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(i) The area \(A\) of the region is given by:
\[ A = \frac{1}{2} \int_{0}^{\pi} r^2 \, d\theta = \frac{1}{2} \int_{0}^{\pi} 4(1 + \cos \theta)^2 \, d\theta \]
\[ A = 2 \int_{0}^{\pi} (1 + 2\cos \theta + \cos^2 \theta) \, d\theta \]
Using the identity \[\cos^2 \theta = \frac{1 + \cos 2\theta}{2}\]:
\[ A = 2 \int_{0}^{\pi} \left(1 + 2\cos \theta + \frac{1}{2} + \frac{1}{2}\cos 2\theta\right) \, d\theta \]
\[ A = 2 \left[ \frac{3}{2}\theta + 2\sin \theta + \frac{1}{4}\sin 2\theta \right]_{0}^{\pi} \]
Substituting the limits:
\[ A = 2 \left( \left(\frac{3}{2}\pi + 0 + 0\right) - (0) \right) = 3\pi \]

(ii) The distance from the initial line is given by \(y = r \sin \theta\):
\[ y = 2(1 + \cos \theta)\sin \theta = 2\sin \theta + 2\sin \theta \cos \theta = 2\sin \theta + \sin 2\theta \]
To maximize \(y\), we find \(\frac{dy}{d\theta} = 0\):
\[ \frac{dy}{d\theta} = 2\cos \theta + 2\cos 2\theta = 0 \]
\[ \cos \theta + (2\cos^2 \theta - 1) = 0 \implies 2\cos^2 \theta + \cos \theta - 1 = 0 \]
Factoring the quadratic:
\[ (2\cos \theta - 1)(\cos \theta + 1) = 0 \]
Since \(0 \le \theta \le \pi\), we have:
\(\cos \theta = \frac{1}{2} \implies \theta = \frac{\pi}{3}\) (giving a maximum)
or \(\cos \theta = -1 \implies \theta = \pi\) (giving \(y = 0\), a minimum).

At \(\theta = \frac{\pi}{3}\):
\(r = 2\left(1 + \cos \frac{\pi}{3}\right) = 2\left(1 + \frac{1}{2}\right) = 3\).

The Cartesian coordinates are:
\[ x = r \cos \theta = 3 \cos \frac{\pi}{3} = \frac{3}{2} \]
\[ y = r \sin \theta = 3 \sin \frac{\pi}{3} = \frac{3\sqrt{3}}{2} \]
Thus, the coordinates are \(\left(\frac{3}{2}, \frac{3\sqrt{3}}{2}\right)\).

(i)
M1: Use of \(\frac{1}{2}\int r^2 \, d\theta\) with correct limits.
A1: Correct expansion of \((1 + \cos \theta)^2\).
M1: Use of double-angle formula for \(\cos^2 \theta\).
A1: Correct integration.
A1: Obtains exact area \(3\pi\).

(ii)
M1: Expresses distance as \(y = r \sin \theta\).
M1: Differentiates and equates \(\frac{dy}{d\theta}\) to 0.
A1: Obtains \(\cos \theta = \frac{1}{2}\) or \(\theta = \frac{\pi}{3}\).
M1: Calculates corresponding \(r\) and converts to Cartesian coordinates.
A1: Correct x-coordinate \(\frac{3}{2}\).
A1: Correct y-coordinate \(\frac{3\sqrt{3}}{2}\).

(i) By algebraic division:
\[ x^2 - x - 2 = (x - 3)(x + 2) + 4 \]
So,
\[ y = x + 2 + \frac{4}{x - 3} \]
Thus, the asymptotes are:
Vertical asymptote: \(x = 3\)
Oblique asymptote: \(y = x + 2\)

(ii) Rearranging the equation of the curve to form a quadratic in \(x\):
\[ y(x - 3) = x^2 - x - 2 \implies x^2 - (y + 1)x + (3y - 2) = 0 \]
Since \(x\) is real, the discriminant must be non-negative:
\[ \Delta \ge 0 \implies (y+1)^2 - 4(1)(3y-2) \ge 0 \]
\[ y^2 + 2y + 1 - 12y + 8 \ge 0 \]
\[ y^2 - 10y + 9 \ge 0 \]
Factoring this inequality:
\[ (y-1)(y-9) \ge 0 \]
This holds when \(y \le 1\) or \(y \ge 9\).
Therefore, there are no points on \(C\) for which \(1 < y < 9\).

(iii) Stationary points occur where \(y = 1\) and \(y = 9\).
When \(y = 1\):
\(x^2 - 2x + 1 = 0 \implies x = 1\). So, \((1, 1)\) is a local maximum.
When \(y = 9\):
\(x^2 - 10x + 25 = 0 \implies x = 5\). So, \((5, 9)\) is a local minimum.

Intersections with axes:
With x-axis (\(y = 0\)):
\(x^2 - x - 2 = 0 \implies (x-2)(x+1) = 0 \implies (2, 0)\) and \((-1, 0)\).
With y-axis (\(x = 0\)):
\(y = \frac{-2}{-3} = \frac{2}{3} \implies (0, \frac{2}{3})\).

The sketch consists of two branches separated by the vertical asymptote \(x = 3\) and approaching the line \(y = x + 2\).

(i)
M1: Attempt at algebraic division or equating coefficients.
A1: Correct vertical asymptote \(x = 3\).
A1: Correct oblique asymptote \(y = x + 2\).

(ii)
M1: Clears fraction to form quadratic equation in \(x\).
A1: Correct quadratic equation.
M1: Uses discriminant \(\Delta \ge 0\).
A1: Deduces range inequality \(y \le 1\) or \(y \ge 9\) hence showing no points exist in \(1 < y < 9\).

(iii)
B1: Correct intercepts with axes: \((2,0)\), \((-1,0)\), and \((0, 2/3)\).
B1: Correct coordinates of turning points \((1,1)\) and \((5,9)\).
B1.7: Correct sketch showing two branches approaching asymptotes accurately.

(i) Since \(\mathbf{v}\) is an eigenvector of \(\mathbf{M}\) with eigenvalue \(\lambda\), we have:
\[ \mathbf{M} \mathbf{v} = \lambda \mathbf{v} \]
\[ \begin{pmatrix} a & 1 & 2 \\ 0 & b & -1 \\ 1 & 0 & 3 \end{pmatrix} \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} = \lambda \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} \]
Evaluating the matrix-vector product on the left-hand side:
\[ \begin{pmatrix} a(1) + 1(-1) + 2(1) \\ 0(1) + b(-1) - 1(1) \\ 1(1) + 0(-1) + 3(1) \end{pmatrix} = \begin{pmatrix} a + 1 \\ -b - 1 \\ 4 \end{pmatrix} \]
Thus, we have the system of equations:
1) \(a + 1 = \lambda\)
2) \(-b - 1 = -\lambda\)
3) \(4 = \lambda\)

From equation (3), \(\lambda = 4\).
Substituting \(\lambda = 4\) into (1):
\(a + 1 = 4 \implies a = 3\).

Substituting \(\lambda = 4\) into (2):
\(-b - 1 = -4 \implies b = 3\).

So, \(a = 3\), \(b = 3\), and \(\lambda = 4\).

(ii) With these values, \(\mathbf{M} = \begin{pmatrix} 3 & 1 & 2 \\ 0 & 3 & -1 \\ 1 & 0 & 3 \end{pmatrix}\).

