2023 Cambridge IAL Mathematics - Further (9231) 模擬試題連答案詳解

(i) Let \(\frac{4r + 2}{r(r+1)(r+2)} = \frac{A}{r} + \frac{B}{r+1} + \frac{C}{r+2}\).
Multiplying both sides by \(r(r+1)(r+2)\) gives:
\(4r + 2 = A(r+1)(r+2) + Br(r+2) + Cr(r+1)\).
- Set \(r = 0\): \(2 = A(1)(2) \implies A = 1\).
- Set \(r = -1\): \(-2 = B(-1)(1) \implies B = 2\).
- Set \(r = -2\): \(-6 = C(-2)(-1) \implies C = -3\).
Thus, the partial fraction decomposition is:
\(\frac{1}{r} + \frac{2}{r+1} - \frac{3}{r+2}\).

(ii) We can rewrite the general term as:
\(\left(\frac{1}{r} - \frac{1}{r+1}\right) + 3\left(\frac{1}{r+1} - \frac{1}{r+2}\right)\).
Letting \(r\) range from \(1\) to \(n\):
\(S_n = \sum_{r=1}^{n} \left[ \left(\frac{1}{r} - \frac{1}{r+1}\right) + 3\left(\frac{1}{r+1} - \frac{1}{r+2}\right) \right]\)
\(S_n = \left[ \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \dots + \left(\frac{1}{n} - \frac{1}{n+1}\right) \right] + 3\left[ \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \dots + \left(\frac{1}{n+1} - \frac{1}{n+2}\right) \right]\).
By the method of differences, almost all terms cancel:
\(S_n = \left(1 - \frac{1}{n+1}\right) + 3\left(\frac{1}{2} - \frac{1}{n+2}\right)\)
\(S_n = 1 - \frac{1}{n+1} + \frac{3}{2} - \frac{3}{n+2}\)
\(S_n = \frac{5}{2} - \frac{n+2 + 3(n+1)}{(n+1)(n+2)} = \frac{5}{2} - \frac{4n + 5}{(n+1)(n+2)}\).

(iii) As \(n \to \infty\), \(\frac{4n + 5}{(n+1)(n+2)} \to 0\).
Therefore, the sum to infinity is:
\(S_{\infty} = \frac{5}{2}\).

(i)
M1: Attempt to express in partial fractions with three terms.
A1: Correct values for any two of \(A\), \(B\), \(C\).
A1: Correct overall partial fractions.

(ii)
M1: Writing the terms in a form suitable for differences (e.g. splitting into two difference sequences).
M1: Writing out the first few and last few terms to show cancellation.
A1: Correct cancellation showing remaining terms.
A1: Showing algebraic steps to simplify to the given expression.

(iii)
B1: Stating that \(\lim_{n \to \infty} \frac{4n + 5}{(n+1)(n+2)} = 0\) and obtaining \(\frac{5}{2}\).

(i) Let \(P(n)\) be the statement that \(f(n) = 3^{2n+2} - 8n - 9\) is divisible by 64.
- **Base case:** For \(n = 1\):
\(f(1) = 3^4 - 8(1) - 9 = 81 - 17 = 64\).
Since 64 is divisible by 64, \(P(1)\) is true.

- **Inductive step:** Assume that \(P(k)\) is true for some positive integer \(k\). That is,
\(3^{2k+2} - 8k - 9 = 64M\) for some integer \(M\).
Thus, \(3^{2k+2} = 64M + 8k + 9\).

Now, we consider \(P(k+1)\):
\(f(k+1) = 3^{2(k+1)+2} - 8(k+1) - 9\)
\(f(k+1) = 3^{2k+4} - 8k - 8 - 9\)
\(f(k+1) = 9 \cdot 3^{2k+2} - 8k - 17\).

Using the inductive hypothesis to substitute for \(3^{2k+2}\):
\(f(k+1) = 9(64M + 8k + 9) - 8k - 17\)
\(f(k+1) = 9(64M) + 72k + 81 - 8k - 17\)
\(f(k+1) = 9(64M) + 64k + 64\)
\(f(k+1) = 64(9M + k + 1)\).

Since \(M\) and \(k\) are integers, \(9M + k + 1\) is also an integer. Hence, \(f(k+1)\) is divisible by 64. Thus, \(P(k)\) implies \(P(k+1)\).

- **Conclusion:** Since \(P(1)\) 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) By substituting \(n = 10\) into the proven expression:
\(3^{2(10)+2} - 8(10) - 9 = 3^{22} - 89\) is divisible by 64.
So, \(3^{22} - 89 = 64K\) for some integer \(K\).
Therefore, \(3^{22} - 80 = 64K + 9\).
This shows that the remainder when \(3^{22} - 80\) is divided by 64 is 9.

B1: Verifying the base case \(n=1\).
M1: Stating the inductive hypothesis clearly.
M1: Considering the expression for \(n=k+1\).
A1: Substituting the inductive hypothesis correctly into the expression.
M1: Factoring out 64 from the simplified expression.
A1: Completing the algebraic proof and showing \(f(k+1) = 64(9M+k+1)\).
A1: Providing a complete and correct logical concluding statement.
M1: Recognizing that \(3^{22} - 89\) is divisible by 64 using \(n=10\).
A1: Deduce the remainder is 9.

(i) \(\mathbf{M}\) is singular when its determinant is zero:
\(\det(\mathbf{M}) = a(a-1) - (2)(1) = a^2 - a - 2 = 0\)
\((a-2)(a+1) = 0 \implies a = 2\) or \(a = -1\).

(ii) For \(a = 3\), \(\mathbf{M} = \begin{pmatrix} 3 & 2 \\ 1 & 2 \end{pmatrix}\).
Let \(\begin{pmatrix} x' \\ y'
\end{pmatrix} = \begin{pmatrix} 3 & 2 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}\).
This gives the system of equations:
1) \(x' = 3x + 2y\)
2) \(y' = x + 2y\)

We are given the line \(y = 2x + 1\). Substitute this into (1) and (2):
\(x' = 3x + 2(2x+1) = 7x + 2 \implies x = \frac{x'-2}{7}\)
\(y' = x + 2(2x+1) = 5x + 2\).

Substitute the expression for \(x\) into the equation for \(y'\):
\(y' = 5\left(\frac{x'-2}{7}\right) + 2\)
\(7y' = 5(x'-2) + 14\)
\(7y' = 5x' + 4\).

Thus, the equation of the image line is \(7y = 5x + 4\).

(iii) The area of triangle \(OAB\) is:
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}\).

The determinant of \(\mathbf{M}\) when \(a=3\) is \(3(2) - 2(1) = 4\).

Using the property that the area of the image is \(|\det(\mathbf{M})| \times \text{original area}\):
\(\text{Area of image} = 4 \times \frac{1}{2} = 2\).

(i)
M1: Setting the determinant of \(\mathbf{M}\) to 0.
A1: Finding correct values \(a = 2\) and \(a = -1\).

(ii)
M1: Writing transformation equations for \(x'\) and \(y'\) with \(a = 3\).
M1: Substituting \(y = 2x+1\) to express \(x'\) and \(y'\) in terms of \(x\).
M1: Eliminating \(x\) to find the relationship between \(y'\) and \(x'\).
A1: Finding correct equation \(7y = 5x + 4\).

(iii)
B1: Stating original area is \(\frac{1}{2}\).
M1: Using the determinant as the area scale factor.
A1: Correctly calculating the image area as 2.

