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

(i) To find the direction of the common perpendicular to both lines, we compute the cross product of their direction vectors:
\[ \mathbf{n} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} \times \begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 3 \\ 1 & 1 & -2 \end{vmatrix} = \begin{pmatrix} -1 \\ 7 \\ 3 \end{pmatrix} \]

The magnitude of \( \mathbf{n} \) is:
\[ |\mathbf{n}| = \sqrt{(-1)^2 + 7^2 + 3^2} = \sqrt{59} \]

Let \( A(1, 2, -1) \) be a point on \( l_1 \) and \( B(3, 0, 4) \) be a point on \( l_2 \). Then:
\[ \vec{AB} = \begin{pmatrix} 3 - 1 \\ 0 - 2 \\ 4 - (-1) \end{pmatrix} = \begin{pmatrix} 2 \\ -2 \\ 5 \end{pmatrix} \]

The shortest distance \( d \) is the projection of \( \vec{AB} \) onto \( \mathbf{n} \):
\[ d = \frac{|\vec{AB} \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{|2(-1) + (-2)(7) + 5(3)|}{\sqrt{59}} = \frac{|-2 - 14 + 15|}{\sqrt{59}} = \frac{1}{\sqrt{59}} \]

(ii) Let \( P \) and \( Q \) be points on \( l_1 \) and \( l_2 \) respectively such that \( \vec{PQ} \) is perpendicular to both lines.
\[ P = \begin{pmatrix} 1 + 2\lambda \\ 2 - \lambda \\ -1 + 3\lambda \end{pmatrix}, \quad Q = \begin{pmatrix} 3 + \mu \\ \mu \\ 4 - 2\mu \end{pmatrix} \]
\[ \vec{PQ} = \begin{pmatrix} 2 + \mu - 2\lambda \\ -2 + \mu + \lambda \\ 5 - 2\mu - 3\lambda \end{pmatrix} \]

Since \( \vec{PQ} \cdot \mathbf{d}_1 = 0 \):
\[ 2(2 + \mu - 2\lambda) - 1(-2 + \mu + \lambda) + 3(5 - 2\mu - 3\lambda) = 0 \implies 14\lambda + 5\mu = 21 \]

Since \( \vec{PQ} \cdot \mathbf{d}_2 = 0 \):
\[ 1(2 + \mu - 2\lambda) + 1(-2 + \mu + \lambda) - 2(5 - 2\mu - 3\lambda) = 0 \implies 5\lambda + 6\mu = 10 \]

Solving these simultaneous equations yields:
\[ \lambda = \frac{76}{59}, \quad \mu = \frac{35}{59} \]

Substituting \( \lambda = \frac{76}{59} \) into the expression for \( P \):
\[ P = \begin{pmatrix} 1 + \frac{152}{59} \\ 2 - \frac{76}{59} \\ -1 + \frac{228}{59} \end{pmatrix} = \frac{1}{59}\begin{pmatrix} 211 \\ 42 \\ 169 \end{pmatrix} \]

Hence, the equation of the line of shortest distance is:
\[ \mathbf{r} = \frac{1}{59}\begin{pmatrix} 211 \\ 42 \\ 169 \end{pmatrix} + t \begin{pmatrix} -1 \\ 7 \\ 3 \end{pmatrix} \]

M1: For calculating the cross product of the direction vectors.
A1: For correct normal vector \( \mathbf{n} = -\mathbf{i} + 7\mathbf{j} + 3\mathbf{k} \).
M1: For using the projection formula \( \frac{|\vec{AB} \cdot \mathbf{n}|}{|\mathbf{n}|} \).
A1: For obtaining the correct shortest distance \( \frac{1}{\sqrt{59}} \).
M1: For setting up the general vector \( \vec{PQ} \) and applying dot products with direction vectors.
A1: For the two correct linear equations in \( \lambda \) and \( \mu \).
M1: For solving the simultaneous equations.
A1: For correct parameter values \( \lambda = \frac{76}{59} \) and \( \mu = \frac{35}{59} \).
M1: For finding a point on the common perpendicular.
A1.7: For the correct vector equation of the line.

(i) The vertical asymptote occurs where the denominator is zero, so \( x = -1 \). (Since the numerator is \( 2(-1)^2 + 5(-1) - 3 = -6 \neq 0 \), this is a vertical asymptote).
To find the oblique asymptote, we perform algebraic division:
\[ 2x^2 + 5x - 3 = (2x + 3)(x + 1) - 6 \]
Thus, the equation of the curve can be rewritten as:
\[ y = 2x + 3 - \frac{6}{x + 1} \]
As \( x \to \pm\infty \), \( y \to 2x + 3 \).
So, the oblique asymptote is \( y = 2x + 3 \).

(ii) Differentiating \( y \) with respect to \( x \):
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(4x + 5)(x + 1) - (2x^2 + 5x - 3)(1)}{(x + 1)^2} = \frac{4x^2 + 9x + 5 - 2x^2 - 5x + 3}{(x + 1)^2} = \frac{2x^2 + 4x + 8}{(x + 1)^2} = \frac{2(x^2 + 2x + 4)}{(x + 1)^2} \]
For stationary points, \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \implies 2(x^2 + 2x + 4) = 0 \).
However, the discriminant of \( x^2 + 2x + 4 = 0 \) is \( D = 2^2 - 4(1)(4) = -12 < 0 \).
Since there are no real roots for the derivative, there are no stationary points on \( C \).

