2023 Cambridge IAS-Level Mathematics - Further (9231) Practice Paper with Answers

Let the proposition be denoted by \(P(n)\): \(\sum_{r=1}^n (3r-2)(3r+1) = n(3n^2 + 3n - 2)\).

**Base case:** For \(n=1\):
LHS: \((3(1)-2)(3(1)+1) = 1 \times 4 = 4\).
RHS: \(1(3(1)^2 + 3(1) - 2) = 1(3+3-2) = 4\).
Since LHS = RHS, \(P(1)\) is true.

**Inductive step:** Assume that \(P(k)\) is true for some positive integer \(k\), so:
\[\sum_{r=1}^k (3r-2)(3r+1) = k(3k^2 + 3k - 2).\]
We need to show that \(P(k+1)\) is true, i.e.,
\[\sum_{r=1}^{k+1} (3r-2)(3r+1) = (k+1)\left(3(k+1)^2 + 3(k+1) - 2\right).\]
Simplifying the target RHS:
\[(k+1)(3k^2 + 6k + 3 + 3k + 3 - 2) = (k+1)(3k^2 + 9k + 4) = 3k^3 + 12k^2 + 13k + 4.\]
Now, adding the \((k+1)\)-th term to the inductive hypothesis:
\[\sum_{r=1}^{k+1} (3r-2)(3r+1) = k(3k^2 + 3k - 2) + \left(3(k+1)-2\right)\left(3(k+1)+1\right)\]
\[= 3k^3 + 3k^2 - 2k + (3k+1)(3k+4)\]
\[= 3k^3 + 3k^2 - 2k + 9k^2 + 15k + 4\]
\[= 3k^3 + 12k^2 + 13k + 4.\]
This is identical to the target RHS. Thus, if \(P(k)\) is true, then \(P(k+1)\) is also true.

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

Base case verified for n = 1: [1 mark]
State the inductive hypothesis clearly: [1 mark]
Add the (k+1)-th term to the sum of k terms: [3 marks]
Perform algebraic expansion and simplification correctly: [4 marks]
Conclude the proof with a complete mathematical statement: [1.7 marks]

(a) The determinant of \(\mathbf{A}\) represents the area scale factor of the transformation \(T\).
\[\det(\mathbf{A}) = (1)(4) - (k)(2) = 4 - 2k.\]
Since the area is multiplied by a factor of 6, we have:
\[|4 - 2k| = 6.\]
This yields two cases:
Case 1: \(4 - 2k = 6 \implies 2k = -2 \implies k = -1\).
Case 2: \(4 - 2k = -6 \implies 2k = 10 \implies k = 5\).
So the two possible values of \(k\) are \(-1\) and \(5\).

(b) For the negative value, \(k = -1\), the matrix is \(\mathbf{A} = \begin{pmatrix} 1 & -1 \\ 2 & 4 \end{pmatrix}\).
Let \(y = mx\) be an invariant line through the origin.
A point on the line \((x, mx)\) is mapped to \((X, Y)\), where:
\[\begin{pmatrix} X \\ Y \end{pmatrix} = \begin{pmatrix} 1 & -1 \\ 2 & 4 \end{pmatrix} \begin{pmatrix} x \\ mx \end{pmatrix} = \begin{pmatrix} x - mx \\ 2x + 4mx \end{pmatrix}.\]
Since the line is invariant, \((X, Y)\) must also lie on the line, so \(Y = mX\):
\[2x + 4mx = m(x - mx).\]
Since this holds for any point on the line (excluding \(x = 0\)), we divide by \(x\):
\[2 + 4m = m(1 - m) \implies 2 + 4m = m - m^2 \implies m^2 + 3m + 2 = 0.\]
Factoring the quadratic equation gives:
\[(m + 1)(m + 2) = 0 \implies m = -1 \text{ or } m = -2.\]
Thus, the invariant lines through the origin are \(y = -x\) and \(y = -2x\).

(a) Identify that the determinant corresponds to the area scale factor: [1 mark]
Formulate the absolute value equation |4 - 2k| = 6: [1 mark]
Solve to get k = -1: [1.5 marks]
Solve to get k = 5: [1.5 marks]
(b) Set up the matrix multiplication for the invariant line through the origin: [1.5 marks]
Establish the relation Y = mX and eliminate x to form the quadratic equation: [1.5 marks]
Form the quadratic equation m^2 + 3m + 2 = 0: [1.2 marks]
Solve to get m = -1 and m = -2, giving the final equations: [1.5 marks]

(a) From the cubic equation \(x^3 + 3x - 1 = 0\), the relations between roots and coefficients are:
\[\sum \alpha = 0,\]
\[\sum \alpha\beta = 3,\]
\[\alpha\beta\gamma = 1.\]
Using the algebraic identity:
\[\alpha^2 + \beta^2 + \gamma^2 = \left(\sum \alpha\right)^2 - 2\sum \alpha\beta = 0^2 - 2(3) = -6.\]

