2023 AQA A-Level Further Mathematics 7367 模擬試題連答案詳解

For an \(n \times n\) matrix \(\mathbf{M}\) and scalar \(c\), the determinant property states that \(\det(c\mathbf{M}) = c^n \det(\mathbf{M})\). Here, \(\mathbf{A}\) is a \(3 \times 3\) matrix, so its inverse \(\mathbf{A}^{-1}\) is also a \(3 \times 3\) matrix. Therefore, \(\det(2\mathbf{A}^{-1}) = 2^3 \det(\mathbf{A}^{-1}) = 8 \det(\mathbf{A}^{-1})\). Since \(\det(\mathbf{A}^{-1}) = \frac{1}{\det(\mathbf{A})} = \frac{1}{k}\), we obtain \(\det(2\mathbf{A}^{-1}) = \frac{8}{k}\).

B1: Correctly selects option B.

A tangent perpendicular to the initial line occurs when \(\frac{\mathrm{d}x}{\mathrm{d}\theta} = 0\) (provided \(\frac{\mathrm{d}y}{\mathrm{d}\theta} \neq 0\)). Since \(x = r \cos\theta = a(1 + \sin\theta)\cos\theta\), differentiating with respect to \(\theta\) gives: \(\frac{\mathrm{d}x}{\mathrm{d}\theta} = a(-\sin\theta(1 + \sin\theta) + \cos^2\theta) = a(-\sin\theta - \sin^2\theta + 1 - \sin^2\theta) = a(1 - \sin\theta - 2\sin^2\theta)\). Setting this to zero: \(2\sin^2\theta + \sin\theta - 1 = 0 \implies (2\sin\theta - 1)(\sin\theta + 1) = 0\). This yields \(\sin\theta = \frac{1}{2}\) or \(\sin\theta = -1\). Out of the given options, \(\theta = \frac{\pi}{6}\) satisfies \(\sin\theta = \frac{1}{2}\).

B1: Correctly selects option C.

Using the exponential definitions of hyperbolic functions: \(\sinh x = \frac{e^x - e^{-x}}{2}\) and \(\cosh x = \frac{e^x + e^{-x}}{2}\). Substituting these into the equation: \(3\left(\frac{e^x - e^{-x}}{2}\right) - \left(\frac{e^x + e^{-x}}{2}\right) = 1\). Multiplying both sides by 2 gives: \(3(e^x - e^{-x}) - (e^x + e^{-x}) = 2 \implies 2e^x - 4e^{-x} = 2 \implies e^x - 2e^{-x} - 1 = 0\). Multiplying by \(e^x\) and letting \(u = e^x\) yields: \(u^2 - u - 2 = 0 \implies (u-2)(u+1) = 0\). Since \(e^x > 0\), we must have \(u = 2\), which gives \(x = \ln 2\).

B1: Correctly selects option A.

Using the exponential definition of the hyperbolic cosine function, we have: \(\cosh(x) = \frac{e^x + e^{-x}}{2}\). Let \(x = \ln(3 + 2\sqrt{2})\). Then: \(e^x = e^{\ln(3 + 2\sqrt{2})} = 3 + 2\sqrt{2}\). To find \(e^{-x}\), we take the reciprocal: \(e^{-x} = \frac{1}{3 + 2\sqrt{2}}\). Rationalising the denominator gives: \(e^{-x} = \frac{1}{3 + 2\sqrt{2}} \times \frac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}} = \frac{3 - 2\sqrt{2}}{9 - 8} = 3 - 2\sqrt{2}\). Substituting these back into the definition of \(\cosh(x)\): \(\cosh(\ln(3 + 2\sqrt{2})) = \frac{(3 + 2\sqrt{2}) + (3 - 2\sqrt{2})}{2} = \frac{6}{2} = 3\). This corresponds to option A.

B1: Obtains the correct option A.

(a) To find a reduction formula for \( I_n = \int_{0}^{1} x^n e^{-x} \, dx \), use integration by parts:
Let \( u = x^n \implies \frac{du}{dx} = n x^{n-1} \)
Let \( \frac{dv}{dx} = e^{-x} \implies v = -e^{-x} \)

Applying the integration by parts formula:
\[ I_n = \left[ -x^n e^{-x} \right]_0^1 - \int_{0}^{1} -n x^{n-1} e^{-x} \, dx \]
\[ I_n = \left( -1^n e^{-1} - 0 \right) + n \int_{0}^{1} x^{n-1} e^{-x} \, dx \]
\[ I_n = -\frac{1}{e} + n I_{n-1} \]
\[ I_n = n I_{n-1} - \frac{1}{e} \]

(b) First find \( I_0 \):
\[ I_0 = \int_{0}^{1} e^{-x} \, dx = \left[ -e^{-x} \right]_0^1 = -e^{-1} - (-e^0) = 1 - \frac{1}{e} \]

Now apply the reduction formula successively:
For \( n = 1 \):
\[ I_1 = 1 \cdot I_0 - \frac{1}{e} = \left(1 - \frac{1}{e}\right) - \frac{1}{e} = 1 - \frac{2}{e} \]

For \( n = 2 \):
\[ I_2 = 2 I_1 - \frac{1}{e} = 2\left(1 - \frac{2}{e}\right) - \frac{1}{e} = 2 - \frac{4}{e} - \frac{1}{e} = 2 - \frac{5}{e} \]

For \( n = 3 \):
\[ I_3 = 3 I_2 - \frac{1}{e} = 3\left(2 - \frac{5}{e}\right) - \frac{1}{e} = 6 - \frac{15}{e} - \frac{1}{e} = 6 - \frac{16}{e} \]

(a)
- M1: Applies integration by parts with appropriate choices of \( u \) and \( v \).
- A1: Obtains \( \frac{du}{dx} = n x^{n-1} \) and \( v = -e^{-x} \).
- M1: Evaluates the boundary term \( \left[ -x^n e^{-x} \right]_0^1 \) to get \( -e^{-1} \) or \( -\frac{1}{e} \).
- A1: Completes the proof to show \( I_n = n I_{n-1} - \frac{1}{e} \) clearly.

(b)
- B1: Correctly calculates \( I_0 = 1 - \frac{1}{e} \) (or equivalent).
- M1: Applies the reduction formula to find \( I_1 \) in terms of \( e \).
- M1: Applies the reduction formula to find \( I_2 \) and then \( I_3 \).
- A1: Obtains the correct exact final answer \( 6 - \frac{16}{e} \).

(a) If \( \lambda = 3 \) is an eigenvalue, then the determinant of \( \mathbf{M} - 3\mathbf{I} \) must be zero:
\[ \mathbf{M} - 3\mathbf{I} = \begin{pmatrix} k-3 & 1 & 1 \\ 0 & -1 & -1 \\ 3 & 1 & -2 \end{pmatrix} \]
\[ \det(\mathbf{M} - 3\mathbf{I}) = (k-3)[(-1)(-2) - (1)(-1)] - 1[0(-2) - 3(-1)] + 1[0(1) - 3(-1)] = 0 \]
\[ (k-3)(2 + 1) - 3 + 3 = 0 \]
\[ 3(k-3) = 0 \implies k = 3 \]

(b) With \( k = 3 \), the characteristic equation is \( \det(\mathbf{M} - \lambda\mathbf{I}) = 0 \):
\[ \begin{vmatrix} 3-\lambda & 1 & 1 \\ 0 & 2-\lambda & -1 \\ 3 & 1 & 1-\lambda \end{vmatrix} = 0 \]
Expanding along the first column:
\[ (3-\lambda) \left[ (2-\lambda)(1-\lambda) - (-1) \right] + 3 \left[ -1 - (2-\lambda) \right] = 0 \]
\[ (3-\lambda)(\lambda^2 - 3\lambda + 2 + 1) + 3(\lambda - 3) = 0 \]
\[ (3-\lambda)(\lambda^2 - 3\lambda + 3) - 3(3-\lambda) = 0 \]
Factorizing out \( (3-\lambda) \):
\[ (3-\lambda)(\lambda^2 - 3\lambda + 3 - 3) = 0 \]
\[ (3-\lambda)(\lambda^2 - 3\lambda) = 0 \]
\[ \lambda(\lambda-3)^2 = 0 \]
So the eigenvalues are \( \lambda = 3 \) (repeated) and \( \lambda = 0 \).
Thus, the other eigenvalues are \( 3 \) and \( 0 \).

(c) For \( \lambda = 0 \), solve \( \mathbf{M}\mathbf{v} = \mathbf{0} \):
\[ \begin{pmatrix} 3 & 1 & 1 \\ 0 & 2 & -1 \\ 3 & 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \]
From the second row: \( 2y - z = 0 \implies z = 2y \).
From the first row: \( 3x + y + z = 0 \implies 3x + y + 2y = 0 \implies 3x + 3y = 0 \implies x = -y \).
Letting \( y = 1 \), we get the eigenvector:
\[ \mathbf{v} = \begin{pmatrix} -1 \\ 1 \\ 2 \end{pmatrix} \]
To normalize, divide by its magnitude \( \sqrt{(-1)^2 + 1^2 + 2^2} = \sqrt{6} \):
\[ \hat{\mathbf{v}} = \pm \frac{1}{\sqrt{6}} \begin{pmatrix} -1 \\ 1 \\ 2 \end{pmatrix} \]

(a)
- M1: Sets up the matrix \( \mathbf{M} - 3\mathbf{I} \).
- M1: Evaluates the determinant of \( \mathbf{M} - 3\mathbf{I} \) and equates to zero.
- A1: Solves to obtain \( k = 3 \) with clear steps.

(b)
- M1: Sets up the determinant equation for the characteristic polynomial with \( k = 3 \).
- A1: Correctly expands to get the cubic equation, e.g., \( \lambda(\lambda-3)^2 = 0 \) or equivalent.
- A1: Identifies the other eigenvalues as \( 3 \) and \( 0 \).

(c)
- M1: Sets up the system of equations for \( \lambda = 0 \) and finds a non-zero eigenvector, e.g. \( \begin{pmatrix} -1 \\ 1 \\ 2 \end{pmatrix} \).
- A1: Correctly normalizes the vector to get \( \pm \frac{1}{\sqrt{6}} \begin{pmatrix} -1 \\ 1 \\ 2 \end{pmatrix} \).

