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(i) Let \(P(n)\) be the statement that \(7^{2n} + 16n - 1\) is divisible by 64. For \(n = 1\), \(f(1) = 7^2 + 16(1) - 1 = 64\), which is divisible by 64, so \(P(1)\) is true. Assume \(P(k)\) is true for some positive integer \(k\), so \(7^{2k} + 16k - 1 = 64m\) for some integer \(m\). Then for \(n = k+1\): \(f(k+1) = 7^{2(k+1)} + 16(k+1) - 1 = 49 \cdot 7^{2k} + 16k + 15 = 49(64m - 16k + 1) + 16k + 15 = 49(64m) - 784k + 49 + 16k + 15 = 49(64m) - 768k + 64 = 64(49m - 12k + 1)\). Since \(49m - 12k + 1\) is an integer, \(f(k+1)\) is divisible by 64. Thus, if \(P(k)\) is true, then \(P(k+1)\) is true. By mathematical induction, \(P(n)\) is true for all positive integers \(n\). (ii) We have \(7^{2n+2} + 16n + 15 = 49(7^{2n}) + 16n + 15\). From (i), \(7^{2n} \equiv 1 - 16n \pmod{64}\). Therefore, \(49(7^{2n}) + 16n + 15 \equiv 49(1 - 16n) + 16n + 15 = 49 - 784n + 16n + 15 = 64 - 768n = 64(1 - 12n)\). Since \(64(1 - 12n)\) is a multiple of 64, the remainder when \(7^{2n+2} + 16n + 15\) is divided by 64 is 0.

(i) B1: Verify base case n = 1. M1: State inductive hypothesis and write expression for f(k+1). A1: Correctly substitute 7^(2k) = 64m - 16k + 1 or equivalent. M1: Factorise out 64. A1: Complete proof with clear concluding statement. (ii) M1: Express the given term in relation to f(n). A1: Correctly show that the expression is a multiple of 64 and conclude that the remainder is 0.

(i) Let \(u_r = \frac{A}{(2r-1)^2} + \frac{B}{(2r+1)^2}\). Then \(A(2r+1)^2 + B(2r-1)^2 = 4r\). Substituting \(r = 1/2\) gives \(4A = 2 \Rightarrow A = 1/2\). Substituting \(r = -1/2\) gives \(4B = -2 \Rightarrow B = -1/2\). Thus, \(u_r = \frac{1}{2(2r-1)^2} - \frac{1}{2(2r+1)^2}\). Using the method of differences: \(\sum_{r=1}^n u_r = \sum_{r=1}^n \left( \frac{1}{2(2r-1)^2} - \frac{1}{2(2r+1)^2} \right) = \left( \frac{1}{2} - \frac{1}{18} \right) + \left( \frac{1}{18} - \frac{1}{50} \right) + \dots + \left( \frac{1}{2(2n-1)^2} - \frac{1}{2(2n+1)^2} \right) = \frac{1}{2} - \frac{1}{2(2n+1)^2}\). (ii) As \(n \to \infty\), \(\frac{1}{2(2n+1)^2} \to 0\), so \(\sum_{r=1}^{\infty} u_r = 1/2\). (iii) We have \(\sum_{r=N}^{\infty} u_r = \sum_{r=1}^{\infty} u_r - \sum_{r=1}^{N-1} u_r = \frac{1}{2} - \left( \frac{1}{2} - \frac{1}{2(2(N-1)+1)^2} \right) = \frac{1}{2(2N-1)^2}\). We want \(\frac{1}{2(2N-1)^2} < 10^{-4} \Rightarrow (2N-1)^2 > 5000 \Rightarrow 2N-1 > \sqrt{5000} \approx 70.71 \Rightarrow 2N > 71.71 \Rightarrow N > 35.85\). Since \(N\) is an integer, the smallest value of \(N\) is 36.

(i) M1: Set up partial fractions. A1: Obtain correct partial fractions. M1: Use method of differences. A1: Complete showing the given sum. (ii) B1: Identify sum to infinity is 1/2. (iii) M1: Express sum from N to infinity as 1/(2(2N-1)^2). M1: Set up inequality and solve for N. A1: Correctly obtain N = 36.

(i) From the given equation: \(\sum \alpha = 3/2\), \(\sum \alpha\beta = 2\), and \(\alpha\beta\gamma = 5/2\). Thus, \(\sum \alpha^2 = (\sum \alpha)^2 - 2\sum \alpha\beta = (3/2)^2 - 2(2) = -7/4\). \(\sum \alpha^2\beta^2 = (\sum \alpha\beta)^2 - 2\alpha\beta\gamma\sum \alpha = 2^2 - 2(5/2)(3/2) = -7/2\). \(\alpha^2\beta^2\gamma^2 = (\alpha\beta\gamma)^2 = 25/4\). (ii) The cubic equation with roots \(\alpha^2, \beta^2, \gamma^2\) is given by \(y^3 - (\sum \alpha^2)y^2 + (\sum \alpha^2\beta^2)y - \alpha^2\beta^2\gamma^2 = 0\). Substituting the values: \(y^3 + \frac{7}{4}y^2 - \frac{7}{2}y - \frac{25}{4} = 0\). Multiplying by 4 to get integer coefficients: \(4y^3 + 7y^2 - 14y - 25 = 0\). (iii) Since \(\alpha, \beta, \gamma\) are roots of the original equation: \(2\alpha^3 - 3\alpha^2 + 4\alpha - 5 = 0\). Summing over the three roots: \(2\sum \alpha^3 - 3\sum \alpha^2 + 4\sum \alpha - 15 = 0 \Rightarrow 2\sum \alpha^3 - 3(-7/4) + 4(3/2) - 15 = 0 \Rightarrow 2\sum \alpha^3 + 21/4 + 6 - 15 = 0 \Rightarrow 2\sum \alpha^3 = 15/4 \Rightarrow \sum \alpha^3 = 15/8\).

(i) B1: Correctly state sum, product of roots from the original equation. M1: Find \sum \alpha^2. A1: Find \sum \alpha^2\beta^2 and \alpha^2\beta^2\gamma^2. (ii) M1: Formulate the cubic equation using the found values. A1: Correctly obtain integer coefficients. (iii) M1: Use relation 2\sum \alpha^3 - 3\sum \alpha^2 + 4\sum \alpha - 15 = 0. A1: Solve for \sum \alpha^3 to get 15/8.

(i) By algebraic division: \(y = \frac{(x-2)^2 + 1}{x-2} = x - 2 + \frac{1}{x-2}\). As \(x \to \pm\infty\), \(y \to x - 2\), so the oblique asymptote is \(y = x - 2\). The denominator is zero when \(x = 2\), so the vertical asymptote is \(x = 2\). (ii) Differentiating \(y = x - 2 + (x-2)^{-1}\) gives \(\frac{dy}{dx} = 1 - \frac{1}{(x-2)^2}\). Setting \(\frac{dy}{dx} = 0\) gives \((x-2)^2 = 1 \Rightarrow x = 3\) or \(x = 1\). When \(x = 3\), \(y = 2\). When \(x = 1\), \(y = -2\). The turning points are \((3, 2)\) and \((1, -2)\). (iii) Rearranging the equation: \(y(x-2) = x^2 - 4x + 5 \Rightarrow x^2 - (y+4)x + 2y + 5 = 0\). For real \(x\), the discriminant \(D \ge 0 \Rightarrow (y+4)^2 - 4(2y+5) \ge 0 \Rightarrow y^2 - 4 \ge 0 \Rightarrow y \ge 2\) or \(y \le -2\). Thus, there are no points on the curve with \(-2 < y < 2\). (iv) The curve has two branches: one in \(x > 2\) with a local minimum at \((3, 2)\), and one in \(x < 2\) with a local maximum at \((1, -2)\). The y-intercept is at \((0, -2.5)\).

