2024 Cambridge IAL Mathematics - Further (9231) Practice Paper with Answers

(i) Since \(\alpha + \beta + \gamma = -p\), we have:
\(-p = -2 \implies p = 2\).

We know that:
\(\alpha^2 + \beta^2 + \gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha)\)
\(14 = (-2)^2 - 2q\)
\(14 = 4 - 2q\)
\(2q = -10 \implies q = -5\).

To find \(r\), we use the identity:
\(\alpha^3 + \beta^3 + \gamma^3 - 3\alpha\beta\gamma = (\alpha+\beta+\gamma)(\alpha^2+\beta^2+\gamma^2 - (\alpha\beta+\beta\gamma+\gamma\alpha))\)
\(-20 - 3(-r) = (-2)(14 - (-5))\)
\(-20 + 3r = -2(19)\)
\(-20 + 3r = -38\)
\(3r = -18 \implies r = -6\).

(ii) Let \(S_k = \alpha^k + \beta^k + \gamma^k\). Using Newton's sums for the cubic equation \(x^3 + 2x^2 - 5x - 6 = 0\):
\(S_4 + p S_3 + q S_2 + r S_1 = 0\)
\(S_4 + 2(-20) - 5(14) - 6(-2) = 0\)
\(S_4 - 40 - 70 + 12 = 0\)
\(S_4 = 98\).

(iii) Let \(y = x^2\). The original equation is \(x^3 + 2x^2 - 5x - 6 = 0\).
Rearranging to group even and odd powers of \(x\):
\(x(x^2 - 5) = -2x^2 + 6\)
\(x(y - 5) = -2y + 6\)

Squaring both sides:
\(x^2(y-5)^2 = (-2y+6)^2\)
\(y(y^2 - 10y + 25) = 4y^2 - 24y + 36\)
\(y^3 - 10y^2 + 25y = 4y^2 - 24y + 36\)
\(y^3 - 14y^2 + 49y - 36 = 0\).

(i)
B1: State \(p = 2\).
M1: Use formula for sum of product pairs \(q\).
A1: Obtain \(q = -5\).
M1: Use identity for sum of cubes with substitution of values.
A1: Obtain \(r = -6\).

(ii)
M1: Formulate the Newton's sum relation for \(S_4\).
M1: Substitute the values of \(p, q, r, S_1, S_2, S_3\) into the relation.
A1: Obtain \(S_4 = 98\).

(iii)
M1: Express the original equation in terms of \(y = x^2\) and isolate \(x\).
M1: Square both sides to eliminate \(x\).
A1: Obtain the correct simplified cubic equation: \(y^3 - 14y^2 + 49y - 36 = 0\) (allow any variable).

(i) The vertical asymptotes occur where the denominator is zero:
\(x^2 - 4 = 0 \implies x = 2\) and \(x = -2\).
As \(x \to \pm\infty\), \(y \to \frac{1}{1} = 1\), so the horizontal asymptote is \(y = 1\).

(ii) Rewrite the equation of the curve as:
\(y(x^2 - 4) = x^2 + 6x + 9\)
\((y - 1)x^2 - 6x - (4y + 9) = 0\)

For real values of \(x\), the discriminant of this quadratic in \(x\) must be non-negative (\(\Delta \ge 0\)):
\(\Delta = (-6)^2 - 4(y - 1)(-(4y + 9)) \ge 0\)
\(36 + 4(y - 1)(4y + 9) \ge 0\)
\(36 + 4(4y^2 + 5y - 9) \ge 0\)
\(36 + 16y^2 + 20y - 36 \ge 0\)
\(16y^2 + 20y \ge 0\)
\(4y(4y + 5) \ge 0\)

This inequality is satisfied when \(y \le -\frac{5}{4}\) or \(y \ge 0\).
Hence, there are no real values of \(x\) for which \(-\frac{5}{4} < y < 0\).

(iii) Intersections with axes:
When \(x = 0\), \(y = -\frac{9}{4}\).
When \(y = 0\), \(x + 3 = 0 \implies x = -3\).

Stationary points occur at the boundary of the range of \(y\):
At \(y = 0\):
\(-x^2 - 6x - 9 = 0 \implies -(x+3)^2 = 0 \implies x = -3\). So there is a local minimum at \((-3, 0)\).

At \(y = -\frac{5}{4}\):
\((-\frac{5}{4} - 1)x^2 - 6x - (4(-\frac{5}{4}) + 9) = 0\)
\(-\frac{9}{4}x^2 - 6x - 4 = 0\)
\(9x^2 + 24x + 16 = 0 \implies (3x+4)^2 = 0 \implies x = -\frac{4}{3}\). So there is a local maximum at \((-\frac{4}{3}, -\frac{5}{4})\).

Intersection with horizontal asymptote:
\(1 = \frac{x^2 + 6x + 9}{x^2 - 4} \implies x^2 - 4 = x^2 + 6x + 9 \implies 6x = -13 \implies x = -\frac{13}{6}\).

The sketch consists of three branches:
1. For \(x > 2\): The curve lies above \(y=1\) and decreases from \(+\infty\) towards \(y=1\).
2. For \(-2 < x < 2\): The curve has a local maximum at \((-\frac{4}{3}, -\frac{5}{4})\), passes through \((0, -2.25)\), and goes to \(-\infty\) as \(x \to 2^-\), and to \(-\infty\) as \(x \to -2^+\).
3. For \(x < -2\): The curve comes from \(y = 1\) (for \(x \to -\infty\)), crosses \(y=1\) at \(x = -\frac{13}{6}\), reaches a minimum at \((-3, 0)\), and goes to \(+\infty\) as \(x \to -2^-\).

(i)
B1: Identify vertical asymptotes \(x = 2\) and \(x = -2\).
B1: Identify horizontal asymptote \(y = 1\).

(ii)
M1: Clear attempt to rearrange the equation into a quadratic in \(x\).
A1: Obtain the correct quadratic \((y-1)x^2 - 6x - (4y+9) = 0\).
M1: Apply the discriminant condition \(\Delta \ge 0\).
A1: Correctly solve the inequality to find the excluded interval \(-\frac{5}{4} < y < 0\).

(iii)
B1: Identify stationary points \((-3, 0)\) and \((-\frac{4}{3}, -\frac{5}{4})\) and axial intercepts.
B1: Sketch the right-hand branch correctly relative to the asymptotes.
B1: Sketch the middle branch showing the local maximum correctly.
B1: Sketch the left-hand branch showing the minimum on the x-axis and crossing the horizontal asymptote.

(i) Let \(u_r = \frac{4r}{4r^4 + 1}\). Using the hint, we factor the denominator:
\(4r^4 + 1 = (2r^2 - 2r + 1)(2r^2 + 2r + 1)\).

We express \(u_r\) as a difference of two terms:
Let \(V_r = \frac{1}{2r^2 - 2r + 1}\).
Then \(V_{r+1} = \frac{1}{2(r+1)^2 - 2(r+1) + 1} = \frac{1}{2(r^2 + 2r + 1) - 2r - 2 + 1} = \frac{1}{2r^2 + 2r + 1}\).

Now, compute \(V_r - V_{r+1}\):
\(V_r - V_{r+1} = \frac{1}{2r^2 - 2r + 1} - \frac{1}{2r^2 + 2r + 1} = \frac{(2r^2 + 2r + 1) - (2r^2 - 2r + 1)}{(2r^2 - 2r + 1)(2r^2 + 2r + 1)}\)
\(V_r - V_{r+1} = \frac{4r}{4r^4 + 1} = u_r\).

Now, sum from \(r = 1\) to \(n\):
\(\sum_{r=1}^{n} u_r = \sum_{r=1}^{n} (V_r - V_{r+1})\)
\(= (V_1 - V_2) + (V_2 - V_3) + \dots + (V_n - V_{n+1})\)
\(= V_1 - V_{n+1}\)

Since \(V_1 = \frac{1}{2(1)^2 - 2(1) + 1} = 1\), and \(V_{n+1} = \frac{1}{2n^2 + 2n + 1}\), we have:
\(\sum_{r=1}^{n} u_r = 1 - \frac{1}{2n^2 + 2n + 1}\)
\(= \frac{2n^2 + 2n + 1 - 1}{2n^2 + 2n + 1} = \frac{2n(n+1)}{2n^2 + 2n + 1}\).

(ii) The sum of the infinite series is:
\(\lim_{n \to \infty} \sum_{r=1}^{n} u_r = \lim_{n \to \infty} \frac{2n^2 + 2n}{2n^2 + 2n + 1} = 1\).

(iii) We require:
\(\frac{2n^2 + 2n}{2n^2 + 2n + 1} > 0.99\)
\(2n^2 + 2n > 0.99(2n^2 + 2n + 1)\)
\(2n^2 + 2n > 1.98n^2 + 1.98n + 0.99\)
\(0.02n^2 + 0.02n - 0.99 > 0\)
Multiplying by 50:
\(n^2 + n - 49.5 > 0\)

Solving the quadratic equation \(n^2 + n - 49.5 = 0\):
\(n = \frac{-1 \pm \sqrt{1 - 4(1)(-49.5)}}{2} = \frac{-1 \pm \sqrt{199}}{2}\)
Since \(\sqrt{199} \approx 14.1067\):
\(n \approx \frac{-1 + 14.1067}{2} \approx 6.55\).

Since \(n\) must be an integer, the smallest value is \(n = 7\).