(i) From the given cubic equation, we have:
- \(\sum \alpha = 3\)
- \(\sum \alpha\beta = 5\)
- \(\alpha\beta\gamma = 2\)

To find \(\alpha^2 + \beta^2 + \gamma^2\):
\(\sum \alpha^2 = (\sum \alpha)^2 - 2\sum \alpha\beta = 3^2 - 2(5) = 9 - 10 = -1\).

Since \(\alpha, \beta, \gamma\) are roots of the cubic equation, they satisfy:
\(x^3 - 3x^2 + 5x - 2 = 0\).
Summing over the three roots:
\(\sum \alpha^3 - 3\sum \alpha^2 + 5\sum \alpha - 6 = 0\).

Substitute the known values:
\(\sum \alpha^3 - 3(-1) + 5(3) - 6 = 0\)
\(\sum \alpha^3 + 3 + 15 - 6 = 0\)
\(\sum \alpha^3 = -12\).

(ii) Let \(y = x^2\) represent the roots of the new equation. Then \(x = \sqrt{y}\).
Substitute into the rewritten original equation:
\(x(x^2 + 5) = 3x^2 + 2\)
\(\sqrt{y}(y + 5) = 3y + 2\).

Square both sides:
\(y(y + 5)^2 = (3y + 2)^2\)
\(y(y^2 + 10y + 25) = 9y^2 + 12y + 4\)
\(y^3 + 10y^2 + 25y = 9y^2 + 12y + 4\)
\(y^3 + y^2 + 13y - 4 = 0\).

Alternatively, we can compute the coefficients using symmetric functions:
- New sum of roots: \(\sum \alpha^2 = -1\).
- New sum of products in pairs:
\(\sum \alpha^2\beta^2 = (\sum \alpha\beta)^2 - 2\alpha\beta\gamma \sum \alpha = 5^2 - 2(2)(3) = 25 - 12 = 13\).
- New product of roots:
\(\alpha^2\beta^2\gamma^2 = (\alpha\beta\gamma)^2 = 2^2 = 4\).

The required cubic equation is:
\(y^3 - (\sum \alpha^2)y^2 + (\sum \alpha^2\beta^2)y - \alpha^2\beta^2\gamma^2 = 0\)
\(y^3 - (-1)y^2 + 13y - 4 = 0\)
\(y^3 + y^2 + 13y - 4 = 0\).

(i)
B1: Correctly stating the relations \(\sum \alpha = 3\), \(\sum \alpha\beta = 5\), and \(\alpha\beta\gamma = 2\).
M1: Finding \(\sum \alpha^2 = -1\) using the algebraic identity.
M1: Setting up the recurrence relation for the cubic sum.
A1: Obtaining \(\sum \alpha^3 = -12\).

(ii)
M1: Attempting the substitution \(y = x^2\) or evaluating symmetric sums for \(\alpha^2, \beta^2, \gamma^2\).
M1: Correctly squaring both sides of the substituted equation or correctly calculating \(\sum \alpha^2\beta^2\).
A1: Finding the constant term of the new cubic equation.
A1: Completing the algebra to obtain the final equation \(y^3 + y^2 + 13y - 4 = 0\).

(i) To find a normal vector to the plane, we first find two vectors in the plane:
\(\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = (\mathbf{i} + 3\mathbf{j} - 2\mathbf{k}) - (2\mathbf{i} - \mathbf{j} + \mathbf{k}) = -\mathbf{i} + 4\mathbf{j} - 3\mathbf{k}\).
\(\overrightarrow{AC} = \mathbf{c} - \mathbf{a} = (3\mathbf{i} + \mathbf{j} + 2\mathbf{k}) - (2\mathbf{i} - \mathbf{j} + \mathbf{k}) = \mathbf{i} + 2\mathbf{j} + \mathbf{k}\).

Now, calculate the cross product \(\overrightarrow{AB} \times \overrightarrow{AC}\):
\(\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 4 & -3 \\ 1 & 2 & 1 \end{vmatrix}\)
\(= \mathbf{i}(4(1) - (-3)(2)) - \mathbf{j}((-1)(1) - (-3)(1)) + \mathbf{k}((-1)(2) - 4(1))\)
\(= 10\mathbf{i} - 2\mathbf{j} - 6\mathbf{k}\).

Dividing by 2 to obtain a simpler vector, we can use:
\(\mathbf{n} = 5\mathbf{i} - \mathbf{j} - 3\mathbf{k}\).

(ii) Using the point \(A(2, -1, 1)\), the constant \(d\) is given by:
\(d = \mathbf{a}
cdot \mathbf{n} = (2\mathbf{i} - \mathbf{j} + \mathbf{k})
cdot (5\mathbf{i} - \mathbf{j} - 3\mathbf{k}) = 2(5) + (-1)(-1) + 1(-3) = 10 + 1 - 3 = 8\).

So the vector equation of the plane is:
\(\mathbf{r}
cdot (5\mathbf{i} - \mathbf{j} - 3\mathbf{k}) = 8\).

(iii) The shortest distance from point \(D(4, 2, -1)\) with position vector \(\mathbf{d} = 4\mathbf{i} + 2\mathbf{j} - \mathbf{k}\) is:
\(\text{Distance} = \frac{|\mathbf{d}
cdot \mathbf{n} - d|}{|\mathbf{n}|}\).

Calculate \(\mathbf{d}
cdot \mathbf{n}\):
\(\mathbf{d}
cdot \mathbf{n} = (4\mathbf{i} + 2\mathbf{j} - \mathbf{k})
cdot (5\mathbf{i} - \mathbf{j} - 3\mathbf{k}) = 4(5) + 2(-1) + (-1)(-3) = 20 - 2 + 3 = 21\).

Calculate the magnitude of \(\mathbf{n}\):
\(|\mathbf{n}| = \sqrt{5^2 + (-1)^2 + (-3)^2} = \sqrt{25 + 1 + 9} = \sqrt{35}\).

Thus, the shortest distance is:
\(\text{Distance} = \frac{|21 - 8|}{\sqrt{35}} = \frac{13}{\sqrt{35}}\).

(i)
M1: Correctly finding vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\).
M1: Attempting the cross product of \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\).
A1: Finding the correct normal vector (or any scalar multiple).

(ii)
M1: Using the formula \(\mathbf{r}
cdot \mathbf{n} = \mathbf{a}
cdot \mathbf{n}\).
A1: Finding the correct equation \(\mathbf{r}
cdot (5\mathbf{i} - \mathbf{j} - 3\mathbf{k}) = 8\).

(iii)
M1: Recalling the shortest distance formula.
M1: Computing the dot product \(\mathbf{d}
cdot \mathbf{n}\).
A1: Correctly calculating \(|\mathbf{n}| = \sqrt{35}\).
A1: Obtaining the correct final distance \(\frac{13}{\sqrt{35}}\).

(i) The curve is a cardioid symmetric about the line \(\theta = \frac{\pi}{2}\) (the y-axis), with a cusp at the pole since \(r = 0\) when \(\theta = -\frac{\pi}{2}\).

(ii) The area \(A\) is given by:
\(A = \frac{1}{2} \int_{-\pi}^{\pi} r^2 \, d\theta = \frac{1}{2} \int_{-\pi}^{\pi} 4(1 + \sin\theta)^2 \, d\theta = 2 \int_{-\pi}^{\pi} (1 + 2\sin\theta + \sin^2\theta) \, d\theta\).

