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

Using the standard sum formulae: \\ \sum_{r=1}^{n} r^2 = \frac{1}{6}n(n+1)(2n+1) \\ and \\ \sum_{r=1}^{n} r = \frac{1}{2}n(n+1) \\ \\ We can write: \\ \sum_{r=1}^{n} r(3r-1) = 3\sum_{r=1}^{n} r^2 - \sum_{r=1}^{n} r \\ = 3 \left[ \frac{1}{6}n(n+1)(2n+1) \right] - \frac{1}{2}n(n+1) \\ = \frac{1}{2}n(n+1)(2n+1) - \frac{1}{2}n(n+1) \\ \\ Factorising out \frac{1}{2}n(n+1): \\ = \frac{1}{2}n(n+1) \left[ (2n+1) - 1 \right] \\ = \frac{1}{2}n(n+1)(2n) \\ = n^2(n+1) \\ \\ as required.

M1: Attempts to split the summation and substitute the standard formulae for \(\sum r^2\) and \(\sum r\). \\ A1: Correct unsimplified algebraic expression, e.g., \(\frac{3}{6}n(n+1)(2n+1) - \frac{1}{2}n(n+1)\). \\ M1: Factorises out at least \(\frac{1}{2}n(n+1)\) to simplify the expression. \\ A1: Fully correct completion to show the given result \(n^2(n+1)\) with no errors.

Let \(f(n) = 3^{2n} + 7\). \\ \\ **Step 1: Base case** \\ For \(n = 1\), \(f(1) = 3^2 + 7 = 16\), which is divisible by 8 (since \(16 = 8 \times 2\)). Thus, the statement is true for \(n = 1\). \\ \\ **Step 2: Inductive hypothesis** \\ Assume the statement is true for \(n = k\), where \(k\) is a positive integer. \\ That is, \(f(k) = 3^{2k} + 7 = 8m\) for some integer \(m\). \\ \\ **Step 3: Inductive step** \\ For \(n = k+1\): \\ \(f(k+1) = 3^{2(k+1)} + 7 = 3^{2k+2} + 7 = 9 \cdot 3^{2k} + 7\) \\ We can rewrite this in terms of \(f(k)\): \\ \(f(k+1) = 9(3^{2k} + 7) - 56 = 9(8m) - 56 = 8(9m - 7)\) \\ Since \(9m - 7\) is an integer, \(f(k+1)\) is divisible by 8. \\ \\ **Step 4: Conclusion** \\ Since the statement is true for \(n = 1\), and if it is true for \(n = k\) then it is also true for \(n = k+1\), the statement is true for all positive integers \(n\) by mathematical induction.

B1: Verifies the statement for \(n = 1\) with clear working. \\ M1: Makes a clear assumption for \(n = k\) and writes down an expression for \(f(k+1)\). \\ M1: Attempts to manipulate the expression for \(f(k+1)\) to relate it to the induction hypothesis or to show divisibility by 8. \\ A1: Obtains a correct simplified expression showing divisibility by 8, such as \(8(9m-7)\) or \(8(3^{2k} + m)\). \\ A1: Complete and logical conclusion containing all key steps: true for \(n=1\), if true for \(n=k\) then true for \(n=k+1\), and therefore true for all positive integers \(n\) by induction.

From the given equation, the sum of the roots is \(\alpha + \beta = 2\) and the product is \(\alpha\beta = \dfrac{5}{3}\). Let the new roots be \(\phi = \alpha + \dfrac{2}{\beta}\) and \(\psi = \beta + \dfrac{2}{\alpha}\). The sum of the new roots is \(S = \phi + \psi = \alpha + \beta + \dfrac{2(\alpha+\beta)}{\alpha\beta}\). Substituting the known values gives \(S = 2 + \dfrac{2(2)}{\frac{5}{3}} = 2 + \dfrac{12}{5} = \dfrac{22}{5}\). The product of the new roots is \(P = \phi\psi = \alpha\beta + 4 + \dfrac{4}{\alpha\beta}\). Substituting the known values gives \(P = \dfrac{5}{3} + 4 + \dfrac{4}{\frac{5}{3}} = \dfrac{5}{3} + 4 + \dfrac{12}{5} = \dfrac{121}{15}\). The new quadratic equation is \(x^2 - Sx + P = 0\), which is \(x^2 - \dfrac{22}{5}x + \dfrac{121}{15} = 0\). Multiplying by 15 gives the integer coefficient form: \(15x^2 - 66x + 121 = 0\).

M1: Identifies sum and product of original roots \(\alpha+\beta=2\) and \(\alpha\beta=\frac{5}{3}\). M1: Expresses the sum of new roots \(S\) in terms of \(\alpha+\beta\) and \(\alpha\beta\). A1: Evaluates \(S = \frac{22}{5}\). M1: Expresses the product of new roots \(P\) in terms of \(\alpha\beta\). A1: Evaluates \(P = \frac{121}{15}\). M1: Uses their \(S\) and \(P\) in \(x^2 - Sx + P = 0\). A1: Obtains \(15x^2 - 66x + 121 = 0\).

Multiply the numerator and denominator of the right hand side by \(2 - \text{i}\): \(\dfrac{15 - 5\text{i}}{2 + \text{i}} = \dfrac{(15 - 5\text{i})(2 - \text{i})}{(2 + \text{i})(2 - \text{i})} = \dfrac{30 - 15\text{i} - 10\text{i} - 5}{5} = 5 - 5\text{i}\). Substitute \(z = x + \text{i}y\) and \(z^* = x - \text{i}y\) into the left hand side: \(4(x + \text{i}y) - 3(x - \text{i}y) = x + 7\text{i}y\). Equating real and imaginary parts: \(x = 5\) and \(7y = -5\), which gives \(y = -\dfrac{5}{7}\). Thus, \(z = 5 - \dfrac{5}{7}\text{i}\).

M1: Multiplies RHS by conjugate of denominator. A1: Obtains \(5 - 5\text{i}\). M1: Substitutes \(z\) and \(z^*\) on LHS. A1: Obtains \(x + 7\text{i}y\). M1: Equates real and imaginary parts. A1: Obtains \(z = 5 - \frac{5}{7}\text{i}\).

Expanding the term: \((3r - 1)(3r + 2) = 9r^2 + 3r - 2\). Summing this: \(\sum_{r=1}^n (9r^2 + 3r - 2) = 9\sum r^2 + 3\sum r - 2n\). Substituting standard formulas: \(9\dfrac{n(n+1)(2n+1)}{6} + 3\dfrac{n(n+1)}{2} - 2n = \dfrac{3n(n+1)(2n+1)}{2} + \dfrac{3n(n+1)}{2} - 2n\). Factoring out \(\dfrac{n}{2}\): \(\dfrac{n}{2} [3(2n^2 + 3n + 1) + 3n + 3 - 4] = \dfrac{n}{2} [6n^2 + 12n + 2] = n(3n^2 + 6n + 1)\).

M1: Expands general term to \(9r^2 + 3r - 2\). M1: Uses standard formulas for sum of \(r^2\) and sum of \(r\). A1: Correct expression \(\frac{3n(n+1)(2n+1)}{2} + \frac{3n(n+1)}{2} - 2n\). M1: Factors out \(\frac{n}{2}\) or equivalent common factor. A1: Simplifies bracket to \(6n^2 + 12n + 2\). A1: Reaches the printed result \(n(3n^2 + 6n + 1)\) with no errors.

(a) We find the sign of \(f(x)\) at the midpoint of \([2.5, 3]\), which is \(2.75\). \(f(2.5) = -4\), \(f(3) = 5\), and \(f(2.75) = 2(2.75)^3 - 5(2.75)^2 - 4 = -0.21875\). Since \(f(2.75) < 0\) and \(f(3) > 0\), the root \(\alpha\) lies in \([2.75, 3]\). (b) Differentiating \(f(x)\) gives \(f'(x) = 6x^2 - 10x\). Evaluating at \(x_0 = 2.8\): \(f(2.8) = 0.704\) and \(f'(2.8) = 19.04\). Using the Newton-Raphson formula: \(x_1 = 2.8 - \dfrac{f(2.8)}{f'(2.8)} = 2.8 - \dfrac{0.704}{19.04} \approx 2.763025\). To 3 decimal places, the root is \(2.763\).

Part (a): M1: Evaluates \(f(2.75)\). A1: Concludes with interval \([2.75, 3]\). Part (b): M1: Finds derivative \(f'(x) = 6x^2 - 10x\). M1: Evaluates \(f(2.8)\) and \(f'(2.8)\). M1: Uses Newton-Raphson formula. A1: Reaches \(2.763\) to 3 d.p.

(a) Since the coefficients are real, complex roots must occur in conjugate pairs, so \(z_2 = -1 - 2\text{i}\) is also a root. Let the third root be \(\alpha\). The product of the roots is \(z_1 z_2 \alpha = -15\). We have \(z_1 z_2 = (-1 + 2\text{i})(-1 - 2\text{i}) = 5\), so \(5\alpha = -15 \implies \alpha = -3\). (b) The sum of the roots is \(-p\), so \(-p = (-1+2\text{i}) + (-1-2\text{i}) - 3 = -5 \implies p = 5\). The sum of products of roots in pairs is \(q\), so \(q = z_1 z_2 + \alpha(z_1 + z_2) = 5 + (-3)(-2) = 11\). Thus \(p = 5, q = 11\).

Part (a): M1: States \(-1 - 2\text{i}\). M1: Uses product of roots to form equation for third root. A1: Identifies third root is \(-3\). Part (b): M1: Uses sum of roots formula for \(p\). A1: Correctly finds \(p = 5\). M1: Uses pairwise product of roots formula for \(q\). A1: Correctly finds \(q = 11\).

