2023 AQA AS-Level Further Mathematics 7366 模擬試題連答案詳解

An invariant line through the origin exists if and only if the matrix \(\mathbf{M}\) has at least one real eigenvalue. Therefore, the transformation has no invariant lines through the origin if and only if \(\mathbf{M}\) has no real eigenvalues.

We find the characteristic equation by solving \(\det(\mathbf{M} - \lambda \mathbf{I}) = 0\):
\[\det\begin{pmatrix} k-1-\lambda & 2 \\ -2 & 2-\lambda \end{pmatrix} = 0\]
\[(k-1-\lambda)(2-\lambda) - (2)(-2) = 0\]
\[\lambda^2 - (k+1)\lambda + 2(k-1) + 4 = 0\]
\[\lambda^2 - (k+1)\lambda + (2k+2) = 0\]

For the matrix to have no real eigenvalues, the discriminant \(\Delta\) of this quadratic equation must be strictly negative:
\[\Delta = (k+1)^2 - 4(1)(2k+2) < 0\]
\[k^2 + 2k + 1 - 8k - 8 < 0\]
\[k^2 - 6k - 7 < 0\]
\[(k-7)(k+1) < 0\]

Solving this inequality gives:
\[-1 < k < 7\]

Hence, the correct range of values is \(-1 < k < 7\).

M1: Realises that no invariant lines through the origin corresponds to no real eigenvalues, and sets up the characteristic equation to find the discriminant.
A1: Correctly simplifies the discriminant and solves the inequality to choose option B.

The half-line \(H\) starts at \((2, 1)\) and extends in the direction of \(\frac{\pi}{4}\). The line containing \(H\) has the equation:
\[y - 1 = \tan\left(\frac{\pi}{4}\right)(x - 2) \implies y = x - 1 \quad \text{for } x > 2\]

This line can be written in general form as \(x - y - 1 = 0\).

The circle \(C\) has center \((a, 0)\) and radius \(r = 2\).
For \(C\) to be tangent to this line, the perpendicular distance from \((a,0)\) to the line must equal the radius:
\[\frac{|a - 0 - 1|}{\sqrt{1^2 + (-1)^2}} = 2 \implies \frac{|a - 1|}{\sqrt{2}} = 2 \implies |a - 1| = 2\sqrt{2}\]

Since \(a\) is real, this gives two possible values:
\[a = 1 + 2\sqrt{2} \quad \text{or} \quad a = 1 - 2\sqrt{2}\]

Now, we must check which of these values yields a point of tangency on the half-line \(H\) (i.e., with \(x > 2\)).
The line of centers is perpendicular to the tangent line and passes through \((a,0)\), so its equation is:
\[y - 0 = -1(x - a) \implies y = a - x\]

Finding the intersection (the point of tangency \(P(x_0, y_0)\)):
\[x - 1 = a - x \implies 2x = a + 1 \implies x_0 = \frac{a + 1}{2}\]

For \(P\) to lie on \(H\), we must have \(x_0 > 2\):
- If \(a = 1 + 2\sqrt{2}\):
\[x_0 = \frac{1 + 2\sqrt{2} + 1}{2} = 1 + \sqrt{2} \approx 2.414 > 2 \quad \text{(Valid)}\]
- If \(a = 1 - 2\sqrt{2}\):
\[x_0 = \frac{1 - 2\sqrt{2} + 1}{2} = 1 - \sqrt{2} \approx -0.414 \le 2 \quad \text{(Invalid)}\]

Thus, the only valid value of \(a\) is \(1 + 2\sqrt{2}\).

M1: Applies the perpendicular distance formula from the center of the circle to the line containing the half-line to find the potential values of \(a\), and tests the validity of the tangency point.
A1: Correctly selects option A.

We use the hyperbolic identity \[\cosh^2 x - \sinh^2 x = 1 \implies \cosh^2 x = 1 + \sinh^2 x\] to rewrite the equation in terms of \(\sinh x\):
\[2(1 + \sinh^2 x) - 5\sinh x - 5 = 0\]
\[2\sinh^2 x - 5\sinh x - 3 = 0\]

Factorising this quadratic equation:
\[(2\sinh x + 1)(\sinh x - 3) = 0\]

This gives two cases:
1) \(\sinh x = 3\)
Using the exponential definition of \(\sinh x\):
\[\frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2} = 3 \implies \mathrm{e}^x - \mathrm{e}^{-x} = 6\]
Multiplying by \(\mathrm{e}^x\):
\[\mathrm{e}^{2x} - 6\mathrm{e}^x - 1 = 0\]
Solving this quadratic in \(\mathrm{e}^x\):
\[\mathrm{e}^x = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(-1)}}{2} = \frac{6 \pm \sqrt{40}}{2} = 3 \pm \sqrt{10}\]
Since \(\mathrm{e}^x > 0\) for any real \(x\), we reject the negative root \(3 - \sqrt{10} < 0\). Thus, \(\mathrm{e}^x = 3 + \sqrt{10}\).

2) \(\sinh x = -\frac{1}{2}\)
\[\frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2} = -\frac{1}{2} \implies \mathrm{e}^x - \mathrm{e}^{-x} = -1 \implies \mathrm{e}^{2x} + \mathrm{e}^x - 1 = 0\]
\[\mathrm{e}^x = \frac{-1 \pm \sqrt{5}}{2}\]
Rejecting the negative root gives \(\mathrm{e}^x = \frac{\sqrt{5} - 1}{2}\).

Comparing with the options, \(3 + \sqrt{10}\) is a possible value of \(\mathrm{e}^x\).

M1: Substitutes \(\cosh^2 x = 1 + \sinh^2 x\), solves the quadratic in \(\sinh x\), and sets up the quadratic equation for \(\mathrm{e}^x\).
A1: Identifies the valid positive root for \(\mathrm{e}^x\) and matches it to option A.

To find \(|z|\), we can solve the quadratic equation to find the roots: \(z = \frac{6 \pm \sqrt{36 - 4 \times 13}}{2} = 3 \pm 2\mathrm{i}\). Now we calculate the modulus of the root: \(|z| = \sqrt{3^2 + 2^2} = \sqrt{13}\). Alternatively, since the quadratic has real coefficients, the roots are complex conjugates, so the product of the roots is \(z z^* = |z|^2 = \frac{c}{a} = 13\), which immediately gives \(|z| = \sqrt{13}\).

