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

(a) Starting from the right-hand side:
\[ \frac{1}{6}\left( \frac{1}{(3r-2)(3r+1)} - \frac{1}{(3r+1)(3r+4)} \right) = \frac{1}{6}\left( \frac{(3r+4) - (3r-2)}{(3r-2)(3r+1)(3r+4)} \right) \]
\[ = \frac{1}{6}\left( \frac{6}{(3r-2)(3r+1)(3r+4)} \right) = \frac{1}{(3r-2)(3r+1)(3r+4)} \]

(b) Using the method of differences:
\[ \sum_{r=1}^n \frac{1}{(3r-2)(3r+1)(3r+4)} = \frac{1}{6}\sum_{r=1}^n \left( \frac{1}{(3r-2)(3r+1)} - \frac{1}{(3r+1)(3r+4)} \right) \]
Writing out the terms:
\[ r=1: \quad \frac{1}{1 \cdot 4} - \frac{1}{4 \cdot 7} \]
\[ r=2: \quad \frac{1}{4 \cdot 7} - \frac{1}{7 \cdot 10} \]
\[ \vdots \]
\[ r=n: \quad \frac{1}{(3n-2)(3n+1)} - \frac{1}{(3n+1)(3n+4)} \]
Summing and cancelling intermediate terms gives:
\[ S_n = \frac{1}{6}\left( \frac{1}{4} - \frac{1}{(3n+1)(3n+4)} \right) = \frac{1}{24} - \frac{1}{6(3n+1)(3n+4)} \]
Alternatively expressed as a single fraction:
\[ S_n = \frac{(3n+1)(3n+4) - 4}{24(3n+1)(3n+4)} = \frac{9n^2 + 15n}{24(3n+1)(3n+4)} = \frac{n(3n+5)}{8(3n+1)(3n+4)} \]

(c) As \(n \to \infty\), \(\frac{1}{6(3n+1)(3n+4)} \to 0\).
Therefore, the sum to infinity is:
\[ S_\infty = \frac{1}{24} \]

(d) We have:
\[ \sum_{r=n+1}^{2n} \frac{1}{(3r-2)(3r+1)(3r+4)} = S_{2n} - S_n \]
\[ = \left( \frac{1}{24} - \frac{1}{6(6n+1)(6n+4)} \right) - \left( \frac{1}{24} - \frac{1}{6(3n+1)(3n+4)} \right) \]
\[ = \frac{1}{6}\left( \frac{1}{(3n+1)(3n+4)} - \frac{1}{(6n+1)(6n+4)} \right) \]
\[ = \frac{1}{6}\left( \frac{(6n+1)(6n+4) - (3n+1)(3n+4)}{(3n+1)(3n+4)(6n+1)(6n+4)} \right) \]
Expanding the numerator:
\[ (36n^2 + 30n + 4) - (9n^2 + 15n + 4) = 27n^2 + 15n = 3n(9n+5) \]
Thus:
\[ \sum_{r=n+1}^{2n} \frac{1}{(3r-2)(3r+1)(3r+4)} = \frac{3n(9n+5)}{6(3n+1)(3n+4)(6n+1)(6n+4)} = \frac{n(9n+5)}{2(3n+1)(3n+4)(6n+1)(6n+4)} \]

(a)
M1: Puts the terms on the RHS over a common denominator or subtracts fractions with correct common denominator.
A1 (AG): Simplifies numerator to 6 and cancels factor of \(\frac{1}{6}\) to achieve the given result with all intermediate steps shown clearly.

(b)
M1: Expresses the summation in difference form and shows list of terms to illustrate cancellation (at least first term, second term, and final \(n\)-th term).
A1: Obtains \(\frac{1}{6}\left( \frac{1}{4} - \frac{1}{(3n+1)(3n+4)} \right)\) or equivalent unsimplified expression.
A1: Correct simplified expression in terms of \(n\), e.g. \(\frac{1}{24} - \frac{1}{6(3n+1)(3n+4)}\) or \(\frac{n(3n+5)}{8(3n+1)(3n+4)}\).

(c)
B1: State \(\frac{1}{24}\) (allow FT from candidate's non-zero constant term in part (b)).

(d)
M1: Uses \(S_{2n} - S_n\) or applies the difference method between limits \(n+1\) and \(2n\).
A1: Correct single algebraic fraction: \(\frac{n(9n+5)}{2(3n+1)(3n+4)(6n+1)(6n+4)}\) or expanded equivalent.

For \( n = 1 \):
\[ \begin{pmatrix} 1+6(1) & -9(1) \\ 4(1) & 1-6(1) \end{pmatrix} = \begin{pmatrix} 7 & -9 \\ 4 & -5 \end{pmatrix} = \mathbf{M} \]
So the result is true for \( n = 1 \).

Assume that the result is true for \( n = k \), where \( k \) is a positive integer, so
\[ \mathbf{M}^k = \begin{pmatrix} 1+6k & -9k \\ 4k & 1-6k \end{pmatrix} \]

Then for \( n = k + 1 \):
\[ \mathbf{M}^{k+1} = \mathbf{M}^k \mathbf{M} = \begin{pmatrix} 1+6k & -9k \\ 4k & 1-6k \end{pmatrix} \begin{pmatrix} 7 & -9 \\ 4 & -5 \end{pmatrix} \]

Multiplying the matrices:
- Top-left entry: \( (1+6k)(7) + (-9k)(4) = 7 + 42k - 36k = 7 + 6k = 1 + 6(k+1) \)
- Top-right entry: \( (1+6k)(-9) + (-9k)(-5) = -9 - 54k + 45k = -9 - 9k = -9(k+1) \)
- Bottom-left entry: \( (4k)(7) + (1-6k)(4) = 28k + 4 - 24k = 4 + 4k = 4(k+1) \)
- Bottom-right entry: \( (4k)(-9) + (1-6k)(-5) = -36k - 5 + 30k = -5 - 6k = 1 - 6(k+1) \)

Thus,
\[ \mathbf{M}^{k+1} = \begin{pmatrix} 1+6(k+1) & -9(k+1) \\ 4(k+1) & 1-6(k+1) \end{pmatrix} \]

Therefore, if the result is true for \( n = k \), it is also true for \( n = k + 1 \).
Since the result is true for \( n = 1 \), by mathematical induction the result is true for all positive integers \( n \).

B1: Verifies the base case \( n = 1 \) by evaluating the given formula and showing it equals \( \mathbf{M} \).

B1: States inductive hypothesis clearly (e.g. "Assume true for \( n = k \)").

M1: Considers \( \mathbf{M}^{k+1} = \mathbf{M}^k \mathbf{M} \) (or \( \mathbf{M}\mathbf{M}^k \)) using the induction hypothesis.

A1: Multiplies matrices correctly showing unsimplified or partially simplified entries for all 4 positions.

A1: Correctly rearranges each entry into the specific form for \( k+1 \), i.e., \( \begin{pmatrix} 1+6(k+1) & -9(k+1) \\ 4(k+1) & 1-6(k+1) \end{pmatrix} \).

A1: Complete and convincing concluding statement asserting truth for \( n = 1 \), truth for \( n = k \implies n = k + 1 \), and conclusion for all positive integers \( n \) (dependent on all previous marks earned).

(a) An invariant line through the origin has the form \(y = mx\) (or \(x = 0\)).
Consider a point \(\begin{pmatrix} x \\ mx \end{pmatrix}\) on the line:
\[\begin{pmatrix} X \\ Y \end{pmatrix} = \begin{pmatrix} 5 & -2 \\ 4 & -1 \end{pmatrix}\begin{pmatrix} x \\ mx \end{pmatrix} = \begin{pmatrix} (5 - 2m)x \\ (4 - m)x \end{pmatrix}.\]
Since the line is invariant, the point \((X, Y)\) lies on \(Y = mX\), so
\[(4 - m)x = m(5 - 2m)x.\]
Assuming \(x \neq 0\):
\[4 - m = 5m - 2m^2 \implies 2m^2 - 6m + 4 = 0 \implies m^2 - 3m + 2 = 0.\]
Solving the quadratic equation gives
\[(m - 1)(m - 2) = 0 \implies m = 1 \text{ or } m = 2.\]
Testing \(x = 0\):
\[\begin{pmatrix} 5 & -2 \\ 4 & -1 \end{pmatrix}\begin{pmatrix} 0 \\ y \end{pmatrix} = \begin{pmatrix} -2y \\ -y \end{pmatrix},\]
which does not lie on \(x = 0\) unless \(y = 0\). Thus \(x = 0\) is not invariant.
Therefore, the invariant lines through the origin are \(y = x\) and \(y = 2x\).

(b) A point \(\begin{pmatrix} x \\ y \end{pmatrix}\) is an invariant point if \(\mathbf{M}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}\):
\[\begin{pmatrix} 5 & -2 \\ 4 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}\]
\[\implies 5x - 2y = x \implies 4x = 2y \implies y = 2x,\]
and
\[4x - y = y \implies 4x = 2y \implies y = 2x.\]
Both equations reduce to \(y = 2x\).
Thus, every point on the line \(y = 2x\) maps directly to itself, so \(y = 2x\) is a line of invariant points.

(c)(i) The combined transformation "\(\mathbf{M}\) followed by \(\mathbf{N}\)" is represented by the product \(\mathbf{NM}\):
\[\mathbf{NM} = \begin{pmatrix} 2 & 1 \\ -1 & 3 \end{pmatrix}\begin{pmatrix} 5 & -2 \\ 4 & -1 \end{pmatrix} = \begin{pmatrix} 2(5) + 1(4) & 2(-2) + 1(-1) \\ -1(5) + 3(4) & -1(-2) + 3(-1) \end{pmatrix} = \begin{pmatrix} 14 & -5 \\ 7 & -1 \end{pmatrix}.\]

(c)(ii) The determinant of the transformation matrix gives the area scale factor:
\[\det(\mathbf{NM}) = (14)(-1) - (-5)(7) = -14 + 35 = 21.\]
(Alternatively, \(\det(\mathbf{NM}) = \det(\mathbf{N})\det(\mathbf{M}) = (6 - (-1))( -5 - (-8) ) = 7 \times 3 = 21\).)
\[\text{Area of image} = |\det(\mathbf{NM})| \times \text{Area of } T = 21 \times 4 = 84.\]

(a)
M1: Sets up the transformation of \(\begin{pmatrix} x \\ mx \end{pmatrix}\) using matrix multiplication and substitutes into \(Y = mX\).
A1: Obtains correct quadratic equation in \(m\), e.g. \(2m^2 - 6m + 4 = 0\) or \(m^2 - 3m + 2 = 0\).
M1: Solves their 3-term quadratic equation for \(m\).
A1: States the correct equations \(y = x\) and \(y = 2x\) (or \(y - x = 0\), \(y - 2x = 0\)).

(b)
M1: Sets \(\mathbf{M}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}\) (or tests \(\begin{pmatrix} x \\ 2x \end{pmatrix}\) in \(\mathbf{M}\)).
A1: Shows clearly that \(y = 2x\) gives \(\mathbf{M}\begin{pmatrix} x \\ 2x \end{pmatrix} = \begin{pmatrix} x \\ 2x \end{pmatrix}\) for all \(x\) and concludes it is a line of invariant points.

(c)(i)
M1: Multiplies matrices in correct order \(\mathbf{NM}\).
A1: Obtains \(\begin{pmatrix} 14 & -5 \\ 7 & -1 \end{pmatrix}\).

(c)(ii)
B1: Correctly evaluates area scale factor \(|\det(\mathbf{NM})| = 21\) and multiplies by \(4\) to obtain \(84\).

From the given cubic equation \(x^3 - 3x^2 + 2x - 5 = 0\):
\[\Sigma \alpha = \alpha + \beta + \gamma = 3\]
\[\Sigma \alpha\beta = \alpha\beta + \beta\gamma + \gamma\alpha = 2\]
\[\alpha\beta\gamma = 5\]

(a)(i)
\[\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\]
\[\alpha^2 + \beta^2 + \gamma^2 = 3^2 - 2(2) = 9 - 4 = 5\]

(a)(ii)
Since each root satisfies the original cubic equation:
\[x^3 = 3x^2 - 2x + 5\]
Summing over all three roots:
\[\Sigma \alpha^3 = 3\Sigma \alpha^2 - 2\Sigma \alpha + 5(3)\]
\[\Sigma \alpha^3 = 3(5) - 2(3) + 15 = 15 - 6 + 15 = 24\]

(b)
Rearrange the equation to isolate terms involving odd powers of \(x\):
\[x^3 + 2x = 3x^2 + 5\]
\[x(x^2 + 2) = 3x^2 + 5\]
Squaring both sides:
\[x^2(x^2 + 2)^2 = (3x^2 + 5)^2\]
Substitute \(y = x^2\):
\[y(y + 2)^2 = (3y + 5)^2\]
\[y(y^2 + 4y + 4) = 9y^2 + 30y + 25\]
\[y^3 + 4y^2 + 4y = 9y^2 + 30y + 25\]
\[y^3 - 5y^2 - 26y - 25 = 0\]

(c)
Using the cubic equation found in part (b), whose roots are \(y_1 = \alpha^2\), \(y_2 = \beta^2\), \(y_3 = \gamma^2\):
\[\Sigma y = \alpha^2 + \beta^2 + \gamma^2 = 5\]
\[\Sigma y_1 y_2 = \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 = -26\]
Then:
\[\alpha^4 + \beta^4 + \gamma^4 = y_1^2 + y_2^2 + y_3^2 = (\Sigma y)^2 - 2\Sigma y_1 y_2\]
\[\alpha^4 + \beta^4 + \gamma^4 = 5^2 - 2(-26) = 25 + 52 = 77\]

(Alternatively, using the recurrence relation from the original equation: \(\Sigma \alpha^4 = 3\Sigma \alpha^3 - 2\Sigma \alpha^2 + 5\Sigma \alpha = 3(24) - 2(5) + 5(3) = 72 - 10 + 15 = 77\).)

(a)(i)
M1: Uses the identity \(\Sigma \alpha^2 = (\Sigma \alpha)^2 - 2\Sigma \alpha\beta\) with \(\Sigma \alpha = 3\) and \(\Sigma \alpha\beta = 2\).
A1: Obtains 5.

(a)(ii)
M1: Uses a valid method to find \(\Sigma \alpha^3\), e.g., the recurrence relation \(\Sigma \alpha^3 = 3\Sigma \alpha^2 - 2\Sigma \alpha + 15\) or the algebraic identity.
A1: Obtains 24.

(b)
M1: Rearranges the original equation to separate odd and even powers of \(x\) and squares both sides.
M1: Expands both sides correctly in terms of \(y\).
A1: Obtains \(y^3 - 5y^2 - 26y - 25 = 0\) (allow all terms on one side equal to 0).

(c)
M1: Uses \(\Sigma (\alpha^2)^2 = (\Sigma \alpha^2)^2 - 2\Sigma \alpha^2\beta^2\) with values from part (b), OR sets up \(\Sigma \alpha^4 = 3\Sigma \alpha^3 - 2\Sigma \alpha^2 + 5\Sigma \alpha\).
M1: Substitutes numerical values correctly into the chosen relationship.
A1: Obtains 77 (cwo).

