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

(a)
Using common denominators on the right hand side:
\(\frac{4r-3}{r(r+1)} - \frac{4r+1}{(r+1)(r+2)} = \frac{(4r-3)(r+2) - r(4r+1)}{r(r+1)(r+2)}\)
\(= \frac{(4r^2 + 5r - 6) - (4r^2 + r)}{r(r+1)(r+2)} = \frac{4r-6}{r(r+1)(r+2)} = u_r\).

(b)
Let \(f(r) = \frac{4r-3}{r(r+1)}\). Then \(u_r = f(r) - f(r+1)\).
\(\sum_{r=1}^n u_r = (f(1) - f(2)) + (f(2) - f(3)) + \dots + (f(n) - f(n+1)) = f(1) - f(n+1)\).
Since \(f(1) = \frac{4(1)-3}{1(2)} = \frac{1}{2}\) and \(f(n+1) = \frac{4n+1}{(n+1)(n+2)}\),
\(\sum_{r=1}^n u_r = \frac{1}{2} - \frac{4n+1}{(n+1)(n+2)}\).

(c)
As \(n \to \infty\), \(\frac{4n+1}{(n+1)(n+2)} \to 0\).
Therefore, \(\sum_{r=1}^{\infty} u_r = \frac{1}{2}\).

(d)
The second part of the sum is a geometric series:
\(\sum_{r=1}^n \left(\frac{1}{3}\right)^r = \frac{\frac{1}{3}\left(1 - (\frac{1}{3})^n\right)}{1 - \frac{1}{3}} = \frac{1}{2}\left(1 - \left(\frac{1}{3}\right)^n\right)\).
Thus, the total sum is:
\(\sum_{r=1}^n \left( u_r + \left(\frac{1}{3}\right)^r \right) = \frac{1}{2} - \frac{4n+1}{(n+1)(n+2)} + \frac{1}{2} - \frac{1}{2}\left(\frac{1}{3}\right)^n = 1 - \frac{4n+1}{(n+1)(n+2)} - \frac{1}{2}\left(\frac{1}{3}\right)^n\).
As \(n \to \infty\), both \(\frac{4n+1}{(n+1)(n+2)}\) and \(\frac{1}{2}\left(\frac{1}{3}\right)^n\) tend to \(0\), so the limit is \(1\).

(a) M1: For algebraic attempt to combine fractions. A1: Correct expansion of numerator. A1: Fully correct simplification leading to the LHS.
(b) M1: Recognising the difference form \(f(r) - f(r+1)\). M1: Systematic cancellation of intermediate terms. A1: Correct expression for \(f(1)\). A1: Correct final expression in terms of \(n\).
(c) B1: State \(\frac{1}{2}\) following correct work.
(d) M1: Applying the GP sum formula to \(\sum (1/3)^r\). A1: Correct expression for the combined sum. A1: Correct limit of \(1\).

(a)
Eigenvalues are found by solving \(\det(M - \lambda I) = 0\):
\(\begin{vmatrix} 3-\lambda & -1 & 1 \\ -1 & 5-\lambda & -1 \\ 1 & -1 & 3-\lambda \end{vmatrix} = 0\).
Subtracting Row 3 from Row 1 gives:
\(\begin{vmatrix} 2-\lambda & 0 & \lambda-2 \\ -1 & 5-\lambda & -1 \\ 1 & -1 & 3-\lambda \end{vmatrix} = (2-\lambda) \begin{vmatrix} 1 & 0 & -1 \\ -1 & 5-\lambda & -1 \\ 1 & -1 & 3-\lambda \end{vmatrix} = 0\).
Adding Column 1 to Column 3:
\((2-\lambda) \begin{vmatrix} 1 & 0 & 0 \\ -1 & 5-\lambda & -2 \\ 1 & -1 & 4-\lambda \end{vmatrix} = (2-\lambda)[(5-\lambda)(4-\lambda) - 2] = (2-\lambda)(\lambda^2 - 9\lambda + 18) = (2-\lambda)(\lambda-3)(\lambda-6) = 0\).
Thus, the eigenvalues are \(\lambda = 2, 3, 6\).

(b)
For \(\lambda = 2\):
\(\begin{pmatrix} 1 & -1 & 1 \\ -1 & 3 & -1 \\ 1 & -1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies x - y + z = 0\) and \(-x + 3y - z = 0\).
Adding these gives \(2y = 0 \implies y = 0 \implies x + z = 0\).
An eigenvector is \(\mathbf{e}_1 = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}\).

For \(\lambda = 3\):
\(\begin{pmatrix} 0 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies x - y = 0\) and \(-y + z = 0\).
An eigenvector is \(\mathbf{e}_2 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\).

For \(\lambda = 6\):
\(\begin{pmatrix} -3 & -1 & 1 \\ -1 & -1 & -1 \\ 1 & -1 & -3 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies x + y + z = 0\) and \(-3x - y + z = 0\).
Solving yields \(x = z\) and \(y = -2x\).
An eigenvector is \(\mathbf{e}_3 = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}\).

(c)
To form the orthogonal matrix \(P\), normalize the eigenvectors:
\(\mathbf{u}_1 = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}\), \(\mathbf{u}_2 = \frac{1}{\sqrt{3}}\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\), \(\mathbf{u}_3 = \frac{1}{\sqrt{6}}\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}\).
Thus, \(P = \begin{pmatrix} 1/\sqrt{2} & 1/\sqrt{3} & 1/\sqrt{6} \\ 0 & 1/\sqrt{3} & -2/\sqrt{6} \\ -1/\sqrt{2} & 1/\sqrt{3} & 1/\sqrt{6} \end{pmatrix}\) and \(D = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 6 \end{pmatrix}\).

(a) M1: Set up characteristic equation \(\det(M - \lambda I) = 0\). A2: Correct characteristic equation. A1: Find eigenvalues \(2, 3, 6\).
(b) M1: Set up systems of equations to find eigenvectors. A2: Correctly identify any two eigenvectors. A1: Correctly identify the third eigenvector.
(c) B1: Correct diagonal matrix \(D\). M1: Normalising the eigenvectors. A1: Correct orthogonal matrix \(P\).

(a)
Base case: For \(n=1\),
\(M^1 = \begin{pmatrix} 2(1)+1 & -4(1) \\ 1 & 1-2(1) \end{pmatrix} = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}\), which is true.

Inductive step: Assume the statement is true for \(n = k\), i.e.,
\(M^k = \begin{pmatrix} 2k+1 & -4k \\ k & 1-2k \end{pmatrix}\).
We must show it is true for \(n = k+1\):
\(M^{k+1} = M^k M = \begin{pmatrix} 2k+1 & -4k \\ k & 1-2k \end{pmatrix} \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}\)
\(= \begin{pmatrix} 3(2k+1) - 4k & -4(2k+1) + 4k \\ 3k + (1-2k) & -4k - (1-2k) \end{pmatrix}\)
\(= \begin{pmatrix} 6k+3-4k & -8k-4+4k \\ k+1 & -2k-1 \end{pmatrix}\)
\(= \begin{pmatrix} 2(k+1)+1 & -4(k+1) \\ k+1 & 1-2(k+1) \end{pmatrix}\).
This is of the same form with \(n = k+1\).
Since the base case is true and the inductive step holds, the statement is true for all positive integers \(n\) by mathematical induction.

(b)
To find \((M^n)^{-1}\), first find the determinant:
\(\det(M^n) = (2n+1)(1-2n) - (-4n)(n) = 1 - 4n^2 + 4n^2 = 1\).
Using the formula for the inverse of a \(2 \times 2\) matrix:
\((M^n)^{-1} = \frac{1}{1} \begin{pmatrix} 1-2n & 4n \\ -n & 2n+1 \end{pmatrix} = \begin{pmatrix} 1-2n & 4n \\ -n & 2n+1 \end{pmatrix}\).

(c)
For \(n = 10\),
\(M^{10} = \begin{pmatrix} 21 & -40 \\ 10 & -19 \end{pmatrix}\).
Applying this transformation to the point \((1, 2)\):
\(\begin{pmatrix} 21 & -40 \\ 10 & -19 \end{pmatrix} \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 21 - 80 \\ 10 - 38 \end{pmatrix} = \begin{pmatrix} -59 \\ -28 \end{pmatrix}\).
So the image coordinates are \((-59, -28)\).

(a) B1: Verify the base case \(n=1\). M1: State the inductive hypothesis. M1: Attempt to multiply \(M^k M\). A2: Correct algebraic expansion and factorization showing the form for \(n=k+1\). B1: Clear concluding statement.
(b) M1: Calculate determinant of \(M^n\). A1: Show determinant is 1. A1: Correctly write down the inverse matrix.
(c) M1: Calculate the matrix \(M^{10}\) and perform the multiplication. A1: Correct coordinates \((-59, -28)\).

