2023 Cambridge IAL Mathematics - Further (9231) 模擬試題連答案詳解

(a)
Let \(H_n\) be the statement: \(\frac{\mathrm{d}^n}{\mathrm{d}x^n} \left( x \mathrm{e}^{2x} \right) = (2x + n) 2^{n-1} \mathrm{e}^{2x}\).

**Base case**: For \(n = 1\):
\(\text{LHS} = \frac{\mathrm{d}}{\mathrm{d}x} \left( x \mathrm{e}^{2x} \right) = \mathrm{e}^{2x} + 2x \mathrm{e}^{2x} = (2x + 1) \mathrm{e}^{2x}\).
\(\text{RHS} = (2x + 1) 2^0 \mathrm{e}^{2x} = (2x + 1) \mathrm{e}^{2x}\).
Since LHS = RHS, \(H_1\) is true.

**Inductive step**: Assume \(H_k\) is true for some positive integer \(k\):
\[ \frac{\mathrm{d}^k}{\mathrm{d}x^k} \left( x \mathrm{e}^{2x} \right) = (2x + k) 2^{k-1} \mathrm{e}^{2x} \]
We wish to show that \(H_{k+1}\) is true, i.e.,
\[ \frac{\mathrm{d}^{k+1}}{\mathrm{d}x^{k+1}} \left( x \mathrm{e}^{2x} \right) = (2x + k + 1) 2^k \mathrm{e}^{2x} \]
Differentiating the expression for \(H_k\) with respect to \(x\):
\[ \frac{\mathrm{d}^{k+1}}{\mathrm{d}x^{k+1}} \left( x \mathrm{e}^{2x} \right) = \frac{\mathrm{d}}{\mathrm{d}x} \left[ (2x + k) 2^{k-1} \mathrm{e}^{2x} \right] \]
Using the product rule:
\[ = 2^{k-1} \left[ 2 \mathrm{e}^{2x} + (2x + k) \cdot 2\mathrm{e}^{2x} \right] \]
\[ = 2^{k-1} \cdot 2 \mathrm{e}^{2x} \left[ 1 + 2x + k \right] \]
\[ = (2x + k + 1) 2^k \mathrm{e}^{2x} \]
This matches the statement \(H_{k+1}\).
Since the base case is true and \(H_k \implies H_{k+1}\), the statement \(H_n\) is true for all positive integers \(n\) by mathematical induction.

(b)
Using the formula with \(n = 10\) and \(x = 0\):
\[ \left. \frac{\mathrm{d}^{10}}{\mathrm{d}x^{10}} \left( x \mathrm{e}^{2x} \right) \right|_{x=0} = (2(0) + 10) 2^9 \mathrm{e}^0 = 10 \cdot 512 \cdot 1 = 5120 \]

M1: Verifies the formula for the base case n = 1.
A1: Correctly completes the base case.
M1: Assumes formula is true for n = k.
M1: Differentiates the assumed formula for n = k.
A1: Correct algebraic working for product rule application.
A1: Factorises the derivative correctly to match n = k+1 form.
A1: Provides a clear concluding induction statement.
M1: Substitutes n = 10 and x = 0 into the proven formula.
A1: Obtains the correct numerical answer 5120.

(a)
Let the invariant line through the origin be \(y = mx\).
Any point on this line has coordinates of the form \(\begin{pmatrix} x \\ mx \end{pmatrix}\).
Under the transformation \(\mathbf{M}\):
\[ \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix} \begin{pmatrix} x \\ mx \end{pmatrix} = \begin{pmatrix} (2 + 3m)x \\ (1 + 4m)x \end{pmatrix} \]
For the line to be invariant, the image point must also lie on the line \(y = mx\):
\[ (1 + 4m)x = m(2 + 3m)x \]
Since \(x \neq 0\) on the line:
\[ 1 + 4m = 2m + 3m^2 \implies 3m^2 - 2m - 1 = 0 \]
\[ (3m + 1)(m - 1) = 0 \]
Thus, \(m = 1\) or \(m = -\frac{1}{3}\).
The invariant lines are \(y = x\) and \(y = -\frac{1}{3}x\).

(b)
Let the coordinates of a point on the original line be \((x, y)\) and its image be \((X, Y)\).
\[ \begin{pmatrix} X \\ Y \end{pmatrix} = \mathbf{M} \begin{pmatrix} x \\ y \end{pmatrix} \implies \begin{pmatrix} x \\ y \end{pmatrix} = \mathbf{M}^{-1} \begin{pmatrix} X \\ Y \end{pmatrix} \]
The determinant of \(\mathbf{M}\) is \(2(4) - 3(1) = 5\).
The inverse matrix is:
\[ \mathbf{M}^{-1} = \frac{1}{5} \begin{pmatrix} 4 & -3 \\ -1 & 2 \end{pmatrix} \]
So:
\[ x = \frac{4X - 3Y}{5} \]
\[ y = \frac{-X + 2Y}{5} \]
Substitute these into the equation of the line \(y = 2x + 1\):
\[ \frac{-X + 2Y}{5} = 2 \left( \frac{4X - 3Y}{5} \right) + 1 \]
Multiply the entire equation by 5:
\[ -X + 2Y = 2(4X - 3Y) + 5 \]
\[ -X + 2Y = 8X - 6Y + 5 \]
\[ 8Y = 9X + 5 \implies Y = \frac{9}{8}X + \frac{5}{8} \]
Thus, the equation of the image line is \(y = \frac{9}{8}x + \frac{5}{8}\).

M1: Applies matrix multiplication to the point (x, mx)^T.
M1: Sets up the equation y' = m x' for the image coordinates.
A1: Simplifies to obtain the quadratic equation 3m^2 - 2m - 1 = 0.
M1: Solves the quadratic equation for m.
A1: Obtains both correct equations: y = x and y = -1/3 x.
M1: Recognizes the need to find the inverse matrix M^-1.
A1: Correctly calculates M^-1.
M1: Expresses x and y in terms of image coordinates X and Y.
M1: Substitutes expressions into y = 2x + 1.
A1: Simplifies to the final correct equation y = 9/8 x + 5/8 (or equivalent).

(a)
From the relation between roots and coefficients:
\[ \alpha + \beta + \gamma = 3 \]
\[ \alpha\beta + \beta\gamma + \gamma\alpha = 5 \]
\[ \alpha\beta\gamma = 2 \]
First, find \(\alpha^2 + \beta^2 + \gamma^2\):
\[ \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) = 3^2 - 2(5) = 9 - 10 = -1 \]
Since \(\alpha, \beta, \gamma\) are roots of the cubic equation, we have:
\[ \alpha^3 - 3\alpha^2 + 5\alpha - 2 = 0 \]
\[ \beta^3 - 3\beta^2 + 5\beta - 2 = 0 \]
\[ \gamma^3 - 3\gamma^2 + 5\gamma - 2 = 0 \]
Summing these three equations:
\[ (\alpha^3 + \beta^3 + \gamma^3) - 3(\alpha^2 + \beta^2 + \gamma^2) + 5(\alpha + \beta + \gamma) - 6 = 0 \]
Substitute the known values:
\[ (\alpha^3 + \beta^3 + \gamma^3) - 3(-1) + 5(3) - 6 = 0 \]
\[ (\alpha^3 + \beta^3 + \gamma^3) + 3 + 15 - 6 = 0 \]
\[ \alpha^3 + \beta^3 + \gamma^3 + 12 = 0 \implies \alpha^3 + \beta^3 + \gamma^3 = -12 \]

(b)
Let \(y = x^2\), so \(x = \sqrt{y}\). Substituting into the cubic equation:
\[ y\sqrt{y} - 3y + 5\sqrt{y} - 2 = 0 \]
\[ \sqrt{y}(y + 5) = 3y + 2 \]
Squaring both sides:
\[ y(y+5)^2 = (3y+2)^2 \]
\[ y(y^2 + 10y + 25) = 9y^2 + 12y + 4 \]
\[ y^3 + 10y^2 + 25y = 9y^2 + 12y + 4 \]
\[ y^3 + y^2 + 13y - 4 = 0 \]
Thus, a cubic equation with roots \(\alpha^2, \beta^2, \gamma^2\) is \(y^3 + y^2 + 13y - 4 = 0\).

B1: States correct values for sum of roots (3) and sum of product of roots (5).
M1: Correctly applies formula for sum of squares of roots.
A1: Finds sum of squares of roots to be -1.
M1: Sets up the sum of cubic equations to find the sum of cubes.
A1: Correctly substitutes the values into the relation.
A1: Obtains -12 with clear, rigorous working.
M1: Introduces substitution y = x^2 and rearranges to square both sides.
A1: Correctly squares the equation.
M1: Expands both sides correctly.
A1: Formulates the correct cubic equation in y.

