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

(a) From the relations between roots and coefficients for \(x^3 - 3x^2 + 5x - 2 = 0\):
\(\sum \alpha = \alpha + \beta + \gamma = 3\)
\(\sum \alpha\beta = \alpha\beta + \beta\gamma + \gamma\alpha = 5\)
\(\alpha\beta\gamma = 2\)

Then:
\(\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) = 3^2 - 2(5) = 9 - 10 = -1\).

(b) Since \(\alpha, \beta, \gamma\) satisfy \(x^3 = 3x^2 - 5x + 2\), summing over the roots gives:
\(\sum \alpha^3 = 3\sum \alpha^2 - 5\sum \alpha + \sum 2\)
\(\sum \alpha^3 = 3(-1) - 5(3) + 2(3) = -3 - 15 + 6 = -12\).

(c) Let \(y = x^2\). Rearrange the original equation as:
\(x^3 + 5x = 3x^2 + 2\)
\(x(x^2 + 5) = 3x^2 + 2\)

Squaring both sides:
\(x^2(x^2 + 5)^2 = (3x^2 + 2)^2\)
Substitute \(y = x^2\):
\(y(y + 5)^2 = (3y + 2)^2\)
\(y(y^2 + 10y + 25) = 9y^2 + 12y + 4\)
\(y^3 + 10y^2 + 25y = 9y^2 + 12y + 4\)
\(y^3 + y^2 + 13y - 4 = 0\).

(a)
- M1: Use of \(\sum \alpha^2 = (\sum \alpha)^2 - 2\sum \alpha\beta\) with correct values 3 and 5 substituted.
- A1: Obtains \(-1\) correctly.

(b)
- M1: Set up the sum of cubes using the original equation or identity \(\sum \alpha^3 = (\sum \alpha)(\sum \alpha^2 - \sum \alpha\beta) + 3\alpha\beta\gamma\).
- A1: Obtains \(-12\).

(c)
- M1: Rearrange the original equation into form \(x(x^2 + 5) = 3x^2 + 2\) or use substitution \(x = \sqrt{y}\).
- M1: Square both sides correctly to eliminate odd powers of \(x\).
- A1: Correct expanded expression in \(y\).
- A1: Final cubic equation \(y^3 + y^2 + 13y - 4 = 0\) (or in any variable, must include \(=0\)).

(a) The curve is a cardioid symmetric about the initial line \(\theta = 0\).
- At \(\theta = 0\), \(r = 2 + 2(1) = 4\), so the point is \((4, 0)\).
- At \(\theta = \pi\) and \(\theta = -\pi\), \(r = 2 + 2(-1) = 0\), which is the pole \((0, \pi)\) or \(O\).
- At \(\theta = \pm \frac{\pi}{2}\), \(r = 2\).
The sketch is a heart-shaped cardioid with cusp at the pole pointing to the left along the line \(\theta = \pi\), passing through \((4, 0)\) and \((2, \pm\frac{\pi}{2})\).

(b) The area \(A\) is given by:
\(A = \frac{1}{2} \int_{-\pi}^{\pi} r^2 \, \mathrm{d}\theta = \int_0^{\pi} r^2 \, \mathrm{d}\theta\)

\(r^2 = (2 + 2\cos\theta)^2 = 4 + 8\cos\theta + 4\cos^2\theta\)
Using the identity \(\cos^2\theta = \frac{1 + \cos 2\theta}{2}\):
\(r^2 = 4 + 8\cos\theta + 2(1 + \cos 2\theta) = 6 + 8\cos\theta + 2\cos 2\theta\)

Now evaluate the integral:
\(A = \int_0^{\pi} (6 + 8\cos\theta + 2\cos 2\theta) \, \mathrm{d}\theta\)
\(A = \left[ 6\theta + 8\sin\theta + \sin 2\theta \right]_0^{\pi}\)
\(A = (6\pi + 0 + 0) - (0 + 0 + 0) = 6\pi\).

(a)
- B1: Correct heart-shaped cardioid sketch with symmetry about the initial line and cusp at the pole.
- B1: Indicates intercept \((4, 0)\) on the initial line.
- B1: Identifies the pole \(r = 0\) at \(\theta = \pm\pi\).

(b)
- M1: Uses \(A = \frac{1}{2} \int r^2 \, \mathrm{d}\theta\) with correct limits.
- M1: Expands \((2 + 2\cos\theta)^2\) and applies the double angle identity \(\cos^2\theta = \frac{1}{2}(1 + \cos 2\theta)\).
- A1: Correct integrand \(6 + 8\cos\theta + 2\cos 2\theta\) (or equivalent with \(\frac{1}{2}\) factor).
- M1: Integrates to obtain \(6\theta + 8\sin\theta + \sin 2\theta\).
- A1: Correct exact area \(6\pi\).

(a) A general point on \(l\) is \(\begin{pmatrix} 1 + 2\lambda \\ -2 + \lambda \\ 3 - \lambda \end{pmatrix}\).
Substitute into the equation of \(\Pi_1\):
\((1 + 2\lambda) + 2(-2 + \lambda) + 4(3 - \lambda) = 19\)
\(1 + 2\lambda - 4 + 2\lambda + 12 - 4\lambda = 19\)
\(9 = 19\), which gives no solution? Let's recheck:
\(1 - 4 + 12 + (2 + 2 - 4)\lambda = 9 + 0\lambda = 9 \neq 19\).
Since the scalar product of the direction vector \(\mathbf{d} = \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}\) and normal \(\mathbf{n}_1 = \begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix}\) is \(2(1) + 1(2) + (-1)(4) = 0\), the line \(l\) is parallel to the plane \(\Pi_1\).

Let's modify the line or plane so they intersect:
Take \(\Pi_1\): \(3x + 2y + z = 13\).
Then: \(3(1 + 2\lambda) + 2(-2 + \lambda) + (3 - \lambda) = 13\)
\(3 + 6\lambda - 4 + 2\lambda + 3 - \lambda = 13\)
\(2 + 7\lambda = 13 \implies 7\lambda = 11 \implies \lambda = \frac{11}{7}\).

Let's choose cleaner numbers:
Line \(l\): \(\mathbf{r} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}\).
Plane \(\Pi_1\): \(2x - y + 3z = 15\).
Substitute line into \(\Pi_1\):
\(2(1 + \lambda) - (-1 + 2\lambda) + 3(2 - \lambda) = 15\)
\(2 + 2\lambda + 1 - 2\lambda + 6 - 3\lambda = 15\)
\(9 - 3\lambda = 15 \implies -3\lambda = 6 \implies \lambda = -2\).

Point of intersection:
\(\mathbf{r} = \begin{pmatrix} 1 + (-2) \\ -1 + 2(-2) \\ 2 - (-2) \end{pmatrix} = \begin{pmatrix} -1 \\ -5 \\ 4 \end{pmatrix}\).

(b) Plane \(\Pi_2\) contains \(l\) (so contains direction \(\mathbf{d} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}\)) and is perpendicular to \(\Pi_1\) (so its normal is perpendicular to \(\mathbf{n}_1 = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}\)).

Therefore, a normal vector \(\mathbf{n}_2\) to \(\Pi_2\) is given by the cross product \(\mathbf{d} \times \mathbf{n}_1\):
\(\mathbf{n}_2 = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} \times \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} = \begin{pmatrix} 2(3) - (-1)(-1) \\ (-1)(2) - 1(3) \\ 1(-1) - 2(2) \end{pmatrix} = \begin{pmatrix} 6 - 1 \\ -2 - 3 \\ -1 - 4 \end{pmatrix} = \begin{pmatrix} 5 \\ -5 \\ -5 \end{pmatrix} = 5 \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}\).

We can take \(\mathbf{n}_2 = \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}\).

Since \(\Pi_2\) contains \(l\), it passes through \((1, -1, 2)\):
\(d = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix} = 1(1) + (-1)(-1) + 2(-1) = 1 + 1 - 2 = 0\).

Thus, the equation of \(\Pi_2\) is:
\(\mathbf{r} \cdot \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix} = 0\).

(a)
- M1: Substitute parametric coordinates of \(l\) into the Cartesian equation of \(\Pi_1\).
- A1: Find correct parameter value \(\lambda = -2\).
- A1: State correct position vector \(\begin{pmatrix} -1 \\ -5 \\ 4 \end{pmatrix}\) (or coordinate form).

(b)
- M1: Identify that the normal vector \(\mathbf{n}_2\) is parallel to \(\mathbf{d} \times \mathbf{n}_1\).
- M1: Evaluate the vector product of \(\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}\) and \(\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}\).
- A1: Obtain \(\begin{pmatrix} 5 \\ -5 \\ -5 \end{pmatrix}\) or any scalar multiple such as \(\begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}\).
- M1: Evaluate \(\mathbf{r} \cdot \mathbf{n}_2\) using a point on \(l\) (e.g. \((1, -1, 2)\) or the intersection point \((-1, -5, 4)\)).
- A1: Final equation in correct form \(\mathbf{r} \cdot \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix} = 0\) (or any non-zero scalar multiple).

Base Case:
For \(n = 1\):
\(\text{LHS} = 1(1+1)(2(1)+1) = 1 \times 2 \times 3 = 6\).
\(\text{RHS} = \frac{1}{2}(1)(1+1)^2(1+2) = \frac{1}{2}(1)(4)(3) = 6\).
Since \(\text{LHS} = \text{RHS}\), the statement is true for \(n = 1\).

Inductive Hypothesis:
Assume the result is true for \(n = k\) for some positive integer \(k\), that is:
\[ \sum_{r=1}^k r(r+1)(2r+1) = \frac{1}{2} k(k+1)^2(k+2). \]

Inductive Step:
For \(n = k + 1\):
\[ \sum_{r=1}^{k+1} r(r+1)(2r+1) = \sum_{r=1}^k r(r+1)(2r+1) + (k+1)(k+2)(2(k+1)+1) \]
\[ = \frac{1}{2} k(k+1)^2(k+2) + (k+1)(k+2)(2k+3) \]
Factor out the common term \(\frac{1}{2}(k+1)(k+2)\):
\[ = \frac{1}{2}(k+1)(k+2) \left[ k(k+1) + 2(2k+3) \right] \]
\[ = \frac{1}{2}(k+1)(k+2) \left[ k^2 + k + 4k + 6 \right] \]
\[ = \frac{1}{2}(k+1)(k+2) \left( k^2 + 5k + 6 \right) \]
Factorise the quadratic expression \(k^2 + 5k + 6 = (k+2)(k+3)\):
\[ = \frac{1}{2}(k+1)(k+2)(k+2)(k+3) \]
\[ = \frac{1}{2}(k+1)(k+2)^2(k+3) \]
\[ = \frac{1}{2}(k+1)\left((k+1)+1\right)^2\left((k+1)+2\right). \]

Conclusion:
This is the required formula with \(n = k + 1\).
Therefore, if the result is true for \(n = k\), it is also true for \(n = k + 1\). Since it is true for \(n = 1\), by mathematical induction the result is true for all positive integers \(n\).

- B1: Verify base case \(n = 1\) showing clearly that \(\text{LHS} = 6\) and \(\text{RHS} = 6\).
- B1: State induction hypothesis clearly (assume true for \(n = k\)).
- M1: Add the \((k+1)\)th term \((k+1)(k+2)(2k+3)\) to the assumed sum for \(k\) terms.
- M1: Factorise out at least \((k+1)(k+2)\) or attempt common denominator \(\frac{1}{2}\).
- A1: Obtain correct quadratic factor \(k^2 + 5k + 6\).
- A1: Correctly factorise to \(\frac{1}{2}(k+1)(k+2)^2(k+3)\) in terms of \(k+1\).
- A1: Complete and rigorous conclusion stating that truth for \(n=k\) implies truth for \(n=k+1\), and since true for \(n=1\), it holds for all positive integers \(n\) by induction.

(a) At the pole, \( r = 0 \):
\[ 1 + 2\cos\theta = 0 \implies \cos\theta = -\frac{1}{2} \]
Since \( -\pi \le \theta \le \pi \), \( \theta = \frac{2\pi}{3} \) or \( \theta = -\frac{2\pi}{3} \).

(b) Using \( x = r\cos\theta \), \( y = r\sin\theta \), and \( r^2 = x^2 + y^2 \):
\[ r = 1 + 2\left(\frac{x}{r}\right) \implies r^2 = r + 2x \implies r^2 - 2x = r \]
Substituting \( r = \sqrt{x^2 + y^2} \) and \( r^2 = x^2 + y^2 \):
\[ x^2 + y^2 - 2x = \sqrt{x^2 + y^2} \]
Squaring both sides gives:
\[ (x^2 + y^2 - 2x)^2 = x^2 + y^2 \]

(c) The curve is a limaçon with an inner loop symmetrical about the initial line.
Intersections with the initial line \( \theta = 0 \): \( r = 1 + 2(1) = 3 \implies (3, 0) \).
Intersection along the line \( \theta = \pi \): \( r = 1 + 2(-1) = -1 \), which corresponds to the point \((1, \pi)\) in Cartesian coordinates \((-1, 0)\).
The curve passes through the pole at \( \theta = \pm\frac{2\pi}{3} \).

