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(i) For the quartic equation \(x^4 - 2x^3 + 3x^2 - 4x + 5 = 0\), the sum of the roots is:
\(\sum \alpha = 2\)
and the sum of the products of roots in pairs is:
\(\sum \alpha\beta = 3\)

We know that:
\(\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = \left(\sum \alpha\right)^2 - 2\sum \alpha\beta\)
\(\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = (2)^2 - 2(3) = 4 - 6 = -2\).

(ii) Let \(S_n = \sum \alpha^n\). We use Newton's relations:
\(S_n - p_1 S_{n-1} + p_2 S_{n-2} - p_3 S_{n-3} + n p_n = 0\)
where \(p_1 = 2\), \(p_2 = 3\), \(p_3 = 4\), \(p_4 = 5\).

For \(n = 3\):
\(S_3 - p_1 S_2 + p_2 S_1 - 3p_3 = 0\)
Substitute the known values:
\(S_3 - 2(-2) + 3(2) - 3(4) = 0\)
\(S_3 + 4 + 6 - 12 = 0\)
\(S_3 - 2 = 0 \implies S_3 = 2\).

So \(\alpha^3 + \beta^3 + \gamma^3 + \delta^3 = 2\).

(iii) Substituting \(y = \frac{1}{x} \implies x = \frac{1}{y}\) into the original equation:
\(\left(\frac{1}{y}\right)^4 - 2\left(\frac{1}{y}\right)^3 + 3\left(\frac{1}{y}\right)^2 - 4\left(\frac{1}{y}\right) + 5 = 0\)
Multiplying by \(y^4\):
\(5y^4 - 4y^3 + 3y^2 - 2y + 1 = 0\).

Dividing by 5:
\(y^4 - \frac{4}{5}y^3 + \frac{3}{5}y^2 - \frac{2}{5}y + \frac{1}{5} = 0\).

Let \(T_2 = \sum y^2 = \sum \frac{1}{\alpha^2}\).
Using the identity for the sum of squares on the roots of this equation:
\(T_2 = \left(\sum y\right)^2 - 2\sum y_i y_j\)
From the coefficients of the \(y\)-equation:
\[ \sum y = \frac{4}{5}, \quad \sum y_i y_j = \frac{3}{5} \]
Therefore:
\(T_2 = \left(\frac{4}{5}\right)^2 - 2\left(\frac{3}{5}\right) = \frac{16}{25} - \frac{6}{5} = -\frac{14}{25}\).

(i) [3.5 marks]
M1: Use formula for \(\sum \alpha^2\) with correct substitution of sum of roots and pairwise products.
A1: Complete and correct algebraic working to obtain \(-2\).

(ii) [3.5 marks]
M1: Use Newton's relation for \(S_3\) or another valid identity.
A1: Substitute correct values of \(S_1\), \(S_2\), and coefficients.
A1: Obtain final answer 2.

(iii) [3.71 marks]
M1: Substitute \(y = 1/x\) and correctly find the polynomial equation in \(y\).
M1: State or use \(\sum y = 4/5\) and \(\sum y_i y_j = 3/5\) in the identity for sum of squares.
A1: Obtain \(-\frac{14}{25}\) (or \(-0.56\)) as the final answer.

(i) Expressing \(y\) by performing polynomial division:
\(y = x + 1 + \frac{4}{x - 2}\).
Thus, the asymptotes are:
Vertical asymptote: \(x = 2\)
Oblique asymptote: \(y = x + 1\).

(ii) To find the stationary points, differentiate \(y\):
\(\frac{dy}{dx} = 1 - \frac{4}{(x-2)^2}\).
Setting \(\frac{dy}{dx} = 0 \implies (x-2)^2 = 4 \implies x-2 = \pm 2\).
- If \(x - 2 = 2 \implies x = 4\), then \(y = 4 + 1 + \frac{4}{2} = 7\).
- If \(x - 2 = -2 \implies x = 0\), then \(y = 0 + 1 + \frac{4}{-2} = -1\).

Thus, the stationary points are \((4, 7)\) and \((0, -1)\).

To determine the nature:
\(\frac{d^2y}{dx^2} = \frac{8}{(x-2)^3}\).
- At \(x = 4\): \(\frac{d^2y}{dx^2} = 1 > 0 \implies (4, 7)\) is a local minimum.
- At \(x = 0\): \(\frac{d^2y}{dx^2} = -1 < 0 \implies (0, -1)\) is a local maximum.

(iii) The curve has no intersections with the \(x\)-axis because \(x^2 - x + 2 = 0\) has no real roots. The \(y\)-intercept is at \((0, -1)\), which is also a stationary point. The sketch consists of two branches separated by the vertical asymptote \(x = 2\) and bounded by the oblique asymptote \(y = x + 1\).

(i) [3 marks]
M1: Attempt algebraic division to express the function in the form \(y = ax + b + \frac{c}{x-2}\).
A1: Find vertical asymptote \(x = 2\).
A1: Find oblique asymptote \(y = x + 1\).

(ii) [4.71 marks]
M1: Differentiate to find \(\frac{dy}{dx}\).
M1: Set \(\frac{dy}{dx} = 0\) and solve for \(x\).
A1: Correct coordinates \((4, 7)\) and \((0, -1)\).
A1: Correctly justify nature (local min for \((4,7)\) and local max for \((0,-1)\)) using second derivative or sign table.

(iii) [3 marks]
B1: Sketch shows both branches with correct asymptotic behavior.
B1: Label asymptotes \(x = 2\) and \(y = x + 1\).
B1: Label coordinates of stationary points \((4,7)\) and \((0,-1)\).

(i) Let \(f(r) = \frac{1}{r^2}\). Then:
\(f(r) - f(r+1) = \frac{1}{r^2} - \frac{1}{(r+1)^2} = \frac{(r+1)^2 - r^2}{r^2(r+1)^2} = \frac{2r+1}{r^2(r+1)^2} = u_r\).
So \(u_r = \frac{1}{r^2} - \frac{1}{(r+1)^2}\).

(ii) Using the method of differences:
\(\sum_{r=1}^{n} u_r = \sum_{r=1}^{n} \left( \frac{1}{r^2} - \frac{1}{(r+1)^2} \right)\)
\(= \left(1 - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{9}\right) + \dots + \left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right)\).

All intermediate terms cancel out, leaving:
\(\sum_{r=1}^{n} u_r = 1 - \frac{1}{(n+1)^2}\).

(iii) As \(n \to \infty\), \(\frac{1}{(n+1)^2} \to 0\).
Therefore, \(\sum_{r=1}^{\infty} u_r = 1\).

(iv) We require:
\(1 - \frac{1}{(N+1)^2} > 0.999 \implies \frac{1}{(N+1)^2} < 0.001\)
\((N+1)^2 > 1000 \implies N+1 > \sqrt{1000} \approx 31.62 \implies N > 30.62\).

Since \(N\) must be an integer, the least value of \(N\) is 31.

(i) [3 marks]
M1: Set up partial fractions or identify \(f(r) = 1/r^2\).
A1: Show algebraic verification clearly to arrive at the difference form.

(ii) [3.71 marks]
M1: Write out the first few terms and last term to show the cancellation pattern.
A1: Correctly simplify to obtain \(1 - \frac{1}{(n+1)^2}\).

(iii) [1 mark]
B1: State \(1\) by taking limit of sum as \(n \to \infty\).

(iv) [3 marks]
M1: Set up inequality \(1 - \frac{1}{(N+1)^2} > 0.999\) and solve for \(N\).
A1: Find boundary value \(30.62\).
A1: State final integer value \(31\).

(i) Expanding the determinant along the first column:
\(\det(\mathbf{M}) = 2(3(1) - (-1)(1)) - 0 + 1((-1)(-1) - 3(1))\)
\(= 2(4) + 1(-2) = 6\).

(ii) Find the matrix of cofactors, \(\mathbf{C}\):
\(C_{11} = +(3(1) - (-1)(1)) = 4\)
\(C_{12} = -(0(1) - (-1)(1)) = -1\)
\(C_{13} = +(0(1) - 3(1)) = -3\)

\(C_{21} = -((-1)(1) - 1(1)) = 2\)
\(C_{22} = +(2(1) - 1(1)) = 1\)
\(C_{23} = -(2(1) - (-1)(1)) = -3\)

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

Thus, \(\mathbf{C} = \begin{pmatrix} 4 & -1 & -3 \\ 2 & 1 & -3 \\ -2 & 2 & 6 \end{pmatrix}\).