B1: Correctly chooses option B (or identifies \(|z| = \sqrt{13}\)).

Since the coefficients of the cubic equation \(z^3 - 7z^2 + kz - 13 = 0\) are real, the complex roots must occur in conjugate pairs.

Therefore, \(z_2 = 3 + 2\mathrm{i}\) is also a root of the equation.

The third root, \(z_3\), must be real.

Using the product of the roots of a cubic equation:
\[ z_1 z_2 z_3 = -\frac{d}{a} \]
\[ (3 - 2\mathrm{i})(3 + 2\mathrm{i}) z_3 = -(-13) \]
\[ (3^2 + 2^2) z_3 = 13 \]
\[ 13 z_3 = 13 \implies z_3 = 1 \]

Using the sum of the products of roots in pairs to find \(k\):
\[ k = z_1 z_2 + z_2 z_3 + z_3 z_1 \]
\[ k = 13 + (3 + 2\mathrm{i})(1) + (3 - 2\mathrm{i})(1) \]
\[ k = 13 + 3 + 2\mathrm{i} + 3 - 2\mathrm{i} = 19 \]

Alternative method:
Substitute \(z = 3-2\mathrm{i}\) directly into the equation:
\[ (3-2\mathrm{i})^3 - 7(3-2\mathrm{i})^2 + k(3-2\mathrm{i}) - 13 = 0 \]
Using \((3-2\mathrm{i})^2 = 9 - 12\mathrm{i} - 4 = 5 - 12\mathrm{i}\)
And \((3-2\mathrm{i})^3 = (5-12\mathrm{i})(3-2\mathrm{i}) = 15 - 10\mathrm{i} - 36\mathrm{i} - 24 = -9 - 46\mathrm{i}\)

So:
\[ (-9 - 46\mathrm{i}) - 7(5 - 12\mathrm{i}) + k(3 - 2\mathrm{i}) - 13 = 0 \]
\[ -9 - 46\mathrm{i} - 35 + 84\mathrm{i} + 3k - 2k\mathrm{i} - 13 = 0 \]

Equating real parts:
\[ -9 - 35 - 13 + 3k = 0 \implies 3k - 57 = 0 \implies k = 19 \]

M1: Recognises that \(3 + 2\mathrm{i}\) is also a root, or attempts to substitute \(z = 3-2\mathrm{i}\) into the cubic equation.
M1: Uses the product of roots to find the third root is 1, OR correctly expands both \((3-2\mathrm{i})^2\) and \((3-2\mathrm{i})^3\).
A1: Obtains \(k = 19\) with no errors in working.

Since the matrix \(\mathbf{M}\) is singular, its determinant must be equal to zero.

\[ \det(\mathbf{M}) = (k+2)(k-3) - (3)(-2) = 0 \]

Expand the terms:
\[ k^2 - 3k + 2k - 6 + 6 = 0 \]
\[ k^2 - k = 0 \]

Factorise:
\[ k(k-1) = 0 \]

Therefore, the possible values of \(k\) are \(k = 0\) and \(k = 1\).

M1: Sets the determinant of \(\mathbf{M}\) to 0, i.e., \((k+2)(k-3) - (3)(-2) = 0\).
A1: Correctly expands to obtain the quadratic equation \(k^2 - k = 0\) (or equivalent).
A1: Correctly solves to find both \(k = 0\) and \(k = 1\).

Using the standard summation formulae:
\[ \sum_{r=1}^{n} r(2r-1) = \sum_{r=1}^{n} (2r^2 - r) = 2\sum_{r=1}^{n} r^2 - \sum_{r=1}^{n} r \]

Substitute the standard formulae:
\[ \sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6} \]
\[ \sum_{r=1}^{n} r = \frac{n(n+1)}{2} \]

So,
\[ \sum_{r=1}^{n} r(2r-1) = 2\left[\frac{n(n+1)(2n+1)}{6}\right] - \frac{n(n+1)}{2} \]
\[ = \frac{n(n+1)(2n+1)}{3} - \frac{n(n+1)}{2} \]

Factorise out the common factor \(\frac{n(n+1)}{6}\):
\[ = \frac{n(n+1)}{6} \left[ 2(2n+1) - 3 \right] \]
\[ = \frac{n(n+1)}{6} \left[ 4n + 2 - 3 \right] \]
\[ = \frac{n(n+1)(4n-1)}{6} \]

M1: Splits the summation into \(2\sum r^2 - \sum r\) and substitutes the standard formulae for \(\sum r^2\) and \(\sum r\).
M1: Attempts to factorise out a common factor, such as \(\frac{n(n+1)}{6}\) or \(\frac{n(n+1)}{2}\).
A1: Obtains the fully factorised form \(\frac{n(n+1)(4n-1)}{6}\) or equivalent.

A matrix is singular if and only if its determinant is equal to zero. First, we find the determinant of \( \mathbf{A} \): \( \det(\mathbf{A}) = (k+3)(k) - (2)(5) = k^2 + 3k - 10 \). Setting the determinant to zero for a singular matrix: \( k^2 + 3k - 10 = 0 \). Factorising the quadratic equation: \( (k+5)(k-2) = 0 \). This gives the solutions: \( k = -5 \) or \( k = 2 \).

M1: Attempts to find the determinant of \( \mathbf{A} \) and sets it equal to 0, resulting in a quadratic equation. A1: Obtains the correct quadratic equation \( k^2 + 3k - 10 = 0 \) and attempts to solve it by factorising or using the quadratic formula. A0.5: Finds both correct values: \( k = -5 \) and \( k = 2 \).

First, we find \( z^2 \): \( z^2 = (3 - 2\mathrm{i})^2 = 9 - 12\mathrm{i} - 4 = 5 - 12\mathrm{i} \). Next, we find the complex conjugate of \( z \): \( z^* = 3 + 2\mathrm{i} \). Now, we compute \( w \): \( w = z^2 + z^* = (5 - 12\mathrm{i}) + (3 + 2\mathrm{i}) = (5 + 3) + (-12 + 2)\mathrm{i} = 8 - 10\mathrm{i} \).

