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(i) A general point on the line \(y = x\) can be written as \(\begin{pmatrix} t \\ t \end{pmatrix}\).

Under the transformation represented by \(\mathbf{M}\), its image is:
\[ \mathbf{M} \begin{pmatrix} t \\ t \end{pmatrix} = \begin{pmatrix} a & 2 \\ -1 & b \end{pmatrix} \begin{pmatrix} t \\ t \end{pmatrix} = \begin{pmatrix} (a+2)t \\ (b-1)t \end{pmatrix} \]

Since this image point lies on the line \(4x + 3y = 0\), we have:
\[ 4(a+2)t + 3(b-1)t = 0 \]

Since this holds for all \(t \ne 0\), we obtain:
\[ 4a + 8 + 3b - 3 = 0 \implies 4a + 3b = -5 \]

We are also given that the determinant of \(\mathbf{M}\) is \(-1\):
\[ \det(\mathbf{M}) = a(b) - 2(-1) = ab + 2 = -1 \implies ab = -3 \]

From \(ab = -3\), we substitute \(b = -\frac{3}{a}\) into the linear equation:
\[ 4a + 3\left(-\frac{3}{a}\right) = -5 \implies 4a - \frac{9}{a} = -5 \]

Multiplying by \(a\) gives the quadratic equation:
\[ 4a^2 + 5a - 9 = 0 \implies (4a + 9)(a - 1) = 0 \]

Since \(a > 0\), we choose \(a = 1\).

Then, we find \(b = -\frac{3}{1} = -3\).

(ii) The matrix \(\mathbf{M}\) is:
\[ \mathbf{M} = \begin{pmatrix} 1 & 2 \\ -1 & -3 \end{pmatrix} \]

To find the image of the point \((2, -1)\), we compute:
\[ \begin{pmatrix} 1 & 2 \\ -1 & -3 \end{pmatrix} \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 1(2) + 2(-1) \\ -1(2) - 3(-1) \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \]

Thus, the image is the point \((0, 1)\).

M1: Sets up the image of a general point \((t, t)^T\) on the line \(y = x\).
M1: Substitutes the coordinates of the image point into the equation of the line \(4x + 3y = 0\) to obtain a relation between \(a\) and \(b\).
A1: Obtains the correct linear equation \(4a + 3b = -5\).
M1: Uses the determinant condition to find \(ab = -3\) and solves the resulting quadratic equation in \(a\).
A1: Selects the positive value \(a = 1\) and finds the corresponding value \(b = -3\).
M1: Multiplies \(\mathbf{M}\) by the column vector \(\begin{pmatrix} 2 \\ -1 \end{pmatrix}\).
A1: Obtains the correct image point \((0, 1)\).

(i) To find the points of intersection, we set the two polar equations equal:
\[ 3 \cos \theta = 1 + \cos \theta \implies 2 \cos \theta = 1 \implies \cos \theta = \frac{1}{2} \]

Since \(-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}\), we find:
\[ \theta = \pm \frac{\pi}{3} \]

Substituting \(\theta = \pm \frac{\pi}{3}\) into \(r = 3 \cos \theta\):
\[ r = 3 \cos\left(\pm \frac{\pi}{3}\right) = 3 \left(\frac{1}{2}\right) = \frac{3}{2} \]

So the polar coordinates of the points of intersection are \(\left(\frac{3}{2}, \frac{\pi}{3}\right)\) and \(\left(\frac{3}{2}, -\frac{\pi}{3}\right)\).

(ii) The region lies inside \(C_1\) but outside \(C_2\) for \(-\frac{\pi}{3} \le \theta \le \frac{\pi}{3}\).
Using symmetry, the area \(A\) is given by:
\[ A = 2 \times \frac{1}{2} \int_{0}^{\pi/3} (r_1^2 - r_2^2) \, \mathrm{d}\theta \]
\[ A = \int_{0}^{\pi/3} \left( (3 \cos \theta)^2 - (1 + \cos \theta)^2 \right) \, \mathrm{d}\theta \]
\[ A = \int_{0}^{\pi/3} \left( 9 \cos^2 \theta - (1 + 2 \cos \theta + \cos^2 \theta) \right) \, \mathrm{d}\theta \]
\[ A = \int_{0}^{\pi/3} \left( 8 \cos^2 \theta - 2 \cos \theta - 1 \right) \, \mathrm{d}\theta \]

Using the double-angle identity \(\cos^2 \theta = \frac{1 + \cos 2\theta}{2}\):
\[ 8 \cos^2 \theta = 4(1 + \cos 2\theta) = 4 + 4 \cos 2\theta \]

Substituting this back into the integral:
\[ A = \int_{0}^{\pi/3} \left( 4 + 4 \cos 2\theta - 2 \cos \theta - 1 \right) \, \mathrm{d}\theta = \int_{0}^{\pi/3} \left( 3 + 4 \cos 2\theta - 2 \cos \theta \right) \, \mathrm{d}\theta \]

Now, integrating term-by-term:
\[ A = \left[ 3\theta + 2 \sin 2\theta - 2 \sin \theta \right]_{0}^{\pi/3} \]

Evaluating at the limits:
\[ A = \left( 3\left(\frac{\pi}{3}\right) + 2 \sin\left(\frac{2\pi}{3}\right) - 2 \sin\left(\frac{\pi}{3}\right) \right) - (0) \]
\[ A = \pi + 2\left(\frac{\sqrt{3}}{2}\right) - 2\left(\frac{\sqrt{3}}{2}\right) = \pi \]

So the area of the region is \(\pi\).

M1: Sets the two polar equations equal and solves for \(\cos \theta\).
A1: Obtains the correct polar coordinates \((1.5, \pm \pi/3)\) or equivalent.
M1: Uses the correct area formula \(\frac{1}{2} \int (r_1^2 - r_2^2) \, \mathrm{d}\theta\) with appropriate limits (e.g. from \(-\pi/3\) to \(\pi/3\), or twice from \(0\) to \(\pi/3\)).
M1: Expands the integrand correctly and applies the identity for \(\cos^2 \theta\).
A1: Obtains the correct simplified integrand, e.g., \(3 + 4 \cos 2\theta - 2 \cos \theta\).
M1: Integrates correctly to get \(3\theta + 2 \sin 2\theta - 2 \sin \theta\) and substitutes the limits.
A1: Obtains the exact final area of \(\pi\).

(i) The vertical asymptote of \(C\) is found where the denominator is zero:
\[ x - 1 = 0 \implies x = 1 \]

To find the oblique asymptote, we perform algebraic division on the rational expression:
\[ x^2 - 3x + 6 = x(x-1) - 2x + 6 = x(x-1) - 2(x-1) + 4 = (x-2)(x-1) + 4 \]

So we can rewrite \(y\) as:
\[ y = x - 2 + \frac{4}{x - 1} \]

As \(x \to \pm\infty\), the term \(\frac{4}{x - 1} \to 0\).
Therefore, the oblique asymptote is:
\[ y = x - 2 \]

(ii) To find the range of \(y\), we rearrange the equation into a quadratic in \(x\):
\[ y(x - 1) = x^2 - 3x + 6 \implies yx - y = x^2 - 3x + 6 \]
\[ x^2 - (3 + y)x + (6 + y) = 0 \]

Since \(x\) is a real number, the discriminant \(\Delta\) of this quadratic equation must be non-negative:
\[ \Delta = [-(3 + y)]^2 - 4(1)(6 + y) \ge 0 \]
\[ y^2 + 6y + 9 - 24 - 4y \ge 0 \]
\[ y^2 + 2y - 15 \ge 0 \]

We solve this inequality by factoring:
\[ (y + 5)(y - 3) \ge 0 \]

This gives the set of values that \(y\) can take as:
\[ y \le -5 \quad \text{or} \quad y \ge 3 \]

B1: Correctly identifies the vertical asymptote as \(x = 1\).
M1: Rewrites the algebraic expression using polynomial division or equivalent.
A1: Correctly identifies the oblique asymptote as \(y = x - 2\).
M1: Rearranges the original equation to form a quadratic equation in \(x\).
M1: Sets the discriminant of the quadratic equation to be greater than or equal to zero.
A1: Simplifies to the quadratic inequality \(y^2 + 2y - 15 \ge 0\) (or equivalent).
A1: Obtains the correct range of \(y \le -5\) or \(y \ge 3\).

(a) The curve is a dimpled limacon that does not pass through the pole. It is symmetric about the initial line. Key points to mark on the sketch: - Intersection with the initial line (\(\theta = 0\)): \((5a, 0)\) - Intersection with the line \(\theta = \pi\): \((a, \pi)\) - Intersections with \(\theta = \frac{\pi}{2}\) and \(\theta = \frac{3\pi}{2}\): \((3a, \frac{\pi}{2})\) and \((3a, \frac{3\pi}{2})\). (b) The area \(A\) is given by: \(A = \frac{1}{2} \int_{0}^{2\pi} r^2 \mathrm{d}\theta = \frac{1}{2} a^2 \int_{0}^{2\pi} (3 + 2\cos\theta)^2 \mathrm{d}\theta = \frac{1}{2} a^2 \int_{0}^{2\pi} (9 + 12\cos\theta + 4\cos^2\theta) \mathrm{d}\theta\). Using the identity \(\cos^2\theta = \frac{1}{2}(1 + \cos 2\theta)\): \(A = \frac{1}{2} a^2 \int_{0}^{2\pi} (9 + 12\cos\theta + 2 + 2\cos 2\theta) \mathrm{d}\theta = \frac{1}{2} a^2 \int_{0}^{2\pi} (11 + 12\cos\theta + 2\cos 2\theta) \mathrm{d}\theta = \frac{1}{2} a^2 \left[ 11\theta + 12\sin\theta + \sin 2\theta \right]_0^{2\pi} = \frac{1}{2} a^2 (22\pi) = 11\pi a^2\). (c) A tangent parallel to \(\theta = \frac{\pi}{2}\) occurs when \(\frac{\mathrm{d}x}{\mathrm{d}\theta} = 0\), provided \(\frac{\mathrm{d}y}{\mathrm{d}\theta} \neq 0\). Using \(x = r\cos\theta\): \(x = a(3 + 2\cos\theta)\cos\theta = a(3\cos\theta + 2\cos^2\theta)\). Then \(\frac{\mathrm{d}x}{\mathrm{d}\theta} = a(-3\sin\theta - 4\cos\theta\sin\theta) = -a\sin\theta(3 + 4\cos\theta)\). Setting \(\frac{\mathrm{d}x}{\mathrm{d}\theta} = 0\): Either \(\sin\theta = 0 \implies \theta = 0, \pi\). At \(\theta = 0\), \(r = 5a\), giving \((5a, 0)\). At \(\theta = \pi\), \(r = a\), giving \((a, \pi)\). Or \(3 + 4\cos\theta = 0 \implies \cos\theta = -\frac{3}{4}\). Since \(\cos\theta < 0\), there are two solutions in \([0, 2\pi)\): \(\theta = \pi - \arccos(\frac{3}{4})\) and \(\theta = \pi + \arccos(\frac{3}{4})\). For these angles, \(r = a(3 + 2(-\frac{3}{4})) = \frac{3}{2}a\). So the points are: \((5a, 0)\), \((a, \pi)\), \((\frac{3}{2}a, \pi - \arccos(\frac{3}{4}))\), and \((\frac{3}{2}a, \pi + \arccos(\frac{3}{4}))\).

