2024 Cambridge IAL Mathematics (9709) 模擬試題連答案詳解

For the curve to lie entirely above the \(x\)-axis, two conditions must be met:

1) The coefficient of \(x^2\) must be positive, which gives:
\[ k > 0 \]

2) The curve must not intersect or touch the \(x\)-axis, meaning the discriminant of \(kx^2 + 4x + (k-3) = 0\) must be strictly less than zero:
\[ \Delta = b^2 - 4ac < 0 \]
\[ 4^2 - 4(k)(k-3) < 0 \]
\[ 16 - 4k^2 + 12k < 0 \]

Divide the inequality by \(-4\) and reverse the inequality sign:
\[ k^2 - 3k - 4 > 0 \]

Factorising the quadratic expression:
\[ (k-4)(k+1) > 0 \]

The critical values are \(k = 4\) and \(k = -1\).
Solving the inequality gives:
\[ k > 4 \quad \text{or} \quad k < -1 \]

Combining this result with the condition \(k > 0\), we discard the region \(k < -1\).

Thus, the set of values is \(k > 4\).

M1: State or imply that \(k > 0\) and discriminant \(\Delta < 0\).
A1: Formulate the discriminant expression correctly: \(16 - 4k(k-3)\).
M1: Set the discriminant \(< 0\) and form a quadratic inequality, e.g., \(k^2 - 3k - 4 > 0\).
A1: Find critical values \(k = 4\) and \(k = -1\).
M1: Correctly solve the quadratic inequality to get \(k > 4\) or \(k < -1\).
A0.8: Combine with the condition \(k > 0\) to obtain the final unique range \(k > 4\).

(a) To find \(\mathrm{f}^{-1}(x)\), let \(y = 3 - \frac{4}{x+2}\):
\[ 3 - y = \frac{4}{x+2} \]
\[ x + 2 = \frac{4}{3-y} \]
\[ x = \frac{4}{3-y} - 2 = \frac{4 - 2(3-y)}{3-y} = \frac{2y - 2}{3-y} \]

So \(\mathrm{f}^{-1}(x) = \frac{2x - 2}{3-x}\).

The domain of \(\mathrm{f}^{-1}\) is the range of \(\mathrm{f}\).
Since \(x > -2\), we have \(x+2 > 0\), which implies \(\frac{4}{x+2} > 0\).
Therefore, \(\mathrm{f}(x) = 3 - \frac{4}{x+2} < 3\).
Thus, the domain of \(\mathrm{f}^{-1}\) is \(x < 3\).

(b) To solve \(\mathrm{ff}(x) = \frac{5}{3}\), let \(\mathrm{f}(x) = y\).
Then \(\mathrm{f}(y) = \frac{5}{3} \implies y = \mathrm{f}^{-1}\left(\frac{5}{3}\right)\).
Using the expression found in part (a) (since \(\frac{5}{3} < 3\)):
\[ y = \frac{2(5/3) - 2}{3 - 5/3} = \frac{10/3 - 6/3}{9/3 - 5/3} = \frac{4/3}{4/3} = 1 \]

Now solve \(\mathrm{f}(x) = 1\):
\[ 3 - \frac{4}{x+2} = 1 \]
\[ \frac{4}{x+2} = 2 \]
\[ x + 2 = 2 \implies x = 0 \]

Since \(0 > -2\), this is a valid solution.

(a)
M1: Attempt to make \(x\) the subject of \(y = 3 - \frac{4}{x+2}\).
A1: Obtain correct expression \(\mathrm{f}^{-1}(x) = \frac{2x-2}{3-x}\) or equivalent.
B1.8: State correct domain \(x < 3\) (accept \((-\infty, 3)\)).

(b)
M1: Set up equation \(\mathrm{f}(y) = \frac{5}{3}\) and attempt to find \(y\), or expand \(\mathrm{f}(\mathrm{f}(x)) = \frac{5}{3}\).
A1: Obtain \(y = 1\) (or equivalent intermediate stage).
M1: Solve \(\mathrm{f}(x) = 1\) to find \(x\).
A1: Correctly obtain \(x = 0\).