M1: Correct method to expand \( z^2 \) with at least three terms correct, or states the correct complex conjugate \( z^* = 3 + 2\mathrm{i} \). A1: Obtains \( z^2 = 5 - 12\mathrm{i} \). A0.5: Obtains the final correct complex number \( 8 - 10\mathrm{i} \).

Express the given term with a common denominator: \( \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\beta\gamma + \alpha\gamma + \alpha\beta}{\alpha\beta\gamma} \). From the relations between roots and coefficients for the cubic equation \( ax^3 + bx^2 + cx + d = 0 \): \( \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = \frac{4}{2} = 2 \) and \( \alpha\beta\gamma = -\frac{d}{a} = -\frac{-3}{2} = \frac{3}{2} \). Substituting these values into our expression gives: \( \frac{2}{3/2} = 2 \times \frac{2}{3} = \frac{4}{3} \).

M1: Expresses the target expression with a common denominator of \( \alpha\beta\gamma \) or identifies the required root relationship formulas. A1: Obtains both correct values for the root relationships: \( \sum \alpha\beta = 2 \) and \( \alpha\beta\gamma = \frac{3}{2} \). A0.5: Correctly simplifies the division to obtain the final exact value of \( \frac{4}{3} \).

For \(\mathbf{A}\) to be singular, we must have \(\det(\mathbf{A}) = 0\). The determinant is calculated as: \(\det(\mathbf{A}) = k(k-5) - (3)(2) = k^2 - 5k - 6\). Setting this to zero gives: \(k^2 - 5k - 6 = 0\). Factoring the quadratic expression, we get: \((k-6)(k+1) = 0\). Thus, the values of \(k\) are \(k = 6\) and \(k = -1\).

M1: Sets the determinant of \(\mathbf{A}\) equal to zero, obtaining a quadratic equation in \(k\). A1: Correctly simplifies the determinant to obtain \(k^2 - 5k - 6 = 0\). A0.5: Solves the quadratic equation to find both correct values: \(k = 6\) and \(k = -1\).

To express \(z\) in the form \(a + b\mathrm{i}\), multiply both the numerator and the denominator by the complex conjugate of the denominator, which is \(1 + 2\mathrm{i}\): \(z = \frac{(5 + \mathrm{i})(1 + 2\mathrm{i})}{(1 - 2\mathrm{i})(1 + 2\mathrm{i})}\). The denominator simplifies to: \(1^2 + (-2)^2 = 1 + 4 = 5\). The numerator simplifies to: \(5(1) + 10\mathrm{i} + \mathrm{i} + 2\mathrm{i}^2 = 5 + 11\mathrm{i} - 2 = 3 + 11\mathrm{i}\). Therefore, \(z = \frac{3 + 11\mathrm{i}}{5} = \frac{3}{5} + \frac{11}{5}\mathrm{i}\).

M1: Multiplies the numerator and denominator by the complex conjugate of the denominator, \(1 + 2\mathrm{i}\). A1: Correctly expands either the numerator to \(3 + 11\mathrm{i}\) or the denominator to \(5\). A0.5: Obtains the correct final expression: \(\frac{3}{5} + \frac{11}{5}\mathrm{i}\) or \(0.6 + 2.2\mathrm{i}\).

From the relation between roots and coefficients of a cubic equation: \(\sum \alpha = \alpha + \beta + \gamma = -\frac{b}{a} = -\frac{-5}{2} = 2.5\). \(\sum \alpha\beta = \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = \frac{4}{2} = 2\). Using the algebraic identity: \(\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\). Substituting the values, we get: \(\alpha^2 + \beta^2 + \gamma^2 = (2.5)^2 - 2(2) = 6.25 - 4 = 2.25\).

B1: States both correct values of the sum of roots \(\sum \alpha = 2.5\) and the sum of products of roots pairwise \(\sum \alpha\beta = 2\). M1: Applies the identity \(\alpha^2 + \beta^2 + \gamma^2 = (\sum \alpha)^2 - 2\sum \alpha\beta\). A0.5: Obtains the correct final value of \(2.25\) (or \(\frac{9}{4}\)).

For the matrix \(\mathbf{M}\) to be singular, its determinant must be zero.

\(\det(\mathbf{M}) = a(a-1) - 3(a+2) = 0\)

Expanding the terms:

\(a^2 - a - 3a - 6 = 0\)

\(a^2 - 4a - 6 = 0\)

Using the quadratic formula to solve for \(a\):

\(a = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-6)}}{2(1)}\)

\(a = \frac{4 \pm \sqrt{16 + 24}}{2} = \frac{4 \pm \sqrt{40}}{2}\)

Since \(\sqrt{40} = 2\sqrt{10}\):

\(a = \frac{4 \pm 2\sqrt{10}}{2} = 2 \pm \sqrt{10}\)

M1: Equates the determinant of \(\mathbf{M}\) to zero to form a quadratic equation in \(a\).

A1: Obtains the correct simplified quadratic equation \(a^2 - 4a - 6 = 0\).

A0.5: Correctly solves the quadratic equation to find both values: \(a = 2 \pm \sqrt{10}\) (or equivalent simplified surd form).

Since the quadratic equation has real coefficients, its non-real complex roots must occur in conjugate pairs.

Therefore, the second root is:

\(z_2 = 3 + 2\mathrm{i}\)

Using the relationships between the roots and coefficients of a quadratic equation:

\(-p = z_1 + z_2 = (3 - 2\mathrm{i}) + (3 + 2\mathrm{i}) = 6 \implies p = -6\)

\(q = z_1 z_2 = (3 - 2\mathrm{i})(3 + 2\mathrm{i}) = 3^2 - (2\mathrm{i})^2 = 9 + 4 = 13\)

Thus, \(p = -6\) and \(q = 13\).

M1: Recognises that the other root is the complex conjugate \(3 + 2\mathrm{i}\) and attempts to find either the sum or product of the roots (or substitutes \(3 - 2\mathrm{i}\) into the equation and equates real and imaginary parts).

A1: Correctly determines \(p = -6\).

A0.5: Correctly determines \(q = 13\).

