2025 Cambridge IAS-Level Mathematics - Further (9231) Practice Paper with Answers

A shear parallel to the \( y \)-axis with the \( y \)-axis invariant has the general matrix form \( \mathbf{S} = \begin{pmatrix} 1 & 0 \\ k & 1 \end{pmatrix} \). Given that the point \( (1, 0) \) is mapped to \( (1, 3) \): \( \begin{pmatrix} 1 & 0 \\ k & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ k \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \end{pmatrix} \). Thus, \( k = 3 \), and the shear matrix is \( \mathbf{S} = \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix} \). The matrix representing a reflection in the line \( y = -x \) is \( \mathbf{R} = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} \). Since the shear is followed by the reflection, the matrix \( \mathbf{M} \) representing the combined transformation is \( \mathbf{M} = \mathbf{R} \mathbf{S} = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix} = \begin{pmatrix} 0(1) + (-1)(3) & 0(0) + (-1)(1) \\ -1(1) + 0(3) & -1(0) + 0(1) \end{pmatrix} = \begin{pmatrix} -3 & -1 \\ -1 & 0 \end{pmatrix} \).

M1: State or use the correct matrix for the shear parallel to the \( y \)-axis, \( \mathbf{S} = \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix} \). M1: State the correct matrix for the reflection in \( y = -x \), \( \mathbf{R} = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} \). M1: Realise that the combined matrix is \( \mathbf{R}\mathbf{S} \) in the correct order of multiplication. A0.66: Obtain the correct final matrix \( \mathbf{M} = \begin{pmatrix} -3 & -1 \\ -1 & 0 \end{pmatrix} \).

The area scale factor of a transformation represented by a \( 2 \times 2 \) matrix \( \mathbf{A} \) is given by \( | \det \mathbf{A} | \). Here, the area scale factor is \( 60 / 12 = 5 \). The determinant of \( \mathbf{A} \) is: \( \det \mathbf{A} = (a)(3) - (2)(-1) = 3a + 2 \). Since \( a > 0 \), we have \( 3a + 2 > 0 \), so \( 3a + 2 = 5 \implies 3a = 3 \implies a = 1 \). With \( a = 1 \), the matrix is \( \mathbf{A} = \begin{pmatrix} 1 & 2 \\ -1 & 3 \end{pmatrix} \). The determinant is \( \det \mathbf{A} = 5 \). The inverse matrix \( \mathbf{A}^{-1} \) is given by: \( \mathbf{A}^{-1} = \frac{1}{5} \begin{pmatrix} 3 & -2 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 0.6 & -0.4 \\ 0.2 & 0.2 \end{pmatrix} \).

M1: Use the relation that the area scale factor is equal to \( | \det \mathbf{A} | \) to write \( 3a + 2 = 5 \). A1: Solve for \( a \) to obtain \( a = 1 \). M1: Use the standard formula for the inverse of a \( 2 \times 2 \) matrix with their value of \( a \). A0.66: Obtain the correct inverse matrix \( \mathbf{A}^{-1} = \begin{pmatrix} 0.6 & -0.4 \\ 0.2 & 0.2 \end{pmatrix} \) or equivalent.

For part (i): We evaluate the product \( \mathbf{P}\mathbf{Q} = \begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix} \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix} = \begin{pmatrix} 2(3) + 1(-5) & 2(-1) + 1(2) \\ 5(3) + 3(-5) & 5(-1) + 3(2) \end{pmatrix} = \begin{pmatrix} 6 - 5 & -2 + 2 \\ 15 - 15 & -5 + 6 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \mathbf{I} \). Since \( \mathbf{P}\mathbf{Q} = \mathbf{I} \), \( \mathbf{Q} \) is the inverse of \( \mathbf{P} \). For part (ii): We are given the matrix equation: \( \mathbf{P}\mathbf{X}\mathbf{P} = \mathbf{D} \), where \( \mathbf{D} = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} \). Pre-multiplying by \( \mathbf{P}^{-1} \) and post-multiplying by \( \mathbf{P}^{-1} \) gives: \( \mathbf{X} = \mathbf{P}^{-1} \mathbf{D} \mathbf{P}^{-1} \). Since \( \mathbf{P}^{-1} = \mathbf{Q} \), we have \( \mathbf{X} = \mathbf{Q} \mathbf{D} \mathbf{Q} \). First, calculate \( \mathbf{Q} \mathbf{D} = \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 3 & -2 \\ -5 & 4 \end{pmatrix} \). Now, calculate \( \mathbf{X} = (\mathbf{Q} \mathbf{D}) \mathbf{Q} = \begin{pmatrix} 3 & -2 \\ -5 & 4 \end{pmatrix} \dots \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix} = \begin{pmatrix} 3(3) + (-2)(-5) & 3(-1) + (-2)(2) \\ -5(3) + 4(-5) & -5(-1) + 4(2) \end{pmatrix} = \begin{pmatrix} 19 & -7 \\ -35 & 13 \end{pmatrix} \).

M1: Show \( \mathbf{P}\mathbf{Q} = \mathbf{I} \) by matrix multiplication (or equivalent verification). M1: Rearrange the matrix equation to express \( \mathbf{X} \) as \( \mathbf{P}^{-1}\mathbf{D}\mathbf{P}^{-1} \) or \( \mathbf{Q}\mathbf{D}\mathbf{Q} \). M1: Perform one step of multiplication (e.g. \( \mathbf{Q}\mathbf{D} \) or \( \mathbf{D}\mathbf{Q} \) ) correctly. A0.66: Obtain the correct final matrix \( \mathbf{X} = \begin{pmatrix} 19 & -7 \\ -35 & 13 \end{pmatrix} \).

Let \(P(n)\) be the statement that \(5^{2n} + 24n - 1\) is divisible by \(48\).

**Base case:**
For \(n = 1\):
\(5^{2(1)} + 24(1) - 1 = 25 + 24 - 1 = 48\)
Since \(48\) is divisible by \(48\), \(P(1)\) is true.

**Inductive step:**
Assume that \(P(k)\) is true for some positive integer \(k\). That is,
\(5^{2k} + 24k - 1 = 48M\) for some integer \(M\).
This can be rewritten as:
\(5^{2k} = 48M - 24k + 1\)

We want to show that \(P(k+1)\) is true, i.e., \(5^{2(k+1)} + 24(k+1) - 1\) is divisible by \(48\).

Consider the expression for \(n = k+1\):
\(5^{2(k+1)} + 24(k+1) - 1 = 5^{2k+2} + 24k + 24 - 1\)
\(= 25 \cdot 5^{2k} + 24k + 23\)

Substituting \(5^{2k} = 48M - 24k + 1\) into the expression:
\(= 25(48M - 24k + 1) + 24k + 23\)
\(= 25 \cdot 48M - 600k + 25 + 24k + 23\)
\(= 25 \cdot 48M - 576k + 48\)

Factor out \(48\) from the terms:
\(= 48(25M - 12k + 1)\)

Since \(M\) and \(k\) are integers, \(25M - 12k + 1\) is an integer. Thus, \(5^{2(k+1)} + 24(k+1) - 1\) is divisible by \(48\), which means \(P(k+1)\) is true.

**Conclusion:**
Since \(P(1)\) is true, and \(P(k) \implies P(k+1)\), by mathematical induction, \(P(n)\) is true for all positive integers \(n\).

**B1:** Verify the base case \(n=1\) clearly showing that \(5^2 + 24(1) - 1 = 48\) is divisible by 48.

**M1:** State the induction hypothesis for \(n=k\) clearly (e.g. \(5^{2k} + 24k - 1 = 48M\)).

**M1:** Attempt to analyze the expression for \(n = k+1\), substituting \(5^{2k+2} = 25 \cdot 5^{2k}\) or equivalent manipulation.

**A1:** Correct substitution and algebraic expansion to obtain \(25 \cdot 48M - 576k + 48\) or an equivalent algebraic expression.

**A1:** Correctly factorize out \(48\) to show that the expression is \(48(25M - 12k + 1)\) (or equivalent form), showing divisibility by 48.

**A1:** Give a complete, logical conclusion stating that since the base case is true and the inductive step holds, the statement is true for all positive integers \(n\) by mathematical induction.

Let the roots of the equation be \(\alpha\), \(\beta\), \(\gamma\) and \(\delta\).
From the relationship between roots and coefficients:
\(\sum \alpha = \alpha + \beta + \gamma + \delta = -\frac{-3}{2} = \frac{3}{2}\)
\(\sum \alpha \beta = \alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta = \frac{4}{2} = 2\)
We use the identity:
\(\sum \alpha^2 = \left(\sum \alpha\right)^2 - 2\sum \alpha\beta\)
Substitute the values into the identity:
\(\sum \alpha^2 = \left(\frac{3}{2}\right)^2 - 2(2) = \frac{9}{4} - 4 = -\frac{7}{4} = -1.75\).

