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

To show the identity, we combine the fractions on the right-hand side over a common denominator: \(\frac{1}{2r^2 - 2r + 1} - \frac{1}{2r^2 + 2r + 1} = \frac{(2r^2 + 2r + 1) - (2r^2 - 2r + 1)}{(2r^2 - 2r + 1)(2r^2 + 2r + 1)}\). The numerator simplifies to: \(2r^2 + 2r + 1 - (2r^2 - 2r + 1) = 4r\). The denominator is expanded as: \((2r^2 + 1 - 2r)(2r^2 + 1 + 2r) = (2r^2 + 1)^2 - (2r)^2 = 4r^4 + 4r^2 + 1 - 4r^2 = 4r^4 + 1\). Thus, the identity is shown. Next, we use the method of differences: \(\sum_{r=1}^{n} \frac{4r}{4r^4 + 1} = \sum_{r=1}^{n} \left( \frac{1}{2r^2 - 2r + 1} - \frac{1}{2r^2 + 2r + 1} \right)\). Listing the terms: for \(r=1\): \(\frac{1}{1} - \frac{1}{5}\), for \(r=2\): \(\frac{1}{5} - \frac{1}{13}\), for \(r=3\): \(\frac{1}{13} - \frac{1}{25}\), ..., for \(r=n\): \(\frac{1}{2n^2 - 2n + 1} - \frac{1}{2n^2 + 2n + 1}\). Summing these terms, all intermediate terms cancel, leaving: \(S_n = 1 - \frac{1}{2n^2 + 2n + 1}\). As \(n \to \infty\), \(\frac{1}{2n^2 + 2n + 1} \to 0\). Therefore, the sum to infinity is \(1\).

M1: For attempting to combine the fractions on the right-hand side over a common denominator. A1: For obtaining a numerator of \(4r\). A1: For correctly expanding the denominator to \(4r^4 + 1\) to complete the proof. M1: For applying the method of differences, showing the first few terms and the general term. A1: For identifying which terms cancel. A1: For the correct sum of \(n\) terms, \(1 - \frac{1}{2n^2 + 2n + 1}\) (or equivalent). B1: For stating that the sum to infinity is \(1\), with a clear explanation that the limit of the fractional term is \(0\) 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 the statement is true for \( n = k \), where \( k \) is a positive integer. That is, assume \( f(k) = 5^{2k} + 24k - 1 = 48m \) for some integer \( m \). We need to show that the statement is true for \( n = k + 1 \), i.e., \( f(k+1) = 5^{2(k+1)} + 24(k+1) - 1 \) is divisible by 48. Consider \( f(k+1) = 5^{2k+2} + 24k + 24 - 1 = 25(5^{2k}) + 24k + 23 \). Substituting \( 5^{2k} = 48m - 24k + 1 \), we get: \( f(k+1) = 25(48m - 24k + 1) + 24k + 23 = 25 \times 48m - 600k + 25 + 24k + 23 = 25 \times 48m - 576k + 48 \). Since \( 576 = 12 \times 48 \), we can factor out 48: \( f(k+1) = 48(25m - 12k + 1) \). Since \( m \) and \( k \) are integers, \( 25m - 12k + 1 \) is an integer, so \( f(k+1) \) is divisible by 48. Conclusion: Since the base case is true and the inductive step holds, by the principle of mathematical induction, \( 5^{2n} + 24n - 1 \) is divisible by 48 for all positive integers \( n \).

B1: Verify the base case \( n = 1 \) is true, showing \( f(1) = 48 \). M1: State the inductive hypothesis (assume true for \( n = k \)) and set up the expression for \( n = k + 1 \). M1: Use a valid algebraic method to relate \( f(k+1) \) to \( f(k) \) or to substitute \( 5^{2k} = 48m - 24k + 1 \). A1: Obtain a correct unsimplified expression showing divisibility, e.g., \( 25 \times 48m - 576k + 48 \). A1: Correctly factor out 48 to show \( f(k+1) = 48(25m - 12k + 1) \) or equivalent, showing \( 576 = 12 \times 48 \). A1: Provide a complete and correct concluding statement of induction.

(a) From the given cubic equation \(x^3 - 2x^2 + 4x - 3 = 0\), we write down the relations between the roots and coefficients:
\(\sum \alpha = 2\)
\(\sum \alpha\beta = 4\)
\(\alpha\beta\gamma = 3\)

First, calculate \(\sum \alpha^2\):
\(\sum \alpha^2 = (\sum \alpha)^2 - 2\sum \alpha\beta = 2^2 - 2(4) = 4 - 8 = -4\)

Since \(\alpha\), \(\beta\), and \(\gamma\) are roots of the cubic equation, they satisfy:
\(\alpha^3 - 2\alpha^2 + 4\alpha - 3 = 0\)
\(\beta^3 - 2\beta^2 + 4\beta - 3 = 0\)
\(\gamma^3 - 2\gamma^2 + 4\gamma - 3 = 0\)

Summing these three equations yields:
\(\sum \alpha^3 - 2\sum \alpha^2 + 4\sum \alpha - 9 = 0\)

Substitute the known values into this sum:
\(\sum \alpha^3 - 2(-4) + 4(2) - 9 = 0\)
\(\sum \alpha^3 + 8 + 8 - 9 = 0\)
\(\sum \alpha^3 + 7 = 0\)
\(\sum \alpha^3 = -7\)

(b) Let \(y = x^2\) be the roots of the new equation. Thus, \(x = \sqrt{y}\).

Rearranging the original cubic equation to group odd and even powers of \(x\):
\(x^3 + 4x = 2x^2 + 3\)
\(x(x^2 + 4) = 2x^2 + 3\)

Substituting \(x = \sqrt{y}\) and \(x^2 = y\):
\(\sqrt{y}(y + 4) = 2y + 3\)

Squaring both sides of the equation:
\(y(y + 4)^2 = (2y + 3)^2\)
\(y(y^2 + 8y + 16) = 4y^2 + 12y + 9\)

Expanding and simplifying:
\(y^3 + 8y^2 + 16y = 4y^2 + 12y + 9\)
\(y^3 + 4y^2 + 4y - 9 = 0\)

(Alternatively, by symmetric functions:
\(S_1 = \alpha^2 + \beta^2 + \gamma^2 = -4\)
\(S_2 = \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 = (\sum \alpha\beta)^2 - 2\alpha\beta\gamma(\sum \alpha) = 4^2 - 2(3)(2) = 16 - 12 = 4\)
\(S_3 = \alpha^2\beta^2\gamma^2 = (\alpha\beta\gamma)^2 = 3^2 = 9\)

The required equation is \(y^3 - S_1 y^2 + S_2 y - S_3 = 0\), which gives \(y^3 + 4y^2 + 4y - 9 = 0\))

(a)
M1: For finding the sum of the roots squared, \(\sum \alpha^2\), or for utilizing the cubic identity for \(\sum \alpha^3\).
A1: For showing \(\sum \alpha^2 = -4\).
A1: For obtaining \(\sum \alpha^3 = -7\).

