2022 AQA A-Level Further Mathematics 7367 Practice Paper with Answers

An invariant point \(\begin{pmatrix} x \\ y \end{pmatrix}\) under the transformation matrix \(\mathbf{M}\) must satisfy the equation:
\(\mathbf{M}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}\)

This gives:
\(\begin{pmatrix} 5 & -2 \\ 6 & -2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}\)

Multiplying out the matrix yields the system of equations:
1) \(5x - 2y = x \implies 4x - 2y = 0 \implies y = 2x\)
2) \(6x - 2y = y \implies 6x - 3y = 0 \implies y = 2x\)

Both equations simplify to \(y = 2x\). Therefore, the line of invariant points is \(y = 2x\).

B1: Correctly identifies the line of invariant points as \(y = 2x\).

Using the logarithmic definition for inverse hyperbolic sine:
\(\operatorname{arsinh}(x) = \ln\left(x + \sqrt{x^2 + 1}\right)\)

Substituting \(x = \frac{3}{4}\):
\(\operatorname{arsinh}\left(\frac{3}{4}\right) = \ln\left(\frac{3}{4} + \sqrt{\left(\frac{3}{4}\right)^2 + 1}\right)\)
\(= \ln\left(\frac{3}{4} + \sqrt{\frac{9}{16} + 1}\right)\)
\(= \ln\left(\frac{3}{4} + \sqrt{\frac{25}{16}}\right)\)
\(= \ln\left(\frac{3}{4} + \frac{5}{4}\right)\)
\(= \ln\left(\frac{8}{4}\right) = \ln(2)\)

B1: Correctly obtains the exact value \\ln(2).

We are given a cubic equation \(x^3 + px^2 + qx + r = 0\) with roots \(\alpha\), \(\beta\), and \(\gamma\).

From the relations between roots and coefficients of a polynomial:
\(\sum \alpha = \alpha + \beta + \gamma = -p\)
\(\sum \alpha\beta = \alpha\beta + \beta\gamma + \gamma\alpha = q\)
\(\alpha\beta\gamma = -r\)

We are given:
\(\alpha + \beta + \gamma = -3\)
\(\alpha^2 + \beta^2 + \gamma^2 = 17\)

We can use the algebraic identity:
\(\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\)

Substituting the known values into this identity:
\(17 = (-3)^2 - 2q\)
\(17 = 9 - 2q\)
\(2q = 9 - 17\)
\(2q = -8\)
\(q = -4\)

Hence, the correct option is A.

B1: Correctly identifies and applies the identity \(\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2\sum\alpha\beta\) to find \(q = -4\).

We want to find the exact value of \(\sinh(2\ln 2)\).

**Method 1: Using the definition of \(\sinh x\)**
Using the definition \(\sinh x = \frac{e^x - e^{-x}}{2}\):
\(\sinh(2\ln 2) = \sinh(\ln(2^2)) = \sinh(\ln 4)\)

Applying the definition:
\(\sinh(\ln 4) = \frac{e^{\ln 4} - e^{-\ln 4}}{2}\)

Since \(e^{\ln 4} = 4\) and \(e^{-\ln 4} = \frac{1}{4}\):
\(\sinh(\ln 4) = \frac{4 - \frac{1}{4}}{2} = \frac{\frac{15}{4}}{2} = \frac{15}{8}\)

**Method 2: Using double-angle identities**
Using the identity \(\sinh(2x) = 2\sinh x \cosh x\):
\(\sinh(2\ln 2) = 2\sinh(\ln 2)\cosh(\ln 2)\)

Calculating \(\sinh(\ln 2)\) and \(\cosh(\ln 2)\):
\(\sinh(\ln 2) = \frac{2 - \frac{1}{2}}{2} = \frac{3}{4}\)
\(\cosh(\ln 2) = \frac{2 + \frac{1}{2}}{2} = \frac{5}{4}\)

Therefore:
\(\sinh(2\ln 2) = 2 \left(\frac{3}{4}\right)\left(\frac{5}{4}\right) = \frac{30}{16} = \frac{15}{8}\)

Hence, the correct option is A.

B1: Correctly evaluates \(\sinh(2\ln 2) = \frac{15}{8}\) using either the exponential definition of hyperbolic sine or hyperbolic identities.

**(a)**
**Base step:** For \(n=1\):
\((\cos\theta + \mathrm{i}\sin\theta)^1 = \cos(1\theta) + \mathrm{i}\sin(1\theta) = \cos\theta + \mathrm{i}\sin\theta\).
So the statement is true for \(n=1\).

**Inductive step:** Assume the statement is true for \(n=k\), where \(k \ge 1\) is an integer.
That is, \((\cos\theta + \mathrm{i}\sin\theta)^k = \cos(k\theta) + \mathrm{i}\sin(k\theta)\).
We want to show it is true for \(n=k+1\):
\((\cos\theta + \mathrm{i}\sin\theta)^{k+1} = (\cos\theta + \mathrm{i}\sin\theta)^k (\cos\theta + \mathrm{i}\sin\theta)\)
\(= (\cos k\theta + \mathrm{i}\sin k\theta)(\cos\theta + \mathrm{i}\sin\theta)\)
\(= (\cos k\theta\cos\theta - \sin k\theta\sin\theta) + \mathrm{i}(\sin k\theta\cos\theta + \cos k\theta\sin\theta)\)
Using the trigonometric addition formulae:
\(\cos(A+B) = \cos A\cos B - \sin A\sin B\) and \(\sin(A+B) = \sin A\cos B + \cos A\sin B\),
\(= \cos(k\theta + \theta) + \mathrm{i}\sin(k\theta + \theta)\)
\(= \cos((k+1)\theta) + \mathrm{i}\sin((k+1)\theta)\).

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

**(b)**
By de Moivre's theorem, \(\cos(5\theta) + \mathrm{i}\sin(5\theta) = (\cos\theta + \mathrm{i}\sin\theta)^5\).
Using the Binomial Theorem to expand:
\((\cos\theta + \mathrm{i}\sin\theta)^5 = \cos^5\theta + 5\mathrm{i}\cos^4\theta\sin\theta - 10\cos^3\theta\sin^2\theta - 10\mathrm{i}\cos^2\theta\sin^3\theta + 5\cos\theta\sin^4\theta + \mathrm{i}\sin^5\theta\).

Equating the imaginary parts:
\(\sin(5\theta) = 5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta\).

Using \(\cos^2\theta = 1 - \sin^2\theta\):
\(\sin(5\theta) = 5(1-\sin^2\theta)^2\sin\theta - 10(1-\sin^2\theta)\sin^3\theta + \sin^5\theta\)
\(= 5(1 - 2\sin^2\theta + \sin^4\theta)\sin\theta - 10\sin^3\theta + 10\sin^5\theta + \sin^5\theta\)
\(= 5\sin\theta - 10\sin^3\theta + 5\sin^5\theta - 10\sin^3\theta + 11\sin^5\theta\)
\(= 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta\).

**(c)**
Let \(\sin(5\theta) = 0\).
For \(0 < \theta < \pi\), \(5\theta\) can be \(\pi, 2\pi, 3\pi, 4\pi\), so \(\theta = \frac{\pi}{5}, \frac{2\pi}{5}, \frac{3\pi}{5}, \frac{4\pi}{5}\).
From (b), \(\sin(5\theta) = \sin\theta(16\sin^4\theta - 20\sin^2\theta + 5) = 0\).
Since \(0 < \theta < \pi\), \(\sin\theta \neq 0\).
Thus, the equation \(16x^4 - 20x^2 + 5 = 0\) has roots \(x = \sin\theta\) where \(\theta \in \left\{\frac{\pi}{5}, \frac{2\pi}{5}, \frac{3\pi}{5}, \frac{4\pi}{5}\right\}\).
The positive values of these are \(\sin\left(\frac{\pi}{5}\right)\) and \(\sin\left(\frac{2\pi}{5}\right)\) (since \(\sin\left(\frac{3\pi}{5}\right) = \sin\left(\frac{2\pi}{5}\right)\) and \(\sin\left(\frac{4\pi}{5}\right) = \sin\left(\frac{\pi}{5}\right)\)).
Thus, the exact positive roots of \(16x^4 - 20x^2 + 5 = 0\) are \(\sin\left(\frac{\pi}{5}\right)\) and \(\sin\left(\frac{2\pi}{5}\right)\).

**(a)**
**M1**: Verifies the base case \(n=1\).
**M1**: Sets up inductive hypothesis assuming the statement is true for \(n=k\).
**M1**: Multiplies the expression for \(n=k\) by \((\cos\theta + \mathrm{i}\sin\theta)\) to progress towards \(n=k+1\).
**A1**: Employs trigonometric compound angle identities correctly to obtain \(\cos((k+1)\theta) + \mathrm{i}\sin((k+1)\theta)\).
**A1**: Provides a clear and complete inductive proof conclusion.

**(b)**
**M1**: Applies Binomial expansion to expand \((\cos\theta + \mathrm{i}\sin\theta)^5\).
**A1**: Correctly expands with correct powers of \(\mathrm{i}\).
**M1**: Equates imaginary parts to find the expression for \(\sin(5\theta)\).
**M1**: Applies the Pythagorean identity \(\cos^2\theta = 1 - \sin^2\theta\) to rewrite the expression solely in terms of \(\sin\theta\).
**A1**: Achieves the given identity through correct algebraic simplification.

**(c)**
**M1**: Solves \(\sin(5\theta) = 0\) to find possible values of \(\theta\) and links roots \(x\) to \(\sin\theta\).
**A1**: Identifies the unique positive roots as \(\sin\left(\frac{\pi}{5}\right)\) and \(\sin\left(\frac{2\pi}{5}\right)\).

