2026 Edexcel IAL Further Mathematics (YFM01) Practice Paper with Answers

(a) From the given equation \(2x^2 - 5x + 4 = 0\), we have:
\(\alpha + \beta = \frac{5}{2}\)
\(\alpha\beta = \frac{4}{2} = 2\)

Using the identity:
\(\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)\)
\(\alpha^3 + \beta^3 = \left(\frac{5}{2}\right)^3 - 3(2)\left(\frac{5}{2}\right) = \frac{125}{8} - 15 = \frac{5}{8}\)

(b) Let the new roots be \(u = \alpha^2 + \frac{2}{\beta}\) and \(v = \beta^2 + \frac{2}{\alpha}\).
Since \(\alpha\beta = 2\), we have \(\frac{2}{\beta} = \alpha\) and \(\frac{2}{\alpha} = \beta\).
Thus:
\(u = \alpha^2 + \alpha\)
\(v = \beta^2 + \beta\)

Sum of new roots:
\(u + v = (\alpha^2 + \beta^2) + (\alpha + \beta)\)
Since \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(\frac{5}{2}\right)^2 - 2(2) = \frac{25}{4} - 4 = \frac{9}{4}\),
we have:
\(u + v = \frac{9}{4} + \frac{5}{2} = \frac{19}{4}\)

Product of new roots:
\(uv = (\alpha^2 + \alpha)(\beta^2 + \beta) = \alpha^2\beta^2 + \alpha^2\beta + \alpha\beta^2 + \alpha\beta = (\alpha\beta)^2 + \alpha\beta(\alpha + \beta) + \alpha\beta\)
\(uv = 2^2 + 2\left(\frac{5}{2}\right) + 2 = 4 + 5 + 2 = 11\)

The new quadratic equation is:
\(x^2 - (u + v)x + uv = 0\)
\(x^2 - \frac{19}{4}x + 11 = 0\)
Multiplying by 4 to get integer coefficients:
\(4x^2 - 19x + 44 = 0\)

(a)
M1: Attempt to write down \(\alpha + \beta\) and \(\alpha\beta\) (at least one correct).
M1: Use the correct algebraic identity \(\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)\).
A1: Correct value of \(\frac{5}{8}\) (or 0.625).
(b)
M1: Realise \(\frac{2}{\beta} = \alpha\) and \(\frac{2}{\alpha} = \beta\) using \(\alpha\beta = 2\), or express the sum and product directly in terms of \(\alpha\) and \(\beta\).
M1: Substitute values to find the sum of the new roots, \(u + v = \frac{19}{4}\).
A1: Find the product of the new roots, \(uv = 11\).
A1: Form the quadratic equation \(4x^2 - 19x + 44 = 0\) (or any integer multiple thereof, must include '= 0').

(a) Evaluate \(\text{f}(x)\) at the interval endpoints:
\(\text{f}(2.8) = 2(2.8)^3 - 5(2.8)^2 - 4(2.8) + 6 = 43.904 - 39.2 - 11.2 + 6 = -0.496\)
\(\text{f}(2.9) = 2(2.9)^3 - 5(2.9)^2 - 4(2.9) + 6 = 48.778 - 42.05 - 11.6 + 6 = 1.128\)
Since there is a change of sign between \(\text{f}(2.8) < 0\) and \(\text{f}(2.9) > 0\), and \(\text{f}(x)\) is continuous, \(\alpha\) lies in the interval \([2.8, 2.9]\).

(b) First derivative:
\(\text{f}'(x) = 6x^2 - 10x - 4\)
At \(x_0 = 2.8\):
\(\text{f}'(2.8) = 6(2.8)^2 - 10(2.8) - 4 = 47.04 - 28 - 4 = 15.04\)
Apply Newton-Raphson:
\(x_1 = x_0 - \frac{\text{f}(x_0)}{\text{f}'(x_0)} = 2.8 - \frac{-0.496}{15.04} \approx 2.8 + 0.0329787 = 2.833\) (to 3 d.p.)

(c) Using linear interpolation on \([2.8, 2.9]\):
\(\frac{x - 2.8}{2.9 - x} =
\frac{|\text{f}(2.8)|}{|\text{f}(2.9)|} = \frac{0.496}{1.128}\)
\(x = 2.8 + \frac{0.496}{0.496 + 1.128} \times (2.9 - 2.8) = 2.8 + \frac{0.496}{1.624} \times 0.1 \approx 2.8 + 0.03054187 = 2.831\) (to 3 d.p.)

(a)
M1: Attempt to evaluate both \(\text{f}(2.8)\) and \(\text{f}(2.9)\).
A1: Obtain \(\text{f}(2.8) = -0.496\) and \(\text{f}(2.9) = 1.128\), and conclude with a reference to sign change and continuity.
(b)
M1: Differentiate \(\text{f}(x)\) to find \(\text{f}'(x)\).
M1: Correct substitution of \(x_0 = 2.8\) into the Newton-Raphson formula.
A1: Correct answer \(2.833\) (must be 3 d.p.).
(c)
M1: Set up a correct linear interpolation equation or ratio.
M1: Solve the equation for the approximation.
A1: Correct answer \(2.831\) (must be 3 d.p.).

(a) Since the coefficients of the cubic equation are real, the complex roots must occur in conjugate pairs.
Therefore, the second root is \(z_2 = 3 - 2\text{i}\).

(b) Let the third root be \(z_3\).
The product of the roots is given by the constant term of the cubic equation:
\(z_1 z_2 z_3 = -(-26) = 26\)
\((3 + 2\text{i})(3 - 2\text{i}) z_3 = 26\)
\((9 + 4) z_3 = 26 \implies 13 z_3 = 26 \implies z_3 = 2\)

(c) The sum of the roots is:
\(z_1 + z_2 + z_3 = -a\)
\((3 + 2\text{i}) + (3 - 2\text{i}) + 2 = 8 \implies -a = 8 \implies a = -8\)

The sum of the pairwise products of the roots is:
\(z_1 z_2 + z_2 z_3 + z_3 z_1 = b\)
\(13 + 2(3 - 2\text{i}) + 2(3 + 2\text{i}) = b\)
\(13 + 6 - 4\text{i} + 6 + 4\text{i} = b \implies b = 25\)

Thus, \(a = -8\) and \(b = 25\).

(a)
B1: Correct conjugate root \(3 - 2\text{i}\).
(b)
M1: State or use the relationship between the product of the three roots and the constant term of the cubic.
M1: Correct calculation of the product \((3+2\text{i})(3-2\text{i}) = 13\).
A1: Correct value \(z_3 = 2\).
(c)
M1: Use the sum of roots relation to form an equation for \(a\).
A1: Correct value \(a = -8\).
M1: Use the sum of products pairwise relation to form an equation for \(b\) (or substitute \(z=2\) into the original equation).
A1: Correct value \(b = 25\).

(a) Differentiating \(y^2 = 12x\) implicitly with respect to \(x\):
\(2y \frac{\text{d}y}{\text{d}x} = 12 \implies \frac{\text{d}y}{\text{d}x} =
\frac{6}{y}\)
At the point \(P(3t^2, 6t)\):
\(\frac{\text{d}y}{\text{d}x} = \frac{6}{6t} = \frac{1}{t}\)
So the gradient of the normal is \(-t\).
The equation of the normal at \(P\) is:
\(y - 6t = -t(x - 3t^2)\)
\(y - 6t = -tx + 3t^3\)
\(y + tx = 6t + 3t^3\) (as required).

(b) The normal cuts the \(x\)-axis at \(N\), where \(y = 0\).
Substitute \(y = 0\) into the normal equation:
\(tx = 6t + 3t^3\)
Since \(t \ne 0\), divide by \(t\):
\(x = 6 + 3t^2\)
So the coordinates of \(N\) are \((6 + 3t^2, 0)\).
The coordinates of the origin \(O\) are \((0, 0)\).
The coordinates of the midpoint of \(ON\) are:
\(\left(\frac{0 + 6 + 3t^2}{2}, 0\right) = \left(3 + \frac{3}{2}t^2, 0\right)\)

(a)
M1: Attempt to differentiate \(y^2 = 12x\) to find \(\frac{\text{d}y}{\text{d}x}\).
A1: Correct gradient of tangent at \(P\) is \(\frac{1}{t}\).
M1: State that the gradient of the normal is \(-t\) and write down a linear equation through \(P\) with this gradient.
A1*: Fully correct derivation leading to the given equation of the normal.
(b)
M1: Set \(y = 0\) in the normal equation to find the \(x\)-coordinate of \(N\).
A1: Correct coordinates of \(N(6 + 3t^2, 0)\).
A1: Correct coordinates of the midpoint \((3 + 1.5t^2, 0)\).