(b)
M1: For rearranging the cubic equation to isolate odd powers of \(x\) (or for expressing \(S_2\) in terms of symmetric sums).
M1: For squaring both sides and substituting \(y = x^2\) (or for evaluating \(S_2 = 4\) correctly using the identity).
A1: For expanding and simplifying correctly (or for obtaining \(S_3 = 9\) correctly).
A1: For collecting terms into a standard polynomial form.
A1: For the final correct equation \(y^3 + 4y^2 + 4y - 9 = 0\) (accept any variable except \(x\), must include "= 0").

(i) First, check if the direction vectors \(\mathbf{d}_1 = \mathbf{i} + 2\mathbf{j} - \mathbf{k}\) and \(\mathbf{d}_2 = 2\mathbf{i} - \mathbf{j} + 3\mathbf{k}\) are parallel. Since there is no scalar \(k\) such that \(\mathbf{d}_1 = k\mathbf{d}_2\) (as \(2/1 \neq -1/2\)), the lines are not parallel.

Next, test for intersection by setting the coordinates equal:
1) \(1 + \lambda = 2 + 2\mu \Rightarrow \lambda - 2\mu = 1\)
2) \(-2 + 2\lambda = 1 - \mu \Rightarrow 2\lambda + \mu = 3\)
3) \(3 - \lambda = 3\mu \Rightarrow \lambda + 3\mu = 3\)

From (1) and (2):
Multiplying (2) by 2 gives \(4\lambda + 2\mu = 6\).
Adding this to (1) gives \(5\lambda = 7 \Rightarrow \lambda = 1.4\).
Substituting back into (2) gives \(2(1.4) + \mu = 3 \Rightarrow \mu = 0.2\).

Check these values in (3):
\(\lambda + 3\mu = 1.4 + 3(0.2) = 2.0 \neq 3\).

Since the system of equations has no consistent solution, the lines do not intersect. Since they are not parallel and do not intersect, they are skew.

(ii) The common perpendicular direction \(\mathbf{n}\) is given by \(\mathbf{d}_1 \times \mathbf{d}_2\):
\(\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -1 \\ 2 & -1 & 3 \end{vmatrix} = \mathbf{i}(6 - 1) - \mathbf{j}(3 - (-2)) + \mathbf{k}(-1 - 4) = 5\mathbf{i} - 5\mathbf{j} - 5\mathbf{k}\).

We can use the simplified normal vector \(\mathbf{n}_0 = \mathbf{i} - \mathbf{j} - \mathbf{k}\).

Let \(A(1, -2, 3)\) be a point on \(l_1\) and \(B(2, 1, 0)\) be a point on \(l_2\).
The vector connecting these points is \(\vec{AB} = (2-1)\mathbf{i} + (1 - (-2))\mathbf{j} + (0-3)\mathbf{k} = \mathbf{i} + 3\mathbf{j} - 3\mathbf{k}\).

The shortest distance \(d\) is the projection of \(\vec{AB}\) onto \(\mathbf{n}_0\):
\(d = \frac{|\vec{AB} \cdot \mathbf{n}_0|}{|\mathbf{n}_0|} = \frac{|1(1) + 3(-1) + (-3)(-1)|}{\sqrt{1^2 + (-1)^2 + (-1)^2}} = \frac{|1 - 3 + 3|}{\sqrt{3}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}\).

(iii) The normal to the plane \(\Pi\) must be perpendicular to both \(l_1\) and \(l_2\), so we can use \(\mathbf{n}_0 = \mathbf{i} - \mathbf{j} - \mathbf{k}\) (or any non-zero scalar multiple).
Since \(\Pi\) contains \(l_1\), it contains the point \(A(1, -2, 3)\).

The equation of the plane is:
\(1(x - 1) - 1(y + 2) - 1(z - 3) = 0\)
\(x - 1 - y - 2 - z + 3 = 0\)
\(x - y - z = 0\).

(i)
- M1: Show direction vectors are not parallel (e.g., stating they are not scalar multiples of each other).
- M1: Attempt to solve the simultaneous equations for \(\lambda\) and \(\mu\) using two components.
- A1: Find correct values (e.g., \(\lambda = 1.4, \mu = 0.2\)) and show inconsistency with the third component, concluding the lines are skew.

(ii)
- M1: Find the cross product of the two direction vectors.
- A1: Obtain a correct normal vector (e.g., \(5\mathbf{i} - 5\mathbf{j} - 5\mathbf{k}\) or \(\mathbf{i} - \mathbf{j} - \mathbf{k}\)).
- M1: Compute \(\vec{AB}\) and apply the projection formula \(d = \frac{|\vec{AB} \cdot \mathbf{n}|}{|\mathbf{n}|}\).
- A1: Correctly calculate the shortest distance as \(\frac{1}{\sqrt{3}}\) or \(\frac{\sqrt{3}}{3}\) (accept decimal equivalent of 0.577).

(iii)
- M1: Use the cross product vector from (ii) as the normal vector and write down an equation of the form \(x - y - z = D\) using a point on \(l_1\).
- A1: Obtain the correct Cartesian equation \(x - y - z = 0\).

(i) A matrix is singular when its determinant is equal to zero.
\(\det(\mathbf{M}) = k(k-3) - (2)(5) = k^2 - 3k - 10\)
Setting the determinant to zero:
\(k^2 - 3k - 10 = 0\)
\((k-5)(k+2) = 0\)
Hence, the two values of \(k\) are \(k = 5\) and \(k = -2\).

(ii) For \(k = 4\), the determinant is:
\(\det(\mathbf{M}) = 4^2 - 3(4) - 10 = 16 - 12 - 10 = -6\)
The area scale factor of the transformation is given by \(|\det(\mathbf{M})| = |-6| = 6\).
Therefore, the area of the image shape is:
\(\text{Area} = 7 \times 6 = 42\).

(iii) For \(k = 1\), the matrix is \(\mathbf{M} = \begin{pmatrix} 1 & 2 \\ 5 & -2 \end{pmatrix}\).
Let the equation of an invariant line through the origin be \(y = mx\).
A general point on this line is of the form \(\begin{pmatrix} x \\ mx \end{pmatrix}\).
Its image under the transformation is:
\(\begin{pmatrix} X \\ Y \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 5 & -2 \end{pmatrix} \begin{pmatrix} x \\ mx \end{pmatrix} = \begin{pmatrix} x + 2mx \\ 5x - 2mx \end{pmatrix} = \begin{pmatrix} x(1+2m) \\ x(5-2m) \end{pmatrix}\)
Since the line is invariant, the point \((X, Y)\) must also lie on the line \(Y = mX\):
\(x(5-2m) = m \cdot x(1+2m)\)
For \(x \neq 0\), we can divide both sides by \(x\):
\(5-2m = m(1+2m)\)
\(5-2m = m + 2m^2\)
\(2m^2 + 3m - 5 = 0\)
\((2m + 5)(m - 1) = 0\)
This gives \(m = -\frac{5}{2}\) or \(m = 1\).
Thus, the equations of the invariant lines through the origin are:
\(y = -\frac{5}{2}x\) and \(y = x\).

