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

(i) We have \(\sum \alpha = 2\) and \(\sum \alpha\beta = 3\). Then \(\sum \alpha^2 = (\sum \alpha)^2 - 2\sum \alpha\beta = 2^2 - 2(3) = 4 - 6 = -2\). (ii) Let \(S_k = \alpha^k + \beta^k + \gamma^k + \delta^k\). Using Newton's relations for the equation \(x^4 - p_1 x^3 + p_2 x^2 - p_3 x + p_4 = 0\) where \(p_1 = 2, p_2 = 3, p_3 = 4, p_4 = 5\): we have \(S_1 = 2\), and \(S_2 - 2 S_1 + 2(3) = 0 \Rightarrow S_2 = -2\). Then \(S_3 - 2 S_2 + 3 S_1 - 3(4) = 0 \Rightarrow S_3 - 2(-2) + 3(2) - 12 = 0 \Rightarrow S_3 + 4 + 6 - 12 = 0 \Rightarrow S_3 = 2\). (iii) Let \(y = \frac{1}{x^2}\), which gives \(x^2 = \frac{1}{y}\). From the quartic equation, we group the even and odd powers: \(x^4 + 3x^2 + 5 = 2x^3 + 4x = 2x(x^2 + 2)\). Squaring both sides: \((x^4 + 3x^2 + 5)^2 = 4x^2(x^2 + 2)^2\). Substituting \(x^2 = \frac{1}{y}\): \(( \frac{1}{y^2} + \frac{3}{y} + 5 )^2 = \frac{4}{y} ( \frac{1}{y} + 2 )^2\). Multiplying both sides by \(y^4\): \((1 + 3y + 5y^2)^2 = 4y(1 + 2y)^2\). Expanding both sides: \(25y^4 + 30y^3 + 19y^2 + 6y + 1 = 4y(4y^2 + 4y + 1) = 16y^3 + 16y^2 + 4y\). Subtracting the right-hand side from the left-hand side gives: \(25y^4 + 14y^3 + 3y^2 + 2y + 1 = 0\).

M1: Use the correct identity for \(\sum \alpha^2\). A1: Obtain -2. M1: Set up the Newton's relation for \(S_3\) or use alternative algebraic expansions. M1: Correctly substitute \(S_1\) and \(S_2\). A1: Obtain 2. M1: Attempt substitution \(y = 1/x^2\) or equivalent. M1: Rearrange the quartic to group even/odd powers and square. A1: Correctly expand LHS. A1: Correctly expand RHS. A1: Obtain final equation \(25y^4 + 14y^3 + 3y^2 + 2y + 1 = 0\).

(i) By algebraic division, we can write \(y = x - 1 + \frac{1}{x - 2}\). As \(x \to 2\), \(y \to \pm\infty\), so the vertical asymptote is \(x = 2\). As \(x \to \pm\infty\), \(y \to x - 1\), so the oblique asymptote is \(y = x - 1\). (ii) Rearranging \(y = \frac{x^2 - 3x + 3}{x - 2}\) gives \(y(x - 2) = x^2 - 3x + 3\), which simplifies to the quadratic equation \(x^2 - (y + 3)x + (2y + 3) = 0\). For \(x\) to be real, the discriminant must be non-negative: \((y + 3)^2 - 4(2y + 3) \ge 0 \Rightarrow y^2 + 6y + 9 - 8y - 12 \ge 0 \Rightarrow y^2 - 2y - 3 \ge 0 \Rightarrow (y - 3)(y + 1) \ge 0\). This inequality is satisfied when \(y \ge 3\) or \(y \le -1\). Thus, \(y\) cannot take any value in the open interval \(-1 < y < 3\). (iii) For \(y = 3\), we have \(x^2 - 6x + 9 = 0 \Rightarrow (x - 3)^2 = 0 \Rightarrow x = 3\), giving the local minimum point \((3, 3)\). For \(y = -1\), we have \(x^2 - 2x + 1 = 0 \Rightarrow (x - 1)^2 = 0 \Rightarrow x = 1\), giving the local maximum point \((1, -1)\). The \(y\)-intercept occurs when \(x = 0\), giving \(y = -1.5\). There are no \(x\)-intercepts as \(x^2 - 3x + 3 = 0\) has no real roots. The sketch consists of two branches separated by the vertical asymptote \(x = 2\), with one branch containing the local minimum \((3, 3)\) approaching the asymptotes, and the other branch containing the local maximum \((1, -1)\) and \(y\)-intercept \((0, -1.5)\).

