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

(i) From the given cubic equation:
\(\sum \alpha = 4\)
\(\sum \alpha\beta = 2\)
\(\alpha\beta\gamma = 1\)

Using the identity:
\(\sum \alpha^2 = (\sum \alpha)^2 - 2\sum \alpha\beta = 4^2 - 2(2) = 16 - 4 = 12\).

To find \(\sum \alpha^3\), we use the relation \(\alpha^3 = 4\alpha^2 - 2\alpha + 1\). Summing this over all three roots:
\(\sum \alpha^3 = 4\sum \alpha^2 - 2\sum \alpha + 3 = 4(12) - 2(4) + 3 = 48 - 8 + 3 = 43\).

(ii) Let \(y = x^2\), so \(x = \sqrt{y}\).
Substitute this into the original cubic equation:
\(y\sqrt{y} - 4y + 2\sqrt{y} - 1 = 0\)
\(\sqrt{y}(y + 2) = 4y + 1\)

Squaring both sides:
\(y(y + 2)^2 = (4y + 1)^2\)
\(y(y^2 + 4y + 4) = 16y^2 + 8y + 1\)
\(y^3 + 4y^2 + 4y = 16y^2 + 8y + 1\)
\(y^3 - 12y^2 - 4y - 1 = 0\)

(iii) The roots of the new equation are \(y_1 = \alpha^2, y_2 = \beta^2, y_3 = \gamma^2\).
We want to find \(\sum \alpha^4 = \sum y^2\).
Using the identity for the roots of the equation in \(y\):
\(\sum y^2 = (\sum y)^2 - 2\sum y_1 y_2\)

From the equation \(y^3 - 12y^2 - 4y - 1 = 0\):
\(\sum y = 12\)
\ \sum y_1 y_2 = -4\)

Therefore,
\(\sum y^2 = 12^2 - 2(-4) = 144 + 8 = 152\).
So, \(\sum \alpha^4 = 152\).

(i)
- M1: Attempt to find \(\sum \alpha^2\) using the standard identity.
- A1: Correct value of 12.
- M1: Use of \(\alpha^3 = 4\alpha^2 - 2\alpha + 1\) (or equivalent cubic formula) to find \(\sum \alpha^3\).
- A1: Correct value of 43.

(ii)
- M1: Substitute \(x = \sqrt{y}\) and rearrange to isolate the square root terms.
- M1: Square both sides and expand correctly.
- A1: Obtain the correct cubic equation in \(y\) (or any variable) with integer coefficients.

(iii)
- M1: Express \(\sum \alpha^4\) as \(\sum y^2\) and use the sum/product of roots of the new equation.
- A1ft: Correct substitution of their values from (ii).
- A1: Correct final answer 152.

(i) Combining the terms on the right-hand side over a common denominator:
\(\frac{1}{4}\left(\frac{(2r+3) - (2r-1)}{(2r-1)(2r+1)(2r+3)}\right) =
\frac{1}{4}\left(\frac{4}{(2r-1)(2r+1)(2r+3)}\right) =
\frac{1}{(2r-1)(2r+1)(2r+3)}\).
This completes the proof.

(ii) Let \(u_r = \frac{1}{(2r-1)(2r+1)}\). Then the summand is \(\frac{1}{4}(u_r - u_{r+1})\).
Using the method of differences:
\(\sum_{r=1}^n \frac{1}{(2r-1)(2r+1)(2r+3)} = \frac{1}{4} \sum_{r=1}^n (u_r - u_{r+1})\)
\(= \frac{1}{4} [ (u_1 - u_2) + (u_2 - u_3) + \dots + (u_n - u_{n+1}) ]\)
\(= \frac{1}{4} (u_1 - u_{n+1})\).

Since \(u_1 = \frac{1}{(1)(3)} = \frac{1}{3}\) and \(u_{n+1} = \frac{1}{(2n+1)(2n+3)}\),
the sum is:
\(\frac{1}{4} \left( \frac{1}{3} - \frac{1}{(2n+1)(2n+3)} \right) = \frac{1}{12} - \frac{1}{4(2n+1)(2n+3)}\).

(iii) As \(n \to \infty\), the term \(\frac{1}{4(2n+1)(2n+3)} \to 0\).
Therefore, the sum to infinity is \(\frac{1}{12}\).

(i)
- M1: Attempt to combine the RHS fraction with a common denominator.
- A1: Show clearly that the numerator simplifies to 4, cancelling out with the outer coefficient to yield LHS.

(ii)
- M1: Write out the first few terms of the series using the difference form to show the telescoping behavior.
- A1: Identify the remaining terms after cancellation, i.e., \(u_1\) and \(u_{n+1}\).
- M1: Substitute \(r=1\) into \(u_1\) and \(r=n+1\) into \(u_{n+1}\).
- A1: Correctly simplified expression: \(\frac{1}{12} - \frac{1}{4(2n+1)(2n+3)}\) (or equivalent single fraction).

(iii)
- M1: State that \(\lim_{n \to \infty} \frac{1}{(2n+1)(2n+3)} = 0\).
- A1: Deduce the limit is \(\frac{1}{12}\).

(i) A matrix is singular when its determinant is zero.
\(\det(\mathbf{A}) = 1(k - 1) - 2(2k - 1) + 1(2 - 1)\)
\(= k - 1 - 4k + 2 + 1 = 2 - 3k\).
Set \(\det(\mathbf{A}) = 0 \implies 2 - 3k = 0 \implies k = \frac{2}{3}\).

