2026 Cambridge IAL Mathematics - Further (9231) 模拟试题及答案详解

(a) Combining the fractions over a common denominator:
\[ \frac{1}{r(r+2)} - \frac{1}{(r+1)(r+3)} = \frac{(r+1)(r+3) - r(r+2)}{r(r+1)(r+2)(r+3)} \]
\[ = \frac{(r^2 + 4r + 3) - (r^2 + 2r)}{r(r+1)(r+2)(r+3)} = \frac{2r+3}{r(r+1)(r+2)(r+3)} \]

(b) Let \( f(r) = \frac{1}{r(r+2)} \). Then the sum is:
\[ \sum_{r=1}^n \left( f(r) - f(r+1) \right) \]
Writing out the terms:
\[ r=1: \quad f(1) - f(2) \]
\[ r=2: \quad f(2) - f(3) \]
\[ \vdots \]
\[ r=n: \quad f(n) - f(n+1) \]
Summing and cancelling intermediate terms gives:
\[ S_n = f(1) - f(n+1) = \frac{1}{1(3)} - \frac{1}{(n+1)(n+3)} = \frac{1}{3} - \frac{1}{(n+1)(n+3)} \]
Expressing as a single fraction:
\[ S_n = \frac{(n+1)(n+3) - 3}{3(n+1)(n+3)} = \frac{n^2 + 4n + 3 - 3}{3(n+1)(n+3)} = \frac{n(n+4)}{3(n+1)(n+3)} \]

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

(d) The sum from \( r = n+1 \) to \( \infty \) is:
\[ S_{\infty} - S_n = \frac{1}{3} - \left( \frac{1}{3} - \frac{1}{(n+1)(n+3)} \right) = \frac{1}{(n+1)(n+3)} \]

(a)
- M1: For putting fractions over a common denominator \( r(r+1)(r+2)(r+3) \) and expanding the numerator.
- A1: For obtaining \( r^2+4r+3 - (r^2+2r) = 2r+3 \) correctly with all working shown.

(b)
- M1: For using the method of differences, showing clear cancellation of intermediate terms.
- A1: For obtaining \( \frac{1}{3} - \frac{1}{(n+1)(n+3)} \) or equivalent.
- A1: For simplifying to \( \frac{n(n+4)}{3(n+1)(n+3)} \) or \( \frac{n^2+4n}{3(n+1)(n+3)} \).

(c)
- B1: For \( \frac{1}{3} \) correctly deduced from \( n \to \infty \).

(d)
- B1: For \( \frac{1}{(n+1)(n+3)} \) OE.

From the given cubic equation \(x^3 - 3x^2 + 2x + 5 = 0\), we have:
\[\sum \alpha = \alpha + \beta + \gamma = 3\]
\[\sum \alpha\beta = \alpha\beta + \beta\gamma + \gamma\alpha = 2\]
\[\alpha\beta\gamma = -5\]

(a)
Using the identity for the sum of squares:
\[\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\]
\[\alpha^2 + \beta^2 + \gamma^2 = 3^2 - 2(2) = 9 - 4 = 5\]

(b)
Since \(\alpha, \beta, \gamma\) satisfy the original equation, \(x^3 = 3x^2 - 2x - 5\).
Summing over the three roots:
\[\alpha^3 + \beta^3 + \gamma^3 = 3(\alpha^2 + \beta^2 + \gamma^2) - 2(\alpha + \beta + \gamma) - 5(3)\]
Substituting the known values:
\[\alpha^3 + \beta^3 + \gamma^3 = 3(5) - 2(3) - 15 = 15 - 6 - 15 = -6\]

(c)
Since \(\alpha + \beta + \gamma = 3\), the required roots can be written as:
\[\beta + \gamma - \alpha = (\alpha + \beta + \gamma) - 2\alpha = 3 - 2\alpha\]
\[\gamma + \alpha - \beta = 3 - 2\beta\]
\[\alpha + \beta - \gamma = 3 - 2\gamma\]
Using the substitution \(y = 3 - 2x\), we get \(x = \frac{3-y}{2}\).
Substitute \(x = \frac{3-y}{2}\) into the original equation:
\[\left(\frac{3-y}{2}\right)^3 - 3\left(\frac{3-y}{2}\right)^2 + 2\left(\frac{3-y}{2}\right) + 5 = 0\]
Multiply through by \(8\):
\[(3-y)^3 - 6(3-y)^2 + 8(3-y) + 40 = 0\]
Expanding each term:
\[(27 - 27y + 9y^2 - y^3) - 6(9 - 6y + y^2) + (24 - 8y) + 40 = 0\]
\[(27 - 27y + 9y^2 - y^3) - (54 - 36y + 6y^2) + (24 - 8y) + 40 = 0\]
Collecting like terms:
\[-y^3 + (9 - 6)y^2 + (-27 + 36 - 8)y + (27 - 54 + 24 + 40) = 0\]
\[-y^3 + 3y^2 + y + 37 = 0\]
\[y^3 - 3y^2 - y - 37 = 0\]

(a) [2 marks]
- M1: Uses \(\sum \alpha^2 = (\sum \alpha)^2 - 2\sum \alpha\beta\) with \(\sum \alpha = 3\) and \(\sum \alpha\beta = 2\).
- A1: Obtains \(5\).

(b) [2 marks]
- M1: Sets up a valid method for \(\sum \alpha^3\), e.g. \(\sum \alpha^3 = 3\sum \alpha^2 - 2\sum \alpha - 15\) or the cubic expansion identity.
- A1: Obtains \(-6\).

(c) [3 marks]
- M1: Recognises the relation \(\beta + \gamma - \alpha = 3 - 2\alpha\) and substitutes \(x = \frac{3-y}{2}\) into the original cubic equation (or calculates new elementary symmetric sums).
- M1: Clears fractions and expands the terms correctly.
- A1: Obtains \(y^3 - 3y^2 - y - 37 = 0\) (or any equivalent equation with integer coefficients and "= 0").

(a)
Differentiating \(y = x^2 \text{e}^{3x}\) with respect to \(x\) using the product rule:
\[\frac{\text{d}y}{\text{d}x} = 2x\text{e}^{3x} + 3x^2\text{e}^{3x} = (3x^2 + 2x)\text{e}^{3x}\]
Differentiating again with respect to \(x\):
\[\frac{\text{d}^2 y}{\text{d}x^2} = (6x + 2)\text{e}^{3x} + 3(3x^2 + 2x)\text{e}^{3x} = (9x^2 + 12x + 2)\text{e}^{3x}\]

(b)
**Base case (\(n=1\)):**
Using the given formula with \(n=1\):
\[\frac{\text{d}^1 y}{\text{d}x^1} = 3^{1-2} \left(9x^2 + 6(1)x + 1(0)\right) \text{e}^{3x} = 3^{-1}(9x^2 + 6x)\text{e}^{3x} = (3x^2 + 2x)\text{e}^{3x}\]
This matches the result for \(\frac{\text{d}y}{\text{d}x}\) found in part (a). Hence, the statement is true for \(n = 1\).

Inductive hypothesis:
Assume that the statement is true for \(n = k\), where \(k\) is a positive integer, so:
\[\frac{\text{d}^k y}{\text{d}x^k} = 3^{k-2} \left(9x^2 + 6kx + k(k-1)\right) \text{e}^{3x}\]

Inductive step:
Differentiating both sides with respect to \(x\) using the product rule:
\[\frac{\text{d}^{k+1} y}{\text{d}x^{k+1}} = \frac{\text{d}}{\text{d}x}\left[3^{k-2} \left(9x^2 + 6kx + k(k-1)\right) \text{e}^{3x}\right]\]
\[= 3^{k-2} (18x + 6k)\text{e}^{3x} + 3 \cdot 3^{k-2} \left(9x^2 + 6kx + k(k-1)\right)\text{e}^{3x}\]
\[= 3^{k-1} (6x + 2k)\text{e}^{3x} + 3^{k-1} \left(9x^2 + 6kx + k^2 - k\right)\text{e}^{3x}\]
\[= 3^{k-1} \left[ 9x^2 + (6k + 6)x + (k^2 + k) \right] \text{e}^{3x}\]
\[= 3^{(k+1)-2} \left[ 9x^2 + 6(k+1)x + (k+1)k \right] \text{e}^{3x}\]
Since \((k+1)k = (k+1)((k+1)-1)\), this is of the form:
\[3^{(k+1)-2} \left(9x^2 + 6(k+1)x + (k+1)((k+1)-1)\right) \text{e}^{3x}\]

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

(a)
M1: Applies the product rule correctly to find \(\frac{\text{d}y}{\text{d}x}\).
A1: Obtains correct expressions for both \(\frac{\text{d}y}{\text{d}x} = (3x^2 + 2x)\text{e}^{3x}\) and \(\frac{\text{d}^2 y}{\text{d}x^2} = (9x^2 + 12x + 2)\text{e}^{3x}\).

(b)
B1: Verifies the base case \(n = 1\) correctly using the given formula and comparing with \(\frac{\text{d}y}{\text{d}x}\).
M1: States the inductive hypothesis clearly for \(n = k\).
M1: Differentiates the assumed \(k\)-th derivative with respect to \(x\) using the product rule.
A1: Obtains a correct unsimplified expression for \(\frac{\text{d}^{k+1}y}{\text{d}x^{k+1}}\).
A1: Correctly rearranges the expression into the exact required form for \(n = k + 1\).
A1: Gives a complete, logically sound conclusion linking the base case, the step from \(n = k\) to \(n = k + 1\), and mathematical induction.

(a) For invariant points, $\mathbf{M}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}$:
$$\begin{pmatrix} 4 & 1 \\ 6 & 3 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}$$
This gives the system of equations:
$$4x + y = x \implies 3x + y = 0$$
$$6x + 3y = y \implies 6x + 2y = 0 \implies 3x + y = 0$$
Both equations reduce to $y = -3x$.
Therefore, the set of invariant points is the line $y = -3x$ (or points of the form $(t, -3t)$ for $t \in \mathbb{R}$).

(b) The determinant of $\mathbf{M}$ is:
$$\det(\mathbf{M}) = (4)(3) - (1)(6) = 12 - 6 = 6$$
The area scale factor is $|\det(\mathbf{M})| = 6$.
$$\text{Area of image triangle} = 6 \times 5 = 30\text{ cm}^2$$

(c) Let a point $(x, y)$ lie on the line $y = mx + c$.
Its image $(X, Y)$ is given by:
$$\begin{pmatrix} X \\ Y \end{pmatrix} = \begin{pmatrix} 4 & 1 \\ 6 & 3 \end{pmatrix}\begin{pmatrix} x \\ mx + c \end{pmatrix} = \begin{pmatrix} (4 + m)x + c \\ (6 + 3m)x + 3c \end{pmatrix}$$
For the line to be invariant, $(X, Y)$ must satisfy $Y = mX + c$:
$$(6 + 3m)x + 3c = m[(4 + m)x + c] + c$$
$$(6 + 3m)x + 3c = (m^2 + 4m)x + mc + c$$
Equating coefficients of $x$:
$$m^2 + 4m = 6 + 3m \implies m^2 + m - 6 = 0 \implies (m - 2)(m + 3) = 0$$
Equating constant terms:
$$mc + c = 3c \implies (m - 2)c = 0$$
For this condition to hold for all $c \in \mathbb{R}$, we require $m - 2 = 0 \implies m = 2$.
When $m = 2$, the quadratic equation is also satisfied ($2^2 + 2 - 6 = 0$).
Thus, $y = 2x + c$ is a family of parallel invariant lines for any $c \in \mathbb{R}$, with $m = 2$.

(d) From the quadratic equation for $m$, the other root is $m = -3$.
Substituting $m = -3$ into the constant equation:
$$(-3 - 2)c = 0 \implies -5c = 0 \implies c = 0$$
Thus, $y = -3x$ is an invariant line.

Now consider vertical lines of the form $x = k$:
$$X = 4k + y$$
Since $X$ depends on $y$, it cannot be constant for all points on $x = k$.
Hence, there are no invariant lines of the form $x = k$.
Therefore, the only other invariant line under $T$ is $y = -3x$.