M1 for stating both \(\sum \alpha = 1.5\) and \(\sum \alpha\beta = 2\) (or showing equivalent work with coefficients).
M1 for substituting these values into the correct identity \(\sum \alpha^2 = (\sum \alpha)^2 - 2\sum \alpha\beta\).
A1 for the correct final value of \(-1.75\) (or \(-\frac{7}{4}\)).

Let \(y = \frac{1}{x}\), which gives \(x = \frac{1}{y}\).
Substitute \(x = \frac{1}{y}\) into the given equation:
\(\left(\frac{1}{y}\right)^4 + 2\left(\frac{1}{y}\right)^2 - 3\left(\frac{1}{y}\right) + 1 = 0\)
Multiply the entire equation by \(y^4\) to eliminate the denominators:
\(1 + 2y^2 - 3y^3 + y^4 = 0\)
Rearranging the terms in descending powers of \(y\):
\(y^4 - 3y^3 + 2y^2 + 1 = 0\)
Any other variable name (such as \(x\)) is acceptable, e.g., \(x^4 - 3x^3 + 2x^2 + 1 = 0\).

M1 for substituting \(x = \frac{1}{y}\) (or equivalent substitution method).
M1 for multiplying by \(y^4\) to clear the fractional terms.
A1 for obtaining the correct equation \(y^4 - 3y^3 + 2y^2 + 1 = 0\) (or equivalent using another variable) with integer coefficients.

Let \(f(x) = x^4 - 2x^3 + x^2 + 3x - 1\).
Since \(\alpha\), \(\beta\), \(\gamma\) and \(\delta\) are the roots of \(f(x) = 0\), we can write:
\(f(x) = (x-\alpha)(x-\beta)(x-\gamma)(x-\delta)\)
To find the value of \((1-\alpha)(1-\beta)(1-\gamma)(1-\delta)\), substitute \(x = 1\) into the expression:
\(f(1) = (1-\alpha)(1-\beta)(1-\gamma)(1-\delta)\)
Evaluating \(f(1)\):
\(f(1) = 1^4 - 2(1)^3 + 1^2 + 3(1) - 1 = 1 - 2 + 1 + 3 - 1 = 2\).

Alternatively, expanding the expression gives:
\(1 - \sum \alpha + \sum \alpha\beta - \sum \alpha\beta\gamma + \alpha\beta\gamma\delta\)
From the coefficients of the quartic equation:
\(\sum \alpha = 2\)
\(\sum \alpha\beta = 1\)
\(\sum \alpha\beta\gamma = -3\)
\(\alpha\beta\gamma\delta = -1\)
Substituting these values:
\(1 - 2 + 1 - (-3) + (-1) = 2\).

M1 for recognizing that the expression can be evaluated by substituting \(x = 1\) into the factored form of the quartic, or for expanding the product in terms of symmetric sums.
M1 for evaluating the expression with \(x = 1\) or for correctly identifying all required symmetric sum values from the coefficients of the quartic.
A1 for obtaining the correct value of \(2\).

Let the general term be \(u_r = \frac{2r+1}{r^2(r+1)^2} = \frac{1}{r^2} - \frac{1}{(r+1)^2}\).

Writing out the terms of the sum, we get:
For \(r = 1\): \(\frac{1}{1^2} - \frac{1}{2^2}\)
For \(r = 2\): \(\frac{1}{2^2} - \frac{1}{3^2}\)
For \(r = 3\): \(\frac{1}{3^2} - \frac{1}{4^2}\)
...
For \(r = n\): \(\frac{1}{n^2} - \frac{1}{(n+1)^2}\)

Summing these terms from \(r = 1\) to \(n\):
\(\sum_{r=1}^{n} \left( \frac{1}{r^2} - \frac{1}{(r+1)^2} \right) = \left(1 - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{9}\right) + \dots + \left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right)\)

Cancelling out terms telescope-fashion, we are left with:
\(S_n = 1 - \frac{1}{(n+1)^2}\)

To find the sum to infinity, we take the limit as \(n \to \infty\):
\(\lim_{n \to \infty} S_n = \lim_{n \to \infty} \left( 1 - \frac{1}{(n+1)^2} \right) = 1\)

M1: For applying the method of differences and listing the terms of the series showing telescopic cancellation.
A1: For obtaining the correct simplified expression \(1 - \frac{1}{(n+1)^2}\) for the sum of the first \(n\) terms.
A1: For taking the limit as \(n \to \infty\) and correctly finding the sum to infinity is \(1\).

First, verify the identity:
\(\frac{1}{2}\left( \frac{1}{r(r+1)} - \frac{1}{(r+1)(r+2)} \right) = \frac{1}{2} \left( \frac{(r+2) - r}{r(r+1)(r+2)} \right) = \frac{1}{2} \left( \frac{2}{r(r+1)(r+2)} \right) = \frac{1}{r(r+1)(r+2)}\).

Let \(f(r) = \frac{1}{r(r+1)}\). The sum is \(\frac{1}{2} \sum_{r=1}^n (f(r) - f(r+1))\).

Listing the terms:
For \(r=1\): \(\frac{1}{2} \left( \frac{1}{2} - \frac{1}{6} \right)\)
For \(r=2\): \(\frac{1}{2} \left( \frac{1}{6} - \frac{1}{12} \right)\)
...
For \(r=n\): \(\frac{1}{2} \left( \frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} \right)\)

Adding these equations, the intermediate terms cancel out:
\(S_n = \frac{1}{2} \left( \frac{1}{1(2)} - \frac{1}{(n+1)(n+2)} \right) = \frac{1}{4} - \frac{1}{2(n+1)(n+2)}\).

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

M1: For writing down the series terms using the given difference form to show cancellation.
A1: For obtaining the correct sum of \(n\) terms as \(\frac{1}{4} - \frac{1}{2(n+1)(n+2)}\).
A1: For stating the correct limit as \(n \to \infty\) to get \(\frac{1}{4}\).

Let the terms of the summation be written using the identity:
\(\sum_{r=1}^n \frac{2}{r(r+2)} = \sum_{r=1}^n \left( \frac{1}{r} - \frac{1}{r+2} \right)\).

Write down the terms of the sum to observe the cancellation:
For \(r = 1\): \(1 - \frac{1}{3}\)
For \(r = 2\): \(\frac{1}{2} - \frac{1}{4}\)
For \(r = 3\): \(\frac{1}{3} - \frac{1}{5}\)
...
For \(r = n-1\): \(\frac{1}{n-1} - \frac{1}{n+1}\)
For \(r = n\): \(\frac{1}{n} - \frac{1}{n+2}\)

Adding these terms, we see that terms from \(\frac{1}{3}\) to \(\frac{1}{n}\) cancel out.
We are left with:
\(S_n = 1 + \frac{1}{2} - \frac{1}{n+1} - \frac{1}{n+2} = \frac{3}{2} - \frac{1}{n+1} - \frac{1}{n+2}\).

Taking the limit as \(n \to \infty\):
\(\lim_{n\to\infty} S_n = \frac{3}{2} - 0 - 0 = \frac{3}{2}\).

M1: For expanding the series and showing the partial cancellation where two positive terms and two negative terms remain.
A1: For the correct expression of \(S_n\), which is \(\frac{3}{2} - \frac{1}{n+1} - \frac{1}{n+2}\) (or equivalent single fraction).
A1: For taking the limit to find the sum to infinity as \(\frac{3}{2}\).

Let \(\mathbf{a}_1 = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}\) and \(\mathbf{a}_2 = \begin{pmatrix} 2 \\ 2 \\ 1 \end{pmatrix}\). The direction vectors are \(\mathbf{d}_1 = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}\) and \(\mathbf{d}_2 = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}\). The common perpendicular vector \(\mathbf{n}\) is given by the vector product: \(\mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2 = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix} \times \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 1(2) - 3(-1) \\ 3(1) - 2(2) \\ 2(-1) - 1(1) \end{pmatrix} = \begin{pmatrix} 5 \\ -1 \\ -3 \end{pmatrix}\). The magnitude of \(\mathbf{n}\) is: \(|\mathbf{n}| = \sqrt{5^2 + (-1)^2 + (-3)^2} = \sqrt{35}\). The vector connecting the points on the two lines is: \(\mathbf{a}_2 - \mathbf{a}_1 = \begin{pmatrix} 2 - 1 \\ 2 - 0 \\ 1 - (-1) \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}\). The shortest distance \(d\) is: \(d = \frac{|(\mathbf{a}_2 - \mathbf{a}_1) \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{|1(5) + 2(-1) + 2(-3)|}{\sqrt{35}} = \frac{|5 - 2 - 6|}{\sqrt{35}} = \frac{3}{\sqrt{35}} \approx 0.507\).

