2023 Cambridge IAS-Level Mathematics - Further (9231) 模擬試題連答案詳解

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

**Base case:**
For \(n=1\):
LHS \(= (3(1)-2)2^{1-1} = 1 \cdot 2^0 = 1\).
RHS \(= (3(1)-5)2^1 + 5 = -2(2) + 5 = -4 + 5 = 1\).
Since LHS = RHS, the statement \(P(1)\) is true.

**Inductive step:**
Assume \(P(k)\) is true for some positive integer \(k\), so
\[ \sum_{r=1}^{k} (3r-2)2^{r-1} = (3k-5)2^k + 5. \]

Now consider the sum for \(n=k+1\):
\[ \sum_{r=1}^{k+1} (3r-2)2^{r-1} = \left[ \sum_{r=1}^{k} (3r-2)2^{r-1} \right] + (3(k+1)-2)2^{(k+1)-1} \]
\[ = (3k-5)2^k + 5 + (3k+1)2^k \]
\[ = 2^k [ (3k-5) + (3k+1) ] + 5 \]
\[ = 2^k [ 6k - 4 ] + 5 \]
\[ = 2^k \cdot 2(3k-2) + 5 \]
\[ = (3k-2)2^{k+1} + 5 \]
\[ = (3(k+1)-5)2^{k+1} + 5. \]

This is the form of the formula with \(n = k+1\).

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

M1: Verify the base case \(n=1\).
M1: State inductive hypothesis for \(n=k\).
M1: Attempt to find the sum for \(n=k+1\) by adding the \((k+1)\)-th term to the assumed sum.
A1: Correctly factor out \(2^k\) or \(2^{k+1}\) from the terms.
A1: Correct algebraic simplification to obtain \((3k-2)2^{k+1} + 5\).
A1: Express the simplified form in terms of \(k+1\), i.e., \((3(k+1)-5)2^{k+1} + 5\).
A1: Complete the proof with a clear concluding statement referencing mathematical induction.

Let \(\begin{pmatrix} x' \\ y' \end{pmatrix}\) be the image of \(\begin{pmatrix} x \\ y \end{pmatrix}\) under the transformation \(\mathbf{A}\).

We have:
\[ \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ 1 & 4 \end{pmatrix} \begin{pmatrix} x \\ mx+c \end{pmatrix} = \begin{pmatrix} 2x - (mx+c) \\ x + 4(mx+c) \end{pmatrix} \]

This gives the parametric equations:
1) \(x' = (2-m)x - c \implies x = \frac{x'+c}{2-m}\) (assuming \(m \neq 2\))
2) \(y' = (1+4m)x + 4c\)

Substitute \(x\) into the equation for \(y'\):
\[ y' = (1+4m)\frac{x'+c}{2-m} + 4c \]
\[ y' = \left( \frac{1+4m}{2-m} \right) x' + \frac{c(1+4m)}{2-m} + 4c \]

We are given that the image line is \(y' = 2x' - 5\). Comparing coefficients:

For the gradient:
\[ \frac{1+4m}{2-m} = 2 \implies 1+4m = 4-2m \implies 6m = 3 \implies m = \frac{1}{2} \]

For the intercept:
\[ \frac{c(1+4m)}{2-m} + 4c = -5 \]

Substituting \(m = \frac{1}{2}\):
\[ \frac{c(1+2)}{2-0.5} + 4c = -5 \implies \frac{3c}{1.5} + 4c = -5 \implies 2c + 4c = -5 \implies 6c = -5 \implies c = -\frac{5}{6} \]

M1: Write down the matrix multiplication \(\mathbf{A}\begin{pmatrix} x \\ mx+c \end{pmatrix}\).
A1: Correct expressions for \(x'\) and \(y'\) in terms of \(x, m, c\).
M1: Express \(x\) in terms of \(x'\).
A1: Obtain the equation of the image line in terms of \(x'\) and \(y'\).
M1: Set the coefficient of \(x'\) equal to 2 and solve for \(m\).
A1: Correct value of \(m = 1/2\).
M1: Set the constant term equal to \(-5\) and substitute \(m\).
A1: Correct value of \(c = -5/6\).