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

Thus, \(\mathbf{M}^{-1} = \frac{1}{6} \begin{pmatrix} 4 & 2 & -2 \\ -1 & 1 & 2 \\ -3 & -3 & 6 \end{pmatrix}\).

(iii) The system can be written as \(\mathbf{M} \mathbf{X} = \mathbf{B}\), where:
\(\mathbf{X} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}\), \(\mathbf{B} = \begin{pmatrix} -1 \\ 5 \\ 2 \end{pmatrix}\).

Multiplying by \(\mathbf{M}^{-1}\):
\(\mathbf{X} = \mathbf{M}^{-1} \mathbf{B} = \frac{1}{6} \begin{pmatrix} 4 & 2 & -2 \\ -1 & 1 & 2 \\ -3 & -3 & 6 \end{pmatrix} \begin{pmatrix} -1 \\ 5 \\ 2 \end{pmatrix} = \frac{1}{6} \begin{pmatrix} 2 \\ 10 \\ 0 \end{pmatrix} = \begin{pmatrix} 1/3 \\ 5/3 \\ 0 \end{pmatrix}\).

So \(x = \frac{1}{3}\), \(y = \frac{5}{3}\), \(z = 0\).

(i) [2 marks]
M1: Attempt expansion of a \(3 \times 3\) determinant.
A1: Correct value of \(6\).

(ii) [5 marks]
M1: Find at least 4 correct cofactors.
A1: Complete and correct cofactor matrix \(\mathbf{C}\).
M1: Transpose the cofactor matrix.
A1: Correct adjugate matrix.
A1: Correct inverse matrix including division by \(\det(\mathbf{M})\).

(iii) [3.71 marks]
M1: Formulate the equation \(\mathbf{X} = \mathbf{M}^{-1}\mathbf{B}\).
M1: Multiply \(\mathbf{M}^{-1}\) by \(\mathbf{B}\).
A1: Correct values for \(x\), \(y\), and \(z\).

(i) The curve \(C\) is the upper half of a cardioid. It starts at \(r = 2a\) when \(\theta = 0\), passes through \(r = a\) when \(\theta = \frac{\pi}{2}\), and terminates at the pole \(r = 0\) when \(\theta = \pi\).

(ii) The area \(A\) is given by:
\(A = \frac{1}{2} \int_{0}^{\pi} r^2 \, d\theta = \frac{1}{2} a^2 \int_{0}^{\pi} (1 + \cos\theta)^2 \, d\theta\)
\(= \frac{1}{2} a^2 \int_{0}^{\pi} \left(1 + 2\cos\theta + \cos^2\theta\right) \, d\theta\)
\(= \frac{1}{2} a^2 \int_{0}^{\pi} \left(\frac{3}{2} + 2\cos\theta + \frac{1}{2}\cos 2\theta\right) \, d\theta\)
\(= \frac{1}{2} a^2 \left[ \frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta \right]_{0}^{\pi}\)
\(= \frac{1}{2} a^2 \left( \frac{3\pi}{2} \right) = \frac{3\pi a^2}{4}\).

(iii) The distance from the initial line is \(y = r\sin\theta = a\sin\theta(1+\cos\theta)\).
The tangent is parallel to the initial line when \(\frac{dy}{d\theta} = 0\):
\(\frac{dy}{d\theta} = a(\cos\theta + \cos 2\theta) = a(2\cos^2\theta + \cos\theta - 1) = 0\).
This factorizes to:
\((2\cos\theta - 1)(\cos\theta + 1) = 0\).
Since \(0 \le \theta \le \pi\), we find:
\(\cos\theta = \frac{1}{2} \implies \theta = \frac{\pi}{3}\).

Substituting \(\theta = \frac{\pi}{3}\) into \(r = a(1+\cos\theta)\):
\(r = a(1 + 0.5) = \frac{3}{2}a\).

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

(i) [2 marks]
B1: Correct shape starting at \((2a,0)\) and terminating at the pole.
B1: Upper half of cardioid correctly represented with no portion below the initial line.

(ii) [5 marks]
M1: Use polar area formula \(\frac{1}{2} \int r^2 d\theta\).
M1: Expand \((1+\cos\theta)^2\) and use double angle identity.
A1: Correctly integrated expression.
M1: Substitute limits \(0\) and \(\pi\).
A1: Correct area \(\frac{3\pi a^2}{4}\).

(iii) [3.71 marks]
M1: Set up expression for \(y = r\sin\theta\).
M1: Differentiate \(y\) and set \(\frac{dy}{d\theta} = 0\).
A1: Solve quadratic to obtain \(\theta = \frac{\pi}{3}\).
A1: Obtain the polar coordinates \(\left(\frac{3}{2}a, \frac{\pi}{3}\right)\).

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

To check for intersection, we write components:
\(1 + 2\lambda = 2 + \mu \implies 2\lambda - \mu = 1 \quad (1)\)
\(2 - \lambda = -1 + \mu \implies \lambda + \mu = 3 \quad (2)\)
\(-1 + 3\lambda = 1 - 2\mu \implies 3\lambda + 2\mu = 2 \quad (3)\)

Adding (1) and (2) yields \(3\lambda = 4 \implies \lambda = 4/3\).
Then \(\mu = 3 - 4/3 = 5/3\).

Checking in (3):
\(3(4/3) + 2(5/3) = 4 + 10/3 = 22/3 \ne 2\).
Thus, the system is inconsistent, and the lines do not intersect. Since they are also not parallel, they are skew.

(ii) A vector perpendicular to both lines is:
\(\mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2 = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} \times \begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix} = \begin{pmatrix} 2 - 3 \\ 3 - (-4) \\ 2 - (-1) \end{pmatrix} = \begin{pmatrix} -1 \\ 7 \\ 3 \end{pmatrix}\).

Let \(A(1, 2, -1)\) be a point on \(l_1\) and \(B(2, -1, 1)\) on \(l_2\).
\(\mathbf{AB} = \begin{pmatrix} 2-1 \\ -1-2 \\ 1-(-1) \end{pmatrix} = \begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix}\).

The shortest distance is:
\(d = \frac{|\mathbf{AB} \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{|1(-1) + (-3)(7) + 2(3)|}{\sqrt{(-1)^2 + 7^2 + 3^2}} = \frac{|-1 - 21 + 6|}{\sqrt{59}} = \frac{16}{\sqrt{59}}\).

(iii) The plane contains \(l_1\), so it contains point \((1, 2, -1)\). It is parallel to \(l_2\), so its normal is \(\mathbf{n} = \begin{pmatrix} -1 \\ 7 \\ 3 \end{pmatrix}\).

Its equation is:
\(-x + 7y + 3z = -1(1) + 7(2) + 3(-1) = 10\)
or
\(x - 7y - 3z = -10\).

(i) [3 marks]
B1: Explain why the lines are not parallel.
M1: Write 3 component equations and attempt to solve two for \(\lambda\) and \(\mu\).
A1: Show contradiction in the third equation and conclude that they are skew.

(ii) [5 marks]
M1: Attempt vector cross product of direction vectors.
A1: Correct normal vector \(\begin{pmatrix} -1 \\ 7 \\ 3 \end{pmatrix}\).
M1: Find a vector between a point on each line.
M1: Apply the shortest distance formula \(d = \frac{|\mathbf{AB} \cdot \mathbf{n}|}{|\mathbf{n}|}\).
A1: Correct distance \(\frac{16}{\sqrt{59}}\).

(iii) [2.71 marks]
M1: Use normal vector and point on \(l_1\) to establish the plane equation.
A1: Correct Cartesian equation in requested form, e.g., \(x - 7y - 3z = -10\).

Let \(P(n)\) be the proposition that \(\mathbf{A}^n = \begin{pmatrix} 2n+1 & 4n \\ -n & 1-2n \end{pmatrix}\).