Step 1: Find the complementary function (CF).
The auxiliary equation is:
\[ m^2 - 4m + 4 = 0 \]
\[ (m-2)^2 = 0 \implies m = 2 \text{ (repeated root)} \]
Therefore, the complementary function is:
\[ y_c = (A + Bx)e^{2x} \]

Step 2: Find the particular integral (PI).
Since \( e^{2x} \) and \( xe^{2x} \) are already in the CF, we must try a PI of the form:
\[ y_p = C x^2 e^{2x} \]
Differentiating \( y_p \):
\[ \frac{dy_p}{dx} = 2Cx e^{2x} + 2Cx^2 e^{2x} = 2C(x + x^2)e^{2x} \]
\[ \frac{d^2 y_p}{dx^2} = 2C(1 + 2x)e^{2x} + 4C(x + x^2)e^{2x} = 2C(1 + 4x + 2x^2)e^{2x} \]
Substitute into the differential equation:
\[ 2C(1 + 4x + 2x^2)e^{2x} - 4 \left[ 2C(x + x^2)e^{2x} \right] + 4 C x^2 e^{2x} = e^{2x} \]
Divide through by \( e^{2x} \):
\[ 2C + 8Cx + 4Cx^2 - 8Cx - 8Cx^2 + 4Cx^2 = 1 \]
\[ 2C = 1 \implies C = \frac{1}{2} \]
So the particular integral is \( y_p = \frac{1}{2} x^2 e^{2x} \).

Step 3: State the general solution.
\[ y = (A + Bx)e^{2x} + \frac{1}{2} x^2 e^{2x} \]

Step 4: Use the boundary conditions to find \( A \) and \( B \).
At \( x = 0 \), \( y = 1 \):
\[ 1 = (A + 0)e^0 + 0 \implies A = 1 \]

Now differentiate the general solution to use the second condition:
\[ \frac{dy}{dx} = B e^{2x} + 2(A + Bx)e^{2x} + x e^{2x} + x^2 e^{2x} \]
At \( x = 0 \), \( \frac{dy}{dx} = 3 \):
\[ 3 = B e^0 + 2(A + 0)e^0 + 0 + 0 \]
\[ 3 = B + 2A \]
Since \( A = 1 \):
\[ 3 = B + 2 \implies B = 1 \]

Therefore, the particular solution is:
\[ y = (1 + x + \frac{1}{2}x^2)e^{2x} \]

- M1: Formulates the auxiliary equation \( m^2 - 4m + 4 = 0 \) and finds the root \( m=2 \).
- A1: Correctly writes down the complementary function \( y_c = (A + Bx)e^{2x} \).
- M1: Recognizes the correct form of the particular integral \( y_p = C x^2 e^{2x} \) and differentiates it.
- A1: Substitutes into the differential equation and solves to find \( C = \frac{1}{2} \).
- A1: Combines to write the correct general solution \( y = (A + Bx)e^{2x} + \frac{1}{2} x^2 e^{2x} \).
- M1: Applies the initial condition \( y(0) = 1 \) to find \( A = 1 \).
- M1: Differentiates the general solution and applies \( y'(0) = 3 \) to find \( B \).
- A1: Obtains the correct final particular solution \( y = (1 + x + \frac{1}{2}x^2)e^{2x} \).

(a) By de Moivre's theorem:
$$\cos(5\theta) + i\sin(5\theta) = (\cos\theta + i\sin\theta)^5$$
Using the binomial expansion:
$$(\cos\theta + i\sin\theta)^5 = \cos^5\theta + 5i\cos^4\theta\sin\theta - 10\cos^3\theta\sin^2\theta - 10i\cos^2\theta\sin^3\theta + 5\cos\theta\sin^4\theta + i\sin^5\theta$$
Equating imaginary parts:
$$\sin(5\theta) = 5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta$$
Using the identity $\cos^2\theta = 1 - \sin^2\theta$:
$$\sin(5\theta) = 5(1-\sin^2\theta)^2\sin\theta - 10(1-\sin^2\theta)\sin^3\theta + \sin^5\theta$$
$$\sin(5\theta) = 5(1-2\sin^2\theta+\sin^4\theta)\sin\theta - 10\sin^3\theta + 10\sin^5\theta + \sin^5\theta$$
$$\sin(5\theta) = 5\sin\theta - 10\sin^3\theta + 5\sin^5\theta - 10\sin^3\theta + 11\sin^5\theta$$
$$\sin(5\theta) = 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta \quad \text{(as required)}$$

(b) We want to solve $16x^4 - 20x^2 + 5 = 0$.
Since $x
eq 0$ is clear, we can multiply by $x$ to get:
$$16x^5 - 20x^3 + 5x = 0 \quad (x \neq 0)$$
Let $x = \sin\theta$. The equation becomes:
$$\sin(5\theta) = 0 \quad \text{where} \quad \sin\theta \neq 0$$
The general solution to $\sin(5\theta) = 0$ is $5\theta = k\pi \implies \theta = \frac{k\pi}{5}$ for $k \in \mathbb{Z}$.
Since $x
eq 0$, $k$ cannot be a multiple of $5$.
This gives distinct values for $x = \sin\theta$ when $k = 1, 2, -1, -2$:
$$x = \sin\left(\frac{\pi}{5}\right), \quad x = \sin\left(\frac{2\pi}{5}\right), \quad x = \sin\left(-\frac{\pi}{5}\right) = -\sin\left(\frac{\pi}{5}\right), \quad x = \sin\left(-\frac{2\pi}{5}\right) = -\sin\left(\frac{2\pi}{5}\right)$$
So the exact roots of the quartic equation are:
$$x = \pm\sin\left(\frac{\pi}{5}\right) \quad \text{and} \quad x = \pm\sin\left(\frac{2\pi}{5}\right)$$

(a)
* **M1**: Uses de Moivre's theorem to write $\cos(5\theta) + i\sin(5\theta) = (\cos\theta + i\sin\theta)^5$ or equivalent.
* **M1**: Expands $(\cos\theta + i\sin\theta)^5$ using the binomial theorem.
* **A1**: Equates the imaginary parts to obtain $\sin(5\theta) = 5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta$.
* **M1**: Substitutes $\cos^2\theta = 1 - \sin^2\theta$ into their expression.
* **A1**: Fully simplifies to show the given identity with no errors.

(b)
* **M1**: Relates the equation $16x^4 - 20x^2 + 5 = 0$ to $16x^5 - 20x^3 + 5x = 0$ and sets $x = \sin\theta$ to get $\sin(5\theta) = 0$.
* **M1**: Solves $\sin(5\theta) = 0$ to find $\theta = \frac{k\pi}{5}$ and identifies that $k$ must not be a multiple of $5$.
* **A1**: Correctly identifies the four distinct roots as $x = \pm\sin\left(\frac{\pi}{5}\right)$ and $x = \pm\sin\left(\frac{2\pi}{5}\right)$.

(a) Let a point on the line $y = 3x + c$ be $(x, 3x+c)$.
Under the transformation $T$, this point maps to $(x', y')$ given by:
$$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} x \\ 3x+c \end{pmatrix}$$
$$x' = 2x + (3x + c) = 5x + c$$
$$y' = 3x + 4(3x + c) = 15x + 4c$$
For the line to be invariant, $(x', y')$ must lie on the line $y' = 3x' + c$.
Substitute $x'$ and $y'$ into this line equation:
$$15x + 4c = 3(5x + c) + c$$
$$15x + 4c = 15x + 3c + c = 15x + 4c$$
Since this equation holds identically for all $x$ and $c$, the line $y = 3x + c$ is indeed invariant under $T$ for any real constant $c$.

(b) Let the line be $y = mx + c$. A point $(x, mx+c)$ maps to $(x', y')$:
$$\begin{pmatrix} x' \\ y'
\end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} x \\ mx+c \end{pmatrix} = \begin{pmatrix} (2+m)x + c \\ (3+4m)x + 4c \end{pmatrix}$$
For this line to be invariant, we require $y' = mx' + c$ for all $x$:
$$(3+4m)x + 4c = m((2+m)x + c) + c$$
$$(3+4m)x + 4c = (2m+m^2)x + (m+1)c$$
Equating coefficients of $x$:
$$3+4m = 2m+m^2 \implies m^2 - 2m - 3 = 0 \implies (m-3)(m+1) = 0$$
So $m = 3$ or $m = -1$.
Equating the constant terms:
$$4c = (m+1)c$$
If $m = 3$:
$$4c = 4c \quad \text{which is true for all } c \in \mathbb{R}$$
This yields the family $y = 3x + c$ already shown in part (a).
If $m = -1$:
$$4c = 0 \implies c = 0$$
This yields the line $y = -x$.
Thus, the only other invariant line of the form $y = mx + c$ is $y = -x$.

(a)
* **M1**: Sets up the matrix multiplication $\begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} x \\ 3x+c \end{pmatrix}$.
* **A1**: Obtains correct expressions for $x'$ and $y'$: $x' = 5x + c$ and $y' = 15x + 4c$.
* **M1**: Substitutes these expressions into the line equation $y' = 3x' + c$.
* **A1**: Correctly verifies that $15x + 4c = 15x + 4c$ holds for all $x$, and states the conclusion.

(b)
* **M1**: Sets up the mapping of $(x, mx+c)$ to $(x', y')$ and substitutes into $y' = mx' + c$ to form an equation in $m$ and $c$.
* **M1**: Compares coefficients of $x$ to get $m^2 - 2m - 3 = 0$ and solves to find $m = 3$ and $m = -1$.
* **M1**: Compares constant terms to find $4c = (m+1)c$ and deduces that for $m = -1$, $c$ must be $0$.
* **A1**: Concludes that the only other invariant line of the form $y = mx+c$ is $y = -x$.