(iii) Intercepts:
With \( y \)-axis: \( x = 0 \implies y = -3 \). So \( (0, -3) \).
With \( x \)-axis: \( y = 0 \implies 2x^2 + 5x - 3 = 0 \implies (2x - 1)(x + 3) = 0 \implies x = \frac{1}{2} \) or \( x = -3 \). So \( (0.5, 0) \) and \( (-3, 0) \).

Sketching \( C \):
Draw asymptotes \( x = -1 \) and \( y = 2x + 3 \).
Plot the intercepts \( (0, -3) \), \( (0.5, 0) \), and \( (-3, 0) \).
Sketch two branches: one in the upper-left region (passing through \( (-3,0) \)) and one in the lower-right region (passing through \( (0,-3) \) and \( (0.5,0) \)), both strictly increasing since \( \frac{\mathrm{d}y}{\mathrm{d}x} > 0 \).

M1: For identifying the vertical asymptote \( x = -1 \).
M1: For executing algebraic division to find the linear term.
A1: For obtaining \( y = 2x + 3 \).
M1: For differentiating the quotient correctly.
A1: For simplifying to \( \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2(x^2 + 2x + 4)}{(x+1)^2} \).
A1: For demonstrating that the discriminant of the numerator is negative and concluding there are no stationary points.
B1: For finding the \( y \)-intercept \( (0, -3) \).
B1: For finding both \( x \)-intercepts \( (-3, 0) \) and \( (0.5, 0) \).
B2.7: For drawing a fully correct sketch showing both branches, labeled asymptotes, and clearly marked intersection points.

(i) We apply integration by parts to \( I_n = \int_{0}^{\frac{\pi}{2}} x^n \cos x \, \mathrm{d}x \).
Let \( u = x^n \implies \mathrm{d}u = n x^{n-1} \, \mathrm{d}x \).
Let \( \mathrm{d}v = \cos x \, \mathrm{d}x \implies v = \sin x \).

\[ I_n = \left[ x^n \sin x \right]_{0}^{\frac{\pi}{2}} - n \int_{0}^{\frac{\pi}{2}} x^{n-1} \sin x \, \mathrm{d}x \]
Since \( \sin(\frac{\pi}{2}) = 1 \) and \( \sin(0) = 0 \), this simplifies to:
\[ I_n = \left(\frac{\pi}{2}\right)^n - n \int_{0}^{\frac{\pi}{2}} x^{n-1} \sin x \, \mathrm{d}x \]

Now, we apply integration by parts again to the remaining integral.
Let \( u = x^{n-1} \implies \mathrm{d}u = (n-1)x^{n-2} \, \mathrm{d}x \).
Let \( \mathrm{d}v = \sin x \, \mathrm{d}x \implies v = -\cos x \).

\[ \int_{0}^{\frac{\pi}{2}} x^{n-1} \sin x \, \mathrm{d}x = \left[ -x^{n-1} \cos x \right]_{0}^{\frac{\pi}{2}} + (n-1) \int_{0}^{\frac{\pi}{2}} x^{n-2} \cos x \, \mathrm{d}x \]
Since \( \cos(\frac{\pi}{2}) = 0 \) and the lower limit is \( 0 \) (for \( n \ge 2 \)), the boundary term vanishes.
Thus,
\[ \int_{0}^{\frac{\pi}{2}} x^{n-1} \sin x \, \mathrm{d}x = (n-1) I_{n-2} \]

Substituting this back into the expression for \( I_n \) yields:
\[ I_n = \left(\frac{\pi}{2}\right)^n - n(n-1)I_{n-2} \]

(ii) We first compute \( I_0 \):
\[ I_0 = \int_{0}^{\frac{\pi}{2}} \cos x \, \mathrm{d}x = [\sin x]_{0}^{\frac{\pi}{2}} = 1 \]

Using the reduction formula for \( n = 2 \):
\[ I_2 = \left(\frac{\pi}{2}\right)^2 - 2(1)I_0 = \frac{\pi^2}{4} - 2 \]

Now using the reduction formula for \( n = 4 \):
\[ I_4 = \left(\frac{\pi}{2}\right)^4 - 4(3)I_2 = \frac{\pi^4}{16} - 12\left(\frac{\pi^2}{4} - 2\right) = \frac{\pi^4}{16} - 3\pi^2 + 24 \]

M1: For applying integration by parts once to \( I_n \).
A1: For obtaining the correct intermediate boundary term and integral.
M1: For applying integration by parts a second time to the integral involving \( \sin x \).
A1: For correctly evaluating the boundary terms for the second integration by parts.
A1: For obtaining the final correct reduction formula \( I_n = \left(\frac{\pi}{2}\right)^n - n(n-1)I_{n-2} \).
B1: For finding \( I_0 = 1 \).
M1: For using the reduction formula to calculate \( I_2 \).
A1: For \( I_2 = \frac{\pi^2}{4} - 2 \).
M1: For using the reduction formula to find \( I_4 \) in terms of \( I_2 \).
A1.7: For the correct exact answer \( \frac{\pi^4}{16} - 3\pi^2 + 24 \).