M1: For finding the vector product of the direction vectors. A1: For obtaining the correct perpendicular direction vector. M1: For using the shortest distance formula with correct substitution. A0.66: For the correct exact distance or decimal equivalent to 3 s.f.

Let \(\mathbf{d} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}\) be the direction vector of the line, and \(\mathbf{n} = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}\) be the normal vector to the plane. The angle \(\theta\) between the line and the plane satisfies \(\sin \theta = \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}| |\mathbf{n}|}\). Calculating the dot product: \(\mathbf{d} \cdot \mathbf{n} = 1(2) + 2(-1) + (-1)(1) = -1\). Calculating the magnitudes: \(|\mathbf{d}| = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{6}\) and \(|\mathbf{n}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{6}\). Thus, \(\sin \theta = \frac{|-1|}{\sqrt{6}\sqrt{6}} = \frac{1}{6}\). Solving for \(\theta\) gives \(\theta = \arcsin\left(\frac{1}{6}\right) \approx 9.594^\circ\), which is \(9.6^\circ\) correct to 1 decimal place.

M1: For calculating the dot product and magnitudes of the direction and normal vectors. A1: For obtaining \(\sin \theta = \frac{1}{6}\) (or \(\cos \phi = \frac{1}{6}\) for the angle with the normal). M1: For applying the correct inverse trigonometric relation to find the acute angle. A0.66: For obtaining \(9.6^\circ\).

Let \(A = (1, 1, 2)\) be a point on the line, and let \(\mathbf{d} = \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}\) be the direction vector of the line. The vector from \(A\) to \(P\) is \(\vec{AP} = \begin{pmatrix} 3 - 1 \\ 0 - 1 \\ 1 - 2 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \\ -1 \end{pmatrix}\). The perpendicular distance \(D\) is given by \(D = \frac{|\vec{AP} \times \mathbf{d}|}{|\mathbf{d}|}\). First, find the cross product: \(\vec{AP} \times \mathbf{d} = \begin{pmatrix} 2 \\ -1 \\ -1 \end{pmatrix} \times \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix} = \begin{pmatrix} -2 - 2 \\ -(4 - (-1)) \\ -4 - (-1) \end{pmatrix} = \begin{pmatrix} -4 \\ -5 \\ -3 \end{pmatrix}\). Next, find the magnitudes: \(|\vec{AP} \times \mathbf{d}| = \sqrt{(-4)^2 + (-5)^2 + (-3)^2} = \sqrt{50} = 5\sqrt{2}\), and \(|\mathbf{d}| = \sqrt{1^2 + (-2)^2 + 2^2} = 3\). Thus, the perpendicular distance is \(D = \frac{5\sqrt{2}}{3} \approx 2.36\).

M1: For finding a vector from a point on the line to the given point P. A1: For finding the vector product of this vector with the direction vector of the line. M1: For dividing the magnitude of the vector product by the magnitude of the direction vector. A0.66: For the correct exact value or decimal equivalent to 3 s.f.

To find the vertical asymptote, we set the denominator of the rational function to zero: \(x - 2 = 0 \implies x = 2\). To find the oblique asymptote, we perform algebraic division on the rational expression: \(3x^2 - 5x + 1 = (3x + 1)(x - 2) + 3\). Thus, \(y = 3x + 1 + \frac{3}{x - 2}\). As \(x \to \pm\infty\), the term \(\frac{3}{x - 2} \to 0\). Therefore, the oblique asymptote is \(y = 3x + 1\).

M1: Attempt to find vertical asymptote by setting denominator to 0, and attempt algebraic division to find oblique asymptote. A1: Correctly identify the oblique asymptote \(y = 3x + 1\). A0.8: Correctly identify the vertical asymptote \(x = 2\).

Equating the function to \(y\), we have \(y = \frac{x^2 - 2x + 5}{x - 1}\). Multiplying by \(x-1\) gives \(y(x - 1) = x^2 - 2x + 5 \implies x^2 - (2 + y)x + (5 + y) = 0\). For \(x\) to be real, the discriminant of this quadratic equation in \(x\) must be greater than or equal to zero: \(\Delta = [-(2 + y)]^2 - 4(1)(5 + y) \ge 0\). Simplifying this expression: \((4 + 4y + y^2) - 20 - 4y \ge 0 \implies y^2 - 16 \ge 0\). This inequality is satisfied when \(y \le -4\) or \(y \ge 4\). Therefore, the values of \(y\) that the function cannot take are those in the interval \(-4 < y < 4\).

M1: Set up the equation as a quadratic in \(x\) and find an expression for the discriminant in terms of \(y\). A1: Correctly obtain the inequality \(y^2 - 16 \ge 0\) for real \(x\). A0.8: Correctly deduce that the set of values that the function cannot take is \(-4 < y < 4\).

To find the stationary points, we differentiate \(y = \frac{x^2 + 3}{x - 1}\) with respect to \(x\) using the quotient rule: \(\frac{dy}{dx} = \frac{2x(x - 1) - (x^2 + 3)(1)}{(x - 1)^2} = \frac{2x^2 - 2x - x^2 - 3}{(x - 1)^2} = \frac{x^2 - 2x - 3}{(x - 1)^2}\). Setting \(\frac{dy}{dx} = 0\) for stationary points gives \(x^2 - 2x - 3 = 0 \implies (x - 3)(x + 1) = 0\), which yields \(x = 3\) or \(x = -1\). Substituting \(x = 3\) back into the original curve equation: \(y = \frac{3^2 + 3}{3 - 1} = \frac{12}{2} = 6\), giving the point \((3, 6)\). Substituting \(x = -1\): \(y = \frac{(-1)^2 + 3}{-1 - 1} = \frac{4}{-2} = -2\), giving the point \((-1, -2)\).

M1: Differentiate the rational function using the quotient rule and equate the numerator to zero to find the \(x\)-coordinates. A1: Correctly find both \(x\)-values, \(x = 3\) and \(x = -1\). A0.8: Correctly find the corresponding \(y\)-coordinates to state the stationary points \((-1, -2)\) and \((3, 6)\).

Since the curve has a vertical asymptote at \(x = -2\), the denominator must be zero at this value, so \(-2 + c = 0 \implies c = 2\). This gives the equation \(y = \frac{ax^2 + bx}{x + 2}\). We perform algebraic division to find the oblique asymptote: \(ax^2 + bx = ax(x+2) - 2ax + bx = ax(x+2) + (b-2a)x = ax(x+2) + (b-2a)(x+2) - 2(b-2a)\). Thus, \(y = ax + (b - 2a) - \frac{2(b - 2a)}{x + 2}\). The oblique asymptote is given by the linear part, so \(y = ax + (b - 2a)\). Comparing this to the given oblique asymptote \(y = 2x - 4\), we get \(a = 2\) and \(b - 2a = -4\). Substituting \(a = 2\) into the second equation: \(b - 2(2) = -4 \implies b - 4 = -4 \implies b = 0\). Thus, \(a = 2, b = 0, c = 2\).

M1: Identify \(c = 2\) from the vertical asymptote and attempt algebraic division to express the oblique asymptote in terms of \(a\) and \(b\). A1: Set up the simultaneous equations \(a = 2\) and \(b - 2a = -4\). A0.8: Solve to obtain the correct values \(a = 2, b = 0\).

To find the points of intersection, we set the equations equal to each other: \(x - 2 = \frac{2x - 3}{x - 1}\). Multiplying both sides by \(x - 1\) (where \(x \neq 1\)): \((x - 2)(x - 1) = 2x - 3 \implies x^2 - 3x + 2 = 2x - 3\). Rearranging into standard quadratic form: \(x^2 - 5x + 5 = 0\). Using the quadratic formula to solve for \(x\): \(x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(5)}}{2(1)} = \frac{5 \pm \sqrt{25 - 20}}{2} = \frac{5 \pm \sqrt{5}}{2}\). To find the corresponding \(y\)-coordinates, we substitute \(x\) back into the simpler linear equation \(y = x - 2\): For \(x_1 = \frac{5 + \sqrt{5}}{2}\), \(y_1 = \frac{5 + \sqrt{5}}{2} - 2 = \frac{1 + \sqrt{5}}{2}\). For \(x_2 = \frac{5 - \sqrt{5}}{2}\), \(y_2 = \frac{5 - \sqrt{5}}{2} - 2 = \frac{1 - \sqrt{5}}{2}\). Thus, the exact coordinates are \(\left(\frac{5 + \sqrt{5}}{2}, \frac{1 + \sqrt{5}}{2}\right)\) and \(\left(\frac{5 - \sqrt{5}}{2}, \frac{1 - \sqrt{5}}{2}\right)\).