First, find the determinant of \(\mathbf{M}\):
\[ \det(\mathbf{M}) = 3(3(3) - 0(-1)) - 1(0(3) - 1(-1)) + 2(0(0) - 1(3)) \]
\[ \det(\mathbf{M}) = 3(9) - 1(1) + 2(-3) = 27 - 1 - 6 = 20 \]

Next, find the matrix of cofactors, \(\mathbf{C}\):
\[ C_{11} = 9, \quad C_{12} = -(0 - (-1)) = -1, \quad C_{13} = 0 - 3 = -3 \]
\[ C_{21} = -(3 - 0) = -3, \quad C_{22} = 9 - 2 = 7, \quad C_{23} = -(0 - 1) = 1 \]
\[ C_{31} = -1 - 6 = -7, \quad C_{32} = -(-3 - 0) = 3, \quad C_{33} = 9 - 0 = 9 \]
\[ \mathbf{C} = \begin{pmatrix} 9 & -1 & -3 \\ -3 & 7 & 1 \\ -7 & 3 & 9 \end{pmatrix} \]

The adjugate matrix is the transpose of \(\mathbf{C}\):
\[ \mathbf{adj}(\mathbf{M}) = \mathbf{C}^T = \begin{pmatrix} 9 & -3 & -7 \\ -1 & 7 & 3 \\ -3 & 1 & 9 \end{pmatrix} \]

Thus, the inverse is:
\[ \mathbf{M}^{-1} = \frac{1}{20} \begin{pmatrix} 9 & -3 & -7 \\ -1 & 7 & 3 \\ -3 & 1 & 9 \end{pmatrix} \]

(i)
M1: Sets up the matrix equation \(\mathbf{M}\mathbf{v} = \lambda \mathbf{v}\).
A1: Calculates the correct matrix-vector product.
A1: Identifies \(\lambda = 4\).
A1: Solves for \(a = 3\).
A1: Solves for \(b = 3\).

(ii)
M1: Evaluates the determinant of \(\mathbf{M}\).
A1: Obtains \(\det(\mathbf{M}) = 20\).
M1: Finds the cofactors of \(\mathbf{M}\) (at least 4 correct for method).
A1: Obtains correct cofactor matrix.
M1: Transposes the cofactor matrix to find the adjugate.
A0.7: Correctly writes the final inverse matrix \(\mathbf{M}^{-1}\).

(i) Let \(u = \sqrt{e^x - 1}\).
Then \(u^2 = e^x - 1 \implies e^x = u^2 + 1\).
Differentiating with respect to \(u\):
\[ e^x \frac{dx}{du} = 2u \implies dx = \frac{2u}{u^2 + 1} \, du \]

Now, change the integration limits:
When \(x = 0\), \(u = \sqrt{e^0 - 1} = 0\).
When \(x = \ln 2\), \(u = \sqrt{e^{\ln 2} - 1} = \sqrt{2 - 1} = 1\).

Substitute into the integral:
\[ \int_{0}^{\ln 2} \sqrt{e^x - 1} \, dx = \int_{0}^{1} u \cdot \frac{2u}{u^2 + 1} \, du = \int_{0}^{1} \frac{2u^2}{u^2 + 1} \, du \]
\[ = 2 \int_{0}^{1} \left(1 - \frac{1}{u^2 + 1}\right) \, du \]
\[ = 2 \left[ u - \arctan u \right]_{0}^{1} \]
\[ = 2 \left( (1 - \arctan 1) - (0) \right) = 2 \left(1 - \frac{\pi}{4}\right) = 2 - \frac{\pi}{2} \]
This completes the proof.

(ii) We use the same substitution \(u = \sqrt{e^x - 1}\), where \(e^{-x} = \frac{1}{u^2 + 1}\) and \(dx = \frac{2u}{u^2 + 1} \, du\):
\[ \int_{0}^{\ln 2} e^{-x} \sqrt{e^x - 1} \, dx = \int_{0}^{1} \left(\frac{1}{u^2 + 1}\right) u \left(\frac{2u}{u^2 + 1}\right) \, du \]
\[ = \int_{0}^{1} \frac{2u^2}{(u^2 + 1)^2} \, du \]

To integrate this, we use integration by parts. Let:
\(p = u \implies dp = du\)
\(dq = \frac{2u}{(u^2 + 1)^2} \, du \implies q = -\frac{1}{u^2 + 1}\)

Applying the integration by parts formula \(\int p \, dq = pq - \int q \, dp\):
\[ \int \frac{2u^2}{(u^2 + 1)^2} \, du = -\frac{u}{u^2 + 1} - \int \left(-\frac{1}{u^2 + 1}\right) \, du \]
\[ = -\frac{u}{u^2 + 1} + \arctan u \]

Evaluating between the limits \(0\) and \(1\):
\[ \left[ -\frac{u}{u^2 + 1} + \arctan u \right]_{0}^{1} = \left(-\frac{1}{2} + \arctan 1\right) - (0) \]
\[ = -\frac{1}{2} + \frac{\pi}{4} = \frac{\pi}{4} - \frac{1}{2} \]

(i)
M1: Substitutes \(u = \sqrt{e^x - 1}\) and obtains \(dx = \frac{2u}{u^2 + 1} \, du\).
M1: Correctly transforms the limits to \(0\) and \(1\).
A1: Obtains the integrand in terms of \(u\): \(\frac{2u^2}{u^2 + 1}\).
M1: Integrates by rewriting \(\frac{2u^2}{u^2 + 1}\) as \(2\left(1 - \frac{1}{u^2+1}\right)\).
A1: Obtains the correct final answer \(2 - \frac{\pi}{2}\).

(ii)
M1: Sets up the integral with the substitution, obtaining \(\int_{0}^{1} \frac{2u^2}{(u^2+1)^2} \, du\).
M1: Applies integration by parts with appropriate choice of parts.
A1: Correctly evaluates \(q = -\frac{1}{u^2+1}\).
A1: Obtains the correct antiderivative \(-\frac{u}{u^2+1} + \arctan u\).
M1: Substitutes limits correctly.
A0.7: Obtains the exact final answer \(\frac{\pi}{4} - \frac{1}{2}\).

First, find the complementary function (CF) by solving the auxiliary equation:
\[ m^2 + 4m + 5 = 0 \]
\[ (m + 2)^2 + 1 = 0 \implies m = -2 \pm i \]
So the CF is:
\[ y_{CF} = e^{-2x}(A \cos x + B \sin x) \]

Next, find the particular integral (PI). Since the right-hand side is \(10e^{-x}\), we try:
\[ y_{PI} = C e^{-x} \]
\[ \frac{dy_{PI}}{dx} = -C e^{-x} \]
\[ \frac{d^2 y_{PI}}{dx^2} = C e^{-x} \]
Substituting these into the differential equation:
\[ C e^{-x} + 4(-C e^{-x}) + 5(C e^{-x}) = 10e^{-x} \]
\[ (C - 4C + 5C)e^{-x} = 10e^{-x} \]
\[ 2C e^{-x} = 10e^{-x} \implies C = 5 \]
Thus, the PI is:
\[ y_{PI} = 5e^{-x} \]

The general solution is:
\[ y = e^{-2x}(A \cos x + B \sin x) + 5e^{-x} \]

Now apply the initial conditions to find the constants \(A\) and \(B\):
At \(x = 0\), \(y = 1\):
\[ 1 = e^{0}(A \cos 0 + B \sin 0) + 5e^{0} \]
\[ 1 = A + 5 \implies A = -4 \]

