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**(a)**
Let \((t, 2t)^T\) be a general point on the line \(y = 2x\).
Its image under the transformation represented by \(\mathbf{M}\) is:
\[\begin{pmatrix} 3 & k \\ -2 & -4 \end{pmatrix} \begin{pmatrix} t \\ 2t \end{pmatrix} = \begin{pmatrix} 3t + 2kt \\ -2t - 8t \end{pmatrix} = \begin{pmatrix} (3 + 2k)t \\ -10t \end{pmatrix}\]
For \(y = 2x\) to be an invariant line, the image point must also lie on \(y = 2x\) for all \(t\):
\[-10t = 2(3 + 2k)t\]
Since \(t \neq 0\):
\[-10 = 6 + 4k \implies 4k = -16 \implies k = -4\]

**(b)**
With \(k = -4\), the matrix is \(\mathbf{M} = \begin{pmatrix} 3 & -4 \\ -2 & -4 \end{pmatrix}\).
Let \(y = mx\) be an invariant line through the origin.
A general point on this line is \((t, mt)^T\). Its image is:
\[\begin{pmatrix} 3 & -4 \\ -2 & -4 \end{pmatrix} \begin{pmatrix} t \\ mt \end{pmatrix} = \begin{pmatrix} (3 - 4m)t \\ (-2 - 4m)t \end{pmatrix}\]
Since the image must lie on \(y = mx\):
\[-2 - 4m = m(3 - 4m)\]
\[-2 - 4m = 3m - 4m^2\]
\[4m^2 - 7m - 2 = 0\]
\[(4m + 1)(m - 2) = 0\]
Since \(m = 2\) is the line from part (a), the other invariant line corresponds to \(m = -\frac{1}{4}\).
Thus, the equation of the other invariant line is \(y = -\frac{1}{4}x\) (or \(x + 4y = 0\)).

**(c)**
With \(k = 1\), \(\mathbf{M} = \begin{pmatrix} 3 & 1 \\ -2 & -4 \end{pmatrix}\).

**Method 1: Using the inverse matrix**
Let \((x, y)\) be a point on the line \(y = 3x - 1\) and \((X, Y)\) be its image.
\[\begin{pmatrix} X \\ Y \end{pmatrix} = \begin{pmatrix} 3 & 1 \\ -2 & -4 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}\]
Find the inverse of \(\mathbf{M}\):
\[\det(\mathbf{M}) = 3(-4) - 1(-2) = -10\]
\[\mathbf{M}^{-1} = -\frac{1}{10} \begin{pmatrix} -4 & -1 \\ 2 & 3 \end{pmatrix} = \begin{pmatrix} 0.4 & 0.1 \\ -0.2 & -0.3 \end{pmatrix}\]
So:
\[x = 0.4X + 0.1Y\]
\[y = -0.2X - 0.3Y\]
Substitute these into \(y = 3x - 1\):
\[-0.2X - 0.3Y = 3(0.4X + 0.1Y) - 1\]
\[-0.2X - 0.3Y = 1.2X + 0.3Y - 1\]
\[-0.6Y = 1.4X - 1\]
Multiply by \(-10\):
\[6Y = -14X + 10 \implies Y = -\frac{7}{3}X + \frac{5}{3}\]
Thus, the equation of \(l\) is \(y = -\frac{7}{3}x + \frac{5}{3}\).

**Method 2: Using two points**
Choose two points on the line \(y = 3x - 1\), for example, \(A(0, -1)\) and \(B(1, 2)\).
Find their images under \(\mathbf{M}\):
\[\mathbf{M} \begin{pmatrix} 0 \\ -1 \end{pmatrix} = \begin{pmatrix} 3 & 1 \\ -2 & -4 \end{pmatrix} \begin{pmatrix} 0 \\ -1 \end{pmatrix} = \begin{pmatrix} -1 \\ 4 \end{pmatrix}\]
\[\mathbf{M} \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 3 & 1 \\ -2 & -4 \end{pmatrix} \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 5 \\ -10 \end{pmatrix}\]
So the image line passes through \((-1, 4)\) and \(5, -10)\).
The gradient of this line is:
\[p = \frac{-10 - 4}{5 - (-1)} = \frac{-14}{6} = -\frac{7}{3}\]
The equation of the line is:
\[y - 4 = -\frac{7}{3}(x + 1) \implies y = -\frac{7}{3}x + \frac{5}{3}\]

**(a)**
* **M1**: For multiplying \(\mathbf{M}\) by \(\begin{pmatrix} t \\ 2t \end{pmatrix}\) (or equivalent) and setting up an equation for \(k\).
* **A1**: For obtaining a correct equation in \(k\), e.g., \(-10 = 2(3 + 2k)\).
* **A1**: For \(k = -4\).

**(b)**
* **M1**: For setting up the equation for the invariant line using \(k = -4\), e.g., \(-2-4m = m(3-4m)\).
* **A1**: For the correct quadratic \(4m^2 - 7m - 2 = 0\) (or equivalent).
* **A1**: For \(y = -\frac{1}{4}x\) (or \(x + 4y = 0\); ignore \(y=2x\) if also given).

**(c)**
* **M1**: For a valid method to find the transformation of the line, e.g., finding \(\mathbf{M}^{-1}\) or transforming two points on the line.
* **A1**: For correctly finding \(\mathbf{M}^{-1} = \begin{pmatrix} 0.4 & 0.1 \\ -0.2 & -0.3 \end{pmatrix}\) (or equivalent fraction form), OR for correctly finding the images of two distinct points (e.g., \((-1, 4)\) and \((5, -10)\)).
* **M1**: For substituting the expressions for \(x\) and \(y\) into \(y = 3x - 1\), OR for finding the gradient of the line connecting the two image points.
* **A1**: For a correct intermediate equation, e.g., \(-0.2X - 0.3Y = 1.2X + 0.3Y - 1\) OR gradient \(p = -\frac{7}{3}\).
* **A1**: For \(y = -\frac{7}{3}x + \frac{5}{3}\) (accept equivalent fractions, but must be in the form \(y = px+q\)).

**Step 1: Base Case**
For \(n=1\):
\(\mathbf{A}^1 = \begin{pmatrix} 2(1)+1 & -4(1) \\ 1 & 1-2(1) \end{pmatrix} = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}\), which is equal to \(\mathbf{A}\).
Therefore, the statement is true for \(n=1\).

**Step 2: Inductive Hypothesis**
Assume that the statement is true for \(n=k\), where \(k\) is a positive integer.
That is, \(\mathbf{A}^k = \begin{pmatrix} 2k+1 & -4k \\ k & 1-2k \end{pmatrix}\).

**Step 3: Inductive Step**
We need to show that the statement is true for \(n=k+1\).
\(\mathbf{A}^{k+1} = \mathbf{A}^k \mathbf{A}\)
\(\mathbf{A}^{k+1} = \begin{pmatrix} 2k+1 & -4k \\ k & 1-2k \end{pmatrix} \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}\)
\(\mathbf{A}^{k+1} = \begin{pmatrix} (2k+1)(3) + (-4k)(1) & (2k+1)(-4) + (-4k)(-1) \\ (k)(3) + (1-2k)(1) & (k)(-4) + (1-2k)(-1) \end{pmatrix}\)
\(\mathbf{A}^{k+1} = \begin{pmatrix} 6k+3 - 4k & -8k-4 + 4k \\ 3k + 1 - 2k & -4k - 1 + 2k \end{pmatrix}\)
\(\mathbf{A}^{k+1} = \begin{pmatrix} 2k+3 & -4k-4 \\ k+1 & -2k-1 \end{pmatrix}\)
\(\mathbf{A}^{k+1} = \begin{pmatrix} 2(k+1)+1 & -4(k+1) \\ k+1 & 1-2(k+1) \end{pmatrix}\)

This is of the same form as the original statement with \(n\) replaced by \(k+1\).

**Conclusion**
Since the statement is true for \(n=1\), and if it is true for \(n=k\) then it is also true for \(n=k+1\), the statement is true for all positive integers \(n\) by mathematical induction.

**B1**: Verifies the base case \(n=1\).
**M1**: Assumes the statement is true for \(n=k\).
**M1**: Multiplies \(\mathbf{A}^k\) by \(\mathbf{A}\) (or vice versa).
**A1**: Obtains correct elements in the multiplied matrix: \(\begin{pmatrix} 2k+3 & -4k-4 \\ k+1 & -2k-1 \end{pmatrix}\).
**A1**: Correctly factors/simplifies to show the form for \(n=k+1\).
**B1**: Gives a complete, mathematically sound concluding statement of induction.

Let the roots of \(x^3 + 3x^2 - x + 2 = 0\) be \(\alpha, \beta, \gamma\).

(a) Method 1: Substitution
Let \(y = x^2\), which gives \(x = \sqrt{y}\).
Substituting \(x = \sqrt{y}\) into the original cubic equation:
\((\sqrt{y})^3 + 3(\sqrt{y})^2 - \sqrt{y} + 2 = 0\)
\(y\sqrt{y} + 3y - \sqrt{y} + 2 = 0\)
\(\sqrt{y}(y - 1) = -(3y + 2)\)
Squaring both sides:
\(y(y - 1)^2 = (3y + 2)^2\)
\(y(y^2 - 2y + 1) = 9y^2 + 12y + 4\)
\(y^3 - 2y^2 + y = 9y^2 + 12y + 4\)
\(y^3 - 11y^2 - 11y - 4 = 0\) as required.

Method 2: Root Relationships
From the original equation, we have:
\(\Sigma \alpha = -3\)
\(\Sigma \alpha\beta = -1\)
\(\alpha\beta\gamma = -2\)
For the new equation with roots \(u_1 = \alpha^2, u_2 = \beta^2, u_3 = \gamma^2\):
\(\Sigma \alpha^2 = (\Sigma \alpha)^2 - 2\Sigma \alpha\beta = (-3)^2 - 2(-1) = 9 + 2 = 11\)
\(\Sigma \alpha^2\beta^2 = (\Sigma \alpha\beta)^2 - 2\alpha\beta\gamma(\Sigma \alpha) = (-1)^2 - 2(-2)(-3) = 1 - 12 = -11\)
\(\alpha^2\beta^2\gamma^2 = (\alpha\beta\gamma)^2 = (-2)^2 = 4\)
The required cubic equation is \(y^3 - (\Sigma \alpha^2)y^2 + (\Sigma \alpha^2\beta^2)y - \alpha^2\beta^2\gamma^2 = 0\), which gives:
\(y^3 - 11y^2 - 11y - 4 = 0\).

(b) Let \(S_n = \alpha^n + \beta^n + \gamma^n\).
Using the recurrence relation for \(x^3 + 3x^2 - x + 2 = 0\):
\(S_3 + 3S_2 - S_1 + 6 = 0\)
We know \(S_1 = \Sigma \alpha = -3\) and \(S_2 = \Sigma \alpha^2 = 11\).
Substituting these values in:
\(S_3 + 3(11) - (-3) + 6 = 0\)
\(S_3 + 33 + 3 + 6 = 0\)
\(S_3 = -42\).
Alternatively, using the identity:
\(\alpha^3 + \beta^3 + \gamma^3 = (\alpha+\beta+\gamma)(\alpha^2+\beta^2+\gamma^2 - (\alpha\beta+\beta\gamma+\gamma\alpha)) + 3\alpha\beta\gamma\)
\(\alpha^3 + \beta^3 + \gamma^3 = (-3)(11 - (-1)) + 3(-2) = -3(12) - 6 = -42\).

(c) We want to find \(\alpha^{-2} + \beta^{-2} + \gamma^{-2} = \frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2}\).
This can be written as:
\(\frac{\alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2}{\alpha^2\beta^2\gamma^2}\)
From the equation in (a), we know that:
\(\alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 = -11\)
\(\alpha^2\beta^2\gamma^2 = 4\)
Thus, the required sum is:
\(\frac{-11}{4} = -2.75\).

Alternatively, let \(w = \frac{1}{y}\). The equation for \(w\) is:
\(1 - 11w - 11w^2 - 4w^3 = 0 \implies 4w^3 + 11w^2 + 11w - 1 = 0\)
The sum of the roots is \(\Sigma w = -\frac{11}{4}\).

(a)
M1: For substituting \(y = x^2\) (or \(x = \sqrt{y}\)) and attempting to isolate the square root terms, OR for identifying the values of \(\Sigma \alpha = -3\), \(\Sigma \alpha\beta = -1\) and \(\alpha\beta\gamma = -2\).
A1: For squaring and expanding to obtain a polynomial in \(y\), OR for finding \(\Sigma \alpha^2 = 11\).
M1: For simplifying the expression to cubic form, OR for using the correct identity to find \(\Sigma \alpha^2\beta^2 = -11\).
A1: For completing the proof to obtain \(y^3 - 11y^2 - 11y - 4 = 0\) with no errors.

(b)
M1: For using the recurrence relation \(S_3 + 3S_2 - S_1 + 6 = 0\) with their \(S_1\) and \(S_2\), OR for using the identity \(\alpha^3+\beta^3+\gamma^3 = (\Sigma \alpha)(\Sigma \alpha^2 - \Sigma \alpha\beta) + 3\alpha\beta\gamma\).
A1: For obtaining the correct value of \(-42\).

(c)
M1: For writing \(\alpha^{-2} + \beta^{-2} + \gamma^{-2} = \frac{\alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2}{\alpha^2\beta^2\gamma^2}\) OR substituting \(y = 1/w\) to get \(4w^3 + 11w^2 + 11w - 1 = 0\).
A1: For substituting the correct values of \(-11\) and \(4\) into the fraction, OR for obtaining the correct coefficients for the equation in \(w\).
A1: For the final answer \(-\frac{11}{4}\) (or \(-2.75\)).

First, we show the algebraic identity by simplifying the right-hand side:
\[ \frac{1}{2(2r-1)^2} - \frac{1}{2(2r+1)^2} = \frac{(2r+1)^2 - (2r-1)^2}{2(2r-1)^2(2r+1)^2} \]
\[ = \frac{(4r^2+4r+1) - (4r^2-4r+1)}{2(2r-1)^2(2r+1)^2} \]
\[ = \frac{8r}{2(2r-1)^2(2r+1)^2} = \frac{4r}{(2r-1)^2(2r+1)^2} \]
This confirms the identity.