(a)
Starting with the RHS:
\[ \frac{1}{r^2 - r + 1} - \frac{1}{r^2 + r + 1} = \frac{(r^2 + r + 1) - (r^2 - r + 1)}{(r^2 - r + 1)(r^2 + r + 1)} \]
\[ = \frac{2r}{(r^2 + 1)^2 - r^2} = \frac{2r}{r^4 + 2r^2 + 1 - r^2} = \frac{2r}{r^4 + r^2 + 1} = \text{LHS} \]

(b)
Using the method of differences:
Let \(f(r) = \frac{1}{r^2 - r + 1}\). Then \(f(r+1) = \frac{1}{(r+1)^2 - (r+1) + 1} = \frac{1}{r^2 + r + 1}\).
We have:
\[ \sum_{r=1}^n \frac{2r}{r^4 + r^2 + 1} = \sum_{r=1}^n \left( f(r) - f(r+1) \right) \]
\[ = \left( f(1) - f(2) \right) + \left( f(2) - f(3) \right) + \dots + \left( f(n) - f(n+1) \right) \]
\[ = f(1) - f(n+1) \]
Since \(f(1) = \frac{1}{1^2 - 1 + 1} = 1\), and \(f(n+1) = \frac{1}{n^2 + n + 1}\), we get:
\[ \sum_{r=1}^n \frac{2r}{r^4 + r^2 + 1} = 1 - \frac{1}{n^2 + n + 1} \]

(c)
The sum to infinity is:
\[ S_{\infty} = \lim_{n \to \infty} \left( 1 - \frac{1}{n^2 + n + 1} \right) = 1 \]
The sum from \(r=N+1\) to \(\infty\) is:
\[ \sum_{r=N+1}^{\infty} \frac{2r}{r^4 + r^2 + 1} = S_{\infty} - S_N = 1 - \left( 1 - \frac{1}{N^2 + N + 1} \right) = \frac{1}{N^2 + N + 1} \]
We require:
\[ \frac{1}{N^2 + N + 1} < 0.001 \implies N^2 + N + 1 > 1000 \implies N^2 + N - 999 > 0 \]
Using the quadratic formula to find the boundary value:
\[ N = \frac{-1 + \sqrt{1 + 4(999)}}{2} = \frac{-1 + \sqrt{3997}}{2} \approx 31.11 \]
Testing integer values:
- If \(N = 31\): \(31^2 + 31 + 1 = 993 \le 1000\)
- If \(N = 32\): \(32^2 + 32 + 1 = 1057 > 1000\)
Thus, the smallest integer is \(N = 32\).

M1: Attempts to find a common denominator for the RHS.
A1: Correctly expands denominator to r^4 + r^2 + 1.
A1: Fully completes the algebraic proof of LHS = RHS.
M1: Applies the difference method by writing out a few terms of the series.
A1: Demonstrates terms cancelling clearly.
A1: Correctly identifies remaining terms f(1) and f(n+1).
A1: Arrives at the given formula for S_n.
M1: Expresses the infinite tail sum in terms of N.
A1: Formulates the inequality N^2 + N + 1 > 1000.
M1: Solves the quadratic inequality or tests integers near the root.
A1: Concludes with the correct smallest integer N = 32.

(a)
The curve is a cardioid-like shape with a dimple. Key points on the sketch:
- On \(\theta = 0\), \(r = 5\).
- On \(\theta = \pm\pi/2\), \(r = 3\).
- On \(\theta = \pi\), \(r = 1\).
- Symmetrical about the initial line.

(b)
The area \(A\) is:
\[ A = \frac{1}{2} \int_{-\pi}^{\pi} r^2 \mathrm{d}\theta = \int_{0}^{\pi} (3 + 2\cos\theta)^2 \mathrm{d}\theta \]
\[ A = \int_{0}^{\pi} (9 + 12\cos\theta + 4\cos^2\theta) \mathrm{d}\theta \]
Using \(\cos^2\theta = \frac{1 + \cos 2\theta}{2\mu}\):
\[ A = \int_{0}^{\pi} (11 + 12\cos\theta + 2\cos 2\theta) \mathrm{d}\theta \]
\[ A = \left[ 11\theta + 12\sin\theta + \sin 2\theta \right]_0^{\pi} \]
\[ A = 11\pi \]

(c)
The tangent is perpendicular to the initial line when \(\frac{\mathrm{d}x}{\mathrm{d}\theta} = 0\):
\[ x = r\cos\theta = (3 + 2\cos\theta)\cos\theta = 3\cos\theta + 2\cos^2\theta \]
\[ \frac{\mathrm{d}x}{\mathrm{d}\theta} = -3\sin\theta - 4\cos\theta\sin\theta = -\sin\theta(3 + 4\cos\theta) = 0 \]
- \(\sin\theta = 0 \implies \theta = 0, \pi\).
Points: \((5, 0)\) and \((1, \pi)\).
- \(\cos\theta = -\frac{3}{4} \implies r = 3 + 2\left(-\frac{3}{4}\right) = \frac{3}{2}\).
Points: \(\left(\frac{3}{2}, \pm \arccos\left(-\frac{3}{4}\right)\right)\).

G1: Correct cardioid-like shape (dimpled, no loop, symmetrical).
G1: Correct intercepts labeled on the graph (at (5,0), (3, ±pi/2), (1, pi)).
M1: Uses the correct polar area integral formula.
M1: Expands the integrand correctly.
M1: Uses the double angle identity for cos^2(theta).
A1: Integrates correctly.
A1: Obtains the correct final area of 11*pi.
M1: Formulates x = r cos(theta) and differentiates with respect to theta.
A1: Correctly factors the derivative as -sin(theta)(3 + 4cos(theta)).
A1: Correctly identifies the polar coordinates of all four points.

(a)
The vertical asymptote is found where the denominator is zero:
\[ x = 3 \]
By algebraic division:
\[ y = \frac{x^2 - x - 2}{x - 3} = x + 2 + \frac{4}{x - 3} \]
As \(x \to \pm\infty\), \(\frac{4}{x-3} \to 0\).
So the oblique asymptote is:
\[ y = x + 2 \]

(b)
Differentiating \(y = x + 2 + \frac{4}{x - 3}\) with respect to \(x\):
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = 1 - \frac{4}{(x-3)^2} \]
Setting the derivative to 0:
\[ (x-3)^2 = 4 \implies x-3 = \pm 2 \]
- If \(x = 5\), then \(y = 5 + 2 + \frac{4}{2} = 9\).
- If \(x = 1\), then \(y = 1 + 2 + \frac{4}{-2} = 1\).
Thus, the stationary points are \((1, 1)\) and \((5, 9)\).

(c)
- Intercepts with axes:
- \(x\)-intercepts (\(y=0\)): \(x^2 - x - 2 = 0 \implies (x-2)(x+1) = 0 \implies x = 2, -1\). Intercepts: \((-1, 0)\), \((2, 0)\).
- \(y\)-intercept (\(x=0\)): \(y = \frac{2}{3}\). Intercept: \((0, 2/3)\).
- Sketch shows the curve with asymptotes \(x = 3\) and \(y = x + 2\), local maximum at \((1, 1)\), local minimum at \((5, 9)\), and correct intercepts.

B1: States the vertical asymptote is x = 3.
M1: Uses algebraic division or equivalent method to find oblique asymptote.
A1: Correctly identifies oblique asymptote as y = x + 2.
M1: Differentiates the curve's equation correctly.
M1: Sets derivative to 0 and solves for x.
A1: Obtains x = 1 and x = 5.
A1: Coordinates of stationary points are (1, 1) and (5, 9).
G1: Correctly plots asymptotes as dashed lines with equations.
G1: Correctly draws both branches of the curve with correct intercepts.
G1: Labels stationary points at correct locations.

(a)
Direction vectors of \(l_1\) and \(l_2\) are:
\[ \mathbf{d}_1 = \mathbf{i} + 2\mathbf{j} - \mathbf{k}, \quad \mathbf{d}_2 = 2\mathbf{i} - \mathbf{j} + 3\mathbf{k} \]
A common normal vector is given by \(\mathbf{d}_1 \times \mathbf{d}_2\):
\[ \mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -1 \\ 2 & -1 & 3 \end{vmatrix} = 5\mathbf{i} - 5\mathbf{j} - 5\mathbf{k} \]
We can use the simplified normal vector \(\mathbf{n}_0 = \mathbf{i} - \mathbf{j} - \mathbf{k}\).
Let \(A(1, -1, 2)\) be a point on \(l_1\) and \(B(2, 3, 0)\) be a point on \(l_2\).
\[ \vec{AB} = \mathbf{i} + 4\mathbf{j} - 2\mathbf{k} \]
The shortest distance \(d\) is:
\[ d = \frac{|\vec{AB} \cdot \mathbf{n}_0|}{|\mathbf{n}_0|} = \frac{|(1)(1) + (4)(-1) + (-2)(-1)|}{\sqrt{1^2 + (-1)^2 + (-1)^2}} = \frac{|1 - 4 + 2|}{\sqrt{3}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \approx 0.577 \]

(b)
The plane \(\Pi\) has normal vector \(\mathbf{n}_0 = \mathbf{i} - \mathbf{j} - \mathbf{k}\) and contains \(A(1, -1, 2)\).
\[ 1(x - 1) - 1(y - (-1)) - 1(z - 2) = 0 \implies x - y - z = 0 \]

(c)
The direction of \(l_2\) is \(\mathbf{d}_2 = 2\mathbf{i} - \mathbf{j} + 3\mathbf{k}\).
The normal to \(\Pi'\) is \(\mathbf{n}' = \mathbf{i} + \mathbf{j} + \mathbf{k}\).
The acute angle \(\theta\) satisfies:
\[ \sin\theta = \frac{|\mathbf{d}_2 \cdot \mathbf{n}'|}{|\mathbf{d}_2| |\mathbf{n}'|} = \frac{|2(1) - 1(1) + 3(1)|}{\sqrt{14}\sqrt{3}} = \frac{4}{\sqrt{42}} \approx 0.6172 \]
\[ \theta = \arcsin(0.6172) \approx 38.1^\circ \text{ (or } 0.665 \text{ rad)} \]

M1: Calculates the cross product of the two direction vectors.
A1: Obtains a correct normal vector (e.g. i - j - k).
M1: Obtains a vector between a point on l_1 and a point on l_2.
M1: Projects the vector onto the normal vector to find shortest distance.
A1: Correctly calculates the distance as 1/sqrt(3) or equivalent.
M1: Uses the normal vector and point on l_1 to set up the plane equation.
A1: Expresses the plane equation correctly as x - y - z = 0.
M1: Applies the angle formula sin(theta) = |d_2 . n'| / (|d_2||n'|).
A1: Correctly computes sin(theta) = 4/sqrt(42).
A1: Obtains correct angle 38.1 degrees or 0.665 radians.

