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

(a) Setting up the identity:
\[ \frac{A}{r(r+1)} - \frac{B}{(r+2)(r+3)} = \frac{A(r+2)(r+3) - B(r)(r+1)}{r(r+1)(r+2)(r+3)} = \frac{2r+3}{r(r+1)(r+2)(r+3)} \]
Expanding the numerator:
\[ A(r^2 + 5r + 6) - B(r^2 + r) = (A - B)r^2 + (5A - B)r + 6A \]
Equating coefficients:
\( r^2 \): \( A - B = 0 \implies A = B \)
\( r^0 \): \( 6A = 3 \implies A = \frac{1}{2} \)
Thus, \( B = \frac{1}{2} \).
So,
\[ \frac{2r+3}{r(r+1)(r+2)(r+3)} = \frac{1}{2}\left( \frac{1}{r(r+1)} - \frac{1}{(r+2)(r+3)} \right) \]

(b) Let \( u_r = \frac{1}{2}\left( \frac{1}{r(r+1)} - \frac{1}{(r+2)(r+3)} \right) \).
Writing out the first and last few terms of \( \sum_{r=1}^{n} u_r \):
For \( r = 1 \): \( \frac{1}{2}\left( \frac{1}{1\cdot 2} - \frac{1}{3\cdot 4} \right) \)
For \( r = 2 \): \( \frac{1}{2}\left( \frac{1}{2\cdot 3} - \frac{1}{4\cdot 5} \right) \)
For \( r = 3 \): \( \frac{1}{2}\left( \frac{1}{3\cdot 4} - \frac{1}{5\cdot 6} \right) \)
\( \dots \)
For \( r = n-1 \): \( \frac{1}{2}\left( \frac{1}{(n-1)n} - \frac{1}{(n+1)(n+2)} \right) \)
For \( r = n \): \( \frac{1}{2}\left( \frac{1}{n(n+1)} - \frac{1}{(n+2)(n+3)} \right) \)

Summing all terms, intermediate terms cancel in pairs, leaving:
\[ \sum_{r=1}^{n} u_r = \frac{1}{2}\left[ \frac{1}{2} + \frac{1}{6} - \frac{1}{(n+1)(n+2)} - \frac{1}{(n+2)(n+3)} \right] \]
\[ = \frac{1}{2}\left[ \frac{2}{3} - \frac{1}{n+2}\left( \frac{1}{n+1} + \frac{1}{n+3} \right) \right] \]
\[ = \frac{1}{3} - \frac{1}{2(n+2)}\left( \frac{2n+4}{(n+1)(n+3)} \right) \]
\[ = \frac{1}{3} - \frac{1}{2(n+2)}\cdot \frac{2(n+2)}{(n+1)(n+3)} = \frac{1}{3} - \frac{1}{(n+1)(n+3)} \]

(c) As \( n \to \infty \), \( \frac{1}{(n+1)(n+3)} \to 0 \).
Therefore,
\[ \sum_{r=1}^{\infty} \frac{2r+3}{r(r+1)(r+2)(r+3)} = \frac{1}{3} \]

(d) The required sum is:
\[ \sum_{r=n+1}^{2n} u_r = \sum_{r=1}^{2n} u_r - \sum_{r=1}^{n} u_r \]
\[ = \left( \frac{1}{3} - \frac{1}{(2n+1)(2n+3)} \right) - \left( \frac{1}{3} - \frac{1}{(n+1)(n+3)} \right) \]
\[ = \frac{1}{(n+1)(n+3)} - \frac{1}{(2n+1)(2n+3)} \]
\[ = \frac{(2n+1)(2n+3) - (n+1)(n+3)}{(n+1)(n+3)(2n+1)(2n+3)} \]
\[ = \frac{(4n^2 + 8n + 3) - (n^2 + 4n + 3)}{(n+1)(n+3)(2n+1)(2n+3)} \]
\[ = \frac{3n^2 + 4n}{(n+1)(n+3)(2n+1)(2n+3)} = \frac{n(3n+4)}{(n+1)(n+3)(2n+1)(2n+3)} \]

(a)
- M1: Sets up algebraic identity \( A(r+2)(r+3) - B(r)(r+1) = 2r+3 \) and attempts to solve for \( A \) and \( B \).
- A1: Obtains correct values \( A = \frac{1}{2} \) and \( B = \frac{1}{2} \).

(b)
- M1: Expresses the sum showing at least the first two terms and the last two terms to demonstrate the pattern of cancellation.
- M1: Correctly identifies surviving terms: \( \frac{1}{2}\left( \frac{1}{2} + \frac{1}{6} - \frac{1}{(n+1)(n+2)} - \frac{1}{(n+2)(n+3)} \right) \).
- A1 (AG): Completes full algebraic simplification cleanly and convincingly to arrive at \( \frac{1}{3} - \frac{1}{(n+1)(n+3)} \).

(c)
- B1: States \( \frac{1}{3} \) (must be exact).

(d)
- M1: Uses \( S_{2n} - S_n \) to set up \( \frac{1}{(n+1)(n+3)} - \frac{1}{(2n+1)(2n+3)} \).
- A1: Simplifies correctly to a single fraction \( \frac{n(3n+4)}{(n+1)(n+3)(2n+1)(2n+3)} \) or \( \frac{3n^2+4n}{(n+1)(n+3)(2n+1)(2n+3)} \).

**Step 1: Base case (\(n = 1\))**
For \(n = 1\), the formula gives:
\[ \frac{2(1) + 1}{1} = \frac{3}{1} = 3 \]
Since it is given that \(u_1 = 3\), the statement is true for \(n = 1\).

Step 2: Inductive hypothesis
Assume that the result is true for \(n = k\) for some positive integer \(k\), that is,
\[ u_k = \frac{2k + 1}{k} \]

**Step 3: Inductive step (\(n = k + 1\))**
Using the recurrence relation:
\[ u_{k+1} = \frac{3u_k - 4}{u_k - 1} \]
Substitute the inductive hypothesis \(u_k = \frac{2k + 1}{k}\):
\[ u_{k+1} = \frac{3\left(\frac{2k+1}{k}\right) - 4}{\left(\frac{2k+1}{k}\right) - 1} = \frac{\frac{3(2k+1) - 4k}{k}}{\frac{(2k+1) - k}{k}} \]
Simplifying numerator and denominator:
\[ u_{k+1} = \frac{6k + 3 - 4k}{2k + 1 - k} = \frac{2k + 3}{k + 1} \]
Rewriting in the required form:
\[ u_{k+1} = \frac{2(k+1) + 1}{k + 1} \]

Step 4: Conclusion
Therefore, if the result is true for \(n = k\), it is also true for \(n = k + 1\). Since the result has been shown to be true for \(n = 1\), by mathematical induction it is true for all positive integers \(n\).

B1: Verifies the base case \(n = 1\) correctly by showing \(\frac{2(1)+1}{1} = 3 = u_1\).

M1: States the inductive hypothesis: assumes \(u_k = \frac{2k+1}{k}\) for some positive integer \(k\).

M1: Substitutes the expression for \(u_k\) into the given recurrence relation \(u_{k+1} = \frac{3u_k - 4}{u_k - 1}\).

A1: Correctly simplifies the algebraic fraction to \(\frac{2k+3}{k+1}\).

A1: Expresses the result explicitly in the target form \(\frac{2(k+1)+1}{k+1}\).

A1: (Dependent on all previous marks) Gives a complete, logically coherent conclusion stating that if true for \(n = k\), then true for \(n = k + 1\), and since true for \(n = 1\), the statement is true for all positive integers \(n\) by mathematical induction.

(a) Let a line through the origin have equation \(y = mx\).
Under the transformation \(T\), a general point \(\begin{pmatrix} x \\ mx \end{pmatrix}\) is mapped to \(\begin{pmatrix} X \\ Y \end{pmatrix}\):
\[\begin{pmatrix} X \\ Y \end{pmatrix} = \begin{pmatrix} 1 & 4 \\ 2 & 3 \end{pmatrix} \begin{pmatrix} x \\ mx \end{pmatrix} = \begin{pmatrix} x + 4mx \\ 2x + 3mx \end{pmatrix} = x \begin{pmatrix} 1 + 4m \\ 2 + 3m \end{pmatrix}\]
Since the line is invariant, the image point \((X, Y)\) must also satisfy \(Y = mX\):
\[2 + 3m = m(1 + 4m)\]
\[2 + 3m = m + 4m^2\]
\[4m^2 - 2m - 2 = 0\]
\[2m^2 - m - 1 = 0\]
\[(2m + 1)(m - 1) = 0\]
Thus, \(m = 1\) or \(m = -\frac{1}{2}\).
Checking the vertical line \(x = 0\):
\[\begin{pmatrix} 1 & 4 \\ 2 & 3 \end{pmatrix} \begin{pmatrix} 0 \\ y \end{pmatrix} = \begin{pmatrix} 4y \\ 3y \end{pmatrix}\]
which is not on \(x = 0\) for \(y \neq 0\), so \(x = 0\) is not invariant.
Therefore, the invariant lines through the origin are \(y = x\) and \(y = -\frac{1}{2}x\) (or \(x + 2y = 0\)).

(b) The determinant of \(\mathbf{M}\) is:
\[\det(\mathbf{M}) = (1)(3) - (4)(2) = 3 - 8 = -5\]
The area scale factor of the transformation is \(|\det(\mathbf{M})| = |-5| = 5\).
Therefore, the area of the image triangle is:
\[\text{Area} = 5 \times 6 = 30\text{ cm}^2\]

(c) The matrix representing the composite transformation "\(T\) followed by \(S\)" is \(\mathbf{NM}\):
\[\mathbf{NM} = \begin{pmatrix} 2 & -1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 4 \\ 2 & 3 \end{pmatrix} = \begin{pmatrix} 2(1) + (-1)(2) & 2(4) + (-1)(3) \\ 1(1) + 1(2) & 1(4) + 1(3) \end{pmatrix} = \begin{pmatrix} 0 & 5 \\ 3 & 7 \end{pmatrix}\]
We are given:
\[\begin{pmatrix} 0 & 5 \\ 3 & 7 \end{pmatrix} \begin{pmatrix} p \\ q \end{pmatrix} = \begin{pmatrix} 5 \\ 16 \end{pmatrix}\]
From the first row:
\[5q = 5 \implies q = 1\]
From the second row:
\[3p + 7q = 16 \implies 3p + 7(1) = 16 \implies 3p = 9 \implies p = 3\]
Thus, \(p = 3\) and \(q = 1\).