**(a)**
First, we solve the homogeneous equation:
\[ \frac{\mathrm{d}^2 x}{\mathrm{d} t^2} + 4\frac{\mathrm{d} x}{\mathrm{d} t} + 5x = 0 \]
The auxiliary equation is:
\[ r^2 + 4r + 5 = 0 \]
Solving for \(r\):
\[ (r+2)^2 + 1 = 0 \implies r = -2 \pm \mathrm{i} \]
Therefore, the complementary function (CF) is:
\[ x_c(t) = \mathrm{e}^{-2t}(A\cos t + B\sin t) \]

Next, we find the particular integral (PI). Since the right-hand side is \(10\cos t\), we try:
\[ x_p(t) = P\cos t + Q\sin t \]
Differentiating:
\[ \frac{\mathrm{d} x_p}{\mathrm{d} t} = -P\sin t + Q\cos t \]
\[ \frac{\mathrm{d}^2 x_p}{\mathrm{d} t^2} = -P\cos t - Q\sin t \]

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

By equating coefficients:
For \(\cos t\): \(4P + 4Q = 10 \implies P + Q = \frac{5}{2}\)
For \(\sin t\): \(4Q - 4P = 0 \implies P = Q\)

Solving these simultaneously:
\[ 2P = \frac{5}{2} \implies P = \frac{5}{4}, \quad Q = \frac{5}{4} \]
So the particular integral is:
\[ x_p(t) = \frac{5}{4}\cos t + \frac{5}{4}\sin t \]

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

**(b)**
We are given that at \(t = 0\), \(x = 1\) and \(\frac{\mathrm{d} x}{\mathrm{d} t} = 0\).
Using \(x(0) = 1\):
\[ 1 = \mathrm{e}^{0}(A\cos 0 + B\sin 0) + \frac{5}{4}\cos 0 + \frac{5}{4}\sin 0 \]
\[ 1 = A + \frac{5}{4} \implies A = -\frac{1}{4} \]

Now differentiate the general solution to find \(\frac{\mathrm{d} x}{\mathrm{d} t}\):
\[ \frac{\mathrm{d} x}{\mathrm{d} t} = -2\mathrm{e}^{-2t}(A\cos t + B\sin t) + \mathrm{e}^{-2t}(-A\sin t + B\cos t) - \frac{5}{4}\sin t + \frac{5}{4}\cos t \]
Using \(\frac{\mathrm{d} x}{\mathrm{d} t} = 0\) at \(t = 0\):
\[ 0 = -2(A) + B + \frac{5}{4} \]
Substitute \(A = -\frac{1}{4}\):
\[ 0 = -2\left(-\frac{1}{4}\right) + B + \frac{5}{4} \]
\[ 0 = \frac{1}{2} + B + \frac{5}{4} \implies B + \frac{7}{4} = 0 \implies B = -\frac{7}{4} \]

Thus, the particular solution is:
\[ x(t) = \mathrm{e}^{-2t}\left(-\frac{1}{4}\cos t - \frac{7}{4}\sin t\right) + \frac{5}{4}\cos t + \frac{5}{4}\sin t \]

**(c)**
As \(t \to \infty\), the term \(\mathrm{e}^{-2t} \to 0\).
Therefore, the transient term \(\mathrm{e}^{-2t}\left(-\frac{1}{4}\cos t - \frac{7}{4}\sin t\right) \to 0\).
The long-term displacement of the particle is given by:
\[ x(t) \approx \frac{5}{4}\cos t + \frac{5}{4}\sin t \]
This represents a sustained sinusoidal oscillation with constant amplitude.

**(a)**
**M1**: Sets up auxiliary equation \(r^2 + 4r + 5 = 0\).
**A1**: Resolves auxiliary equation to find complex roots \(r = -2 \pm \mathrm{i}\).
**A1**: Expresses correct complementary function \(x_c = \mathrm{e}^{-2t}(A\cos t + B\sin t)\).
**M1**: Identifies appropriate trial particular integral \(x_p = P\cos t + Q\sin t\) and differentiates.
**M1**: Substitutes into the differential equation and compares coefficients of \(\cos t\) and \(\sin t\).
**A1**: Obtains equations \(4P+4Q=10\) and \(4Q-4P=0\).
**A1**: Solves coefficients correctly to find \(P = \frac{5}{4}\) and \(Q = \frac{5}{4}\).
**B1**: Combines CF and PI to write the general solution.

**(b)**
**M1**: Applies initial value \(x(0) = 1\) to determine \(A\).
**A1**: Finds \(A = -\frac{1}{4}\).
**M1**: Differentiates general solution and applies \(\dot{x}(0) = 0\) to obtain \(B\).
**A1**: Correctly identifies \(B = -\frac{7}{4}\) and writes down the full particular solution.

**(c)**
**B1**: Explains that \(\mathrm{e}^{-2t} \to 0\) as \(t \to \infty\), hence the motion approaches a sustained oscillation modelled by \(x(t) \to \frac{5}{4}\cos t + \frac{5}{4}\sin t\).

(a) Write the system of equations in matrix form \(\mathbf{M}\mathbf{x} = \mathbf{c}\), where \(\mathbf{M} = \begin{pmatrix} 1 & 1 & 2 \\ 2 & a & 1 \\ a & 2 & 1 \end{pmatrix}\). For a unique solution, we require \(\det(\mathbf{M}) \neq 0\). Let us calculate the determinant: \(\det(\mathbf{M}) = 1(a - 2) - 1(2 - a) + 2(4 - a^2) = a - 2 - 2 + a + 8 - 2a^2 = -2a^2 + 2a + 4 = -2(a^2 - a - 2) = -2(a - 2)(a + 1)\). For a unique solution, \(\det(\mathbf{M}) \neq 0 \implies -2(a - 2)(a + 1) \neq 0 \implies a \neq 2\) and \(a \neq -1\).

(b)(i) When \(a = 2\), the second and third equations are: \(2x + 2y + z = b\) and \(2x + 2y + z = 1\). For these equations to be consistent, the left-hand sides are identical, so we must have \(b = 1\).

(b)(ii) For \(a = 2\) and \(b = 1\), the system of equations reduces to: \(1) x + y + 2z = 3\) and \(2) 2x + 2y + z = 1\). Let \(x = t\). Then equation (2) gives \(2t + 2y + z = 1 \implies z = 1 - 2t - 2y\). Substituting this into (1) gives: \(t + y + 2(1 - 2t - 2y) = 3 \implies t + y + 2 - 4t - 4y = 3 \implies -3t - 3y + 2 = 3 \implies -3y = 3t + 1 \implies y = -t - \frac{1}{3}\). Substituting this back into the expression for \(z\): \(z = 1 - 2t - 2(-t - \frac{1}{3}) = 1 - 2t + 2t + \frac{2}{3} = \frac{5}{3}\). Thus, the general solution is: \(x = t\), \(y = -t - \frac{1}{3}\), \(z = \frac{5}{3}\).

(c)(i) When \(a = -1\), the normal vectors to the three planes are \(\mathbf{n}_1 = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}\), \(\mathbf{n}_2 = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}\), and \(\mathbf{n}_3 = \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix}\). Since no two normal vectors are parallel, no two planes are parallel. Since the determinant is zero and \(b \neq 2\), the system is inconsistent. Therefore, the three planes form a triangular prism.

(c)(ii) Since the system of equations is inconsistent, there are no points of intersection common to all three planes. However, any two of the planes will intersect in a line, yielding three parallel lines of intersection (representing the edges of the prism).

(a)
* M1: Attempts to find the determinant of \(\mathbf{M}\) by expanding.
* A1: Obtains the correct simplified quadratic expression \(-2a^2 + 2a + 4\) or equivalent.
* A1: Solves \(\det(\mathbf{M}) = 0\) to find \(a = 2\) and \(a = -1\) and concludes with the correct conditions for a unique solution.

(b)(i)
* M1: Substitutes \(a = 2\) into the equations and compares the second and third equations.
* A1: Deduces \(b = 1\) with clear justification.

(b)(ii)
* M1: Introduces a parameter \(t\) and attempts to solve the reduced system of two equations.
* A1: Obtains the correct expression for one variable in terms of \(t\) (e.g. \(y = -t - \frac{1}{3}\)).
* A1: Obtains the correct full general solution, e.g. \(x = t, y = -t - \frac{1}{3}, z = \frac{5}{3}\).

(c)(i)
* M1: Notes that no two normal vectors are parallel (i.e. no two planes are parallel).
* A1: Identifies the geometric arrangement of the three planes as a triangular prism.

(c)(ii)
* M1: Explains that because the system is inconsistent, there is no common point of intersection for all three planes.
* A1: States that there are three parallel lines representing pairwise intersections between the planes.

(a) By de Moivre's theorem: \(\cos(5\theta) + \mathrm{i}\sin(5\theta) = (\cos\theta + \mathrm{i}\sin\theta)^5\).
Expanding using the binomial theorem: \((\cos\theta + \mathrm{i}\sin\theta)^5 = \cos^5\theta + 5\mathrm{i}\cos^4\theta\sin\theta - 10\cos^3\theta\sin^2\theta - 10\mathrm{i}\cos^2\theta\sin^3\theta + 5\cos\theta\sin^4\theta + \mathrm{i}\sin^5\theta\).
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 \(\cos^2\theta = 1 - \sin^2\theta\): \(\sin(5\theta) = 5(1 - \sin^2\theta)^2\sin\theta - 10(1 - \sin^2\theta)\sin^3\theta + \sin^5\theta\)
\(= 5(1 - 2\sin^2\theta + \sin^4\theta)\sin\theta - 10\sin^3\theta + 10\sin^5\theta + \sin^5\theta\)
\(= 5\sin\theta - 10\sin^3\theta + 5\sin^5\theta - 10\sin^3\theta + 11\sin^5\theta = 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta\). (proved)

(b) Let \(x = \sin\theta\). Then the equation \(16x^4 - 20x^2 + 5 = 0\) is equivalent to the equation \(16\sin^4\theta - 20\sin^2\theta + 5 = 0\).
Multiplying by \(\sin\theta\) gives \(16\sin^5\theta - 20\sin^3\theta + 5\sin\theta = 0\), which from part (a) is \(\sin(5\theta) = 0\).
We seek non-zero roots, so \(\sin\theta \neq 0\).
\(\sin(5\theta) = 0 \implies 5\theta = k\pi \implies \theta = \frac{k\pi}{5}\) for \(k \in \mathbb{Z}\).
To obtain four distinct non-zero values of \(x = \sin\theta\), we choose \(k = -2, -1, 1, 2\).
This yields: \(x = \sin\left(\frac{\pi}{5}\right)\), \(x = \sin\left(\frac{2\pi}{5}\right)\), \(x = \sin\left(-\frac{\pi}{5}\right) = -\sin\left(\frac{\pi}{5}\right)\), and \(x = \sin\left(-\frac{2\pi}{5}\right) = -\sin\left(\frac{2\pi}{5}\right)\).
So the solutions are \(x = \pm \sin\left(\frac{\pi}{5}\right)\) and \(x = \pm \sin\left(\frac{2\pi}{5}\right)\).

(c) The product of the roots of the quartic equation \(16x^4 - 20x^2 + 5 = 0\) is given by \(\frac{E}{A} = \frac{5}{16}\).
The product of the four roots found in part (b) is: \(\left(\sin\left(\frac{\pi}{5}\right)\right)\left(-\sin\left(\frac{\pi}{5}\right)\right)\left(\sin\left(\frac{2\pi}{5}\right)\right)\left(-\sin\left(\frac{2\pi}{5}\right)\right) = \sin^2\left(\frac{\pi}{5}\right)\sin^2\left(\frac{2\pi}{5}\right)\).
Therefore, \(\sin^2\left(\frac{\pi}{5}\right)\sin^2\left(\frac{2\pi}{5}\right) = \frac{5}{16}\).
Since \(0 < \frac{\pi}{5} < \frac{\pi}{2}\) and \(0 < \frac{2\pi}{5} < \frac{\pi}{2}\), both \(\sin\left(\frac{\pi}{5}\right)\) and \(\sin\left(\frac{2\pi}{5}\right)\) are positive.
Taking the positive square root of both sides gives: \(\sin\left(\frac{\pi}{5}\right)\sin\left(\frac{2\pi}{5}\right) = \sqrt{\frac{5}{16}} = \frac{\sqrt{5}}{4}\). (proved)

(a)
* M1: Applies de Moivre's theorem to express \(\cos(5\theta) + \mathrm{i}\sin(5\theta)\).
* A1: Correctly expands binomial expression and identifies the imaginary part.
* M1: Uses the trigonometric identity \(\cos^2\theta = 1 - \sin^2\theta\) to eliminate \(\cos\theta\).
* A1: Correctly expands and simplifies the algebraic expression.
* A1: Arrives at the given identity with no algebraic errors.