(i) B1: State vertical asymptote x = 2. M1: Attempt algebraic division. A1: State oblique asymptote y = x - 2. (ii) M1: Differentiate and set to 0. A1: Find x values. A1: Find coordinates (3, 2) and (1, -2). (iii) M1: Form quadratic in x and find discriminant. A1: Show y^2 - 4 >= 0. (iv) B1: Sketch the asymptotes. B1: Sketch both branches of the curve with correct shapes and turning points.

(i) \(\det(\mathbf{A}) = 2(2 - 0) - 1(0 - 1) + 3(0 - 1) = 4 + 1 - 3 = 2\). (ii) Matrix of cofactors: \(C_{11} = 2, C_{12} = 1, C_{13} = -1, C_{21} = -2, C_{22} = 1, C_{23} = 1, C_{31} = -2, C_{32} = -2, C_{33} = 2\). Adjoint matrix is \(\mathbf{C}^T = \begin{pmatrix} 2 & -2 & -2 \\ 1 & 1 & -2 \\ -1 & 1 & 2 \end{pmatrix}\). Thus, \(\mathbf{A}^{-1} = \frac{1}{2} \begin{pmatrix} 2 & -2 & -2 \\ 1 & 1 & -2 \\ -1 & 1 & 2 \end{pmatrix} = \begin{pmatrix} 1 & -1 & -1 \\ 0.5 & 0.5 & -1 \\ -0.5 & 0.5 & 1 \end{pmatrix}\). (iii) The system can be written as \(\mathbf{A} \mathbf{x} = \mathbf{b}\), where \(\mathbf{x} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix} 5 \\ 1 \\ 2 \end{pmatrix}\). Then \(\mathbf{x} = \mathbf{A}^{-1} \mathbf{b} = \frac{1}{2} \begin{pmatrix} 2 & -2 & -2 \\ 1 & 1 & -2 \\ -1 & 1 & 2 \end{pmatrix} \begin{pmatrix} 5 \\ 1 \\ 2 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 4 \\ 2 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}\). Thus, \(x = 2, y = 1, z = 0\).

(i) M1: Attempt determinant calculation. A1: Correctly obtain 2. (ii) M1: Calculate cofactors. A1: Transpose cofactors. A1: Divide by determinant to obtain inverse. (iii) M1: Represent system in matrix form. A1: Multiply inverse by constant vector. A1: Correct values for x, y, z.

(i) The curve is a non-dimpled limacon. Max \(r = 3a\) at \(\theta = 0\), min \(r = a\) at \(\theta = \pi\). It is symmetric about the initial line. (ii) Area \(A = \frac{1}{2} \int_0^{2\pi} r^2 d\theta = \frac{1}{2} a^2 \int_0^{2\pi} (2 + \cos\theta)^2 d\theta = \frac{1}{2} a^2 \int_0^{2\pi} (4 + 4\cos\theta + \cos^2\theta) d\theta\). Using \(\cos^2\theta = \frac{1+\cos(2\theta)}{2}\), we have \(A = \frac{1}{2} a^2 \int_0^{2\pi} \left( \frac{9}{2} + 4\cos\theta + \frac{1}{2}\cos(2\theta) \right) d\theta = \frac{1}{2} a^2 \left[ \frac{9}{2}\theta + 4\sin\theta + \frac{1}{4}\sin(2\theta) \right]_0^{2\pi} = \frac{9}{2}\pi a^2\). (iii) Tangent perpendicular to the initial line implies \(\frac{dx}{d\theta} = 0\). Since \(x = r\cos\theta = a(2\cos\theta + \cos^2\theta)\), \(\frac{dx}{d\theta} = a(-2\sin\theta - 2\cos\theta\sin\theta) = -2a\sin\theta(1+\cos\theta) = 0\). This gives \(\sin\theta = 0 \Rightarrow \theta = 0, \pi\) or \(\cos\theta = -1 \Rightarrow \theta = \pi\). For \(\theta = 0\), \(r = 3a \Rightarrow (x, y) = (3a, 0)\). For \(\theta = \pi\), \(r = a \Rightarrow (x, y) = (-a, 0)\).

(i) B1: Correct overall shape. B1: Correct intercepts on initial line. (ii) M1: Use formula 1/2 \int r^2 d\theta. M1: Use double angle identity for \cos^2\theta. A1: Correct integration. A1: Obtain 9/2 \pi a^2. (iii) M1: Express x in terms of \theta and differentiate. A1: Set dx/d\theta = 0 and solve for \theta. A1: Find correct Cartesian coordinates.

(i) Assume the lines intersect. Then \(1 + 2\lambda = 2 + \mu \Rightarrow 2\lambda - \mu = 1\) (1) and \(-2 + \lambda = 1 - \mu \Rightarrow \lambda + \mu = 3\) (2). Adding (1) and (2) gives \(3\lambda = 4 \Rightarrow \lambda = 4/3\). Then \(\mu = 5/3\). Checking the third component: \(z_1 = 3 - \lambda = 5/3\) and \(z_2 = 2\mu = 10/3\). Since \(5/3 \ne 10/3\), the lines do not intersect. The direction vectors are not parallel, so the lines are skew. (ii) The direction of the common perpendicular is \(\mathbf{n} = \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} \times \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 1 \\ -5 \\ -3 \end{pmatrix}\). A vector connecting points on the lines is \(\mathbf{a}_2 - \mathbf{a}_1 = \begin{pmatrix} 2-1 \\ 1-(-2) \\ 0-3 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \\ -3 \end{pmatrix}\). The shortest distance is \(\frac{|(\mathbf{a}_2 - \mathbf{a}_1) \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{|1(1) + 3(-5) + (-3)(-3)|}{\sqrt{1^2 + (-5)^2 + (-3)^2}} = \frac{5}{\sqrt{35}} = \frac{\sqrt{35}}{7}\). (iii) The plane has normal vector \(\mathbf{n} = \begin{pmatrix} 1 \\ -5 \\ -3 \end{pmatrix}\) and passes through \((1, -2, 3)\). Its equation is \(1(x-1) - 5(y+2) - 3(z-3) = 0 \Rightarrow x - 5y - 3z = 2\).

(i) M1: Set up system of equations for intersection. A1: Solve for \lambda and \mu. A1: Show contradiction in z-component and conclude lines are skew. (ii) M1: Find cross product of direction vectors. A1: Correct cross product. M1: Use shortest distance formula. A1: Correctly obtain 5/\sqrt{35} or \sqrt{35}/7. (iii) M1: Use cross product as normal and substitute point on l1. A1: Correct Cartesian equation.

(i) By writing \(1\) in polar form as \(e^{2k\pi i}\), the equation \(z^5 = 1\) becomes \(z^5 = e^{2k\pi i}\).
Using de Moivre's theorem, the five distinct roots are given by \(z = e^{2k\pi i/5}\) for \(k = 0, 1, 2, 3, 4\).

