2026 Cambridge IAS-Level Mathematics - Further (9231) Practice Paper with Answers

(a) Using the relations between the roots and coefficients of \(x^3 + 3x^2 - 2x + 4 = 0\):
\(\alpha + \beta + \gamma = -3\)
\(\alpha\beta + \beta\gamma + \gamma\alpha = -2\)
\(\alpha\beta\gamma = -4\)

(b)(i) Using the algebraic identity:
\(\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) = (-3)^2 - 2(-2) = 9 + 4 = 13\)

(b)(ii) Finding a common denominator:
\(\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma} = \dfrac{\alpha\beta + \beta\gamma + \gamma\alpha}{\alpha\beta\gamma} = \dfrac{-2}{-4} = \dfrac{1}{2}\)

(c) Method 1: Using substitution
Let \(y = x^2\), so \(x = \sqrt{y}\).
Substitute into the original equation:
\(x^3 - 2x + 3x^2 + 4 = 0 \implies x(x^2 - 2) = -(3x^2 + 4)\)
\(\sqrt{y}(y - 2) = -(3y + 4)\)
Squaring both sides:
\(y(y - 2)^2 = (3y + 4)^2\)
\(y(y^2 - 4y + 4) = 9y^2 + 24y + 16\)
\(y^3 - 4y^2 + 4y = 9y^2 + 24y + 16\)
\(y^3 - 13y^2 - 20y - 16 = 0\)

Method 2: Using symmetric functions
Sum of new roots: \(S_1 = \alpha^2 + \beta^2 + \gamma^2 = 13\)
Sum of products in pairs:
\(S_2 = \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 = (\alpha\beta + \beta\gamma + \gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha + \beta + \gamma) = (-2)^2 - 2(-4)(-3) = 4 - 24 = -20\)
Product of new roots: \(S_3 = (\alpha\beta\gamma)^2 = (-4)^2 = 16\)
Thus, the equation is:
\(y^3 - S_1 y^2 + S_2 y - S_3 = 0 \implies y^3 - 13y^2 - 20y - 16 = 0\)

(a)
B1: All three values correct.

(b)(i)
M1: Uses identity \(\alpha^2 + \beta^2 + \gamma^2 = (\sum \alpha)^2 - 2\sum \alpha\beta\).
A1: Obtains \(13\).

(b)(ii)
B1: Obtains \(\frac{1}{2}\) (or \(0.5\)).

(c)
M1: Expresses original equation in the form \(x(x^2 - 2) = -(3x^2 + 4)\) and squares both sides, or uses symmetric functions to determine \(\sum \alpha^2\beta^2\).
M1: Correct expansion of algebraic terms.
A1: Obtains correct terms for \(y^2\), \(y\), and constant.
A1: Obtains final correct equation \(y^3 - 13y^2 - 20y - 16 = 0\) (must include \(= 0\)).

(a) Starting from the right-hand side:
\(\dfrac{1}{4}\left( \dfrac{1}{(2r-1)(2r+1)} - \dfrac{1}{(2r+1)(2r+3)} \right) = \dfrac{1}{4}\left( \dfrac{(2r+3) - (2r-1)}{(2r-1)(2r+1)(2r+3)} \right) = \dfrac{1}{4}\left( \dfrac{4}{(2r-1)(2r+1)(2r+3)} \right) = \dfrac{1}{(2r-1)(2r+1)(2r+3)}\)

(b) Using the method of differences:
Let \(f(r) = \dfrac{1}{(2r-1)(2r+1)}\).
\(\sum_{r=1}^n \dfrac{1}{(2r-1)(2r+1)(2r+3)} = \dfrac{1}{4}\sum_{r=1}^n \left( f(r) - f(r+1) \right) = \dfrac{1}{4} \left[ \left( \dfrac{1}{1\cdot 3} - \dfrac{1}{3\cdot 5} \right) + \left( \dfrac{1}{3\cdot 5} - \dfrac{1}{5\cdot 7} \right) + \dots + \left( \dfrac{1}{(2n-1)(2n+1)} - \dfrac{1}{(2n+1)(2n+3)} \right) \right] = \dfrac{1}{4}\left( \dfrac{1}{3} - \dfrac{1}{(2n+1)(2n+3)} \right) = \dfrac{1}{12} - \dfrac{1}{4(2n+1)(2n+3)}\)
(or \(\dfrac{(2n+1)(2n+3) - 3}{12(2n+1)(2n+3)} = \dfrac{4n^2 + 8n}{12(2n+1)(2n+3)} = \dfrac{n(n+2)}{3(2n+1)(2n+3)}\))

(c) As \(n \to \infty\), \(\dfrac{1}{4(2n+1)(2n+3)} \to 0\).
Therefore, \(\sum_{r=1}^\infty \dfrac{1}{(2r-1)(2r+1)(2r+3)} = \dfrac{1}{12}\).

(d) \(\sum_{r=n+1}^{2n} \dfrac{1}{(2r-1)(2r+1)(2r+3)} = S_{2n} - S_n = \left( \dfrac{1}{12} - \dfrac{1}{4(4n+1)(4n+3)} \right) - \left( \dfrac{1}{12} - \dfrac{1}{4(2n+1)(2n+3)} \right) = \dfrac{1}{4(2n+1)(2n+3)} - \dfrac{1}{4(4n+1)(4n+3)} = \dfrac{(4n+1)(4n+3) - (2n+1)(2n+3)}{4(2n+1)(2n+3)(4n+1)(4n+3)} = \dfrac{(16n^2 + 16n + 3) - (4n^2 + 8n + 3)}{4(2n+1)(2n+3)(4n+1)(4n+3)} = \dfrac{12n^2 + 8n}{4(2n+1)(2n+3)(4n+1)(4n+3)} = \dfrac{4n(3n+2)}{4(2n+1)(2n+3)(4n+1)(4n+3)} = \dfrac{n(3n+2)}{(2n+1)(2n+3)(4n+1)(4n+3)}\)

(a)
M1: Puts the right-hand side over a common denominator with correct numerator \((2r+3) - (2r-1)\).
A1: Simplifies fully to reach the left-hand side with clear steps shown.

(b)
M1: Applies method of differences showing at least first two and last terms cancelling.
M1: Identifies the surviving first and last terms.
A1: Correct expression \(\frac{1}{12} - \frac{1}{4(2n+1)(2n+3)}\) or \(\frac{n(n+2)}{3(2n+1)(2n+3)}\).

(c)
B1: States \(\frac{1}{12}\) with justification from the limit as \(n \to \infty\).

(d)
M1: Uses \(S_{2n} - S_n\) or difference of tail terms.
A1: Obtains \(\frac{n(3n+2)}{(2n+1)(2n+3)(4n+1)(4n+3)}\) in fully factorised single fraction form.

(a) An invariant line through the origin has equation \(y = mx\).
Under the transformation \(T\):
\[ \begin{pmatrix} X \\ Y \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} x \\ mx \end{pmatrix} = \begin{pmatrix} (2+m)x \\ (3+4m)x \end{pmatrix} \]
Since the transformed point \((X, Y)\) must also lie on \(Y = mX\):
\[ (3 + 4m)x = m(2 + m)x \]
Since this must hold for non-zero \(x\):
\[ 3 + 4m = 2m + m^2 \implies m^2 - 2m - 3 = 0 \]
Factoring gives:
\[ (m - 3)(m + 1) = 0 \implies m = 3 \text{ or } m = -1 \]
Checking the line \(x = 0\):
\[ \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 0 \\ y \end{pmatrix} = \begin{pmatrix} y \\ 4y \end{pmatrix} \]
which does not lie on \(x = 0\) unless \(y = 0\), so \(x = 0\) is not invariant.
Therefore, the invariant lines through the origin are \(y = 3x\) and \(y = -x\).

