2024 Cambridge IAL Mathematics - Further (9231) Practice Paper with Answers

(i) A matrix is singular when its determinant is zero. \(\det(\mathbf{M}) = 2(0 - 3) - (-1)(k - (-3)) + 1(1 - 0) = -6 + k + 3 + 1 = k - 2\). Setting \(\det(\mathbf{M}) = 0\) gives \(k = 2\).

(ii) The vector equation of the line is \(\mathbf{r} = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}\). Any point on this line is of the form \(\begin{pmatrix} 1 + 2\lambda \\ \lambda \\ -1 - \lambda \end{pmatrix}\). Transforming this point under \(\mathbf{M}\) with \(k = 4\):
\(\mathbf{M} \mathbf{r} = \begin{pmatrix} 2 & -1 & 1 \\ 1 & 0 & 3 \\ -1 & 1 & 4 \end{pmatrix} \begin{pmatrix} 1 + 2\lambda \\ \lambda \\ -1 - \lambda \end{pmatrix} = \begin{pmatrix} 2(1+2\lambda) - \lambda + (-1-\lambda) \\ (1+2\lambda) + 3(-1-\lambda) \\ -(1+2\lambda) + \lambda + 4(-1-\lambda) \end{pmatrix} = \begin{pmatrix} 1 + 2\lambda \\ -2 - \lambda \\ -5 - 5\lambda \end{pmatrix}\).
Thus, the image line has vector equation \(\mathbf{r} = \begin{pmatrix} 1 \\ -2 \\ -5 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -1 \\ -5 \end{pmatrix}\).

(iii) Let the original coordinates be \((x, y, z)\) and the transformed coordinates be \((x', y', z')\). We have:
\(x' = 2x - y + z\)
\(y' = x + 3z\)
\(z' = -x + y + 4z\)
Substituting these into the image plane equation \(x' - y' + z' = 0\):
\((2x - y + z) - (x + 3z) + (-x + y + 4z) = 0\)
\((2 - 1 - 1)x + (-1 + 1)y + (1 - 3 + 4)z = 0 \implies 2z = 0 \implies z = 0\).

(i) M1: Attempts to find the determinant of \(\mathbf{M}\) in terms of \(k\). A1: Correct determinant \(k - 2\). A1: Correct value \(k = 2\).
(ii) M1: Expresses the line in vector form. M1: Multiplies the matrix by the general point of the line. A1: Correctly performs the matrix multiplication. A1: Obtains a correct simplified position vector. A1: Writes down the final vector equation of the image line.
(iii) M1: Substitutes the transformation equations into the plane equation \(x' - y' + z' = 0\). A1.7: Simplifies correctly to obtain \(z = 0\).

(i) From the relation between roots and coefficients:
\(\alpha + \beta + \gamma = 4\)
\(\alpha\beta + \beta\gamma + \gamma\alpha = 2\)
\(\alpha\beta\gamma = 1\)
Using the identity \(\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\):
\(\alpha^2 + \beta^2 + \gamma^2 = 4^2 - 2(2) = 16 - 4 = 12\).

(ii) Since \(\alpha, \beta, \gamma\) are roots of the cubic equation, they satisfy:
\(\alpha^3 - 4\alpha^2 + 2\alpha - 1 = 0\)
\(\beta^3 - 4\beta^2 + 2\beta - 1 = 0\)
\(\gamma^3 - 4\gamma^2 + 2\gamma - 1 = 0\)
Summing these equations:
\((\alpha^3 + \beta^3 + \gamma^3) - 4(\alpha^2 + \beta^2 + \gamma^2) + 2(\alpha + \beta + \gamma) - 3 = 0\)
\((\alpha^3 + \beta^3 + \gamma^3) - 4(12) + 2(4) - 3 = 0\)
\(\alpha^3 + \beta^3 + \gamma^3 - 48 + 8 - 3 = 0 \implies \alpha^3 + \beta^3 + \gamma^3 = 43\).

(iii) Let \(y = x^2\), so \(x = \sqrt{y}\).
Substitute \(x = \sqrt{y}\) into the original cubic equation:
\(y\sqrt{y} - 4y + 2\sqrt{y} - 1 = 0 \implies \sqrt{y}(y + 2) = 4y + 1\).
Squaring both sides:
\(y(y + 2)^2 = (4y + 1)^2\)
\(y(y^2 + 4y + 4) = 16y^2 + 8y + 1\)
\(y^3 + 4y^2 + 4y = 16y^2 + 8y + 1\)
\(y^3 - 12y^2 - 4y - 1 = 0\).

(i) B1: Correctly states the sum and product of roots. M1: Applies the identity for \(\sum \alpha^2\). A1: Obtains 12.
(ii) M1: Uses the cubic relation for the roots. M1: Forms the sum equation. A1.7: Correctly calculates \(\sum \alpha^3 = 43\).
(iii) M1: Substitutes \(x = \sqrt{y}\) or equivalent method. M1: Rearranges to isolate square roots. M1: Squares both sides. A1: Correctly simplifies to \(y^3 - 12y^2 - 4y - 1 = 0\).

(i) The direction vectors of the lines \(AB\) and \(CD\) are:
\(\vec{AB} = \begin{pmatrix} 1 \\ -3 \\ 4 \end{pmatrix}\) and \(\vec{CD} = \begin{pmatrix} 3 \\ -1 \\ -1 \end{pmatrix}\).
A normal vector \(\mathbf{n}\) to both lines is:
\(\mathbf{n} = \vec{AB} \times \vec{CD} = \begin{pmatrix} 1 \\ -3 \\ 4 \end{pmatrix} \times \begin{pmatrix} 3 \\ -1 \\ -1 \end{pmatrix} = \begin{pmatrix} 7 \\ 13 \\ 8 \end{pmatrix}\).
We also find the vector \(\vec{AC} = \begin{pmatrix} -1 \\ -1 \\ 3 \end{pmatrix}\).
The shortest distance \(d\) is the projection of \(\vec{AC}\) onto \(\mathbf{n}\):
\(d = \frac{|\vec{AC} \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{|-1(7) - 1(13) + 3(8)|}{\sqrt{7^2 + 13^2 + 8^2}} = \frac{4}{\sqrt{49 + 169 + 64}} = \frac{4}{\sqrt{282}}\).

(ii) The normal vector \(\mathbf{n}_1\) to the plane \(ABC\) is:
\(\mathbf{n}_1 = \vec{AB} \times \vec{AC} = \begin{pmatrix} 1 \\ -3 \\ 4 \end{pmatrix} \times \begin{pmatrix} -1 \\ -1 \\ 3 \end{pmatrix} = \begin{pmatrix} -5 \\ -7 \\ -4 \end{pmatrix}\), or we can use \(\begin{pmatrix} 5 \\ 7 \\ 4 \end{pmatrix}\).
The vector \(\vec{AD} = \begin{pmatrix} 2 \\ -2 \\ 2 \end{pmatrix}\).
The normal vector \(\mathbf{n}_2\) to the plane \(ABD\) is:
\(\mathbf{n}_2 = \vec{AB} \times \vec{AD} = \begin{pmatrix} 1 \\ -3 \\ 4 \end{pmatrix} \times \begin{pmatrix} 2 \\ -2 \\ 2 \end{pmatrix} = \begin{pmatrix} 2 \\ 6 \\ 4 \end{pmatrix}\), or we can use \(\begin{pmatrix} 1 \\ 3 \\ 2 \end{pmatrix}\).
The angle \(\theta\) between the two planes satisfies:
\(\cos\theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|} = \frac{|5(1) + 7(3) + 4(2)|}{\sqrt{5^2+7^2+4^2}\sqrt{1^2+3^2+2^2}} = \frac{34}{\sqrt{90}\sqrt{14}} = \frac{34}{6\sqrt{35}} = \frac{17}{3\sqrt{35}}\).
Thus, \(\theta = \arccos\left(\frac{17}{3\sqrt{35}}\right) \approx 16.7^\circ\) (or \(0.292\) radians).