(i) From the given equation, we have:
\(\sum \alpha = 3\)
\(\sum \alpha\beta = 5\)
\(\alpha\beta\gamma = 4\)

We calculate \(\alpha^2 + \beta^2 + \gamma^2\):
\[ \alpha^2 + \beta^2 + \gamma^2 = (\sum \alpha)^2 - 2\sum \alpha\beta = 3^2 - 2(5) = 9 - 10 = -1. \]

To find \(\sum \alpha^3\), we use the relation \(\alpha^3 - 3\alpha^2 + 5\alpha - 4 = 0\). Summing over the three roots:
\[ \sum \alpha^3 - 3\sum \alpha^2 + 5\sum \alpha - 12 = 0 \]
\[ \sum \alpha^3 - 3(-1) + 5(3) - 12 = 0 \]
\[ \sum \alpha^3 + 3 + 15 - 12 = 0 \implies \sum \alpha^3 = -6. \]

(ii) Let the roots of the new equation be \(u = \frac{1}{\alpha^2}\), \(v = \frac{1}{\beta^2}\), \(w = \frac{1}{\gamma^2}\).

First, find \(u+v+w\):
\[ u+v+w = \frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} = \frac{\alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2}{\alpha^2\beta^2\gamma^2} \]
We know \(\alpha^2\beta^2\gamma^2 = (\alpha\beta\gamma)^2 = 16\).
For the numerator:
\[ \sum \alpha^2\beta^2 = (\sum \alpha\beta)^2 - 2\alpha\beta\gamma\sum \alpha = 5^2 - 2(4)(3) = 25 - 24 = 1. \]
Thus, \(u+v+w = \frac{1}{16}\).

Second, find \(uv+vw+wu\):
\[ uv+vw+wu = \frac{1}{\alpha^2\beta^2} + \frac{1}{\beta^2\gamma^2} + \frac{1}{\gamma^2\alpha^2} = \frac{\alpha^2 + \beta^2 + \gamma^2}{\alpha^2\beta^2\gamma^2} = \frac{-1}{16}. \]

Third, find \(uvw\):
\[ uvw = \frac{1}{\alpha^2\beta^2\gamma^2} = \frac{1}{16}. \]

The cubic equation is given by:
\[ y^3 - (u+v+w)y^2 + (uv+vw+wu)y - uvw = 0 \]
\[ y^3 - \frac{1}{16}y^2 - \frac{1}{16}y - \frac{1}{16} = 0 \]
Multiplying by 16 to obtain integer coefficients:
\[ 16y^3 - y^2 - y - 1 = 0. \]

M1: State correct values of sum of roots, sum of products, and product of roots.
M1: Use formula for \(\sum \alpha^2\) and substitute values.
A1: Correct value \(-1\).
M1: Use the cubic equation to find \(\sum \alpha^3\).
A1: Correct value \(-6\).
M1: Calculate \(\sum \alpha^2\beta^2\) using correct formula.
A1: Correct value of \(1\).
M1: Find expressions for new coefficients in terms of \(\alpha, \beta, \gamma\).
A1: Correct values for new coefficients.
A1: Correct final cubic equation with integer coefficients.

(i) Combining the fractions on the LHS:
\[ \frac{1}{(2r-1)(2r+1)(2r+3)} - \frac{1}{(2r+1)(2r+3)(2r+5)} \]
\[ = \frac{(2r+5) - (2r-1)}{(2r-1)(2r+1)(2r+3)(2r+5)} \]
\[ = \frac{6}{(2r-1)(2r+1)(2r+3)(2r+5)}. \]

(ii) Let \(f(r) = \frac{1}{(2r-1)(2r+1)(2r+3)}\). Then from part (i):
\[ \frac{1}{(2r-1)(2r+1)(2r+3)(2r+5)} = \frac{1}{6}[f(r) - f(r+1)] \]

Summing from \(r=1\) to \(n\):
\[ \sum_{r=1}^n \frac{1}{(2r-1)(2r+1)(2r+3)(2r+5)} = \frac{1}{6} \sum_{r=1}^n [f(r) - f(r+1)] \]
\[ = \frac{1}{6} [f(1) - f(2) + f(2) - f(3) + \dots + f(n) - f(n+1)] \]
\[ = \frac{1}{6} [f(1) - f(n+1)] \]

We have:
\[ f(1) = \frac{1}{1 \cdot 3 \cdot 5} = \frac{1}{15} \]
\[ f(n+1) = \frac{1}{(2n+1)(2n+3)(2n+5)} \]

Thus, the sum is:
\[ \frac{1}{6} \left[ \frac{1}{15} - \frac{1}{(2n+1)(2n+3)(2n+5)} \right] = \frac{1}{90} - \frac{1}{6(2n+1)(2n+3)(2n+5)} \]

(iii) As \(n \to \infty\), \(\frac{1}{6(2n+1)(2n+3)(2n+5)} \to 0\).
Therefore, the sum to infinity is \(\frac{1}{90}\).

