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

(i) Combining the terms on the right-hand side over a common denominator: \(\frac{1}{r} + \frac{1}{r+1} - \frac{2}{r+2} = \frac{(r+1)(r+2) + r(r+2) - 2r(r+1)}{r(r+1)(r+2)}\). Expanding the numerator: \(r^2 + 3r + 2 + r^2 + 2r - 2r^2 - 2r = 3r + 2\). Therefore, \(\frac{1}{r} + \frac{1}{r+1} - \frac{2}{r+2} = \frac{3r+2}{r(r+1)(r+2)}\). (ii) Let \(f(r) = \frac{1}{r} + \frac{2}{r+1}\). Then \(f(r) - f(r+1) = \left(\frac{1}{r} + \frac{2}{r+1}\right) - \left(\frac{1}{r+1} + \frac{2}{r+2}\right) = \frac{1}{r} + \frac{1}{r+1} - \frac{2}{r+2} = u_r\). Using the method of differences: \(\sum_{r=1}^n u_r = \sum_{r=1}^n [f(r) - f(r+1)] = f(1) - f(n+1)\). Since \(f(1) = 1 + \frac{2}{2} = 2\) and \(f(n+1) = \frac{1}{n+1} + \frac{2}{n+2}\), we have \(S_n = 2 - \frac{1}{n+1} - \frac{2}{n+2} = 2 - \frac{3n+4}{(n+1)(n+2)} = \frac{2(n^2+3n+2) - (3n+4)}{(n+1)(n+2)} = \frac{2n^2+3n}{(n+1)(n+2)} = \frac{n(2n+3)}{(n+1)(n+2)}\). (iii) As \(n \to \infty\), \(\frac{1}{n+1} \to 0\) and \(\frac{2}{n+2} \to 0\). Thus, \(\sum_{r=1}^{\infty} u_r = 2\).

(i) M1: For attempting to combine the terms on the RHS over a common denominator. A1: For correct algebraic expansion of the numerator. A1: For obtaining \(3r+2\) in the numerator and completing the proof. (ii) M1: For expressing the terms of the sum to show the cancellation (telescoping) pattern, or identifying \(f(r) - f(r+1)\). A1: For correctly finding the remaining non-cancelled terms: \(2 - \frac{1}{n+1} - \frac{2}{n+2}\). A1: For fully correct algebraic working leading to the given expression. (iii) B1: For writing down the correct sum to infinity, 2, with some reference to the limit as \(n \to \infty\).

Let \(f(n) = 5^{2n} + 24n - 1\).

**Base case:**
For \(n = 1\):
\(f(1) = 5^{2(1)} + 24(1) - 1 = 25 + 24 - 1 = 48\).
Since 48 is divisible by 48, the statement is true for \(n = 1\).

**Inductive step:**
Assume that the statement is true for \(n = k\), where \(k\) is a positive integer.
That is, \(f(k) = 5^{2k} + 24k - 1 = 48M\) for some integer \(M\).
This can be rewritten as:
\(5^{2k} = 48M - 24k + 1\).

Now consider the expression for \(n = k+1\):
\(f(k+1) = 5^{2(k+1)} + 24(k+1) - 1\)
\(f(k+1) = 5^{2k} \cdot 5^2 + 24k + 24 - 1\)
\(f(k+1) = 25(5^{2k}) + 24k + 23\)

Substitute the inductive hypothesis \(5^{2k} = 48M - 24k + 1\) into the expression:
\(f(k+1) = 25(48M - 24k + 1) + 24k + 23\)
\(f(k+1) = 25(48M) - 600k + 25 + 24k + 23\)
\(f(k+1) = 25(48M) - 576k + 48\)

Since \(576 = 12 \times 48\), we can factor out 48 from each term:
\(f(k+1) = 48(25M - 12k + 1)\).

Since \(M\) and \(k\) are integers, \(25M - 12k + 1\) is also an integer, which means \(f(k+1)\) is divisible by 48.

**Conclusion:**
Since the base case \(n=1\) is true, and the assumption that the statement is true for \(n=k\) implies that it is also true for \(n=k+1\), the statement is true by mathematical induction for all positive integers \(n\).

**B1**: Verify the base case \(n = 1\) to show \(f(1) = 48\), and state that it is divisible by 48.
**M1**: State a clear inductive hypothesis, assuming that the result is true for some positive integer \(k\) (e.g., \(5^{2k} + 24k - 1 = 48M\)).
**M1**: Consider \(f(k+1)\) and attempt to express it using the inductive hypothesis.
**A1**: Obtain a correct algebraic expansion, such as \(25(48M) - 576k + 48\) or equivalent.
**A1**: Factor out 48 correctly to show \(f(k+1) = 48(25M - 12k + 1)\).
**B1**: Give a complete and accurate final conclusion linking \(n=1\), the inductive step, and the principle of mathematical induction.

For (i): From the given cubic equation, the sum of roots is \(\Sigma \alpha = 2\) and the sum of products of roots taken two at a time is \(\Sigma \alpha \beta = 4\). Using the identity \(\Sigma \alpha^2 = (\Sigma \alpha)^2 - 2 \Sigma \alpha \beta\), we get \(\Sigma \alpha^2 = 2^2 - 2(4) = 4 - 8 = -4\). For (ii): Since \(\alpha\), \(\beta\), and \(\gamma\) are roots of the equation, they satisfy \(x^3 = 2x^2 - 4x + 3\). Summing this equation over all three roots gives \
\(\Sigma \alpha^3 = 2 \Sigma \alpha^2 - 4 \Sigma \alpha + 3(3)\). Substituting the values, we obtain \(\Sigma \alpha^3 = 2(-4) - 4(2) + 9 = -8 - 8 + 9 = -7\). For (iii): Let \(y = x^2\), which means \(x = \sqrt{y}\). Substituting this into the original cubic equation gives \(y\sqrt{y} - 2y + 4\sqrt{y} - 3 = 0\). Rearranging gives \(\sqrt{y}(y + 4) = 2y + 3\). Squaring both sides yields \(y(y + 4)^2 = (2y + 3)^2\). Expanding this equation gives \(y(y^2 + 8y + 16) = 4y^2 + 12y + 9\), which simplifies to \(y^3 + 8y^2 + 16y = 4y^2 + 12y + 9\). Rearranging terms, we obtain the new cubic equation \(y^3 + 4y^2 + 4y - 9 = 0\).

(i) M1: For attempting to use the identity \(\Sigma \alpha^2 = (\Sigma \alpha)^2 - 2 \Sigma \alpha \beta\) with correct values. A1: For obtaining \(-4\). (ii) M1: For establishing a correct relation to find \(\Sigma \alpha^3\) (either using the cubic recurrence or algebraic identity). M1: For substituting the values of \(\Sigma \alpha^2\) and \(\Sigma \alpha\) into their relation. A1: For obtaining \(-7\). (iii) M1: For substituting \(x = \sqrt{y}\) into the cubic equation and separating the radical terms, or for finding the required coefficients \(S_1, S_2, S_3\) of the new equation. M1: For completing the algebraic expansion or successfully finding the symmetric sum \(\Sigma \alpha^2\beta^2 = 4\). A1: For obtaining the correct equation \(y^3 + 4y^2 + 4y - 9 = 0\).

(i) Let \(A\) be the point \((2, 0, 3)\) on the line \(l\). A vector in the plane is:
\[\vec{AP} = \mathbf{p} - \mathbf{a} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} - \begin{pmatrix} 2 \\ 0 \\ 3 \end{pmatrix} = \begin{pmatrix} -1 \\ 2 \\ -4 \end{pmatrix}.\]

The direction vector of the line is \(\mathbf{u} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}\).