(a) Using the exponential definitions \(\cosh x = \frac{e^x + e^{-x}}{2}\) and \(\sinh x = \frac{e^x - e^{-x}}{2}\):

\(3 \left(\frac{e^x + e^{-x}}{2}\right) - \left(\frac{e^x - e^{-x}}{2}\right) = k\)

\(\implies \frac{3}{2}e^x + \frac{3}{2}e^{-x} - \frac{1}{2}e^x + \frac{1}{2}e^{-x} = k\)

\(\implies e^x + 2e^{-x} = k\)

Multiplying by \(e^x\) gives:

\(e^{2x} - ke^x + 2 = 0\)

Let \(y = e^x\). Since \(x \in \mathbb{R}\), we must have \(y > 0\).

For the quadratic equation \(y^2 - ky + 2 = 0\) to have real roots, the discriminant must be non-negative:

\(\Delta = k^2 - 8 \ge 0 \implies k \ge 2\sqrt{2}\) or \(k \le -2\sqrt{2}\).

If \(k \le -2\sqrt{2}\), the sum of the roots is \(k < 0\). Since the product of the roots is \(2 > 0\), both roots of the quadratic must be negative. Thus, there is no positive root for \(y\), so no real solution for \(x\) exists.

If \(k \ge 2\sqrt{2}\), the sum of the roots is \(k > 0\) and the product is \(2 > 0\), so both roots are positive. This yields real values for \(x\).

Thus, a real solution for \(x\) exists if and only if \(k \ge 2\sqrt{2}\).

(b) Substituting \(k = 3\) (which satisfies \(3 \ge 2\sqrt{2}\)):

\(e^{2x} - 3e^x + 2 = 0\)

\(\implies (e^x - 1)(e^x - 2) = 0\)

Thus, \(e^x = 1 \implies x = 0\) (or \(\ln 1\)), and \(e^x = 2 \implies x = \ln 2\).

(a)
M1: Expresses \(\cosh x\) and \(\sinh x\) in terms of exponentials.
A1: Obtains the correct quadratic equation in \(e^x\), i.e., \(e^{2x} - ke^x + 2 = 0\).
M1: Applies the discriminant condition \(\Delta \ge 0\) to find the critical values \(\pm 2\sqrt{2}\).
A1: Correctly reasons why the case \(k \le -2\sqrt{2}\) must be rejected because \(e^x\) must be strictly positive.

(b)
M1: Substitutes \(k=3\) and factors the resulting quadratic equation in \(e^x\).
A1: Obtains the roots \(e^x = 1\) and \(e^x = 2\).
A1: Gives the final solutions as \(x = 0\) and \(x = \ln 2\) in logarithmic form.

(a) Since \(A\) is an upper triangular matrix, its eigenvalues are the entries on the leading diagonal, which are \(\lambda_1 = 1\), \(\lambda_2 = 2\), and \(\lambda_3 = 3\).

To find the eigenvector \(\mathbf{v}_1 = \begin{pmatrix} x \\ y \\ z \end{pmatrix}\) corresponding to \(\lambda_1 = 1\):
\((A - I)\mathbf{v}_1 = \mathbf{0} \implies \begin{pmatrix} 0 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies y=0, z=0\).
So a corresponding eigenvector is \(\mathbf{v}_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\).

To find the eigenvector \(\mathbf{v}_2 = \begin{pmatrix} x \\ y \\ z \end{pmatrix}\) corresponding to \(\lambda_2 = 2\):
\((A - 2I)\mathbf{v}_2 = \mathbf{0} \implies \begin{pmatrix} -1 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies z=0, -x+y=0 \implies x=y\).
So a corresponding eigenvector is \(\mathbf{v}_2 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}\).

To find the eigenvector \(\mathbf{v}_3 = \begin{pmatrix} x \\ y \\ z \end{pmatrix}\) corresponding to \(\lambda_3 = 3\):
\((A - 3I)\mathbf{v}_3 = \mathbf{0} \implies \begin{pmatrix} -2 & 1 & 1 \\ 0 & -1 & 1 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies -y+z=0 \implies y=z\), and \(-2x+y+z=0 \implies 2x = 2y \implies x=y\).
So a corresponding eigenvector is \(\mathbf{v}_3 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\).

(b) From part (a), we can write:
\(D = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}\) and \(P = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}\).

To find \(P^{-1}\), we use row operations on \(\begin{pmatrix} P & | & I \end{pmatrix}\):
\(P^{-1} = \begin{pmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{pmatrix}\).

Now, we compute \(A^n = P D^n P^{-1}\):
\(A^n = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1^n & 0 & 0 \\ 0 & 2^n & 0 \\ 0 & 0 & 3^n \end{pmatrix} \begin{pmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{pmatrix}\)

\(= \begin{pmatrix} 1 & 2^n & 3^n \\ 0 & 2^n & 3^n \\ 0 & 0 & 3^n \end{pmatrix} \begin{pmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{pmatrix}\)

\(= \begin{pmatrix} 1 & 2^n - 1 & 3^n - 2^n \\ 0 & 2^n & 3^n - 2^n \\ 0 & 0 & 3^n \end{pmatrix}\).

(a)
M1: Identifies the eigenvalues from the diagonal entries of the triangular matrix.
A1: Correctly finds the eigenvector for \(\lambda_1 = 1\).
A1: Correctly finds the eigenvector for \(\lambda_2 = 2\).
A1: Correctly finds the eigenvector for \(\lambda_3 = 3\).

(b)
M1: Forms matrices \(P\) and \(D\) from the eigenvectors and eigenvalues.
M1: Calculates the inverse matrix \(P^{-1}\) correctly.
M1: Sets up the matrix multiplication for \(P D^n P^{-1}\).
A1: Displays correct intermediate matrix \(P D^n\).
A1: Completes the multiplication to obtain the given expression for \(A^n\) rigorously.

(a) Given \(f(x) = \ln(1+\sinh x)\):

At \(x=0\), \(f(0) = \ln(1+0) = 0\).

First derivative:
\(f'(x) = \frac{\cosh x}{1+\sinh x}\)

At \(x=0\):
\(f'(0) = \frac{\cosh 0}{1+\sinh 0} = \frac{1}{1} = 1\).

Second derivative (using the quotient rule):
\(f''(x) = \frac{\sinh x(1+\sinh x) - \cosh^2 x}{(1+\sinh x)^2} = \frac{\sinh x + \sinh^2 x - (1+\sinh^2 x)}{(1+\sinh x)^2} = \frac{\sinh x - 1}{(1+\sinh x)^2}\)

At \(x=0\):
\(f''(0) = \frac{0-1}{(1)^2} = -1\).

Third derivative (differentiating the relation \(f''(x)(1+\sinh x)^2 = \sinh x - 1\)):
\(f'''(x)(1+\sinh x)^2 + 2f''(x)(1+\sinh x)\cosh x = \cosh x\)

At \(x=0\):
\(f'''(0)(1)^2 + 2f''(0)(1)(1) = 1\)

\(\implies f'''(0) + 2(-1) = 1 \implies f'''(0) = 3\).

(b) The Maclaurin series formula is:
\(f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots\)

Substituting the values found in part (a):
\(f(x) = 0 + 1x + \frac{-1}{2}x^2 + \frac{3}{6}x^3 + \dots = x - \frac{1}{2}x^2 + \frac{1}{2}x^3\).

(a)
M1: Applies chain rule to differentiate \(\ln(1+\sinh x)\).
A1: Finds \(f'(0) = 1\).
M1: Applies quotient rule or alternative method to find \(f''(x)\).
A1: Finds \(f''(0) = -1\).
M1: Differentiates again to set up an expression for \(f'''(x)\).
A1: Finds \(f'''(0) = 3\).

(b)
M1: Recalls and applies the Maclaurin series formula up to \(x^3\).
A1: Obtains the correct expansion \(x - \frac{1}{2}x^2 + \frac{1}{2}x^3\).

(a) We apply integration by parts to \(I_n\):

Let \(u = x^n\) and \(dv = \sin x \, dx\), which gives \(du = n x^{n-1} \, dx\) and \(v = -\cos x\).

\(I_n = \left[-x^n \cos x\right]_0^{\frac{\pi}{2}} + n \int_0^{\frac{\pi}{2}} x^{n-1} \cos x \, dx\)

Since \(\cos(\frac{\pi}{2}) = 0\) and \(0^n = 0\) for \(n \ge 2\), the boundary term vanishes:

\(I_n = n \int_0^{\frac{\pi}{2}} x^{n-1} \cos x \, dx\)

Apply integration by parts again to this new integral:

Let \(u = x^{n-1}\) and \(dv = \cos x \, dx\), which gives \(du = (n-1)x^{n-2} \, dx\) and \(v = \sin x\).