M1: Use common denominator to combine the fractions in (i).
A1: Obtain the given numerator 6 and complete the proof of (i).
M1: Set up the method of differences using the result of (i).
A1: Correctly write the general term in terms of \(f(r) - f(r+1)\).
M1: Show the cancellation of intermediate terms.
A1: Correctly evaluate \(f(1) = 1/15\).
A1: Correct sum in terms of \(n\).
M1: Take the limit as \(n \to \infty\).
A1: Correct sum to infinity of \(1/90\).

(i) The curve is a lima\u00e7on with an inner loop. It is symmetrical about the initial line \(\theta = 0\).
At \(\theta = 0\), \(r = 3a\).
At \(\theta = \pi/2\), \(r = a\).
At \(\theta = 2\pi/3\), \(r = 0\).
At \(\theta = \pi\), \(r = -a\).
The sketch should show a larger outer loop and an inner loop lying to the left of the pole, passing through the origin at \(\theta = 2\pi/3\) and \(\theta = 4\pi/3\).

(ii) The inner loop corresponds to values of \(\theta\) where \(r \le 0\).
\(1 + 2\cos\theta \le 0 \implies \cos\theta \le -\frac{1}{2} \implies \frac{2\pi}{3} \le \theta \le \frac{4\pi}{3}\).

The area \(A\) of the inner loop is:
\[ A = \frac{1}{2} \int_{2\pi/3}^{4\pi/3} r^2 d\theta \]
By symmetry:
\[ A = \int_{2\pi/3}^{\pi} a^2 (1 + 2\cos\theta)^2 d\theta \]
\[ = a^2 \int_{2\pi/3}^{\pi} (1 + 4\cos\theta + 4\cos^2\theta) d\theta \]
Using \(\cos^2\theta = \frac{1}{2}(1 + \cos 2\theta)\):
\[ 1 + 4\cos\theta + 4\cos^2\theta = 1 + 4\cos\theta + 2(1 + \cos 2\theta) = 3 + 4\cos\theta + 2\cos 2\theta \]

Now integrate:
\[ A = a^2 \left[ 3\theta + 4\sin\theta + \sin 2\theta \right]_{2\pi/3}^{\pi} \]
At the upper limit \(\theta = \pi\):
\[ 3\pi + 4\sin\pi + \sin 2\pi = 3\pi \]
At the lower limit \(\theta = 2\pi/3\):
\[ 3\left(\frac{2\pi}{3}\right) + 4\sin\left(\frac{2\pi}{3}\right) + \sin\left(\frac{4\pi}{3}\right) = 2\pi + 4\left(\frac{\sqrt{3}}{2}\right) - \frac{\sqrt{3}}{2} = 2\pi + \frac{3\sqrt{3}}{2} \]

Subtracting the limits:
\[ A = a^2 \left[ 3\pi - \left(2\pi + \frac{3\sqrt{3}}{2}\right) \right] = a^2\left(\pi - \frac{3\sqrt{3}}{2}\right). \]

B1: Sketch showing correct overall shape (lima\u00e7on with inner loop).
B1: Correct intercepts labeled/noted (e.g., origin crossings, x-axis intercepts at \(3a\) and \(-a\)).
M1: Find the correct limits for the inner loop by setting \(r = 0\).
A1: Obtain \(\theta = 2\pi/3\) and \(4\pi/3\).
M1: State the correct integral formula for polar area.
M1: Use \(\cos^2\theta\) identity to rewrite integrand.
A1: Correctly integrated expression: \(3\theta + 4\sin\theta + \sin 2\theta\).
M1: Substitute limits \(2\pi/3\) and \(\pi\) (or equivalent symmetry limits).
A1: Show clear steps in evaluating trigonometric functions.
A1: Correct final simplified area expression \(a^2\left(\pi - \frac{3\sqrt{3}}{2}\right)\).