The normal vector \(\mathbf{n}\) is given by the cross product \(\mathbf{u} \times \vec{AP}\):
\[\mathbf{n} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} \times \begin{pmatrix} -1 \\ 2 \\ -4 \end{pmatrix} = \begin{pmatrix} (-1)(-4) - (2)(2) \\ (2)(-1) - (1)(-4) \\ (1)(2) - (-1)(-1) \end{pmatrix} = \begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix}.\]

The equation of the plane is \(\mathbf{r} \cdot \mathbf{n} = \mathbf{p} \cdot \mathbf{n}\):
\[\mathbf{r} \cdot \begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix} = 0 + 4 - 1 = 3.\]

Thus, the vector equation of the plane is:
\[\mathbf{r} \cdot \begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix} = 3.\]

(ii) The Cartesian equation of the plane \(\Pi\) is \(2y + z - 3 = 0\).
The perpendicular distance from \(Q(3, 4, 1)\) to the plane is:
\[d = \frac{|0(3) + 2(4) + 1(1) - 3|}{\sqrt{0^2 + 2^2 + 1^2}} = \frac{|8 + 1 - 3|}{\sqrt{5}} = \frac{6}{\sqrt{5}} = \frac{6\sqrt{5}}{5}.\]

(iii) The direction vector of the line \(l\) is \(\mathbf{d}_1 = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}\).
The direction vector of the line \(m\) is given by \(\vec{PQ}\):
\[\vec{PQ} = \begin{pmatrix} 3-1 \\ 4-2 \\ 1-(-1) \end{pmatrix} = \begin{pmatrix} 2 \\ 2 \\ 2 \end{pmatrix},\]
which is parallel to \(\mathbf{d}_2 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\).

The angle \(\theta\) between the lines is given by:
\[\cos \theta = \frac{|\mathbf{d}_1 \cdot \mathbf{d}_2|}{|\mathbf{d}_1| |\mathbf{d}_2|} = \frac{|1(1) - 1(1) + 2(1)|}{\sqrt{1^2 + (-1)^2 + 2^2} \sqrt{1^2 + 1^2 + 1^2}} = \frac{2}{\sqrt{6}\sqrt{3}} = \frac{2}{\sqrt{18}} = \frac{2}{3\sqrt{2}} = \frac{\sqrt{2}}{3}.\]

\[\theta = \arccos\left(\frac{\sqrt{2}}{3}\right) \approx 61.9^\circ.\]

**Part (i)**
* **M1**: Find a connecting vector between \(P\) and a point on \(l\) (e.g. \(\vec{AP} = \begin{pmatrix} -1 \\ 2 \\ -4 \end{pmatrix}\)).
* **M1**: Calculate the vector cross product of the direction vector of the line and the connecting vector.
* **A1**: Obtain a correct normal vector, e.g., \(\mathbf{n} = \begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix}\).
* **A1**: Formulate the correct vector equation of the plane: \(\mathbf{r} \cdot \begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix} = 3\).

**Part (ii)**
* **M1**: Express the plane equation in Cartesian form, \(2y + z - 3 = 0\), or use an equivalent vector projection method.
* **M1**: Correctly substitute \(Q(3, 4, 1)\) into the distance formula.
* **A1**: Obtain the correct exact distance \(\frac{6\sqrt{5}}{5}\) (accept \(\frac{6}{\sqrt{5}}\)).

**Part (iii)**
* **M1**: Determine the direction vector of the line \(m\) and apply the dot product formula to find the cosine of the angle.
* **A1**: Obtain \(61.9^\circ\) (accept \(1.08\) radians).

(a) The matrix for a shear parallel to the \(y\)-axis with shear factor \(k\) is:
\(\mathbf{S} = \begin{pmatrix} 1 & 0 \\\\ k & 1 \end{pmatrix}\)
The matrix for a reflection in the line \(y = x\) is:
\(\mathbf{R} = \begin{pmatrix} 0 & 1 \\\\ 1 & 0 \end{pmatrix}\)
Since the transformation \(T\) is the shear followed by the reflection, the matrix representing \(T\) is:
\(\mathbf{M} = \mathbf{R}\mathbf{S} = \begin{pmatrix} 0 & 1 \\\\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\\\ k & 1 \end{pmatrix} = \begin{pmatrix} k & 1 \\\\ 1 & 0 \end{pmatrix}\)

(b) Let \(y = mx\) be an invariant line under \(T\). A point \((x, mx)^T\) on this line is mapped to:
\(\begin{pmatrix} X' \\\\ Y' \end{pmatrix} = \mathbf{M} \begin{pmatrix} x \\\\ mx \end{pmatrix} = \begin{pmatrix} k & 1 \\\\ 1 & 0 \end{pmatrix} \begin{pmatrix} x \\\\ mx \end{pmatrix} = \begin{pmatrix} kx + mx \\\\ x \end{pmatrix}\)
For the line to be invariant, the image point must lie on the line \(y = mx\), so \(Y' = m X'\):
\(x = m(kx + mx) \implies x = (mk + m^2)x\)
For non-zero points on the line (where \(x \neq 0\)), we divide by \(x\):
\(1 = mk + m^2 \implies m^2 + km - 1 = 0\)
Using the quadratic formula to solve for \(m\):
\(m = \frac{-k \pm \sqrt{k^2 - 4(1)(-1)}}{2} = \frac{-k \pm \sqrt{k^2 + 4}}{2}\)
The discriminant of this quadratic is \(\Delta = k^2 + 4\).
Since \(k^2 \ge 0\) for all real \(k\), we have \(\Delta \ge 4 > 0\) for all real \(k\).
Since the discriminant is strictly positive, there are always two distinct real values for \(m\). Thus, there are always two distinct real invariant lines through the origin for any real value of \(k\).

(c) The area of an image under a transformation represented by \(\mathbf{M}\) is scaled by \(|\det(\mathbf{M})|\).
Calculating the determinant:
\(\det(\mathbf{M}) = k(0) - (1)(1) = -1\)
The area scale factor is \(|\det(\mathbf{M})| = |-1| = 1\).
Since the area scale factor is exactly 1, the area of any shape is invariant under the transformation \(T\) for all real values of \(k\).

(d) The area of the original triangle \(P\) with vertices \(O(0, 0)\), \(A(2, 1)\), and \(B(1, 4)\) is:
\(\text{Area}(P) = \frac{1}{2} |x_A y_B - x_B y_A| = \frac{1}{2} |2(4) - 1(1)| = 3.5\)
The determinant of the transformation matrix \(\mathbf{N}\) is:
\(\det(\mathbf{N}) = 2(-2) - 5(1) = -9\)
The area of the image triangle \(Q\) is:
\(\text{Area}(Q) = |\det(\mathbf{N})| \times \text{Area}(P) = |-9| \times 3.5 = 31.5\)

Part (a) [3 marks]:
M1: For writing down the correct matrix for either the shear \(\mathbf{S} = \begin{pmatrix} 1 & 0 \\\\ k & 1 \end{pmatrix}\) or the reflection \(\mathbf{R} = \begin{pmatrix} 0 & 1 \\\\ 1 & 0 \end{pmatrix}\).
M1: For multiplying the matrices in the correct order: \(\mathbf{M} = \mathbf{R}\mathbf{S}\) (shear first, then reflection).
A1: For obtaining the correct matrix \(\mathbf{M} = \begin{pmatrix} k & 1 \\\\ 1 & 0 \end{pmatrix}\).

Part (b) [5 marks]:
M1: For setting up the invariant line relation, e.g., \(\mathbf{M} \begin{pmatrix} x \\\\ mx \end{pmatrix} = \begin{pmatrix} X' \\\\ Y' \end{pmatrix}\) and setting \(Y' = mX'\).
M1: For deriving the quadratic equation \(m^2 + km - 1 = 0\).
A1: For finding the correct expressions for the gradient: \(m = \frac{-k \pm \sqrt{k^2+4}}{2}\).
M1: For identifying the discriminant \(\Delta = k^2 + 4\) (or equivalent analysis).
A1: For a complete and rigorous argument showing that \(\Delta \ge 4 > 0\) for all real \(k\), thereby proving the existence of two distinct real invariant lines.

Part (c) [3 marks]:
M1: For stating that the area of the image is scaled by \(|\det(\mathbf{M})|\).
M1: For calculating \(\det(\mathbf{M}) = -1\).
A1: For concluding that since \(|\det(\mathbf{M})| = 1\), the area is invariant for all real \(k\).

Part (d) [4 marks]:
M1: For calculating the area of the original triangle \(P\) as 3.5 (or \(\frac{7}{2}\)) using a valid method.
M1: For calculating \(\det(\mathbf{N}) = -9\).
M1: For multiplying the area of \(P\) by \(|\det(\mathbf{N})|\).
A1: For obtaining the correct final area of 31.5 (or \(\frac{63}{2}\)).

