2023 Cambridge IAS-Level Mathematics - Further (9231) 模擬試題連答案詳解

Let \(P(n)\) be the statement that \(5^{2n} + 24n - 1\) is divisible by 48.

**Base step:**
For \(n = 1\):
\(5^{2(1)} + 24(1) - 1 = 25 + 24 - 1 = 48\), which is divisible by 48.
Thus, \(P(1)\) is true.

**Induction step:**
Assume that \(P(k)\) is true for some positive integer \(k\).
That is, \(5^{2k} + 24k - 1 = 48M\) for some integer \(M\).

Now consider \(P(k+1)\):
\(5^{2(k+1)} + 24(k+1) - 1 = 5^{2k+2} + 24k + 24 - 1\)
\(= 25 \cdot 5^{2k} + 24k + 23\).

From the inductive hypothesis, we can express \(5^{2k} = 48M - 24k + 1\).
Substituting this into the expression for \(P(k+1)\):
\(25(48M - 24k + 1) + 24k + 23\)
\(= 25(48M) - 600k + 25 + 24k + 23\)
\(= 48(25M) - 576k + 48\)
\(= 48(25M - 12k + 1)\).

Since \(M\) and \(k\) are integers, \(25M - 12k + 1\) is also an integer.
Therefore, \(5^{2(k+1)} + 24(k+1) - 1\) is divisible by 48.

This shows that if \(P(k)\) is true, then \(P(k+1)\) is also true. Since \(P(1)\) is true and \(P(k) \implies P(k+1)\), by mathematical induction \(P(n)\) is true for all positive integers \(n\).

M1: Verify the base case \(n = 1\) showing it equals 48.
A1: Conclude \(P(1)\) is true.
M1: State the induction hypothesis for \(n = k\).
M1: Write expression for \(n = k+1\) and attempt to substitute the induction hypothesis.
A1: Correctly expand and simplify to get \(25(48M) - 576k + 48\) or an equivalent expression.
M1: Factorize out 48 from the simplified expression to show divisibility.
A1: Correctly show that the term is \(48 \times \text{integer}\).
A1: Clear concluding statement showing understanding of mathematical induction.

(a) Starting with the right-hand side:
\[ \frac{r-1}{r(r+1)} - \frac{r}{(r+1)(r+2)} = \frac{(r-1)(r+2) - r(r)}{r(r+1)(r+2)} \]
\[ = \frac{r^2 + r - 2 - r^2}{r(r+1)(r+2)} = \frac{r-2}{r(r+1)(r+2)} \]
which is equal to the left-hand side.

(b) Using the method of differences:
\[ \sum_{r=1}^n \frac{r-2}{r(r+1)(r+2)} = \sum_{r=1}^n \left( \frac{r-1}{r(r+1)} - \frac{r}{(r+1)(r+2)} \right) \]
Let \(f(r) = \frac{r-1}{r(r+1)}\). The sum telescopes as follows:
\[ (f(1) - f(2)) + (f(2) - f(3)) + \dots + (f(n) - f(n+1)) = f(1) - f(n+1) \]
\[ = \frac{0}{1(2)} - \frac{n}{(n+1)(n+2)} = -\frac{n}{(n+1)(n+2)} \]

(c) The sum from \(r=3\) to infinity can be written as:
\[ \sum_{r=3}^\infty u_r = \lim_{n\to\infty} \sum_{r=3}^n u_r \]
We know that:
\[ \sum_{r=3}^n u_r = \sum_{r=1}^n u_r - u_1 - u_2 \]
We calculate the first two terms:
\[ u_1 = \frac{1-2}{1(2)(3)} = -\frac{1}{6} \]
\[ u_2 = \frac{2-2}{2(3)(4)} = 0 \]
Thus:
\[ \sum_{r=3}^n u_r = -\frac{n}{(n+1)(n+2)} - \left(-\frac{1}{6}\right) - 0 = \frac{1}{6} - \frac{n}{(n+1)(n+2)} \]
Taking the limit as \(n \to \infty\):
\[ \lim_{n\to\infty} \left( \frac{1}{6} - \frac{n}{(n+1)(n+2)} \right) = \frac{1}{6} - 0 = \frac{1}{6} \]

(a)
M1: Combine fractions on RHS over a common denominator.
A1: Show algebra clearly leading to LHS.

(b)
M1: Apply the difference method showing cancellation of terms.
A1: Correctly identify the remaining terms.
A1: Simplify to obtain \(-\frac{n}{(n+1)(n+2)}\).

(c)
M1: Relate the sum from \(r=3\) to the sum from \(r=1\) by subtracting \(u_1\) and \(u_2\).
A1: Find \(u_1 = -1/6\) and \(u_2 = 0\).
M1: Take the limit of the resulting expression as \(n \to \infty\).
A1: Obtain the final correct answer of \(1/6\).