(a) The determinant of \(\mathbf{M}\) is:
\(\det(\mathbf{M}) = (2a)(a) - (a-1)(-3) = 2a^2 + 3(a-1) = 2a^2 + 3a - 3\)
We are given \(\det(\mathbf{M}) = 6\):
\(2a^2 + 3a - 3 = 6 \implies 2a^2 + 3a - 9 = 0\)
Factorising the quadratic equation:
\((2a - 3)(a + 3) = 0\)
Thus, the possible values of \(a\) are \(a = 1.5\) and \(a = -3\).

(b) Using the negative value of \(a\), we have \(a = -3\).
The matrix \(\mathbf{M}\) becomes:
\(\mathbf{M} = \begin{pmatrix} 2(-3) & -3-1 \\ -3 & -3 \end{pmatrix} = \begin{pmatrix} -6 & -4 \\ -3 & -3 \end{pmatrix}\)
The determinant is \(\det(\mathbf{M}) = 6\).
The inverse matrix \(\mathbf{M}^{-1}\) is:
\(\mathbf{M}^{-1} = \frac{1}{6} \begin{pmatrix} -3 & -(-4) \\ -(-3) & -6 \end{pmatrix} = \frac{1}{6} \begin{pmatrix} -3 & 4 \\ 3 & -6 \end{pmatrix} = \begin{pmatrix} -0.5 & \frac{2}{3} \\ 0.5 & -1 \end{pmatrix}\)

(a)
M1: Correct expression for \(\det(\mathbf{M})\) in terms of \(a\).
M1: Set \(\det(\mathbf{M}) = 6\) to form a quadratic equation in \(a\).
A1: Both correct values \(a = 1.5\) and \(a = -3\).
(b)
B1: State the correct matrix \(\mathbf{M} = \begin{pmatrix} -6 & -4 \\ -3 & -3 \end{pmatrix}\).
M1: Apply the formula for the inverse of a \(2 \times 2\) matrix (swapping leading diagonal elements and negating other elements).
M1: Correctly divide by the determinant (which is 6).
A2: Correct inverse matrix \(\begin{pmatrix} -0.5 & \frac{2}{3} \\ 0.5 & -1 \end{pmatrix}\) (deduct 1 mark for minor arithmetic errors).

(a) Expanding the summand:
\(\sum_{r=1}^{n} r(r + 3) = \sum_{r=1}^{n} (r^2 + 3r) = \sum_{r=1}^{n} r^2 + 3\sum_{r=1}^{n} r\)
Using standard formulae:
\(\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n+1)(2n+1)\)
\(\sum_{r=1}^{n} r = \frac{1}{2}n(n+1)\)
So,
\(\sum_{r=1}^{n} r(r + 3) = \frac{1}{6}n(n+1)(2n+1) + 3 \left[\frac{1}{2}n(n+1)\right]\)
Factor out \(\frac{1}{6}n(n+1)\):
\(= \frac{1}{6}n(n+1) [ (2n+1) + 9 ] =
\frac{1}{6}n(n+1) (2n + 10)\)
\(= \frac{1}{3}n(n+1)(n+5)\) (as required).

(b) Note that:
\(\sum_{r=n+1}^{2n} r(r + 3) = \sum_{r=1}^{2n} r(r + 3) - \sum_{r=1}^{n} r(r+3)\)
Using the result from (a):
\(= \frac{1}{3}(2n)(2n+1)(2n+5) - \frac{1}{3}n(n+1)(n+5)\)
Factor out \(\frac{1}{3}n\):
\(= \frac{1}{3}n [ 2(2n+1)(2n+5) - (n+1)(n+5) ]\)
\(=
\frac{1}{3}n [ 2(4n^2 + 12n + 5) - (n^2 + 6n + 5) ]\)
\(= \frac{1}{3}n [ 8n^2 + 24n + 10 - n^2 - 6n - 5 ]\)
\(= \frac{1}{3}n(7n^2 + 18n + 5)\) (as required).

(a)
M1: Expand to \(\sum r^2 + 3\sum r\) and substitute standard summation formulae.
M1: Attempt to factorise out at least \(\frac{1}{6}n(n+1)\).
A1: Simplify the brackets to \((2n+10)\).
A1*: Fully correct working leading to the given expression.
(b)
M1: Use the identity \(\sum_{r=n+1}^{2n} = \sum_{r=1}^{2n} - \sum_{r=1}^{n}\).
M1: Substitute the formula from (a) into both terms and attempt to factor out \(\frac{1}{3}n\).
A1*: Fully correct algebraic simplification leading to the final given result.

Let \(P(n)\) be the statement \(\sum_{r=1}^{n} (3r - 2)2^r = (3n - 5)2^{n+1} + 10\).

**Base Case:** \(n=1\)
LHS: \((3(1) - 2)2^1 = 1 \times 2 = 2\)
RHS: \((3(1) - 5)2^{1+1} + 10 = (-2)(4) + 10 = -8 + 10 = 2\)
Since LHS = RHS, \(P(1)\) is true.

**Inductive Step:**
Assume \(P(k)\) is true for some positive integer \(k\). That is:
\(\sum_{r=1}^{k} (3r - 2)2^r = (3k - 5)2^{k+1} + 10\)
We must show that \(P(k+1)\) is true:
\(\sum_{r=1}^{k+1} (3r - 2)2^r = (3(k+1) - 5)2^{k+2} + 10 = (3k - 2)2^{k+2} + 10\)

Now:
\(\sum_{r=1}^{k+1} (3r - 2)2^r = \sum_{r=1}^{k} (3r - 2)2^r + [3(k+1) - 2]2^{k+1}\)
\(= (3k - 5)2^{k+1} + 10 + (3k + 1)2^{k+1}\) (by inductive hypothesis)
\(= [ (3k - 5) + (3k + 1) ] 2^{k+1} + 10\) (factorising out \(2^{k+1}\))
\(= (6k - 4)2^{k+1} + 10\)
\(= 2(3k - 2)2^{k+1} + 10\)
\(= (3k - 2)2^{k+2} + 10\)

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

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

B1: Show that the formula holds for \(n=1\).
M1: State the inductive hypothesis clearly (assume true for \(n=k\)) and write an expression for \(n=k+1\).
M1: Substitute the inductive hypothesis into the sum of \(k+1\) terms.
A1: Factor out \(2^{k+1}\) to obtain \((6k-4)2^{k+1} + 10\).
A1: Correctly manipulate this to \((3k-2)2^{k+2} + 10\).
C1: Provide a complete and clear conclusion stating that the result is true for all positive integers \(n\) by mathematical induction.

(a) The matrix representing reflection in \(y = x\) is:
\(\mathbf{U} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\)
The matrix representing a rotation of \(90^\circ\) anticlockwise about the origin is:
\(\mathbf{V} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\)
Since \(W\) is \(U\) followed by \(V\), we have:
\(\mathbf{W} = \mathbf{V}\mathbf{U} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}\)

(b) The coordinates of the vertices of the image triangle \(T'\) are found by multiplying the coordinate matrix of \(T\) by \(\mathbf{W}\):
\(\mathbf{W}\mathbf{T} = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 2 & 5 & 5 \\ 1 & 1 & 3 \end{pmatrix} = \begin{pmatrix} -2 & -5 & -5 \\ 1 & 1 & 3 \end{pmatrix}\)
So, the vertices of the image triangle \(T'\) are \(A'(-2, 1)\), \(B'(-5, 1)\), and \(C'(-5, 3)\).

(c) The matrix \(\mathbf{W} = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}\) represents a reflection in the \(y\)-axis (or the line \(x = 0\)).

(a)
B1: Correct matrix \(\mathbf{U}\) representing reflection in \(y=x\).
B1: Correct matrix \(\mathbf{V}\) representing rotation by \(90^\circ\).
M1: Evaluate the product \(\mathbf{V}\mathbf{U}\) in the correct order.
A1: Correct matrix \(\mathbf{W} = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}\).
(b)
M1: Attempt to multiply \(\mathbf{W}\) by the coordinate matrix of the triangle.
A1: Correct vertices \(A'(-2, 1)\), \(B'(-5, 1)\), and \(C'(-5, 3)\).
(c)
M1: Identify the type of transformation as a reflection.
A1: Correctly state that the line of reflection is the \(y\)-axis (or \(x = 0\)).

