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(a) Equating the expressions for \(y\):
\(x^2 - x - 0.5 = mx - 1\)
\(x^2 - (m+1)x + 0.5 = 0\)

For the line to meet the curve at a single point, the discriminant of this quadratic equation must be zero:
\(\Delta = [-(m+1)]^2 - 4(1)(0.5) = 0\)
\(m^2 + 2m + 1 - 2 = 0 \implies m^2 + 2m - 1 = 0\). (This is shown).

(b) Solving the quadratic equation \(m^2 + 2m - 1 = 0\) using the quadratic formula:
\(m = \frac{-2 \pm \sqrt{2^2 - 4(1)(-1)}}{2} = \frac{-2 \pm \sqrt{8}}{2} = -1 \pm \sqrt{2}\)

Since the discriminant of this quadratic equation in \(m\) is \(8 > 0\), there are two distinct real values of \(m\):
\(m_1 = -1 + \sqrt{2}\) and \(m_2 = -1 - \sqrt{2}\).

To prove the tangent lines are perpendicular, we find the product of their gradients:
\(m_1 \times m_2 = (-1 + \sqrt{2})(-1 - \sqrt{2}) = (-1)^2 - (\sqrt{2})^2 = 1 - 2 = -1\).

Since the product of their gradients is \(-1\), the two tangent lines are perpendicular.

M1: Equating curve and line and rearranging into a standard quadratic equation.
A1: Calculating the discriminant and setting it to 0.
A1: Showing the given equation \(m^2 + 2m - 1 = 0\).
M1: Solving the quadratic equation in \(m\) to find two distinct real values.
A1: Correctly stating the values \(m_1 = -1 + \sqrt{2}\) and \(m_2 = -1 - \sqrt{2}\).
M1: Multiplying the two gradients to show the product is \(-1\).
A1: Clear conclusion that the lines are perpendicular.

(a) Completing the square for \(f(x)\):
\(f(x) = x^2 - 4x + 5 = (x-2)^2 - 4 + 5 = (x-2)^2 + 1\).
Since \((x-2)^2 \ge 0\) for all \(x \in \mathbb{R}\), the minimum value of \(f(x)\) is 1 (occurring at \(x = 2\)).

Completing the square for \(g(x)\):
\(g(x) = -x^2 + 2x + c = -(x^2 - 2x) + c = -[(x-1)^2 - 1] + c = -(x-1)^2 + c + 1\).
Since \(-(x-1)^2 \le 0\) for all \(x \in \mathbb{R}\), the maximum value of \(g(x)\) is \(c+1\) (occurring at \(x = 1\)).

(b) If \(c < 0\), then \(c+1 < 1\).
Since \(f(x) \ge 1\) for all \(x\) and \(g(x) \le c+1\) for all \(x\), we have:
\(f(x) \ge 1 > c+1 \ge g(x)\).
This means \(f(x) > g(x)\) for all \(x \in \mathbb{R}\), so the graphs of \(y = f(x)\) and \(y = g(x)\) can never intersect.

To check the converse: 'If the graphs of \(y = f(x)\) and \(y = g(x)\) never intersect, then \(c < 0\).'
Let's find the intersection condition:
\(x^2 - 4x + 5 = -x^2 + 2x + c \implies 2x^2 - 6x + (5-c) = 0\).
They do not intersect if the discriminant \(\Delta < 0\):
\(\Delta = (-6)^2 - 4(2)(5-c) = 36 - 40 + 8c = 8c - 4 < 0 \implies c < 0.5\).
Thus, the graphs do not intersect for any \(c < 0.5\).
Since the graphs do not intersect for, say, \(c = 0.2\) (which is not less than 0), the converse is false.

M1: Completing the square for \(f(x)\) and identifying the minimum value as 1.
A1: Completing the square for \(g(x)\) and identifying the maximum value as \(c+1\).
M1: Stating that \(c < 0 \implies c+1 < 1\).
A1: Showing the inequality \(f(x) \ge 1 > c+1 \ge g(x)\) leads to no intersection.
M1: Setting up the intersection equation and finding the discriminant.
A1: Deducing the non-intersection condition \(c < 0.5\).
A1: Concluding that the converse is false with a valid reason or counterexample.

(a) LHS \(= \frac{\sin \theta}{1 - \cos \theta} - \frac{1 - \cos \theta}{\sin \theta} = \frac{\sin^2\theta - (1 - \cos\theta)^2}{\sin\theta(1 - \cos\theta)}\)
\(= \frac{\sin^2\theta - (1 - 2\cos\theta + \cos^2\theta)}{\sin\theta(1 - \cos\theta)}\)
\(= \frac{\sin^2\theta + \cos^2\theta - 1 + 2\cos\theta - 2\cos^2\theta}{\sin\theta(1 - \cos\theta)}\) (using \(\sin^2\theta + \cos^2\theta = 1\))
\(= \frac{1 - 1 + 2\cos\theta - 2\cos^2\theta}{\sin\theta(1 - \cos\theta)} = \frac{2\cos\theta(1 - \cos\theta)}{\sin\theta(1 - \cos\theta)}\)
\(= \frac{2\cos\theta}{\sin\theta} = \frac{2}{\tan\theta} =\) RHS. (This is shown).

(b) Substituting the identity into the given equation:
\(\left(\frac{2}{\tan \theta}\right)^2 + 4\left(\frac{2}{\tan \theta}\right) + 5 = 0\)
Let \(u = \frac{2}{\tan \theta}\). The equation becomes:
\(u^2 + 4u + 5 = 0\).
This is a quadratic equation in \(u\). We check the discriminant of this quadratic:
\(\Delta = b^2 - 4ac = 4^2 - 4(1)(5) = 16 - 20 = -4\).
Since the discriminant is less than zero, there are no real solutions for \(u\).
Since \(u\) must be a real number for any real value of \(\theta\), it follows that the original equation has no real solutions for \(\theta\).

M1: Combining fractions with a common denominator.
A1: Expanding \((1 - \cos\theta)^2\) correctly.
M1: Using the identity \(\sin^2\theta + \cos^2\theta = 1\) and factoring the numerator.
A1: Cancelling terms to obtain \(\frac{2}{\tan\theta}\).
M1: Substituting the identity into the second equation and letting \(u = \frac{2}{\tan\theta}\).
A1: Forming the quadratic \(u^2 + 4u + 5 = 0\).
A1: Calculating the discriminant and explaining why the negative discriminant means there are no real solutions.

(a) Let the first term of the AP be \(T_1 = a\) and the common difference be \(d\).
The 3rd term is \(T_3 = a + 2d\).
The 9th term is \(T_9 = a + 8d\).
Since \(T_1\), \(T_3\), and \(T_9\) are the first three terms of a geometric progression, the common ratio must be constant:
\(\frac{T_3}{T_1} = \frac{T_9}{T_3} \implies (T_3)^2 = T_1 \times T_9\)
\((a + 2d)^2 = a(a + 8d)\)
Expanding both sides:
\(a^2 + 4ad + 4d^2 = a^2 + 8ad\)
Subtracting \(a^2\) and \(4ad\) from both sides:
\(4d^2 = 4ad\)
Since \(d \neq 0\), we can divide both sides by \(4d\):
\(d = a\). (This is shown).

(b) The formula for the sum of the first \(n\) terms of an arithmetic progression is:
\(S_n = \frac{n}{2}[2a + (n-1)d]\).
Substituting \(d = a\) into the formula:
\(S_n = \frac{n}{2}[2a + (n-1)a]\)
\(S_n = \frac{n}{2}[2a + an - a]\)
\(S_n = \frac{n}{2}[an + a]\)
\(S_n = \frac{n}{2}a(n+1) = \frac{1}{2}an(n+1)\). (This is shown).

M1: Writing expressions for the 1st, 3rd, and 9th terms of the AP in terms of \(a\) and \(d\).
M1: Setting up the GP relation \((a+2d)^2 = a(a+8d)\).
A1: Expanding and simplifying to \(4d^2 = 4ad\).
A1: Explaining why division by \(4d\) is valid (\(d \neq 0\)) to get \(d = a\).
M1: Recalling the formula for \(S_n\) of an AP.
M1: Substituting \(d = a\) into the formula.
A1: Showing clear algebraic steps to arrive at the final given expression.

(a) Differentiating \(y = 2x^3 - 9x^2 + 12x + k\) with respect to \(x\):
\(\frac{dy}{dx} = 6x^2 - 18x + 12\).
At stationary points, \(\frac{dy}{dx} = 0\):
\(6x^2 - 18x + 12 = 0 \implies 6(x^2 - 3x + 2) = 0 \implies 6(x-1)(x-2) = 0\).
So the stationary points occur at \(x = 1\) and \(x = 2\).
When \(x = 1\): \(y = 2(1)^3 - 9(1)^2 + 12(1) + k = 5 + k\).
When \(x = 2\): \(y = 2(2)^3 - 9(2)^2 + 12(2) + k = 16 - 36 + 24 + k = 4 + k\).
Thus, the stationary points are \((1, 5+k)\) and \((2, 4+k)\).

(b) At the point \(P\) where \(x = 3\):
\(\frac{dy}{dx} = 6(3)^2 - 18(3) + 12 = 54 - 54 + 12 = 12\).
The gradient of the tangent at \(P\) is 12.
For any other point on \(C\) to have a tangent parallel to this, we must have \(\frac{dy}{dx} = 12\):
\(6x^2 - 18x + 12 = 12 \implies 6x^2 - 18x = 0 \implies 6x(x-3) = 0\).
This gives \(x = 3\) (which is point \(P\)) or \(x = 0\) (which is point \(Q\)).
Thus, \(Q\) is the only other point, and its \(x\)-coordinate is \(0\).
Let's find the \(y\)-coordinates:
At \(P\) (\(x = 3\)): \(y_P = 2(3)^3 - 9(3)^2 + 12(3) + k = 9+k\). So \(P = (3, 9+k)\).
At \(Q\) (\(x = 0\)): \(y_Q = 2(0)^3 - 9(0)^2 + 12(0) + k = k\). So \(Q = (0, k)\).

The midpoint of \(PQ\) is:
\(M = \left(\frac{3+0}{2}, \frac{9+k+k}{2}\right) = (1.5, 4.5+k)\).
Now we check if \(M\) lies on \(C\). Substitute \(x = 1.5 = \frac{3}{2}\) into the equation of \(C\):
\(y = 2\left(\frac{3}{2}\right)^3 - 9\left(\frac{3}{2}\right)^2 + 12\left(\frac{3}{2}\right) + k\)
\(y = 2\left(\frac{27}{8}\right) - 9\left(\frac{9}{4}\right) + 18 + k = \frac{27}{4} - \frac{81}{4} + 18 + k\)
\(y = -\frac{54}{4} + 18 + k = -13.5 + 18 + k = 4.5 + k\).
Since the calculated \(y\)-value matches the \(y\)-coordinate of \(M\), the midpoint of \(PQ\) indeed lies on \(C\).

B1: Finding \(\frac{dy}{dx}\) correctly.
M1: Setting \(\frac{dy}{dx} = 0\) and solving.
A1: Finding the two stationary points \((1, 5+k)\) and \((2, 4+k)\).
B1: Finding the gradient at \(x=3\) is 12.
M1: Setting \(\frac{dy}{dx} = 12\) and finding the other point \(x=0\).
M1: Stating coordinates of \(P\) and \(Q\), and calculating the midpoint \(M(1.5, 4.5+k)\).
A1: Showing that \(x = 1.5\) yields \(y = 4.5+k\) on the curve \(C\).

