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

(i) Let \(P(n)\) be the statement that \(7^{2n} + 16n - 1\) is divisible by 64.

**Base step:**
For \(n = 1\), \(7^2 + 16(1) - 1 = 49 + 16 - 1 = 64\), which is divisible by 64. Thus, \(P(1)\) is true.

**Inductive step:**
Assume \(P(k)\) is true for some positive integer \(k\). That is, \(7^{2k} + 16k - 1 = 64m\) for some integer \(m\).

We need to show that \(P(k+1)\) is true, i.e., \(7^{2(k+1)} + 16(k+1) - 1\) is divisible by 64.

\[7^{2(k+1)} + 16(k+1) - 1 = 7^{2k+2} + 16k + 16 - 1\]
\[= 49 \cdot 7^{2k} + 16k + 15\]

From our induction hypothesis, we have \(7^{2k} = 64m - 16k + 1\). Substituting this in:

\[49(64m - 16k + 1) + 16k + 15 = 49 \cdot 64m - 784k + 49 + 16k + 15\]
\[= 49 \cdot 64m - 768k + 64\]
\[= 64(49m - 12k + 1)\]

Since \(m\) and \(k\) are integers, \(49m - 12k + 1\) is an integer. Thus, \(7^{2(k+1)} + 16(k+1) - 1\) is divisible by 64, so \(P(k+1)\) is true.

**Conclusion:**
By mathematical induction, \(7^{2n} + 16n - 1\) is divisible by 64 for all positive integers \(n\).

(ii) From part (i), we can write \(7^{2n} = 64A - 16n + 1\) for some integer \(A\).

Consider the expression:
\[7^{2n+1} + 112n + 57 = 7(7^{2n}) + 112n + 57\]
\[= 7(64A - 16n + 1) + 112n + 57\]
\[= 7 \cdot 64A - 112n + 7 + 112n + 57\]
\[= 7 \cdot 64A + 64\]
\[= 64(7A + 1)\]

Since \(A\) is an integer, \(7A + 1\) is an integer. Therefore, \(7^{2n+1} + 112n + 57\) is divisible by 64 for all positive integers \(n\).

(i)
- M1: For verifying the statement for \(n = 1\).
- M1: For formulating the induction hypothesis \(7^{2k} + 16k - 1 = 64m\).
- A1: For expressing the term for \(n = k+1\) as \(49 \cdot 7^{2k} + 16k + 15\).
- M1: For substituting the induction hypothesis to eliminate \(7^{2k}\).
- A1: For obtaining a simplified expression of the form \(64(49m - 12k + 1)\).
- A1: For a complete and coherent mathematical induction proof with a concluding statement.

(ii)
- M1: For rewriting \(7^{2n+1}\) as \(7 \cdot 7^{2n}\).
- M1: For using the result of part (i) to substitute for \(7^{2n}\).
- A1.7: For simplifying to \(64(7A+1)\) and drawing the final conclusion.

(i) We expand the difference \(f(r) - f(r+1)\):
\[f(r) - f(r+1) = \frac{4r+A}{2r(r+1)} - \frac{4(r+1)+A}{2(r+1)(r+2)}\]
\[= \frac{(4r+A)(r+2) - r(4r+4+A)}{2r(r+1)(r+2)}\]
\[= \frac{(4r^2 + 8r + Ar + 2A) - (4r^2 + 4r + Ar)}{2r(r+1)(r+2)}\]
\[= \frac{4r + 2A}{2r(r+1)(r+2)}\]
\[= \frac{2r + A}{r(r+1)(r+2)}\]

Comparing this with \(\frac{2r+3}{r(r+1)(r+2)}\), we find \(A = 3\).

Thus, \(f(r) = \frac{4r+3}{2r(r+1)}\).

(ii) Using the method of differences:
\[\sum_{r=1}^{n} \frac{2r+3}{r(r+1)(r+2)} = \sum_{r=1}^{n} (f(r) - f(r+1))\]
\[= (f(1) - f(2)) + (f(2) - f(3)) + \dots + (f(n) - f(n+1))\]
\[= f(1) - f(n+1)\]

We calculate \(f(1)\):
\[f(1) = \frac{4(1)+3}{2(1)(2)} = \frac{7}{4}\]

And write down \(f(n+1)\):
\[f(n+1) = \frac{4(n+1)+3}{2(n+1)(n+2)} = \frac{4n+7}{2(n+1)(n+2)}\]

Thus, the sum of the series is:
\[\sum_{r=1}^{n} \frac{2r+3}{r(r+1)(r+2)} = \frac{7}{4} - \frac{4n+7}{2(n+1)(n+2)}\]

(iii) As \(n \to \infty\), \(\frac{4n+7}{2(n+1)(n+2)} \to 0\).

Therefore,
\[\sum_{r=1}^{\infty} \frac{2r+3}{r(r+1)(r+2)} = \frac{7}{4}\]

(iv) The sum from \(n+1\) to infinity is:
\[\sum_{r=n+1}^{\infty} \frac{2r+3}{r(r+1)(r+2)} = \sum_{r=1}^{\infty} \frac{2r+3}{r(r+1)(r+2)} - \sum_{r=1}^{n} \frac{2r+3}{r(r+1)(r+2)}\]
\[= \frac{7}{4} - \left( \frac{7}{4} - \frac{4n+7}{2(n+1)(n+2)} \right) = \frac{4n+7}{2(n+1)(n+2)}\]

(i)
- M1: For expanding \(f(r) - f(r+1)\) over a common denominator.
- A1: For obtaining \(\frac{2r+A}{r(r+1)(r+2)}\) and comparing with the given term.
- A1: For concluding \(A = 3\) and giving \(f(r) = \frac{4r+3}{2r(r+1)}\).

(ii)
- M1: For writing out terms to show how cancellation occurs.
- A1: For identifying the non-cancelling terms as \(f(1) - f(n+1)\).
- M1: For calculating \(f(1) = \frac{7}{4}\).
- A1: For obtaining \(\frac{7}{4} - \frac{4n+7}{2(n+1)(n+2)}\).

(iii)
- M1: For taking the limit as \(n \to \infty\).
- A0.7: For stating \(\frac{7}{4}\).

(iv)
- M1: For recognizing the relationship between the finite sum, the sum to infinity, and the required tail sum.
- A1: For obtaining \(\frac{4n+7}{2(n+1)(n+2)}\).

(i) From the relation between the coefficients and roots of \(x^3 - 3x^2 + 5x - 2 = 0\):
\[\sum \alpha = 3\]
\[\sum \alpha\beta = 5\]
\[\alpha\beta\gamma = 2\]

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

(ii) Since \(\alpha, \beta, \gamma\) are roots of the cubic equation:
\[\alpha^3 - 3\alpha^2 + 5\alpha - 2 = 0\]
\[\beta^3 - 3\beta^2 + 5\beta - 2 = 0\]
\[\gamma^3 - 3\gamma^2 + 5\gamma - 2 = 0\]

Summing these equations, we get:
\[\sum \alpha^3 - 3 \sum \alpha^2 + 5 \sum \alpha - 6 = 0\]
\[\sum \alpha^3 = 3 \sum \alpha^2 - 5 \sum \alpha + 6 = 3(-1) - 5(3) + 6 = -3 - 15 + 6 = -12\]

(iii) Let \(y = x^2\), so \(x = \sqrt{y}\). Substituting into the cubic equation:
\[y^{3/2} - 3y + 5y^{1/2} - 2 = 0\]
\[y^{1/2}(y + 5) = 3y + 2\]

Squaring both sides:
\[y(y+5)^2 = (3y+2)^2\]
\[y(y^2 + 10y + 25) = 9y^2 + 12y + 4\]
\[y^3 + 10y^2 + 25y = 9y^2 + 12y + 4\]
\[y^3 + y^2 + 13y - 4 = 0\]

(iv) The roots of the cubic equation \(y^3 + y^2 + 13y - 4 = 0\) are \(\alpha^2, \beta^2, \gamma^2\).
We wish to find:
\[\alpha^{-2} + \beta^{-2} + \gamma^{-2} = \frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} = \frac{\alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2}{\alpha^2\beta^2\gamma^2}\]

Using the coefficients from the equation in \(y\):
Sum of product of roots in pairs: \(\alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 = 13\)
Product of roots: \(\alpha^2\beta^2\gamma^2 = 4\)

Thus:
\[\alpha^{-2} + \beta^{-2} + \gamma^{-2} = \frac{13}{4}\]

(i)
- M1: For stating and using the relation \(\alpha^2+\beta^2+\gamma^2 = (\sum \alpha)^2 - 2\sum\alpha\beta\).
- A1: For obtaining \(-1\).

(ii)
- M1: For summing the cubic identities for the roots to set up \(\sum \alpha^3 - 3\sum\alpha^2 + 5\sum\alpha - 6 = 0\).
- M1: For substituting the values of \(\sum\alpha^2\) and \(\sum\alpha\).
- A1: For obtaining \(-12\).