\(\int_0^{\frac{\pi}{2}} x^{n-1} \cos x \, dx = \left[x^{n-1} \sin x\right]_0^{\frac{\pi}{2}} - (n-1)\int_0^{\frac{\pi}{2}} x^{n-2} \sin x \, dx\)

\(= \left(\frac{\pi}{2}\right)^{n-1} - (n-1)I_{n-2}\)

Substituting this back into the expression for \(I_n\):

\(I_n = n \left[ \left(\frac{\pi}{2}\right)^{n-1} - (n-1)I_{n-2} \right] = n \left(\frac{\pi}{2}\right)^{n-1} - n(n-1)I_{n-2}\).

(b) We first calculate \(I_0\):

\(I_0 = \int_0^{\frac{\pi}{2}} \sin x \, dx = \left[-\cos x\right]_0^{\frac{\pi}{2}} = 1\).

Using the reduction formula with \(n = 2\):

\(I_2 = 2 \left(\frac{\pi}{2}\right)^{1} - 2(1)I_0 = \pi - 2(1) = \pi - 2\).

Now using the reduction formula with \(n = 4\):

\(I_4 = 4 \left(\frac{\pi}{2}\right)^{3} - 4(3)I_2 = \frac{\pi^3}{2} - 12(\pi - 2) = \frac{\pi^3}{2} - 12\pi + 24\).

(a)
M1: Applies integration by parts to \(I_n\) with appropriate choices for \(u\) and \(dv\).
A1: Correctly shows that the boundary terms vanish, yielding the intermediate integral.
M1: Applies integration by parts a second time on \(\int x^{n-1}\cos x \, dx\).
A1: Correctly evaluates the boundary term \(\left(\frac{\pi}{2}\right)^{n-1}\).
A1: Combines the results to obtain the given reduction formula.

(b)
M1: Evaluates the base case \(I_0 = 1\).
M1: Applies the reduction formula to find \(I_2\).
A1: Applies the reduction formula to find \(I_4\) in terms of \(I_2\).
A1: Obtains the correct exact simplified value \(\frac{\pi^3}{2} - 12\pi + 24\).

(a) By de Moivre's theorem, we have:

\(\cos 5\theta + i \sin 5\theta = (\cos \theta + i \sin \theta)^5\)

Expanding the right-hand side using the binomial theorem:

\((\cos \theta + i \sin \theta)^5 = \cos^5 \theta + 5i \cos^4 \theta \sin \theta - 10 \cos^3 \theta \sin^2 \theta - 10i \cos^2 \theta \sin^3 \theta + 5 \cos \theta \sin^4 \theta + i \sin^5 \theta\)

Equating the real parts on both sides:

\(\cos 5\theta = \cos^5 \theta - 10 \cos^3 \theta \sin^2 \theta + 5 \cos \theta \sin^4 \theta\)

Substitute \(\sin^2 \theta = 1 - \cos^2 \theta\):

\(\cos 5\theta = \cos^5 \theta - 10 \cos^3 \theta (1 - \cos^2 \theta) + 5 \cos \theta (1 - \cos^2 \theta)^2\)

\(= \cos^5 \theta - 10 \cos^3 \theta + 10 \cos^5 \theta + 5 \cos \theta (1 - 2\cos^2 \theta + \cos^4 \theta)\)

\(= 11 \cos^5 \theta - 10 \cos^3 \theta + 5 \cos \theta - 10 \cos^3 \theta + 5 \cos^5 \theta\)

\(= 16 \cos^5 \theta - 20 \cos^3 \theta + 5 \cos \theta\).

(b) Let \(\cos 5\theta = 1\).

The solutions to this equation are \(5\theta = 2k\pi \implies \theta = \frac{2k\pi}{5}\) for \(k = 0, 1, 2, 3, 4\).

This correspond to the roots \(x = \cos\left(\frac{2k\pi}{5}\right)\) of the polynomial equation:

\(16x^5 - 20x^3 + 5x - 1 = 0\)

Since \(x = \cos 0 = 1\) is a root, we can divide by \(x - 1\):

\(16x^5 - 20x^3 + 5x - 1 = (x - 1)(16x^4 + 16x^3 - 4x^2 - 4x + 1) = 0\)

Notice that the quartic term is a perfect square:

\(16x^4 + 16x^3 - 4x^2 - 4x + 1 = (4x^2 + 2x - 1)^2 = 0\)

So the other roots must satisfy the quadratic equation:

\(4x^2 + 2x - 1 = 0\)

Using the quadratic formula:

\(x = \frac{-2 \pm \sqrt{4 - 4(4)(-1)}}{8} = \frac{-2 \pm \sqrt{20}}{8} = \frac{-1 \pm \sqrt{5}}{4\vphantom{2}}\)

Since \(0 < \frac{2\pi}{5} < \frac{\pi}{2}\), the value of \(\cos\left(\frac{2\pi}{5}\right)\) must be positive.

Thus, \(\cos\left(\frac{2\pi}{5}\right) = \frac{\sqrt{5}-1}{4\vphantom{2}}\).

(a)
M1: Writes down the binomial expansion of \((\cos\theta + i\sin\theta)^5\).
A1: Equates the real parts to obtain the expression for \(\cos 5\theta\).
M1: Uses the identity \(\sin^2\theta = 1 - \cos^2\theta\) to eliminate \(\sin\theta\).
A1: Correctly expands and simplifies terms.
A1: Reaches the target formula with no errors shown.

(b)
M1: Identifies the roots of \(\cos 5\theta = 1\) as \(x = \cos(2k\pi/5)\).
M1: Divides the polynomial by \(x-1\) to get the quartic equation.
M1: Factorises the quartic as a perfect square of a quadratic, \(4x^2 + 2x - 1 = 0\).
A1: Solves the quadratic equation to get \(\frac{-1 \pm \sqrt{5}}{4}\).
A1: Explains why the positive root is chosen and completes the proof.

First, we find the complementary function (CF) by solving the auxiliary equation:

\(m^2 + 2m + 5 = 0 \implies (m+1)^2 + 4 = 0 \implies m = -1 \pm 2i\)

Thus, the CF is:

\(y_c = e^{-x}(A \cos 2x + B \sin 2x)\)

Next, we find the particular integral (PI). Since \(-1+i\) is not a root of the auxiliary equation, we try:

\(y_p = e^{-x}(C \cos x + D \sin x)\)

Let \(u = C \cos x + D \sin x\), so \(y_p = e^{-x} u\).

\(y_p' = e^{-x}(u' - u)\)

\(y_p'' = e^{-x}(u'' - 2u' + u)\)

Substituting these into the differential equation:

\(e^{-x}(u'' - 2u' + u) + 2e^{-x}(u' - u) + 5e^{-x} u = 10 e^{-x} \cos x\)

\(\implies u'' + 4u = 10 \cos x\)

Since \(u = C \cos x + D \sin x\), we have \(u'' = -C \cos x - D \sin x\).

So,

\((-C \cos x - D \sin x) + 4(C \cos x + D \sin x) = 10 \cos x\)

\(\implies 3C \cos x + 3D \sin x = 10 \cos x\)

Equating coefficients yields:

\(3C = 10 \implies C = \frac{10}{3}\) and \(D = 0\).

Thus, the PI is:

\(y_p = \frac{10}{3} e^{-x} \cos x\)

The general solution is:

\(y = e^{-x}(A \cos 2x + B \sin 2x) + \frac{10}{3} e^{-x} \cos x\)

We now use the initial conditions to find \(A\) and \(B\).

At \(x = 0\), \(y = 4\):

\(4 = A + \frac{10}{3} \implies A = \frac{2}{3}\)

Now, differentiate \(y\):

\(\frac{dy}{dx} = -e^{-x}(A \cos 2x + B \sin 2x) + e^{-x}(-2A \sin 2x + 2B \cos 2x) - \frac{10}{3} e^{-x} \cos x - \frac{10}{3} e^{-x} \sin x\)

At \(x = 0\), \(\frac{dy}{dx} = 0\):

\(0 = -A + 2B - \frac{10}{3}\)

Substitute \(A = \frac{2}{3}\):

\(0 = -\frac{2}{3} + 2B - \frac{10}{3} \implies 2B = 4 \implies B = 2\).

Therefore, the particular solution is:

\(y = e^{-x}\left(\frac{2}{3} \cos 2x + 2 \sin 2x\right) + \frac{10}{3} e^{-x} \cos x\).

M1: Solves the auxiliary equation correctly to find the complex roots.
A1: Obtains the correct complementary function.
M1: Proposes a correct form for the particular integral, \(y_p = e^{-x}(C\cos x + D\sin x)\).
A1: Evaluates the derivatives and substitutes into the DE to form a simplified system for \(C\) and \(D\).
A1: Finds the correct values \(C = 10/3\) and \(D = 0\).
M1: Sets up the general solution and applies \(y(0)=4\) to find \(A\).
A1: Correctly finds \(A = 2/3\).
M1: Differentiates the general solution and applies \(y'(0)=0\).
A1: Correctly finds \(B = 2\).
A1: Writes down the final particular solution in its complete form.

(a) The formula for the surface area \(S\) of revolution about the \(x\)-axis is:

\(S = 2\pi \int_{x_1}^{x_2} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\)

For \(y = \cosh x\), we have \(\frac{dy}{dx} = \sinh x\).

