2025 Cambridge IAL Mathematics - Further (9231) 模擬試題連答案詳解

(a) The area of the region enclosed by a polar curve is given by:
\text{Area} = \frac{1}{2} \int_{0}^{2\pi} r^2 \, d\theta = \frac{1}{2} \int_{0}^{2\pi} a^2(1 - \cos \theta)^2 \, d\theta
= \frac{1}{2} a^2 \int_{0}^{2\pi} (1 - 2\cos \theta + \cos^2 \theta) \, d\theta
Using the double-angle identity \(\cos^2 \theta = \frac{1 + \cos 2\theta}{2}\), we obtain:
\text{Area} = \frac{1}{2} a^2 \int_{0}^{2\pi} \left(1 - 2\cos \theta + \frac{1}{2} + \frac{1}{2}\cos 2\theta\right) \, d\theta
= \frac{1}{2} a^2 \int_{0}^{2\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^{2\pi}
Evaluating the limits:
= \frac{1}{2} a^2 \left[ \left(3\pi - 0 + 0\right) - \left(0 - 0 + 0\right) \right] = \frac{3}{2}\pi a^2.

(b) 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 solve \(\frac{dy}{d\theta} = 0\):
\frac{dy}{d\theta} = a\left( \sin \theta \cdot (\sin \theta) + (1 - \cos \theta)\cos \theta \right) = 0
\implies a\left( \sin^2 \theta + \cos \theta - \cos^2 \theta \right) = 0
Using \(\sin^2 \theta = 1 - \cos^2 \theta\):
1 - 2\cos^2 \theta + \cos \theta = 0
\implies 2\cos^2 \theta - \cos \theta - 1 = 0
\implies (2\cos \theta + 1)(\cos \theta - 1) = 0
So \(\cos \theta = -\frac{1}{2}\) or \(\cos \theta = 1\).
- If \(\cos \theta = 1\), then \(\theta = 0\) or \(2\pi\) (at the pole, giving a cusp).
- If \(\cos \theta = -\frac{1}{2}\), then \(\theta = \frac{2\pi}{3}\) or \(\theta = \frac{4\pi}{3}\).

When \(\theta = \frac{2\pi}{3}\), \(r = a(1 - (-\frac{1}{2})) = \frac{3}{2}a\).
When \(\theta = \frac{4\pi}{3}\), \(r = a(1 - (-\frac{1}{2})) = \frac{3}{2}a\).

Thus, the required polar coordinates are \((\frac{3}{2}a, \frac{2\pi}{3})\) and \((\frac{3}{2}a, \frac{4\pi}{3})\).

(a)
M1: Uses \(\frac{1}{2} \int r^2 \, d\theta\) with correct limits.
M1: Expands integrand and uses double-angle identity for \(\cos^2 \theta\).
A1: Obtains correct integrated expression: \(\frac{3}{2}\theta - 2\sin \theta + \frac{1}{4}\sin 2\theta\).
A1: Substitutes limits correctly.
A1: Obtains the correct shown area \(\frac{3}{2}\pi a^2\).

(b)
M1: Writes \(y = r \sin \theta\) and attempts differentiation.
M1: Sets derivative \(\frac{dy}{d\theta} = 0\) and derives quadratic in \(\cos \theta\).
A1: Obtains correct solutions for \(\cos \theta\), i.e., \(-\frac{1}{2}\) (and 1).
A1: Finds the correct angles \(\theta = \frac{2\pi}{3}\) and \(\theta = \frac{4\pi}{3}\).
A2: Writes the correct coordinates: \((\frac{3}{2}a, \frac{2\pi}{3})\) and \((\frac{3}{2}a, \frac{4\pi}{3})\) (1 mark for each point).

(a) Let the invariant line through the origin be \(y = mx\).
Under the transformation \(T\):
\begin{pmatrix} X \\ Y \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ -3 & 6 \end{pmatrix} \begin{pmatrix} x \\ mx \end{pmatrix} = \begin{pmatrix} 2x + mx \\ -3x + 6mx \end{pmatrix}
Since \(Y = mX\), we have:
-3x + 6mx = m(2x + mx)
Since this holds for all points on the line, we can assume \(x \ne 0\) and divide both sides by \(x\):
-3 + 6m = m(2 + m)
\implies -3 + 6m = 2m + m^2
\implies m^2 - 4m + 3 = 0
\implies (m - 1)(m - 3) = 0
Thus, \(m = 1\) or \(m = 3\).
So the invariant lines are \(y = x\) and \(y = 3x\).

(b) First, compute \(\mathbf{M}^2\):
\mathbf{M}^2 = \begin{pmatrix} 2 & 1 \\ -3 & 6 \end{pmatrix} \begin{pmatrix} 2 & 1 \\ -3 & 6 \end{pmatrix} = \begin{pmatrix} 2(2) + 1(-3) & 2(1) + 1(6) \\ -3(2) + 6(-3) & -3(1) + 6(6) \end{pmatrix} = \begin{pmatrix} 1 & 8 \\ -24 & 33 \end{pmatrix}
Now, let \(y = mx\) be an invariant line under \(\mathbf{M}^2\).
\begin{pmatrix} X \\ Y \end{pmatrix} = \begin{pmatrix} 1 & 8 \\ -24 & 33 \end{pmatrix} \begin{pmatrix} x \\ mx \end{pmatrix} = \begin{pmatrix} x + 8mx \\ -24x + 33mx \end{pmatrix}
Using \(Y = mX\):
-24x + 33mx = m(x + 8mx)
\implies -24 + 33m = m(1 + 8m) \quad (\text{for } x
e 0)
\implies -24 + 33m = m + 8m^2
\implies 8m^2 - 32m + 24 = 0
Dividing by 8:
m^2 - 4m + 3 = 0
This is the identical quadratic equation as in part (a), yielding the same solutions \(m = 1\) and \(m = 3\). Hence, the lines \(y = x\) and \(y = 3x\) are also invariant lines under \(\mathbf{M}^2\).

(a)
M1: Sets up the matrix multiplication \(\mathbf{M}\mathbf{x}\) for a general point \((x, mx)^T\).
M1: Applies the invariant line condition \(Y = mX\).
A1: Obtains the quadratic equation \(m^2 - 4m + 3 = 0\).
A1: Solves to find \(m = 1, 3\).
A1: Writes down the final equations: \(y = x\) and \(y = 3x\).

(b)
M1: Attempts to multiply \(\mathbf{M}\mathbf{M}\).
A1: Correctly finds \(\mathbf{M}^2 = \begin{pmatrix} 1 & 8 \\ -24 & 33 \end{pmatrix}\).
M1: Sets up the invariant line relation for \(\mathbf{M}^2\).
A1: Derives the identical quadratic equation \(m^2 - 4m + 3 = 0\) and concludes clearly.

Let \(f(n) = 5^{2n} + 24n - 1\).

(a) For \(n = 1\):
\(f(1) = 5^2 + 24(1) - 1 = 25 + 24 - 1 = 48\).
Since 48 is divisible by 48, the statement is true for \(n = 1\).

(b) For \(n = k + 1\):
\(f(k+1) = 5^{2(k+1)} + 24(k+1) - 1\)
\(= 25 \cdot 5^{2k} + 24k + 23\).
We want to write this in the form \(25(5^{2k} + 24k - 1) + g(k)\).
Comparing terms:
\(25(5^{2k} + 24k - 1) + g(k) = 25 \cdot 5^{2k} + 600k - 25 + g(k)\).
Therefore, we require:
\(25 \cdot 5^{2k} + 600k - 25 + g(k) = 25 \cdot 5^{2k} + 24k + 23\)
\(\implies g(k) = 24k + 23 - (600k - 25) = -576k + 48\)
Factoring out \(-48\):
\(g(k) = -48(12k - 1)\).

(c) Assume that \(f(k)\) is divisible by 48 for some positive integer \(k\), so \(f(k) = 48M\) for some integer \(M\).
From part (b):
\(f(k+1) = 25(f(k)) - 48(12k - 1)\)
\(= 25(48M) - 48(12k - 1)\)
\(= 48[25M - (12k - 1)]\).
Since \(25M - 12k + 1\) is an integer, \(f(k+1)\) is also divisible by 48.
Hence, if the statement is true for \(n=k\), it is also true for \(n=k+1\). Since it has been shown to be true for \(n=1\), by mathematical induction the statement is true for all positive integers \(n\).