(i) M1: Finds direction vectors for both lines. M1: Finds cross product of direction vectors. A1: Correct normal vector \(\begin{pmatrix} 7 \\ 13 \\ 8 \end{pmatrix}\). M1: Finds \(\vec{AC}\) or another vector connecting the two lines. M1: Uses the projection formula. A0.7: Correct shortest distance \(\frac{4}{\sqrt{282}}\).
(ii) M1: Computes normal vector to plane \(ABC\). M1: Computes normal vector to plane \(ABD\). M1: Uses dot product formula to find the angle. A1: Correctly calculates \(\cos\theta\). A1: Correct final angle in degrees or radians.

(i) The vertical asymptote occurs where the denominator is zero, so \(x = 3\).
By algebraic division:
\(y = \frac{x^2 - x - 2}{x-3} = x + 2 + \frac{4}{x-3}\).
As \(x \to \pm\infty\), \(\frac{4}{x-3} \to 0\), so the oblique asymptote is \(y = x + 2\).

(ii) Differentiating \(y\):
\(\frac{dy}{dx} = 1 - \frac{4}{(x-3)^2}\).
Setting \(\frac{dy}{dx} = 0\) gives:
\(\frac{4}{(x-3)^2} = 1 \implies (x-3)^2 = 4 \implies x-3 = \pm 2\).
This yields \(x = 5\) or \(x = 1\).
When \(x = 5\), \(y = 5 + 2 + \frac{4}{2} = 9\).
When \(x = 1\), \(y = 1 + 2 + \frac{4}{-2} = 1\).
The turning points are \((1, 1)\) and \((5, 9)\).

(iii) Intercepts:
- With the \(y\)-axis: when \(x=0\), \(y = \frac{-2}{-3} = \frac{2}{3}\).
- With the \(x\)-axis: when \(y=0\), \(x^2 - x - 2 = 0 \implies (x-2)(x+1) = 0\), so \(x = 2\) or \(x = -1\).
The sketch should show:
- The two branches of the curve.
- Asymptotes \(x = 3\) and \(y = x + 2\).
- Turning points at \((1, 1)\) and \((5, 9)\).
- Intercepts at \((0, \frac{2}{3})\), \((-1, 0)\), and \((2, 0)\).

(i) B1: Correct vertical asymptote \(x = 3\). M1: Performs division to find the oblique asymptote. A1: Correct oblique asymptote \(y = x + 2\).
(ii) M1: Obtains the first derivative of \(y\). M1: Sets the derivative to 0 and solves for \(x\). A1: Correct \(x\)-coordinates. A1: Correct turning points \((1, 1)\) and \((5, 9)\).
(iii) B1: Correct intercepts with axes. B1: Correct asymptotes drawn. B1.7: Correct shape of both branches with turning points labeled.

(i) The curve is a dimpled lima\u00e7on, symmetric about the initial line, with \(r\) ranging from \(3\) (at \(\theta = 0\)) to \(1\) (at \(\theta = \pi\)).

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

(iii) Let \(x = r\cos\theta = (2+\cos\theta)\cos\theta = 2\cos\theta + \cos^2\theta\) and \(y = r\sin\theta = (2+\cos\theta)\sin\theta = 2\sin\theta + \frac{1}{2}\sin(2\theta)\).
At \(\theta = \frac{\pi}{3}\):
\(x = \frac{5}{4}\), \(y = \frac{5\sqrt{3}}{4}\).
Now find derivatives with respect to \(\theta\):
\(\frac{dx}{d\theta} = -2\sin\theta - \sin(2\theta)\)
\(\frac{dy}{d\theta} = 2\cos\theta + \cos(2\theta)\)
At \(\theta = \frac{\pi}{3}\):
\(\frac{dx}{d\theta} = -2\left(\frac{\sqrt{3}}{2}\right) - \frac{\sqrt{3}}{2} = -\frac{3\sqrt{3}}{2}\)
\(\frac{dy}{d\theta} = 2\left(\frac{1}{2}\right) - \frac{1}{2} = \frac{1}{2}\)
\(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{1/2}{-3\sqrt{3}/2} = -\frac{1}{3\sqrt{3}} = -\frac{\sqrt{3}}{9}\).
The equation of the tangent line is:
\(y - \frac{5\sqrt{3}}{4} = -\frac{\sqrt{3}}{9}\left(x - \frac{5}{4}\right)\)
Multiplying by \(36\):
\(36y - 45\sqrt{3} = -4\sqrt{3}x + 5\sqrt{3} \implies 4\sqrt{3}x + 36y = 50\sqrt{3}\)
Dividing by \(2\sqrt{3}\):
\(2x + 6\sqrt{3}y = 25\).

(i) B1: Symmetry about the initial line. B1: Correct values of \(r\) at key points \(0, \pi/2, \pi\). B1: Smooth closed loop without crossing the pole.
(ii) M1: Sets up the area integral with correct limits. M1: Expands the integrand. M1: Uses the double angle identity. A1: Integrates correctly. A0.7: Correct final area of \(\frac{9\pi}{2}\).
(iii) M1: Parametrizes \(x\) and \(y\) in terms of \(\theta\). M1: Computes \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\). A1: Finds the gradient of the tangent and the point, simplifying to \(2x + 6\sqrt{3}y = 25\).

(i) Expanding the right-hand side:
\((2r^2 - 2r + 1)(2r^2 + 2r + 1) = (2r^2 + 1 - 2r)(2r^2 + 1 + 2r) = (2r^2 + 1)^2 - (2r)^2 = 4r^4 + 4r^2 + 1 - 4r^2 = 4r^4 + 1\).

(ii) We have:
\(\frac{4r}{4r^4 + 1} = \frac{A}{2r^2 - 2r + 1} + \frac{B}{2r^2 + 2r + 1}\)
\(4r = A(2r^2 + 2r + 1) + B(2r^2 - 2r + 1)\)
Equating coefficients:
For \(r^2\): \(2A + 2B = 0 \implies B = -A\).
For \(r\): \(2A - 2B = 4 \implies 4A = 4 \implies A = 1\), which means \(B = -1\).
For constant term: \(A + B = 0\), which is consistent.
So \(\frac{4r}{4r^4 + 1} = \frac{1}{2r^2 - 2r + 1} - \frac{1}{2r^2 + 2r + 1}\).

(iii) Using the method of differences:
Let \(f(r) = \frac{1}{2r^2 - 2r + 1}\).
Then \(f(r+1) = \frac{1}{2(r+1)^2 - 2(r+1) + 1} = \frac{1}{2r^2 + 2r + 1}\).
\(\sum_{r=1}^n \frac{4r}{4r^4 + 1} = \sum_{r=1}^n (f(r) - f(r+1)) = f(1) - f(n+1)\)
\(f(1) = \frac{1}{2 - 2 + 1} = 1\).
\(f(n+1) = \frac{1}{2n^2 + 2n + 1}\).
So \(\sum_{r=1}^n \frac{4r}{4r^4 + 1} = 1 - \frac{1}{2n^2 + 2n + 1}\).

(iv) The sum to infinity is:
\(\lim_{n \to \infty} \left( 1 - \frac{1}{2n^2 + 2n + 1} \right) = 1\).

(i) M1: Expands the RHS. A1: Correctly simplifies to show equality.
(ii) M1: Sets up the partial fraction equation. M1: Solves for coefficients \(A\) and \(B\). A1: Correct partial fractions.
(iii) M1: Uses the method of differences. M1: Recognizes the cancellation of terms. A1: Correctly simplifies the remaining terms. A0.7: Correct algebraic form \(1 - \frac{1}{2n^2 + 2n + 1}\).
(iv) M1: Applies the limit as \(n \to \infty\). A1: Correct sum to infinity of 1.