Then,

\(1 + \left(\frac{dy}{dx}\right)^2 = 1 + \sinh^2 x = \cosh^2 x\)

Since \(\cosh x > 0\) for all real \(x\), we have \(\sqrt{1 + \left(\frac{dy}{dx}\right)^2} = \cosh x\).

Substituting this back into the formula:

\(S = 2\pi \int_0^{\ln 2} \cosh x \cdot \cosh x \, dx = 2\pi \int_0^{\ln 2} \cosh^2 x \, dx\).

(b) We use the double angle hyperbolic identity \(\cosh^2 x = \frac{\cosh 2x + 1}{2}\):

\(S = 2\pi \int_0^{\ln 2} \left( \frac{\cosh 2x + 1}{2} \right) \, dx = \pi \int_0^{\ln 2} (\cosh 2x + 1) \, dx\)

Integrating term by term:

\(S = \pi \left[ \frac{\sinh 2x}{2} + x \right]_0^{\ln 2} = \pi \left( \frac{\sinh(2\ln 2)}{2} + \ln 2 \right)\)

We simplify the term \(\sinh(2\ln 2)\) using \(2\ln 2 = \ln 4\):

\(\sinh(\ln 4) = \frac{e^{\ln 4} - e^{-\ln 4}}{2} = \frac{4 - \frac{1}{4}}{2} = \frac{15}{8}\)

Thus,

\(\frac{\sinh(2\ln 2)}{2} = \frac{15}{16}\)

Substituting this back into the equation for \(S\):

\(S = \pi \left( \frac{15}{16} + \ln 2 \right)\).

Here, \(a = \frac{15}{16}\) and \(b = 2\).

(a)
M1: Recalls the formula for the surface area of revolution.
A1: Correctly differentiates \(\cosh x\) and simplifies \(1 + (dy/dx)^2\) using hyperbolic identities.
A1: Integrates the expression cleanly to reach the given integral form.

(b)
M1: Uses the identity \(\cosh^2 x = \frac{1}{2}(\cosh 2x + 1)\).
A1: Integrates correctly to get \(\frac{1}{2}\sinh 2x + x\).
M1: Evaluates the integrated expression at limits \(0\) and \(\ln 2\).
A1: Converts \(\sinh(2\ln 2)\) into a rational number correctly (yielding \(15/8\)).
A1: Simplifies the final term to show that \(S = \pi \left(\frac{15}{16} + \ln 2\right)\).

(a) Let \(z = \cos \theta + i \sin \theta\). Then:

\(2i \sin \theta = z - z^{-1}\)

Using the binomial theorem to expand \((2i \sin \theta)^6\):

\((2i \sin \theta)^6 = (z - z^{-1})^6\)

\(-64 \sin^6 \theta = z^6 - 6z^4 + 15z^2 - 20 + 15z^{-2} - 6z^{-4} + z^{-6}\)

Grouping terms with positive and negative exponents:

\(-64 \sin^6 \theta = (z^6 + z^{-6}) - 6(z^4 + z^{-4}) + 15(z^2 + z^{-2}) - 20\)

Using the identity \(z^n + z^{-n} = 2 \cos n\theta\):

\(-64 \sin^6 \theta = 2 \cos 6\theta - 12 \cos 4\theta + 30 \cos 2\theta - 20\)

Dividing both sides by \(-2\):

\(32 \sin^6 \theta = 10 - 15 \cos 2\theta + 6 \cos 4\theta - \cos 6\theta\)

\(\implies \sin^6 \theta = \frac{1}{32} (10 - 15 \cos 2\theta + 6 \cos 4\theta - \cos 6\theta)\).

(b) We integrate the identity from part (a):

\(\int_0^{\frac{\pi}{2}} \sin^6 \theta \, d\theta = \frac{1}{32} \int_0^{\frac{\pi}{2}} (10 - 15 \cos 2\theta + 6 \cos 4\theta - \cos 6\theta) \, d\theta\)

\(= \frac{1}{32} \left[ 10\theta - \frac{15}{2} \sin 2\theta + \frac{3}{2} \sin 4\theta - \frac{1}{6} \sin 6\theta \right]_0^{\frac{\pi}{2}}\)

Since \(\sin(k\pi) = 0\) for any integer \(k\), the sine terms vanish at both boundaries \(\theta = 0\) and \(\theta = \frac{\pi}{2}\).

Therefore:

\(\int_0^{\frac{\pi}{2}} \sin^6 \theta \, d\theta = \frac{1}{32} \left[ 10\left(\frac{\pi}{2}\right) - 0 \right] = \frac{5\pi}{32}\).

(a)
M1: Expresses \(2i\sin\theta\) in terms of \(z\) and \(z^{-1}\).
A1: Writes down the binomial expansion of \((z-z^{-1})^6\) correctly.
M1: Regroups terms to pair \(z^n\) with \(z^{-n}\).
A1: Substitutes \(2\cos n\theta\) correctly for each pair.
A1: Simplifies the coefficients to obtain the given identity.

(b)
M1: Integrates the trigonometric terms in the identity term-by-term.
A1: Correctly obtains the integrated expression.
M1: Plugs in the limits \(0\) and \(\pi/2\) and notes that all sine terms vanish.
A1: Reaches the correct exact answer of \(\frac{5\pi}{32}\).

Let \(r\) be the distance of the particle from \(O\). The natural length of the string is \(l = 0.8\text{ m}\), and the initial distance is \(r_0 = 1.2\text{ m}\). Since \(r_0 > l\), the string is initially extended by \(x_0 = r_0 - l = 0.4\text{ m}\).

During the motion, angular momentum about \(O\) is conserved. Since the particle is projected perpendicular to \(OA\) with speed \(u\), the angular momentum per unit mass is \(h = r_0 u = 1.2 u\). At any distance \(r\), the transverse component of the velocity is \(v_{\theta} = \frac{1.2 u}{r}\).

By conservation of energy:
\[ \frac{1}{2} m (v_r^2 + v_{\theta}^2) + \frac{\lambda (r - l)^2}{2l} = \frac{1}{2} m u^2 + \frac{\lambda (r_0 - l)^2}{2l} \]

Substituting \(m = 0.6\text{ kg}\), \(\lambda = 24\text{ N}\), and \(l = 0.8\text{ m}\):
\[ 0.3 \left(v_r^2 + \frac{1.44 u^2}{r^2}\right) + 15 (r - 0.8)^2 = 0.3 u^2 + 15 (0.4)^2 \]
\[ 0.3 v_r^2 + \frac{0.432 u^2}{r^2} + 15 (r - 0.8)^2 = 0.3 u^2 + 2.4 \]

For the string not to slacken, the distance \(r\) must remain greater than or equal to the natural length \(l = 0.8\text{ m}\). The boundary condition for this occurs when the minimum distance reached is exactly \(r_{\min} = 0.8\text{ m}\). At this minimum point, the radial velocity \(v_r = 0\).

Substituting \(r = 0.8\) and \(v_r = 0\) into the energy equation:
\[ 0 + \frac{0.432 u^2}{0.64} + 15 (0)^2 = 0.3 u^2 + 2.4 \]
\[ 0.675 u^2 = 0.3 u^2 + 2.4 \]
\[ 0.375 u^2 = 2.4 \]
\[ u^2 = 6.4 \]
\[ u = \sqrt{6.4} \approx 2.53\text{ m s}^{-1} \]

If \(u \ge 2.53\text{ m s}^{-1}\), the minimum distance of the particle from \(O\) will be at least \(0.8\text{ m}\), meaning the string never slackens.

M1: For writing down the conservation of angular momentum: \(v_{\theta} = 1.2u / r\).
M1: For setting up the conservation of energy equation with elastic potential energy.
A1: Correct energy equation with numerical values substituted.
M1: Recognizing that the boundary condition for not slackening is \(v_r = 0\) at \(r = 0.8\text{ m}\).
A1: For substituting \(r = 0.8\) and obtaining the simplified equation \(0.675 u^2 = 0.3 u^2 + 2.4\).
M1: Solving for \(u\).
A1: Finding the correct value \(u = 2.53\text{ m s}^{-1}\) (or \(\frac{4}{5}\sqrt{10}\)).

Let the line of centers of the spheres at the moment of impact be the \(x\)-axis.

Before the collision, the velocity components of \(A\) are:
\(u_{Ax} = u \cos \theta\), \(u_{Ay} = u \sin \theta\).
Since sphere \(B\) is at rest, \(u_{Bx} = 0\), \(u_{By} = 0\).

Since the spheres are smooth, the components of velocity perpendicular to the line of centers (along the \(y\)-axis) do not change:
\(v_{Ay} = u \sin \theta\), \(v_{By} = 0\).