Now, we use the method of differences to find the sum:
\[ S_n = \sum_{r=1}^n \frac{4r}{(2r-1)^2(2r+1)^2} = \sum_{r=1}^n \left( \frac{1}{2(2r-1)^2} - \frac{1}{2(2r+1)^2} \right) \]
Listing the terms:
\[ r=1: \quad \frac{1}{2(1)^2} - \frac{1}{2(3)^2} \]
\[ r=2: \quad \frac{1}{2(3)^2} - \frac{1}{2(5)^2} \]
\[ \dots \]
\[ r=n: \quad \frac{1}{2(2n-1)^2} - \frac{1}{2(2n+1)^2} \]
Summing these terms, all intermediate terms cancel out, leaving:
\[ S_n = \frac{1}{2} - \frac{1}{2(2n+1)^2} \]

For the second part, as \(n \to \infty\), the term \\frac{1}{2(2n+1)^2} \\to 0\\, so the sum to infinity is:
\[ S_{\infty} = \frac{1}{2} = 0.5 \]

We want to find the least integer \(N\) such that \(S_N > 0.4999\):
\[ \frac{1}{2} - \frac{1}{2(2N+1)^2} > 0.4999 \]
\[ 0.5 - 0.4999 > \frac{1}{2(2N+1)^2} \]
\[ 0.0001 > \frac{1}{2(2N+1)^2} \]
\[ 2(2N+1)^2 > 10000 \]
\[ (2N+1)^2 > 5000 \]
\[ 2N+1 > \sqrt{5000} \approx 70.7107 \]
\[ 2N > 69.7107 \]
\[ N > 34.855 \]
Since \(N\) must be an integer, the least integer value is \(N = 35\).

**Part (i) [5 Marks]**
- **M1**: Attempts to combine the RHS of the identity over a common denominator, or uses partial fractions on the LHS.
- **A1**: Completes algebra correctly to show the identity.
- **M1**: Uses the method of differences, writing out at least three terms including the first and the \(n\)-th term to show cancellation.
- **A1**: Correctly identifies cancelling terms and writes down the uncancelled terms.
- **A1**: Obtains correct simplified sum: \(S_n = \frac{1}{2} - \frac{1}{2(2n+1)^2}\).

**Part (ii) [3 Marks]**
- **B1**: Identifies the sum to infinity as \(\frac{1}{2}\) (or 0.5) by taking \(n \to \infty\).
- **M1**: Sets up the inequality \(S_N > 0.4999\) and attempts to solve for \(N\).
- **A1**: Obtains \(N = 35\) as the least integer (must be a single integer, not a range).

**(a) Find the shortest distance between \(l_1\) and \(l_2\)**

First, identify the direction vectors of the two lines:
\[\mathbf{d}_1 = 3\mathbf{i} - 2\mathbf{j} = \begin{pmatrix} 3 \\ -2 \\ 0 \end{pmatrix}\]
\[\mathbf{d}_2 = 2\mathbf{j} - \mathbf{k} = \begin{pmatrix} 0 \\ 2 \\ -1 \end{pmatrix}\]

The common perpendicular vector \(\mathbf{n}_1\) to both lines is given by the cross product \(\mathbf{d}_1 \times \mathbf{d}_2\):
\[\mathbf{n}_1 = \begin{pmatrix} 3 \\ -2 \\ 0 \end{pmatrix} \times \begin{pmatrix} 0 \\ 2 \\ -1 \end{pmatrix} = \begin{pmatrix} (-2)(-1) - (0)(2) \\ (0)(0) - (3)(-1) \\ (3)(2) - (-2)(0) \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\ 6 \end{pmatrix}\]

The magnitude of \(\mathbf{n}_1\) is:
\[|\mathbf{n}_1| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7\]

Choose a point \(A\) on \(l_1\) and a point \(B\) on \(l_2\):
\[A = (1, 1, 1), \quad B = (2, 1, 3)\]

The displacement vector \(\vec{AB}\) is:
\[\vec{AB} = \begin{pmatrix} 2 - 1 \\ 1 - 1 \\ 3 - 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}\]

The shortest distance \(d\) between the skew lines is the projection of \(\vec{AB}\) onto the common perpendicular \(\mathbf{n}_1\):
\[d = \frac{|\vec{AB} \cdot \mathbf{n}_1|}{|\mathbf{n}_1|} = \frac{|(1)(2) + (0)(3) + (2)(6)|}{7} = \frac{|2 + 0 + 12|}{7} = \frac{14}{7} = 2\]

**(b) Find the Cartesian equation of the plane \(\Pi_1\)**

Since \(\Pi_1\) contains \(l_1\) and is parallel to \(l_2\), its normal vector is parallel to the common perpendicular \(\mathbf{n}_1 = 2\mathbf{i} + 3\mathbf{j} + 6\mathbf{k}\).

The equation of the plane is of the form:
\[2x + 3y + 6z = D\]

Since \(\Pi_1\) contains \(l_1\), the point \(A(1, 1, 1)\) lies on \(\Pi_1\):
\[D = 2(1) + 3(1) + 6(1) = 11\]

Thus, the Cartesian equation of \(\Pi_1\) is:
\[2x + 3y + 6z = 11\]

**(c) Find the acute angle between \(\Pi_1\) and \(\Pi_2\)**

The plane \(\Pi_2\) contains \(l_2\), so it contains the point \(B(2, 1, 3)\) and is parallel to the direction vector \(\mathbf{d}_2 = 2\mathbf{j} - \mathbf{k}\).
It also contains the point \(C(3, 1, 4)\).

Find a second vector in the plane \(\Pi_2\):
\[\vec{BC} = \begin{pmatrix} 3 - 2 \\ 1 - 1 \\ 4 - 3 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}\]

The normal vector \(\mathbf{n}_2\) to \(\Pi_2\) is perpendicular to both \(\mathbf{d}_2\) and \vec{BC}:
\[\mathbf{n}_2 = \mathbf{d}_2 \times \vec{BC} = \begin{pmatrix} 0 \\ 2 \\ -1 \end{pmatrix} \times \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} (2)(1) - (-1)(0) \\ (-1)(1) - (0)(1) \\ (0)(0) - (2)(1) \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \\ -2 \end{pmatrix}\]

The magnitude of \(\mathbf{n}_2\) is:
\[|\mathbf{n}_2| = \sqrt{2^2 + (-1)^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3\]

The angle \(\theta\) between the planes \(\Pi_1\) and \(\Pi_2\) is the angle between their normals \(\mathbf{n}_1\) and \(\mathbf{n}_2\):
\[\cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|} = \frac{|(2)(2) + (3)(-1) + (6)(-2)|}{7 \times 3} = \frac{|4 - 3 - 12|}{21} = \frac{|-11|}{21} = \frac{11}{21}\]

\[\theta = \arccos\left(\frac{11}{21}\right) \approx 58.411...^\circ\]

To 1 decimal place, the acute angle is \(58.4^\circ\).

**(a)**
* **M1**: Attempt to find the vector cross product of the two direction vectors \(\mathbf{d}_1\) and \(\mathbf{d}_2\).
* **A1**: Obtain correct common perpendicular vector \(\mathbf{n}_1 = 2\mathbf{i} + 3\mathbf{j} + 6\mathbf{k}\) (or any non-zero multiple).
* **M1**: Find a displacement vector between a point on \(l_1\) and a point on \(l_2\) (e.g. \(\mathbf{i} + 2\mathbf{k}\)).
* **M1**: Substitute their vectors into the formula for the shortest distance \(d = \frac{|\vec{AB} \cdot \mathbf{n}_1|}{|\mathbf{n}_1|}\).
* **A1**: Obtain the correct distance of \(2\) (must be positive).

**(b)**
* **M1**: State or imply that the normal to \(\Pi_1\) is \(\mathbf{n}_1\) and form an equation of the form \(\mathbf{r} \cdot \mathbf{n}_1 = d\).
* **M1**: Substitute the coordinates of a point on \(l_1\) (e.g., \((1, 1, 1)\)) into their plane equation.
* **A1**: Obtain the correct equation \(2x + 3y + 6z = 11\) (or any equivalent Cartesian form).

**(c)**
* **M1**: Find a vector connecting \(C\) to a point on \(l_2\) (e.g., \(\vec{BC}\)) and attempt the cross product of this vector with \(\mathbf{d}_2\) to find the normal vector \(\mathbf{n}_2\) of \(\Pi_2\).
* **A1**: Obtain the correct normal vector \(\mathbf{n}_2 = 2\mathbf{i} - \mathbf{j} - 2\mathbf{k}\) (or any non-zero multiple).
* **M1**: Apply the formula \(\cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|}\) and calculate \(\theta\).
* **A1**: Obtain \(58.4^\circ\) (accept \(1.02\) radians).

(i) To find the vertical asymptote, we set the denominator to zero:
\[ 2x - 4 = 0 \implies x = 2. \]
To find the oblique asymptote, we perform algebraic division on the equation of \( C \):
\[ y = \frac{x^2 - 2x + 9}{2x - 4} = \frac{\frac{1}{2}x(2x - 4) + 9}{2x - 4} = \frac{1}{2}x + \frac{9}{2x - 4}. \]
As \( x \to \pm\infty \), \( \frac{9}{2x - 4} \to 0 \).
Thus, the oblique asymptote is:
\[ y = \frac{1}{2}x. \]

(ii) To find the turning points, we differentiate \( y \) with respect to \( x \). Using the expression \( y = \frac{1}{2}x + 9(2x - 4)^{-1} \):
\[ \frac{\text{d}y}{\text{d}x} = \frac{1}{2} - 9(2x - 4)^{-2} \cdot 2 = \frac{1}{2} - \frac{18}{(2x - 4)^2}. \]
Alternatively, using the quotient rule on \( y = \frac{x^2 - 2x + 9}{2x - 4} \):
\[ \frac{\text{d}y}{\text{d}x} = \frac{(2x - 2)(2x - 4) - 2(x^2 - 2x + 9)}{(2x - 4)^2} = \frac{4x^2 - 12x + 8 - 2x^2 + 4x - 18}{(2x - 4)^2} = \frac{2x^2 - 8x - 10}{(2x - 4)^2}. \]
Setting \( \frac{\text{d}y}{\text{d}x} = 0 \):
\[ 2x^2 - 8x - 10 = 0 \implies x^2 - 4x - 5 = 0 \implies (x - 5)(x + 1) = 0. \]
Thus, \( x = 5 \) or \( x = -1 \).

When \( x = 5 \):
\[ y = \frac{5^2 - 2(5) + 9}{2(5) - 4} = \frac{25 - 10 + 9}{6} = 4. \]

When \( x = -1 \):
\[ y = \frac{(-1)^2 - 2(-1) + 9}{2(-1) - 4} = \frac{1 + 2 + 9}{-6} = -2. \]

Therefore, the coordinates of the turning points are \( (5, 4) \) and \( (-1, -2) \).

(iii) For the sketch:
- Vertical asymptote is \( x = 2 \).
- Oblique asymptote is \( y = \frac{1}{2}x \).
- The \( y \)-intercept is found by setting \( x = 0 \):
\[ y = \frac{9}{-4} = -2.25 \implies (0, -2.25). \]
- There are no \( x \)-intercepts since the numerator \( x^2 - 2x + 9 = (x - 1)^2 + 8 > 0 \) for all real \( x \).
- The turning points are \( (-1, -2) \) (which is a local maximum) and \( (5, 4) \) (which is a local minimum).
- The graph consists of two branches: one to the left of \( x = 2 \) passing through \( (0, -2.25) \) with a local maximum at \( (-1, -2) \), and one to the right of \( x = 2 \) with a local minimum at \( (5, 4) \).

(iv) To find the intersection of \( y = kx \) and \( C \):
\[ kx = \frac{x^2 - 2x + 9}{2x - 4} \implies 2kx^2 - 4kx = x^2 - 2x + 9 \implies (2k - 1)x^2 + (2 - 4k)x - 9 = 0. \]
For the line and the curve to intersect at two distinct points, this quadratic equation must have two distinct real roots, meaning its discriminant \( \Delta > 0 \):
\[ \Delta = (2 - 4k)^2 - 4(2k - 1)(-9) > 0 \]
\[ (2 - 4k)^2 + 36(2k - 1) > 0 \]
\[ 4(1 - 2k)^2 + 36(2k - 1) > 0 \]
\[ 4(4k^2 - 4k + 1) + 72k - 36 > 0 \]
\[ 16k^2 - 16k + 4 + 72k - 36 > 0 \]
\[ 16k^2 + 56k - 32 > 0. \]
Dividing the inequality by 8:
\[ 2k^2 + 7k - 4 > 0 \implies (2k - 1)(k + 4) > 0. \]
Thus, the set of values of \( k \) is \( k < -4 \) or \( k > \frac{1}{2} \).

(i)
B1: State the vertical asymptote \( x = 2 \).
M1: Attempt algebraic division to write \( y \) in the form \( ax + b + \frac{c}{2x-4} \).
A1: Obtain correct oblique asymptote \( y = \frac{1}{2}x \).

(ii)
M1: Attempt differentiation using quotient rule or chain rule (at least one term correct).
A1: Correct derivative, e.g., \( \frac{\text{d}y}{\text{d}x} = \frac{2x^2 - 8x - 10}{(2x-4)^2} \) or equivalent.
M1: Set derivative to zero and solve the resulting quadratic equation to find two \( x \)-values.
A1: Correct coordinates \( (5, 4) \) and \( (-1, -2) \).

(iii)
B1: Draw both asymptotes correctly, labelled with their equations or clearly showing their positions.
B1: Correctly position and label the \( y \)-intercept at \( (0, -2.25) \) and show no intersection with the \( x \)-axis.
B1: Correctly plot and label the turning points \( (5, 4) \) and \( (-1, -2) \).
B1: Draw two smooth, well-positioned branches asymptotically approaching the lines \( x = 2 \) and \( y = \frac{1}{2}x \).