(iv) For \(k = 3\), the matrix is \(\mathbf{M} = \begin{pmatrix} 3 & 2 \\ 5 & 0 \end{pmatrix}\).
Let a point on the original line be \((x, y)\) and its image be \((x', y')\):
\(\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 3 & 2 \\ 5 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}\)
This gives the system of equations:
\(x' = 3x + 2y\)
\(y' = 5x \implies x = \frac{1}{5}y'\)
Substitute \(x\) into the equation for \(x'\):
\(x' = 3\left(\frac{1}{5}y'\right) + 2y \implies 2y = x' - \frac{3}{5}y' \implies y = \frac{1}{2}x' - \frac{3}{10}y'\)
Now, substitute the expressions for \(x\) and \(y\) into the equation of the original line \(y = 3x + 2\):
\(\frac{1}{2}x' - \frac{3}{10}y' = 3\left(\frac{1}{5}y'\right) + 2\)
Multiply the entire equation by 10 to clear denominators:
\(5x' - 3y' = 6y' + 20\)
\(9y' = 5x' - 20\)
\(y' = \frac{5}{9}x' - \frac{20}{9}\)
Thus, the equation of the image line is \(y = \frac{5}{9}x - \frac{20}{9}\) (or in standard form: \(5x - 9y - 20 = 0\)).

(i)
- **M1**: Attempts to find the determinant in terms of \(k\) and sets it equal to 0.
- **M1**: Solves the resulting quadratic equation for \(k\).
- **A1**: Correctly identifies both values \(k = 5\) and \(k = -2\).

(ii)
- **M1**: Substitutes \(k = 4\) to find the determinant of \(\mathbf{M}\).
- **A1**: States that the area scale factor is the absolute value of the determinant, which is \(6\).
- **A1**: Correctly calculates the final image area as \(42\).

(iii)
- **M1**: Multiplies the matrix by \(\begin{pmatrix} x \\ mx \end{pmatrix}\) to find the transformed coordinates.
- **M1**: Uses the relation \(Y = mX\) to set up an equation involving \(m\).
- **A1**: Simplifies to obtain the correct quadratic equation in \(m\), \(2m^2 + 3m - 5 = 0\).
- **M1**: Solves the quadratic equation to find two values for \(m\).
- **A1**: Gives both correct equations: \(y = -\frac{5}{2}x\) and \(y = x\).

(iv)
- **M1**: Expresses \(x\) in terms of \(y'\) using the transformation equations.
- **M1**: Eliminates \(x\) to express \(y\) in terms of \(x'\) and \(y'\).
- **M1**: Substitutes these expressions into the line equation \(y = 3x + 2\).
- **A1**: Obtains the correct simplified equation \(y = \frac{5}{9}x - \frac{20}{9}\) or equivalent form.

(i) The curve \( C \) is a cardioid symmetric about the initial line. For \( \theta = 0 \), \( r = a(1 + 1) = 2a \), so the intersection with the initial line is \( (2a, 0) \). For \( \theta = \pi \), \( r = a(1 - 1) = 0 \), which is the pole. For \( \theta = \pm\frac{\pi}{2} \), \( r = a(1 + 0) = a \), so the intersections with the line \( \theta = \pm\frac{\pi}{2} \) are \( (a, \frac{\pi}{2}) \) and \( (a, -\frac{\pi}{2}) \). (ii) The area \( A \) of the region is given by \( A = \frac{1}{2} \int_{-\pi}^{\pi} r^2 \, d\theta = \frac{1}{2} \int_{-\pi}^{\pi} a^2(1 + \cos\theta)^2 \, d\theta \). Expanding the integrand gives \( A = \frac{1}{2}a^2 \int_{-\pi}^{\pi} (1 + 2\cos\theta + \cos^2\theta) \, d\theta \). Using the identity \( \cos^2\theta = \frac{1}{2}(1 + \cos 2\theta) \), we rewrite this as \( A = \frac{1}{2}a^2 \int_{-\pi}^{\pi} (\frac{3}{2} + 2\cos\theta + \frac{1}{2}\cos 2\theta) \, d\theta \). Integrating each term yields \( A = \frac{1}{2}a^2 \left[ \frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta \right]_{-\pi}^{\pi} = \frac{1}{2}a^2 \left( \frac{3\pi}{2} - \left(-\frac{3\pi}{2}\right) \right) = \frac{3}{2}\pi a^2 \). (iii) The Cartesian \( x \)-coordinate is given by \( x = r\cos\theta = a(1+\cos\theta)\cos\theta = a(\cos\theta + \cos^2\theta) \). The tangent to \( C \) is perpendicular to the initial line when \( \frac{dx}{d\theta} = 0 \) and \( \frac{dy}{d\theta} \neq 0 \). Differentiating \( x \) with respect to \( \theta \) gives \( \frac{dx}{d\theta} = a(-\sin\theta - 2\sin\theta\cos\theta) = -a\sin\theta(1 + 2\cos\theta) \). Setting \( \frac{dx}{d\theta} = 0 \) gives \( \sin\theta = 0 \) or \( \cos\theta = -\frac{1}{2} \). For \( \sin\theta = 0 \), we have \( \theta = 0 \) or \( \theta = \pi \). If \( \theta = 0 \), then \( r = 2a \) and \( \frac{dy}{d\theta} = a(\cos 0 + \cos 0) = 2a \neq 0 \), so \( (2a, 0) \) is a valid point. If \( \theta = \pi \), then \( r = 0 \) and \( \frac{dy}{d\theta} = a(\cos\pi + \cos 2\pi) = 0 \); calculating the limiting gradient using L'Hopital's rule shows \( \lim_{\theta \to \pi} \frac{dy}{dx} = 0 \), which means the tangent at the cusp is horizontal, so this point is rejected. For \( \cos\theta = -\frac{1}{2} \), we have \( \theta = \frac{2\pi}{3} \) or \( \theta = -\frac{2\pi}{3} \). For these values, \( r = a(1 - \frac{1}{2}) = \frac{1}{2}a \), and \( \frac{dy}{d\theta} = a(-\frac{1}{2} - \frac{1}{2}) = -a \neq 0 \), so they are valid points. Thus, the three points where the tangent is perpendicular to the initial line are \( (2a, 0) \), \( (\frac{1}{2}a, \frac{2\pi}{3}) \), and \( (\frac{1}{2}a, -\frac{2\pi}{3}) \).

Part (i): M1: Sketching a closed cardioid-shaped curve symmetric about the initial line with a cusp at the pole. A1: Correctly labelling the intercepts on the initial line: (2a, 0) and the pole. A1: Correctly labelling the intercepts on the line \theta = \pm\frac{\pi}{2}: (a, \pm\frac{\pi}{2}). Part (ii): M1: Attempting to use the polar area formula A = \frac{1}{2} \int r^2 d\theta. M1: Expanding r^2 and substituting the double-angle identity for \cos^2\theta. A1: Correctly integrating the expanded expression. A1: Correctly applying the limits of integration. A1: Obtaining the final exact area of \frac{3}{2}\pi a^2. Part (iii): M1: Writing x = r\cos\theta and differentiating with respect to \theta. A1: Obtaining the correct derivative \frac{dx}{d\theta} = -a\sin\theta(1+2\cos\theta). M1: Setting \frac{dx}{d\theta} = 0 and solving for \theta. A1: Identifying \theta = 0 and finding the point (2a, 0). A2: Identifying \theta = \pm\frac{2\pi}{3} and finding both points (\frac{1}{2}a, \pm\frac{2\pi}{3}) (1 mark for each). B1: Correctly explaining or showing via limits that the pole at \theta = \pi is rejected because the tangent there is horizontal (or because \frac{dy}{dx} \to 0).