(a) For the cubic equation \(2x^3 - 3x^2 + 4x - 5 = 0\), the relations between roots and coefficients are:
\[ \sum \alpha = \frac{3}{2} \]
\[ \sum \alpha\beta = \frac{4}{2} = 2 \]
\[ \alpha\beta\gamma = \frac{5}{2} \]
Using the identity:
\[ \alpha^2 + \beta^2 + \gamma^2 = (\sum \alpha)^2 - 2\sum \alpha\beta = \left(\frac{3}{2}\right)^2 - 2(2) = \frac{9}{4} - 4 = -\frac{7}{4} = -1.75 \]

(b) Since \(\alpha\), \(\beta\), and \(\gamma\) are roots of the cubic equation, we have:
\[ 2\alpha^3 - 3\alpha^2 + 4\alpha - 5 = 0 \]
\[ 2\beta^3 - 3\beta^2 + 4\beta - 5 = 0 \]
\[ 2\gamma^3 - 3\gamma^2 + 4\gamma - 5 = 0 \]
Summing these three equations yields:
\[ 2\sum \alpha^3 - 3\sum \alpha^2 + 4\sum \alpha - 15 = 0 \]
Substitute the values of \(\sum \alpha^2 = -\frac{7}{4}\) and \(\sum \alpha = \frac{3}{2}\):
\[ 2\sum \alpha^3 - 3\left(-\frac{7}{4}\right) + 4\left(\frac{3}{2}\right) - 15 = 0 \]
\[ 2\sum \alpha^3 + \frac{21}{4} + 6 - 15 = 0 \]
\[ 2\sum \alpha^3 + \frac{21}{4} - 9 = 0 \]
\[ 2\sum \alpha^3 = 9 - \frac{21}{4} = \frac{15}{4} \implies \sum \alpha^3 = \frac{15}{8} = 1.875 \]

(c) Let \(y = x + 1\), which means \(x = y - 1\). Substitute this into the original cubic equation:
\[ 2(y-1)^3 - 3(y-1)^2 + 4(y-1) - 5 = 0 \]
Expanding each term:
\[ 2(y^3 - 3y^2 + 3y - 1) - 3(y^2 - 2y + 1) + 4y - 4 - 5 = 0 \]
\[ 2y^3 - 6y^2 + 6y - 2 - 3y^2 + 6y - 3 + 4y - 9 = 0 \]
\[ 2y^3 - 9y^2 + 16y - 14 = 0 \]
Replacing \(y\) with \(x\) (as the standard variable), the required equation is:
\[ 2x^3 - 9x^2 + 16x - 14 = 0 \]

(a)
M1: Identify the sum of roots and sum of product of roots from the equation.
M1: Apply the correct identity for \(\sum \alpha^2\).
A1: Obtain \(-1.75\) (or \(-7/4\)).

(b)
M1: Set up the sum of the cubic equations for the roots.
M1: Substitute previous values of \(\sum \alpha^2\) and \(\sum \alpha\).
A1: Obtain \(1.875\) (or \(15/8\)).

(c)
M1: Use the substitution method by setting \(x = y-1\).
M1: Expand the algebraic expressions correctly.
A1: Group terms together to form the correct cubic equation with integer coefficients.

(a) We perform algebraic long division on the numerator:
\[ 2x^2 + 5x - 3 = (2x + 3)(x + 1) - 6 \]
Thus:
\[ y = 2x + 3 - \frac{6}{x + 1} \]
As \(x \to -1\), \(y \to \pm\infty\), so the vertical asymptote is \(x = -1\).
As \(x \to \pm\infty\), \(\frac{6}{x + 1} \to 0\), so the oblique asymptote is \(y = 2x + 3\).

(b) We differentiate \(y = 2x + 3 - 6(x+1)^{-1}\) with respect to \(x\):
\[ \frac{dy}{dx} = 2 + \frac{6}{(x + 1)^2} \]
Since \((x+1)^2 > 0\) for all \(x \neq -1\), the term \(\frac{6}{(x+1)^2}\) is always positive.
Therefore, \(\frac{dy}{dx} \ge 2 > 0\) for all \(x \neq -1\). Since \(\frac{dy}{dx}\) is never 0, there are no stationary points on \(C\).

(c) For the \(y\)-intercept, set \(x = 0\):
\[ y = \frac{-3}{1} = -3 \implies (0, -3) \]
For the \(x\)-intercepts, set \(y = 0\):
\[ 2x^2 + 5x - 3 = 0 \implies (2x - 1)(x + 3) = 0 \implies x = 1/2 \text{ or } x = -3 \]
So the intersections are \((1/2, 0)\) and \((-3, 0)\).

(d) The equation \(\frac{2x^2 + 5x - 3}{x + 1} = k\) is equivalent to finding the intersection of the horizontal line \(y = k\) with \(C\).
Since \(\frac{dy}{dx} > 0\) everywhere (excluding the asymptote), the function is strictly increasing on the interval \((-\infty, -1)\) and also strictly increasing on the interval \((-1, \infty)\).
On \((-\infty, -1)\), \(y\) increases from \(-\infty\) (as \(x \to -\infty\)) to \(\infty\) (as \(x \to -1^-\)). Thus, \(y=k\) intersects this branch exactly once.
On \((-1, \infty)\), \(y\) increases from \(-\infty\) (as \(x \to -1^+\)) to \(\infty\) (as \(x \to \infty\)). Thus, \(y=k\) intersects this branch exactly once.
Consequently, for any real constant \(k\), there are exactly 2 real roots.