(i) Let \(P(n)\) be the statement: \(\sum_{r=1}^n r \cdot 2^{r-1} = (n-1)2^n + 1\).
Base case: for \(n=1\):
LHS \(= 1 \cdot 2^0 = 1\).
RHS \(= (1-1)2^1 + 1 = 1\).
Since LHS \(=\) RHS, \(P(1)\) is true.

Inductive step: Assume \(P(k)\) is true for some positive integer \(k\), so \(\sum_{r=1}^k r \cdot 2^{r-1} = (k-1)2^k + 1\).
For \(n=k+1\):
\(\sum_{r=1}^{k+1} r \cdot 2^{r-1} = \sum_{r=1}^k r \cdot 2^{r-1} + (k+1)2^k\)
\(= (k-1)2^k + 1 + (k+1)2^k\)
\(= ((k-1) + (k+1))2^k + 1\)
\(= (2k)2^k + 1 = k 2^{k+1} + 1 = ((k+1)-1)2^{k+1} + 1\).
This is of the same form as the formula with \(n\) replaced by \(k+1\).
Thus, if \(P(k)\) is true, then \(P(k+1)\) is true. Since \(P(1)\) is true, by mathematical induction, the statement is true for all positive integers \(n\).

(ii) Let \(Q(n)\) be the statement that \(3^{2n} + 7\) is divisible by \(8\).
Base case: for \(n=1\):
\(3^2 + 7 = 16 = 2 \times 8\), which is divisible by 8. So \(Q(1)\) is true.

Inductive step: Assume \(Q(k)\) is true for some positive integer \(k\), so \(3^{2k} + 7 = 8m\) for some integer \(m\).
For \(n=k+1\):
\(3^{2(k+1)} + 7 = 3^{2k+2} + 7 = 9(3^{2k}) + 7\)
Using the inductive hypothesis, \(3^{2k} = 8m - 7\):
\(9(8m - 7) + 7 = 72m - 63 + 7 = 72m - 56 = 8(9m - 7)\).
Since \(9m - 7\) is an integer, \(3^{2(k+1)} + 7\) is divisible by 8.
Thus, if \(Q(k)\) is true, then \(Q(k+1)\) is true. Since \(Q(1)\) is true, by mathematical induction, the statement is true for all positive integers \(n\).

(i) B1: Verifies the base case \(n=1\). M1: States the inductive hypothesis clearly. M1: Adds the \((k+1)\)-th term to the sum. M1: Factorizes and simplifies. A1: Reaches the correct form for \(n=k+1\). B0.7: Correct concluding logic statement.
(ii) B1: Verifies the base case \(n=1\). M1: States the inductive hypothesis clearly. M1: Expresses the expression for \(n=k+1\) using the hypothesis. A1: Obtains a form clearly divisible by 8. B1: Correct concluding logic statement.

(a) To find the eigenvalues, solve the characteristic equation \(\det(A - \lambda I) = 0\): \(\begin{vmatrix} 3-\lambda & -1 & 1 \\ -1 & 5-\lambda & -1 \\ 1 & -1 & 3-\lambda \end{vmatrix} = 0\). Expanding this determinant gives \((3-\lambda)[(5-\lambda)(3-\lambda) - 1] + 1[-(3-\lambda) + 1] + 1[1 - (5-\lambda)] = 0\), which simplifies to \(\lambda^3 - 11\lambda^2 + 36\lambda - 36 = 0\). Factoring the cubic equation yields \((\lambda-2)(\lambda-3)(\lambda-6) = 0\). Thus, the eigenvalues are \(\lambda = 2, 3, 6\). (b) For \(\lambda = 2\): \((A-2I)\mathbf{v} = \mathbf{0} \implies \begin{pmatrix} 1 & -1 & 1 \\ -1 & 3 & -1 \\ 1 & -1 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\). This yields \(y=0\) and \(x+z=0\), so an eigenvector is \(\begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}\). For \(\lambda = 3\): \((A-3I)\mathbf{v} = \mathbf{0} \implies \begin{pmatrix} 0 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 0 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\). This yields \(x=y=z\), so an eigenvector is \(\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\). For \(\lambda = 6\): \((A-6I)\mathbf{v} = \mathbf{0} \implies \begin{pmatrix} -3 & -1 & 1 \\ -1 & -1 & -1 \\ 1 & -1 & -3 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\). This yields \(x=z\) and \(y=-2x\), so an eigenvector is \(\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}\). (c) The matrix \(P\) is formed by the eigenvectors as columns, and \(D\) is the diagonal matrix of the corresponding eigenvalues: \(P = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & -2 \\ -1 & 1 & 1 \end{pmatrix}\) and \(D = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 6 \end{pmatrix}\).

(a) M1: Setting up characteristic equation and expanding determinant. A1: Obtaining the correct cubic equation. A1: Finding correct eigenvalues 2, 3, 6. (b) M1: Setting up the homogeneous system for one eigenvalue. A2: Finding three correct eigenvectors (1 mark for any two, 2 marks for all three). (c) M1: Stating the relationship between diagonalization and powers. A1.375: Writing down correct P and D matrices consistent with eigenvalues and eigenvectors.

The auxiliary equation is \(2m^2 + 5m + 2 = 0\), which factors to \((2m+1)(m+2) = 0\), giving roots \(m = -2, -\frac{1}{2}\). Thus, the complementary function (CF) is \(y_{CF} = Ae^{-2x} + Be^{-x/2}\). For the particular integral (PI), try \(y_{PI} = p \cos x + q \sin x\). Differentiating twice yields \(\frac{\mathrm{d}y_{PI}}{\mathrm{d}x} = -p \sin x + q \cos x\) and \(\frac{\mathrm{d}^2y_{PI}}{\mathrm{d}x^2} = -p \cos x - q \sin x\). Substituting these into the differential equation: \(2(-p \cos x - q \sin x) + 5(-p \sin x + q \cos x) + 2(p \cos x + q \sin x) = 10 \cos x\). Comparing coefficients: for \(\cos x\): \(-2p + 5q + 2p = 10 \implies 5q = 10 \implies q = 2\). For \(\sin x\): \(-2q - 5p + 2q = 0 \implies -5p = 0 \implies p = 0\). So \(y_{PI} = 2\sin x\). The general solution is \(y = Ae^{-2x} + Be^{-x/2} + 2\sin x\). Applying the initial condition \(y=3\) at \(x=0\) gives \(A + B = 3\). Finding the derivative: \(\frac{\mathrm{d}y}{\mathrm{d}x} = -2Ae^{-2x} - \frac{1}{2}Be^{-x/2} + 2\cos x\). Applying \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) at \(x=0\) gives \(-2A - \frac{1}{2}B + 2 = 0 \implies 4A + B = 4\). Subtracting the two equations: \(3A = 1 \implies A = \frac{1}{3}\). Substituting back: \(B = \frac{8}{3}\). Therefore, the particular solution is \(y = \frac{1}{3}e^{-2x} + \frac{8}{3}e^{-x/2} + 2\sin x\).

M1: Solve the auxiliary equation. A1: Correct complementary function. M1: Define PI form and substitute into DE. A1: Correct values for p and q. A1: Correct general solution. M1: Use initial conditions to set up equations for A and B. A2: Solve equations to get correct values of A and B (1 mark each). A0.375: Correct final particular solution.