(i)
M1: Recognize the factorization of the denominator.
M1: Establish a partial fraction form or difference format \(V_r - V_{r+1}\).
A1: Show clearly that \(V_r - V_{r+1} = \frac{4r}{4r^4 + 1}\).
M1: Write out the first few terms and last few terms to show telescoping cancellation.
A1: Correctly sum to obtain \(V_1 - V_{n+1}\).
M1: Substitute the values for \(V_1\) and \(V_{n+1}\).
A1: Show algebraic equivalence to the required expression \(\frac{2n(n+1)}{2n^2 + 2n + 1}\).

(ii)
B1: State that the limit as \(n \to \infty\) is 1.

(iii)
M1: Set up the inequality \(\frac{2n^2 + 2n}{2n^2 + 2n + 1} > 0.99\).
M1: Formulate and solve the corresponding quadratic inequality.
A1: Conclude \(n = 7\).

(i) For \(\mathbf{M}\) to be singular, its determinant must be 0:
\(\det(\mathbf{M}) = 2(3(2) - 1(0)) - 1(0(2) - 1(1)) + a(0(0) - 3(1))\)
\(= 2(6) - 1(-1) + a(-3)\)
\(= 12 + 1 - 3a = 13 - 3a\).

Setting \(\det(\mathbf{M}) = 0 \implies 13 - 3a = 0 \implies a = \frac{13}{3}\).

(ii) Given \(a = -1\), the determinant is \(\det(\mathbf{M}) = 13 - 3(-1) = 16\).
We find the matrix of cofactors of \(\mathbf{M} = \begin{pmatrix} 2 & 1 & -1 \\ 0 & 3 & 1 \\ 1 & 0 & 2 \end{pmatrix}\):
\(C_{11} = +(6 - 0) = 6\)
\(C_{12} = -(0 - 1) = 1\)
\(C_{13} = +(0 - 3) = -3\)

\(C_{21} = -(2 - 0) = -2\)
\(C_{22} = +(4 - (-1)) = 5\)
\(C_{23} = -(0 - 1) = 1\)

\(C_{31} = +(1 - (-3)) = 4\)
\(C_{32} = -(2 - 0) = -2\)
\(C_{33} = +(6 - 0) = 6\)

So the cofactor matrix is:
\(\mathbf{C} = \begin{pmatrix} 6 & 1 & -3 \\ -2 & 5 & 1 \\ 4 & -2 & 6 \end{pmatrix}\).

The adjugate matrix is \(\mathbf{C}^T\):
\(\mathbf{C}^T = \begin{pmatrix} 6 & -2 & 4 \\ 1 & 5 & -2 \\ -3 & 1 & 6 \end{pmatrix}\).

Thus, the inverse matrix is:
\(\mathbf{M}^{-1} = \frac{1}{16} \begin{pmatrix} 6 & -2 & 4 \\ 1 & 5 & -2 \\ -3 & 1 & 6 \end{pmatrix}\).

(iii) The system can be written as \(\mathbf{M} \mathbf{x} = \mathbf{b}\), where:
\(\mathbf{M} = \begin{pmatrix} 2 & 1 & -1 \\ 0 & 3 & 1 \\ 1 & 0 & 2 \end{pmatrix}\), \(\mathbf{x} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}\), and \(\mathbf{b} = \begin{pmatrix} 4 \\ -1 \\ 5 \end{pmatrix}\).

Then:
\(\mathbf{x} = \mathbf{M}^{-1} \mathbf{b} = \frac{1}{16} \begin{pmatrix} 6 & -2 & 4 \\ 1 & 5 & -2 \\ -3 & 1 & 6 \end{pmatrix} \begin{pmatrix} 4 \\ -1 \\ 5 \end{pmatrix}\)
\(= \frac{1}{16} \begin{pmatrix} 6(4) - 2(-1) + 4(5) \\ 1(4) + 5(-1) - 2(5) \\ -3(4) + 1(-1) + 6(5) \end{pmatrix}\)
\(= \frac{1}{16} \begin{pmatrix} 24 + 2 + 20 \\ 4 - 5 - 10 \\ -12 - 1 + 30 \end{pmatrix} = \frac{1}{16} \begin{pmatrix} 46 \\ -11 \\ 17 \end{pmatrix}\).

Thus, the solution is \(x = \frac{23}{8}\), \(y = -\frac{11}{16}\), \(z = \frac{17}{16}\).

(i)
M1: Attempt to evaluate the determinant of \(\mathbf{M}\) in terms of \(a\).
A1: Obtain \(a = 13/3\).

(ii)
M1: Find the determinant when \(a = -1\) (det = 16).
M1: Show a clear method to find cofactors (at least 4 correct cofactors).
A1: Obtain the correct cofactor matrix.
M1: Transpose the cofactor matrix and multiply by \(1/\det\).
A1: Obtain the correct inverse matrix.

(iii)
M1: Express the system of equations in matrix form \(\mathbf{M}\mathbf{x} = \mathbf{b}\).
M1: Multiply \(\mathbf{M}^{-1}\) by the column vector \(\mathbf{b}\).
A2: Obtain all three values correctly (A1 for any two correct).

(i) The curve is a cardioid symmetric about the line \(\theta = \pi/2\).
- At \(\theta = 0, \pi\), \(r = 2\).
- At \(\theta = \pi/2\), \(r = 4\).
- At \(\theta = 3\pi/2\), \(r = 0\).
The sketch is a heart-shaped curve enclosing the pole, symmetric about the vertical axis.

(ii) The area \(A\) is given by:
\(A = \frac{1}{2} \int_{0}^{\pi} r^2 \, d\theta = \frac{1}{2} \int_{0}^{\pi} 4(1 + \sin \theta)^2 \, d\theta\)
\(= 2 \int_{0}^{\pi} (1 + 2\sin \theta + \sin^2 \theta) \, d\theta\)

Using the double-angle identity \(\sin^2 \theta = \frac{1 - \cos 2\theta}{2\prime}\):
\(A = 2 \int_{0}^{\pi} \left(1 + 2\sin \theta + \frac{1}{2} - \frac{1}{2}\cos 2\theta\right) \, d\theta\)
\(= 2 \int_{0}^{\pi} \left(\frac{3}{2} + 2\sin \theta - \frac{1}{2}\cos 2\theta\right) \, d\theta\)
\(= 2 \left[ \frac{3}{2}\theta - 2\cos \theta - \frac{1}{4}\sin 2\theta \right]_{0}^{\pi}\)
\(= 2 \left[ \left(\frac{3}{2}\pi - 2(-1) - 0\right) - (0 - 2(1) - 0) \right]\)
\(= 2 \left( \frac{3}{2}\pi + 2 + 2 \right) = 3\pi + 8\).

(iii) The tangent is perpendicular to the initial line when \(\frac{dx}{d\theta} = 0\).
\(x = r \cos \theta = 2(1 + \sin \theta) \cos \theta = 2\cos \theta + 2\sin \theta \cos \theta = 2\cos \theta + \sin 2\theta\).

Differentiating with respect to \(\theta\):
\(\frac{dx}{d\theta} = -2\sin \theta + 2\cos 2\theta = 0 \implies \cos 2\theta = \sin \theta\)
\(1 - 2\sin^2 \theta = \sin \theta \implies 2\sin^2 \theta + \sin \theta - 1 = 0\)
\((2\sin \theta - 1)(\sin \theta + 1) = 0\).

Since the point is in the first quadrant, \(0 < \theta < \pi/2\). Thus:
\(\sin \theta = \frac{1}{2} \implies \theta = \frac{\pi}{6}\).

At \(\theta = \frac{\pi}{6}\):
\(r = 2(1 + \sin(\pi/6)) = 2(1 + 1/2) = 3\).

Cartesian coordinates:
\(x = r \cos \theta = 3 \cos(\pi/6) = \frac{3\sqrt{3}}{2}\).
\(y = r \sin \theta = 3 \sin(\pi/6) = \frac{3}{2}\).

So the coordinates are \(\left(\frac{3\sqrt{3}}{2}, \frac{3}{2}\right)\).

(i)
G1: Sketch cardioid shape with correct symmetry.
G1: Correct intercepts \((2, 0)\), \((4, \pi/2)\), \((2, \pi)\).
G1: Clear pass through the pole at \(\theta = 3\pi/2\).

(ii)
M1: State correct area integral formula with correct limits.
M1: Expand the integrand and use identity for \(\sin^2\theta\).
A1: Obtain the correct integrated form.
M1: Evaluate limits correctly.
A1: Obtain final area \(3\pi + 8\).

(iii)
M1: Formulate expression for \(x = r\cos\theta\) and differentiate.
M1: Set \(\frac{dx}{d\theta} = 0\) and solve for \(\theta\) in the first quadrant.
A1: Convert to Cartesian coordinates to obtain \(\left(\frac{3\sqrt{3}}{2}, \frac{3}{2}\right)\).

(i) Direction vectors of \(l_1\) and \(l_2\) are \(\mathbf{d}_1 = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}\) and \(\mathbf{d}_2 = \begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix}\). Since \(\mathbf{d}_1\) is not a scalar multiple of \(\mathbf{d}_2\), the lines are not parallel.