(a)
M1: Perform algebraic division on \(y\).
A1: Correctly identify the vertical asymptote \(x = -1\).
A1: Correctly identify the oblique asymptote \(y = 2x + 3\).

(b)
M1: Attempt to differentiate \(y\).
A1: Find the derivative \(\frac{dy}{dx} = 2 + \frac{6}{(x + 1)^2}\).
A1: Provide a complete and mathematically sound argument showing why \(\frac{dy}{dx} > 0\).

(c)
B1: Find the \(y\)-intercept \((0, -3)\).
B1: Factorize the quadratic numerator to find the \(x\)-intercepts \((-3, 0)\) and \((1/2, 0)\).

(d)
M1: Use the strictly increasing nature of both branches of the curve to explain intersections.
A1: Correctly deduce that there are exactly 2 real roots for any real \(k\).

(a) The area \(A\) is given by:
\[ A = \frac{1}{2} \int_{0}^{\frac{2\pi}{3}} r^2 \, d\theta = \frac{1}{2} \int_{0}^{\frac{2\pi}{3}} 4(1 + \cos\theta)^2 \, d\theta \]
\[ = 2 \int_{0}^{\frac{2\pi}{3}} (1 + 2\cos\theta + \cos^2\theta) \, d\theta \]
Using the double-angle identity \(\cos^2\theta = \frac{1 + \cos 2\theta}{2}\):
\[ 2 \int_{0}^{\frac{2\pi}{3}} \left( \frac{3}{2} + 2\cos\theta + \frac{1}{2}\cos 2\theta \right) \, d\theta \]
\[ = 2 \left[ \frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta \right]_{0}^{\frac{2\pi}{3}} \]
\[ = 2 \left( \frac{3}{2}\left(\frac{2\pi}{3}\right) + 2\sin\left(\frac{2\pi}{3}\right) + \frac{1}{4}\sin\left(\frac{4\pi}{3}\right) - 0 \right) \]
\[ = 2 \left( \pi + 2\left(\frac{\sqrt{3}}{2}\right) + \frac{1}{4}\left(-\frac{\sqrt{3}}{2}\right) \right) \]
\[ = 2 \left( \pi + \sqrt{3} - \frac{\sqrt{3}}{8} \right) = 2 \left( \pi + \frac{7\sqrt{3}}{8} \right) = 2\pi + \frac{7\sqrt{3}}{4} \]

(b) The Cartesian coordinate \(x\) is given by:
\[ x = r\cos\theta = 2(1+\cos\theta)\cos\theta = 2\cos\theta + 2\cos^2\theta \]
The tangent is perpendicular to the initial line when \(\frac{dx}{d\theta} = 0\):
\[ \frac{dx}{d\theta} = -2\sin\theta - 4\cos\theta\sin\theta = -2\sin\theta(1 + 2\cos\theta) \]
Setting \(\frac{dx}{d\theta} = 0\) for \(0 \le \theta \le \pi\):
- \(\sin\theta = 0 \implies \theta = 0\) or \(\theta = \pi\). (But \(\theta = 0\) is excluded by the question, and \(\theta = \pi\) corresponds to \(r=0\), which is the pole).
- \(1 + 2\cos\theta = 0 \implies \cos\theta = -\frac{1}{2} \implies \theta = \frac{2\pi}{3}\).

When \(\theta = \frac{2\pi}{3}\), the value of \(r\) is:
\[ r = 2(1 + \cos\frac{2\pi}{3}) = 2(1 - 1/2) = 1 \]
Thus, the polar coordinates of the point are \((1, \frac{2\pi}{3})\).

(a)
M1: Use correct area formula \(A = \frac{1}{2} \int r^2 \, d\theta\).
M1: Correctly expand the integrand and apply double-angle identity.
A1: Integrate terms correctly to obtain \(\frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta\).
M1: Substitute the limit \(\theta = \frac{2\pi}{3}\) into the integrated expression.
A1: Obtain the exact area \(2\pi + \frac{7\sqrt{3}}{4}\) (or equivalent).

(b)
M1: Write expression for \(x = r\cos\theta\) in terms of \(\theta\).
M1: Differentiate \(x\) and set to 0.
A1: Find \(\theta = \frac{2\pi}{3}\).
A1: Find \(r = 1\) and state coordinates as \((1, \frac{2\pi}{3})\).

(a) Let the direction vectors of \(l_1\) and \(l_2\) be:
\[ \mathbf{d}_1 = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}, \quad \mathbf{d}_2 = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \]
A vector normal to both lines is:
\[ \mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 1 \\ 1 & 1 & 2 \end{vmatrix} = \mathbf{i}(-2-1) - \mathbf{j}(4-1) + \mathbf{k}(2 - (-1)) = \begin{pmatrix} -3 \\ -3 \\ 3 \end{pmatrix} \]
We can simplify this normal vector to \(\mathbf{n}_0 = \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}\).

The magnitude of \(\mathbf{n}_0\) is \(|\mathbf{n}_0| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3}\).