(a) The area \(A_1\) is given by:
\(A_1 = \int_1^a x^{-1/4} dx = \left[ \frac{x^{3/4}}{3/4} \right]_1^a = \left[ \frac{4}{3}x^{3/4} \right]_1^a = \frac{4}{3}(a^{3/4} - 1)\). (This is shown).

The area \(A_2\) is given by:
\(A_2 = \int_a^{a^2} x^{-1/4} dx = \left[ \frac{4}{3}x^{3/4} \right]_a^{a^2} = \frac{4}{3}((a^2)^{3/4} - a^{3/4}) = \frac{4}{3}(a^{3/2} - a^{3/4})\).
Factoring out \(a^{3/4}\):
\(A_2 = a^{3/4} \left[ \frac{4}{3}(a^{3/4} - 1) \right] = a^{3/4} A_1\). (This is shown).

(b) The volume of revolution \(V\) is:
\(V = \pi \int_1^a y^2 dx = \pi \int_1^a (x^{-1/4})^2 dx = \pi \int_1^a x^{-1/2} dx\)
\(V = \pi \left[ \frac{x^{1/2}}{1/2} \right]_1^a = \pi \left[ 2\sqrt{x} \right]_1^a = 2\pi(\sqrt{a} - 1)\).

As \(a \to \infty\), \(\sqrt{a} \to \infty\), and hence the volume \(V\) increases without limit (tends to infinity / diverges).

M1: Integrating \(x^{-1/4}\) correctly to obtain \(\frac{4}{3}x^{3/4}\).
A1: Evaluating \(A_1 = \frac{4}{3}(a^{3/4} - 1)\).
M1: Integrating over the limits \(a\) and \(a^2\) and factoring out \(a^{3/4}\) to prove \(A_2 = a^{3/4} A_1\).
A1: Showing the algebraic steps of the proof clearly.
M1: Setting up the volume integral \(\pi \int y^2 dx\) and integrating \(x^{-1/2}\).
A1: Obtaining \(V = 2\pi(\sqrt{a} - 1)\).
A1: Stating clearly that \(V \to \infty\) as \(a \to \infty\).

(a) Let the sector have radius \(r\) and angle \(\theta\) radians.
The arc length of the sector is \(r\theta\).
The perimeter \(P\) of the sector is:
\(P = 2r + r\theta \implies r\theta = P - 2r\).

The area \(A\) of the sector is:
\(A = \frac{1}{2} r^2 \theta = \frac{1}{2} r (r\theta)\).

Substituting \(r\theta = P - 2r\):
\(A = \frac{1}{2} r (P - 2r) = \frac{1}{2} P r - r^2\). (This is shown).

(b) Here, the perimeter is fixed at \(L\), so \(P = L\).
The area expression becomes:
\(A = \frac{1}{2} L r - r^2\).

To find the maximum area, we complete the square:
\(A = -(r^2 - \frac{1}{2} L r) = -\left[ \left( r - \frac{L}{4} \right)^2 - \frac{L^2}{16} \right] = \frac{L^2}{16} - \left( r - \frac{L}{4} \right)^2\).

Since \(\left( r - \frac{L}{4} \right)^2 \ge 0\), the maximum value of \(A\) is \(\frac{L^2}{16}\), which occurs when \(r = \frac{L}{4}\).

To find the angle \(\theta\) for this maximum area:
We have \(r = \frac{L}{4}\) and the perimeter is \(L\):
\(P = 2r + r\theta \implies L = 2\left(\frac{L}{4}\right) + \left(\frac{L}{4}\right)\theta\)
\(L = \frac{L}{2} + \frac{L}{4}\theta\).

Since \(L > 0\), we divide by \(L\):
\(1 = \frac{1}{2} + \frac{1}{4}\theta \implies \frac{1}{4}\theta = \frac{1}{2} \implies \theta = 2\) radians.

M1: Writing the sector formulas for perimeter \(P = 2r + r\theta\) and area \(A = \frac{1}{2} r^2 \theta\).
M1: Eliminating \(\theta\) from these equations.
A1: Showing the given relation \(A = \frac{1}{2} P r - r^2\).
M1: Substituting \(P = L\) and using completing the square or differentiation to find the maximum.
A1: Proving that the maximum area is indeed \(\frac{L^2}{16}\) at \(r = \frac{L}{4}\).
M1: Setting up an equation to find \(\theta\) at \(r = \frac{L}{4}\).
A1: Showing that \(\theta = 2\) radians.

(a) Grouping the terms:
\(x^2 - 10x + y^2 - 10y + 49 = 0\).
Completing the square for both variables:
\((x-5)^2 - 25 + (y-5)^2 - 25 + 49 = 0\)
\((x-5)^2 + (y-5)^2 - 1 = 0 \implies (x-5)^2 + (y-5)^2 = 1\).
Thus, the center of the circle is \((5, 5)\) and the radius is \(\sqrt{1} = 1\).

(b) Substituting \(y = mx\) into the original circle equation:
\(x^2 + (mx)^2 - 10x - 10(mx) + 49 = 0\)
\((1+m^2)x^2 - 10(1+m)x + 49 = 0\).

For the line to be tangent to the circle, this quadratic equation in \(x\) must have exactly one solution, so its discriminant \(\Delta = 0\):
\(\Delta = [-10(1+m)]^2 - 4(1+m^2)(49) = 0\)
\(100(1+m)^2 - 196(1+m^2) = 0\).
Dividing by 4:
\(25(1 + 2m + m^2) - 49(1 + m^2) = 0\)
\(25 + 50m + 25m^2 - 49 - 49m^2 = 0\)
\(-24m^2 + 50m - 24 = 0\).
Multiplying by \(-1\) and dividing by 2:
\(12m^2 - 25m + 12 = 0\).

Solving this quadratic equation in \(m\):
\((3m - 4)(4m - 3) = 0 \implies m = \frac{4}{3}\) or \(m = \frac{3}{4}\).

Since these are two distinct real values, there are indeed two distinct tangent lines.
To show they are reciprocals:
\(m_1 \times m_2 = \frac{4}{3} \times \frac{3}{4} = 1\).
Since their product is 1, they are reciprocals of each other.

M1: Completing the square for both \(x\) and \(y\) terms.
A1: Expressing the circle in standard form \((x-5)^2 + (y-5)^2 = 1\).
A1: Stating center is \((5, 5)\) and radius is 1.
M1: Substituting \(y = mx\) and setting the discriminant of the resulting quadratic to 0.
A1: Deriving the quadratic equation \(12m^2 - 25m + 12 = 0\).
M1: Solving for \(m\) to get \(\frac{4}{3}\) and \(\frac{3}{4}\).
A1: Confirming the product is 1, proving they are reciprocals.

(a) We complete the square of the quadratic expression for \(y\): \(y = x^2 + 2mx + 2m^2 = (x + m)^2 - m^2 + 2m^2 = (x + m)^2 + m^2\). For all real values of \(x\) and \(m\), \((x+m)^2 \ge 0\). Since \(m \ne 0\), we have \(m^2 > 0\). Therefore, \(y = (x+m)^2 + m^2 > 0\) for all real \(x\). Thus, the curve lies entirely above the \(x\)-axis. (b) To find the intersection of the line \(y = -3\) and the curve \(y = x^2 + 2mx + 2m^2\), we set \(x^2 + 2mx + 2m^2 = -3\). Rearranging into standard quadratic form: \(x^2 + 2mx + (2m^2 + 3) = 0\). For this quadratic equation to have real roots, its discriminant \(\Delta\) must satisfy \(\Delta \ge 0\). We calculate \(\Delta = b^2 - 4ac = (2m)^2 - 4(1)(2m^2 + 3) = 4m^2 - 8m^2 - 12 = -4m^2 - 12 = -4(m^2 + 3)\). Since \(m\) is real, \(m^2 \ge 0\), which means \(m^2 + 3 \ge 3 > 0\). Multiplying by \(-4\) gives \(-4(m^2 + 3) \le -12 < 0\). Since \(\Delta < 0\) for all real values of \(m\), the equation has no real roots, meaning the line \(y = -3\) never intersects the curve for any real value of \(m\).

(a) M1: Attempt to complete the square on \(x^2 + 2mx + 2m^2\). A1: Obtain \((x+m)^2 + m^2\). A1: Correct explanation mentioning \((x+m)^2 \ge 0\) and \(m^2 > 0\) for \(m \ne 0\), hence \(y > 0\). (b) M1: Equate the curve and the line to form a quadratic in \(x\). A1: Correct quadratic equation: \(x^2 + 2mx + 2m^2 + 3 = 0\). M1: Find the discriminant of this quadratic in terms of \(m\). A1: Obtain \(\Delta = -4m^2 - 12\). A1: Show clearly that \(\Delta < 0\) for all real \(m\) and conclude no intersection.

(a) LHS \(= \frac{(1 + \sin\theta)^2 + \cos^2\theta}{\cos\theta(1 + \sin\theta)} = \frac{1 + 2\sin\theta + \sin^2\theta + \cos^2\theta}{\cos\theta(1 + \sin\theta)}\. Using the identity \)\sin^2\theta + \cos^2\theta \equiv 1\), this simplifies to \(\frac{1 + 2\sin\theta + 1}{\cos\theta(1 + \sin\theta)} = \frac{2 + 2\sin\theta}{\cos\theta(1 + \sin\theta)} = \frac{2(1 + \sin\theta)}{\cos\theta(1 + \sin\theta)} = \frac{2}{\cos\theta} = \) RHS. (b) Using the identity from part (a) with \(\theta = 2x\), the equation becomes \(\frac{2}{\cos 2x} = 3\tan 2x\). Since \(\tan 2x = \frac{\sin 2x}{\cos 2x}\), we have \(\frac{2}{\cos 2x} = \frac{3\sin 2x}{\cos 2x}\). Since \(\cos 2x \ne 0\), we can multiply both sides by \(\cos 2x\), yielding \(2 = 3\sin 2x \implies \sin 2x = \frac{2}{3}\). Given \(0^\circ < x < 180^\circ\), we have \(0^\circ < 2x < 360^\circ\). The principal value is \(2x = \sin^{-1}(2/3) \approx 41.81^\circ\). The second value in this range is \(2x = 180^\circ - 41.81^\circ = 138.19^\circ\). Dividing by 2 gives \(x \approx 20.9^\circ\) or \(x \approx 69.1^\circ\) (to 1 d.p.).

(a) M1: Express LHS over a common denominator. A1: Correctly expand the numerator to get \(1 + 2\sin\theta + \sin^2\theta + \cos^2\theta\). M1: Use \(\sin^2\theta + \cos^2\theta = 1\) to simplify the numerator to \(2(1 + \sin\theta)\). A1: Cancel the common factor of \((1 + \sin\theta)\) in numerator and denominator to achieve the RHS. (b) M1: Formulate the equation \(\sin 2x = \frac{2}{3}\) using the identity. A1: Find \(x \approx 20.9^\circ\). A1: Find \(x \approx 69.1^\circ\) and no other values in range.

We can solve the inequality by squaring both sides, since both sides are non-negative: \((2x - 5)^2 > 9(x + 1)^2\). Expanding both sides gives \(4x^2 - 20x + 25 > 9(x^2 + 2x + 1)\), which simplifies to \(4x^2 - 20x + 25 > 9x^2 + 18x + 9\). Rearranging all terms to one side, we obtain the quadratic inequality \(5x^2 + 38x - 16 < 0\). Factorising the quadratic expression yields \((5x - 2)(x + 8) < 0\). The critical values are \(x = \frac{2}{5}\) and \(x = -8\). Since the quadratic expression must be less than zero, the solution set is \(-8 < x < \frac{2}{5}\).

