2023 Cambridge IAL Mathematics - Further (9231) Practice Paper with Answers

For (i): From the cubic equation, we have \(\alpha + \beta + \gamma = 4\) and \(\alpha\beta + \beta\gamma + \gamma\alpha = 2\). Using the identity \(\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\), we substitute the values to get \(4^2 - 2(2) = 16 - 4 = 12\). For (ii): Let \(y = \frac{1}{x+1}\). This gives \(x = \frac{1}{y} - 1 = \frac{1-y}{y}\). Substituting this expression into the original cubic equation: \(\left(\frac{1-y}{y}\right)^3 - 4\left(\frac{1-y}{y}\right)^2 + 2\left(\frac{1-y}{y}\right) - 3 = 0\). Multiplying the entire equation by \(y^3\) to clear the denominators yields: \((1-y)^3 - 4y(1-y)^2 + 2y^2(1-y) - 3y^3 = 0\). Expanding each term: \((1 - 3y + 3y^2 - y^3) - 4y(1 - 2y + y^2) + (2y^2 - 2y^3) - 3y^3 = 0\), which simplifies to \(1 - 3y + 3y^2 - y^3 - 4y + 8y^2 - 4y^3 + 2y^2 - 2y^3 - 3y^3 = 0\). Grouping like terms gives: \(-10y^3 + 13y^2 - 7y + 1 = 0\). Multiplying by \(-1\) to get integer coefficients with a positive leading term gives the final equation: \(10y^3 - 13y^2 + 7y - 1 = 0\).

Part (i): [M1] For stating and using the sum and sum of product of roots: sum = 4, product sum = 2. [A1] For correctly calculating the sum of squares as 12. Part (ii): [M1] For substituting x = (1-y)/y into the cubic equation. [A1] For the correct expansion of (1-y)^3. [A1] For the correct expansion of -4y(1-y)^2. [M1] For collecting and grouping coefficients of y^3, y^2, y, and the constant term. [A1] For obtaining the correct final cubic equation 10y^3 - 13y^2 + 7y - 1 = 0.

For (i): By algebraic division, we rewrite the equation as: \(y = \frac{x^2 - x - 2x + 6}{x - 1} = x - 2 + \frac{4}{x - 1}\). Thus, the vertical asymptote is \(x = 1\) and the oblique asymptote is \(y = x - 2\). For (ii): We find the derivative: \(\frac{dy}{dx} = 1 - \frac{4}{(x-1)^2}\). Setting the derivative to 0 for stationary points: \(1 - \frac{4}{(x-1)^2} = 0 \implies (x-1)^2 = 4 \implies x - 1 = \pm 2\). This gives \(x = 3\) or \(x = -1\). Substituting these back into the original curve equation: when \(x = 3\), \(y = \frac{3^2 - 3(3) + 6}{3 - 1} = 3\). When \(x = -1\), \(y = \frac{(-1)^2 - 3(-1) + 6}{-1 - 1} = -5\). The stationary points are \((3, 3)\) and \((-1, -5)\). For (iii): The curve does not cross the x-axis because the numerator has no real roots (discriminant is \(9 - 24 < 0\)). The y-intercept is at \((0, -6)\). The sketch consists of two branches, one in the region \(x > 1\) with a local minimum at \((3, 3)\), and the other in the region \(x < 1\) with a local maximum at \((-1, -5)\) and crossing the y-axis at \((0, -6)\).

Part (i): [M1] For performing algebraic division to express y in the form x + a + b/(x-1). [A1] For identifying the vertical asymptote x = 1. [A1] For identifying the oblique asymptote y = x - 2. Part (ii): [M1] For differentiating the expression and setting the derivative to zero. [A1] For solving the resulting equation to find x = 3 and x = -1. [A1] For finding the corresponding y-coordinates, giving stationary points (3,3) and (-1,-5). Part (iii): [B1] For sketching two correct branches of the curve asymptotic to x = 1 and y = x - 2. [B1] For showing the coordinates of the stationary points and the y-intercept (0,-6) on the sketch.

For (i): Combining the terms on the right-hand side over a common denominator: \(\frac{3r+1}{r(r+1)} - \frac{3r+4}{(r+1)(r+2)} = \frac{(3r+1)(r+2) - r(3r+4)}{r(r+1)(r+2)}\). Expanding the numerator: \((3r^2 + 7r + 2) - (3r^2 + 4r) = 3r + 2\). Thus, LHS = RHS. For (ii): We write the sum as a telescoping series: \(\sum_{r=1}^{n} \left( f(r) - f(r+1) \right)\) where \(f(r) = \frac{3r+1}{r(r+1)}\). Writing out the terms: \(\left( f(1) - f(2) \right) + \left( f(2) - f(3) \right) + \dots + \left( f(n) - f(n+1) \right) = f(1) - f(n+1)\). Evaluating these expressions: \(f(1) = \frac{3(1)+1}{1(1+1)} = 2\), and \(f(n+1) = \frac{3(n+1)+1}{(n+1)(n+2)} = \frac{3n+4}{(n+1)(n+2)}\). Thus, the sum is \(2 - \frac{3n+4}{(n+1)(n+2)}\). For (iii): Taking the limit as \(n \to \infty\), \(\lim_{n\to\infty} \frac{3n+4}{(n+1)(n+2)} = 0\). Therefore, the sum to infinity is \(2\).

Part (i): [M1] For attempting to combine the terms on the right-hand side over a common denominator. [A1] For showing the algebraic steps clearly to arrive at the left-hand side. Part (ii): [M1] For recognizing the telescoping structure of the sum. [A1] For showing terms cancelling out. [A1] For correctly finding f(1) = 2. [A1] For expressing the final sum as 2 - (3n+4)/((n+1)(n+2)). Part (iii): [B1] For taking the limit as n approaches infinity to obtain 2.

For (i): The matrix is singular when its determinant is 0. \(\det(\mathbf{A}) = k(1(2) - (-1)(1)) - 2(0(2) - (-1)(3)) + 1(0(1) - 1(3)) = 3k - 2(3) + 1(-3) = 3k - 9\). Setting \(\det(\mathbf{A}) = 0 \implies 3k - 9 = 0 \implies k = 3\). For (ii): When \(k = 4\), \(\det(\mathbf{A}) = 3(4) - 9 = 3\). We find the matrix of cofactors: \(C_{11} = 3\), \(C_{12} = -3\), \(C_{13} = -3\), \(C_{21} = -3\), \(C_{22} = 5\), \(C_{23} = 2\), \(C_{31} = -3\), \(C_{32} = 4\), \(C_{33} = 4\). Thus, the cofactor matrix \(\mathbf{C} = \begin{pmatrix} 3 & -3 & -3 \\ -3 & 5 & 2 \\ -3 & 4 & 4 \end{pmatrix}\). Transposing \(\mathbf{C}\) gives \(\mathbf{C}^T = \begin{pmatrix} 3 & -3 & -3 \\ -3 & 5 & 4 \\ -3 & 2 & 4 \end{pmatrix}\). Hence, \(\mathbf{A}^{-1} = \frac{1}{3} \begin{pmatrix} 3 & -3 & -3 \\ -3 & 5 & 4 \\ -3 & 2 & 4 \end{pmatrix}\). For (iii): The system can be written as \(\mathbf{A}\mathbf{x} = \mathbf{b}\), where \(\mathbf{b} = \begin{pmatrix} 1 \\ -2 \\ 5 \end{pmatrix}\). Thus, \(\mathbf{x} = \mathbf{A}^{-1} \mathbf{b} = \frac{1}{3} \begin{pmatrix} 3 & -3 & -3 \\ -3 & 5 & 4 \\ -3 & 2 & 4 \end{pmatrix} \begin{pmatrix} 1 \\ -2 \\ 5 \end{pmatrix} = \frac{1}{3} \begin{pmatrix} 3(1) - 3(-2) - 3(5) \\ -3(1) + 5(-2) + 4(5) \\ -3(1) + 2(-2) + 4(5) \end{pmatrix} = \frac{1}{3} \begin{pmatrix} -6 \\ 7 \\ 13 \end{pmatrix} = \begin{pmatrix} -2 \\ 7/3 \\ 13/3 \end{pmatrix}\). So, \(x = -2\), \(y = 7/3\), and \(z = 13/3\).