(a) From the given quadratic equation \(2x^2 + 5x + 4 = 0\), we can identify the sum and product of the roots:
\(\alpha + \beta = -\frac{5}{2}\)
\(\alpha\beta = \frac{4}{2} = 2\)

We can express the sum of cubes using the identity:
\(\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)\)

Substituting the values:
\(\alpha^3 + \beta^3 = \left(-\frac{5}{2}\right)^3 - 3(2)\left(-\frac{5}{2}\right)\)
\(\alpha^3 + \beta^3 = -\frac{125}{8} + 15 = -\frac{125}{8} + \frac{120}{8} = -\frac{5}{8}\) (or \(-0.625\))

(b) Let the new roots be \(\gamma = \frac{\alpha}{\beta^2}\) and \(\delta = \frac{\beta}{\alpha^2}\).

The sum of the new roots is:
\(\gamma + \delta = \frac{\alpha}{\beta^2} + \frac{\beta}{\alpha^2} = \frac{\alpha^3 + \beta^3}{\alpha^2\beta^2} = \frac{\alpha^3 + \beta^3}{(\alpha\beta)^2}\)

Substituting our known values:
\(\gamma + \delta = \frac{-\frac{5}{8}}{2^2} = \frac{-\frac{5}{8}}{4} = -\frac{5}{32}\)

The product of the new roots is:
\(\gamma\delta = \left(\frac{\alpha}{\beta^2}\right)\left(\frac{\beta}{\alpha^2}\right) = \frac{\alpha\beta}{\alpha^2\beta^2} = \frac{1}{\alpha\beta}\)

Substituting \(\alpha\beta = 2\):
\(\gamma\delta = \frac{1}{2}\)

A quadratic equation with these roots is given by:
\(x^2 - (\gamma + \delta)x + \gamma\delta = 0\)
\(x^2 - \left(-\frac{5}{32}\right)x + \frac{1}{2} = 0\)
\(x^2 + \frac{5}{32}x + \frac{1}{2} = 0\)

Multiplying through by 32 to obtain integer coefficients:
\(32x^2 + 5x + 16 = 0\)

(a)
- B1: For correctly identifying \(\alpha + \beta = -2.5\) and \(\alpha\beta = 2\).
- M1: For using a correct algebraic identity for \(\alpha^3 + \beta^3\), such as \((\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)\).
- A1: For obtaining the correct value of \(-\frac{5}{8}\) or \(-0.625\).

(b)
- M1: For attempting to find the sum of the new roots in terms of \(\alpha^3 + \beta^3\) and \(\alpha\beta\), and substituting their values.
- A1: For finding the correct sum of the roots, \(-\frac{5}{32}\).
- B1: For finding the correct product of the roots, \(\frac{1}{2}\).
- M1: For attempting to form a quadratic equation using \(x^2 - (\text{sum})x + (\text{product}) = 0\).
- A1: For the correct equation \(32x^2 + 5x + 16 = 0\) (or any integer multiple thereof).

(a) Evaluating the function at the boundaries of the interval:
\(\text{f}(1.5) = 3(1.5)^3 - 4(1.5)^2 - 5(1.5) + 3\)
\(\text{f}(1.5) = 3(3.375) - 4(2.25) - 7.5 + 3 = 10.125 - 9 - 7.5 + 3 = -3.375\)

\(\text{f}(2) = 3(2)^3 - 4(2)^2 - 5(2) + 3\)
\(\text{f}(2) = 3(8) - 4(4) - 10 + 3 = 24 - 16 - 10 + 3 = 1\)

Since \(\text{f}(x)\) is continuous on the interval \([1.5, 2]\) and there is a change of sign between \(\text{f}(1.5) < 0\) and \(\text{f}(2) > 0\), there must be a root \(\alpha\) in the interval \([1.5, 2]\).

(b) Using linear interpolation on \([1.5, 2]\):
Let \(x_1 = 1.5\) and \(x_2 = 2\), with corresponding function values \(\text{f}(1.5) = -3.375\) and \(\text{f}(2) = 1\).

Using the linear interpolation formula:
\(\frac{\alpha - 1.5}{2 - \alpha} =
\frac{0 - (-3.375)}{1 - 0}\)
\(\frac{\alpha - 1.5}{2 - \alpha} = 3.375\)

Solving for \(\alpha\):
\(\alpha - 1.5 = 3.375(2 - \alpha)\)
\(\alpha - 1.5 = 6.75 - 3.375\alpha\)
\(4.375\alpha = 8.25\)
\(\alpha = \frac{8.25}{4.375} = \frac{66}{35} \approx 1.885714...\)

To 3 decimal places, the approximation to \(\alpha\) is \(1.886\).

(c) First, find the derivative of \(\text{f}(x)\):
\(\text{f}'(x) = 9x^2 - 8x - 5\)

Evaluating the function and its derivative at the first approximation \(x_1 = 2\):
\(\text{f}(2) = 1\)
\(\text{f}'(2) = 9(2)^2 - 8(2) - 5 = 36 - 16 - 5 = 15\)

Applying the Newton-Raphson formula:
\(x_2 = x_1 - \frac{\text{f}(x_1)}{\text{f}'(x_1)}
= 2 - \frac{1}{15} = \frac{29}{15} \approx 1.93333...\)

To 3 decimal places, the second approximation to \(\alpha\) is \(1.933\).

(a)
- M1: For attempting to evaluate both \(\text{f}(1.5)\) and \(\text{f}(2)\).
- A1: For obtaining \(\text{f}(1.5) = -3.375\) and \(\text{f}(2) = 1\), followed by a valid conclusion mentioning the sign change and continuity of the function.

(b)
- M1: For setting up a correct linear interpolation equation or ratio, e.g., \(\frac{\alpha - 1.5}{2 - \alpha} = 3.375\) or \(\alpha = 1.5 + \frac{3.375}{1 - (-3.375)} \times 0.5\).
- A1: For a correct intermediate simplified equation, e.g., \(4.375\alpha = 8.25\) or equivalent.
- A1: For the correct approximation of \(1.886\) (must be 3 d.p.).

(c)
- B1: For the correct derivative \(\text{f}'(x) = 9x^2 - 8x - 5\).
- M1: For attempting to evaluate \(\text{f}'(2)\) and substituting into the Newton-Raphson formula \(x_2 = 2 - \frac{\text{f}(2)}{\text{f}'(2)}\).
- A1: For the correct approximation of \(1.933\) (must be 3 d.p.).

Multiply both sides by \((x+2)^2(x-1)^2\), where \(x \neq -2\) and \(x \neq 1\):
\[ (2x - 3)(x + 2)(x - 1)^2 < x(x - 1)(x + 2)^2 \]
Factor out common terms:
\[ (x + 2)(x - 1) \left[ (2x - 3)(x - 1) - x(x + 2) \right] < 0 \]
Simplify the quadratic expression inside the bracket:
\[ (2x^2 - 5x + 3) - (x^2 + 2x) = x^2 - 7x + 3 \]
So we have:
\[ (x + 2)(x - 1)(x^2 - 7x + 3) < 0 \]
Find the roots of the quadratic equation \(x^2 - 7x + 3 = 0\):
\[ x = \frac{7 \pm \sqrt{(-7)^2 - 4(1)(3)}}{2} = \frac{7 \pm \sqrt{37}}{2} \]
The critical values are:
\[ x = -2, \quad x = \frac{7 - \sqrt{37}}{2} \approx 0.458, \quad x = 1, \quad x = \frac{7 + \sqrt{37}}{2} \approx 6.542 \]
Testing the sign of \((x + 2)(x - 1)(x^2 - 7x + 3)\) across the intervals:
- For \( x < -2 \): Positive
- For \( -2 < x < \frac{7 - \sqrt{37}}{2} \): Negative
- For \( \frac{7 - \sqrt{37}}{2} < x < 1 \): Positive
- For \( 1 < x < \frac{7 + \sqrt{37}}{2} \): Negative
- For \( x > \frac{7 + \sqrt{37}}{2} \): Positive

Thus, the inequality holds for:
\[ -2 < x < \frac{7 - \sqrt{37}}{2} \quad \text{and} \quad 1 < x < \frac{7 + \sqrt{37}}{2} \]

M1: Multiplies both sides by \((x+2)^2(x-1)^2\) to form a polynomial inequality.
A1: Correctly expands and factors to obtain the inequality \((x+2)(x-1)(x^2-7x+3) < 0\).
M1: Solves the quadratic equation \(x^2-7x+3 = 0\) using the quadratic formula.
A1: Correct critical values \(x = -2\), \(x = 1\), and \(x = \frac{7 \pm \sqrt{37}}{2}\).
M1: Employs a valid method (e.g. number line or test points) to find the intervals.
A1.33: Correct final solution set: \(-2 < x < \frac{7-\sqrt{37}}{2}\) or \(1 < x < \frac{7+\sqrt{37}}{2}\).