Let \(v_{Ax}\) and \(v_{Bx}\) be the velocity components along the \(x\)-axis after the collision.
Applying conservation of linear momentum along the \(x\)-axis:
\[ m v_{Ax} + 2m v_{Bx} = m u \cos \theta \implies v_{Ax} + 2 v_{Bx} = u \cos \theta \quad \text{--- (1)} \]

Applying Newton's law of restitution along the \(x\)-axis:
\[ v_{Bx} - v_{Ax} = e (u \cos \theta - 0) = \frac{2}{3} u \cos \theta \quad \text{--- (2)} \]

Multiply equation (2) by 2:
\[ 2 v_{Bx} - 2 v_{Ax} = \frac{4}{3} u \cos \theta \]

Subtracting this from equation (1) gives:
\[ 3 v_{Ax} = u \cos \theta - \frac{4}{3} u \cos \theta = -\frac{1}{3} u \cos \theta \implies v_{Ax} = -\frac{1}{9} u \cos \theta \]

The initial direction of motion of \(A\) is represented by the vector \(\mathbf{d}_i = (\cos \theta, \sin \theta)\).
The final velocity vector of \(A\) is \(\mathbf{v}_A = \left(-\frac{1}{9} u \cos \theta, u \sin \theta\right)\).
Since the final direction is perpendicular to the initial direction, their dot product must be zero:
\[ \mathbf{d}_i \cdot \mathbf{v}_A = 0 \implies -\frac{1}{9} u \cos^2 \theta + u \sin^2 \theta = 0 \]

Since \(u \neq 0\), we divide by \(u \cos^2 \theta\):
\[ \tan^2 \theta = \frac{1}{9} \]

Since \(\theta\) is an acute angle, \(\tan \theta\) must be positive:
\[ \tan \theta = \frac{1}{3} \]

M1: State the velocity components of \(A\) parallel and perpendicular to the line of centers before and after the collision.
M1: Apply conservation of momentum along the line of centers.
A1: Correct momentum equation: \(v_{Ax} + 2 v_{Bx} = u \cos \theta\).
M1: Apply Newton's law of restitution.
A1: Correct restitution equation and expression for \(v_{Ax} = -\frac{1}{9} u \cos \theta\).
M1: Use the perpendicular condition (dot product or trigonometry of angles) to set up an equation in \(\theta\).
A1: Obtain the correct value \(\tan \theta = \frac{1}{3}\).

Let \(\theta\) be the angle that the string makes with the upward vertical through \(O\). The lowest point of the path corresponds to \(\theta = 180^{\circ}\).

By conservation of energy, taking the lowest point as the reference level for potential energy:
\[ \frac{1}{2} m u^2 = \frac{1}{2} m v^2 + m g L (1 + \cos \theta) \]
\[ v^2 = u^2 - 2 g L (1 + \cos \theta) \quad \text{--- (1)} \]

Resolving forces radially towards the center \(O\):
\[ T + m g \cos \theta = \frac{m v^2}{L} \quad \text{--- (2)} \]
where \(T\) is the tension in the string.

The string becomes slack when \(T = 0\). Let this occur at \(\theta = 60^{\circ}\):
\[ m g \cos 60^{\circ} = \frac{m v^2}{L} \implies v^2 = g L \cos 60^{\circ} = 0.5 g L \]

Substitute \(\theta = 60^{\circ}\) and \(v^2 = 0.5 g L\) into equation (1):
\[ 0.5 g L = u^2 - 2 g L (1 + \cos 60^{\circ}) \]
\[ 0.5 g L = u^2 - 2 g L (1.5) \]
\[ 0.5 g L = u^2 - 3 g L \]
\[ u^2 = 3.5 g L \]
\[ u = \sqrt{3.5 g L} \]

M1: Use conservation of energy to write down an equation relating \(u\), \(v\), and \(\theta\).
A1: Obtain the correct energy equation: \(v^2 = u^2 - 2gL(1 + \cos \theta)\).
M1: Set up the radial equation of motion including tension and the component of gravity.
A1: Correct radial equation: \(T + mg \cos \theta = mv^2 / L\).
M1: Set \(T = 0\) at \(\theta = 60^{\circ}\) to find \(v^2 = 0.5 g L\).
M1: Substitute \(v^2\) and \(\theta = 60^{\circ}\) back into the energy equation.
A1: Correctly solve for \(u\) to get \(u = \sqrt{3.5 g L}\) (or equivalent).

Let \(R\) and \(F\) be the normal and frictional forces acting on the ladder at \(A\) respectively. Let \(S\) be the normal reaction force of the smooth vertical wall acting on the ladder at \(B\).

Let \(x\) be the distance of the man from \(A\) along the ladder when it is on the point of slipping.

Since the ladder is on the point of slipping, the friction force \(F\) at \(A\) is at its maximum value:
\[ F = \mu R = 0.5 R \]

Resolving vertically for equilibrium:
\[ R = W + 2W = 3W \]

Thus, the maximum friction force is:
\[ F = 0.5 (3W) = 1.5W \]

Resolving horizontally for equilibrium:
\[ S = F = 1.5W \]

Taking moments about \(A\) for the ladder in equilibrium:
\[ W (a \cos \alpha) + 2W (x \cos \alpha) = S (2a \sin \alpha) \]

Substitute \(S = 1.5W\) into the moment equation:
\[ W a \cos \alpha + 2W x \cos \alpha = 1.5W (2a \sin \alpha) = 3W a \sin \alpha \]

Divide the entire equation by \(W \cos \alpha\):
\[ a + 2x = 3a \tan \alpha \]

We are given \(\tan \alpha = 1.5\):
\[ a + 2x = 3a (1.5) = 4.5a \]
\[ 2x = 3.5a \]
\[ x = 1.75a \]

Therefore, the man can climb a distance of \(1.75a\) from \(A\) before the ladder begins to slip.

M1: Resolve forces vertically to find the normal reaction at the floor, \(R = 3W\).
M1: Apply the friction condition \(F = \mu R\) at the point of slipping.
A1: Correctly find \(F = 1.5W\) and show \(S = 1.5W\) by resolving horizontally.
M1: Set up a moments equation about \(A\) (or any other point).
A1: Correct moments equation: \(W a \cos \alpha + 2W x \cos \alpha = 2S a \sin \alpha\).
M1: Substitute \(S\) and \(\tan \alpha = 1.5\) into the moments equation.
A1: Solve for \(x\) to get \(x = 1.75a\) (or \(\frac{7}{4}a\)).

The equation of motion of the particle of mass \(m = 0.5\text{ kg}\) is:
\[ m a = -F \implies 0.5 v \frac{dv}{dx} = -\frac{1}{v+1} \]

We separate the variables to find \(x\) in terms of \(v\):
\[ v(v+1) \, dv = -2 \, dx \]
\[ (v^2 + v) \, dv = -2 \, dx \]

Integrating both sides with the initial conditions that \(x = 0\) when \(v = 4\):
\[ \int_{4}^{v} (v^2 + v) \, dv = \int_{0}^{x} -2 \, dx \]
\[ \left[ \frac{1}{3}v^3 + \frac{1}{2}v^2 \right]_{4}^{v} = -2x \]
\[ \left(\frac{1}{3}v^3 + \frac{1}{2}v^2\right) - \left(\frac{1}{3}(4)^3 + \frac{1}{2}(4)^2\right) = -2x \]
\[ \left(\frac{1}{3}v^3 + \frac{1}{2}v^2\right) - \left(\frac{64}{3} + 8\right) = -2x \]
\[ \left(\frac{1}{3}v^3 + \frac{1}{2}v^2\right) - \frac{88}{3} = -2x \]

We want to find the distance \(x\) when the velocity is \(v = 1\text{ m s}^{-1}\). Substituting \(v = 1\) into the equation:
\[ \left(\frac{1}{3}(1)^3 + \frac{1}{2}(1)^2\right) - \frac{88}{3} = -2x \]
\[ \left(\frac{1}{3} + \frac{1}{2}\right) - \frac{88}{3} = -2x \]
\[ \frac{5}{6} - \frac{176}{6} = -2x \]
\[ -\frac{171}{6} = -2x \]
\[ 2x = 28.5 \implies x = 14.25\text{ m} \]

Thus, the distance traveled by the particle is \(14.25\text{ m}\).

M1: Write down the equation of motion using \(a = v \frac{dv}{dx}\).
A1: Correct differential equation: \(0.5 v \frac{dv}{dx} = -\frac{1}{v+1}\).
M1: Separate variables correctly: \((v^2 + v) dv = -2 dx\).
M1: Integrate both sides, including limits or a constant of integration.
A1: Correct integration: \(\frac{1}{3}v^3 + \frac{1}{2}v^2 = -2x + C\) or equivalent with limits.
M1: Use initial conditions \(v = 4\) at \(x = 0\) and substitute \(v = 1\).
A1: Obtain the correct distance \(14.25\text{ m}\) (or \(\frac{57}{4}\text{ m}\)).

The equation of the trajectory of a projectile is given by:
\[ y = x \tan \theta - \frac{g x^2}{2 V^2} (1 + \tan^2 \theta) \]

Given that the projectile passes through \(P(20, 10)\) and \(g = 10\text{ m s}^{-2}\):
\[ 10 = 20 \tan \theta - \frac{10 (20)^2}{2 V^2} (1 + \tan^2 \theta) \]
\[ 10 = 20 \tan \theta - \frac{2000}{V^2} (1 + \tan^2 \theta) \]

Divide the entire equation by 10:
\[ 1 = 2 \tan \theta - \frac{200}{V^2} (1 + \tan^2 \theta) \]

Rearranging this as a quadratic equation in terms of \(\tan \theta\):
\[ \frac{200}{V^2} \tan^2 \theta - 2 \tan \theta + \left(1 + \frac{200}{V^2}\right) = 0 \]

For a real trajectory to exist, the angle \(\theta\) must be real, meaning that this quadratic equation must have real roots for \(\tan \theta\). Therefore, the discriminant \(\Delta\) of the quadratic must be non-negative:
\[ \Delta = B^2 - 4AC \ge 0 \]
\[ (-2)^2 - 4 \left(\frac{200}{V^2}\right) \left(1 + \frac{200}{V^2}\right) \ge 0 \]
\[ 4 - \frac{800}{V^2} - \frac{160000}{V^4} \ge 0 \]

Divide by 4:
\[ 1 - \frac{200}{V^2} - \frac{40000}{V^4} \ge 0 \]

Let \(W = V^2\). The inequality becomes:
\[ 1 - \frac{200}{W} - \frac{40000}{W^2} \ge 0 \implies W^2 - 200 W - 40000 \ge 0 \]

To find the critical value of \(W\), we solve the quadratic equation \(W^2 - 200 W - 40000 = 0\):
\[ W = \frac{200 \pm \sqrt{(-200)^2 - 4(1)(-40000)}}{2} = \frac{200 \pm \sqrt{40000 + 160000}}{2} = \frac{200 \pm \sqrt{200000}}{2} \]
\[ W = 100 \pm 100\sqrt{5} \]

Since \(W = V^2 > 0\), we must have:
\[ V^2 \ge 100(1 + \sqrt{5}) \approx 323.607 \]
\[ V \ge \sqrt{323.607} \approx 17.989\text{ m s}^{-1} \]

Thus, the minimum possible value of \(V\) to 3 significant figures is \(18.0\text{ m s}^{-1}\).