M1: Attempt algebraic division or limits to find asymptotes. A1: State \(x = 2\). A1: State \(y = x - 1\). M1: Rearrange to a quadratic in \(x\). M1: Apply the discriminant condition \(D \ge 0\). A1: Obtain \((y-3)(y+1) \ge 0\). A1: Conclude the interval correctly. B1: Correct coordinates of local minimum and maximum. B1: Correct \(y\)-intercept. B1: Sketch showing the branch for \(x > 2\) with minimum. B1: Sketch showing the branch for \(x < 2\) with maximum and \(y\)-intercept.

(i) Starting with the right-hand side: \(\frac{1}{4} \left( \frac{1}{(2r-1)(2r+1)} - \frac{1}{(2r+1)(2r+3)} \right) = \frac{1}{4} \left( \frac{(2r+3) - (2r-1)}{(2r-1)(2r+1)(2r+3)} \right) = \frac{1}{4} \left( \frac{4}{(2r-1)(2r+1)(2r+3)} \right) = \frac{1}{(2r-1)(2r+1)(2r+3)}\). (ii) Let \(v_r = \frac{1}{(2r-1)(2r+1)}\). Then the sum is \(\frac{1}{4} \sum_{r=1}^n (v_r - v_{r+1}) = \frac{1}{4} (v_1 - v_{n+1})\) because of the telescoping nature of the series. We have \(v_1 = \frac{1}{1 \times 3} = \frac{1}{3}\) and \(v_{n+1} = \frac{1}{(2n+1)(2n+3)}\). Thus, the sum is \(\frac{1}{4} \left( \frac{1}{3} - \frac{1}{(2n+1)(2n+3)} \right) = \frac{1}{4} \left( \frac{(2n+1)(2n+3) - 3}{3(2n+1)(2n+3)} \right) = \frac{4n^2 + 8n}{12(2n+1)(2n+3)} = \frac{n(n+2)}{3(2n+1)(2n+3)}\). (iii) Taking the limit as \(n \to \infty\): \(\sum_{r=1}^{\infty} \frac{1}{(2r-1)(2r+1)(2r+3)} = \lim_{n \to \infty} \frac{1}{4} \left( \frac{1}{3} - \frac{1}{(2n+1)(2n+3)} \right) = \frac{1}{12}\). (iv) We have \(\sum_{r=n+1}^{2n} u_r = \sum_{r=1}^{2n} u_r - \sum_{r=1}^n u_r = \frac{1}{4} \left( \frac{1}{(2n+1)(2n+3)} - \frac{1}{(4n+1)(4n+3)} \right)\). Combining these fractions: \(\frac{(4n+1)(4n+3) - (2n+1)(2n+3)}{4(2n+1)(2n+3)(4n+1)(4n+3)} = \frac{16n^2 + 16n + 3 - (4n^2 + 8n + 3)}{4(2n+1)(2n+3)(4n+1)(4n+3)} = \frac{12n^2 + 8n}{4(2n+1)(2n+3)(4n+1)(4n+3)} = \frac{n(3n+2)}{(2n+1)(2n+3)(4n+1)(4n+3)}\).

M1: Attempt to combine RHS terms over a common denominator. A1: Complete proof of identity. M1: Apply method of differences to telescope the series. A1: Correctly obtain \(\frac{1}{4} (v_1 - v_{n+1})\). A1: Complete the algebraic simplification to reach the given form. M1: Apply limit as \(n \to \infty\). A1: Obtain 1/12. M1: Use sum from n+1 to 2n equals sum from 1 to 2n minus sum from 1 to n. M1: Express as a difference of two algebraic fractions. A1: Obtain the simplified expression.