M1: Set up the equation by equating the line and the curve and form a quadratic equation. A1: Correctly solve the quadratic equation to obtain \(x = \frac{5 \pm \sqrt{5}}{2}\). A0.8: Correctly find both corresponding exact \(y\)-coordinates and express the final coordinates.

The area \(A\) of the region is given by \(A = \frac{1}{2} \int_{0}^{\pi} r^2 \mathrm{d}\theta\). Substitute \(r = 2(1 + \sin \theta)\): \(A = \frac{1}{2} \int_{0}^{\pi} 4(1 + \sin \theta)^2 \mathrm{d}\theta = 2 \int_{0}^{\pi} (1 + 2\sin \theta + \sin^2 \theta) \mathrm{d}\theta\). Using the identity \(\sin^2 \theta = \frac{1 - \cos 2\theta}{2}\), the integrand becomes: \(1 + 2\sin \theta + \frac{1}{2} - \frac{1}{2}\cos 2\theta = \frac{3}{2} + 2\sin \theta - \frac{1}{2}\cos 2\theta\). Now, integrate with respect to \(\theta\): \(A = 2 \left[ \frac{3}{2}\theta - 2\cos \theta - \frac{1}{4}\sin 2\theta \right]_{0}^{\pi}\). Evaluate this at the limits: At \(\theta = \pi\): \(2 \left( \frac{3\pi}{2} - 2(-1) - 0 \right) = 3\pi + 4\). At \(\theta = 0\): \(2 \left( 0 - 2(1) - 0 \right) = -4\). Subtracting the lower limit value gives: \(A = (3\pi + 4) - (-4) = 3\pi + 8\).

M1: Uses area formula \(\frac{1}{2}\int r^2 \mathrm{d}\theta\) with correct limits. M1: Expands and applies the double-angle identity for \(\sin^2 \theta\) correctly. A1: Integrates correctly to get \(2 [\frac{3}{2}\theta - 2\cos \theta - \frac{1}{4}\sin 2\theta]\) or equivalent. A1: Evaluates limits to obtain the correct exact area of \(3\pi + 8\).

A tangent is parallel to the initial line when \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = 0\), where \(y = r\sin \theta\). Substitute \(r = 2(1 + \cos \theta)\): \(y = 2(1 + \cos \theta)\sin \theta = 2\sin \theta + 2\sin \theta \cos \theta = 2\sin \theta + \sin 2\theta\). Differentiate with respect to \(\theta\): \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = 2\cos \theta + 2\cos 2\theta\). Set this derivative to 0: \(2\cos \theta + 2(2\cos^2 \theta - 1) = 0 \implies 2\cos^2 \theta + \cos \theta - 1 = 0\). Solve the quadratic in \(\cos \theta\): \((2\cos \theta - 1)(\cos \theta + 1) = 0\). Since \(0 \le \theta < \pi\), we exclude \(\cos \theta = -1\) (which corresponds to \(\theta = \pi\)). Thus, \(\cos \theta = \frac{1}{2}\), which gives \(\theta = \frac{\pi}{3}\). Find \(r\) at \(\theta = \frac{\pi}{3}\): \(r = 2(1 + \cos \frac{\pi}{3}) = 2(1 + 0.5) = 3\). The polar coordinates of the point are \((3, \frac{\pi}{3})\).

M1: Expresses \(y = r\sin \theta\) in terms of \(\theta\) and differentiates. M1: Sets \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = 0\) and solves the resulting trigonometric quadratic equation. A1: Identifies the correct value \(\theta = \frac{\pi}{3}\) within the given range. A1: Finds \(r = 3\) and states the polar coordinates as \((3, \frac{\pi}{3})\).

One loop of the curve \(C\) corresponds to the interval of \(\theta\) for which \(r \ge 0\) around a petal. Setting \(r = 0\) gives \(\cos(3\theta) = 0 \implies 3\theta = \pm\frac{\pi}{2} \implies \theta = \pm\frac{\pi}{6}\). The area \(A\) of this loop is: \(A = \frac{1}{2} \int_{-\pi/6}^{\pi/6} r^2 \mathrm{d}\theta = \frac{1}{2} \int_{-\pi/6}^{\pi/6} 16\cos^2(3\theta) \mathrm{d}\theta = 8 \int_{-\pi/6}^{\pi/6} \cos^2(3\theta) \mathrm{d}\theta\). Using the identity \(\cos^2(3\theta) = \frac{1 + \cos(6\theta)}{2}\): \(A = 8 \int_{-\pi/6}^{\pi/6} \frac{1 + \cos(6\theta)}{2} \mathrm{d}\theta = 4 \int_{-\pi/6}^{\pi/6} (1 + \cos(6\theta)) \mathrm{d}\theta\). Integrating gives: \(A = 4 \left[ \theta + \frac{1}{6}\sin(6\theta) \right]_{-\pi/6}^{\pi/6} = 4 \left( \left( \frac{\pi}{6} + 0 \right) - \left( -\frac{\pi}{6} + 0 \right) \right) = 4 \left( \frac{\pi}{3} \right) = \frac{4\pi}{3}\,.\)

B1: Correctly identifies the limits of integration for one loop as \(-\frac{\pi}{6}\) and \(\frac{\pi}{6}\) (or equivalent 0 to \(\frac{\pi}{6}\) with doubling). M1: Sets up the area integral with \(r^2\). M1: Uses the double-angle identity \(\cos^2(3\theta) = \frac{1+\cos(6\theta)}{2}\) and integrates correctly. A1: Obtains the exact final area \(\frac{4\pi}{3}\).

Rearrange the polar equation: \(r(2 - \cos\theta) = 6 \implies 2r - r\cos\theta = 6\). Substitute the standard relations \(r\cos\theta = x\) and \(r = \sqrt{x^2+y^2}\): \(2\sqrt{x^2+y^2} - x = 6 \implies 2\sqrt{x^2+y^2} = x + 6\). Square both sides to eliminate the square root: \(4(x^2+y^2) = (x+6)^2 \implies 4x^2 + 4y^2 = x^2 + 12x + 36\). Rearrange the terms: \(3x^2 - 12x + 4y^2 = 36\). Complete the square for the \(x\) terms: \(3(x^2 - 4x) + 4y^2 = 36 \implies 3(x-2)^2 - 12 + 4y^2 = 36 \implies 3(x-2)^2 + 4y^2 = 48\). Divide the entire equation by 48 to obtain the standard ellipse form: \(\frac{(x-2)^2}{16} + \frac{y^2}{12} = 1\). Matching this with the given form, we find the constants: \(h = 2\), \(a^2 = 16 \implies a = 4\), and \(b^2 = 12 \implies b = \sqrt{12} = 2\sqrt{3}\).

M1: Rearranges polar equation and substitutes \(r\cos\theta = x\) and \(r = \sqrt{x^2+y^2}\). M1: Correctly squares both sides to obtain a polynomial equation in \(x\) and \(y\). M1: Completes the square for the quadratic \(x\) terms and groups constants. A1: Obtains the correct Cartesian equation and clearly states the values \(h = 2\), \(a = 4\), and \(b = \sqrt{12}\) (or \(2\sqrt{3}\)).

**Method 1: Using standard series expansions**

The Maclaurin series for \( \mathrm{e}^u \) and \( \cos u \) are:
\( \mathrm{e}^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \dots \)
\( \cos u = 1 - \frac{u^2}{2!} + \dots \)

Substituting \( u = -x \) into the series for \( \mathrm{e}^u \):
\( \mathrm{e}^{-x} = 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \dots \)

Substituting \( u = 2x \) into the series for \( \cos u \):
\( \cos 2x = 1 - \frac{(2x)^2}{2!} + \dots = 1 - 2x^2 + \dots \)

Multiplying these two series and keeping terms up to \( x^3 \):
\( \mathrm{e}^{-x} \cos 2x = \left(1 - x + \frac{1}{2}x^2 - \frac{1}{6}x^3\right)(1 - 2x^2) + \mathcal{O}(x^4) \)
\( = 1(1 - 2x^2) - x(1 - 2x^2) + \frac{1}{2}x^2(1) - \frac{1}{6}x^3(1) + \mathcal{O}(x^4) \)
\( = 1 - 2x^2 - x + 2x^3 + \frac{1}{2}x^2 - \frac{1}{6}x^3 + \mathcal{O}(x^4) \)
\( = 1 - x - \frac{3}{2}x^2 + \frac{11}{6}x^3 \)

**Method 2: Repeated differentiation**

Let \( f(x) = \mathrm{e}^{-x} \cos 2x \). Then:
\( f(0) = \mathrm{e}^0 \cos 0 = 1 \)