Part (i): [M1] For finding the determinant of A in terms of k. [A1] For setting determinant to zero and solving to get k = 3. Part (ii): [M1] For calculating the cofactors of the matrix. [A1] For the correct cofactor matrix or transpose. [A1] For dividing by the determinant to obtain the correct inverse matrix. Part (iii): [M1] For attempting to solve the system using matrix multiplication with the inverse. [A1] For obtaining x = -2. [A1] For obtaining y = 7/3 and z = 13/3.

For (i): The curve is a dimpled limaçon symmetric about the initial line, with \(r = 5\) when \(o = 0\), \(r = 3\) when \(o = \pm \pi/2\), and \(r = 1\) when \(o = \pi\). For (ii): The area \(A\) is given by \(\frac{1}{2} \int_{0}^{2\pi} r^2 d\theta = \frac{1}{2} \int_{0}^{2\pi} (3 + 2\cos\theta)^2 d\theta\). Expanding: \((3 + 2\cos\theta)^2 = 9 + 12\cos\theta + 4\cos^2\theta = 9 + 12\cos\theta + 2(1 + \cos 2\theta) = 11 + 12\cos\theta + 2\cos 2\theta\). Integrating: \(A = \frac{1}{2} \left[ 11\theta + 12\sin\theta + \sin 2\theta \right]_{0}^{2\pi} = \frac{1}{2} (22\pi) = 11\pi\). For (iii): At \(\theta = \frac{\pi}{3}\), \(r = 3 + 2\cos(\pi/3) = 4\). The Cartesian coordinates of the point are \(x = r\cos\theta = 4(1/2) = 2\) and \(y = r\sin\theta = 4(\sqrt{3}/2) = 2\sqrt{3}\). We have \(\frac{dx}{d\theta} = \frac{dr}{d\theta}\cos\theta - r\sin\theta = -2\sin\theta\cos\theta - (3+2\cos\theta)\sin\theta\). At \(\theta = \pi/3\), \(\frac{dx}{d\theta} = -2(\sqrt{3}/2)(1/2) - 4(\sqrt{3}/2) = -\frac{5\sqrt{3}}{2}\). Similarly, \(\frac{dy}{d\theta} = \frac{dr}{d\theta}\sin\theta + r\cos\theta = -2\sin^2\theta + (3+2\cos\theta)\cos\theta\). At \(\theta = \pi/3\), \(\frac{dy}{d\theta} = -2(3/4) + 4(1/2) = 1/2\). Thus, the gradient of the tangent is \(\frac{dy}{dx} = \frac{1/2}{-5\sqrt{3}/2} = -\frac{1}{5\sqrt{3}}\). The equation of the tangent is \(y - 2\sqrt{3} = -\frac{1}{5\sqrt{3}}(x - 2) \implies 5\sqrt{3}y - 30 = -x + 2 \implies x + 5\sqrt{3}y = 32\).

Part (i): [B1] For a correct closed loop symmetric about the initial line, not passing through the pole. [B1] For indicating the key intercepts (5, 3, 1). Part (ii): [M1] For using the polar area formula with correct limits. [M1] For using the double angle formula to rewrite cos^2(theta). [A1] For correct integration. [A1] For getting the area 11*pi. Part (iii): [M1] For finding the Cartesian coordinates at theta = pi/3. [M1] For differentiating x and y with respect to theta. [A1] For finding dy/dx = -1/(5*sqrt(3)). [A1] For writing the Cartesian equation of the tangent line correctly as x + 5*sqrt(3)*y = 32.

For (i): The direction vectors of \(l_1\) and \(l_2\) are \(\mathbf{d}_1 = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}\) and \(\mathbf{d}_2 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\). Since \(\mathbf{d}_1\) is not a scalar multiple of \(\mathbf{d}_2\), the lines are not parallel. To show they do not intersect, we set up simultaneous equations: \(1 + 2\lambda = 3 + \mu \implies 2\lambda - \mu = 2\) (1) and \(2 - \lambda = -1 + \mu \implies \lambda + \mu = 3\) (2). Adding (1) and (2) gives \(3\lambda = 5 \implies \lambda = 5/3\) and \(\mu = 4/3\). Checking these values in the z-coordinates: for \(l_1\), \(z = -1 + 3(5/3) = 4\), and for \(l_2\), \(z = 2 + 1(4/3) = 10/3\). Since \(4 \neq 10/3\), the lines do not intersect. Since they are neither parallel nor intersecting, they are skew. For (ii): The common normal vector to both lines is \(\mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2 = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} \times \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} -1(1) - 3(1) \\ 3(1) - 2(1) \\ 2(1) - (-1)(1) \end{pmatrix} = \begin{pmatrix} -4 \\ 1 \\ 3 \end{pmatrix}\). A vector connecting a point on \(l_1\) to a point on \(l_2\) is \(\mathbf{AB} = \begin{pmatrix} 3 - 1 \\ -1 - 2 \\ 2 - (-1) \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \\ 3 \end{pmatrix}\). The shortest distance \(d\) is the projection of \(\mathbf{AB}\) onto \(\mathbf{n}\): \(d = \frac{|\mathbf{AB} \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{|2(-4) + (-3)(1) + 3(3)|}{\sqrt{(-4)^2 + 1^2 + 3^2}} = \frac{|-8 - 3 + 9|}{\sqrt{26}} = \frac{2}{\sqrt{26}}\).

Part (i): [B1] For showing that direction vectors are not parallel. [M1] For setting up equations for intersection. [A1] For solving parameters from two equations. [A1] For verifying in the third and concluding that the lines are skew. Part (ii): [M1] For finding the cross product of the direction vectors. [A1] For the correct normal vector. [M1] For finding the displacement vector AB. [M1] For using the shortest distance formula (dot product of AB and normal, divided by magnitude of normal). [A1] For the correct shortest distance of 2/sqrt(26) (or equivalent).