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

The shortest distance \(d\) is the projection of \(\mathbf{AB}\) onto \(\mathbf{n}_0\):
\[ d = \frac{|\mathbf{AB} \cdot \mathbf{n}_0|}{|\mathbf{n}_0|} = \frac{|2(1) - 3(1) + 3(-1)|}{\sqrt{3}} = \frac{|2 - 3 - 3|}{\sqrt{3}} = \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3} \approx 2.31 \]

(b) The plane \(\Pi\) contains the line \(l_1\), so it passes through the point \((1, 2, -1)\) and is parallel to both direction vectors \(\mathbf{d}_1\) and \(\mathbf{d}_2\). Therefore, its normal vector is \(\mathbf{n}_0 = \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}\).

The equation of the plane is:
\[ 1(x - 1) + 1(y - 2) - 1(z + 1) = 0 \]
\[ x - 1 + y - 2 - z - 1 = 0 \]
\[ x + y - z = 4 \]

(a)
M1: Attempt to find the vector product of the direction vectors of the two lines.
A1: Obtain a correct normal vector, e.g., \(\begin{pmatrix} -3 \\ -3 \\ 3 \end{pmatrix}\) or \(\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}\).
M1: Find a vector connecting a point on \(l_1\) to a point on \(l_2\).
M1: Apply the correct shortest distance formula.
A1: Obtain the correct shortest distance \(\frac{4\sqrt{3}}{3}\) (or 2.31).

(b)
M1: Use the cross product vector as the normal to the plane.
M1: Write down the equation of the plane using the point \((1, 2, -1)\).
A1: Obtain the correct equation \(x + y - z = 4\).

(a) Find the determinant of \(\mathbf{A}\):
\[ \det(\mathbf{A}) = 2(3) - (-1)(1) = 6 + 1 = 7 \]
The inverse matrix \(\mathbf{A}^{-1}\) is:
\[ \mathbf{A}^{-1} = \frac{1}{7} \begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix} \]

(b) Let \(\begin{pmatrix} x' \\ y' \end{pmatrix} = \mathbf{A} \begin{pmatrix} x \\ y \end{pmatrix}\). Using the inverse matrix:
\[ \begin{pmatrix} x \\ y \end{pmatrix} = \mathbf{A}^{-1} \begin{pmatrix} x' \\ y' \end{pmatrix} = \frac{1}{7} \begin{pmatrix} 3x' + y' \\ -x' + 2y' \end{pmatrix} \]
So:
\[ x = \frac{3x' + y'}{7} \quad \text{and} \quad y = \frac{-x' + 2y'}{7} \]
Substitute these into the equation of the line \(y = 2x + 1\):
\[ \frac{-x' + 2y'}{7} = 2\left( \frac{3x' + y'}{7} \right) + 1 \]
Multiply both sides by 7:
\[ -x' + 2y' = 6x' + 2y' + 7 \]
Subtract \(2y'\) from both sides:
\[ -x' = 6x' + 7 \implies 7x' = -7 \implies x' = -1 \]
Thus, the image of the line is the vertical line \(x = -1\).

(c) Let a line of the form \(y = mx + c\) be invariant. A point \((x, mx+c)\) on this line is mapped to \((x', y')\) where:
\[ x' = 2x - y = 2x - (mx+c) = (2-m)x - c \]
\[ y' = x + 3y = x + 3(mx+c) = (1+3m)x + 3c \]
If this line is invariant, the point \((x', y')\) must also lie on the line \(y' = mx' + c\):
\[ (1+3m)x + 3c = m\left[ (2-m)x - c \right] + c \]
Comparing the coefficients of \(x\) on both sides:
\[ 1 + 3m = m(2 - m) \]
\[ 1 + 3m = 2m - m^2 \implies m^2 + m + 1 = 0 \]
The discriminant of this quadratic equation is:
\[ \Delta = 1^2 - 4(1)(1) = -3 < 0 \]
Since the discriminant is negative, there are no real solutions for \(m\). Therefore, the transformation \(T\) has no invariant lines of the form \(y = mx + c\).

(a)
M1: Find the determinant of matrix \(\mathbf{A}\).
A1: Correctly write \(\mathbf{A}^{-1}\).

(b)
M1: Express original coordinates \(x\) and \(y\) in terms of image coordinates \(x'\) and \(y'\).
M1: Substitute the expressions into the line equation \(y = 2x + 1\).
A1: Correctly simplify the expression to show \(x' = -1\).

(c)
M1: Set up the transformation equations for a point on the line \(y = mx + c\).
M1: Establish a relation between the coefficients of \(x\) to obtain a quadratic equation in \(m\).
A1: Obtain the quadratic equation \(m^2 + m + 1 = 0\).
A1: Calculate the discriminant and show that it has no real solutions.