(b) Let \(y = x^3\). From the original cubic equation, we have:
\[x^3 - 1 = -3x.\]
Cubing both sides:
\[(x^3 - 1)^3 = (-3x)^3 \implies (x^3 - 1)^3 = -27x^3.\]
Substituting \(y = x^3\):
\[(y - 1)^3 = -27y \implies y^3 - 3y^2 + 3y - 1 = -27y.\]
Rearranging gives:
\[y^3 - 3y^2 + 30y - 1 = 0.\]

(c) Since \(\alpha, \beta, \gamma\) are roots of \(x^3 + 3x - 1 = 0\), we can multiply the equation by \(x\) to get:
\[x^4 + 3x^2 - x = 0.\]
Summing this over all three roots:
\[\sum \alpha^4 + 3\sum \alpha^2 - \sum \alpha = 0.\]
Substituting \(\sum \alpha^2 = -6\) and \(\sum \alpha = 0\):
\[\sum \alpha^4 + 3(-6) - 0 = 0 \implies \sum \alpha^4 = 18.\]

(a) State correct sum of roots and sum of products: [1 mark]
Use correct identity to find -6: [1.5 marks]
(b) Rearrange original equation to isolate a term for cubing: [1.5 marks]
Cube both sides and introduce y: [1.5 marks]
Expand and simplify to obtain the correct equation: [2 marks]
(c) Multiply the cubic equation by x and write in sum of powers form: [1.2 marks]
Substitute known values of sum of squares and sum of roots: [1 mark]
Obtain 18: [1 mark]

(a) Combining the terms on the RHS:
\[\frac{2}{r^2 - r + 1} - \frac{2}{r^2 + r + 1} = \frac{2(r^2 + r + 1) - 2(r^2 - r + 1)}{(r^2 - r + 1)(r^2 + r + 1)}\]
\[= \frac{2r^2 + 2r + 2 - 2r^2 + 2r - 2}{((r^2 + 1) - r)((r^2 + 1) + r)} = \frac{4r}{(r^2 + 1)^2 - r^2}\]
\[= \frac{4r}{r^4 + 2r^2 + 1 - r^2} = \frac{4r}{r^4 + r^2 + 1}.\]
This completes the proof.

(b) Let \(f(r) = \frac{2}{r^2 - r + 1}\). Then:
\[f(r+1) = \frac{2}{(r+1)^2 - (r+1) + 1} = \frac{2}{r^2 + 2r + 1 - r - 1 + 1} = \frac{2}{r^2 + r + 1}.\]
Therefore, the summation becomes:
\[\sum_{r=1}^n \frac{4r}{r^4 + r^2 + 1} = \sum_{r=1}^n \left( f(r) - f(r+1) \right)\]
\[= \left( f(1) - f(2) \right) + \left( f(2) - f(3) \right) + \dots + \left( f(n) - f(n+1) \right)\]
\[= f(1) - f(n+1).\]
Evaluating \(f(1)\) and \(f(n+1)\):
\[f(1) = \frac{2}{1^2 - 1 + 1} = 2,\]
\[f(n+1) = \frac{2}{(n+1)^2 - (n+1) + 1} = \frac{2}{n^2 + n + 1}.\]
Thus, the sum is:
\[2 - \frac{2}{n^2 + n + 1} = \frac{2(n^2 + n + 1) - 2}{n^2 + n + 1} = \frac{2n(n+1)}{n^2 + n + 1}.\]

(c) The sum to infinity is the limit as \(n \to \infty\):
\[\lim_{n \to \infty} \frac{2n^2 + 2n}{n^2 + n + 1} = 2.\]

(a) Express RHS over a common denominator: [1 mark]
Expand and simplify numerator to get 4r: [1 mark]
Expand denominator to show it is equal to r^4 + r^2 + 1: [1.5 marks]
(b) Write the series as a telescoping sum: [2 marks]
Show the cancellation of intermediate terms: [1 mark]
Substitute expressions for the first and last term: [1.2 marks]
Simplify to a single algebraic fraction: [1.5 marks]
(c) Correctly take the limit as n approaches infinity to obtain 2: [1.5 marks]

(a) The area \(A\) of the region is:
\[A = \frac{1}{2} \int_{0}^{\pi} r^2 \, \mathrm{d}\theta = \frac{1}{2} \int_{0}^{\pi} 4a^2 (1 + \cos\theta)^2 \, \mathrm{d}\theta = 2a^2 \int_{0}^{\pi} (1 + 2\cos\theta + \cos^2\theta) \, \mathrm{d}\theta.\]
Using the identity \(\cos^2\theta = \frac{1}{2}(1 + \cos 2\theta)\):
\[A = 2a^2 \int_{0}^{\pi} \left(1 + 2\cos\theta + \frac{1}{2} + \frac{1}{2}\cos 2\theta\right) \, \mathrm{d}\theta\]
\[= 2a^2 \int_{0}^{\pi} \left(\frac{3}{2} + 2\cos\theta + \frac{1}{2}\cos 2\theta\right) \, \mathrm{d}\theta\]
\[= 2a^2 \left[ \frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta \right]_{0}^{\pi}\]
\[= 2a^2 \left[\left(\frac{3\pi}{2} + 0 + 0\right) - (0 + 0 + 0)\right] = 3\pi a^2.\]