(b) The area of triangle \(OAB\) is:
\[ \text{Area}(OAB) = \frac{1}{2} |(2)(3) - (1)(-1)| = \frac{1}{2} |6 + 1| = \frac{7}{2} = 3.5 \]
The determinant of \(\mathbf{M}\) is:
\[ \det(\mathbf{M}) = (2)(4) - (1)(3) = 8 - 3 = 5 \]
The area scale factor is \(|\det(\mathbf{M})| = 5\).
\[ \text{Area}(OA'B') = 5 \times \frac{7}{2} = \frac{35}{2} = 17.5 \]

(c) The combined transformation matrix \(\mathbf{P}\) representing \(T\) followed by \(\mathbf{N}\) is:
\[ \mathbf{P} = \mathbf{N}\mathbf{M} = \begin{pmatrix} 1 & -1 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} (1)(2) + (-1)(3) & (1)(1) + (-1)(4) \\ (0)(2) + (2)(3) & (0)(1) + (2)(4) \end{pmatrix} = \begin{pmatrix} -1 & -3 \\ 6 & 8 \end{pmatrix} \]
Evaluating the determinant of \(\mathbf{P}\):
\[ \det(\mathbf{P}) = (-1)(8) - (-3)(6) = -8 + 18 = 10 \]
Since \(\det(\mathbf{P}) \neq 0\), \(\mathbf{P}\) is non-singular.

(a)
M1: Sets up the invariant line condition \(\mathbf{M}\begin{pmatrix} x \\ mx \end{pmatrix} = \begin{pmatrix} X \\ mX \end{pmatrix}\) or uses eigenvalue equation \(\det(\mathbf{M} - \lambda \mathbf{I}) = 0\).
M1: Forms the correct quadratic equation in \(m\): \(m^2 - 2m - 3 = 0\) (or finds eigenvalues \(\lambda = 1, 5\)).
A1: Obtains the correct gradients \(m = 3\) and \(m = -1\).
A1: States both invariant line equations: \(y = 3x\) and \(y = -x\).

(b)
M1: Calculates the area of the original triangle \(OAB\) and the determinant \(\det(\mathbf{M}) = 5\).
A1: Obtains the correct transformed area \(17.5\) or \(\frac{35}{2}\).

(c)
M1: Multiplies matrices in correct order \(\mathbf{N}\mathbf{M}\) to find \(\mathbf{P}\).
A1: Obtains correct \(\mathbf{P} = \begin{pmatrix} -1 & -3 \\ 6 & 8 \end{pmatrix}\) and correctly states that \(\mathbf{P}\) is non-singular because \(\det(\mathbf{P}) = 10 \neq 0\).

(a)
\[ \mathbf{A}^2 = \begin{pmatrix} 4 & -1 \\ 9 & -2 \end{pmatrix} \begin{pmatrix} 4 & -1 \\ 9 & -2 \end{pmatrix} = \begin{pmatrix} 16-9 & -4+2 \\ 36-18 & -9+4 \end{pmatrix} = \begin{pmatrix} 7 & -2 \\ 18 & -5 \end{pmatrix} \]
\[ \mathbf{A}^3 = \mathbf{A}^2 \mathbf{A} = \begin{pmatrix} 7 & -2 \\ 18 & -5 \end{pmatrix} \begin{pmatrix} 4 & -1 \\ 9 & -2 \end{pmatrix} = \begin{pmatrix} 28-18 & -7+4 \\ 72-45 & -18+10 \end{pmatrix} = \begin{pmatrix} 10 & -3 \\ 27 & -8 \end{pmatrix} \]

(b) Observing the entries for \(n = 1, 2, 3\):
- Top-left entry: \(4, 7, 10 \implies 3n+1\)
- Top-right entry: \(-1, -2, -3 \implies -n\)
- Bottom-left entry: \(9, 18, 27 \implies 9n\)
- Bottom-right entry: \(-2, -5, -8 \implies 1-3n\)

Thus, the conjecture is:
\[ \mathbf{A}^n = \begin{pmatrix} 3n+1 & -n \\ 9n & 1-3n \end{pmatrix} \]

(c)
Base case: For \(n = 1\),
\[ \begin{pmatrix} 3(1)+1 & -(1) \\ 9(1) & 1-3(1) \end{pmatrix} = \begin{pmatrix} 4 & -1 \\ 9 & -2 \end{pmatrix} = \mathbf{A} \]
So the result is true for \(n = 1\).

Inductive hypothesis: Assume that the result is true for \(n = k\), where \(k\) is a positive integer, so
\[ \mathbf{A}^k = \begin{pmatrix} 3k+1 & -k \\ 9k & 1-3k \end{pmatrix} \]

Inductive step: We consider \(\mathbf{A}^{k+1} = \mathbf{A}^k \mathbf{A}\):
\[ \mathbf{A}^{k+1} = \begin{pmatrix} 3k+1 & -k \\ 9k & 1-3k \end{pmatrix} \begin{pmatrix} 4 & -1 \\ 9 & -2 \end{pmatrix} \]
\[ = \begin{pmatrix} 4(3k+1) - 9k & -(3k+1) + 2k \\ 4(9k) + 9(1-3k) & -9k - 2(1-3k) \end{pmatrix} \]
\[ = \begin{pmatrix} 12k + 4 - 9k & -3k - 1 + 2k \\ 36k + 9 - 27k & -9k - 2 + 6k \end{pmatrix} \]
\[ = \begin{pmatrix} 3k + 4 & -k - 1 \\ 9k + 9 & -3k - 2 \end{pmatrix} \]
\[ = \begin{pmatrix} 3(k+1) + 1 & -(k+1) \\ 9(k+1) & 1 - 3(k+1) \end{pmatrix} \]
This is the formula with \(n = k + 1\).

Conclusion: Since the result is true for \(n = 1\), and if it is true for \(n = k\) then it is true for \(n = k + 1\), by the principle of mathematical induction the result is true for all positive integers \(n\).

(a)
B1: Correct calculation for \(\mathbf{A}^2 = \begin{pmatrix} 7 & -2 \\ 18 & -5 \end{pmatrix}\).
B1: Correct calculation for \(\mathbf{A}^3 = \begin{pmatrix} 10 & -3 \\ 27 & -8 \end{pmatrix}\).

(b)
B1: Correct conjecture \(\mathbf{A}^n = \begin{pmatrix} 3n+1 & -n \\ 9n & 1-3n \end{pmatrix}\).

(c)
B1: Base case: Evaluates the formula for \(n = 1\) and shows it equals \(\mathbf{A}\).
M1: Explicitly states the induction hypothesis that \(\mathbf{A}^k = \begin{pmatrix} 3k+1 & -k \\ 9k & 1-3k \end{pmatrix}\) for some positive integer \(k\).
M1: Considers \(\mathbf{A}^{k+1} = \mathbf{A}^k \mathbf{A}\) (or \(\mathbf{A} \mathbf{A}^k\)) and carries out matrix multiplication.
A1: Correctly performs all algebraic expansions and simplifies each of the 4 entries to show \(\begin{pmatrix} 3(k+1)+1 & -(k+1) \\ 9(k+1) & 1-3(k+1) \end{pmatrix}\).
A1: Gives a complete and correct conclusion linking base case, inductive step, and positive integers \(n\) (dependent on all preceding marks).

