2025 Cambridge IAS-Level Mathematics - Further (9231) Practice Paper with Answers

(a) Combining the fractions on the left-hand side over a common denominator:
\[ \frac{1}{(2r-1)(2r+1)} - \frac{1}{(2r+1)(2r+3)} = \frac{(2r+3) - (2r-1)}{(2r-1)(2r+1)(2r+3)} \]
\[ = \frac{4}{(2r-1)(2r+1)(2r+3)} \]
This completes the proof.

(b) Using the identity from part (a), we can write:
\[ u_r = \frac{1}{4} \left( \frac{1}{(2r-1)(2r+1)} - \frac{1}{(2r+1)(2r+3)} \right) \]
Summing from \( r = 1 \) to \( n \):
\[ \sum_{r=1}^{n} u_r = \frac{1}{4} \sum_{r=1}^{n} \left( \frac{1}{(2r-1)(2r+1)} - \frac{1}{(2r+1)(2r+3)} \right) \]
This is a telescoping sum:
\[ = \frac{1}{4} \left[ \left( \frac{1}{1 \cdot 3} - \frac{1}{3 \cdot 5} \right) + \left( \frac{1}{3 \cdot 5} - \frac{1}{5 \cdot 7} \right) + \dots + \left( \frac{1}{(2n-1)(2n+1)} - \frac{1}{(2n+1)(2n+3)} \right) \right] \]
All intermediate terms cancel out, leaving:
\[ = \frac{1}{4} \left[ \frac{1}{3} - \frac{1}{(2n+1)(2n+3)} \right] \]
To express this as a single fraction:
\[ = \frac{1}{4} \left[ \frac{(2n+1)(2n+3) - 3}{3(2n+1)(2n+3)} \right] \]
\[ = \frac{1}{4} \left[ \frac{4n^2 + 8n + 3 - 3}{3(2n+1)(2n+3)} \right] \]
\[ = \frac{4n(n+2)}{12(2n+1)(2n+3)} = \frac{n(n+2)}{3(2n+1)(2n+3)} \]

(c) As \( n \to \infty \), the term \( \frac{1}{(2n+1)(2n+3)} \to 0 \).
Therefore,
\[ \sum_{r=1}^{\infty} u_r = \frac{1}{4} \left( \frac{1}{3} \right) = \frac{1}{12} \]

(d) The required sum can be found by taking the difference of two partial sums:
\[ \sum_{r=n+1}^{2n} u_r = \sum_{r=1}^{2n} u_r - \sum_{r=1}^{n} u_r \]
Using the unsimplified form of the sum from part (b):
\[ = \left( \frac{1}{12} - \frac{1}{4(4n+1)(4n+3)} \right) - \left( \frac{1}{12} - \frac{1}{4(2n+1)(2n+3)} \right) \]
\[ = \frac{1}{4(2n+1)(2n+3)} - \frac{1}{4(4n+1)(4n+3)} \]
\[ = \frac{1}{4} \left[ \frac{(4n+1)(4n+3) - (2n+1)(2n+3)}{(2n+1)(2n+3)(4n+1)(4n+3)} \right] \]
Expanding and simplifying the numerator:
\[ (16n^2 + 16n + 3) - (4n^2 + 8n + 3) = 12n^2 + 8n = 4n(3n+2) \]
Thus, the expression becomes:
\[ = \frac{1}{4} \cdot \frac{4n(3n+2)}{(2n+1)(2n+3)(4n+1)(4n+3)} = \frac{n(3n+2)}{(2n+1)(2n+3)(4n+1)(4n+3)} \]

(a)
M1: For combining the two algebraic fractions over a common denominator.
A1: For showing numerator simplifies to 4 and completing the proof correctly.

(b)
M1: For using the result of (a) to write the general term as a difference.
M1: For showing the telescoping effect by listing terms or indicating cancellation clearly.
A1: For obtaining the correct unsimplified sum: \( \frac{1}{4} \left[ \frac{1}{3} - \frac{1}{(2n+1)(2n+3)} \right] \).
A1: For obtaining the fully simplified single fraction \( \frac{n(n+2)}{3(2n+1)(2n+3)} \).

(c)
B1: For finding the limit \( \frac{1}{12} \) (or equivalent).

(d)
M1: For expressing the sum as \( S_{2n} - S_n \) using either the simplified or unsimplified forms.
A1: For obtaining the correct simplified algebraic expression: \( \frac{n(3n+2)}{(2n+1)(2n+3)(4n+1)(4n+3)} \).

(i) A shear parallel to the \(y\)-axis is represented by a matrix of the form \(\begin{pmatrix} 1 & 0 \\ k & 1 \end{pmatrix}\).
Applying this to the point \((2, 1)\):
\[\begin{pmatrix} 1 & 0 \\ k & 1 \end{pmatrix} \begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 2k + 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 7 \end{pmatrix}\]
This gives:
\[2k + 1 = 7 \implies 2k = 6 \implies k = 3\]
Thus, the shear matrix is \(\mathbf{A} = \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix}\).

The reflection in the line \(y = -x\) maps \((1, 0) \to (0, -1)\) and \((0, 1) \to (-1, 0)\), which yields the matrix:
\[\mathbf{B} = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}\]

(ii) The matrix \(\mathbf{C}\) is calculated as:
\[\mathbf{C} = \mathbf{AB} = \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix} \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ -1 & -3 \end{pmatrix}\]
To find the invariant lines through the origin, we let the line have the equation \(y = mx\).
Any point \((x, mx)\) on this line maps to \((x', y')\) which must also lie on the line, meaning \(y' = mx'\):
\[\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ -1 & -3 \end{pmatrix} \begin{pmatrix} x \\ mx \end{pmatrix} = \begin{pmatrix} -mx \\ -x - 3mx \end{pmatrix}\]
Substituting these into \(y' = mx'\):
\[-x - 3mx = m(-mx) \implies -x - 3mx = -m^2 x\]
Since \(x \neq 0\) for points away from the origin on the line, we can divide by \(-x\):
\[1 + 3m = m^2 \implies m^2 - 3m - 1 = 0\]
Using the quadratic formula to solve for \(m\):
\[m = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(-1)}}{2} = \frac{3 \pm \sqrt{13}}{2}\]
Hence, the equations of the invariant lines are:
\[y = \left(\frac{3 + \sqrt{13}}{2}\right)x \quad \text{and} \quad y = \left(\frac{3 - \sqrt{13}}{2}\right)x\]

(iii) (a) Since \((1, -2)\) is mapped to \((8, -2)\) under \(\mathbf{D}\):
\[\begin{pmatrix} p & q \\ 2 & p \end{pmatrix} \begin{pmatrix} 1 \\ -2 \end{pmatrix} = \begin{pmatrix} 8 \\ -2 \end{pmatrix}\]
This gives the system of linear equations:
\[p - 2q = 8\]
\[2 - 2p = -2\]
From the second equation:
\[-2p = -4 \implies p = 2\]
Substituting \(p = 2\) into the first equation:
\[2 - 2q = 8 \implies -2q = 6 \implies q = -3\]
So, \(p = 2\) and \(q = -3\).

(b) The matrix \(\mathbf{D}\) is:
\[\mathbf{D} = \begin{pmatrix} 2 & -3 \\ 2 & 2 \end{pmatrix}\]
The determinant of \(\mathbf{D}\) is:
\[\det \mathbf{D} = (2)(2) - (-3)(2) = 4 + 6 = 10\]
The area scale factor is given by \(|\det \mathbf{D}|\), so the area of the triangle is enlarged by a factor of 10.

(i)
M1: For setting up the general matrix of a shear parallel to the \(y\)-axis and substituting \((2, 1) \to (2, 7)\).
A1: For the correct matrix \(\mathbf{A}\).
B1: For the correct reflection matrix \(\mathbf{B}\).

(ii)
B1: For the correct product matrix \(\mathbf{C} = \mathbf{AB}\).
M1: For using \(y = mx\) to write the transformation equation \(y' = mx'\) in terms of \(m\) and \(x\).
M1: For deriving the quadratic equation \(m^2 - 3m - 1 = 0\).
A1: For both correct invariant line equations in exact form.

(iii) (a)
M1: For setting up the matrix equation or the corresponding system of linear equations.
A1: For finding \(p = 2\).
A1: For finding \(q = -3\).

(iii) (b)
B1: For computing the determinant and stating that the area enlargement factor is 10.

Let \(H_n\) be the statement:
\[ \begin{pmatrix} 5 & -4 \\ 1 & 1 \end{pmatrix}^n = 3^{n-1} \begin{pmatrix} 2n+3 & -4n \\ n & 3-2n \end{pmatrix}. \]

**Base Case:**
For \(n = 1\):
\[ \text{LHS} = \begin{pmatrix} 5 & -4 \\ 1 & 1 \end{pmatrix}^1 = \begin{pmatrix} 5 & -4 \\ 1 & 1 \end{pmatrix} \]
\[ \text{RHS} = 3^{1-1} \begin{pmatrix} 2(1)+3 & -4(1) \\ 1 & 3-2(1) \end{pmatrix} = 3^0 \begin{pmatrix} 5 & -4 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 5 & -4 \\ 1 & 1 \end{pmatrix} \]
Since \( \text{LHS} = \text{RHS} \), the statement \(H_1\) is true.