(a)
M1: Sets up $\begin{pmatrix} 4 & 1 \\ 6 & 3 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}$ and obtains a linear equation in $x$ and $y$.
A1: States $y = -3x$ or equivalent line equation / parametric form $(t, -3t)$. [2]

(b)
M1: Calculates $\det(\mathbf{M}) = 6$ and uses image area $= |\det(\mathbf{M})| \times 5$.
A1: Obtains $30$ (units not required). [2]

(c)
M1: Applies matrix $\mathbf{M}$ to $\begin{pmatrix} x \\ mx + c \end{pmatrix}$ to find $X$ and $Y$ in terms of $x, m, c$.
M1: Substitutes $X$ and $Y$ into $Y = mX + c$ and equates coefficients of $x$ and constant terms.
A1: Obtains $m^2 + m - 6 = 0$ and $(m - 2)c = 0$.
A1: Correctly deduces $m = 2$ so that $(m - 2)c = 0$ holds for all $c \in \mathbb{R}$, and concludes $y = 2x + c$. [4]

(d)
M1: Uses $m = -3$ in $(m - 2)c = 0$ to find $c = 0$.
A1: Obtains $y = -3x$ (or $3x + y = 0$).
B1: Considers $x = k$, shows $X = 4k + y$ is not constant, and concludes no vertical invariant lines exist. [3]

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

A vector perpendicular to both lines is given by the cross product:
\[ \mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 1 \\ 1 & 1 & -2 \end{vmatrix} = \mathbf{i}(2 - 1) - \mathbf{j}(-4 - 1) + \mathbf{k}(2 - (-1)) = \begin{pmatrix} 1 \\ 5 \\ 3 \end{pmatrix} \]

A vector joining a point on \( l_1 \) to a point on \( l_2 \) is:
\[ \vec{AB} = \begin{pmatrix} 3 \\ 0 \\ 4 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 2 \\ -2 \\ 5 \end{pmatrix} \]

The shortest distance \( d \) is the length of the projection of \( \vec{AB} \) onto \( \mathbf{n} \):
\[ d = \frac{|\vec{AB} \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{|2(1) + (-2)(5) + 5(3)|}{\sqrt{1^2 + 5^2 + 3^2}} = \frac{|2 - 10 + 15|}{\sqrt{35}} = \frac{7}{\sqrt{35}} = \frac{\sqrt{35}}{5} \approx 1.18 \]

(b)
Since \( \Pi_1 \) contains \( l_1 \) and is parallel to \( l_2 \), its normal vector is \( \mathbf{n} = \begin{pmatrix} 1 \\ 5 \\ 3 \end{pmatrix} \).
Using the point \( (1, 2, -1) \) on \( l_1 \):
\[ \mathbf{r} \cdot \begin{pmatrix} 1 \\ 5 \\ 3 \end{pmatrix} = 1(1) + 5(2) + 3(-1) = 8 \]
Thus, the Cartesian equation is \( x + 5y + 3z = 8 \).

(c)(i)
The normal to \( \Pi_1 \) is \( \mathbf{n}_1 = \begin{pmatrix} 1 \\ 5 \\ 3 \end{pmatrix} \) with \( |\mathbf{n}_1| = \sqrt{35} \).
The normal to \( \Pi_2 \) is \( \mathbf{n}_2 = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix} \) with \( |\mathbf{n}_2| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{9} = 3 \).

The angle \( \theta \) between the planes satisfies:
\[ \cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|} = \frac{|1(2) + 5(-1) + 3(2)|}{3\sqrt{35}} = \frac{3}{3\sqrt{35}} = \frac{1}{\sqrt{35}} \]
\[ \theta = \arccos\left(\frac{1}{\sqrt{35}}\right) \approx 80.272^\circ \approx 80.3^\circ \]

(c)(ii)
The direction vector \( \mathbf{d} \) of the line of intersection is given by \( \mathbf{n}_1 \times \mathbf{n}_2 \):
\[ \mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 5 & 3 \\ 2 & -1 & 2 \end{vmatrix} = \mathbf{i}(10 - (-3)) - \mathbf{j}(2 - 6) + \mathbf{k}(-1 - 10) = \begin{pmatrix} 13 \\ 4 \\ -11 \end{pmatrix} \]

To find a point lying on both planes, solve the system:
\( x + 5y + 3z = 8 \)
\( 2x - y + 2z = 9 \)

Setting \( y = 1 \):
\( x + 3z = 3 \)
\( 2x + 2z = 10 \implies x + z = 5 \implies z = 5 - x \)
Substituting into the first equation: \( x + 3(5 - x) = 3 \implies 15 - 2x = 3 \implies x = 6 \).
Then \( z = 5 - 6 = -1 \).
So \( (6, 1, -1) \) lies on the line of intersection.

Thus, a vector equation of the line of intersection is:
\[ \mathbf{r} = \begin{pmatrix} 6 \\ 1 \\ -1 \end{pmatrix} + t \begin{pmatrix} 13 \\ 4 \\ -11 \end{pmatrix} \quad (t \in \mathbb{R}) \]

(a)
- M1: For evaluating the vector product of the two direction vectors \( \mathbf{d}_1 \times \mathbf{d}_2 \).
- A1: For obtaining \( \begin{pmatrix} 1 \\ 5 \\ 3 \end{pmatrix} \) or any non-zero scalar multiple.
- M1: For applying the shortest distance formula \( \frac{|(\mathbf{a}_2 - \mathbf{a}_1) \cdot \mathbf{n}|}{|\mathbf{n}|} \).
- A1: For the correct exact distance \( \frac{7}{\sqrt{35}} \) or \( \frac{\sqrt{35}}{5} \) (or \( 1.18 \) to 3 s.f.).

(b)
- M1: For using \( \mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n} \) with their normal vector and a point on \( l_1 \).
- A1: For the correct Cartesian equation \( x + 5y + 3z = 8 \) (or any non-zero integer multiple).

(c)(i)
- M1: For using the scalar product of the two normal vectors: \( \cos\theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|} \).
- A1: For finding \( \cos\theta = \frac{1}{\sqrt{35}} \) (or unsimplified equivalent \( \frac{3}{3\sqrt{35}} \)).
- A1: For \( 80.3^\circ \) (allow \( 1.40 \text{ rad} \)).

(c)(ii)
- M1: For calculating the direction vector using \( \mathbf{n}_1 \times \mathbf{n}_2 \) or by eliminating a variable to find directional ratios.
- A1: For correct direction vector \( \begin{pmatrix} 13 \\ 4 \\ -11 \end{pmatrix} \) (or any non-zero scalar multiple).
- M1: For finding a particular point on the line by setting one coordinate and solving the resulting simultaneous equations.
- A1: For a fully correct vector equation of the line, including ' \( \mathbf{r} = \dots \) '.

(a) Express \( y \) in quotient-remainder form by algebraic division:

\[ x^2 - x - 2 = (x + 2)(x - 3) + 4 \]

\[ y = x + 2 + \frac{4}{x - 3} \]

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 \( y = x + 2 \).

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(b) Rearrange the equation into a quadratic in \( x \):

\[ y(x - 3) = x^2 - x - 2 \]

\[ x^2 - (1 + y)x + (3y - 2) = 0 \]

Since \( x \) is real, the discriminant \( \Delta \ge 0 \):

\[ [-(1 + y)]^2 - 4(1)(3y - 2) \ge 0 \]

\[ 1 + 2y + y^2 - 12y + 8 \ge 0 \]

\[ y^2 - 10y + 9 \ge 0 \]

\[ (y - 1)(y - 9) \ge 0 \]

This gives \( y \le 1 \) or \( y \ge 9 \).

Hence, there are no points on \( C \) for which \( 1 < y < 9 \).

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(c) At the stationary points, the quadratic in \( x \) has a repeated root (i.e. \( \Delta = 0 \)):

For \( y = 1 \):
\[ x^2 - 2x + 1 = 0 \implies (x - 1)^2 = 0 \implies x = 1 \]

For \( y = 9 \):
\[ x^2 - 10x + 25 = 0 \implies (x - 5)^2 = 0 \implies x = 5 \]

Therefore, the coordinates of the stationary points are \( (1, 1) \) and \( (5, 9) \).

*(Alternatively, using \( \frac{\mathrm{d}y}{\mathrm{d}x} = 1 - \frac{4}{(x-3)^2} = 0 \implies (x-3)^2 = 4 \implies x = 1 \text{ or } 5 \).)*

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(d) Axes intercepts for \( C \):

- When \( y = 0 \): \( x^2 - x - 2 = 0 \implies (x - 2)(x + 1) = 0 \implies x = -1 \) and \( x = 2 \).
- When \( x = 0 \): \( y = \frac{-2}{-3} = \frac{2}{3} \).

The sketch consists of two separate branches:
1. A left branch containing the maximum turning point at \( (1, 1) \), passing through \( (-1, 0) \), \( \left(0, \frac{2}{3}\right) \), and \( (2, 0) \), and approaching asymptotes \( y = x + 2 \) as \( x \to -\infty \) and \( x = 3 \) as \( x \to 3^- \).
2. A right branch for \( x > 3 \) containing the minimum turning point at \( (5, 9) \), approaching asymptotes \( x = 3 \) as \( x \to 3^+ \) and \( y = x + 2 \) as \( x \to \infty \).

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(e) For the curve \( y^2 = f(x) \), real values of \( y \) exist only where \( f(x) \ge 0 \):

- On the interval \( [-1, 2] \), \( f(x) \ge 0 \), producing a symmetrical closed loop intersecting the \( x \)-axis at \( (-1, 0) \) and \( (2, 0) \), intersecting the \( y \)-axis at \( \left(0, \pm \sqrt{\frac{2}{3}}\right) \), with maximum and minimum points at \( (1, 1) \) and \( (1, -1) \).
- For \( x > 3 \), \( f(x) \ge 9 > 0 \), producing two symmetrical branches for \( x > 3 \) asymptotic to \( x = 3 \), with turning points at \( (5, 3) \) and \( (5, -3) \).

(a)
- M1: Attempt algebraic division or express in form \( Ax + B + \frac{C}{x - 3} \)
- A1: State oblique asymptote \( y = x + 2 \)
- A1: State vertical asymptote \( x = 3 \)

(b)
- M1: Rearrange to a quadratic in \( x \) and apply condition \( \Delta \ge 0 \) for real \( x \) (or equivalent calculus method finding range)
- A1: Obtain correct simplified inequality \( y^2 - 10y + 9 \ge 0 \) or \( (y - 1)(y - 9) \ge 0 \)
- A1: Correct conclusion clearly stating \( y \le 1 \text{ or } y \ge 9 \), and concluding there are no points for \( 1 < y < 9 \)

(c)
- B1: Stationary point \( (1, 1) \)
- B1: Stationary point \( (5, 9) \)

(d)
- B1: Two correctly shaped branches in the correct regions relative to asymptotes
- B1: Both asymptotes \( x = 3 \) and \( y = x + 2 \) drawn as dashed lines and clearly labelled
- B1: All axes intercepts clearly marked: \( (-1, 0) \), \( (2, 0) \), and \( \left(0, \frac{2}{3}\right) \) alongside turning points \( (1, 1) \) and \( (5, 9) \)

(e)
- B1: Correct closed loop for \( -1 \le x \le 2 \) symmetric about the \( x \)-axis, with intercepts \( (-1, 0) \) and \( (2, 0) \)
- B1: Two symmetric branches for \( x > 3 \) showing asymptote \( x = 3 \) and turning points at \( (5, \pm 3) \)

(a)
Equating the two expressions for \(r\):
\[ a(1 + \cos\theta) = 3a\cos\theta \]
\[ 1 + \cos\theta = 3\cos\theta \implies 2\cos\theta = 1 \implies \cos\theta = \frac{1}{2} \]
Since \(-\frac{1}{2}\pi \le \theta \le \frac{1}{2}\pi\), the solutions are:
\[ \theta = \frac{\pi}{3} \quad \text{or} \quad \theta = -\frac{\pi}{3} \]
Substituting back into the equation for \(r\):
\[ r = 3a\cos\left(\pm\frac{\pi}{3}\right) = 3a\left(\frac{1}{2}\right) = \frac{3}{2}a \]
Thus, the points of intersection are \(\left(\frac{3}{2}a, \frac{\pi}{3}\right)\) and \(\left(\frac{3}{2}a, -\frac{\pi}{3}\right)\).

(b)
- \(C_1\) is a cardioid with a cusp at the pole (where \(\theta = \pi\)), passing through \((2a, 0)\) on the initial line and \((a, \frac{\pi}{2})\), \((a, -\frac{\pi}{2})\).
- \(C_2\) is a circle of diameter \(3a\) passing through the pole, with centre at \((\frac{3}{2}a, 0)\) in Cartesian coordinates, extending from \(\theta = -\frac{\pi}{2}\) to \(\theta = \frac{\pi}{2}\).
- The curves intersect at \(\left(\frac{3}{2}a, \pm\frac{\pi}{3}\right)\) and touch at the pole.
- Both curves are symmetrical about the initial line, \(\theta = 0\).

(c)
The perpendicular distance \(y\) of a point on \(C_1\) from the initial line is given by:
\[ y = r\sin\theta = a(1 + \cos\theta)\sin\theta = a(\sin\theta + \sin\theta\cos\theta) = a\left(\sin\theta + \frac{1}{2}\sin 2\theta\right) \]
To find the maximum distance in the upper half-plane (\(0 < \theta < \pi\)), differentiate with respect to \(\theta\):
\[ \frac{\mathrm{d}y}{\mathrm{d}\theta} = a(\cos\theta + \cos 2\theta) = a(2\cos^2\theta + \cos\theta - 1) \]
Setting \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = 0\):
\[ (2\cos\theta - 1)(\cos\theta + 1) = 0 \]
For \(0 < \theta < \pi\), \(\cos\theta + 1 \ne 0\), so \(2\cos\theta - 1 = 0 \implies \cos\theta = \frac{1}{2} \implies \theta = \frac{\pi}{3}\).

Substituting \(\theta = \frac{\pi}{3}\) into the expression for \(y\):
\[ y_{\max} = a\left(1 + \frac{1}{2}\right)\sin\left(\frac{\pi}{3}\right) = \frac{3}{2}a \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{4}a \]

(d)
By symmetry about the initial line, the total area \(A\) common to both curves is twice the area in the upper half-plane.
For \(0 \le \theta \le \frac{\pi}{3}\), \(C_1\) lies inside \(C_2\) since \(a(1+\cos\theta) \le 3a\cos\theta\).
For \(\frac{\pi}{3} \le \theta \le \frac{\pi}{2}\), \(C_2\) lies inside \(C_1\) since \(3a\cos\theta \le a(1+\cos\theta)\).