(iv)
M1: Equate \( kx \) to the curve's equation and rearrange to form a standard quadratic equation in \( x \).
M1: State and apply the condition for two distinct real roots, \( \Delta > 0 \), on their quadratic equation.
A1: Obtain the correct ranges \( k < -4 \) or \( k > \frac{1}{2} \) (or equivalent interval notation).

(a) At \(\theta = 0\), \(r = 2(1+1) = 4\), so the intersection with the initial line is \((4, 0)\).
At \(\theta = \frac{\pi}{2}\), \(r = 2(1+0) = 2\), so the intersection with the line \(\theta = \frac{\pi}{2}\) is \((2, \frac{\pi}{2})\).
At \(\theta = \pi\), \(r = 2(1-1) = 0\), which is the pole \((0, \pi)\).
The sketch is a half-cardioid starting at \((4,0)\) in the upper half-plane, passing through \((2, \frac{\pi}{2})\), and ending at the pole.

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

(c) 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 find the points where the tangent is parallel to the initial line, we set \(\frac{dy}{d\theta} = 0\):
\(\frac{dy}{d\theta} = 2\cos\theta + 2\cos 2\theta = 0\)
\(2\cos\theta + 2(2\cos^2\theta - 1) = 0\)
\(4\cos^2\theta + 2\cos\theta - 2 = 0\)
\(2(2\cos\theta - 1)(\cos\theta + 1) = 0\)
This gives:
\(\cos\theta = \frac{1}{2} \implies \theta = \frac{\pi}{3}\) (since \(0 \le \theta \le \pi\))
\(\cos\theta = -1 \implies \theta = \pi\) (which is the pole, excluded by the question).
For \(\theta = \frac{\pi}{3}\):
\(r = 2\left(1 + \cos\frac{\pi}{3}\right) = 2\left(1 + \frac{1}{2}\right) = 3\).
Thus, the polar coordinates of the point are \((3, \frac{\pi}{3})\).

(a) [4 marks]
- B1: Correct overall shape of a half-cardioid in the upper half-plane.
- B1: Correct intersection at \((4, 0)\) shown on the sketch or clearly labeled.
- B1: Correct intersection at \((2, \frac{\pi}{2})\) shown or clearly labeled.
- B1: Correct ending at the pole \((0, \pi)\).

(b) [6 marks]
- M1: Uses area formula \(A = \frac{1}{2} \int r^2 d\theta\) with limits \(0\) and \(\pi\).
- A1: Obtains \(2 \int_{0}^{\pi} (1 + 2\cos\theta + \cos^2\theta) d\theta\).
- M1: Uses double-angle identity \(\cos^2\theta = \frac{1+\cos 2\theta}{2}\) to rewrite integrand.
- M1: Integrates term-by-term (at least two terms integrated correctly).
- A1: Obtains correct integrated expression \(2 \left[ \frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta \right]\).
- A1: Obtains final correct area of \(3\pi\).

(c) [6 marks]
- M1: Sets up the equation \(y = r\sin\theta\).
- M1: Differentiates \(y\) with respect to \(\theta\) using product rule or double-angle identity.
- A1: Obtains correct derivative \(\frac{dy}{d\theta} = 2\cos\theta + 2\cos 2\theta\) (or equivalent quadratic in \(\cos\theta\)).
- M1: Sets \(\frac{dy}{d\theta} = 0\) and solves the resulting quadratic in \(\cos\theta\).
- A1: Identifies \(\theta = \frac{\pi}{3}\) (accepting \(r = 3\) implicitly here).
- A1: Expresses final answer as coordinates \((3, \frac{\pi}{3})\) (must be in polar form \((r, \theta)\)).

Method 1: Using standard series

We know the standard Maclaurin series for \(\sinh x\) and \(\ln(1 + u)\):
\(\sinh x = x + \frac{x^3}{6} + \mathcal{O}(x^5)\)
\(\ln(1 + u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \mathcal{O}(u^4)\)

Substituting \(u = \sinh x\) into the expansion of \(\ln(1 + u)\):
\(\ln(1 + \sinh x) = \left(x + \frac{x^3}{6}\right) - \frac{1}{2}\left(x + \frac{x^3}{6}\right)^2 + \frac{1}{3}\left(x + \frac{x^3}{6}\right)^3 - \dots\)

Now we expand each term, keeping only powers of \(x\) up to \(x^3\):
\(\ln(1 + \sinh x) = \left(x + \frac{x^3}{6}\right) - \frac{1}{2}\left(x^2 + \dots\right) + \frac{1}{3}\left(x^3 + \dots\right)\)
\(\ln(1 + \sinh x) = x - \frac{1}{2}x^2 + \left(\frac{1}{6} + \frac{1}{3}\right)x^3\)
\(\ln(1 + \sinh x) = x - \frac{1}{2}x^2 + \frac{1}{2}x^3\)

Method 2: By successive differentiation

Let \(f(x) = \ln(1 + \sinh x)\). Then:
\(f(0) = \ln(1 + 0) = 0\)

First derivative:
\(f'(x) = \frac{\cosh x}{1 + \sinh x} \implies f'(0) = \frac{1}{1} = 1\)

Second derivative:
\(f''(x) = \frac{\sinh x(1 + \sinh x) - \cosh^2 x}{(1 + \sinh x)^2} = \frac{\sinh x - 1}{(1 + \sinh x)^2} \implies f''(0) = \frac{-1}{1} = -1\)

Third derivative:
\(f'''(x) = \frac{\cosh x(1 + \sinh x)^2 - 2(1 + \sinh x)\cosh x(\sinh x - 1)}{(1 + \sinh x)^4} = \frac{3\cosh x - \cosh x\sinh x}{(1 + \sinh x)^3} \implies f'''(0) = 3\)

Applying Maclaurin's formula:
\(f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots\)

\(f(x) = x - \frac{1}{2}x^2 + \frac{1}{2}x^3\)

M1: For expressing \(\sinh x\) or \(f'(x)\) correctly and evaluating at \(x = 0\).
M1: For expressing \(\ln(1 + u)\) or \(f''(x)\) correctly and evaluating at \(x = 0\).
M1: For substituting and expanding up to the \(x^3\) term, or finding \(f'''(0) = 3\).
A1: For obtaining the correct quadratic approximation: \(x - \frac{1}{2}x^2\).
A1: For obtaining the correct cubic term, giving the final series: \(x - \frac{1}{2}x^2 + \frac{1}{2}x^3\).

**(a)**

First, we find the derivative of \( y = \frac{1}{3} \cosh(3x) \):
\[ \frac{dy}{dx} = \sinh(3x) \]

Using the formula for arc length:
\[ s = \int_{0}^{\ln 2} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \]

Substitute \( \frac{dy}{dx} \):
\[ s = \int_{0}^{\ln 2} \sqrt{1 + \sinh^2(3x)} \, dx \]

Since \( 1 + \sinh^2(3x) = \cosh^2(3x) \) and \( \cosh(3x) > 0 \):
\[ s = \int_{0}^{\ln 2} \cosh(3x) \, dx \]

Integrating gives:
\[ s = \left[ \frac{1}{3} \sinh(3x) \right]_{0}^{\ln 2} = \frac{1}{3} \sinh(3\ln 2) - \frac{1}{3} \sinh(0) \]

Since \( 3\ln 2 = \ln(2^3) = \ln 8 \):
\[ \sinh(\ln 8) = \frac{e^{\ln 8} - e^{-\ln 8}}{2} = \frac{8 - \frac{1}{8}}{2} = \frac{63}{16} \]

Therefore, the arc length is:
\[ s = \frac{1}{3} \times \frac{63}{16} = \frac{21}{16} \]


**(b)**

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

The area of the surface of revolution \( S \) generated by rotating the curve about the \( x \)-axis is:
\[ S = 2\pi \int_{0}^{\ln 3} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \]

Substitute \( y \) and \( \frac{dy}{dx} \):
\[ S = 2\pi \int_{0}^{\ln 3} \cosh x \sqrt{1 + \sinh^2 x} \, dx = 2\pi \int_{0}^{\ln 3} \cosh^2 x \, dx \]

Using the identity \( \cosh^2 x = \frac{1}{2}(\cosh(2x) + 1) \):
\[ S = 2\pi \int_{0}^{\ln 3} \frac{1}{2}(\cosh(2x) + 1) \, dx = \pi \int_{0}^{\ln 3} (\cosh(2x) + 1) \, dx \]

Integrating:
\[ S = \pi \left[ \frac{1}{2} \sinh(2x) + x \right]_{0}^{\ln 3} \]

At the upper limit \( x = \ln 3 \):
\[ 2x = 2\ln 3 = \ln 9 \]
\[ \sinh(\ln 9) = \frac{e^{\ln 9} - e^{-\ln 9}}{2} = \frac{9 - \frac{1}{9}}{2} = \frac{80}{18} = \frac{40}{9} \]

So,
\[ \frac{1}{2} \sinh(2\ln 3) + \ln 3 = \frac{20}{9} + \ln 3 \]

At the lower limit \( x = 0 \):
\[ \frac{1}{2} \sinh(0) + 0 = 0 \]

Thus, the exact surface area is:
\[ S = \pi \left(\frac{20}{9} + \ln 3\right) \]

**(a)**
* **M1**: Differentiate \( y \) to find \( \frac{dy}{dx} = \sinh(3x) \).
* **M1**: Apply the arc length formula and use the hyperbolic identity \( 1 + \sinh^2(3x) = \cosh^2(3x) \) to simplify the integrand.
* **A1**: Obtain the correct integrated expression \( \left[ \frac{1}{3} \sinh(3x) \right]_0^{\ln 2} \) or equivalent.
* **A1**: Correctly substitute the limits and evaluate to obtain \( \frac{21}{16} \) (or equivalent fraction).

**(b)**
* **M1**: Set up the correct surface area integral: \( S = 2\pi \int_{0}^{\ln 3} \cosh^2 x \, dx \).
* **M1**: Use the double-angle hyperbolic identity \( \cosh^2 x = \frac{1}{2}(\cosh(2x) + 1) \) to rewrite the integrand.
* **A1**: Integrate to obtain \( \pi \left[ \frac{1}{2} \sinh(2x) + x \right] \) with correct limits.
* **A1**: Evaluate to obtain the exact value \( \pi \left(\frac{20}{9} + \ln 3\right) \) (accept \( a = \frac{20}{9} \), \( b = 3 \)).

(i) Differentiating the equation implicitly with respect to \( x \):
\[ \frac{\mathrm{d}}{\mathrm{d}x}\left(\mathrm{e}^{2x-y}\right) + \frac{\mathrm{d}}{\mathrm{d}x}\left(\ln(y^2 - 3x)\right) = \frac{\mathrm{d}}{\mathrm{d}x}(2x - y + 1) \]

Using the chain rule, product rule, and quotient rule where appropriate:
\[ \left(2 - \frac{\mathrm{d}y}{\mathrm{d}x}\right)\mathrm{e}^{2x-y} + \frac{2y\frac{\mathrm{d}y}{\mathrm{d}x} - 3}{y^2 - 3x} = 2 - \frac{\mathrm{d}y}{\mathrm{d}x} \]

Substitute the coordinates of the point \( (1, 2) \):
\[ \left(2 - \frac{\mathrm{d}y}{\mathrm{d}x}\right)\mathrm{e}^{2(1)-2} + \frac{2(2)\frac{\mathrm{d}y}{\mathrm{d}x} - 3}{2^2 - 3(1)} = 2 - \frac{\mathrm{d}y}{\mathrm{d}x} \]

Since \( \mathrm{e}^0 = 1 \) and \( 2^2 - 3(1) = 1 \), this simplifies to:
\[ 2 - \frac{\mathrm{d}y}{\mathrm{d}x} + 4\frac{\mathrm{d}y}{\mathrm{d}x} - 3 = 2 - \frac{\mathrm{d}y}{\mathrm{d}x} \]
\[ 3\frac{\mathrm{d}y}{\mathrm{d}x} - 1 = 2 - \frac{\mathrm{d}y}{\mathrm{d}x} \]
\[ 4\frac{\mathrm{d}y}{\mathrm{d}x} = 3 \implies \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3}{4} \]

(ii) To find \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} \) at \( (1, 2) \), we differentiate the first-derivative equation with respect to \( x \):
\[ \frac{\mathrm{d}}{\mathrm{d}x}\left[\left(2 - \frac{\mathrm{d}y}{\mathrm{d}x}\right)\mathrm{e}^{2x-y}\right] + \frac{\mathrm{d}}{\mathrm{d}x}\left[\frac{2y\frac{\mathrm{d}y}{\mathrm{d}x} - 3}{y^2 - 3x}\right] = \frac{\mathrm{d}}{\mathrm{d}x}\left[2 - \frac{\mathrm{d}y}{\mathrm{d}x}\right] \]

We differentiate each term and evaluate at \( x = 1 \), \( y = 2 \), and \( \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3}{4} \):

For the first term, using the product rule:
\[ \frac{\mathrm{d}}{\mathrm{d}x}\left[\left(2 - \frac{\mathrm{d}y}{\mathrm{d}x}\right)\mathrm{e}^{2x-y}\right] = -\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\mathrm{e}^{2x-y} + \left(2 - \frac{\mathrm{d}y}{\mathrm{d}x}\right)^2\mathrm{e}^{2x-y} \]
Evaluating at the point:
\[ -\frac{\mathrm{d}^2y}{\mathrm{d}x^2}(1) + \left(2 - \frac{3}{4}\right)^2(1) = -\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + \frac{25}{16} \]

For the second term, let \( u = 2y\frac{\mathrm{d}y}{\mathrm{d}x} - 3 \) and \( v = y^2 - 3x \):
At \( (1, 2) \) with \( \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3}{4} \):
\[ u = 2(2)\left(\frac{3}{4}\right) - 3 = 0 \]
\[ v = 2^2 - 3(1) = 1 \]
Differentiating \( u \) and \( v \):
\[ \frac{\mathrm{d}u}{\mathrm{d}x} = 2\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 + 2y\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \]
At the point:
\[ \frac{\mathrm{d}u}{\mathrm{d}x} = 2\left(\frac{3}{4}\right)^2 + 2(2)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{9}{8} + 4\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \]
Since \( u = 0 \) at this point, the quotient rule derivative \( \frac{u'v - uv'}{v^2} \) simplifies to:
\[ \frac{u'}{v} = \frac{9}{8} + 4\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \]

For the RHS term:
\[ \frac{\mathrm{d}}{\mathrm{d}x}\left[2 - \frac{\mathrm{d}y}{\mathrm{d}x}\right] = -\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \]

Equating the differentiated parts:
\[ -\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + \frac{25}{16} + \frac{9}{8} + 4\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \]
\[ 3\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + \frac{43}{16} = -\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \]
\[ 4\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -\frac{43}{16} \implies \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -\frac{43}{64} \]

Part (i):
M1: For attempt at implicit differentiation of the first term using chain rule (must obtain \( (2 - y')\mathrm{e}^{2x-y} \) or equivalent).
M1: For attempt at implicit differentiation of the second term (must use quotient rule or chain/product rule to obtain \( \frac{2yy' - 3}{y^2 - 3x} \)).
A1: Correctly differentiated equation (either LHS/RHS or rearranged).
A1: Substituting \( (1, 2) \) and solving to obtain \( \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3}{4} \) (or equivalent fraction/decimal).