First, find the complementary function (CF) by solving the auxiliary equation:
\[ m^2 + 4m + 5 = 0 \]
\[ m = \frac{-4 \pm \sqrt{16 - 20}}{2} = -2 \pm \mathrm{i} \]
So, the CF is:
\[ y_c = \mathrm{e}^{-2x}(A \cos x + B \sin x) \]

Next, find the particular integral (PI) by trying \( y_p = P \cos x + Q \sin x \).
Then:
\[ y_p' = -P \sin x + Q \cos x \]
\[ y_p'' = -P \cos x - Q \sin x \]

Substituting these into the differential equation:
\[ (-P \cos x - Q \sin x) + 4(-P \sin x + Q \cos x) + 5(P \cos x + Q \sin x) = 10\cos x \]
\[ (4P + 4Q)\cos x + (-4P + 4Q)\sin x = 10\cos x \]

Equating coefficients:
\[ 4P + 4Q = 10 \implies P + Q = 2.5 \]
\[ -4P + 4Q = 0 \implies P = Q \]
Thus, \( 2P = 2.5 \implies P = 1.25 \) and \( Q = 1.25 \).
So, the PI is:
\[ y_p = \frac{5}{4}(\cos x + \sin x) \]

The general solution is:
\[ y = \mathrm{e}^{-2x}(A \cos x + B \sin x) + \frac{5}{4}(\cos x + \sin x) \]

Now, apply the initial condition \( y(0) = 0 \):
\[ 0 = A + \frac{5}{4} \implies A = -\frac{5}{4} \]

Differentiating the general solution to find \( y' \):
\[ y' = -2\mathrm{e}^{-2x}(A \cos x + B \sin x) + \mathrm{e}^{-2x}(-A \sin x + B \cos x) + \frac{5}{4}(-\sin x + \cos x) \]

Apply the initial condition \( y'(0) = 2 \):
\[ 2 = -2A + B + \frac{5}{4} \]
Substituting \( A = -\frac{5}{4} \):
\[ 2 = -2\left(-\frac{5}{4}\right) + B + \frac{5}{4} \]
\[ 2 = \frac{5}{2} + B + \frac{5}{4} = \frac{15}{4} + B \]
\[ B = 2 - \frac{15}{4} = -\frac{7}{4} \]

Therefore, the particular solution is:
\[ y = \mathrm{e}^{-2x}\left(-\frac{5}{4}\cos x - \frac{7}{4}\sin x\right) + \frac{5}{4}(\cos x + \sin x) \]

M1: For writing the auxiliary equation and finding the roots.
A1: For correct complementary function (CF).
M1: For stating a trial particular integral (PI) form and finding derivatives.
M1: For substituting into the differential equation and equating coefficients.
A1: For correct values of P and Q.
A1: For correct general solution.
M1: For applying initial conditions to find the constant A.
M1: For differentiating the general solution and applying initial conditions to find the constant B.
A1: For the correct particular solution.

(a) To find the eigenvalues, we solve the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \):
\[ \det\begin{pmatrix} 2-\lambda & 1 & 0 \\ 0 & 3-\lambda & 1 \\ 0 & 0 & 2-\lambda \end{pmatrix} = 0 \]
Since the matrix is upper triangular, the determinant is the product of the diagonal entries:
\[ (2-\lambda)^2(3-\lambda) = 0 \]
This yields the eigenvalues \( \lambda = 2 \) (multiplicity 2) and \( \lambda = 3 \).

(b) For \( \lambda = 3 \), solve \( (\mathbf{A} - 3\mathbf{I})\mathbf{v} = \mathbf{0} \):
\[ \begin{pmatrix} -1 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \]
From the second and third rows, \( z = 0 \).
From the first row, \( -x + y = 0 \implies x = y \).
So the eigenvector corresponding to \( \lambda = 3 \) is any non-zero multiple of \( \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \).

(c) For \( \lambda = 2 \), solve \( (\mathbf{A} - 2\mathbf{I})\mathbf{v} = \mathbf{0} \):
\[ \begin{pmatrix} 0 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \]
From the first row, \( y = 0 \).
From the second row, \( y + z = 0 \implies z = 0 \).
Since \( x \) is free, the general vector is \( \begin{pmatrix} x \\ 0 \\ 0 \end{pmatrix} \).
A basis for this eigenspace is \( \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \).

The algebraic multiplicity of \( \lambda = 2 \) is 2, but the geometric multiplicity (the dimension of its eigenspace) is only 1. Since the geometric multiplicity is strictly less than the algebraic multiplicity, the matrix \( \mathbf{A} \) is defective (it does not possess 3 linearly independent eigenvectors and hence cannot be diagonalised).

Part (a):
M1: For setting up the determinant equation \( \det(\mathbf{A} - \lambda\mathbf{I}) = 0 \).
A1: For deriving the characteristic equation and showing eigenvalues are 2 and 3.
Part (b):
M1: For setting up the linear system for \( \lambda = 3 \).
A1: For solving to find the eigenvector \( [1, 1, 0]^{\mathrm{T}} \).
Part (c):
M1: For setting up the linear system for \( \lambda = 2 \).
A1: For correctly identifying that \( y=0 \) and \( z=0 \) with \( x \) free, yielding the basis vector.
A1: For explaining that since the geometric multiplicity of \( \lambda = 2 \) is 1, which is less than its algebraic multiplicity of 2, the matrix is defective.