**Base Case: n = 1**
\(\mathbf{A}^1 = \begin{pmatrix} 2(1)+1 & 4(1) \\ -1 & 1-2(1) \end{pmatrix} = \begin{pmatrix} 3 & 4 \\ -1 & -1 \end{pmatrix}\).
This matches \(\mathbf{A}\), so \(P(1)\) is true.

**Inductive Step**
Assume that \(P(k)\) is true for some positive integer \(k\). That is:
\(\mathbf{A}^k = \begin{pmatrix} 2k+1 & 4k \\ -k & 1-2k \end{pmatrix}\).

We must prove that \(P(k+1)\) is true, i.e.:
\(\mathbf{A}^{k+1} = \begin{pmatrix} 2(k+1)+1 & 4(k+1) \\ -(k+1) & 1-2(k+1) \end{pmatrix} = \begin{pmatrix} 2k+3 & 4k+4 \\ -k-1 & -2k-1 \end{pmatrix}\).

Using the induction hypothesis:
\(\mathbf{A}^{k+1} = \mathbf{A}^k \mathbf{A} = \begin{pmatrix} 2k+1 & 4k \\ -k & 1-2k \end{pmatrix} \begin{pmatrix} 3 & 4 \\ -1 & -1 \end{pmatrix}\)

Now compute the matrix multiplication:
- Element (1,1):
\((2k+1)(3) + (4k)(-1) = 6k + 3 - 4k = 2k + 3 = 2(k+1) + 1\).
- Element (1,2):
\((2k+1)(4) + (4k)(-1) = 8k + 4 - 4k = 4k + 4 = 4(k+1)\).
- Element (2,1):
\((-k)(3) + (1-2k)(-1) = -3k - 1 + 2k = -k - 1 = -(k+1)\).
- Element (2,2):
\((-k)(4) + (1-2k)(-1) = -4k - 1 + 2k = -2k - 1 = 1 - 2(k+1)\).

Thus, \(\mathbf{A}^{k+1} = \begin{pmatrix} 2(k+1)+1 & 4(k+1) \\ -(k+1) & 1-2(k+1) \end{pmatrix}\).

Hence, if \(P(k)\) is true, then \(P(k+1)\) is true. Since \(P(1)\) is true, by mathematical induction, the result is true for all positive integers \(n\).

[10.71 marks total]
B1: Verify base case \(n=1\) clearly.
M1: State inductive hypothesis \(P(k)\) clearly.
M1: Establish equation \(\mathbf{A}^{k+1} = \mathbf{A}^k \mathbf{A}\) and substitute the hypothesis.
M1: Perform matrix multiplication (at least two elements done correctly).
A1: Correct element (1,1) shown in terms of \(k+1\).
A1: Correct element (1,2) shown in terms of \(k+1\).
A1: Correct element (2,1) shown in terms of \(k+1\).
A1: Correct element (2,2) shown in terms of \(k+1\).
A1: Complete, logical conclusion citing the principle of mathematical induction.

We know that \(\cosh x = \frac{e^x + e^{-x}}{2}\) and \(\sinh x = \frac{e^x - e^{-x}}{2}\).

(i) Substitute these into the equation:
\(5\left(\frac{e^x + e^{-x}}{2}\right) - \left(\frac{e^x - e^{-x}}{2}\right) = 7\)

Multiply through by 2:
\(5(e^x + e^{-x}) - (e^x - e^{-x}) = 14\)
\(4e^x + 6e^{-x} = 14\)

Divide by 2:
\(2e^x + 3e^{-x} = 7\)

Multiply by \(e^x\):
\(2e^{2x} - 7e^x + 3 = 0\) (as required).

(ii) Let \(u = e^x\). The quadratic equation is:
\(2u^2 - 7u + 3 = 0\)

Factorizing gives:
\((2u - 1)(u - 3) = 0\)

So \(u = \frac{1}{2}\) or \(u = 3\).

Since \(u = e^x\):
\(e^x = \frac{1}{2} \implies x = \ln\left(\frac{1}{2}\right) = -\ln 2\)

\(e^x = 3 \implies x = \ln 3\).

M1: Substitute definitions of \(\cosh x\) and \(\sinh x\) in terms of exponentials.
A1: Correctly simplify to \(2e^x + 3e^{-x} = 7\) or equivalent.
A1: Show the given quadratic equation clearly.
M1: Factorize the quadratic in \(e^x\).
A1: Obtain \(e^x = \frac{1}{2}\) and \(e^x = 3\).
A1: Express solutions in the form \(\ln a\) as requested.

(i) Find eigenvalues of \(\mathbf{B}\) by solving \(\det(\mathbf{B} - \lambda \mathbf{I}) = 0\):
\(\begin{vmatrix} 1-\lambda & 1 & 3 \\ 1 & 5-\lambda & 1 \\ 3 & 1 & 1-\lambda \end{vmatrix} = 0\)

Row 1 - Row 3:
\(\begin{vmatrix} -2-\lambda & 0 & 2+\lambda \\ 1 & 5-\lambda & 1 \\ 3 & 1 & 1-\lambda \end{vmatrix} = 0\)

Factor out \(\lambda+2\) from Row 1:
\(-(\lambda+2) \begin{vmatrix} 1 & 0 & -1 \\ 1 & 5-\lambda & 1 \\ 3 & 1 & 1-\lambda \end{vmatrix} = 0\)

Col 3 + Col 1:
\(-(\lambda+2) \begin{vmatrix} 1 & 0 & 0 \\ 1 & 5-\lambda & 2 \\ 3 & 1 & 4-\lambda \end{vmatrix} = 0\)

\(-(\lambda+2) [ (5-\lambda)(4-\lambda) - 2 ] = -(\lambda+2)(\lambda^2 - 9\lambda + 18) = -(\lambda+2)(\lambda-3)(\lambda-6) = 0\)

Eigenvalues are \(\lambda_1 = -2\), \(\lambda_2 = 3\), \(\lambda_3 = 6\).

(ii) Find eigenvectors:
For \(\lambda = -2\):
\(\begin{pmatrix} 3 & 1 & 3 \\ 1 & 7 & 1 \\ 3 & 1 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies y = 0, x + z = 0 \implies \mathbf{v}_1 = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}\).
Normalizing: \(\mathbf{u}_1 = \begin{pmatrix} 1/\sqrt{2} \\ 0 \\ -1/\sqrt{2} \end{pmatrix}\).

For \(\lambda = 3\):
\(\begin{pmatrix} -2 & 1 & 3 \\ 1 & 2 & 1 \\ 3 & 1 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies z = x, y = -x \implies \mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}\).
Normalizing: \(\mathbf{u}_2 = \begin{pmatrix} 1/\sqrt{3} \\ -1/\sqrt{3} \\ 1/\sqrt{3} \end{pmatrix}\).

For \(\lambda = 6\):
\(\begin{pmatrix} -5 & 1 & 3 \\ 1 & -5 & 1 \\ 3 & 1 & -5 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\).

Subtract Row 1 from Row 3:
\(8x - 8z = 0 \implies z = x\).

Substitute \(z=x\) into Row 1: \(-5x + y + 3x = 0 \implies y = 2x \implies \mathbf{v}_3 = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}\).
Normalizing: \(\mathbf{u}_3 = \begin{pmatrix} 1/\sqrt{6} \\ 2/\sqrt{6} \\ 1/\sqrt{6} \end{pmatrix}\).

(iii) The columns of \(\mathbf{P}\) are the normalized eigenvectors, and the diagonal elements of \(\mathbf{D}\) are the corresponding eigenvalues:
\(\mathbf{P} = \begin{pmatrix} 1/\sqrt{2} & 1/\sqrt{3} & 1/\sqrt{6} \\ 0 & -1/\sqrt{3} & 2/\sqrt{6} \\ -1/\sqrt{2} & 1/\sqrt{3} & 1/\sqrt{6} \end{pmatrix}\)
\(\mathbf{D} = \begin{pmatrix} -2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 6 \end{pmatrix}\).

M1: Formulate the characteristic equation \(\det(\mathbf{B} - \lambda \mathbf{I}) = 0\).
A1: Obtain the correct eigenvalues \(\lambda = -2, 3, 6\).
M1: Attempt to find eigenvectors for at least two eigenvalues.
A1: Obtain correct eigenvectors (un-normalized).
M1: Normalize the three eigenvectors.
A1: Correctly state normalized eigenvectors.
B1: Write down \(\mathbf{D}\) corresponding to the order of columns in \(\mathbf{P}\).
B1: Write down the correct orthogonal matrix \(\mathbf{P}\).