(a) First, find $\frac{dy}{dx}$:
$$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{1}{2}x^2 - \frac{1}{4}\ln x\right) = x - \frac{1}{4x}$$
Now, compute $1 + \left(\frac{dy}{dx}\right)^2$:
$$1 + \left(\frac{dy}{dx}\right)^2 = 1 + \left(x - \frac{1}{4x}\right)^2 = 1 + x^2 - 2(x)\left(\frac{1}{4x}\right) + \frac{1}{16x^2}$$
$$= 1 + x^2 - \frac{1}{2} + \frac{1}{16x^2} = x^2 + \frac{1}{2} + \frac{1}{16x^2}$$
Observe that this is a perfect square:
$$x^2 + \frac{1}{2} + \frac{1}{16x^2} = \left(x + \frac{1}{4x}\right)^2$$
Since $1 \le x \le e$, $x + \frac{1}{4x} > 0$, we have:
$$\sqrt{1 + \left(\frac{dy}{dx}\right)^2} = \sqrt{\left(x + \frac{1}{4x}\right)^2} = x + \frac{1}{4x} \quad \text{(as required)}$$

(b) The length of the curve $s$ is given by:
$$s = \int_{1}^{e} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx = \int_{1}^{e} \left(x + \frac{1}{4x}\right) dx$$
$$s = \left[ \frac{1}{2}x^2 + \frac{1}{4}\ln x \right]_{1}^{e}$$
$$s = \left(\frac{1}{2}e^2 + \frac{1}{4}\ln e\right) - \left(\frac{1}{2}(1)^2 + \frac{1}{4}\ln 1\right)$$
$$s = \left(\frac{1}{2}e^2 + \frac{1}{4}\right) - \left(\frac{1}{2}\right) = \frac{1}{2}e^2 - \frac{1}{4}$$

(c) The curved surface area $S$ generated by rotating about the $y$-axis is given by:
$$S = 2\pi \int_{1}^{e} x \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx$$
$$S = 2\pi \int_{1}^{e} x \left(x + \frac{1}{4x}\right) dx = 2\pi \int_{1}^{e} \left(x^2 + \frac{1}{4}\right) dx$$
$$S = 2\pi \left[ \frac{1}{3}x^3 + \frac{1}{4}x \right]_{1}^{e}$$
$$S = 2\pi \left[ \left(\frac{1}{3}e^3 + \frac{1}{4}e\right) - \left(\frac{1}{3} + \frac{1}{4}\right) \right]$$
$$S = 2\pi \left[ \frac{1}{3}e^3 + \frac{1}{4}e - \frac{7}{12} \right] = \pi \left( \frac{2}{3}e^3 + \frac{1}{2}e - \frac{7}{6} \right)$$

(a)
* **B1**: Differentiates correctly to find $\frac{dy}{dx} = x - \frac{1}{4x}$.
* **M1**: Expands $1 + \left(\frac{dy}{dx}\right)^2$ to obtain $x^2 + \frac{1}{2} + \frac{1}{16x^2}$.
* **A1**: Shows clearly that this simplifies to $\left(x + \frac{1}{4x}\right)^2$ and takes the square root to obtain the given result.

(b)
* **M1**: Uses the arc length formula with limits $1$ and $e$ and integrates $\left(x + \frac{1}{4x}\right)$ to obtain $\frac{1}{2}x^2 + \frac{1}{4}\ln x$.
* **A1**: Obtains the correct exact value $\frac{1}{2}e^2 - \frac{1}{4}$ or equivalent.

(c)
* **M1**: Recalls and applies the surface area formula for rotation about the $y$-axis: $S = 2\pi \int x \, ds$.
* **M1**: Integrates $x^2 + \frac{1}{4}$ to get $\frac{1}{3}x^3 + \frac{1}{4}x$ and substitutes limits.
* **A1**: Obtains the correct exact area $\pi \left( \frac{2}{3}e^3 + \frac{1}{2}e - \frac{7}{6} \right)$ or any exact equivalent.

(a) Let \( \frac{5}{(x^2+1)(x+2)} = \frac{Ax+B}{x^2+1} + \frac{C}{x+2} \). Multiplying both sides by the denominator gives \( 5 = (Ax+B)(x+2) + C(x^2+1) \). Substituting \( x = -2 \) gives \( 5 = C((-2)^2 + 1) \implies 5 = 5C \implies C = 1 \). Comparing coefficients of \( x^2 \) gives \( 0 = A + C \implies A = -1 \). Comparing the constant terms gives \( 5 = 2B + C \implies 5 = 2B + 1 \implies B = 2 \). Therefore, \( f(x) = \frac{2-x}{x^2+1} + \frac{1}{x+2} \). (b) We write the improper integral as \( \lim_{t \to \infty} \int_{0}^{t} \left( \frac{2}{x^2+1} - \frac{x}{x^2+1} + \frac{1}{x+2} \right) dx \). Integrating term by term gives \( \left[ 2 \arctan(x) - \frac{1}{2} \ln(x^2+1) + \ln(x+2) \right]_{0}^{t} \). Combining the logarithmic terms gives \( \left[ 2 \arctan(x) + \ln\left(\frac{x+2}{\sqrt{x^2+1}}\right) \right]_{0}^{t} \). Evaluating at the upper limit \( t \) and taking the limit: as \( t \to \infty \), \( 2 \arctan(t) \to 2\left(\frac{\pi}{2}\right) = \pi \), and \( \lim_{t \to \infty} \ln\left(\frac{t+2}{\sqrt{t^2+1}}\right) = \lim_{t \to \infty} \ln\left(\frac{1+2/t}{\sqrt{1+1/t^2}}\right) = \ln(1) = 0 \). Evaluating at the lower limit \( x = 0 \) gives \( 2\arctan(0) + \ln\left(\frac{2}{1}\right) = \ln(2) \). Thus, the exact value of the improper integral is \( \pi - \ln(2) \).

(a) M1: Sets up the correct form of partial fractions and attempts to find the constants. A1: Correctly finds at least two constants (e.g. \( C = 1 \) and \( A = -1 \)). A1: Obtains the fully correct partial fractions \( \frac{2-x}{x^2+1} + \frac{1}{x+2} \). (b) M1: Integrates to find terms of the form \( p \arctan(x) + q \ln(x^2+1) + r \ln(x+2) \). A1: Correctly integrates to get \( 2 \arctan(x) - \frac{1}{2} \ln(x^2+1) + \ln(x+2) \). M1: Combines the logarithmic terms into a single fraction to allow limit evaluation as \( t \to \infty \). A1: Clearly shows that \( \lim_{t \to \infty} 2\arctan(t) = \pi \) and \( \lim_{t \to \infty} \ln\left(\frac{t+2}{\sqrt{t^2+1}}\right) = 0 \). A1: Obtains the final exact answer of \( \pi - \ln(2) \).

(a) Let \( z = x + iy \). Substituting this into the locus equation: \( \left| (x-3) + i(y-4) \right| = 2 \left| x + i(y-1) \right| \). Squaring both sides: \( (x-3)^2 + (y-4)^2 = 4[x^2 + (y-1)^2] \). Expanding both sides: \( x^2 - 6x + 9 + y^2 - 8y + 16 = 4x^2 + 4(y^2 - 2y + 1) \implies x^2 - 6x + 25 + y^2 - 8y = 4x^2 + 4y^2 - 8y + 4 \). Rearranging and grouping terms: \( 3x^2 + 3y^2 + 6x - 21 = 0 \). Dividing through by 3 gives: \( x^2 + y^2 + 2x - 7 = 0 \). Completing the square for \( x \): \( (x+1)^2 - 1 + y^2 - 7 = 0 \implies (x+1)^2 + y^2 = 8 \). This is the equation of a circle with centre \( (-1, 0) \) and radius \( \sqrt{8} = 2\sqrt{2} \). (b) The distance from the origin to the centre of the circle is \( \sqrt{(-1)^2 + 0^2} = 1 \). Since the radius of the circle is \( 2\sqrt{2} \approx 2.83 \), which is greater than 1, the origin lies inside the circle. The maximum distance from the origin to any point \( z \) on the circle is given by the distance from the origin to the centre plus the radius of the circle. Thus, the maximum value of \( \left| z \right| \) is \( 1 + 2\sqrt{2} \).

(a) M1: Substitutes \( z = x + iy \) and squares both sides of the modulus equation. M1: Correctly expands both sides. A1: Simplifies to get a correct equation in \( x \) and \( y \) (e.g., \( 3x^2 + 3y^2 + 6x - 21 = 0 \) or equivalent). M1: Completes the square to get the standard form of a circle's equation. A1: Correctly identifies the centre as \( (-1, 0) \) (or \( -1 \)) and the radius as \( \sqrt{8} \) or \( 2\sqrt{2} \). (b) M1: Realises that the maximum distance is the sum of the distance from the origin to the centre and the radius. A1: Calculates the distance from the origin to the centre as 1. A1: Obtains the correct final exact answer of \( 1 + 2\sqrt{2} \).

(a) The coefficient matrix is \( \mathbf{M} = \begin{pmatrix} 1 & -1 & 2 \\ 3 & 1 & k \\ 1 & 3 & -k \end{pmatrix} \). The determinant of \( \mathbf{M} \) is \( \det(\mathbf{M}) = 1[1(-k) - 3(k)] - (-1)[3(-k) - 1(k)] + 2[3(3) - 1(1)] = -4k - 4k + 16 = 16 - 8k \). A unique solution exists if and only if \( \det(\mathbf{M}) \neq 0 \). Thus, \( 16 - 8k \neq 0 \implies k \neq 2 \). (b)(i) When \( k = 2 \), the equations are: (1) \( x - y + 2z = 4 \), (2) \( 3x + y + 2z = 2 \), (3) \( x + 3y - 2z = 1 \). Adding equations (1) and (3) yields: \( 2x + 2y = 5 \). Subtracting equation (1) from equation (2) yields: \( 2x + 2y = -2 \). Because the equations \( 2x + 2y = 5 \) and \( 2x + 2y = -2 \) are contradictory, there are no solutions. Thus, the system is inconsistent. (b)(ii) The normal vectors of the three planes are \( \mathbf{n}_1 = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} \), \( \mathbf{n}_2 = \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix} \), and \( \mathbf{n}_3 = \begin{pmatrix} 1 \\ 3 \\ -2 \end{pmatrix} \). Since no two normal vectors are parallel (none are scalar multiples of each other), no two planes are parallel. As the system is inconsistent and no two planes are parallel, the three planes must form a triangular prism.

(a) M1: Attempts to find the determinant of the 3x3 coefficient matrix. A1: Correctly calculates the determinant as \( 16 - 8k \) (or equivalent). A1: Explains that a unique solution exists when \( \det(\mathbf{M}) \neq 0 \), and correctly concludes \( k \neq 2 \). (b)(i) M1: Substitutes \( k = 2 \) and attempts to eliminate one of the variables from two different pairs of equations. A1: Successfully obtains two contradictory equations (such as \( 2x + 2y = 5 \) and \( 2x + 2y = -2 \)). A1: Concludes that because the equations are contradictory, there are no solutions, meaning the system is inconsistent. (b)(ii) M1: States or shows that no two planes are parallel by examining their normal vectors. A1: Correctly identifies that the three planes form a triangular prism.