(i) By algebraic division:
\[ 2x^2 + 3x + 9 = (2x+1)(x+1) + 8 \]
So:
\[ y = 2x + 1 + \frac{8}{x+1} \]

Thus, the asymptotes are:
- Vertical asymptote: \(x = -1\)
- Oblique asymptote: \(y = 2x + 1\)

(ii) Differentiating \(y\):
\[ \frac{dy}{dx} = 2 - \frac{8}{(x+1)^2} \]

For stationary points, set \(\frac{dy}{dx} = 0\):
\[ 2 - \frac{8}{(x+1)^2} = 0 \implies (x+1)^2 = 4 \implies x+1 = \pm 2 \]
So \(x = 1\) or \(x = -3\).

When \(x = 1\):
\[ y = \frac{2(1)^2 + 3(1) + 9}{1+1} = \frac{14}{2} = 7 \]
Stationary point: \((1, 7)\).

When \(x = -3\):
\[ y = \frac{2(-3)^2 + 3(-3) + 9}{-3+1} = \frac{18-9+9}{-2} = -9 \]
Stationary point: \((-3, -9)\).

(iii) Y-intercept: When \(x = 0\), \(y = 9\).
X-intercepts: Set \(2x^2 + 3x + 9 = 0\). The discriminant is \(3^2 - 4(2)(9) = -63 < 0\), so there are no x-intercepts.
Sketch details:
- Two branches separated by the vertical asymptote \(x = -1\).
- For \(x > -1\), the curve has a local minimum at \((1, 7)\) and passes through \((0, 9)\).
- For \(x < -1\), the curve has a local maximum at \((-3, -9)\).
- Asymptotes \(x = -1\) and \(y = 2x + 1\) are drawn as dashed lines.

M1: Use algebraic division to express \(y\) in the form \(ax + b + \frac{k}{x+c}\).
A1: Correct vertical asymptote \(x = -1\).
A1: Correct oblique asymptote \(y = 2x + 1\).
M1: Differentiate to find \(\frac{dy}{dx}\).
A1: Correct derivative.
M1: Set \(\frac{dy}{dx} = 0\) and solve for \(x\).
A1: Correct coordinates \((1, 7)\) and \((-3, -9)\).
B1: Correct shape with two distinct branches.
B1: Correctly positioned relative to asymptotes and stationary points labeled.

(i) The direction vectors of \(l_1\) and \(l_2\) are \(\mathbf{d_1} = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}\) and \(\mathbf{d_2} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}\).

A normal vector \(\mathbf{n}\) to both lines is:
\[ \mathbf{n} = \mathbf{d_1} \times \mathbf{d_2} = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix} \times \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 1(2) - 3(-1) \\ 3(1) - 2(2) \\ 2(-1) - 1(1) \end{pmatrix} = \begin{pmatrix} 5 \\ -1 \\ -3 \end{pmatrix} \]

A vector connecting a point on \(l_1\), \(A(1, 2, -1)\), and a point on \(l_2\), \(B(2, -1, 4)\), is:
\[ \mathbf{AB} = \begin{pmatrix} 2-1 \\ -1-2 \\ 4-(-1)
\end{pmatrix} = \begin{pmatrix} 1 \\ -3 \\ 5 \end{pmatrix} \]

The shortest distance \(d\) is the projection of \(\mathbf{AB}\) onto \(\mathbf{n}\):
\[ d = \frac{|\mathbf{AB} \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{|1(5) + (-3)(-1) + 5(-3)|}{\sqrt{5^2 + (-1)^2 + (-3)^2}} = \frac{|5 + 3 - 15|}{\sqrt{25 + 1 + 9}} = \frac{7}{\sqrt{35}} = \frac{\sqrt{35}}{5} \approx 1.18 \]

(ii) Let \(P\) be a general point on \(l_1\) and \(Q\) be a general point on \(l_2\):
\[ \mathbf{p} = \begin{pmatrix} 1+2\lambda \\ 2+\lambda \\ -1+3\lambda \end{pmatrix}, \quad \mathbf{q} = \begin{pmatrix} 2+\mu \\ -1-\mu \\ 4+2\mu \end{pmatrix} \]

The vector \(\mathbf{PQ}\) is:
\[ \mathbf{PQ} = \mathbf{q} - \mathbf{p} = \begin{pmatrix} 1+\mu-2\lambda \\ -3-\mu-\lambda \\ 5+2\mu-3\lambda \end{pmatrix} \]