(i) Let \(f(x) = \ln(1 + \sin x)\).
At \(x = 0\): \(f(0) = \ln(1 + 0) = 0\).

Using the chain rule:
\(f'(x) = \frac{\cos x}{1 + \sin x}\)
At \(x = 0\): \(f'(0) = \frac{1}{1} = 1\).

Using the quotient rule:
\(f''(x) = \frac{-\sin x(1 + \sin x) - \cos x(\cos x)}{(1 + \sin x)^2} = \frac{-\sin x - \sin^2 x - \cos^2 x}{(1 + \sin x)^2} = \frac{-\sin x - 1}{(1 + \sin x)^2} = -\frac{1}{1 + \sin x}\)
At \(x = 0\): \(f''(0) = -1\).

Now find the third derivative:
\(f'''(x) = \frac{d}{dx} \left[ -(1 + \sin x)^{-1} \right] = (1 + \sin x)^{-2} \cos x = \frac{\cos x}{(1 + \sin x)^2}\)
At \(x = 0\): \(f'''(0) = \frac{1}{1^2} = 1\).

(ii) The Maclaurin series expansion is given by:
\(f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots\)

Substitute the values:
\(f(x) = 0 + (1)x + \frac{-1}{2}x^2 + \frac{1}{6}x^3 + \dots = x - \frac{1}{2}x^2 + \frac{1}{6}x^3\).

M1: Correctly differentiate \(f(x)\) to find \(f'(x)\).
A1: Show \(f'(0) = 1\).
M1: Correctly differentiate \(f'(x)\) to find \(f''(x)\) and simplify.
A1: Show \(f''(0) = -1\).
M1: Correctly differentiate \(f''(x)\) to find \(f'''(x)\).
A1: Show \(f'''(0) = 1\).
M1: Apply the Maclaurin series formula with their derivatives.
A1: Obtain the correct expansion \(x - \frac{1}{2}x^2 + \frac{1}{6}x^3\).

(i) We use integration by parts for \(I_n = \int_0^1 x^n e^{-2x} \, dx\).
Let \(u = x^n \implies du = n x^{n-1} \, dx\)
Let \(dv = e^{-2x} \, dx \implies v = -\frac{1}{2} e^{-2x}\)

Applying integration by parts:
\(I_n = \left[ -\frac{1}{2} x^n e^{-2x} \right]_0^1 - \int_0^1 \left( -\frac{1}{2} e^{-2x} \right) (n x^{n-1}) \, dx\)

\(I_n = \left( -\frac{1}{2}(1)^n e^{-2} - 0 \right) + \frac{n}{2} \int_0^1 x^{n-1} e^{-2x} \, dx\)

\(I_n = -\frac{1}{2e^2} + \frac{n}{2} I_{n-1}\) (as required).

(ii) First, calculate \(I_0\):
\(I_0 = \int_0^1 e^{-2x} \, dx = \left[ -\frac{1}{2} e^{-2x} \right]_0^1 = -\frac{1}{2} e^{-2} - \left( -\frac{1}{2} \right) = \frac{1}{2} - \frac{1}{2e^2}\).

Now, use the reduction formula for \(n = 1\):
\(I_1 = -\frac{1}{2e^2} + \frac{1}{2} I_0 = -\frac{1}{2e^2} + \frac{1}{2}\left( \frac{1}{2} - \frac{1}{2e^2} \right) = \frac{1}{4} - \frac{3}{4e^2}\).

Now, use the reduction formula for \(n = 2\):
\(I_2 = -\frac{1}{2e^2} + I_1 = -\frac{1}{2e^2} + \left( \frac{1}{4} - \frac{3}{4e^2} \right) = \frac{1}{4} - \frac{5}{4e^2}\).

M1: Set up integration by parts with appropriate choice of \(u\) and \(dv\).
A1: Obtain the correct integrated term \(\left[ -\frac{1}{2} x^n e^{-2x} \right]_0^1\) and integral term.
A1: Correctly substitute limits and establish the given reduction formula.
M1: Evaluate \(I_0\) correctly.
M1: Calculate \(I_1\) using the reduction formula.
A1: Calculate \(I_2\) using the reduction formula.
A1: Correctly simplify the final answer to \(\frac{1}{4} - \frac{5}{4e^2}\).

First, solve the homogeneous equation \(\frac{d^2 y}{dx^2} + 2 \frac{dy}{dx} + 5 y = 0\).
The characteristic equation is:
\(m^2 + 2m + 5 = 0\)

\(m = \frac{-2 \pm \sqrt{4 - 20}}{2} = -1 \pm 2i\).

So the complementary function is:
\(y_{CF} = e^{-x} (A \cos 2x + B \sin 2x)\).

Next, find a particular integral of the form \(y_{PI} = k e^{-x}\):
\(\frac{dy}{dx} = -k e^{-x}\)

\(\frac{d^2 y}{dx^2} = k e^{-x}\)

Substitute into the differential equation:
\(k e^{-x} + 2(-k e^{-x}) + 5(k e^{-x}) = 10 e^{-x}\)

\(4k e^{-x} = 10 e^{-x} \implies k = \frac{5}{2}\).

So the general solution is:
\(y = e^{-x} (A \cos 2x + B \sin 2x) + \frac{5}{2} e^{-x}\).

Apply the initial conditions to find \(A\) and \(B\):
When \(x = 0, y = \frac{7}{2}\):
\(\frac{7}{2} = e^0 (A \cos 0 + B \sin 0) + \frac{5}{2} e^0 \implies \frac{7}{2} = A + \frac{5}{2} \implies A = 1\).

Now differentiate the general solution to find \(\frac{dy}{dx}\):
\(y = e^{-x} (\cos 2x + B \sin 2x + \frac{5}{2})\)

\(\frac{dy}{dx} = -e^{-x} (\cos 2x + B \sin 2x + \frac{5}{2}) + e^{-x} (-2 \sin 2x + 2B \cos 2x)\)

When \(x = 0, \frac{dy}{dx} = 0\):
\(0 = -(1 + 0 + \frac{5}{2}) + (0 + 2B)\)

\(0 = -\frac{7}{2} + 2B \implies B = \frac{7}{4}\).

Thus, the particular solution is:
\(y = e^{-x} \left(\cos 2x + \frac{7}{4} \sin 2x + \frac{5}{2}\right)\).

M1: Find the auxiliary equation and solve for roots.
A1: Formulate the correct complementary function.
M1: Propose a particular integral of the form \(k e^{-x}\) and substitute into the DE.
A1: Determine \(k = 2.5\).
M1: Substitute \(x=0\) and \(y=\frac{7}{2}\) to find \(A\).
A1: Obtain \(A = 1\).
M1: Differentiate the general solution and substitute \(x=0, y'=0\) to find \(B\).
A1: Obtain \(B = \frac{7}{4\)}.
A1: Write down the correct final particular solution.

(i) By de Moivre's theorem:
\(\cos 5\theta + i \sin 5\theta = (\cos \theta + i \sin \theta)^5\)

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

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

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

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

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

(ii) Let \(\cos 5\theta = 0\).
The solutions in the interval \(0 \le \theta \le \pi\) are:
\(5\theta = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}, \frac{9\pi}{2} \implies \theta = \frac{\pi}{10}, \frac{3\pi}{10}, \frac{\pi}{2}, \frac{7\pi}{10}, \frac{9\pi}{10}\).

So \(x = \cos \theta\) are the roots of \(16x^5 - 20x^3 + 5x = 0\).
Since \(\cos(\frac{\pi}{2}) = 0\), \(x = 0\) is a root.
Factorizing \(x\) out:
\(x(16x^4 - 20x^2 + 5) = 0\).
The non-zero roots must satisfy \(16x^4 - 20x^2 + 5 = 0\).
These non-zero roots are:
\(x = \cos(\frac{\pi}{10}), \cos(\frac{3\pi}{10}), \cos(\frac{7\pi}{10}), \cos(\frac{9\pi}{10})\).
Since \(\cos(\frac{9\pi}{10}) = -\cos(\frac{\pi}{10})\) and \(\cos(\frac{7\pi}{10}) = -\cos(\frac{3\pi}{10})\), the roots are indeed \(\pm \cos(\frac{\pi}{10})\) and \(\pm \cos(\frac{3\pi}{10})\).

