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

The area \(A\) of the region is given by the polar area formula:
\(A = \frac{1}{2} \int_{\pi/6}^{\pi/3} r^2 \, d\theta\)

Substitute \(r = 4 \cos^2\theta\):
\(A = \frac{1}{2} \int_{\pi/6}^{\pi/3} 16 \cos^4\theta \, d\theta = 8 \int_{\pi/6}^{\pi/3} \cos^4\theta \, d\theta\)

We express \(\cos^4\theta\) in terms of multiple angles using double-angle identities:
\(\cos^2\theta = \frac{1 + \cos 2\theta}{2}\)
\(\cos^4\theta = \left(\frac{1 + \cos 2\theta}{2}\right)^2 = \frac{1}{4}(1 + 2\cos 2\theta + \cos^2 2\theta)\)

Since \(\cos^2 2\theta = \frac{1 + \cos 4\theta}{2}\), we have:
\(\cos^4\theta = \frac{1}{4}\left(1 + 2\cos 2\theta + \frac{1 + \cos 4\theta}{2}\right) = \frac{3}{8} + \frac{1}{2}\cos 2\theta + \frac{1}{8}\cos 4\theta\)

Thus, the integrand is:
\(8\cos^4\theta = 3 + 4\cos 2\theta + \cos 4\theta\)

Now integrate this expression:
\(\int (3 + 4\cos 2\theta + \cos 4\theta) \, d\theta = 3\theta + 2\sin 2\theta + \frac{1}{4}\sin 4\theta\)

Evaluate at the upper limit \(\theta = \frac{\pi}{3}\):
\(3\left(\frac{\pi}{3}\right) + 2\sin\left(\frac{2\pi}{3}\right) + \frac{1}{4}\sin\left(\frac{4\pi}{3}\right) = \pi + 2\left(\frac{\sqrt{3}}{2}\right) + \frac{1}{4}\left(-\frac{\sqrt{3}}{2}\right) = \pi + \sqrt{3} - \frac{\sqrt{3}}{8} = \pi + \frac{7\sqrt{3}}{8}\)

Evaluate at the lower limit \(\theta = \frac{\pi}{6}\):
\(3\left(\frac{\pi}{6}\right) + 2\sin\left(\frac{\pi}{3}\right) + \frac{1}{4}\sin\left(\frac{2\pi}{3}\right) = \frac{\pi}{2} + 2\left(\frac{\sqrt{3}}{2}\right) + \frac{1}{4}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{2} + \sqrt{3} + \frac{\sqrt{3}}{8} = \frac{\pi}{2} + \frac{9\sqrt{3}}{8}\)

Subtract the lower limit value from the upper limit value:
\(A = \left(\pi + \frac{7\sqrt{3}}{8}\right) - \left(\frac{\pi}{2} + \frac{9\sqrt{3}}{8}\right) = \frac{\pi}{2} - \frac{\sqrt{3}}{4}\)

M1: Use area formula \(A = \frac{1}{2} \int r^2 \, d\theta\) with correct limits.
A1: Obtain the integral \(8 \int_{\pi/6}^{\pi/3} \cos^4\theta \, d\theta\).
M1: Use double-angle identity to express \(\cos^4\theta\) in terms of \(\cos 2\theta\) and \(\cos 4\theta\).
A1: Correctly obtain \(8\cos^4\theta = 3 + 4\cos 2\theta + \cos 4\theta\).
M1: Integrate correctly to get \(3\theta + 2\sin 2\theta + \frac{1}{4}\sin 4\theta\).
M1: Substitute limits \(\frac{\pi}{3}\) and \(\frac{\pi}{6}\) and subtract.
A1: Correctly show the given result \(\frac{\pi}{2} - \frac{\sqrt{3}}{4}\).

(a) Let \(y = x^2\), which implies \(x = y^{1/2}\). Substituting this into the cubic equation:
\(y^{3/2} + 3y^{1/2} - 1 = 0 \implies y^{1/2}(y + 3) = 1\)

Squaring both sides of the equation:
\(y(y + 3)^2 = 1 \implies y(y^2 + 6y + 9) = 1 \implies y^3 + 6y^2 + 9y - 1 = 0\)

Thus, the cubic equation is \(y^3 + 6y^2 + 9y - 1 = 0\).

(b) The roots of the new equation are \(y_1 = \alpha^2\), \(y_2 = \beta^2\), and \(y_3 = \gamma^2\).
We want to find the value of:
\(\alpha^4 + \beta^4 + \gamma^4 = y_1^2 + y_2^2 + y_3^2\)

Using the identity for the sum of squares of roots:
\(y_1^2 + y_2^2 + y_3^2 = (y_1 + y_2 + y_3)^2 - 2(y_1 y_2 + y_2 y_3 + y_3 y_1)\)

From our cubic equation in part (a), we have:
\(y_1 + y_2 + y_3 = -6\)

\(y_1 y_2 + y_2 y_3 + y_3 y_1 = 9\)

Therefore:
\(\alpha^4 + \beta^4 + \gamma^4 = (-6)^2 - 2(9) = 36 - 18 = 18\)

(a)
M1: Use substitution \(y = x^2\) or compute symmetric sums.
M1: Rearrange and square to eliminate fractional powers.
A1: Correctly expand to get \(y(y+3)^2 = 1\) or equivalent.
A1: Final correct equation \(y^3 + 6y^2 + 9y - 1 = 0\) (or any variable name, with integer coefficients).

(b)
M1: Relate \(\alpha^4 + \beta^4 + \gamma^4\) to the roots of the new equation as \(\sum y_1^2 = (\sum y_1)^2 - 2 \sum y_1 y_2\) (or use recurrence relation on the original cubic equation).
M1: Substitute the values of the sum and sum of products of roots of the new equation (i.e., \(-6\) and \(9\)).
A1: Obtain the correct value of \(18\).

(a) Start with the right-hand side:
\(\frac{1}{r^2 - r + 1} - \frac{1}{r^2 + r + 1} = \frac{(r^2 + r + 1) - (r^2 - r + 1)}{(r^2 - r + 1)(r^2 + r + 1)}\)

Simplify the numerator:
\((r^2 + r + 1) - (r^2 - r + 1) = 2r\)

Simplify the denominator:
\((r^2 - r + 1)(r^2 + r + 1) = ((r^2 + 1) - r)((r^2 + 1) + r) = (r^2 + 1)^2 - r^2 = r^4 + 2r^2 + 1 - r^2 = r^4 + r^2 + 1\)

So the expression becomes:
\(\frac{2r}{r^4 + r^2 + 1}\) (as required).