Since \(\sin\theta\) is an odd function, \(\int_{-\pi}^{\pi} \sin\theta \, d\theta = 0\).
Using the double-angle identity \(\sin^2\theta = \frac{1 - \cos 2\theta}{2}\):
\(A = 2 \int_{-\pi}^{\pi} \left(1 + \frac{1 - \cos 2\theta}{2}\right) \, d\theta = 2 \int_{-\pi}^{\pi} \left(\frac{3}{2} - \frac{1}{2}\cos 2\theta\right) \, d\theta\)
\(A = 2 \left[ \frac{3}{2}\theta - \frac{1}{4}\sin 2\theta \right]_{-\pi}^{\pi} = 2 \left( \frac{3\pi}{2} - \left(-\frac{3\pi}{2}\right) - 0 \right) = 6\pi\).

(iii) The tangent is parallel to the initial line when \(\frac{dy}{d\theta} = 0\) (provided \(\frac{dx}{d\theta} \ne 0\)), where \(y = r \sin\theta\):
\(y = 2(1 + \sin\theta)\sin\theta = 2\sin\theta + 2\sin^2\theta\).

Differentiating with respect to \(\theta\):
\(\frac{dy}{d\theta} = 2\cos\theta + 4\sin\theta\cos\theta = 2\cos\theta(1 + 2\sin\theta) = 0\).

This gives \(\cos\theta = 0\) or \(\sin\theta = -\frac{1}{2}\).
- From \(\cos\theta = 0\), we get \(\theta = \frac{\pi}{2}\) (the value \(\theta = -\frac{\pi}{2}\) yields \(r=0\) which is a cusp, where both derivatives are 0).
For \(\theta = \frac{\pi}{2}\): \(r = 2(1 + 1) = 4\).
Cartesian coordinates: \(x = 4\cos(\pi/2) = 0\), \(y = 4\sin(\pi/2) = 4\). So, \((0, 4)\).

- From \(\sin\theta = -\frac{1}{2}\), we get \(\theta = -\frac{\pi}{6}\) or \(\theta = -\frac{5\pi}{6}\).
In both cases, \(r = 2(1 - 1/2) = 1\).
For \(\theta = -\frac{\pi}{6}\): \(x = 1\cos(-\pi/6) = \frac{\sqrt{3}}{2}\), \(y = 1\sin(-\pi/6) = -\frac{1}{2}\). Point is \(\left(\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)\).
For \(\theta = -\frac{5\pi}{6}\): \(x = 1\cos(-5\pi/6) = -\frac{\sqrt{3}}{2}\), \(y = 1\sin(-5\pi/6) = -\frac{1}{2}\). Point is \(\left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)\).

(i)
B1: Cardioid shape drawn in correct orientation.
B1: Showing intercept at \((0, 4)\) and cusp at pole.

(ii)
M1: Setting up correct integral for area.
M1: Expressing \(\sin^2\theta\) using double angle formula.
A1: Correct integration.
A1: Finding the area of \(6\pi\).

(iii)
M1: Expressing \(y\) in terms of \(\theta\) and finding \(\frac{dy}{d\theta}\).
M1: Solving \(\frac{dy}{d\theta} = 0\).
A1: Identifying the angles \(\theta = \frac{\pi}{2}, -\frac{\pi}{6}, -\frac{5\pi}{6}\).
A1: Finding all three correct Cartesian coordinate points.

(i) Perform algebraic division to express the equation of \(C\) in a more helpful form:
\(y = \frac{x(x - 1) + 4}{x - 1} = x + \frac{4}{x - 1}\).
- As \(x \to 1\), \(y \to \pm\infty\), so the vertical asymptote is \(x = 1\).
- As \(x \to \pm\infty\), \(\frac{4}{x-1} \to 0\), so the oblique asymptote is \(y = x\).

(ii) Differentiate \(y\) with respect to \(x\):
\(\frac{dy}{dx} = 1 - \frac{4}{(x - 1)^2}\).

To find stationary points, set \(\frac{dy}{dx} = 0\):
\(1 - \frac{4}{(x-1)^2} = 0 \implies (x-1)^2 = 4\)
\(x - 1 = 2\) or \(x - 1 = -2\)
\(x = 3\) or \(x = -1\).

- When \(x = 3\): \(y = 3 + \frac{4}{2} = 5\). Point is \((3, 5)\).
- When \(x = -1\): \(y = -1 + \frac{4}{-2} = -3\). Point is \((-1, -3)\).

(iii) Intersections with axes:
- When \(x = 0\), \(y = -4\). The y-intercept is \((0, -4)\).
- When \(y = 0\), \(x^2 - x + 4 = 0\). The discriminant is \((-1)^2 - 4(1)(4) = -15 < 0\), so there are no real roots and no x-intercepts.

The graph consists of two branches:
- One branch is to the left of \(x=1\), passing through \((0, -4)\) with a local maximum at \((-1, -3)\).
- The other branch is to the right of \(x=1\), with a local minimum at \((3, 5)\).
Both branches asymptotically approach the lines \(x = 1\) and \(y = x\).

(i)
M1: Attempting algebraic division of the rational expression.
A1: Correctly identifying vertical asymptote \(x = 1\).
A1: Correctly identifying oblique asymptote \(y = x\).

(ii)
M1: Finding \(\frac{dy}{dx}\).
M1: Solving \(\frac{dy}{dx} = 0\) for \(x\).
A1: Correct coordinates for the first stationary point \((3, 5)\).
A1: Correct coordinates for the second stationary point \((-1, -3)\).

(iii)
B1: Sketching asymptotes \(x=1\) and \(y=x\) and showing the y-intercept at \((0, -4)\).
B1: Drawing two branches with correct shapes, with local maximum and minimum in the correct relative quadrants.

Using the identity \( \cosh^2 x - \sinh^2 x = 1 \), we have \( \cosh^2 x = 1 + \sinh^2 x \). Substituting this into the equation gives \( 3(1 + \sinh^2 x) + 4\sinh x = 7 \), which simplifies to \( 3\sinh^2 x + 4\sinh x - 4 = 0 \). Let \( y = \sinh x \). The quadratic equation is \( 3y^2 + 4y - 4 = 0 \), which factors to \( (3y - 2)(y + 2) = 0 \). Thus, \( \sinh x = \frac{2}{3} \) or \( \sinh x = -2 \). Using the logarithmic form \( \sinh^{-1} y = \ln(y + \sqrt{y^2 + 1}) \), we get: For \( y = \frac{2}{3} \), \( x = \ln\left(\frac{2}{3} + \sqrt{\frac{4}{9} + 1}\right) = \ln\left(\frac{2+\sqrt{13}}{3}\right) \). For \( y = -2 \), \( x = \ln\left(-2 + \sqrt{4 + 1}\right) = \ln(\sqrt{5} - 2) \).

M1 for using the identity \( \cosh^2 x - \sinh^2 x = 1 \). A1 for obtaining the correct quadratic in \( \sinh x \). M1 for solving the quadratic equation. A1 for finding both correct values of \( \sinh x \). M1 for using the logarithmic formula for \( \sinh^{-1} y \). A1 for each correct logarithmic form (2 marks total). A0.375 for final correct presentation of both exact values.

