2025 Cambridge IAL Mathematics - Further (9231) 模擬試題連答案詳解

(i) Starting from the right-hand side: \(\frac{1}{r^3} - \frac{1}{(r+1)^3} = \frac{(r+1)^3 - r^3}{r^3(r+1)^3}\). Expanding the numerator using the binomial expansion: \((r+1)^3 - r^3 = (r^3 + 3r^2 + 3r + 1) - r^3 = 3r^2 + 3r + 1\). Thus, \(\frac{1}{r^3} - \frac{1}{(r+1)^3} = \frac{3r^2+3r+1}{r^3(r+1)^3}\) (as required). (ii) Let \(u_r = \frac{3r^2+3r+1}{r^3(r+1)^3} = \frac{1}{r^3} - \frac{1}{(r+1)^3}\). Using the method of differences: \(\sum_{r=1}^{n} u_r = \sum_{r=1}^{n} \left( \frac{1}{r^3} - \frac{1}{(r+1)^3} \right)\). Writing out the terms of the sum: For \(r=1\): \(1 - \frac{1}{2^3}\); For \(r=2\): \(\frac{1}{2^3} - \frac{1}{3^3}\); For \(r=3\): \(\frac{1}{3^3} - \frac{1}{4^3}\); ... For \(r=n\): \(\frac{1}{n^3} - \frac{1}{(n+1)^3}\). Summing these terms, all intermediate terms cancel: \(\sum_{r=1}^{n} u_r = 1 - \frac{1}{(n+1)^3}\). (iii) As \(n \to \infty\), the term \(\frac{1}{(n+1)^3} \to 0\). Therefore, the sum to infinity is: \(\sum_{r=1}^{\infty} \frac{3r^2+3r+1}{r^3(r+1)^3} = 1\). (iv) We want to find the smallest integer \(N\) such that: \(\sum_{r=N}^{\infty} \frac{3r^2+3r+1}{r^3(r+1)^3} < 10^{-6}\). We can write the sum from \(r=N\) to \(\infty\) as: \(\sum_{r=N}^{\infty} u_r = \lim_{M \to \infty} \sum_{r=N}^{M} \left( \frac{1}{r^3} - \frac{1}{(r+1)^3} \right) = \lim_{M \to \infty} \left( \frac{1}{N^3} - \frac{1}{(M+1)^3} \right) = \frac{1}{N^3}\). We set up the inequality: \(\frac{1}{N^3} < 10^{-6} \implies N^3 > 10^6 \implies N > 100\). Since \(N\) must be an integer, the smallest value is \(N = 101\).

(i) M1: Attempt to combine the RHS over a common denominator. A1: Correctly expand the numerator to obtain the LHS. (ii) M1: For writing down the list of terms showing cancelling structure. M1: For identifying that the sum simplifies to the first and last terms. A1: Correct simplified expression: \(1 - \frac{1}{(n+1)^3}\). (iii) B1: State 1 (follow through on their part (ii)). (iv) M1: Correctly express the sum from \(N\) to infinity as \(\frac{1}{N^3}\) and set up the inequality \(\frac{1}{N^3} < 10^{-6}\). A1: Deduce \(N > 100\) and state \(N = 101\) as the smallest integer.

(i) From the given equation \(x^3 + 2x^2 - 3x + 5 = 0\), the relations between the roots and the coefficients are:
\(\sum \alpha = -2\)
\(\sum \alpha\beta = -3\)
\(\alpha\beta\gamma = -5\)

We know that:
\(\alpha^2 + \beta^2 + \gamma^2 = (\sum \alpha)^2 - 2\sum \alpha\beta\)
Substituting the values:
\(\alpha^2 + \beta^2 + \gamma^2 = (-2)^2 - 2(-3) = 4 + 6 = 10\) (as required).

(ii) We can find the new equation using two main methods.

Method 1: Substitution
Let \(y = x^2\), which implies \(x = y^{1/2}\).
Substituting this into the original equation:
\((y^{1/2})^3 + 2(y^{1/2})^2 - 3y^{1/2} + 5 = 0\)
\(y^{3/2} + 2y - 3y^{1/2} + 5 = 0\)

Group the terms with fractional powers:
\(y^{3/2} - 3y^{1/2} = -(2y + 5)\)
\(y^{1/2}(y - 3) = -(2y + 5)\)

Square both sides:
\(y(y - 3)^2 = (2y + 5)^2\)
\(y(y^2 - 6y + 9) = 4y^2 + 20y + 25\)
\(y^3 - 6y^2 + 9y = 4y^2 + 20y + 25\)
\(y^3 - 10y^2 - 11y - 25 = 0\)

Method 2: Symmetric Functions
Let the new roots be \(A = \alpha^2\), \(B = \beta^2\), and \(C = \gamma^2\).
The new cubic equation is \(y^3 - S_1 y^2 + S_2 y - S_3 = 0\).

From part (i):
\(S_1 = A + B + C = 10\)

For the sum of products of the new roots:
\(S_2 = AB + BC + CA = \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 = (\sum \alpha\beta)^2 - 2\alpha\beta\gamma(\sum \alpha)\)
\(S_2 = (-3)^2 - 2(-5)(-2) = 9 - 20 = -11\)

For the product of the new roots:
\(S_3 = ABC = (\alpha\beta\gamma)^2 = (-5)^2 = 25\)

Substituting these values into the general cubic form:
\(y^3 - 10y^2 - 11y - 25 = 0\)

Part (i):
- M1: For stating \(\sum \alpha = -2\) and \(\sum \alpha\beta = -3\) or using them correctly in the identity.
- A1: For complete and correct substitution showing the result \(10\).

Part (ii):
- M1: For attempting to use a valid method (either substitution \(y=x^2\) or finding symmetric sums of the roots).
- M1: For algebraically correct steps to eliminate fractional powers, e.g., squaring to obtain \(y(y-3)^2 = (2y+5)^2\), OR for using the identity \(\sum \alpha^2\beta^2 = (\sum \alpha\beta)^2 - 2\alpha\beta\gamma(\sum \alpha)\).
- A1: For obtaining \(S_3 = 25\) (or the correct constant term in the substitution method).
- A1: For obtaining \(S_2 = -11\) (or the correct coefficient of \(y\) in the substitution method).
- A1: For the final correct cubic equation: \(y^3 - 10y^2 - 11y - 25 = 0\) (accept any variable name).

Let \(f(n) = 8^n + 6^{2n-1}\). For \(n = 1\), \(f(1) = 8^1 + 6^{2(1)-1} = 8 + 6 = 14\), which is divisible by 14. Thus, the statement is true for \(n = 1\). Assume that the statement is true for \(n = k\), where \(k\) is a positive integer. That is, assume \(f(k) = 8^k + 6^{2k-1} = 14M\) for some integer \(M\). We now consider the case \(n = k + 1\): \(f(k+1) = 8^{k+1} + 6^{2(k+1)-1} = 8 \cdot 8^k + 6^{2k+1} = 8 \cdot 8^k + 36 \cdot 6^{2k-1}\). Substituting \(8^k = 14M - 6^{2k-1}\) from our inductive hypothesis, we obtain: \(f(k+1) = 8(14M - 6^{2k-1}) + 36 \cdot 6^{2k-1} = 112M - 8 \cdot 6^{2k-1} + 36 \cdot 6^{2k-1} = 112M + 28 \cdot 6^{2k-1} = 14(8M + 2 \cdot 6^{2k-1})\). Since \(M\) and \(k\) are positive integers, \(8M + 2 \cdot 6^{2k-1}\) is an integer, meaning \(f(k+1)\) is divisible by 14. Therefore, since the base case is true and the truth of the statement for \(n = k\) implies its truth for \(n = k + 1\), by the principle of mathematical induction, the statement is true for all positive integers \(n\).

B1: Verifies the base case \(n = 1\) showing \(f(1) = 14\) and stating it is divisible by 14. M1: States the inductive hypothesis clearly, assuming \(8^k + 6^{2k-1} = 14M\) for some integer \(M\). A1: Writes down the correct expression for \(f(k+1)\) in terms of \(8^k\) and \(6^{2k-1}\). M1: Integrates the inductive hypothesis into the expression for \(f(k+1)\). A1: Obtains a correct simplified expression showing a factor of 14, such as \(14(8M + 2 \cdot 6^{2k-1})\). A1: Explains why the expression inside the brackets is an integer, showing \(f(k+1)\) is divisible by 14. B1: Provides a complete and logical concluding statement.

(i) Under the transformation represented by \(M\), we have: \\\\begin{pmatrix} X \\\\ Y \\\\end{pmatrix} = \\\\begin{pmatrix} 2 & 1 \\\\ 3 & 4 \\\\end{pmatrix} \\\\begin{pmatrix} x \\\\ y \\\\end{pmatrix} \\\\implies X = 2x + y, \\\\ Y = 3x + 4y\). For an invariant line of the form \(y = mx + c\), the image point \((X, Y)\) must lie on the line \(Y = mX + c\) whenever \((x, y)\) lies on \(y = mx + c\). Substituting \(y = mx + c\) and \(Y = mX + c\) into the transformation equations: \(3x + 4(mx + c) = m(2x + mx + c) + c\). Expanding and grouping terms gives: \((3 + 4m)x + 4c = (2m + m^2)x + mc + c\). Equating the coefficients of \(x\): \(3 + 4m = 2m + m^2 \\\implies m^2 - 2m - 3 = 0 \\\implies (m - 3)(m + 1) = 0\), which gives \(m = 3\) or \(m = -1\). Equating the constant terms: \(4c = mc + c \\\implies (3 - m)c = 0\). For \(m = 3\), this equation becomes \(0 = 0\), which is true for all real values of \(c\). Thus, \(y = 3x + c\) (where \(c\) is any real constant) represents a family of invariant lines. For \(m = -1\), this equation becomes \(4c = 0 \\\implies c = 0\). Thus, \(y = -x\) is an invariant line. (ii) For an invariant point \((x, y)\), we must have \\\\begin{pmatrix} 2 & 1 \\\\ 3 & 4 \\\\end{pmatrix} \\\\begin{pmatrix} x \\\\ y \\\\end{pmatrix} = \\\\begin{pmatrix} x \\\\ y \\\\end{pmatrix}\). This yields the system of equations: \(2x + y = x \\\implies x + y = 0\) and \(3x + 4y = y \\\implies 3x + 3y = 0 \\\implies x + y = 0\). Both equations represent the same line. Thus, the set of invariant points is the line of points satisfying \(y = -x\) (or points of the form \((t, -t)\) for \(t \\\in \\\mathbb{R}\)).

(i) M1: For writing the transformation equations and substituting \(y = mx + c\) and \(Y = mX + c\). M1: For equating the coefficients of \(x\) to form a quadratic in \(m\). A1: For obtaining \(m = 3\) and \(m = -1\). M1: For equating the constant terms to get \((3-m)c = 0\) or equivalent. A1: For correctly identifying the family of lines \(y = 3x + c\) (where \(c\) is any real constant). A1: For correctly identifying the line \(y = -x\). (ii) M1: For setting up the matrix equation for invariant points: \(M\\\mathbf{v} = \\\mathbf{v}\). M1: For expanding this into linear equations in \(x\) and \(y\). A1: For obtaining the final line of invariant points \(y = -x\) (or equivalent form).

(a)
- \( C_1 \) is the upper half of a cardioid, starting at \( (2a, 0) \) on the initial line, passing through \( (a, \frac{\pi}{2}) \), and terminating at the pole \( (0, \pi) \).
- \( C_2 \) is the upper half of a circle of diameter \( 3a \) centered on the initial line, starting at \( (3a, 0) \) and terminating at the pole at \( \theta = \frac{\pi}{2} \).
- They intersect in the first quadrant where \( a(1+\cos\theta) = 3a\cos\theta \implies \cos\theta = \frac{1}{2} \implies \theta = \frac{\pi}{3} \).

(b)
The region in the first quadrant common to both curves is bounded by \( C_1 \) from \( \theta = 0 \) to \( \theta = \frac{\pi}{3} \), and by \( C_2 \) from \( \theta = \frac{\pi}{3} \) to \( \theta = \frac{\pi}{2} \).

