2026 Cambridge IAS-Level Mathematics - Further (9231) 模擬試題連答案詳解

(a) Let \(\dfrac{3}{(3r-1)(3r+2)} = \dfrac{A}{3r-1} + \dfrac{B}{3r+2}\).
\(3 = A(3r+2) + B(3r-1)\)
Setting \(r = -\frac{2}{3} \implies 3 = -3B \implies B = -1\).
Setting \(r = \frac{1}{3} \implies 3 = 3A \implies A = 1\).
Thus, \(\dfrac{3}{(3r-1)(3r+2)} = \dfrac{1}{3r-1} - \dfrac{1}{3r+2}\).

(b) From part (a), \(\dfrac{1}{(3r-1)(3r+2)} = \dfrac{1}{3}\left(\dfrac{1}{3r-1} - \dfrac{1}{3r+2}\right)\).
Using the method of differences:
\(\sum_{r=1}^{n}\dfrac{1}{(3r-1)(3r+2)} = \dfrac{1}{3}\sum_{r=1}^{n}\left(\dfrac{1}{3r-1} - \dfrac{1}{3r+2}\right)\)
\(= \dfrac{1}{3}\left[ \left(\dfrac{1}{2} - \dfrac{1}{5}\right) + \left(\dfrac{1}{5} - \dfrac{1}{8}\right) + \dots + \left(\dfrac{1}{3n-1} - \dfrac{1}{3n+2}\right) \right]\)
\(= \dfrac{1}{3}\left(\dfrac{1}{2} - \dfrac{1}{3n+2}\right)\)
\(= \dfrac{1}{3}\left(\dfrac{(3n+2) - 2}{2(3n+2)}\right)\)
\(= \dfrac{1}{3}\left(\dfrac{3n}{2(3n+2)}\right) = \dfrac{n}{2(3n+2)}\).

(c) Method 1 (Using sums from \(r=1\)):
\(\sum_{r=n}^{2n}\dfrac{1}{(3r-1)(3r+2)} = \sum_{r=1}^{2n}\dfrac{1}{(3r-1)(3r+2)} - \sum_{r=1}^{n-1}\dfrac{1}{(3r-1)(3r+2)}\)
\(= \dfrac{2n}{2(3(2n)+2)} - \dfrac{n-1}{2(3(n-1)+2)}\)
\(= \dfrac{2n}{2(6n+2)} - \dfrac{n-1}{2(3n-1)}\)
\(= \dfrac{n}{2(3n+1)} - \dfrac{n-1}{2(3n-1)}\)
\(= \dfrac{n(3n-1) - (n-1)(3n+1)}{2(3n+1)(3n-1)}\)
\(= \dfrac{(3n^2 - n) - (3n^2 - 2n - 1)}{2(9n^2 - 1)}\)
\(= \dfrac{n+1}{2(9n^2 - 1)}\)

Method 2 (Using differences directly):
\(\sum_{r=n}^{2n}\dfrac{1}{(3r-1)(3r+2)} = \dfrac{1}{3}\left( \dfrac{1}{3n-1} - \dfrac{1}{3(2n)+2} \right)\)
\(= \dfrac{1}{3}\left( \dfrac{1}{3n-1} - \dfrac{1}{2(3n+1)} \right)\)
\(= \dfrac{1}{3}\left( \dfrac{2(3n+1) - (3n-1)}{2(3n-1)(3n+1)} \right)\)
\(= \dfrac{1}{3}\left( \dfrac{3n+3}{2(3n-1)(3n+1)} \right) = \dfrac{n+1}{2(3n-1)(3n+1)}\).

(d) As \(n \to \infty\), the degree of the denominator is 2 and the numerator is 1:
\(\lim_{n \to \infty}\dfrac{n+1}{2(9n^2 - 1)} = 0\).

(a)
B1: Correct partial fractions \(\frac{1}{3r-1} - \frac{1}{3r+2}\).

(b)
M1: Expresses general term in difference form with factor \(\frac{1}{3}\) and lists at least first two and last terms showing cancellation.
A1: Correct unsimplified expression in terms of \(n\), e.g. \(\frac{1}{3}\left(\frac{1}{2} - \frac{1}{3n+2}\right)\).
A1 (AG): Fully correct algebraic steps with no omissions leading to \(\frac{n}{2(3n+2)}\).

(c)
M1: Writes the sum as \(S_{2n} - S_{n-1}\) or applies the difference method between limits \(n\) and \(2n\).
A1: Correct single fraction \(\frac{n+1}{2(3n-1)(3n+1)}\) or \(\frac{n+1}{2(9n^2-1)}\) or \(\frac{n+1}{18n^2-2}\).

(d)
B1: States \(0\) (allow follow through from part (c) if correctly evaluated).

From the given cubic equation \(x^3 - 3x^2 + 4x - 2 = 0\), the relations between roots and coefficients are:
\[\sum \alpha = \alpha + \beta + \gamma = 3\]
\[\sum \alpha\beta = \alpha\beta + \beta\gamma + \gamma\alpha = 4\]
\[\alpha\beta\gamma = 2\]

(a)(i)
Using the identity for the sum of squares:
\[\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\]
\[\alpha^2 + \beta^2 + \gamma^2 = 3^2 - 2(4) = 9 - 8 = 1\]

(a)(ii)
Since \(\alpha, \beta, \gamma\) satisfy the equation \(x^3 = 3x^2 - 4x + 2\):
\[\alpha^3 = 3\alpha^2 - 4\alpha + 2\]
\[\beta^3 = 3\beta^2 - 4\beta + 2\]
\[\gamma^3 = 3\gamma^2 - 4\gamma + 2\]
Summing these three equations gives:
\[\sum \alpha^3 = 3\sum \alpha^2 - 4\sum \alpha + 3(2)\]
Substitute the known values:
\[\sum \alpha^3 = 3(1) - 4(3) + 6 = 3 - 12 + 6 = -3\]

(b)
Rearrange the original equation by grouping odd and even powers of \(x\):
\[x^3 + 4x = 3x^2 + 2 \implies x(x^2 + 4) = 3x^2 + 2\]
Squaring both sides:
\[x^2(x^2 + 4)^2 = (3x^2 + 2)^2\]
Let \(y = x^2\):
\[y(y + 4)^2 = (3y + 2)^2\]
\[y(y^2 + 8y + 16) = 9y^2 + 12y + 4\]
\[y^3 + 8y^2 + 16y = 9y^2 + 12y + 4\]
\[y^3 - y^2 + 4y - 4 = 0\]

(a)(i)
M1: Uses \(\sum \alpha^2 = (\sum \alpha)^2 - 2\sum \alpha\beta\) with \(\sum \alpha = 3\) and \(\sum \alpha\beta = 4\).
A1: Obtains 1.

(a)(ii)
M1: Uses the relation \(\sum \alpha^3 = 3\sum \alpha^2 - 4\sum \alpha + 3(2)\) or equivalent algebraic identity with their value of \(\sum \alpha^2\).
A1: Obtains \(-3\).

(b)
M1: Rearranges the original equation to separate odd and even powers, e.g. \(x(x^2 + 4) = 3x^2 + 2\), and squares both sides.
DM1: Substitutes \(y = x^2\) and expands both sides correctly.
A1: Obtains \(y^3 - y^2 + 4y - 4 = 0\) (accept \(p = -1, q = 4, r = -4\)).

(a)
**Base step (\(n = 1\)):**
For \(n = 1\), \(u_1 = \frac{5}{2} = 2.5\).
Since \(2 < 2.5 < 3\), the statement is true for \(n = 1\).

Inductive step:
Assume that the statement is true for \(n = k\) for some positive integer \(k\), so
\[ 2 < u_k < 3. \]
Since \(u_k > 0\), squaring gives
\[ 4 < u_k^2 < 9. \]
Adding \(6\) to all parts of the inequality:
\[ 10 < u_k^2 + 6 < 15. \]
Dividing by \(5\):
\[ \frac{10}{5} < \frac{u_k^2 + 6}{5} < \frac{15}{5} \implies 2 < u_{k+1} < 3. \]
Hence, if the statement is true for \(n = k\), it is also true for \(n = k + 1\).

Conclusion:
Since the statement is true for \(n = 1\) and, whenever it is true for \(n = k\), it is also true for \(n = k + 1\), by mathematical induction \(2 < u_n < 3\) for all positive integers \(n\).

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(b)
**Base step (\(n = 1\)):**
For \(n = 1\), \(u_1 = \frac{5}{2} = 2.5\).
Using the recurrence relation,
\[ u_2 = \frac{\left(\frac{5}{2}\right)^2 + 6}{5} = \frac{\frac{25}{4} + 6}{5} = \frac{\frac{49}{4}}{5} = \frac{49}{20} = 2.45. \]
Since \(u_1 = 2.5 > 2.45 = u_2\), the statement is true for \(n = 1\).

Inductive step:
Assume that the statement is true for \(n = k\) for some positive integer \(k\), so \(u_k > u_{k+1}\).
We consider the difference \(u_{k+1} - u_{k+2}\):
\[ u_{k+1} - u_{k+2} = \frac{u_k^2 + 6}{5} - \frac{u_{k+1}^2 + 6}{5} = \frac{u_k^2 - u_{k+1}^2}{5} = \frac{(u_k - u_{k+1})(u_k + u_{k+1})}{5}. \]
From the inductive hypothesis, \(u_k - u_{k+1} > 0\).
From part (a), \(u_k > 2\) and \(u_{k+1} > 2\), which means \(u_k + u_{k+1} > 4 > 0\).
Since both \((u_k - u_{k+1})\) and \((u_k + u_{k+1})\) are strictly positive, it follows that
\[ u_{k+1} - u_{k+2} > 0 \implies u_{k+1} > u_{k+2}. \]
Hence, if the statement is true for \(n = k\), it is also true for \(n = k + 1\).