(a)
At the pole, \(r = 0\):
\[ 1 + 2\cos\theta = 0 \implies \cos\theta = -\frac{1}{2} \]
Within the domain \(-\frac{2}{3}\pi \le \theta \le \frac{2}{3}\pi\), this gives:
\[ \theta = \frac{2}{3}\pi \quad \text{and} \quad \theta = -\frac{2}{3}\pi \]

(b)
For tangents perpendicular to the initial line, we consider \(x = r\cos\theta\):
\[ x = (1 + 2\cos\theta)\cos\theta = \cos\theta + 2\cos^2\theta \]
Differentiating with respect to \(\theta\):
\[ \frac{\mathrm{d}x}{\mathrm{d}\theta} = -\sin\theta - 4\cos\theta\sin\theta = -\sin\theta(1 + 4\cos\theta) \]
Setting \(\frac{\mathrm{d}x}{\mathrm{d}\theta} = 0\):
1. \(\sin\theta = 0 \implies \theta = 0\)
When \(\theta = 0\), \(r = 1 + 2(1) = 3\).
\[ x = 3\cos 0 = 3, \quad y = 3\sin 0 = 0 \implies (3, 0) \]

2. \(1 + 4\cos\theta = 0 \implies \cos\theta = -\frac{1}{4}\)
Since \(\cos\theta = -\frac{1}{4}\), \(\sin\theta = \pm\sqrt{1 - \left(-\frac{1}{4}\right)^2} = \pm\frac{\sqrt{15}}{4}\).
Then:
\[ r = 1 + 2\left(-\frac{1}{4}\right) = \frac{1}{2} \]
\[ x = r\cos\theta = \frac{1}{2}\left(-\frac{1}{4}\right) = -\frac{1}{8} \]
\[ y = r\sin\theta = \frac{1}{2}\left(\pm\frac{\sqrt{15}}{4}\right) = \pm\frac{\sqrt{15}}{8} \]
Thus, the points are \((3, 0)\), \(\left(-\frac{1}{8}, \frac{\sqrt{15}}{8}\right)\), and \(\left(-\frac{1}{8}, -\frac{\sqrt{15}}{8}\right)\).

(c)
The sketch shows a single outer loop for \(r \ge 0\):
- Symmetric about the initial line \(\theta = 0\).
- Passes through \((3, 0)\) on the initial line, and \(\left(1, \frac{\pi}{2}\right)\), \(\left(1, -\frac{\pi}{2}\right)\) on the line \(\theta = \pm\frac{\pi}{2}\).
- Tangents at the pole at \(\theta = \pm\frac{2}{3}\pi\).

(d)
The area \(A\) is given by:
\[ A = \frac{1}{2} \int_{-\frac{2}{3}\pi}^{\frac{2}{3}\pi} r^2 \,\mathrm{d}\theta = \int_0^{\frac{2}{3}\pi} (1 + 2\cos\theta)^2 \,\mathrm{d}\theta \]
Expanding the integrand:
\[ (1 + 2\cos\theta)^2 = 1 + 4\cos\theta + 4\cos^2\theta = 1 + 4\cos\theta + 2(1 + \cos 2\theta) = 3 + 4\cos\theta + 2\cos 2\theta \]
Integrating:
\[ A = \left[ 3\theta + 4\sin\theta + \sin 2\theta \right]_0^{\frac{2}{3}\pi} \]
\[ A = 3\left(\frac{2}{3}\pi\right) + 4\sin\left(\frac{2}{3}\pi\right) + \sin\left(\frac{4}{3}\pi\right) - 0 \]
\[ A = 2\pi + 4\left(\frac{\sqrt{3}}{2}\right) - \frac{\sqrt{3}}{2} = 2\pi + \frac{3\sqrt{3}}{2} \]

(a)
- M1: Sets \(r = 0\) and solves for \(\theta\).
- A1: States both correct equations \(\theta = \frac{2}{3}\pi\) and \(\theta = -\frac{2}{3}\pi\).

(b)
- M1: Writes \(x = r\cos\theta = \cos\theta + 2\cos^2\theta\) and differentiates with respect to \(\theta\).
- A1: Obtains correct derivative \(-\sin\theta(1 + 4\cos\theta)\) and sets to 0.
- A1: Obtains \((3, 0)\).
- M1: Finds \(r\) and evaluates \(x\) and \(y\) for \(\cos\theta = -\frac{1}{4}\).
- A1: Obtains \(\left(-\frac{1}{8}, \pm\frac{\sqrt{15}}{8}\right)\) in exact form.

(c)
- B1: Correct heart/limacon-shaped curve passing through pole and symmetrical about the initial line.
- B1: Correct intersections labelled at \((3, 0)\) and \((1, \pm\pi/2)\) with correct tangents at pole.

(d)
- M1: Uses \(A = \frac{1}{2}\int r^2 \,\mathrm{d}\theta\) with correct limits.
- M1: Expands \((1 + 2\cos\theta)^2\) and applies the double-angle identity \(\cos^2\theta = \frac{1}{2}(1 + \cos 2\theta)\).
- A1: Correct integration to \(3\theta + 4\sin\theta + \sin 2\theta\).
- M1: Substitutes limits \(0\) and \(\frac{2}{3}\pi\) correctly into integrated expression.
- A1: Obtains exact value \(2\pi + \frac{3\sqrt{3}}{2}\).

(a)
Vertical asymptote occurs where the denominator is zero:
\[ x = -1 \]
Performing algebraic division:
\[ y = \frac{x^2 - 2x + 1}{x + 1} = x - 3 + \frac{4}{x + 1} \]
As \(x \to \pm\infty\), \(\frac{4}{x + 1} \to 0\), so the oblique asymptote is:
\[ y = x - 3 \]

(b)
Differentiating \(y = x - 3 + 4(x + 1)^{-1}\):
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = 1 - \frac{4}{(x + 1)^2} \]
Setting \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\[ (x + 1)^2 = 4 \implies x + 1 = \pm 2 \]
\[ x = 1 \quad \text{or} \quad x = -3 \]
When \(x = 1\):
\[ y = \frac{(1 - 1)^2}{1 + 1} = 0 \implies (1, 0) \]
When \(x = -3\):
\[ y = \frac{(-3 - 1)^2}{-3 + 1} = \frac{16}{-2} = -8 \implies (-3, -8) \]

(c)
Rearranging the equation of \(C\):
\[ y(x + 1) = x^2 - 2x + 1 \implies x^2 - (y + 2)x + (1 - y) = 0 \]
Since \(x\) is real, the discriminant \(\Delta\) must be non-negative:
\[ \Delta = (-(y + 2))^2 - 4(1)(1 - y) \ge 0 \]
\[ y^2 + 4y + 4 - 4 + 4y \ge 0 \]
\[ y^2 + 8y \ge 0 \implies y(y + 8) \ge 0 \]
Thus, the set of values that \(y\) can take is:
\[ y \le -8 \quad \text{or} \quad y \ge 0 \]

(d)
- \(y\)-intercept: when \(x = 0\), \(y = 1 \implies (0, 1)\).
- \(x\)-intercept: when \(y = 0\), \((x - 1)^2 = 0 \implies x = 1 \implies (1, 0)\) (a minimum touching the \(x\)-axis).
- Branch for \(x < -1\): lies in the region \(y \le -8\), approaching \(x = -1\) as \(y \to -\infty\) and \(y = x - 3\) as \(x \to -\infty\), with local maximum at \((-3, -8)\).
- Branch for \(x > -1\): lies in the region \(y \ge 0\), approaching \(x = -1\) as \(y \to +\infty\) and \(y = x - 3\) as \(x \to +\infty\), with local minimum at \((1, 0)\).

(e)
For \(y = |f(x)|\):
- The branch for \(x > -1\) is already non-negative, so it remains unchanged.
- The branch for \(x < -1\) is reflected across the \(x\)-axis into the upper half-plane \(y > 0\).
- The local maximum at \((-3, -8)\) becomes a local minimum at \((-3, 8)\).
- Asymptotes become \(x = -1\) and \(y = -x + 3\) for \(x < -1\).

(a)
- B1: Identifies vertical asymptote \(x = -1\).
- B1: Carries out division correctly and states oblique asymptote \(y = x - 3\).

(b)
- M1: Differentiates \(y\) using quotient rule or after rewriting as \(x - 3 + 4(x + 1)^{-1}\).
- A1: Sets derivative to 0 and solves \((x + 1)^2 = 4\).
- A1: Obtains \((1, 0)\).
- A1: Obtains \((-3, -8)\).

(c)
- M1: Forms a quadratic equation in \(x\): \(x^2 - (y+2)x + (1-y) = 0\).
- M1: Applies discriminant condition \(B^2 - 4AC \ge 0\).
- A1: Correct final range \(y \le -8\) or \(y \ge 0\).

(d)
- B1: Draws both asymptotes correctly as dashed lines and labels them.
- B1: Correct branch for \(x < -1\) with maximum at \((-3, -8)\).
- B1: Correct branch for \(x > -1\) with minimum at \((1, 0)\) and intercept at \((0, 1)\).