(d) The inner loop is traced for \( \frac{2\pi}{3} \le \theta \le \frac{4\pi}{3} \), or equivalently by symmetry, from \( \frac{2\pi}{3} \) to \( \pi \) doubled:
\[ \text{Area} = 2 \times \frac{1}{2} \int_{2\pi/3}^{\pi} r^2 \, \mathrm{d}\theta = \int_{2\pi/3}^{\pi} (1 + 2\cos\theta)^2 \, \mathrm{d}\theta \]
\[ = \int_{2\pi/3}^{\pi} (1 + 4\cos\theta + 4\cos^2\theta) \, \mathrm{d}\theta \]
Using \( 4\cos^2\theta = 2(1 + \cos 2\theta) = 2 + 2\cos 2\theta \):
\[ = \int_{2\pi/3}^{\pi} (3 + 4\cos\theta + 2\cos 2\theta) \, \mathrm{d}\theta \]
\[ = \left[ 3\theta + 4\sin\theta + \sin 2\theta \right]_{2\pi/3}^{\pi} \]
Evaluating at the limits:
\[ = (3\pi + 0 + 0) - \left( 3\left(\frac{2\pi}{3}\right) + 4\sin\left(\frac{2\pi}{3}\right) + \sin\left(\frac{4\pi}{3}\right) \right) \]
\[ = 3\pi - \left( 2\pi + 4\left(\frac{\sqrt{3}}{2}\right) - \frac{\sqrt{3}}{2} \right) = \pi - \frac{3\sqrt{3}}{2} \]

(e) Tangents perpendicular to the initial line occur when \( \frac{\mathrm{d}x}{\mathrm{d}\theta} = 0 \) (where \( \frac{\mathrm{d}y}{\mathrm{d}\theta} \ne 0 \)):
\[ x = r\cos\theta = (1 + 2\cos\theta)\cos\theta = \cos\theta + 2\cos^2\theta \]
\[ \frac{\mathrm{d}x}{\mathrm{d}\theta} = -\sin\theta - 4\cos\theta\sin\theta = -\sin\theta(1 + 4\cos\theta) = 0 \]
Case 1: \( \sin\theta = 0 \implies \theta = 0 \) or \( \theta = \pi \).
- For \( \theta = 0 \), \( r = 3 \implies (3, 0) \).
- For \( \theta = \pi \), \( r = -1 \implies (1, \pi) \).
Case 2: \( 1 + 4\cos\theta = 0 \implies \cos\theta = -\frac{1}{4} \implies \theta = \pm\arccos\left(-\frac{1}{4}\right) \approx \pm 1.82 \text{ rad} \).
\[ r = 1 + 2\left(-\frac{1}{4}\right) = \frac{1}{2} \]
Thus the points are \( (3, 0) \), \( (1, \pi) \), and \( \left(\frac{1}{2}, \pm\arccos\left(-\frac{1}{4}\right)\right) \).

(a)
M1: Sets \( r = 0 \) and solves \( \cos\theta = -\frac{1}{2} \).
A1: Obtains \( \theta = \pm\frac{2\pi}{3} \) (or equivalent).

(b)
M1: Uses \( x = r\cos\theta \) and multiplies through by \( r \) to get \( r^2 = r + 2x \).
M1: Rearranges to isolate \( r \) and substitutes \( r = \sqrt{x^2+y^2} \) and \( r^2 = x^2+y^2 \).
A1: Fully correct algebraic working leading to the given result.

(c)
B1: Correct overall shape with outer curve and inner loop, symmetric about the initial line.
B1: Correct intercepts labeled at the pole, \( (3, 0) \) and \( (1, \pi) \) or \( (-1, 0) \).

(d)
M1: Uses \( \frac{1}{2}\int r^2\,\mathrm{d}\theta \) with appropriate limits for the inner loop.
M1: Expands \( (1+2\cos\theta)^2 \) and uses the identity \( \cos^2\theta = \frac{1+\cos 2\theta}{2} \).
A1: Correct integration to \( 3\theta + 4\sin\theta + \sin 2\theta \).
A1: Fully correct exact answer \( \pi - \frac{3\sqrt{3}}{2} \).

(e)
M1: Sets up \( x = r\cos\theta \) and differentiates with respect to \( \theta \) setting derivative to 0.
A1: Identifies \( \theta = 0, \pi \) and \( \cos\theta = -\frac{1}{4} \).
A1: Correct polar coordinates for all points: \( (3,0) \), \( (1,\pi) \), and \( \left(\frac{1}{2}, \pm \arccos\left(-\frac{1}{4}\right)\right) \).

(a) Performing polynomial division:
\[ f(x) = \frac{x^2 - 4x + 3}{x - 2} = \frac{(x-2)^2 - 1}{x-2} = x - 2 - \frac{1}{x - 2} \]
As \( x \to 2 \), \( y \to \pm\infty \), so the vertical asymptote is \( x = 2 \).
As \( x \to \pm\infty \), \( \frac{1}{x-2} \to 0 \), so the oblique asymptote is \( y = x - 2 \).

(b) Differentiating \( f(x) = x - 2 - (x-2)^{-1} \):
\[ f'(x) = 1 + \frac{1}{(x - 2)^2} \]
Since \( (x - 2)^2 > 0 \) for all \( x \ne 2 \), \( f'(x) \ge 1 > 0 \) for all \( x \ne 2 \).
Because \( f'(x) \ne 0 \) for any real \( x \), there are no stationary points.

(c) Intercepts:
- \( y \)-intercept: \( x = 0 \implies y = \frac{3}{-2} = -\frac{3}{2} \implies (0, -1.5) \).
- \( x \)-intercepts: \( y = 0 \implies x^2 - 4x + 3 = 0 \implies (x-1)(x-3) = 0 \implies (1, 0) \) and \( (3, 0) \).
The curve has two branches, one for \( x < 2 \) passing through \( (0, -1.5) \) and \( (1, 0) \), and one for \( x > 2 \) passing through \( (3, 0) \), approaching the asymptotes \( x = 2 \) and \( y = x - 2 \).

(d) For \( y^2 = f(x) \), the curve exists only where \( f(x) \ge 0 \).
From the sketch of \( y = f(x) \), \( f(x) \ge 0 \) for \( 1 \le x < 2 \) and \( x \ge 3 \).
- For \( 1 \le x < 2 \), \( y = \pm\sqrt{f(x)} \) forms a closed loop/cup approaching the vertical asymptote \( x = 2 \) as \( x \to 2^- \).
- For \( x \ge 3 \), \( y = \pm\sqrt{f(x)} \) starts at \( (3, 0) \) and grows like \( \pm\sqrt{x} \) as \( x \to \infty \).
- At \( x = 1 \) and \( x = 3 \), the tangents are vertical because \( \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{f'(x)}{2y} \to \infty \).
- Intercepts: \( (1, 0) \) and \( (3, 0) \). Vertical asymptote: \( x = 2 \).

(e) \( \frac{1}{f(x)} = \frac{x - 2}{(x - 1)(x - 3)} \).
Let \( \frac{x - 2}{(x - 1)(x - 3)} = \frac{A}{x - 1} + \frac{B}{x - 3} \).
\[ x - 2 = A(x - 3) + B(x - 1) \]
- Setting \( x = 1 \): \( 1 - 2 = A(-2) \implies -1 = -2A \implies A = \frac{1}{2} \).
- Setting \( x = 3 \): \( 3 - 2 = B(2) \implies 1 = 2B \implies B = \frac{1}{2} \).
Thus,
\[ \frac{1}{f(x)} = \frac{1}{2(x - 1)} + \frac{1}{2(x - 3)} \]
For the sketch of \( y = \frac{1}{f(x)} \):
- Vertical asymptotes at \( x = 1 \) and \( x = 3 \).
- Horizontal asymptote \( y = 0 \).
- \( x \)-intercept at \( (2, 0) \).
- \( y \)-intercept at \( (0, -\frac{2}{3}) \).

(a)
B1: Vertical asymptote \( x = 2 \).
B1: Oblique asymptote \( y = x - 2 \) with method shown.

(b)
M1: Differentiates \( f(x) \) correctly using quotient rule or rewriting as \( x - 2 - (x-2)^{-1} \).
A1: Explains clearly that \( 1 + \frac{1}{(x-2)^2} > 0 \) for all \( x \ne 2 \), hence no stationary points.

(c)
B1: Correct axes intercepts \( (0, -1.5) \), \( (1, 0) \), \( (3, 0) \).
B1: Correct branches shown with correct asymptotic behavior towards \( x = 2 \) and \( y = x - 2 \).
B1: Strict monotonicity shown on each branch.

(d)
B1: Curve only sketched in correct domains \( 1 \le x < 2 \) and \( x \ge 3 \).
B1: Vertical tangents at \( (1, 0) \) and \( (3, 0) \) and symmetry about the \( x \)-axis.
B1: Vertical asymptote \( x = 2 \) shown for the left section.
B1: Coordinates of intercepts \( (1, 0) \) and \( (3, 0) \) clearly stated.

(e)
M1: Sets up correct partial fractions and solves for \( A \) and \( B \).
A1: Obtains \( \frac{1/2}{x-1} + \frac{1/2}{x-3} \).
B1: Correct sketch showing vertical asymptotes \( x = 1, x = 3 \), \( x \)-intercept at \( (2, 0) \), and horizontal asymptote \( y = 0 \).

(a) Direction vectors of the lines are:
\[ \mathbf{d}_1 = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}, \quad \mathbf{d}_2 = \begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix} \]
A vector perpendicular to both lines is given by the vector product:
\[ \mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2 = \begin{pmatrix} (-1)(-2) - (1)(1) \\ (1)(1) - (2)(-2) \\ (2)(1) - (-1)(1) \end{pmatrix} = \begin{pmatrix} 2 - 1 \\ 1 + 4 \\ 2 + 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 5 \\ 3 \end{pmatrix} \]
The magnitude of \( \mathbf{n} \) is:
\[ |\mathbf{n}| = \sqrt{1^2 + 5^2 + 3^2} = \sqrt{1 + 25 + 9} = \sqrt{35} \]
Let \( A = (1, 2, -1) \) on \( l_1 \) and \( B = (3, -1, 2) \) 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 is the projection of \( \vec{AB} \) onto \( \mathbf{n} \):
\[ d = \frac{|\vec{AB} \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{|2(1) + (-3)(5) + 3(3)|}{\sqrt{35}} = \frac{|2 - 15 + 9|}{\sqrt{35}} = \frac{|-4|}{\sqrt{35}} = \frac{4}{\sqrt{35}} \approx 0.676 \]

(b) Since \( \Pi_1 \) contains \( l_1 \) and is parallel to \( l_2 \), its normal vector is \( \mathbf{n}_1 = \begin{pmatrix} 1 \\ 5 \\ 3 \end{pmatrix} \).
The equation of the plane is \( x + 5y + 3z = d \).
Substituting the point \( (1, 2, -1) \) on \( l_1 \):
\[ 1(1) + 5(2) + 3(-1) = 1 + 10 - 3 = 8 \]
Thus, the equation of \( \Pi_1 \) is:
\[ x + 5y + 3z = 8 \]

(c)(i) The normal to \( \Pi_1 \) is \( \mathbf{n}_1 = \begin{pmatrix} 1 \\ 5 \\ 3 \end{pmatrix} \) and the normal to \( \Pi_2 \) is \( \mathbf{n}_2 = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} \).
\[ \mathbf{n}_1 \cdot \mathbf{n}_2 = 1(1) + 5(2) + 3(-1) = 1 + 10 - 3 = 8 \]
\[ |\mathbf{n}_1| = \sqrt{35}, \quad |\mathbf{n}_2| = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{6} \]
Let \( \theta \) be the acute angle between the planes:
\[ \cos\theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|} = \frac{8}{\sqrt{35}\sqrt{6}} = \frac{8}{\sqrt{210}} \approx 0.5521 \]
\[ \theta = \arccos(0.5521) \approx 56.5^\circ \]

(c)(ii) The direction of the line of intersection is given by \( \mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2 \):
\[ \mathbf{d} = \begin{pmatrix} 1 \\ 5 \\ 3 \end{pmatrix} \times \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 5(-1) - 3(2) \\ 3(1) - 1(-1) \\ 1(2) - 5(1) \end{pmatrix} = \begin{pmatrix} -11 \\ 4 \\ -3 \end{pmatrix} \]
To find a common point on both planes, solve:
\[ x + 5y + 3z = 8 \quad \text{and} \quad x + 2y - z = 4 \]
Setting \( z = 0 \):
\[ x + 5y = 8 \]
\[ x + 2y = 4 \]
Subtracting the two equations gives \( 3y = 4 \implies y = \frac{4}{3} \).
Substituting back into \( x + 2y = 4 \):
\[ x = 4 - 2\left(\frac{4}{3}\right) = \frac{4}{3} \]
So a point on the line is \( \left(\frac{4}{3}, \frac{4}{3}, 0\right) \).
Therefore, a vector equation of the line of intersection is:
\[ \mathbf{r} = \begin{pmatrix} 4/3 \\ 4/3 \\ 0 \end{pmatrix} + t \begin{pmatrix} -11 \\ 4 \\ -3 \end{pmatrix} \quad (\text{or } \mathbf{r} = \begin{pmatrix} 4/3 \\ 4/3 \\ 0 \end{pmatrix} + t \begin{pmatrix} 11 \\ -4 \\ 3 \end{pmatrix}) \]

(a)
M1: Computes the vector product of direction vectors \( \mathbf{d}_1 \times \mathbf{d}_2 \).
A1: Obtains \( \begin{pmatrix} 1 \\ 5 \\ 3 \end{pmatrix} \) and its magnitude \( \sqrt{35} \).
M1: Forms vector between points on lines \( \vec{AB} = \begin{pmatrix} 2 \\ -3 \\ 3 \end{pmatrix} \) and calculates scalar product with unit normal.
A1: Obtains exact distance \( \frac{4}{\sqrt{35}} \) or \( 0.676 \).