**Inductive Step:**
Assume that \(H_k\) is true for some positive integer \(k\). That is,
\[ \begin{pmatrix} 5 & -4 \\ 1 & 1 \end{pmatrix}^k = 3^{k-1} \begin{pmatrix} 2k+3 & -4k \\ k & 3-2k \end{pmatrix} \]
We must show that \(H_{k+1}\) is true, i.e.,
\[ \begin{pmatrix} 5 & -4 \\ 1 & 1 \end{pmatrix}^{k+1} = 3^k \begin{pmatrix} 2(k+1)+3 & -4(k+1) \\ k+1 & 3-2(k+1) \end{pmatrix} = 3^k \begin{pmatrix} 2k+5 & -4k-4 \\ k+1 & 1-2k \end{pmatrix} \]
Using the inductive hypothesis:
\[ \begin{pmatrix} 5 & -4 \\ 1 & 1 \end{pmatrix}^{k+1} = \begin{pmatrix} 5 & -4 \\ 1 & 1 \end{pmatrix}^k \begin{pmatrix} 5 & -4 \\ 1 & 1 \end{pmatrix} \]
\[ = 3^{k-1} \begin{pmatrix} 2k+3 & -4k \\ k & 3-2k \end{pmatrix} \begin{pmatrix} 5 & -4 \\ 1 & 1 \end{pmatrix} \]
Now perform the matrix multiplication:
\[ \begin{pmatrix} 2k+3 & -4k \\ k & 3-2k \end{pmatrix} \begin{pmatrix} 5 & -4 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} (2k+3)(5) + (-4k)(1) & (2k+3)(-4) + (-4k)(1) \\ (k)(5) + (3-2k)(1) & (k)(-4) + (3-2k)(1) \end{pmatrix} \]
\[ = \begin{pmatrix} 10k + 15 - 4k & -8k - 12 - 4k \\ 5k + 3 - 2k & -4k + 3 - 2k \end{pmatrix} \]
\[ = \begin{pmatrix} 6k + 15 & -12k - 12 \\ 3k + 3 & 3 - 6k \end{pmatrix} \]
\[ = 3 \begin{pmatrix} 2k + 5 & -4k - 4 \\ k + 1 & 1 - 2k \end{pmatrix} \]
\[ = 3 \begin{pmatrix} 2(k+1)+3 & -4(k+1) \\ k+1 & 3-2(k+1) \end{pmatrix} \]

Multiplying by the scalar factor \(3^{k-1}\):
\[ \begin{pmatrix} 5 & -4 \\ 1 & 1 \end{pmatrix}^{k+1} = 3^{k-1} \cdot 3 \begin{pmatrix} 2(k+1)+3 & -4(k+1) \\ k+1 & 3-2(k+1) \end{pmatrix} \]
\[ = 3^k \begin{pmatrix} 2(k+1)+3 & -4(k+1) \\ k+1 & 3-2(k+1) \end{pmatrix} \]
This is the statement \(H_{k+1}\).

**Conclusion:**
Since \(H_1\) is true, and \(H_k \implies H_{k+1}\), by mathematical induction, \(H_n\) is true for all positive integers \(n\).

**B1**: Verifies the base case \(n = 1\) by showing both LHS and RHS reduce to the same matrix.

**M1**: States the inductive hypothesis \(H_k\) and sets up the product \(\mathbf{M}^{k+1} = \mathbf{M}^k \mathbf{M}\) (or \(\mathbf{M}\mathbf{M}^k\)).

**M1**: Substitutes the inductive hypothesis and attempts matrix multiplication of the two matrices.

**A1**: Obtains at least two correct entries in the unsimplified product matrix (e.g., \(6k+15\) and \(3k+3\)).

**A1**: Obtains a fully correct product matrix, such as \(\begin{pmatrix} 6k + 15 & -12k - 12 \\ 3k + 3 & 3 - 6k \end{pmatrix}\).

**A1**: Pulls out the factor of 3 to show the matrix in the required form, resulting in \(3^k \begin{pmatrix} 2(k+1)+3 & -4(k+1) \\ k+1 & 3-2(k+1) \end{pmatrix}\).

**A1**: Gives a complete and logical conclusion to the proof by induction.

From the given quartic equation \(x^4 + 2x^2 - 3x + 1 = 0\), we identify the coefficients:
\(p = \sum \alpha = 0\)
\(q = \sum \alpha\beta = 2\)
\(r = \sum \alpha\beta\gamma = 3\)
\(s = \alpha\beta\gamma\delta = 1\)

**(i)**
Using the relation for the sum of squares of the roots:
\[\sum \alpha^2 = \left(\sum \alpha\right)^2 - 2\sum \alpha\beta\]
Substitute the values:
\[\sum \alpha^2 = 0^2 - 2(2) = -4\]

**(ii)**
Using Newton's sums or algebraic identities:
\[S_1 = 0\]
\[S_2 = -4\]
For a quartic equation, the relation for \(S_3\) is:
\[S_3 - p S_2 + q S_1 - 3r = 0\]
Substituting the known values:
\[S_3 - 0(-4) + 2(0) - 3(3) = 0 \implies S_3 = 9\]

Alternatively, using the identity:
\[\left(\sum \alpha\right)\left(\sum \alpha^2 - \sum \alpha\beta\right) = \sum \alpha^3 - 3\sum \alpha\beta\gamma\]
Since \(\sum \alpha = 0\):
\[0 = \sum \alpha^3 - 3(3) \implies \sum \alpha^3 = 9\]

**(iii)**
We want to find \(\sum \alpha^2 \beta^2\).
Consider the expansion:
\[\left(\sum \alpha\beta\right)^2 = \sum \alpha^2\beta^2 + 2\sum \alpha^2\beta\gamma + 6\alpha\beta\gamma\delta\]
To find \(\sum \alpha^2\beta\gamma\), we use:
\[\left(\sum \alpha\right)\left(\sum \alpha\beta\gamma\right) = \sum \alpha^2\beta\gamma + 4\alpha\beta\gamma\delta\]
Since \(\sum \alpha = 0\):
\[0 = \sum \alpha^2\beta\gamma + 4(1) \implies \sum \alpha^2\beta\gamma = -4\]
Substitute these into the first expansion:
\[2^2 = \sum \alpha^2\beta^2 + 2(-4) + 6(1)\]
\[4 = \sum \alpha^2\beta^2 - 8 + 6\]
\[4 = \sum \alpha^2\beta^2 - 2 \implies \sum \alpha^2\beta^2 = 6\]

Alternatively, using the general formula:
\[\sum \alpha^2\beta^2 = q^2 - 2pr + 2s\]
Substituting the values:
\[\sum \alpha^2\beta^2 = 2^2 - 2(0)(3) + 2(1) = 4 + 2 = 6\]

**(i)**
* **M1**: For using the correct identity \(\sum \alpha^2 = (\sum \alpha)^2 - 2\sum \alpha\beta\) with substitution of \(\sum \alpha = 0\) and \(\sum \alpha\beta = 2\).
* **A1**: For obtaining the correct answer \(-4\).

**(ii)**
* **M1**: For attempting to use Newton's sum formula \(S_3 - p S_2 + q S_1 - 3r = 0\) or a valid algebraic identity (e.g. expressing \(\sum \alpha^3\) in terms of known symmetric sums).
* **M1**: For substituting the correct values (e.g. \(p=0\), \(q=2\), \(r=3\), \(S_1=0\), \(S_2=-4\)).
* **A1**: For obtaining the correct answer \(9\).

**(iii)**
* **M1**: For stating or using a valid relation for \(\sum \alpha^2 \beta^2\), such as \((\sum \alpha\beta)^2 = \sum \alpha^2\beta^2 + 2\sum \alpha^2\beta\gamma + 6\alpha\beta\gamma\delta\) or the formula \(q^2 - 2pr + 2s\).
* **M1**: For finding \(\sum \alpha^2\beta\gamma = -4\) (or showing its derivation if the general formula is not assumed without proof), and substituting values correctly.
* **A1**: For obtaining the correct answer \(6\).

