2026 Cambridge IAS-Level Mathematics - Further (9231) 模擬試題連答案詳解

(a) Starting from the right-hand side:
\[ \frac{1}{(2r - 1)(2r + 1)} - \frac{1}{(2r + 1)(2r + 3)} = \frac{(2r + 3) - (2r - 1)}{(2r - 1)(2r + 1)(2r + 3)} \]
\[ = \frac{2r + 3 - 2r + 1}{(2r - 1)(2r + 1)(2r + 3)} = \frac{4}{(2r - 1)(2r + 1)(2r + 3)} \]

(b) Using the result from part (a):
\[ \sum_{r=1}^{n} \frac{1}{(2r - 1)(2r + 1)(2r + 3)} = \frac{1}{4} \sum_{r=1}^{n} \left( \frac{1}{(2r - 1)(2r + 1)} - \frac{1}{(2r + 1)(2r + 3)} \right) \]
Listing terms:
\[ r = 1: \quad \frac{1}{1 \times 3} - \frac{1}{3 \times 5} \]
\[ r = 2: \quad \frac{1}{3 \times 5} - \frac{1}{5 \times 7} \]
\[ \vdots \]
\[ r = n: \quad \frac{1}{(2n - 1)(2n + 1)} - \frac{1}{(2n + 1)(2n + 3)} \]
Summing these terms, all intermediate terms cancel, leaving:
\[ \sum_{r=1}^{n} \frac{1}{(2r - 1)(2r + 1)(2r + 3)} = \frac{1}{4} \left( \frac{1}{3} - \frac{1}{(2n + 1)(2n + 3)} \right) \]
\[ = \frac{1}{4} \left( \frac{(2n + 1)(2n + 3) - 3}{3(2n + 1)(2n + 3)} \right) \]
\[ = \frac{1}{4} \left( \frac{4n^2 + 8n + 3 - 3}{3(2n + 1)(2n + 3)} \right) \]
\[ = \frac{1}{4} \left( \frac{4n(n + 2)}{3(2n + 1)(2n + 3)} \right) = \frac{n(n + 2)}{3(2n + 1)(2n + 3)} \]

(c) As \(n \to \infty\), \(\frac{1}{(2n + 1)(2n + 3)} \to 0\), so:
\[ \sum_{r=1}^{\infty} \frac{1}{(2r - 1)(2r + 1)(2r + 3)} = \frac{1}{4} \left( \frac{1}{3} - 0 \right) = \frac{1}{12} \]

(d) Using \(S_n = \sum_{r=1}^{n} \frac{1}{(2r - 1)(2r + 1)(2r + 3)}\):
\[ \sum_{r=n+1}^{2n} \frac{1}{(2r - 1)(2r + 1)(2r + 3)} = S_{2n} - S_n \]
\[ = \frac{1}{4} \left( \frac{1}{3} - \frac{1}{(4n + 1)(4n + 3)} \right) - \frac{1}{4} \left( \frac{1}{3} - \frac{1}{(2n + 1)(2n + 3)} \right) \]
\[ = \frac{1}{4} \left( \frac{1}{(2n + 1)(2n + 3)} - \frac{1}{(4n + 1)(4n + 3)} \right) \]

(a)
- M1: For combining fractions over a common denominator.
- A1: For complete and convincing working leading to the given result (AG).

(b)
- M1: For expressing the sum using the difference of terms and writing out at least the first two and the last terms to show cancellation.
- M1: For obtaining \(\frac{1}{4}\left(\frac{1}{3} - \frac{1}{(2n+1)(2n+3)}\right)\).
- A1: For correct algebraic simplification to the given form \(\frac{n(n+2)}{3(2n+1)(2n+3)}\) (AG).

(c)
- B1: For \(\frac{1}{12}\) (or exact equivalent).

(d)
- M1: For using \(S_{2n} - S_n\) or equivalent difference of sums.
- A1: For correct expression in terms of \(n\), e.g., \(\frac{1}{4}\left[\frac{1}{(2n+1)(2n+3)} - \frac{1}{(4n+1)(4n+3)}\right]\) or \(\frac{n(6n+5)}{2(2n+1)(2n+3)(4n+1)(4n+3)}\).

Step 1: Base case

For \(n = 1\):
\[\text{RHS} = \begin{pmatrix} 1 + 4(1) & -4(1) \[2pt] 4(1) & 1 - 4(1) \end{pmatrix} = \begin{pmatrix} 5 & -4 \[2pt] 4 & -3 \end{pmatrix} = \mathbf{A} = \mathbf{A}^1.\]
Hence, the result is true for \(n = 1\).

Step 2: Inductive hypothesis

Assume that the result is true for \(n = k\), where \(k\) is a positive integer. That is,
\[\mathbf{A}^k = \begin{pmatrix} 1 + 4k & -4k \[2pt] 4k & 1 - 4k \end{pmatrix}.\]

Step 3: Inductive step

We consider \(n = k + 1\):
\[\mathbf{A}^{k+1} = \mathbf{A}^k \mathbf{A} = \begin{pmatrix} 1 + 4k & -4k \[2pt] 4k & 1 - 4k \end{pmatrix} \begin{pmatrix} 5 & -4 \[2pt] 4 & -3 \end{pmatrix}.\]

Multiplying the matrices:
\[\mathbf{A}^{k+1} = \begin{pmatrix} (1 + 4k)(5) + (-4k)(4) & (1 + 4k)(-4) + (-4k)(-3) \[2pt] (4k)(5) + (1 - 4k)(4) & (4k)(-4) + (1 - 4k)(-3) \end{pmatrix}.\]

Expanding each element:
- Top-left entry: \(5 + 20k - 16k = 5 + 4k = 1 + 4(k + 1)\)
- Top-right entry: \(-4 - 16k + 12k = -4 - 4k = -4(k + 1)\)
- Bottom-left entry: \(20k + 4 - 16k = 4 + 4k = 4(k + 1)\)
- Bottom-right entry: \(-16k - 3 + 12k = -3 - 4k = 1 - 4(k + 1)\)

Thus,
\[\mathbf{A}^{k+1} = \begin{pmatrix} 1 + 4(k+1) & -4(k+1) \[2pt] 4(k+1) & 1 - 4(k+1) \end{pmatrix}.\]

Step 4: Conclusion

Since the result is true for \(n = 1\), and if it is true for \(n = k\) then it is also true for \(n = k + 1\), the result is true for all positive integers \(n\) by mathematical induction.

B1: Verifies the base case \(n = 1\) correctly by substituting \(n = 1\) into the given formula and showing it equals \(\mathbf{A}\).

B1: States inductive hypothesis clearly (e.g., "Assume true for \(n = k\)").

M1: Sets up product \(\mathbf{A}^{k+1} = \mathbf{A}^k \mathbf{A}\) (or \(\mathbf{A}\mathbf{A}^k\)) using the assumed form for \(\mathbf{A}^k\) and attempts matrix multiplication.

A1: Obtains a correct unsimplified matrix with all 4 entries correctly evaluated in terms of \(k\).

A1: Correctly and clearly simplifies all 4 entries to show the explicit form corresponding to \(n = k + 1\).

A1: Gives a complete, correct, and convincing conclusion linking the base case, the inductive step, and the principle of mathematical induction.

(a)
A shear parallel to the \(x\)-axis has matrix representation of the form:
\[\mathbf{A} = \begin{pmatrix} 1 & s \\ 0 & 1 \end{pmatrix}\]
Given that \(\mathbf{A}\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 6 \\ 1 \end{pmatrix}\):
\[\begin{pmatrix} 1 + s \\ 1 \end{pmatrix} = \begin{pmatrix} 6 \\ 1 \end{pmatrix} \implies s = 5\]
Thus,
\[\mathbf{A} = \begin{pmatrix} 1 & 5 \\ 0 & 1 \end{pmatrix}\]

For \(T_2\), the rotation matrix is:
\[\mathbf{B} = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}\]
Since \(\tan \theta = \frac{4}{3}\) with \(0 < \theta < \frac{1}{2}\pi\), we have \(\cos \theta = \frac{3}{5}\) and \(\sin \theta = \frac{4}{5}\).
Thus,
\[\mathbf{B} = \begin{pmatrix} \frac{3}{5} & -\frac{4}{5} \[1mm] \frac{4}{5} & \frac{3}{5} \end{pmatrix} = \frac{1}{5}\begin{pmatrix} 3 & -4 \\ 4 & 3 \end{pmatrix}\]

(b)
The combined transformation \(T\) is \(T_1\) followed by \(T_2\), so:
\[\mathbf{C} = \mathbf{B}\mathbf{A} = \frac{1}{5}\begin{pmatrix} 3 & -4 \\ 4 & 3 \end{pmatrix}\begin{pmatrix} 1 & 5 \\ 0 & 1 \end{pmatrix} = \frac{1}{5}\begin{pmatrix} 3(1) - 4(0) & 3(5) - 4(1) \\ 4(1) + 3(0) & 4(5) + 3(1) \end{pmatrix} = \frac{1}{5}\begin{pmatrix} 3 & 11 \\ 4 & 23 \end{pmatrix}\]
\[\mathbf{C} = \begin{pmatrix} \frac{3}{5} & \frac{11}{5} \[1mm] \frac{4}{5} & \frac{23}{5} \end{pmatrix}\]

(c)
Let an invariant line through the origin have equation \(y = mx\).
Under the transformation \(T\), a point \(\begin{pmatrix} x \\ mx \end{pmatrix}\) is mapped to \(\begin{pmatrix} X \\ Y \end{pmatrix}\) where \(Y = mX\):
\[\begin{pmatrix} X \\ Y \end{pmatrix} = \frac{1}{5}\begin{pmatrix} 3 & 11 \\ 4 & 23 \end{pmatrix}\begin{pmatrix} x \\ mx \end{pmatrix} = \frac{x}{5}\begin{pmatrix} 3 + 11m \\ 4 + 23m \end{pmatrix}\]
Since \(Y = mX\):
\[\frac{x}{5}(4 + 23m) = m \left(\frac{x}{5}(3 + 11m)\right)\]
For non-zero points on the line, we divide by \(\frac{x}{5}\):
\[4 + 23m = 3m + 11m^2\]
\[11m^2 - 20m - 4 = 0\]
Factorising:
\[(11m + 2)(m - 2) = 0\]
\[m = 2 \quad \text{or} \quad m = -\frac{2}{11}\]

Checking if the line \(x = 0\) is invariant:
\[\mathbf{C}\begin{pmatrix} 0 \\ y \end{pmatrix} = \frac{y}{5}\begin{pmatrix} 11 \\ 23 \end{pmatrix}\]
Since \(X = \frac{11}{5}y \neq 0\) for \(y \neq 0\), \(x = 0\) is not invariant.