Part (ii):
M1: For differentiating \( (2 - y')\mathrm{e}^{2x-y} \) using the product rule.
M1: For differentiating \( \frac{2yy' - 3}{y^2 - 3x} \) using the quotient rule.
A1: Correct expressions for the derivatives of individual terms (either as formulas or with values substituted).
M1: Substituting \( x = 1 \), \( y = 2 \), and \( \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3}{4} \) and attempting to solve for \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} \).
A1: Correct value of \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -\frac{43}{64} \) (or exact equivalent, e.g., \( -0.671875 \)).

*(i) Find the general solution:*

First, find the complementary function (CF) by solving the auxiliary equation:
\[ m^2 - 6m + 9 = 0 \]
\[ (m - 3)^2 = 0 \implies m = 3 \quad \text{(repeated root)} \]

Therefore, the complementary function is:
\[ y_c = (A + Bx) e^{3x} \]

Next, find the particular integral (PI). Try a particular solution of the form:
\[ y_p = p \sin x + q \cos x \]
Differentiating this gives:
\[ \frac{\mathrm{d}y_p}{\mathrm{d}x} = p \cos x - q \sin x \]
\[ \frac{\mathrm{d}^2 y_p}{\mathrm{d}x^2} = -p \sin x - q \cos x \]

Substitute these into the original differential equation:
\[ (-p \sin x - q \cos x) - 6(p \cos x - q \sin x) + 9(p \sin x + q \cos x) = 50 \sin x \]
Group terms by \( \sin x \) and \( \cos x \):
\[ (8p + 6q) \sin x + (8q - 6p) \cos x = 50 \sin x \]

Equating coefficients of \( \sin x \) and \( \cos x \):
1) \( 8p + 6q = 50 \implies 4p + 3q = 25 \)
2) \( 8q - 6p = 0 \implies 3p = 4q \implies p = \frac{4}{3}q \)

Substitute \( p \) into the first equation:
\[ 4\left(\frac{4}{3}q\right) + 3q = 25 \implies \frac{25}{3}q = 25 \implies q = 3 \]
Then,
\[ p = 4 \]

Thus, the particular integral is:
\[ y_p = 4 \sin x + 3 \cos x \]

The general solution is:
\[ y = (A + Bx) e^{3x} + 4 \sin x + 3 \cos x \]

*(ii) Find the particular solution:*

Apply the initial condition \( y = 1 \) when \( x = 0 \):
\[ 1 = (A + B(0)) e^0 + 4 \sin(0) + 3 \cos(0) \]
\[ 1 = A + 3 \implies A = -2 \]

To apply the second initial condition, find the derivative of the general solution:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = B e^{3x} + 3(A + Bx) e^{3x} + 4 \cos x - 3 \sin x \]

Apply the condition \( \frac{\mathrm{d}y}{\mathrm{d}x} = 2 \) when \( x = 0 \):
\[ 2 = B e^0 + 3(A) e^0 + 4 \cos(0) - 3 \sin(0) \]
\[ 2 = B + 3A + 4 \]
Substitute \( A = -2 \):
\[ 2 = B - 6 + 4 \implies B = 4 \]

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

**Part (i) [7 marks]:**
* **M1**: Attempting to solve the auxiliary equation \( m^2 - 6m + 9 = 0 \).
* **A1**: Finding the repeated root \( m = 3 \) and stating the correct complementary function \( y_c = (A + Bx) e^{3x} \).
* **M1**: Choosing a correct form for the particular integral \( y_p = p \sin x + q \cos x \) and finding derivatives.
* **A1**: Substituting and equating coefficients to obtain simultaneous equations (e.g., \( 8p + 6q = 50 \) and \( 8q - 6p = 0 \)).
* **A1**: Correctly solving for \( p = 4 \) and \( q = 3 \).
* **A1**: Writing the correct general solution (must include arbitrary constants).

**Part (ii) [3 marks]:**
* **M1**: Applying the first boundary condition \( y = 1 \) at \( x = 0 \) to find \( A \).
* **A1**: Finding \( A = -2 \) correctly.
* **M1**: Differentiating the general solution and applying \( y' = 2 \) at \( x = 0 \) to find \( B \).
* **A1**: Finding \( B = 4 \) and stating the correct final particular solution \( y = (4x - 2) e^{3x} + 4 \sin x + 3 \cos x \).

**(a)**
By De Moivre's theorem:
\[\cos(5\theta) + i\sin(5\theta) = (\cos\theta + i\sin\theta)^5\]
Expanding the right-hand side using the binomial theorem:
\[(\cos\theta + i\sin\theta)^5 = \cos^5\theta + 5i\cos^4\theta\sin\theta + 10i^2\cos^3\theta\sin^2\theta + 10i^3\cos^2\theta\sin^3\theta + 5i^4\cos\theta\sin^4\theta + i^5\sin^5\theta\]
Using the powers of \(i\) (\(i^2 = -1\), \(i^3 = -i\), \(i^4 = 1\), \(i^5 = i\)):
\[(\cos\theta + i\sin\theta)^5 = \cos^5\theta + 5i\cos^4\theta\sin\theta - 10\cos^3\theta\sin^2\theta - 10i\cos^2\theta\sin^3\theta + 5\cos\theta\sin^4\theta + i\sin^5\theta\]
Equating the real parts from both sides of the equation:
\[\cos(5\theta) = \cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta\]
Substitute \(\sin^2\theta = 1 - \cos^2\theta\):
\[\cos(5\theta) = \cos^5\theta - 10\cos^3\theta(1 - \cos^2\theta) + 5\cos\theta(1 - \cos^2\theta)^2\]
\[\cos(5\theta) = \cos^5\theta - 10\cos^3\theta + 10\cos^5\theta + 5\cos\theta(1 - 2\cos^2\theta + \cos^4\theta)\]
\[\cos(5\theta) = 11\cos^5\theta - 10\cos^3\theta + 5\cos\theta - 10\cos^3\theta + 5\cos^5\theta\]
\[\cos(5\theta) = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta\]

**(b)**
Given the equation:
\[32x^5 - 40x^3 + 10x - \sqrt{3} = 0\]
Divide by 2:
\[16x^5 - 20x^3 + 5x = \frac{\sqrt{3}}{2}\]
Let \(x = \cos\theta\). Using the identity from part (a), the equation simplifies to:
\[\cos(5\theta) = \frac{\sqrt{3}}{2}\]
Since we require \(x = \cos\alpha\) where \(0 < \alpha < \pi\), we find the values of \(\theta\) such that \(0 < \theta < \pi\).

The general solution for \(5\theta\) is:
\[5\theta = 2k\pi \pm \frac{\pi}{6} \quad \text{for } k \in \mathbb{Z}\]

Setting up the values within the interval \(0 < 5\theta < 5\pi\):
- For \(k = 0\):
- \(5\theta = \frac{\pi}{6} \implies \theta = \frac{\pi}{30}\)
- For \(k = 1\):
- \(5\theta = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6} \implies \theta = \frac{11\pi}{30}\)
- \(5\theta = 2\pi + \frac{\pi}{6} = \frac{13\pi}{6} \implies \theta = \frac{13\pi}{30}\)
- For \(k = 2\):
- \(5\theta = 4\pi - \frac{\pi}{6} = \frac{23\pi}{6} \implies \theta = \frac{23\pi}{30}\)
- \(5\theta = 4\pi + \frac{\pi}{6} = \frac{25\pi}{6} \implies \theta = \frac{25\pi}{30} = \frac{5\pi}{6}\)

Thus, the five roots are:
\[x = \cos\left(\frac{\pi}{30}\right), \cos\left(\frac{11\pi}{30}\right), \cos\left(\frac{13\pi}{30}\right), \cos\left(\frac{23\pi}{30}\right), \cos\left(\frac{5\pi}{6}\right)\]

**(a)**
- **M1**: State De Moivre's theorem: \(\cos(5\theta) + i\sin(5\theta) = (\cos\theta + i\sin\theta)^5\).
- **A1**: Correctly expand the right-hand side using the binomial theorem.
- **M1**: Equate real parts and substitute \(\sin^2\theta = 1 - \cos^2\theta\).
- **A1**: Fully correct algebraic derivation leading to the given expression \(\cos(5\theta) = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta\).

**(b)**
- **M1**: Divide the given equation by 2 and substitute \(x = \cos\theta\).
- **M1**: Deduce the equation \(\cos(5\theta) = \frac{\sqrt{3}}{2}\).
- **M1**: Identify the general form of the solutions: \(5\theta = 2k\pi \pm \frac{\pi}{6}\).
- **A2**: Find the 5 distinct values of \(\theta\) in the interval \((0, \pi)\): \(\frac{\pi}{30}\) , \(\frac{11\pi}{30}\), \(\frac{13\pi}{30}\), \(\frac{23\pi}{30}\), \(\frac{5\pi}{6}\). (Award 1 mark for any 3 correct angles, 2 marks for all 5 correct angles).
- **A1**: Correctly state all five roots in the required form: \(x = \cos\alpha\).

(a) The derivative of \(f(x)\) is \(f'(x) = -\frac{2x}{(1+x^2)^2}\). For \(0 < x \le 1\), \(f'(x) < 0\), which shows that \(f(x) = \frac{1}{1+x^2}\) is strictly decreasing on \([0, 1]\). We divide the interval \([0, 1]\) into \(n\) equal subintervals, each of width \(\Delta x = \frac{1}{n}\). The right-endpoint Riemann sum, representing an under-estimate of the area under the curve, is \(R_n = \sum_{r=1}^n \frac{1}{n} f\left(\frac{r}{n}\right) = \sum_{r=1}^n \frac{1}{n} \frac{1}{1 + (r/n)^2} = \sum_{r=1}^n \frac{n}{n^2 + r^2}\). The left-endpoint Riemann sum, representing an over-estimate of the area under the curve, is \(L_n = \sum_{r=0}^{n-1} \frac{1}{n} f\left(\frac{r}{n}\right) = \sum_{r=0}^{n-1} \frac{1}{n} \frac{1}{1 + (r/n)^2} = \sum_{r=0}^{n-1} \frac{n}{n^2 + r^2}\). The exact area under the curve is \(\int_0^1 \frac{1}{1+x^2} \, dx = [\arctan(x)]_0^1 = \frac{\pi}{4}\). Since the function is strictly decreasing, we have \(R_n < \int_0^1 f(x) \, dx < L_n\), which yields \(\sum_{r=1}^{n} \frac{n}{n^2 + r^2} < \frac{\pi}{4} < \sum_{r=0}^{n-1} \frac{n}{n^2 + r^2}\).

(b) We rewrite the general term of the sum by factoring out \(n^3\) from the denominator: \(\sum_{r=1}^n \frac{n^2}{(n^2 + r^2)^{3/2}} = \sum_{r=1}^n \frac{n^2}{\left(n^2\left(1 + (r/n)^2\right)\right)^{3/2}} = \sum_{r=1}^n \frac{n^2}{n^3 \left(1 + (r/n)^2\right)^{3/2}} = \frac{1}{n} \sum_{r=1}^n \frac{1}{\left(1 + (r/n)^2\right)^{3/2}}\). As \(n \to \infty\), this Riemann sum converges to the definite integral \(\int_0^1 \frac{1}{(1+x^2)^{3/2}} \, dx\). To evaluate this integral, we use the substitution \(x = \tan\theta\), which gives \(dx = \sec^2\theta \, d\theta\). The limits of integration change as follows: when \(x=0\), \(\theta=0\); when \(x=1\), \(\theta=\frac{\pi}{4}\). Substituting these into the integral gives \(\int_0^{\frac{\pi}{4}} \frac{\sec^2\theta}{(\sec^2\theta)^{3/2}} \, d\theta = \int_0^{\frac{\pi}{4}} \frac{\sec^2\theta}{\sec^3\theta} \, d\theta = \int_0^{\frac{\pi}{4}} \cos\theta \, d\theta = [\sin\theta]_0^{\frac{\pi}{4}} = \sin\left(\frac{\pi}{4}\right) - \sin(0) = \frac{\sqrt{2}}{2}\).

Part (a):
M1: States or shows that \(f(x)\) is strictly decreasing on the interval \([0, 1]\) (e.g., via differentiation).
M1: Identifies the width of each subinterval as \(1/n\) and sets up the left-endpoint and right-endpoint Riemann sums.
A1: Evaluates the exact integral as \(\frac{\pi}{4}\).
A1: Combines the inequalities correctly to prove the given result.