(a) A vector perpendicular to both lines is given by the cross product of their direction vectors:
\[ \mathbf{d}_1 \times \mathbf{d}_2 = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} \times \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} = \begin{pmatrix} (-1)(-1) - (2)(1) \\ (2)(2) - (1)(-1) \\ (1)(1) - (-1)(2) \end{pmatrix} = \begin{pmatrix} -1 \\ 5 \\ 3 \end{pmatrix}. \]

The magnitude of this vector is:
\[ |\mathbf{d}_1 \times \mathbf{d}_2| = \sqrt{(-1)^2 + 5^2 + 3^2} = \sqrt{1 + 25 + 9} = \sqrt{35}. \]

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

The shortest distance \(d\) between the skew lines is:
\[ d = \frac{|\mathbf{AB} \cdot (\mathbf{d}_1 \times \mathbf{d}_2)|}{|\mathbf{d}_1 \times \mathbf{d}_2|} = \frac{|3(-1) + (-3)(5) + 4(3)|}{\sqrt{35}} = \frac{|-3 - 15 + 12|}{\sqrt{35}} = \frac{|-6|}{\sqrt{35}} = \frac{6}{\sqrt{35}} = \frac{6\sqrt{35}}{35} \approx 1.01. \]

(b) The plane \(\Pi_1\) is parallel to both \(\mathbf{d}_1\) and \(\mathbf{d}_2\), so a normal vector to \(\Pi_1\) is:
\[ \mathbf{n} = \begin{pmatrix} -1 \\ 5 \\ 3 \end{pmatrix}. \]

Since \(\Pi_1\) contains \(l_1\), it passes through the point \((1, 2, -1)\). Thus:
\[ p = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 5 \\ 3 \end{pmatrix} = 1(-1) + 2(5) + (-1)(3) = -1 + 10 - 3 = 6. \]

Hence, the equation of \(\Pi_1\) is:
\[ \mathbf{r} \cdot \begin{pmatrix} -1 \\ 5 \\ 3 \end{pmatrix} = 6 \quad \left(\text{or } \mathbf{r} \cdot \begin{pmatrix} 1 \\ -5 \\ -3 \end{pmatrix} = -6\right). \]

(c) The normal vector to \(\Pi_1\) is \(\mathbf{n}_1 = \begin{pmatrix} -1 \\ 5 \\ 3 \end{pmatrix}\) and the normal vector to \(\Pi_2\) is \(\mathbf{n}_2 = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}\).

Calculate the magnitudes:
\[ |\mathbf{n}_1| = \sqrt{35}, \]
\[ |\mathbf{n}_2| = \sqrt{2^2 + (-1)^2 + 3^2} = \sqrt{4 + 1 + 9} = \sqrt{14}. \]

Calculate the scalar product:
\[ \mathbf{n}_1 \cdot \mathbf{n}_2 = (-1)(2) + (5)(-1) + (3)(3) = -2 - 5 + 9 = 2. \]

The angle \(\theta\) between the two planes satisfies:
\[ \cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|} = \frac{2}{\sqrt{35}\sqrt{14}} = \frac{2}{\sqrt{490}} = \frac{2}{7\sqrt{10}}. \]

\[ \theta = \arccos\left(\frac{2}{7\sqrt{10}}\right) \approx \arccos(0.09035) \approx 84.8^\circ. \]

(a)
M1: Attempt vector product of direction vectors \(\begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} \times \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}\)
A1: Correct normal vector \(\begin{pmatrix} -1 \\ 5 \\ 3 \end{pmatrix}\) (or any non-zero scalar multiple)
M1: Correct method for shortest distance using \(\frac{|(\mathbf{b} - \mathbf{a}) \cdot \mathbf{n}|}{|\mathbf{n}|}\)
A1: Obtain \(\frac{6}{\sqrt{35}}\) or \(\frac{6\sqrt{35}}{35}\) or \(1.01\) (CWO)

(b)
B1: State or imply normal vector \(\mathbf{n} = \begin{pmatrix} -1 \\ 5 \\ 3 \end{pmatrix}\) or multiple
M1: Substitute point \((1, 2, -1)\) into \(\mathbf{r} \cdot \mathbf{n} = p\)
A1: Correct equation in required form \(\mathbf{r} \cdot \begin{pmatrix} -1 \\ 5 \\ 3 \end{pmatrix} = 6\) (or any integer multiple)

(c)
B1: Identify normal vector of \(\Pi_2\) as \(\mathbf{n}_2 = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}\)
M1: Correct use of scalar product formula \(\cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|}\)
A1: Correct numerical expression for \(\cos \theta\), e.g. \(\frac{2}{\sqrt{490}}\) or \(\frac{2}{7\sqrt{10}}\) (or \(0.09035...\))
A1: \(84.8^\circ\) (accept 1.48 radians; awrt 84.8)

(a)
Dividing the numerator by the denominator:
\[ \frac{x^2 + x + 2}{x + 2} = x - 1 + \frac{4}{x + 2} \]
As \( x \to -2 \), \( y \to \pm\infty \), so the vertical asymptote is \( x = -2 \).
As \( x \to \pm\infty \), \( \frac{4}{x+2} \to 0 \), so the oblique asymptote is \( y = x - 1 \).

(b)
Rearranging the equation of \( C \):
\[ y(x + 2) = x^2 + x + 2 \implies x^2 + (1 - y)x + (2 - 2y) = 0 \]
Since \( x \) is real, the discriminant of this quadratic equation must be non-negative:
\[ \Delta = (1 - y)^2 - 4(1)(2 - 2y) \ge 0 \]
\[ 1 - 2y + y^2 - 8 + 8y \ge 0 \]
\[ y^2 + 6y - 7 \ge 0 \]
\[ (y + 7)(y - 1) \ge 0 \]
Thus, \( y \le -7 \) or \( y \ge 1 \).
Hence, \( y \) cannot take any value in the open interval \( -7 < y < 1 \).

The stationary points occur where the discriminant is zero:
For \( y = 1 \):
\[ x^2 + (1 - 1)x + (2 - 2) = 0 \implies x^2 = 0 \implies x = 0 \]
So one stationary point is \( (0, 1) \).

For \( y = -7 \):
\[ x^2 + (1 - (-7))x + (2 - 2(-7)) = 0 \implies x^2 + 8x + 16 = 0 \implies (x + 4)^2 = 0 \implies x = -4 \]
So the other stationary point is \( (-4, -7) \).

(c)
- Vertical asymptote at \( x = -2 \).
- Oblique asymptote at \( y = x - 1 \).
- \( y \)-intercept: when \( x = 0 \), \( y = 1 \).
- \( x \)-intercepts: \( x^2 + x + 2 = 0 \) has no real roots, so no intersections with the \( x \)-axis.
- Branch 1 (for \( x > -2 \)): passes through \( (0, 1) \) as a local minimum, approaching \( x = -2 \) as \( y \to +\infty \) and approaching \( y = x - 1 \) from above as \( x \to +\infty \).
- Branch 2 (for \( x < -2 \)): has a local maximum at \( (-4, -7) \), approaching \( x = -2 \) as \( y \to -\infty \) and approaching \( y = x - 1 \) from below as \( x \to -\infty \).

(d)
For \( y^2 = f(x) = \frac{x^2 + x + 2}{x + 2} \):
- The graph exists only where \( f(x) \ge 0 \). Since \( x^2 + x + 2 > 0 \) for all real \( x \), \( f(x) \ge 0 \iff x + 2 > 0 \iff x > -2 \). No part of the graph exists for \( x \le -2 \).
- The graph is symmetrical about the \( x \)-axis.
- Vertical asymptote is \( x = -2 \).
- Since \( f(x) \ge 1 \) for \( x > -2 \), there are no \( x \)-intercepts.
- The stationary point of \( f(x) \) at \( (0, 1) \) gives two stationary points on \( y^2 = f(x) \) at \( (0, 1) \) and \( (0, -1) \).
- As \( x \to +\infty \), \( y \approx \pm\sqrt{x - 1} \).

(a)
- M1: Uses algebraic division or inspection to write \( y = x - 1 + \frac{4}{x + 2} \)
- A1: States vertical asymptote \( x = -2 \)
- A1: States oblique asymptote \( y = x - 1 \)

(b)
- M1: Rearranges to form a quadratic equation in \( x \) and sets discriminant \( \Delta \ge 0 \) (or uses differentiation and sets \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \))
- A1: Obtains \( (y + 7)(y - 1) \ge 0 \) and correctly concludes \( -7 < y < 1 \) is not possible (AG)
- M1: Finds the corresponding \( x \)-coordinates when \( y = 1 \) and \( y = -7 \)
- A1: Correct stationary points \( (0, 1) \) and \( (-4, -7) \)

(c)
- B1: Both asymptotes drawn as dashed lines and correctly labelled (or with intercepts \( (0, -1) \), \( (1, 0) \) shown)
- B1: Right-hand branch correct: in upper-right region with local minimum at \( (0, 1) \) and correct asymptotic behaviour
- B1: Left-hand branch correct: in lower-left region with local maximum at \( (-4, -7) \) and correct asymptotic behaviour

(d)
- B1: Graph drawn exclusively for \( x > -2 \) with no curve to the left of \( x = -2 \)
- B1: Correct shape showing symmetry in the \( x \)-axis and asymptotic to \( x = -2 \)
- B1: Stationary points correctly located and labelled at \( (0, 1) \) and \( (0, -1) \)
- B1: States vertical asymptote \( x = -2 \)

(a) The curve is a cardioid symmetric about the initial line \(\theta = 0\).
\(r\) is greatest when \(\cos\theta = 1 \implies \theta = 0\), giving \(r = 2a\). The polar coordinates of this point are \((2a, 0)\).
At the pole, \(r = 0 \implies 1 + \cos\theta = 0 \implies \theta = \pm\pi\). The curve forms a sharp point (cusp) at the pole, where the tangent is along the line \(\theta = \pi\).

(b) Using the Cartesian conversion \(x = r\cos\theta\):
\[ x = a(1 + \cos\theta)\cos\theta = a(\cos\theta + \cos^2\theta) \]
Tangents perpendicular to the initial line occur when \(\frac{\mathrm{d}x}{\mathrm{d}\theta} = 0\) (and \(\frac{\mathrm{d}y}{\mathrm{d}\theta} \ne 0\) or determined by limit):
\[ \frac{\mathrm{d}x}{\mathrm{d}\theta} = a(-\sin\theta - 2\cos\theta\sin\theta) = -a\sin\theta(1 + 2\cos\theta) \]
Setting \(\frac{\mathrm{d}x}{\mathrm{d}\theta} = 0\):
- \(\sin\theta = 0 \implies \theta = 0\) or \(\theta = \pm\pi\).
- For \(\theta = 0\): \(r = 2a\), \(x = 2a\cos 0 = 2a\), \(y = 2a\sin 0 = 0\), giving the point \((2a, 0)\).
- (Note: at \(\theta = \pm\pi\), \(r = 0\), which is the cusp where the tangent is horizontal along \(y = 0\)).
- \(1 + 2\cos\theta = 0 \implies \cos\theta = -\frac{1}{2} \implies \theta = \pm\frac{2}{3}\pi\).
- When \(\cos\theta = -\frac{1}{2}\), \(r = a\left(1 - \frac{1}{2}\right) = \frac{1}{2}a\).
- \(x = r\cos\theta = \left(\frac{1}{2}a\right)\left(-\frac{1}{2}\right) = -\frac{1}{4}a\).
- \(y = r\sin\theta = \left(\frac{1}{2}a\right)\left(\pm\frac{\sqrt{3}}{2}\right) = \pm\frac{\sqrt{3}}{4}a\).

Thus, the Cartesian coordinates are \((2a, 0)\), \(\left(-\frac{1}{4}a, \frac{\sqrt{3}}{4}a\right)\), and \(\left(-\frac{1}{4}a, -\frac{\sqrt{3}}{4}a\right)\).

(c) The area \(A\) enclosed by \(C_1\) is given by:
\[ A = \frac{1}{2}\int_{-\pi}^{\pi} r^2\,\mathrm{d}\theta = \int_0^\pi a^2(1 + \cos\theta)^2\,\mathrm{d}\theta \]
\[ A = a^2 \int_0^\pi (1 + 2\cos\theta + \cos^2\theta)\,\mathrm{d}\theta \]
Using \(\cos^2\theta = \frac{1 + \cos 2\theta}{2}\):
\[ A = a^2 \int_0^\pi \left(\frac{3}{2} + 2\cos\theta + \frac{1}{2}\cos 2\theta\right)\,\mathrm{d}\theta \]
\[ A = a^2 \left[ \frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta \right]_0^\pi = a^2\left(\frac{3}{2}\pi\right) = \frac{3}{2}\pi a^2 \]

(d) First find the points of intersection between \(C_1\) and \(C_2\):
\[ a(1 + \cos\theta) = 3a\cos\theta \implies 2\cos\theta = 1 \implies \cos\theta = \frac{1}{2} \implies \theta = \pm\frac{1}{3}\pi \]
For \(0 \le \theta \le \frac{1}{3}\pi\), \(3a\cos\theta \ge a(1 + \cos\theta)\), so \(C_1\) is the inner boundary.
For \(\frac{1}{3}\pi \le \theta \le \frac{1}{2}\pi\), \(a(1 + \cos\theta) \ge 3a\cos\theta\), so \(C_2\) is the inner boundary.
By symmetry about the initial line, the common area is:
\[ A_{\text{common}} = 2 \left( \frac{1}{2}\int_0^{\pi/3} a^2(1+\cos\theta)^2\,\mathrm{d}\theta + \frac{1}{2}\int_{\pi/3}^{\pi/2} 9a^2\cos^2\theta\,\mathrm{d}\theta \right) \]
Evaluate the first integral:
\[ \int_0^{\pi/3} a^2\left(\frac{3}{2} + 2\cos\theta + \frac{1}{2}\cos 2\theta\right)\,\mathrm{d}\theta = a^2\left[ \frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta \right]_0^{\pi/3} \]
\[ = a^2\left( \frac{\pi}{2} + 2\left(\frac{\sqrt{3}}{2}\right) + \frac{1}{4}\left(\frac{\sqrt{3}}{2}\right) \right) = a^2\left( \frac{\pi}{2} + \frac{9\sqrt{3}}{8} \right) \]
Evaluate the second integral:
\[ 9a^2 \int_{\pi/3}^{\pi/2} \frac{1+\cos 2\theta}{2}\,\mathrm{d}\theta = \frac{9a^2}{2}\left[ \theta + \frac{1}{2}\sin 2\theta \right]_{\pi/3}^{\pi/2} \]
\[ = \frac{9a^2}{2}\left( \frac{\pi}{2} - \frac{\pi}{3} - \frac{\sqrt{3}}{4} \right) = \frac{9a^2}{2}\left( \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right) = a^2\left( \frac{3\pi}{4} - \frac{9\sqrt{3}}{8} \right) \]
Summing the two integrals:
\[ A_{\text{common}} = a^2\left( \frac{\pi}{2} + \frac{9\sqrt{3}}{8} + \frac{3\pi}{4} - \frac{9\sqrt{3}}{8} \right) = \frac{5}{4}\pi a^2 \]

(a)
B1: Correct heart-shaped cardioid sketch, symmetrical about the initial line, with positive orientation.
B1: States polar coordinates of greatest \(r\) as \((2a, 0)\).
B1: Identifies cusp at the pole / tangent \(\theta = \pm\pi\).