Therefore, the equations of the invariant lines through the origin are:
\[y = 2x \quad \text{and} \quad y = -\frac{2}{11}x \quad (\text{or } 2x + 11y = 0)\]

(a) [3 marks]
- B1: State correct matrix \(\mathbf{A} = \begin{pmatrix} 1 & 5 \\ 0 & 1 \end{pmatrix}\).
- M1: Use right-angled triangle or trigonometric identities to find \(\cos \theta = \frac{3}{5}\) and \(\sin \theta = \frac{4}{5}\), and substitute into rotation matrix form.
- A1: Obtain correct matrix \(\mathbf{B} = \begin{pmatrix} \frac{3}{5} & -\frac{4}{5} \\ \frac{4}{5} & \frac{3}{5} \end{pmatrix}\) (oe, e.g. with scalar factor \(\frac{1}{5}\) outside).

(b) [2 marks]
- M1: Correct order of matrix multiplication \(\mathbf{C} = \mathbf{BA}\) applied to their \(\mathbf{A}\) and \(\mathbf{B}\).
- A1: Correct matrix \(\mathbf{C} = \begin{pmatrix} \frac{3}{5} & \frac{11}{5} \\ \frac{4}{5} & \frac{23}{5} \end{pmatrix}\) or \(\frac{1}{5}\begin{pmatrix} 3 & 11 \\ 4 & 23 \end{pmatrix}\).

(c) [4 marks]
- M1: Set up equation for invariant line \(\mathbf{C}\begin{pmatrix} x \\ mx \end{pmatrix} = \begin{pmatrix} X \\ mX \end{pmatrix}\) or equate \(\frac{Y}{X} = m\).
- A1: Obtain correct quadratic equation in \(m\), e.g. \(11m^2 - 20m - 4 = 0\).
- M1: Valid method to solve their 3-term quadratic equation in \(m\).
- A1: State both correct equations: \(y = 2x\) and \(y = -\frac{2}{11}x\) (oe).

From the given equation \( x^3 + 2x^2 - 4x + 3 = 0 \):

\( \alpha + \beta + \gamma = -2 \)

\( \alpha\beta + \beta\gamma + \gamma\alpha = -4 \)

\( \alpha\beta\gamma = -3 \)

(a)(i)
\( \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) \)

\( = (-2)^2 - 2(-4) = 4 + 8 = 12 \)

(a)(ii)
Since \( \alpha, \beta, \gamma \) are roots of \( x^3 + 2x^2 - 4x + 3 = 0 \):

\( \alpha^3 = -2\alpha^2 + 4\alpha - 3 \)

\( \beta^3 = -2\beta^2 + 4\beta - 3 \)

\( \gamma^3 = -2\gamma^2 + 4\gamma - 3 \)

Summing these three equations:

\( \sum \alpha^3 = -2\sum \alpha^2 + 4\sum \alpha - 3(3) \)

\( \sum \alpha^3 = -2(12) + 4(-2) - 9 = -24 - 8 - 9 = -41 \)

(a)(iii)
\( \frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} = \frac{\beta^2\gamma^2 + \gamma^2\alpha^2 + \alpha^2\beta^2}{(\alpha\beta\gamma)^2} \)

We have:
\( \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 = (\alpha\beta + \beta\gamma + \gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha + \beta + \gamma) \)

\( = (-4)^2 - 2(-3)(-2) = 16 - 12 = 4 \)

Therefore,
\( \frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} = \frac{4}{(-3)^2} = \frac{4}{9} \)

(b)
Method 1: Using substitution
Let \( y = x^2 \), so \( x = \sqrt{y} \).

Rearranging the original equation to separate odd and even powers of \( x \):
\( x^3 - 4x = -2x^2 - 3 \implies x(x^2 - 4) = -(2x^2 + 3) \)

Squaring both sides:
\( x^2(x^2 - 4)^2 = (2x^2 + 3)^2 \)

Substituting \( y = x^2 \):
\( y(y - 4)^2 = (2y + 3)^2 \)

\( y(y^2 - 8y + 16) = 4y^2 + 12y + 9 \)

\( y^3 - 8y^2 + 16y = 4y^2 + 12y + 9 \)

\( y^3 - 12y^2 + 4y - 9 = 0 \)

Method 2: Using symmetric sums
For the roots \( \alpha^2, \beta^2, \gamma^2 \):
- Sum of roots: \( \sum \alpha^2 = 12 \)
- Sum of products in pairs: \( \sum \alpha^2\beta^2 = 4 \)
- Product of roots: \( (\alpha\beta\gamma)^2 = (-3)^2 = 9 \)

The equation is \( y^3 - (\sum \alpha^2)y^2 + (\sum \alpha^2\beta^2)y - (\alpha\beta\gamma)^2 = 0 \)

\( y^3 - 12y^2 + 4y - 9 = 0 \)

(a)(i)
- M1: Uses the identity \( \sum \alpha^2 = (\sum \alpha)^2 - 2\sum \alpha\beta \) with stated values \( \sum \alpha = -2 \) and \( \sum \alpha\beta = -4 \).
- A1: Obtains \( 12 \).

(a)(ii)
- M1: Sets up a valid method to find \( \sum \alpha^3 \), e.g., substituting into the cubic equation \( \sum \alpha^3 = -2\sum \alpha^2 + 4\sum \alpha - 9 \) or using \( \sum \alpha^3 - 3\alpha\beta\gamma = (\sum \alpha)(\sum \alpha^2 - \sum \alpha\beta) \).
- A1 (AG): Fully correct working leading to \( -41 \) without omitted steps.

(a)(iii)
- M1: Expresses sum over common denominator \( (\alpha\beta\gamma)^2 \) and evaluates \( \sum \alpha^2\beta^2 = (\sum \alpha\beta)^2 - 2\alpha\beta\gamma(\sum \alpha) \).
- A1: Obtains \( \frac{4}{9} \) (must be exact).

(b)
- M1: Uses substitution \( y = x^2 \) by rearranging the equation into the form \( x(x^2 - 4) = -(2x^2 + 3) \) and squaring, OR sets up the polynomial using roots \( \sum \alpha^2, \sum \alpha^2\beta^2, (\alpha\beta\gamma)^2 \).
- M1: Correctly expands the squared terms \( y(y^2 - 8y + 16) = 4y^2 + 12y + 9 \) or states all three coefficients correctly from prior parts.
- A1: Any two of \( p = -12 \), \( q = 4 \), \( r = -9 \) correct.
- A1: Fully correct equation \( y^3 - 12y^2 + 4y - 9 = 0 \) including \( = 0 \).

(a)
Direction vectors of \(l_1\) and \(l_2\) are \(\mathbf{d}_1 = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}\) and \(\mathbf{d}_2 = \begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix}\).

A vector perpendicular to both lines is
\[ \mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 1 \\ 1 & 1 & -2 \end{vmatrix} = \begin{pmatrix} (-1)(-2) - (1)(1) \\ (1)(1) - (2)(-2) \\ (2)(1) - (-1)(1) \end{pmatrix} = \begin{pmatrix} 1 \\ 5 \\ 3 \end{pmatrix} \]

Let \(\mathbf{a}_1 = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}\) on \(l_1\) and \(\mathbf{a}_2 = \begin{pmatrix} 3 \\ 1 \\ 4 \end{pmatrix}\) on \(l_2\).
\[ \mathbf{a}_2 - \mathbf{a}_1 = \begin{pmatrix} 3 - 1 \\ 1 - 2 \\ 4 - (-1) \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \\ 5 \end{pmatrix} \]

The shortest distance \(d\) between the skew lines is
\[ d = \frac{|(\mathbf{a}_2 - \mathbf{a}_1) \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{\left| \begin{pmatrix} 2 \\ -1 \\ 5 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 5 \\ 3 \end{pmatrix} \right|}{\sqrt{1^2 + 5^2 + 3^2}} = \frac{|2(1) + (-1)(5) + 5(3)|}{\sqrt{1 + 25 + 9}} = \frac{|12|}{\sqrt{35}} = \frac{12}{\sqrt{35}} = \frac{12\sqrt{35}}{35} \approx 2.03 \]

(b)
The plane \(\Pi_1\) is perpendicular to \(\mathbf{n} = \begin{pmatrix} 1 \\ 5 \\ 3 \end{pmatrix}\) and contains the point \((1, 2, -1)\).

The equation of \(\Pi_1\) is
\[ \mathbf{r} \cdot \begin{pmatrix} 1 \\ 5 \\ 3 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 5 \\ 3 \end{pmatrix} = 1(1) + 2(5) + (-1)(3) = 8 \]

Hence, the Cartesian equation of \(\Pi_1\) is
\[ x + 5y + 3z = 8 \]

(c)(i)
The normal to \(\Pi_1\) is \(\mathbf{n}_1 = \begin{pmatrix} 1 \\ 5 \\ 3 \end{pmatrix}\) and the normal to \(\Pi_2\) is \(\mathbf{n}_2 = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}\).

The direction of the line of intersection is
\[ \mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 5 & 3 \\ 1 & 2 & -1 \end{vmatrix} = \begin{pmatrix} 5(-1) - 3(2) \\ 3(1) - 1(-1) \\ 1(2) - 5(1) \end{pmatrix} = \begin{pmatrix} -11 \\ 4 \\ -3 \end{pmatrix} \sim \begin{pmatrix} 11 \\ -4 \\ 3 \end{pmatrix} \]

To find a common point, solve the system:
\[ x + 5y + 3z = 8 \]
\[ x + 2y - z = 7 \]
Subtracting gives \(3y + 4z = 1\). Setting \(z = 1\) yields \(3y = -3 \implies y = -1\).
Substituting back into the second equation: \(x + 2(-1) - 1 = 7 \implies x = 10\).
A point on the line is \((10, -1, 1)\).