(ii) The sum of the roots of the equation \(z^5 - 1 = 0\) is \(0\), so:
\[1 + e^{2\pi i/5} + e^{4\pi i/5} + e^{6\pi i/5} + e^{8\pi i/5} = 0.\]
Taking the real part of this sum:
\[1 + \cos\frac{2\pi}{5} + \cos\frac{4\pi}{5} + \cos\frac{6\pi}{5} + \cos\frac{8\pi}{5} = 0.\]
Using the symmetry of the cosine function, \(\cos\frac{6\pi}{5} = \cos\frac{4\pi}{5}\) and \(\cos\frac{8\pi}{5} = \cos\frac{2\pi}{5}\).
Therefore, \(1 + 2\cos\frac{2\pi}{5} + 2\cos\frac{4\pi}{5} = 0\), which simplifies to:
\[\cos\frac{2\pi}{5} + \cos\frac{4\pi}{5} = -\frac{1}{2}.\]

(iii) Let \(x = \cos\frac{2\pi}{5}\). Since \(\cos\frac{4\pi}{5} = 2\cos^2\frac{2\pi}{5} - 1 = 2x^2 - 1\), the equation becomes:
\[x + 2x^2 - 1 = -\frac{1}{2} \implies 2x^2 + x - \frac{1}{2} = 0 \implies 4x^2 + 2x - 1 = 0.\]
Solving this using the quadratic formula:
\[x = \frac{-2 \pm \sqrt{4 - 4(4)(-1)}}{8} = \frac{-2 \pm \sqrt{20}}{8} = \frac{-1 \pm \sqrt{5}}{4}.\]
Since \(\frac{2\pi}{5}\) lies in the first quadrant, \(\cos\frac{2\pi}{5} > 0\).
Thus, \(\cos\frac{2\pi}{5} = \frac{\sqrt{5}-1}{4}\).

(i) [2 marks]
- M1: For writing \(1 = e^{2k\pi i}\) and taking fifth roots.
- A1: For listing the correct roots for \(k = 0, 1, 2, 3, 4\).

(ii) [3.375 marks]
- M1: For setting the sum of the roots equal to 0.
- A1: For equating the real parts and using symmetry \(\cos(6\pi/5) = \cos(4\pi/5)\) and \(\cos(8\pi/5) = \cos(2\pi/5)\).
- A1.375: For obtaining the final relation \(\cos\frac{2\pi}{5} + \cos\frac{4\pi}{5} = -\frac{1}{2}\).

(iii) [4 marks]
- M1: For substituting the identity \(\cos(4\pi/5) = 2\cos^2(2\pi/5) - 1\).
- M1: For obtaining the quadratic equation \(4x^2 + 2x - 1 = 0\).
- A1: For solving the quadratic equation to get \(\frac{-1 \pm \sqrt{5}}{4}\).
- A1: For justifying the positive sign and giving the final exact value.

(i) Using the exponential definitions of hyperbolic cosine and hyperbolic sine:
\(\cosh x = \frac{e^x + e^{-x}}{2}\) and \(\sinh x = \frac{e^x - e^{-x}}{2}\).
Then,
\[\cosh^2 x - \sinh^2 x = \left(\frac{e^x + e^{-x}}{2}\right)^2 - \left(\frac{e^x - e^{-x}}{2}\right)^2\]
\[= \frac{e^{2x} + 2 + e^{-2x}}{4} - \frac{e^{2x} - 2 + e^{-2x}}{4} = \frac{4}{4} = 1.\]

(ii) Substitute \(\cosh^2 x = 1 + \sinh^2 x\) into the equation:
\[3(1 + \sinh^2 x) - 8 \sinh x = 3\]
\[3 \sinh^2 x - 8 \sinh x = 0 \implies \sinh x (3 \sinh x - 8) = 0.\]
This gives two cases:
- Case 1: \(\sinh x = 0 \implies x = 0\).
- Case 2: \(\sinh x = \frac{8}{3}\). Using the logarithmic definition of \(\text{arsinh } u = \ln(u + \sqrt{u^2 + 1})\):
\[x = \ln\left(\frac{8}{3} + \sqrt{\frac{64}{9} + 1}\right) = \ln\left(\frac{8 + \sqrt{73}}{3}\right).\]

(iii) To find the stationary points:
\(\frac{dy}{dx} = 6\cosh x \sinh x - 8\cosh x = 2\cosh x (3\sinh x - 4)\).
Setting \(\frac{dy}{dx} = 0\):
Since \(\cosh x \ge 1 > 0\) for all real \(x\), we must have \(3\sinh x - 4 = 0 \implies \sinh x = \frac{4}{3}\).
This gives \(x = \ln\left(\frac{4}{3} + \sqrt{\frac{16}{9} + 1}\right) = \ln 3\).
When \(x = \ln 3\), \(\sinh x = \frac{4}{3}\) and \(\cosh x = \frac{5}{3}\), so:
\[y = 3\left(\frac{25}{9}\right) - 8\left(\frac{4}{3}\right) = -\frac{7}{3}.\]
So the minimum point is \((\ln 3, -\frac{7}{3})\).
The \(y\)-intercept is at \(x = 0 \implies y = 3\).
The curve is a smooth, continuous U-shaped curve passing through \((0, 3)\) with a minimum at \((\ln 3, -\frac{7}{3})\).

(i) [1 mark]
- B1: For correct algebraic proof of the identity using exponential definitions.

(ii) [4.375 marks]
- M1: For substituting \(\cosh^2 x = 1 + \sinh^2 x\).
- A1: For obtaining \(\sinh x(3\sinh x - 8) = 0\).
- A1: For finding \(x = 0\) as one solution.
- A1.375: For using the correct log formula to find the second solution \(x = \ln\left(\frac{8+\sqrt{73}}{3}\right)\).

(iii) [4 marks]
- M1: For differentiating to find \(\frac{dy}{dx}\).
- A1: For finding the stationary point at \(x = \ln 3, y = -\frac{7}{3}\).
- B1: For finding the \(y\)-intercept at \((0, 3)\).
- B1: For a correctly sketched curve with minimum in the first quadrant and correct \(y\)-intercept.

(i) The characteristic equation is \(\det(A - \lambda I) = 0\):
\[\det \begin{pmatrix} 2-\lambda & 2 & -1 \\ 0 & 3-\lambda & 0 \\ -1 & 2 & 2-\lambda \end{pmatrix} = 0.\]
Expanding along the second row:
\[(3-\lambda) [(2-\lambda)^2 - 1] = 0 \implies -(\lambda-1)(\lambda-3)^2 = 0.\]
So the eigenvalues are \(\lambda = 1\) and \(\lambda = 3\) (of multiplicity 2).

(ii) For \(\lambda = 1\):
\[(A - I)\mathbf{v} = 0 \implies \begin{pmatrix} 1 & 2 & -1 \\ 0 & 2 & 0 \\ -1 & 2 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}.\]
From the second row, \(2y = 0 \implies y = 0\).
From the first row, \(x - z = 0 \implies x = z\).
An eigenvector is \(\mathbf{v}_1 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}\).

For \(\lambda = 3\):
\[(A - 3I)\mathbf{v} = 0 \implies \begin{pmatrix} -1 & 2 & -1 \\ 0 & 0 & 0 \\ -1 & 2 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}.\]
This gives \(-x + 2y - z = 0 \implies z = 2y - x\).
We can choose two linearly independent eigenvectors:
If \(y = 1, x = 0 \implies z = 2\), giving \(\mathbf{v}_2 = \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}\).
If \(y = 0, x = 1 \implies z = -1\), giving \(\mathbf{v}_3 = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}\).

(iii) We define:
\[D = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{pmatrix} \quad \text{and} \quad P = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 2 & -1 \end{pmatrix}.\]
Finding \(P^{-1}\):
\(\det(P) = 1(-1) + 1(-1) = -2\).
The cofactor matrix of \(P\) is:
\[C = \begin{pmatrix} -1 & 0 & -1 \\ 2 & -2 & -2 \\ -1 & 0 & 1 \end{pmatrix}.\]
Taking the transpose and dividing by \(\det(P)\):
\[P^{-1} = \frac{1}{2} \begin{pmatrix} 1 & -2 & 1 \\ 0 & 2 & 0 \\ 1 & 2 & -1 \end{pmatrix}.\]

(i) [3.375 marks]
- M1: For setting up \(\text{det}(A - Ι\lambda) = 0\).
- A1: For expanding and finding the characteristic equation.
- A1.375: For stating the correct eigenvalues \(\lambda = 1, 3\).