If the lines intersect, there must be values of \(\lambda\) and \(\mu\) such that:
1) \(1 + 2\lambda = 3 + \mu \implies 2\lambda - \mu = 2\)
2) \(2 - \lambda = -1 + \mu \implies \lambda + \mu = 3\)
3) \(-1 + 3\lambda = 2 - 2\mu \implies 3\lambda + 2\mu = 3\)

From (1) and (2), adding them gives \(3\lambda = 5 \implies \lambda = \frac{5}{3}\).
Then \(\mu = 3 - \frac{5}{3} = \frac{4}{3}\).
Substituting into (3):
\(3\left(\frac{5}{3}\right) + 2\left(\frac{4}{3}\right) = 5 + \frac{8}{3} = \frac{23}{3} \neq 3\).

Since the system has no solution, the lines do not intersect. Because they are neither parallel nor intersecting, they are skew.

(ii) The shortest distance \(d\) is given by \(d = \frac{|(\mathbf{a}_2 - \mathbf{a}_1) \cdot \mathbf{n}|}{|\mathbf{n}|}\), where \(\mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2\).
\(\mathbf{n} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} \times \begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix} = \begin{pmatrix} (-1)(-2) - (3)(1) \\ (3)(1) - (2)(-2) \\ (2)(1) - (-1)(1) \end{pmatrix} = \begin{pmatrix} -1 \\ 7 \\ 3
\end{pmatrix}\).

\(\mathbf{a}_2 - \mathbf{a}_1 = \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \\ 3 \end{pmatrix}\).

\((\mathbf{a}_2 - \mathbf{a}_1) \cdot \mathbf{n} = 2(-1) - 3(7) + 3(3) = -2 - 21 + 9 = -14\).
\(|\mathbf{n}| = \sqrt{(-1)^2 + 7^2 + 3^2} = \sqrt{1 + 49 + 9} = \sqrt{59}\).

Thus, \(d = \frac{|-14|}{\sqrt{59}} = \frac{14}{\sqrt{59}}\).

(iii) Let \(P\) be on \(l_1\) and \(Q\) on \(l_2\) such that \(\overrightarrow{PQ}\) is parallel to \(\mathbf{n}\).
\(\overrightarrow{PQ} = \mathbf{r}_2(\mu) - \mathbf{r}_1(\lambda) = \begin{pmatrix} 2 + \mu - 2\lambda \\ -3 + \mu + \lambda \\ 3 - 2\mu - 3\lambda \end{pmatrix} = k \begin{pmatrix} -1 \\ 7 \\ 3 \end{pmatrix}\).

Solving the three component equations:
1) \(2 + \mu - 2\lambda = -k \implies \mu = 2\lambda - k - 2\)
2) \(-3 + \mu + \lambda = 7k \implies 3\lambda - k - 5 = 7k \implies 3\lambda = 8k + 5\)
3) \(3 - 2\mu - 3\lambda = 3k \implies 3 - 2(2\lambda - k - 2) - 3\lambda = 3k \implies 7 - 7\lambda + 2k = 3k \implies \lambda = 1 - \frac{k}{7}\).

Equating expressions for \(\lambda\):
\(\frac{8k+5}{3} = 1 - \frac{k}{7} \implies 56k + 35 = 21 - 3k \implies 59k = -14 \implies k = -\frac{14}{59}\).

Thus, \(\lambda = 1 - \frac{-14/59}{7} = \frac{61}{59}\).

The position vector of \(P\) is:
\(\mathbf{p} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} + \frac{61}{59} \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} = \frac{1}{59} \begin{pmatrix} 181 \\ 57 \\ 124 \end{pmatrix}\).

A vector equation of the line of shortest distance is:
\(\mathbf{r} = \frac{1}{59} \begin{pmatrix} 181 \\ 57 \\ 124 \end{pmatrix} + t \begin{pmatrix} -1 \\ 7 \\ 3 \end{pmatrix}\).

(i)
B1: Explain that direction vectors are not parallel.
M1: Write down 3 simultaneous equations for intersection.
A1: Show that they are inconsistent.
A1: Conclude they are skew.

(ii)
M1: Find the cross product of the direction vectors.
A1: Correct vector \(\mathbf{n}\).
M1: Apply correct formula for shortest distance.
A1: Obtain \(\frac{14}{\sqrt{59}}\).

(iii)
M1: Set up vector equation \(\overrightarrow{PQ} = k\mathbf{n}\).
M1: Solve for parameter \(\lambda\) or \(\mu\).
A1: Correct vector equation.

(i) Let the statement be \(P(n): \frac{d^n}{dx^n} (x e^{2x}) = 2^{n-1} (2x + n) e^{2x}\).

**Base Case: \(n = 1\)**
LHS: \(\frac{d}{dx} (x e^{2x}) = e^{2x} + 2x e^{2x} = (2x + 1) e^{2x}\).
RHS: \(2^{1-1} (2x + 1) e^{2x} = (2x + 1) e^{2x}\).
Since LHS = RHS, \(P(1)\) is true.

**Inductive Step:**
Assume that \(P(k)\) is true for some positive integer \(k\):
\(\frac{d^k}{dx^k} (x e^{2x}) = 2^{k-1} (2x + k) e^{2x}\).

We want to show that \(P(k+1)\) is true:
\(\frac{d^{k+1}}{dx^{k+1}} (x e^{2x}) = \frac{d}{dx} \left[ 2^{k-1} (2x + k) e^{2x} \right]\)
\(= 2^{k-1} \frac{d}{dx} \left[ (2x + k) e^{2x} \right]\)

Applying the product rule:
\(= 2^{k-1} \left[ 2 e^{2x} + (2x + k) (2 e^{2x}) \right]\)
\(= 2^{k-1} \cdot 2 e^{2x} \left[ 1 + (2x + k) \right]\)
\(= 2^k (2x + k + 1) e^{2x}\).

This is of the same form as the original formula with \(n\) replaced by \(k+1\). Thus, if \(P(k)\) is true, then \(P(k+1)\) is also true.

Since \(P(1)\) is true and \(P(k) \implies P(k+1)\), by mathematical induction \(P(n)\) is true for all positive integers \(n\).

(ii) The Maclaurin series of \(f(x)\) up to \(x^3\) is:
\(f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!} x^2 + \frac{f'''(0)}{3!} x^3 + \dots\)

Given \(f(x) = x e^{2x} \implies f(0) = 0\).

Using the formula proved in part (i), the \(n\)-th derivative evaluated at \(x = 0\) is:
\(f^{(n)}(0) = 2^{n-1}(2(0) + n) e^0 = n 2^{n-1}\).

Thus:
- For \(n=1\): \(f'(0) = 1 \cdot 2^0 = 1\).
- For \(n=2\): \(f''(0) = 2 \cdot 2^1 = 4\).
- For \(n=3\): \(f'''(0) = 3 \cdot 2^2 = 12\).

Substitute these into the Maclaurin series:
\(f(x) = 0 + 1x + \frac{4}{2} x^2 + \frac{12}{6} x^3 + \dots\)
\(= x + 2x^2 + 2x^3 + \dots\)

(i)
B1: Verify base case \(n=1\).
M1: State inductive hypothesis clearly.
M1: Write down derivative for \(n = k+1\) as derivative of \(n=k\) term.
M1: Apply product rule correctly.
A1: Factorize terms correctly.
A1: Complete the algebra to match formula for \(n = k+1\).
B1: Proper conclusion stating mathematical induction principles.

(ii)
M1: State general Maclaurin expansion formula.
M1: Use derivative formula to find \(f'(0)\), \(f''(0)\), and \(f'''(0)\).
A1: Obtain correct values for derivatives.
A1: Formulate final series \(x + 2x^2 + 2x^3\).

Using de Moivre's theorem, we have:
\[ \cos(5\theta) + \mathrm{i}\sin(5\theta) = (\cos \theta + \mathrm{i} \sin \theta)^5 \]
Expanding the right-hand side 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 the real parts:
\[ \cos(5\theta) = \cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta \]
Using \(\sin^2\theta = 1 - \cos^2\theta\):
\[ \cos(5\theta) = \cos^5\theta - 10\cos^3\theta(1 - \cos^2\theta) + 5\cos\theta(1 - \cos^2\theta)^2 \]
\[ \cos(5\theta) = \cos^5\theta - 10\cos^3\theta + 10\cos^5\theta + 5\cos\theta(1 - 2\cos^2\theta + \cos^4\theta) \]
\[ \cos(5\theta) = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta \]
Now let \(x = \cos\theta\). The equation becomes:
\[ 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta = \frac{1}{2} \]
\[ \cos(5\theta) = \frac{1}{2} \]
This yields:
\[ 5\theta = \pm \frac{\pi}{3} + 2k\pi \quad \text{for } k \in \mathbb{Z} \]
\[ \theta = \pm \frac{\pi}{15} + \frac{2k\pi}{5} \]
For \(k = 0, 1, 2, 3, 4\), the five distinct real values of \(\cos\theta\) in the interval \([0, \pi]\) are:
\[ \cos\left(\frac{\pi}{15}\right), \, \cos\left(\frac{\pi}{3}\right), \, \cos\left(\frac{7\pi}{15}\right), \, \cos\left(\frac{11\pi}{15}\right), \, \cos\left(\frac{13\pi}{15}\right) \]

M1: Use de Moivre's theorem to express \(\cos(5\theta) + \mathrm{i}\sin(5\theta)\).
A1: Correctly expand the binomial expression and equate real parts.
M1: Substitute \(\sin^2\theta = 1 - \cos^2\theta\) and simplify.
A1: Obtain the correct given identity.
M1: Relate the polynomial equation to \(\cos(5\theta) = \frac{1}{2}\).
A1: Solve to find the general solutions for \(\theta\).
A2.4: Correctly state all five distinct roots of the polynomial in terms of cosines (1 mark for finding at least three correct, 2.4 marks for all five).

