2025 Cambridge IAL Mathematics - Further (9231) Practice Paper with Answers

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

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

**Inductive Step:** Assume that \(P(k)\) is true for some positive integer \(k\). That is,
\[ \frac{\text{d}^k}{\text{d}x^k} (x e^{2x}) = 2^{k-1}(2x+k)e^{2x} \]
We must show that \(P(k+1)\) is true, i.e.,
\[ \frac{\text{d}^{k+1}}{\text{d}x^{k+1}} (x e^{2x}) = 2^k(2x+k+1)e^{2x} \]

Differentiating the expression for the \(k\)-th derivative with respect to \(x\):
\[ \frac{\text{d}^{k+1}}{\text{d}x^{k+1}} (x e^{2x}) = \frac{\text{d}}{\text{d}x} \left[ 2^{k-1}(2x+k)e^{2x} \right] \]
Applying the product rule:
\[ = 2^{k-1} \left[ 2e^{2x} + (2x+k) \cdot 2e^{2x} \right] \]
\[ = 2^{k-1} \cdot 2e^{2x} \left[ 1 + 2x + k \right] \]
\[ = 2^k (2x + k + 1)e^{2x} \]
This is the statement \(P(k+1)\).

Thus, \(P(1)\) is true, and if \(P(k)\) is true then \(P(k+1)\) is true. By mathematical induction, \(P(n)\) is true for all positive integers \(n\).

(ii) For \(n=10\), the tenth derivative is:
\[ \frac{\text{d}^{10}}{\text{d}x^{10}} (x e^{2x}) = 2^9 (2x + 10) e^{2x} \]

Substituting \(x = -\frac{1}{2}\):
\[ 2^9 \left(2\left(-\frac{1}{2}\right) + 10\right) e^{2(-1/2)} = 512 (-1 + 10) e^{-1} = 512(9)e^{-1} = 4608 e^{-1} \]

(i)
M1: Establishes base case \(n=1\) and shows LHS = RHS.
M1: States inductive hypothesis \(P(k)\) clearly.
M1: Sets up the derivative for \(P(k+1)\) from \(P(k)\).
M1: Applies product rule correctly to differentiate \(2^{k-1}(2x+k)e^{2x}\).
A1: Factors out \(2^k e^{2x}\) correctly.
A1: Obtains the correct expression \(2^k(2x+k+1)e^{2x}\).
A1: Provides a complete and clear concluding statement of mathematical induction.

(ii)
M1: Substitutes \(n=10\) into the formula.
M1: Substitutes \(x = -\frac{1}{2}\) into the derivative formula.
A1.7: Simplifies to obtain the exact final value \(4608 e^{-1}\) (or \(\frac{4608}{e}\)).

(i) From the given cubic equation, we have:
\(\sum \alpha = -2\)
\(\sum \alpha\beta = -3\)
\(\alpha\beta\gamma = -4\)

We know that:
\[ \sum \alpha^2 = (\sum \alpha)^2 - 2\sum \alpha\beta = (-2)^2 - 2(-3) = 4 + 6 = 10 \]

For \(\sum \alpha^3\), we use the relation \(\alpha^3 + 2\alpha^2 - 3\alpha + 4 = 0\). Summing over the three roots:
\[ \sum \alpha^3 + 2\sum \alpha^2 - 3\sum \alpha + 12 = 0 \]

Substituting the known values:
\[ \sum \alpha^3 + 2(10) - 3(-2) + 12 = 0 \]
\[ \sum \alpha^3 + 20 + 6 + 12 = 0 \implies \sum \alpha^3 = -38 \]

(ii) Let the roots of the new equation be \(u = \alpha\beta\), \(v = \beta\gamma\), and \(w = \gamma\alpha\).
We find the coefficients of the cubic equation \(y^3 - S_1 y^2 + S_2 y - S_3 = 0\).

\[ S_1 = u + v + w = \alpha\beta + \beta\gamma + \gamma\alpha = -3 \]

\[ S_2 = uv + vw + wu = \alpha^2\beta\gamma + \alpha\beta^2\gamma + \alpha\beta\gamma^2 = \alpha\beta\gamma(\alpha + \beta + \gamma) = (-4)(-2) = 8 \]

\[ S_3 = uvw = (\alpha\beta\gamma)^2 = (-4)^2 = 16 \]

Thus, the cubic equation is:
\[ y^3 - (-3)y^2 + 8y - 16 = 0 \implies y^3 + 3y^2 + 8y - 16 = 0 \]

(i)
B1: Identifies \(\sum \alpha = -2\), \(\sum \alpha\beta = -3\), and \(\alpha\beta\gamma = -4\).
M1: Uses identity \(\sum \alpha^2 = (\sum \alpha)^2 - 2\sum \alpha\beta\).
A1: Obtains \(\sum \alpha^2 = 10\).
M1: Uses the cubic relation to sum \(\alpha^3\).
A1: Obtains \(\sum \alpha^3 = -38\).

(ii)
M1: Identifies \(S_1 = \sum \alpha\beta = -3\).
M1: Expresses \(S_2 = uv + vw + wu\) as \(\alpha\beta\gamma(\sum \alpha)\).
A1: Computes \(S_2 = 8\).
M1: Expresses \(S_3 = uvw\) as \((\alpha\beta\gamma)^2\) and computes \(S_3 = 16\).
A1.7: Writes down the final equation \(y^3 + 3y^2 + 8y - 16 = 0\).

(i) Starting with the RHS:
\[ [\ln(r) - \ln(r+1)] - [\ln(r+1) - \ln(r+2)] = \ln(r) - 2\ln(r+1) + \ln(r+2) \]
Using the laws of logarithms:
\[ = \ln(r) + \ln(r+2) - \ln((r+1)^2) = \ln\left(\frac{r(r+2)}{(r+1)^2}\right) \]
This matches the LHS.

(ii) Using the method of differences with \(f(r) = \ln(r) - \ln(r+1)\):
\[ \sum_{r=1}^n \ln\left(\frac{r(r+2)}{(r+1)^2}\right) = \sum_{r=1}^n [f(r) - f(r+1)] \]
This is a telescoping series:
\[ = [f(1) - f(2)] + [f(2) - f(3)] + \dots + [f(n) - f(n+1)] \]
\[ = f(1) - f(n+1) \]
Since \(f(1) = \ln(1) - \ln(2) = -\ln(2)\), and \(f(n+1) = \ln(n+1) - \ln(n+2)\):
\[ \sum_{r=1}^n \ln\left(\frac{r(r+2)}{(r+1)^2}\right) = (\ln(1) - \ln(2)) - (\ln(n+1) - \ln(n+2)) \]
\[ = -\ln(2) - \ln(n+1) + \ln(n+2) = \ln(n+2) - \ln(2(n+1)) = \ln\left(\frac{n+2}{2(n+1)}\right) \]

(iii) To find the sum to infinity, we take the limit as \(n \to \infty\):
\[ \lim_{n \to \infty} \sum_{r=1}^n \ln\left(\frac{r(r+2)}{(r+1)^2}\right) = \lim_{n \to \infty} \ln\left(\frac{n+2}{2n+2}\right) \]
Since \(\frac{n+2}{2n+2} = \frac{1 + 2/n}{2 + 2/n} \to \frac{1}{2}\) as \(n \to \infty\):
\[ = \ln\left(\frac{1}{2}\right) = -\ln(2) \]

(i)
M1: Applies logarithm laws to expand or combine terms.
A1: Shows LHS = RHS clearly.

(ii)
M1: Expresses the summation in the difference form \(\sum [f(r) - f(r+1)]\).
M1: Writes out the first few terms and the last few terms to show cancelling.
A1: Obtains the unsimplified sum \(f(1) - f(n+1)\).
M1: Substitutes the expressions for \(f(1)\) and \(f(n+1)\).
A1: Simplifies to the final form \(\ln\left(\frac{n+2}{2(n+1)}\right)\).

(iii)
M1: Identifies the limit of \(\frac{n+2}{2n+2}\) as \(n \to \infty\).
A1: Finds the limit is \(\frac{1}{2}\).
A1.7: Concludes with \(\ln\left(\frac{1}{2}\right)\) or \(-\ln(2)\).

(i) To find the shortest distance between two skew lines, we find a vector normal to both lines:
\[ \mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -1 \\ 2 & -1 & 3 \end{vmatrix}
= \mathbf{i}(6 - 1) - \mathbf{j}(3 - (-2)) + \mathbf{k}(-1 - 4) = 5\mathbf{i} - 5\mathbf{j} - 5\mathbf{k} \]
We can use the simplified normal vector \(\mathbf{n}' = \mathbf{i} - \mathbf{j} - \mathbf{k}\).