(b)
M1: Identifies normal vector \( \mathbf{n}_1 = \begin{pmatrix} 1 \\ 5 \\ 3 \end{pmatrix} \) and uses point \( (1, 2, -1) \).
A1: Correct substitution \( 1(1) + 5(2) + 3(-1) = 8 \).
A1: Final equation \( x + 5y + 3z = 8 \).

(c)(i)
M1: Uses scalar product formula \( \cos\theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|} \).
A1: Evaluates to \( \cos\theta = \frac{8}{\sqrt{210}} \).
A1: Obtains \( 56.5^\circ \) (or 0.986 rad).

(c)(ii)
M1: Finds direction vector of line via \( \mathbf{n}_1 \times \mathbf{n}_2 \).
A1: Correct direction vector \( \begin{pmatrix} -11 \\ 4 \\ -3 \end{pmatrix} \) (or any scalar multiple).
M1: Sets one coordinate to 0 (or another constant) to find a particular point on both planes.
A1: Correct coordinates of a point on the line (e.g. \( (4/3, 4/3, 0) \) or \( (0, 20/11, -4/11) \)).
A1: Correct vector equation of the line of intersection in standard form \( \mathbf{r} = \mathbf{a} + t\mathbf{b} \).

(a) By de Moivre's theorem,
\(\cos 5\theta + \mathrm{i}\sin 5\theta = (\cos\theta + \mathrm{i}\sin\theta)^5\).

Expanding using 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 \(\sin^2\theta = 1 - \cos^2\theta\):
\(\cos 5\theta = \cos^5\theta - 10\cos^3\theta(1 - \cos^2\theta) + 5\cos\theta(1 - \cos^2\theta)^2\)
\(\cos 5\theta = \cos^5\theta - 10\cos^3\theta + 10\cos^5\theta + 5\cos\theta(1 - 2\cos^2\theta + \cos^4\theta)\)
\(\cos 5\theta = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta\).

(b) The equation is \(16x^5 - 20x^3 + 5x = \frac{1}{\sqrt{2}}\).
Letting \(x = \cos\theta\), this becomes:
\(\cos 5\theta = \frac{1}{\sqrt{2}} = \cos\left(\frac{\pi}{4}\right)\).

Thus, \(5\theta = 2k\pi \pm \frac{\pi}{4}\) for \(k \in \mathbb{Z}\).
Choosing values of \(k\) to obtain 5 distinct values of \(\cos\theta\) in \([0, \pi]\):
- \(k = 0 \implies 5\theta = \frac{\pi}{4} \implies \theta = \frac{\pi}{20}\)
- \(k = 1 \implies 5\theta = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4} \implies \theta = \frac{7\pi}{20}\)
- \(k = 1 \implies 5\theta = 2\pi + \frac{\pi}{4} = \frac{9\pi}{4} \implies \theta = \frac{9\pi}{20}\)
- \(k = 2 \implies 5\theta = 4\pi - \frac{\pi}{4} = \frac{15\pi}{4} \implies \theta = \frac{15\pi}{20} = \frac{3\pi}{4}\)
- \(k = 2 \implies 5\theta = 4\pi + \frac{\pi}{4} = \frac{17\pi}{4} \implies \theta = \frac{17\pi}{20}\)

Therefore, the 5 roots are:
\(x = \cos\left(\frac{\pi}{20}\right), \cos\left(\frac{7\pi}{20}\right), \cos\left(\frac{9\pi}{20}\right), \cos\left(\frac{3\pi}{4}\right) = -\frac{1}{\sqrt{2}}, \cos\left(\frac{17\pi}{20}\right)\).

(a)
- M1: Uses de Moivre's theorem and expands \((\cos\theta + \mathrm{i}\sin\theta)^5\) using binomial theorem.
- M1: Equates real parts and applies \(\sin^2\theta = 1 - \cos^2\theta\).
- A1: Obtains correct expression \(16\cos^5\theta - 20\cos^3\theta + 5\cos\theta\).

(b)
- M1: Relates equation to \(\cos 5\theta = \frac{1}{\sqrt{2}}\) by setting \(x = \cos\theta\).
- A1: Solves for at least 3 correct values of \(\theta\).
- A1.5: States all 5 distinct roots correctly in exact trigonometric form (accept \(-\frac{1}{\sqrt{2}}\) or \(\cos\frac{3\pi}{4}\)).

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

\[I_n = \left[ -\frac{1}{2} x^n \mathrm{e}^{-2x} \right]_0^1 - \int_0^1 \left(-\frac{1}{2}\mathrm{e}^{-2x}\right) n x^{n-1} \, \mathrm{d}x\]
\[I_n = \left( -\frac{1}{2}\mathrm{e}^{-2} - 0 \right) + \frac{n}{2}\int_0^1 x^{n-1}\mathrm{e}^{-2x}\,\mathrm{d}x\]
\[I_n = \frac{n}{2} I_{n-1} - \frac{1}{2\mathrm{e}^2}\]

(b) First evaluate \(I_0\):
\[I_0 = \int_0^1 \mathrm{e}^{-2x}\,\mathrm{d}x = \left[ -\frac{1}{2}\mathrm{e}^{-2x} \right]_0^1 = -\frac{1}{2}\mathrm{e}^{-2} - \left(-\frac{1}{2}\right) = \frac{1}{2} - \frac{1}{2\mathrm{e}^2}\]

Now use the reduction formula:
\[I_1 = \frac{1}{2} I_0 - \frac{1}{2\mathrm{e}^2} = \frac{1}{2}\left(\frac{1}{2} - \frac{1}{2\mathrm{e}^2}\right) - \frac{1}{2\mathrm{e}^2} = \frac{1}{4} - \frac{3}{4\mathrm{e}^2}\]

\[I_2 = \frac{2}{2} I_1 - \frac{1}{2\mathrm{e}^2} = I_1 - \frac{1}{2\mathrm{e}^2} = \frac{1}{4} - \frac{3}{4\mathrm{e}^2} - \frac{1}{2\mathrm{e}^2} = \frac{1}{4} - \frac{5}{4\mathrm{e}^2}\]

\[I_3 = \frac{3}{2} I_2 - \frac{1}{2\mathrm{e}^2} = \frac{3}{2}\left(\frac{1}{4} - \frac{5}{4\mathrm{e}^2}\right) - \frac{1}{2\mathrm{e}^2} = \frac{3}{8} - \frac{15}{8\mathrm{e}^2} - \frac{4}{8\mathrm{e}^2} = \frac{3}{8} - \frac{19}{8\mathrm{e}^2} = \frac{3\mathrm{e}^2 - 19}{8\mathrm{e}^2}\]

(a)
- M1: Applies integration by parts with appropriate choices for \(u\) and \(\mathrm{d}v\).
- M1: Correctly evaluates the boundary term at limits \(0\) and \(1\).
- A1.5: Reaches the given reduction formula with fully correct working.

(b)
- B1: Finds correct \(I_0 = \frac{1}{2} - \frac{1}{2\mathrm{e}^2}\).
- M1: Applies the reduction formula iteratively from \(I_0\) to \(I_3\).
- A1: Obtains exact value \(\frac{3\mathrm{e}^2 - 19}{8\mathrm{e}^2}\) or \(\frac{3}{8} - \frac{19}{8\mathrm{e}^2}\).

(a) Let \(y = \operatorname{arcosh} x\) for \(x \ge 1\), so \(y \ge 0\) and \(x = \cosh y\).
Using the definition of \(\cosh y\):
\[x = \frac{\mathrm{e}^y + \mathrm{e}^{-y}}{2}\]
\[2x = \mathrm{e}^y + \frac{1}{\mathrm{e}^y}\]
Multiplying by \(\mathrm{e}^y\):
\[(\mathrm{e}^y)^2 - 2x\mathrm{e}^y + 1 = 0\]

Using the quadratic formula to solve for \(\mathrm{e}^y\):
\[\mathrm{e}^y = \frac{2x \pm \sqrt{4x^2 - 4}}{2} = x \pm \sqrt{x^2 - 1}\]
Since \(y \ge 0\), we must have \(\mathrm{e}^y \ge 1\).
For \(x \ge 1\), \(x - \sqrt{x^2 - 1} \le 1\) (with equality only at \(x = 1\)), while \(x + \sqrt{x^2 - 1} \ge 1\).
Therefore, \(\mathrm{e}^y = x + \sqrt{x^2 - 1}\).
Taking the natural logarithm of both sides:
\[y = \operatorname{arcosh} x = \ln\left(x + \sqrt{x^2 - 1}\right)\]

(b) Express \(\cosh x\) and \(\sinh x\) in exponential form:
\[5\left(\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right) + 3\left(\frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right) = 4\]
\[\frac{8\mathrm{e}^x + 2\mathrm{e}^{-x}}{2} = 4\]
\[4\mathrm{e}^x + \mathrm{e}^{-x} = 4\]

Multiply by \(\mathrm{e}^x\):
\[4(\mathrm{e}^x)^2 - 4\mathrm{e}^x + 1 = 0\]
\[(2\mathrm{e}^x - 1)^2 = 0\]
\[2\mathrm{e}^x - 1 = 0 \implies \mathrm{e}^x = \frac{1}{2}\]
\[x = \ln\left(\frac{1}{2}\right)\]

(a)
- M1: Sets \(x = \frac{\mathrm{e}^y + \mathrm{e}^{-y}}{2}\) and forms quadratic equation in \(\mathrm{e}^y\).
- M1: Solves quadratic equation using the quadratic formula.
- A1.5: Justifies the choice of the positive root using \(y \ge 0 \implies \mathrm{e}^y \ge 1\) and deduces logarithmic form.

(b)
- M1: Substitutes exponential forms for \(\cosh x\) and \(\sinh x\) to form a quadratic in \(\mathrm{e}^x\).
- M1: Solves the quadratic equation for \(\mathrm{e}^x\).
- A1: Obtains \(x = \ln\left(\frac{1}{2}\right)\) (or \(-\ln 2\)).

(a) Let \(C = \sum_{r=0}^\infty \left(\frac{1}{2}\right)^r \cos(r\theta)\) and \(S = \sum_{r=0}^\infty \left(\frac{1}{2}\right)^r \sin(r\theta)\).
Then:
\[C + \mathrm{i}S = \sum_{r=0}^\infty \left(\frac{1}{2}\right)^r (\cos(r\theta) + \mathrm{i}\sin(r\theta)) = \sum_{r=0}^\infty \left(\frac{1}{2}\mathrm{e}^{\mathrm{i}\theta}\right)^r\]

Let \(z = \frac{1}{2}\mathrm{e}^{\mathrm{i}\theta}\). Since \(|z| = \frac{1}{2} < 1\), this is a convergent infinite geometric series with first term \(a = 1\) and common ratio \(z\):
\[C + \mathrm{i}S = \frac{1}{1 - \frac{1}{2}\mathrm{e}^{\mathrm{i}\theta}} = \frac{2}{2 - \mathrm{e}^{\mathrm{i}\theta}} = \frac{2}{(2 - \cos\theta) - \mathrm{i}\sin\theta}\]

Multiplying numerator and denominator by the complex conjugate \((2 - \cos\theta) + \mathrm{i}\sin\theta\):
\[C + \mathrm{i}S = \frac{2\left((2 - \cos\theta) + \mathrm{i}\sin\theta\right)}{(2 - \cos\theta)^2 + \sin^2\theta}\]
\[C + \mathrm{i}S = \frac{4 - 2\cos\theta + 2\mathrm{i}\sin\theta}{4 - 4\cos\theta + \cos^2\theta + \sin^2\theta} = \frac{4 - 2\cos\theta + 2\mathrm{i}\sin\theta}{5 - 4\cos\theta}\]

Equating real parts gives:
\[C = \sum_{r=0}^\infty \left(\frac{1}{2}\right)^r \cos(r\theta) = \frac{4 - 2\cos\theta}{5 - 4\cos\theta}\]

(b) Setting \(\theta = \frac{\pi}{3}\):
\[\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\]
Substituting into the formula from part (a):
\[\sum_{r=0}^\infty \left(\frac{1}{2}\right)^r \cos\left(\frac{r\pi}{3}\right) = \frac{4 - 2\left(\frac{1}{2}\right)}{5 - 4\left(\frac{1}{2}\right)} = \frac{4 - 1}{5 - 2} = \frac{3}{3} = 1\]

(a)
- M1: Considers \(C + \mathrm{i}S\) and writes as a geometric progression in \(z = \frac{1}{2}\mathrm{e}^{\mathrm{i}\theta}\).
- M1: Applies sum to infinity formula \(S_\infty = \frac{a}{1 - z}\).
- M1: Multiplies numerator and denominator by conjugate \((2 - \cos\theta) + \mathrm{i}\sin\theta\).
- A1: Equates real parts to arrive at the given expression.

(b)
- M1: Substitutes \(\theta = \frac{\pi}{3}\) and uses \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\).
- A1.5: Obtains final simplified answer \(1\).

(a) For the complementary function, solve the auxiliary equation:
\[m^2 + 4m + 13 = 0 \implies (m+2)^2 + 9 = 0 \implies m = -2 \pm 3\mathrm{i}.\]
Thus, \(y_{CF} = \mathrm{e}^{-2x}(A\cos 3x + B\sin 3x)\).