(a) - B1: Correct cardiod/limacon shape (dimpled, symmetric about the initial line, not passing through the pole). - B1: Coordinates \((5a, 0)\) and \((a, \pi)\) clearly labeled. - B1: Coordinates \((3a, \frac{\pi}{2})\) and \((3a, \frac{3\pi}{2})\) clearly labeled. (b) - M1: Uses correct area formula \(A = \frac{1}{2} \int r^2 \mathrm{d}\theta\). - M1: Expands integrand and applies \(\cos^2\theta = \frac{1}{2}(1 + \cos 2\theta)\). - A1: Integrates correctly to get \(\frac{1}{2} a^2 [11\theta + 12\sin\theta + \sin 2\theta]\) (condone missing limits). - A1: Evaluates limits \(0\) and \(2\pi\) correctly to show \(11\pi a^2\) with no errors. (c) - M1: Expresses \(x = r\cos\theta\) in terms of \(\theta\) and differentiates to find \(\frac{\mathrm{d}x}{\mathrm{d}\theta}\). - M1: Sets \(\frac{\mathrm{d}x}{\mathrm{d}\theta} = 0\) and factors to find equations for \(\theta\). - A1: Identifies \(\theta = 0\) and \(\theta = \pi\) and finds points \((5a, 0)\) and \((a, \pi)\). - M1: Solves \(\cos\theta = -\frac{3}{4}\) to find \(\theta = \pi \pm \arccos(\frac{3}{4})\). - A1: Correctly evaluates \(r = \frac{3}{2}a\) and lists the remaining polar coordinates: \((\frac{3}{2}a, \pi \pm \arccos(\frac{3}{4}))\).

(a) To find the eigenvalues of \(\mathbf{A}\), we solve the characteristic equation \(\det(\mathbf{A} - \lambda\mathbf{I}) = 0\): \(\det \begin{pmatrix} 3-\lambda & -1 & 1 \\ -1 & 5-\lambda & -1 \\ 1 & -1 & 3-\lambda \end{pmatrix} = 0\). Expanding the determinant: \((3-\lambda) [ (5-\lambda)(3-\lambda) - 1 ] + 1 [ -(3-\lambda) + 1 ] + 1 [ 1 - (5-\lambda) ] = 0 \implies (3-\lambda) [ \lambda^2 - 8\lambda + 14 ] + (\lambda - 2) + (\lambda - 4) = 0 \implies 3\lambda^2 - 24\lambda + 42 - \lambda^3 + 8\lambda^2 - 14\lambda + 2\lambda - 6 = 0 \implies -\lambda^3 + 11\lambda^2 - 36\lambda + 36 = 0 \implies \lambda^3 - 11\lambda^2 + 36\lambda - 36 = 0\). Testing \(\lambda = 2\): \(2^3 - 11(2^2) + 36(2) - 36 = 8 - 44 + 72 - 36 = 0\). So \(\lambda = 2\) is an eigenvalue. Dividing by \(\lambda - 2\) yields \(\lambda^2 - 9\lambda + 18 = 0\), which factors as \((\lambda-3)(\lambda-6) = 0\). Thus, the eigenvalues are \(\lambda = 2, 3, 6\). (b) For \(\lambda = 2\): \(\begin{pmatrix} 1 & -1 & 1 \\ -1 & 3 & -1 \\ 1 & -1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\). From row 1, \(x - y + z = 0\). From row 1 + row 2, \(2y = 0 \implies y = 0\). Thus, \(x = -z\). An eigenvector is \(\mathbf{v}_1 = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}\). For \(\lambda = 3\): \(\begin{pmatrix} 0 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\). From row 1, \(y = z\). From row 3, \(x = y\). Thus, \(x = y = z\). An eigenvector is \(\mathbf{v}_2 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\). For \(\lambda = 6\): \(\begin{pmatrix} -3 & -1 & 1 \\ -1 & -1 & -1 \\ 1 & -1 & -3 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\). Adding row 1 and 3 gives \(-2x - 2y - 2z = 0 \implies x + y + z = 0\). Adding row 2 and 3 gives \(-2y - 4z = 0 \implies y = -2z\). Then \(x - 2z + z = 0 \implies x = z\). An eigenvector is \(\mathbf{v}_3 = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}\). These eigenvectors are mutually orthogonal. (c) To construct an orthogonal matrix \(\mathbf{P}\), we normalize each of the orthogonal eigenvectors: \(\hat{\mathbf{v}}_1 = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}\), \(\hat{\mathbf{v}}_2 = \frac{1}{\sqrt{3}} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\), \(\hat{\mathbf{v}}_3 = \frac{1}{\sqrt{6}} \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}\). Thus, we have: \(\mathbf{P} = \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} \\ 0 & \frac{1}{\sqrt{3}} & -\frac{2}{\sqrt{6}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} \end{pmatrix}\) and \(\mathbf{D} = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 6 \end{pmatrix}\).

(a) - M1: Formulates the characteristic equation \(\det(\mathbf{A} - \lambda\mathbf{I}) = 0\). - A1: Expands to obtain the correct cubic equation \(\lambda^3 - 11\lambda^2 + 36\lambda - 36 = 0\). - B1: Shows that \(\lambda = 2\) is a root. - M1: Factorises the cubic expression. - A1: Obtains the eigenvalues \(2, 3, 6\). (b) - M1: Solves \((\mathbf{A} - 2\mathbf{I})\mathbf{v} = \mathbf{0}\) to find eigenvector for \(\lambda = 2\). - A1: Finds correct eigenvector \(\begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}\) (or scalar multiple). - A1: Finds correct eigenvectors \(\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\) and \(\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}\) (or scalar multiples). - B1: Confirms/explains that they are mutually orthogonal. (c) - B1: States correct diagonal matrix \(\mathbf{D}\) with eigenvalues \(2, 3, 6\) on the main diagonal. - M1: Normalises the three eigenvectors from part (b). - A1: Construct the correct orthogonal matrix \(\mathbf{P}\) ensuring column order matches the diagonal entries of \(\mathbf{D}\).

(a) The curve \(C\) is a dimpled limaçon without an internal loop (since \(3 > 2\)). It is symmetrical about the initial line \(\theta = 0\). The distance from the pole is \(5a\) at \(\theta = 0\), \(3a\) at \(\theta = \pm\frac{\pi}{2}\), and \(a\) at \(\theta = \pi\).

(b) The area \(A\) of the region enclosed by \(C\) is given by:
\[A = \frac{1}{2} \int_{-\pi}^{\pi} r^2 \, \mathrm{d}\theta = \int_{0}^{\pi} r^2 \, \mathrm{d}\theta\]
\[r^2 = a^2(3 + 2\cos\theta)^2 = a^2(9 + 12\cos\theta + 4\cos^2\theta)\]
Using the identity \(\cos^2\theta = \frac{1}{2}(1 + \cos 2\theta)\):
\[r^2 = a^2(9 + 12\cos\theta + 2(1 + \cos 2\theta)) = a^2(11 + 12\cos\theta + 2\cos 2\theta)\]
Integrating this expression:
\[A = \frac{1}{2} a^2 \int_{-\pi}^{\pi} (11 + 12\cos\theta + 2\cos 2\theta) \, \mathrm{d}\theta\]
\[A = \frac{1}{2} a^2 \left[ 11\theta + 12\sin\theta + \sin 2\theta \right]_{-\pi}^{\pi}\]
\[A = \frac{1}{2} a^2 (11\pi - (-11\pi)) = 11\pi a^2\]

(c) The Cartesian coordinate \(x\) is given by:
\[x = r\cos\theta = a(3 + 2\cos\theta)\cos\theta = a(3\cos\theta + 2\cos^2\theta)\]
Tangents perpendicular to the initial line occur when \(\frac{\mathrm{d}x}{\mathrm{d}\theta} = 0\):
\[\frac{\mathrm{d}x}{\mathrm{d}\theta} = a(-3\sin\theta - 4\cos\theta\sin\theta) = -a\sin\theta(3 + 4\cos\theta) = 0\]
This gives \(\sin\theta = 0\) or \(\cos\theta = -\frac{3}{4}\).
- For \(\sin\theta = 0\), the values of \(\theta\) in the interval \(-\pi < \theta \le \pi\) are \(\theta = 0\) and \(\theta = \pi\).
- If \(\theta = 0\), then \(r = a(3 + 2(1)) = 5a\). The polar coordinates are \((5a, 0)\).
- If \(\theta = \pi\), then \(r = a(3 + 2(-1)) = a\). The polar coordinates are \((a, \pi)\).
- For \(\cos\theta = -\frac{3}{4}\), we have \(\theta = \pm \arccos\left(-\frac{3}{4}\right)\).
- If \(\cos\theta = -\frac{3}{4}\), then \(r = a\left(3 + 2\left(-\frac{3}{4}\right)\right) = \frac{3}{2}a\). The polar coordinates are \(\left(\frac{3}{2}a, \pm \arccos\left(-\frac{3}{4}\right)\right)\).

(Checking that \(\frac{\mathrm{d}y}{\mathrm{d}\theta} \neq 0\) at these points confirms they are all valid non-cusp points of vertical tangency.)

(a)
- **M1**: For sketching a closed, non-self-intersecting curve symmetric about the initial line.
- **A1**: Correct cardioid-like shape with a dimple at \(\theta = \pi\).
- **A1**: Intercepts clearly labelled or indicated at \((5a, 0)\), \((3a, \pm\frac{\pi}{2})\), and \((a, \pi)\).

(b)
- **M1**: Use of correct area formula \(\frac{1}{2}\int r^2 \, \mathrm{d}\theta\) with appropriate limits.
- **M1**: Expansion of \((3 + 2\cos\theta)^2\) and use of double-angle identity for \(\cos^2\theta\).
- **A1**: Correct linearized integrand \(11 + 12\cos\theta + 2\cos 2\theta\).
- **M1**: Correct integration to obtain terms \(11\theta\), \(12\sin\theta\), and \(\sin 2\theta\).
- **A1**: Correct evaluation of limits to get final area of \(11\pi a^2\).

(c)
- **M1**: Expressing \(x = r\cos\theta\) in terms of \(\theta\) and differentiating with respect to \(\theta\).
- **A1**: Correct expression for \(\frac{\mathrm{d}x}{\mathrm{d}\theta}\), equivalent to \(-a\sin\theta(3 + 4\cos\theta)\).
- **M1**: Setting \(\frac{\mathrm{d}x}{\mathrm{d}\theta} = 0\) and solving for \(\theta\).
- **A1**: Identifying \(\theta = 0\) and \(\theta = \pi\) and finding the corresponding polar coordinates: \((5a, 0)\) and \((a, \pi)\).
- **A1**: Identifying \(\cos\theta = -\frac{3}{4}​\) and finding the corresponding radial distance \(r = \frac{3}{2}a\).
- **A2**: Awarding full marks for stating the remaining coordinates as \(\left(\frac{3}{2}a, \pm \arccos\left(-\frac{3}{4}\right)\right)\) or equivalent decimal representations (approximately \(\pm 2.42\) radians). Deduct 1 mark if \(\pm\) is omitted or only one of the two solutions is written.

(a) The determinant of \(\mathbf{M}\) is:
\[ \det(\mathbf{M}) = (3)(0) - (2)(2) = -4 \]
The area scale factor of the transformation is given by \(|\det(\mathbf{M})| = |-4| = 4\).
Therefore, the area of the image of the triangle is:
\[ \text{Area} = 5 \times 4 = 20 \]

(b) Let \(y = mx\) be an invariant line through the origin. Under the transformation represented by \(\mathbf{M}\), a point \((x, mx)\) on the line is mapped to \((X, Y)\), where:
\[ \begin{pmatrix} X \\ Y \end{pmatrix} = \begin{pmatrix} 3 & 2 \\ 2 & 0 \end{pmatrix} \begin{pmatrix} x \\ mx \end{pmatrix} = \begin{pmatrix} (3 + 2m)x \\ 2x \end{pmatrix} \]
Since the line is invariant, the point \((X, Y)\) must also lie on the line \(y = mx\), meaning \(Y = mX\):
\[ 2x = m(3 + 2m)x \]
Since this relation must hold for all points on the line (where \(x \neq 0\)):
\[ 2 = 3m + 2m^2 \implies 2m^2 + 3m - 2 = 0 \]
Solving this quadratic equation:
\[ (2m - 1)(m + 2) = 0 \implies m = \frac{1}{2} \quad \text{or} \quad m = -2 \]
Thus, the equations of the invariant lines through the origin are \(y = \frac{1}{2}x\) and \(y = -2x\).