(b)
* M1: Identifies the link between the quartic equation and part (a) by setting \(x = \sin\theta\).
* A1: Solves \(\sin(5\theta) = 0\) to find \(\theta = \frac{k\pi}{5}\).
* M1: Correctly selects appropriate values of \(k\) to produce four distinct roots.
* A1: States the exact roots clearly as \(x = \pm \sin\left(\frac{\pi}{5}\right)\) and \(x = \pm \sin\left(\frac{2\pi}{5}\right)\).

(c)
* M1: Applies Vieta's formulas to state that the product of the roots is \(\frac{5}{16}\).
* A1: Sets up the equation \(\sin^2\left(\frac{\pi}{5}\right)\sin^2\left(\frac{2\pi}{5}\right) = \frac{5}{16}\).
* A1: Resolves the signs correctly (explaining that both sines are positive) to conclude that \(\sin\left(\frac{\pi}{5}\right)\sin\left(\frac{2\pi}{5}\right) = \frac{\sqrt{5}}{4}\).

**Part (a)**
Using the product rule for $y = x^2 \text{e}^{3x}$:
$$\frac{\text{d}y}{\text{d}x} = 2x \text{e}^{3x} + 3x^2 \text{e}^{3x} = \text{e}^{3x}(3x^2 + 2x)$$

Differentiating again using the product rule:
$$\frac{\text{d}^2 y}{\text{d}x^2} = (6x+2)\text{e}^{3x} + 3(3x^2 + 2x)\text{e}^{3x}$$
$$\frac{\text{d}^2 y}{\text{d}x^2} = \text{e}^{3x}(9x^2 + 12x + 2)$$

**Part (b)**
Let $P(n)$ be the proposition that:
$$\frac{\text{d}^n}{\text{d}x^n}(x^2 \text{e}^{3x}) = 3^{n-2} \text{e}^{3x} \big(9x^2 + 6nx + n(n-1)\big)$$

**Base Case: $n = 1$**
LHS: $\frac{\text{d}}{\text{d}x}(x^2 \text{e}^{3x}) = \text{e}^{3x}(3x^2 + 2x)$
RHS for $n = 1$:
$$3^{1-2}\text{e}^{3x}\big(9x^2 + 6(1)x + 1(1-1)\big) = \frac{1}{3}\text{e}^{3x}(9x^2 + 6x) = \text{e}^{3x}(3x^2 + 2x)$$
Since LHS = RHS, $P(1)$ is true.

**Inductive Step**
Assume that $P(k)$ is true for some positive integer $k$. That is:
$$\frac{\text{d}^k}{\text{d}x^k}(x^2 \text{e}^{3x}) = 3^{k-2} \text{e}^{3x} \big(9x^2 + 6kx + k(k-1)\big)$$

We must show that this implies $P(k+1)$ is true, i.e.:
$$\frac{\text{d}^{k+1}}{\text{d}x^{k+1}}(x^2 \text{e}^{3x}) = 3^{k-1} \text{e}^{3x} \big(9x^2 + 6(k+1)x + (k+1)k\big)$$

Now, differentiating the expression for the $k$-th derivative:
$$\frac{\text{d}^{k+1}}{\text{d}x^{k+1}}(x^2 \text{e}^{3x}) = \frac{\text{d}}{\text{d}x} \left[ 3^{k-2} \text{e}^{3x} \big(9x^2 + 6kx + k(k-1)\big) \right]$$
Applying the product rule:
$$= 3^{k-2} \cdot 3\text{e}^{3x}\big(9x^2 + 6kx + k(k-1)\big) + 3^{k-2}\text{e}^{3x}\big(18x + 6k\big)$$
Factor out $3^{k-2} \text{e}^{3x}$:
$$= 3^{k-2}\text{e}^{3x} \left[ 3\big(9x^2 + 6kx + k(k-1)\big) + (18x + 6k) \right]$$
Expand inside the square brackets:
$$= 3^{k-2}\text{e}^{3x} \left[ 27x^2 + 18kx + 3k^2 - 3k + 18x + 6k \right]$$
$$= 3^{k-2}\text{e}^{3x} \left[ 27x^2 + (18k+18)x + 3k^2 + 3k \right]$$
Factor out a $3$ from the brackets:
$$= 3^{k-2} \cdot 3 \text{e}^{3x} \left[ 9x^2 + 6(k+1)x + k(k+1) \right]$$
$$= 3^{k-1} \text{e}^{3x} \left[ 9x^2 + 6(k+1)x + (k+1)k \right]$$
This is the statement $P(k+1)$.

**Conclusion**
Since $P(1)$ is true, and if $P(k)$ is true then $P(k+1)$ is also true, by mathematical induction the result is true for all positive integers $n \ge 1$.

- **Part (a):**
- **M1:** Applies product rule to find the first derivative.
- **A1:** Correct first derivative: $\text{e}^{3x}(3x^2 + 2x)$ (or equivalent).
- **A1:** Correct second derivative: $\text{e}^{3x}(9x^2 + 12x + 2)$ (or equivalent).

- **Part (b):**
- **B1:** Verifies the base case $n = 1$ correctly.
- **M1:** States the induction hypothesis clearly (assuming $P(k)$ is true for some positive integer $k$).
- **M1:** Recognizes that $\frac{\text{d}^{k+1}}{\text{d}x^{k+1}} = \frac{\text{d}}{\text{d}x}\left(\frac{\text{d}^k}{\text{d}x^k}\right)$.
- **M1:** Differentiates the assumed formula using the product rule.
- **A1:** Obtains $3^{k-2}\text{e}^{3x} [ 3(9x^2 + 6kx + k(k-1)) + (18x + 6k) ]$ or equivalent.
- **A1:** Collects terms inside brackets to get $27x^2 + 18(k+1)x + 3k(k+1)$.
- **A1:** Completes the algebraic manipulation to factor out $3$ and obtain the correct expression for $n = k+1$.
- **R1:** Provides a complete and clear conclusion stating that $P(1)$ is true and $P(k) \implies P(k+1)$, hence true for all integers $n \ge 1$ by induction.

**Part (a)**
We find the determinant by expanding along the third row (or any other method):
$$\det(\mathbf{M}) = a\big((-2)(-2) - 1(1)\big) - (-3)\big(1(-2) - 1(3)\big) + 0$$
$$\det(\mathbf{M}) = a(4 - 1) + 3(-2 - 3) = 3a - 15$$

For $\mathbf{M}$ to be singular, we require $\det(\mathbf{M}) = 0$:
$$3a - 15 = 0 \implies a = 5$$

**Part (b)**
For $a = 2$, we have:
$$\mathbf{M} = \begin{pmatrix} 1 & -2 & 1 \\ 3 & 1 & -2 \\ 2 & -3 & 0 \end{pmatrix}$$
$$\det(\mathbf{M}) = 3(2) - 15 = -9$$

Next, we find the matrix of cofactors, $\mathbf{C}$:
- $C_{11} = +(0 - 6) = -6$
- $C_{12} = -(0 - (-4)) = -4$
- $C_{13} = +(-9 - 2) = -11$
- $C_{21} = -(0 - (-3)) = -3$
- $C_{22} = +(0 - 2) = -2$
- $C_{23} = -(-3 - (-4)) = -1$
- $C_{31} = +(4 - 1) = 3$
- $C_{32} = -(-2 - 3) = 5$
- $C_{33} = +(1 - (-6)) = 7$

So,
$$\mathbf{C} = \begin{pmatrix} -6 & -4 & -11 \\ -3 & -2 & -1 \\ 3 & 5 & 7 \end{pmatrix}$$

The transpose of the cofactor matrix (adjugate) is:
$$\mathbf{C}^{\text{T}} = \begin{pmatrix} -6 & -3 & 3 \\ -4 & -2 & 5 \\ -11 & -1 & 7 \end{pmatrix}$$

Hence, the inverse matrix is:
$$\mathbf{M}^{-1} = \frac{1}{-9} \begin{pmatrix} -6 & -3 & 3 \\ -4 & -2 & 5 \\ -11 & -1 & 7 \end{pmatrix} = \frac{1}{9} \begin{pmatrix} 6 & 3 & -3 \\ 4 & 2 & -5 \\ 11 & 1 & -7 \end{pmatrix}$$

**Part (c)(i)**
With $a = 5$, the system of equations is:
1) $x - 2y + z = b$
2) $3x + y - 2z = 1$
3) $5x - 3y = 3$

Notice that if we multiply equation (1) by $2$ and add equation (2):
$$2(x - 2y + z) + (3x + y - 2z) = 2b + 1$$
$$5x - 3y = 2b + 1$$

For the system to be consistent, this must match equation (3), which is $5x - 3y = 3$.
Therefore:
$$2b + 1 = 3 \implies 2b = 2 \implies b = 1$$

**Part (c)(ii)**
Using $b = 1$, the equations are:
$x - 2y + z = 1$
$5x - 3y = 3 \implies y = \frac{5}{3}x - 1$

Let $x = 3\lambda$. Then:
$$y = 5\lambda - 1$$
Substitute $x = 3\lambda$ and $y = 5\lambda - 1$ into equation (1):
$$3\lambda - 2(5\lambda - 1) + z = 1$$
$$-7\lambda + 2 + z = 1 \implies z = 7\lambda - 1$$

Thus, the general solution is:
$$\mathbf{r} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ -1 \\ -1 \end{pmatrix} + \lambda \begin{pmatrix} 3 \\ 5 \\ 7 \end{pmatrix}$$

- **Part (a):**
- **M1:** Attempts to find the determinant of $\mathbf{M}$ in terms of $a$ (at least one row expansion term correct).
- **A1:** Obtains $\det(\mathbf{M}) = 3a - 15$ (or equivalent).
- **A1:** Sets $\det(\mathbf{M}) = 0$ and shows $a = 5$ with clear reasoning.

