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

(a) From the given cubic equation, we have:
\(\alpha + \beta + \gamma = \frac{3}{2}\)
\(\alpha\beta + \beta\gamma + \gamma\alpha = 2\)
\(\alpha\beta\gamma = \frac{5}{2}\)

We know that:
\(\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\)
\(\alpha^2 + \beta^2 + \gamma^2 = \left(\frac{3}{2}\right)^2 - 2(2) = \frac{9}{4} - 4 = -\frac{7}{4}\).

(b) To find the equation for \(u = x^2\), we rewrite the original equation as:
\(2x^3 + 4x = 3x^2 + 5 \implies 2x(x^2 + 2) = 3x^2 + 5\)

Squaring both sides:
\(4x^2(x^2 + 2)^2 = (3x^2 + 5)^2\)

Substitute \(u = x^2\):
\(4u(u + 2)^2 = (3u + 5)^2\)
\(4u(u^2 + 4u + 4) = 9u^2 + 30u + 25\)
\(4u^3 + 16u^2 + 16u = 9u^2 + 30u + 25\)
\(4u^3 + 7u^2 - 14u - 25 = 0\).

(c) Let the sum of the squares of the roots be \(S = \alpha^2 + \beta^2 + \gamma^2 = -\frac{7}{4}\).
The roots of the new equation are \(w = S - u = -\frac{7}{4} - u\).
So, \(u = -w - \frac{7}{4}\).

Substitute \(u\) into the equation from part (b):
\(4\left(-w - \frac{7}{4}\right)^3 + 7\left(-w - \frac{7}{4}\right)^2 - 14\left(-w - \frac{7}{4}\right) - 25 = 0\)

Multiplying by \(-1\) and expanding:
\(4\left(w + \frac{7}{4}\right)^3 - 7\left(w + \frac{7}{4}\right)^2 - 14\left(w + \frac{7}{4}\right) + 25 = 0\)

Expanding each term:
\(4\left(w^3 + \frac{21}{4}w^2 + \frac{147}{16}w + \frac{343}{64}\right) - 7\left(w^2 + \frac{7}{2}w + \frac{49}{16}\right) - 14\left(w + \frac{7}{4}\right) + 25 = 0\)
\(4w^3 + 21w^2 + \frac{147}{4}w + \frac{343}{16} - 7w^2 - \frac{49}{2}w - \frac{343}{16} - 14w - \frac{49}{2} + 25 = 0\)

Combine the coefficients:
- \(w^3\) term: \(4\)
- \(w^2\) term: \(21 - 7 = 14\)
- \(w\) term: \(\frac{147}{4} - \frac{98}{4} - \frac{56}{4} = -\frac{7}{4}\)
- Constant term: \(-\frac{49}{2} + 25 = \frac{1}{2}\)

So the equation in \(w\) is:
\(4w^3 + 14w^2 - \frac{7}{4}w + \frac{1}{2} = 0\)

Multiplying by 4 to obtain integer coefficients:
\(16w^3 + 56w^2 - 7w + 2 = 0\).

(a)
- M1: For stating the relations \(\sum \alpha = 3/2\) and \(\sum \alpha\beta = 2\) and using the identity for \(\sum \alpha^2\).
- A1: For the correct value of \(-\frac{7}{4}\).

(b)
- M1: For rearranging the cubic equation to separate odd and even powers of \(x\).
- M1: For squaring both sides.
- M1: For substituting \(u = x^2\) and expanding.
- A1: For obtaining the correct cubic equation \(4u^3 + 7u^2 - 14u - 25 = 0\).

(c)
- M1: For relating the new roots to the old roots using \(w = -\frac{7}{4} - u\).
- M1: For substituting \(u = -w - \frac{7}{4}\) into the cubic in (b).
- A1: For a correct expansion with fractional coefficients.
- A1.7: For the correct final integer-coefficient cubic equation: \(16w^3 + 56w^2 - 7w + 2 = 0\).

(a) Performing algebraic division:
\(y = \frac{x^2 - 2x + 9}{x - 1} = \frac{(x - 1)^2 + 8}{x - 1} = x - 1 + \frac{8}{x - 1}\).

- As \(x \to 1\), \(y \to \pm \infty\), so the vertical asymptote is \(x = 1\).
- As \(x \to \pm \infty\), \(\frac{8}{x - 1} \to 0\), so the oblique asymptote is \(y = x - 1\).

(b) Differentiating \(y = x - 1 + 8(x - 1)^{-1}\) with respect to \(x\):
\(\frac{\mathrm{d}y}{\mathrm{d}x} = 1 - \frac{8}{(x - 1)^2}\)

Setting \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\(1 = \frac{8}{(x - 1)^2} \implies (x - 1)^2 = 8 \implies x - 1 = \pm 2\sqrt{2}\)
So \(x = 1 \pm 2\sqrt{2}\).

Now find the corresponding \(y\)-values:
\(y = (1 \pm 2\sqrt{2}) - 1 + \frac{8}{\pm 2\sqrt{2}} = \pm 2\sqrt{2} \pm 2\sqrt{2} = \pm 4\sqrt{2}\).

Thus, the stationary points are:
\((1 + 2\sqrt{2}, 4\sqrt{2})\) (local minimum) and \((1 - 2\sqrt{2}, -4\sqrt{2})\) (local maximum).

(c) Intersections with axes:
- With y-axis: When \(x = 0\), \(y = \frac{9}{-1} = -9\). So the y-intercept is \((0, -9)\).
- With x-axis: \(x^2 - 2x + 9 = 0\) has discriminant \(D = (-2)^2 - 4(1)(9) = -32 < 0\), so there are no real roots. The curve does not cross the x-axis.

Sketch the asymptotes \(x = 1\) and \(y = x - 1\). Draw two branches: one in the upper-right region with local minimum at \((1 + 2\sqrt{2}, 4\sqrt{2})\), and the other in the lower-left region passing through \((0, -9)\) with local maximum at \((1 - 2\sqrt{2}, -4\sqrt{2})\).

(a)
- M1: For performing algebraic division or equivalent process to write \(y\) in the form \(ax+b + \frac{c}{x-1}\).
- A1: For the vertical asymptote \(x = 1\).
- A1: For the oblique asymptote \(y = x - 1\).

(b)
- M1: For finding \(\frac{\mathrm{d}y}{\mathrm{d}x}\) and setting it to 0.
- A1: For finding the correct x-coordinates \(x = 1 \pm 2\sqrt{2}\).
- A1: For finding the correct y-coordinates \(y = \pm 4\sqrt{2}\).
- A0.7: For clearly identifying the complete coordinates of the stationary points.

(c)
- B1: For correctly drawn and labeled asymptotes \(x = 1\) and \(y = x - 1\).
- B1: For showing the correct y-intercept at \((0, -9)\) and no x-intercepts.
- B2: For sketching the two branches of the curve in correct quadrants with stationary points correctly positioned relative to asymptotes.

(a) Let us factorize the denominator:
\(4r^4 + 1 = (2r^2 + 1)^2 - 4r^2 = (2r^2 - 2r + 1)(2r^2 + 2r + 1)\).

Let \(f(r) = \frac{1}{2r^2 - 2r + 1}\).
Then \(f(r+1) = \frac{1}{2(r+1)^2 - 2(r+1) + 1} = \frac{1}{2r^2 + 2r + 1}\).

Consider the difference:
\(f(r) - f(r+1) = \frac{1}{2r^2 - 2r + 1} - \frac{1}{2r^2 + 2r + 1}\)
\(= \frac{(2r^2 + 2r + 1) - (2r^2 - 2r + 1)}{(2r^2 - 2r + 1)(2r^2 + 2r + 1)} = \frac{4r}{4r^4 + 1}\).

Using the method of differences:
\(\sum_{r=1}^{n} \frac{4r}{4r^4 + 1} = \sum_{r=1}^{n} [f(r) - f(r+1)]\)
\(= [f(1) - f(2)] + [f(2) - f(3)] + \dots + [f(n) - f(n+1)]\)
\(= f(1) - f(n+1)\).

Since \(f(1) = \frac{1}{2(1)^2 - 2(1) + 1} = 1\) and \(f(n+1) = \frac{1}{2n^2 + 2n + 1}\),
we have:
\(\sum_{r=1}^{n} \frac{4r}{4r^4 + 1} = 1 - \frac{1}{2n^2 + 2n + 1} = \frac{2n^2 + 2n}{2n^2 + 2n + 1} = \frac{2n(n+1)}{2n^2 + 2n + 1}\).

(b) The sum to infinity \(S_{\infty}\) is:
\(S_{\infty} = \lim_{n \to \infty} \frac{2n(n+1)}{2n^2 + 2n + 1} = 1\).

The difference between the sum to infinity and the sum of the first \(n\) terms is:
\(S_{\infty} - S_n = 1 - \frac{2n^2 + 2n}{2n^2 + 2n + 1} = \frac{1}{2n^2 + 2n + 1}\).