The vector \(\vec{AB}\) connecting point \(A(1, -2, 3)\) on \(l_1\) to point \(B(2, 1, -1)\) on \(l_2\) is:
\[ \vec{AB} = (2-1)\mathbf{i} + (1 - (-2))\mathbf{j} + (-1 - 3)\mathbf{k} = \mathbf{i} + 3\mathbf{j} - 4\mathbf{k} \]

The shortest distance \(d\) is the projection of \(\vec{AB}\) onto the direction of the normal vector \(\mathbf{n}'\):
\[ d = \frac{|\vec{AB} \cdot \mathbf{n}'|}{|\mathbf{n}'|} = \frac{|(1)(1) + (3)(-1) + (-4)(-1)|}{\sqrt{1^2 + (-1)^2 + (-1)^2}}
= \frac{|1 - 3 + 4|}{\sqrt{3}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3} \]

(ii) The plane \(\Pi\) contains \(l_1\) and is parallel to \(l_2\).
Thus, the normal to the plane is parallel to \(\mathbf{d}_1 \times \mathbf{d}_2\), so we can use \(\mathbf{n} = \mathbf{i} - \mathbf{j} - \mathbf{k}\).

Since \(\Pi\) contains \(l_1\), it contains the point \(A(1, -2, 3)\).
The Cartesian equation of the plane is:
\[ 1(x - 1) - 1(y + 2) - 1(z - 3) = 0 \implies x - 1 - y - 2 - z + 3 = 0 \implies x - y - z = 0 \]

(i)
M1: Finds the vector product of direction vectors \(\mathbf{d}_1 \times \mathbf{d}_2\).
A1: Obtains \(5\mathbf{i} - 5\mathbf{j} - 5\mathbf{k}\) or any scalar multiple like \(\mathbf{i} - \mathbf{j} - \mathbf{k}\).
M1: Computes the vector \(\vec{AB} = \mathbf{i} + 3\mathbf{j} - 4\mathbf{k}\).
M1: Applies the formula for the shortest distance \(d = \frac{|\vec{AB} \cdot \mathbf{n}|}{|\mathbf{n}|}\).
A1: Obtains numerator \(2\).
A1: Obtains correct final exact distance \(\frac{2\sqrt{3}}{3}\) (or \(\frac{2}{\sqrt{3}}\)).

(ii)
M1: Recognises that the normal vector to the plane is the same as the common perpendicular \(\mathbf{i} - \mathbf{j} - \mathbf{k}\).
M1: Uses the coordinates of point \(A\) (or any point on \(l_1\)) in the plane equation.
A2.7: Arrives at the correct simplified Cartesian equation \(x - y - z = 0\).

(i) A matrix is singular if its determinant is zero.
\[ \det(\mathbf{M}) = k(k+3) - (2)(-1) = k^2 + 3k + 2 \]
Setting the determinant to zero:
\[ k^2 + 3k + 2 = 0 \implies (k+1)(k+2) = 0 \]
Thus, \(\mathbf{M}\) is singular when \(k = -1\) or \(k = -2\).

(ii) For \(k = 1\):
\[ \det(\mathbf{M}) = 1^2 + 3(1) + 2 = 6 \]
The area of the original triangle \(T\) with vertices \((0,0)\), \((2,0)\), and \((1,3)\) is:
\[ \text{Area}(T) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 3 = 3 \]

The area of the image triangle \(T'\) is given by:
\[ \text{Area}(T') = |\det(\mathbf{M})| \times \text{Area}(T) = 6 \times 3 = 18 \]

(iii) When \(k = 1\), the matrix is:
\[ \mathbf{M} = \begin{pmatrix} 1 & 2 \\ -1 & 4 \end{pmatrix} \]

Since \(\det(\mathbf{M}) = 6\), the inverse is:
\[ \mathbf{M}^{-1} = \frac{1}{6} \begin{pmatrix} 4 & -2 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} \frac{2}{3} & -\frac{1}{3} \\ \frac{1}{6} & \frac{1}{6} \end{pmatrix} \]

(i)
M1: Expresses determinant of \(\mathbf{M}\) in terms of \(k\).
M1: Sets determinant to zero and solves the quadratic equation.
A1: Finds correct values \(k = -1\) and \(k = -2\).

(ii)
M1: Calculates the determinant for \(k = 1\) (value is 6).
M1: Calculates the area of the original triangle \(T\) (value is 3).
M1: Uses the relationship \(\text{Area}(T') = |\det(\mathbf{M})| \times \text{Area}(T)\).
A1.7: Obtains area of \(T'\) as 18.

(iii)
M1: Applies the standard 2x2 inverse formula with correct signs/determinant.
A1: Correctly swaps elements and negates off-diagonals.
A1: Expresses the inverse clearly, e.g., \(\begin{pmatrix} 2/3 & -1/3 \\ 1/6 & 1/6 \end{pmatrix}\).

(i) The curve \(r = 2(1+\cos\theta)\) is a cardioid.
For \(\theta = 0\), \(r = 4\).
For \(\theta = \frac{\pi}{2}\), \(r = 2\).
For \(\theta = \pi\), \(r = 0\).
The sketch should show a smooth cardioid shape in the upper half-plane starting at \((4,0)\) in Cartesian coordinates, passing through \((0,2)\) on the positive y-axis, and terminating at the origin.

(ii) The region in the first quadrant enclosed by \(C\) and \(\theta = \frac{\pi}{2}\) is bounded by \(\theta = 0\) and \(\theta = \frac{\pi}{2}\).
The area \(A\) is given by:
\[ A = \frac{1}{2} \int_{0}^{\pi/2} r^2 \text{d}\theta = \frac{1}{2} \int_{0}^{\pi/2} [2(1 + \cos\theta)]^2 \text{d}\theta = 2 \int_{0}^{\pi/2} (1 + 2\cos\theta + \cos^2\theta) \text{d}\theta \]

Using the identity \(\cos^2\theta = \frac{1 + \cos 2\theta}{2\dots}\):
\[ = 2 \int_{0}^{\pi/2} \left( 1 + 2\cos\theta + \frac{1}{2} + \frac{1}{2}\cos 2\theta \right) \text{d}\theta \]

\[ = 2 \int_{0}^{\pi/2} \left( \frac{3}{2} + 2\cos\theta + \frac{1}{2}\cos 2\theta \right) \text{d}\theta \]

\[ = 2 \left[ \frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta \right]_{0}^{\pi/2} \]

Evaluating at the upper and lower limits:
\[ = 2 \left( \left[ \frac{3\pi}{4} + 2\sin\left(\frac{\pi}{2}\right) + \frac{1}{4}\sin(\pi) \right] - [0] \right) = 2 \left( \frac{3\pi}{4} + 2(1) + 0 \right) = \frac{3\pi}{2} + 4 \]

(i)
B1: Identifies the correct coordinates for key points, e.g., \((4, 0)\), \((2, \pi/2)\), \((0, \pi)\).
B2: A correctly drawn cardioid-like curve in the upper half-plane with a cusp at the pole.

(ii)
M1: Writes the correct integral for polar area with limits \(0\) to \(\frac{\pi}{2}\).
M1: Substitutes \(r = 2(1 + \cos\theta)\) and expands \(r^2\) correctly.
M1: Uses double angle identity \(\cos^2\theta = \frac{1+\cos 2\theta}{2}\).
A1: Integrates terms correctly to get \(\frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta\).
M1: Substitutes the limits \(0\) and \(\frac{\pi}{2}\) correctly.
A2.7: Arrives at the correct simplified exact area \(\frac{3\pi}{2} + 4\).

(i) To find the vertical asymptote, we find where the denominator is zero:
\[ x - 3 = 0 \implies x = 3 \]

To find the oblique asymptote, we perform algebraic division on the rational function:
\[ \frac{x^2 - x - 2}{x - 3} = \frac{x(x-3) + 2(x-3) + 4}{x - 3} = x + 2 + \frac{4}{x - 3} \]
As \(x \to \pm\infty\), \(\frac{4}{x-3} \to 0\).
Thus, the oblique asymptote is:
\[ y = x + 2 \]

(ii) To find the stationary points, we differentiate \(y = x + 2 + \frac{4}{x-3}\) with respect to \(x\):
\[ \frac{\text{d}y}{\text{d}x} = 1 - \frac{4}{(x-3)^2} \]
Setting \(\frac{\text{d}y}{\text{d}x} = 0\):
\[ 1 - \frac{4}{(x-3)^2} = 0 \implies (x-3)^2 = 4 \implies x - 3 = \pm 2 \]
So:
\(x = 5\) or \(x = 1\)

We find the corresponding \(y\)-coordinates:
For \(x = 5\):
\[ y = \frac{5^2 - 5 - 2}{5 - 3} = \frac{18}{2} = 9 \]

For \(x = 1\):
\[ y = \frac{1^2 - 1 - 2}{1 - 3} = \frac{-2}{-2} = 1 \]

Thus, the stationary points are \((5, 9)\) and \((1, 1)\).