M1: Attempt to square both sides and expand correctly. A1: Obtain a correct quadratic inequality, e.g., \(5x^2 + 38x - 16 < 0\). M1: Solve the quadratic inequality to find critical values. A1: State the final correct range \(-8 < x < \frac{2}{5}\).

Using the laws of logarithms, we write the left-hand side as a single logarithm: \(\ln(2y + 3) - \ln(y^2) = \ln(2)\), which gives \(\ln\left(\frac{2y + 3}{y^2}\right) = \ln(2)\). Taking exponentials of both sides, we get \(\frac{2y + 3}{y^2} = 2\), which rearranges to the quadratic equation \(2y^2 - 2y - 3 = 0\). Using the quadratic formula, we find the roots: \(y = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(2)(-3)}}{2(2)} = \frac{2 \pm \sqrt{28}}{4} = \frac{1 \pm \sqrt{7}}{2}\). Since \(\ln(y)\) is defined only for \(y > 0\), and because \(\frac{1 - \sqrt{7}}{2} < 0\), we must discard the negative root. Thus, the only valid solution is \(y = \frac{1 + \sqrt{7}}{2}\).

M1: Apply log laws to combine the terms on the left-hand side into a single logarithm. A1: Obtain the equation \(\frac{2y+3}{y^2} = 2\). M1: Solve the quadratic equation to find the roots. A1: Obtain both roots \(y = \frac{1 \pm \sqrt{7}}{2}\). A1: State that \(y = \frac{1 + \sqrt{7}}{2}\) is the only solution and explain why the negative root is rejected.

We apply the quotient rule to differentiate \(y = \frac{e^{2x}}{\cos x}\). Let \(u = e^{2x}\) and \(v = \cos x\). Then \(u' = 2e^{2x}\) and \(v' = -\sin x\). Applying the quotient rule: \(\frac{dy}{dx} = \frac{2e^{2x}\cos x - e^{2x}(-\sin x)}{\cos^2 x} = \frac{e^{2x}(2\cos x + \sin x)}{\cos^2 x}\). For a stationary point, \(\frac{dy}{dx} = 0\). Since \(e^{2x} \neq 0\) and \(\cos x \neq 0\) in the given domain, we solve \(2\cos x + \sin x = 0\). Dividing by \(\cos x\) gives \(\tan x = -2\). Since \(-\frac{\pi}{2} < x < \frac{\pi}{2}\), the exact solution is \(x = -\tan^{-1}(2)\).

M1: Apply the quotient rule or product rule to differentiate. A1: Correctly find the derivative of the numerator as \(2e^{2x}\cos x + e^{2x}\sin x\). M1: Set \(\frac{dy}{dx} = 0\) and obtain a trigonometric equation. A1: Solve to find \(\tan x = -2\). A1: State the exact value of \(x\) as \(-\tan^{-1}(2)\).

We use the identity \(2\cos^2 x = 1 + \cos(2x)\) to rewrite the integrand: \(\int_0^{\frac{\pi}{4}} (1 + \cos(2x) + e^{2x}) \, dx\). Integrating term-by-term, we get: \(\left[ x + \frac{1}{2}\sin(2x) + \frac{1}{2}e^{2x} \right]_0^{\frac{\pi}{4}}\). Substituting the upper and lower limits: \(\left( \frac{\pi}{4} + \frac{1}{2}\sin\left(\frac{\pi}{2}\right) + \frac{1}{2}e^{\frac{\pi}{2}} \right) - \left( 0 + \frac{1}{2}\sin(0) + \frac{1}{2}e^0 \right) = \left( \frac{\pi}{4} + \frac{1}{2}(1) + \frac{1}{2}e^{\frac{\pi}{2}} \right) - \left( 0 + 0 + \frac{1}{2} \right) = \frac{\pi}{4} + \frac{1}{2}e^{\frac{\pi}{2}}\).

M1: Rewrite the integrand using the correct double-angle identity. A1: Correct expression \(1 + \cos(2x) + e^{2x}\). M1: Integrate to get \(x + \frac{1}{2}\sin(2x) + \frac{1}{2}e^{2x}\). M1: Substitute limits correctly. A1: Obtain the exact value \%\frac{\pi}{4} + \frac{1}{2}e^{\frac{\pi}{2}}\%.

(i) Let \(f(x) = x^3 + 2x - 7\). Evaluating at the boundaries: \(f(1.5) = (1.5)^3 + 2(1.5) - 7 = 3.375 + 3 - 7 = -0.625 < 0\), and \(f(2.0) = (2.0)^3 + 2(2.0) - 7 = 8 + 4 - 7 = 5 > 0\). Since there is a change of sign over the interval, a real root \(\alpha\) exists between \(1.5\) and \(2.0\).

(ii) Using the iterative formula with \(x_1 = 1.6\):
\(x_2 = (7 - 2(1.6))^{\frac{1}{3}} = (3.8)^{\frac{1}{3}} \approx 1.56049\)
\(x_3 = (7 - 2(1.56049))^{\frac{1}{3}} \approx 1.57127\)
\(x_4 = (7 - 2(1.57127))^{\frac{1}{3}} \approx 1.56834\)
\(x_5 = (7 - 2(1.56834))^{\frac{1}{3}} \approx 1.56914\)
\(x_6 = (7 - 2(1.56914))^{\frac{1}{3}} \approx 1.56892\)
\(x_7 = (7 - 2(1.56892))^{\frac{1}{3}} \approx 1.56898\)
These values converge to \(1.569\) when rounded to \(3\) decimal places. Therefore, \(\alpha \approx 1.569\).

M1: Evaluate the function at the two endpoints. A1: State the correct values and conclude there is a sign change. M1: Correctly calculate the value of \(x_2\). A1: Calculate subsequent iterations up to \(x_6\) or \(x_7\) correct to 5 decimal places. A1: State the final converged root as \(1.569\).

Using the double-angle identity \(\sin(2\theta) = 2\sin\theta\cos\theta\), we rewrite the equation: \(3(2\sin\theta\cos\theta) = 2\cos\theta \implies 6\sin\theta\cos\theta - 2\cos\theta = 0\). Factorising this expression gives: \(2\cos\theta(3\sin\theta - 1) = 0\). This yields two cases:

1) \(\cos\theta = 0 \implies \theta = 90^\circ, 270^\circ\).

2) \(3\sin\theta - 1 = 0 \implies \sin\theta = \frac{1}{3}\). The principal value is \(\theta = \sin^{-1}(1/3) \approx 19.47^\circ\). The second quadrant solution is \(180^\circ - 19.47^\circ \approx 160.53^\circ\). Hence, \(\theta \approx 19.5^\circ, 160.5^\circ\).

Combining all solutions in order gives \(\theta = 19.5^\circ, 90.0^\circ, 160.5^\circ, 270.0^\circ\).

M1: Apply the double-angle formula for \(\sin(2\theta)\). A1: Correctly factorise the equation to get \(2\cos\theta(3\sin\theta-1)=0\). M1: Solve \(\cos\theta = 0\) in the interval. A1: State \(\theta = 90^\circ, 270^\circ\). M1: Solve \(\sin\theta = 1/3\) in the interval. A1: State \(\theta = 19.5^\circ\) and \(160.5^\circ\).

(i) Since \((x-2)\) is a factor, \(P(2) = 0\). Substituting \(x=2\): \(2(8) + 4a + 2b - 6 = 0 \implies 4a + 2b = -10 \implies 2a + b = -5\). Since dividing by \(x+1\) gives remainder \(-9\), \(P(-1) = -9\). Substituting \(x=-1\): \(2(-1) + a - b - 6 = -9 \implies a - b = -1\). Adding the two equations: \(3a = -6 \implies a = -2\). From \(a - b = -1\), we get \(-2 - b = -1 \implies b = -1\).

(ii) Substituting the values of \(a\) and \(b\) gives \(P(x) = 2x^3 - 2x^2 - x - 6\). Since \((x-2)\) is a factor, we can divide \(P(x)\) by \((x-2)\): \(2x^3 - 2x^2 - x - 6 = (x-2)(2x^2 + 2x + 3)\). We check if the quadratic factor \(2x^2 + 2x + 3\) can be factorised. Its discriminant is \(\Delta = 2^2 - 4(2)(3) = 4 - 24 = -20 < 0\). Since the discriminant is negative, there are no further real linear factors. Thus, the completely factorised form of \(P(x)\) is \((x-2)(2x^2 + 2x + 3)\).

M1: Set up a linear equation using \(P(2) = 0\). M1: Set up a linear equation using \(P(-1) = -9\). A1: Solve the system to get \(a = -2\) and \(b = -1\). M1: Attempt polynomial division or coefficient matching. A1: Obtain the quadratic factor \(2x^2 + 2x + 3\). A1: Verify that the quadratic has no real roots and state the final factorised form.

First, find the y-coordinate at \(x = 0\). Substituting \(x = 0\) gives \(y e^{0} - 0^2 + 3y^2 = 4\), which simplifies to \(3y^2 + y - 4 = 0\). Factorising this quadratic equation, we get \((3y + 4)(y - 1) = 0\). Since \(y > 0\), we must have \(y = 1\). Thus, the point of interest is \((0, 1)\). Next, we differentiate the curve equation implicitly with respect to \(x\): \(e^{2x} \frac{dy}{dx} + 2y e^{2x} - 2x + 6y \frac{dy}{dx} = 0\). Substituting the coordinates \(x = 0\) and \(y = 1\) into this derivative equation gives: \((1)\frac{dy}{dx} + 2(1)(1) - 0 + 6(1)\frac{dy}{dx} = 0\), which simplifies to \(7 \frac{dy}{dx} + 2 = 0\). This gives the gradient of the tangent as \(\frac{dy}{dx} = -\frac{2}{7}\). Finally, using the point-slope formula for the tangent line: \(y - 1 = -\frac{2}{7}(x - 0)\), which simplifies to \(2x + 7y - 7 = 0\).

M1: Substitute \(x = 0\) and solve the quadratic to find \(y = 1\) (ignore negative root). M1: Apply product rule to differentiate \(y e^{2x}\). A1: Correct implicit differentiation of the entire equation. M1: Substitute \(x = 0\) and \(y = 1\) to find the gradient \(\frac{dy}{dx} = -\frac{2}{7}\). M1: Form the equation of the tangent using their gradient and point. A1: Correct final equation in the form \(2x + 7y - 7 = 0\) or any equivalent integer-coefficient form.

We use integration by parts, \(\int u \, dv = uv - \int v \, du\). Let \(u = x\) and \(dv = e^{-2x} \, dx\). This gives \(du = dx\) and \(v = -\frac{1}{2} e^{-2x}\). Applying the formula, we obtain: \(\int x e^{-2x} \, dx = -\frac{1}{2} x e^{-2x} - \int -\frac{1}{2} e^{-2x} \, dx = -\frac{1}{2} x e^{-2x} - \frac{1}{4} e^{-2x}\). Now, we evaluate this expression between the limits \(0\) and \(\ln 2\). For the upper limit \(x = \ln 2\): \(-\frac{1}{2}(\ln 2)e^{-2\ln 2} - \frac{1}{4}e^{-2\ln 2} = -\frac{1}{2}(\ln 2)\left(\frac{1}{4}\right) - \frac{1}{4}\left(\frac{1}{4}\right) = -\frac{1}{8}\ln 2 - \frac{1}{16}\). For the lower limit \(x = 0\): \(-\frac{1}{2}(0)e^{0} - \frac{1}{4}e^{0} = -\frac{1}{4}\). Subtracting the lower limit value from the upper limit value gives: \(\left(-\frac{1}{8}\ln 2 - \frac{1}{16}\right) - \left(-\frac{1}{4}\right) = \frac{3}{16} - \frac{1}{8}\ln 2\).