(i) By carrying out algebraic division or expressing the numerator in terms of \( (x - 2) \):
\( 3x^2 - 7x + 14 = 3x(x - 2) - x + 14 = 3x(x - 2) - (x - 2) + 12 = (3x - 1)(x - 2) + 12 \)
So we can write the equation of \( C \) as:
\( y = 3x - 1 + \frac{12}{x - 2} \)
As \( x \to \pm \infty \), \( \frac{12}{x - 2} \to 0 \), which gives the oblique asymptote as:
\( y = 3x - 1 \)
The denominator is zero when \( x = 2 \), and the numerator is non-zero, which gives the vertical asymptote as:
\( x = 2 \)

(ii) Differentiating the equation of \( C \) with respect to \( x \):
\( \frac{\mathrm{d}y}{\mathrm{d}x} = 3 - \frac{12}{(x - 2)^2} \)
Setting the derivative to \( 0 \) to find stationary points:
\( 3 - \frac{12}{(x - 2)^2} = 0 \implies (x - 2)^2 = 4 \implies x - 2 = \pm 2 \)
This gives:
\( x = 4 \) or \( x = 0 \)
Substituting these values back into the equation of \( C \):
For \( x = 4 \): \( y = 3(4) - 1 + \frac{12}{4 - 2} = 17 \)
For \( x = 0 \): \( y = 3(0) - 1 + \frac{12}{0 - 2} = -7 \)
So the stationary points of \( C \) are \( (0, -7) \) and \( (4, 17) \).

(iii) To sketch the curve:
- Draw the vertical asymptote \( x = 2 \) and the oblique asymptote \( y = 3x - 1 \) as dashed lines.
- Find the \( y \)-intercept by putting \( x = 0 \), which gives \( y = -7 \). This is also a local maximum.
- Find any \( x \)-intercepts by solving \( 3x^2 - 7x + 14 = 0 \). Since the discriminant \( \Delta = (-7)^2 - 4(3)(14) = 49 - 168 = -119 < 0 \), there are no \( x \)-intercepts.
- The left branch lies in the region \( x < 2 \), has a local maximum at \( (0, -7) \), and is asymptotic to \( y = 3x - 1 \) and \( x = 2 \).
- The right branch lies in the region \( x > 2 \), has a local minimum at \( (4, 17) \), and is asymptotic to \( x = 2 \) and \( y = 3x - 1 \).

(iv) To find the intersections of the line \( y = kx - 1 \) and the curve \( C \):
\( kx - 1 = \frac{3x^2 - 7x + 14}{x - 2} \)
For \( x \neq 2 \):
\( (kx - 1)(x - 2) = 3x^2 - 7x + 14 \)
\( kx^2 - 2kx - x + 2 = 3x^2 - 7x + 14 \)
\( (k - 3)x^2 + (6 - 2k)x - 12 = 0 \)

Case 1: \( k \neq 3 \)
This is a quadratic equation. For there to be no intersection, the discriminant must be strictly less than zero:
\( \Delta = (6 - 2k)^2 - 4(k - 3)(-12) < 0 \)
\( [2(3 - k)]^2 + 48(k - 3) < 0 \)
\( 4(k - 3)^2 + 48(k - 3) < 0 \)
\( 4(k - 3)(k - 3 + 12) < 0 \)
\( 4(k - 3)(k + 9) < 0 \)
This gives the range of \( k \) as:
\( -9 < k < 3 \)

Case 2: \( k = 3 \)
When \( k = 3 \), the equation simplifies to:
\( 0x^2 + 0x - 12 = 0 \implies -12 = 0 \)
This equation has no real solutions. (Geometrically, the line \( y = 3x - 1 \) is the oblique asymptote of the curve \( C \), which the curve never intersects.) Thus, there are no intersections when \( k = 3 \).

Combining both cases, the set of values of \( k \) is \( -9 < k \le 3 \).

(i)
- M1: Attempt algebraic long division or equating coefficients to express the rational function in the form \( y = px + q + \frac{r}{x-2} \).
- A1: Obtain correct oblique asymptote \( y = 3x - 1 \).
- B1: State correct vertical asymptote \( x = 2 \).

(ii)
- M1: Correctly differentiate the function (either using the quotient rule or from the simplified form).
- A1: Equate the derivative to \( 0 \) and solve for \( x \) to find \( x = 0 \) and \( x = 4 \).
- A1: Find the corresponding \( y \)-coordinates \( -7 \) and \( 17 \).
- A1: Express final answers clearly as coordinates: \( (0, -7) \) and \( (4, 17) \).

(iii)
- B1: Draw both asymptotes correctly as dashed lines and label them with their equations.
- B1: Sketch the left branch with the correct shape, showing the local maximum at \( (0, -7) \) on the \( y \)-axis, approaching both asymptotes.
- B1: Sketch the right branch with the correct shape, showing the local minimum at \( (4, 17) \), approaching both asymptotes.
- B1: Ensure there are no \( x \)-axis intersections and label all key points clearly.

(iv)
- M1: Equate the line and the curve and rearrange into a standard quadratic-like form in \( x \).
- M1: Use the discriminant condition \( \Delta < 0 \) on their quadratic expression.
- A1: Solve the discriminant inequality to find the interval \( -9 < k < 3 \).
- A1: Analyze the case \( k = 3 \) correctly to conclude that \( k = 3 \) is included, and state the final range as \( -9 < k \le 3 \) (or equivalent interval notation).

To find the value of \(k\) for which the system does not have a unique solution, we set the determinant of the coefficient matrix to zero:
\[
\det\begin{pmatrix} 1 & 2 & -1 \\ 2 & 1 & 3 \\ 1 & -1 & k \end{pmatrix} = 0
\]
Evaluating the determinant:
\[
1(k - (-3)) - 2(2k - 3) - 1(-2 - 1) = 0
\]
\[
(k + 3) - (4k - 6) + 3 = 0
\]
\[
12 - 3k = 0 \implies k = 4
\]
With \(k = 4\), the system becomes:
1) \(x + 2y - z = 3\)
2) \(2x + y + 3z = 4\)
3) \(x - y + 4z = 1\)