(a) A matrix is singular if its determinant is zero. \(\det(\mathbf{M}) = (k+2)(k-5) - (3)(-4) = k^2 - 3k - 10 + 12 = k^2 - 3k + 2\). Set the determinant to zero: \(k^2 - 3k + 2 = 0 \implies (k-1)(k-2) = 0\). So the matrix is singular when \(k = 1\) or \(k = 2\). (b) When \(k = 3\), \(\mathbf{M} = \begin{pmatrix} 5 & 3 \\ -4 & -2 \end{pmatrix}\). The determinant is \(\det(\mathbf{M}) = 5(-2) - 3(-4) = 2\). The inverse is \(\mathbf{M}^{-1} = \dfrac{1}{2} \begin{pmatrix} -2 & -3 \\ 4 & 5 \end{pmatrix} = \begin{pmatrix} -1 & -1.5 \\ 2 & 2.5 \end{pmatrix}\).

Part (a): M1: Attempts determinant of \(\mathbf{M}\) in terms of \(k\). A1: Obtains \(k^2 - 3k + 2 = 0\). A1: Correctly finds \(k = 1\) and \(k = 2\). Part (b): M1: Obtains correct matrix for \(k=3\). M1: Applies correct formula for inverse. A1: Obtains \(\begin{pmatrix} -1 & -1.5 \\ 2 & 2.5 \end{pmatrix}\) or equivalent.

(a) The equation of the rectangular hyperbola is \(y = \dfrac{c^2}{x} = c^2 x^{-1}\). Differentiation with respect to \(x\) gives: \(\frac{\text{d}y}{\text{d}x} = -c^2 x^{-2} = -\frac{c^2}{x^2}\). At the point \(P\left(cp, \dfrac{c}{p}\right)\), the gradient of the tangent is: \(m_T = -\frac{c^2}{(cp)^2} = -\frac{1}{p^2}\). Therefore, the gradient of the normal to \(H\) at \(P\) is: \(m_N = -\frac{1}{m_T} = p^2\). The equation of the normal is: \(y - \frac{c}{p} = p^2(x - cp)\). Multiplying both sides by \(p\) to clear the fraction: \(py - c = p^3(x - cp) \implies py - c = p^3 x - cp^4\). Rearranging to the required form: \(p^3 x - py = c(p^4 - 1)\). (b) Since \(Q\left(cq, \dfrac{c}{q}\right)\) lies on the normal, we can substitute its coordinates into the equation of the normal: \(p^3 (cq) - p\left(\frac{c}{q}\right) = c(p^4 - 1)\). Since \(c \neq 0\), we can divide through by \(c\): \(p^3 q - \frac{p}{q} = p^4 - 1\). Multiplying through by \(q\): \(p^3 q^2 - p = q(p^4 - 1) \implies p^3 q^2 - (p^4 - 1)q - p = 0\). Expanding the middle term to factorize: \(p^3 q^2 - p^4 q + q - p = 0 \implies p^3 q(q - p) + 1(q - p) = 0 \implies (q - p)(p^3 q + 1) = 0\). Since \(P\) and \(Q\) are distinct points, \(q \neq p\). Thus, we must have: \(p^3 q + 1 = 0 \implies p^3 q = -1\). (c) The line has equation \(y = x\). Substituting \(y = x\) into the equation of the normal: \(p^3 x - px = c(p^4 - 1) \implies x(p^3 - p) = c(p^2 - 1)(p^2 + 1) \implies xp(p^2 - 1) = c(p^2 - 1)(p^2 + 1)\). Since \(p^2 \neq 1\), we have \(p^2 - 1 \neq 0\). We can divide both sides by \(p^2 - 1\): \(xp = c(p^2 + 1)\). Since \(p \neq 0\), we can divide by \(p\): \(x = \frac{c(p^2 + 1)}{p}\). Since \(y = x\) for any point on the line, the \(y\)-coordinate of \(R\) is also \(\frac{c(p^2+1)}{p}\). Hence, the coordinates of \(R\) are: \(\left( \frac{c(p^2+1)}{p}, \frac{c(p^2+1)}{p} \right)\).

(a) M1: Attempts to differentiate \(y = \dfrac{c^2}{x}\) or uses parametric differentiation to find \(\dfrac{\text{d}y}{\text{d}x}\). A1: Correctly finds \(\dfrac{\text{d}y}{\text{d}x} = -\dfrac{1}{p^2}\) at \(P\). B1: States that the gradient of the normal is \(p^2\) (negative reciprocal of their tangent gradient). M1: Substitutes their normal gradient and coordinates of \(P\) into the straight line equation. A1: Fully correct algebraic steps to show the given equation. No errors seen. (b) M1: Substitutes \(x = cq\) and \(y = \dfrac{c}{q}\) into the equation of the normal. A1: Obtains a correct simplified equation in terms of \(p\), \(q\), and \(c\). M1: Factorizes the quadratic in \(q\) to obtain \((q - p)(p^3 q + 1) = 0\) or equivalent. A0.5: States clearly that since \(q \neq p\), then \(p^3 q = -1\). (c) M1: Substitutes \(y = x\) into the equation of the normal. A1: Factored form shown, e.g., \(xp(p^2 - 1) = c(p^2 - 1)(p^2 + 1)\). A1: Divides through by \(p(p^2 - 1)\) giving the correct coordinates of \(R\).

(a) Let \(H_n\) be the statement that \(\sum_{r=1}^{n} r(r+1)\left(\frac{1}{2}\right)^r = 8 - \dfrac{n^2 + 5n + 8}{2^n}\). Base case: For \(n = 1\): \(\text{LHS} = 1(1+1)\left(\frac{1}{2}\right)^1 = 1\) and \(\text{RHS} = 8 - \frac{1^2 + 5(1) + 8}{2^1} = 8 - 7 = 1\). Since \(\text{LHS} = \text{RHS}\), the statement \(H_1\) is true. Inductive step: Assume that the statement \(H_k\) is true for some positive integer \(k\), so \(\sum_{r=1}^{k} r(r+1)\left(\frac{1}{2}\right)^r = 8 - \frac{k^2 + 5k + 8}{2^k}\). For \(n = k+1\): \(\sum_{r=1}^{k+1} r(r+1)\left(\frac{1}{2}\right)^r = \left[ \sum_{r=1}^{k} r(r+1)\left(\frac{1}{2}\right)^r \right] + (k+1)(k+2)\left(\frac{1}{2}\right)^{k+1} = 8 - \frac{k^2 + 5k + 8}{2^k} + \frac{k^2 + 3k + 2}{2^{k+1}}\). Expressing over a common denominator of \(2^{k+1}\): \(= 8 - \left( \frac{2(k^2 + 5k + 8) - (k^2 + 3k + 2)}{2^{k+1}} \right) = 8 - \left( \frac{2k^2 + 10k + 16 - k^2 - 3k - 2}{2^{k+1}} \right) = 8 - \frac{k^2 + 7k + 14}{2^{k+1}}\). This matches the target expression for \(n = k+1\) since \((k+1)^2 + 5(k+1) + 8 = k^2 + 2k + 1 + 5k + 5 + 8 = k^2 + 7k + 14\). Conclusion: Since the statement is true for \(n = 1\), and if it is true for \(n = k\) then it is also true for \(n = k+1\), by mathematical induction the statement is true for all positive integers \(n\). (b) We want to evaluate: \(\sum_{r=5}^{10} r(r+1)\left(\frac{1}{2}\right)^r = \sum_{r=1}^{10} r(r+1)\left(\frac{1}{2}\right)^r - \sum_{r=1}^{4} r(r+1)\left(\frac{1}{2}\right)^r\). Using the formula proved in (a): For \(n = 10\): \(\sum_{r=1}^{10} r(r+1)\left(\frac{1}{2}\right)^r = 8 - \frac{10^2 + 5(10) + 8}{2^{10}} = 8 - \frac{158}{1024} = 8 - \frac{79}{512} = \frac{4017}{512}\). For \(n = 4\): \(\sum_{r=1}^{4} r(r+1)\left(\frac{1}{2}\right)^r = 8 - \frac{4^2 + 5(4) + 8}{2^4} = 8 - \frac{44}{16} = 8 - \frac{11}{4} = \frac{21}{4}\). Therefore, \(\sum_{r=5}^{10} r(r+1)\left(\frac{1}{2}\right)^r = \frac{4017}{512} - \frac{21}{4} = \frac{4017 - 2688}{512} = \frac{1329}{512}\).

(a) B1: Verifies the formula holds for \(n=1\) by showing both sides equal 1. M1: Formulates the inductive hypothesis and writes \(\sum_{r=1}^{k+1} = \sum_{r=1}^{k} + u_{k+1}\) using the assumed formula. M1: Combines the algebraic terms over a common denominator of \(2^{k+1}\). A1: Obtains the correct numerator \(2(k^2 + 5k + 8) - (k^2 + 3k + 2)\) with correct signs. A1: Simplifies the numerator successfully to \(k^2 + 7k + 14\). M1: Shows clearly that the target formula for \(n=k+1\) expands to the same numerator. A1: Correct conclusion including all essential steps. (b) M1: Recognizes and applies the relationship \(\sum_{r=5}^{10} = \sum_{r=1}^{10} - \sum_{r=1}^{4}\). M1: Evaluates both sums using the formula derived in part (a). A1: Obtains both values \(\dfrac{4017}{512}\) and \(\dfrac{21}{4}\) correctly. A1: Simplifies the subtraction to obtain the final fraction \(\dfrac{1329}{512}\).