(a)
M1: Sets up the matrix equation for points on \(y = mx\) and substitutes \(Y = mX\) into \(\mathbf{M}\binom{x}{mx} = \binom{X}{Y}\).
A1: Obtains a correct 3-term quadratic equation in \(m\), e.g. \(2m^2 - m - 1 = 0\) or \(4m^2 - 2m - 2 = 0\).
M1: Solves their 3-term quadratic equation in \(m\).
A1: Obtains both correct equations \(y = x\) and \(y = -\frac{1}{2}x\) (or \(x + 2y = 0\)).

(b)
M1: Calculates \(\det(\mathbf{M}) = -5\) and applies the area scale factor \(|\det(\mathbf{M})|\) to multiply by 6.
A1: Obtains correct area \(30\) (or \(30\text{ cm}^2\)).

(c)
M1: Multiplies matrices in correct order \(\mathbf{NM}\) or uses \(\mathbf{M}\binom{p}{q} = \mathbf{N}^{-1}\binom{5}{16}\).
A1: Obtains correct composite matrix \(\begin{pmatrix} 0 & 5 \\ 3 & 7 \end{pmatrix}\) or sets up correct linear system \(5q = 5\) and \(3p + 7q = 16\).
A1: Correct values \(p = 3\) and \(q = 1\).

(a)
From the relations between the roots and coefficients of \(x^3 - 3x^2 + x - 2 = 0\):
\[\alpha + \beta + \gamma = -\frac{-3}{1} = 3\]
\[\alpha\beta + \beta\gamma + \gamma\alpha = \frac{1}{1} = 1\]
\[\alpha\beta\gamma = -\frac{-2}{1} = 2\]

(b)
Using the identity for the sum of squares:
\[\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\]
\[\alpha^2 + \beta^2 + \gamma^2 = 3^2 - 2(1) = 9 - 2 = 7\]

Since each root satisfies the original equation:
\[\alpha^3 - 3\alpha^2 + \alpha - 2 = 0\]
\[\beta^3 - 3\beta^2 + \beta - 2 = 0\]
\[\gamma^3 - 3\gamma^2 + \gamma - 2 = 0\]
Summing these three equations gives:
\[(\alpha^3 + \beta^3 + \gamma^3) - 3(\alpha^2 + \beta^2 + \gamma^2) + (\alpha + \beta + \gamma) - 3(2) = 0\]
\[\alpha^3 + \beta^3 + \gamma^3 = 3(7) - 3 + 6 = 21 - 3 + 6 = 24\]

(c)
Let \(y = x^2\), so \(x = \sqrt{y}\).
Rearranging the original cubic equation:
\[x^3 + x = 3x^2 + 2\]
\[x(x^2 + 1) = 3x^2 + 2\]
Squaring both sides:
\[x^2(x^2 + 1)^2 = (3x^2 + 2)^2\]
Substituting \(y = x^2\):
\[y(y + 1)^2 = (3y + 2)^2\]
\[y(y^2 + 2y + 1) = 9y^2 + 12y + 4\]
\[y^3 + 2y^2 + y = 9y^2 + 12y + 4\]
\[y^3 - 7y^2 - 11y - 4 = 0\]

(d)(i)
The roots of \(y^3 - 7y^2 - 11y - 4 = 0\) are \(y_1 = \alpha^2, y_2 = \beta^2, y_3 = \gamma^2\).
From the coefficients:
\[y_1 y_2 + y_2 y_3 + y_3 y_1 = \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 = -11\]
\[y_1 y_2 y_3 = \alpha^2\beta^2\gamma^2 = 4\]
Therefore,
\[\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} = \frac{\alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2}{\alpha^2\beta^2\gamma^2} = -\frac{11}{4}\]

(d)(ii)
\[\alpha^4 + \beta^4 + \gamma^4 = y_1^2 + y_2^2 + y_3^2\]
\[y_1^2 + y_2^2 + y_3^2 = (y_1 + y_2 + y_3)^2 - 2(y_1 y_2 + y_2 y_3 + y_3 y_1)\]
\[\alpha^4 + \beta^4 + \gamma^4 = 7^2 - 2(-11) = 49 + 22 = 71\]

(a) B1: States all three correct values (\(\sum \alpha = 3\), \(\sum \alpha\beta = 1\), \(\alpha\beta\gamma = 2\)).

(b)
M1: Applies \(\sum \alpha^2 = (\sum \alpha)^2 - 2\sum \alpha\beta\).
A1: Obtains \(\alpha^2 + \beta^2 + \gamma^2 = 7\).
B1: Shows \(\sum \alpha^3 = 24\) with clear, complete algebraic steps (e.g. using the recurrence relation \(\sum \alpha^3 - 3\sum \alpha^2 + \sum \alpha - 6 = 0\)).

(c)
M1: Rearranges equation to isolate odd powers and squares both sides, or uses an equivalent algebraic substitution method.
M1: Expands both sides correctly in terms of \(y\).
A1: Obtains \(y^3 - 7y^2 - 11y - 4 = 0\) (allow any single variable, must be simplified with integer coefficients).

(d)(i)
B1: Obtains \(-\dfrac{11}{4}\) or \(-2.75\).

(d)(ii)
M1: Uses \((\sum \alpha^2)^2 - 2\sum \alpha^2\beta^2\) or the recurrence relation \(\sum \alpha^4 = 3\sum \alpha^3 - \sum \alpha^2 + 2\sum \alpha\).
A1: Obtains 71.

(a)
Direction vector of \( l_1 \): \( \mathbf{d}_1 = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix} \)
Direction vector of \( l_2 \): \( \mathbf{d}_2 = \begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix} \)

A vector perpendicular to both lines is:
\[ \mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 2 \\ 1 & 2 & -2 \end{vmatrix} = \mathbf{i}(2 - 4) - \mathbf{j}(-4 - 2) + \mathbf{k}(4 - (-1)) = \begin{pmatrix} -2 \\ 6 \\ 5 \end{pmatrix} \]

Magnitude of \( \mathbf{n} \):
\[ |\mathbf{n}| = \sqrt{(-2)^2 + 6^2 + 5^2} = \sqrt{4 + 36 + 25} = \sqrt{65} \]

Vector joining a point on \( l_1 \), \( A(1, 2, -1) \), to a point on \( l_2 \), \( B(3, -1, 4) \):
\[ \vec{AB} = \begin{pmatrix} 3 - 1 \\ -1 - 2 \\ 4 - (-1) \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix} \]

Shortest distance \( d \):
\[ d = \frac{|\vec{AB} \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{|2(-2) + (-3)(6) + 5(5)|}{\sqrt{65}} = \frac{|-4 - 18 + 25|}{\sqrt{65}} = \frac{3}{\sqrt{65}} = \frac{3\sqrt{65}}{65} \approx 0.372 \]

(b)
The normal to plane \( \Pi_1 \) is parallel to \( \mathbf{n} = \begin{pmatrix} 2 \\ -6 \\ -5 \end{pmatrix} \).
Since \( \Pi_1 \) contains \( l_1 \), it contains the point \( (1, 2, -1) \):
\[ 2(1) - 6(2) - 5(-1) = 2 - 12 + 5 = -5 \]
Hence, the Cartesian equation of \( \Pi_1 \) is:
\[ 2x - 6y - 5z = -5 \quad \text{or} \quad -2x + 6y + 5z = 5 \]

(c)
Normal vector to \( \Pi_1 \): \( \mathbf{n}_1 = \begin{pmatrix} 2 \\ -6 \\ -5 \end{pmatrix} \), with \( |\mathbf{n}_1| = \sqrt{65} \).
Normal vector to \( \Pi_2 \): \( \mathbf{n}_2 = \begin{pmatrix} 3 \\ -2 \\ 1 \end{pmatrix} \), with \( |\mathbf{n}_2| = \sqrt{3^2 + (-2)^2 + 1^2} = \sqrt{14} \).

Let \( \theta \) be the acute angle between the planes:
\[ \cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|} = \frac{|2(3) + (-6)(-2) + (-5)(1)|}{\sqrt{65}\sqrt{14}} = \frac{|6 + 12 - 5|}{\sqrt{910}} = \frac{13}{\sqrt{910}} \]
\[ \theta = \arccos\left(\frac{13}{\sqrt{910}}\right) \approx \arccos(0.43095) \approx 64.5^\circ \]

(d)
The direction vector of the line of intersection is given by:
\[ \mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -6 & -5 \\ 3 & -2 & 1 \end{vmatrix} = \mathbf{i}(-6 - 10) - \mathbf{j}(2 - (-15)) + \mathbf{k}(-4 - (-18)) = \begin{pmatrix} -16 \\ -17 \\ 14 \end{pmatrix} \]
(or any non-zero scalar multiple, e.g., \( 16\mathbf{i} + 17\mathbf{j} - 14\mathbf{k} \)).

To find a point on the line of intersection, set \( x = 1 \):
From \( \Pi_1 \): \( 2(1) - 6y - 5z = -5 \implies 6y + 5z = 7 \)
From \( \Pi_2 \): \( 3(1) - 2y + z = 4 \implies -2y + z = 1 \implies z = 1 + 2y \)

Substitute \( z = 1 + 2y \) into \( 6y + 5z = 7 \):
\[ 6y + 5(1 + 2y) = 7 \implies 16y = 2 \implies y = \frac{1}{8} \]
Then \( z = 1 + 2\left(\frac{1}{8}\right) = \frac{5}{4} \).

A point on the line is \( \begin{pmatrix} 1 \\ \frac{1}{8} \\ \frac{5}{4} \end{pmatrix} \).

Thus, a vector equation of the line of intersection is:
\[ \mathbf{r} = \begin{pmatrix} 1 \\ \frac{1}{8} \\ \frac{5}{4} \end{pmatrix} + t\begin{pmatrix} 16 \\ 17 \\ -14 \end{pmatrix} \]

(a)
- M1: Calculates vector product of direction vectors \( \mathbf{d}_1 \times \mathbf{d}_2 \) (allow one sign error).
- A1: Obtains correct normal vector \( \pm(-2\mathbf{i} + 6\mathbf{j} + 5\mathbf{k}) \) and its magnitude \( \sqrt{65} \).
- M1: Applies correct formula for shortest distance \( \frac{|\vec{AB} \cdot \mathbf{n}|}{|\mathbf{n}|} \).
- A1: Obtains \( \frac{3}{\sqrt{65}} \) or \( \frac{3\sqrt{65}}{65} \) or \( 0.372 \) (3 s.f.).