For (i): Let \(P(n)\) be the statement. Base Case: For \(n = 1\), \(\mathbf{M}^1 = \begin{pmatrix} 2(1)+1 & -4(1) \\ 1 & 1-2(1) \end{pmatrix} = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}\), which is true. Inductive Step: Assume \(P(k)\) is true for some positive integer \(k\), i.e., \(\mathbf{M}^k = \begin{pmatrix} 2k + 1 & -4k \\ k & 1 - 2k \end{pmatrix}\). We need to show that \(P(k+1)\) is true. Indeed, \(\mathbf{M}^{k+1} = \mathbf{M}^k \mathbf{M} = \begin{pmatrix} 2k+1 & -4k \\ k & 1-2k \end{pmatrix} \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}\). Performing matrix multiplication: Top-left element: \((2k+1)(3) + (-4k)(1) = 6k + 3 - 4k = 2k + 3 = 2(k+1) + 1\). Top-right element: \((2k+1)(-4) + (-4k)(-1) = -8k - 4 + 4k = -4(k+1)\). Bottom-left element: \(k(3) + (1-2k)(1) = 3k + 1 - 2k = k + 1\). Bottom-right element: \(k(-4) + (1-2k)(-1) = -4k - 1 + 2k = 1 - 2(k+1)\). Since the elements match the formula for \(n = k+1\), \(P(k+1)\) is true. By the principle of mathematical induction, the result holds for all positive integers \(n\). For (ii): Substituting \(n = 25\): \(\mathbf{M}^{25} = \begin{pmatrix} 2(25) + 1 & -4(25) \\ 25 & 1 - 2(25) \end{pmatrix} = \begin{pmatrix} 51 & -100 \\ 25 & -49 \end{pmatrix}\). For (iii): The leading diagonal elements of \(\mathbf{M}^n\) are \(2n+1\) and \(1-2n\). Their sum is \((2n+1) + (1-2n) = 2\), which is a constant and thus independent of \(n\).

Part (i): [B1] For verifying the base case n = 1. [M1] For assuming the induction hypothesis for n = k. [M1] For attempting to find M^(k+1) by multiplying M^k and M. [A1] For correctly expanding and simplifying all four elements. [A1] For writing a coherent concluding statement about proof by induction. Part (ii): [B1] For correctly writing the matrix M^25. Part (iii): [M1] For identifying and summing the diagonal elements 2n+1 and 1-2n. [A1] For showing that the sum equals 2, which is independent of n.

Using the exponential definitions of hyperbolic functions: \( \cosh x = \frac{e^x + e^{-x}}{2} \) and \( \sinh x = \frac{e^x - e^{-x}}{2} \). Substitute these into the given equation: \( 3\left(\frac{e^x + e^{-x}}{2}\right) + 5\left(\frac{e^x - e^{-x}}{2}\right) = 4 \). Multiplying both sides by 2 yields: \( 3(e^x + e^{-x}) + 5(e^x - e^{-x}) = 8 \). Simplifying this gives: \( 8e^x - 2e^{-x} = 8 \), which divides by 2 to become: \( 4e^x - e^{-x} = 4 \). Let \( u = e^x \) (with \( u > 0 \) since \( e^x \) is always positive). The equation becomes: \( 4u - \frac{1}{u} = 4 \). Multiplying by \( u \) gives the quadratic equation: \( 4u^2 - 4u - 1 = 0 \). Using the quadratic formula, we obtain: \( u = \frac{4 \pm \sqrt{(-4)^2 - 4(4)(-1)}}{2(4)} = \frac{4 \pm \sqrt{16 + 16}}{8} = \frac{4 \pm 4\sqrt{2}}{8} = \frac{1 \pm \sqrt{2}}{2} \). Since \( u > 0 \), we reject the negative root because \( \sqrt{2} \approx 1.414 \) and thus \( 1 - \sqrt{2} < 0 \). Therefore, \( u = \frac{1 + \sqrt{2}}{2} \). Since \( u = e^x \), we have \( e^x = \frac{1 + \sqrt{2}}{2} \), which gives the final solution: \( x = \ln\left(\frac{1 + \sqrt{2}}{2}\right) \).

M1: Substitute the exponential definitions of hyperbolic functions. A1: Simplify to a linear combination of exponential terms such as 8e^x - 2e^{-x} = 8. M1: Substitute u = e^x to obtain a quadratic equation. A1: Obtain 4u^2 - 4u - 1 = 0. M1: Apply the quadratic formula. A1: Obtain u = (1 +/- sqrt(2))/2. M1: Show a reason for rejecting the negative root. A1.375: State the final exact answer in the requested logarithmic form.

Let \( f(x) = \ln(\cosh x) \). First, find the values of \( f(x) \) and its derivatives at \( x = 0 \): \( f(0) = \ln(\cosh 0) = \ln(1) = 0 \). By the chain rule, \( f'(x) = \frac{\sinh x}{\cosh x} = \tanh x \), so \( f'(0) = \tanh 0 = 0 \). The second derivative is \( f''(x) = \text{sech}^2 x \), so \( f''(0) = \text{sech}^2 0 = 1 \). The third derivative is \( f'''(x) = -2\text{sech}^2 x \tanh x \), so \( f'''(0) = 0 \). The fourth derivative is found using the product/chain rules: \( f^{(4)}(x) = -2\text{sech}^4 x + 4\text{sech}^2 x \tanh^2 x \), so \( f^{(4)}(0) = -2(1)^4 + 0 = -2 \). The Maclaurin series expansion is: \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \dots \). Substituting the values: \( f(x) = 0 + 0 + \frac{1}{2}x^2 + 0 - \frac{2}{24}x^4 = \frac{1}{2}x^2 - \frac{1}{12}x^4 \). Now, integrate this series from 0 to 0.5: \( \int_{0}^{\frac{1}{2}} \ln(\cosh x) \, dx \approx \int_{0}^{\frac{1}{2}} \left( \frac{1}{2}x^2 - \frac{1}{12}x^4 \right) \, dx = \left[ \frac{1}{6}x^3 - \frac{1}{60}x^5 \right]_0^{\frac{1}{2}} = \left( \frac{1}{6}\left(\frac{1}{8}\right) - \frac{1}{60}\left(\frac{1}{32}\right) \right) - 0 = \frac{1}{48} - \frac{1}{1920} = \frac{40 - 1}{1920} = \frac{39}{1920} = \frac{13}{640} \).

M1: Compute the first and second derivatives of f(x). A1: Obtain f'(0) = 0 and f''(0) = 1. M1: Compute the third and fourth derivatives of f(x). A1: Obtain f'''(0) = 0 and f^{(4)}(0) = -2. A1: Correctly write the Maclaurin expansion as 1/2 x^2 - 1/12 x^4. M1: Integrate the polynomial term-by-term. A1: Obtain 1/6 x^3 - 1/60 x^5. M1: Substitute the upper limit 1/2 and lower limit 0. A1.375: Obtain the simplified fraction 13/640.

We apply integration by parts to \( I_n = \int_{1}^{e} (\ln x)^n \, dx \). Let \( u = (\ln x)^n \) and \( \frac{dv}{dx} = 1 \). Then \( \frac{du}{dx} = n(\ln x)^{n-1} \cdot \frac{1}{x} \) and \( v = x \). Applying the integration by parts formula: \( I_n = \left[ x(\ln x)^n \right]_1^e - \int_{1}^{e} x \cdot \left( n(\ln x)^{n-1} \cdot \frac{1}{x} \right) \, dx \). Evaluating the boundary term: at \( x = e \), \( e(\ln e)^n = e(1)^n = e \); at \( x = 1 \), \( 1(\ln 1)^n = 0 \) (for \( n \ge 1 \)). The remaining integral simplifies to \( n \int_{1}^{e} (\ln x)^{n-1} \, dx = n I_{n-1} \). Hence, \( I_n = e - n I_{n-1} \). To find \( I_3 \), we first calculate \( I_0 = \int_{1}^{e} 1 \, dx = [x]_1^e = e - 1 \). Then: \( I_1 = e - 1 \cdot I_0 = e - (e - 1) = 1 \). \( I_2 = e - 2 \cdot I_1 = e - 2(1) = e - 2 \). \( I_3 = e - 3 \cdot I_2 = e - 3(e - 2) = e - 3e + 6 = 6 - 2e \).