We require:
\(\frac{1}{2n^2 + 2n + 1} < 10^{-4} \implies 2n^2 + 2n + 1 > 10000\)
\(2n^2 + 2n - 9999 > 0\).

Solving the quadratic equation \(2n^2 + 2n - 9999 = 0\):
\(n = \frac{-2 \pm \sqrt{4 - 4(2)(-9999)}}{4} = \frac{-2 \pm \sqrt{79996}}{4}\).

Using the positive root:
\(n \approx \frac{-2 + 282.836}{4} = 70.21\).

Since \(n\) must be a positive integer, the least value of \(n\) is \(71\).

(a)
- M1: For algebraic factorization of \(4r^4 + 1\).
- A1: For finding correct partial fractions of the form \(\frac{1}{2r^2-2r+1} - \frac{1}{2r^2+2r+1}\).
- A1: For identifying the terms as \(f(r) - f(r+1)\).
- M1: For writing out the sum and showing the cancellation of intermediate terms.
- M1: For obtaining \(f(1) - f(n+1)\).
- A1: For substituting the values \(f(1) = 1\) and \(f(n+1) = \frac{1}{2n^2+2n+1}\).
- A1: For obtaining the required simplified fraction.

(b)
- B1: For the sum to infinity \(1\).
- M1: For setting up the inequality \(\frac{1}{2n^2 + 2n + 1} < 10^{-4}\).
- A1.7: For solving the inequality to find the least integer \(n = 71\).

(a) A matrix is singular if and only if its determinant is zero.
\(\det(\mathbf{M}) = 1((0)(1) - (1)(-1)) - 2((2)(1) - (1)(1)) + a((2)(-1) - (0)(1))\)
\(\det(\mathbf{M}) = 1(1) - 2(1) + a(-2) = 1 - 2 - 2a = -1 - 2a\).

Setting \(\det(\mathbf{M}) = 0\):
\(-1 - 2a = 0 \implies a = -\frac{1}{2}\).

(b) For \(a = 3\), \(\det(\mathbf{M}) = -1 - 2(3) = -7\).
The matrix is \(\mathbf{M} = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 0 & 1 \\ 1 & -1 & 1 \end{pmatrix}\).

We find the matrix of cofactors, \(\mathbf{C}\):
\(C_{11} = +(0 - (-1)) = 1\)
\(C_{12} = -(2 - 1) = -1\)
\(C_{13} = +(-2 - 0) = -2\)
\(C_{21} = -(2 - (-3)) = -5\)
\(C_{22} = +(1 - 3) = -2\)
\(C_{23} = -(-1 - 2) = 3\)
\(C_{31} = +(2 - 0) = 2\)
\(C_{32} = -(1 - 6) = 5\)
\(C_{33} = +(0 - 4) = -4\)

So \(\mathbf{C} = \begin{pmatrix} 1 & -1 & -2 \\ -5 & -2 & 3 \\ 2 & 5 & -4 \end{pmatrix}\).

Taking the transpose to get the adjugate matrix:
\(\mathbf{C}^T = \begin{pmatrix} 1 & -5 & 2 \\ -1 & -2 & 5 \\ -2 & 3 & -4 \end{pmatrix}\).

Then:
\(\mathbf{M}^{-1} = \frac{1}{\det(\mathbf{M})} \mathbf{C}^T = -\frac{1}{7} \begin{pmatrix} 1 & -5 & 2 \\ -1 & -2 & 5 \\ -2 & 3 & -4 \end{pmatrix} = \frac{1}{7} \begin{pmatrix} -1 & 5 & -2 \\ 1 & 2 & -5 \\ 2 & -3 & 4 \end{pmatrix}\).

(c) The system of equations can be written in matrix form as:
\(\mathbf{M} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 4 \\ 11 \\ 5 \end{pmatrix}\).

Multiplying both sides by \(\mathbf{M}^{-1}\):
\(\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{7} \begin{pmatrix} -1 & 5 & -2 \\ 1 & 2 & -5 \\ 2 & -3 & 4 \end{pmatrix} \begin{pmatrix} 4 \\ 11 \\ 5 \end{pmatrix}\)
\(\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{7} \begin{pmatrix} -1(4) + 5(11) - 2(5) \\ 1(4) + 2(11) - 5(5) \\ 2(4) - 3(11) + 4(5) \end{pmatrix} = \frac{1}{7} \begin{pmatrix} -4 + 55 - 10 \\ 4 + 22 - 25 \\ 8 - 33 + 20 \end{pmatrix} = \frac{1}{7} \begin{pmatrix} 41 \\ 1 \\ -5 \end{pmatrix}\).

So, \(x = \frac{41}{7}\), \(y = \frac{1}{7}\), \(z = -\frac{5}{7}\).

(a)
- M1: For finding the determinant in terms of \(a\).
- A1: For obtaining the correct value of \(a = -1/2\).

(b)
- B1: For the determinant value \(-7\).
- M1: For finding at least four correct cofactors.
- M1: For transpose of a cofactor matrix.
- A2: For the complete correct inverse matrix \(\mathbf{M}^{-1}\) (deduct 1 mark for minor calculation errors).

(c)
- M1: For representing the system of equations as a matrix equation and using \(\mathbf{M}^{-1}\).
- A2.7: For correct values of \(x\), \(y\), and \(z\) (A1 for one correct, A2 for two, A2.7 for all three).

(a) The curve is a cardioid symmetric about the line \(\theta = \frac{\pi}{2}\).
- For \(\theta = 0\), \(r = 2\), yielding polar coordinates \((2, 0)\).
- For \(\theta = \pi\), \(r = 2\), yielding polar coordinates \((2, \pi)\).
- For \(\theta = -\frac{\pi}{2}\), \(r = 0\), which represents the pole.
- For \(\theta = \frac{\pi}{2}\), \(r = 4\).

(b) The area \(A\) is given by:
\(A = \frac{1}{2} \int_{0}^{\pi/2} r^2 \mathrm{d}\theta = \frac{1}{2} \int_{0}^{\pi/2} 4(1 + \sin\theta)^2 \mathrm{d}\theta = 2 \int_{0}^{\pi/2} (1 + 2\sin\theta + \sin^2\theta) \mathrm{d}\theta\).

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

(c) We express \(x\) and \(y\) parametrically:
\(x = r \cos\theta = 2(1 + \sin\theta)\cos\theta = 2\cos\theta + \sin 2\theta\)
\(y = r \sin\theta = 2(1 + \sin\theta)\sin\theta = 2\sin\theta + 1 - \cos 2\theta\)

Differentiate with respect to \(\theta\):
\(\frac{\mathrm{d}x}{\mathrm{d}\theta} = -2\sin\theta + 2\cos 2\theta\)
\(\frac{\mathrm{d}y}{\mathrm{d}\theta} = 2\cos\theta + 2\sin 2\theta\)

At \(\theta = \frac{\pi}{6}\):
\(\frac{\mathrm{d}x}{\mathrm{d}\theta} = -2\sin\left(\frac{\pi}{6}\right) + 2\cos\left(\frac{\pi}{3}\right) = -2\left(\frac{1}{2}\right) + 2\left(\frac{1}{2}\right) = 0\).

Since \(\frac{\mathrm{d}x}{\mathrm{d}\theta} = 0\) and \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = 2\cos\left(\frac{\pi}{6}\right) + 2\sin\left(\frac{\pi}{3}\right) = 2\sqrt{3} \ne 0\), the tangent is vertical.

The equation of the vertical tangent is \(x = c\), where \(c\) is the \(x\)-coordinate at \(\theta = \frac{\pi}{6}\):
\(x = 2\left(1 + \sin\left(\frac{\pi}{6}\right)\right)\cos\left(\frac{\pi}{6}\right) = 2\left(\frac{3}{2}\right)\left(\frac{\sqrt{3}}{2}\right) = \frac{3\sqrt{3}}{2}\).

So the cartesian equation is \(x = \frac{3\sqrt{3}}{2}\).

(a)
- B1: For a correct cardioid shape.
- B1: For the correct orientation (symmetric about \(\theta = \pi/2\) with the cusp at the pole).
- B1: For labeling the intercept \((2, 0)\) and the pole.

(b)
- M1: For using the area formula \(A = \frac{1}{2}\int r^2 \mathrm{d}\theta\) with correct limits.
- M1: For using \(\sin^2\theta = \frac{1-\cos 2\theta}{2}\) to integrate.
- A1: For correct integration.
- A1.7: For correct final area \(\frac{3\pi}{2} + 4\).

(c)
- M1: For finding \(\frac{\mathrm{d}x}{\mathrm{d}\theta}\) and evaluating it at \(\theta = \pi/6\).
- M1: For recognizing that the tangent is vertical since \(\frac{\mathrm{d}x}{\mathrm{d}\theta} = 0\).
- A1: For the correct cartesian equation \(x = \frac{3\sqrt{3}}{2}\).