Thus:
\[ A = 2 \left[ \frac{1}{2}\int_0^{\pi/3} [a(1+\cos\theta)]^2\,\mathrm{d}\theta + \frac{1}{2}\int_{\pi/3}^{\pi/2} [3a\cos\theta]^2\,\mathrm{d}\theta \right] \]
\[ A = a^2 \int_0^{\pi/3} (1 + 2\cos\theta + \cos^2\theta)\,\mathrm{d}\theta + 9a^2 \int_{\pi/3}^{\pi/2} \cos^2\theta\,\mathrm{d}\theta \]
Using \(\cos^2\theta = \frac{1 + \cos 2\theta}{2}\):

First integral:
\[ \int_0^{\pi/3} \left( \frac{3}{2} + 2\cos\theta + \frac{1}{2}\cos 2\theta \right)\,\mathrm{d}\theta = \left[ \frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta \right]_0^{\pi/3} \]
\[ = \frac{3}{2}\left(\frac{\pi}{3}\right) + 2\left(\frac{\sqrt{3}}{2}\right) + \frac{1}{4}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{2} + \sqrt{3} + \frac{\sqrt{3}}{8} = \frac{\pi}{2} + \frac{9\sqrt{3}}{8} \]

Second integral:
\[ 9\int_{\pi/3}^{\pi/2} \frac{1 + \cos 2\theta}{2}\,\mathrm{d}\theta = \frac{9}{2}\left[ \theta + \frac{1}{2}\sin 2\theta \right]_{\pi/3}^{\pi/2} \]
\[ = \frac{9}{2}\left( \left(\frac{\pi}{2} + 0\right) - \left(\frac{\pi}{3} + \frac{\sqrt{3}}{4}\right) \right) = \frac{9}{2}\left(\frac{\pi}{6} - \frac{\sqrt{3}}{4}\right) = \frac{3\pi}{4} - \frac{9\sqrt{3}}{8} \]

Summing the two contributions:
\[ A = a^2 \left( \left(\frac{\pi}{2} + \frac{9\sqrt{3}}{8}\right) + \left(\frac{3\pi}{4} - \frac{9\sqrt{3}}{8}\right) \right) = a^2 \left(\frac{\pi}{2} + \frac{3\pi}{4}\right) = \frac{5}{4}\pi a^2 \]

(a)
- M1: Sets \(a(1+\cos\theta) = 3a\cos\theta\) and solves for \(\cos\theta\).
- A1: Finds \(\theta = \pm\frac{\pi}{3}\).
- A1: Gives correct coordinates \(\left(\frac{3}{2}a, \frac{\pi}{3}\right)\) and \(\left(\frac{3}{2}a, -\frac{\pi}{3}\right)\).

(b)
- B1: Correct sketch of cardioid \(C_1\) passing through \((2a, 0)\) and cusp at the pole.
- B1: Correct sketch of circle \(C_2\) passing through the pole with diameter along the initial line.
- B1: Correct relative positioning showing two intersection points in quadrants 1 and 4.
- B1: States the line of symmetry is the initial line (or \(\theta = 0\)).

(c)
- M1: Expresses \(y = r\sin\theta = a(1+\cos\theta)\sin\theta\) and differentiates with respect to \(\theta\).
- M1: Uses \(\cos 2\theta = 2\cos^2\theta - 1\) to obtain a quadratic in \(\cos\theta\) and solves for \(\cos\theta\).
- A1: Identifies \(\theta = \frac{\pi}{3}\) (or \(\cos\theta = \frac{1}{2}\)).
- A1: Obtains exact maximum distance \(\frac{3\sqrt{3}}{4}a\).

(d)
- M1: Applies the area formula \(\frac{1}{2}\int r^2\,\mathrm{d}\theta\) with symmetry, correctly splitting the range into \([0, \frac{\pi}{3}]\) using \(C_1\) and \([\frac{\pi}{3}, \frac{\pi}{2}]\) using \(C_2\).
- M1: Uses the double angle identity \(\cos^2\theta = \frac{1+\cos 2\theta}{2}\) on both integrands.
- A1: Correct integration of both parts.
- M1: Substitutes limits correctly into both integrated parts.
- A1: Obtains exact total area \(\frac{5}{4}\pi a^2\).

(a) The matrix of coefficients is
\[ \mathbf{A} = \begin{pmatrix} 1 & 2 & -1 \\ 2 & 3 & 1 \\ 3 & 4 & a \end{pmatrix} \]
The system does not have a unique solution when \(\det \mathbf{A} = 0\).
Evaluating the determinant:
\[ \begin{aligned} \det \mathbf{A} &= 1(3a - 4) - 2(2a - 3) - 1(8 - 9) \\ &= 3a - 4 - 4a + 6 + 1 \\ &= 3 - a \end{aligned} \]
Setting \(\det \mathbf{A} = 0\) gives:
\[ 3 - a = 0 \implies a = 3 \]

(b) When \(a = 3\), consider the augmented matrix:
\[ \begin{pmatrix} 1 & 2 & -1 & | & 3 \\ 2 & 3 & 1 & | & 5 \\ 3 & 4 & 3 & | & b \end{pmatrix} \]
Applying row operations:
\(R_2 \to R_2 - 2R_1\):
\[ \begin{pmatrix} 0 & -1 & 3 & | & -1 \end{pmatrix} \]
\(R_3 \to R_3 - 3R_1\):
\[ \begin{pmatrix} 0 & -2 & 6 & | & b - 9 \end{pmatrix} \]
\(R_3 \to R_3 - 2R_2\):
\[ \begin{pmatrix} 0 & 0 & 0 & | & (b - 9) - 2(-1) \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 & | & b - 7 \end{pmatrix} \]
For the system to be consistent, the bottom row must be all zeros:
\[ b - 7 = 0 \implies b = 7 \]
To find the general solution, let \(z = \lambda\), where \(\lambda \in \mathbb{R}\).
From \(R_2\):
\[ -y + 3\lambda = -1 \implies y = 1 + 3\lambda \]
From \(R_1\):
\[ x + 2(1 + 3\lambda) - \lambda = 3 \implies x + 2 + 5\lambda = 3 \implies x = 1 - 5\lambda \]
Thus, the general solution is
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} -5 \\ 3 \\ 1 \end{pmatrix}, \quad \lambda \in \mathbb{R} \]

(c) When \(a = 3\) and \(b = 10\), the system is inconsistent (since \(b \ne 7\)). Since no two of the normal vectors \(\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}\), \(\begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix}\), and \(\begin{pmatrix} 3 \\ 4 \\ 3 \end{pmatrix}\) are parallel, no two planes are parallel. Therefore, the three planes form a triangular prism (intersecting pairwise in three distinct parallel lines).

(a)
M1: Sets the determinant of the coefficient matrix equal to zero, or uses equivalent Gaussian elimination.
A1: Obtains \(a = 3\).

(b)
M1: Applies row operations on the augmented matrix with \(a = 3\) to find the condition for consistency.
A1: Obtains \(b = 7\).
A1: Correct general solution expressed in terms of a single parameter (e.g. \(x = 1 - 5\lambda, y = 1 + 3\lambda, z = \lambda\) or vector form).

(c)
B1: States that the three planes form a triangular prism (or planes intersect pairwise in three parallel lines / no common point with no planes parallel).

(a)
To find the eigenvalues of \(\mathbf{A}\), we solve the characteristic equation \(\det(\mathbf{A} - \lambda \mathbf{I}) = 0\):
\[ \begin{vmatrix} 1 - \lambda & 1 & 2 \\ 0 & 2 - \lambda & 2 \\ -1 & 1 & 4 - \lambda \end{vmatrix} = 0 \]
Expanding along the first column:
\[ (1 - \lambda) \left[ (2 - \lambda)(4 - \lambda) - 2 \right] - (-1) \left[ 2 - 2(2 - \lambda) \right] = 0 \]
\[ (1 - \lambda)(\lambda^2 - 6\lambda + 6) + (2 - 4 + 2\lambda) = 0 \]
\[ (1 - \lambda)(\lambda^2 - 6\lambda + 6) + 2(\lambda - 1) = 0 \]
\[ (1 - \lambda)(\lambda^2 - 6\lambda + 6 - 2) = 0 \]
\[ (1 - \lambda)(\lambda^2 - 6\lambda + 8) = 0 \]
\[ (1 - \lambda)(\lambda - 2)(\lambda - 4) = 0 \]
Thus, the eigenvalues are \(\lambda_1 = 1\), \(\lambda_2 = 2\), and \(\lambda_3 = 4\).

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

For \(\lambda = 2\):
\[ (\mathbf{A} - 2\mathbf{I})\mathbf{e}_2 = \begin{pmatrix} -1 & 1 & 2 \\ 0 & 0 & 2 \\ -1 & 1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \]
From the second row, \(2z = 0 \implies z = 0\).
From the first row, \(-x + y = 0 \implies x = y\).
Setting \(x = 1\), we obtain the eigenvector \(\mathbf{e}_2 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}\).

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

(c)
Let \(\mathbf{P}\) be the matrix of eigenvectors and \(\mathbf{D}\) the diagonal matrix of eigenvalues:
\[ \mathbf{P} = \begin{pmatrix} 1 & 1 & 1 \\ -2 & 1 & 1 \\ 1 & 0 & 1 \end{pmatrix}, \quad \mathbf{D} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{pmatrix} \]
We find \(\mathbf{P}^{-1}\):
\[ \det(\mathbf{P}) = 1(1 - 0) - 1(-2 - 1) + 1(0 - 1) = 1 + 3 - 1 = 3 \]
The adjugate matrix of \(\mathbf{P}\) is:
\[ \operatorname{adj}(\mathbf{P}) = \begin{pmatrix} 1 & -1 & 0 \\ 3 & 0 & -3 \\ -1 & 1 & 3 \end{pmatrix} \]
Hence,
\[ \mathbf{P}^{-1} = \frac{1}{3} \begin{pmatrix} 1 & -1 & 0 \\ 3 & 0 & -3 \\ -1 & 1 & 3 \end{pmatrix} \]
Now compute \(\mathbf{A}^n = \mathbf{P} \mathbf{D}^n \mathbf{P}^{-1}\):
\[ \mathbf{P} \mathbf{D}^n = \begin{pmatrix} 1 & 1 & 1 \\ -2 & 1 & 1 \\ 1 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2^n & 0 \\ 0 & 0 & 4^n \end{pmatrix} = \begin{pmatrix} 1 & 2^n & 4^n \\ -2 & 2^n & 4^n \\ 1 & 0 & 4^n \end{pmatrix} \]
Multiplying by \(\mathbf{P}^{-1}\):
\[ \mathbf{A}^n = \frac{1}{3} \begin{pmatrix} 1 & 2^n & 4^n \\ -2 & 2^n & 4^n \\ 1 & 0 & 4^n \end{pmatrix} \begin{pmatrix} 1 & -1 & 0 \\ 3 & 0 & -3 \\ -1 & 1 & 3 \end{pmatrix} \]
\[ \mathbf{A}^n = \frac{1}{3} \begin{pmatrix} 1 + 3(2^n) - 4^n & 4^n - 1 & 3(4^n - 2^n) \\ -2 + 3(2^n) - 4^n & 2 + 4^n & 3(4^n - 2^n) \\ 1 - 4^n & 4^n - 1 & 3(4^n) \end{pmatrix} \]

(a) [3 marks]
- M1: Sets up \(\det(\mathbf{A} - \lambda \mathbf{I}) = 0\) and attempts expansion to find the characteristic polynomial.
- A1: Obtains correct characteristic equation, e.g., \((1 - \lambda)(\lambda - 2)(\lambda - 4) = 0\) or \(-\lambda^3 + 7\lambda^2 - 14\lambda + 8 = 0\).
- A1: States all three correct eigenvalues: \(\lambda = 1, 2, 4\).

(b) [3 marks]
- M1: Substitutes at least one eigenvalue into \((\mathbf{A} - \lambda \mathbf{I})\mathbf{v} = \mathbf{0}\) and sets up a system of linear equations.
- A1: Obtains two correct eigenvectors.
- A1: Obtains all three correct eigenvectors (or any non-zero scalar multiples).

(c) [3 marks]
- B1: Correctly states \(\mathbf{P}\) and \(\mathbf{D}\) consistent with eigenvectors and eigenvalues from parts (a) and (b).
- M1: Finds \(\mathbf{P}^{-1}\) correctly and attempts the product \(\mathbf{P}\mathbf{D}^n\mathbf{P}^{-1}\).
- A1: Fully correct matrix \(\mathbf{A}^n\) in terms of \(n\).