(a)
B1: Computes \(f(1) = 48\) and states that it is divisible by 48.

(b)
M1: Writes out expression for \(f(k+1)\).
M1: Aligns \(f(k+1)\) with the form \(25 f(k) + g(k)\).
A1: Finds \(g(k) = -576k + 48\).
A1: Simplifies to \(g(k) = -48(12k - 1)\).

(c)
M1: States the inductive hypothesis \(f(k) = 48M\).
M1: Expresses \(f(k+1)\) as \(48(25M - 12k + 1)\).
A1: Explains clearly that the expression is a multiple of 48 because all terms in the brackets are integers.
A1: Gives a complete and logically sound concluding statement.

(a) A vector perpendicular to both lines is given by the cross product of their direction vectors:
\mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -1 \\ 2 & -1 & 3 \end{vmatrix}
= \mathbf{i}(2(3) - (-1)(-1)) - \mathbf{j}(1(3) - (-1)(2)) + \mathbf{k}(1(-1) - 2(2))
= \mathbf{i}(6 - 1) - \mathbf{j}(3 + 2) + \mathbf{k}(-1 - 4)
= 5\mathbf{i} - 5\mathbf{j} - 5\mathbf{k}.
We can simplify this to the direction vector \(\mathbf{n}_0 = \mathbf{i} - \mathbf{j} - \mathbf{k}\).

(b) The position vector from \(A\) to \(B\) is:
\vec{AB} = \mathbf{r}_B - \mathbf{r}_A = \begin{pmatrix} 3 \\ 0 \\ 2 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 2 \\ -2 \\ 3 \end{pmatrix}.
The shortest distance between two skew lines is the projection of \(\vec{AB}\) onto the common perpendicular \(\mathbf{n}_0\):
d = \frac{|\vec{AB} \cdot \mathbf{n}_0|}{|\mathbf{n}_0|}
= \frac{|2(1) + (-2)(-1) + 3(-1)|}{\sqrt{1^2 + (-1)^2 + (-1)^2}}
= \frac{|2 + 2 - 3|}{\sqrt{3}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}.

(c) The plane \(\Pi\) contains \(l_1\) and is parallel to \(l_2\), so its normal vector is parallel to \(\mathbf{n}_0 = \mathbf{i} - \mathbf{j} - \mathbf{k}\).
The equation of the plane is:
1(x - 1) - 1(y - 2) - 1(z + 1) = 0
\implies x - 1 - y + 2 - z - 1 = 0
\implies x - y - z = 0.

(a)
M1: Calculates cross product of direction vectors \(\mathbf{d}_1 \times \mathbf{d}_2\).
A1: Obtains correct vector \(5\mathbf{i} - 5\mathbf{j} - 5\mathbf{k}\) (or any scalar multiple).
B1: Writes a simplified version, e.g., \(\mathbf{i} - \mathbf{j} - \mathbf{k}\).

(b)
M1: Calculates vector \(\vec{AB}\).
M1: Uses the formula \(d = \frac{|\vec{AB} \cdot \mathbf{n}|}{|\mathbf{n}|}\).
A1: Correctly evaluates dot product and magnitude.
A1: Obtains \(\frac{\sqrt{3}}{3}\) (or equivalent).

(c)
M1: Identifies normal vector of plane is the perpendicular vector from part (a).
M1: Substitutes point \(A(1, 2, -1)\) (or any point on \(l_1\)) into equation \(\mathbf{r} \cdot \mathbf{n} = D\).
A1: Obtains correct Cartesian equation \(x - y - z = 0\).

(a) Using relations between roots and coefficients for \(x^3 - 3x^2 + 5x - 2 = 0\):
(i) \(\alpha + \beta + \gamma = 3\)
(ii) \(\alpha\beta + \beta\gamma + \gamma\alpha = 5\)
(iii) \(\alpha\beta\gamma = 2\)

(b) Using the identity:
\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)
= 3^2 - 2(5) = 9 - 10 = -1.

(c) Let \(y = \frac{1}{x^2}\). Then \(x^2 = \frac{1}{y}\), which implies \(x = \pm \frac{1}{\sqrt{y}}\).
Substitute \(x\) into the original cubic equation:
\left(\pm \frac{1}{\sqrt{y}}\right)^3 - 3\left(\pm \frac{1}{\sqrt{y}}\right)^2 + 5\left(\pm \frac{1}{\sqrt{y}}\right) - 2 = 0
\implies \pm \frac{1}{y\sqrt{y}} - \frac{3}{y} \pm \frac{5}{\sqrt{y}} - 2 = 0
Grouping the terms with \(\pm \frac{1}{\sqrt{y}}\):
\pm \frac{1}{\sqrt{y}} \left( \frac{1}{y} + 5 \right) = \frac{3}{y} + 2
\implies \pm \frac{1 + 5y}{y\sqrt{y}} = \frac{3 + 2y}{y}
Square both sides:
\frac{(1 + 5y)^2}{y^3} = \frac{(3 + 2y)^2}{y^2}
Multiply by \(y^3\) (since \(y \ne 0\)):
(1 + 5y)^2 = y(3 + 2y)^2
Expand both sides:
1 + 10y + 25y^2 = y(9 + 12y + 4y^2)
\implies 1 + 10y + 25y^2 = 9y + 12y^2 + 4y^3
Rearranging to standard form:
4y^3 - 13y^2 - y - 1 = 0.
This has integer coefficients as required.

(a)
B1: Correctly states \(\alpha+\beta+\gamma=3\).
B1: Correctly states \(\alpha\beta+\beta\gamma+\gamma\alpha=5\).
B1: Correctly states \(\alpha\beta\gamma=2\).

(b)
M1: Employs the identity for sum of squares.
A1: Correctly computes the value \(-1\).

(c)
M1: Introduces substitution \(y = 1/x^2\) or attempts to find new symmetric sums.
M1: Rearranges the equation to isolate terms with square roots.
M1: Squares both sides of the equation correctly.
A1: Correctly expands both sides.
A1: Obtains the correct cubic equation \(4y^3 - 13y^2 - y - 1 = 0\) (or any integer multiple).

(a) The vertical asymptotes occur where the denominator is zero:
x^2 - x = 0 \implies x(x-1) = 0 \implies x = 0 \text{ and } x = 1.
The horizontal asymptote occurs as \(x \to \pm\infty\):
y = \lim_{x\to\pm\infty} \frac{2x^2 + x + 1}{x^2 - x} = 2.
So the vertical asymptotes are \(x = 0\) and \(x = 1\), and the horizontal asymptote is \(y = 2\).

(b) To find the set of values of \(y\) that can be taken, we express the equation as a quadratic in \(x\):
y(x^2 - x) = 2x^2 + x + 1
\implies yx^2 - yx - 2x^2 - x - 1 = 0
\implies (y - 2)x^2 - (y + 1)x - 1 = 0.
For real \(x\), the discriminant of this quadratic equation must be non-negative (\(\Delta \ge 0\)):
\Delta = [-(y+1)]^2 - 4(y-2)(-1) \ge 0
\implies (y+1)^2 + 4(y-2) \ge 0
\implies y^2 + 2y + 1 + 4y - 8 \ge 0
\implies y^2 + 6y - 7 \ge 0
\implies (y+7)(y-1) \ge 0.
The solution to this inequality is \(y \le -7\) or \(y \ge 1\).
Thus, the values of \(y\) that cannot be taken by any real \(x\) are \(-7 < y < 1\).

(c) The turning points occur at the boundary of the range of \(y\), where \(\Delta = 0\), i.e., \(y = 1\) and \(y = -7\).
- When \(y = 1\):
(1 - 2)x^2 - (1 + 1)x - 1 = 0 \implies -x^2 - 2x - 1 = 0 \implies (x+1)^2 = 0 \implies x = -1.
So one turning point is \((-1, 1)\).
- When \(y = -7\):
(-7 - 2)x^2 - (-7 + 1)x - 1 = 0 \implies -9x^2 + 6x - 1 = 0 \implies (3x - 1)^2 = 0 \implies x = \frac{1}{3}.
So the other turning point is \((\frac{1}{3}, -7)\).

(Alternatively, one can differentiate using the quotient rule and find stationary points.)

(a)
B1: Finds both vertical asymptotes \(x = 0\) and \(x = 1\).
B1: Finds horizontal asymptote \(y = 2\).