- **Part (b):**
- **M1:** Evaluates $\det(\mathbf{M}) = -9$ for $a = 2$.
- **M1:** Attempts to find the cofactors (at least 4 correct cofactors).
- **A1:** Correct cofactor matrix or its transpose.
- **M1:** Transposes and divides by the determinant.
- **A1:** Fully correct inverse matrix $\mathbf{M}^{-1} = \frac{1}{9} \begin{pmatrix} 6 & 3 & -3 \\ 4 & 2 & -5 \\ 11 & 1 & -7 \end{pmatrix}$.

- **Part (c)(i):**
- **M1:** Attempts to eliminate one variable (e.g., $z$) from two equations, or uses row reduction.
- **A1:** Shows $b = 1$ for consistency.

- **Part (c)(ii):**
- **M1:** Sets up a parameter (e.g., $x = 3\lambda$) and expresses $y$ and $z$ in terms of this parameter.
- **A1:** Expresses the general solution in the required vector form, e.g., $\mathbf{r} = \begin{pmatrix} 0 \\ -1 \\ -1 \end{pmatrix} + \lambda \begin{pmatrix} 3 \\ 5 \\ 7 \end{pmatrix}$ (or any valid equivalent vector equation).

(a) The auxiliary equation is \(r^2 + 4r + 13 = 0\). Solving this quadratic equation: \(r = \frac{-4 \pm \sqrt{16 - 52}}{2} = -2 \pm 3i\). Therefore, the complementary function (CF) is: \(y_c = e^{-2x}(A \cos(3x) + B \sin(3x))\) where \(A\) and \(B\) are arbitrary constants. (b) To find the particular integral, we use \(y_p = x u(x)\) where \(u(x) = e^{-2x}(C \cos(3x) + D \sin(3x))\). We know that \(u(x)\) satisfies the homogeneous equation \(u'' + 4u' + 13u = 0\). Differentiating \(y_p\): \(y_p' = u + x u'\) and \(y_p'' = 2u' + x u''\). Substituting these into the differential equation gives: \(y_p'' + 4y_p' + 13y_p = 2u' + x u'' + 4(u + x u') + 13 x u = 2u' + 4u + x(u'' + 4u' + 13u)\). Since \(u'' + 4u' + 13u = 0\), this simplifies to: \(y_p'' + 4y_p' + 13y_p = 2u' + 4u\). We differentiate \(u(x)\): \(u' = -2e^{-2x}(C \cos(3x) + D \sin(3x)) + 3e^{-2x}(-C \sin(3x) + D \cos(3x)) = e^{-2x}[(-2C + 3D)\cos(3x) - (3C + 2D)\sin(3x)]\). Thus: \(2u' + 4u = 2e^{-2x}[(-2C + 3D)\cos(3x) - (3C + 2D)\sin(3x)] + 4e^{-2x}[C \cos(3x) + D \sin(3x)] = e^{-2x} [ (-4C + 6D + 4C)\cos(3x) + (-6C - 4D + 4D)\sin(3x) ] = e^{-2x} [ 6D \cos(3x) - 6C \sin(3x) ]\). We require \(2u' + 4u = e^{-2x}\sin(3x)\), so: \(6D \cos(3x) - 6C \sin(3x) = \sin(3x)\). Equating coefficients: \(6D = 0 \implies D = 0\) and \(-6C = 1 \implies C = -\frac{1}{6}\). Thus, \(y_p = -\frac{1}{6} x e^{-2x} \cos(3x)\). (c) The general solution is: \(y = e^{-2x}(A \cos(3x) + B \sin(3x)) - \frac{1}{6} x e^{-2x} \cos(3x)\). Using the initial condition \(y(0) = 0\): \(0 = 1(A \cdot 1 + 0) - 0 \implies A = 0\). So, \(y = B e^{-2x} \sin(3x) - \frac{1}{6} x e^{-2x} \cos(3x)\). Differentiating this: \(\frac{\mathrm{d} y}{\mathrm{d} x} = B e^{-2x}(3\cos(3x) - 2\sin(3x)) - \frac{1}{6} e^{-2x} \cos(3x) - \frac{1}{6} x \frac{\mathrm{d}}{\mathrm{d} x}(e^{-2x}\cos(3x))\). Using the condition \(\frac{\mathrm{d} y}{\mathrm{d} x} = 1\) at \(x=0\): \(1 = B(3 - 0) - \frac{1}{6}(1) - 0 \implies 1 = 3B - \frac{1}{6} \implies 3B = \frac{7}{6} \implies B = \frac{7}{18}\). Thus, the particular solution is: \(y = \frac{7}{18} e^{-2x} \sin(3x) - \frac{1}{6} x e^{-2x} \cos(3x)\).

(a) M1: For writing down the auxiliary equation \(r^2 + 4r + 13 = 0\) and attempting to solve it. A1: For finding correct roots \(r = -2 \pm 3i\). A1: For writing down the correct complementary function \(y_c = e^{-2x}(A \cos(3x) + B \sin(3x))\) (allow any constant names). (b) M1: Attempt to differentiate \(y_p = x e^{-2x}(C \cos(3x) + D \sin(3x))\) using product rule. A1: Correct first derivative \(y_p'\) (or equivalent simplified expression). M1: Attempt to find second derivative \(y_p''\). A1: Correctly substitutes \(y_p\), \(y_p'\) and \(y_p''\) into the differential equation and simplifies. M1: Compares coefficients of \(e^{-2x}\cos(3x)\) and \(e^{-2x}\sin(3x)\) to form equations for \(C\) and \(D\). A1: Obtains \(C = -\frac{1}{6}\) and \(D = 0\) (or equivalent). (c) M1: Uses the condition \(y(0) = 0\) to find \(A = 0\). M1: Differentiates their general solution and uses \(\frac{\mathrm{d}y}{\mathrm{d}x} = 1\) at \(x=0\) to find \(B\). A1: Correct particular solution \(y = \frac{7}{18} e^{-2x} \sin(3x) - \frac{1}{6} x e^{-2x} \cos(3x)\) (or any equivalent form).

(a) By de Moivre's theorem: \(\cos(5\theta) + i\sin(5\theta) = (\cos\theta + i\sin\theta)^5\). Using the binomial theorem: \((\cos\theta + i\sin\theta)^5 = \cos^5\theta + 5i\cos^4\theta\sin\theta - 10\cos^3\theta\sin^2\theta - 10i\cos^2\theta\sin^3\theta + 5\cos\theta\sin^4\theta + i\sin^5\theta\). Equating real and imaginary parts: \(\cos(5\theta) = \cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta\) and \(\sin(5\theta) = 5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta\). Since \(\tan(5\theta) = \frac{\sin(5\theta)}{\cos(5\theta)}\): \(\tan(5\theta) = \frac{5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta}{\cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta}\). Dividing both numerator and denominator by \(\cos^5\theta\): \(\tan(5\theta) = \frac{5\tan\theta - 10\tan^3\theta + \tan^5\theta}{1 - 10\tan^2\theta + 5\tan^4\theta}\). (b) If we set \(\tan(5\theta) = 0\), we have \(5\theta = k\pi\) for \(k \in \mathbb{Z}\), which gives \(\theta = \frac{k\pi}{5}\). From the expression in part (a), \(\tan(5\theta) = 0\) implies: \(\frac{5\tan\theta - 10\tan^3\theta + \tan^5\theta}{1 - 10\tan^2\theta + 5\tan^4\theta} = 0\), which means \(5\tan\theta - 10\tan^3\theta + \tan^5\theta = 0\) provided the denominator is not zero. Factorising this expression gives \(\tan\theta (\tan^4\theta - 10\tan^2\theta + 5) = 0\). For \(k = 1, 2, 3, 4\), we have \(\theta = \frac{\pi}{5}, \frac{2\pi}{5}, \frac{3\pi}{5}, \frac{4\pi}{5}\). For these values, \(\tan\theta \neq 0\). Therefore, \(t = \tan\theta\) must satisfy: \(t^4 - 10t^2 + 5 = 0\). The four roots are: \(t_1 = \tan\left(\frac{\pi}{5}\right)\), \(t_2 = \tan\left(\frac{2\pi}{5}\right)\), \(t_3 = \tan\left(\frac{3\pi}{5}\right) = -\tan\left(\frac{2\pi}{5}\right)\), \(t_4 = \tan\left(\frac{4\pi}{5}\right) = -\tan\left(\frac{\pi}{5}\right)\). Hence, the roots of the equation \(t^4 - 10t^2 + 5 = 0\) are indeed \(\pm \tan\left(\frac{\pi}{5}\right)\) and \(\pm \tan\left(\frac{2\pi}{5}\right)\). (c) For a quartic equation of the form \(t^4 + b t^3 + c t^2 + d t + e = 0\), the product of the roots is \(e\). For the equation \(t^4 - 10t^2 + 5 = 0\), the product of the roots is \(5\). Thus, \(\left(\tan\left(\frac{\pi}{5}\right)\right) \left(-\tan\left(\frac{\pi}{5}\right)\right) \left(\tan\left(\frac{2\pi}{5}\right)\right) \left(-\tan\left(\frac{2\pi}{5}\right)\right) = \tan^2\left(\frac{\pi}{5}\right) \tan^2\left(\frac{2\pi}{5}\right) = 5\). Taking the square root of both sides gives \(\tan\left(\frac{\pi}{5}\right) \tan\left(\frac{2\pi}{5}\right) = \pm\sqrt{5}\). Since both \(\frac{\pi}{5}\) and \(\frac{2\pi}{5}\) lie in the first quadrant, their tangents are positive. Thus, the product must be positive: \(\tan\left(\frac{\pi}{5}\right) \tan\left(\frac{2\pi}{5}\right) = \sqrt{5}\).

(a) M1: Writes down the expansion of \((\cos\theta + i\sin\theta)^5\) using the binomial theorem. A1: Correctly simplifies powers of \(i\) to find the real and imaginary parts. M1: Uses \(\tan(5\theta) = \frac{\sin(5\theta)}{\cos(5\theta)}\). M1: Divides numerator and denominator by \(\cos^5\theta\). A1: Fully correct proof leading to the given expression. (b) M1: Equates \(\tan(5\theta) = 0\) and identifies the roots \(\theta = \frac{k\pi}{5}\). M1: Relates \(\tan(5\theta) = 0\) to the algebraic equation \(5t - 10t^3 + t^5 = 0\). A1: Explains why \(t = 0\) is discarded, yielding \(t^4 - 10t^2 + 5 = 0\). A1: Uses symmetry/trigonometric identities to show the roots are \(\pm\tan\left(\frac{\pi}{5}\right)\) and \(\pm\tan\left(\frac{2\pi}{5}\right)\). (c) M1: States that the product of the roots of the quartic equation is 5. M1: Sets up the equation \(\tan^2\left(\frac{\pi}{5}\right) \tan^2\left(\frac{2\pi}{5}\right) = 5\). A1: Deduces that \(\tan\left(\frac{\pi}{5}\right) \tan\left(\frac{2\pi}{5}\right) = \sqrt{5}\) and justifies the positive sign.

