2024 Cambridge IAS-Level Mathematics - Further (9231) Practice Paper with Answers

(a) A line of invariant points exists if there are non-trivial solutions to \(\mathbf{M}\mathbf{x} = \mathbf{x}\), which can be written as \((\mathbf{M} - \mathbf{I})\mathbf{x} = \mathbf{0}\).
This requires the determinant of \(\mathbf{M} - \mathbf{I}\) to be zero:
\[ \det(\mathbf{M} - \mathbf{I}) = \det \begin{pmatrix} a - 1 & 2 \\ 3 & a - 2 \end{pmatrix} = 0 \]
\[ (a - 1)(a - 2) - 6 = 0 \]
\[ a^2 - 3a - 4 = 0 \]
\[ (a - 4)(a + 1) = 0 \]
So the two values of \(a\) are \(a = 4\) and \(a = -1\).

If \(a = 4\), the system \((\mathbf{M} - \mathbf{I})\mathbf{x} = \mathbf{0}\) becomes:
\[ \begin{pmatrix} 3 & 2 \\ 3 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \]
which gives the line of invariant points: \(3x + 2y = 0\) (or \(y = -\frac{3}{2}x\)).

If \(a = -1\), the system becomes:
\[ \begin{pmatrix} -2 & 2 \\ 3 & -3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \]
which gives the line of invariant points: \(y = x\) (or \(x - y = 0\)).

(b) A general point \(\mathbf{u}\) on the line \(l\) has coordinates:
\[ \mathbf{u} = \begin{pmatrix} 1 + 2\lambda \\ -2 + \lambda \\ 3 - \lambda \end{pmatrix} \]
The image \(\mathbf{u}'\) of this point under the transformation represented by \(\mathbf{A}\) is given by:
\[ \mathbf{u}' = \mathbf{A}\mathbf{u} = \begin{pmatrix} 1 & -1 & 2 \\ 2 & 1 & 1 \\ 0 & 3 & -1 \end{pmatrix} \begin{pmatrix} 1 + 2\lambda \\ -2 + \lambda \\ 3 - \lambda \end{pmatrix} \]
\[ \mathbf{u}' = \begin{pmatrix} 1(1+2\lambda) - 1(-2+\lambda) + 2(3-\lambda) \\ 2(1+2\lambda) + 1(-2+\lambda) + 1(3-\lambda) \\ 0(1+2\lambda) + 3(-2+\lambda) - 1(3-\lambda) \end{pmatrix} \]
\[ \mathbf{u}' = \begin{pmatrix} 1 + 2\lambda + 2 - \lambda + 6 - 2\lambda \\ 2 + 4\lambda - 2 + \lambda + 3 - \lambda \\ -6 + 3\lambda - 3 + \lambda \end{pmatrix} = \begin{pmatrix} 9 - \lambda \\ 3 + 4\lambda \\ -9 + 4\lambda \end{pmatrix} \]
Separating the constant and parameter parts, we obtain the vector equation of the image line:
\[ \mathbf{r} = \begin{pmatrix} 9 \\ 3 \\ -9 \end{pmatrix} + t \begin{pmatrix} -1 \\ 4 \\ 4 \end{pmatrix} \]
(where \(t\) is a real parameter, replacing \(\lambda\)).

Alternatively, we can find the image of two points on \(l\):
For \(\lambda = 0\), the point is \(\mathbf{p}_1 = \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix}\), with image:
\[ \mathbf{A}\mathbf{p}_1 = \begin{pmatrix} 1 & -1 & 2 \\ 2 & 1 & 1 \\ 0 & 3 & -1 \end{pmatrix} \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix} = \begin{pmatrix} 9 \\ 3 \\ -9 \end{pmatrix} \]
For \(\lambda = 1\), the point is \(\mathbf{p}_2 = \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix}\), with image:
\[ \mathbf{A}\mathbf{p}_2 = \begin{pmatrix} 1 & -1 & 2 \\ 2 & 1 & 1 \\ 0 & 3 & -1 \end{pmatrix} \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 8 \\ 7 \\ -5 \end{pmatrix} \]
The direction vector of the image line is:
\[ \mathbf{A}\mathbf{p}_2 - \mathbf{A}\mathbf{p}_1 = \begin{pmatrix} 8 \\ 7 \\ -5 \end{pmatrix} - \begin{pmatrix} 9 \\ 3 \\ -9 \end{pmatrix} = \begin{pmatrix} -1 \\ 4 \\ 4 \end{pmatrix} \]
Thus, the vector equation of the image line is:
\[ \mathbf{r} = \begin{pmatrix} 9 \\ 3 \\ -9 \end{pmatrix} + t \begin{pmatrix} -1 \\ 4 \\ 4 \end{pmatrix} \]

(a)
- **M1**: Sets up the condition for a line of invariant points, \(\det(\mathbf{M} - \mathbf{I}) = 0\).
- **A1**: Correctly solves the quadratic equation to find \(a = 4\) and \(a = -1\).
- **A1**: Correctly finds the equation of the line for \(a = 4\), which is \(3x + 2y = 0\) (or any equivalent form).
- **A1**: Correctly finds the equation of the line for \(a = -1\), which is \(y = x\) (or any equivalent form).

(b)
- **M1**: Attempts to find the image of a general point on the line \(l\), or the images of two distinct points on \(l\).
- **A1**: Finds the image of one point on the line correctly, e.g., \(\begin{pmatrix} 9 \\ 3 \\ -9 \end{pmatrix}\).
- **A1**: Finds the image of a second point on the line correctly, e.g., \(\begin{pmatrix} 8 \\ 7 \\ -5 \end{pmatrix}\) (or writes down the general image vector in terms of \(\lambda\) correctly).
- **M1**: Finds a direction vector of the image line by subtracting two image points or by extracting the coefficient of \(\lambda\).
- **A1**: Obtains a correct direction vector, e.g., \(\begin{pmatrix} -1 \\ 4 \\ 4 \end{pmatrix}\) (or any non-zero scalar multiple).
- **A1**: Writes the final answer in a correct vector equation format (including \(\mathbf{r} = \dots\) and a parameter): \(\mathbf{r} = \begin{pmatrix} 9 \\ 3 \\ -9 \end{pmatrix} + t \begin{pmatrix} -1 \\ 4 \\ 4 \end{pmatrix}\).

Let \( H(n) \) be the statement that \( 7^n + 4^n + 1 \) is divisible by 6. For \( n = 1 \): \( 7^1 + 4^1 + 1 = 12 \), which is divisible by 6, so the base case holds. Assume that the statement is true for \( n = k \), so \( 7^k + 4^k + 1 = 6m \) for some integer \( m \). For \( n = k+1 \), we have \( 7^{k+1} + 4^{k+1} + 1 = 7 \cdot 7^k + 4 \cdot 4^k + 1 \). Substituting \( 7^k = 6m - 4^k - 1 \) yields \( 7(6m - 4^k - 1) + 4 \cdot 4^k + 1 = 42m - 3 \cdot 4^k - 6 = 6(7m - 1) - 3 \cdot 4^k \). Since \( k \ge 1 \), \( 3 \cdot 4^k = 3 \cdot 2 \cdot 2 \cdot 4^{k-1} = 6(2 \cdot 4^{k-1}) \), which is a multiple of 6. Thus, the expression is equal to \( 6(7m - 1 - 2 \cdot 4^{k-1}) \), which is divisible by 6. Therefore, if the statement is true for \( n = k \), it is also true for \( n = k+1 \). Since the base case is true, by mathematical induction the statement is true for all positive integers \( n \).

**B1**: Show that the base case \( n=1 \) is true: \( 7^1 + 4^1 + 1 = 12 \), which is divisible by 6. **M1**: State the inductive hypothesis clearly: assume \( 7^k + 4^k + 1 = 6m \) for some integer \( m \). **M1**: Write down the expression for \( n=k+1 \) and attempt to substitute the inductive hypothesis. **A1**: Obtain a correct simplified expression, such as \( 42m - 3 \cdot 4^k - 6 \) or equivalent. **A1**: Justify why the remaining term (e.g. \( 3 \cdot 4^k \)) is divisible by 6. **A1**: Complete the proof with a correct concluding statement.

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

(i) Using the identity for the sum of squares:
\(\alpha^2 + \beta^2 + \gamma^2 = (\sum \alpha)^2 - 2\sum \alpha\beta\)
\(\alpha^2 + \beta^2 + \gamma^2 = 3^2 - 2(5) = 9 - 10 = -1\)

(ii) Since \(\alpha\), \(\beta\) and \(\gamma\) are roots of the cubic equation, we have:
\(\alpha^3 - 3\alpha^2 + 5\alpha - 2 = 0\)
Summing this relation over the three roots yields:
\(\sum \alpha^3 - 3\sum \alpha^2 + 5\sum \alpha - 6 = 0\)
Substitute the known values:
\(\sum \alpha^3 - 3(-1) + 5(3) - 6 = 0\)
\(\sum \alpha^3 + 3 + 15 - 6 = 0\)
\(\sum \alpha^3 + 12 = 0 \implies \alpha^3 + \beta^3 + \gamma^3 = -12\)

Alternatively, using the identity:
\(\sum \alpha^3 - 3\alpha\beta\gamma = (\sum \alpha)(\sum \alpha^2 - \sum \alpha\beta)\)
\(\sum \alpha^3 - 3(2) = 3(-1 - 5)\)
\(\sum \alpha^3 - 6 = -18 \implies \sum \alpha^3 = -12\).

(b) Let \(y = x^2 - 1\), which means \(x^2 = y + 1\).
Rearranging the original cubic equation to separate the terms with odd and even powers of \(x\):
\(x^3 + 5x = 3x^2 + 2\)
\(x(x^2 + 5) = 3x^2 + 2\)
Squaring both sides:
\(x^2(x^2 + 5)^2 = (3x^2 + 2)^2\)
Substitute \(x^2 = y + 1\) into this equation:
\((y + 1)(y + 1 + 5)^2 = (3(y + 1) + 2)^2\)
\((y + 1)(y + 6)^2 = (3y + 5)^2\)
Expand both sides:
\((y + 1)(y^2 + 12y + 36) = 9y^2 + 30y + 25\)
\(y^3 + 12y^2 + 36y + y^2 + 12y + 36 = 9y^2 + 30y + 25\)
\(y^3 + 13y^2 + 48y + 36 = 9y^2 + 30y + 25\)
Simplifying by collecting all terms to one side:
\(y^3 + 4y^2 + 18y + 11 = 0\)

Part (a):
- M1: For identifying the elementary symmetric sums: \(\sum \alpha = 3\), \(\sum \alpha\beta = 5\), and \(\alpha\beta\gamma = 2\).
- A1: For obtaining \(\alpha^2 + \beta^2 + \gamma^2 = -1\).
- M1: For applying a valid method to find \(\sum \alpha^3\) (such as the cubic recurrence relation or the standard algebraic identity).
- A1: For obtaining \(\alpha^3 + \beta^3 + \gamma^3 = -12\).

Part (b) (using substitution method):
- M1: For setting up the substitution \(y = x^2 - 1\) (or equivalent) and isolating the odd/even powers of \(x\).
- M1: For squaring both sides to produce an equation in terms of \(x^2\) only.
- M1: For substituting \(x^2 = y + 1\) into the squared equation.
- M1: For expanding both sides of the substituted equation correctly.
- A1: For combining terms to obtain the correct cubic expression \(y^3 + 4y^2 + 18y + 11\).
- A1: For writing the final answer as a complete equation, i.e., \(y^3 + 4y^2 + 18y + 11 = 0\) (accept any variable name instead of \(y\)).

Part (b) (alternative method using root relations):
- M1: For finding \(\sum (\alpha^2 - 1) = \sum \alpha^2 - 3 = -4\).
- M1: For expressing \(\sum (\alpha^2 - 1)(\beta^2 - 1) = \sum \alpha^2\beta^2 - 2\sum \alpha^2 + 3\).
- M1: For finding \(\sum \alpha^2\beta^2 = (\sum \alpha\beta)^2 - 2\alpha\beta\gamma\sum\alpha = 25 - 12 = 13\).
- M1: For calculating \(\sum (\alpha^2 - 1)(\beta^2 - 1) = 13 - 2(-1) + 3 = 18\).
- A1: For calculating the constant term from the product: \((\alpha^2 - 1)(\beta^2 - 1)(\gamma^2 - 1) = 4 - 13 - 1 - 1 = -11\).
- A1: For writing the correct final equation \(y^3 + 4y^2 + 18y + 11 = 0\).

To show the identity, we start with the right-hand side (RHS):
\[ \frac{1}{(3r-2)(3r+1)} - \frac{1}{(3r+1)(3r+4)} \]
Combining the fractions over a common denominator:
\[ = \frac{(3r+4) - (3r-2)}{(3r-2)(3r+1)(3r+4)} \]
\[ = \frac{6}{(3r-2)(3r+1)(3r+4)} \]
which matches the left-hand side (LHS).