(iii)
- M1: For making the substitution \(y = x^2\) (or equivalent algebraic substitution) and grouping terms.
- M1: For squaring to eliminate fractional powers.
- A1.7: For obtaining the correct equation \(y^3 + y^2 + 13y - 4 = 0\) (accept any variable name).

(iv)
- M1: For rewriting the target sum in terms of symmetric functions of \(\alpha^2, \beta^2, \gamma^2\).
- A1: For obtaining \(\frac{13}{4}\).

(i) Performing algebraic division on \(y\):
\[2x^2 + 3x - 2 = 2x(x-1) + 5x - 2 = 2x(x-1) + 5(x-1) + 3\]

Thus, we can express the equation as:
\[y = 2x + 5 + \frac{3}{x-1}\]

As \(x \to 1\), \(y \to \pm\infty\), giving the vertical asymptote:
\[x = 1\]

As \(x \to \pm\infty\), \(\frac{3}{x-1} \to 0\), giving the oblique asymptote:
\[y = 2x + 5\]

(ii) Differentiating \(y\) with respect to \(x\):
\[\frac{dy}{dx} = 2 - \frac{3}{(x-1)^2}\]

To find the stationary points, we set \(\frac{dy}{dx} = 0\):
\[2 - \frac{3}{(x-1)^2} = 0 \implies (x-1)^2 = \frac{3}{2} \implies x - 1 = \pm \sqrt{\frac{3}{2}} = \pm \frac{\sqrt{6}}{2}\]

So, the x-coordinates are:
\[x = 1 \pm \frac{\sqrt{6}}{2}\]

Substitute these into the expression \(y = 2x + 5 + \frac{3}{x-1}\):

For \(x = 1 + \frac{\sqrt{6}}{2}\):
\[y = 2\left(1 + \frac{\sqrt{6}}{2}\right) + 5 + \frac{3}{\sqrt{6}/2} = 2 + \sqrt{6} + 5 + \sqrt{6} = 7 + 2\sqrt{6}\]

For \(x = 1 - \frac{\sqrt{6}}{2}\):
\[y = 2\left(1 - \frac{\sqrt{6}}{2}\right) + 5 - \frac{3}{\sqrt{6}/2} = 2 - \sqrt{6} + 5 - \sqrt{6} = 7 - 2\sqrt{6}\]

Therefore, the stationary points are:
\[\left(1 + \frac{\sqrt{6}}{2}, 7 + 2\sqrt{6}\right) \quad \text{and} \quad \left(1 - \frac{\sqrt{6}}{2}, 7 - 2\sqrt{6}\right)\]

(iii) Intersections with the axes:
- When \(x = 0\), \(y = \frac{-2}{-1} = 2\).
- When \(y = 0\), \(2x^2 + 3x - 2 = 0 \implies (2x-1)(x+2) = 0 \implies x = 1/2\) or \(x = -2\).

The sketch consists of two distinct branches:
- One branch in the upper right, staying above the oblique asymptote \(y=2x+5\) and to the right of \(x=1\), with a local minimum at \(\left(1+\frac{\sqrt{6}}{2}, 7+2\sqrt{6}\right) \approx (2.22, 11.90)\).
- One branch in the lower left, staying below the oblique asymptote and to the left of \(x=1\), passing through \((-2,0)\), \((0,2)\), and \((1/2,0)\), with a local maximum at \(\left(1-\frac{\sqrt{6}}{2}, 7-2\sqrt{6}\right) \approx (-0.22, 2.10)\).

(i)
- M1: For carrying out division or showing equivalent algebra to rewrite the fractional function.
- A1: For vertical asymptote \(x = 1\).
- A1: For oblique asymptote \(y = 2x + 5\).

(ii)
- M1: For differentiating the expression and setting \(\frac{dy}{dx} = 0\).
- A1: For solving to find \(x = 1 \pm \frac{\sqrt{6}}{2}\) (any equivalent exact form).
- M1: For substituting \(x\) values back to find corresponding \(y\) coordinates.
- A1.7: For both correct exact coordinates: \(\left(1 + \frac{\sqrt{6}}{2}, 7 + 2\sqrt{6}\right)\) and \(\left(1 - \frac{\sqrt{6}}{2}, 7 - 2\sqrt{6}\right)\).

(iii)
- G1: For two branches of the curve with correct asymptotic behavior.
- G1: For drawing and labelling both asymptotes \(x=1\) and \(y=2x+5\).
- G1: For indicating key coordinates (intercepts and stationary points) correctly.

(i) The curve is the upper half of a cardioid.
At \(\theta = 0\), \(r = 2a\).
At \(\theta = \pi/2\), \(r = a\).
At \(\theta = \pi\), \(r = 0\).
The sketch should show a smooth closed curve starting at \((2a, 0)\), passing through \((a, \pi/2)\), and ending at the pole \((0, \pi)\).

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

Using the identity \(\cos^2\theta = \frac{1 + \cos 2\theta}{2}\):
\[A = \frac{1}{2} a^2 \int_{0}^{\pi} \left( 1 + 2\cos\theta + \frac{1 + \cos 2\theta}{2} \right) \, d\theta\]
\[= \frac{1}{2} a^2 \int_{0}^{\pi} \left( \frac{3}{2} + 2\cos\theta + \frac{1}{2}\cos 2\theta \right) \, d\theta\]
\[= \frac{1}{2} a^2 \left[ \frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta \right]_{0}^{\pi}\]
\[= \frac{1}{2} a^2 \left[ \left( \frac{3}{2}\pi + 0 + 0 \right) - (0) \right] = \frac{3}{8}\pi a^2\]

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

Setting \(\frac{dx}{d\theta} = 0\):
Since \(0 < \theta < \pi\), \(\sin\theta \ne 0\).
Thus, \(1 + 2\cos\theta = 0 \implies \cos\theta = -\frac{1}{2}\).

Since \(0 \le \theta \le \pi\), we find:
\[\theta = \frac{2\pi}{3}\]

Substituting \(\theta = \frac{2\pi}{3}\) into the equation for \(r\):
\[r = a\left(1 + \cos\frac{2\pi}{3}\right) = a\left(1 - \frac{1}{2}\right) = \frac{1}{2}a\]

So, the polar coordinates of the point are \(\left(\frac{1}{2}a, \frac{2\pi}{3}\right)\).

(i)
- G1: For drawing the correct upper half of a cardioid, starting on the initial line and ending at the pole.
- G1: For indicating coordinates at \(\theta = 0\) and \(\theta = \pi/2\).

(ii)
- M1: For using the area formula \(A = \frac{1}{2}\int r^2 \, d\theta\).
- M1: For expanding \((1+\cos\theta)^2\) and utilizing the double-angle identity for \(\cos^2\theta\).
- A1: For performing the integration correctly.
- A1.7: For substituting limits correctly and establishing the given area \(\frac{3}{8}\pi a^2\).

(iii)
- M1: For setting up the equation for \(x = r\cos\theta\).
- M1: For differentiating \(x\) with respect to \(\theta\) and setting \(\frac{dx}{d\theta} = 0\).
- A1: For finding \(\theta = \frac{2\pi}{3}\).
- A1: For finding \(r = \frac{1}{2}a\) and stating the coordinates as \(\left(\frac{1}{2}a, \frac{2\pi}{3}\right)\).

(i) The normal vector \(\mathbf{n}_1\) to the plane \(\Pi_1\) is perpendicular to the directions of both \(l_1\) and \(l_2\):
\[\mathbf{d}_1 = \mathbf{i} + 2\mathbf{j} -
\mathbf{k}\]
\[\mathbf{d}_2 = 2\mathbf{i} - \mathbf{j} + \mathbf{k}\]

Using the vector product:
\[\mathbf{n}_1 = \mathbf{d}_1 \times \mathbf{d}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -1 \\ 2 & -1 & 1 \end{vmatrix}\]
\[= \mathbf{i}(2 - 1) - \mathbf{j}(1 - (-2)) + \mathbf{k}(-1 - 4) = \mathbf{i} - 3\mathbf{j} - 5\mathbf{k}\]

Since \(\Pi_1\) contains \(l_1\), it must contain the point \(A(1, -1, 2)\) on \(l_1\). The equation of the plane is:
\[1(x-1) - 3(y+1) - 5(z-2) = 0\]
\[x - 3y - 5z + 6 = 0 \implies x - 3y - 5z = -6\]

(ii) Let \(B(2, 3, 0)\) be a point on \(l_2\). Then the vector \(\vec{AB}\) is:
\[\vec{AB} = (2-1)\mathbf{i} + (3 - (-1))\mathbf{j} + (0-2)\mathbf{k} = \mathbf{i} + 4\mathbf{j} - 2\mathbf{k}\]