Since \(\mathbf{PQ}\) is perpendicular to both \(\mathbf{d_1}\) and \(\mathbf{d_2}\):
1) \(\mathbf{PQ} \cdot \mathbf{d_1} = 0 \implies 2(1+\mu-2\lambda) + 1(-3-\mu-\lambda) + 3(5+2\mu-3\lambda) = 0\)
\[ \implies 2 + 2\mu - 4\lambda - 3 - \mu - \lambda + 15 + 6\mu - 9\lambda = 0 \implies 14 + 7\mu - 14\lambda = 0 \implies 2\lambda - \mu = 2 \]

2) \(\mathbf{PQ} \cdot \mathbf{d_2} = 0 \implies 1(1+\mu-2\lambda) - 1(-3-\mu-\lambda) + 2(5+2\mu-3\lambda) = 0\)
\[ \implies 1 + \mu - 2\lambda + 3 + \mu + \lambda + 10 + 4\mu - 6\lambda = 0 \implies 14 + 6\mu - 7\lambda = 0 \implies 7\lambda - 6\mu = 14 \]

Solving these simultaneously:
From the first equation, \(\mu = 2\lambda - 2\).
Substitute into the second:
\[ 7\lambda - 6(2\lambda - 2) = 14 \implies 7\lambda - 12\lambda + 12 = 14 \implies -5\lambda = 2 \implies \lambda = -0.4 \]
Then \(\mu = 2(-0.4) - 2 = -2.8\).

Substitute \(\lambda = -0.4\) into \(P\):
\[ P = (1 + 2(-0.4), 2 + (-0.4), -1 + 3(-0.4)) = (0.2, 1.6, -2.2) \]

Substitute \(\mu = -2.8\) into \(Q\):
\[ Q = (2 + (-2.8), -1 - (-2.8), 4 + 2(-2.8)) = (-0.8, 1.8, -1.6) \]

M1: Find the cross product of the direction vectors.
A1: Correct normal vector \(\begin{pmatrix} 5 \\ -1 \\ -3 \end{pmatrix}\).
M1: Find a vector connecting any point on \(l_1\) to any point on \(l_2\).
M1: Use the projection formula for shortest distance.
A1: Correct shortest distance of \(\frac{\sqrt{35}}{5}\).
M1: Set up expressions for general points \(P\) and \(Q\).
M1: Write down \(\mathbf{PQ} \cdot \mathbf{d_1} = 0\) and \(\mathbf{PQ} \cdot \mathbf{d_2} = 0\) to form two equations.
A1: Obtain correct simultaneous equations in \(\lambda\) and \(\mu\).
M1: Solve for \(\lambda\) and \(\mu\).
A1: Correct coordinates of \(P\) and \(Q\).

To find the eigenvalues, solve the characteristic equation \(\det(\mathbf{A} - \lambda \mathbf{I}) = 0\). This gives \(\begin{vmatrix} 3-\lambda & -1 & 1 \\ -1 & 5-\lambda & -1 \\ 1 & -1 & 3-\lambda \end{vmatrix} = 0\), which expands to \(-\lambda^3 + 11\lambda^2 - 36\lambda + 36 = 0\). Factoring this polynomial gives \(-(\lambda - 2)(\lambda - 3)(\lambda - 6) = 0\). Thus, the eigenvalues are \(\lambda = 2, 3, 6\). Next, find the corresponding eigenvectors. For \(\lambda = 2\), the system \((\mathbf{A} - 2\mathbf{I})\mathbf{v} = \mathbf{0}\) leads to \(x - y + z = 0\) and \(-x + 3y - z = 0\), giving the eigenvector \(\mathbf{v}_1 = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}\). Normalizing gives \(\mathbf{u}_1 = \begin{pmatrix} \frac{1}{\sqrt{2}} \\ 0 \\ -\frac{1}{\sqrt{2}} \end{pmatrix}\). For \(\lambda = 3\), the system \((\mathbf{A} - 3\mathbf{I})\mathbf{v} = \mathbf{0}\) yields the eigenvector \(\mathbf{v}_2 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\). Normalizing gives \(\mathbf{u}_2 = \begin{pmatrix} \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \end{pmatrix}\). For \(\lambda = 6\), the system \((\mathbf{A} - 6\mathbf{I})\mathbf{v} = \mathbf{0}\) yields the eigenvector \(\mathbf{v}_3 = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}\). Normalizing gives \(\mathbf{u}_3 = \begin{pmatrix} \frac{1}{\sqrt{6}} \\ -\frac{2}{\sqrt{6}} \\ \frac{1}{\sqrt{6}} \end{pmatrix}\). The matrix \(\mathbf{P}\) is formed by the normalized eigenvectors as columns: \(\mathbf{P} = \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} \\ 0 & \frac{1}{\sqrt{3}} & -\frac{2}{\sqrt{6}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} \end{pmatrix}\). The diagonal matrix \(\mathbf{D}\) contains the eigenvalues on the diagonal: \(\mathbf{D} = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 6 \end{pmatrix}\).