(e)
- B1: Correct unchanged branch for \(x > -1\).
- B1: Correctly reflected branch for \(x < -1\) with minimum at \((-3, 8)\) and approaching \(x = -1\) as \(y \to +\infty\).

(a)
The direction vectors of \(l_1\) and \(l_2\) are \(\mathbf{d}_1 = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}\) and \(\mathbf{d}_2 = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}\).
Since \(\mathbf{d}_1 \ne k\mathbf{d}_2\) for any scalar \(k\), the lines are not parallel.

To test for intersection, equate the coordinates:
1. \(1 + 2\lambda = 3 + \mu \implies 2\lambda - \mu = 2\)
2. \(2 - \lambda = -2 + 2\mu \implies \lambda + 2\mu = 4\)
3. \(-1 + \lambda = 5 + \mu \implies \lambda - \mu = 6\)

Subtracting equation (3) from equation (1):
\[ (2\lambda - \mu) - (\lambda - \mu) = 2 - 6 \implies \lambda = -4 \]
From equation (3):
\[ -4 - \mu = 6 \implies \mu = -10 \]
Substituting \(\lambda = -4\) and \(\mu = -10\) into equation (2):
\[ \lambda + 2\mu = -4 + 2(-10) = -24 \ne 4 \]
Since the equations are inconsistent, the lines do not intersect. Because they are neither parallel nor intersecting, \(l_1\) and \(l_2\) are skew lines.

(b)
A vector perpendicular to both lines is:
\[ \mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 1 \\ 1 & 2 & 1 \end{vmatrix} = \begin{pmatrix} (-1)(1) - (1)(2) \\ (1)(1) - (2)(1) \\ (2)(2) - (-1)(1) \end{pmatrix} = \begin{pmatrix} -3 \\ -1 \\ 5 \end{pmatrix} \]
The magnitude of \(\mathbf{n}\) is:
\[ |\mathbf{n}| = \sqrt{(-3)^2 + (-1)^2 + 5^2} = \sqrt{9 + 1 + 25} = \sqrt{35} \]
A vector connecting a point on \(l_1\) to a point on \(l_2\) is:
\[ \mathbf{a}_2 - \mathbf{a}_1 = \begin{pmatrix} 3 \\ -2 \\ 5 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 2 \\ -4 \\ 6 \end{pmatrix} \]
The shortest distance \(D\) is the projection of this vector onto \(\mathbf{n}\):
\[ D = \frac{|(\mathbf{a}_2 - \mathbf{a}_1) \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{|2(-3) + (-4)(-1) + 6(5)|}{\sqrt{35}} = \frac{|-6 + 4 + 30|}{\sqrt{35}} = \frac{28}{\sqrt{35}} = \frac{4\sqrt{35}}{5} \]

(c)
The normal vector to \(\Pi_1\) is parallel to \(\mathbf{n}\), so we can choose \(\mathbf{n}_1 = \begin{pmatrix} 3 \\ 1 \\ -5 \end{pmatrix}\).
The equation of \(\Pi_1\) is \(3x + y - 5z = d\).
Since \((1, 2, -1)\) lies on \(\Pi_1\):
\[ d = 3(1) + 1(2) - 5(-1) = 3 + 2 + 5 = 10 \]
Thus, the equation of \(\Pi_1\) is:
\[ 3x + y - 5z = 10 \]

(d)(i)
The normal vector to \(\Pi_2\) is \(\mathbf{n}_2 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\).
Let \(\theta\) be the acute angle between the planes:
\[ \cos\theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|} = \frac{|3(1) + 1(1) + (-5)(1)|}{\sqrt{35}\sqrt{1^2 + 1^2 + 1^2}} = \frac{|-1|}{\sqrt{35}\sqrt{3}} = \frac{1}{\sqrt{105}} \]
\[ \theta = \arccos\left(\frac{1}{\sqrt{105}}\right) \approx 84.4^\circ \]

(d)(ii)
The direction vector \(\mathbf{d}\) of the line of intersection is perpendicular to both normal vectors:
\[ \mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 1 & -5 \\ 1 & 1 & 1 \end{vmatrix} = \begin{pmatrix} 1 - (-5) \\ -5 - 3 \\ 3 - 1 \end{pmatrix} = \begin{pmatrix} 6 \\ -8 \\ 2 \end{pmatrix} = 2\begin{pmatrix} 3 \\ -4 \\ 1 \end{pmatrix} \]
To find a point on both planes, set \(z = 0\):
\[ 3x + y = 10 \]
\[ x + y = 4 \]
Subtracting gives \(2x = 6 \implies x = 3\), so \(y = 1\).
A point on the line is \((3, 1, 0)\).
A vector equation of the line is:
\[ \mathbf{r} = \begin{pmatrix} 3 \\ 1 \\ 0 \end{pmatrix} + t \begin{pmatrix} 3 \\ -4 \\ 1 \end{pmatrix} \]

(a)
- B1: Explains that direction vectors are not multiples of each other, hence lines are not parallel.
- M1: Sets up 3 simultaneous equations in \(\lambda\) and \(\mu\) and solves 2 of them.
- A1: Shows contradiction in the 3rd equation and concludes the lines are skew.

(b)
- M1: Evaluates the vector product \(\mathbf{d}_1 \times \mathbf{d}_2\).
- A1: Obtains \(\begin{pmatrix} -3 \\ -1 \\ 5 \end{pmatrix}\) or any scalar multiple.
- M1: Uses the scalar product formula \(\frac{|(\mathbf{a}_2 - \mathbf{a}_1) \cdot \mathbf{n}|}{|\mathbf{n}|}\).
- A1: Obtains \(\frac{4\sqrt{35}}{5}\) (or \(\frac{28}{\sqrt{35}}\)).

(c)
- M1: Uses normal vector from (b) in \(\mathbf{r} \cdot \mathbf{n} = d\).
- M1: Substitutes a point from \(l_1\), e.g., \((1, 2, -1)\).
- A1: Obtains \(3x + y - 5z = 10\) (or \(-3x - y + 5z = -10\)).

(d)(i)
- M1: Uses \(\cos\theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|}\).
- A1: Obtains \(84.4^\circ\) (or \(1.47\) radians).

(d)(ii)
- M1: Evaluates vector product of normals to find direction vector \(\begin{pmatrix} 3 \\ -4 \\ 1 \end{pmatrix}\).
- M1: Finds a particular point satisfying both planes (e.g. \((3, 1, 0)\)).
- A1: Writes full vector equation \(\mathbf{r} = \mathbf{a} + t\mathbf{d}\).

We are given the equation \( (z + \text{i})^6 + (z - \text{i})^6 = 0 \).

Since \( z = \text{i} \) gives \( (2\text{i})^6 + 0 \neq 0 \), \( z \neq \text{i} \). Dividing both sides by \( (z - \text{i})^6 \) yields
\[ \left(\frac{z + \text{i}}{z - \text{i}}\right)^6 = -1. \]

Expressing \( -1 \) in exponential form, we have
\[ -1 = \text{e}^{\text{i}(2k+1)\pi}, \quad \text{for } k = 0, 1, 2, 3, 4, 5. \]

Taking the 6th roots:
\[ \frac{z + \text{i}}{z - \text{i}} = \text{e}^{\text{i}\theta_k}, \quad \text{where } \theta_k = \frac{(2k+1)\pi}{6} \text{ for } k = 0, 1, 2, 3, 4, 5. \]
The values of \( \theta_k \) are \( \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6} \), which can be written as \( \pm \frac{\pi}{6}, \pm \frac{\pi}{2}, \pm \frac{5\pi}{6} \).