(i) The direction vectors of \(l_1\) and \(l_2\) are \(\mathbf{u} = \mathbf{i} + 2\mathbf{j} - \mathbf{k}\) and \(\mathbf{v} = 2\mathbf{i} - \mathbf{j} + 3\mathbf{k}\).
A vector perpendicular to both lines is:
\(\mathbf{n} = \mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -1 \\ 2 & -1 & 3 \end{vmatrix} = \mathbf{i}(6 - 1) - \mathbf{j}(3 + 2) + \mathbf{k}(-1 - 4) = 5\mathbf{i} - 5\mathbf{j} - 5\mathbf{k}\).
We can use the simplified normal vector \(\mathbf{n}_0 = \mathbf{i} - \mathbf{j} - \mathbf{k}\).
Points on the lines are \(A(1, 1, 2)\) on \(l_1\) and \(B(3, 1, 1)\) on \(l_2\).
The vector connecting these points is \(\vec{AB} = (3 - 1)\mathbf{i} + (1 - 1)\mathbf{j} + (1 - 2)\mathbf{k} = 2\mathbf{i} - \mathbf{k}\).
The shortest distance \(d\) is the projection of \(\vec{AB}\) onto the common perpendicular:
\(d = \frac{|\vec{AB} \cdot \mathbf{n}_0|}{|\mathbf{n}_0|} = \frac{|(2)(1) + (0)(-1) + (-1)(-1)|}{\sqrt{1^2 + (-1)^2 + (-1)^2}} = \frac{2 + 1}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}\).

(ii) The normal vector to the plane \(\Pi\) is \(\mathbf{n}_0 = \mathbf{i} - \mathbf{j} - \mathbf{k}\).
Since \(\Pi\) contains \(l_1\), it contains the point \(A(1, 1, 2)\).
Its Cartesian equation is:
\(1(x - 1) - 1(y - 1) - 1(z - 2) = 0 \implies x - y - z = -2\).

(iii) The direction vector of the line \(l_3\) is \(\mathbf{w} = 4\mathbf{i} + 2\mathbf{j} - \mathbf{k}\).
The normal to the plane \(\Pi\) is \(\mathbf{n}_0 = \mathbf{i} - \mathbf{j} - \mathbf{k}\).
Let \(\theta\) be the acute angle between the line and the plane:
\(\sin\theta = \frac{|\mathbf{w} \cdot \mathbf{n}_0|}{|\mathbf{w}| |\mathbf{n}_0|}\)
\(\mathbf{w} \cdot \mathbf{n}_0 = 4(1) + 2(-1) + (-1)(-1) = 4 - 2 + 1 = 3\).
\(|\mathbf{w}| = \sqrt{4^2 + 2^2 + (-1)^2} = \sqrt{21}\).
\(|\mathbf{n}_0| = \sqrt{3}\).
Thus, \(\sin\theta = \frac{3}{\sqrt{21}\sqrt{3}} = \frac{3}{3\sqrt{7}} = \frac{1}{\sqrt{7}}\).
\(\theta = \sin^{-1}\left(\frac{1}{\sqrt{7}}\right) \approx 22.208^\circ\).
To 1 decimal place, the angle is \(22.2^\circ\).

(i)
M1: Attempt to find the vector product of the direction vectors of \(l_1\) and \(l_2\).
A1: Obtain a correct normal vector (e.g., \(\mathbf{i} - \mathbf{j} - \mathbf{k}\) or any scalar multiple).
M1: Use the shortest distance formula \(\frac{|\vec{AB} \cdot \mathbf{n}|}{|\mathbf{n}|}\) with correct coordinates of a point from each line.
A1: Obtain correct distance \(\sqrt{3}\) (or equivalent exact form, e.g., \(\frac{3}{\sqrt{3}}\)).

(ii)
M1: Use the normal vector found in part (i) to form the equation \(x - y - z = D\).
M1: Substitute a point on \(l_1\) (e.g. \((1, 1, 2)\)) into their plane equation to determine \(D\).
A1: Obtain a correct Cartesian equation, e.g., \(x - y - z = -2\) (or any equivalent form like \(-x + y + z = 2\)).

(iii)
B1: State a correct direction vector for \(l_3\), e.g., \(4\mathbf{i} + 2\mathbf{j} - \mathbf{k}\).
M1: Use the correct formula \(\sin\theta = \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}\) to relate the line direction and plane normal.
A1: Obtain \(\sin\theta = \frac{1}{\sqrt{7}}\) (or exact equivalent).
A1: Compute \(22.2^\circ\) (accept only answers rounding to \(22.2^\circ\); award max 3/4 if answer is given in radians as 0.388 without explicit conversion shown).

(i) The curve \(r = a(3 + 2\cos\theta)\) is a dimpled limaon, symmetrical about the initial line.
At \(\theta = 0\), \(r = a(3 + 2(1)) = 5a\). The point is \((5a, 0)\).
At \(\theta = \pi\), \(r = a(3 + 2(-1)) = a\). The point is \((a, \pi)\).
At \(\theta = \frac{1}{2}\pi\), \(r = a(3 + 2(0)) = 3a\). The point is \((3a, \frac{1}{2}\pi)\).
At \(\theta = \frac{3}{2}\pi\), \(r = a(3 + 2(0)) = 3a\). The point is \((3a, \frac{3}{2}\pi)\).

(ii) The area \(A\) is given by:
\(A = \frac{1}{2} \int_{0}^{2\pi} r^2 \, d\theta = \frac{1}{2} \int_{0}^{2\pi} a^2(3 + 2\cos\theta)^2 \, d\theta\)
\(A = \frac{1}{2}a^2 \int_{0}^{2\pi} (9 + 12\cos\theta + 4\cos^2\theta) \, d\theta\)
Using the double-angle identity \(\cos^2\theta = \frac{1}{2}(1 + \cos 2\theta)\):
\(A = \frac{1}{2}a^2 \int_{0}^{2\pi} (9 + 12\cos\theta + 2(1 + \cos 2\theta)) \, d\theta\)
\(A = \frac{1}{2}a^2 \int_{0}^{2\pi} (11 + 12\cos\theta + 2\cos 2\theta) \, d\theta\)
\(A = \frac{1}{2}a^2 \left[ 11\theta + 12\sin\theta + \sin 2\theta \right]_{0}^{2\pi}\)
\(A = \frac{1}{2}a^2 \left( (22\pi + 0 + 0) - (0) \right) = 11\pi a^2\).

(iii) The Cartesian coordinate \(x\) is given by:
\(x = r\cos\theta = a(3 + 2\cos\theta)\cos\theta = a(3\cos\theta + 2\cos^2\theta)\).
Tangents perpendicular to the initial line occur when \(\frac{dx}{d\theta} = 0\):
\(\frac{dx}{d\theta} = a(-3\sin\theta - 4\cos\theta\sin\theta) = -a\sin\theta(3 + 4\cos\theta)\).
Setting \(\frac{dx}{d\theta} = 0\) gives:
Either \(\sin\theta = 0 \implies \theta = 0\) or \(\theta = \pi\) (within \(0 \le \theta < 2\pi\)).
- For \(\theta = 0\), \(r = 5a\), giving the point \((5a, 0)\).
- For \(\theta = \pi\), \(r = a\), giving the point \((a, \pi)\).
Or \(3 + 4\cos\theta = 0 \implies \cos\theta = -\frac{3}{4} = -0.75\).
This yields two values for \(\theta\) in the interval \(0 \le \theta < 2\pi\):
\(\theta = \arccos(-0.75) \approx 2.42\) radians,
\(\theta = 2\pi - \arccos(-0.75) \approx 3.86\) radians.
For both of these values, \(r = a(3 + 2(-0.75)) = 1.5a\).
This gives the points \((1.5a, 2.42)\) and \((1.5a, 3.86)\).
(Note that at all four points, \(\frac{dy}{d\theta} \neq 0\), so these are indeed vertical tangents.)

Part (i):
B1: Sketch of a closed, dimpled loop symmetrical about the initial line, not passing through the pole (in the correct orientation).
B1: Correctly states \((5a, 0)\) and \((a, \pi)\).
B1: Correctly states \((3a, \frac{1}{2}\pi)\) and \((3a, \frac{3}{2}\pi)\).

Part (ii):
M1: Applies the area formula \(A = \frac{1}{2}\int r^2 \, d\theta\) with correct limits.
A1: Obtains \(\frac{1}{2}a^2 \int_{0}^{2\pi} (9 + 12\cos\theta + 4\cos^2\theta) \, d\theta\).
M1: Employs the identity \(\cos^2\theta = \frac{1}{2}(1 + \cos 2\theta)\) to transform the integrand.
A1: Obtains the correct integrated expression \(11\theta + 12\sin\theta + \sin 2\theta\) (or equivalent).
A1: Correctly evaluates to get \(11\pi a^2\).

Part (iii):
M1: Expresses \(x\) as \(r\cos\theta\) in terms of \(\theta\).
A1: Correctly differentiates to find \(\frac{dx}{d\theta} = -a\sin\theta(3+4\cos\theta)\).
M1: Sets \(\frac{dx}{d\theta} = 0\) and solves for \(\theta\).
A1: Identifies \((5a, 0)\) and \((a, \pi)\).
M1: Solves \(\cos\theta = -0.75\) to find at least one value of \(\theta\) in the interval.
A1: Obtains the angles \(\theta \approx 2.42\) and \(\theta \approx 3.86\).
A1: Obtains the radial distance \(r = 1.5a\) and states both points: \((1.5a, 2.42)\) and \((1.5a, 3.86)\) (or with \(\frac{3}{2}a\)).