Let's find the eigenvalues by solving the characteristic equation \(\det(\mathbf{A} - \lambda \mathbf{I}) = 0\):
\[ \det \begin{pmatrix} 1-\lambda & 1 & -2 \\ -1 & 2-\lambda & 1 \\ 0 & 1 & -1-\lambda \end{pmatrix} = 0 \]
Expanding along the first column:
\[ (1-\lambda) [(2-\lambda)(-1-\lambda) - 1] - (-1) [1(-1-\lambda) - 1(-2)] = 0 \]
\[ (1-\lambda) [\lambda^2 - \lambda - 3] + [1 - \lambda] = 0 \]
\[ (1-\lambda) [\lambda^2 - \lambda - 2] = 0 \]
\[ (1-\lambda)(\lambda - 2)(\lambda + 1) = 0 \]
So, the eigenvalues are \(\lambda_1 = 1\), \(\lambda_2 = 2\), \(\lambda_3 = -1\).

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

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

For \(\lambda = -1\):
\[ \begin{pmatrix} 2 & 1 & -2 \\ -1 & 3 & 1 \\ 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \]
From the third row: \(y = 0\). From the first row: \(x = z\).
So the eigenvector is \(\mathbf{v}_3 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}\).

Thus, we can define:
\[ \mathbf{D} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -1 \end{pmatrix}, \quad \mathbf{P} = \begin{pmatrix} 3 & 1 & 1 \\ 2 & 3 & 0 \\ 1 & 1 & 1 \end{pmatrix} \]

M1: Attempt to find the characteristic equation \(\det(\mathbf{A}-\lambda\mathbf{I})=0\).
A1: Obtain the correct cubic equation \((1-\lambda)(\lambda-2)(\lambda+1)=0\).
A1: State eigenvalues \(1\), \(2\), \(-1\).
M1: Set up the homogeneous systems of equations to find eigenvectors.
A2: Find three correct eigenvectors (1 mark for any two correct, 2 marks for all three).
M1: Set up the diagonalisation matrices relation.
A1.4: State correct matrix \(\mathbf{D}\) and the corresponding non-singular matrix \(\mathbf{P}\) (order must match).

(i) Using integration by parts:
Let \( u = x^n \implies du = n x^{n-1} \, dx \).
Let \( dv = e^{2x} \, dx \implies v = \frac{1}{2} e^{2x} \).
Then:
\[ I_n = \left[ \frac{1}{2} x^n e^{2x} \right]_0^1 - \int_0^1 \frac{1}{2} n x^{n-1} e^{2x} \, dx \]
\[ I_n = \frac{1}{2} e^2 - \frac{n}{2} I_{n-1} \]
Multiply through by 2:
\[ 2I_n = e^2 - nI_{n-1} \implies 2I_n + nI_{n-1} = e^2 \]

(ii) Find \( I_0 \):
\[ I_0 = \int_0^1 e^{2x} \, dx = \left[ \frac{1}{2} e^{2x} \right]_0^1 = \frac{1}{2}(e^2 - 1) \]
Now apply the recurrence relation:
For \( n = 1 \):
\[ 2I_1 + I_0 = e^2 \implies 2I_1 = e^2 - \left(\frac{1}{2}e^2 - \frac{1}{2}\right) = \frac{1}{2}e^2 + \frac{1}{2} \implies I_1 = \frac{1}{4}e^2 + \frac{1}{4} \]
For \( n = 2 \):
\[ 2I_2 + 2I_1 = e^2 \implies 2I_2 = e^2 - 2\left(\frac{1}{4}e^2 + \frac{1}{4}\right) = \frac{1}{2}e^2 - \frac{1}{2} \implies I_2 = \frac{1}{4}e^2 - \frac{1}{4} \]
For \( n = 3 \):
\[ 2I_3 + 3I_2 = e^2 \implies 2I_3 = e^2 - 3\left(\frac{1}{4}e^2 - \frac{1}{4}\right) = \frac{1}{4}e^2 + \frac{3}{4} \implies I_3 = \frac{1}{8}e^2 + \frac{3}{8} \]

M1: Apply integration by parts with appropriate choice of functions.
A1: Correctly apply the limits \(0\) and \(1\) to the boundary term.
A1: Establish the recurrence formula \(2I_n + nI_{n-1} = e^2\).
M1: Calculate \(I_0\) correctly.
M1: Use the relation to calculate \(I_1\).
A1: Obtain correct value for \(I_1\).
M1: Use the relation to calculate \(I_2\).
A1: Obtain correct value for \(I_2\).
A1.4: Use the relation to obtain the exact value of \(I_3\) as \(\frac{1}{8}e^2 + \frac{3}{8}\).

First, solve the homogeneous equation:
\[ \frac{d^2y}{dx^2} - 4\frac{dy}{dx} + 4y = 0 \]
The auxiliary equation is:
\[ m^2 - 4m + 4 = 0 \implies (m-2)^2 = 0 \]
The roots are repeated, \( m = 2 \).
The complementary function is:
\[ y_c = (A + Bx)e^{2x} \]

Next, find the particular integral. Assume:
\[ y_p = C\sin x + D\cos x \]
Then:
\[ y_p' = C\cos x - D\sin x \]
\[ y_p'' = -C\sin x - D\cos x \]
Substitute into the original differential equation:
\[ (-C\sin x - D\cos x) - 4(C\cos x - D\sin x) + 4(C\sin x + D\cos x) = 9\sin x + 13\cos x \]
Grouping coefficients:
\[ (-C + 4D + 4C)\sin x + (-D - 4C + 4D)\cos x = 9\sin x + 13\cos x \]
\[ (3C + 4D)\sin x + (3D - 4C)\cos x = 9\sin x + 13\cos x \]
Equating coefficients:
1) \( 3C + 4D = 9 \)
2) \( -4C + 3D = 13 \)

Solving this system:
Multiply (1) by 3: \( 9C + 12D = 27 \)
Multiply (2) by 4: \( -16C + 12D = 52 \)
Subtracting the second equation from the first:
\[ 25C = -25 \implies C = -1 \]
Substitute \( C = -1 \) into (1):
\[ -3 + 4D = 9 \implies 4D = 12 \implies D = 3 \]
Thus, the particular integral is:
\[ y_p = -\sin x + 3\cos x \]
The general solution is the sum of the complementary function and the particular integral:
\[ y = (A + Bx)e^{2x} - \sin x + 3\cos x \]

M1: Write down the correct auxiliary equation and solve it.
A1: State the correct complementary function \(y_c = (A+Bx)e^{2x}\).
M1: Assume a particular integral of the form \(y_p = C\sin x + D\cos x\).
M1: Differentiate the assumed PI twice and substitute into the differential equation.
A1: Set up the correct simultaneous equations for \(C\) and \(D\).
M1: Solve the simultaneous equations to find \(C\) and \(D\).
A1: Obtain \(C = -1\) and \(D = 3\).
A1.4: Combine \(y_c\) and \(y_p\) to state the general solution.

(i) Using exponential definitions:
\[ \cosh x = \frac{e^x + e^{-x}}{2}, \quad \sinh x = \frac{e^x - e^{-x}}{2} \]
Then:
\[ \cosh^2 x - \sinh^2 x = \left(\frac{e^x + e^{-x}}{2}\right)^2 - \left(\frac{e^x - e^{-x}}{2}\right)^2 \]
\[ = \frac{e^{2x} + 2 + e^{-2x}}{4} - \frac{e^{2x} - 2 + e^{-2x}}{4} = \frac{4}{4} = 1 \]
Also:
\[ 2\cosh^2 x - 1 = 2\left(\frac{e^{2x} + 2 + e^{-2x}}{4}\right) - 1 = \frac{e^{2x} + 2 + e^{-2x}}{2} - 1 = \frac{e^{2x} + e^{-2x}}{2} = \cosh(2x) \]

(ii) Substitute \(\cosh(2x) = 2\cosh^2 x - 1\) into \(\cosh(2x) - 5\cosh x + 3 = 0\):
\[ (2\cosh^2 x - 1) - 5\cosh x + 3 = 0 \]
\[ 2\cosh^2 x - 5\cosh x + 2 = 0 \]
Let \(u = \cosh x\):
\[ 2u^2 - 5u + 2 = 0 \implies (2u-1)(u-2) = 0 \implies u = \frac{1}{2} \text{ or } u = 2 \]
Since \(\cosh x \ge 1\) for all real \(x\), we discard \(u = \frac{1}{2}\).
Thus, \(\cosh x = 2\).
Using the logarithmic formula for \(\cosh^{-1} z\):
\[ x = \pm \cosh^{-1}(2) = \pm \ln(2 + \sqrt{2^2 - 1}) = \pm \ln(2 + \sqrt{3}) \]

M1: Substitute exponential definitions of hyperbolic functions.
A1: Prove \(\cosh^2 x - \sinh^2 x = 1\) algebraically.
A1: Prove \(\cosh(2x) = 2\cosh^2 x - 1\) algebraically.
M1: Form a quadratic equation in \(\cosh x\).
A1: Correctly solve the quadratic to get \(\cosh x = 2\) or \(\frac{1}{2}\).
M1: Reject \(\cosh x = \frac{1}{2}\) stating a correct reason (\(\cosh x \ge 1\)).
A2.4: Find both values \(x = \pm\ln(2+\sqrt{3})\) in logarithmic form.