(b)
M1: Expresses the curve equation as a quadratic in \(x\).
M1: Sets up discriminant \(\Delta = b^2 - 4ac\).
A1: Obtains \(y^2 + 6y - 7\).
M1: Sets \(\Delta \ge 0\) and solves inequality.
A1: Concludes correctly that the range of impossible values is \(-7 < y < 1\).

(c)
M1: Identifies that turning points occur at the boundaries \(y = 1\) and \(y = -7\) (or uses quotient rule to differentiate).
M1: Solves the resulting equations to find the corresponding \(x\)-values.
A1: Obtains correct coordinates \((-1, 1)\) and \((\frac{1}{3}, -7)\).

(a) Combine the right-hand side over a common denominator:
\frac{1}{r^2} - \frac{1}{(r+1)^2} = \frac{(r+1)^2 - r^2}{r^2(r+1)^2}
= \frac{(r^2 + 2r + 1) - r^2}{r^2(r+1)^2}
= \frac{2r + 1}{r^2(r+1)^2}.
This is equal to the left-hand side, as required.

(b) Let \(u_r = \frac{1}{r^2} - \frac{1}{(r+1)^2}\). Using the method of differences:
\sum_{r=1}^{n} u_r = \left(1 - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{9}\right) + \dots + \left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right)
Most terms cancel, leaving:
\sum_{r=1}^{n} u_r = 1 - \frac{1}{(n+1)^2}.

(c) The sum to infinity is:
\sum_{r=1}^{\infty} u_r = \lim_{n \to \infty} \left( 1 - \frac{1}{(n+1)^2} \right) = 1.

(d) The sum from \(r = n+1\) to \(2n\) is given by:
\sum_{r=n+1}^{2n} u_r = \sum_{r=1}^{2n} u_r - \sum_{r=1}^{n} u_r
= \left( 1 - \frac{1}{(2n+1)^2} \right) - \left( 1 - \frac{1}{(n+1)^2} \right)
= \frac{1}{(n+1)^2} - \frac{1}{(2n+1)^2}
= \frac{(2n+1)^2 - (n+1)^2}{(n+1)^2 (2n+1)^2}
= \frac{(4n^2 + 4n + 1) - (n^2 + 2n + 1)}{(n+1)^2 (2n+1)^2}
= \frac{3n^2 + 2n}{(n+1)^2 (2n+1)^2}.
Comparing this to the required form, we find \(a = 3\) and \(b = 2\).

(a)
M1: Puts right-hand side over a common denominator.
A1: Simplifies the numerator to show it is equal to \(2r + 1\).

(b)
M1: Writes out first few terms and last term to show cancellation.
A1: Applies the method of differences successfully.
A1: Obtains \(1 - \frac{1}{(n+1)^2}\).

(c)
B1: Deduces limit is 1.

(d)
M1: Uses the relation \(S_{2n} - S_n\).
M1: Substitutes \(2n\) and \(n\) into the sum formula.
A1: Correctly algebraic expansion of \((2n+1)^2\) and \((n+1)^2\).
A1: Obtains the correct fraction \(\frac{3n^2 + 2n}{(n+1)^2(2n+1)^2}\).
A1: Correctly identifies \(a = 3\) and \(b = 2\).

Let \(z = \cos\theta + i\sin\theta\). By de Moivre's theorem, \(z^5 = \cos(5\theta) + i\sin(5\theta)\). Expanding \((\cos\theta + i\sin\theta)^5\) using the binomial theorem gives the imaginary part as: \(\sin(5\theta) = 5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta\). Substituting \(\cos^2\theta = 1 - \sin^2\theta\), we get: \(\sin(5\theta) = 5(1 - \sin^2\theta)^2\sin\theta - 10(1 - \sin^2\theta)\sin^3\theta + \sin^5\theta = 5(1 - 2\sin^2\theta + \sin^4\theta)\sin\theta - 10\sin^3\theta + 10\sin^5\theta + \sin^5\theta = 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta\). To solve \(16x^4 - 20x^2 + 5 = 0\), multiply by \(x\) (since \(x \neq 0\) as \(x=0\) is not a solution) to get \(16x^5 - 20x^3 + 5x = 0\). Substituting \(x = \sin\theta\), this equation becomes \(\sin(5\theta) = 0\), which has solutions \(5\theta = k\pi\) for integers \(k\). This gives \(\theta = \frac{k\pi}{5}\). Since \(x \neq 0\), we exclude \(k = 0\) and multiples of 5. The unique positive and negative values for \(x\) are given by \(k = 1, 2, -1, -2\), which yield \(x = \pm \sin\left(\frac{\pi}{5}\right)\) and \(x = \pm \sin\left(\frac{2\pi}{5}\right)\).

M1: Apply de Moivre's theorem and expand binomially to find the imaginary part. A1: Correct expansion. M1: Substitute \(\cos^2\theta = 1 - \sin^2\theta\) and simplify. A1: Correct proof of the identity. M1: Recognize the connection to the quartic equation by introducing \(x = \sin\theta\). M1: Solve \(\sin(5\theta) = 0\) to obtain \(\theta = \frac{k\pi}{5}\). A1: Correct non-zero roots identified. A1.3: Clearly state all four distinct roots in the correct form.

(a) Substituting \(\cosh x = \frac{e^x + e^{-x}}{2}\) and \(\sinh x = \frac{e^x - e^{-x}}{2}\) into the equation gives: \(3\left(\frac{e^x + e^{-x}}{2}\right) - 2\left(\frac{e^x - e^{-x}}{2}\right) = k\). Let \(u = e^x\). Then, \(3(u + u^{-1}) - 2(u - u^{-1}) = 2k\). Simplifying this yields \(u + 5u^{-1} = 2k\). Multiplying both sides by \(u\) (since \(u > 0\)) gives the quadratic equation \(u^2 - 2ku + 5 = 0\). (b) For \(x\) to be a real solution, \(u = e^x\) must be a real, positive root of the quadratic equation. For real roots, the discriminant must be non-negative: \((-2k)^2 - 4(1)(5) \ge 0 \implies 4k^2 - 20 \ge 0 \implies k^2 \ge 5\). Thus, \(k \ge \sqrt{5}\) or \(k \le -\sqrt{5}\). Additionally, since the product of the roots is \(5 > 0\), both roots have the same sign. For the roots to be positive (as \(u = e^x > 0\)), their sum must be positive: \(2k > 0 \implies k > 0\). Therefore, we discard \(k \le -\sqrt{5}\). The set of values is \(k \ge \sqrt{5}\).

M1: Substitute exponential definitions of hyperbolic functions. A1: Correctly obtain the expression in terms of \(u\) and \(u^{-1}\). M1: Rearrange to the form \(u^2 - 2ku + 5 = 0\). A1: Correct proof of part (a). M1: Set the discriminant \(\ge 0\) and solve for \(k\). A1: Obtain \(k^2 \ge 5\). M1: Express the condition that the roots of the quadratic must be positive (i.e., sum of roots \(2k > 0\)). A1.3: Conclude with the final correct range \(k \ge \sqrt{5}\).

This is a first-order linear differential equation with integrating factor \(I(x) = e^{\int \tanh x \, dx}\). Since \(\int \tanh x \, dx = \ln(\cosh x)\), the integrating factor is \(I(x) = e^{\ln(\cosh x)} = \cosh x\). Multiplying the differential equation by \(\cosh x\) gives: \(\cosh x \frac{dy}{dx} + y \sinh x = 2\sinh x \cosh x\), which simplifies to \(\frac{d}{dx}(y \cosh x) = \sinh(2x)\). Integrating both sides with respect to \(x\) gives: \(y \cosh x = \int \sinh(2x) \, dx = \frac{1}{2}\cosh(2x) + C\). Using the initial condition \(y = 3\) when \(x = 0\): \(3 \cosh(0) = \frac{1}{2}\cosh(0) + C \implies 3 = \frac{1}{2} + C \implies C = \frac{5}{2}\). Thus, \(y \cosh x = \frac{1}{2}\cosh(2x) + \frac{5}{2}\). Using the identity \(\cosh(2x) = 2\cosh^2 x - 1\), this becomes: \(y \cosh x = \frac{1}{2}(2\cosh^2 x - 1) + \frac{5}{2} = \cosh^2 x + 2\). Dividing by \(\cosh x\), we obtain the particular solution: \(y = \cosh x + \frac{2}{\cosh x} = \cosh x + 2\text{sech } x\).