The area \( A \) is given by:
\[ A = \frac{1}{2} \int_{0}^{\frac{\pi}{3}} [a(1+\cos\theta)]^2 \, d\theta + \frac{1}{2} \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} [3a\cos\theta]^2 \, d\theta \]

First integral:
\[ I_1 = a^2 \int_{0}^{\frac{\pi}{3}} (1 + 2\cos\theta + \cos^2\theta) \, d\theta \]
\[ = a^2 \int_{0}^{\frac{\pi}{3}} \left( \frac{3}{2} + 2\cos\theta + \frac{1}{2}\cos 2\theta \right) \, d\theta \]
\[ = a^2 \left[ \frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta \right]_{0}^{\frac{\pi}{3}} \]
\[ = a^2 \left( \frac{\pi}{2} + 2\left(\frac{\sqrt{3}}{2}\right) + \frac{1}{4}\left(\frac{\sqrt{3}}{2}\right) - 0 \right) \]
\[ = a^2 \left( \frac{\pi}{2} + \frac{9\sqrt{3}}{8} \right) \]

Second integral:
\[ I_2 = 9a^2 \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \cos^2\theta \, d\theta \]
\[ = 9a^2 \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \left( \frac{1}{2} + \frac{1}{2}\cos 2\theta \right) \, d\theta \]
\[ = \frac{9a^2}{2} \left[ \theta + \frac{1}{2}\sin 2\theta \right]_{\frac{\pi}{3}}^{\frac{\pi}{2}} \]
\[ = \frac{9a^2}{2} \left( \left(\frac{\pi}{2} + 0\right) - \left(\frac{\pi}{3} + \frac{\sqrt{3}}{4}\right) \right) \]
\[ = \frac{9a^2}{2} \left( \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right) = a^2 \left( \frac{3\pi}{4} - \frac{9\sqrt{3}}{8} \right) \]

Adding the two components:
\[ A = \frac{1}{2} I_1 + \frac{1}{2} I_2 \]
\[ = \frac{1}{2} a^2 \left( \frac{\pi}{2} + \frac{9\sqrt{3}}{8} + \frac{3\pi}{4} - \frac{9\sqrt{3}}{8} \right) \]
\[ = \frac{1}{2} a^2 \left( \frac{5\pi}{4} \right) = \frac{5\pi a^2}{8} \]

(c)
The Cartesian coordinate \( x \) on \( C_1 \) is given by:
\[ x = r \cos \theta = a(1 + \cos \theta)\cos \theta = a(\cos \theta + \cos^2 \theta) \]

The tangent is perpendicular to the initial line when \( \frac{dx}{d\theta} = 0 \):
\[ \frac{dx}{d\theta} = a(-\sin \theta - 2\cos\theta\sin\theta) = -a\sin\theta(1 + 2\cos\theta) = 0 \]

Since \( 0 < \theta < \pi \), we have \( \sin \theta \neq 0 \).
Thus, \( 1 + 2\cos\theta = 0 \implies \cos\theta = -\frac{1}{2} \).
Within the interval, this gives \( \theta = \frac{2\pi}{3} \).

Substituting this back into the equation for \( C_1 \):
\[ r = a\left(1 + \cos\frac{2\pi}{3}\right) = a\left(1 - \frac{1}{2}\right) = \frac{1}{2}a \]

Thus, the polar coordinates of the point are \( \left(\frac{1}{2}a, \frac{2\pi}{3}\right) \).

(a)
- B1: Correct sketch of the upper half-cardioid \( C_1 \) showing starting at \( (2a, 0) \) and ending at the pole.
- B1: Correct sketch of the upper semicircle \( C_2 \) starting at \( (3a, 0) \) and ending at the pole.
- B1: Clearly indicating the pole and the crossing points of both curves on the initial line.

(b)
- M1: Equating \( C_1 \) and \( C_2 \) to find the correct intersection angle \( \theta = \frac{\pi}{3} \).
- M1: Stating the correct split-integral formula for the area.
- A1: Correct integration of \( (1+\cos\theta)^2 \) with double-angle formula.
- A1: Correct integration of \( \cos^2\theta \) with double-angle formula.
- M1: Applying limits \( 0 \) to \( \frac{\pi}{3} \) and \( \frac{\pi}{3} \) to \( \frac{\pi}{2} \) correctly.
- A1: Showing cancellation of \( \sqrt{3} \) terms to arrive at \( \frac{5\pi a^2}{8} \).

(c)
- M1: Writing down \( x = r\cos\theta \) for \( C_1 \).
- A1: Correctly differentiating to find \( \frac{dx}{d\theta} = -a\sin\theta(1+2\cos\theta) \).
- M1: Setting \( \frac{dx}{d\theta} = 0 \) and identifying \( \cos\theta = -\frac{1}{2} \) as the only solution in the interval.
- A1: Obtaining the correct polar coordinates \( \left(\frac{1}{2}a, \frac{2\pi}{3}\right) \).

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

Now assume that \(l_1\) and \(l_2\) intersect. Equating the coordinates:
1) \(1 + 2\lambda = 3 - \mu \implies 2\lambda + \mu = 2\)
2) \(2 + \lambda = -1 + 2\mu \implies \lambda - 2\mu = -3\)
3) \(-1 + 3\lambda = 2 + \mu \implies 3\lambda - \mu = 3\)

Solving (1) and (2):
Multiply (2) by 2: \(2\lambda - 4\mu = -6\)
Subtracting this from (1) gives \(5\mu = 8 \implies \mu = 1.6\).
Then, \(\lambda = 2(1.6) - 3 = 0.2\).

Substituting these values into (3):
\text{LHS} = 3(0.2) - 1.6 = -1.0\\
\text{RHS} = 3\\
Since \(\text{LHS} \neq \text{RHS}\), the system of equations is inconsistent. Therefore, the lines do not intersect.
Since they are neither parallel nor intersecting, \(l_1\) and \(l_2\) are skew.

**(b)**
The common perpendicular vector \(\mathbf{n}\) to both lines is:
\[\mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2 = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix} \times \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 1(1) - 3(2) \\ 3(-1) - 2(1) \\ 2(2) - 1(-1) \end{pmatrix} = \begin{pmatrix} -5 \\ -5 \\ 5 \end{pmatrix}\]
We can use the simplified normal vector \(\mathbf{n}' = \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}\).

A vector connecting a point on each line is:
\[\mathbf{a}_2 - \mathbf{a}_1 = \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \\ 3 \end{pmatrix}\]

The shortest distance \(d\) is the projection of this vector onto the normal vector:
\[d = \frac{|(\mathbf{a}_2 - \mathbf{a}_1) \cdot \mathbf{n}'|}{||\mathbf{n}'||} = \frac{|2(1) - 3(1) + 3(-1)|}{\sqrt{1^2 + 1^2 + (-1)^2}} = \frac{|2 - 3 - 3|}{\sqrt{3}} = \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3} \approx 2.31\]

**(c)**
The normal to the plane \(\Pi_1\) is \(\mathbf{n}' = \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}\).
Since \(\Pi_1\) contains \(l_1\), it contains the point \((1, 2, -1)\).

The equation of the plane is:
\[\mathbf{r} \cdot \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} = 1(1) + 2(1) - 1(-1) = 4\]

Thus, the Cartesian equation of \(\Pi_1\) is:
\[x + y - z = 4\]

**(d)**
Since \(\Pi_2\) is perpendicular to \(l_1\), its normal vector is the direction vector of \(l_1\), i.e., \(\mathbf{d}_1 = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}\).
Since it contains the point \(A(1, 0, 1)\), the equation of \(\Pi_2\) is:
\[2x + y + 3z = 2(1) + 1(0) + 3(1) = 5\]

Any point on \(l_2\) can be written parametrically as:
\[x = 3 - \mu, \quad y = -1 + 2\mu, \quad z = 2 + \mu\]

Substituting these expressions into the equation of \(\Pi_2\):
\[2(3 - \mu) + (-1 + 2\mu) + 3(2 + \mu) = 5\]
\[6 - 2\mu - 1 + 2\mu + 6 + 3\mu = 5\]
\[11 + 3\mu = 5 \implies 3\mu = -6 \implies \mu = -2\]

Substituting \(\mu = -2\) back into the parametric equations of \(l_2\):
\[x = 3 - (-2) = 5\]
\[y = -1 + 2(-2) = -5\]
\[z = 2 + (-2) = 0\]

Hence, the coordinates of the point of intersection are \((5, -5, 0)\).

**(a)**
* **B1**: Explains clearly why the lines are not parallel (e.g., showing direction vectors are not scalar multiples).
* **M1**: Sets up the three simultaneous equations in \(\lambda\) and \(\mu\).
* **A1**: Solves a pair of equations to find values for both parameters (e.g., \(\lambda = 0.2\), \(\mu = 1.6\)).
* **A1**: Shows that these values do not satisfy the third equation and concludes that the lines are skew.

**(b)**
* **M1**: Attempts to find the cross product of the two direction vectors.
* **A1**: Obtains a correct normal vector, e.g., \(\begin{pmatrix} -5 \\ -5 \\ 5 \end{pmatrix}\) or any non-zero multiple.
* **M1**: Uses a correct formula for shortest distance, using their normal vector and a vector connecting the two lines.
* **A1**: Obtains \(\frac{4}{\sqrt{3}}\) (or \(\frac{4\sqrt{3}}{3}\) or 2.31).

**(c)**
* **M1**: Employs the normal vector from (b) to write the equation of the plane in the form \(\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}\).
* **M1**: Substitutes a point from \(l_1\) (such as \((1, 2, -1)\)) to evaluate the constant.
* **A1**: Arrives at the correct Cartesian equation: \(x + y - z = 4\) (or any equivalent form).

**(d)**
* **B1**: Identifies the normal vector of \(\Pi_2\) as \(\begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}\).
* **A1**: Finds the correct equation for \(\Pi_2\): \(2x + y + 3z = 5\).
* **M1**: Substitutes the parametric form of \(l_2\) into their equation for \(\Pi_2\).
* **A1**: Solves for the parameter correctly to find \(\mu = -2\).
* **A1**: Obtains the correct coordinates \((5, -5, 0)\).

(a) Express the algebraic fraction in quotient-remainder form using division:
\[ y = \frac{x(x - 1) - (x - 1) + 4}{x - 1} = x - 1 + \frac{4}{x - 1} \]
As $x \to \pm\infty$, $\frac{4}{x - 1} \to 0$, which gives the oblique asymptote:
\[ y = x - 1 \]
As $x \to 1$, $y \to \pm\infty$, which gives the vertical asymptote:
\[ x = 1 \]

(b) Differentiate $y$ with respect to $x$:
\[ \frac{dy}{dx} = 1 - \frac{4}{(x-1)^2} \]
Setting $\frac{dy}{dx} = 0$:
\[ 1 - \frac{4}{(x-1)^2} = 0 \implies (x-1)^2 = 4 \implies x - 1 = \pm 2 \]
This yields two critical values:
- For $x = 3$: $y = 3 - 1 + \frac{4}{2} = 4$. The stationary point is $(3, 4)$.
- For $x = -1$: $y = -1 - 1 + \frac{4}{-2} = -4$. The stationary point is $(-1, -4)$.

To determine their nature, find the second derivative:
\[ \frac{d^2y}{dx^2} = \frac{8}{(x-1)^3} \]
- At $x = 3$: $\frac{d^2y}{dx^2} = \frac{8}{8} = 1 > 0$, so $(3, 4)$ is a local minimum.
- At $x = -1$: $\frac{d^2y}{dx^2} = \frac{8}{-8} = -1 < 0$, so $(-1, -4)$ is a local maximum.

(c) Sketch the graph with the following features:
- Dashed lines representing asymptotes at $x = 1$ and $y = x - 1$.
- $y$-intercept at $(0, -5)$, found by setting $x = 0$ in the equation of $C$.
- Local minimum plotted and labeled at $(3, 4)$.
- Local maximum plotted and labeled at $(-1, -4)$.
- Two branches drawn: one in the upper-right region asymptotic to $x=1$ and $y=x-1$, and one in the lower-left region asymptotic to $x=1$ and $y=x-1$, passing through $(0, -5)$.

(d) The inequality $\left| \frac{x^2 - 2x + 5}{x - 1} \right| < 5$ is equivalent to:
\[ -5 < \frac{x^2 - 2x + 5}{x - 1} < 5 \]
First, solve for boundary values:
- Case 1: $\frac{x^2 - 2x + 5}{x - 1} = 5 \implies x^2 - 2x + 5 = 5x - 5 \implies x^2 - 7x + 10 = 0 \implies (x-2)(x-5) = 0 \implies x = 2$ or $x = 5$.
- Case 2: $\frac{x^2 - 2x + 5}{x - 1} = -5 \implies x^2 - 2x + 5 = -5x + 5 \implies x^2 + 3x = 0 \implies x(x+3) = 0 \implies x = 0$ or $x = -3$.