M1: Substitute coordinates of \(P\) and \(g = 10\) into the trajectory equation.
A1: Correctly obtain the trajectory equation in \(\tan \theta\): \(1 = 2 \tan \theta - \frac{200}{V^2}(1+\tan^2\theta)\).
M1: Rearrange into standard quadratic form: \(A \tan^2 \theta + B \tan \theta + C = 0\).
M1: State and apply the discriminant condition \(\Delta \ge 0\) for real roots.
A1: Correct inequality for \(V^2\) or \(W\): \(W^2 - 200 W - 40000 \ge 0\) (or equivalent).
M1: Solve the quadratic inequality for \(V^2\) and find the positive root.
A1: State the final answer as \(18.0\) (3 s.f.).

We choose a coordinate system with the \(x'\)-axis pointing up the line of greatest slope of the plane, and the \(y'\)-axis perpendicular to the plane.

The initial velocity of the particle has components:
\[ u_{x'} = 20 \cos 30^{\circ} = 10\sqrt{3}\text{ m s}^{-1} \]
\[ u_{y'} = 20 \sin 30^{\circ} = 10\text{ m s}^{-1} \]

The acceleration due to gravity has components:
\[ a_{x'} = -g \sin 30^{\circ} = -10 (0.5) = -5\text{ m s}^{-2} \]
\[ a_{y'} = -g \cos 30^{\circ} = -10 \left(\frac{\sqrt{3}}{2}\right) = -5\sqrt{3}\text{ m s}^{-2} \]

The equations of motion for the coordinates of the projectile are:
\[ x'(t) = 10\sqrt{3} t - \frac{5}{2} t^2 \]
\[ y'(t) = 10 t - \frac{5\sqrt{3}}{2} t^2 \]

The particle strikes the plane when \(y'(t) = 0\) for \(t > 0\):
\[ 10 t - \frac{5\sqrt{3}}{2} t^2 = 0 \implies t = \frac{20}{5\sqrt{3}} = \frac{4}{\sqrt{3}}\text{ seconds} \]

Now we substitute this time of flight \(t\) into the equation for \(x'\) to find the range \(R\) up the plane:
\[ R = x'\left(\frac{4}{\sqrt{3}}\right) = 10\sqrt{3} \left(\frac{4}{\sqrt{3}}\right) - \frac{5}{2} \left(\frac{4}{\sqrt{3}}\right)^2 \]
\[ R = 40 - \frac{5}{2} \left(\frac{16}{3}\right) \]
\[ R = 40 - \frac{40}{3} = \frac{80}{3}\text{ m} \]

So the distance from \(O\) to the point where the particle strikes the plane is \(\frac{80}{3}\text{ m}\) (or approximately \(26.7\text{ m}\)).

M1: Resolve the initial velocity parallel and perpendicular to the inclined plane.
A1: Correct components: \(u_{x'} = 10\sqrt{3}\), \(u_{y'} = 10\).
M1: Resolve gravity parallel and perpendicular to the inclined plane.
A1: Correct acceleration components: \(a_{x'} = -5\), \(a_{y'} = -5\sqrt{3}\).
M1: Set \(y'(t) = 0\) and solve for the time of flight \(t = 4/\sqrt{3}\).
M1: Substitute the time of flight into the \(x'(t)\) expression.
A1: Find the correct range \(\frac{80}{3}\text{ m}\) (or \(26.7\text{ m}\)).

(i) For \(x < 0\), \(F(x) = 0\).

For \(0 \le x < 2\):
\[F(x) = \int_0^x \frac{1}{8}t \, dt = \left[ \frac{t^2}{16} \right]_0^x = \frac{x^2}{16}\]

At \(x = 2\), \(F(2) = \frac{4}{16} = 0.25\).

For \(2 \le x \le 4\):
\[F(x) = 0.25 + \int_2^x \frac{3}{8}(4-t) \, dt = 0.25 + \frac{3}{8}\left[ 4t - \frac{t^2}{2} \right]_2^x\]
\[F(x) = 0.25 + \frac{3}{8}\left( \left(4x - \frac{x^2}{2}\right) - (8 - 2) \right) = 0.25 + \frac{3}{2}x - \frac{3}{16}x^2 - 2.25 = \frac{3}{2}x - \frac{3}{16}x^2 - 2\]

For \(x > 4\), \(F(x) = 1\).

Thus, the cumulative distribution function is:
\[F(x) = \begin{cases} 0 & x < 0, \\ \frac{x^2}{16} & 0 \le x < 2, \\ \frac{3}{2}x - \frac{3}{16}x^2 - 2 & 2 \le x \le 4, \\ 1 & x > 4. \end{cases}\]

(ii) Since \(F(2) = 0.25 < 0.5\), the median \(m\) must lie in the interval \(2 \le m \le 4\).
Set \(F(m) = 0.5\):
\[\frac{3}{2}m - \frac{3}{16}m^2 - 2 = 0.5\]
Multiplying by 16 and rearranging gives:
\[3m^2 - 24m + 40 = 0\]
Applying the quadratic formula:
\[m = \frac{24 \pm \sqrt{(-24)^2 - 4(3)(40)}}{2(3)} = \frac{24 \pm \sqrt{576 - 480}}{6} = 4 \pm \frac{\sqrt{96}}{6} = 4 \pm \frac{2\sqrt{6}}{3}\]
Since \(m \le 4\), we have:
\[m = 4 - \frac{2\sqrt{6}}{3} \approx 2.37\]

(i)
M1: Integrates \(f(t)\) in the interval \(0 \le x < 2\).
A1: Correct expression \(\frac{x^2}{16}\) for this interval.
M1: Integrates \(f(t)\) in the interval \(2 \le x \le 4\) and adds \(F(2)\).
A1: Correct piecewise function with all branches specified (including \(x < 0\) and \(x > 4\)).

(ii)
M1: Realises that median lies in \(2 \le x \le 4\) and equates expression to \(0.5\).
M1: Obtains and solves a correct quadratic equation for \(m\).
A1: Selects the correct root and rounds to 2.37 (3 s.f.).

Let \(D\) be the difference in jump distance, \(D = \text{After} - \text{Before}\).

Calculate the differences for each student:
- A: \(1.99 - 1.85 = +0.14\)
- B: \(1.91 - 1.94 = -0.03\)
- C: \(1.88 - 1.76 = +0.12\)
- D: \(2.21 - 2.10 = +0.11\)
- E: \(1.83 - 1.88 = -0.05\)
- F: \(2.13 - 2.05 = +0.08\)
- G: \(2.05 - 1.90 = +0.15\)
- H: \(1.84 - 1.82 = +0.02\)

Order the non-zero absolute differences and assign ranks:
1. \(|+0.02| = 0.02\) \(\rightarrow\) Rank 1 (+)
2. \(|-0.03| = 0.03\) \(\rightarrow\) Rank 2 (-)
3. \(|-0.05| = 0.05\) \(\rightarrow\) Rank 3 (-)
4. \(|+0.08| = 0.08\) \(\rightarrow\) Rank 4 (+)
5. \(|+0.11| = 0.11\) \(\rightarrow\) Rank 5 (+)
6. \(|+0.12| = 0.12\) \(\rightarrow\) Rank 6 (+)
7. \(|+0.14| = 0.14\) \(\rightarrow\) Rank 7 (+)
8. \(|+0.15| = 0.15\) \(\rightarrow\) Rank 8 (+)

Sum of negative ranks: \(T_- = 2 + 3 = 5\).
Sum of positive ranks: \(T_+ = 1 + 4 + 5 + 6 + 7 + 8 = 31\).

The test statistic is \(T = \min(T_-, T_+) = 5\).

Hypotheses:
\(H_0\): Median difference in jump distances is zero (no improvement).
\(H_1\): Median difference in jump distances is greater than zero (improvement).
This is a one-tailed test with \(n = 8\).

At the 5% significance level, the critical value for a one-tailed Wilcoxon signed-rank test with \(n = 8\) is 5.
Since \(T \le 5\) (i.e., \(5 \le 5\)), we reject \(H_0\).

There is sufficient evidence at the 5% significance level to conclude that the training program improves standing long jump distances.