(a) To show the lines are skew, we show they are not parallel and do not intersect.

1. Direction vectors are \(\mathbf{d}_1 = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}\) and \(\mathbf{d}_2 = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}\).
Since \(\mathbf{d}_1\) is not a scalar multiple of \(\mathbf{d}_2\), the lines are not parallel.

2. Equating coordinates to find an intersection:
- x-component: \(1 + 2\lambda = 4 + \mu \implies 2\lambda - \mu = 3\) (1)
- y-component: \(3 - \lambda = 1 + 2\mu \implies \lambda + 2\mu = 2\) (2)
- z-component: \(-1 + 2\lambda = 3 + 2\mu \implies 2\lambda - 2\mu = 4\) (3)

From (1) and (3), subtracting (3) from (1) gives \(\mu = -1\).
Substituting \(\mu = -1\) into (1) gives \(2\lambda + 1 = 3 \implies \lambda = 1\).
Now test \(\lambda = 1, \mu = -1\) in (2):
\(\lambda + 2\mu = 1 + 2(-1) = -1 \ne 2\).
Since the equations are inconsistent, the lines do not intersect. Therefore, the lines are skew.

(b) The direction of the common perpendicular is given by \(\mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2\):
\(\mathbf{n} = \det\begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 2 \\ 1 & 2 & 2 \end{pmatrix} = \mathbf{i}(-2 - 4) - \mathbf{j}(4 - 2) + \mathbf{k}(4 - (-1)) = -6\mathbf{i} - 2\mathbf{j} + 5\mathbf{k}\).

Its magnitude is:
\(|\mathbf{n}| = \sqrt{(-6)^2 + (-2)^2 + 5^2} = \sqrt{36 + 4 + 25} = \sqrt{65}\).

Let \(A(1, 3, -1)\) be a point on \(L_1\) and \(B(4, 1, 3)\) be a point on \(L_2\).
\(\vec{AB} = \begin{pmatrix} 3 \\ -2 \\ 4 \end{pmatrix}\).

The shortest distance \(d\) is the projection of \(\vec{AB}\) onto \(\mathbf{n}\):
\(d = \frac{|\vec{AB} \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{|3(-6) + (-2)(-2) + 4(5)|}{\sqrt{65}} = \frac{|-18 + 4 + 20|}{\sqrt{65}} = \frac{6}{\sqrt{65}}\).

(c) The plane has normal vector \(\mathbf{n} = \begin{pmatrix} -6 \\ -2 \\ 5 \end{pmatrix}\) and contains the point \((1, 3, -1)\).

Its equation is:
\(-6(x - 1) - 2(y - 3) + 5(z + 1) = 0\)
\(-6x + 6 - 2y + 6 + 5z + 5 = 0\)
\(-6x - 2y + 5z + 17 = 0\)

Multiplying by \(-1\):
\(6x + 2y - 5z = 17\).

(a)
- B1: For explaining why the lines are not parallel.
- M1: For writing down the coordinate equations and attempting to solve them.
- M1: For showing that the values obtained do not satisfy the third equation.
- A1: For concluding that the lines are skew.

(b)
- M1: For calculating the cross product of the two direction vectors.
- A1: For obtaining \(-6\mathbf{i} - 2\mathbf{j} + 5\mathbf{k}\).
- A1.7: For evaluating the shortest distance to get \(\frac{6}{\sqrt{65}}\).

(c)
- M1: For using the normal vector in a plane equation.
- A2: For obtaining the correct plane equation \(6x + 2y - 5z = 17\) (or any equivalent form).

(a) Let \(P(n)\) be the statement:
\[ \sum_{r=1}^{n} r(r+1)(r+2) = \frac{1}{4}n(n+1)(n+2)(n+3) \]

**Base Case:** For \(n = 1\):
LHS \(= 1(2)(3) = 6\).
RHS \(= \frac{1}{4}(1)(2)(3)(4) = 6\).
Since LHS \(=\) RHS, \(P(1)\) is true.

**Inductive Step:** Assume \(P(k)\) is true for some positive integer \(k\), so:
\[ \sum_{r=1}^{k} r(r+1)(r+2) = \frac{1}{4}k(k+1)(k+2)(k+3) \]

We must show that \(P(k+1)\) is true, i.e.:
\[ \sum_{r=1}^{k+1} r(r+1)(r+2) = \frac{1}{4}(k+1)(k+2)(k+3)(k+4) \]

Now:
\(\sum_{r=1}^{k+1} r(r+1)(r+2) = \sum_{r=1}^{k} r(r+1)(r+2) + (k+1)(k+2)(k+3)\)
\(= \frac{1}{4}k(k+1)(k+2)(k+3) + (k+1)(k+2)(k+3)\)
\(= (k+1)(k+2)(k+3) \left( \frac{1}{4}k + 1 \right)\)
\(= \frac{1}{4}(k+1)(k+2)(k+3)(k+4)\).

This is indeed \(P(k+1)\).
Therefore, by mathematical induction, the statement \(P(n)\) is true for all positive integers \(n\).

(b) We can write:
\(\sum_{r=n+1}^{2n} r(r+1)(r+2) = \sum_{r=1}^{2n} r(r+1)(r+2) - \sum_{r=1}^{n} r(r+1)(r+2)\)

Using the formula from part (a):
\(\sum_{r=1}^{2n} r(r+1)(r+2) = \frac{1}{4}(2n)(2n+1)(2n+2)(2n+3) = n(n+1)(2n+1)(2n+3)\).
\(\sum_{r=1}^{n} r(r+1)(r+2) = \frac{1}{4}n(n+1)(n+2)(n+3)\).

Subtracting the two sums:
\(\sum_{r=n+1}^{2n} r(r+1)(r+2) = n(n+1)(2n+1)(2n+3) - \frac{1}{4}n(n+1)(n+2)(n+3)\)
\(= \frac{1}{4}n(n+1) \left[ 4(2n+1)(2n+3) - (n+2)(n+3) \right]\)
\(= \frac{1}{4}n(n+1) \left[ 4(4n^2 + 8n + 3) - (n^2 + 5n + 6) \right]\)
\(= \frac{1}{4}n(n+1) \left[ 16n^2 + 32n + 12 - n^2 - 5n - 6 \right]\)
\(= \frac{1}{4}n(n+1) \left[ 15n^2 + 27n + 6 \right]\)
\(= \frac{3}{4}n(n+1) \left[ 5n^2 + 9n + 2 \right]\).

(a)
- B1: For verifying the base case \(n=1\).
- M1: For stating the inductive hypothesis and setting up the sum for \(n=k+1\).
- M1: For factoring out the common term \((k+1)(k+2)(k+3)\).
- A1: For obtaining \(\frac{1}{4}(k+1)(k+2)(k+3)(k+4)\).
- A2: For a complete and mathematically sound inductive conclusion.

(b)
- M1: For using \(S_{2n} - S_n\).
- A1: For the correct expression for \(S_{2n}\).
- M1: For factoring out \(\frac{1}{4}n(n+1)\).
- A1.7: For expanding and factoring the remaining quadratic to obtain the required result \(\frac{3}{4}n(n+1)(5n^2 + 9n + 2)\).

(i) We write
\( I_n = \int_{0}^{\ln 2} \tanh^{n-2} x \, \tanh^2 x \, dx \).
Using the identity \( \tanh^2 x = 1 - \text{sech}^2 x \), we have
\( I_n = \int_{0}^{\ln 2} \tanh^{n-2} x \, (1 - \text{sech}^2 x) \, dx = \int_{0}^{\ln 2} \tanh^{n-2} x \, dx - \int_{0}^{\ln 2} \tanh^{n-2} x \, \text{sech}^2 x \, dx \).
This gives
\( I_n = I_{n-2} - \left[ \frac{\tanh^{n-1} x}{n-1} \right]_{0}^{\ln 2} \).
Since \( \tanh(\ln 2) = \frac{e^{\ln 2} - e^{-\ln 2}}{e^{\ln 2} + e^{-\ln 2}} = \frac{2 - 1/2}{2 + 1/2} = \frac{3}{5} \) and \( \tanh(0) = 0 \), we obtain
\( I_n = I_{n-2} - \frac{1}{n-1} \left( \frac{3}{5} \right)^{n-1} \).

(ii) For \( n = 3 \),
\( I_3 = I_1 - \frac{1}{2} \left( \frac{3}{5} \right)^2 = I_1 - \frac{9}{50} \).
We calculate \( I_1 \):
\( I_1 = \int_{0}^{\ln 2} \tanh x \, dx = \left[ \ln(\cosh x) \right]_{0}^{\ln 2} \).
Since \( \cosh(\ln 2) = \frac{2 + 1/2}{2} = \frac{5}{4} \) and \( \cosh(0) = 1 \),
\( I_1 = \ln\left(\frac{5}{4}\right) - \ln(1) = \ln\left(\frac{5}{4}\right) \).
Thus, \( I_3 = -\frac{9}{50} + \ln\left(\frac{5}{4}\right) \).