First, find the eigenvalues of \( \mathbf{A} \). The characteristic equation is \( \det(\mathbf{A} - \lambda\mathbf{I}) = 0 \), which gives \( (3-\lambda)(-\lambda) - (-2) = \lambda^2 - 3\lambda + 2 = 0 \). The eigenvalues are \( \lambda = 1 \) and \( \lambda = 2 \). For \( \lambda = 1 \), we solve \( (\mathbf{A} - \mathbf{I})\mathbf{v} = 0 \), leading to \( 2x - y = 0 \), so an eigenvector is \( \begin{pmatrix} 1 \\ 2 \end{pmatrix} \). For \( \lambda = 2 \), we solve \( (\mathbf{A} - 2\mathbf{I})\mathbf{v} = 0 \), leading to \( x - y = 0 \), so an eigenvector is \( \begin{pmatrix} 1 \\ 1 \end{pmatrix} \). Thus, \( \mathbf{P} = \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix} \) and \( \mathbf{D} = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} \). The inverse of \( \mathbf{P} \) is \( \mathbf{P}^{-1} = \frac{1}{-1}\begin{pmatrix} 1 & -1 \\ -2 & 1 \end{pmatrix} = \begin{pmatrix} -1 & 1 \\ 2 & -1 \end{pmatrix} \). Using \( \mathbf{A}^n = \mathbf{P}\mathbf{D}^n\mathbf{P}^{-1} \), we have \( \mathbf{A}^n = \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} 1^n & 0 \\ 0 & 2^n \end{pmatrix} \begin{pmatrix} -1 & 1 \\ 2 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 2^n \\ 2 & 2^n \end{pmatrix} \begin{pmatrix} -1 & 1 \\ 2 & -1 \end{pmatrix} = \begin{pmatrix} 2^{n+1} - 1 & 1 - 2^n \\ 2^{n+1} - 2 & 2 - 2^n \end{pmatrix} \).

M1 for setting up and solving the characteristic equation. A1 for finding eigenvalues 1 and 2. M1 for finding the corresponding eigenvectors. A1 for forming correct matrices \( \mathbf{P} \) and \( \mathbf{D} \). M1 for finding the inverse matrix \( \mathbf{P}^{-1} \). M1 for setting up the multiplication \( \mathbf{P}\mathbf{D}^n\mathbf{P}^{-1} \). A1.375 for carrying out the matrix multiplication correctly to obtain the final form.

Let \( f(x) = \ln(1 + \sin x) \), so \( f(0) = \ln 1 = 0 \). Differentiating gives \( f'(x) = \frac{\cos x}{1 + \sin x} \), so \( f'(0) = 1 \). Differentiating again using the quotient rule: \( f''(x) = \frac{-\sin x(1+\sin x) - \cos^2 x}{(1+\sin x)^2} = \frac{-\sin x - (\sin^2 x + \cos^2 x)}{(1+\sin x)^2} = \frac{-(1+\sin x)}{(1+\sin x)^2} = -(1+\sin x)^{-1} \), so \( f''(0) = -1 \). Differentiating a third time: \( f'''(x) = (1+\sin x)^{-2}\cos x \), so \( f'''(0) = 1 \). Using Maclaurin's formula \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots \), we obtain \( f(x) = x - \frac{1}{2}x^2 + \frac{1}{6}x^3 \).

M1 for finding \( f'(x) \). A1 for correct value \( f'(0) = 1 \). M1 for finding \( f''(x) \). A1 for simplifying \( f''(x) \) and finding \( f''(0) = -1 \). M1 for finding \( f'''(x) \). A1 for finding \( f'''(0) = 1 \). M1 for substituting values into Maclaurin's series formula. A1.375 for obtaining the correct final series.

Using integration by parts with \( u = x^n \) and \( dv = e^{-x} \, dx \), we have \( du = n x^{n-1} \, dx \) and \( v = -e^{-x} \). Thus, \( I_n = [-x^n e^{-x}]_0^1 + \int_0^1 n x^{n-1} e^{-x} \, dx = -e^{-1} + n I_{n-1} \). This establishes the reduction formula. Next, we find the base case: \( I_0 = \int_0^1 e^{-x} \, dx = [-e^{-x}]_0^1 = 1 - e^{-1} \). Using the reduction formula: \( I_1 = I_0 - e^{-1} = 1 - 2e^{-1} \). Then, \( I_2 = 2 I_1 - e^{-1} = 2(1 - 2e^{-1}) - e^{-1} = 2 - 5e^{-1} \). Finally, \( I_3 = 3 I_2 - e^{-1} = 3(2 - 5e^{-1}) - e^{-1} = 6 - 16e^{-1} \).

M1 for applying integration by parts with correct assignment of \( u \) and \( dv \). A1 for correct integration by parts expression. A1 for deriving the reduction formula \( I_n = n I_{n-1} - e^{-1} \). M1 for calculating \( I_0 \). A1 for finding \( I_1 \). M1 for finding \( I_2 \). M1 for finding \( I_3 \). A1.375 for the correct exact value of \( I_3 \).

By de Moivre's theorem, \( \cos(5\theta) + i\sin(5\theta) = (\cos\theta + i\sin\theta)^5 \). Expanding this using the binomial theorem and equating the imaginary parts gives \( \sin(5\theta) = 5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta \). Substituting \( \cos^2\theta = 1 - \sin^2\theta \), we obtain \( \sin(5\theta) = 5(1-\sin^2\theta)^2\sin\theta - 10(1-\sin^2\theta)\sin^3\theta + \sin^5\theta = 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta \). To find the non-zero roots of \( 16x^4 - 20x^2 + 5 = 0 \), we multiply by \( x \) to get \( 16x^5 - 20x^3 + 5x = 0 \). Letting \( x = \sin\theta \), this equation is equivalent to \( \sin(5\theta) = 0 \). Thus, \( 5\theta = m\pi \) for integers \( m \). For non-zero roots of the degree 4 polynomial, we choose the four distinct positive and negative values \( \theta = \pm\frac{\pi}{5} \) and \( \theta = \pm\frac{2\pi}{5} \). Thus, the roots are \( x = \pm\sin\left(\frac{\pi}{5}\right) \) and \( x = \pm\sin\left(\frac{2\pi}{5}\right) \).

M1 for expanding \( (\cos\theta + i\sin\theta)^5 \) using the binomial theorem. A1 for identifying the imaginary part corresponding to \( \sin(5\theta) \). M1 for substituting \( \cos^2\theta = 1 - \sin^2\theta \). A1 for the correct identity \( \sin(5\theta) = 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta \). M1 for establishing the connection between the polynomial equation and \( \sin(5\theta) = 0 \). A1 for identifying the appropriate angles \( \theta = \frac{m\pi}{5} \). A2.375 for presenting all four roots in the requested form.

First, solve the auxiliary equation \( m^2 + 2m + 5 = 0 \) to find the complementary function (CF). The roots are \( m = -1 \pm 2i \), giving \( y_c = e^{-x}(A\cos 2x + B\sin 2x) \). Next, seek a particular integral (PI) of the form \( y_p = C\cos x + D\sin x \). Differentiating, we obtain \( y_p' = -C\sin x + D\cos x \) and \( y_p'' = -C\cos x - D\sin x \). Substituting into the differential equation: \( (-C\cos x - D\sin x) + 2(-C\sin x + D\cos x) + 5(C\cos x + D\sin x) = 10\sin x \). Simplifying gives \( (4C + 2D)\cos x + (4D - 2C)\sin x = 10\sin x \). Equating coefficients: \( 4C + 2D = 0 \Rightarrow D = -2C \) and \( 4D - 2C = 10 \Rightarrow -8C - 2C = 10 \Rightarrow C = -1 \), which yields \( D = 2 \). Hence, the PI is \( y_p = -\cos x + 2\sin x \). The general solution is \( y = y_c + y_p = e^{-x}(A\cos 2x + B\sin 2x) - \cos x + 2\sin x \).