B1: States correct hypotheses \(H_0\) and \(H_1\).
M1: Calculates all 8 differences correctly.
M1: Ranks absolute differences correctly without any tied rank issues.
M1: Calculates sum of positive ranks (31) and sum of negative ranks (5).
A1: Identifies test statistic \(T = 5\).
B1: States correct critical value of 5 for a one-tailed test with \(n = 8\) at 5% significance level.
A1: Correctly compares \(T\) with critical value and draws the conclusion in context.

(i) Mean \(\lambda = \frac{(0 \times 38) + (1 \times 35) + (2 \times 18) + (3 \times 7) + (4 \times 2)}{100} = \frac{0 + 35 + 36 + 21 + 8}{100} = \frac{100}{100} = 1.0\).

Using \(\lambda = 1.0\), the Poisson probabilities \(P(X = k) = \frac{e^{-1} 1^k}{k!}\) are calculated:
- \(E_0 = 100 \times e^{-1} \approx 36.79\)
- \(E_1 = 100 \times e^{-1} \approx 36.79\)
- \(E_2 = 100 \times \frac{e^{-1}}{2} \approx 18.39\)
- \(E_3 = 100 \times \frac{e^{-1}}{6} \approx 6.13\)
- \(E_{\ge 4} = 100 \times \left(1 - e^{-1}\left(1 + 1 + \frac{1}{2} + \frac{1}{6}\right)\right) = 100 \times \left(1 - \frac{8}{3}e^{-1}\right) \approx 1.90\).

(ii) Since \(E_{\ge 4} = 1.90 < 5\), we must combine this class with the \(x = 3\) class. Let the combined class be \(x \ge 3\):
- Combined Observed Frequency \(= 7 + 2 + 0 = 9\).
- Combined Expected Frequency \(= 6.13 + 1.90 = 8.03\).

The classes to test are:
- \(x = 0\): \(O = 38\), \(E = 36.79\)
- \(x = 1\): \(O = 35\), \(E = 36.79\)
- \(x = 2\): \(O = 18\), \(E = 18.39\)
- \(x \ge 3\): \(O = 9\), \(E = 8.03\)

Calculate \(\chi^2 = \sum \frac{(O - E)^2}{E}\):
- \(x = 0\): \(\frac{(38 - 36.79)^2}{36.79} = 0.0398\)
- \(x = 1\): \(\frac{(35 - 36.79)^2}{36.79} = 0.0871\)
- \(x = 2\): \(\frac{(18 - 18.39)^2}{18.39} = 0.0083\)
- \(x \ge 3\): \(\frac{(9 - 8.03)^2}{8.03} = 0.1172\)

Sum of contributions: \\ \(\chi^2 \approx 0.0398 + 0.0871 + 0.0083 + 0.1172 = 0.2524 \approx 0.252\) (3 s.f.).

Degrees of freedom: \(d.f. = \text{number of classes} - 1 - \text{number of estimated parameters} = 4 - 1 - 1 = 2\).

Critical value for \(\chi^2_2\) at 5% level is \(5.991\).
Since \(0.2524 < 5.991\), we fail to reject \(H_0\).
There is insufficient evidence to suggest that the Poisson distribution is not a suitable model.

(i)
M1: Calculates the sample mean correctly as 1.0.
A1: Shows or explains the calculation of expected frequencies with \(\lambda = 1.0\) to match the given values.

(ii)
M1: Recognises the need to combine classes because \(E_{\ge 4} < 5\) and correctly pools \(x=3\) and \(x\ge 4\).
A1: Obtains correct pooled observed (9) and expected (8.03) values.
M1: Calculates the \(\chi^2\) test statistic values correctly.
A1: Correct value \(\chi^2 \approx 0.252\).
B1: Identifies 2 degrees of freedom and states correct critical value of 5.991.
A1: Makes correct comparison and final conclusion in context.

(i) Rewrite \(G_X(t) = (3 - 2t)^{-1}\).
First derivative:
\[G_X'(t) = -1(3 - 2t)^{-2} \times (-2) = 2(3 - 2t)^{-2}\]
The mean is:
\[E(X) = G_X'(1) = 2(3 - 2)^{-2} = 2\]

Second derivative:
\[G_X''(t) = 2 \times (-2)(3 - 2t)^{-3} \times (-2) = 8(3 - 2t)^{-3}\]
At \(t = 1\):
\[G_X''(1) = 8(3 - 2)^{-3} = 8\]

The variance of \(X\) is:
\[\text{Var}(X) = G_X''(1) + G_X'(1) - (G_X'(1))^2 = 8 + 2 - 2^2 = 6\]

(ii) Since \(X\) and \(Y\) are independent, the PGF of \(W = X + Y\) is:
\[G_W(t) = G_X(t) \cdot G_Y(t) = \frac{(0.4 + 0.6t)^2}{3 - 2t}\]

We find \(P(W = 1)\) using the derivative formula \(P(W = 1) = G_W'(0)\):
\[G_W'(t) = \frac{(3 - 2t) \cdot 2(0.4 + 0.6t)(0.6) - (0.4 + 0.6t)^2 \cdot (-2)}{(3 - 2t)^2}\]
Substitute \(t = 0\):
\[G_W'(0) = \frac{3 \cdot 2(0.4)(0.6) + 2(0.4)^2}{3^2} = \frac{3 \cdot 0.48 + 2(0.16)}{9} = \frac{1.44 + 0.32}{9} = \frac{1.76}{9} = \frac{44}{225} \approx 0.196\]

(i)
M1: Differentiates \(G_X(t)\) to find \(G_X'(t)\) and evaluates at \(t=1\).
A1: Obtains Mean = 2.
M1: Differentiates again to find \(G_X''(t)\) and uses \(\text{Var}(X) = G_X''(1) + G_X'(1) - [G_X'(1)]^2\).
A1: Obtains Variance = 6.

(ii)
M1: Identifies \(G_W(t) = G_X(t) \cdot G_Y(t)\) and writes expression.
M1: Uses differentiation \(G_W'(0)\) or series expansion coefficient of \(t\).
A1: Obtains exact probability of \(\frac{44}{225}\) or 0.196 (3 s.f.).

(i) The pooled estimate of the population variance, \(s_p^2\), is:
\[s_p^2 = \frac{(n_A - 1)s_A^2 + (n_B - 1)s_B^2}{n_A + n_B - 2}\]
\[s_p^2 = \frac{(10 - 1)(12.8) + (12 - 1)(15.4)}{10 + 12 - 2} = \frac{9(12.8) + 11(15.4)}{20} = \frac{115.2 + 169.4}{20} = 14.23\]

(ii) State the hypotheses:
\(H_0: \mu_A = \mu_B\)
\(H_1: \mu_A > \mu_B\)
where \(\mu_A\) and \(\mu_B\) are the population mean yields for Fertilizers A and B respectively.

This is a one-tailed two-sample t-test with \(d.f. = 10 + 12 - 2 = 20\).

Calculate the test statistic:
\[t = \frac{\bar{x}_A - \bar{x}_B}{\sqrt{s_p^2 \left(\frac{1}{n_A} + \frac{1}{n_B}\right)}}\]
\[t = \frac{45.2 - 41.5}{\sqrt{14.23 \left(\frac{1}{10} + \frac{1}{12}\right)}} = \frac{3.7}{\sqrt{14.23 \times \frac{11}{60}}} = \frac{3.7}{\sqrt{2.60883}} \approx \frac{3.7}{1.6152} \approx 2.291 \approx 2.29\]

The critical value of \(t\) for a one-tailed test with 20 degrees of freedom at the 5% level is \(1.725\).
Since \(t = 2.29 > 1.725\), we reject \(H_0\).

There is sufficient evidence at the 5% significance level to conclude that the mean crop yield using Fertilizer A is greater than the mean crop yield using Fertilizer B.

(i)
M1: Uses the correct formula for the pooled estimate of population variance.
A1: Obtains 14.23.

(ii)
B1: Correctly states hypotheses \(H_0\) and \(H_1\).
M1: Applies the correct two-sample t-test statistic formula.
A1: Calculates test statistic \(t \approx 2.29\).
B1: Identifies 20 degrees of freedom and states the correct critical value of 1.725.
A1: Concludes correctly by rejecting \(H_0\) in context.

(i) For \(x \ge 0\), the cumulative distribution function is:
\[F(x) = \int_0^x t e^{-t} \, dt\]

Use integration by parts, let \(u = t\) and \(v' = e^{-t}\), so \(u' = 1\) and \(v = -e^{-t}\):
\[F(x) = \left[ -t e^{-t} \right]_0^x - \int_0^x -e^{-t} \, dt\]
\[F(x) = -x e^{-x} + \left[ -e^{-t} \right]_0^x = -x e^{-x} - e^{-x} - (-1)\]
\[F(x) = 1 - (x + 1)e^{-x}\]

(ii) The probability that \(X\) is greater than 2 is:
\[P(X > 2) = 1 - F(2)\]
\[P(X > 2) = 1 - \left( 1 - (2 + 1)e^{-2} \right) = 3e^{-2} \approx 3 \times 0.135335 = 0.406005 \approx 0.406\]

(i)
M1: Sets up the integral for \(F(x)\).
M1: Integrates by parts correctly.
A1: Shows step-by-step substitution of limits to arrive at \(1 - (x + 1)e^{-x}\).

(ii)
M1: Identifies that \(P(X > 2) = 1 - F(2)\).
A1: Substitutes \(x = 2\) into the CDF correctly.
A1: Obtains 0.406 (3 s.f.).