By definition, if \(y = \operatorname{artanh}(x)\), then \(x = \tanh(y) = \frac{e^{2y}-1}{e^{2y}+1}\).

Setting \(x = \frac{1}{2}\):
\[\frac{1}{2} = \frac{e^{2y}-1}{e^{2y}+1}\]
\[e^{2y}+1 = 2e^{2y}-2\]
\[e^{2y} = 3\]
\[2y = \ln(3) \implies y = \frac{1}{2}\ln(3)\]

Thus, \(\operatorname{artanh}\left(\frac{1}{2}\right) = \frac{1}{2}\ln(3)\), which corresponds to option A.

B1: Correctly identifies option A.

The standard matrix representing a reflection in the line \(y = x\tan\theta\) is:
\[\mathbf{R} = \begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}\]

Comparing this to the given matrix \(\mathbf{M}\):
\[\cos 2\theta = -0.8 \quad \text{and} \quad \sin 2\theta = 0.6\]

We can find \(\tan 2\theta\):
\[\tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta} = \frac{0.6}{-0.8} = -0.75 = -\frac{3}{4}\]

Using the identity \(\tan 2\theta = \frac{2\tan\theta}{1-\tan^2\theta}\) and letting \(m = \tan\theta\):
\[-\frac{3}{4} = \frac{2m}{1-m^2}\]
\[-3(1-m^2) = 8m \implies 3m^2 - 8m - 3 = 0\]
\[(3m+1)(m-3) = 0\]

Since \(\sin 2\theta > 0\) and \(\cos 2\theta < 0\), the angle \(2\theta\) lies in the second quadrant, meaning \(\theta\) lies in the first quadrant (specifically, between \(45^\circ\) and \(90^\circ\)). Therefore, \(m = \tan\theta\) must be positive, which gives \(m = 3\).

This corresponds to option B.

B1: Correctly identifies option B.

The matrix representing a reflection in the line \(y = x\tan\theta\) is given by:
\[ \mathbf{M} = \begin{pmatrix} \cos 2\theta & \sin 2\theta \\\\\\\\ \sin 2\theta & -\cos 2\theta \end{pmatrix} \]
Here, the line is \(y = 3x\), so \(\tan\theta = 3\).
Using the double-angle identity for cosine:
\[ a = \cos 2\theta = \frac{1 - \tan^2\theta}{1 + \tan^2\theta} \]
Substituting \(\tan\theta = 3\):
\[ a = \frac{1 - 3^2}{1 + 3^2} = \frac{1 - 9}{1 + 9} = \frac{-8}{10} = -0.8 \]

B1: Correctly calculates \(a = -0.8\) using the reflection matrix formulas or equivalent geometric argument.

The equation \(|z - (3 + 4\mathrm{i})| = 2\) represents a circle in the Argand diagram with centre \(C(3, 4)\) and radius \(r = 2\).

The distance from the origin \(O(0,0)\) to the centre \(C\) is:
\[ |3 + 4\mathrm{i}| = \sqrt{3^2 + 4^2} = 5 \]

The maximum distance from the origin to any point \(z\) on the circle is the distance to the centre plus the radius:
\[ |z|_{\text{max}} = 5 + 2 = 7 \]

B1: Obtains the correct maximum value of \(7\) by finding the distance from the origin to the centre of the circle and adding the radius.

For part (a):
Let \( n = 1 \). Then \( \mathbf{M}^1 = \begin{pmatrix} 2(1)+1 & -4(1) \\ 1 & 1-2(1) \end{pmatrix} = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix} \), which is equal to the given matrix. Thus, the statement is true for \( n = 1 \).
Assume the statement is true for \( n = k \), where \( k \) is a positive integer, so that \( \mathbf{M}^k = \begin{pmatrix} 2k+1 & -4k \\ k & 1-2k \end{pmatrix} \).
We consider \( n = k + 1 \):
\( \mathbf{M}^{k+1} = \mathbf{M}^k \mathbf{M} = \begin{pmatrix} 2k+1 & -4k \\ k & 1-2k \end{pmatrix} \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix} \)
Multiplying the matrices:
Top-left element: \( (2k+1)(3) + (-4k)(1) = 6k + 3 - 4k = 2(k+1) + 1 \)
Top-right element: \( (2k+1)(-4) + (-4k)(-1) = -8k - 4 + 4k = -4(k+1) \)
Bottom-left element: \( (k)(3) + (1-2k)(1) = 3k + 1 - 2k = k+1 \)
Bottom-right element: \( (k)(-4) + (1-2k)(-1) = -4k - 1 + 2k = 1 - 2(k+1) \)
Thus, \( \mathbf{M}^{k+1} = \begin{pmatrix} 2(k+1)+1 & -4(k+1) \\ k+1 & 1-2(k+1) \end{pmatrix} \).
Since the statement is true for \( n = 1 \), and if true for \( n = k \) it is also true for \( n = k+1 \), the statement is true for all positive integers \( n \) by mathematical induction.

For part (b):
An invariant point \( \begin{pmatrix} x \\ y \end{pmatrix} \) under the transformation represented by \( \mathbf{M} \) must satisfy:
\( \mathbf{M} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix} \)
\( \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix} \)
This yields the equations:
\( 3x - 4y = x \implies 2x - 4y = 0 \implies y = \frac{1}{2}x \)
\( x - y = y \implies x - 2y = 0 \implies y = \frac{1}{2}x \)
Both equations simplify to the same relation. Therefore, the Cartesian equation of the line of invariant points is \( y = \frac{1}{2}x \).

Part (a):
M1: Evaluates \( \mathbf{M}^1 \) and shows it matches the given matrix.
M1: Formulates a clear induction hypothesis assuming the statement is true for \( n = k \).
M1: Attempts to multiply their assumed matrix \( \mathbf{M}^k \) by \( \mathbf{M} \) (or vice versa).
A1: Shows correct matrix multiplication with algebraic steps for all four elements.
A1: Successfully factors expressions to show the form for \( n = k+1 \).
A1: Writes a complete, coherent mathematical induction conclusion statement.

Part (b):
M1: Sets up the matrix equation \( \mathbf{M}\mathbf{r} = \mathbf{r} \).
A1: Obtains at least one correct simplified linear equation from the system.
A1: Correctly deduces the line of invariant points as \( y = \frac{1}{2}x \) or equivalent.

First, we find the complementary function (CF) of the homogeneous equation:
The auxiliary equation is \( m^2 - 4m + 4 = 0 \implies (m-2)^2 = 0 \), giving a repeated root of \( m = 2 \).
So, the CF is \( y_c = (A + Bx)\mathrm{e}^{2x} \).

Next, we find the particular integral (PI). Since the terms \( \mathrm{e}^{2x} \) and \( x\mathrm{e}^{2x} \) are already present in the CF, we must try a trial PI of the form:
\( y_p = C x^2 \mathrm{e}^{2x} \)
Differentiating \( y_p \):
\( y_p' = 2Cx\mathrm{e}^{2x} + 2Cx^2\mathrm{e}^{2x} \)
\( y_p'' = 2C\mathrm{e}^{2x} + 4Cx\mathrm{e}^{2x} + 4Cx\mathrm{e}^{2x} + 4Cx^2\mathrm{e}^{2x} = (2C + 8Cx + 4Cx^2)\mathrm{e}^{2x} \)
Substituting these into the original differential equation:
\( (2C + 8Cx + 4Cx^2)\mathrm{e}^{2x} - 4(2Cx + 2Cx^2)\mathrm{e}^{2x} + 4(Cx^2\mathrm{e}^{2x}) = 6\mathrm{e}^{2x} \)
Simplifying the coefficient on the left-hand side:
\( 2C + 8Cx + 4Cx^2 - 8Cx - 8Cx^2 + 4Cx^2 = 2C \)
Thus, \( 2C = 6 \implies C = 3 \).
The particular integral is \( y_p = 3x^2\mathrm{e}^{2x} \).

The general solution is:
\( y = (A + Bx + 3x^2)\mathrm{e}^{2x} \)

Now, we apply the initial conditions to find the constants \( A \) and \( B \).
At \( x = 0 \), \( y = 1 \):
\( 1 = (A + 0 + 0)\mathrm{e}^0 \implies A = 1 \)

Differentiating the general solution to apply the second condition:
\( y' = (B + 6x)\mathrm{e}^{2x} + 2(A + Bx + 3x^2)\mathrm{e}^{2x} \)
At \( x = 0 \), \( y' = 4 \):
\( 4 = (B + 0)\mathrm{e}^0 + 2(A + 0 + 0)\mathrm{e}^0 \implies 4 = B + 2A \)
Since \( A = 1 \):
\( 4 = B + 2 \implies B = 2 \)

Therefore, the particular solution is:
\( y = (1 + 2x + 3x^2)\mathrm{e}^{2x} \)

M1: Solves auxiliary equation to find the repeated root \( m = 2 \).
A1: Obtains the correct CF: \( (A + Bx)\mathrm{e}^{2x} \).
M1: Identifies the correct form of the PI: \( y_p = C x^2 \mathrm{e}^{2x} \) due to the repeated roots.
M1: Differentiates the trial PI twice and substitutes into the differential equation.
A1: Obtains the correct value of the parameter \( C = 3 \).
A1: States the correct general solution: \( y = (A + Bx + 3x^2)\mathrm{e}^{2x} \).
M1: Applies the initial condition \( y(0) = 1 \) to obtain \( A = 1 \).
M1: Correctly differentiates their general solution and applies \( y'(0) = 4 \).
A1: Finds \( B = 2 \).
A1: Concludes with the correct particular solution \( y = (1 + 2x + 3x^2)\mathrm{e}^{2x} \).

**(a)**
First, we establish the base case for \( n = 1 \).
LHS = \( (2(1)-1)3^{1-1} = 1 \cdot 3^0 = 1 \).
RHS = \( (1-1)3^1 + 1 = 0 + 1 = 1 \).
Since LHS = RHS, the statement is true for \( n = 1 \).

Next, assume the statement is true for \( n = k \), where \( k \) is a positive integer:
\(\sum_{r=1}^{k} (2r-1)3^{r-1} = (k-1)3^k + 1\)

Now, we consider the case \( n = k+1 \):
\(\sum_{r=1}^{k+1} (2r-1)3^{r-1} = \sum_{r=1}^{k} (2r-1)3^{r-1} + (2(k+1)-1)3^{(k+1)-1}\)
Substituting the induction hypothesis:
\( = (k-1)3^k + 1 + (2k+1)3^k \)
Factor out \( 3^k \):
\( = 3^k [ (k-1) + (2k+1) ] + 1 \)
\( = 3^k [ 3k ] + 1 \)
\( = k \cdot 3^{k+1} + 1 \)
\( = ((k+1)-1)3^{k+1} + 1 \)

This is of the required form for \( n = k+1 \).