(b)
M1: Expresses \(x = r\cos\theta = a(\cos\theta + \cos^2\theta)\).
M1: Differentiates \(x\) with respect to \(\theta\) and sets \(\frac{\mathrm{d}x}{\mathrm{d}\theta} = 0\).
A1: Obtains \(-a\sin\theta(1+2\cos\theta) = 0\) leading to \(\cos\theta = -\frac{1}{2}\) and \(\sin\theta = 0\).
A1: Finds \((2a, 0)\).
M1: Evaluates \(r\), \(x\), and \(y\) for \(\cos\theta = -\frac{1}{2}\).
A1: Obtains \(\left(-\frac{1}{4}a, \frac{\sqrt{3}}{4}a\right)\) and \(\left(-\frac{1}{4}a, -\frac{\sqrt{3}}{4}a\right)\).

(c)
M1: Uses area formula \(\frac{1}{2}\int r^2\,\mathrm{d}\theta\) with appropriate symmetry or limits \(-\pi\) to \(\pi\).
M1: Expands \((1+\cos\theta)^2\) and applies the double angle formula \(\cos^2\theta = \frac{1}{2}(1+\cos 2\theta)\).
A1: Correct integration to get \(a^2\left[\frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta\right]\).
A1: Obtains \(\frac{3}{2}\pi a^2\) (CAO).

(d)
M1: Sets \(a(1+\cos\theta) = 3a\cos\theta\) to find intersection angle \(\theta = \pm\frac{\pi}{3}\).
M1: Sets up area expression as sum of two integrals: \(\int_0^{\pi/3} r_1^2\,\mathrm{d}\theta + \int_{\pi/3}^{\pi/2} r_2^2\,\mathrm{d}\theta\) (or total area of circle minus outer region).
A1: Correctly integrates both parts with correct limits: \(a^2\left(\frac{\pi}{2} + \frac{9\sqrt{3}}{8}\right)\) and \(a^2\left(\frac{3\pi}{4} - \frac{9\sqrt{3}}{8}\right)\).
A1: Obtains exact answer \(\frac{5}{4}\pi a^2\) (CWO).

(a)
Differentiating \(y = \mathrm{e}^{\operatorname{arsinh} x}\) with respect to \(x\):
\[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{\sqrt{1+x^2}}\,\mathrm{e}^{\operatorname{arsinh} x} = \frac{y}{\sqrt{1+x^2}}\]
Multiplying by \(\sqrt{1+x^2}\):
\[\sqrt{1+x^2}\,\frac{\mathrm{d}y}{\mathrm{d}x} = y\]
Differentiating both sides with respect to \(x\) using the product rule:
\[\frac{x}{\sqrt{1+x^2}}\,\frac{\mathrm{d}y}{\mathrm{d}x} + \sqrt{1+x^2}\,\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = \frac{\mathrm{d}y}{\mathrm{d}x}\]
Multiplying through by \(\sqrt{1+x^2}\):
\[x\frac{\mathrm{d}y}{\mathrm{d}x} + (1+x^2)\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = \sqrt{1+x^2}\,\frac{\mathrm{d}y}{\mathrm{d}x}\]
Since \(\sqrt{1+x^2}\,\frac{\mathrm{d}y}{\mathrm{d}x} = y\), we substitute to get:
\[(1+x^2)\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + x\frac{\mathrm{d}y}{\mathrm{d}x} - y = 0\]

(b)
At \(x = 0\):
\(y(0) = \mathrm{e}^0 = 1\)
\(y'(0) = \frac{y(0)}{\sqrt{1+0}} = 1\)
From the differential equation:
\(y''(0) + 0 - y(0) = 0 \implies y''(0) = y(0) = 1\)

Differentiating \((1+x^2)y'' + xy' - y = 0\) with respect to \(x\):
\[(1+x^2)y''' + 2xy'' + xy'' + y' - y' = 0 \implies (1+x^2)y''' + 3xy'' = 0\]
At \(x = 0\):
\[y'''(0) + 0 = 0 \implies y'''(0) = 0\]

Differentiating again with respect to \(x\):
\[(1+x^2)y^{(4)} + 2xy''' + 3xy''' + 3y'' = 0 \implies (1+x^2)y^{(4)} + 5xy''' + 3y'' = 0\]
At \(x = 0\):
\[y^{(4)}(0) + 0 + 3y''(0) = 0 \implies y^{(4)}(0) = -3(1) = -3\]

Using Maclaurin's formula:
\[y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \frac{y^{(4)}(0)}{4!}x^4 + \dots\]
\[y = 1 + x + \frac{1}{2!}x^2 + 0 + \frac{-3}{4!}x^4 = 1 + x + \frac{1}{2}x^2 - \frac{1}{8}x^4\]

(a)
M1: Obtains \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{y}{\sqrt{1+x^2}}\) or equivalent, and applies the product/quotient rule or differentiates after squaring.
A1: Completes differentiation and algebraic manipulation correctly to show \((1+x^2)y'' + xy' - y = 0\) (AG).

(b)
M1: Evaluates \(y(0)=1\), \(y'(0)=1\), and \(y''(0)=1\).
M1: Differentiates the given differential equation successively to obtain \(y'''(0)=0\) and \(y^{(4)}(0)=-3\).
A1: Obtains correct Maclaurin's series \(1 + x + \frac{1}{2}x^2 - \frac{1}{8}x^4\).

(a) For \( n \ge 2 \), write the integral as
\[ I_n = \int_0^1 x^{n-1} \cdot \left(x\sqrt{1 - x^2}\right) \mathrm{d}x. \]
Let \( u = x^{n-1} \) and \( \frac{\mathrm{d}v}{\mathrm{d}x} = x(1 - x^2)^{\frac{1}{2}} \).
Then \( \frac{\mathrm{d}u}{\mathrm{d}x} = (n - 1)x^{n-2} \) and \( v = -\frac{1}{3}(1 - x^2)^{\frac{3}{2}} \).

Integrating by parts:
\[ I_n = \left[ -\frac{1}{3}x^{n-1}(1 - x^2)^{\frac{3}{2}} \right]_0^1 + \frac{n - 1}{3} \int_0^1 x^{n-2}(1 - x^2)^{\frac{3}{2}} \, \mathrm{d}x. \]
For \( n \ge 2 \), \( n - 1 \ge 1 \), so the boundary term vanishes at both \( x = 1 \) and \( x = 0 \):
\[ \left[ -\frac{1}{3}x^{n-1}(1 - x^2)^{\frac{3}{2}} \right]_0^1 = 0 - 0 = 0. \]

Now rewrite \( (1 - x^2)^{\frac{3}{2}} = (1 - x^2)\sqrt{1 - x^2} \):
\[ I_n = \frac{n - 1}{3} \int_0^1 x^{n-2}(1 - x^2)\sqrt{1 - x^2} \, \mathrm{d}x \]
\[ I_n = \frac{n - 1}{3} \left( \int_0^1 x^{n-2}\sqrt{1 - x^2} \, \mathrm{d}x - \int_0^1 x^n \sqrt{1 - x^2} \, \mathrm{d}x \right) \]
\[ I_n = \frac{n - 1}{3} (I_{n-2} - I_n). \]

Multiplying by 3:
\[ 3I_n = (n - 1)I_{n-2} - (n - 1)I_n \]
\[ (3 + n - 1)I_n = (n - 1)I_{n-2} \]
\[ (n + 2)I_n = (n - 1)I_{n-2}. \]

(b) First evaluate \( I_0 \):
\[ I_0 = \int_0^1 \sqrt{1 - x^2} \, \mathrm{d}x. \]
Using the substitution \( x = \sin\theta \), \( \mathrm{d}x = \cos\theta \, \mathrm{d}\theta \):
\[ I_0 = \int_0^{\frac{\pi}{2}} \cos^2\theta \, \mathrm{d}\theta = \int_0^{\frac{\pi}{2}} \frac{1 + \cos 2\theta}{2} \, \mathrm{d}\theta = \left[ \frac{\theta}{2} + \frac{\sin 2\theta}{4} \right]_0^{\frac{\pi}{2}} = \frac{\pi}{4}. \]
(Alternatively, recognising this as the area of a quarter circle of radius 1 gives \( I_0 = \frac{\pi}{4} \).)

Using the reduction formula with \( n = 2 \):
\[ (2 + 2)I_2 = (2 - 1)I_0 \implies 4I_2 = I_0 \implies I_2 = \frac{1}{4}I_0 = \frac{\pi}{16}. \]

Using the reduction formula with \( n = 4 \):
\[ (4 + 2)I_4 = (4 - 1)I_2 \implies 6I_4 = 3I_2 \implies I_4 = \frac{1}{2}I_2 = \frac{1}{2} \left(\frac{\pi}{16}\right) = \frac{\pi}{32}. \]

(a)
- M1: Attempts integration by parts with \( u = x^{n-1} \) and \( v' = x(1-x^2)^{\frac{1}{2}} \) (or equivalent splitting).
- A1: Obtains correct expression \( \left[ -\frac{1}{3}x^{n-1}(1-x^2)^{\frac{3}{2}} \right]_0^1 + \frac{n-1}{3}\int_0^1 x^{n-2}(1-x^2)^{\frac{3}{2}}\,\mathrm{d}x \) and correctly evaluates the boundary term to 0.
- M1: Uses \( (1-x^2)^{\frac{3}{2}} = (1-x^2)\sqrt{1-x^2} \) to split the integral into \( I_{n-2} - I_n \).
- A1: (AG) Rearranges completely and convincingly to obtain \( (n+2)I_n = (n-1)I_{n-2} \).

(b)
- B1: Evaluates \( I_0 = \frac{\pi}{4} \) with sufficient working shown or correct geometric justification.
- M1: Applies the reduction formula with \( n=2 \) and \( n=4 \) to express \( I_4 \) in terms of \( I_0 \).
- A1: Obtains exact value \( \frac{\pi}{32} \) (or \( \frac{1}{32}\pi \)).

(a)
Using the identity \(\sinh^2 x = \frac{1}{2}(\cosh 2x - 1)\):
\[ \sinh^4 x = \left(\frac{\cosh 2x - 1}{2}\right)^2 = \frac{1}{4}\left(\cosh^2 2x - 2\cosh 2x + 1\right) \]
Using the double-angle identity \(\cosh^2 2x = \frac{1}{2}(\cosh 4x + 1)\):
\[ \sinh^4 x = \frac{1}{4}\left(\frac{\cosh 4x + 1}{2} - 2\cosh 2x + 1\right) = \frac{1}{8}\cosh 4x - \frac{1}{2}\cosh 2x + \frac{3}{8} \]
Integrating with respect to \(x\):
\[ \int \sinh^4 x \,\mathrm{d}x = \int \left(\frac{1}{8}\cosh 4x - \frac{1}{2}\cosh 2x + \frac{3}{8}\right)\,\mathrm{d}x = \frac{1}{32}\sinh 4x - \frac{1}{4}\sinh 2x + \frac{3}{8}x + c \]

(b)
The given first order linear differential equation is in standard form:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} + (\coth x) y = 16 \sinh^3 x \]
Find the integrating factor \(I(x)\):
\[ I(x) = \mathrm{e}^{\int \coth x \,\mathrm{d}x} = \mathrm{e}^{\ln(\sinh x)} = \sinh x \quad (\text{since } x > 0, \sinh x > 0) \]
Multiply the differential equation by \(I(x)\):
\[ \frac{\mathrm{d}}{\mathrm{d}x}(y \sinh x) = 16 \sinh^4 x \]
Integrate both sides with respect to \(x\), using the result from part (a):
\[ y \sinh x = 16 \left( \frac{1}{32}\sinh 4x - \frac{1}{4}\sinh 2x + \frac{3}{8}x \right) + C \]
\[ y \sinh x = \frac{1}{2}\sinh 4x - 4\sinh 2x + 6x + C \]

Substitute the initial condition \(x = \ln 2\) and \(y = 3\ln 2\):
\[ \mathrm{e}^{\ln 2} = 2, \quad \mathrm{e}^{-\ln 2} = \frac{1}{2} \implies \sinh(\ln 2) = \frac{2 - \frac{1}{2}}{2} = \frac{3}{4} \]
\[ \sinh(2\ln 2) = \sinh(\ln 4) = \frac{4 - \frac{1}{4}}{2} = \frac{15}{8} \]
\[ \sinh(4\ln 2) = \sinh(\ln 16) = \frac{16 - \frac{1}{16}}{2} = \frac{255}{32} \]

Substitute these values into the integrated equation:
\[ (3\ln 2)\left(\frac{3}{4}\right) = \frac{1}{2}\left(\frac{255}{32}\right) - 4\left(\frac{15}{8}\right) + 6\ln 2 + C \]
\[ \frac{9}{4}\ln 2 = \frac{255}{64} - \frac{60}{8} + 6\ln 2 + C \]
\[ \frac{9}{4}\ln 2 = -\frac{225}{64} + 6\ln 2 + C \]
\[ C = \frac{225}{64} - \frac{15}{4}\ln 2 \]

Thus, the particular solution is:
\[ y = \frac{\frac{1}{2}\sinh 4x - 4\sinh 2x + 6x + \frac{225}{64} - \frac{15}{4}\ln 2}{\sinh x} \]

(a)
M1: Expresses \(\sinh^2 x\) in terms of \(\cosh 2x\) and expands \(\left(\frac{\cosh 2x - 1}{2}\right)^2\).
M1: Uses \(\cosh^2 2x = \frac{1}{2}(\cosh 4x + 1)\) to express \(\sinh^4 x\) in linear terms of \(\cosh 4x\) and \(\cosh 2x\).
A1: Obtains \(\sinh^4 x = \frac{1}{8}\cosh 4x - \frac{1}{2}\cosh 2x + \frac{3}{8}\) correctly.
A1 (AG): Integrates each term correctly with respect to \(x\) to reach the printed answer with constant \(c\).

(b)
M1: Identifies and computes the integrating factor \(I(x) = \mathrm{e}^{\int \coth x \,\mathrm{d}x} = \sinh x\).
M1: Multiplies through and writes in the form \(\frac{\mathrm{d}}{\mathrm{d}x}(y \sinh x) = 16 \sinh^4 x\).
A1: Correct integration using part (a) leading to \(y \sinh x = \frac{1}{2}\sinh 4x - 4\sinh 2x + 6x + C\).
M1: Calculates values of \(\sinh(\ln 2)\), \(\sinh(2\ln 2)\), and \(\sinh(4\ln 2)\) using exponential definitions.
M1: Substitutes \(x = \ln 2\) and \(y = 3\ln 2\) into their general solution to find \(C\).
A1: Obtains the correct value of \(C = \frac{225}{64} - \frac{15}{4}\ln 2\) (or exact equivalent).
A1: Expresses \(y\) explicitly in the form \(y = f(x)\).