The shortest distance \(d\) is the length of the projection of \(\vec{AB}\) onto the normal vector \(\mathbf{n}_1\):
\[d = \frac{|\vec{AB} \cdot \mathbf{n}_1|}{|\mathbf{n}_1|}\]
\[\vec{AB} \cdot \mathbf{n}_1 = 1(1) + 4(-3) + (-2)(-5) = 1 - 12 + 10 = -1\]
\[|\mathbf{n}_1| = \sqrt{1^2 + (-3)^2 + (-5)^2} = \sqrt{1 + 9 + 25} = \sqrt{35}\]

Thus:
\[d = \frac{|-1|}{\sqrt{35}} = \frac{1}{\sqrt{35}}\]

(iii) Let \(\theta\) be the acute angle between \(l_1\) and the plane \(\Pi_2\), which has normal vector \(\mathbf{n}_2 = 2\mathbf{i} - \mathbf{j} + 2\mathbf{k}\). The angle satisfies:
\[\sin\theta = \frac{|\mathbf{d}_1 \cdot \mathbf{n}_2|}{|\mathbf{d}_1| |\mathbf{n}_2|}\]

We compute the dot product and magnitudes:
\[\mathbf{d}_1 \cdot \mathbf{n}_2 = 1(2) + 2(-1) - 1(2) = 2 - 2 - 2 = -2\]
\[|\mathbf{d}_1| = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{6}\]
\[|\mathbf{n}_2| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{9} = 3\]

Thus:
\[\sin\theta = \frac{|-2|}{3\sqrt{6}} = \frac{2}{3\sqrt{6}}\]
\[\theta = \sin^{-1}\left(\frac{2}{3\sqrt{6}}\right) = \sin^{-1}\left(\frac{\sqrt{6}}{9}\right) \approx 15.8^{\circ} \quad (\text{or } 0.276 \text{ radians})\]

(i)
- M1: For calculating the cross product of the direction vectors of the lines.
- A1: For obtaining the normal vector \(\mathbf{i} - 3\mathbf{j} - 5\mathbf{k}\).
- M1: For substituting the point \((1, -1, 2)\) to find the plane constant.
- A1: For stating \(x - 3y - 5z = -6\) (or any scalar multiple).

(ii)
- M1: For finding a vector between any point on \(l_1\) and any point on \(l_2\).
- M1: For applying the projection formula using their normal vector.
- A1.7: For obtaining \(\frac{1}{\sqrt{35}}\).

(iii)
- M1: For utilizing the formula \(\sin\theta = \frac{|\mathbf{d} \cdot \mathbf{n}_2|}{|\mathbf{d}||\mathbf{n}_2|}\).
- A1: For calculating \(\sin\theta = \frac{2}{3\sqrt{6}}\).
- A1: For evaluating \(\theta \approx 15.8^{\circ}\) or \(0.276\) radians.

(i) Expanding the determinant of \(\mathbf{M}\) along the first row:
\[\det(\mathbf{M}) = 1\begin{vmatrix} k & 1 \\ -1 & 2 \end{vmatrix} - 2\begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} - 1\begin{vmatrix} 2 & k \\ 1 & -1 \end{vmatrix}\]
\[= 1(2k - (-1)) - 2(4 - 1) - 1(-2 - k)\]
\[= 2k + 1 - 6 + 2 + k\]
\[= 3k - 3\]

(ii) A system of linear equations does not have a unique solution if and only if the determinant of the coefficient matrix is equal to 0:
\[3k - 3 = 0 \implies k = 1\]

(iii) For \(k = 1\), the system becomes:
\[1) \quad x + 2y - z = 2\]
\[2) \quad 2x + y + z = 1\]
\[3) \quad x - y + 2z = a\]

We perform row operations to analyze consistency:
Subtract \(2 \times\) Equation (1) from Equation (2):
\[(2x + y + z) - 2(x + 2y - z) = 1 - 4 \implies -3y + 3z = -3 \implies y - z = 1\]

Subtract Equation (1) from Equation (3):
\[(x - y + 2z) - (x + 2y - z) = a - 2 \implies -3y + 3z = a - 2 \implies -3(y - z) = a - 2\]

For the system to be consistent, these two equations must be compatible. Substituting \(y - z = 1\) into the second relation:
\[-3(1) = a - 2 \implies a = -1\]

When \(a = -1\), the equations are consistent. To find the general solution, we let \(z = t\) (where \(t \in \mathbb{R}\)).

From \(y - z = 1\), we obtain:
\[y = t + 1\]

Substitute both into Equation (1):
\[x + 2(t + 1) - t = 2 \implies x + t + 2 = 2 \implies x = -t\]

So the general solution is:
\[\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -t \\ t + 1 \\ t \end{pmatrix}\]

(i)
- M1: For calculating the determinant of a 3x3 matrix.
- A1.7: For simplifying the expression to \(3k - 3\).

(ii)
- M1: For setting their determinant equal to 0.
- A1: For identifying \(k = 1\).

(iii)
- M1: For carrying out a systematic elimination process on the system with \(k=1\).
- A1: For obtaining the simplified equations, such as \(y - z = 1\) and \(-3(y-z) = a-2\).
- A1: For finding \(a = -1\).
- M1: For introducing a parameter \(t\) to find the general solution.
- A1: For finding \(x = -t\) and \(y = t + 1\).
- A1: For writing the final vector solution in a clear form.

The auxiliary equation is:
\[ m^2 + 4m + 5 = 0 \]
Solving this gives:
\[ m = \frac{-4 \pm \sqrt{16 - 20}}{2} = -2 \pm \mathrm{i} \]
Thus, the complementary function (CF) is:
\[ y_c = \mathrm{e}^{-2x}(A\cos x + B\sin x) \]
For the particular integral (PI), let \(y_p = k \mathrm{e}^{-x}\).
Then:
\[ \frac{\mathrm{d} y_p}{\mathrm{d} x} = -k \mathrm{e}^{-x} \quad \text{and} \quad \frac{\mathrm{d}^2 y_p}{\mathrm{d} x^2} = k \mathrm{e}^{-x} \]
Substituting these into the differential equation:
\[ k \mathrm{e}^{-x} + 4(-k \mathrm{e}^{-x}) + 5(k \mathrm{e}^{-x}) = 10 \mathrm{e}^{-x} \]
\[ 2k \mathrm{e}^{-x} = 10 \mathrm{e}^{-x} \implies k = 5 \]
So the general solution is:
\[ y = \mathrm{e}^{-2x}(A\cos x + B\sin x) + 5\mathrm{e}^{-x} \]
Using the boundary conditions:
When \(x = 0\), \(y = 3\):
\[ 3 = A + 5 \implies A = -2 \]
Differentiating the general solution:
\[ \frac{\mathrm{d} y}{\mathrm{d} x} = -2\mathrm{e}^{-2x}(A\cos x + B\sin x) + \mathrm{e}^{-2x}(-A\sin x + B\cos x) - 5\mathrm{e}^{-x} \]
When \(x = 0\), \(\frac{\mathrm{d} y}{\mathrm{d} x} = 0\):
\[ 0 = -2A + B - 5 \]
Substituting \(A = -2\):
\[ 0 = 4 + B - 5 \implies B = 1 \]
Thus, the particular solution is:
\[ y = \mathrm{e}^{-2x}(\sin x - 2\cos x) + 5\mathrm{e}^{-x} \]

M1: Formulates the auxiliary equation and solves to find roots \(-2 \pm \mathrm{i}\).
A1: Writes down the correct Complementary Function.
M1: States a correct trial form for the Particular Integral and differentiates.
A1: Finds the correct Particular Integral \(y_p = 5\mathrm{e}^{-x}\).
A1: States the general solution.
M1: Applies the initial condition \(y(0) = 3\) to find \(A\).
M1: Differentiates the general solution and applies \(\frac{\mathrm{d}y}{\mathrm{d}x}(0) = 0\).
A1: Finds \(A = -2\) and \(B = 1\).
A1: States the correct final particular solution.