*(a)*
To find the eigenvalues of \(\mathbf{A}\), we solve the characteristic equation \(\det(\mathbf{A} - \lambda \mathbf{I}) = 0\):
\[ \det\begin{pmatrix} 1-\lambda & 1 & 2 \\ 0 & 2-\lambda & 2 \\ -1 & 1 & 3-\lambda \end{pmatrix} = 0 \]
Expanding along the first column:
\[ (1-\lambda) \left[ (2-\lambda)(3-\lambda) - 2 \right] - 1 \left[ 2 - 2(2-\lambda) \right] = 0 \]
\[ (1-\lambda)(\lambda^2 - 5\lambda + 4) - (2\lambda - 2) = 0 \]
\[ (1-\lambda)(\lambda-1)(\lambda-4) + 2(1-\lambda) = 0 \]
Factoring out \((1-\lambda)\):
\[ (1-\lambda) \left[ (\lambda-1)(\lambda-4) + 2 \right] = 0 \]
\[ (1-\lambda)(\lambda^2 - 5\lambda + 6) = 0 \]
\[ (1-\lambda)(\lambda-2)(\lambda-3) = 0 \]
This shows that \(\lambda = 1\) is indeed an eigenvalue, and the other two eigenvalues are \(\lambda = 2\) and \(\lambda = 3\).

*(b)*
To find an eigenvector for \(\lambda = 3\), we solve \((\mathbf{A} - 3\mathbf{I})\mathbf{v} = \mathbf{0}\):
\[ \begin{pmatrix} -2 & 1 & 2 \\ 0 & -1 & 2 \\ -1 & 1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \]
From the third row:
\[ -x + y = 0 \implies x = y \]
From the second row:
\[ -y + 2z = 0 \implies y = 2z \]
Thus, \(x = 2z\) and \(y = 2z\). Setting \(z = 1\) gives an eigenvector:
\[ \mathbf{v} = \begin{pmatrix} 2 \\ 2 \\ 1 \end{pmatrix} \]
The magnitude of this vector is:
\[ |\mathbf{v}| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{9} = 3 \]
Therefore, the normalised eigenvector is:
\[ \pm \frac{1}{3} \begin{pmatrix} 2 \\ 2 \\ 1 \end{pmatrix} = \pm \begin{pmatrix} 2/3 \\ 2/3 \\ 1/3 \end{pmatrix} \]

*(a)*
- **M1**: Sets up characteristic equation \(\det(\mathbf{A} - \lambda\mathbf{I}) = 0\).
- **A1**: Correctly expands the determinant in terms of \(\lambda\).
- **A1**: Demonstrates that \(\lambda = 1\) is an eigenvalue (e.g. by factoring out \((1-\lambda)\) or showing that \(\det(\mathbf{A}-\mathbf{I})=0\)).
- **A1**: Finds other two eigenvalues \(\lambda = 2\) and \(\lambda = 3\).

*(b)*
- **M1**: Sets up equations for eigenvector corresponding to \(\lambda = 3\).
- **A1**: Solves the system of equations to find a non-zero eigenvector, e.g. \(\begin{pmatrix} 2 \\ 2 \\ 1 \end{pmatrix}\).
- **M1**: Calculates the magnitude of their eigenvector.
- **A1**: Correctly obtains the normalised eigenvector \(\pm \frac{1}{3} \begin{pmatrix} 2 \\ 2 \\ 1 \end{pmatrix}\).

*(a)*
Using the product rule to differentiate with respect to \(t\):
\[ \frac{dx}{dt} = e^t \cos t - e^t \sin t = e^t(\cos t - \sin t) \]
\[ \frac{dy}{dt} = e^t \sin t + e^t \cos t = e^t(\sin t + \cos t) \]
Now, squaring each term:
\[ \left(\frac{dx}{dt}\right)^2 = e^{2t}(\cos^2 t - 2\sin t \cos t + \sin^2 t) = e^{2t}(1 - 2\sin t \cos t) \]
\[ \left(\frac{dy}{dt}\right)^2 = e^{2t}(\sin^2 t + 2\sin t \cos t + \cos^2 t) = e^{2t}(1 + 2\sin t \cos t) \]
Adding these together yields:
\[ \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = e^{2t}(1 - 2\sin t \cos t + 1 + 2\sin t \cos t) = 2e^{2t} \]
which is the required expression.

*(b)*
The length of a parametric curve \(s\) is given by:
\[ s = \int_{0}^{\pi} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]
Substituting the result from part (a):
\[ s = \int_{0}^{\pi} \sqrt{2e^{2t}} \, dt = \int_{0}^{\pi} \sqrt{2} e^t \, dt \]
Evaluating the integral:
\[ s = \left[ \sqrt{2} e^t \right]_{0}^{\pi} = \sqrt{2}(e^{\pi} - e^0) = \sqrt{2}(e^{\pi} - 1) \]

*(a)*
- **M1**: Differentiates \(x\) and \(y\) with respect to \(t\) using the product rule.
- **A1**: Obtains correct expressions for both \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).
- **M1**: Squares both derivatives.
- **A1**: Correctly expands and simplifies using \(\sin^2 t + \cos^2 t = 1\) to show the given result.

*(b)*
- **M1**: Uses the arc length formula with limits \(0\) and \(\pi\).
- **A1**: Correctly simplifies the integrand to \(\sqrt{2} e^t\).
- **M1**: Integrates to get \(\sqrt{2} e^t\).
- **A1**: Obtains the correct exact length \(\sqrt{2}(e^{\pi} - 1)\).

*(a)*
Let \(z = \cos \theta + \mathrm{i} \sin \theta\).
Using De Moivre's theorem, we have:
\[ z - \frac{1}{z} = 2\mathrm{i}\sin \theta \quad \text{and} \quad z^n - \frac{1}{z^n} = 2\mathrm{i}\sin n\theta \]
Then:
\[ (2\mathrm{i}\sin \theta)^5 = \left(z - \frac{1}{z}\right)^5 \]
Using the binomial expansion:
\[ 32\mathrm{i}\sin^5 \theta = z^5 - 5z^3 + 10z - \frac{10}{z} + \frac{5}{z^3} - \frac{1}{z^5} \]
Grouping terms with reciprocal powers:
\[ 32\mathrm{i}\sin^5 \theta = \left(z^5 - \frac{1}{z^5}\right) - 5\left(z^3 - \frac{1}{z^3}\right) + 10\left(z - \frac{1}{z}\right) \]
Substituting the sine identities:
\[ 32\mathrm{i}\sin^5 \theta = 2\mathrm{i}\sin 5\theta - 5(2\mathrm{i}\sin 3\theta) + 10(2\mathrm{i}\sin \theta) \]
Dividing both sides by \(32\mathrm{i}\):
\[ \sin^5 \theta = \frac{1}{16}\sin 5\theta - \frac{5}{16}\sin 3\theta + \frac{5}{8}\sin \theta \]
So \(a = \frac{1}{16}\), \(b = -\frac{5}{16}\), \(c = \frac{5}{8}\).

*(b)*
We integrate the expression term-by-term:
\[ \int_0^{\pi/3} \sin^5 \theta \, d\theta = \int_0^{\pi/3} \left( \frac{1}{16}\sin 5\theta - \frac{5}{16}\sin 3\theta + \frac{5}{8}\sin \theta \right) \, d\theta \]
\[ = \left[ -\frac{1}{80}\cos 5\theta + \frac{5}{48}\cos 3\theta - \frac{5}{8}\cos \theta \right]_0^{\pi/3} \]
Evaluating at the upper limit \(\theta = \frac{\pi}{3}\):
\[ -\frac{1}{80}\cos\left(\frac{5\pi}{3}\right) + \frac{5}{48}\cos\pi - \frac{5}{8}\cos\left(\frac{\pi}{3}\right) = -\frac{1}{80}\left(\frac{1}{2}\right) + \frac{5}{48}(-1) - \frac{5}{8}\left(\frac{1}{2}\right) \]
\[ = -\frac{1}{160} - \frac{5}{48} - \frac{5}{16} = -\frac{3}{480} - \frac{50}{480} - \frac{150}{480} = -\frac{203}{480} \]
Evaluating at the lower limit \(\theta = 0\):
\[ -\frac{1}{80}\cos 0 + \frac{5}{48}\cos 0 - \frac{5}{8}\cos 0 = -\frac{1}{80} + \frac{5}{48} - \frac{5}{8} \]
\[ = -\frac{6}{480} + \frac{50}{480} - \frac{300}{480} = -\frac{256}{480} \]
Subtracting the lower limit value from the upper limit value:
\[ -\frac{203}{480} - \left(-\frac{256}{480}\right) = \frac{53}{480} \]

*(a)*
- **M1**: Uses \(2\mathrm{i}\sin\theta = z - z^{-1}\) or equivalent exponential form.
- **M1**: Applies binomial expansion to \((z - z^{-1})^5\).
- **A1**: Correctly groups powers to obtain terms in \(z^n - z^{-n}\).
- **M1**: Substitutes back to sine functions using \(z^n - z^{-n} = 2\mathrm{i}\sin(n\theta)\).
- **A1**: Obtains correct coefficients \(a = \frac{1}{16}\), \(b = -\frac{5}{16}\), and \(c = \frac{5}{8}\).

*(b)*
- **M1**: Integrates the trigonometric terms correctly (allow minor coefficient or sign errors).
- **A1**: Obtains the correct integrated function: \(-\frac{1}{80}\cos 5\theta + \frac{5}{48}\cos 3\theta - \frac{5}{8}\cos \theta\).
- **A1**: Evaluates at both limits correctly and subtracts to obtain \(\frac{53}{480}\).

By letting \(z = x + iy\), the equation \(|z - 3| = 2|z - 3i|\) becomes \(|x - 3 + iy| = 2|x + i(y - 3)|\). Squaring both sides yields \((x-3)^2 + y^2 = 4(x^2 + (y-3)^2)\). Expanding this gives \(x^2 - 6x + 9 + y^2 = 4x^2 + 4y^2 - 24y + 36\). Rearranging leads to \(3x^2 + 3y^2 + 6x - 24y + 27 = 0\). Dividing by 3, we get \(x^2 + y^2 + 2x - 8y + 9 = 0\). Completing the square results in \((x+1)^2 - 1 + (y-4)^2 - 16 + 9 = 0\), which simplifies to \((x+1)^2 + (y-4)^2 = 8\). This represents a circle with centre \((-1, 4)\) and radius \(\sqrt{8} = 2\sqrt{2}\).