M1: State a correct choice of \(u\) and \(dv\) and find \(du\) and \(v\). A1: Correctly apply the integration by parts formula to get \(-\frac{1}{2} x e^{-2x} + \frac{1}{2} \int e^{-2x} \, dx\). A1: Complete integration to get \(-\frac{1}{2} x e^{-2x} - \frac{1}{4} e^{-2x}\). M1: Substitute limits \(\ln 2\) and \(0\) correctly, using \(e^{-2\ln 2} = \frac{1}{4}\). A1: Obtain correct upper limit value or lower limit value. A1: Obtain the final exact answer \(\frac{3}{16} - \frac{1}{8}\ln 2\).

To express \(w\) in the standard form \(x + \mathrm{i}y\), multiply the numerator and the denominator by the complex conjugate of the denominator, which is \(2 + \mathrm{i}\): \(w = \frac{(1 + 7\mathrm{i})(2 + \mathrm{i})}{(2 - \mathrm{i})(2 + \mathrm{i})} = \frac{2 + \mathrm{i} + 14\mathrm{i} - 7}{4 + 1} = \frac{-5 + 15\mathrm{i}}{5} = -1 + 3\mathrm{i}\). Now, find the modulus of \(w\): \(|w| = \sqrt{(-1)^2 + 3^2} = \sqrt{10}\). Next, calculate the argument of \(w\). Since \(w = -1 + 3\mathrm{i}\) lies in the second quadrant: \(\arg(w) = \pi - \arctan(3) \approx 3.14159 - 1.24905 = 1.89254\) radians. Rounded to 3 significant figures, the argument is \(1.89\) radians.

M1: Multiply numerator and denominator by \(2 + \mathrm{i}\) to realise the denominator. A1: Obtain \(w = -1 + 3\mathrm{i}\). M1: Calculate the modulus using \(\sqrt{x^2 + y^2}\). A1: Obtain exact modulus \(\sqrt{10}\). M1: Use correct quadrant-specific trigonometry to find the argument, \(\pi - \arctan(3)\). A1: Obtain argument \(1.89\) radians.

The direction vectors of the lines are \(\mathbf{d}_1 = \mathbf{i} + 2\mathbf{j} - 2\mathbf{k}\) and \(\mathbf{d}_2 = 2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}\). First, calculate the dot product of the direction vectors: \(\mathbf{d}_1 \cdot \mathbf{d}_2 = (1)(2) + (2)(-3) + (-2)(6) = 2 - 6 - 12 = -16\). Next, calculate the magnitudes of each vector: \(|\mathbf{d}_1| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{9} = 3\) and \(|\mathbf{d}_2| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{49} = 7\). Let \(\theta\) be the acute angle between the two directions. Then: \(\cos \theta = \frac{|\mathbf{d}_1 \cdot \mathbf{d}_2|}{|\mathbf{d}_1||\mathbf{d}_2|} = \frac{|-16|}{3 \times 7} = \frac{16}{21}\). Finally, solve for the acute angle: \(\theta = \arccos\left(\frac{16}{21}\right) \approx 40.367^\circ\). Correct to 1 decimal place, \(\theta = 40.4^\circ\).

M1: Identify the correct direction vectors \(\mathbf{d}_1\) and \(\mathbf{d}_2\). M1: Calculate the scalar product of their direction vectors. M1: Correctly calculate the magnitude of both direction vectors. M1: Apply the cosine angle formula. A1: Obtain \(\cos \theta = \frac{16}{21}\) or equivalent. A1: Obtain \(40.4^\circ\) (allow answers rounding to 40.4).

Separate the variables to set up the integration: \(\int \frac{1}{y^2 + 1} \, dy = \int \frac{1}{x^2} \, dx\). Integrating both sides gives: \(\arctan(y) = -\frac{1}{x} + C\), where \(C\) is the constant of integration. Apply the initial condition \(y = 0\) when \(x = 1\): \(\arctan(0) = -\frac{1}{1} + C \implies 0 = -1 + C \implies C = 1\). Thus, the particular solution satisfies: \(\arctan(y) = 1 - \frac{1}{x}\). Taking the tangent of both sides to express \(y\) in terms of \(x\) yields: \(y = \tan\left(1 - \frac{1}{x}\right)\).

M1: Separate variables and set up both integrals correctly. A1: Correctly integrate LHS to obtain \(\arctan(y)\). A1: Correctly integrate RHS to obtain \(-\frac{1}{x}\). M1: Substitute \(x = 1\) and \(y = 0\) into an integrated equation with a constant of integration. A1: Find \(C = 1\). A1: Rearrange and express the final answer explicitly as \(y = \tan\left(1 - \frac{1}{x}\right)\).

First, differentiate \(x\) and \(y\) with respect to \(t\): \(\frac{dx}{dt} = 3 - 3\cos(3t)\) and \(\frac{dy}{dt} = 3\sin(3t)\). Using the parametric differentiation chain rule, we find \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3\sin(3t)}{3(1 - \cos(3t))} = \frac{\sin(3t)}{1 - \cos(3t)}\). Now, substitute \(t = \frac{\pi}{9}\) into this expression: The angle term is \(3t = 3\left(\frac{\pi}{9}\right) = \frac{\pi}{3}\). Evaluating the trigonometric ratios: \(\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\) and \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\). Thus, the gradient is: \(\frac{dy}{dx} = \frac{\sqrt{3}/2}{1 - 1/2} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}\).

M1: Differentiate \(x\) with respect to \(t\). M1: Differentiate \(y\) with respect to \(t\). A1: Correct derivatives for both, showing chain rule was used. M1: Use \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\). A1: Correct simplified derivative in terms of \(t\). M1: Substitute \(t = \frac{\pi}{9}\) and use exact trig values. A1: Obtain the exact value \(\sqrt{3}\).

Let \(\frac{3x + 5}{(x + 1)(2x + 1)} = \frac{A}{x + 1} + \frac{B}{2x + 1}\). Thus: \(3x + 5 = A(2x + 1) + B(x + 1)\). Set \(x = -1\) to find \(A\): \(3(-1) + 5 = A(2(-1) + 1) \implies 2 = -A \implies A = -2\). Set \(x = -\frac{1}{2}\) to find \(B\): \(3\left(-\frac{1}{2}\right) + 5 = B\left(-\frac{1}{2} + 1\right) \implies \frac{7}{2} = \frac{1}{2}B \implies B = 7\). Therefore, \(\frac{3x + 5}{(x + 1)(2x + 1)} = \frac{7}{2x + 1} - \frac{2}{x + 1}\). Now, integrate this from \(0\) to \(2\): \(\int_{0}^{2} \left( \frac{7}{2x + 1} - \frac{2}{x + 1} \right) \, dx = \left[ \frac{7}{2} \ln|2x + 1| - 2 \ln|x + 1| \right]_{0}^{2}\). Substitute the upper limit \(x = 2\): \(\left(\frac{7}{2} \ln(5) - 2 \ln(3)\right)\). Substitute the lower limit \(x = 0\): \(\left(\frac{7}{2} \ln(1) - 2 \ln(1)\right) = 0\). Subtracting the lower limit from the upper limit, the exact value is: \(\frac{7}{2} \ln(5) - 2 \ln(3)\).

M1: Formulate partial fractions equation and attempt to solve for \(A\) and \(B\). A1: Correctly find \(A = -2\) and \(B = 7\). M1: Integrate partial fractions to obtain log forms. A1: Correct integration with correct coefficients: \(\frac{7}{2}\ln|2x+1| - 2\ln|x+1|\). M1: Substitute limits \(2\) and \(0\) into their integrated expression. A1: Obtain the exact final answer \(\frac{7}{2}\ln 5 - 2\ln 3\) (or any single logarithm equivalent).

Let \(z = x + \mathrm{i}y\). Substituting this into the first equation, \(|x + \mathrm{i}(y - 2)| = |(x - 4) + \mathrm{i}y|\). Squaring both sides: \(x^2 + (y - 2)^2 = (x - 4)^2 + y^2 \implies x^2 + y^2 - 4y + 4 = x^2 - 8x + 16 + y^2\). Simplifying this yields: \(8x - 4y = 12 \implies y = 2x - 3\). The second equation is \(\arg(z - 2) = \frac{3\pi}{4}\). Since \(z - 2 = (x - 2) + \mathrm{i}y\), this represents a ray starting from the point \(2\) on the real axis with an angle of \(135^\circ\). This requires: \(y = -(x - 2) = 2 - x\), with the constraints \(x - 2 < 0 \implies x < 2\) and \(y > 0\). Now solve the system of equations for \(x\) and \(y\): \(2x - 3 = 2 - x \implies 3x = 5 \implies x = \frac{5}{3}\). Since \(x = \frac{5}{3} < 2\), this is valid. Substitute \(x = \frac{5}{3}\) back into \(y = 2 - x\): \(y = 2 - \frac{5}{3} = \frac{1}{3}\). Since \(y = \frac{1}{3} > 0\), this is also valid. Therefore, the complex number is \(z = \frac{5}{3} + \frac{1}{3}\mathrm{i}\).

M1: Substitute \(z = x + \mathrm{i}y\) into the modulus equation and expand. A1: Correct linear equation \(y = 2x - 3\) (or equivalent). M1: Interpret the argument equation as a straight line. A1: Correct equation \(y = 2 - x\) with the half-line constraints. M1: Solve the simultaneous equations for \(x\) and \(y\). A1: Obtain \(z = \frac{5}{3} + \frac{1}{3}\mathrm{i}\).