(i) We write:
\[ I_n = \int_0^{\ln 2} \tanh^{n-2} x \tanh^2 x \, \mathrm{d} x \]
Since \(\tanh^2 x = 1 - \operatorname{sech}^2 x\):
\[ I_n = \int_0^{\ln 2} \tanh^{n-2} x (1 - \operatorname{sech}^2 x) \, \mathrm{d} x \]
\[ I_n = \int_0^{\ln 2} \tanh^{n-2} x \, \mathrm{d} x - \int_0^{\ln 2} \tanh^{n-2} x \operatorname{sech}^2 x \, \mathrm{d} x \]
The first term is \(I_{n-2}\).
For the second term, we use the substitution \(u = \tanh x\), which gives \(\mathrm{d} u = \operatorname{sech}^2 x \, \mathrm{d} x\).
When \(x = 0\), \(u = 0\).
When \(x = \ln 2\),
\[ u = \tanh(\ln 2) = \frac{\mathrm{e}^{\ln 2} - \mathrm{e}^{-\ln 2}}{\mathrm{e}^{\ln 2} + \mathrm{e}^{-\ln 2}} = \frac{2 - 1/2}{2 + 1/2} = \frac{3}{5} \]
So,
\[ \int_0^{\ln 2} \tanh^{n-2} x \operatorname{sech}^2 x \, \mathrm{d} x = \int_0^{3/5} u^{n-2} \, \mathrm{d} u = \left[ \frac{u^{n-1}}{n-1} \right]_0^{3/5} = \frac{1}{n-1}\left(\frac{3}{5}\right)^{n-1} \]
Therefore,
\[ I_n = I_{n-2} - \frac{1}{n-1}\left(\frac{3}{5}\right)^{n-1} \]

(ii) For \(n = 3\):
\[ I_3 = I_1 - \frac{1}{2}\left(\frac{3}{5}\right)^2 = I_1 - \frac{9}{50} \]
We evaluate \(I_1\):
\[ I_1 = \int_0^{\ln 2} \tanh x \, \mathrm{d} x = \left[ \ln(\cosh x) \right]_0^{\ln 2} \]
Since \(\cosh(\ln 2) = \frac{\mathrm{e}^{\ln 2} + \mathrm{e}^{-\ln 2}}{2} = \frac{2 + 1/2}{2} = \frac{5}{4}\) and \(\cosh(0) = 1\):
\[ I_1 = \ln\left(\frac{5}{4}\right) - \ln 1 = \ln\left(\frac{5}{4}\right) \]
Thus,
\[ I_3 = \ln\left(\frac{5}{4}\right) - \frac{9}{50} \]

M1: Splits \(\tanh^n x\) as \(\tanh^{n-2} x (1 - \operatorname{sech}^2 x)\).
A1: Separates into two integrals and identifies \(I_{n-2}\).
M1: Applies substitution \(u = \tanh x\) to the second integral.
A1: Correctly evaluates the limits as \(0\) and \(\frac{3}{5}\).
A1: Obtains the correct reduction formula.
M1: Applies the reduction formula for \(n=3\).
M1: Integrates \(\tanh x\) and evaluates the limits correctly to find \(I_1 = \ln(5/4)\).
A1: Combines to get the correct final answer.

(i) By De Moivre's theorem:
\[ \cos(5\theta) + \mathrm{i}\sin(5\theta) = (\cos\theta + \mathrm{i}\sin\theta)^5 \]
Expanding using the binomial theorem:
\[ (\cos\theta + \mathrm{i}\sin\theta)^5 = \cos^5\theta + 5\mathrm{i}\cos^4\theta\sin\theta - 10\cos^3\theta\sin^2\theta - 10\mathrm{i}\cos^2\theta\sin^3\theta + 5\cos\theta\sin^4\theta + \mathrm{i}\sin^5\theta \]
Equating imaginary parts:
\[ \sin(5\theta) = 5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta \]
Using \(\cos^2\theta = 1 - \sin^2\theta\):
\[ \sin(5\theta) = 5(1 - \sin^2\theta)^2\sin\theta - 10(1 - \sin^2\theta)\sin^3\theta + \sin^5\theta \]
\[ = 5(1 - 2\sin^2\theta + \sin^4\theta)\sin\theta - 10\sin^3\theta + 10\sin^5\theta + \sin^5\theta \]
\[ = 5\
sin\theta - 10\sin^3\theta + 5\sin^5\theta - 10\sin^3\theta + 10\sin^5\theta + \sin^5\theta \]
\[ = 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta \]

(ii) Let \(x = \sin\theta\). Then the equation \(16x^4 - 20x^2 + 5 = 0\) becomes:
\[ 16\sin^4\theta - 20\sin^2\theta + 5 = 0 \]
Multiply both sides by \(\sin\theta\) (assuming \(\sin\theta \ne 0\)):
\[ 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta = 0 \]
which is:
\[ \sin(5\theta) = 0 \]
For \(\sin\theta \ne 0\), the solutions to \(\sin(5\theta) = 0\) are:
\[ 5\theta = k\pi \implies \theta = \frac{k\pi}{5} \quad \text{for } k \in \mathbb{Z}, k \text{ not a multiple of } 5 \]
Choosing \(k = 1, 2, -1, -2\), we obtain the four distinct roots:
\[ x = \sin\left(\frac{\pi}{5}\right), \sin\left(\frac{2\pi}{5}\right), \sin\left(-\frac{\pi}{5}\right) = -\sin\left(\frac{\pi}{5}\right), \sin\left(-\frac{2\pi}{5}\right) = -\sin\left(\frac{2\pi}{5}\right) \]
Thus, the roots of the equation are:
\[ x = \pm\sin\left(\frac{\pi}{5}\right), \pm\sin\left(\frac{2\pi}{5}\right) \]

M1: Uses De Moivre's theorem and expands \((\cos\theta + \mathrm{i}\sin\theta)^5\).
A1: Equates the imaginary parts correctly.
M1: Substitutes \(\cos^2\theta = 1 - \sin^2\theta\).
A1: Obtains the given expression for \(\sin(5\theta)\).
M1: Substitutes \(x = \sin\theta\) and multiplies by \(\sin\theta\) to form the connection with \(\sin(5\theta) = 0\).
M1: Solves \(\sin(5\theta) = 0\) to find the values of \(\theta\).
A1: Identifies the appropriate values of \(k\) to produce 4 distinct non-zero roots.
A1: Writes down the roots clearly as \(\pm\sin(\frac{\pi}{5})\) and \(\pm\sin(\frac{2\pi}{5})\).

(i) To find the eigenvalues, solve \(\det(\mathbf{A} - \lambda \mathbf{I}) = 0\):
\[ \det \begin{pmatrix} 3-\lambda & -1 & 1 \\ -1 & 5-\lambda & -1 \\ 1 & -1 & 3-\lambda \end{pmatrix} = 0 \]
Expanding along the first row:
\[ (3-\lambda)[(5-\lambda)(3-\lambda) - 1] + 1[-(3-\lambda) + 1] + 1[1 - (5-\lambda)] = 0 \]
\[ (3-\lambda)(\lambda^2 - 8\lambda + 14) + (\lambda - 2) + (\lambda - 4) = 0 \]
\[ (3-\lambda)(\lambda^2 - 8\lambda + 14) + 2\lambda - 6 = 0 \]
\[ (3-\lambda)(\lambda^2 - 8\lambda + 14) - 2(3-\lambda) = 0 \]
\[ (3-\lambda)(\lambda^2 - 8\lambda + 12) = 0 \]
\[ (3-\lambda)(\lambda-2)(\lambda-6) = 0 \]
The eigenvalues are \(\lambda = 2, 3, 6\).

(ii) The largest eigenvalue is \(\lambda = 6\).
The corresponding eigenvector \(\mathbf{x}\) satisfies \((\mathbf{A} - 6\mathbf{I})\mathbf{x} = \mathbf{0}\):
\[ \begin{pmatrix} -3 & -1 & 1 \\ -1 & -1 & -1 \\ 1 & -1 & -3 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \]
From row 2: \(x + y + z = 0 \implies y = -x - z\).
From row 1: \(-3x - y + z = 0 \implies -3x - (-x-z) + z = 0 \implies -2x + 2z = 0 \implies x = z\).
Then \(y = -2x\).
Choosing \(x=1\), an eigenvector is \(\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}\).
Normalizing this vector:
\[ \text{Magnitude} = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{6} \]
The normalized eigenvector is \(\frac{1}{\sqrt{6}}\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}\).

(iii) For \(\lambda = 2\):
\[ \begin{pmatrix} 1 & -1 & 1 \\ -1 & 3 & -1 \\ 1 & -1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies y = 0 \text{ and } x = -z \]
Normalized eigenvector is \(\frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}\).
For \(\lambda = 3\):
\[ \begin{pmatrix} 0 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies x = y = z \]
Normalized eigenvector is \(\frac{1}{\sqrt{3}}\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\).
Thus, the orthogonal matrix \(\mathbf{P}\) (formed by the normalized eigenvectors) is:
\[ \mathbf{P} = \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} \\ 0 & \frac{1}{\sqrt{3}} & -\frac{2}{\sqrt{6}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} \end{pmatrix} \]
And the diagonal matrix \(\mathbf{D}\) of eigenvalues is:
\[ \mathbf{D} = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 6 \end{pmatrix} \]

M1: Sets up the characteristic equation \(\det(\mathbf{A}-\lambda\mathbf{I}) = 0\).
A1: Expands the determinant and simplifies to a cubic.
A1: Finds the correct eigenvalues \(2, 3, 6\).
M1: Uses \(\lambda = 6\) to set up the system of equations for the eigenvector.
A1: Finds a correct eigenvector \(\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}\) or any multiple.
A1: Normalizes the eigenvector correctly.
M1: Finds normalized eigenvectors for the other two eigenvalues.
A1: States a correct orthogonal matrix \(\mathbf{P}\).
A1: States the corresponding diagonal matrix \(\mathbf{D}\).