**Part (i)**

We rewrite the equation by algebraic division:
\[ y = \frac{x^2 - 2x + 5}{x - 1} = \frac{(x-1)^2 + 4}{x - 1} = x - 1 + \frac{4}{x - 1} \]
As \( x \to 1 \), \( y \to \pm\infty \), so the vertical asymptote is \( x = 1 \).
As \( x \to \pm\infty \), \( \frac{4}{x - 1} \to 0 \), so the oblique asymptote is \( y = x - 1 \).

**Part (ii)**

Differentiating \( y = x - 1 + 4(x-1)^{-1} \) with respect to \( x \):
\[ \frac{dy}{dx} = 1 - \frac{4}{(x-1)^2} \]
Setting \( \frac{dy}{dx} = 0 \) to find the stationary points:
\[ 1 - \frac{4}{(x-1)^2} = 0 \implies (x-1)^2 = 4 \implies x - 1 = \pm 2 \]
This gives \( x = 3 \) or \( x = -1 \).
Substituting back into the equation of \( C \):
- For \( x = 3 \): \( y = \frac{3^2 - 2(3) + 5}{3 - 1} = 4 \)
- For \( x = -1 \): \( y = \frac{(-1)^2 - 2(-1) + 5}{-1 - 1} = -4 \)

Thus, the turning points are \( (3, 4) \) and \( (-1, -4) \).

**Part (iii)**

- **Intercepts**:
When \( x = 0 \), \( y = -5 \), so the y-intercept is \( (0, -5) \).
For \( y = 0 \), \( x^2 - 2x + 5 = 0 \). Since the discriminant is \( (-2)^2 - 4(1)(5) = -16 < 0 \), there are no x-intercepts.
- **Graph**:
Draw the asymptotes \( x = 1 \) and \( y = x - 1 \) as dashed lines.
The right-hand branch (\( x > 1 \)) lies entirely above the oblique asymptote, with a local minimum at \( (3, 4) \).
The left-hand branch (\( x < 1 \)) lies entirely below the oblique asymptote, with a local maximum at \( (-1, -4) \) and crossing the y-axis at \( (0, -5) \).

**Part (iv)**

The inequality \( \left| \frac{x^2 - 2x + 5}{x - 1} \right| \ge 5 \) corresponds to finding where the graph of \( C \) lies either at or above the line \( y = 5 \), or at or below the line \( y = -5 \).

We find the critical boundaries by solving \( y = 5 \) and \( y = -5 \):
- **Case 1**: \( \frac{x^2 - 2x + 5}{x - 1} = 5 \implies x^2 - 2x + 5 = 5x - 5 \implies x^2 - 7x + 10 = 0 \implies (x-2)(x-5) = 0 \), giving \( x = 2 \) and \( x = 5 \).
- **Case 2**: \( \frac{x^2 - 2x + 5}{x - 1} = -5 \implies x^2 - 2x + 5 = -5x + 5 \implies x^2 + 3x = 0 \implies x(x+3) = 0 \), giving \( x = -3 \) and \( x = 0 \).

By referencing the sketch of \( C \):
- The left-hand branch (\( x < 1 \)) has a local maximum of \( -4 \). The values where \( y \le -5 \) are \( x \le -3 \) and \( 0 \le x < 1 \).
- The right-hand branch (\( x > 1 \)) has a local minimum of \( 4 \). The values where \( y \ge 5 \) are \( 1 < x \le 2 \) and \( x \ge 5 \).

Combining these, the complete set of values is:
\[ x \le -3 \quad \text{or} \quad 0 \le x < 1 \quad \text{or} \quad 1 < x \le 2 \quad \text{or} \quad x \ge 5 \]

**Part (i)**
- **B1**: State \( x = 1 \) as the vertical asymptote.
- **B1**: State \( y = x - 1 \) as the oblique asymptote.

**Part (ii)**
- **M1**: Attempt differentiation using quotient rule or by rewriting the equation and using the power rule.
- **A1**: Correct derivative expression, e.g., \( \frac{dy}{dx} = 1 - \frac{4}{(x-1)^2} \) or \( \frac{x^2 - 2x - 3}{(x-1)^2} \).
- **M1**: Equate the derivative to zero and solve the resulting quadratic to find two values for \( x \).
- **A1**: Correctly state both turning point coordinates: \( (3, 4) \) and \( (-1, -4) \).

**Part (iii)**
- **B1**: Both asymptotes drawn and labeled.
- **B1**: Two-branched curve in correct regions with turning points in correct quadrants relative to asymptotes.
- **B1**: Correct y-intercept \( (0, -5) \) shown and no x-intercepts on the graph.

**Part (iv)**
- **M1**: Set up and attempt to solve both \( \frac{x^2 - 2x + 5}{x - 1} = 5 \) and \( \frac{x^2 - 2x + 5}{x - 1} = -5 \).
- **A1**: Obtain correct critical boundaries \( x = 2, 5 \) and \( x = -3, 0 \).
- **M1**: Use graphical behavior or algebraic sign testing to identify regions where the inequality holds.
- **A1**: Find correct outer intervals: \( x \le -3 \) and \( x \ge 5 \).
- **A1**: Find correct inner intervals: \( 0 \le x < 1 \) and \( 1 < x \le 2 \) (must show exclusion of the vertical asymptote \( x = 1 \)).

The system does not have a unique solution if and only if the determinant of the coefficient matrix is equal to zero. Let \(A\) be the coefficient matrix: \(A = \begin{pmatrix} 1 & 2 & -1 \\ 2 & 5 & k \\ 3 & 7 & 1 \end{pmatrix}\). We compute the determinant of \(A\): \(\det(A) = 1(5 - 7k) - 2(2 - 3k) - 1(14 - 15) = 5 - 7k - 4 + 6k + 1 = 2 - k\). For the system to not have a unique solution, we must have: \(2 - k = 0\), which gives \(k = 2\).

M1: For attempting to find the determinant of the coefficient matrix and setting it to 0 (or equivalent row reduction). A1: For obtaining the correct equation \(2 - k = 0\) (or equivalent). A1: For the correct value \(k = 2\).

(i) Differentiating both parametric equations with respect to \( t \) gives \(\frac{dx}{dt} = \frac{1}{1+t}\) and \(\frac{dy}{dt} = \frac{1}{1+t^2}\). Using the parametric differentiation formula: \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{1/(1+t^2)}{1/(1+t)} = \frac{1+t}{1+t^2}\). (ii) To find the second derivative, we use the chain rule: \(\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{dt}{dx}\). Differentiating \(\frac{dy}{dx}\) with respect to \( t \) using the quotient rule: \(\frac{d}{dt}\left(\frac{1+t}{1+t^2}\right) = \frac{(1)(1+t^2) - (1+t)(2t)}{(1+t^2)^2} = \frac{1 + t^2 - 2t - 2t^2}{(1+t^2)^2} = \frac{1 - 2t - t^2}{(1+t^2)^2}\). Since \(\frac{dt}{dx} = 1+t\), we have: \(\frac{d^2y}{dx^2} = \frac{1-2t-t^2}{(1+t^2)^2} \cdot (1+t) = \frac{(1-2t-t^2)(1+t)}{(1+t^2)^2}\). (iii) For the Maclaurin's series, we evaluate the function and its derivatives at \( x = 0 \). Since \( x = \ln(1+t) \), \( x = 0 \implies t = 0 \). At \( t = 0 \): \( y(0) = \tan^{-1}(0) = 0 \); \( y'(0) = \frac{1+0}{1+0} = 1 \); \( y''(0) = \frac{(1-0-0)(1+0)}{(1+0)^2} = 1 \). To find \( y'''(0) \), we differentiate \(\frac{d^2y}{dx^2}\) with respect to \( t \) and multiply by \(\frac{dt}{dx}\): Let \( u = \frac{1-t-3t^2-t^3}{(1+t^2)^2} \). By the quotient rule: \(\frac{du}{dt} = \frac{(-1-6t-3t^2)(1+t^2)^2 - (1-t-3t^2-t^3) \cdot 4t(1+t^2)}{(1+t^2)^4}\). At \( t = 0 \), this simplifies to \(\left(\frac{du}{dt}\right)_{t=0} = \frac{(-1)(1) - 0}{1} = -1\). Since \(\left(\frac{dt}{dx}\right)_{t=0} = 1\), we have \( y'''(0) = -1 \cdot 1 = -1 \). Substituting these into Maclaurin's expansion formula: \( y = y(0) + x y'(0) + \frac{x^2}{2!} y''(0) + \frac{x^3}{3!} y'''(0) = 0 + x(1) + \frac{x^2}{2}(1) + \frac{x^3}{6}(-1) = x + \frac{1}{2}x^2 - \frac{1}{6}x^3 \).