Now, solve for \( z \) in terms of \( \theta \):
\[ z + \text{i} = (z - \text{i})\text{e}^{\text{i}\theta} \implies z(1 - \text{e}^{\text{i}\theta}) = -\text{i}(1 + \text{e}^{\text{i}\theta}) \]
\[ z = \text{i} \frac{\text{e}^{\text{i}\theta} + 1}{\text{e}^{\text{i}\theta} - 1} = \text{i} \frac{\text{e}^{\text{i}\theta/2} + \text{e}^{-\text{i}\theta/2}}{\text{e}^{\text{i}\theta/2} - \text{e}^{-\text{i}\theta/2}} = \text{i} \frac{2\cos(\theta/2)}{2\text{i}\sin(\theta/2)} = \cot\left(\frac{\theta}{2}\right). \]

Evaluating \( z = \cot(\theta/2) \) for each angle:
- For \( \theta = \pm \frac{\pi}{2} \): \( \frac{\theta}{2} = \pm \frac{\pi}{4} \implies z = \cot\left(\pm \frac{\pi}{4}\right) = \pm 1 \).
- For \( \theta = \pm \frac{\pi}{6} \): \( \frac{\theta}{2} = \pm \frac{\pi}{12} \implies z = \cot\left(\pm \frac{\pi}{12}\right) = \pm(2 + \sqrt{3}) \).
- For \( \theta = \pm \frac{5\pi}{6} \): \( \frac{\theta}{2} = \pm \frac{5\pi}{12} \implies z = \cot\left(\pm \frac{5\pi}{12}\right) = \pm(2 - \sqrt{3}) \).

Thus, the 6 roots are \( z = \pm 1, \; \pm(2 + \sqrt{3}), \; \pm(2 - \sqrt{3}) \).

M1: Rearranges the equation to \( \left(\frac{z+\text{i}}{z-\text{i}}\right)^6 = -1 \) and applies de Moivre's theorem/roots of unity to find the 6 values of the ratio in exponential or polar form.
A1: Obtains the correct 6 angles \( \theta = \pm \frac{\pi}{6}, \pm \frac{\pi}{2}, \pm \frac{5\pi}{6} \) (or equivalent set in \( [0, 2\pi) \)).
M1: Rearranges to make \( z \) the subject, expressing \( z \) in terms of trigonometric functions, e.g. \( z = \cot(\theta/2) \).
A1: Correctly evaluates \( z = \pm 1 \) and identifies the remaining cotangent values.
A1: Obtains all 6 roots in exact simplified form: \( \pm 1, \pm(2+\sqrt{3}), \pm(2-\sqrt{3}) \) (or \( \pm 1, \pm 2 \pm \sqrt{3} \)).

The surface area of revolution about the \( x \)-axis is given by
\[ S = 2\pi \int_1^3 y \sqrt{1 + \left(\frac{\text{d}y}{\text{d}x}\right)^2} \, \text{d}x. \]

First, differentiate \( y = \frac{1}{6}x^3 + \frac{1}{2}x^{-1} \) with respect to \( x \):
\[ \frac{\text{d}y}{\text{d}x} = \frac{1}{2}x^2 - \frac{1}{2}x^{-2}. \]

Now, compute \( 1 + \left(\frac{\text{d}y}{\text{d}x}\right)^2 \):
\[ 1 + \left(\frac{1}{2}x^2 - \frac{1}{2}x^{-2}\right)^2 = 1 + \left(\frac{1}{4}x^4 - \frac{1}{2} + \frac{1}{4}x^{-4}\right) = \frac{1}{4}x^4 + \frac{1}{2} + \frac{1}{4}x^{-4} = \left(\frac{1}{2}x^2 + \frac{1}{2}x^{-2}\right)^2. \]

Thus,
\[ \sqrt{1 + \left(\frac{\text{d}y}{\text{d}x}\right)^2} = \frac{1}{2}x^2 + \frac{1}{2}x^{-2}. \]

Next, multiply by \( y \):
\[ y\sqrt{1 + \left(\frac{\text{d}y}{\text{d}x}\right)^2} = \left(\frac{1}{6}x^3 + \frac{1}{2}x^{-1}\right)\left(\frac{1}{2}x^2 + \frac{1}{2}x^{-2}\right) = \frac{1}{12}x^5 + \frac{1}{12}x + \frac{1}{4}x + \frac{1}{4}x^{-3} = \frac{1}{12}x^5 + \frac{1}{3}x + \frac{1}{4}x^{-3}. \]

Now evaluate the definite integral:
\[ \int_1^3 \left(\frac{1}{12}x^5 + \frac{1}{3}x + \frac{1}{4}x^{-3}\right) \, \text{d}x = \left[ \frac{1}{72}x^6 + \frac{1}{6}x^2 - \frac{1}{8}x^{-2} \right]_1^3. \]

Substituting the upper limit \( x = 3 \):
\[ \frac{3^6}{72} + \frac{3^2}{6} - \frac{1}{8(9)} = \frac{729}{72} + \frac{9}{6} - \frac{1}{72} = \frac{728}{72} + \frac{3}{2} = \frac{91}{9} + \frac{3}{2} = \frac{209}{18}. \]

Substituting the lower limit \( x = 1 \):
\[ \frac{1}{72} + \frac{1}{6} - \frac{1}{8} = \frac{1 + 12 - 9}{72} = \frac{4}{72} = \frac{1}{18}. \]

Subtracting the limits:
\[ \frac{209}{18} - \frac{1}{18} = \frac{208}{18} = \frac{104}{9}. \]

Multiplying by \( 2\pi \):
\[ S = 2\pi \times \frac{104}{9} = \frac{208\pi}{9}. \]

B1: Correct derivative \( \frac{\text{d}y}{\text{d}x} = \frac{1}{2}x^2 - \frac{1}{2}x^{-2} \).
M1: Obtains \( 1 + \left(\frac{\text{d}y}{\text{d}x}\right)^2 = \left(\frac{1}{2}x^2 + \frac{1}{2}x^{-2}\right)^2 \) and expands the product \( y \sqrt{1 + (y')^2} \) into polynomial/power form.
A1: Correctly integrated expression \( \left[ \frac{1}{72}x^6 + \frac{1}{6}x^2 - \frac{1}{8}x^{-2} \right] \).
M1: Applies limits \( 1 \) and \( 3 \) correctly to their integrated expression and multiplies by \( 2\pi \).
A1: Obtains exact answer \( \frac{208\pi}{9} \) (or exact equivalent fraction).

(a) Using integration by parts with \(u = x^n\) and \(\frac{\mathrm{d}v}{\mathrm{d}x} = \sin x\):
\[ \frac{\mathrm{d}u}{\mathrm{d}x} = n x^{n-1}, \quad v = -\cos x \]
\[ I_n = \left[-x^n \cos x\right]_0^{\frac{1}{2}\pi} + \int_0^{\frac{1}{2}\pi} n x^{n-1} \cos x \, \mathrm{d}x \]
Since \(\cos\left(\frac{1}{2}\pi\right) = 0\) and \(0^n = 0\) for \(n \ge 2\), the boundary term vanishes:
\[ I_n = n \int_0^{\frac{1}{2}\pi} x^{n-1} \cos x \, \mathrm{d}x \]
Apply integration by parts a second time with \(u = x^{n-1}\) and \(\frac{\mathrm{d}v}{\mathrm{d}x} = \cos x\):
\[ \frac{\mathrm{d}u}{\mathrm{d}x} = (n-1) x^{n-2}, \quad v = \sin x \]
\[ \int_0^{\frac{1}{2}\pi} x^{n-1} \cos x \, \mathrm{d}x = \left[x^{n-1} \sin x\right]_0^{\frac{1}{2}\pi} - \int_0^{\frac{1}{2}\pi} (n-1) x^{n-2} \sin x \, \mathrm{d}x \]
Evaluating the boundary term:
\[ \left(\frac{1}{2}\pi\right)^{n-1} \sin\left(\frac{1}{2}\pi\right) - 0 = \left(\frac{1}{2}\pi\right)^{n-1} \]
Substituting back:
\[ I_n = n \left[ \left(\frac{1}{2}\pi\right)^{n-1} - (n-1) I_{n-2} \right] = n\left(\frac{1}{2}\pi\right)^{n-1} - n(n-1)I_{n-2} \]

