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

For (i): LHS = \(3\left(\frac{(r+1) - (r-1)}{(r-1)(r+1)}\right) = 3\left(\frac{2}{r^2-1}\right) = \frac{6}{r^2-1} = \text{RHS}\). For (ii): Let \(u_r = 3\left(\frac{1}{r-1} - \frac{1}{r+1}\right)\). Then \(\sum_{r=2}^{n} u_r = 3 \sum_{r=2}^{n} \left(\frac{1}{r-1} - \frac{1}{r+1}\right) = 3 \left[ \left(1 - \frac{1}{3}\right) + \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \dots + \left(\frac{1}{n-2} - \frac{1}{n}\right) + \left(\frac{1}{n-1} - \frac{1}{n+1}\right) \right]\). Cancelling intermediate terms leaves only the first two positive terms and the last two negative terms: \(3 \left[ 1 + \frac{1}{2} - \frac{1}{n} - \frac{1}{n+1} \right] = 3 \left[ \frac{3}{2} - \frac{2n+1}{n(n+1)} \right] = \frac{9n(n+1) - 6(2n+1)}{2n(n+1)} = \frac{9n^2 - 3n - 6}{2n(n+1)} = \frac{3(3n^2 - n - 2)}{2n(n+1)}\). For (iii): As \(n \to \infty\), \(\frac{1}{n} \to 0\) and \(\frac{1}{n+1} \to 0\). Thus, the sum to infinity is \(3 \left(1 + \frac{1}{2}\right) = \frac{9}{2}\).

M1: Algebraic manipulation to prove identity. A1: Correctly shown. M1: Listing terms of the series and identifying terms that cancel. A1: Obtaining the unsimplified remaining terms: \(3(1 + 1/2 - 1/n - 1/(n+1))\). M1: Combining into a single fraction. A1: Fully correct simplified single fraction. B1.33: Correctly finding limit as \(n \to \infty\) as \(9/2\) (or 4.5).

For (i): The sketch is the upper half of a cardioid, starting at \((a, 0)\), peaking at \((2a, \pi/2)\), and ending at \((a, \pi)\). For (ii): Area = \(\frac{1}{2} \int_{0}^{\pi} r^2 \mathrm{d}\theta = \frac{1}{2} a^2 \int_{0}^{\pi} (1 + \sin\theta)^2 \mathrm{d}\theta = \frac{1}{2} a^2 \int_{0}^{\pi} (1 + 2\sin\theta + \sin^2\theta) \mathrm{d}\theta\). Using \(\sin^2\theta = \frac{1-\cos 2\theta}{2}\), we get \(\frac{1}{2} a^2 \int_{0}^{\pi} \left( \frac{3}{2} + 2\sin\theta - \frac{1}{2}\cos 2\theta \right) \mathrm{d}\theta = \frac{1}{2} a^2 \left[ \frac{3}{2}\theta - 2\cos\theta - \frac{1}{4}\sin 2\theta \right]_0^{\pi} = \frac{1}{2} a^2 \left[ \left( \frac{3\pi}{2} + 2 - 0 \right) - (0 - 2 - 0) \right] = \frac{1}{2} a^2 \left( \frac{3\pi}{2} + 4 \right) = a^2 \left( \frac{3\pi}{4} + 2 \right)\).

B1: Sketch showing correct shape in the first quadrant. B1: Sketch showing correct symmetry and coordinates/endpoints in the second quadrant. M1: Setting up correct area integral \(\frac{1}{2}\int r^2 \mathrm{d}\theta\) with correct limits. M1: Expanding and using double-angle identity to integrate \(\sin^2\theta\). A1: Correct integration. M1: Substituting limits correctly. A1.33: Correct simplified area: \(a^2(\frac{3\pi}{4} + 2)\).

For (i): From the coefficients of the cubic equation, \(\alpha + \beta + \gamma = -2\), \(\alpha\beta + \beta\gamma + \gamma\alpha = -3\), and \(\alpha\beta\gamma = -4\). For (ii): Using the algebraic identity, \(\alpha^2 + \beta^2 + \gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = (-2)^2 - 2(-3) = 4 + 6 = 10\). For (iii): Let the new equation be \(y^3 - Sy^2 + Ty - P = 0\). We have \(S = \alpha^2+\beta^2+\gamma^2 = 10\). For the product of the roots, \(P = (\alpha\beta\gamma)^2 = (-4)^2 = 16\). For the sum of the pairwise products, \(T = \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 = (\alpha\beta+\beta\gamma+\gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha+\beta+\gamma) = (-3)^2 - 2(-4)(-2) = 9 - 16 = -7\). Thus, the new equation is \(y^3 - 10y^2 - 7y - 16 = 0\). (Alternative Method: substitute \(x = y^{1/2}\) to get \(y^{3/2} + 2y - 3y^{1/2} + 4 = 0 \implies y^{1/2}(y-3) = -(2y+4)\). Squaring both sides yields \(y(y-3)^2 = (2y+4)^2 \implies y(y^2-6y+9) = 4y^2+16y+16 \implies y^3-10y^2-7y-16=0\).)

B1.33: Correctly stating all three sum/product relations. M1: Using the expansion of \((\alpha+\beta+\gamma)^2\) to express the sum of squares. A1: Correctly showing 10. M1: Calculating product \(P = 16\). M1: Calculating pairwise sum \(T = -7\). A1: Formulating the final cubic equation in \(y\) (or any other variable). (Alternative Method: M1: Substitute \(x = y^{1/2}\). M1: Isolate radical terms. A1: Square both sides. A1: Correctly expand and simplify to get \(y^3 - 10y^2 - 7y - 16 = 0\).)

(a) The curve is the upper half of a cardioid.
- At \( \theta = 0 \), \( r = 2a \).
- At \( \theta = \frac{\pi}{2} \), \( r = a \).
- At \( \theta = \pi \), \( r = 0 \).
- The curve is smooth, starting at \( (2a, 0) \) in Cartesian coordinates, looping through \( (0, a) \), and ending at the origin.