B1: Correctly identifies the circle with centre \((-1, 4)\) and radius \(2\sqrt{2}\).

For a matrix to be singular, its determinant must be zero. The determinant of \(A\) is given by \(\det(A) = k(-k - 4) - 2(3k - 2) + 1(6 + 1) = -k^2 - 10k + 11\). Setting the determinant to zero gives \(k^2 + 10k - 11 = 0\), which factorises to \((k+11)(k-1) = 0\). Therefore, the matrix is singular when \(k = 1\) or \(k = -11\).

B1: Correctly identifies the values of \(k\) as \(1\) and \(-11\).

Using the identity \(\sinh^2 x = \frac{\cosh 2x - 1}{2}\), the integral becomes \(\int_{0}^{\ln 2} \frac{\cosh 2x - 1}{2} \, \mathrm{d}x = \left[ \frac{\sinh 2x}{4} - \frac{x}{2} \right]_{0}^{\ln 2}\). Since \(\sinh(2\ln 2) = \sinh(\ln 4) = \frac{e^{\ln 4} - e^{-\ln 4}}{2} = \frac{4 - 0.25}{2} = \frac{15}{8}\), evaluating at the upper limit yields \(\frac{15}{32} - \frac{\ln 2}{2}\). Evaluating at the lower limit of 0 yields 0. Thus, the exact value of the integral is \(\frac{15 - 16\ln 2}{32}\).

B1: Correctly evaluates the integral to obtain \(\frac{15 - 16\ln 2}{32}\).

To find the arc length \(s\), we use the formula:

\[s = \int_{x_1}^{x_2} \sqrt{1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2} \mathrm{d}x\]

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

\[\frac{\mathrm{d}y}{\mathrm{d}x} = 4 \cdot \frac{1}{4}\sinh\left(\frac{x}{4}\right) = \sinh\left(\frac{x}{4}\right)\]

Next, we simplify the integrand using the hyperbolic identity \(1 + \sinh^2(\theta) = \cosh^2(\theta)\):

\[\sqrt{1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2} = \sqrt{1 + \sinh^2\left(\frac{x}{4}\right)} = \sqrt{\cosh^2\left(\frac{x}{4}\right)} = \cosh\left(\frac{x}{4}\right)\]

Now, we set up and evaluate the definite integral:

\[s = \int_{0}^{4\ln(3)} \cosh\left(\frac{x}{4}\right) \mathrm{d}x = \left[ 4\sinh\left(\frac{x}{4}\right) \right]_{0}^{4\ln(3)}\]

Evaluating at the limits:

\[s = 4\sinh\left(\frac{4\ln(3)}{4}\right) - 4\sinh(0) = 4\sinh(\ln(3)) - 0\]

Using the exponential definition of \(\sinh(\theta)\):

\[\sinh(\ln(3)) = \frac{e^{\ln(3)} - e^{-\ln(3)}}{2} = \frac{3 - \frac{1}{3}}{2} = \frac{\frac{8}{3}}{2} = \frac{4}{3}\]

Therefore, the arc length is:

\[s = 4 \times \frac{4}{3} = \frac{16}{3}\]

M1: Identifies the correct arc length formula, differentiates \(y\) correctly to get \(\sinh(x/4)\), and uses the identity \(1 + \sinh^2(\theta) = \cosh^2(\theta)\) to simplify the integrand.
A1: Integrates correctly and substitutes limits using the definition of \(\sinh\) to obtain the exact value of \(\frac{16}{3}\).

To find the arc length \(s\), we use the formula:

\[s = \int_{x_1}^{x_2} \sqrt{1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2} \mathrm{d}x\]

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

\[\frac{\mathrm{d}y}{\mathrm{d}x} = 4 \cdot \frac{1}{4}\sinh\left(\frac{x}{4}\right) = \sinh\left(\frac{x}{4}\right)\]

Next, we simplify the integrand using the hyperbolic identity \(1 + \sinh^2(\theta) = \cosh^2(\theta)\):

\[\sqrt{1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2} = \sqrt{1 + \sinh^2\left(\frac{x}{4}\right)} = \sqrt{\cosh^2\left(\frac{x}{4}\right)} = \cosh\left(\frac{x}{4}\right)\]

Now, we set up and evaluate the definite integral:

\[s = \int_{0}^{4\ln(3)} \cosh\left(\frac{x}{4}\right) \mathrm{d}x = \left[ 4\sinh\left(\frac{x}{4}\right) \right]_{0}^{4\ln(3)}\]

Evaluating at the limits:

\[s = 4\sinh\left(\frac{4\ln(3)}{4}\right) - 4\sinh(0) = 4\sinh(\ln(3)) - 0\]

Using the exponential definition of \(\sinh(\theta)\):

\[\sinh(\ln(3)) = \frac{e^{\ln(3)} - e^{-\ln(3)}}{2} = \frac{3 - \frac{1}{3}}{2} = \frac{\frac{8}{3}}{2} = \frac{4}{3}\]

Therefore, the arc length is:

\[s = 4 \times \frac{4}{3} = \frac{16}{3}\]

M1: Identifies the correct arc length formula, differentiates \(y\) correctly to get \(\sinh(x/4)\), and uses the identity \(1 + \sinh^2(\theta) = \cosh^2(\theta)\) to simplify the integrand.
A1: Integrates correctly and substitutes limits using the definition of \(\sinh\) to obtain the exact value of \(\frac{16}{3}\).

(a) Let \(w = z^2\). The equation becomes \(w^2 + 8w + 64 = 0\).
Using the quadratic formula:
\(w = \frac{-8 \pm \sqrt{64 - 256}}{2} = -4 \pm 4\sqrt{3}i\).

Expressing \(w\) in polar form:
\(|w| = \sqrt{(-4)^2 + (4\sqrt{3})^2} = \sqrt{16 + 48} = 8\).
\(\arg(w) = \pi - \arctan(\sqrt{3}) = \frac{2\pi}{3}\) (for \(w = -4 + 4\sqrt{3}i\)) and \(-\frac{2\pi}{3}\) (for \(w = -4 - 4\sqrt{3}i\)).

Thus, \(z^2 = 8 e^{i \frac{2\pi}{3}}\) or \(z^2 = 8 e^{-i \frac{2\pi}{3}}\).

Taking the square root:
For \(z^2 = 8 e^{i \frac{2\pi}{3}}\), the roots are \(z = \pm \sqrt{8} e^{i \frac{\pi}{3}} = 2\sqrt{2} e^{i \frac{\pi}{3}}\) and \(2\sqrt{2} e^{-i \frac{2\pi}{3}}\).
For \(z^2 = 8 e^{-i \frac{2\pi}{3}}\), the roots are \(z = \pm \sqrt{8} e^{-i \frac{\pi}{3}} = 2\sqrt{2} e^{-i \frac{\pi}{3}}\) and \(2\sqrt{2} e^{i \frac{2\pi}{3}}\).

So the four roots are:
\(z_1 = 2\sqrt{2} e^{i \frac{\pi}{3}}\), \(z_2 = 2\sqrt{2} e^{i \frac{2\pi}{3}}\), \(z_3 = 2\sqrt{2} e^{-i \frac{\pi}{3}}\), and \(z_4 = 2\sqrt{2} e^{-i \frac{2\pi}{3}}\).

(b) Converting the roots to Cartesian form:
\(z_1 = 2\sqrt{2} (\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}) = \sqrt{2} + i\sqrt{6}\)
\(z_2 = 2\sqrt{2} (\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}) = -\sqrt{2} + i\sqrt{6}\)
\(z_3 = -\sqrt{2} - i\sqrt{6}\)
\(z_4 = \sqrt{2} - i\sqrt{6}\)

The vertices on the Argand diagram are at the Cartesian coordinates \((\sqrt{2}, \sqrt{6})\), \((-\sqrt{2}, \sqrt{6})\), \((-\sqrt{2}, -\sqrt{6})\), and \((\sqrt{2}, -\sqrt{6})\).
These coordinates have horizontal sides parallel to the real axis (at \(y = \pm\sqrt{6}\)) and vertical sides parallel to the imaginary axis (at \(x = \pm\sqrt{2}\)), confirming they form a rectangle.

The width of the rectangle is \(\sqrt{2} - (-\sqrt{2}) = 2\sqrt{2}\).
The height of the rectangle is \(\sqrt{6} - (-\sqrt{6}) = 2\sqrt{6}\).

Area = \(2\sqrt{2} \times 2\sqrt{6} = 4\sqrt{12} = 8\sqrt{3}\).

**Part (a)**
* **M1**: Substitutes \(w = z^2\) and correctly solves the quadratic equation to get \(w = -4 \pm 4\sqrt{3}i\).
* **A1**: Finds correct modulus \(8\) and arguments \(\pm \frac{2\pi}{3}\).
* **M1**: Uses de Moivre's theorem or square roots of a complex number to find the roots of \(z^2 = w\).
* **A1**: Correctly identifies the four roots as \(2\sqrt{2} e^{i \frac{\pi}{3}}\), \(2\sqrt{2} e^{i \frac{2\pi}{3}}\), \(2\sqrt{2} e^{-i \frac{\pi}{3}}\), and \(2\sqrt{2} e^{-i \frac{2\pi}{3}}\).

**Part (b)**
* **M1**: Converts the four roots to Cartesian coordinates.
* **A1**: Correctly obtains the coordinates as \((\pm\sqrt{2}, \pm\sqrt{6})\).
* **M1**: Recognises the shape as a rectangle and calculates its dimensions as \(2\sqrt{2}\) and \(2\sqrt{6}\).
* **A1**: Obtains the correct exact area of \(8\sqrt{3}\).

(a) The characteristic equation is defined by \(\det(\mathbf{M} - \lambda \mathbf{I}) = 0\).
\(\det \begin{pmatrix} 2 - \lambda & k \\ 3 & -2 - \lambda \end{pmatrix} = 0\)
\((2 - \lambda)(-2 - \lambda) - 3k = 0\)
\(\lambda^2 - 4 - 3k = 0\)

Since \(\lambda = 4\) is an eigenvalue, we substitute \(\lambda = 4\) into the characteristic equation:
\(4^2 - 4 - 3k = 0\)
\(16 - 4 - 3k = 0\)
\(12 - 3k = 0 \implies k = 4\).

(b) Substituting \(k = 4\) back into the characteristic equation:
\(\lambda^2 - 16 = 0\)
\(\lambda^2 = 16 \implies \lambda = \pm 4\).