Conclusion:
Since the statement is true for \(n = 1\) and, whenever it is true for \(n = k\), it is also true for \(n = k + 1\), by mathematical induction \(u_n > u_{n+1}\) for all positive integers \(n\).

(a)
- B1: Verify base case \(n = 1\): states \(u_1 = 2.5\) and verifies \(2 < 2.5 < 3\).
- M1: State inductive hypothesis \(2 < u_k < 3\) and attempt to bound \(u_{k+1} = \frac{u_k^2 + 6}{5}\).
- A1: Correctly deduce \(4 < u_k^2 < 9 \implies 10 < u_k^2 + 6 < 15 \implies 2 < u_{k+1} < 3\).
- A1: Complete and convincing conclusion mentioning true for \(n=1\), inductive step (if true for \(k\) then true for \(k+1\)), and induction for all positive integers \(n\).

(b)
- B1: Calculate \(u_2 = \frac{49}{20} = 2.45\) and show \(u_1 > u_2\) for the base case \(n = 1\).
- M1: Assume \(u_k > u_{k+1}\) and express \(u_{k+1} - u_{k+2}\) using the recurrence relation.
- A1: Obtain \(\frac{(u_k - u_{k+1})(u_k + u_{k+1})}{5}\) and justify why this is positive using \(u_k - u_{k+1} > 0\) and \(u_k + u_{k+1} > 0\) (from part (a)).
- A1: Complete and convincing conclusion mentioning true for \(n=1\), inductive step, and induction for all positive integers \(n\).

(a)
An invariant line through the origin has equation \(y = mx\).
Under the transformation:
\[ \begin{pmatrix} 4 & 2 \\ 1 & 5 \end{pmatrix} \begin{pmatrix} x \\ mx \end{pmatrix} = \begin{pmatrix} 4x + 2mx \\ x + 5mx \end{pmatrix} = \begin{pmatrix} x' \\ y' \end{pmatrix} \]
Since the image point \((x', y')\) also lies on the line \(y = mx\), we have \(y' = mx'\):
\[ x(1 + 5m) = m \cdot x(4 + 2m) \]
Assuming \(x \neq 0\):
\[ 1 + 5m = 4m + 2m^2 \]
\[ 2m^2 - m - 1 = 0 \]
\[ (2m + 1)(m - 1) = 0 \]
Hence, \(m = 1\) or \(m = -\frac{1}{2}\).

The equations of the invariant lines through the origin are \(y = x\) and \(y = -\frac{1}{2}x\).

(b)
Consider a general line \(y = mx + c\).
Under the transformation:
\[ \begin{pmatrix} 4 & 2 \\ 1 & 5 \end{pmatrix} \begin{pmatrix} x \\ mx + c \end{pmatrix} = \begin{pmatrix} (4 + 2m)x + 2c \\ (1 + 5m)x + 5c \end{pmatrix} = \begin{pmatrix} x' \\ y' \end{pmatrix} \]
Setting \(y' = mx' + c\):
\[ (1 + 5m)x + 5c = m\big((4 + 2m)x + 2c\big) + c \]
\[ (1 + 5m)x + 5c = (4m + 2m^2)x + 2mc + c \]
Equating the coefficients of \(x\) gives \(2m^2 - m - 1 = 0\), so \(m = 1\) or \(m = -\frac{1}{2}\).

Equating the constant terms:
\[ 5c = 2mc + c \implies 4c - 2mc = 0 \implies 2c(2 - m) = 0 \]
- For \(m = 1\): \(2c(1) = 0 \implies c = 0\).
- For \(m = -\frac{1}{2}\): \(2c\left(\frac{5}{2}\right) = 0 \implies 5c = 0 \implies c = 0\).

In both cases, \(c = 0\), so there are no invariant lines that do not pass through the origin.

(c) (i)
\[ \det(\mathbf{B}) = 1\begin{vmatrix} 1 & 1 \\ -1 & 3 \end{vmatrix} - k\begin{vmatrix} 3 & 1 \\ 2 & 3 \end{vmatrix} + 2\begin{vmatrix} 3 & 1 \\ 2 & -1 \end{vmatrix} \]
\[ \det(\mathbf{B}) = 1(3 - (-1)) - k(9 - 2) + 2(-3 - 2) = 4 - 7k - 10 = -7k - 6 \]
\(\mathbf{B}\) is singular when \(\det(\mathbf{B}) = 0\):
\[ -7k - 6 = 0 \implies k = -\frac{6}{7} \]

(c) (ii)
For \(k = 0\), \(\det(\mathbf{B}) = -6\).
\[ \mathbf{B} = \begin{pmatrix} 1 & 0 & 2 \\ 3 & 1 & 1 \\ 2 & -1 & 3 \end{pmatrix} \]
The matrix of cofactors is:
\[ C_{11} = +(3 - (-1)) = 4, \quad C_{12} = -(9 - 2) = -7, \quad C_{13} = +(-3 - 2) = -5 \]
\[ C_{21} = -(0 - (-2)) = -2, \quad C_{22} = +(3 - 4) = -1, \quad C_{23} = -(-1 - 0) = 1 \]
\[ C_{31} = +(0 - 2) = -2, \quad C_{32} = -(1 - 6) = 5, \quad C_{33} = +(1 - 0) = 1 \]
\[ \mathbf{C} = \begin{pmatrix} 4 & -7 & -5 \\ -2 & -1 & 1 \\ -2 & 5 & 1 \end{pmatrix} \]
The adjugate matrix is the transpose of the cofactor matrix:
\[ \text{adj}(\mathbf{B}) = \mathbf{C}^{\mathrm{T}} = \begin{pmatrix} 4 & -2 & -2 \\ -7 & -1 & 5 \\ -5 & 1 & 1 \end{pmatrix} \]
Therefore,
\[ \mathbf{B}^{-1} = \frac{1}{\det(\mathbf{B})}\text{adj}(\mathbf{B}) = -\frac{1}{6}\begin{pmatrix} 4 & -2 & -2 \\ -7 & -1 & 5 \\ -5 & 1 & 1 \end{pmatrix} = \begin{pmatrix} -\frac{2}{3} & \frac{1}{3} & \frac{1}{3} \[4pt] \frac{7}{6} & \frac{1}{6} & -\frac{5}{6} \[4pt] \frac{5}{6} & -\frac{1}{6} & -\frac{1}{6} \end{pmatrix} \]

(a)
- M1: Sets up the matrix equation \(\begin{pmatrix} 4 & 2 \\ 1 & 5 \end{pmatrix} \begin{pmatrix} x \\ mx \end{pmatrix} = \begin{pmatrix} x' \\ mx' \end{pmatrix}\) and forms a relation between \(x'\) and \(y'\).
- A1: Obtains correct quadratic equation in \(m\), e.g. \(2m^2 - m - 1 = 0\).
- M1: Solves quadratic equation for \(m\).
- A1: Both lines correctly stated: \(y = x\) and \(y = -\frac{1}{2}x\) (or equivalent).

(b)
- M1: Applies matrix transformation to \(\begin{pmatrix} x \\ mx + c \end{pmatrix}\) and equates to \(mX + c\) to form equations in \(m\) and \(c\).
- A1: Shows clearly that \(c(4 - 2m) = 0\) leads to \(c = 0\) for both \(m = 1\) and \(m = -\frac{1}{2}\), concluding that no invariant lines exist with \(c \neq 0\).

(c)(i)
- M1: Correct method for expanding determinant of a \(3 \times 3\) matrix with at most one sign error.
- A1: Correct determinant \(-7k - 6\) (or \(-6 - 7k\)) and states \(k = -\frac{6}{7}\).

(c)(ii)
- M1: Finds at least 4 correct cofactors for \(k = 0\).
- A1: Obtains complete correct cofactor matrix or adjugate matrix.
- A1: Correct inverse matrix \(\mathbf{B}^{-1} = -\frac{1}{6}\begin{pmatrix} 4 & -2 & -2 \\ -7 & -1 & 5 \\ -5 & 1 & 1 \end{pmatrix}\) (oe, fractions may be inside matrix).

(a)
Direction vectors of \(l_1\) and \(l_2\) are \(\mathbf{d}_1 = \begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix}\) and \(\mathbf{d}_2 = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}\).

A common perpendicular direction is:
\[ \mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -2 \\ 2 & -1 & 1 \end{vmatrix} = \begin{pmatrix} 2(1) - (-2)(-1) \\ - (1(1) - (-2)(2)) \\ 1(-1) - 2(2) \end{pmatrix} = \begin{pmatrix} 0 \\ -5 \\ -5 \end{pmatrix} = -5 \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} \]

A vector connecting a point on \(l_1\), \(A(2, -1, 3)\), to a point on \(l_2\), \(B(4, 1, -1)\), is:
\[ \vec{AB} = \begin{pmatrix} 4 - 2 \\ 1 - (-1) \\ -1 - 3 \end{pmatrix} = \begin{pmatrix} 2 \\ 2 \\ -4 \end{pmatrix} \]

The shortest distance is given by the formula:
\[ d = \frac{|\vec{AB} \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{\left| \begin{pmatrix} 2 \\ 2 \\ -4 \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} \right|}{\sqrt{0^2 + 1^2 + 1^2}} = \frac{|0 + 2 - 4|}{\sqrt{2}} = \frac{|-2|}{\sqrt{2}} = \sqrt{2} \approx 1.41 \]

(b)
The normal to \(\Pi_1\) is perpendicular to both \(\mathbf{d}_1\) and \(\mathbf{d}_2\), so \(\mathbf{n}_1 = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\).

Since \(\Pi_1\) contains \(l_1\), the point \((2, -1, 3)\) lies on \(\Pi_1\):
\[ \mathbf{r} \cdot \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = 0(2) + 1(-1) + 1(3) = 2 \]

Thus, the Cartesian equation is \(y + z = 2\).