Rewrite the differential equation in standard linear form by dividing throughout by \(x\):
\[ \frac{\mathrm{d}y}{\mathrm{d}x} + \frac{2}{x}y = \frac{\ln x}{x} \]

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

Multiply the differential equation by the integrating factor:
\[ x^2 \frac{\mathrm{d}y}{\mathrm{d}x} + 2xy = x\ln x \]
\[ \frac{\mathrm{d}}{\mathrm{d}x}\left(x^2 y\right) = x\ln x \]

Integrate both sides with respect to \(x\):
\[ x^2 y = \int x\ln x\,\mathrm{d}x \]

Use integration by parts with \(u = \ln x\) and \(\frac{\mathrm{d}v}{\mathrm{d}x} = x\), so \(\frac{\mathrm{d}u}{\mathrm{d}x} = \frac{1}{x}\) and \(v = \frac{1}{2}x^2\):
\[ \int x\ln x\,\mathrm{d}x = \frac{1}{2}x^2\ln x - \int \frac{1}{2}x^2 \cdot \frac{1}{x}\,\mathrm{d}x = \frac{1}{2}x^2\ln x - \int \frac{1}{2}x\,\mathrm{d}x = \frac{1}{2}x^2\ln x - \frac{1}{4}x^2 + c \]

Thus, the general solution is:
\[ x^2 y = \frac{1}{2}x^2\ln x - \frac{1}{4}x^2 + c \]

Apply the initial condition \(y = \frac{3}{4}\) when \(x = 1\):
\[ (1)^2 \left(\frac{3}{4}\right) = \frac{1}{2}(1)^2\ln(1) - \frac{1}{4}(1)^2 + c \]
\[ \frac{3}{4} = 0 - \frac{1}{4} + c \implies c = 1 \]

Substitute \(c = 1\) into the equation and solve for \(y\):
\[ x^2 y = \frac{1}{2}x^2\ln x - \frac{1}{4}x^2 + 1 \]
\[ y = \frac{1}{2}\ln x - \frac{1}{4} + \frac{1}{x^2} \]

M1: Divide by \(x\) to express the differential equation in the standard form \(\frac{\mathrm{d}y}{\mathrm{d}x} + P(x)y = Q(x)\).

M1: Correct method for finding the integrating factor \(\mathrm{e}^{\int \frac{2}{x}\,\mathrm{d}x}\).

A1: Obtain correct integrating factor \(x^2\).

M1: Set up the integrated form \(x^2 y = \int x\ln x\,\mathrm{d}x\).

M1: Apply integration by parts to \(\int x\ln x\,\mathrm{d}x\).

A1: Obtain correct integrated expression \(\frac{1}{2}x^2\ln x - \frac{1}{4}x^2 (+ c)\).

A1: Substitute \(x = 1, y = \frac{3}{4}\) to find \(c = 1\) and state the correct explicit solution \(y = \frac{1}{2}\ln x - \frac{1}{4} + \frac{1}{x^2}\) (or equivalent single fraction \(y = \frac{2x^2\ln x - x^2 + 4}{4x^2}\)).

(a)
Differentiating \(x\) and \(y\) with respect to \(t\):
\[ \frac{\mathrm{d}x}{\mathrm{d}t} = 1 - \operatorname{sech}^2 t = \tanh^2 t \]
\[ \frac{\mathrm{d}y}{\mathrm{d}t} = -\operatorname{sech} t \tanh t \]

Using the chain rule for parametric differentiation:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{\mathrm{d}x}{\mathrm{d}t}} = \frac{-\operatorname{sech} t \tanh t}{\tanh^2 t} = -\frac{\operatorname{sech} t}{\tanh t} = -\frac{1/\cosh t}{\sinh t/\cosh t} = -\frac{1}{\sinh t} = -\operatorname{cosech} t \]

To find the second derivative:
\[ \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right) \div \frac{\mathrm{d}x}{\mathrm{d}t} \]
\[ \frac{\mathrm{d}}{\mathrm{d}t}(-\operatorname{cosech} t) = \operatorname{cosech} t \coth t = \frac{\cosh t}{\sinh^2 t} \]
\[ \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = \frac{\frac{\cosh t}{\sinh^2 t}}{\tanh^2 t} = \frac{\frac{\cosh t}{\sinh^2 t}}{\frac{\sinh^2 t}{\cosh^2 t}} = \frac{\cosh^3 t}{\sinh^4 t} \]

(b)
The arc length \(s\) is given by:
\[ s = \int_0^{\ln 3} \sqrt{\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2} \, \mathrm{d}t \]
\[ \left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2 = \tanh^4 t + \operatorname{sech}^2 t \tanh^2 t = \tanh^2 t (\tanh^2 t + \operatorname{sech}^2 t) = \tanh^2 t \]
Since \(t \ge 0\), \(\sqrt{\tanh^2 t} = \tanh t\).

\[ s = \int_0^{\ln 3} \tanh t \, \mathrm{d}t = \left[ \ln(\cosh t) \right]_0^{\ln 3} \]
At \(t = \ln 3\):
\[ \cosh(\ln 3) = \frac{\mathrm{e}^{\ln 3} + \mathrm{e}^{-\ln 3}}{2} = \frac{3 + \frac{1}{3}}{2} = \frac{5}{3} \]
At \(t = 0\):
\[ \cosh(0) = 1 \]
\[ s = \ln\left(\frac{5}{3}\right) - \ln(1) = \ln\left(\frac{5}{3}\right) \]

(c)
The surface area \(S\) of revolution about the \(x\)-axis is:
\[ S = 2\pi \int_0^{\ln 3} y \sqrt{\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2} \, \mathrm{d}t = 2\pi \int_0^{\ln 3} \operatorname{sech} t \tanh t \, \mathrm{d}t \]
\[ S = 2\pi \left[ -\operatorname{sech} t \right]_0^{\ln 3} \]
\[ \operatorname{sech}(\ln 3) = \frac{1}{\cosh(\ln 3)} = \frac{3}{5}, \quad \operatorname{sech}(0) = 1 \]
\[ S = 2\pi \left( -\frac{3}{5} - (-1) \right) = 2\pi \left( \frac{2}{5} \right) = \frac{4\pi}{5} \]

(a)
- M1: Finds \(\frac{\mathrm{d}x}{\mathrm{d}t} = \tanh^2 t\) (or \(1 - \operatorname{sech}^2 t\)) and \(\frac{\mathrm{d}y}{\mathrm{d}t} = -\operatorname{sech} t \tanh t\).
- A1: Correctly shows \(\frac{\mathrm{d}y}{\mathrm{d}x} = -\operatorname{cosech} t\) with clear algebraic steps.
- M1: Uses \(\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right) \div \frac{\mathrm{d}x}{\mathrm{d}t}\).
- A1: Obtains \(\frac{\cosh^3 t}{\sinh^4 t}\) (or equivalent single fraction in \(\sinh t\) and \(\cosh t\)).

(b)
- M1: Simplifies \(\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2\) to \(\tanh^2 t\) using \(\tanh^2 t + \operatorname{sech}^2 t = 1\).
- M1: Integrates \(\tanh t\) to \(\ln(\cosh t)\) and substitutes limits \(0\) and \(\ln 3\).
- A1: Obtains exact answer \(\ln\left(\frac{5}{3}\right)\).

(c)
- M1: Sets up correct integral for surface area: \(2\pi \int_0^{\ln 3} \operatorname{sech} t \tanh t \, \mathrm{d}t\).
- M1: Integrates \(\operatorname{sech} t \tanh t\) to \(-\operatorname{sech} t\) and applies limits correctly.
- A1: Obtains exact answer \(\frac{4\pi}{5}\).

(a) Consider the interval \( [1, n+1] \) divided into \( n \) subintervals \( [r, r+1] \) for \( r = 1, 2, \dots, n \), each of width \( 1 \).
Since \( f(x) = \dfrac{1}{\sqrt{2x-1}} \) is a strictly decreasing function for \( x \ge 1 \), on each subinterval \( [r, r+1] \), the curve lies below the rectangle with height \( f(r) = \dfrac{1}{\sqrt{2r-1}} \).

The total area of these \( n \) rectangles is greater than the area under the curve:
\[ \sum_{r=1}^n 1 \times f(r) > \int_1^{n+1} \frac{1}{\sqrt{2x-1}} \, \mathrm{d}x. \]

Evaluating the definite integral:
\[ \int_1^{n+1} (2x - 1)^{-\frac{1}{2}} \, \mathrm{d}x = \left[ \sqrt{2x - 1} \right]_1^{n+1} = \sqrt{2(n+1) - 1} - \sqrt{2(1) - 1} = \sqrt{2n + 1} - 1. \]

Thus,
\[ \sum_{r=1}^n \frac{1}{\sqrt{2r - 1}} > \sqrt{2n + 1} - 1. \]

(b) Consider the interval \( [1, n] \) divided into \( n - 1 \) subintervals \( [r-1, r] \) for \( r = 2, 3, \dots, n \).
On each subinterval \( [r-1, r] \), the rectangle of width \( 1 \) and height \( f(r) = \dfrac{1}{\sqrt{2r-1}} \) lies completely below the curve \( y = f(x) \).

Therefore, the total area of these rectangles is less than the area under the curve from \( x = 1 \) to \( x = n \):
\[ \sum_{r=2}^n f(r) < \int_1^n \frac{1}{\sqrt{2x-1}} \, \mathrm{d}x. \]

Evaluating the integral:
\[ \int_1^n \frac{1}{\sqrt{2x-1}} \, \mathrm{d}x = \left[ \sqrt{2x-1} \right]_1^n = \sqrt{2n - 1} - 1. \]

So,
\[ \sum_{r=2}^n \frac{1}{\sqrt{2r - 1}} < \sqrt{2n - 1} - 1. \]

Adding \( f(1) = \dfrac{1}{\sqrt{2(1)-1}} = 1 \) to both sides gives:
\[ \sum_{r=1}^n \frac{1}{\sqrt{2r - 1}} = 1 + \sum_{r=2}^n \frac{1}{\sqrt{2r - 1}} < 1 + (\sqrt{2n - 1} - 1) = \sqrt{2n - 1}. \]

(c) Setting \( n = 60 \) in both inequalities:
From (a):
\[ \sum_{r=1}^{60} \frac{1}{\sqrt{2r - 1}} > \sqrt{2(60) + 1} - 1 = \sqrt{121} - 1 = 11 - 1 = 10. \]

From (b):
\[ \sum_{r=1}^{60} \frac{1}{\sqrt{2r - 1}} < \sqrt{2(60) - 1} = \sqrt{119}. \]

Since \( 100 < 119 < 121 \), we have \( 10 < \sqrt{119} < 11 \).

Combining the two bounds:
\[ 10 < \sum_{r=1}^{60} \frac{1}{\sqrt{2r - 1}} < \sqrt{119} < 11. \]

Therefore, the integer part of the sum is \( 10 \).

(a)
- M1: Shows or sketches left-endpoint rectangles of width 1 on \( [1, n+1] \) lying above the curve \( y = f(x) \), expressing total area as \( \sum_{r=1}^n f(r) \).
- M1: Correct method to integrate \( \int_1^{n+1} (2x-1)^{-\frac{1}{2}} \, \mathrm{d}x \).
- A1: Fully correct integration \( [\sqrt{2x-1}]_1^{n+1} = \sqrt{2n+1} - 1 \) and complete justification leading to the given inequality.

(b)
- M1: Shows or sketches right-endpoint rectangles of width 1 on \( [1, n] \) lying below the curve \( y = f(x) \).
- M1: Sets up \( \sum_{r=2}^n f(r) < \int_1^n \frac{1}{\sqrt{2x-1}} \, \mathrm{d}x \) and integrates correctly to obtain \( \sqrt{2n-1} - 1 \).
- A1: Adds \( f(1) = 1 \) to complete the rigorous proof that \( \sum_{r=1}^n \frac{1}{\sqrt{2r-1}} < \sqrt{2n-1} \).

(c)
- M1: Applies \( n = 60 \) to both derived inequalities.
- A1: Obtains explicit bounds \( 10 < \sum_{r=1}^{60} \frac{1}{\sqrt{2r-1}} < \sqrt{119} \) (or notes \( \sqrt{119} \approx 10.909 \)).
- A1: Concludes that the integer part is 10 with clear justification that \( 10 < \text{Sum} < 11 \).