(a)
From the coefficients of the cubic equation \(x^3 + 3x^2 - x + 2 = 0\):
(i) \(\alpha + \beta + \gamma = -3\)
(ii) \(\alpha\beta + \beta\gamma + \gamma\alpha = -1\)
(iii) \(\alpha\beta\gamma = -2\)

(b)
Using the identity \(\alpha^2 + \beta^2 + \gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha)\):
\(\alpha^2 + \beta^2 + \gamma^2 = (-3)^2 - 2(-1) = 9 + 2 = 11\).

(c)
Let \(y = x + 1\), which means \(x = y - 1\).
Substitute \(x = y - 1\) into the original equation:
\((y-1)^3 + 3(y-1)^2 - (y-1) + 2 = 0\)
\((y^3 - 3y^2 + 3y - 1) + 3(y^2 - 2y + 1) - y + 1 + 2 = 0\)
\(y^3 - 3y^2 + 3y - 1 + 3y^2 - 6y + 3 - y + 3 = 0\)
\(y^3 - 4y + 5 = 0\).
This is the required cubic equation.

(d)
\(\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} = \frac{\alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2}{\alpha^2\beta^2\gamma^2}\).
We know \(\alpha^2\beta^2\gamma^2 = (\alpha\beta\gamma)^2 = (-2)^2 = 4\).
For the numerator, we use:
\(\alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 = (\alpha\beta + \beta\gamma + \gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha + \beta + \gamma)\)
\(= (-1)^2 - 2(-2)(-3) = 1 - 12 = -11\).
Thus,
\(\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} = \frac{-11}{4} = -2.75\).

(a) B2: All three correct (allow B1 for any two correct).
(b) M1: Use correct algebraic identity. A1: Obtain 11.
(c) M1: Substitute \(x = y - 1\). M1: Expand terms. A1: Simplify to get \(y^3 - 4y + 5 = 0\) (or equivalent variable).
(d) M1: Write expression with a common denominator. M1: Use correct identity for the numerator. A1: Correct value of \(-11\) for numerator. A1: Correct final fraction or decimal \(-2.75\).

(a)
First, check if the lines are parallel. The direction vectors are \(\mathbf{d}_1 = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}\) and \(\mathbf{d}_2 = \begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix}\). Since \(\mathbf{d}_1 \neq k\mathbf{d}_2\), they are not parallel.
Next, test for intersection:
\(1 + 2\lambda = 2 + \mu \implies 2\lambda - ̅\mu = 1 \quad (1)\)
\(2 - \lambda = -1 + \mu \implies \lambda + \mu = 3 \quad (2)\)
\(-1 + 3\lambda = 4 - 2\mu \implies 3\lambda + 2\mu = 5 \quad (3)\)

From (1) and (2), adding them gives \(3\lambda = 4 \implies \lambda = \frac{4}{3}\).
Substituting into (2) gives \(\mu = 3 - \frac{4}{3} = \frac{5}{3}\).
Substituting these into (3) yields LHS \(= 3(\frac{4}{3}) + 2(\frac{5}{3}) = 4 + \frac{10}{3} = \frac{22}{3} \neq 5\).
Since the equations are inconsistent, the lines do not intersect. Since they are neither parallel nor intersecting, they are skew.

(b)
The common perpendicular vector \(\mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2\):
\(\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 3 \\ 1 & 1 & -2 \end{vmatrix} = \mathbf{i}(2-3) - \mathbf{j}(-4-3) + \mathbf{k}(2 - (-1)) = \begin{pmatrix} -1 \\ 7 \\ 3 \end{pmatrix}\).
Its magnitude is \(|\mathbf{n}| = \sqrt{(-1)^2 + 7^2 + 3^2} = \sqrt{59}\).
Let \(\vec{AB}\) be the vector connecting a point on each line:
\(\vec{AB} = \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 1 \\ -3 \\ 5
\end{pmatrix}\).
The shortest distance is the projection of \(\vec{AB}\) onto \(\mathbf{n}\):
\(d = \frac{|\vec{AB} \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{|1(-1) + (-3)(7) + 5(3)|}{\sqrt{59}} = \frac{|-1 - 21 + 15|}{\sqrt{59}} = \frac{7}{\sqrt{59}} \approx 0.911\).

(c)
The normal vector to the plane containing \(l_1\) and parallel to \(l_2\) is the cross product \(\mathbf{n} = \begin{pmatrix} -1 \\ 7 \\ 3 \end{pmatrix}\) (or its negative \(\begin{pmatrix} 1 \\ -7 \\ -3 \end{pmatrix}\)).
The plane passes through the point on \(l_1\), which is \((1, 2, -1)\).
The equation of the plane is:
\(1(x - 1) - 7(y - 2) - 3(z + 1) = 0\)
\(x - 1 - 7y + 14 - 3z - 3 = 0\)
\(x - 7y - 3z = -10\).

(a) B1: Show that directions are not parallel. M1: Set up system of equations for intersection. A1: Solve for parameters using two equations. A1: Show inconsistency in the third equation and conclude skew.
(b) M1: Calculate cross product of the two direction vectors. A1: Correct normal vector. M1: Calculate projection of a position vector difference onto the normal. A1: Correct shortest distance \(\frac{7}{\sqrt{59}}\) or \(0.911\).
(c) M1: Recognize the normal vector is the same as the perpendicular to both lines. M1: Substitute \((1, 2, -1)\) into plane equation. A1: Correct equation in cartesian form.

(a)
Sketching \(r = 1 + 2\cos\theta\):
- At \(\theta = 0\), \(r = 3\).
- At \(\theta = \pi/2\), \(r = 1\).
- \(r = 0\) when \(\cos\theta = -1/2 \implies \theta = \frac{2\pi}{3}\) or \(\frac{4
\pi}{3}\).
- Inside the range \(\frac{2\pi}{3} < \theta < \frac{4\pi}{3}\), \(r < 0\), which produces an inner loop.
- At \(\theta = \pi\), \(r = -1\) (plotted at the pole along \(\theta = 0\) direction, distance 1).

(b)
The inner loop is traced for \(\frac{2\pi}{3} \le \theta \le \frac{4\pi}{3}\).
Using symmetry, we can integrate from \(\frac{2\pi}{3}\) to \(\pi\) and multiply by 2:
\(A = 2 \times \frac{1}{2} \int_{2\pi/3}^{\pi} (1 + 2\cos\theta)^2 \, d\theta = \int_{2\pi/3}^{\pi} (1 + 4\cos\theta + 4\cos^2\theta) \, d\theta\)
Using the identity \(4\cos^2\theta = 2(1 + \cos 2\theta) = 2 + 2\cos 2\theta\):
\(A = \int_{2\pi/3}^{\pi} (3 + 4\cos\theta + 2\cos 2\theta) \, d\theta\)
\(= \left[ 3\theta + 4\sin\theta + \sin 2\theta \right]_{2\pi/3}^{\pi}\)
\(= (3\pi + 0 + 0) - \left(2\pi + 4\sin\left(\frac{2\pi}{3}\right) + \sin\left(\frac{4\pi}{3}\right)\right)\)
\(= 3\pi - \left(2\pi + 4\left(\frac{\sqrt{3}}{2}\right) - \frac{\sqrt{3}}{2}\right)\)
\(= \pi - \frac{3\sqrt{3}}{2}\).

(c)
To find the gradient of the tangent, we use parametric differentiation:
\(x = r\cos\theta = (1 + 2\cos\theta)\cos\theta = \cos\theta + 2\cos^2\theta\)
\(y = r\sin\theta = (1 + 2\cos\theta)\sin\theta = \sin\theta + \sin 2\theta\)
Now differentiate with respect to \(\theta\):
\(\frac{dx}{d\theta} = -\sin\theta - 4\cos\theta\sin\theta = -\sin\theta - 2\sin 2\theta\)
\(\frac{dy}{d\theta} = \cos\theta + 2\cos 2\theta\)
At \(\theta = \frac{\pi}{2}\):
\(\frac{dx}{d\theta} = -\sin(\pi/2) - 2\sin(\pi) = -1\)
\(\frac{dy}{d\theta} = \cos(\pi/2) + 2\cos(\pi) = -2\)
So gradient \(m = \frac{dy/d\theta}{dx/d\theta} = \frac{-2}{-1} = 2\).
At \(\theta = \frac{\pi}{2}\), \(r = 1\), so \(x = 0, y = 1\).
Using \(y - y_1 = m(x - x_1)\):
\(y - 1 = 2(x - 0) \implies y = 2x + 1\).

(a) B1: Correct overall shape with inner and outer loops. B1: Indicated points of intersection on axes. B1: Tangents at pole correctly oriented at \(\theta = \frac{2\pi}{3}, \frac{4\pi}{3}\).
(b) M1: Set up correct integral with limits \([2\pi/3, \pi]\) or \([2\pi/3, 4\pi/3]\). M1: Expand integrand and use double-angle identity. A1: Correct integration. M1: Correct substitution of limits. A1: Correct exact area of \(\pi - \frac{3\sqrt{3}}{2}\).
(c) M1: Express \(x\) and \(y\) in terms of \(\theta\). M1: Evaluate derivatives at \(\theta = \frac{\pi}{2}\). A1: Correct equation \(y = 2x + 1\).