For the particular integral, try \(y_{PI} = px + q + c\mathrm{e}^{-2x}\).
Then:
\[y'_{PI} = p - 2c\mathrm{e}^{-2x}\]
\[y''_{PI} = 4c\mathrm{e}^{-2x}\]
Substitute into the differential equation:
\[(4c\mathrm{e}^{-2x}) + 4(p - 2c\mathrm{e}^{-2x}) + 13(px + q + c\mathrm{e}^{-2x}) = 26x + 21 + 18\mathrm{e}^{-2x}\]
\[13px + (4p + 13q) + (4 - 8 + 13)c\mathrm{e}^{-2x} = 26x + 21 + 18\mathrm{e}^{-2x}\]
\[13px + (4p + 13q) + 9c\mathrm{e}^{-2x} = 26x + 21 + 18\mathrm{e}^{-2x}.\]
Equating coefficients:
\(13p = 26 \implies p = 2\)
\(4(2) + 13q = 21 \implies 13q = 13 \implies q = 1\)
\(9c = 18 \implies c = 2\).

Therefore, \(y_{PI} = 2x + 1 + 2\mathrm{e}^{-2x}\).
The general solution is:
\[y = \mathrm{e}^{-2x}(A\cos 3x + B\sin 3x) + 2x + 1 + 2\mathrm{e}^{-2x}.\]

(b) When \(x = 0\), \(y = 2\):
\[2 = \mathrm{e}^0(A\cos 0 + B\sin 0) + 2(0) + 1 + 2\mathrm{e}^0 \implies 2 = A + 1 + 2 \implies A = -1.\]
Differentiating \(y\):
\[y' = -2\mathrm{e}^{-2x}(A\cos 3x + B\sin 3x) + \mathrm{e}^{-2x}(-3A\sin 3x + 3B\cos 3x) + 2 - 4\mathrm{e}^{-2x}.\]
When \(x = 0\), \(y' = 3\):
\[3 = -2(A) + 3B + 2 - 4 = -2(-1) + 3B - 2 = 3B \implies 3B = 3 \implies B = 1.\]
Hence, the particular solution is:
\[y = \mathrm{e}^{-2x}(-\cos 3x + \sin 3x) + 2x + 1 + 2\mathrm{e}^{-2x}\]
(or equivalently \(y = \mathrm{e}^{-2x}(2 - \cos 3x + \sin 3x) + 2x + 1\)).

(c) As \(x \to \infty\), \(\mathrm{e}^{-2x} \to 0\) and \(\mathrm{e}^{-2x}(-\cos 3x + \sin 3x) \to 0\).
Therefore, \(y \approx 2x + 1\).

(a)
- M1: Forms and solves auxiliary equation \(m^2 + 4m + 13 = 0\).
- A1: Obtains correct complex conjugate roots \(m = -2 \pm 3\mathrm{i}\).
- A1: Correct form for complementary function \(y_{CF} = \mathrm{e}^{-2x}(A\cos 3x + B\sin 3x)\).
- M1: Uses suitable trial particular integral \(y = px + q + c\mathrm{e}^{-2x}\), differentiates twice and substitutes.
- A1: Finds correct values \(p = 2\) and \(q = 1\).
- A1: Finds correct value \(c = 2\).
- A1: Combines to state general solution \(y = y_{CF} + y_{PI}\).

(b)
- M1: Substitutes \(x = 0, y = 2\) into general solution to find \(A\).
- A1: Obtains \(A = -1\).
- M1: Differentiates general solution correctly using product rule and uses \(x = 0, y' = 3\).
- A1: Obtains \(B = 1\) and states correct particular solution.

(c)
- B1: States \(y \approx 2x + 1\) (or \(y = 2x + 1\)).

(a) Use the substitution \(u = \ln x\), so \(\mathrm{d}u = \frac{1}{x}\,\mathrm{d}x\).
\[\int_2^n \frac{1}{x \ln x}\,\mathrm{d}x = [\ln(\ln x)]_2^n = \ln(\ln n) - \ln(\ln 2).\]

(b) The function \(f(x) = \frac{1}{x \ln x}\) is strictly decreasing for \(x \ge 2\) because \(x\) and \(\ln x\) are both positive and strictly increasing.
Consider rectangles of width 1 on the interval \([2, n]\).
- Lower sum (rectangles under the curve, using right-hand endpoints \(x = 3, 4, \dots, n\)):
\[\text{Area} = 1 \cdot f(3) + 1 \cdot f(4) + \dots + 1 \cdot f(n) = \sum_{r=3}^n \frac{1}{r \ln r}.\]
Since the rectangles lie entirely below the curve \(y = f(x)\):
\[\sum_{r=3}^n \frac{1}{r \ln r} < \int_2^n \frac{1}{x \ln x}\,\mathrm{d}x.\]
- Upper sum (rectangles above the curve, using left-hand endpoints \(x = 2, 3, \dots, n-1\)):
\[\text{Area} = 1 \cdot f(2) + 1 \cdot f(3) + \dots + 1 \cdot f(n-1) = \sum_{r=2}^{n-1} \frac{1}{r \ln r}.\]
Since the rectangles cover the entire area under the curve:
\[\int_2^n \frac{1}{x \ln x}\,\mathrm{d}x < \sum_{r=2}^{n-1} \frac{1}{r \ln r}.\]
Substituting the integral from part (a) yields:
\[\sum_{r=3}^n \frac{1}{r \ln r} < \ln(\ln n) - \ln(\ln 2) < \sum_{r=2}^{n-1} \frac{1}{r \ln r}.\]

(c) Replacing \(n\) with \(n+1\) in the right-hand inequality of part (b):
\[\ln(\ln(n+1)) - \ln(\ln 2) < \sum_{r=2}^{(n+1)-1} \frac{1}{r \ln r} = \sum_{r=2}^n \frac{1}{r \ln r}.\]
Rearranging gives:
\[\sum_{r=2}^n \frac{1}{r \ln r} > \ln(\ln(n+1)) - \ln(\ln 2).\]

(d) Let \(S_n = \sum_{r=2}^n \frac{1}{r \ln r}\).
From part (c), \(S_n > \ln(\ln(n+1)) - \ln(\ln 2)\).
As \(n \to \infty\), \(\ln(n+1) \to \infty\), and therefore \(\ln(\ln(n+1)) \to \infty\).
Since \(S_n\) is bounded below by an expression tending to infinity, \(\lim_{n\to\infty} S_n = \infty\).
Hence, the series \(\sum_{r=2}^\infty \frac{1}{r \ln r}\) diverges.

(a)
- M1: Recognises form \(\int \frac{f'(x)}{f(x)}\,\mathrm{d}x\) or uses substitution \(u = \ln x\).
- A1: Fully correct integration with limits substituted to obtain \(\ln(\ln n) - \ln(\ln 2)\).

(b)
- B1: States that \(f(x)\) is a decreasing function for \(x \ge 2\).
- M1: Considers lower sum of rectangles of unit width with heights at right-hand endpoints \(r = 3, \dots, n\).
- A1: Establishes \(\sum_{r=3}^n \frac{1}{r \ln r} < \int_2^n \frac{1}{x \ln x}\,\mathrm{d}x\).
- M1: Considers upper sum of rectangles of unit width with heights at left-hand endpoints \(r = 2, \dots, n-1\).
- A1: Establishes \(\int_2^n \frac{1}{x \ln x}\,\mathrm{d}x < \sum_{r=2}^{n-1} \frac{1}{r \ln r}\) and concludes the combined inequality.

(c)
- M1: Replaces \(n\) by \(n+1\) in the upper bound inequality.
- A1: Fully justifies \(\sum_{r=2}^n \frac{1}{r \ln r} > \ln(\ln(n+1)) - \ln(\ln 2)\).

(d)
- M1: Considers behaviour of lower bound as \(n \to \infty\).
- A1: Identifies that \(\ln(\ln(n+1)) \to \infty\).
- A1: Concludes correctly with sound reasoning that the series diverges.

(a) The characteristic equation is \(\det(\mathbf{A} - \lambda\mathbf{I}) = 0\):
\[\begin{vmatrix} -\lambda & 2 & 1 \\ -1 & 3-\lambda & 1 \\ -3 & 3 & 4-\lambda \end{vmatrix} = 0.\]
Expanding across the first row:
\[-\lambda [(3-\lambda)(4-\lambda) - 3] - 2 [-1(4-\lambda) - (-3)(1)] + 1 [-3 - (-3)(3-\lambda)] = 0\]
\[-\lambda [\lambda^2 - 7\lambda + 9] - 2 [\lambda - 1] + 1 [6 - 3\lambda] = 0\]
\[-\lambda^3 + 7\lambda^2 - 9\lambda - 2\lambda + 2 + 6 - 3\lambda = 0\]
\[-\lambda^3 + 7\lambda^2 - 14\lambda + 8 = 0 \implies \lambda^3 - 7\lambda^2 + 14\lambda - 8 = 0.\]
Factoring:
\[(\lambda - 1)(\lambda^2 - 6\lambda + 8) = (\lambda - 1)(\lambda - 2)(\lambda - 4) = 0.\]
Thus, the eigenvalues are \(\lambda_1 = 1\), \(\lambda_2 = 2\), and \(\lambda_3 = 4\).

(b) For \(\lambda = 1\):
\[\begin{pmatrix} -1 & 2 & 1 \\ -1 & 2 & 1 \\ -3 & 3 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}.\]
From row 3: \(-x + y + z = 0\). From row 1: \(-x + 2y + z = 0\). Subtracting gives \(y = 0\), which leads to \(x = z\).
An eigenvector is \(\mathbf{e}_1 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}\).

For \(\lambda = 2\):
\[\begin{pmatrix} -2 & 2 & 1 \\ -1 & 1 & 1 \\ -3 & 3 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}.\]
Row 2 minus Row 1 gives \(x - y = 0 \implies x = y\). Substituting into \(-x + y + z = 0\) gives \(z = 0\).
An eigenvector is \(\mathbf{e}_2 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}\).

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

(c) Form \(\mathbf{P}\) from the eigenvectors and \(\mathbf{D}\) from the eigenvalues:
\[\mathbf{P} = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 2 \end{pmatrix}, \quad \mathbf{D} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{pmatrix}.\]

(d) First compute \(\mathbf{P}^{-1}\):
\(\det \mathbf{P} = 1(2-0) - 1(0-1) + 1(0-1) = 2 + 1 - 1 = 2\).
\[\mathbf{P}^{-1} = \frac{1}{2} \begin{pmatrix} 2 & -2 & 0 \\ 1 & 1 & -1 \\ -1 & 1 & 1 \end{pmatrix}.\]
Now calculate \(\mathbf{A}^n = \mathbf{P}\mathbf{D}^n\mathbf{P}^{-1}\):
\[\mathbf{P}\mathbf{D}^n = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 2 \end{pmatrix} \begin{pmatrix} 1^n & 0 & 0 \\ 0 & 2^n & 0 \\ 0 & 0 & 4^n \end{pmatrix} = \begin{pmatrix} 1 & 2^n & 4^n \\ 0 & 2^n & 4^n \\ 1 & 0 & 2\cdot 4^n \end{pmatrix}.\]
Multiplying by \(\mathbf{P}^{-1}\):
\[\mathbf{A}^n = \begin{pmatrix} 1 & 2^n & 4^n \\ 0 & 2^n & 4^n \\ 1 & 0 & 2\cdot 4^n \end{pmatrix} \frac{1}{2} \begin{pmatrix} 2 & -2 & 0 \\ 1 & 1 & -1 \\ -1 & 1 & 1 \end{pmatrix}\]
\[= \frac{1}{2} \begin{pmatrix} 2 + 2^n - 4^n & -2 + 2^n + 4^n & -2^n + 4^n \\ 2^n - 4^n & 2^n + 4^n & -2^n + 4^n \\ 2 - 2\cdot 4^n & -2 + 2\cdot 4^n & 2\cdot 4^n \end{pmatrix}.\]

(a)
- M1: Sets up characteristic determinant \(\det(\mathbf{A} - \lambda\mathbf{I}) = 0\).
- M1: Expands determinant to obtain a cubic polynomial in \(\lambda\).
- A1: Obtains \(-\lambda^3 + 7\lambda^2 - 14\lambda + 8 = 0\) (or factorised form).
- A1: Solves correctly to obtain \(\lambda = 1, 2, 4\).

(b)
- M1: Sets up \((\mathbf{A} - \lambda\mathbf{I})\mathbf{v} = \mathbf{0}\) for at least one eigenvalue.
- A1: Correct eigenvector for \(\lambda = 1\) (any non-zero scalar multiple of \([1, 0, 1]^T\)).
- A1: Correct eigenvector for \(\lambda = 2\) (any non-zero scalar multiple of \([1, 1, 0]^T\)).
- A1: Correct eigenvector for \(\lambda = 4\) (any non-zero scalar multiple of \([1, 1, 2]^T\)).

(c)
- B1: States correct \(\mathbf{P}\) corresponding to their order of eigenvectors.
- B1: States correct \(\mathbf{D}\) matching the order in \(\mathbf{P}\).

(d)
- M1: Finds \(\mathbf{P}^{-1}\) correctly.
- M1: Uses \(\mathbf{A}^n = \mathbf{P}\mathbf{D}^n\mathbf{P}^{-1}\) with \(\mathbf{D}^n = \operatorname{diag}(1, 2^n, 4^n)\).
- A1: Obtains fully correct expression for \(\mathbf{A}^n\).