(c) To find the image of the line \(L: y = 2x + 1\), we can express the original coordinates \(x\) and \(y\) in terms of the image coordinates \(X\) and \(Y\) using the inverse matrix \(\mathbf{M}^{-1}\):
\[ \mathbf{M}^{-1} = \frac{1}{-4} \begin{pmatrix} 0 & -2 \\ -2 & 3 \end{pmatrix} = \begin{pmatrix} 0 & \frac{1}{2} \\ \frac{1}{2} & -\frac{3}{4} \end{pmatrix} \]
Thus,
\[ \begin{pmatrix} x \\ y
\end{pmatrix} = \mathbf{M}^{-1} \begin{pmatrix} X \\ Y \end{pmatrix} = \begin{pmatrix} 0 & \frac{1}{2} \\ \frac{1}{2} & -\frac{3}{4} \end{pmatrix} \begin{pmatrix} X \\ Y \end{pmatrix} \]
This gives:
\[ x = \frac{1}{2}Y \quad \text{and} \quad y = \frac{1}{2}X - \frac{3}{4}Y \]
Substituting these into the equation of the line \(y = 2x + 1\):
\[ \frac{1}{2}X - \frac{3}{4}Y = 2\left(\frac{1}{2}Y\right) + 1 \]
\[ \frac{1}{2}X - \frac{3}{4}Y = Y + 1 \]
Multiply the entire equation by 4 to clear the fractions:
\[ 2X - 3Y = 4Y + 4 \implies 7Y = 2X - 4 \]
Dividing by 7 to write it in the requested form:
\[ Y = \frac{2}{7}X - \frac{4}{7} \]
Hence, the equation of the image of \(L\) is \(y = \frac{2}{7}x - \frac{4}{7}\).

(Alternative Method: We can choose two points on \(L\), say \(A(0, 1)\) and \(B(1, 3)\). Under \(T\):
\[ A'(x', y') = \mathbf{M} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 0 \end{pmatrix} \implies A'(2, 0) \]
\[ B'(x', y') = \mathbf{M} \begin{pmatrix} 1 \\ 3 \end{pmatrix} = \begin{pmatrix} 9 \\ 2 \end{pmatrix} \implies B'(9, 2) \]
The equation of the line passing through \((2, 0)\) and \((9, 2)\) has gradient \(m' = \frac{2 - 0}{9 - 2} = \frac{2}{7}\).
Using the point-slope form: \(y - 0 = \frac{2}{7}(x - 2) \implies y = \frac{2}{7}x - \frac{4}{7}\).)

(a)
- **M1**: Correct calculation of the determinant of \(\mathbf{M}\).
- **A1**: Stating that \(\det(\mathbf{M}) = -4\), indicating the area scale factor is 4.
- **A1**: Finding the correct image area of 20.

(b)
- **M1**: Setting up the equation for the transformation of a general point \((x, mx)\) on the invariant line.
- **M1**: Equating the coordinates using \(Y = mX\) to derive an equation in \(m\).
- **A1**: Establishing the correct quadratic equation \(2m^2 + 3m - 2 = 0\).
- **M1**: Solving the quadratic equation for \(m\).
- **A1**: Obtaining both correct line equations \(y = \frac{1}{2}x\) and \(y = -2x\).

(c)
- **M1**: Attempting to find the inverse matrix \(\mathbf{M}^{-1}\).
- **A1**: Showing the correct inverse matrix \(\mathbf{M}^{-1} = \begin{pmatrix} 0 & \frac{1}{2} \\ \frac{1}{2} & -\frac{3}{4} \end{pmatrix}\).
- **M1**: Writing correct equations for \(x\) and \(y\) in terms of the image coordinates \(X\) and \(Y\).
- **A1**: Correctly obtaining \(x = \frac{1}{2}Y\) and \(y = \frac{1}{2}X - \frac{3}{4}Y\).
- **M1**: Substituting these expressions into \(y = 2x + 1\).
- **A2**: Finding the final simplified equation \(y = \frac{2}{7}x - \frac{4}{7}\). (Award **A1** for any correct unsimplified linear equation such as \(2x - 7y = 4\).)

*(Alternatively, for the two-point method: **M1** choosing two distinct points on the line; **A1** transforming the first point correctly; **M1** transforming the second point correctly; **A1** calculating the correct gradient of the image line; **M1** using the line equation formula; **A2** for final simplified equation.)*

(a) To find the eigenvalues, we solve the characteristic equation \(\det(\mathbf{A} - \lambda\mathbf{I}) = 0\):

\(\begin{vmatrix} 3-\lambda & -1 & 0 \\ -1 & 2-\lambda & -1 \\ 0 & -1 & 3-\lambda \end{vmatrix} = 0\)

Expanding along the first row:

\((3-\lambda)[(2-\lambda)(3-\lambda) - 1] - (-1)[(-1)(3-\lambda) - 0] = 0\)

\((3-\lambda)(\lambda^2 - 5\lambda + 5) + (\lambda - 3) = 0\)

Factor out \((3-\lambda)\):

\((3-\lambda)(\lambda^2 - 5\lambda + 5 - 1) = 0\)

\((3-\lambda)(\lambda^2 - 5\lambda + 4) = 0\)

\((3-\lambda)(\lambda-1)(\lambda-4) = 0\)

Thus, the eigenvalues of \(\mathbf{A}\) are \(\lambda = 1, 3, 4\).

(b) For the eigenvalue \(\lambda = 4\), we solve the system \((\mathbf{A} - 4\mathbf{I})\mathbf{v} = \mathbf{0}\):

\(\begin{pmatrix} -1 & -1 & 0 \\ -1 & -2 & -1 \\ 0 & -1 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\)

From the first row, we get \(-x - y = 0 \implies y = -x\).
From the third row, we get \(-y - z = 0 \implies z = -y = x\).
Setting \(x = 1\) yields the eigenvector \(\begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}\).

(c) Since \(\mathbf{v} = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}\) is an eigenvector corresponding to the eigenvalue \(\lambda = 4\), we have:

\(\mathbf{A}^n \mathbf{v} = 4^n \mathbf{v} = 4^n \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}\).

(a)
M1: For attempting to find the characteristic equation by setting \(\det(\mathbf{A} - \lambda\mathbf{I}) = 0\).
A1: For obtaining a correct simplified cubic equation, such as \((3-\lambda)(\lambda^2 - 5\lambda + 4) = 0\).
A1: For verifying the eigenvalues are 1, 3, and 4.

(b)
M1: For attempting to solve \((\mathbf{A} - 4\mathbf{I})\mathbf{v} = \mathbf{0}\).
A1: For finding a correct eigenvector, e.g., \(\begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}\) (or any non-zero constant scalar multiple).

(c)
B1: For writing down the correct expression \(4^n \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}\).

(a) We use integration by parts with \(u = x^n\) and \(\frac{dv}{dx} = (1-x)^{1/2}\).
Then \(\frac{du}{dx} = n x^{n-1}\) and \(v = -\frac{2}{3}(1-x)^{3/2}\).

Applying integration by parts:

\(I_n = \left[ -\frac{2}{3} x^n (1-x)^{3/2} \right]_0^1 + \frac{2}{3} n \int_0^1 x^{n-1} (1-x)^{3/2} \, dx\)

Evaluating the boundary term, it vanishes at both limits because \(1-x = 0\) at \(x=1\), and \(x^n = 0\) at \(x=0\) (since \(n \ge 1\)).

Thus,

\(I_n = \frac{2}{3} n \int_0^1 x^{n-1} (1-x) \sqrt{1-x} \, dx\)

\(I_n = \frac{2}{3} n \int_0^1 (x^{n-1} \sqrt{1-x} - x^n \sqrt{1-x}) \, dx\)

\(I_n = \frac{2}{3} n (I_{n-1} - I_n)\)

Now we solve for \(I_n\):

\(I_n \left(1 + \frac{2}{3} n\right) = \frac{2}{3} n I_{n-1}\)

\(I_n \left(\frac{2n+3}{3}\right) = \frac{2n}{3} I_{n-1}\)

\(I_n = \frac{2n}{2n+3} I_{n-1}\).

(b) We first evaluate \(I_0\):

\(I_0 = \int_0^1 \sqrt{1-x} \, dx = \left[ -\frac{2}{3}(1-x)^{3/2} \right]_0^1 = 0 - \left( -\frac{2}{3} \right) = \frac{2}{3}\).

Using the reduction formula:

For \(n = 1\):

\(I_1 = \frac{2(1)}{2(1)+3} I_0 = \frac{2}{5} \left(\frac{2}{3}\right) = \frac{4}{15}\).

For \(n = 2\):

\(I_2 = \frac{2(2)}{2(2)+3} I_1 = \frac{4}{7} \left(\frac{4}{15}\right) = \frac{16}{105}\).

(a)
M1: For employing integration by parts with appropriate choices of \(u\) and \(dv/dx\).
A1: For obtaining the correct non-integral term and integral term, and showing that the boundary term vanishes.
M1: For expressing \((1-x)^{3/2}\) as \((1-x)\sqrt{1-x}\) and splitting the integral to form recurrence relation terms.
A1: For completing the algebraic steps to establish the given recurrence relation.

(b)
B1: For finding the correct value of \(I_0 = \frac{2}{3}\).
B1: For applying the reduction formula twice to obtain \(I_2 = \frac{16}{105}\).

(a) Let \(z = \cos \theta + i \sin \theta\).
By de Moivre's Theorem, \(z^n + z^{-n} = 2 \cos n\theta\).

Specifically, for \(n = 1\), we have \(z + z^{-1} = 2 \cos \theta\).
Raising this to the 5th power:

\((2 \cos \theta)^5 = (z + z^{-1})^5\)

Using the binomial theorem, we expand the right-hand side:

\(32 \cos^5 \theta = z^5 + 5z^3 + 10z + 10z^{-1} + 5z^{-3} + z^{-5}\)

Grouping terms with reciprocal powers together:

\(32 \cos^5 \theta = (z^5 + z^{-5}) + 5(z^3 + z^{-3}) + 10(z + z^{-1})\)

Using the relation \(z^n + z^{-n} = 2 \cos n\theta\), we substitute these back:

\(32 \cos^5 \theta = 2 \cos 5\theta + 5(2 \cos 3\theta) + 10(2 \cos \theta)\)

\(32 \cos^5 \theta = 2 \cos 5\theta + 10 \cos 3\theta + 20 \cos \theta\)

Dividing by 32 yields:

\[ \cos^5 \theta = \frac{1}{16} (\cos 5\theta + 5\cos 3\theta + 10\cos \theta) \]

(b) Using the identity derived in (a), we integrate:

\(\int_0^{\pi/2} \cos^5 \theta \, d\theta = \frac{1}{16} \int_0^{\pi/2} (\cos 5\theta + 5\cos 3\theta + 10\cos \theta) \, d\theta\)

\(= \frac{1}{16} \left[ \frac{1}{5}\sin 5\theta + \frac{5}{3}\sin 3\theta + 10\sin \theta \right]_0^{\pi/2}\)

At \(\theta = \pi/2\):

\(\sin\left(\frac{5\pi}{2}\right) = 1\), \(\sin\left(\frac{3\pi}{2}\right) = -1\), and \(\sin\left(\frac{\pi}{2}\right) = 1\).

At \(\theta = 0\), all terms are 0.

Substituting these values in:

\(\frac{1}{16} \left( \frac{1}{5} - \frac{5}{3} + 10 \right) = \frac{1}{16} \left( \frac{3 - 25 + 150}{15} \right) = \frac{1}{16} \left( \frac{128}{15} \right) = \frac{8}{15}\).

