2025 Cambridge IAL Mathematics - Further (9231) Practice Paper with Answers

(i) From the relation between roots and coefficients:
\[ \sum \alpha = \frac{3}{2} \quad \text{and} \quad \sum \alpha\beta = 2 \]
We know that:
\[ \alpha^2 + \beta^2 + \gamma^2 = \left(\sum \alpha\right)^2 - 2\sum \alpha\beta = \left(\frac{3}{2}\right)^2 - 2(2) = \frac{9}{4} - 4 = -\frac{7}{4} \]

(ii) Since \(\alpha, \beta, \gamma\) satisfy the original cubic equation, we have:
\[ 2\alpha^3 - 3\alpha^2 + 4\alpha - 5 = 0 \]
Summing this identity for \(\alpha, \beta, \gamma\) yields:
\[ 2\sum \alpha^3 - 3\sum \alpha^2 + 4\sum \alpha - 15 = 0 \]
Substitute the known values of \(\sum \alpha^2\) and \(\sum \alpha\):
\[ 2\sum \alpha^3 - 3\left(-\frac{7}{4}\right) + 4\left(\frac{3}{2}\right) - 15 = 0 \]
\[ 2\sum \alpha^3 + \frac{21}{4} + 6 - 15 = 0 \]
\[ 2\sum \alpha^3 = 9 - \frac{21}{4} = \frac{15}{4} \implies \sum \alpha^3 = \frac{15}{8} \]

(iii) Let \(w = x^2\), which means \(x = w^{1/2}\). Rearranging the original cubic equation to isolate odd powers of \(x\):
\[ 2x^3 + 4x = 3x^2 + 5 \]
\[ x(2x^2 + 4) = 3x^2 + 5 \]
Substituting \(x = w^{1/2}\):
\[ w^{1/2}(2w + 4) = 3w + 5 \]
Squaring both sides:
\[ w(2w + 4)^2 = (3w + 5)^2 \]
\[ w(4w^2 + 16w + 16) = 9w^2 + 30w + 25 \]
\[ 4w^3 + 16w^2 + 16w = 9w^2 + 30w + 25 \]
\[ 4w^3 + 7w^2 - 14w - 25 = 0 \]

(i)
M1: For using \((\sum \alpha)^2 - 2\sum \alpha\beta\).
A1: Correct substitution of values.
A1: Correct simplification to show \(-\frac{7}{4}\).

(ii)
M1: Establish the relation \(2\sum \alpha^3 - 3\sum \alpha^2 + 4\sum \alpha - 15 = 0\).
A1: Substitute known values correctly.
A1: Correctly solve to obtain \(\sum \alpha^3 = \frac{15}{8}\).

(iii)
M1: Express the original cubic with odd powers isolated.
M1: Substitute \(x = w^{1/2}\) and square both sides.
A1: Correct expanded equation before simplification.
A1: Final correct cubic equation with integer coefficients.

(i) Vertical asymptotes occur where the denominator is zero:
\[ x^2 - 4 = 0 \implies x = 2 \quad \text{and} \quad x = -2 \]
Horizontal asymptotes are found by taking the limit as \(x \to \pm\infty\):
\[ y = \lim_{x \to \pm\infty} \frac{3x^2 + 2x - 1}{x^2 - 4} = 3 \]
So the asymptotes are \(x = 2\), \(x = -2\), and \(y = 3\).

(ii) Rearranging the curve equation to find the range of \(y\):
\[ y(x^2 - 4) = 3x^2 + 2x - 1 \implies (y-3)x^2 - 2x - (4y - 1) = 0 \]
Since \(x\) is real, the discriminant of this quadratic in \(x\) must be non-negative (\(\Delta \ge 0\)):
\[ \Delta = (-2)^2 - 4(y-3)(-(4y-1)) \ge 0 \]
\[ 4 + 4(y-3)(4y-1) \ge 0 \]
\[ 1 + 4y^2 - 13y + 3 \ge 0 \implies 4y^2 - 13y + 4 \ge 0 \]
The roots of \(4y^2 - 13y + 4 = 0\) are:
\[ y = \frac{13 \pm \sqrt{13^2 - 4(4)(4)}}{8} = \frac{13 \pm \sqrt{105}}{8} \]
The inequality is satisfied when \(y \le \frac{13 - \sqrt{105}}{8}\) or \(y \ge \frac{13 + \sqrt{105}}{8}\).
Thus, the forbidden interval for \(y\) is \(\left(\frac{13 - \sqrt{105}}{8}, \frac{13 + \sqrt{105}}{8}\right)\).
The length of this interval is:
\[ L = \frac{13 + \sqrt{105}}{8} - \frac{13 - \sqrt{105}}{8} = \frac{2\sqrt{105}}{8} = \frac{\sqrt{105}}{4} \]

(iii) Intercepts:
With \(y\)-axis: when \(x = 0\), \(y = \frac{-1}{-4} = \frac{1}{4}\).
With \(x\)-axis: when \(y = 0\), \(3x^2 + 2x - 1 = 0 \implies (3x-1)(x+1) = 0 \implies x = \frac{1}{3}\) or \(x = -1\).
Intersection with the asymptote \(y = 3\):
\[ 3 = \frac{3x^2 + 2x - 1}{x^2 - 4} \implies 3x^2 - 12 = 3x^2 + 2x - 1 \implies 2x = -11 \implies x = -5.5 \]
The sketch consists of three branches with vertical asymptotes \(x = \pm 2\) and horizontal asymptote \(y = 3\). The left branch crosses the horizontal asymptote at \((-5.5, 3)\). The middle branch contains the intercepts \((-1, 0)\), \((0, 0.25)\), and \((1/3, 0)\).

(i)
M1: Equating denominator to zero to find vertical asymptotes.
A1: Correct equations for vertical asymptotes: \(x = 2\) and \(x = -2\).
A1: Correct horizontal asymptote \(y = 3\).

(ii)
M1: Rearrange to form a quadratic in \(x\).
M1: Apply condition for real roots (\(\Delta \ge 0\)).
A1: Correct quadratic inequality in \(y\): \(4y^2 - 13y + 4 \ge 0\).
A1: Identify critical values of \(y\).
A1: Compute difference to show \(L = \frac{\sqrt{105}}{4}\).

(iii)
B1: Sketch showing three branches with correct asymptotic behavior.
B1: Identify coordinate axis intercepts correctly on the graph.
B1: Identify intersection with the asymptote at \((-5.5, 3)\).