M1 for setting up and solving the auxiliary equation. A1 for finding the correct roots \( -1 \pm 2i \). A1 for the correct complementary function. M1 for stating a trial particular integral of the form \( C\cos x + D\sin x \). M1 for substituting this trial solution into the differential equation. A1 for finding the correct coefficients \( C = -1 \) and \( D = 2 \). A1 for stating the correct particular integral. A1.375 for writing the final correct general solution.

We use the arc length formula \( S = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \). First, find the derivative: \( \frac{dy}{dx} = \frac{-\sin x}{\cos x} = -\tan x \). Thus, \( 1 + \left(\frac{dy}{dx}\right)^2 = 1 + \tan^2 x = \sec^2 x \). Since \( \sec x > 0 \) for \( x \in [0, \frac{\pi}{3}] \), the integrand is \( \sqrt{\sec^2 x} = \sec x \). The arc length is \( S = \int_0^{\pi/3} \sec x \, dx = [\ln|\sec x + \tan x|]_0^{\pi/3} \). Evaluating this at the limits: \( S = \ln|\sec(\frac{\pi}{3}) + \tan(\frac{\pi}{3})| - \ln|\sec(0) + \tan(0)| = \ln(2 + \sqrt{3}) - \ln(1 + 0) = \ln(2 + \sqrt{3}) \).

M1 for differentiating \( y = \ln(\cos x) \). A1 for obtaining \( \frac{dy}{dx} = -\tan x \). M1 for calculating \( 1 + (\frac{dy}{dx})^2 \) and simplifying. A1 for obtaining \( \sec^2 x \). M1 for setting up the correct arc length integral. M1 for integrating \( \sec x \) to get \( \ln|\sec x + \tan x| \). A1.375 for evaluating the limits to obtain the exact value \( \ln(2+\sqrt{3}) \).

We can rewrite the integrand as \( \cosh^3 x = \cosh x \cosh^2 x = \cosh x (1 + \sinh^2 x) \). Let \( u = \sinh x \), which gives \( du = \cosh x \, dx \). We also change the limits of integration: when \( x = 0 \), \( u = \sinh 0 = 0 \); when \( x = \ln 2 \), \( u = \sinh(\ln 2) = \frac{e^{\ln 2} - e^{-\ln 2}}{2} = \frac{2 - 1/2}{2} = \frac{3}{4} \). The integral becomes \( \int_0^{3/4} (1 + u^2) \, du = \left[ u + \frac{u^3}{3} \right]_0^{3/4} = \frac{3}{4} + \frac{1}{3}\left(\frac{27}{64}\right) = \frac{3}{4} + \frac{9}{64} = \frac{48 + 9}{64} = \frac{57}{64} \).

M1 for rewriting the integrand \( \cosh^3 x \) as \( \cosh x (1 + \sinh^2 x) \). A1 for setting up the substitution with correct derivative. M1 for finding the transformed limits of integration \( u = 0 \) and \( u = 3/4 \). A1 for the correct simplified integral \( \int_0^{3/4} (1+u^2) \, du \). M1 for integrating. A1 for substituting the limits. A1.375 for obtaining the exact value \( \frac{57}{64} \).

Let \(O\) be the center of the sphere and let \(A\) be the lowest point. At a general position where the radius to the particle makes an angle \(\theta\) with the upward vertical, the height of the particle above \(A\) is given by:
\[h = a + a\cos\theta = a(1 + \cos\theta)\]

By Conservation of Energy, the kinetic energy at this point is:
\[\frac{1}{2}mv^2 = \frac{1}{2}mu^2 - mgh\]
\[v^2 = u^2 - 2gh = 3ag - 2ag(1 + \cos\theta) = ag(1 - 2\cos\theta)\]

The equation of motion along the inward normal to the sphere is:
\[R + mg\cos\theta = \frac{mv^2}{a}\]
where \(R\) is the normal reaction.

Substitute the expression for \(v^2\):
\[R = \frac{m(ag(1 - 2\cos\theta))}{a} - mg\cos\theta = mg(1 - 3\cos\theta)\]

The particle leaves the surface of the sphere when \(R = 0\):
\[1 - 3\cos\theta = 0 \implies \cos\theta = \frac{1}{3}\]

M1: Apply Conservation of Energy to express \(v^2\) in terms of \(\theta\).
A1: Correct equation for \(v^2 = ag(1 - 2\cos\theta)\).
M1: Write down the radial equation of motion including normal reaction \(R\) and weight component.
A1: Correct radial equation \(R + mg\cos\theta = \frac{mv^2}{a}\).
M1: Set \(R = 0\) to find the condition for leaving the surface.
A1: Solve for \(\cos\theta\) to obtain \(\cos\theta = \frac{1}{3}\).

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

Substitute the given values \(x = 12\), \(y = 8\), \(U = 20\), and \(g = 10\):
\[8 = 12\tan\alpha - \frac{10 \times 12^2}{2 \times 20^2}(1 + \tan^2\alpha)\]
\[8 = 12\tan\alpha - \frac{1440}{800}(1 + \tan^2\alpha)\]
\[8 = 12\tan\alpha - 1.8(1 + \tan^2\alpha)\]

Multiply the equation by 5 to clear the decimals:
\[40 = 60\tan\alpha - 9(1 + \tan^2\alpha)\]
\[9\tan^2\alpha - 60\tan\alpha + 49 = 0\]

Let \(t = \tan\alpha\). This quadratic equation in \(t\) is:
\[9t^2 - 60t + 49 = 0\]

Using Vieta's formulas, the sum of the roots \(t_1 + t_2\) is:
\[t_1 + t_2 = \frac{60}{9} = \frac{20}{3}\]

M1: State or use the standard trajectory equation in terms of \(\tan\alpha\).
A1: Substitute the coordinates \((12, 8)\) and parameters \(U = 20, g = 10\) correctly.
M1: Rearrange the resulting equation into a quadratic form in \(\tan\alpha\).
A1: Correct quadratic equation \(9\tan^2\alpha - 60\tan\alpha + 49 = 0\) (or equivalent).
M1: Use Vieta's formulas or solve the quadratic to find the sum of the roots.
A1: Obtain the correct sum \(\frac{20}{3}\).

Let \(x\) be the extension of the string. The particle reaches its maximum speed when the net force on it is zero (equilibrium position).

At equilibrium:
\[T = mg\]
\[\frac{\lambda x_0}{L} = mg \implies \frac{30 x_0}{1.2} = 2 \times 10 = 20\]
\[25 x_0 = 20 \implies x_0 = 0.8\text{ m}\]

At this point, the total descent of the particle is \(h = L + x_0 = 1.2 + 0.8 = 2.0\text{ m}\).

Using conservation of energy from the release point \(O\) to the point of maximum speed:
\[\text{Loss of GPE} = \text{Gain in KE} + \text{Gain in EPE}\]
\[mgh = \frac{1}{2}mv_{\max}^2 + \frac{\lambda x_0^2}{2L}\]
\[2 \times 10 \times 2.0 = \frac{1}{2} \times 2 \times v_{\max}^2 + \frac{30 \times 0.8^2}{2 \times 1.2}\]
\[40 = v_{\max}^2 + \frac{19.2}{2.4}\]
\[40 = v_{\max}^2 + 8\]
\[v_{\max}^2 = 32 \implies v_{\max} = \sqrt{32} = 4\sqrt{2}\text{ m/s}\]

M1: Identify that maximum speed occurs at the equilibrium position where acceleration is zero.
A1: Calculate the extension at equilibrium \(x_0 = 0.8\text{ m}\).
M1: Formulate the conservation of energy equation relating GPE, KE, and EPE.
A1: Correctly substitute values into the energy equation.
M1: Simplify the energy equation to find \(v_{\max}^2\).
A1: Obtain the exact maximum speed as \(4\sqrt{2}\).