Subtracting equation (3) from equation (1):
\[
(x + 2y - z) - (x - y + 4z) = 3 - 1 \implies 3y - 5z = 2
\]
Let \(z = \lambda\), where \(\lambda\) is a real parameter. Then:
\[
3y - 5\lambda = 2 \implies y = \frac{2}{3} + \frac{5}{3}\lambda
\]
Substituting \(y\) and \(z\) back into equation (1):
\[
x + 2\left(\frac{2}{3} + \frac{5}{3}\lambda\right) - \lambda = 3
\]
\[
x + \frac{4}{3} + \frac{10}{3}\lambda - \lambda = 3
\]
\[
x = 3 - \frac{4}{3} - \frac{7}{3}\lambda = \frac{5}{3} - \frac{7}{3}\lambda
\]
Checking consistency with equation (2):
\[
2\left(\frac{5}{3} - \frac{7}{3}\lambda\right) + \left(\frac{2}{3} + \frac{5}{3}\lambda\right) + 3\lambda = \frac{10}{3} - \frac{14}{3}\lambda + \frac{2}{3} + \frac{5}{3}\lambda + 3\lambda = 4
\]
which is consistent. Thus, the general solution is:
\[
x = \frac{5}{3} - \frac{7}{3}\lambda, \quad y = \frac{2}{3} + \frac{5}{3}\lambda, \quad z = \lambda
\]

M1: For setting the determinant of the coefficient matrix to 0 (or using row reduction to find the condition for no unique solution).
A1: For obtaining \(k = 4\).
M1: For parameterising the system (e.g., setting \(z = \lambda\)) and attempting to find \(x\) and \(y\) in terms of \(\lambda\).
A1: For obtaining the correct general solution (or any equivalent parameterized form, e.g., using integer parameters like \(z = 3t-1\)).

Let \(w = -8 + 8\sqrt{3}\text{i}\). 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 of the Argand diagram, its argument is \(\arg(w) = \pi - \arctan\left(\frac{8\sqrt{3}}{8}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}\). We can write \(w\) in exponential form with a general argument: \(w = 16e^{\text{i}\left(\frac{2\pi}{3} + 2k\pi\right)}\) for \(k \in \mathbb{Z}\). Applying de Moivre's theorem to find the fourth roots: \(z = w^{1/4} = 16^{1/4} e^{\text{i}\left(\frac{2\pi/3 + 2k\pi}{4}\right)} = 2e^{\text{i}\left(\frac{\pi}{6} + \frac{k\pi}{2}\right)}\). To find the four distinct roots in the interval \(-\pi < \theta \le \pi\), we choose \(k = 0, 1, -1, -2\): For \(k = 0\): \(z_0 = 2e^{\text{i}\pi/6}\). For \(k = 1\): \(z_1 = 2e^{\text{i}(\pi/6 + \pi/2)} = 2e^{2\text{i}\pi/3}\). For \(k = -1\): \(z_2 = 2e^{\text{i}(\pi/6 - \pi/2)} = 2e^{-\text{i}\pi/3}\). For \(k = -2\): \(z_3 = 2e^{\text{i}(\pi/6 - \pi)} = 2e^{-5\text{i}\pi/6}\).

M1: Attempts to find the modulus and argument of \(-8 + 8\sqrt{3}\text{i}\). A1: Correctly expresses the complex number in exponential form as \(16e^{\text{i}\frac{2\pi}{3}}\). M1: Applies de Moivre's theorem to obtain an expression for the fourth roots incorporating a general argument. A2: Gives all four correct roots in the required form and range (A1 if only two are correct).

Let .

When , .

First derivative:
Using the chain rule,

At , .

Second derivative:
Using the quotient rule,

At , .

Third derivative:
Writing ,

At , .

Alternative method using implicit differentiation:
From , we have:

Differentiating implicitly with respect to :

At : .

Differentiating again:

At : .

Alternative method using standard series expansions:
We know that

and


Substituting :

Expanding terms up to :




Maclaurin's series formula:

M1: For correctly finding the first derivative (or finding the standard expansion of sin x).
A1: For evaluating the first derivative at 0 to get 1.
M1: For correctly finding the second derivative (or applying the standard expansion of ln(1+u)).
A1: For evaluating the second derivative at 0 to get -1.
M1: For finding the third derivative at 0 to get 1 (or correctly simplifying the combined terms up to x^3).
A1: For the final correct expansion x - 1/2 x^2 + 1/6 x^3.

To find the general solution, we first solve the homogeneous equation:
\[ \frac{\text{d}^2 y}{\text{d}x^2} + 2 \frac{\text{d}y}{\text{d}x} + 5y = 0 \]
The auxiliary equation is:
\[ m^2 + 2m + 5 = 0 \]
Solving for \( m \):
\[ m = \frac{-2 \pm \sqrt{4 - 20}}{2} = -1 \pm 2i \]
The complementary function (CF) is:
\[ y_c = e^{-x} (A \cos 2x + B \sin 2x) \]
For the particular integral (PI), we try a solution of the form:
\[ y_p = k e^{-x} \]
Differentiating twice with respect to \( x \) gives:
\[ \frac{\text{d}y_p}{\text{d}x} = -k e^{-x} \quad \text{and} \quad \frac{\text{d}^2 y_p}{\text{d}x^2} = k e^{-x} \]
Substituting these into the original differential equation:
\[ k e^{-x} + 2(-k e^{-x}) + 5(k e^{-x}) = 10 e^{-x} \]
\[ 4k e^{-x} = 10 e^{-x} \implies k = \frac{5}{2} \]
Thus, the general solution is:
\[ y = e^{-x} (A \cos 2x + B \sin 2x) + \frac{5}{2} e^{-x} \]
Next, we apply the initial conditions to find the particular solution.
When \( x = 0 \), \( y = 3 \):
\[ 3 = e^0 (A \cos 0 + B \sin 0) + \frac{5}{2} e^0 \]
\[ 3 = A + \frac{5}{2} \implies A = \frac{1}{2} \]
To find \( B \), we differentiate the general solution:
\[ y = e^{-x} \left( A \cos 2x + B \sin 2x + \frac{5}{2} \right) \]
\[ \frac{\text{d}y}{\text{d}x} = -e^{-x} \left( A \cos 2x + B \sin 2x + \frac{5}{2} \right) + e^{-x} (-2A \sin 2x + 2B \cos 2x) \]
When \( x = 0 \), \( \frac{\text{d}y}{\text{d}x} = -1 \):
\[ -1 = -1 \left( A + \frac{5}{2} \right) + 2B \]
\[ -1 = -\left(\frac{1}{2} + \frac{5}{2}\right) + 2B \]
\[ -1 = -3 + 2B \implies 2B = 2 \implies B = 1 \]
Hence, the particular solution is:
\[ y = e^{-x} \left( \frac{1}{2} \cos 2x + \sin 2x + \frac{5}{2} \right) \]

M1: Attempt to write down and solve the auxiliary equation.
A1: Obtain correct complex roots \( m = -1 \pm 2i \).
A1: Formulate the correct complementary function (CF).
M1: Substitute a particular integral of the form \( y = k e^{-x} \) into the differential equation.
A1: Obtain \( k = \frac{5}{2} \).
A2: Combine CF and PI to write the general solution (award 1 mark if a minor error is present in either CF or PI but they are correctly summed).
M1: Apply the initial condition \( y = 3 \) at \( x = 0 \) to find \( A \).
M1: Differentiate the general solution and apply the initial condition \( \frac{\text{d}y}{\text{d}x} = -1 \) at \( x = 0 \).
A1: Obtain the correct particular solution with \( A = \frac{1}{2} \) and \( B = 1 \).