M1: Find the correct integrating factor \(\cosh x\). A1: Correctly write the ODE in the form \(\frac{d}{dx}(y \cosh x) = 2\sinh x \cosh x\). M1: Integrate the right-hand side. A1: Obtain \(y \cosh x = \frac{1}{2}\cosh(2x) + C\) (or equivalent). M1: Apply the boundary condition \(y(0) = 3\) to find \(C\). A1: Obtain \(C = 2.5\). M1: Use hyperbolic identities to simplify the expression. A1.3: Correct final solution in the form \(y = \cosh x + 2\text{sech } x\).

(a) To show the reduction formula, we use integration by parts on \(I_n = \int_1^e (\ln x)^n \cdot 1 \, dx\). Let \(u = (\ln x)^n\) and \(dv = dx\). Then \(du = n(\ln x)^{n-1} \cdot \frac{1}{x} \, dx\) and \(v = x\). Applying the integration by parts formula: \(I_n = \left[ x (\ln x)^n \right]_1^e - \int_1^e x \cdot n (\ln x)^{n-1} \frac{1}{x} \, dx\). Evaluating the boundary term: \(\left[ e (\ln e)^n - 1 (\ln 1)^n \right] = e(1)^n - 0 = e\). The integral term simplifies to: \(n \int_1^e (\ln x)^{n-1} \, dx = n I_{n-1}\). Thus, \(I_n = e - n I_{n-1}\). (b) To find \(I_3\), we first evaluate \(I_0\): \(I_0 = \int_1^e 1 \, dx = [x]_1^e = e - 1\). Using the reduction formula: \(I_1 = e - 1 \cdot I_0 = e - (e - 1) = 1\). \(I_2 = e - 2 \cdot I_1 = e - 2(1) = e - 2\). \(I_3 = e - 3 \cdot I_2 = e - 3(e - 2) = e - 3e + 6 = 6 - 2e\).

M1: Apply integration by parts with appropriate choices of \(u\) and \(v\). A1: Correctly differentiate \(u\) and integrate \(dv\). A1: Evaluate boundary limits correctly. A1: Deduce the reduction formula \(I_n = e - n I_{n-1}\). M1: Evaluate the base case \(I_0\) or \(I_1\). M1: Use the reduction formula iteratively to find \(I_3\). A1: Correctly calculate \(I_2\). A1.3: Obtain the final exact value \(6 - 2e\).

(a) Differentiating \(y = e^{\arcsin x}\) with respect to \(x\) gives \(y' = \frac{e^{\arcsin x}}{\sqrt{1-x^2}} = \frac{y}{\sqrt{1-x^2}}\). Squaring both sides yields \((y')^2(1-x^2) = y^2\). Differentiating both sides implicitly with respect to \(x\) gives \(2y'y''(1-x^2) - 2x(y')^2 = 2yy'\). Dividing by \(2y'\) (since \(y' \neq 0\)) gives \((1-x^2)y'' - x y' - y = 0\). (b) Applying Leibniz's theorem to differentiate each term \(n\) times: For \((1-x^2)y''\): \(D^n((1-x^2)y'') = (1-x^2)y^{(n+2)} + n(-2x)y^{(n+1)} + \frac{n(n-1)}{2}(-2)y^{(n)} = (1-x^2)y^{(n+2)} - 2nxy^{(n+1)} - n(n-1)y^{(n)}\). For \(-xy'\): \(D^n(-xy') = -[x y^{(n+1)} + n(1)y^{(n)}] = -xy^{(n+1)} - ny^{(n)}\). For \(-y\): \(D^n(-y) = -y^{(n)}\). Summing these terms and collecting coefficients: \((1-x^2)y^{(n+2)} - (2n+1)xy^{(n+1)} - (n(n-1) + n + 1)y^{(n)} = 0\), which simplifies to \((1-x^2)y^{(n+2)} - (2n+1)xy^{(n+1)} - (n^2+1)y^{(n)} = 0\). (c) At \(x = 0\): \(y(0) = e^0 = 1\). From \(y'∑\sqrt{1-x^2} = y\), \(y'(0) = y(0) = 1\). From the original ODE: \(y''(0) - 0 - y(0) = 0 \implies y''(0) = 1\). Using the recurrence relation from part (b) with \(n=1\) at \(x=0\): \(y'''(0) - (1^2+1)y'(0) = 0 \implies y'''(0) = 2y'(0) = 2\). The Maclaurin series is \(y(x) = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \dots = 1 + x + \frac{1}{2}x^2 + \frac{1}{3}x^3 + \dots\).

M1: Correctly differentiate to find \(y'\). A1: Show \((1-x^2)y'' - xy' - y = 0\). M1: Apply Leibniz's theorem to each term. A1: Correct differentiation of \((1-x^2)y''\). A1: Correct differentiation of \(-xy'\). A1: Establish the recurrence relation. M1: Find \(y(0)\), \(y'(0)\), \(y''(0)\), and use the recurrence to find \(y'''(0)\). A1.3: Correctly express the Maclaurin series up to the \(x^3\) term.

(a) Since \(\mathbf{A}\) is upper triangular, its eigenvalues are the diagonal entries: \(\lambda_1 = 2\), \(\lambda_2 = 3\), \(\lambda_3 = 4\). Thus, \(\mathbf{D} = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{pmatrix}\). For \(\lambda_1 = 2\): \(\mathbf{A} - 2\mathbf{I} = \begin{pmatrix} 0 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 2 \end{pmatrix}\). Solving \((\mathbf{A}-2\mathbf{I})\mathbf{v} = \mathbf{0}\) gives \(y+z=0, 2z=0 \implies y=0, z=0\). An eigenvector is \(\mathbf{v}_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\). For \(\lambda_2 = 3\): \(\mathbf{A} - 3\mathbf{I} = \begin{pmatrix} -1 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 1 \end{pmatrix}\). Solving \((\mathbf{A}-3\mathbf{I})\mathbf{v} = \mathbf{0}\) gives \(z=0, -x+y+z=0 \implies x=y\). An eigenvector is \(\mathbf{v}_2 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}\). For \(\lambda_3 = 4\): \(\mathbf{A} - 4\mathbf{I} = \begin{pmatrix} -2 & 1 & 1 \\ 0 & -1 & 1 \\ 0 & 0 & 0 \end{pmatrix}\). Solving \((\mathbf{A}-4\mathbf{I})\mathbf{v} = \mathbf{0}\) gives \(-y+z=0 \implies y=z\), and \(-2x+y+z=0 \implies x=z\). An eigenvector is \(\mathbf{v}_3 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\). Thus, \(\mathbf{P} = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}\). (b) To find \(\mathbf{P}^{-1}\), we solve \(\mathbf{P}\mathbf{x} = \mathbf{y}\): \(x_1 + x_2 + x_3 = y_1\), \(x_2 + x_3 = y_2\), \(x_3 = y_3\). This yields \(x_2 = y_2 - y_3\) and \(x_1 = y_1 - y_2\). Hence, \(\mathbf{P}^{-1} = \begin{pmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{pmatrix}\). (c) Since \(\mathbf{A}^n = \mathbf{P}\mathbf{D}^n\mathbf{P}^{-1}\): \(\mathbf{P}\mathbf{D}^n = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 2^n & 0 & 0 \\ 0 & 3^n & 0 \\ 0 & 0 & 4^n \end{pmatrix} = \begin{pmatrix} 2^n & 3^n & 4^n \\ 0 & 3^n & 4^n \\ 0 & 0 & 4^n \end{pmatrix}\). Then \(\mathbf{A}^n = \begin{pmatrix} 2^n & 3^n & 4^n \\ 0 & 3^n & 4^n \\ 0 & 0 & 4^n \end{pmatrix} \begin{pmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 2^n & 3^n - 2^n & 4^n - 3^n \\ 0 & 3^n & 4^n - 3^n \\ 0 & 0 & 4^n \end{pmatrix}\).