By comparing with the sketched graph of $C$:
- For $x \in (-3, 0)$, the curve lies strictly between $y = -5$ and $y = 0$, so $|y| < 5$.
- For $x \in (2, 5)$, the curve lies strictly between the minimum $y = 4$ and $y = 5$, so $|y| < 5$.

Thus, the solution set is:
\[ -3 < x < 0 \quad \text{or} \quad 2 < x < 5 \]

*(a)*
**M1**: Attempting algebraic division or writing the equation in the form $y = Ax + B + \frac{C}{x-1}$.
**A1**: Correct oblique asymptote $y = x - 1$.
**A1**: Correct vertical asymptote $x = 1$.

*(b)*
**M1**: Correct method to differentiate $y$ (using quotient rule or from the simplified form).
**M1**: Setting the derivative equal to zero and attempting to solve for $x$.
**A1**: Finding correct coordinates $(3, 4)$ and $(-1, -4)$.
**M1**: Finding the second derivative and substituting $x$-values to check nature (or using first-derivative test).
**A1**: Concluding correctly that $(3, 4)$ is a local minimum and $(-1, -4)$ is a local maximum.

*(c)*
**B1**: Drawing correct dashed lines for asymptotes $x = 1$ and $y = x - 1$.
**B1**: Correctly labeling the $y$-intercept at $(0, -5)$.
**B1**: Labeling correct coordinates for stationary points $(3, 4)$ and $(-1, -4)$.
**B1**: Correctly shaped branches approaching both asymptotes without crossing them.

*(d)*
**M1**: Setting up and solving the equation $y = 5$ to get $x = 2$ and $x = 5$.
**M1**: Setting up and solving the equation $y = -5$ to get $x = 0$ and $x = -3$.
**A1**: Deducing the correct range from the graph or intervals: $-3 < x < 0$ or $2 < x < 5$ (allow equivalent notation).

First, we express the complex number \(w = -16\sqrt{3} + 16i\) in exponential form.

The modulus is:
\[|w| = \sqrt{(-16\sqrt{3})^2 + 16^2} = \sqrt{768 + 256} = \sqrt{1024} = 32\]

Since \(w\) lies in the second quadrant, its argument is:
\[\theta = \pi - \arctan\left(\frac{16}{16\sqrt{3}}\right) = \pi - \frac{\pi}{6} = \frac{5\pi}{6}\]

Thus, we can write the equation as:
\[z^5 = 32 e^{i\left(\frac{5\pi}{6} + 2k\pi\right)}\quad \text{for } k \in \mathbb{Z}\]

Applying de Moivre's theorem to find the fifth roots gives:
\[z = 32^{1/5} e^{i \left(\frac{\frac{5\pi}{6} + 2k\pi}{5}\right)} = 2 e^{i \left(\frac{\pi}{6} + \frac{2k\pi}{5}\right)}\]

We choose five consecutive integers for \(k\) to obtain the roots in the principal range \(-\pi < \theta \le \pi\), for example, \(k = -2, -1, 0, 1, 2\):

- For \(k = 0\): \(z_0 = 2 e^{i \frac{\pi}{6}}\)
- For \(k = 1\): \(z_1 = 2 e^{i \left(\frac{\pi}{6} + \frac{2\pi}{5}\right)} = 2 e^{i \frac{17\pi}{30}}\)
- For \(k = 2\): \(z_2 = 2 e^{i \left(\frac{\pi}{6} + \frac{4\pi}{5}\right)} = 2 e^{i \frac{29\pi}{30}}\)
- For \(k = -1\): \(z_{-1} = 2 e^{i \left(\frac{\pi}{6} - \frac{2\pi}{5}\right)} = 2 e^{-i \frac{7\pi}{30}}\)
- For \(k = -2\): \(z_{-2} = 2 e^{i \left(\frac{\pi}{6} - \frac{4\pi}{5}\right)} = 2 e^{-i \frac{19\pi}{30}}\)

All of these arguments lie within the interval \((-\pi, \pi]\).

M1: For finding the correct modulus \(r = 32\) and argument \(\theta = \frac{5\pi}{6}\) of \(-16\sqrt{3} + 16i\).
A1: For applying de Moivre's theorem to obtain the general root form \(z = 2 e^{i \left(\frac{\pi}{6} + \frac{2k\pi}{5}\right)}\).
M1: For attempting to find five distinct roots by substituting five different integer values of \(k\).
A1: For obtaining at least three correct roots with arguments in the range \((-\pi, \pi]\).
A1: For finding all five correct roots with arguments in the range \((-\pi, \pi]\): \(2e^{i\frac{\pi}{6}}\), \(2e^{i\frac{17\pi}{30}}\), \(2e^{i\frac{29\pi}{30}}\), \(2e^{-i\frac{7\pi}{30}}\), \(2e^{-i\frac{19\pi}{30}}\).

(a) Let \( u = x^n \implies \frac{du}{dx} = n x^{n-1} \) and \( \frac{dv}{dx} = (1-x)^{1/2} \implies v = -\frac{2}{3}(1-x)^{3/2} \). Using integration by parts: \( I_n = \left[ -\frac{2}{3} x^n (1-x)^{3/2} \right]_0^1 + \frac{2}{3} n \int_{0}^{1} x^{n-1} (1-x)^{3/2} dx \). For \( n \ge 1 \), the boundary term evaluated at both limits is 0. Thus, \( I_n = \frac{2n}{3} \int_{0}^{1} x^{n-1} (1-x)\sqrt{1-x} \, dx = \frac{2n}{3} \left( \int_{0}^{1} x^{n-1}\sqrt{1-x} \, dx - \int_{0}^{1} x^n\sqrt{1-x} \, dx \right) \). This gives \( I_n = \frac{2n}{3}(I_{n-1} - I_n) \). Multiplying by 3: \( 3I_n = 2n I_{n-1} - 2n I_n \implies (2n+3)I_n = 2n I_{n-1} \implies I_n = \frac{2n}{2n+3} I_{n-1} \). (b) We calculate \( I_0 = \int_{0}^{1} \sqrt{1-x} \, dx = \left[ -\frac{2}{3}(1-x)^{3/2} \right]_0^1 = \frac{2}{3} \). Using the reduction formula: \( I_1 = \frac{2}{5} I_0 = \frac{2}{5} \times \frac{2}{3} = \frac{4}{15} \). Then, \( I_2 = \frac{4}{7} I_1 = \frac{4}{7} \times \frac{4}{15} = \frac{16}{105} \).

Part (a): M1: For applying integration by parts to \( I_n \). A1: For obtaining \( I_n = \frac{2n}{3} \int_{0}^{1} x^{n-1} (1-x)^{3/2} dx \) with the boundary term vanishing. M1: For expressing \( (1-x)^{3/2} \) as \( (1-x)\sqrt{1-x} \) and obtaining an equation in terms of \( I_n \) and \( I_{n-1} \). A1: For fully correct algebraic work leading to the given reduction formula. Part (b): M1: For finding \( I_0 \) (or \( I_1 \)) by direct integration. M1: For using the reduction formula to express \( I_2 \) in terms of \( I_0 \) (or \( I_1 \)). A1: For the correct exact value of \( I_2 = \frac{16}{105} \).

We use de Moivre's theorem: \[ \cos 5\theta + i\sin 5\theta = (\cos\theta + i\sin\theta)^5 \] Expanding the right-hand side using the binomial theorem: \[ (\cos\theta + i\sin\theta)^5 = \cos^5\theta + 5i\cos^4\theta\sin\theta - 10\cos^3\theta\sin^2\theta - 10i\cos^2\theta\sin^3\theta + 5\cos\theta\sin^4\theta + i\sin^5\theta \] Equating the imaginary parts of both sides: \[ \sin 5\theta = 5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta \] For \(\sin\theta \neq 0\), we divide both sides by \(\sin\theta\): \[ \frac{\sin 5\theta}{\sin\theta} = 5\cos^4\theta - 10\cos^2\theta\sin^2\theta + \sin^4\theta \] Using the identity \(\sin^2\theta = 1 - \cos^2\theta\): \[ \frac{\sin 5\theta}{\sin\theta} = 5\cos^4\theta - 10\cos^2\theta(1 - \cos^2\theta) + (1 - \cos^2\theta)^2 \] \[ = 5\cos^4\theta - 10\cos^2\theta + 10\cos^4\theta + 1 - 2\cos^2\theta + \cos^4\theta \] \[ = 16\cos^4\theta - 12\cos^2\theta + 1 \] This completes the proof of the identity. To solve \(\frac{\sin 5\theta}{\sin\theta} = 1\): \[ 16\cos^4\theta - 12\cos^2\theta + 1 = 1 \] \[ 16\cos^4\theta - 12\cos^2\theta = 0 \] \[ 4\cos^2\theta(4\cos^2\theta - 3) = 0 \] Since \(0 < \theta < \pi\), we have \(\cos\theta = 0\) or \(\cos\theta = \pm\frac{\sqrt{3}}{2}\). This gives: For \(\cos\theta = 0 \implies \theta = \frac{\pi}{2}\), for \(\cos\theta = \frac{\sqrt{3}}{2} \implies \theta = \frac{\pi}{6}\), and for \(\cos\theta = -\frac{\sqrt{3}}{2} \implies \theta = \frac{5\pi}{6}\). Thus, the solutions in the given interval are \(\theta = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}\).

M1: Use de Moivre's theorem and attempt binomial expansion of \((\cos\theta + i\sin\theta)^5\). A1: Obtain correct imaginary part for \(\sin 5\theta\). M1: Divide by \(\sin\theta\) and substitute \(\sin^2\theta = 1 - \cos^2\theta\) to obtain an expression in terms of \(\cos\theta\) only. A1: Correctly show the given identity. M1: Set the expression to 1, factorise, and attempt to solve for \(\cos\theta\). A1: Obtain all three correct solutions \(\theta = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}\) and no others in the interval.

(a) Consider the curve \( y = f(x) = \frac{1}{\sqrt{x}} \) for \( x > 0 \). Since \( f'(x) = -\frac{1}{2}x^{-3/2} < 0 \, \), the function \( f(x) \) is strictly decreasing.
By drawing rectangles of width 1 and height \( f(r) \) on each interval \( [r, r+1] \) for \( r = 1, 2, \dots, n \), the area of these rectangles is greater than the area under the curve \( y = f(x) \) over \( [1, n+1] \).
Thus,
\[ \sum_{r=1}^n f(r) \cdot 1 > \int_1^{n+1} f(x) \, dx \]
\[ \sum_{r=1}^n \frac{1}{\sqrt{r}} > \int_1^{n+1} x^{-1/2} \, dx \]
\[ \sum_{r=1}^n \frac{1}{\sqrt{r}} > \left[ 2\sqrt{x} \right]_1^{n+1} = 2\sqrt{n+1} - 2. \]

(b) Consider the interval \( [r-1, r] \) for \( r = 2, 3, \dots, n \). Since \( f(x) \) is strictly decreasing, the minimum value on each subinterval occurs at the right-hand endpoint, \( x = r \).
Thus, the area of the rectangle of width 1 and height \( f(r) \) is less than the area under the curve over \( [r-1, r] \).
\[ f(r) \cdot 1 < \int_{r-1}^r f(x) \, dx \]
Summing both sides from \( r = 2 \) to \( n \):
\[ \sum_{r=2}^n \frac{1}{\sqrt{r}} < \sum_{r=2}^n \int_{r-1}^r x^{-1/2} \, dx = \int_1^n x^{-1/2} \, dx \]
Evaluating the integral:
\[ \sum_{r=2}^n \frac{1}{\sqrt{r}} < \left[ 2\sqrt{x} \right]_1^n = 2\sqrt{n} - 2 \]
Adding the first term \( 1 \) (for \( r = 1 \)) to both sides:
\[ \sum_{r=1}^n \frac{1}{\sqrt{r}} < 1 + 2\sqrt{n} - 2 = 2\sqrt{n} - 1. \]

(c) Substituting \( n = 100 \) into both inequalities:
From part (a):
\[ \sum_{r=1}^{100} \frac{1}{\sqrt{r}} > 2\sqrt{101} - 2 \]
Since \( 101 > 100 \), we have \( \sqrt{101} > 10 \), which gives:
\[ 2\sqrt{101} - 2 > 2(10) - 2 = 18. \]
From part (b):
\[ \sum_{r=1}^{100} \frac{1}{\sqrt{r}} < 2\sqrt{100} - 1 = 2(10) - 1 = 19. \]
Combining these bounds:
\[ 18 < \sum_{r=1}^{100} \frac{1}{\sqrt{r}} < 19. \]
Thus, the unique integer is \( k = 18 \).