(ii) For \(k = 2\), \(\det(\mathbf{A}) = 2 - 3(2) = -4\).
Next, we find the matrix of cofactors, \(\mathbf{C}\):
\(C_{11} = 1(2) - 1(1) = 1\)
\(C_{12} = -(2(2) - 1(1)) = -3\)
\(C_{13} = 2(1) - 1(1) = 1\)
\(C_{21} = -(2(2) - 1(1)) = -3\)
\(C_{22} = 1(2) - 1(1) = 1\)
\(C_{23} = -(1(1) - 2(1)) = 1\)
\(C_{31} = 2(1) - 1(1) = 1\)
\(C_{32} = -(1(1) - 2(1)) = 1\)
\(C_{33} = 1(1) - 2(2) = -3\)

Thus, \(\mathbf{C} = \begin{pmatrix} 1 & -3 & 1 \\ -3 & 1 & 1 \\ 1 & 1 & -3 \end{pmatrix}\).
Since \(\mathbf{C}\) is symmetric, the adjugate matrix is \(\mathbf{adj}(\mathbf{A}) = \mathbf{C}^T = \mathbf{C}\).
Therefore, the inverse matrix is:
\(\mathbf{A}^{-1} = \frac{1}{\det(\mathbf{A})} \mathbf{adj}(\mathbf{A}) = -\frac{1}{4} \begin{pmatrix} 1 & -3 & 1 \\ -3 & 1 & 1 \\ 1 & 1 & -3 \end{pmatrix} = \begin{pmatrix} -1/4 & 3/4 & -1/4 \\ 3/4 & -1/4 & -1/4 \\ -1/4 & -1/4 & 3/4 \end{pmatrix}\).

(iii) The system of equations can be written as \(\mathbf{A} \mathbf{x} = \mathbf{b}\), where
\(\mathbf{A} = \begin{pmatrix} 1 & 2 & 1 \\ 2 & 1 & 1 \\ 1 & 1 & 2 \end{pmatrix}\), \(\mathbf{x} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}\), and \(\mathbf{b} = \begin{pmatrix} 4 \\ 5 \\ 3 \end{pmatrix}\).

Thus, \(\mathbf{x} = \mathbf{A}^{-1} \mathbf{b}\):
\(\begin{pmatrix} x \\ y \\ z \end{pmatrix} = -\frac{1}{4} \begin{pmatrix} 1 & -3 & 1 \\ -3 & 1 & 1 \\ 1 & 1 & -3 \end{pmatrix} \begin{pmatrix} 4 \\ 5 \\ 3 \end{pmatrix}\)
\(= -\frac{1}{4} \begin{pmatrix} 4 - 15 + 3 \\ -12 + 5 + 3 \\ 4 + 5 - 9 \end{pmatrix} = -\frac{1}{4} \begin{pmatrix} -8 \\ -4 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}\).
So the solution is \(x = 2\), \(y = 1\), \(z = 0\).

(i)
- M1: Attempt to find the determinant of \(\mathbf{A}\) in terms of \(k\).
- A1: Obtain the correct determinant expression \(2 - 3k\).
- A1: Equate to 0 and solve to find \(k = 2/3\).

(ii)
- M1: Calculate \(\det(\mathbf{A}) = -4\) for \(k=2\).
- M2: Attempt to find the cofactors of the matrix (at least 6 elements correct for M1, all correct for M2).
- M1: Transpose the cofactor matrix (noting symmetry means it remains unchanged) and divide by the determinant.
- A1: Correct inverse matrix.

(iii)
- M1: Recognize the system can be solved using \(\mathbf{x} = \mathbf{A}^{-1} \mathbf{b}\).
- A1: Set up the matrix multiplication correctly.
- A1: Obtain the final unique values of \(x=2, y=1, z=0\).

(i) The area \(A\) of the polar region is given by:
\(A = \frac{1}{2} \int_0^{\pi} r^2 \mathrm{d}\theta = \frac{1}{2} \int_0^{\pi} a^2(1 + \cos\theta)^2 \mathrm{d}\theta\)
\(= \frac{a^2}{2} \int_0^{\pi} (1 + 2\cos\theta + \cos^2\theta) \mathrm{d}\theta\)

Using the identity \(\cos^2\theta = \frac{1}{2}(1 + \cos 2\theta)\):
\(A = \frac{a^2}{2} \int_0^{\pi} \left( 1 + 2\cos\theta + \frac{1}{2} + \frac{1}{2}\cos 2\theta \right) \mathrm{d}\theta\)
\(= \frac{a^2}{2} \int_0^{\pi} \left( \frac{3}{2} + 2\cos\theta + \frac{1}{2}\cos 2\theta \right) \mathrm{d}\theta\)
\(= \frac{a^2}{2} \left[ \frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta \right]_0^{\pi}\)
\(= \frac{a^2}{2} \left[ \left(\frac{3\pi}{2} + 0 + 0\right) - (0 + 0 + 0) \right] = \frac{3\pi a^2}{4}\).

(ii) A tangent to \(C\) is parallel to the initial line when \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = 0\) (provided \(\frac{\mathrm{d}x}{\mathrm{d}\theta} \ne 0\)).
Since \(y = r \sin\theta\):
\(y = a(1 + \cos\theta)\sin\theta = a\sin\theta + a\sin\theta\cos\theta = a\sin\theta + \frac{a}{2}\sin 2\theta\).

Differentiating with respect to \(\theta\):
\(\frac{\mathrm{d}y}{\mathrm{d}\theta} = a\cos\theta + a\cos 2\theta\).
Set \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = 0\):
\(\cos\theta + \cos 2\theta = 0\)
\(\cos\theta + 2\cos^2\theta - 1 = 0\)
\(2\cos^2\theta + \cos\theta - 1 = 0\)
\((2\cos\theta - 1)(\cos\theta + 1) = 0\).