(a)
Differentiating the given equation \( \mathrm{e}^{2y} + 2xy - x^2 = 1 \) implicitly with respect to \( x \):
\[ 2\mathrm{e}^{2y}\frac{\mathrm{d}y}{\mathrm{d}x} + 2y + 2x\frac{\mathrm{d}y}{\mathrm{d}x} - 2x = 0 \]
Dividing by 2 gives:
\[ (\mathrm{e}^{2y} + x)\frac{\mathrm{d}y}{\mathrm{d}x} + y - x = 0 \]
Substituting \( x = 0 \) and \( y = 0 \):
\[ (\mathrm{e}^0 + 0)\frac{\mathrm{d}y}{\mathrm{d}x} + 0 - 0 = 0 \implies 1\cdot \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \implies \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \]

(b)
Differentiating \( (\mathrm{e}^{2y} + x)y' + y - x = 0 \) with respect to \( x \):
\[ \frac{\mathrm{d}}{\mathrm{d}x}\left[(\mathrm{e}^{2y} + x)y'\right] + y' - 1 = 0 \]
\[ \left(2\mathrm{e}^{2y}(y') + 1\right)y' + (\mathrm{e}^{2y} + x)y'' + y' - 1 = 0 \]
\[ 2\mathrm{e}^{2y}(y')^2 + (\mathrm{e}^{2y} + x)y'' + 2y' - 1 = 0 \]
Substituting \( x = 0 \), \( y = 0 \), and \( y' = 0 \):
\[ 2(1)(0)^2 + (1 + 0)y'' + 2(0) - 1 = 0 \]
\[ y'' - 1 = 0 \implies \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 1 \]

(c)
Differentiating \( 2\mathrm{e}^{2y}(y')^2 + (\mathrm{e}^{2y} + x)y'' + 2y' - 1 = 0 \) implicitly with respect to \( x \):
\[ 4\mathrm{e}^{2y}(y')^3 + 4\mathrm{e}^{2y}y' y'' + (2\mathrm{e}^{2y}y' + 1)y'' + (\mathrm{e}^{2y} + x)y''' + 2y'' = 0 \]
Substituting \( x = 0 \), \( y = 0 \), \( y' = 0 \), and \( y'' = 1 \):
\[ 0 + 0 + (0 + 1)(1) + (1 + 0)y''' + 2(1) = 0 \]
\[ 1 + y''' + 2 = 0 \implies y''' = -3 \]
So \( \frac{\mathrm{d}^3y}{\mathrm{d}x^3} = -3 \) at \( (0,0) \).

The Maclaurin series for \( y \) is:
\[ y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \dots \]
\[ y = 0 + 0x + \frac{1}{2}x^2 + \frac{-3}{6}x^3 = \frac{1}{2}x^2 - \frac{1}{2}x^3 \]

(a)
- M1: Differentiates implicitly with respect to \( x \), using the chain rule for \( \mathrm{e}^{2y} \) and the product rule for \( 2xy \).
- A1: Substitutes \( x=0, y=0 \) correctly to show \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \) (AG, fully justified).

(b)
- M1: Applies product and chain rules to differentiate the first-order implicit equation with respect to \( x \).
- M1: Substitutes \( x = 0, y = 0, y' = 0 \) into an expression involving \( y'' \).
- A1: Obtains \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 1 \).

(c)
- M1: Differentiates again with respect to \( x \) to form an expression containing \( y''' \) and substitutes known values \( x=0, y=0, y'=0, y''=1 \).
- A1: Obtains \( \frac{\mathrm{d}^3y}{\mathrm{d}x^3} = -3 \).
- A1: Obtains correct Maclaurin series \( y = \frac{1}{2}x^2 - \frac{1}{2}x^3 \) (or equivalent form).

(a) Consider the curve \( y = \frac{1}{\sqrt{x}} \), which is strictly decreasing for \( x > 0 \).

For the lower bound, consider rectangles of width 1 on the intervals \( [r, r+1] \) for \( r = 1, 2, \dots, N \), each with height \( f(r) = \frac{1}{\sqrt{r}} \).
Since \( f(x) \) is decreasing, the rectangle over \( [r, r+1] \) lies entirely above the curve on that interval. Therefore, the area of the rectangle is greater than the area under the curve:
\[ \frac{1}{\sqrt{r}} > \int_r^{r+1} \frac{1}{\sqrt{x}}\,\mathrm{d}x. \]
Summing from \( r = 1 \) to \( N \):
\[ \sum_{r=1}^N \frac{1}{\sqrt{r}} > \sum_{r=1}^N \int_r^{r+1} \frac{1}{\sqrt{x}}\,\mathrm{d}x = \int_1^{N+1} \frac{1}{\sqrt{x}}\,\mathrm{d}x = \left[ 2\sqrt{x} \right]_1^{N+1} = 2\sqrt{N+1} - 2. \]

For the upper bound, consider rectangles of width 1 on the intervals \( [r-1, r] \) for \( r = 2, 3, \dots, N \), each with height \( f(r) = \frac{1}{\sqrt{r}} \).
Since \( f(x) \) is decreasing, each rectangle lies entirely beneath the curve on \( [r-1, r] \):
\[ \frac{1}{\sqrt{r}} < \int_{r-1}^r \frac{1}{\sqrt{x}}\,\mathrm{d}x. \]
Summing from \( r = 2 \) to \( N \):
\[ \sum_{r=2}^N \frac{1}{\sqrt{r}} < \sum_{r=2}^N \int_{r-1}^r \frac{1}{\sqrt{x}}\,\mathrm{d}x = \int_1^N \frac{1}{\sqrt{x}}\,\mathrm{d}x = \left[ 2\sqrt{x} \right]_1^N = 2\sqrt{N} - 2. \]
Adding the first term \( \frac{1}{\sqrt{1}} = 1 \) to both sides:
\[ \sum_{r=1}^N \frac{1}{\sqrt{r}} < 1 + 2\sqrt{N} - 2 = 2\sqrt{N} - 1. \]
Hence,
\[ 2\sqrt{N+1} - 2 < \sum_{r=1}^N \frac{1}{\sqrt{r}} < 2\sqrt{N} - 1. \]

(b) Substitute \( N = 10\,000 \) into the inequality from part (a):
Lower bound:
\[ 2\sqrt{10\,001} - 2 \approx 2(100.0049999) - 2 = 198.0099998 > 198 \]
Upper bound:
\[ 2\sqrt{10\,000} - 1 = 2(100) - 1 = 199 \]
Thus,
\[ 198 < 198.01 < \sum_{r=1}^{10\,000} \frac{1}{\sqrt{r}} < 199. \]
Therefore, the integer part of the sum is \( 198 \).

(c) Rewrite the sum in the form of a Riemann sum:
\[ \sum_{r=1}^n \frac{n}{n^2 + 3r^2} = \sum_{r=1}^n \frac{1}{n} \cdot \frac{1}{1 + 3\left(\frac{r}{n}\right)^2}. \]
Taking the limit as \( n \to \infty \), this converges to the definite integral:
\[ \lim_{n \to \infty} \sum_{r=1}^n \frac{1}{n} \cdot \frac{1}{1 + 3\left(\frac{r}{n}\right)^2} = \int_0^1 \frac{1}{1 + 3x^2}\,\mathrm{d}x. \]
Evaluating the integral:
\[ \int_0^1 \frac{1}{1 + (\sqrt{3}x)^2}\,\mathrm{d}x = \left[ \frac{1}{\sqrt{3}} \arctan(\sqrt{3}x) \right]_0^1 = \frac{1}{\sqrt{3}} \arctan(\sqrt{3}) - 0 = \frac{1}{\sqrt{3}} \cdot \frac{\pi}{3} = \frac{\pi\sqrt{3}}{9}. \]

(a)
M1: Suitable sketch or clear formulation showing upper rectangles of width 1 over $[r, r+1]$ and stating/showing $\sum_{r=1}^N \frac{1}{\sqrt{r}} > \int_1^{N+1} x^{-1/2}\,\mathrm{d}x$
A1: Correct evaluation of $\int_1^{N+1} x^{-1/2}\,\mathrm{d}x = 2\sqrt{N+1}-2$ leading to the lower bound (AG)
M1: Suitable sketch or clear formulation showing lower rectangles of width 1 over $[r-1, r]$ for $r=2, \dots, N$ and adding $f(1)=1$
A1: Correct evaluation of $1 + \int_1^N x^{-1/2}\,\mathrm{d}x = 2\sqrt{N}-1$ leading convincingly to the upper bound (AG)

(b)
M1: Substitute $N = 10\,000$ into both bounds
A1: Correct values: lower bound $\approx 198.01$ and upper bound $= 199$
A1: Deduce that the integer part is 198 with rigorous justification

(c)
M1: Rewrite the sum in the form $\frac{1}{n} \sum_{r=1}^n f\left(\frac{r}{n}\right)$ and recognise the limit as $\int_0^1 \frac{1}{1+3x^2}\,\mathrm{d}x$
M1: Integrate correctly to obtain $\left[ \frac{1}{\sqrt{3}} \arctan(x\sqrt{3}) \right]_0^1$
A1: Correct exact value $\frac{\pi\sqrt{3}}{9}$ or $\frac{\pi}{3\sqrt{3}}$

(a)
To find the complementary function, consider the auxiliary equation:
\[ m^2 + 2m + 5 = 0 \]
\[ (m + 1)^2 + 4 = 0 \implies m = -1 \pm 2\mathrm{i} \]
Thus, the complementary function is:
\[ y_c = \mathrm{e}^{-x}(A \cos 2x + B \sin 2x) \]

For the particular integral, try:
\[ y_p = p \cos 2x + q \sin 2x \]
\[ \frac{\mathrm{d} y_p}{\mathrm{d} x} = -2p \sin 2x + 2q \cos 2x \]
\[ \frac{\mathrm{d}^2 y_p}{\mathrm{d} x^2} = -4p \cos 2x - 4q \sin 2x \]

Substitute into the differential equation:
\[ (-4p \cos 2x - 4q \sin 2x) + 2(-2p \sin 2x + 2q \cos 2x) + 5(p \cos 2x + q \sin 2x) = 17 \cos 2x \]
\[ (p + 4q)\cos 2x + (-4p + q)\sin 2x = 17 \cos 2x \]

Equating coefficients:
\[ p + 4q = 17 \]
\[ -4p + q = 0 \implies q = 4p \]

Substituting \(q = 4p\) into the first equation:
\[ p + 4(4p) = 17 \implies 17p = 17 \implies p = 1 \]
\[ q = 4(1) = 4 \]

So \(y_p = \cos 2x + 4 \sin 2x\).

The general solution is:
\[ y = \mathrm{e}^{-x}(A \cos 2x + B \sin 2x) + \cos 2x + 4 \sin 2x \]

(b)
Using the initial condition \(x = 0, y = 1\):
\[ 1 = \mathrm{e}^0 (A \cos 0 + B \sin 0) + \cos 0 + 4 \sin 0 \implies 1 = A + 1 \implies A = 0 \]

Now, differentiating \(y = B \mathrm{e}^{-x}\sin 2x + \cos 2x + 4 \sin 2x\):
\[ \frac{\mathrm{d} y}{\mathrm{d} x} = B(-\mathrm{e}^{-x}\sin 2x + 2\mathrm{e}^{-x}\cos 2x) - 2\sin 2x + 8\cos 2x \]

Using the condition \(x = 0, \frac{\mathrm{d} y}{\mathrm{d} x} = 0\):
\[ 0 = B(0 + 2) - 0 + 8 \implies 2B + 8 = 0 \implies B = -4 \]

Thus, the particular solution is:
\[ y = -4\mathrm{e}^{-x}\sin 2x + \cos 2x + 4 \sin 2x \]

(c)
As \(x \to \infty\), \(\mathrm{e}^{-x} \to 0\), so the term \(-4\mathrm{e}^{-x}\sin 2x \to 0\).
Therefore, \(y \approx \cos 2x + 4 \sin 2x\).

Expressing in the form \(R \cos(2x - \alpha) = R \cos 2x \cos \alpha + R \sin 2x \sin \alpha\):
\[ R \cos \alpha = 1 \quad \text{and} \quad R \sin \alpha = 4 \]
\[ R = \sqrt{1^2 + 4^2} = \sqrt{17} \]
\[ \tan \alpha = 4 \implies \alpha = \arctan(4) \approx 1.33 \text{ radians} \]

So as \(x \to \infty\), \(y \approx \sqrt{17} \cos(2x - 1.33)\).

(a)
* M1: Sets up auxiliary equation \(m^2 + 2m + 5 = 0\) and attempts to solve.
* A1: Obtains roots \(-1 \pm 2\mathrm{i}\) and states correct complementary function \(\mathrm{e}^{-x}(A \cos 2x + B \sin 2x)\).
* M1: Uses trial particular integral of form \(p \cos 2x + q \sin 2x\), differentiates twice, and substitutes into the differential equation.
* M1: Equates coefficients of \(\cos 2x\) and \(\sin 2x\) to form a system of simultaneous equations in \(p\) and \(q\).
* A1: Solves correctly to obtain \(p = 1\) and \(q = 4\).
* A1: States the complete general solution \(y = y_c + y_p\).

(b)
* M1: Applies boundary conditions \(y(0) = 1\) and \(y'(0) = 0\) to their general solution containing two arbitrary constants.
* A1: Obtains \(A = 0\) and \(B = -4\), giving the particular solution \(y = -4\mathrm{e}^{-x}\sin 2x + \cos 2x + 4\sin 2x\).

(c)
* B1: Identifies that \(\mathrm{e}^{-x} \to 0\) as \(x \to \infty\) so that \(y \to \cos 2x + 4\sin 2x\).
* B1: Obtains exact \(R = \sqrt{17}\) and \(\alpha = 1.33\) (or \(\arctan 4\)), stating \(y \approx \sqrt{17}\cos(2x - 1.33)\).