(i) We first find the derivatives of \(x\) and \(y\) with respect to \(t\):
\[\frac{dx}{dt} = 1 - \text{sech}^2 t = \tanh^2 t\]
\[\frac{dy}{dt} = -\text{sech } t \tanh t\]
Now, using the chain rule:
\[\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-\text{sech } t \tanh t}{\tanh^2 t} = -\frac{\text{sech } t}{\tanh t}\]
Using the definitions \(\text{sech } t = \frac{1}{\cosh t}\) and \(\tanh t = \frac{\sinh t}{\cosh t}\), we obtain:
\[\frac{dy}{dx} = -\frac{1}{\sinh t} = -\text{cosech } t\]

(ii) We compute the sum of the squares of the derivatives:
\[\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = (\tanh^2 t)^2 + (-\text{sech } t \tanh t)^2\]
\[= \tanh^4 t + \text{sech}^2 t \tanh^2 t\]
Factor out \(\tanh^2 t\):
\[= \tanh^2 t (\tanh^2 t + \text{sech}^2 t)\]
Since \(\cosh^2 t - \sinh^2 t = 1\), dividing by \(\cosh^2 t\) gives \(1 - \tanh^2 t = \text{sech}^2 t\), which implies \(\tanh^2 t + \text{sech}^2 t = 1\).
Thus,
\[\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = \tanh^2 t (1) = \tanh^2 t\]

(iii) The arc length \(s\) is given by the formula:
\[s = \int_{0}^{\ln 3} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\]
Substituting the result from part (ii):
\[s = \int_{0}^{\ln 3} \sqrt{\tanh^2 t} \, dt\]
Since \(t \ge 0\), \(\tanh t \ge 0\), so \ \sqrt{\tanh^2 t} = \tanh t\).
\[s = \int_{0}^{\ln 3} \tanh t \, dt\]
\[= \left[ \ln(\cosh t) \right]_{0}^{\ln 3}\]
\[= \ln(\cosh(\ln 3)) - \ln(\cosh 0)\]
Since \(\cosh 0 = 1\), \(\ln(\cosh 0) = \ln 1 = 0\).
Now, evaluate \(\cosh(\ln 3)\):
\[\cosh(\ln 3) = \frac{e^{\ln 3} + e^{-\ln 3}}{2} = \frac{3 + \frac{1}{3}}{2} = \frac{\frac{10}{3}}{2} = \frac{5}{3}\]
Thus, the exact arc length is:
\[s = \ln\left(\frac{5}{3}\right)\]

**Part (i)**
* **M1**: For differentiating \(x\) and \(y\) with respect to \(t\).
* **A1**: For obtaining correct derivatives: \(\frac{dx}{dt} = \tanh^2 t\) (or \(1 - \text{sech}^2 t\)) and \(\frac{dy}{dt} = -\text{sech } t \tanh t\).
* **A1**: For correctly dividing and simplifying to \(-\text{cosech } t\) or \(-\frac{1}{\sinh t}\).

**Part (ii)**
* **M1**: For squaring and adding the expressions for \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).
* **M1**: For factoring out \(\tanh^2 t\).
* **A1**: For correctly applying the hyperbolic identity \(\tanh^2 t + \text{sech}^2 t = 1\) to show the given result.

**Part (iii)**
* **M1**: For setting up the arc length integral with the correct limits and integrand \(\tanh t\).
* **A1**: For integrating \(\tanh t\) to obtain \(\ln(\cosh t)\).
* **M1**: For correctly evaluating \(\cosh(\ln 3)\) using the exponential definition of \(\cosh\).
* **A1**: For obtaining the final exact arc length of \(\ln\left(\frac{5}{3}\right)\).

(i) Using the compound angle formula for hyperbolic cosine:
\[ \cosh(3x) = \cosh(2x + x) = \cosh 2x \cosh x + \sinh 2x \sinh x \]
Using the double angle identities \(\cosh 2x = 2\cosh^2 x - 1\) and \(\sinh 2x = 2\sinh x \cosh x\):
\[ \cosh(3x) = (2\cosh^2 x - 1)\cosh x + (2\sinh x \cosh x)\sinh x \]
\[ \cosh(3x) = 2\cosh^3 x - \cosh x + 2\sinh^2 x \cosh x \]
Using the identity \(\sinh^2 x = \cosh^2 x - 1\):
\[ \cosh(3x) = 2\cosh^3 x - \cosh x + 2(\cosh^2 x - 1)\cosh x \]
\[ \cosh(3x) = 2\cosh^3 x - \cosh x + 2\cosh^3 x - 2\cosh x \]
\[ \cosh(3x) = 4\cosh^3 x - 3\cosh x \]
Rearranging this formula gives:
\[ 4\cosh^3 x = \cosh 3x + 3\cosh x \implies \cosh^3 x = \frac{1}{4}(\cosh 3x + 3\cosh x) \]

(ii) Using the identity proven in part (i):
\[ \int_{0}^{\ln 2} \cosh^3 x \, \mathrm{d}x = \frac{1}{4} \int_{0}^{\ln 2} (\cosh 3x + 3\cosh x) \, \mathrm{d}x \]
\[ = \frac{1}{4} \left[ \frac{1}{3}\sinh 3x + 3\sinh x \right]_{0}^{\ln 2} \]
Evaluating the upper limit \(x = \ln 2\):
\[ \sinh(\ln 2) = \frac{e^{\ln 2} - e^{-\ln 2}}{2} = \frac{2 - \frac{1}{2}}{2} = \frac{3}{4} \]
\[ \sinh(3\ln 2) = \sinh(\ln 8) = \frac{e^{\ln 8} - e^{-\ln 8}}{2} = \frac{8 - \frac{1}{8}}{2} = \frac{63}{16} \]
Substituting these values in:
\[ \frac{1}{4} \left[ \frac{1}{3}\left(\frac{63}{16}\right) + 3\left(\frac{3}{4}\right) \right] = \frac{1}{4} \left( \frac{21}{16} + \frac{9}{4} \right) = \frac{1}{4} \left( \frac{21 + 36}{16} \right) = \frac{57}{64} \]
At the lower limit \(x = 0\), both terms are zero because \(\sinh 0 = 0\).
Thus, the exact value is \(\frac{57}{64}\).

(iii) Express \(5\cosh x - \sinh x = 7\) using exponential definitions:
\[ 5\left(\frac{e^x + e^{-x}}{2}\right) - \left(\frac{e^x - e^{-x}}{2}\right) = 7 \]
Multiply through by 2:
\[ 5(e^x + e^{-x}) - (e^x - e^{-x}) = 14 \]
\[ 4e^x + 6e^{-x} = 14 \]
Divide by 2:
\[ 2e^x + 3e^{-x} = 7 \]
Multiply by \(e^x\):
\[ 2e^{2x} - 7e^x + 3 = 0 \]
Let \(u = e^x\):
\[ 2u^2 - 7u + 3 = 0 \]
\[ (2u - 1)(u - 3) = 0 \]
So \(u = \frac{1}{2}\) or \(u = 3\).
Since \(e^x > 0\), both solutions are valid:
\[ e^x = \frac{1}{2} \implies x = \ln\left(\frac{1}{2}\right) = -\ln 2 \]
\[ e^x = 3 \implies x = \ln 3 \]

**Part (i)**
* **M1**: Use compound angle formula to expand \(\cosh(2x+x)\).
* **A1**: Use correct double angle formulae for \(\cosh 2x\) and \(\sinh 2x\) to write expression in terms of \(\cosh x\) and \(\sinh x\).
* **A1**: Substitute \(\sinh^2 x = \cosh^2 x - 1\) and simplify to obtain \(\cosh 3x = 4\cosh^3 x - 3\cosh x\).
* **A1**: Rearrange correctly to complete the proof of the identity.