(i) Since \(T_1\) is a shear parallel to the \(y\)-axis with the \(y\)-axis invariant, its matrix \(A\) is of the form \(\begin{pmatrix} 1 & 0 \\ k & 1 \end{pmatrix}\). Since \((1, 0)\) maps to \((1, 3)\), we have \(k = 3\), so \(A = \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix}\). (ii) The matrix \(B\) for \(T_2\) is \(\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}\). The composite transformation is \(T_1\) followed by \(T_2\), so the matrix is \(C = B A\). \(C = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix} = \begin{pmatrix} \cos\theta - 3\sin\theta & -\sin\theta \\ \sin\theta + 3\cos\theta & \cos\theta \end{pmatrix}\). Comparing with the given matrix \(C\), we have \(\cos\theta = \frac{1}{2}\) and \(\sin\theta = \frac{\sqrt{3}}{2}\). Since \(0 < \theta < \pi\), this gives \(\theta = \frac{\pi}{3}\) (or \(60^\circ\)). (iii) The determinant of \(C\) is \(\det(C) = \det(B)\det(A) = 1 \times 1 = 1\). Thus, \(C^{-1} = \begin{pmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{3 + \sqrt{3}}{2} & \frac{1 - 3\sqrt{3}}{2} \end{pmatrix}\). Since \(C = B A\), we have \(C^{-1} = A^{-1} B^{-1}\), which represents \(B^{-1\)} followed by \(A^{-1}\). \(B^{-1} = \begin{pmatrix} \cos(-\frac{\pi}{3}) & -\sin(-\frac{\pi}{3}) \\ \sin(-\frac{\pi}{3}) & \cos(-\frac{\pi}{3}) \end{pmatrix}\) represents a clockwise rotation through \(\frac{\pi}{3}\) about the origin. \(A^{-1} = \begin{pmatrix} 1 & 0 \\ -3 & 1 \end{pmatrix}\) represents a shear parallel to the \(y\)-axis with the \(y\)-axis invariant, mapping \((1, 0)\) to \((1, -3)\). Therefore, \(C^{-1}\) is a rotation about the origin through angle \(\frac{\pi}{3}\) clockwise, followed by a shear parallel to the \(y\)-axis with invariant line the \(y\)-axis, mapping \((1, 0)\) to \((1, -3)\).

B1: Write down correct diagonal of \(A\). B1: State correct bottom-left entry of \(A\). M1: Write down correct general rotation matrix \(B\) in terms of \(\theta\). M1: Perform matrix multiplication \(B A\). A1: Compare elements to get trig equations. A1: Solve for \(\theta = \pi/3\) or \(60^\circ\). M1: Find inverse of matrix \(C\). A1: Correctly write \(C^{-1}\). M1: Express \(C^{-1} = A^{-1} B^{-1}\) and identify the two component matrices. A1: Correctly describe the clockwise rotation. A1: Correctly describe the shear and state the correct order.

(i) The curve \(C\) is a closed limaon symmetric about the initial line. At \(\theta = 0\), \(r = 3a\); at \(\theta = \pm \frac{\pi}{2}\), \(r = 2a\); at \(\theta = \pi\), \(r = a\). (ii) The area \(A\) enclosed is \(A = \frac{1}{2} \int_{-\pi}^{\pi} r^2 \, d\theta = \int_{0}^{\pi} r^2 \, d\theta = a^2 \int_{0}^{\pi} (4 + 4\cos\theta + \cos^2\theta) \, d\theta = a^2 \int_{0}^{\pi} \left(4 + 4\cos\theta + \frac{1 + \cos 2\theta}{2}\right) d\theta = a^2 \int_{0}^{\pi} \left(\frac{9}{2} + 4\cos\theta + \frac{1}{2\cos 2\theta}\right) d\theta = a^2 \left[ \frac{9}{2}\theta + 4\sin\theta + \frac{1}{4}\sin 2\theta \right]_0^{\pi} = \frac{9\pi a^2}{2}\). (iii) The Cartesian coordinate \(x\) is given by \(x = r\cos\theta = a(2\cos\theta + \cos^2\theta)\). For the tangent to be perpendicular to the initial line, we need \(\frac{dx}{d\theta} = 0\): \(\frac{dx}{d\theta} = a(-2\sin\theta - 2\sin\theta\cos\theta) = -2a\sin\theta(1 + \cos\theta) = 0\). This gives \(\sin\theta = 0\) or \(\cos\theta = -1\). In the range \(-\pi < \theta \le \pi\), this yields \(\theta = 0\) and \(\theta = \pi\). At \(\theta = 0\), \(r = 3a\), so Cartesian coordinates are \((3a, 0)\). At \(\theta = \pi\), \(r = a\), so Cartesian coordinates are \((-a, 0)\). (At both points, \(\frac{dy}{d\theta} \neq 0\), so these are indeed perpendicular tangents.)

B1: Correct shape (limaon symmetric about initial line, not passing through pole). B1: Correct intercept on initial line \((3a)\) and \(\theta=\pi\) \((a)\). B1: Correct intercepts at \(\theta = \pm \pi/2\) \((2a)\). M1: Set up area integral \(\frac{1}{2}\int r^2 \, d\theta\). M1: Use identity \(\cos^2\theta = \frac{1+\cos 2\theta}{2}\). A1: Correct integration. A1: Obtain \(\frac{9\pi a^2}{2}\). M1: Use \(x = r\cos\theta\) and differentiate. A1: Set \(\frac{dx}{d\theta} = 0\) and find \(\theta = 0, \pi\). A1: Determine Cartesian coordinates \((3a, 0)\) and \((-a, 0)\).