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

(c) The y-coordinate is given by \( y = r\sin\theta = a(1 + \cos\theta)\sin\theta \).
For the tangent to be parallel to the initial line, we need \( \frac{dy}{d\theta} = 0 \).
\( \frac{dy}{d\theta} = a\left(-\sin^2\theta + \cos\theta(1+\cos\theta)\right) \)
\( \frac{dy}{d\theta} = a\left(-\sin^2\theta + \cos\theta + \cos^2\theta\right) \)
Using \( \sin^2\theta = 1 - \cos^2\theta \):
\( \frac{dy}{d\theta} = a\left(2\cos^2\theta + \cos\theta - 1\right) \)
Setting \( \frac{dy}{d\theta} = 0 \) gives:
\( (2\cos\theta - 1)(\cos\theta + 1) = 0 \)
Since \( \theta \ne \pi \), we have \( \cos\theta = \frac{1}{2} \).
Since \( 0 \le \theta \le \pi \) and \( \theta \ne 0 \), we find \( \theta = \frac{\pi}{3} \).
At \( \theta = \frac{\pi}{3} \):
\( r = a\left(1 + \cos\frac{\pi}{3}\right) = \frac{3}{2}a \).
Now find the Cartesian coordinates:
\( x = r\cos\theta = \frac{3}{2}a \cos\frac{\pi}{3} = \frac{3}{4}a \).
\( y = r\sin\theta = \frac{3}{2}a \sin\frac{\pi}{3} = \frac{3\sqrt{3}}{4}a \).
Thus, the Cartesian coordinates are \( \left(\frac{3}{4}a, \frac{3\sqrt{3}}{4}a\right) \).

M1: For a correct sketch showing a closed loop in the upper half plane starting at \( (2a,0) \) and ending at the origin.
A1: For a correctly shaped half-cardioid, showing tangent at the origin along the negative x-axis.

M1: For using the area formula \( \frac{1}{2} \int r^2 \, d\theta \) with correct limits.
M1: For expanding \( (1+\cos\theta)^2 \) and using double-angle formula for \( \cos^2\theta \).
A1: For correct integration of all terms.
A1: For obtaining \( \frac{3}{4}\pi a^2 \) (or equivalent fraction).

M1: For writing \( y = r\sin\theta \) and differentiating with respect to \( \theta \).
M1: For using trigonometric identity to form a quadratic in \( \cos\theta \) and equating to 0.
A1: For finding \( \cos\theta = \frac{1}{2} \) (or \( \theta = \frac{\pi}{3} \)).
M1: For calculating both \( x = r\cos\theta \) and \( y = r\sin\theta \) using their value of \( \theta \).
A1: For both correct coordinates: \( x = \frac{3}{4}a \), \( y = \frac{3\sqrt{3}}{4}a \).

(a) From the given equation, the sum of roots and sum of product of roots in pairs are:
\( \alpha + \beta + \gamma = \frac{3}{2} \)
\( \alpha\beta + \beta\gamma + \gamma\alpha = \frac{4}{2} = 2 \)
We use the identity:
\( \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) \)
\( \alpha^2 + \beta^2 + \gamma^2 = \left(\frac{3}{2}\right)^2 - 2(2) = \frac{9}{4} - 4 = -\frac{7}{4} \).

(b) Since \( \alpha \), \( \beta \), and \( \gamma \) are roots of the cubic equation, they satisfy:
\( 2\alpha^3 - 3\alpha^2 + 4\alpha - 1 = 0 \implies 2\alpha^3 = 3\alpha^2 - 4\alpha + 1 \)
Summing this relation for all three roots, we obtain:
\( 2(\alpha^3 + \beta^3 + \gamma^3) = 3(\alpha^2 + \beta^2 + \gamma^2) - 4(\alpha + \beta + \gamma) + 3 \)
Let \( S_k = \alpha^k + \beta^k + \gamma^k \). Then:
\( 2S_3 = 3S_2 - 4S_1 + 3 \)
\( 2S_3 = 3\left(-\frac{7}{4}\right) - 4\left(\frac{3}{2}\right) + 3 \)
\( 2S_3 = -\frac{21}{4} - 6 + 3 = -\frac{33}{4} \)
\( S_3 = -\frac{33}{8} \).

(c) Let \( y = x^2 \), which means \( x = y^{1/2} \). Substituting this into the cubic equation:
\( 2y^{3/2} - 3y + 4y^{1/2} - 1 = 0 \)
Rearranging terms to separate those with fractional powers:
\( 2y^{3/2} + 4y^{1/2} = 3y + 1 \)
\( y^{1/2}(2y + 4) = 3y + 1 \)
Squaring both sides:
\( y(2y + 4)^2 = (3y + 1)^2 \)
\( y(4y^2 + 16y + 16) = 9y^2 + 6y + 1 \)
\( 4y^3 + 16y^2 + 16y = 9y^2 + 6y + 1 \)
\( 4y^3 + 7y^2 + 10y - 1 = 0 \).
Since all coefficients are integers, the required equation is \( 4y^3 + 7y^2 + 10y - 1 = 0 \) (or using any other variable like \( w \)).

B1: For stating \( \alpha + \beta + \gamma = \frac{3}{2} \) and \( \alpha\beta + \beta\gamma + \gamma\alpha = 2 \).
M1: For using the identity \( \sum \alpha^2 = (\sum \alpha)^2 - 2\sum\alpha\beta \).
A1: For showing the given result \( -\frac{7}{4} \) clearly with no errors.

M1: For using the original cubic equation to write \( 2S_3 - 3S_2 + 4S_1 - 3 = 0 \) (or equivalent method, e.g. using the identity \( S_3 - 3\alpha\beta\gamma = S_1(S_2 - \sum\alpha\beta) \)).
M1: For substituting their values of \( S_1, S_2 \) (and product of roots \( \alpha\beta\gamma = \frac{1}{2} \) if using the identity).
A1: For obtaining \( -\frac{33}{8} \) (or equivalent fraction).

M1: For rearranging the cubic equation to group terms with odd and even powers of \( x \).
M1: For squaring both sides to eliminate fractional powers of \( y \) (or using \( y = x^2 \) substitutions in an algebraic system).
A1: For expanding both sides correctly, obtaining \( y(4y^2 + 16y + 16) \) and \( 9y^2 + 6y + 1 \).
M1: For collecting terms and writing in standard form.
A1: For the correct final equation \( 4y^3 + 7y^2 + 10y - 1 = 0 \) (accept any variable, must have integer coefficients).

**(i)**
To find the vertical asymptotes, set the denominator to zero:
\[x^2 - 4 = 0 \implies x = 2 \text{ and } x = -2\]
Since the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is given by the ratio of the leading coefficients:
\[y = \lim_{x \to \pm\infty} \frac{2x^2 + 3x - 5}{x^2 - 4} = 2\]
Thus, the asymptotes are \(x = 2\), \(x = -2\), and \(y = 2\).

**(ii)**
Using the quotient rule to find the derivative \(y'\):
\[y' = \frac{(4x+3)(x^2-4) - (2x^2+3x-5)(2x)}{(x^2-4)^2}\]
\[y' = \frac{(4x^3 - 16x + 3x^2 - 12) - (4x^3 + 6x^2 - 10x)}{(x^2-4)^2}\]
\[y' = \frac{-3x^2 - 6x - 12}{(x^2-4)^2} = \frac{-3(x^2 + 2x + 4)}{(x^2-4)^2}\]
Completing the square for the quadratic term in the numerator:
\[x^2 + 2x + 4 = (x+1)^2 + 3 > 0 \quad \text{for all } x \in \mathbb{R}\]
Since the numerator is always strictly negative and the denominator is positive for all \(x \neq \pm 2\), we have \(y' < 0\) everywhere. Thus, \(y' \neq 0\) and \(C\) has no stationary points.