M1: State the eigenvalues of \(\mathbf{A}\). A1: Correct \(\mathbf{D}\). M1: Compute eigenvectors for each eigenvalue. A1: Correct \(\mathbf{P}\). M1: Method to invert matrix \(\mathbf{P}\). A1: Correct \(\mathbf{P}^{-1}\). M1: Use \(\mathbf{A}^n = \mathbf{P}\mathbf{D}^n\mathbf{P}^{-1}\). A1.3: Obtain correct matrix expression for \(\mathbf{A}^n\).

(a) Let \(S_n = \sum_{r=1}^n e^{i(2r-1)\theta}\). This is a geometric series with first term \(a = e^{i\theta}\) and common ratio \(r = e^{2i\theta}\). Using the sum formula for a geometric series: \(S_n = e^{i\theta} \frac{1 - e^{2in\theta}}{1 - e^{2i\theta}}\) Factoring out the half-angles from numerator and denominator: \(S_n = e^{i\theta} \frac{e^{in\theta}(e^{-in\theta} - e^{in\theta})}{e^{i\theta}(e^{-i\theta} - e^{i\theta})} = e^{in\theta} \frac{-2i\sin(n\theta)}{-2i\sin\theta} = e^{in\theta} \frac{\sin(n\theta)}{\sin\theta}\). Since \(e^{in\theta} = \cos(n\theta) + i\sin(n\theta)\), we have: \(S_n = \frac{(\cos(n\theta) + i\sin(n\theta))\sin(n\theta)}{\sin\theta}\). Taking the real part: \(\sum_{r=1}^n \cos((2r-1)\theta) = \text{Re}(S_n) = \frac{\cos(n\theta)\sin(n\theta)}{\sin\theta} = \frac{\sin(2n\theta)}{2\sin\theta}\). (b) Taking the imaginary part of \(S_n\): \(\sum_{r=1}^n \sin((2r-1)\theta) = \text{Im}(S_n) = \frac{\sin(n\theta)\sin(n\theta)}{\sin\theta} = \frac{\sin^2(n\theta)}{\sin\theta}\).

M1: Identify the series as geometric and state first term and common ratio. A1: Correct expression using the geometric sum formula. M1: Use half-angle factoring/Euler's formula to simplify. A1: Obtain \(S_n = e^{in\theta} \frac{\sin(n\theta)}{\sin\theta}\). M1: Expand \(e^{in\theta}\) and equate real parts. A1: Correct proof of part (a). M1: Equate imaginary parts to find the sine series sum. A1.3: Obtain correct final simplified expression \(\frac{\sin^2(n\theta)}{\sin\theta}\).

To find the general solution, we first solve the auxiliary equation for the homogeneous equation: \(m^2 - 4m + 4 = 0 \implies (m-2)^2 = 0\). This gives a repeated root \(m = 2\). The complementary function (CF) is \(y_c = (A + Bx)e^{2x}\). Since the term \(e^{2x}\) and \(xe^{2x}\) are already in the CF, we must try a particular integral (PI) of the form \(y_p = C x^2 e^{2x}\). Differentiating \(y_p\): \(y_p' = 2Cx e^{2x} + 2Cx^2 e^{2x} = 2C(x^2 + x)e^{2x}\), \(y_p'' = 2C(2x + 1)e^{2x} + 4C(x^2 + x)e^{2x} = 2C(2x^2 + 4x + 1)e^{2x}\). Substituting into the differential equation: \(2C(2x^2 + 4x + 1)e^{2x} - 8C(x^2 + x)e^{2x} + 4Cx^2 e^{2x} = 8e^{2x}\). Factoring out \(e^{2x}\): \(C(4x^2 + 8x + 2 - 8x^2 - 8x + 4x^2) = 8 \implies 2C = 8 \implies C = 4\). Thus, the particular integral is \(y_p = 4x^2 e^{2x}\). The general solution is \(y = y_c + y_p = (A + Bx)e^{2x} + 4x^2 e^{2x} = (A + Bx + 4x^2)e^{2x}\).

M1: Write and solve the auxiliary equation. A1: Correct repeated roots \(m=2\). A1: Correct complementary function \(y_c = (A + Bx)e^{2x}\). M1: State the appropriate form of the particular integral \(y_p = C x^2 e^{2x}\). M1: Differentiate \(y_p\) twice. A1: Correct derivatives. M1: Substitute back into the ODE and solve for \(C\). A1: Obtain \(C = 4\). A1.3: Write the correct general solution.

Using the equation of the trajectory:
\(y = x \tan\theta - \frac{g x^2}{2 u^2}(1 + \tan^2\theta)\)

Given \(u = 4\sqrt{5}\), \(x = 4\), \(y = 2\), and \(g = 10\):
\(2 = 4 \tan\theta - \frac{10 \times 16}{2 \times (16 \times 5)}(1 + \tan^2\theta)\)

\(2 = 4 \tan\theta - \frac{160}{160}(1 + \tan^2\theta)\)

\(2 = 4 \tan\theta - 1 - \tan^2\theta\)

\(\tan^2\theta - 4\tan\theta + 3 = 0\). (This completes part (i))

For part (ii), we solve the quadratic equation:
\((\tan\theta - 1)(\tan\theta - 3) = 0 \implies \tan\theta = 1 \text{ or } \tan\theta = 3\).

The horizontal distance is given by \(x = u \cos\theta \cdot t \implies t = \frac{x}{u \cos\theta}\).

Since \(\cos^2\theta = \frac{1}{1 + \tan^2\theta}\):

Case 1: \(\tan\theta = 1\)
\(\cos^2\theta = \frac{1}{2} \implies \cos\theta = \frac{1}{\sqrt{2}}\)
\(t_1 = \frac{4}{4\sqrt{5} \times \frac{1}{\sqrt{2}}} = \sqrt{\frac{2}{5}}\text{ s}\).

Case 2: \(\tan\theta = 3\)
\(\cos^2\theta = \frac{1}{10} \implies \cos\theta = \frac{1}{\sqrt{10}}\)
\(t_2 = \frac{4}{4\sqrt{5} \times \frac{1}{\sqrt{10}}} = \sqrt{2}\text{ s}\).

M1: Use the trajectory equation with appropriate substitutions.
A1: Correct simplification to obtain the given quadratic equation.
M1: Solve the quadratic to find two values of \(\tan\theta\).
A1: Correct values \(\tan\theta = 1\) and \(\tan\theta = 3\).
M1: Use the relation \(t = \frac{x}{u\cos\theta}\) and trigonometric identities to find \(t\).
A1: Obtain \(t_1 = \sqrt{\frac{2}{5}}\).
A1: Obtain \(t_2 = \sqrt{2}\).

(i) Let \(x\) be the maximum extension. At this point, the speed of the particle is 0.
By conservation of energy:
Loss in GPE = Gain in EPE
\(mg(a + x) = \frac{\lambda x^2}{2a}\)

Substituting \(\lambda = 4mg\):
\(mg(a + x) = \frac{4mg x^2}{2a} = \frac{2mg x^2}{a}\)

Divide by \(mg\):
\(a + x = \frac{2x^2}{a} \implies 2x^2 - ax - a^2 = 0\)

\((2x + a)(x - a) = 0\)

Since \(x > 0\), the maximum extension is \(x = a\).

(ii) The maximum speed occurs when the acceleration is zero, i.e., at the equilibrium position where the tension equals the weight of the particle:
\(T = mg \implies \frac{4mg e}{a} = mg \implies e = \frac{1}{4}a\)

Let \(v\) be the maximum speed. By conservation of energy at this position:
Loss in GPE = Gain in KE + Gain in EPE
\(mg(a + e) = \frac{1}{2}mv^2 + \frac{4mg e^2}{2a}\)

Substituting \(e = \frac{1}{4}a\):
\(mg\left(\frac{5}{4}a\right) = \frac{1}{2}mv^2 + \frac{2mg (a^2/16)}{a}\)

\(\frac{5}{4}mga = \frac{1}{2}mv^2 + \frac{1}{8}mga\)

\(\frac{1}{2}mv^2 = \left(\frac{5}{4} - \frac{1}{8}\right)mga = \frac{9}{8}mga\)

\(v^2 = \frac{9}{4}ga \implies v = \frac{3}{2}\sqrt{ga}\).

M1: Set up the energy conservation equation for maximum extension.
A1: Correct quadratic equation \(2x^2 - ax - a^2 = 0\).
A1: Solve to find the extension \(x = a\).
M1: Identify that maximum speed occurs at equilibrium and find the extension \(e = \frac{1}{4}a\).
M1: Set up the energy conservation equation at the equilibrium position.
A1: Correctly substitute \(e\) to find \(\frac{1}{2}mv^2 = \frac{9}{8}mga\).
A1: Obtain \(v = \frac{3}{2}\sqrt{ga}\).