In the interval \(0 < \theta < \pi\):
\(\cos\theta = \frac{1}{2} \implies \theta = \frac{\pi}{3}\)
(The other solution \(\cos\theta = -1 \implies \theta = \pi\) is at the boundary).

When \(\theta = \frac{\pi}{3}\):
\(r = a\left(1 + \cos\frac{\pi}{3}\right) = a\left(1 + \frac{1}{2}\right) = \frac{3}{2}a\).

So the polar coordinates of the point are \(\left(\frac{3}{2}a, \frac{\pi}{3}\right)\).

(i)
- M1: Use of standard polar area formula \(\frac{1}{2} \int r^2 \mathrm{d}\theta\) with correct limits.
- M1: Correctly expand \((1 + \cos\theta)^2\).
- M1: Use identity \(\cos^2\theta = \frac{1}{2}(1 + \cos 2\theta)\) to integrate.
- A1: Correct integration.
- A1: Obtain final exact area \(\frac{3\pi a^2}{4}\).

(ii)
- M1: Use \(y = r\sin\theta\) to express \(y\) in terms of \(\theta\).
- M1: Differentiate \(y\) with respect to \(\theta\).
- M1: Equate \(\frac{\mathrm{d}y}{\mathrm{d}\theta}\) to 0 and obtain a quadratic equation in \(\cos\theta\).
- A1: Solve correctly for \(\cos\theta = 1/2\) and find \(\theta = \pi/3\).
- M1: Find the corresponding value of \(r\).
- A1: Write down the correct coordinates as \(\left(\frac{3}{2}a, \frac{\pi}{3}\right)\).

(i) To show they are skew, we first show they are not parallel.
The direction vectors are \(\mathbf{d}_1 = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}\) and \(\mathbf{d}_2 = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}\). Since one is not a scalar multiple of the other, they are not parallel.

Now we test for intersection by setting components equal:
1) \(1 + \lambda = 2 + 2\mu \implies \lambda - 2\mu = 1\)
2) \(-\lambda = 4 + \mu \implies \lambda + \mu = -4\)
3) \(3 + 2\lambda = -1 + 3\mu \implies 2\lambda - 3\mu = -4\)

Subtracting equation (2) from equation (1):
\(-3\mu = 5 \implies \mu = -\frac{5}{3}\).
Substitute into (2):
\(\lambda = -4 - \left(-\frac{5}{3}\right) = -\frac{7}{3}\).

Now substitute these values into equation (3):
LHS: \(2\left(-\frac{7}{3}\right) - 3\left(-\frac{5}{3}\right) = -\frac{14}{3} + \frac{15}{3} = \frac{1}{3}\).
RHS: \(-4\).
Since LHS \(\ne\) RHS, the lines do not intersect. Since they are not parallel and do not intersect, they are skew.

(ii) The common perpendicular direction is given by \(\mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2\):
\(\mathbf{n} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} \times \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix} = \begin{pmatrix} -3 - 2 \\ -(3 - 4) \\ 1 - (-2)
\end{pmatrix} = \begin{pmatrix} -5 \\ 1 \\ 3 \end{pmatrix}\).

Let \(P(1, 0, 3)\) be a point on \(L_1\) and \(Q(2, 4, -1)\) be a point on \(L_2\).
\(\vec{PQ} = \begin{pmatrix} 2 \\ 4 \\ -1 \end{pmatrix} - \begin{pmatrix} 1 \\ 0 \\ 3 \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \\ -4 \end{pmatrix}\).

The shortest distance \(d\) is the projection of \(\vec{PQ}\) onto \(\mathbf{n}\):
\(d = \frac{|\vec{PQ} \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{|1(-5) + 4(1) + (-4)(3)|}{\sqrt{(-5)^2 + 1^2 + 3^2}} = \frac{|-5 + 4 - 12|}{\sqrt{35}} = \frac{13}{\sqrt{35}}\).

(iii) The plane \(\Pi\) contains \(L_1\) and is parallel to \(L_2\). Therefore, the normal vector to the plane is \(\mathbf{n} = \begin{pmatrix} -5 \\ 1 \\ 3 \end{pmatrix}\) (or any scalar multiple).
Using the point \((1, 0, 3)\) in the equation \(\mathbf{n} \cdot \mathbf{r} = \mathbf{n} \cdot \mathbf{r}_0\):
\(-5x + y + 3z = -5(1) + 1(0) + 3(3)\)
\(-5x + y + 3z = 4\)
or \(5x - y - 3z + 4 = 0\).

(i)
- B1: Correctly state that direction vectors are not parallel.
- M1: Set up simultaneous equations to find an intersection.
- A1: Solve equations for \(\lambda\) and \(\mu\).
- A1: Show a contradiction when substituting into the third equation, and conclude they are skew.

(ii)
- M1: Attempt to find the vector product of direction vectors.
- A1: Correct normal vector \(\begin{pmatrix} -5 \\ 1 \\ 3 \end{pmatrix}\) (or equivalent).
- M1: Apply the shortest distance formula \(d = \frac{|\vec{PQ} \cdot \mathbf{n}|}{|\mathbf{n}|}\).
- A1: Correct final distance \(\frac{13}{\sqrt{35}}\).

(iii)
- M1: Use the normal vector found in (ii) to set up plane equation.
- A1: Substitute a point from \(L_1\) to find the constant.
- A1: Correct Cartesian equation.

(i) By expanding and performing algebraic division:
\(y = \frac{x^2 - 2x + 1}{x + 1}\)

Dividing \(x^2 - 2x + 1\) by \(x + 1\):
\(x^2 - 2x + 1 = (x - 3)(x + 1) + 4\)
Thus, \(y = x - 3 + \frac{4}{x + 1}\).

As \(x \to -1\), \(y \to \pm \infty\), so the vertical asymptote is \(x = -1\).
As \(x \to \pm\infty\), \(y \to x - 3\), so the oblique asymptote is \(y = x - 3\).

