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

(a)

We set up the identity:
\[\dfrac{1}{(2r-1)(2r+1)(2r+3)} = \dfrac{A}{(2r-1)(2r+1)} + \dfrac{B}{(2r+1)(2r+3)}\]
Multiplying both sides by \((2r-1)(2r+1)(2r+3)\):
\[1 = A(2r+3) + B(2r-1)\]
\[1 = 2(A+B)r + (3A-B)\]
Equating coefficients of \(r\) and the constant terms:
\[A + B = 0 \implies B = -A\]
\[3A - B = 1 \implies 3A - (-A) = 4A = 1 \implies A = \frac{1}{4}, \quad B = -\frac{1}{4}\]
Thus,
\[\dfrac{1}{(2r-1)(2r+1)(2r+3)} = \dfrac{1}{4(2r-1)(2r+1)} - \dfrac{1}{4(2r+1)(2r+3)}\]

(b)

Using the result from part (a):
\[\sum_{r=1}^{n} \dfrac{1}{(2r-1)(2r+1)(2r+3)} = \dfrac{1}{4} \sum_{r=1}^{n} \left( \dfrac{1}{(2r-1)(2r+1)} - \dfrac{1}{(2r+1)(2r+3)} \right)\]
Writing out the first few terms and the last term:
\[r = 1: \quad \dfrac{1}{1 \times 3} - \dfrac{1}{3 \times 5}\]
\[r = 2: \quad \dfrac{1}{3 \times 5} - \dfrac{1}{5 \times 7}\]
\[\vdots\]
\[r = n: \quad \dfrac{1}{(2n-1)(2n+1)} - \dfrac{1}{(2n+1)(2n+3)}\]
Summing the telescoping series:
\[\sum_{r=1}^{n} \dfrac{1}{(2r-1)(2r+1)(2r+3)} = \dfrac{1}{4} \left( \dfrac{1}{3} - \dfrac{1}{(2n+1)(2n+3)} \right)\]
\[= \dfrac{1}{4} \left( \dfrac{(2n+1)(2n+3) - 3}{3(2n+1)(2n+3)} \right)\]
\[= \dfrac{4n^2 + 8n + 3 - 3}{12(2n+1)(2n+3)} = \dfrac{4n(n+2)}{12(2n+1)(2n+3)} = \dfrac{n(n+2)}{3(2n+1)(2n+3)}\]

(c)

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

(d)

Using the method of differences from \(r = n\) to \(r = 2n\):
\[\sum_{r=n}^{2n} \dfrac{1}{(2r-1)(2r+1)(2r+3)} = \dfrac{1}{4} \left( \dfrac{1}{(2n-1)(2n+1)} - \dfrac{1}{(2(2n)+1)(2(2n)+3)} \right)\]
\[= \dfrac{1}{4} \left( \dfrac{1}{(2n-1)(2n+1)} - \dfrac{1}{(4n+1)(4n+3)} \right)\]
\[= \dfrac{1}{4} \left( \dfrac{(4n+1)(4n+3) - (2n-1)(2n+1)}{(2n-1)(2n+1)(4n+1)(4n+3)} \right)\]
Expanding the numerator:
\[(16n^2 + 16n + 3) - (4n^2 - 1) = 12n^2 + 16n + 4 = 4(3n^2 + 4n + 1) = 4(n+1)(3n+1)\]
Substituting back:
\[\dfrac{1}{4} \left( \dfrac{4(n+1)(3n+1)}{(2n-1)(2n+1)(4n+1)(4n+3)} \right) = \dfrac{(n+1)(3n+1)}{(2n-1)(2n+1)(4n+1)(4n+3)}\]

(a)
- M1: Sets up partial fractions identity \(1 = A(2r+3) + B(2r-1)\) and solves for \(A\) and \(B\).
- A1: Obtains correct values \(A = \frac{1}{4}\) and \(B = -\frac{1}{4}\) (or writes correctly with a factor of \(\frac{1}{4}\)).

(b)
- M1: Demonstrates the method of differences clearly by writing out at least two leading terms and the final term to show cancellation.
- M1: Correctly evaluates the remaining uncancelled terms: \(\frac{1}{4}\left(\frac{1}{3} - \frac{1}{(2n+1)(2n+3)}\right)\).
- A1 (AG): Combines over a common denominator and simplifies fully with intermediate steps shown to obtain \(\frac{n(n+2)}{3(2n+1)(2n+3)}\).

(c)
- B1: Obtains \(\frac{1}{12}\) by taking the limit as \(n \to \infty\).

(d)
- M1: Expresses the sum as \(S_{2n} - S_{n-1}\) or directly identifies remaining uncancelled terms \(\frac{1}{4}\left(\frac{1}{(2n-1)(2n+1)} - \frac{1}{(4n+1)(4n+3)}\right)\).
- A1: Simplifies to the single fraction \(\frac{(n+1)(3n+1)}{(2n-1)(2n+1)(4n+1)(4n+3)}\) (or numerator expanded as \(3n^2+4n+1\)).

Base case:
For \(n = 1\):
\[ \mathbf{A}^1 = \begin{pmatrix} 1 + 3(1) & -1 \\ 9(1) & 1 - 3(1) \end{pmatrix} = \begin{pmatrix} 4 & -1 \\ 9 & -2 \end{pmatrix} = \mathbf{A} \]
Hence the result is true for \(n = 1\).

Inductive hypothesis:
Assume that the result is true for \(n = k\), where \(k\) is a positive integer, so that
\[ \mathbf{A}^k = \begin{pmatrix} 1 + 3k & -k \\ 9k & 1 - 3k \end{pmatrix}. \]

Inductive step:
We consider \(n = k + 1\):
\[ \mathbf{A}^{k+1} = \mathbf{A}^k \mathbf{A} = \begin{pmatrix} 1 + 3k & -k \\ 9k & 1 - 3k \end{pmatrix} \begin{pmatrix} 4 & -1 \\ 9 & -2 \end{pmatrix} \]

Multiplying the matrices:
- Row 1, Column 1: \(4(1 + 3k) - 9k = 4 + 12k - 9k = 4 + 3k = 1 + 3(k + 1)\)
- Row 1, Column 2: \(-(1 + 3k) - 2(-k) = -1 - 3k + 2k = -1 - k = -(k + 1)\)
- Row 2, Column 1: \(4(9k) + 9(1 - 3k) = 36k + 9 - 27k = 9 + 9k = 9(k + 1)\)
- Row 2, Column 2: \(-(9k) - 2(1 - 3k) = -9k - 2 + 6k = -2 - 3k = 1 - 3(k + 1)\)

Thus,
\[ \mathbf{A}^{k+1} = \begin{pmatrix} 1 + 3(k+1) & -(k+1) \\ 9(k+1) & 1 - 3(k+1) \end{pmatrix}. \]

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

B1: Verify base case \(n = 1\) correctly showing equality with \(\mathbf{A}\).
M1: State inductive hypothesis: Assume result is true for \(n = k\).
M1: Consider \(\mathbf{A}^{k+1} = \mathbf{A}^k \mathbf{A}\) (or \(\mathbf{A}\mathbf{A}^k\)) and attempt matrix multiplication using the assumed expression for \(\mathbf{A}^k\).
A1: Obtain at least two correct unsimplified or simplified entries in the product matrix.
A1: Complete algebraic simplification to show all four entries in the form corresponding to \(n = k + 1\).
A1: Full, correct conclusion statement including: "true for \(n = k \implies\) true for \(n = k + 1\)", "true for \(n = 1\)", and "true for all positive integers \(n\) by mathematical induction" (dependent on all previous marks earned).

(a)
For the shear $A$ with the $y$-axis invariant, the matrix has the form $\begin{pmatrix} 1 & 0 \\ k & 1 \end{pmatrix}$.
Since $(1, 0)$ is transformed to $(1, 3)$:
\[ \begin{pmatrix} 1 & 0 \\ k & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ k \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \end{pmatrix} \implies k = 3 \]
Thus, the matrix representing transformation $A$ is $\mathbf{A} = \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix}$.

For the reflection $B$ in the line $y = -x$:
\[ (1, 0) \mapsto (0, -1) \quad \text{and} \quad (0, 1) \mapsto (-1, 0) \]
Thus, the matrix representing transformation $B$ is $\mathbf{B} = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}$.

Since $T$ is $A$ followed by $B$, the combined matrix is given by:
\[ \mathbf{M} = \mathbf{B}\mathbf{A} = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix} = \begin{pmatrix} -3 & -1 \\ -1 & 0 \end{pmatrix} \]

(b)
An invariant point $(x, y)$ satisfies $\mathbf{M}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}$:
\[ \begin{pmatrix} -3 & -1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix} \]
This yields the system of equations:
\[ -3x - y = x \implies 4x + y = 0 \]
\[ -x = y \implies x + y = 0 \]
Subtracting the second equation from the first gives $3x = 0 \implies x = 0$, which then gives $y = 0$.

Hence, the only invariant point is $(0, 0)$.

(c)
First, consider vertical lines of the form $x = k$:
\[ \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} -3 & -1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} k \\ y \end{pmatrix} = \begin{pmatrix} -3k - y \\ -k \end{pmatrix} \]
For this line to be invariant, $x'$ must be constant and equal to $k$, but $-3k - y$ depends on $y$. Thus, there are no vertical invariant lines.

Next, consider non-vertical lines of the form $y = mx + c$:
\[ \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} -3 & -1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} x \\ mx + c \end{pmatrix} = \begin{pmatrix} -3x - (mx + c) \\ -x \end{pmatrix} = \begin{pmatrix} -(3 + m)x - c \\ -x \end{pmatrix} \]
Substituting into $y' = mx' + c$:
\[ -x = m\left( -(3 + m)x - c \right) + c \]
\[ -x = -m(3 + m)x + c(1 - m) \]
Equating coefficients of $x$:
\[ -1 = -m(3 + m) \implies m^2 + 3m - 1 = 0 \]
Using the quadratic formula:
\[ m = \frac{-3 \pm \sqrt{3^2 - 4(1)(-1)}}{2} = \frac{-3 \pm \sqrt{13}}{2} \]
Equating the constant terms:
\[ 0 = c(1 - m) \]
Since $m = \frac{-3 \pm \sqrt{13}}{2}
eq 1$, it follows that $c = 0$.

Thus, the invariant lines of $T$ are:
\[ y = \left(\frac{-3 + \sqrt{13}}{2}\right)x \quad \text{and} \quad y = \left(\frac{-3 - \sqrt{13}}{2}\right)x \]

(a)
- B1: State or use the correct matrix for $A$, $\mathbf{A} = \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix}$, or matrix for $B$, $\mathbf{B} = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}$.
- M1: Correct order of matrix multiplication $\mathbf{M} = \mathbf{BA}$ attempted.
- A1: Correct product $\mathbf{M} = \begin{pmatrix} -3 & -1 \\ -1 & 0 \end{pmatrix}$.

(b)
- M1: Set up the matrix equation $\mathbf{M}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}$ and form a pair of linear equations.
- A1: Deduce that $x = 0, y = 0$ and state that the origin $(0, 0)$ is the only invariant point.

(c)
- M1: Apply the matrix transformation to $\begin{pmatrix} x \\ mx + c \end{pmatrix}$ or consider $y = mx$.
- M1: Substitute transformed coordinates into $y' = mx' + c$ and equate coefficients of $x$ to obtain a quadratic in $m$.
- A1: Solve $m^2 + 3m - 1 = 0$ to find $m = \frac{-3 \pm \sqrt{13}}{2}$.
- A1: Justify that $c = 0$ and state the final equations $y = \left(\frac{-3 + \sqrt{13}}{2}\right)x$ and $y = \left(\frac{-3 - \sqrt{13}}{2}\right)x$ (allow equivalent exact expressions).

(a) From the given cubic equation \(x^3 - 2x^2 + 3x - 5 = 0\):
\[\sum \alpha = 2, \quad \sum \alpha\beta = 3, \quad \alpha\beta\gamma = 5\]
Thus, \(S_1 = \alpha + \beta + \gamma = 2\).

For \(S_2\):
\[S_2 = \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) = 2^2 - 2(3) = 4 - 6 = -2\]

For \(S_3\), since \(\alpha, \beta, \gamma\) satisfy the equation, summing over the roots gives:
\[S_3 - 2S_2 + 3S_1 - 5(3) = 0\]
\[S_3 = 2S_2 - 3S_1 + 15 = 2(-2) - 3(2) + 15 = -4 - 6 + 15 = 5\]

(b) Using the identity for symmetric sums:
\[\alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 = (\alpha\beta + \beta\gamma + \gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha + \beta + \gamma)\]
\[= 3^2 - 2(5)(2) = 9 - 20 = -11\]

(c) Method 1 (Using coefficients of the transformed equation in \(w = x^2\)):
Let \(w\) be the roots \(\alpha^2, \beta^2, \gamma^2\).
Sum of roots: \(\sum \alpha^2 = S_2 = -2\)
Sum of product of pairs: \(\sum \alpha^2\beta^2 = -11\)
Product of roots: \(\alpha^2\beta^2\gamma^2 = (\alpha\beta\gamma)^2 = 5^2 = 25\)

Thus, the cubic equation with roots \(\alpha^2, \beta^2, \gamma^2\) is:
\[w^3 - (\sum \alpha^2)w^2 + (\sum \alpha^2\beta^2)w - (\alpha\beta\gamma)^2 = 0\]
\[w^3 - (-2)w^2 + (-11)w - 25 = 0 \implies w^3 + 2w^2 - 11w - 25 = 0\]

To find the equation with roots \(y = \frac{1}{w}\), substitute \(w = \frac{1}{y}\):
\[\left(\frac{1}{y}\right)^3 + 2\left(\frac{1}{y}\right)^2 - 11\left(\frac{1}{y}\right) - 25 = 0\]
Multiplying by \(y^3\):
\[1 + 2y - 11y^2 - 25y^3 = 0 \implies 25y^3 + 11y^2 - 2y - 1 = 0\]

Method 2 (Using direct substitution and rearrangement):
From \(x^3 + 3x = 2x^2 + 5\), squaring both sides gives:
\[x^2(x^2 + 3)^2 = (2x^2 + 5)^2\]
Let \(y = \frac{1}{x^2} \implies x^2 = \frac{1}{y}\):
\[\frac{1}{y}\left(\frac{1}{y} + 3\right)^2 = \left(\frac{2}{y} + 5\right)^2\]
\[\frac{1}{y}\cdot \frac{(1+3y)^2}{y^2} = \frac{(2+5y)^2}{y^2}\]
\[\frac{1 + 6y + 9y^2}{y^3} = \frac{4 + 20y + 25y^2}{y^2}\]
\[1 + 6y + 9y^2 = y(4 + 20y + 25y^2) = 4y + 20y^2 + 25y^3\]
\[25y^3 + 11y^2 - 2y - 1 = 0\]

(a)
B1: State \(S_1 = 2\)
M1: Use \(S_2 = (\sum \alpha)^2 - 2\sum\alpha\beta\) (or equivalent valid method)
A1: Obtain \(S_2 = -2\)
A1: Obtain \(S_3 = 5\) via \(S_3 - 2S_2 + 3S_1 - 15 = 0\) or standard formula

(b)
M1: Use identity \(\sum \alpha^2\beta^2 = (\sum\alpha\beta)^2 - 2\alpha\beta\gamma(\sum\alpha)\)
A1: Obtain \(-11\)

(c)
M1: Find polynomial equation with roots \(\alpha^2, \beta^2, \gamma^2\) (e.g. \(w^3 + 2w^2 - 11w - 25 = 0\)) OR rearrange original equation and square: \(x^2(x^2+3)^2 = (2x^2+5)^2\)
A1: Correct intermediate cubic in \(w\) or expression in \(x^2\)
M1: Substitute \(w = \frac{1}{y}\) or \(x^2 = \frac{1}{y}\) to obtain an equation in \(y\)
A1: Obtain \(25y^3 + 11y^2 - 2y - 1 = 0\) (or \(-25y^3 - 11y^2 + 2y + 1 = 0\), must be an equation with integer coefficients)

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

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

Let \(A = (1, 2, -1)\) be a point on \(l_1\) and \(B = (3, -1, 2)\) be a point on \(l_2\).
\[ \vec{AB} = \begin{pmatrix} 3 - 1 \\ -1 - 2 \\ 2 - (-1) \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \\ 3 \end{pmatrix} \]

The shortest distance \(d\) between the skew lines is
\[ d = \frac{|\vec{AB} \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{\left| 2(-1) + (-3)(3) + 3(5) \right|}{\sqrt{(-1)^2 + 3^2 + 5^2}} = \frac{|-2 - 9 + 15|}{\sqrt{1 + 9 + 25}} = \frac{4}{\sqrt{35}} = \frac{4\sqrt{35}}{35} \approx 0.676 \]

(b) Since \(\Pi_1\) contains \(l_1\) and is parallel to \(l_2\), a normal to \(\Pi_1\) is \(\mathbf{n}_1 = \begin{pmatrix} -1 \\ 3 \\ 5 \end{pmatrix}\) (or \(\begin{pmatrix} 1 \\ -3 \\ -5 \end{pmatrix}\)).