(i) Let \(\theta\) be the angle that the radius to the particle makes with the downward vertical.
By conservation of energy:
\(\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mga(1 - \cos\theta)\)

Since \(u^2 = 4ga\):
\(2mga = \frac{1}{2}mv^2 + mga(1 - \cos\theta)\)

\(v^2 = 4ga - 2ga(1 - \cos\theta) = 2ga(1 + \cos\theta)\).

The radial equation of motion is:
\(R - mg\cos\theta = \frac{mv^2}{a}\)

where \(R\) is the normal contact force.
Substituting \(v^2\):
\(R = mg\cos\theta + \frac{m}{a}(2ga(1 + \cos\theta)) = mg(3\cos\theta + 2)\).

The particle leaves the surface when \(R = 0\):
\(3\cos\theta + 2 = 0 \implies \cos\theta = -\frac{2}{3}\).

Substitute \(\cos\theta = -\frac{2}{3}\) into the expression for \(v^2\):
\(v^2 = 2ga\left(1 - \frac{2}{3}\right) = \frac{2}{3}ga \implies v = \sqrt{\frac{2}{3}ga}\).

(ii) The height above the level of \(O\) is given by:
\(h = -a\cos\theta\).

Since \(\cos\theta = -\frac{2}{3}\):
\(h = -a\left(-\frac{2}{3}\right) = \frac{2}{3}a\).

M1: Apply conservation of energy to find \(v^2\) in terms of \(\theta\).
A1: Obtain \(v^2 = 2ga(1 + \cos\theta)\).
M1: Write down the radial equation of motion.
A1: Express the normal contact force \(R\) in terms of \(\cos\theta\).
M1: Set \(R = 0\) to find the angle \(\theta\).
A1: Correctly find \(\cos\theta = -\frac{2}{3}\) and use it to show \(v = \sqrt{\frac{2}{3}ga}\).
A1: Deduce that the height above \(O\) is \(\frac{2}{3}a\).

The equation of motion is:
\(m \frac{dv}{dt} = -v(1+v) \implies 0.5 \frac{dv}{dt} = -v(1+v)\).

(i) To find the distance \(x\), we use \(a = v \frac{dv}{dx}\):
\(0.5 v \frac{dv}{dx} = -v(1+v) \implies \frac{dv}{dx} = -2(1+v)\).

Separating variables:
\(\int_{3}^{1} \frac{1}{1+v} dv = \int_{0}^{x} -2 dx\)

\([\ln(1+v)]_{3}^{1} = -2x\)

\(\ln 2 - \ln 4 = -2x \implies \ln(0.5) = -2x \implies 2x = \ln 2 \implies x = \frac{1}{2}\ln 2\text{ m}\).

(ii) To find the time \(t\), we use \(\frac{dv}{dt}\):
\(\frac{dv}{dt} = -2v(1+v)\).

Separating variables:
\(\int_{3}^{1} \frac{1}{v(1+v)} dv = \int_{0}^{t} -2 dt\)

Using partial fractions:
\(\int_{3}^{1} \left(\frac{1}{v} - \frac{1}{1+v}\right) dv = -2t\)

\(\left[\ln\left(\frac{v}{1+v}\right)\right]_{3}^{1} = -2t\)

\(\ln\left(\frac{1}{2}\right) - \ln\left(\frac{3}{4}\right) = -2t\)

\(\ln\left(\frac{1/2}{3/4}\right) = -2t \implies \ln\left(\frac{2}{3}\right) = -2t \implies 2t = \ln(1.5) \implies t = \frac{1}{2}\ln(1.5)\text{ s}\).

M1: Express acceleration as \(v \frac{dv}{dx}\) and set up the differential equation.
A1: Separate variables and integrate correctly.
A1: Obtain \(x = \frac{1}{2}\ln 2\).
M1: Express acceleration as \(\frac{dv}{dt}\) and separate variables.
M1: Use partial fractions to integrate \(\frac{1}{v(1+v)}\).
A1: Apply the limits 3 and 1 correctly.
A1: Obtain \(t = \frac{1}{2}\ln(1.5)\).

(i) Let \(R\) be the normal reaction of the floor at \(A\), and \(F\) be the frictional force at \(A\). Let \(N\) be the normal reaction of the wall at \(B\).
Since the wall is smooth, there is no vertical frictional force at \(B\).

Resolving vertically:
\(R = W\)

Resolving horizontally:
\(F = N\)

Since the rod is in limiting equilibrium, the frictional force at \(A\) is at its maximum:
\(F = \mu R = \mu W\)

Therefore, \(N = \mu W\).

Taking moments about \(A\):
The clockwise moment due to the weight is \(W \times d\cos\theta\).
The anticlockwise moment due to the wall reaction is \(N \times 2a\sin\theta\).

Equating moments:
\(W d \cos\theta = N (2a \sin\theta)\)

Substitute \(N = \mu W\):
\(W d \cos\theta = \mu W (2a \sin\theta)\)

Divide both sides by \(W\cos\theta\):
\(d = 2a\mu\tan\theta\).

(ii) Substituting the given values:
\(a = 1.5\), \(\mu = 0.3\), and \(\theta = 60^\circ\):
\(d = 2(1.5)(0.3)\tan(60^\circ) = 0.9\sqrt{3} \approx 1.5588\text{ m}\).

To 3 significant figures, \(d = 1.56\text{ m}\).

M1: Resolve forces vertically to find \(R = W\).
M1: Apply limiting friction condition \(F = \mu R\) and resolve horizontally to find \(N = \mu W\).
M1: Take moments about \(A\) and set up the equation.
A1: Correct moments equation: \(W d \cos\theta = 2 N a \sin\theta\).
A1: Complete proof of \(d = 2a\mu\tan\theta\).
M1: Substitute given numerical values into the formula.
A1: Obtain \(d = 1.56\text{ m}\).

(i) Let \(u\) be the speed of the sphere before the collision.
The component of velocity parallel to the wall is:
\(u_x = u \cos 30^\circ = u \frac{\sqrt{3}}{2}\).

The component of velocity perpendicular to the wall is:
\(u_y = u \sin 30^\circ = \frac{1}{2}u\).

Since the wall is smooth, the component of velocity parallel to the wall is unchanged after the collision:
\(v_x = u_x = u \frac{\sqrt{3}}{2}\).

The component of velocity perpendicular to the wall after the collision is given by Newton's law of restitution:
\(v_y = e u_y = \frac{2}{3} \left(\frac{1}{2}u\right) = \frac{1}{3}u\).

Let \(\alpha\) be the angle that the direction of motion of the sphere makes with the wall after the collision.
\(\tan\alpha = \frac{v_y}{v_x} = \frac{\frac{1}{3}u}{u \frac{\sqrt{3}}{2}} = \frac{2}{3\sqrt{3}} = \frac{2\sqrt{3}}{9}\).

\(\alpha = \arctan\left(\frac{2\sqrt{3}}{9}\right) \approx 21.05^\circ \approx 21.1^\circ\).

(ii) The initial kinetic energy is:
\(E_i = \frac{1}{2} m u^2\).

The final kinetic energy is:
\(E_f = \frac{1}{2} m (v_x^2 + v_y^2) = \frac{1}{2} m \left( \left(\frac{\sqrt{3}}{2}u\right)^2 + \left(\frac{1}{3}u\right)^2 \right) = \frac{1}{2} m u^2 \left( \frac{3}{4} + \frac{1}{9} \right) = \frac{1}{2} m u^2 \left(\frac{31}{36}\right)\).

The loss in kinetic energy is:
\(\Delta E = E_i - E_f = \frac{1}{2} m u^2 \left(1 - \frac{31}{36}\right) = \frac{5}{72} m u^2\).

The fraction of initial kinetic energy lost is:
\(\frac{\Delta E}{E_i} = \frac{\frac{5}{72} m u^2}{\frac{1}{2} m u^2} = \frac{5}{36}\).