(i) Putting the terms on the right-hand side over a common denominator:
\[ \frac{2}{r} - \frac{4}{r+1} + \frac{2}{r+2} = 2 \left( \frac{1}{r} - \frac{2}{r+1} + \frac{1}{r+2} \right) \]
\[ = 2 \left( \frac{(r+1)(r+2) - 2r(r+2) + r(r+1)}{r(r+1)(r+2)} \right) \]
\[ = 2 \left( \frac{r^2 + 3r + 2 - 2r^2 - 4r + r^2 + r}{r(r+1)(r+2)} \right) \]
\[ = 2 \left( \frac{2}{r(r+1)(r+2)} \right) = \frac{4}{r(r+1)(r+2)} \]

(ii) Let \(V_r = \frac{2}{r} - \frac{2}{r+1}\). Then:
\[ V_r - V_{r+1} = \left( \frac{2}{r} - \frac{2}{r+1} \right) - \left( \frac{2}{r+1} - \frac{2}{r+2} \right) = \frac{2}{r} - \frac{4}{r+1} + \frac{2}{r+2} \]
So the general term is \(u_r = V_r - V_{r+1}\).
The sum \(\sum_{r=1}^n u_r\) is a telescoping series:
\[ \sum_{r=1}^n u_r = (V_1 - V_2) + (V_2 - V_3) + \dots + (V_n - V_{n+1}) = V_1 - V_{n+1} \]
\[ V_1 = \frac{2}{1} - \frac{2}{2} = 1 \]
\[ V_{n+1} = \frac{2}{n+1} - \frac{2}{n+2} = \frac{2(n+2) - 2(n+1)}{(n+1)(n+2)} = \frac{2}{(n+1)(n+2)} \]
Hence,
\[ \sum_{r=1}^{n} \frac{4}{r(r+1)(r+2)} = 1 - \frac{2}{(n+1)(n+2)} \]

(iii) The infinite sum starting from \(r=3\) is:
\[ \sum_{r=3}^{\infty} u_r = \lim_{N \to \infty} \sum_{r=3}^{N} (V_r - V_{r+1}) = V_3 - \lim_{N \to \infty} V_{N+1} \]
\[ V_3 = \frac{2}{3} - \frac{2}{4} = \frac{2}{3} - \frac{1}{2} = \frac{1}{6} \]
Since \(\lim_{N \to \infty} V_{N+1} = 0\), the sum is:
\[ \sum_{r=3}^{\infty} \frac{4}{r(r+1)(r+2)} = \frac{1}{6} \]

(i)
M1: Attempt to write the RHS with a common denominator.
A1: Correctly show that the numerator simplifies to 4.

(ii)
M1: Express terms in a difference form, e.g., \(V_r - V_{r+1}\).
A1: Write out the terms of the series showing cancellation of intermediate terms.
A1: Identify surviving terms as \(V_1 - V_{n+1}\).
A1: Correctly find \(V_1 = 1\).
A1: Correctly simplify \(V_{n+1}\) to \(\frac{2}{(n+1)(n+2)}\).

(iii)
M1: Recognize the relation between \(\sum_{r=3}^{\infty} u_r\) and the telescoping sum.
M1: Express the sum as \(\sum_{r=1}^{\infty} u_r - u_1 - u_2\) or as \(V_3\).
A1: Obtain the correct numerical terms to subtract or calculate \(V_3\).
A1: Deduce the final answer \(\frac{1}{6}\).

(i) For \(\mathbf{M}\) to be singular, its determinant must be zero:
\[ \det(\mathbf{M}) = a(a^2 - 1) - 1(a - 1) + 1(1 - a) \]
\[ = a(a-1)(a+1) - (a-1) - (a-1) \]
\[ = (a-1)[a(a+1) - 2] = (a-1)(a^2 + a - 2) \]
\[ = (a-1)^2(a+2) \]
Setting \(\det(\mathbf{M}) = 0\), we find \(a = 1\) or \(a = -2\).

(ii) When \(a = 2\):
\[ \mathbf{M} = \begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{pmatrix} \]
\[ \det(\mathbf{M}) = (2-1)^2(2+2) = 4 \]
Next, we calculate the cofactors of \(\mathbf{M}\):
\[ C_{11} = 2(2) - 1(1) = 3, \quad C_{12} = -(1(2) - 1(1)) = -1, \quad C_{13} = 1(1) - 2(1) = -1 \]
By symmetry:
\[ \mathbf{C} = \begin{pmatrix} 3 & -1 & -1 \\ -1 & 3 & -1 \\ -1 & -1 & 3 \end{pmatrix} \]
Since the cofactor matrix is symmetric, the transpose \(\mathbf{C}^T = \mathbf{C}\).
\[ \mathbf{M}^{-1} = \frac{1}{\det(\mathbf{M})} \mathbf{C}^T = \frac{1}{4} \begin{pmatrix} 3 & -1 & -1 \\ -1 & 3 & -1 \\ -1 & -1 & 3 \end{pmatrix} \]

(iii) The system can be written as \(\mathbf{M}\mathbf{X} = \mathbf{B}\), where:
\[ \mathbf{X} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \quad \text{and} \quad \mathbf{B} = \begin{pmatrix} 3 \\ 0 \\ 9 \end{pmatrix} \]
Thus:
\[ \mathbf{X} = \mathbf{M}^{-1} \mathbf{B} = \frac{1}{4} \begin{pmatrix} 3 & -1 & -1 \\ -1 & 3 & -1 \\ -1 & -1 & 3 \end{pmatrix} \begin{pmatrix} 3 \\ 0 \\ 9 \end{pmatrix} \]
\[ \mathbf{X} = \frac{1}{4} \begin{pmatrix} 3(3) - 1(0) - 1(9) \\ -1(3) + 3(0) - 1(9) \\ -1(3) - 1(0) + 3(9) \end{pmatrix} = \frac{1}{4} \begin{pmatrix} 0 \\ -12 \\ 24 \end{pmatrix} = \begin{pmatrix} 0 \\ -3 \\ 6 \end{pmatrix} \]
So \(x = 0, y = -3, z = 6\).

(i)
M1: Write down the correct determinant expansion.
A1: Factorise the determinant to get \((a-1)^2(a+2)\).
A1: Correct values of \(a = 1\) and \(a = -2\).

(ii)
M1: Evaluate \(\det(\mathbf{M}) = 4\) for \(a = 2\).
M1: Compute cofactor values (at least 3 correct).
A1: Correct cofactor matrix.
A1: Correctly obtain the transpose.
A1: Correct inverse matrix \(\mathbf{M}^{-1}\).

(iii)
M1: State and use the matrix multiplication \(\mathbf{X} = \mathbf{M}^{-1} \mathbf{B}\).
A1: Perform the multiplication correctly.
A1: Correct values \(x = 0\), \(y = -3\), \(z = 6\).

Let \(P(n)\) be the statement:
\[ \sum_{r=1}^n r \cdot 2^{r-1} = (n-1)2^n + 1 \]

**Base case:** For \(n = 1\):
\[ \text{LHS} = 1 \cdot 2^{1-1} = 1 \cdot 2^0 = 1 \]
\[ \text{RHS} = (1-1)2^1 + 1 = 0 + 1 = 1 \]
Since \(\text{LHS} = \text{RHS}\), \(P(1)\) is true.

**Inductive step:** Assume that \(P(k)\) is true for some positive integer \(k\), so:
\[ \sum_{r=1}^k r \cdot 2^{r-1} = (k-1)2^k + 1 \]
We want to show that \(P(k+1)\) is true, i.e.:
\[ \sum_{r=1}^{k+1} r \cdot 2^{r-1} = k \cdot 2^{k+1} + 1 \]
Starting with the left-hand side for \(n = k+1\):
\[ \sum_{r=1}^{k+1} r \cdot 2^{r-1} = \sum_{r=1}^k r \cdot 2^{r-1} + (k+1)2^k \]
Substitute the induction hypothesis:
\[ = (k-1)2^k + 1 + (k+1)2^k \]
Group the terms involving \(2^k\):
\[ = 2^k [ (k-1) + (k+1) ] + 1 \]
\[ = 2^k [ 2k ] + 1 \]
\[ = k \cdot 2^{k+1} + 1 \]
This matches the RHS for \(n = k+1\).

Thus, if \(P(k)\) is true, then \(P(k+1)\) is also true. Since \(P(1)\) is true, by mathematical induction, the statement is true for all positive integers \(n\).