(b) Let \(f(r) = \frac{1}{r^2 - r + 1}\). Then \(f(r+1) = \frac{1}{(r+1)^2 - (r+1) + 1} = \frac{1}{r^2 + 2r + 1 - r - 1 + 1} = \frac{1}{r^2 + r + 1}\).

Thus, the summation is a telescoping sum:
\(\sum_{r=1}^{n} \frac{2r}{r^4 + r^2 + 1} = \sum_{r=1}^{n} (f(r) - f(r+1))\)

Expanding the sum:
\(= (f(1) - f(2)) + (f(2) - f(3)) + \dots + (f(n) - f(n+1))\)

All middle terms cancel, leaving:
\(= f(1) - f(n+1)\)

Calculate \(f(1)\):
\(f(1) = \frac{1}{1^2 - 1 + 1} = 1\)

Calculate \(f(n+1)\):
\(f(n+1) = \frac{1}{n^2 + n + 1}\)

Thus, the sum \(S_n\) is:
\(S_n = 1 - \frac{1}{n^2 + n + 1} = \frac{n^2 + n + 1 - 1}{n^2 + n + 1} = \frac{n(n+1)}{n^2 + n + 1}\).

(c) The sum to infinity is defined as \(\lim_{n \to \infty} S_n\):
\(S_{\infty} = \lim_{n \to \infty} \left(1 - \frac{1}{n^2 + n + 1}\right)\)

As \(n \to \infty\), \(\frac{1}{n^2 + n + 1} \to 0\).

Thus:
\(S_{\infty} = 1 - 0 = 1\).

(a)
M1: Express the RHS as a single fraction with a common denominator.
A1: Correctly simplify the numerator to \(2r\) and denominator to \(r^4+r^2+1\) to show equivalence.

(b)
M1: Express the sum in telescoping form using the result from (a).
M1: Show the cancellation of intermediate terms, leaving \(f(1) - f(n+1)\).
A1: Obtain the correct sum expression \(1 - \frac{1}{n^2+n+1}\) or \(\frac{n(n+1)}{n^2+n+1}\).

(c)
M1: Apply the limit as \(n \to \infty\) to their expression for \(S_n\).
A1: Obtain the correct sum to infinity of \(1\).

(a) The direction vectors of \(l_1\) and \(l_2\) are \(\mathbf{d}_1 = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}\) and \(\mathbf{d}_2 = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}\) respectively.

A vector perpendicular to both lines is:
\[\mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2 = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} \times \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix} = \begin{pmatrix} (-1)(3) - (1)(1) \\ (1)(2) - (1)(3) \\ (1)(1) - (-1)(2) \end{pmatrix} = \begin{pmatrix} -4 \\ -1 \\ 3 \end{pmatrix}\]

The magnitude of \(\mathbf{n}\) is:
\[|\mathbf{n}| = \sqrt{(-4)^2 + (-1)^2 + 3^2} = \sqrt{26}\]

Points on \(l_1\) and \(l_2\) are \(A(1, 0, 2)\) and \(B(2, 1, -1)\) respectively.
The vector \(\overrightarrow{AB}\) is:
\[\overrightarrow{AB} = \begin{pmatrix} 2 - 1 \\ 1 - 0 \\ -1 - 2 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ -3 \end{pmatrix}\]

The shortest distance \(d\) between the two lines is the projection of \(\overrightarrow{AB}\) onto the common perpendicular \(\mathbf{n}\):
\[d = \frac{|\overrightarrow{AB} \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{\left| \begin{pmatrix} 1 \\ 1 \\ -3 \end{pmatrix} \cdot \begin{pmatrix} -4 \\ -1 \\ 3 \end{pmatrix} \right|}{\sqrt{26}} = \frac{|-4 - 1 - 9|}{\sqrt{26}} = \frac{14}{\sqrt{26}} = \frac{7\sqrt{26}}{13} \approx 2.75\]

(b) The plane \(\Pi\) is parallel to \(l_1\) and contains \(l_2\). Hence, its normal vector is perpendicular to both \(\mathbf{d}_1\) and \(\mathbf{d}_2\). We can use \(\mathbf{n}_0 = \begin{pmatrix} 4 \\ 1 \\ -3 \end{pmatrix}\) as the normal vector.

Since \(\Pi\) contains \(l_2\), it passes through \((2, 1, -1)\).
The Cartesian equation of \(\Pi\) is:
\[4(x - 2) + 1(y - 1) - 3(z + 1) = 0 \implies 4x + y - 3z = 12\]

(c) Let \(\theta\) be the acute angle between \(l_1\) and \(\Pi'\). The direction of \(l_1\) is \(\mathbf{d}_1 = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}\) and the normal to \(\Pi'\) is \(\mathbf{n}' = \begin{pmatrix} 2 \\ -1 \\ -1 \end{pmatrix}\).

\[\sin \theta = \frac{|\mathbf{d}_1 \cdot \mathbf{n}'|}{|\mathbf{d}_1| |\mathbf{n}'|} = \frac{|1(2) + (-1)(-1) + 1(-1)|}{\sqrt{1^2 + (-1)^2 + 1^2} \sqrt{2^2 + (-1)^2 + (-1)^2}} = \frac{|2 + 1 - 1|}{\sqrt{3}\sqrt{6}} = \frac{2}{\sqrt{18}} = \frac{\sqrt{2}}{3}\]

Thus,
\[\theta = \arcsin\left(\frac{\sqrt{2}}{3}\right) \approx 28.1^\circ \text{ (or } 0.491 \text{ radians)}\]

(a)
M1: For attempting to find the cross product of the direction vectors.
A1: For correct normal vector \(\begin{pmatrix} -4 \\ -1 \\ 3 \end{pmatrix}\) (or any multiple).
M1: For finding the vector \(\overrightarrow{AB}\) connecting any point on \(l_1\) to any point on \(l_2\).
M1: For applying the projection formula \(d = \frac{|\overrightarrow{AB} \cdot \mathbf{n}|}{|\mathbf{n}|}\).
A1: For evaluating the dot product correctly to get \(|-14|\) or \(14\).
A1: For the correct exact distance \(\frac{7\sqrt{26}}{13}\) (or 3 s.f. equivalent \(2.75\)).