(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 imaginary parts:
\[ \sin 5\theta = 5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta \]
Factoring out $\sin\theta$:
\[ \sin 5\theta = \sin\theta \left( 5\cos^4\theta - 10\cos^2\theta\sin^2\theta + \sin^4\theta \right) \]
Using $\cos^2\theta = 1 - \sin^2\theta$:
\[ \sin 5\theta = \sin\theta \left( 5(1 - \sin^2\theta)^2 - 10(1 - \sin^2\theta)\sin^2\theta + \sin^4\theta \right) \]
\[ = \sin\theta \left( 5(1 - 2\sin^2\theta + \sin^4\theta) - 10\sin^2\theta + 10\sin^4\theta + \sin^4\theta \right) \]
\[ = \sin\theta \left( 5 - 10\sin^2\theta + 5\sin^4\theta - 10\sin^2\theta + 10\sin^4\theta + \sin^4\theta \right) \]
\[ = \sin\theta \left( 16\sin^4\theta - 20\sin^2\theta + 5 \right) \]

(b) Consider the equation $\sin 5\theta = 0$, which gives $5\theta = k\pi \implies \theta = \frac{k\pi}{5}$ for $k \in \mathbb{Z}$.
From part (a), $\sin 5\theta = \sin\theta(16\sin^4\theta - 20\sin^2\theta + 5) = 0$.
For $k = \pm 1, \pm 2$, $\sin\theta
eq 0$, so $\theta = \pm \frac{\pi}{5}, \pm \frac{2\pi}{5}$ satisfy:
\[ 16\sin^4\theta - 20\sin^2\theta + 5 = 0 \]
Letting $x = \sin\theta$, since $\sin\left(-\frac{\pi}{5}\right) = -\sin\left(\frac{\pi}{5}\right)$ and $\sin\left(-\frac{2\pi}{5}\right) = -\sin\left(\frac{2\pi}{5}\right)$, the four distinct roots of the quartic equation $16x^4 - 20x^2 + 5 = 0$ are:
\[ x = \pm \sin\left(\frac{\pi}{5}\right), \quad x = \pm \sin\left(\frac{2\pi}{5}\right) \]

(c) Let $y = x^2$. The equation becomes $16y^2 - 20y + 5 = 0$, with roots $y_1 = \sin^2\left(\frac{\pi}{5}\right)$ and $y_2 = \sin^2\left(\frac{2\pi}{5}\right)$.
The product of the roots is:
\[ y_1 y_2 = \sin^2\left(\frac{\pi}{5}\right) \sin^2\left(\frac{2\pi}{5}\right) = \frac{5}{16} \]
Since $0 < \frac{\pi}{5} < \frac{2\pi}{5} < \pi$, both $\sin\left(\frac{\pi}{5}\right) > 0$ and $\sin\left(\frac{2\pi}{5}\right) > 0$. Taking the positive square root:
\[ \sin\left(\frac{\pi}{5}\right) \sin\left(\frac{2\pi}{5}\right) = \sqrt{\frac{5}{16}} = \frac{\sqrt{5}}{4} \]

Solving $16y^2 - 20y + 5 = 0$ using the quadratic formula:
\[ y = \frac{20 \pm \sqrt{(-20)^2 - 4(16)(5)}}{2(16)} = \frac{20 \pm \sqrt{400 - 320}}{32} = \frac{20 \pm \sqrt{80}}{32} = \frac{20 \pm 4\sqrt{5}}{32} = \frac{5 \pm \sqrt{5}}{8} = \frac{10 \pm 2\sqrt{5}}{16} \]
Since $0 < \frac{\pi}{5} < \frac{2\pi}{5} < \frac{\pi}{2}$, the sine function is strictly increasing on this interval, so $\sin\left(\frac{\pi}{5}\right) < \sin\left(\frac{2\pi}{5}\right)$, which means $\sin^2\left(\frac{\pi}{5}\right)$ must be the smaller root:
\[ \sin^2\left(\frac{\pi}{5}\right) = \frac{10 - 2\sqrt{5}}{16} \]
Taking the positive square root gives:
\[ \sin\left(\frac{\pi}{5}\right) = \frac{\sqrt{10 - 2\sqrt{5}}}{4} \]

(a)
- M1: Applies de Moivre's theorem to $(\cos\theta + \mathrm{i}\sin\theta)^5$ and attempts binomial expansion.
- A1: Correct expansion and equates imaginary parts to obtain $\sin 5\theta = 5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta$.
- M1: Uses $\cos^2\theta = 1 - \sin^2\theta$ to express all terms purely in terms of $\sin\theta$.
- A1: Fully correct simplification leading convincingly to the given result (AG).

(b)
- M1: Sets $\sin 5\theta = 0$ and identifies $\theta = \frac{k\pi}{5}$ for non-zero $\sin\theta$.
- A1: Fully justifies that the four roots are $x = \pm \sin\left(\frac{\pi}{5}\right)$ and $x = \pm \sin\left(\frac{2\pi}{5}\right)$ by symmetry/odd property of $\sin$.

(c)
- B1: Uses the product of the roots of the quadratic in $y$ to deduce $\sin\left(\frac{\pi}{5}\right)\sin\left(\frac{2\pi}{5}\right) = \frac{\sqrt{5}}{4}$.
- M1: Applies quadratic formula to solve $16y^2 - 20y + 5 = 0$ obtaining $y = \frac{10 \pm 2\sqrt{5}}{16}$ (or $\frac{5 \pm \sqrt{5}}{8}$).
- M1: Justifies the selection of the smaller root for $\sin^2\left(\frac{\pi}{5}\right)$ using the monotonic increasing property of $\sin\theta$ on $\left(0, \frac{\pi}{2}\right)$.
- A1: Concludes with $\sin\left(\frac{\pi}{5}\right) = \frac{\sqrt{10 - 2\sqrt{5}}}{4}$ (AG) with all steps clearly set out.

(a)
To find the eigenvalues, solve \(\det(\mathbf{A} - \lambda\mathbf{I}) = 0\):
\[\det\begin{pmatrix} 3-\lambda & 0 & 1 \\ 2 & 5-\lambda & 1 \\ 2 & 0 & 2-\lambda \end{pmatrix} = 0\]
Expanding along the second column:
\[(5-\lambda)\left[(3-\lambda)(2-\lambda) - 2\right] = 0\]
\[(5-\lambda)(\lambda^2 - 5\lambda + 4) = 0\]
\[(5-\lambda)(\lambda - 1)(\lambda - 4) = 0\]
Thus, the eigenvalues are \(\lambda = 1, 4, 5\).

For \(\lambda = 1\):
\[\begin{pmatrix} 2 & 0 & 1 \\ 2 & 4 & 1 \\ 2 & 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\]
\(2x + z = 0 \implies z = -2x\)
\(2x + 4y + z = 0 \implies 4y = 0 \implies y = 0\)
Eigenvector: \(\mathbf{e}_1 = \begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix}\) (or any non-zero scalar multiple).

For \(\lambda = 4\):
\[\begin{pmatrix} -1 & 0 & 1 \\ 2 & 1 & 1 \\ 2 & 0 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\]
\(-x + z = 0 \implies z = x\)
\(2x + y + z = 0 \implies 3x + y = 0 \implies y = -3x\)
Eigenvector: \(\mathbf{e}_2 = \begin{pmatrix} 1 \\ -3 \\ 1 \end{pmatrix}\) (or any non-zero scalar multiple).

For \(\lambda = 5\):
\[\begin{pmatrix} -2 & 0 & 1 \\ 2 & 0 & 1 \\ 2 & 0 & -3 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\]
From rows 1 and 2: \(-2x + z = 0\) and \(2x + z = 0 \implies x = 0, z = 0\), with \(y\) arbitrary.
Eigenvector: \(\mathbf{e}_3 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\) (or any non-zero scalar multiple).

(b)
Forming \(\mathbf{P}\) from the eigenvectors and \(\mathbf{D}\) from the corresponding eigenvalues:
\[\mathbf{P} = \begin{pmatrix} 1 & 1 & 0 \\ 0 & -3 & 1 \\ -2 & 1 & 0 \end{pmatrix}, \quad \mathbf{D} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 5 \end{pmatrix}\]
*(Note: Any permutation of columns is acceptable provided the order in \(\mathbf{P}\) matches that in \(\mathbf{D}\).)*

(c)
The characteristic equation is:
\[-\lambda^3 + 10\lambda^2 - 29\lambda + 20 = 0 \implies \lambda^3 - 10\lambda^2 + 29\lambda - 20 = 0\]
By the Cayley-Hamilton theorem, \(\mathbf{A}\) satisfies its own characteristic equation:
\[\mathbf{A}^3 - 10\mathbf{A}^2 + 29\mathbf{A} - 20\mathbf{I} = \mathbf{0}\]
Multiplying by \(\mathbf{A}^{-1}\):
\[\mathbf{A}^2 - 10\mathbf{A} + 29\mathbf{I} - 20\mathbf{A}^{-1} = \mathbf{0}\]
\[20\mathbf{A}^{-1} = \mathbf{A}^2 - 10\mathbf{A} + 29\mathbf{I}\]
\[\mathbf{A}^{-1} = \frac{1}{20}\mathbf{A}^2 - \frac{1}{2}\mathbf{A} + \frac{29}{20}\mathbf{I}\]
Thus \(p = \frac{1}{20}\), \(q = -\frac{1}{2}\), \(r = \frac{29}{20}\).

(d)
Express the vector \(\mathbf{v} = \begin{pmatrix} 2 \\ -3 \\ -1 \end{pmatrix}\) as a linear combination of the eigenvectors:
\[\begin{pmatrix} 2 \\ -3 \\ -1 \end{pmatrix} = a\begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix} + b\begin{pmatrix} 1 \\ -3 \\ 1 \end{pmatrix} + c\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\]
\[a + b = 2\]
\[-3b + c = -3\]
\[-2a + b = -1\]
Subtracting the third equation from the first: \(3a = 3 \implies a = 1\).
Then \(b = 2 - 1 = 1\) and \(c = -3 + 3(1) = 0\).
So \(\begin{pmatrix} 2 \\ -3 \\ -1 \end{pmatrix} = \mathbf{e}_1 + \mathbf{e}_2\).

Applying \(\mathbf{A}^n\):
\[\mathbf{A}^n \begin{pmatrix} 2 \\ -3 \\ -1 \end{pmatrix} = \mathbf{A}^n \mathbf{e}_1 + \mathbf{A}^n \mathbf{e}_2 = (1)^n \mathbf{e}_1 + (4)^n \mathbf{e}_2\]
\[= 1 \begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix} + 4^n \begin{pmatrix} 1 \\ -3 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 + 4^n \\ -3 \cdot 4^n \\ -2 + 4^n \end{pmatrix}\]

(a)
- M1: Sets \(\det(\mathbf{A} - \lambda\mathbf{I}) = 0\) and expands the determinant correctly.
- A1: Obtains characteristic equation \((5-\lambda)(\lambda-1)(\lambda-4) = 0\) (or expanded equivalent) and states eigenvalues \(\lambda = 1, 4, 5\).
- M1: Sets up \((\mathbf{A} - \lambda\mathbf{I})\mathbf{x} = \mathbf{0}\) for at least one eigenvalue.
- A1: Correct eigenvector for \(\lambda = 1\) (e.g. \([1, 0, -2]^T\) or non-zero multiple).
- A1: Correct eigenvector for \(\lambda = 4\) (e.g. \([1, -3, 1]^T\) or non-zero multiple).
- A1: Correct eigenvector for \(\lambda = 5\) (e.g. \([0, 1, 0]^T\) or non-zero multiple).

(b)
- B1: Writes a non-singular matrix \(\mathbf{P}\) whose columns are their eigenvectors.
- B1: Writes the diagonal matrix \(\mathbf{D}\) with eigenvalues in corresponding order.

(c)
- M1: States the Cayley-Hamilton theorem result \(\mathbf{A}^3 - 10\mathbf{A}^2 + 29\mathbf{A} - 20\mathbf{I} = \mathbf{0}\).
- M1: Multiplies through by \(\mathbf{A}^{-1}\) and rearranges to make \(\mathbf{A}^{-1}\) the subject.
- A1: Correct values \(p = \frac{1}{20}, q = -\frac{1}{2}, r = \frac{29}{20}\) (or exact fractions).

(d)
- M1: Attempts to express \([2, -3, -1]^T\) as a linear combination of the eigenvectors.
- A1: Correctly identifies \([2, -3, -1]^T = \mathbf{e}_1 + \mathbf{e}_2\).
- A1: Evaluates \(\mathbf{A}^n \mathbf{v} = 1^n \mathbf{e}_1 + 4^n \mathbf{e}_2\) to reach the final column vector \(\begin{pmatrix} 1 + 4^n \\ -3 \cdot 4^n \\ -2 + 4^n \end{pmatrix}\).

(a) Let \(O\) be the centre of the common circular base, and let the axis of symmetry be the \(x\)-axis with the positive direction pointing into the cone.

For the solid hemisphere of radius \(r\):
\[ V_1 = \frac{2}{3}\pi r^3, \quad \bar{x}_1 = -\frac{3}{8}r \]

For the solid cone of radius \(r\) and height \(h = 2r\):
\[ V_2 = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi r^2 (2r) = \frac{2}{3}\pi r^3 \]
The centre of mass of the cone is at a distance of \(\frac{1}{4}h = \frac{1}{4}(2r) = \frac{1}{2}r\) from its base, so:
\[ \bar{x}_2 = \frac{1}{2}r \]

The total volume of \(S\) is:
\[ V = V_1 + V_2 = \frac{2}{3}\pi r^3 + \frac{2}{3}\pi r^3 = \frac{4}{3}\pi r^3 \]

Taking moments about the plane of the common circular base:
\[ V \bar{x} = V_1 \bar{x}_1 + V_2 \bar{x}_2 \]
\[ \left(\frac{4}{3}\pi r^3\right)\bar{x} = \left(\frac{2}{3}\pi r^3\right)\left(-\frac{3}{8}r\right) + \left(\frac{2}{3}\pi r^3\right)\left(\frac{1}{2}r\right) \]
\[ 2\bar{x} = -\frac{3}{8}r + \frac{1}{2}r = \frac{1}{8}r \]
\[ \bar{x} = \frac{1}{16}r \]
Thus, the distance of the centre of mass from the common circular base is \(\frac{1}{16}r\).

(b) When \(S\) is suspended freely from \(A\), the line joining \(A\) to the centre of mass \(G\) is vertical.

In the right-angled triangle \(AOG\):
- \(OA = r\) (along the circular base, perpendicular to the axis of symmetry)
- \(OG = \bar{x} = \frac{1}{16}r\) (along the axis of symmetry)

Let \(\theta\) be the acute angle between the axis of symmetry (line \(OG\)) and the vertical (line \(AG\)):
\[ \tan\theta = \frac{OA}{OG} = \frac{r}{\frac{1}{16}r} = 16 \]
\[ \theta = \arctan(16) \approx 86.4236^\circ \approx 86.4^\circ \]

(a)
B1: State or use correct volume \(\frac{2}{3}\pi r^3\) and position of centre of mass \(\frac{3}{8}r\) from base for hemisphere.
B1: State or use correct volume \(\frac{2}{3}\pi r^3\) and position of centre of mass \(\frac{1}{2}r\) from base for cone.
M1: Formulate a correct moments equation about the common circular base with their volumes and distances.
A1 (AG): Fully correct algebraic working leading to the given answer \(\bar{x} = \frac{1}{16}r\).

(b)
M1: Recognise that the vertical passes through the suspension point \(A\) and the centre of mass \(G\).
M1: Use trigonometry in right-angled triangle \(AOG\) with lengths \(r\) and \(\frac{1}{16}r\) to set up an equation for the angle with the vertical (e.g. \(\tan\theta = 16\) or \(\tan\phi = \frac{1}{16}\)).
A1: Obtain \(86.4^\circ\) (accept \(1.51\) radians; must be to 1 d.p. for degrees or 3 s.f. for radians).

(a) Let \(M\) be the midpoint of \(AB\). Since the triangle \(ABP\) is isosceles with \(AP = BP = 0.5\text{ m}\) and \(AB = 0.6\text{ m}\), the centre of the circular path is \(M\), and \(AM = MB = 0.3\text{ m}\).