(a) From de Moivre's theorem:
\[ \cos(5\theta) + \mathrm{i}\sin(5\theta) = (\cos\theta + \mathrm{i}\sin\theta)^5 \]
Expanding using the Binomial theorem:
\[ (\cos\theta + \mathrm{i}\sin\theta)^5 = \cos^5\theta + 5\mathrm{i}\cos^4\theta\sin\theta - 10\cos^3\theta\sin^2\theta - 10\mathrm{i}\cos^2\theta\sin^3\theta + 5\cos\theta\sin^4\theta + \mathrm{i}\sin^5\theta \]
Equating the real parts:
\[ \cos(5\theta) = \cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta \]
Using the identity \( \sin^2\theta = 1 - \cos^2\theta \):
\[ \cos(5\theta) = \cos^5\theta - 10\cos^3\theta(1-\cos^2\theta) + 5\cos\theta(1-\cos^2\theta)^2 \]
\[ \cos(5\theta) = \cos^5\theta - 10\cos^3\theta + 10\cos^5\theta + 5\cos\theta(1 - 2
\cos^2\theta + \cos^4\theta) \]
\[ \cos(5\theta) = 11\cos^5\theta - 10\cos^3\theta + 5\cos\theta - 10\cos^3\theta + 5\cos^5\theta \]
\[ \cos(5\theta) = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta \]

(b) Consider the equation:
\[ 32x^5 - 40x^3 + 10x - \sqrt{3} = 0 \]
Divide by 2:
\[ 16x^5 - 20x^3 + 5x = \frac{\sqrt{3}}{2} \]
Letting \( x = \cos\theta \), this becomes:
\[ \cos(5\theta) = \frac{\sqrt{3}}{2} \]
Thus, the general solution for \( 5\theta \) is:
\[ 5\theta = 2k\pi \pm \frac{\pi}{6} \quad \text{for } k \in \mathbb{Z} \]
\[ \theta = \frac{2k\pi}{5} \pm \frac{\pi}{30} \]
Evaluating distinct values of \( \theta \) in the interval \( [0, \pi] \):
- For \( k=0, + \): \( \theta_1 = \frac{\pi}{30} \)
- For \( k=1, - \): \( \theta_2 = \frac{11\pi}{30} \)
- For \( k=1, + \): \( \theta_3 = \frac{13\pi}{30} \)
- For \( k=2, - \): \( \theta_4 = \frac{23\pi}{30} \)
- For \( k=2, + \): \( \theta_5 = \frac{25\pi}{30} = \frac{5\pi}{6} \)

Thus, the five roots are:
\[ x = \cos\left(\frac{\pi}{30}\right), \cos\left(\frac{11\pi}{30}\right), \cos\left(\frac{13\pi}{30}\right), \cos\left(\frac{23\pi}{30}\right), \cos\left(\frac{5\pi}{6}\right) \]

Part (a):
M1: For stating de Moivre's theorem for exponent 5.
M1: For binomially expanding and equating real parts.
M1: For substituting \( \sin^2\theta = 1 - \cos^2\theta \).
A1: For simplifying to the required identity.
Part (b):
M1: For rearranging the polynomial equation into the identity form.
M1: For setting up the trigonometric equation \( \cos(5\theta) = \frac{\sqrt{3}}{2} \).
A1: For finding correct general values of \( 5\theta \).
A1: For listing the five distinct roots in terms of cosines correctly.

(a) Recall that:
\[ \cosh x = \frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2} \quad \text{and} \quad \sinh x = \frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2} \]
Substitute these definitions into the equation:
\[ 4\left(\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right) + 3\left(\frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right) = 5 \]
Using the substitution \( u = \mathrm{e}^x \) (and thus \( u^{-1} = \mathrm{e}^{-x} \)):
\[ 2(u + u^{-1}) + \frac{3}{2}(u - u^{-1}) = 5 \]
Multiply both sides by 2:
\[ 4(u + u^{-1}) + 3(u - u^{-1}) = 10 \]
\[ 4u + \frac{4}{u} + 3u - \frac{3}{u} = 10 \]
\[ 7u + \frac{1}{u} = 10 \]
Multiply through by \( u \) (since \( u > 0 \)):
\[ 7u^2 - 10u + 1 = 0 \]

(b) Solve \( 7u^2 - 10u + 1 = 0 \) using the quadratic formula:
\[ u = \frac{10 \pm \sqrt{(-10)^2 - 4(7)(1)}}{2(7)} \]
\[ u = \frac{10 \pm \sqrt{100 - 28}}{14} = \frac{10 \pm \sqrt{72}}{14} \]
\[ u = \frac{10 \pm 6\sqrt{2}}{14} = \frac{5 \pm 3\sqrt{2}}{7} \]
Since \( 3\sqrt{2} \approx 4.24 \), both roots \( \frac{5 + 3\sqrt{2}}{7} \) and \( \frac{5 - 3\sqrt{2}}{7} \) are positive and thus valid values for \( u = \mathrm{e}^x \).
Taking the natural logarithm of both sides:
\[ x = \ln\left(\frac{5 \pm 3\sqrt{2}}{7}\right) \]

Part (a):
M1: For expressing hyperbolic functions in terms of exponentials.
M1: For substituting \( u = \mathrm{e}^x \) correctly.
M1: For clearing the fraction and multiplying by \( u \).
A1: For obtaining \( 7u^2 - 10u + 1 = 0 \).
Part (b):
M1: For solving the quadratic equation using the quadratic formula.
A1: For obtaining correct simplified values of \( u \).
M1: For stating that both values are valid as they are positive.
A1: For the final correct answers in terms of natural logarithms.