Next, we use this identity to evaluate the sum:
\[ \sum_{r=1}^n \frac{1}{(3r-2)(3r+1)(3r+4)} = \frac{1}{6} \sum_{r=1}^n \left( \frac{1}{(3r-2)(3r+1)} - \frac{1}{(3r+1)(3r+4)} \right) \]
Let \( f(r) = \frac{1}{(3r-2)(3r+1)} \). The sum can be written as:
\[ \frac{1}{6} \sum_{r=1}^n \left( f(r) - f(r+1) \right) \]
Writing out the terms of this telescoping series:
\[ = \frac{1}{6} \left[ \left( f(1) - f(2) \right) + \left( f(2) - f(3) \right) + \dots + \left( f(n) - f(n+1) \right) \right] \]
\[ = \frac{1}{6} \left[ f(1) - f(n+1) \right] \]
We calculate the values for \( f(1) \) and \( f(n+1) \):
\[ f(1) = \frac{1}{(1)(4)} = \frac{1}{4} \]
\[ f(n+1) = \frac{1}{(3n+1)(3n+4)} \]
So the sum \( S_n \) is:
\[ S_n = \frac{1}{6} \left( \frac{1}{4} - \frac{1}{(3n+1)(3n+4)} \right) = \frac{1}{24} - \frac{1}{6(3n+1)(3n+4)} \]
We simplify this to a single algebraic fraction:
\[ S_n = \frac{(3n+1)(3n+4) - 4}{24(3n+1)(3n+4)} \]
\[ S_n = \frac{9n^2 + 15n + 4 - 4}{24(3n+1)(3n+4)} \]
\[ S_n = \frac{3n(3n+5)}{24(3n+1)(3n+4)} \]
\[ S_n = \frac{n(3n+5)}{8(3n+1)(3n+4)} \]

Finally, for the sum to infinity, we take the limit as \( n \to \infty \):
\[ S_{\infty} = \lim_{n \to \infty} S_n = \lim_{n \to \infty} \left( \frac{1}{24} - \frac{1}{6(3n+1)(3n+4)} \right) = \frac{1}{24} \]

**M1**: For combining the RHS of the identity over a common denominator.
**A1**: For obtaining the correct LHS numerator of 6 and completing the verification.

**M1**: For expressing the series as a sum of differences using the identity.
**M1**: For showing the telescoping cancellation of terms, leaving only the first and last terms.
**A1**: For finding the correct unsimplified sum \( S_n = \frac{1}{6} \left( \frac{1}{4} - \frac{1}{(3n+1)(3n+4)} \right) \).
**A1**: For simplifying \( S_n \) correctly to the final form \( \frac{n(3n+5)}{8(3n+1)(3n+4)} \).

**M1**: For considering the limit of \( S_n \) as \( n \to \infty \).
**A1**: For the correct value \( \frac{1}{24} \).

To show the identity, we start with the right-hand side (RHS):
\[ \frac{1}{(3r-2)(3r+1)} - \frac{1}{(3r+1)(3r+4)} \]
Combining the fractions over a common denominator:
\[ = \frac{(3r+4) - (3r-2)}{(3r-2)(3r+1)(3r+4)} \]
\[ = \frac{6}{(3r-2)(3r+1)(3r+4)} \]
which matches the left-hand side (LHS).

Next, we use this identity to evaluate the sum:
\[ \sum_{r=1}^n \frac{1}{(3r-2)(3r+1)(3r+4)} = \frac{1}{6} \sum_{r=1}^n \left( \frac{1}{(3r-2)(3r+1)} - \frac{1}{(3r+1)(3r+4)} \right) \]
Let \( f(r) = \frac{1}{(3r-2)(3r+1)} \). The sum can be written as:
\[ \frac{1}{6} \sum_{r=1}^n \left( f(r) - f(r+1) \right) \]
Writing out the terms of this telescoping series:
\[ = \frac{1}{6} \left[ \left( f(1) - f(2) \right) + \left( f(2) - f(3) \right) + \dots + \left( f(n) - f(n+1) \right) \right] \]
\[ = \frac{1}{6} \left[ f(1) - f(n+1) \right] \]
We calculate the values for \( f(1) \) and \( f(n+1) \):
\[ f(1) = \frac{1}{(1)(4)} = \frac{1}{4} \]
\[ f(n+1) = \frac{1}{(3n+1)(3n+4)} \]
So the sum \( S_n \) is:
\[ S_n = \frac{1}{6} \left( \frac{1}{4} - \frac{1}{(3n+1)(3n+4)} \right) = \frac{1}{24} - \frac{1}{6(3n+1)(3n+4)} \]
We simplify this to a single algebraic fraction:
\[ S_n = \frac{(3n+1)(3n+4) - 4}{24(3n+1)(3n+4)} \]
\[ S_n = \frac{9n^2 + 15n + 4 - 4}{24(3n+1)(3n+4)} \]
\[ S_n = \frac{3n(3n+5)}{24(3n+1)(3n+4)} \]
\[ S_n = \frac{n(3n+5)}{8(3n+1)(3n+4)} \]

Finally, for the sum to infinity, we take the limit as \( n \to \infty \):
\[ S_{\infty} = \lim_{n \to \infty} S_n = \lim_{n \to \infty} \left( \frac{1}{24} - \frac{1}{6(3n+1)(3n+4)} \right) = \frac{1}{24} \]

**M1**: For combining the RHS of the identity over a common denominator.
**A1**: For obtaining the correct LHS numerator of 6 and completing the verification.

**M1**: For expressing the series as a sum of differences using the identity.
**M1**: For showing the telescoping cancellation of terms, leaving only the first and last terms.
**A1**: For finding the correct unsimplified sum \( S_n = \frac{1}{6} \left( \frac{1}{4} - \frac{1}{(3n+1)(3n+4)} \right) \).
**A1**: For simplifying \( S_n \) correctly to the final form \( \frac{n(3n+5)}{8(3n+1)(3n+4)} \).

**M1**: For considering the limit of \( S_n \) as \( n \to \infty \).
**A1**: For the correct value \( \frac{1}{24} \).

(i) To sketch \(r = a(7 + 3\cos\theta)\), we note that for all \(\theta\), \(r > 0\) since the minimum value of \(7 + 3\cos\theta\) is \(4\). The curve is symmetric about the initial line because \(\cos(-\theta) = \cos\theta\). Evaluating at key angles: when \(\theta = 0\), \(r = 10a\); when \(\theta = \pi\), \(r = 4a\); when \(\theta = \pm\frac{\pi}{2}\), \(r = 7a\). Thus, the curve is a smooth, closed, egg-like convex shape with no dimple or cusp, intersecting the initial line at \((10a, 0)\) and \((4a, \pi)\), and intersecting \(\theta = \frac{\pi}{2}\) at \((7a, \frac{\pi}{2})\) and \((7a, -\frac{\pi}{2})\). (ii) The distance from the initial line is \(y = r\sin\theta = a(7 + 3\cos\theta)\sin\theta = a(7\sin\theta + 3\sin\theta\cos\theta) = a(7\sin\theta + \frac{3}{2}\sin 2\theta)\). For the tangent to be parallel to the initial line, we set \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = 0\). Differentiating: \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = a(7\cos\theta + 3\cos 2\theta) = a(7\cos\theta + 3(2\cos^2\theta - 1)) = a(6\cos^2\theta + 7\cos\theta - 3)\). Setting \(6\cos^2\theta + 7\cos\theta - 3 = 0\) and factoring yields \((2\cos\theta + 3)(3\cos\theta - 1) = 0\). Since \(\cos\theta\) must lie in the range \([-1, 1]\), we discard the root \(\cos\theta = -1.5\). This leaves \(3\cos\theta - 1 = 0 \implies \cos\theta = \frac{1}{3}\). The corresponding values of \(\theta\) in \((-\pi, \pi]\) are \(\theta = \pm\arccos(\frac{1}{3})\). Substituting \(\cos\theta = \frac{1}{3}\) back into the polar equation gives \(r = a(7 + 3(\frac{1}{3})) = 8a\). Thus, the polar coordinates of the points of contact are \((8a, \arccos(\frac{1}{3}))\) and \((8a, -\arccos(\frac{1}{3}))\). (iii) The area \(A\) of the region enclosed by \(C\) is given by \(A = \frac{1}{2}\int_{-\pi}^{\pi} r^2 \mathrm{d}\theta = \frac{a^2}{2}\int_{-\pi}^{\pi} (7 + 3\cos\theta)^2 \mathrm{d}\theta = \frac{a^2}{2}\int_{-\pi}^{\pi} (49 + 42\cos\theta + 9\cos^2\theta) \mathrm{d}\theta\). Using the identity \(\cos^2\theta = \frac{1}{2}(1 + \cos 2\theta)\), the integrand becomes: \(49 + 42\cos\theta + \frac{9}{2}(1 + \cos 2\theta) = \frac{107}{2} + 42\cos\theta + \frac{9}{2}\cos 2\theta\). Integrating each term with respect to \(\theta\): \(A = \frac{a^2}{2} \left[ \frac{107}{2}\theta + 42\sin\theta + \frac{9}{4}\sin 2\theta \right]_{-\pi}^{\pi}\). Evaluating between the limits: \(A = \frac{a^2}{2} \left[ \left( \frac{107\pi}{2} + 0 + 0 \right) - \left( -\frac{107\pi}{2} + 0 + 0 \right) \right] = \frac{a^2}{2} (107\pi) = \frac{107}{2}\pi a^2\).

Part (i): [3 marks] B1: Correctly shaped closed curve (convex, symmetric about the initial line, with no dimple or loop). B1: Correct polar coordinates for intersections with the initial line: \((10a, 0)\) and \((4a, \pi)\). B1: Correct polar coordinates for intersections with the line \(\theta = \pi/2\): \((7a, \pi/2)\) and \((7a, -\pi/2)\). Part (ii): [5 marks] M1: Expresses \(y = r\sin\theta\) in terms of \(\theta\). M1: Differentiates \(y\) and equates \(\frac{\mathrm{d}y}{\mathrm{d}\theta}\) to zero to obtain a quadratic equation in \(\cos\theta\). A1: Correctly factors and solves the quadratic equation, obtaining \(\cos\theta = 1/3\) and showing the rejection of \(\cos\theta = -1.5\). M1: Substitutes \(\cos\theta = 1/3\) into the polar equation to find \(r = 8a\). A1: Correctly states both pairs of polar coordinates: \((8a, \pm\arccos(1/3))\) (allow equivalent exact expressions). Part (iii): [5 marks] M1: States and uses the polar area formula \(\frac{1}{2}\int_{-\pi}^{\pi} r^2 \mathrm{d}\theta\) (or \(\int_{0}^{\pi} r^2 \mathrm{d}\theta\)). M1: Correctly expands \((7+3\cos\theta)^2\) and applies the identity \(\cos^2\theta = \frac{1}{2}(1+\cos 2\theta)\). A1: Obtains the correct integrated expression: \(\frac{107}{2}\theta + 42\sin\theta + \frac{9}{4}\sin 2\theta\) (or equivalent). M1: Substitutes the limits correctly. A1: Obtains the final exact area \(\frac{107}{2}\pi a^2\) (or \(53.5\pi a^2\)).

(i) By algebraic division:
\[ y = \frac{x(x - 1) - 2(x - 1) + 4}{x - 1} = x - 2 + \frac{4}{x - 1} \]
As \( x \to \pm\infty \), \\frac{4}{x - 1} \\to 0 \), so the oblique asymptote is \( y = x - 2 \).
The denominator is zero when \( x = 1 \) and the numerator is non-zero, so the vertical asymptote is \( x = 1 \).

(ii) Differentiating \( y = x - 2 + 4(x - 1)^{-1} \) with respect to \( x \):
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = 1 - \frac{4}{(x - 1)^2} \]
Setting \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \):
\[ 1 - \frac{4}{(x - 1)^2} = 0 \implies (x - 1)^2 = 4 \implies x - 1 = \pm 2 \]
This gives \( x = 3 \) or \( x = -1 \).
For \( x = 3 \): \( y = 3 - 2 + \frac{4}{3 - 1} = 3 \).
For \( x = -1 \): \( y = -1 - 2 + \frac{4}{-1 - 1} = -5 \).
Thus, the stationary points are \( (3, 3) \) and \( (-1, -5) \).

(iii) The sketch of \( C \) should feature:
- Two branches of the curve (one for \( x > 1 \) with a local minimum at \( (3, 3) \), and one for \( x < 1 \) with a local maximum at \( (-1, -5) \)).
- The vertical asymptote \( x = 1 \) and oblique asymptote \( y = x - 2 \) drawn as dashed lines.
- The \( y \)-intercept labeled at \( (0, -6) \).

(iv) Method 1 (Graphical/Range):
From the sketch, the local maximum value is \( -5 \) and the local minimum value is \( 3 \).
Therefore, the range of the function is \( y \le -5 \) and \( y \ge 3 \).
For the line \( y = k \) to have no intersection with the curve \( C \) (and thus no real roots to the equation), \( k \) must lie strictly between these two regions:
\[ -5 < k < 3 \]

Method 2 (Algebraic):
Rearranging the equation:
\[ x^2 - 3x + 6 = k(x - 1) \implies x^2 - (3 + k)x + (6 + k) = 0 \]
For no real roots, the discriminant \( \Delta < 0 \):
\[ (3 + k)^2 - 4(6 + k) < 0 \]
\[ k^2 + 6k + 9 - 24 - 4k < 0 \]
\[ k^2 + 2k - 15 < 0 \]
\[ (k + 5)(k - 3) < 0 \]
This yields \( -5 < k < 3 \).

(i)
M1: Attempt algebraic division or express the fraction in the form \( ax + b + \frac{c}{x-1} \).
A1: Correct oblique asymptote: \( y = x - 2 \).
A1: Correct vertical asymptote: \( x = 1 \).

(ii)
M1: Attempt differentiation of the function (using quotient rule or from the simplified form).
A1: Correct derivative, e.g., \( \frac{\mathrm{d}y}{\mathrm{d}x} = 1 - \frac{4}{(x-1)^2} \) or \( \frac{x^2 - 2x - 3}{(x-1)^2} \).
M1: Set \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \) and solve the resulting quadratic equation for \( x \).
A1: Correct stationary points: \( (3, 3) \) and \( (-1, -5) \).