To integrate \(\int_{0}^{\frac{1}{2}\pi} x \cos^2 x \,\mathrm{d}x\), we first use the trigonometric double-angle identity:

\[\cos^2 x = \frac{1}{2}(1 + \cos 2x)\]

Substituting this into the integral, we get:

\[\int_{0}^{\frac{1}{2}\pi} x \cos^2 x \,\mathrm{d}x = \frac{1}{2} \int_{0}^{\frac{1}{2}\pi} (x + x \cos 2x) \,\mathrm{d}x\]

We can split this into two separate integrals:

\[\frac{1}{2} \int_{0}^{\frac{1}{2}\pi} x \,\mathrm{d}x + \frac{1}{2} \int_{0}^{\frac{1}{2}\pi} x \cos 2x \,\mathrm{d}x\]

First, evaluate \(\frac{1}{2} \int_{0}^{\frac{1}{2}\pi} x \,\mathrm{d}x\):

\[\frac{1}{2} \left[ \frac{1}{2}x^2 \right]_{0}^{\frac{1}{2}\pi} = \frac{1}{4} \left( \left(\frac{\pi}{2}\right)^2 - 0 \right) = \frac{\pi^2}{16}\]

Second, evaluate \(\int x \cos 2x \,\mathrm{d}x\) using integration by parts with \(u = x\) and \(\frac{\mathrm{d}v}{\mathrm{d}x} = \cos 2x\):

\[\frac{\mathrm{d}u}{\mathrm{d}x} = 1 \quad \text{and} \quad v = \frac{1}{2}\sin 2x\]

Using the integration by parts formula \(\int u \frac{\mathrm{d}v}{\mathrm{d}x} \,\mathrm{d}x = uv - \int v \frac{\mathrm{d}u}{\mathrm{d}x} \,\mathrm{d}x\):

\[\int x \cos 2x \,\mathrm{d}x = \frac{1}{2}x \sin 2x - \int \frac{1}{2}\sin 2x \,\mathrm{d}x = \frac{1}{2}x \sin 2x + \frac{1}{4}\cos 2x\]

Now, apply the limits from \(0\) to \(\frac{1}{2}\pi\):

\[\left[ \frac{1}{2}x \sin 2x + \frac{1}{4}\cos 2x \right]_{0}^{\frac{1}{2}\pi} = \left( \frac{1}{2}\left(\frac{\pi}{2}\right)\sin\pi + \frac{1}{4}\cos\pi \right) - \left( 0 + \frac{1}{4}\cos 0 \right)\]

Since \(\sin\pi = 0\), \(\cos\pi = -1\), and \(\cos 0 = 1\):

\[= \left( 0 - \frac{1}{4} \right) - \left( \frac{1}{4} \right) = -\frac{1}{2}\]

We multiply this part by the front coefficient \(\frac{1}{2}\):

\[\frac{1}{2} \left( -\frac{1}{2} \right) = -\frac{1}{4}\]

Combining both parts, we obtain:

\[\frac{\pi^2}{16} - \frac{1}{4}\]

M1: Use the correct double-angle identity to express \(\cos^2 x\) in terms of \(\cos 2x\).
A1: Obtain the correct integrand \(\frac{1}{2}(x + x \cos 2x)\).
M1: Apply integration by parts correctly to \(\int x \cos 2x \,\mathrm{d}x\) to obtain the form \(ax \sin 2x + b \cos 2x\).
A1: Obtain the correct integrated expression \(\frac{1}{2}x \sin 2x + \frac{1}{4}\cos 2x\).
M1: Integrate \(x\) to \(\frac{1}{2}x^2\) and correctly substitute limits \(0\) and \(\frac{1}{2}\pi\) into their integrated terms.
A1: Obtain the final exact answer \(\frac{\pi^2}{16} - \frac{1}{4}\) (or equivalent single fraction \(\frac{\pi^2 - 4}{16}\)).

(a)
To find the modulus of \(w = -2 + 2\sqrt{3}\mathrm{i}\):

\[|w| = \sqrt{(-2)^2 + (2\sqrt{3})^2} = \sqrt{4 + 12} = \sqrt{16} = 4\]

To find the argument of \(w\), we identify that \(w\) lies in the second quadrant because the real part is negative and the imaginary part is positive:

\[\arg(w) = \pi - \arctan\left(\frac{2\sqrt{3}}{2}\right) = \pi - \arctan(\sqrt{3}) = \pi - \frac{\pi}{3} = \frac{2}{3}\pi\]

(b)
We represent \(w\) in polar form:

\[w = 4 \mathrm{e}^{\mathrm{i}\frac{2}{3}\pi}\]

Let \(z = r\mathrm{e}^{\mathrm{i}\theta}\) be a square root of \(w\), so \(z^2 = w\):

\[r^2 \mathrm{e}^{\mathrm{i}2\theta} = 4 \mathrm{e}^{\mathrm{i}\frac{2}{3}\pi}\]

Taking the square root of the modulus gives \(r = 2\).

For the argument:

\[2\theta = \frac{2}{3}\pi + 2k\pi \implies \theta = \frac{1}{3}\pi \quad (\text{for } k=0)\]

\[\theta = -\frac{2}{3}\pi \quad (\text{for } k=-1)\]

Now we convert the polar roots back to Cartesian form \(a + \mathrm{i}b\):

For \(\theta = \frac{1}{3}\pi\):

\[z_1 = 2\left(\cos\frac{\pi}{3} + \mathrm{i}\sin\frac{\pi}{3}\right) = 2\left(\frac{1}{2} + \mathrm{i}\frac{\sqrt{3}}{2}\right) = 1 + \sqrt{3}\mathrm{i}\]

For \(\theta = -\frac{2}{3}\pi\):

\[z_2 = 2\left(\cos\left(-\frac{2}{3}\pi\right) + \mathrm{i}\sin\left(-\frac{2}{3}\pi\right)\right) = 2\left(-\frac{1}{2} - \mathrm{i}\frac{\sqrt{3}}{2}\right) = -1 - \sqrt{3}\mathrm{i}\]

Thus, the two square roots are \(1 + \sqrt{3}\mathrm{i}\) and \(-1 - \sqrt{3}\mathrm{i}\).

Part (a):
M1: For a correct method to find the modulus or argument (e.g., using arctan).
A1: Obtain modulus 4 and argument \(\frac{2}{3}\pi\) (or \(120^\circ\) or equivalent radian form \(2.09\)).

Part (b):
M1: Set up a method to solve for square roots (either using polar representation \(r^2=4\) and \(2\theta = \arg(w)\) or algebraic representation \((a+\mathrm{i}b)^2 = -2+2\sqrt{3}\mathrm{i}\)).
A1: Identify the correct modulus of the roots is \(2\) and the arguments are \(\frac{1}{3}\pi\) and \(-\frac{2}{3}\pi\) (or obtain \(a^2=1\) and \(b^2=3\)).
A1: State the first square root \(1 + \sqrt{3}\mathrm{i}\) (or exact equivalent).
A1: State the second square root \(-1 - \sqrt{3}\mathrm{i}\) (or exact equivalent).

(a)
If the lines \(l_1\) and \(l_2\) intersect, there must exist values of \(\lambda\) and \(\mu\) that satisfy the vector equation:

\[\begin{pmatrix} 3 + \lambda \\ 3 - 2\lambda \\ -2 + 3\lambda \end{pmatrix} = \begin{pmatrix} 3 + 2\mu \\ a + \mu \\ 5 - \mu \end{pmatrix}\]

Equating the \(x\) and \(z\) components:

\[3 + \lambda = 3 + 2\mu \implies \lambda = 2\mu \quad (1)\]

\[-2 + 3\lambda = 5 - \mu \quad (2)\]

Substitute equation (1) into (2):

\[-2 + 3(2\mu) = 5 - \mu \implies -2 + 6\mu = 5 - \mu \implies 7\mu = 7 \implies \mu = 1\]

Using equation (1), \(\lambda = 2(1) = 2\).

Now, substitute \(\lambda = 2\) and \(\mu = 1\) into the equation for the \(y\) components:

\[3 - 2\lambda = a + \mu \implies 3 - 2(2) = a + 1 \implies -1 = a + 1 \implies a = -2\]

To find the position vector of the point of intersection, substitute \(\lambda = 2\) into the equation of \(l_1\):

\[\mathbf{r} = \begin{pmatrix} 3 + 2 \\ 3 - 4 \\ -2 + 6 \end{pmatrix} = \begin{pmatrix} 5 \\ -1 \\ 4
\end{pmatrix}\]

Thus, the position vector of the point of intersection is \(5\mathbf{i} - \mathbf{j} + 4\mathbf{k}\).

(b)
The direction vectors of the lines are:

\[\mathbf{d}_1 = \mathbf{i} - 2\mathbf{j} + 3\mathbf{k} \quad \text{and} \quad \mathbf{d}_2 = 2\mathbf{i} + \mathbf{j} - \mathbf{k}\]

Let \(\theta\) be the angle between the lines. We find the scalar product and magnitudes:

\[\mathbf{d}_1 \cdot \mathbf{d}_2 = (1)(2) + (-2)(1) + (3)(-1) = 2 - 2 - 3 = -3\]

\[|\mathbf{d}_1| = \sqrt{1^2 + (-2)^2 + 3^2} = \sqrt{14}\]

\[|\mathbf{d}_2| = \sqrt{2^2 + 1^2 + (-1)^2} = \sqrt{6}\]

Using the cosine formula for the acute angle:

\[\cos\theta = \frac{|\mathbf{d}_1 \cdot \mathbf{d}_2|}{|\mathbf{d}_1||\mathbf{d}_2|} = \frac{|-3|}{\sqrt{14}\sqrt{6}} = \frac{3}{\sqrt{84}}\]

\[\theta = \arccos\left(\frac{3}{\sqrt{84}}\right) \approx 70.893^\circ\]

Rounding to 1 decimal place, we get \(\theta = 70.9^\circ\).

Part (a):
M1: Equate components of \(l_1\) and \(l_2\) to form simultaneous equations in \(\lambda\) and \(\mu\).
A1: Solve equations to find \(\mu = 1\) and \(\lambda = 2\).
A1: Show that \(a = -2\) and obtain the correct position vector of the intersection as \(5\mathbf{i} - \mathbf{j} + 4\mathbf{k}\) (or equivalent).

Part (b):
M1: Use the scalar product formula \(\cos\theta = \frac{|\mathbf{u}\cdot\mathbf{v}|}{|\mathbf{u}||\mathbf{v}|}\) with correct direction vectors.
A1: Calculate correct scalar product (\(-3\) or \(3\)) and correct magnitudes (\(\sqrt{14}\) and \(\sqrt{6}\)).
A1: Obtain correct final acute angle of \(70.9^\circ\) (accept \(1.24\) radians).

At time \(t\), the velocity \(v\) is found by integrating the acceleration:

\(v(t) = \int a \, dt = \int (6 - 2t) \, dt = 6t - t^2 + C\)

Since \(v = 7\) when \(t = 0\), we have:

\(7 = 6(0) - (0)^2 + C \implies C = 7\)

Thus, \(v(t) = 6t - t^2 + 7\).

(i) The particle is at rest when \(v = 0\):

\(6t - t^2 + 7 = 0 \implies t^2 - 6t - 7 = 0 \implies (t - 7)(t + 1) = 0\)

Since \(t \ge 0\), we choose \(t = 7\text{ s}\).

(ii) The velocity is maximum when the acceleration \(a = 0\):

\(6 - 2t = 0 \implies t = 3\text{ s}\)

To find the displacement, we integrate the velocity function:

\(s(t) = \int v(t) \, dt = \int (6t - t^2 + 7) \, dt = 3t^2 - \frac{1}{3}t^3 + 7t + D\)

Since \(s(0) = 0\) (displacement is measured from \(O\)), we have \(D = 0\).

Substituting \(t = 3\):

\(s(3) = 3(3)^2 - \frac{1}{3}(3)^3 + 7(3) = 27 - 9 + 21 = 39\text{ m}\)

(i)
M1: Integrate \(a\) to find \(v\) and attempt to find \(C\).
A1: Obtain correct velocity function \(v(t) = 6t - t^2 + 7\).
M1: Set \(v = 0\) and solve the quadratic equation for \(t\).
A1: Correctly state \(t = 7\text{ s}\) (rejecting the negative root).

(ii)
M1: Set \(a = 0\) to find the time \(t = 3\) at maximum velocity, and integrate \(v(t)\) to find displacement.
A1: Correct integration with constant of integration \(D = 0\).
A1: Obtain displacement of \(39\text{ m}\).

Let \(T\) be the tension in the string and \(a\) be the acceleration of the system. Since \(B\) is heavier and hanging vertically, block \(A\) moves up the inclined plane while \(B\) moves downwards.