(iii) Let \(u = x^2\). The equation becomes:
\(16u^2 - 20u + 5 = 0\)

Using the quadratic formula:
\(u = \frac{20 \pm \sqrt{(-20)^2 - 4(16)(5)}}{32} = \frac{20 \pm \sqrt{400 - 320}}{32} = \frac{20 \pm \sqrt{80}}{32} = \frac{20 \pm 4\sqrt{5}}{32} = \frac{5 \pm \sqrt{5}}{8}\).

Since \(0 < \frac{\pi}{10} < \frac{3\pi}{10} < \frac{\pi}{2}\), we have \(\cos(\frac{\pi}{10}) > \cos(\frac{3\pi}{10})\), which means \(\cos^2(\frac{\pi}{10}) > \cos^2(\frac{3\pi}{10})\).
Therefore, \(\cos^2(\frac{\pi}{10})\) must be the larger root:
\(\cos^2(\frac{\pi}{10}) = \frac{5 + \sqrt{5}}{8}\).

M1: Use de Moivre's theorem to expand \((\cos\theta + i\sin\theta)^5\).
A1: Express \(\cos 5\theta\) in terms of \(\cos\theta\) and \(\sin\theta\).
M1: Substitute \(\sin^2\theta = 1-\cos^2\theta\) to convert fully to \(\cos\theta\).
A1: Show the given identity clearly.
M1: Set \(\cos 5\theta = 0\) and find the values of \(\theta\).
A1: Identify the corresponding roots of \(16x^4 - 20x^2 + 5 = 0\) using symmetry of \(\cos\theta\).
M1: Solve the quadratic equation in \(x^2\).
A1: Justify choosing the positive/larger root for \(\cos^2(\frac{\pi}{10})\).
A1: State the final exact value \(\frac{5 + \sqrt{5}}{8}\).

(i) The arc length formula is:
\(s = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\)

For \(y = \ln(\cos x)\):
\(\frac{dy}{dx} = \frac{-\sin x}{\cos x} = -\tan x\)

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

Since \(\sec x > 0\) for \(0 \le x \le \frac{\pi}{3}\):
\(s = \int_0^{\pi/3} \sec x \, dx = \left[ \ln|\sec x + \tan x| \right]_0^{\pi/3}\)

\(s = \ln\left| \sec(\frac{\pi}{3}) + \tan(\frac{\pi}{3}) \right| - \ln|\sec 0 + \tan 0|\)

\(s = \ln(2 + \sqrt{3}) - \ln(1 + 0) = \ln(2 + \sqrt{3})\).

(ii) The surface area formula is:
\(S = 2\pi \int_a^b y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\)

For \(y = \cosh x\):
\(\frac{dy}{dx} = \sinh x\)

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

So:
\(S = 2\pi \int_0^{\ln 2} \cosh x \sqrt{\cosh^2 x} \, dx = 2\pi \int_0^{\ln 2} \cosh^2 x \, dx\)

Using \(\cosh^2 x = \frac{1 + \cosh 2x}{2}\):
\(S = \pi \int_0^{\ln 2} (1 + \cosh 2x) \, dx = \pi \left[ x + \frac{1}{2} \sinh 2x \right]_0^{\ln 2} = \pi [ x + \sinh x \cosh x ]_0^{\ln 2}\)

At \(x = \ln 2\):
\(\sinh(\ln 2) = \frac{e^{\ln 2} - e^{-\ln 2}}{2} = \frac{2 - 1/2}{2} = \frac{3}{4}\)

\(\cosh(\ln 2) = \frac{e^{\ln 2} + e^{-\ln 2}}{2} = \frac{2 + 1/2}{2} = \frac{5}{4}\)

At \(x = 0\):
\(\sinh 0 = 0, \cosh 0 = 1\)

So:
\(S = \pi \left( \ln 2 + \frac{3}{4} \cdot \frac{5}{4} \right) - \pi(0) = \pi \left( \ln 2 + \frac{15}{16} \right)\).

M1: Differentiate \(y = \ln(\cos x)\).
A1: Obtain the correct arc length integrand \(\sec x\).
A1: Integrate and substitute limits to get \(\ln(2 + \sqrt{3})\).
M1: State the correct surface area formula and substitute \(y = \cosh x\).
A1: Show that the integrand simplifies to \(\cosh^2 x\).
M1: Integrate \(\cosh^2 x\) using the identity \(\cosh^2 x = \frac{1+\cosh 2x}{2}\).
A1: Correctly evaluate \(\sinh(\ln 2)\) and \(\cosh(\ln 2)\).
A1: Obtain the final exact area \(\pi \left( \ln 2 + \frac{15}{16} \right)\).

(i) From the first equation:
\(4y = \frac{dx}{dt} + x \implies y = \frac{1}{4} \frac{dx}{dt} + \frac{1}{4} x\)

Differentiate with respect to \(t\):
\(\frac{dy}{dt} = \frac{1}{4} \frac{d^2 x}{dt^2} + \frac{1}{4} \frac{dx}{dt}\)

Substitute \(y\) and \(\frac{dy}{dt}\) into the second equation:
\(\frac{1}{4} \frac{d^2 x}{dt^2} + \frac{1}{4} \frac{dx}{dt} = -x - 5 \left( \frac{1}{4} \frac{dx}{dt} + \frac{1}{4} x \right)\)

Multiply the entire equation by 4:
\(\frac{d^2 x}{dt^2} + \frac{dx}{dt} = -4x - 5 \frac{dx}{dt} - 5x\)

\(\frac{d^2 x}{dt^2} + 6 \frac{dx}{dt} + 9x = 0\) (as required).

(ii) The auxiliary equation for \(x\) is:
\(m^2 + 6m + 9 = 0 \implies (m+3)^2 = 0 \implies m = -3\) (repeated root).

So the general solution for \(x\) is:
\(x = (A + Bt) e^{-3t}\)

To find \(y\), substitute \(x\) and \(\frac{dx}{dt}\) into \(y = \frac{1}{4} \left( \frac{dx}{dt} + x \right)\):
\(\frac{dx}{dt} = B e^{-3t} - 3(A + Bt) e^{-3t} = (B - 3A - 3Bt) e^{-3t}\)

Then:
\(y = \frac{1}{4} [ (B - 3A - 3Bt) e^{-3t} + (A + Bt) e^{-3t} ] = \frac{1}{4} (B - 2A - 2Bt) e^{-3t}\).

(iii) Given \(x = 2\) and \(y = 1\) at \(t = 0\):
From \(x(0) = 2 \implies A = 2\).

From \(y(0) = 1 \implies \frac{1}{4} (B - 2A) = 1 \implies B - 4 = 4 \implies B = 8\).

Thus, the particular solutions are:
\(x = (2 + 8t) e^{-3t}\)

\(y = \frac{1}{4} (8 - 4 - 16t) e^{-3t} = (1 - 4t) e^{-3t}\).

M1: Differentiate the first equation to express \(\frac{dy}{dt}\) in terms of \(x\) derivatives.
A1: Correctly substitute \(y\) and \(\frac{dy}{dt}\) into the second equation.
A1: Obtain the correct second-order ODE for \(x\).
M1: Solve the auxiliary equation for \(x\).
A1: Obtain the correct general solution for \(x\).
M1: Express \(y\) in terms of \(t\).
A1: Obtain the correct general solution for \(y\).
M1: Use initial conditions to find constants \(A\) and \(B\).
A1: Obtain the correct particular solutions.

We use Newton's second law of motion, \(F = m a\), and write acceleration in terms of \(v\) and \(x\) as \(a = v \frac{dv}{dx}\).