First derivative:
\( f'(x) = -\mathrm{e}^{-x} \cos 2x - 2\mathrm{e}^{-x} \sin 2x = -\mathrm{e}^{-x}(\cos 2x + 2\sin 2x) \)
\( f'(0) = -1(1 + 0) = -1 \)

Second derivative:
\( f''(x) = \mathrm{e}^{-x}(\cos 2x + 2\sin 2x) - \mathrm{e}^{-x}(-2\sin 2x + 4\cos 2x) = \mathrm{e}^{-x}(-3\cos 2x + 4\sin 2x) \)
\( f''(0) = 1(-3 + 0) = -3 \)

Third derivative:
\( f'''(x) = -\mathrm{e}^{-x}(-3\cos 2x + 4\sin 2x) + \mathrm{e}^{-x}(6\sin 2x + 8\cos 2x) = \mathrm{e}^{-x}(11\cos 2x + 2\sin 2x) \)
\( f'''(0) = 1(11 + 0) = 11 \)

Substituting into the Maclaurin series formula:
\( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots \)
\( f(x) = 1 - x - \frac{3}{2}x^2 + \frac{11}{6}x^3 \)

M1: For using standard series of \( \mathrm{e}^{-x} \) and \( \cos 2x \) up to at least \( x^3 \) terms, or for calculating \( f'(x) \) and \( f''(x) \) correctly.
A1: For correct expansions of both series: \( \mathrm{e}^{-x} \approx 1 - x + \frac{1}{2}x^2 - \frac{1}{6}x^3 \) and \( \cos 2x \approx 1 - 2x^2 \); or for correct derivatives \( f'(0) = -1 \) and \( f''(0) = -3 \).
M1: For multiplying the two series correctly to obtain at least 3 terms, or for calculating \( f'''(x) \) and finding \( f'''(0) = 11 \).
A1: For obtaining the terms \( 1 - x - \frac{3}{2}x^2 \) (or equivalent fraction format).
A1: For obtaining the final term \( \frac{11}{6}x^3 \).

To find the arc length \( s \), we use the formula:
\( s = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \)

First, we find the derivative \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \sinh(2x) \)

Now, we substitute this into the arc length integrand:
\( 1 + \left(\frac{dy}{dx}\right)^2 = 1 + \sinh^2(2x) = \cosh^2(2x) \)

Thus, the integrand is:
\( \sqrt{\cosh^2(2x)} = \cosh(2x) \) (since \( \cosh(2x) > 0 \) for all real \( x \)).

We integrate this from \( x = 0 \) to \( x = \ln 2 \):
\( s = \int_{0}^{\ln 2} \cosh(2x) \, dx = \left[ \frac{1}{2}\sinh(2x) \right]_{0}^{\ln 2} \)

Evaluating at the upper limit:
\( 2x = 2\ln 2 = \ln 4 \)
\( \sinh(\ln 4) = \frac{e^{\ln 4} - e^{-\ln 4}}{2} = \frac{4 - \frac{1}{4}}{2} = \frac{15}{8} \)

Evaluating at the lower limit:
\( \sinh(0) = 0 \)

Therefore, the arc length is:
\( s = \frac{1}{2} \left( \frac{15}{8} \right) - 0 = \frac{15}{16} \).

M1: Differentiates \( y \) to obtain \( \frac{dy}{dx} = \sinh(2x) \) and substitutes into the arc length formula.
A1: Correctly simplifies the integrand to \( \cosh(2x) \).
M1: Integrates to get \( \frac{1}{2}\sinh(2x) \) and substitutes limits \( 0 \) and \( \ln 2 \) using exponential definitions or direct hyperbolic values.
A1: Obtains the correct exact answer \( \frac{15}{16} \) (or exact decimal equivalent \( 0.9375 \)).

To find the arc length \( s \) of a parametric curve, we use the formula:
\( s = \int_{t_1}^{t_2} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \)

First, we find the derivatives with respect to \( t \):
\( \frac{dx}{dt} = 1 - \text{sech}^2 t = \tanh^2 t \)
\( \frac{dy}{dt} = -\text{sech } t \tanh t \)

Next, we calculate the sum of the squares of the derivatives:
\( \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = (\tanh^2 t)^2 + (-\text{sech } t \tanh t)^2 \)
\( = \tanh^4 t + \text{sech}^2 t \tanh^2 t \)
\( = \tanh^2 t (\tanh^2 t + \text{sech}^2 t) \)

Using the identity \( 1 - \tanh^2 t = \text{sech}^2 t \), we have:
\( \tanh^2 t + \text{sech}^2 t = 1 \)

So,
\( \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = \tanh^2 t \)

Since \( t \ge 0 \), \( \tanh t \ge 0 \), so the integrand simplifies to:
\( \sqrt{\tanh^2 t} = \tanh t \)

We integrate this from \( t = 0 \) to \( t = \ln 3 \):
\( s = \int_{0}^{\ln 3} \tanh t \, dt = [\ln(\cosh t)]_{0}^{\ln 3} \)

Evaluating the limits:
At \( t = \ln 3 \):
\( \cosh(\ln 3) = \frac{e^{\ln 3} + e^{-\ln 3}}{2} = \frac{3 + \frac{1}{3}}{2} = \frac{5}{3} \)

At \( t = 0 \):
\( \cosh 0 = 1 \)

Therefore, the exact arc length is:
\( s = \ln\left(\frac{5}{3}\right) - \ln(1) = \ln\left(\frac{5}{3}\right) \).

M1: Differentiates both \( x \) and \( y \) correctly with respect to \( t \).
A1: Simplifies the expression \( \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 \) to \( \tanh^2 t \) using appropriate hyperbolic identities.
M1: Integrates \( \tanh t \) to get \( \ln(\cosh t) \) and substitutes the limits \( 0 \) and \( \ln 3 \).
A1: Obtains the correct exact answer \( \ln\left(\frac{5}{3}\right) \) (or equivalent form such as \( \ln 5 - \ln 3 \)).

When \(x = 0\), we substitute this into the equation of the curve to find \(y\): \(\ln(0+y) + 0 = 1 \implies \ln y = 1 \implies y = e\). Thus, the point of interest is \((0, e)\). Differentiating both sides of the equation \(\ln(x+y) + xy = 1\) implicitly with respect to \(x\) yields: \(\frac{1}{x+y}\left(1 + \frac{\text{d}y}{\text{d}x}\right) + y + x\frac{\text{d}y}{\text{d}x} = 0\). Substituting the coordinates \(x = 0\) and \(y = e\) into this equation: \(\frac{1}{e}\left(1 + \frac{\text{d}y}{\text{d}x}\right) + e + 0 = 0\). Multiplying through by \(e\): \(1 + \frac{\text{d}y}{\text{d}x} + e^2 = 0 \implies \frac{\text{d}y}{\text{d}x} = -(e^2 + 1)\).

M1: Substitute \(x=0\) to find the corresponding value of \(y = e\). M1: Perform implicit differentiation with respect to \(x\) on \(\ln(x+y)\) using the chain rule. A1: Correctly obtain the fully differentiated equation \(\frac{1}{x+y}\left(1 + \frac{\text{d}y}{\text{d}x}\right) + y + x\frac{\text{d}y}{\text{d}x} = 0\). A1.5: Substitute \(x=0\) and \(y=e\) and solve to obtain the final exact value \(-(e^2 + 1)\).

Differentiating \(x^2 y + y^3 = 10\) implicitly with respect to \(x\): \(2xy + x^2 \frac{\text{d}y}{\text{d}x} + 3y^2 \frac{\text{d}y}{\text{d}x} = 0\). Substituting \(x = 1\) and \(y = 2\): \(2(1)(2) + 1^2 \frac{\text{d}y}{\text{d}x} + 3(2^2) \frac{\text{d}y}{\text{d}x} = 0 \implies 4 + 13 \frac{\text{d}y}{\text{d}x} = 0 \implies \frac{\text{d}y}{\text{d}x} = -\frac{4}{13}\). Differentiating the first derivative equation \(2xy + (x^2 + 3y^2)\frac{\text{d}y}{\text{d}x} = 0\) with respect to \(x\) again: \(2y + 2x \frac{\text{d}y}{\text{d}x} + (2x + 6y \frac{\text{d}y}{\text{d}x})\frac{\text{d}y}{\text{d}x} + (x^2 + 3y^2)\frac{\text{d}^2 y}{\text{d}x^2} = 0\), which simplifies to \(2y + 4x \frac{\text{d}y}{\text{d}x} + 6y \left(\frac{\text{d}y}{\text{d}x}\right)^2 + (x^2 + 3y^2)\frac{\text{d}^2 y}{\text{d}x^2} = 0\). Substituting \(x = 1\), \(y = 2\), and \(\frac{\text{d}y}{\text{d}x} = -\frac{4}{13}\): \(2(2) + 4(1)\left(-\frac{4}{13}\right) + 6(2)\left(-\frac{4}{13}\right)^2 + (1^2 + 3(2^2))\frac{\text{d}^2 y}{\text{d}x^2} = 0 \implies 4 - \frac{16}{13} + \frac{192}{169} + 13 \frac{\text{d}^2 y}{\text{d}x^2} = 0 \implies \frac{660}{169} + 13 \frac{\text{d}^2 y}{\text{d}x^2} = 0 \implies \frac{\text{d}^2 y}{\text{d}x^2} = -\frac{660}{2197}\).