**Conclusion:** If the statement is true for \( n = k \), then it is true for \( n = k+1 \). Since it is true for \( n = 1 \), the statement is true for all positive integers \( n \) by mathematical induction.

**(b)**
Using the result from part (a):
\(\sum_{r=5}^{20} (2r-1)3^{r-1} = \sum_{r=1}^{20} (2r-1)3^{r-1} - \sum_{r=1}^{4} (2r-1)3^{r-1}\)
\( = \left[ (20-1)3^{20} + 1 \right] - \left[ (4-1)3^4 + 1 \right] \)
\( = 19 \cdot 3^{20} + 1 - (3 \cdot 3^4 + 1) \)
\( = 19 \cdot 3^{20} + 1 - (3^5 + 1) \)
\( = 19 \cdot 3^{20} - 243 \)
Thus, \( a = 19 \) and \( b = -243 \).

**Part (a): [6 Marks]**
* **M1**: Correctly verifies the base case \( n = 1 \), showing LHS = RHS = 1.
* **M1**: States the inductive hypothesis clearly: assume the result holds for \( n = k \).
* **M1**: Attempts to write the sum for \( n = k+1 \) by adding the \( (k+1) \)-th term to the assumed sum.
* **A1**: Simplifies the algebraic expression to obtain \( 3^k(3k) + 1 \).
* **A1**: Reaches the final correct simplified expression \( k \cdot 3^{k+1} + 1 \) and writes it in the form \( ((k+1)-1)3^{k+1} + 1 \).
* **R1**: Provides a complete, logical conclusion addressing the inductive steps (If true for \( k \) then true for \( k+1 \), true for \( n=1 \) so true for all positive integers \( n \)).

**Part (b): [4 Marks]**
* **M1**: Recognizes and writes the required sum as \( \sum_{r=1}^{20} - \sum_{r=1}^{4} \).
* **M1**: Substitutes \( n=20 \) and \( n=4 \) into the formula from part (a).
* **A1**: Simplifies correctly to \( [19 \cdot 3^{20} + 1] - [244] \) or equivalent expression.
* **A1**: Correctly identifies \( a=19 \) and \( b=-243 \) to give the final answer: \( 19 \cdot 3^{20} - 243 \).

**(a)**
First, we find the Complementary Function (CF) by solving the auxiliary equation:
\( m^2 - 4m + 4 = 0 \)
\( (m-2)^2 = 0 \)
This gives a repeated root \( m = 2 \).
Therefore, the Complementary Function is:
\( y_c = (A + Bx)\text{e}^{2x} \)

Next, we find the Particular Integral (PI). Since both \( \text{e}^{2x} \) and \( x\text{e}^{2x} \) are terms in the CF, we must try a PI of the form:
\( y_p = Cx^2\text{e}^{2x} \)

We differentiate \( y_p \) with respect to \( x \):
\( \frac{\text{d}y_p}{\text{d}x} = 2Cx\text{e}^{2x} + 2Cx^2\text{e}^{2x} = 2C(x + x^2)\text{e}^{2x} \)
\( \frac{\text{d}^2 y_p}{\text{d}x^2} = 2C(1 + 2x)\text{e}^{2x} + 4C(x + x^2)\text{e}^{2x} = 2C(1 + 4x + 2x^2)\text{e}^{2x} \)

Substitute these into the original differential equation:
\( 2C(1 + 4x + 2x^2)\text{e}^{2x} - 4\left[2C(x + x^2)\text{e}^{2x}\right] + 4\left[Cx^2\text{e}^{2x}\right] = 6\text{e}^{2x} \)

Divide through by \( \text{e}^{2x} \):
\( 2C(1 + 4x + 2x^2) - 8C(x + x^2) + 4Cx^2 = 6 \)
\( 2C + 8Cx + 4Cx^2 - 8Cx - 8Cx^2 + 4Cx^2 = 6 \)
\( 2C = 6 \implies C = 3 \)

Thus, the Particular Integral is \( y_p = 3x^2\text{e}^{2x} \).

The general solution is \( y = y_c + y_p \):
\( y = (A + Bx)\text{e}^{2x} + 3x^2\text{e}^{2x} \)

**(b)**
We use the boundary conditions to find the constants \( A \) and \( B \).
Using \( y = 2 \) at \( x = 0 \):
\( 2 = (A + B(0))\text{e}^{0} + 3(0)^2\text{e}^{0} \implies A = 2 \)

Now we differentiate the general solution to find \( \frac{\text{d}y}{\text{d}x} \):
\( y = (2 + Bx + 3x^2)\text{e}^{2x} \)
\( \frac{\text{d}y}{\text{d}x} = (B + 6x)\text{e}^{2x} + 2(2 + Bx + 3x^2)\text{e}^{2x} \)

Using \( \frac{\text{d}y}{\text{d}x} = 5 \) at \( x = 0 \):
\( 5 = (B + 0)\text{e}^{0} + 2(2 + 0)\text{e}^{0} \)
\( 5 = B + 4 \implies B = 1 \)

Therefore, the particular solution is:
\( y = (2 + x + 3x^2)\text{e}^{2x} \)

**Part (a): [7 Marks]**
* **M1**: Sets up and solves the auxiliary equation \( m^2 - 4m + 4 = 0 \) to find the repeated root \( m = 2 \).
* **A1**: Writes down the correct Complementary Function \( y_c = (A+Bx)\text{e}^{2x} \).
* **M1**: Identifies the correct form for the Particular Integral as \( y_p = Cx^2\text{e}^{2x} \).
* **M1**: Correctly applies the product rule to find the first derivative of their proposed \( y_p \).
* **A1**: Obtains correct first and second derivatives \( y_p' \) and \( y_p'' \).
* **M1**: Substitutes derivatives into the differential equation and solves for \( C \).
* **A1**: Obtains the correct general solution: \( y = (A + Bx)\text{e}^{2x} + 3x^2\text{e}^{2x} \).

**Part (b): [3 Marks]**
* **M1**: Applies \( y = 2 \) when \( x = 0 \) to find \( A = 2 \).
* **M1**: Correctly differentiates the general solution and substitutes \( x = 0 \) and \( \frac{\text{d}y}{\text{d}x} = 5 \) to find \( B = 1 \).
* **A1**: States the correct final particular solution: \( y = (2 + x + 3x^2)\text{e}^{2x} \).

**(a)**

**Step 1: Base case**
When \(n = 1\):
LHS \(= \sum_{r=1}^{1} \frac{4r}{3^r} = \frac{4(1)}{3^1} = \frac{4}{3}\)
RHS \(= 3 - \frac{2(1)+3}{3^1} = 3 - \frac{5}{3} = \frac{4}{3}\)
Since \(\text{LHS} = \text{RHS}\), the statement is true for \(n = 1\).

**Step 2: Inductive hypothesis**
Assume that the statement is true for \(n = k\), where \(k \geq 1\) is an integer. That is,
\[ S_k = 3 - \frac{2k+3}{3^k} \]

**Step 3: Inductive step**
We wish to show that the statement holds for \(n = k+1\). That is, we want to prove:
\[ S_{k+1} = 3 - \frac{2(k+1)+3}{3^{k+1}} = 3 - \frac{2k+5}{3^{k+1}} \]
We can write:
\[ S_{k+1} = S_k + \frac{4(k+1)}{3^{k+1}} \]
Using the inductive hypothesis, substitute \(S_k\):
\[ S_{k+1} = 3 - \frac{2k+3}{3^k} + \frac{4k+4}{3^{k+1}} \]
Expressing the fractions over a common denominator of \(3^{k+1}\):
\[ S_{k+1} = 3 - \frac{3(2k+3)}{3^{k+1}} + \frac{4k+4}{3^{k+1}} \]
\[ S_{k+1} = 3 - \frac{3(2k+3) - (4k+4)}{3^{k+1}} \]
\[ S_{k+1} = 3 - \frac{6k+9 - 4k-4}{3^{k+1}} \]
\[ S_{k+1} = 3 - \frac{2k+5}{3^{k+1}} \]
\[ S_{k+1} = 3 - \frac{2(k+1)+3}{3^{k+1}} \]
This is the required form for \(n = k+1\).

**Step 4: Conclusion**
Since the statement is true for \(n = 1\), and if it is true for \(n = k\) then it is also true for \(n = k+1\), then by mathematical induction the statement is true for all positive integers \(n\).

**(b)**
The sum to infinity is defined as \(\lim_{n \to \infty} S_n\).
As \(n \to \infty\), the linear term in the numerator grows much slower than the exponential term in the denominator, so:
\[ \lim_{n \to \infty} \frac{2n+3}{3^n} = 0 \]
Therefore,
\[ \sum_{r=1}^{\infty} \frac{4r}{3^r} = \lim_{n \to \infty} \left( 3 - \frac{2n+3}{3^n} \right) = 3 \]

**(a)**
* **M1:** Evaluates \(S_1\) and shows that \(\text{LHS} = \text{RHS} = \frac{4}{3}\).
* **A1:** Concludes clearly that the base case \(n=1\) holds.
* **M1:** Formulates the inductive hypothesis clearly by stating "assume true for \(n=k\)" or equivalent, and writes down the assumed formula for \(S_k\).
* **M1:** Expresses the relationship \(S_{k+1} = S_k + \frac{4(k+1)}{3^{k+1}}\) (or equivalent).
* **M1:** Substitutes the assumption and attempts to combine the terms over a common denominator of \(3^{k+1}\).
* **A1:** Correctly simplifies the numerator to obtain \(2k+5\).
* **A1:** Re-writes the final fraction in the inductive form \(3 - \frac{2(k+1)+3}{3^{k+1}}\).
* **A1:** Provides a complete and clear inductive conclusion mentioning that the result holds for \(n=1\), and if it holds for \(n=k\) then it holds for \(n=k+1\), hence it holds for all integers \(n \geq 1\). (All previous 7 marks must be awarded to achieve this mark).

**(b)**
* **M1:** Applies the limit as \(n \to \infty\) and states or demonstrates that \(\frac{2n+3}{3^n} \to 0\).
* **A1:** Obtains the exact value of \(3\).

**(a)**
Using integration by parts:
\[ \int u \, \mathrm{d}v = uv - \int v \, \mathrm{d}u \]
Let:
\[ u = x^n \implies \mathrm{d}u = n x^{n-1} \, \mathrm{d}x \]
\[ \mathrm{d}v = e^{-2x} \, \mathrm{d}x \implies v = -\frac{1}{2} e^{-2x} \]
Applying this to \(I_n\):
\[ I_n = \left[ -\frac{1}{2} x^n e^{-2x} \right]_{0}^{1} - \int_{0}^{1} \left( -\frac{1}{2} e^{-2x} \right) \left( n x^{n-1} \right) \mathrm{d}x \]
Now, evaluate the boundary term \(\left[ -\frac{1}{2} x^n e^{-2x} \right]_{0}^{1}\):
- At the upper limit \(x = 1\): \(-\frac{1}{2}(1)^n e^{-2(1)} = -\frac{1}{2e^2}\)
- At the lower limit \(x = 0\) (since \(n \geq 1\)): \(-\frac{1}{2}(0)^n e^0 = 0\)

Thus, the boundary term is \(-\frac{1}{2e^2}\).