A vector equation of the line is
\[ \mathbf{r} = \begin{pmatrix} 10 \\ -1 \\ 1 \end{pmatrix} + t \begin{pmatrix} 11 \\ -4 \\ 3 \end{pmatrix} \]

(c)(ii)
The acute angle \(\theta\) between the planes is given by
\[ \cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|} = \frac{|1(1) + 5(2) + 3(-1)|}{\sqrt{1^2 + 5^2 + 3^2}\sqrt{1^2 + 2^2 + (-1)^2}} = \frac{8}{\sqrt{35}\sqrt{6}} = \frac{8}{\sqrt{210}} \]
\[ \theta = \arccos\left(\frac{8}{\sqrt{210}}\right) \approx 56.5^\circ \]

(a)
M1: Evaluates the vector product of the two direction vectors \(\begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix} \times \begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix}\).
A1: Obtains \(\begin{pmatrix} 1 \\ 5 \\ 3 \end{pmatrix}\) or any non-zero scalar multiple.
M1: Applies the shortest distance formula \(\frac{|(\mathbf{a}_2 - \mathbf{a}_1) \cdot \mathbf{n}|}{|\mathbf{n}|}\) using their normal and difference of position vectors.
A1: Obtains \(\frac{12}{\sqrt{35}}\) (or \(\frac{12\sqrt{35}}{35}\), or awrt \(2.03\)).

(b)
M1: Uses their normal from part (a) and a point on \(l_1\) to form the plane equation \(\mathbf{r} \cdot \mathbf{n} = \mathbf{a}_1 \cdot \mathbf{n}\).
M1: Evaluates scalar product correctly.
A1: Obtains \(x + 5y + 3z = 8\) (or any integer multiple).

(c)(i)
M1: Finds the vector product of the two normals or solves the two plane equations simultaneously to obtain the line's direction and a valid point.
A1: Obtains a fully correct vector equation, e.g. \(\mathbf{r} = \begin{pmatrix} 10 \\ -1 \\ 1 \end{pmatrix} + t \begin{pmatrix} 11 \\ -4 \\ 3 \end{pmatrix}\) (accept any equivalent form/point).

(c)(ii)
M1: Uses the scalar product of the normal vectors \(\cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|}\).
A1: Obtains \(56.5^\circ\) (or \(0.986\) radians; allow awrt \(56.5^\circ\)).

(a)
By polynomial division or rewriting the numerator:
\[ \frac{x^2 - 3x + 6}{x - 2} = \frac{x(x - 2) - 1(x - 2) + 4}{x - 2} = x - 1 + \frac{4}{x - 2} \]
Therefore, the vertical asymptote is \( x = 2 \) and the oblique asymptote is \( y = x - 1 \).

(b)
Rearranging the equation of \( C \):
\[ y(x - 2) = x^2 - 3x + 6 \implies x^2 - (y + 3)x + (2y + 6) = 0 \]
Since \( x \) is real, the discriminant of this quadratic equation in \( x \) must be greater than or equal to zero:
\[ \Delta = [-(y + 3)]^2 - 4(1)(2y + 6) \ge 0 \]
\[ y^2 + 6y + 9 - 8y - 24 \ge 0 \]
\[ y^2 - 2y - 15 \ge 0 \]
\[ (y - 5)(y + 3) \ge 0 \]
This inequality holds for \( y \le -3 \) or \( y \ge 5 \).
Hence, there are no points on \( C \) for which \( -3 < y < 5 \).

(c)
The stationary points occur where \( \Delta = 0 \):
- For \( y = 5 \): \( x = \frac{5 + 3}{2} = 4 \), giving the local minimum at \( (4, 5) \).
- For \( y = -3 \): \( x = \frac{-3 + 3}{2} = 0 \), giving the local maximum at \( (0, -3) \).

Axial intercepts:
- y-intercept: When \( x = 0 \), \( y = -3 \), which is the point \( (0, -3) \).
- x-intercepts: Setting \( x^2 - 3x + 6 = 0 \) gives \( \Delta_x = 9 - 24 = -15 < 0 \), so there are no intersections with the x-axis.

Sketch of \( C \):
- Asymptotes \( x = 2 \) and \( y = x - 1 \) drawn as dashed lines.
- Branch for \( x < 2 \) lies entirely below \( y = -3 \), passing through the maximum at \( (0, -3) \), asymptotic to \( y = x - 1 \) as \( x \to -\infty \) and to \( x = 2 \) as \( x \to 2^- \).
- Branch for \( x > 2 \) lies entirely above \( y = 5 \), passing through the minimum at \( (4, 5) \), asymptotic to \( x = 2 \) as \( x \to 2^+ \) and to \( y = x - 1 \) as \( x \to +\infty \).

(d)
For the curve \( y^2 = f(x) \):
- The curve is only defined where \( f(x) \ge 0 \). From parts (b) and (c), \( f(x) \ge 0 \) only for \( x > 2 \) (where \( f(x) \ge 5 \)). There is no part of the curve for \( x < 2 \).
- The curve is symmetrical about the x-axis.
- Vertical asymptote is \( x = 2 \).
- The turning points are at \( (4, \sqrt{5}) \) (local minimum) and \( (4, -\sqrt{5}) \) (local maximum).
- As \( x \to \infty \), \( y^2 \approx x - 1 \implies y \approx \pm \sqrt{x - 1} \) (parabolic shape branches, no linear oblique asymptotes).
- There are no intersections with either coordinate axis.

(e)
The inequality \( |f(x)| \le 6 \) is equivalent to \( -6 \le f(x) \le 6 \).

Case 1: \( x > 2 \)
Here, \( f(x) \ge 5 > -6 \) is always satisfied. We only need \( f(x) \le 6 \):
\[ \frac{x^2 - 3x + 6}{x - 2} \le 6 \]
Since \( x - 2 > 0 \):
\[ x^2 - 3x + 6 \le 6(x - 2) \implies x^2 - 9x + 18 \le 0 \implies (x - 3)(x - 6) \le 0 \]
\[ 3 \le x \le 6 \]

Case 2: \( x < 2 \)
Here, \( f(x) \le -3 < 6 \) is always satisfied. We only need \( f(x) \ge -6 \):
\[ \frac{x^2 - 3x + 6}{x - 2} \ge -6 \]
Since \( x - 2 < 0 \), multiplying by \( (x - 2) \) reverses the inequality:
\[ x^2 - 3x + 6 \le -6(x - 2) \implies x^2 - 3x + 6 \le -6x + 12 \implies x^2 + 3x - 6 \le 0 \]
Solving \( x^2 + 3x - 6 = 0 \) gives:
\[ x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-6)}}{2} = \frac{-3 \pm \sqrt{33}}{2} \]
Thus:
\[ \frac{-3 - \sqrt{33}}{2} \le x \le \frac{-3 + \sqrt{33}}{2} \]
(Note that \( \frac{-3 + \sqrt{33}}{2} \approx 1.37 < 2 \), so this interval lies entirely in \( x < 2 \).)

Combining both cases, the set of values of \( x \) is:
\[ \frac{-3 - \sqrt{33}}{2} \le x \le \frac{-3 + \sqrt{33}}{2} \quad \text{or} \quad 3 \le x \le 6 \]

(a)
- M1: For carrying out algebraic division or expressing in form \( y = x + p + \frac{q}{x - 2} \).
- A1: For both equations correct: \( x = 2 \) and \( y = x - 1 \).

(b)
- M1: For forming a quadratic equation in \( x \): \( x^2 - (y + 3)x + (2y + 6) = 0 \).
- M1: For using the condition \( \Delta \ge 0 \) for real roots: \( (y + 3)^2 - 4(2y + 6) \ge 0 \).
- A1: For obtaining \( (y - 5)(y + 3) \ge 0 \) and providing a clear concluding argument that \( y \le -3 \) or \( y \ge 5 \), so \( -3 < y < 5 \) has no points on \( C \) (AG).

(c)
- B1: For drawing both asymptotes correctly (dashed lines with equations or marked intercepts) and showing two distinct branches.
- B1: For correctly positioning the local maximum at \( (0, -3) \) and local minimum at \( (4, 5) \).
- B1: For correct curvature and asymptotic approach, with the y-intercept at \( (0, -3) \) clearly shown/labelled and no x-intercepts.

(d)
- B1: For a graph defined only for \( x > 2 \) and demonstrating reflectional symmetry across the x-axis.
- B1: For showing the vertical asymptote \( x = 2 \) with \( y \to \pm \infty \) as \( x \to 2^+ \), and showing parabolic growth without linear oblique asymptotes as \( x \to \infty \).
- B1: For labelling the stationary points at \( (4, \sqrt{5}) \) and \( (4, -\sqrt{5}) \).

(e)
- M1: For setting up the two quadratic inequalities corresponding to \( f(x) \le 6 \) (for \( x > 2 \)) and \( f(x) \ge -6 \) (for \( x < 2 \)).
- A1: For obtaining \( 3 \le x \le 6 \).
- A1: For obtaining \( \frac{-3 - \sqrt{33}}{2} \le x \le \frac{-3 + \sqrt{33}}{2} \) (exact form required).

(a) For \(C_1\), setting \(\theta = 0\) gives \(r = 2(1 + 1) = 4\), so the point on the initial line is \((4, 0)\). At \(\theta = \pi\), \(r = 0\) (the pole).
For \(C_2\), setting \(\theta = 0\) gives \(r = 6\cos 0 = 6\), so the point on the initial line is \((6, 0)\). At \(\theta = \frac{1}{2}\pi\), \(r = 0\) (the pole).
To find the intersection of \(C_1\) and \(C_2\):
\[2(1 + \cos\theta) = 6\cos\theta \implies 2 = 4\cos\theta \implies \cos\theta = \frac{1}{2}.\]
Since \(0 \le \theta \le \frac{1}{2}\pi\), \(\theta = \frac{1}{3}\pi\).
When \(\theta = \frac{1}{3}\pi\), \(r = 6\cos\left(\frac{1}{3}\pi\right) = 3\).
Thus, the non-pole point of intersection is \(\left(3, \frac{1}{3}\pi\right)\).
The sketch shows the top half of a cardioid for \(C_1\) starting at \((4, 0)\) and looping into the pole at \(\theta = \pi\), and a semicircle for \(C_2\) above the initial line with diameter from \((0,0)\) to \((6,0)\).