(b) For \(n = 0\):
\[ I_0 = \int_0^{\frac{1}{2}\pi} \sin x \, \mathrm{d}x = \left[-\cos x\right]_0^{\frac{1}{2}\pi} = 0 - (-1) = 1 \]
Using the reduction formula for \(n = 2\):
\[ I_2 = 2\left(\frac{1}{2}\pi\right)^1 - 2(1)I_0 = \pi - 2(1) = \pi - 2 \]
Using the reduction formula for \(n = 4\):
\[ I_4 = 4\left(\frac{1}{2}\pi\right)^3 - 4(3)I_2 = 4 \cdot \frac{\pi^3}{8} - 12(\pi - 2) = \frac{1}{2}\pi^3 - 12\pi + 24 \]

(a)
- M1: Applies integration by parts to \(I_n\) with correct parts to reach \(\pm [x^n \cos x]_0^{\frac{1}{2}\pi} + n\int x^{n-1}\cos x\,\mathrm{d}x\).
- A1: Correctly shows boundary term is 0 and obtains \(n \int_0^{\frac{1}{2}\pi} x^{n-1} \cos x \, \mathrm{d}x\).
- M1: Applies integration by parts a second time to \(\int x^{n-1}\cos x\,\mathrm{d}x\).
- A1 (AG): Fully correct intermediate steps leading convincingly to \(I_n = n\left(\frac{1}{2}\pi\right)^{n-1} - n(n-1)I_{n-2}\).

(b)
- B1: Evaluates \(I_0 = 1\) correctly.
- M1: Applies the reduction formula with \(n = 2\) using their \(I_0\).
- A1: Obtains \(I_2 = \pi - 2\).
- M1: Applies the reduction formula with \(n = 4\) using their \(I_2\).
- A1: Obtains exact value \(\frac{1}{2}\pi^3 - 12\pi + 24\) (or exact equivalent expanded form).

(a) Starting from the definition \(\cosh x = \frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\):
\[ 2\cosh^2 x - 1 = 2\left(\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right)^2 - 1 = 2\left(\frac{\mathrm{e}^{2x} + 2 + \mathrm{e}^{-2x}}{4}\right) - 1 = \frac{\mathrm{e}^{2x} + 2 + \mathrm{e}^{-2x} - 2}{2} = \frac{\mathrm{e}^{2x} + \mathrm{e}^{-2x}}{2} = \cosh 2x \]

(b) Substitute \(\cosh 2x = 2\cosh^2 x - 1\) into the equation:
\[ 3(2\cosh^2 x - 1) - 7\cosh x = 2 \implies 6\cosh^2 x - 7\cosh x - 5 = 0 \]
Factorising the quadratic:
\[ (2\cosh x + 1)(3\cosh x - 5) = 0 \]
Since \(\cosh x \ge 1\) for all real \(x\), \(\cosh x = -\frac{1}{2}\) has no real solutions.

Thus \(\cosh x = \frac{5}{3}\).

Using the logarithmic form of \(\operatorname{arcosh} u = \ln(u + \sqrt{u^2 - 1})\):
\[ x = \pm \operatorname{arcosh}\left(\frac{5}{3}\right) = \pm \ln\left(\frac{5}{3} + \sqrt{\frac{25}{9} - 1}\right) = \pm \ln\left(\frac{5}{3} + \frac{4}{3}\right) = \pm \ln 3 \]

(c) Let \(u = \frac{2x}{1+x^2}\). Then \(y = \operatorname{artanh} u\).
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{1-u^2} \cdot \frac{\mathrm{d}u}{\mathrm{d}x} \]
\[ 1 - u^2 = 1 - \frac{4x^2}{(1+x^2)^2} = \frac{(1+x^2)^2 - 4x^2}{(1+x^2)^2} = \frac{(1-x^2)^2}{(1+x^2)^2} \]
\[ \frac{\mathrm{d}u}{\mathrm{d}x} = \frac{2(1+x^2) - 2x(2x)}{(1+x^2)^2} = \frac{2 - 2x^2}{(1+x^2)^2} = \frac{2(1-x^2)}{(1+x^2)^2} \]
Therefore,
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(1+x^2)^2}{(1-x^2)^2} \cdot \frac{2(1-x^2)}{(1+x^2)^2} = \frac{2}{1-x^2} \]
Since \(\frac{\mathrm{d}}{\mathrm{d}x}(\operatorname{artanh} x) = \frac{1}{1-x^2}\), integrating gives \(y = 2\operatorname{artanh} x + C\).
When \(x = 0\), \(y = \operatorname{artanh}(0) = 0\), so \(C = 0\).
Hence \(y = 2\operatorname{artanh} x\).

(a)
- M1: Expresses \(2\cosh^2 x - 1\) in terms of exponential functions and expands.
- A1 (AG): Correct algebraic completion to obtain \(\frac{\mathrm{e}^{2x}+\mathrm{e}^{-2x}}{2} = \cosh 2x\).

(b)
- M1: Uses identity to form a 3-term quadratic equation in \(\cosh x\).
- A1: Obtains \(6\cosh^2 x - 7\cosh x - 5 = 0\) and correctly rejects \(\cosh x = -\frac{1}{2}\).
- M1: Applies logarithmic formula for \(\operatorname{arcosh}\left(\frac{5}{3}\right)\).
- A1: Obtains \(x = \ln 3\) and \(x = -\ln 3\) (or \(x = \pm \ln 3\)).

(c)
- M1: Applies chain rule to differentiate \(\operatorname{artanh} u\) and evaluates \(\frac{\mathrm{d}u}{\mathrm{d}x}\).
- A1: Obtains \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2}{1-x^2}\) in simplified form.
- A1: Deduces \(y = 2\operatorname{artanh} x\) with adequate justification (e.g. constant evaluated at \(x=0\)).

(a) Let \(x = 2\tan\theta\), then \(\mathrm{d}x = 2\sec^2\theta \, \mathrm{d}\theta\).
Also \(\sqrt{x^2+4} = \sqrt{4\tan^2\theta + 4} = 2\sec\theta\).

Substituting into the integral:
\[ \int \frac{1}{(4\tan^2\theta + 1)(2\sec\theta)} \cdot 2\sec^2\theta \, \mathrm{d}\theta = \int \frac{\sec\theta}{4\tan^2\theta + 1} \, \mathrm{d}\theta \]
Multiply numerator and denominator by \(\cos^2\theta\):
\[ \int \frac{\cos\theta}{4\sin^2\theta + \cos^2\theta} \, \mathrm{d}\theta = \int \frac{\cos\theta}{4\sin^2\theta + (1 - \sin^2\theta)} \, \mathrm{d}\theta = \int \frac{\cos\theta}{3\sin^2\theta + 1} \, \mathrm{d}\theta \]
Let \(u = \sin\theta\), so \(\mathrm{d}u = \cos\theta \, \mathrm{d}\theta\):
\[ \int \frac{1}{3u^2 + 1} \, \mathrm{d}u = \frac{1}{\sqrt{3}} \arctan(u\sqrt{3}) + c = \frac{1}{\sqrt{3}} \arctan(\sqrt{3}\sin\theta) + c \]
From \(\tan\theta = \frac{x}{2}\), we have \(\sin\theta = \frac{x}{\sqrt{x^2+4}}\).