M1: Set up integration by parts with correct choices for u and dv. A1: Correctly differentiate u to find du. A1: State the integration by parts expression. A1: Evaluate the boundary limits to establish the reduction formula. M1: Determine the base case value of I_0 or I_1. A1: Establish I_0 = e - 1 or I_1 = 1. M1: Successively apply the reduction formula to find I_3. A1.375: Arrive at the final simplified value 6 - 2e.

From de Moivre's theorem, we have \( \cos(5\theta) + i\sin(5\theta) = (\cos\theta + i\sin\theta)^5 \). Expanding the right-hand side using the Binomial Theorem: \( (\cos\theta + i\sin\theta)^5 = \cos^5\theta + 5i\cos^4\theta\sin\theta - 10\cos^3\theta\sin^2\theta - 10i\cos^2\theta\sin^3\theta + 5\cos\theta\sin^4\theta + i\sin^5\theta \). Equating the imaginary parts: \( \sin(5\theta) = 5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta \). Substitute \( \cos^2\theta = 1 - \sin^2\theta \) into this expression: \( \sin(5\theta) = 5(1 - \sin^2\theta)^2\sin\theta - 10(1 - \sin^2\theta)\sin^3\theta + \sin^5\theta = 5(1 - 2\sin^2\theta + \sin^4\theta)\sin\theta - 10(\sin^3\theta - \sin^5\theta) + \sin^5\theta = 5\sin\theta - 10\sin^3\theta + 5\sin^5\theta - 10\sin^3\theta + 10\sin^5\theta + \sin^5\theta = 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta \). For \( \sin(5\theta) = 0 \), the roots are \( 5\theta = k\pi \implies \theta = \frac{k\pi}{5} \) for \( k \in \mathbb{Z} \). Setting \( x = \sin\theta \), the roots of \( 16x^5 - 20x^3 + 5x = 0 \) are \( \sin\left(-\frac{2\pi}{5}\right), \sin\left(-\frac{\pi}{5}\right), \sin(0), \sin\left(\frac{\pi}{5}\right), \sin\left(\frac{2\pi}{5}\right) \). Removing the root \( x = \sin(0) = 0 \) leaves the quartic equation \( 16x^4 - 20x^2 + 5 = 0 \) with roots \( \pm\sin\left(\frac{\pi}{5}\right) \) and \( \pm\sin\left(\frac{2\pi}{5}\right) \). By Vieta's formulas, the product of the four roots of the quartic equation \( 16x^4 - 20x^2 + 5 = 0 \) is equal to \( \frac{\text{constant term}}{\text{coefficient of } x^4} = \frac{5}{16} \). Since the roots are \( x_1 = \sin\left(\frac{\pi}{5}\right) \), \( x_2 = -\sin\left(\frac{\pi}{5}\right) \), \( x_3 = \sin\left(\frac{2\pi}{5}\right) \), and \( x_4 = -\sin\left(\frac{2\pi}{5}\right) \), their product is \( x_1 x_2 x_3 x_4 = \sin^2\left(\frac{\pi}{5}\right)\sin^2\left(\frac{2\pi}{5}\right) \). Hence, \( \sin^2\left(\frac{\pi}{5}\right)\sin^2\left(\frac{2\pi}{5}\right) = \frac{5}{16} \).

M1: Expand (cos(\theta) + i sin(\theta))^5 and identify imaginary part. A1: Correct expression for sin(5\theta) using powers of cos(\theta) and sin(\theta). M1: Apply cos^2(\theta) = 1 - sin^2(\theta) to eliminate cosine terms. A1: Correctly show that sin(5\theta) = 16sin^5(\theta) - 20sin^3(\theta) + 5sin(\theta). M1: Solve sin(5\theta) = 0 to establish the roots of the corresponding 5th-degree equation. A1: Correctly deduce the roots of the quartic 16x^4 - 20x^2 + 5 = 0. M1: Apply Vieta's product of roots formula to the quartic. A1.375: Obtain the exact product 5/16.

First, find the complementary function (CF) by solving the auxiliary equation: \( m^2 + 4m + 5 = 0 \). Solving for \( m \) using the quadratic formula: \( m = \frac{-4 \pm \sqrt{16 - 20}}{2} = -2 \pm i \). The complementary function is: \( y_c = e^{-2x}(A\cos x + B\sin x) \). Second, find a particular integral (PI) for the non-homogeneous equation. We try a solution of the form: \( y_p = ke^{-x} \). Differentiating: \( y_p' = -ke^{-x} \) and \( y_p'' = ke^{-x} \). Substituting these into the differential equation: \( ke^{-x} + 4(-ke^{-x}) + 5(ke^{-x}) = 10e^{-x} \implies 2ke^{-x} = 10e^{-x} \implies k = 5 \). So the particular integral is: \( y_p = 5e^{-x} \). The general solution is: \( y = e^{-2x}(A\cos x + B\sin x) + 5e^{-x} \). Now use the initial conditions: \( y(0) = 7 \implies 7 = e^0(A\cos 0 + B\sin 0) + 5e^0 \implies 7 = A + 5 \implies A = 2 \). Differentiating the general solution: \( \frac{dy}{dx} = -2e^{-2x}(A\cos x + B\sin x) + e^{-2x}(-A\sin x + B\cos x) - 5e^{-x} \). At \( x = 0 \), \( \frac{dy}{dx} = -3 \): \( -3 = -2(A\cos 0 + B\sin 0) + (-A\sin 0 + B\cos 0) - 5 \implies -3 = -2A + B - 5 \). Substitute \( A = 2 \): \( -3 = -2(2) + B - 5 \implies -3 = B - 9 \implies B = 6 \). Therefore, the particular solution is: \( y = e^{-2x}(2\cos x + 6\sin x) + 5e^{-x} \).

M1: Set up the auxiliary equation and solve for m. A1: Obtain roots -2 +/- i. A1: State the complementary function. M1: Set up and solve for the particular integral using y_p = k e^{-x}. A1: Obtain y_p = 5e^{-x}. M1: Combine CF and PI and apply the condition y(0) = 7 to find A. A1: Find A = 2. M1: Differentiate the general solution and apply y'(0) = -3 to find B. A1.375: Find B = 6 and write the final correct particular solution.

To find the eigenvalues of \( \mathbf{A} \), we solve \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \): \( \det\begin{pmatrix} 2-\lambda & 1 & 0 \\ 0 & 3-\lambda & 0 \\ 0 & 2 & 4-\lambda \end{pmatrix} = (4-\lambda)(2-\lambda)(3-\lambda) = 0 \). Thus, the eigenvalues are \( \lambda_1 = 2 \), \( \lambda_2 = 3 \), \( \lambda_3 = 4 \). Next, we find the eigenvectors: For \( \lambda_1 = 2 \): \( \mathbf{A} - 2\mathbf{I} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 2 & 2 \end{pmatrix} \implies y=0, z=0 \), so an eigenvector is \( \mathbf{v}_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \). For \( \lambda_2 = 3 \): \( \mathbf{A} - 3\mathbf{I} = \begin{pmatrix} -1 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 2 & 1 \end{pmatrix} \implies x = y, z = -2y \). Choosing \( y = 1 \) gives \( \mathbf{v}_2 = \begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix} \). For \( \lambda_3 = 4 \): \( \mathbf{A} - 4\mathbf{I} = \begin{pmatrix} -2 & 1 & 0 \\ 0 & -1 & 0 \\ 0 & 2 & 0 \end{pmatrix} \implies y=0, x=0 \), so an eigenvector is \( \mathbf{v}_3 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \). We define the matrices \( \mathbf{P} = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & -2 & 1 \end{pmatrix} \) and \( \mathbf{D} = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{pmatrix} \). Calculating \( \mathbf{P}^{-1} \): \( \mathbf{P}^{-1} = \begin{pmatrix} 1 & -1 & 0 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \end{pmatrix} \). Now compute \( \mathbf{A}^n = \mathbf{P} \mathbf{D}^n \mathbf{P}^{-1} \): \( \mathbf{A}^n = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & -2 & 1 \end{pmatrix} \begin{pmatrix} 2^n & 0 & 0 \\ 0 & 3^n & 0 \\ 0 & 0 & 4^n \end{pmatrix} \begin{pmatrix} 1 & -1 & 0 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \end{pmatrix} = \begin{pmatrix} 2^n & 3^n & 0 \\ 0 & 3^n & 0 \\ 0 & -2 \cdot 3^n & 4^n \end{pmatrix} \begin{pmatrix} 1 & -1 & 0 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \end{pmatrix} = \begin{pmatrix} 2^n & 3^n - 2^n & 0 \\ 0 & 3^n & 0 \\ 0 & 2(4^n - 3^n) & 4^n \end{pmatrix} \).