(iii) Intercepts with the axes:
For \(x = 0\), \(y = \frac{-2}{-3} = \frac{2}{3}\).
For \(y = 0\), \(x^2 - x - 2 = 0 \implies (x-2)(x+1) = 0 \implies x = 2\) or \(x = -1\).

The curve \(C\) has two branches:
- One branch in the region \(x > 3\), with a local minimum at \((5,9)\), asymptotic to \(x=3\) and \(y=x+2\).
- Another branch in the region \(x < 3\), with a local maximum at \((1,1)\), passing through \((0, 2/3)\), \((2,0)\), and \((-1,0)\), asymptotic to \(x=3\) and \(y=x+2\).

(i)
M1: Finds vertical asymptote by setting denominator to zero.
A1: Obtains \(x = 3\).
M1: Uses long division or algebraic manipulation to find oblique asymptote.
A1: Obtains \(y = x + 2\).

(ii)
M1: Differentiates the equation of the curve.
A1: Obtains correct derivative \(\frac{\text{d}y}{\text{d}x} = 1 - \frac{4}{(x-3)^2}\).
M1: Sets derivative to zero and solves for \(x\).
A1: Obtains \(x = 5\) and \(x = 1\).
A0.7: Correctly calculates \(y\)-coordinates to give \((5, 9)\) and \((1, 1)\).

(iii)
B1: Shows the asymptotes \(x=3\) and \(y=x+2\) correctly, and labels the intercepts with axes.
B1: Draws two correct branches of the hyperbola-like curve with correct local maximum/minimum positions.

(i) To find the eigenvalues, we solve the characteristic equation \( \det(A - \lambda I) = 0 \):
\( \det \begin{pmatrix} 2-\lambda & 2 & -1 \\ 1 & 3-\lambda & -1 \\ -1 & -2 & 2-\lambda \end{pmatrix} = 0 \)
Expanding along the first row:
\( (2-\lambda)[(3-\lambda)(2-\lambda) - 2] - 2[(2-\lambda) - 1] - 1[-2 + (3-\lambda)] = 0 \)
\( (2-\lambda)(\lambda^2 - 5\lambda + 4) - 2(1-\lambda) - (1-\lambda) = 0 \)
\( (2-\lambda)(\lambda-1)(\lambda-4) + 3(\lambda-1) = 0 \)
\( (\lambda-1) [ (2-\lambda)(\lambda-4) + 3 ] = 0 \)
\( (\lambda-1)(-\lambda^2 + 6\lambda - 5) = 0 \)
\( -(\lambda-1)^2(\lambda-5) = 0 \)
So the eigenvalues are \( \lambda = 1, 1, 5 \).

(ii) For the repeated eigenvalue \( \lambda = 1 \):
\( (A - I)\mathbf{v} = 0 \implies \begin{pmatrix} 1 & 2 & -1 \\ 1 & 2 & -1 \\ -1 & -2 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \)
This yields the single equation \( x + 2y - z = 0 \).
We can choose two linearly independent eigenvectors:
If \( y = 0 \), we get \( \mathbf{v}_1 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} \).
If \( z = 0 \), we get \( \mathbf{v}_2 = \begin{pmatrix} -2 \\ 1 \\ 0 \end{pmatrix} \).

For the eigenvalue \( \lambda = 5 \):
\( (A - 5I)\mathbf{v} = 0 \implies \begin{pmatrix} -3 & 2 & -1 \\ 1 & -2 & -1 \\ -1 & -2 & -3 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \)
From the first two rows:
\( -3x + 2y - z = 0 \)
\( x - 2y - z = 0 \)
Subtracting the equations gives \( -4x + 4y = 0 \implies x = y \).
Substituting back gives \( z = -x \).
So an eigenvector is \( \mathbf{v}_3 = \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} \).

(iii) Let \( P \) be the diagonalising matrix:
\( P = \begin{pmatrix} 1 & -2 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & -1 \end{pmatrix} \)
Finding the inverse of \( P \):
\( P^{-1} = \frac{1}{4} \begin{pmatrix} 1 & 2 & 3 \\ -1 & 2 & 1 \\ 1 & 2 & -1 \end{pmatrix} \)
Using \( A^n = P D^n P^{-1} \), where \( D^n = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 5^n \end{pmatrix} \):
\( A^n = \begin{pmatrix} 1 & -2 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & -1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 5^n \end{pmatrix} P^{-1} \)
\( = \begin{pmatrix} 1 & -2 & 5^n \\ 0 & 1 & 5^n \\ 1 & 0 & -5^n \end{pmatrix} \frac{1}{4} \begin{pmatrix} 1 & 2 & 3 \\ -1 & 2 & 1 \\ 1 & 2 & -1 \end{pmatrix} \)
\( = \frac{1}{4} \begin{pmatrix} 3 + 5^n & -2 + 2 \cdot 5^n & 1 - 5^n \\ -1 + 5^n & 2 + 2 \cdot 5^n & 1 - 5^n \\ 1 - 5^n & 2 - 2 \cdot 5^n & 3 + 5^n \end{pmatrix} \)

M1: Set up characteristic equation \( \det(A - \lambda I) = 0 \).
A1: Find correct eigenvalues \( \lambda = 1, 1, 5 \).
M1: Substitute \( \lambda = 1 \) to find eigenvectors.
A1: Obtain two correct linearly independent eigenvectors for \( \lambda = 1 \).
A1: Obtain a correct eigenvector for \( \lambda = 5 \).
M1: Construct matrices \( P \) and find \( P^{-1} \).
A1: Correct inverse matrix \( P^{-1} \).
M1: Perform the multiplication \( P D^n P^{-1} \).
A1: Correct final expression for \( A^n \).

First find the complementary function (CF):
Auxiliary equation: \( m^2 + 4m + 5 = 0 \implies m = -2 \pm \mathrm{i} \).
\( y_{CF} = \mathrm{e}^{-2x} (A \cos x + B \sin x) \).

Now find the particular integral (PI):
Try \( y_{PI} = \mathrm{e}^{-x} (C \cos x + D \sin x) \).
Differentiating:
\( y'_{PI} = \mathrm{e}^{-x} [ (D-C)\cos x - (C+D)\sin x ] \)
\( y''_{PI} = \mathrm{e}^{-x} [ -2D\cos x + 2C\sin x ] \).

Substitute into the differential equation:
\( \mathrm{e}^{-x} [ -2D\cos x + 2C\sin x + 4(D-C)\cos x - 4(C+D)\sin x + 5C\cos x + 5D\sin x ] = 10 \mathrm{e}^{-x} \cos x \)
Grouping coefficients of \( \cos x \) and \( \sin x \):
\( C + 2D = 10 \)
\( -2C + D = 0 \implies D = 2C \).
Substituting \( D = 2C \) yields \( 5C = 10 \implies C = 2 \) and \( D = 4 \).
So, \( y_{PI} = \mathrm{e}^{-x} (2\cos x + 4\sin x) \).

The general solution is:
\( y = \mathrm{e}^{-2x} (A \cos x + B \sin x) + \mathrm{e}^{-x} (2 \cos x + 4 \sin x) \).

Apply initial conditions at \( x = 0 \):
\( y(0) = 2 \implies 2 = A + 2 \implies A = 0 \).
So \( y = B \mathrm{e}^{-2x} \sin x + \mathrm{e}^{-x} (2 \cos x + 4 \sin x) \).

Differentiating this particular form:
\( y' = B \mathrm{e}^{-2x} (\cos x - 2\sin x) + \mathrm{e}^{-x} [ 2 \cos x - 6 \sin x ] \).
Using \( y'(0) = 0 \):
\( 0 = B + 2 \implies B = -2 \).

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

M1: Solve the auxiliary equation to find complex roots.
A1: Correct complementary function.
M1: Set up the correct form for the particular integral.
A1: Correctly differentiate the proposed PI.
M1: Equate coefficients to solve for constants.
A1: Find correct values \( C = 2 \) and \( D = 4 \).
M1: Apply initial conditions to the general solution.
A1: Find \( A = 0 \) and \( B = -2 \).
A1: Correct final expression of the particular solution.

(i) By de Moivre's theorem:
\( \cos 5\theta + \mathrm{i} \sin 5\theta = (\cos \theta + \mathrm{i} \sin \theta)^5 \)
Expanding using the Binomial theorem:
\( \cos 5\theta + \mathrm{i} \sin 5\theta = \cos^5 \theta + 5\mathrm{i} \cos^4 \theta \sin \theta - 10 \cos^3 \theta \sin^2 \theta - 10\mathrm{i} \cos^2 \theta \sin^3 \theta + 5 \cos \theta \sin^4 \theta + \mathrm{i} \sin^5 \theta \)
Equating the imaginary and real parts:
\( \sin 5\theta = 5\cos^4 \theta \sin \theta - 10\cos^2 \theta \sin^3 \theta + \sin^5 \theta \)
\( \cos 5\theta = \cos^5 \theta - 10\cos^3 \theta \sin^2 \theta + 5\cos \theta \sin^4 \theta \)
Dividing \( \sin 5\theta \) by \( \cos 5\theta \) and dividing numerator and denominator by \( \cos^5 \theta \):
\( \tan 5\theta = \frac{5\tan \theta - 10\tan^3 \theta + \tan^5 \theta}{1 - 10\tan^2 \theta + 5\tan^4 \theta} \).