Part (i): M1: For calculating both \( dx/dt \) and \( dy/dt \) and applying the parametric division formula. A1: For obtaining \(\frac{dy}{dx} = \frac{1+t}{1+t^2}\). Part (ii): M1: For applying the parametric chain rule for the second derivative. M1: For applying the quotient rule to differentiate \(\frac{dy}{dx}\) with respect to \( t \). A1: For successfully showing the given result with clear algebraic steps. Part (iii): M1: For finding \( t=0 \) when \( x=0 \) and evaluating \( y'(0) = 1 \) and \( y''(0) = 1 \). M1: For evaluating \( y'''(0) = -1 \) by a correct differentiation of the second derivative or via an equivalent valid expansion method. A1: For the correct series expansion \( y = x + \frac{1}{2}x^2 - \frac{1}{6}x^3 \).

**(i)** We start by using integration by parts on \( I_n = \int_0^1 1 \cdot (1-x^2)^n \mathrm{d}x \).

Let \( u = (1-x^2)^n \) and \( \frac{\mathrm{d}v}{\mathrm{d}x} = 1 \).

Then \( \frac{\mathrm{d}u}{\mathrm{d}x} = -2nx(1-x^2)^{n-1} \) and \( v = x \).

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

For \( n \ge 1 \), the boundary term evaluated at both 1 and 0 is 0:
\( \left[ x(1-x^2)^n \right]_0^1 = 1(0) - 0 = 0 \)

Thus, we have:
\( I_n = 2n \int_0^1 x^2 (1-x^2)^{n-1} \mathrm{d}x \)

We rewrite \( x^2 \) as \( 1 - (1-x^2) \):
\( I_n = 2n \int_0^1 (1 - (1-x^2)) (1-x^2)^{n-1} \mathrm{d}x \)
\( I_n = 2n \left[ \int_0^1 (1-x^2)^{n-1} \mathrm{d}x - \int_0^1 (1-x^2)^n \mathrm{d}x \right] \)
\( I_n = 2n (I_{n-1} - I_n) \)

Now, solve for \( I_n \):
\( I_n = 2n I_{n-1} - 2n I_n \)
\( I_n(1 + 2n) = 2n I_{n-1} \)
\( I_n = \frac{2n}{2n+1} I_{n-1} \) (as required).

**(ii)** We first evaluate \( I_0 \):
\( I_0 = \int_0^1 1 \mathrm{d}x = [x]_0^1 = 1 \)

Using the reduction formula successively:
For \( n = 1 \):
\( I_1 = \frac{2(1)}{2(1)+1} I_0 = \frac{2}{3}(1) = \frac{2}{3} \)

For \( n = 2 \):
\( I_2 = \frac{2(2)}{2(2)+1} I_1 = \frac{4}{5} \left(\frac{2}{3}\right) = \frac{8}{15} \)

For \( n = 3 \):
\( I_3 = \frac{2(3)}{2(3)+1} I_2 = \frac{6}{7} \left(\frac{8}{15}\right) = \frac{16}{35} \)

**(i)**
* **M1**: For attempting integration by parts on \( I_n \) with \( u = (1-x^2)^n \) and \( v' = 1 \).
* **A1**: For evaluating the boundary term to 0 and obtaining \( I_n = 2n \int_0^1 x^2 (1-x^2)^{n-1} \mathrm{d}x \).
* **M1**: For substituting \( x^2 = 1 - (1-x^2) \) to split the integral.
* **A1**: For obtaining the relation in terms of \( I_{n-1} \) and \( I_n \): \( I_n = 2n (I_{n-1} - I_n) \).
* **A1**: For completing the algebraic steps to show the given reduction formula correctly.

**(ii)**
* **B1**: For correctly evaluating \( I_0 = 1 \) (or finding \( I_1 = \frac{2}{3} \) directly by integration).
* **M1**: For applying the reduction formula successively at least twice.
* **A1**: For finding \( I_2 = \frac{8}{15} \) (or equivalent form).
* **A1**: For the correct final exact value \( I_3 = \frac{16}{35} \).

Let \(x = e^t\), which implies \(t = \ln x\) and \(\frac{dt}{dx} = \frac{1}{x}\).

By the chain rule:
\[\frac{dy}{dx} = \frac{dy}{dt} \frac{dt}{dx} = \frac{1}{x} \frac{dy}{dt}\]

Differentiating again with respect to \(x\) using the product rule:
\[\frac{d^2 y}{dx^2} = \frac{d}{dx}\left( \frac{1}{x} \frac{dy}{dt} \right) = -\frac{1}{x^2} \frac{dy}{dt} + \frac{1}{x} \frac{d}{dx}\left(\frac{dy}{dt}\right)\]
\[\frac{d^2 y}{dx^2} = -\frac{1}{x^2} \frac{dy}{dt} + \frac{1}{x} \left(\frac{d^2 y}{dt^2} \frac{dt}{dx}\right) = -\frac{1}{x^2} \frac{dy}{dt} + \frac{1}{x^2} \frac{d^2 y}{dt^2}\]

Multiplying by \(x^2\) gives:
\[x^2 \frac{d^2 y}{dx^2} = \frac{d^2 y}{dt^2} - \frac{dy}{dt}\]

Substitute these expressions, along with \(x \frac{dy}{dx} = \frac{dy}{dt}\) and \(\ln x = t\), into the original differential equation:
\[\left(\frac{d^2 y}{dt^2} - \frac{dy}{dt}\right) + 3\left(\frac{dy}{dt}\right) + y = 4t\]
\[\frac{d^2 y}{dt^2} + 2\frac{dy}{dt} + y = 4t\]
This completes the proof of the transformation.

To find the general solution of the transformed differential equation, we first solve the auxiliary equation for the complementary function (CF):
\[m^2 + 2m + 1 = 0 \implies (m+1)^2 = 0\]
This has a repeated real root \(m = -1\), so the CF is:
\[y_c = (At + B)e^{-t}\]

For the particular integral (PI), we try a linear form \(y_p = kt + c\).
Then \(\frac{dy_p}{dt} = k\) and \(\frac{d^2 y_p}{dt^2} = 0\).
Substituting these into the transformed ODE:
\[0 + 2(k) + (kt + c) = 4t\]
\[kt + (2k + c) = 4t\]
Comparing coefficients:
\[k = 4\]
\[2k + c = 0 \implies c = -2(4) = -8\]
So the PI is \(y_p = 4t - 8\).

The general solution in terms of \(t\) is:
\[y = (At + B)e^{-t} + 4t - 8\]

Now, apply the initial conditions to find the constants \(A\) and \(B\).
At \(x = 1\), we have \(t = \ln(1) = 0\) and \(y = -4\):
\[-4 = (A(0) + B)e^{0} + 4(0) - 8 \implies -4 = B - 8 \implies B = 4\]

To apply the condition for \(\frac{dy}{dx}\), we use the relation \(\frac{dy}{dt} = x \frac{dy}{dx}\).
At \(x = 1\) where \(\frac{dy}{dx} = 5\), we must have:
\[\frac{dy}{dt} = (1)(5) = 5\]

Differentiating the general solution with respect to \(t\):
\[\frac{dy}{dt} = A e^{-t} - (At + B)e^{-t} + 4\]
At \(t = 0\):
\[5 = A(1) - (0 + B)(1) + 4 \implies 5 = A - B + 4\]
Substituting \(B = 4\):
\[5 = A - 4 + 4 \implies A = 5\]

Hence, the particular solution in terms of \(t\) is:
\[y = (5t + 4)e^{-t} + 4t - 8\]

Finally, substituting back \(t = \ln x\) and \(e^{-t} = \frac{1}{x}\), we obtain the particular solution in terms of \(x\):
\[y = \frac{5 \ln x + 4}{x} + 4 \ln x - 8\]

**Part (i)**
* **M1**: Apply the chain rule to obtain the correct expression for the first derivative: \(\frac{dy}{dx} = \frac{1}{x} \frac{dy}{dt}\).
* **M1**: Differentiate again with respect to \(x\) using the product rule and chain rule to obtain the second derivative: \(\frac{d^2 y}{dx^2} = \frac{1}{x^2} \left( \frac{d^2 y}{dt^2} - \frac{dy}{dt} \right)\).
* **A1**: Show both derivative transformations correctly.
* **A1**: Substitute into the given differential equation and show convincing algebraic steps to arrive at the given target equation: \(\frac{d^2 y}{dt^2} + 2\frac{dy}{dt} + y = 4t\).

**Part (ii)**
* **M1**: Form the auxiliary equation \(m^2 + 2m + 1 = 0\) and obtain the complementary function: \(y_c = (At + B)e^{-t}\).
* **M1**: Use a trial particular integral of the form \(y_p = kt + c\) to find the constants \(k = 4\) and \(c = -8\).
* **A1**: State the correct general solution: \(y = (At + B)e^{-t} + 4t - 8\).
* **M1**: Use the condition \(y = -4\) at \(x = 1\) (equivalent to \(t = 0\)) to find \(B = 4\).
* **M1**: Correctly relate derivatives (e.g., using \(\frac{dy}{dt} = x \frac{dy}{dx}\) to set \(\frac{dy}{dt} = 5\) at \(t = 0\)) and solve for \(A\) to obtain \(A = 5\).
* **A1**: State the final correct particular solution in terms of \(x\): \(y = \frac{5 \ln x + 4}{x} + 4 \ln x - 8\) (or equivalent form such as \(y = x^{-1}(5 \ln x + 4) + 4 \ln x - 8\)).