(i) The curve is a dimpled limacon which is symmetrical about the initial line.
Key points are:
At \(\theta = 0\), \(r = 3a\).
At \(\theta = \pi/2\), \(r = 2a\).
At \(\theta = \pi\), \(r = a\).
At \(\theta = 3\pi/2\), \(r = 2a\).

(ii) The area \(A\) is given by:
\[ A = \frac{1}{2} \int_0^{2\pi} r^2 \, d\theta = \frac{1}{2} a^2 \int_0^{2\pi} (2 + \cos \theta)^2 \, d\theta \]
\[ A = \frac{1}{2} a^2 \int_0^{2\pi} (4 + 4\cos\theta + \cos^2\theta) \, d\theta \]
Use \(\cos^2\theta = \frac{1 + \cos(2\theta)}{2}\):
\[ A = \frac{1}{2} a^2 \int_0^{2\pi} \left(4 + 4\cos\theta + \frac{1}{2} + \frac{1}{2}\cos(2\theta)\right) \, d\theta \]
\[ A = \frac{1}{2} a^2 \int_0^{2\pi} \left(\frac{9}{2} + 4\cos\theta + \frac{1}{2}\cos(2\theta)\right) \, d\theta \]
Integrating:
\[ A = \frac{1}{2} a^2 \left[ \frac{9}{2}\theta + 4\sin\theta + \frac{1}{4}\sin(2\theta) \right]_0^{2\pi} \]
Substitute the limits:
\[ A = \frac{1}{2} a^2 \left( \left(\frac{9}{2}(2\pi) + 0 + 0\right) - 0 \right) = \frac{9}{2}\pi a^2 \]

B2: Correct sketch of a closed polar curve showing correct intercepts \(3a, 2a, a\).
M1: State the correct integral formula for polar area.
A1: Correctly expand the integrand to \(4 + 4\cos\theta + \cos^2\theta\).
M1: Apply the double-angle trigonometric identity to simplify \(\cos^2\theta\).
A1: Perform correct term-by-term integration.
A2.4: Apply the limits of integration and simplify to get \(\frac{9}{2}\pi a^2\).

(i) Let \(w = -8 - 8\sqrt{3}\mathrm{i}\).
Find the modulus and argument of \(w\):
\[ |w| = \sqrt{(-8)^2 + (-8\sqrt{3})^2} = \sqrt{64 + 192} = \sqrt{256} = 16 \]
Since \(w\) lies in the third quadrant:
\[ \arg(w) = -\pi + \tan^{-1}\left(\frac{-8\sqrt{3}}{-8}\right) = -\pi + \frac{\pi}{3} = -\frac{2\pi}{3} \]
Thus, \(w = 16 e^{-\frac{2\pi}{3}\mathrm{i}}\).
Using \(z^4 = w\), we let \(z = R e^{\mathrm{i}\phi}\):
\[ R = 16^{1/4} = 2 \]
\[ 4\phi = -\frac{2\pi}{3} + 2k\pi \implies \phi = -\frac{\pi}{6} + \frac{k\pi}{2} \quad \text{for } k=0, 1, 2, 3 \]
The four roots in the range \((-\pi, \pi]\) are:
- \(k=0 \implies z_0 = 2 e^{-\frac{\pi}{6}\mathrm{i}}\)
- \(k=1 \implies z_1 = 2 e^{\frac{\pi}{3}\mathrm{i}}\)
- \(k=2 \implies z_2 = 2 e^{\frac{5\pi}{6}\mathrm{i}}\)
- \(k=3 \implies z_3 = 2 e^{\mathrm{i}\left(-\frac{\pi}{6} + \frac{3\pi}{2}\right)} = 2 e^{\frac{4\pi}{3}\mathrm{i}} \equiv 2 e^{-\frac{2\pi}{3}\mathrm{i}}

(ii) These four roots represent vertices of a square inscribed in a circle of radius 2.
The vertices are separated by angles of \)\frac{\pi}{2}\).
The area of this square is:
\[ \text{Area} = 4 \times \left(\frac{1}{2} \times 2 \times 2 \sin\left(\frac{\pi}{2}\right)\right) = 8 \]

M1: Find modulus and argument of the complex number \(w\).
A1: Express \(w\) correctly as \(16 e^{-\frac{2\pi}{3}\mathrm{i}}\).
M1: Use de Moivre's theorem to find \(R\) and the general form of \(\phi\).
A2: Find all four roots in correct exponential form (1 mark for any two correct, 2 marks for all four).
B1: Sketch the roots as a square on a circle of radius 2 on an Argand diagram.
M1: Attempt to compute the area of the square.
A1.4: Obtain the correct area of \(8\).

Divide both sides of the differential equation by \(x\) to obtain the standard linear form:
\[ \frac{dy}{dx} + \left(2 + \frac{1}{x}\right)y = e^{-2x} \]
Identify \(P(x) = 2 + \frac{1}{x}\).
The integrating factor \(\mu(x)\) is:
\[ \mu(x) = e^{\int \left(2 + \frac{1}{x}\right) dx} = e^{2x + \ln x} = x e^{2x} \]
Multiplying the standard equation by the integrating factor \(x e^{2x}\):
\[ x e^{2x} \frac{dy}{dx} + (2x + 1)e^{2x} y = x \]
Which simplifies to:
\[ \frac{d}{dx} \left( y \cdot x e^{2x} \right) = x \]
Integrate both sides with respect to \(x\):
\[ y \cdot x e^{2x} = \int x \, dx \]
\[ y \cdot x e^{2x} = \frac{1}{2} x^2 + C \]
Use the boundary condition \(y = 3\) when \(x = 1\):
\[ 3 \cdot (1) e^{2} = \frac{1}{2} (1)^2 + C \implies C = 3e^2 - \frac{1}{2} \]
Thus:
\[ y \cdot x e^{2x} = \frac{1}{2} x^2 + 3e^2 - \frac{1}{2} \]
Solving for \(y\):
\[ y = \frac{\frac{1}{2}x^2 + 3e^2 - \frac{1}{2}}{x e^{2x}} = \left( \frac{1}{2}x + \frac{3e^2 - \frac{1}{2}}{x} \right) e^{-2x} \]

M1: Rewrite the equation in the standard form \(\frac{dy}{dx} + P(x)y = Q(x)\).
M1: State the correct integral expression for the integrating factor.
A1: Correctly calculate the integrating factor \(x e^{2x}\).
M1: Perform the integration of the right-hand side of the differential equation.
A1: State the general equation \(y \cdot x e^{2x} = \frac{1}{2} x^2 + C\).
M1: Substitute \(x=1\), \(y=3\) to find the constant \(C\).
A1: Obtain the correct value of \(C = 3e^2 - \frac{1}{2}\).
A1.4: Solve for \(y\) and express the final answer in the correct form.

Using the equation of trajectory:

$$y = x \tan\theta - \frac{gx^2}{2U^2}(1 + \tan^2\theta)$$

Substitute the given values $x = 16$, $y = 4.8$, $U = 20$, and $g = 10$:

$$4.8 = 16 \tan\theta - \frac{10 \times 16^2}{2 \times 20^2}(1 + \tan^2\theta)$$

$$4.8 = 16 \tan\theta - 3.2(1 + \tan^2\theta)$$

Divide the entire equation by $1.6$:

$$3 = 10 \tan\theta - 2(1 + \tan^2\theta)$$

$$3 = 10 \tan\theta - 2 - 2 \tan^2\theta$$

$$2 \tan^2\theta - 10 \tan\theta + 5 = 0$$

Using the quadratic formula to solve for $\tan\theta$:

$$\tan\theta = \frac{10 \pm \sqrt{(-10)^2 - 4(2)(5)}}{2(2)} = \frac{10 \pm \sqrt{60}}{4} = \frac{5 \pm \sqrt{15}}{2}$$

This gives two values:
$$\tan\theta_1 \approx 4.43649 \implies \theta_1 \approx 77.3^{\circ}$$
$$\tan\theta_2 \approx 0.56351 \implies \theta_2 \approx 29.4^{\circ}$$

Thus, the two possible projection angles are $\theta = 29.4^{\circ}$ and $\theta = 77.3^{\circ}$.

M1: For using the equation of trajectory with given values.
A1: For obtaining a correct simplified quadratic in $\tan\theta$, such as $2\tan^2\theta - 10\tan\theta + 5 = 0$.
M1: For attempting to solve the quadratic equation.
A1: For one correct angle (e.g., $29.4^{\circ}$).
A1: For the second correct angle (e.g., $77.3^{\circ}$).

Let the maximum extension of the string from its natural length be $x$.

The initial state of the particle is at distance $2l$ from $A$, so the initial extension is $x_0 = 2l - l = l$.

The initial elastic potential energy (EPE) is:
$$\text{EPE}_i = \frac{\lambda x_0^2}{2l} = \frac{\lambda l^2}{2l} = \frac{1}{2}\lambda l$$

The initial kinetic energy (KE) of the particle is:
$$\text{KE}_i = \frac{1}{2} m u^2$$

At the maximum distance from $A$, the velocity of the particle is zero, so the final kinetic energy is $\text{KE}_f = 0$.