(ii) Setting \( \tan 5\theta = 0 \) gives \( 5\theta = k\pi \implies \theta = \frac{k\pi}{5} \) for \( k \in \mathbb{Z} \).
For \( k = 1, 2, 3, 4 \), we have \( \tan \theta \neq 0 \).
Thus, the numerator of \( \tan 5\theta \) must be 0:
\( \tan \theta ( \tan^4 \theta - 10\tan^2 \theta + 5 ) = 0 \)
Since \( \tan \theta \neq 0 \), this yields \( t^4 - 10t^2 + 5 = 0 \) where \( t = \tan \theta \).
These four roots correspond to \( \theta = \frac{\pi}{5}, \frac{2\pi}{5}, \frac{3\pi}{5}, \frac{4\pi}{5} \).
Since \( \tan \frac{4\pi}{5} = -\tan \frac{\pi}{5} \) and \( \tan \frac{3\pi}{5} = -\tan \frac{2\pi}{5} \), the roots are indeed \( \pm \tan \frac{\pi}{5} \) and \( \pm \tan \frac{2\pi}{5} \).

(iii) For the quartic equation \( t^4 - 10t^2 + 5 = 0 \), the product of the roots is given by the constant term:
\( t_1 t_2 t_3 t_4 = 5 \)
Substituting the roots:
\( \left(\tan \frac{\pi}{5}\right)\left(-\tan \frac{\pi}{5}\right)\left(\tan \frac{2\pi}{5}\right)\left(-\tan \frac{2\pi}{5}\right) = 5 \)
\( \implies \tan^2 \frac{\pi}{5} \tan^2 \frac{2\pi}{5} = 5 \).

M1: Expand the binomial expression to relate trigonometric terms.
A1: Obtain correct formulas for \( \sin 5\theta \) and \( \cos 5\theta \).
A1: Deduce correct expression for \( \tan 5\theta \).
M1: Equate \( \tan 5\theta = 0 \) to find the angle solutions.
A1: Identify the quartic equation in \( t \).
A1: Complete proof of the four root expressions.
M1: Use the product of roots formula for a quartic equation.
A1: Obtain the final exact value of 5.

(i) Using the identity \( \cosh 2x = 2\sinh^2 x + 1 \):
\( 2\sinh^2 x + 1 - 5\sinh x = 13 \implies 2\sinh^2 x - 5\sinh x - 12 = 0 \)
Factorising the quadratic equation:
\( (2\sinh x + 3)(\sinh x - 4) = 0 \)
So \( \sinh x = 4 \) or \( \sinh x = -\frac{3}{2} \).
Using the logarithmic formula \( \sinh^{-1} y = \ln(y + \sqrt{y^2 + 1}) \):
For \( \sinh x = 4 \):
\( x = \ln(4 + \sqrt{16 + 1}) = \ln(4 + \sqrt{17}) \).
For \( \sinh x = -\frac{3}{2} \):
\( x = \ln\left(-\frac{3}{2} + \sqrt{\frac{9}{4} + 1}\right) = \ln\left(\frac{\sqrt{13} - 3}{2}\right) \).

(ii) Let \( y = \tanh^{-1} x \implies x = \tanh y = \frac{\mathrm{e}^y - \mathrm{e}^{-y}}{\mathrm{e}^y + \mathrm{e}^{-y}} = \frac{\mathrm{e}^{2y} - 1}{\mathrm{e}^{2y} + 1} \).
Rearranging to solve for \( \mathrm{e}^{2y} \):
\( x(\mathrm{e}^{2y} + 1) = \mathrm{e}^{2y} - 1 \implies x\mathrm{e}^{2y} + x = \mathrm{e}^{2y} - 1 \)
\( \mathrm{e}^{2y}(1 - x) = 1 + x \implies \mathrm{e}^{2y} = \frac{1+x}{1-x} \).
Taking the natural logarithm of both sides:
\( 2y = \ln\left(\frac{1+x}{1-x}\right) \implies y = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right) \).

M1: Use hyperbolic double-angle identity to obtain a quadratic in \( \sinh x \).
A1: Factorise quadratic correctly.
A1: Find correct values for \( \sinh x \).
M1: Apply logarithmic form of \( \sinh^{-1} \).
A1: Obtain first correct logarithmic answer.
A1: Obtain second correct logarithmic answer.
M1: Write \( \tanh y \) in exponential form.
A1: Rearrange the equation to isolate \( \mathrm{e}^{2y} \).
A1: Correctly conclude the proof.

(i) Integrating by parts on \( I_n = \int_1^{\mathrm{e}} (\ln x)^n \cdot 1 \, \mathrm{d}x \):
Let \( u = (\ln x)^n \implies \mathrm{d}u = n(\ln x)^{n-1} \frac{1}{x} \, \mathrm{d}x \).
Let \( \mathrm{d}v = 1 \, \mathrm{d}x \implies v = x \).
Then:
\( I_n = \left[ x(\ln x)^n \right]_1^{\mathrm{e}} - \int_1^{\mathrm{e}} x \cdot n(\ln x)^{n-1} \frac{1}{x} \, \mathrm{d}x \)
\( = \left( \mathrm{e}(\ln \mathrm{e})^n - 1(\ln 1)^n \right) - n \int_1^{\mathrm{e}} (\ln x)^{n-1} \, \mathrm{d}x \)
Since \( \ln \mathrm{e} = 1 \) and \( \ln 1 = 0 \):
\( I_n = \mathrm{e} - n I_{n-1} \).

(ii) To find \( I_3 \), we first calculate \( I_0 \):
\( I_0 = \int_1^{\mathrm{e}} 1 \, \mathrm{d}x = \mathrm{e} - 1 \).
Applying the reduction formula successively:
\( I_1 = \mathrm{e} - 1 \cdot I_0 = \mathrm{e} - (\mathrm{e} - 1) = 1 \).
\( I_2 = \mathrm{e} - 2 I_1 = \mathrm{e} - 2 \).
\( I_3 = \mathrm{e} - 3 I_2 = \mathrm{e} - 3(\mathrm{e} - 2) = 6 - 2\mathrm{e} \).

M1: Attempt integration by parts with appropriate choice of \( u \) and \( v' \).
A1: Correct derivatives and antiderivatives.
M1: Apply limits of integration correctly.
A1: Establish reduction formula.
M1: Find correct value of \( I_0 \).
A1: Successively calculate \( I_1 \) and \( I_2 \).
A1: Deduce exact value of \( I_3 \) in the form \( 6 - 2\mathrm{e} \).

(i) First find the derivatives with respect to \( t \):
\( \frac{\mathrm{d}x}{\mathrm{d}t} = -6\cos^2 t \sin t \)
\( \frac{\mathrm{d}y}{\mathrm{d}t} = 6\sin^2 t \cos t \).
Then:
\( \left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2 = 36\cos^4 t \sin^2 t + 36\sin^4 t \cos^2 t \)
\( = 36\sin^2 t \cos^2 t (\cos^2 t + \sin^2 t) = 36\sin^2 t \cos^2 t \).
For \( 0 \le t \le \frac{\pi}{2} \), we have \( \sin t \cos t \ge 0 \):
\( \sqrt{\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2} = 6\sin t \cos t = 3\sin 2t \).
Arc length:
\( s = \int_0^{\pi/2} 3\sin 2t \, \mathrm{d}t = \left[ -\frac{3}{2}\cos 2t \right]_0^{\pi/2} = -\frac{3}{2}(-1 - 1) = 3 \).

(ii) Area of surface of revolution about the \( x \)-axis:
\( S = \int_0^{\pi/2} 2\pi y \sqrt{\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2} \, \mathrm{d}t \)
\( = 2\pi \int_0^{\pi/2} (2\sin^3 t)(6\sin t \cos t) \, \mathrm{d}t \)
\( = 24\pi \int_0^{\pi/2} \sin^4 t \cos t \, \mathrm{d}t \).
Using substitution \( u = \sin t \implies \mathrm{d}u = \cos t \, \mathrm{d}t \):
\( S = 24\pi \int_0^1 u^4 \, \mathrm{d}u = 24\pi \left[ \frac{u^5}{5} \right]_0^1 = \frac{24\pi}{5} \).

M1: Find derivatives \( \frac{\mathrm{d}x}{\mathrm{d}t} \) and \( \frac{\mathrm{d}y}{\mathrm{d}t} \).
A1: Show that the sum of squares simplifies to \( 36\sin^2 t \cos^2 t \).
M1: Set up the correct integral for arc length.
A1: Correctly integrate and find arc length is 3.
M1: Write down the correct formula for surface area.
A1: Correctly substitute parametric equations into the surface area formula.
M1: Perform the integration using substitution.
A1: Correctly evaluate the limits and conclude with \( \frac{24\pi}{5} \).