Differentiate the general solution to use the second condition:
\[ \frac{dy}{dx} = -2e^{-2x}(A \cos x + B \sin x) + e^{-2x}(-A \sin x + B \cos x) - 5e^{-x} \]
At \(x = 0\), \(\frac{dy}{dx} = 0\):
\[ 0 = -2(A) + B - 5 \]
Substitute \(A = -4\):
\[ 0 = -2(-4) + B - 5 \]
\[ 0 = 8 + B - 5 \implies B = -3 \]

Thus, the particular solution is:
\[ y = e^{-2x}(-4 \cos x - 3 \sin x) + 5e^{-x} \]

M1: Forms and attempts to solve auxiliary equation.
A1: Correct roots \(-2 \pm i\).
A1: Correct CF \(e^{-2x}(A \cos x + B \sin x)\).
M1: Forms appropriate PI form \(y_{PI} = C e^{-x}\), differentiates, and substitutes.
A1: Correctly finds \(C = 5\).
A1: States general solution.
M1: Applies \(y = 1\) when \(x = 0\) to find \(A\).
A1: Correctly obtains \(A = -4\).
M1: Differentiates general solution correctly (requires product rule).
A1: Applies \(\frac{dy}{dx} = 0\) when \(x = 0\) to find \(B\).
A0.7: Correct particular solution: \(y = e^{-2x}(-4 \cos x - 3 \sin x) + 5e^{-x}\).

(i) Combining the terms on the right-hand side over a common denominator:
\[ \frac{1}{r^2} - \frac{1}{(r+1)^2} = \frac{(r+1)^2 - r^2}{r^2(r+1)^2} \]
\[ = \frac{(r^2 + 2r + 1) - r^2}{r^2(r+1)^2} = \frac{2r+1}{r^2(r+1)^2} \]
This completes the proof.

(ii) Using the identity from part (i) and the method of differences:
\[ \sum_{r=1}^{n} \frac{2r+1}{r^2(r+1)^2} = \sum_{r=1}^{n} \left( \frac{1}{r^2} - \frac{1}{(r+1)^2} \right) \]
Listing the first few terms and the last term:
\[ r=1: \quad \frac{1}{1^2} - \frac{1}{2^2} \]
\[ r=2: \quad \frac{1}{2^2} - \frac{1}{3^2} \]
\[ r=3: \quad \frac{1}{3^2} - \frac{1}{4^2} \]
\[ \dots \]
\[ r=n: \quad \frac{1}{n^2} - \frac{1}{(n+1)^2} \]
Summing these, all intermediate terms cancel, leaving:
\[ \sum_{r=1}^{n} \frac{2r+1}{r^2(r+1)^2} = 1 - \frac{1}{(n+1)^2} \]

(iii) As \(n \to \infty\), the term \(\frac{1}{(n+1)^2} \to 0\).
Therefore, the sum to infinity is:
\[ \sum_{r=1}^{\infty} \frac{2r+1}{r^2(r+1)^2} = 1 \]

(iv) We want to find the least value of \(N\) such that:
\[ \sum_{r=N}^{\infty} \frac{2r+1}{r^2(r+1)^2} < 0.001 \]
The sum from \(r=N\) to \(\infty\) is given by:
\[ \sum_{r=N}^{\infty} \left( \frac{1}{r^2} - \frac{1}{(r+1)^2} \right) = \frac{1}{N^2} \]
We require:
\[ \frac{1}{N^2} < 0.001 \implies N^2 > 1000 \]
Since \(\sqrt{1000} \approx 31.62\), and \(N\) must be a positive integer:
\[ N \ge 32 \]
Thus, the least value of \(N\) is \(32\).

(i)
M1: Expresses \(\frac{1}{r^2} - \frac{1}{(r+1)^2}\) with a common denominator.
A1: Correctly simplifies numerator to show identity.

(ii)
M1: Writes out the sum using the difference form.
A1: Demonstrates cancelling of intermediate terms.
A1: Obtains the correct sum expression \(1 - \frac{1}{(n+1)^2}\).

(iii)
M1: Takes the limit as \(n \to \infty\).
A0.7: Correctly states sum to infinity is \(1\).

(iv)
M1: Obtains the sum from \(N\) to infinity as \(\frac{1}{N^2}\).
M1: Sets up the inequality \(\frac{1}{N^2} < 0.001\).
A1: Solves to find \(N > 31.62\).
A1: States the least integer value \(N = 32\).

(i) From the given cubic equation \(x^3 - 3x^2 + 5x - 2 = 0\), we have:
\(\Sigma_1 = \alpha + \beta + \gamma = 3\)
\(\Sigma_2 = \alpha\beta + \beta\gamma + \gamma\alpha = 5\)
\(\Sigma_3 = \alpha\beta\gamma = 2\)

(a) \(\alpha + \beta + \gamma = 3\).

(b) Using the identity:
\[ \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) \]
\[ = 3^2 - 2(5) = 9 - 10 = -1 \]

(c) Since \(\alpha\), \(\beta\), and \(\gamma\) are roots of \(x^3 - 3x^2 + 5x - 2 = 0\), they satisfy:
\(\alpha^3 - 3\alpha^2 + 5\alpha - 2 = 0\)
\(\beta^3 - 3\beta^2 + 5\beta - 2 = 0\)
\(\gamma^3 - 3\gamma^2 + 5\gamma - 2 = 0\)
Summing these equations yields:
\[ (\alpha^3 + \beta^3 + \gamma^3) - 3(\alpha^2 + \beta^2 + \gamma^2) + 5(\alpha + \beta + \gamma) - 6 = 0 \]
Substitute the values from (a) and (b):
\[ (\alpha^3 + \beta^3 + \gamma^3) - 3(-1) + 5(3) - 6 = 0 \]
\[ (\alpha^3 + \beta^3 + \gamma^3) + 3 + 15 - 6 = 0 \]
\[ \alpha^3 + \beta^3 + \gamma^3 + 12 = 0 \implies \alpha^3 + \beta^3 + \gamma^3 = -12 \]

(ii) Let \(y = x^2\), which means \(x = \sqrt{y}\). Substitute this into the original cubic equation:
\[ (\sqrt{y})^3 - 3(\sqrt{y})^2 + 5\sqrt{y} - 2 = 0 \]
\[ y\sqrt{y} - 3y + 5\sqrt{y} - 2 = 0 \]
Rearrange to isolate the square root terms:
\[ \sqrt{y}(y + 5) = 3y + 2 \]
Square both sides of the equation:
\[ y(y + 5)^2 = (3y + 2)^2 \]
\[ y(y^2 + 10y + 25) = 9y^2 + 12y + 4 \]
\[ y^3 + 10y^2 + 25y = 9y^2 + 12y + 4 \]
Rearrange into standard form:
\[ y^3 + y^2 + 13y - 4 = 0 \]

(i)
B1: States \(\alpha + \beta + \gamma = 3\).
M1: Uses identity \(\alpha^2 + \beta^2 + \gamma^2 = (\Sigma_1)^2 - 2\Sigma_2\).
A1: Correctly evaluates \(\alpha^2 + \beta^2 + \gamma^2 = -1\).
M1: Sets up the linear relation using the original cubic equation for the sum of cubes.
A1.7: Correctly evaluates \(\alpha^3 + \beta^3 + \gamma^3 = -12\).

(ii)
M1: Uses substitution \(y = x^2\) or alternative roots method.
M1: Rearranges equation to separate radical terms.
M1: Squares both sides of the equation.
A1: Expands correctly.
A1: Obtains the correct cubic equation \(y^3 + y^2 + 13y - 4 = 0\) (or any integer multiple thereof).