Therefore, the other eigenvalue of \(\mathbf{M}\) is \(\lambda = -4\).

(c) To find an eigenvector for \(\lambda = 4\):
\(\begin{pmatrix} 2 & 4 \\ 3 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = 4 \begin{pmatrix} x \\ y \end{pmatrix}\)
From the first row: \(2x + 4y = 4x \implies 2x = 4y \implies x = 2y\).
Choosing \(y = 1\), an eigenvector is \(\begin{pmatrix} 2 \\ 1 \end{pmatrix}\).

To find an eigenvector for \(\lambda = -4\):
\(\begin{pmatrix} 2 & 4 \\ 3 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = -4 \begin{pmatrix} x \\ y \end{pmatrix}\)
From the first row: \(2x + 4y = -4x \implies 6x = -4y \implies 3x = -2y\).
Choosing \(y = 3\), an eigenvector is \(\begin{pmatrix} -2 \\ 3 \end{pmatrix}\).

**Part (a)**
* **M1**: Sets up the determinant equation \(\det(\mathbf{M} - 4\mathbf{I}) = 0\) or equivalents.
* **A1**: Shows clearly that \(k = 4\).

**Part (b)**
* **M1**: Sets up characteristic equation with \(k = 4\).
* **A1**: Obtains the second eigenvalue \(\lambda = -4\).

**Part (c)**
* **M1**: Formulates the matrix equation \(\mathbf{M}\mathbf{v} = 4\mathbf{v}\) or equivalent.
* **A1**: Obtains a correct eigenvector for \(\lambda = 4\), e.g., \(\begin{pmatrix} 2 \\ 1 \end{pmatrix}\) or any scalar multiple.
* **M1**: Formulates the matrix equation \(\mathbf{M}\mathbf{v} = -4\mathbf{v}\) or equivalent.
* **A1**: Obtains a correct eigenvector for \(\lambda = -4\), e.g., \(\begin{pmatrix} -2 \\ 3 \end{pmatrix}\) or any scalar multiple.

We can evaluate this integral using hyperbolic substitution.
Let \(x = 3 \sinh u\).
Then \(dx = 3 \cosh u \, du\).

Now find the limits of integration:
When \(x = 0\), \(3 \sinh u = 0 \implies u = 0\).
When \(x = 3\), \(3 \sinh u = 3 \implies \sinh u = 1 \implies u = \operatorname{arsinh}(1) = \ln(1 + \sqrt{2})\).

Substitute these into the integral:
\(\int_{0}^{\ln(1+\sqrt{2})} \frac{9\sinh^2 u}{\sqrt{9 + 9\sinh^2 u}} \cdot 3\cosh u \, du\)

Since \(\sqrt{9 + 9\sinh^2 u} = \sqrt{9(1 + \sinh^2 u)} = 3\cosh u\), the integral simplifies to:
\(\int_{0}^{\ln(1+\sqrt{2})} 9\sinh^2 u \, du\)

Using the hyperbolic identity \ \sinh^2 u = \frac{\cosh(2u) - 1}{2}\):
\(\int_{0}^{\ln(1+\sqrt{2})} \frac{9}{2}(\cosh(2u) - 1) \, du = \left[ \frac{9}{4}\sinh(2u) - \frac{9}{2}u \right]_{0}^{\ln(1+\sqrt{2})}\)

Using the identity \(\sinh(2u) = 2\sinh u\cosh u\):
When \(u = \ln(1+\sqrt{2})\), \\sinh u = 1\) and \(\cosh u = \sqrt{1+\sinh^2 u} = \sqrt{2}\).
Thus, \(\sinh(2u) = 2(1)(\sqrt{2}) = 2\sqrt{2}\).

Substituting the limits:
At upper limit: \(\frac{9}{4}(2\sqrt{2}) - \frac{9}{2}\ln(1+\sqrt{2}) = \frac{9\sqrt{2}}{2} - \frac{9}{2}\ln(1+\sqrt{2})\).
At lower limit: \(0\).

This gives:
\(\int_{0}^{3} \frac{x^2}{\sqrt{9 + x^2}} \, dx = \frac{9}{2}\sqrt{2} - \frac{9}{2}\ln(1+\sqrt{2})\).

Hence, \(a = \frac{9}{2}\) and \(b = -\frac{9}{2}\).

* **M1**: Uses substitution \(x = 3 \sinh u\) (or equivalent trigonometric substitution) and correctly calculates \(dx\).
* **M1**: Changes the limits of integration to \(u = 0\) and \(u = \ln(1+\sqrt{2})\).
* **A1**: Simplifies the integrand correctly to \(9\sinh^2 u\).
* **M1**: Employs the identity \(\sinh^2 u = \frac{\cosh(2u)-1}{2}\) to integrate.
* **A1**: Obtains the correct integrated function \(\frac{9}{4}\sinh(2u) - \frac{9}{2}u\).
* **M1**: Uses \(\sinh(2u) = 2\sinh u\cosh u\) to evaluate \(\sinh(2u)\) at the upper limit.
* **A1**: Obtains \(a = \frac{9}{2}\).
* **A1**: Obtains \(b = -\frac{9}{2}\).

**(a)**
We are given the transformation:
\(w = \frac{z - 2\mathrm{i}}{z + 4}

We express \)z\) in terms of \(w\):
\(w(z + 4) = z - 2\mathrm{i}\)
\(wz + 4w = z - 2\mathrm{i}\)
\(z(w - 1) = -4w - 2\mathrm{i}\)
\(z = \frac{-4w - 2\mathrm{i}}{w - 1}\)

We are given the equation of the locus of \(z\):
\(|z + 2| = 2\)

Substituting \(z\) into this equation:

\(\left| \frac{-4w - 2\mathrm{i}}{w - 1} + 2 \right| = 2\)

Simplify the expression inside the modulus:

\(\frac{-4w - 2\mathrm{i} + 2(w - 1)}{w - 1} = \frac{-2w - 2 - 2\mathrm{i}}{w - 1}\)

So,

\(\left| \frac{-2(w + 1 + \mathrm{i})}{w - 1} \right| = 2\)

\(\frac{2|w + 1 + \mathrm{i}|}{|w - 1|} = 2\)

Dividing both sides by 2:

\(|w + 1 + \mathrm{i}| = |w - 1|\)

This equation is of the form \(|w - w_1| = |w - w_2|\), which represents the perpendicular bisector of the line segment joining \(w_1 = -1 - \mathrm{i}\) and \(w_2 = 1\). Therefore, the image is indeed a straight line.

**(b)**
Let \(w = u + \mathrm{i}v\). Substituting this into \(|w + 1 + \mathrm{i}| = |w - 1|\):

\(|(u + 1) + \mathrm{i}(v + 1)| = |(u - 1) + \mathrm{i}v|\)

Squaring both sides:

\((u + 1)^2 + (v + 1)^2 = (u - 1)^2 + v^2\)

Expand the terms:

\(u^2 + 2u + 1 + v^2 + 2v + 1 = u^2 - 2u + 1 + v^2\)

Subtract \(u^2 + v^2\) from both sides:

\(2u + 2v + 2 = -2u + 1\)

Rearranging to the form \(au + bv + c = 0\):

\(4u + 2v + 1 = 0\)

**(a)**
* **M1**: Attempts to rearrange the transformation equation to make \(z\) the subject.
* **A1**: Correctly obtains \(z = \frac{-4w - 2\mathrm{i}}{w - 1}\) (or an equivalent form).
* **M1**: Substitutes their expression for \(z\) into \(|z + 2| = 2\).
* **A1**: Simplifies the numerator correctly to obtain \(-2w - 2 - 2\mathrm{i}\) or equivalent.
* **A1**: Reaches the form \(|w + 1 + \mathrm{i}| = |w - 1|\) and states that this represents a straight line (perpendicular bisector).

**(b)**
* **M1**: Substitutes \(w = u + \mathrm{i}v\) into their locus equation from part (a).
* **A1**: Correctly expands the squared terms, e.g., \((u + 1)^2 + (v + 1)^2 = (u - 1)^2 + v^2\).
* **A1**: Correctly simplifies to obtain the final equation \(4u + 2v + 1 = 0\).

**(a)**
An invariant point \(\begin{pmatrix} x \\ y \end{pmatrix}\) satisfies the equation:

\(\mathbf{M} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}

\)\begin{pmatrix} 2 & 3 \\ 3 & -6 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}\)

This gives the simultaneous equations:
1) \(2x + 3y = x \implies x + 3y = 0\)
2) \(3x - 6y = y \implies 3x - 7y = 0\)

From (1), \(x = -3y\). Substituting into (2):
\(3(-3y) - 7y = 0 \implies -16y = 0 \implies y = 0\)

Since \(y = 0\), then \(x = 0\).
Thus, the only invariant point is the origin \((0,0)\).

**(b)**
An invariant line through the origin has the equation \(y = mx\).
Under the transformation \(T\), any point \((x, mx)\) on this line is mapped to a point \((X, Y)\) which must also lie on the line, so \(Y = mX\).

\(\begin{pmatrix} X \\ Y \end{pmatrix} = \begin{pmatrix} 2 & 3 \\ 3 & -6 \end{pmatrix} \begin{pmatrix} x \\ mx \end{pmatrix} = \begin{pmatrix} 2x + 3mx \\ 3x - 6mx \end{pmatrix}\)

Thus:
\(X = (2 + 3m)x\)
\(Y = (3 - 6m)x\)

Since \(Y = mX\), we have:
\((3 - 6m)x = m(2 + 3m)x\)

Since this must hold for any point on the line, we can assume \(x \neq 0\):
\(3 - 6m = m(2 + 3m)\)
\(3 - 6m = 2m + 3m^2\)
\(3m^2 + 8m - 3 = 0\)

Factorising the quadratic:
\((3m - 1)(m + 3) = 0\)

This gives:
\(m = \frac{1}{3}\) or \(m = -3\)

Therefore, the equations of the invariant lines are:
\(y = \frac{1}{3}x\) and \(y = -3x\)

**(a)**
* **M1**: Sets up the system of equations \(\mathbf{M}\mathbf{r} = \mathbf{r}\) to find invariant points.
* **A1**: Solves the system correctly to show that \(x = 0\) and \(y = 0\) is the unique solution.