(i) Let us test if the lines intersect by equating components: \(1 + 2\lambda = 2 + \mu \Rightarrow 2\lambda - \mu = 1\) (1), \(2 - \lambda = -1 + \mu \Rightarrow \lambda + \mu = 3\) (2), \-1 + 3\lambda = -\mu \Rightarrow 3\lambda + \mu = 1\) (3). Adding (1) and (2) gives \(3\lambda = 4 \Rightarrow \lambda = \frac{4}{3}\), which yields \(\mu = \frac{5}{3}\). Substituting these into (3) gives \(3(\frac{4}{3}) + \frac{5}{3} = 4 + \frac{5}{3} = \frac{17}{3} \neq 1\). The equations are inconsistent, so the lines do not intersect. Since the direction vectors are not scalar multiples of each other, the lines are not parallel. Hence, they are skew. (ii) The common perpendicular direction is \(\mathbf{n} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} \times \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} = \begin{pmatrix} -2 \\ 5 \\ 3 \end{pmatrix}\). The vector connecting points on each line is \(\mathbf{a}_1 - \mathbf{a}_2 = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} - \begin{pmatrix} 2 \\ -1 \\ 0 \end{pmatrix} = \begin{pmatrix} -1 \\ 3 \\ -1 \end{pmatrix}\). The shortest distance is \(d = \frac{|(\mathbf{a}_1 - \mathbf{a}_2) \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{|(-1)(-2) + (3)(5) + (-1)(3)|}{\sqrt{(-2)^2+5^2+3^2}} = \frac{14}{\sqrt{38}} = \frac{7\sqrt{38}}{19}\). (iii) The plane \(\Pi\) contains \(l_1\), so it passes through \((1, 2, -1)\). It is parallel to \(l_2\), so its normal is \(\mathbf{n} = \begin{pmatrix} -2 \\ 5 \\ 3 \end{pmatrix}\). The equation of the plane is \(-2x + 5y + 3z = -2(1) + 5(2) + 3(-1) = 5\), which simplifies to \(2x - 5y - 3z + 5 = 0\).

M1: Equate components to set up system of equations. A1: Solve for \(\lambda\) and \(\mu\) and show inconsistency. A1: State that direction vectors are not parallel. A1: Conclude lines are skew. M1: Attempt cross product of direction vectors. A1: Obtain the correct normal vector. M1: Apply the shortest distance formula. A1: Obtain \(\frac{7\sqrt{38}}{19}\) or equivalent. M1: Use normal vector and point on \(l_1\) to form plane equation. A1: Substitute coordinates of point correctly. A1: Obtain \(2x - 5y - 3z + 5 = 0\).

(i) Base case: for \(n = 1\), LHS: \(M^1 = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}\). RHS: \(\begin{pmatrix} 2(1)+1 & -4(1) \\ 1 & 1-2(1) \end{pmatrix} = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}\). LHS = RHS, so the statement is true for \(n = 1\). Inductive step: Assume the statement is true for \(n = m\), i.e., \(M^m = \begin{pmatrix} 2m+1 & -4m \\ m & 1-2m \end{pmatrix}\). We must show it is true for \(n = m+1\): \(M^{m+1} = M^m M = \begin{pmatrix} 2m+1 & -4m \\ m & 1-2m \end{pmatrix} \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix} = \begin{pmatrix} (2m+1)(3) + (-4m)(1) & (2m+1)(-4) + (-4m)(-1) \\ (m)(3) + (1-2m)(1) & (m)(-4) + (1-2m)(-1) \end{pmatrix} = \begin{pmatrix} 6m + 3 - 4m & -8m - 4 + 4m \\ 3m + 1 - 2m & -4m - 1 + 2m \end{pmatrix} = \begin{pmatrix} 2(m+1)+1 & -4(m+1) \\ m+1 & 1-2(m+1) \end{pmatrix}\). This is the required form for \(n = m+1\). Since both the base case and inductive step are satisfied, the statement is true for all positive integers \(n\) by mathematical induction. (ii) The determinant of \(M^n\) is \(\det(M^n) = (2n+1)(1-2n) - (-4n)(n) = 1 - 4n^2 + 4n^2 = 1\). The inverse is \((M^n)^{-1} = \begin{pmatrix} 1-2n & 4n \\ -n & 2n+1 \end{pmatrix}\). We compare this with \(\begin{pmatrix} 2k+1 & -4k \\ k & 1-2k \end{pmatrix}\). The bottom-left entry gives \(k = -n\). Testing other entries with \(k = -n\): \(2k+1 = 1-2n\) (matches), \(-4k = 4n\) (matches), \(1-2k = 2n+1\) (matches). Thus, \((M^n)^{-1}\) can indeed be expressed in this form with the function \(k = -n\).