(a)
Vertical asymptote is \(x = -1\).
Using polynomial division:
\(\frac{2x^2 + 3x - 2}{x+1} = \frac{2x(x+1) + (x-2)}{x+1} = 2x + 1 - \frac{3}{x+1}\).
As \(x \to \pm\infty\), the fractional part tends to \(0\), so the oblique asymptote is \(y = 2x + 1\).

(b)
Differentiating \(y\):
\(\frac{dy}{dx} = \frac{(4x+3)(x+1) - (2x^2+3x-2)(1)}{(x+1)^2} = \frac{4x^2 + 7x + 3 - 2x^2 - 3x + 2}{(x+1)^2} = \frac{2x^2 + 4x + 5}{(x+1)^2}\).
To find stationary points, set \(\frac{dy}{dx} = 0 \implies 2x^2 + 4x + 5 = 0\).
The discriminant is \(b^2 - 4ac = 4^2 - 4(2)(5) = 16 - 40 = -24 < 0\).
Since there are no real solutions to the numerator, there are no stationary points. Since the numerator is always positive, the gradient is positive for all \(x \neq -1\).

(c)
Intercepts:
- y-intercept: When \(x = 0\), \(y = -2\).
- x-intercepts: When \(y = 0 \implies 2x^2 + 3x - 2 = 0 \implies (2x - 1)(x + 2) = 0\), so \(x = 0.5\) and \(x = -2\).
Sketch features: Two branches, strictly increasing, asymptotic to \(x = -1\) and \(y = 2x + 1\).

(d)
Set \(y = 2x + k\) equal to the curve:
\(2x + k = \frac{2x^2 + 3x - 2}{x+1}\)
\((2x + k)(x + 1) = 2x^2 + 3x - 2\)
\(2x^2 + (2+k)x + k = 2x^2 + 3x - 2\)
\((k-1)x = -k-2\).
If \(k \neq 1\), we get a unique solution: \(x = \frac{-k-2}{k-1}\).
If \(k = 1\), the equation reduces to \(0x = -3\), which has no solutions.
Thus, the only value of \(k\) for which the line does not intersect \(C\) is \(k = 1\).

(a) B1: Correct vertical asymptote. M1: Attempt polynomial division. A1: Correct oblique asymptote.
(b) M1: Use quotient rule to differentiate. A1: Obtain correct numerator \(2x^2 + 4x + 5\). A1: Show discriminant is negative to conclude no stationary points.
(c) B1: Correct intercepts labeled. B1: Both branches drawn with correct behavior near asymptotes. B1: Fully correct sketch with asymptotes clearly shown.
(d) M1: Equate and form linear equation in \(x\). A1: Correct conclusion that \(k = 1\).

(i) The characteristic equation is \(\det(\mathbf{A} - \lambda \mathbf{I}) = 0\).
\(\det \begin{pmatrix} 3-\lambda & 1 & -1 \\ 1 & 3-\lambda & -1 \\ -1 & -1 & 3-\lambda \end{pmatrix} = 0\)
Expanding along the first row:
\((3-\lambda)[(3-\lambda)^2 - 1] - 1[(3-\lambda) - 1] - 1[-1 + (3-\lambda)] = 0\)
\((3-\lambda)(\lambda^2 - 6\lambda + 8) - 2(2-\lambda) = 0\)
\((3-\lambda)(\lambda-2)(\lambda-4) + 2(\lambda-2) = 0\)
\((\lambda-2)[(3-\lambda)(\lambda-4) + 2] = 0\)
\((\lambda-2)[-\lambda^2 + 7\lambda - 10] = 0\)
\(-(\lambda-2)^2(\lambda-5) = 0\)
So the eigenvalues are \(\lambda = 2\) (repeated) and \(\lambda = 5\).

(ii) For \(\lambda = 5\):
\(\begin{pmatrix} -2 & 1 & -1 \\ 1 & -2 & -1 \\ -1 & -1 & -2 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\)
Subtracting row 2 from row 1 gives \(-3x + 3y = 0 \implies x = y\).
Row 3 gives \(-x - y - 2z = 0 \implies -2x - 2z = 0 \implies z = -x\).
Thus, an eigenvector is \(\mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}\).

For \(\lambda = 2\):
\(\begin{pmatrix} 1 & 1 & -1 \\ 1 & 1 & -1 \\ -1 & -1 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies x + y - z = 0\).
We need two mutually orthogonal eigenvectors from this plane, which must also be orthogonal to \(\mathbf{v}_1\).
Let \(\mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}\). This satisfies \(x + y - z = 0\) and \(\mathbf{v}_2 \cdot \mathbf{v}_1 = 1(1) - 1(1) + 0 = 0\).
Let the third eigenvector be \(\mathbf{v}_3 = \begin{pmatrix} a \\ b \\ c \end{pmatrix}\). For \(\mathbf{v}_3\) to be orthogonal to \(\mathbf{v}_2\), we have \(a - b = 0 \implies a = b\).
Since it must lie in the plane, \(a + b - c = 0 \implies 2a - c = 0 \implies c = 2a\).
Choosing \(a = 1\), we get \(\mathbf{v}_3 = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}\).
This is orthogonal to \(\mathbf{v}_1\) since \(1(1) + 1(1) + 2(-1) = 0\).
Thus, the three mutually orthogonal eigenvectors are \(\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}\), \(\begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}\), and \(\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}\).

(iii) Normalising the eigenvectors:
\(\mathbf{u}_1 = \frac{1}{\sqrt{3}}\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}\), \(\mathbf{u}_2 = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}\), \(\mathbf{u}_3 = \frac{1}{\sqrt{6}}\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}\).
Thus, an orthogonal matrix \(\mathbf{P}\) is
\(\mathbf{P} = \begin{pmatrix} 1/\sqrt{3} & 1/\sqrt{2} & 1/\sqrt{6} \\ 1/\sqrt{3} & -1/\sqrt{2} & 1/\sqrt{6} \\ -1/\sqrt{3} & 0 & 2/\sqrt{6} \end{pmatrix}\)
and the corresponding diagonal matrix is
\(\mathbf{D} = \begin{pmatrix} 5 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix}\).

(i) M1: Attempts determinant expansion of \(\det(\mathbf{A} - \lambda \mathbf{I}) = 0\).
A1: Correct characteristic equation \((\lambda-2)^2(\lambda-5) = 0\).
A1: Obtains \(\lambda = 5\).
A1: Obtains \(\lambda = 2\) (repeated).

(ii) M1: Attempts to find eigenvector for \(\lambda = 5\).
A1: Correct eigenvector \(\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}\) (or any scalar multiple).
M1: Attempts to find two orthogonal eigenvectors for the repeated eigenvalue \(\lambda = 2\).
A1: Correct orthogonal pair, e.g., \(\begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}\) and \(\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}\).

(iii) B1: Correct diagonal matrix \(\mathbf{D}\) corresponding to their order of columns in \(\mathbf{P}\).
B0.375: Correct normalized columns in \(\mathbf{P}\).

(i) \(y = \sec x \implies \frac{\mathrm{d}y}{\mathrm{d}x} = \sec x \tan x = y \tan x\).
Differentiating again using the product rule:
\(\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = \frac{\mathrm{d}y}{\mathrm{d}x}\tan x + y\sec^2 x = (y\tan x)\tan x + y(y^2) = y\tan^2 x + y^3\).
Since \(\tan^2 x = \sec^2 x - 1 = y^2 - 1\), we have:
\(\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = y(y^2 - 1) + y^3 = 2y^3 - y\).

(ii) Evaluating derivatives at \(x = 0\):
\(y(0) = \sec 0 = 1\).
\(y'(0) = 1 \cdot \tan 0 = 0\).
\(y''(0) = 2(1)^3 - 1 = 1\).
Differentiating \(y'' = 2y^3 - y\):
\(y''' = 6y^2 y' - y'\).
At \(x = 0\), since \(y'(0) = 0\), we have \(y'''(0) = 0\).
Differentiating again:
\(y^{(4)} = 12y(y')^2 + 6y^2 y'' - y''\).
At \(x = 0\):
\(y^{(4)}(0) = 12(1)(0)^2 + 6(1)^2(1) - 1 = 5\).
The Maclaurin series formula is:
\(y(x) = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \frac{y^{(4)}(0)}{4!}x^4 + \dots\)
Substituting the values:
\(y(x) = 1 + 0x + \frac{1}{2}x^2 + 0x^3 + \frac{5}{24}x^4 + \dots = 1 + \frac{1}{2}x^2 + \frac{5}{24}x^4\).

