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

(a) We express the terms using partial fractions: \(\frac{3}{(3r-1)(3r+2)} = \frac{1}{3r-1} - \frac{1}{3r+2}\).

Summing from \(r=1\) to \(n\):
\(S_n = \sum_{r=1}^n \left( \frac{1}{3r-1} - \frac{1}{3r+2} \right) = \left( \frac{1}{2} - \frac{1}{5} \right) + \left( \frac{1}{5} - \frac{1}{8} \right) + \dots + \left( \frac{1}{3n-1} - \frac{1}{3n+2} \right)\).

This telescopes to:
\(S_n = \frac{1}{2} - \frac{1}{3n+2} = \frac{3n+2-2}{2(3n+2)} = \frac{3n}{2(3n+2)}\).

(b) The required sum can be written as:
\(\sum_{r=n+1}^{3n} \frac{1}{(3r-1)(3r+2)} = \frac{1}{3} \sum_{r=n+1}^{3n} \frac{3}{(3r-1)(3r+2)} = \frac{1}{3} (S_{3n} - S_n)\).

Substituting our result from (a):
\(\frac{1}{3} \left( \frac{3(3n)}{2(3(3n)+2)} - \frac{3n}{2(3n+2)} \right) = \frac{1}{3} \left( \frac{9n}{2(9n+2)} - \frac{3n}{2(3n+2)} \right) = \frac{n}{2} \left( \frac{3}{9n+2} - \frac{1}{3n+2} \right)\).

Combining the fractions:
\(\frac{n}{2} \left( \frac{3(3n+2) - (9n+2)}{(9n+2)(3n+2)} \right) = \frac{n}{2} \left( \frac{9n+6-9n-2}{(9n+2)(3n+2)} \right) = \frac{n}{2} \left( \frac{4}{(9n+2)(3n+2)} \right) = \frac{2n}{(9n+2)(3n+2)}\).

(c) Expanding the summand:
\((3r-1)(3r+2) = 9r^2 + 3r - 2\).

Now, applying the summation formulae:
\(\sum_{r=1}^n (9r^2 + 3r - 2) = 9 \sum_{r=1}^n r^2 + 3 \sum_{r=1}^n r - 2 \sum_{r=1}^n 1\)
\(= 9 \cdot \frac{n(n+1)(2n+1)}{6} + 3 \cdot \frac{n(n+1)}{2} - 2n\)
\(= \frac{3}{2}n(n+1)(2n+1) + \frac{3}{2}n(n+1) - 2n\)
\(= n \left[ \frac{3}{2}(n+1)(2n+1) + \frac{3}{2}(n+1) - 2 \right]\)
\(= \frac{n}{2} \left[ 3(2n^2 + 3n + 1) + 3(n+1) - 4 \right]\)
\(= \frac{n}{2} [ 6n^2 + 9n + 3 + 3n + 3 - 4 ]\)
\(= \frac{n}{2} [ 6n^2 + 12n + 2 ] = n(3n^2 + 6n + 1)\).

(a) M1 for attempting to use partial fractions or a difference form. A1 for correct difference expression. M1 for writing out terms to show cancelling. A1 for correct final simplified expression.

(b) M1 for expressing the sum as \(\frac{1}{3}(S_{3n} - S_n)\). M1 for substituting the formula from part (a) correctly. A1 for the correct simplified expression in terms of \(n\).

(c) M1 for expanding the algebraic term and splitting the sum. M1 for substituting the standard formulae for sum of squares and sum of integers. M1 for algebraic factorization. A1 for completing the proof to show \(n(3n^2 + 6n + 1)\).

(a) For \(\mathbf{M}\) to be singular, its determinant must be zero:
\(\det(\mathbf{M}) = 1(-2) - 2(k) = -2 - 2k = 0 \implies k = -1\).

(b) For \(k = 2\), \(\mathbf{M} = \begin{pmatrix} 1 & 2 \\ 2 & -2 \end{pmatrix}\).
Under the transformation, a point \((x, mx)\) on the line \(y = mx\) is mapped to \((x', y')\):
\(\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 2 & -2 \end{pmatrix} \begin{pmatrix} x \\ mx \end{pmatrix} = \begin{pmatrix} (1+2m)x \\ (2-2m)x \end{pmatrix}\).

Since the line is invariant, we must have \(y' = mx'\):
\((2-2m)x = m(1+2m)x\).

Assuming \(x \neq 0\), we obtain:
\(2 - 2m = m + 2m^2 \implies 2m^2 + 3m - 2 = 0\).

Factorising this quadratic equation:
\((2m-1)(m+2) = 0 \implies m = \frac{1}{2} \text{ or } m = -2\).

(c) For \(k = 2\), \(\det(\mathbf{M}) = 1(-2) - 2(2) = -6\).
\(\mathbf{M}^{-1} = -\frac{1}{6} \begin{pmatrix} -2 & -2 \\ -2 & 1 \end{pmatrix} = \frac{1}{6} \begin{pmatrix} 2 & 2 \\ 2 & -1 \end{pmatrix}\).