To find the surface area of revolution about the \(x\)-axis, we use the formula \(S = 2\pi \int_{a}^{b} y \sqrt{1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2} \mathrm{d}x\). First, differentiate \(y = \cosh x\) with respect to \(x\) to get \(\frac{\mathrm{d}y}{\mathrm{d}x} = \sinh x\). Then \(\sqrt{1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2} = \sqrt{1 + \sinh^2 x} = \cosh x\). The integral becomes \(S = 2\pi \int_{0}^{\ln 3} \cosh^2 x \mathrm{d}x\). Using the identity \(\cosh^2 x = \frac{1}{2}(\cosh 2x + 1)\), we get \(S = \pi \int_{0}^{\ln 3} (\cosh 2x + 1) \mathrm{d}x = \pi \left[ \frac{1}{2}\sinh 2x + x \right]_{0}^{\ln 3}\). Evaluating this at the upper limit \(x = \ln 3\): \(\sinh(2\ln 3) = \sinh(\ln 9) = \frac{e^{\ln 9} - e^{-\ln 9}}{2} = \frac{9 - 1/9}{2} = \frac{40}{9}\). Thus \(\frac{1}{2}\sinh(2\ln 3) = \frac{20}{9}\). At the lower limit \(x = 0\), the value is \(0\). Therefore, the surface area is \(S = \pi \left( \frac{20}{9} + \ln 3 \right)\).

M1: For finding \(\frac{\mathrm{d}y}{\mathrm{d}x} = \sinh x\) and setting up the surface area integral. A1: For obtaining \(S = 2\pi \int_{0}^{\ln 3} \cosh^2 x \mathrm{d}x\). M1: For using the identity \(\cosh^2 x = \frac{1}{2}(\cosh 2x + 1)\) to integrate. A1: For the correct integration: \(\pi \left[ \frac{1}{2}\sinh 2x + x \right]\). M1: For substituting the limit \(\ln 3\) and calculating \(\sinh(2\ln 3)\). A1: For completing the proof to show \(\pi \left( \frac{20}{9} + \ln 3 \right)\).

First, we find the complementary function (CF) by solving the auxiliary equation: \(m^2 - 4m + 4 = 0 \implies (m-2)^2 = 0\), which has a repeated root \(m = 2\). Thus, the CF is \(y_{CF} = (Ax + B)e^{2x}\). For the particular integral (PI), since both \(e^{2x}\) and \(xe^{2x}\) are terms in the CF, we try a solution of the form \(y_{PI} = C x^2 e^{2x}\). Differentiating, we find: \(\frac{\mathrm{d}y}{\mathrm{d}x} = C(2x e^{2x} + 2x^2 e^{2x})\) and \(\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = C(2e^{2x} + 8xe^{2x} + 4x^2e^{2x})\). Substituting these into the differential equation gives: \(C(2e^{2x} + 8xe^{2x} + 4x^2e^{2x}) - 4C(2xe^{2x} + 2x^2e^{2x}) + 4C x^2 e^{2x} = 8e^{2x}\). Simplifying this, we get: \(2C e^{2x} = 8e^{2x} \implies C = 4\). So, the general solution is \(y = (4x^2 + Ax + B)e^{2x}\). To find the particular solution, we use the initial conditions. At \(x = 0, y = 1 \implies B = 1\). Differentiating the general solution: \(\frac{\mathrm{d}y}{\mathrm{d}x} = (8x + A)e^{2x} + 2(4x^2 + Ax + B)e^{2x}\). At \(x = 0, \frac{\mathrm{d}y}{\mathrm{d}x} = A + 2B = 3 \implies A + 2(1) = 3 \implies A = 1\). Thus, the particular solution is \(y = (4x^2 + x + 1)e^{2x}\).

M1: For solving the auxiliary equation. A1: For the correct CF: \(y_{CF} = (Ax + B)e^{2x}\). M1: For trying a PI of the form \(y_{PI} = Cx^2e^{2x}\). A1: For finding \(C = 4\). A1: For the correct general solution. M1: For applying \(y(0)=1\) to find \(B\). M1: For differentiating the general solution and applying the second boundary condition to find \(A\). A1: For the correct particular solution.

(a) The area \(A\) is given by \(A = \frac{1}{2} \int_{0}^{\pi/2} r^2 \mathrm{d}\theta = \frac{1}{2} \int_{0}^{\pi/2} 4(1 + \cos\theta)^2 \mathrm{d}\theta = 2 \int_{0}^{\pi/2} (1 + 2\cos\theta + \cos^2\theta) \mathrm{d}\theta\). Using the identity \(\cos^2\theta = \frac{1}{2}(1+\cos 2\theta)\), we have: \(A = 2 \int_{0}^{\pi/2} \left(\frac{3}{2} + 2\cos\theta + \frac{1}{2}\cos 2\theta\right) \mathrm{d}\theta = 2 \left[ \frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta \right]_{0}^{\pi/2} = 2 \left( \frac{3\pi}{4} + 2 + 0 - 0 \right) = \frac{3\pi}{2} + 4\). (b) For the tangent to be parallel to the initial line, \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = 0\). Since \(y = r\sin\theta = 2(1+\cos\theta)\sin\theta = 2\sin\theta + \sin 2\theta\), we have \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = 2\cos\theta + 2\cos 2\theta\). Setting this to 0: \(\cos\theta + 2\cos^2\theta - 1 = 0 \implies (2\cos\theta - 1)(\cos\theta + 1) = 0\). In the interval \(0 \le \theta \le \pi\), this gives \(\cos\theta = \frac{1}{2} \implies \theta = \frac{\pi}{3}\) (or \(\cos\theta = -1 \implies \theta = \pi\), which is the pole). At \(\theta = \frac{\pi}{3}\), \(r = 2(1 + 1/2) = 3\). The Cartesian coordinates are \(x = r\cos\theta = 3\cos(\pi/3) = \frac{3}{2}\) and \(y = r\sin\theta = 3\sin(\pi/3) = \frac{3\sqrt{3}}{2}\).

M1: For setting up the area integral with correct limits. M1: For using the identity for \(\cos^2\theta\). A1: For correct integration. A1: For obtaining \(\frac{3\pi}{2} + 4\). M1: For writing \(y = r\sin\theta\) and differentiating. A1: For solving \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = 0\) to get \(\theta = \frac{\pi}{3}\). M1: For finding the corresponding value of \(r\). A1: For finding Cartesian coordinates \(\left(\frac{3}{2}, \frac{3\sqrt{3}}{2}\right)\).

(a) By algebraic division, we rewrite the equation as: \(y = \frac{x^2 + x + 2}{x - 1} = x + 2 + \frac{4}{x - 1}\). As \(x \to 1\), \(y \to \infty\), so the vertical asymptote is \(x = 1\). As \(x \to \pm\infty\), \(\frac{4}{x-1} \to 0\), so the oblique asymptote is \(y = x + 2\). (b) To find the stationary points, we differentiate \(y = x + 2 + 4(x-1)^{-1}\): \(\frac{\mathrm{d}y}{\mathrm{d}x} = 1 - \frac{4}{(x-1)^2}\). Setting this to 0: \((x-1)^2 = 4 \implies x - 1 = \pm 2 \implies x = 3\) or \(x = -1\). If \(x = 3\), \(y = 3 + 2 + \frac{4}{2} = 7\). If \(x = -1\), \(y = -1 + 2 + \frac{4}{-2} = -1\). To determine the nature, we find the second derivative: \(\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = \frac{8}{(x-1)^3}\). For \(x = 3\), \(\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = 1 > 0 \implies (3, 7)\) is a local minimum. For \(x = -1\), \(\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = -1 < 0 \implies (-1, -1)\) is a local maximum.