(b)
M1: For realizing the normal to \(\Pi\) is the same as the common perpendicular from part (a).
M1: For substituting the coordinates of \((2, 1, -1)\) (or another point on \(l_2\)) into the plane equation formula.
A1: For expanding and gathering terms correctly.
A1: For the correct Cartesian equation \(4x + y - 3z = 12\).

(c)
M1: For identifying the correct direction vector of the line and normal vector of the plane.
M1: For writing down the formula \(\sin \theta = \frac{|\mathbf{d}_1 \cdot \mathbf{n}'|}{|\mathbf{d}_1| |\mathbf{n}'|}\) (or equivalent cosine of complementary angle).
A1: For finding the correct value of \(\sin \theta = \frac{\sqrt{2}}{3}\).
A1: For the correct angle in degrees or radians.

(a) Since \(r = a(3 + 2\cos\theta)\), and \(a > 0\), we observe that \(r > 0\) for all \(\theta\) (as the minimum value of \(r\) is \(a\) when \(\theta = \pi\)). The curve is a closed, convex limaçon containing the pole with no dimple or loop.

The intercepts with the initial line \(\theta = 0\) and \(\theta = \pi\) are:
When \(\theta = 0\), \(r = a(3 + 2) = 5a\) \(\implies (5a, 0)\).
When \(\theta = \pi\), \(r = a(3 - 2) = a\) \(\implies (a, \pi)\).

(b) The area \(A\) enclosed by the curve \(C\) is given by:
\[A = \frac{1}{2} \int_{0}^{2\pi} r^2 \, d\theta = \frac{1}{2} a^2 \int_{0}^{2\pi} (3 + 2 \cos \theta)^2 \, d\theta\]
\[(3 + 2 \cos \theta)^2 = 9 + 12 \cos \theta + 4 \cos^2 \theta\]

Using the identity \(\cos^2 \theta = \frac{1 + \cos 2\theta}{2}\):
\[(3 + 2 \cos \theta)^2 = 9 + 12 \cos \theta + 2(1 + \cos 2\theta) = 11 + 12 \cos \theta + 2 \cos 2\theta\]

Integrating this expression:
\[A = \frac{1}{2} a^2 \left[ 11\theta + 12 \sin \theta + \sin 2\theta \right]_{0}^{2\pi} = \frac{1}{2} a^2 (22\pi - 0) = 11\pi a^2\]

(c) Tangents to the curve are perpendicular to the initial line when \(\frac{dx}{d\theta} = 0\) and \(\frac{dy}{d\theta} \ne 0\).

Using \(x = r \cos \theta\):
\[x = a(3 + 2 \cos \theta) \cos \theta = a(3 \cos \theta + 2 \cos^2 \theta)\]
\[\frac{dx}{d\theta} = a(-3 \sin \theta - 4 \sin \theta \cos \theta) = -a \sin \theta (3 + 4 \cos \theta)\]

Setting \(\frac{dx}{d\theta} = 0\):
\[\sin \theta = 0 \quad \text{or} \quad \cos \theta = -0.75\]

Case 1: \(\sin \theta = 0 \implies \theta = 0\) or \(\theta = \pi\).
For \(\theta = 0\), \(r = 5a\). Point is \((5a, 0)\).
For \(\theta = \pi\), \(r = a\). Point is \((a, \pi)\).

Case 2: \(\cos \theta = -0.75 \implies \theta = \arccos(-0.75) \approx 2.42\) rad and \(2\pi - \arccos(-0.75) \approx 3.86\) rad.
For both of these angles:
\[r = a(3 + 2(-0.75)) = 1.5a\]
So the points are \(\left(1.5a, 2.42\right)\) and \(\left(1.5a, 3.86\right)\).

Checking \(\frac{dy}{d\theta} = a(4\cos^2 \theta + 3\cos \theta - 2)\) shows that \(\frac{dy}{d\theta} \ne 0\) at all four of these locations, meaning they are indeed valid vertical tangents.

(a)
M1: For a sketch of a closed convex loop that does not pass through the pole and is symmetric about the initial line.
A1: For labeling the correct general cardioid-like shape.
A1: For identifying intercepts on the initial line at \((5a, 0)\) and \((a, \pi)\).

(b)
M1: For using the area formula \(A = \frac{1}{2} \int r^2 \, d\theta\) with correct limits.
M1: For expanding the integrand to \(9 + 12\cos\theta + 4\cos^2\theta\).
M1: For using the double-angle identity for \(\cos^2\theta\) to rewrite the integrand.
A1: For getting the correct simplified integrand \(11 + 12\cos\theta + 2\cos 2\theta\).
A1: For the correct integration result \(11\pi a^2\).

(c)
M1: For expressing \(x\) as \(r \cos\theta\).
M1: For differentiating \(x\) with respect to \(\theta\).
A1: For setting derivative to zero and obtaining \(\sin\theta = 0\) and \(\cos\theta = -0.75\).
A1: For finding the points \((5a, 0)\) and \((a, \pi)\).
A1: For finding the points \((1.5a, 2.42)\) and \((1.5a, 3.86)\) (allow exact forms with \(\arccos\)).

(a) The denominator is zero when \(x = 2\). At \(x = 2\), the numerator is \(2^2 + 2 - 5 = 1 \ne 0\), so there is a vertical asymptote at \(x = 2\).

Using algebraic division:
\[\frac{x^2 + x - 5}{x - 2} = \frac{x(x - 2) + 3x - 5}{x - 2} = x + 3 + \frac{1}{x - 2}\]

As \(x \to \pm\infty\), the fractional term \(\frac{1}{x-2} \to 0\).
Therefore, the oblique asymptote is \(y = x + 3\).

(b) Differentiating \(y = x + 3 + (x - 2)^{-1}\):
\[\frac{dy}{dx} = 1 - \frac{1}{(x - 2)^2}\]

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

This gives \(x = 3\) or \(x = 1\).
When \(x = 3\), \(y = 3 + 3 + 1 = 7\). Point is \((3, 7)\).
When \(x = 1\), \(y = 1 + 3 - 1 = 3\). Point is \((1, 3)\).