(ii) [3 marks]
- M1: For setting up equations to find eigenvectors.
- A1: For finding the eigenvector for \(\lambda = 1\).
- A1: For finding two linearly independent eigenvectors for \(\lambda = 3\).

(iii) [3 marks]
- B1: For stating correct \(D\) and \(P\).
- M1: For a valid method to find the inverse matrix \(P^{-1}\).
- A1: For the correct inverse matrix.

(i) For \(y = \arcsin x\), we have:
\[\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} \implies (1-x^2)\left(\frac{dy}{dx}\right)^2 = 1.\]
Differentiating both sides with respect to \(x\):
\[-2x\left(\frac{dy}{dx}\right)^2 + (1-x^2) \cdot 2\left(\frac{dy}{dx}\right)\frac{d^2y}{dx^2} = 0.\]
Dividing through by \(2\frac{dy}{dx}\):
\[(1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} = 0.\]

(ii) Let \(u = y^{(2)}\), \(v = 1-x^2\). Applying Leibniz's theorem to the term \((1-x^2)y^{(2)}\):
\[[(1-x^2)y^{(2)}]^{(n)} = y^{(n+2)}(1-x^2) - 2nx y^{(n+1)} - n(n-1) y^{(n)}.\]
Let \(u = y^{(1)}\), \(v = -x\). Applying Leibniz's theorem to the term \(-x y^{(1)}\):
\[[-x y^{(1)}]^{(n)} = -x y^{(n+1)} - n y^{(n)}.\]
Summing these two parts:
\[(1-x^2)y^{(n+2)} - (2n+1)x y^{(n+1)} - (n(n-1) + n) y^{(n)} = 0.\]
Since \(n(n-1) + n = n^2\), we obtain:
\[(1-x^2)y^{(n+2)} - (2n+1)x y^{(n+1)} - n^2 y^{(n)} = 0.\]

(iii) At \(x = 0\):
\(y(0) = 0\), \(y^{(1)}(0) = 1\), \(y^{(2)}(0) = 0\).
From the recurrence relation at \(x = 0\):
\(y^{(n+2)}(0) = n^2 y^{(n)}(0)\).
Since \(y^{(2)}(0) = 0\), all even-order derivatives at \(x = 0\) are zero.
For \(n = 1\): \(y^{(3)}(0) = 1^2 y^{(1)}(0) = 1\).
For \(n = 3\): \(y^{(5)}(0) = 3^2 y^{(3)}(0) = 9\).
The Maclaurin series up to \(x^5\) is:
\[y(x) \approx y(0) + y^{(1)}(0) x + \frac{y^{(3)}(0)}{3!} x^3 + \frac{y^{(5)}(0)}{5!} x^5 = x + \frac{1}{6} x^3 + \frac{3}{40} x^5.\]

(i) [3.375 marks]
- M1: For finding \(\frac{dy}{dx}\) and arranging terms.
- A1: For correct differentiation to get second derivative.
- A1.375: For showing the identity clearly.

(ii) [3 marks]
- M1: For applying Leibniz's theorem to both parts.
- A1: For correct expansion of both terms.
- A1: For simplifying to get the required relation.

(iii) [3 marks]
- M1: For evaluating \(y(0)\), \(y^{(1)}(0)\), and \(y^{(2)}(0)\).
- A1: For using the recurrence relation to find \(y^{(3)}(0) = 1\) and \(y^{(5)}(0) = 9\).
- A1: For writing down the correct Maclaurin series.

(i) Using integration by parts on \(I_n = \int_0^1 x^n e^{-x} \, dx\):
Let \(u = x^n \implies du = n x^{n-1} \, dx\).
Let \(dv = e^{-x} \, dx \implies v = -e^{-x}\).
Then,
\[I_n = \left[ -x^n e^{-x} \right]_0^1 - \int_0^1 -n x^{n-1} e^{-x} \, dx\]
\[= -e^{-1} + n \int_0^1 x^{n-1} e^{-x} \, dx = n I_{n-1} - e^{-1}.\]

(ii) For \(n = 0\):
\[I_0 = \int_0^1 e^{-x} \, dx = \left[ -e^{-x} \right]_0^1 = 1 - e^{-1}.\]

(iii) Using the recurrence relation:
\[I_1 = 1(I_0) - e^{-1} = 1 - 2e^{-1}.\]
\[I_2 = 2 I_1 - e^{-1} = 2(1 - 2e^{-1}) - e^{-1} = 2 - 5e^{-1}.\]
\[I_3 = 3 I_2 - e^{-1} = 3(2 - 5e^{-1}) - e^{-1} = 6 - 16e^{-1}.\]
\[I_4 = 4 I_3 - e^{-1} = 4(6 - 16e^{-1}) - e^{-1} = 24 - 65e^{-1}.\]

(i) [4 marks]
- M1: For attempting integration by parts.
- A1: For correct choice of \(u\) and \(dv\).
- A1: For finding the boundary term \(-e^{-1}\).
- A1: For obtaining the recurrence relation.

(ii) [1.375 marks]
- B1.375: For the correct integration and evaluation of \(I_0 = 1 - e^{-1}\).

(iii) [4 marks]
- M1: For calculating \(I_1\) using the recurrence relation.
- A1: For calculating \(I_2\).
- A1: For calculating \(I_3\).
- A1: For finding the correct value of \(I_4\).

First find the complementary function (CF) by solving the auxiliary equation:
\[m^2 + 4m + 5 = 0 \implies m = \frac{-4 \pm \sqrt{16-20}}{2} = -2 \pm i.\]
Since the roots are complex, the CF is:
\[y_{CF} = e^{-2x} (A \cos x + B \sin x).\]
Next, find the particular integral (PI). Try \(y = k e^{-x}\):
\[\frac{dy}{dx} = -k e^{-x}, \quad \frac{d^2y}{dx^2} = k e^{-x}.\]
Substitute into the differential equation:
\[k e^{-x} + 4(-k e^{-x}) + 5k e^{-x} = 10e^{-x} \implies 2k e^{-x} = 10e^{-x} \implies k = 5.\]
So the PI is \(y_{PI} = 5e^{-x}\), giving the general solution:
\[y = e^{-2x} (A \cos x + B \sin x) + 5e^{-x}.\]
Apply the initial conditions:
At \(x = 0\), \(y = 3\):
\[3 = A + 5 \implies A = -2.\]
Differentiate the general solution to find \(\frac{dy}{dx}\):
\[\frac{dy}{dx} = -2e^{-2x}(A\cos x + B\sin x) + e^{-2x}(-A\sin x + B\cos x) - 5e^{-x}.\]
At \(x = 0\), \(\frac{dy}{dx} = 0\):
\[0 = -2A + B - 5.\]
Substituting \(A = -2\):
\[0 = -2(-2) + B - 5 \implies B = 1.\]
Therefore, the particular solution is:
\[y = e^{-2x}(\sin x - 2\cos x) + 5e^{-x}.\]

- [3 marks] for CF:
- M1: For setting up and solving the auxiliary equation.
- A1: For finding correct roots \(m = -2 \pm i\).
- A1: For writing the correct CF.
- [3 marks] for PI:
- M1: For trying \(y = ke^{-x}\) and finding its derivatives.
- A1: For substituting and finding \(k = 5\).
- A1: For stating the general solution.
- [3.375 marks] for Particular Solution:
- M1: For using \(y(0) = 3\) to find \(A = -2\).
- M1: For differentiating \(y\) using the product rule.
- A1: For using \(y'(0) = 0\) to find \(B = 1\).
- A0.375: For stating the final particular solution.