(i) Starting from the right-hand side:
\[ 2\cosh^2 x - 1 = 2 \left( \frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2} \right)^2 - 1 \]
\[ = 2 \left( \frac{\mathrm{e}^{2x} + 2 + \mathrm{e}^{-2x}}{4} \right) - 1 \]
\[ = \frac{\mathrm{e}^{2x} + 2 + \mathrm{e}^{-2x}}{2} - 1 \]
\[ = \frac{\mathrm{e}^{2x} + \mathrm{e}^{-2x}}{2} = \cosh(2x) \]
Hence proved.

(ii) Using the identity from part (i) to rewrite the equation:
\[ (2\cosh^2 x - 1) - 5\cosh x + 3 = 0 \]
\[ 2\cosh^2 x - 5\cosh x + 2 = 0 \]
Let \(u = \cosh x\).
\[ 2u^2 - 5u + 2 = 0 \implies (2u - 1)(u - 2) = 0 \]
Since \(\cosh x \ge 1\) for all real \(x\), we reject \(u = \frac{1}{2}\).
Thus, \(\cosh x = 2\).
Using the logarithmic form of \(\operatorname{arcosh} z = \ln(z + \sqrt{z^2 - 1})\):
\[ x = \pm \operatorname{arcosh}(2) \]
\[ x = \pm \ln(2 + \sqrt{2^2 - 1}) \]
\[ x = \pm \ln(2 + \sqrt{3}) \]
Here \(a = 2\) and \(b = 3\).

M1: Expresses \(\cosh x\) in exponential form.
A1: Expands the squared term correctly.
A1: Completes the proof to reach \(\cosh(2x)\).
M1: Uses the identity to form a quadratic equation in \(\cosh x\).
A1: Solves the quadratic to find \(\cosh x = 2\) and rejects \(\cosh x = \frac{1}{2}\).
M1: Employs the logarithmic formula for \(\operatorname{arcosh}\).
A1: Simplifies the argument of the logarithm to find \(a = 2\) and \(b = 3\).
A1: Correctly expresses the final solutions as \(\pm \ln(2 + \sqrt{3})\).

(i) If \(y = \arctan x\), then
\[ \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{1}{1+x^2} \implies (1+x^2)\frac{\mathrm{d} y}{\mathrm{d} x} = 1 \]

(ii) Let \(u = \frac{\mathrm{d} y}{\mathrm{d} x}\) and \(v = 1 + x^2\).
Applying Leibniz's theorem to find the \(n\)-th derivative of \(uv\):
\[ \frac{\mathrm{d}^n}{\mathrm{d} x^n}(uv) = u^{(n)}v + n u^{(n-1)}v' + \frac{n(n-1)}{2} u^{(n-2)}v'' + \dots \]
Since \(v = 1+x^2\), we have \(v' = 2x\), \(v'' = 2\), and \(v^{(r)} = 0\) for \(r \ge 3\).
Therefore:
\[ \frac{\mathrm{d}^n}{\mathrm{d} x^n}\left((1+x^2)\frac{\mathrm{d} y}{\mathrm{d} x}\right) = (1+x^2)\frac{\mathrm{d}^{n+1} y}{\mathrm{d} x^{n+1}} + 2nx\frac{\mathrm{d}^n y}{\mathrm{d} x^n} + n(n-1)\frac{\mathrm{d}^{n-1} y}{\mathrm{d} x^{n-1}} \]
Since the \(n\)-th derivative of the right-hand side (which is \(1\)) is \(0\) for \(n \ge 1\), we obtain:
\[ (1+x^2)\frac{\mathrm{d}^{n+1} y}{\mathrm{d} x^{n+1}} + 2nx\frac{\mathrm{d}^n y}{\mathrm{d} x^n} + n(n-1)\frac{\mathrm{d}^{n-1} y}{\mathrm{d} x^{n-1}} = 0 \]

(iii) Let \(y^{(k)}(0)\) denote the value of the \(k\)-th derivative at \(x = 0\).
At \(x = 0\):
\(y(0) = \arctan(0) = 0\).
From part (i), at \(x = 0\):
\[ y^{(1)}(0) = 1 \]
From part (ii), substituting \(x = 0\):
\[ y^{(n+1)}(0) + n(n-1)y^{(n-1)}(0) = 0 \implies y^{(n+1)}(0) = -n(n-1)y^{(n-1)}(0) \]
For \(n = 1\): \(y^{(2)}(0) = 0\).
For \(n = 2\): \(y^{(3)}(0) = -2(1)y^{(1)}(0) = -2\).
For \(n = 3\): \(y^{(4)}(0) = -3(2)y^{(2)}(0) = 0\).
For \(n = 4\): \(y^{(5)}(0) = -4(3)y^{(3)}(0) = -12(-2) = 24\).

The Maclaurin's series expansion is:
\[ y(x) = y(0) + y^{(1)}(0)x + \frac{y^{(2)}(0)}{2!}x^2 + \frac{y^{(3)}(0)}{3!}x^3 + \frac{y^{(4)}(0)}{4!}x^4 + \frac{y^{(5)}(0)}{5!}x^5 + \dots \]
\[ y(x) = x - \frac{2}{6}x^3 + \frac{24}{120}x^5 = x - \frac{1}{3}x^3 + \frac{1}{5}x^5 \]

M1: Differentiates \(\arctan x\) to obtain \(\frac{1}{1+x^2}\).
A1: Rearranges to the given form.
M1: Correctly applies Leibniz's theorem to a product.
A1: Identifies \(v=1+x^2\), \(v'=2x\), \(v''=2\) and zero for higher terms.
A1: Shows the required Leibniz recurrence relation.
M1: Uses the recurrence relation at \(x = 0\) to compute successive derivatives.
A1: Obtains \(y'(0) = 1\), \(y''(0) = 0\), \(y'''(0) = -2\), \(y^{(4)}(0) = 0\), \(y^{(5)}(0) = 24\).
M1: Substitutes derivatives into the Maclaurin's series formula.
A1: Yields \(x - \frac{1}{3}x^3 + \frac{1}{5}x^5\).

First, find the derivative \(\frac{\mathrm{d} y}{\mathrm{d} x}\):
\[ y = \frac{1}{8} x^4 + \frac{1}{4} x^{-2} \]
\[ \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{1}{2} x^3 - \frac{1}{2} x^{-3} \]
Next, compute \(1 + \left(\frac{\mathrm{d} y}{\mathrm{d} x}\right)^2\):
\[ 1 + \left(\frac{\mathrm{d} y}{\mathrm{d} x}\right)^2 = 1 + \left(\frac{1}{2} x^3 - \frac{1}{2} x^{-3}\right)^2 \]
\[ = 1 + \left(\frac{1}{4} x^6 - \frac{1}{2} + \frac{1}{4} x^{-6}\right) \]
\[ = \frac{1}{4} x^6 + \frac{1}{2} + \frac{1}{4} x^{-6} \]
\[ = \left(\frac{1}{2} x^3 + \frac{1}{2} x^{-3}\right)^2 \]
The length of the arc, \(s\), is given by:
\[ s = \int_1^2 \sqrt{1 + \left(\frac{\mathrm{d} y}{\mathrm{d} x}\right)^2} \, \mathrm{d} x \]
\[ s = \int_1^2 \left(\frac{1}{2} x^3 + \frac{1}{2} x^{-3}\right) \, \mathrm{d} x \]
Integrate:
\[ s = \left[ \frac{1}{8} x^4 - \frac{1}{4} x^{-2} \right]_1^2 \]
Evaluate at the limits:
At \(x = 2\):
\[ \frac{1}{8}(16) - \frac{1}{4}\left(\frac{1}{4}\right) = 2 - \frac{1}{16} = \frac{31}{16} \]
At \(x = 1\):
\[ \frac{1}{8}(1) - \frac{1}{4}(1) = \frac{1}{8} - \frac{1}{4} = -\frac{1}{8} \]
Subtract the two values:
\[ s = \frac{31}{16} - \left(-\frac{1}{8}\right) = \frac{31}{16} + \frac{2}{16} = \frac{33}{16} \]

M1: Differentiates \(y\) with respect to \(x\).
A1: Obtains the correct expression for \(\frac{\mathrm{d}y}{\mathrm{d}x}\).
M1: Expresses \(1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2\) and expands correctly.
A1: Simplifies \(1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2\) to a perfect square.
M1: Sets up the correct arc length integral.
A1: Integrates the terms correctly.
M1: Substitutes the limits \(1\) and \(2\) into the integrated expression.
A1: Correctly simplifies the final answer to \(\frac{33}{16}\).