Let \(\theta\) be the angle that each string makes with the horizontal plane through \(M\).
In triangle \(AMP\):
\[ \sin\theta = \frac{AM}{AP} = \frac{0.3}{0.5} = 0.6 \]
Let \(T_A\) and \(T_B\) be the tensions in strings \(AP\) and \(BP\) respectively.
Resolving forces vertically for particle \(P\):
\[ T_A \sin\theta - T_B \sin\theta - mg = 0 \]
\[ (T_A - T_B) \sin\theta = mg \]
Using \(m = 0.3\text{ kg}\), \(g = 10\text{ m s}^{-2}\), and \(\sin\theta = 0.6\):
\[ (T_A - T_B)(0.6) = 0.3 \times 10 = 3 \]
\[ T_A - T_B = \frac{3}{0.6} = 5\text{ N} \]
Thus, the tension in \(AP\) is \(5\text{ N}\) greater than the tension in \(BP\).

(b) The radius of the horizontal circle is \(r = MP\):
\[ r = \sqrt{0.5^2 - 0.3^2} = \sqrt{0.25 - 0.09} = \sqrt{0.16} = 0.4\text{ m} \]
Also, \(\cos\theta = \frac{r}{AP} = \frac{0.4}{0.5} = 0.8\).

Applying Newton's second law horizontally towards the centre \(M\):
\[ T_A \cos\theta + T_B \cos\theta = m r \omega^2 \]
\[ (T_A + T_B)(0.8) = 0.3 \times 0.4 \times 8^2 \]
\[ 0.8(T_A + T_B) = 0.12 \times 64 = 7.68 \]
\[ T_A + T_B = \frac{7.68}{0.8} = 9.6\text{ N} \]
We now have the simultaneous equations:
1) \(T_A - T_B = 5\)
2) \(T_A + T_B = 9.6\)

Adding (1) and (2):
\[ 2T_A = 14.6 \implies T_A = 7.3\text{ N} \]
Subtracting (1) from (2):
\[ 2T_B = 4.6 \implies T_B = 2.3\text{ N} \]
So, the tension in string \(AP\) is \(7.3\text{ N}\) and the tension in string \(BP\) is \(2.3\text{ N}\).

(a)
- M1: Resolves forces vertically for \(P\) to obtain an equation relating \(T_A\), \(T_B\), \(m\), \(g\), and \(\sin\theta\).
- A1 (AG): Fully correct derivation showing \(T_A - T_B = 5\text{ N}\) with \(\sin\theta = 0.6\) clearly shown.

(b)
- M1: Applies Newton's second law horizontally towards the centre of the circle using \(mr\omega^2\) with \(r = 0.4\) and \(\cos\theta = 0.8\).
- M1: Solves the system of two linear equations in \(T_A\) and \(T_B\).
- A1: Obtains \(T_A = 7.3\text{ N}\) and \(T_B = 2.3\text{ N}\) (both correct).

(a) For the upward motion, taking the upward displacement as \(x\):
Newton's second law gives:
\[ 0.5 v \frac{\mathrm{d}v}{\mathrm{d}x} = -0.5g - 0.05v^2 \]
Taking \(g = 10\text{ ms}^{-2}\):
\[ 0.5 v \frac{\mathrm{d}v}{\mathrm{d}x} = -5 - 0.05v^2 = -0.05(100 + v^2) \]
\[ v \frac{\mathrm{d}v}{\mathrm{d}x} = -0.1(100 + v^2) = -\frac{v^2 + 100}{10} \]

Separating variables:
\[ \int \frac{v}{v^2 + 100} \, \mathrm{d}v = -\frac{1}{10} \int \mathrm{d}x \]
\[ \frac{1}{2}\ln(v^2 + 100) = -\frac{x}{10} + C \]

Using the initial condition \(x = 0\) when \(v = 20\):
\[ \frac{1}{2}\ln(20^2 + 100) = C \implies C = \frac{1}{2}\ln(500) \]

Thus:
\[ \frac{x}{10} = \frac{1}{2}\ln(500) - \frac{1}{2}\ln(v^2 + 100) = \frac{1}{2}\ln\left(\frac{500}{v^2 + 100}\right) \]
\[ x = 5\ln\left(\frac{500}{v^2 + 100}\right) \]

At the maximum height \(H\), the speed is \(v = 0\):
\[ H = 5\ln\left(\frac{500}{100}\right) = 5\ln 5\text{ m} \]

(b) For the upward motion in terms of time \(t\):
\[ 0.5 \frac{\mathrm{d}v}{\mathrm{d}t} = -5 - 0.05v^2 \implies \frac{\mathrm{d}v}{\mathrm{d}t} = -\frac{v^2 + 100}{10} \]

Separating variables from \(t = 0, v = 20\) to \(t = T, v = 0\):
\[ \int_0^T \mathrm{d}t = -\int_{20}^0 \frac{10}{v^2 + 100} \, \mathrm{d}v = \int_0^{20} \frac{10}{v^2 + 100} \, \mathrm{d}v \]
\[ T = \left[ \frac{10}{10}\arctan\left(\frac{v}{10}\right) \right]_0^{20} = \arctan(2) - \arctan(0) = \arctan(2) \]
\[ T \approx 1.11\text{ s} \text{ (3 s.f.)} \]

(c) For the downward motion, measuring displacement \(y\) downwards from the highest point:
\[ 0.5 v \frac{\mathrm{d}v}{\mathrm{d}y} = 0.5g - 0.05v^2 = 5 - 0.05v^2 = 0.05(100 - v^2) \]
\[ v \frac{\mathrm{d}v}{\mathrm{d}y} = 0.1(100 - v^2) = \frac{100 - v^2}{10} \]

Separating variables:
\[ \int \frac{v}{100 - v^2} \, \mathrm{d}v = \frac{1}{10} \int \mathrm{d}y \]
\[ -\frac{1}{2}\ln(100 - v^2) = \frac{y}{10} + D \]

At the highest point, \(y = 0\) and \(v = 0\):
\[ -\frac{1}{2}\ln(100) = D \]

So:
\[ \frac{y}{10} = \frac{1}{2}\ln\left(\frac{100}{100 - v^2}\right) \implies y = 5\ln\left(\frac{100}{100 - v^2}\right) \]

When \(P\) returns to \(O\), the total distance fallen is \(y = 5\ln 5\):
\[ 5\ln 5 = 5\ln\left(\frac{100}{100 - v^2}\right) \]
\[ 5 = \frac{100}{100 - v^2} \implies 100 - v^2 = 20 \implies v^2 = 80 \]
\[ v = \sqrt{80} = 4\sqrt{5}\text{ ms}^{-1} \]
Thus \(k = 4\).

(a)
- M1: Applies Newton's second law vertically upwards using \(a = v\frac{\mathrm{d}v}{\mathrm{d}x}\) with two forces (weight and resistance opposing motion).
- A1: Separates variables correctly to reach \(\int \frac{v}{v^2 + 100} \, \mathrm{d}v = -\frac{1}{10} \int \mathrm{d}x\) (or equivalent with limits).
- M1: Integrates to obtain a logarithmic term and uses \(v = 20\) when \(x = 0\) (or uses definite limits).
- A1 (AG): Sets \(v = 0\) and deduces the given expression \(5\ln 5\) with no errors seen.

(b)
- M1: Sets up \(0.5\frac{\mathrm{d}v}{\mathrm{d}t} = -5 - 0.05v^2\), separates variables, and integrates to the form \(k\arctan\left(\frac{v}{10}\right)\).
- A1: Obtains \(T = \arctan(2)\) or awrt \(1.11\text{ s}\).

(c)
- M1: Applies Newton's second law for the downward motion using \(a = v\frac{\mathrm{d}v}{\mathrm{d}y}\) with weight and resistance opposing each other, and separates variables.
- A1: Integrates correctly and applies initial condition \(v = 0\) when \(y = 0\) to obtain \(y = 5\ln\left(\frac{100}{100 - v^2}\right)\) (or equivalent).
- A1: Equates \(y\) to \(5\ln 5\) and correctly determines \(v = 4\sqrt{5}\text{ ms}^{-1}\) (or \(k = 4\)).

(a)
Let components of velocity parallel and perpendicular to the barrier be denoted by subscripts \(\parallel\) and \(\perp\) respectively.

From \(\tan \alpha = \frac{4}{3}\), we have \(\sin \alpha = \frac{4}{5}\) and \(\cos \alpha = \frac{3}{5}\).

Before impact:
\(u_\parallel = u \cos \alpha = \frac{3}{5}u\)
\(u_\perp = u \sin \alpha = \frac{4}{5}u\)

Since the barrier is smooth, the component of velocity parallel to the barrier is unchanged:
\(v_\parallel = u_\parallel = \frac{3}{5}u\)

Perpendicular to the barrier, using Newton's experimental law:
\(v_\perp = e u_\perp = \frac{1}{2}\left(\frac{4}{5}u\right) = \frac{2}{5}u\)

The speed \(v\) of the sphere immediately after impact is:
\(v = \sqrt{v_\parallel^2 + v_\perp^2} = \sqrt{\left(\frac{3}{5}u\right)^2 + \left(\frac{2}{5}u\right)^2} = \sqrt{\frac{9 + 4}{25}u^2} = \frac{\sqrt{13}}{5}u\)

(b)
Let \(\beta\) be the acute angle that the direction of motion makes with the barrier after impact.
\(\tan \beta = \frac{v_\perp}{v_\parallel} = \frac{\frac{2}{5}u}{\frac{3}{5}u} = \frac{2}{3}\)

So \(\beta = \arctan\left(\frac{2}{3}\right) \approx 33.690^\circ\).
Also \(\alpha = \arctan\left(\frac{4}{3}\right) \approx 53.130^\circ\).

The angle of deflection \(\theta\) is:
\(\theta = \alpha + \beta\)
Using the tangent addition formula:
\(\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = \frac{\frac{4}{3} + \frac{2}{3}}{1 - \left(\frac{4}{3}\right)\left(\frac{2}{3}\right)} = \frac{2}{1 - \frac{8}{9}} = \frac{2}{\frac{1}{9}} = 18\)

Therefore,
\(\theta = \arctan(18) \approx 86.820^\circ = 86.8^\circ\) (to 1 d.p.)

(c)
Loss in kinetic energy:
\(\Delta E_k = \frac{1}{2}m u^2 - \frac{1}{2}m v^2\)
\(\Delta E_k = \frac{1}{2}m u^2 - \frac{1}{2}m\left(\frac{\sqrt{13}}{5}u\right)^2 = \frac{1}{2}m u^2 \left(1 - \frac{13}{25}\right) = \frac{1}{2}m u^2 \left(\frac{12}{25}\right) = \frac{6}{25}m u^2\)

(a)
M1: Resolving velocity components parallel and perpendicular to the barrier, applying Newton's experimental law perpendicular to the barrier (e.g. \(v_\parallel = \frac{3}{5}u\), \(v_\perp = \frac{1}{2} \times \frac{4}{5}u = \frac{2}{5}u\)).
M1: Correct use of Pythagoras to determine the magnitude of final velocity: \(v = \sqrt{v_\parallel^2 + v_\perp^2}\).
A1: Obtain \(v = \frac{\sqrt{13}}{5}u\) (or state \(k = 13\)).

(b)
M1: Finding the angle \(\beta\) after impact via \(\tan \beta = \frac{v_\perp}{v_\parallel} = \frac{2}{3}\) (or finding \(\beta = 33.7^\circ\)).
M1: Expressing the angle of deflection as \(\alpha + \beta\) (or evaluating \(\tan(\alpha + \beta)\) using compound angle formula).
A1: Correct deflection angle \(86.8^\circ\) (accept \(1.52\) radians).

(c)
M1: Correct method for kinetic energy loss, e.g. \(\frac{1}{2}m u^2 - \frac{1}{2}m v^2\) or \(\frac{1}{2}m(u_\perp^2 - v_\perp^2)\).
A1: Correct exact expression \(\frac{6}{25}mu^2\) (or \(0.24mu^2\)).

(a)
Let the gravitational potential energy be zero at the lowest point $A$.
When the string makes an angle $\theta$ with the upward vertical, the height of $P$ above $A$ is
\[ h = a + a\cos\theta = a(1 + \cos\theta) \]
By conservation of mechanical energy:
\[ \frac{1}{2}mv^2 + mgh = \frac{1}{2}mu^2 \]
\[ \frac{1}{2}mv^2 + mg a(1 + \cos\theta) = \frac{1}{2}m\left(\frac{7}{2}ga\right) \]
\[ v^2 + 2ga(1 + \cos\theta) = \frac{7}{2}ga \]
\[ v^2 = \frac{7}{2}ga - 2ga - 2ga\cos\theta = ga\left(\frac{3}{2} - 2\cos\theta\right) \]

Resolving forces along the string towards the centre $O$:
\[ T + mg\cos\theta = \frac{mv^2}{a} \]
\[ T = \frac{m}{a}\left[ga\left(\frac{3}{2} - 2\cos\theta\right)\right] - mg\cos\theta \]
\[ T = mg\left(\frac{3}{2} - 2\cos\theta - \cos\theta\right) = mg\left(\frac{3}{2} - 3\cos\theta\right) = 3mg\left(\frac{1}{2} - \cos\theta\right) \]

(b)
The string becomes slack when $T = 0$:
\[ 3mg\left(\frac{1}{2} - \cos\theta\right) = 0 \implies \cos\theta = \frac{1}{2} \]
\[ \theta = \frac{\pi}{3} \text{ (or } 60^\circ\text{)} \]

(c)
At the point where the string becomes slack (let this point be $B$):
\[ v^2 = ga\left(\frac{3}{2} - 2\left(\frac{1}{2}\right)\right) = \frac{1}{2}ga \]
The height of $B$ above $A$ is:
\[ h_B = a\left(1 + \cos\frac{\pi}{3}\right) = a\left(1 + \frac{1}{2}\right) = \frac{3}{2}a \]
Since the string makes an angle of $60^\circ$ with the upward vertical, the velocity vector is perpendicular to the string and makes an angle of $60^\circ$ with the horizontal.
The vertical component of the velocity at $B$ is:
\[ v_y = v\sin 60^\circ = \sqrt{\frac{1}{2}ga}\left(\frac{\sqrt{3}}{2}\right) \]
After the string becomes slack, $P$ moves freely under gravity as a projectile. The additional vertical height $h'$ gained is:
\[ h' = \frac{v_y^2}{2g} = \frac{\frac{1}{2}ga \cdot \frac{3}{4}}{2g} = \frac{3}{16}a \]
Therefore, the greatest vertical height reached by $P$ above $A$ is:
\[ H = h_B + h' = \frac{3}{2}a + \frac{3}{16}a = \frac{27}{16}a \]

(a)
- M1: Uses conservation of energy with correct potential energy expression $mga(1+\cos\theta)$ (or equivalent).
- A1: Fully correct derivation leading to $v^2 = ga\left(\frac{3}{2} - 2\cos\theta\right)$ (AG).
- M1: Sets up radial equation of motion $T + mg\cos\theta = \frac{mv^2}{a}$.
- A1: Correct expression for $T$, e.g. $T = mg\left(\frac{3}{2} - 3\cos\theta\right)$ or $3mg\left(\frac{1}{2} - \cos\theta\right)$.

(b)
- M1: Sets $T = 0$ to find $\cos\theta$.
- A1: Obtains $\theta = \frac{\pi}{3}$ or $60^\circ$ (or $1.05\text{ rad}$).