(b) For the tangent to be parallel to the initial line, we require \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = 0\), where \(y = r\sin\theta\).
\[y = 2a(1 + \cos\theta)\sin\theta = 2a(\sin\theta + \sin\theta\cos\theta) = 2a\sin\theta + a\sin 2\theta.\]
Now, differentiate with respect to \(\theta\):
\[\frac{\mathrm{d}y}{\mathrm{d}\theta} = 2a\cos\theta + 2a\cos 2\theta.\]
Setting \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = 0\) and dividing by \(2a\):
\[\cos\theta + \cos 2\theta = 0 \implies \cos\theta + 2\cos^2\theta - 1 = 0.\]
This factorizes to:
\[(2\cos\theta - 1)(\cos\theta + 1) = 0.\]
Since we exclude the pole (where \(r = 0 \implies \theta = \pi\), which gives \(\cos\theta = -1\)), we must have:
\[2\cos\theta - 1 = 0 \implies \cos\theta = \frac{1}{2}.\]
For \(0 \le \theta \le \pi\), this gives \(\theta = \frac{\pi}{3}\).

Substitute \(\theta = \frac{\pi}{3}\) back into the polar equation:
\[r = 2a\left(1 + \cos\frac{\pi}{3}\right) = 2a\left(1 + \frac{1}{2}\right) = 3a.\]
Thus, the polar coordinates of the point are \((3a, \frac{\pi}{3})\).

(a) State the correct area integral formula: [1 mark]
Substitute for r and expand correctly: [1.5 marks]
Use double-angle identity for cos^2\theta correctly: [1.5 marks]
Integrate and apply the limits [0, \pi]: [1 mark]
Obtain the area of 3\pi a^2: [1 mark]
(b) Express y correctly in terms of \theta: [1 mark]
Differentiate and set dy/d\theta = 0: [1.5 marks]
Solve the trigonometric quadratic equation: [1.2 marks]
Obtain correct coordinates (3a, \pi/3): [1 mark]

(a) A vertical asymptote occurs where the denominator is zero:
\[x - 2 = 0 \implies x = 2.\]
By division:
\[\frac{x^2 - x + 2}{x - 2} = x + 1 + \frac{4}{x - 2}.\]
As \(x \to \pm\infty\), the fraction \(\frac{4}{x-2} \to 0\), so the slant asymptote is:
\[y = x + 1.\]

(b) To find the stationary points, we differentiate \(y = x + 1 + 4(x-2)^{-1}\):
\[\frac{\mathrm{d}y}{\mathrm{d}x} = 1 - \frac{4}{(x-2)^2}.\]
Setting \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\[(x-2)^2 = 4 \implies x - 2 = \pm 2.\]
This gives \(x = 0\) or \(x = 4\).
- At \(x = 0\), \(y = \frac{0 - 0 + 2}{0 - 2} = -1\), giving the point \((0, -1)\).
- At \(x = 4\), \(y = \frac{16 - 4 + 2}{4 - 2} = 7\), giving the point \((4, 7)\).
To find their nature, calculate the second derivative:
\[\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{8}{(x-2)^3}.\]
- For \(x = 0\), \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{8}{(-2)^3} = -1 < 0\), so \((0, -1)\) is a local maximum.
- For \(x = 4\), \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{8}{2^3} = 1 > 0\), so \((4, 7)\) is a local minimum.

(c) Setting \(y = 0\) to find intersections with the \(x\)-axis:
\[\frac{x^2 - x + 2}{x - 2} = 0 \implies x^2 - x + 2 = 0.\]
The discriminant of this quadratic is:
\[\Delta = (-1)^2 - 4(1)(2) = 1 - 8 = -7.\]
Since the discriminant is negative, there are no real roots. Therefore, the curve \(C\) does not cross the \(x\)-axis.

(a) Correctly identify vertical asymptote x = 2: [1 mark]
Perform algebraic division: [1.5 marks]
Obtain slant asymptote y = x + 1: [1 mark]
(b) Differentiate to find dy/dx: [1.5 marks]
Set dy/dx = 0 and solve to find x = 0 and x = 4: [1.2 marks]
Find the corresponding y-values for both points: [1.5 marks]
Use second derivative or alternative valid test to determine the nature: [1.5 marks]
(c) Set y = 0 and compute the discriminant of the numerator: [1 mark]
Conclude correctly that there are no real roots because the discriminant is negative: [0.5 marks]

(a) The direction vector of \(l_1\) is \(\mathbf{d}_1 = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}\) and the direction vector of \(l_2\) is \(\mathbf{d}_2 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\).