Using the point \((1, 2, -1)\) on \(l_1\):
\[ \mathbf{r} \cdot \begin{pmatrix} 1 \\ -3 \\ -5 \end{pmatrix} = 1(1) - 3(2) - 5(-1) = 1 - 6 + 5 = 0 \]

Thus, a Cartesian equation of \(\Pi_1\) is \(x - 3y - 5z = 0\).

(c)(i) The plane \(\Pi_2\) contains \(l_2\) (direction \(\mathbf{d}_2\)) and is perpendicular to \(\Pi_1\) (normal \(\mathbf{n}_1\)). Therefore, a normal \(\mathbf{n}_2\) to \(\Pi_2\) is given by
\[ \mathbf{n}_2 = \mathbf{n}_1 \times \mathbf{d}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 3 & 5 \\ 1 & 2 & -1 \end{vmatrix} = \begin{pmatrix} 3(-1) - 5(2) \\ -\big((-1)(-1) - 5(1)\big) \\ (-1)(2) - 3(1) \end{vmatrix} = \begin{pmatrix} -13 \\ 4 \\ -5 \end{pmatrix} \]

Taking \(\mathbf{n}_2 = \begin{pmatrix} 13 \\ -4 \\ 5 \end{pmatrix}\) and using the point \((3, -1, 2)\) on \(l_2\):
\[ 13(x) - 4(y) + 5(z) = 13(3) - 4(-1) + 5(2) = 39 + 4 + 10 = 53 \]

Thus, a Cartesian equation of \(\Pi_2\) is \(13x - 4y + 5z = 53\).

(c)(ii) Let \(\theta\) be the acute angle between the line \(l_1\) and the plane \(\Pi_2\). Then
\[ \sin\theta = \frac{|\mathbf{d}_1 \cdot \mathbf{n}_2|}{|\mathbf{d}_1| |\mathbf{n}_2|} \]

We compute:
\[ \mathbf{d}_1 \cdot \mathbf{n}_2 = 2(13) + (-1)(-4) + 1(5) = 26 + 4 + 5 = 35 \]
\[ |\mathbf{d}_1| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{6} \]
\[ |\mathbf{n}_2| = \sqrt{13^2 + (-4)^2 + 5^2} = \sqrt{169 + 16 + 25} = \sqrt{210} \]
\[ |\mathbf{d}_1| |\mathbf{n}_2| = \sqrt{6 \times 210} = \sqrt{1260} = 6\sqrt{35} \]

\[ \sin\theta = \frac{35}{6\sqrt{35}} = \frac{\sqrt{35}}{6} \approx 0.98601 \]
\[ \theta = \arcsin\left(\frac{\sqrt{35}}{6}\right) \approx 80.414^\circ \approx 80.4^\circ \]

(a)
M1: Correct method to find the vector product of the direction vectors \(\mathbf{d}_1\) and \(\mathbf{d}_2\).
A1: Correct vector \(\begin{pmatrix} -1 \\ 3 \\ 5 \end{pmatrix}\) (or scalar multiple) and correct vector \(\vec{AB}\).
A1: Correct shortest distance \(\frac{4}{\sqrt{35}}\) or \(\frac{4\sqrt{35}}{35}\) or awrt \(0.676\).

(b)
M1: Sets up the plane equation \(\mathbf{r} \cdot \mathbf{n}_1 = d\) using their normal from (a) and substituting a point on \(l_1\).
A1: Correct Cartesian equation: \(x - 3y - 5z = 0\) or \(-x + 3y + 5z = 0\).

(c)(i)
M1: Vector product of their normal to \(\Pi_1\) and the direction vector of \(l_2\) (or equivalent method).
A1: Correct normal vector \(\begin{pmatrix} 13 \\ -4 \\ 5 \end{pmatrix}\) (or scalar multiple).
A1: Correct Cartesian equation: \(13x - 4y + 5z = 53\) (or integer multiple).

(c)(ii)
M1: Uses \(\sin\theta = \frac{|\mathbf{d}_1 \cdot \mathbf{n}_2|}{|\mathbf{d}_1| |\mathbf{n}_2|}\) with their direction and normal vectors.
M1: Correctly evaluates dot product and both vector magnitudes.
A1: Correct acute angle \(80.4^\circ\) (accept \(80.41^\circ\) or \(1.40\) rad).

(a)
Dividing the numerator by the denominator:
\[ x^2 - x - 2 = (x - 3)(x + 2) + 4 \]
\[ y = x + 2 + \frac{4}{x - 3} \]
As \( x \to 3 \), \( y \to \pm\infty \), so the vertical asymptote is \( x = 3 \).
As \( x \to \pm\infty \), \( \frac{4}{x - 3} \to 0 \), so the oblique asymptote is \( y = x + 2 \).

(b)
Rearranging the equation of \( C \) into a quadratic in \( x \):
\[ y(x - 3) = x^2 - x - 2 \]
\[ x^2 - (y + 1)x + (3y - 2) = 0 \]
Since \( x \) is real, the discriminant must satisfy \( \Delta \ge 0 \):
\[ [-(y + 1)]^2 - 4(1)(3y - 2) \ge 0 \]
\[ y^2 + 2y + 1 - 12y + 8 \ge 0 \]
\[ y^2 - 10y + 9 \ge 0 \]
\[ (y - 1)(y - 9) \ge 0 \]
This inequality holds for \( y \le 1 \) or \( y \ge 9 \).
Hence, there are no points on \( C \) for which \( 1 < y < 9 \).

At the stationary points, \( \Delta = 0 \):
- When \( y = 1 \):
\[ x^2 - 2x + 1 = 0 \implies (x - 1)^2 = 0 \implies x = 1 \]
Stationary point is \( (1, 1) \) (local maximum).
- When \( y = 9 \):
\[ x^2 - 10x + 25 = 0 \implies (x - 5)^2 = 0 \implies x = 5 \]
Stationary point is \( (5, 9) \) (local minimum).

(c)
- Intersection with the \( y \)-axis: when \( x = 0 \), \( y = \frac{-2}{-3} = \frac{2}{3} \), giving \( \left(0, \frac{2}{3}\right) \).
- Intersections with the \( x \)-axis: when \( y = 0 \), \( x^2 - x - 2 = 0 \implies (x - 2)(x + 1) = 0 \), giving \( (-1, 0) \) and \( (2, 0) \).
- The sketch consists of two branches:
1. For \( x < 3 \), the curve passes through \( (-1, 0) \), \( \left(0, \frac{2}{3}\right) \), reaches a local maximum at \( (1, 1) \), passes through \( (2, 0) \), and approaches \( x = 3 \) downwards (\( y \to -\infty \)) and \( y = x + 2 \) as \( x \to -\infty \).
2. For \( x > 3 \), the curve approaches \( x = 3 \) from above (\( y \to +\infty \)), reaches a local minimum at \( (5, 9) \), and approaches \( y = x + 2 \) as \( x \to +\infty \).

(d)
To sketch \( y = |f(x)| \):
- Reflect the portion of the curve where \( y < 0 \) across the \( x \)-axis.
- The region \( x < -1 \) has \( f(x) < 0 \), so it reflects upwards, approaching \( y = -x - 2 \) as \( x \to -\infty \).
- The region \( 2 < x < 3 \) has \( f(x) < 0 \), so it reflects upwards, approaching \( +\infty \) as \( x \to 3^- \).
- Cusps / sharp points occur on the \( x \)-axis at \( (-1, 0) \) and \( (2, 0) \).
- The asymptotes are \( x = 3 \), \( y = x + 2 \) (for \( x > 0 \)), and \( y = -x - 2 \) (for \( x < 0 \)).

(e)
The inequality \( |f(x)| \le 9 \) corresponds to \( -9 \le f(x) \le 9 \).
1. Solving \( f(x) \le 9 \):
\[ \frac{x^2 - x - 2}{x - 3} \le 9 \iff \frac{x^2 - 10x + 25}{x - 3} \le 0 \iff \frac{(x - 5)^2}{x - 3} \le 0 \]
Since \( (x - 5)^2 \ge 0 \) for all \( x \), this holds when \( x < 3 \) or \( x = 5 \).

2. Solving \( f(x) \ge -9 \):
\[ \frac{x^2 - x - 2}{x - 3} \ge -9 \iff \frac{x^2 + 8x - 29}{x - 3} \ge 0 \]
The roots of \( x^2 + 8x - 29 = 0 \) are:
\[ x = \frac{-8 \pm \sqrt{64 - 4(1)(-29)}}{2} = \frac{-8 \pm \sqrt{180}}{2} = -4 \pm 3\sqrt{5} \]
Testing the signs of the expression across intervals with critical values \( -4 - 3\sqrt{5} \), \( -4 + 3\sqrt{5} \approx 2.708 \), and \( 3 \):
\[ -4 - 3\sqrt{5} \le x \le -4 + 3\sqrt{5} \quad \text{or} \quad x > 3 \]

3. Combining both conditions:
- For \( x < 3 \): \( -4 - 3\sqrt{5} \le x \le -4 + 3\sqrt{5} \)
- For \( x > 3 \): \( x = 5 \)

Thus, the solution set is \( -4 - 3\sqrt{5} \le x \le -4 + 3\sqrt{5} \) and \( x = 5 \).

(a)
- B1: For vertical asymptote \( x = 3 \).
- B1: For oblique asymptote \( y = x + 2 \) (must be obtained from correct algebraic division or inspection).

(b)
- M1: Forms a quadratic equation in \( x \), e.g. \( x^2 - (y + 1)x + (3y - 2) = 0 \) (or applies differentiation setting \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \)).
- M1: Uses discriminant condition \( \Delta \ge 0 \) (or solves \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \) for \( x \) and finds corresponding \( y \)-values).
- A1: Accurately shows \( (y - 1)(y - 9) \ge 0 \) and states that no points exist for \( 1 < y < 9 \) (AG, fully justified).
- A1: States both pairs of coordinates correctly: \( (1, 1) \) and \( (5, 9) \).

(c)
- B1: Both branches correctly positioned relative to asymptotes \( x = 3 \) and \( y = x + 2 \).
- B1: Asymptotes shown as dashed lines and local maximum/minimum correctly indicated in their respective quadrants.
- B1: Axis intercepts \( (-1, 0) \), \( (2, 0) \), and \( \left(0, \frac{2}{3}\right) \) clearly labelled.

(d)
- B1: Correctly reflects the sections below the \( x \)-axis so that the curve is strictly non-negative with sharp corners/cusps at \( (-1, 0) \) and \( (2, 0) \).
- B1: Correct asymptotic behaviour shown (approaching \( x = 3 \) from both sides to \( +\infty \), and oblique asymptotes).

(e)
- M1: Formulates the inequality \( -9 \le f(x) \le 9 \) and uses algebra or graphical insight to establish relevant equations/inequalities.
- M1: Solves \( x^2 + 8x - 29 = 0 \) to find \( x = -4 \pm 3\sqrt{5} \) and identifies the isolated solution \( x = 5 \) from \( f(x) = 9 \).
- A1: Fully correct final set of values: \( -4 - 3\sqrt{5} \le x \le -4 + 3\sqrt{5} \) and \( x = 5 \) (or equivalent interval notation \( [-4 - 3\sqrt{5}, -4 + 3\sqrt{5}] \cup \{5\} \)).

(a)
Since \(1 + \cos 2\theta = 2\cos^2\theta\), \(r = 2a\cos^2\theta\ge 0\) for all \(\theta\in[-\frac{1}{2}\pi, \frac{1}{2}\pi]\).
The curve consists of a single closed loop symmetric about the initial line \(\theta = 0\), located in the half-plane \(x \ge 0\).
- At \(\theta = 0\), \(r = 2a\), giving the point furthest from the pole: \((2a, 0)\).
- At the pole \(r = 0 \implies 1 + \cos 2\theta = 0 \implies 2\theta = \pm\pi \implies \theta = \pm\frac{1}{2}\pi\).
Thus, the tangents at the pole are \(\theta = \frac{1}{2}\pi\) and \(\theta = -\frac{1}{2}\pi\).

(b)
The distance of any point \((r, \theta)\) from the line \(\theta = \frac{1}{2}\pi\) (the \(y\)-axis) is given by \(x = r\cos\theta\).
Substituting \(r = a(1 + \cos 2\theta) = 2a\cos^2\theta\):
\[ x = 2a\cos^2\theta \cdot \cos\theta = 2a\cos^3\theta \]
For \(-\frac{1}{2}\pi \le \theta \le \frac{1}{2}\pi\), \(0 \le \cos\theta \le 1\).
The maximum of \(\cos\theta\) is \(1\), occurring at \(\theta = 0\).
Therefore, the greatest distance is \(2a(1)^3 = 2a\).

(c)
The tangent is parallel to the initial line when \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = 0\) with \(\frac{\mathrm{d}x}{\mathrm{d}\theta} \ne 0\).
\[ y = r\sin\theta = 2a\cos^2\theta\sin\theta \]
Differentiating with respect to \(\theta\):
\[ \frac{\mathrm{d}y}{\mathrm{d}\theta} = 2a\left[ 2\cos\theta(-\sin\theta)\sin\theta + \cos^2\theta\cos\theta \right] = 2a\cos\theta(\cos^2\theta - 2\sin^2\theta) \]
Using \(\cos^2\theta = 1 - \sin^2\theta\):
\[ \frac{\mathrm{d}y}{\mathrm{d}\theta} = 2a\cos\theta(1 - 3\sin^2\theta) \]
Setting \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = 0\):
- \(\cos\theta = 0 \implies \theta = \pm\frac{1}{2}\pi\) (corresponds to the pole where \(r = 0\)).
- \(1 - 3\sin^2\theta = 0 \implies \sin^2\theta = \frac{1}{3} \implies \sin\theta = \pm\frac{1}{\sqrt{3}}\).
Since \(\theta \in \left(-\frac{1}{2}\pi, \frac{1}{2}\pi\right)\), \(\cos\theta > 0\), so:
\[ \cos\theta = \sqrt{1 - \frac{1}{3}} = \sqrt{\frac{2}{3}} \]
The value of \(r\) at these points is:
\[ r = 2a\cos^2\theta = 2a\left(\frac{2}{3}\right) = \frac{4}{3}a \]
Now finding the Cartesian coordinates \((x, y)\):
\[ x = r\cos\theta = \frac{4}{3}a \cdot \sqrt{\frac{2}{3}} = \frac{4\sqrt{6}}{9}a \]
\[ y = r\sin\theta = \frac{4}{3}a \left(\pm\frac{1}{\sqrt{3}}\right) = \pm\frac{4\sqrt{3}}{9}a \]
Thus, the points are \(\left(\frac{4\sqrt{6}}{9}a, \, \frac{4\sqrt{3}}{9}a\right)\) and \(\left(\frac{4\sqrt{6}}{9}a, \, -\frac{4\sqrt{3}}{9}a\right)\).

(d)(i)
At intersection points of \(C_1\) and \(C_2\):
\[ 2a\cos^2\theta = a \implies \cos^2\theta = \frac{1}{2} \]
Since \(-\frac{1}{2}\pi \le \theta \le \frac{1}{2}\pi\), \(\cos\theta = \frac{1}{\sqrt{2}} \implies \theta = \pm\frac{\pi}{4}\).
Since \(r = a\), the polar coordinates are \(\left(a, \frac{\pi}{4}\right)\) and \(\left(a, -\frac{\pi}{4}\right)\).