B1: Verify base case \(n = 1\). M1: State inductive hypothesis for \(n = m\) and write down the expression for \(M^{m+1} = M^m M\). M1: Perform matrix multiplication. A1: Simplify matrix elements. A1: Structure the final matrix clearly in terms of \(m+1\) and state the conclusion. M1: Calculate the determinant of \(M^n\) (which is 1). A1: Correctly write down \((M^n)^{-1}\). M1: Compare entries with the template to find \(k\). A1: Correctly state \(k = -n\) and show all other entries match.

First, we find the complementary function (CF) by solving the auxiliary equation:
\[m^2 + 2m + 5 = 0\]
Using the quadratic formula:
\[m = \frac{-2 \pm \sqrt{4 - 20}}{2} = -1 \pm 2i\]
This gives the CF:
\[y_{CF} = e^{-x}(A\cos 2x + B\sin 2x)\]

Next, we find the particular integral (PI). Since the right-hand side is \(10e^{-x}\cos x\) and \(-1 \pm i\) is not a root of the auxiliary equation, we propose a PI of the form:
\[y_{PI} = e^{-x}(C\cos x + D\sin x)\]
We differentiate \(y_{PI}\):
\[y_{PI}' = e^{-x}\left((D - C)\cos x - (C + D)\sin x\right)\]
\[y_{PI}'' = e^{-x}\left(-2D\cos x + 2C\sin x\right)\]

Substituting these derivatives into the original differential equation:
\[y_{PI}'' + 2y_{PI}' + 5y_{PI} = e^{-x}\left[(-2D + 2(D-C) + 5C)\cos x + (2C - 2(C+D) + 5D)\sin x\right] = 10e^{-x}\cos x\]
\[e^{-x}(3C\cos x + 3D\sin x) = 10e^{-x}\cos x\]
Comparing coefficients:
\[3C = 10 \implies C = \frac{10}{3}\]
\[3D = 0 \implies D = 0\]
Thus, the PI is:
\[y_{PI} = \frac{10}{3}e^{-x}\cos x\]

Combining the CF and the PI, we obtain the general solution:
\[y = e^{-x}(A\cos 2x + B\sin 2x) + \frac{10}{3}e^{-x}\cos x\]

M1: For writing down and solving the auxiliary equation.
A1: For correct roots of the auxiliary equation.
A1: For correct complementary function.
M1: For proposing a particular integral of the form \(e^{-x}(C\cos x + D\sin x)\).
M1: For differentiating the PI and substituting it into the differential equation.
A1: For obtaining correct values of \(C = \frac{10}{3}\) and \(D = 0\).
A1: For the final general solution combining the CF and PI.

Since the matrix \(\mathbf{A}\) is upper triangular, its eigenvalues are simply the diagonal elements:
\[\lambda_1 = 2, \quad \lambda_2 = 3, \quad \lambda_3 = 4\]

To find the eigenvector corresponding to \(\lambda_1 = 2\), we solve \((\mathbf{A} - 2\mathbf{I})\mathbf{v}_1 = \mathbf{0}\):
\[\begin{pmatrix} 0 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\]
From the last row, \(2z = 0 \implies z = 0\). From the first row, \(y + z = 0 \implies y = 0\). \(x\) can be any non-zero scalar. Setting \(x = 1\), we get:
\[\mathbf{v}_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\]

To find the eigenvector corresponding to \(\lambda_2 = 3\), we solve \((\mathbf{A} - 3\mathbf{I})\mathbf{v}_2 = \mathbf{0}\):
\[\begin{pmatrix} -1 & 1 & 1 \\ 0 & 0 & 2 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\]
From the second and third rows, \(2z = 0 \implies z = 0\). From the first row, \(-x + y + z = 0 \implies x = y\). Setting \(x = 1\), we get:
\[\mathbf{v}_2 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}\]

To find the eigenvector corresponding to \(\lambda_3 = 4\), we solve \((\mathbf{A} - 4\mathbf{I})\mathbf{v}_3 = \mathbf{0}\):
\[\begin{pmatrix} -2 & 1 & 1 \\ 0 & -1 & 2 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\]
From the second row, \(-y + 2z = 0 \implies y = 2z\). From the first row, \(-2x + y + z = 0 \implies -2x + 2z + z = 0 \implies 2x = 3z\). Setting \(z = 2\) yields \(y = 4\) and \(x = 3\). Thus:
\[\mathbf{v}_3 = \begin{pmatrix} 3 \\ 4 \\ 2 \end{pmatrix}\]