(ii) To find the stationary points, we differentiate \(y\) with respect to \(x\):
\(\frac{\mathrm{d}y}{\mathrm{d}x} = 1 - \frac{4}{(x+1)^2}\).
Set \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\(1 - \frac{4}{(x+1)^2} = 0 \implies (x+1)^2 = 4\)
\(x + 1 = \pm 2\).

This gives two solutions:
- For \(x + 1 = 2 \implies x = 1\).
The corresponding \(y\)-value is \(y = \frac{(1-1)^2}{1+1} = 0\).
- For \(x + 1 = -2 \implies x = -3\).
The corresponding \(y\)-value is \(y = \frac{(-3-1)^2}{-3+1} = \frac{16}{-2} = -8\).

So the coordinates of the stationary points are \((1, 0)\) (local minimum) and \((-3, -8)\) (local maximum).

(iii) To sketch the curve:
- Intersections with axes:
When \(y = 0 \implies x = 1\), so \((1,0)\) is a touch-point on the x-axis.
When \(x = 0 \implies y = 1\), so the y-intercept is \((0,1)\).
- Plot asymptotes \(x = -1\) and \(y = x - 3\).
- Plot the stationary points at \((1,0)\) and \((-3,-8)\).
- Sketch the two branches of the hyperbola-like curve approaching the asymptotes.

(i)
- B1: State the vertical asymptote \(x = -1\).
- M1: Perform division to write the curve in the form \(y = ax + b + \frac{c}{x+1}\).
- A1: State the oblique asymptote \(y = x - 3\).

(ii)
- M1: Differentiate \(y\) with respect to \(x\).
- M1: Equate derivative to 0 and attempt to solve for \(x\).
- A1: Correct \(x\)-values (1 and \(-3\)).
- A1: Correct coordinates \((1, 0)\) and \((-3, -8)\).

(iii)
- B1: Correctly drawn and labeled asymptotes \(x=-1\) and \(y=x-3\).
- B1: Correct coordinates of intercepts \((1,0)\) and \((0,1)\) and stationary points shown.
- B1: Correctly shaped curve showing both branches approaching the asymptotes.

Let \(P(n)\) be the statement:
\(\mathbf{A}^n = \begin{pmatrix} 2n+1 & -4n \\ n & 1-2n \end{pmatrix}\).

**Base Case:**
For \(n = 1\):
LHS: \(\mathbf{A}^1 = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}\).

RHS: \(\begin{pmatrix} 2(1)+1 & -4(1) \\ 1 & 1-2(1) \end{pmatrix} = \begin{pmatrix} 3 & -4 \\ 1 & -1
\end{pmatrix}\).

Since LHS = RHS, \(P(1)\) is true.

**Inductive Step:**
Assume that \(P(k)\) is true for some positive integer \(k\). That is,
\(\mathbf{A}^k = \begin{pmatrix} 2k+1 & -4k \\ k & 1-2k \end{pmatrix}\).

We must show that this implies \(P(k+1)\) is true, where:
\(\mathbf{A}^{k+1} = \begin{pmatrix} 2(k+1)+1 & -4(k+1) \\ k+1 & 1-2(k+1) \end{pmatrix} = \begin{pmatrix} 2k+3 & -4k-4 \\ k+1 & -2k-1 \end{pmatrix}\).

Using matrix multiplication:
\(\mathbf{A}^{k+1} = \mathbf{A}^k \mathbf{A}\)
\(= \begin{pmatrix} 2k+1 & -4k \\ k & 1-2k \end{pmatrix} \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}\)

Calculate the elements of the product matrix:
- Top-Left element:
\((2k+1)(3) + (-4k)(1) = 6k + 3 - 4k = 2k + 3\)
- Top-Right element:
\((2k+1)(-4) + (-4k)(-1) = -8k - 4 + 4k = -4k - 4\)
- Bottom-Left element:
\(k(3) + (1-2k)(1) = 3k + 1 - 2k = k + 1\)
- Bottom-Right element:
\(k(-4) + (1-2k)(-1) = -4k - 1 + 2k = -2k - 1\)

Thus,
\(\mathbf{A}^{k+1} = \begin{pmatrix} 2k+3 & -4k-4 \\ k+1 & -2k-1 \end{pmatrix}\)
\(= \begin{pmatrix} 2(k+1)+1 & -4(k+1) \\ k+1 & 1-2(k+1) \end{pmatrix}\).

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 \(P(n)\) is true for all positive integers \(n\).

- B1: Verify the base case \(n=1\) correctly.
- M1: State the inductive hypothesis clearly for \(n=k\).
- M2: Attempt to find \(\mathbf{A}^{k+1}\) by multiplying \(\mathbf{A}^k\) by \(\mathbf{A}\) (or vice-versa).
- A2: Correct algebraic expansions of all 4 entries of the product (A1 for at least 2 correct entries).
- A1: Factorize and show the terms explicitly matching the required form for \(n = k+1\).
- Q1: Provide a complete and rigorous inductive conclusion linking the base case, the inductive step, and the final statement.