M1: Set up characteristic equation and find eigenvalues. A1: Correct eigenvalues 2, 3, 6. M1: Find eigenvectors for each eigenvalue. A1: Correct eigenvectors. M1: Normalize eigenvectors. A1: Correct matrix P. A1: Correct matrix D.

First, find the complementary function (CF) by solving the auxiliary equation \(m^2 + 2m + 5 = 0\), which yields \(m = -1 \pm 2\mathrm{i}\). Thus, \(y_{CF} = \mathrm{e}^{-x}(A\cos(2x) + B\sin(2x))\). Next, find the particular integral (PI) of the form \(y_{PI} = p\cos x + q\sin x\). Differentiating gives \(y' = -p\sin x + q\cos x\) and \(y'' = -p\cos x - q\sin x\). Substituting these into the differential equation yields \((-p + 2q + 5p)\cos x + (-q - 2p + 5q)\sin x = 10\cos x\). Equating coefficients gives \(4p + 2q = 10\) and \(-2p + 4q = 0\). Solving these gives \(p = 2\) and \(q = 1\). So, the general solution is \(y = \mathrm{e}^{-x}(A\cos(2x) + B\sin(2x)) + 2\cos x + \sin x\). Apply the initial condition \(y(0) = 2\) to get \(2 = A + 2 \implies A = 0\). Now differentiate \(y = B\mathrm{e}^{-x}\sin(2x) + 2\cos x + \sin x\) to get \(y' = B\mathrm{e}^{-x}(2\cos(2x) - \sin(2x)) - 2\sin x + \cos x\). Apply \(y'(0) = 0\) to get \(0 = 2B + 1 \implies B = -\frac{1}{2}\). Therefore, the particular solution is \(y = -\frac{1}{2}\mathrm{e}^{-x}\sin(2x) + 2\cos x + \sin x\).

M1: Solve the auxiliary equation. A1: Obtain correct CF. M1: Postulate PI of correct form and differentiate. A1: Correctly solve for p and q. M1: Set up general solution and use y(0) = 2. A1: Obtain A = 0. M1: Differentiate GS and use y'(0) = 0. A1: Obtain B = -0.5. A1: Correct final expression.

Using integration by parts on \(I_n = \int_1^{\mathrm{e}} 1 \cdot (\ln x)^n \mathrm{d}x\), let \(u = (\ln x)^n\) and \(\mathrm{d}v = \mathrm{d}x\). Then \(\mathrm{d}u = \frac{n(\ln x)^{n-1}}{x}\mathrm{d}x\) and \(v = x\). This gives \(I_n = [x(\ln x)^n]_1^{\mathrm{e}} - \int_1^{\mathrm{e}} n(\ln x)^{n-1} \mathrm{d}x\). Evaluating the boundary term yields \(\mathrm{e}(\ln\mathrm{e})^n - 1(\ln 1)^n = \mathrm{e}\). Thus, \(I_n = \mathrm{e} - n I_{n-1}\). To find \(I_4\), first calculate \(I_0 = \int_1^{\mathrm{e}} 1 \mathrm{d}x = \mathrm{e} - 1\). Then: \(I_1 = \mathrm{e} - 1(I_0) = \mathrm{e} - (\mathrm{e}-1) = 1\), \(I_2 = \mathrm{e} - 2(I_1) = \mathrm{e} - 2\), \(I_3 = \mathrm{e} - 3(I_2) = \mathrm{e} - 3(\mathrm{e}-2) = 6 - 2\mathrm{e}\), \(I_4 = \mathrm{e} - 4(I_3) = \mathrm{e} - 4(6-2\mathrm{e}) = 9\mathrm{e} - 24\).