(a) The curve is defined for \(-\frac{3\pi}{4} \le \theta \le \frac{3\pi}{4}\). Since \(\cos(-\theta) = \cos\theta\), the curve is symmetrical about the initial line.
- When \(\theta = 0\), \(r = a(1 + \sqrt{2})\). Thus, the point of intersection with the initial line is \((a(1+\sqrt{2}), 0)\).
- When \(\theta = \pm\frac{3\pi}{4}\), \(r = a(1 + \sqrt{2}(-\frac{1}{\sqrt{2}})) = 0\), which represents the pole.
- The tangents to the curve at the pole are \(\theta = \pm\frac{3\pi}{4}\).
- Since \(r \ge 0\) for this entire domain, the sketch consists of a single closed loop starting at the pole, reaching its maximum distance at \(r = a(1+\sqrt{2})\) along the initial line, and returning symmetrically to the pole.

(b) By symmetry, the area \(A\) enclosed by \(C\) is given by:
\(A = 2 \times \frac{1}{2} \int_{0}^{\frac{3\pi}{4}} r^2 \, \mathrm{d}\theta = \int_{0}^{\frac{3\pi}{4}} a^2(1 + \sqrt{2}\cos\theta)^2 \, \mathrm{d}\theta\)

Expanding the integrand:
\((1 + \sqrt{2}\cos\theta)^2 = 1 + 2\sqrt{2}\cos\theta + 2\cos^2\theta\)

Using the double-angle identity \(2\cos^2\theta = 1 + \cos 2\theta\):
\((1 + \sqrt{2}\cos\theta)^2 = 1 + 2\sqrt{2}\cos\theta + 1 + \cos 2\theta = 2 + 2\sqrt{2}\cos\theta + \cos 2\theta\)

Now integrate with respect to \(\theta\):
\(A = a^2 \left[ 2\theta + 2\sqrt{2}\sin\theta + \frac{1}{2}\sin 2\theta \right]_{0}^{\frac{3\pi}{4}}\)

Evaluating at the upper limit \(\theta = \frac{3\pi}{4}\):
- \(2\left(\frac{3\pi}{4}\right) = \frac{3\pi}{2}\)
- \(2\sqrt{2}\sin\left(\frac{3\pi}{4}\right) = 2\sqrt{2}\left(\frac{1}{\sqrt{2}}\right) = 2\)
- \ (\frac{1}{2}\sin\left(2 \cdot \frac{3\pi}{4}\right) = \frac{1}{2}\sin\left(\frac{3\pi}{2}\right) = -\frac{1}{2}\)

Evaluating at the lower limit \(\theta = 0\) yields \(0\).

Thus, the area is:
\(A = a^2 \left( \frac{3\pi}{2} + 2 - \frac{1}{2} \right) = a^2 \left( \frac{3\pi}{2} + \frac{3}{2} \right) = \frac{3}{2}a^2(\pi + 1)\)

(c) The tangent is perpendicular to the initial line when \(\frac{\mathrm{d}x}{\mathrm{d}\theta} = 0\) (provided \(\frac{\mathrm{d}y}{\mathrm{d}\theta} \neq 0\)).

Using \(x = r\cos\theta\):
\(x = a(1 + \sqrt{2}\cos\theta)\cos\theta = a\cos\theta + a\sqrt{2}\cos^2\theta\)

Differentiating with respect to \(\theta\):
\(\frac{\mathrm{d}x}{\mathrm{d}\theta} = -a\sin\theta - 2\sqrt{2}a\cos\theta\sin\theta = -a\sin\theta(1 + 2\sqrt{2}\cos\theta)\)

Setting \(\frac{\mathrm{d}x}{\mathrm{d}\theta} = 0\):
- Case 1: \(\sin\theta = 0\)
Since \(-\frac{3\pi}{4} \le \theta \le \frac{3\pi}{4}\), this gives \(\theta = 0\).
When \(\theta = 0\), \(r = a(1+\sqrt{2})\), which yields the Cartesian point:
\(x = a(1+\sqrt{2})\cos(0) = a(1+\sqrt{2})\)
\(y = a(1+\sqrt{2})\sin(0) = 0\)
Point: \((a(1+\sqrt{2}), 0)\)

- Case 2: \(1 + 2\sqrt{2}\cos\theta = 0 \implies \cos\theta = -\frac{1}{2\sqrt{2}} = -\frac{\sqrt{2}}{4}\)
Since \(-\frac{\sqrt{2}}{2} \le -\frac{\sqrt{2}}{4} < 1\), this value lies within the range of \(\cos\theta\) for the domain.
At this value:
\(r = a\left(1 + \sqrt{2}\left(-\frac{\sqrt{2}}{4}\right)\right) = a\left(1 - \frac{2}{4}\right) = \frac{1}{2}a\)

The \(x\)-coordinate is:
\(x = r\cos\theta = \left(\frac{1}{2}a\right)\left(-\frac{\sqrt{2}}{4}\right) = -\frac{\sqrt{2}}{8}a\)

For the \(y\)-coordinate, we use \(\sin^2\theta = 1 - \cos^2\theta\):
\(\sin^2\theta = 1 - \left(-\frac{\sqrt{2}}{4}\right)^2 = 1 - \frac{2}{16} = \frac{7}{8}\)
\(\sin\theta = \pm\sqrt{\frac{7}{8}} = \pm\frac{\sqrt{14}}{4}\)

Thus, the \(y\)-coordinates are:
\(y = r\sin\theta = \left(\frac{1}{2}a\right)\left(\pm\frac{\sqrt{14}}{4}\right) = \pm\frac{\sqrt{14}}{8}a\)

Checking that \(\frac{\mathrm{d}y}{\mathrm{d}\theta} \neq 0\) at these points confirms they are indeed vertical tangents.

The required Cartesian coordinates are:
\((a(1+\sqrt{2}), 0)\) and \(\left(-\frac{\sqrt{2}}{8}a, \pm\frac{\sqrt{14}}{8}a\right)\)

(a)
- B1: Correct closed-loop shape, symmetric about the initial line, with tangents at the pole clearly approaching \(\theta = \pm\frac{3\pi}{4}\).
- B1: Intercept on the initial line labeled or stated as \(a(1+\sqrt{2})\) (or equivalent).
- B1: Passes through the pole at \(\theta = \pm\frac{3\pi}{4}\).

(b)
- M1: For stating the area formula with correct limits or utilizing symmetry: \(\frac{1}{2}\int r^2 \, \mathrm{d}\theta\).
- M1: For correctly expanding \(r^2\) and substituting the identity \(2\cos^2\theta = 1 + \cos 2\theta\).
- A1: For correct integration to obtain \(2\theta + 2\sqrt{2}\sin\theta + \frac{1}{2}\sin 2\theta\) (condone missing factor of \(a^2\) or \(\frac{1}{2}\) temporarily).
- M1: For substituting the correct limits \(0\) and \(\frac{3\pi}{4}\) into their integrated expression.
- A1: For obtaining the exact given answer: \(\frac{3}{2}a^2(\pi + 1)\) with no errors seen.

(c)
- M1: For using \(x = r\cos\theta\) and differentiating with respect to \(\theta\).
- A1: For obtaining correct derivative \(\frac{\mathrm{d}x}{\mathrm{d}\theta} = -a\sin\theta(1 + 2\sqrt{2}\cos\theta)\).
- M1: For setting \(\frac{\mathrm{d}x}{\mathrm{d}\theta} = 0\) and identifying the two key cases: \(\sin\theta = 0\) and \(\cos\theta = -\frac{\sqrt{2}}{4}\).
- A1: For finding the point \((a(1+\sqrt{2}), 0)\).
- M1: For substituting \(\cos\theta = -\frac{\sqrt{2}}{4}\) to find \(r = \frac{1}{2}a\).
- M1: For calculating the corresponding \(x\) and \(y\) coordinates using \(\sin\theta = \pm\sqrt{1 - \cos^2\theta}\).
- A1: For both coordinates: \(\left(-\frac{\sqrt{2}}{8}a, \pm\frac{\sqrt{14}}{8}a\right)\) (accept decimal approximations only if exact values are clearly derived first, but exact form is expected).