Part (a):
* **M1**: For identifying a correct set of rectangles and setting up the inequality \( \sum_{r=1}^n \frac{1}{\sqrt{r}} > \int_1^{n+1} \frac{1}{\sqrt{x}} \, dx \) (or equivalent explanation/diagram).
* **M1**: For evaluating the integral to get \( [2\sqrt{x}]_1^{n+1} = 2\sqrt{n+1} - 2 \).
* **A1**: For obtaining the given result strictly and clearly.

Part (b):
* **M1**: For setting up the inequality for \( r \ge 2 \): \( \sum_{r=2}^n \frac{1}{\sqrt{r}} < \int_1^n \frac{1}{\sqrt{x}} \, dx \).
* **M1**: For evaluating the integral to \( 2\sqrt{n} - 2 \) and adding the term for \( r = 1 \).
* **A1**: For obtaining the given upper bound \( 2\sqrt{n} - 1 \) with clear steps.

Part (c):
* **M1**: For substituting \( n = 100 \) into the lower bound and demonstrating that \( 2\sqrt{101} - 2 > 18 \).
* **M1**: For substituting \( n = 100 \) into the upper bound to get \( 19 \).
* **A1**: For concluding that the unique integer is \( k = 18 \).

First, we find the complementary function (CF) by solving the auxiliary equation:
\[ m^2 + 4m + 5 = 0 \]
Using the quadratic formula:
\[ m = \frac{-4 \pm \sqrt{16 - 20}}{2} = -2 \pm i \]
Since the roots are complex, the complementary function is:
\[ y_c = e^{-2x} (A \cos x + B \sin x) \]

Next, we find the particular integral (PI). Since the right-hand side is \(10e^{-x} + 8 \cos x + 16 \sin x\), we propose a trial particular integral of the form:
\[ y_p = C e^{-x} + D \cos x + E \sin x \]
Differentiating this trial solution, we get:
\[ y_p' = -C e^{-x} - D \sin x + E \cos x \]
\[ y_p'' = C e^{-x} - D \cos x - E \sin x \]
Substituting these expressions into the original differential equation:
\[ (C e^{-x} - D \cos x - E \sin x) + 4(-C e^{-x} - D \sin x + E \cos x) + 5(C e^{-x} + D \cos x + E \sin x) = 10 e^{-x} + 8 \cos x + 16 \sin x \]
Grouping like terms:
\[ (C - 4C + 5C)e^{-x} + (-D + 4E + 5D)\cos x + (-E - 4D + 5E)\sin x = 10 e^{-x} + 8 \cos x + 16 \sin x \]
\[ 2C e^{-x} + (4D + 4E)\cos x + (4E - 4D)\sin x = 10 e^{-x} + 8 \cos x + 16 \sin x \]
Comparing coefficients on both sides:
- For \(e^{-x}\):
\[ 2C = 10 \implies C = 5 \]
- For \(\cos x\):
\[ 4D + 4E = 8 \implies D + E = 2 \]
- For \(\sin x\):
\[ 4E - 4D = 16 \implies -D + E = 4 \]
Adding the two trigonometric equations gives:
\[ 2E = 6 \implies E = 3 \]
Subtracting the first from the second gives:
\[ 2D = -2 \implies D = -1 \]
So the particular integral is:
\[ y_p = 5e^{-x} - \cos x + 3 \sin x \]

The general solution is the sum of the complementary function and the particular integral:
\[ y = e^{-2x} (A \cos x + B \sin x) + 5e^{-x} - \cos x + 3 \sin x \]

Now, we apply the initial conditions to find the constants \(A\) and \(B\).
Using \(y = 4\) when \(x = 0\):
\[ 4 = e^0 (A \cos 0 + B \sin 0) + 5e^0 - \cos 0 + 3 \sin 0 \]
\[ 4 = A + 5 - 1 \implies A = 0 \]

Thus, the solution simplifies to:
\[ y = B e^{-2x} \sin x + 5e^{-x} - \cos x + 3 \sin x \]

Differentiating this with respect to \(x\):
\[ \frac{dy}{dx} = -2B e^{-2x} \sin x + B e^{-2x} \cos x - 5e^{-x} + \sin x + 3 \cos x \]
Using \(\frac{dy}{dx} = 0\) when \(x = 0\):
\[ 0 = B - 5 + 3 \implies B = 2 \]

Therefore, the particular solution is:
\[ y = 2 e^{-2x} \sin x + 5e^{-x} - \cos x + 3 \sin x \]

- **M1**: For setting up and solving the auxiliary equation \(m^2 + 4m + 5 = 0\).
- **A1**: For obtaining the correct complementary function: \(y_c = e^{-2x}(A\cos x + B\sin x)\).
- **M1**: For proposing a trial particular integral of the form \(y_p = C e^{-x} + D \cos x + E \sin x\) and substituting it into the differential equation.
- **A1**: For finding \(C = 5\).
- **A1**: For finding \(D = -1\) and \(E = 3\).
- **A1ft**: For writing the correct general solution as the sum of their CF and PI.
- **M1**: For applying \(y = 4\) at \(x = 0\) to find \(A\).
- **A1**: For obtaining \(A = 0\).
- **M1**: For differentiating their general solution and applying \(\frac{dy}{dx} = 0\) at \(x = 0\) to find \(B\).
- **A1**: For obtaining \(B = 2\) and writing the final particular solution: \(y = 2e^{-2x} \sin x + 5e^{-x} - \cos x + 3 \sin x\).

For (a): We differentiate \( x \) and \( y \) with respect to \( t \): \( \frac{\mathrm{d}x}{\mathrm{d}t} = 3\cosh^2 t \sinh t - 3\sinh t = 3\sinh t (\cosh^2 t - 1) = 3\sinh^3 t \) and \( \frac{\mathrm{d}y}{\mathrm{d}t} = 6\sinh t \cosh t \). Thus, by the chain rule, \( \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t} = \frac{6\sinh t \cosh t}{3\sinh^3 t} = 2\cosh t \sinh^{-2} t \). For (b): Using \( \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = \frac{\mathrm{d}/\mathrm{d}t(\mathrm{d}y/\mathrm{d}x)}{\mathrm{d}x/\mathrm{d}t} \), we first differentiate \( \frac{\mathrm{d}y}{\mathrm{d}x} \) with respect to \( t \): \( \frac{\mathrm{d}}{\mathrm{d}t}(2\cosh t \sinh^{-2} t) = 2\sinh t \sinh^{-2} t - 4\cosh^2 t \sinh^{-3} t = \frac{2\sinh^2 t - 4\cosh^2 t}{\sinh^3 t} \). Using \( \sinh^2 t = \cosh^2 t - 1 \), this simplifies to \( \frac{2(\cosh^2 t - 1) - 4\cosh^2 t}{\sinh^3 t} = -\frac{2(\cosh^2 t + 1)}{\sinh^3 t} \). Dividing by \( \frac{\mathrm{d}x}{\mathrm{d}t} = 3\sinh^3 t \) gives \( \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = -\frac{2(\cosh^2 t + 1)}{3\sinh^6 t} \). For (c): At \( t = \ln 3 \), we have \( \cosh(\ln 3) = \frac{5}{3} \) and \( \sinh(\ln 3) = \frac{4}{3} \). The coordinates are \( x = \left(\frac{5}{3}\right)^3 - 3\left(\frac{5}{3}\right) = -\frac{10}{27} \) and \( y = 3\left(\frac{4}{3}\right)^2 = \frac{16}{3} \). The gradient of the tangent is \( m = \frac{2(5/3)}{(4/3)^2} = \frac{15}{8} \). The equation of the tangent is \( y - \frac{16}{3} = \frac{15}{8}\left(x + \frac{10}{27}\right) \). At \( x = 0 \), \( y = \frac{16}{3} + \frac{15}{8}\left(\frac{10}{27}\right) = \frac{217}{36} \). For (d): The arc length is \( s = \int_{0}^{\ln 3} \sqrt{(\mathrm{d}x/\mathrm{d}t)^2 + (\mathrm{d}y/\mathrm{d}t)^2} \mathrm{d}t \). Here, \( (\mathrm{d}x/\mathrm{d}t)^2 + (\mathrm{d}y/\mathrm{d}t)^2 = 9\sinh^6 t + 36\sinh^2 t \cosh^2 t = 9\sinh^2 t (\sinh^4 t + 4\cosh^2 t) \). Since \( \cosh^2 t = 1 + \sinh^2 t \), we have \( \sinh^4 t + 4\sinh^2 t + 4 = (\sinh^2 t + 2)^2 \). The integrand is \( 3\sinh t (\sinh^2 t + 2) = 3\sinh t \cosh^2 t + 3\sinh t \). The integral is \( [\cosh^3 t + 3\cosh t]_{0}^{\ln 3} = \left(\frac{125}{27} + 5\right) - (1 + 3) = \frac{260}{27} - 4 = \frac{152}{27} \).

For (a): M1: differentiate x and y with respect to t. A1: correct dx/dt and dy/dt. A1: correct division and simplification to show given result. For (b): M1: apply quotient/product rule to dy/dx. A1: correct derivative with respect to t. M1: divide by dx/dt and apply hyperbolic identity. A1: obtain given expression for second derivative. For (c): M1: calculate cosh(ln 3) and sinh(ln 3) to find coordinates x and y. A1: correct values of x, y, and gradient m. M1: find tangent equation and substitute x = 0. A1: correct y-coordinate of 217/36. For (d): M1: substitute derivatives into the arc length formula. M1: simplify the term inside the square root to a perfect square. A1: integrate to get cosh^3 t + 3cosh t. A1: substitute limits to obtain the exact value 152/27.

(i) The differential equation is in the first-order linear form \(\frac{dy}{dx} + P(x)y = Q(x)\), where \(P(x) = \frac{2x}{1+x^2}\) and \(Q(x) = \frac{3x}{\sqrt{1+x^2}}\).

The integrating factor is given by:
\[ I(x) = e^{\int P(x) \, dx} = e^{\int \frac{2x}{1+x^2} \, dx} = e^{\ln(1+x^2)} = 1+x^2 \]

Multiplying the entire differential equation by the integrating factor \(1+x^2\):
\[ (1+x^2)\frac{dy}{dx} + 2xy = 3x\sqrt{1+x^2} \]

\[ \frac{d}{dx} [y(1+x^2)] = 3x(1+x^2)^{1/2} \]

Integrating both sides with respect to \(x\):
\[ y(1+x^2) = \int 3x(1+x^2)^{1/2} \, dx \]

Using the substitution \(u = 1+x^2\), we have \(du = 2x \, dx\), which gives:
\[ y(1+x^2) = (1+x^2)^{3/2} + C \]

Dividing both sides by \(1+x^2\) yields the general solution:
\[ y = \sqrt{1+x^2} + \frac{C}{1+x^2} \]

(ii) Substituting the initial condition \(y = 3\) when \(x = 0\):
\[ 3 = \sqrt{1+0^2} + \frac{C}{1+0^2} \implies 3 = 1 + C \implies C = 2 \]

The particular solution is:
\[ y = \sqrt{1+x^2} + \frac{2}{1+x^2} \]

(iii) Substituting \(x = \sqrt{3}\) into the particular solution:
\[ y = \sqrt{1 + (\sqrt{3})^2} + \frac{2}{1 + (\sqrt{3})^2} = \sqrt{4} + \frac{2}{4} = 2 + 0.5 = 2.5 \text{ (or } \frac{5}{2}\text{)} \]

(i)
* **M1**: Attempt to find the integrating factor by integrating \(P(x)\).
* **A1**: Obtain the correct integrating factor \(1+x^2\).
* **M1**: Multiply through by their integrating factor and express LHS in the form \(\frac{d}{dx}[y \cdot I(x)]\).
* **M1**: Attempt to integrate the RHS, of the form \(kx(1+x^2)^{1/2}\).
* **A1**: Correctly obtain \((1+x^2)^{3/2}\).
* **B1**: Include the constant of integration \(C\).
* **A1**: Express the general solution in the form \(y = \sqrt{1+x^2} + \frac{C}{1+x^2}\) or equivalent.

(ii)
* **M1**: Substitute \(x = 0, y = 3\) into their general solution to find \(C\).
* **A1**: Correct particular solution \(y = \sqrt{1+x^2} + \frac{2}{1+x^2}\).