A common perpendicular vector \(\mathbf{n}\) to both lines is:
\[\mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 3 \\ 1 & 1 & 1 \end{vmatrix}\]
\[= \mathbf{i}(-1 - 3) - \mathbf{j}(2 - 3) + \mathbf{k}(2 - (-1)) = \begin{pmatrix} -4 \\ 1 \\ 3 \end{pmatrix}.\]
Let \(A(1, 2, -1)\) be a point on \(l_1\) and \(B(2, 0, 3)\) be a point on \(l_2\).
The vector \(\mathbf{AB}\) is:
\[\mathbf{AB} = \begin{pmatrix} 2 - 1 \\ 0 - 2 \\ 3 - (-1) \end{pmatrix} = \begin{pmatrix} 1 \\ -2 \\ 4 \end{pmatrix}.\]
The shortest distance \(d\) is the projection of \(\mathbf{AB}\) onto the common normal \(\mathbf{n}\):
\[d = \frac{|\mathbf{AB} \cdot \mathbf{n}|}{|\mathbf{n}|}\]
Calculate \(\mathbf{AB} \cdot \mathbf{n}\):
\[\mathbf{AB} \cdot \mathbf{n} = (1)(-4) + (-2)(1) + (4)(3) = -4 - 2 + 12 = 6.\]
Calculate the magnitude of \(\mathbf{n}\):
\[|\mathbf{n}| = \sqrt{(-4)^2 + 1^2 + 3^2} = \sqrt{26}.\]
Thus, the shortest distance is:
\[d = \frac{6}{\sqrt{26}} = \frac{3\sqrt{26}}{13} \approx 1.18.\]

(b) The plane \(\Pi\) is parallel to both direction vectors \(\mathbf{d}_1\) and \(\mathbf{d}_2\), so its normal vector is parallel to \(\mathbf{n} = \begin{pmatrix} -4 \\ 1 \\ 3 \end{pmatrix}\). Let's use \(\begin{pmatrix} 4 \\ -1 \\ -3 \end{pmatrix}\).

The equation of \(\Pi\) is of the form:
\[4x - y - 3z = D.\]
Since \(\Pi\) contains \(l_1\), it must contain the point \((1, 2, -1)\). Substituting this point:
\[D = 4(1) - 2 - 3(-1) = 4 - 2 + 3 = 5.\]
Thus, the Cartesian equation of the plane \(\Pi\) is:
\[4x - y - 3z = 5.\]

(a) Calculate the cross product of the direction vectors: [2 marks]
Find a vector between points on the two lines: [1.5 marks]
Apply the projection formula for the shortest distance: [1.5 marks]
Compute dot product and magnitude correctly: [1.5 marks]
Obtain 6/\sqrt{26} (or 1.18): [1 mark]
(b) Identify the normal vector of the plane: [1 mark]
Substitute point (1, 2, -1) to find the constant: [1.2 marks]
State the correct Cartesian equation: [1 mark]

(i) Using the definitions of hyperbolic functions, we substitute \( \cosh x = \frac{e^x + e^{-x}}{2} \) and \( \sinh x = \frac{e^x - e^{-x}}{2} \) into the equation to obtain: \( 4\left(\frac{e^x + e^{-x}}{2}\right) - 2\left(\frac{e^x - e^{-x}}{2}\right) = 3.5 \). Simplifying terms gives \( 2(e^x + e^{-x}) - (e^x - e^{-x}) = 3.5 \), which reduces to \( e^x + 3e^{-x} = 3.5 \). Multiplying the entire equation by \( 2e^x \) yields \( 2e^{2x} + 6 = 7e^x \), which can be rearranged to the quadratic form \( 2e^{2x} - 7e^x + 6 = 0 \). Solving this quadratic in terms of \( e^x \) by factoring gives \( (2e^x - 3)(e^x - 2) = 0 \). This yields \( e^x = 1.5 \) or \( e^x = 2 \). Taking natural logarithms on both sides, we find the exact values of \( x \) to be \( x = \ln 1.5 \) or \( x = \ln 2 \). (ii) To evaluate the integral \( \int_0^{\ln 3} \frac{1}{4\cosh x - 2\sinh x} \, \mathrm{d}x \), we rewrite the denominator using the exponential form derived in part (i): \( 4\cosh x - 2\sinh x = e^x + 3e^{-x} \). The integral becomes \( \int_0^{\ln 3} \frac{1}{e^x + 3e^{-x}} \, \mathrm{d}x = \int_0^{\ln 3} \frac{e^x}{e^{2x} + 3} \, \mathrm{d}x \). We use the substitution \( u = e^x \), so \( \mathrm{d}u = e^x \mathrm{d}x \). The limits of integration change from \( x=0 \) to \( u=1 \), and from \( x=\ln 3 \) to \( u=3 \). This gives \( \int_1^3 \frac{1}{u^2 + 3} \, \mathrm{d}u = \left[ \frac{1}{\sqrt{3}}\arctan\left(\frac{u}{\sqrt{3}}\right) \right]_1^3 \). Evaluating at the limits yields \( \frac{1}{\sqrt{3}}\left(\arctan(\sqrt{3}) - \arctan\left(\frac{1}{\sqrt{3}}\right)\right) = \frac{1}{\sqrt{3}}\left(\frac{\pi}{3} - \frac{\pi}{6}\right) = \frac{\pi}{6\sqrt{3}} = \frac{\pi\sqrt{3}}{18} \).

For part (i): M1 for substituting exponential definitions of hyperbolic functions. A1 for correct simplification to a quadratic in terms of e^x. A1 for solving the quadratic and finding the correct exact values of x. For part (ii): M1 for rewriting the integrand in terms of exponentials. M1 for correct substitution u = e^x including transforming the integration limits. M1 for integrating to find the arctan form. A1 for evaluating the limits and finding the exact value in simplified form.