(a) Differentiating \(y = \mathrm{e}^{\arcsin x}\) with respect to \(x\) gives \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{e}^{\arcsin x}}{\sqrt{1-x^2}} = \frac{y}{\sqrt{1-x^2}}\). Squaring both sides and rearranging: \((1-x^2)\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = y^2\). Differentiating both sides implicitly with respect to \(x\): \(-2x\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 + (1-x^2) \cdot 2\frac{\mathrm{d}y}{\mathrm{d}x}\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2y\frac{\mathrm{d}y}{\mathrm{d}x}\). Since \(\frac{\mathrm{d}y}{\mathrm{d}x} \neq 0\), dividing by \(2\frac{\mathrm{d}y}{\mathrm{d}x}\) yields \((1-x^2)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - x\frac{\mathrm{d}y}{\mathrm{d}x} - y = 0\). (b) Evaluating at \(x=0\): \(y(0) = 1\). From the first derivative, \(y'(0) = \frac{1}{\sqrt{1-0}} = 1\). From the differential equation: \(1 \cdot y''(0) - 0 - y(0) = 0 \implies y''(0) = 1\). Differentiating the differential relation once more with respect to \(x\): \((1-x^2)y''' - 2x y'' - y' - x y'' - y' = 0 \implies (1-x^2)y''' - 3x y'' - 2y' = 0\). Evaluating this at \(x=0\) gives \(1 \cdot y'''(0) - 0 - 2y'(0) = 0 \implies y'''(0) = 2(1) = 2\). The Maclaurin series is \(y(x) = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \dots\) Substituting values: \(y(x) = 1 + x + \frac{1}{2}x^2 + \frac{1}{3}x^3 + \dots\).

(a) M1: Differentiate y to obtain y'. M1: Square both sides and perform implicit differentiation. A1: Simplify and divide by y' to arrive at the given relation. (b) B1: State values of y(0) and y'(0). M1: Use relation to find y''(0). M1: Differentiate relation to find expression for y'''. A1: Find y'''(0) = 2. A1.375: Correct Maclaurin series up to x^3 term.

Let \(x = \sinh u\), then \(\mathrm{d}x = \cosh u \mathrm{d}u\). The limits of integration change as follows: when \(x = 0\), \(u = 0\); when \(x = 1\), \(u = \sinh^{-1}(1) = \ln(1+\sqrt{1^2+1}) = \ln(1+\sqrt{2})\). Substituting these into the integral gives \(\int_0^{\ln(1+\sqrt{2})} \frac{\sinh^2 u}{\sqrt{\sinh^2 u + 1}} \cosh u \mathrm{d}u\). Since \(\sqrt{\sinh^2 u + 1} = \cosh u\), the integral simplifies to \(\int_0^{\ln(1+\sqrt{2})} \sinh^2 u \mathrm{d}u\). Using the identity \(\sinh^2 u = \frac{1}{2}(\cosh 2u - 1)\), we integrate: \(\int_0^{\ln(1+\sqrt{2})} \frac{1}{2}(\cosh 2u - 1) \mathrm{d}u = \left[ \frac{1}{4}\sinh 2u - \frac{1}{2}u \right]_0^{\ln(1+\sqrt{2})}\). For the upper limit \(u = \sinh^{-1}(1)\), we have \(\sinh 2u = 2\sinh u \cosh u = 2(1)\sqrt{1+\sinh^2 u} = 2\sqrt{2}\). The value of the integral is \(\frac{1}{4}(2\sqrt{2}) - \frac{1}{2}\ln(1+\sqrt{2}) - 0 = \frac{1}{2}\sqrt{2} - \frac{1}{2}\ln(1+\sqrt{2})\). Thus \(a = \frac{1}{2}\) and \(b = -\frac{1}{2}\).

M1: Set up correct substitution and find dx. B1: Change limits correctly. M1: Use hyperbolic identity to simplify the integrand. M1: Use double angle identity for sinh^2(u). A1: Correctly integrate to get expression. M1: Substitute upper limit using double-angle formula for sinh(2u). A2.375: Final exact answer in correct form (1 mark for a, 1.375 marks for b).

(a) Express \(\cosh x\) and \(\sinh x\) in exponential terms: \(\cosh x = \frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\) and \(\sinh x = \frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\). Substitute these into the given equation: \(4\left(\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right) + \frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2} = 4\). Simplify to: \(2\mathrm{e}^x + 2\mathrm{e}^{-x} + \frac{1}{2}\mathrm{e}^x - \frac{1}{2}\mathrm{e}^{-x} = 4 \implies \frac{5}{2}\mathrm{e}^x + \frac{3}{2}\mathrm{e}^{-x} = 4\). Multiply by 2: \(5\mathrm{e}^x + 3\mathrm{e}^{-x} = 8\). Multiply by \(\mathrm{e}^x\): \(5\mathrm{e}^{2x} - 8\mathrm{e}^x + 3 = 0\). (b) Factor the quadratic equation in \(\mathrm{e}^x\): \((5\mathrm{e}^x - 3)(\mathrm{e}^x - 1) = 0\). This gives \(\mathrm{e}^x = \frac{3}{5}\) or \(\mathrm{e}^x = 1\). Taking natural logarithms: \(x = \ln\left(\frac{3}{5}\right)\) or \(x = \ln(1) = 0\). Both are valid real solutions.

(a) M1: Substitute exponential definitions of cosh and sinh. A1: Correctly simplify to a linear relation in e^x and e^{-x}. A1: Correctly obtain the quadratic equation. (b) M1: Factorize the quadratic equation in e^x. A1: Obtain correct values for e^x. A2.375: Obtain correct exact solutions x = 0 and x = ln(3/5) (1.1875 marks each).

(a) Apply integration by parts to \(I_n = \int_0^1 x^n e^{2x} \mathrm{d}x\). Let \(u = x^n \implies \mathrm{d}u = n x^{n-1} \mathrm{d}x\), and let \(\mathrm{d}v = e^{2x} \mathrm{d}x \implies v = \frac{1}{2}e^{2x}\). Then \(I_n = \left[ \frac{1}{2}x^n e^{2x} \right]_0^1 - \int_0^1 \frac{1}{2}n x^{n-1} e^{2x} \mathrm{d}x\). Evaluating the boundary term: \(\left[ \frac{1}{2}x^n e^{2x} \right]_0^1 = \frac{1}{2}e^2\) (since \(n \geq 1\), the term is 0 at \(x=0\)). So \(I_n = \frac{1}{2}e^2 - \frac{n}{2}I_{n-1}\). Multiplying by 2: \(2I_n = e^2 - n I_{n-1}\). (b) First find \(I_0 = \int_0^1 e^{2x} \mathrm{d}x = \left[ \frac{1}{2}e^{2x} \right]_0^1 = \frac{1}{2}e^2 - \frac{1}{2}\). Using the recurrence relation: For \(n = 1\): \(2I_1 = e^2 - I_0 = e^2 - (\frac{1}{2}e^2 - \frac{1}{2}) = \frac{1}{2}e^2 + \frac{1}{2} \implies I_1 = \frac{1}{4}e^2 + \frac{1}{4}\). For \(n = 2\): \(2I_2 = e^2 - 2I_1 = e^2 - 2(\frac{1}{4}e^2 + \frac{1}{4}) = \frac{1}{2}e^2 - \frac{1}{2} \implies I_2 = \frac{1}{4}e^2 - \frac{1}{4}\). For \(n = 3\): \(2I_3 = e^2 - 3I_2 = e^2 - 3(\frac{1}{4}e^2 - \frac{1}{4}) = \frac{1}{4}e^2 + \frac{3}{4} \implies I_3 = \frac{1}{8}e^2 + \frac{3}{8}\).

(a) M1: Use integration by parts. A1: Correct boundary term and integral term. A1: Correct recurrence relation shown. (b) B1: Correctly evaluate I_0. M1: Use recurrence relation successively to find I_1, I_2, I_3. A2.375: Correct exact value of I_3 (with intermediate values correct).