(d)(ii)
By symmetry, the required area is:
\[ A = 2 \times \frac{1}{2} \int_0^{\pi/4} \left(r_1^2 - r_2^2\right) \mathrm{d}\theta = \int_0^{\pi/4} \left(4a^2\cos^4\theta - a^2\right) \mathrm{d}\theta \]
Express \(\cos^4\theta\) using multiple angles:
\[ \cos^4\theta = \left(\frac{1+\cos 2\theta}{2}\right)^2 = \frac{1 + 2\cos 2\theta + \cos^2 2\theta}{4} = \frac{1 + 2\cos 2\theta + \frac{1+\cos 4\theta}{2}}{4} = \frac{3 + 4\cos 2\theta + \cos 4\theta}{8} \]
Thus:
\[ 4a^2\cos^4\theta = \frac{a^2}{2}(3 + 4\cos 2\theta + \cos 4\theta) \]
Integrating:
\[ \int_0^{\pi/4} 4a^2\cos^4\theta \, \mathrm{d}\theta = \frac{a^2}{2} \left[ 3\theta + 2\sin 2\theta + \frac{1}{4}\sin 4\theta \right]_0^{\pi/4} \]
Evaluating at the limits \(\theta = \frac{\pi}{4}\) and \(\theta = 0\):
\[ = \frac{a^2}{2}\left( \frac{3\pi}{4} + 2\sin\frac{\pi}{2} + \frac{1}{4}\sin\pi \right) = \frac{a^2}{2}\left(\frac{3\pi}{4} + 2\right) = a^2\left(\frac{3\pi}{8} + 1\right) \]
Next, integrate the circular sector term:
\[ \int_0^{\pi/4} a^2 \, \mathrm{d}\theta = a^2 \cdot \frac{\pi}{4} = \frac{\pi a^2}{4} \]
Subtracting:
\[ A = a^2\left(\frac{3\pi}{8} + 1\right) - \frac{\pi a^2}{4} = a^2\left(\frac{\pi}{8} + 1\right) = \frac{1}{8}a^2(\pi + 8) \]

(a)
- B1: Correct sketch of a single smooth closed loop, symmetric about the initial line, completely contained within the half-plane \(-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}\).
- B1: Stating or clearly labelling point furthest from pole as \((2a, 0)\).
- B1: Stating equations of tangents at the pole as \(\theta = \pm\frac{1}{2}\pi\) (or \(\theta = \frac{\pi}{2}, -\frac{\pi}{2}\)).

(b)
- M1: Writes distance from \(y\)-axis as \(x = r\cos\theta\) and substitutes \(r = a(1+\cos 2\theta)\) or \(2a\cos^2\theta\).
- A1: Accurately shows intermediate step \(2a\cos^2\theta\cos\theta = 2a\cos^3\theta\) (AG).
- A1: States maximum distance is \(2a\) with clear justification that \(\max(\cos\theta) = 1\) at \(\theta = 0\).

(c)
- M1: Sets up expression for \(y = r\sin\theta = 2a\cos^2\theta\sin\theta\).
- M1: Uses product/chain rule to find \(\frac{\mathrm{d}y}{\mathrm{d}\theta}\).
- A1: Correct simplified derivative \(2a\cos\theta(1 - 3\sin^2\theta)\) or equivalent.
- M1: Sets derivative to 0 and solves for non-zero \(\sin\theta\) and \(\cos\theta\).
- A1: Correct values \(\sin\theta = \pm\frac{1}{\sqrt{3}}\) and \(\cos\theta = \sqrt{\frac{2}{3}}\).
- A1: Correct exact Cartesian coordinates \(\left(\frac{4\sqrt{6}}{9}a, \, \pm\frac{4\sqrt{3}}{9}a\right)\).

(d)(i)
- M1: Equates equations \(2a\cos^2\theta = a\) and solves for \(\theta\).
- A1: Both polar coordinates correctly stated: \(\left(a, \frac{\pi}{4}\right)\) and \(\left(a, -\frac{\pi}{4}\right)\).

(d)(ii)
- M1: Uses polar area formula \(\frac{1}{2}\int (r_1^2 - r_2^2)\,\mathrm{d}\theta\) with correct integration limits.
- M1: Expands \(\cos^4\theta\) using double/multiple angle identities and integrates correctly.
- A1: Obtains exact final area \(\frac{1}{8}a^2(\pi + 8)\) or \(a^2\left(\frac{\pi}{8} + 1\right)\).

(a) Differentiating \(y = \ln(1 + \sinh 2x)\) with respect to \(x\):
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2\cosh 2x}{1 + \sinh 2x} \]
Multiplying both sides by \(1 + \sinh 2x\):
\[ (1 + \sinh 2x)\frac{\mathrm{d}y}{\mathrm{d}x} = 2\cosh 2x \]
Differentiating implicitly with respect to \(x\) using the product rule:
\[ (1 + \sinh 2x)\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + 2\cosh 2x \frac{\mathrm{d}y}{\mathrm{d}x} = 4\sinh 2x \]

(b) Differentiating the relation with respect to \(x\):
\[ (1 + \sinh 2x)\frac{\mathrm{d}^3 y}{\mathrm{d}x^3} + 2\cosh 2x \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + 2\cosh 2x \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + 4\sinh 2x \frac{\mathrm{d}y}{\mathrm{d}x} = 8\cosh 2x \]
\[ (1 + \sinh 2x)\frac{\mathrm{d}^3 y}{\mathrm{d}x^3} + 4\cosh 2x \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + 4\sinh 2x \frac{\mathrm{d}y}{\mathrm{d}x} = 8\cosh 2x \]

Evaluating the function and derivatives at \(x = 0\):
\[ y(0) = \ln(1 + \sinh 0) = \ln 1 = 0 \]
\[ y'(0) = \frac{2\cosh 0}{1 + \sinh 0} = \frac{2(1)}{1 + 0} = 2 \]
From (a):
\[ (1 + 0)y''(0) + 2(1)(2) = 4(0) \implies y''(0) = -4 \]
From the third derivative:
\[ (1 + 0)y'''(0) + 4(1)(-4) + 4(0)(2) = 8(1) \implies y'''(0) - 16 = 8 \implies y'''(0) = 24 \]

Applying Maclaurin's series formula:
\[ y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \dots \]
\[ y = 0 + 2x + \frac{-4}{2}x^2 + \frac{24}{6}x^3 = 2x - 2x^2 + 4x^3 \]

(a)
M1: For differentiating \(y\) using the chain rule and writing in the form \((1 + \sinh 2x)y' = 2\cosh 2x\) (or applying the quotient rule correctly to obtain \(y''\)).
A1: For completing the differentiation using the product rule to obtain the given equation correctly with all steps clearly shown (AG).

(b)
M1: For differentiating the equation from part (a) implicitly with respect to \(x\).
M1: For substituting \(x = 0\) to determine the values of \(y(0)\), \(y'(0)\), \(y''(0)\), and \(y'''(0)\).
A1: For the correct series \(2x - 2x^2 + 4x^3\) (or \(y = 2x - 2x^2 + 4x^3\)).

(a) Write the integral as
\[ I_n = \int_0^1 x^{n-1} \cdot x (1 - x^2)^{\frac{1}{2}} \, \mathrm{d}x. \]
Let \( u = x^{n-1} \) and \( \frac{\mathrm{d}v}{\mathrm{d}x} = x (1 - x^2)^{\frac{1}{2}} \).
Then \( \frac{\mathrm{d}u}{\mathrm{d}x} = (n - 1)x^{n-2} \) and \( v = -\frac{1}{3}(1 - x^2)^{\frac{3}{2}} \).

Applying integration by parts:
\[ I_n = \left[ -\frac{1}{3} x^{n-1} (1 - x^2)^{\frac{3}{2}} \right]_0^1 + \frac{n - 1}{3} \int_0^1 x^{n-2} (1 - x^2)^{\frac{3}{2}} \, \mathrm{d}x. \]
For \( n \ge 2 \), the boundary term evaluates to
\[ -\frac{1}{3}(1)(0)^{\frac{3}{2}} - \left(-\frac{1}{3}(0)(1)^{\frac{3}{2}}\right) = 0. \]

Since \( (1 - x^2)^{\frac{3}{2}} = (1 - x^2)\sqrt{1 - x^2} \), we have:
\[ I_n = \frac{n - 1}{3} \int_0^1 x^{n-2} (1 - x^2) \sqrt{1 - x^2} \, \mathrm{d}x \]
\[ I_n = \frac{n - 1}{3} \left( \int_0^1 x^{n-2}\sqrt{1 - x^2} \, \mathrm{d}x - \int_0^1 x^n \sqrt{1 - x^2} \, \mathrm{d}x \right) \]
\[ I_n = \frac{n - 1}{3} (I_{n-2} - I_n). \]
Multiplying through by 3 gives:
\[ 3 I_n = (n - 1)I_{n-2} - (n - 1)I_n \]
\[ (3 + n - 1) I_n = (n - 1) I_{n-2} \]
\[ (n + 2) I_n = (n - 1) I_{n-2}. \]

(b) First evaluate \( I_0 \):
\[ I_0 = \int_0^1 \sqrt{1 - x^2} \, \mathrm{d}x. \]
Using the substitution \( x = \sin\theta \), \( \mathrm{d}x = \cos\theta \, \mathrm{d}\theta \):
\[ I_0 = \int_0^{\frac{\pi}{2}} \cos^2\theta \, \mathrm{d}\theta = \int_0^{\frac{\pi}{2}} \frac{1 + \cos 2\theta}{2} \, \mathrm{d}\theta = \left[ \frac{\theta}{2} + \frac{\sin 2\theta}{4} \right]_0^{\frac{\pi}{2}} = \frac{\pi}{4}. \]

Using the reduction formula with \( n = 2 \):
\[ (2 + 2) I_2 = (2 - 1) I_0 \implies 4 I_2 = I_0 \implies I_2 = \frac{1}{4} \left(\frac{\pi}{4}\right) = \frac{\pi}{16}. \]

Using the reduction formula with \( n = 4 \):
\[ (4 + 2) I_4 = (4 - 1) I_2 \implies 6 I_4 = 3 I_2 \implies I_4 = \frac{1}{2} I_2 = \frac{1}{2} \left(\frac{\pi}{16}\right) = \frac{\pi}{32}. \]

(a)
- M1: For splitting integrand as \( x^{n-1} \cdot x\sqrt{1 - x^2} \) (or equivalent) and attempting integration by parts.
- A1: For correct integrated part \( \left[ -\frac{1}{3} x^{n-1} (1 - x^2)^{\frac{3}{2}} \right]_0^1 \) and showing it vanishes at both limits for \( n \ge 2 \).
- M1: For rewriting \( (1 - x^2)^{\frac{3}{2}} \) as \( (1 - x^2)\sqrt{1 - x^2} \) and expressing the integral in terms of \( I_{n-2} \) and \( I_n \).
- A1: For completely correct working leading to the given result \( (n + 2) I_n = (n - 1) I_{n-2} \) (AG).

(b)
- B1: For evaluating \( I_0 = \frac{\pi}{4} \) (via substitution or recognized area of a quarter unit circle).
- M1: For using the reduction formula to express \( I_4 \) in terms of \( I_2 \) and \( I_2 \) in terms of \( I_0 \).
- A1: For the exact value \( \frac{\pi}{32} \).

(a)
Given \(z = y^{-1} = \frac{1}{y}\), differentiating with respect to \(x\) gives:
\[ \frac{\mathrm{d}z}{\mathrm{d}x} = -y^{-2} \frac{\mathrm{d}y}{\mathrm{d}x} \implies \frac{\mathrm{d}y}{\mathrm{d}x} = -y^2 \frac{\mathrm{d}z}{\mathrm{d}x} \]
Substitute this into the original differential equation:
\[ -y^2 \frac{\mathrm{d}z}{\mathrm{d}x} + 2y \tanh x = y^2 \sinh 2x \]
Dividing the entire equation by \(-y^2\):
\[ \frac{\mathrm{d}z}{\mathrm{d}x} - 2y^{-1} \tanh x = -\sinh 2x \]
Since \(z = y^{-1}\), this becomes:
\[ \frac{\mathrm{d}z}{\mathrm{d}x} - 2z \tanh x = -\sinh 2x \]

(b)
The integrating factor is:
\[ I(x) = \exp\left(\int -2\tanh x \,\mathrm{d}x\right) = \exp\left(-2\ln(\cosh x)\right) = (\cosh x)^{-2} = \operatorname{sech}^2 x \]
Multiplying the differential equation by \(I(x)\):
\[ \operatorname{sech}^2 x \frac{\mathrm{d}z}{\mathrm{d}x} - 2z \operatorname{sech}^2 x \tanh x = -\sinh 2x \operatorname{sech}^2 x \]
\[ \frac{\mathrm{d}}{\mathrm{d}x}\left(z \operatorname{sech}^2 x\right) = -\sinh 2x \operatorname{sech}^2 x \]
Using the identity \(\sinh 2x = 2\sinh x \cosh x\):
\[ -\sinh 2x \operatorname{sech}^2 x = -2\sinh x \cosh x \cdot \frac{1}{\cosh^2 x} = -2\frac{\sinh x}{\cosh x} = -2\tanh x \]
Integrating both sides with respect to \(x\):
\[ z \operatorname{sech}^2 x = \int -2\tanh x \,\mathrm{d}x = -2\ln(\cosh x) + c \]
Multiplying through by \(\cosh^2 x\):
\[ z = \cosh^2 x \left(c - 2\ln(\cosh x)\right) \]

(c)
When \(x = 0\), \(y = 1 \implies z = \frac{1}{1} = 1\).
Substituting these values:
\[ 1 = \cosh^2(0)\left(c - 2\ln(\cosh 0)\right) \implies 1 = 1(c - 2\ln 1) \implies c = 1 \]
Thus:
\[ z = \cosh^2 x \left(1 - 2\ln(\cosh x)\right) \]
Since \(y = \frac{1}{z}\):
\[ y = \frac{1}{\cosh^2 x \left(1 - 2\ln(\cosh x)\right)} \quad \text{or} \quad y = \frac{\operatorname{sech}^2 x}{1 - 2\ln(\cosh x)} \]

(d)
The denominator is zero when:
\[ 1 - 2\ln(\cosh x) = 0 \implies \ln(\cosh x) = \frac{1}{2} \implies \cosh x = \mathrm{e}^{1/2} = \sqrt{\mathrm{e}} \]
Using the logarithmic form of \(\operatorname{arcosh} u = \ln\left(u + \sqrt{u^2 - 1}\right)\) for \(u \ge 1\):
\[ x = \operatorname{arcosh}\left(\sqrt{\mathrm{e}}\right) = \ln\left(\sqrt{\mathrm{e}} + \sqrt{(\sqrt{\mathrm{e}})^2 - 1}\right) = \ln\left(\sqrt{\mathrm{e}} + \sqrt{\mathrm{e} - 1}\right) \]

(a)
- M1: Differentiates \(z = y^{-1}\) to find \(\frac{\mathrm{d}z}{\mathrm{d}x} = -y^{-2}\frac{\mathrm{d}y}{\mathrm{d}x}\) and substitutes into the equation.
- A1: Divides through by \(-y^2\) and correctly obtains the given form with no omitted steps (AG).

(b)
- M1: Attempts to find the integrating factor \(\mathrm{e}^{\int -2\tanh x \,\mathrm{d}x}\).
- A1: Obtains \(\operatorname{sech}^2 x\) or \(\frac{1}{\cosh^2 x}\).
- M1: Uses \(\sinh 2x = 2\sinh x \cosh x\) to simplify the right-hand side to \(-2\tanh x\) (or equivalent).
- M1: Integrates the RHS to obtain \(-2\ln(\cosh x)\) (condone omission of \(+c\) at this step).
- A1: Obtains correct general solution \(z = \cosh^2 x(c - 2\ln(\cosh x))\) (must include constant of integration).

(c)
- M1: Substitutes \(x = 0, y = 1\) (or \(z = 1\)) into their expression for \(z\) to find \(c\).
- A1: Obtains correct expression for \(y\) in terms of \(x\), e.g. \(y = \frac{1}{\cosh^2 x (1 - 2\ln(\cosh x))}\).

(d)
- M1: Sets the denominator to 0 and correctly finds \(\cosh x = \sqrt{\mathrm{e}}\).
- A1: Applies inverse hyperbolic formula to deduce \(x = \ln\left(\sqrt{\mathrm{e}} + \sqrt{\mathrm{e} - 1}\right)\) (accept \(k = \mathrm{e} - 1\)).