To find the Maclaurin series expansion, we compute the value of the function and its first three derivatives at \( x = 0 \):

1) At \( x = 0 \):
\[ y(0) = \ln(1 + 0) = 0 \]

2) First derivative:
\[ y' = \frac{\cos x}{1 + \sin x} \]
At \( x = 0 \):
\[ y'(0) = \frac{1}{1 + 0} = 1 \]

3) Second derivative using the quotient rule:
\[ y'' = \frac{(-\sin x)(1 + \sin x) - \cos x(\cos x)}{(1 + \sin x)^2} = \frac{-\sin x - \sin^2 x - \cos^2 x}{(1 + \sin x)^2} \]
Using the identity \( \sin^2 x + \cos^2 x = 1 \):
\[ y'' = \frac{-(1 + \sin x)}{(1 + \sin x)^2} = -\frac{1}{1 + \sin x} = -(1 + \sin x)^{-1} \]
At \( x = 0 \):
\[ y''(0) = -1 \]

4) Third derivative:
\[ y''' = \frac{\mathrm{d}}{\mathrm{d}x} \left( -(1 + \sin x)^{-1} \right) = (1 + \sin x)^{-2} \cos x = \frac{\cos x}{(1 + \sin x)^2} \]
At \( x = 0 \):
\[ y'''(0) = \frac{1}{1^2} = 1 \]

The Maclaurin series is given by:
\[ y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \dots \]
Substitute the calculated values:
\[ y = 0 + 1x + \frac{-1}{2}x^2 + \frac{1}{6}x^3 + \dots \]
\[ y = x - \frac{1}{2}x^2 + \frac{1}{6}x^3 \]

M1: For finding the first derivative \( y' \).
A1: For finding \( y'(0) = 1 \).
M1: For finding the second derivative \( y'' \) using quotient or chain rule.
A1: For simplifying \( y'' \) to \( -(1 + \sin x)^{-1} \) (or equivalent) and finding \( y''(0) = -1 \).
M1: For finding the third derivative \( y''' \).
A1: For finding \( y'''(0) = 1 \).
M1: For stating and applying the Maclaurin series formula.
A1: For obtaining the correct expansion \( x - \frac{1}{2}x^2 + \frac{1}{6}x^3 \).

Let \( u = \mathrm{e}^x \), so \( \mathrm{d}x = \frac{\mathrm{d}u}{u} \).
The limits of integration transform as follows:
- When \( x = 0 \), \( u = 1 \).
- When \( x = \ln 2 \), \( u = 2 \).

Express the denominator in terms of \( u \):
\[ 2\cosh x - 1 = 2\left(\frac{u + u^{-1}}{2}\right) - 1 = u + \frac{1}{u} - 1 = \frac{u^2 - u + 1}{u} \]

Substitute into the integral:
\[ I = \int_{1}^{2} \frac{1}{\frac{u^2 - u + 1}{u}} \cdot \frac{\mathrm{d}u}{u} = \int_{1}^{2} \frac{1}{u^2 - u + 1} \mathrm{d}u \]

Complete the square in the denominator:
\[ u^2 - u + 1 = \left(u - \frac{1}{2}\right)^2 + \frac{3}{4} \]

Now, integrate using the standard form \( \int \frac{1}{w^2 + a^2} \mathrm{d}w = \frac{1}{a} \arctan\left(\frac{w}{a}\right) \) where \( a = \frac{\sqrt{3}}{2} \):
\[ I = \left[ \frac{2}{\sqrt{3}} \arctan\left( \frac{u - \frac{1}{2}}{\frac{\sqrt{3}}{2}} \right) \right]_{1}^{2} = \left[ \frac{2}{\sqrt{3}} \arctan\left( \frac{2u - 1}{\sqrt{3}} \right) \right]_{1}^{2} \]

Evaluate at the upper and lower limits:
- At \( u = 2 \):
\[ \frac{2}{\sqrt{3}} \arctan\left( \frac{3}{\sqrt{3}} \right) = \frac{2}{\sqrt{3}} \arctan(\sqrt{3}) = \frac{2}{\sqrt{3}} \left( \frac{\pi}{3} \right) \]
- At \( u = 1 \):
\[ \frac{2}{\sqrt{3}} \arctan\left( \frac{1}{\sqrt{3}} \right) = \frac{2}{\sqrt{3}} \left( \frac{\pi}{6} \right) \]

Subtracting the two values:
\[ I = \frac{2}{\sqrt{3}} \left( \frac{\pi}{3} - \frac{\pi}{6} \right) = \frac{2}{\sqrt{3}} \left( \frac{\pi}{6} \right) = \frac{\pi}{3\sqrt{3}} = \frac{\pi\sqrt{3}}{9} \]

M1: For using the substitution \( u = \mathrm{e}^x \) or equivalent.
A1: For transforming the limits to 1 and 2.
M1: For expressing \( 2\cosh x - 1 \) in terms of \( u \).
A1: For deriving the simplified integral \( \int_{1}^{2} \frac{1}{u^2 - u + 1} \mathrm{d}u \).
M1: For completing the square of the quadratic denominator.
A1: For integrating to a form involving \( \arctan \).
M1: For substituting the limits of integration correctly.
A1: For simplifying the final result to \( \frac{\pi\sqrt{3}}{9} \) (or equivalent exact form).