To find the intersection with the horizontal asymptote \(y = 2\):
\[2 = \frac{2x^2 + 3x - 5}{x^2 - 4} \implies 2x^2 - 8 = 2x^2 + 3x - 5 \implies 3x = -3 \implies x = -1\]
So, the coordinates of the intersection point are \((-1, 2)\).

**(iii)**
- **Intercepts with axes:**
- When \(y = 0\), \(2x^2 + 3x - 5 = 0 \implies (2x+5)(x-1) = 0 \implies x = -2.5\) and \(x = 1\).
- When \(x = 0\), \(y = \frac{-5}{-4} = 1.25\).
- **Asymptotes:** \(x = -2\), \(x = 2\), and \(y = 2\).
- **Branches:**
- For \(x < -2\), the branch starts near \(y=2\) (from below), passes through \((-2.5, 0)\), and decreases to \(-\infty\) as \(x \to -2^-\).
- For \(-2 < x < 2\), the branch decreases from \(+\infty\) near \(x = -2^+\), passes through \((-1, 2)\), \((0, 1.25)\), and \((1, 0)\), and decreases to \(-\infty\) as \(x \to 2^-\).
- For \(x > 2\), the branch decreases from \(+\infty\) near \(x = 2^+\) towards \(y = 2\) as \(x \to +\infty\).

**(iv)**
We solve \(-2 < \frac{2x^2+3x-5}{x^2-4} < 2\).

*First Inequality:* \(\frac{2x^2+3x-5}{x^2-4} < 2\)
\[\frac{2x^2+3x-5 - 2(x^2-4)}{x^2-4} < 0 \implies \frac{3x+3}{x^2-4} < 0 \implies \frac{3(x+1)}{(x-2)(x+2)} < 0\]
Critical values: \(x = -2, -1, 2\).
Testing intervals yields: \(x < -2\) or \(-1 < x < 2\).

*Second Inequality:* \(\frac{2x^2+3x-5}{x^2-4} > -2\)
\[\frac{2x^2+3x-5 + 2(x^2-4)}{x^2-4} > 0 \implies \frac{4x^2+3x-13}{x^2-4} > 0\]
For the numerator, the roots of \(4x^2+3x-13 = 0\) are:
\[x = \frac{-3 \pm \sqrt{217}}{8} \quad (x_1 \approx -2.22, \quad x_2 \approx 1.47)\]
Critical values: \(x = -2, 2\) and \(x_1, x_2\).
Testing intervals yields: \(x < x_1\) or \(-2 < x < x_2\) or \(x > 2\).

Combining both inequalities (finding the intersection of the two solution sets):
- For \(x < -2\), both hold when \(x < \frac{-3-\sqrt{217}}{8}\).
- For \(-2 < x < 2\), both hold when \(-1 < x < \frac{-3+\sqrt{217}}{8}\).
- For \(x > 2\), there is no common interval.

So the final solution is \(x < \frac{-3-\sqrt{217}}{8}\) or \(-1 < x < \frac{-3+\sqrt{217}}{8}\).

**(i)**
- **M1**: For setting the denominator to 0 to obtain vertical asymptotes.
- **A1**: Both vertical asymptotes \(x = 2\) and \(x = -2\) correct.
- **B1**: Horizontal asymptote \(y = 2\) correct.

**(ii)**
- **M1**: For attempting differentiation using the quotient or product rule.
- **A1**: Correct simplified derivative \(y' = \frac{-3(x^2+2x+4)}{(x^2-4)^2}\).
- **A1**: Showing \(x^2+2x+4 = (x+1)^2 + 3 > 0\) (or using discriminant) and concluding no stationary points.
- **B1**: Setting \(y = 2\) and solving to get \(x = -1\).
- **B0.5**: Correct coordinates \((-1, 2)\).

**(iii)**
- **B1**: Correct shape of the 3 branches with asymptotes drawn.
- **B1**: Intersection with horizontal asymptote at \((-1, 2)\) correctly shown.
- **B1**: Intersections with x-axis at \((-2.5, 0)\) and \((1, 0)\) correctly shown.
- **B1**: Intersection with y-axis at \((0, 1.25)\) correctly shown.

**(iv)**
- **M1**: For translating the absolute value inequality into two critical boundary inequalities.
- **A1**: Finding the critical values \(x = -1\) and \(x = \frac{-3 \pm \sqrt{217}}{8}\) (or 3 s.f. decimals \(-2.22\) and \(1.47\)).
- **M1**: Attempting to find overlapping regions via testing intervals or graphical analysis.
- **A1**: Correct final range of values (accept decimals): \(x < -2.22\) or \(-1 < x < 1.47\).

**(i)**
To show they are skew, we check for intersection. Equating components:
1) \(1 + 2\lambda = 2 + \mu \implies 2\lambda -
\mu = 1\)
2) \(-1 + \lambda = 3 - 2\mu \implies \lambda + 2\mu = 4\)
3) \(2 - \lambda = \mu \implies \lambda + \mu = 2\)

From (1) and (3), adding the equations gives:
\(3\lambda = 3 \implies \lambda = 1\), which then gives \
(\mu = 1\).
Substitute \(\lambda = 1, \mu = 1\) into (2):
\(1 + 2(1) = 3 \neq 4\).
Since the equations are inconsistent, the lines do not intersect. Since their direction vectors are not parallel (\(\begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} \neq k \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}\)), the lines are skew.

To find the shortest distance, find a common perpendicular vector \(\mathbf{n}\):
\(\mathbf{n} = \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} \times \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} = \begin{pmatrix} 1(1) - (-1)(-2) \\ (-1)(1) - 2(1) \\ 2(-2) - 1(1) \end{pmatrix} = \begin{pmatrix} -1 \\ -3 \\ -5 \end{pmatrix} \propto \begin{pmatrix} 1 \\ 3 \\ 5 \end{pmatrix}\).

Let \(A(1, -1, 2)\) be on \(l_1\) and \(B(2, 3, 0)\) be on \(l_2\).
\(\vec{AB} = \begin{pmatrix} 2-1 \\ 3 - (-1) \\ 0-2 \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix}\).