To solve the inequality
\[ \frac{2x - 3}{x + 1} > \frac{x + 1}{x - 2} \]
we first subtract the right-hand side from both sides to form a single fraction:
\[ \frac{2x - 3}{x + 1} - \frac{x + 1}{x - 2} > 0 \]
Finding a common denominator gives:
\[ \frac{(2x - 3)(x - 2) - (x + 1)^2}{(x + 1)(x - 2)} > 0 \]
Expanding and simplifying the numerator:
\[ (2x - 3)(x - 2) = 2x^2 - 7x + 6 \]
\[ (x + 1)^2 = x^2 + 2x + 1 \]
\[ (2x^2 - 7x + 6) - (x^2 + 2x + 1) = x^2 - 9x + 5 \]
So we have:
\[ \frac{x^2 - 9x + 5}{(x + 1)(x - 2)} > 0 \]
The critical values are the roots of the numerator and the roots of the denominator (the poles):
For the denominator:
\[ x = -1 \quad \text{and} \quad x = 2 \]
For the numerator, solving \( x^2 - 9x + 5 = 0 \) using the quadratic formula:
\[ x = \frac{9 \pm \sqrt{(-9)^2 - 4(1)(5)}}{2} = \frac{9 \pm \sqrt{61}}{2} \]
Letting \( x_1 = \frac{9-\sqrt{61}}{2} \approx 0.595 \) and \( x_2 = \frac{9+\sqrt{61}}{2} \approx 8.405 \), the critical values in ascending order are:
\[ -1 < \frac{9-\sqrt{61}}{2} < 2 < \frac{9+\sqrt{61}}{2} \]
Testing the sign of the function \( f(x) = \frac{x^2 - 9x + 5}{(x + 1)(x - 2)} \) in each of the five intervals yields:
- For \( x < -1 \), \( f(x) > 0 \) (True)
- For \( -1 < x < \frac{9-\sqrt{61}}{2} \), \( f(x) < 0 \) (False)
- For \( \frac{9-\sqrt{61}}{2} < x < 2 \), \( f(x) > 0 \) (True)
- For \( 2 < x < \frac{9+\sqrt{61}}{2} \), \( f(x) < 0 \) (False)
- For \( x > \frac{9+\sqrt{61}}{2} \), \( f(x) > 0 \) (True)

Thus, the solution set is:
\[ x < -1 \quad \text{or} \quad \frac{9-\sqrt{61}}{2} < x < 2 \quad \text{or} \quad x > \frac{9+\sqrt{61}}{2} \]

**[5 Marks]**
* **M1**: Multiplies by \( (x+1)^2(x-2)^2 \) or uses a common denominator to obtain a single fraction with a quadratic numerator.
* **A1**: Obtains the correct quadratic numerator or equation \( x^2 - 9x + 5 = 0 \).
* **A1**: Correctly identifies all four critical values: \( -1 \), \( 2 \), and \( \frac{9 \pm \sqrt{61}}{2} \) (or decimal equivalents to at least 3 significant figures, which are \( 0.595 \) and \( 8.41 \)).
* **M1**: Employs a valid method (such as testing values in intervals or sketching a graph) to determine the correct regions.
* **A1**: Fully correct final inequality: \( x < -1 \) or \( \frac{9-\sqrt{61}}{2} < x < 2 \) or \( x > \frac{9+\sqrt{61}}{2} \) (accept equivalent interval or set-theoretic notation).

First, we express \( z \) in terms of \( w \):
\[ w(z + 3) = z - \mathrm{i} \]
\[ wz + 3w = z - \mathrm{i} \]
\[ wz - z = -3w - \mathrm{i} \]
\[ z(w - 1) = -3w - \mathrm{i} \implies z = \frac{3w + \mathrm{i}}{1 - w} \]
Since \( |z| = 2 \), we substitute our expression for \( z \):
\[ \left| \frac{3w + \mathrm{i}}{1 - w} \right| = 2 \]
\[ |3w + \mathrm{i}| = 2|1 - w| \]
Let \( w = u + \mathrm{i}v \), where \( u, v \in \mathbb{R} \):
\[ |3(u + \mathrm{i}v) + \mathrm{i}| = 2|1 - (u + \mathrm{i}v)| \]
\[ |3u + \mathrm{i}(3v + 1)| = 2|(1 - u) - \mathrm{i}v| \]
Squaring both sides of the equation yields:
\[ (3u)^2 + (3v + 1)^2 = 4 \left( (1 - u)^2 + (-v)^2 \right) \]
\[ 9u^2 + 9v^2 + 6v + 1 = 4(1 - 2u + u^2 + v^2) \]
\[ 9u^2 + 9v^2 + 6v + 1 = 4 - 8u + 4u^2 + 4v^2 \]
Rearranging terms to form the standard form of a circle equation:
\[ 5u^2 + 5v^2 + 8u + 6v - 3 = 0 \]
Dividing the equation by 5:
\[ u^2 + v^2 + \frac{8}{5}u + \frac{6}{5}v - \frac{3}{5} = 0 \]
Completing the square for both \( u \) and \( v \):
\[ \left( u + \frac{4}{5} \right)^2 - \frac{16}{25} + \left( v + \frac{3}{5} \right)^2 - \frac{9}{25} - \frac{15}{25} = 0 \]
\[ \left( u + \frac{4}{5} \right)^2 + \left( v + \frac{3}{5} \right)^2 = \frac{40}{25} \]
\[ \left( u + \frac{4}{5} \right)^2 + \left( v + \frac{3}{5} \right)^2 = \frac{8}{5} \]
This is indeed the equation of a circle in the \( w \)-plane with:
- Center: \( \left(-\frac{4}{5}, -\frac{3}{5}\right) \) or as a complex number \( -\frac{4}{5} - \frac{3}{5}\mathrm{i} \)
- Exact Radius: \( \sqrt{\frac{8}{5}} = \frac{2\sqrt{2}}{\sqrt{5}} = \frac{2\sqrt{10}}{5} \)

**[6 Marks]**
* **M1**: Attempts to rearrange the transformation formula to make \( z \) the subject.
* **A1**: Obtains a correct expression for \( z \) in terms of \( w \), e.g., \( z = \frac{3w + \mathrm{i}}{1 - w} \) or equivalent.
* **M1**: Substitutes this expression into \( |z| = 2 \) to obtain \( |3w + \mathrm{i}| = 2|1 - w| \) or equivalent.
* **M1**: Substitutes \( w = u + \mathrm{i}v \) (or \( x + \mathrm{i}y \)), squares both sides, and expands.
* **A1**: Obtains a correct Cartesian equation for the circle, such as \( 5u^2 + 5v^2 + 8u + 6v - 3 = 0 \).
* **A1**: Correctly completes the square to state the center \( -\frac{4}{5} - \frac{3}{5}\mathrm{i} \) (or coordinates \( \left(-\frac{4}{5}, -\frac{3}{5}\right) \)) and the exact radius \( \frac{2\sqrt{10}}{5} \) (or equivalent simplified exact form, e.g., \( \sqrt{1.6} \)).

(a) Starting with the right-hand side:
\[ \frac{1}{2} \left[ \frac{2r+1}{r(r+1)} - \frac{2r+3}{(r+1)(r+2)} \right] = \frac{1}{2} \left[ \frac{(2r+1)(r+2) - r(2r+3)}{r(r+1)(r+2)} \right] \]
\[ = \frac{1}{2} \left[ \frac{(2r^2 + 5r + 2) - (2r^2 + 3r)}{r(r+1)(r+2)} \right] = \frac{1}{2} \left[ \frac{2r + 2}{r(r+1)(r+2)} \right] \]
\[ = \frac{1}{2} \left[ \frac{2(r+1)}{r(r+1)(r+2)} \right] = \frac{1}{r(r+2)} \quad \text{as required.} \]

(b) Let \(V_r = \frac{2r+1}{r(r+1)}\). Using the result from (a):
\[ \sum_{r=1}^n \frac{1}{r(r+2)} = \frac{1}{2} \sum_{r=1}^n (V_r - V_{r+1}) \]
Writing out the terms of the sum:
\[ r=1: \quad \frac{1}{2}(V_1 - V_2) \]
\[ r=2: \quad \frac{1}{2}(V_2 - V_3) \]
\[ \dots \]
\[ r=n: \quad \frac{1}{2}(V_n - V_{n+1}) \]
Summing these terms results in a telescoping series where intermediate terms cancel:
\[ \sum_{r=1}^n \frac{1}{r(r+2)} = \frac{1}{2} (V_1 - V_{n+1}) \]
Since \(V_1 = \frac{2(1)+1}{1(2)} = \frac{3}{2}\) and \(V_{n+1} = \frac{2(n+1)+1}{(n+1)(n+2)} = \frac{2n+3}{(n+1)(n+2)}\), we have:
\[ \sum_{r=1}^n \frac{1}{r(r+2)} = \frac{1}{2} \left[ \frac{3}{2} - \frac{2n+3}{(n+1)(n+2)} \right] \]
\[ = \frac{1}{4} \left[ 3 - \frac{2(2n+3)}{(n+1)(n+2)} \right] = \frac{1}{4} \left[ \frac{3(n^2+3n+2) - (4n+6)}{(n+1)(n+2)} \right] \]
\[ = \frac{1}{4} \left[ \frac{3n^2 + 9n + 6 - 4n - 6}{(n+1)(n+2)} \right] = \frac{3n^2+5n}{4(n+1)(n+2)} = \frac{n(3n+5)}{4(n+1)(n+2)} \quad \text{as required.} \]

(c) As \(n \to \infty\):
\[ \lim_{n \to \infty} \frac{n(3n+5)}{4(n+1)(n+2)} = \lim_{n \to \infty} \frac{3n^2+5n}{4n^2+12n+8} = \frac{3}{4} \]

(a)
- M1: Attempts to combine RHS fractions over a common denominator of \(r(r+1)(r+2)\) or \(r(r+2)\).
- A1: Correctly expands numerator to \(2r^2 + 5r + 2 - 2r^2 - 3r\) or simplifies it to \(2r + 2\).
- A1*: Fully correct proof with no errors, showing clearly the cancellation of \((r+1)\) to reach \(\frac{1}{r(r+2)}\).