(a) Combine the fractions on the right-hand side:
\[ \text{RHS} = \frac{1}{r^2} - \frac{1}{(r+1)^2} = \frac{(r+1)^2 - r^2}{r^2(r+1)^2} = \frac{r^2 + 2r + 1 - r^2}{r^2(r+1)^2} = \frac{2r + 1}{r^2(r+1)^2} = \text{LHS} \]
(b) Applying the method of differences:
\[ \sum_{r=1}^{n} \left( \frac{1}{r^2} - \frac{1}{(r+1)^2} \right) = \left(1 - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{9}\right) + \dots + \left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right) \]
All intermediate terms cancel, leaving:
\[ 1 - \frac{1}{(n+1)^2} \]
(c) The sum from \(r=n\) to infinity is:
\[ \sum_{r=n}^{\infty} \frac{2r+1}{r^2(r+1)^2} = \lim_{k \to \infty} \sum_{r=n}^{k} \left( \frac{1}{r^2} - \frac{1}{(r+1)^2} \right) \]
\[ = \lim_{k \to \infty} \left[ \left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right) + \dots + \left(\frac{1}{k^2} - \frac{1}{(k+1)^2}\right) \]
\[ = \lim_{k \to \infty} \left[ \frac{1}{n^2} - \frac{1}{(k+1)^2} \right] = \frac{1}{n^2} \]

M1 (part a): Combines RHS terms into a single fraction.
A1 (part a): Correctly completes algebraic proof to show identity holds.
M1 (part b): Writes down several terms to show the telescoping cancellations.
A1 (part b): Correctly simplifies the series to \(1 - \frac{1}{(n+1)^2}\).
M1 (part c): Formulates the sum from \(r=n\) to infinity using limits or subtraction of partial sums.
A1.33 (part c): Correctly simplifies to obtain \(\frac{1}{n^2}\).

First, divide through by \(x^2 + 1\) to express the equation in standard form:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} + \frac{3x}{x^2 + 1}y = \frac{x}{(x^2 + 1)^3} \]
This is a first-order linear differential equation with \(P(x) = \frac{3x}{x^2 + 1}\).
The integrating factor is:
\[ I(x) = \mathrm{e}^{\int \frac{3x}{x^2 + 1} \mathrm{d}x} = \mathrm{e}^{\frac{3}{2}\ln(x^2 + 1)} = (x^2 + 1)^{3/2} \]
Multiply the differential equation by this integrating factor:
\[ (x^2 + 1)^{3/2}\frac{\mathrm{d}y}{\mathrm{d}x} + 3x(x^2 + 1)^{1/2}y = \frac{x}{(x^2 + 1)^{3/2}} \]
\[ \frac{\mathrm{d}}{\mathrm{d}x}\left[ y(x^2 + 1)^{3/2} \right] = x(x^2 + 1)^{-3/2} \]
Integrate both sides with respect to \(x\):
\[ y(x^2 + 1)^{3/2} = \int x(x^2 + 1)^{-3/2} \mathrm{d}x \]
Let \(u = x^2 + 1\), then \(\mathrm{d}u = 2x\mathrm{d}x\):
\[ \int x(x^2 + 1)^{-3/2} \mathrm{d}x = \frac{1}{2}\int u^{-3/2}\mathrm{d}u = -u^{-1/2} + C = -(x^2 + 1)^{-1/2} + C \]
Therefore,
\[ y(x^2 + 1)^{3/2} = C - \frac{1}{(x^2 + 1)^{1/2}} \]
Dividing both sides by \((x^2 + 1)^{3/2}\) gives the general solution:
\[ y = \frac{C}{(x^2 + 1)^{3/2}} - \frac{1}{(x^2 + 1)^2} \]

M1: Rearranges into standard first-order linear form.
M1: Evaluates the integrating factor integral.
A1: Correct integrating factor \((x^2+1)^{3/2}\).
M1: Integrates the LHS to get \(y \cdot I(x)\).
M1: Correct method for integrating the RHS (e.g., substitution).
A1: Correctly integrates RHS to obtain \(-(x^2+1)^{-1/2} + C\).
A1.33: Correct general solution in terms of \(y\).

Rearrange the transformation equation to express \(z\) in terms of \(w\):
\[ w(z + 2) = z - \mathrm{i} \implies wz + 2w = z - \mathrm{i} \implies z(w - 1) = -2w - ̔\mathrm{i} \]
\[ z = \frac{2w + \mathrm{i}}{1 - w} \]
Let \(w = u + \mathrm{i}v\). Since \(z\) lies on the imaginary axis, \(\text{Re}(z) = 0\), which implies \(z + z^* = 0\):
\[ \frac{2w + \mathrm{i}}{1 - w} + \frac{2w^* - \mathrm{i}}{1 - w^*} = 0 \]
Multiply both sides by \((1 - w)(1 - w^*)\):
\[ (2w + \mathrm{i})(1 - w^*) + (2w^* - \mathrm{i})(1 - w) = 0 \]
Expand the expression:
\[ 2w - 2|w|^2 + \mathrm{i} - \mathrm{i}w^* + 2w^* - 2|w|^2 - \mathrm{i} + \mathrm{i}w = 0 \]
\[ -4|w|^2 + 2(w + w^*) + \mathrm{i}(w - w^*) = 0 \]
Using \(|w|^2 = u^2 + v^2\), \(w + w^* = 2u\), and \(\mathrm{i}(w - w^*) = -2v\):
\[ -4(u^2 + v^2) + 4u - 2v = 0 \]
Divide by \(-4\):
\[ u^2 + v^2 - u + \frac{1}{2}v = 0 \]
Complete the square for both variables:
\[ \left(u - \frac{1}{2}\right)^2 + \left(v + \frac{1}{4}\right)^2 = \frac{1}{4} + \frac{1}{16} = \frac{5}{16} \]
This equation represents a circle in the \(w\)-plane.
The center is at \(u = \frac{1}{2}\) and \(v = -\frac{1}{4}\), representing the complex number \(\frac{1}{2} - \frac{1}{4}\mathrm{i}\).
The radius is \(\sqrt{\frac{5}{16}} = \frac{\sqrt{5}}{4}\).

M1: Rearranges the transformation equation to find \(z\) in terms of \(w\).
A1: Correct expression \(z = \frac{2w+\mathrm{i}}{1-w}\).
M1: Applies the boundary condition \(\text{Re}(z) = 0\).
A1: Obtains a correct algebraic relation in terms of \(u\) and \(v\).
M1: Completes the square to find center and radius.
A1: Correct center of \(\frac{1}{2} - \frac{1}{4}\mathrm{i}\).
A1.33: Correct radius of \(\frac{\sqrt{5}}{4}\).

First, find the complementary function (CF) by solving:
\[ m^2 - 4m + 4 = 0 \implies (m - 2)^2 = 0 \implies m = 2 \text{ (repeated root)} \]
So the CF is:
\[ y_c = (A + Bx)\mathrm{e}^{2x} \]
Since both \(\mathrm{e}^{2x}\) and \(x\mathrm{e}^{2x}\) are present in the CF, try a particular integral (PI) of the form:
\[ y_p = C x^2 \mathrm{e}^{2x} \]
Differentiate \(y_p\):
\[ \frac{\mathrm{d}y_p}{\mathrm{d}x} = 2C(x + x^2)\mathrm{e}^{2x} \]
\[ \frac{\mathrm{d}^2y_p}{\mathrm{d}x^2} = 2C(1 + 4x + 2x^2)\mathrm{e}^{2x} \]
Substitute into the differential equation:
\[ 2C(1 + 4x + 2x^2)\mathrm{e}^{2x} - 8C(x + x^2)\mathrm{e}^{2x} + 4Cx^2\mathrm{e}^{2x} = 9\mathrm{e}^{2x} \]
\[ C\mathrm{e}^{2x} [ 2 + 8x + 4x^2 - 8x - 8x^2 + 4x^2 ] = 9\mathrm{e}^{2x} \]
\[ 2C\mathrm{e}^{2x} = 9\mathrm{e}^{2x} \implies C = \frac{9}{2} \]
Thus, the general solution is:
\[ y = y_c + y_p = (A + Bx + \frac{9}{2}x^2)\mathrm{e}^{2x} \]

M1: Forms and solves auxiliary equation.
A1: Correct complementary function \(y_c = (A + Bx)\mathrm{e}^{2x}\).
M1: Identifies the correct form for the particular integral: \(y_p = C x^2\mathrm{e}^{2x}\).
M1: Differentiates \(y_p\) twice using the product rule.
A1: Obtains correct derivatives.
M1: Substitutes derivatives into the DE to solve for \(C\).
A1.33: Correct general solution.