For block \(A\) on the inclined plane:
- The normal contact force is \(R = m_A g \cos\alpha = 3(10)(0.8) = 24\text{ N}\).
- The frictional force is \(F = \mu R = 0.25 \times 24 = 6\text{ N}\).
- The component of weight acting down the incline is \(m_A g \sin\alpha = 3(10)(0.6) = 18\text{ N}\).

The equation of motion for block \(A\) up the plane is:
\(T - m_A g \sin\alpha - F = m_A a\)
\(T - 18 - 6 = 3a \implies T - 24 = 3a\) [Equation 1]

For particle \(B\) hanging vertically:
- The downward force is its weight \(m_B g = 5(10) = 50\text{ N}\).

The equation of motion for particle \(B\) is:
\(m_B g - T = m_B a\)
\(50 - T = 5a\) [Equation 2]

Adding Equation 1 and Equation 2:
\((T - 24) + (50 - T) = 3a + 5a\)
\(26 = 8a \implies a = 3.25\text{ m s}^{-2}\)

Substituting \(a = 3.25\) back into Equation 2:
\(T = 50 - 5(3.25) = 50 - 16.25 = 33.75\text{ N}\)

M1: Resolve forces perpendicular to the plane to find \(R = 24\text{ N}\) and use \(F = \mu R\) to obtain \(F = 6\text{ N}\).
M1: Apply Newton's second law to block \(A\) along the plane, accounting for tension, gravity, and friction.
A1: Correct equation for \(A\): \(T - 24 = 3a\).
M1: Apply Newton's second law to particle \(B\).
A1: Correct equation for \(B\): \(50 - T = 5a\).
M1: Solve the simultaneous equations to find acceleration and tension.
A1: Correct acceleration \(a = 3.25\text{ m s}^{-2}\) and tension \(T = 33.75\text{ N}\).

(i) Let the direction of \(P\)'s initial motion be the positive direction.
Initially, \(u_P = 8\text{ m s}^{-1}\) and \(u_Q = -4\text{ m s}^{-1}\) (since they move towards each other).
After the collision, \(P\)'s velocity is \(v_P = -2\text{ m s}^{-1}\).

Using the conservation of linear momentum:
\(m_P u_P + m_Q u_Q = m_P v_P + m_Q v_Q\)
\(0.4(8) + 0.6(-4) = 0.4(-2) + 0.6 v_Q\)
\(3.2 - 2.4 = -0.8 + 0.6 v_Q\)
\(0.8 = -0.8 + 0.6 v_Q\)
\(1.6 = 0.6 v_Q \implies v_Q = \frac{1.6}{0.6} = \frac{8}{3} \approx 2.67\text{ m s}^{-1}\)

Since \(v_Q\) is positive, \(Q\) now moves in the direction of \(P\)'s initial motion, meaning \(Q\)'s direction of motion is reversed.

(ii) The impulse \(I\) exerted by \(P\) on \(Q\) is equal to the change in momentum of \(Q\):
\(I = m_Q(v_Q - u_Q) = 0.6 \left(\frac{8}{3} - (-4)\right) = 0.6 \left(\frac{20}{3}\right) = 4\text{ N s}\)

Alternatively, using \(P\)'s change in momentum:
\(|I| = |m_P(v_P - u_P)| = |0.4(-2 - 8)| = |-4| = 4\text{ N s}\)

(i)
M1: Set up the equation for the conservation of linear momentum, using opposite signs for the initial velocities.
A1: Substitute correct values into the momentum equation.
M1: Solve for the final velocity of \(Q\).
A1: Obtain speed of \(2.67\text{ m s}^{-1}\) (or \(\frac{8}{3}\)) and clearly state that the direction of motion is reversed.

(ii)
M1: Apply the formula for impulse, \(I = m\Delta v\), to either particle.
A1: Calculate the change in momentum correctly.
A1: Correctly state that the magnitude of the impulse is \(4\text{ N s}\).

The forces acting on the block are:
- Weight: \(W = mg = 12 \times 10 = 120\text{ N}\).
- Normal reaction force perpendicular to the plane: \(R = mg \cos 30^\circ = 120 \cos 30^\circ = 60\sqrt{3}\text{ N}\).
- Friction force acting along the plane: \(F\), where \(F \le F_{\text{max}} = \mu R = 0.4 \times 60\sqrt{3} = 24\sqrt{3}\text{ N}\).
- The applied force \(P\) acting up the plane.

We consider the two extreme cases of limiting equilibrium:

Case 1: The block is on the point of slipping down the plane. In this case, friction \(F\) acts up the plane to oppose motion.
Resolving forces parallel to the plane:
\(P_{\text{min}} + F_{\text{max}} = mg \sin 30^\circ\)
\(P_{\text{min}} + 24\sqrt{3} = 120(0.5) = 60\)
\(P_{\text{min}} = 60 - 24\sqrt{3} \approx 18.43 \approx 18.4\text{ N}\)

Case 2: The block is on the point of slipping up the plane. In this case, friction \(F\) acts down the plane.
Resolving forces parallel to the plane:
\(P_{\text{max}} = mg \sin 30^\circ + F_{\text{max}}\)
\(P_{\text{max}} = 60 + 24\sqrt{3} \approx 101.57 \approx 102\text{ N}\)

Therefore, for equilibrium to be maintained, \(P\) must satisfy:
\(60 - 24\sqrt{3} \le P \le 60 + 24\sqrt{3}\)
Which is \(18.4 \le P \le 102\) (to 3 s.f.).

M1: Resolve forces perpendicular to the plane to find \(R = 120 \cos 30^\circ\).
A1: Correctly calculate \(R = 60\sqrt{3}\text{ N}\) and find \(F_{\text{max}} = 24\sqrt{3}\text{ N}\).
M1: Consider the case where the block is on the point of sliding down, setting up \(P + F = mg\sin 30^\circ\).
A1: Obtain \(P_{\text{min}} = 60 - 24\sqrt{3} \approx 18.4\text{ N}\).
M1: Consider the case where the block is on the point of sliding up, setting up \(P = mg\sin 30^\circ + F\).
A1: Obtain \(P_{\text{max}} = 60 + 24\sqrt{3} \approx 102\text{ N}\).
A1: Express the final range of values correctly as \(18.4 \le P \le 102\).

(i) The engine power is \(P = 24000\text{ W}\).
At a speed of \(v = 15\text{ m s}^{-1}\), the driving force \(D\) of the car is:
\(D = \frac{P}{v} = \frac{24000}{15} = 1600\text{ N}\)

The forces opposing the motion are:
- Resistance force \(R = 350\text{ N}\)
- Component of gravity down the incline: \(mg \sin\theta = 1200 \times 10 \times 0.05 = 600\text{ N}\)

Applying Newton's second law along the incline:
\(D - mg \sin\theta - R = m a\)
\(1600 - 600 - 350 = 1200 a\)
\(650 = 1200 a \implies a = \frac{650}{1200} = \frac{13}{24} \approx 0.542\text{ m s}^{-2}\)

(ii) The maximum speed is achieved when the acceleration is zero, meaning the driving force \(D_{\text{max}}\) balances the opposing forces:
\(D_{\text{max}} = mg \sin\theta + R = 600 + 350 = 950\text{ N}\)

Using the relationship \(P = D_{\text{max}} v_{\text{max}}\):
\(v_{\text{max}} = \frac{P}{D_{\text{max}}} = \frac{24000}{950} = \frac{480}{19} \approx 25.3\text{ m s}^{-1}\)

(i)
M1: Use \(P = D v\) to find the driving force \(D = 1600\text{ N}\).
M1: Formulate Newton's second law, accounting for the driving force, gravity component down the incline, and the resistance force.
A1: Correct gravity component \(600\text{ N}\).
A1: Obtain acceleration \(a = 0.542\text{ m s}^{-2}\) (or \(\frac{13}{24}\)).

(ii)
M1: Recognize that max speed occurs when \(a = 0\) and hence the driving force equals opposing forces.
A1: Correctly calculate the total opposing force \(950\text{ N}\).
A1: Calculate maximum speed \(v_{\text{max}} \approx 25.3\text{ m s}^{-1}\) (or \(\frac{480}{19}\)).

(i) During the first stage of constant acceleration:
\(v = u + a t_1 = 0 + 0.8(10) = 8\text{ m s}^{-1}\)
This is the maximum speed of the cyclist.

(ii) Let us find the distance travelled in each of the three stages:
- Stage 1 (Acceleration):
\(s_1 = \frac{1}{2} a t_1^2 = \frac{1}{2} (0.8)(10)^2 = 40\text{ m}\)
- Stage 2 (Constant Speed):
The cyclist travels at \(8\text{ m s}^{-1}\) for \(T\text{ s}\).
\(s_2 = 8T\)
- Stage 3 (Deceleration):
The cyclist decelerates from \(8\text{ m s}^{-1}\) to rest.
The duration of this stage is \(t_3 = 50 - 10 - T = 40 - T\text{ s}\).
The distance travelled during deceleration is:
\(s_3 = \frac{u_3 + v_3}{2} t_3 = \frac{8 + 0}{2} (40 - T) = 4(40 - T) = 160 - 4T\)

The total distance is the sum of the distances of the three stages:
\(s_{\text{total}} = s_1 + s_2 + s_3 = 320\)
\(40 + 8T + 160 - 4T = 320\)
\(200 + 4T = 320\)
\(4T = 120 \implies T = 30\)

(iii) The duration of the final stage is:
\(t_3 = 40 - T = 40 - 30 = 10\text{ s}\)

The deceleration \(d\) is given by:
\(d = \frac{u_3 - v_3}{t_3} = \frac{8 - 0}{10} = 0.8\text{ m s}^{-2}\)

(i)
M1: Use \(v = u + at\) to find maximum speed.
A1: Correctly state max speed as \(8\text{ m s}^{-1}\).

(ii)
M1: Calculate the distance for Stage 1, \(s_1 = 40\text{ m}\).
M1: Set up expressions for distances in Stage 2 (\(8T\)) and Stage 3 (\(4(40-T)\)) in terms of \(T\).
A1: Formulate the total distance equation: \(40 + 8T + 160 - 4T = 320\).
A1: Solve for \(T = 30\).

(iii)
M1: Calculate the duration of the third stage, \(t_3 = 10\text{ s}\), and use it to find acceleration/deceleration.
A1: Correct deceleration magnitude of \(0.8\text{ m s}^{-2}\).

(i) Let us find the forces acting on the particle as it moves up the plane:
- Component of gravity acting down the plane: \(W_x = mg \sin 30^\circ = 2 \times 10 \times 0.5 = 10\text{ N}\).
- Normal reaction force: \(R = mg \cos 30^\circ = 2 \times 10 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}\text{ N}\).
- Since the motion is up the plane, the friction force \(F\) acts down the plane:
\(F = \mu R = 0.3 \times 10\sqrt{3} = 3\sqrt{3}\text{ N} \approx 5.196\text{ N}\).

Using Newton's second law along the plane, the deceleration \(a\) is given by:
\(mg \sin 30^\circ + F = m a\)
\(10 + 3\sqrt{3} = 2a \implies a = 5 + 1.5\sqrt{3} \approx 7.598\text{ m s}^{-2}\)

Now, using \(v^2 = u^2 - 2as\) where \(v = 0\) and \(u = 12\):
\(0 = 12^2 - 2(7.598)s\)
\(144 = 15.196 s \implies s = \frac{144}{15.196} \approx 9.48\text{ m}\) (to 3 s.f.).