(a)
M1: For stating \(z^n + z^{-n} = 2 \cos n\theta\) and setting up \((2\cos\theta)^5 = (z+z^{-1})^5\).
M1: For expanding \((z+z^{-1})^5\) binomial expansion correctly.
A1: For grouping reciprocal pairs of powers of \(z\) correctly.
A1: For substituting \(2\cos n\theta\) and completing the proof to show the given identity.

(b)
M1: For integrating the trigonometric terms correctly to get \(\frac{1}{5}\sin 5\theta + \frac{5}{3}\sin 3\theta + 10\sin \theta\).
A1: For substituting limits correctly and obtaining \(\frac{8}{15}\).

To find the general solution, we first find the complementary function (CF) by solving the auxiliary equation:

\(m^2 + 4m + 5 = 0\)

Using the quadratic formula:

\(m = \frac{-4 \pm \sqrt{16 - 20}}{2} = -2 \pm i\)

This yields the complementary function:

\(y_{cf} = e^{-2x}(A \cos x + B \sin x)\)

Next, we find a particular integral (PI) by trying a function of the form \(y_{pi} = k e^{-x}\).

Then, \(\frac{dy_{pi}}{dx} = -k e^{-x}\) and \(\frac{d^2 y_{pi}}{dx^2} = k e^{-x}\).

Substituting these into the differential equation:

\(k e^{-x} + 4(-k e^{-x}) + 5(k e^{-x}) = 10 e^{-x}\)

\(2k e^{-x} = 10 e^{-x} \implies k = 5\)

So, the particular integral is \(y_{pi} = 5 e^{-x}\).

The general solution of the differential equation is:

\(y = e^{-2x}(A \cos x + B \sin x) + 5e^{-x}\)

Now we use the initial conditions to determine the constants \(A\) and \(B\).

At \(x = 0\), \(y = 6\):

\(6 = e^0(A \cos 0 + B \sin 0) + 5e^0 \implies 6 = A + 5 \implies A = 1\)

To use the second condition, we differentiate \(y\):

\(\frac{dy}{dx} = -2e^{-2x}(A \cos x + B \sin x) + e^{-2x}(-A \sin x + B \cos x) - 5e^{-x}\)

At \(x = 0\), \(\frac{dy}{dx} = -7\):

\(-7 = -2A + B - 5\)

Substitute \(A = 1\):

\(-7 = -2(1) + B - 5 \implies -7 = B - 7 \implies B = 0\)

Thus, the particular solution is:

\(y = e^{-2x} \cos x + 5e^{-x}\).

M1: For writing down the auxiliary equation and solving it to find roots \(m = -2 \pm i\) and the corresponding CF.
M1: For attempting to find a particular integral of the form \(y_{pi} = k e^{-x}\) and determining \(k = 5\).
A1: For stating the general solution \(y = e^{-2x}(A \cos x + B \sin x) + 5e^{-x}\).
M1: For substituting the first boundary condition \(y(0) = 6\) to find \(A = 1\).
M1: For differentiating the general solution and using \(\frac{dy}{dx}(0) = -7\) to find \(B\).
A1: For stating the correct particular solution \(y = e^{-2x} \cos x + 5e^{-x}\).

(i) The characteristic equation of \(\mathbf{A}\) is given by \(\det(\mathbf{A} - \lambda\mathbf{I}) = 0\):
\((a-
\lambda)(c-\lambda) - b^2 = 0 \implies \lambda^2 - (a+c)\lambda + ac - b^2 = 0\).
The discriminant of this quadratic equation is \(\Delta = (a+c)^2 - 4(ac-b^2) = a^2 + 2ac + c^2 - 4ac + 4b^2 = (a-c)^2 + 4b^2\).
Since \(a, b, c\) are real, \((a-c)^2 \ge 0\) and \(4b^2 \ge 0\), so \(\Delta \ge 0\). This guarantees that the eigenvalues \(\lambda_1\) and \(\lambda_2\) are always real.

(ii) If \(b \neq 0\), then \(4b^2 > 0\). Since \((a-c)^2 \ge 0\), we have \(\Delta = (a-c)^2 + 4b^2 > 0\). Hence, the roots \(\lambda_1\) and \(\lambda_2\) are distinct real numbers.
Let \(\mathbf{v}_1\) and \(\mathbf{v}_2\) be eigenvectors corresponding to \(\lambda_1\) and \(\lambda_2\), so \(\mathbf{A}\mathbf{v}_1 = \lambda_1\mathbf{v}_1\) and \(\mathbf{A}\mathbf{v}_2 = \lambda_2\mathbf{v}_2\).
Taking the transpose of the first relation gives \(\mathbf{v}_1^T \mathbf{A}^T = \lambda_1 \mathbf{v}_1^T\). Since \(\mathbf{A}\) is symmetric, \(\mathbf{A}^T = \mathbf{A}\), so \(\mathbf{v}_1^T \mathbf{A} = \lambda_1 \mathbf{v}_1^T\).
Multiplying this on the right by \(\mathbf{v}_2\) gives \(\mathbf{v}_1^T \mathbf{A} \mathbf{v}_2 = \lambda_1 \mathbf{v}_1^T \mathbf{v}_2\).
Multiplying the relation \(\mathbf{A}\mathbf{v}_2 = \lambda_2\mathbf{v}_2\) on the left by \(\mathbf{v}_1^T\) gives \(\mathbf{v}_1^T \mathbf{A} \mathbf{v}_2 = \lambda_2 \mathbf{v}_1^T \mathbf{v}_2\).
Equating the two expressions for \(\mathbf{v}_1^T \mathbf{A} \mathbf{v}_2\) yields \(\lambda_1 \mathbf{v}_1^T \mathbf{v}_2 = \lambda_2 \mathbf{v}_1^T \mathbf{v}_2 \implies (\lambda_1 - \lambda_2) \mathbf{v}_1^T \mathbf{v}_2 = 0\).
Since \(\lambda_1 \neq \lambda_2\), it follows that \(\mathbf{v}_1^T \mathbf{v}_2 = 0\), proving that \(\mathbf{v}_1\) and \(\mathbf{v}_2\) are orthogonal.

(iii) Normalizing the eigenvectors, let \(\mathbf{u}_1 = \frac{\mathbf{v}_1}{\|\mathbf{v}_1\|}\) and \(\mathbf{u}_2 = \frac{\mathbf{v}_2}{\|\mathbf{v}_2\|}\).
Since \(\mathbf{u}_1\) and \(\mathbf{u}_2\) are orthogonal, \(\mathbf{u}_1^T \mathbf{u}_2 = \mathbf{u}_2^T \mathbf{u}_1 = 0\). Because they are normalized, \(\mathbf{u}_1^T \mathbf{u}_1 = \mathbf{u}_2^T \mathbf{u}_2 = 1\).
Let \(\mathbf{P} = (\mathbf{u}_1 \ \mathbf{u}_2)\). Then:
\(\mathbf{P}^T \mathbf{P} = \begin{pmatrix} \mathbf{u}_1^T \\ \mathbf{u}_2^T \end{pmatrix} \begin{pmatrix} \mathbf{u}_1 & \mathbf{u}_2 \end{pmatrix} = \begin{pmatrix} \mathbf{u}_1^T \mathbf{u}_1 & \mathbf{u}_1^T \mathbf{u}_2 \\ \mathbf{u}_2^T \mathbf{u}_1 & \mathbf{u}_2^T \mathbf{u}_2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \mathbf{I\}}.
Thus \)\mathbf{P}\) is an orthogonal matrix.
Now, \(\mathbf{A}\mathbf{P} = \mathbf{A}(\mathbf{u}_1 \ \mathbf{u}_2) = (\mathbf{A}\mathbf{u}_1 \ \mathbf{A}\mathbf{u}_2) = (\lambda_1 \mathbf{u}_1 \ \lambda_2 \mathbf{u}_2)\).
Premultiplying by \(\mathbf{P}^T\):
\(\mathbf{P}^T \mathbf{A} \mathbf{P} = \begin{pmatrix} \mathbf{u}_1^T \\ \mathbf{u}_2^T \end{pmatrix} (\lambda_1 \mathbf{u}_1 \ \lambda_2 \mathbf{u}_2) = \begin{pmatrix} \lambda_1 \mathbf{u}_1^T \mathbf{u}_1 & \lambda_2 \mathbf{u}_1^T \mathbf{u}_2 \\ \lambda_1 \mathbf{u}_2^T \mathbf{u}_1 & \lambda_2 \mathbf{u}_2^T \mathbf{u}_2 \end{pmatrix} = \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix}\).

(i) M1: Setting up the characteristic equation and expanding to obtain the quadratic equation in \(\lambda\).
A1: Finding the discriminant \(\Delta = (a-c)^2 + 4b^2\) or equivalent.
A1: Stating that \((a-c)^2 \ge 0\) and \(4b^2 \ge 0\) for real \(a, b, c\).
A1: Concluding that \(\Delta \ge 0\) and hence eigenvalues are real.

(ii) M1: Showing that \(b \neq 0 \implies \Delta > 0\) hence distinct eigenvalues.
M1: Writing the equations \(\mathbf{A}\mathbf{v}_1 = \lambda_1\mathbf{v}_1\) and \(\mathbf{A}\mathbf{v}_2 = \lambda_2\mathbf{v}_2\).
M1: Taking the transpose of the first relation and using the symmetry of \(\mathbf{A}\) (\(\mathbf{A}^T = \mathbf{A}\)).
A1: Equating \(\mathbf{v}_1^T \mathbf{A} \mathbf{v}_2\) from both expressions to obtain \((\lambda_1 - \lambda_2)\mathbf{v}_1^T\mathbf{v}_2 = 0\).
A1: Explaining that since \(\lambda_1 \neq \lambda_2\), \(\mathbf{v}_1^T\mathbf{v}_2 = 0\), which proves orthogonality.

(iii) M1: Setting up the matrix product \(\mathbf{P}^T \mathbf{P}\) using column vectors \(\mathbf{u}_1, \mathbf{u}_2\).
A1: Evaluating \(\mathbf{P}^T \mathbf{P} = \mathbf{I}\) and concluding \(\mathbf{P}\) is orthogonal.
M1: Expressing \(\mathbf{A}\mathbf{P}\) as \((\lambda_1 \mathbf{u}_1 \ \lambda_2 \mathbf{u}_2)\) and performing the multiplication \(\mathbf{P}^T \mathbf{A} \mathbf{P}\).
A0.75: Concluding that the result is the diagonal matrix \(\mathbf{D}\).

(i) By de Moivre's theorem, we have \(\cos(5\theta) + \mathrm{i}\sin(5\theta) = (\cos\theta + \mathrm{i}\sin\theta)^5\).
Expanding the right-hand side 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 the imaginary parts of both sides gives:
\(\sin(5\theta) = 5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta\).
We substitute \(\cos^2\theta = 1 - \sin^2\theta\) to express the relation entirely in terms of \(\sin\theta\):
\(\sin(5\theta) = 5(1-\sin^2\theta)^2\sin\theta - 10(1-\sin^2\theta)\sin^3\theta + \sin^5\theta\)
\(= 5(1 - 2
\sin^2\theta + \sin^4\theta)\sin\theta - 10(\sin^3\theta - \sin^5\theta) + \sin^5\theta\)
\(= (5\sin\theta - 10\sin^3\theta + 5\sin^5\theta) - (10\sin^3\theta - 10\sin^5\theta) + \sin^5\theta\)
\(= 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta\)
\(= \sin\theta(16\sin^4\theta - 20\sin^2\theta + 5)\).