(a) For \( n \ge 2 \), write \( I_n = \int_0^1 x^{n-1} \cdot \frac{x}{\sqrt{2 - x^2}} \, \mathrm{d}x \).
Let \( u = x^{n-1} \implies \frac{\mathrm{d}u}{\mathrm{d}x} = (n-1)x^{n-2} \) and \( \frac{\mathrm{d}v}{\mathrm{d}x} = \frac{x}{\sqrt{2 - x^2}} \implies v = -\sqrt{2 - x^2} \).
Applying integration by parts:
\[ I_n = \left[ -x^{n-1}\sqrt{2 - x^2} \right]_0^1 - \int_0^1 (n-1)x^{n-2}\left(-\sqrt{2 - x^2}\right) \, \mathrm{d}x \]
Since \( n \ge 2 \), at \( x = 0 \) the term \( 0^{n-1}\sqrt{2} = 0 \), and at \( x = 1 \) the term is \( -(1)^{n-1}\sqrt{2 - 1} = -1 \).
\[ I_n = -1 + (n-1)\int_0^1 x^{n-2}\sqrt{2 - x^2} \, \mathrm{d}x \]
Rewrite the integrand:
\[ x^{n-2}\sqrt{2 - x^2} = x^{n-2} \cdot \frac{2 - x^2}{\sqrt{2 - x^2}} = \frac{2x^{n-2}}{\sqrt{2 - x^2}} - \frac{x^n}{\sqrt{2 - x^2}} \]
Thus,
\[ I_n = -1 + 2(n-1)I_{n-2} - (n-1)I_n \]
\[ n I_n = 2(n-1)I_{n-2} - 1 \]

(b) First evaluate \( I_0 \):
\[ I_0 = \int_0^1 \frac{1}{\sqrt{2 - x^2}} \, \mathrm{d}x = \left[ \arcsin\left(\frac{x}{\sqrt{2}}\right) \right]_0^1 = \arcsin\left(\frac{1}{\sqrt{2}}\right) - 0 = \frac{\pi}{4} \]
Using the reduction formula with \( n = 2 \):
\[ 2 I_2 = 2(1)I_0 - 1 = 2\left(\frac{\pi}{4}\right) - 1 = \frac{\pi}{2} - 1 \implies I_2 = \frac{\pi}{4} - \frac{1}{2} \]
Using the reduction formula with \( n = 4 \):
\[ 4 I_4 = 2(3)I_2 - 1 = 6\left(\frac{\pi}{4} - \frac{1}{2}\right) - 1 = \frac{3\pi}{2} - 3 - 1 = \frac{3\pi}{2} - 4 \]
\[ I_4 = \frac{3\pi}{8} - 1 \]

(c) The volume of revolution \( V \) about the \( x \)-axis is:
\[ V = \pi \int_0^1 y^2 \, \mathrm{d}x = \pi \int_0^1 \left( \frac{x^2}{(2 - x^2)^{1/4}} \right)^2 \, \mathrm{d}x = \pi \int_0^1 \frac{x^4}{\sqrt{2 - x^2}} \, \mathrm{d}x = \pi I_4 \]
Using the value of \( I_4 \) from part (b):
\[ V = \pi \left( \frac{3\pi}{8} - 1 \right) = \frac{3\pi^2}{8} - \pi \]

(a)
M1: Expresses integrand as \( x^{n-1} \cdot \frac{x}{\sqrt{2-x^2}} \) and applies integration by parts to obtain \( \pm x^{n-1}\sqrt{2-x^2} \pm \int (n-1)x^{n-2}\sqrt{2-x^2} \, \mathrm{d}x \).
A1: Evaluates limits correctly to get \( -1 \) and writes integral term as \( +(n-1)\int_0^1 x^{n-2}\sqrt{2-x^2} \, \mathrm{d}x \).
M1: Uses \( \sqrt{2-x^2} = \frac{2-x^2}{\sqrt{2-x^2}} \) to express integral in terms of \( I_{n-2} \) and \( I_n \).
A1: Fully correct argument leading to the given reduction formula \( n I_n = 2(n-1)I_{n-2} - 1 \).

(b)
B1: Obtains \( I_0 = \left[\arcsin(x/\sqrt{2})\right]_0^1 = \frac{\pi}{4} \).
M1: Applies the reduction formula to calculate \( I_2 \) and subsequently \( I_4 \).
A1: Obtains exact value \( I_4 = \frac{3\pi}{8} - 1 \) (or equivalent single fraction \( \frac{3\pi - 8}{8} \)).

(c)
M1: Uses standard volume of revolution formula \( V = \pi \int_0^1 y^2 \, \mathrm{d}x \).
M1: Squares \( y \) and identifies the resulting integral as \( \pi I_4 \).
A1: Obtains exact value \( \frac{3\pi^2}{8} - \pi \) or \( \pi\left(\frac{3\pi}{8} - 1\right) \).

(a) By de Moivre’s theorem,
\[ \cos 5\theta + \mathrm{i}\sin 5\theta = (\cos\theta + \mathrm{i}\sin\theta)^5 \]
Expanding by the binomial theorem:
\[ (\cos\theta + \mathrm{i}\sin\theta)^5 = \cos^5\theta + 5\mathrm{i}\cos^4\theta\sin\theta - 10\cos^3\theta\sin^2\theta - 10\mathrm{i}\cos^2\theta\sin^3\theta + 5\cos\theta\sin^4\theta + \mathrm{i}\sin^5\theta \]
Equating real parts:
\[ \cos 5\theta = \cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta \]
Using $\sin^2\theta = 1 - \cos^2\theta$:
\[ \cos 5\theta = \cos^5\theta - 10\cos^3\theta(1 - \cos^2\theta) + 5\cos\theta(1 - \cos^2\theta)^2 \]
\[ \cos 5\theta = \cos^5\theta - 10\cos^3\theta + 10\cos^5\theta + 5\cos\theta(1 - 2\cos^2\theta + \cos^4\theta) \]
\[ \cos 5\theta = 11\cos^5\theta - 10\cos^3\theta + 5\cos\theta - 10\cos^3\theta + 5\cos^5\theta \]
\[ \cos 5\theta = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta \]

(b) The equation $32x^5 - 40x^3 + 10x - 1 = 0$ can be rewritten as:
\[ 2(16x^5 - 20x^3 + 5x) = 1 \implies 16x^5 - 20x^3 + 5x = \frac{1}{2} \]
Letting $x = \cos\theta$, from part (a) this becomes:
\[ \cos 5\theta = \frac{1}{2} \]
The general solution is:
\[ 5\theta = 2n\pi \pm \frac{\pi}{3} = \frac{(6n \pm 1)\pi}{3} \implies \theta = \frac{(6n \pm 1)\pi}{15} \]
Choosing values of $n$ to generate 5 distinct values of $\theta \in [0, \pi]$:
- $n = 0, +$: $\theta = \frac{\pi}{15}$
- $n = 1, -$: $\theta = \frac{5\pi}{15}$
- $n = 1, +$: $\theta = \frac{7\pi}{15}$
- $n = 2, -$: $\theta = \frac{11\pi}{15}$
- $n = 2, +$: $\theta = \frac{13\pi}{15}$
Since $x = \cos\theta$ and the 5 values of $\theta$ lie in $[0, \pi]$, these yield 5 distinct values of $x$. As the polynomial has degree 5, all roots are given by:
\[ x = \cos\left(\frac{k\pi}{15}\right), \quad k \in \{1, 5, 7, 11, 13\} \]

(c) (i) $\cos\left(\frac{5\pi}{15}\right) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$.

(ii) For the polynomial $32x^5 - 40x^3 + 10x - 1 = 0$, the product of the five roots is:
\[ \prod_{i=1}^5 x_i = -\frac{-1}{32} = \frac{1}{32} \]
Therefore,
\[ \cos\left(\frac{\pi}{15}\right)\cos\left(\frac{5\pi}{15}\right)\cos\left(\frac{7\pi}{15}\right)\cos\left(\frac{11\pi}{15}\right)\cos\left(\frac{13\pi}{15}\right) = \frac{1}{32} \]
Substituting $\cos\left(\frac{5\pi}{15}\right) = \frac{1}{2}$:
\[ \cos\left(\frac{\pi}{15}\right)\cos\left(\frac{7\pi}{15}\right)\cos\left(\frac{11\pi}{15}\right)\cos\left(\frac{13\pi}{15}\right) = \frac{1/32}{1/2} = \frac{1}{16} \]
Using $\cos(\pi - \alpha) = -\cos\alpha$:
\[ \cos\left(\frac{11\pi}{15}\right) = \cos\left(\pi - \frac{4\pi}{15}\right) = -\cos\left(\frac{4\pi}{15}\right) \]
\[ \cos\left(\frac{13\pi}{15}\right) = \cos\left(\pi - \frac{2\pi}{15}\right) = -\cos\left(\frac{2\pi}{15}\right) \]
Thus,
\[ \cos\left(\frac{11\pi}{15}\right)\cos\left(\frac{13\pi}{15}\right) = \left(-\cos\left(\frac{4\pi}{15}\right)\right)\left(-\cos\left(\frac{2\pi}{15}\right)\right) = \cos\left(\frac{2\pi}{15}\right)\cos\left(\frac{4\pi}{15}\right) \]
Substituting this back gives:
\[ \cos\left(\frac{\pi}{15}\right)\cos\left(\frac{2\pi}{15}\right)\cos\left(\frac{4\pi}{15}\right)\cos\left(\frac{7\pi}{15}\right) = \frac{1}{16} \]

(a) [4 marks]
- M1: Applies de Moivre's theorem to express $\cos 5\theta + \mathrm{i}\sin 5\theta = (\cos\theta + \mathrm{i}\sin\theta)^5$ and uses binomial expansion with correct coefficients (1, 5, 10, 10, 5, 1).
- M1: Equates real parts to obtain $\cos 5\theta = \cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta$.
- M1: Uses $\sin^2\theta = 1 - \cos^2\theta$ to express all terms purely in terms of $\cos\theta$.
- A1: Correctly simplifies with all intermediate algebraic steps clearly shown to reach the given result (AG).

(b) [3 marks]
- M1: Relates the equation $32x^5 - 40x^3 + 10x - 1 = 0$ to $2\cos 5\theta = 1$ (or $\cos 5\theta = \frac{1}{2}$) via substitution $x = \cos\theta$.
- M1: Solves $\cos 5\theta = \frac{1}{2}$ to find values of $5\theta = 2n\pi \pm \frac{\pi}{3}$ and determines corresponding values of $\theta$.
- A1: Identifies the five distinct values of $k \in \{1, 5, 7, 11, 13\}$ and explains/justifies why these give all five roots.

(c)(i) [1 mark]
- B1: States $\frac{1}{2}$ (or exact equivalent).

(c)(ii) [3 marks]
- M1: Uses the product of roots formula $\prod x_i = \frac{1}{32}$ for the given degree 5 polynomial.
- A1: Obtains $\cos\left(\frac{\pi}{15}\right)\cos\left(\frac{7\pi}{15}\right)\cos\left(\frac{11\pi}{15}\right)\cos\left(\frac{13\pi}{15}\right) = \frac{1}{16}$.
- A1: Uses $\cos\left(\frac{11\pi}{15}\right) = -\cos\left(\frac{4\pi}{15}\right)$ and $\cos\left(\frac{13\pi}{15}\right) = -\cos\left(\frac{2\pi}{15}\right)$ to deduce the final target expression with full justification.

(a) Given \(z = y^2\), differentiate with respect to \(x\):
\[ \frac{\mathrm{d}z}{\mathrm{d}x} = 2y \frac{\mathrm{d}y}{\mathrm{d}x} \]
Differentiating again with respect to \(x\) using the product rule:
\[ \frac{\mathrm{d}^2z}{\mathrm{d}x^2} = 2y \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 2\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 \]
Substitute these into the left-hand side of the linear ODE:
\[ \frac{\mathrm{d}^2z}{\mathrm{d}x^2} + 4\frac{\mathrm{d}z}{\mathrm{d}x} + 13z = \left(2y \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 2\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2\right) + 4\left(2y \frac{\mathrm{d}y}{\mathrm{d}x}\right) + 13y^2 \]
\[ = 2y \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 2\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 + 8y \frac{\mathrm{d}y}{\mathrm{d}x} + 13y^2 = 26x + 21 \]
Thus,
\[ \frac{\mathrm{d}^2z}{\mathrm{d}x^2} + 4\frac{\mathrm{d}z}{\mathrm{d}x} + 13z = 26x + 21. \]

(b) Auxiliary equation for the complementary function:
\[ m^2 + 4m + 13 = 0 \implies (m + 2)^2 + 9 = 0 \implies m = -2 \pm 3\mathrm{i} \]
So the complementary function is:
\[ z_{\text{CF}} = \mathrm{e}^{-2x}(A\cos 3x + B\sin 3x) \]
For the particular integral, try \(z = px + q\):
\[ \frac{\mathrm{d}z}{\mathrm{d}x} = p, \quad \frac{\mathrm{d}^2z}{\mathrm{d}x^2} = 0 \]
Substitute into the differential equation:
\[ 0 + 4p + 13(px + q) = 26x + 21 \]
\[ 13px + (4p + 13q) = 26x + 21 \]
Equating coefficients:
\[ 13p = 26 \implies p = 2 \]
\[ 4(2) + 13q = 21 \implies 13q = 13 \implies q = 1 \]
So the particular integral is \(z_{\text{PI}} = 2x + 1\).