(i) Since \( \omega \) is a 7th root of unity and \( \omega \neq 1 \):
\( 1 + \omega + \dots + \omega^6 = \frac{\omega^7 - 1}{\omega - 1} = 0 \implies \sum_{k=1}^{6} \omega^k = -1 \).

(ii) Sum:
\( p + q = (\omega + \omega^2 + \omega^4) + (\omega^3 + \omega^5 + \omega^6) = \sum_{k=1}^{6} \omega^k = -1 \).
Product:
\( pq = (\omega + \omega^2 + \omega^4)(\omega^3 + \omega^5 + \omega^6) \)
\( = \omega^4 + \omega^6 + \omega^7 + \omega^5 + \omega^7 + \omega^8 + \omega^7 + \omega^9 + \omega^{10} \)
Using \( \omega^7 = 1 \), \( \omega^8 = \omega \), \( \omega^9 = \omega^2 \), \( \omega^{10} = \omega^3 \):
\( pq = \omega^4 + \omega^6 + 1 + \omega^5 + 1 + \omega + 1 + \omega^2 + \omega^3 \)
\( = 3 + (\omega + \omega^2 + \dots + \omega^6) = 3 - 1 = 2 \).

(iii) Since \( p+q = -1 \) and \( pq = 2 \), \( p \) and \( q \) are roots of the quadratic equation:
\( z^2 + z + 2 = 0 \)
Solving this gives:
\( z = \frac{-1 \pm \mathrm{i}\sqrt{7}}{2} \).
Notice that:
\( \operatorname{Im}(p) = \sin \frac{2\pi}{7} + \sin \frac{4\pi}{7} + \sin \frac{8\pi}{7} \).
Since \( \sin \frac{8\pi}{7} = -\sin \frac{\pi}{7} \), the sum is:
\( \sin \frac{2\pi}{7} + \sin \frac{4\pi}{7} - \sin \frac{\pi}{7} \).
Because \( \sin \frac{4\pi}{7} > \sin \frac{\pi}{7} \) and \( \sin \frac{2\pi}{7} > 0 \), \( \operatorname{Im}(p) > 0 \).
Thus, \( p = \frac{-1 + \mathrm{i}\sqrt{7}}{2} \).
Equating the imaginary parts:
\( \sin \frac{2\pi}{7} + \sin \frac{4\pi}{7} + \sin \frac{8\pi}{7} = \frac{\sqrt{7}}{2} \).

M1: Apply formula for sum of geometric progression.
A1: Complete the proof that the sum is -1.
M1: Write down sum \( p+q \) and product \( pq \).
A1: Simplify \( pq \) using the root of unity property.
A1: Find correct product \( pq = 2 \).
M1: Construct the quadratic equation \( z^2 + z + 2 = 0 \).
A1: Solve the quadratic equation correctly.
M1: Analyse the sign of the imaginary part of \( p \).
A1: Complete the proof to get \( \frac{\sqrt{7}}{2} \).

(i) Differentiating \( u = y^{-2} \) with respect to \( x \) using the chain rule:
\( \frac{\mathrm{d}u}{\mathrm{d}x} = -2y^{-3} \frac{\mathrm{d}y}{\mathrm{d}x} \implies y^{-3}\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{1}{2}\frac{\mathrm{d}u}{\mathrm{d}x} \).
Divide the original differential equation by \( x y^3 \):
\( y^{-3} \frac{\mathrm{d}y}{\mathrm{d}x} + \frac{1}{x} y^{-2} = x \).
Substituting \( y^{-3}\frac{\mathrm{d}y}{\mathrm{d}x} \) and \( y^{-2} = u \):
\( -\frac{1}{2}\frac{\mathrm{d}u}{\mathrm{d}x} + \frac{1}{x} u = x \).
Multiplying by \( -2 \):
\( \frac{\mathrm{d}u}{\mathrm{d}x} - \frac{2}{x} u = -2x \).

(ii) The integrating factor \( I \) is:
\( I = \mathrm{e}^{\int -\frac{2}{x} \, \mathrm{d}x} = \mathrm{e}^{-2\ln x} = x^{-2} = \frac{1}{x^2} \).
Multiplying the differential equation by \( \frac{1}{x^2} \):
\( \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{u}{x^2}\right) = -\frac{2}{x} \).
Integrating both sides with respect to \( x \):
\( \frac{u}{x^2} = -2\ln x + C \implies u = x^2(C - 2\ln x) \).

(iii) Since \( u = y^{-2} = \frac{1}{y^2} \):
\( \frac{1}{y^2} = x^2(C - 2\ln x) \).
Substituting \( x = 1 \) and \( y = 1 \):
\( 1 = 1(C - 0) \implies C = 1 \).
So, \( y^2 = \frac{1}{x^2(1 - 2\ln x)} \).
Since \( y > 0 \):
\( y = \frac{1}{x \sqrt{1 - 2\ln x}} \).

M1: Differentiate \( u = y^{-2} \) using the chain rule.
M1: Divide the original ODE by \( x y^3 \) and substitute.
A1: Correctly show the reduced linear ODE.
M1: Find the correct integrating factor.
A1: Correct general form of the integrated expression.
A1: Correctly solve for \( u \) with a constant of integration.
M1: Re-substitute to find \( y^2 \) and apply the initial condition.
A1: Find \( C = 1 \).
A1: Deduce the correct particular solution for \( y \).

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

Substituting \(x = 10\), \(y = 5\), \(V = 14\) and \(g = 9.8\):
\[5 = 10\tan\alpha - \frac{9.8 \times 10^2(1+\tan^2\alpha)}{2 \times 14^2}\]

Simplify the fractional coefficient:
\[\frac{9.8 \times 100}{2 \times 196} = \frac{980}{392} = 2.5\]

Substituting this back into the equation:
\[5 = 10\tan\alpha - 2.5(1+\tan^2\alpha)\]
\[5 = 10\tan\alpha - 2.5 - 2.5\tan^2\alpha\]
\[2.5\tan^2\alpha - 10\tan\alpha + 7.5 = 0\]

Dividing the entire equation by \(2.5\) yields:
\[\tan^2\alpha - 4\tan\alpha + 3 = 0\]

(ii) Solving this quadratic equation for \(\tan\alpha\):
\[(\tan\alpha - 1)(\tan\alpha - 3) = 0\]
\[\tan\alpha = 1 \implies \alpha = 45.0^\circ\]
\[\tan\alpha = 3 \implies \alpha = \arctan(3) \approx 71.6^\circ\]

(i)
M1: Use of the trajectory equation with appropriate substitutions.
A1: Correct substitution of values into the formula.
M1: Simplification of the quadratic equation terms.
A1: Final verification of the given quadratic equation.

(ii)
M1: Solving the quadratic equation to find two values of \(\tan\alpha\).
A1: Finding the first angle \(\alpha = 45.0^\circ\).
A1: Finding the second angle \(\alpha \approx 71.6^\circ\) (accept 71.6).

Let \(\theta\) be the angle that the radius to the particle makes with the downward vertical. By conservation of energy:
\[\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mga(1 - \cos\theta)\]
\[v^2 = u^2 - 2ga(1 - \cos\theta) = kga - 2ga + 2ga\cos\theta = ga(k - 2 + 2\cos\theta)\]

The radial equation of motion for the particle is:
\[T - mg\cos\theta = \frac{mv^2}{a}\]
\[T = mg\cos\theta + mg(k - 2 + 2\cos\theta) = mg(k - 2 + 3\cos\theta)\]

The particle leaves the surface when the tension \(T = 0\). Let this occur at \(\theta = \theta_0\):
\[\cos\theta_0 = \frac{2-k}{3}\]

At this point of departure, the velocity is \(v_0\) where:
\[v_0^2 = -ga\cos\theta_0\]

After leaving the surface, the particle is a projectile starting at position \((a\sin\theta_0, -a\cos\theta_0)\) relative to the center of the sphere. The velocity components are \(v_0\cos\theta_0\) horizontally and \(v_0\sin\theta_0\) vertically. For the projectile to pass through the center \((0,0)\):
\[a\sin\theta_0 + v_0\cos\theta_0 t = 0 \implies t = -\frac{a\sin\theta_0}{v_0\cos\theta_0}\]

Substituting this into the vertical equation of motion:
\[-a\cos\theta_0 + v_0\sin\theta_0\left(-\frac{a\sin\theta_0}{v_0\cos\theta_0}\right) - \frac{1}{2}gt^2 = 0\]
\[-a\cos\theta_0 - a\frac{\sin^2\theta_0}{\cos\theta_0} - \frac{1}{2}g\left(-\frac{a\sin\theta_0}{v_0\cos\theta_0}\right)^2 = 0\]