M1: Resolve velocity components parallel and perpendicular to the wall before collision.
A1: Correct components: \(u_x = \frac{\sqrt{3}}{2}u\) and \(u_y = \frac{1}{2}u\).
M1: Apply Newton's law of restitution to find the component perpendicular to the wall after collision.
A1: Express \(\tan\alpha = \frac{2}{3\sqrt{3}}\) and find \(\alpha = 21.1^\circ\).
M1: Set up expressions for initial and final kinetic energies.
A1: Correctly calculate the final kinetic energy as \(\frac{31}{72} m u^2\).
A1: Obtain the fraction of lost kinetic energy as \(\frac{5}{36}\).

Let \(x\) represent the extension of the string beyond its natural length \(l\).

When \(x > 0\) (string is taut):
The tension in the string is given by Hooke's Law:
\(T = \frac{\lambda x}{l} = \frac{2mgx}{l}\).

The equation of motion for the particle is:
\(m \ddot{x} = -T \implies m \ddot{x} = -\frac{2mgx}{l} \implies \ddot{x} = -\frac{2g}{l}x\).

This is the equation of Simple Harmonic Motion of the form \(\ddot{x} = -\omega^2 x\), where \(\omega = \sqrt{\frac{2g}{l}}\).

The particle is released from \(OA = 2l\), which corresponds to an initial extension \(x = l\). Thus, the amplitude of the SHM is \(l\).
The time \(t_1\) taken for the particle to move from \(x = l\) to \(x = 0\) (where the string first becomes slack) is a quarter of a full period:
\(t_1 = \frac{1}{4} T_{SHM} = \frac{1}{4} \left(\frac{2\pi}{\omega}\right) = \frac{\pi}{2\omega} = \frac{\pi}{2} \sqrt{\frac{l}{2g}} = \frac{\pi}{4}\sqrt{\frac{2l}{g}}\).

At \(x = 0\) (distance \(l\) from \(O\)), the speed of the particle is at its maximum:
\(v_{max} = \omega A = \sqrt{\frac{2g}{l}} \times l = \sqrt{2gl}\).

For the remaining distance \(l\) to \(O\), the string is slack, and no force acts on the particle. It travels at a constant speed of \(v_{max}\).
The time \(t_2\) taken to travel this distance of \(l\) is:
\(t_2 = \frac{l}{v_{max}} = \frac{l}{\sqrt{2gl}} = \sqrt{\frac{l}{2g}}\).

The total time taken to reach \(O\) is:
\(t = t_1 + t_2 = \frac{\pi}{4}\sqrt{\frac{2l}{g}} + \sqrt{\frac{l}{2g}}\).

Since \(\sqrt{\frac{l}{2g}} = \frac{1}{2}\sqrt{\frac{2l}{g}}\), we have:
\(t = \frac{\pi}{4}\sqrt{\frac{2l}{g}} + \frac{2}{4}\sqrt{\frac{2l}{g}} = \frac{\pi + 2}{4}\sqrt{\frac{2l}{g}}\).

M1: Show the motion is Simple Harmonic and state \(\omega = \sqrt{\frac{2g}{l}}\).
A1: Find the time \(t_1\) to reach the natural length position as \(\frac{\pi}{4}\sqrt{\frac{2l}{g}}\).
M1: Find the maximum speed of the particle as it reaches the natural length.
A1: Correct maximum speed \(v = \sqrt{2gl}\).
M1: Calculate the time \(t_2\) taken for the uniform motion of length \(l\).
A1: Correct \(t_2 = \sqrt{\frac{l}{2g}}\).
A1: Obtain the total time \(t = \frac{\pi + 2}{4}\sqrt{\frac{2l}{g}}\).

First, we state the hypotheses:
\(H_0: \mu = 12.0\)
\(H_1: \mu > 12.0\)

Next, calculate the sample mean \(\bar{x}\) and unbiased variance estimate \(s^2\):
\(\sum x = 12.4 + 14.1 + 10.8 + 15.2 + 13.5 + 11.9 + 14.8 + 13.3 = 106\)
\(\bar{x} = \frac{106}{8} = 13.25\)

\(\sum x^2 = 12.4^2 + 14.1^2 + 10.8^2 + 15.2^2 + 13.5^2 + 11.9^2 + 14.8^2 + 13.3^2 = 1420.04\)

\(s^2 = \frac{1}{7} \left( 1420.04 - \frac{106^2}{8} \right) = \frac{1}{7} (1420.04 - 1404.5) = \frac{15.54}{7} = 2.22\)

Calculate the test statistic \(t\):
\(t = \frac{\bar{x} - \mu_0}{\sqrt{s^2 / n}} = \frac{13.25 - 12.0}{\sqrt{2.22 / 8}} = \frac{1.25}{\sqrt{0.2775}} = \frac{1.25}{0.52678} \approx 2.373\)

Determine the critical value:
Degrees of freedom \(\nu = n - 1 = 7\).
From the \(t\)-distribution table, for a one-tailed test at the 5% significance level with 7 degrees of freedom, the critical value is \(t_7(0.05) = 1.895\).

Conclusion:
Since the test statistic \(t = 2.373 > 1.895\), we reject the null hypothesis \(H_0\). There is sufficient evidence at the 5% significance level to support the manufacturer's claim that the mean lifetime is greater than 12.0 thousand hours.

M1: State correct hypotheses \(H_0\) and \(H_1\).
A1: Calculate correct sample mean \(\bar{x} = 13.25\).
M1: Calculate correct unbiased variance estimate \(s^2 = 2.22\).
M1: Calculate the test statistic \(t \approx 2.37\).
A1: Identify correct critical value \(t_7(0.05) = 1.895\).
M1: Compare \(t\) value with the critical value and make a decision to reject \(H_0\).
A1.3: Conclude in context, stating that the manufacturer's claim is supported.

First, state the hypotheses:
\(H_0: \text{Median difference is zero (the course does not improve memory scores)}\)
\(H_1: \text{Median difference is greater than zero (the course improves memory scores)}\)

Calculate the differences \(d = \text{After} - \text{Before}\):
\begin{array}{r|ccccccccc}
\text{Participant} & A & B & C & D & E & F & G & H & I \\
\hline
\text{Difference } d & +6 & -2 & +9 & +4 & +7 & -1 & +11 & -3 & +12 \\
|d| & 6 & 2 & 9 & 4 & 7 & 1 & 11 & 3 & 12 \\
\text{Rank of } |d| & 5 & 2 & 7 & 4 & 6 & 1 & 8 & 3 & 9 \\
\text{Signed Rank} & +5 & -2 & +7 & +4 & +6 & -1 & +8 & -3 & +9 \\
\end{array}

Sum of positive ranks:
\(W_+ = 5 + 7 + 4 + 6 + 8 + 9 = 39\)

Sum of negative ranks:
\(W_- = 2 + 1 + 3 = 6\)

Our test statistic \(W = \min(W_+, W_-) = 6\).

For a one-tailed Wilcoxon signed-rank test with \(n = 9\) at the 5% level of significance, the critical value from the tables is \(8\).

Since \(W = 6 \le 8\), we reject the null hypothesis \(H_0\). There is significant evidence at the 5% level to conclude that the memory enhancement course improves scores.

M1: State correct null and alternative hypotheses.
A1: Calculate all differences \(d = \text{After} - \text{Before}\) correctly.
M1: Rank absolute values of differences correctly.
A1: Assign correct signs to the ranks and calculate sum of negative ranks \(W_- = 6\) and positive ranks \(W_+ = 39\).
M1: State the test statistic \(W = 6\).
A1: Identify the correct critical value of 8 for \(n=9\) at 5% significance.
M1: Compare \(W\) with critical value correctly.
A1.3: Conclude in context, rejecting \(H_0\).

First, state the hypotheses:
\(H_0:\) Category of book purchased and type of music played are independent.
\(H_1:\) Category of book purchased and type of music played are not independent (there is an association).

Calculate the row and column totals:
Row totals: Classical = 80, Rock = 100, Grand Total \(N = 180\).
Column totals: Fiction = 60, Non-Fiction = 60, Academic = 60.

Calculate the expected frequencies \(E = \frac{\text{Row Total} \times \text{Col Total}}{N}\):
Since all column totals are equal (60), the expected frequencies for each row will be equal:
For Classical (Row 1): \(E = \frac{80 \times 60}{180} = 26.67\) for each cell.
For Rock (Row 2): \(E = \frac{100 \times 60}{180} = 33.33\) for each cell.

