left(-\frac{1}{4} + \sqrt{\frac{17}{16}}\right) = \ln\left(\frac{\sqrt{17}-1}{4}\right) \)

M1: Substitution of \( \cosh(2x) = 1 + 2\sinh^2 x \) to form a quadratic in \( \sinh x \). \\ A1: Correct quadratic equation \( 4\sinh^2 x - 7\sinh x - 2 = 0 \). \\ M1: Solving the quadratic to find \( \sinh x = 2 \) and \( \sinh x = -\frac{1}{4} \). \\ M1: Attempting to use the logarithmic form of arsinh on at least one of their solutions. \\ A1: Both final exact solutions correct and fully simplified: \( \ln(2 + \sqrt{5}) \) and \( \ln\left(\frac{\sqrt{17}-1}{4}\right) \).

The line passing through \( P(0, 4) \) with gradient \( m \) has equation \( y = mx + 4 \). \\ Since this line is tangent to the hyperbola, we can use the tangency condition \( c^2 = a^2 m^2 - b^2 \) where \( a^2 = 25 \), \( b^2 = 9 \), and \( c = 4 \): \\ \( 4^2 = 25m^2 - 9 \implies 16 = 25m^2 - 9 \implies 25m^2 = 25 \implies m = \pm 1 \). \\ Therefore, the equations of the tangents are \( y = x + 4 \) and \( y = -x + 4 \). \\ To find the point of contact for the tangent \( y = x + 4 \), we substitute this into the equation of \( H \): \\ \( \frac{x^2}{25} - \frac{(x+4)^2}{9} = 1 \implies 9x^2 - 25(x^2 + 8x + 16) = 225 \implies -16x^2 - 200x - 400 = 225 \implies 16x^2 + 200x + 625 = 0 \) \\ This factors as \( (4x + 25)^2 = 0 \implies x = -\frac{25}{4} \). \\ The corresponding \( y \)-coordinate is \( y = -\frac{25}{4} + 4 = -\frac{9}{4} \). \\ To find the point of contact for the tangent \( y = -x + 4 \), we substitute into the equation of \( H \): \\ \( \frac{x^2}{25} - \frac{(-x+4)^2}{9} = 1 \implies 9x^2 - 25(x^2 - 8x + 16) = 225 \implies 16x^2 - 200x + 625 = 0 \) \\ This factors as \( (4x - 25)^2 = 0 \implies x = \frac{25}{4} \). \\ The corresponding \( y \)-coordinate is \( y = -\frac{25}{4} + 4 = -\frac{9}{4} \). \\ Thus, the points of contact are \( A\left(\frac{25}{4}, -\frac{9}{4}\right) \) and \( B\left(-\frac{25}{4}, -\frac{9}{4}\right) \).

M1: Set up the equation of the tangent as \( y = mx + 4 \) and apply the tangency condition \( c^2 = a^2m^2 - b^2 \) (or substitute and set discriminant to zero). \\ A1: Solve for the gradients to find \( m = \pm 1 \). \\ M1: Substitute at least one tangent equation back into the hyperbola equation to find the \( x \)-coordinate of the point of contact. \\ A1: Correctly find \( x = \pm \frac{25}{4} \). \\ A1: Correctly find the coordinates of both points of contact: \( \left(\frac{25}{4}, -\frac{9}{4}\right) \) and \( \left(-\frac{25}{4}, -\frac{9}{4}\right) \).

First, write the hyperbolic functions in exponential form: \\ \( 5\cosh x + 3\sinh x = 5\left(\frac{e^x + e^{-x}}{2}\right) + 3\left(\frac{e^x - e^{-x}}{2}\right) = 4e^x + e^{-x} \) \\ The integral becomes: \\ \( \int_{0}^{\ln 2} \frac{1}{4e^x + e^{-x}} \, dx = \int_{0}^{\ln 2} \frac{e^x}{4e^{2x} + 1} \, dx \) \\ Use the substitution \( u = e^x \), so \( du = e^x \, dx \). \\ Change the limits: \\ When \( x = 0 \), \( u = e^0 = 1 \). \\ When \( x = \ln 2 \), \( u = e^{\ln 2} = 2 \). \\ The integral is now: \\ \( \int_{1}^{2} \frac{1}{4u^2 + 1} \, du \) \\ This integrates to: \\ \( \left[ \frac{1}{2}\arctan(2u) \right]_{1}^{2} = \frac{1}{2}\arctan(4) - \frac{1}{2}\arctan(2) = \frac{1}{2}\left(\arctan(4) - \arctan(2)\right) \) \\ Using the identity \( \arctan A - \arctan B = \arctan\left(\frac{A-B}{1+AB}\right) \): \\ \( \arctan(4) - \arctan(2) = \arctan\left(\frac{4-2}{1 + 4 \times 2}\right) = \arctan\left(\frac{2}{9}\right) \) \\ Thus, the exact value is \( \frac{1}{2}\arctan\left(\frac{2}{9}\right) \).