To find the coordinates of \(P\), we solve \(\mathbf{M}\mathbf{x}_P = \begin{pmatrix} -4 \\ 10 \end{pmatrix}\):
\(\mathbf{x}_P = \mathbf{M}^{-1} \begin{pmatrix} -4 \\ 10 \end{pmatrix} = \frac{1}{6} \begin{pmatrix} 2 & 2 \\ 2 & -1 \end{pmatrix} \begin{pmatrix} -4 \\ 10 \end{pmatrix} = \frac{1}{6} \begin{pmatrix} -8+20 \\ -8-10 \end{pmatrix} = \frac{1}{6} \begin{pmatrix} 12 \\ -18 \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \end{pmatrix}\).

So, the coordinates of \(P\) are \((2, -3)\).

(a) M1 for setting determinant to 0. A1 for \(k = -1\).

(b) M1 for setting up the matrix equation with \((x, mx)^T\). M1 for deriving the quadratic equation in \(m\). A1 for correct quadratic \(2m^2 + 3m - 2 = 0\). A1 for \(m = 1/2\), A1 for \(m = -2\).

(c) M1 for finding the determinant of \\mathbf{M} when \(k = 2\). A1 for correct inverse matrix. M1 for multiplying \(\mathbf{M}^{-1}\) by the image vector. A1 for obtaining the coordinates \((2, -3)\).

(a) Let \(f(n) = 5^{2n-1} + 2^{2n-1}\).

Step 1: Base case \(n=1\):
\(f(1) = 5^1 + 2^1 = 7\), which is divisible by 7. Thus, the statement is true for \(n=1\).

Step 2: Inductive hypothesis:
Assume the statement is true for \(n=k\), where \(k\) is a positive integer. That is, \(f(k) = 5^{2k-1} + 2^{2k-1} = 7M\) for some integer \(M\).

Step 3: Inductive step:
We examine \(f(k+1)\):
\(f(k+1) = 5^{2(k+1)-1} + 2^{2(k+1)-1} = 5^{2k+1} + 2^{2k+1}\)
\(= 5^2 \cdot 5^{2k-1} + 2^2 \cdot 2^{2k-1}\)
\(= 25 \cdot 5^{2k-1} + 4 \cdot 2^{2k-1}\)
\(= 21 \cdot 5^{2k-1} + 4 \cdot 5^{2k-1} + 4 \cdot 2^{2k-1}\)
\(= 21 \cdot 5^{2k-1} + 4(5^{2k-1} + 2^{2k-1})\)
\(= 21 \cdot 5^{2k-1} + 4(7M)\)
\(= 7(3 \cdot 5^{2k-1} + 4M)\).

Since \(3 \cdot 5^{2k-1} + 4M\) is an integer, \(f(k+1)\) is divisible by 7. Therefore, if the statement is true for \(n=k\), it is also true for \(n=k+1\).

By mathematical induction, the statement is true for all positive integers \(n\).

(b) Let \(\mathbf{A} = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}\).

Step 1: Base case \(n=1\):
LHS = \(\mathbf{A}^1 = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}\).
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 \(n=1\).

Step 2: Inductive hypothesis:
Assume the statement is true for \(n=k\):
\(\mathbf{A}^k = \begin{pmatrix} 2k+1 & -4k \\ k & 1-2k \end{pmatrix}\).

Step 3: Inductive step:
We evaluate \(\mathbf{A}^{k+1} = \mathbf{A}^k \mathbf{A}\):
\(\mathbf{A}^{k+1} = \begin{pmatrix} 2k+1 & -4k \\ k & 1-2k \end{pmatrix} \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}\)
\(= \begin{pmatrix} 3(2k+1) - 4k & -4(2k+1) + 4k \\ 3k + 1 - 2k & -4k - (1-2k) \end{pmatrix}\)
\(= \begin{pmatrix} 6k+3-4k & -8k-4+4k \\ k+1 & -4k-1+2k \end{pmatrix}\)
\(= \begin{pmatrix} 2k+3 & -4k-4 \\ k+1 & -2k-1 \end{pmatrix}\)
\(= \begin{pmatrix} 2(k+1)+1 & -4(k+1) \\ k+1 & 1-2(k+1) \end{pmatrix}\).

This is the required form for \(n=k+1\). By mathematical induction, the statement is true for all positive integers \(n\).

(a) B1 for testing the base case \(n=1\) correctly. M1 for assuming true for \(n=k\). M1 for expressing \(f(k+1)\) in terms of \(5^{2k-1}\) and \(2^{2k-1}\). A1 for obtaining an expression clearly divisible by 7. A1 for a complete and logical conclusion.

(b) B1 for checking the base case \(n=1\). M1 for inductive hypothesis and setting up \(\mathbf{A}^{k+1} = \mathbf{A}^k \mathbf{A}\). M1 for correct matrix multiplication. A1 for simplifying the matrix elements into the required form. A1 for a complete and logical induction conclusion.

(a) From the coefficients of the cubic equation:
\(\sum \alpha = 3\)
\(\sum \alpha\beta = 5\)
\(\alpha\beta\gamma = -2\)

Using the identity:
\(\alpha^2 + \beta^2 + \gamma^2 = (\sum \alpha)^2 - 2\sum \alpha\beta\)
\(\alpha^2 + \beta^2 + \gamma^2 = 3^2 - 2(5) = 9 - 10 = -1\).