(i) By de Moivre's theorem:
\[\cos 4\theta + i \sin 4\theta = (\cos\theta + i\sin\theta)^4.\]
Expanding the right-hand side using the Binomial Theorem:
\[(\cos\theta + i\sin\theta)^4 = \cos^4\theta + 4i\cos^3\theta\sin\theta - 6\cos^2\theta\sin^2\theta - 4i\cos\theta\sin^3\theta + \sin^4\theta.\]
Equating the imaginary parts:
\[\sin 4\theta = 4\cos^3\theta\sin\theta - 4\cos\theta\sin^3\theta.\]
Factoring out \(\cos\theta\) and substituting \(\cos^2\theta = 1 - \sin^2\theta\):
\[\sin 4\theta = \cos\theta (4(1-\sin^2\theta)\sin\theta - 4\sin^3\theta)\]
\[= \cos\theta (4\sin\theta - 4\sin^3\theta - 4\sin^3\theta) = \cos\theta (4\sin\theta - 8\sin^3\theta).\]

(ii) We are given the equation:
\[\sin(4\theta) = 2\sin(2\theta).\]
We know that \(\sin(2\theta) = 2\sin\theta\cos\theta\), so \(2\sin(2\theta) = 4\sin\theta\cos\theta\).
Using the identity from part (i):
\[\cos\theta(4\sin\theta - 8\sin^3\theta) = 4\sin\theta\cos\theta.\]
Rearranging and factoring:
\[\cos\theta(4\sin\theta - 8\sin^3\theta - 4\sin\theta) = 0 \implies -8\cos\theta\sin^3\theta = 0.\]
This implies that either \(\cos\theta = 0\) or \(\sin\theta = 0\).
In the interval \(0 \le \theta \le \pi\):
- \(\sin\theta = 0 \implies \theta = 0, \pi\).
- \(\cos\theta = 0 \implies \theta = \frac{\pi}{2}\).
Thus, the solutions are \(\theta = 0, \frac{\pi}{2}, \pi\).

(i) [4.375 marks]
- M1: For applying de Moivre's theorem.
- A1: For correct binomial expansion.
- A1: For equating imaginary parts.
- A1.375: For replacing \(\cos^2\theta\) with \(1-\sin^2\theta\) and simplifying.

(ii) [5 marks]
- M1: For substituting \(\sin 2\theta = 2\sin\theta\cos\theta\).
- M1: For substituting the expression for \(\sin 4\theta\) from part (i).
- A1: For simplifying to \(\cos\theta\sin^3\theta = 0\).
- A1: For identifying \(\sin\theta = 0\) gives \(\theta = 0, \pi\).
- A1: For identifying \(\cos\theta = 0\) gives \(\theta = \frac{\pi}{2}\).

(i) For the curve \(y = \ln(\cos x)\):
\[\frac{dy}{dx} = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x.\]
The arc length \(s\) is given by:
\[s = \int_{0}^{\alpha} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx = \int_{0}^{\alpha} \sqrt{1 + (-\tan x)^2} \, dx = \int_{0}^{\alpha} \sqrt{\sec^2 x} \, dx.\]
Since \(0 \le x \le \alpha < \frac{\pi}{2}\), \(\sec x > 0\), so:
\[s = \int_{0}^{\alpha} \sec x \, dx = [\ln(\sec x + \tan x)]_0^\alpha\]
\[= \ln(\sec \alpha + \tan \alpha) - \ln(\sec 0 + \tan 0) = \ln(\sec \alpha + \tan \alpha).\]

(ii) For the curve \(y = \cosh x\):
\[\frac{dy}{dx} = \sinh x.\]
The surface area \(S\) is given by:
\[S = 2\pi \int_{0}^{\ln 2} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx = 2\pi \int_{0}^{\ln 2} \cosh x \sqrt{1 + \sinh^2 x} \, dx = 2\pi \int_{0}^{\ln 2} \cosh^2 x \, dx.\]
Using the identity \(\cosh^2 x = \frac{\cosh(2x) + 1}{2\text{}}\):
\[S = 2\pi \int_{0}^{\ln 2} \frac{\cosh(2x) + 1}{2} \, dx = \pi \left[ \frac{\sinh(2x)}{2} + x \right]_{0}^{\ln 2}\]
\[= \pi \left( \frac{\sinh(2\ln 2)}{2} + \ln 2 \right) - 0.\]
Since \(\sinh(2\ln 2) = \sinh(\ln 4) = \frac{4 - 1/4}{2} = \frac{15}{8}\), we have:
\[S = \pi \left( \frac{15}{16} + \ln 2 \right).\]

(i) [4.375 marks]
- M1: For differentiating \(y = \ln(\cos x)\) to get \(\frac{dy}{dx} = -\tan x\).
- A1: For simplifying \(1 + (\frac{dy}{dx})^2 = \sec^2 x\).
- M1: For integrating \(\sec x\) to get \(\ln(\sec x + \tan x)\).
- A1.375: For substituting limits correctly to obtain \(\ln(\sec\alpha + \tan\alpha)\).

(ii) [5 marks]
- M1: For setting up the surface area integral.
- A1: For simplifying the integrand to \(\cosh^2 x\).
- M1: For using the identity \(\cosh^2 x = \frac{\cosh(2x)+1}{2}\) to integrate.
- A1: For correct evaluation of the limits.
- A1: For obtaining the final answer \(\pi\left(\frac{15}{16} + \ln 2\right)\).

Using the equation of the trajectory in terms of the horizontal range \(R = 12\text{ m}\): \(y = x \tan\theta \left(1 - \frac{x}{R}\right)\). Substituting \(x = 6\) and \(y = 3\) gives: \(3 = 6 \tan\theta \left(1 - \frac{6}{12}\right) = 3 \tan\theta \implies \tan\theta = 1\). Since \(\theta\) is acute, \(\theta = 45^\circ\). To find \(u\), we use the range formula: \(R = \frac{u^2 \sin 2\theta}{g}\). Substituting \(R = 12\), \(\theta = 45^\circ\), and \(g = 10\) gives: \(12 = \frac{u^2 \sin 90^\circ}{10} \implies u^2 = 120 \implies u = \sqrt{120} = 2\sqrt{30}\text{ m s}^{-1}\}.

M1: For using the trajectory equation \(y = x \tan\theta(1 - x/R)\) or equivalent.
A1: For substituting the values to show \(\tan\theta = 1\).
A1: For concluding \(\theta = 45^\circ\).
M1: For using the range formula with \(g = 10\).
A1: For obtaining \(u^2 = 120\).
A1: For the correct exact value of \(u = 2\sqrt{30}\).

First, find the equilibrium extension \(x_0\): \(T_0 = mg \implies \frac{\lambda x_0}{l} = mg \implies \frac{\lambda x_0}{0.8} = 0.4 \times 10 \implies \lambda x_0 = 3.2\). The particle is pulled down \(a = 0.3\text{ m}\) from the equilibrium position. Since \(a < x_0\), the motion is SHM about the equilibrium position with amplitude \(a = 0.3\text{ m}\). The maximum speed of SHM is given by: \(v_{\max} = \omega a\). Given \(v_{\max} = 1.5\text{ m s}^{-1}\) and \(a = 0.3\text{ m}\): \(1.5 = \omega \times 0.3 \implies \omega = 5\text{ rad s}^{-1}\). The angular frequency is \(\omega^2 = \frac{\lambda}{m l}\). Substituting the values: \(5^2 = \frac{\lambda}{0.4 \times 0.8} \implies 25 = \frac{\lambda}{0.32} \implies \lambda = 8\text{ N}\).

M1: For writing the equilibrium equation.
A1: For obtaining \(\lambda x_0 = 3.2\).
M1: For using \(v_{\max} = \omega a\) to find \(\omega\).
A1: For finding \(\omega = 5\).
M1: For using \(\omega^2 = \frac{\lambda}{ml}\).
A1: For finding \(\lambda = 8\).
B1: For verifying that \(x_0 > a\) so that the string remains taut.