(iii)
B1: Sketch both branches of the curve in the correct quadrants relative to the asymptotes.
B1: Draw and label both asymptotes with their correct equations.
B1: Label the stationary points \( (3, 3) \), \( (-1, -5) \) and the \( y \)-intercept \( (0, -6) \).

(iv)
M1: Formulate a quadratic equation in \( x \) and find its discriminant, or utilize the stationary values from part (ii) to set up an inequality.
M1: Correctly solve the resulting quadratic inequality in \( k \) (e.g., \( k^2 + 2k - 15 < 0 \)).
A1: Obtain the correct interval: \( -5 < k < 3 \) (or equivalent interval notation).

(a) To find the shortest distance between the skew lines, we first find a vector perpendicular to both direction vectors, \(\mathbf{u} = 2\mathbf{i} + \mathbf{j}\) and \(\mathbf{v} = \mathbf{j} + \mathbf{k}\):

\(\mathbf{n} = \mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix} = \mathbf{i}(1-0) - \mathbf{j}(2-0) + \mathbf{k}(2-0) = \mathbf{i} - 2\mathbf{j} + 2\mathbf{k}\)

Now, we find its magnitude:

\(|\mathbf{n}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{9} = 3\)

Let \(A(1, 4, -1)\) be a point on \(l_1\) and \(B(2, 1, 3)\) be a point on \(l_2\). The displacement vector \(\mathbf{AB}\) is:

\(\mathbf{AB} = (2-1)\mathbf{i} + (1-4)\mathbf{j} + (3 - (-1))\mathbf{k} = \mathbf{i} - 3\mathbf{j} + 4\mathbf{k}\)

The shortest distance \(d\) is the projection of \(\mathbf{AB}\) onto the common perpendicular \(\mathbf{n}\):

\(d = \frac{|\mathbf{AB} \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{|(1)(1) + (-3)(-2) + (4)(2)|}{3} = \frac{|1 + 6 + 8|}{3} = \frac{15}{3} = 5\)

(b) The plane \(\Pi_1\) contains \(l_1\) and is parallel to \(l_2\), so its normal is parallel to the common perpendicular \(\mathbf{n}_1 = \mathbf{i} - 2\mathbf{j} + 2\mathbf{k}\).

Since \(\Pi_1\) contains \(l_1\), it passes through the point \(A(1, 4, -1)\).

The equation of the plane is:

\(1(x-1) - 2(y-4) + 2(z+1) = 0\)

\(x - 2y + 2z - 1 + 8 + 2 = 0 \implies x - 2y + 2z = -9\)

(c) The plane \(\Pi_2\) contains \(l_1\) and is perpendicular to \(\Pi_1\). Thus, the normal to \(\Pi_2\), which we call \(\mathbf{n}_2\), is perpendicular to the direction vector of \(l_1\) (\(\mathbf{u} = 2\mathbf{i} + \mathbf{j}\)) and the normal of \(\Pi_1\) (\(\mathbf{n}_1 = \mathbf{i} - 2\mathbf{j} + 2\mathbf{k}\)):

\(\mathbf{n}_2 = \mathbf{u} \times \mathbf{n}_1 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & 0 \\ 1 & -2 & 2 \end{vmatrix} = \mathbf{i}(2-0) - \mathbf{j}(4-0) + \mathbf{k}(-4-1) = 2\mathbf{i} - 4\mathbf{j} - 5\mathbf{k}\)

Since \(\Pi_2\) contains \(l_1\), it also passes through \(A(1, 4, -1)\). The equation of \(\Pi_2\) is:

\(2(x-1) - 4(y-4) - 5(z+1) = 0\)

\(2x - 4y - 5z - 2 + 16 - 5 = 0 \implies 2x - 4y - 5z = -9\)

(d) The angle \(\theta\) between the line \(l_2\) (with direction \(\mathbf{v} = \mathbf{j} + \mathbf{k}\)) and the plane \(\Pi_2\) (with normal \(\mathbf{n}_2 = 2\mathbf{i} - 4\mathbf{j} - 5\mathbf{k}\)) satisfies:

\(\sin\theta = \frac{|\mathbf{v} \cdot \mathbf{n}_2|}{|\mathbf{v}| |\mathbf{n}_2|}\)

First, calculate the scalar product and magnitudes:

\(\mathbf{v} \cdot \mathbf{n}_2 = (0)(2) + (1)(-4) + (1)(-5) = -9\)

\(|\mathbf{v}| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2}\)

\(|\mathbf{n}_2| = \sqrt{2^2 + (-4)^2 + (-5)^2} = \sqrt{4 + 16 + 25} = \sqrt{45} = 3\sqrt{5}\)

Thus:

\(\sin\theta = \frac{|-9|}{\sqrt{2} \times 3\sqrt{5}} = \frac{9}{3\sqrt{10}} = \frac{3}{\sqrt{10}}\)

\(\theta = \arcsin\left(\frac{3}{\sqrt{10}}\right) \approx 71.565^\circ\)

To the nearest \(0.1^\circ\), \(\theta = 71.6^\circ\).

(a)
M1: For attempting the cross product of the direction vectors of \(l_1\) and \(l_2\).
A1: For obtaining correct normal vector \(\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}\) (or scalar multiple).
M1: For finding a vector connecting a point on each line, e.g., \(\mathbf{AB} = \mathbf{i} - 3\mathbf{j} + 4\mathbf{k}\), and attempting the dot product with their normal vector.
M1: For using the correct projection formula \(d = \frac{|\mathbf{AB} \cdot \mathbf{n}|}{|\mathbf{n}|}\).
A1: For obtaining the correct distance of 5.

(b)
M1: For identifying that the normal vector of \(\Pi_1\) is parallel to the common perpendicular found in part (a).
M1: For substituting a point on \(l_1\) (e.g. \((1, 4, -1)\)) into the equation \(\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}\).
A1: For obtaining the correct Cartesian equation \(x - 2y + 2z = -9\) (or equivalent).

(c)
M1: For realizing that the normal of \(\Pi_2\) is perpendicular to the direction vector of \(l_1\) and the normal of \(\Pi_1\).
M1: For attempting the cross product of \(2\mathbf{i} + \mathbf{j}\) and \(\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}\).
A1: For obtaining the correct normal vector \(2\mathbf{i} - 4\mathbf{j} - 5\mathbf{k}\) (or scalar multiple).
A1: For obtaining the correct Cartesian equation \(2x - 4y - 5z = -9\) (or equivalent).

(d)
M1: For attempting to use the formula \(\sin\theta = \frac{|\mathbf{v} \cdot \mathbf{n}_2|}{|\mathbf{v}| |\mathbf{n}_2|}\) with their direction vector of \(l_2\) and normal of \(\Pi_2\).
A1: For finding the correct value of \(\sin\theta = \frac{3}{\sqrt{10}}\) (or equivalent).
A1: For obtaining the correct angle \(71.6^\circ\) (allow \(1.25\) radians if specified, otherwise degrees as standard).

Substitute the exponential definitions of \(\cosh x\) and \(\sinh x\):
\(\cosh x = \frac{e^x + e^{-x}}{2}\), \(\sinh x = \frac{e^x - e^{-x}}{2}\).

Substitute these into the given equation:
\(3\left(\frac{e^x + e^{-x}}{2}\right) - 2\left(\frac{e^x - e^{-x}}{2}\right) = 3\).

Multiply by 2 to clear the denominators:
\(3(e^x + e^{-x}) - 2(e^x - e^{-x}) = 6\).

Expand and collect terms:
\(e^x + 5e^{-x} = 6\).

Multiply by \(e^x\) to form a quadratic equation:
\((e^x)^2 - 6e^x + 5 = 0\).

Letting \(u = e^x\), we have:
\(u^2 - 6u + 5 = 0 \implies (u - 1)(u - 5) = 0\).

Thus \(u = 1\) or \(u = 5\).

Since \(u = e^x\):
For \(u = 1\), \(e^x = 1 \implies x = 0\).
For \(u = 5\), \(e^x = 5 \implies x = \ln 5\).

M1: For substituting the exponential definitions of \(\cosh x\) and \(\sinh x\) into the equation.
A1: For obtaining a correct quadratic equation in terms of \(e^x\), such as \((e^x)^2 - 6e^x + 5 = 0\).
M1: For solving their quadratic equation to find two positive values for \(e^x\).
A1: For both correct exact solutions \(x = 0\) and \(x = \ln 5\).

Substitute the exponential definitions of \(\cosh x\) and \(\sinh x\):
\(\cosh x = \frac{e^x + e^{-x}}{2}\), \(\sinh x = \frac{e^x - e^{-x}}{2}\).

Substitute these into the given equation:
\(3\left(\frac{e^x + e^{-x}}{2}\right) - 2\left(\frac{e^x - e^{-x}}{2}\right) = 3\).

Multiply by 2 to clear the denominators:
\(3(e^x + e^{-x}) - 2(e^x - e^{-x}) = 6\).

Expand and collect terms:
\(e^x + 5e^{-x} = 6\).

Multiply by \(e^x\) to form a quadratic equation:
\((e^x)^2 - 6e^x + 5 = 0\).

Letting \(u = e^x\), we have:
\(u^2 - 6u + 5 = 0 \implies (u - 1)(u - 5) = 0\).

Thus \(u = 1\) or \(u = 5\).

Since \(u = e^x\):
For \(u = 1\), \(e^x = 1 \implies x = 0\).
For \(u = 5\), \(e^x = 5 \implies x = \ln 5\).

M1: For substituting the exponential definitions of \(\cosh x\) and \(\sinh x\) into the equation.
A1: For obtaining a correct quadratic equation in terms of \(e^x\), such as \((e^x)^2 - 6e^x + 5 = 0\).
M1: For solving their quadratic equation to find two positive values for \(e^x\).
A1: For both correct exact solutions \(x = 0\) and \(x = \ln 5\).

Let \(y = x^2 \cosh(2x)\). We define \(u = \cosh(2x)\) and \(v = x^2\). By Leibniz's theorem, the \(n\)th derivative of a product is given by \(y^{(n)} = \sum_{r=0}^n \binom{n}{r} u^{(n-r)} v^{(r)}\). The derivatives of \(v = x^2\) are: \(v' = 2x\), \(v'' = 2\), and \(v^{(r)} = 0\) for all \(r \ge 3\). Applying this to the 8th derivative (\(n = 8\)), we get: \(y^{(8)} = \binom{8}{0} u^{(8)} v + \binom{8}{1} u^{(7)} v' + \binom{8}{2} u^{(6)} v''\). Substituting the derivatives of \(v\): \(y^{(8)} = x^2 u^{(8)} + 8(u^{(7)})(2x) + 28(u^{(6)})(2) = x^2 u^{(8)} + 16x u^{(7)} + 56 u^{(6)}\). Evaluating at \(x = 0\), we see that the first two terms vanish: \(y^{(8)}(0) = 56 u^{(6)}(0)\). To find \(u^{(6)}\), we differentiate \(u = \cosh(2x)\) successively: \(u' = 2\sinh(2x)\), \(u'' = 4\cosh(2x)\), \(u''' = 8\sinh(2x)\), and in general, the even derivatives of \(u\) are \(u^{(2k)} = 2^{2k}\cosh(2x)\). For \(2k = 6\), we have \(u^{(6)} = 2^6\cosh(2x) = 64\cosh(2x)\). At \(x = 0\), this gives \(u^{(6)}(0) = 64\cosh(0) = 64\). Therefore, \(y^{(8)}(0) = 56 \times 64 = 3584\).

M1: State or apply Leibniz's theorem for the 8th derivative of the product. A1: Obtain the simplified expression \(y^{(8)} = x^2 u^{(8)} + 16x u^{(7)} + 56 u^{(6)}\) (or equivalent). M1: Substitute \(x = 0\) to obtain \(y^{(8)}(0) = 56 u^{(6)}(0)\). M1: Successively differentiate \(u = \cosh(2x)\) to find the form of the 6th derivative. A1: Find \(u^{(6)}(0) = 64\). A1: Compute the final correct value of 3584.

Let \(y = x^2 \cosh(2x)\). We define \(u = \cosh(2x)\) and \(v = x^2\). By Leibniz's theorem, the \(n\)th derivative of a product is given by \(y^{(n)} = \sum_{r=0}^n \binom{n}{r} u^{(n-r)} v^{(r)}\). The derivatives of \(v = x^2\) are: \(v' = 2x\), \(v'' = 2\), and \(v^{(r)} = 0\) for all \(r \ge 3\). Applying this to the 8th derivative (\(n = 8\)), we get: \(y^{(8)} = \binom{8}{0} u^{(8)} v + \binom{8}{1} u^{(7)} v' + \binom{8}{2} u^{(6)} v''\). Substituting the derivatives of \(v\): \(y^{(8)} = x^2 u^{(8)} + 8(u^{(7)})(2x) + 28(u^{(6)})(2) = x^2 u^{(8)} + 16x u^{(7)} + 56 u^{(6)}\). Evaluating at \(x = 0\), we see that the first two terms vanish: \(y^{(8)}(0) = 56 u^{(6)}(0)\). To find \(u^{(6)}\), we differentiate \(u = \cosh(2x)\) successively: \(u' = 2\sinh(2x)\), \(u'' = 4\cosh(2x)\), \(u''' = 8\sinh(2x)\), and in general, the even derivatives of \(u\) are \(u^{(2k)} = 2^{2k}\cosh(2x)\). For \(2k = 6\), we have \(u^{(6)} = 2^6\cosh(2x) = 64\cosh(2x)\). At \(x = 0\), this gives \(u^{(6)}(0) = 64\cosh(0) = 64\). Therefore, \(y^{(8)}(0) = 56 \times 64 = 3584\).