(b)
- M1: Sets up equation of plane using normal vector from (a) and point \( (1, 2, -1) \).
- A1: Obtains \( 2x - 6y - 5z = -5 \) or \( -2x + 6y + 5z = 5 \) (or equivalent integer Cartesian equation).

(c)
- M1: Uses scalar product of normals divided by product of their magnitudes.
- A1: Obtains \( 64.5^\circ \) (allow \( 1.12 \) rad; must be 1 d.p. for degrees or 3 s.f. for radians).

(d)
- M1: Computes vector product of the normal vectors of \( \Pi_1 \) and \( \Pi_2 \) (or equivalent method to find line direction).
- A1: Obtains correct direction vector (e.g. \( 16\mathbf{i} + 17\mathbf{j} - 14\mathbf{k} \) or any non-zero multiple).
- A1: Finds a valid point lying on both planes and states a complete vector equation of the line \( \mathbf{r} = \mathbf{a} + t\mathbf{d} \).

(a)
Vertical asymptote occurs where the denominator is zero:
\[ x - 1 = 0 \implies x = 1. \]
To find the oblique asymptote, express \(y\) in quotient-remainder form by algebraic division:
\[ \frac{2x^2 - 3x + 3}{x - 1} = \frac{2x(x - 1) - (x - 1) + 2}{x - 1} = 2x - 1 + \frac{2}{x - 1}. \]
As \(x \to \pm\infty\), \(\frac{2}{x - 1} \to 0\), so the oblique asymptote is:
\[ y = 2x - 1. \]

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

Stationary points occur at the boundary values of \(y\):
- For \(y = -3\):
\[ 2x^2 - (-3 + 3)x + (-3 + 3) = 0 \implies 2x^2 = 0 \implies x = 0. \]
Thus, \((0, -3)\) is a stationary point (local maximum).

- For \(y = 5\):
\[ 2x^2 - (5 + 3)x + (5 + 3) = 0 \implies 2x^2 - 8x + 8 = 0 \implies 2(x - 2)^2 = 0 \implies x = 2. \]
Thus, \((2, 5)\) is a stationary point (local minimum).

(c)
- Vertical asymptote: \(x = 1\).
- Oblique asymptote: \(y = 2x - 1\) (intercepts at \((0, -1)\) and \((0.5, 0)\)).
- Left branch: defined for \(x < 1\), passes through the maximum point and \(y\)-intercept at \((0, -3)\), approaching \(y = 2x - 1\) as \(x \to -\infty\) and \(y \to -\infty\) as \(x \to 1^-\).
- Right branch: defined for \(x > 1\), has a minimum turning point at \((2, 5)\), approaching \(y \to +\infty\) as \(x \to 1^+\) and \(y = 2x - 1\) as \(x \to +\infty\).
- No \(x\)-intercepts since \(2x^2 - 3x + 3 = 0\) has discriminant \(9 - 24 = -15 < 0\).

(d)
For \(y^2 = f(x)\), real values of \(y\) exist only where \(f(x) \ge 0\).
From part (b), \(f(x) \ge 5\) for \(x > 1\) and \(f(x) \le -3\) for \(x < 1\). Therefore, the graph exists only for \(x > 1\).
- The curve is symmetrical about the \(x\)-axis: \(y = \pm\sqrt{f(x)}\).
- Vertical asymptote: \(x = 1\) (as \(x \to 1^+\), \(y \to \pm\infty\)).
- Stationary points: \((2, \sqrt{5})\) and \((2, -\sqrt{5})\).
- For large positive \(x\), \(y \approx \pm\sqrt{2x}\).

(a)
- B1: State vertical asymptote \(x = 1\).
- M1: Carry out algebraic division or equate coefficients to express \(y\) in the form \(Ax + B + \frac{C}{x-1}\).
- A1: State oblique asymptote \(y = 2x - 1\).

(b)
- M1: Multiply out and form a 3-term quadratic equation in \(x\): \(2x^2 - (y + 3)x + (y + 3) = 0\).
- M1: Set up condition for real roots using discriminant \(b^2 - 4ac \ge 0\).
- A1: Factorise correctly to obtain \((y + 3)(y - 5) \ge 0\) and conclude that \(y \le -3\) or \(y \ge 5\), so no points exist for \(-3 < y < 5\) (AG).
- M1: Substitute \(y = -3\) and \(y = 5\) into the quadratic equation to find the corresponding \(x\)-values (or differentiate and solve \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\)).
- A1: State both stationary points correctly: \((0, -3)\) and \((2, 5)\).

(c)
- B1: Correct shape of two separate branches approaching the asymptotes correctly.
- B1: Both asymptotes \(x = 1\) and \(y = 2x - 1\) clearly drawn as dashed lines and correctly positioned.
- B1: Stationary points \((0, -3)\) and \((2, 5)\) labeled and the \(y\)-intercept \((0, -3)\) clearly shown.

(d)
- B1: Curve sketched only for \(x > 1\) and symmetrical about the \(x\)-axis.
- B1: Vertical asymptote \(x = 1\) drawn and labeled with both branches tending to \(\pm\infty\) as \(x \to 1^+\).
- B1: Stationary points labeled at \((2, \sqrt{5})\) and \((2, -\sqrt{5})\).

(a) Equating the two expressions for \(r\) for \(r > 0\):
\[ a(1 + \sin\theta) = 3a\sin\theta \implies 2\sin\theta = 1 \implies \sin\theta = \frac{1}{2} \]
For \(0 \le \theta \le \pi\), \(\theta = \frac{\pi}{6}\) or \(\theta = \frac{5\pi}{6}\).
When \(\theta = \frac{\pi}{6}\) or \(\frac{5\pi}{6}\), \(r = a\left(1 + \frac{1}{2}\right) = \frac{3}{2}a\).
At the pole \(r = 0\):
For \(C_1\), \(r = 0\) when \(\theta = -\frac{\pi}{2}\).
For \(C_2\), \(r = 0\) when \(\theta = 0\) and \(\theta = \pi\).
Thus the pole is also a common point.
The points of intersection are \(\left(\frac{3}{2}a, \frac{\pi}{6}\right)\), \(\left(\frac{3}{2}a, \frac{5\pi}{6}\right)\), and the pole \((0, \theta)\).

(b) \(C_1\) is a cardioid with a cusp at the pole pointing downwards along \(\theta = -\frac{\pi}{2}\), passing through \((a, 0)\), \(\left(2a, \frac{\pi}{2}\right)\), and \((a, \pi)\).
\(C_2\) is a circle of diameter \(3a\) lying entirely in the upper half-plane \(0 \le \theta \le \pi\), with diameter on the line \(\theta = \frac{\pi}{2}\) passing through the pole and \(\left(3a, \frac{\pi}{2}\right)\).
The two curves intersect at \(\left(\frac{3}{2}a, \frac{\pi}{6}\right)\), \(\left(\frac{3}{2}a, \frac{5\pi}{6}\right)\), and at the pole.

(c) For a tangent parallel to the initial line, \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = 0\) with \(\frac{\mathrm{d}x}{\mathrm{d}\theta} \ne 0\).
\[ y = r\sin\theta = a(1 + \sin\theta)\sin\theta = a(\sin\theta + \sin^2\theta) \]
\[ \frac{\mathrm{d}y}{\mathrm{d}\theta} = a(\cos\theta + 2\sin\theta\cos\theta) = a\cos\theta(1 + 2\sin\theta) \]
Setting \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = 0\):
- \(\cos\theta = 0 \implies \theta = \frac{\pi}{2}\) or \(\theta = -\frac{\pi}{2}\).
- \(1 + 2\sin\theta = 0 \implies \sin\theta = -\frac{1}{2} \implies \theta = -\frac{\pi}{6}\) or \(\theta = -\frac{5\pi}{6}\).

Now check \(x = r\cos\theta = a(1+\sin\theta)\cos\theta\):
\[ \frac{\mathrm{d}x}{\mathrm{d}\theta} = a(-\sin\theta + \cos 2\theta) = -a(2\sin\theta - 1)(\sin\theta + 1) \]
- At \(\theta = \frac{\pi}{2}\): \(\frac{\mathrm{d}x}{\mathrm{d}\theta} = -2a \ne 0\), \(r = 2a\). Polar coordinate is \(\left(2a, \frac{\pi}{2}\right)\).
- At \(\theta = -\frac{\pi}{2}\): \(\frac{\mathrm{d}x}{\mathrm{d}\theta} = 0\), which corresponds to the cusp at the pole where the tangent is perpendicular to the initial line.
- At \(\theta = -\frac{\pi}{6}\): \(\frac{\mathrm{d}x}{\mathrm{d}\theta} = a \ne 0\), \(r = a\left(1 - \frac{1}{2}\right) = \frac{1}{2}a\). Polar coordinate is \(\left(\frac{1}{2}a, -\frac{\pi}{6}\right)\).
- At \(\theta = -\frac{5\pi}{6}\): \(\frac{\mathrm{d}x}{\mathrm{d}\theta} = a \ne 0\), \(r = a\left(1 - \frac{1}{2}\right) = \frac{1}{2}a\). Polar coordinate is \(\left(\frac{1}{2}a, -\frac{5\pi}{6}\right)\).

Thus, the points are \(\left(2a, \frac{\pi}{2}\right)\), \(\left(\frac{1}{2}a, -\frac{\pi}{6}\right)\), and \(\left(\frac{1}{2}a, -\frac{5\pi}{6}\right)\).