(i) Differentiating \( y = \arctan(2x) \) with respect to \( x \) gives \( \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2}{1 + 4x^2} \). Multiplying by the denominator yields \( (1 + 4x^2)\frac{\mathrm{d}y}{\mathrm{d}x} = 2 \). Differentiating this product with respect to \( x \) gives \( (1 + 4x^2)\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + 8x\frac{\mathrm{d}y}{\mathrm{d}x} = 0 \). (ii) We apply Leibniz's theorem to the equation. For the first term, let \( u = y^{(2)} \) and \( v = 1 + 4x^2 \), giving \( ( (1+4x^2)y^{(2)} )^{(n)} = (1+4x^2)y^{(n+2)} + n(8x)y^{(n+1)} + \frac{n(n-1)}{2}(8)y^{(n)} \). For the second term, let \( u = y^{(1)} \) and \( v = 8x \), giving \( ( 8xy^{(1)} )^{(n)} = 8xy^{(n+1)} + n(8)y^{(n)} \). Summing these two terms and setting them to 0, we have \( (1+4x^2)y^{(n+2)} + [8nx + 8x]y^{(n+1)} + [4n(n-1) + 8n]y^{(n)} = 0 \). Simplifying the coefficients yields \( (1+4x^2)y^{(n+2)} + 8x(n+1)y^{(n+1)} + 4n(n+1)y^{(n)} = 0 \). Evaluating this at \( x = 0 \) eliminates the middle term and leaves \( y^{(n+2)}(0) + 4n(n+1)y^{(n)}(0) = 0 \). (iii) Evaluating the derivatives at \( x = 0 \): \( y(0) = 0 \), and from \( y' = 2(1+4x^2)^{-1} \) we have \( y'(0) = 2 \). From the equation in part (i), \( y''(0) = 0 \). Using the recurrence relation: For \( n=1 \), \( y^{(3)}(0) = -4(1)(2)y^{(1)}(0) = -8(2) = -16 \). For \( n=2 \), \( y^{(4)}(0) = -4(2)(3)y^{(2)}(0) = 0 \). For \( n=3 \), \( y^{(5)}(0) = -4(3)(4)y^{(3)}(0) = -48(-16) = 768 \). Substituting these values into Maclaurin's series expansion: \( y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y^{(3)}(0)}{3!}x^3 + \frac{y^{(4)}(0)}{4!}x^4 + \frac{y^{(5)}(0)}{5!}x^5 + \dots = 2x - \frac{16}{6}x^3 + \frac{768}{120}x^5 = 2x - \frac{8}{3}x^3 + \frac{32}{5}x^5 \).

For part (i): M1 for differentiating y once. M1 for differentiating a second time to get the required differential equation. A1 for correct verification. For part (ii): M1 for applying Leibniz's theorem to (1+4x^2)y''. M1 for applying Leibniz's theorem to 8xy'. A1 for combining and simplifying the algebraic coefficients. A1 for substituting x = 0 to get the correct recurrence relation. For part (iii): M1 for finding the third and fifth derivatives at x = 0 using the recurrence relation. M1 for substituting into the Maclaurin series formula. A1 for obtaining the correct series expansion.

(i) We use integration by parts for \( I_n = \int_0^1 x^n e^{-2x} \, \mathrm{d}x \). Let \( u = x^n \implies \mathrm{d}u = n x^{n-1} \mathrm{d}x \), and let \( \mathrm{d}v = e^{-2x} \mathrm{d}x \implies v = -\frac{1}{2} e^{-2x} \). Then, \( I_n = \left[ -\frac{1}{2} x^n e^{-2x} \right]_0^1 + \frac{n}{2} \int_0^1 x^{n-1} e^{-2x} \, \mathrm{d}x \). Evaluating the boundary term gives \( -\frac{1}{2}(1)^n e^{-2} - 0 = -\frac{1}{2}e^{-2} \). Thus, we have \( I_n = -\frac{1}{2}e^{-2} + \frac{n}{2} I_{n-1} \). Multiplying both sides by 2 yields the required reduction formula: \( 2I_n = n I_{n-1} - e^{-2} \). (ii) We first compute \( I_0 = \int_0^1 e^{-2x} \, \mathrm{d}x = \left[ -\frac{1}{2} e^{-2x} \right]_0^1 = \frac{1}{2}(1 - e^{-2}) \). Using the reduction formula for successive values of \( n \): For \( n=1 \): \( 2I_1 = I_0 - e^{-2} = \frac{1}{2}(1 - e^{-2}) - e^{-2} = \frac{1}{2} - \frac{3}{2}e^{-2} \implies I_1 = \frac{1}{4} - \frac{3}{4}e^{-2} \). For \( n=2 \): \( 2I_2 = 2I_1 - e^{-2} = 2\left(\frac{1}{4} - \frac{3}{4}e^{-2}\right) - e^{-2} = \frac{1}{2} - \frac{5}{2}e^{-2} \implies I_2 = \frac{1}{4} - \frac{5}{4}e^{-2} \). For \( n=3 \): \( 2I_3 = 3I_2 - e^{-2} = 3\left(\frac{1}{4} - \frac{5}{4}e^{-2}\right) - e^{-2} = \frac{3}{4} - \frac{15}{4}e^{-2} - e^{-2} = \frac{3}{4} - \frac{19}{4}e^{-2} \implies I_3 = \frac{3}{8} - \frac{19}{8}e^{-2} \).