M1: State or apply Leibniz's theorem for the 8th derivative of the product. A1: Obtain the simplified expression \(y^{(8)} = x^2 u^{(8)} + 16x u^{(7)} + 56 u^{(6)}\) (or equivalent). M1: Substitute \(x = 0\) to obtain \(y^{(8)}(0) = 56 u^{(6)}(0)\). M1: Successively differentiate \(u = \cosh(2x)\) to find the form of the 6th derivative. A1: Find \(u^{(6)}(0) = 64\). A1: Compute the final correct value of 3584.

We use the formula for the surface area of revolution about the $x$-axis:
$$S = 2\pi \int_{a}^{b} y \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt$$

First, find the derivatives of $x$ and $y$ with respect to $t$:
$$\frac{dx}{dt} = e^t \cos t - e^t \sin t = e^t(\cos t - \sin t)$$
$$\frac{dy}{dt} = e^t \sin t + e^t \cos t = e^t(\sin t + \cos t)$$

Next, calculate the term under the square root:
$$\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = e^{2t}(\cos^2 t - 2\sin t\cos t + \sin^2 t) + e^{2t}(\sin^2 t + 2\sin t\cos t + \cos^2 t)$$
$$= e^{2t}(1 - 2\sin t\cos t + 1 + 2\sin t\cos t) = 2e^{2t}$$

Thus, the element of arc length is:
$$\frac{ds}{dt} = \sqrt{2e^{2t}} = \sqrt{2} e^t$$

Substitute $y$ and $\frac{ds}{dt}$ into the surface area formula:
$$S = 2\pi \int_{0}^{\pi/2} (e^t \sin t)(\sqrt{2} e^t) dt = 2\sqrt{2}\pi \int_{0}^{\pi/2} e^{2t} \sin t dt$$

To evaluate the integral $I = \int e^{2t} \sin t dt$, we use integration by parts:
Let $u = \sin t$ and $dv = e^{2t} dt$, so $du = \cos t dt$ and $v = \frac{1}{2}e^{2t}$.
$$I = \frac{1}{2}e^{2t} \sin t - \frac{1}{2} \int e^{2t} \cos t dt$$

Apply integration by parts again to the second integral:
Let $u = \cos t$ and $dv = e^{2t} dt$, so $du = -\sin t dt$ and $v = \frac{1}{2}e^{2t}$.
$$\int e^{2t} \cos t dt = \frac{1}{2}e^{2t} \cos t - \int \frac{1}{2}e^{2t} (-\sin t) dt = \frac{1}{2}e^{2t} \cos t + \frac{1}{2}I$$

Substitute this back into the expression for $I$:
$$I = \frac{1}{2}e^{2t} \sin t - \frac{1}{2}\left(\frac{1}{2}e^{2t} \cos t + \frac{1}{2}I\right)$$
$$I = \frac{1}{2}e^{2t} \sin t - \frac{1}{4}e^{2t} \cos t - \frac{1}{4}I$$
$$\frac{5}{4}I = \frac{1}{4}e^{2t}(2\sin t - \cos t) \implies I = \frac{1}{5}e^{2t}(2\sin t - \cos t)$$

Now, evaluate the limits from $0$ to $\frac{1}{2}\pi$:
$$\left[\frac{1}{5}e^{2t}(2\sin t - \cos t)\right]_{0}^{\pi/2} = \frac{1}{5}e^{\pi}(2\sin(\pi/2) - \cos(\pi/2)) - \frac{1}{5}e^{0}(2\sin(0) - \cos(0))$$
$$= \frac{1}{5}e^{\pi}(2(1) - 0) - \frac{1}{5}(1)(0 - 1)$$
$$= \frac{2}{5}e^{\pi} + \frac{1}{5} = \frac{2e^{\pi} + 1}{5}$$

Finally, multiply by the constant $2\sqrt{2}\pi$ to find the surface area $S$:
$$S = 2\sqrt{2}\pi \left(\frac{2e^{\pi} + 1}{5}\right) = \frac{2\sqrt{2}\pi}{5}(2e^{\pi} + 1)$$"

**M1**: Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$ using the product rule.
**A1**: Obtain correct derivatives: $\frac{dx}{dt} = e^t(\cos t - \sin t)$ and $\frac{dy}{dt} = e^t(\sin t + \cos t)$.
**M1**: Substitute into $\sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2}$ and simplify.
**A1**: Obtain $\sqrt{2}e^t$ (or equivalent exact simplified form).
**M1**: Set up correct integral for surface area: $S = 2\pi \int_{0}^{\pi/2} (e^t \sin t)(\sqrt{2} e^t) dt$.
**M1**: Use integration by parts twice to integrate $e^{2t} \sin t$ (or state and apply the standard formula correctly).
**A1**: Obtain correct integrated term $\frac{1}{5}e^{2t}(2\sin t - \cos t)$ and evaluate limits correctly to get $\frac{2e^\pi + 1}{5}$.
**A1**: Obtain the final exact surface area $\frac{2\sqrt{2}\pi}{5}(2e^\pi + 1)$ (or any algebraically equivalent form).

**Part (i)**
To find the characteristic equation of \( A \), we set \( \det(A - \lambda I) = 0 \):
\[ \det(A - \lambda I) = \begin{vmatrix} 2-\lambda & 1 & 0 \\ 0 & 1-\lambda & -1 \\ 0 & 2 & 4-\lambda \end{vmatrix} = 0 \]
Expanding along the first column:
\[ (2-\lambda) \left[ (1-\lambda)(4-\lambda) - (-1)(2) \right] = 0 \]
\[ (2-\lambda) \left[ \lambda^2 - 5\lambda + 4 + 2 \right] = 0 \]
\[ (2-\lambda)(\lambda^2 - 5\lambda + 6) = 0 \]
\[ 2\lambda^2 - 10\lambda + 12 - \lambda^3 + 5\lambda^2 - 6\lambda = 0 \]
\[ -\lambda^3 + 7\lambda^2 - 16\lambda + 12 = 0 \]
Multiplying by \(-1\), we obtain the characteristic equation:
\[ \lambda^3 - 7\lambda^2 + 16\lambda - 12 = 0 \]
By the Cayley-Hamilton Theorem, any square matrix satisfies its own characteristic equation. Replacing \( \lambda \) with \( A \) and the constant term \( 12 \) with \( 12I \) yields:
\[ A^3 - 7A^2 + 16A - 12I = 0 \]

**Part (ii)**
Since \( \det(A) = 12 \ne 0 \), the inverse matrix \( A^{-1} \) exists. Multiplying the relation from (i) by \( A^{-1} \):
\[ A^{-1}(A^3 - 7A^2 + 16A - 12I) = 0 \]
\[ A^2 - 7A + 16I - 12A^{-1} = 0 \]
Rearranging for \( A^{-1} \):
\[ 12A^{-1} = A^2 - 7A + 16I \]
\[ A^{-1} = \frac{1}{12}A^2 - \frac{7}{12}A + \frac{4}{3}I \]
Thus, the constants are \( p = \frac{1}{12} \), \( q = -\frac{7}{12} \), and \( r = \frac{4}{3} \).

**Part (iii)**
From the Cayley-Hamilton relation, we can write:
\[ A^3 = 7A^2 - 16A + 12I \]
Multiplying this entire equation by \( A \):
\[ A^4 = 7A^3 - 16A^2 + 12A \]
Now, substitute the expression for \( A^3 \) back into this equation:
\[ A^4 = 7(7A^2 - 16A + 12I) - 16A^2 + 12A \]
\[ A^4 = 49A^2 - 112A + 84I - 16A^2 + 12A \]
\[ A^4 = (49 - 16)A^2 + (-112 + 12)A + 84I \]
\[ A^4 = 33A^2 - 100A + 84I \]
Thus, the constants are \( a = 33 \), \( b = -100 \), and \( c = 84 \).

**Part (i)**
* **M1**: For attempting to find the determinant \( \det(A - \lambda I) = 0 \).
* **A1**: For obtaining the correct characteristic equation \( \lambda^3 - 7\lambda^2 + 16\lambda - 12 = 0 \).
* **B1**: For clearly stating that by the Cayley-Hamilton Theorem, the matrix \( A \) satisfies its characteristic equation to show \( A^3 - 7A^2 + 16A - 12I = 0 \).

**Part (ii)**
* **M1**: For multiplying the equation by \( A^{-1} \) and attempting to isolate \( A^{-1} \).
* **A1**: For obtaining the correct expression \( A^{-1} = \frac{1}{12}A^2 - \frac{7}{12}A + \frac{4}{3}I \) (or equivalent values of \( p, q, r \)).

**Part (iii)**
* **M1**: For multiplying the relation by \( A \) to get \( A^4 = 7A^3 - 16A^2 + 12A \).
* **A1**: For obtaining the correct expression for \( A^4 \) in terms of \( A^3 \).
* **M1**: For substituting \( A^3 = 7A^2 - 16A + 12I \) into the expression for \( A^4 \).
* **A1**: For obtaining the correct final expression \( A^4 = 33A^2 - 100A + 84I \) (or equivalent values of \( a, b, c \)).

To find the complementary function, we solve the auxiliary equation: \(m^2 + 4m + 4 = 0\), which gives \((m+2)^2 = 0\), so \(m = -2\) (repeated root). Thus, the complementary function is \(y_c = (A + Bx)e^{-2x}\). For the particular integral, we consider the two terms on the right-hand side separately. For the term \(8e^{-2x}\), since \(e^{-2x}\) and \(xe^{-2x}\) are already in the complementary function, we try a particular integral of the form \(y_{p1} = Cx^2 e^{-2x}\). Differentiating this gives \(y_{p1}' = C(2x - 2x^2)e^{-2x}\) and \(y_{p1}'' = C(2 - 8x + 4x^2)e^{-2x}\). Substituting these into the differential equation: \(C(2 - 8x + 4x^2)e^{-2x} + 4C(2x - 2x^2)e^{-2x} + 4Cx^2 e^{-2x} = 8e^{-2x}\), which simplifies to \(2Ce^{-2x} = 8e^{-2x}\), hence \(C = 4\). For the term \(25\cos x\), we try a particular integral of the form \(y_{p2} = D\cos x + E\sin x\). Differentiating gives \(y_{p2}' = -D\sin x + E\cos x\) and \(y_{p2}'' = -D\cos x - E\sin x\). Substituting these into the differential equation: \((-D\cos x - E\sin x) + 4(-D\sin x + E\cos x) + 4(D\cos x + E\sin x) = 25\cos x\). Equating coefficients of \(\cos x\) and \(\sin x\) gives: \(3D + 4E = 25\) and \(-4D + 3E = 0\). Solving these simultaneous equations yields \(D = 3\) and \(E = 4\). Thus, the general solution is: \(y = (A + Bx)e^{-2x} + 4x^2 e^{-2x} + 3\cos x + 4\sin x\). Using the initial condition \(y = 3\) when \(x = 0\): \(3 = A + 3\), which gives \(A = 0\). Using the initial condition \(\frac{dy}{dx} = 5\) when \(x = 0\): First differentiate \(y = Bxe^{-2x} + 4x^2 e^{-2x} + 3\cos x + 4\sin x\) to get \(\frac{dy}{dx} = Be^{-2x} - 2Bxe^{-2x} + 8xe^{-2x} - 8x^2 e^{-2x} - 3\sin x + 4\cos x\). Substituting \(x = 0\) and \(\frac{dy}{dx} = 5\) gives \(5 = B + 4\), so \(B = 1\). Therefore, the particular solution is \(y = (x + 4x^2)e^{-2x} + 3\cos x + 4\sin x\).

M1: Attempt to solve the auxiliary equation and find the complementary function. A1: Correct complementary function \(y_c = (A+Bx)e^{-2x}\). M1: Correct form for the first part of the particular integral \(y_{p1} = Cx^2 e^{-2x}\), differentiate and substitute. A1: Correct coefficient \(C = 4\). M1: Correct form for the second part of the particular integral \(y_{p2} = D\cos x + E\sin x\), differentiate and substitute. A1: Correct coefficients \(D = 3\) and \(E = 4\). A1: Correct general solution \(y = (A+Bx)e^{-2x} + 4x^2 e^{-2x} + 3\cos x + 4\sin x\). M1: Use initial condition \(y(0)=3\) to find \(A = 0\). M1: Differentiate and use initial condition \(y'(0)=5\) to find \(B = 1\). A1: Correct particular solution \(y = (x + 4x^2)e^{-2x} + 3\cos x + 4\sin x\).