(a) To find the vertical asymptotes, we set the denominator equal to zero:
\[ x^2 - x - 2 = 0 \implies (x - 2)(x + 1) = 0 \implies x = 2 \text{ or } x = -1 \]
Thus, the vertical asymptotes are \( x = -1 \) and \( x = 2 \).

To find the horizontal asymptote, we examine the behavior as \( x \to \pm\infty \):
\[ y = \frac{2 + \frac{5}{x} - \frac{3}{x^2}}{1 - \frac{1}{x} - \frac{2}{x^2}} \]
As \( x \to \pm\infty \), \( y \to \frac{2}{1} = 2 \).
Thus, the horizontal asymptote is \( y = 2 \).

(b) We differentiate \( y \) with respect to \( x \) using the quotient rule:
\[ \frac{dy}{dx} = \frac{(4x + 5)(x^2 - x - 2) - (2x^2 + 5x - 3)(2x - 1)}{(x^2 - x - 2)^2} \]
First, expand the terms in the numerator:
\[ (4x + 5)(x^2 - x - 2) = 4x^3 - 4x^2 - 8x + 5x^2 - 5x - 10 = 4x^3 + x^2 - 13x - 10 \]
\[ (2x^2 + 5x - 3)(2x - 1) = 4x^3 - 2x^2 + 10x^2 - 5x - 6x + 3 = 4x^3 + 8x^2 - 11x + 3 \]
Subtracting these two expressions gives:
\[ (4x^3 + x^2 - 13x - 10) - (4x^3 + 8^2 - 11x + 3) = -7x^2 - 2x - 13 \]
So,
\[ \frac{dy}{dx} = \frac{-(7x^2 + 2x + 13)}{(x^2 - x - 2)^2} \]
For stationary points, \( \frac{dy}{dx} = 0 \implies 7x^2 + 2x + 13 = 0 \).
Checking the discriminant of this quadratic equation:
\[ \Delta = 2^2 - 4(7)(13) = 4 - 364 = -360 < 0 \]
Since the discriminant is negative, there are no real roots. Therefore, \( C \) has no stationary points.

(c) To find the \( y \)-intercept, we set \( x = 0 \):
\[ y = \frac{-3}{-2} = \frac{3}{2} \implies \left(0, \frac{3}{2}\right) \]

To find the \( x \)-intercepts, we set \( y = 0 \):
\[ 2x^2 + 5x - 3 = 0 \implies (2x - 1)(x + 3) = 0 \implies x = \frac{1}{2} \text{ or } x = -3 \]
Thus, the \( x \)-intercepts are \( (-3, 0) \) and \( \left(\frac{1}{2}, 0\right) \).

(d) Sketching the curve:
- Draw dashed lines for the asymptotes: \( x = -1 \), \( x = 2 \), and \( y = 2 \).
- Plot the intercepts: \( (-3, 0) \), \( (0, 1.5) \), and \( (0.5, 0) \).
- The curve consists of three branches:
1. For \( x < -1 \): The curve approaches the horizontal asymptote \( y = 2 \) from below as \( x \to -\infty \), passes downwards through \( (-3, 0) \), and goes to \( -\infty \) as \( x \to -1^- \).
2. For \( -1 < x < 2 \): The curve starts at \( +\infty \) near \( x = -1^+ \), decreases through \( (0, 1.5) \), crosses \( y = 2 \) at \( x = -\frac{1}{7} \), passes through \( (0.5, 0) \), and goes to \( -\infty \) as \( x \to 2^- \).
3. For \( x > 2 \): The curve decreases from \( +\infty \) near \( x = 2^+ \) towards \( y = 2 \) as \( x \to +\infty \).

**Part (a)**
- **M1**: For attempting to solve the denominator equal to 0 for vertical asymptotes.
- **A1**: For both vertical asymptotes \( x = -1 \) and \( x = 2 \).
- **B1**: For the horizontal asymptote \( y = 2 \).

**Part (b)**
- **M1**: For attempting differentiation of \( y \) with respect to \( x \) using the quotient or product rule.
- **A1**: For obtaining the correct simplified numerator \( -7x^2 - 2x - 13 \) (or equivalent).
- **M1**: For setting the derivative to 0 and analyzing the discriminant (or completing the square) of the resulting quadratic.
- **A1**: For showing \( \Delta = -360 < 0 \) (or equivalent completed square form) and correctly concluding there are no stationary points.

**Part (c)**
- **B1**: For the \( y \)-intercept \( (0, 1.5) \) or \( \left(0, \frac{3}{2}\right) \).
- **M1**: For setting the numerator equal to zero and attempting to solve the quadratic equation \( 2x^2 + 5x - 3 = 0 \).
- **A1**: For both \( x \)-intercepts: \( (-3, 0) \) and \( \left(\frac{1}{2}, 0\right) \).

**Part (d)**
- **G1**: For drawing all three asymptotes correctly with dashed lines, labeled with their equations.
- **G1**: For a correct branch for \( x < -1 \) passing through \( (-3, 0) \).
- **G1**: For a correct branch for \( -1 < x < 2 \) passing through \( (0, 1.5) \) and \( (0.5, 0) \).
- **G1**: For a correct branch for \( x > 2 \) approaching \( y = 2 \) from above.
- **G1**: For clearly labeling all intercepts with their coordinates on the sketch.

Since \(2\) is an eigenvalue of \(\mathbf{M}\), the determinant of \(\mathbf{M} - 2\mathbf{I}\) must be zero:
\[\det(\mathbf{M} - 2\mathbf{I}) = \begin{vmatrix} -1 & 3 & 1 \\ 0 & 0 & k \\ 1 & -1 & 1 \end{vmatrix} = 0\]
Expanding the determinant along the second row:
\[-k \begin{vmatrix} -1 & 3 \\ 1 & -1 \end{vmatrix} = -k(1 - 3) = 2k = 0 \implies k = 0\]

With \(k = 0\), the characteristic equation is given by:
\[\det(\mathbf{M} - \lambda\mathbf{I}) = \begin{vmatrix} 1-\lambda & 3 & 1 \\ 0 & 2-\lambda & 0 \\ 1 & -1 & 3-\lambda \end{vmatrix} = 0\]
Expanding along the second row:
\[(2 - \lambda) \left[ (1 - \lambda)(3 - \lambda) - 1 \right] = 0\]
\[(2 - \lambda)(\lambda^2 - 4\lambda + 2) = 0\]

The other two eigenvalues are the solutions to the quadratic equation \(\lambda^2 - 4\lambda + 2 = 0\):
\[\lambda = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(2)}}{2} = \frac{4 \pm \sqrt{8}}{2} = 2 \pm \sqrt{2}\]

M1: For setting up the equation \(\det(\mathbf{M} - 2\mathbf{I}) = 0\) and attempting to solve for \(k\).
A1: For obtaining \(k = 0\).
M1: For writing down the characteristic equation with \(k=0\) and factoring out \((2-\lambda)\).
A1: For correctly identifying the other two eigenvalues as \(2 \pm \sqrt{2}\) (accept exact equivalents).

Since \(2\) is an eigenvalue of \(\mathbf{M}\), the determinant of \(\mathbf{M} - 2\mathbf{I}\) must be zero:
\[\det(\mathbf{M} - 2\mathbf{I}) = \begin{vmatrix} -1 & 3 & 1 \\ 0 & 0 & k \\ 1 & -1 & 1 \end{vmatrix} = 0\]
Expanding the determinant along the second row:
\[-k \begin{vmatrix} -1 & 3 \\ 1 & -1 \end{vmatrix} = -k(1 - 3) = 2k = 0 \implies k = 0\]

With \(k = 0\), the characteristic equation is given by:
\[\det(\mathbf{M} - \lambda\mathbf{I}) = \begin{vmatrix} 1-\lambda & 3 & 1 \\ 0 & 2-\lambda & 0 \\ 1 & -1 & 3-\lambda \end{vmatrix} = 0\]
Expanding along the second row:
\[(2 - \lambda) \left[ (1 - \lambda)(3 - \lambda) - 1 \right] = 0\]
\[(2 - \lambda)(\lambda^2 - 4\lambda + 2) = 0\]

The other two eigenvalues are the solutions to the quadratic equation \(\lambda^2 - 4\lambda + 2 = 0\):
\[\lambda = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(2)}}{2} = \frac{4 \pm \sqrt{8}}{2} = 2 \pm \sqrt{2}\]

M1: For setting up the equation \(\det(\mathbf{M} - 2\mathbf{I}) = 0\) and attempting to solve for \(k\).
A1: For obtaining \(k = 0\).
M1: For writing down the characteristic equation with \(k=0\) and factoring out \((2-\lambda)\).
A1: For correctly identifying the other two eigenvalues as \(2 \pm \sqrt{2}\) (accept exact equivalents).