(ii) The equation \(\sin(5\theta) = 0\) has solutions given by \(5\theta = k\pi \implies \theta = \frac{k\pi}{5}\) for \(k \in \mathbb{Z}\).
For \(k = 1, 2, 3, 4\), the corresponding values of \(\theta\) are \(\frac{\pi}{5}, \frac{2\pi}{5}, \frac{3\pi}{5}, \frac{4\pi}{5}\).
Since \(\sin\theta \neq 0\) for these values of \(\theta\), we must have:
\(16\sin^4\theta - 20\sin^2\theta + 5 = 0\).
Let \(x = \sin^2\theta\). Then this becomes \(16x^2 - 20x + 5 = 0\).
Notice that:
\(\sin^2\left(\frac{4\pi}{5}\right) = \sin^2\left(\pi - \frac{\pi}{5}\right) = \sin^2\left(\frac{\pi}{5}\right)\),
\(\sin^2\left(\frac{3\pi}{5}\right) = \sin^2\left(\pi - \frac{2\pi}{5}\right) = \sin^2\left(\frac{2\pi}{5}\right)\).
Thus, the four values of \(\theta\) yield exactly two distinct values for \(x = \sin^2\theta\), which are \(\sin^2\left(\frac{\pi}{5}\right)\) and \(\sin^2\left(\frac{2\pi}{5}\right)\).
Since these values satisfy the equation, they are the roots of the quadratic equation \(16x^2 - 20x + 5 = 0\).

(iii) By Vieta's relations, for a quadratic equation \(Ax^2 + Bx + C = 0\), the product of the roots is \(\frac{C}{A}\) and the sum of the roots is \(-\frac{B}{A}\).
For \(16x^2 - 20x + 5 = 0\):
The product of the roots is \(\sin^2\left(\frac{\pi}{5}\right)\sin^2\left(\frac{2\pi}{5}\right) = \frac{5}{16}\).
The sum of the roots is \(\sin^2\left(\frac{\pi}{5}\right) + \sin^2\left(\frac{2\pi}{5}\right) = -\frac{-20}{16} = \frac{5}{4}\).

(i) M1: Applying de Moivre's theorem to write \(\cos(5\theta) + \mathrm{i}\sin(5\theta) = (\cos\theta + \mathrm{i}\sin\theta)^5\).
M1: Expanding the right-hand side using the binomial theorem.
A1: Correctly equating the imaginary parts to obtain \(\sin(5\theta) = 5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta\).
M1: Substituting \(\cos^2\theta = 1 - \sin^2\theta\) and expanding.
A1: Completing the algebraic simplification to reach \(\sin\theta(16\sin^4\theta - 20\sin^2\theta + 5)\).

(ii) M1: Solving \(\sin(5\theta) = 0\) to get \(\theta = \frac{k
\pi}{5}\).
A1: Identifying the four key non-zero roots for \(k = 1, 2, 3, 4\).
M1: Making the substitution \(x = \sin^2\theta\) to form the quadratic equation \(16x^2 - 20x + 5 = 0\).
A1.75: Explaining how the four values of \(\theta\) collapse into two distinct values for \(\sin^2\theta\) due to symmetry (\(\sin(\pi - \phi) = \sin\phi\)), confirming they are indeed the roots of the quadratic equation.

(iii) M1: Using Vieta's formulas for the product of roots on \(16x^2 - 20x + 5 = 0\).
A1: Deduating that the product is \(\frac{5}{16}\).
A1: Finding the sum of the roots as \(\frac{5}{4}\).

(i) We write the integrand as \(\operatorname{sech}^n x = \operatorname{sech}^{n-2} x \operatorname{sech}^2 x\).
We apply integration by parts to \(\int_0^a \operatorname{sech}^{n-2} x \operatorname{sech}^2 x \ \mathrm{d}x\):
Let \(u = \operatorname{sech}^{n-2} x \implies \frac{\mathrm{d}u}{\mathrm{d}x} = -(n-2)\operatorname{sech}^{n-2} x \tanh x\).
Let \(\frac{\mathrm{d}v}{\mathrm{d}x} = \operatorname{sech}^2 x \implies v = \tanh x\).
Using the integration by parts formula \(\int u v' \mathrm{d}x = [uv] - \int u' v \mathrm{d}x\):
\(I_n = \left[ \operatorname{sech}^{n-2} x \tanh x \right]_0^a - \int_0^a -(n-2)\operatorname{sech}^{n-2} x \tanh^2 x \ \mathrm{d}x\)
\(I_n = \operatorname{sech}^{n-2} a \tanh a + (n-2)\int_0^a \operatorname{sech}^{n-2} x \tanh^2 x \ \mathrm{d}x\) (since \(\tanh 0 = 0\)).
Using the identity \(\tanh^2 x = 1 - \operatorname{sech}^2 x\):
\(I_n = \operatorname{sech}^{n-2} a \tanh a + (n-2)\int_0^a \operatorname{sech}^{n-2} x (1 - \operatorname{sech}^2 x) \ \mathrm{d}x\)
\(I_n = \operatorname{sech}^{n-2} a \tanh a + (n-2)\int_0^a \operatorname{sech}^{n-2} x \ \mathrm{d}x - (n-2)\int_0^a \operatorname{sech}^n x \ \mathrm{d}x\)
\(I_n = \operatorname{sech}^{n-2} a \tanh a + (n-2)I_{n-2} - (n-2)I_n\).
Rearranging to group the \(I_n\) terms:
\(I_n + (n-2)I_n = \tanh a \operatorname{sech}^{n-2} a + (n-2)I_{n-2}\)
\((n-1)I_n = \tanh a \operatorname{sech}^{n-2} a + (n-2)I_{n-2}\).

(ii) Using the fundamental hyperbolic identity \(\cosh^2 a - \sinh^2 a = 1\), we divide by \(\cosh^2 a\) to obtain:
\(1 - \tanh^2 a = \operatorname{sech}^2 a\).
Given \(\tanh a = \frac{4}{5}\):
\(
\operatorname{sech}^2 a = 1 - \left(\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{9}{25}\).
Since \(\operatorname{sech} x = \frac{1}{\cosh x} > 0\) for all real \(x\), we have:
\(\operatorname{sech} a = \frac{3}{5}\).

(iii) We want to find \(I_4 = \int_0^a \operatorname{sech}^4 x \ \mathrm{d}x\).
Using the reduction formula with \(n = 4\):
\(3I_4 = \tanh a \operatorname{sech}^2 a + 2I_2\).
First, we compute \(I_2 = \int_0^a \operatorname{sech}^2 x \ \mathrm{d}x = [\tanh x]_0^a = \tanh a = \frac{4}{5}\).
Now substitute the known values \(\tanh a = \frac{4}{5}\) and \(\operatorname{sech} a = \frac{3}{5}\) into the reduction formula:
\(3I_4 = \left(\frac{4}{5}\right)\left(\frac{3}{5}\right)^2 + 2\left(\frac{4}{5}\right)\)
\(3I_4 = \frac{4}{5}\left(\frac{9}{25}\right) + \frac{8}{5} = \frac{36}{125} + \frac{200}{125} = \frac{236}{125}\).
Therefore,
\(I_4 = \frac{236}{375}\).

(i) M1: Splitting the integrand into \(\operatorname{sech}^{n-2} x \operatorname{sech}^2 x\) and attempting integration by parts.
A1: Correctly differentiating \(\operatorname{sech}^{n-2} x\) to get \(-(n-2)\operatorname{sech}^{n-2} x \tanh x\).
A1: Correctly integrating \(\operatorname{sech}^2 x\) to get \(\tanh x\).
M1: Substituting the limits and evaluating the boundary term correctly.
M1: Using \(\tanh^2 x = 1 - \operatorname{sech}^2 x\) to express the remaining integral in terms of \(I_{n-2}\) and \(I_n\).
A1: Rearranging terms to establish the final reduction formula.

(ii) M1: Recalling and using the identity \(1 - \tanh^2 a = \operatorname{sech}^2 a\).
A1: Substituting \(\tanh a = \frac{4}{5}\) to find \(\operatorname{sech}^2 a = \frac{9}{25}\).
A0.75: Justifying that \(\operatorname{sech} a > 0\) and concluding \(\operatorname{sech} a = \frac{3}{5}\).

(iii) M1: Applying the reduction formula with \(n = 4\) to write \(3I_4\) in terms of \(\tanh a\), \(\operatorname{sech} a\), and \(I_2\).
A1: Finding \(I_2 = \frac{4}{5}\).
M1: Substituting the numeric values into the equation for \(3I_4\).
A1: Correctly calculating the final exact fraction \(I_4 = \frac{236}{375}\).

(i) We are given \(y = u \mathrm{e}^{kx}\). Differentiating with respect to \(x\) using the product rule:
\(\frac{\mathrm{d} y}{\mathrm{d} x} =
\frac{\mathrm{d} u}{\mathrm{d} x} \mathrm{e}^{kx} + k u \mathrm{e}^{kx} = \left(\frac{\mathrm{d} u}{\mathrm{d} x} + k u\right)\mathrm{e}^{kx}\).
Differentiating once more with respect to \(x\) using the product rule:
\(\frac{\mathrm{d}^2 y}{\mathrm{d} x^2} = \left(\frac{\mathrm{d}^2 u}{\mathrm{d} x^2} + k \frac{\mathrm{d} u}{\mathrm{d} x}\right)\mathrm{e}^{kx} + k\left(\frac{\mathrm{d} u}{\mathrm{d} x} + k u\right)\mathrm{e}^{kx} = \left(\frac{\mathrm{d}^2 u}{\mathrm{d} x^2} + 2k \frac{\mathrm{d} u}{\mathrm{d} x} + k^2 u\right)\mathrm{e}^{kx}\).
Substituting these expressions for \(y\), \(\frac{\mathrm{d} y}{\mathrm{d} x}\), and \(\frac{\mathrm{d}^2 y}{\mathrm{d} x^2}\) into the original differential equation:
\(\left(\frac{\mathrm{d}^2 u}{\mathrm{d} x^2} + 2k \frac{\mathrm{d} u}{\mathrm{d} x} + k^2 u\right)\mathrm{e}^{kx} - 2k \left(\frac{\mathrm{d} u}{\mathrm{d} x} + k u\right)\mathrm{e}^{kx} + k^2 u \mathrm{e}^{kx} = \mathrm{e}^{kx}\).
Since \(\mathrm{e}^{kx} \neq 0\) for all \(x\), we can divide both sides by \(\mathrm{e}^{kx}\):
\(\frac{\mathrm{d}^2 u}{\mathrm{d} x^2} + 2k \frac{\mathrm{d} u}{\mathrm{d} x} + k^2 u - 2k \frac{\mathrm{d} u}{\mathrm{d} x} - 2k^2 u + k^2 u = 1\).
Combining like terms:
\(\frac{\mathrm{d}^2 u}{\mathrm{d} x^2} + (2k - 2k)\frac{\mathrm{d} u}{\mathrm{d} x} + (k^2 - 2k^2 + k^2)u = 1 \implies \frac{\mathrm{d}^2 u}{\mathrm{d} x^2} = 1\).
This completes the proof.

(ii) To find the general solution, we integrate \(\frac{\mathrm{d}^2 u}{\mathrm{d} x^2} = 1\) twice with respect to \(x\):
First integration: \(\frac{\mathrm{d} u}{\mathrm{d} x} = x + A\), where \(A\) is an arbitrary constant.
Second integration: \(u = \frac{1}{2}x^2 + Ax + B\), where \(B\) is another arbitrary constant.
Since \(y = u \mathrm{e}^{kx}\), we substitute \(u\) back to get:
\(y = \left(\frac{1}{2}x^2 + Ax + B\right)\mathrm{e}^{kx}\).

(iii) We use the initial conditions \(y = 1\) and \(\frac{\mathrm{d} y}{\mathrm{d} x} = k\) at \(x = 0\) to determine the constants \(A\) and \(B\).
Using \(y(0) = 1\):
\(1 = \left(\frac{1}{2}(0)^2 + A(0) + B\right)\mathrm{e}^{0} \implies 1 = B \implies B = 1\).
Our solution becomes:
\(y = \left(\frac{1}{2}x^2 + Ax + 1\right)\mathrm{e}^{kx}\).
Next, we find \(\frac{\mathrm{d} y}{\mathrm{d} x}\) using the product rule:
\(\frac{\mathrm{d} y}{\mathrm{d} x} = (x + A)\mathrm{e}^{kx} + k\left(\frac{1}{2}x^2 + Ax + 1\right)\mathrm{e}^{kx}\).
Using \(\frac{\mathrm{d} y}{\mathrm{d} x} = k\) at \(x = 0\):
\(k = (0 + A)\mathrm{e}^{0} + k\left(0 + 0 + 1\right)\mathrm{e}^{0} \implies k = A + k \implies A = 0\).
Substituting \(A = 0\) and \(B = 1\) back into the solution:
\(y = \left(1 + \frac{1}{2}x^2\right)\mathrm{e}^{kx}\).
This completes the proof.