The height of \(P\) above the lowest point \(A\) is \(h = r(1+\cos\theta)\). By conservation of mechanical energy: \(\frac{1}{2} m u^2 = \frac{1}{2} m v^2 + mgh \implies u^2 = v^2 + 2gr(1+\cos\theta)\). The radial equation of motion is \(R + mg \cos\theta = \frac{m v^2}{r}\). Contact is lost when \(R = 0 \implies v^2 = gr \cos\theta\). Substituting \(v^2\) into the energy equation: \(u^2 = gr\cos\theta + 2gr(1+\cos\theta) = gr(2 + 3\cos\theta)\). Substituting \(g = 10\), \(r = 0.8\), and \(\cos\theta = \frac{1}{3}\) gives: \(u^2 = 10 \times 0.8 \left(2 + 3 \times \frac{1}{3}\right) = 24 \implies u = \sqrt{24} = 2\sqrt{6}\text{ m s}^{-1}\).

M1: For expressing the height \(h = r(1+\cos\theta)\).
M1: For writing the energy conservation equation.
A1: For obtaining \(u^2 = v^2 + 2gr(1+\cos\theta)\).
M1: For writing the radial equation of motion and setting \(R = 0\).
A1: For obtaining \(v^2 = gr\cos\theta\).
M1: For substituting \(v^2\) and evaluating \(u^2 = 24\).
A1: For finding \(u = 2\sqrt{6}\).

The equation of motion is: \(m v \frac{dv}{dx} = 10 - 2v^2 \implies 0.5 v \frac{dv}{dx} = 10 - 2v^2 \implies v \frac{dv}{dx} = 4(5 - v^2)\). Separating variables and integrating from \(x=0, v=0\) to \(x=x, v=2\): \(\int_0^2 \frac{v}{5 - v^2} dv = \int_0^x 4 dx\). This evaluates to: \(\left[ -\frac{1}{2} \ln(5-v^2) \right]_0^2 = 4x \implies \frac{1}{2} \ln 5 = 4x \implies x = \frac{1}{8} \ln 5\).

M1: For using Newton's second law with acceleration as \(v \frac{dv}{dx}\).
A1: For obtaining the correct differential equation.
M1: For separating variables and integrating.
A1: For integrating to get \(-\frac{1}{2}\ln(5-v^2)\) or equivalent.
M1: For substituting the limits correctly.
A1: For obtaining \(\frac{1}{2} \ln 5 = 4x\).
A1: For concluding \(x = \frac{1}{8} \ln 5\) (and hence \(k = 8\)).

Let the line of centers be the x-axis. Before impact, \(\mathbf{u}_A = (u \cos 30^\circ, u \sin 30^\circ)\) and \(\mathbf{u}_B = (0, 0)\). After impact, the components of velocity perpendicular to the line of centers are unchanged: \(v_{Ay} = \frac{u}{2}\). For the components along the line of centers, conservation of momentum gives: \(m \left(\frac{u\sqrt{3}}{2}\right) = m v_{Ax} + 2m v_{Bx} \implies v_{Ax} + 2 v_{Bx} =
\frac{u\sqrt{3}}{2}\). Newton's law of restitution gives: \(v_{Bx} - v_{Ax} = e \left(\frac{u\sqrt{3}}{2}\right) = \frac{u\sqrt{3}}{4}\). Solving these simultaneously yields \(v_{Ax} = 0\). Therefore, after the collision, the velocity of \(A\) is entirely perpendicular to the line of centers. The final direction of motion is at \(90^\circ\) to the line of centers. The angle of deflection is \(90^\circ - 30^\circ = 60^\circ\).

B1: For resolving initial velocities along and perpendicular to the line of centers.
M1: For applying conservation of momentum along the line of centers.
A1: For the equation \(v_{Ax} + 2v_{Bx} = \frac{u\sqrt{3}}{2}\).
M1: For applying Newton's law of restitution.
A1: For the equation \(v_{Bx} - v_{Ax} = \frac{u\sqrt{3}}{4}\).
A1: For finding \(v_{Ax} = 0\).
A1: For calculating the deflection angle of \(60^\circ\).

The angle of the string with the horizontal rod \(AB\) is \(90^\circ - 30^\circ = 60^\circ\). Taking moments about the hinge \(A\): \(T \sin 60^\circ \times 2 = W \times 1 \implies 2 T \left(\frac{\sqrt{3}}{2}\right) = W \implies T = \frac{W}{\sqrt{3}}\). Let \(H\) and \(V\) be the horizontal and vertical forces exerted by the hinge at \(A\). Resolving horizontally: \(H - T \cos 60^\circ = 0 \implies H = \frac{W}{2\sqrt{3}}\). Resolving vertically: \(V + T \sin 60^\circ - W = 0 \implies V = W - \frac{W}{2} = \frac{W}{2}\). The magnitude of the force at the hinge \(A\) is: \(R = \sqrt{H^2 + V^2} = \sqrt{\frac{W^2}{12} + \frac{W^2}{4}} = \sqrt{\frac{W^2}{3}} = \frac{W}{\sqrt{3}}\).

M1: For identifying that the angle of the string with the horizontal is \(60^\circ\).
M1: For taking moments about \(A\).
A1: For finding \(T = \frac{W}{\sqrt{3}}\).
M1: For resolving forces horizontally to find \(H\).
M1: For resolving forces vertically to find \(V\).
A1: For obtaining \(H = \frac{W}{2\sqrt{3}}\) and \(V = \frac{W}{2}\).
A1: For finding the resultant force magnitude \(R = \frac{W}{\sqrt{3}}\) (or \(\frac{W\sqrt{3}}{3}\)).

The radius of the horizontal circle is \(r = l \sin 45^\circ = 0.4\sqrt{2} \times \frac{1}{\sqrt{2}} = 0.4\text{ m}\). Resolving vertically: \(T \cos 45^\circ + R \sin 45^\circ = mg\). Resolving horizontally towards the center: \(T \sin 45^\circ - R \cos 45^\circ = \frac{m v^2}{r}\). When the particle is on the point of losing contact, the normal reaction is \(R = 0\). Therefore, \(T \cos 45^\circ = mg\) and \(T \sin 45^\circ = \frac{m v^2}{r}\). Since \(\sin 45^\circ = \cos 45^\circ\), we equate these: \(mg = \frac{m v^2}{r} \implies v^2 = gr\). Substituting \(g = 10\) and \(r = 0.4\) yields: \(v^2 = 10 \times 0.4 = 4 \implies v = 2\text{ m s}^{-1}\).

M1: For finding the circle radius \(r = 0.4\text{ m}\).
M1: For writing the vertical force resolution equation.
A1: For the correct vertical equation.
M1: For writing the horizontal equation of motion.
A1: For the correct horizontal equation.
M1: For setting \(R = 0\) and solving for \(v^2 = gr\).
A1: For finding \(v = 2\).