To find the complementary function, we solve the auxiliary equation: \(m^2 + 4m + 4 = 0\), which gives \((m+2)^2 = 0\), so \(m = -2\) (repeated root). Thus, the complementary function is \(y_c = (A + Bx)e^{-2x}\). For the particular integral, we consider the two terms on the right-hand side separately. For the term \(8e^{-2x}\), since \(e^{-2x}\) and \(xe^{-2x}\) are already in the complementary function, we try a particular integral of the form \(y_{p1} = Cx^2 e^{-2x}\). Differentiating this gives \(y_{p1}' = C(2x - 2x^2)e^{-2x}\) and \(y_{p1}'' = C(2 - 8x + 4x^2)e^{-2x}\). Substituting these into the differential equation: \(C(2 - 8x + 4x^2)e^{-2x} + 4C(2x - 2x^2)e^{-2x} + 4Cx^2 e^{-2x} = 8e^{-2x}\), which simplifies to \(2Ce^{-2x} = 8e^{-2x}\), hence \(C = 4\). For the term \(25\cos x\), we try a particular integral of the form \(y_{p2} = D\cos x + E\sin x\). Differentiating gives \(y_{p2}' = -D\sin x + E\cos x\) and \(y_{p2}'' = -D\cos x - E\sin x\). Substituting these into the differential equation: \((-D\cos x - E\sin x) + 4(-D\sin x + E\cos x) + 4(D\cos x + E\sin x) = 25\cos x\). Equating coefficients of \(\cos x\) and \(\sin x\) gives: \(3D + 4E = 25\) and \(-4D + 3E = 0\). Solving these simultaneous equations yields \(D = 3\) and \(E = 4\). Thus, the general solution is: \(y = (A + Bx)e^{-2x} + 4x^2 e^{-2x} + 3\cos x + 4\sin x\). Using the initial condition \(y = 3\) when \(x = 0\): \(3 = A + 3\), which gives \(A = 0\). Using the initial condition \(\frac{dy}{dx} = 5\) when \(x = 0\): First differentiate \(y = Bxe^{-2x} + 4x^2 e^{-2x} + 3\cos x + 4\sin x\) to get \(\frac{dy}{dx} = Be^{-2x} - 2Bxe^{-2x} + 8xe^{-2x} - 8x^2 e^{-2x} - 3\sin x + 4\cos x\). Substituting \(x = 0\) and \(\frac{dy}{dx} = 5\) gives \(5 = B + 4\), so \(B = 1\). Therefore, the particular solution is \(y = (x + 4x^2)e^{-2x} + 3\cos x + 4\sin x\).

M1: Attempt to solve the auxiliary equation and find the complementary function. A1: Correct complementary function \(y_c = (A+Bx)e^{-2x}\). M1: Correct form for the first part of the particular integral \(y_{p1} = Cx^2 e^{-2x}\), differentiate and substitute. A1: Correct coefficient \(C = 4\). M1: Correct form for the second part of the particular integral \(y_{p2} = D\cos x + E\sin x\), differentiate and substitute. A1: Correct coefficients \(D = 3\) and \(E = 4\). A1: Correct general solution \(y = (A+Bx)e^{-2x} + 4x^2 e^{-2x} + 3\cos x + 4\sin x\). M1: Use initial condition \(y(0)=3\) to find \(A = 0\). M1: Differentiate and use initial condition \(y'(0)=5\) to find \(B = 1\). A1: Correct particular solution \(y = (x + 4x^2)e^{-2x} + 3\cos x + 4\sin x\).

To find the complementary function, we solve the auxiliary equation: \(m^2 + 4m + 4 = 0\), which gives \((m+2)^2 = 0\), so \(m = -2\) (repeated root). Thus, the complementary function is \(y_c = (A + Bx)e^{-2x}\). For the particular integral, we consider the two terms on the right-hand side separately. For the term \(8e^{-2x}\), since \(e^{-2x}\) and \(xe^{-2x}\) are already in the complementary function, we try a particular integral of the form \(y_{p1} = Cx^2 e^{-2x}\). Differentiating this gives \(y_{p1}' = C(2x - 2x^2)e^{-2x}\) and \(y_{p1}'' = C(2 - 8x + 4x^2)e^{-2x}\). Substituting these into the differential equation: \(C(2 - 8x + 4x^2)e^{-2x} + 4C(2x - 2x^2)e^{-2x} + 4Cx^2 e^{-2x} = 8e^{-2x}\), which simplifies to \(2Ce^{-2x} = 8e^{-2x}\), hence \(C = 4\). For the term \(25\cos x\), we try a particular integral of the form \(y_{p2} = D\cos x + E\sin x\). Differentiating gives \(y_{p2}' = -D\sin x + E\cos x\) and \(y_{p2}'' = -D\cos x - E\sin x\). Substituting these into the differential equation: \((-D\cos x - E\sin x) + 4(-D\sin x + E\cos x) + 4(D\cos x + E\sin x) = 25\cos x\). Equating coefficients of \(\cos x\) and \(\sin x\) gives: \(3D + 4E = 25\) and \(-4D + 3E = 0\). Solving these simultaneous equations yields \(D = 3\) and \(E = 4\). Thus, the general solution is: \(y = (A + Bx)e^{-2x} + 4x^2 e^{-2x} + 3\cos x + 4\sin x\). Using the initial condition \(y = 3\) when \(x = 0\): \(3 = A + 3\), which gives \(A = 0\). Using the initial condition \(\frac{dy}{dx} = 5\) when \(x = 0\): First differentiate \(y = Bxe^{-2x} + 4x^2 e^{-2x} + 3\cos x + 4\sin x\) to get \(\frac{dy}{dx} = Be^{-2x} - 2Bxe^{-2x} + 8xe^{-2x} - 8x^2 e^{-2x} - 3\sin x + 4\cos x\). Substituting \(x = 0\) and \(\frac{dy}{dx} = 5\) gives \(5 = B + 4\), so \(B = 1\). Therefore, the particular solution is \(y = (x + 4x^2)e^{-2x} + 3\cos x + 4\sin x\).

M1: Attempt to solve the auxiliary equation and find the complementary function. A1: Correct complementary function \(y_c = (A+Bx)e^{-2x}\). M1: Correct form for the first part of the particular integral \(y_{p1} = Cx^2 e^{-2x}\), differentiate and substitute. A1: Correct coefficient \(C = 4\). M1: Correct form for the second part of the particular integral \(y_{p2} = D\cos x + E\sin x\), differentiate and substitute. A1: Correct coefficients \(D = 3\) and \(E = 4\). A1: Correct general solution \(y = (A+Bx)e^{-2x} + 4x^2 e^{-2x} + 3\cos x + 4\sin x\). M1: Use initial condition \(y(0)=3\) to find \(A = 0\). M1: Differentiate and use initial condition \(y'(0)=5\) to find \(B = 1\). A1: Correct particular solution \(y = (x + 4x^2)e^{-2x} + 3\cos x + 4\sin x\).

To find the complementary function, we solve the auxiliary equation: \(m^2 + 4m + 4 = 0\), which gives \((m+2)^2 = 0\), so \(m = -2\) (repeated root). Thus, the complementary function is \(y_c = (A + Bx)e^{-2x}\). For the particular integral, we consider the two terms on the right-hand side separately. For the term \(8e^{-2x}\), since \(e^{-2x}\) and \(xe^{-2x}\) are already in the complementary function, we try a particular integral of the form \(y_{p1} = Cx^2 e^{-2x}\). Differentiating this gives \(y_{p1}' = C(2x - 2x^2)e^{-2x}\) and \(y_{p1}'' = C(2 - 8x + 4x^2)e^{-2x}\). Substituting these into the differential equation: \(C(2 - 8x + 4x^2)e^{-2x} + 4C(2x - 2x^2)e^{-2x} + 4Cx^2 e^{-2x} = 8e^{-2x}\), which simplifies to \(2Ce^{-2x} = 8e^{-2x}\), hence \(C = 4\). For the term \(25\cos x\), we try a particular integral of the form \(y_{p2} = D\cos x + E\sin x\). Differentiating gives \(y_{p2}' = -D\sin x + E\cos x\) and \(y_{p2}'' = -D\cos x - E\sin x\). Substituting these into the differential equation: \((-D\cos x - E\sin x) + 4(-D\sin x + E\cos x) + 4(D\cos x + E\sin x) = 25\cos x\). Equating coefficients of \(\cos x\) and \(\sin x\) gives: \(3D + 4E = 25\) and \(-4D + 3E = 0\). Solving these simultaneous equations yields \(D = 3\) and \(E = 4\). Thus, the general solution is: \(y = (A + Bx)e^{-2x} + 4x^2 e^{-2x} + 3\cos x + 4\sin x\). Using the initial condition \(y = 3\) when \(x = 0\): \(3 = A + 3\), which gives \(A = 0\). Using the initial condition \(\frac{dy}{dx} = 5\) when \(x = 0\): First differentiate \(y = Bxe^{-2x} + 4x^2 e^{-2x} + 3\cos x + 4\sin x\) to get \(\frac{dy}{dx} = Be^{-2x} - 2Bxe^{-2x} + 8xe^{-2x} - 8x^2 e^{-2x} - 3\sin x + 4\cos x\). Substituting \(x = 0\) and \(\frac{dy}{dx} = 5\) gives \(5 = B + 4\), so \(B = 1\). Therefore, the particular solution is \(y = (x + 4x^2)e^{-2x} + 3\cos x + 4\sin x\).

M1: Attempt to solve the auxiliary equation and find the complementary function. A1: Correct complementary function \(y_c = (A+Bx)e^{-2x}\). M1: Correct form for the first part of the particular integral \(y_{p1} = Cx^2 e^{-2x}\), differentiate and substitute. A1: Correct coefficient \(C = 4\). M1: Correct form for the second part of the particular integral \(y_{p2} = D\cos x + E\sin x\), differentiate and substitute. A1: Correct coefficients \(D = 3\) and \(E = 4\). A1: Correct general solution \(y = (A+Bx)e^{-2x} + 4x^2 e^{-2x} + 3\cos x + 4\sin x\). M1: Use initial condition \(y(0)=3\) to find \(A = 0\). M1: Differentiate and use initial condition \(y'(0)=5\) to find \(B = 1\). A1: Correct particular solution \(y = (x + 4x^2)e^{-2x} + 3\cos x + 4\sin x\).

To find the complementary function, we solve the auxiliary equation: \(m^2 + 4m + 4 = 0\), which gives \((m+2)^2 = 0\), so \(m = -2\) (repeated root). Thus, the complementary function is \(y_c = (A + Bx)e^{-2x}\). For the particular integral, we consider the two terms on the right-hand side separately. For the term \(8e^{-2x}\), since \(e^{-2x}\) and \(xe^{-2x}\) are already in the complementary function, we try a particular integral of the form \(y_{p1} = Cx^2 e^{-2x}\). Differentiating this gives \(y_{p1}' = C(2x - 2x^2)e^{-2x}\) and \(y_{p1}'' = C(2 - 8x + 4x^2)e^{-2x}\). Substituting these into the differential equation: \(C(2 - 8x + 4x^2)e^{-2x} + 4C(2x - 2x^2)e^{-2x} + 4Cx^2 e^{-2x} = 8e^{-2x}\), which simplifies to \(2Ce^{-2x} = 8e^{-2x}\), hence \(C = 4\). For the term \(25\cos x\), we try a particular integral of the form \(y_{p2} = D\cos x + E\sin x\). Differentiating gives \(y_{p2}' = -D\sin x + E\cos x\) and \(y_{p2}'' = -D\cos x - E\sin x\). Substituting these into the differential equation: \((-D\cos x - E\sin x) + 4(-D\sin x + E\cos x) + 4(D\cos x + E\sin x) = 25\cos x\). Equating coefficients of \(\cos x\) and \(\sin x\) gives: \(3D + 4E = 25\) and \(-4D + 3E = 0\). Solving these simultaneous equations yields \(D = 3\) and \(E = 4\). Thus, the general solution is: \(y = (A + Bx)e^{-2x} + 4x^2 e^{-2x} + 3\cos x + 4\sin x\). Using the initial condition \(y = 3\) when \(x = 0\): \(3 = A + 3\), which gives \(A = 0\). Using the initial condition \(\frac{dy}{dx} = 5\) when \(x = 0\): First differentiate \(y = Bxe^{-2x} + 4x^2 e^{-2x} + 3\cos x + 4\sin x\) to get \(\frac{dy}{dx} = Be^{-2x} - 2Bxe^{-2x} + 8xe^{-2x} - 8x^2 e^{-2x} - 3\sin x + 4\cos x\). Substituting \(x = 0\) and \(\frac{dy}{dx} = 5\) gives \(5 = B + 4\), so \(B = 1\). Therefore, the particular solution is \(y = (x + 4x^2)e^{-2x} + 3\cos x + 4\sin x\).

M1: Attempt to solve the auxiliary equation and find the complementary function. A1: Correct complementary function \(y_c = (A+Bx)e^{-2x}\). M1: Correct form for the first part of the particular integral \(y_{p1} = Cx^2 e^{-2x}\), differentiate and substitute. A1: Correct coefficient \(C = 4\). M1: Correct form for the second part of the particular integral \(y_{p2} = D\cos x + E\sin x\), differentiate and substitute. A1: Correct coefficients \(D = 3\) and \(E = 4\). A1: Correct general solution \(y = (A+Bx)e^{-2x} + 4x^2 e^{-2x} + 3\cos x + 4\sin x\). M1: Use initial condition \(y(0)=3\) to find \(A = 0\). M1: Differentiate and use initial condition \(y'(0)=5\) to find \(B = 1\). A1: Correct particular solution \(y = (x + 4x^2)e^{-2x} + 3\cos x + 4\sin x\).