(c)(i)
Let \(\theta\) be the angle between \(l_1\) and \(\Pi_2\).
The normal to \(\Pi_2\) is \(\mathbf{n}_2 = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}\).

\[ \sin \theta = \frac{|\mathbf{d}_1 \cdot \mathbf{n}_2|}{|\mathbf{d}_1| |\mathbf{n}_2|} \]

Calculate the magnitudes and dot product:
\[ |\mathbf{d}_1| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{9} = 3 \]
\[ |\mathbf{n}_2| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{9} = 3 \]
\[ \mathbf{d}_1 \cdot \mathbf{n}_2 = 1(2) + 2(-1) + (-2)(2) = 2 - 2 - 4 = -4 \]

\[ \sin \theta = \frac{|-4|}{3 \times 3} = \frac{4}{9} \]
\[ \theta = \arcsin\left(\frac{4}{9}\right) \approx 26.3877^\circ = 26.4^\circ \]

(c)(ii)
The direction of the line of intersection is perpendicular to both normal vectors \(\mathbf{n}_1\) and \(\mathbf{n}_2\):
\[ \mathbf{d}_3 = \mathbf{n}_1 \times \mathbf{n}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 1 & 1 \\ 2 & -1 & 2 \end{vmatrix} = \begin{pmatrix} 1(2) - 1(-1) \\ -(0(2) - 1(2)) \\ 0(-1) - 1(2) \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} \]

To find a point lying on both planes:
\[ y + z = 2 \quad \text{and} \quad 2x - y + 2z = 5 \]
Setting \(y = 1\) gives \(z = 1\), and from the second equation \(2x - 1 + 2(1) = 5 \implies 2x = 4 \implies x = 2\).
So \((2, 1, 1)\) is a point on the line.

Therefore, a vector equation of the line is:
\[ \mathbf{r} = \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} + t \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} \]

(a) [4 marks]
- M1: Attempts vector product of direction vectors \(\begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix} \times \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}\).
- A1: Obtains correct normal vector, e.g., \(\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\) or \(\begin{pmatrix} 0 \\ -5 \\ -5 \end{pmatrix}\).
- M1: Evaluates \(\frac{|(\mathbf{b} - \mathbf{a}) \cdot \mathbf{n}|}{|\mathbf{n}|}\) using a vector connecting the two lines and their common normal.
- A1: Correct exact distance \(\sqrt{2}\) (or AWRT 1.41).

(b) [2 marks]
- M1: Uses normal vector from (a) with point \((2, -1, 3)\) to find plane constant \(d\).
- A1: Obtains \(y + z = 2\) (or \(0x + y + z = 2\)).

(c)(i) [3 marks]
- M1: Sets up \(\sin \theta = \frac{|\mathbf{d}_1 \cdot \mathbf{n}_2|}{|\mathbf{d}_1||\mathbf{n}_2|}\) with \(\mathbf{d}_1 = \begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix}\) and \(\mathbf{n}_2 = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}\).
- A1: Obtains \(\sin \theta = \frac{4}{9}\) (or \(\cos(90^\circ - \theta) = \frac{4}{9}\)).
- A1: Obtains \(26.4^\circ\) (accept AWRT 26.4, or 0.461 radians).

(c)(ii) [4 marks]
- M1: Computes vector product of the two normal vectors \(\mathbf{n}_1 \times \mathbf{n}_2\) (or solves system in terms of a parameter).
- A1: Obtains correct direction vector \(\begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix}\) (or any non-zero scalar multiple).
- M1: Finds a particular point satisfying both plane equations.
- A1: Correct vector equation in form \(\mathbf{r} = \mathbf{p} + t \mathbf{d}\), e.g. \(\mathbf{r} = \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} + t \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix}\) or \(\mathbf{r} = \begin{pmatrix} 7/2 \\ 2 \\ 0 \end{pmatrix} + t \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix}\).

(a)
Rewriting the equation of \(C\) using polynomial division:
\[ y = \frac{2x(x - 1) - 3(x - 1) + 2}{x - 1} = 2x - 3 + \frac{2}{x - 1} \]
- The vertical asymptote is given by the zero of the denominator: \(x = 1\).
- As \(x \to \pm\infty\), \(\frac{2}{x - 1} \to 0\), so the oblique asymptote is \(y = 2x - 3\).

(b)
Differentiating \(y\) with respect to \(x\):
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = 2 - \frac{2}{(x - 1)^2} \]
Set \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\[ 2 = \frac{2}{(x - 1)^2} \implies (x - 1)^2 = 1 \implies x - 1 = \pm 1 \]
Thus, \(x = 0\) or \(x = 2\).
- For \(x = 0\): \(y = \frac{5}{-1} = -5\). Coordinates: \((0, -5)\).
- For \(x = 2\): \(y = \frac{2(4) - 5(2) + 5}{2 - 1} = 3\). Coordinates: \((2, 3)\).

To determine the nature, find the second derivative:
\[ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{4}{(x - 1)^3} \]
- At \(x = 0\): \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{4}{(-1)^3} = -4 < 0\), so \((0, -5)\) is a local maximum.
- At \(x = 2\): \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{4}{(1)^3} = 4 > 0\), so \((2, 3)\) is a local minimum.

(c)
Intercepts:
- \(y\)-intercept: when \(x = 0\), \(y = -5\), so \((0, -5)\).
- \(x\)-intercepts: \(2x^2 - 5x + 5 = 0\) has discriminant \(\Delta = (-5)^2 - 4(2)(5) = 25 - 40 = -15 < 0\), so there are no real roots and no intersections with the \(x\)-axis.

Sketch of \(C\):
- Two separate smooth branches.
- Left branch (for \(x < 1\)): approaches \(y = 2x - 3\) as \(x \to -\infty\), reaches local maximum at \((0, -5)\) (which is also the \(y\)-intercept), and goes to \(-\infty\) as \(x \to 1^-\).
- Right branch (for \(x > 1\)): comes from \(+\infty\) as \(x \to 1^+\), reaches local minimum at \((2, 3)\), and approaches \(y = 2x - 3\) from above as \(x \to +\infty\).
- Asymptotes \(x = 1\) and \(y = 2x - 3\) shown as dashed lines.

(d)
For \(y = \frac{1}{f(x)} = \frac{x - 1}{2x^2 - 5x + 5}\):
- Asymptotes: Since the denominator \(2x^2 - 5x + 5 > 0\) for all \(x \in \mathbb{R}\), there are no vertical asymptotes. As \(x \to \pm\infty\), \(y \to 0\), so the horizontal asymptote is \(y = 0\) (the \(x\)-axis).
- Intercepts:
- When \(x = 0\), \(y = \frac{1}{-5} = -\frac{1}{5}\), giving \(\left(0, -\frac{1}{5}\right)\).
- When \(y = 0\), \(x - 1 = 0 \implies x = 1\), giving \((1, 0)\).
- Stationary points: The reciprocal transformation maps:
- Local max \((0, -5)\) of \(f(x)\) to local min \(\left(0, -\frac{1}{5}\right)\) of \(\frac{1}{f(x)}\).
- Local min \((2, 3)\) of \(f(x)\) to local max \(\left(2, \frac{1}{3}\right)\) of \(\frac{1}{f(x)}\).
- Sketch: A single continuous curve approaching \(y = 0\) from below as \(x \to -\infty\), with a local minimum at \(\left(0, -\frac{1}{5}\right)\), passing through \((1, 0)\), reaching a local maximum at \(\left(2, \frac{1}{3}\right)\), and approaching \(y = 0\) from above as \(x \to +\infty\).

(a)
- B1: State vertical asymptote \(x = 1\).
- B1: Carry out division or equate coefficients to obtain oblique asymptote \(y = 2x - 3\).

(b)
- M1: Differentiate using quotient rule or by differentiating the decomposed form \(2x - 3 + 2(x - 1)^{-1}\), and set to zero.
- A1: Obtain \(x = 0\) and \(x = 2\) with correct coordinates \((0, -5)\) and \((2, 3)\).
- M1: Use a valid method (e.g. second derivative, sign of first derivative, or nature of asymptotes) to determine the nature of both stationary points.
- A1: Fully correct justification identifying \((0, -5)\) as a local maximum and \((2, 3)\) as a local minimum.

(c)
- B1: Correctly draw and label both asymptotes \(x = 1\) and \(y = 2x - 3\).
- B1: Show two correctly positioned branches with correct asymptotic behavior and correct curvature near stationary points.
- B1: Clearly indicate the \(y\)-intercept at \((0, -5)\) and show no intersection with the \(x\)-axis.

(d)
- B1: State horizontal asymptote \(y = 0\) and state/show no vertical asymptotes.
- B1: State stationary points \(\left(0, -\frac{1}{5}\right)\) (min) and \(\left(2, \frac{1}{3}\right)\) (max).
- B1: State intercepts \(\left(0, -\frac{1}{5}\right)\) and \((1, 0)\).
- B1: Correct continuous curve drawn, showing correct behavior as \(x \to \pm\infty\) and correct turning points.

(a)
Using the relations \(x = r\cos\theta\), \(y = r\sin\theta\), and \(r^2 = x^2 + y^2\):
\[ r = 2\sqrt{3}\cos\theta \implies r^2 = 2\sqrt{3}r\cos\theta \]
\[ x^2 + y^2 = 2\sqrt{3}x \]
\[ x^2 - 2\sqrt{3}x + y^2 = 0 \]
Completing the square in \(x\):
\[ (x - \sqrt{3})^2 - 3 + y^2 = 0 \implies (x - \sqrt{3})^2 + y^2 = 3 \]
Thus \(a = \sqrt{3}\).