(b)
- M1: Writes the sum in terms of \(V_r - V_{r+1}\) using part (a) and lists at least 3 terms to show cancellation.
- A1: Correctly identifies the remaining terms as \(\frac{1}{2}(V_1 - V_{n+1})\).
- A1: Correctly substitutes \(V_1 = \frac{3}{2}\) and \(V_{n+1} = \frac{2n+3}{(n+1)(n+2)}\).
- dM1: (Dependent on the first M1) Puts the expression over a common denominator of \(4(n+1)(n+2)\).
- A1*: Fully correct algebraic simplification leading to the given answer.

(c)
- B1: States \(\frac{3}{4}\) or 0.75.

(a) Substituting \(x = 1\) and \(y = 2\) into the given equation:
\[ 2 \frac{\mathrm{d}y}{\mathrm{d}x} + 2(1)^2 = 2^2 \implies 2 \frac{\mathrm{d}y}{\mathrm{d}x} + 2 = 4 \implies \frac{\mathrm{d}y}{\mathrm{d}x} = 1 \]

(b) Differentiating the differential equation implicitly with respect to \(x\):
\[ y \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 + 4x = 2y \frac{\mathrm{d}y}{\mathrm{d}x} \]
Substituting \(x = 1\), \(y = 2\), and \(\frac{\mathrm{d}y}{\mathrm{d}x} = 1\):
\[ 2 \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 1^2 + 4(1) = 2(2)(1) \implies 2 \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 5 = 4 \implies \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -0.5 \]

(c) Differentiating again with respect to \(x\):
\[ \frac{\mathrm{d}}{\mathrm{d}x} \left( y y'' + (y')^2 + 4x \right) = \frac{\mathrm{d}}{\mathrm{d}x} (2y y') \]
\[ y y''' + y' y'' + 2y' y'' + 4 = 2 (y')^2 + 2y y'' \]
\[ y y''' + 3y' y'' + 4 = 2 (y')^2 + 2y y'' \]
Substituting \(x = 1\), \(y = 2\), \(y' = 1\), and \(y'' = -0.5\):
\[ 2 y''' + 3(1)(-0.5) + 4 = 2(1)^2 + 2(2)(-0.5) \]
\[ 2 y''' - 1.5 + 4 = 2 - 2 \implies 2 y''' + 2.5 = 0 \implies y''' = -1.25 \]

(d) The Taylor series expansion of \(y\) about \(x = 1\) is:
\[ y(x) \approx y(1) + y'(1)(x-1) + \frac{y''(1)}{2!}(x-1)^2 + \frac{y'''(1)}{3!}(x-1)^3 \]
Substituting the values found:
\[ y(x) \approx 2 + 1(x-1) + \frac{-0.5}{2}(x-1)^2 + \frac{-1.25}{6}(x-1)^3 \]
\[ y(x) \approx 2 + (x-1) - \frac{1}{4}(x-1)^2 - \frac{5}{24}(x-1)^3 \]

(a)
- B1: Correctly identifies \(\frac{\mathrm{d}y}{\mathrm{d}x} = 1\).

(b)
- M1: Differentiates implicitly with respect to \(x\) (must use product rule on \(y \frac{\mathrm{d}y}{\mathrm{d}x}\)).
- A1: Correctly solves to find \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -0.5\) (or \(-\frac{1}{2}\)).

(c)
- M1: Differentiates again with respect to \(x\), using product and chain rules.
- A1: Correctly obtains the equation \(y y''' + 3y' y'' + 4 = 2(y')^2 + 2y y''\) (or equivalent).
- A1: Correctly solves to find \(\frac{\mathrm{d}^3y}{\mathrm{d}x^3} = -1.25\) (or \(-\frac{5}{4}\)).

(d)
- M1: Applies the Taylor series formula using their evaluated derivatives.
- A1: Correct quadratic term \(-\frac{1}{4}(x-1)^2\).
- A1: Correct cubic term \(-\frac{5}{24}(x-1)^3\) leading to the final expression \(2 + (x-1) - \frac{1}{4}(x-1)^2 - \frac{5}{24}(x-1)^3\).

(a) Given \(u = y^{-2}\), differentiating with respect to \(x\) using the chain rule gives:
\[ \frac{\mathrm{d}u}{\mathrm{d}x} = -2y^{-3} \frac{\mathrm{d}y}{\mathrm{d}x} \implies \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{1}{2}y^3 \frac{\mathrm{d}u}{\mathrm{d}x} \]
Substituting this expression into the original differential equation:
\[ -\frac{1}{2}y^3 \frac{\mathrm{d}u}{\mathrm{d}x} + \frac{1}{x}y = x^2 y^3 \]
Divide through by \(-\frac{1}{2}y^3\):
\[ \frac{\mathrm{d}u}{\mathrm{d}x} - \frac{2}{x}y^{-2} = -2x^2 \]
Since \(u = y^{-2}\), we obtain:
\[ \frac{\mathrm{d}u}{\mathrm{d}x} - \frac{2}{x} u = -2x^2 \quad \text{as required.} \]

(b) This is a first order linear differential equation in \(u\). The integrating factor is:
\[ I(x) = \exp\left( \int -\frac{2}{x} \mathrm{d}x \right) = \exp(-2\ln x) = x^{-2} \]
Multiplying the differential equation by this integrating factor:
\[ x^{-2}\frac{\mathrm{d}u}{\mathrm{d}x} - 2x^{-3} u = -2 \implies \frac{\mathrm{d}}{\mathrm{d}x}(u x^{-2}) = -2 \]
Integrating both sides with respect to \(x\):
\[ u x^{-2} = \int -2 \mathrm{d}x \implies u x^{-2} = -2x + C \]
Multiplying by \(x^2\) to solve for \(u\):
\[ u = Cx^2 - 2x^3 \]

(c) Since \(u = y^{-2} = \frac{1}{y^2}\):
\[ \frac{1}{y^2} = Cx^2 - 2x^3 \implies y^2 = \frac{1}{Cx^2 - 2x^3} \]

(a)
- M1: Differentiates \(u = y^{-2}\) with respect to \(x\) to obtain \(\frac{\mathrm{d}u}{\mathrm{d}x} = k y^{-3} \frac{\mathrm{d}y}{\mathrm{d}x}\).
- A1: Correct derivative \(\frac{\mathrm{d}u}{\mathrm{d}x} = -2 y^{-3} \frac{\mathrm{d}y}{\mathrm{d}x}\).
- M1: Substitutes \(\frac{\mathrm{d}y}{\mathrm{d}x}\) and \(u = y^{-2}\) into the original differential equation.
- A1*: Simplifies correctly to show the given equation.

(b)
- M1: Finds the integrating factor \(I(x) = \exp\left( \int -\frac{2}{x} \mathrm{d}x \right)\).
- A1: Correct integrating factor \(x^{-2}\).
- M1: Integrates to find \(u \cdot I(x) = \int -2 \mathrm{d}x\) (must include a constant of integration).
- A1: Correct general solution for \(u\) in terms of \(x\), e.g., \(u = Cx^2 - 2x^3\).

(c)
- B1: Correct general solution \(y^2 = \frac{1}{Cx^2 - 2x^3}\) (or equivalent).

(a) Since \( x = e^t \), we have \( t = \ln x \) and \( \frac{dt}{dx} = \frac{1}{x} = e^{-t} \).
Using the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{dt} \frac{dt}{dx} = \frac{1}{x} \frac{dy}{dt} \]
Differentiating again with respect to \( x \):
\[ \frac{d^2 y}{dx^2} = \frac{d}{dx}\left(\frac{1}{x} \frac{dy}{dt}\right) = -\frac{1}{x^2} \frac{dy}{dt} + \frac{1}{x} \frac{d}{dx}\left(\frac{dy}{dt}\right) = -\frac{1}{x^2} \frac{dy}{dt} + \frac{1}{x} \left(\frac{d^2 y}{dt^2} \frac{dt}{dx}\right) = -\frac{1}{x^2} \frac{dy}{dt} + \frac{1}{x^2} \frac{d^2 y}{dt^2} \]
Therefore:
\[ x \frac{dy}{dx} = \frac{dy}{dt} \quad \text{and} \quad x^2 \frac{d^2 y}{dx^2} = \frac{d^2 y}{dt^2} - \frac{dy}{dt} \]
Substituting these expressions into the original differential equation:
\[ \left(\frac{d^2 y}{dt^2} - \frac{dy}{dt}\right) - 4\left(\frac{dy}{dt}\right) + 6y = 12(e^t)^5 \]
\[ \frac{d^2 y}{dt^2} - 5\frac{dy}{dt} + 6y = 12 e^{5t} \quad \text{(as required)} \]

(b) To solve the transformed equation, we first solve the homogeneous part using the auxiliary equation:
\[ m^2 - 5m + 6 = 0 \implies (m-2)(m-3) = 0 \implies m = 2, 3 \]
The complementary function (CF) is:
\[ y_{\text{CF}} = A e^{2t} + B e^{3t} \]
For the particular integral (PI), try \( y_{\text{PI}} = k e^{5t} \). Then \( \frac{dy}{dt} = 5k e^{5t} \) and \( \frac{d^2 y}{dt^2} = 25k e^{5t} \).
Substituting these into the differential equation:
\[ 25k e^{5t} - 5(5k e^{5t}) + 6k e^{5t} = 12e^{5t} \implies 6k e^{5t} = 12 e^{5t} \implies k = 2 \]
The general solution in terms of \( t \) is:
\[ y = A e^{2t} + B e^{3t} + 2 e^{5t} \]
Substituting back \( x = e^t \):
\[ y = A x^2 + B x^3 + 2 x^5 \]