The general solution for \(z\) is:
\[ z = \mathrm{e}^{-2x}(A\cos 3x + B\sin 3x) + 2x + 1 \]
Since \(z = y^2\), the general solution for \(y\) is:
\[ y = \pm \sqrt{\mathrm{e}^{-2x}(A\cos 3x + B\sin 3x) + 2x + 1} \]

(c) When \(x = 0\), \(y = 2 \implies z(0) = 2^2 = 4\).
From the general solution for \(z\):
\[ z(0) = A + 1 = 4 \implies A = 3 \]
Also, \(\frac{\mathrm{d}z}{\mathrm{d}x} = 2y \frac{\mathrm{d}y}{\mathrm{d}x}\), so when \(x = 0\), \(y = 2\) and \(\frac{\mathrm{d}y}{\mathrm{d}x} = -1\):
\[ z'(0) = 2(2)(-1) = -4 \]
Differentiating \(z(x)\):
\[ z'(x) = -2\mathrm{e}^{-2x}(A\cos 3x + B\sin 3x) + \mathrm{e}^{-2x}(-3A\sin 3x + 3B\cos 3x) + 2 \]
At \(x = 0\):
\[ z'(0) = -2A + 3B + 2 = -2(3) + 3B + 2 = -4 + 3B \]
Equating to \(-4\):
\[ -4 + 3B = -4 \implies 3B = 0 \implies B = 0 \]
Since \(y(0) = 2 > 0\), we take the positive square root:
\[ y = \sqrt{3\mathrm{e}^{-2x}\cos 3x + 2x + 1} \]

(a)
M1: Differentiates \(z = y^2\) twice with respect to \(x\) using the chain and product rules.
A1: Obtains correct expressions \(\frac{\mathrm{d}z}{\mathrm{d}x} = 2y\frac{\mathrm{d}y}{\mathrm{d}x}\) and \(\frac{\mathrm{d}^2z}{\mathrm{d}x^2} = 2y\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 2\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2\).
A1: Fully substitutes into the given equation to convincingly obtain the printed result.

(b)
M1: Forms and solves the auxiliary equation \(m^2 + 4m + 13 = 0\).
A1: Obtains the correct complementary function \(\mathrm{e}^{-2x}(A\cos 3x + B\sin 3x)\).
M1: Attempts a particular integral of the form \(z = px + q\), calculates derivatives and substitutes into the ODE.
A1: Finds correct values \(p = 2\) and \(q = 1\).
A1: Writes correct general solution for \(z\).
A1: Expresses the general solution for \(y\) in terms of \(x\) including \(\pm\).

(c)
M1: Uses \(x = 0, y = 2\) to find \(z(0) = 4\) and solves for \(A\).
M1: Uses \(z'(0) = 2y(0)y'(0) = -4\) and differentiates the general solution of \(z\) to form an equation in \(B\).
A1: Obtains \(A = 3\) and \(B = 0\).
A1: States the correct explicit particular solution with the positive square root chosen.

(a) Since the particle moves in a horizontal circle, there is no vertical acceleration.
Resolving forces vertically:
\[ T\cos\theta = mg \]
Given \(m = 0.6\text{ kg}\), \(g = 10\text{ m s}^{-2}\), and \(\cos\theta = 0.8\):
\[ T(0.8) = 0.6 \times 10 = 6 \]
\[ T = \frac{6}{0.8} = 7.5\text{ N} \]

(b) Since \(\cos\theta = 0.8\), we have:
\[ \sin\theta = \sqrt{1 - 0.8^2} = 0.6 \]
The radius \(r\) of the horizontal circle is:
\[ r = l\sin\theta = 2 \times 0.6 = 1.2\text{ m} \]
Applying Newton's second law horizontally towards the centre of the circle:
\[ T\sin\theta = \frac{mv^2}{r} \]
Substitute the known values:
\[ 7.5 \times 0.6 = \frac{0.6 \times v^2}{1.2} \]
\[ 4.5 = 0.5 v^2 \]
\[ v^2 = \frac{4.5}{0.5} = 9 \]
\[ v = 3\text{ m s}^{-1} \]

(a)
M1: Resolving forces vertically to form an equation involving \(T\), \(\cos\theta\), and \(mg\).
A1: Correct tension \(T = 7.5\text{ N}\).

(b)
M1: Finding the radius \(r = 1.2\text{ m}\) (or \(2\sin\theta\)) and applying \(F = \frac{mv^2}{r}\) horizontally with \(T\sin\theta\).
A1: Correct speed \(v = 3\text{ m s}^{-1}\).

(a)
Set up a Cartesian coordinate system with the origin at $O$, the $x$-axis along $OA$, and the $y$-axis along $OC$.

1. Areas and Ratios:
- Area of rectangle $OABC$: $A_1 = 6a \times 4a = 24a^2$
- Area of removed triangle $ADE$: $A_2 = \frac{1}{2} \times 3a \times 3a = \frac{9}{2}a^2$
- Area of lamina $L$: $A = 24a^2 - \frac{9}{2}a^2 = \frac{39}{2}a^2$

Ratio of areas (Rectangle : Triangle : Lamina) = $16 : 3 : 13$.

2. Centres of Mass of Components:
- Rectangle $OABC$: $(x_1, y_1) = (3a, 2a)$
- Triangle $ADE$ has vertices at $(6a, 0)$, $(6a, 3a)$, and $(3a, 0)$:
\[ x_2 = \frac{6a + 6a + 3a}{3} = 5a, \quad y_2 = \frac{0 + 3a + 0}{3} = a \]

3. **Moments about $OC$ (finding $\bar{x}$):**
\[ \frac{39}{2}a^2 \bar{x} = 24a^2(3a) - \frac{9}{2}a^2(5a) \]
\[ 39 \bar{x} = 48(3a) - 9(5a) = 144a - 45a = 99a \]
\[ \bar{x} = \frac{99}{39}a = \frac{33}{13}a \quad \text{(as required)} \]

4. **Moments about $OA$ (finding $\bar{y}$):**
\[ \frac{39}{2}a^2 \bar{y} = 24a^2(2a) - \frac{9}{2}a^2(a) \]
\[ 39 \bar{y} = 48(2a) - 9(a) = 96a - 9a = 87a \]
\[ \bar{y} = \frac{87}{39}a = \frac{29}{13}a \]

---

(b)
When the lamina is suspended freely from $O$, the line joining $O$ to the centre of mass $G(\bar{x}, \bar{y})$ hangs vertically downward.

Let $\theta$ be the angle that $OA$ makes with the vertical line $OG$:
\[ \tan \theta = \frac{\bar{y}}{\bar{x}} = \frac{\frac{29}{13}a}{\frac{33}{13}a} = \frac{29}{33} \]
\[ \theta = \arctan\left(\frac{29}{33}\right) \approx 41.3^\circ \quad (\text{or } 0.721 \text{ rad}) \]

(a)
- M1: For calculating the correct areas of the rectangle, triangle, and lamina (or correct mass/area ratios: $16 : 3 : 13$).
- B1: For correct centre of mass of the removed triangular portion: $(5a, a)$.
- A1: For forming a correct moment equation about $OC$ and obtaining the given result $\bar{x} = \frac{33}{13}a$ with full working shown.
- A1: For finding the correct distance from $OA$: $\bar{y} = \frac{29}{13}a$ (or exact equivalent).

(b)
- M1: For recognizing that $OG$ is vertical in equilibrium and stating $\tan \theta = \frac{\bar{y}}{\bar{x}}$ or $\tan(90^\circ - \theta) = \frac{\bar{x}}{\bar{y}}$.
- M1: For substituting their $\bar{x}$ and $\bar{y}$ into $\tan \theta = \frac{29}{33}$.
- A1: For $\theta = 41.3^\circ$ (or $41.3$ to 1 d.p., or $0.721$ radians to 3 s.f.).

(a) Let \(v\) be the speed of the particle when the string makes an angle \(\alpha\) with the upward vertical.

Taking the lowest point as the reference level for gravitational potential energy, the height of \(P\) above the lowest point is \(a + a\cos\alpha\).

By conservation of energy:
\[ \frac{1}{2}m u^2 = \frac{1}{2}m v^2 + mg(a + a\cos\alpha) \]

Substitute \(u^2 = \frac{7}{2}ga\):
\[ \frac{1}{2}m\left(\frac{7}{2}ga\right) = \frac{1}{2}m v^2 + mga(1 + \cos\alpha) \]
\[ \frac{7}{4}ga = \frac{1}{2}v^2 + ga + ga\cos\alpha \]
\[ v^2 = \frac{3}{2}ga - 2ga\cos\alpha \]

Applying Newton's second law along the inward radial direction towards \(O\):
\[ T + mg\cos\alpha = \frac{mv^2}{a} \]
\[ T = \frac{m}{a}\left(\frac{3}{2}ga - 2ga\cos\alpha\right) - mg\cos\alpha = mg\left(\frac{3}{2} - 3\cos\alpha\right) \]

The string becomes slack when \(T = 0\):
\[ mg\left(\frac{3}{2} - 3\cos\alpha\right) = 0 \implies 3\cos\alpha = \frac{3}{2} \implies \cos\alpha = \frac{1}{2} \]

(b) When the string becomes slack, \(\cos\alpha = \frac{1}{2}\), so \(\sin^2\alpha = 1 - \left(\frac{1}{2}\right)^2 = \frac{3}{4}\).

The speed \(v\) at this instant is given by:
\[ v^2 = \frac{3}{2}ga - 2ga\left(\frac{1}{2}\right) = \frac{1}{2}ga \]

The velocity of \(P\) is perpendicular to the string, so it makes an angle \(\alpha\) with the horizontal. The vertical component of velocity is:
\[ v_y = v\sin\alpha \]
\[ v_y^2 = v^2\sin^2\alpha = \left(\frac{1}{2}ga\right)\left(\frac{3}{4}\right) = \frac{3}{8}ga \]

After the string becomes slack, \(P\) moves freely under gravity as a projectile. The additional height gained above the point where the string becomes slack is:
\[ h_{\text{extra}} = \frac{v_y^2}{2g} = \frac{\frac{3}{8}ga}{2g} = \frac{3}{16}a \]

The height of \(P\) above the level of \(O\) when the string becomes slack is:
\[ h_0 = a\cos\alpha = \frac{1}{2}a \]

Therefore, the greatest height reached by \(P\) above the level of \(O\) is:
\[ H_{\max} = h_0 + h_{\text{extra}} = \frac{1}{2}a + \frac{3}{16}a = \frac{11}{16}a \]

(a)
- M1: For applying the principle of conservation of energy to relate the initial speed \(u\) to the speed \(v\) at angle \(\alpha\).
- A1: For obtaining a correct expression for \(v^2\), e.g. \(v^2 = \frac{3}{2}ga - 2ga\cos\alpha\).
- M1: For resolving forces radially towards \(O\) to obtain \(T + mg\cos\alpha = \frac{mv^2}{a}\) and setting \(T = 0\).
- A1: For correct completion to the given result \(\cos\alpha = \frac{1}{2}\) with no errors seen.

(b)
- M1: For finding the vertical component of velocity \(v_y = v\sin\alpha\) (or equivalent projectile energy method) to find the additional height.
- A1: For obtaining the correct additional height \(\frac{3}{16}a\) or correct \(v_y^2 = \frac{3}{8}ga\).
- A1: For the correct final answer \(\frac{11}{16}a\) (or \(0.6875a\)).

(a) Let \(v_A\) and \(v_B\) be the velocities of \(A\) and \(B\) immediately after the collision, measured in the direction of the initial motion of \(A\) (towards the wall).

By conservation of linear momentum:
\[ m u = m v_A + 3m v_B \implies v_A + 3v_B = u \]

By Newton's experimental law of restitution:
\[ v_B - v_A = e(u - 0) = eu \]

Adding the two equations:
\[ 4v_B = u + eu = u(1 + e) \implies v_B = \frac{1}{4}u(1 + e) \]

Substituting \(v_B\) into the restitution equation:
\[ v_A = v_B - eu = \frac{1}{4}u(1 + e) - eu = \frac{1}{4}u(1 - 3e) \]

Thus, the velocity of \(A\) is \(\frac{1}{4}u(1 - 3e)\) towards the wall and the velocity of \(B\) is \(\frac{1}{4}u(1 + e)\) towards the wall.

(b) After the collision, sphere \(B\) moves towards the wall with speed \(v_B = \frac{1}{4}u(1 + e)\).

Upon colliding with the wall, \(B\) rebounds with speed \(w_B\) away from the wall, given by:
\[ w_B = \frac{1}{2}v_B = \frac{1}{8}u(1 + e) \]

Now consider the motion of \(A\) and \(B\):
- If \(e \le \frac{1}{3}\), then \(v_A \ge 0\), meaning \(A\) is moving towards the wall (or is at rest). Since \(B\) is moving away from the wall, a second collision will definitely occur.
- If \(e > \frac{1}{3}\), then \(v_A < 0\), meaning \(A\) is moving away from the wall with speed \(\frac{1}{4}u(3e - 1)\).

For \(B\) to catch up with \(A\) and produce a second collision, the speed of \(B\) must be greater than the speed of \(A\):
\[ w_B > |v_A| \]
\[ \frac{1}{8}u(1 + e) > \frac{1}{4}u(3e - 1) \]
\[ 1 + e > 2(3e - 1) \]
\[ 1 + e > 6e - 2 \implies 5e < 3 \implies e < \frac{3}{5} \]

Since \(0 < e \le 1\), the complete range of values of \(e\) for which a second collision occurs is \(0 < e < \frac{3}{5}\).

(a)
M1: Applies conservation of linear momentum to the collision between \(A\) and \(B\).
M1: Applies Newton's experimental law of restitution to the collision.
A1: Obtains both correct velocities: \(v_A = \frac{1}{4}u(1 - 3e)\) and \(v_B = \frac{1}{4}u(1 + e)\).

(b)
B1: States or finds the speed of \(B\) after rebounding from the wall as \(\frac{1}{8}u(1 + e)\).
M1: Sets up a valid inequality comparing the speeds of \(A\) and \(B\) away from the wall for \(e > \frac{1}{3}\), or considers both cases \(e \le \frac{1}{3}\) and \(e > \frac{1}{3}\).
A1: Correct final range \(0 < e < \frac{3}{5}\) (or \(0 < e < 0.6\)).