Using Newton's second law, the equation of motion is:
\[F_{\text{net}} = ma \implies 1200 - 2v^2 = 1000 v \frac{dv}{dx}\]

Separate the variables to integrate for the distance \(x\):
\[1000 v \frac{dv}{dx} = 2(600 - v^2)\]
\[500 v \frac{dv}{dx} = 600 - v^2\]
\[dx = \frac{500 v}{600 - v^2} dv\]

Integrate between the limits \(v = 10\) and \(v = 20\):
\[x = \int_{10}^{20} \frac{500 v}{600 - v^2} dv\]

Let \(u = 600 - v^2 \implies du = -2v\,dv \implies v\,dv = -\frac{1}{2}\,du\).
When \(v = 10\), \(u = 500\).
When \(v = 20\), \(u = 200\).

\[x = \int_{500}^{200} \frac{500 \left(-\frac{1}{2}\right)}{u} du = -250 [\ln u]_{500}^{200}\]
\[x = -250 (\ln 200 - \ln 500) = 250 \ln\left(\frac{500}{200}\right) = 250 \ln(2.5)\text{ m}\]

M1: Set up the differential equation of motion using \(a = v \frac{dv}{dx}\).
A1: Correct differential equation \(1000 v \frac{dv}{dx} = 1200 - 2v^2\).
M1: Separate variables and set up the definite integral with appropriate limits.
A1: Correctly perform the integration to get a form involving natural logarithms.
M1: Apply the integration limits \(10\) and \(20\) correctly.
A1: Obtain the final simplified exact distance \(250 \ln(2.5)\) (or equivalent).

Let \(R_A\) and \(F_A\) be the normal reaction and frictional force at the ground \(A\).
Let \(R_B\) and \(F_B\) be the normal reaction and frictional force at the wall \(B\).

Resolving forces:
Horizontal equilibrium: \(F_A = R_B\)
Vertical equilibrium: \(R_A + F_B = W\)

At the point of slipping, both contacts are in limiting equilibrium:
\[F_A = \mu R_A\]
\[F_B = \mu R_B\]

Using these relations:
\[\mu R_A = R_B \implies R_A = \frac{R_B}{\mu}\]
Substitute into the vertical equation:
\[\frac{R_B}{\mu} + \mu R_B = W \implies R_B \left(\frac{1 + \mu^2}{\mu}\right) = W \implies R_B = \frac{\mu W}{1 + \mu^2}\]
Consequently:
\[F_B = \frac{\mu^2 W}{1 + \mu^2}\]

Taking moments about \(A\):
\[W a \cos\theta = R_B (2a \sin\theta) + F_B (2a \cos\theta)\]
\[W \cos\theta = 2 R_B \sin\theta + 2 F_B \cos\theta\]

Substitute the expressions for \(R_B\) and \(F_B\):
\[W \cos\theta = 2 \left(\frac{\mu W}{1 + \mu^2}\right) \sin\theta + 2 \left(\frac{\mu^2 W}{1 + \mu^2}\right) \cos\theta\]
\[\cos\theta = \frac{2\mu}{1 + \mu^2} \sin\theta + \frac{2\mu^2}{1 + \mu^2} \cos\theta\]
\[(1 + \mu^2) \cos\theta = 2\mu \sin\theta + 2\mu^2 \cos\theta\]
\[(1 - \mu^2) \cos\theta = 2\mu \sin\theta \implies \tan\theta = \frac{1 - \mu^2}{2\mu}\]

Given \( ̅\mu = \frac{1}{3}\):
\[\tan\theta = \frac{1 - (1/3)^2}{2(1/3)} = \frac{8/9}{2/3} = \frac{4}{3}\]

M1: Write horizontal and vertical force equilibrium equations.
A1: Correctly apply limiting friction conditions \(F_A = \mu R_A\) and \(F_B = \mu R_B\).
M1: Solve the system of forces to express normal reactions in terms of \(W\) and \(\mu\).
M1: Formulate the moments equation about one end of the ladder (e.g., \(A\)).
A1: Correctly substitute force expressions into the moment equation.
M1: Derivation of the relation \(\tan\theta = \frac{1 - \mu^2}{2\mu}\).
A1: Correctly calculate \(\tan\theta = \frac{4}{3}\).

Let \(v_A\) and \(v_B\) be the velocities of \(A\) and \(B\) after the collision in the direction of the initial velocity \(u\).

By Conservation of Linear Momentum:
\[3m u = 3m v_A + m v_B \implies 3v_A + v_B = 3u\]

By Newton's Law of Restitution:
\[v_B - v_A = eu = \frac{1}{3}u\]

Solving these equations simultaneously:
Multiply the second equation by 3:
\[3v_B - 3v_A = u\]
Add this to the momentum equation:
\[4v_B = 4u \implies v_B = u\]
Then, find \(v_A\):
\[v_A = v_B - \frac{1}{3}u = u - \frac{1}{3}u = \frac{2}{3}u\]

The total kinetic energy before the collision is:
\[K_i = \frac{1}{2}(3m)u^2 = \frac{3}{2}mu^2\]

The total kinetic energy after the collision is:
\[K_f = \frac{1}{2}(3m)v_A^2 + \frac{1}{2}m v_B^2 = \frac{3}{2}m\left(\frac{2}{3}u\right)^2 + \frac{1}{2}m(u)^2\]
\[K_f = \frac{3}{2}m\left(\frac{4}{9}u^2\right) + \frac{1}{2}mu^2 = \frac{2}{3}mu^2 + \frac{1}{2}mu^2 = \frac{7}{6}mu^2\]

The ratio of the kinetic energies is:
\[\frac{K_f}{K_i} = \frac{\frac{7}{6}mu^2}{\frac{3}{2}mu^2} = \frac{7}{6} \times \frac{2}{3} = \frac{7}{9}\]

M1: Write down the conservation of linear momentum equation.
A1: Correct momentum equation \(3v_A + v_B = 3u\).
M1: Write down the restitution equation with \(e = \frac{1}{3}\).
A1: Solve the simultaneous equations to obtain \(v_B = u\) and \(v_A = \frac{2}{3}u\).
M1: Express the total kinetic energy before and after the collision.
A1: Correct expressions for \(K_i = \frac{3}{2}mu^2\) and \(K_f = \frac{7}{6}mu^2\).
A1: Calculate the final ratio as \ \frac{7}{9}\.

The radius of the circle is \(r = 1.0\text{ m}\), which is greater than the natural length of the string \(L = 0.8\text{ m}\).
Thus, the extension \(x\) of the string is:
\[x = r - L = 1.0 - 0.8 = 0.2\text{ m}\]

Using Hooke's Law, the tension \(T\) in the string is:
\[T = \frac{\lambda x}{L} = \frac{20 \times 0.2}{0.8} = 5\text{ N}\]

This tension provides the necessary centripetal force for the circular motion:
\[T = mr\omega^2\]
\[5 = 0.2 \times 1.0 \times \omega^2\]
\[5 = 0.2 \omega^2\]
\[\omega^2 = 25 \implies \omega = 5\text{ rad/s}\]

M1: Calculate the extension of the string \(x = r - L\).
A1: Obtain \(x = 0.2\text{ m}\).
M1: Apply Hooke's Law to find the tension \(T\).
A1: Correctly find \(T = 5\text{ N}\).
M1: Relate the tension to the centripetal force equation \(T = mr\omega^2\).
A1: Solve for the angular speed to get \(\omega = 5\text{ rad/s}\).