(b) Using \(x = r\cos\theta\) and \(r^2 = x^2 + y^2\):
\[r = 6\cos\theta \implies r^2 = 6r\cos\theta \implies x^2 + y^2 = 6x.\]
Completing the square gives:
\[(x - 3)^2 + y^2 = 9,\]
so \(a = 3\).

(c) For \(C_1\), \(x = r\cos\theta = 2(1 + \cos\theta)\cos\theta = 2\cos\theta + 2\cos^2\theta\).
Differentiating with respect to \(\theta\):
\[\frac{\mathrm{d}x}{\mathrm{d}\theta} = -2\sin\theta - 4\cos\theta\sin\theta = -2\sin\theta(1 + 2\cos\theta).\]
For tangents perpendicular to the initial line, \(\frac{\mathrm{d}x}{\mathrm{d}\theta} = 0\):
\(-2\sin\theta(1 + 2\cos\theta) = 0\).
Within \(0 \le \theta \le \pi\):
- \(\sin\theta = 0 \implies \theta = 0\) or \(\theta = \pi\).
- \(1 + 2\cos\theta = 0 \implies \cos\theta = -\frac{1}{2} \implies \theta = \frac{2}{3}\pi\).

We check \(y = r\sin\theta = 2\sin\theta + 2\sin\theta\cos\theta = 2\sin\theta + \sin 2\theta\):
\[\frac{\mathrm{d}y}{\mathrm{d}\theta} = 2\cos\theta + 2\cos 2\theta = 2(\cos\theta + 2\cos^2\theta - 1) = 2(2\cos\theta - 1)(\cos\theta + 1).\]
- At \(\theta = 0\): \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = 2(1)(2) = 4 \ne 0\), \(r = 4\). So \(\frac{\mathrm{d}y}{\mathrm{d}x} \to \infty\), giving a valid vertical tangent at \((4, 0)\).
- At \(\theta = \frac{2}{3}\pi\): \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = 2(-2)\left(\frac{1}{2}\right) = -2 \ne 0\), \(r = 2\left(1 - \frac{1}{2}\right) = 1\). So \(\frac{\mathrm{d}y}{\mathrm{d}x} \to \infty\), giving a valid vertical tangent at \(\left(1, \frac{2}{3}\pi\right)\).
- At \(\theta = \pi\): \(r = 0\) and \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = 0\); as \(\theta \to \pi\), \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2(2\cos\theta - 1)(\cos\theta + 1)}{-2\sin\theta(2\cos\theta + 1)} = \frac{-(2\cos\theta - 1)\cos(\theta/2)}{\sin(\theta/2)(2\cos\theta + 1)} \to 0\), so the tangent at the pole is the initial line (horizontal), not perpendicular.

Thus, the points are \((4, 0)\) and \(\left(1, \frac{2}{3}\pi\right)\).

(d) For \(0 \le \theta \le \frac{1}{3}\pi\), \(2(1+\cos\theta) \le 6\cos\theta\), so the boundary is given by \(C_1\).
For \(\frac{1}{3}\pi \le \theta \le \frac{1}{2}\pi\), \(6\cos\theta \le 2(1+\cos\theta)\), so the boundary is given by \(C_2\).

The area \(A\) is:
\[A = \frac{1}{2}\int_0^{\pi/3} [2(1 + \cos\theta)]^2 \,\mathrm{d}\theta + \frac{1}{2}\int_{\pi/3}^{\pi/2} (6\cos\theta)^2 \,\mathrm{d}\theta.\]

Calculating the first integral:
\[I_1 = 2\int_0^{\pi/3} (1 + 2\cos\theta + \cos^2\theta) \,\mathrm{d}\theta = 2\int_0^{\pi/3} \left(1 + 2\cos\theta + \frac{1 + \cos 2\theta}{2}\right) \,\mathrm{d}\theta\]
\[= \int_0^{\pi/3} (3 + 4\cos\theta + \cos 2\theta) \,\mathrm{d}\theta = \left[ 3\theta + 4\sin\theta + \frac{1}{2}\sin 2\theta \right]_0^{\pi/3}\]
\[= 3\left(\frac{\pi}{3}\right) + 4\sin\left(\frac{\pi}{3}\right) + \frac{1}{2}\sin\left(\frac{2\pi}{3}\right) = \pi + 4\left(\frac{\sqrt{3}}{2}\right) + \frac{1}{2}\left(\frac{\sqrt{3}}{2}\right) = \pi + \frac{9\sqrt{3}}{4}.\]

Calculating the second integral:
\[I_2 = 18\int_{\pi/3}^{\pi/2} \cos^2\theta \,\mathrm{d}\theta = 9\int_{\pi/3}^{\pi/2} (1 + \cos 2\theta) \,\mathrm{d}\theta = 9\left[ \theta + \frac{1}{2}\sin 2\theta \right]_{\pi/3}^{\pi/2}\]
\[= 9\left[\left(\frac{\pi}{2} + 0\right) - \left(\frac{\pi}{3} + \frac{\sqrt{3}}{4}\right)\right] = 9\left(\frac{\pi}{6} - \frac{\sqrt{3}}{4}\right) = \frac{3\pi}{2} - \frac{9\sqrt{3}}{4}.\]

Summing the two parts:
\[A = I_1 + I_2 = \left(\pi + \frac{9\sqrt{3}}{4}\right) + \left(\frac{3\pi}{2} - \frac{9\sqrt{3}}{4}\right) = \frac{5\pi}{2}.\]

(a)
B1: Correct shape for \(C_1\) (upper half of cardioid starting on initial line and entering pole tangentially at \(\theta = \pi\)) and correct shape for \(C_2\) (semicircle in first quadrant passing through pole).
B1: Correct intersections with initial line clearly indicated: \((4, 0)\) for \(C_1\) and \((6, 0)\) for \(C_2\).
M1: Equates equations \(2(1+\cos\theta) = 6\cos\theta\) to find \(\theta\).
A1: Obtains \(\left(3, \frac{1}{3}\pi\right)\) or exact equivalent.

(b)
M1: Multiplies by \(r\) and substitutes \(r^2 = x^2 + y^2\) and \(r\cos\theta = x\).
A1: Obtains \((x - 3)^2 + y^2 = 9\) (or states \(a = 3\)).

(c)
M1: Sets up \(x = r\cos\theta = 2\cos\theta + 2\cos^2\theta\) and differentiates with respect to \(\theta\).
M1: Sets \(\frac{\mathrm{d}x}{\mathrm{d}\theta} = 0\) and solves for \(\theta\).
A1: Obtains \(\theta = 0\) and \(\theta = \frac{2}{3}\pi\).
M1: Calculates corresponding \(r\) values for both angles.
A1: Obtains \((4, 0)\) and \(\left(1, \frac{2}{3}\pi\right)\), and correctly identifies/rejects the pole \(\theta = \pi\).

(d)
M1: Recognises the need to split the area into two integrals at \(\theta = \frac{1}{3}\pi\) using \(\frac{1}{2}\int r^2 \,\mathrm{d}\theta\).
M1: Uses \(\cos^2\theta = \frac{1}{2}(1 + \cos 2\theta)\) to integrate \([2(1+\cos\theta)]^2\).
A1: Obtains correct integrated expression \(\left[3\theta + 4\sin\theta + \frac{1}{2}\sin 2\theta\right]\) and evaluates at limits \(0\) to \(\frac{\pi}{3}\) to give \(\pi + \frac{9\sqrt{3}}{4}\).
M1: Uses double angle formula to integrate \((6\cos\theta)^2\).
A1: Obtains \(9\left[\theta + \frac{1}{2}\sin 2\theta\right]\) and evaluates between \(\frac{\pi}{3}\) and \(\frac{\pi}{2}\) to give \(\frac{3\pi}{2} - \frac{9\sqrt{3}}{4}\).
A1: Combines both areas to obtain exact value \(\frac{5\pi}{2}\).

Let \( y = \ln(1 + \operatorname{arsinh} x) \). When \( x = 0 \), \( y(0) = \ln(1 + 0) = 0 \).

Differentiating with respect to \( x \):
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{1 + \operatorname{arsinh} x} \cdot \frac{1}{\sqrt{1+x^2}} = (1 + \operatorname{arsinh} x)^{-1}(1+x^2)^{-\frac{1}{2}} \]
When \( x = 0 \):
\[ y'(0) = (1+0)^{-1}(1+0)^{-\frac{1}{2}} = 1 \]

Differentiating a second time using the product rule:
\[ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -(1+\operatorname{arsinh} x)^{-2}\left(\frac{1}{\sqrt{1+x^2}}\right)(1+x^2)^{-\frac{1}{2}} + (1+\operatorname{arsinh} x)^{-1}\left(-\frac{1}{2}\right)(1+x^2)^{-\frac{3}{2}}(2x) \]
\[ = -\frac{1}{(1+\operatorname{arsinh} x)^2(1+x^2)} - \frac{x}{(1+\operatorname{arsinh} x)(1+x^2)^{\frac{3}{2}}} \]
When \( x = 0 \):
\[ y''(0) = -\frac{1}{(1)^2(1)} - 0 = -1 \]

Differentiating a third time:
Let \( T_1 = -(1+\operatorname{arsinh} x)^{-2}(1+x^2)^{-1} \) and \( T_2 = -x(1+\operatorname{arsinh} x)^{-1}(1+x^2)^{-\frac{3}{2}} \).
\[ \frac{\mathrm{d}T_1}{\mathrm{d}x} = 2(1+\operatorname{arsinh} x)^{-3}(1+x^2)^{-\frac{1}{2}}(1+x^2)^{-1} + (1+\operatorname{arsinh} x)^{-2}(1+x^2)^{-2}(2x) \]
At \( x = 0 \), \( \left.\frac{\mathrm{d}T_1}{\mathrm{d}x}\right|_{x=0} = 2(1)(1)(1) + 0 = 2 \).