The final EPE at maximum extension $x$ is:
$$\text{EPE}_f = \frac{\lambda x^2}{2l}$$

By conservation of energy:
$$\text{EPE}_f + \text{KE}_f = \text{EPE}_i + \text{KE}_i$$

$$\frac{\lambda x^2}{2l} = \frac{1}{2}\lambda l + \frac{1}{2} m u^2$$

Multiply by $\frac{2l}{\lambda}$:
$$x^2 = l^2 + \frac{m l u^2}{\lambda}$$

$$x = \sqrt{l^2 + \frac{m l u^2}{\lambda}}$$

The maximum distance of the particle from $A$ is the natural length plus the maximum extension:
$$\text{Maximum distance} = l + x = l + \sqrt{l^2 + \frac{m l u^2}{\lambda}}$$

M1: For writing down the correct expression for initial EPE.
M1: For stating the conservation of energy equation with initial KE and EPE and final EPE.
A1: For obtaining a correct energy equation, e.g., $\frac{\lambda x^2}{2l} = \frac{1}{2}\lambda l + \frac{1}{2} m u^2$.
M1: For solving for the extension $x$ in terms of the given parameters.
A1: For the correct expression for the maximum extension $x = \sqrt{l^2 + \frac{mlu^2}{\lambda}}$.
A1: For obtaining the final maximum distance by adding $l$ to $x$.

Let $v$ be the speed of $P$ and $\theta$ be the angle that $OP$ makes with the upward vertical when the string becomes slack.

Applying conservation of energy between the lowest point and the point where the string becomes slack:
$$\frac{1}{2} m u^2 = \frac{1}{2} m v^2 + mg(a + a\cos\theta)$$

Substitute $u = 4.5$, $a = 0.6$, and $g = 10$:
$$v^2 = 4.5^2 - 2(10)(0.6)(1 + \cos\theta)$$
$$v^2 = 20.25 - 12(1 + \cos\theta)$$
$$v^2 = 8.25 - 12\cos\theta \quad \text{--- (Equation 1)}$$

Using Newton's second law along the radial direction towards the center $O$:
$$T + mg\cos\theta = \frac{mv^2}{a}$$

Since the string is slack, $T = 0$:
$$mg\cos\theta = \frac{mv^2}{a} \implies v^2 = ga\cos\theta$$

Substitute $g = 10$ and $a = 0.6$:
$$v^2 = 6\cos\theta \quad \text{--- (Equation 2)}$$

Equating (1) and (2):
$$6\cos\theta = 8.25 - 12\cos\theta$$
$$18\cos\theta = 8.25 \implies \cos\theta = \frac{8.25}{18} = \frac{11}{24}$$

Substitute $\cos\theta = \frac{11}{24}$ back into Equation 2:
$$v^2 = 6 \left(\frac{11}{24}\right) = 2.75$$
$$v = \sqrt{2.75} \approx 1.66 \text{ m s}^{-1} \text{ (to 3 s.f.)}$$

M1: For using conservation of energy to relate the initial speed and the speed at angle $\theta$.
A1: For obtaining a correct expression for $v^2$ in terms of $\cos\theta$.
M1: For setting up the radial equation of motion and applying the condition for slackness ($T = 0$).
A1: For obtaining $v^2 = ga\cos\theta$ (or $v^2 = 6\cos\theta$).
M1: For solving the simultaneous equations to find $\cos\theta$.
A1: For finding $\cos\theta = \frac{11}{24}$ (or $0.458$).
A1: For calculating the final speed $v \approx 1.66 \text{ m s}^{-1}$.

(i) The resistive force per unit mass is $\frac{4}{2x+3}$, so the equation of motion is:
$$a = v \frac{\mathrm{d}v}{\mathrm{d}x} = -\frac{4}{2x+3}$$

Separating variables and integrating:
$$\int v \,\mathrm{d}v = -\int \frac{4}{2x+3} \,\mathrm{d}x$$
$$\frac{1}{2}v^2 = -2\ln(2x+3) + C$$

Using the boundary condition $v=4$ when $x=0$:
$$\frac{1}{2}(4)^2 = -2\ln(3) + C \implies 8 = -2\ln 3 + C \implies C = 8 + 2\ln 3$$

Thus:
$$\frac{1}{2}v^2 = -2\ln(2x+3) + 8 + 2\ln 3 = 8 - 2\ln\left(\frac{2x+3}{3}\right)$$
$$v^2 = 16 - 4\ln\left(\frac{2x+3}{3}\right)$$

(ii) The particle comes to rest when $v = 0$:
$$16 - 4\ln\left(\frac{2x+3}{3}\right) = 0$$
$$\ln\left(\frac{2x+3}{3}\right) = 4$$
$$\frac{2x+3}{3} = e^4$$
$$2x+3 = 3e^4 \implies x = 1.5(e^4 - 1) \approx 80.4\text{ m}$$

M1: For using $a = v \frac{\mathrm{d}v}{\mathrm{d}x}$ and setting up the differential equation.
M1: For integrating both sides to obtain log and $v^2$ terms.
A1: For the correct integration $\frac{1}{2}v^2 = -2\ln(2x+3) + C$.
M1: For substituting $v=4$ and $x=0$ to find the constant $C$.
A1: For the correct expression $v^2 = 16 - 4\ln\left(\frac{2x+3}{3}\right)$.
M1: For setting $v^2 = 0$ and solving for $x$ in terms of $e$.
A1: For finding $x \approx 80.4$ (or $80.4\text{ m}$).

Let $A$ be the foot of the ladder on the ground and $B$ be the top against the wall.
Let $R$ be the vertical normal reaction of the ground at $A$, $F$ be the friction force at $A$, and $N$ be the horizontal normal reaction of the wall at $B$.

For vertical equilibrium:
$$R = Mg + 2Mg = 3Mg$$

At the point of slipping, the friction force reaches its limiting value:
$$F = \mu R = 0.4(3Mg) = 1.2Mg$$

For horizontal equilibrium:
$$N = F = 1.2Mg$$

Let the person be at a distance $x$ along the ladder from $A$. Taking moments about $A$:
$$N(2L \sin 60^{\circ}) = Mg(L \cos 60^{\circ}) + 2Mg(x \cos 60^{\circ})$$

Substitute the known values:
$$1.2Mg \left(2L \frac{\sqrt{3}}{2}\right) = Mg \left(L \frac{1}{2}\right) + 2Mg \left(x \frac{1}{2}\right)$$

Divide the entire equation by $Mg$:
$$1.2\sqrt{3} L = 0.5L + x$$

$$x = (1.2\sqrt{3} - 0.5) L$$

$$1.2\sqrt{3} - 0.5 \approx 1.2(1.73205) - 0.5 = 2.07846 - 0.5 = 1.57846$$

Thus, $x \approx 1.58L$.

M1: For resolution of vertical forces to obtain $R = 3Mg$.
M1: For using limiting friction $F = \mu R$ to find $F = 1.2Mg$.
A1: For horizontal equilibrium giving $N = 1.2Mg$.
M1: For taking moments about $A$ with terms for ladder weight and person's weight.
A1: For the correct moments equation, e.g., $N(2L \sin 60^{\circ}) = Mg(L \cos 60^{\circ}) + 2Mg(x \cos 60^{\circ})$.
M1: For solving the moments equation for $x$.
A1: For the correct answer $x \approx 1.58L$.

(i) Let $v_A$ and $v_B$ be the velocities of $A$ and $B$ after their collision.
By conservation of momentum:
$$m u = m v_A + 2m v_B \implies u = v_A + 2v_B \quad \text{--- (Equation 1)}$$

By Newton's law of restitution:
$$v_B - v_A = e u \quad \text{--- (Equation 2)}$$

Adding Equation 1 and Equation 2:
$$3v_B = (1+e)u \implies v_B = \frac{1+e}{3} u$$

Subtracting Equation 2 from $v_B$:
$$v_A = v_B - e u = \frac{1-2e}{3} u$$

(ii) Let $w_B$ and $w_C$ be the velocities of $B$ and $C$ after their collision.
By conservation of momentum:
$$2m v_B = 2m w_B + 3m w_C \implies 2v_B = 2w_B + 3w_C \quad \text{--- (Equation 3)}$$

By Newton's law of restitution:
$$w_C - w_B = e v_B \implies 3w_C - 3w_B = 3e v_B \quad \text{--- (Equation 4)}$$

Subtracting Equation 4 from Equation 3:
$$5w_B = (2-3e)v_B \implies w_B = \frac{2-3e}{5} v_B$$

Substitute $v_B = \frac{1+e}{3} u$:
$$w_B = \frac{(2-3e)(1+e)}{15} u$$

(iii) For there to be no further collision between $A$ and $B$, we require $v_A \le w_B$:
$$\frac{1-2e}{3} u \le \frac{(2-3e)(1+e)}{15} u$$

Since $u > 0$, we can divide by $u$ and multiply by 15:
$$5(1-2e) \le (2-3e)(1+e)$$
$$5 - 10e \le 2 - e - 3e^2$$
$$3e^2 - 9e + 3 \le 0 \implies e^2 - 3e + 1 \le 0$$

The roots of $e^2 - 3e + 1 = 0$ are:
$$e = \frac{3 \pm \sqrt{9-4}}{2} = \frac{3 \pm \sqrt{5}}{2}$$

Since $e \le 1$ and $\frac{3+\sqrt{5}}{2} \approx 2.62$, the range of $e$ for no second collision is:
$$\frac{3-\sqrt{5}}{2} \le e \le 1$$

M1: For using conservation of momentum and restitution for the first collision.
A1: For correct velocities $v_A = \frac{1-2e}{3}u$ and $v_B = \frac{1+e}{3}u$.
M1: For using conservation of momentum and restitution for the second collision.
A1: For the correct velocity $w_B = \frac{(2-3e)(1+e)}{15}u$.
M1: For setting up the inequality $v_A \le w_B$ and simplifying to a quadratic in $e$.
A1: For the quadratic inequality $e^2 - 3e + 1 \le 0$.
A1: For finding the correct range $\frac{3-\sqrt{5}}{2} \le e \le 1$.