(iii) We approximate the integral:
\(\int_0^{0.5} \sec x \;\mathrm{d}x \approx \int_0^{0.5} \left(1 + \frac{1}{2}x^2 + \frac{5}{24}x^4\right) \mathrm{d}x\)
\(= \left[ x + \frac{1}{6}x^3 + \frac{1}{24}x^5 \right]_0^{0.5}\)
\(= 0.5 + \frac{1}{6}(0.125) + \frac{1}{24}(0.03125)\)
\(= 0.5 + 0.0208333 + 0.0013021 = 0.5221354\)
To 4 decimal places, the value is \(0.5221\).

(i) M1: Finds first derivative of \(\sec x\) in terms of \(y\) and \(\tan x\).
M1: Differentiates again using product/chain rule.
A1: Replaces \(\tan^2 x\) with \(y^2 - 1\) to obtain the given expression.

(ii) B1: Identifies \(y(0)=1\), \(y'(0)=0\), and \(y''(0)=1\).
M1: Differentiates to find expressions for \(y'''\) and \(y^{(4)}\).
A1: Correct values \(y'''(0) = 0\) and \(y^{(4)}(0) = 5\).
A1.375: Correct Maclaurin series: \(1 + \frac{1}{2}x^2 + \frac{5}{24}x^4\).

(iii) M1: Integrates the Maclaurin series term by term.
A1: Evaluates at the limits and obtains \(0.5221\) (correct to 4 d.p.).

(i) First, we find the derivative:
\(\frac{\mathrm{d}y}{\mathrm{d}x} = x - \frac{1}{4x}\).
Then,
\(1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = 1 + \left(x - \frac{1}{4x}\right)^2 = 1 + x^2 - 2(x)\left(\frac{1}{4x}\right) + \frac{1}{16x^2}\)
\(= 1 + x^2 - \frac{1}{2} + \frac{1}{16x^2} = x^2 + \frac{1}{2} + \frac{1}{16x^2}\)
\(= \left(x + \frac{1}{4x}\right)^2\).

(ii) The arc length \(s\) is given by:
\(s = \int_1^{\mathrm{e}} \sqrt{1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2} \;\mathrm{d}x = \int_1^{\mathrm{e}} \left(x + \frac{1}{4x}\right) \;\mathrm{d}x\)
\(= \left[ \frac{1}{2}x^2 + \frac{1}{4}\ln x \right]_1^{\mathrm{e}}\)
\(= \left(\frac{1}{2}\mathrm{e}^2 + \frac{1}{4}\ln \mathrm{e}\right) - \left(\frac{1}{2}(1)^2 + \frac{1}{4}\ln 1\right)\)
\(= \frac{1}{2}\mathrm{e}^2 + \frac{1}{4} - \frac{1}{2}\)
\(= \frac{1}{2}\mathrm{e}^2 - \frac{1}{4}\).

(iii) The surface area \(S_y\) of revolution about the \(y\)-axis is:
\(S_y = 2\pi \int_1^{\mathrm{e}} x \sqrt{1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2} \;\mathrm{d}x\)
\(= 2\pi \int_1^{\mathrm{e}} x \left(x + \frac{1}{4x}\right) \;\mathrm{d}x\)
\(= 2\pi \int_1^{\mathrm{e}} \left(x^2 + \frac{1}{4}\right) \;\mathrm{d}x\)
\(= 2\pi \left[ \frac{1}{3}x^3 + \frac{1}{4}x \right]_1^{\mathrm{e}}\)
\(= 2\pi \left[ \left(\frac{1}{3}\mathrm{e}^3 + \frac{1}{4}\mathrm{e}\right) - \left(\frac{1}{3} + \frac{1}{4}\right) \right]\)
\(= 2\pi \left( \frac{1}{3}\mathrm{e}^3 + \frac{1}{4}\mathrm{e} - \frac{7}{12} \right)\)
\(= \pi \left( \frac{2}{3}\mathrm{e}^3 + \frac{1}{2}\mathrm{e} - \frac{7}{6} \right)\).

(i) M1: Differentiates \(y\) correctly.
M1: Expands the squared term and adds 1.
A1: Completes the square to show the desired expression.

(ii) M1: Sets up the arc length integral.
A1: Correct integration.
A1.375: Correct exact simplified value \(\frac{1}{2}\mathrm{e}^2 - \frac{1}{4}\).

(iii) M1: Uses the correct formula for surface area about the \(y\)-axis, \(2\pi \int x \;\mathrm{d}s\).
M1: Integrates \(2\pi(x^2 + \frac{1}{4})\).
A1: Obtains exact value \(\pi \left(\frac{2}{3}\mathrm{e}^3 + \frac{1}{2}\mathrm{e} - \frac{7}{6}\right)\).

(i) The auxiliary equation is:
\(m^2 - 4m + 4 = 0 \implies (m-2)^2 = 0\).
Since we have a repeated root \(m = 2\), the complementary function is:
\(y_c = (Ax + B)\mathrm{e}^{2x}\).

(ii) Since \(\mathrm{e}^{2x}\) and \(x\mathrm{e}^{2x}\) are already terms in the complementary function, we try a particular integral of the form:
\(y_p = C x^2 \mathrm{e}^{2x}\).
Differentiating:
\(y_p' = C(2x \mathrm{e}^{2x} + 2x^2 \mathrm{e}^{2x}) = 2C(x + x^2)\mathrm{e}^{2x}\)
\(y_p'' = 2C(1 + 2x)\mathrm{e}^{2x} + 4C(x + x^2)\mathrm{e}^{2x} = 2C(1 + 4x + 2x^2)\mathrm{e}^{2x}\).
Substituting these into the differential equation:
\(2C(1 + 4x + 2x^2)\mathrm{e}^{2x} - 4[2C(x + x^2)\mathrm{e}^{2x}] + 4[C x^2 \mathrm{e}^{2x}] = \mathrm{e}^{2x}\)
\(C\mathrm{e}^{2x}[ (2 + 8x + 4x^2) - (8x + 8x^2) + 4x^2 ] = \mathrm{e}^{2x}\)
\(2C\mathrm{e}^{2x} = \mathrm{e}^{2x} \implies 2C = 1 \implies C = \frac{1}{2}\).
Thus, the particular integral is \(y_p = \frac{1}{2}x^2\mathrm{e}^{2x}\).
The general solution is:
\(y = (Ax + B)\mathrm{e}^{2x} + \frac{1}{2}x^2\mathrm{e}^{2x}\).

(iii) Using boundary conditions at \(x = 0\):
\(y = 1 \implies 1 = B\).
Now we differentiate the general solution:
\(y' = A\mathrm{e}^{2x} + 2(Ax + B)\mathrm{e}^{2x} + x\mathrm{e}^{2x} + x^2\mathrm{e}^{2x}\).
At \(x = 0\), \(y' = 3\):
\(3 = A + 2B \implies 3 = A + 2(1) \implies A = 1\).
Thus, the particular solution is:
\(y = (x + 1)\mathrm{e}^{2x} + \frac{1}{2}x^2\mathrm{e}^{2x} = \left(\frac{1}{2}x^2 + x + 1\right)\mathrm{e}^{2x}\).

(i) M1: Solves auxiliary equation \(m^2 - 4m + 4 = 0\).
A1.375: Correct complementary function \(y_c = (Ax + B)\mathrm{e}^{2x}\).

(ii) M1: Suggests correct form of particular integral \(y_p = C x^2 \mathrm{e}^{2x}\).
M1: Differentiates and substitutes into the ODE.
A1: Solves for \(C = \frac{1}{2}\).
A1: Correct general solution.

(iii) M1: Uses \(y(0) = 1\) to find \(B = 1\).
M1: Differentiates general solution and uses \(y'(0) = 3\) to find \(A\).
A1: Obtains \(y = \left(\frac{1}{2}x^2 + x + 1\right)\mathrm{e}^{2x}\).

(i) By de Moivre's theorem, \(\cos(5\theta) + \mathrm{i}\sin(5\theta) = (\cos\theta + \mathrm{i}\sin\theta)^5\).
Using the binomial expansion:
\((\cos\theta + \mathrm{i}\sin\theta)^5 = \cos^5\theta + 5\mathrm{i}\cos^4\theta\sin\theta - 10\cos^3\theta\sin^2\theta - 10\mathrm{i}\cos^2\theta\sin^3\theta + 5\cos\theta\sin^4\theta + \mathrm{i}\sin^5\theta\).
Equating the imaginary parts:
\(\sin(5\theta) = 5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta\).
Since \(\cos^2\theta = 1 - \sin^2\theta\):
\(\sin(5\theta) = 5(1 - \sin^2\theta)^2\sin\theta - 10(1 - \sin^2\theta)\sin^3\theta + \sin^5\theta\)
\(= 5(1 - 2\sin^2\theta + \sin^4\theta)\sin\theta - 10(\sin^3\theta - \sin^5\theta) + \sin^5\theta\)
\(= 5\sin\theta - 10\sin^3\theta + 5\sin^5\theta - 10\sin^3\theta + 10\sin^5\theta + \sin^5\theta\)
\(= 5\sin\theta - 20\sin^3\theta + 16\sin^5\theta\).