For \( T_2 = -x g(x) \) where \( g(x) = (1+\operatorname{arsinh} x)^{-1}(1+x^2)^{-\frac{3}{2}} \):
\[ \frac{\mathrm{d}T_2}{\mathrm{d}x} = -g(x) - x g'(x) \]
At \( x = 0 \), \( \left.\frac{\mathrm{d}T_2}{\mathrm{d}x}\right|_{x=0} = -g(0) - 0 = -(1)^{-1}(1)^{-\frac{3}{2}} = -1 \).

Therefore:
\[ y'''(0) = 2 + (-1) = 1 \]

The Maclaurin series expansion is:
\[ y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \dots \]
\[ y = 0 + (1)x + \frac{-1}{2}x^2 + \frac{1}{6}x^3 = x - \frac{1}{2}x^2 + \frac{1}{6}x^3 \]

M1: Differentiates \( y \) correctly using the chain rule and the standard derivative \( \frac{\mathrm{d}}{\mathrm{d}x}(\operatorname{arsinh} x) = \frac{1}{\sqrt{1+x^2}} \).
A1: Obtains \( y'(0) = 1 \) and correctly applies the product/chain rule to obtain \( y''(0) = -1 \).
M1: Differentiates again to obtain an expression for \( \frac{\mathrm{d}^3y}{\mathrm{d}x^3} \) and evaluates at \( x = 0 \).
A1: Obtains \( y'''(0) = 1 \).
A1: Correctly substitutes derivatives into the Maclaurin formula to obtain \( y = x - \frac{1}{2}x^2 + \frac{1}{6}x^3 \) (or equivalent simplified form).

(a) Using integration by parts on \( I_n = \int_0^1 x^n \sqrt{1-x} \, \mathrm{d}x \):
Let \( u = x^n \implies \frac{\mathrm{d}u}{\mathrm{d}x} = n x^{n-1} \)
and \( \frac{\mathrm{d}v}{\mathrm{d}x} = (1-x)^{\frac{1}{2}} \implies v = -\frac{2}{3}(1-x)^{\frac{3}{2}} \).

Applying integration by parts:
\[ I_n = \left[ -\frac{2}{3} x^n (1-x)^{\frac{3}{2}} \right]_0^1 - \int_0^1 \left( -\frac{2}{3}(1-x)^{\frac{3}{2}} \right) \left( n x^{n-1} \right) \, \mathrm{d}x \]

Evaluating the boundary term:
\[ \left[ -\frac{2}{3} x^n (1-x)^{\frac{3}{2}} \right]_0^1 = 0 - 0 = 0 \quad (\text{for } n \ge 1). \]

Thus,
\[ I_n = \frac{2n}{3} \int_0^1 x^{n-1} (1-x)^{\frac{3}{2}} \, \mathrm{d}x. \]

Write \( (1-x)^{\frac{3}{2}} = (1-x)\sqrt{1-x} \):
\[ I_n = \frac{2n}{3} \int_0^1 x^{n-1} (1-x) \sqrt{1-x} \, \mathrm{d}x \]
\[ I_n = \frac{2n}{3} \left( \int_0^1 x^{n-1} \sqrt{1-x} \, \mathrm{d}x - \int_0^1 x^n \sqrt{1-x} \, \mathrm{d}x \right) \]
\[ I_n = \frac{2n}{3} (I_{n-1} - I_n). \]

Multiply through by \( 3 \):
\[ 3 I_n = 2n I_{n-1} - 2n I_n \]
\[ (2n + 3)I_n = 2n I_{n-1}. \]

(b) First find \( I_0 \):
\[ I_0 = \int_0^1 \sqrt{1-x} \, \mathrm{d}x = \left[ -\frac{2}{3}(1-x)^{\frac{3}{2}} \right]_0^1 = 0 - \left(-\frac{2}{3}\right) = \frac{2}{3}. \]

Using the reduction formula \( I_n = \frac{2n}{2n+3} I_{n-1} \):
For \( n = 1 \):
\[ I_1 = \frac{2(1)}{2(1)+3} I_0 = \frac{2}{5} \left( \frac{2}{3} \right) = \frac{4}{15}. \]

For \( n = 2 \):
\[ I_2 = \frac{2(2)}{2(2)+3} I_1 = \frac{4}{7} \left( \frac{4}{15} \right) = \frac{16}{105}. \]

For \( n = 3 \):
\[ I_3 = \frac{2(3)}{2(3)+3} I_2 = \frac{6}{9} I_2 = \frac{2}{3} \left( \frac{16}{105} \right) = \frac{32}{315}. \]

(a)
- M1: For attempting integration by parts by setting \( u = x^n \) and \( \frac{\mathrm{d}v}{\mathrm{d}x} = (1-x)^{1/2} \).
- A1: For correct intermediate expression \( \left[ -\frac{2}{3} x^n (1-x)^{3/2} \right]_0^1 + \frac{2n}{3} \int_0^1 x^{n-1} (1-x)^{3/2} \, \mathrm{d}x \) with the limits evaluated to zero.
- M1: For splitting \( (1-x)^{3/2} = (1-x)\sqrt{1-x} \) and expressing the integral in terms of \( I_n \) and \( I_{n-1} \).
- A1 (AG): For fully convincing algebraic progression to obtain \( (2n+3)I_n = 2n I_{n-1} \) with no steps suppressed.

(b)
- B1: For finding the correct initial value \( I_0 = \frac{2}{3} \).
- M1: For applying the reduction formula successively to find \( I_1 \), \( I_2 \), and \( I_3 \).
- A1: For the correct exact value \( I_3 = \frac{32}{315} \).

(a) Using the hyperbolic identity \(\cosh^2 x = 1 + \sinh^2 x\):
\[ \int \cosh^3 x \, \mathrm{d}x = \int \cosh x (1 + \sinh^2 x) \, \mathrm{d}x = \int (\cosh x + \sinh^2 x \cosh x) \, \mathrm{d}x \]
Integrating directly (or using substitution \(u = \sinh x\), \(\mathrm{d}u = \cosh x \, \mathrm{d}x\)):
\[ \int (\cosh x + \sinh^2 x \cosh x) \, \mathrm{d}x = \sinh x + \frac{1}{3}\sinh^3 x + c \]

(b) The given differential equation is first-order linear in standard form:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} + P(x)y = Q(x) \quad \text{where } P(x) = 2\tanh x, \; Q(x) = \cosh x \]
The integrating factor \(I(x)\) is:
\[ I(x) = \exp\left(\int 2\tanh x \, \mathrm{d}x\right) = \exp(2\ln(\cosh x)) = \exp(\ln(\cosh^2 x)) = \cosh^2 x \]
Multiplying the differential equation by \(I(x)\):
\[ \cosh^2 x \frac{\mathrm{d}y}{\mathrm{d}x} + 2y\sinh x \cosh x = \cosh^3 x \]
\[ \frac{\mathrm{d}}{\mathrm{d}x}\left(y \cosh^2 x\right) = \cosh^3 x \]
Integrating both sides with respect to \(x\) and using the result from part (a):
\[ y \cosh^2 x = \int \cosh^3 x \, \mathrm{d}x = \sinh x + \frac{1}{3}\sinh^3 x + c \]
Dividing through by \(\cosh^2 x\):
\[ y = \frac{\sinh x + \frac{1}{3}\sinh^3 x + c}{\cosh^2 x} = \tanh x \operatorname{sech} x + \frac{1}{3}\tanh x \sinh x + c\operatorname{sech}^2 x \]

(c) Using the initial condition \(y(0) = 0\):
\[ 0 = \frac{\sinh(0) + \frac{1}{3}\sinh^3(0) + c}{\cosh^2(0)} = \frac{0 + 0 + c}{1^2} \implies c = 0 \]
Thus, the particular solution is:
\[ y = \frac{\sinh x + \frac{1}{3}\sinh^3 x}{\cosh^2 x} \]
When \(x = \ln 3\):
\[ \mathrm{e}^x = 3, \quad \mathrm{e}^{-x} = \frac{1}{3} \]
\[ \sinh(\ln 3) = \frac{3 - \frac{1}{3}}{2} = \frac{8/3}{2} = \frac{4}{3} \]
\[ \cosh(\ln 3) = \frac{3 + \frac{1}{3}}{2} = \frac{10/3}{2} = \frac{5}{3} \]
Substitute these values into \(y\):
\[ \sinh(\ln 3) + \frac{1}{3}\sinh^3(\ln 3) = \frac{4}{3} + \frac{1}{3}\left(\frac{4}{3}\right)^3 = \frac{4}{3} + \frac{64}{81} = \frac{108 + 64}{81} = \frac{172}{81} \]
\[ \cosh^2(\ln 3) = \left(\frac{5}{3}\right)^2 = \frac{25}{9} = \frac{225}{81} \]
Therefore:
\[ y = \frac{\frac{172}{81}}{\frac{225}{81}} = \frac{172}{225} \]

(a)
M1: Use identity \(\cosh^2 x = 1 + \sinh^2 x\) or substitute \(u = \sinh x\).
A1: Split into integrable forms \(\int (\cosh x + \sinh^2 x \cosh x) \, \mathrm{d}x\) or \(\int (1 + u^2) \, \mathrm{d}u\).
A1: Fully correct derivation leading convincingly to \(\sinh x + \frac{1}{3}\sinh^3 x + c\) (AG).

(b)
M1: State and attempt to evaluate the integrating factor \(I = \mathrm{e}^{\int 2\tanh x \, \mathrm{d}x}\).
A1: Obtain \(I = \cosh^2 x\).
M1: Write as \(\frac{\mathrm{d}}{\mathrm{d}x}(y \cosh^2 x) = \cosh^3 x\) and attempt integration using (a).
A1: Obtain \(y \cosh^2 x = \sinh x + \frac{1}{3}\sinh^3 x + c\).
A1: Express \(y\) explicitly as a function of \(x\).

(c)
M1: Substitute \(x = 0, y = 0\) into candidate solution to find \(c = 0\).
M1: Evaluate \(\sinh(\ln 3) = \frac{4}{3}\) and \(\cosh(\ln 3) = \frac{5}{3}\) using exponential definitions.
A1: Correct final simplified fraction \(\frac{172}{225}\).