The shortest distance \(d\) is the projection of \(\vec{AB}\) onto \(\mathbf{n}\):
\[d = \frac{|\vec{AB} \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{|1(1) + 4(3) - 2(5)|}{\sqrt{1^2+3^2+5^2}} = \frac{|1+12-10|}{\sqrt{35}} = \frac{3}{\sqrt{35}} \approx 0.507\]

**(ii)**
The plane \(\Pi\) contains \(l_1\) and is parallel to \(l_2\), so its normal vector is perpendicular to both lines. Thus, \(\mathbf{n} = \begin{pmatrix} 1 \\ 3 \\ 5 \end{pmatrix}\).
Since \(\Pi\) contains \(l_1\), it contains the point \((1, -1, 2)\).
Equation of \(\Pi\):
\[1(x - 1) + 3(y + 1) + 5(z - 2) = 0 \implies x + 3y + 5z = 8\]

**(iii)**
Let \(\Pi_2\) be the plane containing \(l_1\) and the point \(P(2, 3, 0)\).
Since it contains \(l_1\), it contains \(A(1, -1, 2)\) and has direction \(\begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}\).
Another vector in the plane is \(\vec{AP} = \begin{pmatrix} 2-1 \\ 3-(-1) \\ 0-2 \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix}\).

The normal vector to \(\Pi_2\) is:
\[\mathbf{n}_2 = \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} \times \begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix} = \begin{pmatrix} 1(-2) - (-1)(4) \\ (-1)(1) - 2(-2) \\ 2(4) - 1(1) \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\ 7
\end{pmatrix}\]

Let \(\theta\) be the acute angle between \(\Pi\) and \(\Pi_2\):
\[\cos \theta = \frac{|\mathbf{n} \cdot \mathbf{n}_2|}{|\mathbf{n}| |\mathbf{n}_2|} = \frac{|1(2) + 3(3) + 5(7)|}{\sqrt{35} \sqrt{2^2+3^2+7^2}} = \frac{2+9+35}{\sqrt{35}\sqrt{62}} = \frac{46}{\sqrt{2170}} \approx 0.9875\]
\[\theta = \arccos(0.9875) \approx 9.1^\circ \quad \text{(or } 0.158 \text{ radians)}\]

**(iv)**
A general point \(R\) on \(l_1\) has coordinates \((1+2\lambda, -1+\lambda, 2-\lambda)\).
\[\vec{QR} = \begin{pmatrix} 1+2\lambda - 5 \\ -1+\lambda - (-2) \\ 2-\lambda - 4 \end{pmatrix} = \begin{pmatrix} 2\lambda - 4 \\ \lambda + 1 \\ -\lambda - 2 \end{pmatrix}\]
For \(R\) to be the closest point to \(Q\), \(\vec{QR}\) must be perpendicular to the direction of \(l_1\):
\[\vec{QR} \cdot \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} = 0 \implies 2(2\lambda - 4) + 1(\lambda + 1) - 1(-\lambda - 2) = 0\]
\[4\lambda - 8 + \lambda + 1 + \lambda + 2 = 0 \implies 6\lambda - 5 = 0 \implies \lambda = \frac{5}{6}\]
Substituting \(\lambda = \frac{5}{6}\) into the coordinates of \(R\):
\[x = 1 + 2\left(\frac{5}{6}\right) = \frac{8}{3}\]
\[y = -1 + \frac{5}{6} = -\frac{1}{6}\]
\[z = 2 - \frac{5}{6} = \frac{7}{6}\]
Thus, the closest point is \((\frac{8}{3}, -\frac{1}{6}, \frac{7}{6})\).

**(i)**
- **M1**: For attempting to solve the system of equations for intersection and demonstrating inconsistency.
- **A1**: Correctly concluding the lines are skew (including noting non-parallel directions).
- **M1**: For finding the cross product of the direction vectors of \(l_1\) and \(l_2\).
- **A1**: Correct projection formula and substitution.
- **A0.5**: Correct shortest distance of \(\frac{3}{\sqrt{35}}\).

**(ii)**
- **M1**: For identifying that the normal to \(\Pi\) is the vector \(\mathbf{n} = \begin{pmatrix} 1 \\ 3 \\ 5 \end{pmatrix}\) and using point \((1, -1, 2)\).
- **A2**: Correct Cartesian equation: \(x + 3y + 5z = 8\) (deduct 1 mark if not in \(ax+by+cz=d\) form).

**(iii)**
- **M1**: Finding the second direction vector in \(\Pi_2\) (i.e., \(\vec{AP}\)) and attempting the cross product to find normal \(\mathbf{n}_2\).
- **A1**: Correct normal vector \(\mathbf{n}_2 = \begin{pmatrix} 2 \\ 3 \\ 7 \end{pmatrix}\) (or scalar multiple).
- **M1**: Using the scalar product of normals to find \(\cos \theta\).
- **A1**: Correct angle of \(9.1^\circ\) or \(0.158\) rad.

**(iv)**
- **M1**: Expressing \(\vec{QR}\) in terms of parameter \(\lambda\).
- **M1**: Applying the perpendicularity condition \(\vec{QR} \cdot \mathbf{d}_1 = 0\) and solving for \(\lambda\).
- **A1**: Finding \(\lambda = \frac{5}{6}\).
- **A1**: Finding the correct coordinates: \((\frac{8}{3}, -\frac{1}{6}, \frac{7}{6})\) (or equivalent decimals).

We use the double-angle identity for hyperbolic cosine:
\( \cosh(2x) = 1 + 2\sinh^2 x \)

Substitute this into the given equation:
\( 2(1 + 2\sinh^2 x) - 7\sinh x = 4 \)
\( 2 + 4\sinh^2 x - 7\sinh x = 4 \)

Rearranging terms gives a quadratic equation in \( \sinh x \):
\( 4\sinh^2 x - 7\sinh x - 2 = 0 \)

Factorising this quadratic equation:
\( (4\sinh x + 1)(\sinh x - 2) = 0 \)

This yields two possible values for \( \sinh x \):
\( \sinh x = 2 \quad \text{or} \quad \sinh x = -\frac{1}{4} \)

Using the logarithmic definition of inverse hyperbolic sine, \( \text{arsinh } u = \ln\left(u + \sqrt{u^2 + 1}\right) \):

For \( \sinh x = 2 \):
\( x = \ln\left(2 + \sqrt{2^2 + 1}\right) = \ln(2 + \sqrt{5}) \)

For \( \sinh x = -\frac{1}{4} \):
\( x = \ln\left(-\frac{1}{4} + \sqrt{\left(-\frac{1}{4}\right)^2 + 1}\right) = \ln\left(-\frac{1}{4} + \sqrt{\frac{17}{16}}\right) = \ln\left(\frac{\sqrt{17}-1}{4}\right) \)

M1: Use the identity \( \cosh(2x) = 1 + 2\sinh^2 x \) to form an equation in terms of \( \sinh x \) only.
A1: Obtain the correct quadratic equation \( 4\sinh^2 x - 7\sinh x - 2 = 0 \).
M1: Solve the quadratic equation to find two values of \( \sinh x \) (e.g., \( 2 \) and \( -0.25 \)).
M1: Apply the logarithmic definition of \( \text{arsinh } u \) to at least one of their values.
A1: Obtain the correct exact value \( x = \ln(2 + \sqrt{5}) \).
A1: Obtain the correct exact value \( x = \ln\left(\frac{\sqrt{17}-1}{4}\right) \).