(ii) The equation is \(16x^5 - 20x^3 + 5x = 0\).
Letting \(x = \sin\theta\), the equation becomes \(\sin(5\theta) = 0\).
This occurs when \(5\theta = k\pi \implies \theta = \frac{k\pi}{5}\) for integers \(k\).
Distinct values of \(x\) are obtained for \(k = -2, -1, 0, 1, 2\).
These yield the 5 roots:
\(x = 0\),
\(x = \pm \sin\frac{\pi}{5}\),
\(x = \pm \sin\frac{2\pi}{5}\).

(iii) The non-zero roots of the equation satisfy \(16x^4 - 20x^2 + 5 = 0\).
The product of the roots of this quartic equation is given by \(\frac{\text{constant term}}{\text{leading coefficient}} = \frac{5}{16}\).
The roots are \(\pm\sin\frac{\pi}{5}\) and \(\pm\sin\frac{2\pi}{5}\).
Therefore:
\(\left(\sin\frac{\pi}{5}\right)\left(-\text{sin}\frac{\pi}{5}\right)\left(\sin\frac{2\pi}{5}\right)\left(-\text{sin}\frac{2\pi}{5}\right) = \frac{5}{16}\)
\(\sin^2\frac{\pi}{5} \sin^2\frac{2\pi}{5} = \frac{5}{16}\).
Since both \(\sin\frac{\pi}{5}\) and \(\sin\frac{2\pi}{5}\) are positive (as they are in the first quadrant):
\(\sin\frac{\pi}{5} \sin\frac{2\pi}{5} = \sqrt{\frac{5}{16}} = \frac{\sqrt{5}}{4}\).

(i) M1: Uses de Moivre's theorem to write the expansion.
A1: Obtains the correct imaginary part expression.
M1: Uses \(\cos^2\theta = 1 - \sin^2\theta\) to eliminate \(\cos\theta\).
A1.375: Correct algebraic simplification to the given identity.

(ii) M1: Relates the polynomial equation to \(\sin(5\theta) = 0\).
A1: Solves for \(\theta = \frac{k\pi}{5}\).
A1: Lists the five unique roots: \(0, \pm\sin\frac{\pi}{5}, \pm\sin\frac{2\pi}{5}\).

(iii) M1: Identifies the equation for non-zero roots as \(16x^4 - 20x^2 + 5 = 0\) and uses product of roots formula.
A1: Completes the proof by taking the positive square root.

(i) From the definitions, \(\cosh x = \frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\) and \(\sinh x = \frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\).
Thus,
\(\cosh^2 x - \sinh^2 x = \left(\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right)^2 - \left(\frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right)^2\)
\(= \frac{\mathrm{e}^{2x} + 2 + \mathrm{e}^{-2x}}{4} - \frac{\mathrm{e}^{2x} - 2 + \mathrm{e}^{-2x}}{4} = \frac{4}{4} = 1\).

(ii) Substitute definitions into \(3\cosh x + \sinh x = 5\):
\(3\left(\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right) + \frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2} = 5\)
\(\frac{4\mathrm{e}^x + 2\mathrm{e}^{-x}}{2} = 5 \implies 2\mathrm{e}^x + \mathrm{e}^{-x} = 5\).
Multiplying by \(\mathrm{e}^x\):
\(2\mathrm{e}^{2x} - 5\mathrm{e}^x + 1 = 0\).
Letting \(u = \mathrm{e}^x\), we solve the quadratic equation \(2u^2 - 5u + 1 = 0\):
\(u = \frac{5 \pm \sqrt{25 - 8}}{4} = \frac{5 \pm \sqrt{17}}{4}\).
Since \(u > 0\) for all real \(x\), and both values are positive, we have:
\(x = \ln\left(\frac{5 \pm \sqrt{17}}{4}\right)\).

(iii) Expanding the integrand:
\((3\cosh x + \sinh x)^2 = 9\cosh^2 x + 6\cosh x \sinh x + \sinh^2 x\).
Using hyperbolic identities:
\(9\cosh^2 x + \sinh^2 x = 9\left(\frac{\cosh(2x) + 1}{2}\right) + \left(\frac{\cosh(2x) - 1}{2}\right) = 5\cosh(2x) + 4\),
and \(6\cosh x \sinh x = 3\sinh(2x)\).
So the integral becomes:
\(\int_0^{\ln 2} [5\cosh(2x) + 3\sinh(2x) + 4] \;\mathrm{d}x = \left[ \frac{5}{2}\sinh(2x) + \frac{3}{2}\cosh(2x) + 4x \right]_0^{\ln 2}\).
At \(x = \ln 2\), \(\mathrm{e}^{2x} = 4\) and \(\mathrm{e}^{-2x} = \frac{1}{4}\).
\(\sinh(2\ln 2) = \frac{4 - 1/4}{2} = \frac{15}{8}\) and \(\cosh(2\ln 2) = \frac{4 + 1/4}{2} = \frac{17}{8}\).
Value at upper limit:
\(\frac{5}{2}\left(\frac{15}{8}\right) + \frac{3}{2}\left(\frac{17}{8}\right) + 4\ln 2 = \frac{75 + 51}{16} + 4\ln 2 = \frac{63}{8} + 4\ln 2\).
Value at lower limit \(x = 0\):
\(\frac{5}{2}\sinh(0) + \frac{3}{2}\cosh(0) + 0 = \frac{3}{2}\).
Subtracting:
\(\left(\frac{63}{8} + 4\ln 2\right) - \frac{3}{2} = \frac{51}{8} + 4\ln 2\).

(i) M1: Substitutes the exponential definitions of \(\sinh x\) and \(\cosh x\).
A1: Expands correctly and completes the algebraic proof.

(ii) M1: Rewrites equation in terms of exponentials.
A1: Obtains quadratic in \(\mathrm{e}^x\).
M1: Uses quadratic formula.
A1.375: Obtains exact logarithmic answers \(x = \ln\left(\frac{5 \pm \sqrt{17}}{4}\right)\).

(iii) M1: Expands the integrand and uses double-angle identities.
A1: Integrates terms correctly.
A1: Obtains exact value \(\frac{51}{8} + 4\ln 2\).

(i) Using integration by parts for \(I_n = \int_0^1 x^n e^{-x} \;\mathrm{d}x\):
Let \(u = x^n \implies \mathrm{d}u = n x^{n-1}\mathrm{d}x\).
Let \(\mathrm{d}v = e^{-x}\mathrm{d}x \implies v = -e^{-x}\).
Then,
\(I_n = \left[ -x^n e^{-x} \right]_0^1 + \int_0^1 n x^{n-1} e^{-x} \;\mathrm{d}x\)
\(= -e^{-1} + 0 + n \int_0^1 x^{n-1} e^{-x} \;\mathrm{d}x\)
\(= n I_{n-1} - \frac{1}{\mathrm{e}}\).

(ii) First find \(I_0\):
\(I_0 = \int_0^1 e^{-x} \;\mathrm{d}x = \left[ -e^{-x} \right]_0^1 = 1 - \frac{1}{\mathrm{e}}\).
Now apply the reduction formula:
\(I_1 = 1 \cdot I_0 - \frac{1}{\mathrm{e}} = 1 - \frac{2}{\mathrm{e}}\).
\(I_2 = 2 I_1 - \frac{1}{\mathrm{e}} = 2\left(1 - \frac{2}{\mathrm{e}}\right) - \frac{1}{\mathrm{e}} = 2 - \frac{5}{\mathrm{e}}\).
\(I_3 = 3 I_2 - \frac{1}{\mathrm{e}} = 3\left(2 - \frac{5}{\mathrm{e}}\right) - \frac{1}{\mathrm{e}} = 6 - \frac{16}{\mathrm{e}}\).

(iii) The volume of the solid of revolution \(V\) is:
\(V = \pi \int_0^1 y^2 \;\mathrm{d}x = \pi \int_0^1 \left(x^{3/2} e^{-x/2}\right)^2 \;\mathrm{d}x\)
\(= \pi \int_0^1 x^3 e^{-x} \;\mathrm{d}x\)
\(= \pi I_3\).
Substituting the value of \(I_3\) from part (ii):
\(V = \pi \left(6 - \frac{16}{\mathrm{e}}\right)\).

(i) M1: Uses integration by parts.
A1: Correct boundary evaluation \(-e^{-1}\).
A1.375: Correctly relates remaining integral to \(I_{n-1}\) and completes proof.

(ii) M1: Evaluates base case \(I_0\) correctly.
M1: Applies reduction formula iteratively.
A1: Obtains \(I_3 = 6 - \frac{16}{\mathrm{e}}\).

(iii) M1: Sets up correct volume of revolution formula \(V = \pi \int y^2 \;\mathrm{d}x\).
A1: Recognises the integral as \(I_3\).
A1: State final exact volume \(\pi \left(6 - \frac{16}{\mathrm{e}}\right)\).