(i) First, solve the homogeneous equation \( \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + 4 \frac{\mathrm{d}y}{\mathrm{d}x} + 4y = 0 \).
The auxiliary equation is:
\[ m^2 + 4m + 4 = 0 \implies (m+2)^2 = 0 \implies m = -2 \text{ (repeated root)} \]
Therefore, the complementary function (CF) is:
\[ y_c = (A + Bx)e^{-2x} \]

Next, find a particular integral (PI) of the form:
\[ y_p = p \cos x + q \sin x \]
First and second derivatives of \( y_p \) are:
\[ \frac{\mathrm{d}y_p}{\mathrm{d}x} = -p \sin x + q \cos x \]
\[ \frac{\mathrm{d}^2 y_p}{\mathrm{d}x^2} = -p \cos x - q \sin x \]

Substitute these into the original differential equation:
\[ (-p \cos x - q \sin x) + 4(-p \sin x + q \cos x) + 4(p \cos x + q \sin x) = 25 \cos x \]
Combine coefficients of \( \cos x \) and \( \sin x \):
\[ (3p + 4q) \cos x + (-4p + 3q) \sin x = 25 \cos x \]

This gives the system of linear equations:
\[ 3p + 4q = 25 \]
\[ -4p + 3q = 0 \implies q = \frac{4}{3}p \]

Substitute \( q \) into the first equation:
\[ 3p + 4\left(\frac{4}{3}p\right) = 25 \implies \frac{25}{3}p = 25 \implies p = 3 \]
Then \( q = 4 \).
So, the particular integral is \( y_p = 3 \cos x + 4 \sin x \).

The general solution is:
\[ y = (A + Bx)e^{-2x} + 3 \cos x + 4 \sin x \]

(ii) Apply initial conditions: \( y = 3 \) at \( x = 0 \):
\[ 3 = A e^0 + 3 \cos 0 + 4 \sin 0 \implies 3 = A + 3 \implies A = 0 \]

Now find the derivative of the general solution:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = B e^{-2x} - 2(A + Bx)e^{-2x} - 3 \sin x + 4 \cos x \]
Since \( A = 0 \):
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = (B - 2Bx)e^{-2x} - 3 \sin x + 4 \cos x \]

Apply \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \) at \( x = 0 \):
\[ 0 = B + 4 \implies B = -4 \]

Thus, the particular solution is:
\[ y = -4x e^{-2x} + 3 \cos x + 4 \sin x \]

M1: For solving the auxiliary equation to get \( m = -2 \).
A1: For the correct CF \( (A + Bx)e^{-2x} \).
M1: For proposing a PI of the form \( p \cos x + q \sin x \) and finding derivatives.
M1: For substituting into the DE and setting up simultaneous equations.
A1: For finding the correct coefficients \( p = 3 \) and \( q = 4 \).
A1.7: For stating the correct general solution.
M1: For using the initial condition \( y(0) = 3 \) to find \( A \).
A1: For finding \( A = 0 \).
M1: For differentiating the general solution and using \( y'(0) = 0 \) to find \( B \).
A1: For finding \( B = -4 \) and writing the final particular solution.

(i) The curve \( r = a(1 + \cos \theta) \) is a cardioid symmetrical about the initial line \( \theta = 0 \). At \( \theta = 0 \), \( r = 2a \); at \( \theta = \frac{\pi}{2} \), \( r = a \); at \( \theta = \pi \), \( r = 0 \).

(ii) The area \( A \) enclosed by \( C \) is:
\[ A = \frac{1}{2} \int_{0}^{2\pi} r^2 \, \mathrm{d}\theta = \frac{1}{2} a^2 \int_{0}^{2\pi} (1 + \cos \theta)^2 \, \mathrm{d}\theta \]
\[ A = \frac{1}{2} a^2 \int_{0}^{2\pi} (1 + 2\cos \theta + \cos^2 \theta) \, \mathrm{d}\theta \]

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

(iii) A tangent is perpendicular to the initial line when \( \frac{\mathrm{d}x}{\mathrm{d}\theta} = 0 \).
\[ x = r \cos \theta = a(1 + \cos \theta)\cos \theta = a(\cos \theta + \cos^2 \theta) \]
\[ \frac{\mathrm{d}x}{\mathrm{d}\theta} = a(-\sin \theta - 2\cos \theta \sin \theta) = -a\sin \theta(1 + 2\cos \theta) = 0 \]

This gives two sets of solutions:
1) \( \sin \theta = 0 \implies \theta = 0 \) or \( \theta = \pi \).
At \( \theta = 0 \), \( r = 2a \). The point is \( (2a, 0) \).
At \( \theta = \pi \), \( r = 0 \) (this is the pole, which is excluded by the question).

2) \( 1 + 2\cos \theta = 0 \implies \cos \theta = -\frac{1}{2} \implies \theta = \frac{2\pi}{3} \) or \( \theta = \frac{4\pi}{3} \).
At \( \theta = \frac{2\pi}{3} \), \( r = a(1 - 0.5) = \frac{1}{2}a \). The point is \( (\frac{1}{2}a, \frac{2\pi}{3}) \).
At \( \theta = \frac{4\pi}{3} \), \( r = \frac{1}{2}a \). The point is \( (\frac{1}{2}a, \frac{4\pi}{3}) \).

Thus, the points are \( (2a, 0) \), \( (\frac{1}{2}a, \frac{2\pi}{3}) \), and \( (\frac{1}{2}a, \frac{4\pi}{3}) \).

B3: For a correct cardioid sketch: symmetric, passing through correct intercepts, correct heart shape.
M1: For setting up the area integral with correct limits.
M1: For using double-angle identity to integrate \( \cos^2 \theta \).
A1: For correct integration.
A1.7: For obtaining \( \frac{3}{2}\pi a^2 \) after correctly substituting limits.
M1: For expressing \( x = r\cos\theta \) and differentiating to find \( \frac{\mathrm{d}x}{\mathrm{d}\theta} \).
A1: For getting \( \sin\theta(1 + 2\cos\theta) = 0 \).
A1: For finding the point \( (2a, 0) \).
A1: For finding \( (\frac{1}{2}a, \frac{2\pi}{3}) \) and \( (\frac{1}{2}a, \frac{4\pi}{3}) \).