To find the complementary function, we solve the auxiliary equation: \(m^2 + 4m + 4 = 0\), which gives \((m+2)^2 = 0\), so \(m = -2\) (repeated root). Thus, the complementary function is \(y_c = (A + Bx)e^{-2x}\). For the particular integral, we consider the two terms on the right-hand side separately. For the term \(8e^{-2x}\), since \(e^{-2x}\) and \(xe^{-2x}\) are already in the complementary function, we try a particular integral of the form \(y_{p1} = Cx^2 e^{-2x}\). Differentiating this gives \(y_{p1}' = C(2x - 2x^2)e^{-2x}\) and \(y_{p1}'' = C(2 - 8x + 4x^2)e^{-2x}\). Substituting these into the differential equation: \(C(2 - 8x + 4x^2)e^{-2x} + 4C(2x - 2x^2)e^{-2x} + 4Cx^2 e^{-2x} = 8e^{-2x}\), which simplifies to \(2Ce^{-2x} = 8e^{-2x}\), hence \(C = 4\). For the term \(25\cos x\), we try a particular integral of the form \(y_{p2} = D\cos x + E\sin x\). Differentiating gives \(y_{p2}' = -D\sin x + E\cos x\) and \(y_{p2}'' = -D\cos x - E\sin x\). Substituting these into the differential equation: \((-D\cos x - E\sin x) + 4(-D\sin x + E\cos x) + 4(D\cos x + E\sin x) = 25\cos x\). Equating coefficients of \(\cos x\) and \(\sin x\) gives: \(3D + 4E = 25\) and \(-4D + 3E = 0\). Solving these simultaneous equations yields \(D = 3\) and \(E = 4\). Thus, the general solution is: \(y = (A + Bx)e^{-2x} + 4x^2 e^{-2x} + 3\cos x + 4\sin x\). Using the initial condition \(y = 3\) when \(x = 0\): \(3 = A + 3\), which gives \(A = 0\). Using the initial condition \(\frac{dy}{dx} = 5\) when \(x = 0\): First differentiate \(y = Bxe^{-2x} + 4x^2 e^{-2x} + 3\cos x + 4\sin x\) to get \(\frac{dy}{dx} = Be^{-2x} - 2Bxe^{-2x} + 8xe^{-2x} - 8x^2 e^{-2x} - 3\sin x + 4\cos x\). Substituting \(x = 0\) and \(\frac{dy}{dx} = 5\) gives \(5 = B + 4\), so \(B = 1\). Therefore, the particular solution is \(y = (x + 4x^2)e^{-2x} + 3\cos x + 4\sin x\).

M1: Attempt to solve the auxiliary equation and find the complementary function. A1: Correct complementary function \(y_c = (A+Bx)e^{-2x}\). M1: Correct form for the first part of the particular integral \(y_{p1} = Cx^2 e^{-2x}\), differentiate and substitute. A1: Correct coefficient \(C = 4\). M1: Correct form for the second part of the particular integral \(y_{p2} = D\cos x + E\sin x\), differentiate and substitute. A1: Correct coefficients \(D = 3\) and \(E = 4\). A1: Correct general solution \(y = (A+Bx)e^{-2x} + 4x^2 e^{-2x} + 3\cos x + 4\sin x\). M1: Use initial condition \(y(0)=3\) to find \(A = 0\). M1: Differentiate and use initial condition \(y'(0)=5\) to find \(B = 1\). A1: Correct particular solution \(y = (x + 4x^2)e^{-2x} + 3\cos x + 4\sin x\).

To find the complementary function, we solve the auxiliary equation: \(m^2 + 4m + 4 = 0\), which gives \((m+2)^2 = 0\), so \(m = -2\) (repeated root). Thus, the complementary function is \(y_c = (A + Bx)e^{-2x}\). For the particular integral, we consider the two terms on the right-hand side separately. For the term \(8e^{-2x}\), since \(e^{-2x}\) and \(xe^{-2x}\) are already in the complementary function, we try a particular integral of the form \(y_{p1} = Cx^2 e^{-2x}\). Differentiating this gives \(y_{p1}' = C(2x - 2x^2)e^{-2x}\) and \(y_{p1}'' = C(2 - 8x + 4x^2)e^{-2x}\). Substituting these into the differential equation: \(C(2 - 8x + 4x^2)e^{-2x} + 4C(2x - 2x^2)e^{-2x} + 4Cx^2 e^{-2x} = 8e^{-2x}\), which simplifies to \(2Ce^{-2x} = 8e^{-2x}\), hence \(C = 4\). For the term \(25\cos x\), we try a particular integral of the form \(y_{p2} = D\cos x + E\sin x\). Differentiating gives \(y_{p2}' = -D\sin x + E\cos x\) and \(y_{p2}'' = -D\cos x - E\sin x\). Substituting these into the differential equation: \((-D\cos x - E\sin x) + 4(-D\sin x + E\cos x) + 4(D\cos x + E\sin x) = 25\cos x\). Equating coefficients of \(\cos x\) and \(\sin x\) gives: \(3D + 4E = 25\) and \(-4D + 3E = 0\). Solving these simultaneous equations yields \(D = 3\) and \(E = 4\). Thus, the general solution is: \(y = (A + Bx)e^{-2x} + 4x^2 e^{-2x} + 3\cos x + 4\sin x\). Using the initial condition \(y = 3\) when \(x = 0\): \(3 = A + 3\), which gives \(A = 0\). Using the initial condition \(\frac{dy}{dx} = 5\) when \(x = 0\): First differentiate \(y = Bxe^{-2x} + 4x^2 e^{-2x} + 3\cos x + 4\sin x\) to get \(\frac{dy}{dx} = Be^{-2x} - 2Bxe^{-2x} + 8xe^{-2x} - 8x^2 e^{-2x} - 3\sin x + 4\cos x\). Substituting \(x = 0\) and \(\frac{dy}{dx} = 5\) gives \(5 = B + 4\), so \(B = 1\). Therefore, the particular solution is \(y = (x + 4x^2)e^{-2x} + 3\cos x + 4\sin x\).

M1: Attempt to solve the auxiliary equation and find the complementary function. A1: Correct complementary function \(y_c = (A+Bx)e^{-2x}\). M1: Correct form for the first part of the particular integral \(y_{p1} = Cx^2 e^{-2x}\), differentiate and substitute. A1: Correct coefficient \(C = 4\). M1: Correct form for the second part of the particular integral \(y_{p2} = D\cos x + E\sin x\), differentiate and substitute. A1: Correct coefficients \(D = 3\) and \(E = 4\). A1: Correct general solution \(y = (A+Bx)e^{-2x} + 4x^2 e^{-2x} + 3\cos x + 4\sin x\). M1: Use initial condition \(y(0)=3\) to find \(A = 0\). M1: Differentiate and use initial condition \(y'(0)=5\) to find \(B = 1\). A1: Correct particular solution \(y = (x + 4x^2)e^{-2x} + 3\cos x + 4\sin x\).

To find the complementary function, we solve the auxiliary equation: \(m^2 + 4m + 4 = 0\), which gives \((m+2)^2 = 0\), so \(m = -2\) (repeated root). Thus, the complementary function is \(y_c = (A + Bx)e^{-2x}\). For the particular integral, we consider the two terms on the right-hand side separately. For the term \(8e^{-2x}\), since \(e^{-2x}\) and \(xe^{-2x}\) are already in the complementary function, we try a particular integral of the form \(y_{p1} = Cx^2 e^{-2x}\). Differentiating this gives \(y_{p1}' = C(2x - 2x^2)e^{-2x}\) and \(y_{p1}'' = C(2 - 8x + 4x^2)e^{-2x}\). Substituting these into the differential equation: \(C(2 - 8x + 4x^2)e^{-2x} + 4C(2x - 2x^2)e^{-2x} + 4Cx^2 e^{-2x} = 8e^{-2x}\), which simplifies to \(2Ce^{-2x} = 8e^{-2x}\), hence \(C = 4\). For the term \(25\cos x\), we try a particular integral of the form \(y_{p2} = D\cos x + E\sin x\). Differentiating gives \(y_{p2}' = -D\sin x + E\cos x\) and \(y_{p2}'' = -D\cos x - E\sin x\). Substituting these into the differential equation: \((-D\cos x - E\sin x) + 4(-D\sin x + E\cos x) + 4(D\cos x + E\sin x) = 25\cos x\). Equating coefficients of \(\cos x\) and \(\sin x\) gives: \(3D + 4E = 25\) and \(-4D + 3E = 0\). Solving these simultaneous equations yields \(D = 3\) and \(E = 4\). Thus, the general solution is: \(y = (A + Bx)e^{-2x} + 4x^2 e^{-2x} + 3\cos x + 4\sin x\). Using the initial condition \(y = 3\) when \(x = 0\): \(3 = A + 3\), which gives \(A = 0\). Using the initial condition \(\frac{dy}{dx} = 5\) when \(x = 0\): First differentiate \(y = Bxe^{-2x} + 4x^2 e^{-2x} + 3\cos x + 4\sin x\) to get \(\frac{dy}{dx} = Be^{-2x} - 2Bxe^{-2x} + 8xe^{-2x} - 8x^2 e^{-2x} - 3\sin x + 4\cos x\). Substituting \(x = 0\) and \(\frac{dy}{dx} = 5\) gives \(5 = B + 4\), so \(B = 1\). Therefore, the particular solution is \(y = (x + 4x^2)e^{-2x} + 3\cos x + 4\sin x\).

M1: Attempt to solve the auxiliary equation and find the complementary function. A1: Correct complementary function \(y_c = (A+Bx)e^{-2x}\). M1: Correct form for the first part of the particular integral \(y_{p1} = Cx^2 e^{-2x}\), differentiate and substitute. A1: Correct coefficient \(C = 4\). M1: Correct form for the second part of the particular integral \(y_{p2} = D\cos x + E\sin x\), differentiate and substitute. A1: Correct coefficients \(D = 3\) and \(E = 4\). A1: Correct general solution \(y = (A+Bx)e^{-2x} + 4x^2 e^{-2x} + 3\cos x + 4\sin x\). M1: Use initial condition \(y(0)=3\) to find \(A = 0\). M1: Differentiate and use initial condition \(y'(0)=5\) to find \(B = 1\). A1: Correct particular solution \(y = (x + 4x^2)e^{-2x} + 3\cos x + 4\sin x\).

To find the complementary function, we solve the auxiliary equation: \(m^2 + 4m + 4 = 0\), which gives \((m+2)^2 = 0\), so \(m = -2\) (repeated root). Thus, the complementary function is \(y_c = (A + Bx)e^{-2x}\). For the particular integral, we consider the two terms on the right-hand side separately. For the term \(8e^{-2x}\), since \(e^{-2x}\) and \(xe^{-2x}\) are already in the complementary function, we try a particular integral of the form \(y_{p1} = Cx^2 e^{-2x}\). Differentiating this gives \(y_{p1}' = C(2x - 2x^2)e^{-2x}\) and \(y_{p1}'' = C(2 - 8x + 4x^2)e^{-2x}\). Substituting these into the differential equation: \(C(2 - 8x + 4x^2)e^{-2x} + 4C(2x - 2x^2)e^{-2x} + 4Cx^2 e^{-2x} = 8e^{-2x}\), which simplifies to \(2Ce^{-2x} = 8e^{-2x}\), hence \(C = 4\). For the term \(25\cos x\), we try a particular integral of the form \(y_{p2} = D\cos x + E\sin x\). Differentiating gives \(y_{p2}' = -D\sin x + E\cos x\) and \(y_{p2}'' = -D\cos x - E\sin x\). Substituting these into the differential equation: \((-D\cos x - E\sin x) + 4(-D\sin x + E\cos x) + 4(D\cos x + E\sin x) = 25\cos x\). Equating coefficients of \(\cos x\) and \(\sin x\) gives: \(3D + 4E = 25\) and \(-4D + 3E = 0\). Solving these simultaneous equations yields \(D = 3\) and \(E = 4\). Thus, the general solution is: \(y = (A + Bx)e^{-2x} + 4x^2 e^{-2x} + 3\cos x + 4\sin x\). Using the initial condition \(y = 3\) when \(x = 0\): \(3 = A + 3\), which gives \(A = 0\). Using the initial condition \(\frac{dy}{dx} = 5\) when \(x = 0\): First differentiate \(y = Bxe^{-2x} + 4x^2 e^{-2x} + 3\cos x + 4\sin x\) to get \(\frac{dy}{dx} = Be^{-2x} - 2Bxe^{-2x} + 8xe^{-2x} - 8x^2 e^{-2x} - 3\sin x + 4\cos x\). Substituting \(x = 0\) and \(\frac{dy}{dx} = 5\) gives \(5 = B + 4\), so \(B = 1\). Therefore, the particular solution is \(y = (x + 4x^2)e^{-2x} + 3\cos x + 4\sin x\).

M1: Attempt to solve the auxiliary equation and find the complementary function. A1: Correct complementary function \(y_c = (A+Bx)e^{-2x}\). M1: Correct form for the first part of the particular integral \(y_{p1} = Cx^2 e^{-2x}\), differentiate and substitute. A1: Correct coefficient \(C = 4\). M1: Correct form for the second part of the particular integral \(y_{p2} = D\cos x + E\sin x\), differentiate and substitute. A1: Correct coefficients \(D = 3\) and \(E = 4\). A1: Correct general solution \(y = (A+Bx)e^{-2x} + 4x^2 e^{-2x} + 3\cos x + 4\sin x\). M1: Use initial condition \(y(0)=3\) to find \(A = 0\). M1: Differentiate and use initial condition \(y'(0)=5\) to find \(B = 1\). A1: Correct particular solution \(y = (x + 4x^2)e^{-2x} + 3\cos x + 4\sin x\).