(b) Since \(\alpha, \beta, \gamma\) are roots of \(x^3 - 3x^2 + 5x + 2 = 0\), they satisfy the equation:
\(\alpha^3 - 3\alpha^2 + 5\alpha + 2 = 0\)
\(\beta^3 - 3\beta^2 + 5\beta + 2 = 0\)
\(\gamma^3 - 3\gamma^2 + 5\gamma + 2 = 0\)

Summing these three equations yields:
\(\sum \alpha^3 - 3\sum \alpha^2 + 5\sum \alpha + 6 = 0\).

Substituting \(\sum \alpha^2 = -1\) and \(\sum \alpha = 3\):
\(\sum \alpha^3 - 3(-1) + 5(3) + 6 = 0\)
\(\sum \alpha^3 + 3 + 15 + 6 = 0\)
\(\sum \alpha^3 + 24 = 0 \implies \sum \alpha^3 = -24\).

(c) Let \(y = x^2\), so \(x = \sqrt{y}\).
Substituting this into the cubic equation:
\((\sqrt{y})^3 - 3y + 5\sqrt{y} + 2 = 0\)
\(y\sqrt{y} + 5\sqrt{y} = 3y - 2\)
\(\sqrt{y}(y + 5) = 3y - 2\).

Squaring both sides:
\(y(y+5)^2 = (3y-2)^2\)
\(y(y^2 + 10y + 25) = 9y^2 - 12y + 4\)
\(y^3 + 10y^2 + 25y = 9y^2 - 12y + 4\)
\(y^3 + y^2 + 37y - 4 = 0\).

So, the required cubic equation is \(y^3 + y^2 + 37y - 4 = 0\).

(a) B1 for stating the values of the sum of roots and sum of products of roots. M1 for the identity for \(\sum \alpha^2\). A1 for obtaining \(-1\).

(b) M1 for using the fact that \(\alpha, \beta, \gamma\) satisfy the cubic equation. M1 for setting up the sum equation. A1 for substituting known values. A1 for completing the proof to show \(-24\).

(c) M1 for substituting \(x = \sqrt{y}\) into the equation. M1 for separating terms into rational and irrational parts. M1 for squaring and expanding both sides correctly. A1 for the correct final cubic equation \(y^3 + y^2 + 37y - 4 = 0\) (or equivalent).

(a) The direction vectors of \(l_1\) and \(l_2\) are \(\mathbf{d}_1 = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}\) and \(\mathbf{d}_2 = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\) respectively.

We find a vector perpendicular to both directions using the cross product:
\(\mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2 = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} \times \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1(1) - 0(1) \\ 0(0) - 2(1) \\ 2(1) - 1(0) \end{pmatrix} = \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}\).

Let \(A(1, 2, 1)\) be a point on \(l_1\) and \(B(4, 1, 3)\) be a point on \(l_2\).
\(\vec{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} 4 - 1 \\ 1 - 2 \\ 3 - 1 \end{pmatrix} = \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix}\).

The shortest distance \(d\) is the magnitude of the projection of \(\vec{AB}\) onto the common normal \(\mathbf{n}\):
\(d = \frac{|\vec{AB} \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{\left| \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix} \right|}{\sqrt{1^2 + (-2)^2 + 2^2}} = \frac{|3(1) + (-1)(-2) + 2(2)|}{\sqrt{1 + 4 + 4}} = \frac{|3 + 2 + 4|}{3} = \frac{9}{3} = 3\).

(b) Since \(\Pi\) contains \(l_1\) and is parallel to \(l_2\), its normal vector is \(\mathbf{n} = \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}\).

Using the point \(A(1, 2, 1)\) which lies on \(l_1\) (and hence on \(\Pi\)):
\(1(x - 1) - 2(y - 2) + 2(z - 1) = 0\)
\(x - 1 - 2y + 4 + 2z - 2 = 0\)
\(x - 2y + 2z + 1 = 0 \implies x - 2y + 2z = -1\).

(a) M1 for attempting cross product of \(\mathbf{d}_1\) and \(\mathbf{d}_2\). A1 for obtaining \(\mathbf{n} = (1, -2, 2)^T\). M1 for finding the vector \(\vec{AB}\). M1 for using the shortest distance formula projection. A1 for calculating the dot product as 9. A1 for obtaining final distance \(3\).

(b) M1 for using the normal vector found in part (a). M1 for using the coordinates of a point on \(l_1\) in the plane equation. A1 for obtaining the equation in the correct form \(x - 2y + 2z = -1\) (or any integer multiple).

(a) The curve is a cardioid, symmetric about the initial line \(\theta = 0\), with a cusp at the pole. Key coordinates are:
- \(\theta = 0 \implies r = 4\)
- \(\theta = \frac{\pi}{2} \implies r = 2\)
- \(\theta = \pi \implies r = 0\)
- \(\theta = \frac{3\pi}{2} \implies r = 2\)
- \(\theta = 2\pi \implies r = 4\)

(b) The area \(A\) is given by:
\(A = \frac{1}{2} \int_0^{2\pi} r^2 \, d\theta = \frac{1}{2} \int_0^{2\pi} 4(1 + \cos\theta)^2 \, d\theta = 2 \int_0^{2\pi} (1 + 2\cos\theta + \cos^2\theta) \, d\theta\).