Let $v$ be the velocity of the particle $P$ at angle $\theta$ to the upward vertical.

Using conservation of energy:
$$\frac{1}{2} m V^2 - mga = \frac{1}{2} m v^2 + mga\cos\theta$$

Multiply by $\frac{2}{m}$:
$$V^2 - 2ga = v^2 + 2ga\cos\theta$$

Substitute $V^2 = 3ga$:
$$3ga - 2ga = v^2 + 2ga\cos\theta \implies v^2 = ga(1 - 2\cos\theta) \quad \text{--- (Equation 1)}$$

The normal force $R$ acts radially inwards towards the center $O$. The radial equation of motion is:
$$R + mg\cos\theta = \frac{mv^2}{a}$$

Since $P$ loses contact with the sphere, $R = 0$:
$$mg\cos\theta = \frac{mv^2}{a} \implies v^2 = ga\cos\theta \quad \text{--- (Equation 2)}$$

Equating Equation 1 and Equation 2:
$$ga\cos\theta = ga(1 - 2\cos\theta)$$
$$\cos\theta = 1 - 2\cos\theta \implies 3\cos\theta = 1 \implies \cos\theta = \frac{1}{3}$$

The height of $P$ above the level of the center $O$ is given by $a\cos\theta$:
$$\text{Height} = a\cos\theta = \frac{1}{3}a$$

M1: For setting up the energy conservation equation with correct gravitational potential energy terms.
A1: For obtaining $v^2 = ga(1 - 2\cos\theta)$ using $V^2 = 3ga$.
M1: For setting up the radial equation of motion with $R = 0$.
A1: For obtaining $v^2 = ga\cos\theta$.
M1: For solving the two equations to find $\cos\theta$.
A1: For $\cos\theta = \frac{1}{3}$.
A1: For expressing the height above $O$ as $\frac{1}{3}a$.

Let \(\mu_1\) and \(\mu_2\) be the population mean recovery times for Group 1 and Group 2, respectively.

**Step 1: State hypotheses**
\(H_0: \mu_2 - \mu_1 = 0\) (or \(\mu_1 = \mu_2\))
\(H_1: \mu_2 - \mu_1 > 0\) (or \(\mu_2 > \mu_1\))

**Step 2: Calculate the pooled sample variance \(s_p^2\)**
\(s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1 + n_2 - 2}\)
\(s_p^2 = \frac{(8-1)(3.2) + (10-1)(4.1)}{8 + 10 - 2} = \frac{7(3.2) + 9(4.1)}{16} = \frac{22.4 + 36.9}{16} = \frac{59.3}{16} = 3.70625\)

**Step 3: Calculate the test statistic \(t\)**
\(SE = \sqrt{s_p^2 \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} = \sqrt{3.70625 \left(\frac{1}{8} + \frac{1}{10}\right)} = \sqrt{3.70625 \times 0.225} \approx 0.91318\)
\(t = \frac{\bar{x}_2 - \bar{x}_1}{SE} = \frac{14.8 - 12.5}{0.91318} = \frac{2.3}{0.91318} \approx 2.519\)

**Step 4: Find degrees of freedom and the critical value**
Degrees of freedom \(df = 8 + 10 - 2 = 16\).
From the \(t\)-distribution table, the critical value for a one-tailed test at the 5% level of significance with 16 degrees of freedom is \(t_{16}(0.05) = 1.746\).

**Step 5: Decision and Conclusion**
Since the calculated test statistic \(t = 2.52 > 1.746\), we reject \(H_0\).
There is significant evidence at the 5% level of significance to suggest that the mean recovery time for Group 2 is greater than that for Group 1.

B1: Formulates null and alternative hypotheses correctly.
M1: Standard formula for pooled variance used with correct values.
A1: Obtains \(s_p^2 = 3.71\) (or 3.70625).
M1: Standard error and t-statistic formula applied correctly.
A1: Obtains \(t = 2.52\) (or 2.519).
B1: States correct degrees of freedom (16) and critical value (1.746).
M1: Makes a correct comparison between the calculated test statistic and the critical value.
A1: Rejects \(H_0\) and provides a fully correct contextual conclusion.

**Step 1: State hypotheses**
Let \(D = \text{After} - \text{Before}\).
\(H_0\): The median difference in performance scores is zero.
\(H_1\): The median difference in performance scores is positive (training improved scores).

**Step 2: Calculate differences and omit zeros**
Employee differences (\(D\)):
\(A: +3\), \(B: -1\), \(C: +4\), \(D: +6\), \(E: 0\), \(F: +5\), \(G: -1\), \(H: +5\), \(I: +4\)
Exclude employee \(E\) as the difference is 0. The effective sample size is \(n = 8\).

**Step 3: Rank absolute differences**
Non-zero absolute differences: \(1, 1, 3, 4, 4, 5, 5, 6\).
Rank allocation:
- Absolute value 1: ranks 1 and 2 \(\implies\) mean rank \(1.5\) (for \(B, G\))
- Absolute value 3: rank 3 (for \(A\))
- Absolute value 4: ranks 4 and 5 \(\implies\) mean rank \(4.5\) (for \(C, I\))
- Absolute value 5: ranks 6 and 7 \(\implies\) mean rank \(6.5\) (for \(F, H\))
- Absolute value 6: rank 8 (for \(D\))

**Step 4: Sum positive and negative ranks**
Negative ranks: \(B (-1) \to 1.5\), \(G (-1) \to 1.5\). Sum \(W_- = 1.5 + 1.5 = 3\).
Positive ranks: \(A (+3) \to 3\), \(C (+4) \to 4.5\), \(D (+6) \to 8\), \(F (+5) \to 6.5\), \(H (+5) \to 6.5\), \(I (+4) \to 4.5\). Sum \(W_+ = 3 + 4.5 + 8 + 6.5 + 6.5 + 4.5 = 33\).

**Step 5: Determine test statistic and decision**
The test statistic is \(T = \min(W_+, W_-) = 3\).
For a one-tailed Wilcoxon signed-rank test with \(n = 8\) at the 5% level of significance, the critical value is 5.
Since \(T = 3 \le 5\), we reject \(H_0\).
Therefore, there is sufficient evidence at the 5% level of significance to suggest that the training program improved performance scores.

B1: Formulates null and alternative hypotheses clearly.
M1: Calculates the differences correctly.
B1: Excludes zero difference, identifying \(n = 8\).
M1: Correctly ranks the absolute differences.
A1: Correctly handles tied ranks (averaging them).
M1: Computes the rank sums \(W_- = 3\) and \(W_+ = 33\), and identifies the test statistic \(T = 3\).
B1: States correct critical value of 5.
A1: Compares \(3 \le 5\), rejects \(H_0\), and gives a correct contextual conclusion.

**Step 1: State hypotheses**
\(H_0\): Preference for media format and age group are independent.
\(H_1\): There is an association between preference for media format and age group.

**Step 2: Compute marginal totals**
Row Totals:
- Young: \(15 + 45 + 20 = 80\)
- Middle: \(30 + 35 + 15 = 80\)
- Senior: \(45 + 20 + 15 = 80\)

Column Totals:
- Print: \(15 + 30 + 45 = 90\)
- Digital: \(45 + 35 + 20 = 100\)
- Audio: \(20 + 15 + 15 = 50\)

Grand Total \(N = 240\).

**Step 3: Calculate expected values \(E = \frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}}\)**
Since all row totals are 80, the expected values are identical across the rows for each column:
- Print Expected: \(80 \times 90 / 240 = 30\)
- Digital Expected: \(80 \times 100 / 240 = 33.333\) (or \(\frac{100}{3}\))
- Audio Expected: \(80 \times 50 / 240 = 16.667\) (or \(\frac{50}{3}\))

**Step 4: Calculate the test statistic \(\chi^2 = \sum \frac{(O-E)^2}{E}\)**
For Print:
- Young: \(\frac{(15-30)^2}{30} = 7.5\)
- Middle: \(\frac{(30-30)^2}{30} = 0\)
- Senior: \(\frac{(45-30)^2}{30} = 7.5\)

For Digital:
- Young: \(\frac{(45-33.333)^2}{33.333} \approx 4.0833\)
- Middle: \(\frac{(35-33.333)^2}{33.333} \approx 0.0833\)
- Senior: \(\frac{(20-33.333)^2}{33.333} \approx 5.3333\)

For Audio:
- Young: \(\frac{(20-16.667)^2}{16.667} \approx 0.6667\)
- Middle: \(\frac{(15-16.667)^2}{16.667} \approx 0.1667\)
- Senior: \(\frac{(15-16.667)^2}{16.667} \approx 0.1667\)

Summing all contributions:
\(\chi^2 = 7.5 + 0 + 7.5 + 4.0833 + 0.0833 + 5.3333 + 0.6667 + 0.1667 + 0.1667 = 25.5\)

**Step 5: Compare with critical value**
Degrees of freedom \(df = (r-1)(c-1) = (3-1)(3-1) = 4\).
At the 1% significance level, the critical value \(\chi^2_4(0.01) = 13.277\).
Since \(25.5 > 13.277\), we reject \(H_0\).
There is highly significant evidence at the 1% level to conclude that there is an association between age group and preferred media format.