B1: Identify correct eigenvalues \(\lambda = 2, 3, 4\) from the diagonal.
M1: Setup equations for finding the eigenvector for \(\lambda = 2\).
A1: Correctly find \(\mathbf{v}_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\) (or scalar multiple).
M1: Setup equations for finding the eigenvector for \(\lambda = 3\).
A1: Correctly find \(\mathbf{v}_2 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}\) (or scalar multiple).
M1: Setup equations for finding the eigenvector for \(\lambda = 4\).
A1: Correctly find \(\mathbf{v}_3 = \begin{pmatrix} 3 \\ 4 \\ 2 \end{pmatrix}\) (or scalar multiple).

(a) We write the integrand as \(\sin^n x = \sin^{n-1} x \sin x\) and apply integration by parts:
Let \(u = \sin^{n-1} x \implies du = (n-1)\sin^{n-2} x \cos x \, dx\)
Let \(dv = \sin x \, dx \implies v = -\cos x\)
Applying the formula \(\int u \, dv = uv - \int v \, du\):
\[I_n = \left[-\sin^{n-1} x \cos x\right]_{0}^{\pi/2} + (n-1)\int_{0}^{\pi/2} \sin^{n-2} x \cos^2 x \, dx\]
Since \(\cos(\pi/2) = 0\) and \(\sin(0) = 0\), the boundary term is 0.
Substituting \(\cos^2 x = 1 - \sin^2 x\):
\[I_n = (n-1)\int_{0}^{\pi/2} \sin^{n-2} x (1 - \sin^2 x) \, dx = (n-1)\left(I_{n-2} - I_n\right)\]
\[I_n = (n-1)I_{n-2} - (n-1)I_n \implies n I_n = (n-1)I_{n-2}\]
Thus, we have:
\[I_n = \frac{n-1}{n} I_{n-2}\]

(b) Let the given integral be \(J = \int_{0}^{\pi/2} \sin^5 x \cos^2 x \, dx\).
Using \(\cos^2 x = 1 - \sin^2 x\):
\[J = \int_{0}^{\pi/2} \sin^5 x (1 - \sin^2 x) \, dx = \int_{0}^{\pi/2} (\sin^5 x - \sin^7 x) \, dx = I_5 - I_7\]
From the reduction formula, \(I_7 = \frac{6}{7} I_5\), so:
\[J = I_5 - \frac{6}{7} I_5 = \frac{1}{7} I_5\]
Now, we find \(I_5\):
\[I_5 = \frac{4}{5} I_3\]
\[I_3 = \frac{2}{3} I_1\]
Since \(I_1 = \int_{0}^{\pi/2} \sin x \, dx = \left[-\cos x\right]_{0}^{\pi/2} = 1\), we have:
\[I_3 = \frac{2}{3} \cdot 1 = \frac{2}{3}\]
\[I_5 = \frac{4}{5} \cdot \frac{2}{3} = \frac{8}{15}\]
Thus, the value of the integral is:
\[J = \frac{1}{7} \cdot \frac{8}{15} = \frac{8}{105}\]

(a)
M1: For expressing \(\sin^n x\) as \(\sin^{n-1} x \sin x\) and choosing parts.
A1: For correct integration by parts resulting in \((n-1)\int_{0}^{\pi/2} \sin^{n-2} x \cos^2 x \, dx\).
M1: For using \(\cos^2 x = 1 - \sin^2 x\) to set up a relation in terms of \(I_n\) and \(I_{n-2}\).
A1: For successfully showing the given reduction formula.

(b)
M1: For expressing the target integral in terms of \(I_5\) and \(I_7\).
M1: For expressing \(I_7\) in terms of \(I_5\).
M1: For computing \(I_5\) using the reduction formula iteratively down to \(I_1\).
A1: For finding the correct value of \(I_5 = \frac{8}{15}\).
A1: For the correct final answer of \(\frac{8}{105}\).