Let \(y' = \frac{\mathrm{d}y}{\mathrm{d}x}\), \(y'' = \frac{\mathrm{d}^2y}{\mathrm{d}x^2}\), and \(y''' = \frac{\mathrm{d}^3y}{\mathrm{d}x^3}\).
Given that at \(x = 0\), \(y = 1\).
Using the differential equation:
\[ y' - 2xy = \cos x \implies y' = 2xy + \cos x \]
At \(x = 0\):
\[ y'(0) = 2(0)(1) + \cos(0) = 1 \]
Differentiate the differential equation with respect to \(x\):
\[ y'' - 2y - 2xy' = -\sin x \]
At \(x = 0\):
\[ y''(0) - 2(1) - 2(0)(1) = -\sin(0) \implies y''(0) = 2 \]
Differentiate again with respect to \(x\):
\[ y''' - 2y' - 2y' - 2xy'' = -\cos x \implies y''' - 4y' - 2xy'' = -\cos x \]
At \(x = 0\):
\[ y'''(0) - 4(1) - 2(0)(2) = -\cos(0) \implies y'''(0) - 4 = -1 \implies y'''(0) = 3 \]
The Maclaurin series expansion for \(y\) is:
\[ y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \dots \]
Substitute the values:
\[ y = 1 + (1)x + \frac{2}{2}x^2 + \frac{3}{6}x^3 + \dots \]
\[ y = 1 + x + x^2 + \frac{1}{2}x^3 + \dots \]

M1: Uses the DE and boundary conditions to calculate \(y'(0)\).
A1: Correct value \(y'(0) = 1\).
M1: Differentiates the DE to find an expression for \(y''\).
A1: Correctly evaluates \(y''(0) = 2\).
M1: Differentiates a second time using the product rule.
A1: Correctly evaluates \(y'''(0) = 3\).
A1.33: Obtains the correct series expansion up to the term in \(x^3\).

(a) The sketch is a limacon symmetric about the initial line \(\theta = 0\) with no inner loop. It has coordinates at \((5a, 0)\), \((3a, \pi/2)\), \((a, \pi)\), and \((3a, 3\pi/2)\).
(b) The area is:
\[ A = \frac{1}{2}\int_{0}^{2
\pi} r^2 \mathrm{d}\theta = \frac{1}{2}a^2\int_{0}^{2\pi} (3 + 2\cos\theta)^2 \mathrm{d}\theta \]
\[ (3 + 2\cos\theta)^2 = 9 + 12\cos\theta + 4\cos^2\theta \]
Substitute \(\cos^2\theta = \frac{1 + \cos 2\theta}{2}\):
\[ 9 + 12\cos\theta + 4\left(\frac{1 + \cos 2\theta}{2}\right) = 11 + 12\cos\theta + 2\cos 2\theta \]
Integrate from \(0\) to \(2\pi\):
\[ A = \frac{1}{2}a^2 \int_{0}^{2\pi} (11 + 12\cos\theta + 2\cos 2\theta) \mathrm{d}\theta \]
\[ A = \frac{1}{2}a^2 \left[ 11\theta + 12\sin\theta + \sin 2\theta \right]_0^{2\pi} \]
\[ A = \frac{1}{2}a^2 \left( 22\pi + 0 + 0 - 0 \right) = 11\pi a^2 \]

M1 (part a): Draws a correct closed, loop-free limacon symmetric about the initial line.
A1 (part a): Clearly labels key intercepts on the initial and vertical lines.
M1 (part b): Applies the polar area formula \(\frac{1}{2}\int r^2 \mathrm{d}\theta\) with correct limits.
A1 (part b): Expands the integrand correctly.
M1 (part b): Integrates the expression using double-angle identity.
A1.33 (part b): Correctly arrives at \(11\pi a^2\).

(a) Using 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 the real parts:
\[ \cos 5\theta = \cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta
\sin^4\theta \]
Substitute \(\sin^2\theta = 1 - \cos^2\theta\):
\[ \cos 5\theta = \cos^5\theta - 10\cos^3\theta(1 - \cos^2\theta) + 5\cos\theta(1 - \cos^2\theta)^2 \]
\[ \cos 5\theta = \cos^5\theta - 10\cos^3\theta + 10\cos^5\theta + 5\cos\theta(1 - 2\cos^2\theta + \cos^4\theta) \]
\[ \cos 5\theta = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta \]
(b) Let \(x = \cos\theta\). Multiplying the equation \(16x^4 - 20x^2 + 5 = 0\) by \(x\) (since \(x \neq 0\)):
\[ 16x^5 - 20x^3 + 5x = 0 \implies 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta = 0 \]
Using the identity from part (a):
\[ \cos 5\theta = 0 \]
Solve for \(\theta\):
\[ 5\theta = \frac{\pi}{2} + k\pi \implies \theta = \frac{(2k+1)\pi}{10} \]
Since \(x = \cos\theta \neq 0\), we omit the solution \(\theta = \pi/2\) (when \(k=2\)).
The other angles in the range \([0, \pi]\) correspond to \(k=0, 1, 3, 4\):
- \(k=0 \implies \theta = \frac{\pi}{10} \implies x = \cos\frac{\pi}{10} \approx 0.951\)
- \(k=1 \implies \theta = \frac{3\pi}{10} \implies x = \cos\frac{3\pi}{10} \approx 0.588\)
- \(k=3 \implies \theta = \frac{7\pi}{10} \implies x = \cos\frac{7\pi}{10} \approx -0.588\)
- \(k=4 \implies \theta = \frac{9\pi}{10} \implies x = \cos\frac{9\pi}{10} \approx -0.951\)
Thus, the four roots are \(\pm 0.951\) and \(\pm 0.588\).

M1 (part a): Applies de Moivre's theorem and expands binomially.
A1 (part a): Equates real parts correctly.
M1 (part a): Uses basic trigonometric identities to rewrite entirely in terms of \(\cos\theta\).
A1 (part a): Fully correct proof.
M1 (part b): Relates the algebraic equation to the trigonometric identity.
A1 (part b): Solves \(\cos 5\theta = 0\) to find the key angles.
A1.33 (part b): Calculates all four roots to 3 decimal places.

Part (a): Move all terms to one side of the inequality: \(\frac{2x+3}{x-1} - \frac{x}{x+3} > 0\). Combine the fractions over a common denominator: \(\frac{(2x+3)(x+3) - x(x-1)}{(x-1)(x+3)} > 0\). Expand and simplify the numerator: \((2x+3)(x+3) - x(x-1) = 2x^2 + 9x + 9 - (x^2 - x) = x^2 + 10x + 9\). Factorise the numerator: \(x^2 + 10x + 9 = (x+1)(x+9)\). Thus, the inequality can be written as: \(\frac{(x+1)(x+9)}{(x-1)(x+3)} > 0\). This is in the required form with \(a = -1\) and \(b = -9\) (or vice versa). Part (b): Identify the critical values where the numerator or denominator is zero: \(x = -9, -3, -1, 1\). Test the intervals or use a sign table to find where the expression is positive: For \(x > 1\), the expression is positive. For \(-1 < x < 1\), the expression is negative. For \(-3 < x < -1\), the expression is positive. For \(-9 < x < -3\), the expression is negative. For \(x < -9\), the expression is positive. Therefore, the set of values for which the inequality holds is \(x < -9\) or \(-3 < x < -1\) or \(x > 1\).

Part (a): [M1] for attempting to subtract the fractions and forming a single fraction with denominator \((x-1)(x+3)\). [A1] for obtaining a correct quadratic numerator \(x^2 + 10x + 9\). [A1] for factorising to obtain \((x+1)(x+9)\) and identifying \(a = -1\) and \(b = -9\). Part (b): [B1] for identifying all four critical values: \(-9, -3, -1, 1\). [M1] for using a valid method (e.g., sign table, graph, testing points) to determine the signs in the intervals. [A1] for any two correct intervals. [A1] for all three correct intervals. [A1] for writing the final answer using correct inequality symbols and correct logical connectives (e.g., 'or').

(a) Using the identity \(R\cosh(x - \alpha) = R\cosh x \cosh \alpha - R\sinh x \sinh \alpha\), we compare coefficients with \(5\cosh x - 3\sinh x\):
\(R\cosh \alpha = 5\) (1)
\(R\sinh \alpha = 3\) (2)

Squaring and subtracting:
\(R^2(\cosh^2 \alpha - \sinh^2 \alpha) = 25 - 9 \implies R^2 = 16 \implies R = 4\) (since \(R > 0\)).