Part (b):
M1: Formulates the sum by factoring out \(n^3\) from the denominator to obtain the term \(\frac{1}{n}\) and a function of \(\frac{r}{n}\).
A1: Identifies the limit of the sum as the definite integral \(\int_0^1 \frac{1}{(1+x^2)^{3/2}} \, dx\).
M1: Uses the trigonometric substitution \(x = \tan\theta\) and finds \(dx = \sec^2\theta \, d\theta\).
M1: Successfully updates the limits of integration to \(0\) and \(\frac{\pi}{4}\) and simplifies the integrand to \(\cos\theta\).
A1: Integrates to get \(\sin\theta\).
A1: Evaluates to find the exact limit of \(\frac{\sqrt{2}}{2}\) (accept \(\frac{1}{\sqrt{2}}\)).

To solve the first-order linear differential equation, we first write it in standard form by dividing through by \( x + 1 \) (since \( x > -1 \), \( x + 1 \neq 0 \)):
\[ \frac{dy}{dx} + \frac{2}{x+1}y = \cos x \]

Next, we find the integrating factor, \( I(x) \):
\[ I(x) = e^{\int \frac{2}{x+1} \, dx} = e^{2\ln(x+1)} = (x+1)^2 \]

Multiplying both sides of the standard form by the integrating factor yields:
\[ (x+1)^2 \frac{dy}{dx} + 2(x+1)y = (x+1)^2 \cos x \]
\[ \frac{d}{dx} \left[ y(x+1)^2 \right] = (x+1)^2 \cos x \]

Now, we integrate both sides with respect to \( x \):
\[ y(x+1)^2 = \int (x+1)^2 \cos x \, dx \]

We evaluate the right-hand side using integration by parts:
Let \( u = (x+1)^2 \) and \( dv = \cos x \, dx \).
Then \( du = 2(x+1) \, dx \) and \( v = \sin x \).
\[ \int (x+1)^2 \cos x \, dx = (x+1)^2 \sin x - \int 2(x+1) \sin x \, dx \]

Apply integration by parts again to the remaining integral:
Let \( u_1 = 2(x+1) \) and \( dv_1 = \sin x \, dx \).
Then \( du_1 = 2 \, dx \) and \( v_1 = -\cos x \).
\[ \int 2(x+1) \sin x \, dx = -2(x+1)\cos x - \int -2\cos x \, dx = -2(x+1)\cos x + 2\sin x \]

Substituting this back into our primary integration equation gives:
\[ y(x+1)^2 = (x+1)^2 \sin x - \left( -2(x+1)\cos x + 2\sin x \right) + C \]
\[ y(x+1)^2 = (x+1)^2 \sin x + 2(x+1)\cos x - 2\sin x + C \]

Using the initial condition \( y = 1 \) when \( x = 0 \):
\[ 1(0+1)^2 = (0+1)^2 \sin 0 + 2(0+1)\cos 0 - 2\sin 0 + C \]
\[ 1 = 2 + C \implies C = -1 \]

Thus, the solution is:
\[ y(x+1)^2 = (x+1)^2 \sin x + 2(x+1)\cos x - 2\sin x - 1 \]

Dividing through by \( (x+1)^2 \):
\[ y = \sin x + \frac{2\cos x}{x+1} - \frac{2\sin x + 1}{(x+1)^2} \]

M1: Divides the given equation by \( (x+1) \) to obtain the standard form: \( \frac{dy}{dx} + \frac{2}{x+1}y = \cos x \).
M1: Obtains the correct integrating factor, showing the step \( e^{\int \frac{2}{x+1}\,dx} \) to get \( (x+1)^2 \).
A1: Correctly writes the differential equation in derivative form: \( \frac{d}{dx}[y(x+1)^2] = (x+1)^2\cos x \).
M1: Applies integration by parts to the RHS once, obtaining a form like \( (x+1)^2 \sin x - \int 2(x+1)\sin x \, dx \).
M1: Applies integration by parts a second time to complete the integration.
A1: Obtains the correct integrated expression with an arbitrary constant: \( y(x+1)^2 = (x+1)^2 \sin x + 2(x+1)\cos x - 2\sin x + C \).
M1: Uses the initial condition \( x = 0, y = 1 \) to attempt to find \( C \).
A1: Correctly identifies \( C = -1 \).
A1: Expresses the final particular solution explicitly in the form \( y = f(x) \) as \( y = \sin x + \frac{2\cos x}{x+1} - \frac{2\sin x + 1}{(x+1)^2} \) (or any algebraically equivalent form, such as \( y = \frac{(x+1)^2 \sin x + 2(x+1)\cos x - 2\sin x - 1}{(x+1)^2} \)).

(i) To find the eigenvalues, we solve the characteristic equation \(\det(A - \lambda I) = 0\):
\[ \begin{vmatrix} 1 - \lambda & 1 & 0 \\ -1 & 3 - \lambda & 1 \\ -1 & 1 & 2 - \lambda \end{vmatrix} = 0 \]
Expanding along the first row:
\[ (1-\lambda) \left[ (3-\lambda)(2-\lambda) - 1 \right] - 1 \left[ -1(2-\lambda) - (-1) \right] = 0 \]
\[ (1-\lambda) (\lambda^2 - 5\lambda + 5) - (\lambda - 1) = 0 \]
\[ (1-\lambda) (\lambda^2 - 5\lambda + 5) + (1-\lambda) = 0 \]
\[ (1-\lambda)(\lambda^2 - 5\lambda + 6) = 0 \]
\[ (1-\lambda)(\lambda - 2)(\lambda - 3) = 0 \]
Thus, the eigenvalues of \( A \) are \( \lambda = 1, 2, 3 \).

(ii) For each eigenvalue, we find a non-zero eigenvector \(\mathbf{v} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}\) satisfying \((A - \lambda I)\mathbf{v} = \mathbf{0}\):

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

- For \(\lambda = 2\):
\[ \begin{pmatrix} -1 & 1 & 0 \\ -1 & 1 & 1 \\ -1 & 1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \]
From row 1, \( -x + y = 0 \implies x = y \).
From row 2, \( -x + y + z = 0 \implies z = 0 \).
An eigenvector is \( \mathbf{e}_2 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \).

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

(iii) Using the eigenvectors in the order corresponding to the eigenvalues \( 1, 2, 3 \):
\[ P = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 1 & 0 & 1 \end{pmatrix} \quad \text{and} \quad D = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix} \]

(iv) To find \( P^{-1} \):
\[ \det(P) = 1(1 - 0) - 1(0 - 2) + 1(0 - 1) = 1 + 2 - 1 = 2 \]
The matrix of cofactors of \( P \) is:
\[ C = \begin{pmatrix} 1 & 2 & -1 \\ -1 & 0 & 1 \\ 1 & -2 & 1 \end{pmatrix} \]
Thus, the adjugate matrix of \( P \) is:
\[ \text{adj}(P) = C^T = \begin{pmatrix} 1 & -1 & 1 \\ 2 & 0 & -2 \\ -1 & 1 & 1 \end{pmatrix} \]
This gives:
\[ P^{-1} = \frac{1}{2} \begin{pmatrix} 1 & -1 & 1 \\ 2 & 0 & -2 \\ -1 & 1 & 1 \end{pmatrix} \]
Now, compute \( A^n = P D^n P^{-1} \):
\[ P D^n = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 1 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2^n & 0 \\ 0 & 0 & 3^n \end{pmatrix} = \begin{pmatrix} 1 & 2^n & 3^n \\ 0 & 2^n & 2 \cdot 3^n \\ 1 & 0 & 3^n \end{pmatrix} \]
Finally, compute \( P D^n P^{-1} \):
\[ A^n = \frac{1}{2} \begin{pmatrix} 1 & 2^n & 3^n \\ 0 & 2^n & 2 \cdot 3^n \\ 1 & 0 & 3^n \end{pmatrix} \begin{pmatrix} 1 & -1 & 1 \\ 2 & 0 & -2 \\ -1 & 1 & 1 \end{pmatrix} \]
\[ = \frac{1}{2} \begin{pmatrix} 1 + 2 \cdot 2^n - 3^n & -1 + 0 + 3^n & 1 - 2 \cdot 2^n + 3^n \\ 0 + 2 \cdot 2^n - 2 \cdot 3^n & 0 + 0 + 2 \cdot 3^n & 0 - 2 \cdot 2^n + 2 \cdot 3^n \\ 1 + 0 - 3^n & -1 + 0 + 3^n & 1 + 0 + 3^n \end{pmatrix} \]
\[ = \frac{1}{2} \begin{pmatrix} 1 + 2^{n+1} - 3^n & 3^n - 1 & 1 - 2^{n+1} + 3^n \\ 2^{n+1} - 2 \cdot 3^n & 2 \cdot 3^n & 2 \cdot 3^n - 2^{n+1} \\ 1 - 3^n & 3^n - 1 & 1 + 3^n \end{pmatrix} \]
which is the required expression.

**(i)**
* **M1**: Sets up characteristic equation \(\det(A - \lambda I) = 0\).
* **A1**: Expands determinant correctly to obtain \((1-\lambda)(\lambda^2 - 5
\lambda + 6) = 0\) (or any equivalent cubic form).
* **A2**: Finds correct eigenvalues \(\lambda = 1, 2, 3\) (Award 1 mark for any two correct eigenvalues).

**(ii)**
* **M1**: For a valid attempt to solve \((A - \lambda I)\mathbf{v} = \mathbf{0}\) for at least one of their eigenvalues.
* **A1**: Correct eigenvector for \(\lambda = 1\) (or any non-zero scalar multiple).
* **A1**: Correct eigenvector for \(\lambda = 2\) (or any non-zero scalar multiple).
* **A1**: Correct eigenvector for \(\lambda = 3\) (or any non-zero scalar multiple).

**(iii)**
* **B1**: Writes down consistent matrices \(P\) and \(D\) based on their eigenvalues and eigenvectors.

**(iv)**
* **M1**: Attempts to find the inverse matrix \(P^{-1}\) using determinants and cofactors.
* **A1**: Obtains correct \(P^{-1}\).
* **M1**: Sets up the product \(P D^n P^{-1}\).
* **A1**: Correctly computes the product \(P D^n\).
* **A1**: Completes the matrix multiplication correctly to obtain the given expression for \(A^n\).

Let \( T \) be the tension in the string, \( L = 0.8\text{ m} \) be the length of the string, and \( \theta = 60^\circ \) be the angle with the vertical.

Resolving forces vertically for equilibrium:
\[ T \cos \theta = mg \]

Applying Newton's second law along the radial direction towards the center of the circular path:
\[ T \sin \theta = m r \omega^2 \]
where the radius of the path is \( r = L \sin \theta \). This gives:
\[ T \sin \theta = m (L \sin \theta) \omega^2 \implies T = m L \omega^2 \]

Substituting \( T \) into the vertical equation:
\[ (m L \omega^2) \cos \theta = mg \]
\[ \omega^2 = \frac{g}{L \cos \theta} \]

Substitute the given values \( g = 10 \text{ m/s}^2 \), \( L = 0.8 \text{ m} \), and \( \theta = 60^\circ \):
\[ \omega^2 = \frac{10}{0.8 \cos 60^\circ} = \frac{10}{0.8 \times 0.5} = \frac{10}{0.4} = 25 \]
\[ \omega = 5 \text{ rad s}^{-1} \]

M1: For resolving forces vertically to find \( T \cos 60^\circ = mg \) and applying Newton's second law horizontally to obtain \( T \sin 60^\circ = m (0.8 \sin 60^\circ) \omega^2 \) (or equivalent).
A1: For eliminating \( T \) to obtain a correct equation in terms of \( \omega \), such as \( \omega^2 = \frac{10}{0.8 \cos 60^\circ} \).
A1: For obtaining the correct angular speed of \( 5 \) (or \( 5 \text{ rad s}^{-1} \)).

First, we calculate the constant frictional force acting on the particle during its motion. The normal reaction force is \(R = mg = 0.2 \times 10 = 2\text{ N}\). The maximum frictional force is \(F = \mu R = 0.5 \times 2 = 1\text{ N}\). The motion of the particle consists of three distinct stages. Stage 1: From \(A\) to the first natural length position \(N_1\). The natural length of the string is \(l = 0.5\text{ m}\), so the initial extension of the string is \(x_i = 1.1 - 0.5 = 0.6\text{ m}\). The initial elastic potential energy is \(E_i = \frac{\lambda x_i^2}{2l} = \frac{10 \times 0.6^2}{2 \times 0.5} = 3.6\text{ J}\). The distance from \(A\) to \(N_1\) is \(0.6\text{ m}\), so the work done against friction during this stage is \(W_1 = F \times d_1 = 1 \times 0.6 = 0.6\text{ J}\). By conservation of energy, the kinetic energy at \(N_1\) is \(KE_{N1} = E_i - W_1 = 3.6 - 0.6 = 3.0\text{ J}\). Stage 2: From \(N_1\) to the second natural length position \(N_2\) on the other side of \(O\). The string is slack over this distance of \(N_1 N_2 = 2l = 1.0\text{ m}\). The work done against friction during this stage is \(W_2 = F \times 1.0 = 1.0\text{ J}\). The kinetic energy of the particle as it passes \(N_2\) is \(KE_{N2} = KE_{N1} - W_2 = 3.0 - 1.0 = 2.0\text{ J}\). Stage 3: From \(N_2\) to the point of instantaneous rest \(B\). Let the maximum extension beyond \(N_2\) be \(x'\). The work done against friction is \(W_3 = F \times x' = x'\). The final elastic potential energy is \(E_f = \frac{\lambda (x')^2}{2l} = \frac{10 (x')^2}{2 \times 0.5} = 10(x')^2\). Applying the work-energy principle: \(KE_{N2} = E_f + W_3 \implies 2 = 10(x')^2 + x'\), which simplifies to the quadratic equation: \(10(x')^2 + x' - 2 = 0\). Factoring this equation gives: \((5x' - 2)(2x' + 1) = 0\). Since the extension \(x'\) must be positive, we find \(x' = 0.4\text{ m}\). Thus, the distance of the point of instantaneous rest from \(O\) is \(ON_2 + x' = 0.5 + 0.4 = 0.9\text{ m}\).