(b)
- \(C_1\) is a circle of radius \(\sqrt{3}\) with centre \((\sqrt{3}, 0)\) in Cartesian coordinates, passing through the pole \(O\). On the initial line \(\theta = 0\), \(r = 2\sqrt{3}\).
- \(C_2\) is a cardioid symmetric about the line \(\theta = \tfrac{1}{2}\pi\):
- Meets the initial line (\(\theta = 0\)) at \(r = 2\).
- Meets the line \(\theta = \pi\) at \(r = 2\).
- Meets the line \(\theta = \tfrac{1}{2}\pi\) at \(r = 4\).
- Passes through the pole at \(\theta = -\tfrac{1}{2}\pi\).

(c)
Setting the two polar equations equal for \(r > 0\):
\[ 2\sqrt{3}\cos\theta = 2 + 2\sin\theta \implies \sqrt{3}\cos\theta - \sin\theta = 1 \]
Using the harmonic identity \(R\cos(\theta + \alpha) = R\cos\theta\cos\alpha - R\sin\theta\sin\alpha\):
\[ R = \sqrt{(\sqrt{3})^2 + (-1)^2} = 2, \quad \tan\alpha = \frac{1}{\sqrt{3}} \implies \alpha = \frac{\pi}{6} \]
\[ 2\cos\left(\theta + \frac{\pi}{6}\right) = 1 \implies \cos\left(\theta + \frac{\pi}{6}\right) = \frac{1}{2} \]
For \(-\tfrac{1}{2}\pi \leqslant \theta \leqslant \tfrac{1}{2}\pi\):
\[ \theta + \frac{\pi}{6} = \frac{\pi}{3} \implies \theta = \frac{\pi}{6} \]
When \(\theta = \frac{\pi}{6}\):
\[ r = 2\sqrt{3}\cos\left(\frac{\pi}{6}\right) = 2\sqrt{3}\left(\frac{\sqrt{3}}{2}\right) = 3 \]
Thus, the polar coordinates are \(\left(3, \frac{\pi}{6}\right)\).

(d)
In the first quadrant \(0 \leqslant \theta \leqslant \tfrac{1}{2}\pi\), the boundary of the common region consists of:
- \(C_2\) for \(0 \leqslant \theta \leqslant \frac{\pi}{6}\)
- \(C_1\) for \(\frac{\pi}{6} \leqslant \theta \leqslant \frac{\pi}{2}\)

The area is given by \(A = A_1 + A_2\), where:
\[ A_1 = \frac{1}{2}\int_0^{\pi/6} (2 + 2\sin\theta)^2 \,\mathrm{d}\theta = 2\int_0^{\pi/6} (1 + 2\sin\theta + \sin^2\theta) \,\mathrm{d}\theta \]
Using \(\sin^2\theta = \frac{1 - \cos 2\theta}{2}\):
\[ 1 + 2\sin\theta + \sin^2\theta = \frac{3}{2} + 2\sin\theta - \frac{1}{2}\cos 2\theta \]
\[ A_1 = 2\left[ \frac{3}{2}\theta - 2\cos\theta - \frac{1}{4}\sin 2\theta \right]_0^{\pi/6} \]
Evaluating at \(\theta = \frac{\pi}{6}\):
\[ \frac{3}{2}\left(\frac{\pi}{6}\right) - 2\cos\left(\frac{\pi}{6}\right) - \frac{1}{4}\sin\left(\frac{\pi}{3}\right) = \frac{\pi}{4} - \sqrt{3} - \frac{\sqrt{3}}{8} = \frac{\pi}{4} - \frac{9\sqrt{3}}{8} \]
Evaluating at \(\theta = 0\):
\[ 0 - 2(1) - 0 = -2 \]
\[ A_1 = 2\left(\frac{\pi}{4} - \frac{9\sqrt{3}}{8} + 2\right) = \frac{\pi}{2} - \frac{9\sqrt{3}}{4} + 4 \]

Now, for \(A_2\):
\[ A_2 = \frac{1}{2}\int_{\pi/6}^{\pi/2} (2\sqrt{3}\cos\theta)^2 \,\mathrm{d}\theta = 6\int_{\pi/6}^{\pi/2} \cos^2\theta \,\mathrm{d}\theta = 3\int_{\pi/6}^{\pi/2} (1 + \cos 2\theta) \,\mathrm{d}\theta \]
\[ A_2 = 3\left[ \theta + \frac{1}{2}\sin 2\theta \right]_{\pi/6}^{\pi/2} \]
Evaluating at \(\theta = \frac{\pi}{2}\):
\[ \frac{\pi}{2} + \frac{1}{2}\sin(\pi) = \frac{\pi}{2} \]
Evaluating at \(\theta = \frac{\pi}{6}\):
\[ \frac{\pi}{6} + \frac{1}{2}\sin\left(\frac{\pi}{3}\right) = \frac{\pi}{6} + \frac{\sqrt{3}}{4} \]
\[ A_2 = 3\left( \frac{\pi}{2} - \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right) = 3\left( \frac{\pi}{3} - \frac{\sqrt{3}}{4} \right) = \pi - \frac{3\sqrt{3}}{4} \]

Total area \(A\):
\[ A = A_1 + A_2 = \left(\frac{\pi}{2} - \frac{9\sqrt{3}}{4} + 4\right) + \left(\pi - \frac{3\sqrt{3}}{4}\right) = \frac{3}{2}\pi - 3\sqrt{3} + 4 \]

(a) [3 marks]
- M1: Multiplies by \(r\) to obtain \(r^2 = 2\sqrt{3}r\cos\theta\) and uses standard substitutions \(r^2 = x^2 + y^2\) and \(x = r\cos\theta\).
- A1: Obtains \(x^2 - 2\sqrt{3}x + y^2 = 0\).
- A1: Completes the square correctly to obtain \((x - \sqrt{3})^2 + y^2 = 3\) and states \(a = \sqrt{3}\).

(b) [4 marks]
- B1: Correct sketch of circle \(C_1\) passing through the pole with centre on the initial line.
- B1: Correct sketch of cardioid \(C_2\) symmetrical about \(\theta = \frac{1}{2}\pi\) with cusp at the pole.
- B1: Correct intercepts labelled: \((2\sqrt{3}, 0)\) for \(C_1\); \((2, 0)\), \((2, \pi)\), and \((4, \frac{1}{2}\pi)\) for \(C_2\).
- B1: Curves sketched in correct relative positions (intersecting in the first and fourth quadrants).

(c) [3 marks]
- M1: Equates equations \(2\sqrt{3}\cos\theta = 2 + 2\sin\theta\) and expresses in the form \(R\cos(\theta + \alpha) = c\) or equivalent method.
- DM1: Solves for \(\theta\) within the valid interval \(-\frac{1}{2}\pi \leqslant \theta \leqslant \frac{1}{2}\pi\).
- A1: Obtains \(\left(3, \frac{\pi}{6}\right)\) (both \(r\) and \(\theta\) correct).

(d) [6 marks]
- M1: Sets up the area integral splitting at \(\theta = \frac{\pi}{6}\): \(\frac{1}{2}\int_0^{\pi/6} r_2^2 \,\mathrm{d}\theta + \frac{1}{2}\int_{\pi/6}^{\pi/2} r_1^2 \,\mathrm{d}\theta\).
- M1: Uses standard double angle identities \(\sin^2\theta = \frac{1 - \cos 2\theta}{2}\) and \(\cos^2\theta = \frac{1 + \cos 2\theta}{2}\) to integrate both expressions.
- A1: Correct integration of \((2 + 2\sin\theta)^2\), obtaining \(2\left[\frac{3}{2}\theta - 2\cos\theta - \frac{1}{4}\sin 2\theta\right]\) or equivalent.
- A1: Correct integration of \((2\sqrt{3}\cos\theta)^2\), obtaining \(3\left[\theta + \frac{1}{2}\sin 2\theta\right]\) or equivalent.
- M1: Applies all limits \(0\) to \(\frac{\pi}{6}\) and \(\frac{\pi}{6}\) to \(\frac{\pi}{2}\) correctly.
- A1: Obtains the exact final answer \(\frac{3}{2}\pi - 3\sqrt{3} + 4\) (or identifies \(p = \frac{3}{2}, q = -3, s = 4\)).

(a)
The system fails to have a unique solution when the determinant of the coefficient matrix \(\mathbf{M}\) is zero:
\[ \det(\mathbf{M}) = \begin{vmatrix} 1 & 2 & -1 \\ 2 & 5 & a \\ 3 & 8 & -1 \end{vmatrix} \]
Expanding along the first row:
\[ \det(\mathbf{M}) = 1\bigl(5(-1) - 8a\bigr) - 2\bigl(2(-1) - 3a\bigr) + (-1)\bigl(2(8) - 3(5)\bigr) \]
\[ = (-5 - 8a) - 2(-2 - 3a) - (16 - 15) \]
\[ = -5 - 8a + 4 + 6a - 1 = -2a - 2 \]
Setting \(\det(\mathbf{M}) = 0\):
\[ -2a - 2 = 0 \implies a = -1 \]
Thus, the system does not have a unique solution if and only if \(a = -1\).

(b)(i)
Substitute \(a = -1\) into the augmented matrix:
\[ \begin{pmatrix} 1 & 2 & -1 & | & 1 \\ 2 & 5 & -1 & | & 3 \\ 3 & 8 & -1 & | & b \end{pmatrix} \]
Apply row operations:
- \(R_2 \to R_2 - 2R_1\): \(\begin{pmatrix} 0 & 1 & 1 & | & 1 \end{pmatrix}\)
- \(R_3 \to R_3 - 3R_1\): \(\begin{pmatrix} 0 & 2 & 2 & | & b - 3 \end{pmatrix}\)

Now apply \(R_3 \to R_3 - 2R_2\):
\[ \begin{pmatrix} 0 & 0 & 0 & | & (b - 3) - 2(1) \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 & | & b - 5 \end{pmatrix} \]
For consistency, the system requires \(b - 5 = 0 \implies b = 5\).