We find the eigenvalues of \( \mathbf{A} \) by solving the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \):
\( \det\begin{pmatrix} 3-\lambda & -1 & 1 \\\\ 0 & 2-\lambda & 0 \\\\ 1 & -1 & 3-\lambda \end{pmatrix} = 0 \)

Expanding along the second row:
\( (2-\lambda)\det\begin{pmatrix} 3-\lambda & 1 \\\\ 1 & 3-\lambda \end{pmatrix} = 0 \)
\( (2-\lambda)\left((3-\lambda)^2 - 1\right) = 0 \)
\( (2-\lambda)(\lambda^2 - 6\lambda + 8) = 0 \)
\( (2-\lambda)(\lambda-2)(\lambda-4) = 0 \)

Thus, the eigenvalues are \( \lambda = 2 \) (with multiplicity 2) and \( \lambda = 4 \).

Next, we find the eigenvectors.

For \( \lambda = 4 \):
\( (\mathbf{A} - 4\mathbf{I})\mathbf{v} = \mathbf{0} \implies \begin{pmatrix} -1 & -1 & 1 \\\\ 0 & -2 & 0 \\\\ 1 & -1 & -1 \end{pmatrix} \begin{pmatrix} x \\\\ y \\\\ z \end{pmatrix} = \begin{pmatrix} 0 \\\\ 0 \\\\ 0 \end{pmatrix} \)
From the second row, \( -2y = 0 \implies y = 0 \).
From the first row, \( -x + z = 0 \implies x = z \).
So an eigenvector for \( \lambda = 4 \) is \( \mathbf{v}_1 = \begin{pmatrix} 1 \\\\ 0 \\\\ 1 \end{pmatrix} \).

For \( \lambda = 2 \):
\( (\mathbf{A} - 2\mathbf{I})\mathbf{v} = \mathbf{0} \implies \begin{pmatrix} 1 & -1 & 1 \\\\ 0 & 0 & 0 \\\\ 1 & -1 & 1 \end{pmatrix} \begin{pmatrix} x \\\\ y \\\\ z \end{pmatrix} = \begin{pmatrix} 0 \\\\ 0 \\\\ 0 \end{pmatrix} \)
This yields the single independent equation:
\( x - y + z = 0 \)
We can choose any two linearly independent vectors satisfying this plane equation.
Choosing \( y = 1, z = 0 \) gives \( x = 1 \), so \( \mathbf{v}_2 = \begin{pmatrix} 1 \\\\ 1 \\\\ 0 \end{pmatrix} \).
Choosing \( y = 0, z = 1 \) gives \( x = -1 \), so \( \mathbf{v}_3 = \begin{pmatrix} -1 \\\\ 0 \\\\ 1 \end{pmatrix} \).

We construct \( \mathbf{D} \) from the eigenvalues and \( \mathbf{P} \) from the corresponding eigenvectors in the same order:
\( \mathbf{D} = \begin{pmatrix} 2 & 0 & 0 \\\\ 0 & 2 & 0 \\\\ 0 & 0 & 4 \end{pmatrix} \quad \text{and} \quad \mathbf{P} = \begin{pmatrix} 1 & -1 & 1 \\\\ 1 & 0 & 0 \\\\ 0 & 1 & 1 \end{pmatrix} \)

Other valid orders of eigenvalues and their corresponding eigenvectors, or scalar multiples of the eigenvectors, are also correct.

M1: Set up the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \).
A1: Obtain the correct eigenvalues \( \lambda = 2, 2, 4 \).
A1: Find a correct eigenvector corresponding to \( \lambda = 4 \).
M1: Use the equation \( x - y + z = 0 \) to find two independent eigenvectors for \( \lambda = 2 \).
A1: Obtain two correct linearly independent eigenvectors for \( \lambda = 2 \).
A1: Correctly state a consistent pair of matrices \( \mathbf{D} \) and \( \mathbf{P} \).

Let \( z = \cos\theta + \text{i}\sin\theta \).
Then \( z - z^{-1} = 2\text{i}\sin\theta \) and \( z^n - z^{-n} = 2\text{i}\sin(n\theta) \).

Using the binomial theorem on \( (2\text{i}\sin\theta)^5 = (z - z^{-1})^5 \):
\( 32\text{i}\sin^5\theta = z^5 - 5z^3 + 10z - 10z^{-1} + 5z^{-3} - z^{-5} \)

Grouping terms with reciprocal powers:
\( 32\text{i}\sin^5\theta = (z^5 - z^{-5}) - 5(z^3 - z^{-3}) + 10(z - z^{-1}) \)

Substitute the identity \( z^n - z^{-n} = 2\text{i}\sin(n\theta) \):
\( 32\text{i}\sin^5\theta = 2\text{i}\sin(5\theta) - 10\text{i}\sin(3\theta) + 20\text{i}\sin\theta \)

Dividing through by \( 32\text{i} \) yields the desired identity:
\( \sin^5\theta = \frac{1}{16}\sin(5\theta) - \frac{5}{16}\sin(3\theta) + \frac{5}{8}\sin\theta \)

Now, integrate this expression over the interval \( \left[0, \frac{\pi}{3}\right] \):
\( \int_{0}^{\frac{\pi}{3}} \sin^5 \theta \, \mathrm{d}\theta = \int_{0}^{\frac{\pi}{3}} \left( \frac{1}{16}\sin(5\theta) - \frac{5}{16}\sin(3\theta) + \frac{5}{8}\sin\theta \right) \mathrm{d}\theta \)
\( = \left[ -\frac{1}{80}\cos(5\theta) + \frac{5}{48}\cos(3\theta) - \frac{5}{8}\cos\theta \right]_{0}^{\frac{\pi}{3}} \)

Evaluating at the upper limit \( \theta = \frac{\pi}{3} \):
\( -\frac{1}{80}\cos\left(\frac{5\pi}{3}\right) + \frac{5}{48}\cos(\pi) - \frac{5}{8}\cos\left(\frac{\pi}{3}\right) = -\frac{1}{160} - \frac{5}{48} - \frac{5}{16} = -\frac{203}{480} \)

Evaluating at the lower limit \( \theta = 0 \):
\( -\frac{1}{80}\cos(0) + \frac{5}{48}\cos(0) - \frac{5}{8}\cos(0) = -\frac{1}{80} + \frac{5}{48} - \frac{5}{8} = -\frac{256}{480} \)

Subtracting the lower limit value from the upper limit value:
\( -\frac{203}{480} - \left(-\frac{256}{480}\right) = \frac{53}{480} \)

M1: Express \( \sin\theta \) in terms of \( z \) and \( z^{-1} \) and expand \( (z - z^{-1})^5 \) binomially.
A1: Correctly group terms to prove the identity \( \sin^5\theta = \frac{1}{16}\sin(5\theta) - \frac{5}{16}\sin(3\theta) + \frac{5}{8}\sin\theta \).
M1: Integrate the trigonometric terms (showing understanding that \( \int \sin(k\theta) \, \mathrm{d}\theta = -\frac{1}{k}\cos(k\theta) \)).
A1ft: Correct integrated expression: \( -\frac{1}{80}\cos(5\theta) + \frac{5}{48}\cos(3\theta) - \frac{5}{8}\cos\theta \).
M1: Substitute the limits \( \frac{\pi}{3} \) and \( 0 \) correctly.
A1: Obtain the exact value \( \frac{53}{480} \).