(d) Total area enclosed by \(C_1\):
\[ A_1 = \frac{1}{2} \int_{-\pi}^{\pi} r_1^2\,\mathrm{d}\theta = \frac{1}{2}a^2 \int_{-\pi}^{\pi} (1 + 2\sin\theta + \sin^2\theta)\,\mathrm{d}\theta \]
\[ = \frac{1}{2}a^2 \int_{-\pi}^{\pi} \left( \frac{3}{2} + 2\sin\theta - \frac{1}{2}\cos 2\theta \right)\,\mathrm{d}\theta = \frac{1}{2}a^2 \left[ \frac{3}{2}\theta - 2\cos\theta - \frac{1}{4}\sin 2\theta \right]_{-\pi}^{\pi} = \frac{1}{2}a^2 (3\pi) = \frac{3}{2}\pi a^2 \]

The region enclosed by both \(C_1\) and \(C_2\) consists of \(C_2\) for \(\theta \in \left[0, \frac{\pi}{6}\right]\) and \(\theta \in \left[\frac{5\pi}{6}, \pi\right]\), and \(C_1\) for \(\theta \in \left[\frac{\pi}{6}, \frac{5\pi}{6}\right]\).
By symmetry:
\[ A_{\text{both}} = 2 \times \frac{1}{2} \int_0^{\frac{\pi}{6}} (3a\sin\theta)^2\,\mathrm{d}\theta + 2 \times \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} a^2(1 + \sin\theta)^2\,\mathrm{d}\theta \]
\[ \int_0^{\frac{\pi}{6}} 9a^2\sin^2\theta\,\mathrm{d}\theta = \frac{9a^2}{2} \left[ \theta - \frac{1}{2}\sin 2\theta \right]_0^{\frac{\pi}{6}} = \frac{9a^2}{2}\left( \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right) = a^2\left(\frac{3\pi}{4} - \frac{9\sqrt{3}}{8}\right) \]
\[ a^2 \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \left( \frac{3}{2} + 2\sin\theta - \frac{1}{2}\cos 2\theta \right)\,\mathrm{d}\theta = a^2 \left[ \frac{3}{2}\theta - 2\cos\theta - \frac{1}{4}\sin 2\theta \right]_{\frac{\pi}{6}}^{\frac{\pi}{2}} \]
\[ = a^2 \left( \frac{3\pi}{4} - \left(\frac{\pi}{4} - \sqrt{3} - \frac{\sqrt{3}}{8}\right) \right) = a^2 \left( \frac{\pi}{2} + \frac{9\sqrt{3}}{8} \right) \]
\[ A_{\text{both}} = a^2\left(\frac{3\pi}{4} - \frac{9\sqrt{3}}{8} + \frac{\pi}{2} + \frac{9\sqrt{3}}{8}\right) = \frac{5}{4}\pi a^2 \]

Therefore, the area inside \(C_1\) but outside \(C_2\) is:
\[ A = A_1 - A_{\text{both}} = \frac{3}{2}\pi a^2 - \frac{5}{4}\pi a^2 = \frac{1}{4}\pi a^2 \]

(a)
M1: Set \(a(1 + \sin\theta) = 3a\sin\theta\) and solve for \(\sin\theta\).
A1: Obtain non-pole intersection points \(\left(\frac{3}{2}a, \frac{\pi}{6}\right)\) and \(\left(\frac{3}{2}a, \frac{5\pi}{6}\right)\).
B1: State the pole as an intersection point (or \((0, \theta)\)).

(b)
B1: Cardioid \(C_1\) sketched with correct orientation, cusp at the pole pointing in negative vertical direction, symmetric about \(\theta = \frac{\pi}{2}\).
B1: Circle \(C_2\) sketched entirely in upper half-plane with diameter on \(\theta = \frac{\pi}{2}\) passing through the pole.
B1: Clear intersections in the 1st and 2nd quadrants and at the pole, correctly positioned relative to each other.

(c)
M1: Use \(y = r\sin\theta = a(\sin\theta + \sin^2\theta)\) and differentiate with respect to \(\theta\).
A1: Obtain \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = a\cos\theta(1 + 2\sin\theta)\) and equate to 0.
M1: Solve for angles in the domain: \(\theta = \frac{\pi}{2}\), \(\theta = -\frac{\pi}{6}\), \(\theta = -\frac{5\pi}{6}\) (and discard/justify \(\theta = -\frac{\pi}{2}\)).
A1: Calculate correct \(r\) values: \(r = 2a\) and \(r = \frac{1}{2}a\).
A1: Give all three points in correct polar form \((r, \theta)\).

(d)
M1: Use formula \(\frac{1}{2}\int r^2\,\mathrm{d}\theta\) for sector area on either curve.
M1: Express \(\sin^2\theta\) as \(\frac{1}{2}(1 - \cos 2\theta)\) to integrate.
A1: Correct total area of \(C_1 = \frac{3}{2}\pi a^2\).
M1: Valid method to find overlapping area (or integrate \(r_1^2 - r_2^2\) for outer region directly).
A1: Correct integration and evaluation of trigonometric limits leading to \(\frac{5}{4}\pi a^2\) for common area (or directly \(\frac{1}{4}\pi a^2\)).
A1: Final exact answer \(\frac{1}{4}\pi a^2\) correctly established.

(a) Equating the two expressions for \(r\) for \(r > 0\):
\[ a(1 + \sin\theta) = 3a\sin\theta \implies 2\sin\theta = 1 \implies \sin\theta = \frac{1}{2} \]
For \(0 \le \theta \le \pi\), \(\theta = \frac{\pi}{6}\) or \(\theta = \frac{5\pi}{6}\).
When \(\theta = \frac{\pi}{6}\) or \(\frac{5\pi}{6}\), \(r = a\left(1 + \frac{1}{2}\right) = \frac{3}{2}a\).
At the pole \(r = 0\):
For \(C_1\), \(r = 0\) when \(\theta = -\frac{\pi}{2}\).
For \(C_2\), \(r = 0\) when \(\theta = 0\) and \(\theta = \pi\).
Thus the pole is also a common point.
The points of intersection are \(\left(\frac{3}{2}a, \frac{\pi}{6}\right)\), \(\left(\frac{3}{2}a, \frac{5\pi}{6}\right)\), and the pole \((0, \theta)\).

(b) \(C_1\) is a cardioid with a cusp at the pole pointing downwards along \(\theta = -\frac{\pi}{2}\), passing through \((a, 0)\), \(\left(2a, \frac{\pi}{2}\right)\), and \((a, \pi)\).
\(C_2\) is a circle of diameter \(3a\) lying entirely in the upper half-plane \(0 \le \theta \le \pi\), with diameter on the line \(\theta = \frac{\pi}{2}\) passing through the pole and \(\left(3a, \frac{\pi}{2}\right)\).
The two curves intersect at \(\left(\frac{3}{2}a, \frac{\pi}{6}\right)\), \(\left(\frac{3}{2}a, \frac{5\pi}{6}\right)\), and at the pole.

(c) For a tangent parallel to the initial line, \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = 0\) with \(\frac{\mathrm{d}x}{\mathrm{d}\theta} \ne 0\).
\[ y = r\sin\theta = a(1 + \sin\theta)\sin\theta = a(\sin\theta + \sin^2\theta) \]
\[ \frac{\mathrm{d}y}{\mathrm{d}\theta} = a(\cos\theta + 2\sin\theta\cos\theta) = a\cos\theta(1 + 2\sin\theta) \]
Setting \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = 0\):
- \(\cos\theta = 0 \implies \theta = \frac{\pi}{2}\) or \(\theta = -\frac{\pi}{2}\).
- \(1 + 2\sin\theta = 0 \implies \sin\theta = -\frac{1}{2} \implies \theta = -\frac{\pi}{6}\) or \(\theta = -\frac{5\pi}{6}\).

Now check \(x = r\cos\theta = a(1+\sin\theta)\cos\theta\):
\[ \frac{\mathrm{d}x}{\mathrm{d}\theta} = a(-\sin\theta + \cos 2\theta) = -a(2\sin\theta - 1)(\sin\theta + 1) \]
- At \(\theta = \frac{\pi}{2}\): \(\frac{\mathrm{d}x}{\mathrm{d}\theta} = -2a \ne 0\), \(r = 2a\). Polar coordinate is \(\left(2a, \frac{\pi}{2}\right)\).
- At \(\theta = -\frac{\pi}{2}\): \(\frac{\mathrm{d}x}{\mathrm{d}\theta} = 0\), which corresponds to the cusp at the pole where the tangent is perpendicular to the initial line.
- At \(\theta = -\frac{\pi}{6}\): \(\frac{\mathrm{d}x}{\mathrm{d}\theta} = a \ne 0\), \(r = a\left(1 - \frac{1}{2}\right) = \frac{1}{2}a\). Polar coordinate is \(\left(\frac{1}{2}a, -\frac{\pi}{6}\right)\).
- At \(\theta = -\frac{5\pi}{6}\): \(\frac{\mathrm{d}x}{\mathrm{d}\theta} = a \ne 0\), \(r = a\left(1 - \frac{1}{2}\right) = \frac{1}{2}a\). Polar coordinate is \(\left(\frac{1}{2}a, -\frac{5\pi}{6}\right)\).

Thus, the points are \(\left(2a, \frac{\pi}{2}\right)\), \(\left(\frac{1}{2}a, -\frac{\pi}{6}\right)\), and \(\left(\frac{1}{2}a, -\frac{5\pi}{6}\right)\).

(d) Total area enclosed by \(C_1\):
\[ A_1 = \frac{1}{2} \int_{-\pi}^{\pi} r_1^2\,\mathrm{d}\theta = \frac{1}{2}a^2 \int_{-\pi}^{\pi} (1 + 2\sin\theta + \sin^2\theta)\,\mathrm{d}\theta \]
\[ = \frac{1}{2}a^2 \int_{-\pi}^{\pi} \left( \frac{3}{2} + 2\sin\theta - \frac{1}{2}\cos 2\theta \right)\,\mathrm{d}\theta = \frac{1}{2}a^2 \left[ \frac{3}{2}\theta - 2\cos\theta - \frac{1}{4}\sin 2\theta \right]_{-\pi}^{\pi} = \frac{1}{2}a^2 (3\pi) = \frac{3}{2}\pi a^2 \]

The region enclosed by both \(C_1\) and \(C_2\) consists of \(C_2\) for \(\theta \in \left[0, \frac{\pi}{6}\right]\) and \(\theta \in \left[\frac{5\pi}{6}, \pi\right]\), and \(C_1\) for \(\theta \in \left[\frac{\pi}{6}, \frac{5\pi}{6}\right]\).
By symmetry:
\[ A_{\text{both}} = 2 \times \frac{1}{2} \int_0^{\frac{\pi}{6}} (3a\sin\theta)^2\,\mathrm{d}\theta + 2 \times \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} a^2(1 + \sin\theta)^2\,\mathrm{d}\theta \]
\[ \int_0^{\frac{\pi}{6}} 9a^2\sin^2\theta\,\mathrm{d}\theta = \frac{9a^2}{2} \left[ \theta - \frac{1}{2}\sin 2\theta \right]_0^{\frac{\pi}{6}} = \frac{9a^2}{2}\left( \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right) = a^2\left(\frac{3\pi}{4} - \frac{9\sqrt{3}}{8}\right) \]
\[ a^2 \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \left( \frac{3}{2} + 2\sin\theta - \frac{1}{2}\cos 2\theta \right)\,\mathrm{d}\theta = a^2 \left[ \frac{3}{2}\theta - 2\cos\theta - \frac{1}{4}\sin 2\theta \right]_{\frac{\pi}{6}}^{\frac{\pi}{2}} \]
\[ = a^2 \left( \frac{3\pi}{4} - \left(\frac{\pi}{4} - \sqrt{3} - \frac{\sqrt{3}}{8}\right) \right) = a^2 \left( \frac{\pi}{2} + \frac{9\sqrt{3}}{8} \right) \]
\[ A_{\text{both}} = a^2\left(\frac{3\pi}{4} - \frac{9\sqrt{3}}{8} + \frac{\pi}{2} + \frac{9\sqrt{3}}{8}\right) = \frac{5}{4}\pi a^2 \]

Therefore, the area inside \(C_1\) but outside \(C_2\) is:
\[ A = A_1 - A_{\text{both}} = \frac{3}{2}\pi a^2 - \frac{5}{4}\pi a^2 = \frac{1}{4}\pi a^2 \]

(a)
M1: Set \(a(1 + \sin\theta) = 3a\sin\theta\) and solve for \(\sin\theta\).
A1: Obtain non-pole intersection points \(\left(\frac{3}{2}a, \frac{\pi}{6}\right)\) and \(\left(\frac{3}{2}a, \frac{5\pi}{6}\right)\).
B1: State the pole as an intersection point (or \((0, \theta)\)).