(i) The assumptions required are:
- The two samples are independent random samples.
- The parent populations from which the samples are drawn are normally distributed.
- The two populations have equal variances.

(ii) For Group A:
\(n_A = 6\)
\(\sum x = 78.4\), \(\bar{x}_A = 13.067\)
\(\sum x^2 = 12.4^2 + 14.1^2 + 11.8^2 + 13.5^2 + 12.9^2 + 13.7^2 = 1028.16\)
\(s_A^2 = \frac{1}{5} \left( 1028.16 - \frac{78.4^2}{6} \right) = \frac{3.7333}{5} = 0.7467\)

For Group B:
\(n_B = 7\)
\(\sum y = 100.4\), \(\bar{y}_B = 14.343\)
\(\sum y^2 = 14.2^2 + 13.8^2 + 15.1^2 + 14.6^2 + 13.5^2 + 14.9^2 + 14.3^2 = 1442.0\)
\(s_B^2 = \frac{1}{6} \left( 1442.0 - \frac{100.4^2}{7} \right) = \frac{1.9771}{6} = 0.3295\)

The pooled variance estimate is:
\(s_p^2 = \frac{(n_A - 1)s_A^2 + (n_B - 1)s_B^2}{n_A + n_B - 2} = \frac{5(0.7467) + 6(0.3295)}{11} = \frac{3.7333 + 1.9771}{11} = 0.5191\) (or \(0.519\) to 3 s.f.).

(iii) \(H_0: \mu_A = \mu_B\)
\(H_1: \mu_A \neq \mu_B\)

Standard error of the difference:
\(\text{SE} = \sqrt{s_p^2 \left(\frac{1}{6} + \frac{1}{7}\right)} = \sqrt{0.5191 \times 0.3095} = 0.4008\)

Test statistic:
\(t = \frac{\bar{x}_A - \bar{y}_B}{\text{SE}} = \frac{13.067 - 14.343}{0.4008} = -3.18\)

Degrees of freedom: \(v = 6 + 7 - 2 = 11\).
Critical value for two-tailed test at 5% significance level is \(t_{11}(0.025) = 2.201\).

Since \(|-3.18| > 2.201\), we reject \(H_0\).
There is sufficient evidence at the 5% level of significance to suggest that there is a difference in the mean scores of the two populations.

(i) B1: For stating normal distribution of populations. B1: For stating independence of samples and equal variances.
(ii) M1: For attempting sample variance calculation for both groups. A1: For correct sample variances (0.747 and 0.330). A1: For correct pooled variance of 0.519.
(iii) B1: For correct hypotheses. M1: For calculating the SE and t-statistic. A1: For t = -3.18. B1: For correct critical value of 2.201. A1: For correct rejection and conclusion in context.

(i) Mean \(\lambda = \frac{0(36) + 1(40) + 2(19) + 3(4) + 4(1)}{100} = \frac{40 + 38 + 12 + 4}{100} = 0.94\).

Under a Poisson distribution with \(\lambda = 0.94\):
\(P(X = k) = \frac{e^{-0.94} 0.94^k}{k!}\)

Expected frequencies (\(E_k = 100 \times P(X=k)\)):
\(E_0 = 100 e^{-0.94} = 39.06\)
\(E_1 = 100 e^{-0.94} \times 0.94 = 36.72\)
\(E_2 = 100 e^{-0.94} \times \frac{0.94^2}{2} = 17.26\)

Since expected frequencies must be at least 5, we group the remaining classes:
\(E_{\ge 3} = 100 - (39.06 + 36.72 + 17.26) = 6.96\).
Since \(6.96 > 5\), the final classes are:
- \(x = 0\): Observed = 36, Expected = 39.06
- \(x = 1\): Observed = 40, Expected = 36.72
- \(x = 2\): Observed = 19, Expected = 17.26
- \(x \ge 3\): Observed = 5, Expected = 6.96

(ii) \(H_0\): Poisson distribution fits the data.
\(H_1\): Poisson distribution does not fit the data.

Calculate the test statistic \(\chi^2 = \sum \frac{(O-E)^2}{E}\):
- \(x=0\): \(\frac{(36 - 39.06)^2}{39.06} = 0.2397\)
- \(x=1\): \(\frac{(40 - 36.72)^2}{36.72} = 0.2930\)
- \(x=2\): \(\frac{(19 - 17.26)^2}{17.26} = 0.1754\)
- \(x \ge 3\): \(\frac{(5 - 6.96)^2}{6.96} = 0.5520\)

\(\chi^2 = 0.2397 + 0.2930 + 0.1754 + 0.5520 = 1.26\) (to 3 s.f.)

Degrees of freedom:
Number of classes after combining = 4.
Number of estimated parameters = 1 (mean \(\lambda\)).
\(df = 4 - 1 - 1 = 2\).

Critical value of \(\chi^2\) with 2 degrees of freedom at the 5% level is 5.991.

Since \(1.26 < 5.991\), we fail to reject \(H_0\).
There is insufficient evidence to suggest that the Poisson distribution is not a suitable model.

(i) M1: For calculating the mean as 0.94. M1: For formula of Poisson probabilities. A1: For E0 and E1 correct. A1: For E2 and pooling E>=3 correct. A1: For identifying that no further combining is needed since E>=3 is 6.96.
(ii) B1: For hypotheses. M1: For calculating the chi-square sum. A1: For chi-square value of 1.26. B1: For correct degrees of freedom (2) and critical value (5.991). A1: For correct final conclusion.

(i) Using the property \(G_X(1) = 1\):
\(k(1 + 2(1) + 3(1)^2)^2 = 1 \implies k(6)^2 = 1 \implies 36k = 1 \implies k = \frac{1}{36}\).

To find \(P(X=2)\), we expand \(G_X(t)\):
\(G_X(t) = \frac{1}{36}(1 + 2t + 3t^2)(1 + 2t + 3t^2)\)
\(G_X(t) = \frac{1}{36}(1 + 4t + 10t^2 + 12t^3 + 9t^4)\).

The coefficient of \(t^2\) is the probability \(P(X=2)\):
\(P(X=2) = \frac{10}{36} = \frac{5}{18}\) (or \(0.278\)).

(ii) The first derivative is:
\(G_X'(t) = \frac{1}{36} \cdot 2(1 + 2t + 3t^2)(2 + 6t) = \frac{1}{18}(1 + 2t + 3t^2)(2 + 6t)\).

\(E(X) = G_X'(1) = \frac{1}{18}(1 + 2 + 3)(2 + 6) = \frac{1}{18}(6)(8) = \frac{48}{18} = \frac{8}{3}\) (or \(2.67\)).

The second derivative is:
\(G_X''(t) = \frac{1}{18} \left[ (2+6t)(2+6t) + (1+2t+3t^2)(6) \right]\).

\(G_X''(1) = \frac{1}{18} \left[ 8^2 + 6(6) \right] = \frac{1}{18}(64 + 36) = \frac{100}{18} = \frac{50}{9}\).

\(\text{Var}(X) = G_X''(1) + G_X'(1) - [G_X'(1)]^2\)
\(\text{Var}(X) = \frac{50}{9} + \frac{8}{3} - \left(\frac{8}{3}\right)^2 = \frac{50}{9} + \frac{24}{9} - \frac{64}{9} = \frac{10}{9}\) (or \(1.11\)).