M1: Calculate the constant frictional force \(F = 1\text{ N}\) using \(F = \mu R\). A1: Correctly calculate the initial elastic potential energy \(E_i = 3.6\text{ J}\). M1: Set up the energy equation for the first stage of motion to find the kinetic energy at \(N_1\). A1: Obtain the correct kinetic energy at the second natural length position, \(KE_{N2} = 2.0\text{ J}\) (or equivalent). M1: Formulate the energy equation for the final stage with variable extension \(x'\). A1: Obtain a correct quadratic equation in terms of \(x'\), such as \(10(x')^2 + x' - 2 = 0\). M1: Solve their quadratic equation to find the positive root \(x' = 0.4\text{ m}\). A1: Conclude with the correct final distance of \(0.9\text{ m}\) from \(O\).

For part (i):
Using Newton's second law, the equation of motion is:
\(m \frac{dv}{dt} = -0.5(v + v^3)\)
Given that \(m = 0.5\text{ kg}\), we have:
\(0.5 \frac{dv}{dt} = -0.5(v + v^3) \implies \frac{dv}{dt} = -v(1+v^2)\)
Separating the variables gives:
\(\int \frac{1}{v(1+v^2)} dv = \int -1 dt\)
Using partial fractions:
\(\frac{1}{v(1+v^2)} = \frac{1}{v} - \frac{v}{1+v^2}\)
Integrating both sides:
\(\ln v - \frac{1}{2}\ln(1+v^2) = -t + C\)
Using the initial condition \(v = 3\) when \(t = 0\):
\(C = \ln 3 - \frac{1}{2}\ln(10)\)
When \(v = 1\):
\(\ln 1 - \frac{1}{2}\ln 2 = -t + \ln 3 - \frac{1}{2}\ln 10\)
\(t = \ln 3 - \frac{1}{2}\ln 10 + \frac{1}{2}\ln 2 = \ln 3 - \frac{1}{2}\ln 5 = \ln\left(\frac{3}{\sqrt{5}}\right)\text{ seconds}\).

For part (ii):
We write the acceleration as \(v \frac{dv}{dx}\). The equation of motion becomes:
\(0.5 v \frac{dv}{dx} = -0.5(v + v^3)\)
Since \(v \neq 0\), we can divide by \(0.5v\):
\ \frac{dv}{dx} = -(1+v^2)\)
Separating variables:
\(\int \frac{1}{1+v^2} dv = \int -1 dx\)
Integrating both sides:
\(\arctan v = -x + C'\)
Using the initial condition \(x = 0\) when \(v = 3\):
\(C' = \arctan 3\)
When \(v = 1\):
\(\arctan 1 = -x + \arctan 3\)
\(x = \arctan 3 - \arctan 1 = \arctan 3 - \frac{\pi}{4}\text{ metres}\).

Part (i):
M1: For writing the equation of motion in the form \(0.5 \frac{dv}{dt} = -0.5(v+v^3)\) and separating variables.
M1: For using partial fractions correctly to integrate \(\int \frac{1}{v(1+v^2)} dv\).
A1: For correct integration with limits or constant of integration: \(\ln v - \frac{1}{2}\ln(1+v^2) = -t + C\).
A1: For obtaining the correct time in the form \(\ln k\), which is \(\ln\left(\frac{3}{\sqrt{5}}\right)\) or equivalent exact form.

Part (ii):
M1: For using \(a = v \frac{dv}{dx}\) to set up the differential equation \(v \frac{dv}{dx} = -v(1+v^2)\).
M1: For separating variables and recognizing the standard integral for \(\arctan v\).
A1: For correct integration with limits or constant of integration: \(\arctan v = -x + C\).
A1: For obtaining the correct distance \(\arctan 3 - \frac{\pi}{4}\) m.

Let \(B\) be the origin \((0,0)\), with the \(x\)-axis along the edge \(BC\) and the \(y\)-axis along the edge \(BA\).

The coordinates of the vertices of the uniform triangular lamina are:
\(B(0,0)\), \(C(4a,0)\), and \(A(0,3a)\).

Since the lamina is uniform, its center of mass \(G(\bar{x}, \bar{y})\) is located at the centroid of the triangle:
\[ \bar{x} = \frac{0 + 4a + 0}{3} = \frac{4}{3}a \]
\[ \bar{y} = \frac{0 + 0 + 3a}{3} = a \]
So, \(G = \left(\frac{4}{3}a, a\right)\).

The lamina is on the point of toppling about \(B\) when the vertical line through the center of mass \(G\) passes through the lowest point of contact, which is \(B\).

The angle that the vertical line makes with the normal to the inclined plane (the \(y\)-axis) is \(\theta\). Therefore, at the point of toppling, the angle that the line \(BG\) makes with the \(y\)-axis is also \(\theta\).

Using the coordinates of \(G\):
\[ \tan\theta = \frac{\bar{x}}{\bar{y}} = \frac{\frac{4}{3}a}{a} = \frac{4}{3} \]

M1: For finding the coordinates of the center of mass of the lamina relative to \(B\) (or finding the perpendicular distances of the center of mass from \(AB\) and \(BC\)).
A1: Correct coordinates or distances: \(\bar{x} = \frac{4}{3}a\) and \(\bar{y} = a\).
M1: For identifying that the line of action of the weight must pass through the pivot point \(B\) at the point of toppling, and setting up the equation \\tan\theta = \frac{\bar{x}}{\bar{y}}\.
A1: Correct value \(\tan\theta = \frac{4}{3}\) (or equivalent fraction, or \(1.33\)).

Let \(B\) be the origin \((0,0)\), with the \(x\)-axis along the edge \(BC\) and the \(y\)-axis along the edge \(BA\).

The coordinates of the vertices of the uniform triangular lamina are:
\(B(0,0)\), \(C(4a,0)\), and \(A(0,3a)\).

Since the lamina is uniform, its center of mass \(G(\bar{x}, \bar{y})\) is located at the centroid of the triangle:
\[ \bar{x} = \frac{0 + 4a + 0}{3} = \frac{4}{3}a \]
\[ \bar{y} = \frac{0 + 0 + 3a}{3} = a \]
So, \(G = \left(\frac{4}{3}a, a\right)\).

The lamina is on the point of toppling about \(B\) when the vertical line through the center of mass \(G\) passes through the lowest point of contact, which is \(B\).

The angle that the vertical line makes with the normal to the inclined plane (the \(y\)-axis) is \(\theta\). Therefore, at the point of toppling, the angle that the line \(BG\) makes with the \(y\)-axis is also \(\theta\).

Using the coordinates of \(G\):
\[ \tan\theta = \frac{\bar{x}}{\bar{y}} = \frac{\frac{4}{3}a}{a} = \frac{4}{3} \]

M1: For finding the coordinates of the center of mass of the lamina relative to \(B\) (or finding the perpendicular distances of the center of mass from \(AB\) and \(BC\)).
A1: Correct coordinates or distances: \(\bar{x} = \frac{4}{3}a\) and \(\bar{y} = a\).
M1: For identifying that the line of action of the weight must pass through the pivot point \(B\) at the point of toppling, and setting up the equation \\tan\theta = \frac{\bar{x}}{\bar{y}}\.
A1: Correct value \(\tan\theta = \frac{4}{3}\) (or equivalent fraction, or \(1.33\)).

Let \(\theta\) be the angle that the radius \(OQ\) makes with the upward vertical, where \(O\) is the center of the cylinder. By conservation of energy from \(A\) to \(Q\): \(\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mga(1 + \cos\theta)\). Since \(u^2 = kga\), we have \(v^2 = ga(k - 2 - 2\cos\theta)\) (Equation 1). At the point \(Q\), the particle loses contact, so the normal reaction is zero. Resolving radially inwards: \(mg\cos\theta = \frac{mv^2}{a}\), which gives \(v^2 = ga\cos\theta\) (Equation 2). Equating these two expressions gives \(\cos\theta = \frac{k-2}{3}\) (Equation 3). Let \(O\) be the origin. The coordinates of \(Q\) are \((a\sin\theta, a\cos\theta)\). The horizontal and vertical components of the velocity at \(Q\) are \(u_x = -v\cos\theta\) and \(u_y = v\sin\theta\). Taking \(t=0\) at \(Q\), the coordinates of the projectile at time \(t\) are \(x(t) = a\sin\theta - (v\cos\theta)t\) and \(y(t) = a\cos\theta + (v\sin\theta)t - \frac{1}{2}gt^2\). Since the path passes through \(A(0, -a)\), we find the time of flight \(T\) when \(x(T) = 0\), which gives \(T = \frac{a\sin\theta}{v\cos\theta}\). Substituting \(T\) into the vertical coordinate equation: \(-a = a\cos\theta + v\sin\theta\left(\frac{a\sin\theta}{v\cos\theta}\right) - \frac{1}{2}g\left(\frac{a\sin\theta}{v\cos\theta}\right)^2\). This simplifies to \(-a = \frac{a}{\cos\theta} - \frac{ga^2\sin^2\theta}{2v^2\cos^2\theta}\). Dividing by \(a\) and substituting \(v^2 = ga\cos\theta\) yields \(-1 = \frac{1}{\cos\theta} - \frac{\sin^2\theta}{2\cos^3\theta}\). Multiplying by \(2\cos^3\theta\) and substituting \(\sin^2\theta = 1 - \cos^2\theta\) gives \(2\cos^3\theta + 3\cos^2\theta - 1 = 0\), which factors as \((2\cos\theta - 1)(\cos\theta + 1)^2 = 0\). Since \(0 < \cos\theta < 1\), we must have \(\cos\theta = 0.5\). Substituting this into Equation 3 gives \(0.5 = \frac{k-2}{3}\), which solves to \(k = 3.5\).

M1: Apply conservation of energy between the lowest point and the point of losing contact to find \(v^2\). A1: Correct expression for \(v^2 = ga(k - 2 - 2\cos\theta)\). M1: Apply Newton's second law radially at the point of losing contact with \(R=0\) to find \(v^2 = ga\cos\theta\). M1: Set up projectile equations of motion from \(Q\) and find the time of flight \(T\) to reach \(x=0\). A1: Correct expression for \(T = \frac{a\sin\theta}{v\cos\theta}\). M1: Substitute \(T\) into the vertical displacement equation with final vertical position \(y = -a\). A1: Obtain the trigonometric equation \(2\cos^3\theta + 3\cos^2\theta - 1 = 0\) and solve to find \(\cos\theta = 0.5\). A1: Obtain the correct value of \(k = 3.5\) (or \(7/2\)).

(i) Let \(v_A\) and \(v_B\) be the velocities of \(A\) and \(B\) after the collision along the line of centres.
By conservation of momentum along the line of centres:
\[ m u \cos \theta = m v_A + 2m v_B \implies u \cos \theta = v_A + 2v_B \quad \text{(1)} \]
By Newton's law of restitution along the line of centres:
\[ v_B - v_A = e (u \cos \theta - 0) = e u \cos \theta \quad \text{(2)} \]
From (2), we have \(v_B = v_A + e u \cos \theta\). Substituting this into (1) yields:
\[ u \cos \theta = v_A + 2(v_A + e u \cos \theta) = 3v_A + 2e u \cos \theta \]
\[ 3v_A = u \cos \theta (1 - 2e) \implies v_A = \frac{1}{3} u (1 - 2e) \cos \theta \]
Since the spheres are smooth, the component of velocity perpendicular to the line of centres is unchanged, which is \(u \sin \theta\).

(ii) Let the unit vector along the line of centres (in the direction of the initial component of velocity of \(A\)) be \(\mathbf{i}\) and the unit vector perpendicular to it be \(\mathbf{j}\).
The initial velocity vector of \(A\) is:
\[ \mathbf{u}_A = (u \cos \theta) \mathbf{i} + (u \sin \theta) \mathbf{j} \]
The final velocity vector of \(A\) is:
\[ \mathbf{v}_A = \left[ \frac{1}{3} u (1 - 2e) \cos \theta \right] \mathbf{i} + (u \sin \theta) \mathbf{j} \]
Since the direction of motion is deflected through \(90^\circ\), the vectors \(\mathbf{u}_A\) and \(\mathbf{v}_A\) are perpendicular, so their dot product is zero:
\[ \mathbf{u}_A \cdot \mathbf{v}_A = 0 \implies (u \cos \theta) \left( \frac{1}{3} u (1 - 2e) \cos \theta \right) + (u \sin \theta)^2 = 0 \]
\[ \frac{1}{3} u^2 (1 - 2e) \cos^2 \theta + u^2 \sin^2 \theta = 0 \]
Dividing both sides by \(u^2 \cos^2 \theta\) (since \(u \neq 0\) and \(\cos \theta \neq 0\) for an oblique collision):
\[ \frac{1}{3}(1 - 2e) + \tan^2 \theta = 0 \implies \tan^2 \theta = \frac{2e - 1}{3} \]

(iii) For this deflection to be possible, we must have \(\tan^2 \theta > 0\) (since \(\theta = 0\) is a direct collision and \(\theta = \frac{\pi}{2}\) would result in no collision).
Thus:
\[ \frac{2e - 1}{3} > 0 \implies 2e - 1 > 0 \implies e > \frac{1}{2} \]
Since the coefficient of restitution \(e\) cannot exceed \(1\), the range of possible values for \(e\) is:
\[ \frac{1}{2} < e \le 1 \]

Part (i)
* **M1**: For using conservation of momentum along the line of centres to obtain a linear equation in \(v_A\) and \(v_B\).
* **M1**: For using Newton's law of restitution along the line of centres to obtain another equation in \(v_A\) and \(v_B\).
* **A1**: For finding the correct component along the line of centres, \(\frac{1}{3} u (1 - 2e) \cos \theta\), and stating the perpendicular component is \(u \sin \theta\).

Part (ii)
* **M1**: For using a valid condition for perpendicularity of initial and final velocities (e.g., using the dot product \(\mathbf{u}_A \cdot \mathbf{v}_A = 0\) or trigonometry).
* **M1**: For substituting the expressions for the components and simplifying.
* **A1**: For obtaining the given result \(\tan^2 \theta = \frac{2e - 1}{3}\) with clear and rigorous working.

Part (iii)
* **M1**: For stating or using \(\tan^2 \theta > 0\) to find the lower bound \(e > \frac{1}{2}\).
* **A1**: For the correct complete range \(\frac{1}{2} < e \le 1\) (accept \(\frac{1}{2} < e < 1\)).