(b)
B1: Cardioid \(C_1\) sketched with correct orientation, cusp at the pole pointing in negative vertical direction, symmetric about \(\theta = \frac{\pi}{2}\).
B1: Circle \(C_2\) sketched entirely in upper half-plane with diameter on \(\theta = \frac{\pi}{2}\) passing through the pole.
B1: Clear intersections in the 1st and 2nd quadrants and at the pole, correctly positioned relative to each other.

(c)
M1: Use \(y = r\sin\theta = a(\sin\theta + \sin^2\theta)\) and differentiate with respect to \(\theta\).
A1: Obtain \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = a\cos\theta(1 + 2\sin\theta)\) and equate to 0.
M1: Solve for angles in the domain: \(\theta = \frac{\pi}{2}\), \(\theta = -\frac{\pi}{6}\), \(\theta = -\frac{5\pi}{6}\) (and discard/justify \(\theta = -\frac{\pi}{2}\)).
A1: Calculate correct \(r\) values: \(r = 2a\) and \(r = \frac{1}{2}a\).
A1: Give all three points in correct polar form \((r, \theta)\).

(d)
M1: Use formula \(\frac{1}{2}\int r^2\,\mathrm{d}\theta\) for sector area on either curve.
M1: Express \(\sin^2\theta\) as \(\frac{1}{2}(1 - \cos 2\theta)\) to integrate.
A1: Correct total area of \(C_1 = \frac{3}{2}\pi a^2\).
M1: Valid method to find overlapping area (or integrate \(r_1^2 - r_2^2\) for outer region directly).
A1: Correct integration and evaluation of trigonometric limits leading to \(\frac{5}{4}\pi a^2\) for common area (or directly \(\frac{1}{4}\pi a^2\)).
A1: Final exact answer \(\frac{1}{4}\pi a^2\) correctly established.

Method 1: By successive differentiation

Let \( f(x) = \ln(\cosh 2x) \).

1. First derivative:
\[ f'(x) = \frac{2\sinh 2x}{\cosh 2x} = 2\tanh 2x \]
At \( x = 0 \):
\[ f(0) = \ln(\cosh 0) = \ln 1 = 0, \quad f'(0) = 2\tanh 0 = 0 \]

2. Second derivative:
\[ f''(x) = 4\operatorname{sech}^2 2x \]
At \( x = 0 \):
\[ f''(0) = 4\operatorname{sech}^2 0 = 4 \]

3. Third derivative:
\[ f'''(x) = 4 \cdot 2\operatorname{sech} 2x \cdot (-\operatorname{sech} 2x \tanh 2x \cdot 2) = -16\operatorname{sech}^2 2x \tanh 2x \]
At \( x = 0 \):
\[ f'''(0) = 0 \]

4. Fourth derivative:
\[ f^{(4)}(x) = -16 \left[ -4\operatorname{sech}^2 2x \tanh^2 2x + 2\operatorname{sech}^4 2x \right] = 64\operatorname{sech}^2 2x \tanh^2 2x - 32\operatorname{sech}^4 2x \]
At \( x = 0 \):
\[ f^{(4)}(0) = 0 - 32(1) = -32 \]

5. Applying Maclaurin's theorem:
\[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \dots \]
\[ f(x) = 0 + 0 + \frac{4}{2}x^2 + 0 - \frac{32}{24}x^4 + \dots = 2x^2 - \frac{4}{3}x^4 \]

---

Method 2: Using standard expansions

Using the standard series for \( \cosh u \):
\[ \cosh 2x = 1 + \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} + \dots = 1 + 2x^2 + \frac{2}{3}x^4 + \dots \]

Using the standard series \( \ln(1 + u) = u - \frac{1}{2}u^2 + \dots \) with \( u = 2x^2 + \frac{2}{3}x^4 \):
\[ \ln(\cosh 2x) = \left(2x^2 + \frac{2}{3}x^4\right) - \frac{1}{2}\left(2x^2 + \frac{2}{3}x^4\right)^2 + \dots \]
\[ = 2x^2 + \frac{2}{3}x^4 - \frac{1}{2}(4x^4) + \dots = 2x^2 - \frac{4}{3}x^4 \]

Method 1 (Differentiation):
* M1: For differentiating to find \( f'(x) \) and \( f''(x) \) using the chain rule.
* A1: For correctly evaluating \( f'(0) = 0 \) and \( f''(0) = 4 \).
* M1: For using the product/chain rule to find \( f'''(x) \) and \( f^{(4)}(x) \).
* A1: For correctly evaluating \( f'''(0) = 0 \) and \( f^{(4)}(0) = -32 \).
* A1: For the correct final expansion \( 2x^2 - \frac{4}{3}x^4 \).

Method 2 (Standard series):
* M1: For using the expansion of \( \cosh 2x = 1 + 2x^2 + \frac{2}{3}x^4 + \dots \)
* M1: For using the expansion of \( \ln(1 + u) = u - \frac{1}{2}u^2 + \dots \) with \( u = 2x^2 + \dots \)
* A1: For correct combined expression \( \left(2x^2 + \frac{2}{3}x^4\right) - \frac{1}{2}(2x^2)^2 \).
* A1: For correct coefficients before simplification.
* A1: For the final simplified result \( 2x^2 - \frac{4}{3}x^4 \).

(a)
Apply integration by parts with \( u = x^n \) and \( \frac{\mathrm{d}v}{\mathrm{d}x} = \cos x \):
\[ \frac{\mathrm{d}u}{\mathrm{d}x} = n x^{n-1}, \quad v = \sin x \]
\[ I_n = \left[ x^n \sin x \right]_0^{\frac{1}{2}\pi} - n \int_0^{\frac{1}{2}\pi} x^{n-1} \sin x \, \mathrm{d}x \]

Evaluating the limit term:
\[ \left[ x^n \sin x \right]_0^{\frac{1}{2}\pi} = \left(\frac{\pi}{2}\right)^n \sin\left(\frac{\pi}{2}\right) - 0 = \left(\frac{\pi}{2}\right)^n \]

So,
\[ I_n = \left(\frac{\pi}{2}\right)^n - n \int_0^{\frac{1}{2}\pi} x^{n-1} \sin x \, \mathrm{d}x \]

Apply integration by parts to the remaining integral with \( u = x^{n-1} \) and \( \frac{\mathrm{d}v}{\mathrm{d}x} = \sin x \):
\[ \frac{\mathrm{d}u}{\mathrm{d}x} = (n-1) x^{n-2}, \quad v = -\cos x \]
\[ \int_0^{\frac{1}{2}\pi} x^{n-1} \sin x \, \mathrm{d}x = \left[ -x^{n-1} \cos x \right]_0^{\frac{1}{2}\pi} - \int_0^{\frac{1}{2}\pi} (n-1) x^{n-2} (-\cos x) \, \mathrm{d}x \]

Evaluating the boundary term:
\[ \left[ -x^{n-1} \cos x \right]_0^{\frac{1}{2}\pi} = -\left(\frac{\pi}{2}\right)^{n-1}\cos\left(\frac{\pi}{2}\right) - 0 = 0 \]

Thus,
\[ \int_0^{\frac{1}{2}\pi} x^{n-1} \sin x \, \mathrm{d}x = (n-1) \int_0^{\frac{1}{2}\pi} x^{n-2} \cos x \, \mathrm{d}x = (n-1) I_{n-2} \]

Substituting this back into the expression for \( I_n \):
\[ I_n = \left(\frac{\pi}{2}\right)^n - n(n-1)I_{n-2} \]

(b)
First, calculate \( I_0 \):
\[ I_0 = \int_0^{\frac{1}{2}\pi} \cos x \, \mathrm{d}x = \left[ \sin x \right]_0^{\frac{1}{2}\pi} = 1 - 0 = 1 \]

Using the reduction formula with \( n = 2 \):
\[ I_2 = \left(\frac{\pi}{2}\right)^2 - 2(1)I_0 = \frac{\pi^2}{4} - 2(1) = \frac{\pi^2}{4} - 2 \]

Using the reduction formula with \( n = 4 \):
\[ I_4 = \left(\frac{\pi}{2}\right)^4 - 4(3)I_2 = \frac{\pi^4}{16} - 12\left(\frac{\pi^2}{4} - 2\right) \]
\[ I_4 = \frac{\pi^4}{16} - 3\pi^2 + 24 \]

Hence, \( a = \frac{1}{16} \), \( b = -3 \), \( c = 24 \).

(a)
M1: Attempt integration by parts on \( I_n \) with \( u = x^n \) and \( v' = \cos x \).
A1: Obtain \( \left(\frac{\pi}{2}\right)^n - n \int_0^{\frac{1}{2}\pi} x^{n-1} \sin x \, \mathrm{d}x \) with correct limits evaluated.
M1: Attempt integration by parts a second time on \( \int_0^{\frac{1}{2}\pi} x^{n-1} \sin x \, \mathrm{d}x \).
A1: Correct completion showing all intermediate steps leading convincingly to the given relation (AG).

(b)
B1: State or find \( I_0 = 1 \).
M1: Apply the reduction formula to find \( I_2 \) and subsequently \( I_4 \).
A1: Obtain \( \frac{1}{16}\pi^4 - 3\pi^2 + 24 \) (or clearly state \( a = \frac{1}{16}, b = -3, c = 24 \)).