Multiplying by \(\cos\theta_0\) and simplifying:
\[-a - \frac{ga^2\sin^2\theta_0}{2v_0^2\cos\theta_0} = 0\]

Substitute \(v_0^2 = -ga\cos\theta_0\):
\[1 - \frac{\sin^2\theta_0}{2\cos^2\theta_0} = 0 \implies 2\cos^2\theta_0 = \sin^2\theta_0\]
\[2\cos^2\theta_0 = 1 - \cos^2\theta_0 \implies 3\cos^2\theta_0 = 1\]

Since the particle leaves the surface in the upper half of the sphere, \(\cos\theta_0 < 0\):
\[\cos\theta_0 = -\frac{1}{\sqrt{3}}\]

Equating the expressions for \(\cos\theta_0\):
\[\frac{2-k}{3} = -\frac{1}{\sqrt{3}} \implies k = 2 + \sqrt{3}\]

M1: Expresses conservation of energy to find \(v^2\) in terms of \(\theta\) and \(k\).
M1: Applies the radial equation of motion and sets the tension \(T = 0\) to find \(\cos\theta_0\).
A1: Obtains \(\cos\theta_0 = \frac{2-k}{3}\).
M1: Sets up projectile equations of motion relative to the center of the sphere.
M1: Eliminates \(t\) and uses \(v_0^2 = -ga\cos\theta_0\) to find a relation in terms of \(\theta_0\) only.
A1: Correctly simplifies to obtain \(3\cos^2\theta_0 = 1\).
A1: Solves for \(k\) to obtain \(k = 2 + \sqrt{3}\).

Let \(R_A\) and \(F_A\) be the vertical normal reaction and friction force at the ground \(A\) respectively.
Let \(R_B\) be the normal reaction force at the smooth wall \(B\).
Let \(x\) be the distance from \(A\) that the man climbs.

Resolving vertically for equilibrium:
\[R_A = mg + 2mg = 3mg\]

At the point of slipping, the friction force is at its maximum value:
\[F_A = \mu R_A = \frac{\sqrt{3}}{9} (3mg) = \frac{\sqrt{3}}{3}mg\]

Resolving horizontally for equilibrium:
\[R_B = F_A = \frac{\sqrt{3}}{3}mg\]

Taking moments about \(A\):
\[R_B \times 2a\sin(60^\circ) = mg \times a\cos(60^\circ) + 2mg \times x\cos(60^\circ)\]

Substitute the values for the trigonometric functions:
\[R_B \times 2a \frac{\sqrt{3}}{2} = mg \times a \frac{1}{2} + 2mg \times x \frac{1}{2}\]
\[R_B \times a\sqrt{3} = \frac{1}{2}mg(a + 2x)\]

Now substitute \(R_B = \frac{\sqrt{3}}{3}mg\):
\[\left(\frac{\sqrt{3}}{3}mg\right) a\sqrt{3} = \frac{1}{2}mg(a + 2x)\]
\[mga = \frac{1}{2}mg(a + 2x)\]
\[a = \frac{1}{2}(a + 2x)\]
\[2a = a + 2x \implies 2x = a \implies x = \frac{1}{2}a\]

M1: Resolves forces vertically for the system to find \(R_A\).
A1: Finds \(R_A = 3mg\).
M1: Applies the slipping condition \(F_A = \mu R_A\).
M1: Takes moments about \(A\) for the ladder.
A1: Obtains the correct moment equation containing \(x\).
M1: Integrates horizontal equilibrium and the slipping condition to solve for \(x\).
A1: Obtains \(x = \frac{1}{2}a\).

(i) The forces acting on the particle along the direction of motion are friction \(F_{\text{friction}} = \mu mg\) and air resistance \(F_{\text{air}} = mkv^2\).

The equation of motion is:
\[m v \frac{dv}{dx} = -\mu mg - mkv^2\]
\[v \frac{dv}{dx} = -(\mu g + kv^2)\]

Separating variables:
\[\frac{v}{\mu g + kv^2} dv = -dx\]

Integrating from the initial state \(x = 0, v = u\) to the stopping state \(x = s, v = 0\):
\[\int_u^0 \frac{v}{\mu g + kv^2} dv = -\int_0^s dx\]
\[\left[ \frac{1}{2k} \ln(\mu g + kv^2) \right]_u^0 = -s\]
\[\frac{1}{2k}\ln(\mu g) - \frac{1}{2k}\ln(\mu g + ku^2) = -s\]
\[s = \frac{1}{2k}\ln\left(\frac{\mu g + ku^2}{\mu g}\right) = \frac{1}{2k}\ln\left(1 + \frac{ku^2}{\mu g}\right)\]

(ii) Given values: \(u = 14\), \(\mu = 0.4\), \(k = 0.05\), \(g = 9.8\).
First calculate the inner fraction:
\[\mu g = 0.4 \times 9.8 = 3.92\]
\[ku^2 = 0.05 \times 14^2 = 0.05 \times 196 = 9.8\]
\[\frac{ku^2}{\mu g} = \frac{9.8}{3.92} = 2.5\]

Substitute these into the expression for \(s\):
\[s = \frac{1}{2 \times 0.05} \ln(1 + 2.5) = 10 \ln(3.5)\]
\[s \approx 10 \times 1.25276 = 12.5276\text{ m}\]

To 3 significant figures, \(s = 12.5\text{ m}\).

(i)
M1: Writes down the equation of motion using \(v \frac{dv}{dx}\) with correct signs.
M1: Separates variables and sets up correct limits for integration.
A1: Correct integration to get the log terms.
A1: Completes the algebraic steps to show the given formula.

(ii)
M1: Substitutes values into the formula.
A1: Calculates the correct value for the argument of the logarithm, i.e., \(3.5\).
A1: Calculates the final distance correctly to 3 significant figures.

(i) Let \(x_1\) be the distance \(AP\) and \(x_2\) be the distance \(PB\) when \(P\) is in equilibrium. We have:
\[x_1 + x_2 = 3a\]

Let the extensions in the two strings be \(e_1\) and \(e_2\). Since both strings are taut:
\[e_1 = x_1 - a\text{ and } e_2 = x_2 - a\]
\[e_1 + e_2 = x_1 + x_2 - 2a = 3a - 2a = a\]

In equilibrium, the tensions in the two strings must be equal:
\[T_1 = T_2 \implies \frac{\lambda_1 e_1}{a} = \frac{\lambda_2 e_2}{a} \implies mg e_1 = 2mg e_2 \implies e_1 = 2e_2\]

Using \(e_1 + e_2 = a\):
\[2e_2 + e_2 = a \implies 3e_2 = a \implies e_2 = \frac{1}{3}a \text{ and } e_1 = \frac{2}{3}a\]

The distance of \(P\) from \(A\) in equilibrium is:
\[AP = a + e_1 = a + \frac{2}{3}a = \frac{5}{3}a\]

(ii) Suppose \(P\) is displaced by a distance \(x\) from its equilibrium position towards \(B\).
The new extension of the first string is \(\frac{2}{3}a + x\) and the new extension of the second string is \(\frac{1}{3}a - x\).

Applying Newton's second law for the motion of \(P\):
\[m\ddot{x} = T_2 - T_1\]
\[m\ddot{x} = \frac{2mg}{a}\left(\frac{1}{3}a - x\right) - \frac{mg}{a}\left(\frac{2}{3}a + x\right)\]
\[m\ddot{x} = \frac{2}{3}mg - \frac{2mg}{a}x - \frac{2}{3}mg - \frac{mg}{a}x\]
\[m\ddot{x} = -\frac{3mg}{a}x \implies \ddot{x} = -\frac{3g}{a}x\]

This is the standard equation for simple harmonic motion, \(\ddot{x} = -\omega^2 x\), with:
\[\omega^2 = \frac{3g}{a}\]

The period \(T\) of the motion is:
\[T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{a}{3g}}\]

(i)
M1: Relates the total distance to the extensions of both strings.
M1: Equates the equilibrium tensions \(T_1 = T_2\).
A1: Obtains the correct equilibrium position distance \(\frac{5}{3}a\).

(ii)
M1: Expresses the tensions \(T_1, T_2\) after a displacement \(x\).
M1: Formulates the equation of motion \(m\ddot{x} = T_2 - T_1\).
A1: Simplifies to the SHM form \(\ddot{x} = -\frac{3g}{a}x\).
A1: Identifies \(\omega\) and states the correct period \(T = 2\pi\sqrt{\frac{a}{3g}}\).