(b)(ii)
When \(a = -1\) and \(b = 5\), the equations are consistent with infinitely many solutions. Since the normal vectors \((1, 2, -1)\), \((2, 5, -1)\), and \((3, 8, -1)\) are not parallel, the three planes intersect in a single common straight line (they form a sheaf of planes).

(b)(iii)
When \(a = -1\) and \(b \neq 5\), the system is inconsistent. Since no two normal vectors are parallel (no two planes are parallel or coincident), the three planes form a triangular prism (they intersect in pairs in three mutually parallel, distinct lines).

(a)
- M1: Attempts determinant of the \(3 \times 3\) matrix and sets to 0 (or uses elimination to obtain a row of zeros in the coefficient matrix).
- A1: Correct expansion showing \(\det(\mathbf{M}) = -2a - 2 = 0 \implies a = -1\) with clear working.

(b)(i)
- M1: Sets up augmented matrix with \(a = -1\) and performs valid row operations to eliminate variables in the third row.
- A1: Obtains \(b - 5 = 0 \implies b = 5\).

(b)(ii)
- B1: States that the three planes intersect in a common straight line / form a sheaf of planes.

(b)(iii)
- B1: States that the planes form a triangular prism / intersect in three distinct parallel lines (with no common point).

(a)
The characteristic equation is given by $\det(\mathbf{A} - \lambda \mathbf{I}) = 0$:
$$\begin{vmatrix} 1-\lambda & 0 & 1 \\ 0 & 1-\lambda & 1 \\ 0 & -4 & 6-\lambda \end{vmatrix} = 0$$

Expanding along the first column:
$$(1-\lambda) \left[ (1-\lambda)(6-\lambda) - (1)(-4) \right] = 0$$
$$(1-\lambda)(\lambda^2 - 7\lambda + 6 + 4) = 0$$
$$(1-\lambda)(\lambda^2 - 7\lambda + 10) = 0$$
$$(1-\lambda)(\lambda - 2)(\lambda - 5) = 0$$

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

(b)
For $\lambda = 1$:
$$\begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 1 \\ 0 & -4 & 5 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies z = 0, \; -4y = 0 \implies y = 0$$
$x$ can take any non-zero value, so an eigenvector is $\mathbf{e}_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$.

For $\lambda = 2$:
$$\begin{pmatrix} -1 & 0 & 1 \\ 0 & -1 & 1 \\ 0 & -4 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies -x + z = 0, \; -y + z = 0 \implies x = y = z$$
Choosing $z = 1$, an eigenvector is $\mathbf{e}_2 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$.

For $\lambda = 5$:
$$\begin{pmatrix} -4 & 0 & 1 \\ 0 & -4 & 1 \\ 0 & -4 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies -4x + z = 0, \; -4y + z = 0 \implies z = 4x = 4y$$
Choosing $x = 1$, an eigenvector is $\mathbf{e}_3 = \begin{pmatrix} 1 \\ 1 \\ 4 \end{pmatrix}$.

(c)
Forming the matrices:
$$\mathbf{P} = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 4 \end{pmatrix}, \quad \mathbf{D} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 5 \end{pmatrix}$$

To find $\mathbf{P}^{-1}$:
$$\det(\mathbf{P}) = 1(4 - 1) - 1(0) + 1(0) = 3$$
The matrix of cofactors of $\mathbf{P}$ is:
$$\mathbf{C} = \begin{pmatrix} 3 & 0 & 0 \\ -3 & 4 & -1 \\ 0 & -1 & 1 \end{pmatrix}$$
Transposing gives the adjugate matrix:
$$\operatorname{adj}(\mathbf{P}) = \begin{pmatrix} 3 & -3 & 0 \\ 0 & 4 & -1 \\ 0 & -1 & 1 \end{pmatrix} \implies \mathbf{P}^{-1} = \frac{1}{3} \begin{pmatrix} 3 & -3 & 0 \\ 0 & 4 & -1 \\ 0 & -1 & 1 \end{pmatrix}$$

Now, $\mathbf{A}^n = \mathbf{P}\mathbf{D}^n\mathbf{P}^{-1}$:
$$\mathbf{P}\mathbf{D}^n = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 4 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2^n & 0 \\ 0 & 0 & 5^n \end{pmatrix} = \begin{pmatrix} 1 & 2^n & 5^n \\ 0 & 2^n & 5^n \\ 0 & 2^n & 4 \cdot 5^n \end{pmatrix}$$

Multiplying by $\mathbf{P}^{-1}$:
$$\mathbf{A}^n = \frac{1}{3} \begin{pmatrix} 1 & 2^n & 5^n \\ 0 & 2^n & 5^n \\ 0 & 2^n & 4 \cdot 5^n \end{pmatrix} \begin{pmatrix} 3 & -3 & 0 \\ 0 & 4 & -1 \\ 0 & -1 & 1 \end{pmatrix}$$
$$\mathbf{A}^n = \frac{1}{3} \begin{pmatrix} 3 & -3 + 4(2^n) - 5^n & -2^n + 5^n \\ 0 & 4(2^n) - 5^n & -2^n + 5^n \\ 0 & 4(2^n) - 4(5^n) & -2^n + 4(5^n) \end{pmatrix}$$

(a)
- M1: Sets up characteristic equation $\det(\mathbf{A} - \lambda \mathbf{I}) = 0$ and attempts cofactor expansion.
- A1: Obtains correct factorised polynomial $(1-\lambda)(\lambda-2)(\lambda-5) = 0$ (or expanded $-\lambda^3 + 8\lambda^2 - 17\lambda + 10 = 0$).
- A1: States correct eigenvalues $\lambda = 1, 2, 5$.

(b)
- M1: Sets up $(\mathbf{A} - \lambda\mathbf{I})\mathbf{x} = \mathbf{0}$ for at least one of their eigenvalues and attempts to solve for the components.
- A1: Finds any two correct eigenvectors (or non-zero scalar multiples).
- A1: Finds all three correct eigenvectors.

(c)
- M1: States $\mathbf{P}$ and $\mathbf{D}$ consistently with their eigenvalues and eigenvectors, and correctly finds $\mathbf{P}^{-1}$.
- DM1: Sets up and attempts the matrix multiplication $\mathbf{P}\mathbf{D}^n\mathbf{P}^{-1}$. Dependent on previous M1.
- A1: Obtains fully correct expression for $\mathbf{A}^n$ in terms of $n$ (any correct equivalent factorised or single matrix form).

Rewrite the differential equation in standard linear form by dividing throughout by \(x\) (for \(x > 0\)):
\[ \frac{\mathrm{d}y}{\mathrm{d}x} + \left(3 + \frac{2}{x}\right)y = 4\mathrm{e}^{-x} \]

Find the integrating factor, \(I(x)\):
\[ I(x) = \exp\left( \int \left(3 + \frac{2}{x}\right) \mathrm{d}x \right) = \exp(3x + 2\ln x) = \mathrm{e}^{3x} \cdot \mathrm{e}^{\ln(x^2)} = x^2 \mathrm{e}^{3x} \]

Multiply the differential equation by \(I(x)\):
\[ \frac{\mathrm{d}}{\mathrm{d}x}\left(y x^2 \mathrm{e}^{3x}\right) = 4\mathrm{e}^{-x} \cdot x^2 \mathrm{e}^{3x} = 4x^2 \mathrm{e}^{2x} \]

Integrate both sides with respect to \(x\):
\[ y x^2 \mathrm{e}^{3x} = \int 4x^2 \mathrm{e}^{2x} \mathrm{d}x \]

Using integration by parts:
Let \(u = 4x^2\) and \(v' = \mathrm{e}^{2x}\), so \(u' = 8x\) and \(v = \frac{1}{2}\mathrm{e}^{2x}\):
\[ \int 4x^2 \mathrm{e}^{2x} \mathrm{d}x = 4x^2\left(\frac{1}{2}\mathrm{e}^{2x}\right) - \int 8x\left(\frac{1}{2}\mathrm{e}^{2x}\right) \mathrm{d}x = 2x^2 \mathrm{e}^{2x} - \int 4x \mathrm{e}^{2x} \mathrm{d}x \]

Applying integration by parts a second time on \(\int 4x \mathrm{e}^{2x} \mathrm{d}x\) with \(u = 4x\) and \(v' = \mathrm{e}^{2x}\):
\[ \int 4x \mathrm{e}^{2x} \mathrm{d}x = 4x\left(\frac{1}{2}\mathrm{e}^{2x}\right) - \int 4\left(\frac{1}{2}\mathrm{e}^{2x}\right) \mathrm{d}x = 2x \mathrm{e}^{2x} - 2 \cdot \frac{1}{2}\mathrm{e}^{2x} = (2x - 1)\mathrm{e}^{2x} \]

Thus:
\[ y x^2 \mathrm{e}^{3x} = 2x^2 \mathrm{e}^{2x} - (2x - 1)\mathrm{e}^{2x} + C = (2x^2 - 2x + 1)\mathrm{e}^{2x} + C \]

Using the initial condition \(y = \mathrm{e}^{-1}\) when \(x = 1\):
\[ (\mathrm{e}^{-1})(1)^2 \mathrm{e}^{3(1)} = (2(1)^2 - 2(1) + 1)\mathrm{e}^{2(1)} + C \]
\[ \mathrm{e}^2 = (1)\mathrm{e}^2 + C \implies C = 0 \]

Therefore:
\[ y x^2 \mathrm{e}^{3x} = (2x^2 - 2x + 1)\mathrm{e}^{2x} \]
\[ y = \frac{2x^2 - 2x + 1}{x^2} \mathrm{e}^{-x} = \left(2 - \frac{2}{x} + \frac{1}{x^2}\right) \mathrm{e}^{-x} \]

M1: Write the differential equation in the form \(\frac{\mathrm{d}y}{\mathrm{d}x} + P(x)y = Q(x)\) and attempt to find the integrating factor \(\mathrm{e}^{\int P(x)\,\mathrm{d}x}\).
A1: Obtain correct integrating factor \(x^2 \mathrm{e}^{3x}\).
M1: Multiply by their integrating factor and recognize LHS as \(\frac{\mathrm{d}}{\mathrm{d}x}(y \cdot I(x))\).
M1: Attempt integration by parts on \(\int 4x^2 \mathrm{e}^{2x}\,\mathrm{d}x\) at least once.
A1: Obtain fully correct integral \((2x^2 - 2x + 1)\mathrm{e}^{2x} + C\).
M1: Substitute \(x = 1\) and \(y = \mathrm{e}^{-1}\) into their general solution including constant \(C\) to determine \(C\).
A1: Obtain \(y = \frac{2x^2 - 2x + 1}{x^2} \mathrm{e}^{-x}\) or \(y = \left(2 - \frac{2}{x} + \frac{1}{x^2}\right) \mathrm{e}^{-x}\).