Let \( w = -8 + 8\sqrt{3}\mathrm{i} \).

First, we find the modulus and argument of \( w \):
\( |w| = \sqrt{(-8)^2 + (8\sqrt{3})^2} = \sqrt{64 + 192} = \sqrt{256} = 16 \).

Since \( w \) lies in the second quadrant, its argument is:
\( \arg(w) = \pi - \arctan\left(\frac{8\sqrt{3}}{8}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \).

Thus, \( w = 16 \mathrm{e}^{\mathrm{i}\frac{2\pi}{3}} \).

To solve \( z^4 = w \), we write \( z^4 = 16 \mathrm{e}^{\mathrm{i}\left(\frac{2\pi}{3} + 2k\pi\right)} \) for \( k \in \mathbb{Z} \).

Using de Moivre's theorem, the roots are given by:
\( z = 16^{1/4} \mathrm{e}^{\mathrm{i}\left(\frac{2\pi/3 + 2k\pi}{4}\right)} = 2 \mathrm{e}^{\mathrm{i}\left(\frac{\pi}{6} + \frac{k\pi}{2}\right)} \).

We choose four consecutive values of \( k \) to obtain the roots in the range \( -\pi < \theta \le \pi \):
- For \( k = 0 \): \( z_1 = 2 \mathrm{e}^{\mathrm{i}\frac{\pi}{6}} \)
- For \( k = 1 \): \( z_2 = 2 \mathrm{e}^{\mathrm{i}\frac{2\pi}{3}} \)
- For \( k = -1 \): \( z_3 = 2 \mathrm{e}^{-\mathrm{i}\frac{\pi}{3}} \)
- For \( k = -2 \): \( z_4 = 2 \mathrm{e}^{-\mathrm{i}\frac{5\pi}{6}} \)

Therefore, the roots are \( 2 \mathrm{e}^{\mathrm{i}\pi/6} \), \( 2 \mathrm{e}^{\mathrm{i}2\pi/3} \), \( 2 \mathrm{e}^{-\mathrm{i}\pi/3} \), and \( 2 \mathrm{e}^{-\mathrm{i}5\pi/6} \).

**M1**: For finding the modulus of \( -8 + 8\sqrt{3}\mathrm{i} \) to get 16, and finding the argument to get \( \frac{2\pi}{3} \) (or equivalent).

**A1**: For the correct exponential form \( 16\mathrm{e}^{\mathrm{i}2\pi/3} \).

**M1**: For using de Moivre's theorem to state the general expression for the fourth roots, dividing the argument (including the general term \( 2k\pi \)) by 4.

**A1**: For obtaining the correct modulus of 2 for the roots and finding at least two correct arguments.

**A1**: For correctly stating all four roots in the requested form with arguments strictly within the range \( (-\pi, \pi] \).

For (i): The first derivative is \( f'(x) = \cosh x \mathrm{e}^{\sinh x} \). Substituting \( x = 0 \) gives \( f'(0) = \cosh(0) \mathrm{e}^{0} = 1 \). Using the product rule, the second derivative is \( f''(x) = (\sinh x + \cosh^2 x) \mathrm{e}^{\sinh x} \). Substituting \( x = 0 \) gives \( f''(0) = (0 + 1^2) \cdot 1 = 1 \). Applying the product rule again, the third derivative is \( f'''(x) = (\cosh x + 2\cosh x\sinh x) \mathrm{e}^{\sinh x} + (\sinh x + \cosh^2 x)\cosh x \mathrm{e}^{\sinh x} \). Substituting \( x = 0 \) gives \( f'''(0) = (1 + 0) \cdot 1 + (0 + 1) \cdot 1 \cdot 1 = 2 \). For (ii): Knowing that \( f(0) = 1 \), we substitute these values into the Maclaurin series formula: \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots \). This yields \( f(x) = 1 + x + \frac{1}{2}x^2 + \frac{2}{6}x^3 = 1 + x + \frac{1}{2}x^2 + \frac{1}{3}x^3 \).

M1: For correctly finding \( f'(x) \) and evaluating \( f'(0) = 1 \). M1: For applying the product rule to obtain a correct form of \( f''(x) \). A1: For obtaining \( f''(0) = 1 \). A1: For obtaining \( f'''(0) = 2 \) with clear supporting work. M1: For substituting the values of \( f(0) \), \( f'(0) \), \( f''(0) \), and \( f'''(0) \) into the standard Maclaurin's series formula. A1: For obtaining the correct expansion \( 1 + x + \frac{1}{2}x^2 + \frac{1}{3}x^3 \).

To find the particular solution of the differential equation:
\[ \frac{d^2 y}{dx^2} + 2 \frac{dy}{dx} + 2y = 5 \cos x \]

**Step 1: Find the complementary function (CF)**
Consider the homogeneous equation:
\[ \frac{d^2 y}{dx^2} + 2 \frac{dy}{dx} + 2y = 0 \]

The auxiliary equation is:
\[ m^2 + 2m + 2 = 0 \]

Solving for \( m \):
\[ m = \frac{-2 \pm \sqrt{2^2 - 4(1)(2)}}{2} = \frac{-2 \pm \sqrt{-4}}{2} = -1 \pm i \]

Since the roots are complex conjugate \( -1 \pm i \), the complementary function is:
\[ y_c = e^{-x} (A \cos x + B \sin x) \]
where \( A \) and \( B \) are arbitrary constants.

**Step 2: Find the particular integral (PI)**
Since the right-hand side is \( 5 \cos x \), we try a particular integral of the form:
\[ y_p = P \cos x + Q \sin x \]

Differentiating this expression gives:
\[ \frac{dy_p}{dx} = -P \sin x + Q \cos x \]
\[ \frac{d^2 y_p}{dx^2} = -P \cos x - Q \sin x \]

Substituting these into the original differential equation:
\[ (-P \cos x - Q \sin x) + 2(-P \sin x + Q \cos x) + 2(P \cos x + Q \sin x) = 5 \cos x \]

Grouping the \( \cos x \) and \( \sin x \) terms:
\[ (-P + 2Q + 2P)\cos x + (-Q - 2P + 2Q)\sin x = 5 \cos x \]
\[ (P + 2Q)\cos x + (Q - 2P)\sin x = 5 \cos x \]

Equating the coefficients on both sides:
1) \( P + 2Q = 5 \)
2) \( Q - 2P = 0 \implies Q = 2P \)

Substitute \( Q = 2P \) into the first equation:
\[ P + 2(2P) = 5 \implies 5P = 5 \implies P = 1 \]
Then:
\[ Q = 2(1) = 2 \]

Thus, the particular integral is:
\[ y_p = \cos x + 2 \sin x \]

**Step 3: State the general solution**
The general solution is the sum of the complementary function and the particular integral:
\[ y = e^{-x} (A \cos x + B \sin x) + \cos x + 2 \sin x \]

**Step 4: Apply the initial conditions**
We are given that \( y = 2 \) when \( x = 0 \):
\[ 2 = e^{0} (A \cos 0 + B \sin 0) + \cos 0 + 2 \sin 0 \]
\[ 2 = A + 1 \implies A = 1 \]

Next, differentiate the general solution to apply the second initial condition:
\[ \frac{dy}{dx} = -e^{-x}(A \cos x + B \sin x) + e^{-x}(-A \sin x + B \cos x) - \sin x + 2 \cos x \]

We are given that \( \frac{dy}{dx} = 0 \) when \( x = 0 \):
\[ 0 = -e^{0}(A \cos 0 + B \sin 0) + e^{0}(-A \sin 0 + B \cos 0) - \sin 0 + 2 \cos 0 \]
\[ 0 = -A + B + 2 \]