(i) Divide the equation by \(x\) to put it in standard form:
\(\frac{\mathrm{d}y}{\mathrm{d}x} + \frac{2}{x}y = \frac{\sin x}{x^2}\).
The integrating factor is:
\(I = \mathrm{e}^{\int \frac{2}{x} \mathrm{d}x} = \mathrm{e}^{2\ln x} = x^2\).
Multiplying both sides by \(x^2\):
\(x^2 \frac{\mathrm{d}y}{\mathrm{d}x} + 2xy = \sin x\)
\(\frac{\mathrm{d}}{\mathrm{d}x}(x^2 y) = \sin x\).
Integrating both sides with respect to \(x\):
\(x^2 y = \int \sin x \;\mathrm{d}x = -\cos x + C\).
So the general solution is:
\(y = \frac{C - \cos x}{x^2}\).

(ii) Using the boundary condition \(y(\pi) = \frac{2}{\pi^2}\):
\(\frac{2}{\pi^2} = \frac{C - \cos \pi}{\pi^2} = \frac{C - (-1)}{\pi^2}\)
\(2 = C + 1 \implies C = 1\).
Thus, the particular solution is:
\(y = \frac{1 - \cos x}{x^2}\).

(iii) We find the limit of \(y\) as \(x \to 0^+\):
\(\lim_{x\to 0^+} \frac{1 - \cos x}{x^2}\).
As \(x \to 0\), the limit is of the form \(\frac{0}{0}\). Applying L'Hopital's Rule:
\(\lim_{x\to 0^+} \frac{\frac{\mathrm{d}}{\mathrm{d}x}(1 - \cos x)}{\frac{\mathrm{d}}{\mathrm{d}x}(x^2)} = \lim_{x\to 0^+} \frac{\sin x}{2x}\).
Since \(\lim_{x\to 0} \frac{\sin x}{x} = 1\), we have:
\(\lim_{x\to 0^+} \frac{\sin x}{2x} = \frac{1}{2}\).
Alternatively, using the Taylor series \(\cos x = 1 - \frac{1}{2}x^2 + O(x^4)\):
\(y = \frac{1 - (1 - \frac{1}{2}x^2 + O(x^4))}{x^2} = \frac{1}{2} + O(x^2)\),
which clearly approaches \(\frac{1}{2}\) as \(x \to 0^+\).

(i) M1: Divides by \(x\) and finds the integrating factor \(x^2\).
M1: Multiplies by integrating factor and writes LHS as product derivative.
A1: Integrates \(\sin x\) correctly with constant of integration.
A1.375: Obtains general solution \(y = \frac{C - \cos x}{x^2}\).

(ii) M1: Substitutes boundary conditions to solve for \(C\).
A1: Obtains \(C = 1\) and states particular solution \(y = \frac{1 - \cos x}{x^2}\).

(iii) M1: Recognises indeterminate form \(0/0\) and applies L'Hopital's rule or Taylor series expansion.
A1: Obtains first derivative limit \(\frac{\sin x}{2x}\) or series expansion.
A1: Correctly evaluates the limit as \(\frac{1}{2}\).

Let the equation of the trajectory be: \( y = x \tan \theta - \frac{g x^2}{2 V^2 \cos^2 \theta} \). Given that \( \tan \theta = \frac{4}{3} \), we have \( \cos^2 \theta = \frac{1}{1 + \tan^2 \theta} = \frac{1}{1 + 16/9} = \frac{9}{25} \). Substituting the given values \( x = 30 \), \( y = 20 \), and \( g = 10 \) into the equation: \( 20 = 30 \left(\frac{4}{3}\right) - \frac{10 (30)^2}{2 V^2 (9/25)} \) which simplifies to \( 20 = 40 - \frac{9000}{\frac{18}{25}V^2} \). This further simplifies to \( 20 = 40 - \frac{12500}{V^2} \). Rearranging this gives \( \frac{12500}{V^2} = 20 \), so \( V^2 = 625 \). Taking the positive square root, we get \( V = 25 \).

M1: For using the projectile trajectory equation with \( \tan \theta \) and \( \cos^2 \theta \). A1: For correctly identifying that \( \cos^2 \theta = \frac{9}{25} \). M1: For substituting the values \( x = 30 \), \( y = 20 \), and \( g = 10 \) into the equation. A1: For simplifying the expression to the form \( 20 = 40 - \frac{12500}{V^2} \). M1: For solving the equation to find \( V^2 = 625 \). A1.14: For the correct final answer \( V = 25 \).

Let \(\theta\) be the angle the string makes with the downward vertical. The height of the particle above its lowest point is given by \( h = L(1 - \cos\theta) \). Since \( h = 1.5L \), we have \( L(1 - \cos\theta) = 1.5L \implies \cos\theta = -0.5 \). By conservation of energy, \( \frac{1}{2} m u^2 = \frac{1}{2} m v^2 + m g h \), which gives \( v^2 = u^2 - 2gh = u^2 - 3gL \). The radial equation of motion for the particle towards \(O\) is \( T - mg\cos\theta = \frac{mv^2}{L} \). When the string becomes slack, the tension \( T = 0 \), so \( -mg\cos\theta = \frac{mv^2}{L} \implies -g\cos\theta = \frac{v^2}{L} \). Substituting \( \cos\theta = -0.5 \) and \( v^2 = u^2 - 3gL \) yields \( 0.5g = \frac{u^2 - 3gL}{L} \implies 0.5gL = u^2 - 3gL \implies u^2 = 3.5gL = \frac{7gL}{2} \). Thus, \( u = \sqrt{\frac{7gL}{2}} \).

M1: For using \( h = L(1 - \cos\theta) \) with \( 1.5L \) to find \( \cos\theta = -0.5 \). M1: For setting up the conservation of energy equation. A1: For obtaining the correct expression for speed \( v^2 = u^2 - 3gL \). M1: For setting up the radial equation of motion and identifying that \( T = 0 \) when the string is slack. A1: For the equation \( -g\cos\theta = \frac{v^2}{L} \). M1: For substituting and solving for \( u^2 \). A1.14: For the correct exact answer \( u = \sqrt{\frac{7gL}{2}} \).

Using Newton's second law: \( F = ma \implies -(v^2 + 4v) = 0.5 v \frac{dv}{dx} \). Since \( v > 0 \) during the motion, we can divide by \( v \) to get \( -(v + 4) = 0.5 \frac{dv}{dx} \). Separating the variables gives \( \frac{dv}{v + 4} = -2 dx \). Integrating both sides using the boundary conditions \( v = 10 \) when \( x = 0 \): \( \int_{10}^{v} \frac{1}{v+4} dv = \int_{0}^{x} -2 dx \implies [\ln(v+4)]_{10}^{v} = -2x \implies \ln\left(\frac{v+4}{14}\right) = -2x \). The particle comes to rest when \( v = 0 \). Substituting this in gives: \( \ln\left(\frac{4}{14}\right) = -2x \implies \ln\left(\frac{2}{7}\right) = -2x \implies 2x = -\ln\left(\frac{2}{7}\right) = \ln\left(\frac{7}{2}\right) = \ln 3.5 \). Therefore, the distance travelled is \( x = \frac{1}{2} \ln 3.5 \) m.

M1: For using \( a = v \frac{dv}{dx} \) in Newton's second law. A1: For obtaining the simplified differential equation \( -(v + 4) = 0.5 \frac{dv}{dx} \). M1: For separating the variables and attempting to integrate. A1: For the correct integrated form \( \ln(v+4) = -2x + C \) or equivalent definite integral. M1: For applying the initial condition \( v = 10 \) when \( x = 0 \) to find \( C \). M1: For setting \( v = 0 \) to find the total distance. A1.14: For the correct final exact answer \( \frac{1}{2} \ln 3.5 \) (or equivalent rational fraction form).

Let \(x\) be the maximum extension of the string. At the point of maximum extension, the speed of the particle is zero, so it is instantaneously at rest. By conservation of energy, the loss in gravitational potential energy equals the gain in elastic potential energy. The total vertical distance fallen is \( L + x = 0.5 + x \). Loss in gravitational potential energy: \( E_{gp} = mg(L + x) = 2 \times 10 \times (0.5 + x) = 10 + 20x \). Gain in elastic potential energy: \( E_{ep} = \frac{\lambda x^2}{2L} = \frac{80 x^2}{2 \times 0.5} = 80 x^2 \). Equating the energies: \( 80 x^2 = 10 + 20x \implies 8 x^2 - 2 x - 1 = 0 \). Factoring the quadratic equation gives \( (4x + 1)(2x - 1) = 0 \). Since the extension \(x\) must be positive, we find \( x = 0.5 \) m.

M1: For writing a conservation of energy equation relating GPE loss and EPE gain. M1: For expressing the loss in gravitational potential energy as \( mg(L+x) \). A1: For obtaining \( 10 + 20x \) for GPE loss. M1: For expressing elastic potential energy as \( \frac{\lambda x^2}{2L} \). A1: For obtaining \( 80x^2 \) for EPE. M1: For setting up and attempting to solve the quadratic equation. A1.14: For the correct positive solution \( x = 0.5 \) m.