(iii)
* **B1**: State the correct exact value \(2.5\) or \(\frac{5}{2}\).

(i) To find the eigenvalues of \(A\), we solve the characteristic equation \(\det(A - \lambda I) = 0\):
\(\det \begin{pmatrix} -\lambda & 2 & 1 \\ -3 & 5-\lambda & 3 \\ -2 & 2 & 3-\lambda \end{pmatrix} = 0\)

Expanding along the first row:
\(-\lambda ((5-\lambda)(3-\lambda) - 6) - 2 (-3(3-\lambda) + 6) + 1 (-6 - 2(5-\lambda)) = 0\)
\(-\lambda (\lambda^2 - 8\lambda + 9) - 2 (3\lambda - 3) - 2\lambda + 4 = 0\)
\(-\lambda^3 + 8\lambda^2 - 9\lambda - 6\lambda + 6 - 2\lambda + 4 = 0\)
\(-\lambda^3 + 8\lambda^2 - 17\lambda + 10 = 0\)
\(\lambda^3 - 8\lambda^2 + 17\lambda - 10 = 0\)

Testing integer factors of 10, we find \(\lambda = 1\) is a root since \(1 - 8 + 17 - 10 = 0\).
Factorising by polynomial division gives:
\((\lambda - 1)(\lambda^2 - 7\lambda + 10) = 0\)
\((\lambda - 1)(\lambda - 2)(\lambda - 5) = 0\)

Thus, the eigenvalues are \(\lambda = 1\), \(\lambda = 2\), and \(\lambda = 5\).

(ii) For each eigenvalue, we find the corresponding eigenvector by solving \((A - \lambda I)\mathbf{v} = \mathbf{0}\).

For \(\lambda = 1\):
\(\begin{pmatrix} -1 & 2 & 1 \\ -3 & 4 & 3 \\ -2 & 2 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies y = 0, x = z\)
Eigenvector: \(\mathbf{v}_1 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}\).

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

For \(\lambda = 5\):
\(\begin{pmatrix} -5 & 2 & 1 \\ -3 & 0 & 3 \\ -2 & 2 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies x = z, y = 2x\)
Eigenvector: \(\mathbf{v}_3 = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}\).

(iii) We construct the transition matrix \(P\) using the eigenvectors as columns, and \(D\) as the diagonal matrix of eigenvalues:
\(P = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 1 & 0 & 1 \end{pmatrix}\), \(D = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 5 \end{pmatrix}\).

To find \(P^{-1}\), we use the cofactor method:
\(\det(P) = 1(1-0) - 1(0-2) + 1(0-1) = 2\).

The matrix of cofactors \(C\) is:
\(C = \begin{pmatrix} 1 & 2 & -1 \\ -1 & 0 & 1 \\ 1 & -2 & 1 \end{pmatrix}\).

Taking the transpose of \(C\):
\(C^T = \begin{pmatrix} 1 & -1 & 1 \\ 2 & 0 & -2 \\ -1 & 1 & 1 \end{pmatrix}\).

Thus, \(P^{-1} = \frac{1}{2} \begin{pmatrix} 1 & -1 & 1 \\ 2 & 0 & -2 \\ -1 & 1 & 1 \end{pmatrix}\).

(iv) Using \(A^n = P D^n P^{-1}\), we have:
\(A^n = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 1 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1^n & 0 & 0 \\ 0 & 2^n & 0 \\ 0 & 0 & 5^n \end{pmatrix} \left[ \frac{1}{2} \begin{pmatrix} 1 & -1 & 1 \\ 2 & 0 & -2 \\ -1 & 1 & 1 \end{pmatrix} \right]\)
\(A^n = \frac{1}{2} \begin{pmatrix} 1 & 2^n & 5^n \\ 0 & 2^n & 2 \cdot 5^n \\ 1 & 0 & 5^n \end{pmatrix} \begin{pmatrix} 1 & -1 & 1 \\ 2 & 0 & -2 \\ -1 & 1 & 1 \end{pmatrix}\)

Multiplying these matrices gives:
- Row 1, Col 1: \(1 + 2(2^n) - 5^n = 1 + 2^{n+1} - 5^n\)
- Row 1, Col 2: \(-1 + 5^n = 5^n - 1\)
- Row 1, Col 3: \(1 - 2(2^n) + 5^n = 1 - 2^{n+1} + 5^n\)
- Row 2, Col 1: \(2(2^n) - 2(5^n) = 2^{n+1} - 2 \cdot 5^n\)
- Row 2, Col 2: \(2(5^n)\)
- Row 2, Col 3: \(-2(2^n) + 2(5^n) = 2 \cdot 5^n - 2^{n+1}\)
- Row 3, Col 1: \(1 - 5^n\)
- Row 3, Col 2: \(-1 + 5^n = 5^n - 1\)
- Row 3, Col 3: \(1 + 5^n\)

Combining these elements into the matrix gives the required result.

Part (i):
M1: Attempt to expand \(\det(A - \lambda I)\) with at least two correct algebraic terms.
A1: Correctly obtain the cubic equation \(\lambda^3 - 8\lambda^2 + 17\lambda - 10 = 0\).
M1: Establish a first root and factorise the cubic polynomial to a product of linear terms.
A1: Correctly state eigenvalues \(\lambda = 1, 2, 5\).

Part (ii):
M1: Set up system of linear equations \((A - \lambda I)\mathbf{v} = \mathbf{0}\) for at least one of their eigenvalues.
A1: Find one correct eigenvector (e.g., \(\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}\) corresponding to \(\lambda = 1\)).
A1: Find a second correct eigenvector (e.g., \(\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}\) corresponding to \(\lambda = 2\)).
A1: Find a third correct eigenvector (e.g., \(\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}\) corresponding to \(\lambda = 5\)).

Part (iii):
B1: State correct diagonal matrix \(D\) and corresponding transition matrix \(P\) matching their eigenvectors.
M1: Standard method to find the inverse of their 3x3 matrix \(P\) (cofactor method or row reductions).
A1: Obtain correct inverse matrix \(P^{-1} = \frac{1}{2} \begin{pmatrix} 1 & -1 & 1 \\ 2 & 0 & -2 \\ -1 & 1 & 1 \end{pmatrix}\).

Part (iv):
M1: Formulate the product \(P D^n P^{-1}\) using their matrices and execute matrix multiplication of at least one row/column combination.
A1: Fully correct multiplication leading to the given expression for \(A^n\) (must contain clear intermediate steps).

Using Newton's second law of motion along the line of motion:

\[ m a = F_{\text{driving}} - F_{\text{resistance}} \]

Since we require the relationship between speed \(v\) and distance \(x\), we use the acceleration in the form \(a = v \frac{\mathrm{d}v}{\mathrm{d}x}\). Substituting \(m = 0.5\), the driving force of \(4\text{ N}\), and the resistance of \(0.2v^2\text{ N}\), we get:

\[ 0.5 v \frac{\mathrm{d}v}{\mathrm{d}x} = 4 - 0.2 v^2 \]

Multiply the entire equation by \(2\) to simplify:

\[ v \frac{\mathrm{d}v}{\mathrm{d}x} = 8 - 0.4 v^2 = 0.4(20 - v^2) \]

Separating the variables gives:

\[ \frac{v}{20 - v^2} \mathrm{d}v = 0.4 \mathrm{d}x \]

Integrating both sides with respect to their corresponding variables, using the boundary condition that \(x = 0\) when \(v = 2\), and finding the distance \(x\) when \(v = 4\):

\[ \int_{2}^{4} \frac{v}{20 - v^2} \mathrm{d}v = \int_{0}^{x} 0.4 \mathrm{d}x \]

Let \(u = 20 - v^2\), so \(\mathrm{d}u = -2v \mathrm{d}v\). When \(v = 2\), \(u = 16\); when \(v = 4\), \(u = 4\). The integral becomes:

\[ \int_{16}^{4} -\frac{1}{2u} \mathrm{d}u = \left[ -\frac{1}{2} \ln|u| \right]_{16}^{4} = -\frac{1}{2} \ln 4 - \left( -\frac{1}{2} \ln 16 \right) \]
\[ = \frac{1}{2} (\ln 16 - \ln 4) = \frac{1}{2} \ln\left(\frac{16}{4}\right) = \frac{1}{2} \ln 4 \]

Since \(\ln 4 = \ln(2^2) = 2\ln 2\), the LHS simplifies to:

\[ \frac{1}{2} (2\ln 2) = \ln 2 \]

Now, integrating the RHS:

\[ \int_{0}^{x} 0.4 \mathrm{d}x = 0.4x \]

Equating the LHS and RHS:

\[ 0.4x = \ln 2 \]
\[ x = \frac{\ln 2}{0.4} = \frac{5}{2} \ln 2 \]

M1: Attempts to use Newton's second law with \(a = v \frac{\mathrm{d}v}{\mathrm{d}x}\) and sets up the differential equation.
M1: Separates variables and attempts integration of both sides.
A1: Obtains a correct integrated expression (e.g., \(-1.25 \ln(20 - v^2)\) or equivalent on the LHS, and \(x\) on the RHS).
M1: Applies the limits (from \(v=2\) to \(v=4\) and \(x=0\) to \(x\)) correctly or evaluates the constant of integration.
A1: Correctly simplifies the final answer to the required form, obtaining \(\frac{5}{2} \ln 2\) (or \(2.5 \ln 2\)).

We use the standard equation of the trajectory of a projectile:
\(y = x \tan\theta - \frac{g x^2}{2 u^2}(1 + \tan^2\theta)\)

We are given:
- Launch speed \(u = 20\text{ m s}^{-1}\)
- Acceleration due to gravity \(g = 10\text{ m s}^{-2}\)
- Horizontal distance \(x = 20\text{ m}\)
- To clear the wall, we require the height of the projectile \(y \ge 10\text{ m}\) at this distance.

Substituting these values into the trajectory equation gives:
\(10 \le 20 \tan\theta - \frac{10(20)^2}{2(20)^2}(1 + \tan^2\theta)\)

Simplifying the coefficient of the second term:
\(\frac{10(20)^2}{2(20)^2} = 5\)

So, the inequality becomes:
\(10 \le 20 \tan\theta - 5(1 + \tan^2\theta)\)
\(10 \le 20 \tan\theta - 5 - 5\tan^2\theta\)

Rearranging all terms to one side:
\(5\tan^2\theta - 20\tan\theta + 15 \le 0\)

Dividing the entire inequality by 5:
\(\tan^2\theta - 4\tan\theta + 3 \le 0\)

Factoring the quadratic expression:
\((\tan\theta - 1)(\tan\theta - 3) \le 0\)

This yields the range for \(\tan\theta\):
\(1 \le \tan\theta \le 3\)

Now, we find the corresponding values of \(\theta\) for \(0^\circ < \theta < 90^\circ\):
- \(\tan\theta \ge 1 \implies \theta \ge \arctan(1) = 45.0^\circ\)
- \(\tan\theta \le 3 \implies \theta \le \arctan(3) \approx 71.565^\circ\)

Thus, the set of possible values for \(\theta\) is:
\(45.0^\circ \le \theta \le 71.6^\circ\)

M1: For using the projectile trajectory equation \(y = x \tan\theta - \frac{gx^2}{2u^2}(1+\tan^2\theta)\) (or equivalent method using components to eliminate time).
A1: For correct substitution of \(u = 20\), \(g = 10\), \(x = 20\), and \(y \ge 10\).
M1: For simplifying the inequality to a quadratic form in \(\tan\theta\).
A1: For obtaining \(\tan^2\theta - 4\tan\theta + 3 \le 0\) and correctly showing \(1 \le \tan\theta \le 3\).
M1: For taking the inverse tangent of both boundaries to find \(\theta\).
A1: For the final correct range: \(45.0^\circ \le \theta \le 71.6^\circ\) (accept \(45^\circ \le \theta \le 71.6^\circ\)).

Let \(R\) be the normal reaction of the cone on the particle, and let \(F\) be the frictional force.
Since \(\tan\alpha = \frac{3}{4}\), we have \(\sin\alpha = \frac{3}{5}\) and \(\cos\alpha = \frac{4}{5}\).