Since \( A = 1 \):
\[ 0 = -1 + B + 2 \implies B = -1 \]

**Step 5: Write down the final particular solution**
Substituting \( A = 1 \) and \( B = -1 \) back into the general solution:
\[ y = e^{-x} (\cos x - \sin x) + \cos x + 2 \sin x \]

- **M1**: For setting up and attempting to solve the auxiliary equation \( m^2 + 2m + 2 = 0 \).
- **A1**: For finding the correct roots \( m = -1 \pm i \).
- **A1**: For stating the correct complementary function \( y_c = e^{-x}(A \cos x + B \sin x) \).
- **M1**: For assuming a particular integral of the form \( y_p = P \cos x + Q \sin x \) and substituting its derivatives into the differential equation.
- **A1**: For finding the correct constants \( P = 1 \) and \( Q = 2 \) (leading to the correct general solution).
- **M1**: For substituting the boundary condition \( y(0) = 2 \) into their general solution to find \( A \).
- **A1**: For obtaining \( A = 1 \).
- **M1**: For differentiating their general solution using the product rule.
- **M1**: For substituting \( \frac{dy}{dx}(0) = 0 \) and their value of \( A \) to find \( B \).
- **A1**: For obtaining \( B = -1 \) and stating the correct final particular solution \( y = e^{-x}(\cos x - \sin x) + \cos x + 2 \sin x \).

**(i)**
First, find the derivatives of \(x\) and \(y\) with respect to \(t\):
\[ \frac{dx}{dt} = \sec t - \cos t \]
\[ \frac{dy}{dt} = -\sin t \]

Next, calculate the term under the square root for the arc length formula:
\[ \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = (\sec t - \cos t)^2 + (-\sin t)^2 \]
\[ = \sec^2 t - 2\sec t\cos t + \cos^2 t + \sin^2 t \]
Since \(\sec t \cos t = 1\), this simplifies to:
\[ = \sec^2 t - 2 + 1 = \sec^2 t - 1 = \tan^2 t \]

Since \(0 \le t \le \frac{\pi}{3}\), we have \(\tan t \ge 0\), so:
\[ \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} = \tan t \]

The arc length \(s\) is given by:
\[ s = \int_{0}^{\pi/3} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt = \int_{0}^{\pi/3} \tan t \, dt \]
\[ = \left[ \ln|\sec t| \right]_{0}^{\pi/3} \]
\[ = \ln(\sec(\pi/3)) - \ln(\sec(0)) \]
Since \(\sec(\pi/3) = 2\) and \(\sec(0) = 1\):
\[ s = \ln 2 - \ln 1 = \ln 2 \]

**(ii)**
The surface area \(S\) of the surface generated by rotating the arc about the \(x\)-axis is given by:
\[ S = 2\pi \int_{0}^{\pi/3} y \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]
Substitute \(y = \cos t\) and the square root term from part (i):
\[ S = 2\pi \int_{0}^{\pi/3} \cos t \tan t \, dt \]
Since \(\cos t \tan t = \sin t\):
\[ S = 2\pi \int_{0}^{\pi/3} \sin t \, dt \]
\[ = 2\pi \left[ -\cos t \right]_{0}^{\pi/3} \]
\[ = 2\pi \left( -\cos\left(\frac{\pi}{3}\right) - (-\cos(0)) \right) \]
\[ = 2\pi \left( -\frac{1}{2} + 1 \right) = \pi \]

**(i)**
- **M1**: Differentiate both \(x\) and \(y\) correctly with respect to \(t\) (allow minor slips).
- **M1**: Substitute into \(\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2\) and simplify to \(\tan^2 t\) using trigonometric identities.
- **A1**: Obtain \(\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} = \tan t\).
- **M1**: Integrate \(\tan t\) to get \(\ln(\sec t)\) (or equivalent form) and substitute limits \(0\) and \(\frac{\pi}{3}\).
- **A1**: Obtain the exact arc length \(\ln 2\).

**(ii)**
- **M1**: State or use the correct formula for the surface area of revolution about the \(x\)-axis: \(2\pi \int y \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\).
- **M1**: Substitute \(y = \cos t\) and the result from part (i) to obtain \(2\pi \int_{0}^{\pi/3} \cos t \tan t \, dt\).
- **A1**: Simplify the integrand to \(\sin t\).
- **M1**: Integrate \(\sin t\) and substitute the limits \(0\) and \(\frac{\pi}{3}\) correctly.
- **A1**: Obtain the exact surface area \(\pi\).

**(i)**

First, find the complementary function (CF) by solving the auxiliary equation:
\[ m^2 - 4 = 0 \implies m = \pm 2 \]
Thus, the CF is:
\[ y_c = A \cosh(2x) + B \sinh(2x) \]
(or \(C e^{2x} + D e^{-2x}\)).

Next, find the particular integral (PI). Since \(\cosh(2x)\) is a solution of the homogeneous equation, we must use the form:
\[ y_p = x(K \cosh(2x) + L \sinh(2x)) \]

Differentiate twice with respect to \(x\):
\[ y_p' = K \cosh(2x) + L \sinh(2x) + 2x(K \sinh(2x) + L \cosh(2x)) \]
\[ y_p'' = 2(K \sinh(2x) + L \cosh(2x)) + 2(K \sinh(2x) + L \cosh(2x)) + 4x(K \cosh(2x) + L \sinh(2x)) \]
\[ y_p'' = 4K \sinh(2x) + 4L \cosh(2x) + 4y_p \]

Substitute \(y_p\) and \(y_p''\) into the original differential equation:
\[ (4K \sinh(2x) + 4L \cosh(2x) + 4y_p) - 4y_p = 8 \cosh(2x) \]
\[ 4K \sinh(2x) + 4L \cosh(2x) = 8 \cosh(2x) \]

Equating coefficients of \(\sinh(2x)\) and \(\cosh(2x)\):
\[ 4K = 0 \implies K = 0 \]
\[ 4L = 8 \implies L = 2 \]

So the PI is:
\[ y_p = 2x \sinh(2x) \]

Thus, the general solution is:
\[ y = A \cosh(2x) + B \sinh(2x) + 2x \sinh(2x) \]
(or \(y = C e^{2x} + D e^{-2x} + 2x \sinh(2x)\)).

**(ii)**

We are given the initial conditions \(y = 1\) and \(\frac{\text{d} y}{\text{d} x} = 0\) at \(x = 0\).

Using \(y = 1\) at \(x = 0\):
\[ 1 = A \cosh(0) + B \sinh(0) + 0 \implies A = 1 \]

Now, differentiate the general solution:
\[ \frac{\text{d} y}{\text{d} x} = 2A \sinh(2x) + 2B \cosh(2x) + 2 \sinh(2x) + 4x \cosh(2x) \]

Using \(\frac{\text{d} y}{\text{d} x} = 0\) at \(x = 0\):
\[ 0 = 0 + 2B + 0 + 0 \implies B = 0 \]

Therefore, the particular solution is:
\[ y = \cosh(2x) + 2x \sinh(2x) \]

**(iii)**

To find stationary points, we set \(\frac{\text{d} y}{\text{d} x} = 0\):
\[ \frac{\text{d} y}{\text{d} x} = 2 \sinh(2x) + 2 \sinh(2x) + 4x \cosh(2x) = 4(\sinh(2x) + x \cosh(2x)) = 0 \]
\[ \sinh(2x) + x \cosh(2x) = 0 \]

Clearly, if \(x = 0\), we have \(\sinh(0) + 0 \cdot \cosh(0) = 0\), which satisfies the equation. The corresponding \(y\)-value is:
\[ y = \cosh(0) + 0 = 1 \]
So \((0, 1)\) is a stationary point.

Now we show there are no other stationary points:
- If \(x > 0\), then \(\sinh(2x) > 0\) and \(x \cosh(2x) > 0\), hence \(\sinh(2x) + x \cosh(2x) > 0\).
- If \(x < 0\), then \(\sinh(2x) < 0\) and \(x \cosh(2x) < 0\), hence \(\sinh(2x) + x \cosh(2x) < 0\).
Therefore, \(x = 0\) is the unique solution to \(\frac{\text{d} y}{\text{d} x} = 0\).