For part (i): M1 for setting up integration by parts with correct choice of u and v. A1 for correct evaluation of the boundary term. M1 for identifying the remaining integral as I_{n-1}. A1 for completing the algebraic steps to show the given reduction formula. For part (ii): B1 for correct initial integration to find I_0. M1 for calculating I_1 using the formula. M1 for calculating I_2 using the formula. M1 for calculating I_3. A1 for the final correct exact value of I_3 in terms of e.

(i) Let \( z = \cos \theta + i \sin \theta \). By de Moivre's theorem, \( z^n - z^{-n} = 2i \sin(n\theta) \) and \( z - z^{-1} = 2i \sin \theta \). We expand \( (2i \sin \theta)^5 \) using the binomial theorem: \( (2i \sin \theta)^5 = (z - z^{-1})^5 = z^5 - 5z^3 + 10z - 10z^{-1} + 5z^{-3} - z^{-5} \). Grouping symmetric terms: \( 32i \sin^5 \theta = (z^5 - z^{-5}) - 5(z^3 - z^{-3}) + 10(z - z^{-1}) \). Substituting back the sine representations: \( 32i \sin^5 \theta = 2i \sin(5\theta) - 10i \sin(3\theta) + 20i \sin \theta \). Dividing both sides by \( 32i \) yields: \( \sin^5 \theta = \frac{1}{16}(\sin 5\theta - 5\sin 3\theta + 10\sin \theta) \). (ii) Using the identity from part (i) to integrate: \( \int_0^{\frac{\pi}{2}} \sin^5 \theta \, \mathrm{d}\theta = \frac{1}{16} \int_0^{\frac{\pi}{2}} (\sin 5\theta - 5\sin 3\theta + 10\sin \theta) \, \mathrm{d}\theta = \frac{1}{16} \left[ -\frac{1}{5}\cos 5\theta + \frac{5}{3}\cos 3\theta - 10\cos \theta \right]_0^{\frac{\pi}{2}} \). Since \( \cos\left(\frac{5\pi}{2}\right) = 0 \), \( \cos\left(\frac{3\pi}{2}\right) = 0 \), and \( \cos\left(\frac{\pi}{2}\right) = 0 \), the upper limit evaluates to \( 0 \). Evaluating at the lower limit \( \theta = 0 \): \( \frac{1}{16} \left( 0 - \left( -\frac{1}{5} + \frac{5}{3} - 10 \right) \right) = -\frac{1}{16} \left( -\frac{3}{15} + \frac{25}{15} - \frac{150}{15} \right) = -\frac{1}{16} \left( -\frac{128}{15} \right) = \frac{8}{15} \).

For part (i): M1 for stating the expressions for z^n - z^{-n} and z - z^{-1}. M1 for writing down the binomial expansion of (z - z^{-1})^5. M1 for grouping terms into pairs of the form (z^k - z^{-k}). M1 for expressing these terms back in terms of sine. A1 for obtaining the correct identity. For part (ii): M1 for setting up the integration of the trigonometric terms. M1 for integrating each sine term correctly. M1 for substituting the limits of integration. A1 for evaluating the result to get the exact value 8/15.

(i) The characteristic equation of \( \mathbf{A} \) is \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). This is \( \det\begin{pmatrix} 2-\lambda & 1 & 0 \\ 0 & 3-\lambda & 0 \\ 0 & 2 & 1-\lambda \end{pmatrix} = 0 \). Expanding along the first column gives \( (2-\lambda)(3-\lambda)(1-\lambda) = 0 \). The eigenvalues are therefore \( \lambda_1 = 1 \), \( \lambda_2 = 2 \), and \( \lambda_3 = 3 \). (ii) To find the eigenvectors: For \( \lambda = 1 \): Solve \( (\mathbf{A} - \mathbf{I})\mathbf{v} = \mathbf{0} \implies \begin{pmatrix} 1 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 2 & 0 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \). This gives \( x+y = 0 \) and \( y=0 \), which implies \( x=0, y=0 \), and \( z \) is free. An eigenvector is \( \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \). For \( \lambda = 2 \): Solve \( (\mathbf{A} - 2\mathbf{I})\mathbf{v} = \mathbf{0} \implies \begin{pmatrix} 0 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 2 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \). This gives \( y = 0 \) and \( 2y - z = 0 \implies z = 0 \), and \( x \) is free. An eigenvector is \( \mathbf{v}_2 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \). For \( \lambda = 3 \): Solve \( (\mathbf{A} - 3\mathbf{I})\mathbf{v} = \mathbf{0} \implies \begin{pmatrix} -1 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 2 & -2 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \). This gives \( -x+y = 0 \) and \( 2y-2z=0 \implies x=y=z \). An eigenvector is \( \mathbf{v}_3 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \). (iii) We define \( \mathbf{P} = \begin{pmatrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 1 & 0 & 1 \end{pmatrix} \) and \( \mathbf{D} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix} \). To find \( \mathbf{P}^{-1} \), we calculate the determinant of \( \mathbf{P} \) which is \( 1 \), and find the inverse matrix: \( \mathbf{P}^{-1} = \begin{pmatrix} 0 & -1 & 1 \\ 1 & -1 & 0 \\ 0 & 1 & 0 \end{pmatrix} \). Then \( \mathbf{A}^n = \mathbf{P}\mathbf{D}^n\mathbf{P}^{-1} \). Carrying out the multiplication: \( \mathbf{P}\mathbf{D}^n = \begin{pmatrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 1 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1^n & 0 & 0 \\ 0 & 2^n & 0 \\ 0 & 0 & 3^n \end{pmatrix} = \begin{pmatrix} 0 & 2^n & 3^n \\ 0 & 0 & 3^n \\ 1 & 0 & 3^n \end{pmatrix} \). Next, \( \mathbf{A}^n = \begin{pmatrix} 0 & 2^n & 3^n \\ 0 & 0 & 3^n \\ 1 & 0 & 3^n \end{pmatrix} \begin{pmatrix} 0 & -1 & 1 \\ 1 & -1 & 0 \\ 0 & 1 & 0 \end{pmatrix} = \begin{pmatrix} 2^n & 3^n - 2^n & 0 \\ 0 & 3^n & 0 \\ 0 & 3^n - 1 & 1 \end{pmatrix} \).