Dividing (2) by (1):
\(\tanh \alpha = \frac{3}{5}\)

Using the logarithmic form of \(\text{artanh}(x)\):
\(\alpha = \frac{1}{2}\ln\left(\frac{1 + 3/5}{1 - 3/5}\right) = \frac{1}{2}\ln(4) = \ln 2\).

Thus, \(5\cosh x - 3\sinh x = 4\cosh(x - \ln 2)\).

(b) Using the result from part (a):
\(4\cosh(x - \ln 2) = 7 \implies \cosh(x - \ln 2) = \frac{7}{4}\).

Let \(u = x - \ln 2\), then \(u = \pm\text{arcosh}\left(\frac{7}{4}\right)\).
Using the formula \(\text{arcosh}(y) = \ln(y + \sqrt{y^2-1})\):
\(\text{arcosh}\left(\frac{7}{4}\right) = \ln\left(\frac{7}{4} + \sqrt{\frac{49}{16} - 1}\right) = \ln\left(\frac{7 + \sqrt{33}}{4}\right)\).

Therefore, \(x - \ln 2 = \pm\ln\left(\frac{7 + \sqrt{33}}{4}\right)\).

For the positive case:
\(x = \ln 2 + \ln\left(\frac{7 + \sqrt{33}}{4}\right) = \ln\left(\frac{2(7 + \sqrt{33})}{4}\right) = \ln\left(\frac{7 + \sqrt{33}}{2}\right)\).

For the negative case, using the fact that \(-\ln(A) = \ln(1/A)\):
\(x = \ln 2 - \ln\left(\frac{7 + \sqrt{33}}{4}\right) = \ln 2 + \ln\left(\frac{4}{7 + \sqrt{33}}\right) = \ln 2 + \ln\left(\frac{7 - \sqrt{33}}{4}\right) = \ln\left(\frac{7 - \sqrt{33}}{2}\right)\).

Thus, \(x = \ln\left(\frac{7 \pm \sqrt{33}}{2}\right)\).

(a)
M1: Equates coefficients of \(\cosh x\) and \(\sinh x\) to establish equations in \(R\) and \(\alpha\).
A1: Correctly finds \(R = 4\).
M1: Uses \(\tanh \alpha = 3/5\) to find \(\alpha\) in logarithmic form.
A1: Obtains \(\alpha = \ln 2\).

(b)
M1: Sets \(\cosh(x - \ln 2) = 7/4\) and applies inverse hyperbolic cosine.
A1: Obtains \(x - \ln 2 = \pm\ln\left(\frac{7 + \sqrt{33}}{4}\right)\) or equivalent correct form.
M1: Adds \(\ln 2\) and correctly applies laws of logarithms to combine terms.
A1: Reaches final exact answer \(x = \ln\left(\frac{7 \pm \sqrt{33}}{2}\right)\).

(a) Differentiating the equation of the ellipse implicitly with respect to \(x\):
\(\frac{2x}{16} + \frac{2y}{9}\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{9x}{16y}\).

At the point \(P(4\cos\theta, 3\sin\theta)\):
\(\frac{dy}{dx} = -\frac{9(4\cos\theta)}{16(3\sin\theta)} = -\frac{3\cos\theta}{4\sin\theta}\).

Therefore, the gradient of the normal is \(\frac{4\sin\theta}{3\cos\theta}\).
The equation of the normal is:
\(y - 3\sin\theta = \frac{4\sin\theta}{3\cos\theta}(x - 4\cos\theta)\).

Multiplying by \(3\cos\theta\):
\(3y\cos\theta - 9\sin\theta\cos\theta = 4x\sin\theta - 16\sin\theta\cos\theta\)
\(4x\sin\theta - 3y\cos\theta = 7\sin\theta\cos\theta\).

(b) For \(\theta = \frac{\pi}{6}\), \(\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\) and \(\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\).
Substituting these values into the normal equation:
\(4x\left(\frac{1}{2}\right) - 3y\left(\frac{\sqrt{3}}{2}\right) = 7\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)\)
\(2x - \frac{3\sqrt{3}}{2}y = \frac{7\sqrt{3}}{4}\).

At the point \(A\) (where \(y = 0\)):
\(2x = \frac{7\sqrt{3}}{4} \implies x = \frac{7\sqrt{3}}{8}\).
So \(A\) is the point \(\left(\frac{7\sqrt{3}}{8}, 0\right)\).

At the point \(B\) (where \(x = 0\)):
\(-\frac{3\sqrt{3}}{2}y = \frac{7\sqrt{3}}{4} \implies y = -\frac{7}{6}\).
So \(B\) is the point \(\left(0, -\frac{7}{6}\right)\).

The area of triangle \(OAB\) is:
\(\text{Area} = \frac{1}{2} \times OA \times OB = \frac{1}{2} \times \left(\frac{7\sqrt{3}}{8}\right) \times \left|-\frac{7}{6}\right| = \frac{49\sqrt{3}}{96}\).

(a)
M1: Differentiates implicitly or uses parametric differentiation to find \(\frac{dy}{dx}\).
A1: Obtains gradient of tangent \(-\frac{3\cos\theta}{4\sin\theta}\).
M1: Correctly finds gradient of normal and writes the equation of the normal at \(P\).
A1*: Fully correct rearrangement to lead to the given equation.

(b)
M1: Substitutes \(\theta = \frac{\pi}{6}\) into the equation of the normal.
M1: Finds the coordinates of the intercepts \(A\) and \(B\).
M1: Applies the triangle area formula \(\frac{1}{2} \times \text{base} \times \text{height}\).
A1: Correct exact area of \(\frac{49\sqrt{3}}{96}\).

(a) Let \(y = \text{artanh}(\sin x)\). Then \(\tanh y = \sin x\).
Differentiating implicitly with respect to \(x\):
\(\text{sech}^2 y \frac{dy}{dx} = \cos x\).

Using the identity \(\text{sech}^2 y = 1 - \tanh^2 y\):
\(\text{sech}^2 y = 1 - \sin^2 x = \cos^2 x\).

Therefore,
\(\cos^2 x \frac{dy}{dx} = \cos x\).

Since \(-\frac{\pi}{2} < x < \frac{\pi}{2}\), \(\cos x \ne 0\), we can divide by \(\cos^2 x\):
\(\frac{dy}{dx} = \frac{\cos x}{\cos^2 x} = \sec x\).

(b) Let \(u = \text{arcosh}(\sec x)\). Then \(\cosh u = \sec x\).
Differentiating implicitly with respect to \(x\):
\(\sinh u \frac{du}{dx} = \sec x \tan x\).

We know that \(\sinh^2 u = \cosh^2 u - 1 = \sec^2 x - 1 = \tan^2 x\).
Since \(0 < x < \frac{\pi}{2}\), \(\sec x > 1\) so \(u > 0\) and \(\sinh u > 0\).
Also, in this interval, \(\tan x > 0\).
Hence, \(\sinh u = \tan x\).

Substituting this back into the derivative equation:
\(\tan x \frac{du}{dx} = \sec x \tan x\).

Since \(\tan x \ne 0\) for \(0 < x < \frac{\pi}{2}\), we can divide by \(\tan x\):
\(\frac{du}{dx} = \sec x\).

(a)
M1: Writes relation as \(\tanh y = \sin x\) and differentiates implicitly.
M1: Uses standard identity \(\text{sech}^2 y = 1 - \tanh^2 y\) to express LHS in terms of \(x\).
A1*: Reaches the given result \(\sec x\) with clear, mathematically sound steps.

(b)
M1: Writes relation as \(\cosh u = \sec x\) and differentiates.
A1: Correctly differentiates RHS to obtain \(\sec x \tan x\).
M1: Replaces \(\sinh u\) with \(\pm \sqrt{\sec^2 x - 1}\) and justifies choice of positive sign.
A1*: Completes the proof to obtain \(\sec x\) clearly.

(a) We express \(\tanh^n x\) as \(\tanh^{n-2} x \tanh^2 x\):
\(I_n = \int_0^{\ln 2} \tanh^{n-2} x (1 - \text{sech}^2 x) \, dx\)
\(I_n = \int_0^{\ln 2} \tanh^{n-2} x \, dx - \int_0^{\ln 2} \tanh^{n-2} x \text{sech}^2 x \, dx\)
\(I_n = I_{n-2} - \left[ \frac{\tanh^{n-1} x}{n-1} \right]_0^{\ln 2}\).