(a) When \(x = 0\):
\[ y^3 + 3(0)^2y - 4\mathrm{e}^{0} + 3 = 0 \implies y^3 - 4 + 3 = 0 \implies y^3 = 1 \implies y = 1. \]
Differentiating the given equation implicitly with respect to \(x\):
\[ 3y^2 \frac{\mathrm{d}y}{\mathrm{d}x} + 6xy + 3x^2 \frac{\mathrm{d}y}{\mathrm{d}x} - 8\mathrm{e}^{2x} = 0. \]
Substituting \(x = 0\) and \(y = 1\):
\[ 3(1)^2 \frac{\mathrm{d}y}{\mathrm{d}x} + 6(0)(1) + 3(0)^2 \frac{\mathrm{d}y}{\mathrm{d}x} - 8\mathrm{e}^{0} = 0 \]
\[ 3\frac{\mathrm{d}y}{\mathrm{d}x} - 8 = 0 \implies \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{8}{3}. \]

(b) Differentiating implicitly with respect to \(x\) again:
\[ \frac{\mathrm{d}}{\mathrm{d}x}\left(3y^2 \frac{\mathrm{d}y}{\mathrm{d}x}\right) = 6y\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 + 3y^2 \frac{\mathrm{d}^2y}{\mathrm{d}x^2}, \]
\[ \frac{\mathrm{d}}{\mathrm{d}x}(6xy) = 6y + 6x\frac{\mathrm{d}y}{\mathrm{d}x}, \]
\[ \frac{\mathrm{d}}{\mathrm{d}x}\left(3x^2 \frac{\mathrm{d}y}{\mathrm{d}x}\right) = 6x\frac{\mathrm{d}y}{\mathrm{d}x} + 3x^2 \frac{\mathrm{d}^2y}{\mathrm{d}x^2}, \]
\[ \frac{\mathrm{d}}{\mathrm{d}x}\left(-8\mathrm{e}^{2x}\right) = -16\mathrm{e}^{2x}. \]
Combining all terms gives:
\[ 6y\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 + 3(y^2 + x^2)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 6y + 12x\frac{\mathrm{d}y}{\mathrm{d}x} - 16\mathrm{e}^{2x} = 0. \]
Substituting \(x = 0\), \(y = 1\), and \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{8}{3}\):
\[ 6(1)\left(\frac{8}{3}\right)^2 + 3(1^2 + 0^2)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 6(1) + 12(0)\left(\frac{8}{3}\right) - 16\mathrm{e}^0 = 0 \]
\[ 6\left(\frac{64}{9}\right) + 3\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 6 - 16 = 0 \]
\[ \frac{128}{3} - 10 + 3\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 0 \]
\[ \frac{98}{3} + 3\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 0 \implies \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -\frac{98}{9}. \]

(c) Using the Maclaurin series formula:
\[ y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \dots \]
\[ y = 1 + \frac{8}{3}x + \frac{1}{2}\left(-\frac{98}{9}\right)x^2 = 1 + \frac{8}{3}x - \frac{49}{9}x^2. \]

(a)
- B1: Find \(y = 1\) when \(x = 0\).
- M1: Differentiate implicitly with respect to \(x\) (at least two terms correct including product rule on \(3x^2y\)).
- A1: Obtain \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{8}{3}\) (oe exact form).

(b)
- M1: Differentiate implicitly again with respect to \(x\), applying the product/chain rule to terms involving \(\frac{\mathrm{d}y}{\mathrm{d}x}\).
- A1: Correct second derivative expression (or correctly differentiated equation before rearrangement).
- M1: Substitute \(x = 0\), \(y = 1\), and their \(\frac{\mathrm{d}y}{\mathrm{d}x}\) into the second derivative equation.
- A1: Obtain \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -\frac{98}{9}\) (or \(-10\frac{8}{9}\)).

(c)
- B1 FT: Correct Maclaurin series up to \(x^2\): \(1 + \frac{8}{3}x - \frac{49}{9}x^2\), follow through on their \(y'(0)\) and \(y''(0)\).

(a)
Since \( f(x) = \dfrac{1}{x^2 + 4} \) is strictly decreasing for \( x > 0 \), for any integer \( r \geqslant 1 \) and for \( x \in [r, r+1] \):
\[ f(r+1) < f(x) < f(r) \]
Integrating across the interval \([r, r+1]\) of width 1:
\[ \int_{r}^{r+1} f(r+1) \, \mathrm{d}x < \int_{r}^{r+1} f(x) \, \mathrm{d}x < \int_{r}^{r+1} f(r) \, \mathrm{d}x \]
\[ \frac{1}{(r+1)^2 + 4} < \int_{r}^{r+1} \frac{1}{x^2 + 4} \, \mathrm{d}x < \frac{1}{r^2 + 4} \]

Summing the right-hand inequality \( \int_{r}^{r+1} \frac{1}{x^2 + 4} \, \mathrm{d}x < \frac{1}{r^2 + 4} \) from \( r = 1 \) to \( n \):
\[ \sum_{r=1}^{n} \int_{r}^{r+1} \frac{1}{x^2 + 4} \, \mathrm{d}x < \sum_{r=1}^{n} \frac{1}{r^2 + 4} \implies \int_{1}^{n+1} \frac{1}{x^2 + 4} \, \mathrm{d}x < \sum_{r=1}^{n} \frac{1}{r^2 + 4} \]

Summing the left-hand inequality \( \frac{1}{(r+1)^2 + 4} < \int_{r}^{r+1} \frac{1}{x^2 + 4} \, \mathrm{d}x \) from \( r = 1 \) to \( n - 1 \):
\[ \sum_{r=1}^{n-1} \frac{1}{(r+1)^2 + 4} < \sum_{r=1}^{n-1} \int_{r}^{r+1} \frac{1}{x^2 + 4} \, \mathrm{d}x \implies \sum_{k=2}^{n} \frac{1}{k^2 + 4} < \int_{1}^{n} \frac{1}{x^2 + 4} \, \mathrm{d}x \]
Adding \( f(1) = \dfrac{1}{1^2 + 4} = \dfrac{1}{5} \) to both sides:
\[ \sum_{r=1}^{n} \frac{1}{r^2 + 4} < \frac{1}{5} + \int_{1}^{n} \frac{1}{x^2 + 4} \, \mathrm{d}x \]
Combining both results yields the required inequality:
\[ \int_{1}^{n+1} \frac{1}{x^2 + 4} \, \mathrm{d}x < \sum_{r=1}^{n} \frac{1}{r^2 + 4} < \frac{1}{5} + \int_{1}^{n} \frac{1}{x^2 + 4} \, \mathrm{d}x \]

(b)
Evaluate the integral:
\[ \int_{1}^{n} \frac{1}{x^2 + 4} \, \mathrm{d}x = \left[ \frac{1}{2} \arctan\left(\frac{x}{2}\right) \right]_1^n = \frac{1}{2}\arctan\left(\frac{n}{2}\right) - \frac{1}{2}\arctan\left(\frac{1}{2}\right) \]
Taking the limit as \( n \to \infty \):
\[ \lim_{n \to \infty} \int_{1}^{n} \frac{1}{x^2 + 4} \, \mathrm{d}x = \frac{1}{2}\left(\frac{\pi}{2}\right) - \frac{1}{2}\arctan\left(\frac{1}{2}\right) = \frac{\pi}{4} - \frac{1}{2}\arctan\left(\frac{1}{2}\right) \]
Since \( \lim_{n \to \infty} \int_{1}^{n+1} \frac{1}{x^2 + 4} \, \mathrm{d}x = \frac{\pi}{4} - \frac{1}{2}\arctan\left(\frac{1}{2}\right) \), letting \( S = \sum_{r=1}^{\infty} \frac{1}{r^2 + 4} \):
\[ \frac{\pi}{4} - \frac{1}{2}\arctan\left(\frac{1}{2}\right) < S \leqslant \frac{1}{5} + \frac{\pi}{4} - \frac{1}{2}\arctan\left(\frac{1}{2}\right) \]
(or with strict upper inequality \( < \)).

(c)
Rewrite the general term in terms of \( \frac{r}{n} \):
\[ \sum_{r=1}^{n} \frac{n}{n^2 + 4r^2} = \sum_{r=1}^{n} \frac{n}{n^2\left(1 + 4\left(\frac{r}{n}\right)^2\right)} = \frac{1}{n} \sum_{r=1}^{n} \frac{1}{1 + 4\left(\frac{r}{n}\right)^2} \]
Taking the limit as \( n \to \infty \), this is the Riemann sum for the integral:
\[ \int_{0}^{1} \frac{1}{1 + 4x^2} \, \mathrm{d}x \]
Evaluating the integral:
\[ \int_{0}^{1} \frac{1}{1 + (2x)^2} \, \mathrm{d}x = \left[ \frac{1}{2}\arctan(2x) \right]_0^1 = \frac{1}{2}\arctan(2) - 0 = \frac{1}{2}\arctan(2) \]

(a)
- M1: Uses areas of rectangles or sets up interval inequality \( f(r+1) < f(x) < f(r) \) for decreasing function \( f(x) \).
- A1: Correctly shows the lower bound \( \int_{1}^{n+1} \frac{1}{x^2+4}\,\mathrm{d}x < \sum_{r=1}^{n} \frac{1}{r^2+4} \) with clear summation limits.
- M1: Sets up the upper sum for \( \sum_{r=2}^n \frac{1}{r^2+4} < \int_1^n \frac{1}{x^2+4}\,\mathrm{d}x \) and adds \( f(1) = \frac{1}{5} \).
- A1: Fully correct intermediate working leading convincingly to the given double inequality (AG).

(b)
- M1: Integrates correctly to obtain \( \left[ \frac{1}{2}\arctan\left(\frac{x}{2}\right) \right] \) and applies limits \( 1 \) and \( n \).
- A1: Obtains exact limit \( \frac{\pi}{4} - \frac{1}{2}\arctan\left(\frac{1}{2}\right) \).
- A1: States correct bounds for \( S \) using the limit obtained (allow either strict or non-strict inequalities).

(c)
- M1: Expresses the sum in the form \( \frac{1}{n}\sum f\left(\frac{r}{n}\right) \) with \( f(x) = \frac{1}{1+4x^2} \).
- M1: Identifies the integral as \( \int_{0}^{1} \frac{1}{1+4x^2} \, \mathrm{d}x \) and integrates to \( \left[ \frac{1}{2}\arctan(2x) \right]_0^1 \).
- A1: Obtains exact value \( \frac{1}{2}\arctan(2) \) (or equivalent single arctan expression).

(a)
First, find the complementary function by setting the homogeneous equation to zero:
\[ m^2 + 4m + 5 = 0 \]
Solving the auxiliary equation:
\[ (m + 2)^2 + 1 = 0 \implies m = -2 \pm \mathrm{i} \]
Thus, the complementary function is:
\[ y_{\text{CF}} = \mathrm{e}^{-2x}(A\cos x + B\sin x) \]

For the particular integral, try:
\[ y = p\cos 2x + q\sin 2x \]
\[ \frac{\mathrm{d} y}{\mathrm{d} x} = -2p\sin 2x + 2q\cos 2x \]
\[ \frac{\mathrm{d}^2 y}{\mathrm{d} x^2} = -4p\cos 2x - 4q\sin 2x \]

Substitute these into the differential equation:
\[ (-4p\cos 2x - 4q\sin 2x) + 4(-2p\sin 2x + 2q\cos 2x) + 5(p\cos 2x + q\sin 2x) = -15\cos 2x - 10\sin 2x \]
\[ (p + 8q)\cos 2x + (-8p + q)\sin 2x = -15\cos 2x - 10\sin 2x \]

Equating coefficients:
\[ p + 8q = -15 \]
\[ -8p + q = -10 \]

From the second equation, \( q = 8p - 10 \). Substituting into the first equation:
\[ p + 8(8p - 10) = -15 \implies 65p - 80 = -15 \implies 65p = 65 \implies p = 1 \]
\[ q = 8(1) - 10 = -2 \]

Thus, the particular integral is \( y_{\text{PI}} = \cos 2x - 2\sin 2x \).

The general solution is:
\[ y = \mathrm{e}^{-2x}(A\cos x + B\sin x) + \cos 2x - 2\sin 2x \]

(b)
Applying initial conditions:
When \( x = 0 \), \( y = 2 \):
\[ 2 = \mathrm{e}^{0}(A\cos 0 + B\sin 0) + \cos 0 - 2\sin 0 \implies 2 = A + 1 \implies A = 1 \]

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

When \( x = 0 \), \( \frac{\mathrm{d} y}{\mathrm{d} x} = 0 \):
\[ 0 = -2(A) + B - 4 \]
Using \( A = 1 \):
\[ 0 = -2(1) + B - 4 \implies B = 6 \]

Hence, the particular solution is:
\[ y = \mathrm{e}^{-2x}(\cos x + 6\sin x) + \cos 2x - 2\sin 2x \]

(a)
- M1: State and solve the auxiliary equation \( m^2 + 4m + 5 = 0 \).
- A1: Correct roots \( m = -2 \pm \mathrm{i} \) and complementary function \( y_{\text{CF}} = \mathrm{e}^{-2x}(A\cos x + B\sin x) \).
- B1: State correct trial form for particular integral \( y = p\cos 2x + q\sin 2x \).
- M1: Differentiate twice and substitute into the differential equation.
- M1: Equate coefficients of \( \cos 2x \) and \( \sin 2x \) to form a pair of simultaneous equations.
- A1: Obtain \( p = 1 \) and \( q = -2 \).
- A1: State correct general solution \( y = y_{\text{CF}} + y_{\text{PI}} \).

(b)
- M1: Use \( x = 0, y = 2 \) in general solution to find \( A \).
- M1: Differentiate general solution with product rule and use \( x = 0, y' = 0 \) to find \( B \).
- A1: Correct particular solution \( y = \mathrm{e}^{-2x}(\cos x + 6\sin x) + \cos 2x - 2\sin 2x \) (CWO).

(a) Since \(\omega^7 = 1\) and \(z^7 - 1 = (z - 1)(1 + z + z^2 + z^3 + z^4 + z^5 + z^6)\), and \(\omega \ne 1\), we have:
\[ 1 + \omega + \omega^2 + \omega^3 + \omega^4 + \omega^5 + \omega^6 = 0. \]

(b) Sum of \(\alpha\) and \(\beta\):
\[ \alpha + \beta = \omega + \omega^2 + \omega^4 + \omega^3 + \omega^5 + \omega^6 = \sum_{k=1}^6 \omega^k = -1. \]
Product of \(\alpha\) and \(\beta\):
\[ \begin{aligned} \alpha\beta &= (\omega + \omega^2 + \omega^4)(\omega^3 + \omega^5 + \omega^6) \\ &= \omega(\omega^3 + \omega^5 + \omega^6) + \omega^2(\omega^3 + \omega^5 + \omega^6) + \omega^4(\omega^3 + \omega^5 + \omega^6) \\ &= (\omega^4 + \omega^6 + \omega^7) + (\omega^5 + \omega^7 + \omega^8) + (\omega^7 + \omega^9 + \omega^{10}). \end{aligned} \]
Using \(\omega^7 = 1\), we simplify the powers:
\[ \omega^8 = \omega, \quad \omega^9 = \omega^2, \quad \omega^{10} = \omega^3. \]
Thus,
\[ \alpha\beta = (\omega^4 + \omega^6 + 1) + (\omega^5 + 1 + \omega) + (1 + \omega^2 + \omega^3) = 3 + (\omega + \omega^2 + \omega^3 + \omega^4 + \omega^5 + \omega^6). \]
Since \(\sum_{k=1}^6 \omega^k = -1\),
\[ \alpha\beta = 3 + (-1) = 2. \]

(c) A quadratic equation with roots \(\alpha\) and \(\beta\) is given by:
\[ z^2 - (\alpha + \beta)z + \alpha\beta = 0 \implies z^2 - (-1)z + 2 = 0 \implies z^2 + z + 2 = 0. \]

(d) Solving the quadratic equation \(z^2 + z + 2 = 0\):
\[ z = \frac{-1 \pm \sqrt{1 - 4(1)(2)}}{2} = \frac{-1 \pm \mathrm{i}\sqrt{7}}{2}. \]
Using Euler's formula, \(\omega^k = \cos\left(\frac{2k\pi}{7}\right) + \mathrm{i}\sin\left(\frac{2k\pi}{7}\right)\), the imaginary part of \(\alpha\) is:
\[ \operatorname{Im}(\alpha) = \sin\left(\frac{2\pi}{7}\right) + \sin\left(\frac{4\pi}{7}\right) + \sin\left(\frac{8\pi}{7}\right). \]
Since \(\sin\left(\frac{8\pi}{7}\right) = \sin\left(\pi + \frac{\pi}{7}\right) = -\sin\left(\frac{\pi}{7}\right)\), we have:
\[ \operatorname{Im}(\alpha) = \sin\left(\frac{2\pi}{7}\right) + \sin\left(\frac{4\pi}{7}\right) - \sin\left(\frac{\pi}{7}\right). \]
Since \(\frac{2\pi}{7}, \frac{4\pi}{7} \in (0, \pi)\), \(\sin\left(\frac{2\pi}{7}\right) > 0\) and \(\sin\left(\frac{4\pi}{7}\right) = \sin\left(\frac{3\pi}{7}\right) > \sin\left(\frac{\pi}{7}\right) > 0\), so \(\operatorname{Im}(\alpha) > 0\).
Therefore, \(\alpha = \frac{-1 + \mathrm{i}\sqrt{7}}{2}\), which gives:
\[ \sin\left(\frac{2\pi}{7}\right) + \sin\left(\frac{4\pi}{7}\right) - \sin\left(\frac{\pi}{7}\right) = \operatorname{Im}(\alpha) = \frac{\sqrt{7}}{2}. \]

(a) B1: For factorising \(z^7 - 1 = (z - 1)(1 + z + \dots + z^6)\) or using the sum of a GP with \(\omega \ne 1\) to establish the given result.