To determine the nature, find the second derivative:
\[\frac{d^2y}{dx^2} = \frac{2}{(x - 2)^3}\]

At \(x = 3\), \(\frac{d^2y}{dx^2} = 2 > 0 \implies (3, 7)\) is a local minimum.
At \(x = 1\), \(\frac{d^2y}{dx^2} = -2 < 0 \implies (1, 3)\) is a local maximum.

(c) For the y-intercept, set \(x = 0\):
\[y = \frac{-5}{-2} = 2.5 \implies (0, 2.5)\]

For the x-intercepts, set \(y = 0\):
\[x^2 + x - 5 = 0 \implies x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-5)}}{2} = \frac{-1 \pm \sqrt{21}}{2}\]

The intercepts are \(\left(\frac{-1 - \sqrt{21}}{2}, 0\right) \approx (-2.79, 0)\) and \(\left(\frac{-1 + \sqrt{21}}{2}, 0\right) \approx (1.79, 0)\).

(d) The sketch should feature:
- The two asymptotes \(x = 2\) and \(y = x + 3\) represented as dashed lines.
- A left branch with a local maximum at \((1, 3)\), crossing the y-axis at \((0, 2.5)\) and the x-axis at \(\approx -2.79\) and \(\approx 1.79\).
- A right branch with a local minimum at \((3, 7)\) residing fully in the region where \(x > 2\) and \(y > x+3\).

(a)
A1: Correct vertical asymptote \(x = 2\).
M1: For attempting algebraic division or equivalent.
A1: Correct oblique asymptote \(y = x + 3\).

(b)
M1: For differentiating the function correctly.
M1: For setting the derivative to zero and solving for \(x\).
A1: For identifying stationary points \((3, 7)\) and \((1, 3)\).
A1: For evaluating second derivative or using a sign table to correctly identify \((3, 7)\) as a local minimum and \((1, 3)\) as a local maximum.

(c)
A1: Correct y-intercept \((0, 2.5)\).
A1: Correct x-intercepts \(\frac{-1 \pm \sqrt{21}}{2}\) (exact or decimal equivalents).

(d)
G1: For two distinct and correctly placed branches.
G1: For showing both asymptotes as dashed lines with equations.
G1: For labeling the coordinates of the stationary points.
G1: For labeling all three axis intercepts.

(a) Let \(\theta\) be the angle that the reflection line \(y = 2x\) makes with the positive x-axis. Thus, \(\tan \theta = 2\).
The general matrix representing a reflection in the line \(y = x \tan \theta\) is:
\[\mathbf{A} = \begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}\]

Using trigonometric double-angle formulas:
\[\cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \frac{1 - 2^2}{1 + 2^2} = -\frac{3}{5}\]
\[\sin 2\theta = \frac{2 \tan \theta}{1 + \tan^2 \theta} = \frac{2(2)}{1 + 2^2} = \frac{4}{5}\]

Substituting these values yields:
\[\mathbf{A} = \begin{pmatrix} -3/5 & 4/5 \\ 4/5 & 3/5 \end{pmatrix} = \frac{1}{5}\begin{pmatrix} -3 & 4 \\ 4 & 3 \end{pmatrix}\]

(b) The combined transformation \(T_2\) followed by \(T_1\) is represented by the product \(\mathbf{C} = \mathbf{A}\mathbf{B}\):
\[\mathbf{C} = \frac{1}{5}\begin{pmatrix} -3 & 4 \\ 4 & 3 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix}\]
\[\mathbf{C} = \frac{1}{5}\begin{pmatrix} -3(1) + 4(3) & -3(2) + 4(2) \\ 4(1) + 3(3) & 4(2) + 3(2) \end{pmatrix} = \frac{1}{5}\begin{pmatrix} 9 & 2 \\ 13 & 14 \end{pmatrix}\]

(c) Let the equation of the invariant line be \(y = mx + c\).
Under \(T_2\), a point \((x, y)\) is mapped to \((x', y')\) where:
\[x' = x + 2y\]
\[y' = 3x + 2y\]

Substituting \(y = mx + c\) into these equations:
\[x' = x + 2(mx + c) = (1 + 2m)x + 2c\]
\[y' = 3x + 2(mx + c) = (3 + 2m)x + 2c\]

For the line to be invariant, \(y' = mx' + c\) must hold for all \(x\):
\[(3 + 2m)x + 2c = m\left((1 + 2m)x + 2c\right) + c\]
\[(3 + 2m)x + 2c = (m + 2m^2)x + (2mc + c)\]

Equating coefficients of \(x\):
\[3 + 2m = m + 2m^2 \implies 2m^2 - m - 3 = 0\]
\[(2m - 3)(m + 1) = 0 \implies m = \frac{3}{2} \quad \text{or} \quad m = -1\]

Equating the constant terms:
\[2c = 2mc + c \implies (2m - 1)c = 0\]

Since \(m = \frac{3}{2}\) or \(m = -1\), we have \(2m - 1 \ne 0\), which forces \(c = 0\).

Thus, the invariant lines are:
\[y = \frac{3}{2}x \quad \text{and} \quad y = -x\]

(a)
M1: For stating the reflection matrix form in terms of \(2\theta\) (or mapping basis vectors).
M1: For using \(\tan \theta = 2\) to find \(\cos 2\theta\) and \(\sin 2\theta\).
A1: For correctly evaluating \(\cos 2\theta = -0.6\) and \(\sin 2\theta = 0.8\).
A1: For showing the final matrix factorized with \(1/5\).

(b)
M1: For setting up the matrix multiplication as \(\mathbf{A}\mathbf{B}\) (order is critical).
A1: For correct intermediate calculations.
A1: For the correct matrix \(\mathbf{C} = \frac{1}{5}\begin{pmatrix} 9 & 2 \\ 13 & 14 \end{pmatrix}\).

(c)
M1: For setting up the mapping of the general point on the line \(y = mx + c\).
M1: For substituting \(y = mx + c\) into equations for \(x'\) and \(y'\).
M1: For substituting these into \(y' = mx' + c\) and equating the coefficients of \(x\).
A1: For finding the correct quadratic in \(m\): \(2m^2 - m - 3 = 0\).
A1: For solving to find \(m = 1.5\) and \(m = -1\).
A1: For equating constants to show \(c = 0\), leading to final lines \(y = 1.5x\) and \(y = -x\).