(a) Apply integration by parts to \(I_n = \int_0^{\frac{\pi}{2}} x^n \cos x\,\mathrm{d}x\):
Let \(u = x^n\) and \(v' = \cos x\), so \(u' = n x^{n-1}\) and \(v = \sin x\).
\[I_n = [x^n \sin x]_0^{\frac{\pi}{2}} - n \int_0^{\frac{\pi}{2}} x^{n-1} \sin x\,\mathrm{d}x\]
\[= \left(\frac{\pi}{2}\right)^n \sin\left(\frac{\pi}{2}\right) - 0 - n \int_0^{\frac{\pi}{2}} x^{n-1} \sin x\,\mathrm{d}x = \left(\frac{\pi}{2}\right)^n - n \int_0^{\frac{\pi}{2}} x^{n-1} \sin x\,\mathrm{d}x.\]
Apply integration by parts to \(\int_0^{\frac{\pi}{2}} x^{n-1} \sin x\,\mathrm{d}x\):
Let \(u = x^{n-1}\) and \(v' = \sin x\), so \(u' = (n-1)x^{n-2}\) and \(v = -\cos x\).
\[\int_0^{\frac{\pi}{2}} x^{n-1} \sin x\,\mathrm{d}x = [-x^{n-1}\cos x]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} -(n-1)x^{n-2}\cos x\,\mathrm{d}x\]
\[= \left(0 - 0\right) + (n-1) \int_0^{\frac{\pi}{2}} x^{n-2} \cos x\,\mathrm{d}x = (n-1) I_{n-2}.\]
Substituting back into the expression for \(I_n\):
\[I_n = \left(\frac{\pi}{2}\right)^n - n(n-1)I_{n-2}.\]

(b) For \(n = 0\):
\[I_0 = \int_0^{\frac{\pi}{2}} \cos x\,\mathrm{d}x = [\sin x]_0^{\frac{\pi}{2}} = 1 - 0 = 1.\]
For \(n = 1\):
\[I_1 = \int_0^{\frac{\pi}{2}} x \cos x\,\mathrm{d}x = [x \sin x + \cos x]_0^{\frac{\pi}{2}} = \left(\frac{\pi}{2}(1) + 0\right) - (0 + 1) = \frac{\pi}{2} - 1.\]

(c) Using the reduction formula for \(n = 2\):
\[I_2 = \left(\frac{\pi}{2}\right)^2 - 2(1)I_0 = \frac{\pi^2}{4} - 2(1) = \frac{\pi^2}{4} - 2.\]
Using the reduction formula for \(n = 4\):
\[I_4 = \left(\frac{\pi}{2}\right)^4 - 4(3)I_2 = \frac{\pi^4}{16} - 12\left(\frac{\pi^2}{4} - 2\right) = \frac{\pi^4}{16} - 3\pi^2 + 24.\]

(d) Use the half-angle identity \(\sin^2\left(\frac{x}{2}\right) = \frac{1 - \cos x}{2}\):
\[\int_0^{\frac{\pi}{2}} x^2 \sin^2\left(\frac{x}{2}\right)\,\mathrm{d}x = \int_0^{\frac{\pi}{2}} x^2 \left(\frac{1 - \cos x}{2}\right)\,\mathrm{d}x\]
\[= \frac{1}{2} \int_0^{\frac{\pi}{2}} x^2\,\mathrm{d}x - \frac{1}{2} \int_0^{\frac{\pi}{2}} x^2 \cos x\,\mathrm{d}x\]
\[= \frac{1}{2} \left[\frac{x^3}{3}\right]_0^{\frac{\pi}{2}} - \frac{1}{2} I_2\]
\[= \frac{1}{2} \left(\frac{\pi^3}{24}\right) - \frac{1}{2} \left(\frac{\pi^2}{4} - 2\right) = \frac{\pi^3}{48} - \frac{\pi^2}{8} + 1.\]

(a)
- M1: Applies integration by parts once to \(I_n\) with appropriate choice of \(u\) and \(v'\).
- A1: Obtains \(I_n = \left(\frac{\pi}{2}\right)^n - n \int_0^{\frac{\pi}{2}} x^{n-1} \sin x\,\mathrm{d}x\).
- M1: Applies integration by parts to \(\int_0^{\frac{\pi}{2}} x^{n-1} \sin x\,\mathrm{d}x\).
- A1: Obtains \((n-1)I_{n-2}\) with boundary terms evaluating correctly to \(0\).
- A1: Completes proof to show \(I_n = \left(\frac{\pi}{2}\right)^n - n(n-1)I_{n-2}\).

(b)
- B1: Obtains \(I_0 = 1\).
- B1: Obtains \(I_1 = \frac{\pi}{2} - 1\).

(c)
- M1: Uses reduction formula with \(n = 2\) to find \(I_2\).
- A1: Obtains \(I_2 = \frac{\pi^2}{4} - 2\).
- A1: Uses reduction formula with \(n = 4\) to find \(I_4 = \frac{\pi^4}{16} - 3\pi^2 + 24\).

(d)
- M1: Uses identity \(\sin^2(x/2) = \frac{1 - \cos x}{2}\) to split the integral into \(\frac{1}{2}\int x^2\,\mathrm{d}x - \frac{1}{2}I_2\).
- A1: Evaluates correctly to \(\frac{\pi^3}{48} - \frac{\pi^2}{8} + 1\).

Let the line of centres be the $x$-axis and the perpendicular direction in the plane of motion be the $y$-axis.

Immediately before impact:
For sphere $A$: $u_x = u\cos 30^\circ = \frac{\sqrt{3}}{2}u$, $u_y = u\sin 30^\circ = \frac{1}{2}u$.
For sphere $B$: $u_{Bx} = 0$, $u_{By} = 0$.

Let the velocities after impact be $(v_{Ax}, v_{Ay})$ for $A$ and $(v_{Bx}, v_{By})$ for $B$.
Since the spheres are smooth, there is no impulse perpendicular to the line of centres:
$v_{Ay} = u_y = \frac{1}{2}u$ and $v_{By} = 0$.

(a) Conservation of linear momentum along the line of centres:
$m u_x + m(0) = m v_{Ax} + m v_{Bx} \implies v_{Ax} + v_{Bx} = \frac{\sqrt{3}}{2}u$

Newton's experimental law of restitution along the line of centres:
$v_{Bx} - v_{Ax} = e(u_x - 0) = \frac{1}{2}\left(\frac{\sqrt{3}}{2}u\right) = \frac{\sqrt{3}}{4}u$

Adding the two equations:
$2 v_{Bx} = \frac{\sqrt{3}}{2}u + \frac{\sqrt{3}}{4}u = \frac{3\sqrt{3}}{4}u \implies v_{Bx} = \frac{3\sqrt{3}}{8}u$.
Since $v_{By} = 0$, the speed of $B$ after impact is $v_B = \frac{3\sqrt{3}}{8}u$.

(b) Subtracting the equations gives:
$2 v_{Ax} = \frac{\sqrt{3}}{2}u - \frac{\sqrt{3}}{4}u = \frac{\sqrt{3}}{4}u \implies v_{Ax} = \frac{\sqrt{3}}{8}u$.

The speed of $A$ after impact is:
$v_A = \sqrt{v_{Ax}^2 + v_{Ay}^2} = \sqrt{\left(\frac{\sqrt{3}}{8}u\right)^2 + \left(\frac{1}{2}u\right)^2} = u\sqrt{\frac{3}{64} + \frac{1}{4}} = u\sqrt{\frac{3+16}{64}} = \frac{\sqrt{19}}{8}u$.

(c) Let $\theta$ be the angle that the velocity of $A$ makes with the line of centres after impact:
$\tan \theta = \frac{v_{Ay}}{v_{Ax}} = \frac{\frac{1}{2}u}{\frac{\sqrt{3}}{8}u} = \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3} \approx 2.3094$.
Thus $\theta = \arctan\left(\frac{4}{\sqrt{3}}\right) \approx 66.5867^\circ$.

The angle of deflection is $\theta - 30^\circ = 66.5867^\circ - 30^\circ = 36.6^\circ$ (or $21.8^\circ$ if measured relative to another axis; here deflecting from $30^\circ$ gives $66.59^\circ - 30^\circ = 36.6^\circ$).

M1: Set up conservation of linear momentum and Newton's experimental law parallel to the line of centres.
A1: Obtain $v_{Bx} = \frac{3\sqrt{3}}{8}u$ and state speed of $B$.
M1: Find $v_{Ax} = \frac{\sqrt{3}}{8}u$ and apply Pythagoras with unchanged perpendicular component $v_{Ay} = \frac{1}{2}u$.
A1: Correctly show $v_A = \frac{\sqrt{19}}{8}u$ with all steps shown clearly.
M1: Calculate new angle $\theta = \arctan\left(\frac{v_{Ay}}{v_{Ax}}\right)$.
A1: Correct angle of deflection $36.6^\circ$ (or $0.639\text{ rad}$). Accept $36.59^\circ$.

(a) Let $\theta$ be the angle that the string makes with the upward vertical. The height of $P$ above $A$ is $h = l(1 + \cos\theta)$.
By conservation of mechanical energy from $A$ to the point where the string becomes slack:
$\frac{1}{2}m u^2 = \frac{1}{2}m v^2 + mgh$
$\frac{1}{2}m\left(\frac{7}{2}gl\right) = \frac{1}{2}m v^2 + mgl(1 + \cos\theta)$
$v^2 = \frac{7}{2}gl - 2gl(1 + \cos\theta) = \frac{3}{2}gl - 2gl\cos\theta$.

The radial equation of motion towards $O$ is:
$T + mg\cos\theta = \frac{mv^2}{l}$.
When the string becomes slack, $T = 0$:
$mg\cos\theta = \frac{m}{l}\left(\frac{3}{2}gl - 2gl\cos\theta\right)$
$\cos\theta = \frac{3}{2} - 2\cos\theta \implies 3\cos\theta = \frac{3}{2} \implies \cos\theta = \frac{1}{2}$.
Thus $\theta = 60^\circ$.

(b) At this instant:
$v^2 = \frac{3}{2}gl - 2gl\left(\frac{1}{2}\right) = \frac{1}{2}gl$, so $v = \sqrt{\frac{1}{2}gl}$.
The height above $A$ at the instant the string becomes slack is:
$h_1 = l(1 + \cos 60^\circ) = l\left(1 + \frac{1}{2}\right) = \frac{3}{2}l$.

The particle now moves freely under gravity as a projectile. The velocity $v$ makes an angle of $\theta = 60^\circ$ with the horizontal.
The vertical component of velocity is $v_y = v\sin 60^\circ = \sqrt{\frac{1}{2}gl}\left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{2\sqrt{2}}\sqrt{gl}$.

The additional height reached during projectile motion is:
$h_2 = \frac{v_y^2}{2g} = \frac{\frac{3}{8}gl}{2g} = \frac{3}{16}l$.

Therefore, the greatest height reached above the level of $A$ is:
$H = h_1 + h_2 = \frac{3}{2}l + \frac{3}{16}l = \frac{27}{16}l = \frac{54}{32}l$.
Wait, let us re-evaluate the angle of velocity with the horizontal:
The string makes an angle $\theta = 60^\circ$ with the upward vertical, so the tangent to the circle (direction of $v$) makes an angle $\theta = 60^\circ$ with the horizontal. Thus $v_y = v\sin 60^\circ$, giving $v_y^2 = \frac{1}{2}gl \cdot \frac{3}{4} = \frac{3}{8}gl$, so $h_2 = \frac{3}{16}l$.
Total height $= \frac{3}{2}l + \frac{3}{16}l = \frac{27}{16}l$.

M1: Apply conservation of energy to express $v^2$ in terms of $\theta$.
M1: Use Newton's second law along the string and set $T = 0$ to find $\cos\theta$.
A1: Obtain $\cos\theta = \frac{1}{2}$, giving $\theta = 60^\circ$ (or $\frac{\pi}{3}$ radians).
M1: Calculate $v^2 = \frac{1}{2}gl$ and determine the vertical velocity component $v_y = v\sin 60^\circ$.
M1: Use $h_2 = \frac{v_y^2}{2g}$ to find the additional height gained.
A1: Correctly calculate the total greatest height $H = \frac{27}{16}l$.

(a) The equation of motion for $P$ using $a = v\frac{dv}{dx}$ is:
$m v\frac{dv}{dx} = -m k(v + v^3)$
$v\frac{dv}{dx} = -k v(1 + v^2)$
$\frac{1}{1 + v^2}\,dv = -k\,dx$

Integrating both sides between $v = u$ ($x = 0$) and $v = \frac{1}{2}u$ ($x = X$):
$\int_{u}^{u/2} \frac{1}{1 + v^2}\,dv = -k \int_{0}^{X} dx$
$[\arctan v]_{u}^{u/2} = -k X$
$\arctan\left(\frac{u}{2}\right) - \arctan(u) = -k X$
$k X = \arctan(u) - \arctan\left(\frac{u}{2}\right)$

Using the compound angle identity $\arctan A - \arctan B = \arctan\left(\frac{A - B}{1 + AB}\right)$:
$k X = \arctan\left(\frac{u - \frac{u}{2}}{1 + u\left(\frac{u}{2}\right)}\right) = \arctan\left(\frac{\frac{u}{2}}{1 + \frac{u^2}{2}}\right) = \arctan\left(\frac{u}{2 + u^2}\right)$

Hence,
$X = \frac{1}{k}\arctan\left(\frac{u}{2 + u^2}\right)$.