Using the identity \(\cos^2\theta = \frac{1 + \cos 2\theta}{2}\):
\(A = 2 \int_0^{2\pi} \left( 1 + 2\cos\theta + \frac{1}{2} + \frac{1}{2}\cos 2\theta \right) \, d\theta = 2 \int_0^{2\pi} \left( \frac{3}{2} + 2\cos\theta + \frac{1}{2}\cos 2\theta \right) \, d\theta\)
\(= 2 \left[ \frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta \right]_0^{2\pi}\)
\(= 2 \left( \frac{3}{2}(2\pi) - 0 \right) = 6\pi\).

(c) We represent the \(y\)-coordinate as:
\(y = r\sin\theta = 2(1 + \cos\theta)\sin\theta = 2\sin\theta + 2\sin\theta\cos\theta = 2\sin\theta + \sin 2\theta\).

For a tangent parallel to the initial line, we find \(\frac{dy}{d\theta} = 0\):
\(\frac{dy}{d\theta} = 2\cos\theta + 2\cos 2\theta = 0 \implies \cos\theta + (2\cos^2\theta - 1) = 0\)
\(2\cos^2\theta + \cos\theta - 1 = 0 \implies (2\cos\theta - 1)(\cos\theta + 1) = 0\).

This gives \(\cos\theta = \frac{1}{2}\) or \(\cos\theta = -1\).

Since we exclude the pole (where \(r = 0 \implies \theta = \pi\)), we reject \(\cos\theta = -1\).
Thus, \(\cos\theta = \frac{1}{2} \implies \theta = \frac{\pi}{3}\) or \(\theta = \frac{5\pi}{3}\).

At \(\theta = \frac{\pi}{3}\), \(r = 2(1 + \cos\frac{\pi}{3}) = 3\).
At \(\theta = \frac{5\pi}{3}\), \(r = 2(1 + \cos\frac{5
\pi}{3}) = 3\).

The required polar coordinates are \((3, \frac{\pi}{3})\) and \((3, \frac{5\pi}{3})\).

(a) B1 for correct shape (cardioid with cusp at origin). B1 for showing key intercepts \(r=4\) at \(\theta=0\), \(r=2\) at \(\theta = \pm \pi/2\).

(b) M1 for correct area integral formula. M1 for expanding and using double angle identity. A1 for correct integrated expression. A1 for final area \(6\pi\).

(c) M1 for using \(y = r\sin\theta\) to express \(y\) in terms of \(\theta\). M1 for differentiating and setting \(\frac{dy}{d\theta} = 0\). A1 for finding \(\cos\theta = 1/2\). A1 for correct polar coordinates \((3, \frac{\pi}{3})\) and \((3, \frac{5\pi}{3})\).

(a) Express the curve by algebraic division:
\(y = \frac{x(x-2) - 2x + 8}{x-2} = \frac{x(x-2) - 2(x-2) + 4}{x-2} = x - 2 + \frac{4}{x - 2}\).

- As \(x \to 2\), \(y \to \pm\infty\), so the vertical asymptote is \(x = 2\).
- As \(x \to \pm\infty\), \(y \to x - 2\), so the oblique asymptote is \(y = x - 2\).

(b) Differentiating \(y = x - 2 + 4(x-2)^{-1}\):
\(\frac{dy}{dx} = 1 - \frac{4}{(x-2)^2}\).

For stationary points, set \(\frac{dy}{dx} = 0\):
\(1 - \frac{4}{(x-2)^2} = 0 \implies (x-2)^2 = 4 \implies x-2 = \pm 2\).

This gives:
- \(x = 4 \implies y = 4 - 2 + \frac{4}{2} = 4\).
- \(x = 0 \implies y = 0 - 2 + \frac{4}{-2} = -4\).

So the stationary points are \((4, 4)\) and \((0, -4)\).

(c) To sketch \(C\):
- Plot the asymptotes \(x = 2\) and \(y = x - 2\).
- Plot the stationary points: \((4, 4)\) is a local minimum, and \((0, -4)\) is a local maximum.
- The \(y\)-axis intercept is at \((0, -4)\) (which is also the local maximum).
- No \(x\)-axis intercepts exist since the numerator \(x^2 - 4x + 8 = 0\) has no real solutions (discriminant \(16 - 32 = -16 < 0\)).
- Sketch the two separate branches of the curve asymptotic to the lines.

(a) M1 for attempting algebraic division. A1 for vertical asymptote \(x = 2\). A1 for oblique asymptote \(y = x - 2\).

(b) M1 for differentiating the function. A1 for setting derivative to 0 and solving for \(x\). A1 for \(x=0\) and \(x=4\). A1 for correct coordinates \((4, 4)\) and \((0, -4)\).

(c) B1 for drawing the asymptotes correctly. B1 for plotting the stationary points and the \(y\)-intercept. B2 for sketching both branches approaching the asymptotes correctly.