M1: For rewriting \(y\) in the form \(ax+b+\frac{c}{x-d}\) by division. A1: For vertical asymptote \(x = 1\). A1: For oblique asymptote \(y = x + 2\). M1: For finding \(\frac{\mathrm{d}y}{\mathrm{d}x}\). A1: For setting to 0 and finding \(x = 3, -1\). A1: For obtaining coordinates \((3, 7)\) and \((-1, -1)\). M1: For checking the second derivative. A1: For correct determination of nature.

First, we find the eigenvalues of \(A\) by solving \(\det(A - \lambda I) = 0\): \(\begin{vmatrix} 1-\lambda & 0 & 2 \\ 0 & 3-\lambda & 0 \\ 2 & 0 & 1-\lambda \end{vmatrix} = (3-\lambda)[(1-\lambda)^2 - 4] = (3-\lambda)(\lambda^2-2\lambda-3) = -(\lambda-3)^2(\lambda+1) = 0\). The eigenvalues are \(\lambda = -1\) and \(\lambda = 3\) (with multiplicity 2). For \(\lambda = -1\), we solve \((A + I)\mathbf{v} = \mathbf{0}\): \(\begin{pmatrix} 2 & 0 & 2 \\ 0 & 4 & 0 \\ 2 & 0 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies y = 0, x+z=0\). An eigenvector is \(\begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}\). Normalizing, we get \(\mathbf{u}_1 = \begin{pmatrix} 1/\sqrt{2} \\ 0 \\ -1/\sqrt{2} \end{pmatrix}\). For \(\lambda = 3\), we solve \((A - 3I)\mathbf{v} = \mathbf{0}\): \(\begin{pmatrix} -2 & 0 & 2 \\ 0 & 0 & 0 \\ 2 & 0 & -2 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies x - z = 0\). We can choose two mutually orthogonal unit vectors in this eigenspace, for example: \(\mathbf{u}_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\) and \(\mathbf{u}_3 = \begin{pmatrix} 1/\sqrt{2} \\ 0 \\ 1/\sqrt{2} \end{pmatrix}\). These are also orthogonal to \(\mathbf{u}_1\). Thus, we can define \(P = \begin{pmatrix} 1/\sqrt{2} & 0 & 1/\sqrt{2} \\ 0 & 1 & 0 \\ -1/\sqrt{2} & 0 & 1/\sqrt{2} \end{pmatrix}\) and \(D = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{pmatrix}\).

M1: For setting up characteristic equation. A1: For eigenvalues \(\lambda = -1, 3, 3\). M1: For finding the eigenvector for \(\lambda = -1\). A1: For normalizing this eigenvector. M1: For finding two orthogonal eigenvectors for \(\lambda = 3\). A1: For normalizing both eigenvectors. M1: For constructing \(P\) and \(D\). A1: For correct \(P\) and \(D\) matching each other.

We express the hyperbolic functions in terms of exponential functions: \(\cosh x = \frac{e^x + e^{-x}}{2}\) and \(\sinh x = \frac{e^x - e^{-x}}{2}\). Substituting these into the equation, we get: \(3 \left( \frac{e^x + e^{-x}}{2} \right) - 2 \left( \frac{e^x - e^{-x}}{2} \right) = 3\). Multiplying both sides by 2: \(3(e^x + e^{-x}) - 2(e^x - e^{-x}) = 6 \implies e^x + 5e^{-x} = 6\). Multiplying both sides by \(e^x\), we obtain a quadratic equation in terms of \(e^x\): \(e^{2x} - 6e^x + 5 = 0\). Factoring this quadratic, we get: \((e^x - 1)(e^x - 5) = 0\). Therefore, \(e^x = 1 \implies x = \ln 1 = 0\), or \(e^x = 5 \implies x = \ln 5\).

M1: For expressing \(\cosh x\) and \(\sinh x\) in exponentials. A1: For simplifying to get \(e^x + 5e^{-x} = 6\). M1: For converting to a quadratic equation in \(e^x\). A1: For the correct quadratic equation. M1: For solving the quadratic equation. A1: For finding \(x = 0\) and \(x = \ln 5\).

(a) We apply integration by parts to \(I_n = \int_{0}^{1} x^n e^{-x} \mathrm{d}x\) by setting \(u = x^n\) and \(\frac{\mathrm{d}v}{\mathrm{d}x} = e^{-x}\). Then \(\frac{\mathrm{d}u}{\mathrm{d}x} = n x^{n-1}\) and \(v = -e^{-x}\). Thus, \(I_n = \left[ -x^n e^{-x} \right]_{0}^{1} - \int_{0}^{1} \left( -n x^{n-1} e^{-x} \right) \mathrm{d}x = -1^n e^{-1} - 0 + n \int_{0}^{1} x^{n-1} e^{-x} \mathrm{d}x = n I_{n-1} - e^{-1}\). (b) First, compute \(I_0\): \(I_0 = \int_{0}^{1} e^{-x} \mathrm{d}x = \left[ -e^{-x} \right]_{0}^{1} = 1 - e^{-1}\). Now, apply the reduction formula step-by-step: \(I_1 = 1 \cdot I_0 - e^{-1} = 1 - e^{-1} - e^{-1} = 1 - 2e^{-1}\); \(I_2 = 2 \cdot I_1 - e^{-1} = 2(1 - 2e^{-1}) - e^{-1} = 2 - 5e^{-1}\); \(I_3 = 3 \cdot I_2 - e^{-1} = 3(2 - 5e^{-1}) - e^{-1} = 6 - 15e^{-1} - e^{-1} = 6 - 16e^{-1}\).

M1: For applying integration by parts. A1: For correct intermediate boundary term and integral term. A1: For establishing the reduction formula. M1: For calculating \(I_0\). M1: For finding \(I_1\). A1: For finding \(I_2\). M1: For finding \(I_3\). A1: For the final exact answer \(6 - 16e^{-1}\).

Let \(w = -8 + 8\sqrt{3}\mathrm{i}\). First, express \(w\) in polar form. The modulus of \(w\) is \(|w| = \sqrt{(-8)^2 + (8\sqrt{3})^2} = \sqrt{64 + 192} = \sqrt{256} = 16\). The argument of \(w\) is in the second quadrant: \(\arg(w) = \pi - \arctan\left(\frac{8\sqrt{3}}{8}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}\). Thus, \(w = 16 e^{\mathrm{i} \left( \frac{2\pi}{3} + 2k\pi \right)}\). By de Moivre's theorem, \(z = w^{1/4} = 2 e^{\mathrm{i} \left( \frac{\pi}{6} + \frac{k\pi}{2} \right)}\) for \(k = 0, 1, 2, 3\). We evaluate for each value of \(k\): For \(k = 0\): \(z_0 = 2 \left( \cos\frac{\pi}{6} + \mathrm{i}\sin\frac{\pi}{6} \right) = 2 \left( \frac{\sqrt{3}}{2} + \frac{1}{2}\mathrm{i} \right) = \sqrt{3} + \mathrm{i}\). For \(k = 1\): \(z_1 = 2 \left( \cos\frac{2\pi}{3} + \mathrm{i}\sin\frac{2\pi}{3} \right) = 2 \left( -\frac{1}{2} + \frac{\sqrt{3}}{2}\mathrm{i} \right) = -1 + \sqrt{3}\mathrm{i}\). For \(k = 2\): \(z_2 = 2 \left( \cos\frac{7\pi}{6} + \mathrm{i}\sin\frac{7\pi}{6} \right) = 2 \left( -\frac{\sqrt{3}}{2} - \frac{1}{2}\mathrm{i} \right) = -\sqrt{3} - \mathrm{i}\). For \(k = 3\): \(z_3 = 2 \left( \cos\frac{5\pi}{3} + \mathrm{i}\sin\frac{5\pi}{3} \right) = 2 \left( \frac{1}{2} - \frac{\sqrt{3}}{2}\mathrm{i} \right) = 1 - \sqrt{3}\mathrm{i}\).