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

\[ \begin{vmatrix} 2-\lambda & 1 & -1 \\ 0 & 3-\lambda & 1 \\ 0 & 4 & 3-\lambda \end{vmatrix} = 0 \]

Expanding along the first column:

\[ (2-\lambda) \left[ (3-\lambda)^2 - 4 \right] = 0 \]
\[ (2-\lambda)(\lambda^2 - 6\lambda + 5) = 0 \]
\[ (2-\lambda)(\lambda-1)(\lambda-5) = 0 \]

So the eigenvalues are \(\lambda_1 = 1\), \(\lambda_2 = 2\), and \(\lambda_3 = 5\).

The largest eigenvalue is \(\lambda = 5\). To find the corresponding eigenvector, we solve \((\mathbf{A} - 5\mathbf{I})\mathbf{v} = \mathbf{0}\):

\[ \begin{pmatrix} -3 & 1 & -1 \\ 0 & -2 & 1 \\ 0 & 4 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \]

From the second row, we have:
\[ -2y + z = 0 \implies z = 2y \]

From the first row, we have:
\[ -3x + y - z = 0 \implies -3x + y - 2y = 0 \implies -3x - y = 0 \implies y = -3x \]

Setting \(x = 1\), we obtain \(y = -3\) and \(z = -6\).

Thus, a suitable eigenvector is \(\begin{pmatrix} 1 \\ -3 \\ -6 \end{pmatrix}\) (or any non-zero scalar multiple).

M1: Sets up the determinant equation \(\det(\mathbf{A} - \lambda\mathbf{I}) = 0\) and attempts expansion.
A1: Correctly finds the characteristic equation and obtains eigenvalues \(1, 2, 5\).
M1: Sets up the system of equations \((\mathbf{A} - 5\mathbf{I})\mathbf{v} = \mathbf{0}\) for the largest eigenvalue \(\lambda = 5\).
M1: Solves the resulting system to find relationships between components.
A1.33: Obtains a correct eigenvector (e.g., \([1, -3, -6]^T\) or any scalar multiple).

We use the hyperbolic substitution \(x = \sinh u\), which gives \(\mathrm{d}x = \cosh u \, \mathrm{d}u\).

Changing the limits of integration:
- When \(x = 0\), \(u = \operatorname{arsinh}(0) = 0\).
- When \(x = 1\), \(u = \operatorname{arsinh}(1) = \ln(1+\sqrt{2})\).

The integral becomes:
\[ \int_{0}^{\ln(1+\sqrt{2})} \frac{\sinh^2 u}{\cosh u} \cosh u \, \mathrm{d}u = \int_{0}^{\ln(1+\sqrt{2})} \sinh^2 u \, \mathrm{d}u \]

Using the identity \(\sinh^2 u = \frac{\cosh 2u - 1}{2}\):
\[ \int_{0}^{\ln(1+\sqrt{2})} \frac{\cosh 2u - 1}{2} \, \mathrm{d}u = \left[ \frac{\sinh 2u}{4} - \frac{u}{2} \right]_{0}^{\ln(1+\sqrt{2})} \]

Using the identity \(\sinh 2u = 2\sinh u \cosh u = 2\sinh u \sqrt{1+\sinh^2 u}\):
At \(u = \ln(1+\sqrt{2})\) (where \(\sinh u = 1\)), we have:
\[ \sinh 2u = 2(1)\sqrt{1+1} = 2\sqrt{2} \]

Evaluating the limits:
\[ \left( \frac{2\sqrt{2}}{4} - \frac{\ln(1+\sqrt{2})}{2} \right) - (0) = \frac{\sqrt{2} - \ln(1+\sqrt{2})}{2} \]

M1: Applies a valid method, such as substitution \(x = \sinh u\) (or integration by parts).
A1: Correctly simplifies the integrand (e.g., to \(\sinh^2 u\)).
M1: Uses a double-angle identity or equivalent technique to integrate correctly.
M1: Standardizes the limits and correctly applies the logarithmic representation of \(\operatorname{arsinh}(1)\).
A1.33: Obtains the correct exact answer: \(\frac{\sqrt{2} - \ln(1+\sqrt{2})}{2}\).

First, we find the complementary function (CF) by solving the auxiliary equation:
\[ m^2 + 4m + 4 = 0 \implies (m+2)^2 = 0 \implies m = -2 \text{ (repeated root)} \]

Thus, the CF is:
\[ y_c = (Ax + B)\mathrm{e}^{-2x} \]

Next, we find the particular integral (PI). Since \(-2\) is a double root of the auxiliary equation and the RHS is \(8\mathrm{e}^{-2x}\), we try a PI of the form:
\[ y_p = C x^2 \mathrm{e}^{-2x} \]

Differentiating \(y_p\):
\[ y_p' = C(2x - 2x^2)\mathrm{e}^{-2x} \]
\[ y_p'' = C(2 - 8x + 4x^2)\mathrm{e}^{-2x} \]

Substituting these into the original ODE:
\[ C\mathrm{e}^{-2x} \left[ (2 - 8x + 4x^2) + 4(2x - 2x^2) + 4x^2 \right] = 8\mathrm{e}^{-2x} \]
\[ C\mathrm{e}^{-2x} (2) = 8\mathrm{e}^{-2x} \implies 2C = 8 \implies C = 4 \]

So the PI is \(y_p = 4x^2 \mathrm{e}^{-2x}\). This gives the general solution:
\[ y = (Ax + B)\mathrm{e}^{-2x} + 4x^2 \mathrm{e}^{-2x} = (4x^2 + Ax + B)\mathrm{e}^{-2x} \]

Now we apply the initial conditions:
1. At \(x = 0\), \(y = 3\):
\[ 3 = B \]

2. Differentiating the general solution:
\[ y' = (8x + A)\mathrm{e}^{-2x} - 2(4x^2 + Ax + B)\mathrm{e}^{-2x} \]
At \(x = 0\), \(y' = -2\):
\[ -2 = A - 2B \implies -2 = A - 6 \implies A = 4 \]

Therefore, the particular solution is:
\[ y = (4x^2 + 4x + 3)\mathrm{e}^{-2x} \]

M1: Solves the auxiliary equation to find the correct CF form \(y_c = (Ax+B)\mathrm{e}^{-2x}\).
A1: Identifies the correct PI form \(y_p = C x^2 \mathrm{e}^{-2x}\), differentiates, substitutes, and finds \(C = 4\).
M1: Formulates the general solution and uses \(y(0)=3\) to determine \(B=3\).
M1: Correctly differentiates the general solution and uses \(y'(0)=-2\) to determine \(A=4\).
A1.33: States the correct particular solution \(y = (4x^2 + 4x + 3)\mathrm{e}^{-2x}\).