(i) Let \( (x, y) \) be a point on the line \( y = 2x + 1 \), and let its image under \( \mathbf{M} \) be \( (X, Y) \).
\[ \begin{pmatrix} X \\ Y \end{pmatrix} = \mathbf{M} \begin{pmatrix} x \\ y \end{pmatrix} \implies \begin{pmatrix} x \\ y \end{pmatrix} = \mathbf{M}^{-1} \begin{pmatrix} X \\ Y \end{pmatrix} \]

First, find \( \mathbf{M}^{-1} \):
\[ \det(\mathbf{M}) = (3)(2) - (2)(1) = 4 \]
\[ \mathbf{M}^{-1} = \frac{1}{4} \begin{pmatrix} 2 & -2 \\ -1 & 3 \end{pmatrix} \]

This gives:
\[ x = \frac{1}{4}(2X - 2Y) = \frac{1}{2}(X - Y) \]
\[ y = \frac{1}{4}(-X + 3Y) \]

Substitute these expressions into \( y = 2x + 1 \):
\[ \frac{1}{4}(-X + 3Y) = 2\left( \frac{1}{2}(X - Y) \right) + 1 \]
\[ \frac{1}{4}(-X + 3Y) = X - Y + 1 \]
Multiply both sides by 4:
\[ -X + 3Y = 4X - 4Y + 4 \]
\[ 7Y = 5X + 4 \implies Y = \frac{5}{7}X + \frac{4}{7} \]

Thus, the equation of the image line is \( y = \frac{5}{7}x + \frac{4}{7} \).

(ii) Let \( y = mx \) be an invariant line through the origin.
Then, the point \( (x, mx) \) is mapped to \( (X, mX) \) on the same line.
\[ \begin{pmatrix} X \\ Y \end{pmatrix} = \begin{pmatrix} 3 & 2 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} x \\ mx \end{pmatrix} = \begin{pmatrix} 3x + 2mx \\ x + 2mx \end{pmatrix} \]

Since \( Y = mX \):
\[ x + 2mx = m(3x + 2mx) \]
\[ x(1 + 2m) = x(3m + 2m^2) \]

Since \( x \neq 0 \) for points other than the origin:
\[ 2m^2 + m - 1 = 0 \]
\[ (2m - 1)(m + 1) = 0 \]

This gives \( m = \frac{1}{2} \) or \( m = -1 \).

Therefore, the equations of the invariant lines are \( y = \frac{1}{2}x \) and \( y = -x \).

M1: For calculating the determinant and writing the correct inverse matrix \( \mathbf{M}^{-1} \).
A1: For obtaining \( x = \frac{1}{2}(X-Y) \) and \( y = \frac{1}{4}(-X+3Y) \).
M1: For substituting the expressions for \( x \) and \( y \) into \( y = 2x + 1 \).
A1.7: For simplifying to the correct equation \( y = \frac{5}{7}x + \frac{4}{7} \).
M1: For setting up the transformation of a point \( (x, mx) \).
M1: For substituting the transformed coordinates into the invariant line equation \( Y = mX \).
A1: For formulating the quadratic equation \( 2m^2 + m - 1 = 0 \).
M1: For factoring or solving the quadratic equation.
A2: For obtaining both correct equations: \( y = \frac{1}{2}x \) and \( y = -x \) (1 mark for each).

(i) We start from the right-hand side and combine into a single fraction:
\[ \text{RHS} = \frac{1}{8} \left[ \frac{(2r + 1)^2 - (2r - 1)^2}{(2r - 1)^2(2r + 1)^2} \right] \]
Expand the terms in the numerator:
\[ (2r+1)^2 - (2r-1)^2 = (4r^2 + 4r + 1) - (4r^2 - 4r + 1) = 8r \]
Substitute this back:
\[ \text{RHS} = \frac{1}{8} \left[ \frac{8r}{(2r - 1)^2(2r + 1)^2} \right] = \frac{r}{(2r - 1)^2(2r + 1)^2} = \text{LHS} \]

(ii) Using the identity from part (i):
\[ S_n = \frac{1}{8} \sum_{r=1}^n \left( \frac{1}{(2r - 1)^2} - \frac{1}{(2r + 1)^2} \right) \]
Writing out the terms of this telescoping series:
\[ r=1: \quad \frac{1}{1^2} - \frac{1}{3^2} \]
\[ r=2: \quad \frac{1}{3^2} - \frac{1}{5^2} \]
\[ \dots \]
\[ r=n: \quad \frac{1}{(2n - 1)^2} - \frac{1}{(2n + 1)^2} \]
Summing these terms leads to a cascade cancellation where all but the first and last terms vanish:
\[ S_n = \frac{1}{8} \left( 1 - \frac{1}{(2n + 1)^2} \right) \]
Simplify the algebraic expression:
\[ S_n = \frac{1}{8} \left( \frac{(2n + 1)^2 - 1}{(2n + 1)^2} \right) = \frac{1}{8} \left( \frac{4n^2 + 4n}{(2n + 1)^2} \right) = \frac{1}{8} \left( \frac{4n(n + 1)}{(2n + 1)^2} \right) = \frac{n(n + 1)}{2(2n + 1)^2} \]