M1: For finding modulus is 16. A1: For finding argument is \(\frac{2\pi}{3}\). M1: For taking the 4th root of the modulus to get 2. M1: For obtaining the argument of the roots as \(\frac{\pi}{6} + \frac{k\pi}{2}\). A1: For listing the four arguments. M1: For converting to Cartesian form. A1: For two correct Cartesian roots. A1: For the remaining two correct Cartesian roots.

The equation of the trajectory of the projectile is given by: \(y = x \tan\alpha - \frac{g x^2}{2 u^2} (1 + \tan^2\alpha)\). Substituting \(g = 10\), \(u = 20\), and \(x = 20\), we obtain: \(y = 20 \tan\alpha - \frac{10 \times 400}{2 \times 400} (1 + \tan^2\alpha) = 20 \tan\alpha - 5(1 + \tan^2\alpha)\). For the projectile to pass over the wall, we require \(y \ge 5\) when \(x = 20\): \(20 \tan\alpha - 5 - 5 \tan^2\alpha \ge 5\), which simplifies to \(5 \tan^2\alpha - 20 \tan\alpha + 10 \le 0\). Dividing by 5 yields \(\tan^2\alpha - 4 \tan\alpha + 2 \le 0\). Solving the quadratic equation \(\tan^2\alpha - 4 \tan\alpha + 2 = 0\) using the quadratic formula: \(\tan\alpha = \frac{4 \pm \sqrt{16 - 8}}{2} = 2 \pm \sqrt{2}\). Therefore, the range of possible values for \(\tan\alpha\) is \(2 - \sqrt{2} \le \tan\alpha \le 2 + \sqrt{2}\).

M1: For attempting to use the trajectory equation with given values of \(u\), \(x\), and \(g\). A1: For obtaining a correct quadratic expression in terms of \(\tan\alpha\). M1: For setting up the inequality \(y \ge 5\). A1: For obtaining the simplified inequality \(\tan^2\alpha - 4 \tan\alpha + 2 \le 0\). M1: For solving the quadratic equation. A1: For writing the final correct inequality range.

Let \(l = 1.2\text{ m}\) be the natural length and \(x = 0.8\text{ m}\) be the maximum extension of the string. The total distance fallen by the particle from rest to the point of maximum extension is \(d = l + x = 1.2 + 0.8 = 2.0\text{ m}\). By conservation of energy, the loss in gravitational potential energy of the particle is equal to the gain in elastic potential energy of the string: \(m g d = \frac{\lambda x^2}{2 l}\). Substituting the known values: \(0.5 \times 10 \times 2.0 = \frac{\lambda \times 0.8^2}{2 \times 1.2}\), which gives \(10 = \frac{0.64 \lambda}{2.4}\). Simplifying the fraction: \(10 = \frac{4}{15} \lambda\), hence \(\lambda = \frac{150}{4} = 37.5\).

M1: For finding the total vertical distance fallen \(d = 2.0\text{ m}\). A1: For correctly expressing the loss in gravitational potential energy as \(10\text{ J}\). M1: For attempting to use the formula for elastic potential energy \(\frac{\lambda x^2}{2l}\). A1: For correctly expressing the gain in elastic potential energy as \(\frac{4}{15}\lambda\). M1: For equating loss in GPE to gain in EPE. A1: For obtaining the final correct value \(\lambda = 37.5\).

Let \(\theta = 60^\circ\) be the angle between \(OP\) and the upward vertical at the instant the string becomes slack. The tension in the string is \(T = 0\). Resolving the forces along the radial direction: \(T + m g \cos\theta = \frac{m v^2}{r}\), where \(v\) is the speed of the particle at this instant and \(r = 1.4\text{ m}\). Since \(T = 0\), we have \(m g \cos 60^\circ = \frac{m v^2}{r}\), which simplifies to \(v^2 = g r \cos 60^\circ = 10 \times 1.4 \times 0.5 = 7\). Next, we use conservation of energy between the point of projection and the slack point. The height above the lowest point is \(h = r + r \cos 60^\circ = 1.4 (1 + 0.5) = 2.1\text{ m}\). By energy conservation: \(\frac{1}{2} m u^2 = \frac{1}{2} m v^2 + m g h\), which gives \(u^2 = v^2 + 2 g h\). Substituting the known values: \(u^2 = 7 + 2 \times 10 \times 2.1 = 7 + 42 = 49\). Thus, \(u = 7\).

M1: For resolving forces radially with \(T = 0\). A1: For deriving \(v^2 = g r \cos\theta\). A1: For calculating the correct value \(v^2 = 7\). M1: For expressing the height in terms of the angle: \(h = r(1 + \cos 60^\circ)\). A1: For obtaining the correct height \(2.1\text{ m}\). M1: For using conservation of energy to link \(u\) and \(v\). A1: For obtaining the correct speed \(u = 7\).

By Newton's second law, \(m \frac{\mathrm{d}v}{\mathrm{d}t} = -R\). Given \(m = 0.5\) and \(R = 2v + 1\), we have: \(0.5 \frac{\mathrm{d}v}{\mathrm{d}t} = -(2v + 1)\), which can be rewritten as \(\frac{\mathrm{d}v}{\mathrm{d}t} = -2(2v + 1)\). Separating the variables gives: \(\frac{\mathrm{d}v}{2v + 1} = -2 \mathrm{d}t\). Integrating both sides from \(t=0\) (where \(v=4\)) to \(t\) (where \(v=1\)): \(\int_{4}^{1} \frac{1}{2v + 1} \mathrm{d}v = \int_{0}^{t} -2 \mathrm{d}t\). Evaluating the integrals: \(\left[ \frac{1}{2} \ln(2v + 1) \right]_{4}^{1} = -2t\), which becomes \(\frac{1}{2} (\ln 3 - \ln 9) = -2t\). Since \(\ln 9 = 2 \ln 3\), this simplifies to \(\frac{1}{2} (-\ln 3) = -2t\), which yields \(t = \frac{1}{4} \ln 3\).

M1: For applying Newton's second law to set up the differential equation. A1: For obtaining the correct differential equation \(\frac{\mathrm{d}v}{\mathrm{d}t} = -2(2v+1)\). M1: For separating variables and integrating. A1: For obtaining a correct logarithmic integral term \(\frac{1}{2}\ln(2v+1)\). M1: For applying the limits 4 and 1. A1: For simplifying the expression to find \(t\) explicitly. A1: For obtaining the exact answer \(\frac{1}{4}\ln 3\) (or equivalent).