(i) Using the definitions of hyperbolic functions:
\(\cosh x = \frac{e^x + e^{-x}}{2}\) and \(\sinh x = \frac{e^x - e^{-x}}{2}\).

Substitute these into the expression:
\[ f(x) = 3 \left(\frac{e^x + e^{-x}}{2}\right) + 5 \left(\frac{e^x - e^{-x}}{2}\right) \]
\[ f(x) = \frac{3e^x + 3e^{-x} + 5e^x - 5e^{-x}}{2} \]
\[ f(x) = \frac{8e^x - 2e^{-x}}{2} = 4e^x - e^{-x} \]

(ii) Now we evaluate the integral:
\[ I = \int_0^{\ln 2} \frac{1}{4e^x - e^{-x}} \text{d}x = \int_0^{\ln 2} \frac{e^x}{4e^{2x} - 1} \text{d}x \]

Using the substitution \( u = e^x \), we have \( \text{d}u = e^x \text{d}x \).
When \( x = 0 \), \( u = e^0 = 1 \).
When \( x = \ln 2 \), \( u = e^{\ln 2} = 2 \).

The integral becomes:
\[ I = \int_1^2 \frac{1}{4u^2 - 1} \text{d}u \]

Using partial fractions:
\[ \frac{1}{4u^2 - 1} = \frac{1}{(2u-1)(2u+1)} = \frac{1}{2} \left( \frac{1}{2u-1} - \frac{1}{2u+1} \right) \]

So:
\[ I = \left[ \frac{1}{4} \ln|2u-1| - \frac{1}{4} \ln|2u+1| \right]_1^2 = \left[ \frac{1}{4} \ln \left| \frac{2u-1}{2u+1} \right| \right]_1^2 \]

Substitute the upper limit \( u = 2 \):
\[ \frac{1}{4} \ln \left( \frac{3}{5} \right) \]

Substitute the lower limit \( u = 1 \):
\[ \frac{1}{4} \ln \left( \frac{1}{3} \right) \]

Subtract the limits:
\[ I = \frac{1}{4} \ln \left( \frac{3}{5} \right) - \frac{1}{4} \ln \left( \frac{1}{3} \right) = \frac{1}{4} \ln \left( \frac{3/5}{1/3} \right) = \frac{1}{4} \ln \left( \frac{9}{5} \right) \]

(i)
- **M1**: Substituting exponential definitions of \(\cosh x\) and \(\sinh x\).
- **A1**: Correctly simplifying to \(4e^x - e^{-x}\).

(ii)
- **M1**: Applying substitution \(u = e^x\) to convert the integral.
- **A1**: Finding correct new integration limits \(u = 1\) and \(u = 2\).
- **M1**: Performing partial fraction decomposition or utilizing standard formula \(\int \frac{1}{u^2-a^2} \text{d}u\).
- **A1**: Obtaining the correct integrated form, e.g., \(\frac{1}{4} \ln \left| \frac{2u-1}{2u+1} \right|\).
- **M1**: Substituting limits and using logarithm properties to combine the terms.
- **A1**: Reaching the final simplified exact value of \(\frac{1}{4} \ln \frac{9}{5}\).

(i) We solve the characteristic equation \(\det(\mathbf{A} - \lambda \mathbf{I}) = 0\):
\[ \begin{vmatrix} 2-\lambda & 1 & -1 \\ 0 & 3-\lambda & 0 \\ 0 & 1 & 1-\lambda \end{vmatrix} = 0 \]

Expanding along the first column:
\[ (2-\lambda) \begin{vmatrix} 3-\lambda & 0 \\ 1 & 1-\lambda \end{vmatrix} = (2-\lambda)(3-\lambda)(1-\lambda) = 0 \]
Thus, the eigenvalues are \( \lambda = 1 \), \( \lambda = 2 \), and \( \lambda = 3 \).

(ii) For \( \lambda = 1 \):
\[ (\mathbf{A} - \mathbf{I})\mathbf{x} = \mathbf{0} \implies \begin{pmatrix} 1 & 1 & -1 \\ 0 & 2 & 0 \\ 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \]
This gives \( y = 0 \) and \( x - z = 0 \implies x = z \). An eigenvector is:
\[ \mathbf{e}_1 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} \]

For \( \lambda = 2 \):
\[ (\mathbf{A} - 2\mathbf{I})\mathbf{x} = \mathbf{0} \implies \begin{pmatrix} 0 & 1 & -1 \\ 0 & 1 & 0 \\ 0 & 1 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \]
This gives \( y = 0 \) and \( y - z = 0 \implies z = 0 \), with \( x \) free. An eigenvector is:
\[ \mathbf{e}_2 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \]

For \( \lambda = 3 \):
\[ (\mathbf{A} - 3\mathbf{I})\mathbf{x} = \mathbf{0} \implies \begin{pmatrix} -1 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 1 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \]
From the third row, \( y = 2z \).
From the first row, \( -x + y - z = 0 \implies -x + 2z - z = 0 \implies x = z \).
Setting \( z = 1 \), we have \( x = 1, y = 2 \). An eigenvector is:
\[ \mathbf{e}_3 = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} \]

(iii) We construct the matrices:
\[ \mathbf{P} = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 0 & 2 \\ 1 & 0 & 1 \end{pmatrix}, \quad \mathbf{D} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix} \]

To find \( \mathbf{P}^{-1} \), we calculate the determinant and adjugate matrix:
\[ \det(\mathbf{P}) = 1(0 - 0) - 1(0 - 2) + 1(0 - 0) = 2 \]
The cofactor matrix \( \mathbf{C} \) is:
\[ \mathbf{C} = \begin{pmatrix} 0 & 2 & 0 \\ -1 & 0 & 1 \\ 2 & -2 & 0 \end{pmatrix} \]

The adjugate is:
\[ \mathbf{C}^T = \begin{pmatrix} 0 & -1 & 2 \\ 2 & 0 & -2 \\ 0 & 1 & 0 \end{pmatrix} \]

Thus:
\[ \mathbf{P}^{-1} = \frac{1}{2} \begin{pmatrix} 0 & -1 & 2 \\ 2 & 0 & -2 \\ 0 & 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -0.5 & 1 \\ 1 & 0 & -1 \\ 0 & 0.5 & 0 \end{pmatrix} \]

(i)
- **M1**: Attempting to find characteristic equation \( \det(\mathbf{A}-\lambda \mathbf{I}) = 0 \).
- **A1**: Finding eigenvalues \( 1, 2, 3 \).