Divide the given differential equation by \(y^3\):
\[ y^{-3} \frac{\mathrm{d} y}{\mathrm{d} x} + y^{-2} = \mathrm{e}^{2x} \]
Given the substitution \(u = y^{-2}\):
\[ \frac{\mathrm{d} u}{\mathrm{d} x} = -2 y^{-3} \frac{\mathrm{d} y}{\mathrm{d} x} \implies y^{-3} \frac{\mathrm{d} y}{\mathrm{d} x} = -\frac{1}{2} \frac{\mathrm{d} u}{\mathrm{d} x} \]
Substituting these into the equation:
\[ -\frac{1}{2} \frac{\mathrm{d} u}{\mathrm{d} x} + u = \mathrm{e}^{2x} \]
Multiply by \(-2\):
\[ \frac{\mathrm{d} u}{\mathrm{d} x} - 2u = -2 \mathrm{e}^{2x} \]
This is a first-order linear differential equation. The integrating factor is:
\[ I(x) = \mathrm{e}^{\int -2 \, \mathrm{d} x} = \mathrm{e}^{-2x} \]
Multiply the differential equation by \( \mathrm{e}^{-2x} \):
\[ \frac{\mathrm{d}}{\mathrm{d} x}\left(u \mathrm{e}^{-2x}\right) = -2 \]
Integrate both sides with respect to \(x\):
\[ u \mathrm{e}^{-2x} = -2x + C \]
Multiply by \(\mathrm{e}^{2x}\):
\[ u = (C - 2x)\mathrm{e}^{2x} \]
Since \(u = y^{-2}\):
\[ y^{-2} = (C - 2x)\mathrm{e}^{2x} \implies y^2 = \frac{\mathrm{e}^{-2x}}{C - 2x} \]
Using the boundary condition: when \(x = 0\), \(y = 1\):
\[ 1^2 = \frac{\mathrm{e}^0}{C - 0} \implies 1 = \frac{1}{C} \implies C = 1 \]
Therefore, the particular solution is:
\[ y^2 = \frac{\mathrm{e}^{-2x}}{1 - 2x} \]

M1: Divides by \(y^3\) and prepares the equation for substitution.
M1: Differentiates \(u = y^{-2}\) to find \(\frac{\mathrm{d}u}{\mathrm{d}x}\) in terms of \(y\) and \(\frac{\mathrm{d}y}{\mathrm{d}x}\).
A1: Obtains the correct linear differential equation in \(u\).
M1: Finds the correct integrating factor \(\mathrm{e}^{-2x}\).
A1: Integrates both sides correctly to obtain \(u\mathrm{e}^{-2x} = -2x + C\).
A1: Writes the solution in terms of \(y\).
M1: Applies the initial condition \(y(0) = 1\) to find \(C\).
A1: Finds \(C = 1\).
A1: States the correct final relation \(y^2 = \frac{\mathrm{e}^{-2x}}{1 - 2x}\).

Let the coordinates of the particle at the given instant be \((d, 15)\). The trajectory equation of a projectile is given by:
\(y = x \tan \alpha - \frac{g x^2}{2 u^2 \cos^2 \alpha}\)
Thus at this instant:
\(15 = d \tan \alpha - \frac{g d^2}{2 u^2 \cos^2 \alpha}\) [1]

The horizontal and vertical components of the velocity at any time \(t\) are:
\(v_x = u \cos \alpha\)
\(v_y = u \sin \alpha - gt\)
The direction of motion \(\beta\) satisfies:
\(\tan \beta = \frac{v_y}{v_x} = \tan \alpha - \frac{gt}{u \cos \alpha}\)
Since \(d = u t \cos \alpha\), we have \(t = \frac{d}{u \cos \alpha}\). Substituting this into the equation for \(\tan \beta\):
\(\tan \beta = \tan \alpha - \frac{gd}{u^2 \cos^2 \alpha} \implies \frac{gd}{u^2 \cos^2 \alpha} = \tan \alpha - \tan \beta\) [2]

Substituting [2] into [1]:
\(15 = d \tan \alpha - \frac{d}{2}(\tan \alpha - \tan \beta)\)
\(15 = \frac{d}{2}(\tan \alpha + \tan \beta)\)
Given \(\tan \alpha = 2\) and \(\tan \beta = 0.5\):
\(15 = \frac{d}{2}(2 + 0.5)\)
\(15 = 1.25 d \implies d = 12\)

M1: Attempt to use the components of velocity to relate vertical and horizontal positions or write the trajectory equation.
A1: Correct equation for \(y\) in terms of \(d\), \(u\), and \(\alpha\).
M1: Express \(\tan \beta\) in terms of velocity components and substitute \(t = \frac{d}{u \cos \alpha}\).
A1: Obtain the relation \(\frac{gd}{u^2 \cos^2 \alpha} = \tan \alpha - \tan \beta\).
M1: Substitute this relation into the trajectory equation to eliminate \(u\).
A1: Obtain the simplified relation \(y = \frac{d}{2}(\tan \alpha + \tan \beta)\).
A1: Substitute the given values and solve to obtain \(d = 12\).

Let \(L = 0.8\text{ m\)} be the natural length of the string and \(x = 0.4\text{ m\)} be the maximum extension. At the point of maximum extension, the particle is instantaneously at rest. By the principle of conservation of energy, the loss in gravitational potential energy of the particle is equal to the gain in elastic potential energy of the string. The total distance the particle slides down the slope is:
\(d = L + x = 0.8 + 0.4 = 1.2\text{ m\)}
The vertical height descended by the particle is:
\(h = d \sin \theta = 1.2 \times 0.6 = 0.72\text{ m\)}
The loss in gravitational potential energy is:
\(E_{gp} = m g h = 2 \times 10 \times 0.72 = 14.4\text{ J\)}
The gain in elastic potential energy of the string at maximum extension is:
\(E_{ep} = \frac{\lambda x^2}{2 L} = \frac{\lambda (0.4)^2}{2 \times 0.8} = \frac{0.16 \lambda}{1.6} = 0.1 \lambda\text{ J\)}
Equating the two energies:
\(14.4 = 0.1 \lambda \implies \lambda = 144\)

M1: State or imply that the total distance moved along the slope is \(0.8 + 0.4 = 1.2\text{ m}\).
M1: Calculate the vertical height descended using \(h = d \sin \theta\).
A1: Obtain \(h = 0.72\text{ m}\).
M1: Formulate the loss of gravitational potential energy \(mgh\) and calculate its value.
A1: Obtain \(14.4\text{ J}\).
M1: Formulate the elastic potential energy in terms of \(\lambda\).
A1: Obtain \(0.1 \lambda\).
A1: Equate and solve for \(\lambda = 144\).

Let \(a = 1.4\text{ m\)} be the length of the string and \(y = 2.1\text{ m\)} be the height at which the string becomes slack. The angle \(\phi\) that the string makes with the upward vertical satisfies:
\(\cos \phi = \frac{y - a}{a} = \frac{2.1 - 1.4}{1.4} = 0.5\)
At the point where the string becomes slack, the tension \(T = 0\). The radial equation of motion towards the center \(O\) is:
\(T + mg \cos \phi = \frac{mv^2}{a}\)
Setting \(T = 0\):
\(v^2 = a g \cos \phi = 1.4 \times 10 \times 0.5 = 7\text{ m}^2\text{ s}^{-2}\)
By the conservation of energy between the lowest point and the point where the string becomes slack:
\(\frac{1}{2} m u^2 = \frac{1}{2} m v^2 + m g y\)
Dividing by \(\frac{1}{2} m\):
\(u^2 = v^2 + 2 g y\)
Substitute the values:
\(u^2 = 7 + 2(10)(2.1) = 7 + 42 = 49\)
Thus, \(u = \sqrt{49} = 7\)

M1: Use geometry to find \(\cos \phi\) where \(\phi\) is the angle with the upward vertical.
A1: Obtain \(\cos \phi = 0.5\).
M1: Apply Newton's second law radially with \(T = 0\) to find \(v^2\).
A1: Obtain \(v^2 = 7\).
M1: Apply conservation of energy between the launch point and the slack point.
A1: Obtain \(u^2 = v^2 + 2gy\).
A1: Solve to find \(u = 7\).

The equation of motion of the particle is:
\(m v \frac{dv}{dx} = \frac{36}{x^2} - 4\)
Given \(m = 2\text{ kg\)}:
\(2 v \frac{dv}{dx} = \frac{36}{x^2} - 4 \implies v \frac{dv}{dx} = \frac{18}{x^2} - 2\)
Integrating both sides with respect to \(x\):
\(\int v \, dv = \int \left( 18x^{-2} - 2 \right) dx\)
\(\frac{1}{2} v^2 = -\frac{18}{x} - 2x + C\)
Using the initial condition that at \(x = 3\), \(v = 4\):
\(\frac{1}{2} (4)^2 = -\frac{18}{3} - 2(3) + C \implies 8 = -6 - 6 + C \implies C = 20\)
So the velocity equation is:
\(\frac{1}{2} v^2 = -\frac{18}{x} - 2x + 20\)
The particle reaches its maximum distance from \(O\) when its velocity \(v = 0\):
\(0 = -\frac{18}{x} - 2x + 20 \implies 2x + \frac{18}{x} - 20 = 0 \implies 2x^2 - 20x + 18 = 0\)
Dividing by 2:
\(x^2 - 10x + 9 = 0 \implies (x - 1)(x - 9) = 0\)
This gives \(x = 1\) or \(x = 9\).
Since the particle starts at \(x = 3\) and is moving in the direction of increasing \(x\) (since initial velocity is positive), the maximum distance reached is \(x = 9\text{ m\)}.