(a)
Differentiating \(x\) and \(y\) with respect to \(t\):
\[ \frac{\mathrm{d}x}{\mathrm{d}t} = 3\cosh^2 t \sinh t \]
\[ \frac{\mathrm{d}y}{\mathrm{d}t} = 3\sinh^2 t \cosh t \]

Using the chain rule for parametric differentiation:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{\mathrm{d}x}{\mathrm{d}t}} = \frac{3\sinh^2 t \cosh t}{3\cosh^2 t \sinh t} = \frac{\sinh t}{\cosh t} = \tanh t \]

(b)
Using \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)}{\frac{\mathrm{d}x}{\mathrm{d}t}}\):
\[ \frac{\mathrm{d}}{\mathrm{d}t}\left(\tanh t\right) = \operatorname{sech}^2 t \]

Thus,
\[ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{\operatorname{sech}^2 t}{3\cosh^2 t \sinh t} = \frac{1}{3\cosh^4 t \sinh t} = \frac{1}{3}\operatorname{sech}^4 t \operatorname{cosech} t \]

(c)
The arc length \(s\) is given by
\[ s = \int_0^{\ln(1+\sqrt{2})} \sqrt{\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2} \,\mathrm{d}t \]

We compute the integrand:
\[ \left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2 = 9\cosh^4 t \sinh^2 t + 9\sinh^4 t \cosh^2 t \]
\[ = 9\sinh^2 t \cosh^2 t (\cosh^2 t + \sinh^2 t) \]
Using the identities \(2\sinh t \cosh t = \sinh 2t\) and \(\cosh^2 t + \sinh^2 t = \cosh 2t\):
\[ = 9 \left(\frac{1}{2}\sinh 2t\right)^2 \cosh 2t = \frac{9}{4} \sinh^2(2t) \cosh(2t) \]

Since \(t \ge 0\), taking the square root gives:
\[ \sqrt{\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2} = \frac{3}{2}\sinh(2t) \sqrt{\cosh(2t)} \]

Integrating with respect to \(t\):
\[ s = \int_0^{\ln(1+\sqrt{2})} \frac{3}{2}\sinh(2t) (\cosh 2t)^{\frac{1}{2}} \,\mathrm{d}t = \left[ \frac{3}{2} \cdot \frac{1}{2} \cdot \frac{2}{3} (\cosh 2t)^{\frac{3}{2}} \right]_0^{\ln(1+\sqrt{2})} = \left[ \frac{1}{2}(\cosh 2t)^{\frac{3}{2}} \right]_0^{\ln(1+\sqrt{2})} \]

At the upper limit \(t = \ln(1+\sqrt{2})\):
\[ \mathrm{e}^{2t} = (1+\sqrt{2})^2 = 3 + 2\sqrt{2} \]
\[ \mathrm{e}^{-2t} = \frac{1}{3+2\sqrt{2}} = 3 - 2\sqrt{2} \]
\[ \cosh(2t) = \frac{(3+2\sqrt{2}) + (3-2\sqrt{2})}{2} = 3 \]

At the lower limit \(t = 0\):
\[ \cosh(0) = 1 \]

Hence,
\[ s = \frac{1}{2}\left(3^{\frac{3}{2}} - 1^{\frac{3}{2}}\right) = \frac{1}{2}(3\sqrt{3} - 1) \]

(a)
M1: Differentiates \(x\) and \(y\) with respect to \(t\) using the chain rule and forms \(\frac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t}\).
A1: Obtains \(\tanh t\) in simplified form.

(b)
M1: Differentiates their \(\frac{\mathrm{d}y}{\mathrm{d}x}\) with respect to \(t\) to get \(\operatorname{sech}^2 t\).
M1: Divides by \(\frac{\mathrm{d}x}{\mathrm{d}t}\) (correct parametric second derivative formula).
A1: Fully correct derivation leading to the given result \(\frac{1}{3}\operatorname{sech}^4 t \operatorname{cosech} t\) without missing steps (AG).

(c)
M1: Correctly sets up the arc length integral with \(\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2\).
M1: Factorises expression and applies hyperbolic identities (\(\cosh^2 t + \sinh^2 t = \cosh 2t\) and \(2\sinh t \cosh t = \sinh 2t\)) to express under the square root.
A1: Integrates correctly to obtain \(\frac{1}{2}(\cosh 2t)^{\frac{3}{2}}\).
M1: Evaluates \(\cosh(2\ln(1+\sqrt{2})) = 3\) correctly.
A1: Obtains exact value \(\frac{1}{2}(3\sqrt{3} - 1)\) (or equivalent single fraction \(\frac{3\sqrt{3}-1}{2}\)).

(a) Using the standard integral or the substitution \(x = 3\sinh u\):
\[ \int \frac{1}{\sqrt{x^2 + 9}} \, \mathrm{d}x = \operatorname{arsinh}\left(\frac{x}{3}\right) = \ln\left(\frac{x}{3} + \sqrt{\left(\frac{x}{3}\right)^2 + 1}\right) = \ln\left(\frac{x + \sqrt{x^2 + 9}}{3}\right) \]
Applying the limits from \(x = 0\) to \(x = N\):
\[ \int_{0}^{N} \frac{1}{\sqrt{x^2 + 9}} \, \mathrm{d}x = \left[ \ln\left(\frac{x + \sqrt{x^2 + 9}}{3}\right) \right]_0^N = \ln\left(\frac{N + \sqrt{N^2 + 9}}{3}\right) - \ln\left(\frac{0 + \sqrt{9}}{3}\right) = \ln\left(\frac{N + \sqrt{N^2 + 9}}{3}\right) - \ln 1 = \ln\left(\frac{N + \sqrt{N^2 + 9}}{3}\right) \]

(b) The function \(f(x) = \frac{1}{\sqrt{x^2 + 9}}\) is strictly decreasing for \(x \ge 0\).
Consider rectangles of width \(1\) with bases on the intervals \([r-1, r]\) for \(r = 1, 2, \dots, n\), where the height of each rectangle is evaluated at the right-hand endpoint, \(x = r\).
The height of the \(r\)th rectangle is \(f(r) = \frac{1}{\sqrt{r^2 + 9}}\).
Since \(f(x)\) is decreasing, each rectangle lies entirely below the curve on \([r-1, r]\):
\[ \text{Area of rectangle} = 1 \times f(r) = \frac{1}{\sqrt{r^2 + 9}} < \int_{r-1}^{r} \frac{1}{\sqrt{x^2 + 9}} \, \mathrm{d}x \]
Summing both sides from \(r = 1\) to \(n\):
\[ \sum_{r=1}^{n} \frac{1}{\sqrt{r^2 + 9}} < \sum_{r=1}^{n} \int_{r-1}^{r} \frac{1}{\sqrt{x^2 + 9}} \, \mathrm{d}x = \int_{0}^{n} \frac{1}{\sqrt{x^2 + 9}} \, \mathrm{d}x \]
Using the result from part (a) with \(N = n\):
\[ \sum_{r=1}^{n} \frac{1}{\sqrt{r^2 + 9}} < \ln\left(\frac{n + \sqrt{n^2 + 9}}{3}\right) \]

(c) Consider rectangles of width \(1\) on the intervals \([r, r+1]\) for \(r = 1, 2, \dots, n\), where the height of each rectangle is evaluated at the left-hand endpoint, \(x = r\).
The height of the \(r\)th rectangle is \(f(r) = \frac{1}{\sqrt{r^2 + 9}}\).
Since \(f(x)\) is decreasing, each rectangle lies entirely above the curve on \([r, r+1]\):
\[ \text{Area of rectangle} = 1 \times f(r) = \frac{1}{\sqrt{r^2 + 9}} > \int_{r}^{r+1} \frac{1}{\sqrt{x^2 + 9}} \, \mathrm{d}x \]
Summing both sides from \(r = 1\) to \(n\):
\[ \sum_{r=1}^{n} \frac{1}{\sqrt{r^2 + 9}} > \sum_{r=1}^{n} \int_{r}^{r+1} \frac{1}{\sqrt{x^2 + 9}} \, \mathrm{d}x = \int_{1}^{n+1} \frac{1}{\sqrt{x^2 + 9}} \, \mathrm{d}x \]
Evaluating the definite integral:
\[ \int_{1}^{n+1} \frac{1}{\sqrt{x^2 + 9}} \, \mathrm{d}x = \left[ \ln\left(x + \sqrt{x^2 + 9}\right) \right]_1^{n+1} = \ln\left(n + 1 + \sqrt{(n+1)^2 + 9}\right) - \ln\left(1 + \sqrt{1^2 + 9}\right) \]
\[ = \ln\left(\frac{n + 1 + \sqrt{n^2 + 2n + 1 + 9}}{1 + \sqrt{10}}\right) = \ln\left(\frac{n + 1 + \sqrt{n^2 + 2n + 10}}{1 + \sqrt{10}}\right) \]
Hence,
\[ \sum_{r=1}^{n} \frac{1}{\sqrt{r^2 + 9}} > \ln\left(\frac{n + 1 + \sqrt{n^2 + 2n + 10}}{1 + \sqrt{10}}\right) \]

(a)
- M1: Integrates \(\frac{1}{\sqrt{x^2 + 9}}\) to \(\operatorname{arsinh}\left(\frac{x}{3}\right)\) or \(\ln(x + \sqrt{x^2 + 9})\).
- A1: Applies limits \(0\) and \(N\) correctly, showing \(\ln(3/3) = \ln 1 = 0\), and obtains the given logarithmic form accurately (AG).