**Case 1: Minimum angular speed \(\omega_1\) (impending motion down the slope)**
In this case, the frictional force \(F\) acts up the slope.
Resolving vertically:
\[R \sin\alpha + F \cos\alpha = mg\]
Since the motion is limiting, \(F = \mu R = \frac{1}{3}R\). Substituting this and the trigonometric values:
\[R\left(\frac{3}{5}\right) + \frac{1}{3}R\left(\frac{4}{5}\right) = mg\]
\[R\left(\frac{9}{15} + \frac{4}{15}\right) = mg \implies \frac{13}{15}R = mg \implies R = \frac{15}{13}mg\]

Resolving horizontally towards the center of the circle:
\[R \cos\alpha - F \sin\alpha = m r \omega_1^2\]
\[R\left(\frac{4}{5}\right) - \frac{1}{3}R\left(\frac{3}{5}\right) = m r \omega_1^2\]
\[R\left(\frac{4}{5} - \frac{1}{5}\right) = m r \omega_1^2 \implies \frac{3}{5}R = m r \omega_1^2\]

Substituting \(R = \frac{15}{13}mg\):
\[\frac{3}{5}\left(\frac{15}{13}mg\right) = m r \omega_1^2 \implies \frac{9}{13}g = r \omega_1^2 \implies \omega_1 = \sqrt{\frac{9g}{13r}}\]

**Case 2: Maximum angular speed \(\omega_2\) (impending motion up the slope)**
In this case, the frictional force \(F\) acts down the slope.
Resolving vertically:
\[R \sin\alpha - F \cos\alpha = mg\]
Substituting \(F = \frac{1}{3}R\):
\[R\left(\frac{3}{5}\right) - \frac{1}{3}R\left(\frac{4}{5}\right) = mg\]
\[R\left(\frac{9}{15} - \frac{4}{15}\right) = mg \implies \frac{5}{15}R = mg \implies R = 3mg\]

Resolving horizontally towards the center of the circle:
\[R \cos\alpha + F \sin\alpha = m r \omega_2^2\]
\[R\left(\frac{4}{5}\right) + \frac{1}{3}R\left(\frac{3}{5}\right) = m r \omega_2^2\]
\[R\left(\frac{4}{5} + \frac{1}{5}\right) = m r \omega_2^2 \implies R = m r \omega_2^2\]

Substituting \(R = 3mg\):
\[3mg = m r \omega_2^2 \implies \omega_2 = \sqrt{\frac{3g}{r}}\]

Thus, the set of values for \(\omega\) is:
\[\sqrt{\frac{9g}{13r}} \le \omega \le \sqrt{\frac{3g}{r}}\]

- **M1**: Resolves forces vertically for the case of minimum speed, including both a normal reaction component and a friction component.
- **M1**: Resolves forces horizontally for the case of minimum speed, equating to the centripetal force \(m r \omega^2\).
- **A1**: Correctly obtains the minimum angular speed \(\omega_1 = \sqrt{\frac{9g}{13r}}\).
- **M1**: Resolves forces vertically for the case of maximum speed, including both a normal reaction component and a friction component with appropriate sign changes.
- **M1**: Resolves forces horizontally for the case of maximum speed, equating to the centripetal force.
- **A1**: Correctly obtains the maximum angular speed \(\omega_2 = \sqrt{\frac{3g}{r}}\).
- **A1**: Correctly states the final range of \(\omega\) in a combined inequality: \(\sqrt{\frac{9g}{13r}} \le \omega \le \sqrt{\frac{3g}{r}}\).

(i) Let the center of the common circular face be the origin \(O\), and let the axis of symmetry be the \(z\)-axis, with the hemisphere lying in the region \(z > 0\) and the cylinder in \(z < 0\).

The volume of the cylinder is:
\(V_c = \pi r^2 h = \pi (3a)^2 (4a) = 36\pi a^3\)
Its center of mass is at:
\(z_c = -2a\)

The volume of the hemisphere is:
\(V_h = \frac{2}{3}\pi r^3 = \frac{2}{3}\pi (3a)^3 = 18\pi a^3\)
Its center of mass is at:
\(z_h = \frac{3}{8}r = \frac{3}{8}(3a) = \frac{9}{8}a\)

Let \(\bar{z}\) be the \(z\)-coordinate of the center of mass of the compound solid:
\(\bar{z} = \frac{V_c z_c + V_h z_h}{V_c + V_h}\)
\(\bar{z} = \frac{36\pi a^3 (-2a) + 18\pi a^3 \left(\frac{9}{8}a\right)}{36\pi a^3 + 18\pi a^3}\)
\(\bar{z} = \frac{-72a + \frac{81}{4}a}{54} = \frac{-\frac{207}{4}a}{54} = -\frac{23}{24}a\)

Thus, the distance of the center of mass from the common face is \(\frac{23}{24}a\).

(ii) Let \(A\) be the point of suspension on the rim of the common face. In equilibrium, the center of mass \(G\) lies vertically below \(A\), so the line \(AG\) is vertical.

The axis of symmetry of the cylinder is along the line \(OG\).

The angle that the axis of symmetry makes with the vertical is \(\theta = \angle OGA\).

Since the axis of symmetry is perpendicular to the common face, and \(OA\) is a radius of the common face, \(\angle AOG = 90^\circ\).

In the right-angled triangle \(AOG\):
\(\tan \theta = \frac{OA}{OG} = \frac{3a}{\frac{23}{24}a} = \frac{72}{23}\)
\(\theta = \arctan\left(\frac{72}{23}\right) \approx 72.28^\circ \approx 72.3^\circ\)

(i)
M1: For finding the volumes of the cylinder and hemisphere (or showing they are in the ratio 2:1).
A1: For correct positions of the individual centers of mass from the common face (\(-2a\) and \(\frac{9}{8}a\)).
M1: For setting up the moments equation about the common face.
A1: For obtaining the given result \(\frac{23}{24}a\) convincingly.

(ii)
M1: For identifying that the angle \(\theta\) satisfies \(\tan \theta = \frac{r}{d}\), where \(r\) is the radius of the common face and \(d\) is the distance of the center of mass from the common face.
A1: For substituting the correct values to get \(\tan \theta = \frac{72}{23}\).
A1: For calculating the angle \(\theta \approx 72.3^\circ\) (accept only 72.3).

Let the final position of the particle be at a distance of \(2.0\text{ m}\) down the slope from \(O\).

First, we calculate the normal reaction force \(R\) and the frictional force \(F\):
\[ R = mg \cos \theta = 5 \times 10 \times 0.8 = 40\text{ N} \]
\[ F = \mu R = 0.125 \times 40 = 5\text{ N} \]

The work done against friction, \(W_f\), as the particle moves a distance \(d = 2.0\text{ m}\) is:
\[ W_f = F \times d = 5 \times 2.0 = 10\text{ J} \]

Next, we calculate the gravitational potential energy (GPE) lost by the particle. Taking the final position as the reference level for GPE:
\[ \text{Loss in GPE} = mgd \sin \theta = 5 \times 10 \times 2.0 \times 0.6 = 60\text{ J} \]

At the final position, the extension of the elastic string is:
\[ x = 2.0 - 1.2 = 0.8\text{ m} \]

The elastic potential energy (EPE) gained by the string is:
\[ \text{Gain in EPE} = \frac{\lambda x^2}{2l} = \frac{37.5 \times 0.8^2}{2 \times 1.2} = \frac{37.5 \times 0.64}{2.4} = 10\text{ J} \]

Let \(v\) be the speed of the particle at the final position. The kinetic energy (KE) of the particle is:
\[ \text{KE} = \frac{1}{2} mv^2 = \frac{1}{2} \times 5 \times v^2 = 2.5v^2 \]

Applying the principle of conservation of energy:
\[ \text{Loss in GPE} = \text{Gain in EPE} + \text{KE} + W_f \]
\[ 60 = 10 + 2.5v^2 + 10 \]
\[ 60 = 20 + 2.5v^2 \]
\[ 2.5v^2 = 40 \]
\[ v^2 = 16 \]
\[ v = 4\text{ m/s} \]

M1: For calculating the normal reaction \(R = 40\text{ N}\) and finding the friction force \(F = 5\text{ N}\) or the work done against friction \(10\text{ J}\).
A1: Correct work done against friction of \(10\text{ J}\).
M1: For using Hooke's Law energy formula to find the elastic potential energy \(EPE = 10\text{ J}\).
A1: Correct EPE value.
M1: For setting up the complete work-energy equation of the form \(GPE_{\text{loss}} = EPE_{\text{gain}} + KE_{\text{gain}} + W_{\text{friction}}\).
A1: For obtaining the correct speed of \(4\text{ m/s}\).

Let the line of centres at the moment of impact be the $x$-axis, and the direction perpendicular to it be the $y$-axis.

(i) Before impact, the velocity of $A$ is:
$\mathbf{u}_A = (u \cos\alpha) \mathbf{i} + (u \sin\alpha) \mathbf{j}$
and the velocity of $B$ is $\mathbf{u}_B = 0$.

Let the velocity of $A$ after impact be:
$\mathbf{v}_A = v_{Ax} \mathbf{i} + v_{Ay} \mathbf{j}$
and the velocity of $B$ after impact be:
$\mathbf{v}_B = v_{Bx} \mathbf{i} + v_{By} \mathbf{j}$

Since the spheres are smooth, the component of velocity perpendicular to the line of centres does not change:
$v_{Ay} = u \sin\alpha$
$v_{By} = 0$

By conservation of linear momentum along the line of centres:
$m(u \cos\alpha) + 3m(0) = m v_{Ax} + 3m v_{Bx} \implies u \cos\alpha = v_{Ax} + 3 v_{Bx}$ --- (1)

By Newton's law of restitution:
$v_{Bx} - v_{Ax} = e(u \cos\alpha - 0) = e u \cos\alpha$ --- (2)

Multiplying (2) by 3 and subtracting from (1) gives:
$u \cos\alpha - 3e u \cos\alpha = 4 v_{Ax} \implies v_{Ax} = \frac{u \cos\alpha (1 - 3e)}{4}$

Since the direction of motion of $A$ after collision is perpendicular to its direction before collision, the dot product of their velocities is zero:
$\mathbf{u}_A \cdot \mathbf{v}_A = 0 \implies (u \cos\alpha) v_{Ax} + (u \sin\alpha) v_{Ay} = 0$

Substitute $v_{Ax}$ and $v_{Ay}$:
$u \cos\alpha \left( \frac{u \cos\alpha (1 - 3e)}{4} \right) + (u \sin\alpha)(u \sin\alpha) = 0$
$\frac{u^2 \cos^2\alpha (1 - 3e)}{4} + u^2 \sin^2\alpha = 0$

Dividing by $u^2 \cos^2\alpha$:
$\frac{1 - 3e}{4} + \tan^2\alpha = 0 \implies 1 - 3e + 4\tan^2\alpha = 0$
$3e = 1 + 4\tan^2\alpha \implies e = \frac{1}{3}(1 + 4\tan^2\alpha)$ (as required).

(ii) Given $\tan\alpha = \frac{1}{2}$, we have:
$e = \frac{1}{3} \left( 1 + 4 \left( \frac{1}{4} \right) \right) = \frac{2}{3}$
Since $\tan\alpha = \frac{1}{2}$, we have $\cos\alpha = \frac{2}{\sqrt{5}}$.

Substitute these into $v_{Ax}$:
$v_{Ax} = \frac{u \left( \frac{2}{\sqrt{5}} \right) \left( 1 - 3 \left(\frac{2}{3}\right) \right)}{4} = \frac{u \left( \frac{2}{\sqrt{5}} \right) (-1)}{4} = -\frac{u}{2\sqrt{5}}$

Using (2):
$v_{Bx} = v_{Ax} + e u \cos\alpha = -\frac{u}{2\sqrt{5}} + \frac{2}{3} u \left( \frac{2}{\sqrt{5}} \right) = u \left( -\frac{1}{2\sqrt{5}} + \frac{4}{3\sqrt{5}} \right) = \frac{5}{6\sqrt{5}} u = \frac{\sqrt{5}}{6} u$

Since $v_{By} = 0$, the speed of $B$ after the collision is $v_{Bx} = \frac{\sqrt{5}}{6} u$.

Part (i):
M1: For writing a correct equation for conservation of momentum along the line of centres.
M1: For writing a correct equation for Newton's law of restitution along the line of centres.
A1: For finding a correct expression for the final component of $A$'s velocity along the line of centres, $v_{Ax}$, in terms of $u$, $\alpha$, and $e$.
M1: For using the condition that the initial and final directions of motion of $A$ are perpendicular (e.g., using $\mathbf{u}_A \cdot \mathbf{v}_A = 0$ or equivalent tangent relation).
M1: For eliminating $u$ and expressing the relation in terms of $\tan\alpha$.
A1: For completing the algebraic proof to show $e = \frac{1}{3}(1 + 4\tan^2\alpha)$ convincingly.