Therefore,
\[ \int \frac{1}{(x^2 + 1)\sqrt{x^2 + 4}} \, \mathrm{d}x = \frac{1}{\sqrt{3}} \arctan\left(\frac{x\sqrt{3}}{\sqrt{x^2 + 4}}\right) + c \]

(b) Rearrange the differential equation into standard linear form:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} + \frac{2x}{x^2+1}y = \frac{1}{(x^2+1)^2\sqrt{x^2+4}} \]
The integrating factor is:
\[ I(x) = \mathrm{e}^{\int \frac{2x}{x^2+1}\,\mathrm{d}x} = \mathrm{e}^{\ln(x^2+1)} = x^2 + 1 \]
Multiplying through by the integrating factor:
\[ \frac{\mathrm{d}}{\mathrm{d}x}\left[(x^2+1)y\right] = \frac{1}{(x^2+1)\sqrt{x^2+4}} \]
Integrating both sides using the result from part (a):
\[ (x^2+1)y = \frac{1}{\sqrt{3}} \arctan\left(\frac{x\sqrt{3}}{\sqrt{x^2+4}}\right) + C \]
Using the initial condition \(y(0) = 0\):
\[ (0 + 1)(0) = \frac{1}{\sqrt{3}} \arctan(0) + C \implies C = 0 \]
Hence,
\[ y = \frac{1}{\sqrt{3}(x^2+1)} \arctan\left(\frac{x\sqrt{3}}{\sqrt{x^2+4}}\right) \]

(a)
- M1: Uses substitution \(x = 2\tan\theta\) with \(\mathrm{d}x = 2\sec^2\theta\,\mathrm{d}\theta\) and \(\sqrt{x^2+4} = 2\sec\theta\).
- A1: Transforms integral correctly to \(\int \frac{\sec\theta}{4\tan^2\theta+1}\,\mathrm{d}\theta\).
- M1: Converts integrand to terms in \(\sin\theta\) and \(\cos\theta\) to get \(\int \frac{\cos\theta}{3\sin^2\theta+1}\,\mathrm{d}\theta\).
- A1: Integrates correctly using standard form to obtain \(\frac{1}{\sqrt{3}}\arctan(u\sqrt{3})\).
- A1 (AG): Replaces \(\sin\theta\) with \(\frac{x}{\sqrt{x^2+4}}\) with full reasoning shown.

(b)
- M1: Finds the correct integrating factor \(\mathrm{e}^{\int \frac{2x}{x^2+1}\,\mathrm{d}x} = x^2+1\).
- M1: Recognises the integral on RHS and uses the result from part (a) to write \((x^2+1)y = \frac{1}{\sqrt{3}}\arctan\left(\frac{x\sqrt{3}}{\sqrt{x^2+4}}\right) + C\).
- M1: Uses \(x=0, y=0\) to evaluate arbitrary constant \(C\).
- A1: Obtains \(y = \frac{1}{\sqrt{3}(x^2+1)} \arctan\left(\frac{x\sqrt{3}}{\sqrt{x^2+4}}\right)\).

(a) Auxiliary equation: \(m^2 + 4m + 13 = 0\).
Solving gives \(m = \frac{-4 \pm \sqrt{16 - 52}}{2} = -2 \pm 3\text{i}\).
Complementary function: \(y_c = \text{e}^{-2x}\big(A\cos(3x) + B\sin(3x)\big)\).

For the particular integral, try \(y_p = Cx + D + E\text{e}^{-x}\).
Then \(y_p' = C - E\text{e}^{-x}\) and \(y_p'' = E\text{e}^{-x}\).
Substitute into the ODE:
\(E\text{e}^{-x} + 4(C - E\text{e}^{-x}) + 13(Cx + D + E\text{e}^{-x}) = 26x - 5 + 34\text{e}^{-x}\)
\(13Cx + (4C + 13D) + (1 - 4 + 13)E\text{e}^{-x} = 26x - 5 + 34\text{e}^{-x}\)
\(13Cx + (4C + 13D) + 10E\text{e}^{-x} = 26x - 5 + 34\text{e}^{-x}\)

Equating coefficients:
\(13C = 26 \implies C = 2\)
\(4(2) + 13D = -5 \implies 13D = -13 \implies D = -1\)
\(10E = 34 \implies E = \frac{17}{5} = 3.4\) [Wait, let's verify: \(E - 4E + 13E = 10E = 34\text{e}^{-x}\), giving \(E = \frac{17}{5}\); or if RHS was \(20\text{e}^{-x}\), \(E=2\). For \(34\text{e}^{-x}\), \(E = 3.4\). Let us use exact fraction \(E = \frac{17}{5}\)].
General solution:
\[y = \text{e}^{-2x}\big(A\cos(3x) + B\sin(3x)\big) + 2x - 1 + \frac{17}{5}\text{e}^{-x}.\]

(b) Using initial condition \(y(0) = 3\):
\(3 = A - 1 + \frac{17}{5} \implies A = 4 - \frac{17}{5} = \frac{3}{5}\).

Differentiating \(y\):
\(y' = -2\text{e}^{-2x}\big(A\cos(3x) + B\sin(3x)\big) + \text{e}^{-2x}\big(-3A\sin(3x) + 3B\cos(3x)\big) + 2 - \frac{17}{5}\text{e}^{-x}\).

Using \(y'(0) = 1\):
\(1 = -2A + 3B + 2 - \frac{17}{5}\)
\(1 = -2\left(\frac{3}{5}\right) + 3B + 2 - \frac{17}{5} = -\frac{6}{5} - \frac{17}{5} + 2 + 3B = -\frac{23}{5} + \frac{10}{5} + 3B = -\frac{13}{5} + 3B\)
\(3B = 1 + \frac{13}{5} = \frac{18}{5} \implies B = \frac{6}{5}\).

Particular solution:
\[y = \text{e}^{-2x}\left(\frac{3}{5}\cos(3x) + \frac{6}{5}\sin(3x)\right) + 2x - 1 + \frac{17}{5}\text{e}^{-x}.\]

(c) As \(x \to \infty\), \(\text{e}^{-2x} \to 0\) and \(\text{e}^{-x} \to 0\).
Therefore, \(y \approx 2x - 1\) (i.e. \(y\) approaches the oblique line \(y = 2x - 1\)).

(a)
M1: Forms auxiliary equation and solves for complex roots.
A1: Correct complementary function.
M1: Uses trial particular integral of form \(Cx + D + E\text{e}^{-x}\) and differentiates.
M1: Substitutes into DE and equates coefficients.
A1: Correct values of \(C, D, E\).
A1: States full general solution \(y = y_c + y_p\).

(b)
M1: Substitutes \(x=0, y=3\) to find \(A\).
M1: Differentiates general solution using product rule.
M1: Substitutes \(x=0, y'=1\) to find \(B\).
A1: Fully correct particular solution.

(c)
B1: Identifies exponential terms tend to 0.
B1: States that \(y \to 2x - 1\) as \(x \to \infty\).

(a) Consider the decreasing function \(f(x) = \frac{1}{2x+1}\) for \(x \ge 0\).
On each interval \([r-1, r]\) for \(r = 1, 2, \dots, n\), the lower rectangle has height \(f(r) = \frac{1}{2r+1}\) and width 1.
The total area of these inscribed rectangles is \(\sum_{r=1}^{n} \frac{1}{2r+1}\).
Since \(f(x)\) is strictly decreasing, the area of these rectangles is strictly less than the area under the curve:
\[\sum_{r=1}^{n} \frac{1}{2r+1} < \int_{0}^{n} \frac{1}{2x+1}\,\mathrm{d}x.\]
Evaluating the integral:
\[\int_{0}^{n} \frac{1}{2x+1}\,\mathrm{d}x = \left[ \frac{1}{2}\ln(2x+1) \right]_{0}^{n} = \frac{1}{2}\ln(2n+1) - 0 = \frac{1}{2}\ln(2n+1).\]
Thus, \(\sum_{r=1}^{n} \frac{1}{2r+1} < \frac{1}{2}\ln(2n+1)\).

(b) On each interval \([r-1, r]\), the upper (circumscribed) rectangle has height \(f(r-1) = \frac{1}{2(r-1)+1} = \frac{1}{2r-1}\) and width 1.
The total area of these circumscribed rectangles is \(\sum_{r=1}^{n} \frac{1}{2r-1}\).
Since the curve lies below these rectangles:
\[\sum_{r=1}^{n} \frac{1}{2r-1} > \int_{0}^{n} \frac{1}{2x+1}\,\mathrm{d}x = \frac{1}{2}\ln(2n+1).\]

(c) From part (b), we have:
\(\sum_{r=1}^{n} \frac{1}{2r-1} = 1 + \sum_{r=1}^{n-1} \frac{1}{2r+1} > \frac{1}{2}\ln(2n+1)\),
which gives
\(\sum_{r=1}^{n} \frac{1}{2r+1} = \left(\sum_{r=1}^{n-1} \frac{1}{2r+1}\right) + \frac{1}{2n+1} > \frac{1}{2}\ln(2n-1) - 1 + \frac{1}{2n+1}\).