**Part (ii)**
* **M1**: For integrating \(\cosh 3x\) and \(\cosh x\) to get \(\frac{1}{3}\sinh 3x\) and \(\sinh x\).
* **M1**: For evaluating \(\sinh(\ln 2)\) and \(\sinh(3\ln 2)\) (or \(\sinh(\ln 8)\)) in terms of fractions.
* **A1**: Correctly substituting values into the integrated expression.
* **A1**: Obtaining the final exact value of \(\frac{57}{64}\).

**Part (iii)**
* **M1**: Substituting exponential forms of \(\cosh x\) and \(\sinh x\) into the given equation.
* **A1**: Obtaining the simplified equation \(2e^x + 3e^{-x} = 7\) (or equivalent).
* **M1**: Forming a quadratic equation in \(e^x\), namely \(2e^{2x} - 7e^x + 3 = 0\).
* **M1**: Solving the quadratic equation to find the values of \(e^x\).
* **A1**: Obtaining both correct values \(e^x = \frac{1}{2}\) and \(e^x = 3\).
* **A1**: Obtaining final logarithmic forms \(x = -\ln 2\) (or \(\ln(0.5)\)) and \(x = \ln 3\).

**(i)**
To find the eigenvalues of \(\mathbf{A}\), we solve the characteristic equation \(\det(\mathbf{A} - \lambda \mathbf{I}) = 0\):
\[ \begin{vmatrix} 2-\lambda & 2 & -1 \\ 2 & -1-\lambda & 2 \\ -1 & 2 & 2-\lambda \end{vmatrix} = 0 \]
Expanding along the first row:
\[ (2-\lambda) [ (-1-\text{\(\lambda\)})(2-\lambda) - 4 ] - 2 [ 2(2-\lambda) - (-2) ] - 1 [ 4 - (-1-\lambda) ] = 0 \]
\[ (2-\lambda) (\lambda^2 - \lambda - 6) - 2 (6 - 2\lambda) - (5 + \lambda) = 0 \]
\[ (-\lambda^3 + 3\lambda^2 + 4\lambda - 12) + (4\lambda - 12) + (\lambda - 5) = 0 \]
\[ -\lambda^3 + 3\lambda^2 + 9\lambda - 27 = 0 \]
\[ \lambda^3 - 3\lambda^2 - 9\lambda + 27 = 0 \]
\[ \lambda^2(\lambda - 3) - 9(\lambda - 3) = 0 \]
\[ (\lambda^2 - 9)(\lambda - 3) = 0 \]
\[ (\lambda - 3)^2(\lambda + 3) = 0 \]
Thus, the eigenvalues are \(\lambda = 3\) (repeated) and \(\lambda = -3\).

**(ii)**
For \(\lambda = -3\), we solve \((\mathbf{A} + 3\mathbf{I})\mathbf{v} = \mathbf{0}\):
\[ \begin{pmatrix} 5 & 2 & -1 \\ 2 & 2 & 2 \\ -1 & 2 & 5 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \]
From row 2, \(x + y + z = 0\). Subtracting row 3 from row 1 gives \(6x - 6z = 0 \implies x = z\). Substituting into \(x+y+z=0\) gives \(y = -2x\).
Thus, an eigenvector is \(\mathbf{e}_1 = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}\).

For \(\lambda = 3\), we solve \((\mathbf{A} - 3\mathbf{I})\mathbf{v} = \mathbf{0}\):
\[ \begin{pmatrix} -1 & 2 & -1 \\ 2 & -4 & 2 \\ -1 & 2 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \]
This reduces to the single equation \(x - 2y + z = 0\).
Any vector satisfying this equation is orthogonal to \(\mathbf{e}_1\).
Let us choose one non-zero vector satisfying \(x - 2y + z = 0\), say \(\mathbf{e}_2 = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}\).
To find a third eigenvector \(\mathbf{e}_3\) in this eigenspace that is orthogonal to \(\mathbf{e}_2\), we require:
\[ \mathbf{e}_3 \cdot \mathbf{e}_2 = 0 \implies x - z = 0 \implies x = z \]
Substituting \(x = z\) into \(x - 2y + z = 0\) gives \(2x - 2y = 0 \implies x = y\).
So we can choose \(\mathbf{e}_3 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\).
Thus, a set of three mutually orthogonal eigenvectors is:
\[ \left\{ \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \right\} \]
(Accept any scalar multiples of these vectors, as long as they remain mutually orthogonal.)

**(iii)**
To form the orthogonal matrix \(\mathbf{P}\), we normalize each of the eigenvectors found in part (ii):
- For \(\mathbf{e}_1\), the magnitude is \(\sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{6}\).
- For \(\mathbf{e}_2\), the magnitude is \(\sqrt{1^2 + 0^2 + (-1)^2} = \sqrt{2}\).
- For \(\mathbf{e}_3\), the magnitude is \(\sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}\).

Thus, we have:
\[ \mathbf{P} = \begin{pmatrix} \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} \\ -\frac{2}{\sqrt{6}} & 0 & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} \end{pmatrix} \]
and the corresponding diagonal matrix \(\mathbf{D}\) is:
\[ \mathbf{D} = \begin{pmatrix} -3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{pmatrix} \]
(Other consistent orderings of columns in \(\mathbf{P}\) and diagonal elements in \(\mathbf{D}\) are acceptable.)

To find \(\mathbf{A}^5\), we note that:
\[ \mathbf{D}^5 = \begin{pmatrix} (-3)^5 & 0 & 0 \\ 0 & 3^5 & 0 \\ 0 & 0 & 3^5 \end{pmatrix} = \begin{pmatrix} -243 & 0 & 0 \\ 0 & 243 & 0 \\ 0 & 0 & 243 \end{pmatrix} = 81 \mathbf{D} \]
Therefore:
\[ \mathbf{A}^5 = \mathbf{P} \mathbf{D}^5 \mathbf{P}^{\mathrm{T}} = 81 \mathbf{P} \mathbf{D} \mathbf{P}^{\mathrm{T}} = 81 \mathbf{A} \]
Evaluating this product:
\[ \mathbf{A}^5 = 81 \begin{pmatrix} 2 & 2 & -1 \\ 2 & -1 & 2 \\ -1 & 2 & 2 \end{pmatrix} = \begin{pmatrix} 162 & 162 & -81 \\ 162 & -81 & 162 \\ -81 & 162 & 162 \end{pmatrix} \]

**(i)**
- **M1**: Sets up determinant equation \(\det(\mathbf{A}-\lambda\mathbf{I}) = 0\) and attempts expansion.
- **A1**: Obtains correct characteristic equation \(\lambda^3 - 3\lambda^2 - 9\lambda + 27 = 0\) (or equivalent).
- **M1**: Factorises and solves the cubic equation.
- **A1**: Correctly identifies eigenvalues as \(\lambda = 3\) (repeated) and \(\lambda = -3\).