(a) Let the interval \( [0, 1] \) be divided into \( n \) subintervals of equal width \( \delta x = \frac{1}{n} \). The partition points are \( x_r = \frac{r}{n} \) for \( r = 0, 1, 2, \dots, n \).

Since \( f(x) = \frac{1}{1+x^2} \) is strictly decreasing on \( [0, 1] \) (as \( f'(x) = -\frac{2x}{(1+x^2)^2} < 0 \) for \( x > 0 \)), on each subinterval \( [x_{r-1}, x_r] \), the minimum value of \( f(x) \) is at the right endpoint \( x_r \) and the maximum value is at the left endpoint \( x_{r-1} \).

The sum of the areas of the lower (inscribed) rectangles is:
\[ L_n = \sum_{r=1}^n f(x_r) \delta x = \sum_{r=1}^n \frac{1}{1 + (r/n)^2} \frac{1}{n} = \sum_{r=1}^n \frac{n}{n^2 + r^2} \]

The sum of the areas of the upper (circumscribed) rectangles is:
\[ U_n = \sum_{r=1}^n f(x_{r-1}) \delta x = \sum_{r=0}^{n-1} f(x_r) \delta x = \sum_{r=0}^{n-1} \frac{n}{n^2 + r^2} \]

Since \( f(x) \) is continuous and strictly decreasing, the area under the curve \( y = f(x) \) from \( x = 0 \) to \( x = 1 \) satisfies:
\[ L_n < \int_0^1 f(x) \, dx < U_n \]

Evaluating the definite integral:
\[ \int_0^1 \frac{1}{1+x^2} \, dx = \left[ \arctan x \right]_0^1 = \arctan 1 - \arctan 0 = \frac{\pi}{4} \]

Substituting this back, we obtain the required inequality:
\[ \sum_{r=1}^n \frac{n}{n^2 + r^2} < \frac{\pi}{4} < \sum_{r=0}^{n-1} \frac{n}{n^2 + r^2} \]

(b) We rewrite the given sum as:
\[ \sum_{r=n+1}^{2n} \frac{n}{n^2 + r^2} = \sum_{r=n+1}^{2n} \frac{1}{1 + (r/n)^2} \frac{1}{n} \]

As \( n \to \infty \), this Riemann sum converges to the definite integral of \( f(x) = \frac{1}{1+x^2} \).

The lower limit of integration is:
\[ \lim_{n \to \infty} \frac{n+1}{n} = 1 \]

The upper limit of integration is:
\[ \lim_{n \to \infty} \frac{2n}{n} = 2 \]

Thus, the limit of the sum is:
\[ L = \int_1^2 \frac{1}{1+x^2} \, dx = \left[ \arctan x \right]_1^2 = \arctan 2 - \arctan 1 \]

Since \( \arctan 1 = \frac{\pi}{4} \), this is \( \arctan 2 - \frac{\pi}{4} \). Using the identity for \( \tan(A - B) \):
\[ \tan(L) = \tan(\arctan 2 - \arctan 1) = \frac{2 - 1}{1 + (2)(1)} = \frac{1}{3} \]

Since \( 0 < L < \frac{\pi}{2} \), we have:
\[ L = \arctan\left(\frac{1}{3}\right) \]

Thus, the limit is \( \arctan\left(\frac{1}{3}\right) \), with \( k = \frac{1}{3} \).

Part (a):
- M1: For dividing the interval \( [0, 1] \) into \( n \) subintervals of width \( \frac{1}{n} \) and using the partition points \( x_r = \frac{r}{n} \).
- M1: For identifying that \( f(x) = \frac{1}{1+x^2} \) is decreasing, so lower and upper rectangle sums bound the integral.
- A1: For deriving correct algebraic expressions for the lower sum \( L_n \) and upper sum \( U_n \).
- B1: For evaluating \( \int_0^1 \frac{1}{1+x^2} \, dx = \frac{\pi}{4} \).
- A1: For completing the proof of the inequality with appropriate justification.

Part (b):
- M1: For expressing the sum as \( \frac{1}{n} \sum_{r=n+1}^{2n} \frac{1}{1+(r/n)^2} \).
- M1: For determining the limits of integration as \( 1 \) and \( 2 \).
- A1: For writing the limit of the sum as the definite integral \( \int_1^2 \frac{1}{1+x^2} \, dx \).
- A1: For evaluating the integral to get \( \arctan 2 - \arctan 1 \) (or equivalent).
- M1: For applying the tangent subtraction identity to simplify \( \arctan 2 - \arctan 1 \).
- A1: For obtaining the correct final value \( \arctan\left(\frac{1}{3}\right) \) (or stating \( k = \frac{1}{3} \)).

(i) To find the eigenvalues, we solve the characteristic equation \(\det(A - \lambda I) = 0\):
\[\det \begin{pmatrix} 1-\lambda & 1 & 2 \\ 0 & 2-\lambda & 2 \\ 0 & 1 & 3-\lambda \end{pmatrix} = 0\]
Expanding along the first column:
\[(1-\lambda) \left[ (2-\lambda)(3-\lambda) - 2 \right] = 0\]
\[(1-\lambda)(\lambda^2 - 5\lambda + 4) = 0\]
\[(1-\lambda)(\lambda-1)(\lambda-4) = 0\]
So the eigenvalues are \(\lambda = 1\) (repeated) and \(\lambda = 4\).

(ii) For \(\lambda = 1\):
\[(A - I)\mathbf{v} = \mathbf{0} \implies \begin{pmatrix} 0 & 1 & 2 \\ 0 & 1 & 2 \\ 0 & 1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\]
This gives the single equation \(y + 2z = 0\).
Choosing \(x = 1, z = 0\) gives the eigenvector \(\mathbf{v}_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\).
Choosing \(x = 0, z = 1\) gives \(y = -2\), yielding the eigenvector \(\mathbf{v}_2 = \begin{pmatrix} 0 \\ -2 \\ 1 \end{pmatrix}\).
These two eigenvectors are linearly independent.

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

(iii) The diagonalizing matrix \(P\) is formed by placing the eigenvectors as columns, and \(D\) contains the corresponding eigenvalues on the diagonal:
\[P = \begin{pmatrix} 1 & 0 & 1 \\ 0 & -2 & 1 \\ 0 & 1 & 1 \end{pmatrix}, \quad D = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 4 \end{pmatrix}\]

(i)
M1: Setting up characteristic equation \(\det(A - \lambda I) = 0\) and expanding.
A1: Correct characteristic cubic polynomial, e.g., \((1-\lambda)(\lambda-1)(\lambda-4) = 0\).
A1: Correct eigenvalues \(\lambda = 1, 1, 4\).

(ii)
M1: Substituting \(\lambda = 1\) into \((A - \lambda I)\mathbf{v} = 0\) and attempting to solve for eigenvectors.
A1: Finding two linearly independent eigenvectors for \(\lambda = 1\), e.g., \(\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\) and \(\begin{pmatrix} 0 \\ -2 \\ 1 \end{pmatrix}\) (or any valid linear combinations).
M1: Substituting \(\lambda = 4\) into \((A - \lambda I)\mathbf{v} = 0\) and attempting to solve.
A1: Finding a correct eigenvector for \(\lambda = 4\), e.g., \(\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\).

(iii)
M1: Constructing matrix \(P\) with columns as eigenvectors and \(D\) as corresponding diagonal entries.
A1: Correct matrices \(P\) and \(D\) (allowing for consistent column permutations).

(i) We find the characteristic equation \(\det(B - \lambda I) = 0\):
\[\det \begin{pmatrix} 3-\lambda & 1 & -1 \\ 1 & 3-\lambda & -1 \\ -1 & -1 & 5-\lambda \end{pmatrix} = 0\]
Expanding along the first row:
\[(3-\lambda) \left[ (3-\lambda)(5-\lambda) - 1 \right] - 1 \left[ (5-\lambda) - 1 \right] - 1 \left[ -1 - (-(3-\lambda)) \right] = 0\]
\[(3-\lambda)(\lambda^2 - 8\lambda + 14) - (4-\lambda) - (\lambda-2) = 0\]
\[(3-\lambda)(\lambda^2 - 8\lambda + 14) - 2(3-\lambda) = 0\]
\[(3-\lambda)(\lambda^2 - 8\lambda + 12) = 0\]
\[(3-\lambda)(\lambda-2)(\lambda-6) = 0\]
Thus, the eigenvalues of \(B\) are indeed 2, 3, and 6.

(ii) Next, we find the corresponding eigenvectors.