Substituting the given mass and force:
\(2 v \frac{dv}{dx} = \frac{4}{3v+1}\)

We separate the variables to integrate:
\(2v(3v+1) \, dv = 4 \, dx\)
\((3v^2 + v) \, dv = 2 \, dx\)

Integrating both sides with respect to their variables:
\(\int (3v^2 + v) \, dv = \int 2 \, dx\)

\(v^3 + \frac{1}{2}v^2 = 2x + C\)

Using the initial conditions, \(v = 1\) when \(x = 0\):
\(1^3 + \frac{1}{2}(1)^2 = 2(0) + C \implies C = 1.5\)

So, the equation relating \(v\) and \(x\) is:
\(v^3 + \frac{1}{2}v^2 = 2x + 1.5\)

Substitute \(v = 2\) to find \(x\):
\(2^3 + \frac{1}{2}(2)^2 = 2x + 1.5\)
\(8 + 2 = 2x + 1.5\)
\(10 = 2x + 1.5\)
\(2x = 8.5 \implies x = 4.25\)

M1: Set up the differential equation using \(F = m v \frac{dv}{dx}\).
A1: Correct equation: \(2v \frac{dv}{dx} = \frac{4}{3v+1}\).
M1: Separate variables and integrate both sides.
A1: Correct integration to find \(v^3 + \frac{1}{2}v^2 = 2x + C\) (or equivalent definite integral).
M1: Apply the initial condition \(v = 1\) at \(x = 0\) to evaluate the constant of integration.
A1: Obtain the correct equation relating \(x\) and \(v\).
A1.14: Solve for \(x\) to obtain \(4.25\) (or \(\frac{17}{4}\)).

The string becomes slack when its extension \(x\) becomes zero, meaning its length is \(1.0\text{ m}\).

We apply the principle of conservation of energy between the release point \(B\) and the point where the string first becomes slack.

At point \(B\):
Extension \(x = AB - L_0 = 1.5 - 1.0 = 0.5\text{ m}\).
Elastic Potential Energy (EPE) = \(\frac{\lambda x^2}{2 L_0} = \frac{40 \times 0.5^2}{2 \times 1.0} = \frac{40 \times 0.25}{2.0} = 5\text{ J}\).
Kinetic Energy (KE) = 0 (since it is released from rest).

When the string becomes slack:
Extension \(x = 0\), so EPE = 0.
Kinetic Energy (KE) = \(\frac{1}{2} m v^2 = \frac{1}{2} (0.4) v^2 = 0.2 v^2\).

Since the horizontal table is smooth, total mechanical energy is conserved:
\(\text{Initial EPE} + \text{Initial KE} = \text{Final EPE} + \text{Final KE}\)
\(5 + 0 = 0 + 0.2 v^2\)
\(0.2 v^2 = 5\)
\(v^2 = 25 \implies v = 5\text{ m/s}\)

M1: Calculate the initial extension of the string \(x = 0.5\text{ m}\).
M1: Use Hooke's Law EPE formula \(\frac{\lambda x^2}{2l}\).
A1: Obtain initial elastic potential energy as \(5\text{ J}\).
M1: State that elastic potential energy is completely converted into kinetic energy at the slack point.
A1: Formulate the energy conservation equation: \(0.2 v^2 = 5\).
M1: Solve for \(v^2\).
A1.14: Correctly find \(v = 5\text{ m/s}\).

Let \(v\) be the speed of \(P\) at the point where the string becomes slack.

Using conservation of energy between the lowest point of motion and the point where the string makes an angle \(\theta\) with the upward vertical:
Vertical height gained, \(h = r(1 + \cos\theta) = 1.0 \times \left(1 + \frac{1}{6}\right) = \frac{7}{6}\text{ m}\).

Energy conservation equation:
\(\frac{1}{2} m u^2 = \frac{1}{2} m v^2 + m g h \implies u^2 = v^2 + 2 g h\)
\(u^2 = v^2 + 2(10)\left(\frac{7}{6}\right) = v^2 + \frac{70}{3}\) (Equation 1)

At the point where the string becomes slack, the tension \(T = 0\).
Resolving forces radially towards the center \(O\):
\(T + m g \cos\theta = \frac{m v^2}{r}\)
Since \(T = 0\):
\(m g \cos\theta = \frac{m v^2}{r} \implies v^2 = g r \cos\theta\)
Substitute \(g=10\), \(r=1.0\), and \(\cos\theta = \frac{1}{6}\):
\(v^2 = 10 \times 1.0 \times \frac{1}{6} = \frac{5}{3}\)

Substitute \(v^2 = \frac{5}{3}\) into Equation 1:
\(u^2 = \frac{5}{3} + \frac{70}{3} = \frac{75}{3} = 25\)
\(u = 5\text{ m/s}\)

M1: Use conservation of energy between the lowest point and the point where the string becomes slack.
A1: Find the vertical height gained \(h = r(1+\cos\theta) = \frac{7}{6}\text{ m}\).
A1: Set up the energy equation correctly: \(u^2 = v^2 + \frac{70}{3}\).
M1: Resolve forces radially and set tension \(T = 0\).
A1: Obtain \(v^2 = g r \cos\theta = \frac{5}{3}\).
M1: Substitute \(v^2\) into the energy equation.
A1.14: Correctly determine \(u = 5\text{ m/s}\).

Let \(d\) be the distance the man climbs from \(A\) along the ladder.

First, resolve forces vertically for the system (ladder + man):
\(R_A = W_{\text{ladder}} + W_{\text{man}} = 15g + 75g = 90g = 900\text{ N}\).

Since slipping is about to occur, the friction force \(F_A\) at the ground is limiting:
\(F_A = \mu R_A = 0.5 \times 900 = 450\text{ N}\).

Resolve forces horizontally:
\(R_B = F_A = 450\text{ N}\) (where \(R_B\) is the normal reaction of the wall at \(B\)).

Take moments about point \(A\) for the ladder in equilibrium:
\(R_B (6 \sin\alpha) = W_{\text{ladder}} (3 \cos\alpha) + W_{\text{man}} (d \cos\alpha)\)

Divide the entire moments equation by \(\cos\alpha\):
\(R_B (6 \tan\alpha) = 3 W_{\text{ladder}} + d W_{\text{man}}\)

Given \(\tan\alpha = \frac{4}{3}\), \(W_{\text{ladder}} = 150\text{ N}\), \(W_{\text{man}} = 750\text{ N}\), and \(R_B = 450\text{ N}\):
\(450 \times 6 \times \frac{4}{3} = 3(150) + d(750)\)
\(450 \times 8 = 450 + 750d\)
\(3600 = 450 + 750d\)
\(750d = 3150 \implies d = 4.2\text{ m}\)

M1: Resolve forces vertically to find normal reaction \(R_A = 900\text{ N}\).
A1: Determine maximum frictional force \(F_A = 450\text{ N}\).
M1: Relate horizontal reaction at the wall to friction: \(R_B = 450\text{ N}\).
M1: Take moments about \(A\) (or any other point) with appropriate components.
A1: Correct moments equation: \(R_B (6 \tan\alpha) = 3 W_{\text{ladder}} + d W_{\text{man}}\).
M1: Substitute known values and simplify.
A1.14: Find the distance \(d = 4.2\text{ m}\).

Let the line of centers at impact be the x-axis, with \(A\) moving in the positive x-direction.

Initial velocity components:
For \(A\):
\(u_{Ax} = 5u \cos\theta = 3u\)
\(u_{Ay} = 5u \sin\theta = 4u\)
For \(B\):
\(u_{Bx} = -5u\)
\(u_{By} = 0\)

Since the spheres are smooth, the components of velocity perpendicular to the line of centers are unchanged by the collision:
\(v_{Ay} = u_{Ay} = 4u\)
\(v_{By} = u_{By} = 0\)

Along the line of centers (x-direction), we apply conservation of linear momentum:
\(m u_{Ax} + m u_{Bx} = m v_{Ax} + m v_{Bx}\)
\(m(3u) + m(-5u) = m v_{Ax} + m v_{Bx}\)
\(v_{Ax} + v_{Bx} = -2u\) (Equation 1)

We also apply Newton's law of restitution along the line of centers:
\(v_{Bx} - v_{Ax} = e(u_{Ax} - u_{Bx})\)
\(v_{Bx} - v_{Ax} = 0.5(3u - (-5u)) = 0.5(8u) = 4u\) (Equation 2)

To find \(v_{Ax}\), subtract Equation 2 from Equation 1:
\(2v_{Ax} = -2u - 4u = -6u \implies v_{Ax} = -3u\)

The components of velocity of \(A\) after collision are \(v_{Ax} = -3u\) and \(v_{Ay} = 4u\).