To find the eigenvalues, solve the characteristic equation \(\det(A - \lambda I) = 0\):
\[\begin{vmatrix} 2-\lambda & 2 & -1 \\ 1 & 3-\lambda & -1 \\ -1 & -2 & 2-\lambda \end{vmatrix} = 0.\]
Expanding the determinant along the first row:
\[(2-\lambda)[(3-\lambda)(2-\lambda) - 2] - 2[1(2-\lambda) - 1] - 1[-2 - (-(3-\lambda))] = 0.\]
\[(2-\lambda)(\lambda^2 - 5\lambda + 4) - 2(1-\lambda) - (1-\lambda) = 0.\]
\[(2-\lambda)(\lambda-1)(\lambda-4) + 3(\lambda-1) = 0.\]
\[(\lambda-1)[(2-\lambda)(\lambda-4) + 3] = 0.\]
\[(\lambda-1)(-\lambda^2 + 6\lambda - 5) = 0 \implies -(\lambda-1)^2(\lambda-5) = 0.\]
Thus, the eigenvalues are \(\lambda = 1\) (repeated) and \(\lambda = 5\).

For \(\lambda = 5\):
\[(A - 5I)\mathbf{v} = \begin{pmatrix} -3 & 2 & -1 \\ 1 & -2 & -1 \\ -1 & -2 & -3 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}.\]
From the second row, \(x - 2y - z = 0 \implies x = 2y + z\).
Substitute into the first row: \(-3(2y+z) + 2y - z = 0 \implies -4y - 4z = 0 \implies y = -z\).
This yields \(x = -z\).
Thus, an eigenvector is \(\mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}\).

For \(\lambda = 1\):
\[(A - I)\mathbf{v} = \begin{pmatrix} 1 & 2 & -1 \\ 1 & 2 & -1 \\ -1 & -2 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}.\]
This gives the single equation \(x + 2y - z = 0 \implies z = x + 2y\).
Choosing \(y = 0\) and \(x = 1\) gives \(\mathbf{v}_2 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}\).
Choosing \(x = 0\) and \(y = 1\) gives \(\mathbf{v}_3 = \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}\).

Since we have three linearly independent eigenvectors, we can form the matrices \(P\) and \(D\):
\[P = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 2 & -1 \end{pmatrix}, \quad D = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 5 \end{pmatrix}.\]

M1: Set up characteristic equation and attempt expansion.
A1: Correct characteristic equation/eigenvalues \(\lambda = 1, 1, 5\).
M1: Substitute \(\lambda = 5\) to find corresponding eigenvector.
A1: Correct eigenvector \(\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}\).
M1: Substitute \(\lambda = 1\) and write equation of plane of eigenvectors.
A1: Two linearly independent eigenvectors for \(\lambda = 1\).
B1: Write correct matrix \(P\) (columns must match the order in \(D\)).
B1: Write correct diagonal matrix \(D\).

By de Moivre's theorem, we have:
\[\cos(5\theta) + i\sin(5\theta) = (\cos\theta + i\sin\theta)^5.\]
Expanding the RHS 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 real parts:
\[\cos(5\theta) = \cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta.\]
Substitute \(\sin^2\theta = 1 - \cos^2\theta\):
\[\cos(5\theta) = \cos^5\theta - 10\cos^3\theta(1-\cos^2\theta) + 5\cos\theta(1-\cos^2\theta)^2.\]
\[\cos(5\theta) = \cos^5\theta - 10\cos^3\theta + 10\cos^5\theta + 5\cos\theta(1 - 2\cos^2\theta + \cos^4\theta).\]
\[\cos(5\theta) = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta.\]

To find the roots of \(16x^5 - 20x^3 + 5x = 0\), we substitute \(x = \cos\theta\):
\[16\cos^5\theta - 20\cos^3\theta + 5\cos\theta = 0 \implies \cos(5\theta) = 0.\]
Solving \(\cos(5\theta) = 0\) yields:
\[5\theta = \frac{\pi}{2} + k\pi \implies \theta = \frac{\pi}{10} + \frac{k\pi}{5}.\]
For \(k = -2, -1, 0, 1, 2\), we get:
\[\theta = -\frac{3\pi}{10}, -\frac{\pi}{10}, \frac{\pi}{10}, \frac{3\pi}{10}, \frac{\pi}{2}.\]
The corresponding values of \(x = \cos\theta\) are:
\[x = \cos\left(\frac{\pi}{2}\right) = 0,\quad x = \cos\left(\pm\frac{\pi}{10}\right) = \cos\left(\frac{\pi}{10}\right), \quad x = \cos\left(\pm\frac{3\pi}{10}\right) = \cos\left(\frac{3\pi}{10}\right).\]
Since \(\cos(-\theta) = \cos\theta\), we obtain the distinct roots:
\[x = 0, \pm \cos\left(\frac{\pi}{10}\right), \pm \cos\left(\frac{3\pi}{10}\right).\]

M1: State de Moivre's theorem and write \((\cos\theta + i\sin\theta)^5\).
A1: Correct binomial expansion.
M1: Equate real parts and substitute \(\sin^2\theta = 1 - \cos^2\theta\).
A1: Obtain the correct expression for \(\cos(5\theta)\).
M1: Set \(x = \cos\theta\) and relate to \(\cos(5\theta) = 0\).
A1: Solve \(\cos(5\theta) = 0\) to find five values for \(\theta\).
A1: Deduce the 5 distinct roots of the given polynomial in terms of cosine.