M1: Find eigenvalues by solving characteristic equation. A1: Correctly state eigenvalues 2, 3, 4. M1: Find corresponding eigenvectors for each eigenvalue. A1: Correctly find all three eigenvectors. M1: Construct P and compute P^{-1}. A1: Obtain correct inverse matrix P^{-1}. M1: State general diagonalized power formula A^n = P D^n P^{-1}. A1.375: Perform matrix multiplication to successfully derive the final expression.

(a) The derivative is \( \frac{dy}{dx} = \sinh(2x) \). Thus, the integrand for the arc length is \( \sqrt{1 + \left(\frac{dy}{dx}\right)^2} = \sqrt{1 + \sinh^2(2x)} = \cosh(2x) \). The arc length \( s \) is: \( s = \int_{0}^{\ln 3} \cosh(2x) \, dx = \left[ \frac{1}{2}\sinh(2x) \right]_0^{\ln 3} = \frac{1}{2}\sinh(2\ln 3) \). Since \( 2\ln 3 = \ln 9 \): \( \sinh(\ln 9) = \frac{e^{\ln 9} - e^{-\ln 9}}{2} = \frac{9 - 1/9}{2} = \frac{80}{18} = \frac{40}{9} \). Therefore, \( s = \frac{1}{2} \left(\frac{40}{9}\right) = \frac{20}{9} \). (b) The surface area of revolution \( S \) is given by: \( S = 2\pi \int_{0}^{\ln 3} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx = 2\pi \int_{0}^{\ln 3} \left(\frac{1}{2}\cosh(2x)\right) \cosh(2x) \, dx = \pi \int_{0}^{\ln 3} \cosh^2(2x) \, dx \). Using the identity \( \cosh^2(2x) = \frac{1 + \cosh(4x)}{2} \): \( S = \pi \int_{0}^{\ln 3} \frac{1 + \cosh(4x)}{2} \, dx = \frac{\pi}{2} \left[ x + \frac{1}{4}\sinh(4x) \right]_0^{\ln 3} = \frac{\pi}{2} \left( \ln 3 + \frac{1}{4}\sinh(4\ln 3) \right) \). Since \( 4\ln 3 = \ln 81 \): \( \sinh(\ln 81) = \frac{e^{\ln 81} - e^{-\ln 81}}{2} = \frac{81 - 1/81}{2} = \frac{6560}{162} = \frac{3280}{81} \). Hence, \( S = \frac{\pi}{2} \left( \ln 3 + \frac{1}{4}\left(\frac{3280}{81}\right) \right) = \frac{\pi}{2} \left( \ln 3 + \frac{820}{81} \right) = \pi \left( \frac{1}{2}\ln 3 + \frac{410}{81} \right) \).

Part (a): M1: Differentiate y and establish standard integrand form. A1: Simplify integrand to cosh(2x). M1: Integrate to 1/2 sinh(2x) and apply limits. A1: Correct answer 20/9. Part (b): M1: Set up the correct surface area integral formula. A1: Simplify to \pi \int cosh^2(2x) dx. M1: Use double angle hyperbolic identity. A1: Correctly integrate to obtain \pi/2 [x + 1/4 sinh(4x)]. A1.375: Substitute limits and obtain exact final answer \pi(410/81 + 1/2 \ln 3).

First, rewrite the differential equation in standard linear form: \( \frac{dy}{dx} + \left(2 + \frac{1}{x}\right)y = \frac{e^{-2x}}{x} \). The integrating factor \( I(x) \) is: \( I(x) = e^{\int \left(2 + \frac{1}{x}\right) \, dx} = e^{2x + \ln x} = e^{2x} \cdot x = x e^{2x} \). Multiplying both sides of the standard form by \( I(x) \): \( x e^{2x} \frac{dy}{dx} + (2x + 1)e^{2x} y = 1 \implies \frac{d}{dx} \left( x e^{2x} y \right) = 1 \). Integrating both sides with respect to \( x \): \( x e^{2x} y = x + C \). Thus, the general solution is: \( y = e^{-2x} \left(1 + \frac{C}{x}\right) \). Now apply the boundary condition \( y = 2 \) when \( x = 1 \): \( 2 = e^{-2} \left(1 + \frac{C}{1}\right) \implies 2e^2 = 1 + C \implies C = 2e^2 - 1 \). Therefore, the particular solution is: \( y = e^{-2x} \left(1 + \frac{2e^2 - 1}{x}\right) \).

M1: Write the equation in standard form. A1: Correctly identify the coefficient function P(x) = 2 + 1/x. M1: Find integrating factor using exponential formula. A1: Correctly find integrating factor x e^{2x}. M1: Formulate derivative of product and integrate both sides. A1: Correctly find general solution y = e^{-2x}(1 + C/x). M1: Apply condition y(1) = 2 to find constant C. A1.375: State correct particular solution.

Using the horizontal equation of motion, \(x = u t \cos\theta\), we have \(t = \frac{x}{u \cos\theta}\). Substituting this into the vertical equation of motion, \(y = u t \sin\theta - \frac{1}{2}gt^2\), gives the trajectory equation \(y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}\). Given \(\tan\theta = \frac{3}{4}\), we find \(\cos^2\theta = \frac{1}{1+\tan^2\theta} = \frac{16}{25}\). Substituting the coordinates of the wall \((8, 1)\) and \(g = 10\text{ m s}^{-2}\) into the trajectory equation, we get \(1 = 8\left(\frac{3}{4}\right) - \frac{10 \times 64}{2u^2 \times \frac{16}{25}}\). This simplifies to \(1 = 6 - \frac{500}{u^2}\), which gives \(\frac{500}{u^2} = 5\). Solving for \(u\), we obtain \(u^2 = 100\), hence \(u = 10\).

M1: State or derive the equation of the trajectory \(y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}\) or equivalent.
A1: Correctly calculate \(\cos^2\theta = \frac{16}{25}\).
M1: Substitute the coordinates \((8, 1)\) into the trajectory equation.
A1: Obtain the correct unsimplified equation \(1 = 6 - \frac{500}{u^2}\).
M1: Solve the equation for \(u^2\).
A1: Correctly find \(u = 10\).