M1: Set up the equation of motion using \(m v \frac{dv}{dx} = F_{net}\).
A1: Obtain \(2 v \frac{dv}{dx} = \frac{36}{x^2} - 4\).
M1: Integrate the differential equation with respect to \(x\).
A1: Obtain \(\frac{1}{2} v^2 = -\frac{18}{x} - 2x + C\).
M1: Apply the initial conditions to find the constant of integration \(C\).
A1: Obtain \(C = 20\).
M1: Set \(v = 0\) and solve the resulting quadratic equation in \(x\).
A1: Identify \(x = 9\) as the maximum distance with physical reasoning.

Let \(R\) be the normal reaction of the ground on the rod at \(A\), and \(F\) be the frictional force at \(A\). Let \(N\) be the normal reaction of the wall on the rod at \(B\). At the maximum value of \(P\), the rod is on the point of slipping towards the wall. Thus, the friction force \(F\) acts away from the wall (to the left) and is at its limiting value:
\(F = \mu R\), where \(\mu = 0.4\).
Resolving forces vertically:
\(R - 120 = 0 \implies R = 120\text{ N\)}
Thus, the maximum friction force is:
\(F = 0.4 \times 120 = 48\text{ N\)}
Resolving forces horizontally:
\(P - F - N = 0 \implies P = N + 48\)

To find \(N\), we take moments about \(A\) for the rod in equilibrium: The weight of \(120\text{ N\)} acts downwards at the midpoint of the rod (at a distance of \(2\text{ m\)} from \(A\)). The horizontal normal reaction \(N\) acts at \(B\) (at a distance of \(4\text{ m\)} from \(A\)).
Taking moments about \(A\):
\(120 \times (2 \cos 60^\circ) = N \times (4 \sin 60^\circ)\)
\(120 \times 1 = N \times 4 \left( \frac{\sqrt{3}}{2} \right)\)
\(120 = 2\sqrt{3} N \implies N = \frac{60}{\sqrt{3}} = 20\sqrt{3}\text{ N\)}
Substituting \(N\) back into the horizontal force equation:
\(P = 20\sqrt{3} + 48 = 48 + 20\sqrt{3}\)

M1: State or imply \(R = 120\text{ N}\) by resolving vertically.
A1: State that at maximum \(P\), friction is limiting and directed away from the wall, so \(F = 0.4 \times 120 = 48\text{ N}\).
M1: Resolve forces horizontally to obtain \(P = N + F\).
M1: Take moments about a point (e.g., \(A\)) to find \(N\).
A1: Set up the correct moment equation, e.g., \(120 \cos 60^\circ \times 2 = N \sin 60^\circ \times 4\).
A1: Solve for \(N = 20\sqrt{3}\).
A1: State the final answer as \(48 + 20\sqrt{3}\).

Let the line of centres of the spheres at the instant of collision be the \(x\)-axis. The initial velocity of \(A\) has components:
\(u_{Ax} = u \cos 30^\circ = \frac{\sqrt{3}}{2} u\)
\(u_{Ay} = u \sin 30^\circ = \frac{1}{2} u\)
Since sphere \(B\) is initially at rest:
\(u_{Bx} = 0\), \(u_{By} = 0\)
Since the contact is smooth, there is no impulse perpendicular to the line of centres, so the \(y\)-components of velocity are unchanged:
\(v_{Ay} = u_{Ay} = \frac{1}{2} u\)
\(v_{By} = 0\)

For the motion along the line of centres (\(x\)-axis):
Conservation of linear momentum:
\(m u_{Ax} + 2m u_{Bx} = m v_{Ax} + 2m v_{Bx} \implies v_{Ax} + 2 v_{Bx} = \frac{\sqrt{3}}{2} u\) [1]
Using Newton's law of restitution with \(e = 0.5\):
\(v_{Bx} - v_{Ax} = e(u_{Ax} - u_{Bx}) \implies v_{Bx} - v_{Ax} = 0.5 \left( \frac{\sqrt{3}}{2} u - 0 \right) = \frac{\sqrt{3}}{4} u\) [2]

Multiplying equation [2] by 2:
\(2 v_{Bx} - 2 v_{Ax} = \frac{\sqrt{3}}{2} u\) [3]
Subtracting [3] from [1]:
\(3 v_{Ax} = 0 \implies v_{Ax} = 0\)

So the velocity of \(A\) after the collision has components \((0, \frac{1}{2} u)\), which means its direction of motion is perpendicular to the line of centres (at \(90^\circ\) to it). The angle of deflection of \(A\) is the angle between its initial and final directions of motion:
\(\theta_{\text{deflected}} = 90^\circ - 30^\circ = 60^\circ\)

M1: Resolve initial velocity of \(A\) along and perpendicular to the line of centres.
A1: Obtain \(u_{Ax} = \frac{\sqrt{3}}{2} u\) and \(u_{Ay} = \frac{1}{2} u\).
M1: Set up the conservation of momentum equation along the line of centres.
A1: Obtain \(v_{Ax} + 2 v_{Bx} = \frac{\sqrt{3}}{2} u\).
M1: Set up the Newton's law of restitution equation along the line of centres.
A1: Obtain \(v_{Bx} - v_{Ax} = \frac{\sqrt{3}}{4} u\).
A1: Solve the system to show \(v_{Ax} = 0\).
A1: Find the angle of deflection \(90^\circ - 30^\circ = 60^\circ\).

(i) For \(0 \le x \le 2\), the cumulative distribution function is \(F(x) = \int_0^x \frac{3}{8}t^2 \, dt = [\frac{1}{8}t^3]_0^x = \frac{1}{8}x^3\). Hence, \(F(x) = 0\) for \(x < 0\), \(F(x) = \frac{1}{8}x^3\) for \(0 \le x \le 2\), and \(F(x) = 1\) for \(x > 2\). (ii) For \(0 \le y \le 4\), the cumulative distribution function of \(Y\) is given by \(G(y) = P(Y \le y) = P(X^2 \le y) = P(X \le \sqrt{y}) = F(\sqrt{y}) = \frac{1}{8}(\sqrt{y})^3 = \frac{1}{8}y^{3/2}\). Differentiating \(G(y)\) with respect to \(y\) gives the probability density function: \(g(y) = G'(y) = \frac{3}{16}y^{1/2}\) for \(0 \le y \le 4\), and \(0\) otherwise. (iii) Let \(m\) be the median of \(Y\). We solve \(G(m) = 0.5 \implies \frac{1}{8}m^{3/2} = 0.5 \implies m^{3/2} = 4 \implies m = 4^{2/3} = (2^2)^{2/3} = 2^{4/3}\).

(i) M1 for integrating \(f(x)\) to find \(F(x)\). A1 for correct piecewise CDF with ranges. (ii) M1 for writing \(P(Y \le y) = P(X \le \sqrt{y})\). A1 for \(G(y) = \frac{1}{8}y^{3/2}\). M1 for differentiating \(G(y)\) to get \(g(y)\). A1 for correct \(g(y)\) and domain \(0 \le y \le 4\). (iii) M1 for setting \(G(m) = 0.5\) and solving for \(m\). A1.33 for showing \(m = 2^{4/3}\) clearly.

(i) Since \(G_X(1) = 1\), we have \(k(2 + 1 + 1)^2 = 1 \implies 16k = 1 \implies k = \frac{1}{16}\). (ii) Expanding \(G_X(t)\): \(G_X(t) = \frac{1}{16}(2 + t + t^2)^2 = \frac{1}{16}(4 + 4t + 5t^2 + 2t^3 + t^4)\). Comparing coefficients of \(t^r\) with \(P(X = r)\): \(P(X = 0) = \frac{4}{16} = \frac{1}{4}\), \(P(X = 1) = \frac{4}{16} = \frac{1}{4}\), \(P(X = 2) = \frac{5}{16}\), \(P(X = 3) = \frac{2}{16} = \frac{1}{8}\), \(P(X = 4) = \frac{1}{16}\). (iii) Differentiating \(G_X(t)\): \(G'_X(t) = \frac{2}{16}(2 + t + t^2)(1 + 2t) = \frac{1}{8}(2 + t + t^2)(1 + 2t)\). Thus, \(E(X) = G'_X(1) = \frac{1}{8}(4)(3) = 1.5\). Differentiating again: \(G''_X(t) = \frac{1}{8}[(1 + 2t)^2 + 2(2 + t + t^2)]\). Thus, \(G''_X(1) = \frac{1}{8}[3^2 + 2(4)] = \frac{17}{8} = 2.125\). The variance is \(Var(X) = G''_X(1) + G'_X(1) - (G'_X(1))^2 = 2.125 + 1.5 - 1.5^2 = 3.625 - 2.25 = 1.375\).