(b) B1: For correctly stating \(\alpha + \beta = \sum_{k=1}^6 \omega^k = -1\).
M1: For expanding the product \(\alpha\beta\) to obtain 9 terms.
M1: For reducing powers of \(\omega\) using \(\omega^7 = 1\) to express the product in terms of \(3 + \sum_{k=1}^6 \omega^k\).
A1: For fully correct working leading to \(\alpha\beta = 2\) (AG).

(c) M1: For using \(z^2 - (\text{sum})z + (\text{product}) = 0\).
A1: For \(z^2 + z + 2 = 0\) (or any integer multiple thereof).

(d) M1: For solving the quadratic equation to obtain roots \(\frac{-1 \pm \mathrm{i}\sqrt{7}}{2}\).
M1: For considering \(\operatorname{Im}(\alpha)\) and simplifying \(\sin\left(\frac{8\pi}{7}\right) = -\sin\left(\frac{\pi}{7}\right)\).
A1: For justifying \(\operatorname{Im}(\alpha) > 0\) and concluding \(\sin\left(\frac{2\pi}{7}\right) + \sin\left(\frac{4\pi}{7}\right) - \sin\left(\frac{\pi}{7}\right) = \frac{\sqrt{7}}{2}\) correctly.

(a)
To find the eigenvalues of \(\mathbf{A}\), solve the characteristic equation \(\det(\lambda\mathbf{I} - \mathbf{A}) = 0\):
\[ \det \begin{pmatrix} \lambda - 2 & -1 & 1 \\ -3 & \lambda & 1 \\ -3 & 1 & \lambda \end{pmatrix} = 0 \]
Expanding along the first row:
\[ (\lambda - 2)(\lambda^2 - 1) - (-1)(-3\lambda - (-3)) + 1(-3 - (-3\lambda)) = 0 \]
\[ (\lambda - 2)(\lambda^2 - 1) + (-3\lambda + 3) + (3\lambda - 3) = 0 \]
\[ (\lambda - 2)(\lambda - 1)(\lambda + 1) = 0 \]
\[ \lambda^3 - 2\lambda^2 - \lambda + 2 = 0 \]
Thus, the eigenvalues are \(\lambda_1 = 1\), \(\lambda_2 = -1\), \(\lambda_3 = 2\).

(b)
For \(\lambda_1 = 1\):
\[ (\mathbf{A} - \mathbf{I})\mathbf{v} = \begin{pmatrix} 1 & 1 & -1 \\ 3 & -1 & -1 \\ 3 & -1 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \]
From \(x + y - z = 0\) and \(3x - y - z = 0\), adding the two gives \(4x - 2z = 0 \implies z = 2x\).
Then \(y = z - x = x\). Choosing \(x = 1\), an eigenvector is \(\mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}\).

For \(\lambda_2 = -1\):
\[ (\mathbf{A} + \mathbf{I})\mathbf{v} = \begin{pmatrix} 3 & 1 & -1 \\ 3 & 1 & -1 \\ 3 & -1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \]
From \(3x + y - z = 0\) and \(3x - y + z = 0\), adding gives \(6x = 0 \implies x = 0\).
Then \(y = z\). Choosing \(y = 1\), an eigenvector is \(\mathbf{v}_2 = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\).

For \(\lambda_3 = 2\):
\[ (\mathbf{A} - 2\mathbf{I})\mathbf{v} = \begin{pmatrix} 0 & 1 & -1 \\ 3 & -2 & -1 \\ 3 & -1 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \]
From the first equation, \(y = z\). Substituting into \(3x - 2y - z = 0\) gives \(3x - 3y = 0 \implies x = y\).
Choosing \(x = 1\), an eigenvector is \(\mathbf{v}_3 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\).

(c)
Let \(\mathbf{P} = \begin{pmatrix} 1 & 0 & 1 \\ 1 & 1 & 1 \\ 2 & 1 & 1 \end{pmatrix}\) and \(\mathbf{D} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 2 \end{pmatrix}\).

Find \(\mathbf{P}^{-1}\):
\(\det(\mathbf{P}) = 1(1 - 1) - 0(1 - 2) + 1(1 - 2) = -1\).
The matrix of cofactors is:
\[ \mathbf{C} = \begin{pmatrix} 0 & 1 & -1 \\ 1 & -1 & -1 \\ -1 & 0 & 1 \end{pmatrix} \]
\[ \mathbf{P}^{-1} = \frac{1}{-1}\mathbf{C}^{\mathrm{T}} = \begin{pmatrix} 0 & -1 & 1 \\ -1 & 1 & 0 \\ 1 & 1 & -1 \end{pmatrix} \]
Then \(\mathbf{A}^n = \mathbf{P}\mathbf{D}^n\mathbf{P}^{-1}\):
\[ \mathbf{A}^n = \begin{pmatrix} 1 & 0 & 1 \\ 1 & 1 & 1 \\ 2 & 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & (-1)^n & 0 \\ 0 & 0 & 2^n \end{pmatrix} \begin{pmatrix} 0 & -1 & 1 \\ -1 & 1 & 0 \\ 1 & 1 & -1 \end{pmatrix} \]
\[ \mathbf{P}\mathbf{D}^n = \begin{pmatrix} 1 & 0 & 2^n \\ 1 & (-1)^n & 2^n \\ 2 & (-1)^n & 2^n \end{pmatrix} \]
Multiplying by \(\mathbf{P}^{-1}\):
\[ \mathbf{A}^n = \begin{pmatrix} 1 & 0 & 2^n \\ 1 & (-1)^n & 2^n \\ 2 & (-1)^n & 2^n \end{pmatrix} \begin{pmatrix} 0 & -1 & 1 \\ -1 & 1 & 0 \\ 1 & 1 & -1 \end{pmatrix} \]
\[ \mathbf{A}^n = \begin{pmatrix} 2^n & 2^n - 1 & 1 - 2^n \\ 2^n - (-1)^n & 2^n + (-1)^n - 1 & 1 - 2^n \\ 2^n - (-1)^n & 2^n + (-1)^n - 2 & 2 - 2^n \end{pmatrix} \]

(d)
By the Cayley-Hamilton theorem, \(\mathbf{A}\) satisfies its characteristic equation:
\[ \mathbf{A}^3 - 2\mathbf{A}^2 - \mathbf{A} + 2\mathbf{I} = \mathbf{0} \]
\[ \implies \mathbf{A}^3 = 2\mathbf{A}^2 + \mathbf{A} - 2\mathbf{I} \]
Multiplying both sides by \(\mathbf{A}\):
\[ \mathbf{A}^4 = 2\mathbf{A}^3 + \mathbf{A}^2 - 2\mathbf{A} \]
Substitute \(\mathbf{A}^3\) into the expression:
\[ \mathbf{A}^4 = 2(2\mathbf{A}^2 + \mathbf{A} - 2\mathbf{I}) + \mathbf{A}^2 - 2\mathbf{A} = 4\mathbf{A}^2 + 2\mathbf{A} - 4\mathbf{I} + \mathbf{A}^2 - 2\mathbf{A} = 5\mathbf{A}^2 - 4\mathbf{I} \]
To find \(\mathbf{A}^4\), either use \(\mathbf{A}^n\) with \(n=4\) or calculate \(5\mathbf{A}^2 - 4\mathbf{I}\):
\[ \mathbf{A}^2 = \begin{pmatrix} 2 & 1 & -1 \\ 3 & 0 & -1 \\ 3 & -1 & 0 \end{pmatrix}\begin{pmatrix} 2 & 1 & -1 \\ 3 & 0 & -1 \\ 3 & -1 & 0 \end{pmatrix} = \begin{pmatrix} 4 & 3 & -3 \\ 3 & 4 & -3 \\ 3 & 3 & -2 \end{pmatrix} \]
\[ \mathbf{A}^4 = 5\begin{pmatrix} 4 & 3 & -3 \\ 3 & 4 & -3 \\ 3 & 3 & -2 \end{pmatrix} - 4\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 16 & 15 & -15 \\ 15 & 16 & -15 \\ 15 & 15 & -14 \end{pmatrix} \]

(a)
- M1: Sets up characteristic determinant \(\det(\lambda\mathbf{I} - \mathbf{A}) = 0\) and attempts expansion.
- A1: Obtains correct cubic equation \(\lambda^3 - 2\lambda^2 - \lambda + 2 = 0\).
- M1: Factorises cubic equation to find the roots.
- A1: Correct eigenvalues \(\lambda = 1, -1, 2\).

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

(c)
- B1: Writes down compatible \(\mathbf{P}\) and \(\mathbf{D}\) (order of columns in \(\mathbf{P}\) matching order of eigenvalues in \(\mathbf{D}\)).
- M1: Correct method to find \(\mathbf{P}^{-1}\) (determinant + adjugate/cofactors).
- M1: Multiplies \(\mathbf{P}\mathbf{D}^n\mathbf{P}^{-1}\) algebraically in terms of \(n\).
- A1: Fully correct intermediate matrix multiplication leading convincingly to the given result (AG).

(d)
- M1: States the Cayley-Hamilton equation \(\mathbf{A}^3 - 2\mathbf{A}^2 - \mathbf{A} + 2\mathbf{I} = \mathbf{0}\) and multiplies by \(\mathbf{A}\).
- A1: Replaces \(\mathbf{A}^3\) to show \(\mathbf{A}^4 = 5\mathbf{A}^2 - 4\mathbf{I}\) clearly with no missing steps.
- A1: Correct numerical matrix \(\mathbf{A}^4\).

(a) Let \(\rho\) be the uniform density of the material.

For the solid cone:
\[ \text{Volume } V_1 = \frac{1}{3}\pi a^2 (3a) = \pi a^3 \]
The centre of mass of a solid cone of height \(h\) is at a distance \(\frac{1}{4}h\) from its base. Thus, its distance from the base is:
\[ y_1 = \frac{1}{4}(3a) = \frac{3}{4}a \]

For the solid hemisphere:
\[ \text{Volume } V_2 = \frac{2}{3}\pi a^3 \]
The centre of mass of a solid hemisphere of radius \(a\) is at a distance \(\frac{3}{8}a\) from its base. Measuring upwards from the base:
\[ y_2 = -\frac{3}{8}a \]

Total volume of the composite body:
\[ V = V_1 + V_2 = \pi a^3 + \frac{2}{3}\pi a^3 = \frac{5}{3}\pi a^3 \]

Taking moments about the plane of the base:
\[ V \bar{y} = V_1 y_1 + V_2 y_2 \]
\[ \left(\frac{5}{3}\pi a^3\right) \bar{y} = (\pi a^3)\left(\frac{3}{4}a\right) + \left(\frac{2}{3}\pi a^3\right)\left(-\frac{3}{8}a\right) \]
\[ \frac{5}{3}\bar{y} = \frac{3}{4}a - \frac{1}{4}a = \frac{1}{2}a \]
\[ \bar{y} = \frac{3}{5} \times \frac{1}{2}a = \frac{3}{10}a \]

(b) Let \(O\) be the centre of the common circular base and \(G\) be the centre of mass of the composite body.
\(G\) lies on the axis of symmetry of the body inside the cone, so \(OG = \bar{y} = \frac{3}{10}a\).

Since \(P\) is on the circumference of the common base, the line segment \(OP\) lies in the plane of the base and has length \(OP = a\).

The axis of symmetry is perpendicular to the plane of the base, so \(\angle POG = 90^\circ\).

When the body is suspended freely from \(P\), the line \(PG\) is vertical in equilibrium.

Let \(\alpha\) be the angle that the plane of the base makes with the vertical. The line \(OP\) lies in the base, so \(\alpha = \angle OPG\).

In the right-angled triangle \(POG\):
\[ \tan \alpha = \frac{OG}{OP} = \frac{\frac{3}{10}a}{a} = \frac{3}{10} = 0.3 \]
\[ \alpha = \arctan(0.3) \approx 16.699^\circ = 16.7^\circ \]

(a)
- M1: Calculates the volumes (or masses) of the cone and hemisphere, obtaining ratio \(3 : 2\) or \(\pi a^3\) and \(\frac{2}{3}\pi a^3\).
- B1: States correct positions of the individual centres of mass from the base: \(\frac{3}{4}a\) (cone) and \(\frac{3}{8}a\) (hemisphere).
- M1: Forms a correct moments equation about the plane of the base using their volumes and centres of mass.
- A1 (AG): Simplifies correctly to obtain \(\bar{y} = \frac{3}{10}a\) with no omitted steps.

(b)
- M1: Recognises that the line joining the suspension point \(P\) to the centre of mass \(G\) is vertical.
- M1: Uses trigonometry on right-angled triangle \(POG\) to relate the angle to \(OG\) and \(OP\), e.g., \(\tan \alpha = \frac{OG}{OP}\) or \(\tan\theta = \frac{OP}{OG}\).
- A1: Obtains \(16.7^\circ\) (or \(0.291\) rad). (Accept 16.7, allow rounding from 16.699...)

1. Determine the geometry of the system:
In \(\triangle ABP\), the side lengths are \(AP = 1.0\text{ m}\), \(BP = 0.75\text{ m}\), and \(AB = 1.25\text{ m}\).
Since \(1.0^2 + 0.75^2 = 1 + 0.5625 = 1.5625 = 1.25^2\), \(\triangle ABP\) is a right-angled triangle with \(\angle APB = 90^\circ\).

Let \(C\) be the centre of the circular path on \(AB\), and let \(r\) be the radius of the circle (the perpendicular distance from \(P\) to \(AB\)):
\[ \text{Area}(\triangle ABP) = \frac{1}{2} \times 1.0 \times 0.75 = \frac{1}{2} \times 1.25 \times r \implies r = \frac{0.75}{1.25} = 0.6\text{ m} \]

Let \(\theta\) be the angle \(AP\) makes with the vertical \(AB\), and \(\phi\) be the angle \(BP\) makes with the vertical \(AB\):
\[ \cos\theta = \frac{1.0}{1.25} = 0.8, \quad \sin\theta = \frac{0.6}{1.0} = 0.6 \]
\[ \cos\phi = \frac{0.75}{1.25} = 0.6, \quad \sin\phi = \frac{0.6}{0.75} = 0.8 \]

2. Resolve vertically for equilibrium:
Let \(T_A\) and \(T_B\) be the tensions in \(AP\) and \(BP\) respectively.
\[ T_A \cos\theta - T_B \cos\phi - mg = 0 \]
\[ 0.8 T_A - 0.6 T_B = 0.5 \times 10 = 5 \implies 4 T_A - 3 T_B = 25 \quad \text{--- (1)} \]

3. **Apply Newton's Second Law horizontally towards the centre \(C\):**
\[ T_A \sin\theta + T_B \sin\phi = m r \omega^2 \]
\[ 0.6 T_A + 0.8 T_B = 0.5 \times 0.6 \times 5^2 = 7.5 \implies 3 T_A + 4 T_B = 37.5 \quad \text{--- (2)} \]

4. Solve the simultaneous equations:
Multiplying (1) by 3 and (2) by 4:
\[ 12 T_A - 9 T_B = 75 \]
\[ 12 T_A + 16 T_B = 150 \]
Subtracting the first from the second:
\[ 25 T_B = 75 \implies T_B = 3.0\text{ N} \]

M1: Valid method to find the radius of the horizontal circle \(r = 0.6\text{ m}\) and correct trigonometric ratios for the angles with the vertical.
A1: Correct vertical equation of equilibrium \(0.8 T_A - 0.6 T_B = 5\) (or equivalent).
M1: Correct structure for horizontal equation of motion with centripetal acceleration: \(T_A \sin\theta + T_B \sin\phi = m r \omega^2\).
A1: Correct horizontal equation \(0.6 T_A + 0.8 T_B = 7.5\) (or equivalent).
A1: Correct value for tension \(T_B = 3.0\text{ N}\).