(a) Differentiating the equation implicitly with respect to \(x\):
\[ 2\cosh(2y)\frac{\mathrm{d}y}{\mathrm{d}x} + \left(\mathrm{e}^y + x\mathrm{e}^y\frac{\mathrm{d}y}{\mathrm{d}x}\right) - \sin(x+y)\left(1 + \frac{\mathrm{d}y}{\mathrm{d}x}\right) = 0. \]
Substitute \(x = 0\) and \(y = 0\):
\[ 2\cosh(0)\frac{\mathrm{d}y}{\mathrm{d}x} + \mathrm{e}^0 + 0 - \sin(0)\left(1 + \frac{\mathrm{d}y}{\mathrm{d}x}\right) = 0 \]
\[ 2(1)\frac{\mathrm{d}y}{\mathrm{d}x} + 1 - 0 = 0 \implies \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{1}{2}. \]

(b) Differentiating the first derivative equation implicitly with respect to \(x\):
\[ \left(4\sinh(2y)\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 + 2\cosh(2y)\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\right) + \left(\mathrm{e}^y\frac{\mathrm{d}y}{\mathrm{d}x} + \mathrm{e}^y\frac{\mathrm{d}y}{\mathrm{d}x} + x\mathrm{e}^y\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 + x\mathrm{e}^y\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\right) - \left(\cos(x+y)\left(1 + \frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 + \sin(x+y)\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\right) = 0. \]
Substitute \(x = 0\), \(y = 0\), and \(\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{1}{2}\):
\[ 4\sinh(0)\left(-\frac{1}{2}\right)^2 + 2\cosh(0)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 2\mathrm{e}^0\left(-\frac{1}{2}\right) + 0 - \cos(0)\left(1 - \frac{1}{2}\right)^2 - \sin(0)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 0 \]
\[ 0 + 2(1)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 1 - 1\left(\frac{1}{2}\right)^2 - 0 = 0 \]
\[ 2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 1 - \frac{1}{4} = 0 \implies 2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{5}{4} \implies \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{5}{8}. \]

(c) Using the Maclaurin series expansion:
\[ y = y(0) + x y'(0) + \frac{x^2}{2!} y''(0) + \dots \]
Given \(y(0) = 0\), \(y'(0) = -\frac{1}{2}\), and \(y''(0) = \frac{5}{8}\):
\[ y = 0 + \left(-\frac{1}{2}\right)x + \frac{1}{2}\left(\frac{5}{8}\right)x^2 = -\frac{1}{2}x + \frac{5}{16}x^2. \]

(a)
M1: Differentiates implicitly with respect to \(x\), using the chain rule on \(\sinh(2y)\) and \(\cos(x+y)\) and the product rule on \(x\mathrm{e}^y\).
A1: Obtains a fully correct first derivative equation.
A1: Substitutes \(x=0, y=0\) to obtain \(\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{1}{2}\).

(b)
M1: Differentiates the implicit expression a second time with respect to \(x\), applying product/chain rules systematically.
A1: Correctly substitutes \(x = 0\), \(y = 0\), and \(y' = -\frac{1}{2}\) into an unsimplified or simplified second derivative expression.
A1: (AG) Correctly simplifies without missing steps to achieve \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{5}{8}\).

(c)
M1: Applies the Maclaurin expansion formula \(y = y(0) + x y'(0) + \frac{x^2}{2}y''(0)\).
A1: Obtains \(y = -\frac{1}{2}x + \frac{5}{16}x^2\) (or exact equivalent).

(a) Let \( f(x) = \frac{1}{\sqrt{2x-1}} \). For \( x \ge 1 \), \( f'(x) = -(2x-1)^{-\frac{3}{2}} < 0 \), so \( f(x) \) is strictly decreasing.

Consider rectangles constructed on the intervals \([r, r+1]\) for \( r = 1, 2, \dots, n \), each of width \( 1 \) and height \( f(r) = \frac{1}{\sqrt{2r-1}} \).

Since \( f(x) \) is strictly decreasing, each rectangle of height \( f(r) \) lies above the curve \( y = f(x) \) on \([r, r+1]\).

The total area of these \( n \) rectangles is
\[ \text{Area} = \sum_{r=1}^n \left( \frac{1}{\sqrt{2r-1}} \times 1 \right) = \sum_{r=1}^n \frac{1}{\sqrt{2r-1}}. \]

The area under the curve between \( x = 1 \) and \( x = n+1 \) is
\[ \int_1^{n+1} \frac{1}{\sqrt{2x-1}}\,\mathrm{d}x = \left[ \sqrt{2x-1} \right]_1^{n+1} = \sqrt{2(n+1)-1} - \sqrt{2(1)-1} = \sqrt{2n+1} - 1. \]

Since the sum of the areas of the rectangles exceeds the area under the curve,
\[ \sum_{r=1}^n \frac{1}{\sqrt{2r-1}} > \sqrt{2n+1} - 1. \]

(b) Now consider rectangles constructed on the intervals \([r-1, r]\) for \( r = 2, 3, \dots, n \), each of width \( 1 \) and height \( f(r) = \frac{1}{\sqrt{2r-1}} \).

Since \( f(x) \) is strictly decreasing, each rectangle of height \( f(r) \) lies entirely below the curve on \([r-1, r]\).

The total area of these \( n-1 \) rectangles is
\[ \sum_{r=2}^n \frac{1}{\sqrt{2r-1}}. \]

The area under the curve between \( x = 1 \) and \( x = n \) is
\[ \int_1^n \frac{1}{\sqrt{2x-1}}\,\mathrm{d}x = \left[ \sqrt{2x-1} \right]_1^n = \sqrt{2n-1} - 1. \]

Since the rectangles lie below the curve,
\[ \sum_{r=2}^n \frac{1}{\sqrt{2r-1}} < \sqrt{2n-1} - 1. \]

Adding the first term \( \frac{1}{\sqrt{2(1)-1}} = 1 \) to both sides yields
\[ \sum_{r=1}^n \frac{1}{\sqrt{2r-1}} < \sqrt{2n-1}. \]

(c) Combining the results of parts (a) and (b), for \( n \ge 2 \):
\[ \sqrt{2n+1} - 1 < \sum_{r=1}^n \frac{1}{\sqrt{2r-1}} < \sqrt{2n-1}. \]

Dividing throughout by \( \sqrt{n} \):
\[ \frac{\sqrt{2n+1} - 1}{\sqrt{n}} < \frac{1}{\sqrt{n}} \sum_{r=1}^n \frac{1}{\sqrt{2r-1}} < \frac{\sqrt{2n-1}}{\sqrt{n}}. \]

Taking limits as \( n \to \infty \):
\[ \lim_{n \to \infty} \frac{\sqrt{2n+1} - 1}{\sqrt{n}} = \lim_{n \to \infty} \left( \sqrt{2 + \frac{1}{n}} - \frac{1}{\sqrt{n}} \right) = \sqrt{2}, \]
\[ \lim_{n \to \infty} \frac{\sqrt{2n-1}}{\sqrt{n}} = \lim_{n \to \infty} \sqrt{2 - \frac{1}{n}} = \sqrt{2}. \]

By the Sandwich (Squeeze) Theorem,
\[ \lim_{n \to \infty} \frac{1}{\sqrt{n}} \sum_{r=1}^n \frac{1}{\sqrt{2r-1}} = \sqrt{2}. \]

(a)
B1: Sketch of \( y = (2x-1)^{-1/2} \) for \( x \ge 1 \) showing a strictly decreasing curve and clearly indicating rectangles extending above the curve.
M1: Evaluates the integral \( \int_1^{n+1} (2x-1)^{-1/2} \,\mathrm{d}x = [\sqrt{2x-1}]_1^{n+1} \).
A1: Obtains \( \sqrt{2n+1} - 1 \) correctly with correct limits.
A1: Completes the justification with a clear statement comparing the sum of rectangle areas to the integral (AG).

(b)
M1: Considers lower/inscribed rectangles on intervals \([r-1, r]\) of width 1 and height \( f(r) \) for \( r = 2, \dots, n \).
M1: Integrates between \( 1 \) and \( n \) to obtain \( \sqrt{2n-1} - 1 \).
A1: Concludes \( \sum_{r=1}^n \frac{1}{\sqrt{2r-1}} < \sqrt{2n-1} \) by adding \( 1 \) to both sides (AG).

(c)
M1: Sets up the inequality \( \frac{\sqrt{2n+1}-1}{\sqrt{n}} < \frac{1}{\sqrt{n}}\sum_{r=1}^n \frac{1}{\sqrt{2r-1}} < \frac{\sqrt{2n-1}}{\sqrt{n}} \).
M1: Evaluates the limit of both bounding expressions as \( n \to \infty \).
A1: States final answer \( \sqrt{2} \) with reference to the sandwich/squeeze theorem.

(a)

To find the complementary function, consider the auxiliary equation:
\[ m^2 + 6m + 9 = 0 \implies (m + 3)^2 = 0 \]
This gives a repeated root $m = -3$.
Therefore, the complementary function is:
\[ y_{\text{CF}} = (A + Bx)\mathrm{e}^{-3x} \]

For the particular integral, since $\mathrm{e}^{-3x}$ and $x\mathrm{e}^{-3x}$ appear in the complementary function, we try a particular integral of the form:
\[ y_{\text{PI}} = C x^2 \mathrm{e}^{-3x} + P\cos x + Q\sin x \]

Let $y_1 = C x^2 \mathrm{e}^{-3x}$:
\[ y_1' = C(2x - 3x^2)\mathrm{e}^{-3x} \]
\[ y_1'' = C(2 - 6x - 6x + 9x^2)\mathrm{e}^{-3x} = C(2 - 12x + 9x^2)\mathrm{e}^{-3x} \]
Substituting into the differential equation:
\[ y_1'' + 6y_1' + 9y_1 = C\left[(2 - 12x + 9x^2) + 6(2x - 3x^2) + 9x^2\right]\mathrm{e}^{-3x} = 2C\mathrm{e}^{-3x} \]
Setting $2C\mathrm{e}^{-3x} = 18\mathrm{e}^{-3x}$ gives:
\[ 2C = 18 \implies C = 9 \]

Let $y_2 = P\cos x + Q\sin x$:
\[ y_2' = -P\sin x + Q\cos x \]
\[ y_2'' = -P\cos x - Q\sin x \]
Substituting into the differential equation:
\[ y_2'' + 6y_2' + 9y_2 = (-P + 6Q + 9P)\cos x + (-Q - 6P + 9Q)\sin x \]
\[ = (8P + 6Q)\cos x + (-6P + 8Q)\sin x \]
Equating coefficients with $50\sin x$:
\[ 8P + 6Q = 0 \implies P = -\frac{3}{4}Q \]
\[ -6P + 8Q = 50 \implies -6\left(-\frac{3}{4}Q\right) + 8Q = 50 \implies \frac{25}{2}Q = 50 \implies Q = 4 \]
Then $P = -\frac{3}{4}(4) = -3$.