M1: Splitting the integral into \( \tanh^{n-2} x (1 - \text{sech}^2 x) \).
A1: Integrating the term \( \tanh^{n-2} x \text{sech}^2 x \) correctly to obtain \( \frac{\tanh^{n-1} x}{n-1} \).
A1: Correct substitution of limits \( \ln 2 \) and \( 0 \) showing \( \tanh(\ln 2) = \frac{3}{5} \), and completing the proof.
M1: Using the reduction formula with \( n=3 \).
M1: Integrating \( \tanh x \) to obtain \( \ln(\cosh x) \) and evaluating at limits.
A1: Final correct value of \( I_3 \) in the specified form.

(i) Differentiating \( y = \arcsin(2x) \):
\( \frac{dy}{dx} = \frac{2}{\sqrt{1 - 4x^2}} \implies (1 - 4x^2) \left(\frac{dy}{dx}\right)^2 = 4 \).
Differentiating implicitly with respect to \( x \):
\( -8x \left(\frac{dy}{dx}\right)^2 + (1 - 4x^2) \cdot 2\frac{dy}{dx}\frac{d^2 y}{dx^2} = 0 \).
Dividing by \( 2\frac{dy}{dx} \) gives
\( (1 - 4x^2) \frac{d^2 y}{dx^2} - 4x \frac{dy}{dx} = 0 \).

(ii) Applying Leibniz's theorem to \( (1-4x^2) y^{(2)} - 4x y^{(1)} = 0 \):
For the first term, we get
\( y^{(n+2)}(1-4x^2) + n y^{(n+1)}(-8x) + \frac{n(n-1)}{2} y^{(n)}(-8) \).
For the second term, we get
\( -4x y^{(n+1)} - 4n y^{(n)} \).
Adding these together:
\( (1-4x^2) y^{(n+2)} - (8n + 4)x y^{(n+1)} - [4n(n-1) + 4n] y^{(n)} = 0 \),
\( (1-4x^2) y^{(n+2)} - 4(2n+1)x y^{(n+1)} - 4n^2 y^{(n)} = 0 \).

(iii) When \( x = 0 \):
\( y(0) = \arcsin(0) = 0 \).
\( y'(0) = 2 \).
From the original equation, \( y''(0) - 0 = 0 \implies y''(0) = 0 \).
For \( n = 1 \) at \( x = 0 \):
\( y^{(3)}(0) - 4 y^{(1)}(0) = 0 \implies y^{(3)}(0) = 4(2) = 8 \).
Thus, the Maclaurin series up to \( x^3 \) is:
\( y(x) = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 = 2x + \frac{4}{3}x^3 \).

M1: Differentiating \( y \) and rearranging to a form without fractions.
A1: Differentiating again to obtain the given second order differential relation.
M1: Expressing the \( n \)-th derivative of the terms using Leibniz's theorem.
A1: Correct expansion of both terms with binomial coefficients.
A1: Simplifying algebraic terms to match the target equation.
M1: Evaluating \( y(0) \), \( y'(0) \), and \( y''(0) \).
M1: Using the recurrence relation from part (ii) with \( n=1 \) to find \( y'''(0) \).
A1: Correctly writing the Maclaurin series up to the \( x^3 \) term.

(i) We substitute \( \cosh^2 x = 1 + \sinh^2 x \):
\( 3(1 + \sinh^2 x) - 5\sinh x - 5 = 0 \implies 3\sinh^2 x - 5\sinh x - 2 = 0 \).
Factoring the quadratic equation:
\( (3\sinh x + 1)(\sinh x - 2) = 0 \).
So \( \sinh x = -\frac{1}{3} \) or \( \sinh x = 2 \).
Using the formula \( \text{arsinh}(y) = \ln(y + \sqrt{y^2 + 1}) \):
For \( \sinh x = 2 \), \( x = \ln(2 + \sqrt{5}) \).
For \( \sinh x = -\frac{1}{3} \), \( x = \ln\left(-\frac{1}{3} + \sqrt{\frac{1}{9} + 1}\right) = \ln\left(\frac{-1 + \sqrt{10}}{3}\right) \).

(ii) We integrate using inverse hyperbolic sine substitution:
\( \int \frac{1}{\sqrt{4x^2 + 9}} \, dx = \frac{1}{2} \int \frac{1}{\sqrt{x^2 + (3/2)^2}} \, dx = \frac{1}{2} \text{arsinh}\left(\frac{2x}{3}\right) \).
Evaluating from \( 0 \) to \( 2 \):
\( \frac{1}{2} \left[ \text{arsinh}\left(\frac{4}{3}\right) - \text{arsinh}(0) \right] = \frac{1}{2} \ln\left(\frac{4}{3} + \sqrt{\frac{16}{9} + 1}\right) = \frac{1}{2} \ln\left(\frac{4}{3} + \frac{5}{3}\right) = \frac{1}{2}\ln 3 \).

M1: Using the identity \( \cosh^2 x = 1 + \sinh^2 x \).
A1: Solving the quadratic in \( \sinh x \) to get \( \sinh x = 2 \) and \( \sinh x = -1/3 \).
M1: Using the logarithmic form of \( \text{arsinh} \) to find both values of \( x \).
A1: Correctly writing both final logarithmic forms.
M1: Standard integration leading to \( \frac{1}{2}\text{arsinh}(\frac{2x}{3}) \).
M1: Substituting the limits \( 0 \) and \( 2 \).
A1: Obtaining the exact simplified logarithmic value of \( \frac{1}{2}\ln 3 \).

(i) Using de Moivre's theorem:
\( \cos(5\theta) + i\sin(5\theta) = (\cos\theta + i\sin\theta)^5 \).
Equating the imaginary parts:
\( \sin(5\theta) = 5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta \)
\( = 5(1-\sin^2\theta)^2\sin\theta - 10(1-\sin^2\theta)\sin^3\theta + \sin^5\theta \)
\( = 5(1 - 2\sin^2\theta + \sin^4\theta)\sin\theta - 10(\sin^3\theta - \sin^5\theta) + \sin^5\theta \)
\( = 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta \).

(ii) If \( \sin(5\theta) = 0 \), then \( 5\theta = k\pi \implies \theta = \frac{k\pi}{5} \) for \( k \in \mathbb{Z} \).
Since \( \sin(5\theta) = \sin\theta(16\sin^4\theta - 20\sin^2\theta + 5) \), if \( \sin\theta \neq 0 \) (i.e., \( k \) is not a multiple of 5), then \( 16\sin^4\theta - 20\sin^2\theta + 5 = 0 \).
For \( k = 1, 2, 3, 4 \), we get distinct non-zero values of \( \sin\theta \), which are \( \sin\left(\frac{\pi}{5}\right), \sin\left(\frac{2\pi}{5}\right), \sin\left(\frac{3\pi}{5}\right) = \sin\left(\frac{2\pi}{5}\right), \sin\left(\frac{4\pi}{5}\right) = \sin\left(\frac{\pi}{5}\right) \).
Considering negative orientations (from \( k = -1, -2 \)), we obtain the roots as \( \pm\sin\left(\frac{\pi}{5}\right) \) and \( \pm\sin\left(\frac{2\pi}{5}\right) \).

(iii) Let \( y = x^2 \). The quartic \( 16x^4 - 20x^2 + 5 = 0 \) becomes \( 16y^2 - 20y + 5 = 0 \).
The roots of this quadratic in \( y \) are \( y_1 = \sin^2\left(\frac{\pi}{5}\right) \) and \( y_2 = \sin^2\left(\frac{2\pi}{5}\right) \).
By the product of roots formula:
\( y_1 y_2 = \frac{5}{16} \).
Thus, \( \sin^2\left(\frac{\pi}{5}\right) \sin^2\left(\frac{2\pi}{5}\right) = \frac{5}{16} \).

M1: Attempting to expand \( (\cos\theta + i\sin\theta)^5 \) and extract the imaginary part.
A1: Correct expression in terms of \( \cos\theta \) and \( \sin\theta \).
A1: Substituting \( \cos^2\theta = 1 - \sin^2\theta \) and correctly simplifying to get \( 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta \).
M1: Setting \( \sin(5\theta) = 0 \) and identifying solutions \( \theta = k\pi/5 \).
A1: Explaining how the non-zero roots map to the roots of the quartic \( 16x^4 - 20x^2 + 5 = 0 \).
M1: Using a substitution \( y = x^2 \) to obtain the quadratic equation \( 16y^2 - 20y + 5 = 0 \).
A1: Recognizing the product of roots \( y_1 y_2 \) gives the required exact value \( 5/16 \).