The maximum speed of the particle occurs when the net force is zero, which is when the upward tension in the string equals the downward gravitational force. Thus, \(T = mg\), where \(mg = 2 \times 10 = 20\text{ N}\). By Hooke's Law, \(T = \frac{\lambda x}{L}\), where \(L = 0.4\text{ m}\) and \(\lambda = 10\text{ N}\). This gives \(\frac{10x}{0.4} = 20 \implies x = 0.8\text{ m}\). Let \(v\) be the maximum speed. By conservation of energy from the release point \(A\) to the point of maximum speed: Loss in gravitational potential energy = Gain in kinetic energy + Gain in elastic potential energy. The total vertical distance fallen is \(L + x = 0.4 + 0.8 = 1.2\text{ m}\). Therefore, \(mg(L+x) = \frac{1}{2}mv^2 + \frac{\lambda x^2}{2L}\). Substituting the known values gives \(2 \times 10 \times 1.2 = \frac{1}{2} \times 2 \times v^2 + \frac{10 \times 0.8^2}{2 \times 0.4}\). This simplifies to \(24 = v^2 + 8\), which yields \(v^2 = 16\). Thus, the maximum speed is \(v = 4\text{ m s}^{-1}\).

M1: Identify that maximum speed occurs at the equilibrium position where \(T = mg\).
A1: Calculate the extension \(x = 0.8\text{ m}\).
M1: Apply the principle of conservation of energy.
A1: Correctly express the gravitational potential energy loss as \(24\text{ J}\).
A1: Correctly express the elastic potential energy gain as \(8\text{ J}\).
M1: Formulate and solve the equation \(24 = v^2 + 8\).
A1: Obtain \(v = 4\).

Since \(6^2 + 8^2 = 10^2\), the triangle \(APB\) is right-angled at \(P\). Let \(\alpha\) be the angle the upper string \(AP\) makes with the vertical wall, and let \(\beta\) be the angle the lower string \(BP\) makes with the vertical wall. We have \(\cos\alpha = 0.6\), \(\sin\alpha = 0.8\), \(\cos\beta = 0.8\), and \(\sin\beta = 0.6\). The radius \(r\) of the horizontal circle is \(r = AP \sin\alpha = 6 \times 0.8 = 4.8\text{ m}\). Let \(T_1\) and \(T_2\) be the tensions in the upper and lower strings respectively. Resolving vertically: \(T_1 \cos\alpha - T_2 \cos\beta = mg \implies 0.6 T_1 - 0.8 T_2 = 30\). Resolving horizontally: \(T_1 \sin\alpha + T_2 \sin\beta = \frac{mv^2}{r} \implies 0.8 T_1 + 0.6 T_2 = \frac{3v^2}{4.8} = \frac{5v^2}{8}\). To eliminate \(T_1\), multiply the vertical equation by \(0.8\) and the horizontal equation by \(0.6\): \(0.48 T_1 - 0.64 T_2 = 24\) and \(0.48 T_1 + 0.36 T_2 = \frac{3v^2}{8}\). Subtracting these equations gives \(T_2 = \frac{3v^2}{8} - 24 = 3\left(\frac{v^2}{8} - 8\right)\). For the lower string to remain taut, we must have \(T_2 \ge 0\), which requires \(\frac{v^2}{8} - 8 \ge 0 \implies v^2 \ge 64 \implies v \ge 8\). Thus, the minimum value of \(v\) is \(8\).

M1: Recognize that \(\triangle APB\) is right-angled and find the trigonometric ratios of the angles.
A1: Calculate the radius of the circle \(r = 4.8\text{ m}\).
M1: Set up the vertical force balance equation.
A1: Set up the horizontal force balance equation.
M1: Eliminate \(T_1\) to solve for \(T_2\) in terms of \(v^2\).
A1: Obtain the correct expression for \(T_2\).
A1: Set \(T_2 \ge 0\) and obtain the minimum speed \(v = 8\).

Using Newton's second law, \(m a = F\), where \(a = v \frac{dv}{dx}\), we have \(0.5 v \frac{dv}{dx} = -(v + 2v^2)\). Dividing both sides by \(v\) (since \(v > 0\)) gives \(0.5 \frac{dv}{dx} = -(1 + 2v)\). Separating the variables, we get \(\frac{1}{1+2v} dv = -2 dx\). Integrating both sides gives \(\int \frac{1}{1+2v} dv = -2 \int dx \implies \frac{1}{2} \ln(1+2v) = -2x + C\). Using the initial condition \(x = 0\) when \(v = 4\), we find \(C = \frac{1}{2} \ln(9) = \ln(3)\). Therefore, \(\frac{1}{2} \ln(1+2v) = -2x + \ln(3)\). To find the distance \(x\) when \(v = 1\), we substitute \(v = 1\) into the equation: \(\frac{1}{2} \ln(3) = -2x + \ln(3) \implies 2x = \ln(3) - \frac{1}{2} \ln(3) = \frac{1}{2} \ln(3) \implies x = \frac{1}{4} \ln(3)\).

M1: Set up the differential equation \(m v \frac{dv}{dx} = - (v + 2v^2)\).
A1: Correctly separate variables to get \(\frac{1}{1+2v} dv = -2 dx\).
M1: Integrate both sides to get \(\frac{1}{2} \ln(1+2v) = -2x + C\).
A1: Find the correct constant of integration \(C = \ln(3)\) or equivalent.
M1: Substitute \(v = 1\) into the integrated equation.
A1: Solve for \(x\) and express the final answer in terms of \(\ln(3)\).
A1: Obtain the correct final answer \(x = \frac{1}{4}\ln(3)\).

Let \(R_A\) be the normal reaction force at the ground, \(F\) be the friction force at the ground, and \(R_B\) be the normal reaction force at the wall. Resolving forces vertically: \(R_A = 15g + 75g = 900\text{ N}\). Since the ladder is on the point of slipping, the friction force is at its maximum: \(F = \mu R_A = 0.4 \times 900 = 360\text{ N}\). Resolving forces horizontally: \(R_B = F = 360\text{ N}\). Taking moments about \(A\): \(15g \times 2 \cos\theta + 75g \times d \cos\theta = R_B \times 4 \sin\theta\). Dividing both sides by \(\cos\theta\) gives \(30g + 75g d = 4 R_B \tan\theta\). Substituting \(g = 10\text{ m s}^{-2}\), \(R_B = 360\text{ N}\), and \(\tan\theta = \frac{4}{3}\): \(300 + 750d = 4 \times 360 \times \frac{4}{3} = 1920\). This simplifies to \(750d = 1620\), which yields \(d = 2.16\text{ m}\).

M1: Resolve forces vertically to find \(R_A\).
A1: Obtain \(R_A = 900\text{ N}\).
A1: Use \(F = \mu R_A\) to find \(F = 360\text{ N}\) and resolve horizontally to find \(R_B = 360\text{ N}\).
M1: Set up the moment equation about \(A\).
A1: Substitute \(\tan\theta = 4/3\) and other values correctly.
M1: Solve the linear equation for \(d\).
A1: Obtain \(d = 2.16\).