(i) Let the line of centres be the \(x\)-axis. The velocity of \(A\) before collision is:
\[\mathbf{u}_A = \begin{pmatrix} u\cos\alpha \\ u\sin\alpha \end{pmatrix}\]

Since the spheres are smooth, the component of velocity of \(A\) perpendicular to the line of centres is unchanged after collision:
\[v_{Ay} = u\sin\alpha\]

Let \(v_{Ax}\) and \(v_{Bx}\) be the components of the velocities of \(A\) and \(B\) after collision parallel to the line of centres.
By conservation of momentum parallel to the line of centres:
\[m u\cos\alpha = m v_{Ax} + 2m v_{Bx} \implies u\cos\alpha = v_{Ax} + 2v_{Bx}\]

By Newton's law of restitution:
\[v_{Bx} - v_{Ax} = e(u\cos\alpha - 0) = e u\cos\alpha\]

Substituting \(v_{Bx} = v_{Ax} + e u\cos\alpha\) into the momentum equation:
\[u\cos\alpha = v_{Ax} + 2(v_{Ax} + e u\cos\alpha)\]
\[u\cos\alpha = 3v_{Ax} + 2e u\cos\alpha \implies v_{Ax} = \frac{1}{3}u\cos\alpha (1 - 2e)\]

So the components of the velocity of \(A\) after collision are:
Parallel component: \(v_{Ax} = \frac{1}{3}u\cos\alpha (1 - 2e)\)
Perpendicular component: \(v_{Ay} = u\sin\alpha\)

(ii) The direction of motion of \(A\) is perpendicular to its initial direction if the dot product of their velocity vectors is zero:
\[\mathbf{u}_A \cdot \mathbf{v}_A = 0\]
\[\begin{pmatrix} u\cos\alpha \\ u\sin\alpha \end{pmatrix} \cdot \begin{pmatrix} \frac{1}{3}u\cos\alpha (1 - 2e) \\ u\sin\alpha \end{pmatrix} = 0\]
\[\frac{1}{3}u^2\cos^2\alpha (1 - 2e) + u^2\sin^2\alpha = 0\]

Dividing by \(u^2\cos^2\alpha\) (where \(u \ne 0, \cos\alpha \ne 0\)):
\[\frac{1}{3}(1 - 2e) + \tan^2\alpha = 0\]
\[\tan^2\alpha = -\frac{1}{3}(1 - 2e) = \frac{2e - 1}{3}\]

(i)
M1: Applies conservation of momentum along the line of centres.
M1: Applies Newton's law of restitution along the line of centres.
A1: Correctly expresses the parallel component \(v_{Ax}\).
A1: Correctly states that the perpendicular component is \(u\sin\alpha\).

(ii)
M1: Expresses the perpendicularity condition using the dot product (or geometry).
M1: Obtains a trigonometric relation in terms of \(\tan\alpha\).
A1: Successfully completes the algebraic steps to show the given relation.

(i) Let \(\theta\) be the angular displacement of the particle from the highest point. By conservation of energy:
\[\frac{1}{2}mu^2 + mga = \frac{1}{2}mv^2 + mga\cos\theta\]
\[v^2 = u^2 + 2ga(1 - \cos\theta)\]

The equation of motion along the normal to the surface, directed towards the center, is:
\[mg\cos\theta - R = \frac{mv^2}{a}\]

Substituting the expression for \(v^2\):
\[R = mg\cos\theta - \frac{m}{a}\left[u^2 + 2ga(1 - \cos\theta)\right]\]
\[R = mg\cos\theta - \frac{mu^2}{a} - 2mg + 2mg\cos\theta = mg(3\cos\theta - 2) - \frac{mu^2}{a}\]

The particle leaves the surface of the hemisphere when the reaction force \(R = 0\):
\[mg(3\cos\theta - 2) - \frac{mu^2}{a} = 0 \implies 3\cos\theta - 2 = \frac{u^2}{ga}\]
\[\cos\theta = \frac{2}{3} + \frac{u^2}{3ga}\]

(ii) Given \(u = \frac{1}{2}\sqrt{ga}\), we have \(u^2 = \frac{1}{4}ga\).
Substituting this into the equation for \(\cos\theta\):
\[\cos\theta = \frac{2}{3} + \frac{\frac{1}{4}ga}{3ga} = \frac{2}{3} + \frac{1}{12} = \frac{8}{12} + \frac{1}{12} = \frac{9}{12} = \frac{3}{4}\]

The vertical distance \(h\) of the particle below the highest point of the hemisphere is:
\[h = a - a\cos\theta = a\left(1 - \frac{3}{4}\right) = \frac{1}{4}a\]

(i)
M1: Expresses conservation of energy correctly to find \(v^2\).
M1: Applies Newton's second law along the radial direction to find \(R\).
M1: Sets \(R=0\) as the condition for leaving the surface.
A1: Derives the given expression for \(\cos\theta\).

(ii)
M1: Substitutes \(u = \frac{1}{2}\sqrt{ga}\) into the equation for \(\cos\theta\).
A1: Finds \(\cos\theta = \frac{3}{4}\).
A1: Obtains the vertical distance \(h = \frac{1}{4}a\).

First, we determine the range of the transformed random variable \(Y = \frac{8}{X^3}\).
Since \(1 \le X \le 2\):
When \(X = 1\), \(Y = 8\).
When \(X = 2\), \(Y = 1\).
Thus, the range of \(Y\) is \(1 \le Y \le 8\).

Now, we find the cumulative distribution function (CDF) of \(Y\), denoted by \(F_Y(y)\):
\[ F_Y(y) = P(Y \le y) = P\left(\frac{8}{X^3} \le y\right) \]
\[ = P\left(X^3 \ge \frac{8}{y}\right) = P\left(X \ge \frac{2}{y^{1/3}}\right) \]
\[ = \int_{2 y^{-1/3}}^{2} \frac{3}{7}x^2 \, dx \]
\[ = \left[ \frac{1}{7}x^3 \right]_{2 y^{-1/3}}^{2} = \frac{1}{7} \left( 8 - \frac{8}{y} \right) = \frac{8}{7}\left(1 - y^{-1}\right) \]

To find the probability density function (PDF) \(g(y)\), we differentiate the CDF with respect to \(y\):
\[ g(y) = F'_Y(y) = \frac{d}{dy}\left( \frac{8}{7} - \frac{8}{7y} \right) = \frac{8}{7y^2} \]

Thus, the PDF is:
\[ g(y) = \begin{cases} \frac{8}{7y^2} & 1 \le y \le 8 \\ 0 & \text{otherwise} \end{cases} \]

M1: Find the correct range for Y, which is [1, 8].
M1: Write down the probability expression for the CDF of Y in terms of X.
A1: Obtain the correct integration limits and set up the integral for the CDF of Y.
A1: Correctly evaluate the CDF to get F_Y(y) = 8/7 * (1 - 1/y).
M1: Differentiate the CDF with respect to y to find the PDF.
A1: State the final PDF with the correct domain.

(i) Using the property that \(G_X(1) = 1\):
\[ G_X(1) = \frac{k}{4 - 3(1)} = 1 \implies \frac{k}{1} = 1 \implies k = 1 \]

(ii) The PGF of \(Y = 2X + 1\) is given by:
\[ G_Y(t) = E(t^{2X+1}) = t E((t^2)^X) = t G_X(t^2) \]
Substituting the expression for \(G_X(t)\):
\[ G_Y(t) = t \left( \frac{1}{4 - 3t^2} \right) = \frac{t}{4 - 3t^2} \]

(iii) To find \(\text{Var}(X)\), we use derivatives of \(G_X(t) = (4-3t)^{-1}\):
First derivative:
\[ G'_X(t) = -1(4-3t)^{-2}(-3) = 3(4-3t)^{-2} \]
\[ E(X) = G'_X(1) = 3(4-3)^{-2} = 3 \]

Second derivative:
\[ G''_X(t) = -6(4-3t)^{-3}(-3) = 18(4-3t)^{-3} \]
\[ G''_X(1) = 18(4-3)^{-3} = 18 \]

Using the variance formula for PGFs:
\[ \text{Var}(X) = G''_X(1) + G'_X(1) - [G'_X(1)]^2 \]
\[ \text{Var}(X) = 18 + 3 - 3^2 = 21 - 9 = 12 \]

(i) B1: Correctly apply G_X(1) = 1 to find k.
(ii) M1: Apply the property of PGF transformation G_Y(t) = t * G_X(t^2).
A1: Obtain the correct simplified expression for G_Y(t).
(iii) M1: Differentiate G_X(t) once and substitute t = 1 to find E(X).
M1: Differentiate G_X(t) a second time and evaluate at t = 1.
M1: Use the variance formula Var(X) = G''(1) + G'(1) - [G'(1)]^2.
A1: Correctly calculate Var(X) = 12.