(iii) As \( n \to \infty \), the term \( \frac{1}{(2n + 1)^2} \to 0 \).
Therefore:
\[ S_{\infty} = \lim_{n \to \infty} S_n = \frac{1}{8}(1 - 0) = \frac{1}{8} \]

M1: For combining the terms on the RHS into a single algebraic fraction.
A1: For correctly expanding the numerator and showing it equals \( 8r \).
A1: For establishing the identity clearly.
M1: For writing down the first few terms and the last term of the series to show the telescoping pattern.
A1: For obtaining the unsimplified sum \( \frac{1}{8} \left( 1 - \frac{1}{(2n + 1)^2} \right) \).
M1: For bringing the sum to a single simplified fraction.
A1.7: For finding \( S_n = \frac{n(n+1)}{2(2n+1)^2} \).
M1: For evaluating the limit as \( n \to \infty \).
A2: For obtaining the correct sum to infinity \( \frac{1}{8} \).

First, we find the complementary function (CF) by solving the auxiliary equation:
\(m^2 + 4m + 4 = 0 \Rightarrow (m+2)^2 = 0 \Rightarrow m = -2\) (repeated root).
Thus, the CF is:
\(y_{CF} = (Ax + B)\text{e}^{-2x}\).

Next, we find the particular integral (PI) for the right-hand side, \(8\text{e}^{-2x} + 12x\).
Since both \(\text{e}^{-2x}\) and \(x\text{e}^{-2x}\) are present in the CF, our trial PI for the exponential term must be of the form \(C x^2 \text{e}^{-2x}\).
For the linear term, the trial PI is \(Dx + E\).
So the complete trial PI is:
\(y_p = C x^2 \text{e}^{-2x} + Dx + E\).

Let's find the derivatives of the exponential part \(y_1 = C x^2 \text{e}^{-2x}\):
\(y_1' = C(2x - 2x^2)\text{e}^{-2x}\)
\(y_1'' = C(2 - 8x + 4x^2)\text{e}^{-2x}\)
Substituting \(y_1\) into the left-hand side of the differential equation:
\(y_1'' + 4y_1' + 4y_1 = C\text{e}^{-2x} [ (2 - 8x + 4x^2) + 4(2x - 2x^2) + 4x^2 ] = 2C\text{e}^{-2x}\).
Setting this equal to \(8\text{e}^{-2x}\) gives:
\(2C = 8 \Rightarrow C = 4\).

Now, for the linear part \(y_2 = Dx + E\):
\(y_2' = D\)
\(y_2'' = 0\)
Substituting into the differential equation:
\(0 + 4D + 4(Dx + E) = 12x \Rightarrow 4Dx + (4D + 4E) = 12x\).
Equating coefficients:
\(4D = 12 \Rightarrow D = 3\)
\(4D + 4E = 0 \Rightarrow 12 + 4E = 0 \Rightarrow E = -3\).

Therefore, the particular integral is:
\(y_p = 4x^2\text{e}^{-2x} + 3x - 3\).

The general solution is the sum of the CF and the PI:
\(y = (Ax + B)\text{e}^{-2x} + 4x^2\text{e}^{-2x} + 3x - 3\).

M1: Set up and solve the auxiliary equation.
A1: Correct CF: \((Ax + B)\text{e}^{-2x}\).
M1: Identify correct form of PI: \(C x^2 \text{e}^{-2x} + Dx + E\).
M1: Differentiate the trial PI and substitute into the DE.
A1: Obtain \(C = 4\).
A1: Obtain \(D = 3\) and \(E = -3\).
A1.375: Combine CF and PI to write the general solution.

Divide the given differential equation by \( x \) to express it in standard linear form:
\[ \frac{\text{d}y}{\text{d}x} + \left(2 + \frac{1}{x}\right)y = 3\text{e}^{-2x} \]

Find the integrating factor (IF):
\[ I = \exp\left(\int \left(2 + \frac{1}{x}\right)\text{d}x\right) = \exp(2x + \ln x) = x\text{e}^{2x} \]

Multiply both sides of the standard equation by the integrating factor:
\[ x\text{e}^{2x}\frac{\text{d}y}{\text{d}x} + (2x + 1)\text{e}^{2x}y = 3x \]
\[ \frac{\text{d}}{\text{d}x} \left( y x \text{e}^{2x} \right) = 3x \]

Integrate both sides with respect to \( x \):
\[ y x \text{e}^{2x} = \int 3x \text{d}x = \frac{3}{2}x^2 + C \]

Apply the boundary condition \( y = 2 \) when \( x = 1 \):
\[ (2)(1)\text{e}^{2(1)} = \frac{3}{2}(1)^2 + C \Rightarrow 2\text{e}^2 = \frac{3}{2} + C \Rightarrow C = 2\text{e}^2 - \frac{3}{2} \]

Substitute \( C \) back into the equation:
\[ y x \text{e}^{2x} = \frac{3}{2}x^2 + 2\text{e}^2 - \frac{3}{2} \]

Divide by \( x\text{e}^{2x} \) to solve for \( y \):
\[ y = \text{e}^{-2x} \left( \frac{3}{2}x + \frac{2\text{e}^2 - 1.5}{x} \right) = \text{e}^{-2x} \left( \frac{3x^2 + 4\text{e}^2 - 3}{2x} \right) \]

M1: Divide by \(x\) to find the standard first-order linear form.
A1: Correct integrating factor \( I = x\text{e}^{2x} \).
M1: Write the LHS as a perfect derivative and integrate the RHS.
A1: Correct integration: \( y x \text{e}^{2x} = \frac{3}{2}x^2 + C \).
M1: Substitute boundary conditions \( y = 2 \) and \( x = 1 \) to find \( C \).
A1: Obtain \( C = 2\text{e}^2 - \frac{3}{2} \).
A2.375: Solve for \( y \) explicitly and simplify to the final expression.