B1: Verify base case \(n = 1\) showing LHS = RHS = 1.
B1: Clearly state inductive hypothesis \(P(k)\).
M1: Express sum of \(k+1\) terms as sum of \(k\) terms plus the \((k+1)\)-th term.
A1: Substitute the inductive hypothesis.
M1: Factorise and group terms with \(2^k\).
A1: Show correct intermediate step \(2^k [2k] + 1\).
A1: Obtain the correct RHS for \(n = k+1\).
B1: Clear concluding statement referencing the principle of mathematical induction.

(i) The curve \(r = a(2 + \cos \theta)\) is a dimpled limacon. It is symmetric about the initial line \(\theta = 0\).
- At \(\theta = 0\), \(r = 3a\).
- At \(\theta = \pm \pi/2\), \(r = 2a\).
- At \(\theta = \pm \pi\), \(r = a\).
The sketch is a single closed loop containing the origin, with no loop-within-a-loop since \(2 > 1\).

(ii) The area \(A\) is:
\[ A = \frac{1}{2} \int_{-\pi}^{\pi} r^2 \, \mathrm{d}\theta = \int_{0}^{\pi} a^2 (2 + \cos \theta)^2 \, \mathrm{d}\theta \]
\[ = a^2 \int_{0}^{\pi} (4 + 4\cos \theta + \cos^2 \theta) \, \mathrm{d}\theta \]
Using the double-angle identity \[ \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \]:
\[ = a^2 \int_{0}^{\pi} \left(4 + 4\cos \theta + \frac{1}{2} + \frac{1}{2}\cos 2\theta\right) \, \mathrm{d}\theta \]
\[ = a^2 \int_{0}^{\pi} \left(\frac{9}{2} + 4\cos \theta + \frac{1}{2}\cos 2\theta\right) \, \mathrm{d}\theta \]
\[ = a^2 \left[ \frac{9}{2}\theta + 4\sin \theta + \frac{1}{4}\sin 2\theta \right]_{0}^{\pi} \]
\[ = a^2 \left( \frac{9\pi}{2} + 0 + 0 - 0 \right) = \frac{9\pi a^2}{2} \]

(iii) When \(\theta = \frac{\pi}{3}\):
\[ r = a\left(2 + \cos \frac{\pi}{3}\right) = a\left(2 + \frac{1}{2}\right) = \frac{5}{2}a \]
Using conversion to Cartesian coordinates:
\[ x = r \cos \theta = \frac{5}{2}a \cos \frac{\pi}{3} = \frac{5}{2}a \left(\frac{1}{2}\right) = \frac{5}{4}a \]
\[ y = r \sin \theta = \frac{5}{2}a \sin \frac{\pi}{3} = \frac{5}{2}a \left(\frac{\sqrt{3}}{2}\right) = \frac{5\sqrt{3}}{4}a \]
So the coordinates are \(\left(\frac{5}{4}a, \frac{5\sqrt{3}}{4}a\right)\).

(i)
B1: Sketch of a closed loop symmetric about the initial line.
B1: Label correct coordinates or values of intercepts (\(a\), \(2a\), \(3a\)).

(ii)
M1: Use area formula \(\frac{1}{2}\int r^2 \mathrm{d}\theta\) with appropriate limits.
M1: Correctly expand and use \(\cos^2 \theta = \frac{1+\cos 2\theta}{2}\).
A1: Correct integration.
A1: Correct substitution of limits.
A1: Final correct area \(\frac{9\pi a^2}{2}\).

(iii)
M1: Calculate the value of \(r\) at \(\theta = \frac{\pi}{3}\).
M1: Apply formulas \(x = r\cos\theta\) and \(y = r\sin\theta\).
A1: Correct coordinates \(\left(\frac{5}{4}a, \frac{5\sqrt{3}}{4}a\right)\).

(i) Let \(\mathbf{d}_1 = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}\) and \(\mathbf{d}_2 = \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}\) be the direction vectors of \(l_1\) and \(l_2\).
A vector \(\mathbf{n}\) perpendicular to both lines is:
\[ \mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2 = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} \times \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} = \begin{pmatrix} (-1)(-1) - (2)(1) \\ (2)(2) - (1)(-1) \\ (1)(1) - (-1)(2) \end{pmatrix} = \begin{pmatrix} -1 \\ 5 \\ 3 \end{pmatrix} \]
The magnitude of \(\mathbf{n}\) is:
\[ |\mathbf{n}| = \sqrt{(-1)^2 + 5^2 + 3^2} = \sqrt{35} \]
Let \(A = (2, 1, -1)\) be a point on \(l_1\) and \(B = (1, 4, 2)\) be a point on \(l_2\). Then:
\[ \vec{AB} = \begin{pmatrix} 1 - 2 \\ 4 - 1 \\ 2 - (-1) \end{pmatrix} = \begin{pmatrix} -1 \\ 3 \\ 3 \end{pmatrix} \]
The shortest distance \(d\) is the projection of \(\vec{AB}\) on \(\mathbf{n}\):
\[ d = \frac{|\vec{AB} \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{|(-1)(-1) + 3(5) + 3(3)|}{\sqrt{35}} = \frac{1 + 15 + 9}{\sqrt{35}} = \frac{25}{\sqrt{35}} = \frac{5\sqrt{35}}{7} \approx 4.23 \]

(ii) The plane \(\Pi\) containing \(l_1\) and parallel to \(l_2\) must have normal vector \(\mathbf{n} = \begin{pmatrix} -1 \\ 5 \\ 3 \end{pmatrix}\).
Therefore, the equation of the plane is of the form:
\[ -x + 5y + 3z = D \]
Substituting the point \((2, 1, -1)\) on \(l_1\):
\[ D = -2 + 5(1) + 3(-1) = 0 \]
So the equation is \(-x + 5y + 3z = 0\), or equivalently:
\[ x - 5y - 3z = 0 \]

(i)
M1: Attempt to find cross product of direction vectors.
A1: Correct normal vector \(\begin{pmatrix} -1 \\ 5 \\ 3 \end{pmatrix}\).
M1: Find a vector connecting any point on \(l_1\) to any point on \(l_2\).
M1: Apply correct shortest distance projection formula.
A1: Evaluate numerator correctly to 25.
A1: Correct shortest distance \(\frac{5\sqrt{35}}{7}\) (or 3 s.f. equivalent 4.23).

(ii)
M1: Use normal vector \(\mathbf{n}\) found in (i).
M1: Set up the plane equation \(ax + by + cz = D\) and substitute a point on \(l_1\).
A1: Find \(D = 0\).
A1: Correct plane equation \(x - 5y - 3z = 0\).

1. Using the exponential definitions:
\(\cosh \theta = \frac{e^\theta + e^{-\theta}}{2}\) and \(\sinh \theta = \frac{e^\theta - e^{-\theta}}{2}\).

Then:
\(\cosh^2 \theta - \sinh^2 \theta = \left(\frac{e^\theta + e^{-\theta}}{2}\right)^2 - \left(\frac{e^\theta - e^{-\theta}}{2}\right)^2\)
\(= \frac{e^{2\theta} + 2 + e^{-2\theta}}{4} - \frac{e^{2\theta} - 2 + e^{-2\theta}}{4}\)
\(= \frac{4}{4} = 1\).

2. Substituting the exponential definitions into the given equation:
\(5 \left(\frac{e^{2x} + e^{-2x}}{2}\right) - 3 \left(\frac{e^{2x} - e^{-2x}}{2}\right) = 5\)

Multiply the entire equation by 2:
\(5(e^{2x} + e^{-2x}) - 3(e^{2x} - e^{-2x}) = 10\)
\(2e^{2x} + 8e^{-2x} = 10\)

Divide by 2:
\(e^{2x} + 4e^{-2x} = 5\)

Multiply by \(e^{2x}\):
\(e^{4x} - 5e^{2x} + 4 = 0\)

Let \(u = e^{2x}\):
\(u^2 - 5u + 4 = 0\)
\((u - 4)(u - 1) = 0\)

Thus, \(u = 4\) or \(u = 1\).