Thus, the speed of \(A\) after the collision is:
\(V_A = \sqrt{v_{Ax}^2 + v_{Ay}^2} = \sqrt{(-3u)^2 + (4u)^2} = \sqrt{9u^2 + 16u^2} = 5u\)

M1: Resolve initial velocities parallel and perpendicular to the line of centers.
A1: Identify that the perpendicular component of \(A\)'s velocity remains \(4u\).
M1: Apply conservation of momentum along the line of centers to get \(v_{Ax} + v_{Bx} = -2u\).
A1: Apply Newton's law of restitution correctly to get \(v_{Bx} - v_{Ax} = 4u\).
M1: Solve the simultaneous equations for the parallel velocity component of \(A\).
A1: Obtain \(v_{Ax} = -3u\).
A1.14: Calculate the final speed of \(A\) as \(5u\).

From \(\tan\theta = \frac{3}{4}\), we get:
\(\sin\theta = \frac{3}{5} = 0.6\)
\(\cos\theta = \frac{4}{5} = 0.8\)

Using the horizontal equation of motion for a projectile:
\(x = V \cos\theta \times T\)
\(80 = V (0.8) T \implies V T = 100\) (Equation 1)

Using the vertical equation of motion:
\(y = V \sin\theta \times T - \frac{1}{2} g T^2\)
\(40 = V (0.6) T - 5 T^2\) (Equation 2)

Substitute \(V T = 100\) into Equation 2:
\(40 = 0.6 (100) - 5 T^2\)
\(40 = 60 - 5 T^2\)
\(5 T^2 = 20\)
\(T^2 = 4 \implies T = 2\text{ s}\) (since \(T > 0\))

Substitute \(T = 2\) back into Equation 1:
\(V (2) = 100 \implies V = 50\text{ m/s}\)

M1: Determine \(\sin\theta = 0.6\) and \(\cos\theta = 0.8\).
M1: Establish horizontal motion equation \(80 = V (0.8) T\).
A1: Deduce the relation \(VT = 100\).
M1: Establish vertical motion equation \(40 = V (0.6) T - 5 T^2\).
A1: Substitute \(VT = 100\) and simplify to find \(5 T^2 = 20\).
M1: Solve for \(T\) to find \(T = 2\).
A1.14: Substitute \(T = 2\) to find \(V = 50\).

Let \(T\) be the tension in the string, so \(T = 10\text{ N}\).

Resolving vertically for the particle in circular motion (no vertical acceleration):
\(T \cos\theta = m g\)
\(10 \cos\theta = 0.6 \times 10 = 6\)
\(\cos\theta = 0.6\)

Using the trigonometric identity, we find \(\sin\theta\):
\(\sin\theta = \sqrt{1 - 0.6^2} = 0.8\)

The radius \(r\) of the horizontal circle is:
\(r = L \sin\theta = 1.5 \times 0.8 = 1.2\text{ m}\)

Resolving horizontally towards the center of the circle:
\(T \sin\theta = \frac{m v^2}{r}\)

Substitute the known values:
\(10 \times 0.8 = \frac{0.6 \times v^2}{1.2}\)
\(8 = 0.5 v^2\)
\(v^2 = 16 \implies v = 4\text{ m/s}\)

M1: Resolve forces vertically for the conical pendulum to set up \(T \cos\theta = mg\).
A1: Find \(\cos\theta = 0.6\) and thus \(\sin\theta = 0.8\).
M1: Formulate the horizontal radius \(r = L \sin\theta\).
A1: Calculate \(r = 1.2\text{ m}\).
M1: Set up the equation for centripetal force: \(T \sin\theta = \frac{m v^2}{r}\).
A1: Substitute values to obtain the equation \(8 = 0.5 v^2\).
A1.14: Solve for \(v\) to obtain \(4\text{ m/s}\).

The probability generating function of \( W = 2X + Y \) is given by: \( G_W(t) = \mathrm{E}(t^W) = \mathrm{E}(t^{2X+Y}) = \mathrm{E}((t^2)^X) \mathrm{E}(t^Y) = G_X(t^2)G_Y(t) \). Substituting the given PGFs: \( G_W(t) = (0.3 + 0.7t^2)^2 e^{1.5(t-1)} \). To find the mean and variance, we find the first derivative: \( G'_W(t) = 2(0.3 + 0.7t^2)(1.4t) e^{1.5(t-1)} + 1.5(0.3 + 0.7t^2)^2 e^{1.5(t-1)} = [2.8t(0.3 + 0.7t^2) + 1.5(0.3 + 0.7t^2)^2] e^{1.5(t-1)} \). Evaluating at \( t = 1 \): \( \mathrm{E}(W) = G'_W(1) = [2.8(1)(1) + 1.5(1)^2] e^0 = 2.8 + 1.5 = 4.3 \). To find \( \mathrm{Var}(W) \), we use \( \mathrm{Var}(W) = G''_W(1) + G'_W(1) - [G'_W(1)]^2 \). Let \( u(t) = 2.8t(0.3 + 0.7t^2) + 1.5(0.3 + 0.7t^2)^2 \) and \( v(t) = e^{1.5(t-1)} \). We have \( u(1) = 4.3 \) and \( v'(1) = 1.5 \). Differentiating \( u(t) \): \( u'(t) = 2.8(0.3 + 0.7t^2) + 3.92t^2 + 4.2t(0.3 + 0.7t^2) \). At \( t = 1 \), \( u'(1) = 2.8 + 3.92 + 4.2 = 10.92 \). Thus, \( G''_W(1) = u'(1)v(1) + u(1)v'(1) = 10.92(1) + 4.3(1.5) = 17.37 \). Finally, \( \mathrm{Var}(W) = 17.37 + 4.3 - 4.3^2 = 3.18 \).

M1: For using the key property \( G_W(t) = G_X(t^2)G_Y(t) \). A1: Correct expression for \( G_W(t) \). M1: Attempt to differentiate \( G_W(t) \) using the product rule. A1: Correct derivative and evaluation at \( t=1 \) to get \( \mathrm{E}(W) = 4.3 \). M1: Attempt to find second derivative \( G''_W(1) \). A1: Correct value of \( G''_W(1) = 17.37 \). A1: Correct final value of \( \mathrm{Var}(W) = 3.18 \).

For \( 0 \le x \le 4 \), the cumulative distribution function is: \( \mathrm{F}(x) = \int_0^x \frac{3}{64} (4t^2 - t^3) \, dt = \frac{3}{64} \left[ \frac{4}{3}t^3 - \frac{1}{4}t^4 \right]_0^x = \frac{3}{64} \left( \frac{4}{3}x^3 - \frac{1}{4}x^4 \right) = \frac{x^3}{16} - \frac{3x^4}{256} \). To find \( \mathrm{P}(1 < X < 3) \), we evaluate: \( \mathrm{P}(1 < X < 3) = \mathrm{F}(3) - \mathrm{F}(1) \). Here, \( \mathrm{F}(3) = \frac{27}{16} - \frac{3(81)}{256} = \frac{432 - 243}{256} = \frac{189}{256} \). \( \mathrm{F}(1) = \frac{1}{16} - \frac{3}{256} = \frac{16 - 3}{256} = \frac{13}{256} \). Therefore, \( \mathrm{P}(1 < X < 3) = \frac{189}{256} - \frac{13}{256} = \frac{176}{256} = \frac{11}{16} = 0.6875 \).

M1: Integrates the PDF to find \( \mathrm{F}(x) \). A1: Obtains \( \mathrm{F}(x) = \frac{x^3}{16} - \frac{3x^4}{256} \) with limits. M1: Uses \( \mathrm{F}(3) - \mathrm{F}(1) \) or integrates the PDF from 1 to 3. A1: Obtains \( \mathrm{F}(3) = \frac{189}{256} \). A1: Obtains \( \mathrm{F}(1) = \frac{13}{256} \). A1: Correct final probability of \( \frac{11}{16} \) (or 0.6875).