First, find the complementary function (CF) by solving the auxiliary equation:
\( m^2 + 4m + 5 = 0 \implies m = -2 \pm i \).
So the CF is:
\( y_c = e^{-2x} (A\cos x + B\sin x) \).

Next, find the particular integral (PI). Since \( e^{-2x} \) is not a term of the CF, we try the form:
\( y_p = C e^{-2x} + Dx + E \).
Differentiating once and twice:
\( \frac{dy_p}{dx} = -2C e^{-2x} + D \),
\( \frac{d^2 y_p}{dx^2} = 4C e^{-2x} \).
Substituting into the differential equation:
\( 4C e^{-2x} + 4(-2C e^{-2x} + D) + 5(C e^{-2x} + Dx + E) = 10e^{-2x} + 5x \),
\( C e^{-2x} + 5Dx + (4D + 5E) = 10e^{-2x} + 5x \).
Comparing coefficients:
For \( e^{-2x} \): \( C = 10 \).
For \( x \): \( 5D = 5 \implies D = 1 \).
For constant: \( 4D + 5E = 0 \implies 4(1) + 5E = 0 \implies E = -\frac{4}{5} \).

So the PI is \( y_p = 10e^{-2x} + x - \frac{4}{5} \).
The general solution is the sum of the CF and PI:
\( y = e^{-2x}(A\cos x + B\sin x) + 10e^{-2x} + x - \frac{4}{5} \).

M1: Setting up and solving the auxiliary equation.
A1: Correct complementary function.
M1: Attempting a particular integral of the form \( C e^{-2x} + Dx + E \).
M1: Differentiating and substituting into the differential equation.
A1: Correctly finding \( C = 10 \).
A1: Correctly finding \( D = 1 \) and \( E = -4/5 \).
A1: Summing CF and PI to write the general solution.

(i) We solve the characteristic equation \( \det(\mathbf{A} - \lambda\mathbf{I}) = 0 \):
\( \det\begin{pmatrix} 3-\lambda & -1 & 1 \\ -1 & 5-\lambda & -1 \\ 1 & -1 & 3-\lambda \end{pmatrix} = 0 \)
\( (3-\lambda)[(5-\lambda)(3-\lambda) - 1] + 1[-(3-\lambda) + 1] + 1[1 - (5-\lambda)] = 0 \)
\( (3-\lambda)(\lambda^2 - 8\lambda + 14) + (\lambda - 2) + (\lambda - 4) = 0 \)
\( (3-\lambda)(\lambda^2 - 8\lambda + 12) = 0 \)
\( (3-\lambda)(\lambda-2)(\lambda-6) = 0 \).
Thus, the eigenvalues are indeed \( 2, 3, 6 \).

(ii) For \( \lambda = 2 \):
\( \begin{pmatrix} 1 & -1 & 1 \\ -1 & 3 & -1 \\ 1 & -1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies y = 0, z = -x \).
Eigenvector is \( \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} \), unit eigenvector: \( \mathbf{v}_1 = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} \).

For \( \lambda = 3 \):
\( \begin{pmatrix} 0 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies x = y = z \).
Eigenvector is \( \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \), unit eigenvector: \( \mathbf{v}_2 = \frac{1}{\sqrt{3}} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \).

For \( \lambda = 6 \):
\( \begin{pmatrix} -3 & -1 & 1 \\ -1 & -1 & -1 \\ 1 & -1 & -3 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies x = z, y = -2x \).
Eigenvector is \( \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} \), unit eigenvector: \( \mathbf{v}_3 = \frac{1}{\sqrt{6}} \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} \).

(iii) We construct \( \mathbf{P} \) using the unit eigenvectors as columns:
\( \mathbf{P} = \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} \\ 0 & \frac{1}{\sqrt{3}} & -\frac{2}{\sqrt{6}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} \end{pmatrix} \).
The diagonal matrix with the corresponding eigenvalues is:
\( \mathbf{D} = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 6 \end{pmatrix} \).

M1: Writing down the characteristic equation \( \det(\mathbf{A} - \lambda\mathbf{I}) = 0 \).
A1: Showing that it factorizes to find the eigenvalues 2, 3, 6.
M1: Attempting to find eigenvectors for each eigenvalue.
A1: Finding the correct direction for all three eigenvectors.
A1: Normalizing the eigenvectors to have unit length.
M1: Constructing matrix \( \mathbf{P} \) with the unit eigenvectors.
A1: Writing down the correct \( \mathbf{P} \) and corresponding \( \mathbf{D} \).

(i) First, find the derivatives:
\( \frac{dx}{dt} = e^t \cos t - e^t \sin t = e^t (\cos t - \sin t) \),
\( \frac{dy}{dt} = e^t \sin t + e^t \cos t = e^t (\sin t + \cos t) \).
Now calculate \( \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 \):
\( e^{2t} [(\cos t - \sin t)^2 + (\sin t + \cos t)^2] = e^{2t} [2(\cos^2 t + \sin^2 t)] = 2e^{2t} \).
So the arc length is:
\( S = \int_{0}^{\pi} \sqrt{2e^{2t}} \, dt = \sqrt{2} \int_{0}^{\pi} e^t \, dt = \sqrt{2} [e^t]_{0}^{\pi} = \sqrt{2}(e^\pi - 1) \).

(ii) The surface area \( A \) is given by:
\( A = 2\pi \int_{0}^{\pi} y \sqrt{ \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 } \, dt = 2\pi \int_{0}^{\pi} e^t \sin t (\sqrt{2} e^t) \, dt = 2\sqrt{2}\pi \int_{0}^{\pi} e^{2t} \sin t \, dt \).
Using the integration formula:
\( \int e^{2t} \sin t \, dt = \frac{e^{2t}(2\sin t - \cos t)}{5} \).
Evaluating this from \( 0 \) to \( \pi \):
\( \left[ \frac{e^{2t}(2\sin t - \cos t)}{5} \right]_{0}^{\pi} = \frac{e^{2\pi}(0 - (-1))}{5} - \frac{e^0(0 - 1)}{5} = \frac{e^{2\pi} + 1}{5} \).
Thus, the surface area is:
\( A = 2\sqrt{2}\pi \left( \frac{e^{2\pi} + 1}{5} \right) = \frac{2\sqrt{2}\pi}{5}(e^{2\pi} + 1) \).

M1: Finding \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).
A1: Simplifying \( \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 \) to \( 2e^{2t} \).
A1: Correct integration and limits to establish the arc length.
M1: Using the formula for surface area of revolution.
M1: Integrating \( e^{2t} \sin t \) using integration by parts (twice or via standard formula).
A1: Finding the correct integral \( \frac{e^{2t}(2\sin t - \cos t)}{5} \).
A1: Substituting limits correctly to get the final exact surface area.

(i) The characteristic equation is:
\( \det(\mathbf{M} - \lambda \mathbf{I}) = \det\begin{pmatrix} 4-\lambda & -2 \\ 1 & 1-\lambda \end{pmatrix} = (4-\lambda)(1-\lambda) + 2 = \lambda^2 - 5\lambda + 6 = 0 \).
So the eigenvalues are \( \lambda = 2 \) and \( \lambda = 3 \).

For \( \lambda = 2 \):
\( \begin{pmatrix} 2 & -2 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies x = y \).
Eigenvector is \( \mathbf{u}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix} \).

For \( \lambda = 3 \):
\( \begin{pmatrix} 1 & -2 \\ 1 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies x = 2y \).
Eigenvector is \( \mathbf{u}_2 = \begin{pmatrix} 2 \\ 1 \end{pmatrix} \).

(ii) Let \( \mathbf{Q} = \begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix} \) and \( \mathbf{D} = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix} \).
The inverse of \( \mathbf{Q} \) is:
\( \mathbf{Q}^{-1} = \frac{1}{1 - 2}\begin{pmatrix} 1 & -2 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} -1 & 2 \\ 1 & -1 \end{pmatrix} \).
We use \( \mathbf{M}^n = \mathbf{Q} \mathbf{D}^n \mathbf{Q}^{-1} \):
\( \mathbf{M}^n = \begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 2^n & 0 \\ 0 & 3^n \end{pmatrix} \begin{pmatrix} -1 & 2 \\ 1 & -1 \end{pmatrix} \)
\( = \begin{pmatrix} 2^n & 2 \cdot 3^n \\ 2^n & 3^n \end{pmatrix} \begin{pmatrix} -1 & 2 \\ 1 & -1 \end{pmatrix} \)
\( = \begin{pmatrix} -2^n + 2 \cdot 3^n & 2 \cdot 2^n - 2 \cdot 3^n \\ -2^n + 3^n & 2 \cdot 2^n - 3^n \end{pmatrix} \).

M1: Finding the characteristic equation.
A1: Finding correct eigenvalues 2 and 3.
M1: Finding an eigenvector for \( \lambda = 2 \).
M1: Finding an eigenvector for \( \lambda = 3 \).
M1: Finding the transition matrix \( \mathbf{Q} \) and its inverse.
M1: Setting up \( \mathbf{M}^n = \mathbf{Q} \mathbf{D}^n \mathbf{Q}^{-1} \).
A1: Correctly completing matrix multiplication to find the general entry.