If \(e^{2x} = 4\):
\(2x = \ln 4 = 2\ln 2 \implies x = \ln 2\).

If \(e^{2x} = 1\):
\(2x = 0 \implies x = 0\).

The solutions in exact logarithmic form are \(x = 0\) and \(x = \ln 2\).

M1: Substituting exponential forms into the identity.
A1: Correct expansion and completion of proof.
M1: Substituting exponential forms into the equation.
A1: Simplifying to obtain a quadratic in \(e^{2x}\).
M1: Solving the quadratic equation for \(e^{2x}\).
A1: Finding \(x = 0\).
A1: Finding \(x = \ln 2\).

We have \(f(x) = \ln(1 + \sinh x)\).
At \(x = 0\), \(f(0) = \ln(1 + 0) = 0\).

Differentiating with respect to \(x\):
\(f'(x) = \frac{\cosh x}{1 + \sinh x}\)
At \(x = 0\), \(f'(0) = \frac{1}{1} = 1\).

Differentiating again:
\(f''(x) = \frac{\sinh x(1 + \sinh x) - \cosh^2 x}{(1 + \sinh x)^2} = \frac{\sinh x + \sinh^2 x - \cosh^2 x}{(1 + \sinh x)^2}\)
Since \(\cosh^2 x - \sinh^2 x = 1\), this simplifies to:
\(f''(x) = \frac{\sinh x - 1}{(1 + \sinh x)^2}\)
At \(x = 0\), \(f''(0) = \frac{-1}{1} = -1\).

Differentiating a third time:
\(f'''(x) = \frac{\cosh x(1 + \sinh x)^2 - 2(1+\sinh x)\cosh x(\sinh x - 1)}{(1 + \sinh x)^4} = \frac{\cosh x(1 + \sinh x) - 2\cosh x(\sinh x - 1)}{(1 + \sinh x)^3}\)
\(f'''(x) = \frac{3\cosh x - \cosh x\sinh x}{(1 + \sinh x)^3}\)
At \(x = 0\), \(f'''(0) = \frac{3(1) - 0}{1} = 3\).

The Maclaurin series is given by:
\(f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots\)
Substituting the values:
\(f(x) = x - \frac{1}{2}x^2 + \frac{3}{6}x^3 + \dots = x - \frac{1}{2}x^2 + \frac{1}{2}x^3 + \dots\)

The first three non-zero terms are \(x - \frac{1}{2}x^2 + \frac{1}{2}x^3\).

M1: Correctly differentiating to find \(f'(x)\) and evaluating at \(x = 0\).
A1: Finding \(f'(0) = 1\).
M1: Correctly differentiating to find \(f''(x)\) using quotient rule and evaluating at \(x = 0\).
A1: Finding \(f''(0) = -1\).
M1: Correctly differentiating to find \(f'''(x)\) and evaluating at \(x = 0\).
A1: Finding \(f'''(0) = 3\).
M1: Using the Maclaurin expansion formula with their derivative values.
A1: Correctly obtaining \(x - \frac{1}{2}x^2 + \frac{1}{2}x^3\).

1. Let \(x = a \tanh \theta\). Then \(dx = a \text{sech}^2 \theta \, d\theta\).
Substituting into the integral:
\(\int \frac{a \text{sech}^2 \theta}{a^2 - a^2 \tanh^2 \theta} \, d\theta = \int \frac{a \text{sech}^2 \theta}{a^2(1 - \tanh^2 \theta)} \, d\theta\)
Using the identity \(1 - \tanh^2 \theta = \text{sech}^2 \theta\):
\(\int \frac{a \text{sech}^2 \theta}{a^2 \text{sech}^2 \theta} \, d\theta = \frac{1}{a} \int 1 \, d\theta = \frac{1}{a} \theta + C\)

Since \(\tanh \theta = \frac{x}{a}\), we have:
\(\frac{e^\theta - e^{-\theta}}{e^\theta + e^{-\theta}} = \frac{x}{a} \implies \frac{e^{2\theta}-1}{e^{2\theta}+1} = \frac{x}{a}\)
\(a(e^{2\theta} - 1) = x(e^{2\theta} + 1) \implies e^{2\theta}(a-x) = a+x \implies e^{2\theta} = \frac{a+x}{a-x}\)
\(2\theta = \ln\left|\frac{a+x}{a-x}\right| \implies \theta = \frac{1}{2}\ln\left|\frac{a+x}{a-x}\right|\)
Thus, \(\int \frac{1}{a^2 - x^2} \, dx = \frac{1}{2a} \ln\left|\frac{a+x}{a-x}\right| + C\).

2. For the definite integral:
\(\int_{0}^{\sqrt{3}} \frac{1}{12 - 3x^2} \, dx = \frac{1}{3} \int_{0}^{\sqrt{3}} \frac{1}{4 - x^2} \, dx\)
Using the formula with \(a = 2\):
\(\frac{1}{3} \left[ \frac{1}{4} \ln\left|\frac{2+x}{2-x}\right| \right]_{0}^{\sqrt{3}} = \frac{1}{12} \left( \ln\left(\frac{2+\sqrt{3}}{2-\sqrt{3}}\right) - \ln(1) \right)\)
\(= \frac{1}{12} \ln\left( \frac{(2+\sqrt{3})^2}{(2-\sqrt{3})(2+\sqrt{3})} \right) = \frac{1}{12} \ln((2+\sqrt{3})^2)\)
\(= \frac{2}{12} \ln(2+\sqrt{3}) = \frac{1}{6} \ln(2+\sqrt{3})\).

M1: Using substitution \(x = a \tanh \theta\) and finding \(dx\).
A1: Simplifying the integrand using the hyperbolic identity.
M1: Integrating and obtaining \(\frac{1}{a}\theta\).
M1: Expressing \(\theta\) in terms of logarithms.
A1: Completing the proof.
M1: Factoring out \(\frac{1}{3}\) from the integral and applying the formula with \(a=2\).
M1: Substituting the limits \(0\) and \(\sqrt{3}\).
A1: Rationalizing and simplifying the log expression to obtain \(\frac{1}{6} \ln(2+\sqrt{3})\).

1. By de Moivre's theorem, \(\cos 5\theta + i \sin 5\theta = (\cos \theta + i \sin \theta)^5\).
Using binomial expansion:
\(\cos 5\theta = \cos^5 \theta - 10 \cos^3 \theta \sin^2 \theta + 5 \cos \theta \sin^4 \theta\)

Substitute \(\sin^2 \theta = 1 - \cos^2 \theta\):
\(\cos 5\theta = \cos^5 \theta - 10 \cos^3 \theta (1 - \cos^2 \theta) + 5 \cos \theta (1 - \cos^2 \theta)^2\)
\(= \cos^5 \theta - 10 \cos^3 \theta + 10 \cos^5 \theta + 5 \cos \theta (1 - 2\cos^2 \theta + \cos^4 \theta)\)
\(= 11 \cos^5 \theta - 10 \cos^3 \theta + 5 \cos \theta - 10 \cos^3 \theta + 5 \cos^5 \theta\)
\(= 16 \cos^5 \theta - 20 \cos^3 \theta + 5 \cos \theta\).