(c) Using the conditions \( y = 5 \) and \( \frac{dy}{dx} = 11 \) at \( x = 1 \):
From \( y = A x^2 + B x^3 + 2 x^5 \):
\[ 5 = A(1)^2 + B(1)^3 + 2(1)^5 \implies A + B = 3 \]
Differentiating the general solution with respect to \( x \):
\[ \frac{dy}{dx} = 2Ax + 3Bx^2 + 10x^4 \]
Substituting \( x = 1 \) and \( \frac{dy}{dx} = 11 \):
\[ 11 = 2A + 3B + 10 \implies 2A + 3B = 1 \]
Solving the system \( A + B = 3 \) and \( 2A + 3B = 1 \):
From the first equation, \( A = 3 - B \). Substituting this into the second equation:
\[ 2(3 - B) + 3B = 1 \implies 6 + B = 1 \implies B = -5 \]
Then \( A = 3 - (-5) = 8 \).
The particular solution is:
\[ y = 8x^2 - 5x^3 + 2x^5 \]

Part (a):
M1: Correctly applies the chain rule to find \( \frac{dy}{dx} \) in terms of \( \frac{dy}{dt} \) and \( x \).
M1: Attempts to differentiate \( \frac{dy}{dx} \) with respect to \( x \) using the product rule or quotient rule.
A1: Correctly derives \( \frac{d^2 y}{dx^2} \) in terms of derivatives with respect to \( t \) and \( x \).
M1: Substitutes the expressions for \( x^2 \frac{d^2 y}{dx^2} \) and \( x \frac{dy}{dx} \) into the given differential equation.
A1*: Fully correct simplification to reach the given differential equation with no errors shown.

Part (b):
M1: Formulates and solves the auxiliary equation to find the roots.
A1: Writes the correct complementary function (CF).
M1: Substitutes a trial PI of the correct form \( k e^{5t} \) into the ODE and solves for \( k \).
A1: Obtains the correct general solution in terms of \( x \).

Part (c):
M1: Uses the condition \( y = 5 \) at \( x = 1 \) to establish an equation in \( A \) and \( B \).
M1: Differentiates their general solution and uses \( \frac{dy}{dx} = 11 \) at \( x = 1 \) to find a second equation in \( A \) and \( B \).
A1: Solves the equations to find the correct constants \( A = 8, B = -5 \) and writes the correct particular solution.

(a) Given the substitution \( z = y^{-2} \).
Differentiating both sides with respect to \( x \) using the chain rule:
\[ \frac{dz}{dx} = -2 y^{-3} \frac{dy}{dx} \implies y^{-3} \frac{dy}{dx} = -\frac{1}{2} \frac{dz}{dx} \]
Now, we divide the original differential equation by \( y^3 \):
\[ y^{-3} \frac{dy}{dx} + \frac{2}{x} y^{-2} = 3x^2 \]
Substituting \( y^{-3} \frac{dy}{dx} = -\frac{1}{2} \frac{dz}{dx} \) and \( y^{-2} = z \) into this equation gives:
\[ -\frac{1}{2} \frac{dz}{dx} + \frac{2}{x} z = 3x^2 \]
Multiplying the entire equation by \( -2 \):
\[ \frac{dz}{dx} - \frac{4}{x} z = -6x^2 \quad \text{(as required)} \]

(b) The equation is a first-order linear differential equation in \( z \) of the form \( \frac{dz}{dx} + P(x)z = Q(x) \), where \( P(x) = -\frac{4}{x} \) and \( Q(x) = -6x^2 \).
The integrating factor is:
\[ I(x) = e^{\int -\frac{4}{x} \, dx} = e^{-4\ln x} = x^{-4} \]
Multiplying the differential equation by this integrating factor:
\[ x^{-4} \frac{dz}{dx} - 4x^{-5} z = -6x^{-2} \implies \frac{d}{dx} (z x^{-4}) = -6x^{-2} \]
Integrating both sides with respect to \( x \):
\[ z x^{-4} = \int -6x^{-2} \, dx = 6x^{-1} + C \]
Multiplying both sides by \( x^4 \):
\[ z = 6x^3 + C x^4 \]
Since \( z = y^{-2} = \frac{1}{y^2} \), we have:
\[ y^2 = \frac{1}{6x^3 + C x^4} \]

(c) Using the condition \( y = \frac{1}{2} \) at \( x = 1 \):
\[ \left(\frac{1}{2}\right)^2 = \frac{1}{6(1)^3 + C(1)^4} \implies \frac{1}{4} = \frac{1}{6 + C} \]
Solving for \( C \):
\[ 6 + C = 4 \implies C = -2 \]
Thus, the particular solution is:
\[ y^2 = \frac{1}{6x^3 - 2x^4} \]

Part (a):
M1: Differentiates \( z = y^{-2} \) with respect to \( x \) to find \( \frac{dz}{dx} \) in terms of \( y \) and \( \frac{dy}{dx} \).
A1: Obtains the correct expression \( y^{-3}\frac{dy}{dx} = -\frac{1}{2}\frac{dz}{dx} \).
M1: Divides the original ODE by \( y^3 \) and substitutes expressions in \( z \).
A1*: Fully justifies the steps to arrive at the given target differential equation.

Part (b):
M1: Finds the correct form of the integrating factor \( I(x) = e^{\int -\frac{4}{x}\,dx} \).
A1: Obtains \( I(x) = x^{-4} \).
M1: Multiplies by the integrating factor and integrates \( -6x^{-2} \) with respect to \( x \).
A1: Obtains \( z x^{-4} = 6x^{-1} + C \).
A1: Substitutes back \( z = y^{-2} \) and writes \( y^2 \) as a function of \( x \).

Part (c):
M1: Substitutes \( y = 1/2 \) and \( x = 1 \) into their general solution.
A1: Correctly calculates \( C = -2 \).
A1: Writes down the final particular solution in a correct form, e.g., \( y^2 = \frac{1}{6x^3 - 2x^4} \) or \( y^2 = \frac{1}{2x^3(3 - x)} \).

(a) Given the substitution \( y = u e^{-2x} \).
Differentiating with respect to \( x \) using the product rule:
\[ \frac{dy}{dx} = \frac{du}{dx} e^{-2x} - 2u e^{-2x} = \left( \frac{du}{dx} - 2u \right) e^{-2x} \]
Differentiating once more with respect to \( x \):
\[ \frac{d^2 y}{dx^2} = \left( \frac{d^2 u}{dx^2} - 2\frac{du}{dx} \right) e^{-2x} - 2\left( \frac{du}{dx} - 2u \right) e^{-2x} = \left( \frac{d^2 u}{dx^2} - 4\frac{du}{dx} + 4u \right) e^{-2x} \]
Substituting \( y \), \( \frac{dy}{dx} \), and \( \frac{d^2 y}{dx^2} \) into the given differential equation:
\[ \left( \frac{d^2 u}{dx^2} - 4\frac{du}{dx} + 4u \right) e^{-2x} + 4\left( \frac{du}{dx} - 2u \right) e^{-2x} + 4u e^{-2x} = e^{-2x} \ln x \]
Factorizing out the common exponential term \( e^{-2x} \):
\[ e^{-2x} \left( \frac{d^2 u}{dx^2} - 4\frac{du}{dx} + 4u + 4\frac{du}{dx} - 8u + 4u \right) = e^{-2x} \ln x \]
Simplifying the term inside the brackets:
\[ e^{-2x} \frac{d^2 u}{dx^2} = e^{-2x} \ln x \]
Since \( e^{-2x} \neq 0 \), dividing both sides by \( e^{-2x} \) yields:
\[ \frac{d^2 u}{dx^2} = \ln x \quad \text{(as required)} \]

(b) Integrating \( \frac{d^2 u}{dx^2} = \ln x \) with respect to \( x \) using integration by parts:
Let \( I_1 = \int \ln x \, dx \). Choosing \( u_{\text{by parts}} = \ln x \) and \( v_{\text{by parts}}' = 1 \):
\[ I_1 = x \ln x - \int x \left( \frac{1}{x} \right) dx = x \ln x - x + A \]
So:
\[ \frac{du}{dx} = x \ln x - x + A \]
Integrating again with respect to \( x \) to find \( u \):
\[ u = \int (x \ln x - x + A) \, dx = \int x \ln x \, dx - \frac{1}{2}x^2 + Ax + B \]
To evaluate \( \int x \ln x \, dx \), we use integration by parts with \( u_{\text{by parts}} = \ln x \) and \( v_{\text{by parts}}' = x \):
\[ \int x \ln x \, dx = \frac{1}{2}x^2 \ln x - \int \frac{1}{2}x^2 \left( \frac{1}{x} \right) dx = \frac{1}{2}x^2 \ln x - \frac{1}{4}x^2 \]
Thus:
\[ u = \left( \frac{1}{2}x^2 \ln x - \frac{1}{4}x^2 \right) - \frac{1}{2}x^2 + Ax + B \]
\[ u = \frac{1}{2}x^2 \ln x - \frac{3}{4}x^2 + Ax + B \]
Since \( y = u e^{-2x} \), the general solution is:
\[ y = e^{-2x} \left( \frac{1}{2}x^2 \ln x - \frac{3}{4}x^2 + Ax + B \right) \]

(c) Using the conditions at \( x = 1 \):
When \( x = 1 \), \( y = 0 \):
\[ 0 = e^{-2} \left( \frac{1}{2}(1)^2 \ln(1) - \frac{3}{4}(1)^2 + A(1) + B \right) \implies -\frac{3}{4} + A + B = 0 \implies A + B = \frac{3}{4} \]
Now, we have \( \frac{dy}{dx} = e^{-2x} \left( \frac{du}{dx} - 2u \right) \).
At \( x = 1 \), since \( y = 0 \implies u = 0 \).
Thus, at \( x = 1 \), \( \frac{dy}{dx} = e^{-2} \frac{du}{dx} \).
Given \( \frac{dy}{dx} = 0 \) at \( x = 1 \), we must have \( \frac{du}{dx} = 0 \) at \( x = 1 \).
Using \( \frac{du}{dx} = x \ln x - x + A \):
\[ 0 = 1 \ln(1) - 1 + A \implies A = 1 \]
Substituting \( A = 1 \) back into \( A + B = \frac{3}{4} \):
\[ 1 + B = \frac{3}{4} \implies B = -\frac{1}{4} \]
So, the constants of integration are \( A = 1 \) and \( B = -\frac{1}{4} \).