**(ii)**
- **M1**: Correct process to find an eigenvector for \(\lambda = -3\).
- **A1**: Obtains \(\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}\) (or any non-zero multiple).
- **M1**: Processes \(\lambda = 3\) to find two orthogonal eigenvectors in the 2D eigenspace.
- **A1**: Correctly states a complete set of three mutually orthogonal eigenvectors, e.g. \(\left\{ \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \right\}\).

**(iii)**
- **B1**: States a correct orthogonal matrix \(\mathbf{P}\) with normalized columns and the corresponding diagonal matrix \(\mathbf{D}\) (consistent ordering).
- **M1**: Uses relation \(\mathbf{A}^5 = \mathbf{P} \mathbf{D}^5 \mathbf{P}^{\mathrm{T}}\) or identifies \(\mathbf{A}^2 = 9\mathbf{I}\) to simplify power calculations.
- **A1**: Obtains correct final matrix \(\mathbf{A}^5 = \begin{pmatrix} 162 & 162 & -81 \\ 162 & -81 & 162 \\ -81 & 162 & 162 \end{pmatrix}\).

(a) We first rewrite the given sum in a form that reveals the Riemann sum structure:

\(\lim_{n \to \infty} \sum_{r=1}^n \frac{3n + r}{9n^2 + r^2} = \lim_{n \to \infty} \sum_{r=1}^n \frac{n\left(3 + \frac{r}{n}\right)}{n^2\left(9 + \left(\frac{r}{n}\right)^2\right)} = \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^n \frac{3 + \frac{r}{n}}{9 + \left(\frac{r}{n}\right)^2}\)

This is the Riemann sum for the function \(f(x) = \frac{3+x}{9+x^2}\) over the interval \([0, 1]\). Thus, the limit is equal to the definite integral:

\(\int_0^1 \frac{3+x}{9+x^2} \, dx\)

We can split this integral into two parts:

\(\int_0^1 \frac{3}{9+x^2} \, dx + \int_0^1 \frac{x}{9+x^2} \, dx\)

Evaluating each part:

\(\int_0^1 \frac{3}{9+x^2} \, dx = \left[ \arctan\left(\frac{x}{3}\right) \right]_0^1 = \arctan\left(\frac{1}{3}\right)\)

\(\int_0^1 \frac{x}{9+x^2} \, dx = \left[ \frac{1}{2}\ln(9+x^2) \right]_0^1 = \frac{1}{2}\ln(10) - \frac{1}{2}\ln(9) = \frac{1}{2}\ln\left(\frac{10}{9}\right)\)

Combining these, we get the final exact value:

\(\arctan\left(\frac{1}{3}\right) + \frac{1}{2}\ln\left(\frac{10}{9}\right)\)

(b) First, we find the derivative of \(y\) with respect to \(x\):

\(\frac{dy}{dx} = x - \frac{1}{4x}\)

Now, we set up the integrand for the arc length, \(\sqrt{1 + \left(\frac{dy}{dx}\right)^2}\):

\(1 + \left(\frac{dy}{dx}\right)^2 = 1 + \left(x - \frac{1}{4x}\right)^2 = 1 + x^2 - \frac{1}{2} + \frac{1}{16x^2} = x^2 + \frac{1}{2} + \frac{1}{16x^2} = \left(x + \frac{1}{4x}\right)^2\)

Since \(x \ge 1\), the term \(x + \frac{1}{4x}\) is positive, so:

\(\sqrt{1 + \left(\frac{dy}{dx}\right)^2} = x + \frac{1}{4x}\)

The arc length \(s\) is given by:

\(s = \int_1^2 \left(x + \frac{1}{4x}\right) \, dx = \left[ \frac{1}{2}x^2 + \frac{1}{4}\ln x \right]_1^2\)

Substituting the limits:

\(s = \left(2 + \frac{1}{4}\ln 2\right) - \left(\frac{1}{2} + 0\right) = \frac{3}{2} + \frac{1}{4}\ln 2\)

(c) The formula for the area of the surface generated by rotating a curve about the \(y\)-axis is:

\(S_y = 2\pi \int_{x_1}^{x_2} x \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\)

Substituting the expression for \(\sqrt{1 + \left(\frac{dy}{dx}\right)^2}\) from part (b):

\(S_y = 2\pi \int_1^2 x \left(x + \frac{1}{4x}\right) \, dx = 2\pi \int_1^2 \left(x^2 + \frac{1}{4}\right) \, dx\)

Integrating this expression:

\(S_y = 2\pi \left[ \frac{1}{3}x^3 + \frac{1}{4}x \right]_1^2\)

Evaluating at the limits:

At \(x=2\): \(\frac{8}{3} + \frac{2}{4} = \frac{19}{6}\)

At \(x=1\): \(\frac{1}{3} + \frac{1}{4} = \frac{7}{12}\)

Subtracting the lower limit value from the upper limit value:

\(\frac{19}{6} - \frac{7}{12} = \frac{38 - 7}{12} = \frac{31}{12}\)

Therefore, the surface area is:

\(S_y = 2\pi \left(\frac{31}{12}\right) = \frac{31\pi}{6}\)

Part (a) [5 Marks]:

M1: For factoring out \(1/n\) and rewriting the sum in terms of \(r/n\).

A1: For writing down the correct definite integral \(\int_0^1 \frac{3+x}{9+x^2} \, dx\) (including correct limits).

A1: For integrating the first term to get \(\arctan(x/3)\).

A1: For integrating the second term to get \(\frac{1}{2}\ln(9+x^2)\).

A1: For substituting limits correctly to obtain the exact value \(\arctan\left(\frac{1}{3}\right) + \frac{1}{2}\ln\left(\frac{10}{9}\right)\).

Part (b) [5 Marks]:

B1: For finding the correct derivative \(\frac{dy}{dx} = x - \frac{1}{4x}\).

M1: For simplifying \(1 + \left(\frac{dy}{dx}\right)^2\) and expressing it as a perfect square.

A1: For obtaining the simplified square root as \(x + \frac{1}{4x}\).

M1: For integrating to find \(\frac{1}{2}x^2 + \frac{1}{4}\ln x\).

A1: For substituting the limits correctly to get \(\frac{3}{2} + \frac{1}{4}\ln 2\).

Part (c) [5 Marks]:

M1: For using the correct formula for surface area of revolution about the \(y\)-axis, \(S_y = 2\pi \int x \, ds\).

A1: For simplifying the integrand to \(2\pi \left(x^2 + \frac{1}{4}\right)\).

M1: For performing the integration to get \(\frac{1}{3}x^3 + \frac{1}{4}x\).

M1: For correctly substituting the limits \(1\) and \(2\) into their integrated expression.

A1: For obtaining the exact final answer \(\frac{31\pi}{6}\).