For \(\lambda = 2\):
\[(B-2I)\mathbf{v} = \mathbf{0} \implies \begin{pmatrix} 1 & 1 & -1 \\ 1 & 1 & -1 \\ -1 & -1 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\]
From row 1 and row 3, we easily find \(z=0\) and \(x+y=0\). An eigenvector is \(\begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}\).
Normalizing gives: \(\mathbf{u}_1 = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}\).

For \(\lambda = 3\):
\[(B-3I)\mathbf{v} = \mathbf{0} \implies \begin{pmatrix} 0 & 1 & -1 \\ 1 & 0 & -1 \\ -1 & -1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\]
From rows 1 and 2, \(y=z\) and \(x=z\). An eigenvector is \(\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\).
Normalizing gives: \(\mathbf{u}_2 = \frac{1}{\sqrt{3}} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\).

For \(\lambda = 6\):
\[(B-6I)\mathbf{v} = \mathbf{0} \implies \begin{pmatrix} -3 & 1 & -1 \\ 1 & -3 & -1 \\ -1 & -1 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\]
From row 3, \(z = -x-y\). Substituting into row 1 gives \(-3x+y - (-x-y) = 0 \implies -2x+2y = 0 \implies x=y\). This leads to \(z = -2x\). An eigenvector is \(\begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix}\).
Normalizing gives: \(\mathbf{u}_3 = \frac{1}{\sqrt{6}} \begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix}\).

Since \(B\) is symmetric, these normalized eigenvectors are mutually orthogonal. Thus, the orthogonal matrix \(Q\) is:
\[Q = \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} \\ 0 & \frac{1}{\sqrt{3}} & -\frac{2}{\sqrt{6}} \end{pmatrix}\]
And the diagonal matrix \(D\) is:
\[D = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 6 \end{pmatrix}\]

(iii) The eigenvalues of \(B^{-1}\) are the reciprocals of the eigenvalues of \(B\), which are \(\frac{1}{2}, \frac{1}{3}, \frac{1}{6}\).
The eigenvalues of \(B^3\) are the cubes of the eigenvalues of \(B\), which are \(2^3 = 8\), \(3^3 = 27\), and \(6^3 = 216\).

(i)
M1: Sets up characteristic determinant and attempts expansion.
A1: Achieves a correct simplified form of the cubic equation, e.g., \((3-\lambda)(\lambda^2 - 8\lambda + 12) = 0\).
A1: Correctly verifies that eigenvalues are 2, 3, and 6.

(ii)
M1: Attempts to find eigenvectors for at least two eigenvalues.
A1: Finds correct eigenvectors for all three eigenvalues.
M1: Normalizes the three eigenvectors.
A1: Provides a correct orthogonal matrix \(Q\) (with columns as normalized eigenvectors).
A1: Provides the corresponding diagonal matrix \(D\) consistent with the column ordering of \(Q\).

(iii)
B1: Correctly states eigenvalues of \(B^{-1}\) as \(\frac{1}{2}, \frac{1}{3}, \frac{1}{6}\).
B1: Correctly states eigenvalues of \(B^3\) as \(8, 27, 216\).

**(a)**

The given differential equation is a first-order linear differential equation of the form \(\frac{\mathrm{dy}}{\mathrm{dx}} + P(x)y = Q(x)\), with \(P(x) = \tanh x\) and \(Q(x) = \frac{1}{\sqrt{9 + \sinh^2 x}}\).

First, find the integrating factor \(I(x)\):

\(I(x) = e^{\int \tanh x \, \mathrm{dx}} = e^{\ln(\cosh x)} = \cosh x\) (since \(\cosh x > 0\) for all real \(x\)).

Multiply the differential equation by the integrating factor:

\(\cosh x \frac{\mathrm{dy}}{\mathrm{dx}} + y \sinh x = \frac{\cosh x}{\sqrt{9 + \sinh^2 x}}\)

\(\frac{\mathrm{d}}{\mathrm{dx}}(y \cosh x) = \frac{\cosh x}{\sqrt{9 + \sinh^2 x}}\)

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

\(y \cosh x = \int \frac{\cosh x}{\sqrt{9 + \sinh^2 x}} \, \mathrm{dx}\)

To evaluate the integral, use the substitution \(u = \sinh x\), which gives \(\mathrm{du} = \cosh x \, \mathrm{dx\)}:

\(\int \frac{1}{\sqrt{9 + u^2}} \, \mathrm{du} = \operatorname{arsinh}\left(\frac{u}{3}\right) + C = \operatorname{arsinh}\left(\frac{\sinh x}{3}\right) + C\)

Thus, the general solution is:

\(y = \frac{1}{\cosh x} \left( \operatorname{arsinh}\left(\frac{\sinh x}{3}\right) + C \right)\)

**(b)**

Apply the initial condition \(y = 0\) when \(x = 0\):

\(0 = \frac{1}{\cosh 0} \left( \operatorname{arsinh}(0) + C \right)\)

Since \(\cosh 0 = 1\) and \(\operatorname{arsinh}(0) = 0\), we find:

\(C = 0\)

So the particular solution is:

\(y = \frac{\operatorname{arsinh}\left(\frac{\sinh x}{3}\right)}{\cosh x}\)

Using the logarithmic definition of the inverse hyperbolic sine function, \(\operatorname{arsinh}(z) = \ln\left(z + \sqrt{z^2 + 1}\right)\):

\(\operatorname{arsinh}\left(\frac{\sinh x}{3}\right) = \ln\left( \frac{\sinh x}{3} + \sqrt{\frac{\sinh^2 x}{9} + 1} \right) = \ln\left( \frac{\sinh x + \sqrt{\sinh^2 x + 9}}{3} \right)\)

Thus, the particular solution can be written as:

\(y = \frac{\ln\left(\frac{\sinh x + \sqrt{\sinh^2 x + 9}}{3}\right)}{\cosh x}\)

where \(f(x) = \frac{\sinh x + \sqrt{\sinh^2 x + 9}}{3}\).

**(c)**

Substitute \(x = 0\) and \(y = 0\) directly into the original differential equation:

\(\frac{\mathrm{dy}}{\mathrm{dx}} + (0)\tanh(0) = \frac{1}{\sqrt{9 + \sinh^2(0)}}\)

\(\frac{\mathrm{dy}}{\mathrm{dx}} + 0 = \frac{1}{\sqrt{9 + 0}} = \frac{1}{3}\)

Therefore, at \(x = 0\), \(\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{3}\).

**(a)**
M1: Attempt to find the integrating factor using \(e^{\int P(x) \mathrm{dx}}\).
A1: Obtain the correct integrating factor \(I(x) = \cosh x\).
M1: Formulate \(\frac{\mathrm{d}}{\mathrm{dx}}(y I(x)) = Q(x) I(x)\) and set up the integration.
M1: Use a valid substitution (e.g., \(u = \sinh x\)) or direct standard formula to integrate the right-hand side.
A1: Obtain the correct integral \(\operatorname{arsinh}\left(\frac{\sinh x}{3}\right)\) (or equivalent form).
A1: Include a constant of integration \(C\).
A1: State the final explicit solution for \(y\).

**(b)**
M1: Use the initial condition \(y(0) = 0\) to show that \(C = 0\).
M1: Apply the logarithmic definition of the inverse hyperbolic sine function.
A1: Deduce the correct form for \(f(x)\).

**(c)**
M1: Use the original differential equation with \(x = 0\) and \(y = 0\) (or differentiate the particular solution using the quotient/product rule).
A1: Show that \(\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{3}\).