Let \( \eta \) be the population median difference (After - Before). State the hypotheses: \( H_0: \eta = 0 \) and \( H_1: \eta < 0 \) (one-tailed test). Next, order the absolute values of the differences and assign ranks: Absolute differences: \( 1, 2, 3, 4, 5, 7, 8, 10, 12, 15 \). Corresponding ranks: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Now sum the ranks for positive differences: Ranks of positive differences (+1, +3, +5) are 1, 3, 5 respectively. Thus, \( T^+ = 1 + 3 + 5 = 9 \). The sum of negative ranks is \( T^- = 2 + 4 + 6 + 7 + 8 + 9 + 10 = 46 \). The test statistic is \( T = \min(T^+, T^-) = 9 \). For a one-tailed test with \( n = 10 \) at the 5% level, the critical value from the Wilcoxon signed-rank table is 10. Since \( T = 9 \le 10 \), we reject \( H_0 \). There is sufficient evidence to conclude that reaction times are reduced after drinking coffee.

B1: States correct hypotheses \( H_0 \) and \( H_1 \) with median difference identified. M1: Orders the absolute differences and assigns ranks 1 to 10. A1: Identifies the correct ranks for positive differences: 1, 3, 5. A1: Calculates the sum of positive ranks \( T^+ = 9 \). M1: Compares test statistic \( T = 9 \) with the critical value of 10. A1: Concludes to reject \( H_0 \) with a contextually correct conclusion.

\( H_0 \): Preferred study method and gender are independent. \( H_1 \): Preferred study method and gender are associated. Row totals: Male = 120, Female = 130. Column totals: Online = 60, In-person = 100, Hybrid = 90. Grand total = 250. Expected frequencies are calculated as \( \frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}} \): Male-Online: \( 28.8 \), Male-In-person: \( 48.0 \), Male-Hybrid: \( 43.2 \); Female-Online: \( 31.2 \), Female-In-person: \( 52.0 \), Female-Hybrid: \( 46.8 \). Calculating \( \chi^2 = \sum \frac{(O-E)^2}{E} \): Male-Online: \( (35-28.8)^2 / 28.8 = 1.3347 \); Male-In-person: \( (45-48.0)^2 / 48.0 = 0.1875 \); Male-Hybrid: \( (40-43.2)^2 / 43.2 = 0.2370 \); Female-Online: \( (25-31.2)^2 / 31.2 = 1.2321 \); Female-In-person: \( (55-52.0)^2 / 52.0 = 0.1731 \); Female-Hybrid: \( (50-46.8)^2 / 46.8 = 0.2188 \). Summing these terms gives \( \chi^2 = 3.3832 \). The degrees of freedom are \( (2-1) \times (3-1) = 2 \). The critical value at 5% significance with 2 d.f. is 5.991. Since \( 3.3832 < 5.991 \), we fail to reject \( H_0 \). There is insufficient evidence to suggest an association.

B1: States correct hypotheses. M1: Calculates expected values. A1: All expected values correct. M1: Applies the \( \chi^2 \) test statistic formula. A1: Correctly calculates \( \chi^2 \approx 3.38 \). B1: Correct degrees of freedom (2) and critical value (5.991). A1: Correct non-rejection conclusion in context.

The range of \( X \) is \( [0, 1] \). When \( x=0 \), \( y=1 \); and when \( x=1 \), \( y=1/4 \). Since \( y = \frac{1}{(x+1)^2} \) is decreasing on \( [0, 1] \), the range of \( Y \) is \( \frac{1}{4} \le y \le 1 \). The cumulative distribution function of \( Y \) is: \( \mathrm{F}_Y(y) = \mathrm{P}(Y \le y) = \mathrm{P}\left(\frac{1}{(X+1)^2} \le y\right) = \mathrm{P}\left(X+1 \ge y^{-1/2}\right) = \mathrm{P}\left(X \ge y^{-1/2} - 1\right) = 1 - \mathrm{F}_X(y^{-1/2} - 1) \). Since \( \mathrm{F}_X(x) = x^2 \) for \( 0 \le x \le 1 \), we have: \( \mathrm{F}_Y(y) = 1 - (y^{-1/2} - 1)^2 = 1 - (y^{-1} - 2y^{-1/2} + 1) = 2y^{-1/2} - y^{-1} \). Differentiating with respect to \( y \) gives the PDF of \( Y \): \( g(y) = \frac{d}{dy}(2y^{-1/2} - y^{-1}) = -y^{-3/2} + y^{-2} = y^{-2} - y^{-3/2} \) for \( \frac{1}{4} \le y \le 1 \), and 0 otherwise.

B1: Correct range \( \frac{1}{4} \le y \le 1 \). M1: Correct probability statement relating \( Y \) to \( X \). M1: Finds the CDF of \( X \) as \( x^2 \). A1: Expresses the CDF of \( Y \) correctly as \( 2y^{-1/2} - y^{-1} \). M1: Differentiates the CDF of \( Y \). A1: Obtains \( y^{-2} - y^{-3/2} \). A1: Fully specifies the piecewise PDF.

First, we find the pooled estimate of the common variance: \( s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2} = \frac{7 \times 12.4 + 9 \times 15.1}{8+10-2} = \frac{86.8 + 135.9}{16} = 13.91875 \). The standard error of the difference is: \( \mathrm{SE} = \sqrt{s_p^2 \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} = \sqrt{13.91875 \left(\frac{1}{8} + \frac{1}{10}\right)} = \sqrt{13.91875 \times 0.225} \approx 1.7697 \). The degrees of freedom is \( 8 + 10 - 2 = 16 \). The critical value for a two-tailed 95% confidence interval with 16 degrees of freedom is \( t_{16}(0.025) = 2.120 \). The difference in sample means is \( 45.2 - 41.5 = 3.7 \). The margin of error is \( 2.120 \times 1.7697 = 3.7517 \). The confidence interval is \( 3.7 \pm 3.7517 \), which yields \( [-0.05, 7.45] \) to 2 decimal places.

M1: Formula for pooled estimate of variance applied. A1: Correct pooled variance \( s_p^2 = 13.91875 \). M1: Standard error of difference calculation. A1: Correct standard error \( \approx 1.7697 \). B1: Identifies the correct critical value \( t = 2.120 \). M1: Construction of confidence interval. A1: Correct final interval \( [-0.05, 7.45] \) (accept answers rounding to these values).

(a) The PGF of \( S \) is given by \( G_S(t) = G_N(G_X(t)) \). Substituting \( G_X(t) \) into \( G_N(t) \): \( G_S(t) = \frac{0.6(0.5t + 0.5t^2)}{1 - 0.4(0.5t + 0.5t^2)} = \frac{0.3(t + t^2)}{1 - (0.2t + 0.2t^2)} \). Multiplying both numerator and denominator by 10 gives: \( G_S(t) = \frac{3t(1+t)}{10 - 2t - 2t^2} \). (b) To find \( \mathrm{P}(S = 3) \), we expand \( G_S(t) \) using the binomial series: \( G_S(t) = 0.3(t + t^2) [1 - (0.2t + 0.2t^2)]^{-1} \). Using the expansion \( (1-u)^{-1} = 1 + u + u^2 + \dots \): \( [1 - (0.2t + 0.2t^2)]^{-1} = 1 + (0.2t + 0.2t^2) + (0.2t + 0.2t^2)^2 + \dots = 1 + 0.2t + 0.24t^2 + 0.08t^3 + \dots \). Now, multiplying by \( 0.3t + 0.3t^2 \): \( G_S(t) = (0.3t + 0.3t^2)(1 + 0.2t + 0.24t^2 + 0.08t^3 + \dots) \). The coefficient of \( t^3 \) is: \( (0.3 \times 0.24) + (0.3 \times 0.2) = 0.072 + 0.06 = 0.132 \). Hence, \( \mathrm{P}(S = 3) = 0.132 \).

M1: States or uses \( G_S(t) = G_N(G_X(t)) \). A1: Correct substitution. A1: Clearly simplifies to show the given result. M1: Recognizes that \( \mathrm{P}(S=3) \) is the coefficient of \( t^3 \). M1: Expands the denominator binomial series up to at least \( t^2 \). A1: Correctly identifies all contributions to the \( t^3 \) term. A1: Obtains \( 0.132 \).