Let the velocity of the sphere just before impact be resolved into components parallel and perpendicular to the wall. The component parallel to the wall is \(u_{\parallel} = u \cos 60^\circ = 0.5u\), and the component perpendicular to the wall is \(u_{\perp} = u \sin 60^\circ = \frac{\sqrt{3}}{2} u\). After the collision, the component parallel to the wall remains unchanged: \(v_{\parallel} = 0.5u\). The component perpendicular to the wall is reversed and reduced by the coefficient of restitution \(e = 0.5\): \(v_{\perp} = e u_{\perp} = 0.5 \left(\frac{\sqrt{3}}{2}u\right) = \frac{\sqrt{3}}{4}u\). The final kinetic energy is \(K_f = \frac{1}{2}m(v_{\parallel}^2 + v_{\perp}^2) = \frac{1}{2}m\left( \frac{1}{4}u^2 + \frac{3}{16}u^2 \right) = \frac{7}{16} \left(\frac{1}{2}mu^2\right)\). The initial kinetic energy is \(K_i = \frac{1}{2}mu^2\). The loss in kinetic energy is \(\Delta K = K_i - K_f = \left(1 - \frac{7}{16}\right) \left(\frac{1}{2}mu^2\right) = \frac{9}{16} \left(\frac{1}{2}mu^2\right)\). Thus, the fractional loss of kinetic energy is \(\frac{9}{16}\).

M1: Determine the components of the initial velocity parallel and perpendicular to the wall.
A1: Correctly apply the conservation of momentum parallel to the wall to get \(v_{\parallel} = 0.5u\).
M1: Apply the coefficient of restitution perpendicular to the wall to get \(v_{\perp} = \frac{\sqrt{3}}{4}u\).
A1: Calculate the final velocity squared as \(v^2 = \frac{7}{16}u^2\).
M1: Write expressions for initial and final kinetic energies.
A1: Calculate the energy loss.
A1: Find the correct fractional loss of \(\frac{9}{16}\).

Let \(\theta\) be the angle the radius to the particle makes with the upward vertical. By conservation of energy, the kinetic energy at angle \(\theta\) is equal to the initial kinetic energy plus the loss in gravitational potential energy: \(\frac{1}{2}mv^2 = \frac{1}{2}mu^2 + mg(a - a\cos\theta)\). Substituting \(u = \sqrt{0.25 ag}\) gives \(\frac{1}{2}mv^2 = \frac{1}{2}m(0.25 ag) + mga(1 - \cos\theta)\), which simplifies to \(v^2 = 0.25 ag + 2ag(1 - \cos\theta) = 2.25 ag - 2ag\cos\theta\). The equation of motion in the radial direction is \(mg\cos\theta - R = \frac{mv^2}{a}\), where \(R\) is the normal reaction. The particle leaves the surface of the sphere when \(R = 0\), which gives \(v^2 = ag\cos\theta\). Equating the two expressions for \(v^2\), we have \(ag\cos\theta = 2.25 ag - 2ag\cos\theta \implies 3\cos\theta = 2.25 \implies \cos\theta = 0.75 = \frac{3}{4}\).

M1: Set up the conservation of energy equation.
A1: Correctly substitute \(u^2\) and obtain \(v^2\) in terms of \(\theta\).
A1: Correctly simplify to \(v^2 = 2.25 ag - 2ag\cos\theta\).
M1: Set up the radial equation of motion \(mg\cos\theta - R = \frac{mv^2}{a}\).
M1: Use the condition \(R = 0\) for leaving the surface.
A1: Equate the two expressions to get \(3\cos\theta = 2.25\).
A1: Find \(\cos\theta = \frac{3}{4}\).

(i) For \(1 \le x \le 2\):
\[
F(x) = \int_1^x \frac{4}{15}u^3 \, du = \left[ \frac{u^4}{15} \right]_1^x = \frac{x^4 - 1}{15}
\]
For \(x < 1\), \(F(x) = 0\), and for \(x > 2\), \(F(x) = 1\).

(ii) Let \(G(y)\) be the cumulative distribution function of \(Y\).
For \(1 \le y \le 4\):
\[
G(y) = P(Y \le y) = P(X^2 \le y) = P(X \le \sqrt{y}) = F(\sqrt{y}) = \frac{(\sqrt{y})^4 - 1}{15} = \frac{y^2 - 1}{15}
\]
Differentiating to find the probability density function \(g(y)\):
\[
g(y) = G'(y) = \frac{2}{15}y \quad \text{for } 1 \le y \le 4
\]
and \(g(y) = 0\) otherwise.

(iii) Since \(\sqrt{Y} = X\):
\[
E(\sqrt{Y}) = E(X) = \int_1^2 x \cdot \frac{4}{15}x^3 \, dx = \frac{4}{15} \int_1^2 x^4 \, dx = \frac{4}{15} \left[ \frac{x^5}{5} \right]_1^2 = \frac{4}{75}(32 - 1) = \frac{124}{75}
\]

M1: Integrates \(f(x)\) with correct limits to find \(F(x)\).
A1: Correct \(F(x)\) with specified domain.
M1: Expresses \(P(Y \le y)\) in terms of \(X\) and uses \(F(x)\).
A1: Correctly derives CDF of \(Y\).
M1: Differentiates CDF to find PDF.
A1: Correctly states \(g(y)\) with its correct domain.
M1: Sets up the integral for \(E(X)\) or integrates \(\sqrt{y}g(y)\).
A1: Obtains the correct exact fraction \(\frac{124}{75}\).

Let \(H_0\): Median difference in reaction times is zero.
Let \(H_1\): Median difference in reaction times is not zero.

Calculate the differences \(d_i = \text{Before} - \text{After}\):
\[
\begin{array}{lcccccccc}
\text{Participant} & A & B & C & D & E & F & G & H \\
\text{Difference } (d_i) & +8 & +5 & -3 & +12 & -2 & +10 & -4 & +9
\end{array}
\]
Rank the absolute differences \(|d_i|\):
\[
\begin{array}{lcccccccc}
\text{Participant} & E & C & G & B & A & H & F & D \\
|d_i| & 2 & 3 & 4 & 5 & 8 & 9 & 10 & 12 \\
\text{Rank} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\text{Sign} & - & - & - & + & + & + & + & +
\end{array}
\]
Calculate the sum of the positive ranks (\(T^+\)) and negative ranks (\(T^-\)):
\[
T^- = 1 + 2 + 3 = 6
\]
\[
T^+ = 4 + 5 + 6 + 7 + 8 = 30
\]
The test statistic \(T\) is \(\min(T^+, T^-) = 6\).
For \(n = 8\) at the 5% level of significance for a two-tailed test, the critical value is 4.
Since \(T = 6 > 4\), we fail to reject \(H_0\). There is insufficient evidence at the 5% level to conclude that there is a significant difference in reaction times before and after consuming the beverage.

H1: Formulates correct null and alternative hypotheses.
M1: Calculates correct differences for all pairs.
M1: Ranks the absolute differences correctly (no ties in this dataset).
M1: Identifies signs and calculates sum of positive and negative ranks.
A1: Correctly identifies \(T = 6\).
B1: States correct critical value of 4 for \(n = 8\), two-tailed at 5%.
M1: Makes a correct comparison (reject if \(T \le 4\)).
A1: Draws correct non-contextual and contextual conclusion.

Let \(H_0\): The population phenotypes follow the 9:3:3:1 Mendelian ratio.
Let \(H_1\): The population phenotypes do not follow the Mendelian ratio.