(b)
- B1: Draws a clear sketch showing a decreasing curve for \(x \ge 0\) with inscribed (lower/right-endpoint) rectangles of unit width from \(x = 0\) to \(x = n\).
- M1: States the inequality for the area of a single rectangle \(\frac{1}{\sqrt{r^2 + 9}} < \int_{r-1}^{r} \frac{1}{\sqrt{x^2 + 9}}\,\mathrm{d}x\) or sums over \(r = 1\dots n\).
- A1: Fully justifies the summation to obtain \(\sum_{r=1}^n \frac{1}{\sqrt{r^2 + 9}} < \int_{0}^n \frac{1}{\sqrt{x^2 + 9}}\,\mathrm{d}x\) and completes the proof to the given result (AG).

(c)
- B1: Draws or indicates circumscribed (upper/left-endpoint) rectangles on intervals \([r, r+1]\) for \(r = 1, \dots, n\).
- M1: Forms the sum of rectangle areas and states the inequality \(\sum_{r=1}^{n} \frac{1}{\sqrt{r^2 + 9}} > \int_{1}^{n+1} \frac{1}{\sqrt{x^2 + 9}}\,\mathrm{d}x\).
- M1: Evaluates the integral using limits \(1\) and \(n+1\).
- A1: Expands \((n+1)^2 + 9 = n^2 + 2n + 10\) and combines the logarithms into a single quotient to achieve the given result accurately (AG).

(a)
Apply integration by parts to \(I_n = \int_0^1 x^n (1-x)^{\frac{1}{2}} \,\mathrm{d}x\):
Let \(u = x^n \implies \frac{\mathrm{d}u}{\mathrm{d}x} = n x^{n-1}\)
and \(\frac{\mathrm{d}v}{\mathrm{d}x} = (1-x)^{\frac{1}{2}} \implies v = -\frac{2}{3}(1-x)^{\frac{3}{2}}\).

Then
\[I_n = \left[ -\frac{2}{3} x^n (1-x)^{\frac{3}{2}} \right]_0^1 - \int_0^1 \left(-\frac{2}{3}(1-x)^{\frac{3}{2}}\right) n x^{n-1} \,\mathrm{d}x.\]
Evaluating the boundary term:
\[\left[ -\frac{2}{3} x^n (1-x)^{\frac{3}{2}} \right]_0^1 = 0 - 0 = 0 \quad \text{for } n \ge 1.\]
Thus,
\[I_n = \frac{2n}{3} \int_0^1 x^{n-1} (1-x)^{\frac{3}{2}} \,\mathrm{d}x.\]
Rewrite \((1-x)^{\frac{3}{2}} = (1-x)\sqrt{1-x}\):
\[I_n = \frac{2n}{3} \int_0^1 x^{n-1}(1-x)\sqrt{1-x} \,\mathrm{d}x = \frac{2n}{3} \int_0^1 \left(x^{n-1}\sqrt{1-x} - x^n\sqrt{1-x}\right) \,\mathrm{d}x.\]
Express in terms of \(I_{n-1}\) and \(I_n\):
\[I_n = \frac{2n}{3} (I_{n-1} - I_n).\]
Multiply through by 3:
\[3 I_n = 2n I_{n-1} - 2n I_n \implies (2n + 3) I_n = 2n I_{n-1}.\]

(b)
\[I_0 = \int_0^1 (1-x)^{\frac{1}{2}} \,\mathrm{d}x = \left[ -\frac{2}{3}(1-x)^{\frac{3}{2}} \right]_0^1 = 0 - \left(-\frac{2}{3}\right) = \frac{2}{3}.\]

(c)
Using the reduction formula \(I_n = \frac{2n}{2n+3} I_{n-1}\):
- For \(n = 1\):
\[I_1 = \frac{2(1)}{2(1)+3} I_0 = \frac{2}{5} \times \frac{2}{3} = \frac{4}{15}.\]
- For \(n = 2\):
\[I_2 = \frac{2(2)}{2(2)+3} I_1 = \frac{4}{7} \times \frac{4}{15} = \frac{16}{105}.\]
- For \(n = 3\):
\[I_3 = \frac{2(3)}{2(3)+3} I_2 = \frac{6}{9} \times \frac{16}{105} = \frac{2}{3} \times \frac{16}{105} = \frac{32}{315}.\]

(a)
M1: Sets up integration by parts with \(u = x^n\) and \(\frac{\mathrm{d}v}{\mathrm{d}x} = (1-x)^{\frac{1}{2}}\).
A1: Correct integrated part and integral: \(\left[-\frac{2}{3}x^n (1-x)^{\frac{3}{2}}\right]_0^1 + \frac{2n}{3}\int_0^1 x^{n-1}(1-x)^{\frac{3}{2}}\,\mathrm{d}x\).
B1: Correctly shows/justifies that the boundary term vanishes at both limits.
M1: Writes \((1-x)^{\frac{3}{2}} = (1-x)(1-x)^{\frac{1}{2}}\) to split into integrals corresponding to \(I_{n-1}\) and \(I_n\).
A1: Rearranges convincingly to obtain \((2n+3)I_n = 2n I_{n-1}\) with no missing steps (AG).

(b)
B1: Correct value \(I_0 = \frac{2}{3}\).

(c)
M1: Applies reduction formula to find \(I_1\).
A1: \(I_1 = \frac{4}{15}\).
M1: Applies reduction formula successively for \(n = 2\) and \(n = 3\).
A1: Obtains exact simplified fraction \(I_3 = \frac{32}{315}\).

(a)
By de Moivre's theorem,
\[\cos 5\theta + \mathrm{i}\sin 5\theta = (\cos\theta + \mathrm{i}\sin\theta)^5\]
Expanding using the binomial theorem:
\[(\cos\theta + \mathrm{i}\sin\theta)^5 = \cos^5\theta + 5\mathrm{i}\cos^4\theta\sin\theta - 10\cos^3\theta\sin^2\theta - 10\mathrm{i}\cos^2\theta\sin^3\theta + 5\cos\theta\sin^4\theta + \mathrm{i}\sin^5\theta\]
Equating real parts:
\[\cos 5\theta = \cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta\]
Substitute \(\sin^2\theta = 1 - \cos^2\theta\):
\begin{align*}
\cos 5\theta &= \cos^5\theta - 10\cos^3\theta(1 - \cos^2\theta) + 5\cos\theta(1 - \cos^2\theta)^2 \\
&= \cos^5\theta - 10\cos^3\theta + 10\cos^5\theta + 5\cos\theta(1 - 2\cos^2\theta + \cos^4\theta) \\
&= 11\cos^5\theta - 10\cos^3\theta + 5\cos\theta - 10\cos^3\theta + 5\cos^5\theta \\
&= 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta.
\end{align*}

(b)
Consider the equation \(\cos 5\theta = 0\).
From part (a), this gives:
\[16\cos^5\theta - 20\cos^3\theta + 5\cos\theta = 0 \implies \cos\theta\left(16\cos^4\theta - 20\cos^2\theta + 5\right) = 0.\]
The solutions to \(\cos 5\theta = 0\) for \(0 < \theta < \pi\) are:
\[5\theta = \frac{(2k+1)\pi}{2} \implies \theta = \frac{\pi}{10}, \frac{3\pi}{10}, \frac{5\pi}{10}, \frac{7\pi}{10}, \frac{9\pi}{10}.\]
For \(\theta = \frac{5\pi}{10} = \frac{\pi}{2}\), \(\cos\left(\frac{\pi}{2}\right) = 0\), which corresponds to the factor \(\cos\theta = 0\).