(a) Let \(u = \cosh x\), then \(\mathrm{d}u = \sinh x \, \mathrm{d}x\).
Using the identity \(\sinh^2 x = \cosh^2 x - 1 = u^2 - 1\), the integral becomes:
\[ \int \frac{\sinh^3 x}{1 + \cosh^2 x} \, \mathrm{d}x = \int \frac{\sinh^2 x}{1 + \cosh^2 x} \sinh x \, \mathrm{d}x = \int \frac{u^2 - 1}{u^2 + 1} \, \mathrm{d}u \]
Rewriting the integrand:
\[ \frac{u^2 - 1}{u^2 + 1} = \frac{(u^2 + 1) - 2}{u^2 + 1} = 1 - \frac{2}{u^2 + 1} \]
Integrating with respect to \(u\):
\[ \int \left(1 - \frac{2}{u^2 + 1}\right) \mathrm{d}u = u - 2\arctan u + c \]
Substituting back \(u = \cosh x\):
\[ = \cosh x - 2\arctan(\cosh x) + c \]

(b) The differential equation is linear and in standard form:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} + P(x)y = Q(x), \quad \text{where } P(x) = \tanh x \]
The integrating factor \(I(x)\) is:
\[ I(x) = \mathrm{e}^{\int \tanh x \, \mathrm{d}x} = \mathrm{e}^{\ln(\cosh x)} = \cosh x \]
Multiplying both sides of the differential equation by \(\cosh x\):
\[ \cosh x \frac{\mathrm{d}y}{\mathrm{d}x} + y \sinh x = \frac{\sinh^3 x}{1 + \cosh^2 x} \]
\[ \frac{\mathrm{d}}{\mathrm{d}x}(y \cosh x) = \frac{\sinh^3 x}{1 + \cosh^2 x} \]
Integrating both sides with respect to \(x\), using the result from part (a):
\[ y \cosh x = \cosh x - 2\arctan(\cosh x) + C \]
Thus, the general solution is:
\[ y = 1 - \frac{2\arctan(\cosh x)}{\cosh x} + \frac{C}{\cosh x} \quad \text{or} \quad y = 1 - 2\operatorname{sech} x \arctan(\cosh x) + C\operatorname{sech} x \]

(c) Using the initial condition \(y = 1 - \frac{1}{2}\pi\) when \(x = 0\):
\[ \left(1 - \frac{1}{2}\pi\right)\cosh(0) = \cosh(0) - 2\arctan(\cosh 0) + C \]
Since \(\cosh(0) = 1\) and \(\arctan(1) = \frac{\pi}{4}\):
\[ 1 - \frac{1}{2}\pi = 1 - 2\left(\frac{\pi}{4}\right) + C \]
\[ 1 - \frac{1}{2}\pi = 1 - \frac{1}{2}\pi + C \implies C = 0 \]
Therefore, the particular solution is:
\[ y = 1 - \frac{2\arctan(\cosh x)}{\cosh x} \]
To find the limiting value as \(x \to \infty\):
As \(x \to \infty\), \(\cosh x \to \infty\), so \(\arctan(\cosh x) \to \frac{\pi}{2}\) and \(\frac{1}{\cosh x} \to 0\).
Hence:
\[ \lim_{x \to \infty} y = 1 - 2 \left(\frac{\pi}{2}\right) (0) = 1 \]

(a)
- M1: Uses the substitution \(u = \cosh x\), \(\mathrm{d}u = \sinh x \, \mathrm{d}x\), and identity \(\sinh^2 x = u^2 - 1\).
- A1: Obtains the correct integrand in terms of \(u\): \(\int \frac{u^2 - 1}{u^2 + 1} \, \mathrm{d}u\).
- M1: Decomposes the rational expression to \(1 - \frac{2}{u^2 + 1}\) and integrates to obtain \(u - 2\arctan u\).
- A1 (AG): Correctly reaches \(\cosh x - 2\arctan(\cosh x) + c\) with all steps clearly shown.

(b)
- M1: Finds the integrating factor \(\mathrm{e}^{\int \tanh x \, \mathrm{d}x} = \cosh x\).
- A1: Expresses the differential equation in the form \(\frac{\mathrm{d}}{\mathrm{d}x}(y \cosh x) = \frac{\sinh^3 x}{1 + \cosh^2 x}\).
- M1: Integrates both sides using part (a) and includes an arbitrary constant of integration \(C\).
- A1: Obtains the correct general solution, e.g., \(y \cosh x = \cosh x - 2\arctan(\cosh x) + C\) or explicit in \(y\).

(c)
- M1: Substitutes \(x = 0\) and \(y = 1 - \frac{1}{2}\pi\) into their general solution to evaluate \(C\).
- A1: Obtains \(C = 0\) and states the particular solution \(y = 1 - \frac{2\arctan(\cosh x)}{\cosh x}\) (or equivalent).
- B1: States the limiting value \(1\) with valid justification (e.g. observing \(\frac{\arctan(\cosh x)}{\cosh x} \to 0\) as \(x \to \infty\)).

(a)
Differentiating the equation \(y^2 \cosh x + x \sinh y = 4\) implicitly with respect to \(x\):
\[ \frac{\mathrm{d}}{\mathrm{d}x}\left(y^2 \cosh x\right) + \frac{\mathrm{d}}{\mathrm{d}x}\left(x \sinh y\right) = 0 \]
\[ 2y \frac{\mathrm{d}y}{\mathrm{d}x} \cosh x + y^2 \sinh x + \sinh y + x \cosh y \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \]

At the point \((0, 2)\), we have \(x = 0\), \(y = 2\), \(\cosh 0 = 1\), and \(\sinh 0 = 0\):
\[ 2(2)\frac{\mathrm{d}y}{\mathrm{d}x}(1) + 2^2(0) + \sinh 2 + 0 \cdot \cosh 2 \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \]
\[ 4\frac{\mathrm{d}y}{\mathrm{d}x} + \sinh 2 = 0 \]
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{1}{4}\sinh 2 \]

(b)
We differentiate the first derivative equation with respect to \(x\):
\[ 2y y' \cosh x + y^2 \sinh x + \sinh y + x y' \cosh y = 0 \]

Differentiating term by term:
- \(\frac{\mathrm{d}}{\mathrm{d}x}\left(2y y' \cosh x\right) = 2(y')^2 \cosh x + 2y y'' \cosh x + 2y y' \sinh x\)
- \(\frac{\mathrm{d}}{\mathrm{d}x}\left(y^2 \sinh x\right) = 2y y' \sinh x + y^2 \cosh x\)
- \(\frac{\mathrm{d}}{\mathrm{d}x}\left(\sinh y\right) = y' \cosh y\)
- \(\frac{\mathrm{d}}{\mathrm{d}x}\left(x y' \cosh y\right) = y' \cosh y + x y'' \cosh y + x (y')^2 \sinh y\)

Combining all terms gives:
\[ 2(y')^2 \cosh x + 2y y'' \cosh x + 4y y' \sinh x + y^2 \cosh x + 2y' \cosh y + x y'' \cosh y + x (y')^2 \sinh y = 0 \]

Substituting \(x = 0\), \(y = 2\), and \(y' = -\frac{1}{4}\sinh 2\):
\[ 2\left(-\frac{1}{4}\sinh 2\right)^2(1) + 2(2)y''(1) + 4(2)\left(-\frac{1}{4}\sinh 2\right)(0) + 2^2(1) + 2\left(-\frac{1}{4}\sinh 2\right)\cosh 2 + 0 + 0 = 0 \]
\[ 2\left(\frac{1}{16}\sinh^2 2\right) + 4y'' + 4 - \frac{1}{2}\sinh 2 \cosh 2 = 0 \]
\[ \frac{1}{8}\sinh^2 2 + 4y'' + 4 - \frac{1}{2}\sinh 2 \cosh 2 = 0 \]
\[ 4y'' = -4 + \frac{1}{2}\sinh 2 \cosh 2 - \frac{1}{8}\sinh^2 2 \]
\[ y'' = -1 + \frac{1}{8}\sinh 2 \cosh 2 - \frac{1}{32}\sinh^2 2 \]

*(Alternatively: \(y'' = -1 + \frac{1}{16}\sinh 4 - \frac{1}{32}\sinh^2 2\))*

(a)
- M1: Applies product rule and chain rule to differentiate both terms implicitly with respect to \(x\).
- A1: Obtains fully correct expression for the first derivative, e.g. \(2y y' \cosh x + y^2 \sinh x + \sinh y + x y' \cosh y = 0\).
- A1: Substitutes \(x = 0, y = 2\) correctly to obtain \(\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{1}{4}\sinh 2\).

(b)
- M1: Differentiates \(2y y' \cosh x\) using the product rule to produce an expression involving \(y''\) and \((y')^2\).
- M1: Differentiates the remaining terms of the first-order implicit equation with respect to \(x\).
- A1: Obtains a completely correct second-derivative equation before substitution.
- M1: Substitutes \(x = 0\), \(y = 2\), and their \(y' = -\frac{1}{4}\sinh 2\) into the second-derivative equation.
- A1: Obtains \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -1 + \frac{1}{8}\sinh 2 \cosh 2 - \frac{1}{32}\sinh^2 2\) or any exact equivalent form (e.g. \(-1 + \frac{1}{16}\sinh 4 - \frac{1}{32}\sinh^2 2\)).