Let \(\mu_A\) and \(\mu_B\) represent the population mean scores of Group A and Group B respectively.
Hypotheses:
\(H_0: \mu_A = \mu_B\)
\(H_1: \mu_A > \mu_B\)

We calculate the sample statistics:
Group A:
\(n_A = 7\)
\(\sum x_A = 536\)
\(\bar{x}_A = \frac{536}{7} \approx 76.5714\)
\(\sum x_A^2 = 41260\)
\(s_A^2 = \frac{1}{6}\left(41260 - \frac{536^2}{7}\right) \approx 35.8095\)

Group B:
\(n_B = 6\)
\(\sum x_B = 424\)
\(\bar{x}_B = \frac{424}{6} \approx 70.6667\)
\(\sum x_B^2 = 30086\)
\(s_B^2 = \frac{1}{5}\left(30086 - \frac{424^2}{6}\right) \approx 24.6667\)

Pooled variance estimate \(s_p^2\):
\s_p^2 = \frac{(n_A - 1)s_A^2 + (n_B - 1)s_B^2}{n_A + n_B - 2} = \frac{6(35.8095) + 5(24.6667)}{11} = \frac{214.8571 + 123.3333}{11} \approx 30.7446

Test statistic \(t\):
\[ t = \frac{\bar{x}_A - \bar{x}_B}{\sqrt{s_p^2 \left(\frac{1}{n_A} + \frac{1}{n_B}\right)}} = \frac{76.5714 - 70.6667}{\sqrt{30.7446 \left(\frac{1}{7} + \frac{1}{6}\right)}} \approx \frac{5.9047}{\sqrt{9.5161}} \approx 1.914 \]

Degrees of freedom \(\nu = 7 + 6 - 2 = 11\).
At the 5% level of significance for a one-tailed test with \(df = 11\), the critical value of \(t\) is \(1.796\).
Since the calculated value \(t = 1.914 > 1.796\), we reject \(H_0\).
There is sufficient evidence at the 5% level of significance to conclude that the mean score of Group A is greater than that of Group B.

B1: State both null and alternative hypotheses correctly.
M1: Calculate the sample means and unbiased sample variances for both groups.
A1: Obtain s_A^2 ≈ 35.81 and s_B^2 ≈ 24.67 (or equivalent accuracy).
M1: Compute the pooled variance estimate s_p^2 ≈ 30.74.
M1: Calculate the correct test statistic t ≈ 1.91.
B1: Correctly identify degrees of freedom as 11 and state the critical value 1.796.
A1: Make a consistent comparison (1.914 > 1.796) and draw a correct contextual conclusion.

Hypotheses:
\(H_0\): There is no difference in the reaction times before and after the training.
\(H_1\): There is a difference in the reaction times before and after the training (two-tailed test).

First, we list the differences, find their absolute values, and assign ranks from smallest to largest:

1. Difference: \(0.5\), Absolute value: \(0.5\), Rank: 1, Sign: +
2. Difference: \(-1.1\), Absolute value: \(1.1\), Rank: 2, Sign: -
3. Difference: \(1.2\), Absolute value: \(1.2\), Rank: 3, Sign: +
4. Difference: \(-2.1\), Absolute value: \(2.1\), Rank: 4, Sign: -
5. Difference: \(-3.5\), Absolute value: \(3.5\), Rank: 5, Sign: -
6. Difference: \(-4.2\), Absolute value: \(4.2\), Rank: 6, Sign: -
7. Difference: \(4.8\), Absolute value: \(4.8\), Rank: 7, Sign: +
8. Difference: \(5.3\), Absolute value: \(5.3\), Rank: 8, Sign: +
9. Difference: \(6.2\), Absolute value: \(6.2\), Rank: 9, Sign: +
10. Difference: \(7.1\), Absolute value: \(7.1\), Rank: 10, Sign: +

Sum of positive ranks \(W_+\):
\[ W_+ = 1 + 3 + 7 + 8 + 9 + 10 = 38 \]

Sum of negative ranks \(W_-\):
\[ W_- = 2 + 4 + 5 + 6 = 17 \]

The test statistic \(T\) is the smaller of these two sums, so \(T = 17\).

For a two-tailed test at the 5% significance level with \(n = 10\), the critical value for the Wilcoxon signed-rank test is 8.

Since \(T = 17 > 8\), we fail to reject the null hypothesis. There is insufficient evidence at the 5% level of significance to show a difference in reaction times before and after the training.

B1: State the hypotheses clearly, indicating a two-tailed test.
M1: Calculate the absolute differences and rank them correctly.
A1: Identify the correct ranks and their corresponding signs.
M1: Calculate the sum of positive ranks (38) and sum of negative ranks (17).
A1: Identify the test statistic T = 17.
B1: State the correct critical value of 8 for n = 10 at 5% significance.
A1: Compare the test statistic to the critical value and conclude correctly in context.

Hypotheses:
\(H_0\): Exercise preference is independent of age group.
\(H_1\): Exercise preference is associated with age group.

First, we calculate the row, column, and grand totals:
Row Totals:
- Under 30: \(45 + 35 + 20 = 100\)
- 30 and Over: \(35 + 25 + 40 = 100\)

Column Totals:
- Aerobic: \(45 + 35 = 80\)
- Strength: \(35 + 25 = 60\)
- Flexibility: \(20 + 40 = 60\)
Grand Total \(N = 200\).

Expected Frequencies \(E = \frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}}\):
For both row groups, because the row totals are equal (100):
- Expected Aerobic: \(\frac{100 \times 80}{200} = 40\)
- Expected Strength: \(\frac{100 \times 60}{200} = 30\)
- Expected Flexibility: \(\frac{100 \times 60}{200} = 30\)

Now, calculate the test statistic \(\chi^2 = \sum \frac{(O - E)^2}{E}\):
\[ \chi^2 = \frac{(45 - 40)^2}{40} + \frac{(35 - 30)^2}{30} + \frac{(20 - 30)^2}{30} + \frac{(35 - 40)^2}{40} + \frac{(25 - 30)^2}{30} + \frac{(40 - 30)^2}{30} \]
\[ = \frac{25}{40} + \frac{25}{30} + \frac{100}{30} + \frac{25}{40} + \frac{25}{30} + \frac{100}{30} \]
\[ = 0.625 + 0.8333 + 3.3333 + 0.625 + 0.8333 + 3.3333 = 9.5833 \]

Degrees of freedom \(df = (r-1)(c-1) = (2-1)(3-1) = 2\).
At the 5% significance level with \(df = 2\), the critical value of \(\chi^2\) is \(5.991\).

Since \(\chi^2 = 9.583 > 5.991\), we reject the null hypothesis \(H_0\).
There is sufficient evidence at the 5% significance level to conclude that exercise preference is associated with age group.

B1: State null and alternative hypotheses correctly.
M1: Calculate the expected frequencies for each cell.
A1: Correctly compute expected frequencies (40, 30, 30 for both rows).
M1: Compute the individual term contributions for the chi-squared statistic.
A1: Find the correct calculated value of chi-squared ≈ 9.58 (or 9.583).
B1: State correct degrees of freedom (2) and critical value (5.991).
A1: Compare and state a correct decision to reject H_0 with a consistent conclusion.

First, we calculate the sample mean \(\bar{x}\) and unbiased variance estimate \(s^2\):
\[ \bar{x} = \frac{502 + 498 + 505 + 497 + 501 + 503 + 496 + 502}{8} = \frac{4004}{8} = 500.5 \text{ g} \]

We calculate \(\sum (x_i - \bar{x})^2\) using deviation values:
Deviations \(d_i = x_i - 500\) are \(2, -2, 5, -3, 1, 3, -4, 2\).
\[ \sum d_i = 4 \implies \bar{x} = 500 + \frac{4}{8} = 500.5 \]
\[ \sum d_i^2 = 4 + 4 + 25 + 9 + 1 + 9 + 16 + 4 = 72 \]
\[ s^2 = \frac{1}{n-1} \left( \sum d_i^2 - \frac{(\sum d_i)^2}{n} \right) = \frac{1}{7} \left( 72 - \frac{16}{8} \right) = \frac{1}{7}(70) = 10 \]
So \(s = \sqrt{10} \approx 3.1623\).

Since the population variance is unknown and the sample size is small (\(n = 8\)), we use the \(t\)-distribution with \(n-1 = 7\) degrees of freedom.
For a 95% confidence interval, the critical value \(t_7(0.025) = 2.365\).

The margin of error is:
\[ E = t_{7} \times \frac{s}{\sqrt{n}} = 2.365 \times \frac{\sqrt{10}}{\sqrt{8}} = 2.365 \times \sqrt{1.25} \approx 2.365 \times 1.1180 = 2.644 \]

Therefore, the 95% confidence interval is:
\[ 500.5 \pm 2.644 \]
\[ [497.856, \ 503.144] \]
Rounding to 4 significant figures / 1 decimal place:
\[ [497.9, \ 503.1] \]

M1: Calculate the sample mean (500.5).
M1: Set up calculation for unbiased sample variance s^2.
A1: Correctly calculate s^2 = 10 (or s ≈ 3.162).
B1: Identify the correct degrees of freedom as 7 and use the t-distribution.
B1: State the correct critical t-value of 2.365.
M1: Evaluate the margin of error (2.365 * sqrt(10) / sqrt(8)).
A1: Give the correct final confidence interval [497.9, 503.1] (accept 498 to 503).