(ii)
- **M1**: Setting up equations to find eigenvectors for each eigenvalue.
- **A2**: Finding three correct eigenvectors (1 mark for any two correct, 2 marks for all three correct).

(iii)
- **B1**: Identifying a valid matrix \( \mathbf{P} \) and its corresponding diagonal matrix \( \mathbf{D} \).
- **M1**: An appropriate method to calculate the inverse of a \( 3 \times 3 \) matrix (e.g., adjugate method or row operations).
- **A1**: Finding the correct inverse matrix \( \mathbf{P}^{-1} \).

(i) The auxiliary equation is:
\[ m^2 - 4m + 4 = 0 \implies (m-2)^2 = 0 \]
This has a repeated root \( m = 2 \).
Thus, the complementary function is:
\[ y_c = (A + Bx)e^{2x} \]

(ii) The right-hand side is \( 6e^{2x} + 25\cos x \).
Since 2 is a double root of the auxiliary equation, the trial solution for the exponential term must be \( C x^2 e^{2x} \).
For the trigonometric term, the trial solution is \( D\cos x + E\sin x \).
So, we let:
\[ y_p = C x^2 e^{2x} + D\cos x + E\sin x \]

First, let's find the derivatives for \( y_{p1} = C x^2 e^{2x} \):
\[ y'_{p1} = 2Cx e^{2x} + 2Cx^2 e^{2x} \]
\[ y''_{p1} = 2C e^{2x} + 8Cx e^{2x} + 4Cx^2 e^{2x} \]

Substituting \( y_{p1} \) into the LHS:
\[ y''_{p1} - 4y'_{p1} + 4y_{p1} = e^{2x} (2C + 8Cx + 4Cx^2 - 8Cx - 8Cx^2 + 4Cx^2) = 2C e^{2x} \]
We set this equal to the first term on the RHS: \( 2C e^{2x} = 6e^{2x} \implies C = 3 \).

Now, let's find the derivatives for \( y_{p2} = D\cos x + E\sin x \):
\[ y'_{p2} = -D\sin x + E\cos x \]
\[ y''_{p2} = -D\cos x - E\sin x \]

Substituting \( y_{p2} \) into the LHS:
\[ (-D\cos x - E\sin x) - 4(-D\sin x + E\cos x) + 4(D\cos x + E\sin x) = 25\cos x \]
Grouping like terms:
\[ (3D - 4E)\cos x + (4D + 3E)\sin x = 25\cos x \]

This gives the simultaneous equations:
1) \( 3D - 4E = 25 \)
2) \( 4D + 3E = 0 \implies D = -\frac{3}{4}E \)

Substitute (2) into (1):
\[ 3\left(-\frac{3}{4}E\right) - 4E = 25 \implies -\frac{25}{4}E = 25 \implies E = -4 \]
Then \( D = -\frac{3}{4}(-4) = 3 \).

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

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

Using the first initial condition \( y(0) = 3 \):
\[ 3 = A(1) + 0 + 3(1) - 0 \implies A = 0 \]

So the solution simplifies to:
\[ y = Bx e^{2x} + 3x^2 e^{2x} + 3\cos x - 4\sin x \]

Now find the derivative \( \frac{\text{d}y}{\text{d}x} \):
\[ \frac{\text{d}y}{\text{d}x} = B e^{2x} + 2Bx e^{2x} + 6x e^{2x} + 6x^2 e^{2x} - 3\sin x - 4\cos x \]

Using the second initial condition \( \frac{\text{d}y}{\text{d}x} = -2 \) when \( x = 0 \):
\[ -2 = B(1) + 0 + 0 + 0 - 0 - 4(1) \implies B - 4 = -2 \implies B = 2 \]

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

(i)
- **M1**: Reaching correct auxiliary equation quadratic form.
- **A1**: Finding correct complementary function \( y_c = (A + Bx)e^{2x} \).

(ii)
- **M1**: Setting up correct forms of particular integrals: \( C x^2 e^{2x} \) and \( D\cos x + E\sin x \).
- **M1**: Substituting \( y_{p1} \) or \( y_{p2} \) and differentiating to set up equations.
- **A1**: Finding \( C = 3 \).
- **A1**: Finding \( D = 3 \) and \( E = -4 \).

(iii)
- **M1**: Stating general solution and substituting \( x = 0 \) and \( y = 3 \) to find \( A \).
- **M1**: Differentiating general solution and substituting \( x = 0 \) and \( y' = -2 \) to find \( B \).
- **A1**: Writing down the correct final particular solution \( y = (2x + 3x^2)e^{2x} + 3\cos x - 4\sin x \).

(a) The auxiliary equation is \(m^2 + 2m + 5 = 0\). Solving this quadratic equation gives:
\[m = \frac{-2 \pm \sqrt{4 - 20}}{2} = -1 \pm 2i\]
Thus, the complementary function (CF) is:
\[y_{CF} = e^{-x}(A\cos(2x) + B\sin(2x))\]

For the particular integral (PI), we propose:
\[y_{PI} = p\sin(2x) + q\cos(2x) + r e^{-x}\]
Differentiating:
\[y_{PI}' = 2p\cos(2x) - 2q\sin(2x) - r e^{-x}\]
\[y_{PI}'' = -4p\sin(2x) - 4q\cos(2x) + r e^{-x}\]

Substituting these into the original differential equation:
\[(-4p\sin(2x) - 4q\cos(2x) + r e^{-x}) + 2(2p\cos(2x) - 2q\sin(2x) - r e^{-x}) + 5(p\sin(2x) + q\cos(2x) + r e^{-x}) = 17\sin(2x) + 8e^{-x}\]

Grouping terms:
\[(p - 4q)\sin(2x) + (4p + q)\cos(2x) + 4re^{-x} = 17\sin(2x) + 8e^{-x}\]

Equating coefficients:
For \(e^{-x}\): \(4r = 8 \implies r = 2\).
For \(\cos(2x)\): \(4p + q = 0 \implies q = -4p\).
For \(\sin(2x)\): \(p - 4q = 17 \implies p - 4(-4p) = 17 \implies 17p = 17 \implies p = 1\), which gives \(q = -4\).