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

Next, we find the particular integral (PI). Since the right-hand side is \( 10e^{-x} \), we assume a particular integral of the form:
\[ y_p = ke^{-x} \]
Differentiating this gives:
\[ \frac{dy_p}{dx} = -ke^{-x} \quad \text{and} \quad \frac{d^2y_p}{dx^2} = ke^{-x} \]
Substituting these into the original differential equation:
\[ ke^{-x} + 4(-ke^{-x}) + 5ke^{-x} = 10e^{-x} \]
\[ (1 - 4 + 5)ke^{-x} = 10e^{-x} \implies 2k = 10 \implies k = 5 \]
Thus, the particular integral is:
\[ y_p = 5e^{-x} \]

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

Now we apply the boundary conditions:
1. When \( x = 0 \), \( y = 3 \):
\[ 3 = e^{0}(A\cos 0 + B\sin 0) + 5e^{0} \implies 3 = A + 5 \implies A = -2 \]

2. We differentiate the general solution to apply the second boundary condition:
\[ \frac{dy}{dx} = -2e^{-2x}(A\cos x + B\sin x) + e^{-2x}(-A\sin x + B\cos x) - 5e^{-x} \]
When \( x = 0 \), \( \frac{dy}{dx} = 0 \):
\[ 0 = -2(A) + B - 5 \]
Since \( A = -2 \):
\[ 0 = -2(-2) + B - 5 \implies 0 = 4 + B - 5 \implies B = 1 \]

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

M1: Set up and solve the auxiliary equation \( m^2+4m+5=0 \).
A1: Correct complementary function \( y_c = e^{-2x}(A\cos x + B\sin x) \).
M1: Assume \( y_p = ke^{-x} \) and substitute to solve for \( k \).
A1: Correctly obtain \( k = 5 \).
B1: State general solution \( y = e^{-2x}(A\cos x + B\sin x) + 5e^{-x} \).
M1: Apply condition \( y(0) = 3 \) to find \( A \).
A1: Obtain \( A = -2 \).
M1: Differentiate and apply condition \( y'(0) = 0 \) to find \( B \).
A1.375: Obtain \( B = 1 \) and write down the final particular solution.

(i) To find the eigenvalues, we solve the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \):
\[ \det\begin{pmatrix} 3-\lambda & 1 & 1 \\ 2 & 4-\lambda & 2 \\ 1 & 1 & 3-\lambda \end{pmatrix} = 0 \]
Row reductions or direct expansion yields:
\[ (3-\lambda)[(4-\lambda)(3-\lambda) - 2] - 1[2(3-\lambda) - 2] + 1[2 - (4-\lambda)] = 0 \]
\[ (3-\lambda)(\lambda^2 - 7\lambda + 10) - (4 - 2\lambda) + (\lambda - 2) = 0 \]
\[ (3-\lambda)(\lambda-2)(\lambda-5) + 3(\lambda-2) = 0 \]
Factoring out \( (\lambda-2) \):
\[ (\lambda-2) [ (3-\lambda)(\lambda-5) + 3 ] = (\lambda-2) ( -\lambda^2 + 8\lambda - 12 ) = -(\lambda-2)^2(\lambda-6) = 0 \]
So the eigenvalues are \( \lambda = 2 \) (repeated twice) and \( \lambda = 6 \).

(ii) For \( \lambda = 2 \):
\[ \mathbf{A} - 2\mathbf{I} = \begin{pmatrix} 1 & 1 & 1 \\ 2 & 2 & 2 \\ 1 & 1 & 1 \end{pmatrix} \]
This reduces to the single equation \( x + y + z = 0 \). Two linearly independent eigenvectors can be chosen by setting free variables:
If \( y = -1, z = 0 \), then \( x = 1 \), giving \( \mathbf{v}_1 = \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} \).
If \( y = 0, z = -1 \), then \( x = 1 \), giving \( \mathbf{v}_2 = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} \).

For \( \lambda = 6 \):
\[ \mathbf{A} - 6\mathbf{I} = \begin{pmatrix} -3 & 1 & 1 \\ 2 & -2 & 2 \\ 1 & 1 & -3 \end{pmatrix} \]
From Row 2: \( x - y + z = 0 \implies y = x + z \).
Substituting this into Row 1: \( -3x + (x+z) + z = 0 \implies -2x + 2z = 0 \implies x = z \).
Then \( y = 2x \). Setting \( x = 1 \), we obtain \( \mathbf{v}_3 = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} \).

(iii) The transition matrix \( \mathbf{P} \) is formed by using the eigenvectors as columns, and \( \mathbf{D} \) is the diagonal matrix of the corresponding eigenvalues:
\[ \mathbf{P} = \begin{pmatrix} 1 & 1 & 1 \\ -1 & 0 & 2 \\ 0 & -1 & 1 \end{pmatrix}, \quad \mathbf{D} = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 6 \end{pmatrix} \]

M1: Set up and write down the characteristic equation determinant.
A2: Correctly solve to find the eigenvalues 2, 2, 6 (1 mark for showing root 6, 1 mark for showing root 2 with multiplicity 2).
M1: Substitute \( \lambda = 2 \) to set up the system \( (\mathbf{A}-2\mathbf{I})\mathbf{v}=\mathbf{0} \).
A2: Correctly identify two linearly independent eigenvectors for \( \lambda = 2 \).
M1: Substitute \( \lambda = 6 \) to set up the system \( (\mathbf{A}-6\mathbf{I})\mathbf{v}=\mathbf{0} \).
A1: Correctly find an eigenvector for \( \lambda = 6 \).
A1.375: State correct matrices \( \mathbf{P} \) and \( \mathbf{D} \) in a mutually consistent order.