(i) To find the constant \(k\), we integrate the probability density function over its range and set the integral equal to 1:
\[ \int_{0}^{4} kx^2(4-x) \, dx = 1 \]
\[ k \int_{0}^{4} (4x^2 - x^3) \, dx = 1 \]
\[ k \left[ \frac{4}{3}x^3 - \frac{1}{4}x^4 \right]_{0}^{4} = 1 \]
\[ k \left( \frac{256}{3} - 64 \right) = 1 \]
\[ k \left( \frac{64}{3} \right) = 1 \implies k = \frac{3}{64} \]

(ii) For \(0 \le x \le 4\), the cumulative distribution function \(\text{F}(x)\) is:
\[ \text{F}(x) = \int_{0}^{x} \frac{3}{64}(4t^2 - t^3) \, dt \]
\[ \text{F}(x) = \frac{3}{64} \left[ \frac{4}{3}t^3 - \frac{1}{4}t^4 \right]_{0}^{x} = \frac{1}{16}x^3 - \frac{3}{256}x^4 \]

(iii) The mean \(\text{E}(X)\) is given by:
\[ \text{E}(X) = \int_{0}^{4} x f(x) \, dx = \frac{3}{64} \int_{0}^{4} (4x^3 - x^4) \, dx \]
\[ \text{E}(X) = \frac{3}{64} \left[ x^4 - \frac{1}{5}x^5 \right]_{0}^{4} = \frac{3}{64} \left( 256 - \frac{1024}{5} \right) = \frac{3}{64} \times \frac{256}{5} = 2.4 \]

Now, we calculate the probability \(\text{P}(X > 2.4)\):
\[ \text{P}(X > 2.4) = 1 - \text{F}(2.4) = 1 - \left( \frac{1}{16}(2.4)^3 - \frac{3}{256}(2.4)^4 \right) \]
\[ \text{F}(2.4) = 0.864 - 0.3888 = 0.4752 \]
\[ \text{P}(X > 2.4) = 1 - 0.4752 = 0.5248 \approx 0.525 \text{ (to 3 s.f.)} \]

(i) M1: For attempting to integrate \(f(x)\) and equate to 1.
A1: Correct integration and showing \(k = \frac{3}{64}\) with clear steps.

(ii) M1: For integrating from 0 to \(x\).
A1: Correct CDF expression \(\frac{1}{16}x^3 - \frac{3}{256}x^4\).

(iii) M1: For correct method to find the mean \(\text{E}(X)\).
A1: Correct mean of 2.4.
M1: For evaluating \(1 - \text{F}(2.4)\).
A1: Correct probability of 0.525 (accept 0.5248).

(i) Let \(d = \text{Speed after} - \text{Speed before}\).
\(H_0\): The median difference in typing speeds is zero (the training program does not increase speed).
\(H_1\): The median difference in typing speeds is greater than zero (the training program increases speed).

(ii) Calculate the difference \(d\) for each assistant:
- A: \(49 - 45 = +4\)
- B: \(51 - 52 = -1\)
- C: \(46 - 38 = +8\)
- D: \(55 - 48 = +7\)
- E: \(58 - 60 = -2\)
- F: \(51 - 41 = +10\)
- G: \(61 - 55 = +6\)
- H: \(55 - 50 = +5\)

Rank the absolute differences \(|d|\):
- \(|d| = 1\) (Assistant B): Rank 1, sign \(-\)
- \(|d| = 2\) (Assistant E): Rank 2, sign \(-\)
- \(|d| = 4\) (Assistant A): Rank 3, sign \(+\)
- \(|d| = 5\) (Assistant H): Rank 4, sign \(+\)
- \(|d| = 6\) (Assistant G): Rank 5, sign \(+\)
- \(|d| = 7\) (Assistant D): Rank 6, sign \(+\)
- \(|d| = 8\) (Assistant C): Rank 7, sign \(+\)
- \(|d| = 10\) (Assistant F): Rank 8, sign \(+\)

Sum of positive ranks:
\[ W_+ = 3 + 4 + 5 + 6 + 7 + 8 = 33 \]

Sum of negative ranks:
\[ W_- = 1 + 2 = 3 \]

The test statistic \(T\) is the minimum of \(W_+\) and \(W_-\), so \(T = 3\).

(iii) For a one-tailed test with \(n = 8\) at the 5% significance level, the critical value is 5.
Since \(T = 3 \le 5\), we reject the null hypothesis \(H_0\).
There is sufficient evidence at the 5% significance level to suggest that the training program increases typing speed.

(i) B1: For stating both hypotheses correctly, specifying the median difference.

(ii) M1: For calculating the differences for all 8 assistants.
M1: For ranking the absolute differences correctly.
A1: For correct positive and negative rank sums (33 and 3).
A1: For identifying the test statistic \(T = 3\).

(iii) B1: For stating the correct critical value of 5.
M1: For comparing the test statistic with the critical value.
A1: For drawing the correct conclusion in context.

(i) \(H_0\): There is no association between age group and preferred music genre.
\(H_1\): There is an association between age group and preferred music genre.

(ii) Row totals: Under 30 = 100, 30 and Over = 100.
Column totals: Classical = 40, Rock = 80, Pop = 80.
Grand total = 200.
Expected frequencies are calculated as \(E = \frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}}\):
- Under 30, Classical: \(\frac{100 \times 40}{200} = 20\)
- Under 30, Rock: \(\frac{100 \times 80}{200} = 40\)
- Under 30, Pop: \(\frac{100 \times 80}{200} = 40\)
- 30 and Over, Classical: \(\frac{100 \times 40}{200} = 20\)
- 30 and Over, Rock: \(\frac{100 \times 80}{200} = 40\)
- 30 and Over, Pop: \(\frac{100 \times 80}{200} = 40\)

Expected frequencies matrix:
\[ \begin{array}{l|ccc} \text{Age Group} & \text{Classical} & \text{Rock} & \text{Pop} \\\\ \hline \text{Under 30} & 20 & 40 & 40 \\\\ \text{30 and Over} & 20 & 40 & 40 \end{array} \]

(iii) The test statistic \(\chi^2 = \sum \frac{(O - E)^2}{E}\):
\[ \chi^2 = \frac{(15-20)^2}{20} + \frac{(45-40)^2}{40} + \frac{(40-40)^2}{40} + \frac{(25-20)^2}{20} + \frac{(35-40)^2}{40} + \frac{(40-40)^2}{40} \]
\[ \chi^2 = \frac{25}{20} + \frac{25}{40} + 0 + \frac{25}{20} + \frac{25}{40} + 0 \]
\[ \chi^2 = 1.25 + 0.625 + 1.25 + 0.625 = 3.75 \]

(iv) Degrees of freedom \(df = (r-1)(c-1) = (2-1)(3-1) = 2\).
At the 5% significance level, the critical value of \(\chi^2\) with 2 degrees of freedom is 5.991.
Since \(3.75 < 5.991\), we fail to reject \(H_0\).
There is insufficient evidence at the 5% level to conclude that there is an association between age group and preferred music genre.

(i) B1: For correctly stating both hypotheses.

(ii) M1: For attempting to calculate expected frequencies.
A1: For all correct expected frequencies.

(iii) M1: For attempting to sum \(\frac{(O-E)^2}{E}\).
A1: For correct contributions.
A1: For \(\chi^2 = 3.75\).

(iv) B1: For correct degrees of freedom (2) and identifying the critical value 5.991.
A1: For making the correct non-rejection decision and conclusion in context.

(i) Since \(G_X(1) = 1\):
\[ k(1 + 2(1) + 3(1)^2)^2 = 1 \]
\[ k(6)^2 = 1 \implies 36k = 1 \implies k = \frac{1}{36} \]

(ii) Expand the expression \((1 + 2t + 3t^2)^2\):
\[ (1 + 2t + 3t^2)(1 + 2t + 3t^2) = 1(1 + 2t + 3t^2) + 2t(1 + 2t + 3t^2) + 3t^2(1 + 2t + 3t^2) \]
\[ = (1 + 2t + 3t^2) + (2t + 4t^2 + 6t^3) + (3t^2 + 6t^3 + 9t^4) \]
\[ = 1 + 4t + 10t^2 + 12t^3 + 9t^4 \]

Dividing by 36, we get:
\[ G_X(t) = \frac{1}{36} + \frac{1}{9}t + \frac{5}{18}t^2 + \frac{1}{3}t^3 + \frac{1}{4}t^4 \]

The probability distribution of \(X\) is:
- \(\text{P}(X=0) = \frac{1}{36}\)
- \(\text{P}(X=1) = \frac{1}{9}\)
- \(\text{P}(X=2) = \frac{5}{18}\)
- \(\text{P}(X=3) = \frac{1}{3}\)
- \(\text{P}(X=4) = \frac{1}{4}\)
For all other values of \(r\), \(\text{P}(X=r) = 0\).