(a) We write the integrand as:
\[I_n = \int_0^1 x^{n-1} \cdot \left( x e^{-x^2} \right) \, dx.\]
Let \(u = x^{n-1}\), which gives \(du = (n-1)x^{n-2} \, dx\).
Let \(dv = x e^{-x^2} \, dx\), which integrates to \(v = -\frac{1}{2}e^{-x^2}\).
Applying integration by parts:
\[I_n = \left[ -\frac{1}{2}x^{n-1}e^{-x^2} \right]_0^1 - \int_0^1 \left( -\frac{1}{2}e^{-x^2} \right) (n-1)x^{n-2} \, dx.\]
Since \(n \ge 2\), evaluating the first term at \(x = 1\) and \(x = 0\) gives:
\[\left[ -\frac{1}{2}x^{n-1}e^{-x^2} \right]_0^1 = -\frac{1}{2e} - 0 = -\frac{1}{2e}.\]
Thus:
\[I_n = -\frac{1}{2e} + \frac{n-1}{2} \int_0^1 x^{n-2} e^{-x^2} \, dx = \frac{n-1}{2}I_{n-2} - \frac{1}{2e}.\]

(b) To find \(I_5\), we use the reduction formula:
\[I_5 = 2 I_3 - \frac{1}{2e}.\]
\[I_3 = I_1 - \frac{1}{2e}.\]
We calculate \(I_1\):
\[I_1 = \int_0^1 x e^{-x^2} \, dx = \left[ -\frac{1}{2} e^{-x^2} \right]_0^1 = -\frac{1}{2e} + \frac{1}{2}.\]
Now substitute \(I_1\) back to find \(I_3\):
\[I_3 = \left(\frac{1}{2} - \frac{1}{2e}\right) - \frac{1}{2e} = \frac{1}{2} - \frac{1}{e}.\]
Substitute \(I_3\) to find \(I_5\):
\[I_5 = 2 \left(\frac{1}{2} - \frac{1}{e}\right) - \frac{1}{2e} = 1 - \frac{2}{e} - \frac{1}{2e} = 1 - \frac{5}{2e}.\]

M1: Write integrand in a form suitable for integration by parts.
A1: Correct derivatives/integrals \(u\) and \(v\).
M1: Apply integration by parts formula.
A1: Correctly evaluate the boundary term.
A1: Obtain the reduction formula.
M1: Relate \(I_5\) to \(I_3\) and \(I_1\).
A1: Compute \(I_1\) correctly.
A1: Evaluate \(I_5\) to get \(1 - \frac{5}{2e}\).

First, find the complementary function (CF) by solving the auxiliary equation:
\[m^2 + 4m + 5 = 0 \implies (m+2)^2 + 1 = 0 \implies m = -2 \pm i.\]
Thus, the complementary function is:
\[y_{CF} = e^{-2x}(A\cos x + B\sin x).\]

Since the RHS of the differential equation, \(10e^{-2x}\cos x\), is a part of the complementary function, the particular integral (PI) must be of the form:
\[y = x e^{-2x}(C\cos x + D\sin x).\]
Let \(w = e^{-2x}(C\cos x + D\sin x)\). Then \(y = xw\).
\[\frac{dy}{dx} = w + x\frac{dw}{dx},\]
\[\frac{d^2 y}{dx^2} = 2\frac{dw}{dx} + x\frac{d^2 w}{dx^2}.\]
Substitute these into the differential equation:
\[\left(2\frac{dw}{dx} + x\frac{d^2 w}{dx^2}\right) + 4\left(w + x\frac{dw}{dx}\right) + 5xw = 10e^{-2x}\cos x.\]
\[x\left(\frac{d^2 w}{dx^2} + 4\frac{dw}{dx} + 5w\right) + 2\frac{dw}{dx} + 4w = 10e^{-2x}\cos x.\]
Since \(w\) satisfies the homogeneous equation, the term in parentheses is 0. Thus, we are left with:
\[2\frac{dw}{dx} + 4w = 10e^{-2x}\cos x.\]
Now find \(\frac{dw}{dx}\):
\[\frac{dw}{dx} = -2e^{-2x}(C\cos x + D\sin x) + e^{-2x}(-C\sin x + D\cos x) = e^{-2x}((D - 2C)\cos x - (C + 2D)\sin x).\]
Substitute \(\frac{dw}{dx}\) and \(w\) into the LHS of our equation:
\[2e^{-2x}((D - 2C)\cos x - (C + 2D)\sin x) + 4e^{-2x}(C\cos x + D\sin x) = 10e^{-2x}\cos x.\]
Simplifying inside the brackets:
\[e^{-2x}((2D - 4C + 4C)\cos x + (-2C - 4D + 4D)\sin x) = 10e^{-2x}\cos x,\]
\[e^{-2x}(2D\cos x - 2C\sin x) = 10e^{-2x}\cos x.\]
Comparing coefficients:
\[2D = 10 \implies D = 5,\]
\[-2C = 0 \implies C = 0.\]
Thus, the particular integral is:
\[y_{PI} = 5x e^{-2x}\sin x.\]
The general solution is:
\[y = e^{-2x}(A\cos x + B\sin x) + 5x e^{-2x}\sin x.\]

M1: Set up auxiliary equation and solve for roots.
A1: Correct complementary function.
M1: Identify appropriate trial particular integral \(y_{PI} = x e^{-2x}(C\cos x + D\sin x)\).
M1: Differentiate the trial particular integral.
A1: Substitute derivatives back and simplify.
M1: Compare coefficients to solve for \(C\) and \(D\).
A1: Correct particular integral.
A1: Correct general solution with constants of integration.