First, find the derivative \( \frac{\text{d}y}{\text{d}x} \):
\[ \frac{\text{d}y}{\text{d}x} = \sinh(2x) \]

The formula for arc length \( s \) is:
\[ s = \int_{a}^{b} \sqrt{1 + \left(\frac{\text{d}y}{\text{d}x}\right)^2} \text{d}x \]

Substitute the derivative into the integrand:
\[ 1 + \left(\frac{\text{d}y}{\text{d}x}\right)^2 = 1 + \sinh^2(2x) = \cosh^2(2x) \]

Since \( \cosh(2x) \ge 1 > 0 \) for all real \( x \), the integrand simplifies to:
\[ \sqrt{\cosh^2(2x)} = \cosh(2x) \]

Now, integrate \( \cosh(2x) \) from \( x = 0 \) to \( x = \ln 3 \):
\[ s = \int_{0}^{\ln 3} \cosh(2x) \text{d}x = \left[ \frac{1}{2}\sinh(2x) \right]_{0}^{\ln 3} \]
\[ s = \frac{1}{2}\sinh(2\ln 3) - \frac{1}{2}\sinh(0) \]
\[ s = \frac{1}{2}\sinh(\ln 9) - 0 \]

Using the exponential definition of \(
\sinh(\theta) = \frac{\text{e}^{\theta} - \text{e}^{-\theta}}{2} \):
\[ \sinh(\ln 9) = \frac{\text{e}^{\ln 9} - \text{e}^{-\ln 9}}{2} = \frac{9 - \frac{1}{9}}{2} = \frac{\frac{80}{9}}{2} = \frac{40}{9} \]

Thus, the arc length is:
\[ s = \frac{1}{2} \left(\frac{40}{9}\right) = \frac{20}{9} \]

M1: Differentiate \( y \) correctly to obtain \( \frac{\text{d}y}{\text{d}x} = \sinh(2x) \).
M1: Substitute into the arc length formula integrand \( \sqrt{1 + y'^2} \).
A1: Simplify the integrand to \( \cosh(2x) \) using the identity \( 1 + \sinh^2(u) = \cosh^2(u) \).
M1: Integrate \( \cosh(2x) \) to get \( \frac{1}{2}\sinh(2x) \).
M1: Substitute the limits of integration \( 0 \) and \( \ln 3 \).
A1: Express \( \sinh(2\ln 3) \) in exponential form.
A2.375: Evaluate to obtain the final exact value \( \frac{20}{9} \).

(a) To show the reduction formula, we apply integration by parts to \( I_n = \int_{0}^{1} x^n \text{e}^{-x} \text{d}x \).
Let \( u = x^n \) and \( \text{d}v = \text{e}^{-x}\text{d}x \).
Then \( \text{d}u = n x^{n-1} \text{d}x \) and \( v = -\text{e}^{-x} \).

Applying the integration by parts formula:
\[ I_n = \left[ -x^n \text{e}^{-x} \right]_{0}^{1} - \int_{0}^{1} (-\text{e}^{-x})(n x^{n-1}) \text{d}x \]
\[ I_n = \left( -(1)^n \text{e}^{-1} - 0 \right) + n \int_{0}^{1} x^{n-1} \text{e}^{-x} \text{d}x \]
\[ I_n = -\text{e}^{-1} + n I_{n-1} \]
\[ I_n = n I_{n-1} - \text{e}^{-1} \]

(b) First, compute the base case \( I_0 \):
\[ I_0 = \int_{0}^{1} \text{e}^{-x} \text{d}x = \left[ -\text{e}^{-x} \right]_{0}^{1} = -\text{e}^{-1} - (-\text{e}^0) = 1 - \text{e}^{-1} \]

Now, apply the reduction formula successively:
For \( n = 1 \):
\[ I_1 = 1 I_0 - \text{e}^{-1} = (1 - \text{e}^{-1}) - \text{e}^{-1} = 1 - 2\text{e}^{-1} \]

For \( n = 2 \):
\[ I_2 = 2 I_1 - \text{e}^{-1} = 2(1 - 2\text{e}^{-1}) - \text{e}^{-1} = 2 - 5\text{e}^{-1} \]

For \( n = 3 \):
\[ I_3 = 3 I_2 - \text{e}^{-1} = 3(2 - 5\text{e}^{-1}) - \text{e}^{-1} = 6 - 15\text{e}^{-1} - \text{e}^{-1} = 6 - 16\text{e}^{-1} \]

Part (a) [4 marks]:
M1: Correct choice of \( u \) and \( \text{d}v \) for integration by parts.
A1: Correct intermediate boundary evaluation \( -\text{e}^{-1} \).
A1: Correct term \( n \int x^{n-1}\text{e}^{-x}\text{d}x \).
A1: Complete the proof clearly.

Part (b) [5.375 marks]:
M1: Find \( I_0 = 1 - \text{e}^{-1} \).
A1: Apply formula to get \( I_1 = 1 - 2\text{e}^{-1} \).
A1: Apply formula to get \( I_2 = 2 - 5\text{e}^{-1} \).
A2.375: Correctly find \( I_3 = 6 - 16\text{e}^{-1} \).