From the coefficients of the cubic equation \(2x^3 - 5x^2 + 4x - 3 = 0\), we have:

\(\alpha\beta + \beta\gamma + \gamma\alpha = \frac{4}{2} = 2\)

\(\alpha\beta\gamma = -\frac{-3}{2} = \frac{3}{2}\)

We want to find the value of:

\(\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\beta\gamma + \alpha\gamma + \alpha\beta}{\alpha\beta\gamma}\)

Substituting the known values into this expression:

\(\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{2}{\frac{3}{2}} = 2 \times \frac{2}{3} = \frac{4}{3}\)

M1: Expresses the sum of the reciprocals with a common denominator, \(\frac{\alpha\beta + \beta\gamma + \gamma\alpha}{\alpha\beta\gamma}\), and attempts to find the required root products from the coefficients.

A1: Obtains the correct values for \(\alpha\beta + \beta\gamma + \gamma\alpha = 2\) and \(\alpha\beta\gamma = \frac{3}{2}\).

A0.5: Correctly evaluates the expression to get \(\frac{4}{3}\) (or equivalent fraction).

Since \(\mathbf{M}\) is singular, its determinant must be zero: \(\det(\mathbf{M}) = (k-2)(k+2) - (3)(4) = 0\). Expanding the terms, we get: \(k^2 - 4 - 12 = 0 \implies k^2 - 16 = 0 \implies k^2 = 16\). Thus, the possible values of \(k\) are: \(k = 4\) or \(k = -4\).

M1: Setting the determinant of \(\mathbf{M}\) equal to zero, i.e., \((k-2)(k+2) - 12 = 0\). A1: Correctly simplifying to a quadratic equation, e.g., \(k^2 - 16 = 0\). A0.5: Finding both correct values \(k = 4\) and \(k = -4\) (accept \(k = \pm 4\)).

Method 1: Since the coefficients of the cubic equation are real, complex roots must occur in conjugate pairs. Thus, \(z_2 = 2 + \mathrm{i}\) is also a root. Let the third root be \(\alpha\). Using the sum of roots: \((2 - \mathrm{i}) + (2 + \mathrm{i}) + \alpha = 3 \implies 4 + \alpha = 3 \implies \alpha = -1\). Using the product of roots: \((2 - \mathrm{i})(2 + \mathrm{i})\alpha = -b \implies (4 - \mathrm{i}^2)(-1) = -b \implies 5(-1) = -b \implies b = 5\). Using the sum of pairwise products: \((2 - \mathrm{i})(2 + \mathrm{i}) + (2 - \mathrm{i})\alpha + (2 + \mathrm{i})\alpha = a \implies 5 - 4 = a \implies a = 1\). Method 2: Direct substitution of \(z = 2 - \mathrm{i}\) into the equation yields \((2 - \mathrm{i})^3 - 3(2 - \mathrm{i})^2 + a(2 - \mathrm{i}) + b = 0\). Expanding terms: \((2 - 11\mathrm{i}) + (-9 + 12\mathrm{i}) + a(2 - \mathrm{i}) + b = 0 \implies (2a + b - 7) + (1 - a)\mathrm{i} = 0\). Equating real and imaginary parts to zero gives \(1 - a = 0 \implies a = 1\) and \(2(1) + b - 7 = 0 \implies b = 5\).

M1: Recognising that the complex conjugate \(2 + \mathrm{i}\) is also a root, or attempting direct substitution of \(2 - \mathrm{i}\) into the equation. A1: Correctly determining the third root \(\alpha = -1\), or correctly simplifying the substitution to obtain the system of equations \(2a + b - 7 = 0\) and \(1 - a = 0\). A0.5: Correctly finding \(a = 1\) and \(b = 5\).

We split the summation using standard formulae: \(\sum_{r=1}^{20} (3r^2 - 2r) = 3\sum_{r=1}^{20} r^2 - 2\sum_{r=1}^{20} r\). Using the standard formulae \(\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n+1)(2n+1)\) and \(\sum_{r=1}^{n} r = \frac{1}{2}n(n+1)\) with \(n = 20\): \(\sum_{r=1}^{20} r^2 = \frac{1}{6}(20)(21)(41) = 2870\) and \(\sum_{r=1}^{20} r = \frac{1}{2}(20)(21) = 210\). Substituting these values back: \(3(2870) - 2(210) = 8610 - 420 = 8190\).

M1: Splitting the summation and identifying the correct standard formulae for \(\sum r^2\) and \(\sum r\). A1: Correct evaluation of both \(\sum_{r=1}^{20} r^2 = 2870\) and \(\sum_{r=1}^{20} r = 210\) (or substituting these values correctly into the expression: \(3(2870) - 2(210)\)). A0.5: Obtains the final answer 8190.

For a matrix to be singular, its determinant must be equal to zero. Therefore, \(\det(\mathbf{M}) = a(a+5) - 3(-2) = 0\). Expanding and simplifying this gives \(a^2 + 5a + 6 = 0\). Factoring the quadratic expression, we get \((a+2)(a+3) = 0\). Thus, the possible values of \(a\) are \(a = -2\) and \(a = -3\).

M1: Sets the determinant of \(\mathbf{M}\) to 0, i.e., \(a(a+5) - (3)(-2) = 0\). A1: Forms the correct quadratic equation \(a^2 + 5a + 6 = 0\). A0.5: Solves the quadratic equation correctly to find both values \(a = -2\) and \(a = -3\).

Let \(z = x + \mathrm{i}y\).

Substituting this into the given equation:
\[|x + \mathrm{i}y - 4 - 3\mathrm{i}| = |x + \mathrm{i}y + 2 - \mathrm{i}|\]
\[|(x - 4) + \mathrm{i}(y - 3)| = |(x + 2) + \mathrm{i}(y - 1)|\]

Squaring both sides:
\[(x - 4)^2 + (y - 3)^2 = (x + 2)^2 + (y - 1)^2\]

Expanding the terms:
\[x^2 - 8x + 16 + y^2 - 6y + 9 = x^2 + 4x + 4 + y^2 - 2y + 1\]

Simplifying by cancelling \(x^2\) and \(y^2\) on both sides:
\[-8x - 6y + 25 = 4x - 2y + 5\]
\[12x + 4y = 20\]
\[3x + y = 5 \implies y = -3x + 5\]

This is a straight line with \(m = -3\) and \(c = 5\).