(a) Let \(y = \cosh^{-1} x\). Then \(x = \cosh y = \frac{e^y + e^{-y}}{2}\) for \(y \ge 0\).
Multiply by \(2e^y\):
\[2x e^y = (e^y)^2 + 1 \implies (e^y)^2 - 2x e^y + 1 = 0.\]
Solving this quadratic in \(e^y\):
\[e^y = \frac{2x \pm \sqrt{4x^2 - 4}}{2} = x \pm \sqrt{x^2 - 1}.\]
Since \(y \ge 0\), we must have \(e^y \ge 1\). Since \(x \ge 1\), \(x - \sqrt{x^2 - 1} \le 1\), so we choose the positive sign:
\[e^y = x + \sqrt{x^2 - 1}.\]
Taking natural logarithms of both sides:
\[y = \ln(x + \sqrt{x^2 - 1}).\]

(b) Using the identity \(\cosh(2\theta) = 1 + 2\sinh^2\theta\), we substitute into the equation:
\[3(1 + 2\sinh^2\theta) - 8\sinh\theta = 5,\]
\[6\sinh^2\theta - 8\sinh\theta - 2 = 0 \implies 3\sinh^2\theta - 4\sinh\theta - 1 = 0.\]
Solving this quadratic in \(\sinh\theta\):
\[\sinh\theta = \frac{4 \pm \sqrt{16 - 4(3)(-1)}}{6} = \frac{4 \pm 2\sqrt{7}}{6} = \frac{2 \pm \sqrt{7}}{3}.\]
We know \(\sinh^{-1} u = \ln(u + \sqrt{u^2 + 1})\).
Let \(u = \sinh\theta\). Then:
\[\theta = \ln\left(u + \sqrt{u^2 + 1}\right).\]
Since \(3u^2 - 4u - 1 = 0 \implies u^2 = \frac{4u+1}{3}\), we have:
\[u^2 + 1 = \frac{4u+4}{3} = \frac{4(u+1)}{3}.\]
For \(u = \frac{2 \pm \sqrt{7}}{3}\), we have \(u+1 = \frac{5 \pm \sqrt{7}}{3}\).
Hence, \(u^2 + 1 = \frac{20 \pm 4\sqrt{7}}{9}\), giving \(\sqrt{u^2+1} = \frac{\sqrt{20 \pm 4\sqrt{7}}}{3}\).
Thus, the solutions are:
\[\theta = \ln\left( \frac{2 \pm \sqrt{7} + \sqrt{20 \pm 4\sqrt{7}}}{3} \right).\]

M1: Set \(x = \cosh y\) and form quadratic equation in \(e^y\).
A1: Solve quadratic for \(e^y\).
A1: Justify choice of sign and take logs.
M1: Use double-angle identity for hyperbolic cosine.
A1: Form correct quadratic in \(\sinh\theta\).
A1: Solve for \(\sinh\theta\).
M1: Apply logarithmic form of \(\sinh^{-1}\).
A1: Express solutions in exact logarithmic forms.

Let \(f(x) = \ln(1 + \sin x)\). Then \(f(0) = \ln(1 + 0) = 0\).
Find the first derivative:
\[f'(x) = \frac{\cos x}{1 + \sin x} \implies f'(0) = \frac{1}{1} = 1.\]
Find the second derivative:
\[f''(x) = \frac{(-\sin x)(1 + \sin x) - \cos x(\cos x)}{(1 + \sin x)^2} = \frac{-\sin x - (\sin^2 x + \cos^2 x)}{(1 + \sin x)^2} = \frac{-(1 + \sin x)}{(1 + \sin x)^2} = -\frac{1}{1 + \sin x}.\]
Thus, \(f''(0) = -\frac{1}{1+0} = -1\).
Find the third derivative:
\[f'''(x) = \frac{d}{dx} \left[ -(1 + \sin x)^{-1} \right] = (1 + \sin x)^{-2} \cos x = \frac{\cos x}{(1 + \sin x)^2}.\]
Thus, \(f'''(0) = \frac{1}{1^2} = 1\).
By Maclaurin's theorem:
\[f(x) \approx f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3\]
\[f(x) \approx 0 + (1)x + \frac{-1}{2}x^2 + \frac{1}{6}x^3 = x - \frac{1}{2}x^2 + \frac{1}{6}x^3.\]

B1: Correctly find \(f(0)\).
M1: Differentiate to find \(f'(x)\).
A1: Correctly find \(f'(0) = 1\).
M1: Differentiate to find \(f''(x)\) and simplify.
A1: Correctly find \(f''(0) = -1\).
M1: Differentiate to find \(f'''(x)\).
A1: Correctly find \(f'''(0) = 1\).
A1: Correct Maclaurin series expansion.

(a) To find the eigenvalues, solve the characteristic equation \(\det(M - \lambda I) = 0\):
\[\begin{vmatrix} 5-\lambda & -6 \\ 3 & -4-\lambda \end{vmatrix} = 0 \implies (5-\lambda)(-4-\lambda) + 18 = 0,\]
\[\lambda^2 - \lambda - 2 = 0 \implies (\lambda - 2)(\lambda + 1) = 0.\]
The eigenvalues are \(\lambda_1 = 2\) and \(\lambda_2 = -1\).