M1: Apply integration by parts with correct u and v. A1: Show boundary evaluation correctly. A1: Derive reduction formula. B1: Obtain correct base case I_0. M1: Use reduction formula iteratively. A1: Correct values for I_1 and I_2. A1: Correct value for I_3. A1: Correct final exact value for I_4.

(i) Let \(y = \cosh^{-1} x\), then \(x = \cosh y = \frac{\mathrm{e}^y + \mathrm{e}^{-y}}{2}\). This gives \(\mathrm{e}^{2y} - 2x\mathrm{e}^y + 1 = 0\). Solving this quadratic in \(\mathrm{e}^y\) using the quadratic formula yields \(\mathrm{e}^y = x \pm \sqrt{x^2-1}\). Since \(y \ge 0\) for the principal value, \(\mathrm{e}^y \ge 1\), so we take the positive root. Taking the natural logarithm gives \(y = \ln(x + \sqrt{x^2-1})\). (ii) Substitute \(\cosh(2\theta) = 2\cosh^2\theta - 1\) into the equation: \(2\cosh^2\theta - 1 - 3\cosh\theta = 1 \implies 2\cosh^2\theta - 3\cosh\theta - 2 = 0\). Factoring gives \((2\cosh\theta + 1)(\cosh\theta - 2) = 0\). Since \(\cosh\theta \ge 1\), the only valid solution is \(\cosh\theta = 2\). Thus, \(\theta = \cosh^{-1}(2) = \ln(2 + \sqrt{2^2-1}) = \ln(2+\sqrt{3})\).

M1: Express cosh in terms of exponentials. A1: Formulate the quadratic equation in e^y. A1: Solve the quadratic and justify choice of sign. M1: Use identity cosh(2x) = 2cosh^2(x) - 1. A1: Obtain correct quadratic in cosh. A1: Solve quadratic and reject invalid root. A1: Apply logarithmic formula to obtain exact answer.

Differentiating \(y = \arctan x\) gives \(y' = \frac{1}{1+x^2}\), so \((1+x^2)y' = 1\). Apply Leibniz's theorem to differentiate \((1+x^2)y' = 1\) exactly \(n\) times. Let \(u = y'\) and \(v = 1+x^2\). Then \((uv)^{(n)} = u^{(n)}v + n u^{(n-1)}v' + \frac{n(n-1)}{2}u^{(n-2)}v'' + 0 = y^{(n+1)}(1+x^2) + n y^{(n)}(2x) + \frac{n(n-1)}{2}y^{(n-1)}(2) = 0\), which simplifies to \((1+x^2)y^{(n+1)} + 2nxy^{(n)} + n(n-1)y^{(n-1)} = 0\). Evaluating this at \(x = 0\) gives \(y^{(n+1)}(0) = -n(n-1)y^{(n-1)}(0)\). We know \(y(0) = 0\) and \(y'(0) = 1\). For \(n=1\): \(y''(0) = 0\). For \(n=2\): \(y^{(3)}(0) = -2(1)y'(0) = -2\). For \(n=3\): \(y^{(4)}(0) = 0\). For \(n=4\): \(y^{(5)}(0) = -4(3)y^{(3)}(0) = 24\). The Maclaurin series is \(y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y^{(3)}(0)}{3!}x^3 + \frac{y^{(4)}(0)}{4!}x^4 + \frac{y^{(5)}(0)}{5!}x^5 = x - \frac{2}{6}x^3 + \frac{24}{120}x^5 = x - \frac{1}{3}x^3 + \frac{1}{5}x^5\).

B1: Prove (1+x^2)y' = 1. M1: Set up Leibniz's theorem correctly. A1: Correctly differentiate to obtain the given relation. M1: Evaluate relation at x = 0 to obtain recurrence formula. A1: Correctly calculate y'(0), y''(0) and y'''(0). A1: Correctly calculate fourth and fifth derivatives at x = 0. A1: Construct final Maclaurin series.