Divide both sides of the differential equation by \(x\) to write it in standard form: \(\frac{\mathrm{d}y}{\mathrm{d}x} + \left(1 + \frac{1}{x}\right)y = x\mathrm{e}^{-x}\). The integrating factor is \(I(x) = \exp\left(\int \left(1 + \frac{1}{x}\right) \mathrm{d}x\right) = \exp(x + \ln x) = x\mathrm{e}^x\). Multiplying the standard form equation by the integrating factor gives: \(\frac{\mathrm{d}}{\mathrm{d}x}\left( y \cdot x\mathrm{e}^x \right) = (x\mathrm{e}^{-x})(x\mathrm{e}^x) = x^2\). Integrating both sides with respect to \(x\): \(y \cdot x\mathrm{e}^x = \int x^2 \mathrm{d}x = \frac{1}{3}x^3 + C\). Substitute the initial condition \(x=1, y=2\mathrm{e}^{-1}\): \(2\mathrm{e}^{-1} \cdot (1)\mathrm{e}^1 = \frac{1}{3}(1)^3 + C \implies 2 = \frac{1}{3} + C \implies C = \frac{5}{3}\). Thus, \(y \cdot x\mathrm{e}^x = \frac{1}{3}x^3 + \frac{5}{3}\). Solving for \(y\) yields \(y = \frac{x^3 + 5}{3x\mathrm{e}^x}\).

M1: Put differential equation in standard form. M1: Determine the correct integrating factor. A1: Correctly write the integrated LHS. M1: Integrate RHS. A1: Correct general solution with constant C. M1: Apply initial conditions to find C. A2.375: Correct particular solution in explicit form.

(a) Differentiating \(y\) with respect to \(x\): \(\frac{\mathrm{d}y}{\mathrm{d}x} = 6\cosh x \sinh x - 4\cosh x = 2\cosh x(3\sinh x - 2)\). To find the stationary point, set \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\). Since \(\cosh x \geq 1 > 0\) for all real \(x\), we must have \(3\sinh x - 2 = 0 \implies \sinh x = \frac{2}{3}\). Solving for \(x\): \(x = \sinh^{-1}\left(\frac{2}{3}\right) = \ln\left(\frac{2}{3} + \sqrt{\frac{4}{9} + 1}\right) = \ln\left(\frac{2 + \sqrt{13}}{3}\right)\). Substitute \(\sinh x = \frac{2}{3}\) back into the curve equation to find \(y\). Since \(\cosh^2 x = 1 + \sinh^2 x = 1 + \frac{4}{9} = \frac{13}{9}\): \(y = 3\left(\frac{13}{9}\right) - 4\left(\frac{2}{3}\right) = \frac{13}{3} - \frac{8}{3} = \frac{5}{3}\). So the stationary point is \(\left( \ln\left(\frac{2+\sqrt{13}}{3}\right), \frac{5}{3} \right)\). (b) Find the second derivative: \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 6\cosh^2 x + 6\sinh^2 x - 4\sinh x\). Evaluate at \(\sinh x = \frac{2}{3}\): \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 6\left(\frac{13}{9}\right) + 6\left(\frac{4}{9}\right) - 4\left(\frac{2}{3}\right) = \frac{26}{3} + \frac{8}{3} - \frac{8}{3} = \frac{26}{3}\). Since \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} > 0\), the stationary point is a local minimum.

(a) M1: Differentiate y using chain rule. A1: Set derivative to 0 and find sinh x = 2/3. M1: Find exact x-coordinate in logarithmic form. A1: Find exact y-coordinate. (b) M1: Find second derivative. A1: Substitute sinh x to determine sign. A2.375: State correct conclusion (minimum) with detailed reasoning.

Let \(x\) be the distance traveled by the particle. The equation of motion is:
\[ m v \frac{dv}{dx} = -F \]

Substituting \(m = 0.5\) and \(F = 2(v^2 + 4)\):
\[ 0.5 v \frac{dv}{dx} = -2(v^2 + 4) \]
\[ v \frac{dv}{dx} = -4(v^2 + 4) \]

Separating variables:
\[ \frac{v}{v^2 + 4} dv = -4 dx \]

Integrating both sides:
\[ \int \frac{v}{v^2 + 4} dv = \int -4 dx \]
\[ \frac{1}{2} \ln(v^2 + 4) = -4x + C \]

Using the initial condition \(v = 4\) when \(x = 0\):
\[ \frac{1}{2} \ln(16 + 4) = C \implies C = \frac{1}{2} \ln(20) \]

Thus, the equation relating \(v\) and \(x\) is:
\[ \frac{1}{2} \ln(v^2 + 4) = -4x + \frac{1}{2} \ln(20) \]
\[ 4x = \frac{1}{2} \ln(20) - \frac{1}{2} \ln(v^2 + 4) = \frac{1}{2} \ln\left(\frac{20}{v^2 + 4}\right) \]

Substituting \(v = 2\):
\[ 4x = \frac{1}{2} \ln\left(\frac{20}{2^2 + 4}\right) = \frac{1}{2} \ln\left(\frac{20}{8}\right) = \frac{1}{2} \ln(2.5) \]
\[ x = \frac{1}{8} \ln(2.5) \approx 0.1146 \approx 0.116\text{ m} \]

M1: Formulates the equation of motion using \(v \frac{dv}{dx}\).
A1: Obtains \(v \frac{dv}{dx} = -4(v^2 + 4)\) or equivalent.
M1: Separates variables and attempts integration to obtain logarithmic form.
A1: Integrates correctly to get \(\frac{1}{2} \ln(v^2 + 4) = -4x + C\).
M1: Uses the initial conditions to find \(C\).
M1: Substitutes \(v = 2\) to find \(x\).
A1.14: Obtains the correct distance in exact form \(\frac{1}{8} \ln(2.5)\) or decimal form \(0.116\text{ m}\) (to 3 s.f.).

Let \(e\) be the maximum extension of the string. At the maximum extension, the velocity of the particle is \(0\).

Let the reference level for gravitational potential energy (GPE) be the lowest point of the motion, which is at a distance of \(l + e\) below \(O\), where \(l = 1.0\text{ m}\) is the natural length of the string.

At the release point (at \(O\)):
- Kinetic Energy (KE) = \(0\)
- Elastic Potential Energy (EPE) = \(0\)
- GPE = \(m g (l + e) = 0.4 \times 10 \times (1.0 + e) = 4(1 + e)\)

At the lowest point (at maximum extension \(e\)):
- KE = \(0\)
- GPE = \(0\)
- EPE = \(\frac{\lambda e^2}{2l} = \frac{6 e^2}{2 \times 1.0} = 3e^2\)

By conservation of mechanical energy:
\[ 4(1 + e) = 3e^2 \]
\[ 3e^2 - 4e - 4 = 0 \]

Solving the quadratic equation:
\[ (3e + 2)(e - 2) = 0 \]

Since extension \(e\) must be positive:
\[ e = 2\text{ m} \]

M1: Expresses the loss in gravitational potential energy as \(mg(l+e)\).
M1: Expresses the gain in elastic potential energy as \(\frac{\lambda e^2}{2l}\).
M1: Sets up the equation using conservation of mechanical energy.
A1: Derives the correct quadratic equation in terms of \(e\): \(3e^2 - 4e - 4 = 0\).
M1: Attempts to solve the quadratic equation.
A1.14: Obtains the correct positive root \(e = 2\text{ m}\) with appropriate units.