(a) Note that \( f(x) = \frac{1}{(x+1)^2 + 1} \) is a strictly decreasing function for \( x \ge 0 \).
Consider intervals \( [r-1, r] \) for \( r = 1, 2, \dots, n \).
On each interval \( [r-1, r] \), the lower rectangle has width 1 and height \( f(r) = \frac{1}{r^2 + 2r + 2} \).
The total area of these \( n \) lower rectangles is strictly less than the area under the curve from \( x = 0 \) to \( x = n \):
\[ \sum_{r=1}^n f(r) < \int_0^n f(x) \, \mathrm{d}x \]
Evaluating the integral:
\[ \int_0^n \frac{1}{(x+1)^2 + 1} \, \mathrm{d}x = \left[ \arctan(x+1) \right]_0^n = \arctan(n+1) - \arctan(1) = \arctan(n+1) - \frac{1}{4}\pi \]
Hence,
\[ \sum_{r=1}^n \frac{1}{r^2 + 2r + 2} < \arctan(n+1) - \frac{1}{4}\pi \]

(b) Consider intervals \( [r, r+1] \) for \( r = 1, 2, \dots, n \).
On each interval \( [r, r+1] \), the upper rectangle has width 1 and height \( f(r) = \frac{1}{r^2 + 2r + 2} \).
The total area of these \( n \) upper rectangles is strictly greater than the area under the curve from \( x = 1 \) to \( x = n+1 \):
\[ \sum_{r=1}^n f(r) > \int_1^{n+1} f(x) \, \mathrm{d}x \]
Evaluating the integral:
\[ \int_1^{n+1} \frac{1}{(x+1)^2 + 1} \, \mathrm{d}x = \left[ \arctan(x+1) \right]_1^{n+1} = \arctan(n+2) - \arctan 2 \]
Hence,
\[ \sum_{r=1}^n \frac{1}{r^2 + 2r + 2} > \arctan(n+2) - \arctan 2 \]

(c) The sequence of partial sums \( S_n = \sum_{r=1}^n \frac{1}{r^2 + 2r + 2} \) is strictly increasing since each term is positive. As \( n \to \infty \), \( \arctan(n+1) < \frac{1}{2}\pi \), so \( S_n < \frac{1}{2}\pi - \frac{1}{4}\pi = \frac{1}{4}\pi \). Since \( S_n \) is increasing and bounded above, the infinite series converges to a sum \( S \).
Taking the limit as \( n \to \infty \) in the inequalities from (a) and (b):
\[ \lim_{n \to \infty} (\arctan(n+2) - \arctan 2) \le S \le \lim_{n \to \infty} \left( \arctan(n+1) - \frac{1}{4}\pi \right) \]
Since \( \lim_{n \to \infty} \arctan(n+1) = \lim_{n \to \infty} \arctan(n+2) = \frac{1}{2}\pi \), we have:
\[ \frac{1}{2}\pi - \arctan 2 \le S \le \frac{1}{4}\pi \]

(d) For \( r \ge 2 \), consider the lower rectangles on \( [r-1, r] \) from \( x = 1 \) to \( x = n \):
\[ \sum_{r=2}^n \frac{1}{r^2+2r+2} < \int_1^n \frac{1}{(x+1)^2+1} \, \mathrm{d}x = \left[ \arctan(x+1) \right]_1^n = \arctan(n+1) - \arctan 2 \]
For \( r = 1 \), \( f(1) = \frac{1}{1^2 + 2(1) + 2} = \frac{1}{5} \).
Therefore,
\[ S_n = \frac{1}{5} + \sum_{r=2}^n \frac{1}{r^2+2r+2} < \frac{1}{5} + \arctan(n+1) - \arctan 2 \]
Taking the limit as \( n \to \infty \):
\[ S \le \frac{1}{5} + \frac{1}{2}\pi - \arctan 2 \]
Thus \( a = \frac{1}{5} \) and \( b = 2 \).

(a)
- M1: For setting up the sum of areas of lower rectangles with unit width as \( \sum_{r=1}^n f(r) < \int_0^n f(x) \, \mathrm{d}x \) (or equivalent graphical justification).
- M1: For integrating \( \int \frac{1}{(x+1)^2+1} \, \mathrm{d}x = \arctan(x+1) \) and applying limits \( 0 \) to \( n \).
- A1: For fully correct working with \( \arctan 1 = \frac{1}{4}\pi \) leading to the given result (AG).

(b)
- M1: For setting up the sum of areas of upper rectangles with unit width as \( \sum_{r=1}^n f(r) > \int_1^{n+1} f(x) \, \mathrm{d}x \).
- M1: For applying limits \( 1 \) to \( n+1 \) to \( \arctan(x+1) \).
- A1: For fully correct working leading to the given result (AG).

(c)
- B1: For stating/justifying that the series converges (e.g. sequence of partial sums is increasing and bounded above).
- B1: For taking \( n \to \infty \) using \( \lim_{n \to \infty} \arctan(n+c) = \frac{1}{2}\pi \) to obtain the given inequality.

(d)
- M1: For establishing \( \sum_{r=2}^n f(r) < \int_1^n f(x) \, \mathrm{d}x = \arctan(n+1) - \arctan 2 \) and adding \( f(1) = \frac{1}{5} \).
- A1: For correctly deducing \( a = \frac{1}{5} \) and \( b = 2 \) (or stating \( S \le \frac{1}{5} + \frac{1}{2}\pi - \arctan 2 \)).

(a)

To find the complementary function, consider the auxiliary equation:
\[ m^2 + 4m + 13 = 0 \]
\[ (m + 2)^2 + 9 = 0 \implies m = -2 \pm 3\mathrm{i} \]

Hence, the complementary function is:
\[ y_{\text{CF}} = \mathrm{e}^{-2x}(A\cos 3x + B\sin 3x) \]

For the particular integral, try $y = Px^2 + Qx + R$.
Differentiating with respect to $x$:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = 2Px + Q \]
\[ \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = 2P \]

Substituting these into the differential equation:
\[ 2P + 4(2Px + Q) + 13(Px^2 + Qx + R) = 39x^2 - 2x + 11 \]
\[ 13P x^2 + (8P + 13Q)x + (2P + 4Q + 13R) = 39x^2 - 2x + 11 \]

Equating coefficients:
- For $x^2$: $13P = 39 \implies P = 3$
- For $x$: $8(3) + 13Q = -2 \implies 13Q = -26 \implies Q = -2$
- Constant term: $2(3) + 4(-2) + 13R = 11 \implies 6 - 8 + 13R = 11 \implies 13R = 13 \implies R = 1$

Thus, the particular integral is $y_{\text{PI}} = 3x^2 - 2x + 1$.

The general solution is the sum of the complementary function and the particular integral:
\[ y = \mathrm{e}^{-2x}(A\cos 3x + B\sin 3x) + 3x^2 - 2x + 1 \]

(b)

Apply the initial condition $y = 3$ when $x = 0$:
\[ 3 = \mathrm{e}^0 (A\cos 0 + B\sin 0) + 3(0)^2 - 2(0) + 1 \implies 3 = A + 1 \implies A = 2 \]

Differentiate the general solution with respect to $x$ using the product rule:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = -2\mathrm{e}^{-2x}(A\cos 3x + B\sin 3x) + \mathrm{e}^{-2x}(-3A\sin 3x + 3B\cos 3x) + 6x - 2 \]

Apply the condition $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$ when $x = 0$:
\[ 0 = -2(A) + 3B - 2 \]
Substitute $A = 2$:
\[ 0 = -2(2) + 3B - 2 \implies 3B - 6 = 0 \implies B = 2 \]

Therefore, the particular solution is:
\[ y = \mathrm{e}^{-2x}(2\cos 3x + 2\sin 3x) + 3x^2 - 2x + 1 \]

(a)
- M1: Sets up and solves the auxiliary equation $m^2 + 4m + 13 = 0$.
- A1: Correct roots $m = -2 \pm 3\mathrm{i}$.
- A1: Correct complementary function $y_{\text{CF}} = \mathrm{e}^{-2x}(A\cos 3x + B\sin 3x)$ (allow other equivalent forms or constants).
- M1: Substitutes $y = Px^2 + Qx + R$, with its first and second derivatives, into the differential equation.
- M1: Equates coefficients to find values of $P, Q, R$.
- A1: Correct values: $P = 3, Q = -2, R = 1$.
- A1: Correct general solution $y = \mathrm{e}^{-2x}(A\cos 3x + B\sin 3x) + 3x^2 - 2x + 1$ (must be written as $y = \dots$).

(b)
- M1: Uses $x=0, y=3$ in their general solution to evaluate $A$, and differentiates their general solution using the product rule.
- M1: Substitutes $x=0, y'=0$ and their value of $A$ to obtain an equation in $B$ and solves for $B$.
- A1: Correct particular solution: $y = \mathrm{e}^{-2x}(2\cos 3x + 2\sin 3x) + 3x^2 - 2x + 1$ (or $y = 2\mathrm{e}^{-2x}(\cos 3x + \sin 3x) + 3x^2 - 2x + 1$).

(a) (i) The roots of \(z^7 = 1 = \mathrm{e}^{\mathrm{i} 2k\pi}\) are
\[ z = \mathrm{e}^{\mathrm{i} \frac{2k\pi}{7}}, \quad k = 0, \pm 1, \pm 2, \pm 3 \]

(ii) We have \(z^7 - 1 = (z - 1)(z^6 + z^5 + z^4 + z^3 + z^2 + z + 1)\).
Since \(z = 1\) corresponds to \(k = 0\), the roots of \(z^6 + z^5 + z^4 + z^3 + z^2 + z + 1 = 0\) are the 6 non-real roots \(\mathrm{e}^{\pm \mathrm{i} \frac{2\pi}{7}}, \mathrm{e}^{\pm \mathrm{i} \frac{4\pi}{7}}, \mathrm{e}^{\pm \mathrm{i} \frac{6\pi}{7}}\).
Pairing each root with its complex conjugate gives quadratic factors:
\[ (z - \mathrm{e}^{\mathrm{i}\theta})(z - \mathrm{e}^{-\mathrm{i}\theta}) = z^2 - (\mathrm{e}^{\mathrm{i}\theta} + \mathrm{e}^{-\mathrm{i}\theta})z + 1 = z^2 - 2z\cos\theta + 1 \]
Setting \(\theta = \frac{2\pi}{7}, \frac{4\pi}{7}, \frac{6\pi}{7}\), the polynomial factorises over \(\mathbb{R}\) as:
\[ z^6 + z^5 + z^4 + z^3 + z^2 + z + 1 = \left(z^2 - 2z\cos\frac{2\pi}{7} + 1\right)\left(z^2 - 2z\cos\frac{4\pi}{7} + 1\right)\left(z^2 - 2z\cos\frac{6\pi}{7} + 1\right) \]

(b) (i) The coefficient of \(z^5\) on the right-hand side is obtained by choosing the \(-2z\cos\theta\) term from one quadratic factor and the \(z^2\) term from the other two factors:
\[ \text{Coefficient of } z^5 = -2\cos\frac{2\pi}{7} - 2\cos\frac{4\pi}{7} - 2\cos\frac{6\pi}{7} = -2\left(\cos\frac{2\pi}{7} + \cos\frac{4\pi}{7} + \cos\frac{6\pi}{7}\right) \]
Equating this to the coefficient of \(z^5\) on the left-hand side, which is \(1\):
\[ 1 = -2\left(\cos\frac{2\pi}{7} + \cos\frac{4\pi}{7} + \cos\frac{6\pi}{7}\right) \implies \cos\frac{2\pi}{7} + \cos\frac{4\pi}{7} + \cos\frac{6\pi}{7} = -\frac{1}{2} \]