Now, simplify the remaining integral term:
\[ -\int_{0}^{1} \left( -\frac{1}{2} e^{-2x} \right) \left( n x^{n-1} \right) \mathrm{d}x = \frac{n}{2} \int_{0}^{1} x^{n-1} e^{-2x} \, \mathrm{d}x = \frac{n}{2} I_{n-1} \]
Combining these components yields:
\[ I_n = -\frac{1}{2e^2} + \frac{n}{2} I_{n-1} \]
Rearranging the terms gives:
\[ I_n = \frac{n}{2} I_{n-1} - \frac{1}{2e^2} \]
which is the required reduction formula.

**(b)**
For \(n = 0\):
\[ I_0 = \int_{0}^{1} e^{-2x} \, \mathrm{d}x \]
\[ I_0 = \left[ -\frac{1}{2} e^{-2x} \right]_{0}^{1} \]
\[ I_0 = -\frac{1}{2} e^{-2} - \left( -\frac{1}{2} e^{0} \right) \]
\[ I_0 = \frac{1}{2} - \frac{1}{2e^2} \]

**(c)**
Apply the reduction formula with \(n = 1\):
\[ I_1 = \frac{1}{2} I_0 - \frac{1}{2e^2} \]
Substitute the expression for \(I_0\) from part (b):
\[ I_1 = \frac{1}{2} \left( \frac{1}{2} - \frac{1}{2e^2} \right) - \frac{1}{2e^2} \]
\[ I_1 = \frac{1}{4} - \frac{1}{4e^2} - \frac{1}{2e^2} = \frac{1}{4} - \frac{3}{4e^2} \]

Now, apply the reduction formula with \(n = 2\):
\[ I_2 = \frac{2}{2} I_1 - \frac{1}{2e^2} = I_1 - \frac{1}{2e^2} \]
Substitute the expression for \(I_1\):
\[ I_2 = \left( \frac{1}{4} - \frac{3}{4e^2} \right) - \frac{1}{2e^2} \]
\[ I_2 = \frac{1}{4} - \frac{5}{4e^2} \]

**(a)**
* **M1:** Sets up integration by parts by assigning \(u = x^n\) and \(\mathrm{d}v = e^{-2x} \mathrm{d}x\), and finding \(\mathrm{d}u\) and \(v\).
* **A1:** Obtains correct term \(v = -\frac{1}{2}e^{-2x}\) and writes down the correct integration by parts expression.
* **M1:** Evaluates the limits \(0\) and \(1\) on \(uv\) correctly, noting that the term at \(x=0\) vanishes because \(n \ge 1\).
* **A1:** Replaces \(\int_{0}^{1} x^{n-1} e^{-2x} \, \mathrm{d}x\) with \(I_{n-1}\) and completes the proof to show the given reduction formula with no errors.

**(b)**
* **M1:** Integrates \(e^{-2x}\) to obtain \(-\frac{1}{2}e^{-2x}\) and applies limits \(0\) and \(1\).
* **A1:** Reaches the exact value \(\frac{1}{2} - \frac{1}{2e^2}\) (or equivalent form like \(\frac{e^2 - 1}{2e^2}\)).

**(c)**
* **M1:** Applies the reduction formula with \(n=1\) utilizing their value of \(I_0\).
* **A1:** Simplifies correctly to find \(I_1 = \frac{1}{4} - \frac{3}{4e^2}\) (or equivalent).
* **M1:** Applies the reduction formula with \(n=2\) utilizing their value of \(I_1\).
* **A1:** Achieves the final exact simplified value of \(\frac{1}{4} - \frac{5}{4e^2}\) (or \(\frac{e^2 - 5}{4e^2}\)).

### Part (a)

**Step 1: Base Case**
For \(n = 1\):
\[u_1 = \frac{5^1 + 2^{1+1}}{3} = \frac{5 + 4}{3} = \frac{9}{3} = 3\]
Since this matches the given value of \(u_1 = 3\), the formula is true for \(n = 1\).

**Step 2: Inductive Hypothesis**
Assume the formula is true for some positive integer \(n = k\), where \(k \ge 1\):
\[u_k = \frac{5^k + 2^{k+1}}{3}\]

**Step 3: Inductive Step**
We wish to show that the formula holds for \(n = k+1\), i.e.,
\[u_{k+1} = \frac{5^{k+1} + 2^{k+2}}{3}\]

Using the given recurrence relation:
\[u_{k+1} = 5u_k - 2^{k+1}\]

Substitute the inductive hypothesis into the recurrence relation:
\[u_{k+1} = 5 \left( \frac{5^k + 2^{k+1}}{3} \right) - 2^{k+1}\]

Express with a common denominator of 3:
\[u_{k+1} = \frac{5 \cdot 5^k + 5 \cdot 2^{k+1} - 3 \cdot 2^{k+1}}{3}\]

Simplify the terms in the numerator:
\[u_{k+1} = \frac{5^{k+1} + (5 - 3) \cdot 2^{k+1}}{3}\]
\[u_{k+1} = \frac{5^{k+1} + 2 \cdot 2^{k+1}}{3}\]
\[u_{k+1} = \frac{5^{k+1} + 2^{k+2}}{3}\]

This is the required formula for \(n = k+1\).

**Step 4: Conclusion**
Since the formula is true for \(n = 1\), and if it is true for \(n = k\) then it is also true for \(n = k+1\), it is true for all integers \(n \ge 1\) by mathematical induction.

---

### Part (b)

We need to find the smallest integer \(n\) such that:
\[u_n > 10^8\]
\[\frac{5^n + 2^{n+1}}{3} > 10^8\]
\[5^n + 2^{n+1} > 3 \times 10^8\]

Since \(5^n\) grows much faster than \(2^{n+1}\), we can approximate the relation to find an initial estimate:
\[5^n \approx 3 \times 10^8\]

Taking base-10 logarithms:
\[n \log_{10}(5) \approx \log_{10}(3) + 8\]
\[n \approx \frac{0.4771 + 8}{0.6990} \approx 12.13\]

Now, test integer values around \(12\):

- For \(n = 12\):
\[u_{12} = \frac{5^{12} + 2^{13}}{3} = \frac{244,140,625 + 8,192}{3} = \frac{244,148,817}{3} = 81,382,939\]
Since \(81,382,939 < 10^8\), \(n = 12\) is not large enough.

- For \(n = 13\):
\[u_{13} = \frac{5^{13} + 2^{14}}{3} = \frac{1,220,703,125 + 16,384}{3} = \frac{1,220,719,509}{3} = 406,906,503\]
Since \(406,906,503 > 10^8\), \(n = 13\) satisfies the inequality.

Thus, the smallest integer value of \(n\) is 13.

**Part (a)**
- **B1**: Verifies the base case \(n=1\) is true, showing clear calculations yielding 3.
- **M1**: States inductive hypothesis clearly (assuming the formula is true for \(n=k\)).
- **M1**: Applies the recurrence relation for \(u_{k+1}\) and substitutes the inductive hypothesis.
- **M1**: Combines the terms over a common denominator of 3.
- **A1**: Correctly simplifies the numerator to obtain \(5^{k+1} + 2^{k+2}\) and completes the inductive step.
- **R1**: Gives a complete and logically sound inductive conclusion containing all key phrases (e.g., "true for \(n=1\)", "if true for \(n=k\) then true for \(n=k+1\)", "true for all positive integers \(n\) by induction").

**Part (b)**
- **M1**: Sets up the correct inequality \(5^n + 2^{n+1} > 3 \times 10^8\) or equivalent equation.
- **M1**: Uses a valid method (such as logarithms on the dominant term or systematic trial and improvement) to approximate \(n\).
- **A1**: Evaluates \(u_{12}\) (or shows \(u_{12} \approx 8.14 \times 10^7 < 10^8\)) and \(u_{13}\) (or shows \(u_{13} \approx 4.07 \times 10^8 > 10^8\)).
- **A1**: Correctly identifies \(n = 13\) as the smallest integer.

### Part (a)

Let a general point on the line \(y = x + c\) be represented by \(\begin{pmatrix} x \\ x+c \end{pmatrix}\).

Applying the transformation matrix \(\mathbf{M}\):
\[\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix} \begin{pmatrix} x \\ x+c \end{pmatrix}\]

This gives:
\[x' = 2x + 3(x + c) = 5x + 3c\]
\[y' = x + 4(x + c) = 5x + 4c\]

We must show that the image point \((x', y')\) also lies on the same line \(y' = x' + c\).
Evaluating the right-hand side:
\[x' + c = (5x + 3c) + c = 5x + 4c\]

Since this is equal to \(y'\), the point \((x', y')\) lies on the line \(y' = x' + c\). Thus, the family of lines \(y = x + c\) is invariant under \(T\).

---

### Part (b)

An invariant line through the origin can be written in the form \(y = mx\).
Let a point on this line be \(\begin{pmatrix} x \\ mx \end{pmatrix}\). Under \(T\), it maps to \(\begin{pmatrix} x' \\ y' \end{pmatrix}\) where:
\[x' = 2x + 3mx = (2+3m)x\]
\[y' = x + 4mx = (1+4m)x\]

For the line to be invariant, the image point must satisfy \(y' = mx'\):
\[(1+4m)x = m(2+3m)x\]

Since this must hold for any point on the line, we can divide by \(x\) (for \(x \neq 0\)):
\[1 + 4m = m(2 + 3m)\]
\[1 + 4m = 2m + 3m^2\]
\[3m^2 - 2m - 1 = 0\]

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

This yields two possible values for the gradient \(m\):
\[m = 1 \quad \text{and} \quad m = -\frac{1}{3}\]

The root \(m = 1\) corresponds to the line \(y = x\) (which is a member of the family of lines from part (a) with \(c=0\)).

Therefore, the other invariant line through the origin is:
\[y = -\frac{1}{3}x\]

---

### Part (c)

Consider a general line \(y = mx + c\). Since the gradient of any invariant line must be one of the eigenvalues' associated directions, \(m\) must be either \(1\) or \(-\frac{1}{3}\).

Let's test \(m = -\frac{1}{3}\) with a non-zero vertical intercept \(c\).
Under the transformation:
\[x' = 2x + 3\left(-\frac{1}{3}x + c\right) = x + 3c\]
\[y' = x + 4\left(-\frac{1}{3}x + c\right) = -\frac{1}{3}x + 4c\]

We check if \(y' = -\frac{1}{3}x' + c\):
\[-\frac{1}{3}x' + c = -\frac{1}{3}(x + 3c) + c = -\frac{1}{3}x - c + c = -\frac{1}{3}x\]

For the line to be invariant, we must have \(y' = -\frac{1}{3}x' + c\), which requires:
\[-\frac{1}{3}x + 4c = -\frac{1}{3}x\]
\[4c = 0 \implies c = 0\]

Thus, any invariant line with gradient \(-\frac{1}{3}\) must have \(c = 0\), meaning it must pass through the origin.