(a)
During the upward motion, let \(x\) be the displacement of \(P\) upwards from \(O\). The equation of motion is:
\[ m v \frac{\mathrm{d}v}{\mathrm{d}x} = -mg - 0.005v^2 \]
Substituting \(m = 0.2\) and \(g = 10\):
\[ 0.2 v \frac{\mathrm{d}v}{\mathrm{d}x} = -2 - 0.005v^2 = -\frac{400 + v^2}{200} \]
\[ v \frac{\mathrm{d}v}{\mathrm{d}x} = -\frac{400 + v^2}{40} \]

Separating variables:
\[ \int \frac{40v}{400 + v^2}\, \mathrm{d}v = -\int \mathrm{d}x \]
\[ 20 \ln(400 + v^2) = -x + c \]

When \(x = 0\), \(v = 20\):
\[ c = 20 \ln(400 + 20^2) = 20 \ln 800 \]

Hence:
\[ x = 20 \ln 800 - 20 \ln(400 + v^2) = 20 \ln\left(\frac{800}{400 + v^2}\right) \]

At the maximum height \(H\), \(v = 0\):
\[ H = 20 \ln\left(\frac{800}{400}\right) = 20 \ln 2\text{ m} \]

---

(b)
Using the time-derivative form of acceleration:
\[ m \frac{\mathrm{d}v}{\mathrm{d}t} = -\frac{400 + v^2}{200} \implies \frac{\mathrm{d}v}{\mathrm{d}t} = -\frac{400 + v^2}{40} \]

Separating variables with limits from \(t = 0, v = 20\) to \(t = T, v = 0\):
\[ \int_{20}^0 \frac{40}{400 + v^2}\, \mathrm{d}v = -\int_0^T \mathrm{d}t \]
\[ T = 40 \int_0^{20} \frac{1}{v^2 + 20^2}\, \mathrm{d}v \]
\[ T = 40 \left[ \frac{1}{20} \arctan\left(\frac{v}{20}\right) \right]_0^{20} = 2 \left( \arctan(1) - \arctan(0) \right) \]
\[ T = 2 \times \frac{\pi}{4} = \frac{\pi}{2}\text{ s} \]

---

(c)
For the downward motion, let \(y\) be the distance fallen from the highest point and \(v\) be the downward speed. The equation of motion is:
\[ m v \frac{\mathrm{d}v}{\mathrm{d}y} = mg - 0.005v^2 \]
\[ 0.2 v \frac{\mathrm{d}v}{\mathrm{d}y} = 2 - \frac{v^2}{200} = \frac{400 - v^2}{200} \]
\[ v \frac{\mathrm{d}v}{\mathrm{d}y} = \frac{400 - v^2}{40} \]

Separating variables from \(y = 0, v = 0\) to \(y = 20\ln 2, v = v_1\):
\[ \int_0^{v_1} \frac{40v}{400 - v^2}\, \mathrm{d}v = \int_0^{20\ln 2} \mathrm{d}y \]
\[ \left[ -20 \ln(400 - v^2) \right]_0^{v_1} = 20 \ln 2 \]
\[ -20 \ln(400 - v_1^2) + 20 \ln 400 = 20 \ln 2 \]
\[ \ln\left(\frac{400}{400 - v_1^2}\right) = \ln 2 \]
\[ \frac{400}{400 - v_1^2} = 2 \implies 400 - v_1^2 = 200 \implies v_1^2 = 200 \]
\[ v_1 = \sqrt{200} = 10\sqrt{2} \approx 14.1\text{ m s}^{-1} \]

(a) [4 marks]
- M1: Sets up Newton's second law for upward motion using \(a = v\frac{\mathrm{d}v}{\mathrm{d}x}\) with both weight and resistance opposing motion.
- M1: Separates variables and integrates to obtain a logarithmic term of the form \(k\ln(400+v^2)\).
- M1: Applies initial condition \(x = 0, v = 20\) (or uses definite integration limits \(0\) to \(H\) and \(20\) to \(0\)).
- A1: Obtains given result \(20\ln 2\) convincingly with all working shown.

(b) [2 marks]
- M1: Uses \(a = \frac{\mathrm{d}v}{\mathrm{d}t}\), separates variables and integrates using the standard \(\arctan\) form.
- A1: Obtains \(T = \frac{\pi}{2}\) (or \(\frac{1}{2}\pi\)).

(c) [3 marks]
- M1: Sets up Newton's second law for downward motion with correct signs (weight downwards, resistance upwards).
- M1: Separates variables, integrates to logarithmic form, and equates distance fallen to \(20\ln 2\).
- A1: Obtains \(10\sqrt{2}\text{ m s}^{-1}\) (or exact equivalent, or \(14.1\text{ m s}^{-1}\)).

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

The initial horizontal and vertical velocity components are:
\[ u_x = 25 \cos\theta = 25 \times \frac{3}{5} = 15\text{ m s}^{-1} \]
\[ u_y = 25 \sin\theta = 25 \times \frac{4}{5} = 20\text{ m s}^{-1} \]

The horizontal and vertical displacements from \(O\) at time \(t\) are:
\[ x = 15t \]
\[ y = 20t - \frac{1}{2}(10)t^2 = 20t - 5t^2 \]

The particle strikes the inclined plane when \(y = x\tan\alpha\), where \(\tan\alpha = \frac{3}{4}\):
\[ 20t - 5t^2 = \frac{3}{4}(15t) \]
\[ 20t - 5t^2 = \frac{45}{4}t \]

Since \(t > 0\):
\[ 20 - 5t = \frac{45}{4} \]
\[ 5t = 20 - \frac{45}{4} = \frac{35}{4} \]
\[ t = \frac{7}{4} = 1.75\text{ s} \]

(b) At \(t = 1.75\text{ s}\), the horizontal displacement is:
\[ x = 15 \times 1.75 = 26.25\text{ m} = \frac{105}{4}\text{ m} \]

Given \(\tan\alpha = \frac{3}{4}\), \(\cos\alpha = \frac{4}{5}\).
The distance \(d\) along the inclined plane from \(O\) is:
\[ d = \frac{x}{\cos\alpha} = \frac{105/4}{4/5} = \frac{105}{4} \times \frac{5}{4} = \frac{525}{16} = 32.8125\text{ m} \approx 32.8\text{ m} \]

(c) The velocity components at \(t = 1.75\text{ s}\) are:
\[ v_x = 15\text{ m s}^{-1} \]
\[ v_y = 20 - 10(1.75) = 2.5\text{ m s}^{-1} \]

The speed \(v\) is given by:
\[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{15^2 + 2.5^2} = \sqrt{225 + 6.25} = \sqrt{231.25} = \frac{5\sqrt{37}}{2} \approx 15.2\text{ m s}^{-1} \]

(a)
M1: Writes down expressions for horizontal and vertical displacements \(x = (25\cos\theta)t\) and \(y = (25\sin\theta)t - \frac{1}{2}gt^2\), using \(\sin\theta = \frac{4}{5}\) and \(\cos\theta = \frac{3}{5}\) (or resolves parallel and perpendicular to the plane).
M1: Applies the condition for landing on the plane, \(y = x\tan\alpha\) (or displacement perpendicular to plane equals 0).
A1: Obtains \(t = 1.75\text{ s}\) correctly from fully correct working (AG).

(b)
M1: Uses \(d = \frac{x}{\cos\alpha}\) or \(d = \sqrt{x^2 + y^2}\) with \(x = 15 \times 1.75\).
A1: Obtains \(\frac{525}{16}\text{ m}\) or \(32.8\text{ m}\) (or \(32.8125\text{ m}\)).

(c)
M1: Finds velocity components \(v_x = 15\) and \(v_y = 2.5\) and applies \(v = \sqrt{v_x^2 + v_y^2}\), or uses energy conservation \(v = \sqrt{u^2 - 2gy}\).
A1: Obtains \(15.2\text{ m s}^{-1}\) (accept \(\frac{5\sqrt{37}}{2}\) or \(\sqrt{231.25}\)).

Given: \(m = 2\text{ kg}\), \(l = 1.2\text{ m}\), \(\lambda = 60\text{ N}\), \(\sin\alpha = 0.6\), \(\cos\alpha = 0.8\), \(\mu = 0.2\), \(g = 10\text{ m s}^{-2}\).

The normal reaction force \(R\) is:
\[ R = mg\cos\alpha = 2(10)(0.8) = 16\text{ N} \]
The magnitude of the frictional force while \(P\) is in motion is:
\[ F = \mu R = 0.2 \times 16 = 3.2\text{ N} \]
The component of weight down the plane is:
\[ mg\sin\alpha = 2(10)(0.6) = 12\text{ N} \]

(a) Maximum speed occurs when the acceleration is zero, i.e. when forces parallel to the plane are in equilibrium:
\[ mg\sin\alpha - F - T = 0 \]
\[ 12 - 3.2 - \frac{60x}{1.2} = 0 \implies 8.8 - 50x = 0 \implies x = 0.176\text{ m} \]
Applying the work-energy principle between \(B\) (where \(x = 0, v = 0\)) and this point:
\[ \frac{1}{2}mv_{\text{max}}^2 - 0 = (mg\sin\alpha - F)x - \frac{\lambda x^2}{2l} \]
\[ \frac{1}{2}(2)v_{\text{max}}^2 = (12 - 3.2)(0.176) - \frac{60(0.176)^2}{2(1.2)} \]
\[ v_{\text{max}}^2 = 8.8(0.176) - 25(0.176)^2 = 0.176(8.8 - 4.4) = 0.176 \times 4.4 = 0.7744 \]
\[ v_{\text{max}} = \sqrt{0.7744} = 0.88\text{ m s}^{-1} \]

(b) Let \(d\) be the distance travelled from \(B\) to \(C\). At \(C\), \(v = 0\) and extension \(x = d\).
Using the work-energy principle from \(B\) to \(C\):
\[ 0 - 0 = (mg\sin\alpha - F)d - \frac{\lambda d^2}{2l} \]
\[ 8.8d - 25d^2 = 0 \]
Since \(d > 0\):
\[ d = \frac{8.8}{25} = 0.352\text{ m} \]

(c) At point \(C\), the extension of the string is \(x = 0.352\text{ m}\).
The tension in the string acting up the plane is:
\[ T = \frac{60 \times 0.352}{1.2} = 50 \times 0.352 = 17.6\text{ N} \]
The force opposing motion up the plane consists of the component of weight down the plane and the maximum static friction:
\[ mg\sin\alpha + F_{\text{max}} = 12 + 3.2 = 15.2\text{ N} \]
Since \(T = 17.6\text{ N} > 15.2\text{ N}\), the net upward force along the plane is positive (\(17.6 - 12 - 3.2 = 2.4\text{ N} > 0\)).
Therefore, \(P\) does not remain at rest and begins to move back up the plane.

(a)
- M1: Resolves forces to find normal reaction \(R\) and sets up equilibrium equation \(mg\sin\alpha - \mu R - T = 0\) to find extension at max speed.
- A1: Obtains \(x = 0.176\text{ m}\).
- M1: Sets up energy equation involving change in KE, work done by gravity and friction, and EPE.
- A1: Obtains \(v_{\text{max}} = 0.88\text{ m s}^{-1}\).

(b)
- M1: Applies work-energy equation from release at \(B\) to instantaneous rest at \(C\).
- A1: Obtains correct unsimplified quadratic equation, e.g. \(8.8d - 25d^2 = 0\).
- A1: Obtains \(d = 0.352\text{ m}\) (or \(\frac{44}{125}\text{ m}\)).

(c)
- M1: Calculates the tension at \(C\) using \(T = \frac{\lambda d}{l}\).
- A1: Calculates the maximum resisting force down the slope (\(mg\sin\alpha + \mu R = 15.2\text{ N}\)) or finds the resultant force.
- A1: Clearly compares \(T = 17.6\text{ N}\) with \(15.2\text{ N}\) (or calculates net upward force \(2.4\text{ N} > 0\)) and correctly concludes that \(P\) moves back up the plane.

(a)
First find the sample means:
\[ \bar{x} = \frac{376.8}{8} = 47.1 \]
\[ \bar{y} = \frac{435.0}{10} = 43.5 \]

Calculate the sums of squared deviations:
\[ \sum (x - \bar{x})^2 = \sum x^2 - \frac{(\sum x)^2}{8} = 17765.48 - \frac{376.8^2}{8} = 17765.48 - 17747.28 = 18.20 \]
\[ \sum (y - \bar{y})^2 = \sum y^2 - \frac{(\sum y)^2}{10} = 18942.50 - \frac{435.0^2}{10} = 18942.50 - 18922.50 = 20.00 \]

The pooled estimate of the common variance is:
\[ s_p^2 = \frac{\sum (x - \bar{x})^2 + \sum (y - \bar{y})^2}{n_A + n_B - 2} = \frac{18.20 + 20.00}{8 + 10 - 2} = \frac{38.20}{16} = 2.3875 \text{ (or } \frac{191}{80} \approx 2.39 \text{)} \]

(b)
The number of degrees of freedom is \(8 + 10 - 2 = 16\).
For a 95% confidence interval, the critical value from the \(t\)-distribution table is \(t_{16, 0.025} = 2.120\).