Thus, the four non-zero roots of \(16x^4 - 20x^2 + 5 = 0\) are given by \(x = \cos\theta\) for \(\theta = \frac{\pi}{10}, \frac{3\pi}{10}, \frac{7\pi}{10}, \frac{9\pi}{10}\).
Since \(\cos\left(\frac{9\pi}{10}\right) = -\cos\left(\frac{\pi}{10}\right)\) and \(\cos\left(\frac{7\pi}{10}\right) = -\cos\left(\frac{3\pi}{10}\right)\), the four roots are:
\[x = \pm \cos\left(\frac{\pi}{10}\right) \quad \text{and} \quad x = \pm \cos\left(\frac{3\pi}{10}\right).\]

(c)(i)
The product of the four roots of the quartic \(16x^4 - 20x^2 + 5 = 0\) is:
\[\left(\cos\left(\frac{\pi}{10}\right)\right)\left(-\cos\left(\frac{\pi}{10}\right)\right)\left(\cos\left(\frac{3\pi}{10}\right)\right)\left(-\cos\left(\frac{3\pi}{10}\right)\right) = \frac{5}{16}\]
\[\implies \cos^2\left(\frac{\pi}{10}\right)\cos^2\left(\frac{3\pi}{10}\right) = \frac{5}{16}.\]

(c)(ii)
Let \(u = x^2\). The equation \(16x^4 - 20x^2 + 5 = 0\) becomes the quadratic:
\[16u^2 - 20u + 5 = 0.\]
The roots of this quadratic equation are \(u_1 = \cos^2\left(\frac{\pi}{10}\right)\) and \(u_2 = \cos^2\left(\frac{3\pi}{10}\right)\).
Using the relations between roots and coefficients:
\[u_1 + u_2 = \frac{20}{16} = \frac{5}{4}, \qquad u_1 u_2 = \frac{5}{16}.\]
Then:
\begin{align*}
\sec^2\left(\frac{\pi}{10}\right) + \sec^2\left(\frac{3\pi}{10}\right) &= \frac{1}{\cos^2\left(\frac{\pi}{10}\right)} + \frac{1}{\cos^2\left(\frac{3\pi}{10}\right)} \[6pt]
&= \frac{1}{u_1} + \frac{1}{u_2} \[6pt]
&= \frac{u_1 + u_2}{u_1 u_2} \[6pt]
&= \frac{5/4}{5/16} = \frac{5}{4} \times \frac{16}{5} = 4.
\end{align*}

(a) [4 marks]
- M1: Uses de Moivre's theorem to state \(\cos 5\theta + \mathrm{i}\sin 5\theta = (\cos\theta + \mathrm{i}\sin\theta)^5\) and attempts full binomial expansion.
- A1: Correctly identifies and equates the real part: \(\cos 5\theta = \cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta\).
- M1: Applies \(\sin^2\theta = 1 - \cos^2\theta\) throughout the real part expression.
- A1: Fully simplifies and shows the given result \(\cos 5\theta = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta\) with complete, clear working (AG).

(b) [3 marks]
- M1: Sets \(\cos 5\theta = 0\), factorises out \(\cos\theta\), and finds the general solutions for \(\theta\).
- A1: Identifies the relevant values \(\theta = \frac{\pi}{10}, \frac{3\pi}{10}, \frac{7\pi}{10}, \frac{9\pi}{10}\) and notes that \(\theta = \frac{\pi}{2}\) corresponds to \(\cos\theta = 0\).
- A1: Correctly uses the symmetry \(\cos(\pi - \theta) = -\cos\theta\) to deduce the roots are \(\pm\cos\left(\frac{\pi}{10}\right)\) and \(\pm\cos\left(\frac{3\pi}{10}\right)\).

(c)(i) [1 mark]
- B1: Deduces \(\cos^2\left(\frac{\pi}{10}\right)\cos^2\left(\frac{3\pi}{10}\right) = \frac{5}{16}\) directly from product of roots of the quartic or substitution \(u = x^2\).

(c)(ii) [3 marks]
- M1: Considers substitution \(u = x^2\) yielding \(16u^2 - 20u + 5 = 0\) and identifies sum of roots \(u_1 + u_2 = \frac{5}{4}\).
- M1: Expresses \(\sec^2\left(\frac{\pi}{10}\right) + \sec^2\left(\frac{3\pi}{10}\right)\) as \(\frac{u_1 + u_2}{u_1 u_2}\).
- A1: Obtains exact integer value \(4\).

(a)
Let \(x = \mathrm{e}^t\), so \(t = \ln x\) and \(\frac{\mathrm{d}t}{\mathrm{d}x} = \frac{1}{x} = \mathrm{e}^{-t}\).
By the chain rule:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y}{\mathrm{d}t} \frac{\mathrm{d}t}{\mathrm{d}x} = \frac{1}{x}\frac{\mathrm{d}y}{\mathrm{d}t} \implies x\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y}{\mathrm{d}t} \]
Differentiating again with respect to \(x\):
\[ \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{x}\frac{\mathrm{d}y}{\mathrm{d}t}\right) = -\frac{1}{x^2}\frac{\mathrm{d}y}{\mathrm{d}t} + \frac{1}{x}\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\mathrm{d}y}{\mathrm{d}t}\right) = -\frac{1}{x^2}\frac{\mathrm{d}y}{\mathrm{d}t} + \frac{1}{x}\left(\frac{\mathrm{d}^2 y}{\mathrm{d}t^2}\frac{\mathrm{d}t}{\mathrm{d}x}\right) = \frac{1}{x^2}\left(\frac{\mathrm{d}^2 y}{\mathrm{d}t^2} - \frac{\mathrm{d}y}{\mathrm{d}t}\right) \]
Thus,
\[ x^2\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = \frac{\mathrm{d}^2 y}{\mathrm{d}t^2} - \frac{\mathrm{d}y}{\mathrm{d}t} \]
Substituting these into the original equation:
\[ \left(\frac{\mathrm{d}^2 y}{\mathrm{d}t^2} - \frac{\mathrm{d}y}{\mathrm{d}t}\right) + 5\left(\frac{\mathrm{d}y}{\mathrm{d}t}\right) + 4y = 12t \]
\[ \frac{\mathrm{d}^2 y}{\mathrm{d}t^2} + 4\frac{\mathrm{d}y}{\mathrm{d}t} + 4y = 12t \]

(b)
For the complementary function, the auxiliary equation is:
\[ m^2 + 4m + 4 = 0 \implies (m+2)^2 = 0 \implies m = -2 \text{ (repeated)} \]
So the complementary function is \(y_{\text{CF}} = (A + Bt)\mathrm{e}^{-2t}\).

For the particular integral, try \(y_{\text{PI}} = Ct + D\):
\[ \frac{\mathrm{d}y}{\mathrm{d}t} = C, \quad \frac{\mathrm{d}^2 y}{\mathrm{d}t^2} = 0 \]
Substituting into the differential equation:
\[ 0 + 4C + 4(Ct + D) = 12t \implies 4Ct + (4C + 4D) = 12t \]
Equating coefficients:
\[ 4C = 12 \implies C = 3 \]
\[ 4C + 4D = 0 \implies 4(3) + 4D = 0 \implies D = -3 \]
So \(y_{\text{PI}} = 3t - 3\).

The general solution in terms of \(t\) is:
\[ y = (A + Bt)\mathrm{e}^{-2t} + 3t - 3 \]
Since \(t = \ln x\) and \(\mathrm{e}^{-2t} = x^{-2}\), the general solution in terms of \(x\) is:
\[ y = (A + B\ln x)x^{-2} + 3\ln x - 3 = \frac{A + B\ln x}{x^2} + 3\ln x - 3 \]

Using the initial conditions:
When \(x = 1\), \(y = -1\):
\[ -1 = \frac{A + 0}{1} + 0 - 3 \implies A - 3 = -1 \implies A = 2 \]
Differentiating \(y\) with respect to \(x\):
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = -2x^{-3}(A + B\ln x) + x^{-2}\left(\frac{B}{x}\right) + \frac{3}{x} = \frac{-2A - 2B\ln x + B}{x^3} + \frac{3}{x} \]
When \(x = 1\), \(\frac{\mathrm{d}y}{\mathrm{d}x} = 5\):
\[ 5 = \frac{-2(2) + B}{1} + 3 \implies 5 = -4 + B + 3 \implies B = 6 \]
Thus, the particular solution is:
\[ y = \frac{2 + 6\ln x}{x^2} + 3\ln x - 3 \]

(c)
From the particular solution:
\[ y - 3\ln x = \frac{2 + 6\ln x}{x^2} - 3 \]
As \(x \to \infty\), \(\frac{2}{x^2} \to 0\) and \(\frac{\ln x}{x^2} \to 0\), so:
\[ \lim_{x \to \infty} (y - 3\ln x) = 0 - 3 = -3 \]
Now consider the difference \(y - (3\ln x - 3) = \frac{2 + 6\ln x}{x^2}\).
For all \(x > \mathrm{e}^{-1/3}\), \(2 + 6\ln x > 0\), which means \(\frac{2 + 6\ln x}{x^2} > 0\).
Therefore, \(y > 3\ln x - 3\) for large \(x\), so the curve approaches \(y = 3\ln x - 3\) from above.

(a)
- M1: Uses chain rule to find \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{x}\frac{\mathrm{d}y}{\mathrm{d}t}\) or \(\frac{\mathrm{d}t}{\mathrm{d}x} = \frac{1}{x}\).
- M1: Differentiates again with product/chain rule to express \(\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}\) in terms of derivatives with respect to \(t\).
- A1: Correct expressions: \(x\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y}{\mathrm{d}t}\) and \(x^2\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = \frac{\mathrm{d}^2 y}{\mathrm{d}t^2} - \frac{\mathrm{d}y}{\mathrm{d}t}\).
- A1: Substitutes into the given equation and simplifies to obtain the printed result correctly (AG).

(b)
- M1: States and solves auxiliary equation \(m^2 + 4m + 4 = 0\).
- A1: Correct complementary function \((A + Bt)\mathrm{e}^{-2t}\).
- M1: Uses trial form \(y = Ct + D\) and substitutes into the transformed equation.
- A1: Correct particular integral \(3t - 3\).
- M1: Replaces \(t\) with \(\ln x\) and applies initial condition \(y(1) = -1\) to find \(A\).
- M1: Differentiates \(y(x)\) and applies condition \(y'(1) = 5\) to find \(B\).
- A1: Correct particular solution \(y = \frac{2 + 6\ln x}{x^2} + 3\ln x - 3\) or \(y = (2 + 6\ln x)x^{-2} + 3\ln x - 3\).

(c)
- B1: Uses standard limits \(\frac{\ln x}{x^2} \to 0\) and \(\frac{1}{x^2} \to 0\) as \(x \to \infty\) to show limit is \(-3\).
- B1: Shows that \(y - (3\ln x - 3) = \frac{2 + 6\ln x}{x^2} > 0\) for large \(x\) and concludes that the curve approaches from above.