Part (ii):
M1: For finding the correct value of $e = \frac{2}{3}$ and the value of $\cos\alpha = \frac{2}{\sqrt{5}}$.
M1: For solving for the component of $B$'s velocity along the line of centres, $v_{Bx}$, by substituting $e$ and $\alpha$.
A1: For obtaining the correct speed of $B$ as $\frac{\sqrt{5}}{6} u$ (or exact equivalent, e.g. $\frac{5}{6\sqrt{5}}u$).

For part (a): We use conservation of energy to find the speed \(v\) of the particle at angle \(\theta\). Taking the lowest point \(A\) as the reference level for potential energy: \(\frac{1}{2} m u^2 = \frac{1}{2} m v^2 + mg a(1 - \cos\theta)\). Thus, \(v^2 = u^2 - 2ga(1 - \cos\theta)\). Applying Newton's second law along the normal to the sphere at angle \(\theta\), towards the center \(O\): \(R - mg \cos\theta = \frac{m v^2}{a}\). Substituting \(v^2\) into this equation: \(R = mg \cos\theta + \frac{m}{a}\left(u^2 - 2ga(1 - \cos\theta)\right) = \frac{mu^2}{a} - 2mg + 3mg \cos\theta\). For part (b): The particle leaves the surface when the normal reaction \(R = 0\). Setting \(R = 0\) when \(\cos\theta = -\frac{1}{3}\) gives \(\frac{mu^2}{a} - 2mg + 3mg\left(-\frac{1}{3}\right) = 0\), which simplifies to \(\frac{u^2}{a} - 3g = 0\), leading to \(u = \sqrt{3ga}\). For part (c): Let \(\alpha\) be the angle at which the particle leaves the surface, so \(\cos\alpha = -\frac{1}{3}\). The height of this point of release above the center \(O\) is \(y_0 = -a \cos\alpha = \frac{1}{3}a\). The speed \(v\) at the point of release is given by \(v^2 = u^2 - 2ga(1 - \cos\alpha) = 3ga - 2ga\left(1 - \left(-\frac{1}{3}\right)\right) = \frac{1}{3}ga\). At this point, the velocity of the particle is tangent to the sphere, making an angle of \(\alpha\) with the horizontal. The vertical component of the velocity is \(v_y = v \sin\alpha\). Since \(\sin^2\alpha = 1 - \cos^2\alpha = 1 - \frac{1}{9} = \frac{8}{9}\), we have \(v_y^2 = v^2 \sin^2\alpha = \frac{1}{3}ga \cdot \frac{8}{9} = \frac{8}{27}ga\). The maximum height \(H\) reached by the projectile above the release point is \(H = \frac{v_y^2}{2g} = \frac{\frac{8}{27}ga}{2g} = \frac{4}{27}a\). Therefore, the maximum height of the path above the level of the center \(O\) is \(y_{\text{max}} = y_0 + H = \frac{1}{3}a + \frac{4}{27}a = \frac{13}{27}a\).

For part (a): M1 for writing a correct energy equation to express \(v^2\). M1 for resolving forces radially towards the center of the sphere to obtain \(R - mg\cos\theta = \frac{mv^2}{a}\). A1 for substituting \(v^2\) and simplifying to obtain the given expression for \(R\). For part (b): M1 for setting \(R = 0\). M1 for substituting \(\cos\theta = -\frac{1}{3}\) and solving for \(u^2\) or \(u\). A1 for obtaining \(u = \sqrt{3ga}\). For part (c): B1 for stating or using the height of the release point above \(O\) is \(y_0 = \frac{1}{3}a\). M1 for calculating the speed \(v\) (or \(v^2\)) at the point of release, obtaining \(v^2 = \frac{1}{3}ga\). M1 for finding the vertical velocity component and using it in \(H = \frac{v_y^2}{2g}\) to find the height reached after leaving the sphere. A1 for successfully showing that the total maximum height is \(\frac{13}{27}a\).

Let $\mu_A$ and $\mu_B$ be the population mean scores for Method A and Method B respectively.

**Step 1: State hypotheses**
$$H_0: \mu_A = \mu_B$$
$$H_1: \mu_A > \mu_B$$

**Step 2: Calculate sample statistics**
For Method A ($n_A = 8$):
$$\sum x_A = 72 + 78 + 85 + 69 + 74 + 81 + 75 + 70 = 604$$
$$\bar{x}_A = \frac{604}{8} = 75.5$$
$$\sum x_A^2 = 72^2 + 78^2 + 85^2 + 69^2 + 74^2 + 81^2 + 75^2 + 70^2 = 45816$$
$$\sum (x_A - \bar{x}_A)^2 = 45816 - \frac{604^2}{8} = 45816 - 45602 = 214$$

For Method B ($n_B = 10$):
$$\sum x_B = 66 + 73 + 71 + 68 + 75 + 70 + 64 + 72 + 69 + 67 = 695$$
$$\bar{x}_B = \frac{695}{10} = 69.5$$
$$\sum x_B^2 = 66^2 + 73^2 + 71^2 + 68^2 + 75^2 + 70^2 + 64^2 + 72^2 + 69^2 + 67^2 = 48405$$
$$\sum (x_B - \bar{x}_B)^2 = 48405 - \frac{695^2}{10} = 48405 - 48302.5 = 102.5$$

**Step 3: Calculate pooled estimate of variance**
$$s_p^2 = \frac{\sum (x_A - \bar{x}_A)^2 + \sum (x_B - \bar{x}_B)^2}{n_A + n_B - 2}$$
$$s_p^2 = \frac{214 + 102.5}{8 + 10 - 2} = \frac{316.5}{16} = 19.78125 \approx 19.8$$

**Step 4: Calculate test statistic**
$$t = \frac{\bar{x}_A - \bar{x}_B}{\sqrt{s_p^2 \left(\frac{1}{n_A} + \frac{1}{n_B}\right)}} = \frac{75.5 - 69.5}{\sqrt{19.78125 \left(\frac{1}{8} + \frac{1}{10}\right)}} = \frac{6}{\sqrt{19.78125 \times 0.225}} = \frac{6}{\sqrt{4.45078}} \approx 2.844$$

**Step 5: Determine critical value and make decision**
Degrees of freedom: $
u = 8 + 10 - 2 = 16$.
At the 1% significance level for a one-tailed test, the critical value of $t$ is $t_{16}(0.99) = 2.583$.

Since $t_{\text{calc}} = 2.844 > 2.583$, the test statistic falls in the critical region.
We reject $H_0$.
There is significant evidence at the 1% level to suggest that the mean test score for students taught by Method A is greater than the mean test score for students taught by Method B.

• **B1**: State both hypotheses correctly in terms of population means $\mu_A$ and $\mu_B$ (or clearly defined symbols).
• **B1**: Find both sample means $\bar{x}_A = 75.5$ and $\bar{x}_B = 69.5$.
• **M1**: Use a correct method to find the pooled sample variance (either using sum of squares or sample variances).
• **A1**: Obtain $s_p^2 = 19.8$ (or $19.78$ or exact $\frac{633}{32}$).
• **M1**: Calculate the test statistic using the correct two-sample t-test formula.
• **A1**: Obtain $t = 2.84$ (or $2.844$).
• **A1**: Identify the correct critical value of $2.583$, compare with their test statistic, and state a correct contextual conclusion (reject $H_0$, supporting Method A's higher mean).

(i) To find the cumulative distribution function \(\mathrm{F}(x) = \mathrm{P}(X \le x)\):
For \(x < 0\), \(\mathrm{F}(x) = 0\).
For \(0 \le x \le 2\):
\(\mathrm{F}(x) = \int_0^x \frac{3}{8}t^2 \mathrm{d}t = \left[ \frac{1}{8}t^3 \right]_0^x = \frac{x^3}{8}\).
For \(x > 2\), \(\mathrm{F}(x) = 1\).
Thus, the cumulative distribution function is:
\(\mathrm{F}(x) = \begin{cases} 0 & x < 0 \\\\ \frac{x^3}{8} & 0 \le x \le 2 \\\\ 1 & x > 2 \end{cases}\)

(ii) Since the domain of \(X\) is \(0 < x \le 2\), we find the domain of \(Y = \frac{8}{X^3}\):
As \(x \to 0^+\), \(y \to \infty\).
As \(x = 2\), \(y = 1\).
So, the range of \(Y\) is \(y \ge 1\).
Let \(\mathrm{G}(y)\) be the cumulative distribution function of \(Y\). For \(y \ge 1\):
\(\mathrm{G}(y) = \mathrm{P}(Y \le y) = \mathrm{P}\left(\frac{8}{X^3} \le y\right) = \mathrm{P}\left(X^3 \ge \frac{8}{y}\right) = \mathrm{P}\left(X \ge \frac{2}{y^{1/3}}\right)\)
Using the complement rule:
\(\mathrm{G}(y) = 1 - \mathrm{F}\left(\frac{2}{y^{1/3}}\right) = 1 - \frac{\left(2y^{-1/3}\right)^3}{8} = 1 - \frac{8y^{-1}}{8} = 1 - \frac{1}{y}\) for \(y \ge 1\).
Differentiating \(\mathrm{G}(y)\) with respect to \(y\) to obtain the probability density function \(g(y)\):
\(g(y) = \frac{\mathrm{d}}{\mathrm{d}y}\left(1 - \frac{1}{y}\right) = \frac{1}{y^2}\) for \(y \ge 1\).
Hence:
\(g(y) = \begin{cases} \frac{1}{y^2} & y \ge 1 \\\\ 0 & \text{otherwise} \end{cases}\)

(iii) We find the expectation of \(\sqrt{Y}\):
\(\mathrm{E}(\sqrt{Y}) = \int_1^{\infty} \sqrt{y} \cdot g(y) \mathrm{d}y = \int_1^{\infty} y^{1/2} \cdot y^{-2} \mathrm{d}y = \int_1^{\infty} y^{-3/2} \mathrm{d}y\)
Integrating:
\(\mathrm{E}(\sqrt{Y}) = \left[ -2y^{-1/2} \right]_1^{\infty} = \lim_{b \to \infty} \left( -2b^{-1/2} \right) - (-2 \cdot 1^{-1/2}) = 0 + 2 = 2\).

(i)
M1: Attempting to integrate \(f(x)\) over the interval \([0, x]\).
A1: Correctly obtaining \(\frac{x^3}{8}\) for \(0 \le x \le 2\).
A1: Expressing the fully defined piecewise cumulative distribution function \(\mathrm{F}(x)\).

(ii)
B1: Explaining or stating that the range of \(Y\) is \(y \ge 1\).
M1: Setting up the probability statement \(\mathrm{P}(Y \le y) = \mathrm{P}\left(X \ge \frac{2}{y^{1/3}}\right)\) (or equivalent).
M1: Correctly substituting into the cumulative distribution function: \(1 - \mathrm{F}(2y^{-1/3})\).
A1: Simplifying correctly to obtain \(\mathrm{G}(y) = 1 - \frac{1}{y}\).
A1: Differentiating to obtain the required pdf \(g(y) = \frac{1}{y^2}\) and stating the domain.

(iii)
M1: Setting up the correct integral for the expectation \(\int_1^{\infty} y^{1/2} \frac{1}{y^2} \mathrm{d}y\).
A1: Correctly simplifying the integrand to \(y^{-3/2}\).
M1: Integrating correctly to \(-2y^{-1/2}\) and applying limits.
A1: Reaching the exact final value of 2.

Let \(X\) be the number of germinated seeds in a group of 4.

**Step 1: State the hypotheses.**
- \(H_0\): A binomial distribution \(B(4, p)\) is a suitable model for the number of germinated seeds.
- \(H_1\): A binomial distribution \(B(4, p)\) is not a suitable model for the number of germinated seeds.