Now, evaluating the boundary values:
At \(x = \ln 2\):
\(\tanh(\ln 2) = \frac{e^{\ln 2} - e^{-\ln 2}}{e^{\ln 2} + e^{-\ln 2}} = \frac{2 - 1/2}{2 + 1/2} = \frac{3/2}{5/2} = \frac{3}{5}\).
At \(x = 0\), \(\tanh 0 = 0\).

Thus,
\(I_n = I_{n-2} - \frac{1}{n-1} \left(\frac{3}{5}\right)^{n-1}\).

(b) Using the reduction formula for \(n = 3\):
\(I_3 = I_1 - \frac{1}{2} \left(\frac{3}{5}\right)^2 = I_1 - \frac{9}{50}\).

Now, we compute \(I_1\):
\(I_1 = \int_0^{\ln 2} \tanh x \, dx = \left[ \ln(\cosh x) \right]_0^{\ln 2}\).
Evaluating at the boundaries:
\(\cosh(\ln 2) = \frac{e^{\ln 2} + e^{-\ln 2}}{2} = \frac{2 + 1/2}{2} = \frac{5}{4}\).
\(\cosh 0 = 1\).

So \(I_1 = \ln\left(\frac{5}{4}\right) - \ln 1 = \ln\left(\frac{5}{4}\right)\).

Substituting this back:
\(I_3 = \ln\left(\frac{5}{4}\right) - \frac{9}{50}\).

(a)
M1: Splits \(\tanh^n x\) into \(\tanh^{n-2} x (1 - \text{sech}^2 x)\).
A1: Integrates \(\tanh^{n-2} x \text{sech}^2 x\) to get \(\frac{\tanh^{n-1} x}{n-1}\).
M1: Correctly calculates \(\tanh(\ln 2) = 3/5\) using exponential definitions.
A1*: Substitutes boundaries correctly to show the given reduction formula.

(b)
M1: Applies the reduction formula with \(n=3\).
M1: Integrates \(\tanh x\) to find \(I_1\).
A1: Calculates \(\cosh(\ln 2) = 5/4\).
A1: Reaches the correct final exact answer of \(\ln(5/4) - 9/50\).

(a) The direction vectors of \(L_1\) and \(L_2\) are \(\mathbf{d}_1 = 2\mathbf{i} + \mathbf{j} + 3\mathbf{k}\) and \(\mathbf{d}_2 = \mathbf{i} - \mathbf{j} + 2\mathbf{k}\).
A vector perpendicular to both directions is \(\mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2\):
\[\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & 3 \\ 1 & -1 & 2 \end{vmatrix}\]
\[\mathbf{n} = \mathbf{i}(1(2) - 3(-1)) - \mathbf{j}(2(2) - 3(1)) + \mathbf{k}(2(-1) - 1(1))\]
\[\mathbf{n} = 5\mathbf{i} - \mathbf{j} - 3\mathbf{k}\]

(b) The position vectors of points on \(L_1\) and \(L_2\) are \(\mathbf{a}_1 = \mathbf{i} - \mathbf{k}\) and \(\mathbf{a}_2 = 2\mathbf{i} + 3\mathbf{j} + \mathbf{k}\).
The vector connecting these points is:
\[\mathbf{a}_2 - \mathbf{a}_1 = (2-1)\mathbf{i} + 3\mathbf{j} + (1 - (-1))\mathbf{k} = \mathbf{i} + 3\mathbf{j} + 2\mathbf{k}\]

The shortest distance \(d\) is the magnitude of the projection of \(\mathbf{a}_2 - \mathbf{a}_1\) onto \(\mathbf{n}\):
\[d = \frac{|(\mathbf{a}_2 - \mathbf{a}_1) \cdot \mathbf{n}|}{|\mathbf{n}|}\]

First calculate the dot product:
\[(\mathbf{a}_2 - \mathbf{a}_1) \cdot \mathbf{n} = (1)(5) + (3)(-1) + (2)(-3) = 5 - 3 - 6 = -4\]

Next calculate the magnitude of \(\mathbf{n}\):
\[|\mathbf{n}| = \sqrt{5^2 + (-1)^2 + (-3)^2} = \sqrt{25 + 1 + 9} = \sqrt{35}\]

Thus, the shortest distance is:
\[d = \frac{|-4|}{\sqrt{35}} = \frac{4}{\sqrt{35}} = \frac{4\sqrt{35}}{35}\]

(a)
M1: Attempts cross product of the two direction vectors.
A1: Correct cross product calculations.
A1: Obtains \(5\mathbf{i} - \mathbf{j} - 3\mathbf{k}\) or any scalar multiple.

(b)
M1: Determines vector \(\mathbf{a}_2 - \mathbf{a}_1\) between point on each line.
M1: Applies formula for the shortest distance \(\frac{|(\mathbf{a}_2 - \mathbf{a}_1) \cdot \mathbf{n}|}{|\mathbf{n}|}\).
A1: Calculates dot product as \(-4\) (or \(4\)) and magnitude of normal as \(\sqrt{35}\).
A1: Rationalises denominator to obtain \(\frac{4\sqrt{35}}{35}\).

(a) The characteristic equation is \(\det(\mathbf{M} - \lambda\mathbf{I}) = 0\):
\[\det \begin{pmatrix} 1-\lambda & 0 & 2 \\ 0 & 3-\lambda & 0 \\ 2 & 0 & 1-\lambda \end{pmatrix} = 0\]

Expanding along the second row:
\[(3-\lambda) [(1-\lambda)^2 - 4] = 0\]
\[(3-\lambda) (\lambda^2 - 2\lambda - 3) = 0\]
\[-(\lambda-3)^2(\lambda+1) = 0\]

Thus, the eigenvalues are \(\lambda = -1\) and \(\lambda = 3\) (repeated).

(b) For \(\lambda = -1\):
\[\mathbf{M} - (-1)\mathbf{I} = \begin{pmatrix} 2 & 0 & 2 \\ 0 & 4 & 0 \\ 2 & 0 & 2 \end{pmatrix}\]
Solving \((\mathbf{M} + \mathbf{I})\mathbf{x} = \mathbf{0}\):
\[2x + 2z = 0 \implies z = -x\]
\[4y = 0 \implies y = 0\]

An eigenvector is \(\begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}\).
Normalizing this vector:
\[\mathbf{u}_1 = \frac{1}{\sqrt{1^2+0^2+(-1)^2}}\begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} = \begin{pmatrix} 1/\sqrt{2} \\ 0 \\ -1/\sqrt{2} \end{pmatrix}\]

(c) For the repeated eigenvalue \(\lambda = 3\):
\[\mathbf{M} - 3\mathbf{I} = \begin{pmatrix} -2 & 0 & 2 \\ 0 & 0 & 0 \\ 2 & 0 & -2 \end{pmatrix}\]
Solving \((\mathbf{M} - 3\mathbf{I})\mathbf{x} = \mathbf{0}\):
\[-2x + 2z = 0 \implies x = z\]

So eigenvectors are of the form \(\begin{pmatrix} x \\ y \\ x \end{pmatrix}\).
We must choose two mutually orthogonal eigenvectors in this eigenspace:
Choose \(\mathbf{x}_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\), which has magnitude 1.
Choose \(\mathbf{x}_3 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}\). Normalizing gives \(\mathbf{u}_3 = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 1/\sqrt{2} \\ 0 \\ 1/\sqrt{2} \end{pmatrix}\).

Since \(\mathbf{u}_1\), \(\mathbf{u}_2\), and \(\mathbf{u}_3\) are orthonormal, we construct \(\mathbf{P}\):
\[\mathbf{P} = \begin{pmatrix} 1/\sqrt{2} & 0 & 1/\sqrt{2} \\ 0 & 1 & 0 \\ -1/\sqrt{2} & 0 & 1/\sqrt{2} \end{pmatrix}\]

And the corresponding diagonal matrix \(\mathbf{D}\) is:
\[\mathbf{D} = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{pmatrix}\]

(a)
M1: Sets up characteristic equation \(\det(\mathbf{M} - \lambda\mathbf{I}) = 0\).
A1: Finds correct cubic equation \((\lambda-3)^2(\lambda+1) = 0\) or equivalent.
A1: Finds correct eigenvalues \(-1\) and \(3\).

(b)
M1: Substitutes \(\lambda = -1\) to get system of linear equations.
A1: Obtains the normalized eigenvector \(\frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}\).

(c)
M1: Solves system for \(\lambda = 3\) to get general eigenvector.
M1: Produces two orthogonal normalized eigenvectors for the eigenvalue \(3\).
A1: Constructs a correct orthogonal matrix \(\mathbf{P}\).
A1: States the matching diagonal matrix \(\mathbf{D}\).