To find the complementary function, we solve the auxiliary equation: \(m^2 + 4m + 4 = 0\), which gives \((m+2)^2 = 0\), so \(m = -2\) (repeated root). Thus, the complementary function is \(y_c = (A + Bx)e^{-2x}\). For the particular integral, we consider the two terms on the right-hand side separately. For the term \(8e^{-2x}\), since \(e^{-2x}\) and \(xe^{-2x}\) are already in the complementary function, we try a particular integral of the form \(y_{p1} = Cx^2 e^{-2x}\). Differentiating this gives \(y_{p1}' = C(2x - 2x^2)e^{-2x}\) and \(y_{p1}'' = C(2 - 8x + 4x^2)e^{-2x}\). Substituting these into the differential equation: \(C(2 - 8x + 4x^2)e^{-2x} + 4C(2x - 2x^2)e^{-2x} + 4Cx^2 e^{-2x} = 8e^{-2x}\), which simplifies to \(2Ce^{-2x} = 8e^{-2x}\), hence \(C = 4\). For the term \(25\cos x\), we try a particular integral of the form \(y_{p2} = D\cos x + E\sin x\). Differentiating gives \(y_{p2}' = -D\sin x + E\cos x\) and \(y_{p2}'' = -D\cos x - E\sin x\). Substituting these into the differential equation: \((-D\cos x - E\sin x) + 4(-D\sin x + E\cos x) + 4(D\cos x + E\sin x) = 25\cos x\). Equating coefficients of \(\cos x\) and \(\sin x\) gives: \(3D + 4E = 25\) and \(-4D + 3E = 0\). Solving these simultaneous equations yields \(D = 3\) and \(E = 4\). Thus, the general solution is: \(y = (A + Bx)e^{-2x} + 4x^2 e^{-2x} + 3\cos x + 4\sin x\). Using the initial condition \(y = 3\) when \(x = 0\): \(3 = A + 3\), which gives \(A = 0\). Using the initial condition \(\frac{dy}{dx} = 5\) when \(x = 0\): First differentiate \(y = Bxe^{-2x} + 4x^2 e^{-2x} + 3\cos x + 4\sin x\) to get \(\frac{dy}{dx} = Be^{-2x} - 2Bxe^{-2x} + 8xe^{-2x} - 8x^2 e^{-2x} - 3\sin x + 4\cos x\). Substituting \(x = 0\) and \(\frac{dy}{dx} = 5\) gives \(5 = B + 4\), so \(B = 1\). Therefore, the particular solution is \(y = (x + 4x^2)e^{-2x} + 3\cos x + 4\sin x\).

M1: Attempt to solve the auxiliary equation and find the complementary function. A1: Correct complementary function \(y_c = (A+Bx)e^{-2x}\). M1: Correct form for the first part of the particular integral \(y_{p1} = Cx^2 e^{-2x}\), differentiate and substitute. A1: Correct coefficient \(C = 4\). M1: Correct form for the second part of the particular integral \(y_{p2} = D\cos x + E\sin x\), differentiate and substitute. A1: Correct coefficients \(D = 3\) and \(E = 4\). A1: Correct general solution \(y = (A+Bx)e^{-2x} + 4x^2 e^{-2x} + 3\cos x + 4\sin x\). M1: Use initial condition \(y(0)=3\) to find \(A = 0\). M1: Differentiate and use initial condition \(y'(0)=5\) to find \(B = 1\). A1: Correct particular solution \(y = (x + 4x^2)e^{-2x} + 3\cos x + 4\sin x\).

To find the complementary function, we solve the auxiliary equation: \(m^2 + 4m + 4 = 0\), which gives \((m+2)^2 = 0\), so \(m = -2\) (repeated root). Thus, the complementary function is \(y_c = (A + Bx)e^{-2x}\). For the particular integral, we consider the two terms on the right-hand side separately. For the term \(8e^{-2x}\), since \(e^{-2x}\) and \(xe^{-2x}\) are already in the complementary function, we try a particular integral of the form \(y_{p1} = Cx^2 e^{-2x}\). Differentiating this gives \(y_{p1}' = C(2x - 2x^2)e^{-2x}\) and \(y_{p1}'' = C(2 - 8x + 4x^2)e^{-2x}\). Substituting these into the differential equation: \(C(2 - 8x + 4x^2)e^{-2x} + 4C(2x - 2x^2)e^{-2x} + 4Cx^2 e^{-2x} = 8e^{-2x}\), which simplifies to \(2Ce^{-2x} = 8e^{-2x}\), hence \(C = 4\). For the term \(25\cos x\), we try a particular integral of the form \(y_{p2} = D\cos x + E\sin x\). Differentiating gives \(y_{p2}' = -D\sin x + E\cos x\) and \(y_{p2}'' = -D\cos x - E\sin x\). Substituting these into the differential equation: \((-D\cos x - E\sin x) + 4(-D\sin x + E\cos x) + 4(D\cos x + E\sin x) = 25\cos x\). Equating coefficients of \(\cos x\) and \(\sin x\) gives: \(3D + 4E = 25\) and \(-4D + 3E = 0\). Solving these simultaneous equations yields \(D = 3\) and \(E = 4\). Thus, the general solution is: \(y = (A + Bx)e^{-2x} + 4x^2 e^{-2x} + 3\cos x + 4\sin x\). Using the initial condition \(y = 3\) when \(x = 0\): \(3 = A + 3\), which gives \(A = 0\). Using the initial condition \(\frac{dy}{dx} = 5\) when \(x = 0\): First differentiate \(y = Bxe^{-2x} + 4x^2 e^{-2x} + 3\cos x + 4\sin x\) to get \(\frac{dy}{dx} = Be^{-2x} - 2Bxe^{-2x} + 8xe^{-2x} - 8x^2 e^{-2x} - 3\sin x + 4\cos x\). Substituting \(x = 0\) and \(\frac{dy}{dx} = 5\) gives \(5 = B + 4\), so \(B = 1\). Therefore, the particular solution is \(y = (x + 4x^2)e^{-2x} + 3\cos x + 4\sin x\).

M1: Attempt to solve the auxiliary equation and find the complementary function. A1: Correct complementary function \(y_c = (A+Bx)e^{-2x}\). M1: Correct form for the first part of the particular integral \(y_{p1} = Cx^2 e^{-2x}\), differentiate and substitute. A1: Correct coefficient \(C = 4\). M1: Correct form for the second part of the particular integral \(y_{p2} = D\cos x + E\sin x\), differentiate and substitute. A1: Correct coefficients \(D = 3\) and \(E = 4\). A1: Correct general solution \(y = (A+Bx)e^{-2x} + 4x^2 e^{-2x} + 3\cos x + 4\sin x\). M1: Use initial condition \(y(0)=3\) to find \(A = 0\). M1: Differentiate and use initial condition \(y'(0)=5\) to find \(B = 1\). A1: Correct particular solution \(y = (x + 4x^2)e^{-2x} + 3\cos x + 4\sin x\).

(a) Since \( f(x) = \frac{1}{2x+1} \) is a strictly decreasing function for \( x \ge 0 \):
- For any interval \( [r-1, r] \) where \( r \ge 1 \), the minimum value of \( f(x) \) occurs at the right endpoint, so \( f(r) < f(x) \) for all \( x \in [r-1, r) \). Taking the integral, we get:
\[ f(r) < \int_{r-1}^r f(x) \, dx \]
Summing this inequality from \( r = 1 \) to \( n \):
\[ \sum_{r=1}^n f(r) < \sum_{r=1}^n \int_{r-1}^r f(x) \, dx = \int_0^n f(x) \, dx \]
Substituting \( f(r) = \frac{1}{2r+1} \):
\[ \sum_{r=1}^n \frac{1}{2r+1} < \int_0^n \frac{1}{2x+1} \, dx \]

- Similarly, for any interval \( [r, r+1] \) where \( r \ge 1 \), the maximum value of \( f(x) \) occurs at the left endpoint, so \( f(x) < f(r) \) for all \( x \in (r, r+1] \). Taking the integral, we get:
\[ \int_r^{r+1} f(x) \, dx < f(r) \]
Summing this inequality from \( r = 1 \) to \( n \):
\[ \int_1^{n+1} f(x) \, dx = \sum_{r=1}^n \int_r^{r+1} f(x) \, dx < \sum_{r=1}^n f(r) \]
Substituting \( f(r) = \frac{1}{2r+1} \):
\[ \int_1^{n+1} \frac{1}{2x+1} \, dx < \sum_{r=1}^n \frac{1}{2r+1} \]
Combining both results gives the required inequality:
\[ \int_{1}^{n+1} \frac{1}{2x+1} \, dx < \sum_{r=1}^n \frac{1}{2r+1} < \int_{0}^{n} \frac{1}{2x+1} \, dx \]

(b) First, evaluate the lower bound integral:
\[ \int_1^{n+1} \frac{1}{2x+1} \, dx = \left[ \frac{1}{2} \ln(2x+1) \right]_1^{n+1} = \frac{1}{2} \ln(2n+3) - \frac{1}{2} \ln(3) = \frac{1}{2} \ln \left( \frac{2n+3}{3} \right) \]
Next, evaluate the upper bound integral:
\[ \int_0^n \frac{1}{2x+1} \, dx = \left[ \frac{1}{2} \ln(2x+1) \right]_0^n = \frac{1}{2} \ln(2n+1) - \frac{1}{2} \ln(1) = \frac{1}{2} \ln(2n+1) \]
Substituting these evaluated integrals back into the inequality from part (a) yields:
\[ \frac{1}{2} \ln \left( \frac{2n+3}{3} \right) < \sum_{r=1}^n \frac{1}{2r+1} < \frac{1}{2} \ln(2n+1) \]

(c) The function \( g(x) = \frac{1}{(2x+1)^2} \) is also strictly decreasing for \( x \ge 1 \). Using the same rectangular method for \( r \ge 2 \):
\[ \int_2^{\infty} g(x) \, dx < \sum_{r=2}^{\infty} g(r) < \int_1^{\infty} g(x) \, dx \]
We find the general integral:
\[ \int_k^{\infty} \frac{1}{(2x+1)^2} \, dx = \left[ -\frac{1}{2(2x+1)} \right]_k^{\infty} = 0 - \left( -\frac{1}{2(2k+1)} \right) = \frac{1}{2(2k+1)} \]
For the lower bound (with \( k = 2 \)):
\[ \int_2^{\infty} \frac{1}{(2x+1)^2} \, dx = \frac{1}{2(5)} = \frac{1}{10} \]
For the upper bound (with \( k = 1 \)):
\[ \int_1^{\infty} \frac{1}{(2x+1)^2} \, dx = \frac{1}{2(3)} = \frac{1}{6} \]
Thus, we have:
\[ \frac{1}{10} < \sum_{r=2}^{\infty} \frac{1}{(2r+1)^2} < \frac{1}{6} \]
Now, we add the first term of the sum (for \( r = 1 \)), which is \( T_1 = \frac{1}{(2(1)+1)^2} = \frac{1}{9} \):
\[ \frac{1}{9} + \frac{1}{10} < S < \frac{1}{9} + \frac{1}{6} \]
\[ \frac{19}{90} < S < \frac{5}{18} \]
Hence, the lower bound is \( \frac{19}{90} \) and the upper bound is \( \frac{5}{18} \).

(a)
- M1: Draw/sketch the curve y = 1/(2x+1) for x >= 0 with rectangles of width 1.
- A1: Explain or show that the sum of rectangle areas for r=1 to n on intervals [r-1, r] lies under the curve, hence establishing the upper bound.
- A1: Explain or show that the sum of rectangle areas for r=1 to n on intervals [r, r+1] lies over the curve, hence establishing the lower bound.
- A1: Combine inequalities clearly to obtain the final stated inequality.

(b)
- M1: Integrate 1/(2x+1) to obtain k ln(2x+1) with constant k = 1/2.
- A1: Show evaluation of lower bound integral at n+1 and 1 to obtain 1/2 ln((2n+3)/3).
- A1: Show evaluation of upper bound integral at n and 0 to obtain 1/2 ln(2n+1).
- A1: Conclude the final combined logarithmic inequality.

(c)
- M1: State or use the inequality: integral from 2 to infinity of g(x) dx < sum from r=2 to infinity of g(r) < integral from 1 to infinity of g(x) dx.
- M1: Integrate 1/(2x+1)^2 to obtain -1/(2(2x+1)) with correct constant.
- A1: Show that the lower integral evaluates to 1/10.
- A1: Show that the upper integral evaluates to 1/6.
- M1: Add the first term T_1 = 1/9 to both bounds.
- A1: Obtain the correct lower bound 19/90 and upper bound 5/18 as simplified rational numbers.

**(a)**

First, we divide the given differential equation by $\cos x$ to express it in standard linear form:
$$\frac{\mathrm{d}y}{\mathrm{d}x} + 2y\tan x = \cos^2 x \sin x$$

The integrating factor $I(x)$ is:
$$I(x) = e^{\int 2\tan x \,\mathrm{d}x} = e^{2\ln|\sec x|} = e^{\ln(\sec^2 x)} = \sec^2 x$$

Multiplying both sides of the differential equation by $\sec^2 x$ gives:
$$\sec^2 x \frac{\mathrm{d}y}{\mathrm{d}x} + 2y\sec^2 x \tan x = \cos^2 x \sin x \sec^2 x$$
$$\frac{\mathrm{d}}{\mathrm{d}x}\left(y\sec^2 x\right) = \sin x$$

Integrating both sides with respect to $x$:
$$y\sec^2 x = \int \sin x \,\mathrm{d}x = -\cos x + C$$

Multiplying by $\cos^2 x$ to solve for $y$:
$$y = C\cos^2 x - \cos^3 x$$

**(b)(i)**

We use the boundary condition $y = 0$ when $x = \frac{\pi}{3}$:
$$0 = C\cos^2\left(\frac{\pi}{3}\right) - \cos^3\left(\frac{\pi}{3}\right)$$

Since $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$:
$$0 = C\left(\frac{1}{4}\right) - \frac{1}{8} \implies C = \frac{1}{2}$$

The particular solution is:
$$y = \frac{1}{2}\cos^2 x - \cos^3 x$$

**(b)(ii)**

Let $u = \cos x$. For $-\frac{\pi}{2} < x < \frac{\pi}{2}$, we have $0 < u \le 1$.