Let \(R\) and \(F\) represent the normal reaction and friction force at the ground, and \(N\) be the normal reaction of the wall. Resolving forces vertically: \(R = (20 + 80)g = 100g = 1000\text{ N}\). Resolving forces horizontally: \(F = N\). Taking moments about the base of the ladder (where it contacts the ground) when the man is at the top: \(N \times 6\sin\theta = 20g \times 3\cos\theta + 80g \times 6\cos\theta\). Substituting \(g = 10\): \(6 N \sin\theta = 600 \cos\theta + 4800 \cos\theta = 5400 \cos\theta\). Dividing both sides by \(6 \cos\theta\) yields: \(N \tan\theta = 900\), so \(N = \frac{900}{\tan\theta}\). For no slipping, the friction force must satisfy \(F \le \mu R\), which means \(N \le 0.375 \times 1000 = 375\). Substituting \(N\): \(\frac{900}{\tan\theta} \le 375 \implies \tan\theta \ge \frac{900}{375} = 2.4\). Thus, the minimum value of \(\tan\theta\) is 2.4.

M1: For resolving forces vertically to find \(R\). A1: For finding \(R = 1000\text{ N}\). M1: For taking moments about the base of the ladder. A1: For obtaining the correct moment equation: \(6N\sin\theta = 5400\cos\theta\). M1: For utilizing the condition for no slipping \(F \le \mu R\). A1: For solving the inequality to find \(\tan\theta\). A1: For obtaining the correct minimum value of 2.4.

Let \(u\) be the speed of the sphere immediately before the collision. The component of the velocity parallel to the wall is \(u_{\parallel} = u \cos 60^\circ = 0.5u\), and the component perpendicular to the wall is \(u_{\perp} = u \sin 60^\circ = \frac{\sqrt{3}}{2} u\). After the impact, the parallel component of the velocity remains unchanged: \(v_{\parallel} = 0.5u\). The perpendicular component is reversed and scaled by the coefficient of restitution: \(v_{\perp} = e u_{\perp} = 0.5 \times \frac{\sqrt{3}}{2} u = \frac{\sqrt{3}}{4} u\). The initial kinetic energy is \(E_i = \frac{1}{2} m u^2\). The final kinetic energy is \(E_f = \frac{1}{2} m (v_{\parallel}^2 + v_{\perp}^2) = \frac{1}{2} m \left( (0.5u)^2 + \left(\frac{\sqrt{3}}{4}u\right)^2 \right) = \frac{1}{2} m u^2 \left( \frac{1}{4} + \frac{3}{16} \right) = \frac{7}{16} E_i\). The fraction of the initial kinetic energy lost in the collision is \(1 - \frac{7}{16} = \frac{9}{16} = 0.5625\).

M1: For finding the components of the initial velocity parallel and perpendicular to the wall. A1: For correct components \(0.5u\) and \(\frac{\sqrt{3}}{2}u\). M1: For finding the components of the final velocity. A1: For correct final perpendicular component \(\frac{\sqrt{3}}{4}u\). M1: For seting up the expression for final kinetic energy. A1: For finding final kinetic energy as \(\frac{7}{16}\) of initial kinetic energy. A1: For finding the fractional loss as \(0.5625\) (or \(\frac{9}{16}\)).

Let \(u\) be the speed of projection and \(\theta\) be the angle of projection. The maximum height is given by \(H = \frac{u^2 \sin^2\theta}{2g}\), so \(\frac{u^2 \sin^2\theta}{20} = 7.2 \implies u^2 \sin^2\theta = 144\). The horizontal range is given by \(R = \frac{u^2 \sin 2\theta}{g} = \frac{2 u^2 \sin\theta \cos\theta}{10} = 38.4 \implies u^2 \sin\theta \cos\theta = 192\). Dividing the equation for \(H\) by the equation for \(R\) gives: \(\frac{u^2 \sin^2\theta}{2 u^2 \sin\theta \cos\theta} = \frac{144}{2 \times 192} \implies \frac{\tan\theta}{2} = \frac{144}{384} = \frac{3}{8} \implies \tan\theta = \frac{3}{4}\). Since \(\tan\theta = 0.75\), we find \(\sin\theta = 0.6\). Substituting this into the first equation: \(u^2 (0.6)^2 = 144 \implies 0.36 u^2 = 144 \implies u^2 = 400\). Taking the square root gives \(u = 20\text{ m s}^{-1}\).

M1: For setting up the formula for maximum height \(H\) with given values. A1: For obtaining \(u^2 \sin^2\theta = 144\). M1: For setting up the formula for horizontal range \(R\) with given values. A1: For obtaining \(u^2 \sin\theta \cos\theta = 192\). M1: For dividing the equations to find \(\tan\theta\). A1: For finding \(\tan\theta = 0.75\). A1: For solving to find \(u = 20\).

(i) For \(1 \le x \le 5\), the cumulative distribution function is \(F(x) = \int_1^x \frac{3}{32}(t-1)(5-t) dt = \frac{3}{32} \int_1^x (6t - t^2 - 5) dt = \frac{3}{32} \left[ 3t^2 - \frac{1}{3}t^3 - 5t \right]_1^x = \frac{3}{32} \left[ 3x^2 - \frac{1}{3}x^3 - 5x + \frac{7}{3} \right] = \frac{1}{32}(9x^2 - x^3 - 15x + 7)\). (ii) By the symmetry of the PDF about the line \(x = 3\), \(E(X) = 3\). Alternatively, by integration, \(E(X) = \int_1^5 x f(x) dx = \frac{3}{32} \int_1^5 (6x^2 - x^3 - 5x) dx = \frac{3}{32} \left[ 2x^3 - \frac{1}{4}x^4 - \frac{5}{2}x^2 \right]_1^5 = 3\). (iii) \(P(2 < X < 4) = F(4) - F(2) = \frac{1}{32}(9(16) - 64 - 15(4) + 7) - \frac{1}{32}(9(4) - 8 - 15(2) + 7) = \frac{27}{32} - \frac{5}{32} = \frac{22}{32} = \frac{11}{16} = 0.6875\).

M1: Set up integration of the PDF to find CDF. A1: Correct integration with correct limits or constant of integration. A1: Correct final expression for \(F(x)\). M1: Recognize symmetry or use \(E(X) = \int x f(x) dx\). A1: Correctly calculate \(E(X) = 3\). M1: Evaluate \(F(4) - F(2)\) or equivalent integral. A1: Correctly obtain \(F(4)\) and \(F(2)\). A1: Correct final probability of \(11/16\) or \(0.688\).

Let \(x = \text{before} - \text{after}\). We test H0: Median difference is 0 against H1: Median difference is greater than 0 (one-tailed test). We rank the absolute differences: 0.5 (Rank 1, +), -0.8 (Rank 2, -), -1.5 (Rank 3, -), 1.9 (Rank 4, +), 2.1 (Rank 5, +), -2.5 (Rank 6, -), 3.2 (Rank 7, +), 3.8 (Rank 8, +), 4.0 (Rank 9, +), 4.2 (Rank 10, +). Sum of positive ranks: \(W_+ = 1 + 4 + 5 + 7 + 8 + 9 + 10 = 44\). Sum of negative ranks: \(W_- = 2 + 3 + 6 = 11\). The test statistic is \(T = \min(W_+, W_-) = 11\). From the critical values table for \(n = 10\) at the 5% level (one-tailed), the critical value is 10. Since \(T = 11 > 10\), we do not reject H0. There is insufficient evidence at the 5% significance level to suggest that the energy drink significantly improves running times.

B1: State correct hypotheses (one-tailed). M1: Rank absolute values of differences. A1: Assign correct ranks. M1: Calculate sum of positive and negative ranks. A1: Identify test statistic \(T = 11\). B1: Correct critical value of 10. M1: Correct comparison (T > 10). A1: Conclude in context correctly.