(b) Using $a = \frac{dv}{dt}$, the equation of motion is:
$m\frac{dv}{dt} = -m k(v + v^3) \implies \frac{dv}{dt} = -k v(1 + v^2)$
$\frac{1}{v(1 + v^2)}\,dv = -k\,dt$

Using partial fractions:
$\frac{1}{v(1 + v^2)} = \frac{1}{v} - \frac{v}{1 + v^2}$

Integrating from $t = 0, v = u$ to $t = T, v = \frac{1}{2}u$:
$\int_{u}^{u/2} \left(\frac{1}{v} - \frac{v}{1 + v^2}\right) dv = -k \int_0^T dt$
$\left[ \ln v - \frac{1}{2}\ln(1 + v^2) \right]_{u}^{u/2} = -k T$
$\left( \ln(u/2) - \frac{1}{2}\ln\left(1 + \frac{u^2}{4}\right) \right) - \left( \ln u - \frac{1}{2}\ln(1 + u^2) \right) = -k T$
$\ln\left(\frac{u/2}{u}\right) - \frac{1}{2}\ln\left(\frac{4 + u^2}{4(1 + u^2)}\right) = -k T$
$\ln\left(\frac{1}{2}\right) - \frac{1}{2}\ln\left(\frac{4+u^2}{4(1+u^2)}\right) = -k T$
$-\frac{1}{2}\ln 4 - \frac{1}{2}\ln\left(\frac{4+u^2}{4(1+u^2)}\right) = -k T$
$-\frac{1}{2}\ln\left(\frac{4+u^2}{1+u^2}\right) = -k T$
$T = \frac{1}{2k}\ln\left(\frac{4(1+u^2)}{4+u^2}\right)$.

M1: Set up $v\frac{dv}{dx} = -k(v + v^3)$ and separate variables.
A1: Integrate to get $[\arctan v] = -kx$ with correct limits.
M1: Apply identity for difference of arctangents to establish the given formula for $X$.
M1: Set up $\frac{dv}{dt} = -k v(1 + v^2)$ and apply partial fractions $\frac{1}{v} - \frac{v}{1+v^2}$.
A1: Correctly integrate to obtain $\ln v - \frac{1}{2}\ln(1+v^2)$.
A1: Substitute limits and obtain fully simplified expression $T = \frac{1}{2k}\ln\left(\frac{4(1+u^2)}{4+u^2}\right)$ (or equivalent).

(a) The particle moves in a horizontal circle at a distance $l$ along the cone surface from $V$.
The radius of the circular path is $r = l\sin 30^\circ = \frac{1}{2}l$.
The height above $V$ is $l\cos 30^\circ = \frac{\sqrt{3}}{2}l$.

Forces acting on $P$ are:
- Weight $mg$ acting vertically downwards,
- Tension $T$ acting along the surface towards $V$ (at angle $30^\circ$ to the upward vertical, so pointing downwards and inwards at $30^\circ$ to the vertical),
Wait, since $V$ is downwards, the string extends upwards from $V$ to $P$. Thus the string pulls $P$ towards $V$, which is downwards along the surface at angle $30^\circ$ to the downward vertical.
- Normal reaction $R$ perpendicular to the surface of the cone, directed upwards and inwards at angle $30^\circ$ above the horizontal (i.e. $60^\circ$ to the upward vertical).

Resolving forces vertically:
$R\cos 30^\circ - T\cos 30^\circ - mg = 0 \implies (R - T)\frac{\sqrt{3}}{2} = mg \implies R - T = \frac{2}{\sqrt{3}}mg$

Resolving forces horizontally (towards the centre of the circle):
$R\sin 30^\circ + T\sin 30^\circ = m r \omega^2$
$(R + T)\frac{1}{2} = m\left(\frac{1}{2}l\right)\omega^2 \implies R + T = m l \omega^2$

Subtracting the first equation from the second:
$2T = m l \omega^2 - \frac{2}{\sqrt{3}}mg \implies T = \frac{1}{2}m l \omega^2 - \frac{1}{\sqrt{3}}mg$.
Wait! Let's check the direction of tension: if the particle is on the inner surface and $V$ is at the bottom, $T$ points towards $V$ (downwards along the cone). Then vertical equilibrium gives $R\cos 30^\circ - T\cos 30^\circ = mg$ if $R$ is inwards/upwards.
Wait, if the particle is at rest, $R\cos 30^\circ = mg \implies R = \frac{2}{\sqrt{3}}mg > 0$ and $T=0$. When rotating faster, it tends to slide up, so tension $T$ acts towards $V$ to hold it down!
Then $T = m\left(\frac{1}{2}l\omega^2 - \frac{1}{\sqrt{3}}mg\right)$.
Adding the two equations:
$2R = m l \omega^2 + \frac{2}{\sqrt{3}}mg \implies R = m\left(\frac{1}{\sqrt{3}}g + \frac{1}{2}l\omega^2\right)$.

Wait, let us consider the standard setup where $V$ is above the particle (vertex at top):
If $V$ is fixed at the top (axis vertical), string goes from $V$ down to $P$ along the outer surface of a cone (or inside an inverted cone):
$T$ acts upwards along the generator at $30^\circ$ to the vertical.
$R$ acts perpendicular to the cone (upwards and outwards at $60^\circ$ to the vertical).
Resolving vertically: $T\cos 30^\circ + R\sin 30^\circ = mg \implies T\frac{\sqrt{3}}{2} + R\frac{1}{2} = mg$.
Resolving horizontally: $T\sin 30^\circ - R\cos 30^\circ = m r \omega^2 = m\left(\frac{1}{2}l\right)\omega^2$.
Then:
$\frac{\sqrt{3}}{2}T + \frac{1}{2}R = mg$
$\frac{1}{2}T - \frac{\sqrt{3}}{2}R = \frac{1}{2}m l \omega^2$

Multiplying the first by $\sqrt{3}$ and adding to the second:
$\left(\frac{3}{2} + \frac{1}{2}\right)T = \sqrt{3}mg + \frac{1}{2}m l \omega^2 \implies 2T = \sqrt{3}mg + \frac{1}{2}m l \omega^2 \implies T = m\left(\frac{\sqrt{3}}{2}g + \frac{1}{4}l\omega^2\right)$.
This matches the stated question formula perfectly! So $V$ is at the top (vertex upwards).

(b) To find $R$, multiply the first equation by 1 and the second by $-\sqrt{3}$:
$\left(\frac{1}{2} + \frac{3}{2}\right)R = mg - \sqrt{3}\left(\frac{1}{2}m l \omega^2\right)$
$2R = mg - \frac{\sqrt{3}}{2}m l \omega^2 \implies R = m\left(\frac{1}{2}g - \frac{\sqrt{3}}{4}l\omega^2\right)$.

(c) For the particle to remain in contact with the cone, $R \geqslant 0$:
$\frac{1}{2}g - \frac{\sqrt{3}}{4}l\omega^2 \geqslant 0$
$\frac{\sqrt{3}}{4}l\omega^2 \leqslant \frac{1}{2}g \implies \omega^2 \leqslant \frac{2g}{\sqrt{3}l} \implies \omega \leqslant \sqrt{\frac{2g}{\sqrt{3}l}}$.

M1: State correct radius $r = l\sin 30^\circ = \frac{1}{2}l$ and set up equations of motion resolving horizontally and vertically.
A1: Obtain correct simultaneous equations $\frac{\sqrt{3}}{2}T + \frac{1}{2}R = mg$ and $\frac{1}{2}T - \frac{\sqrt{3}}{2}R = \frac{1}{2}ml\omega^2$.
A1: Eliminate $R$ to correctly obtain $T = m\left(\frac{\sqrt{3}}{2}g + \frac{1}{4}l\omega^2\right)$.
M1: Eliminate $T$ to find $R$.
A1: Correct expression $R = m\left(\frac{1}{2}g - \frac{\sqrt{3}}{4}l\omega^2\right)$.
A1: Set $R \geqslant 0$ and deduce $\omega \leqslant \sqrt{\frac{2g}{\sqrt{3}l}}$ (or equivalent).

Let the wall be along the $x$-axis and the normal to the wall be the $y$-axis.
Immediately before impact, the velocity components are:
$u_x = u\cos\alpha$ (parallel to the wall),
$u_y = u\sin\alpha$ (perpendicular to the wall).

Initial kinetic energy is $E_k = \frac{1}{2}m u^2$.

Since the wall is smooth, the velocity component parallel to the wall is unchanged:
$v_x = u_x = u\cos\alpha$.
By Newton's experimental law perpendicular to the wall:
$v_y = e u_y = \frac{1}{3}u\sin\alpha$.

The kinetic energy after impact is:
$E_k' = \frac{1}{2}m(v_x^2 + v_y^2) = \frac{1}{2}m\left(u^2\cos^2\alpha + \frac{1}{9}u^2\sin^2\alpha\right)$.

The loss in kinetic energy is:
$\Delta E_k = E_k - E_k' = \frac{1}{2}m u^2 - \frac{1}{2}m u^2\left(\cos^2\alpha + \frac{1}{9}\sin^2\alpha\right) = \frac{1}{2}m u^2\left(1 - \cos^2\alpha - \frac{1}{9}\sin^2\alpha\right) = \frac{1}{2}m u^2\left(\frac{8}{9}\sin^2\alpha\right)$.

(a) We are given $\Delta E_k = \frac{2}{9} E_k = \frac{2}{9}\left(\frac{1}{2}m u^2\right)$:
$\frac{1}{2}m u^2\left(\frac{8}{9}\sin^2\alpha\right) = \frac{2}{9}\left(\frac{1}{2}m u^2\right)$
$\frac{8}{9}\sin^2\alpha = \frac{2}{9} \implies \sin^2\alpha = \frac{2}{8} = \frac{1}{4}$.
Wait! If $\sin^2\alpha = \frac{1}{4}$, then $\sin\alpha = \frac{1}{2}$, so $\alpha = 30^\circ$.
Let's check: $\Delta E_k = \frac{1}{2}mu^2 \cdot \frac{8}{9} \sin^2(30^\circ) = \frac{1}{2}mu^2 \cdot \frac{8}{9} \cdot \frac{1}{4} = \frac{2}{9}(\frac{1}{2}mu^2)$.
This gives $\sin\alpha = \frac{1}{2}$ and $\alpha = 30^\circ$.

(b) Since $\alpha = 30^\circ$:
$v_x = u\cos 30^\circ = \frac{\sqrt{3}}{2}u$,
$v_y = \frac{1}{3}u\sin 30^\circ = \frac{1}{3}\left(\frac{1}{2}u\right) = \frac{1}{6}u$.

The speed after impact is:
$v = \sqrt{v_x^2 + v_y^2} = \sqrt{\left(\frac{\sqrt{3}}{2}u\right)^2 + \left(\frac{1}{6}u\right)^2} = u\sqrt{\frac{3}{4} + \frac{1}{36}} = u\sqrt{\frac{27 + 1}{36}} = u\sqrt{\frac{28}{36}} = \frac{\sqrt{7}}{3}u$.

(c) Let $\beta$ be the angle the velocity makes with the wall after impact:
$\tan\beta = \frac{v_y}{v_x} = \frac{\frac{1}{6}u}{\frac{\sqrt{3}}{2}u} = \frac{1}{3\sqrt{3}} = \frac{\sqrt{3}}{9} \approx 0.19245$.
$\beta = \arctan\left(\frac{\sqrt{3}}{9}\right) \approx 10.9^\circ$.

M1: Write components of velocity parallel and perpendicular to the wall after impact ($v_x = u\cos\alpha$, $v_y = \frac{1}{3}u\sin\alpha$).
M1: Form expression for loss in kinetic energy $\frac{1}{2}mu^2(1 - e^2)\sin^2\alpha = \frac{1}{2}mu^2\left(\frac{8}{9}\sin^2\alpha\right)$.
A1: Equate to $\frac{2}{9}E_k$ and solve to find $\sin\alpha = \frac{1}{2}$, giving $\alpha = 30^\circ$.
M1: Calculate $v_x = \frac{\sqrt{3}}{2}u$ and $v_y = \frac{1}{6}u$ and apply Pythagoras.
A1: Correct post-impact speed $v = \frac{\sqrt{7}}{3}u$.
A1: Find $\tan\beta = \frac{v_y}{v_x} = \frac{\sqrt{3}}{9}$ and deduce $\beta = 10.9^\circ$ (or $0.190\text{ rad}$).

(a) Let \(\rho\) be the uniform density of the material.
Mass of the hemisphere: \(M_1 = \rho \times \frac{2}{3}\pi r^3\).
Mass of the cone: \(M_2 = \rho \times \frac{1}{3}\pi r^2 (3r) = \rho \pi r^3\).
Thus, the ratio of masses is \(M_1 : M_2 = 2 : 3\), with total mass proportional to \(5\).

Let the centre of the common base be the origin \(O\), with the axis of symmetry taken as the \(y\)-axis directed into the cone.
Distance of the centre of mass of the hemisphere from \(O\) is \(\bar{y}_1 = -\frac{3}{8}r\).
Distance of the centre of mass of the cone from \(O\) is \(\bar{y}_2 = \frac{1}{4}h = \frac{1}{4}(3r) = \frac{3}{4}r\).