**Step 2: Estimate the parameter \(p\) from the sample.**
First, find the sample mean \(\bar{x}\):
\[ \bar{x} = \frac{(0 \times 6) + (1 \times 20) + (2 \times 38) + (3 \times 28) + (4 \times 8)}{100} = \frac{0 + 20 + 76 + 84 + 32}{100} = \frac{212}{100} = 2.12 \]
Since the mean of a binomial distribution \(B(n, p)\) is given by \(np\) with \(n = 4\):
\[ 4p = 2.12 \implies p = 0.53 \]

**Step 3: Calculate the expected frequencies.**
Using the binomial probability formula \(P(X = x) = \binom{4}{x} (0.53)^x (0.47)^{4-x}\):
- \(E_0 = 100 \times P(X = 0) = 100 \times (0.47)^4 \approx 4.880\)
- \(E_1 = 100 \times P(X = 1) = 100 \times 4 \times 0.53 \times (0.47)^3 \approx 22.010\)
- \(E_2 = 100 \times P(X = 2) = 100 \times 6 \times (0.53)^2 \times (0.47)^2 \approx 37.225\)
- \(E_3 = 100 \times P(X = 3) = 100 \times 4 \times (0.53)^3 \times 0.47 \approx 27.989\)
- \(E_4 = 100 \times P(X = 4) = 100 \times (0.53)^4 \approx 7.890\)

**Step 4: Combine classes with expected frequencies less than 5.**
Since \(E_0 = 4.880 < 5\), we must combine the classes for \(x = 0\) and \(x = 1\):
- Combined Observed (\(x \le 1\)): \(6 + 20 = 26\)
- Combined Expected (\(x \le 1\)): \(4.880 + 22.010 = 26.890\)

This yields the following table for the test statistic calculation:

| Class | Observed (\(O\)) | Expected (\(E\)) | \(\frac{(O-E)^2}{E}\) |
| :---: | :---: | :---: | :---: |
| \(x \le 1\) | 26 | 26.890 | 0.02947 |
| \(x = 2\) | 38 | 37.225 | 0.01613 |
| \(x = 3\) | 28 | 27.989 | 0.00000 |
| \(x = 4\) | 8 | 7.890 | 0.00153 |

**Step 5: Calculate the test statistic \(\chi^2\).**
\[ \chi^2 = \sum \frac{(O-E)^2}{E} \approx 0.02947 + 0.01613 + 0.00000 + 0.00153 = 0.04713 \approx 0.0471 \]

**Step 6: Determine the critical value.**
- Number of classes after combining = 4.
- Since \(p\) was estimated, the number of degrees of freedom is:
\[ \nu = 4 - 1 - 1 = 2 \]
- The critical value at the 5% significance level with \(\nu = 2\) is \(5.991\).

**Step 7: Conclusion.**
Since the calculated test statistic \(\chi^2 \approx 0.0471\) is less than the critical value \(5.991\), we do not reject the null hypothesis \(H_0\).
There is insufficient evidence at the 5% significance level to suggest that the binomial distribution \(B(4, 0.53)\) is not a suitable model for these data.

- **B1**: For correctly stating the null and alternative hypotheses in context.
- **M1**: For finding the mean of the data and estimating \(p = 0.53\).
- **M1**: For calculating binomial probabilities and expected frequencies.
- **A1**: For obtaining the correct expected frequencies and demonstrating the necessity to combine the first two cells (since \(E_0 < 5\)).
- **M1**: For applying the formula \(\sum \frac{(O-E)^2}{E}\) with their combined cells.
- **A1**: For obtaining the correct test statistic \(\chi^2 \approx 0.0471\) (accept range \(0.046\) to \(0.048\)).
- **B1**: For stating the correct number of degrees of freedom (\(\nu = 2\)) and identifying the critical value as \(5.991\).
- **A1**: For comparing their test statistic with the critical value and giving a correct conclusion in context (e.g., "do not reject \(H_0\); the binomial distribution is a suitable model").

First, we find the summary statistics for each sample:

For Process $A$ ($n_1 = 8$):
$$\sum x = 12.4 + 14.1 + 11.8 + 13.5 + 15.0 + 12.9 + 13.2 + 14.3 = 107.2$$
$$\bar{x} = \frac{107.2}{8} = 13.4$$
$$\sum x^2 = 12.4^2 + 14.1^2 + 11.8^2 + 13.5^2 + 15.0^2 + 12.9^2 + 13.2^2 + 14.3^2 = 1444.2$$
$$\sum (x - \bar{x})^2 = \sum x^2 - \frac{(\sum x)^2}{n_1} = 1444.2 - 1436.48 = 7.72$$

For Process $B$ ($n_2 = 10$):
$$\sum y = 15.1 + 13.8 + 16.2 + 14.5 + 15.5 + 13.9 + 14.8 + 15.2 + 16.0 + 14.0 = 149.0$$
$$\bar{y} = \frac{149.0}{10} = 14.9$$
$$\sum y^2 = 15.1^2 + 13.8^2 + 16.2^2 + 14.5^2 + 15.5^2 + 13.9^2 + 14.8^2 + 15.2^2 + 16.0^2 + 14.0^2 = 2226.68$$
$$\sum (y - \bar{y})^2 = \sum y^2 - \frac{(\sum y)^2}{n_2} = 2226.68 - 2220.1 = 6.58$$

Next, we calculate the pooled sample variance, $s_p^2$:
$$s_p^2 = \frac{\sum (x - \bar{x})^2 + \sum (y - \bar{y})^2}{n_1 + n_2 - 2} = \frac{7.72 + 6.58}{8 + 10 - 2} = \frac{14.3}{16} = 0.89375$$

We find the critical value from the $t$-distribution. The degrees of freedom are:
$$\nu = n_1 + n_2 - 2 = 16$$
For a $95\%$ confidence interval, the critical value is:
$$t_{16}(0.975) = 2.120$$

The standard error of the difference of means is:
$$SE = \sqrt{s_p^2 \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} = \sqrt{0.89375 \left(\frac{1}{8} + \frac{1}{10}\right)} = \sqrt{0.20109375} \approx 0.448435$$

The $95\%$ confidence interval for $\mu_B - \mu_A$ is given by:
$$(\bar{y} - \bar{x}) \pm t_{16}(0.975) \times SE$$
$$= (14.9 - 13.4) \pm 2.120 \times 0.448435$$
$$= 1.5 \pm 0.95068$$

This gives the interval:
$$[0.5493, 2.4507]$$

To 3 significant figures, the confidence interval is $[0.549, 2.45]$.

**M1**: For calculating the sample means $\bar{x} = 13.4$ and $\bar{y} = 14.9$ and showing a valid method for calculating sum of squares/variances for both samples.

**A1**: For finding the correct sums of squares: $\sum (x - \bar{x})^2 = 7.72$ and $\sum (y - \bar{y})^2 = 6.58$ (or equivalent values of $s_1^2 \approx 1.10$ and $s_2^2 \approx 0.731$).

**M1**: For using the correct pooled variance formula with their values.

**A1**: For obtaining the correct pooled variance $s_p^2 = 0.89375$ (or $\frac{143}{160}$).

**B1**: For stating the correct critical value $t = 2.120$ (accept $2.12$).

**M1**: For calculating the standard error and combining to form the confidence interval: $(\bar{y} - \bar{x}) \pm t \times SE$.

**A1**: For obtaining the correct interval $[0.549, 2.45]$ (or $0.549 < \mu_B - \mu_A < 2.45$) to 3 significant figures (accept $[0.55, 2.45]$ if 3sf rounding is carried through with $2.12$ instead of $2.120$).

Let $D = \text{After} - \text{Before}$.

**Step 1: State the hypotheses**
Let $\eta$ be the population median difference (After $-$ Before).
$H_0: \eta = 0$
$H_1: \eta > 0$ (one-tailed test)

**Step 2: Calculate differences and ranks**
$$\begin{array}{|l|c|c|c|c|c|c|c|c|}
\hline
\text{Patient} & A & B & C & D & E & F & G & H \\
\hline
\text{Difference, } D & +6 & -2 & +7 & +9 & -2 & +8 & +10 & -1 \\
\hline
|D| & 6 & 2 & 7 & 9 & 2 & 8 & 10 & 1 \\
\hline
\text{Rank of } |D| & 4 & 2.5 & 5 & 7 & 2.5 & 6 & 8 & 1 \\
\hline
\end{array}$$

*Note on ties:* The absolute differences of $2$ for patients $B$ and $E$ share the ranks $2$ and $3$, giving them each an average rank of $\frac{2+3}{2} = 2.5$.

**Step 3: Calculate sum of ranks**
Sum of positive ranks:
$$W_+ = 4 + 5 + 7 + 6 + 8 = 30$$

Sum of negative ranks:
$$W_- = 2.5 + 2.5 + 1 = 6$$

Check: $W_+ + W_- = 30 + 6 = 36$ (which matches $\frac{8 \times 9}{2} = 36$).

**Step 4: Determine test statistic**
The test statistic $T$ is the smaller of $W_+$ and $W_-$:
$$T = 6$$

**Step 5: Critical value and comparison**
For a one-tailed test with $n=8$ at the $5\%$ significance level:
From Wilcoxon signed-rank tables, the critical value is $5$.
(The critical region is $T \le 5$ since $P(T \le 5) \approx 0.0391 < 0.05$ and $P(T \le 6) \approx 0.0547 > 0.05$.)

Since $T = 6 > 5$, the result is not significant.

**Step 6: Conclusion**
Do not reject $H_0$. There is insufficient evidence at the $5\%$ level of significance to suggest that the dietary supplement increases the concentration of the chemical in the blood.

**B1**: State null hypothesis correctly in words or symbols ($H_0: \text{median difference } = 0$).
**B1**: State alternative hypothesis correctly in words or symbols ($H_1: \text{median difference } > 0$, indicating a one-tailed test).
**M1**: Attempt to find differences ($D = \text{After} - \text{Before}$) and rank absolute differences.
**M1**: Handle ties correctly (assigning rank $2.5$ to absolute differences of $2$).
**A1**: Obtain all correct ranks for $|D|$ ($1, 2.5, 2.5, 4, 5, 6, 7, 8$).
**A1**: Calculate correct rank sums $W_+ = 30$ and $W_- = 6$, and state test statistic $T = 6$.
**B1**: State the correct critical value of $5$ (or identify critical region $T \le 5$).
**M1**: Compare the calculated test statistic $T = 6$ with the critical value $5$ (e.g., $6 > 5$).
**A1**: Correctly conclude to accept/not reject $H_0$.
**A1**: Provide a well-phrased final conclusion in context (e.g., insufficient evidence that the supplement increases the concentration of the chemical).

We can expand the probability generating function of \(X\) as follows: \(G_X(t) = \frac{1}{9}t(4 + 4t + t^2) = \frac{1}{9}(4t + 4t^2 + t^3)\). (i) To find the mean, we differentiate \(G_X(t)\) with respect to \(t\): \(G_X'(t) = \frac{1}{9}(4 + 8t + 3t^2)\). The mean is: \(\mathrm{E}(X) = G_X'(1) = \frac{1}{9}(4 + 8 + 3) = \frac{15}{9} = \frac{5}{3}\). To find the variance, we differentiate a second time: \(G_X''(t) = \frac{1}{9}(8 + 6t)\). Evaluating this derivative at \(t = 1\) gives: \(G_X''(1) = \frac{1}{9}(8 + 6) = \frac{14}{9}\). Using the formula for the variance: \(\mathrm{Var}(X) = G_X''(1) + G_X'(1) - [G_X'(1)]^2\), we find: \(\mathrm{Var}(X) = \frac{14}{9} + \frac{15}{9} - \left(\frac{5}{3}\right)^2 = \frac{29}{9} - \frac{25}{9} = \frac{4}{9}\). (ii) From the expansion \(G_X(t) = \frac{4}{9}t + \frac{4}{9}t^2 + \frac{1}{9}t^3\), the probability distribution of \(X\) is: \(P(X=1) = \frac{4}{9}\), \(P(X=2) = \frac{4}{9}\), and \(P(X=3) = \frac{1}{9}\). The event \(X_1 + X_2 = 3\) can occur in two mutually exclusive ways: \((X_1 = 1, X_2 = 2)\) or \((X_1 = 2, X_2 = 1)\). Since the observations are independent: \(P(X_1 + X_2 = 3) = P(X=1)P(X=2) + P(X=2)P(X=1) = \left(\frac{4}{9}\right)\left(\frac{4}{9}\right) + \left(\frac{4}{9}\right)\left(\frac{4}{9}\right) = \frac{16}{81} + \frac{16}{81} = \frac{32}{81} \approx 0.395\).

M1: For differentiating \(G_X(t)\) with respect to \(t\) and evaluating at \(t=1\). A1: For obtaining \(\mathrm{E}(X) = \frac{5}{3}\). M1: For finding \(G_X''(1)\) and applying the variance formula. A1: For obtaining \(\mathrm{Var}(X) = \frac{4}{9}\). M1: For identifying both cases for \(X_1+X_2=3\) and multiplying/adding their probabilities (or finding the coefficient of \(t^3\) in \([G_X(t)]^2\)). A1: For obtaining \(\frac{32}{81}\) (or 0.395).