Part (a):
M1: Correctly applies the product rule to find the first derivative \( \frac{dy}{dx} \).
A1: Correctly applies the product rule to find the second derivative \( \frac{d^2 y}{dx^2} \).
M1: Substitutes \( y \), \( \frac{dy}{dx} \), and \( \frac{d^2 y}{dx^2} \) into the original differential equation.
A1*: Fully simplifies to get the given target equation \( \frac{d^2 u}{dx^2} = \ln x \).

Part (b):
M1: Correctly integrates \( \ln x \) using integration by parts (must show integration by parts process).
A1: Obtains the correct expression for \( \frac{du}{dx} = x \ln x - x + A \).
M1: Applies integration by parts to integrate \( x \ln x \).
A1: Obtains \( u = \frac{1}{2}x^2 \ln x - \frac{3}{4}x^2 + Ax + B \).
A1: Multiplies by \( e^{-2x} \) to write down the correct general solution for \( y \).

Part (c):
M1: Uses \( y = 0 \) at \( x = 1 \) to establish the linear relation \( A + B = 3/4 \).
M1: Connects the boundary conditions on \( \frac{dy}{dx} \) and \( y \) to establish that \( \frac{du}{dx} = 0 \) at \( x = 1 \), and uses this to find \( A \).
A1: Obtains the correct values \( A = 1 \) and \( B = -1/4 \).

Part (a): We have \(x = 5\cosh t\) and \(y = 3\sinh t\). Differentiating both with respect to \(t\): \(\frac{dx}{dt} = 5\sinh t\) and \(\frac{dy}{dt} = 3\cosh t\). Thus, the gradient of the tangent is \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3\cosh t}{5\sinh t}\). The gradient of the normal is the negative reciprocal: \(m = -\frac{5\sinh t}{3\cosh t}\). The equation of the normal at \(P\) is: \(y - 3\sinh t = -\frac{5\sinh t}{3\cosh t}(x - 5\cosh t)\). Multiplying both sides by \(3\cosh t\) yields: \(3y\cosh t - 9\sinh t\cosh t = -5x\sinh t + 25\sinh t\cosh t\). Rearranging terms gives: \(5x\sinh t + 3y\cosh t = 34\sinh t\cosh t\). Part (b): To find \(A\) (intersection with the x-axis), set \(y = 0\): \(5x\sinh t = 34\sinh t\cosh t\). Since \(t > 0\), \(\sinh t \neq 0\), so \(x = \frac{34}{5}\cosh t\), giving \(A\left(\frac{34}{5}\cosh t, 0\right)\). To find \(B\) (intersection with the y-axis), set \(x = 0\): \(3y\cosh t = 34\sinh t\cosh t\). Since \(\cosh t \neq 0\), \(y = \frac{34}{3}\sinh t\), giving \(B\left(0, \frac{34}{3}\sinh t\right)\). The midpoint \(M(x, y)\) of \(AB\) has coordinates: \(x = \frac{1}{2}\left(\frac{34}{5}\cosh t + 0\right) = \frac{17}{5}\cosh t\) and \(y = \frac{1}{2}\left(0 + \frac{34}{3}\sinh t\right) = \frac{17}{3}\sinh t\). So the coordinates of \(M\) are \(M\left(\frac{17}{5}\cosh t, \frac{17}{3}\sinh t\right)\). Part (c): From the coordinates of \(M\), we have \(\cosh t = \frac{5x}{17}\) and \(\sinh t = \frac{3y}{17}\). Using the identity \(\cosh^2 t - \sinh^2 t = 1\), we obtain: \(\left(\frac{5x}{17}\right)^2 - \left(\frac{3y}{17}\right)^2 = 1 \implies \frac{25x^2}{289} - \frac{9y^2}{289} = 1 \implies 25x^2 - 9y^2 = 289\). Thus, the locus of \(M\) is a hyperbola with equation \(25x^2 - 9y^2 = 289\), where \(k = 289\).

Part (a): M1: Attempt to differentiate both \(x\) and \(y\) with respect to \(t\). A1: Correct expression for \(\frac{dy}{dx} = \frac{3\cosh t}{5\sinh t}\). M1: Uses the negative reciprocal gradient to form the equation of the normal. A1*: Correctly completes the algebraic steps to show the given equation. Part (b): M1: Sets \(x=0\) or \(y=0\) to find at least one coordinate of \(A\) or \(B\). A1: Correct coordinates of \(M\left(\frac{17}{5}\cosh t, \frac{17}{3}\sinh t\right)\). Part (c): M1: Rearranges coordinates to express \(\cosh t\) and \(\sinh t\) in terms of \(x\) and \(y\). M1: Uses the identity \(\cosh^2 t - \sinh^2 t = 1\) to eliminate the parameter \(t\). A1: Obtains the correct Cartesian equation and states \(k = 289\).

Part (a): (i) For a hyperbola, \(b^2 = a^2(e^2 - 1)\). Here, \(a^2 = 16\) and \(b^2 = 9\). Substituting these values: \(9 = 16(e^2 - 1) \implies e^2 - 1 = \frac{9}{16} \implies e^2 = \frac{25}{16}\). Since \(e > 1\), we have \(e = \frac{5}{4}\). (ii) The foci are located at \((\pm ae, 0)\). Since \(ae = 4 \times \frac{5}{4} = 5\), the coordinates of the foci are \(S(5, 0)\) and \(S'(-5, 0)\). Part (b): Differentiating \(x = 4\cosh \phi\) and \(y = 3\sinh \phi\) with respect to \(\phi\): \(\frac{dx}{d\phi} = 4\sinh \phi\) and \(\frac{dy}{d\phi} = 3\cosh \phi\). The gradient of the tangent is \(\frac{dy}{dx} = \frac{dy/d\phi}{dx/d\phi} = \frac{3\cosh \phi}{4\sinh \phi}\). The equation of the tangent at \(Q\) is: \(y - 3\sinh \phi = \frac{3\cosh \phi}{4\sinh \phi}(x - 4\cosh \phi)\). Multiplying both sides by \(4\sinh \phi\) yields: \(4y\sinh \phi - 12\sinh^2 \phi = 3x\cosh \phi - 12\cosh^2 \phi \implies 3x\cosh \phi - 4y\sinh \phi = 12(\cosh^2 \phi - \sinh^2 \phi)\). Since \(\cosh^2 \phi - \sinh^2 \phi = 1\), this simplifies to: \(3x\cosh \phi - 4y\sinh \phi = 12\). Part (c): The perpendicular distance from a point \((x_1, y_1)\) to the line \(Ax + By + C = 0\) is given by \(\frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\). Here, the tangent equation is \(3x\cosh \phi - 4y\sinh \phi - 12 = 0\). The distance from \(S(5,0)\) is \(d_1 = \frac{|3(5)\cosh \phi - 4(0)\sinh \phi - 12|}{\sqrt{(3\cosh \phi)^2 + (-4\sinh \phi)^2}} = \frac{|15\cosh \phi - 12|}{\sqrt{9\cosh^2 \phi + 16\sinh^2 \phi}}\). The distance from \(S'(-5,0)\) is \(d_2 = \frac{|3(-5)\cosh \phi - 4(0)\sinh \phi - 12|}{\sqrt{(3\cosh \phi)^2 + (-4\sinh \phi)^2}} = \frac{|-15\cosh \phi - 12|}{\sqrt{9\cosh^2 \phi + 16\sinh^2 \phi}} = \frac{|15\cosh \phi + 12|}{\sqrt{9\cosh^2 \phi + 16\sinh^2 \phi}}\). The product of the distances is: \(d_1 d_2 = \frac{|(15\cosh \phi - 12)(15\cosh \phi + 12)|}{9\cosh^2 \phi + 16\sinh^2 \phi} = \frac{|225\cosh^2 \phi - 144|}{9\cosh^2 \phi + 16\sinh^2 \phi}\). Since \(\cosh \phi \ge 1\), \(225\cosh^2 \phi - 144 > 0\), so the absolute value bars can be removed. Writing the denominator in terms of \(\cosh \phi\) using \(\sinh^2 \phi = \cosh^2 \phi - 1\): \(9\cosh^2 \phi + 16(\cosh^2 \phi - 1) = 25\cosh^2 \phi - 16\). Note that \(225\cosh^2 \phi - 144 = 9(25\cosh^2 \phi - 16)\). Therefore: \(d_1 d_2 = \frac{9(25\cosh^2 \phi - 16)}{25\cosh^2 \phi - 16} = 9\). This product is constant and independent of \(\phi\), and its value is 9.