(a) First, we find the horizontal and vertical components of the initial velocity:
\(u_x = 20 \cos\theta = 20 \times 0.6 = 12\text{ m s}^{-1}\)
\(u_y = 20 \sin\theta = 20 \times 0.8 = 16\text{ m s}^{-1}\)

The horizontal distance to the wall is \(12\text{ m}\). The time \(t_1\) taken to reach the wall is:
\(t_1 = \frac{12}{u_x} = \frac{12}{12} = 1\text{ s}\)

Now, we find the vertical height of \(W\) using the equation of vertical motion:
\(y = u_y t_1 - \frac{1}{2} g t_1^2\)
Taking \(g = 10\text{ m s}^{-2}\):
\(y = 16(1) - 5(1)^2 = 11\text{ m}\)

(b) The vertical component of the velocity immediately before the impact is:
\(v_y = u_y - g t_1 = 16 - 10(1) = 6\text{ m s}^{-1}\)

Since the wall is smooth, the vertical component of velocity is unchanged by the collision. Thus, immediately after the impact, the vertical velocity is still \(6\text{ m s}^{-1}\) upwards.

Let \(t_2\) be the time taken to travel from \(W\) to the floor. Using the vertical displacement equation with the origin at \(W\):
\(-11 = 6 t_2 - 5 t_2^2\)
\(5 t_2^2 - 6 t_2 - 11 = 0\)
\((5 t_2 - 11)(t_2 + 1) = 0\)

Since \(t_2 > 0\), we have \(t_2 = 2.2\text{ s}\).

Since the landing point \(A\) is at a distance of \(5.4\text{ m}\) from \(O\), and the wall is at a distance of \(12\text{ m}\) from \(O\), the horizontal distance traveled by the particle after the rebound is:
\(d_{\text{horizontal}} = 12 - 5.4 = 6.6\text{ m}\)

The horizontal velocity after the impact is \(e u_x = 12e\).
Using the horizontal motion after impact:
\(12e \times t_2 = 6.6\)
\(12e \times 2.2 = 6.6\)
\(26.4e = 6.6 \implies e = 0.25\)

(c) The horizontal velocity immediately after impact is:
\(v_x' = 12(0.25) = 3\text{ m s}^{-1}\)

The vertical velocity immediately after impact is:
\(v_y' = 6\text{ m s}^{-1}\)

The speed immediately after impact is:
\(v' = \sqrt{(v_x')^2 + (v_y')^2} = \sqrt{3^2 + 6^2} = \sqrt{45} = 3\sqrt{5}\text{ m s}^{-1}\)

(a)
- M1: For calculating the components of the initial velocity, \(u_x = 12\) and \(u_y = 16\).
- A1: For finding the correct time to reach the wall, \(t_1 = 1\text{ s}\).
- M1: For substituting \(t_1\) into the vertical motion equation to find the height.
- A1: For obtaining the correct height \(11\text{ m}\).

(b)
- B1: For finding the vertical component of velocity just before the impact, \(v_y = 6\text{ m s}^{-1}\).
- M1: For setting up the quadratic equation for the subsequent time of flight, \(5t_2^2 - 6t_2 - 11 = 0\).
- A1: For finding the correct time of flight, \(t_2 = 2.2\text{ s}\).
- M1: For relating the horizontal distance \(12 - 5.4 = 6.6\) to the horizontal velocity after impact, \(12e \times 2.2 = 6.6\).
- A1: For obtaining \(e = 0.25\).

(c)
- M1: For using Pythagoras' theorem with the horizontal speed \(3\text{ m s}^{-1}\) and vertical speed \(6\text{ m s}^{-1}\).
- A1: For simplifying the speed to \(3\sqrt{5}\text{ m s}^{-1}\).

Let the null and alternative hypotheses be:
\(H_0\): The genetic model is a suitable fit (i.e., the phenotypes occur in the ratio \(4:3:2:1\)).
\(H_1\): The genetic model is not a suitable fit.

The total number of plants is \(N = 74 + 68 + 43 + 15 = 200\).
Under \(H_0\), the expected frequencies \(E\) are calculated using the ratio \(4:3:2:1\) (total parts = 10):
- \(E_A = 200 \times \frac{4}{10} = 80\)
- \(E_B = 200 \times \frac{3}{10} = 60\)
- \(E_C = 200 \times \frac{2}{10} = 40\)
- \(E_D = 200 \times \frac{1}{10} = 20\)

We now compute the test statistic \(\chi^2 = \sum \frac{(O - E)^2}{E}\):
- Category \(A\): \(\frac{(74-80)^2}{80} = \frac{36}{80} = 0.45\)
- Category \(B\): \(\frac{(68-60)^2}{60} = \frac{64}{60} \approx 1.0667\)
- Category \(C\): \(\frac{(43-40)^2}{40} = \frac{9}{40} = 0.225\)
- Category \(D\): \(\frac{(15-20)^2}{20} = \frac{25}{20} = 1.25\)

Summing these values:
\(\chi^2 = 0.45 + 1.0667 + 0.225 + 1.25 \approx 2.992\) (or \(2.99\) to 3 s.f.)

The number of degrees of freedom is \(\nu = k - 1 = 4 - 1 = 3\).
At the 5% significance level, the critical value from the chi-squared distribution table is \(\chi^2_3(0.05) = 7.815\).

Since \(2.992 < 7.815\), we do not reject \(H_0\).
There is insufficient evidence at the 5% significance level to suggest that the genetic model is not a suitable fit. We conclude that the genetic model is a suitable fit for the data.

B1: State both null and alternative hypotheses correctly in context.
M1: Calculate all four expected frequencies correctly (80, 60, 40, 20).
M1: Attempt to calculate the \(\chi^2\) test statistic using the correct formula \(\sum \frac{(O-E)^2}{E}\).
A1: Obtain the correct test statistic value of \(2.99\) (or \(2.992\)).
M1: Identify the correct degrees of freedom \(\nu = 3\) and state the correct critical value of \(7.815\), comparing it to their calculated test statistic.
A1: Formulate a correct conclusion in context, stating that there is insufficient evidence to reject the genetic model (or that the model is a suitable fit).

First, calculate the sample size, mean, and sum of squares of deviations for each sample:

For Regime A:
\(n_A = 8\)
\(\sum x_A = 108 \implies \bar{x}_A = 13.5\)
\(\sum x_A^2 = 1484\)
\(\sum(x_A - \bar{x}_A)^2 = 1484 - \frac{108^2}{8} = 26\)

For Regime B:
\(n_B = 10\)
\(\sum x_B = 106 \implies \bar{x}_B = 10.6\)
\(\sum x_B^2 = 1144\)
\(\sum(x_B - \bar{x}_B)^2 = 1144 - \frac{106^2}{10} = 20.4\)

Now, calculate the pooled estimate of the common variance, \(s_p^2\):

\(s_p^2 = \frac{\sum(x_A - \bar{x}_A)^2 + \sum(x_B - \bar{x}_B)^2}{n_A + n_B - 2}\)
\(s_p^2 = \frac{26 + 20.4}{8 + 10 - 2} = \frac{46.4}{16} = 2.9\)

The number of degrees of freedom is \(\nu = 8 + 10 - 2 = 16\).
For a 95% confidence interval, we use the critical value from the t-distribution:
\(t_{16}(0.975) = 2.120\)

The standard error of the difference between the means is:
\(SE = \sqrt{s_p^2 \left( \frac{1}{n_A} + \frac{1}{n_B} \right)} = \sqrt{2.9 \left( \frac{1}{8} + \frac{1}{10} \right)} = \sqrt{2.9 \times 0.225} = \sqrt{0.6525} \approx 0.80777\)

The 95% confidence interval is:
\((\bar{x}_A - \bar{x}_B) \pm t_{16} \times SE\)
\(= (13.5 - 10.6) \pm 2.120 \times 0.80777\)
\(= 2.9 \pm 1.7125\)

This gives the interval \([1.1875, 4.6125]\).
Rounding to 3 significant figures, we get:
\([1.19, 4.61]\) (or \(1.19 < \mu_A - \mu_B < 4.61\)).

* **M1**: For calculating the sample means \(\bar{x}_A = 13.5\) and \(\bar{x}_B = 10.6\).
* **M1**: For finding the sums of squares of deviations: \(26\) for A and \(20.4\) for B (or equivalent correct calculations of sample variances).
* **A1**: For finding the pooled sample variance \(s_p^2 = 2.9\).
* **B1**: For identifying 16 degrees of freedom and using the correct t-value of 2.120.
* **M1**: For calculating the standard error of the difference \(\sqrt{2.9 \left(\frac{1}{8} + \frac{1}{10}\right)}\).
* **M1**: For substituting values into the confidence interval formula \((\bar{x}_A - \bar{x}_B) \pm t \cdot SE\).
* **A1**: For obtaining the correct interval \([1.19, 4.61]\) (accept \(1.19 < \mu_A - \mu_B < 4.61\)).

(i) For \(0 \le x \le 2\), \(F(x) = \int_0^x \frac{3}{8} t^2 \mathrm{d}t = [\frac{1}{8}t^3]_0^x = \frac{1}{8}x^3\). Thus, the cumulative distribution function of \(X\) is \(F(x) = 0\) for \(x < 0\), \(F(x) = \frac{1}{8}x^3\) for \(0 \le x \le 2\), and \(F(x) = 1\) for \(x > 2\). (ii) Since \(0 < X \le 2\), the range of \(Y = \frac{4}{X^2}\) is \(Y \ge 1\). For \(y \ge 1\), the cumulative distribution function of \(Y\) is \(G(y) = \mathrm{P}(Y \le y) = \mathrm{P}(\frac{4}{X^2} \le y)\). Since \(X > 0\), this is equivalent to \(X^2 \ge \frac{4}{y} \implies X \ge \frac{2}{\sqrt{y}}\). So \(G(y) = \mathrm{P}(X \ge \frac{2}{\sqrt{y}}) = 1 - \mathrm{P}(X < \frac{2}{\sqrt{y}}) = 1 - F(\frac{2}{\sqrt{y}})\). Since \(y \ge 1\), we have \(0 < \frac{2}{\sqrt{y}} \le 2\). Using the CDF of \(X\), we get: \(G(y) = 1 - \frac{1}{8}(\frac{2}{\sqrt{y}})^3 = 1 - \frac{1}{8}(\frac{8}{y^{3/2}}) = 1 - y^{-3/2}\). For \(y < 1\), \(G(y) = 0\). Differentiating with respect to \(y\) for \(y \ge 1\), we obtain the probability density function of \(Y\): \(g(y) = G'(y) = \frac{\mathrm{d}}{\mathrm{d}y}(1 - y^{-3/2}) = \frac{3}{2}y^{-5/2}\). For \(y < 1\), \(g(y) = 0\). Thus, \(g(y) = \begin{cases} \frac{3}{2}y^{-\frac{5}{2}} & y \ge 1 \\ 0 & \text{otherwise} \end{cases}\). (iii) \(\mathrm{E}(\sqrt{Y}) = \int_1^\infty \sqrt{y} \cdot g(y) \mathrm{d}y = \int_1^\infty y^{1/2} (\frac{3}{2}y^{-5/2}) \mathrm{d}y = \int_1^\infty \frac{3}{2}y^{-2} \mathrm{d}y = [-\frac{3}{2}y^{-1}]_1^\infty = 0 - (-\frac{3}{2}) = \frac{3}{2}\).

(i) M1: For integrating the pdf of \(X\) to find \(F(x)\) (limits 0 to \(x\)). A1: For obtaining correct \(F(x) = \frac{1}{8}x^3\) for \(0 \le x \le 2\) (with full definition for all intervals). (ii) M1: For stating or using \(G(y) = \mathrm{P}(Y \le y)\) and attempting to write this in terms of \(X\). M1: For correctly changing the inequality to \(X \ge \frac{2}{\sqrt{y}}\). A1: For obtaining \(G(y) = 1 - y^{-3/2}\) for \(y \ge 1\). M1: For differentiating their \(G(y)\) with respect to \(y\). A1: For correctly obtaining the given pdf \(g(y) = \frac{3}{2}y^{-5/2}\) with the appropriate range \(y \ge 1\). (iii) M1: For setting up the integral \(\int_1^\infty \sqrt{y} \cdot g(y) \mathrm{d}y\). A1: For simplifying to \(\int_1^\infty \frac{3}{2}y^{-2} \mathrm{d}y\) and obtaining the correct antiderivative \(-\frac{3}{2}y^{-1}\). A1: For evaluating the limits correctly to get \(\frac{3}{2}\) (or 1.5).

Let \(M_A\) and \(M_B\) be the median drying times of Brand \(A\) and Brand \(B\) respectively.

**Step 1: State the hypotheses**
\(H_0\): The median drying times of the two brands of paint are equal (\(M_A = M_B\)).
\(H_1\): The median drying times of the two brands of paint are different (\(M_A \neq M_B\)).

**Step 2: Combine and rank the data**
We have sample sizes \(n_1 = 6\) (Brand \(A\)) and \(n_2 = 7\) (Brand \(B\)). Let us rank the combined 13 observations in ascending order:

1. \(12.8\) (Brand \(A\)) \(\rightarrow\) Rank 1
2. \(13.9\) (Brand \(A\)) \(\rightarrow\) Rank 2
3. \(14.2\) (Brand \(A\)) \(\rightarrow\) Rank 3
4. \(14.8\) (Brand \(B\)) \(\rightarrow\) Rank 4
5. \(15.0\) (Brand \(A\)) \(\rightarrow\) Rank 5
6. \(15.8\) (Brand \(B\)) \(\rightarrow\) Rank 6
7. \(16.1\) (Brand \(B\)) \(\rightarrow\) Rank 7
8. \(16.5\) (Brand \(A\)) \(\rightarrow\) Rank 8
9. \(17.2\) (Brand \(B\)) \(\rightarrow\) Rank 9
10. \(18.1\) (Brand \(A\)) \(\rightarrow\) Rank 10
11. \(18.5\) (Brand \(B\)) \(\rightarrow\) Rank 11
12. \(19.5\) (Brand \(B\)) \(\rightarrow\) Rank 12
13. \(20.3\) (Brand \(B\)) \(\rightarrow\) Rank 13

**Step 3: Calculate the sum of ranks**
For the smaller sample (Brand \(A\), \(n_1 = 6\)), the sum of ranks is:
\(W_A = 1 + 2 + 3 + 5 + 8 + 10 = 29\)

For Brand \(B\) (\(n_2 = 7\)), the sum of ranks is:
\(W_B = 4 + 6 + 7 + 9 + 11 + 12 + 13 = 62\)

**Step 4: Calculate the test statistic \(U\)**
Using the formula for the Wilcoxon rank-sum test statistic:
\(U = W_A - \frac{1}{2} n_1(n_1 + 1)\)
\(U = 29 - \frac{1}{2}(6)(7) = 29 - 21 = 8\)

(Alternatively, \(U' = n_1 n_2 - U = 6 \times 7 - 8 = 34\). The test statistic is \(\min(U, U') = 8\).)