M1: Differentiate the original implicit equation with respect to \(x\) to obtain a first-order differential expression. A1: Correctly find \(\frac{\text{d}y}{\text{d}x} = -\frac{4}{13}\) at the point \((1, 2)\). M1: Differentiate implicitly a second time, applying the product and chain rules appropriately. A1: Correctly obtain the second derivative equation in terms of \(x\), \(y\), \(\frac{\text{d}y}{\text{d}x}\), and \(\frac{\text{d}^2y}{\text{d}x^2}\). A0.5: Substitute numerical values and solve to obtain the final exact value of \(-\frac{660}{2197}\).

(i) Let \(y = u e^{-2x}\).
Using the product rule:
\[\frac{dy}{dx} = \frac{du}{dx} e^{-2x} - 2u e^{-2x}\]
\[\frac{d^2 y}{dx^2} = \frac{d^2 u}{dx^2} e^{-2x} - 2\frac{du}{dx} e^{-2x} - 2\frac{du}{dx} e^{-2x} + 4u e^{-2x} = e^{-2x} \left( \frac{d^2 u}{dx^2} - 4\frac{du}{dx} + 4u \right)\]
Substitute these into the left-hand side of equation (I):
\[\frac{d^2 y}{dx^2} + 4\frac{dy}{dx} + 5y = e^{-2x} \left( \frac{d^2 u}{dx^2} - 4\frac{du}{dx} + 4u \right) + 4 e^{-2x} \left( \frac{du}{dx} - 2u \right) + 5 u e^{-2x}\]
\[= e^{-2x} \left( \frac{d^2 u}{dx^2} - 4\frac{du}{dx} + 4u + 4\frac{du}{dx} - 8u + 5u \right)\]
\[= e^{-2x} \left( \frac{d^2 u}{dx^2} + u \right)\]
Since this equals \(10 e^{-x}\), we have:
\[e^{-2x} \left( \frac{d^2 u}{dx^2} + u \right) = 10 e^{-x}\]
Multiplying both sides by \(e^{2x}\) gives:
\[\frac{d^2 u}{dx^2} + u = 10 e^{x}\]

(ii) To solve the homogeneous equation \(\frac{d^2 u}{dx^2} + u = 0\), we write the auxiliary equation:
\[m^2 + 1 = 0 \implies m = \pm i\]
Thus, the complementary function (CF) is:
\[u_{CF} = A \cos x + B \sin x\]
To find the particular integral (PI) for \(\frac{d^2 u}{dx^2} + u = 10 e^x\), we try \(u = k e^x\):
\[\frac{du}{dx} = k e^x \quad \text{and} \quad \frac{d^2 u}{dx^2} = k e^x\]
Substituting into equation (II):
\[k e^x + k e^x = 10 e^x \implies 2k = 10 \implies k = 5\]
So, the particular integral is \(u_{PI} = 5 e^x\).
Thus, the general solution for \(u\) is:
\[u = A \cos x + B \sin x + 5 e^x\]
Using the substitution \(y = u e^{-2x}\), the general solution for \(y\) is:
\[y = e^{-2x} (A \cos x + B \sin x + 5 e^x) = e^{-2x} (A \cos x + B \sin x) + 5 e^{-x}\]

(iii) We are given that \(y = 7\) when \(x = 0\):
\[7 = e^0 (A \cos 0 + B \sin 0) + 5 e^0 \implies 7 = A + 5 \implies A = 2\]
Now differentiate \(y\) with respect to \(x\):
\[\frac{dy}{dx} = -2e^{-2x}(A \cos x + B \sin x) + e^{-2x}(-A \sin x + B \cos x) - 5e^{-x}\]
Substitute \(x = 0\) and \(\frac{dy}{dx} = -9\):
\[-9 = -2(A) + B - 5 \implies -2A + B = -4\]
Since \(A = 2\):
\[-2(2) + B = -4 \implies -4 + B = -4 \implies B = 0\]
Therefore, the particular solution is:
\[y = 2 e^{-2x} \cos x + 5 e^{-x}\]

(i)
M1: For attempting to find \(\frac{dy}{dx}\) and \(\frac{d^2y}{dx^2}\) in terms of \(u\) using the product rule.
A1: Correct expressions for \(\frac{dy}{dx}\) and \(\frac{d^2y}{dx^2}\).
A1: Correctly substitutes into equation (I) and simplifies to obtain the given differential equation (II) with clear working shown.

(ii)
M1: Solves auxiliary equation to find the complementary function for \(u\).
A1: Correct CF: \(u = A\cos x + B\sin x\) (or using other arbitrary constant names).
M1: Substitute form \(u = ke^x\) to find the particular integral.
A1: Correct PI: \(u = 5e^x\), leading to general solution for \(u\).
A1: Correct general solution for \(y\): \(y = e^{-2x}(A\cos x + B\sin x) + 5e^{-x}\) (accept equivalent forms).

(iii)
M1: Applies initial conditions to find the constants \(A\) and \(B\) (must attempt to differentiate general solution of \(y\) to apply \(\frac{dy}{dx} = -9\)).
A1: Correct particular solution: \(y = 2 e^{-2x} \cos x + 5 e^{-x}\).

Let \(w = -8 + 8\sqrt{3}\mathrm{i}\). First, find the modulus and argument of \(w\): \(|w| = \sqrt{(-8)^2 + (8\sqrt{3})^2} = \sqrt{64 + 192} = \sqrt{256} = 16\). Since the real part is negative and the imaginary part is positive, \(w\) lies in the second quadrant: \(\arg(w) = \pi - \arctan\left(\frac{8\sqrt{3}}{8}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}\). Using de Moivre's theorem, we express \(w\) as: \(w = 16\left(\cos\left(\frac{2\pi}{3} + 2k\pi\right) + \mathrm{i}\sin\left(\frac{2\pi}{3} + 2k\pi\right)\right)\) for \(k \in \mathbb{Z}\). Taking the fourth root: \(z = 16^{1/4}\left(\cos\left(\frac{\frac{2\pi}{3} + 2k\pi}{4}\right) + \mathrm{i}\sin\left(\frac{\frac{2\pi}{3} + 2k\pi}{4}\right)\right)\). The modulus is \(r = 16^{1/4} = 2\). The arguments for \(k = 0, 1, -1, -2\) to ensure \(-\pi < \theta \le \pi\) are: - For \(k = 0\): \(\theta_1 = \frac{2\pi}{12} = \frac{\pi}{6}\); - For \(k = 1\): \(\theta_2 = \frac{2\pi/3 + 2\pi}{4} = \frac{8\pi}{12} = \frac{2\pi}{3}\); - For \(k = -1\): \(\theta_3 = \frac{2\pi/3 - 2\pi}{4} = -\frac{4\pi}{12} = -\frac{\pi}{3}\); - For \(k = -2\): \(\theta_4 = \frac{2\pi/3 - 4\pi}{4} = -\frac{10\pi}{12} = -\frac{5\pi}{6}\). Therefore, the roots are: \(2\left(\cos\frac{\pi}{6} + \mathrm{i}\sin\frac{\pi}{6}\right)\), \(2\left(\cos\frac{2\pi}{3} + \mathrm{i}\sin\frac{2\pi}{3}\right)\), \(2\left(\cos\left(-\frac{\pi}{3}\right) + \mathrm{i}\sin\left(-\frac{\pi}{3}\right)\right)\), and \(2\left(\cos\left(-\frac{5\pi}{6}\right) + \mathrm{i}\sin\left(-\frac{5\pi}{6}\right)\right)\).

B1: Correctly find the modulus (16) and principal argument (\(2\pi/3\)) of \(-8 + 8\sqrt{3}\mathrm{i}\). M1: Apply de Moivre's theorem to find the general formula for the fourth root (dividing argument by 4, taking fourth root of modulus). A1: Correct modulus for the roots, \(r = 2\). A1: Correctly determine at least two of the four arguments in the range \((-\pi, \pi]\). A1: Express all four roots correctly in the required form.