2. Let \(x = \cos \theta\). The equation \(16x^5 - 20x^3 + 5x - 1 = 0\) becomes:
\(16 \cos^5 \theta - 20 \cos^3 \theta + 5 \cos \theta - 1 = 0\)
\(\cos 5\theta - 1 = 0 \implies \cos 5\theta = 1\).

The roots are \(\theta = \frac{2k\pi}{5}\) for \(k = 0, 1, 2, 3, 4\).
This gives \(x = \cos\left(\frac{2k\pi}{5}\right)\).
For \(k = 0\), \(x = \cos 0 = 1\).
The roots not equal to 1 must satisfy the remaining factor. Divide the polynomial by \(x - 1\):
\(16x^5 - 20x^3 + 5x - 1 = (x - 1)(16x^4 + 16x^3 - 4x^2 - 4x + 1) = 0\).

Observe that the quartic factor can be written as a perfect square:
\(16x^4 + 16x^3 - 4x^2 - 4x + 1 = (4x^2 + 2x - 1)^2\).

Since \((4x^2 + 2x - 1)^2 = 0\), any root other than 1 must satisfy:
\(4x^2 + 2x - 1 = 0\).

M1: Attempting to expand \((\cos \theta + i \sin \theta)^5\) and equating the real part to \(\cos 5\theta\).
A1: Finding correct expression in terms of \(\cos\) and \(\sin\).
M1: Substituting \(\sin^2 \theta = 1 - \cos^2 \theta\).
A1: Correctly deriving \(\cos 5\theta = 16\cos^5 \theta - 20\cos^3 \theta + 5\cos \theta\).
M1: Relating the polynomial equation to the trigonometric identity and identifying roots as \(x = \cos\left(\frac{2k\pi}{5}\right)\).
M1: Performing polynomial division to factor out \((x - 1)\).
A1: Obtaining the quartic factor \(16x^4 + 16x^3 - 4x^2 - 4x + 1\).
A1: Correctly showing that this quartic is equal to \((4x^2 + 2x - 1)^2\), concluding the proof.

1. Find the complementary function (CF):
Auxiliary equation: \(\lambda^2 + 4\lambda + 4 = 0 \implies (\lambda + 2)^2 = 0 \implies \lambda = -2\) (repeated root).
Thus, \(y_{CF} = (Ax + B)e^{-2x}\).

2. Find the particular integral (PI):
Since both \(e^{-2x}\) and \(x e^{-2x}\) are in the CF, we try \(y_p = C x^2 e^{-2x} + D x^2 + E x + F\).

Let's evaluate \(y_{p1} = C x^2 e^{-2x}\):
\(y_{p1}' = C(2x - 2x^2)e^{-2x}\)
\(y_{p1}'' = C(2 - 8x + 4x^2)e^{-2x}\)
Substituting into LHS:
\(y_{p1}'' + 4y_{p1}' + 4y_{p1} = C(2 - 8x + 4x^2)e^{-2x} + 4C(2x - 2x^2)e^{-2x} + 4C x^2 e^{-2x} = 2C e^{-2x}\)
We require \(2C e^{-2x} = 8 e^{-2x} \implies C = 4\).

Now evaluate \(y_{p2} = D x^2 + E x + F\):
\(y_{p2}' = 2Dx + E\)
\(y_{p2}'' = 2D\)
Substituting into LHS:
\(2D + 4(2Dx + E) + 4(D x^2 + E x + F) = 4x^2\)
\(4D x^2 + (8D + 4E)x + (2D + 4E + 4F) = 4x^2\)
Equating coefficients:
For \(x^2\): \(4D = 4 \implies D = 1\)
For \(x\): \(8D + 4E = 0 \implies 8 + 4E = 0 \implies E = -2\)
For constant: \(2D + 4E + 4F = 0 \implies 2 - 8 + 4F = 0 \implies F = \frac{3}{2}\).

Thus, the PI is \(y_p = 4x^2 e^{-2x} + x^2 - 2x + \frac{3}{2}\).

3. The general solution is:
\(y = (Ax + B)e^{-2x} + 4x^2 e^{-2x} + x^2 - 2x + \frac{3}{2}\).

M1: Setting up and solving the auxiliary equation.
A1: Correct complementary function.
M1: Formulating a suitable particular integral including \(C x^2 e^{-2x}\) and a quadratic.
M1: Differentiating \(C x^2 e^{-2x}\) and substituting to find \(C\).
A1: Finding \(C = 4\).
M1: Substituting the quadratic part to find \(D, E, F\).
A1: Finding \(D = 1, E = -2, F = 1.5\).
A1: Combining to write the complete general solution with arbitrary constants.

1. Eigenvalues of \(\mathbf{A}\):
\(\det(\mathbf{A} - \lambda \mathbf{I}) = \det \begin{pmatrix} 3-\lambda & -1 & 1 \\ -1 & 5-\lambda & -1 \\ 1 & -1 & 3-\lambda \end{pmatrix} = 0\)
Expanding the determinant:
\((3-\lambda)[(5-\lambda)(3-\lambda) - 1] - (-1)[-(3-\lambda) - (-1)] + 1[1 - (5-\lambda)] = 0\)
\((3-\lambda)(\lambda^2 - 8\lambda + 14) + (\lambda - 2) + (\lambda - 4) = 0\)
\(-\lambda^3 + 11\lambda^2 - 36\lambda + 36 = 0\)
Factoring this cubic:
\((\lambda - 2)(\lambda - 3)(\lambda - 6) = 0\)
Thus, the eigenvalues are \(\lambda = 2, 3, 6\).

2. Find normalized eigenvectors:
For \(\lambda = 2\):
\(\begin{pmatrix} 1 & -1 & 1 \\ -1 & 3 & -1 \\ 1 & -1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies y = 0, x + z = 0\).
Eigenvector is \(\begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}\), normalized to \(\begin{pmatrix} 1/\sqrt{2} \\ 0 \\ -1/\sqrt{2} \end{pmatrix}\).

For \(\lambda = 3\):
\(\begin{pmatrix} 0 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies x = y = z\).
Eigenvector is \(\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\), normalized to \(\begin{pmatrix} 1/\sqrt{3} \\ 1/\sqrt{3} \\ 1/\sqrt{3} \end{pmatrix}\).

For \(\lambda = 6\):
\(\begin{pmatrix} -3 & -1 & 1 \\ -1 & -1 & -1 \\ 1 & -1 & -3 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies x = z, y = -2z\).
Eigenvector is \(\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}\), normalized to \(\begin{pmatrix} 1/\sqrt{6} \\ -2/\sqrt{6} \\ 1/\sqrt{6} \end{pmatrix}\).

3. Construct matrices:
\(\mathbf{P} = \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} \\ 0 & \frac{1}{\sqrt{3}} & -\frac{2}{\sqrt{6}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} \end{pmatrix}\)
\(\mathbf{D} = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 6 \end{pmatrix}\)
This choice satisfies \(\mathbf{A} = \mathbf{P}\mathbf{D}\mathbf{P}^{\mathrm{T}}\).

M1: Setting up the characteristic equation.
A1: Correct expansion of determinant to a cubic equation.
A1: Factoring the cubic to confirm eigenvalues 2, 3, and 6.
M1: Finding an eigenvector for one eigenvalue.
M1: Normalizing the eigenvectors.
A1: Correct normalized eigenvector for \(\lambda = 2\).
A1: Correct normalized eigenvector for \(\lambda = 3\).
A1: Correct normalized eigenvector for \(\lambda = 6\).
A1: Stating correct \(\mathbf{P}\) and \(\mathbf{D}\) matching in column/diagonal order.