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

Next, find the particular integral (PI) of the form:
\[ y_p = P \cos x + Q \sin x \]
Differentiating:
\[ y_p' = -P \sin x + Q \cos x \]
\[ y_p'' = -P \cos x - Q \sin x \]

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

Equating coefficients:
\[ 4P + 4Q = 10 \implies P + Q = \frac{5}{2} \]
\[ 4Q - 4P = 0 \implies P = Q \]

Thus, \( 2P = \frac{5}{2} \implies P = \frac{5}{4} \) and \( Q = \frac{5}{4} \).
So the particular integral is:
\[ y_p = \frac{5}{4}(\cos x + \sin x) \]

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

Now use the boundary conditions to find the constants \( A \) and \( B \):
When \( x = 0 \), \( y = 0 \):
\[ 0 = A + \frac{5}{4} \implies A = -\frac{5}{4} \]

Differentiating the general solution:
\[ y' = -2\text{e}^{-2x}(A\cos x + B\sin x) + \text{e}^{-2x}(-A\sin x + B\cos x) + \frac{5}{4}(-\sin x + \cos x) \]
When \( x = 0 \), \( y' = 0 \):
\[ 0 = -2A + B + \frac{5}{4} \]
Substitute \( A = -\frac{5}{4} \):
\[ 0 = -2(-\frac{5}{4}) + B + \frac{5}{4} = \frac{5}{2} + B + \frac{5}{4} = B + \frac{15}{4} \]
\[ B = -\frac{15}{4} \]

Therefore, the particular solution is:
\[ y = -\frac{5}{4} \text{e}^{-2x} (\cos x + 3\sin x) + \frac{5}{4}(\cos x + \sin x) \]

M1: Solves auxiliary equation to find complex roots.
A1: Correct complementary function.
M1: Sets up PI form \( P \cos x + Q \sin x \) and differentiates.
M1: Substitutes into DE and solves for constants \( P \) and \( Q \).
A1: Correct values \( P = Q = \frac{5}{4} \).
A1: Correct general solution.
M1: Applies boundary condition \( y(0) = 0 \) to find \( A \).
M1: Differentiates general solution and applies \( y'(0) = 0 \) to find \( B \).
A1.75: Fully correct particular solution (equivalent forms accepted).

Express the hyperbolic functions in terms of exponential functions:
\[ \cosh x = \frac{\text{e}^x + \text{e}^{-x}}{2} \]
\[ \sinh x = \frac{\text{e}^x - \text{e}^{-x}}{2} \]

Substitute these into the denominator:
\[ 3 \cosh x + 5 \sinh x = 3 \left(\frac{\text{e}^x + \text{e}^{-x}}{2}\right) + 5 \left(\frac{\text{e}^x - \text{e}^{-x}}{2}\right) = \frac{8\text{e}^x - 2\text{e}^{-x}}{2} = 4\text{e}^x - \text{e}^{-x} \]

Thus, the integral becomes:
\[ \int_0^{\ln 2} \frac{1}{4\text{e}^x - \text{e}^{-x}} \, \text{dx} \]

Substitute \( u = \text{e}^x \), so \( \text{d}u = \text{e}^x \, \text{dx} \implies \text{dx} = \frac{\text{d}u}{u} \).
Change the limits of integration:
When \( x = 0 \), \( u = \text{e}^0 = 1 \).
When \( x = \ln 2 \), \( u = \text{e}^{\ln 2} = 2 \).

The integral becomes:
\[ \int_1^2 \frac{1}{4u - u^{-1}} \frac{\text{d}u}{u} = \int_1^2 \frac{1}{4u^2 - 1} \, \text{d}u \]

Use partial fractions or standard integral:
\[ \frac{1}{4u^2 - 1} = \frac{1}{(2u-1)(2u+1)} = \frac{1}{2} \left( \frac{1}{2u-1} - \frac{1}{2u+1} \right) \]

Integrate term-by-term:
\[ \int_1^2 \frac{1}{4u^2 - 1} \, \text{d}u = \left[ \frac{1}{4} \ln |2u-1| - \frac{1}{4} \ln |2u+1| \right]_1^2 = \left[ \frac{1}{4} \ln \left| \frac{2u-1}{2u+1} \right| \right]_1^2 \]

Evaluate at the limits:
Upper limit \( u = 2 \): \( \frac{1}{4} \ln \left( \frac{3}{5} \right) \)
Lower limit \( u = 1 \): \( \frac{1}{4} \ln \left( \frac{1}{3} \right) \)

Subtracting the lower limit from the upper limit:
\[ \frac{1}{4} \ln \left( \frac{3}{5} \right) - \frac{1}{4} \ln \left( \frac{1}{3} \right) = \frac{1}{4} \ln \left( \frac{3/5}{1/3} \right) = \frac{1}{4} \ln \left( \frac{9}{5} \right) \]

M1: Expresses \(\cosh x\) and \(\sinh x\) in terms of exponentials.
A1: Obtains the correct integrand in terms of \(\text{e}^x\).
M1: Uses substitution \(u = \text{e}^x\) (or similar substitution).
A1: Correctly changes limits and gets the integral in terms of \(u\).
M1: Applies partial fractions or the standard integration formula for \(\int \frac{1}{u^2-a^2} \text{d}u\).
A1: Obtains correct antiderivative: \(\frac{1}{4} \ln \left| \frac{2u-1}{2u+1} \right|\).
M1: Substitutes the limits 1 and 2.
A1: Uses logarithm laws to combine the terms.
A1.75: Obtains the final exact value \(\frac{1}{4} \ln \left( \frac{9}{5} \right)\).

(a) We differentiate \( y = \text{e}^{-x} \cos x \) successively:
At \( x = 0 \): \( y = \text{e}^0 \cos 0 = 1 \).

First derivative:
\[ y' = -\text{e}^{-x} \cos x - \text{e}^{-x} \sin x = -\text{e}^{-x} (\cos x + \sin x) \]
At \( x = 0 \): \( y'(0) = -1(1 + 0) = -1 \).