Using the equations of projectile motion:

1. Horizontal motion:
\( x = u t \cos\theta \implies 16 = 20 t \cos\theta \implies t = \frac{16}{20\cos\theta} = \frac{4}{5\cos\theta} \)

2. Vertical motion:
\( y = u t \sin\theta - \frac{1}{2}gt^2 \)

Substituting \( g = 10 \) and the expression for \( t \):
\( 8 = 20 \left(\frac{4}{5\cos\theta}\right)\sin\theta - 5\left(\frac{16}{25\cos^2\theta}\right) \)
\( 8 = 16 \tan\theta - \frac{16}{5}\sec^2\theta \)

Dividing the entire equation by 8:
\( 1 = 2\tan\theta - \frac{2}{5}(1 + \tan^2\theta) \)

Multiplying by 5 and rearranging into a quadratic equation in \( \tan\theta \):
\( 5 = 10\tan\theta - 2 - 2\tan^2\theta \)
\( 2\tan^2\theta - 10\tan\theta + 7 = 0 \)

Solving for \( \tan\theta \) using the quadratic formula:
\( \tan\theta = \frac{10 \pm \sqrt{(-10)^2 - 4(2)(7)}}{2(2)} = \frac{10 \pm \sqrt{100 - 56}}{4} = \frac{10 \pm \sqrt{44}}{4} = \frac{5 \pm \sqrt{11}}{2} \)

This gives two possible values for \( \tan\theta \):
- \( \tan\theta_1 \approx 4.1583 \implies \theta_1 \approx 76.5^\circ \)
- \( \tan\theta_2 \approx 0.8417 \implies \theta_2 \approx 40.1^\circ \)

M1: Attempts to express time \( t \) in terms of \( \theta \) using horizontal motion.
A1: Obtains a correct expression for \( t \), e.g., \( t = \frac{4}{5\cos\theta} \).
M1: Substitutes \( t \) into the vertical motion equation and uses the identity \( \sec^2\theta = 1 + \tan^2\theta \).
A1: Obtains a correct quadratic equation in \( \tan\theta \), e.g., \( 2\tan^2\theta - 10\tan\theta + 7 = 0 \).
M1: Solves the quadratic equation to find two values for \( \tan\theta \).
A1: Gives both angles correctly to 1 decimal place (accept \( 40.1^\circ \) and \( 76.5^\circ \)).

The maximum speed of the particle occurs when its acceleration is zero, which is when the forces acting on it are in equilibrium.

Let \( x \) be the extension of the string at this point.
Tension in the string: \( T = \frac{\lambda x}{l} = \frac{40x}{0.8} = 50x \).
At equilibrium:
\( T = mg \implies 50x = 2(10) \implies 50x = 20 \implies x = 0.4 \text{ m} \).

Let \( v \) be the maximum speed. We use conservation of energy between the release point \( O \) and the point where the speed is maximum.
At the point of maximum speed, the particle has fallen a total vertical distance of:
\( h = l + x = 0.8 + 0.4 = 1.2 \text{ m} \).

Loss in Gravitational Potential Energy (GPE):
\( \Delta \text{GPE} = mgh = 2 \times 10 \times 1.2 = 24 \text{ J} \).

Gain in Kinetic Energy (KE):
\( \Delta \text{KE} = \frac{1}{2}mv^2 = \frac{1}{2}(2)v^2 = v^2 \).

Gain in Elastic Potential Energy (EPE):
\( \Delta \text{EPE} = \frac{\lambda x^2}{2l} = \frac{40 \times 0.4^2}{2 \times 0.8} = \frac{6.4}{1.6} = 4 \text{ J} \).

By conservation of energy:
\( \Delta \text{GPE} = \Delta \text{KE} + \Delta \text{EPE} \)
\( 24 = v^2 + 4 \)
\( v^2 = 20 \implies v = \sqrt{20} = 2\sqrt{5} \approx 4.47 \text{ m s}^{-1} \).

M1: States or implies that maximum speed occurs at the equilibrium position where \( T = mg \).
A1: Calculates the correct extension at maximum speed, \( x = 0.4 \text{ m} \).
M1: Writes a conservation of energy equation relating GPE loss to KE gain and EPE gain.
A1: Correctly evaluates GPE loss as \( mgh = 24 \text{ J} \).
A1: Correctly evaluates EPE gain as \( 4 \text{ J} \).
M1: Solves the energy equation for \( v \).
A1: Obtains the correct maximum speed, \( 4.47 \text{ m s}^{-1} \) (or \( 2\sqrt{5} \text{ m s}^{-1} \)).

Let \( \theta \) be the angle that the radius from the center of the sphere to the particle makes with the upward vertical.

When the particle has fallen a vertical distance \( h \), we have:
\( h = R(1 - \cos\theta) \)

Using conservation of energy:
\( \text{Loss in GPE} = \text{Gain in KE} \)
\( mgh = \frac{1}{2}mv^2 \implies v^2 = 2gh = 2gR(1 - \cos\theta) \)

Considering the radial forces acting on the particle:
\( mg\cos\theta - R_N = \frac{mv^2}{R} \)
where \( R_N \) is the normal reaction force.

The particle loses contact with the sphere when \( R_N = 0 \):
\( mg\cos\theta = \frac{m v^2}{R} \implies g\cos\theta = \frac{2gR(1 - \cos\theta)}{R} \)
\( g\cos\theta = 2g(1 - \cos\theta) \)
\( \cos\theta = 2 - 2\cos\theta \implies 3\cos\theta = 2 \implies \cos\theta = \frac{2}{3} \)

The vertical distance fallen \( h \) is:
\( h = R(1 - \cos\theta) = R\left(1 - \frac{2}{3}\right) = \frac{1}{3}R \).

M1: Applies conservation of energy to find an expression for \( v^2 \) in terms of \( \theta \) or \( h \).
A1: Obtains \( v^2 = 2gR(1 - \cos\theta) \) or equivalent.
M1: Writes down the equation of motion along the normal to the surface of the sphere.
A1: Correctly identifies \( mg\cos\theta - R_N = \frac{mv^2}{R} \).
M1: Sets \( R_N = 0 \) to find the condition for leaving the surface.
A1: Correctly solves for \( \cos\theta = \frac{2}{3} \).
A1: Clearly derives the final vertical distance fallen as \( \frac{1}{3}R \).

Using Newton's second law of motion:
\( m a = -\frac{2}{v + 1} \)

Since \( a = v \frac{\mathrm{d}v}{\mathrm{d}x} \) and \( m = 0.5 \):
\( 0.5 v \frac{\mathrm{d}v}{\mathrm{d}x} = -\frac{2}{v + 1} \)

Separating variables:
\( v(v + 1) \mathrm{d}v = -4 \mathrm{d}x \)
\( (v^2 + v) \mathrm{d}v = -4 \mathrm{d}x \)

Integrating both sides with respect to the given limits from \( x = 0 \) (where \( v = 3 \)) to the required \( x \) (where \( v = 1 \)):
\( \int_{3}^{1} (v^2 + v) \mathrm{d}v = \int_{0}^{x} -4 \mathrm{d}x \)

\( \left[ \frac{1}{3}v^3 + \frac{1}{2}v^2 \right]_{3}^{1} = -4x \)

Evaluating the left-hand side:
\( \left( \frac{1}{3}(1)^3 + \frac{1}{2}(1)^2 \right) - \left( \frac{1}{3}(3)^3 + \frac{1}{2}(3)^2 \right) = -4x \)
\( \left( \frac{1}{3} + \frac{1}{2} \right) - \left( 9 + 4.5 \right) = -4x \)
\( \frac{5}{6} - \frac{27}{2} = -4x \)
\( \frac{5 - 81}{6} = -4x \)
\( -\frac{76}{6} = -4x \implies -\frac{38}{3} = -4x \)

Solving for \( x \):
\( x = \frac{38}{12} = \frac{19}{6} \approx 3.17 \text{ m} \).

M1: Uses Newton's second law with \( a = v \frac{\mathrm{d}v}{\mathrm{d}x} \).
A1: Formulates the differential equation correctly: \( 0.5 v \frac{\mathrm{d}v}{\mathrm{d}x} = -\frac{2}{v + 1} \).
M1: Separates the variables to obtain a integrable form, \( (v^2 + v) \mathrm{d}v = -4 \mathrm{d}x \).
A1: Integrates both sides correctly, including the integration constants or correct definite limits.
M1: Substitutes the boundary conditions \( v = 3 \) at \( x = 0 \) and \( v = 1 \).
A1: Evaluates the algebraic fraction correctly to obtain \( -\frac{38}{3} = -4x \).
A1: Gives the final correct value of \( x = \frac{19}{6} \text{ m} \) or \( 3.17 \text{ m} \).