Taking moments about the common circular base:
\((M_1 + M_2)\bar{y} = M_1\bar{y}_1 + M_2\bar{y}_2\)
\(5\bar{y} = 2\left(-\frac{3}{8}r\right) + 3\left(\frac{3}{4}r\right) = -\frac{3}{4}r + \frac{9}{4}r = \frac{6}{4}r = \frac{3}{2}r\)
\(\bar{y} = \frac{3}{10}r\).
Since \(\bar{y} > 0\), the centre of mass lies inside the cone.

(b) Let \(G\) be the centre of mass of the composite solid, so \(OG = \frac{3}{10}r\) along the axis of symmetry.
Point \(A\) is on the circumference of the base, so \(OA = r\) and \(OA \perp OG\).
When freely suspended from \(A\), the line \(AG\) is vertical.
Let \(\theta\) be the angle between the axis of symmetry \(OG\) and the vertical \(AG\):
\(\tan \theta = \frac{OA}{OG} = \frac{r}{\frac{3}{10}r} = \frac{10}{3}\).
The common circular base is perpendicular to \(OG\), so the angle \(\phi\) between the base and the vertical is:
\(\phi = 90^\circ - \theta\)
\(\tan \phi = \cot \theta = \frac{3}{10} = 0.3\)
\(\phi = \arctan(0.3) \approx 16.699^\circ \approx 16.7^\circ\).

(c) The solid rests on a generator \(VC\), where \(V\) is the cone vertex and \(C\) is a point on the rim of the base.
In \(\triangle VOC\), \(VO = 3r\), \(OC = r\), and \(VC = \sqrt{(3r)^2 + r^2} = r\sqrt{10}\).
Let \(G\) be the centre of mass on \(VO\) with \(OG = 0.3r\).
Position of \(G\) relative to \(C\):
Distance along the generator from \(C\) towards \(V\):
\(x_G = r \sin(\angle OVC) + (3r - 0.3r)\cos(\angle OVC) = r\left(\frac{1}{\sqrt{10}}\right) + 2.7r\left(\frac{3}{\sqrt{10}}\right) = \frac{1 + 8.1}{\sqrt{10}}r = \frac{9.1}{\sqrt{10}}r\) from \(V\), so from \(C\):
\(x_G = VC - \frac{9.1}{\sqrt{10}}r = \frac{10 - 9.1}{\sqrt{10}}r = \frac{0.9}{\sqrt{10}}r\).
Perpendicular distance of \(G\) from the generator \(VC\):
\(y_G = 2.7r \sin(\angle OVC) = 2.7r\left(\frac{1}{\sqrt{10}}\right) = \frac{2.7}{\sqrt{10}}r\).
Wait, checking coordinates with \(C\) as origin: \(x_G = \frac{1.9}{\sqrt{10}}r\) when resolved directly: \(r\cos(\angle VCO) - 0.3r\sin(\angle VCO) = r\left(\frac{1}{\sqrt{10}}\right) + 0.3r\left(\frac{3}{\sqrt{10}}\right) = \frac{1.9}{\sqrt{10}}r\).
At the point of toppling over \(C\), the vertical line through \(G\) passes through \(C\):
\(\tan \alpha = \frac{x_G}{y_G} = \frac{\frac{1.9}{\sqrt{10}}r}{\frac{2.7}{\sqrt{10}}r} = \frac{19}{27}\).
Therefore, \(\alpha = \arctan\left(\frac{19}{27}\right) \approx 35.1^\circ\).

(a)
- M1: Attempts volumes/masses of hemisphere and cone with correct ratio \(2:3\).
- B1: States correct centres of mass for individual parts from the base (\(\frac{3}{8}r\) and \(\frac{3}{4}r\)).
- M1: Takes moments about the base: \(5\bar{y} = 3\left(\frac{3}{4}r\right) - 2\left(\frac{3}{8}r\right)\).
- A1: Obtains \(\bar{y} = \frac{3}{10}r\) and explicitly states it lies inside the cone.

(b)
- M1: Uses right-angled triangle \(AOG\) to find \(\tan \theta = \frac{OA}{OG}\) or \(\tan \phi = \frac{OG}{OA}\).
- M1: Correctly identifies the required angle as \(90^\circ - \theta\) or \(\arctan(0.3)\).
- A1: Obtains \(16.7^\circ\) (or \(0.291\text{ rad}\)).

(c)
- M1: Identifies the condition for toppling over the rim \(C\) and calculates perpendicular and parallel distances of \(G\) from \(VC\).
- M1: Equates \(\tan \alpha\) to the ratio of distances \(\frac{x_G}{y_G}\).
- A1: Obtains \(\alpha = \arctan\left(\frac{19}{27}\right) \approx 35.1^\circ\).

(a) At the equilibrium position \(E\), the tension in the string equals the weight of the particle:
\(T = mg\)
Using Hooke's law, \(T = \frac{\lambda x}{l}\):
\(\frac{6mg x}{l} = mg \implies x = \frac{1}{6}l\).
Therefore, \(AE = l + x = l + \frac{1}{6}l = \frac{7}{6}l\).

(b)(i) At the release point \(C\), \(AC = 2l\), so the extension is \(e_1 = 2l - l = l\).
Elastic potential energy at \(C\):
\(\text{EPE}_C = \frac{\lambda e_1^2}{2l} = \frac{6mg(l)^2}{2l} = 3mgl\).
Taking \(C\) as the datum level for gravitational potential energy (\(\text{GPE}_C = 0\)), the total initial mechanical energy is \(E = 3mgl\).

The string becomes slack when the particle is at distance \(l\) below \(A\), which is a height of \(2l - l = l\) above \(C\).
At this slack point, \(\text{EPE} = 0\) and \(\text{GPE} = mgl\).
Let \(v_s\) be the speed of \(P\) at the slack point:
\(\frac{1}{2}mv_s^2 + mgl = 3mgl \implies \frac{1}{2}mv_s^2 = 2mgl \implies v_s^2 = 4gl\).

Above the slack point, the particle moves freely under gravity.
Let \(H\) be the additional height gained above the slack point before coming to instantaneous rest:
\(0 = v_s^2 - 2gH \implies H = \frac{v_s^2}{2g} = \frac{4gl}{2g} = 2l\).

Since the slack point is at distance \(l\) below \(A\), the height reached above \(A\) is:
\(H - l = 2l - l = l\).
Thus, \(P\) first comes to instantaneous rest at a height \(l\) above \(A\).

(b)(ii) The maximum speed occurs when the acceleration is zero, which is at the equilibrium position \(E\).
At \(E\), \(AE = \frac{7}{6}l\), so the height of \(E\) above \(C\) is \(h_E = 2l - \frac{7}{6}l = \frac{5}{6}l\).
The extension at \(E\) is \(x = \frac{1}{6}l\).

Energy at \(E\):
\(\text{EPE}_E = \frac{6mg\left(\frac{1}{6}l\right)^2}{2l} = \frac{6mg \cdot \frac{1}{36}l^2}{2l} = \frac{1}{12}mgl\).
\(\text{GPE}_E = mg \cdot \frac{5}{6}l = \frac{5}{6}mgl\).

By conservation of energy between \(C\) and \(E\):
\(\frac{1}{2}mv_{\max}^2 + \text{EPE}_E + \text{GPE}_E = 3mgl\)
\(\frac{1}{2}mv_{\max}^2 + \frac{1}{12}mgl + \frac{10}{12}mgl = 3mgl\)
\(\frac{1}{2}mv_{\max}^2 + \frac{11}{12}mgl = 3mgl\)
\(\frac{1}{2}mv_{\max}^2 = \frac{25}{12}mgl\)
\(v_{\max}^2 = \frac{25}{6}gl\)
\(v_{\max} = \frac{5}{\sqrt{6}}\sqrt{gl} = \frac{5\sqrt{6}}{6}\sqrt{gl}\).

(a)
- M1: Applies equilibrium condition \(T = mg\) and uses Hooke's law \(T = \frac{\lambda x}{l}\).
- A1: Obtains \(AE = \frac{7}{6}l\).

(b)(i)
- M1: Calculates correct EPE at \(C\): \(\frac{6mg(l)^2}{2l} = 3mgl\).
- M1: Applies conservation of energy to find the speed at the slack point \(v_s^2 = 4gl\) (or considers direct energy conservation up to the highest point).
- M1: Uses kinematics under gravity \(v^2 = u^2 - 2gH\) to find the height gained from the slack point (\(H = 2l\)).
- M1: Calculates height above \(A\) by subtracting the distance \(l\) from \(A\) to the slack point.
- A1: Fully correct proof showing the height above \(A\) is \(l\).

(b)(ii)
- B1: Identifies that maximum speed occurs at the equilibrium position \(E\).
- M1: Writes expressions for EPE and GPE at \(E\).
- M1: Sets up energy equation: \(\frac{1}{2}mv_{\max}^2 + \frac{1}{12}mgl + \frac{5}{6}mgl = 3mgl\).
- A1: Obtains \(v_{\max} = \frac{5\sqrt{6}}{6}\sqrt{gl}\) (or \(\sqrt{\frac{25}{6}gl}\)).

(a) The necessary assumptions are:
1. The seedling heights for both populations are normally distributed.
2. The two populations have equal variances ($\sigma_x^2 = \sigma_y^2$).

(b) Calculate sample means:
$$\bar{x} = \frac{216.0}{8} = 27.0$$
$$\bar{y} = \frac{295.0}{10} = 29.5$$

Calculate sum of squared deviations for each sample:
$$S_{xx} = \sum x^2 - \frac{(\sum x)^2}{n_x} = 5844.50 - \frac{216.0^2}{8} = 5844.50 - 5832.00 = 12.50$$
$$S_{yy} = \sum y^2 - \frac{(\sum y)^2}{n_y} = 8721.50 - \frac{295.0^2}{10} = 8721.50 - 8702.50 = 19.00$$

Calculate the pooled estimate of the common variance, $s_p^2$:
$$s_p^2 = \frac{S_{xx} + S_{yy}}{n_x + n_y - 2} = \frac{12.50 + 19.00}{8 + 10 - 2} = \frac{31.50}{16} = 1.96875$$

Degrees of freedom: $
u = 8 + 10 - 2 = 16$.
Critical value for $95\%$ confidence interval with 16 degrees of freedom:
$$t_{16}(0.975) = 2.120$$

Standard error of the difference between sample means:
$$\text{SE} = \sqrt{s_p^2 \left(\frac{1}{n_x} + \frac{1}{n_y}\right)} = \sqrt{1.96875 \left(\frac{1}{8} + \frac{1}{10}\right)} = \sqrt{1.96875 \times 0.225} = \sqrt{0.44296875} \approx 0.66556$$

Confidence interval:
$$(\bar{y} - \bar{x}) \pm t_{16} \times \text{SE} = (29.5 - 27.0) \pm 2.120 \times 0.66556$$
$$= 2.5 \pm 1.411 = [1.089, 3.911]$$
To 3 significant figures, the $95\%$ confidence interval is $[1.09, 3.91]$.

(c) Since the entire confidence interval is strictly positive (greater than 0), there is evidence at the $5\%$ significance level that the new organic fertilizer results in a greater mean seedling height than the standard fertilizer.

(a)
B1: States that both populations are normally distributed.
B1: States that both populations have equal variances.

(b)
B1: Correct sample means $\bar{x} = 27.0$ and $\bar{y} = 29.5$.
M1: Correct method for pooled variance $s_p^2 = \frac{12.5 + 19.0}{16}$.
A1: $s_p^2 = 1.96875$ (or $s_p \approx 1.403$).
B1: Correct critical $t$-value $t = 2.120$.
M1: Evaluates $(\bar{y} - \bar{x}) \pm t \sqrt{s_p^2(1/8 + 1/10)}$.
A1: $[1.09, 3.91]$ (accept $[1.089, 3.911]$).

(c)
B1: Concludes that the organic fertilizer is effective / increases mean height because the interval contains only positive values (does not contain 0).

(a) A non-parametric test is preferred because the sample sizes are small and the underlying populations may not be normally distributed (or no assumption of normality is required).

(b) Let $m_A$ and $m_B$ denote the population median battery lives of Model A and Model B respectively.
Hypotheses:
$$H_0: m_A = m_B$$
$$H_1: m_B > m_A$$

Combine the two samples ($N = 6 + 7 = 13$) and arrange in ascending order to assign ranks:
$$\begin{array}{|c|c|c|}\hline \text{Battery Life (hours)} & \text{Model} & \text{Rank} \\hline 13.5 & \text{A} & 1 \\ 14.2 & \text{A} & 2 \\ 15.1 & \text{A} & 3 \\ 15.6 & \text{B} & 4 \\ 15.9 & \text{A} & 5 \\ 16.2 & \text{B} & 6 \\ 16.8 & \text{A} & 7 \\ 17.1 & \text{B} & 8 \\ 17.4 & \text{A} & 9 \\ 17.8 & \text{B} & 10 \\ 18.3 & \text{B} & 11 \\ 18.7 & \text{B} & 12 \\ 19.0 & \text{B} & 13 \\hline \end{array}$$

Calculate the sum of ranks for the smaller sample (Model A, $m = 6$):
$$R_A = 1 + 2 + 3 + 5 + 7 + 9 = 27$$

Calculate the sum of ranks for Model B ($n = 7$):
$$R_B = 4 + 6 + 8 + 10 + 11 + 12 + 13 = 64$$
Check: $R_A + R_B = 27 + 64 = 91 = \frac{13 \times 14}{2}$.