Thus, the particular integral is:
\[y_{PI} = \sin(2x) - 4\cos(2x) + 2e^{-x}\]

Combining the CF and PI, the general solution is:
\[y = e^{-x}(A\cos(2x) + B\sin(2x)) + \sin(2x) - 4\cos(2x) + 2e^{-x}\]

(b) Using the initial condition \(y = 0\) when \(x = 0\):
\[0 = 1(A + 0) + 0 - 4 + 2 \implies A = 2\]

Now, differentiating the general solution:
\[\frac{dy}{dx} = -e^{-x}(A\cos(2x) + B\sin(2x)) + e^{-x}(-2A\sin(2x) + 2B\cos(2x)) + 2\cos(2x) + 8\sin(2x) - 2e^{-x}\]

Using the boundary condition \(\frac{dy}{dx} = 0\) when \(x = 0\):
\[0 = -1(A) + 1(2B) + 2 - 2 \implies 0 = -A + 2B\]
Since \(A = 2\), we have \(2B = 2 \implies B = 1\).

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

(a)
- M1: Attempt to solve the auxiliary equation \(m^2 + 2m + 5 = 0\).
- A1: Find correct roots \(m = -1 \pm 2i\).
- A1: State correct complementary function \(y_{CF} = e^{-x}(A\cos 2x + B\sin 2x)\).
- B1: Identify the correct form of the PI: \(y_{PI} = p\sin 2x + q\cos 2x + r e^{-x}\).
- M1: Differentiate the PI twice and substitute into the DE.
- A1: Obtain correct simplified expression in terms of coefficients, leading to equations for \(p, q, r\).
- A1: Correctly calculate \(r = 2\).
- A1: Correctly calculate \(p = 1\) and \(q = -4\).
- A1: State the general solution correctly.

(b)
- M1: Substitute \(x = 0, y = 0\) into the general solution to obtain a value for \(A\).
- A1: Correctly find \(A = 2\).
- M1: Correctly apply the product rule to differentiate the general solution.
- A1: Substitute \(x = 0, y' = 0\) and obtain \(B = 1\).
- A1: State the correct particular solution.

(a) Let \(y = \sinh^{-1} x\). Then \(\sinh y = x\).
Differentiating both sides with respect to \(x\):
\[\cosh y \frac{dy}{dx} = 1 \implies \frac{dy}{dx} = \frac{1}{\cosh y}\]
Using the identity \(\cosh^2 y - \sinh^2 y = 1\), we obtain \(\cosh y = \sqrt{1 + \sinh^2 y} = \sqrt{1+x^2}\) (since \(\cosh y > 0\) for all real \(y\)).
Thus, \(\frac{dy}{dx} = \frac{1}{\sqrt{x^2+1}}\).

(b) Complete the square for the quadratic terms:
\[4x^2 + 4x + 5 = 4(x^2 + x) + 5 = 4(x + 1/2)^2 + 4 = (2x+1)^2 + 4\]
Let \(u = 2x+1\), then \(du = 2 \, dx \implies dx = \frac{1}{2} du\).
For the limits of integration:
When \(x = 0\), \(u = 1\).
When \(x = 1/2\), \(u = 2\).
Thus, the integral becomes:
\[\frac{1}{2} \int_{1}^{2} \frac{1}{\sqrt{u^2+4}} \, du = \frac{1}{2} \left[ \sinh^{-1}\left(\frac{u}{2}\right) \right]_{1}^{2}\]
Using the logarithmic definition \(\sinh^{-1}(z) = \ln(z + \sqrt{z^2+1})\):
At \(u=2\): \(\sinh^{-1}(1) = \ln(1+\sqrt{2})\).
At \(u=1\): \(\sinh^{-1}(1/2) = \ln\left(\frac{1}{2} + \sqrt{\frac{1}{4}+1}\right) = \ln\left(\frac{1+\sqrt{5}}{2}\right)\).
So the integral evaluates to:
\[\frac{1}{2} \left[ \ln(1+\sqrt{2}) - \ln\left(\frac{1+\sqrt{5}}{2}\right) \right] = \frac{1}{2} \ln\left( \frac{2(1+\sqrt{2})}{1+\sqrt{5}} \right)\]

(c) We express the numerator in terms of the derivative of the denominator's quadratic expression, which is \(8x+4\):
\[x = \frac{1}{8}(8x+4) - \frac{1}{2}\]
So the integral is:
\[\int_{0}^{1/2} \frac{\frac{1}{8}(8x+4) - \frac{1}{2}}{\sqrt{4x^2+4x+5}} \, dx = \frac{1}{8} \int_{0}^{1/2} (8x+4)(4x^2+4x+5)^{-1/2} \, dx - \frac{1}{2} \int_{0}^{1/2} \frac{1}{\sqrt{4x^2+4x+5}} \, dx\]
The first term integrates to:
\[\frac{1}{8} \left[ 2\sqrt{4x^2+4x+5} \right]_{0}^{1/2} = \frac{1}{4} \left[ \sqrt{4(1/4)+4(1/2)+5} - \sqrt{5} \right] = \frac{2\sqrt{2} - \sqrt{5}}{4}\]
The second term is \(-\frac{1}{2} I\), where \(I\) is the integral from part (b).
Thus, the second term is:
\[-\frac{1}{4} \ln\left( \frac{2(1+\sqrt{2})}{1+\sqrt{5}} \right)\]
Combining the terms yields the exact value:
\[\frac{2\sqrt{2} - \sqrt{5}}{4} - \frac{1}{4} \ln \left( \frac{2(1+\sqrt{2})}{1+\sqrt{5}} \right)\]

(a)
- M1: Use substitution \(y = \sinh^{-1}x\) to write \(\sinh y = x\).
- M1: Perform implicit differentiation with respect to \(x\).
- M1: Substitute \(\cosh y = \sqrt{1+\sinh^2 y}\).
- A1: Fully deduce the given expression.

(b)
- B1: Complete the square to get \((2x+1)^2+4\).
- M1: Use substitution \(u = 2x+1\) and adjust limits to \(1\) and \(2\).
- A1: Find the correct integral \(\frac{1}{2}\left[\sinh^{-1}(u/2)\right]\).
- M1: Apply the logarithmic form of inverse hyperbolic sine to the limits.
- A1: Show the given answer with detailed algebraic steps.

(c)
- M1: Split the integrand into a derivative term and a constant term.
- A1: Obtain \(\frac{1}{4}\sqrt{4x^2+4x+5}\) as the antiderivative of the first part.
- A1: Calculate the value of the first part as \(\frac{2\sqrt{2}-\sqrt{5}}{4}\).
- M1: Relate the second part of the split integral to the result of part (b).
- A1: State the final combined exact value.