Combining the inequalities:
\[\frac{\frac{1}{2}\ln(2n+1) - 1}{\ln n} < \frac{1}{\ln n}\sum_{r=1}^{n} \frac{1}{2r+1} < \frac{\frac{1}{2}\ln(2n+1)}{\ln n}.\]
Now evaluate \(\lim_{n \to \infty} \frac{\ln(2n+1)}{\ln n} = \lim_{n \to \infty} \frac{\ln n + \ln(2 + 1/n)}{\ln n} = 1\).
Thus,
\[\lim_{n \to \infty} \frac{\frac{1}{2}\ln(2n+1)}{\ln n} = \frac{1}{2}.\]
By the squeeze theorem, the limit is \(\frac{1}{2}\).

(a)
B1: States that \(f(x) = \frac{1}{2x+1}\) is a decreasing function.
M1: Sets up sum of lower rectangle areas as \(\sum_{r=1}^n \frac{1}{2r+1} \times 1\).
M1: Integrates \(\int_0^n \frac{1}{2x+1}\,\mathrm{d}x\) correctly to get \(\frac{1}{2}[\ln(2x+1)]_0^n\).
A1: Accurately concludes the inequality \(\sum_{r=1}^n \frac{1}{2r+1} < \frac{1}{2}\ln(2n+1)\).

(b)
M1: Identifies upper rectangle heights as \(f(r-1) = \frac{1}{2r-1}\).
M1: Compares sum of upper rectangles to the definite integral \(\int_0^n \frac{1}{2x+1}\,\mathrm{d}x\).
A1: Concludes the inequality \(\sum_{r=1}^n \frac{1}{2r-1} > \frac{1}{2}\ln(2n+1)\).

(c)
M1: Re-indexes or bounds \(\sum_{r=1}^n \frac{1}{2r+1}\) from below using part (b).
M1: Divides both bounds by \(\ln n\).
M1: Takes limit as \(n \to \infty\) of \(\frac{\ln(2n+1)}{\ln n}\).
A1: Obtains final limit \(\frac{1}{2}\) with full justification.

(a) Characteristic equation: \(\det(\mathbf{A} - \lambda\mathbf{I}) = 0\).
\[\det\begin{pmatrix} 1-\lambda & 2 & 0 \\ 0 & 3-\lambda & 0 \\ 2 & -4 & -1-\lambda \end{pmatrix} = 0.\]
Expanding along the first row or third column:
\((-1-\lambda)(1-\lambda)(3-\lambda) = 0 \implies -(\lambda+1)(1-\lambda)(3-\lambda) = 0\).
So the eigenvalues are \(\lambda_1 = -1\), \(\lambda_2 = 1\), \(\lambda_3 = 3\).

For \(\lambda_1 = -1\):
\(\begin{pmatrix} 2 & 2 & 0 \\ 0 & 4 & 0 \\ 2 & -4 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies y = 0, x = 0, z\text{ is free}\).
Eigenvector: \(\mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\).

For \(\lambda_2 = 1\):
\(\begin{pmatrix} 0 & 2 & 0 \\ 0 & 2 & 0 \\ 2 & -4 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies y = 0, 2x - 2z = 0 \implies x = z\).
Eigenvector: \(\mathbf{v}_2 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}\).

For \(\lambda_3 = 3\):
\(\begin{pmatrix} -2 & 2 & 0 \\ 0 & 0 & 0 \\ 2 & -4 & -4 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies x = y, 2x - 4x - 4z = 0 \implies -2x = 4z \implies z = -\frac{1}{2}x\).
Choosing \(x = 2\), \(y = 2\), \(z = -1\).
Eigenvector: \(\mathbf{v}_3 = \begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix}\).

(b) \(\mathbf{P} = \begin{pmatrix} 0 & 1 & 2 \\ 0 & 0 & 2 \\ 1 & 1 & -1 \end{pmatrix}\), \(\mathbf{D} = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{pmatrix}\).

(c) The characteristic equation is \(-(\lambda + 1)(\lambda - 1)(\lambda - 3) = -(\lambda^2 - 1)(\lambda - 3) = -(\lambda^3 - 3\lambda^2 - \lambda + 3) = 0\).
Thus, \(\lambda^3 - 3\lambda^2 - \lambda + 3 = 0\).
By the Cayley–Hamilton theorem, \(\mathbf{A}\) satisfies its own characteristic equation:
\[\mathbf{A}^3 - 3\mathbf{A}^2 - \mathbf{A} + 3\mathbf{I} = \mathbf{0} \implies \mathbf{A}^3 = 3\mathbf{A}^2 + \mathbf{A} - 3\mathbf{I}.\]
Multiplying by \(\mathbf{A}^{-1}\):
\[\mathbf{A}^2 - 3\mathbf{A} - \mathbf{I} + 3\mathbf{A}^{-1} = \mathbf{0} \implies 3\mathbf{A}^{-1} = 3\mathbf{A} + \mathbf{I} - \mathbf{A}^2.\]
We compute \(\mathbf{A}^2\):
\[\mathbf{A}^2 = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 3 & 0 \\ 2 & -4 & -1 \end{pmatrix}\begin{pmatrix} 1 & 2 & 0 \\ 0 & 3 & 0 \\ 2 & -4 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 8 & 0 \\ 0 & 9 & 0 \\ 0 & -4 & 1 \end{pmatrix}.\]
Then:
\[3\mathbf{A} + \mathbf{I} - \mathbf{A}^2 = \begin{pmatrix} 3 & 6 & 0 \\ 0 & 9 & 0 \\ 6 & -12 & -3 \end{pmatrix} + \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} - \begin{pmatrix} 1 & 8 & 0 \\ 0 & 9 & 0 \\ 0 & -4 & 1 \end{pmatrix} = \begin{pmatrix} 3 & -2 & 0 \\ 0 & 1 & 0 \\ 6 & -8 & -3 \end{pmatrix}.\]
Therefore,
\[\mathbf{A}^{-1} = \frac{1}{3}\begin{pmatrix} 3 & -2 & 0 \\ 0 & 1 & 0 \\ 6 & -8 & -3 \end{pmatrix} = \begin{pmatrix} 1 & -\frac{2}{3} & 0 \\ 0 & \frac{1}{3} & 0 \\ 2 & -\frac{8}{3} & -1 \end{pmatrix}.\]

(a)
M1: Sets up and evaluates \(\det(\mathbf{A} - \lambda\mathbf{I}) = 0\).
A1: Obtains eigenvalues \(\lambda = -1, 1, 3\).
M1: Sets up \((\mathbf{A} - \lambda\mathbf{I})\mathbf{v} = \mathbf{0}\) for their eigenvalues.
A1: Correct eigenvector for \(\lambda = -1\) (any non-zero scalar multiple).
A1: Correct eigenvector for \(\lambda = 1\) (any non-zero scalar multiple).
A1: Correct eigenvector for \(\lambda = 3\) (any non-zero scalar multiple).

(b)
B1: Writes correct \(\mathbf{P}\) whose columns are the eigenvectors.
B1: Writes corresponding diagonal matrix \(\mathbf{D}\) with matching eigenvalue order.

(c)
M1: Expands characteristic polynomial and applies Cayley–Hamilton theorem.
A1: Obtains \(\mathbf{A}^3 = 3\mathbf{A}^2 + \mathbf{A} - 3\mathbf{I}\).
M1: Multiplies equation by \(\mathbf{A}^{-1}\) to express \(\mathbf{A}^{-1}\) in terms of \(\mathbf{A}^2, \mathbf{A}, \mathbf{I}\).
A1: Computes \(\mathbf{A}^2\) and evaluates \(\mathbf{A}^{-1}\) correctly.