We test H0: Median test score for Method A = Median test score for Method B against H1: Median test scores are different (two-tailed test). Combine and rank all 11 observations: 65 (B, Rank 1), 68 (A, Rank 2), 69 (B, Rank 3), 72 (B, Rank 4), 74 (A, Rank 5), 75 (B, Rank 6), 78 (B, Rank 7), 80 (B, Rank 8), 81 (A, Rank 9), 85 (A, Rank 10), 92 (A, Rank 11). Let \(n_1 = 5\) (Method A) and \(n_2 = 6\) (Method B). The rank sum for the smaller sample (Method A) is \(R_A = 2 + 5 + 9 + 10 + 11 = 37\). Alternatively, the Mann-Whitney statistic is \(U = n_1 n_2 + \frac{n_1(n_1+1)}{2} - R_A = 30 + 15 - 37 = 8\). For \(n_1 = 5, n_2 = 6\) at the 5% significance level (two-tailed), the critical values for the rank-sum \(R_1\) are 18 and 42 (non-rejection interval [18, 42]), or critical value for \(U\) is 3. Since \(18 \le 37 \le 42\) (or \(8 > 3\)), we do not reject H0. There is insufficient evidence to conclude that there is a difference in the median test scores.

B1: State correct hypotheses (two-tailed). M1: Order and rank all 11 scores. A1: Assign correct ranks. M1: Calculate sum of ranks for Method A (or Method B). A1: Correct rank sum \(R_A = 37\) (or \(R_B = 29\), or \(U = 8\)). B1: Correct critical values (18 and 42, or 3 for U). M1: Correct comparison of statistic with critical values. A1: Correct final conclusion in context.

We test H0: Sport type and injury frequency are independent against H1: There is an association. Row totals: Football: 80, Basketball: 70, Tennis: 50. Total = 200. Column totals: Low: 80, Medium: 75, High: 45. Total = 200. Expected frequencies \(E = \frac{\text{Row Total} \times \text{Col Total}}{200}\): Football: Low=32, Med=30, High=18; Basketball: Low=28, Med=26.25, High=15.75; Tennis: Low=20, Med=18.75, High=11.25. Test statistic: \(\chi^2 = \sum \frac{(O-E)^2}{E} = \frac{(25-32)^2}{32} + \frac{(35-30)^2}{30} + \frac{(20-18)^2}{18} + \frac{(30-28)^2}{28} + \frac{(25-26.25)^2}{26.25} + \frac{(15-15.75)^2}{15.75} + \frac{(25-20)^2}{20} + \frac{(15-18.75)^2}{18.75} + \frac{(10-11.25)^2}{11.25} = 1.53125 + 0.83333 + 0.22222 + 0.14286 + 0.05952 + 0.03571 + 1.25 + 0.75 + 0.13889 \approx 4.964\). Degrees of freedom: \(df = (3-1)(3-1) = 4\). Critical value of \(\chi^2\) with 4 df at 5% significance level is 9.488. Since \(4.964 < 9.488\), we do not reject H0. There is no significant association between sport type and injury frequency.

B1: Correctly state hypotheses. M1: Calculate expected frequencies (at least 3 correct). A1: Correctly calculate all expected frequencies. M1: Calculate \(\sum \frac{(O-E)^2}{E}\). A1: Correctly calculate \(\chi^2 \approx 4.96\). B1: Correct degrees of freedom (4) and critical value (9.488). M1: Compare test statistic with critical value. A1: Correct conclusion in context.

(i) Rewrite \(G_X(t) = t(3-2t)^{-2}\). Differentiating: \(G'_X(t) = (3-2t)^{-2} + 4t(3-2t)^{-3}\). Evaluation at \(t=1\) gives \(E(X) = G'_X(1) = 1^{-2} + 4(1)(1)^{-3} = 5\). (ii) Differentiating again: \(G''_X(t) = 8(3-2t)^{-3} + 24t(3-2t)^{-4}\). Evaluation at \(t=1\) gives \(G''_X(1) = 8 + 24 = 32\). Variance is \(\operatorname{Var}(X) = G''_X(1) + G'_X(1) - (G'_X(1))^2 = 32 + 5 - 25 = 12\). (iii) Since \(Y = X_1 + X_2\) and they are independent, \(G_Y(t) = [G_X(t)]^2 = \frac{t^2}{(3-2t)^4}\). Rewriting, \(G_Y(t) = t^2 \cdot 3^{-4} (1 - \frac{2}{3}t)^{-4} = \frac{t^2}{81} (1 - \frac{2}{3}t)^{-4}\). Expanding: \((1 - \frac{2}{3}t)^{-4} = 1 + (-4)(-\frac{2}{3}t) + \dots = 1 + \frac{8}{3}t + \dots\). Thus, \(G_Y(t) = \frac{t^2}{81} (1 + \frac{8}{3}t + \dots) = \frac{1}{81}t^2 + \frac{8}{243}t^3 + \dots\). The coefficient of \(t^3\) is \(\frac{8}{243}\).

M1: Differentiate \(G_X(t)\) with respect to \(t\). A1: Correctly find \(E(X) = 5\). M1: Differentiate a second time. A1: Correct value of \(G''_X(1) = 32\). M1: Use correct variance formula. A1: Correctly obtain \(\operatorname{Var}(X) = 12\). B1: State correct PGF \(G_Y(t) = \frac{t^2}{(3-2t)^4}\). M1: Expand to find coefficient of \(t^3\). A1: Correct coefficient \(\frac{8}{243}\) (or 0.0329).

We test H0: \(\mu_1 = \mu_2\) against H1: \(\mu_1 < \mu_2\) (one-tailed). Sample means: \(\bar{x}_1 = \frac{144}{8} = 18\) and \(\bar{x}_2 = \frac{210}{10} = 21\). Unbiased estimates of the variances: \(s_1^2 = \frac{1}{7}\left(2652 - \frac{144^2}{8}\right) = \frac{60}{7}\) and \(s_2^2 = \frac{1}{9}\left(4510 - \frac{210^2}{10}\right) = \frac{100}{9}\). Pooled estimate of variance: \(s_p^2 = \frac{7(s_1^2) + 9(s_2^2)}{8 + 10 - 2} = \frac{60 + 100}{16} = 10\). Standard error of difference: \(SE = \sqrt{s_p^2\left(\frac{1}{n_1} + \frac{1}{n_2}\right)} = \sqrt{10\left(\frac{1}{8} + \frac{1}{10}\right)} = \sqrt{2.25} = 1.5\). Test statistic: \(t = \frac{\bar{x}_1 - \bar{x}_2}{SE} = \frac{18 - 21}{1.5} = -2\). Degrees of freedom: \(v = 8 + 10 - 2 = 16\). The critical value of \(t\) for a one-tailed test with 16 df at the 5% significance level is \(-1.746\). Since \(t = -2 < -1.746\), we reject H0. There is significant evidence that the mean time for Group I is shorter than for Group II.

B1: Correctly state hypotheses. M1: Calculate unbiased estimates of variance \(s_1^2\) and \(s_2^2\). A1: Correctly obtain \(s_1^2 = 60/7\) and \(s_2^2 = 100/9\). M1: Calculate pooled variance \(s_p^2\) and standard error. A1: Correctly obtain \(s_p^2 = 10\) and \(SE = 1.5\). M1: Calculate test statistic \(t = -2\). B1: Correct degrees of freedom (16) and critical value (-1.746). M1: Compare test statistic to critical value. A1: Correct conclusion in context.