(iii) First derivative of \(G_X(t)\):
\[ G_X'(t) = \frac{2}{36}(1 + 2t + 3t^2)(2 + 6t) = \frac{1}{18}(1 + 2t + 3t^2)(2 + 6t) \]
\[ \text{E}(X) = G_X'(1) = \frac{1}{18}(1+2+3)(2+6) = \frac{1}{18}(6)(8) = \frac{8}{3} \approx 2.67 \]

Second derivative of \(G_X(t)\):
Using the product rule on \(G_X'(t) = \frac{1}{18}(2+6t)(1+2t+3t^2)\):
\[ G_X''(t) = \frac{1}{18} \left[ 6(1 + 2t + 3t^2) + (2 + 6t)(2 + 6t) \right] \]
\[ G_X''(1) = \frac{1}{18} \left[ 6(6) + (8)^2 \right] = \frac{1}{18}(36 + 64) = \frac{100}{18} = \frac{50}{9} \]

Using the formula for variance:
\[ \text{Var}(X) = G_X''(1) + G_X'(1) - (G_X'(1))^2 \]
\[ \text{Var}(X) = \frac{50}{9} + \frac{8}{3} - \left(\frac{8}{3}\right)^2 = \frac{50}{9} + \frac{24}{9} - \frac{64}{9} = \frac{10}{9} \approx 1.11 \]

(i) B1: Correctly showing \(k = \frac{1}{36}\) using the condition \(G_X(1) = 1\).

(ii) M1: For attempting to expand the quadratic trinomial squared.
A1: Correct coefficients of the expanded polynomial.
A1: Fully stated probability distribution of \(X\).

(iii) M1: For finding the first derivative \(G_X'(t)\) and evaluating at \(t=1\).
A1: Correct \(\text{E}(X) = \frac{8}{3}\).
M1: For finding the second derivative \(G_X''(t)\) and evaluating at \(t=1\) to find \(\text{Var}(X)\).
A1: Correct \(\text{Var}(X) = \frac{10}{9}\).

(i) For Brand A (\(n_1 = 5\)):
\[ \bar{x}_A = \frac{12.4 + 12.1 + 12.5 + 12.8 + 12.2}{5} = \frac{62.0}{5} = 12.4 \]
\[ \sum (x_A - \bar{x}_A)^2 = 0^2 + (-0.3)^2 + 0.1^2 + 0.4^2 + (-0.2)^2 = 0 + 0.09 + 0.01 + 0.16 + 0.04 = 0.3 \]
\[ s_A^2 = \frac{0.3}{4} = 0.075 \]

For Brand B (\(n_2 = 6\)):
\[ \bar{x}_B = \frac{11.8 + 11.5 + 12.0 + 11.6 + 11.9 + 11.4}{6} = \frac{70.2}{6} = 11.7 \]
\[ \sum (x_B - \bar{x}_B)^2 = 0.1^2 + (-0.2)^2 + 0.3^2 + (-0.1)^2 + 0.2^2 + (-0.3)^2 = 0.01 + 0.04 + 0.09 + 0.01 + 0.04 + 0.09 = 0.28 \]
\[ s_B^2 = \frac{0.28}{5} = 0.056 \]

(ii) The pooled estimate of the population variance \(s_p^2\) is:
\[ s_p^2 = \frac{(n_1 - 1)s_A^2 + (n_2 - 1)s_B^2}{n_1 + n_2 - 2} = \frac{4(0.075) + 5(0.056)}{9} = \frac{0.3 + 0.28}{9} = \frac{0.58}{9} \approx 0.06444 \]

(iii) Degrees of freedom \(df = 5 + 6 - 2 = 9\).
For a 95% confidence interval, the critical value \(t\) is \(t_{0.025, 9} = 2.262\).
The standard error is:
\[ \text{SE} = \sqrt{s_p^2 \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} = \sqrt{\frac{0.58}{9} \left(\frac{1}{5} + \frac{1}{6}\right)} = \sqrt{\frac{0.58}{9} \times \frac{11}{30}} = \sqrt{\frac{6.38}{270}} \approx 0.15372 \]

The margin of error is:
\[ \text{ME} = t \times \text{SE} = 2.262 \times 0.15372 \approx 0.34774 \]

The difference in means is \(\bar{x}_A - \bar{x}_B = 12.4 - 11.7 = 0.7\).

The 95% confidence interval is:
\[ 0.7 \pm 0.3477 \implies [0.352, 1.05] \text{ (to 3 s.f.)} \]

(i) B1: Correct means \(\bar{x}_A = 12.4\) and \(\bar{x}_B = 11.7\).
M1: For attempting to calculate the unbiased variances.
A1: Correct unbiased variances \(s_A^2 = 0.075\) and \(s_B^2 = 0.056\).

(ii) B1: Correct pooled variance \(s_p^2 = \frac{0.58}{9} \approx 0.0644\).

(iii) B1: Correct critical value \(t = 2.262\).
M1: Correct method to calculate the standard error.
M1: For attempting the confidence interval formula \((\bar{x}_A - \bar{x}_B) \pm t \times \text{SE}\).
A1: Correct confidence interval \([0.352, 1.05]\).

(i) The cumulative distribution function of \(Y\), \(\text{G}(y)\) for \(y \ge 1\), is defined as:
\[ \text{G}(y) = \text{P}(Y \le y) = \text{P}(e^{X/3} \le y) = \text{P}\left(X \le 3\ln y\right) \]

Integrating the probability density function of \(X\) from 0 to \(3\ln y\):
\[ \text{G}(y) = \int_{0}^{3\ln y} \frac{1}{2}e^{-x/2} \, dx = \left[ -e^{-x/2} \right]_{0}^{3\ln y} \]
\[ \text{G}(y) = -e^{-\frac{3}{2}\ln y} - (-1) = 1 - (e^{\ln y})^{-3/2} = 1 - y^{-3/2} \]
For \(y < 1\), \(\text{G}(y) = 0\).

(ii) The probability density function \(g(y)\) is the derivative of \(\text{G}(y)\) with respect to \(y\):
\[ g(y) = \frac{d}{dy}(1 - y^{-3/2}) = \frac{3}{2}y^{-5/2} \]
Thus, the probability density function \(g(y)\) is:
\[ g(y) = \begin{cases} \frac{3}{2}y^{-5/2} & y \ge 1 \\ 0 & \text{otherwise} \end{cases} \]

(iii) The probability \(\text{P}(2 < Y < 4)\) is:
\[ \text{P}(2 < Y < 4) = \text{G}(4) - \text{G}(2) \]
\[ \text{P}(2 < Y < 4) = (1 - 4^{-3/2}) - (1 - 2^{-3/2}) = 2^{-3/2} - 4^{-3/2} \]
\[ = \frac{1}{2\sqrt{2}} - \frac{1}{8} \approx 0.35355 - 0.125 = 0.22855 \approx 0.229 \text{ (to 3 s.f.)} \]

(i) M1: For setting up \(\text{G}(y) = \text{P}(X \le 3\ln y)\).
M1: For attempting to integrate \(f(x)\).
A1: Correct integration.
A1: Correct CDF expression \(1 - y^{-3/2}\).

(ii) M1: For differentiating the CDF to find the PDF.
A1: Correct PDF and specifying the range \(y \ge 1\).

(iii) M1: For evaluating \(\text{G}(4) - \text{G}(2)\).
A1: Correct probability of 0.229 (or 0.22855).