To find the complex number \(w\) on this locus with the minimum modulus, we find the point on the line \(y = -3x + 5\) closest to the origin.

The line perpendicular to the locus passing through the origin has gradient \(m_{\perp} = -\frac{1}{-3} = \frac{1}{3}\).
Its equation is:
\[y = \frac{1}{3}x\]

Finding the intersection of the two lines:
\[\frac{1}{3}x = -3x + 5\]
\[\frac{10}{3}x = 5 \implies x = 1.5 = \frac{3}{2}\]

Substituting \(x = \frac{3}{2}\) back into the equation of the perpendicular line:
\[y = \frac{1}{3}\left(\frac{3}{2}\right) = \frac{1}{2}\]

Alternatively, we can minimize the square of the modulus \(|z|^2 = x^2 + y^2\):
\[f(x) = x^2 + (-3x + 5)^2 = x^2 + 9x^2 - 30x + 25 = 10x^2 - 30x + 25\]
\[f'(x) = 20x - 30 = 0 \implies x = 1.5\]
\[y = -3(1.5) + 5 = 0.5\]

Thus, the complex number \(w\) is:
\[w = \frac{3}{2} + \frac{1}{2}\mathrm{i}\]

M1: Substitutes \(z = x + \mathrm{i}y\) into the equation or interprets the locus as the perpendicular bisector of the line segment joining \(4 + 3\mathrm{i}\) and \(-2 + \mathrm{i}\).
A1: Obtains a correct unsimplified Cartesian equation, e.g. \((x-4)^2 + (y-3)^2 = (x+2)^2 + (y-1)^2\), or finds midpoint \((1, 2)\) and gradient \(-3\) using the perpendicular bisector method.
A1: Deduces the line equation \(y = -3x + 5\) (or equivalent).
M1: Employs a valid method to find the point of minimum modulus, e.g., finding the intersection of the locus with the perpendicular line \(y = \frac{1}{3}x\), or minimizing \(x^2 + y^2 = 10x^2 - 30x + 25\).
A1: Solves to find \(x = 1.5\) and \(y = 0.5\) (or equivalents).
A1: States the correct complex number \(w = 1.5 + 0.5\mathrm{i}\) (or in equivalent fraction form).

A reflection in the line \(y = mx\) (where \(m = \tan\theta\)) is represented by the matrix:
\[\mathbf{R} = \begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}\]

A stretch parallel to the \(y\)-axis with scale factor \(k\) is represented by the matrix:
\[\mathbf{S} = \begin{pmatrix} 1 & 0 \\ 0 & k \end{pmatrix}\]

Since the reflection is followed by the stretch, the combined transformation matrix \(\mathbf{M}\) is given by:
\[\mathbf{M} = \mathbf{S}\mathbf{R}\]
\[\mathbf{M} = \begin{pmatrix} 1 & 0 \\ 0 & k \end{pmatrix} \begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix} = \begin{pmatrix} \cos 2\theta & \sin 2\theta \\ k\sin 2\theta & -k\cos 2\theta \end{pmatrix}\]

We are given:
\[\mathbf{M} = \begin{pmatrix} -0.8 & 0.6 \\ 1.8 & 2.4 \end{pmatrix}\]

Comparing the top row elements:
\[\cos 2\theta = -0.8\]
\[\sin 2\theta = 0.6\]

Comparing the bottom row elements:
\[k\sin 2\theta = 1.8 \implies k(0.6) = 1.8 \implies k = 3\]
\[-k\cos 2\theta = 2.4 \implies -3(-0.8) = 2.4 \quad \text{(which is consistent)}\]

Now, we find \(m = \tan\theta\). Using double-angle identities:
\[\cos 2\theta = \frac{1 - \tan^2\theta}{1 + \tan^2\theta} = \frac{1 - m^2}{1 + m^2}\]

Setting this equal to \(-0.8\):
\[\frac{1 - m^2}{1 + m^2} = -0.8\]
\[1 - m^2 = -0.8(1 + m^2)\]
\[1 - m^2 = -0.8 - 0.8m^2\]
\[1.8 = 0.2m^2\]
\[m^2 = 9\]

Since \(m > 0\), we have:
\[m = 3\]

(Alternatively, because \(\cos 2\theta = -0.8 < 0\) and \(\sin 2\theta = 0.6 > 0\), the angle \(2\theta\) lies in the second quadrant, so \(\theta\) lies in the first quadrant, confirming \(\tan\theta > 0\).)

Thus, \(k = 3\) and \(m = 3\).

M1: Identifies the matrix representation of a stretch parallel to the \(y\)-axis as \(\begin{pmatrix} 1 & 0 \\ 0 & k \end{pmatrix}\).
M1: Identifies the general matrix representation of a reflection in \(y = mx\) in terms of \(\theta\) or directly in terms of \(m\).
M1: Forms the product \(\mathbf{S}\mathbf{R}\) in the correct order to represent the combined transformation.
A1: Equates coefficients to obtain \(\cos 2\theta = -0.8\), \(\sin 2\theta = 0.6\) and \(k\sin 2\theta = 1.8\) (or equivalent equations in terms of \(m\)).
A1: Solves to find \(k = 3\).
A1: Uses trigonometric double-angle formulae (or equivalent algebraic methods) to find \(m = 3\), with appropriate justification for selecting the positive root.

Method 1: Substitution
Let \(w = x^2\) be the roots of the new equation. This implies \(x = \sqrt{w}\).

Substitute \(x = \sqrt{w}\) into the original cubic equation:
\[2(\sqrt{w})^3 - 5(\sqrt{w})^2 + 4\sqrt{w} - 3 = 0\]
\[2w\sqrt{w} - 5w + 4\sqrt{w} - 3 = 0\]

Rearrange the terms with \(\sqrt{w}\) on one side:
\[\sqrt{w}(2w + 4) = 5w + 3\]

Square both sides of the equation:
\[w(2w + 4)^2 = (5w + 3)^2\]
\[w(4w^2 + 16w + 16) = 25w^2 + 30w + 9\]
\[4w^3 + 16w^2 + 16w = 25w^2 + 30w + 9\]

Collect all terms on one side:
\[4w^3 - 9w^2 - 14w - 9 = 0\]

Method 2: Using Relations between Roots and Coefficients
For the original cubic equation \(2x^3 - 5x^2 + 4x - 3 = 0\):
\[\sum \alpha = \frac{5}{2}\]
\[\sum \alpha\beta = \frac{4}{2} = 2\]
\[\alpha\beta\gamma = \frac{3}{2}\]

Let the new roots be \(u = \alpha^2\), \(v = \beta^2\), \(w = \gamma^2\).