Part (a): M1: Uses the standard hyperbola identity \(b^2 = a^2(e^2 - 1)\) to set up an equation in terms of \(e\). A1: Correct eccentricity \(e = \frac{5}{4}\). A1: Correct coordinates for both foci \(S(5, 0)\) and \(S'(-5, 0)\). Part (b): M1: Differentiates parametric equations to find \(\frac{dx}{d\phi}\) and \(\frac{dy}{d\phi}\). M1: Uses their gradients to form an equation of the tangent line. A1*: Correctly simplifies the tangent equation to the required form using the identity \(\cosh^2 \phi - \sinh^2 \phi = 1\). Part (c): M1: Applies the perpendicular distance formula from at least one of the foci to the tangent. M1: Sets up the product of distances and attempts to express the denominator using \(\sinh^2 \phi = \cosh^2 \phi - 1\). A1: Completes the simplification to show that the product is 9, clearly stating that it is independent of \(\phi\).

(a) Since \( y = \frac{1}{2} \cosh(2x) \), we have \( \frac{dy}{dx} = \sinh(2x) \).
Using the formula for arc length:
\( s = \int_{0}^{\ln 2} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx = \int_{0}^{\ln 2} \sqrt{1 + \sinh^2(2x)} \, dx = \int_{0}^{\ln 2} \cosh(2x) \, dx \)
Integrating gives:
\( s = \left[ \frac{1}{2} \sinh(2x) \right]_{0}^{\ln 2} = \frac{1}{2} \sinh(2 \ln 2) - 0 \)
Since \( \sinh(2\ln 2) = \frac{e^{2\ln 2} - e^{-2\ln 2}}{2} = \frac{4 - 1/4}{2} = \frac{15}{8} \),
we obtain:
\( s = \frac{1}{2} \times \frac{15}{8} = \frac{15}{16} \) as required.

(b) Using the formula for surface area of revolution:
\( S = 2\pi \int_{0}^{\ln 2} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \)
\( S = 2\pi \int_{0}^{\ln 2} \left( \frac{1}{2} \cosh(2x) \right) \cosh(2x) \, dx = \pi \int_{0}^{\ln 2} \cosh^2(2x) \, dx \)
Using the identity \( \cosh^2(2x) = \frac{1 + \cosh(4x)}{2} \):
\( S = \frac{\pi}{2} \int_{0}^{\ln 2} (1 + \cosh(4x)) \, dx = \frac{\pi}{2} \left[ x + \frac{1}{4} \sinh(4x) \right]_{0}^{\ln 2} \)
Evaluating at the limits:
\( S = \frac{\pi}{2} \left( \ln 2 + \frac{1}{4} \sinh(4\ln 2) - 0 \right) \)
Since \( \sinh(4\ln 2) = \frac{e^{4\ln 2} - e^{-4\ln 2}}{2} = \frac{16 - 1/16}{2} = \frac{255}{32} \),
we have:
\( S = \frac{\pi}{2} \left( \ln 2 + \frac{255}{128} \right) = \frac{\pi}{2} \left( \frac{128\ln 2 + 255}{128} \right) = \frac{\pi}{256}(128\ln 2 + 255) \).
So \( k = 256 \), \( a = 128 \), \( b = 255 \).

(a)
M1: Formulates the integral for arc length, finds \( \frac{dy}{dx} = \sinh(2x) \), and uses the hyperbolic identity \( 1 + \sinh^2(2x) = \cosh^2(2x) \).
A1: Integrates correctly to obtain \( \frac{1}{2}\sinh(2x) \).
M1: Substitutes the limit \( x = \ln 2 \) and uses the exponential form of sinh.
A1: Obtains the correct length of \( \frac{15}{16} \) through correct working.

(b)
M1: Formulates the correct integral for the surface area of revolution.
A1: Simplifies the integrand to \( \cosh^2(2x) \).
M1: Uses the hyperbolic identity \( \cosh^2(2x) = \frac{1+\cosh(4x)}{2} \).
A1: Integrates to find \( \frac{1}{2}\left(x + \frac{1}{4}\sinh(4x)\right) \).
M1: Substitutes the upper limit \( x = \ln 2 \) and evaluates \( \sinh(4\ln 2) \) using exponential forms.
M1: Combines terms into a single fraction over a common denominator.
A1: Correct exact final answer in the requested form: \( \frac{\pi}{256}(128 \ln 2 + 255) \).

(a) Let \( \mathbf{d}_1 = \mathbf{i} - \mathbf{j} + 2\mathbf{k} \) and \( \mathbf{d}_2 = 2\mathbf{i} + \mathbf{j} - \mathbf{k} \) be the direction vectors of \( L_1 \) and \( L_2 \).
A vector perpendicular to both is given by their cross product:
\( \mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 2 \\ 2 & 1 & -1 \end{vmatrix} = \mathbf{i}(1 - 2) - \mathbf{j}(-1 - 4) + \mathbf{k}(1 - (-2)) = -\mathbf{i} + 5\mathbf{j} + 3\mathbf{k} \).

(b) Let \( A \) be the point \( (2, 1, -1) \) on \( L_1 \) and \( B \) be the point \( (3, 5, 2) \) on \( L_2 \).
Then \( \mathbf{AB} = \mathbf{b} - \mathbf{a} = (3-2)\mathbf{i} + (5-1)\mathbf{j} + (2 - (-1))\mathbf{k} = \mathbf{i} + 4\mathbf{j} + 3\mathbf{k} \).
The shortest distance between the lines is the projection of \( \mathbf{AB} \) onto \( \mathbf{n} \):
\( d = \frac{|\mathbf{AB} \cdot \mathbf{n}|}{|\mathbf{n}|} \)
\( \mathbf{AB} \cdot \mathbf{n} = (1)(-1) + (4)(5) + (3)(3) = -1 + 20 + 9 = 28 \).
\( |\mathbf{n}| = \sqrt{(-1)^2 + 5^2 + 3^2} = \sqrt{1 + 25 + 9} = \sqrt{35} \).
So \( d = \frac{28}{\sqrt{35}} = \frac{4\sqrt{35}}{5} \).

(c) The plane contains \( L_1 \) and is parallel to \( L_2 \), so its normal vector is \( \mathbf{n} = -\mathbf{i} + 5\mathbf{j} + 3\mathbf{k} \).
Since the plane contains the point \( A(2, 1, -1) \) on \( L_1 \), we have:
\( \mathbf{r} \cdot (-\mathbf{i} + 5\mathbf{j} + 3\mathbf{k}) = (2\mathbf{i} + \mathbf{j} - \mathbf{k}) \cdot (-\mathbf{i} + 5\mathbf{j} + 3\mathbf{k}) = 2(-1) + 1(5) - 1(3) = 0 \).
Thus, the equation of the plane is \( \mathbf{r} \cdot (-\mathbf{i} + 5\mathbf{j} + 3\mathbf{k}) = 0 \).

(a)
M1: Attempts cross product of the two direction vectors.
A1: Correctly calculates two components.
A1: Fully correct vector \(-\mathbf{i} + 5\mathbf{j} + 3\mathbf{k}\) (or any scalar multiple).

(b)
M1: Identifies a point on \( L_1 \) and a point on \( L_2 \) and finds the vector connecting them.
A1: Correct vector \(\mathbf{AB} = \mathbf{i} + 4\mathbf{j} + 3\mathbf{k}\).
M1: Applies the formula for the shortest distance using the dot product with their normal vector.
A1: Correct scalar product of 28 and correct magnitude \(\sqrt{35}\).
A1: Correct exact distance \(\frac{4\sqrt{35}}{5}\).

(c)
M1: Recognises that the normal vector to the plane is perpendicular to both lines, and uses a point on \( L_1 \) to find \( d \).
M1: Calculates \(\mathbf{r} \cdot \mathbf{n}\) using their normal and point.
A1: Correct equation \(\mathbf{r} \cdot (-\mathbf{i} + 5\mathbf{j} + 3\mathbf{k}) = 0\) (or equivalent multiple).

(a) Using integration by parts for \( I_n = \int_{0}^{1} x^n e^{-2x} \, dx \):
Let \( u = x^n \implies \frac{du}{dx} = n x^{n-1} \)
Let \( \frac{dv}{dx} = e^{-2x} \implies v = -\frac{1}{2} e^{-2x} \)
Then:
\( I_n = \left[ -\frac{1}{2} x^n e^{-2x} \right]_{0}^{1} - \int_{0}^{1} \left( -\frac{1}{2} e^{-2x} \right) \left( n x^{n-1} \right) \, dx \)
\( I_n = \left( -\frac{1}{2} e^{-2} - 0 \right) + \frac{n}{2} \int_{0}^{1} x^{n-1} e^{-2x} \, dx \)
\( I_n = -\frac{1}{2} e^{-2} + \frac{n}{2} I_{n-1} \)
Multiplying both sides by 2 gives:
\( 2 I_n = n I_{n-1} - e^{-2} \) (as required).

(b) First find \( I_0 \):
\( I_0 = \int_{0}^{1} e^{-2x} \, dx = \left[ -\frac{1}{2} e^{-2x} \right]_{0}^{1} = -\frac{1}{2} e^{-2} + \frac{1}{2} = \frac{1}{2} - \frac{1}{2} e^{-2} \).
Using the reduction formula with \( n = 1 \):
\( 2 I_1 = I_0 - e^{-2} = \frac{1}{2} - \frac{1}{2} e^{-2} - e^{-2} = \frac{1}{2} - \frac{3}{2} e^{-2} \implies I_1 = \frac{1}{4} - \frac{3}{4} e^{-2} \).
With \( n = 2 \):
\( 2 I_2 = 2 I_1 - e^{-2} = 2 \left( \frac{1}{4} - \frac{3}{4} e^{-2} \right) - e^{-2} = \frac{1}{2} - \frac{3}{2} e^{-2} - e^{-2} = \frac{1}{2} - \frac{5}{2} e^{-2} \implies I_2 = \frac{1}{4} - \frac{5}{4} e^{-2} \).
With \( n = 3 \):
\( 2 I_3 = 3 I_2 - e^{-2} = 3 \left( \frac{1}{4} - \frac{5}{4} e^{-2} \right) - e^{-2} = \frac{3}{4} - \frac{15}{4} e^{-2} - e^{-2} = \frac{3}{4} - \frac{19}{4} e^{-2} \implies I_3 = \frac{3}{8} - \frac{19}{8} e^{-2} \).
Thus \( a = \frac{3}{8} \) and \( b = -\frac{19}{8} \).