First, rewrite the differential equation in standard form by dividing by \( x \) (assuming \( x > 0 \)):
\[ \frac{\mathrm{d}y}{\mathrm{d}x} + \left( 2 + \frac{1}{x} \right)y = \mathrm{e}^{-2x} \]

Identify the integrating factor (IF):
\[ \mathrm{I.F.} = \mathrm{e}^{\int \left( 2 + \frac{1}{x} \right) \mathrm{d}x} = \mathrm{e}^{2x + \ln x} = \mathrm{e}^{2x} \cdot \mathrm{e}^{\ln x} = x\mathrm{e}^{2x} \]

Multiply the standard form differential equation by the integrating factor:
\[ x\mathrm{e}^{2x} \frac{\mathrm{d}y}{\mathrm{d}x} + (2x + 1)\mathrm{e}^{2x}y = x \]
\[ \frac{\mathrm{d}}{\mathrm{d}x} \left( y \cdot x\mathrm{e}^{2x} \right) = x \]

Integrate both sides with respect to \( x \):
\[ y \cdot x\mathrm{e}^{2x} = \int x \mathrm{d}x = \frac{1}{2}x^2 + C \]

Apply the initial condition \( y = 1 \) when \( x = 1 \):
\[ 1 \cdot 1 \cdot \mathrm{e}^{2(1)} = \frac{1}{2}(1)^2 + C \]
\[ \mathrm{e}^2 = \frac{1}{2} + C \implies C = \mathrm{e}^2 - \frac{1}{2} \]

Substitute \( C \) back into the equation:
\[ y \cdot x\mathrm{e}^{2x} = \frac{1}{2}x^2 + \mathrm{e}^2 - \frac{1}{2} \]
Divide by \( x\mathrm{e}^{2x} \) to solve for \( y \):
\[ y = \left( \frac{\frac{1}{2}x^2 + \mathrm{e}^2 - \frac{1}{2}}{x} \right) \mathrm{e}^{-2x} = \left( \frac{x^2 + 2\mathrm{e}^2 - 1}{2x} \right) \mathrm{e}^{-2x} \]

M1: For dividing by \( x \) to write in standard form.
M1: For attempting to find the integrating factor.
A1: For finding the correct integrating factor \( x\mathrm{e}^{2x} \).
M1: For multiplying and writing the equation in derivative form.
A1: For integrating both sides correctly to get \( y \cdot x\mathrm{e}^{2x} = \frac{1}{2}x^2 + C \).
M1: For applying the boundary condition to find the constant \( C \).
A1: For the correct value of \( C \).
A1: For writing the final particular solution in the required form \( y = f(x) \).

(a) A matrix is singular if and only if its determinant is equal to zero:
\[ \det(\mathbf{M}) = k(k+3) - (2)(2) = k^2 + 3k - 4 = 0 \]
Factoring the quadratic equation:
\[ (k + 4)(k - 1) = 0 \implies k = 1 \text{ or } k = -4 \]

(b) For \( k = 1 \), the matrix is:
\[ \mathbf{M} = \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix} \]
Find the eigenvalues \( \lambda \):
\[ \det(\mathbf{M} - \lambda\mathbf{I}) = (1-\lambda)(4-\lambda) - 4 = \lambda^2 - 5\lambda = 0 \]
\[ \lambda(\lambda - 5) = 0 \implies \lambda_1 = 0, \lambda_2 = 5 \]
Thus, the diagonal matrix \( \mathbf{D} \) is:
\[ \mathbf{D} = \begin{pmatrix} 0 & 0 \\ 0 & 5 \end{pmatrix} \]

Find normalized eigenvectors:
1) For \( \lambda_1 = 0 \):
\[ \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies x + 2y = 0 \]
An eigenvector is \( \begin{pmatrix} -2 \\ 1 \end{pmatrix} \). Normalizing this vector:
\[ \mathbf{u}_1 = \frac{1}{\sqrt{5}}\begin{pmatrix} -2 \\ 1 \end{pmatrix} \]

2) For \( \lambda_2 = 5 \):
\[ \begin{pmatrix} -4 & 2 \\ 2 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies 2x - y = 0 \]
An eigenvector is \( \begin{pmatrix} 1 \\ 2 \end{pmatrix} \). Normalizing this vector:
\[ \mathbf{u}_2 = \frac{1}{\sqrt{5}}\begin{pmatrix} 1 \\ 2 \end{pmatrix} \]

Thus, the orthogonal matrix \( \mathbf{P} \) is constructed from the normalized eigenvectors:
\[ \mathbf{P} = \frac{1}{\sqrt{5}}\begin{pmatrix} -2 & 1 \\ 1 & 2 \end{pmatrix} \]

Part (a):
M1: For setting up the determinant of \( \mathbf{M} \) and equating it to zero.
A1: For obtaining \( k^2 + 3k - 4 = 0 \).
A1: For correct values of \( k \).
Part (b):
M1: For finding the eigenvalues of the matrix with \( k=1 \).
A1: For the diagonal matrix \( \mathbf{D} \).
M1: For finding eigenvectors of each eigenvalue.
M1: For normalizing eigenvectors to ensure \( \mathbf{P} \) is orthogonal.
A1: For the correct orthogonal matrix \( \mathbf{P} \).