(iii) Since \(Y = X_1 + X_2\) and they are independent:
\(G_Y(t) = [G_X(t)]^2 = \frac{1}{36^2}(1 + 4t + 10t^2 + 12t^3 + 9t^4)^2\).

We need the coefficient of \(t\) in \(G_Y(t)\).
In the expansion of \((1 + 4t + 10t^2 + \dots)^2\), the linear term is \(2 \times 1 \times 4t = 8t\).

Thus, \(P(Y = 1) = \frac{8}{36^2} = \frac{8}{1296} = \frac{1}{162}\) (or \(0.00617\)).

(i) B1: For verifying k = 1/36. M1: For attempting to expand or find the t^2 term. A1: For P(X=2) = 5/18.
(ii) M1: For differentiating the PGF once. A1: For E(X) = 8/3. M1: For differentiating the PGF a second time. A1: For G''(1) = 50/9. A1: For Var(X) = 10/9.
(iii) M1: For identifying P(Y=1) as the coefficient of t in [G(t)]^2. A1: For P(Y=1) = 1/162.

(i) For \(x < 0\), \(F(x) = 0\).
For \(0 \le x \le 2\):
\(F(x) = \int_0^x \frac{3}{8}u^2 du = \left[ \frac{1}{8}u^3 \right]_0^x = \frac{1}{8}x^3\).

For \(x > 2\), \(F(x) = 1\).

(ii) \(Y = \frac{1}{X^2}\). Since the range of \(X\) is \([0, 2]\), the range of \(Y\) is \(Y \ge \frac{1}{4}\).
For \(y \ge \frac{1}{4}\):
\(G(y) = P(Y \le y) = P\left(\frac{1}{X^2} \le y\right) = P\left(X^2 \ge \frac{1}{y}\right)\)

Since \(X\) is non-negative:
\(G(y) = P\left(X \ge \frac{1}{\sqrt{y}}\right) = 1 - F\left(\frac{1}{\sqrt{y}}\right)\)

Substitute the CDF of \(X\):
\(G(y) = 1 - \frac{1}{8}\left(\frac{1}{\sqrt{y}}\right)^3 = 1 - \frac{1}{8y^{3/2}} = 1 - \frac{1}{8 y \sqrt{y}}\).
For \(y < \frac{1}{4}\), \(G(y) = 0\).

(iii) The PDF of \(Y\), \(g(y) = G'(y)\):
For \(y \ge \frac{1}{4}\):
\(g(y) = \frac{d}{dy}\left(1 - \frac{1}{8}y^{-3/2}\right) = \frac{3}{16}y^{-5/2}\).
For \(y < \frac{1}{4}\), \(g(y) = 0\).

Now calculate \(E(Y)\):
\(E(Y) = \int_{1/4}^{\infty} y \cdot \frac{3}{16}y^{-5/2} dy = \int_{1/4}^{\infty} \frac{3}{16}y^{-3/2} dy\)
\(E(Y) = \left[ -\frac{3}{8}y^{-1/2} \right]_{1/4}^{\infty} = 0 - \left( -\frac{3}{8} \sqrt{4} \right) = \frac{6}{8} = 0.75\).

Alternatively, using the "otherwise" method:
\(E(Y) = E\left(\frac{1}{X^2}\right) = \int_0^2 \frac{1}{x^2} \cdot \frac{3}{8}x^2 dx = \int_0^2 \frac{3}{8} dx = \left[ \frac{3}{8}x \right]_0^2 = 0.75\).

(i) M1: For integrating the PDF of X. A1: For correct CDF with ranges.
(ii) M1: For writing P(Y <= y) as P(X^2 >= 1/y). A1: For converting to 1 - F(1/\sqrt{y}). M1: For substitution of CDF of X. A1: For showing the final algebraic form correctly.
(iii) M1: For differentiating the CDF of Y. A1: For correct PDF of Y. M1: For attempting integration to find expectation (or LOTUS). A1: For E(Y) = 0.75.

(i) Let \(m_D\) be the median of the differences (Yield B \(-\) Yield A).
\(H_0: m_D = 0\) (There is no difference in the median yields of the two fertilizers)
\(H_1: m_D > 0\) (Fertilizer B produces a higher median yield than Fertilizer A)

(ii) We compute the difference \(D = \text{Yield B} - \text{Yield A}\) for each pair:
- Pair 1: \(4.5 - 4.2 = +0.3\)
- Pair 2: \(4.3 - 3.8 = +0.5\)
- Pair 3: \(5.0 - 5.1 = -0.1\)
- Pair 4: \(4.6 - 4.0 = +0.6\)
- Pair 5: \(4.2 - 4.4 = -0.2\)
- Pair 6: \(4.7 - 3.9 = +0.8\)
- Pair 7: \(4.8 - 4.5 = +0.3\)
- Pair 8: \(4.9 - 4.8 = +0.1\)
- Pair 9: \(4.5 - 4.1 = +0.4\)
- Pair 10: \(4.4 - 4.6 = -0.2\)

Next, list the absolute differences and rank them:
- Absolute differences: \(0.1, 0.1, 0.2, 0.2, 0.3, 0.3, 0.4, 0.5, 0.6, 0.8\)
- Rank allocation for ties:
- For \(0.1\) (ranks 1, 2): average rank is \(1.5\).
- For \(0.2\) (ranks 3, 4): average rank is \(3.5\).
- For \(0.3\) (ranks 5, 6): average rank is \(5.5\).
- For \(0.4\): rank is \(7\).
- For \(0.5\): rank is \(8\).
- For \(0.6\): rank is \(9\).
- For \(0.8\): rank is \(10\).

Assign signs to the ranks:
- Positive ranks (where \(D > 0\)):
- Pair 8 (diff \(+0.1\), rank \(1.5\))
- Pair 1 (diff \(+0.3\), rank \(5.5\))
- Pair 7 (diff \(+0.3\), rank \(5.5\))
- Pair 9 (diff \(+0.4\), rank \(7\))
- Pair 2 (diff \(+0.5\), rank \(8\))
- Pair 4 (diff \(+0.6\), rank \(9\))
- Pair 6 (diff \(+0.8\), rank \(10\))
- Sum of positive ranks \(T_+ = 1.5 + 5.5 + 5.5 + 7 + 8 + 9 + 10 = 46.5\).

- Negative ranks (where \(D < 0\)):
- Pair 3 (diff \(-0.1\), rank \(1.5\))
- Pair 5 (diff \(-0.2\), rank \(3.5\))
- Pair 10 (diff \(-0.2\), rank \(3.5\))
- Sum of negative ranks \(T_- = 1.5 + 3.5 + 3.5 = 8.5\).

Test statistic \(T = \min(T_+, T_-) = 8.5\).

(iii) The critical value at the 5% significance level for a one-tailed test with \(n=10\) is 10.
Since \(T = 8.5 \le 10\), we reject \(H_0\).

There is sufficient evidence at the 5% level of significance to conclude that the new organic fertilizer (Fertilizer B) increases the crop yield.

(i) B1: For stating H0. B1: For stating a correct one-tailed H1.
(ii) M1: For calculating the differences for all pairs. M1: For ordering and ranking absolute differences correctly, handling ties properly. A1: For correct ranks. M1: For separating and summing positive and negative ranks. A1: For T+ = 46.5 and T- = 8.5. A1: For T = 8.5.
(iii) M1: For comparing the test statistic T with the critical value of 10. A1: For rejecting H0 and drawing the correct conclusion in context.