For part (i): M1 for setting up det(A - lambda I) = 0. A1 for finding the correct eigenvalues 1, 2, and 3. For part (ii): M1 for solving (A - lambda I)v = 0 for at least one eigenvalue. A2 for finding all three eigenvectors correctly (1 mark if only one or two are correct). For part (iii): M1 for constructing P and D and attempting to find the inverse matrix P^{-1}. A1 for the correct inverse matrix P^{-1}. M1 for the matrix multiplication process of P * D^n * P^{-1}. A1 for the correct simplified expression of A^n.

First, we find the complementary function (CF) by solving the auxiliary equation: \( m^2 - 4m + 4 = 0 \implies (m-2)^2 = 0 \implies m = 2 \) (a repeated real root). Thus, the CF is \( y_c = (Ax + B)e^{2x} \). Next, we find the particular integral (PI). Since \( e^{2x} \) and \( x e^{2x} \) are present in the CF, the PI for the \( 8e^{2x} \) term must be of the form \( C x^2 e^{2x} \). The PI for the polynomial term \( 4x \) is of the form \( Dx + E \). Thus, we assume the PI is \( y_p = C x^2 e^{2x} + Dx + E \). Differentiating \( y_p \): \( y_p' = 2C x e^{2x} + 2C x^2 e^{2x} + D \) and \( y_p'' = 2C e^{2x} + 8C x e^{2x} + 4C x^2 e^{2x} \). Substituting these into the differential equation: \( (2C e^{2x} + 8C x e^{2x} + 4C x^2 e^{2x}) - 4(2C x e^{2x} + 2C x^2 e^{2x} + D) + 4(C x^2 e^{2x} + Dx + E) = 8e^{2x} + 4x \). Simplifying terms yields: \( 2C e^{2x} + 4Dx + 4E - 4D = 8e^{2x} + 4x \). Comparing coefficients: For the \( e^{2x} \) term: \( 2C = 8 \implies C = 4 \). For the \( x \) term: \( 4D = 4 \implies D = 1 \). For the constant term: \( 4E - 4D = 0 \implies E = 1 \). Thus, the PI is \( y_p = 4x^2 e^{2x} + x + 1 \). The general solution is \( y = (Ax + B)e^{2x} + 4x^2 e^{2x} + x + 1 \). We apply the boundary conditions at \( x = 0 \): \( y = 2 \implies 2 = B + 1 \implies B = 1 \). Differentiating the general solution: \( y' = A e^{2x} + 2(Ax + B)e^{2x} + 8x e^{2x} + 8x^2 e^{2x} + 1 \). At \( x = 0 \), \( y' = 4 \implies A + 2B + 1 = 4 \). Substituting \( B = 1 \) yields \( A + 3 = 4 \implies A = 1 \). The particular solution is therefore \( y = (x + 1)e^{2x} + 4x^2 e^{2x} + x + 1 \).

M1 for finding the complementary function (correct auxiliary equation and form). A1 for correct CF: (Ax+B)e^{2x}. M1 for identifying the correct form of the particular integral (Cx^2 e^{2x} + Dx + E). M1 for differentiating the PI and substituting it into the differential equation. A1 for obtaining the correct coefficients C=4, D=1, E=1. A1 for the correct general solution. M1 for applying the boundary condition y(0)=2 to find B. M1 for differentiating the general solution and applying y'(0)=4 to find A. A1 for the correct final particular solution.