To determine the nature of the stationary point, we can evaluate the second derivative:
From the differential equation:
\[ \frac{\text{d}^2 y}{\text{d} x^2} = 4y + 8 \cosh(2x) \]
At \(x = 0\) and \(y = 1\):
\[ \frac{\text{d}^2 y}{\text{d} x^2} = 4(1) + 8 \cosh(0) = 12 > 0 \]
Since the second derivative is positive, the stationary point \((0, 1)\) is a minimum.

**(i)**
M1: Find roots of the auxiliary equation \(m^2 - 4 = 0\).
A1: Obtain correct CF (either in exponential or hyperbolic form).
M1: State a correct form for the PI: \(y_p = x(K \cosh(2x) + L \sinh(2x))\) or equivalent exponential form.
M1: Differentiate the proposed PI twice with respect to \(x\).
M1: Substitute into the differential equation and solve for coefficients \(K\) and \(L\).
A1: Obtain \(K = 0\) and \(L = 2\) (or equivalent for exponential form).
A1: State the correct general solution.

**(ii)**
M1: Substitute \(x = 0, y = 1\) into the general solution to find one constant.
M1: Differentiate the general solution and substitute \(x = 0, y' = 0\) to find the other constant.
A1: Obtain the correct particular solution \(y = \cosh(2x) + 2x \sinh(2x)\).

**(iii)**
M1: Differentiate the particular solution and set \(\frac{\text{d} y}{\text{d} x} = 0\) to obtain \(4(\sinh(2x) + x \cosh(2x)) = 0\).
A1: Identify \(x = 0\) as a solution and state coordinates \((0, 1)\).
A1: Provide a convincing argument showing that \(\sinh(2x) + x \cosh(2x) = 0\) has no solutions for \(x > 0\) and \(x < 0\).
A1: Calculate \(\frac{\text{d}^2 y}{\text{d} x^2} = 12 > 0\) at \(x = 0\) (or use first derivative sign test) to justify that it is a minimum.

**(i)** To find the eigenvalues of \(A\), we solve the characteristic equation \(\det(A - \lambda I) = 0\):

\(\det \begin{pmatrix} 3-\lambda & -1 & 1 \\ -1 & 5-\lambda & -1 \\ 1 & -1 & 3-\lambda \end{pmatrix} = 0\)

Expanding along the first row:

\((3-\lambda) \left[ (5-\lambda)(3-\lambda) - 1 \right] - (-1) \left[ -1(3-\lambda) - (-1) \right] + 1 \left[ 1 - (5-\lambda) \right] = 0\)

Simplifying the terms:

\((3-\lambda)(\lambda^2 - 8\lambda + 14) + (\lambda - 2) + (\lambda - 4) = 0\)

\((3\lambda^2 - 24\lambda + 42 - \lambda^3 + 8\lambda^2 - 14\lambda) + 2\lambda - 6 = 0\)

\(-\lambda^3 + 11\lambda^2 - 36\lambda + 36 = 0\)

We test integer roots. For \(\lambda = 2\):

\(-(2)^3 + 11(2)^2 - 36(2) + 36 = -8 + 44 - 72 + 36 = 0\)

Thus \(\lambda = 2\) is an eigenvalue. Factoring out \(\lambda - 2\):

\(-(\lambda - 2)(\lambda^2 - 9\lambda + 18) = 0\)

\(-(\lambda - 2)(\lambda - 3)(\lambda - 6) = 0\)

So, the eigenvalues of \(A\) are \(\lambda_1 = 2\), \(\lambda_2 = 3\), and \(\lambda_3 = 6\).

**(ii)** Now find eigenvectors for each eigenvalue:

- **For \(\lambda = 2\)**:

\((A - 2I)\mathbf{v} = 0 \implies \begin{pmatrix} 1 & -1 & 1 \\ -1 & 3 & -1 \\ 1 & -1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\)

From row 1, \(x - y + z = 0\) and from row 2, \(-x + 3y - z = 0\). Adding these gives \(2y = 0 \implies y = 0\). Then \(x + z = 0 \implies z = -x\). Thus, a corresponding eigenvector is \(\mathbf{v}_1 = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}\).

- **For \(\lambda = 3\)**:

\((A - 3I)\mathbf{v} = 0 \implies \begin{pmatrix} 0 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\)

From row 1, \(-y + z = 0 \implies y = z\) and from row 3, \(x - y = 0 \implies x = y\). Thus, \(x = y = z\). A corresponding eigenvector is \(\mathbf{v}_2 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\).

- **For \(\lambda = 6\)**:

\((A - 6I)\mathbf{v} = 0 \implies \begin{pmatrix} -3 & -1 & 1 \\ -1 & -1 & -1 \\ 1 & -1 & -3 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\)

From row 2, \(x + y + z = 0 \implies y = -x - z\). From row 1, \(-3x - (-x - z) + z = 0 \implies -2x + 2z = 0 \implies x = z\). Then \(y = -2x\). A corresponding eigenvector is \(\mathbf{v}_3 = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}\).

Using these eigenvectors, we construct \(P\) and \(D\):

\(P = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & -2 \\ -1 & 1 & 1 \end{pmatrix}, \quad D = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 6 \end{pmatrix}\)

*(Note: any consistent ordering of the eigenvalues and eigenvectors is acceptable.)*

**(iii)** To find \(A^{-1}\), we find \(P^{-1}\). Since the columns of \(P\) are orthogonal, we can use the relation \(P^{-1} = D_P^{-1} P^T\) where \(D_P = P^T P = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 6 \end{pmatrix}\), or use the standard adjugate formula:

\(\det(P) = 1(1 - (-2)) - 1(0 - 2) + 1(0 - (-1)) = 6\)

Matrix of cofactors \(C = \begin{pmatrix} 3 & 2 & 1 \\ 0 & 2 & -2 \\ -3 & 2 & 1 \end{pmatrix} \implies P^{-1} = \frac{1}{6} C^T = \frac{1}{6} \begin{pmatrix} 3 & 0 & -3 \\ 2 & 2 & 2 \\ 1 & -2 & 1 \end{pmatrix} = \begin{pmatrix} 1/2 & 0 & -1/2 \\ 1/3 & 1/3 & 1/3 \\ 1/6 & -1/3 & 1/6 \end{pmatrix}\).

Now, compute \(A^{-1} = P D^{-1} P^{-1}\):

\(D^{-1} = \begin{pmatrix} 1/2 & 0 & 0 \\ 0 & 1/3 & 0 \\ 0 & 0 & 1/6 \end{pmatrix}\)

\(P D^{-1} = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & -2 \\ -1 & 1 & 1 \end{pmatrix} \begin{pmatrix} 1/2 & 0 & 0 \\ 0 & 1/3 & 0 \\ 0 & 0 & 1/6 \end{pmatrix} = \begin{pmatrix} 1/2 & 1/3 & 1/6 \\ 0 & 1/3 & -1/3 \\ -1/2 & 1/3 & 1/6 \end{pmatrix}\)

\(A^{-1} = \begin{pmatrix} 1/2 & 1/3 & 1/6 \\ 0 & 1/3 & -1/3 \\ -1/2 & 1/3 & 1/6 \end{pmatrix} \begin{pmatrix} 1/2 & 0 & -1/2 \\ 1/3 & 1/3 & 1/3 \\ 1/6 & -1/3 & 1/6 \end{pmatrix} = \begin{pmatrix} 7/18 & 1/18 & -1/9 \\ 1/18 & 2/9 & 1/18 \\ -1/9 & 1/18 & 7/18 \end{pmatrix} = \frac{1}{18} \begin{pmatrix} 7 & 1 & -2 \\ 1 & 4 & 1 \\ -2 & 1 & 7 \end{pmatrix}\).

**(i)**
- **M1**: For attempting to find the characteristic equation \(\det(A - \lambda I) = 0\).
- **A1**: For obtaining the correct simplified cubic equation \(\lambda^3 - 11\lambda^2 + 36\lambda - 36 = 0\).
- **A1**: For finding one correct eigenvalue (e.g., \(\lambda = 2\)).
- **A1**: For finding the remaining two eigenvalues (\(\lambda = 3, 6\)).