(i) M1: Attempting to differentiate \(y = u \mathrm{e}^{kx}\) with respect to \(x\).
A1: Finding the correct expression for \(\frac{\mathrm{d} y}{\mathrm{d} x}\).
M1: Attempting to differentiate a second time using the product rule.
A1: Finding the correct expression for \(\frac{\mathrm{d}^2 y}{\mathrm{d} x^2}\).
M1: Substituting these derivatives into the original differential equation.
A1: Correctly simplifying and dividing by \(\mathrm{e}^{kx}\) to obtain \(\frac{\mathrm{d}^2 u}{\mathrm{d} x^2} = 1\).

(ii) M1: Integrating \(\frac{\mathrm{d}^2 u}{\mathrm{d} x^2} = 1\) twice with respect to \(x\) (including arbitrary constants).
A1: Correctly finding \(u = \frac{1}{2}x^2 + Ax + B\).
A0.75: Stating the general solution \(y = \left(\frac{1}{2}x^2 + Ax + B\right)\mathrm{e}^{kx}\).

(iii) M1: Applying the initial condition \(y(0) = 1\) to find \(B\).
A1: Correctly obtaining \(B = 1\).
M1: Differentiating the solution and applying \(\frac{\mathrm{d} y}{\mathrm{d} x} = k\) at \(x = 0\) to find \(A\).
A1: Finding \(A = 0\) and concluding with the correct particular solution \(y = \left(1 + \frac{1}{2}x^2\right)\mathrm{e}^{kx}\).

Let \(AP = x\). The tension in string 1 is \(T_1 = \frac{4mg}{2a}(x - 2a) = \frac{2mg}{a}(x - 2a)\) for \(x > 2a\). The tension in string 2 is \(T_2 = \frac{mg}{a}(3a - x)\) for \(x < 3a\). At equilibrium, \(T_1 = T_2\), which gives \(2(x_0 - 2a) = 3a - x_0 \implies 3x_0 = 7a \implies x_0 = \frac{7}{3}a\). The initial position is \(x = 2.5a\). Initial elastic potential energy is \(EPE_i = \frac{4mg(2.5a - 2a)^2}{4a} + \frac{mg(3a - 2.5a)^2}{2a} = \frac{mg(0.25a^2)}{a} + \frac{mg(0.25a^2)}{2a} = \frac{3}{8}mga\). Since the particle is released from rest, the initial kinetic energy is 0. At the equilibrium position \(x_0 = \frac{7}{3}a\), the elastic potential energy is \(EPE_{eq} = \frac{4mg(\frac{7}{3}a - 2a)^2}{4a} + \frac{mg(3a - \frac{7}{3}a)^2}{2a} = \frac{1}{9}mga + \frac{2}{9}mga = \frac{1}{3}mga\). By conservation of energy, \(E_i = EPE_{eq} + KE_{eq} \implies \frac{3}{8}mga = \frac{1}{3}mga + \frac{1}{2}mv^2 \implies \frac{1}{2}mv^2 = \frac{1}{24}mga \implies v^2 = \frac{1}{12}ga \implies v = \sqrt{\frac{ga}{12}} = \frac{1}{6}\sqrt{3ga}\).

M1: Set up equilibrium equation T_1 = T_2. A1: Correctly find equilibrium position x = 7a/3. M1: Calculate initial EPE of both strings. A1: Correctly find initial EPE as 3mga/8. M1: Calculate EPE of both strings at equilibrium. A1: Correctly find EPE at equilibrium as mga/3. M1: Use conservation of energy to find v. A0.5: Correctly find final speed v = \sqrt{ga/12}.

Immediately after the horizontal blow, the speed of \(P\) is \(u = I/m\). When the string makes an angle \(\theta\) with the upward vertical, the height \(h\) of the particle above the lowest point is \(h = L + L\cos\theta\). Since \(\cos\theta = 1/3\), \(h = L(1 + 1/3) = \frac{4}{3}L\). By conservation of energy, \(\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mgh \implies v^2 = u^2 - 2g(\frac{4}{3}L) = u^2 - \frac{8}{3}gL\). The equation of motion along the radial direction is \(T + mg\cos\theta = \frac{mv^2}{L}\). The string becomes slack when \(T = 0\), which gives \(mg\cos\theta = \frac{mv^2}{L} \implies v^2 = gL\cos\theta\). Since \(\cos\theta = 1/3\), \(v^2 = \frac{1}{3}gL\). Equating the two expressions for \(v^2\) gives \(\frac{1}{3}gL = u^2 - \frac{8}{3}gL \implies u^2 = 3gL \implies u = \sqrt{3gL}\). Since \(I = mu\), we have \(I = m\sqrt{3gL}\).

B1: State relation between impulse and initial velocity I = mu. M1: Use conservation of energy to find v^2 in terms of u, g, and L. A1: Correctly find v^2 = u^2 - 8gL/3. M1: Use radial component of Newton's second law. A1: Correctly identify slackness condition T = 0 to get v^2 = gL/3. M2: Solve for u. A0.5: Obtain correct final answer I = m\sqrt{3gL}.

(a) Using Newton's second law, \(m \frac{dv}{dt} = -mv(1 + k^2 v^2) \implies \frac{dv}{dt} = -v(1 + k^2 v^2)\). Separating variables gives \(\int \frac{1}{v(1 + k^2 v^2)} dv = -\int dt\). Using partial fractions, \(\int (\frac{1}{v} - \frac{k^2 v}{1 + k^2 v^2}) dv = -t + C \implies \ln\left(\frac{v}{\sqrt{1 + k^2 v^2}}\right) = -t + C\). At \(t = 0\), \(v = u\), so \(C = \ln\left(\frac{u}{\sqrt{1 + k^2 u^2}}\right)\). This gives \(\frac{v}{\sqrt{1 + k^2 v^2}} = \frac{u}{\sqrt{1 + k^2 u^2}} e^{-t}\). Squaring and rearranging: \(\frac{v^2}{1 + k^2 v^2} = \frac{u^2 e^{-2t}}{1 + k^2 u^2} \implies v^2 = \frac{u^2}{(1 + k^2 u^2)e^{2t} - k^2 u^2}\). Since \(v > 0\), \(v = \frac{u}{\sqrt{(1+k^2 u^2)e^{2t} - k^2 u^2}}\). (b) Using \(v \frac{dv}{dx} = -v(1 + k^2 v^2) \implies \frac{dv}{dx} = -(1 + k^2 v^2)\). Separating variables gives \(\int \frac{1}{1 + k^2 v^2} dv = -\int dx \implies \frac{1}{k}\arctan(kv) = -x + D\). Since \(x = 0\) when \(v = u\), \(D = \frac{1}{k}\arctan(ku)\). Thus, \(x = \frac{1}{k}(\arctan(ku) - \arctan(kv))\). When \(v = \frac{1}{2}u\), \(x = \frac{1}{k}\left(\arctan(ku) - \arctan\left(\frac{1}{2}ku\right)\right)\).

Part (a) [5 marks]: M1: Write down differential equation and separate variables. M1: Use partial fractions correctly. A1: Correct integration with constant of integration. M1: Substitute boundary conditions to find C and eliminate logs. A1: Complete correct algebraic steps to show the required expression. Part (b) [2.5 marks]: M1: Use acceleration in the form v dv/dx and separate variables. A1: Integrate correctly to get x in terms of v. A0.5: Substitute v = u/2 to obtain the correct final distance x.

**Part (i):**
We are given that the resisting force is \(F = v(1+v^2)\). Using Newton's second law:
\[m \frac{\mathrm{d}v}{\mathrm{d}t} = -v(1+v^2)\]
Since \(m = 1\):
\[\frac{\mathrm{d}v}{\mathrm{d}t} = -v(1+v^2)\]
Separating variables gives:
\[\int \frac{1}{v(1+v^2)} \,\mathrm{d}v = -\int \mathrm{d}t\]
Using partial fractions:
\[\frac{1}{v(1+v^2)} = \frac{1}{v} - \frac{v}{1+v^2}\]
Integrating both sides:
\[\int \left(\frac{1}{v} - \frac{v}{1+v^2}\right) \,\mathrm{d}v = -t + C\]
\[\ln v - \frac{1}{2}\ln(1+v^2) = -t + C\]
\[\frac{1}{2}\ln\left(\frac{v^2}{1+v^2}\right) = -t + C\]
Using the initial condition \(v(0) = 3\):
\[C = \frac{1}{2}\ln\left(\frac{9}{10}\right)\]
Thus:
\[-t = \frac{1}{2}\ln\left(\frac{v^2}{1+v^2}\right) - \frac{1}{2}\ln\left(\frac{9}{10}\right) = \frac{1}{2}\ln\left(\frac{10v^2}{9(1+v^2)}\right)\]
When \(v = 1\):
\[-t = \frac{1}{2}\ln\left(\frac{10}{18}\right) = \frac{1}{2}\ln\left(\frac{5}{9}\right)\]
\[t = \frac{1}{2}\ln\left(\frac{9}{5}\right) = \frac{1}{2}\ln(1.8) \approx 0.294\text{ s}.\]

**Part (ii):**
Using \(a = v \frac{\mathrm{d}v}{\mathrm{d}x}\):
\[v \frac{\mathrm{d}v}{\mathrm{d}x} = -v(1+v^2)\]
Since \(v \neq 0\), we can divide by \(v\):
\[\frac{\mathrm{d}v}{\mathrm{d}x} = -(1+v^2)\]
Separating variables:
\[\int \frac{1}{1+v^2} \,\mathrm{d}v = -\int \mathrm{d}x\]
\[\arctan v = -x + C'\]
Using the initial condition \(x = 0\) when \(v = 3\):
\[C' = \arctan 3\]
So:
\[x = \arctan 3 - \arctan v\]
When \(v = 1\):
\[x = \arctan 3 - \arctan 1 = \arctan 3 - \frac{\pi}{4} \approx 0.464\text{ m}.\]

- **Part (i) [5 marks]:**
- M1: Set up the differential equation \(\frac{\mathrm{d}v}{\mathrm{d}t} = -v(1+v^2)\).
- A1: Separate variables and apply partial fractions correctly.
- M1: Integrate to obtain \(\ln v - \frac{1}{2}\ln(1+v^2)\) or equivalent.
- A1: Correctly apply the limits \(v = 3\) to \(v = 1\) (or find the constant of integration).
- A1: Obtain the final exact time \(t = \frac{1}{2}\ln(1.8)\) or \(0.294\text{ s}\).
- **Part (ii) [4.33 marks]:**
- M1: Use \(a = v \frac{\mathrm{d}v}{\mathrm{d}x}\) to set up the differential equation in terms of \(v\) and \(x\).
- A1: Separate variables to get \(\int \frac{1}{1+v^2} \,\mathrm{d}v = -x\).
- M1: Integrate to get \(\arctan v\) and apply limits \(v = 3\) to \(v = 1\).
- A1.33: Obtain the final exact distance \(x = \arctan(3) - \frac{\pi}{4}\) or \(0.464\text{ m}\).