**Step 5: Determine the critical value and make a decision**
From the critical values table for the Wilcoxon rank-sum test with \(n_1 = 6\) and \(n_2 = 7\) at a \(5\%\) significance level (two-tailed), the critical value is \(6\).

Since our calculated value \(U = 8 > 6\), the result is not significant. We fail to reject \(H_0\).

**Step 6: Conclusion**
There is insufficient evidence at the \(5\%\) significance level to suggest that there is a difference in the median drying times of the two brands of quick-drying paint.

**B1**: For stating both \(H_0\) and \(H_1\) correctly in terms of medians or identical distributions (two-tailed).
**M1**: For ordering the combined data and attempting to assign ranks (allow up to 2 ranking errors).
**A1**: For obtaining the correct ranks for Brand \(A\) and Brand \(B\) as listed in the solution.
**M1**: For calculating the sum of ranks for Brand \(A\) (or Brand \(B\)).
**A1**: For obtaining \(W_A = 29\) (or \(W_B = 62\)) and calculating the test statistic \(U = 8\) (or \(U' = 34\)).
**B1**: For stating the correct critical value of \(6\) from the statistical tables.
**M1**: For comparing the calculated test statistic with the critical value (e.g., comparing \(8\) with \(6\)).
**A1**: For a correct conclusion in context, stating that there is no significant evidence of a difference in the median drying times of the two brands.

(i) The population of differences in productivity scores (After \( - \) Before) must be normally distributed.

(ii) Let \(d = Y - X\) represent the difference in productivity scores for each employee.

The individual differences are calculated as follows:
- Employee A: \(76 - 72 = 4\)
- Employee B: \(71 - 65 = 6\)
- Employee C: \(79 - 80 = -1\)
- Employee D: \(79 - 74 = 5\)
- Employee E: \(69 - 61 = 8\)
- Employee F: \(83 - 78 = 5\)
- Employee G: \(70 - 69 = 1\)

The sample size is \(n = 7\).

Let's calculate the sample mean \(\bar{d}\) of the differences:
\[
\bar{d} = \frac{4 + 6 - 1 + 5 + 8 + 5 + 1}{7} = \frac{28}{7} = 4
\]

Let's calculate the sum of squares of differences:
\[
\sum d^2 = 4^2 + 6^2 + (-1)^2 + 5^2 + 8^2 + 5^2 + 1^2 = 16 + 36 + 1 + 25 + 64 + 25 + 1 = 168
\]

The unbiased estimate of the population variance \(s^2\) is:
\[
s^2 = \frac{1}{n-1} \left( \sum d^2 - \frac{(\sum d)^2}{n} \right) = \frac{1}{6} \left( 168 - \frac{28^2}{7} \right) = \frac{1}{6} (168 - 112) = \frac{56}{6} = \frac{28}{3} \approx 9.333
\]

We set up the hypotheses:
- \(H_0: \mu_d = 0\)
- \(H_1: \mu_d > 0\)
where \(\mu_d\) is the population mean difference (After \( - \) Before).

The test statistic is calculated as:
\[
t = \frac{\bar{d} - 0}{\sqrt{\frac{s^2}{n}}} = \frac{4}{\sqrt{\frac{28}{3 \times 7}}} = \frac{4}{\sqrt{\frac{4}{3}}} = \frac{4}{2/\sqrt{3}} = 2\sqrt{3} \approx 3.464
\]

The number of degrees of freedom is \(\nu = n - 1 = 6\).

For a one-tailed test at the 5% significance level with \(6\) degrees of freedom, the critical value from the \(t\)-distribution table is \(t_{crit} = 1.943\).

Comparing the values:
\[
t = 3.464 > 1.943
\]

Since the test statistic is greater than the critical value, we reject the null hypothesis \(H_0\).

There is significant evidence at the 5% level of significance to suggest that using the focus application has led to an increase in productivity.

**Part (i)**
* **B1**: State that the differences in productivity scores must be normally distributed.

**Part (ii)**
* **B1**: Formulate correct hypotheses \(H_0: \mu_d = 0\) and \(H_1: \mu_d > 0\) (or equivalent with defined notation).
* **M1**: Calculate the differences \(d\) and attempt to calculate both the mean \(\bar{d}\) and the unbiased variance estimate \(s^2\).
* **A1**: Obtain \(\bar{d} = 4\) and \(s^2 = \frac{28}{3}\) (or 9.33).
* **M1**: Calculate the paired \(t\)-test statistic using the correct formula \(t = \frac{\bar{d}}{s/\sqrt{n}}\).
* **A1**: Obtain \(t = 2\sqrt{3} \approx 3.46\).
* **M1**: Find the correct critical value \(t_{0.05, 6} = 1.943\) and compare it with the calculated test statistic.
* **A1**: Correctly reject \(H_0\) and conclude in context that there is significant evidence that the application has led to an increase in productivity.

(i) The population of differences in productivity scores (After \( - \) Before) must be normally distributed.

(ii) Let \(d = Y - X\) represent the difference in productivity scores for each employee.

The individual differences are calculated as follows:
- Employee A: \(76 - 72 = 4\)
- Employee B: \(71 - 65 = 6\)
- Employee C: \(79 - 80 = -1\)
- Employee D: \(79 - 74 = 5\)
- Employee E: \(69 - 61 = 8\)
- Employee F: \(83 - 78 = 5\)
- Employee G: \(70 - 69 = 1\)

The sample size is \(n = 7\).

Let's calculate the sample mean \(\bar{d}\) of the differences:
\[
\bar{d} = \frac{4 + 6 - 1 + 5 + 8 + 5 + 1}{7} = \frac{28}{7} = 4
\]

Let's calculate the sum of squares of differences:
\[
\sum d^2 = 4^2 + 6^2 + (-1)^2 + 5^2 + 8^2 + 5^2 + 1^2 = 16 + 36 + 1 + 25 + 64 + 25 + 1 = 168
\]

The unbiased estimate of the population variance \(s^2\) is:
\[
s^2 = \frac{1}{n-1} \left( \sum d^2 - \frac{(\sum d)^2}{n} \right) = \frac{1}{6} \left( 168 - \frac{28^2}{7} \right) = \frac{1}{6} (168 - 112) = \frac{56}{6} = \frac{28}{3} \approx 9.333
\]

We set up the hypotheses:
- \(H_0: \mu_d = 0\)
- \(H_1: \mu_d > 0\)
where \(\mu_d\) is the population mean difference (After \( - \) Before).

The test statistic is calculated as:
\[
t = \frac{\bar{d} - 0}{\sqrt{\frac{s^2}{n}}} = \frac{4}{\sqrt{\frac{28}{3 \times 7}}} = \frac{4}{\sqrt{\frac{4}{3}}} = \frac{4}{2/\sqrt{3}} = 2\sqrt{3} \approx 3.464
\]

The number of degrees of freedom is \(\nu = n - 1 = 6\).

For a one-tailed test at the 5% significance level with \(6\) degrees of freedom, the critical value from the \(t\)-distribution table is \(t_{crit} = 1.943\).

Comparing the values:
\[
t = 3.464 > 1.943
\]

Since the test statistic is greater than the critical value, we reject the null hypothesis \(H_0\).

There is significant evidence at the 5% level of significance to suggest that using the focus application has led to an increase in productivity.

**Part (i)**
* **B1**: State that the differences in productivity scores must be normally distributed.

**Part (ii)**
* **B1**: Formulate correct hypotheses \(H_0: \mu_d = 0\) and \(H_1: \mu_d > 0\) (or equivalent with defined notation).
* **M1**: Calculate the differences \(d\) and attempt to calculate both the mean \(\bar{d}\) and the unbiased variance estimate \(s^2\).
* **A1**: Obtain \(\bar{d} = 4\) and \(s^2 = \frac{28}{3}\) (or 9.33).
* **M1**: Calculate the paired \(t\)-test statistic using the correct formula \(t = \frac{\bar{d}}{s/\sqrt{n}}\).
* **A1**: Obtain \(t = 2\sqrt{3} \approx 3.46\).
* **M1**: Find the correct critical value \(t_{0.05, 6} = 1.943\) and compare it with the calculated test statistic.
* **A1**: Correctly reject \(H_0\) and conclude in context that there is significant evidence that the application has led to an increase in productivity.

(a) Since \(\sum_{r=0}^{\infty} P(X = r) = 1\), we have:
\[\sum_{r=0}^{\infty} k(r + 1)a^r = 1\]
Using the standard binomial series expansion \(\sum_{r=0}^{\infty} (r + 1)x^r = (1 - x)^{-2}\) for \(|x| < 1\), we have:
\[k (1 - a)^{-2} = 1 \implies k = (1 - a)^2\]
Now, the probability generating function \(G_X(t)\) is defined as:
\[G_X(t) = \mathrm{E}(t^X) = \sum_{r=0}^{\infty} P(X = r)t^r = \sum_{r=0}^{\infty} (1 - a)^2 (r + 1)a^r t^r = (1 - a)^2 \sum_{r=0}^{\infty} (r + 1)(at)^r\]
Using the same binomial series expansion with \(x = at\), we get:
\[G_X(t) = \frac{(1 - a)^2}{(1 - at)^2}\]
which converges for \(|at| < 1\), or \(|t| < \frac{1}{a}\).

(b) To find \(\mathrm{E}(X)\), we differentiate \(G_X(t)\) with respect to \(t\):
\[G_X'(t) = \frac{\mathrm{d}}{\mathrm{d}t} \left[ (1 - a)^2 (1 - at)^{-2} \right] = (1 - a)^2 \cdot (-2)(1 - at)^{-3} \cdot (-a) = 2a(1 - a)^2 (1 - at)^{-3}\]
Evaluating at \(t = 1\):
\[\mathrm{E}(X) = G_X'(1) = 2a(1 - a)^2 (1 - a)^{-3} = \frac{2a}{1 - a}\]

To find \(\mathrm{Var}(X)\), we find the second derivative of \(G_X(t)\):
\[G_X''(t) = \frac{\mathrm{d}}{\mathrm{d}t} \left[ 2a(1 - a)^2 (1 - at)^{-3} \right] = 2a(1 - a)^2 \cdot (-3)(1 - at)^{-4} \cdot (-a) = 6a^2(1 - a)^2 (1 - at)^{-4}\]
Evaluating at \(t = 1\):
\[G_X''(1) = 6a^2(1 - a)^2 (1 - a)^{-4} = \frac{6a^2}{(1 - a)^2}\]
Using the variance formula for PGFs:
\[\mathrm{Var}(X) = G_X''(1) + G_X'(1) - [G_X'(1)]^2 = \frac{6a^2}{(1 - a)^2} + \frac{2a}{1 - a} - \left( \frac{2a}{1 - a} \right)^2\]
\[\mathrm{Var}(X) = \frac{6a^2 + 2a(1 - a) - 4a^2}{(1 - a)^2} = \frac{2a^2 + 2a}{(1 - a)^2} = \frac{2a}{(1 - a)^2}\]

(c) Since \(X\) and \(Y\) are independent and identically distributed, the PGF of their sum \(Z = X + Y\) is:
\[G_Z(t) = G_X(t) \cdot G_Y(t) = \left[ \frac{(1 - a)^2}{(1 - at)^2} \right]^2 = \frac{(1 - a)^4}{(1 - at)^4}\]

To find \(P(Z = 2)\), we need to find the coefficient of \(t^2\) in the Maclaurin expansion of \(G_Z(t)\):
\[G_Z(t) = (1 - a)^4 (1 - at)^{-4} = (1 - a)^4 \left[ 1 + (-4)(-at) + \frac{(-4)(-5)}{2!}(-at)^2 + \dots \right]\]
\[G_Z(t) = (1 - a)^4 \left[ 1 + 4at + 10a^2 t^2 + \dots \right]\]
Thus, the coefficient of \(t^2\) is:
\[P(Z = 2) = 10a^2 (1 - a)^4\]

Part (a)
- M1: For equating the sum of the probabilities to 1.
- A1: For showing \(k = (1 - a)^2\) using the binomial expansion of \((1 - a)^{-2}\).
- M1: For setting up the expression for \(G_X(t)\) as \(\sum P(X=r)t^r\).
- A1: For correctly deriving the given expression for \(G_X(t)\).

Part (b)
- M1: For differentiating \(G_X(t)\) to find \(G_X'(t)\) and substituting \(t=1\).
- A1: For obtaining \(\mathrm{E}(X) = \frac{2a}{1-a}\).
- M1: For finding \(G_X''(t)\) and using \(\mathrm{Var}(X) = G_X''(1) + G_X'(1) - [G_X'(1)]^2\).
- A1: For obtaining \(\mathrm{Var}(X) = \frac{2a}{(1-a)^2}\) (fully simplified).

Part (c)
- B1: For finding the correct PGF of \(Z\), \(G_Z(t) = \frac{(1-a)^4}{(1-at)^4}\).
- M1: For attempting to expand \((1 - at)^{-4}\) to find the coefficient of \(t^2\).
- A1: For obtaining \(P(Z=2) = 10a^2(1-a)^4\).