1) Sum of the new roots:
\[\sum \alpha^2 = (\sum \alpha)^2 - 2\sum \alpha\beta = \left(\frac{5}{2}\right)^2 - 2(2) = \frac{25}{4} - 4 = \frac{9}{4}\]

2) Sum of products in pairs:
\[\sum \alpha^2\beta^2 = (\sum \alpha\beta)^2 - 2\alpha\beta\gamma\sum \alpha = 2^2 - 2\left(\frac{3}{2}\right)\left(\frac{5}{2}\right) = 4 - \frac{15}{2} = -\frac{7}{2}\]

3) Product of the roots:
\[\alpha^2\beta^2\gamma^2 = (\alpha\beta\gamma)^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4}\]

Thus, the new cubic equation is of the form:
\[y^3 - \left(\sum \alpha^2\right)y^2 + \left(\sum \alpha^2\beta^2\right)y - \alpha^2\beta^2\gamma^2 = 0\]
\[y^3 - \frac{9}{4}y^2 - \frac{7}{2}y - \frac{9}{4} = 0\]

Multiplying by 4 to obtain integer coefficients:
\[4y^3 - 9y^2 - 14y - 9 = 0\]

Method 1:
M1: Sets \(w = x^2\) (or equivalent) and substitutes \(x = \sqrt{w}\) into the cubic equation.
M1: Rearranges the equation to isolate terms containing \(\sqrt{w}\), e.g., \(\sqrt{w}(2w + 4) = 5w + 3\).
M1: Squares both sides to eliminate the square root.
A1: Correctly expands both sides to obtain \(w(4w^2 + 16w + 16) = 25w^2 + 30w + 9\) (or equivalent).
A1: Simplifies to a cubic equation in standard form.
A1: Correctly expresses the final equation with integer coefficients: \(4w^3 - 9w^2 - 14w - 9 = 0\) (or any variable name like \(x\) or \(y\)).

Method 2:
M1: States the values of \(\sum \alpha\), \(\sum \alpha\beta\) and \(\alpha\beta\gamma\) correctly.
M1: Attempts to find the sum of the new roots \(\sum \alpha^2\) using a correct algebraic identity.
M1: Attempts to find \(\sum \alpha^2\beta^2\) using a correct algebraic identity.
A1: Obtains the correct values: \(\sum \alpha^2 = \frac{9}{4}\), \(\sum \alpha^2\beta^2 = -\frac{7}{2}\), and \(\alpha^2\beta^2\gamma^2 = \frac{9}{4}\).
A1: Formulates the new cubic equation using these values.
A1: Correctly expresses the final equation with integer coefficients: \(4y^3 - 9y^2 - 14y - 9 = 0\) (or equivalent).

(a) The system does not have a unique solution when the determinant of the coefficient matrix is equal to zero.
Let \(\mathbf{A} = \begin{pmatrix} 1 & -2 & 1 \\ 2 & k & -1 \\ 3 & -1 & 2 \end{pmatrix}\).

Calculating the determinant of \(\mathbf{A}\):
\(\det(\mathbf{A}) = 1 \begin{vmatrix} k & -1 \\ -1 & 2 \end{vmatrix} - (-2) \begin{vmatrix} 2 & -1 \\ 3 & 2 \end{vmatrix} + 1 \begin{vmatrix} 2 & k \\ 3 & -1 \end{vmatrix}\)
\(\det(\mathbf{A}) = 1(2k - 1) + 2(4 - (-3)) + 1(-2 - 3k)\)
\(\det(\mathbf{A}) = 2k - 1 + 14 - 2 - 3k\)
\(\det(\mathbf{A}) = 11 - k\)

Setting the determinant to zero:
\(11 - k = 0 \implies k = 11\)

(b) Substitute \(k = 11\) into the system of equations:
(1) \(x - 2y + z = 3\)
(2) \(2x + 11y - z = 1\)
(3) \(3x - y + 2z = 16\)

Eliminate \(z\) from equations (1) and (2) by adding them:
\((x - 2y + z) + (2x + 11y - z) = 3 + 1\)
\(3x + 9y = 4 \implies x + 3y = \frac{4}{3}\)

Eliminate \(z\) from equations (1) and (3) by multiplying (1) by 2 and subtracting from (3):
\((3x - y + 2z) - 2(x - 2y + z) = 16 - 2(3)\)
\(3x - y + 2z - 2x + 4y - 2z = 16 - 6\)
\(x + 3y = 10\)

This gives two contradictory equations:
\(x + 3y = \frac{4}{3}\) and \(x + 3y = 10\).
Since these cannot both be true simultaneously, there is no solution, and the system is **inconsistent**.

(a)
- M1: Attempts to find the determinant of the coefficient matrix in terms of \(k\).
- A1: Obtains \(\det(\mathbf{A}) = 11 - k\) (or equivalent expression).
- A1: Sets determinant to 0 and correctly finds \(k = 11\).

(b)
- M1: Substitutes \(k = 11\) and attempts to eliminate one variable (e.g., \(z\)) from two pairs of equations.
- A1: Correctly obtains two equations in two variables, such as \(x + 3y = \frac{4}{3}\) (or \(3x + 9y = 4\)) and \(x + 3y = 10\).
- R1: Explains that because these equations are contradictory (e.g., \(\frac{4}{3} \neq 10\)), the system has no solutions and is therefore inconsistent.