Let \( R_A \) be the normal reaction force at \( A \), \( F_A \) be the frictional force at \( A \), and \( R_B \) be the normal reaction force at \( B \). Resolving forces vertically: \( R_A = W \). Resolving forces horizontally: \( F_A = R_B \). Taking moments about \( A \): \( R_B \times 2a \sin\theta = W \times a \cos\theta \implies R_B = \frac{W \cos\theta}{2 \sin\theta} = \frac{W}{2 \tan\theta} \). Given \( \tan\theta = \frac{3}{4} \), we find \( R_B = \frac{W}{2 \times (3/4)} = \frac{2W}{3} \). Thus, the frictional force is \( F_A = \frac{2}{3}W \). For the rod to remain in equilibrium, we must have \( F_A \le \mu R_A \). Substituting the known expressions: \( \frac{2}{3}W \le \mu W \implies \mu \ge \frac{2}{3} \). The minimum value of the coefficient of friction is therefore \( \frac{2}{3} \).

M1: For resolving forces vertically to obtain \( R_A = W \). M1: For resolving forces horizontally to relate \( F_A = R_B \). M1: For taking moments about \( A \) (or any other suitable point). A1: For obtaining the correct moment equation: \( R_B \times 2a \sin\theta = W \times a \cos\theta \). A1: For finding the friction force \( F_A = \frac{2}{3}W \). M1: For using the friction inequality \( F_A \le \mu R_A \). A1.14: For obtaining the minimum value of \( \mu = \frac{2}{3} \) (or 0.667).

Let \( v_A \) and \( v_B \) be the velocities of \( A \) and \( B \) after the collision in the direction of \( u \). By conservation of momentum: \( m u = m v_A + 3m v_B \implies u = v_A + 3v_B \). By Newton's law of restitution: \( v_B - v_A = e u \). Adding these two equations: \( 4 v_B = u(1 + e) \implies v_B = \frac{u(1+e)}{4} \). Substituting this back gives: \( v_A = v_B - e u = \frac{u(1-3e)}{4} \). The initial kinetic energy is \( E_i = \frac{1}{2} m u^2 \). The final kinetic energy is \( E_f = \frac{1}{2} m v_A^2 + \frac{1}{2} (3m) v_B^2 = \frac{1}{2} m \left[ \frac{u^2(1-3e)^2}{16} \right] + \frac{3}{2} m \left[ \frac{u^2(1+e)^2}{16} \right] = \frac{mu^2}{32} [ (1 - 6e + 9e^2) + 3(1 + 2e + e^2) ] = \frac{mu^2}{32} [ 4 + 12e^2 ] = \frac{mu^2}{8}(1 + 3e^2) \). The loss of kinetic energy is \( \Delta E = E_i - E_f = \frac{1}{2} m u^2 - \frac{1}{8} m u^2 (1 + 3e^2) = \frac{m u^2}{8} [ 4 - (1 + 3e^2) ] = \frac{3}{8} m u^2 (1 - e^2) \).

M1: For applying conservation of momentum. M1: For applying Newton's law of restitution. A1: For finding the correct expressions for \( v_A \) and \( v_B \) in terms of \( u \) and \( e \). M1: For writing down the formula for the total final kinetic energy. A1: For simplifying the final kinetic energy to \( \frac{1}{8} m u^2 (1 + 3e^2) \). M1: For subtracting the final kinetic energy from the initial kinetic energy. A1.14: For completing the algebraic steps to show \( \frac{3}{8} m u^2 (1 - e^2) \).

Let \(\theta\) be the angle that the radius vector of the particle makes with the upward vertical. The vertical distance fallen by the particle is \( R(1 - \cos\theta) \). By conservation of energy, \( \frac{1}{2} m v^2 = \frac{1}{2} m u^2 + m g R (1 - \cos\theta) \implies v^2 = u^2 + 2gR(1 - \cos\theta) \). Substituting \( u = \frac{1}{2}\sqrt{gR} \) gives \( u^2 = \frac{1}{4}gR \), so \( v^2 = \frac{1}{4}gR + 2gR - 2gR\cos\theta = gR \left( \frac{9}{4} - 2\cos\theta \right) \). The radial equation of motion for the particle while in contact with the sphere is \( m g \cos\theta - R_N = \frac{m v^2}{R} \), where \( R_N \) is the normal reaction force. The particle leaves the surface of the sphere when \( R_N = 0 \), which gives \( g \cos\theta = \frac{v^2}{R} \). Substituting the expression for \( v^2 \): \( g \cos\theta = g \left( \frac{9}{4} - 2\cos\theta \right) \implies \cos\theta = \frac{9}{4} - 2\cos\theta \implies 3\cos\theta = \frac{9}{4} \implies \cos\theta = \frac{3}{4} \).

M1: For applying the principle of conservation of energy to relate the speeds. A1: For the equation \( v^2 = u^2 + 2gR(1 - \cos\theta) \). M1: For substituting \( u^2 = \frac{1}{4}gR \) to obtain \( v^2 = gR\left(\frac{9}{4} - 2\cos\theta\right) \). M1: For writing the radial equation of motion. A1: For identifying that the particle leaves the surface when \( R_N = 0 \), resulting in \( g\cos\theta = \frac{v^2}{R} \). M1: For substituting the expression for \( v^2 \) into the leaving condition. A1.14: For completing the proof to show that \( \cos\theta = \frac{3}{4} \).

Let \(D = \text{After} - \text{Before}\). The differences \(d\) for the students A, B, C, D, E, F, G are 15, 5, 15, -2, 15, 20, 5 respectively. The number of pairs is \(n = 7\). The mean difference is \(\bar{d} = \frac{15 + 5 + 15 - 2 + 15 + 20 + 5}{7} = \frac{73}{7} \approx 10.43\). The sum of squares of differences is \(\sum d^2 = 15^2 + 5^2 + 15^2 + (-2)^2 + 15^2 + 20^2 + 5^2 = 1129\). The sample variance of the differences is \(s^2 = \frac{1}{6} \left( 1129 - \frac{73^2}{7} \right) = \frac{1}{6} \left( 1129 - 761.29 \right) = 61.29\), so the standard deviation is \(s \approx 7.829\). We test \(H_0: \mu_D = 0\) against \(H_1: \mu_D > 0\). Under \(H_0\), the test statistic is \(t = \frac{\bar{d} - 0}{s / \sqrt{n}} = \frac{10.4286}{7.8285 / \sqrt{7}} \approx 3.52\). The critical value from the t-distribution with 6 degrees of freedom at the 5% significance level (one-tailed) is \(1.943\). Since \(3.52 > 1.943\), we reject \(H_0\). There is significant evidence that the program increases reading speed. Assumption: The differences in reading speeds are normally distributed.

M1 for calculating the differences and finding their mean (10.43). A1 for calculating the sample standard deviation (7.83). M1 for stating the null and alternative hypotheses. M1 for calculating the test statistic t (3.52). A1 for identifying the correct critical value of t (1.943) and making a correct comparison. A1 for the correct conclusion in context. B1 for stating the required assumption of normality of differences.

The null hypothesis \(H_0\) is that movie genre preference is independent of age group. The alternative hypothesis \(H_1\) is that there is an association. The row totals are 70 and 80, and the column totals are 54 (Action), 60 (Comedy), and 36 (Drama), with a grand total of 150. Expected frequencies are calculated using \((Row \, Total \times Column \, Total) / 150\). Under 30 expected values: Action = 25.2, Comedy = 28.0, Drama = 16.8. 30 and Over expected values: Action = 28.8, Comedy = 32.0, Drama = 19.2. The test statistic is \(\chi^2 = \sum \frac{(O-E)^2}{E} = \frac{(32-25.2)^2}{25.2} + \frac{(28-28.0)^2}{28.0} + \frac{(10-16.8)^2}{16.8} + \frac{(22-28.8)^2}{28.8} + \frac{(32-32.0)^2}{32.0} + \frac{(26-19.2)^2}{19.2} = 1.8349 + 0 + 2.7524 + 1.6056 + 0 + 2.4083 = 8.60\). The number of degrees of freedom is \((2-1)(3-1) = 2\). The critical value for a chi-squared test with 2 degrees of freedom at the 5% level is \(5.991\). Since \(8.60 > 5.991\), we reject \(H_0\). There is significant evidence at the 5% level to suggest an association between preferred movie genre and age group.

B1 for stating hypotheses correctly. M1 for calculating expected frequencies (at least three correct). A1 for all expected frequencies correct. M1 for calculating the test statistic (8.60). A1 for correct degrees of freedom (2) and critical value (5.991). A1 for comparing test statistic with critical value and making correct decision. A1 for the correct conclusion in context.