The direction vectors of the two lines are \( \mathbf{u}_1 = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} \) and \( \mathbf{u}_2 = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \).

Find a common normal vector \( \mathbf{n} \) perpendicular to both lines using the cross product:
\[ \mathbf{n} = \mathbf{u}_1 \times \mathbf{u}_2 = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} \times \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \]
\[ \mathbf{n} = \begin{pmatrix} (-1)(2) - (3)(1) \\ (3)(1) - (2)(2) \\ (2)(1) - (-1)(1) \end{pmatrix} = \begin{pmatrix} -5 \\ -1 \\ 3 \end{pmatrix} \]

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

Let \( A \) be a point on \( l_1 \), \( (1, 2, -1) \), and \( B \) be a point on \( l_2 \), \( (3, 0, 4) \).
Find the vector \( \vec{AB} \):
\[ \vec{AB} = \mathbf{r}_B - \mathbf{r}_A = \begin{pmatrix} 3 - 1 \\ 0 - 2 \\ 4 - (-1) \end{pmatrix} = \begin{pmatrix} 2 \\ -2 \\ 5 \end{pmatrix} \]

Now, the shortest distance \( d \) is the projection of \( \vec{AB} \) onto the unit normal vector:
\[ d = \frac{|\vec{AB} \cdot \mathbf{n}|}{|\mathbf{n}|} \]
\[ \vec{AB} \cdot \mathbf{n} = \begin{pmatrix} 2 \\ -2 \\ 5 \end{pmatrix} \cdot \begin{pmatrix} -5 \\ -1 \\ 3 \end{pmatrix} = 2(-5) + (-2)(-1) + 5(3) = -10 + 2 + 15 = 7 \]

Thus, the shortest distance is:
\[ d = \frac{|7|}{\sqrt{35}} = \frac{7}{\sqrt{35}} = \frac{\sqrt{35}}{5} \]

M1: Set up the cross product of the direction vectors \( \mathbf{u}_1 \times \mathbf{u}_2 \).
A1: Correct normal vector \( \mathbf{n} = \begin{pmatrix} -5 \\ -1 \\ 3 \end{pmatrix} \).
A1: Correct magnitude \( |\mathbf{n}| = \sqrt{35} \).
M1: Identify any point on \( l_1 \) and \( l_2 \) and calculate \( \vec{AB} \).
A1: Correct vector \( \vec{AB} = \begin{pmatrix} 2 \\ -2 \\ 5 \end{pmatrix} \).
M1: Compute the scalar dot product \( \vec{AB} \cdot \mathbf{n} \).
A2.375: Evaluate the shortest distance and simplify to \( \frac{\sqrt{35}}{5} \) (or equivalent numerical value \( \approx 1.18 \)).

First, find two direction vectors in the plane \(\Pi\):
\[ \vec{PQ} = \begin{pmatrix} 2-1 \\ 3-1 \\ 0-2 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix} \]
\[ \vec{PR} = \begin{pmatrix} -1-1 \\ 2-1 \\ 4-2 \end{pmatrix} = \begin{pmatrix} -2 \\ 1 \\ 2 \end{pmatrix} \]

Calculate a normal vector \( \mathbf{n} \) to the plane \(\Pi\) by finding the cross product \( \vec{PQ} \times \vec{PR} \):
\[ \mathbf{n} = \begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix} \times \begin{pmatrix} -2 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} (2)(2) - (-2)(1) \\ (-2)(-2) - (1)(2) \\ (1)(1) - (2)(-2) \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \\ 5 \end{pmatrix} \]

The Cartesian equation of the plane \(\Pi\) is given by:
\[ 6x + 2y + 5z = D \]
To find \( D \), substitute the coordinates of point \( P(1, 1, 2) \):
\[ 6(1) + 2(1) + 5(2) = 18 \Rightarrow 6x + 2y + 5z - 18 = 0 \]

The perpendicular distance \( d \) from the point \( S(4, -1, 5) \) to the plane \( 6x + 2y + 5z - 18 = 0 \) is:
\[ d = \frac{|6(4) + 2(-1) + 5(5) - 18|}{\sqrt{6^2 + 2^2 + 5^2}} \]
\[ d = \frac{|24 - 2 + 25 - 18|}{\sqrt{36 + 4 + 25}} = \frac{29}{\sqrt{65}} \]

M1: Find two vectors in the plane, e.g., \( \vec{PQ} \) and \( \vec{PR} \).
A1: Correct vectors: \( \vec{PQ} = \begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix} \) and \( \vec{PR} = \begin{pmatrix} -2 \\ 1 \\ 2 \end{pmatrix} \).
M1: Compute the cross product to find the normal vector.
A1: Correct normal vector \( \mathbf{n} = \begin{pmatrix} 6 \\ 2 \\ 5 \end{pmatrix} \).
M1: Form the Cartesian equation of the plane.
A1: Correct equation: \( 6x + 2y + 5z - 18 = 0 \).
M1: Substitute point \( S \) into the perpendicular distance formula.
A1.375: Obtain the exact perpendicular distance \( \frac{29}{\sqrt{65}} \) (or \( \approx 3.60 \)).

(a) To find the asymptotes, perform polynomial division on the equation of \(C\):
\[ 2x^2 + x - 3 = (2x + 5)(x - 2) + 7 \]
So,
\[ y = 2x + 5 + \frac{7}{x - 2} \]

As \( x \to 2 \), \( y \to \pm \infty \), so the vertical asymptote is \( x = 2 \).
As \( x \to \pm \infty \), \( \frac{7}{x - 2} \to 0 \), so the oblique asymptote is \( y = 2x + 5 \).