By de Moivre's theorem: \(\cos 5\theta + \mathrm{i}\sin 5\theta = (\cos\theta + \mathrm{i}\sin\theta)^5\). Using the binomial expansion: \((\cos\theta + \mathrm{i}\sin\theta)^5 = \cos^5\theta + 5\mathrm{i}\cos^4\theta\sin\theta + 10\mathrm{i}^2\cos^3\theta\sin^2\theta + 10\mathrm{i}^3\cos^2\theta\sin^3\theta + 5\mathrm{i}^4\cos\theta\sin^4\theta + \mathrm{i}^5\sin^5\theta\). Substitute the powers of \(\mathrm{i}\): \(= \cos^5\theta + 5\mathrm{i}\cos^4\theta\sin\theta - 10\cos^3\theta\sin^2\theta - 10\mathrm{i}\cos^2\theta\sin^3\theta + 5\cos\theta\sin^4\theta + \mathrm{i}\sin^5\theta\). Equating the imaginary parts of both sides: \(\sin 5\theta = 5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta\). We need to express this solely in terms of \(\sin\theta\), so we use the identity \(\cos^2\theta = 1 - \sin^2\theta\): \(\sin 5\theta = 5(1-\sin^2\theta)^2\sin\theta - 10(1-\sin^2\theta)\sin^3\theta + \sin^5\theta\). Expand each term: \(\sin 5\theta = 5(1 - 2\sin^2\theta + \sin^4\theta)\sin\theta - 10(\sin^3\theta - \sin^5\theta) + \sin^5\theta\) which gives \(\sin 5\theta = 5\sin\theta - 10\sin^3\theta + 5\sin^5\theta - 10\sin^3\theta + 10\sin^5\theta + \sin^5\theta\). Combine like terms to get: \(\sin 5\theta = 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta\).

M1: State de Moivre's relation and expand \((\cos\theta + \mathrm{i}\sin\theta)^5\) binomially. A1: Identify and equate imaginary parts correctly to get \(\sin 5\theta = 5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta\). M1: Substitute \(\cos^2\theta = 1 - \sin^2\theta\) to eliminate all cosine terms. A1: Expand and simplify algebraic steps correctly. A1: Obtain the correct final simplified expression: \(16\sin^5\theta - 20\sin^3\theta + 5\sin\theta\).

Since \(f(x) = \frac{1}{x^2}\) is a decreasing function for \(x \ge 1\), the area of the rectangles lies above the curve: \(\sum_{r=2}^{n} f(r) < \int_{1}^{n} f(x) \, dx\). Evaluating this definite integral: \(\int_{1}^{n} \frac{1}{x^2} \, dx = \left[ -\frac{1}{x} \right]_1^n = 1 - \frac{1}{n}\). Therefore, we have: \(\sum_{r=2}^{n} \frac{1}{r^2} < 1 - \frac{1}{n}\). Adding the first term \(f(1) = 1\) to both sides of the inequality gives: \(\sum_{r=1}^{n} \frac{1}{r^2} < 1 + 1 - \frac{1}{n} = 2 - \frac{1}{n}\). Thus, the upper bound is \(2 - \frac{1}{n}\).

M1: For using a valid comparison between the sum of the series and the integral of \(f(x)\), such as \(\sum_{r=2}^{n} \frac{1}{r^2} < \int_{1}^{n} \frac{1}{x^2} \, dx\). A1: For obtaining the correct integrated expression of \(1 - \frac{1}{n}\). A1: For adding the first term \(1\) to get the final upper bound of \(2 - \frac{1}{n}\).

Since \(f(x) = \frac{1}{2x+1}\) is a decreasing function for \(x \ge 1\), the area of the rectangles lies below the curve: \(\sum_{r=1}^{n} f(r) > \int_{1}^{n+1} f(x) \, dx\). We evaluate the definite integral: \(\int_{1}^{n+1} \frac{1}{2x+1} \, dx = \left[ \frac{1}{2} \ln(2x+1) \right]_1^{n+1} = \frac{1}{2} \ln(2n+3) - \frac{1}{2} \ln 3 = \frac{1}{2} \ln\left(\frac{2n+3}{3}\right)\). Thus, a lower bound for the sum is \(\frac{1}{2} \ln\left(\frac{2n+3}{3}\right)\).

M1: For establishing the inequality \(\sum_{r=1}^{n} f(r) > \int_{1}^{n+1} f(x) \, dx\) with appropriate limits. A1: For performing the integration correctly to obtain \(\left[ \frac{1}{2} \ln(2x+1) \right]\). A1: For correctly evaluating the limits to obtain the lower bound of \(\frac{1}{2} \ln\left(\frac{2n+3}{3}\right)\) or equivalent.

Since \(f(x) = xe^{-x}\) is decreasing for \(x \ge 1\), we have \(\sum_{r=2}^{n} f(r) < \int_{1}^{n} f(x) \, dx\). We integrate \(f(x)\) by parts: Let \(u = x \implies du = dx\), and \(dv = e^{-x} dx \implies v = -e^{-x}\). Then \(\int xe^{-x} \, dx = -xe^{-x} - \int -e^{-x} \, dx = -(x+1)e^{-x}\). Evaluating the integral from \(1\) to \(n\): \(\int_{1}^{n} xe^{-x} \, dx = \left[ -(x+1)e^{-x} \right]_1^n = -(n+1)e^{-n} + 2e^{-1}\). Adding the first term \(f(1) = e^{-1}\) to both sides: \(\sum_{r=1}^{n} re^{-r} < e^{-1} + 2e^{-1} - (n+1)e^{-n} = 3e^{-1} - (n+1)e^{-n}\). Thus, the upper bound is \(3e^{-1} - (n+1)e^{-n}\).

M1: For integrating \(\int xe^{-x} \, dx\) using integration by parts. A1: For obtaining the correct integral value \(2e^{-1} - (n+1)e^{-n}\) with limits \(1\) to \(n\). A1: For adding the first term \(e^{-1}\) to get the correct upper bound of \(3e^{-1} - (n+1)e^{-n}\).

(a) Given the substitution \(u = y^{-2}\), differentiating with respect to \(x\) using the chain rule gives:
\[ \frac{du}{dx} = -2y^{-3} \frac{dy}{dx} \implies y^{-3} \frac{dy}{dx} = -\frac{1}{2} \frac{du}{dx}. \]

Now divide the original differential equation
\[ x \frac{dy}{dx} + 3y = x^2 y^3 \]
by \(y^3\) to obtain:
\[ x y^{-3} \frac{dy}{dx} + 3 y^{-2} = x^2. \]

Substitute \(y^{-3} \frac{dy}{dx} = -\frac{1}{2} \frac{du}{dx}\) and \(u = y^{-2}\):
\[ x \left(-\frac{1}{2} \frac{du}{dx}\right) + 3u = x^2 \]
\[ -\frac{1}{2} x \frac{du}{dx} + 3u = x^2. \]

Dividing both sides by \(-\frac{1}{2}x\) (for \(x \neq 0\)):
\[ \frac{du}{dx} - \frac{6}{x} u = -2x. \]
This is the required form.

(b) To solve the linear differential equation in \(u\):
\[ \frac{du}{dx} - \frac{6}{x} u = -2x, \]
we find the integrating factor \(I(x)\):
\[ I(x) = e^{\int -\frac{6}{x} \\, dx} = e^{-6 \ln x} = x^{-6}. \]

Multiplying both sides of the differential equation by \(x^{-6}\):
\[ x^{-6} \frac{du}{dx} - 6 x^{-7} u = -2 x^{-5} \]
\[ \frac{d}{dx} \left(u x^{-6}\right) = -2 x^{-5}. \]

Integrating both sides with respect to \(x\):
\[ u x^{-6} = \int -2 x^{-5} \\, dx \]
\[ u x^{-6} = \frac{-2}{-4} x^{-4} + C \]
\[ u x^{-6} = \frac{1}{2} x^{-4} + C. \]

Multiplying by \(x^6\) gives:
\[ u = \frac{1}{2} x^2 + C x^6. \]

Substitute back \(u = y^{-2}\):
\[ y^{-2} = \frac{1}{2} x^2 + C x^6. \]

Using the boundary condition \(y = 1\) when \(x = 1\):
\[ 1^{-2} = \frac{1}{2} (1)^2 + C (1)^6 \implies 1 = \frac{1}{2} + C \implies C = \frac{1}{2}. \]

Thus, the equation is:
\[ y^{-2} = \frac{1}{2} x^2 + \frac{1}{2} x^6 = \frac{x^2 + x^6}{2}. \]

Taking the reciprocal to express \(y^2\) in terms of \(x\):
\[ y^2 = \frac{2}{x^2 + x^6} \quad \left(\text{or } y^2 = \frac{2}{x^2(1+x^4)}\right). \]

(a)
- M1: For differentiating \(u = y^{-2}\) to get \(\frac{du}{dx} = k y^{-3} \frac{dy}{dx}\) (where \(k \neq 0\)).
- M1: For substituting \(u\) and \(\frac{du}{dx}\) into the divided form of the given differential equation.
- A1: For obtaining the printed result \(\frac{du}{dx} - \frac{6}{x} u = -2x\) with clear, correct working.