B1: States correct null and alternative hypotheses.
M1: Computes marginal row, column, and grand totals.
A1: Calculates all expected frequencies correctly (can be written in a table or listed).
M1: Applies the correct test statistic formula \(\sum \frac{(O-E)^2}{E}\).
A1: Obtains the correct test statistic of 25.5.
B1: Identifies 4 degrees of freedom and states the correct critical value of 13.28 (or 13.277).
A1: Compares \(\chi^2\) value with critical value, rejects \(H_0\), and writes a correct contextual conclusion.

**(i) Show that \(k = \frac{4}{27}\)**

For \(f(x)\) to be a valid probability density function:
\[ \int_0^3 k x^2 (3 - x) \, dx = 1 \]
\[ k \int_0^3 (3x^2 - x^3) \, dx = k \left[ x^3 - \frac{x^4}{4} \right]_0^3 = 1 \]
\[ k \left( 27 - \frac{81}{4} \right) = k \left( \frac{27}{4} \right) = 1 \implies k = \frac{4}{27}. \] (as required)

**(ii) Find the mean \(E(X)\)**

\[ E(X) = \int_0^3 x f(x) \, dx = \frac{4}{27} \int_0^3 (3x^3 - x^4) \, dx \]
\[ E(X) = \frac{4}{27} \left[ \frac{3x^4}{4} - \frac{x^5}{5} \right]_0^3 = \frac{4}{27} \left( \frac{243}{4} - \frac{243}{5} \right) = \frac{4}{27} \times 243 \left( \frac{1}{4} - \frac{1}{5} \right) \]
\[ E(X) = 36 \left( \frac{1}{20} \right) = 1.8. \]

**(iii) Find \(F(x)\) and show median range**

For \(0 \le x \le 3\):
\[ F(x) = \int_0^x \frac{4}{27} (3t^2 - t^3) \, dt = \frac{4}{27} \left[ t^3 - \frac{t^4}{4} \right]_0^x = \frac{4x^3 - x^4}{27}. \]

To show that the median \(m\) lies between 1.8 and 1.9, we evaluate \(F(x)\) at these points:
\[ F(1.8) = \frac{4(1.8)^3 - (1.8)^4}{27} = \frac{(1.8)^3 (4 - 1.8)}{27} = \frac{5.832 \times 2.2}{27} = \frac{12.8304}{27} \approx 0.4752. \]
Since \(F(1.8) \approx 0.4752 < 0.5\), the median is greater than 1.8.

\[ F(1.9) = \frac{4(1.9)^3 - (1.9)^4}{27} = \frac{(1.9)^3 (4 - 1.9)}{27} = \frac{6.859 \times 2.1}{27} = \frac{14.4039}{27} \approx 0.5335. \]
Since \(F(1.9) \approx 0.5335 > 0.5\), the median is less than 1.9.

Since \(F(1.8) < 0.5 < F(1.9)\), the median of \(X\) lies between 1.8 and 1.9.

(i) [2 marks]
M1: Sets up the integral equal to 1 and integrates successfully.
A1: Completes the proof to show \(k = 4/27\).

(ii) [3 marks]
M1: Sets up correct integral for \(E(X)\).
A1: Performs integration correctly.
A1: Simplifies to obtain 1.8 (or \(9/5\)).

(iii) [3.3 marks]
M1: Sets up integral to find \(F(x)\).
A1: Obtains \(F(x) = \frac{4x^3 - x^4}{27}\).
M1: Calculates both \(F(1.8)\) and \(F(1.9)\).
A1: Shows both values clearly with explanation that \(F(1.8) < 0.5 < F(1.9)\).

**(i) Find \(E(X)\) and \(\text{Var}(X)\)**

Rewrite \(G_X(t) = t(3 - 2t)^{-1}\).
Using the product rule or quotient rule to differentiate:
\[ G_X'(t) = (3 - 2t)^{-1} + t (-1)(3 - 2t)^{-2}(-2) = \frac{1}{3 - 2t} + \frac{2t}{(3 - 2t)^2}. \]
Evaluating at \(t = 1\):
\[ E(X) = G_X'(1) = \frac{1}{1} + \frac{2}{1} = 3. \]

Differentiate again to find \(G_X''(t)\):
\[ G_X''(t) = -1(3-2t)^{-2}(-2) + 2(3-2t)^{-2} + 2t(-2)(3-2t)^{-3}(-2) \]
\[ G_X''(t) = \frac{2}{(3-2t)^2} + \frac{2}{(3-2t)^2} + \frac{8t}{(3-2t)^3} = \frac{4}{(3-2t)^2} + \frac{8t}{(3-2t)^3}. \]
Evaluating at \(t = 1\):
\[ G_X''(1) = 4 + 8 = 12. \]

Calculate Variance:
\[ \text{Var}(X) = G_X''(1) + G_X'(1) - [G_X'(1)]^2 = 12 + 3 - (3)^2 = 15 - 9 = 6. \]

**(ii) Find the PGF of \(W\) and find \(P(W = 3)\)**

Since \(W = X_1 + X_2\) and they are independent:
\[ G_W(t) = [G_X(t)]^2 = \left( \frac{t}{3 - 2t} \right)^2 = \frac{t^2}{(3 - 2t)^2}. \]

To find \(P(W = 3)\), we require the coefficient of \(t^3\) in \(G_W(t)\).
Rewrite \(G_W(t)\):
\[ G_W(t) = t^2 (3 - 2t)^{-2} = t^2 \cdot 3^{-2} \left( 1 - \frac{2}{3}t \right)^{-2} = \frac{t^2}{9} \left( 1 - \frac{2}{3}t \right)^{-2}. \]
Expanding the term using binomial expansion:
\[ \left( 1 - \frac{2}{3}t \right)^{-2} = 1 + (-2)\left(-\frac{2}{3}t\right) + \frac{(-2)(-3)}{2!}\left(-\frac{2}{3}t\right)^2 + \dots = 1 + \frac{4}{3}t + \dots \]
Then:
\[ G_W(t) = \frac{t^2}{9} \left( 1 + \frac{4}{3}t + \dots \right) = \frac{1}{9}t^2 + \frac{4}{27}t^3 + \dots \]
The coefficient of \(t^3\) is \(P(W = 3) = \frac{4}{27}\).

(i) [5 marks]
M1: Differentiates \(G_X(t)\) to find \(G_X'(t)\).
A1: Evaluates at \(t = 1\) to get \(E(X) = 3\).
M1: Differentiates again to find \(G_X''(t)\).
A1: Evaluates at \(t = 1\) to get \(G_X''(1) = 12\).
A1: Correct formula for Variance applied to get 6.

(ii) [3.3 marks]
M1: Formulates \(G_W(t) = [G_X(t)]^2\).
M1: Applies binomial expansion to \((1 - 2t/3)^{-2}\).
A1: Finds the coefficient of \(t^3\) as \(4/27\) (or approx 0.148).

**Step 1: State hypotheses**
\(H_0\): The population distributions of reaction times under both conditions are identical.
\(H_1\): The population distributions of reaction times under both conditions are different.

**Step 2: Combine and rank all observations**
Arrange the combined samples of size \(n_1 = 6\) and \(n_2 = 6\) in ascending order and rank them:
- 1.2 (A) \(\to\) Rank 1
- 1.5 (A) \(\to\) Rank 2
- 1.6 (B) \(\to\) Rank 3
- 1.8 (A) \(\to\) Rank 4
- 2.0 (B) \(\to\) Rank 5
- 2.1 (A) \(\to\) Rank 6
- 2.3 (B) \(\to\) Rank 7
- 2.4 (A) \(\to\) Rank 8
- 2.5 (B) \(\to\) Rank 9
- 2.7 (A) \(\to\) Rank 10
- 2.9 (B) \(\to\) Rank 11
- 3.2 (B) \(\to\) Rank 12

**Step 3: Calculate the rank sums**
Sum of ranks for Condition A:
\(W_A = 1 + 2 + 4 + 6 + 8 + 10 = 31\)

Sum of ranks for Condition B:
\(W_B = 3 + 5 + 7 + 9 + 11 + 12 = 47\)

Check: \(W_A + W_B = 31 + 47 = 78 = \frac{12 \times 13}{2}\) (Correct).

**Step 4: Determine the test statistic**
Using the Mann-Whitney \(U\) statistic:
\(U_A = W_A - \frac{n_1(n_1+1)}{2} = 31 - \frac{6 \times 7}{2} = 31 - 21 = 10\)
\(U_B = 6 \times 6 - 10 = 26\)
So \(U = \min(10, 26) = 10\).

**Step 5: Compare with the critical value**
For a two-tailed Wilcoxon rank-sum test with \(n_1 = 6\) and \(n_2 = 6\) at the 5% significance level, the critical value for \(U\) is 5.
Since \(U = 10 > 5\), we do not reject \(H_0\).

Therefore, there is insufficient evidence at the 5% level of significance to suggest a difference in the population distributions of reaction times under the two conditions.

B1: Correctly states null and alternative hypotheses.
M1: Combines and orders the data.
A1: Correctly ranks all observations.
M1: Finds the sum of ranks for at least one group (e.g., \(W_A = 31\)).
A1: Correctly calculates \(U = 10\) (or equivalent rank-sum test statistic).
M1: Identifies the correct critical region / value (critical value \(U = 5\) or critical rank-sum range \([26, 52]\)).
B1: Makes a correct comparison.
A1: Correctly concludes that \(H_0\) is not rejected and provides the contextual conclusion.