(a) For the upward motion, measuring displacement \( x \) vertically upwards from \( O \):
\[ m v \frac{\mathrm{d}v}{\mathrm{d}x} = -mg - mkv^2 \implies v \frac{\mathrm{d}v}{\mathrm{d}x} = -(g + kv^2) \]
Separating variables:
\[ \int_0^H \mathrm{d}x = -\int_u^0 \frac{v}{g + kv^2} \,\mathrm{d}v = \int_0^u \frac{v}{g + kv^2} \,\mathrm{d}v \]
Integrating:
\[ H = \left[ \frac{1}{2k} \ln(g + kv^2) \right]_0^u = \frac{1}{2k} \left( \ln(g + ku^2) - \ln g \right) = \frac{1}{2k} \ln\left( 1 + \frac{ku^2}{g} \right) \]

(b) For the downward motion, measuring displacement \( x \) vertically downwards from the highest point:
\[ m v \frac{\mathrm{d}v}{\mathrm{d}x} = mg - mkv^2 \implies v \frac{\mathrm{d}v}{\mathrm{d}x} = g - kv^2 \]
Separating variables and integrating from \( x = 0 \) (\( v = 0 \)) to \( x = H \) (\( v = w \)):
\[ \int_0^w \frac{v}{g - kv^2} \,\mathrm{d}v = \int_0^H \mathrm{d}x = H \]
\[ \left[ -\frac{1}{2k} \ln(g - kv^2) \right]_0^w = H \implies -\frac{1}{2k} \ln\left( 1 - \frac{kw^2}{g} \right) = \frac{1}{2k} \ln\left( 1 + \frac{ku^2}{g} \right) \]
\[ \ln\left( 1 - \frac{kw^2}{g} \right)^{-1} = \ln\left( 1 + \frac{ku^2}{g} \right) \implies 1 - \frac{kw^2}{g} = \frac{1}{1 + \frac{ku^2}{g}} = \frac{g}{g + ku^2} \]
\[ \frac{kw^2}{g} = 1 - \frac{g}{g + ku^2} = \frac{ku^2}{g + ku^2} \implies w^2 = \frac{gu^2}{g + ku^2} \]
\[ w = u \sqrt{\frac{g}{g + ku^2}} = \frac{u}{\sqrt{1 + \frac{ku^2}{g}}} \]

(c) For the upward motion, let \( T_{\text{up}} \) be the time taken:
\[ \frac{\mathrm{d}v}{\mathrm{d}t} = -(g + kv^2) \implies T_{\text{up}} = \int_0^u \frac{\mathrm{d}v}{g + kv^2} = \frac{1}{\sqrt{gk}} \left[ \arctan\left( v \sqrt{\frac{k}{g}} \right) \right]_0^u = \frac{1}{\sqrt{gk}} \arctan\left( u \sqrt{\frac{k}{g}} \right) \]
Given \( k = \frac{g}{3u^2} \), we have \( u \sqrt{\frac{k}{g}} = \frac{1}{\sqrt{3}} \):
\[ T_{\text{up}} = \frac{1}{\sqrt{gk}} \arctan\left( \frac{1}{\sqrt{3}} \right) = \frac{\pi}{6\sqrt{gk}} \]

For the downward motion, let \( T_{\text{down}} \) be the time taken:
\[ \frac{\mathrm{d}v}{\mathrm{d}t} = g - kv^2 \implies T_{\text{down}} = \int_0^w \frac{\mathrm{d}v}{g - kv^2} \]
When \( k = \frac{g}{3u^2} \), \( w = u \sqrt{\frac{g}{g + g/3}} = \frac{\sqrt{3}}{2} u \), so \( w \sqrt{\frac{k}{g}} = \frac{\sqrt{3}}{2} u \cdot \frac{1}{\sqrt{3}u} = \frac{1}{2} \).
\[ T_{\text{down}} = \frac{1}{\sqrt{gk}} \int_0^{1/2} \frac{\mathrm{d}y}{1 - y^2} = \frac{1}{2\sqrt{gk}} \left[ \ln\left( \frac{1+y}{1-y} \right) \right]_0^{1/2} = \frac{1}{2\sqrt{gk}} \ln\left( \frac{3/2}{1/2} \right) = \frac{\ln 3}{2\sqrt{gk}} \]

The ratio is:
\[ \frac{T_{\text{down}}}{T_{\text{up}}} = \frac{\frac{\ln 3}{2\sqrt{gk}}}{\frac{\pi}{6\sqrt{gk}}} = \frac{3\ln 3}{\pi} \]

(a)
M1: Sets up equation of motion for upward journey using acceleration in the form \( v \frac{\mathrm{d}v}{\mathrm{d}x} \) with opposing gravity and resistance.
M1: Separates variables and correctly integrates to obtain a logarithmic form.
A1: Applies limits 0 to \( H \) and \( u \) to 0 (or evaluates constant of integration) to obtain the given expression for \( H \) convincingly. [3]

(b)
M1: Sets up equation of motion for downward journey \( v \frac{\mathrm{d}v}{\mathrm{d}x} = g - kv^2 \) and separates variables.
M1: Integrates correctly and equates to \( H \) (or substitutes \( H \) from part (a)).
A1: Correct expression for \( w \), e.g. \( w = u \sqrt{\frac{g}{g+ku^2}} \) or \( \frac{u}{\sqrt{1 + \frac{ku^2}{g}}} \). [3]

(c)
M1: Sets up \( \frac{\mathrm{d}v}{\mathrm{d}t} = -(g + kv^2) \), integrates to an arctan form, and finds \( T_{\text{up}} = \frac{\pi}{6\sqrt{gk}} \).
M1: Sets up \( \frac{\mathrm{d}v}{\mathrm{d}t} = g - kv^2 \), integrates to a logarithmic/artanh form, and finds \( T_{\text{down}} = \frac{\ln 3}{2\sqrt{gk}} \).
A1: Obtains exact ratio \( \frac{3\ln 3}{\pi} \) (or \( \frac{\ln 27}{\pi} \)). [3]

(a)
Given \(\tan \alpha = \frac{4}{3}\), we have \(\cos \alpha = \frac{3}{5}\) and \(\sin \alpha = \frac{4}{5}\).

Components of velocity before impact with \(W_1\):
- Parallel to \(W_1\): \(u_{\parallel} = u \cos \alpha = \frac{3}{5}u\)
- Perpendicular to \(W_1\): \(u_{\perp} = u \sin \alpha = \frac{4}{5}u\)

After impact with \(W_1\):
- Parallel component is unchanged: \(v_{1\parallel} = \frac{3}{5}u\)
- Perpendicular component: \(v_{1\perp} = e_1 u_{\perp} = \frac{1}{2}\left(\frac{4}{5}u\right) = \frac{2}{5}u\)

Speed after first impact:
\[ v_1 = \sqrt{\left(\frac{3}{5}u\right)^2 + \left(\frac{2}{5}u\right)^2} = \frac{u}{5}\sqrt{3^2 + 2^2} = \frac{\sqrt{13}}{5}u \approx 0.721u \]

Let \(\theta\) be the acute angle the path makes with \(W_1\):
\[ \tan \theta = \frac{v_{1\perp}}{v_{1\parallel}} = \frac{\frac{2}{5}u}{\frac{3}{5}u} = \frac{2}{3} \implies \theta = \arctan\left(\frac{2}{3}\right) \approx 33.7^\circ \]

(b)
Since \(W_2\) is perpendicular to \(W_1\), the velocity component parallel to \(W_1\) is perpendicular to \(W_2\), and the component perpendicular to \(W_1\) is parallel to \(W_2\).

Immediately before impact with \(W_2\):
- Component perpendicular to \(W_2\): \(\frac{3}{5}u\)
- Component parallel to \(W_2\): \(\frac{2}{5}u\)

Immediately after impact with \(W_2\):
- Component perpendicular to \(W_2\): \(e \left(\frac{3}{5}u\right) = \frac{3}{5}eu\)
- Component parallel to \(W_2\): \(\frac{2}{5}u\) (unchanged)

Speed after second impact \(v_2\):
\[ v_2 = \sqrt{\left(\frac{3}{5}eu\right)^2 + \left(\frac{2}{5}u\right)^2} = \sqrt{\frac{9e^2 u^2 + 4u^2}{25}} = \frac{u}{5}\sqrt{9e^2 + 4} \]

(c)
Initial kinetic energy:
\[ E_0 = \frac{1}{2}mu^2 \]
Final kinetic energy:
\[ E_2 = \frac{1}{2}m v_2^2 = \frac{1}{2}m \left(\frac{u^2}{25}(9e^2 + 4)\right) \]
Given that the loss in kinetic energy is \(\frac{3}{4}E_0\):
\[ E_2 = E_0 - \frac{3}{4}E_0 = \frac{1}{4}E_0 \]
\[ \frac{1}{2}m \frac{u^2}{25}(9e^2 + 4) = \frac{1}{4}\left(\frac{1}{2}mu^2\right) \]
\[ \frac{9e^2 + 4}{25} = \frac{1}{4} \implies 9e^2 + 4 = 6.25 \implies 9e^2 = 2.25 = \frac{9}{4} \]
\[ e^2 = \frac{1}{4} \implies e = \frac{1}{2} \quad (\text{since } 0 < e \le 1) \]

To find the deflection angle, choose unit vectors \(\mathbf{i}\) along \(W_1\) and \(\mathbf{j}\) perpendicular to \(W_1\) away from the wall:
- Initial velocity vector: \(\mathbf{u} = \frac{3}{5}u\,\mathbf{i} - \frac{4}{5}u\,\mathbf{j}\)
- Velocity vector after first impact: \(\mathbf{v}_1 = \frac{3}{5}u\,\mathbf{i} + \frac{2}{5}u\,\mathbf{j}\)
- Velocity vector after second impact (reversing and scaling the \(\mathbf{i}\)-component with \(e = \frac{1}{2}\)):
\[ \mathbf{v}_2 = -\left(\frac{1}{2} \cdot \frac{3}{5}u\right)\mathbf{i} + \frac{2}{5}u\,\mathbf{j} = -\frac{3}{10}u\,\mathbf{i} + \frac{4}{10}u\,\mathbf{j} = -\frac{1}{2}\left(\frac{3}{5}u\,\mathbf{i} - \frac{4}{5}u\,\mathbf{j}\right) = -\frac{1}{2}\mathbf{u} \]
Since \(\mathbf{v}_2\) is parallel to \(\mathbf{u}\) and in the opposite direction, the angle through which the direction of motion has been deflected is \(180^\circ\) (or \(\pi\) radians).

(a) [3 marks]
- M1: Resolves initial velocity into components parallel and perpendicular to \(W_1\) and applies Newton's experimental law perpendicular to \(W_1\).
- A1: Obtains correct speed \(\frac{\sqrt{13}}{5}u\) (or \(0.721u\)).
- A1: Obtains correct angle \(\theta = \arctan\left(\frac{2}{3}\right) \approx 33.7^\circ\) (or \(0.588\text{ rad}\)).

(b) [2 marks]
- M1: Recognises that components interchange roles for impact with perpendicular wall \(W_2\), applies restitution \(e\) to the normal component, and applies Pythagoras theorem to find the resultant speed.
- A1: Completes convincing algebraic steps to reach the given result \(\frac{u}{5}\sqrt{9e^2 + 4}\) correctly without error (AG).

(c) [3 marks]
- M1: Sets up kinetic energy relation \(v_2^2 = \frac{1}{4}u^2\) or \(\text{Loss} = \frac{3}{4}E_0\).
- A1: Solves correctly for \(e = \frac{1}{2}\).
- A1: Deduces that final velocity is anti-parallel to initial velocity (e.g. \(\mathbf{v}_2 = -\frac{1}{2}\mathbf{u}\)) and concludes the deflection angle is \(180^\circ\) (or \(\pi\) radians).

(a) Let the horizontal plane through the lowest point of the circular path be the reference level for gravitational potential energy.
When the string makes an angle \(\theta\) with the upward vertical, the height of \(P\) above the lowest point is \(h = a + a\cos\theta = a(1 + \cos\theta)\).
By conservation of mechanical energy:
\[ \frac{1}{2}m v^2 + mgh = \frac{1}{2}m u^2 \]
\[ \frac{1}{2}m v^2 + mga(1 + \cos\theta) = \frac{1}{2}m u^2 \]
Multiplying by \(\frac{2}{m}\):
\[ v^2 = u^2 - 2ga(1 + \cos\theta) \]

(b) Resolving forces along the string towards the centre \(O\):
\[ T + mg\cos\theta = \frac{mv^2}{a} \]
Substitute \(v^2 = u^2 - 2ga(1 + \cos\theta)\):
\[ T = \frac{m}{a}\left(u^2 - 2ga(1 + \cos\theta)\right) - mg\cos\theta \]
\[ T = \frac{m}{a}\left(u^2 - 2ga - 3ga\cos\theta\right) \]

(c)(i) The string becomes slack when \(T = 0\):
\[ u^2 - 2ga - 3ga\cos\theta = 0 \]
Given \(u^2 = \frac{7}{2}ga\):
\[ \frac{7}{2}ga - 2ga - 3ga\cos\theta = 0 \]
\[ \frac{3}{2}ga = 3ga\cos\theta \implies \cos\theta = \frac{1}{2} \]

(c)(ii) At the instant the string becomes slack, \(\cos\theta = \frac{1}{2}\) and \(\sin\theta = \frac{\sqrt{3}}{2}\).
The speed \(v_1\) of the particle at this point is:
\[ v_1^2 = \frac{7}{2}ga - 2ga\left(1 + \frac{1}{2}\right) = \frac{7}{2}ga - 3ga = \frac{1}{2}ga \]
The height of \(P\) above \(O\) at this instant is:
\[ y_0 = a\cos\theta = \frac{1}{2}a \]
The particle now moves freely under gravity as a projectile. The velocity vector is perpendicular to the string, making an angle \(\theta\) above the horizontal.
The vertical component of velocity is \(v_y = v_1\sin\theta\).
The additional vertical height \(h_{\text{add}}\) gained to the highest point of the trajectory is:
\[ h_{\text{add}} = \frac{v_y^2}{2g} = \frac{v_1^2\sin^2\theta}{2g} = \frac{\frac{1}{2}ga \cdot \left(\frac{\sqrt{3}}{2}\right)^2}{2g} = \frac{\frac{1}{2}ga \cdot \frac{3}{4}}{2g} = \frac{3}{16}a \]
Therefore, the maximum height above the level of \(O\) is:
\[ H_{\max} = y_0 + h_{\text{add}} = \frac{1}{2}a + \frac{3}{16}a = \frac{11}{16}a \]

(a)
M1: Writes down a valid conservation of energy equation involving kinetic energy and gravitational potential energy with correct height \(a(1 + \cos\theta)\).
A1: Obtains the given expression correctly with fully justified working (AG).

(b)
M1: Uses Newton's second law radially towards \(O\), \(T + mg\cos\theta = \frac{mv^2}{a}\).
A1: Substitutes \(v^2\) and simplifies to \(T = \frac{m}{a}(u^2 - 2ga - 3ga\cos\theta)\) or equivalent.

(c)(i)
M1: Sets \(T = 0\) and substitutes \(u^2 = \frac{7}{2}ga\).
A1: Obtains \(\cos\theta = \frac{1}{2}\).