**Part (i):**
Let \(R_A\) and \(F_A\) be the normal reaction and friction force at the ground \(A\).
Let \(R_B\) and \(F_B\) be the normal reaction and friction force at the wall \(B\).
Since the ladder is on the point of slipping downwards at \(B\), the end \(A\) is on the point of slipping away from the wall (to the left). Thus, the friction force \(F_A\) acts to the right (towards the wall), and \(F_B\) acts upwards.
Since limiting equilibrium is reached at both contacts:
\[F_A = \mu R_A = 0.5 R_A\]
\[F_B = \mu R_B = 0.5 R_B\]

Resolving forces:
Horizontally: \(F_A - R_B = 0 \implies R_B = 0.5 R_A \implies R_A = 2 R_B\)
Vertically: \(R_A + F_B - W = 0 \implies 2 R_B + 0.5 R_B = W \implies 2.5 R_B = W \implies R_B = 0.4 W\).
Then, \(R_A = 0.8 W\) and \(F_B = 0.2 W\).

Taking moments about \(A\):
The clockwise moment is due to the weight: \(W \times a \cos\theta\).
The anticlockwise moments are due to \(R_B\) and \(F_B\):
\[R_B (2a \sin\theta) + F_B (2a \cos\theta)\]
Equating moments:
\[W a \cos\theta = 2 R_B a \sin\theta + 2 F_B a \cos\theta\]
Divide by \(a\):
\[W \cos\theta = 2 R_B \sin\theta + 2 F_B \cos\theta\]
Substitute \(R_B = 0.4 W\) and \(F_B = 0.2 W\):
\[W \cos\theta = 2 (0.4 W) \sin\theta + 2 (0.2 W) \cos\theta\]
Divide by \(W\):
\[\cos\theta = 0.8 \sin\theta + 0.4 \cos\theta \implies 0.6 \cos\theta = 0.8 \sin\theta\]
\[\tan\theta = \frac{0.6}{0.8} = 0.75\].

**Part (ii):**
Let the person of weight \(W\) be at a distance \(x\) from \(A\) along the ladder.
Since the ladder is on the point of slipping:
\[F_A = 0.5 R_A \quad \text{and} \quad F_B = 0.5 R_B\]
Resolving forces:
Horizontally: \(F_A = R_B\)
Vertically: \(R_A + F_B = W_{\text{ladder}} + W_{\text{person}} = 2W\).
Substituting \(F_A = R_B\) and the friction laws:
\[R_B = 0.5 R_A\]
\[R_A + 0.5 R_B = 2W \implies R_A + 0.25 R_A = 2W \implies 1.25 R_A = 2W \implies R_A = 1.6 W\]
This gives:
\[R_B = 0.8 W \quad \text{and} \quad F_B = 0.4 W\]

Taking moments about \(A\):
\[W a \cos\theta + W x \cos\theta = R_B (2a \sin\theta) + F_B (2a \cos\theta)\]
Divide by \(W \cos\theta\):
\[a + x = 2a R_B \tan\theta / W + 2a F_B / W\]
Substitute \( \tan\theta = 0.75 \), \( R_B = 0.8 W \) and \( F_B = 0.4 W \):
\[a + x = 2a (0.8) (0.75) + 2a (0.4)\]
\[a + x = 1.2a + 0.8a = 2a \implies x = a\].
So the person can climb a distance \(a\) (exactly halfway up the ladder) before slipping occurs.

- **Part (i) [5 marks]:**
- M1: Set up horizontal and vertical equations of equilibrium with correct friction directions.
- A1: Show that \(R_A = 2 R_B\) and express \(R_B\) and \(F_B\) in terms of \(W\) (\(R_B = 0.4 W, F_B = 0.2 W\)).
- M1: Set up moments equation about \(A\) (or any other point).
- A1: Correctly substitute force components into the moments equation.
- A1: Show clearly that \(\tan\theta = 0.75\).
- **Part (ii) [4.33 marks]:**
- M1: Write down new vertical force equilibrium equation including the person's weight.
- A1: Calculate the new reactions \(R_A = 1.6 W\) and \(R_B = 0.8 W\).
- M1: Set up the new moments equation about \(A\) containing \(x\).
- A1.33: Substitute the reactions and \(\tan\theta = 0.75\) to obtain \(x = a\).

**Part (i):**
Let \(x\) be the extension of the string.
The particle falls from \(O\), so when the extension is \(x\), the total distance fallen is \(L + x\).
At the point of maximum extension, the velocity of the particle is \(0\), so its kinetic energy is \(0\).
By conservation of mechanical energy, the loss in gravitational potential energy equals the gain in elastic potential energy:
\[\Delta E_k + \Delta E_p + \Delta E_e = 0\]
Loss in GPE = \(mg(L+x)\)
Gain in EPE = \(\frac{\lambda x^2}{2L} = \frac{4mg x^2}{2L} = \frac{2mg x^2}{L}\)
Equating these:
\[mg(L+x) = \frac{2mg x^2}{L}\]
Since \(m \neq 0\) and \(g \neq 0\):
\[L + x = \frac{2x^2}{L} \implies 2x^2 - Lx - L^2 = 0\]
Factorizing this quadratic equation:
\[(2x + L)(x - L) = 0\]
Since the extension \(x\) must be positive, we find \(x = L\).
Thus, the maximum extension is \(L\).

**Part (ii):**
The maximum speed occurs when the acceleration is zero, which is the equilibrium position where the upward tension \(T\) equals the downward gravitational force \(mg\).
Using Hooke's Law:
\[T = \frac{\lambda x_0}{L} = \frac{4mg x_0}{L}\]
Setting \(T = mg\):
\[\frac{4mg x_0}{L} = mg \implies x_0 = \frac{1}{4}L\]
Let \(v\) be the maximum speed. At this position, the extension is \(x_0 = \frac{1}{4}L\), and the total distance fallen is \(L + x_0 = \frac{5}{4}L\).
By conservation of energy:
Loss in GPE = Gain in KE + Gain in EPE
\[mg\left(\frac{5}{4}L\right) = \frac{1}{2}mv^2 + \frac{4mg (L/4)^2}{2L}\]
\[\frac{5}{4}mgL = \frac{1}{2}mv^2 + \frac{4mg L^2 / 16}{2L}\]
\[\frac{5}{4}mgL = \frac{1}{2}mv^2 + \frac{1}{8}mgL\]
\[\frac{1}{2}mv^2 = \left(\frac{5}{4} - \frac{1}{8}\right)mgL = \frac{9}{8}mgL\]
\[v^2 = \frac{9}{4}gL\]
\[v = \frac{3}{2}\sqrt{gL}\].

- **Part (i) [5 marks]:**
- M1: Apply conservation of energy between the release point and the point of maximum extension.
- A1: Write down correct gravitational potential energy expression \(mg(L+x)\).
- A1: Write down correct elastic potential energy expression \(\frac{2mg x^2}{L}\).
- M1: Formulate and solve the quadratic equation \(2x^2 - Lx - L^2 = 0\).
- A1: Deduce that \(x = L\) is the only valid physical solution.
- **Part (ii) [4.33 marks]:**
- M1: State or imply that maximum speed occurs at the equilibrium position (zero acceleration).
- A1: Find the equilibrium extension \(x_0 = \frac{1}{4}L\).
- M1: Apply conservation of energy at \(x_0 = \frac{1}{4}L\).
- A1.33: Solve for \(v\) to obtain \(v = \frac{3}{2}\sqrt{gL}\) or equivalent form.

We define the differences as \(d = \text{Before} - \text{After}\). Let us calculate \(d\) for each athlete:
- Athlete A: \(12.4 - 12.0 = +0.4\)
- Athlete B: \(11.8 - 12.0 = -0.2\)
- Athlete C: \(12.2 - 11.6 = +0.6\)
- Athlete D: \(11.5 - 11.8 = -0.3\)
- Athlete E: \(12.6 - 11.8 = +0.8\)
- Athlete F: \(11.9 - 11.8 = +0.1\)
- Athlete G: \(12.1 - 11.1 = +1.0\)
- Athlete H: \(12.3 - 11.8 = +0.5\)

We rank the absolute differences \(|d|\):
1. \(|d| = 0.1\): rank 1
2. \(|d| = 0.2\): rank 2
3. \(|d| = 0.3\): rank 3
4. \(|d| = 0.4\): rank 4
5. \(|d| = 0.5\): rank 5
6. \(|d| = 0.6\): rank 6
7. \(|d| = 0.8\): rank 7
8. \(|d| = 1.0\): rank 8

Next, we sum the ranks of the positive and negative differences:
- Positive differences: \(+0.4, +0.6, +0.8, +0.1, +1.0, +0.5\). Ranks: \(4, 6, 7, 1, 8, 5\). Sum \(T^+ = 4 + 6 + 7 + 1 + 8 + 5 = 31\).
- Negative differences: \(-0.2, -0.3\). Ranks: \(2, 3\). Sum \(T^- = 2 + 3 = 5\).

The test statistic is \(T = \min(T^+, T^-) = 5\).

Let the hypotheses be:
- \(H_0\): The median difference is 0 (no change in running times).
- \(H_1\): The median difference is greater than 0 (the training programme reduces running times).

For a sample size of \(n = 8\), the critical value for a one-tailed Wilcoxon signed-rank test at the 5% significance level is 5.

Since \(T = 5 \le 5\), we reject \(H_0\).

There is sufficient evidence at the 5% level of significance to suggest that the speed training programme reduces the median running time of athletes.

M1: State appropriate null and alternative hypotheses.
A1: Calculate the correct differences and their absolute values.
M1: Correctly rank the absolute differences (no errors).
A1: Calculate the sum of positive ranks \(T^+ = 31\) and negative ranks \(T^- = 5\).
A1: Identify the correct test statistic \(T = 5\).
B1: State the correct critical value of 5 for \(n=8\) at the 5% level (one-tailed).
M1: Compare \(T\) with the critical value and make a decision to reject \(H_0\).
A0.75: Provide a clear contextual conclusion in terms of the training programme reducing the median running time.

Let \(X \sim \text{Po}(1.75)\). The expected frequencies \(E_i = 100 \times P(X=i)\) are calculated using the Poisson probability function:
- \(P(X=0) = e^{-1.75} \approx 0.1738 \implies E_0 = 17.38\)
- \(P(X=1) = e^{-1.75} \times 1.75 \approx 0.3041 \implies E_1 = 30.41\)
- \(P(X=2) = e^{-1.75} \times \frac{1.75^2}{2} \approx 0.2661 \implies E_2 = 26.61\)
- \(P(X=3) = e^{-1.75} \times \frac{1.75^3}{6} \approx 0.1552 \implies E_3 = 15.52\)
- \(P(X=4) = e^{-1.75} \times \frac{1.75^4}{24} \approx 0.0679 \implies E_4 = 6.79\)
- \(P(X \ge 5) = 1 - (0.1738 + 0.3041 + 0.2661 + 0.1552 + 0.0679) \approx 0.0329 \implies E_{\ge 5} = 3.29\)

Since \(E_{\ge 5} = 3.29 < 5\), we pool the classes for \(x = 4\) and \(x \ge 5\):
- For \(x \ge 4\), Observed frequency \(O = 8 + 2 = 10\)
- Expected frequency \(E = 6.79 + 3.29 = 10.08\)

Now we calculate the contribution to \(\chi^2\):

$$\chi^2 = \sum \frac{(O - E)^2}{E}$$

- \(x = 0\): \(\frac{(16 - 17.38)^2}{17.38} \approx 0.1096\)
- \(x = 1\): \(\frac{(31 - 30.41)^2}{30.41} \approx 0.0114\)
- \(x = 2\): \(\frac{(27 - 26.61)^2}{26.61} \approx 0.0057\)
- \(x = 3\): \(\frac{(16 - 15.52)^2}{15.52} \approx 0.0148\)
- \(x \ge 4\): \(\frac{(10 - 10.08)^2}{10.08} \approx 0.0006\)

Sum of contributions \(\chi^2 \approx 0.1096 + 0.0114 + 0.0057 + 0.0148 + 0.0006 = 0.1421\) (or \(0.143\) to 3 d.p.).

Degrees of freedom \(\nu\):
- Number of classes after pooling = 5.
- We estimated 1 parameter (\(\lambda = 1.75\)).
- Hence, \(\nu = 5 - 1 - 1 = 3\).