(ii) When the particle comes to rest at its highest point, the component of gravity pulling it down the plane is:
\(W_x = mg \sin 30^\circ = 10\text{ N}\)

The maximum possible friction force that can act up the plane to keep it in equilibrium is:
\(F_{\text{max}} = \mu R = 3\sqrt{3} \approx 5.20\text{ N}\)

Since \(W_x > F_{\text{max}}\) (i.e. \(10\text{ N} > 5.20\text{ N}\)), the component of gravity down the plane is greater than the limiting static friction.

Therefore, the particle cannot remain in equilibrium and will slide back down the plane.

(i)
M1: Find the normal force \(R = 10\sqrt{3}\) and apply \(F = \mu R\) to find the friction force.
A1: Correct friction force \(3\sqrt{3}\text{ N}\) (or \(5.20\text{ N}\)).
M1: Set up the equation of motion to find deceleration \(a\).
A1: Correct deceleration value \(a = 5 + 1.5\sqrt{3} \approx 7.60\text{ m s}^{-2}\).
M1: Apply \(v^2 = u^2 - 2as\) to find the stopping distance.
A1: Obtain distance \(s = 9.48\text{ m}\).

(ii)
M1: Compare the weight component down the incline with the maximum limiting friction.
A1: Conclude that the block slides down since \(10 > 5.20\) with appropriate justification.

First, calculate the mean of the coded data \(u\):
\(\bar{u} = \frac{\sum u}{n} = \frac{180}{50} = 3.6\).

Using the coding relation \(u = x - 100\), we find the mean of the original data \(x\):
\(\bar{x} = \bar{u} + 100 = 3.6 + 100 = 103.6\) hours.

Next, calculate the variance of the coded data \(u\):
\(\text{Var}(u) = \frac{\sum u^2}{n} - (\bar{u})^2 = \frac{2450}{50} - (3.6)^2 = 49 - 12.96 = 36.04\).

Since coding by subtracting a constant does not affect the spread (variance or standard deviation), the variance of \(x\) is the same as the variance of \(u\):
\(\text{Var}(x) = 36.04\).

Therefore, the standard deviation of \(x\) is:
\(\text{SD}(x) = \sqrt{36.04} \approx 6.00\) hours (to 3 significant figures).

M1: For calculating the coded mean \(\bar{u} = 3.6\).
A1: For obtaining the correct mean of \(x\) as 103.6.
M1: For using the variance formula for \(u\), \(\frac{\sum u^2}{n} - (\bar{u})^2\).
A1: For obtaining the correct variance of 36.04.
A1: For correctly identifying that standard deviation is unchanged by shift, giving 6.00 (accept 6).

We can solve this using the relationship between the frequency of a class and the area of its block in the histogram.

For the first class \(10 < w \le 15\):
Area of the block = \(\text{width} \times \text{height} = 2 \text{ cm} \times 4.5 \text{ cm} = 9 \text{ cm}^2\).
This block represents a frequency of 18 parcels.
Therefore, 1 \text{ cm}^2\) of area represents:
\(\frac{18}{9} = 2\) parcels.

For the second class \(15 < w \le 25\):
Area of the block = \(\text{width} \times \text{height} = 4 \text{ cm} \times 3 \text{ cm} = 12 \text{ cm}^2\).

Using the area scale, the frequency \(f\) represented by this block is:
\(f = 12 \text{ cm}^2 \times 2 \text{ parcels/cm}^2 = 24\).

M1: For finding the area of the first block (9 cm^2).
M1: For establishing the scale ratio between frequency and area (e.g., 1 cm^2 = 2 parcels or frequency density ratio).
M1: For calculating the area of the second block (12 cm^2).
A1: For obtaining the correct frequency f = 24.

To find the estimated variance, we use the midpoints (\(x\)) of each class interval:
- For \(0 < h \le 10\): midpoint \(x = 5\), \(f = 8\)
- For \(10 < h \le 20\): midpoint \(x = 15\), \(f = 15\)
- For \(20 < h \le 40\): midpoint \(x = 30\), \(f = 22\)
- For \(40 < h \le 50\): midpoint \(x = 45\), \(f = 5\)

First, calculate \(\sum fx\):
\(\sum fx = (8 \times 5) + (15 \times 15) + (22 \times 30) + (5 \times 45) = 40 + 225 + 660 + 225 = 1150\).

The estimated mean \(\bar{x}\) is:
\(\bar{x} = \frac{1150}{50} = 23\).

Next, calculate \(\sum fx^2\):
\(\sum fx^2 = (8 \times 5^2) + (15 \times 15^2) + (22 \times 30^2) + (5 \times 45^2)\)
\(\sum fx^2 = (8 \times 25) + (15 \times 225) + (22 \times 900) + (5 \times 2025) = 200 + 3375 + 19800 + 10125 = 33500\).

The estimated variance is:
\(\text{Variance} = \frac{\sum fx^2}{\sum f} - (\bar{x})^2 = \frac{33500}{50} - 23^2 = 670 - 529 = 141\).

M1: For identifying all correct midpoints (5, 15, 30, 45).
M1: For calculating the estimated mean \(\bar{x} = 23\) using \(\frac{\sum fx}{50}\).
M1: For calculating \(\sum fx^2 = 33500\).
M1: For applying the correct variance formula \(\frac{\sum fx^2}{\sum f} - (\bar{x})^2\).
A1: For obtaining the correct estimated variance of 141.

(i) There are \(n = 15\) ordered data points for Orchard A.
The median is the 8th value, which is 98.
The lower quartile (\(Q_1\)) is the median of the lower 7 values (values 1 to 7: 82, 86, 89, 91, 93, 94, 96). Thus, \(Q_1\) is the 4th value, which is 91 grams.
The upper quartile (\(Q_3\)) is the median of the upper 7 values (values 9 to 15: 100, 102, 105, 107, 109, 112, 118). Thus, \(Q_3\) is the 12th value, which is 107 grams.
The interquartile range (IQR) of Orchard A is:
\(\text{IQR}_A = Q_3 - Q_1 = 107 - 91 = 16\) grams.

(ii) For Orchard B, the interquartile range is:
\(\text{IQR}_B = 108 - 88 = 20\) grams.

Comparing the two interquartile ranges, Orchard A has a smaller interquartile range than Orchard B (16 < 20). Therefore, Orchard A has more consistent apple weights.

M1: For finding the correct lower quartile (91) or upper quartile (107) of Orchard A.
A1: For the correct interquartile range of Orchard A (16 grams).
M1: For calculating the interquartile range of Orchard B (20 grams).
A1: For making the correct comparison and concluding that Orchard A is more consistent because its IQR is smaller.

For the mean under a linear transformation \(z = ay + b\):
\(\bar{z} = a\bar{y} + b\).
Substituting the given values:
\(\bar{z} = 1.5(45) - 10 = 67.5 - 10 = 57.5\).

For the standard deviation under a linear transformation \(z = ay + b\):
\(\sigma_z = |a| \sigma_y\).
Note that adding or subtracting a constant does not affect the standard deviation.
Substituting the given values:
\(\sigma_z = 1.5 \times 8 = 12\).

M1: For applying the mean coding formula \(1.5 \bar{y} - 10\).
A1: For obtaining the correct mean of 57.5.
M1: For applying the scaling factor to standard deviation (\(1.5 \times \sigma_y\)).
A1: For concluding that subtraction has no effect on standard deviation and obtaining the correct standard deviation of 12.

First, calculate the sum of the observations for each group:
\(\sum x_1 = n_1 \times \bar{x}_1 = 40 \times 12.5 = 500\),
\(\sum x_2 = n_2 \times \bar{x}_2 = 60 \times 15.0 = 900\).

The combined sum of the observations is:
\(\sum x = \sum x_1 + \sum x_2 = 500 + 900 = 1400\).

The combined mean is:
\(\bar{x} = \frac{1400}{100} = 14.0\).

Next, find the combined sum of squares:
\(\sum x^2 = \sum x_1^2 + \sum x_2^2 = 6800 + 14200 = 21000\).

The combined variance is:
\(\text{Var}(x) = \frac{\sum x^2}{n} - (\bar{x})^2 = \frac{21000}{100} - (14.0)^2 = 210 - 196 = 14\).

The combined standard deviation is:
\(\text{SD}(x) = \sqrt{14} \approx 3.74\) (to 3 significant figures).

M1: For calculating the sum of observations for both groups (500 and 900).
A1: For obtaining the correct combined mean of 14.0.
M1: For finding the combined sum of squares (21000).
M1: For substituting into the combined variance formula \(\frac{21000}{100} - 14^2\).
A1: For obtaining the correct standard deviation of 3.74 (accept \(\sqrt{14}\)).

(i) The interquartile range (IQR) is calculated as:
\(\text{IQR} = Q_3 - Q_1 = 210 - 120 = 90\) cars.

(ii) To analyze the skewness, we compare the distances between the quartiles and the median:
\(Q_2 - Q_1 = 180 - 120 = 60\),
\(Q_3 - Q_2 = 210 - 180 = 30\).

Since \(Q_2 - Q_1 > Q_3 - Q_2\), the lower 50% of the data below the median is more spread out than the upper 50% above the median. This indicates a longer tail to the left, showing that the distribution is negatively skewed (or skewed to the left).

M1: For calculating the correct IQR of 90.
M1: For evaluating the distance from the lower quartile to the median (60) and from the median to the upper quartile (30).
A1: For concluding that because \(Q_2 - Q_1 > Q_3 - Q_2\) (or equivalent comparison), the distribution is negatively skewed.

Let \( \lambda \) be the daily mean sales of sourdough loaves.

State the hypotheses:
\( H_0: \lambda = 4.2 \) (the daily mean is unchanged, so the 5-day mean is \( 21 \))
\( H_1: \lambda > 4.2 \) (the daily mean has increased)

Under the null hypothesis \( H_0 \), let \( Y \) be the number of loaves sold over a 5-day period.
\( Y \sim \text{Po}(21) \)

Since \( \lambda' = 21 > 15 \), we can use the Normal approximation to the Poisson distribution:
\( Y \approx N(21, 21) \)

We want to find the probability of observing 28 or more loaves, applying a continuity correction:
\( P(Y \ge 28) \approx P(Y_{\text{normal}} > 27.5) \)

Standardize the value:
\( z = \frac{27.5 - 21}{\sqrt{21}} = \frac{6.5}{4.5826} \approx 1.418 \)

For a one-tailed test at the 5% significance level, the critical value of \( z \) is \( 1.645 \).

Since \( 1.418 < 1.645 \) (or the p-value \( 1 - \Phi(1.418) = 0.0781 > 0.05 \)), we do not reject \( H_0 \).

There is insufficient evidence at the 5% significance level to support the bakery owner's claim that the mean daily sales of sourdough loaves has increased.

B1: Correctly state both null and alternative hypotheses (either in terms of daily mean or 5-day mean).
B1: Identify the mean and variance for the 5-day period under the null hypothesis as 21.
M1: Use a Normal approximation with a standardizing formula that includes a continuity correction (evaluating \(27.5\) or \(28.5\)).
A1: Obtain the correct standardized test statistic \( z = 1.418 \) (or 1.42).
M1: Compare the calculated test statistic with the critical value of 1.645 (or compare the calculated probability of 0.0781 with 0.05).
A1: Draw the correct conclusion in context, stating that there is insufficient evidence of an increase.