1. Apply integration by parts to \(I_n = \int_{0}^{1} x^n e^{-x} \, dx\):
Let \(u = x^n \implies du = n x^{n-1} \, dx\),
\(dv = e^{-x} \, dx \implies v = -e^{-x}\).

Then:
\(I_n = \left[ -x^n e^{-x} \right]_{0}^{1} - \int_{0}^{1} -n x^{n-1} e^{-x} \, dx\)
\(I_n = -e^{-1} + n \int_{0}^{1} x^{n-1} e^{-x} \, dx\)
\(I_n = n I_{n-1} - e^{-1}\).

2. First find \(I_0\):
\(I_0 = \int_{0}^{1} e^{-x} \, dx = \left[ -e^{-x} \right]_{0}^{1} = 1 - e^{-1}\).

Apply the reduction formula successively:
\(I_1 = 1 I_0 - e^{-1} = 1 - e^{-1} - e^{-1} = 1 - 2e^{-1}\)
\(I_2 = 2 I_1 - e^{-1} = 2(1 - 2e^{-1}) - e^{-1} = 2 - 5e^{-1}\)
\(I_3 = 3 I_2 - e^{-1} = 3(2 - 5e^{-1}) - e^{-1} = 6 - 16e^{-1}\).

Thus, the exact value of the integral is \(6 - 16e^{-1}\).

M1: Applying integration by parts to \(I_n\).
A1: Correctly evaluating the boundary term \(\left[ -x^n e^{-x} \right]_{0}^{1}\).
A1: Completing the proof of the reduction formula.
M1: Evaluating \(I_0\) correctly.
M1: Successive application of the formula to find \(I_1\) and \(I_2\).
A1: Finding \(I_2 = 2 - 5e^{-1}\).
A1: Finding \(I_3 = 6 - 16e^{-1}\) (or equivalent form \(6 - \frac{16}{e}\)).

Express \(32i\) in polar form:
\(32i = 32 e^{i (\frac{\pi}{2} + 2k\pi)}\), where \(k \in \mathbb{Z}\).

Taking the fifth root of both sides:
\(z = 32^{1/5} e^{i \left( \frac{\pi/2 + 2k\pi}{5} \right)} = 2 e^{i \left( \frac{\pi}{10} + \frac{2k\pi}{5} \right)}\) for \(k = 0, 1, 2, 3, 4\).

We determine the arguments in the range \((-\pi, \pi]\):
- For \(k = 0\): \(\theta_0 = \frac{\pi}{10}\)
- For \(k = 1\): \(\theta_1 = \frac{\pi}{10} + \frac{2\pi}{5} = \frac{\pi}{2}\)
- For \(k = 2\): \(\theta_2 = \frac{\pi}{10} + \frac{4\pi}{5} = \frac{9\pi}{10}\)
- For \(k = 3\) (or \(k = -2\)): \(\theta_3 = \frac{\pi}{10} - \frac{4\pi}{5} = -\frac{7\pi}{10}\)
- For \(k = 4\) (or \(k = -1\)): \(\theta_4 = \frac{\pi}{10} - \frac{2\pi}{5} = -\frac{3\pi}{10}\)

Thus, the five roots are:
\(2 e^{i \frac{\pi}{10}}\), \(2 e^{i \frac{\pi}{2}}\), \(2 e^{i \frac{9\pi}{10}}\), \(2 e^{-i \frac{7\pi}{10}}\), \(2 e^{-i \frac{3\pi}{10}}\).

M1: Writing \(32i\) in exponential form with a general argument.
A1: Correct modulus \(r = 2\).
M1: Applying de Moivre's theorem to find roots.
M1: Correctly generating arguments for five roots.
A1: Finding \(2 e^{i \frac{\pi}{10}}\).
A1: Finding \(2 e^{i \frac{\pi}{2}}\).
A1: Finding \(2 e^{i \frac{9\pi}{10}}\).
A1: Finding \(2 e^{-i \frac{7\pi}{10}}\).
A1: Finding \(2 e^{-i \frac{3\pi}{10}}\).

Let x be the vertical distance of P below O. While x is less than or equal to 1.0 m, the string is slack. When x is greater than 1.0 m, the extension is e = x - 1.0. The tension in the string is T = 20 * (x - 1.0). The equation of motion for P is m * g - T = m * a, which gives 4.0 - 20 * (x - 1.0) = 0.4 * a. The maximum speed occurs when the acceleration is zero: 4.0 - 20 * (x - 1.0) = 0, which gives x - 1.0 = 0.2 m, so x = 1.2 m. By conservation of energy, taking the gravitational potential energy (GPE) reference level at O, the initial total energy is 0. At x = 1.2 m, the loss in GPE is m * g * x = 0.4 * 10 * 1.2 = 4.8 J. The gain in kinetic energy (KE) is 0.5 * m * v^2 = 0.2 * v^2. The gain in elastic potential energy (EPE) is 20 * (0.2)^2 / 2 = 0.4 J. By conservation of energy, we have 4.8 = 0.2 * v^2 + 0.4, which gives 0.2 * v^2 = 4.4, so v^2 = 22. Thus, the maximum speed v is \sqrt{22} \approx 4.69 m s^{-1}.

M1: For expressing tension in terms of extension using Hooke's Law. A1: For finding the extension at maximum speed, e = 0.2 m (or x = 1.2 m). M1: For writing a conservation of energy equation. A1: For correct GPE and EPE terms. M1: For solving the energy equation for v. A1: For the correct maximum speed of 4.69 m s^{-1} (or \sqrt{22}).

Using the trajectory equation in terms of the horizontal range R: y = x * tan(\theta) * (1 - x / R). Here, the horizontal range R is given as 18 m. Since the particle passes through the point at the top of the wall (6, 4), we substitute these coordinates into the equation: 4 = 6 * tan(\theta) * (1 - 6 / 18) = 6 * tan(\theta) * (2 / 3) = 4 * tan(\theta). This simplifies to tan(\theta) = 1, which means \theta = 45^\circ. The range formula is R = u^2 * sin(2\theta) / g. Substituting R = 18, \theta = 45^\circ, and g = 10, we obtain: 18 = u^2 * sin(90^\circ) / 10, which simplifies to u^2 = 180. Thus, u = \sqrt{180} = 6\sqrt{5} \approx 13.4 m s^{-1}.

M1: For using the trajectory equation in terms of range or standard form. A1: For substituting the coordinates of the wall and range correctly. M1: For solving to find tan(\theta) = 1. M1: For using the range formula to relate u and \theta. A1: For obtaining u^2 = 180. A1: For the correct initial speed u = 13.4 (or 6\sqrt{5}).

Let \theta be the angle the string makes with the downward vertical. By conservation of energy, 0.5 * m * u^2 = 0.5 * m * v^2 + m * g * L * (1 - cos(\theta)), which gives v^2 = u^2 - 2 * g * L * (1 - cos(\theta)). Substituting u = 5, L = 0.8, and g = 10, we obtain v^2 = 25 - 16 * (1 - cos(\theta)) = 9 + 16 * cos(\theta). The radial equation of motion is T - m * g * cos(\theta) = m * v^2 / L. The string becomes slack when T = 0, so g * cos(\theta) + v^2 / L = 0. Substituting v^2 and L: 10 * cos(\theta) + (9 + 16 * cos(\theta)) / 0.8 = 0, which simplifies to 8 * cos(\theta) + 9 + 16 * cos(\theta) = 0, so 24 * cos(\theta) = -9, giving cos(\theta) = -0.375. The height of the particle above the lowest point is h = L * (1 - cos(\theta)) = 0.8 * (1 - (-0.375)) = 0.8 * 1.375 = 1.1 m.