(i) By de Moivre's theorem, \( \cos 6\theta + i\sin 6\theta = (\cos\theta + i\sin\theta)^6 \). Equating the real parts: \( \cos 6\theta = \cos^6\theta - \binom{6}{2}\cos^4\theta\sin^2\theta + \binom{6}{4}\cos^2\theta\sin^4\theta - \sin^6\theta \). Replacing \( \sin^2\theta = 1 - \cos^2\theta \), we obtain: \( \cos 6\theta = \cos^6\theta - 15\cos^4\theta(1 - \cos^2\theta) + 15\cos^2\theta(1 - \cos^2\theta)^2 - (1 - \cos^2\theta)^3 \). Expanding and simplifying: \( \cos 6\theta = \cos^6\theta - 15\cos^4\theta + 15\cos^6\theta + 15\cos^2\theta(1 - 2\cos^2\theta + \cos^4\theta) - (1 - 3\cos^2\theta + 3\cos^4\theta - \cos^6\theta) \) which simplifies to \( \cos 6\theta = 32\cos^6\theta - 48\cos^4\theta + 18\cos^2\theta - 1 \). (ii) Let \( x = \cos\theta \). The equation is \( 64x^6 - 96x^4 + 36x^2 - 1 = 0 \). Dividing by 2: \( 32x^6 - 48x^4 + 18x^2 - \frac{1}{2} = 0 \), which can be rewritten as \( 32x^6 - 48x^4 + 18x^2 - 1 = -\frac{1}{2} \). Using the result from (i), this is equivalent to \( \cos 6\theta = -\frac{1}{2} \). The solutions for \( 6\theta \) in the range \( 0 < 6\theta < 6\pi \) are \( 6\theta = \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3}, \frac{10\pi}{3}, \frac{14\pi}{3}, \frac{16\pi}{3} \), giving \( \theta = \frac{\pi}{9}, \frac{2\pi}{9}, \frac{4\pi}{9}, \frac{5\pi}{9}, \frac{7\pi}{9}, \frac{8\pi}{9} \). Since \( \cos\frac{8\pi}{9} = -\cos\frac{\pi}{9} \), \( \cos\frac{7\pi}{9} = -\cos\frac{2\pi}{9} \), and \( \cos\frac{5\pi}{9} = -\cos\frac{4\pi}{9} \), the six roots are indeed \( \pm\cos\frac{\pi}{9} \), \( \pm\cos\frac{2\pi}{9} \), and \( \pm\cos\frac{4\pi}{9} \). (iii) Substituting \( x^2 = 1 - y \) into the equation \( 64(x^2)^3 - 96(x^2)^2 + 36(x^2) - 1 = 0 \) gives \( 64(1-y)^3 - 96(1-y)^2 + 36(1-y) - 1 = 0 \). Expanding this: \( 64(1 - 3y + 3y^2 - y^3) - 96(1 - 2y + y^2) + 36(1-y) - 1 = 0 \), which simplifies to \( 64y^3 - 96y^2 + 36y - 3 = 0 \). Since the roots in \( x \) are \( \pm\cos\frac{\pi}{9}, \pm\cos\frac{2\pi}{9}, \pm\cos\frac{4\pi}{9} \), the roots in \( y = 1 - x^2 = \sin^2\theta \) are \( \sin^2\frac{\pi}{9}, \sin^2\frac{2\pi}{9}, \sin^2\frac{4\pi}{9} \). By Vieta's formulas, the product of the roots is \( \sin^2\frac{\pi}{9}\sin^2\frac{2\pi}{9}\sin^2\frac{4\pi}{9} = \frac{3}{64} \). Taking the positive square root since the sines of these angles in the first quadrant are positive: \( \sin\frac{\pi}{9}\sin\frac{2\pi}{9}\sin\frac{4\pi}{9} = \frac{\sqrt{3}}{8} \). Since \( \cos\frac{\pi}{18} = \sin\frac{4\pi}{9} \), \( \cos\frac{5\pi}{18} = \sin\frac{2\pi}{9} \), and \( \cos\frac{7\pi}{18} = \sin\frac{\pi}{9} \), we obtain \( \cos\left(\frac{\pi}{18}\right)\cos\left(\frac{5\pi}{18}\right)\cos\left(\frac{7\pi}{18}\right) = \frac{\sqrt{3}}{8} \).

Part (i): M1: For using de Moivre's theorem to write \( \cos 6\theta \) as the real part of \( (\cos\theta + i\sin\theta)^6 \). A1: For correct binomial expansion. M1: For substituting \( \sin^2\theta = 1 - \cos^2\theta \) and expanding terms. A1: For obtaining the correct final identity (AG). Part (ii): M1: For relating the given equation to the identity in part (i) to get \( \cos 6\theta = -\frac{1}{2} \). A1: For finding the correct values of \( 6\theta \) or \( \theta \). M1: For showing how the six roots correspond to the positive and negative values of the three cosines. A1: For fully justifying the roots (AG). Part (iii): M1: For applying the substitution \( y = 1-x^2 \) or \( x^2 = 1-y \) to the equation. A1: For obtaining the correct simplified cubic equation in \( y \). M1: For finding the product of roots using Vieta's formulas and relating it to the product of sines. A1: For using the complementary angle identities to finish the proof (AG).

(i) By de Moivre's theorem, \( \cos 6\theta + i\sin 6\theta = (\cos\theta + i\sin\theta)^6 \). Equating the real parts: \( \cos 6\theta = \cos^6\theta - \binom{6}{2}\cos^4\theta\sin^2\theta + \binom{6}{4}\cos^2\theta\sin^4\theta - \sin^6\theta \). Replacing \( \sin^2\theta = 1 - \cos^2\theta \), we obtain: \( \cos 6\theta = \cos^6\theta - 15\cos^4\theta(1 - \cos^2\theta) + 15\cos^2\theta(1 - \cos^2\theta)^2 - (1 - \cos^2\theta)^3 \). Expanding and simplifying: \( \cos 6\theta = \cos^6\theta - 15\cos^4\theta + 15\cos^6\theta + 15\cos^2\theta(1 - 2\cos^2\theta + \cos^4\theta) - (1 - 3\cos^2\theta + 3\cos^4\theta - \cos^6\theta) \) which simplifies to \( \cos 6\theta = 32\cos^6\theta - 48\cos^4\theta + 18\cos^2\theta - 1 \). (ii) Let \( x = \cos\theta \). The equation is \( 64x^6 - 96x^4 + 36x^2 - 1 = 0 \). Dividing by 2: \( 32x^6 - 48x^4 + 18x^2 - \frac{1}{2} = 0 \), which can be rewritten as \( 32x^6 - 48x^4 + 18x^2 - 1 = -\frac{1}{2} \). Using the result from (i), this is equivalent to \( \cos 6\theta = -\frac{1}{2} \). The solutions for \( 6\theta \) in the range \( 0 < 6\theta < 6\pi \) are \( 6\theta = \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3}, \frac{10\pi}{3}, \frac{14\pi}{3}, \frac{16\pi}{3} \), giving \( \theta = \frac{\pi}{9}, \frac{2\pi}{9}, \frac{4\pi}{9}, \frac{5\pi}{9}, \frac{7\pi}{9}, \frac{8\pi}{9} \). Since \( \cos\frac{8\pi}{9} = -\cos\frac{\pi}{9} \), \( \cos\frac{7\pi}{9} = -\cos\frac{2\pi}{9} \), and \( \cos\frac{5\pi}{9} = -\cos\frac{4\pi}{9} \), the six roots are indeed \( \pm\cos\frac{\pi}{9} \), \( \pm\cos\frac{2\pi}{9} \), and \( \pm\cos\frac{4\pi}{9} \). (iii) Substituting \( x^2 = 1 - y \) into the equation \( 64(x^2)^3 - 96(x^2)^2 + 36(x^2) - 1 = 0 \) gives \( 64(1-y)^3 - 96(1-y)^2 + 36(1-y) - 1 = 0 \). Expanding this: \( 64(1 - 3y + 3y^2 - y^3) - 96(1 - 2y + y^2) + 36(1-y) - 1 = 0 \), which simplifies to \( 64y^3 - 96y^2 + 36y - 3 = 0 \). Since the roots in \( x \) are \( \pm\cos\frac{\pi}{9}, \pm\cos\frac{2\pi}{9}, \pm\cos\frac{4\pi}{9} \), the roots in \( y = 1 - x^2 = \sin^2\theta \) are \( \sin^2\frac{\pi}{9}, \sin^2\frac{2\pi}{9}, \sin^2\frac{4\pi}{9} \). By Vieta's formulas, the product of the roots is \( \sin^2\frac{\pi}{9}\sin^2\frac{2\pi}{9}\sin^2\frac{4\pi}{9} = \frac{3}{64} \). Taking the positive square root since the sines of these angles in the first quadrant are positive: \( \sin\frac{\pi}{9}\sin\frac{2\pi}{9}\sin\frac{4\pi}{9} = \frac{\sqrt{3}}{8} \). Since \( \cos\frac{\pi}{18} = \sin\frac{4\pi}{9} \), \( \cos\frac{5\pi}{18} = \sin\frac{2\pi}{9} \), and \( \cos\frac{7\pi}{18} = \sin\frac{\pi}{9} \), we obtain \( \cos\left(\frac{\pi}{18}\right)\cos\left(\frac{5\pi}{18}\right)\cos\left(\frac{7\pi}{18}\right) = \frac{\sqrt{3}}{8} \).

Part (i): M1: For using de Moivre's theorem to write \( \cos 6\theta \) as the real part of \( (\cos\theta + i\sin\theta)^6 \). A1: For correct binomial expansion. M1: For substituting \( \sin^2\theta = 1 - \cos^2\theta \) and expanding terms. A1: For obtaining the correct final identity (AG). Part (ii): M1: For relating the given equation to the identity in part (i) to get \( \cos 6\theta = -\frac{1}{2} \). A1: For finding the correct values of \( 6\theta \) or \( \theta \). M1: For showing how the six roots correspond to the positive and negative values of the three cosines. A1: For fully justifying the roots (AG). Part (iii): M1: For applying the substitution \( y = 1-x^2 \) or \( x^2 = 1-y \) to the equation. A1: For obtaining the correct simplified cubic equation in \( y \). M1: For finding the product of roots using Vieta's formulas and relating it to the product of sines. A1: For using the complementary angle identities to finish the proof (AG).