The critical value for \(\chi^2_3\) at the 5% level is 7.815.

Since \(0.142 \le 7.815\), we fail to reject the null hypothesis.

There is no significant evidence to suggest that the Poisson distribution is not a suitable model for the number of insects per leaf.

M1: State the null and alternative hypotheses clearly.
A1: Calculate all expected frequencies correctly.
M1: Recognize the need to pool classes with expected frequencies less than 5 and pool correctly.
A1: Correctly calculate individual contributions to \(\chi^2\) and find the sum \(\chi^2 \approx 0.142\) (or \(0.143\)).
B1: Determine the correct number of degrees of freedom (\(\nu = 3\)).
B1: Identify the correct critical value of 7.815.
M1: Compare the test statistic to the critical value.
A0.75: State a coherent conclusion in context, noting that the Poisson model is suitable.

Let the difference in fuel efficiency for each van be defined as \(d = \text{After} - \text{Before}\).

The calculated differences \(d_i\) are:
- Van 1: \(29.6 - 28.4 = 1.2\)
- Van 2: \(31.9 - 31.2 = 0.7\)
- Van 3: \(27.1 - 25.5 = 1.6\)
- Van 4: \(29.5 - 29.8 = -0.3\)
- Van 5: \(34.6 - 33.0 = 1.6\)
- Van 6: \(28.5 - 27.1 = 1.4\)

Let us find the sample mean difference \(\bar{d}\) and the sample variance \(s_d^2\):

$$\bar{d} = \frac{1.2 + 0.7 + 1.6 - 0.3 + 1.6 + 1.4}{6} = \frac{6.2}{6} \approx 1.0333$$

$$\sum d^2 = 1.2^2 + 0.7^2 + 1.6^2 + (-0.3)^2 + 1.6^2 + 1.4^2 = 1.44 + 0.49 + 2.56 + 0.09 + 2.56 + 1.96 = 9.10$$

$$s_d^2 = \frac{1}{n-1} \left( \sum d^2 - \frac{(\sum d)^2}{n} \right) = \frac{1}{5} \left( 9.10 - \frac{6.2^2}{6} \right) = \frac{1}{5} (9.10 - 6.4067) = \frac{2.6933}{5} \approx 0.5387$$

$$s_d = \sqrt{0.5387} \approx 0.7339$$

Let the population mean difference be \(\mu_d\). The hypotheses are:
- \(H_0\): \(\mu_d = 0\)
- \(H_1\): \(\mu_d > 0\)

The test statistic is:

$$t = \frac{\bar{d} - 0}{s_d / \sqrt{n}} = \frac{1.0333}{0.7339 / \sqrt{6}} \approx 3.449$$

The number of degrees of freedom is \(\nu = 6 - 1 = 5\).
For a one-tailed test at the 5% level of significance, the critical value of \(t_5\) is 2.015.

Since \(3.449 > 2.015\), we reject \(H_0\).

There is significant evidence at the 5% level of significance that the new fuel additive increases fuel efficiency.

M1: State correct hypotheses with parameter \(\mu_d\) defined clearly.
A1: Calculate differences correctly.
M1: Correctly calculate the sample mean difference \(\bar{d}\) and sample standard deviation \(s_d\).
A1: Obtain \(\bar{d} \approx 1.033\) and \(s_d \approx 0.734\).
M1: Formulate the test statistic \(t = \frac{\bar{d}}{s_d/\sqrt{6}}\) and obtain \(t \approx 3.45\).
B1: Identify the correct critical value of 2.015 from the \(t_5\) distribution.
M1: Compare \(t\) with the critical value and make a correct rejection decision.
A0.75: Give a concluding statement in context.

Let the hypotheses be:
- \(H_0\): The population distributions of lifetimes for Brand A and Brand B are identical (no difference in medians).
- \(H_1\): The population distributions differ (difference in median lifetimes).

We first combine and rank all 11 observations in ascending order, tracking the brand of each observation:
1. 13.8 (Brand B) - Rank 1
2. 14.2 (Brand B) - Rank 2
3. 14.7 (Brand B) - Rank 3
4. 14.9 (Brand A) - Rank 4
5. 15.0 (Brand B) - Rank 5
6. 15.2 (Brand A) - Rank 6
7. 15.5 (Brand B) - Rank 7
8. 16.1 (Brand B) - Rank 8
9. 16.5 (Brand A) - Rank 9
10. 17.8 (Brand A) - Rank 10
11. 18.2 (Brand A) - Rank 11

Let \(n_A = 5\) (the size of the smaller sample) and \(n_B = 6\).
We calculate the sum of the ranks for Brand A (the smaller sample):

$$W = 4 + 6 + 9 + 10 + 11 = 40$$

For \(n_1 = 5\) and \(n_2 = 6\), at the 5% significance level for a two-tailed test, the critical region for \(W\) is:

$$W \le 18 \quad \text{or} \quad W \ge 42$$

Since \(W = 40\) is not in the critical region (\(18 < 40 < 42\)), we do not reject the null hypothesis \(H_0\).

There is insufficient evidence at the 5% level of significance to suggest a difference between the median lifetimes of Brand A and Brand B.

M1: State correct hypotheses for the Wilcoxon rank-sum test.
A1: Combine and rank all 11 observations correctly in ascending order.
M1: Sum the ranks of the smaller sample (Brand A) to find the test statistic \(W\).
A1: Obtain \(W = 40\) (or \(W_B = 26\), or Mann-Whitney \(U = 5\)).
B1: Identify the correct critical bounds (\(18\) and \(42\) for \(W\); or \(U \le 3\) for \(U\)).
M1: Compare the test statistic with the critical bounds and decide to not reject \(H_0\).
A1.75: Give a correct conclusion in context with a statement of uncertainty.

(i) For \( 0 \le x \le 2 \), the cumulative distribution function \( F(x) \) is found by integrating the probability density function:
\[ F(x) = \int_0^x \frac{3}{8}t^2 \mathrm{d}t = \left[ \frac{t^3}{8} \right]_0^x = \frac{x^3}{8} \]

(ii) Since \( 0 < X \le 2 \), the range of \( Y = \frac{8}{X^3} \) is \( Y \ge 1 \).
For \( y \ge 1 \):
\[ G(y) = \mathrm{P}(Y \le y) = \mathrm{P}\left(\frac{8}{X^3} \le y\right) \]
\[ = \mathrm{P}\left(X^3 \ge \frac{8}{y}\right) = \mathrm{P}\left(X \ge \frac{2}{y^{1/3}}\right) \]
\[ = 1 - F\left(\frac{2}{y^{1/3}}\right) = 1 - \frac{\left(\frac{2}{y^{1/3}}\right)^3}{8} = 1 - \frac{8/y}{8} = 1 - \frac{1}{y} \]
To find the probability density function \( g(y) \), we differentiate \( G(y) \):
\[ g(y) = G'(y) = \frac{\mathrm{d}}{\mathrm{d}y}\left(1 - \frac{1}{y}\right) = \frac{1}{y^2} \]
So,
\[ g(y) = \begin{cases} \frac{1}{y^2} & \text{for } y \ge 1, \\ 0 & \text{otherwise.} \end{cases} \]

(iii) The expected value of \( \sqrt{Y} \) is:
\[ \mathrm{E}(\sqrt{Y}) = \int_1^\infty \sqrt{y} \cdot g(y) \mathrm{d}y = \int_1^\infty y^{1/2} \cdot \frac{1}{y^2} \mathrm{d}y = \int_1^\infty y^{-3/2} \mathrm{d}y \]
\[ = \left[ -2y^{-1/2} \right]_1^\infty = \lim_{b \to \infty} \left( -2b^{-1/2} \right) - (-2(1)^{-1/2}) = 0 - (-2) = 2 \]

(i)
- M1: For integrating \( f(x) \) with correct limits \( [0, x] \).
- A1: For correctly showing \( F(x) = \frac{x^3}{8} \).

(ii)
- M1: For attempting to write \( G(y) = \mathrm{P}(Y \le y) \) in terms of \( X \ , leading to \) \mathrm{P}\left(X \ge \frac{2}{y^{1/3}}\right) \).
- M1: For using \( 1 - F\left(\frac{2}{y^{1/3}}\right) \).
- A1: For establishing \( G(y) = 1 - \frac{1}{y} \) and stating range \( y \ge 1 \).
- A1: For differentiating \( G(y) \) to obtain \( g(y) = \frac{1}{y^2} \) (stating range of non-zero density).

(iii)
- M1: For setting up the integral for \( \mathrm{E}(\sqrt{Y}) \) with correct integrand.
- A1: For obtaining the simplified integral \( \int_1^\infty y^{-3/2} \mathrm{d}y \).
- M1: For integrating to obtain \( \left[ -2y^{-1/2} \right] \) and applying limits.
- A0.5: For the correct final exact value of 2.

(i)
\( H_0 \): The number of attempts follows a geometric distribution with \( p = 0.5 \).
\( H_1 \): The number of attempts does not follow a geometric distribution with \( p = 0.5 \).

(ii) Under \( H_0 \), the probability distribution is \( \mathrm{P}(X = k) = (0.5)^k \) for \( k = 1, 2, 3, 4 \) and \( \mathrm{P}(X \ge 5) = (0.5)^4 = 0.0625 \).
Expected frequencies \( E_i = 200 \times \mathrm{P}(X) \):
- For \( X \ge 5 \): \( E_5 = 200 \times 0.0625 = 12.5 \) (as required).
- For \( X = 1 \): \( E_1 = 200 \times 0.5 = 100 \).
- For \( X = 2 \): \( E_2 = 200 \times 0.25 = 50 \).
- For \( X = 3 \): \( E_3 = 200 \times 0.125 = 25 \).
- For \( X = 4 \): \( E_4 = 200 \times 0.0625 = 12.5 \).

(iii) The test statistic is calculated using \( \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \):
- \( X = 1 \): \( \frac{(92-100)^2}{100} = 0.64 \)
- \( X = 2 \): \( \frac{(46-50)^2}{50} = 0.32 \)
- \( X = 3 \): \( \frac{(28-25)^2}{25} = 0.36 \)
- \( X = 4 \): \( \frac{(18-12.5)^2}{12.5} = \frac{30.25}{12.5} = 2.42 \)
- \( X \ge 5 \): \( \frac{(16-12.5)^2}{12.5} = \frac{12.25}{12.5} = 0.98 \)
Summing these values:
\[ \chi^2 = 0.64 + 0.32 + 0.36 + 2.42 + 0.98 = 4.72 \]

(iv) The number of degrees of freedom is \( \nu = c - 1 = 5 - 1 = 4 \) (since no parameters were estimated).
At the 5% significance level, the critical value is \( \chi^2_4(0.05) = 9.488 \).
Since the calculated value of \( 4.72 \) is less than the critical value of \( 9.488 \), we do not reject the null hypothesis \( H_0 \).
Conclusion: There is insufficient evidence at the 5% significance level to suggest that the geometric distribution with \( p = 0.5 \) is not a suitable model. We conclude that the model fits the data adequately.

(i)
- B1: Both hypotheses stated correctly in context (accepting equivalent contextual descriptions).

(ii)
- M1: For calculating \( \mathrm{P}(X \ge 5) = 0.5^4 = 0.0625 \) or equivalent, and showing expected value is 12.5.
- A1.5: For calculating the remaining expected frequencies (0.5 marks each for 100, 50, and 25/12.5 grouped together).

(iii)
- M1: For attempting to calculate individual cell contributions \( \frac{(O-E)^2}{E} \).
- A1: For at least three cell values calculated correctly.
- A1: For the correct test statistic of 4.72.

(iv)
- B1: For stating the critical value of 9.488 (associated with \( \nu = 4 \)).
- M1: For comparing their calculated test statistic with their critical value.
- A1: For a correct contextual conclusion showing understanding of uncertainty (e.g. 'insufficient evidence to reject the model').