(ii) Substituting \(z = 1\) into the identity from (a)(ii):
\[ \text{LHS} = 1 + 1 + 1 + 1 + 1 + 1 + 1 = 7 \]
\[ \text{RHS} = \left(2 - 2\cos\frac{2\pi}{7}\right)\left(2 - 2\cos\frac{4\pi}{7}\right)\left(2 - 2\cos\frac{6\pi}{7}\right) \]
Using the identity \(2 - 2\cos(2\alpha) = 4\sin^2\alpha\):
\[ \text{RHS} = \left(4\sin^2\frac{\pi}{7}\right)\left(4\sin^2\frac{2\pi}{7}\right)\left(4\sin^2\frac{3\pi}{7}\right) = 64\sin^2\frac{\pi}{7}\sin^2\frac{2\pi}{7}\sin^2\frac{3\pi}{7} \]
Thus:
\[ 64\sin^2\frac{\pi}{7}\sin^2\frac{2\pi}{7}\sin^2\frac{3\pi}{7} = 7 \implies \sin^2\frac{\pi}{7}\sin^2\frac{2\pi}{7}\sin^2\frac{3\pi}{7} = \frac{7}{64} \]
Since \(0 < \frac{\pi}{7} < \frac{2\pi}{7} < \frac{3\pi}{7} < \pi\), each sine value is strictly positive, hence:
\[ \sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7} = \frac{\sqrt{7}}{8} \]

(c) Substituting \(z = -1\) into the identity from (a)(ii):
\[ \text{LHS} = (-1)^6 + (-1)^5 + (-1)^4 + (-1)^3 + (-1)^2 + (-1) + 1 = 1 \]
\[ \text{RHS} = \left(2 + 2\cos\frac{2\pi}{7}\right)\left(2 + 2\cos\frac{4\pi}{7}\right)\left(2 + 2\cos\frac{6\pi}{7}\right) \]
Using the identity \(2 + 2\cos(2\alpha) = 4\cos^2\alpha\):
\[ \text{RHS} = \left(4\cos^2\frac{\pi}{7}\right)\left(4\cos^2\frac{2\pi}{7}\right)\left(4\cos^2\frac{3\pi}{7}\right) = 64\cos^2\frac{\pi}{7}\cos^2\frac{2\pi}{7}\cos^2\frac{3\pi}{7} \]
Thus:
\[ 64\cos^2\frac{\pi}{7}\cos^2\frac{2\pi}{7}\cos^2\frac{3\pi}{7} = 1 \implies \cos^2\frac{\pi}{7}\cos^2\frac{2\pi}{7}\cos^2\frac{3\pi}{7} = \frac{1}{64} \]
Since \(0 < \frac{\pi}{7} < \frac{2\pi}{7} < \frac{3\pi}{7} < \frac{\pi}{2}\), all three cosine values are positive, giving:
\[ \cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{3\pi}{7} = \frac{1}{8} \]
Therefore:
\[ \tan\frac{\pi}{7}\tan\frac{2\pi}{7}\tan\frac{3\pi}{7} = \frac{\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7}}{\cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{3\pi}{7}} = \frac{\frac{\sqrt{7}}{8}}{\frac{1}{8}} = \sqrt{7} \]

(a)(i)
- B1: Correct roots listed in exponential form: \(1, \mathrm{e}^{\pm \mathrm{i}\frac{2\pi}{7}}, \mathrm{e}^{\pm \mathrm{i}\frac{4\pi}{7}}, \mathrm{e}^{\pm \mathrm{i}\frac{6\pi}{7}}\) or \(\mathrm{e}^{\mathrm{i}\frac{2k\pi}{7}}\) for \(k=0, \pm 1, \pm 2, \pm 3\).

(a)(ii)
- M1: Identifies the roots of \(z^6 + \dots + 1 = 0\) as the 6 non-real roots and pairs conjugate roots \((z - \mathrm{e}^{\mathrm{i}\theta})(z - \mathrm{e}^{-\mathrm{i}\theta})\).
- M1: Expands the conjugate pair to show \(z^2 - 2z\cos\theta + 1\).
- A1 (AG): Fully correct deduction of the product form with clear reasoning.

(b)(i)
- M1: Expands the product to find the coefficient of \(z^5\) on the RHS in terms of the cosines and equates to the coefficient on the LHS (\(1\)).
- A1 (AG): Correctly concludes \(\cos\frac{2\pi}{7} + \cos\frac{4\pi}{7} + \cos\frac{6\pi}{7} = -\frac{1}{2}\).

(b)(ii)
- M1: Substitutes \(z = 1\) to get \(7 = 64\sin^2\frac{\pi}{7}\sin^2\frac{2\pi}{7}\sin^2\frac{3\pi}{7}\) via a valid trigonometric identity.
- A1 (AG): Justifies taking the positive square root to obtain \(\frac{\sqrt{7}}{8}\).

(c)
- B1: Correctly substitutes \(z = -1\) and finds \(\cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{3\pi}{7} = \frac{1}{8}\) with justification of signs.
- B1: Deduces \(\tan\frac{\pi}{7}\tan\frac{2\pi}{7}\tan\frac{3\pi}{7} = \sqrt{7}\).

(a) The characteristic equation is given by \( \det(A - \lambda I) = 0 \):
\[ \det \begin{pmatrix} -1-\lambda & 3 & 2 \\ -2 & 4-\lambda & 2 \\ -3 & 3 & 4-\lambda \end{pmatrix} = 0 \]
Expanding along the first row:
\[ (-1-\lambda) \left[ (4-\lambda)^2 - 6 \right] - 3 \left[ -2(4-\lambda) - (-6) \right] + 2 \left[ -6 - (-3)(4-\lambda) \right] = 0 \]
\[ (-1-\lambda)(\lambda^2 - 8\lambda + 10) - 3(2\lambda - 2) + 2(3\lambda - 6) = 0 \]
\[ -\lambda^3 + 8\lambda^2 - 10\lambda - \lambda^2 + 8\lambda - 10 - 6\lambda + 6 + 6\lambda - 12 = 0 \]
\[ -\lambda^3 + 7\lambda^2 - 14\lambda + 8 = 0 \implies \lambda^3 - 7\lambda^2 + 14\lambda - 8 = 0 \quad \text{(AG)} \]

Factorising:
\( \lambda = 1 \implies 1 - 7 + 14 - 8 = 0 \), so \( (\lambda - 1) \) is a factor.
\[ (\lambda - 1)(\lambda^2 - 6\lambda + 8) = (\lambda - 1)(\lambda - 2)(\lambda - 4) = 0 \]
Thus, the eigenvalues are \( \lambda = 1, \lambda = 2, \lambda = 4 \).

(b) For \( \lambda = 1 \):
\[ \begin{pmatrix} -2 & 3 & 2 \\ -2 & 3 & 2 \\ -3 & 3 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \]
From row 3: \( -x + y + z = 0 \implies x = y + z \).
Substituting into row 1: \( -2(y+z) + 3y + 2z = 0 \implies y = 0 \implies x = z \).
An eigenvector is \( \mathbf{v}_1 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} \).

For \( \lambda = 2 \):
\[ \begin{pmatrix} -3 & 3 & 2 \\ -2 & 2 & 2 \\ -3 & 3 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \]
From row 2: \( -x + y + z = 0 \implies z = x - y \).
From row 1: \( -3x + 3y + 2(x - y) = 0 \implies -x + y = 0 \implies x = y \implies z = 0 \).
An eigenvector is \( \mathbf{v}_2 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \).

For \( \lambda = 4 \):
\[ \begin{pmatrix} -5 & 3 & 2 \\ -2 & 0 & 2 \\ -3 & 3 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \]
From row 2: \( x = z \).
From row 3: \( x = y \).
Thus \( x = y = z \).
An eigenvector is \( \mathbf{v}_3 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \).

(c) Using the eigenvectors as columns in the corresponding order:
\[ P = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{pmatrix}, \qquad D = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{pmatrix} \]

(d)(i) By the Cayley–Hamilton theorem, \( A \) satisfies its characteristic equation:
\[ A^3 - 7A^2 + 14A - 8I = O \implies A^3 = 7A^2 - 14A + 8I \]
Multiplying by \( A \):
\[ A^4 = 7A^3 - 14A^2 + 8A \]
Substitute \( A^3 \):
\[ A^4 = 7(7A^2 - 14A + 8I) - 14A^2 + 8A \]
\[ A^4 = 49A^2 - 98A + 56I - 14A^2 + 8A = 35A^2 - 90A + 56I \]
Thus \( p = 35, q = -90, r = 56 \).

(d)(ii) Multiplying the Cayley–Hamilton identity \( A^3 - 7A^2 + 14A - 8I = O \) by \( A^{-1} \):
\[ A^2 - 7A + 14I - 8A^{-1} = O \]
\[ 8A^{-1} = A^2 - 7A + 14I \]
\[ A^{-1} = \frac{1}{8}A^2 - \frac{7}{8}A + \frac{7}{4}I \]
Thus \( a = \frac{1}{8}, b = -\frac{7}{8}, c = \frac{7}{4} \).

(a) [4 marks]
- M1: Sets up \( \det(A - \lambda I) = 0 \) and expands determinant of \( 3 \times 3 \) matrix.
- A1 (AG): Correct algebraic expansion leading convincingly to \( \lambda^3 - 7\lambda^2 + 14\lambda - 8 = 0 \).
- M1: Finds one root and factorises the cubic into linear factors.
- A1: Obtains \( \lambda = 1, 2, 4 \).

(b) [3 marks]
- M1: Sets up \( (A - \lambda I)\mathbf{v} = \mathbf{0} \) for at least one eigenvalue and attempts to solve for the components.
- A1: Obtains any two correct non-zero eigenvectors.
- A1: Obtains all three correct eigenvectors (or any non-zero scalar multiples).

(c) [1 mark]
- B1: Writes down matrices \( P \) and \( D \) with columns and diagonal entries in matching corresponding order.

(d)(i) [3 marks]
- M1: States or uses Cayley–Hamilton theorem: \( A^3 = 7A^2 - 14A + 8I \).
- M1: Multiplies by \( A \) and substitutes expression for \( A^3 \).
- A1: Obtains \( A^4 = 35A^2 - 90A + 56I \) (or values \( p = 35, q = -90, r = 56 \)).

(d)(ii) [3 marks]
- M1: Multiplies Cayley–Hamilton equation by \( A^{-1} \).
- M1: Rearranges to make \( A^{-1} \) the subject.
- A1: Obtains \( A^{-1} = \frac{1}{8}A^2 - \frac{7}{8}A + \frac{7}{4}I \) (or values \( a = \frac{1}{8}, b = -\frac{7}{8}, c = \frac{7}{4} \)).