Hence, no other non-origin invariant lines exist under \(T\) other than those with gradient \(1\) (which are \(y = x + c\) as shown in part a).

**Part (a)**
- **M1**: Formulates the transformation of a general point \((x, x+c)\) under \(\mathbf{M}\).
- **M1**: Finds algebraic expressions for \(x' = 5x + 3c\) and \(y' = 5x + 4c\).
- **A1**: Shows clearly that \(y' = x' + c\) by substituting the expressions.
- **A1**: Writes a clear concluding sentence establishing the invariance of \(y = x+c\).

**Part (b)**
- **M1**: Uses the transformation of a point \((x, mx)\) on the line \(y = mx\) to write \(x' = (2+3m)x\) and \(y' = (1+4m)x\).
- **M1**: Forms the equation \(1+4m = m(2+3m)\) and simplifies it to a quadratic equation.
- **A1**: Obtains \(3m^2 - 2m - 1 = 0\) and correctly finds the roots \(m=1\) and \(m=-\frac{1}{3}\).
- **A1**: Selects the other root and states the final equation as \(y = -\frac{1}{3}x\) (or equivalent).

**Part (c)**
- **M1**: Investigates the general line equation \(y = -\frac{1}{3}x + c\) under the transformation and compares constants (e.g. showing that \(4c = 0\) or \(c(1-m)=0\)).
- **A1**: Fully explains that \(c\) must be \(0\) for the gradient \(m = -\frac{1}{3}\), and therefore no other non-origin lines can exist.

(a) By de Moivre's theorem, we have:
\\cos(5\theta) + \mathrm{i}\sin(5\theta) = (\cos\theta + \mathrm{i}\sin\theta)^5\\

Expanding the right-hand side using the binomial theorem:
\(\cos\theta + \mathrm{i}\sin\theta)^5 = \cos^5\theta + 5\mathrm{i}\cos^4\theta\sin\theta - 10\cos^3\theta\sin^2\theta - 10\mathrm{i}\cos^2\theta\sin^3\theta + 5\cos\theta\sin^4\theta + \mathrm{i}\sin^5\theta\\n\nEquating the real parts on both sides:\n\\cos(5\theta) = \cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta\\n\nUsing the identity \)\sin^2\theta = 1 - \cos^2\theta\):
\\\cos(5\theta) = \cos^5\theta - 10\cos^3\theta(1 - \cos^2\theta) + 5\cos\theta(1 - \cos^2\theta)^2\\
\\\cos(5\theta) = \cos^5\theta - 10\cos^3\theta + 10\cos^5\theta + 5\cos\theta(1 - 2\cos^2\theta + \cos^4\theta)\\
\\\cos(5\theta) = 11\cos^5\theta - 10\cos^3\theta + 5\cos\theta - 10\cos^3\theta + 5\cos^5\theta\\
\\\cos(5\theta) = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta\\

(b) Let \(x = \cos\theta\). Substituting this into the given equation yields:
\\16\cos^5\theta - 20\cos^3\theta + 5\cos\theta = \frac{1}{2} \implies \cos(5\theta) = \frac{1}{2}\\

For \(0 \le \theta \le \pi\), we have \(0 \le 5\theta \le 5\pi\).
Solving \(\cos(5\theta) = \frac{1}{2}\) in this interval gives:
\\5\theta = \frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}, \frac{11\pi}{3}, \frac{13\pi}{3}\\
\\\theta = \frac{\pi}{15}, \frac{\pi}{3}, \frac{7\pi}{15}, \frac{11\pi}{15}, \frac{13\pi}{15}\\

Since \(\cos\theta\) is strictly decreasing on \([0, \pi]\), these values provide 5 distinct roots:
\\x = \cos\left(\frac{\pi}{15}\right), \cos\left(\frac{\pi}{3}\right), \cos\left(\frac{7\pi}{15}\right), \cos\left(\frac{11\pi}{15}\right), \cos\left(\frac{13\pi}{15}\right)\\

(c) The polynomial equation is \(16x^5 - 20x^3 + 5x - \frac{1}{2} = 0\).
By Viete's formulas, the product of the roots of a fifth-degree equation is given by \(-\frac{a_0}{a_5}\):
\\x_1 x_2 x_3 x_4 x_5 = -\frac{-1/2}{16} = \frac{1}{32}\\

Substituting the roots found in part (b):
\\\cos\left(\frac{\pi}{15}\right) \cos\left(\frac{\pi}{3}\right) \cos\left(\frac{7\pi}{15}\right) \cos\left(\frac{11\pi}{15}\right) \cos\left(\frac{13\pi}{15}\right) = \frac{1}{32}\\

Since \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\):
\\\frac{1}{2} \cos\left(\frac{\pi}{15}\right) \cos\left(\frac{7\pi}{15}\right) \cos\left(\frac{11\pi}{15}\right) \cos\left(\frac{13\pi}{15}\right) = \frac{1}{32}\\
\\\cos\left(\frac{\pi}{15}\right) \cos\left(\frac{7\pi}{15}\right) \cos\left(\frac{11\pi}{15}\right) \cos\left(\frac{13\pi}{15}\right) = \frac{1}{16}\\

Part (a) [5 marks]:
- M1: Applies de Moivre's theorem to write \(\cos(5\theta) + \mathrm{i}\sin(5\theta) = (\cos\theta + \mathrm{i}\sin\theta)^5\) and begins binomial expansion.
- A1: Obtains the correct expanded real part: \(\cos(5\theta) = \cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta\).
- M1: Uses \(\sin^2\theta = 1 - \cos^2\theta\) to substitute for all \(\sin^2\theta\) terms.
- A1: Expands the algebraic terms correctly.
- A1: Simplifies to successfully show the given expression.

Part (b) [3 marks]:
- M1: Direct substitution of \(x = \cos\theta\) and identifies the equation as \(\cos(5\theta) = \frac{1}{2}\).
- A1: Finds at least 3 correct values of \(5\theta\) or \(\theta\).
- A1: Fully specifies all 5 distinct roots in the correct format with values inside the interval \([0, \pi]\).

Part (c) [2 marks]:
- M1: Applies the product of roots formula to obtain \(\text{Product} = \frac{1}{32}\).
- A1: Uses \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\) to complete the proof cleanly.

(a) To derive the reduction formula, apply integration by parts to \(I_n = \int_0^1 x^n e^{-2x} \, \mathrm{d}x\):
Let \(u = x^n \implies \frac{\mathrm{d}u}{\mathrm{d}x} = n x^{n-1}\\nLet \)\frac{\mathrm{d}v}{\mathrm{d}x} = e^{-2x} \implies v = -\frac{1}{2} e^{-2x}\\\n\nUsing the formula \(\int u \frac{\mathrm{d}v}{\mathrm{d}x} \, \mathrm{d}x = u v - \int v \frac{\mathrm{d}u}{\mathrm{d}x} \, \mathrm{d}x\):
\(I_n = \left[ -\frac{1}{2} x^n e^{-2x} \right]_0^1 - \int_0^1 \left(-\frac{1}{2} e^{-2x}\right) \left(n x^{n-1}\right) \, \mathrm{d}x\\n\)I_n = \left( -\frac{1}{2} (1)^n e^{-2} - 0 \right) + \frac{n}{2} \int_0^1 x^{n-1} e^{-2x} \, \mathrm{d}x\\\n\(I_n = -\frac{1}{2} e^{-2} + \frac{n}{2} I_{n-1}\\n\nMultiplying both sides by 2 yields:\n\)2 I_n = n I_{n-1} - e^{-2}\\\n\n(b) For \(n = 0\):
\(I_0 = \int_0^1 e^{-2x} \, \mathrm{d}x = \left[ -\frac{1}{2} e^{-2x} \right]_0^1\\n\)I_0 = -\frac{1}{2} e^{-2} - \left(-\frac{1}{2} e^{0}\right) = \frac{1}{2} - \frac{1}{2} e^{-2} = \frac{1}{2}(1 - e^{-2})\\\n\n(c) We now apply the reduction formula systematically for \(n = 1, 2, 3\):
For \(n = 1\):
\(2 I_1 = I_0 - e^{-2} = \frac{1}{2}(1 - e^{-2}) - e^{-2} = \frac{1}{2} - \frac{3}{2} e^{-2}\\n\)I_1 = \frac{1}{4} - \frac{3}{4} e^{-2}\\\n\nFor \(n = 2\):
\(2 I_2 = 2 I_1 - e^{-2} = 2\left(\frac{1}{4} - \frac{3}{4} e^{-2}\right) - e^{-2} = \frac{1}{2} - \frac{3}{2} e^{-2} - e^{-2} = \frac{1}{2} - \frac{5}{2} e^{-2}\\n\)I_2 = \frac{1}{4} - \frac{5}{4} e^{-2}\\\n\nFor \(n = 3\):
\(2 I_3 = 3 I_2 - e^{-2} = 3\left(\frac{1}{4} - \frac{5}{4} e^{-2}\right) - e^{-2} = \frac{3}{4} - \frac{15}{4} e^{-2} - e^{-2} = \frac{3}{4} - \frac{19}{4} e^{-2}\\
\)I_3 = \frac{3}{8} - \frac{19}{8} e^{-2}\\

Part (a) [4 marks]:
- M1: Uses integration by parts with correct assignment of \(u\) and \(v\).
- M1: Correctly evaluates the boundary term \(\left[ -\frac{1}{2} x^n e^{-2x} \right]_0^1\).
- A1: Writes down a correct recurrence relation relating \(I_n\) and \(I_{n-1}\).
- A1: Clearly rearranges to show the required equation: \(2 I_n = n I_{n-1} - e^{-2}\).

Part (b) [2 marks]:
- M1: Correctly integrates \(\int_0^1 e^{-2x} \, \mathrm{d}x\).
- A1: Obtains the correct exact value \(\frac{1}{2}(1 - e^{-2})\) or equivalent.

Part (c) [4 marks]:
- M1: Uses the reduction formula with \(n=1\) and their \(I_0\) to find a correct expression for \(I_1\).
- A1: Uses the reduction formula with \(n=2\) to find the correct expression for \(I_2 = \frac{1}{4} - \frac{5}{4}e^{-2}\).
- M1: Uses the reduction formula with \(n=3\) to obtain an equation for \(I_3\).
- A1: Obtains the correct exact value of \(I_3 = \frac{3}{8} - \frac{19}{8} e^{-2}\) (or equivalent single fraction).