To find the roots of \(z^7 = -1\), we write \(-1\) in exponential form:
\[z^7 = e^{i(\pi + 2k\pi)} \quad \text{for } k \in \mathbb{Z}\]
Applying de Moivre's theorem:
\[z_k = e^{i\frac{(2k+1)\pi}{7}}\]
To satisfy \(-\pi < \theta \leq \pi\), we select \(k = -3, -2, -1, 0, 1, 2, 3\). This yields the seven roots:
\[z_{-3} = e^{-5i\pi/7}, \quad z_{-2} = e^{-3i\pi/7}, \quad z_{-1} = e^{-i\pi/7}, \quad z_0 = e^{i\pi/7}, \quad z_1 = e^{3i\pi/7}, \quad z_2 = e^{5i\pi/7}, \quad z_3 = e^{i\pi} = -1\]
We can write these conjugate pairs as:
\[e^{\pm i\pi/7}, \quad e^{\pm 3i\pi/7}, \quad e^{\pm 5i\pi/7}, \quad -1\]

Now, for the polynomial equation \(z^7 + 1 = 0\), the sum of the roots is equal to the coefficient of \(z^6\), which is 0:
\[\sum_{k=-3}^3 z_k = 0\]
\[\left(e^{-5i\pi/7} + e^{5i\pi/7}\right) + \left(e^{-3i\pi/7} + e^{3i\pi/7}\right) + \left(e^{-i\pi/7} + e^{i\pi/7}\right) - 1 = 0\]
Using Euler's relation \(e^{i\theta} + e^{-i\theta} = 2\cos\theta\):
\[2\cos\left(\frac{5\pi}{7}\right) + 2\cos\left(\frac{3\pi}{7}\right) + 2\cos\left(\frac{\pi}{7}\right) - 1 = 0\]
Dividing both sides by 2 and moving 1 to the other side:
\[\cos\left(\frac{\pi}{7}\right) + \cos\left(\frac{3\pi}{7}\right) + \cos\left(\frac{5\pi}{7}\right) = \frac{1}{2}\]
which is the required proof.

M1: For expressing \(-1\) as \(e^{i(\pi + 2k\pi)}\).
M1: For applying de Moivre's theorem to find the roots in the form \(e^{i\theta}\).
A1: For listing the seven correct roots within the specified range.
M1: For stating that the sum of the roots is equal to 0.
M1: For grouping conjugate pairs into terms of the form \(2\cos\theta\).
A1: For completing the proof clearly and logically.

We express the hyperbolic functions in terms of exponential functions:
\[\cosh x = \frac{e^x + e^{-x}}{2}, \quad \sinh x = \frac{e^x - e^{-x}}{2}\]
Substituting these definitions into the given equation:
\[4\left(\frac{e^x + e^{-x}}{2}\right) - 2\left(\frac{e^x - e^{-x}}{2}\right) = 5\]
\[2(e^x + e^{-x}) - (e^x - e^{-x}) = 5\]
\[2e^x + 2e^{-x} - e^x + e^{-x} = 5\]
\[e^x + 3e^{-x} = 5\]

Multiply the entire equation by \(e^x\):
\[(e^x)^2 + 3 = 5e^x \implies (e^x)^2 - 5e^x + 3 = 0\]
Let \(u = e^x\). We solve the quadratic equation:
\[u^2 - 5u + 3 = 0\]
Using the quadratic formula:
\[u = \frac{5 \pm \sqrt{25 - 12}}{2} = \frac{5 \pm \sqrt{13}}{2}\]
Since both roots \(\frac{5 + \sqrt{13}}{2}\) and \(\frac{5 - \sqrt{13}}{2}\) are positive (note that \(\sqrt{13} \approx 3.61 < 5\)), we take the natural logarithm to find the solutions for \(x\):
\[x = \ln\left(\frac{5 \pm \sqrt{13}}{2}\right)\]

M1: For writing \(\cosh x\) and \(\sinh x\) in terms of exponentials.
A1: For a correct simplified equation in \(e^x\) and \(e^{-x}\).
M1: For setting up a quadratic equation in \(e^x\).
A1: For the correct quadratic equation, e.g., \((e^x)^2 - 5e^x + 3 = 0\).
M1: For solving the quadratic equation to find the values of \(e^x\).
A1: For verifying that both roots are positive.
A1: For the final answers in exact logarithmic form.

We can find the Maclaurin's series by repeated differentiation.
First, evaluate \(y\) at \(x = 0\):
\[y(0) = \ln(1 + \sin 0) = \ln(1) = 0\]

Differentiate with respect to \(x\):
\[y' = \frac{\cos x}{1 + \sin x}\]
At \(x = 0\):
\[y'(0) = \frac{\cos 0}{1 + \sin 0} = 1\]

We can write \(y'(1 + \sin x) = \cos x\). Differentiating both sides implicitly:
\[y''(1 + \sin x) + y'\cos x = -\sin x\]
At \(x = 0\):
\[y''(0)(1 + 0) + (1)(1) = 0 \implies y''(0) = -1\]