We define the difference as \(D = \text{After} - \text{Before}\). The differences for the 9 participants are: +7, -2, +15, +12, 0, +13, +15, +10, +12. Participant 5 has a difference of 0 and is excluded, leaving \(n = 8\). The remaining differences are: +7, -2, +15, +12, +13, +15, +10, +12. Sorting the absolute differences: 2, 7, 10, 12, 12, 13, 15, 15. The ranks are: |Difference| = 2: rank 1. |Difference| = 7: rank 2. |Difference| = 10: rank 3. |Difference| = 12 (two of them): average rank 4.5. |Difference| = 13: rank 6. |Difference| = 15 (two of them): average rank 7.5. The only negative difference is -2, which has a rank of 1. Thus, the sum of negative ranks is \(T^- = 1\), and the sum of positive ranks is \(T^+ = 2 + 3 + 4.5 + 4.5 + 6 + 7.5 + 7.5 = 35\). The test statistic is \(T = \min(T^-, T^+) = 1\). Under the null hypothesis \(H_0\): median difference is zero (no increase), against the alternative hypothesis \(H_1\): median difference is greater than zero (plank duration increases). For a one-tailed test with \(n = 8\) at the 5% level, the critical value is \(5\). Since \(T = 1 \le 5\), we reject \(H_0\). There is significant evidence that the training program increases plank duration.

M1 for calculating differences and omitting the zero difference. M1 for ranking the absolute differences correctly, including handling of tied ranks. A1 for correct sum of negative ranks (1) or positive ranks (35). B1 for stating the correct hypotheses. B1 for stating the correct critical value (5) for a one-tailed test with n=8. M1 for comparing the test statistic with the critical value. A1 for the correct conclusion in context.

(i) Since the total probability is 1, we must have \(\int_0^3 f(x) \, dx = 1\). This is split as \(\int_0^2 k(4x-x^2) \, dx + \int_2^3 \frac{4}{3}k(3-x) \, dx = 1\). Calculating the first integral: \(\int_0^2 (4x-x^2) \, dx = \left[ 2x^2 - \frac{1}{3}x^3 \right]_0^2 = 8 - \frac{8}{3} = \frac{16}{3}\). Calculating the second integral: \(\int_2^3 (3-x) \, dx = \left[ 3x - \frac{1}{2}x^2 \right]_2^3 = \left(9 - 4.5\right) - \left(6 - 2\right) = 0.5 = \frac{1}{2}\). Thus, we have \(k \left( \frac{16}{3} + \frac{4}{3} \times \frac{1}{2} \right) = 1 \implies k \left( \frac{16}{3} + \frac{2}{3} \right) = 1 \implies 6k = 1 \implies k = \frac{1}{6}\). (ii) For \(0 \le x \le 2\), the cumulative distribution function is \(F(x) = \int_0^x f(t) \, dt = \int_0^x \frac{1}{6} t(4 - t) \, dt = \frac{1}{6} \left[ 2t^2 - \frac{1}{3}t^3 \right]_0^x = \frac{x^2(6-x)}{18}\). (iii) The probability \(P(X \ge 1.5) = 1 - F(1.5)\). Substituting \(x = 1.5\) into \(F(x)\) gives \(F(1.5) = \frac{1.5^2(6-1.5)}{18} = \frac{2.25 \times 4.5}{18} = \frac{10.125}{18} = 0.5625 = \frac{9}{16}\). Therefore, \(P(X \ge 1.5) = 1 - \frac{9}{16} = \frac{7}{16} = 0.4375\).

M1 for setting up the sum of integrals to equal 1. A1 for correct integration of both parts and demonstrating k = 1/6. M1 for integrating f(t) to find F(x). A1 for obtaining the correct F(x) = x^2(6-x)/18. M1 for expressing P(X >= 1.5) as 1 - F(1.5). A1 for correct calculation of F(1.5) = 9/16. A1 for the final probability of 7/16 or 0.4375.

(i) Using the property of PGFs that \(G_X(1) = 1\), we substitute \(t = 1\) into \(G_X(t)\) to get \(a(2 + 1)^3 = 1 \implies 27a = 1 \implies a = \frac{1}{27}\). (ii) The mean of \(X\) is given by \(E(X) = G'_X(1)\). Differentiating \(G_X(t) = \frac{1}{27}(2+t)^3\) with respect to \(t\), we get \(G'_X(t) = \frac{3}{27}(2+t)^2 = \frac{1}{9}(2+t)^2\). Substituting \(t = 1\) gives \(E(X) = G'_X(1) = \frac{1}{9}(3)^2 = 1\). To find the variance, we first find the second derivative \(G''_X(t) = \frac{2}{9}(2+t)\). Substituting \(t = 1\) gives \(G''_X(1) = \frac{2}{9}(3) = \frac{2}{3}\). The variance is given by \(\text{Var}(X) = G''_X(1) + G'_X(1) - (G'_X(1))^2 = \frac{2}{3} + 1 - 1^2 = \frac{2}{3}\). (iii) The PGF of \(Y = 2X + 1\) is \(G_Y(t) = E(t^Y) = E(t^{2X+1}) = t E((t^2)^X) = t G_X(t^2)\). Thus, \(G_Y(t) = t \left( \frac{1}{27}(2 + t^2)^3 \right) = \frac{t(2+t^2)^3}{27}\).

M1 for using G_X(1) = 1 to show a = 1/27. M1 for differentiating G_X(t) once and substituting t = 1 to find the mean. A1 for correct mean of 1. M1 for differentiating a second time and using the variance formula. A1 for correct variance of 2/3. M1 for using the property G_Y(t) = t * G_X(t^2). A1 for the correct simplified expression.

First, we calculate the sample mean \(\bar{x}\) and the sample variance \(s^2\). The sum of the 10 weights is \(\sum x = 201.0\), so \(\bar{x} = \frac{201.0}{10} = 20.1\). The sum of squared deviations from the mean is \(\sum (x - \bar{x})^2 = (20.2-20.1)^2 + (19.8-20.1)^2 + \dots + (20.1-20.1)^2 = 0.01 + 0.09 + 0.16 + 0.00 + 0.04 + 0.04 + 0.01 + 0.16 + 0.09 + 0.00 = 0.60\). The sample variance is \(s^2 = \frac{1}{9} \times 0.60 = \frac{1}{15} \approx 0.06667\), giving the sample standard deviation \(s \approx 0.2582\). Since the population variance is unknown, we use a t-distribution with \(n-1 = 9\) degrees of freedom. For a 95% confidence level, the critical value \(t_{9, 0.025}\) is \(2.262\). The confidence interval is given by \(\bar{x} \pm t \frac{s}{\sqrt{n}} = 20.1 \pm 2.262 \times \frac{0.2582}{\sqrt{10}} = 20.1 \pm 2.262 \times 0.08165 = 20.1 \pm 0.185\). This gives the confidence interval \([19.915, 20.285]\), which to 2 decimal places is \([19.92, 20.28]\) or to 3 significant figures is \([19.9, 20.3]\).

M1 for finding the sample mean 20.1. M1 for calculating the sample standard deviation (0.2582). A1 for correct values of both. B1 for identifying the correct critical value of t with 9 degrees of freedom (2.262). M1 for calculating the margin of error (0.185). A1 for the correct confidence interval bounds (accept 19.9 to 20.3, or 19.92 to 20.28).

(i) First, determine the range of \(Y\). Since \(0 \le X \le 1\), we have \(0 \le X^2 \le 1\), so \(Y = \frac{1}{X^2}\) takes values in \([1, \infty)\). For \(y \ge 1\), the cumulative distribution function \(F_Y(y)\) is given by \(F_Y(y) = P(Y \le y) = P\left(\frac{1}{X^2} \le y\right) = P\left(X^2 \ge \frac{1}{y}\right) = P\left(X \ge \frac{1}{\sqrt{y}}\right)\) since \(X\) takes only positive values. This can be written as \(1 - F_X\left(\frac{1}{\sqrt{y}}\right)\). For \(0 \le x \le 1\), the cumulative distribution function of \(X\) is \(F_X(x) = \int_0^x 2t \, dt = x^2\). Therefore, \(F_X\left(\frac{1}{\sqrt{y}}\right) = \left(\frac{1}{\sqrt{y}}\right)^2 = \frac{1}{y}\). Hence, \(F_Y(y) = 1 - \frac{1}{y}\) for \(y \ge 1\), and \(F_Y(y) = 0\) for \(y < 1\). (ii) To find the probability density function \(g(y)\), we differentiate \(F_Y(y)\) with respect to \(y\): \(g(y) = F'_Y(y) = \frac{d}{dy}\left(1 - \frac{1}{y}\right) = \frac{1}{y^2}\) for \(y \ge 1\), and \(0\) otherwise.

M1 for identifying the range of Y as y >= 1. M1 for expressing P(Y <= y) in terms of X. A1 for correctly rewriting this as P(X >= y^{-1/2}). M1 for using the CDF of X, which is x^2. A1 for obtaining F_Y(y) = 1 - 1/y for y >= 1. M1 for differentiating F_Y(y) to obtain g(y). A1 for the correct PDF g(y) = 1/y^2 for y >= 1, 0 otherwise.