(c)
- B1: Calculates the height of the slack point above $A$, $h_B = \frac{3}{2}a$, and finds $v^2 = \frac{1}{2}ga$.
- M1: Identifies the angle of projection as $60^\circ$ to the horizontal and calculates the additional height $h' = \frac{v^2\sin^2 60^\circ}{2g}$.
- A1: Correct total height $H = \frac{27}{16}a$ (or $1.6875a$ or $1.69a$).

(a) Before the string becomes taut, \(P\) falls freely under gravity through a vertical distance equal to the natural length \(l = 0.8\text{ m}\).
Using the equation of motion:
\[ v^2 = u^2 + 2gs \]
\[ v^2 = 0 + 2(10)(0.8) = 16 \]
\[ v = \sqrt{16} = 4\text{ m s}^{-1} \]

(b) Let \(x\) be the maximum extension of the string when \(P\) reaches its lowest point. At this point, the speed of \(P\) is momentarily zero.
Taking the initial position at \(O\) as the datum for gravitational potential energy:
Initial energy at \(O\) is \(E_i = 0\).
At the lowest point, the total distance fallen is \(0.8 + x\).
Gravitational potential energy \(= -mg(0.8 + x) = -0.5(10)(0.8 + x) = -5(0.8 + x) = -4 - 5x\).
Elastic potential energy \(= \dfrac{\lambda x^2}{2l} = \dfrac{60 x^2}{2(0.8)} = 37.5 x^2\).
Kinetic energy \(= 0\).

By the conservation of mechanical energy:
\[ 37.5 x^2 - 5x - 4 = 0 \]

Multiply the entire equation by 2:
\[ 75 x^2 - 10x - 8 = 0 \]
\[ (5x - 2)(15x + 4) = 0 \]

Since extension \(x > 0\):
\[ x = \dfrac{2}{5} = 0.4\text{ m} \]

(c) At the lowest point, the extension is \(x = 0.4\text{ m}\).
The tension in the string is:
\[ T = \dfrac{\lambda x}{l} = \dfrac{60(0.4)}{0.8} = 30\text{ N} \]

The weight of the particle is:
\[ W = mg = 0.5(10) = 5\text{ N} \]

Applying Newton's second law vertically upwards:
\[ T - mg = ma \]
\[ 30 - 5 = 0.5 a \]
\[ 25 = 0.5 a \implies a = 50\text{ m s}^{-2} \]

(a)
M1: Uses \(v^2 = 2gl\) or \(\frac{1}{2}mv^2 = mgl\) with \(g = 10\) and \(l = 0.8\).
A1: Obtains \(v = 4\text{ m s}^{-1}\) with correct intermediate working shown (AG).

(b)
M1: Sets up a valid energy equation connecting the release point (or taut position) and the lowest point, including gravitational potential energy and elastic potential energy terms: \(\frac{\lambda x^2}{2l} = mg(l + x)\) or \(\frac{1}{2}mv^2 + mgx = \frac{\lambda x^2}{2l}\).
M1: Solves a 3-term quadratic equation in \(x\).
A1: Obtains \(x = 0.4\text{ m}\) (or \(\frac{2}{5}\text{ m}\)) and rejects the negative root.

(c)
B1: Calculates tension \(T = 30\text{ N}\) and obtains magnitude of acceleration \(a = 50\text{ m s}^{-2}\).

(a) Taking horizontal and vertical axes through \(O\):
Horizontal displacement: \(x = (20\cos 60^\circ)t = 10t\)
Vertical displacement: \(y = (20\sin 60^\circ)t - \frac{1}{2}gt^2 = 10\sqrt{3}t - 5t^2\)

At the point \(A\) on the inclined plane, \(y = x\tan 30^\circ = \frac{x}{\sqrt{3}}\).
Substituting for \(x\) and \(y\):
\[10\sqrt{3}t - 5t^2 = \frac{10t}{\sqrt{3}}\]
Since \(t \neq 0\):
\[5t = 10\sqrt{3} - \frac{10}{\sqrt{3}} = \frac{30 - 10}{\sqrt{3}} = \frac{20}{\sqrt{3}}\]
\[t = \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3}\text{ s}\]

(b) The horizontal distance to \(A\) is:
\[x = 10\left(\frac{4\sqrt{3}}{3}\right) = \frac{40\sqrt{3}}{3}\text{ m}\]
The distance along the inclined plane is:
\[OA = \frac{x}{\cos 30^\circ} = \frac{\frac{40\sqrt{3}}{3}}{\frac{\sqrt{3}}{2}} = \frac{80}{3}\text{ m} \approx 26.7\text{ m}\]

(c) The velocity components at time \(t = \frac{4\sqrt{3}}{3}\text{ s}\) are:
\[v_x = 20\cos 60^\circ = 10\text{ m s}^{-1}\]
\[v_y = 20\sin 60^\circ - gt = 10\sqrt{3} - 10\left(\frac{4\sqrt{3}}{3}\right) = -\frac{10\sqrt{3}}{3}\text{ m s}^{-1}\]

The speed \(v\) of \(P\) is:
\[v = \sqrt{v_x^2 + v_y^2} = \sqrt{10^2 + \left(-\frac{10\sqrt{3}}{3}\right)^2} = \sqrt{100 + \frac{100}{3}} = \sqrt{\frac{400}{3}} = \frac{20\sqrt{3}}{3}\text{ m s}^{-1} \approx 11.5\text{ m s}^{-1}\]

Let \(\theta\) be the acute angle the velocity vector makes below the horizontal:
\[\tan\theta = \frac{|v_y|}{v_x} = \frac{\frac{10\sqrt{3}}{3}}{10} = \frac{1}{\sqrt{3}} \implies \theta = 30^\circ\]
Since the inclined plane makes an angle of \(30^\circ\) above the horizontal and the velocity vector is directed at \(30^\circ\) below the horizontal, the angle between the direction of motion and the inclined plane is:
\[30^\circ + 30^\circ = 60^\circ\]

(a)
M1: Sets up parametric equations for horizontal and vertical displacements, or resolves components of motion parallel and perpendicular to the inclined plane.
M1: Uses the landing condition \(y = x\tan 30^\circ\) (or perpendicular displacement \(= 0\)) to set up an equation in \(t\).
A1 (AG): Obtains \(t = \frac{4\sqrt{3}}{3}\text{ s}\) correctly with full and sufficient intermediate working shown.

(b)
B1: Correct distance \(OA = \frac{80}{3}\text{ m}\) or \(26.7\text{ m}\).

(c)
B1: Correct speed \(v = \frac{20\sqrt{3}}{3}\text{ m s}^{-1}\) or \(11.5\text{ m s}^{-1}\).
B1: Correct angle \(60^\circ\) with the inclined plane.

(a)
The population distribution of differences must be symmetrical (about the median).

(b)
Let \(m_D\) denote the population median of the differences \(D = \text{Before} - \text{After}\).

Null hypothesis: \(H_0: m_D = 0\)
Alternative hypothesis: \(H_1: m_D > 0\)

Calculate the differences \(d = \text{Before} - \text{After}\) and their signs for each station:

- Station \(A\): \(42.1 - 38.4 = +3.7\) \((+)\)
- Station \(B\): \(38.5 - 39.1 = -0.6\) \((-)\)
- Station \(C\): \(55.0 - 49.7 = +5.3\) \((+)\)
- Station \(D\): \(48.2 - 43.0 = +5.2\) \((+)\)
- Station \(E\): \(36.4 - 37.2 = -0.8\) \((-)\)
- Station \(F\): \(61.3 - 54.6 = +6.7\) \((+)\)
- Station \(G\): \(44.8 - 41.5 = +3.3\) \((+)\)
- Station \(H\): \(51.6 - 46.8 = +4.8\) \((+)\)
- Station \(I\): \(39.0 - 36.1 = +2.9\) \((+)\)
- Station \(J\): \(47.5 - 42.1 = +5.4\) \((+)\)
- Station \(K\): \(53.2 - 48.9 = +4.3\) \((+)\)

Number of positive signs \((+)\) is \(9\).
Number of negative signs \((-)\) is \(2\).
Total number of non-zero differences \(n = 11\).

Under \(H_0\), the number of negative differences \(X \sim \text{B}(11, 0.5)\).

We calculate the one-tailed \(p\)-value:
\[ P(X \le 2) = P(X = 0) + P(X = 1) + P(X = 2) \]
\[ P(X \le 2) = \left(\frac{1}{2}\right)^{11} \left[ \binom{11}{0} + \binom{11}{1} + \binom{11}{2} \right] \]
\[ P(X \le 2) = \frac{1}{2048} [1 + 11 + 55] = \frac{67}{2048} \approx 0.0327 \]

Comparison:
Since \(0.0327 < 0.05\) (or \(p < 0.05\)), we reject \(H_0\).

Conclusion:
There is sufficient evidence at the \(5\%\) significance level to suggest that the traffic management scheme has resulted in a reduction in the median \(\text{PM}_{2.5}\) concentration.

(a)
B1: State that the distribution of differences is symmetrical (or symmetric).

(b)
B1: Correctly state both null and alternative hypotheses defining the parameter (e.g. \(H_0: m_D = 0\) versus \(H_1: m_D > 0\), or \(p = 0.5\) versus \(p > 0.5\) / \(p < 0.5\) where \(m_D\) is median of differences).
M1: Determine the correct signs of differences (9 positive, 2 negative).
M1: State or use binomial model \(X \sim \text{B}(11, 0.5)\) and identify the relevant tail probability \(P(X \le 2)\) or \(P(X \ge 9)\).
A1: Calculate correct probability \(P(X \le 2) = \frac{67}{2048} \approx 0.0327\) (accept 0.033).
M1: Valid comparison of their calculated \(p\)-value with \(0.05\) (e.g. \(0.0327 < 0.05\)).
A1: Correct contextual conclusion in non-assertive language (FT their \(p\)-value comparison).

(a)
\(H_0\): There is no association between an employee's age group and their primary mode of transport to work (i.e. age group and mode of transport are independent).
\(H_1\): There is an association between an employee's age group and their primary mode of transport to work (i.e. age group and mode of transport are not independent).

(b)
Row totals:
- Under 30: \(28 + 18 + 14 = 60\)
- 30 to 50: \(24 + 44 + 12 = 80\)
- Over 50: \(18 + 28 + 14 = 60\)
- Total \(= 200\)

Column totals:
- Public transport: \(28 + 24 + 18 = 70\)
- Private car: \(18 + 44 + 28 = 90\)
- Active: \(14 + 12 + 14 = 40\)
- Total \(= 200\)

Expected frequency formula: \(E = \dfrac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}} Expected frequencies table: \[ \begin{array}{|l|c|c|c|} \hline \text{Age group} & \text{Public transport} & \text{Private car} & \text{Active} \\ \hline \text{Under } 30 & \frac{60 \times 70}{200} = 21 & \frac{60 \times 90}{200} = 27 & \frac{60 \times 40}{200} = 12 \\ 30\text{ to } 50 & \frac{80 \times 70}{200} = 28 & \frac{80 \times 90}{200} = 36 & \frac{80 \times 40}{200} = 16 \\ \text{Over } 50 & \frac{60 \times 70}{200} = 21 & \frac{60 \times 90}{200} = 27 & \frac{60 \times 40}{200} = 12 \\ \hline \end{array} \] All expected values are \)\ge 5\), so no pooling of cells is required.

(c)
Test statistic:
\[ \chi^2 = \sum \frac{(O - E)^2}{E} \]
Contributions for each cell:
- Under 30:
- Public: \(\dfrac{(28 - 21)^2}{21} = \dfrac{49}{21} = \dfrac{7}{3} \approx 2.3333\)
- Car: \(\dfrac{(18 - 27)^2}{27} = \dfrac{81}{27} = 3\)
- Active: \(\dfrac{(14 - 12)^2}{12} = \dfrac{4}{12} = \dfrac{1}{3} \approx 0.3333\)
- Row 1 sum \(= \dfrac{17}{3} \approx 5.6667\)

- 30 to 50:
- Public: \(\dfrac{(24 - 28)^2}{28} = \dfrac{16}{28} = \dfrac{4}{7} \approx 0.5714\)
- Car: \(\dfrac{(44 - 36)^2}{36} = \dfrac{64}{36} = \dfrac{16}{9} \approx 1.7778\)
- Active: \(\dfrac{(12 - 16)^2}{16} = \dfrac{16}{16} = 1\)
- Row 2 sum \(= \dfrac{211}{63} \approx 3.3492\)

- Over 50:
- Public: \(\dfrac{(18 - 21)^2}{21} = \dfrac{9}{21} = \dfrac{3}{7} \approx 0.4286\)
- Car: \(\dfrac{(28 - 27)^2}{27} = \dfrac{1}{27} \approx 0.0370\)
- Active: \(\dfrac{(14 - 12)^2}{12} = \dfrac{4}{12} = \dfrac{1}{3} \approx 0.3333\)
- Row 3 sum \(= \dfrac{151}{189} \approx 0.7989\)

Total test statistic:
\[ \chi^2 = \frac{17}{3} + \frac{211}{63} + \frac{151}{189} = \frac{1855}{189} = \frac{265}{27} \approx 9.815 \text{ (or } 9.81\text{ to 3 s.f.)} \]

Degrees of freedom: \(\nu = (r - 1)(c - 1) = (3 - 1)(3 - 1) = 4\).
Critical value at the 5% significance level: \(\chi^2_{4, 0.05} = 9.488\).

Since \(9.815 > 9.488\) (or \(p\text{-value} \approx 0.0436 < 0.05\)), we reject \(H_0\).
There is sufficient evidence at the 5% significance level to suggest that there is an association between employees' age group and their primary mode of transport to work.

(d)
The 'Under 30' age group contributes \(5.67\) (over half of the total \(\chi^2\) value of \(9.815\)), which is the largest contribution.
In this group, considerably more employees use public transport than expected (28 observed vs 21 expected) and fewer use a private car than expected (18 observed vs 27 expected).

(a)
B1: Both hypotheses correctly stated in context (mentioning age group and transport method/mode).

(b)
M1: Correct method shown for calculating expected frequencies (at least two correct products/ratios shown).
A1: All 9 expected frequencies correct (21, 27, 12; 28, 36, 16; 21, 27, 12).

(c)
M1: Correct formula for \(\chi^2\) used, showing at least 3 terms evaluated.
A1: Correct individual contributions calculated (at least 6 correct terms or correct row sums).
A1: Test statistic \(\chi^2 = 9.81\) or \(9.82\) or \(\frac{265}{27}\).
B1: Degrees of freedom \(\nu = 4\) stated or clearly used.
B1: Critical value \(9.488\) correctly stated from MF19 tables.
A1: Correct comparison (\(9.815 > 9.488\)) and correct contextual conclusion in non-assertive language (FT on their \(\chi^2\) and critical value).

(d)
B1: Correctly identifies 'Under 30' age group and makes a correct qualitative remark (e.g. higher public transport use / lower car use than expected).

(a) The required assumptions are:
1. The two parent populations are normally distributed.
2. The two parent populations have equal (common) variances.