(i) Using standard series expansions:
\[ \sinh x = x + \frac{x^3}{6} + O(x^5) \]
Let \( u = \sinh x \). The expansion for \( \ln(1 + u) \) is:
\[ \ln(1 + u) = u - \frac{u^2}{2} + \frac{u^3}{3} - O(u^4) \]
Substituting \( u = x + \frac{x^3}{6} \) into this series up to degree 3:
\[ \ln(1 + \sinh x) = \left( x + \frac{x^3}{6} \right) - \frac{1}{2}\left( x + \frac{x^3}{6} \right)^2 + \frac{1}{3}\left( x + \frac{x^3}{6} \right)^3 + \dots \]
Expanding each term and keeping powers up to \( x^3 \):
\[ = x + \frac{x^3}{6} - \frac{1}{2}(x^2 + O(x^4)) + \frac{1}{3}(x^3 + O(x^5)) \]
\[ = x - \frac{1}{2}x^2 + \left( \frac{1}{6} + \frac{1}{3} \right)x^3 + \dots \]
\[ = x - \frac{1}{2}x^2 + \frac{1}{2}x^3 \]

(ii) Substituting the Maclaurin series expansion of \( \ln(1 + \sinh x) \) into the given limit expression:
\[ \lim_{x \to 0} \frac{2\left(x - \frac{1}{2}x^2 + \frac{1}{2}x^3 + O(x^4)\right) - 2x + x^2}{x^3} \]
\[ = \lim_{x \to 0} \frac{2x - x^2 + x^3 + O(x^4) - 2x + x^2}{x^3} \]
\[ = \lim_{x \to 0} \frac{x^3 + O(x^4)}{x^3} = \lim_{x \to 0} (1 + O(x)) = 1 \]

M1: State or use the standard series for \( \sinh x \).
M1: Set up the substitution of \( u = \sinh x \) into the standard series for \( \ln(1+u) \).
A2: Correctly expand and retain powers up to \( x^3 \) (1 mark for the quadratic term, 1 mark for the cubic term).
A1: Correctly write the final Maclaurin series \( x - \frac{1}{2}x^2 + \frac{1}{2}x^3 \).
M2: Substitute the obtained series into the limit expression.
A1.375: Obtain the correct limit of 1.

(i) We start by writing out the difference expression:
\[ I_n - I_{n-2} = \int_0^{\ln 2} (\tanh^n x - \tanh^{n-2} x) \, dx = \int_0^{\ln 2} \tanh^{n-2} x (\tanh^2 x - 1) \, dx \]
Using the hyperbolic identity \( 1 - \tanh^2 x = \text{sech}^2 x \), we have:
\[ \tanh^2 x - 1 = -\text{sech}^2 x \]
Substituting this into the integral yields:
\[ I_n - I_{n-2} = \int_0^{\ln 2} -\tanh^{n-2} x \, \text{sech}^2 x \, dx \]
Now we use the substitution \( u = \tanh x \), which gives \( du = \text{sech}^2 x \, dx \).
We also change the limits of integration:
- When \( x = 0 \), \( u = \tanh 0 = 0 \).
- When \( x = \ln 2 \), \( u = \tanh(\ln 2) = \frac{e^{\ln 2} - e^{-\ln 2}}{e^{\ln 2} + e^{-\ln 2}} = \frac{2 - 1/2}{2 + 1/2} = \frac{3/2}{5/2} = \frac{3}{5} \).

Now, perform the integration in terms of \( u \):
\[ I_n - I_{n-2} = \int_0^{3/5} -u^{n-2} \, du = \left[ -\frac{u^{n-1}}{n-1} \right]_0^{3/5} = -\frac{1}{n-1} \left(\frac{3}{5}\right)^{n-1} \]
This completes the proof.

(ii) To find \( I_3 \), we set \( n = 3 \) in the recurrence relation:
\[ I_3 - I_1 = -\frac{1}{2} \left(\frac{3}{5}\right)^2 = -\frac{9}{50} \implies I_3 = I_1 - \frac{9}{50} \]
We calculate \( I_1 \):
\[ I_1 = \int_0^{\ln 2} \tanh x \, dx = \int_0^{\ln 2} \frac{\sinh x}{\cosh x} \, dx = \left[ \ln(\cosh x) \right]_0^{\ln 2} \]
Since \( \cosh(\ln 2) = \frac{2 + 1/2}{2} = \frac{5}{4} \) and \( \cosh 0 = 1 \):
\[ I_1 = \ln\left(\frac{5}{4}\right) - \ln 1 = \ln\left(\frac{5}{4}\right) \]
Substituting this value back into the expression for \( I_3 \):
\[ I_3 = \ln\left(\frac{5}{4}\right) - \frac{9}{50} \]

M1: Express \( I_n - I_{n-2} \) as \( \int \tanh^{n-2} x (\tanh^2 x - 1) \, dx \).
M1: Apply the hyperbolic identity \( \tanh^2 x - 1 = -\text{sech}^2 x \).
A1: Substitute \( u = \tanh x \) and calculate \( u = \frac{3}{5} \) when \( x = \ln 2 \).
A1: Perform the integration of \( -u^{n-2} \) correctly.
A1: Obtain the correct given recurrence formula.
M1: Integrate \( \tanh x \) to find \( I_1 \).
A1: Find \( I_1 = \ln(5/4) \).
M1: Substitute \( n = 3 \) and use \( I_1 \) in the formula.
A1.375: Obtain the final exact value \( I_3 = \ln(5/4) - \frac{9}{50} \).

(i) We rewrite the equation using the identity \( \cosh^2 x = 1 + \sinh^2 x \):
\[ 3(1 + \sinh^2 x) + 4\sinh x = 7 \implies 3\sinh^2 x + 4\sinh x - 4 = 0 \]
Let \( y = \sinh x \). The quadratic becomes:
\[ 3y^2 + 4y - 4 = 0 \implies (3y - 2)(y + 2) = 0 \]
So \( \sinh x = \frac{2}{3} \) or \( \sinh x = -2 \).