Second derivative:
\[ y'' = \text{e}^{-x} (\cos x + \sin x) - \text{e}^{-x} (-\sin x + \cos x) = 2\text{e}^{-x} \sin x \]
At \( x = 0 \): \( y''(0) = 2(1)(0) = 0 \).

Third derivative:
\[ y''' = -2\text{e}^{-x} \sin x + 2\text{e}^{-x} \cos x = 2\text{e}^{-x} (\cos x - \sin x) \]
At \( x = 0 \): \( y'''(0) = 2(1)(1 - 0) = 2 \).

(b) The Maclaurin series expansion is:
\[ y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \dots \]
Substituting the values:
\[ y \approx 1 - x + \frac{0}{2}x^2 + \frac{2}{6}x^3 = 1 - x + \frac{1}{3}x^3 \]

(c) Approximating the integral:
\[ \int_0^{0.2} \text{e}^{-x} \cos x \, \text{dx} \approx \int_0^{0.2} \left(1 - x + \frac{1}{3}x^3\right) \, \text{dx} \]

\[ = \left[ x - \frac{x^2}{2} + \frac{x^4}{12} \right]_0^{0.2} \]
\[ = 0.2 - \frac{(0.2)^2}{2} + \frac{(0.2)^4}{12} \]
\[ = 0.2 - \frac{0.04}{2} + \frac{0.0016}{12} \]
\[ = 0.2 - 0.02 + \frac{0.0004}{3} \]
\[ = 0.18 + 0.00013333... \approx 0.18013 \]

M1: Correctly applies product rule to find \( y' \).
A1: Correct values of \( y(0) = 1 \) and \( y'(0) = -1 \).
M1: Correctly differentiates to find \( y'' \).
A1: Correct values of \( y''(0) = 0 \).
M1: Correctly differentiates to find \( y''' \).
A1: Correct value of \( y'''(0) = 2 \).
M1: Uses Maclaurin's formula to write down the expansion.
A1: Correct series \( 1 - x + \frac{1}{3}x^3 \).
M1: Integrates term-by-term.
A1.75: Correctly substitutes limits and calculates the decimal value to 5 d.p. (0.18013).

(a) Using the definition of \( I_n \):
\[ I_n + I_{n-2} = \int_0^{\frac{\pi}{4}} \tan^n x \, \text{dx} + \int_0^{\frac{\pi}{4}} \tan^{n-2} x \, \text{dx} \]
\[ = \int_0^{\frac{\pi}{4}} (\tan^n x + \tan^{n-2} x) \, \text{dx} \]
\[ = \int_0^{\frac{\pi}{4}} \tan^{n-2} x (\tan^2 x + 1) \, \text{dx} \]

Since \( \tan^2 x + 1 = \sec^2 x \), we have:
\[ I_n + I_{n-2} = \int_0^{\frac{\pi}{4}} \tan^{n-2} x \sec^2 x \, \text{dx} \]

Let \( u = \tan x \), then \( \text{d}u = \sec^2 x \, \text{dx} \).
The limits of integration change as follows:
When \( x = 0 \), \( u = \tan 0 = 0 \).
When \( x = \frac{\pi}{4} \), \( u = \tan \frac{\pi}{4} = 1 \).

Thus:
\[ I_n + I_{n-2} = \int_0^1 u^{n-2} \, \text{d}u = \left[ \frac{u^{n-1}}{n-1} \right]_0^1 = \frac{1^{n-1}}{n-1} - 0 = \frac{1}{n-1} \] (for \( n \ge 2 \)).

(b) We want to find \( I_5 \).
Using the reduction formula:
For \( n = 5 \):
\[ I_5 + I_3 = \frac{1}{4} \implies I_5 = \frac{1}{4} - I_3 \]

For \( n = 3 \):
\[ I_3 + I_1 = \frac{1}{2} \implies I_3 = \frac{1}{2} - I_1 \]

We now calculate \( I_1 \):
\[ I_1 = \int_0^{\frac{\pi}{4}} \tan x \, \text{dx} = [ \ln |\sec x| ]_0^{\frac{\pi}{4}} \]
\[ = \ln \left( \sec \frac{\pi}{4} \right) - \ln(\sec 0) \]
Since \( \sec \frac{\pi}{4} = \sqrt{2} \) and \( \sec 0 = 1 \):
\[ I_1 = \ln \sqrt{2} - 0 = \frac{1}{2} \ln 2 \]

Now substitute back:
\[ I_3 = \frac{1}{2} - \frac{1}{2} \ln 2 \]
\[ I_5 = \frac{1}{4} - \left( \frac{1}{2} - \frac{1}{2} \ln 2 \right) = \frac{1}{2} \ln 2 - \frac{1}{4} \]

M1: Combines \( I_n + I_{n-2} \) into a single integral.
M1: Factors out \( \tan^{n-2} x \) and uses the identity \( 1 + \tan^2 x = \sec^2 x \).
M1: Uses substitution \( u = \tan x \) and changes limits.
A1: Completes the integration to show the required reduction formula.
M1: Applies the reduction formula for \( n = 5 \).
A1: Correctly expresses \( I_5 \) in terms of \( I_3 \).
M1: Applies the reduction formula for \( n = 3 \) to express \( I_3 \) in terms of \( I_1 \).
M1: Integrates \( \tan x \) to find \( I_1 \).
A1: Correct value \( I_1 = \frac{1}{2} \ln 2 \).
A1.75: Obtains final exact value \( \frac{1}{2} \ln 2 - \frac{1}{4} \) (accept equivalent exact forms).