(a) Since the coefficients of the cubic equation are real, complex roots must occur in conjugate pairs. Thus, the second root is:
\(z_2 = 1 + 2\mathrm{i}\)

(b) For the cubic equation \(2z^3 - 3z^2 + pz + q = 0\), the sum of the roots is:
\(z_1 + z_2 + z_3 = -\frac{b}{a} = -\frac{-3}{2} = \frac{3}{2}\)

Substitute the known roots \(z_1\) and \(z_2\):
\((1 - 2\mathrm{i}) + (1 + 2\mathrm{i}) + z_3 = \frac{3}{2}\)
\(2 + z_3 = \frac{3}{2}\)
\(z_3 = \frac{3}{2} - 2 = -\frac{1}{2}\) (as required).

(c) Method 1: Product of roots and sum of pairwise products
The product of the roots is:
\(z_1 z_2 z_3 = -\frac{d}{a} = -\frac{q}{2}\)
\((1 - 2\mathrm{i})(1 + 2\mathrm{i})\left(-\frac{1}{2}\right) = -\frac{q}{2}\)
\((1^2 + 2^2)\left(-\frac{1}{2}\right) = -\frac{q}{2}\)
\(5\left(-\frac{1}{2}\right) = -\frac{q}{2} \implies q = 5\)

The sum of products of roots in pairs is:
\(z_1 z_2 + z_2 z_3 + z_3 z_1 = \frac{c}{a} = \frac{p}{2}\)
\(5 + \left(-\frac{1}{2}\right)(1 + 2\mathrm{i}) + \left(-\frac{1}{2}\right)(1 - 2\mathrm{i}) = \frac{p}{2}\)
\(5 - \frac{1}{2}(1 + 2\mathrm{i} + 1 - 2\mathrm{i}) = \frac{p}{2}\)
\(5 - \frac{1}{2}(2) = \frac{p}{2}\)
\(4 = \frac{p}{2} \implies p = 8\)

Method 2: Factoring the cubic
The cubic expression can be written as:
\(2\left(z + \frac{1}{2}\right)(z - (1 - 2\mathrm{i}))(z - (1 + 2\mathrm{i})) = 0\)
\((2z + 1)(z^2 - 2z + 5) = 0\)
\(2z(z^2 - 2z + 5) + 1(z^2 - 2z + 5) = 0\)
\(2z^3 - 4z^2 + 10z + z^2 - 2z + 5 = 0\)
\(2z^3 - 3z^2 + 8z + 5 = 0\)

Comparing coefficients with \(2z^3 - 3z^2 + pz + q = 0\):
\(p = 8\) and \(q = 5\).

(a)
- B1: Correctly writes \(z_2 = 1 + 2\mathrm{i}\).

(b)
- M1: Uses the sum of roots formula \(z_1 + z_2 + z_3 = \frac{3}{2}\) (or substitutes \(z = -\frac{1}{2}\) into the factors).
- A1: Shows that \(z_3 = -\frac{1}{2}\) clearly and convincingly.

(c)
- M1: Uses relations between roots and coefficients (e.g., product of roots \(z_1 z_2 z_3 = -\frac{q}{2}\) or sum of pairwise products) OR expands the product of factors \((2z+1)(z^2-2z+5)\).
- A1: Correctly finds \(q = 5\).
- A1: Correctly finds \(p = 8\).

(a) To find \(\int_0^{\pi/2} x \sin(2x) \, \mathrm{d}x\), use integration by parts:
\(\int u \frac{\mathrm{d}v}{\mathrm{d}x} \, \mathrm{d}x = u v - \int v \frac{\mathrm{d}u}{\mathrm{d}x} \, \mathrm{d}x\)

Let \(u = x \implies \frac{\mathrm{d}u}{\mathrm{d}x} = 1\)
Let \(\frac{\mathrm{d}v}{\mathrm{d}x} = \sin(2x) \implies v = -\frac{1}{2}\cos(2x)\)

Applying the formula:
\(\int x \sin(2x) \, \mathrm{d}x = \left[ -\frac{1}{2}x \cos(2x) \right]_0^{\pi/2} - \int_0^{\pi/2} \left( -\frac{1}{2}\cos(2x) \right) \mathrm{d}x\)
\(= \left[ -\frac{1}{2}x \cos(2x) \right]_0^{\pi/2} + \frac{1}{2} \int_0^{\pi/2} \cos(2x) \, \mathrm{d}x\)
\(= \left[ -\frac{1}{2}x \cos(2x) \right]_0^{\pi/2} + \left[ \frac{1}{4}\sin(2x) \right]_0^{\pi/2}\)
\(= \left[ -\frac{1}{2}x \cos(2x) + \frac{1}{4}\sin(2x) \right]_0^{\pi/2}\)

Now evaluate at the limits:
Upper limit \(x = \frac{\pi}{2}\):
\(-\frac{1}{2}\left(\frac{\pi}{2}\right) \cos(\pi) + \frac{1}{4}\sin(\pi) = -\frac{\pi}{4}(-1) + 0 = \frac{\pi}{4}\)

Lower limit \(x = 0\):
\(-\frac{1}{2}(0)\cos(0) + \frac{1}{4}\sin(0) = 0\)

Thus:
\(\int_0^{\pi/2} x \sin(2x) \, \mathrm{d}x = \frac{\pi}{4} - 0 = \frac{\pi}{4}\) (as required).

(b) The formula for the mean value of a function \(f(x)\) over the interval \([a, b]\) is:
\(\text{Mean Value} = \frac{1}{b - a} \int_a^b f(x) \, \mathrm{d}x\)

Here, \(a = 0\) and \(b = \frac{\pi}{2}\).
\(\text{Mean Value} = \frac{1}{\frac{\pi}{2} - 0} \int_0^{\pi/2} x \sin(2x) \, \mathrm{d}x\)
\(\text{Mean Value} = \frac{2}{\pi} \left( \frac{\pi}{4} \right) = \frac{1}{2}\).

(a)
- M1: Applies integration by parts with \(u = x\) and \(v' = \sin(2x)\) to obtain \(uv = -\frac{1}{2}x\cos(2x)\).
- A1: Obtains the correct integrated term \(uv - \int v \, \mathrm{d}u = -\frac{1}{2}x\cos(2x) + \frac{1}{2}\int \cos(2x) \, \mathrm{d}x\).
- A1: Fully integrates to find \(-\frac{1}{2}x\cos(2x) + \frac{1}{4}\sin(2x)\).
- A1: Evaluates limits correctly to show the result is \(\frac{\pi}{4}\) with no errors.