The test statistic for sample size $m = 6, n = 7$ is:
$$W = R_A = 27$$

From the Wilcoxon rank-sum table for $m = 6, n = 7$, the critical value for a one-tailed test at the $5\%$ significance level is $27$.

Rejection region: Reject $H_0$ if $W \le 27$.

Since $W = 27 \le 27$, we reject $H_0$ at the $5\%$ significance level.

Conclusion: There is significant evidence at the $5\%$ significance level to conclude that the population median battery life of Model B is greater than that of Model A.

(a)
B1: Mentions that the population distribution may not be normal / small sample without normality assumption.

(b)
B1: States both hypotheses correctly with indication of one-tailed direction ($H_0: m_A = m_B$, $H_1: m_B > m_A$).
M1: Combines and ranks all 13 observations in ascending order.
A1: Correct rank assignments for all data points.
M1: Calculates sum of ranks for Model A: $R_A = 1 + 2 + 3 + 5 + 7 + 9 = 27$.
A1: Obtains $W = 27$.
B1: Identifies the correct critical value of $27$ for $m=6, n=7$ at $5\%$ one-tailed level.
M1: Compares test statistic with critical value ($27 \le 27$).
A1: Correct contextual conclusion that there is evidence that Model B has a greater median battery life than Model A.

(a) The necessary assumptions are:
1. The plant heights in both populations are normally distributed.
2. The two populations have equal (common) variance.

(b) First, compute the sample means:
\[ \bar{x} = \frac{180.8}{8} = 22.60 \]
\[ \bar{y} = \frac{248.0}{10} = 24.80 \]

The unbiased estimates of variance are:
\[ s_A^2 = \frac{1}{7} \left( 4092.40 - \frac{180.8^2}{8} \right) = \frac{1}{7}(4092.40 - 4086.08) = \frac{6.32}{7} \approx 0.9029 \]
\[ s_B^2 = \frac{1}{9} \left( 6164.50 - \frac{248.0^2}{10} \right) = \frac{1}{9}(6164.50 - 6150.40) = \frac{14.10}{9} \approx 1.5667 \]

The pooled variance estimate is:
\[ s_p^2 = \frac{(n_A - 1)s_A^2 + (n_B - 1)s_B^2}{n_A + n_B - 2} = \frac{6.32 + 14.10}{8 + 10 - 2} = \frac{20.42}{16} = 1.27625 \approx 1.28 \]

(c) Hypotheses:
Let $\mu_A$ and $\mu_B$ be the population mean heights for standard and enriched compost respectively.
\[ H_0: \mu_B = \mu_A \quad \text{(or } \mu_B - \mu_A = 0\text{)} \]
\[ H_1: \mu_B > \mu_A \quad \text{(or } \mu_B - \mu_A > 0\text{)} \]

The test statistic is:
\[ t = \frac{\bar{y} - \bar{x}}{\sqrt{s_p^2 \left( \frac{1}{n_A} + \frac{1}{n_B} \right)}} = \frac{24.80 - 22.60}{\sqrt{1.27625 \left( \frac{1}{8} + \frac{1}{10} \right)}} = \frac{2.20}{\sqrt{1.27625 \times 0.225}} = \frac{2.20}{\sqrt{0.287156}} \approx 4.105 \]

Degrees of freedom: $
u = 8 + 10 - 2 = 16$.
Critical value for a one-tailed test at the $5\%$ level with 16 degrees of freedom is $t_{16, 0.05} = 1.746$.

Since $t = 4.105 > 1.746$, we reject $H_0$.

Conclusion: There is significant evidence at the $5\%$ level that the enriched compost results in a greater mean tomato plant height.

(a) B1: States that both populations must be normally distributed.
B1: States that both populations must have equal / common variance.

(b) M1: Correct formula for pooled variance used with sample values substituted.
A1: Obtains $s_p^2 = 1.28$ or $\frac{1021}{800}$ (1.27625).

(c) B1: Correctly states $H_0$ and $H_1$ with appropriate symbols/notation.
M1: Calculates test statistic $t$ using the pooled standard error.
A1: Obtains $t = 4.10$ to $4.11$ (or critical region $t > 1.746$).
A1: Correct decision to reject $H_0$ and fully contextualized conclusion.

(a) Using the property \(G_X(1) = 1\):
\[ G_X(1) = \frac{k}{3 - 2(1)^2} = \frac{k}{1} = 1 \implies k = 1. \]

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

Differentiating again using the product/quotient rule:
\[ G''_X(t) = 4(3 - 2t^2)^{-2} + 4t(-2)(3 - 2t^2)^{-3}(-4t) = \frac{4}{(3 - 2t^2)^2} + \frac{32t^2}{(3 - 2t^2)^3}. \]
Evaluating at \(t = 1\):
\[ G''_X(1) = \frac{4}{1^2} + \frac{32(1)^2}{1^3} = 4 + 32 = 36. \]

Using the variance formula \(\operatorname{Var}(X) = G''_X(1) + G'_X(1) - [G'_X(1)]^2\):
\[ \operatorname{Var}(X) = 36 + 4 - 4^2 = 40 - 16 = 24. \]

(c) Since \(Y = X_1 + X_2 + \dots + X_n\) where \(X_i\) are independent copies of \(X\):
\[ G_Y(t) = [G_X(t)]^n = (3 - 2t^2)^{-n} = 3^{-n} \left(1 - \frac{2}{3}t^2\right)^{-n}. \]

Expanding using the binomial series:
\[ \left(1 - \frac{2}{3}t^2\right)^{-n} = 1 + (-n)\left(-\frac{2}{3}t^2\right) + \dots = 1 + \frac{2n}{3}t^2 + \dots \]

Thus,
\[ G_Y(t) = 3^{-n} \left( 1 + \frac{2n}{3}t^2 + \dots \right) = 3^{-n} + \frac{2n}{3^{n+1}}t^2 + \dots \]

The probability \(\mathrm{P}(Y = 2)\) is the coefficient of \(t^2\):
\[ \mathrm{P}(Y = 2) = \frac{2n}{3^{n+1}}. \]

(a) B1: States \(k = 1\) from \(G_X(1) = 1\).

(b) M1: Differentiates \(G_X(t)\) correctly with respect to \(t\) and evaluates at \(t = 1\) to find \(\mathrm{E}(X) = 4\).
A1: Correct value \(\mathrm{E}(X) = 4\).
M1: Finds second derivative \(G''_X(t)\) and evaluates \(G''_X(1) = 36\).
A1: Applies \(\operatorname{Var}(X) = G''_X(1) + G'_X(1) - [G'_X(1)]^2\) correctly to achieve 24.

(c) M1: Applies the product property \(G_Y(t) = [G_X(t)]^n\).
M1: Uses binomial expansion on \((3 - 2t^2)^{-n}\) to identify the term in \(t^2\).
A1: Obtains \(\frac{2n}{3^{n+1}}\) (or equivalent single fraction).

(a) The total number of plants is:
\[ \sum fx = (0 \times 25) + (1 \times 40) + (2 \times 32) + (3 \times 16) + (4 \times 7) = 0 + 40 + 64 + 48 + 28 = 180. \]
The sample mean is \(\bar{x} = \frac{180}{120} = 1.5\).

(b) Fitting \(X \sim \mathrm{Po}(1.5)\):
\[ \mathrm{P}(X = 0) = \mathrm{e}^{-1.5} \approx 0.22313 \implies E_0 = 120 \times 0.22313 = 26.78 \]
\[ \mathrm{P}(X = 1) = 1.5 \mathrm{e}^{-1.5} \approx 0.33470 \implies E_1 = 120 \times 0.33470 = 40.16 \]
\[ \mathrm{P}(X = 2) = \frac{1.5^2}{2}\mathrm{e}^{-1.5} \approx 0.25102 \implies E_2 = 120 \times 0.25102 = 30.12 \]
\[ \mathrm{P}(X = 3) = \frac{1.5^3}{6}\mathrm{e}^{-1.5} \approx 0.12551 \implies E_3 = 120 \times 0.12551 = 15.06 \]
\[ \mathrm{P}(X \ge 4) = 1 - \sum_{x=0}^3 \mathrm{P}(X = x) = 1 - 0.93436 = 0.06564 \implies E_{\ge 4} = 120 \times 0.06564 = 7.88 \]

All expected frequencies are \(\ge 5\), so no cells need to be combined.

(c) Hypotheses:
\(H_0\): A Poisson distribution is a suitable model for the distribution of plants.
\(H_1\): A Poisson distribution is not a suitable model for the distribution of plants.

Calculating the test statistic \(\chi^2 = \sum \frac{(O - E)^2}{E}\):
- \(x = 0\): \(\frac{(25 - 26.776)^2}{26.776} = 0.1178\)
- \(x = 1\): \(\frac{(40 - 40.164)^2}{40.164} = 0.0007\)
- \(x = 2\): \(\frac{(32 - 30.123)^2}{30.123} = 0.1169\)
- \(x = 3\): \(\frac{(16 - 15.062)^2}{15.062} = 0.0584\)
- \(x \ge 4\): \(\frac{(7 - 7.876)^2}{7.876} = 0.0974\)

\[ \chi^2 = 0.1178 + 0.0007 + 0.1169 + 0.0584 + 0.0974 = 0.391 \quad (\text{or } 0.393 \text{ using rounded } E) \]

Degrees of freedom: \(\nu = c - 1 - p = 5 - 1 - 1 = 3\).
Critical value at 5% significance level: \(\chi^2_{3, 0.05} = 7.815\).

Since \(\chi^2_{\text{calc}} = 0.391 < 7.815\), we do not reject \(H_0\).
There is insufficient evidence to suggest that the Poisson distribution is not a suitable model (i.e. the Poisson model is adequate).

(a) B1: Correct calculation showing \(\bar{x} = \frac{180}{120} = 1.5\).

(b) M1: Calculates Poisson probabilities with \(\lambda = 1.5\).
A1: At least two correct expected frequencies to 2 d.p. (or 3 s.f.).
A1: All 5 expected frequencies correct: 26.78, 40.16, 30.12, 15.06, 7.88 (accept 26.8, 40.2, 30.1, 15.1, 7.9).
B1: Notes that each expected frequency is \(\ge 5\), so no pooling is required.

(c) B1: Correctly states \(H_0\) and \(H_1\) in context.
M1: Calculates test statistic \(\sum \frac{(O-E)^2}{E}\) (awarded for correct method with their values).
A1: Obtains \(\chi^2 \approx 0.391\) to \(0.393\) and states degrees of freedom \(\nu = 3\) with critical value \(7.815\).
A1(FT): Correct conclusion in context comparing test statistic with critical value.

(a) For \(0 \le x \le 4\):
\[ F(x) = \int_0^x \frac{3}{32}(4t - t^2) \, \mathrm{d}t = \frac{3}{32} \left[ 2t^2 - \frac{1}{3}t^3 \right]_0^x = \frac{3}{32} \left( 2x^2 - \frac{1}{3}x^3 \right) = \frac{6x^2 - x^3}{32}. \]

(b) Let \(G(y)\) be the cumulative distribution function of \(Y = \sqrt{X}\).
Since \(0 \le X \le 4\), we have \(0 \le Y \le 2\).
For \(0 \le y \le 2\):
\[ G(y) = \mathrm{P}(Y \le y) = \mathrm{P}(\sqrt{X} \le y) = \mathrm{P}(X \le y^2) = F(y^2). \]

Substituting \(x = y^2\) into \(F(x)\):
\[ G(y) = \frac{6(y^2)^2 - (y^2)^3}{32} = \frac{6y^4 - y^6}{32}. \]

Differentiating with respect to \(y\) to find \(g(y)\):
\[ g(y) = G'(y) = \frac{24y^3 - 6y^5}{32} = \frac{6y^3(4 - y^2)}{32} = \frac{3y^3(4 - y^2)}{16}. \]

Thus,
\[ g(y) = \begin{cases} \frac{3}{16}(4y^3 - y^5) & 0 \le y \le 2, \\ 0 & \text{otherwise.} \end{cases} \]

(c) We find \(\mathrm{E}(Y)\):
\[ \mathrm{E}(Y) = \int_0^2 y g(y) \, \mathrm{d}y = \int_0^2 \frac{3}{16}(4y^4 - y^6) \, \mathrm{d}y = \frac{3}{16} \left[ \frac{4y^5}{5} - \frac{y^7}{7} \right]_0^2 \]
\[ = \frac{3}{16} \left( \frac{4(32)}{5} - \frac{128}{7} \right) = \frac{3}{16} \times 128 \left( \frac{1}{5} - \frac{1}{7} \right) = 24 \left( \frac{2}{35} \right) = \frac{48}{35}. \]

(a) M1: Integrates \(f(t)\) from 0 to \(x\).
A1: Obtains \(F(x) = \frac{6x^2 - x^3}{32}\) (or equivalent).

(b) M1: Sets up \(G(y) = \mathrm{P}(\sqrt{X} \le y) = F(y^2)\).
A1: Correct expression for \(G(y) = \frac{6y^4 - y^6}{32}\).
M1: Differentiates \(G(y)\) with respect to \(y\).
A1: Fully correct PDF \(g(y) = \frac{3y^3(4 - y^2)}{16}\) (or \(\frac{3}{16}(4y^3 - y^5)\)) including the correct range \(0 \le y \le 2\) (and 0 otherwise).

(c) M1: Evaluates \(\int_0^2 y g(y) \, \mathrm{d}y\) (or \(\int_0^4 x^{1/2} f(x) \, \mathrm{d}x\)).
A1: Obtains exact value \(\frac{48}{35}\).