The equation of the projectile trajectory is given by:
\[ y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta} = x \tan \theta - \frac{g x^2 (1 + \tan^2 \theta)}{2 u^2} \]

Let \(T = \tan \theta\) and \(K = \frac{g(1+T^2)}{2u^2} = \frac{5(1+T^2)}{u^2}\). The trajectory becomes:
\[ y = x T - K x^2 \]

We are given two points on the trajectory: \((8, 4)\) and \((16, 6)\).

For \((8, 4)\):
\[ 4 = 8 T - 64 K \implies 1 = 2 T - 16 K \quad \text{--- (1)} \]

For \((16, 6)\):
\[ 6 = 16 T - 256 K \implies 3 = 8 T - 128 K \quad \text{--- (2)} \]

Multiply Equation (1) by 8:
\[ 8 = 16 T - 128 K \quad \text{--- (3)} \]

Subtract Equation (2) from (3):
\[ 5 = 8 T \implies T = \frac{5}{8} = 0.625 \]

Now, substitute \(T = 0.625\) into Equation (1):
\[ 1 = 2(0.625) - 16 K \implies 1 = 1.25 - 16 K \implies 16 K = 0.25 \implies K = \frac{1}{64} \]

Since \(K = \frac{5(1+T^2)}{u^2}\):
\[ \frac{1}{64} = \frac{5\left(1 + (5/8)^2\right)}{u^2} \]
\[ u^2 = 320 \left(1 + \frac{25}{64}\right) = 320 \left(\frac{89}{64}\right) = 5 \times 89 = 445 \]
\[ u = \sqrt{445} \approx 21.1\text{ m s}^{-1} \]

M1: Recalls and uses the trajectory equation in terms of \(x\) and \(\tan \theta\).
M1: Sets up two simultaneous equations using coordinates \((8,4)\) and \((16,6)\).
A1: Correctly simplifies equations to a linear system in terms of \(T\) and \(K\).
M1: Eliminates one variable to solve for \(\tan \theta\).
A1: Obtains \(\tan \theta = 0.625\) (or \(5/8\)).
M1: Uses the value of \(\tan \theta\) and the second relation to solve for \(u^2\).
A1.14: Obtains the correct initial speed \(u \approx 21.1\text{ m s}^{-1}\) (or \(\sqrt{445}\)).

(i) For the upward motion, both the weight of the particle and the resistive force act vertically downwards. By Newton's second law:
\[ m a = -m g - R \]
Using \(a = v \frac{dv}{dx}\), \(m = 2\text{ kg}\), \(g = 10\text{ m s}^{-2}\), and \(R = 0.1 v^2\):
\[ 2 v \frac{dv}{dx} = -2(10) - 0.1 v^2 \]
\[ 2 v \frac{dv}{dx} = -(20 + 0.1 v^2) \]
Dividing both sides by 2:
\[ v \frac{dv}{dx} = -(10 + 0.05 v^2) \] (as required).

(ii) At the maximum height \(H\), the speed \(v = 0\). Separating variables:
\[ \frac{v}{10 + 0.05 v^2} dv = -dx \]

Integrating from \(x = 0\) (where \(v = 20\)) to \(x = H\) (where \(v = 0\)):
\[ \int_{20}^{0} \frac{v}{10 + 0.05 v^2} dv = \int_{0}^{H} -dx \]

Using the substitution \(u = 10 + 0.05 v^2\), so \(du = 0.1 v \, dv\):
\[ \left[ 10 \ln(10 + 0.05 v^2) \right]_{20}^{0} = -H \]
\[ 10 \ln(10) - 10 \ln(10 + 0.05 \times 400) = -H \]
\[ 10 \ln(10) - 10 \ln(30) = -H \]
\[ -10 \ln(3) = -H \implies H = 10 \ln(3) \approx 10.986 \approx 11.0\text{ m} \]

M1: Applies Newton's second law with correct force components for upward motion.
A1: Correctly derives the given differential equation \(v \frac{dv}{dx} = -(10 + 0.05 v^2)\).
M1: Separates variables for integration.
A1: Integrates the LHS correctly to get \(10 \ln(10 + 0.05 v^2)\) (or equivalent).
M1: Identifies the limits \(v = 20\) to \(v = 0\) and applies them.
A1.14: Obtains the final height \(H \approx 11.0\text{ m}\) (to 3 s.f.).

(i) The component of the gravitational force acting on the particle down the slope is:
\[ F_g = m g \sin 30^\circ = 0.5 m g \]

Let \(x\) be the distance of the particle from \(O\) down the slope. For \(x > a\), the string is stretched, and the tension is:
\[ T = \frac{4mg(x - a)}{a} \]

The speed of the particle is maximum when its acceleration is zero:
\[ T = F_g \implies \frac{4mg(x - a)}{a} = 0.5 m g \]
\[ 4(x - a) = 0.5 a \implies 4x - 4a = 0.5 a \implies 4x = 4.5 a \implies x = 1.125 a \] (or \(\frac{9}{8}a\)).

(ii) We use conservation of mechanical energy from the release point \(O\) to the point of maximum speed \(x = 1.125 a\).
Let the reference level for GPE be the point of maximum speed.

At \(O\):
- KE = \(0\)
- EPE = \(0\)
- GPE = \(m g x \sin 30^\circ = m g (1.125 a) (0.5) = \frac{9}{16} m g a\)

At the point of maximum speed:
- KE = \(\frac{1}{2} m v_{max}^2\)
- GPE = \(0\)
- EPE = \(\frac{\lambda (x - a)^2}{2a} = \frac{4mg (0.125 a)^2}{2a} = \frac{4mg \left(\frac{1}{8}a\right)^2}{2a} = \frac{1}{32} m g a\)

By conservation of energy:
\[ \frac{9}{16} m g a = \frac{1}{2} m v_{max}^2 + \frac{1}{32} m g a \]
\[ \frac{1}{2} m v_{max}^2 = \left(\frac{18}{32} - \frac{1}{32}\right) m g a = \frac{17}{32} m g a \]
\[ v_{max}^2 = \frac{17}{16} g a \implies v_{max} = \frac{\sqrt{17}}{4} \sqrt{g a} \]

M1: Uses the condition that acceleration is zero at maximum speed.
A1: Finds the correct distance \(x = 1.125a\) (or \(\frac{9}{8}a\)).
M1: Uses conservation of energy with correct GPE loss of \(m g x \sin 30^\circ\).
A1: Obtains the correct GPE term \(\frac{9}{16} mga\).
A1: Obtains the correct EPE term \(\frac{1}{32} mga\).
M1: Solves the energy equation for \(v_{max}\).
A1.14: Obtains \(v_{max} = \frac{\sqrt{17}}{4} \sqrt{g a}\) (or equivalent exact form).

(i) Using the equation of trajectory:
\[ y = x \tan \alpha - \frac{g x^2 (1 + \tan^2 \alpha)}{2 U^2} \]
Substitute \(x = 30\), \(y = 15\), \(U = 30\), and \(g = 10\):
\[ 15 = 30 \tan \alpha - \frac{10 \times 30^2 (1 + \tan^2 \alpha)}{2 \times 30^2} \]
\[ 15 = 30 \tan \alpha - 5(1 + \tan^2 \alpha) \]
\[ 3 = 6 \tan \alpha - (1 + \tan^2 \alpha) \]
\[ \tan^2 \alpha - 6 \tan \alpha + 4 = 0 \]

Solving for \(\tan \alpha\):
\[ \tan \alpha = \frac{6 \pm \sqrt{36 - 16}}{2} = 3 \pm \sqrt{5} \]

Thus:
\[ \tan \alpha_1 = 3 + \sqrt{5} \approx 5.236 \implies \alpha_1 \approx 79.17^\circ \approx 79.2^\circ \]
\[ \tan \alpha_2 = 3 - \sqrt{5} \approx 0.764 \implies \alpha_2 \approx 37.38^\circ \approx 37.4^\circ \]