(b)
- M1: Applies the mean value formula \(\frac{1}{b-a} \int_a^b f(x) \, \mathrm{d}x\) with \(a = 0\) and \(b = \frac{\pi}{2}\).
- A1: Obtains the correct mean value of \(\frac{1}{2}\) (or equivalent constant).

(a) Let a general point on the line \(y=2x\) be \(\begin{pmatrix} x \\ 2x \end{pmatrix}\). Under the transformation \(T\), the image is:
\[ \begin{pmatrix} a & 2 \\\\ 1 & 4 \end{pmatrix} \begin{pmatrix} x \\\\ 2x \end{pmatrix} = \begin{pmatrix} ax + 4x \\\\ x + 8x \end{pmatrix} = \begin{pmatrix} (a+4)x \\\\ 9x \end{pmatrix} \]
Since the line \(y=2x\) is invariant, the image point must also lie on the line \(y=2x\). Therefore:
\[ 9x = 2(a+4)x \]
Since this must hold for all points on the line (where \(x \neq 0\)):
\[ 9 = 2a + 8 \implies 2a = 1 \implies a = \frac{1}{2} \]

(b) Let another invariant line through the origin be \(y = mx\). Points on this line are of the form \(\begin{pmatrix} x \\ mx \end{pmatrix}\). Under \(T\) with \(a = \frac{1}{2}\):
\[ \begin{pmatrix} 0.5 & 2 \\\\ 1 & 4 \end{pmatrix} \begin{pmatrix} x \\\\ mx \end{pmatrix} = \begin{pmatrix} (0.5 + 2m)x \\\\ (1 + 4m)x \end{pmatrix} \]
For the image to lie on the line \(y=mx\), we require:
\[ (1 + 4m)x = m(0.5 + 2m)x \]
Since \(x \neq 0\):
\[ 1 + 4m = 0.5m + 2m^2 \]
\[ 2m^2 - 3.5m - 1 = 0 \implies 4m^2 - 7m - 2 = 0 \]
Factoring this quadratic equation:
\[ (4m + 1)(m - 2) = 0 \]
This gives \(m = 2\) (which is the given line \(y=2x\)) or \(m = -\frac{1}{4}\).

Thus, the equation of the other invariant line is \(y = -\frac{1}{4}x\) (or \(x + 4y = 0\)).

**Part (a)**
* **M1**: Sets up the transformation of a point on the line \(y=2x\), i.e., \(\mathbf{M}\begin{pmatrix} x \\\\ 2x \end{pmatrix}\) or uses the condition \(y' = 2x'\) with \((x', y') = (ax+2y, x+4y)\).
* **M1**: Uses the condition for invariance to set up an equation in terms of \(a\), such as \(9x = 2(ax+4x)\).
* **A1**: Correctly solves to obtain \(a = \frac{1}{2}\) (or \(0.5\)) with all steps clearly shown.

**Part (b)**
* **M1**: Sets up the invariance condition for a general line \(y=mx\) with \(a = \frac{1}{2}\), obtaining an equation in \(m\), such as \(1+4m = m(0.5+2m)\).
* **M1**: Solves the resulting quadratic equation in \(m\), e.g., \(4m^2 - 7m - 2 = 0\), by factoring or quadratic formula.
* **A1**: Identifies \(m = -\frac{1}{4}\) and states the correct equation \(y = -\frac{1}{4}x\) (or equivalent, e.g., \(x + 4y = 0\)).

(a) To find the equation with roots \(y = 2x\), we can use the substitution \(x = \frac{y}{2}\).
Substituting this into the original cubic equation \(x^3 - 3x^2 + 5x - 4 = 0\):
\[ \left(\frac{y}{2}\right)^3 - 3\left(\frac{y}{2}\right)^2 + 5\left(\frac{y}{2}\right) - 4 = 0 \]
\[ \frac{y^3}{8} - \frac{3y^2}{4} + \frac{5y}{2} - 4 = 0 \]
Multiply the entire equation by 8 to clear the denominators:
\[ y^3 - 6y^2 + 20y - 32 = 0 \]
which is the required equation.

(b) Let the roots of the new equation be \(w = 2x - 1\).
Since \(y = 2x\), we have \(w = y - 1\), which means \(y = w + 1\).
Substituting \(y = w + 1\) into the equation from part (a):
\[ (w + 1)^3 - 6(w + 1)^2 + 20(w + 1) - 32 = 0 \]
Expanding each term:
\[ (w^3 + 3w^2 + 3w + 1) - 6(w^2 + 2w + 1) + 20(w + 1) - 32 = 0 \]
\[ w^3 + 3w^2 + 3w + 1 - 6w^2 - 12w - 6 + 20w + 20 - 32 = 0 \]
Grouping the terms together:
\[ w^3 + (3 - 6)w^2 + (3 - 12 + 20)w + (1 - 6 + 20 - 32) = 0 \]
\[ w^3 - 3w^2 + 11w - 17 = 0 \]
(Note: Any variable name such as \(x\) or \(u\) is acceptable, e.g. \(x^3 - 3x^2 + 11x - 17 = 0\).)

**Part (a)**
* **M1**: Uses a valid method, such as substitution \(x = \frac{y}{2}\) or finding the sum/products of the new roots in terms of the original roots (e.g., \(\sum 2\alpha = 2(3) = 6\)).
* **M1**: Substitutes \(x = \frac{y}{2}\) into the cubic equation or calculates all three symmetric sums of the new roots.
* **A1**: Fully correct algebraic steps showing that \(y^3 - 6y^2 + 20y - 32 = 0\).

**Part (b)**
* **M1**: Identifies a suitable substitution to relate the new roots to the results of part (a), e.g., \(w = y - 1 \implies y = w + 1\), or uses root relationships (e.g., \(\sum (2\alpha - 1) = \sum 2\alpha - 3\)).
* **M1**: Substitutes \(y = w + 1\) into the equation from part (a) and attempts to expand, or correctly computes the coefficients using root relationships.
* **A1**: Obtains the correct cubic equation \(w^3 - 3w^2 + 11w - 17 = 0\) (or equivalent with another variable).