(i) Let \(z = \cos\theta + \mathrm{i}\sin\theta\). Then \(2\mathrm{i}\sin\theta = z - z^{-1}\). Expanding \((2\mathrm{i}\sin\theta)^5 = (z - z^{-1})^5\) using the Binomial Theorem gives \(32\mathrm{i}\sin^5\theta = z^5 - 5z^3 + 10z - 10z^{-1} + 5z^{-3} - z^{-5} = (z^5 - z^{-5}) - 5(z^3 - z^{-3}) + 10(z - z^{-1})\). Since \(z^n - z^{-n} = 2\mathrm{i}\sin(n\theta)\), this becomes \(32\mathrm{i}\sin^5\theta = 2\mathrm{i}\sin(5\theta) - 10\mathrm{i}\sin(3\theta) + 20\mathrm{i}\sin\theta\). Dividing by \(32\mathrm{i}\) yields \(\sin^5\theta = \frac{1}{16}\sin(5\theta) - \frac{5}{16}\sin(3\theta) + \frac{5}{8}\sin\theta\). (ii) Integrate: \(\int_0^{\pi/2} \sin^5\theta \mathrm{d}\theta = \left[ -\frac{1}{80}\cos(5\theta) + \frac{5}{48}\cos(3\theta) - \frac{5}{8}\cos\theta \right]_0^{\pi/2}\). At \(\pi/2\), all cosine terms are 0. At 0, the expression evaluates to \(-\frac{1}{80} + \frac{5}{48} - \frac{5}{8} = -\frac{256}{480} = -\frac{8}{15}\). Subtracting this value at the lower limit gives \(0 - (-\frac{8}{15}) = \frac{8}{15}\).

M1: Express sin in terms of z and z^-1. M1: Apply binomial expansion correctly. A1: Correctly group terms. A1: Substitute 2isin(ntheta) back. A1: Obtain correct identity. M1: Integrate the identity term-by-term. A1: Correctly apply limits. A1: Simplify to final fraction.

First, find \(\frac{\mathrm{d}y}{\mathrm{d}x}\) for \(y = \ln(\cos x)\). This gives \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x\). The arc length \(s\) is given by \(s = \int_0^{\pi/3} \sqrt{1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2} \mathrm{d}x\). Substituting \(\frac{\mathrm{d}y}{\mathrm{d}x}\) yields \(s = \int_0^{\pi/3} \sqrt{1 + \tan^2 x} \mathrm{d}x = \int_0^{\pi/3} \sec x \mathrm{d}x\). Integrating \(\sec x\) gives \([\ln(\sec x + \tan x)]_0^{\pi/3}\). Evaluating at the upper limit \(x = \frac{\pi}{3}\): \(\sec(\frac{\pi}{3}) = 2\) and \(\tan(\frac{\pi}{3}) = \sqrt{3}\). Evaluating at the lower limit \(x = 0\): \(\sec 0 = 1\) and \(\tan 0 = 0\). Therefore, \(s = \ln(2 + \sqrt{3}) - \ln(1 + 0) = \ln(2 + \sqrt{3})\).

M1: Differentiate y to obtain dy/dx. A1: Correct derivative -tan x. M1: Substitute dy/dx into the arc length formula. A1: Simplify integrand to sec x. M1: Integrate sec x correctly. A1: Substitute limits. A1: Obtain correct final exact value.

First, rewrite the differential equation in standard form by dividing by \(\cos x\): \(\frac{\mathrm{d}y}{\mathrm{d}x} + y\tan x = \cos^2 x\). The integrating factor is \(I = \mathrm{e}^{\int \tan x \mathrm{d}x} = \mathrm{e}^{\ln(\sec x)} = \sec x\). Multiplying both sides of the standard equation by \(\sec x\) yields \(\frac{\mathrm{d}}{\mathrm{d}x}(y\sec x) = \cos^2 x \sec x = \cos x\). Integrating both sides gives \(y\sec x = \sin x + C\), so \(y = \sin x\cos x + C\cos x\). Applying the initial condition \(y(0) = 2\) yields \(2 = 0 + C(1) \implies C = 2\). Therefore, the particular solution is \(y = \sin x\cos x + 2\cos x\). Substituting \(x = \frac{\pi}{3}\) gives \(y = \sin(\frac{\pi}{3})\cos(\frac{\pi}{3}) + 2\cos(\frac{\pi}{3}) = \frac{\sqrt{3}}{2} \cdot \frac{1}{2} + 2 \cdot \frac{1}{2} = \frac{\sqrt{3}}{4} + 1\).

M1: Rewrite differential equation in standard linear form. A1: Find correct integrating factor sec x. M1: Multiply by integrating factor and integrate. A1: Correct general solution. M1: Apply initial conditions to find C. A1: Obtain C = 2. M1: Substitute x = pi/3 into the particular solution. A1: Correct final exact value.