Let \( R_B \) be the normal reaction force at the floor, \( F_B \) be the frictional force at the floor, and \( R_A \) be the normal reaction force at the smooth wall.

1. Resolving forces vertically:
\( R_B = Mg \)

2. Resolving forces horizontally:
\( F_B = R_A \)

Since the ladder is in limiting equilibrium, the frictional force reaches its maximum value:
\( F_B = \mu R_B = \mu Mg \implies R_A = \mu Mg \)

3. Taking moments about the base of the ladder, \( B \):
\( Mg \times L \cos 60^\circ = R_A \times 2L \sin 60^\circ \)
\( Mg \cos 60^\circ = 2 R_A \sin 60^\circ \)

Substitute \( R_A = \mu Mg \) into the moment equation:
\( Mg \cos 60^\circ = 2 (\mu Mg) \sin 60^\circ \)
\( \cos 60^\circ = 2 \mu \sin 60^\circ \)
\( \mu = \frac{\cos 60^\circ}{2 \sin 60^\circ} = \frac{1}{2\tan 60^\circ} \)

Since \( \tan 60^\circ = \sqrt{3} \):
\( \mu = \frac{1}{2\sqrt{3}} = \frac{\sqrt{3}}{6} \approx 0.289 \).

M1: Resolves forces vertically to find \( R_B = Mg \).
M1: Identifies limiting friction condition \( F_B = \mu R_B \) and resolves horizontally.
A1: Relates the normal wall reaction to friction: \( R_A = \mu Mg \).
M1: Takes moments about point \( B \) (or any other valid point) with correct trigonometric components.
A1: Formulates the correct moment equation, e.g., \( Mg L \cos 60^\circ = R_A 2L \sin 60^\circ \).
M1: Combines the equations to solve for \( \mu \).
A1: Obtains the correct final value of \( \mu = \frac{\sqrt{3}}{6} \) or \( 0.289 \).

Let the line of centers be the \( x \)-axis.

Before collision:
- Velocity components of \( A \):
\( u_{Ax} = 4u \cos 30^\circ = 2\sqrt{3}u \)
\( u_{Ay} = 4u \sin 30^\circ = 2u \)
- Velocity components of \( B \):
\( u_{Bx} = -u \)
\( u_{By} = 0 \)

Since the spheres are smooth, the components of velocity perpendicular to the line of centers are unchanged after collision:
\( v_{Ay} = u_{Ay} = 2u \)

By conservation of momentum parallel to the line of centers:
\( 2m(2\sqrt{3}u) + m(-u) = 2m(v_{Ax}) + m(v_{Bx}) \)
\( 2 v_{Ax} + v_{Bx} = (4\sqrt{3} - 1)u \) --- (Equation 1)

Using Newton's law of restitution along the line of centers:
\( v_{Bx} - v_{Ax} = e(u_{Ax} - u_{Bx}) \)
\( v_{Bx} - v_{Ax} = 0.5(2\sqrt{3}u - (-u)) = (\sqrt{3} + 0.5)u \) --- (Equation 2)

Subtract Equation 2 from Equation 1:
\( 3 v_{Ax} = (4\sqrt{3} - 1)u - (\sqrt{3} + 0.5)u \)
\( 3 v_{Ax} = (3\sqrt{3} - 1.5)u \implies v_{Ax} = (\sqrt{3} - 0.5)u \)

The speed of \( A \) after the collision, \( v_A \), is:
\( v_A = \sqrt{v_{Ax}^2 + v_{Ay}^2} = \sqrt{(\sqrt{3} - 0.5)^2 u^2 + (2u)^2} \)
\( v_A = u \sqrt{(3 - \sqrt{3} + 0.25) + 4} = u \sqrt{7.25 - \sqrt{3}} \approx 2.35u \).

M1: Resolves the initial velocity of \( A \) parallel and perpendicular to the line of centers.
A1: Obtains \( u_{Ax} = 2\sqrt{3}u \) and \( u_{Ay} = 2u \).
M1: Applies conservation of momentum along the line of centers.
A1: Obtains \( 2v_{Ax} + v_{Bx} = (4\sqrt{3} - 1)u \).
M1: Applies the law of restitution along the line of centers.
A1: Obtains \( v_{Bx} - v_{Ax} = (\sqrt{3} + 0.5)u \).
M1: Solves the simultaneous equations to find \( v_{Ax} \) and uses Pythagoras to find the final speed of \( A \).
A1: Obtains the correct final speed \( 2.35u \) (or \( u\sqrt{7.25 - \sqrt{3}} \)).

Let \( \theta \) be the angle of the string \( OP \) from the downward vertical.
When the string makes an angle of \( 60^\circ \) with the upward vertical, the angle with the downward vertical is \( \theta = 180^\circ - 60^\circ = 120^\circ \).

Let \( v \) be the speed of the particle when the string becomes slack.

1. Using conservation of energy:
\( \text{Loss in KE} = \text{Gain in GPE} \)
\( \frac{1}{2} m u^2 - \frac{1}{2} m v^2 = mg L(1 - \cos 120^\circ) \)
\( u^2 - v^2 = 2gL(1 - (-0.5)) = 2gL(1.5) = 3gL \)
\( v^2 = u^2 - 3gL \) --- (Equation 1)

2. Equation of motion along the radial direction towards the center \( O \):
\( T + mg\cos\theta = \frac{m v^2}{L} \)
At \( \theta = 120^\circ \):
\( T + mg\cos 120^\circ = \frac{m v^2}{L} \implies T - 0.5mg = \frac{m v^2}{L} \)

The string becomes slack when the tension \( T = 0 \):
\( -0.5mg = \frac{m v^2}{L} \implies v^2 = 0.5gL \) --- (Equation 2)

3. Equating both expressions for \( v^2 \):
\( 0.5gL = u^2 - 3gL \)
\( u^2 = 3.5gL \)
\( u = \sqrt{3.5gL} = \sqrt{\frac{7gL}{2}} \).

M1: Identifies the position where the string is slack (at \( 120^\circ \) from downward vertical or equivalent).
M1: Applies conservation of energy to write an equation for \( v^2 \).
A1: Correctly obtains \( v^2 = u^2 - 3gL \).
M1: Writes the equation of motion along the radial direction.
A1: Correctly identifies \( T + mg\cos 120^\circ = \frac{mv^2}{L} \).
M1: Sets \( T = 0 \) and solves the system of equations for \( u \).
A1: Obtains the correct final speed \( u = \sqrt{3.5gL} \) (or equivalent form).

First find the cumulative distribution function (CDF) of \(X\), denoted by \(F_X(x)\): For \(0 \le x \le 2\): \(F_X(x) = \int_0^x \frac{3}{8}t^2 dt = [\frac{1}{8}t^3]_0^x = \frac{1}{8}x^3\). The CDF of \(Y\), denoted by \(G(y)\), is defined as \(P(Y \le y)\): \(G(y) = P\left(\frac{8}{X^3} \le y\right)\). Since \(X\) lies in the interval \((0, 2]\), \(Y\) lies in the interval \([1, \infty)\). For \(y \ge 1\): \(G(y) = P\left(X^3 \ge \frac{8}{y}\right) = P\left(X \ge 2y^{-1/3}\right) = 1 - F_X(2y^{-1/3})\). Substitute \(2y^{-1/3}\) into the CDF of \(X\): \(G(y) = 1 - \frac{1}{8}(2y^{-1/3})^3 = 1 - \frac{1}{8}\left(\frac{8}{y}\right) = 1 - \frac{1}{y}\). To find the probability density function (PDF) of \(Y\), differentiate \(G(y)\) with respect to \(y\): \(g(y) = \frac{d}{dy}\left(1 - \frac{1}{y}\right) = \frac{1}{y^2}\) for \(y \ge 1\), and \(0\) otherwise.

M1: Attempt to find the CDF of \(X\) by integration.
A1: Correct CDF \(F_X(x) = \frac{1}{8}x^3\).
M1: Express \(P(Y \le y)\) in terms of \(P(X \ge \dots)\) or equivalent.
A1: Correctly obtain the CDF \(G(y) = 1 - \frac{1}{y}\) for \(y \ge 1\).
M1: Differentiate the CDF \(G(y)\) with respect to \(y\).
A1: Correct PDF \(g(y) = \frac{1}{y^2}\).
A1: Correctly state range \(y \ge 1\).