**(b)**
* **M1**: Uses the invariant line definition to write \(\mathbf{M}\begin{pmatrix} x \\ mx \end{pmatrix} = \begin{pmatrix} X \\ mX \end{pmatrix}\).
* **A1**: Expresses \(X\) and \(Y\) correctly in terms of \(x\) and \(m\): \(X = (2+3m)x\) and \(Y = (3-6m)x\).
* **M1**: Sets up the equation relating \(Y\) and \(X\): \(3-6m = m(2+3m)\).
* **A1**: Obtains the correct quadratic equation \(3m^2 + 8m - 3 = 0\).
* **M1**: Solves the quadratic equation to find values of \(m\).
* **A1**: States the correct equations of the invariant lines: \(y = \frac{1}{3}x\) and \(y = -3x\).

**(a)**
First, find the derivative \(\frac{\mathrm{d}y}{\mathrm{d}x}\):

\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{2}x^2 - \frac{1}{4}\ln x\right) = x - \frac{1}{4x}\)

Now, substitute this into the expression \(1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2\):

\(1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = 1 + \left(x - \frac{1}{4x}\right)^2\)

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

\(= 1 + x^2 - \frac{1}{2} + \frac{1}{16x^2}\)

\(= x^2 + \frac{1}{2} + \frac{1}{16x^2}\)

We recognize that this is a perfect square:

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

Hence, \(1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = \left(x + \frac{1}{4x}\right)^2\).

**(b)**
The formula for the arc length \(s\) of a curve is:

\(s = \int_{x_1}^{x_2} \sqrt{1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2} \,\mathrm{d}x\)

Substituting the limits and the expression from part (a):

\(s = \int_{1}^{4} \sqrt{\left(x + \frac{1}{4x}\right)^2} \,\mathrm{d}x\)

Since \(x > 0\) on the interval \([1, 4]\), we have \(x + \frac{1}{4x} > 0\). Thus:

\(s = \int_{1}^{4} \left(x + \frac{1}{4x}\right) \,\mathrm{d}x\)

Now perform the integration:

\(s = \left[ \frac{1}{2}x^2 + \frac{1}{4}\ln x \right]_{1}^{4}\)

Substitute the upper limit \(x = 4\):

\(\left( \frac{1}{2}(4)^2 + \frac{1}{4}\ln 4 \right) = 8 + \frac{1}{4}\ln(2^2) = 8 + \frac{2}{4}\ln 2 = 8 + \frac{1}{2}\ln 2\)

Substitute the lower limit \(x = 1\):

\(\left( \frac{1}{2}(1)^2 + \frac{1}{4}\ln 1 \right) = \frac{1}{2} + 0 = \frac{1}{2}\)

Subtract the lower limit value from the upper limit value:

\(s = \left(8 + \frac{1}{2}\ln 2\right) - \frac{1}{2} = \frac{15}{2} + \frac{1}{2}\ln 2\)

This is in the form \(a + b\ln 2\) with \(a = \frac{15}{2}\) and \(b = \frac{1}{2}\).

**(a)**
* **M1**: Correctly differentiates the given function to get \(\frac{\mathrm{d}y}{\mathrm{d}x} = x - \frac{1}{4x}\).
* **M1**: Attempts to expand \(1 + \left(x - \frac{1}{4x}\right)^2\).
* **A1**: Correctly simplifies the algebraic expression and shows it factorises to \(\left(x + \frac{1}{4x}\right)^2\).

**(b)**
* **M1**: Quotes the correct formula for arc length and substitutes their expression from part (a).
* **A1**: Reaches the simplified integral \(\int_{1}^{4} \left(x + \frac{1}{4x}\right) \,\mathrm{d}x\).
* **M1**: Integrates correctly to get \(\left[ \frac{1}{2}x^2 + \frac{1}{4}\ln x \right]\) (ignore limits for this mark).
* **M1**: Substitutes limits 4 and 1 into their integrated expression.
* **A1**: Obtains the correct final exact length \(\frac{15}{2} + \frac{1}{2}\ln 2\) (or equivalent exact form, but must be simplified to \(a + b\ln 2\)).

(a) To find the intersections, we set \(2(1 + \cos\theta) = 3\). This gives \(1 + \cos\theta = 1.5\), so \(\cos\theta = 0.5\). In the range \(-\pi \le \theta \le \pi\), this has solutions \(\theta = \frac{\pi}{3}\) and \(\theta = -\frac{\pi}{3}\). Substituting these back gives \(r = 3\). Hence the polar coordinates of the points of intersection are \((3, \frac{\pi}{3})\) and \((3, -\frac{\pi}{3})\). (b) The region inside \(C_1\) but outside \(C_2\) lies between \(\theta = -\frac{\pi}{3}\) and \(\theta = \frac{\pi}{3}\). The area \(A\) is given by: \(A = \frac{1}{2} \int_{-\pi/3}^{\pi/3} \left( r_1^2 - r_2^2 \right) d\theta\). By symmetry: \(A = \int_{0}^{\pi/3} \left[ 4(1+\cos\theta)^2 - 9 \right] d\theta = \int_{0}^{\pi/3} \left[ 4(1 + 2\cos\theta + \cos^2\theta) - 9 \right] d\theta = \int_{0}^{\pi/3} \left[ 4 + 8\cos\theta + 4\left(\frac{1+\cos 2\theta}{2}\right) - 9 \right] d\theta = \int_{0}^{\pi/3} \left[ 8\cos\theta + 2\cos 2\theta - 3 \right] d\theta\). Integrating term by term: \(A = \left[ 8\sin\theta + \sin 2\theta - 3\theta \right]_0^{\pi/3} = \left( 8\sin\frac{\pi}{3} + \sin\frac{2\pi}{3} - 3\left(\frac{\pi}{3}\right) \right) - (0) = 8\left(\frac{\sqrt{3}}{2}\right) + \frac{\sqrt{3}}{2} - \pi = \frac{9\sqrt{3}}{2} - \pi\).

M1: Sets the two equations equal to find \(\cos\theta = 0.5\). A1: Correctly identifies both values of \(\theta = \pm\frac{\pi}{3}\). A1: Writes both polar coordinates correctly. M1: Sets up the area integral with correct limits and integrand form \(\frac{1}{2} \int (r_1^2 - r_2^2) d\theta\). M1: Uses the double angle identity \(\cos^2\theta = \frac{1+\cos 2\theta}{2}\) to express the integrand in an integrable form. A1: Obtains the correct integrated expression \(8\sin\theta + \sin 2\theta - 3\theta\) (or equivalent). M1: Substitutes the limits \(0\) and \(\frac{\pi}{3}\) (or equivalents if using \(-\frac{\pi}{3}\) to \(\frac{\pi}{3}\)) correctly. A1: Obtains the exact area \(\frac{9\sqrt{3}}{2} - \pi\).

(a) The roots of \(z^5 = 1\) are of the form \(e^{2k\pi i/5}\) for integers \(k\). For \(-\pi < \theta \le \pi\), we choose \(k = 0, \pm 1, \pm 2\). This gives the roots: \(1, e^{\frac{2\pi i}{5}}, e^{-\frac{2\pi i}{5}}, e^{\frac{4\pi i}{5}}, e^{-\frac{4\pi i}{5}}\). (b) Since \(w\) is a root of \(z^5 - 1 = 0\), we have \(w^5 - 1 = 0\). Factoring gives \((w-1)(1+w+w^2+w^3+w^4) = 0\). Since \(w = e^{2\pi i/5} \neq 1\), we must have \(1+w+w^2+w^3+w^4 = 0\). (c) Notice that \(w^4 = e^{8\pi i/5} = e^{-2\pi i/5} = w^{-1}\), and \(w^3 = e^{6\pi i/5} = e^{-4\pi i/5} = w^{-2}\). Thus, the equation \(1+w+w^2+w^3+w^4=0\) can be written as \(1 + (w + w^{-1}) + (w^2 + w^{-2}) = 0\). Using Euler's formula, \(w^k + w^{-k} = e^{2k\pi i/5} + e^{-2k\pi i/5} = 2\cos\left(\frac{2k\pi}{5}\right)\). Substituting this in: \(1 + 2\cos\left(\frac{2\pi}{5}\right) + 2\cos\left(\frac{4\pi}{5}\right) = 0\), which simplifies to \(\cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) = -\frac{1}{2}\). (d) Let \(x = \cos\left(\frac{2\pi}{5}\right)\). Then \(\cos\left(\frac{4\pi}{5}\right) = 2\cos^2\left(\frac{2\pi}{5}\right) - 1 = 2x^2 - 1\). Substituting this into our equation: \(x + 2x^2 - 1 = -\frac{1}{2} \implies 2x^2 + x - \frac{1}{2} = 0 \implies 4x^2 + 2x - 1 = 0\). Solving this quadratic: \(x = \frac{-2 \pm \sqrt{4 - 4(4)(-1)}}{8} = \frac{-2 \pm \sqrt{20}}{8} = \frac{-1 \pm \sqrt{5}}{4}\). Since \(\frac{2\pi}{5}\) is an acute angle, \(\cos\left(\frac{2\pi}{5}\right) > 0\). Thus we choose the positive root: \(\cos\left(\frac{2\pi}{5}\right) = \frac{-1 + \sqrt{5}}{4}\).

M1: Identifies the general form of the roots as \(e^{2k\pi i/5}\). A1: Correctly lists all five roots within the required interval. B1: States that \(w^5 - 1 = 0\) and factors to show \(1+w+w^2+w^3+w^4=0\) since \(w \neq 1\). M1: Pairs the terms as \(w + w^4\) and \(w^2 + w^3\) or equivalent. M1: Expresses these pairs as \(2\cos\left(\frac{2\pi}{5}\right)\) and \(2\cos\left(\frac{4\pi}{5}\right)\). A1: Fully completes the proof to show the given identity. M1: Uses the double angle formula to rewrite \(\cos\left(\frac{4\pi}{5}\right)\) as \(2\cos^2\left(\frac{2\pi}{5}\right) - 1\) and forms a quadratic equation. A1: Solves the quadratic and justifies choosing the positive root to obtain \(\frac{-1+\sqrt{5}}{4}\).