Using the logarithmic formula \( \text{arsinh } u = \ln(u + \sqrt{u^2 + 1}) \):
- For \( \sinh x = \frac{2}{3} \):
\[ x = \ln\left( \frac{2}{3} + \sqrt{\frac{4}{9} + 1} \right) = \ln\left( \frac{2}{3} + \frac{\sqrt{13}}{3} \right) = \ln\left( \frac{2 + \sqrt{13}}{3} \right) \]
- For \( \sinh x = -2 \):
\[ x = \ln\left( -2 + \sqrt{(-2)^2 + 1} \right) = \ln\left( \sqrt{5} - 2 \right) \]

(ii) To prove the identity, we use the compound angle formula \( \cosh(A+B) = \cosh A \cosh B + \sinh A \sinh B \):
\[ \cosh(3x) = \cosh(2x + x) = \cosh(2x)\cosh x + \sinh(2x)\sinh x \]
We use the double-angle identities \( \cosh(2x) = 2\cosh^2 x - 1 \) and \( \sinh(2x) = 2\sinh x \cosh x \):
\[ = (2\cosh^2 x - 1)\cosh x + (2\sinh x \cosh x)\sinh x \]
\[ = 2\cosh^3 x - \cosh x + 2\sinh^2 x \cosh x \]
Since \( \sinh^2 x = \cosh^2 x - 1 \):
\[ = 2\cosh^3 x - \cosh x + 2(\cosh^2 x - 1)\cosh x \]
\[ = 2\cosh^3 x - \cosh x + 2\cosh^3 x - 2\cosh x \]
\[ = 4\cosh^3 x - 3\cosh x \]

M1: Substitute \( \cosh^2 x = 1 + \sinh^2 x \) to form a quadratic equation.
A1: Correct quadratic equation \( 3\sinh^2 x + 4\sinh x - 4 = 0 \).
M1: Solve quadratic equation to find values of \( \sinh x \).
A1: Correctly write \( x = \ln\left(\frac{2+\sqrt{13}}{3}\right) \).
A1: Correctly write \( x = \ln(\sqrt{5}-2) \).
M1: For (ii), expand \( \cosh(2x+x) \) using the compound angle identity.
A1: Substitute double-angle formulas for \( \cosh 2x \) and \( \sinh 2x \).
A1.375: Use \( \sinh^2 x = \cosh^2 x - 1 \) and simplify to obtain the identity.

(i) By de Moivre's theorem, we have:
\[ \cos(5\theta) + i\sin(5\theta) = (\cos\theta + i\sin\theta)^5 \]
Expanding the RHS 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 imaginary parts from both sides:
\[ \sin(5\theta) = 5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta \]
Using the identity \( \cos^2\theta = 1 - \sin^2\theta \):
\[ \sin(5\theta) = 5(1 - \sin^2\theta)^2\sin\theta - 10(1 - \sin^2\theta)\sin^3\theta + \sin^5\theta \]
\[ = 5(1 - 2\sin^2\theta + \sin^4\theta)\sin\theta - 10(\sin^3\theta - \sin^5\theta) + \sin^5\theta \]
\[ = 5\sin\theta - 10\sin^3\theta + 5\sin^5\theta - 10\sin^3\theta + 10\sin^5\theta + \sin^5\theta \]
\[ = 5\sin\theta - 20\sin^3\theta + 16\sin^5\theta \]

(ii) We are given the equation:
\[ 16x^5 - 20x^3 + 5x = 0 \]
Let \( x = \sin\theta \). Using the identity from part (i), this becomes:
\[ \sin(5\theta) = 0 \]
This equation has solutions:
\[ 5\theta = k\pi \implies \theta = \frac{k\pi}{5} \quad \text{for } k \in \mathbb{Z} \]
To obtain five distinct roots for \( x \), we choose \( k = -2, -1, 0, 1, 2 \). This gives:
\[ x = \sin\left(-\frac{2\pi}{5}\right), \sin\left(-\frac{\pi}{5}\right), \sin 0, \sin\left(\frac{\pi}{5}\right), \sin\left(\frac{2\pi}{5}\right) \]
Using \( \sin(-\phi) = -\sin\phi \), the roots are:
\[ x = 0, \pm\sin\left(\frac{\pi}{5}\right), \pm\sin\left(\frac{2\pi}{5}\right) \]

M1: Expand \( (\cos\theta + i\sin\theta)^5 \) using the binomial theorem.
A1: Equate imaginary parts to get \( \sin(5\theta) \) in terms of \( \cos\theta \) and \( \sin\theta \).
M1: Substitute \( \cos^2\theta = 1 - \sin^2\theta \) into the expression.
A1: Correct expansion and grouping of terms to show the identity.
M1: Use substitution \( x = \sin\theta \) to relate the equation to \( \sin 5\theta = 0 \).
A1: Solve \( \sin 5\theta = 0 \) to find \( \theta = \frac{k\pi}{5} \).
A2.375: State all 5 distinct roots of \( x \) (1 mark for \( x = 0 \), 1.375 marks for trigonometric representations of the non-zero roots).

This is a first-order linear differential equation of the form \( \frac{dy}{dx} + P(x)y = Q(x) \) with \( P(x) = \cot x \).
First, find the integrating factor \( I(x) \):
\[ I(x) = e^{\int \cot x \, dx} = e^{\ln|\sin x|} = \sin x \]
(since \( 0 < x < \pi \), \( \sin x > 0 \)).