Hence, the particular integral is:
\[ y_{\text{PI}} = 9x^2\mathrm{e}^{-3x} - 3\cos x + 4\sin x \]

The general solution is $y = y_{\text{CF}} + y_{\text{PI}}$:
\[ y = (A + Bx + 9x^2)\mathrm{e}^{-3x} - 3\cos x + 4\sin x \]

(b)

Using the initial condition $y(0) = 1$:
\[ 1 = (A + 0 + 0)\mathrm{e}^{0} - 3\cos 0 + 4\sin 0 \implies 1 = A - 3 \implies A = 4 \]

Differentiating the general solution:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = (B + 18x)\mathrm{e}^{-3x} - 3(A + Bx + 9x^2)\mathrm{e}^{-3x} + 3\sin x + 4\cos x \]

Using the initial condition $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$ when $x = 0$:
\[ 0 = B - 3A + 4 \]
Substituting $A = 4$:
\[ 0 = B - 3(4) + 4 \implies 0 = B - 8 \implies B = 8 \]

Therefore, the particular solution is:
\[ y = (4 + 8x + 9x^2)\mathrm{e}^{-3x} - 3\cos x + 4\sin x \]

(a)
- M1: Sets up and solves the auxiliary equation $m^2 + 6m + 9 = 0$.
- A1: Obtains correct complementary function $(A + Bx)\mathrm{e}^{-3x}$.
- B1: Recognises the form $y = Cx^2\mathrm{e}^{-3x}$ for the exponential part of the PI.
- M1: Differentiates $Cx^2\mathrm{e}^{-3x}$ twice and substitutes into LHS to determine $C$.
- A1: Obtains $C = 9$.
- M1: Uses $P\cos x + Q\sin x$, substitutes into LHS and solves the simultaneous equations for $P$ and $Q$.
- A1: Obtains $P = -3, Q = 4$ and states the full general solution $y = (A + Bx + 9x^2)\mathrm{e}^{-3x} - 3\cos x + 4\sin x$.

(b)
- M1: Uses $x = 0, y = 1$ in their general solution to form an equation in $A$ and finds $A$.
- M1: Differentiates general solution (applying product rule correctly) and uses $x = 0, y' = 0$ to find $B$.
- A1: Obtains $y = (4 + 8x + 9x^2)\mathrm{e}^{-3x} - 3\cos x + 4\sin x$ (or equivalent expanded form).

(a)
Rearranging the given equation:
\[ \left(\frac{z+1}{z-1}\right)^6 = 1 = \mathrm{e}^{\mathrm{i}2k\pi}, \quad \text{for } k = 0, 1, 2, 3, 4, 5. \]
Taking the 6th root:
\[ \frac{z+1}{z-1} = \mathrm{e}^{\mathrm{i}\frac{2k\pi}{6}} = \mathrm{e}^{\mathrm{i}\frac{k\pi}{3}}. \]
For \( k = 0 \), \( \frac{z+1}{z-1} = 1 \implies z+1 = z-1 \), which has no finite solution.

For \( k \in \{1, 2, 3, 4, 5\} \):
\[ z+1 = \mathrm{e}^{\mathrm{i}\frac{k\pi}{3}}(z-1) \implies z\left(1 - \mathrm{e}^{\mathrm{i}\frac{k\pi}{3}}\right) = -\left(1 + \mathrm{e}^{\mathrm{i}\frac{k\pi}{3}}\right) \]
\[ z = \frac{\mathrm{e}^{\mathrm{i}\frac{k\pi}{3}} + 1}{\mathrm{e}^{\mathrm{i}\frac{k\pi}{3}} - 1} = \frac{\mathrm{e}^{\mathrm{i}\frac{k\pi}{6}}\left(\mathrm{e}^{\mathrm{i}\frac{k\pi}{6}} + \mathrm{e}^{-\mathrm{i}\frac{k\pi}{6}}\right)}{\mathrm{e}^{\mathrm{i}\frac{k\pi}{6}}\left(\mathrm{e}^{\mathrm{i}\frac{k\pi}{6}} - \mathrm{e}^{-\mathrm{i}\frac{k\pi}{6}}\right)} = \frac{2\cos\left(\frac{k\pi}{6}\right)}{2\mathrm{i}\sin\left(\frac{k\pi}{6}\right)} = -\mathrm{i}\cot\left(\frac{k\pi}{6}\right) = \mathrm{i}\cot\left(\frac{(6-k)\pi}{6}\right). \]
Alternatively, using the half-angle factoring directly:
\[ z = \frac{\mathrm{e}^{\mathrm{i}\frac{k\pi}{6}} + \mathrm{e}^{-\mathrm{i}\frac{k\pi}{6}}}{\mathrm{e}^{\mathrm{i}\frac{k\pi}{6}} - \mathrm{e}^{-\mathrm{i}\frac{k\pi}{6}}} = \frac{2\cos\left(\frac{k\pi}{6}\right)}{2\mathrm{i}\sin\left(\frac{k\pi}{6}\right)} = -\mathrm{i}\cot\left(\frac{k\pi}{6}\right). \]
For \( k = 3 \), \( \cot\left(\frac{3\pi}{6}\right) = \cot\left(\frac{\pi}{2}\right) = 0 \), giving the root \( z = 0 \).

Thus the non-zero roots correspond to \( k = 1, 2, 4, 5 \), which give \( z = \mathrm{i}\cot\left(\frac{k\pi}{6}\right) \) since \( -\cot\left(\frac{k\pi}{6}\right) = \cot\left(\frac{(6-k)\pi}{6}\right) \).

(b)
Expanding both binomial terms using the Binomial Theorem:
\[ (z+1)^6 = z^6 + 6z^5 + 15z^4 + 20z^3 + 15z^2 + 6z + 1 \]
\[ (z-1)^6 = z^6 - 6z^5 + 15z^4 - 20z^3 + 15z^2 - 6z + 1 \]
Subtracting the two expressions:
\[ (z+1)^6 - (z-1)^6 = 12z^5 + 40z^3 + 12z = 0 \]
Dividing by \( 4z \) for non-zero roots \( (z \ne 0) \):
\[ 3z^4 + 10z^2 + 3 = 0. \]
Let \( w = z^2 \). Then:
\[ 3w^2 + 10w + 3 = 0. \]
The non-zero roots of the original polynomial in \( z \) are:
\[ z = \pm \mathrm{i}\cot\left(\frac{\pi}{6}\right), \quad z = \pm \mathrm{i}\cot\left(\frac{\pi}{3}\right). \]
Thus, the roots of the quadratic in \( w = z^2 \) are:
\[ w_1 = \left(\mathrm{i}\cot\left(\frac{\pi}{6}\right)\right)^2 = -\cot^2\left(\frac{\pi}{6}\right) \]
\[ w_2 = \left(\mathrm{i}\cot\left(\frac{\pi}{3}\right)\right)^2 = -\cot^2\left(\frac{\pi}{3}\right) \]
Using the sum of the roots of the quadratic equation \( 3w^2 + 10w + 3 = 0 \):
\[ w_1 + w_2 = -\frac{10}{3} \implies -\cot^2\left(\frac{\pi}{6}\right) - \cot^2\left(\frac{\pi}{3}\right) = -\frac{10}{3} \]
\[ \cot^2\left(\frac{\pi}{6}\right) + \cot^2\left(\frac{\pi}{3}\right) = \frac{10}{3}. \]

(c)
Let \( S = \sum_{r=1}^{\infty} \left(\frac{1}{3}\right)^r \sin(r\theta) \).
Consider \( C + \mathrm{i}S \):
\[ C + \mathrm{i}S = \sum_{r=1}^{\infty} \left(\frac{1}{3}\right)^r \left(\cos(r\theta) + \mathrm{i}\sin(r\theta)\right) = \sum_{r=1}^{\infty} \left(\frac{1}{3}\mathrm{e}^{\mathrm{i}\theta}\right)^r. \]
This is an infinite geometric series with first term \( a = \frac{1}{3}\mathrm{e}^{\mathrm{i}\theta} \) and common ratio \( q = \frac{1}{3}\mathrm{e}^{\mathrm{i}\theta} \).
Since \( |q| = \left|\frac{1}{3}\mathrm{e}^{\mathrm{i}\theta}\right| = \frac{1}{3} < 1 \), the sum to infinity converges to:
\[ C + \mathrm{i}S = \frac{\frac{1}{3}\mathrm{e}^{\mathrm{i}\theta}}{1 - \frac{1}{3}\mathrm{e}^{\mathrm{i}\theta}} = \frac{\mathrm{e}^{\mathrm{i}\theta}}{3 - \mathrm{e}^{\mathrm{i}\theta}}. \]
Multiplying the numerator and denominator by the complex conjugate of the denominator, \( 3 - \mathrm{e}^{-\mathrm{i}\theta} \):
\[ C + \mathrm{i}S = \frac{\mathrm{e}^{\mathrm{i}\theta}(3 - \mathrm{e}^{-\mathrm{i}\theta})}{(3 - \mathrm{e}^{\mathrm{i}\theta})(3 - \mathrm{e}^{-\mathrm{i}\theta})} = \frac{3\mathrm{e}^{\mathrm{i}\theta} - 1}{9 - 3(\mathrm{e}^{\mathrm{i}\theta} + \mathrm{e}^{-\mathrm{i}\theta}) + 1} = \frac{3(\cos\theta + \mathrm{i}\sin\theta) - 1}{10 - 6\cos\theta}. \]
Equating the real parts:
\[ C = \frac{3\cos\theta - 1}{10 - 6\cos\theta}. \]

(a) [4 marks]
- M1: Replaces the equation with \( \left(\frac{z+1}{z-1}\right)^6 = 1 = \mathrm{e}^{\mathrm{i}2k\pi} \) and takes the 6th root.
- M1: Expresses \( z \) in terms of \( \mathrm{e}^{\mathrm{i}\frac{k\pi}{3}} \) and uses exponential definitions of sine and cosine (half-angle method).
- A1: Correct simplification showing \( z = -\mathrm{i}\cot\left(\frac{k\pi}{6}\right) \) or \( \mathrm{i}\cot\left(\frac{k\pi}{6}\right) \).
- A1: Explains clearly why \( k=0 \) gives no solution, \( k=3 \) gives the zero root, and correctly concludes the form for \( k = 1, 2, 4, 5 \) (AG).