(a) We use integration by parts: \(\int u \, dv = uv - \int v \, du\). Let \(u = x^n\) so that \(du = n x^{n-1} \, dx\). Let \(dv = e^{-2x} \, dx\) so that \(v = -\frac{1}{2} e^{-2x}\). This gives: \(I_n = \left[ -\frac{1}{2} x^n e^{-2x} \right]_{0}^{1} - \int_{0}^{1} \left( -\frac{1}{2} e^{-2x} \right) (n x^{n-1}) \, dx\). Evaluating the boundary term: at \(x=1\), we have \(-\frac{1}{2} (1)^n e^{-2} = -\frac{1}{2} e^{-2}\); at \(x=0\) (since \(n \ge 1\)), the term is \(0\). Thus: \(I_n = -\frac{1}{2} e^{-2} + \frac{n}{2} \int_{0}^{1} x^{n-1} e^{-2x} \, dx\). Since \(I_{n-1} = \int_{0}^{1} x^{n-1} e^{-2x} \, dx\), we have: \(I_n = -\frac{1}{2} e^{-2} + \frac{n}{2} I_{n-1}\). (b) First, find \(I_0\): \(I_0 = \int_{0}^{1} e^{-2x} \, dx = \left[ -\frac{1}{2} e^{-2x} \right]_{0}^{1} = -\frac{1}{2} e^{-2} - \left(-\frac{1}{2}\right) = \frac{1}{2} - \frac{1}{2} e^{-2}\). Next, find \(I_1\) using the reduction formula with \(n=1\): \(I_1 = -\frac{1}{2} e^{-2} + \frac{1}{2} I_0 = -\frac{1}{2} e^{-2} + \frac{1}{2} \left( \frac{1}{2} - \frac{1}{2} e^{-2} \right) = \frac{1}{4} - \frac{3}{4} e^{-2}\). Finally, find \(I_2\) using the reduction formula with \(n=2\): \(I_2 = -\frac{1}{2} e^{-2} + I_1 = -\frac{1}{2} e^{-2} + \frac{1}{4} - \frac{3}{4} e^{-2} = \frac{1}{4} - \frac{5}{4} e^{-2}\).

M1: Applies integration by parts with correct assignment of \(u\) and \(dv\). A1: Correctly integrates to get the term \(-\frac{1}{2} x^n e^{-2x}\) and the integral term \(\frac{n}{2} \int x^{n-1} e^{-2x} \, dx\). M1: Evaluates the limits on the boundary term correctly (noting \(x^n = 0\) at \(x=0\) for \(n \ge 1\)). A1: Obtains the correct reduction relation. M1: Calculates the base case \(I_0\) correctly as \(\frac{1}{2} - \frac{1}{2} e^{-2}\). M1: Applies the reduction formula to find \(I_1\). M1: Applies the reduction formula again to find \(I_2\). A1: Obtains the correct exact final answer \(\frac{1}{4} - \frac{5}{4} e^{-2}\) (or equivalent form).

(a) Rearranging the transformation formula:
\(w(z - \mathrm{i}) = z + \mathrm{i} \implies wz - w\mathrm{i} = z + \mathrm{i}\)
\(wz - z = w\mathrm{i} + \mathrm{i} \implies z(w - 1) = \mathrm{i}(w + 1)\)
\(z = \frac{\mathrm{i}(w+1)}{w-1}\)

(b) Substitute \(w = u + \mathrm{i}v\) into the expression for \(z\):
\(z = \frac{\mathrm{i}(u + \mathrm{i}v + 1)}{u + \mathrm{i}v - 1} = \frac{-v + \mathrm{i}(u+1)}{(u-1) + \mathrm{i}v}\)

Multiply the numerator and the denominator by the conjugate of the denominator, \((u-1) - \mathrm{i}v\):
\(z = \frac{(-v + \mathrm{i}(u+1))((u-1) - \mathrm{i}v)}{(u-1)^2 + v^2}\)
\(z = \frac{-v(u-1) + v(u+1) + \mathrm{i}((u+1)(u-1) + v^2)}{(u-1)^2 + v^2}\)
\(z = \frac{2v + \mathrm{i}(u^2 - 1 + v^2)}{(u-1)^2 + v^2}\)

Thus, the real part of \(z\) is:
\(\operatorname{Re}(z) = \frac{2v}{(u-1)^2 + v^2}\)

Since \(\operatorname{Re}(z) = 1\):
\(\frac{2v}{(u-1)^2 + v^2} = 1 \implies 2v = (u-1)^2 + v^2\)
\((u-1)^2 + v^2 - 2v = 0 \implies (u-1)^2 + (v-1)^2 = 1\)

(c) The Cartesian equation \((u-1)^2 + (v-1)^2 = 1\) describes a circle with centre \((1, 1)\) and radius \(1\).

M1: Attempts to rearrange \(w = \frac{z + \mathrm{i}}{z - \mathrm{i}}\) to make \(z\) the subject.
A1: Obtains \(z = \frac{\mathrm{i}(w+1)}{w-1}\) correctly with all steps shown.
M1: Substitutes \(w = u + \mathrm{i}v\) into the expression for \(z\).
M1: Multiplies numerator and denominator by the conjugate \((u-1) - \mathrm{i}v\) to find the real part.
A1: Correctly identifies \(\operatorname{Re}(z) = \frac{2v}{(u-1)^2 + v^2}\).
A1: Equates to 1 and completes algebra to show \((u-1)^2 + (v-1)^2 = 1\).
B1: States centre is \((1, 1)\) (accept \(1+\mathrm{i}\)).
B1: States radius is \(1\).

(a) The system does not have a unique solution when the determinant of the coefficient matrix is zero:
\(\det \begin{pmatrix} k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{pmatrix} = 0\)
\(k(k^2 - 1) - 1(k - 1) + 1(1 - k) = 0\)
\(k^3 - k - k + 1 + 1 - k = 0\)
\(k^3 - 3k + 2 = 0\)
Factoring this cubic gives:
\((k-1)^2(k+2) = 0\)
Thus, the system does not have a unique solution when \(k = 1\) or \(k = -2\).

(b) When \(k = 1\), the three equations are:
\(x + y + z = 1\)
\(x + y + z = 1\)
\(x + y + z = 1\)
These equations are identical. Therefore, the system is consistent (with infinitely many solutions) and the three planes are coincident.

(c) When \(k = -2\), the equations are:
(1) \(-2x + y + z = 1\)
(2) \(x - 2y + z = -2\)
(3) \(x + y - 2z = 4\)
Adding (1), (2), and (3) together:
\((-2x+x+x) + (y-2y+y) + (z+z-2z) = 1 - 2 + 4\)
\(0 = 3\)
This contradiction shows that there are no solutions, meaning the system is inconsistent.
Since none of the three planes are parallel (the normal vectors \((-2,1,1)\), \(1,-2,1\), and \(1,1,-2\) are not multiples of each other), the planes intersect in pairs to form a triangular prism.

M1: Sets up the determinant of the coefficient matrix and attempts expansion.
A1: Obtains the correct simplified cubic equation \(k^3 - 3k + 2 = 0\).
A1: Correctly solves to find both \(k = 1\) and \(k = -2\).
B1: For \(k=1\), identifies that the equations are identical / system is consistent.
B1: States that the three planes are coincident.
M1: For \(k=-2\), attempts to show inconsistency by adding equations or reducing rows to produce a contradiction.
A1: Explicitly concludes that the system is inconsistent.
B1: Correctly describes the geometric arrangement as a triangular prism.

(a) For the points of intersection, we set the equations equal:
\(2(1 + \cos\theta) = \frac{3}{2\cos\theta}\)
\(4\cos\theta(1 + \cos\theta) = 3\)
\(4\cos^2\theta + 4\cos\theta - 3 = 0\)
\((2\cos\theta - 1)(2\cos\theta + 3) = 0\)
Since \(\cos\theta \ge -1\), we reject the second bracket, leaving:
\(\cos\theta = \frac{1}{2} \implies \theta = \pm \frac{\pi}{3}\)
When \(\theta = \pm \frac{\pi}{3}\), \(r = 2(1 + \frac{1}{2}) = 3\).
Thus, the polar coordinates of the points of intersection are \((3, \frac{\pi}{3})\) and \((3, -\frac{\pi}{3})\).

(b) The region \(R\) is symmetric about the initial line. Its area is given by:
\(\text{Area} = \frac{1}{2} \int_{-\pi/3}^{\pi/3} \left( r_C^2 - r_L^2 \right) \mathrm{d}\theta = \int_{0}^{\pi/3} \left( 4(1+\cos\theta)^2 - \frac{9}{4}\sec^2\theta \right) \mathrm{d}\theta\)
\(\text{Area} = \int_{0}^{\pi/3} \left( 4(1 + 2\cos\theta + \cos^2\theta) - \frac{9}{4}\sec^2\theta \right) \mathrm{d}\theta\)
Using the identity \(\cos^2\theta = \frac{1+\cos 2\theta}{2}\):
\(\text{Area} = \int_{0}^{\pi/3} \left( 6 + 8\cos\theta + 2\cos 2\theta - \frac{9}{4}\sec^2\theta \right) \mathrm{d}\theta\)
Integrating term by term:
\(\text{Area} = \left[ 6\theta + 8\sin\theta +
\sin 2\theta - \frac{9}{4}\tan\theta \right]_{0}^{\pi/3}\)
Substituting the upper limit \(\theta = \frac{\pi}{3}\):
\(\text{Area} = \left( 6\left(\frac{\pi}{3}\right) + 8\sin\left(\frac{\pi}{3}\right) + \sin\left(\frac{2\pi}{3}\right) - \frac{9}{4}\tan\left(\frac{\pi}{3}\right) \right) - 0\)
\(\text{Area} = 2\pi + 8\left(\frac{\sqrt{3}}{2}\right) + \frac{\sqrt{3}}{2} - \frac{9}{4}\sqrt{3}\)
\(\text{Area} = 2\pi + 4\sqrt{3} + \frac{1}{2}\sqrt{3} - \frac{9}{4}\sqrt{3}\)
\(\text{Area} = 2\pi + \frac{9\sqrt{3}}{4}\)

M1: Equates the polar equations and forms a quadratic in \(\cos\theta\).
A1: Obtains the correct intersection coordinates: \((3, \frac{\pi}{3})\) and \((3, -\frac{\pi}{3})\).
M1: Writes down a correct integral for the area of the region using correct limits.
M1: Expands the integrand and uses the double-angle identity for \(\cos^2\theta\).
A1: Correct integration of the expression to obtain \(6\theta + 8\sin\theta + \sin 2\theta - \frac{9}{4}\tan\theta\).
M1: Correctly substitutes the limit \(\frac{\pi}{3}\).
A1: Correctly simplifies the trigonometric exact values.
A1: Obtains the exact area \(2\pi + \frac{9\sqrt{3}}{4}\) or equivalent form.