(i) To find \( k \), we integrate the PDF over the entire range and equate to 1:
\[ \int_{0}^{2} k(4x - x^3) \, dx = 1 \]
\[ k \left[ 2x^2 - \frac{x^4}{4} \right]_{0}^{2} = 1 \]
\[ k \left( 2(4) - \frac{16}{4} \right) = 1 \]
\[ k (8 - 4) = 1 \implies 4k = 1 \implies k = \frac{1}{4} \]

(ii) The expectation is given by:
\[ E(X) = \int_{0}^{2} x f(x) \, dx = \int_{0}^{2} \frac{1}{4}(4x^2 - x^4) \, dx \]
\[ E(X) = \frac{1}{4} \left[ \frac{4x^3}{3} - \frac{x^5}{5} \right]_{0}^{2} \]
\[ E(X) = \frac{1}{4} \left( \frac{32}{3} - \frac{32}{5} \right) = \frac{1}{4} \left( \frac{160 - 96}{15} \right) = \frac{1}{4} \left( \frac{64}{15} \right) = \frac{16}{15} \approx 1.07 \]

(iii) Let \( m \) be the median of \( X \). Then:
\[ \int_{0}^{m} f(x) \, dx = 0.5 \]
\[ \int_{0}^{m} \frac{1}{4}(4x - x^3) \, dx = 0.5 \]
\[ \left[ 2x^2 - \frac{x^4}{4} \right]_{0}^{m} = 2 \]
\[ 2m^2 - \frac{m^4}{4} = 2 \]
\[ m^4 - 8m^2 + 8 = 0 \]

Using the quadratic formula for \( m^2 \):
\[ m^2 = \frac{8 \pm \sqrt{64 - 32}}{2} = 4 \pm 2\sqrt{2} \]

Since \( 0 \le m \le 2 \), we must have \( 0 \le m^2 \le 4 \). Thus:
\[ m^2 = 4 - 2\sqrt{2} \approx 1.1716 \]
\[ m = \sqrt{4 - 2\sqrt{2}} \approx 1.08 \]

Part (i):
M1: Integrate \( f(x) \) from 0 to 2 and equate the result to 1.
A1: Correct integration and showing that \( k = 1/4 \).

Part (ii):
M1: Use the correct formula for \( E(X) = \int x f(x) \, dx \) with limits 0 and 2.
A1: Correct analytical integration and calculation to obtain \( 16/15 \) (or 1.07).

Part (iii):
M1: Equate the cumulative probability up to \( m \) to 0.5.
A1: Form a quadratic equation in terms of \( m^2 \).
A1: Solve the equation and correctly select the root within the valid range to obtain \( 1.08 \).

(i) Unbiased estimate of the population mean \( \bar{x} \):
\[ \bar{x} = \frac{\sum x}{n} = \frac{50150}{100} = 501.5 \text{ grams} \]

Unbiased estimate of the population variance \( s^2 \):
\[ s^2 = \frac{1}{n-1} \left( \sum x^2 - \frac{(\sum x)^2}{n} \right) \]
\[ s^2 = \frac{1}{99} \left( 25150850 - \frac{50150^2}{100} \right) \]
\[ s^2 = \frac{1}{99} (25150850 - 25150225) = \frac{625}{99} \approx 6.3131 \approx 6.31 \text{ grams}^2 \]

(ii) For a 98% confidence interval, the critical value \( z \) is such that \( \Phi(z) = 0.99 \), which gives \( z = 2.326 \).

The confidence interval is given by:
\[ \bar{x} \pm z \times \frac{s}{\sqrt{n}} \]
\[ 501.5 \pm 2.326 \times \frac{\sqrt{6.3131}}{\sqrt{100}} \]
\[ 501.5 \pm 2.326 \times 0.25126 \]
\[ 501.5 \pm 0.5844 \]

This yields the interval:
\[ [500.9, 502.1] \text{ grams} \]

(iii) Yes, it was necessary. The distribution of the population of masses is not stated to be normal. Because the sample size \( n = 100 \) is large, the Central Limit Theorem allows us to assume that the sample mean \( \bar{X} \) is approximately normally distributed.

Part (i):
B1: Calculate the unbiased mean of 501.5.
M1: Apply the correct unbiased variance formula.
A1: Obtain the correct unbiased variance of 6.31 (or 625/99).

Part (ii):
B1: State the correct critical value \( z = 2.326 \).
M1: Substitute values correctly into the confidence interval formula.
A1: Obtain the correct interval limits of [500.9, 502.1] (accept 501 to 502 if 3 sf used).

Part (iii):
B1: Correctly state 'Yes' and refer to the population distribution being unknown and the sample size being large.

(i) We want to find \( P(A > 1.6B) \), which is equivalent to \( P(A - 1.6B > 0) \).

Let \( W = A - 1.6B \).
\[ E(W) = E(A) - 1.6E(B) = 150 - 1.6(100) = -10 \]
\[ \text{Var}(W) = \text{Var}(A) + 1.6^2 \text{Var}(B) = 15^2 + 2.56(10^2) = 225 + 256 = 481 \]

So \( W \sim N(-10, 481) \).

Standardizing:
\[ P(W > 0) = P\left( Z > \frac{0 - (-10)}{\sqrt{481}} \right) = P\left( Z > \frac{10}{21.9317} \right) = P(Z > 0.456) \]
\[ P(W > 0) = 1 - \Phi(0.456) = 1 - 0.6758 = 0.324 \]

(ii) Let \( T \) be the total weight of 4 large and 6 small apples:
\[ T = A_1 + A_2 + A_3 + A_4 + B_1 + B_2 + B_3 + B_4 + B_5 + B_6 \]

\[ E(T) = 4E(A) + 6E(B) = 4(150) + 6(100) = 600 + 600 = 1200 \]
\[ \text{Var}(T) = 4\text{Var}(A) + 6\text{Var}(B) = 4(15^2) + 6(10^2) = 900 + 600 = 1500 \]

So \( T \sim N(1200, 1500) \).

Standardizing to find \( P(T < 1250) \):
\[ P(T < 1250) = P\left( Z < \frac{1250 - 1200}{\sqrt{1500}} \right) = P\left( Z < \frac{50}{38.7298} \right) = P(Z < 1.291) \]
\[ P(T < 1250) = \Phi(1.291) = 0.9017 \approx 0.902 \]

Part (i):
M1: Set up a linear combination variable, finding its expected value.
A1: Calculate correct mean (-10) and variance (481).
M1: Standardize with correct values and evaluate probability.
A1: Obtain correct probability 0.324.

Part (ii):
M1: Find the mean and variance of the sum of 10 independent apples.
A1: Obtain mean = 1200 and variance = 1500.
M1: Standardize with their mean and standard deviation.
A1: Obtain correct probability 0.902.

(i) For a 2-hour period, the mean is \( \lambda = 2.4 \times 2 = 4.8 \).
Let \( X \) be the number of complaints in 2 hours, where \( X \sim \text{Po}(4.8) \).
\[ P(X = 3) = \frac{e^{-4.8} 4.8^3}{3!} \approx 0.1517 \approx 0.152 \]

(ii) For a 1.5-hour period, the mean is \( \lambda = 2.4 \times 1.5 = 3.6 \).
Let \( Y \) be the number of complaints in 1.5 hours, where \( Y \sim \text{Po}(3.6) \).
\[ P(Y > 4) = 1 - P(Y \le 4) = 1 - e^{-3.6} \left( 1 + 3.6 + \frac{3.6^2}{2} + \frac{3.6^3}{6} + \frac{3.6^4}{24} \right) \]
\[ P(Y > 4) = 1 - e^{-3.6} (1 + 3.6 + 6.48 + 7.776 + 6.9984) = 1 - e^{-3.6} (25.8544) \]
\[ P(Y > 4) \approx 1 - 0.7064 = 0.2936 \approx 0.294 \]

(iii) For a 24-hour period, the mean is \( \lambda = 2.4 \times 24 = 57.6 \).
Since \( \lambda = 57.6 > 15 \), we can approximate with a Normal distribution \( W \sim N(57.6, 57.6) \).
Applying continuity correction:
\[ P(\text{Complaints} \ge 50) \approx P(W \ge 49.5) \]
Standardizing:
\[ z = \frac{49.5 - 57.6}{\sqrt{57.6}} = \frac{-8.1}{7.5895} \approx -1.067 \]
\[ P(W \ge 49.5) = P(Z > -1.067) = \Phi(1.067) \approx 0.857 \]

Part (i):
B1: Calculate the scaled mean of 4.8 and find correct probability of 0.152.

Part (ii):
M1: Scale the mean to 3.6 and set up \( 1 - P(Y \le 4) \).
A1: Correct computation to obtain 0.294.

Part (iii):
B1: State the correct mean of 57.6.
M1: Standardize using a Normal distribution with mean and variance 57.6, including a continuity correction (using 49.5).
A1: Obtain the correct standardized value \( z = -1.067 \) (or \( -1.07 \)).
A1: Compute correct final probability of 0.857.

(i) Hypotheses:
\( H_0: p = 0.25 \)
\( H_1: p < 0.25 \)

(ii) Under \( H_0 \), \( X \sim B(20, 0.25) \). We want to find the largest critical value \( c \) such that \( P(X \le c) \le 0.05 \).

Let's calculate the cumulative probabilities:
\[ P(X = 0) = 0.75^{20} \approx 0.00317 \]
\[ P(X = 1) = \binom{20}{1} (0.25)^1 (0.75)^{19} = 20 \times 0.25 \times 0.004228 \approx 0.02114 \]
\[ P(X \le 1) = 0.00317 + 0.02114 = 0.02431 \]

Now check \( X = 2 \):
\[ P(X = 2) = \binom{20}{2} (0.25)^2 (0.75)^{18} = 190 \times 0.0625 \times 0.005638 \approx 0.06695 \]
\[ P(X \le 2) = 0.02431 + 0.06695 = 0.09126 \]

Since \( P(X \le 1) = 0.0243 \le 0.05 \) and \( P(X \le 2) = 0.0913 > 0.05 \), the critical region is:
\( X \le 1 \) (or \( X = {0, 1} \))

(iii) A Type II error occurs when we fail to reject \( H_0 \) given that the true parameter is \( p = 0.10 \).

We fail to reject \( H_0 \) when \( X \ge 2 \).

Under the alternative hypothesis \( p = 0.10 \), \( X \sim B(20, 0.10) \).
\[ P(\text{Type II error}) = P(X \ge 2 \mid p = 0.10) = 1 - P(X \le 1 \mid p = 0.10) \]
\[ P(X = 0) = 0.9^{20} \approx 0.12158 \]
\[ P(X = 1) = 20 \times (0.1)^1 \times (0.9)^{19} \approx 0.27017 \]
\[ P(X \le 1) = 0.12158 + 0.27017 = 0.39175 \]
\[ P(\text{Type II error}) = 1 - 0.39175 = 0.60825 \approx 0.608 \]

Part (i):
B1: Formulate the correct null and alternative hypotheses.

Part (ii):
M1: Evaluate binomial probabilities under the null hypothesis starting from \( X = 0 \).
A1: Identify the correct critical boundary where the cumulative probability does not exceed 0.05, establishing the critical region as \( X \le 1 \).

Part (iii):
B1: Clearly define the condition for a Type II error (finding \( P(X \ge 2) \) when \( p = 0.10 \)).
M1: Set up the calculations for binomial terms \( P(X = 0) \) and \( P(X = 1) \) using \( p = 0.1 \).
A1: Correctly calculate the final probability of Type II error as 0.608.