Differentiating again implicitly:
\[y'''(1 + \sin x) + y''\cos x + y''\cos x - y'\sin x = -\cos x\]
\[y'''(1 + \sin x) + 2y''\cos x - y'\sin x = -\cos x\]
At \(x = 0\):
\[y'''(0)(1) + 2(-1)(1) - (1)(0) = -1 \implies y'''(0) - 2 = -1 \implies y'''(0) = 1\]

Now, substitute the values of the derivatives into the Maclaurin series formula:
\[y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \dots\]
\[y = 0 + 1(x) + \frac{-1}{2}x^2 + \frac{1}{6}x^3 + \dots = x - \frac{1}{2}x^2 + \frac{1}{6}x^3\]

B1: For finding \(y(0) = 0\).
M1: For differentiating \(y\) to find \(y'\).
A1: For finding \(y'(0) = 1\).
M1: For differentiating again to find \(y''\) (using quotient rule, product rule or implicit differentiation).
A1: For finding \(y''(0) = -1\).
M1: For differentiating a third time to find \(y'''\).
A1: For finding \(y'''(0) = 1\).
A1: For constructing the correct final Maclaurin's series up to the term in \(x^3\).

The formula for the arc length \(S\) of a parametric curve is given by:
\[S = \int_{t_1}^{t_2} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\]
First, we compute the derivatives of \(x\) and \(y\) with respect to \(t\):
\[\frac{dx}{dt} = 2t\]
\[\frac{dy}{dt} = t^2 - 1\]

Next, we calculate the term under the square root:
\[\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = (2t)^2 + (t^2 - 1)^2\]
\[= 4t^2 + t^4 - 2t^2 + 1 = t^4 + 2t^2 + 1\]
Notice that this expression is a perfect square:
\[t^4 + 2t^2 + 1 = (t^2 + 1)^2\]

Thus, the integrand is:
\[\sqrt{(t^2 + 1)^2} = t^2 + 1 \quad \text{for } 0 \leq t \leq 2\]

Now, we evaluate the definite integral:
\[S = \int_{0}^{2} (t^2 + 1) \, dt = \left[ \frac{1}{3}t^3 + t \right]_{0}^{2}\]
\[S = \left( \frac{1}{3}(2)^3 + 2 \right) - \left( 0 \right) = \frac{8}{3} + 2 = \frac{14}{3}\]

M1: For differentiating both \(x\) and \(y\) with respect to \(t\).
A1: For correct derivatives \(\frac{dx}{dt} = 2t\) and \(\frac{dy}{dt} = t^2 - 1\).
M1: For setting up the expression \(\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2\).
A1: For simplifying the expression to the perfect square \((t^2 + 1)^2\).
M1: For integrating the simplified function \(t^2 + 1\).
A1: For substituting the limits 0 and 2 correctly.
A1: For the final exact answer of \(\frac{14}{3}\).

Given the substitution \(u = y^2\), we differentiate with respect to \(x\) using the chain rule:
\[\frac{du}{dx} = 2y \frac{dy}{dx}\]

We substitute \(u\) and \(\frac{du}{dx}\) into the given differential equation:
\[\frac{du}{dx} + \frac{2}{x} u = x^3\]
This is a first-order linear differential equation of the form \(\frac{du}{dx} + P(x)u = Q(x)\), where \(P(x) = \frac{2}{x}\) and \(Q(x) = x^3\).

We find the integrating factor, \(I(x)\):
\[I(x) = e^{\int \frac{2}{x} \, dx} = e^{2\ln x} = e^{\ln(x^2)} = x^2\]

Multiplying both sides of the differential equation by \(x^2\):
\[x^2 \frac{du}{dx} + 2xu = x^5\]
\[\frac{d}{dx} (x^2 u) = x^5\]

Now, integrate both sides with respect to \(x\):
\[x^2 u = \int x^5 \, dx\]
\[x^2 u = \frac{1}{6}x^6 + C\]
\[u = \frac{1}{6}x^4 + C x^{-2}\]

Finally, substitute back \(u = y^2\) to get the general solution:
\[y^2 = \frac{1}{6}x^4 + C x^{-2}\]

M1: For applying the substitution \(u = y^2\) and finding the derivative \(\frac{du}{dx} = 2y\frac{dy}{dx}\).
A1: For transforming the differential equation into a first-order linear form in terms of \(u\).
M1: For finding the correct integrating factor, \(I(x) = x^2\).
M1: For multiplying by the integrating factor and expressing the left side as a single derivative.
M1: For integrating both sides with respect to \(x\).
A1: For getting the correct integration term \(\frac{1}{6}x^6 + C\) (including the constant of integration).
A1: For substituting back \(u = y^2\) and obtaining the correct final general solution.