(b)
- M1: For finding the correct integrating factor, \(I(x) = e^{\int -\frac{6}{x} \\, dx} = x^{-6}\).
- M1: For multiplying by their integrating factor and integrating the RHS of the form \(A x^{-5}\) to obtain \(B x^{-4}\).
- A1: For the correct integrated expression: \(u x^{-6} = \frac{1}{2} x^{-4} + C\) (constant of integration must be present).
- M1: For substituting \(u = y^{-2}\) and applying the initial conditions \(x=1, y=1\) to find \(C\).
- A1: For finding \(C = \frac{1}{2}\) (or equivalent value depending on the form of the general solution).
- A1: For expressing the final answer as \(y^2 = \frac{2}{x^2 + x^6}\) or \(y^2 = \frac{2}{x^2(1+x^4)}\).

To find the diagonalisation of \(A\), we first calculate its eigenvalues by solving the characteristic equation \(\det(A - \lambda I) = 0\):
\[\det \begin{pmatrix} 5-\lambda & -4 \\ 2 & -1-\lambda \end{pmatrix} = 0\]
\[(5-\lambda)(-1-\lambda) - (-8) = 0 \implies \lambda^2 - 4\lambda + 3 = 0\]
\[(\lambda - 1)(\lambda - 3) = 0\]
So the eigenvalues are \(\lambda = 1\) and \(\lambda = 3\).

Next, we find the corresponding eigenvectors.
For \(\lambda = 1\):
\[\begin{pmatrix} 4 & -4 \\ 2 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies x = y\]
An eigenvector is \(\mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}\).

For \(\lambda = 3\):
\[\begin{pmatrix} 2 & -4 \\ 2 & -4 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies x = 2y\]
An eigenvector is \(\mathbf{v}_2 = \begin{pmatrix} 2 \\ 1 \end{pmatrix}\).

Thus, we can define the matrices:
\[P = \begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix} \quad \text{and} \quad D = \begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix}\]
Now find the inverse of \(P\):
\[P^{-1} = \frac{1}{(1)(1) - (2)(1)} \begin{pmatrix} 1 & -2 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} -1 & 2 \\ 1 & -1 \end{pmatrix}\]
Using the relation \(A^n = P D^n P^{-1}\), we have:
\[A^n = \begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1^n & 0 \\ 0 & 3^n \end{pmatrix} \begin{pmatrix} -1 & 2 \\ 1 & -1 \end{pmatrix}\]
\[A^n = \begin{pmatrix} 1 & 2 \cdot 3^n \\ 1 & 3^n \end{pmatrix} \begin{pmatrix} -1 & 2 \\ 1 & -1 \end{pmatrix}\]
\[A^n = \begin{pmatrix} -1 + 2 \cdot 3^n & 2 - 2 \cdot 3^n \\ -1 + 3^n & 2 - 3^n \end{pmatrix} = \begin{pmatrix} 2 \cdot 3^n - 1 & 2 - 2 \cdot 3^n \\ 3^n - 1 & 2 - 3^n \end{pmatrix}\]

M1: For setting up and solving \(\det(A - \lambda I) = 0\) to find eigenvalues \(\lambda = 1\) and \(\lambda = 3\).
M1: For finding correct eigenvectors \(\begin{pmatrix} 1 \\ 1 \end{pmatrix}\) and \(\begin{pmatrix} 2 \\ 1 \end{pmatrix}\) (or any scalar multiples).
A1: For writing correct \(P\), \(D\) and finding the inverse matrix \(P^{-1} = \begin{pmatrix} -1 & 2 \\ 1 & -1 \end{pmatrix}\).
A1.66: For multiplying the three matrices \(P D^n P^{-1}\) correctly to obtain the given expression for \(A^n\).

We set up the augmented matrix for the system and perform row operations to find conditions for consistency:
\[\begin{pmatrix} 1 & 2 & -1 & | & 3 \\ 2 & 5 & a & | & 7 \\ 3 & 7 & 2 & | & b \end{pmatrix}\]
Subtract \(2 \times\) Row 1 from Row 2, and \(3 \times\) Row 1 from Row 3:
\[\begin{pmatrix} 1 & 2 & -1 & | & 3 \\ 0 & 1 & a+2 & | & 1 \\ 0 & 1 & 5 & | & b-9 \end{pmatrix}\]
Subtract Row 2 from Row 3:
\[\begin{pmatrix} 1 & 2 & -1 & | & 3 \\ 0 & 1 & a+2 & | & 1 \\ 0 & 0 & 3-a & | & b-10 \end{pmatrix}\]
For the system to have infinitely many solutions, the last row must represent the equation \(0x + 0y + 0z = 0\). This requires:
\[3 - a = 0 \implies a = 3\]
\[b - 10 = 0 \implies b = 10\]
Substituting \(a = 3\) back into the system, the remaining equations are:
\[y + 5z = 1\]
\[x + 2y - z = 3\]
Let \(z = t\), where \(t\) is a real parameter. Then:
\[y = 1 - 5t\]
Substitute \(y\) and \(z\) into the first equation:
\[x + 2(1 - 5t) - t = 3 \implies x + 2 - 11t = 3 \implies x = 11t + 1\]
Thus, the general solution is:
\[x = 11t + 1, \quad y = 1 - 5t, \quad z = t\]

M1: For executing row operations on the augmented matrix to obtain a row-echelon form equivalent to \((3-a)z = b-10\).
A1: For deducing that \(3-a=0\) and \(b-10=0\) which gives \(a=3\) and \(b=10\).
A1: For parameterizing the system with parameter \(t\) (or any other letter) and establishing \(y = 1 - 5t\).
A1.66: For substituting back to find \(x = 11t + 1\) and stating the complete general solution correctly.

First, we find the eigenvalues of \(A\) by solving the characteristic equation \(\det(A - \lambda I) = 0\):
\[\begin{vmatrix} 1-\lambda & 0 & 2 \\ 0 & 3-\lambda & 0 \\ 2 & 0 & 1-\lambda \end{vmatrix} = 0\]
Expanding along the second row gives:
\[(3-\lambda) \left((1-\lambda)^2 - 4\right) = 0 \implies (3-\lambda)(\lambda^2 - 2\lambda - 3) = 0\]
\[(3-\lambda)(\lambda - 3)(\lambda + 1) = 0\]
Thus, the eigenvalues are \(\lambda = -1\) and \(\lambda = 3\) (with multiplicity 2).

Now, we find corresponding normalized eigenvectors.
For \(\lambda = -1\):
\[\begin{pmatrix} 2 & 0 & 2 \\ 0 & 4 & 0 \\ 2 & 0 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies y = 0 \text{ and } x + z = 0\]
An eigenvector is \(\begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}\). Normalizing this vector yields \(\mathbf{u}_1 = \begin{pmatrix} \frac{1}{\sqrt{2}} \\ 0 \\ -\frac{1}{\sqrt{2}} \end{pmatrix}\).

For \(\lambda = 3\):
\[\begin{pmatrix} -2 & 0 & 2 \\ 0 & 0 & 0 \\ 2 & 0 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies x - z = 0\]
We need two orthogonal eigenvectors in this eigenspace. We can choose:
\(\mathbf{v}_2 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}\) and \(\mathbf{v}_3 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\).
These are mutually orthogonal since \(\mathbf{v}_2 \cdot \mathbf{v}_3 = 0\).
Normalizing these yields:
\[\mathbf{u}_2 = \begin{pmatrix} \frac{1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}} \end{pmatrix} \quad \text{and} \quad \mathbf{u}_3 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\]

Thus, the orthogonal matrix \(P\) and diagonal matrix \(D\) are:
\[P = \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1 \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \end{pmatrix} \quad \text{and} \quad D = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{pmatrix}\]
(Note: Permutations of the columns of \(P\) and the diagonal entries of \(D\) are also correct.)

M1: For finding the eigenvalues \(\lambda = -1\) and \(\lambda = 3\) (with multiplicity 2).
A1: For finding three correct, mutually orthogonal eigenvectors.
M1: For normalizing all three eigenvectors to unit length.
A1.66: For constructing the correct orthogonal matrix \(P\) and corresponding diagonal matrix \(D\).