**(ii)**
- **M1**: For setting up \((A - \lambda I)\mathbf{v} = 0\) and attempting to find eigenvectors.
- **A1**: For finding any one correct eigenvector.
- **A1**: For finding all three correct eigenvectors (e.g., \([1, 0, -1]^T\), \([1, 1, 1]^T\), \([1, -2, 1]^T\)).
- **B1**: For writing down a correct \(P\) and \(D\) which match their order of eigenvectors.

**(iii)**
- **M1**: For a complete, correct method to find \(P^{-1}\).
- **M1**: For setting up the matrix multiplication \(P D^{-1} P^{-1}\) or another valid method like Cayley-Hamilton.
- **A1**: For the correct final matrix \(A^{-1} = \frac{1}{18} \begin{pmatrix} 7 & 1 & -2 \\ 1 & 4 & 1 \\ -2 & 1 & 7 \end{pmatrix}\).

(i) Let \( f(x) = \frac{1}{x(\ln x)^2} \). For \( x > 1 \), the first derivative is:
\( f'(x) = -\frac{(\ln x)^2 + 2\ln x}{x^2 (\ln x)^4} \)
Since \( f'(x) < 0 \) for all \( x \ge 2 \), \( f(x) \) is a strictly decreasing function.

For any integer \( r \ge 3 \), on the interval \( [r-1, r] \), the minimum value of \( f(x) \) is \( f(r) \) and the maximum value is \( f(r-1) \).
Therefore, for all \( x \in (r-1, r) \):
\( f(r) < f(x) < f(r-1) \)

Integrating this inequality from \( r-1 \) to \( r \):
\( \int_{r-1}^{r} f(r) \, dx < \int_{r-1}^{r} f(x) \, dx < \int_{r-1}^{r} f(r-1) \, dx \)
Which simplifies to:
\( f(r) < \int_{r-1}^{r} f(x) \, dx < f(r-1) \)

Summing this inequality from \( r = 3 \) to \( N \):
\( \sum_{r=3}^{N} f(r) < \sum_{r=3}^{N} \int_{r-1}^{r} f(x) \, dx < \sum_{r=3}^{N} f(r-1) \)

Evaluating the middle term as a single integral:
\( \sum_{r=3}^{N} \int_{r-1}^{r} f(x) \, dx = \int_{2}^{N} f(x) \, dx \)

And shifting the index of the right-hand sum with \( u = r-1 \):
\( \sum_{r=3}^{N} f(r-1) = \sum_{u=2}^{N-1} f(u) \)

Thus, we obtain:
\( \sum_{r=3}^{N} \frac{1}{r(\ln r)^2} < \int_{2}^{N} \frac{1}{x(\ln x)^2} \, dx < \sum_{r=2}^{N-1} \frac{1}{r(\ln r)^2} \)

(ii) To find \( \int_{2}^{N} \frac{1}{x(\ln x)^2} \, dx \), we use the substitution \( u = \ln x \), which gives \( du = \frac{1}{x} \, dx \).
When \( x = 2 \), \( u = \ln 2 \).
When \( x = N \), \( u = \ln N \).

Substituting these into the integral:
\( \int_{\ln 2}^{\ln N} u^{-2} \, du = \left[ -\frac{1}{u} \right]_{\ln 2}^{\ln N} = \frac{1}{\ln 2} - \frac{1}{\ln N} \)

(iii) From part (i), we have:
\( \sum_{r=3}^{N} f(r) < \int_{2}^{N} f(x) \, dx \)
Adding \( f(2) \) to both sides:
\( \sum_{r=2}^{N} f(r) < f(2) + \int_{2}^{N} f(x) \, dx \)
Substituting the result from part (ii):
\( \sum_{r=2}^{N} f(r) < \frac{1}{2(\ln 2)^2} + \frac{1}{\ln 2} - \frac{1}{\ln N} < \frac{1}{2(\ln 2)^2} + \frac{1}{\ln 2} \)
Since all terms of the series are positive, the sequence of partial sums \( S_N = \sum_{r=2}^{N} f(r) \) is strictly increasing and bounded above by \( \frac{1}{2(\ln 2)^2} + \frac{1}{\ln 2} \). Therefore, the infinite series converges.

Taking the limit as \( N \to \infty \) of the upper bound:
\( S \le \frac{1}{2(\ln 2)^2} + \frac{1}{\ln 2} \approx 2.483 \)

From the inequality in part (i):
\( \int_{2}^{N} f(x) \, dx < \sum_{r=2}^{N-1} f(r) < S \)
Taking the limit as \( N \to \infty \):
\( S \ge \lim_{N \to \infty} \int_{2}^{N} f(x) \, dx = \frac{1}{\ln 2} \approx 1.443 \)

So, the bounds are:
\( \text{Lower Bound} = \frac{1}{\ln 2} \)
\( \text{Upper Bound} = \frac{1}{\ln 2} + \frac{1}{2(\ln 2)^2} \)

(iv) The approximation error \( E \) is:
\( E = S - \sum_{r=2}^{M-1} f(r) = \sum_{r=M}^{\infty} f(r) \)
Using the inequality from the lower rectangles, summing from \( r = M \) to \( N \):
\( \sum_{r=M}^{N} f(r) < \int_{M-1}^{N} f(x) \, dx \)
Taking the limit as \( N \to \infty \):
\( E = \sum_{r=M}^{\infty} f(r) \le \int_{M-1}^{\infty} f(x) \, dx = \left[ -\frac{1}{\ln x} \right]_{M-1}^{\infty} = \frac{1}{\ln(M-1)} \)
Hence, \( E < \frac{1}{\ln(M-1)} \).

To ensure the error is guaranteed to be less than \( 0.5 \), we set the upper bound of the error to be less than or equal to \( 0.5 \):
\( \frac{1}{\ln(M-1)} \le 0.5 \implies \ln(M-1) \ge 2 \implies M-1 \ge e^2 \approx 7.389 \implies M \ge 8.389 \)
Since \( M \) must be an integer, the smallest integer is \( M = 9 \).

**Part (i) [5 marks]**
- **M1**: Establishes that \( f(x) = \frac{1}{x(\ln x)^2} \) is strictly decreasing for \( x \ge 2 \) (e.g., via derivative or clear sign analysis).
- **M1**: Sets up interval inequality \( f(r) < f(x) < f(r-1) \) for \( x \in (r-1, r) \).
- **M1**: Integrates interval inequality to yield \( f(r) < \int_{r-1}^{r} f(x) \, dx < f(r-1) \).
- **A1**: Sums from \( r = 3 \) to \( N \) to show LHS: \( \sum_{r=3}^{N} f(r) < \int_{2}^{N} f(x) \, dx \).
- **A1**: Sums from \( r = 3 \) to \( N \) to show RHS: \( \int_{2}^{N} f(x) \, dx < \sum_{r=2}^{N-1} f(r) \) and completes the full inequality.

**Part (ii) [3 marks]**
- **M1**: Employs a valid substitution method such as \( u = \ln x \).
- **A1**: Obtains correct antiderivative \( -\frac{1}{\ln x} \).
- **A1**: Evaluates limits correctly to find \( \frac{1}{\ln 2} - \frac{1}{\ln N} \).

**Part (iii) [4 marks]**
- **M1**: Formulates \( \sum_{r=2}^{N} f(r) < f(2) + \int_{2}^{N} f(x) \, dx \) and explains convergence using monotonic increasing partial sums bounded above.
- **A1**: Obtains correct limit of the integral as \( \frac{1}{\ln 2} \).
- **A1**: Computes the upper bound correctly: \( \frac{1}{\ln 2} + \frac{1}{2(\ln 2)^2} \) (accept decimal equivalent \( \approx 2.48 \)).
- **A1**: Computes the lower bound correctly: \( \frac{1}{\ln 2} \) (accept decimal equivalent \( \approx 1.44 \)).

**Part (iv) [3 marks]**
- **M1**: Identifies the error bound as \( E \le \int_{M-1}^{\infty} f(x) \, dx \) and integrates to find \( \frac{1}{\ln(M-1)} \).
- **M1**: Formulates inequality \( \frac{1}{\ln(M-1)} \le 0.5 \) and solves for \( M \).
- **A1**: Deduces correct smallest integer \( M = 9 \).