The difference between sample means is:
\[ \bar{x} - \bar{y} = 47.1 - 43.5 = 3.6 \]

The estimated standard error of the difference between means is:
\[ \text{SE} = \sqrt{s_p^2 \left(\frac{1}{n_A} + \frac{1}{n_B}\right)} = \sqrt{2.3875 \left(\frac{1}{8} + \frac{1}{10}\right)} = \sqrt{2.3875 \times 0.225} = \sqrt{0.5371875} \approx 0.73293 \]

The 95% confidence interval is:
\[ (\bar{x} - \bar{y}) \pm t_{16} \times \text{SE} = 3.6 \pm 2.120 \times 0.73293 = 3.6 \pm 1.5538 \]
\[ [2.0462, 5.1538] \implies [2.05, 5.15] \quad \text{(to 3 s.f.)} \]

(c)
Since the confidence interval does not contain 0 (and is entirely positive), it indicates that there is evidence that the population mean drying times differ, specifically that the mean drying time for formulation \(A\) is greater than that for formulation \(B\).

(a)
- M1: For calculating sums of squares and applying the pooled variance formula \(s_p^2 = \frac{S_{xx} + S_{yy}}{n_A + n_B - 2}\) with \(\text{d.f.} = 16\).
- A1: Correct pooled variance \(s_p^2 = 2.3875\) (or \(\frac{191}{80}\) or \(2.39\)).

(b)
- B1: For stating/using the correct critical value \(t_{16} = 2.120\) (or \(2.12\)).
- M1: For correct confidence interval structure: \((\bar{x} - \bar{y}) \pm t \sqrt{s_p^2\left(\frac{1}{8} + \frac{1}{10}\right)}\).
- A1: For correct confidence interval \([2.05, 5.15]\) (or \([2.046, 5.154]\)).

(c)
- B1: For correct interpretation stating that 0 is not in the interval, so there is evidence of a difference in mean drying times (or formulation \(A\) takes longer on average).

(a) Using the property \(G_X(1) = 1\):
\[ G_X(1) = \frac{k}{(4 - 3(1))^2} = \frac{k}{1^2} = k \]
Since \(G_X(1) = 1\), we have \(k = 1\).

(b) We have \(G_X(t) = (4 - 3t)^{-2}\).
Differentiating with respect to \(t\):
\[ G_X'(t) = -2(4 - 3t)^{-3}(-3) = 6(4 - 3t)^{-3} \]
Thus,
\[ \text{E}(X) = G_X'(1) = 6(4 - 3)^{-3} = 6 \]

Differentiating again:
\[ G_X''(t) = 6(-3)(4 - 3t)^{-4}(-3) = 54(4 - 3t)^{-4} \]
Thus,
\[ G_X''(1) = 54(4 - 3)^{-4} = 54 \]

Using the formula for variance:
\[ \text{Var}(X) = G_X''(1) + G_X'(1) - [G_X'(1)]^2 \]
\[ \text{Var}(X) = 54 + 6 - 6^2 = 60 - 36 = 24 \]

(c) The random variable \(Y = X_1 + X_2\), where \(X_1\) and \(X_2\) are independent observations of \(X\).
The probability generating function of \(Y\) is:
\[ G_Y(t) = [G_X(t)]^2 = (4 - 3t)^{-4} = \frac{1}{256}\left(1 - \frac{3}{4}t\right)^{-4} \]
Expanding using the binomial series:
\[ \left(1 - \frac{3}{4}t\right)^{-4} = 1 + (-4)\left(-\frac{3}{4}t\right) + \frac{(-4)(-5)}{2!}\left(-\frac{3}{4}t\right)^2 + \dots = 1 + 3t + 10\left(\frac{9}{16}t^2\right) + \dots \]
\[ = 1 + 3t + \frac{45}{8}t^2 + \dots \]

Therefore,
\[ \text{P}(Y = 2) = \text{coefficient of } t^2 \text{ in } G_Y(t) = \frac{1}{256} \times \frac{45}{8} = \frac{45}{2048} \approx 0.0220 \]

(a)
B1: Equates \(G_X(1) = 1\) and demonstrates \(k = 1\).

(b)
M1: Differentiates \(G_X(t)\) correctly using chain rule to find \(G'_X(t)\).
A1: Evaluates \(G'_X(1)\) to get \(\text{E}(X) = 6\).
M1: Finds \(G''_X(t)\), evaluates at \(t = 1\), and applies \(\text{Var}(X) = G''_X(1) + G'_X(1) - [G'_X(1)]^2\).
A1: Obtains \(\text{Var}(X) = 24\).

(c)
M1: Forms \(G_Y(t) = [G_X(t)]^2 = (4 - 3t)^{-4}\) and attempts to find coefficient of \(t^2\) or uses \(\frac{1}{2!}G''_Y(0)\).
A1: Obtains \(\frac{45}{2048}\) (or exact equivalent, e.g. \(0.0220\) to 3 s.f.).

(a) For \(0 \leqslant x \leqslant 4\):
\[ F(x) = \int_0^x f(t) \,\mathrm{d}t = \int_0^x \frac{3}{64}t^2 \,\mathrm{d}t = \left[ \frac{t^3}{64} \right]_0^x = \frac{x^3}{64} \]

(b) Let \(m\) be the median of \(X\). Then \(F(m) = 0.5\):
\[ \frac{m^3}{64} = \frac{1}{2} \implies m^3 = 32 \]
\[ m = \sqrt[3]{32} = \sqrt[3]{8 \times 4} = 2\sqrt[3]{4} \]

(c)(i) When \(X = 0\), \(Y = \frac{6}{0+2} = 3\).
When \(X = 4\), \(Y = \frac{6}{4+2} = 1\).
Thus, \(Y\) takes values in the interval \([1, 3]\).

For \(1 \leqslant y \leqslant 3\):
\[ G(y) = \mathrm{P}(Y \leqslant y) = \mathrm{P}\left(\frac{6}{X+2} \leqslant y\right) = \mathrm{P}\left(X+2 \geqslant \frac{6}{y}\right) = \mathrm{P}\left(X \geqslant \frac{6}{y} - 2\right) \]
\[ G(y) = 1 - F\left(\frac{6}{y} - 2\right) \]

Substituting into \(F(x)\):
\[ G(y) = 1 - \frac{\left(\frac{6}{y} - 2\right)^3}{64} = 1 - \frac{8\left(\frac{3}{y} - 1\right)^3}{64} = 1 - \frac{1}{8}\left(\frac{3 - y}{y}\right)^3 = 1 - \frac{(3-y)^3}{8y^3} \]

Including all intervals:
\[ G(y) = \begin{cases} 0 & y < 1, \[4pt] 1 - \frac{(3-y)^3}{8y^3} & 1 \leqslant y \leqslant 3, \[4pt] 1 & y > 3. \end{cases} \]

(c)(ii) For \(1 \leqslant y \leqslant 3\), differentiate \(G(y)\):
\[ g(y) = G'(y) = -\frac{\mathrm{d}}{\mathrm{d}y}\left[ \frac{1}{8}\left(\frac{3}{y} - 1\right)^3 \right] = -\frac{3}{8}\left(\frac{3}{y} - 1\right)^2 \left(-\frac{3}{y^2}\right) = \frac{9}{8y^2}\left(\frac{3-y}{y}\right)^2 = \frac{9(3-y)^2}{8y^4} \]

Specifying for all \(y \in \mathbb{R}\):
\[ g(y) = \begin{cases} \frac{9(3-y)^2}{8y^4} & 1 \leqslant y \leqslant 3, \[4pt] 0 & \text{otherwise.} \end{cases} \]

(a)
- M1: For integrating \(f(x)\) between limits \(0\) and \(x\) (or integrating and using \(F(0)=0\) or \(F(4)=1\) to evaluate constant of integration).
- A1: For completely correct working leading to the given result \(F(x) = \frac{x^3}{64}\).

(b)
- M1: Sets \(F(m) = 0.5\) (or \(\frac{1}{2}\)) and attempts to solve for \(m\).
- A1: Correct value in the required form: \(2\sqrt[3]{4}\).

(c)(i)
- M1: States \(\mathrm{P}(Y \leqslant y) = \mathrm{P}\left(X \geqslant \frac{6}{y}-2\right) = 1 - F\left(\frac{6}{y}-2\right)\) with correct change of inequality direction.
- A1: Obtains correct algebraic expression \(1 - \frac{(3-y)^3}{8y^3}\) (or equivalent form such as \(1 - \frac{(6-2y)^3}{64y^3}\)) for \(1 \leqslant y \leqslant 3\).
- A1: States correct piecewise definition of \(G(y)\) including \(0\) for \(y < 1\) and \(1\) for \(y > 3\).

(c)(ii)
- M1: Differentiates their \(G(y)\) with respect to \(y\) using the chain rule or quotient rule.
- A1: Correct derivative \(\frac{9(3-y)^2}{8y^4}\) (or \(\frac{9(y-3)^2}{8y^4}\) or expanded \(\frac{9y^2 - 54y + 81}{8y^4}\)).
- A1: Completely correct probability density function defined piecewise for all real \(y\) including \(0\) otherwise.

(a)
Two necessary assumptions are:
1. The grain yields for both Variety $A$ and Variety $B$ are normally distributed.
2. The two population variances are equal (i.e. $\sigma_A^2 = \sigma_B^2$).

(b)
Let $\mu_A$ and $\mu_B$ denote the population mean yields of Variety $A$ and Variety $B$ respectively.

State hypotheses:
$H_0: \mu_A = \mu_B$ (or $\mu_A - \mu_B = 0$)
$H_1: \mu_A > \mu_B$ (or $\mu_A - \mu_B > 0$)

Calculate sample means:
$$\bar{x}_A = \frac{268.0}{8} = 33.5$$
$$\bar{x}_B = \frac{308.0}{10} = 30.8$$

Calculate sum of squared deviations from the mean for each sample:
$$\sum (x_A - \bar{x}_A)^2 = \sum x_A^2 - \frac{(\sum x_A)^2}{n_A} = 9012.32 - \frac{268.0^2}{8} = 9012.32 - 8978.00 = 34.32$$
$$\sum (x_B - \bar{x}_B)^2 = \sum x_B^2 - \frac{(\sum x_B)^2}{n_B} = 9534.68 - \frac{308.0^2}{10} = 9534.68 - 9486.40 = 48.28$$

Calculate the pooled estimate of common variance, $s_p^2$:
$$s_p^2 = \frac{\sum (x_A - \bar{x}_A)^2 + \sum (x_B - \bar{x}_B)^2}{n_A + n_B - 2} = \frac{34.32 + 48.28}{8 + 10 - 2} = \frac{82.60}{16} = 5.1625$$

Calculate the standard error:
$$\text{s.e.} = \sqrt{s_p^2\left(\frac{1}{n_A} + \frac{1}{n_B}\right)} = \sqrt{5.1625\left(\frac{1}{8} + \frac{1}{10}\right)} = \sqrt{5.1625 \times 0.225} = \sqrt{1.1615625} \approx 1.0778$$

Calculate the test statistic $t$:
$$t = \frac{\bar{x}_A - \bar{x}_B}{\text{s.e.}} = \frac{33.5 - 30.8}{1.077758} = \frac{2.7}{1.077758} \approx 2.505$$

Degrees of freedom: $
u = 8 + 10 - 2 = 16$.
Critical value for a one-tailed test at the $5\%$ significance level with $
u = 16$:
$$t_{\text{crit}} = 1.746$$

Comparison and conclusion:
Since $t = 2.505 > 1.746$, the test statistic falls in the rejection region.
Reject $H_0$.
There is significant evidence at the $5\%$ level to indicate that the population mean yield of Variety $A$ is greater than that of Variety $B$.

(c)
For a $95\%$ confidence interval with $
u = 16$ degrees of freedom, the two-tailed critical value is:
$$t_{16, 0.025} = 2.120$$

The confidence interval for $\mu_A - \mu_B$ is given by:
$$(\bar{x}_A - \bar{x}_B) \pm t_{16, 0.025} \times \text{s.e.}$$
$$2.7 \pm 2.120 \times 1.077758 = 2.7 \pm 2.28485$$
$$[0.415, 4.98]$$ (to 3 significant figures)

(a)
B1: States that both populations are normally distributed.
B1: States that population variances are equal (or common population variance).

(b)
B1: Both hypotheses correctly stated using population parameters $\mu_A$ and $\mu_B$ (one-tailed test).
M1: Correct method to find sample variances / sums of squares and pooled variance $s_p^2$.
A1: Obtains $s_p^2 = 5.1625$ (or $s_p \approx 2.272$).
M1: Correct method to calculate test statistic $t$.
A1: Obtains $t = 2.51$ (accept $2.50$ to $2.51$).
A1: States critical value $1.746$, compares correctly ($2.505 > 1.746$), and gives a correct contextual conclusion in non-assertive language.

(c)
M1: Correct formula for confidence interval $(\bar{x}_A - \bar{x}_B) \pm t \times \text{s.e.}$ using $t = 2.120$.
A1: Correct interval $[0.415, 4.98]$ (accept $[0.415, 4.985]$ or $0.415 < \mu_A - \mu_B < 4.98$).