Now, calculate the test statistic \(\chi^2 = \sum \frac{(O - E)^2}{E}\):
\(\chi^2 = \frac{(40 - 26.67)^2}{26.67} + \frac{(25 - 26.67)^2}{26.67} + \frac{(15 - 26.67)^2}{26.67} + \frac{(20 - 33.33)^2}{33.33} + \frac{(35 - 33.33)^2}{33.33} + \frac{(45 - 33.33)^2}{33.33}\)
\(\chi^2 = \frac{177.69}{26.67} + \frac{1.37}{26.67} + \frac{136.19}{26.67} + \frac{177.69}{33.33} + \frac{2.79}{33.33} + \frac{136.19}{33.33}\)
\(\chi^2 \approx 6.663 + 0.051 + 5.106 + 5.331 + 0.084 + 4.086 = 21.321\) (using more precise intermediate values: \(21.37\)).

Degrees of freedom:
\(v = (2 - 1) \times (3 - 1) = 2\).

At the 1% significance level, the critical value is \(\chi^2_2(0.01) = 9.210\).

Since \(21.37 > 9.210\), we reject \(H_0\). There is highly significant evidence that the category of book purchased is associated with the type of music played.

M1: State correct hypotheses.
A1: Calculate correct expected values (e.g., 26.67 and 33.33).
M1: Apply the formula \(\chi^2 = \sum \frac{(O - E)^2}{E}\) correctly.
A1: Calculate the test statistic value \(\chi^2 \approx 21.37\) (accept 21.3 to 21.4).
M1: State correct degrees of freedom \(v = 2\) and identify critical value \(9.210\).
M1: Compare test statistic to critical value and make a correct rejection decision.
A1.3: Conclude clearly in context.

(i) Since \(G_X(1) = 1\), we have:
\(k(1 + 2(1) + 3(1)^2)^2 = 1 \implies k(6)^2 = 1 \implies 36k = 1 \implies k = \frac{1}{36}\).

(ii) To find \(E(X)\), we compute the derivative \(G_X'(t)\):
\(G_X(t) = \frac{1}{36}(1 + 2t + 3t^2)^2\)
\(G_X'(t) = \frac{2}{36}(1 + 2t + 3t^2)(2 + 6t) = \frac{1}{18}(1 + 2t + 3t^2)(2 + 6t)\)

Evaluate at \(t = 1\):
\(E(X) = G_X'(1) = \frac{1}{18}(1 + 2 + 3)(2 + 6) = \frac{1}{18}(6)(8) = \frac{48}{18} = \frac{8}{3}\).

(iii) To find \(Var(X)\), we first compute \(G_X''(t)\) using the product rule:
\(G_X''(t) = \frac{1}{18} \left[ (2 + 6t)(2 + 6t) + (1 + 2t + 3t^2)(6) \right]\)

Evaluate at \(t = 1\):
\(G_X''(1) = \frac{1}{18} \left[ (8)(8) + (6)(6) \right] = \frac{1}{18} [64 + 36] = \frac{100}{18} = \frac{50}{9}\).

Using the formula for variance:
\(Var(X) = G_X''(1) + G_X'(1) - [G_X'(1)]^2\)
\(Var(X) = \frac{50}{9} + \frac{8}{3} - \left(\frac{8}{3}\right)^2 = \frac{50}{9} + \frac{24}{9} - \frac{64}{9} = \frac{10}{9}\).

M1: Use the property \(G_X(1) = 1\) to find \(k\).
A0.3: Conclude \(k = 1/36\).
M1: Differentiate \(G_X(t)\) with respect to \(t\).
A1: Obtain correct first derivative.
A1: Evaluate \(G_X'(1)\) to find \(E(X) = 8/3\).
M1: Differentiate a second time to obtain \(G_X''(t)\).
A1: Correctly evaluate \(G_X''(1) = 50/9\).
M1: Use \(Var(X) = G_X''(1) + G_X'(1) - [G_X'(1)]^2\) to find variance.
A1: Obtain \(Var(X) = 10/9\).

(i) To find the cumulative distribution function \(F(x)\) for \(0 \le x \le 4\):
\(F(x) = \int_0^x f(t) \, dt = \int_0^x \frac{3}{32}(4t - t^2) \, dt = \frac{3}{32} \left[ 2t^2 - \frac{1}{3}t^3 \right]_0^x = \frac{3}{32} \left(2x^2 - \frac{1}{3}x^3\right)\)

Thus, the complete CDF is:
\(F(x) = \begin{cases} 0 & x < 0 \\ \frac{3}{32} \left(2x^2 - \frac{1}{3}x^3\right) & 0 \le x \le 4 \\ 1 & x > 4 \end{cases}\)

(ii) The probability density function is symmetric about \(x = 2\) because:
\(f(4 - x) = \frac{3}{32}(4(4-x) - (4-x)^2) = \frac{3}{32}(16 - 4x - (16 - 8x + x^2)) = \frac{3}{32}(4x - x^2) = f(x)\).
Therefore, by symmetry, \(E(X) = 2\).

Now find \(E(X^2)\):
\(E(X^2) = \int_0^4 x^2 \cdot f(x) \, dx = \int_0^4 x^2 \cdot \frac{3}{32}(4x - x^2) \, dx = \frac{3}{32} \int_0^4 (4x^3 - x^4) \, dx\)
\(E(X^2) = \frac{3}{32} \left[ x^4 - \frac{1}{5}x^5 \right]_0^4 = \frac{3}{32} \left( 256 - \frac{1024}{5} \right) = \frac{3}{32} \left( \frac{1280 - 1024}{5} \right) = \frac{3}{32} \cdot \frac{256}{5} = \frac{24}{5} = 4.8\)

Calculate the variance:
\(Var(X) = E(X^2) - [E(X)]^2 = 4.8 - 2^2 = 4.8 - 4 = 0.8\).

M1: Integrate \(f(t)\) over the interval \([0, x]\).
A1: Obtain correct expression \(\frac{3}{32}(2x^2 - \frac{1}{3}x^3)\) for \(0 \le x \le 4\).
A1: State complete piecewise definition of \(F(x)\) including boundaries.
B1: State \(E(X) = 2\) with reference to symmetry.
M1: Write down the correct integral for \(E(X^2)\).
A1: Obtain \(E(X^2) = 4.8\) (or \(24/5\)).
M1: Use \(Var(X) = E(X^2) - [E(X)]^2\).
A1.3: Obtain \(Var(X) = 0.8\) (or \(4/5\)).

State the hypotheses:
\(H_0: \mu_A = \mu_B\)
\(H_1: \mu_A > \mu_B\)

Calculate the pooled sample variance \(s_p^2\):
\(s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2} = \frac{5(12.4) + 7(10.8)}{6 + 8 - 2} = \frac{62.0 + 75.6}{12} = \frac{137.6}{12} \approx 11.467\)

Calculate the test statistic \(t\):
\(SE = \sqrt{s_p^2 \left( \frac{1}{n_1} + \frac{1}{n_2} \right)} = \sqrt{11.467 \left( \frac{1}{6} + \frac{1}{8} \right)} = \sqrt{11.467 \times \frac{7}{24}} = \sqrt{3.3444} \approx 1.8288\)

\(t = \frac{\bar{x}_1 - \bar{x}_2}{SE} = \frac{45.2 - 41.5}{1.8288} = \frac{3.7}{1.8288} \approx 2.023\)

Determine the critical value:
Degrees of freedom \(\nu = n_1 + n_2 - 2 = 6 + 8 - 2 = 12\).
For a one-tailed test at the 5% significance level with 12 degrees of freedom, the critical value is \(t_{12}(0.05) = 1.782\).

Conclusion:
Since the calculated test statistic \(t = 2.023 > 1.782\), we reject the null hypothesis \(H_0\). There is significant evidence at the 5% significance level that the mean height of plants in group A is greater than that of group B.

M1: State correct null and alternative hypotheses.
M1: Calculate the pooled sample variance \(s_p^2 = 11.47\).
A1: Obtain the correct standard error of difference \(SE \approx 1.829\).
M1: Compute the test statistic \(t\).
A1: Obtain \(t \approx 2.02\).
B1: Identify correct degrees of freedom (12) and critical value \(t_{12}(0.05) = 1.782\).
M1: Compare calculated \(t\) value with the critical value.
A1.3: Draw correct conclusion in context, rejecting \(H_0\).