(b) To find the stationary points, we find \( \frac{\text{d}y}{\text{d}x} \) and set it to 0.
Using \( y = 2x + 5 + 7(x-2)^{-1} \):
\[ \frac{\text{d}y}{\text{d}x} = 2 - \frac{7}{(x - 2)^2} \]

Setting the derivative to 0:
\[ 2 - \frac{7}{(x - 2)^2} = 0 \Rightarrow (x - 2)^2 = \frac{7}{2} = 3.5 \]
\[ x - 2 = \pm \sqrt{3.5} = \pm \frac{\sqrt{14}}{2} \]
\[ x = 2 \pm \frac{\sqrt{14}}{2} \]

Now substitute these values back into \( y = 2x + 5 + \frac{7}{x - 2} \):
For \( x = 2 + \frac{\sqrt{14}}{2} \):
\[ y = 2\left(2 + \frac{\sqrt{14}}{2}\right) + 5 + \frac{7}{\frac{\sqrt{14}}{2}} = 4 + \sqrt{14} + 5 + \sqrt{14} = 9 + 2\sqrt{14} \]

For \( x = 2 - \frac{\sqrt{14}}{2} \):
\[ y = 2\left(2 - \frac{\sqrt{14}}{2}\right) + 5 - \frac{7}{\frac{\sqrt{14}}{2}} = 4 - \sqrt{14} + 5 - \sqrt{14} = 9 - 2\sqrt{14} \]

So the coordinates of the stationary points are:
\[ \left(2 + \frac{\sqrt{14}}{2}, 9 + 2\sqrt{14}\right) \quad \text{and} \quad \left(2 - \frac{\sqrt{14}}{2}, 9 - 2\sqrt{14}\right) \]

Part (a) [3.375 marks]:
M1: Perform algebraic division of the rational function.
A1: Correct vertical asymptote: \( x = 2 \).
A1.375: Correct oblique asymptote: \( y = 2x + 5 \).

Part (b) [6 marks]:
M1: Differentiate the rational function \( y \).
A1: Correct derivative \( \frac{\text{d}y}{\text{d}x} = 2 - \frac{7}{(x-2)^2} \).
M1: Set derivative to 0 and solve for \( x \).
A1: Correct \( x \)-coordinates: \( x = 2 \pm \frac{\sqrt{14}}{2} \).
M1: Substitute the \( x \)-values back into the curve equation to find \( y \).
A1: Correct stationary points: \( \left(2 \pm \frac{\sqrt{14}}{2}, 9 \pm 2\sqrt{14}\right) \).

(a) We rewrite the expression by performing division:
\[ y = \frac{(x^2 - 4x + 5) - 2}{x^2 - 4x + 5} = 1 - \frac{2}{x^2 - 4x + 5} \]

Complete the square for the quadratic expression in the denominator:
\[ x^2 - 4x + 5 = (x - 2)^2 + 1 \]

Since \( (x-2)^2 \ge 0 \) for all real \( x \), we have:
\[ (x - 2)^2 + 1 \ge 1 \]

This implies that the fraction is positive and bounded:
\[ \frac{2}{x^2 - 4x + 5} > 0 \]

Thus,
\[ y = 1 - \frac{2}{x^2 - 4x + 5} < 1 \]
This proves that \( y \le 1 \) (specifically, \( y < 1 \)) for all real \( x \).

(b) Differentiate \( y = 1 - 2(x^2 - 4x + 5)^{-1} \) using the chain rule:
\[ \frac{\text{d}y}{\text{d}x} = 2(x^2 - 4x + 5)^{-2} \cdot (2x - 4) = \frac{4(x - 2)}{(x^2 - 4x + 5)^2} \]

Setting \( \frac{\text{d}y}{\text{d}x} = 0 \) gives:
\[ 4(x - 2) = 0 \Rightarrow x = 2 \]

When \( x = 2 \):
\[ y = 1 - \frac{2}{2^2 - 4(2) + 5} = 1 - \frac{2}{1} = -1 \]
So the stationary point is \( (2, -1) \).

To find the nature of the stationary point:
Since \( (x^2 - 4x + 5)^2 > 0 \) for all real \( x \), the sign of the derivative is determined solely by the numerator, \( 4(x-2) \).
- For \( x < 2 \), \( \frac{\text{d}y}{\text{d}x} < 0 \).
- For \( x > 2 \), \( \frac{\text{d}y}{\text{d}x} > 0 \).

Since the gradient changes from negative to positive, the stationary point \( (2, -1) \) is a local minimum.

Part (a) [3 marks]:
M1: Rewrite \( y \) as \( 1 - \frac{2}{x^2 - 4x + 5} \).
M1: Complete the square on the denominator to show \( x^2-4x+5 \ge 1 \).
A1: Deduce that \( y < 1 \) and hence \( y \le 1 \).

Part (b) [6.375 marks]:
M1: Differentiate \( y \) using quotient rule or chain rule.
A1.375: Correct derivative \( \frac{\text{d}y}{\text{d}x} = \frac{4(x-2)}{(x^2-4x+5)^2} \).
M1: Set derivative to 0 and find \( x = 2 \).
A1: Obtain the point \( (2, -1) \).
M1: Test the gradient on either side or use the second derivative to determine the nature.
A1: Conclude that \( (2, -1) \) is a local minimum.