(b) First, calculate the sample means:
\[ \bar{x} = \frac{142.0}{10} = 14.2 \]
\[ \bar{y} = \frac{98.4}{8} = 12.3 \]

Calculate the sum of squares of deviations for each sample:
\[ \sum (x - \bar{x})^2 = \sum x^2 - \frac{(\sum x)^2}{n_A} = 2038.36 - \frac{142.0^2}{10} = 2038.36 - 2016.40 = 21.96 \]
\[ \sum (y - \bar{y})^2 = \sum y^2 - \frac{(\sum y)^2}{n_B} = 1228.38 - \frac{98.4^2}{8} = 1228.38 - 1210.32 = 18.06 \]

Calculate the pooled estimate of the common variance, \(s_p^2\):
\[ s_p^2 = \frac{\sum (x - \bar{x})^2 + \sum (y - \bar{y})^2}{n_A + n_B - 2} = \frac{21.96 + 18.06}{10 + 8 - 2} = \frac{40.02}{16} = 2.50125 \]

The degrees of freedom is \(\nu = 10 + 8 - 2 = 16\).
For a 95\% confidence interval with 16 degrees of freedom, the critical value is:
\[ t_{16, 0.025} = 2.120 \]

The standard error of the difference between the sample means is:
\[ \text{SE} = \sqrt{s_p^2\left(\frac{1}{n_A} + \frac{1}{n_B}\right)} = \sqrt{2.50125\left(\frac{1}{10} + \frac{1}{8}\right)} = \sqrt{2.50125 \times 0.225} = \sqrt{0.56278125} \approx 0.750187 \]

The margin of error is:
\[ t \times \text{SE} = 2.120 \times 0.750187 \approx 1.5904 \]

Hence, the 95\% confidence interval for \(\mu_A - \mu_B\) is:
\[ (\bar{x} - \bar{y}) \pm 1.5904 = (14.2 - 12.3) \pm 1.5904 = 1.9 \pm 1.5904 \]
\[ [0.310, 3.49] \quad \text{or} \quad 0.310 < \mu_A - \mu_B < 3.49 \]

(a)
B1: States that both populations (or distributions of yield) are normally distributed.
B1: States that the two populations have equal (or common) variances / standard deviations.

(b)
M1: Correct method for finding the pooled sample variance \(s_p^2\).
A1: Obtains \(s_p^2 = 2.50125\) (or \(\frac{40.02}{16}\)) or standard error \(\approx 0.7502\).
B1: Uses correct critical value \(t = 2.120\) (from \(t_{16}\)).
A1: Correct 95% confidence interval \([0.310, 3.49]\) or \(0.310 < \mu_A - \mu_B < 3.49\) (to 3 s.f.).

(a)
For \(x < 0\), \(\mathrm{F}(x) = 0\).

For \(0 \le x \le 2\):
\[ \mathrm{F}(x) = \int_0^x \frac{1}{4}t \, \mathrm{d}t = \left[ \frac{1}{8}t^2 \right]_0^x = \frac{1}{8}x^2. \]

For \(2 < x \le 4\):
\[ \mathrm{F}(x) = \mathrm{F}(2) + \int_2^x \frac{2}{t^2} \, \mathrm{d}t = \frac{1}{8}(2^2) + \left[ -\frac{2}{t} \right]_2^x = \frac{1}{2} + \left( -\frac{2}{x} - (-1) \right) = \frac{3}{2} - \frac{2}{x}. \]

For \(x > 4\), \(\mathrm{F}(x) = 1\).

Hence, the cumulative distribution function is:
\[ \mathrm{F}(x) = \begin{cases} 0 & x < 0, \\ \frac{1}{8}x^2 & 0 \le x \le 2, \\ \frac{3}{2} - \frac{2}{x} & 2 < x \le 4, \\ 1 & x > 4. \end{cases} \]

(b)
\[ \mathrm{E}(X) = \int_0^2 x\left(\frac{1}{4}x\right) \mathrm{d}x + \int_2^4 x\left(\frac{2}{x^2}\right) \mathrm{d}x \]
\[ = \int_0^2 \frac{1}{4}x^2 \, \mathrm{d}x + \int_2^4 \frac{2}{x} \, \mathrm{d}x \]
\[ = \left[ \frac{1}{12}x^3 \right]_0^2 + \left[ 2\ln x \right]_2^4 \]
\[ = \frac{8}{12} + 2(\ln 4 - \ln 2) = \frac{2}{3} + 2\ln 2 \quad (\text{or } \frac{2}{3} + \ln 4). \]

(c)
Since \(0 < X \le 4\), \(Y = \frac{1}{\sqrt{X}} \ge \frac{1}{\sqrt{4}} = \frac{1}{2}\).

The CDF of \(Y\), \(\mathrm{G}(y)\), is:
\[ \mathrm{G}(y) = \mathrm{P}(Y \le y) = \mathrm{P}\left(\frac{1}{\sqrt{X}} \le y\right) = \mathrm{P}\left(\sqrt{X} \ge \frac{1}{y}\right) = \mathrm{P}\left(X \ge \frac{1}{y^2}\right) = 1 - \mathrm{F}_X\left(\frac{1}{y^2}\right). \]

- When \(2 < X \le 4 \iff \frac{1}{2} \le y < \frac{1}{\sqrt{2}}\):
\[ \mathrm{G}(y) = 1 - \left( \frac{3}{2} - \frac{2}{1/y^2} \right) = 1 - \frac{3}{2} + 2y^2 = 2y^2 - \frac{1}{2}. \]
Differentiating with respect to \(y\):
\[ \mathrm{g}(y) = \mathrm{G}'(y) = 4y. \]

- When \(0 < X \le 2 \iff y \ge \frac{1}{\sqrt{2}}\):
\[ \mathrm{G}(y) = 1 - \frac{1}{8}\left(\frac{1}{y^2}\right)^2 = 1 - \frac{1}{8y^4}. \]
Differentiating with respect to \(y\):
\[ \mathrm{g}(y) = \mathrm{G}'(y) = -\left(-\frac{4}{8y^5}\right) = \frac{1}{2y^5}. \]

Therefore, the probability density function of \(Y\) is:
\[ \mathrm{g}(y) = \begin{cases} 4y & \frac{1}{2} \le y < \frac{1}{\sqrt{2}}, \\ \frac{1}{2y^5} & y \ge \frac{1}{\sqrt{2}}, \\ 0 & \text{otherwise.} \end{cases} \]

(a)
* M1: For integrating \(\frac{1}{4}x\) to find \(\mathrm{F}(x)\) for the interval \(0 \le x \le 2\).
* A1: For \(\frac{1}{8}x^2\) (with limits \(0\) to \(x\)).
* M1: For integrating \(\frac{2}{t^2}\) and correctly applying \(\mathrm{F}(2) = \frac{1}{2}\) or limits \(2\) to \(x\).
* A1: Fully correct piecewise function \(\mathrm{F}(x)\) clearly stating all intervals including \(x < 0\) and \(x > 4\).

(b)
* M1: For applying \(\int x \mathrm{f}(x) \, \mathrm{d}x\) across both non-zero intervals with correct limits.
* A1: For obtaining the correct exact value \(\frac{2}{3} + 2\ln 2\) (or \(\frac{2}{3} + \ln 4\)).

(c)
* M1: For using \(\mathrm{P}(Y \le y) = 1 - \mathrm{F}_X(y^{-2})\) or applying the transformation formula \(g(y) = f(x) |\frac{\mathrm{d}x}{\mathrm{d}y}|\).
* A1: For obtaining either correct branch of \(\mathrm{g}(y)\) with its respective domain.
* A1: For the fully correct PDF with all domains correctly identified: \(\mathrm{g}(y) = 4y\) for \(\frac{1}{2} \le y < \frac{1}{\sqrt{2}}\) and \(\mathrm{g}(y) = \frac{1}{2y^5}\) for \(y \ge \frac{1}{\sqrt{2}}\), \(0\) otherwise.

(a) The differences in recovery times (Before $-$ After) are normally distributed.

(b) Let $d = x - y$ represent the difference in recovery time for an individual athlete (Before $-$ After), and let $\mu_D$ be the population mean difference.

State the hypotheses:
$$H_0: \mu_D = 0$$
$$H_1: \mu_D > 0$$

Calculate the differences $d_i = x_i - y_i$:
$$\begin{array}{|l|c|c|c|c|c|c|c|}
\hline
\text{Athlete} & A & B & C & D & E & F & G \\
\hline
d & 2.4 & -0.8 & 1.5 & 3.1 & 0.6 & 2.0 & 1.7 \\
\hline
\end{array}$$

Summary statistics:
$$\sum d = 2.4 - 0.8 + 1.5 + 3.1 + 0.6 + 2.0 + 1.7 = 10.5$$
$$\bar{d} = \frac{10.5}{7} = 1.5$$

$$\sum d^2 = 2.4^2 + (-0.8)^2 + 1.5^2 + 3.1^2 + 0.6^2 + 2.0^2 + 1.7^2 = 25.51$$

Calculate the unbiased estimate of the population variance $s^2$:
$$s^2 = \frac{1}{n-1} \left( \sum d^2 - \frac{(\sum d)^2}{n} \right) = \frac{1}{6} \left( 25.51 - \frac{10.5^2}{7} \right) = \frac{1}{6} (25.51 - 15.75) = \frac{9.76}{6} \approx 1.6267$$
$$s = \sqrt{1.6267} \approx 1.2754$$

Calculate the test statistic $t$:
$$t = \frac{\bar{d} - 0}{s / \sqrt{n}} = \frac{1.5}{\sqrt{1.6267 / 7}} = \frac{1.5}{0.48206} \approx 3.112$$

Degrees of freedom $
u = 7 - 1 = 6$.

From tables of the $t$-distribution, the critical value for a one-tailed test at the $5\%$ significance level with $6$ degrees of freedom is:
$$t_{\text{crit}} = 1.943$$

Since $t = 3.112 > 1.943$, the test statistic falls in the critical region. Reject $H_0$.

There is sufficient evidence at the $5\%$ significance level to conclude that the training course reduces the mean recovery time of athletes.

(a)
B1: State that the population of differences is normally distributed (allow: recovery times before and after have a bivariate normal distribution).

(b)
B1: State correct hypotheses with parameter defined (e.g., $H_0: \mu_D = 0, H_1: \mu_D > 0$, where $\mu_D$ is the population mean difference).
M1: Attempt to calculate differences and obtain sample mean $\bar{d} = 1.5$.
M1: Correct method for finding the unbiased sample variance $s^2$ (or standard deviation $s$).
A1: Correct value of $s^2 = \frac{9.76}{6} \approx 1.63$ or $s \approx 1.28$.
M1: Calculate test statistic using $t = \frac{\bar{d}}{s/\sqrt{7}}$.
A1: Obtain $t = 3.11$ (accept $3.11$ to $3.12$, or $-3.11$ if defined as After $-$ Before with $H_1: \mu_D < 0$).
A1: Correct critical value $1.943$ (or $-1.943$), correct comparison, and a correctly phrased non-assertive conclusion in context.

(a) Let $\mathrm{P}(X = r) = kr$ for $r = 1, 2, 3, 4$.
Since the sum of probabilities must equal 1:
\[ \sum_{r=1}^{4} kr = k(1 + 2 + 3 + 4) = 10k = 1 \implies k = \frac{1}{10}. \]
Thus, $\mathrm{P}(X = r) = \frac{r}{10}$ for $r = 1, 2, 3, 4$.
The probability generating function of $X$ is:
\[ G_X(t) = \sum_{r=1}^{4} \mathrm{P}(X = r)t^r = \frac{1}{10}t + \frac{2}{10}t^2 + \frac{3}{10}t^3 + \frac{4}{10}t^4 = \frac{1}{10}t(1 + 2t + 3t^2 + 4t^3). \]

(b) Differentiating $G_X(t) = \frac{1}{10}(t + 2t^2 + 3t^3 + 4t^4)$ with respect to $t$:
\[ G'_X(t) = \frac{1}{10}(1 + 4t + 9t^2 + 16t^3). \]
Evaluating at $t = 1$:
\[ \mathrm{E}(X) = G'_X(1) = \frac{1}{10}(1 + 4 + 9 + 16) = \frac{30}{10} = 3. \]

Differentiating again:
\[ G''_X(t) = \frac{1}{10}(4 + 18t + 48t^2). \]
Evaluating at $t = 1$:
\[ G''_X(1) = \frac{1}{10}(4 + 18 + 48) = \frac{70}{10} = 7. \]

Using the formula for variance:
\[ \mathrm{Var}(X) = G''_X(1) + G'_X(1) - [G'_X(1)]^2 = 7 + 3 - 3^2 = 10 - 9 = 1. \]

(c)(i) Since $S = X_1 + X_2 + X_3$, where $X_1, X_2, X_3$ are independent and identically distributed as $X$:
\[ G_S(t) = [G_X(t)]^3 = \frac{1}{1000}t^3(1 + 2t + 3t^2 + 4t^3)^3. \]

(ii) The probability $\mathrm{P}(S = 6)$ is the coefficient of $t^6$ in $G_S(t)$.
This is equivalent to finding $\frac{1}{1000} \times (\text{coefficient of } t^3 \text{ in } (1 + 2t + 3t^2 + 4t^3)^3)$.

Expanding $(1 + (2t + 3t^2 + 4t^3))^3$:
\[ (1 + u)^3 = 1 + 3u + 3u^2 + u^3 \]
where $u = 2t + 3t^2 + 4t^3$.
- The $t^3$ term from $3u$ is $3(4t^3) = 12t^3$.
- The $t^3$ term from $3u^2 = 3(2t + 3t^2 + 4t^3)^2$ is $3 \times 2(2t)(3t^2) = 36t^3$.
- The $t^3$ term from $u^3 = (2t + 3t^2 + 4t^3)^3$ is $(2t)^3 = 8t^3$.

Sum of coefficients of $t^3 = 12 + 36 + 8 = 56$.

Hence,
\[ \mathrm{P}(S = 6) = \frac{56}{1000} = \frac{7}{125} = 0.056. \]

(a)
M1: Sets sum of probabilities $k(1+2+3+4)=1$ to find $k = \frac{1}{10}$.
A1: AG. Correctly substitutes probabilities into definition $\sum \mathrm{P}(X=r)t^r$ and factors out $\frac{1}{10}t$ with no missing steps.

(b)
M1: Differentiates $G_X(t)$ correctly to find $G'_X(t)$ and evaluates at $t=1$.
A1: Obtains $\mathrm{E}(X) = 3$.
M1: Differentiates again to find $G''_X(t)$, evaluates at $t=1$, and correctly uses $\mathrm{Var}(X) = G''_X(1) + G'_X(1) - [G'_X(1)]^2$.
A1: Obtains $\mathrm{Var}(X) = 1$.

(c)(i)
B1: States $[G_X(t)]^3$ or equivalent expression.

(c)(ii)
M1: Identifies $\mathrm{P}(S=6)$ as the coefficient of $t^6$ in $G_S(t)$ (or coefficient of $t^3$ in the cubed bracket) or identifies all valid partitions of 6 into 3 parts from {1, 2, 3, 4}.
M1: Valid algebraic expansion or probabilistic summation method to collect all terms corresponding to $t^6$.
A1: Obtains $\frac{7}{125}$ or $\frac{56}{1000}$ or $0.056$.