Substitute $u$ into the expression for $y$:
$$y = \frac{1}{2}u^2 - u^3$$

Differentiating $y$ with respect to $u$:
$$\frac{\mathrm{d}y}{\mathrm{d}u} = u - 3u^2 = u(1-3u)$$

Setting $\frac{\mathrm{d}y}{\mathrm{d}u} = 0$, and since $u > 0$, we get:
$$1 - 3u = 0 \implies u = \frac{1}{3}$$

To confirm the nature of this stationary point, find the second derivative:
$$\frac{\mathrm{d}^2 y}{\mathrm{d}u^2} = 1 - 6u$$

At $u = \frac{1}{3}$:
$$\frac{\mathrm{d}^2 y}{\mathrm{d}u^2} = 1 - 6\left(\frac{1}{3}\right) = -1 < 0$$

Since the second derivative is negative, $u = \frac{1}{3}$ corresponds to a local maximum.

Now we check the boundary points and the stationary point:
- As $u \to 0^+$, $y \to 0$.
- At $u = 1$, $y = \frac{1}{2}(1)^2 - (1)^3 = -\frac{1}{2}$.
- At $u = \frac{1}{3}$:
$$y = \frac{1}{2}\left(\frac{1}{3}\right)^2 - \left(\frac{1}{3}\right)^3 = \frac{1}{18} - \frac{1}{27} = \frac{1}{54}$$

Thus, the maximum value of $y$ in the interval is $\frac{1}{54}$.

**(a)**
* **M1**: For dividing by $\cos x$ to obtain the standard linear form $\frac{\mathrm{d}y}{\mathrm{d}x} + 2y\tan x = \cos^2 x \sin x$.
* **M1**: For a correct attempt to find the integrating factor, evaluating $\int 2\tan x \,\mathrm{d}x$.
* **A1**: For obtaining the correct integrating factor $\sec^2 x$ (or $\frac{1}{\cos^2 x}$).
* **M1**: For writing the left-hand side as $\frac{\mathrm{d}}{\mathrm{d}x}(y \times \text{IF})$ and integrating the right-hand side correctly.
* **A1**: For the correct integration, leading to $y\sec^2 x = -\cos x + C$.
* **A1**: For expressing $y$ explicitly in terms of $x$ to get the general solution $y = C\cos^2 x - \cos^3 x$.

**(b)(i)**
* **M1**: For substituting $x = \frac{\pi}{3}$ and $y = 0$ into their general solution to find $C$.
* **A1**: For finding $C = \frac{1}{2}$ and stating the correct particular solution $y = \frac{1}{2}\cos^2 x - \cos^3 x$.

**(b)(ii)**
* **M1**: For a valid method to find the maximum value, e.g., substituting $u = \cos x$ and differentiating $y$ with respect to $u$, or differentiating $y$ directly with respect to $x$ and finding critical values.
* **A1**: For showing that the maximum occurs at $\cos x = \frac{1}{3}$ and finding the maximum value of $y = \frac{1}{54}$ with justification (such as checking boundaries or second derivative).

**(a)**

First, we divide the given differential equation by $\cos x$ to express it in standard linear form:
$$\frac{\mathrm{d}y}{\mathrm{d}x} + 2y\tan x = \cos^2 x \sin x$$

The integrating factor $I(x)$ is:
$$I(x) = e^{\int 2\tan x \,\mathrm{d}x} = e^{2\ln|\sec x|} = e^{\ln(\sec^2 x)} = \sec^2 x$$

Multiplying both sides of the differential equation by $\sec^2 x$ gives:
$$\sec^2 x \frac{\mathrm{d}y}{\mathrm{d}x} + 2y\sec^2 x \tan x = \cos^2 x \sin x \sec^2 x$$
$$\frac{\mathrm{d}}{\mathrm{d}x}\left(y\sec^2 x\right) = \sin x$$

Integrating both sides with respect to $x$:
$$y\sec^2 x = \int \sin x \,\mathrm{d}x = -\cos x + C$$

Multiplying by $\cos^2 x$ to solve for $y$:
$$y = C\cos^2 x - \cos^3 x$$

**(b)(i)**

We use the boundary condition $y = 0$ when $x = \frac{\pi}{3}$:
$$0 = C\cos^2\left(\frac{\pi}{3}\right) - \cos^3\left(\frac{\pi}{3}\right)$$

Since $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$:
$$0 = C\left(\frac{1}{4}\right) - \frac{1}{8} \implies C = \frac{1}{2}$$

The particular solution is:
$$y = \frac{1}{2}\cos^2 x - \cos^3 x$$

**(b)(ii)**

Let $u = \cos x$. For $-\frac{\pi}{2} < x < \frac{\pi}{2}$, we have $0 < u \le 1$.

Substitute $u$ into the expression for $y$:
$$y = \frac{1}{2}u^2 - u^3$$

Differentiating $y$ with respect to $u$:
$$\frac{\mathrm{d}y}{\mathrm{d}u} = u - 3u^2 = u(1-3u)$$

Setting $\frac{\mathrm{d}y}{\mathrm{d}u} = 0$, and since $u > 0$, we get:
$$1 - 3u = 0 \implies u = \frac{1}{3}$$

To confirm the nature of this stationary point, find the second derivative:
$$\frac{\mathrm{d}^2 y}{\mathrm{d}u^2} = 1 - 6u$$

At $u = \frac{1}{3}$:
$$\frac{\mathrm{d}^2 y}{\mathrm{d}u^2} = 1 - 6\left(\frac{1}{3}\right) = -1 < 0$$

Since the second derivative is negative, $u = \frac{1}{3}$ corresponds to a local maximum.

Now we check the boundary points and the stationary point:
- As $u \to 0^+$, $y \to 0$.
- At $u = 1$, $y = \frac{1}{2}(1)^2 - (1)^3 = -\frac{1}{2}$.
- At $u = \frac{1}{3}$:
$$y = \frac{1}{2}\left(\frac{1}{3}\right)^2 - \left(\frac{1}{3}\right)^3 = \frac{1}{18} - \frac{1}{27} = \frac{1}{54}$$

Thus, the maximum value of $y$ in the interval is $\frac{1}{54}$.

**(a)**
* **M1**: For dividing by $\cos x$ to obtain the standard linear form $\frac{\mathrm{d}y}{\mathrm{d}x} + 2y\tan x = \cos^2 x \sin x$.
* **M1**: For a correct attempt to find the integrating factor, evaluating $\int 2\tan x \,\mathrm{d}x$.
* **A1**: For obtaining the correct integrating factor $\sec^2 x$ (or $\frac{1}{\cos^2 x}$).
* **M1**: For writing the left-hand side as $\frac{\mathrm{d}}{\mathrm{d}x}(y \times \text{IF})$ and integrating the right-hand side correctly.
* **A1**: For the correct integration, leading to $y\sec^2 x = -\cos x + C$.
* **A1**: For expressing $y$ explicitly in terms of $x$ to get the general solution $y = C\cos^2 x - \cos^3 x$.

**(b)(i)**
* **M1**: For substituting $x = \frac{\pi}{3}$ and $y = 0$ into their general solution to find $C$.
* **A1**: For finding $C = \frac{1}{2}$ and stating the correct particular solution $y = \frac{1}{2}\cos^2 x - \cos^3 x$.

**(b)(ii)**
* **M1**: For a valid method to find the maximum value, e.g., substituting $u = \cos x$ and differentiating $y$ with respect to $u$, or differentiating $y$ directly with respect to $x$ and finding critical values.
* **A1**: For showing that the maximum occurs at $\cos x = \frac{1}{3}$ and finding the maximum value of $y = \frac{1}{54}$ with justification (such as checking boundaries or second derivative).

(a)(i) Using de Moivre's theorem or Euler's formula:
\(z^n = \cos n\theta + \text{i} \sin n\theta\)
\(z^{-n} = \cos(-n\theta) + \text{i} \sin(-n\theta) = \cos n\theta - \text{i} \sin n\theta\).

Subtracting the second equation from the first gives:
\(z^n - z^{-n} = 2\text{i} \sin n\theta\).

(a)(ii) Using the binomial expansion:
\((z - z^{-1})^6 = z^6 - 6z^4 + 15z^2 - 20 + 15z^{-2} - 6z^{-4} + z^{-6}\)
\(= (z^6 + z^{-6}) - 6(z^4 + z^{-4}) + 15(z^2 + z^{-2}) - 20\).

Since \(z^k + z^{-k} = 2 \cos k\theta\), we can substitute these into the expansion:
\((z - z^{-1})^6 = 2 \cos 6\theta - 12 \cos 4\theta + 30 \cos 2\theta - 20\).

From part (a)(i), we know that \(z - z^{-1} = 2\text{i} \sin \theta\). Therefore:
\((z - z^{-1})^6 = (2\text{i} \sin \theta)^6 = 64\text{i}^6 \sin^6 \theta = -64 \sin^6 \theta\).

Equating the two expressions for \((z - z^{-1})^6\):
\(-64 \sin^6 \theta = 2 \cos 6\theta - 12 \cos 4\theta + 30 \cos 2\theta - 20\).

Dividing both sides by \(-64\) yields:
\(\sin^6 \theta = -\frac{1}{32} \cos 6\theta + \frac{3}{16} \cos 4\theta - \frac{15}{32} \cos 2\theta + \frac{5}{16}\).

(b)(i) Given \(I_n = \int_0^{\frac{\pi}{2}} x^n \sin x \\, \text{d}x\).
Using integration by parts with \(u = x^n\) and \(\text{d}v = \sin x \\, \text{d}x\), we have:
\(\text{d}u = n x^{n-1} \\, \text{d}x\) and \(v = -\cos x\).

Applying the integration by parts formula:
\(I_n = \left[-x^n \cos x\right]_0^{\frac{\pi}{2}} + n \int_0^{\frac{\pi}{2}} x^{n-1} \cos x \\, \text{d}x\).

For \(n \ge 2\), the boundary term evaluates to:
\(\left[-(\frac{\pi}{2})^n \cos(\frac{\pi}{2})\right] - \left[-0^n \cos 0\right] = 0 - 0 = 0\).

Thus, \(I_n = n \int_0^{\frac{\pi}{2}} x^{n-1} \cos x \\, \text{d}x\).

We apply integration by parts a second time to this integral, with \(u = x^{n-1}\) and \(\text{d}v = \cos x \\, \text{d}x\):
\(\text{d}u = (n-1) x^{n-2} \\, \text{d}x\) and \(v = \sin x\).

This gives:
\(\int_0^{\frac{\pi}{2}} x^{n-1} \cos x \\, \text{d}x = \left[x^{n-1} \sin x\right]_0^{\frac{\pi}{2}} - (n-1) \int_0^{\frac{\pi}{2}} x^{n-2} \sin x \\, \text{d}x\)
\(= \left(\frac{\pi}{2}\right)^{n-1} \sin(\frac{\pi}{2}) - 0 - (n-1) I_{n-2}\)
\(= \left(\frac{\pi}{2}\right)^{n-1} - (n-1) I_{n-2}\).

Substituting this back into the expression for \(I_n\):
\(I_n = n \left[ \left(\frac{\pi}{2}\right)^{n-1} - (n-1) I_{n-2} \right] = n \left(\frac{\pi}{2}\right)^{n-1} - n(n-1) I_{n-2}\).

(b)(ii) First, compute \(I_0\):
\(I_0 = \int_0^{\frac{\pi}{2}} \sin x \\, \text{d}x = [-\cos x]_0^{\frac{\pi}{2}} = -\cos(\frac{\pi}{2}) + \cos 0 = 1\).

Using the reduction formula with \(n = 2\):
\(I_2 = 2 \left(\frac{\pi}{2}\right)^1 - 2(1) I_0 = \pi - 2(1) = \pi - 2\).

Using the reduction formula with \(n = 4\):
\(I_4 = 4 \left(\frac{\pi}{2}\right)^3 - 4(3) I_2 = 4 \left(\frac{\pi^3}{8}\right) - 12 (\pi - 2) = \frac{\pi^3}{2} - 12\pi + 24\).

(a)(i)
B1: For showing the correct derivation of \(z^n - z^{-n} = 2\text{i} \sin n\theta\).

(a)(ii)
M1: For expanding \((z - z^{-1})^6\) using the binomial theorem.
M1: For pairing the terms and substituting \(z^k + z^{-k} = 2\cos k\theta\).
M1: For identifying that \((z-z^{-1})^6 = -64 \sin^6 \theta\).
A1: For the correct unsimplified equation linking \(\sin^6 \theta\) to multiple angles of cosine.
A1: For dividing by \(-64\) and obtaining the given identity in the exact form requested.

(b)(i)
M1: For integrating by parts once and correctly evaluating the boundary term to 0.
A1: For obtaining the correct expression \(I_n = n \int_0^{\frac{\pi}{2}} x^{n-1} \cos x \\, \text{d}x\).
M1: For integrating by parts a second time and evaluating the boundary term correctly.
A1: For completing the steps to establish the given reduction formula.

(b)(ii)
B1: For finding the correct value of \(I_0 = 1\).
M1: For applying the reduction formula to find \(I_2\).
A1: For obtaining the correct exact value of \(I_2 = \pi - 2\).
A1: For applying the reduction formula a second time and obtaining \(I_4 = \frac{\pi^3}{2} - 12\pi + 24\).