(a) Differentiating \(x\) and \(y\) with respect to \(t\):
\(\frac{dx}{dt} = 1 - \text{sech}^2 t = \tanh^2 t\)
\(\frac{dy}{dt} = -\text{sech} t \tanh t\)

Now compute the sum of the squares:
\(\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = (\tanh^2 t)^2 + (-\text{sech} t \tanh t)^2\)
\(= \tanh^4 t + \text{sech}^2 t \tanh^2 t\)
\(= \tanh^2 t (\tanh^2 t + \text{sech}^2 t)\)

Using the identity \(\tanh^2 t + \text{sech}^2 t = 1\):
\(\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = \tanh^2 t\).

(b) The formula for the arc length \(s\) is:
\(s = \int_0^{\ln 3} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\)
\(s = \int_0^{\ln 3} \sqrt{\tanh^2 t} \, dt\)

Since \(t \ge 0\), \(\tanh t \ge 0\), we have:
\(s = \int_0^{\ln 3} \tanh t \, dt = \left[ \ln(\cosh t) \right]_0^{\ln 3}\)

Evaluating at the upper limit:
\(\cosh(\ln 3) = \frac{e^{\ln 3} + e^{-\ln 3}}{2} = \frac{3 + 1/3}{2} = \frac{5}{3}\)

Evaluating at the lower limit:
\(\cosh 0 = 1\)

Thus:
\(s = \ln(5/3) - \ln 1 = \ln(5/3)\).

(a)
M1: Correctly differentiates \(x\) with respect to \(t\).
M1: Correctly differentiates \(y\) with respect to \(t\).
M1: Squares and adds derivatives, factoring out \(\tanh^2 t\).
A1: Uses identity \(\tanh^2 t + \text{sech}^2 t = 1\).
A1*: Completes verification cleanly to get \(\tanh^2 t\).

(b)
M1: Sets up the arc length integral with the correct limits.
A1: Integrates \(\tanh t\) to get \(\ln(\cosh t)\).
M1: Replaces \(\cosh(\ln 3)\) with its exact rational form \(5/3\).
A1: Reaches the correct final exact value of \(\ln(5/3)\).

(a) Since \(x = a\cosh t\) and \(y = b\sinh t\):
\(\frac{dx}{dt} = a\sinh t\) and \(\frac{dy}{dt} = b\cosh t\).

The gradient of the tangent is:
\(\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{b\cosh t}{a\sinh t}\).

The equation of the tangent at \(P\) is:
\(y - b\sinh t = \frac{b\cosh t}{a\sinh t}(x - a\cosh t)\)

Multiplying both sides by \(a\sinh t\):
\(ay\sinh t - ab\sinh^2 t = bx\cosh t - ab\cosh^2 t\)
\(bx\cosh t - ay\sinh t = ab(\cosh^2 t - \sinh^2 t)\)

Using the identity \(\cosh^2 t - \sinh^2 t = 1\):
\(bx\cosh t - ay\sinh t = ab\).

Dividing by \(ab\):
\(\frac{x\cosh t}{a} - \frac{y\sinh t}{b} = 1\).

(b) The equations of the asymptotes of \(H\) are \(y = \pm \frac{b}{a}x \implies \frac{y}{b} = \pm \frac{x}{a}\).

For the asymptote \(\frac{y}{b} = \frac{x}{a}\), substitute into the tangent equation:
\(\frac{x\cosh t}{a} - \frac{x\sinh t}{a} = 1 \implies \frac{x}{a}(\cosh t - \sinh t) = 1 \implies \frac{x}{a}e^{-t} = 1 \implies x = ae^t\).
Then, \(y = be^t\). Thus, \(Q\) is the point \((ae^t, be^t)\).

For the asymptote \(\frac{y}{b} = -\frac{x}{a}\), substitute into the tangent equation:
\(\frac{x\cosh t}{a} + \frac{x\sinh t}{a} = 1 \implies \frac{x}{a}(\cosh t + \sinh t) = 1 \implies \frac{x}{a}e^{t} = 1 \implies x = ae^{-t}\).
Then, \(y = -be^{-t}\). Thus, \(R\) is the point \((ae^{-t}, -be^{-t})\).

The area of triangle \(OQR\) with vertices \((0,0)\), \(Q(ae^t, be^t)\), and \(R(ae^{-t}, -be^{-t})\) is:
\(\text{Area} = \frac{1}{2}|x_Q y_R - x_R y_Q|\)
\(\text{Area} = \frac{1}{2}|(ae^t)(-be^{-t}) - (ae^{-t})(be^t)| = \frac{1}{2}|-ab - ab| = \frac{1}{2}|-2ab| = ab\).

This is a constant value \(ab\) and is independent of \(t\).

(a)
M1: Uses parametric differentiation to find \(\frac{dy}{dx}\).
A1: Obtains \(\frac{b\cosh t}{a\sinh t}\).
M1: Applies straight line equation formula and identity \(\cosh^2 t - \sinh^2 t = 1\).
A1*: Rearranges correctly to reach the given form.

(b)
M1: Identifies the correct equations of the asymptotes.
M1: Solves the simultaneous equations to find the coordinates of \(Q\) and \(R\).
A1: Obtains \(Q(ae^t, be^t)\) and \(R(ae^{-t}, -be^{-t})\).
M1: Uses the coordinate method for the area of triangle \(OQR\).
A1*: Correctly simplifes the determinant to prove the area is \(ab\).

(a) We can write \( \tanh^n x = \tanh^{n-2} x \tanh^2 x = \tanh^{n-2} x (1 - \text{sech}^2 x) \). Thus, \( I_n = \int_{0}^{\ln 3} \tanh^{n-2} x \ \mathrm{d}x - \int_{0}^{\ln 3} \tanh^{n-2} x \text{sech}^2 x \ \mathrm{d}x \). The first integral is \( I_{n-2} \). For the second integral, using the substitution \( u = \tanh x \), we get \( \int \tanh^{n-2} x \text{sech}^2 x \ \mathrm{d}x = \frac{\tanh^{n-1} x}{n-1} \). Evaluating this at the limits: at \( x = \ln 3 \), \( \tanh(\ln 3) = \frac{e^{\ln 3} - e^{-\ln 3}}{e^{\ln 3} + e^{-\ln 3}} = \frac{3 - 1/3}{3 + 1/3} = \frac{8/3}{10/3} = \frac{4}{5} \). At \( x = 0 \), \( \tanh(0) = 0 \). Thus, \( \left[ \frac{\tanh^{n-1} x}{n-1} \right]_{0}^{\ln 3} = \frac{1}{n-1} \left(\frac{4}{5}\right)^{n-1} \). Substituting this back gives \( I_n = I_{n-2} - \frac{1}{n-1} \left(\frac{4}{5}\right)^{n-1} \). (b) Using the reduction formula with \( n = 3 \), we have \( I_3 = I_1 - \frac{1}{2} \left(\frac{4}{5}\right)^2 = I_1 - \frac{8}{25} \). We calculate \( I_1 \): \( I_1 = \int_{0}^{\ln 3} \tanh x \ \mathrm{d}x = [\ln(\cosh x)]_{0}^{\ln 3} \). Since \( \cosh(\ln 3) = \frac{e^{\ln 3} + e^{-\ln 3}}{2} = \frac{3 + 1/3}{2} = \frac{5}{3} \) and \( \cosh(0) = 1 \), we find \( I_1 = \ln\left(\frac{5}{3}\right) - \ln 1 = \ln\left(\frac{5}{3}\right) \). Therefore, \( I_3 = -\frac{8}{25} + \ln\left(\frac{5}{3}\right) \), with \( a = -\frac{8}{25} \) and \( b = \frac{5}{3} \).

Part (a): M1: Applies the identity \( \tanh^2 x = 1 - \text{sech}^2 x \) and splits the integral into two parts. M1: Integrates \( \tanh^{n-2} x \text{sech}^2 x \) to obtain \( \frac{\tanh^{n-1} x}{n-1} \). A1: Evaluates \( \tanh(\ln 3) = \frac{4}{5} \) correctly. A1*: Achieves the given reduction formula with no errors. Part (b): M1: Applies the reduction formula for \( n = 3 \) to express \( I_3 \) in terms of \( I_1 \). M1: Integrates \( \tanh x \) to get \( \ln(\cosh x) \) and applies the limits. A1: Correctly finds \( I_1 = \ln\left(\frac{5}{3}\right) \). A1: Obtains \( I_3 = -\frac{8}{25} + \ln\left(\frac{5}{3}\right) \).