(c)(ii)
M1: Finds the speed at slackness, \(v_1^2 = \frac{1}{2}ga\), and identifies the initial height above \(O\) as \(a\cos\theta = \frac{1}{2}a\).
M1: Calculates the additional vertical height gained using projectile motion, \(\frac{(v_1\sin\theta)^2}{2g}\) or energy conservation.
A1: Correct final height \(\frac{11}{16}a\) (or \(0.6875a\)).

(a) Let \(x\) be the extension of the string when \(P\) comes instantaneously to rest at its lowest point.

The total distance fallen by \(P\) from \(A\) is \(0.8 + x\).

By conservation of mechanical energy between \(A\) and the lowest point:
\[ \text{Loss in GPE} = \text{Gain in EPE} \]
\[ mg(0.8 + x) = \frac{\lambda x^2}{2l} \]
Substitute \(m = 0.8\), \(g = 10\), \(l = 0.8\), and \(\lambda = 32\):
\[ 0.8 \times 10 \times (0.8 + x) = \frac{32 x^2}{2(0.8)} \]
\[ 8(0.8 + x) = 20x^2 \]
\[ 6.4 + 8x = 20x^2 \implies 20x^2 - 8x - 6.4 = 0 \]
Multiplying by \(5\):
\[ 100x^2 - 40x - 32 = 0 \implies 25x^2 - 10x - 8 = 0 \]
Factoring the quadratic:
\[ (5x - 4)(5x + 2) = 0 \]
Since \(x > 0\), we have \(x = 0.8\text{ m}\).

Therefore, the greatest distance below \(A\) reached by \(P\) is:
\[ 0.8 + 0.8 = 1.6\text{ m} \]

(b) Maximum speed occurs when the acceleration is zero, which is at the equilibrium position where tension equals weight:
\[ T = mg \implies \frac{\lambda e}{l} = mg \]
\[ \frac{32 e}{0.8} = 0.8 \times 10 \implies 40e = 8 \implies e = 0.2\text{ m} \]
The distance fallen from \(A\) to the equilibrium position is \(h = 0.8 + 0.2 = 1.0\text{ m}\).

By conservation of energy between \(A\) and the equilibrium position:
\[ \text{Loss in GPE} = \text{Gain in KE} + \text{Gain in EPE} \]
\[ mgh = \frac{1}{2}mv_{\max}^2 + \frac{\lambda e^2}{2l} \]
\[ 0.8 \times 10 \times 1.0 = \frac{1}{2}(0.8)v_{\max}^2 + \frac{32(0.2)^2}{2(0.8)} \]
\[ 8 = 0.4 v_{\max}^2 + \frac{32 \times 0.04}{1.6} \]
\[ 8 = 0.4 v_{\max}^2 + 0.8 \]
\[ 0.4 v_{\max}^2 = 7.2 \implies v_{\max}^2 = 18 \]
\[ v_{\max} = \sqrt{18} = 3\sqrt{2} \approx 4.24\text{ m s}^{-1} \]

(a)
M1: Sets up an energy equation at the lowest point equating loss in GPE to elastic potential energy \(\frac{\lambda x^2}{2l}\).
A1: Forms a correct 3-term quadratic equation in \(x\) (or total depth \(d\)), e.g. \(20x^2 - 8x - 6.4 = 0\).
A1: Obtains the correct total distance below \(A\) of \(1.6\text{ m}\).

(b)
B1: Identifies that maximum speed occurs at equilibrium and calculates the equilibrium extension \(e = 0.2\text{ m}\).
M1: Sets up a correct energy equation connecting release point \(A\) and the equilibrium position including GPE, KE, and EPE terms.
A1: Obtains \(v_{\max} = 3\sqrt{2}\text{ m s}^{-1}\) or \(4.24\text{ m s}^{-1}\) (allow exact or 3 s.f.).

(a)

Let the direction parallel to and up the inclined plane be the positive \(x'\)-axis, and the direction perpendicular to the plane upwards be the positive \(y'\)-axis.

The acceleration components are:
\[ a_{x'} = -g \sin \alpha, \quad a_{y'} = -g \cos \alpha \]

The initial velocity components relative to these axes are:
\[ u_{x'} = u \cos(\theta - \alpha), \quad u_{y'} = u \sin(\theta - \alpha) \]

At the point of landing \(A\), the perpendicular displacement is zero, \(y' = 0\):
\[ u_{y'} T - \frac{1}{2}g \cos \alpha T^2 = 0 \implies T = \frac{2u\sin(\theta - \alpha)}{g\cos \alpha} \quad (T > 0) \]

Since the particle strikes the plane at right angles, the velocity component parallel to the plane at time \(T\) is zero, \(v_{x'} = 0\):
\[ v_{x'} = u\cos(\theta - \alpha) - g \sin \alpha T = 0 \]

Substituting \(T\):
\[ u\cos(\theta - \alpha) = g \sin \alpha \left( \frac{2u\sin(\theta - \alpha)}{g\cos \alpha} \right) = 2u \tan \alpha \sin(\theta - \alpha) \]
\[ \cot(\theta - \alpha) = 2\tan \alpha \implies \tan(\theta - \alpha) = \frac{1}{2\tan \alpha} \]

Given \(\tan \alpha = \frac{3}{4}\):
\[ \tan(\theta - \alpha) = \frac{1}{2 \left(\frac{3}{4}\right)} = \frac{2}{3} \]

Using the compound angle formula:
\[ \tan \theta = \tan((\theta - \alpha) + \alpha) = \frac{\tan(\theta - \alpha) + \tan \alpha}{1 - \tan(\theta - \alpha)\tan \alpha} = \frac{\frac{2}{3} + \frac{3}{4}}{1 - \left(\frac{2}{3}\right)\left(\frac{3}{4}\right)} = \frac{\frac{17}{12}}{1 - \frac{1}{2}} = \frac{17}{6} \]

(b)

Given \(\tan(\theta - \alpha) = \frac{2}{3}\), we have \(\sin(\theta - \alpha) = \frac{2}{\sqrt{13}}\) and \(\cos(\theta - \alpha) = \frac{3}{\sqrt{13}}\).

Since \(\tan \alpha = \frac{3}{4}\), \(\cos \alpha = \frac{4}{5}\).

With \(u = 20\text{ m s}^{-1}\) and \(g = 10\text{ m s}^{-2}\):
\[ T = \frac{2(20)\left(\frac{2}{\sqrt{13}}\right)}{10\left(\frac{4}{5}\right)} = \frac{\frac{80}{\sqrt{13}}}{8} = \frac{10}{\sqrt{13}}\text{ s} \]

The displacement along the plane is:
\[ OA = u_{x'} T - \frac{1}{2}g \sin \alpha T^2 \]

Using \(g \sin \alpha T = u_{x'}\):
\[ OA = u_{x'} T - \frac{1}{2}u_{x'} T = \frac{1}{2}u_{x'} T = \frac{1}{2} u \cos(\theta - \alpha) T \]
\[ OA = \frac{1}{2} \times 20 \times \frac{3}{\sqrt{13}} \times \frac{10}{\sqrt{13}} = \frac{300}{13}\text{ m} \approx 23.1\text{ m} \]

(a)
- M1: For setting up equations for motion parallel and perpendicular to the inclined plane (or resolving horizontally and vertically) and finding an expression for the time of flight \(T\).
- A1: For obtaining correct time of flight \(T = \frac{2u\sin(\theta - \alpha)}{g\cos \alpha}\) or \(T = \frac{2u\cos \theta(\tan \theta - \tan \alpha)}{g}\).
- M1: For applying the condition that the particle strikes at right angles, \(v_{x'} = 0\) (or using gradients \(m_1 m_2 = -1\)), to form an equation relating \(\tan(\theta - \alpha)\) (or \(\tan \theta\)) to \(\tan \alpha\).
- A1 (AG): For fully correct algebraic working leading to \(\tan \theta = \frac{17}{6}\) with no errors.

(b)
- M1: For a complete method to find the distance \(OA\) along the plane (e.g. using \(x' = \frac{1}{2} u_{x'} T\) or \(x_A / \cos \alpha\)).
- A1: For \(OA = \frac{300}{13}\text{ m}\) (exact) or \(23.1\text{ m}\) (3 s.f.).

(a)
The distribution of the differences is symmetrical.

(b)
Let $m$ be the population median difference in scores, where difference $d = \text{Score before} - \text{Score after}$.

$H_0: m = 0$
$H_1: m > 0$

Calculate the differences $d = \text{Score before} - \text{Score after}$:
- $\text{A}: 62 - 58 = +4$
- $\text{B}: 71 - 73 = -2$
- $\text{C}: 83 - 72 = +11$
- $\text{D}: 65 - 58 = +7$
- $\text{E}: 79 - 63 = +16$
- $\text{F}: 81 - 86 = -5$
- $\text{G}: 74 - 65 = +9$
- $\text{H}: 90 - 76 = +14$
- $\text{I}: 68 - 67 = +1$

Order the absolute differences $|d|$ and assign ranks $1$ to $9$:
- $|d| = 1$ (I) $\rightarrow$ rank $1$
- $|d| = 2$ (B) $\rightarrow$ rank $2$
- $|d| = 4$ (A) $\rightarrow$ rank $3$
- $|d| = 5$ (F) $\rightarrow$ rank $4$
- $|d| = 7$ (D) $\rightarrow$ rank $5$
- $|d| = 9$ (G) $\rightarrow$ rank $6$
- $|d| = 11$ (C) $\rightarrow$ rank $7$
- $|d| = 14$ (H) $\rightarrow$ rank $8$
- $|d| = 16$ (E) $\rightarrow$ rank $9$

Sum of positive ranks:
\[ T_+ = 1 + 3 + 5 + 6 + 7 + 8 + 9 = 39 \]

Sum of negative ranks:
\[ T_- = 2 + 4 = 6 \]

Test statistic $T = \min(T_+, T_-) = 6$.

From tables of the Wilcoxon signed-rank distribution with $n = 9$ at the $5\%$ one-tailed level, the critical value is $8$.

Since $T = 6 \leqslant 8$, reject $H_0$.

There is sufficient evidence at the $5\%$ significance level to conclude that the mindfulness training programme reduces stress levels.

(a)
- B1: State that the population distribution of differences is symmetrical (or symmetric).

(b)
- B1: State both hypotheses correctly in terms of median difference (e.g. $H_0: m = 0$ and $H_1: m > 0$).
- M1: Calculate differences and rank their absolute values correctly.
- A1: Calculate $T_- = 6$ (or $T_+ = 39$) correctly, giving test statistic $T = 6$.
- B1: State the correct critical value of $8$ (from MF19 for $n=9$, $1$-tail $\alpha = 0.05$).
- M1: Compare test statistic with critical value correctly (i.e. $6 \leqslant 8$) and decide to reject $H_0$.
- A1: State a correct contextual conclusion (e.g. evidence that the training programme reduces stress levels). Must not be assertively worded.

(a)
$H_0$: Age group and primary purpose for visiting the library are independent.
$H_1$: Age group and primary purpose for visiting the library are not independent (there is an association).

(b)
Calculate the row and column totals:
- Row totals: $\text{Under } 30 = 28 + 22 + 10 = 60$, $30\text{ to } 50 = 25 + 44 + 21 = 90$, $\text{Over } 50 = 17 + 14 + 19 = 50$.
- Column totals: $\text{Borrowing} = 28 + 25 + 17 = 70$, $\text{Study spaces} = 22 + 44 + 14 = 80$, $\text{Community events} = 10 + 21 + 19 = 50$.
- Grand total: $N = 200$.

Expected frequency $E = \dfrac{\text{Row Total} \times \text{Column Total}}{N}$:
$$\begin{array}{|l|c|c|c|}
\hline
\text{Age group} & \text{Borrowing books} & \text{Using study spaces} & \text{Attending events} \\
\hline
\text{Under } 30 & \frac{60 \times 70}{200} = 21 & \frac{60 \times 80}{200} = 24 & \frac{60 \times 50}{200} = 15 \\
30\text{ to } 50 & \frac{90 \times 70}{200} = 31.5 & \frac{90 \times 80}{200} = 36 & \frac{90 \times 50}{200} = 22.5 \\
\text{Over } 50 & \frac{50 \times 70}{200} = 17.5 & \frac{50 \times 80}{200} = 20 & \frac{50 \times 50}{200} = 12.5 \\
\hline
\end{array}$$
All expected frequencies are $\ge 5$, so no pooling of rows or columns is required.

(c)
Calculate $\chi^2 = \sum \dfrac{(O - E)^2}{E}$ for each of the 9 cells:
- Under 30, Borrowing: $\dfrac{(28 - 21)^2}{21} = \dfrac{49}{21} \approx 2.333$
- Under 30, Study: $\dfrac{(22 - 24)^2}{24} = \dfrac{4}{24} \approx 0.167$
- Under 30, Events: $\dfrac{(10 - 15)^2}{15} = \dfrac{25}{15} \approx 1.667$
- 30 to 50, Borrowing: $\dfrac{(25 - 31.5)^2}{31.5} = \dfrac{42.25}{31.5} \approx 1.341$
- 30 to 50, Study: $\dfrac{(44 - 36)^2}{36} = \dfrac{64}{36} \approx 1.778$
- 30 to 50, Events: $\dfrac{(21 - 22.5)^2}{22.5} = \dfrac{2.25}{22.5} = 0.100$
- Over 50, Borrowing: $\dfrac{(17 - 17.5)^2}{17.5} = \dfrac{0.25}{17.5} \approx 0.014$
- Over 50, Study: $\dfrac{(14 - 20)^2}{20} = \dfrac{36}{20} = 1.800$
- Over 50, Events: $\dfrac{(19 - 12.5)^2}{12.5} = \dfrac{42.25}{12.5} = 3.380$

Summing all contributions:
$$\chi^2 = 2.333 + 0.167 + 1.667 + 1.341 + 1.778 + 0.100 + 0.014 + 1.800 + 3.380 = 12.58$$

Degrees of freedom $
u = (r - 1)(c - 1) = (3 - 1)(3 - 1) = 4$.

Critical value at 5% significance level: $\chi^2_{0.05, 4} = 9.488$.

Since $12.58 > 9.488$, we reject $H_0$.

There is sufficient evidence at the 5% significance level to conclude that there is an association between age group and primary purpose for visiting the library.

(d)
The cell contributing most to the test statistic is Over 50 attending community events, with a contribution of $3.38$.

Interpretation: The observed frequency (19) is notably higher than the expected frequency (12.5), which indicates that members aged over 50 are more likely to visit the library to attend community events than would be expected if purpose were independent of age.

(a)
- B1: Both hypotheses stated correctly in context (e.g. "independent" vs "not independent" or "no association" vs "association").

(b)
- M1: Correct method for finding expected frequencies $\frac{\text{Row Total} \times \text{Col Total}}{\text{Grand Total}}$, showing at least two correct calculations.
- A1: At least 5 expected values correct.
- A1: All 9 expected values correct and exact/terminating decimals ($21, 24, 15, 31.5, 36, 22.5, 17.5, 20, 12.5$).

(c)
- M1: Correct method for $\chi^2$ calculation: $\sum \frac{(O - E)^2}{E}$ with at least 3 terms shown.
- A1: Correct calculated test statistic $\chi^2 \approx 12.58$ or $12.6$.
- B1: Stating correct degrees of freedom $
u = 4$ and critical value $9.488$.
- A1 (FT): Correct comparison of test statistic with critical value and correct non-assertive conclusion in context. (Do not award if hypotheses are contradictory or missing context).

(d)
- B1: Correctly identifies the cell with the largest contribution ("Over 50 / Community events").
- B1: Meaningful contextual interpretation (observing that more older members attend community events than expected under independence).