(i) B1 for setting \(G_X(1) = 1\) and showing \(k = 1/16\). (ii) M1 for expanding the expression. A2 for all correct probabilities (A1 for at least 3 correct). (iii) M1 for differentiating \(G_X(t)\) to find \(G'_X(1)\). A1 for \(E(X) = 1.5\). M1 for finding \(G''_X(1)\). A1 for \(G''_X(1) = 2.125\). M1 for using the variance formula. A0.33 for \(Var(X) = 1.375\).

Let \(D\) be the difference \(\text{Before} - \text{After}\). The hypotheses are: \(H_0\): The median difference is zero (no change in anxiety). \(H_1\): The median difference is greater than zero (anxiety is reduced after therapy). The differences \(D\) for Patients 1 to 8 are: Patient 1: \(24 - 18 = +6\); Patient 2: \(15 - 17 = -2\); Patient 3: \(31 - 22 = +9\); Patient 4: \(20 - 21 = -1\); Patient 5: \(28 - 20 = +8\); Patient 6: \(22 - 15 = +7\); Patient 7: \(19 - 24 = -5\); Patient 8: \(26 - 23 = +3\). The absolute differences in ascending order are: 1, 2, 3, 5, 6, 7, 8, 9. Assigning ranks to these absolute values: 1 (Rank 1, negative), 2 (Rank 2, negative), 3 (Rank 3, positive), 5 (Rank 4, negative), 6 (Rank 5, positive), 7 (Rank 6, positive), 8 (Rank 7, positive), 9 (Rank 8, positive). Sum of negative ranks: \(W_- = 1 + 2 + 4 = 7\). Sum of positive ranks: \(W_+ = 3 + 5 + 6 + 7 + 8 = 29\). The test statistic \(T = \min(W_-, W_+) = 7\). For a one-tailed test with \(n = 8\) at the 5% significance level, the critical value is 5. Since \(T = 7 > 5\), we do not reject \(H_0\). There is insufficient evidence at the 5% level to suggest that the cognitive therapy reduces anxiety.

B1.33 for correctly stating \(H_0\) and \(H_1\) (one-tailed). M1 for calculating the differences for each patient. M2 for ranking the absolute differences correctly. M2 for calculating the sum of positive and negative ranks (\(W_- = 7\), \(W_+ = 29\)). B1 for identifying the correct critical value of 5. A1 for comparing the test statistic with the critical value and stating a correct conclusion in context.

(i) For Type A: \(n_A = 6\), \(\sum x_A = 81\), \(\bar{x}_A = 13.5\), \(\sum x_A^2 = 1099.24\). Sum of squares \(\sum(x_A - \bar{x}_A)^2 = 1099.24 - \frac{81^2}{6} = 5.74\). For Type B: \(n_B = 5\), \(\sum x_B = 59.5\), \(\bar{x}_B = 11.9\), \(\sum x_B^2 = 710.75\). Sum of squares \(\sum(x_B - \bar{x}_B)^2 = 710.75 - \frac{59.5^2}{5} = 2.70\). The pooled estimate of the population variance is: \(s_p^2 = \frac{5.74 + 2.70}{6 + 5 - 2} = \frac{8.44}{9} \approx 0.9378\). (ii) We test \(H_0: \mu_A = \mu_B\) against \(H_1: \mu_A > \mu_B\). The standard error is \(SE = \sqrt{s_p^2 (\frac{1}{n_A} + \frac{1}{n_B})} = \sqrt{0.9378 (\frac{1}{6} + \frac{1}{5})} = \sqrt{0.34385} \approx 0.5864\). The test statistic is: \(t = \frac{\bar{x}_A - \bar{x}_B}{SE} = \frac{13.5 - 11.9}{0.5864} \approx 2.73\). The degrees of freedom is \(\nu = 6 + 5 - 2 = 9\). The critical value for a one-tailed test at the 5% significance level with 9 degrees of freedom is \(1.833\). Since \(t = 2.73 > 1.833\), we reject \(H_0\). There is significant evidence at the 5% level to conclude that the mean breaking strength of Type A fiber is greater than that of Type B fiber.

(i) M1 for finding the sum of squares for both samples. M1 for the pooled variance formula. A1 for \(s_p^2 \approx 0.9378\). (ii) B1 for stating the hypotheses. M1 for calculating the test statistic \(t\). A1 for obtaining \(t \approx 2.73\). B1 for identifying 9 degrees of freedom and the critical value of 1.833. A1.33 for comparing the test statistic with the critical value and making a correct decision with context.

The hypotheses are: \(H_0\): Age group and music genre preference are independent. \(H_1\): Age group and music genre preference are associated. The row and column totals are: Row 1 (Under 30) total = 100, Row 2 (30 and Over) total = 80; Column 1 (Classical) total = 50, Column 2 (Rock) total = 70, Column 3 (Pop) total = 60; Total sample size \(N = 180\). The expected frequencies \(E = \frac{\text{Row Total} \times \text{Column Total}}{N}\) are: Under 30 & Classical: \(27.78\); Under 30 & Rock: \(38.89\); Under 30 & Pop: \(33.33\); 30 and Over & Classical: \(22.22\); 30 and Over & Rock: \(31.11\); 30 and Over & Pop: \(26.67\). Calculating \(\chi^2 = \sum \frac{(O - E)^2}{E}\): Under 30 & Classical: \(\frac{(15 - 27.78)^2}{27.78} = 5.88\); Under 30 & Rock: \(\frac{(45 - 38.89)^2}{38.89} = 0.96\); Under 30 & Pop: \(\frac{(40 - 33.33)^2}{33.33} = 1.33\); 30 and Over & Classical: \(\frac{(35 - 22.22)^2}{22.22} = 7.35\); 30 and Over & Rock: \(\frac{(25 - 31.11)^2}{31.11} = 1.20\); 30 and Over & Pop: \(\frac{(20 - 26.67)^2}{26.67} = 1.67\). Summing these terms: \(\chi^2 = 5.88 + 0.96 + 1.33 + 7.35 + 1.20 + 1.67 = 18.39\). Degrees of freedom: \(\nu = (2 - 1) \times (3 - 1) = 2\). The critical value of \(\chi^2\) with 2 degrees of freedom at the 1% significance level is \(9.210\). Since \(\chi^2 = 18.39 > 9.210\), we reject \(H_0\). There is significant evidence at the 1% level to conclude that there is an association between age group and preferred music genre.

B1.33 for stating the hypotheses correctly. M3 for calculating all 6 expected frequencies (0.5 marks each). M2 for calculating \(\chi^2 = 18.39\). B1 for identifying 2 degrees of freedom and the critical value of 9.210. A1 for comparing with critical value and concluding in context.

(i) For \(x < 0\), \(F(x) = 0\). For \(0 \le x < 1\), \(F(x) = \int_0^x \frac{1}{2} \, dt = \frac{1}{2}x\). For \(x \ge 1\), \(F(x) = F(1) + \int_1^x \frac{1}{2t^2} \, dt = \frac{1}{2} + [-\frac{1}{2t}]_1^x = \frac{1}{2} + \frac{1}{2} - \frac{1}{2x} = 1 - \frac{1}{2x}\). So, the cumulative distribution function is: \(F(x) = 0\) for \(x < 0\), \(F(x) = \frac{1}{2}x\) for \(0 \le x < 1\), and \(F(x) = 1 - \frac{1}{2x}\) for \(x \ge 1\). (ii) The lower quartile \(q_1\) satisfies \(F(q_1) = 0.25\). Since \(F(1) = 0.5\), \(q_1\) lies in \([0, 1)\): \(\frac{1}{2}q_1 = 0.25 \implies q_1 = 0.5\). The upper quartile \(q_3\) satisfies \(F(q_3) = 0.75\). Since \(0.75 > 0.5\), \(q_3\) lies in \([1, \infty)\): \(1 - \frac{1}{2q_3} = 0.75 \implies \frac{1}{2q_3} = 0.25 \implies q_3 = 2\). The interquartile range is \(IQR = q_3 - q_1 = 2 - 0.5 = 1.5\). (iii) The conditional probability is \(P(X > 3 \mid X > 0.5) = \frac{P(X > 3)}{P(X > 0.5)}\). We have \(P(X > 3) = 1 - F(3) = 1 - (1 - \frac{1}{6}) = \frac{1}{6}\), and \(P(X > 0.5) = 1 - F(0.5) = 1 - 0.25 = 0.75 = \frac{3}{4}\). Thus, \(P(X > 3 \mid X > 0.5) = \frac{1/6}{3/4} = \frac{2}{9}\).

(i) M1 for integrating \(f(x)\) in the first interval. M1 for integrating in the second interval. A1 for the fully correct piecewise CDF with ranges. (ii) M1 for setting \(F(q_1) = 0.25\) and solving. A1 for \(q_1 = 0.5\). M1 for setting \(F(q_3) = 0.75\) and solving. A0.33 for \(q_3 = 2\) and finding \(IQR = 1.5\). (iii) M1 for writing the conditional probability formula. A1 for calculating \(2/9\).