(b) [3 marks]
- M1: Uses the Binomial Theorem to expand both \( (z+1)^6 \) and \( (z-1)^6 \), correctly subtracting and dividing by \( 4z \) to obtain \( 3z^4 + 10z^2 + 3 = 0 \).
- M1: Sets up a substitution \( w = z^2 \) and relates the roots of the quadratic equation to \( -\cot^2\left(\frac{\pi}{6}\right) \) and \( -\cot^2\left(\frac{\pi}{3}\right) \).
- A1: Correctly applies the relation for the sum of the roots to deduce \( \cot^2\left(\frac{\pi}{6}\right) + \cot^2\left(\frac{\pi}{3}\right) = \frac{10}{3} \).

(c) [3 marks]
- M1: Considers the geometric series \( C + \mathrm{i}S = \sum_{r=1}^{\infty} \left(\frac{1}{3}\mathrm{e}^{\mathrm{i}\theta}\right)^r \) and correctly applies the sum to infinity formula \( S_\infty = \frac{a}{1-q} \).
- M1: Multiplies numerator and denominator by the complex conjugate \( 3 - \mathrm{e}^{-\mathrm{i}\theta} \) (or uses Cartesian form \( 3 - \cos\theta + \mathrm{i}\sin\theta \)) to realise the denominator.
- A1: Equates the real parts correctly to achieve the given result \( C = \frac{3\cos\theta - 1}{10 - 6\cos\theta} \) (AG).

(a) The characteristic equation of \(\mathbf{A}\) is \(\det(\mathbf{A} - \lambda\mathbf{I}) = 0\):
\[\det\begin{pmatrix} 2-\lambda & 1 & -2 \\ 0 & 3-\lambda & -2 \\ 3 & 1 & -3-\lambda \end{pmatrix} = 0\]
Expanding along the first column:
\[(2-\lambda)[(3-\lambda)(-3-\lambda) - (-2)(1)] + 3[1(-2) - (3-\lambda)(-2)] = 0\]
\[(2-\lambda)(\lambda^2 - 9 + 2) + 3(-2 + 6 - 2\lambda) = 0\]
\[(2-\lambda)(\lambda^2 - 7) + 3(4 - 2\lambda) = 0\]
\[2\lambda^2 - 14 - \lambda^3 + 7\lambda + 12 - 6\lambda = 0\]
\[-\lambda^3 + 2\lambda^2 + \lambda - 2 = 0\]
\[-\lambda^2(\lambda - 2) + 1(\lambda - 2) = -(\lambda^2 - 1)(\lambda - 2) = -(\lambda - 1)(\lambda + 1)(\lambda - 2) = 0\]
Thus, the eigenvalues are \(\lambda = 1, 2, -1\).

(b) For \(\lambda = 1\):
\[\begin{pmatrix} 1 & 1 & -2 \\ 0 & 2 & -2 \\ 3 & 1 & -4 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\]
From the second row, \(2y - 2z = 0 \implies y = z\).
From the first row, \(x + y - 2y = 0 \implies x = y\).
Setting \(y = 1\), an eigenvector is \(\mathbf{e}_1 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\).

For \(\lambda = 2\):
\[\begin{pmatrix} 0 & 1 & -2 \\ 0 & 1 & -2 \\ 3 & 1 & -5 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\]
From the first row, \(y - 2z = 0 \implies y = 2z\).
From the third row, \(3x + 2z - 5z = 0 \implies 3x = 3z \implies x = z\).
Setting \(z = 1\), an eigenvector is \(\mathbf{e}_2 = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}\).

For \(\lambda = -1\):
\[\begin{pmatrix} 3 & 1 & -2 \\ 0 & 4 & -2 \\ 3 & 1 & -2 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\]
From the second row, \(4y - 2z = 0 \implies z = 2y\).
From the first row, \(3x + y - 4y = 0 \implies 3x = 3y \implies x = y\).
Setting \(y = 1\), an eigenvector is \(\mathbf{e}_3 = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}\).

(c) Using the eigenvectors as columns:
\[\mathbf{P} = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{pmatrix}, \quad \mathbf{D} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -1 \end{pmatrix}\]
(Other consistent column orderings of \(\mathbf{P}\) and diagonal entries of \(\mathbf{D}\) are also valid).

(d) The characteristic equation of \(\mathbf{A}\) is \(-\lambda^3 + 2\lambda^2 + \lambda - 2 = 0\).
By the Cayley-Hamilton theorem, \(\mathbf{A}\) satisfies its characteristic equation:
\[-\mathbf{A}^3 + 2\mathbf{A}^2 + \mathbf{A} - 2\mathbf{I} = \mathbf{0}\]
\[\mathbf{A}^3 = 2\mathbf{A}^2 + \mathbf{A} - 2\mathbf{I} \quad \text{[AG]}\]
Multiplying throughout by \(\mathbf{A}^{-1}\):
\[\mathbf{A}^2 = 2\mathbf{A} + \mathbf{I} - 2\mathbf{A}^{-1}\]
\[2\mathbf{A}^{-1} = -\mathbf{A}^2 + 2\mathbf{A} + \mathbf{I}\]
\[\mathbf{A}^{-1} = -\frac{1}{2}\mathbf{A}^2 + \mathbf{A} + \frac{1}{2}\mathbf{I}\]
Hence \(p = -\frac{1}{2}, q = 1, r = \frac{1}{2}\).

(e) First compute \(\mathbf{A}^2\):
\[\mathbf{A}^2 = \begin{pmatrix} 2 & 1 & -2 \\ 0 & 3 & -2 \\ 3 & 1 & -3 \end{pmatrix} \begin{pmatrix} 2 & 1 & -2 \\ 0 & 3 & -2 \\ 3 & 1 & -3 \end{pmatrix} = \begin{pmatrix} -2 & 3 & 0 \\ -6 & 7 & 0 \\ -3 & 3 & 1 \end{pmatrix}\]
Now substitute into the expression for \(\mathbf{A}^{-1}\):
\[\mathbf{A}^{-1} = -\frac{1}{2}\begin{pmatrix} -2 & 3 & 0 \\ -6 & 7 & 0 \\ -3 & 3 & 1 \end{pmatrix} + \begin{pmatrix} 2 & 1 & -2 \\ 0 & 3 & -2 \\ 3 & 1 & -3 \end{pmatrix} + \frac{1}{2}\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\]
\[\mathbf{A}^{-1} = \begin{pmatrix} 1 & -\frac{3}{2} & 0 \\ 3 & -\frac{7}{2} & 0 \\ \frac{3}{2} & -\frac{3}{2} & -\frac{1}{2} \end{pmatrix} + \begin{pmatrix} 2 & 1 & -2 \\ 0 & 3 & -2 \\ 3 & 1 & -3 \end{pmatrix} + \begin{pmatrix} \frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{2} & 0 \\ 0 & 0 & \frac{1}{2} \end{pmatrix}\]
\[\mathbf{A}^{-1} = \begin{pmatrix} \frac{7}{2} & -\frac{1}{2} & -2 \\ 3 & 0 & -2 \\ \frac{9}{2} & -\frac{1}{2} & -3 \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 7 & -1 & -4 \\ 6 & 0 & -4 \\ 9 & -1 & -6 \end{pmatrix}\]

(a)
M1: Set up characteristic equation \(\det(\mathbf{A} - \lambda\mathbf{I}) = 0\) and expand determinant to obtain a cubic equation in \(\lambda\).
A1: Correct cubic equation \(\lambda^3 - 2\lambda^2 - \lambda + 2 = 0\) (or \(-\lambda^3 + 2\lambda^2 + \lambda - 2 = 0\)).
A1: Correct eigenvalues \(\lambda = 1, 2, -1\).

(b)
M1: Attempt to solve \((\mathbf{A} - \lambda\mathbf{I})\mathbf{x} = \mathbf{0}\) for at least one eigenvalue.
A1: Correct eigenvector for \(\lambda = 1\), e.g. \(\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\) or any non-zero multiple.
A1: Correct eigenvector for \(\lambda = 2\), e.g. \(\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}\) or any non-zero multiple.
A1: Correct eigenvector for \(\lambda = -1\), e.g. \(\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}\) or any non-zero multiple.

(c)
B1: State correct matrix \(\mathbf{P}\) and diagonal matrix \(\mathbf{D}\) matching the order of the eigenvectors in part (b).

(d)
B1: Apply Cayley-Hamilton theorem to state that \(\mathbf{A}\) satisfies its characteristic equation to show \(\mathbf{A}^3 = 2\mathbf{A}^2 + \mathbf{A} - 2\mathbf{I}\) convincingly (AG).
M1: Multiply equation by \(\mathbf{A}^{-1}\) and rearrange for \(\mathbf{A}^{-1}\).
A1: Obtain \(\mathbf{A}^{-1} = -\frac{1}{2}\mathbf{A}^2 + \mathbf{A} + \frac{1}{2}\mathbf{I}\) (or explicitly identify \(p = -\frac{1}{2}, q = 1, r = \frac{1}{2}\)).

(e)
M1: Correct method to calculate \(\mathbf{A}^2\).
A1: Correct matrix \(\mathbf{A}^2 = \begin{pmatrix} -2 & 3 & 0 \\ -6 & 7 & 0 \\ -3 & 3 & 1 \end{pmatrix}\).
A1: Fully correct inverse matrix \(\mathbf{A}^{-1} = \begin{pmatrix} \frac{7}{2} & -\frac{1}{2} & -2 \\ 3 & 0 & -2 \\ \frac{9}{2} & -\frac{1}{2} & -3 \end{pmatrix}\) or equivalent.