Calculate the sample means:
\[ \bar{x} = \frac{188.0}{8} = 23.5 \]
\[ \bar{y} = \frac{254.4}{12} = 21.2 \]
\[ \bar{x} - \bar{y} = 23.5 - 21.2 = 2.3 \]

Calculate the sums of squared deviations:
\[ \sum (x - \bar{x})^2 = \sum x^2 - \frac{(\sum x)^2}{n_A} = 4434.50 - \frac{188.0^2}{8} = 4434.50 - 4418.00 = 16.50 \]
\[ \sum (y - \bar{y})^2 = \sum y^2 - \frac{(\sum y)^2}{n_B} = 5418.06 - \frac{254.4^2}{12} = 5418.06 - 5393.28 = 24.78 \]

Calculate the pooled estimate of the common variance, \(s^2\):
\[ s^2 = \frac{\sum (x - \bar{x})^2 + \sum (y - \bar{y})^2}{n_A + n_B - 2} = \frac{16.50 + 24.78}{8 + 12 - 2} = \frac{41.28}{18} = \frac{172}{75} \approx 2.29333 \]

The number of degrees of freedom is \(\nu = 8 + 12 - 2 = 18\).
The critical value from the \(t\)-distribution for a 95% confidence interval with 18 degrees of freedom is:
\[ t_{18}(0.025) = 2.101 \]

Calculate the standard error:
\[ \text{SE} = \sqrt{s^2 \left( \frac{1}{n_A} + \frac{1}{n_B} \right)} = \sqrt{\frac{41.28}{18} \left( \frac{1}{8} + \frac{1}{12} \right)} = \sqrt{\frac{41.28}{18} \times \frac{5}{24}} = \sqrt{\frac{43}{90}} \approx 0.691215 \]

The 95% confidence interval is:
\[ (\bar{x} - \bar{y}) \pm t \times \text{SE} = 2.3 \pm 2.101 \times 0.691215 = 2.3 \pm 1.45224 \]

Lower limit: \(2.3 - 1.45224 = 0.84776 \approx 0.848\)
Upper limit: \(2.3 + 1.45224 = 3.75224 \approx 3.75\)

Hence, the 95% confidence interval is \([0.848, 3.75]\) (or \(0.848 < \mu_A - \mu_B < 3.75\)).

M1: For calculating sample means \(\bar{x} = 23.5\) and \(\bar{y} = 21.2\) and their difference \(\bar{x} - \bar{y} = 2.3\).
M1: For finding the sum of squared deviations \(\sum (x-\bar{x})^2 = 16.50\) and \(\sum (y-\bar{y})^2 = 24.78\) (or unbiased variances \(s_A^2 = 2.357\), \(s_B^2 = 2.253\)).
M1: For finding the pooled variance estimate \(s^2 = \frac{16.50 + 24.78}{18} = 2.2933\dots\) or \(\frac{172}{75}\).
B1: For the correct critical value \(t = 2.101\) (from \(t_{18}\)).
M1: For standard error calculation and CI structure \((\bar{x} - \bar{y}) \pm t \sqrt{s^2\left(\frac{1}{n_A} + \frac{1}{n_B}\right)}\).
A1: For correct confidence interval \([0.848, 3.75]\) or \(0.848 < \mu_A - \mu_B < 3.75\) (accept \([0.847, 3.753]\) due to rounding).

(a) Since \(\text{F}(x)\) is a cumulative distribution function of a continuous random variable, it must be continuous at \(x = 3\):
\[ \lim_{x \to 3^-} \text{F}(x) = \lim_{x \to 3^+} \text{F}(x) \]
\[ \frac{1}{6}(3-1)^2 = 1 - \frac{a}{(3-1)^2} \]
\[ \frac{4}{6} = 1 - \frac{a}{4} \implies \frac{2}{3} = 1 - \frac{a}{4} \]
\[ \frac{a}{4} = \frac{1}{3} \implies a = \frac{4}{3} \]

(b) Differentiating \(\text{F}(x)\) with respect to \(x\):
For \(1 \le x < 3\):
\[ \text{f}(x) = \frac{\mathrm{d}}{\mathrm{d}x}\left[\frac{1}{6}(x-1)^2\right] = \frac{1}{3}(x-1) \]
For \(x \ge 3\):
\[ \text{f}(x) = \frac{\mathrm{d}}{\mathrm{d}x}\left[1 - \frac{4}{3}(x-1)^{-2}\right] = \frac{8}{3}(x-1)^{-3} \]
Thus, the probability density function is:
\[ \text{f}(x) = \begin{cases} \frac{1}{3}(x-1) & 1 \le x < 3, \\ \frac{8}{3}(x-1)^{-3} & x \ge 3, \\ 0 & \text{otherwise}. \end{cases} \]

(c) \(\text{E}(X) = \int_{-\infty}^{\infty} x\,\text{f}(x)\,\mathrm{d}x\):
\[ \text{E}(X) = \int_1^3 x \cdot \frac{1}{3}(x-1)\,\mathrm{d}x + \int_3^{\infty} x \cdot \frac{8}{3}(x-1)^{-3}\,\mathrm{d}x \]
Evaluating the first integral:
\[ \frac{1}{3} \int_1^3 (x^2 - x)\,\mathrm{d}x = \frac{1}{3} \left[ \frac{x^3}{3} - \frac{x^2}{2} \right]_1^3 = \frac{1}{3} \left[ \left(9 - \frac{9}{2}\right) - \left(\frac{1}{3} - \frac{1}{2}\right) \right] = \frac{1}{3} \left[ \frac{9}{2} + \frac{1}{6} \right] = \frac{1}{3} \left( \frac{28}{6} \right) = \frac{14}{9} \]
Evaluating the second integral (using substitution \(u = x-1\), \(x = u+1\)):
\[ \frac{8}{3} \int_2^{\infty} (u+1)u^{-3}\,\mathrm{d}u = \frac{8}{3} \int_2^{\infty} (u^{-2} + u^{-3})\,\mathrm{d}u = \frac{8}{3} \left[ -u^{-1} - \frac{1}{2}u^{-2} \right]_2^{\infty} \]
\[ = \frac{8}{3} \left[ 0 - \left( -\frac{1}{2} - \frac{1}{8} \right) \right] = \frac{8}{3} \left( \frac{5}{8} \right) = \frac{5}{3} \]
Adding the two parts:
\[ \text{E}(X) = \frac{14}{9} + \frac{5}{3} = \frac{14 + 15}{9} = \frac{29}{9} \quad (\approx 3.22) \]

(d) For \(X \ge 1\), \(X - 1 \ge 0\), so \(Y = (X-1)^2 \ge 0\).
For \(y < 0\), \(\text{F}_Y(y) = 0\).
For \(y \ge 0\):
\[ \text{F}_Y(y) = \text{P}(Y \le y) = \text{P}((X-1)^2 \le y) = \text{P}(X - 1 \le \sqrt{y}) = \text{P}(X \le 1 + \sqrt{y}) = \text{F}_X(1 + \sqrt{y}) \]
The boundary \(x = 3\) corresponds to \(1 + \sqrt{y} = 3 \implies \sqrt{y} = 2 \implies y = 4\).
For \(0 \le y < 4\):
\[ \text{F}_Y(y) = \frac{1}{6}\left((1 + \sqrt{y}) - 1\right)^2 = \frac{1}{6}(\sqrt{y})^2 = \frac{1}{6}y \]
For \(y \ge 4\):
\[ \text{F}_Y(y) = 1 - \frac{4/3}{\left((1 + \sqrt{y}) - 1\right)^2} = 1 - \frac{4}{3y} \]
Hence,
\[ \text{F}_Y(y) = \begin{cases} 0 & y < 0, \\ \frac{1}{6}y & 0 \le y < 4, \\ 1 - \frac{4}{3y} & y \ge 4. \end{cases} \]

(a)
B1: Equates the two expressions at \(x = 3\) to show \(a = \frac{4}{3}\) with full algebraic working.

(b)
M1: Differentiates \(\text{F}(x)\) in both non-zero intervals.
A1: Correctly states \(\text{f}(x)\) fully, including intervals and the zero case elsewhere.

(c)
M1: Sets up the sum of two integrals for \(\text{E}(X) = \int x\,\text{f}(x)\,\mathrm{d}x\) with correct limits.
A1: Obtains either \(\frac{14}{9}\) or \(\frac{5}{3}\) correctly from integration.
A1: Obtains \(\frac{29}{9}\) (or exact recurring decimal \(3.\dot{2}\), or \(3.22\) to 3 s.f.) CWO.

(d)
M1: Writes \(\text{P}(Y \le y) = \text{P}(X \le 1 + \sqrt{y})\) or equivalent method connecting \(\text{F}_Y(y)\) and \(\text{F}_X(x)\).
A1: Correct expressions for \(0 \le y < 4\) and \(y \ge 4\).
A1: Fully correct piecewise CDF including the domain \(y < 0\) (or '0 otherwise').

(a)
The population of differences between the paired operating temperatures (Lubricant A $-$ Lubricant B) is assumed to follow a normal distribution.

(b)
Let \(d = \text{Lubricant A} - \text{Lubricant B}\), and let \(\mu_D\) denote the population mean difference.

State the null and alternative hypotheses:
\[ H_0: \mu_D = 0 \]
\[ H_1: \mu_D > 0 \]

Calculate the differences \(d\) for each machine:
\[ \begin{array}{|l|c|c|c|c|c|c|c|} \hline \text{Machine} & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline d & 2.8 & 1.6 & -0.8 & 2.4 & 3.2 & 1.8 & 3.0 \\ \hline \end{array} \]

Calculate the sample mean of the differences:
\[ \sum d = 2.8 + 1.6 - 0.8 + 2.4 + 3.2 + 1.8 + 3.0 = 14.0 \]
\[ \bar{d} = \frac{\sum d}{n} = \frac{14.0}{7} = 2.0 \]

Calculate the unbiased estimate of the population variance \(s^2\):
\[ \sum d^2 = 2.8^2 + 1.6^2 + (-0.8)^2 + 2.4^2 + 3.2^2 + 1.8^2 + 3.0^2 = 7.84 + 2.56 + 0.64 + 5.76 + 10.24 + 3.24 + 9.00 = 39.28 \]
\[ s^2 = \frac{1}{n-1} \left( \sum d^2 - \frac{(\sum d)^2}{n} \right) = \frac{1}{6} \left( 39.28 - \frac{14.0^2}{7} \right) = \frac{1}{6} (39.28 - 28.00) = \frac{11.28}{6} = 1.88 \]
\[ s = \sqrt{1.88} \approx 1.371 \]

Calculate the test statistic:
\[ t = \frac{\bar{d} - 0}{s / \sqrt{n}} = \frac{2.0}{\sqrt{1.88 / 7}} = \frac{2.0}{\sqrt{0.26857}} = \frac{2.0}{0.5182} \approx 3.86 \]

Determine the critical value for a one-tailed test with degrees of freedom \(\nu = n - 1 = 6\) at the 5% significance level:
\[ t_{\text{crit}} = t_{6}(0.05) = 1.943 \]

Comparison and conclusion:
Since the calculated test statistic \(t = 3.86 > 1.943\), the result is significant at the 5% level.
Reject \(H_0\).

There is sufficient evidence at the 5% significance level to suggest that Lubricant B reduces the mean operating temperature of the machinery compared to Lubricant A.

(a)
- B1: States that the differences (in operating temperature) are normally distributed. (Must refer to the differences, not merely the individual distributions of A and B).

(b)
- B1: Correct hypotheses stated with \(\mu_D\) (or \(\mu_A - \mu_B\)) clearly defined: \(H_0: \mu_D = 0\) and \(H_1: \mu_D > 0\).
- B1: Correct calculation of sample mean difference \(\bar{d} = 2.0\).
- M1: Valid method for calculating the unbiased estimate of population variance \(s^2\) or standard deviation \(s\) for the differences.
- A1: Correct value \(s^2 = 1.88\) (or \(s \approx 1.371\)).
- M1: Correct substitution into the test statistic formula \(t = \frac{\bar{d}}{s/\sqrt{7}}\).
- B1: Correct critical value \(t_6(0.05) = 1.943\) (or critical region \(t > 1.943\) / comparison of \(p\)-value \(p \approx 0.0042 < 0.05\)).
- A1: Correct conclusion following a correct comparison (reject \(H_0\)), phrased in context with non-assertive language.

(a)
We differentiate \(\mathrm{G}_X(t) = t(4 - 3t)^{-1}\) with respect to \(t\):
\[ \mathrm{G}'_X(t) = 1 \cdot (4 - 3t)^{-1} + t \cdot (-1)(4 - 3t)^{-2}(-3) = (4 - 3t)^{-1} + 3t(4 - 3t)^{-2} = \frac{4}{(4 - 3t)^2} = 4(4 - 3t)^{-2} \]
Evaluating at \(t = 1\):
\[ \mathrm{E}(X) = \mathrm{G}'_X(1) = 4(4 - 3(1))^{-2} = 4 \]

Differentiating again:
\[ \mathrm{G}''_X(t) = 4(-2)(4 - 3t)^{-3}(-3) = 24(4 - 3t)^{-3} \]
Evaluating at \(t = 1\):
\[ \mathrm{G}''_X(1) = 24(4 - 3(1))^{-3} = 24 \]

Using the formula for variance:
\[ \operatorname{Var}(X) = \mathrm{G}''_X(1) + \mathrm{G}'_X(1) - [\mathrm{G}'_X(1)]^2 = 24 + 4 - 4^2 = 12 \]

(b)
Since \(X_1\) and \(X_2\) are independent:
\[ \mathrm{G}_Z(t) = \mathrm{G}_{X_1}(t) \times \mathrm{G}_{X_2}(t) = [\mathrm{G}_X(t)]^2 = \left(\frac{t}{4 - 3t}\right)^2 = \frac{t^2}{(4 - 3t)^2} \]

(c)(i)
Since \(W = Z + Y\) where \(Z\) and \(Y\) are independent:
\[ \mathrm{G}_W(t) = \mathrm{G}_Z(t) \times \mathrm{G}_Y(t) = \frac{t^2}{(4 - 3t)^2} \times \frac{1}{8}(1 + t)^3 = \frac{t^2(1 + t)^3}{8(4 - 3t)^2} \]

(c)(ii)
\(\mathrm{P}(W = 4)\) is the coefficient of \(t^4\) in the Maclaurin expansion of \(\mathrm{G}_W(t)\).
We expand the terms:
\[ (1 + t)^3 = 1 + 3t + 3t^2 + t^3 \]
\[ (4 - 3t)^{-2} = 4^{-2}\left(1 - \frac{3}{4}t\right)^{-2} = \frac{1}{16}\left(1 + 2\left(\frac{3}{4}t\right) + 3\left(\frac{3}{4}t\right)^2 + \dots\right) = \frac{1}{16}\left(1 + \frac{3}{2}t + \frac{27}{16}t^2 + \dots\right) \]

Thus,
\[ \mathrm{G}_W(t) = \frac{1}{128} t^2 (1 + 3t + 3t^2 + t^3)\left(1 + \frac{3}{2}t + \frac{27}{16}t^2 + \dots\right) \]
The coefficient of \(t^4\) in \(\mathrm{G}_W(t)\) is \(\frac{1}{128}\) multiplied by the coefficient of \(t^2\) in the product \((1 + 3t + 3t^2)\left(1 + \frac{3}{2}t + \frac{27}{16}t^2\right)\):
\[ \text{Coefficient of } t^2 = 1 \times \frac{27}{16} + 3 \times \frac{3}{2} + 3 \times 1 = \frac{27}{16} + \frac{72}{16} + \frac{48}{16} = \frac{147}{16} \]

Therefore,
\[ \mathrm{P}(W = 4) = \frac{1}{128} \times \frac{147}{16} = \frac{147}{2048} \approx 0.0718 \]

(a)
- M1: Attempts first derivative using product/quotient/chain rule on \(\mathrm{G}_X(t)\).
- A1: Correct \(\mathrm{E}(X) = 4\).
- M1: Differentiates again to find \(\mathrm{G}''_X(1)\) and uses \(\operatorname{Var}(X) = \mathrm{G}''(1) + \mathrm{G}'(1) - [\mathrm{G}'(1)]^2\).
- A1: Obtains \(\operatorname{Var}(X) = 12\).

(b)
- B1: Correct expression \(\frac{t^2}{(4 - 3t)^2}\) or \(t^2(4 - 3t)^{-2}\).

(c)(i)
- M1: Uses \(\mathrm{G}_W(t) = \mathrm{G}_Z(t) \times \mathrm{G}_Y(t)\) with their \(\mathrm{G}_Z(t)\).
- A1: Fully correct algebraic justification to the given expression (AG).

(c)(ii)
- M1: Applies binomial expansion to \((4 - 3t)^{-2}\) or \(\left(1 - \frac{3}{4}t\right)^{-2}\) up to the term in \(t^2\).
- M1: Collects all relevant product terms to find the coefficient of \(t^4\) in \(\mathrm{G}_W(t)\).
- A1: Correct exact probability \(\frac{147}{2048}\) (or \(0.0718\) to 3 s.f.).