M1: Expresses the denominator in terms of exponentials to obtain \( 4e^x + e^{-x} \). \\ M1: Employs a substitution \( u = e^x \) to convert the integral to the form \( \int \frac{1}{4u^2+1} \, du \) and attempts to find new limits. \\ A1: Correctly integrates to obtain \( \frac{1}{2}\arctan(2u) \) with correct limits \( 1 \) and \( 2 \). \\ M1: Substitutes the limits to find \( \frac{1}{2}(\arctan 4 - \arctan 2) \). \\ A1: Simplifies using the arctangent subtraction identity to find the final exact form \( \frac{1}{2}\arctan\left(\frac{2}{9}\right) \) (accept \( a=2, b=2, c=9 \)).

(a) First, we find the characteristic equation of \( \mathbf{A} \) by setting \( \det(\mathbf{A} - \lambda\mathbf{I}) = 0 \): \\ \( \det\begin{pmatrix} 2-\lambda & 0 & 1 \\ 0 & 1-\lambda & 0 \\ 1 & 0 & 2-\lambda \end{pmatrix} = 0 \) \\ Expanding along the second row: \\ \( (1-\lambda) \left( (2-\lambda)^2 - 1 \right) = (1-\lambda)(\lambda^2 - 4\lambda + 3) = -\lambda^3 + 5\lambda^2 - 7\lambda + 3 = 0 \) \\ Multiplying by \( -1 \) gives the characteristic equation: \\ \( \lambda^3 - 5\lambda^2 + 7\lambda - 3 = 0 \) \\ By the Cayley-Hamilton theorem, \( \mathbf{A} \) satisfies its own characteristic equation, hence: \\ \( \mathbf{A}^3 - 5\mathbf{A}^2 + 7\mathbf{A} - 3\mathbf{I} = \mathbf{0} \) \\ \\ (b) To find \( \mathbf{A}^{-1} \), we multiply the equation by \( \mathbf{A}^{-1} \): \\ \( \mathbf{A}^2 - 5\mathbf{A} + 7\mathbf{I} - 3\mathbf{A}^{-1} = \mathbf{0} \implies 3\mathbf{A}^{-1} = \mathbf{A}^2 - 5\mathbf{A} + 7\mathbf{I} \) \\ First, calculate \( \mathbf{A}^2 \): \\ \( \mathbf{A}^2 = \begin{pmatrix} 2 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 2 \end{pmatrix} \begin{pmatrix} 2 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 2 \end{pmatrix} = \begin{pmatrix} 5 & 0 & 4 \\ 0 & 1 & 0 \\ 4 & 0 & 5 \end{pmatrix} \) \\ Now evaluate \( 3\mathbf{A}^{-1} \): \\ \( 3\mathbf{A}^{-1} = \begin{pmatrix} 5 & 0 & 4 \\ 0 & 1 & 0 \\ 4 & 0 & 5 \end{pmatrix} - 5\begin{pmatrix} 2 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 2 \end{pmatrix} + 7\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 0 & -1 \\ 0 & 3 & 0 \\ -1 & 0 & 2 \end{pmatrix} \) \\ Thus: \\ \( \mathbf{A}^{-1} = \frac{1}{3}\begin{pmatrix} 2 & 0 & -1 \\ 0 & 3 & 0 \\ -1 & 0 & 2 \end{pmatrix} = \begin{pmatrix} \frac{2}{3} & 0 & -\frac{1}{3} \\ 0 & 1 & 0 \\ -\frac{1}{3} & 0 & \frac{2}{3} \end{pmatrix} \)

(a) \\ M1: Attempting to calculate the characteristic equation \( \det(\mathbf{A} - \lambda\mathbf{I}) = 0 \). \\ A1: Correctly deriving the polynomial equation \( \lambda^3 - 5\lambda^2 + 7\lambda - 3 = 0 \). \\ A1: Citing the Cayley-Hamilton theorem to state that \( \mathbf{A}^3 - 5\mathbf{A}^2 + 7\mathbf{A} - 3\mathbf{I} = \mathbf{0} \). \\ \\ (b) \\ M1: Rearranging the equation to isolate \( \mathbf{A}^{-1} \), calculating \( \mathbf{A}^2 \), and attempting to evaluate the combination \( \mathbf{A}^2 - 5\mathbf{A} + 7\mathbf{I} \). \\ A1: Correctly finding the inverse matrix \( \mathbf{A}^{-1} = \begin{pmatrix} \frac{2}{3} & 0 & -\frac{1}{3} \\ 0 & 1 & 0 \\ -\frac{1}{3} & 0 & \frac{2}{3} \end{pmatrix} \) (either in fraction-factored form or as a full matrix).