(a) To find the eigenvalues, we solve the characteristic equation \(\det(\mathbf{A} - \lambda \mathbf{I}) = 0\). This is given by \(\det \begin{pmatrix} 3-\lambda & -1 & 1 \\ -1 & 5-\lambda & -1 \\ 1 & -1 & 3-\lambda \end{pmatrix} = 0\). Expanding along the first row: \((3-\lambda) [(5-\lambda)(3-\lambda) - 1] - (-1) [ -1(3-\lambda) - (-1) ] + 1 [ 1 - (5-\lambda) ] = 0\). Simplifying inside the brackets: \((3-\lambda)(\lambda^2 - 8\lambda + 14) + (\lambda - 2) + (\lambda - 4) = 0 \implies -\lambda^3 + 11\lambda^2 - 36\lambda + 36 = 0\). Test integer factors of 36: for \(\lambda = 2\), we get \(8 - 44 + 72 - 36 = 0\), so \(\lambda = 2\) is a root. Factoring out \(\lambda - 2\), we get \((\lambda - 2)(\lambda^2 - 9\lambda + 18) = 0 \implies (\lambda - 2)(\lambda - 3)(\lambda - 6) = 0\). The eigenvalues are \(\lambda = 2, 3, 6\). (b) For \(\lambda = 2\): \(\begin{pmatrix} 1 & -1 & 1 \\ -1 & 3 & -1 \\ 1 & -1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\). From Row 1 and Row 2, adding them gives \(2y = 0 \implies y = 0\). Then \(x + z = 0 \implies x = -z\). An eigenvector is \(\mathbf{v}_1 = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}\). For \(\lambda = 3\): \(\begin{pmatrix} 0 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\). From Row 1, \(y = z\). From Row 3, \(x = y\). An eigenvector is \(\mathbf{v}_2 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\). For \(\lambda = 6\): \(\begin{pmatrix} -3 & -1 & 1 \\ -1 & -1 & -1 \\ 1 & -1 & -3 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\). Row 1 minus three times Row 2 gives \(2y + 4z = 0 \implies y = -2z\). Row 3 gives \(x - y - 3z = 0 \implies x - (-2z) - 3z = 0 \implies x = z\). An eigenvector is \(\mathbf{v}_3 = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}\). (c) Choosing columns in order of eigenvalues \(\lambda = 2, 3, 6\), we have \(\mathbf{D} = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 6 \end{pmatrix}\) and \(\mathbf{P} = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & -2 \\ -1 & 1 & 1 \end{pmatrix}\). To find \(\mathbf{P}^{-1}\), we calculate \(\det(\mathbf{P}) = 1(1 - (-2)) - 1(0 - 2) + 1(0 - (-1)) = 3 + 2 + 1 = 6\). The cofactor matrix of \(\mathbf{P}\) is \(\mathbf{C} = \begin{pmatrix} 3 & 2 & 1 \\ 0 & 2 & -2 \\ -3 & 2 & 1 \end{pmatrix}\). The transpose of the cofactor matrix is the adjugate matrix \(\operatorname{adj}(\mathbf{P}) = \begin{pmatrix} 3 & 0 & -3 \\ 2 & 2 & 2 \\ 1 & -2 & 1 \end{pmatrix}\). Therefore, \(\mathbf{P}^{-1} = \frac{1}{6} \begin{pmatrix} 3 & 0 & -3 \\ 2 & 2 & 2 \\ 1 & -2 & 1 \end{pmatrix}\). (d) Using \(\mathbf{A}^n = \mathbf{P} \mathbf{D}^n \mathbf{P}^{-1}\), we find: \(\mathbf{P} \mathbf{D}^n = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & -2 \\ -1 & 1 & 1 \end{pmatrix} \begin{pmatrix} 2^n & 0 & 0 \\ 0 & 3^n & 0 \\ 0 & 0 & 6^n \end{pmatrix} = \begin{pmatrix} 2^n & 3^n & 6^n \\ 0 & 3^n & -2 \cdot 6^n \\ -2^n & 3^n & 6^n \end{pmatrix}\). Now multiplying \(\mathbf{P} \mathbf{D}^n\) by \(\mathbf{P}^{-1}\) yields: \(\mathbf{A}^n = \frac{1}{6} \begin{pmatrix} 2^n & 3^n & 6^n \\ 0 & 3^n & -2 \cdot 6^n \\ -2^n & 3^n & 6^n \end{pmatrix} \begin{pmatrix} 3 & 0 & -3 \\ 2 & 2 & 2 \\ 1 & -2 & 1 \end{pmatrix} = \frac{1}{6} \begin{pmatrix} 3 \cdot 2^n + 2 \cdot 3^n + 6^n & 2 \cdot 3^n - 2 \cdot 6^n & -3 \cdot 2^n + 2 \cdot 3^n + 6^n \\ 2 \cdot 3^n - 2 \cdot 6^n & 2 \cdot 3^n + 4 \cdot 6^n & 2 \cdot 3^n - 2 \cdot 6^n \\ -3 \cdot 2^n + 2 \cdot 3^n + 6^n & 2 \cdot 3^n - 2 \cdot 6^n & 3 \cdot 2^n + 2 \cdot 3^n + 6^n \end{pmatrix}\).

Part (a): [5 marks] M1: Attempt to construct characteristic equation \(\det(\mathbf{A}-\lambda\mathbf{I}) = 0\). A1: Correct determinant expansion leading to the cubic polynomial \(-\lambda^3 + 11\lambda^2 - 36\lambda + 36 = 0\). M1: Use of factor theorem to find one integer root (e.g., testing \(\lambda=2\) or \(\lambda=3\)). M1: Factorisation of the cubic polynomial to a quadratic expression. A1: Correctly identification of all three eigenvalues \(2, 3, 6\). Part (b): [4 marks] M1: Method for finding eigenvector corresponding to \(\lambda=2\) (setting up and solving linear equations). A1: Correct eigenvector for \(\lambda=2\) (or any non-zero multiple). A1: Correct eigenvector for \(\lambda=3\) (or any non-zero multiple). A1: Correct eigenvector for \(\lambda=6\) (or any non-zero multiple). Part (c): [4 marks] B1: Correct diagonal matrix \(\mathbf{D}\) and transforming matrix \(\mathbf{P}\) written down based on eigenvectors from (b). M1: Correct method for finding the determinant and cofactor/adjugate matrix of \(\mathbf{P}\). A1: Correct adjugate matrix \(\operatorname{adj}(\mathbf{P})\) or cofactor matrix. A1: Correct inverse matrix \(\mathbf{P}^{-1} = \frac{1}{6} \begin{pmatrix} 3 & 0 & -3 \\ 2 & 2 & 2 \\ 1 & -2 & 1 \end{pmatrix}\) (or consistent with their choice of \(\mathbf{P}\)). Part (d): [3 marks] M1: Correctly compute the matrix product \(\mathbf{P}\mathbf{D}^n\). M1: Attempt matrix multiplication \((\mathbf{P}\mathbf{D}^n)\mathbf{P}^{-1}\) showing at least two elements correctly worked out in terms of \(n\). A1: Complete and accurate proof showing the target matrix for \(\mathbf{A}^n\).