(ii) The time of flight \(t\) to reach the horizontal distance \(x = 30\) is given by:
\[ t = \frac{x}{U \cos \alpha} = \frac{30}{30 \cos \alpha} = \sec \alpha \]

Since \(\sec^2 \alpha = 1 + \tan^2 \alpha\):
For \(\alpha_1\):
\[ \sec^2 \alpha_1 = 1 + (3 + \sqrt{5})^2 = 1 + (9 + 6\sqrt{5} + 5) = 15 + 6\sqrt{5} \approx 28.416 \]
\[ t_1 = \sqrt{15 + 6\sqrt{5}} \approx 5.331\text{ s} \]

For \(\alpha_2\):
\[ \sec^2 \alpha_2 = 1 + (3 - \sqrt{5})^2 = 1 + (9 - 6\sqrt{5} + 5) = 15 - 6\sqrt{5} \approx 1.584 \]

\[ t_2 = \sqrt{15 - 6\sqrt{5}} \approx 1.258\text{ s} \]

The difference in times of flight is:
\[ \Delta t = t_1 - t_2 = 5.331 - 1.258 \approx 4.07\text{ s} \]

M1: Substitutes the values into the trajectory equation.
A1: Derives the quadratic equation \(\tan^2 \alpha - 6 \tan \alpha + 4 = 0\).
A1: Solves the quadratic to find \(\alpha_1 \approx 79.2^\circ\) and \(\alpha_2 \approx 37.4^\circ\).
M1: Relates the time of flight to the angle of projection, e.g., \(t = \sec \alpha\).
A1: Calculates the time of flight for the first angle: \(t_1 \approx 5.33\text{ s}\).
A1: Calculates the time of flight for the second angle: \(t_2 \approx 1.26\text{ s}\).
A1.14: Obtains the correct time difference \(4.07\text{ s}\) (to 3 s.f.).

(i) The mid-point \(P\) divides the string into two halves, each of natural length \(L\) and modulus of elasticity \(4mg\).
When the particle \(P\) is at a vertical distance \(y\) below the level of \(AB\), the length of each half of the string is:
\[ AP = BP = \sqrt{L^2 + y^2} \]

The extension of each half of the string is \(\sqrt{L^2 + y^2} - L\).

The total elastic potential energy (EPE) of both halves of the string is:
\[ EPE = 2 \times \left[ \frac{\lambda (\text{extension})^2}{2L} \right] = 2 \times \left[ \frac{4mg (\sqrt{L^2+y^2}-L)^2}{2L} \right] = \frac{4mg(\sqrt{L^2+y^2}-L)^2}{L} \]

Let the horizontal line \(AB\) be the reference level for gravitational potential energy (GPE). The GPE is \(-mgy\), and the kinetic energy (KE) is \(\frac{1}{2} m v^2\).

Therefore, the total mechanical energy \(E\) of the system is:
\[ E = \frac{1}{2} m v^2 - mgy + \frac{4mg\left(\sqrt{L^2+y^2}-L\right)^2}{L} \]

(ii) Initially, the particle is released from rest at the mid-point of \(AB\), so at \(y=0\), \(v=0\).
Since \(y=0\), the extension of each half is \(L - L = 0\), meaning the initial EPE is also \(0\). Hence, the initial total energy is:
\[ E_0 = 0 \]

By conservation of mechanical energy, when the particle is at its maximum depth \(y = h\), its speed is \(v = 0\):
\[ -mgh + \frac{4mg\left(\sqrt{L^2+h^2}-L\right)^2}{L} = 0 \]
\[ \frac{4mg\left(\sqrt{L^2+h^2}-L\right)^2}{L} = mgh \]

Dividing both sides by \(mg\) and multiplying by \(L\):
\[ 4\left(\sqrt{L^2 + h^2} - L\right)^2 = h L \] (as required).

M1: Identifies that each half of the string has natural length \(L\) and modulus \(4mg\).
M1: Obtains the extension of one half of the string as \(\sqrt{L^2+y^2}-L\).
A1: Writes a correct expression for the total energy including KE, GPE, and EPE.
M1: Realizes that the initial total mechanical energy of the system is zero.
M1: Sets the velocity to zero at the maximum depth \(y=h\) in the energy conservation equation.
A1: Obtains \(-mgh + \frac{4mg\left(\sqrt{L^2+h^2}-L\right)^2}{L} = 0\).
A1.14: Correctly manipulates the equation to show \(4\left(\sqrt{L^2 + h^2} - L\right)^2 = h L\).

Let \(D = \text{Before} - \text{After}\). Under the null hypothesis, the median difference is zero: \(H_0: M_D = 0\). Under the alternative hypothesis, the median difference is positive: \(H_1: M_D > 0\) (since Before > After indicates a reduction in time). The differences are: A: 4, B: 4, C: -2, D: 7, E: -1, F: 6, G: 6, H: 7. Ranking the absolute differences: |1| (rank 1, sign -), |2| (rank 2, sign -), |4| (ranks 3, 4: average 3.5, sign +), |4| (rank 3.5, sign +), |6| (ranks 5, 6: average 5.5, sign +), |6| (rank 5.5, sign +), |7| (ranks 7, 8: average 7.5, sign +), |7| (rank 7.5, sign +). The sum of negative ranks is \(T^- = 1 + 2 = 3\). The sum of positive ranks is \(T^+ = 3.5 + 3.5 + 5.5 + 5.5 + 7.5 + 7.5 = 33\). The test statistic is \(T = \min(T^+, T^-) = 3\). For \(n = 8\) at the 5% level of significance for a one-tailed test, the critical value is 5. Since \(T = 3 \le 5\), we reject \(H_0\) and conclude that the training program significantly reduces task completion time.

M1: For calculating the correct differences. A1: For correctly assigning ranks (with average ranks for ties). M1: For calculating \(T^- = 3\) and \(T^+ = 33\). A1: For identifying the test statistic \(T = 3\). B1: For stating correct hypotheses \(H_0\) and \(H_1\). M1: For comparing the test statistic with the critical value of 5. A1: For rejecting \(H_0\) and making the correct contextual conclusion.

(a) By definition, the probability generating function of \(X\) is \(G_X(t) = \sum_{r=0}^{\infty} P(X=r) t^r = \sum_{r=0}^{\infty} \frac{2}{3} \left(\frac{t}{3}\right)^r\). This is an infinite geometric series with first term \(\frac{2}{3}\) and common ratio \(\frac{t}{3}\). Thus, \(G_X(t) = \frac{2/3}{1 - t/3} = \frac{2}{3-t}\). The PGF of \(Y = 2X + 1\) is \(G_Y(t) = E(t^Y) = E(t^{2X+1}) = t G_X(t^2) = \frac{2t}{3-t^2}\).

(b) Differentiating \(G_Y(t) = 2t(3-t^2)^{-1}\) gives: \(G_Y'(t) = 2(3-t^2)^{-1} + 2t(-1)(3-t^2)^{-2}(-2t) = \frac{2}{3-t^2} + \frac{4t^2}{(3-t^2)^2}\). Substituting \(t = 1\), \(E(Y) = G_Y'(1) = \frac{2}{2} + \frac{4}{4} = 2\). Differentiating again: \(G_Y''(t) = 4t(3-t^2)^{-2} + 8t(3-t^2)^{-2} + 16t^3(3-t^2)^{-3} = \frac{12t}{(3-t^2)^2} + \frac{16t^3}{(3-t^2)^3}\). Substituting \(t = 1\), \(G_Y''(1) = \frac{12}{4} + \frac{16}{8} = 3 + 2 = 5\). Finally, \(\text{Var}(Y) = G_Y''(1) + G_Y'(1) - [G_Y'(1)]^2 = 5 + 2 - 2^2 = 3\).