For \(\lambda_1 = 2\):
\[M - 2I = \begin{pmatrix} 3 & -6 \\ 3 & -6 \end{pmatrix} \implies x - 2y = 0.\]
An eigenvector is \(\mathbf{v}_1 = \begin{pmatrix} 2 \\ 1 \end{pmatrix}\).

For \(\lambda_2 = -1\):
\[M + I = \begin{pmatrix} 6 & -6 \\ 3 & -3 \end{pmatrix} \implies x - y = 0.\]
An eigenvector is \(\mathbf{v}_2 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}\).

(b) Using diagonalisation, \(M = PDP^{-1}\), where:
\[P = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} \quad \text{and} \quad D = \begin{pmatrix} 2 & 0 \\ 0 & -1 \end{pmatrix}.\]
Find the inverse of \(P\):
\[\det P = 2(1) - 1(1) = 1 \implies P^{-1} = \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}.\]
Since \(M^n = P D^n P^{-1}\):
\[D^n = \begin{pmatrix} 2^n & 0 \\ 0 & (-1)^n \end{pmatrix}.\]
Multiply \(P D^n\):
\[P D^n = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 2^n & 0 \\ 0 & (-1)^n \end{pmatrix} = \begin{pmatrix} 2 \cdot 2^n & (-1)^n \\ 2^n & (-1)^n \end{pmatrix}.\]
Now multiply by \(P^{-1}\):
\[M^n = \begin{pmatrix} 2 \cdot 2^n & (-1)^n \\ 2^n & (-1)^n
\end{pmatrix} \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 2 \cdot 2^n - (-1)^n & -2 \cdot 2^n + 2(-1)^n \\ 2^n - (-1)^n & -2^n + 2(-1)^n \end{pmatrix}.\]
This is the required formula.

M1: Set up determinant and solve quadratic for eigenvalues.
A1: Correct eigenvalues \(\lambda = 2, -1\).
M1: Substitute \(\lambda = 2\) and solve for eigenvector.
A1: Correct eigenvector \(\begin{pmatrix} 2 \\ 1 \end{pmatrix}\).
M1: Substitute \(\lambda = -1\) and solve for eigenvector.
A1: Correct eigenvector \(\begin{pmatrix} 1 \\ 1 \end{pmatrix}\).
M1: Write down matrices \(P\) and \(D^n\) and find \(P^{-1}\).
A1: Perform matrix multiplication \(P D^n P^{-1}\) correctly to yield the given result.

(a) Let \(z = \cos\theta + i\sin\theta\). Then:
\[z - z^{-1} = 2i\sin\theta.\]
By the Binomial Theorem:
\[(2i\sin\theta)^4 = (z - z^{-1})^4\]
\[16\sin^4\theta = z^4 - 4z^2 + 6 - 4z^{-2} + z^{-4}.\]
Grouping matching powers:
\[16\sin^4\theta = (z^4 + z^{-4}) - 4(z^2 + z^{-2}) + 6.\]
Using de Moivre's theorem, \(z^n + z^{-n} = 2\cos(n\theta)\):
\[16\sin^4\theta = 2\cos 4\theta - 8\cos 2\theta + 6.\]
Divide by 16:
\[\sin^4\theta = \frac{1}{8}\cos 4\theta - \frac{1}{2}\cos 2\theta + \frac{3}{8} = \frac{1}{8}(\cos 4\theta - 4\cos 2\theta + 3).\]

(b) Integrate this expression from \(0\) to \(\frac{\pi}{4}\):
\[\int_0^{\frac{\pi}{4}} \sin^4\theta \, d\theta = \frac{1}{8} \int_0^{\frac{\pi}{4}} (\cos 4\theta - 4\cos 2\theta + 3) \, d\theta\]
\[= \frac{1}{8} \left[ \frac{1}{4}\sin 4\theta - 2\sin 2\theta + 3\theta \right]_0^{\frac{\pi}{4}}.\]
Evaluating at the upper limit \(\theta = \frac{\pi}{4}\):
\[\frac{1}{8} \left( \frac{1}{4}\sin\pi - 2\sin\frac{\pi}{2} + 3\left(\frac{\pi}{4}\right) \right) = \frac{1}{8} \left( 0 - 2 + \frac{3\pi}{4} \right) = \frac{3\pi}{32} - \frac{1}{4}.\]
Evaluating at the lower limit \(\theta = 0\) yields \(0\).
Thus, the exact value is:
\[\frac{3\pi}{32} - \frac{1}{4}.\]

M1: Express \(\sin\theta\) in terms of \(z - z^{-1}\).
A1: Apply Binomial expansion to \((z - z^{-1})^4\).
M1: Regroup terms in pairs and apply de Moivre's theorem \(z^n + z^{-n} = 2\cos(n\theta)\).
A1: Correctly show the identity.
M1: Attempt integration of the identity.
A1: Correct antiderivative.
M1: Substitute limits \(0\) and \(\frac{\pi}{4}\).
A1: Obtain the correct exact value.