M1: For applying conservation of energy to express v^2 in terms of cos(\theta). A1: For obtaining v^2 = 9 + 16 * cos(\theta). M1: For resolving forces radially to find tension T. A1: For setting T = 0 to get the slackness condition. M1: For solving the equation to get cos(\theta) = -0.375. A1: For calculating the correct height h = 1.1 m.

Let R be the normal reaction force, r be the radius of the circle, and h be the height above the vertex. From the geometry of the cone, r = h * tan(\alpha). Resolving forces vertically: R * sin(\alpha) = m * g. Resolving forces horizontally: R * cos(\alpha) = m * \omega^2 * r. Dividing these two equations gives tan(\alpha) = g / (\omega^2 * r). Substituting r = h * tan(\alpha) yields tan(\alpha) = g / (\omega^2 * h * tan(\alpha)), which gives h = g / (\omega^2 * tan^2(\alpha)). Substituting g = 10, \omega = 5, and tan(\alpha) = 0.8, we get h = 10 / (25 * 0.64) = 10 / 16 = 0.625 m.

M1: For resolving forces vertically and horizontally. A1: For obtaining R * sin(\alpha) = mg and R * cos(\alpha) = m * \omega^2 * r. M1: For obtaining tan(\alpha) = g / (\omega^2 * r). A1: For using the geometric relationship r = h * tan(\alpha). M1: For substituting to find an equation for h. A1: For finding the correct height h = 0.625 m.

The equation of motion is F = m * a, so 4 / (2v+3) = 0.5 * v * dv/dx. Separating variables gives (2 * v^2 + 3 * v) dv = 8 dx. Integrating both sides yields 2/3 * v^3 + 1.5 * v^2 = 8x + C. Using the initial condition v = 1 when x = 0: 2/3 * 1^3 + 1.5 * 1^2 = C, which gives C = 13/6. Therefore, 2/3 * v^3 + 1.5 * v^2 = 8x + 13/6. Substituting v = 4: 2/3 * 64 + 1.5 * 16 = 8x + 13/6, so 128/3 + 24 = 8x + 13/6, which gives 200/3 = 8x + 13/6. Thus, 8x = 400/6 - 13/6 = 387/6 = 64.5, which gives x = 8.0625 m.

M1: For using Newton's second law with acceleration v * dv/dx. A1: For the correct differential equation 4 / (2v+3) = 0.5 * v * dv/dx. M1: For separating variables and integrating both sides. A1: For the integrated equation with constant C. M1: For using the boundary condition to find C = 13/6. A1: For substituting v = 4 and solving for x. A1: For the correct value x = 8.0625.

The rod has length 2.4 m, and AC = 1.8 m. The length of the string BC is \sqrt{2.4^2 + 1.8^2} = 3.0 m. The angle \theta that the string makes with the horizontal rod AB has sin(\theta) = 1.8 / 3.0 = 0.6 and cos(\theta) = 2.4 / 3.0 = 0.8. Taking moments about A: T * sin(\theta) * 2.4 = 80 * 1.2, which gives T * 0.6 * 2.4 = 96, so 1.44 T = 96, yielding T = 200/3 N. Let H and V be the horizontal and vertical components of the hinge reaction force at A. Resolving horizontally: H = T * cos(\theta) = (200/3) * 0.8 = 160/3 N. Resolving vertically: V + T * sin(\theta) = 80, which gives V = 80 - (200/3) * 0.6 = 40 N. The magnitude of the reaction force at A is R = \sqrt{H^2 + V^2} = \sqrt{(160/3)^2 + 40^2} = \sqrt{25600/9 + 1600} = \sqrt{40000/9} = 200/3 \approx 66.7 N.

M1: For finding the geometry and angle \theta. M1: For taking moments about A to find the tension T. A1: For the correct tension T = 200/3 N. M1: For resolving horizontally and vertically to find H and V. A1: For correct components H = 160/3 N and V = 40 N. M1: For using Pythagoras' theorem to find the magnitude of the reaction. A1: For the correct magnitude of 66.7 N.

Let v_A and v_B be the velocities of A and B after the collision. By conservation of linear momentum: 0.3 * 6 + 0 = 0.3 * v_A + 0.1 * v_B, which gives 3 * v_A + v_B = 18. By Newton's experimental law of restitution: v_B - v_A = 0.5 * 6 = 3. Solving these equations gives v_A = 3.75 m s^{-1} and v_B = 6.75 m s^{-1}. The initial kinetic energy of the system is KE_i = 0.5 * 0.3 * 6^2 = 5.4 J. The final kinetic energy of the system is KE_f = 0.5 * 0.3 * 3.75^2 + 0.5 * 0.1 * 6.75^2 = 2.109375 + 2.278125 = 4.3875 J. The loss of kinetic energy is KE_i - KE_f = 5.4 - 4.3875 = 1.0125 J.

M1: For writing the conservation of momentum equation. A1: For a correct conservation equation. M1: For using Newton's law of restitution. A1: For finding the correct velocities after collision: v_A = 3.75 m s^{-1} and v_B = 6.75 m s^{-1}. M1: For calculating the initial and final kinetic energies. A1: For finding the loss of kinetic energy is 1.0125 J.

(a) For \(x \ge 0\), \(F(x) = \int_0^x \frac{1}{2} t^2 e^{-t} dt\).
Using integration by parts:
\(\int t^2 e^{-t} dt = -t^2 e^{-t} + \int 2t e^{-t} dt = -t^2 e^{-t} - 2t e^{-t} - 2e^{-t}\).
Evaluating from 0 to \(x\):
\(F(x) = \frac{1}{2} \left[ -e^{-x}(x^2 + 2x + 2) - (-2) \right] = 1 - e^{-x}(1 + x + \frac{1}{2}x^2)\).

(b) \(P(X > 2) = 1 - F(2) = e^{-2}(1 + 2 + \frac{1}{2}(4)) = 5e^{-2}\).

(c) \(E[X] = \int_0^\infty \frac{1}{2} x^3 e^{-x} dx = \frac{1}{2} \Gamma(4) = \frac{6}{2} = 3\).

(a) M1 for attempting integration by parts. M1 for correct second integration by parts. A1 for correct final expression of F(x) showing limits substitution.
(b) M1 for using 1 - F(2). A1 for correct exact value 5/e^2.
(c) M1 for setting up the integral for E[X]. A1 for correct mean of 3.

Null hypothesis \(H_0\): Median increase in sleeping time is zero.
Alternative hypothesis \(H_1\): Median increase in sleeping time is greater than zero.

Discard the zero value, leaving sample size \(n = 7\).

Absolute differences and ranks:
0.3 (Rank 1, sign -)
0.5 (Rank 2, sign -)
0.8 (Rank 3, sign +)
1.2 (Rank 4, sign +)
1.5 (Rank 5, sign +)
1.9 (Rank 6, sign +)
2.1 (Rank 7, sign +)

Sum of positive ranks \(W^+ = 3 + 4 + 5 + 6 + 7 = 25\).
Sum of negative ranks \(W^- = 1 + 2 = 3\).

Test statistic \(T = \min(W^+, W^-) = 3\).

For \(n = 7\), at the 5% significance level (one-tailed), the critical value is 3.

Since \(T \le 3\), we reject \(H_0\). There is sufficient evidence at the 5% level to conclude that the therapy increases sleeping time.