Since \(G_X(t)\) is a probability generating function, we must have \(G_X(1) = 1\). Therefore: \(G_X(1) = k(2 + 1 + 2)^2 = k(5)^2 = 25k = 1 \implies k = \frac{1}{25}\). The mean \(E(X)\) is given by \(G'_X(1)\). Differentiating \(G_X(t)\) once: \(G'_X(t) = \frac{2}{25}(2 + t + 2t^2)(1 + 4t)\). At \(t = 1\): \(E(X) = G'_X(1) = \frac{2}{25}(5)(5) = 2\). To find the variance, we use \(Var(X) = G''_X(1) + G'_X(1) - [G'_X(1)]^2\). Differentiating again using the product rule: \(G''_X(t) = \frac{2}{25}\left[ (1+4t)^2 + (2+t+2t^2)(4) \right]\). At \(t = 1\): \(G''_X(1) = \frac{2}{25}\left[ 5^2 + 5(4) \right] = \frac{2}{25}(45) = 3.6\). Hence, the variance is: \(Var(X) = 3.6 + 2 - 2^2 = 1.6\).

M1: Use the property \(G_X(1) = 1\) to establish an equation for \(k\).
A1: Show \(k = \frac{1}{25}\).
M1: Differentiate \(G_X(t)\) once with respect to \(t\).
A1: Correctly evaluate \(E(X) = 2\).
M1: Differentiate \(G'_X(t)\) with respect to \(t\).
A1: Correctly evaluate \(G''_X(1) = 3.6\).
M1: Apply the variance formula \(Var(X) = G''_X(1) + G'_X(1) - [G'_X(1)]^2\).
A1: Correctly obtain \(Var(X) = 1.6\).

Let the differences be \(D = \text{After} - \text{Before}\). The differences for the 10 students are: A: 5, B: -2, C: 8, D: 12, E: 3, F: -1, G: 7, H: 10, I: 4, J: -6. We rank the absolute differences \(|D|\): \([5, 2, 8, 12, 3, 1, 7, 10, 4, 6]\). Sorted absolute differences and their ranks: 1. \(|D| = 1\) (Student F, rank 1), 2. \(|D| = 2\) (Student B, rank 2), 3. \(|D| = 3\) (Student E, rank 3), 4. \(|D| = 4\) (Student I, rank 4), 5. \(|D| = 5\) (Student A, rank 5), 6. \(|D| = 6\) (Student J, rank 6), 7. \(|D| = 7\) (Student G, rank 7), 8. \(|D| = 8\) (Student C, rank 8), 9. \(|D| = 10\) (Student H, rank 9), 10. \(|D| = 12\) (Student D, rank 10). The hypotheses are: \(H_0\): Median difference is 0; \(H_1\): Median difference is greater than 0 (one-tailed test). Sum of negative ranks: \(T^- = 1 + 2 + 6 = 9\). Sum of positive ranks: \(T^+ = 3 + 4 + 5 + 7 + 8 + 9 + 10 = 46\). The test statistic is \(T = \min(T^+, T^-) = 9\). For \(n = 10\), one-tailed test at 5% significance level, the critical value of \(T\) is 10. Since \(T = 9 \le 10\), we reject \(H_0\). There is sufficient evidence at the 5% significance level to conclude that the memory enhancement course has successfully improved test scores.

B1: State correct hypotheses \(H_0\) and \(H_1\) (one-tailed).
M1: Find correct differences and their absolute values.
A1: Assign correct ranks to the absolute differences.
M1: Calculate sum of positive ranks and negative ranks.
A1: Identify test statistic \(T = 9\).
B1: State correct critical value of 10 for \(n = 10\), one-tailed, 5% level.
M1: Compare test statistic with critical value (reject \(H_0\) if \(T \le 10\)).
A1: Correct conclusion in context.

Hypotheses: \(H_0\): Age group and preference are independent; \(H_1\): Age group and preference are associated. Row totals are both 100. Column totals are: Like = 60, Neutral = 70, Dislike = 70. Grand total = 200. Expected frequencies are: For Under 30: Like = \(100 \times 60 / 200 = 30\), Neutral = \(100 \times 70 / 200 = 35\), Dislike = \(100 \times 70 / 200 = 35\). For 30 and Over: Like = 30, Neutral = 35, Dislike = 35. Test statistic \(\chi^2 = \sum \frac{(O - E)^2}{E} = \frac{(40 - 30)^2}{30} + \frac{(30 - 35)^2}{35} + \frac{(30 - 35)^2}{35} + \frac{(20 - 30)^2}{30} + \frac{(40 - 35)^2}{35} + \frac{(40 - 35)^2}{35} = 2 \times \left(\frac{100}{30} + \frac{25}{35} + \frac{25}{35}\right) = 2 \times \left(\frac{10}{3} + \frac{10}{7}\right) = 2 \times \frac{100}{21} = \frac{200}{21} \approx 9.524\). Degrees of freedom: \(df = (2 - 1)(3 - 1) = 2\). At 5% level, critical value of \(\chi^2\) with 2 df is 5.991. Since \(9.524 > 5.991\), we reject the null hypothesis and conclude that preference is associated with age group.

B1: State correct hypotheses.
M1: Calculate expected frequencies for all cells.
A1: Correct expected frequencies (30, 35, 35 for both rows).
M1: Calculate \(\frac{(O-E)^2}{E}\) for each cell.
A1: Correct test statistic \(\chi^2 \approx 9.52\) (or \(9.524\)).
B1: State correct degrees of freedom (2) and critical value (5.991).
M1: Compare test statistic with critical value.
A1: Conclude in context that there is significant evidence of association.

Hypotheses: \(H_0: \mu_1 = \mu_2\), \(H_1: \mu_1 \ne \mu_2\). First calculate the pooled sample variance: \(s_p^2 = \frac{\sum (x - \bar{x})^2 + \sum (y - \bar{y})^2}{n_1 + n_2 - 2} = \frac{42.0 + 54.0}{8 + 10 - 2} = \frac{96.0}{16} = 6.0\). The standard error of the difference of means is: \(SE = \sqrt{s_p^2 \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} = \sqrt{6 \left(\frac{1}{8} + \frac{1}{10}\right)} = \sqrt{6 \times 0.225} = \sqrt{1.35} \approx 1.1619\). The test statistic is: \(t = \frac{\bar{x} - \bar{y}}{SE} = \frac{15.4 - 12.2}{1.1619} = \frac{3.2}{1.1619} \approx 2.754\). Degrees of freedom: \(df = n_1 + n_2 - 2 = 16\). For a two-tailed test at the 5% significance level with 16 degrees of freedom, the critical value of \(t\) is 2.120. Since \(|t| = 2.754 > 2.120\), we reject \(H_0\). There is significant evidence of a difference between the population means.

B1: State correct hypotheses.
M1: Find pooled variance \(s_p^2 = 6.0\).
A1: Correct \(s_p^2\).
M1: Calculate standard error \(SE = \sqrt{1.35} \approx 1.1619\).
A1: Correct test statistic \(t \approx 2.75\) (or \(2.754\)).
B1: State correct degrees of freedom (16) and critical value (2.120).
M1: Compare calculated \(t\) value with critical value.
A1: Conclude in context that there is significant evidence of a difference.

(i) Since \(f(x)\) is a PDF, \(\int_0^4 c(4x - x^2) dx = 1 \implies c \left[ 2x^2 - \frac{1}{3}x^3 \right]_0^4 = 1 \implies c \left( 32 - \frac{64}{3} \right) = 1 \implies c \left( \frac{32}{3} \right) = 1 \implies c = \frac{3}{32}\). (ii) By symmetry of \(f(x)\) about \(x = 2\), the mean is \(E(X) = 2\). We find \(E(X^2)\): \(E(X^2) = \int_0^4 x^2 f(x) dx = \frac{3}{32} \int_0^4 (4x^3 - x^4) dx = \frac{3}{32} \left[ x^4 - \frac{1}{5}x^5 \right]_0^4 = \frac{3}{32} \left( 256 - \frac{1024}{5} \right) = \frac{3}{32} \left( \frac{256}{5} \right) = \frac{24}{5} = 4.8\). The variance is \(Var(X) = E(X^2) - [E(X)]^2 = 4.8 - 2^2 = 4.8 - 4 = 0.8\). (iii) \(P(X > 3) = \int_3^4 \frac{3}{32} (4x - x^2) dx = \frac{3}{32} \left[ 2x^2 - \frac{1}{3}x^3 \right]_3^4 = \frac{3}{32} \left[ \left(32 - \frac{64}{3}\right) - \left(18 - 9\right) \right] = \frac{3}{32} \left[ \frac{32}{3} - 9 \right] = \frac{3}{32} \left( \frac{5}{3} \right) = \frac{5}{32}\).

M1: Set up the integral \(\int_0^4 c(4x - x^2) dx = 1\).
A1: Show \(c = \frac{3}{32}\).
M1: Set up the integral for \(E(X^2)\).
A1: Correctly calculate \(E(X^2) = 4.8\).
B1: State \(E(X) = 2\).
M1: Compute \(Var(X) = E(X^2) - [E(X)]^2\).
A1: Correctly obtain \(Var(X) = 0.8\).
M1: Integrate to find \(P(X > 3)\).
A1: Correct probability \(\frac{5}{32}\).