Multiply both sides of the differential equation by \( \sin x \):
\[ \sin x \frac{dy}{dx} + y \cos x = 2\cos^2 x \sin^2 x \]
\[ \frac{d}{dx}(y \sin x) = 2\cos^2 x \sin^2 x \]

Now rewrite the right-hand side using trigonometric identities to make it integrable:
\[ 2\cos^2 x \sin^2 x = \frac{1}{2}(2\sin x \cos x)^2 = \frac{1}{2}\sin^2(2x) = \frac{1}{2}\left( \frac{1 - \cos 4x}{2} \right) = \frac{1}{4}(1 - \cos 4x) \]

Integrate both sides with respect to \( x \):
\[ y \sin x = \int \frac{1}{4}(1 - \cos 4x) \, dx = \frac{1}{4}x - \frac{1}{16}\sin 4x + C \]

Now apply the boundary condition \( y = \frac{1}{2} \) when \( x = \frac{\pi}{2} \):
\[ \frac{1}{2} \sin\left(\frac{\pi}{2}\right) = \frac{1}{4}\left(\frac{\pi}{2}\right) - \frac{1}{16}\sin(2\pi) + C \]
\[ \frac{1}{2}(1) = \frac{\pi}{8} - 0 + C \implies C = \frac{1}{2} - \frac{\pi}{8} \]

Thus, the particular solution is:
\[ y \sin x = \frac{1}{4}x - \frac{1}{16}\sin 4x + \frac{1}{2} - \frac{\pi}{8} \]
\[ y = \csc x \left( \frac{1}{4}x - \frac{1}{16}\sin 4x + \frac{1}{2} - \frac{\pi}{8} \) \]

M1: Attempt to find the integrating factor.
A1: Correctly obtain IF \( = \sin x \).
M1: Multiply the differential equation by the IF and express LHS as derivative of a product.
M1: Use double-angle identities to rewrite the RHS in an integrable form.
A1: Correctly express RHS as \( \frac{1}{4}(1 - \cos 4x) \).
M1: Integrate both sides with respect to \( x \).
A1: Correct integration with constant: \( y\sin x = \frac{1}{4}x - \frac{1}{16}\sin 4x + C \).
M1: Apply boundary conditions to find the value of \( C \).
A1.375: State the final particular solution for \( y \) in terms of \( x \).

(i) To find the characteristic equation, we solve \( \det(\mathbf{M} - \lambda \mathbf{I}) = 0 \):
\[ \det\begin{pmatrix} 1-\lambda & 2 & 1 \\ 0 & 1-\lambda & 3 \\ 2 & 1 & 1-\lambda \end{pmatrix} = 0 \]
Expanding along the second row:
\[ (1-\lambda) [ (1-\lambda)^2 - 2 ] - 3 [ (1-\lambda) - 4 ] = 0 \]
\[ (1-\lambda) [ \lambda^2 - 2\lambda - 1 ] - 3 [ -\lambda - 3 ] = 0 \]
\[ (\lambda^2 - 2\lambda - 1 - \lambda^3 + 2\lambda^2 + \lambda) + 3\lambda + 9 = 0 \]
\[ -\lambda^3 + 3\lambda^2 + 2\lambda + 8 = 0 \]
Thus, the characteristic equation is:
\[ \lambda^3 - 3\lambda^2 - 2\lambda - 8 = 0 \]

(ii) By the Cayley-Hamilton theorem, \( \mathbf{M} \) satisfies its own characteristic equation:
\[ \mathbf{M}^3 - 3\mathbf{M}^2 - 2\mathbf{M} - 8\mathbf{I} = \mathbf{0} \]
Multiplying through by \( \mathbf{M}^{-1} \):
\[ \mathbf{M}^2 - 3\mathbf{M} - 2\mathbf{I} - 8\mathbf{M}^{-1} = \mathbf{0} \implies 8\mathbf{M}^{-1} = \mathbf{M}^2 - 3\mathbf{M} - 2\mathbf{I} \]
So,
\[ \mathbf{M}^{-1} = \frac{1}{8}(\mathbf{M}^2 - 3\mathbf{M} - 2\mathbf{I}) \]

Let us first calculate \( \mathbf{M}^2 \):
\[ \mathbf{M}^2 = \begin{pmatrix} 1 & 2 & 1 \\ 0 & 1 & 3 \\ 2 & 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 2 & 1 \\ 0 & 1 & 3 \\ 2 & 1 & 1 \end{pmatrix} = \begin{pmatrix} 3 & 5 & 8 \\ 6 & 4 & 6 \\ 4 & 6 & 6 \end{pmatrix} \]

Now, calculate the matrix expression \( \mathbf{M}^2 - 3\mathbf{M} - 2\mathbf{I} \):
\[ \mathbf{M}^2 - 3\mathbf{M} - 2\mathbf{I} = \begin{pmatrix} 3 & 5 & 8 \\ 6 & 4 & 6 \\ 4 & 6 & 6 \end{pmatrix} - \begin{pmatrix} 3 & 6 & 3 \\ 0 & 3 & 9 \\ 6 & 3 & 3 \end{pmatrix} - \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix} = \begin{pmatrix} -2 & -1 & 5 \\ 6 & -1 & -3 \\ -2 & 3 & 1 \end{pmatrix} \]

Thus, the inverse matrix is:
\[ \mathbf{M}^{-1} = \frac{1}{8}\begin{pmatrix} -2 & -1 & 5 \\ 6 & -1 & -3 \\ -2 & 3 & 1 \end{pmatrix} \]

M1: Set up the characteristic equation determinant.
A2: Correctly expand and simplify to find \( \lambda^3 - 3\lambda^2 - 2\lambda - 8 = 0 \) (1 mark for trace and determinant, 1 mark for correct polynomial).
M1: Apply the Cayley-Hamilton theorem to form the matrix equation.
M1: Solve for \( \mathbf{M}^{-1} \) in terms of \( \mathbf{M}^2 \), \( \mathbf{M} \), and \( \mathbf{I} \).
A1: Correctly calculate \( \mathbf{M}^2 \).
M1: Substitute and calculate \( \mathbf{M}^2 - 3\mathbf{M} - 2\mathbf{I} \).
A2.375: Correctly find \( \mathbf{M}^{-1} \) (1 mark for correct scalar factor, 1.375 marks for correct elements of the matrix).