(a)
M1: Applies integration by parts with correct assignment of $u$ and $\frac{dv}{dx}$.
A1: Obtains the correct boundary term: \( \left[ -\frac{1}{2} x^n e^{-2x} \right]_{0}^{1} \).
A1: Obtains the correct integral term: \( \frac{n}{2} \int_{0}^{1} x^{n-1} e^{-2x} \, dx \).
M1: Substitutes the limits into the boundary term.
A1: Completes algebraic steps to show the required reduction relation: \( 2 I_n = n I_{n-1} - e^{-2} \).

(b)
M1: Integrates to find \( I_0 \) correctly.
A1: Correct \( I_0 = \frac{1}{2} - \frac{1}{2} e^{-2} \).
M1: Uses reduction formula once to find \( I_1 \).
M1: Uses reduction formula again to find \( I_2 \).
M1: Uses reduction formula third time to find \( I_3 \).
A1: Correct exact final answer \( I_3 = \frac{3}{8} - \frac{19}{8} e^{-2} \) (or equivalent rational numbers).

(a) The direction of the line of intersection, \( \mathbf{d} \), is perpendicular to the normal vectors of both planes:
\( \mathbf{n}_1 = 2\mathbf{i} - \mathbf{j} + \mathbf{k} \)
\( \mathbf{n}_2 = \mathbf{i} + \mathbf{j} + 2\mathbf{k} \)
\( \mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 1 \\ 1 & 1 & 2 \end{vmatrix} = \mathbf{i}(-2 - 1) - \mathbf{j}(4 - 1) + \mathbf{k}(2 - (-1)) = -3\mathbf{i} - 3\mathbf{j} + 3\mathbf{k} \).
We can scale this to \( \mathbf{d} = \mathbf{i} + \mathbf{j} - \mathbf{k} \).
Now find a point on both planes. Adding the equations of \( \Pi_1 \) and \( \Pi_2 \):
\( 3x + 3z = 12 \implies x + z = 4 \).
Let \( z = 1 \), then \( x = 3 \).
Substituting into \( \Pi_2 \): \( 3 + y + 2 = 9 \implies y = 4 \).
So a point on the line is \( (3, 4, 1) \).
Thus, a vector equation of the line \( L \) is:
\( \mathbf{r} = (3\mathbf{i} + 4\mathbf{j} + \mathbf{k}) + t(\mathbf{i} + \mathbf{j} - \mathbf{k}) \).

(b) Let \( F \) be the foot of the perpendicular from \( P(5, -1, 4) \) to \( L \).
The position vector of \( F \) is \( \mathbf{f} = (3+t)\mathbf{i} + (4+t)\mathbf{j} + (1-t)\mathbf{k} \).
The vector \( \mathbf{PF} \) is:
\( \mathbf{PF} = \mathbf{f} - \mathbf{p} = (t-2)\mathbf{i} + (t+5)\mathbf{j} + (-t-3)\mathbf{k} \).
Since \( \mathbf{PF} \) is perpendicular to \( \mathbf{d} \):
\( \mathbf{PF} \cdot \mathbf{d} = 0 \)
\( (t-2)(1) + (t+5)(1) + (-t-3)(-1) = 0 \)
\( t - 2 + t + 5 + t + 3 = 0 \implies 3t + 6 = 0 \implies t = -2 \).
Substituting \( t = -2 \) back into \( \mathbf{PF} \):
\( \mathbf{PF} = -4\mathbf{i} + 3\mathbf{j} - \mathbf{k} \).
The perpendicular distance is the magnitude of \( \mathbf{PF} \):
\( d_p = \sqrt{(-4)^2 + 3^2 + (-1)^2} = \sqrt{16 + 9 + 1} = \sqrt{26} \).

(a)
M1: Attempts cross product of the normal vectors of the two planes.
A1: Correct cross product vector \(-3\mathbf{i} - 3\mathbf{j} + 3\mathbf{k}\) (or multiple).
M1: Sets up a system of equations to find a common point by setting one variable to a constant (e.g., \( z = 1 \)).
A1: Solves correctly to find a point, e.g., \( (3, 4, 1) \).
M1: Formulates vector equation of the line using their point and direction vector.
A1: Correct vector equation: \( \mathbf{r} = (3\mathbf{i} + 4\mathbf{j} + \mathbf{k}) + t(\mathbf{i} + \mathbf{j} - \mathbf{k}) \) (or equivalent).

(b)
M1: Expresses the vector \( \mathbf{PF} \) from \( P \) to a general point on the line in terms of \( t \).
A1: Correct vector \( \mathbf{PF} = (t-2)\mathbf{i} + (t+5)\mathbf{j} + (-t-3)\mathbf{k} \).
M1: Sets the dot product \( \mathbf{PF} \cdot \mathbf{d} = 0 \) and solves for \( t \).
A1: Obtains \( t = -2 \) and evaluates \( \mathbf{PF} = -4\mathbf{i} + 3\mathbf{j} - \mathbf{k} \).
A1: Finds the correct distance \(\sqrt{26}\).

(a) Using the substitution \( x = 2\sinh \theta \):
\( \frac{dx}{d\theta} = 2\cosh \theta \implies dx = 2\cosh \theta \, d\theta \).
Also, \( \sqrt{x^2 + 4} = \sqrt{4\sinh^2 \theta + 4} = \sqrt{4(\sinh^2 \theta + 1)} = \sqrt{4\cosh^2 \theta} = 2\cosh \theta \), since \( \cosh \theta > 0 \).
Substituting these into the integral:
\( \int \sqrt{x^2 + 4} \, dx = \int (2\cosh \theta)(2\cosh \theta) \, d\theta = 4 \int \cosh^2 \theta \, d\theta \) (as required).

(b) Using the hyperbolic identity \( \cosh^2 \theta = \frac{\cosh 2\theta + 1}{2} \):
\( 4 \int \cosh^2 \theta \, d\theta = 4 \int \frac{\cosh 2\theta + 1}{2} \, d\theta = 2 \int (\cosh 2\theta + 1) \, d\theta = \sinh 2\theta + 2\theta \).
Now find the limits in terms of \( \theta \):
For \( x = 0 \): \( 0 = 2\sinh\theta \implies \theta = 0 \).
For \( x = 3 \): \( 3 = 2\sinh\theta \implies \sinh\theta = \frac{3}{2} \implies \theta = \text{arsinh}\left(\frac{3}{2}\right) = \ln\left(\frac{3 + \sqrt{13}}{2}\right) \).
Evaluate the integrated expression at the limits:
\( \left[ \sinh 2\theta + 2\theta \right]_{0}^{\ln\left(\frac{3+\sqrt{13}}{2}\right)} \).
Since \( \sinh 2\theta = 2\sinh\theta\cosh\theta \):
At \( x = 3 \), \( \sinh\theta = \frac{3}{2} \) and \( \cosh\theta = \sqrt{1 + \left(\frac{3}{2}\right)^2} = \sqrt{\frac{13}{4}} = \frac{\sqrt{13}}{2} \).
So \( \sinh 2\theta = 2 \left(\frac{3}{2}\right) \left(\frac{\sqrt{13}}{2}\right) = \frac{3\sqrt{13}}{2} \).
And \( 2\theta = 2\ln\left(\frac{3 + \sqrt{13}}{2}\right) \).
At the lower limit \( \theta = 0 \), both terms are \( 0 \).
Thus, the definite integral is:
\( \frac{3}{2}\sqrt{13} + 2\ln\left(\frac{3 + \sqrt{13}}{2}\right) \).
So \( p = \frac{3}{2} \) and \( q = 2 \).

(a)
M1: Applies substitution \( x = 2\sinh \theta \) and finds \( dx = 2\cosh \theta \, d\theta \).
M1: Uses \( \cosh^2 \theta - \sinh^2 \theta = 1 \) to simplify \( \sqrt{x^2+4} \) to \( 2\cosh\theta \).
A1: Successfully shows the given integral transforms to \( 4\int \cosh^2\theta \, d\theta \).

(b)
M1: Employs hyperbolic identity \( \cosh^2\theta = \frac{\cosh 2\theta + 1}{2} \).
A1: Integrates correctly to obtain \( \sinh 2\theta + 2\theta \) (or equivalent in terms of \( \theta \)).
M1: Uses identity \( \sinh 2\theta = 2\sinh\theta\cosh\theta \).
M1: Finds correct limit values for \( \theta \) or transforms expression back to \( x \).
A1: Shows that upper limit \( \theta = \ln\left(\frac{3+\sqrt{13}}{2}\right) \) or equivalent logarithmic form.
M1: Evaluates \( \sinh 2\theta \) at the upper limit to obtain \( \frac{3\sqrt{13}}{2} \).
M1: Substitutes both limits and subtracts.
A1: Correct exact final answer \( \frac{3}{2}\sqrt{13} + 2\ln\left(\frac{3 + \sqrt{13}}{2}\right) \) (or equivalent values of \( p \) and \( q \)).