(a) M1: For attempting to find \(G_X(t)\) as a geometric series. A1: For deriving \(G_X(t) = \frac{2}{3-t}\). M1: For using \(G_Y(t) = t G_X(t^2)\) to obtain the given expression.

(b) M1: For correct first derivative of \(G_Y(t)\). A1: For finding \(E(Y) = 2\). M1: For correct second derivative of \(G_Y(t)\). A1: For finding \(G_Y''(1) = 5\). A1: For correct variance \(\text{Var}(Y) = 3\).

Let \(\mu_1\) and \(\mu_2\) be the population means of Species A and Species B. Hypotheses: \(H_0: \mu_1 = \mu_2\) vs \(H_1: \mu_1 \neq \mu_2\). The pooled estimate of the population variance is: \(s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2} = \frac{9(2.45) + 11(1.85)}{20} = \frac{22.05 + 20.35}{20} = 2.12\). The standard error of the difference is: \(SE = \sqrt{s_p^2 \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} = \sqrt{2.12 \left(\frac{1}{10} + \frac{1}{12}\right)} = \sqrt{2.12 \times \frac{11}{60}} \approx 0.6234\). The value of the test statistic is: \(t = \frac{\bar{x}_1 - \bar{x}_2}{SE} = \frac{15.4 - 13.8}{0.6234} \approx 2.567\). The number of degrees of freedom is \(\nu = 10 + 12 - 2 = 20\). For a two-tailed test at the 5% level, the critical value of \(t\) is \(t_{20}(0.025) = 2.086\). Since \(|t| = 2.567 > 2.086\), we reject the null hypothesis \(H_0\).

B1: For stating correct hypotheses \(H_0\) and \(H_1\). M1: For calculating the pooled variance \(s_p^2\). A1: For obtaining \(s_p^2 = 2.12\). M1: For calculating the standard error of difference. A1: For finding \(SE \approx 0.6234\). M1: For calculating the test statistic \(t\). A1: For obtaining \(t \approx 2.567\) (or \(2.57\)). B1: For stating the correct critical value of 2.086. A1: For rejecting \(H_0\) and stating the conclusion in context.

Hypotheses: \(H_0\): No association between preferred exercise type and age group. \(H_1\): There is an association. Calculate Row and Column Totals: Row Totals: Under 30 = 100, 30 & Over = 100. Column Totals: Cardio = 70, Strength = 70, Yoga = 60. Grand Total = 200. Expected frequencies: \(E = \frac{\text{Row Total} \times \text{Col Total}}{\text{Grand Total}}\). Under 30 expected values: Cardio = 35, Strength = 35, Yoga = 30. 30 and Over expected values: Cardio = 35, Strength = 35, Yoga = 30. Calculate \(\sum \frac{(O-E)^2}{E}\): Under 30: \(\frac{(45-35)^2}{35} + 0 + \frac{(20-30)^2}{30} = 2.857 + 0 + 3.333 = 6.19\). 30 and Over: \(\frac{(25-35)^2}{35} + 0 + \frac{(40-30)^2}{30} = 2.857 + 0 + 3.333 = 6.19\). Total \(\chi^2 = 12.38\). Degrees of freedom: \((2-1)(3-1) = 2\). At 1% significance level, critical value of \(\chi^2\) with 2 df is 9.210. Since \(\chi^2 = 12.38 > 9.210\), we reject \(H_0\) and conclude there is an association.

B1: For stating appropriate null and alternative hypotheses. M1: For calculating expected frequencies. A1: For obtaining correct expected frequencies (35, 35, 30 for both rows). M1: For using the test statistic formula. A2: For calculating the correct term values (award 1 mark for at least 3 correct terms). A1: For obtaining \(\chi^2 \approx 12.38\). B1: For identifying 2 degrees of freedom and stating the critical value 9.210. A1: For rejecting \(H_0\) and stating the conclusion in context.

(a) For \(0 \le x \le 2\), \(F(x) = \int_0^x \frac{3}{8} t^2 dt = \frac{1}{8} x^3\). For \(x < 0\), \(F(x) = 0\); for \(x > 2\), \(F(x) = 1\).

(b) Since \(0 \le x \le 2\), the range of \(Y = \frac{4}{X^2}\) is \(y \ge 1\). For \(y \ge 1\), the cumulative distribution function \(G(y)\) is: \(G(y) = P(Y \le y) = P\left(\frac{4}{X^2} \le y\right) = P\left(X^2 \ge \frac{4}{y}\right) = P\left(X \ge \frac{2}{\sqrt{y}}\right) = 1 - F\left(\frac{2}{\sqrt{y}}\right)\). Substituting the expression for \(F(x)\) gives \(G(y) = 1 - \frac{1}{8}\left(\frac{2}{\sqrt{y}}\right)^3 = 1 - y^{-3/2}\). Differentiating \(G(y)\) with respect to \(y\): \(g(y) = G'(y) = \frac{3}{2} y^{-5/2}\) for \(y \ge 1\), and \(0\) otherwise.

(c) \(E(\sqrt{Y}) = \int_1^{\infty} \sqrt{y} \cdot \frac{3}{2} y^{-5/2} dy = \int_1^{\infty} \frac{3}{2} y^{-2} dy = \left[ -\frac{3}{2y} \right]_1^{\infty} = 0 - \left(-\frac{3}{2}\right) = \frac{3}{2}\).

(a) M1: For integrating \(f(x)\). A1: For correct CDF expression with ranges.

(b) M1: For establishing the link between \(P(Y \le y)\) and \(P(X \ge \dots)\). A1: For deriving \(G(y) = 1 - y^{-3/2}\). M1: For differentiating to get the PDF. A1: For showing the given PDF with correct range \(y \ge 1\).

(c) M1: For attempting to integrate \(\sqrt{y} g(y)\) from 1 to infinity. A1: For obtaining the correct exact value of 3/2 (or 1.5).

Let \(M_P\) and \(M_Q\) represent the median yields. \(H_0: M_P = M_Q\) vs \(H_1: M_P > M_Q\). Combine and rank the observations: 9.5 (Q, rank 1), 10.8 (Q, rank 2), 11.2 (Q, rank 3), 11.5 (P, rank 4), 12.0 (Q, rank 5), 12.3 (P, rank 6), 12.9 (P, rank 7), 13.1 (Q, rank 8), 13.6 (P, rank 9), 14.8 (P, rank 10), 15.1 (P, rank 11). Sample sizes: \(n_1 = 6\) (P) and \(n_2 = 5\) (Q). Sum of ranks for the smaller sample (Type Q): \(W = R_Q = 1 + 2 + 3 + 5 + 8 = 19\). (The rank sum for P is \(R_P = 47\)). For \(n_1 = 6\) and \(n_2 = 5\) on a one-tailed test at the 5% significance level, the critical value for the rank sum of the smaller sample is 20. Since \(W = 19 \le 20\), we reject \(H_0\). (Alternative using Mann-Whitney: \(U = 19 - \frac{5 \times 6}{2} = 4\); critical value of \(U\) is 5, since \(4 \le 5\), we reject \(H_0\)). There is sufficient evidence at the 5% significance level to conclude that the median yield of Type P plants is greater than the median yield of Type Q plants.

B1: For stating correct hypotheses \(H_0\) and \(H_1\). M1: For combining and ranking all 11 observations correctly. A1: For identifying the correct ranks of Type Q: 1, 2, 3, 5, 8. M1: For calculating the sum of ranks of the smaller sample \(W = 19\) (or \(U = 4\)). B1: For stating the correct critical value of 20 for \(W\) (or 5 for \(U\)). M1: For comparing the test statistic with the critical value. A1: For rejecting \(H_0\) and drawing the correct contextual conclusion.