(i) Differentiating \( y \) with respect to \( x \) gives \( \frac{\mathrm{d}y}{\mathrm{d}x} = x^2 - \frac{1}{4x^2} \). Squaring this derivative: \( \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = x^4 - 2(x^2)\left(\frac{1}{4x^2}\right) + \frac{1}{16x^4} = x^4 - \frac{1}{2} + \frac{1}{16x^4} \). Adding 1 yields: \( 1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = x^4 + \frac{1}{2} + \frac{1}{16x^4} = \left(x^2 + \frac{1}{4x^2}\right)^2 \). (ii) The arc length \( s \) is given by \( s = \int_1^3 \sqrt{1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2} \, \mathrm{d}x = \int_1^3 \left(x^2 + \frac{1}{4x^2}\right) \, \mathrm{d}x \). Integrating each term: \( \int \left(x^2 + \frac{1}{4}x^{-2}\right) \, \mathrm{d}x = \left[ \frac{1}{3}x^3 - \frac{1}{4x} \right]_1^3 \). Evaluating at the limits: \( \left(9 - \frac{1}{12}\right) - \left(\frac{1}{3} - \frac{1}{4}\right) = \frac{107}{12} - \frac{1}{12} = \frac{106}{12} = \frac{53}{6} \). (iii) The surface area of revolution \( S \) is given by \( S = 2\pi \int_1^3 y \sqrt{1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2} \, \mathrm{d}x = 2\pi \int_1^3 \left(\frac{1}{3}x^3 + \frac{1}{4x}\right)\left(x^2 + \frac{1}{4x^2}\right) \, \mathrm{d}x \). Expanding the integrand: \( \left(\frac{1}{3}x^3 + \frac{1}{4x}\right)\left(x^2 + \frac{1}{4x^2}\right) = \frac{1}{3}x^5 + \frac{1}{12}x + \frac{1}{4}x + \frac{1}{16x^3} = \frac{1}{3}x^5 + \frac{1}{3}x + \frac{1}{16}x^{-3} \). Integrating term by term: \( \int \left(\frac{1}{3}x^5 + \frac{1}{3}x + \frac{1}{16}x^{-3}\right) \, \mathrm{d}x = \left[ \frac{1}{18}x^6 + \frac{1}{6}x^2 - \frac{1}{32x^2} \right]_1^3 \). Evaluating at the limits: At \( x = 3 \): \( \frac{729}{18} + \frac{9}{6} - \frac{1}{288} = 40.5 + 1.5 - \frac{1}{288} = 42 - \frac{1}{288} \). At \( x = 1 \): \( \frac{1}{18} + \frac{1}{6} - \frac{1}{32} = \frac{4}{18} - \frac{1}{32} = \frac{2}{9} - \frac{1}{32} = \frac{55}{288} \). Subtracting these values: \( \left(42 - \frac{1}{288}\right) - \frac{55}{288} = 42 - \frac{56}{288} = 42 - \frac{7}{36} = \frac{1505}{36} \). Thus, \( S = 2\pi \times \frac{1505}{36} = \frac{1505\pi}{18} \).

For part (i): B1 for finding the correct derivative dy/dx. M1 for setting up 1 + (dy/dx)^2. A1 for algebraic simplification to the required square form. For part (ii): M1 for substituting the expression into the arc length formula. M1 for carrying out the integration. A1 for evaluating the integral to get the correct exact value 53/6. For part (iii): M1 for writing down the correct integral for surface area of revolution. M1 for expanding the algebraic integrand correctly. M1 for performing the integration on the expanded terms. A1 for substituting limits and calculating the exact surface area as 1505*pi/18.

First, we rewrite the differential equation in standard linear form by dividing through by \( x \): \( \frac{\mathrm{d}y}{\mathrm{d}x} + \left(1 + \frac{2}{x}\right)y = \frac{e^{-x}}{x} \). The integrating factor \( I \) is given by \( I = e^{\int \left(1 + \frac{2}{x}\right) \, \mathrm{d}x} = e^{x + 2\ln x} = e^x \cdot e^{\ln(x^2)} = x^2 e^x \). Multiplying the standard form of the differential equation by this integrating factor yields: \( \frac{\mathrm{d}}{\mathrm{d}x}\left(x^2 e^x y\right) = x^2 e^x \cdot \frac{e^{-x}}{x} = x \). Integrating both sides with respect to \( x \): \( x^2 e^x y = \int x \, \mathrm{d}x = \frac{1}{2}x^2 + C \). To find the constant \( C \), we apply the initial condition that \( y = 2e^{-1} \) when \( x = 1 \): \( (1)^2 e^1 (2e^{-1}) = \frac{1}{2}(1)^2 + C \implies 2 = \frac{1}{2} + C \implies C = \frac{3}{2} \). Therefore, the equation becomes \( x^2 e^x y = \frac{1}{2}x^2 + \frac{3}{2} \). Dividing by \( x^2 e^x \) to express \( y \) in terms of \( x \): \( y = e^{-x}\left(\frac{1}{2} + \frac{3}{2x^2}\right) = \frac{(x^2 + 3)e^{-x}}{2x^2} \).

M1 for dividing by x to get standard linear form. M1 for setting up the integrating factor integral. A1 for correct integrating factor x^2 e^x. M1 for rewriting the left side as derivative of product of IF and y. A1 for correct simplified RHS after multiplication. M1 for integrating RHS to get x^2 / 2. A1 for the correct general solution including constant C. M1 for using x = 1, y = 2e^{-1} to find C. A1 for obtaining the correct particular solution in the required form.