The total number of offspring is \(N = 160\).
The expected frequencies are calculated as:
* Round & Yellow: \(160 \times \frac{9}{16} = 90\)
* Round & Green: \(160 \times \frac{3}{16} = 30\)
* Wrinkled & Yellow: \(160 \times \frac{3}{16} = 30\)
* Wrinkled & Green: \(160 \times \frac{1}{16} = 10\)

Calculate the \(\chi^2\) test statistic:
\[
\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}
\]
\[
\chi^2 = \frac{(98-90)^2}{90} + \frac{(27-30)^2}{30} + \frac{(25-30)^2}{30} + \frac{(10-10)^2}{10}
\]
\[
\chi^2 = \frac{64}{90} + \frac{9}{30} + \frac{25}{30} + 0 = 0.7111 + 0.3000 + 0.8333 + 0 = 1.8444 \approx 1.84
\]

Degrees of freedom \(\nu = k - 1 = 4 - 1 = 3\).
The critical value of \(\chi^2\) with 3 degrees of freedom at the 5% level of significance is \(7.815\).

Since \(\chi^2_{\text{calc}} = 1.84 < 7.815\), we fail to reject \(H_0\). There is no significant evidence at the 5% level to suggest that the observed phenotypic ratios differ from the Mendelian ratio.

B1: States appropriate null and alternative hypotheses.
M1: Calculates the expected frequencies correctly (90, 30, 30, 10).
M1: Applies the \(\chi^2\) formula correctly with their expected frequencies.
A1: Obtains the correct test statistic value of approximately 1.84.
B1: States correct degrees of freedom (3) and critical value (7.815).
M1: Compares the calculated statistic with the critical value.
A1: Concludes correctly in context (fail to reject \(H_0\)).

(i) Differentiating \(G_X(t)\):
\[
G'_X(t) = \frac{(4-3t)(1) - t(-3)}{(4-3t)^2} = \frac{4}{(4-3t)^2}
\]
Therefore:
\[
E(X) = G'_X(1) = \frac{4}{(4-3(1))^2} = \frac{4}{1} = 4
\]

(ii) Differentiating again:
\[
G''_X(t) = \frac{d}{dt}\left[ 4(4-3t)^{-2} \right] = 4 \cdot (-2)(4-3t)^{-3} \cdot (-3) = \frac{24}{(4-3t)^3}
\]
Evaluating at \(t=1\):
\[
G''_X(1) = \frac{24}{(4-3)^3} = 24
\]
Using the formula for variance:
\[
Var(X) = G''_X(1) + G'_X(1) - [G'_X(1)]^2 = 24 + 4 - 4^2 = 28 - 16 = 12
\]

(iii) Expanding \(G_X(t)\) as a power series:
\[
G_X(t) = \frac{t}{4}\left(1 - \frac{3}{4}t\right)^{-1} = \frac{t}{4} \sum_{r=0}^{\infty} \left(\frac{3}{4}t\right)^r = \sum_{r=0}^{\infty} \frac{1}{4}\left(\frac{3}{4}\right)^r t^{r+1}
\]
So:
\[
P(X = 1) = \frac{1}{4} = 0.25
\]
\[
P(X = 2) = \frac{1}{4}\left(\frac{3}{4}\right) = \frac{3}{16} = 0.1875
\]
Thus:
\[
P(X \ge 3) = 1 - P(X = 1) - P(X = 2) = 1 - \frac{1}{4} - \frac{3}{16} = \frac{9}{16} = 0.5625
\]

M1: Differentiates \(G_X(t)\) using quotient or chain rule.
A1: Obtains correct expression for \(G'_X(t)\).
A1: Evaluates to find \(E(X) = 4\).
M1: Obtains a correct second derivative \(G''_X(t)\).
M1: Correctly applies variance formula \(G''_X(1) + G'_X(1) - (G'_X(1))^2\).
A1: Obtains \(Var(X) = 12\).
M1: Identifies coefficients of \(t\) and \(t^2\) from power series or by successive differentiation.
A1: Reaches the correct value of \(P(X \ge 3) = \frac{9}{16}\).

Let \(H_0\): \(\mu_1 = \mu_2\)
Let \(H_1\): \(\mu_1 \neq \mu_2\)

Calculate the pooled estimate of variance, \(s_p^2\):
\[
s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}
\]
\[
s_p^2 = \frac{5(2) + 7\left(\frac{12}{7}\right)}{6 + 8 - 2} = \frac{10 + 12}{12} = \frac{22}{12} = \frac{11}{6} \approx 1.8333
\]

Calculate the test statistic \(t\):
\[
t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{s_p^2 \left( \frac{1}{n_1} + \frac{1}{n_2} \right)}}
\]
\[
t = \frac{12 - 15}{\sqrt{\frac{11}{6} \left( \frac{1}{6} + \frac{1}{8} \right)}} = \frac{-3}{\sqrt{\frac{11}{6} \cdot \frac{7}{24}}} = \frac{-3}{\sqrt{\frac{77}{144}}} = \frac{-3}{\frac{\sqrt{77}}{12}} = -\frac{36}{\sqrt{77}} \approx -4.103
\]

Degrees of freedom \(\nu = n_1 + n_2 - 2 = 12\).
For a two-tailed test at the 5% significance level with 12 degrees of freedom, the critical value is \(2.179\).

Since \(|t| = 4.103 > 2.179\), we reject \(H_0\).
There is significant evidence at the 5% level to conclude that the population means are different.

B1: Correctly states hypotheses.
M1: Applies the pooled variance formula.
A1: Obtains \(s_p^2 = \frac{11}{6}\) (or 1.83).
M1: Applies the two-sample \(t\)-test statistic formula.
A1: Obtains \(t \approx -4.10\) (or positive equivalent).
B1: Identifies correct degrees of freedom (12) and critical value (2.179).
M1: Compares calculated \(t\) to critical value correctly.
A1: Concludes correctly with context.

First, calculate the sample mean \(\bar{x}\) and unbiased variance \(s^2\):
\[
\sum x = 45.2 + 44.8 + 45.5 + 45.1 + 44.9 + 45.3 + 45.6 + 44.7 + 45.2 + 45.3 = 451.6
\]
\[
\bar{x} = \frac{451.6}{10} = 45.16
\]
\[
\sum x^2 = 45.2^2 + 44.8^2 + 45.5^2 + 45.1^2 + 44.9^2 + 45.3^2 + 45.6^2 + 44.7^2 + 45.2^2 + 45.3^2 = 20395.02
\]
\[
s^2 = \frac{1}{n-1} \left( \sum x^2 - \frac{(\sum x)^2}{n} \right) = \frac{1}{9} \left( 20395.02 - \frac{451.6^2}{10} \right)
\]
\[
s^2 = \frac{1}{9} (20395.02 - 20394.256) = \frac{0.764}{9} \approx 0.084889
\]
\[
s = \sqrt{0.084889} \approx 0.29136
\]

For a 95% confidence interval with \(n-1 = 9\) degrees of freedom using the \(t\)-distribution:
\[
t_{0.025, 9} = 2.262
\]

The margin of error is:
\[
E = t \cdot \frac{s}{\sqrt{n}} = 2.262 \cdot \frac{0.29136}{\sqrt{10}} = 2.262 \cdot 0.092136 \approx 0.2084
\]

Thus, the 95% confidence interval is:
\[
\bar{x} \pm E = 45.16 \pm 0.2084 = [44.95, 45.37] \text{ (to 2 decimal places)}
\]

M1: Calculates sample mean \(\bar{x} = 45.16\).
M1: Calculates unbiased variance \(s^2 \approx 0.0849\).
A1: Correct standard deviation \(s \approx 0.2914\).
B1: Identifies critical value \(t = 2.262\) for 9 degrees of freedom.
M1: Calculates margin of error using \(t \cdot \frac{s}{\sqrt{n}}\).
A1: Correct margin of error \(\approx 0.208\).
A1: Correct lower limit of confidence interval (44.95).
A1: Correct upper limit of confidence interval (45.37).