B1 for stating correct hypotheses.
B1 for discarding zero and correct ranking of absolute differences.
M1 for finding W+ and W- correctly.
A1 for identifying test statistic T = 3.
B1 for stating correct critical value of 3.
M1 for comparing T with the critical value.
A1 for correct conclusion in context.

Null hypothesis \(H_0\): There is no association between preferred music genre and age group.
Alternative hypothesis \(H_1\): There is an association between preferred music genre and age group.

Row totals: Under 30 = 100, 30 and over = 100. Column totals: Classical = 50, Rock = 80, Pop = 70. Grand total = 200.

Expected frequencies:
Under 30: Classical = 25, Rock = 40, Pop = 35
30 and over: Classical = 25, Rock = 40, Pop = 35

\(\chi^2 = \sum \frac{(O - E)^2}{E}\):
Under 30:
Classical: \((15 - 25)^2 / 25 = 4.000\)
Rock: \((45 - 40)^2 / 40 = 0.625\)
Pop: \((40 - 35)^2 / 35 = 0.714\)

30 and over:
Classical: \((35 - 25)^2 / 25 = 4.000\)
Rock: \((35 - 40)^2 / 40 = 0.625\)
Pop: \((30 - 35)^2 / 35 = 0.714\)

\(\chi^2 = 2 \times (4.000 + 0.625 + 0.714) = 10.68\) (3 s.f.)

Degrees of freedom: \(df = (2-1)(3-1) = 2\).
Critical value of \(\chi^2\) at 5% level with 2 df is 5.991.

Since \(10.68 > 5.991\), we reject \(H_0\). There is sufficient evidence at the 5% significance level to conclude that there is an association between preferred music genre and age group.

B1 for stating correct hypotheses.
M1 for calculating correct expected frequencies.
A1 for correct test statistic chi^2 = 10.68.
B1 for identifying correct degrees of freedom (2) and critical value (5.991).
M1 for comparing the test statistic to the critical value.
A1 for correct conclusion in context.

(a) Expanding \(G_X(t)\):
\(G_X(t) = \frac{t}{3}(1 - \frac{2}{3}t)^{-1} = \frac{t}{3} [ 1 + \frac{2}{3}t + \frac{4}{9}t^2 + \dots ] = \frac{1}{3}t + \frac{2}{9}t^2 + \frac{4}{27}t^3 + \dots\)
Therefore, \(P(X=1) = \frac{1}{3}\) and \(P(X=2) = \frac{2}{9}\).

(b) Differentiating \(G_X(t)\):
\(G'_X(t) = \frac{(3-2t) - t(-2)}{(3-2t)^2} = \frac{3}{(3-2t)^2}\)
\(E[X] = G'_X(1) = \frac{3}{1^2} = 3\).

Differentiating again:
\(G''_X(t) = \frac{12}{(3-2t)^3}\)
\(G''_X(1) = 12\).

Using the formula for variance:
\(Var(X) = G''_X(1) + G'_X(1) - [G'_X(1)]^2 = 12 + 3 - 9 = 6\).

(a) M1 for expanding G_X(t) using binomial expansion. A1 for correct P(X=1) and P(X=2).
(b) M1 for finding first derivative G'_X(t). A1 for mean = 3. M1 for finding second derivative G''_X(t). M1 for using Var(X) formula. A1 for variance = 6.

(a) Given \(n = 10\), \(\bar{x} = 12.4\), and \(s = \sqrt{1.44} = 1.2\).
Degrees of freedom \(df = 9\).
For a 95% confidence interval, the critical value \(t_9(0.025) = 2.262\).
Confidence interval: \(\bar{x} \pm t \frac{s}{\sqrt{n}} = 12.4 \pm 2.262 \times \frac{1.2}{\sqrt{10}} = 12.4 \pm 2.262 \times 0.37947 = 12.4 \pm 0.858\).
This gives the interval \([11.54, 13.26]\) (to 2 decimal places).

(b) \(H_0: \mu = 11.5\) vs \(H_1: \mu \neq 11.5\).
Test statistic \(t = \frac{12.4 - 11.5}{1.2 / \sqrt{10}} = 2.372\).
At 10% significance level (two-tailed), the critical value for \(df=9\) is \(t_9(0.05) = 1.833\).
Since \(|2.372| > 1.833\), we reject \(H_0\). There is sufficient evidence at the 10% significance level to conclude that the population mean concentration differs from 11.5 mg/L.

(a) B1 for identifying critical value 2.262. M1 for correct formula of confidence interval. A1 for correct limits of the interval.
(b) B1 for stating correct hypotheses. M1 for calculating test statistic t = 2.372. B1 for critical value 1.833. A1 for correct conclusion in context.

Null hypothesis \(H_0\): The two samples come from populations with the same distribution.
Alternative hypothesis \(H_1\): The two samples come from populations with different distributions.

Combine and rank all 11 observations:
10 (B) - Rank 1
12 (A) - Rank 2
14 (B) - Rank 3
15 (A) - Rank 4
16 (B) - Rank 5
17 (B) - Rank 6
18 (A) - Rank 7
20 (B) - Rank 8
21 (B) - Rank 9
22 (A) - Rank 10
25 (A) - Rank 11

Sum of ranks for A: \(R_1 = 2 + 4 + 7 + 10 + 11 = 34\).
Sum of ranks for B: \(R_2 = 1 + 3 + 5 + 6 + 8 + 9 = 32\).

Calculate \(U_1\) and \(U_2\):
\(U_1 = n_1 n_2 + \frac{n_1(n_1+1)}{2} - R_1 = 5(6) + \frac{5(6)}{2} - 34 = 11\).
\(U_2 = n_1 n_2 + \frac{n_2(n_2+1)}{2} - R_2 = 5(6) + \frac{6(7)}{2} - 32 = 19\).

Test statistic \(U = \min(U_1, U_2) = 11\).

The critical value is 5.

Since \(U = 11 > 5\), we do not reject \(H_0\). There is insufficient evidence at the 10% significance level to conclude that the two populations have different distributions.

B1 for stating correct hypotheses.
M1 for pooling and ranking all observations. A1 for correct sum of ranks R_1 = 34 or R_2 = 32.
M1 for calculating U_1 or U_2. A1 for finding U = 11.
M1 for comparing U to the critical value 5. A1 for correct conclusion in context.

Null hypothesis \(H_0\): \(\mu_1 = \mu_2\)
Alternative hypothesis \(H_1\): \(\mu_1 > \mu_2\) (one-tailed)

Calculate the pooled estimate of the variance \(s_p^2\):
\(s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2} = \frac{7(12.8) + 9(15.4)}{16} = 14.2625\).

Calculate the test statistic \(t\):
\(t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{s_p^2 (\frac{1}{n_1} + \frac{1}{n_2})}} = \frac{45.2 - 41.6}{\sqrt{14.2625 (\frac{1}{8} + \frac{1}{10})}} = \frac{3.6}{\sqrt{14.2625 \times 0.225}} = \frac{3.6}{1.791386} = 2.010\) (3 d.p.)

Degrees of freedom \(df = 16\).
For a one-tailed test at the 5% level, the critical value is \(t_{16}(0.05) = 1.746\).

Since \(t = 2.010 > 1.746\), we reject \(H_0\). There is sufficient evidence at the 5% significance level to conclude that the population mean of the first population is greater than that of the second population.

B1 for stating correct hypotheses.
M1 for calculating the pooled variance s_p^2 = 14.26.
A1 for calculating the standard error or denominator correctly.
M1 for calculating the test statistic t = 2.010.
B1 for identifying df = 16 and critical value 1.746.
M1 for comparing the test statistic with the critical value.
A1 for correct conclusion in context.