(a) The radius \(r\) is the distance between \(C(3, -2)\) and \(P(6, 2)\):
\[ r = \sqrt{(6-3)^2 + (2 - (-2))^2} = \sqrt{3^2 + 4^2} = \sqrt{25} = 5 \]

The equation of the circle is:
\[ (x-3)^2 + (y+2)^2 = r^2 \]
\[ (x-3)^2 + (y+2)^2 = 25 \]

(b) The gradient of the radius \(CP\) is:
\[ m_{CP} = \frac{2 - (-2)}{6 - 3} = \frac{4}{3} \]

The tangent is perpendicular to the radius, so its gradient \(m_t\) is:
\[ m_t = -\frac{3}{4} \]

Using the point \(P(6, 2)\), the equation of the tangent is:
\[ y - 2 = -\frac{3}{4}(x - 6) \]
\[ y = -\frac{3}{4}x + \frac{9}{2} + 2 \]
\[ y = -\frac{3}{4}x + \frac{13}{2} \quad \text{or} \quad 3x + 4y = 26 \]

(c) To find the coordinates of \(Q\), substitute \(x = 2\) into the equation of the tangent:
\[ 3(2) + 4y = 26 \implies 6 + 4y = 26 \implies 4y = 20 \implies y = 5 \]

So \(Q\) is the point \((2, 5)\).

The distance \(CQ\) between \(C(3, -2)\) and \(Q(2, 5)\) is:
\[ CQ = \sqrt{(3-2)^2 + (-2-5)^2} = \sqrt{1^2 + (-7)^2} = \sqrt{1 + 49} = \sqrt{50} = 5\sqrt{2} \]

(a)
M1: Calculate the radius or radius squared using the distance formula.
A1: Correctly write the equation of the circle: \((x-3)^2 + (y+2)^2 = 25\).

(b)
M1: Find the gradient of the radius \(CP\) and use the negative reciprocal for the tangent.
A1: Correct equation of the tangent, e.g., \(3x + 4y = 26\) or \(y = -\frac{3}{4}x + \frac{13}{2}\).

(c)
M1: Substitute \(x = 2\) into the tangent equation to find the \(y\)-coordinate of \(Q\).
A1: Obtain \(Q(2, 5)\).
A0.8: Use the distance formula to find \(CQ = 5\sqrt{2}\) (or \(\sqrt{50}\)).

(a) The perimeter \(P\) of the sector consists of two radii and the arc length:
\[ P = 2r + r\theta \]

Given \(P = 30\):
\[ 2r + r\theta = 30 \implies r\theta = 30 - 2r \implies \theta = \frac{30 - 2r}{r} \]

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

Substitute the expression for \(\theta\):
\[ A = \frac{1}{2}r^2\left(\frac{30 - 2r}{r}\right) = \frac{1}{2}r(30 - 2r) = 15r - r^2 \quad \text{(as required)} \]

(b) Given that the area is \(54\text{ cm}^2\):
\[ 15r - r^2 = 54 \]
\[ r^2 - 15r + 54 = 0 \]

Factorising the quadratic:
\[ (r-6)(r-9) = 0 \]

So the two possible values of \(r\) are \(r = 6\) and \(r = 9\).

If \(r = 6\):
\[ \theta = \frac{30 - 2(6)}{6} = \frac{18}{6} = 3\text{ radians} \]

If \(r = 9\):
\[ \theta = \frac{30 - 2(9)}{9} = \frac{12}{9} = \frac{4}{3}\text{ radians} \]

(a)
M1: State or use the formula for perimeter \(2r + r\theta = 30\) to express \(\theta\) in terms of \(r\).
M1: Substitute into the sector area formula \(A = \frac{1}{2}r^2\theta\).
A0.8: Complete the algebra to show \(A = 15r - r^2\) clearly.

(b)
M1: Set the area expression equal to 54 and form a quadratic equation.
A1: Solve the quadratic to find \(r = 6\) and \(r = 9\).
M1: Substitute both values of \(r\) back to find the corresponding values of \(\theta\).
A1: Obtain \(\theta = 3\) for \(r = 6\), and \(\theta = \frac{4}{3}\) (or 1.33) for \(r = 9\).

(a) Put the LHS over a common denominator:
\[ \frac{\sin\theta}{1 - \cos\theta} + \frac{1 - \cos\theta}{\sin\theta} = \frac{\sin^2\theta + (1 - \cos\theta)^2}{\sin\theta(1 - \cos\theta)} \]

Expand the numerator:
\[ \sin^2\theta + 1 - 2\cos\theta + \cos^2\theta \]

Using the identity \(\sin^2\theta + \cos^2\theta = 1\):
\[ 1 + 1 - 2\cos\theta = 2 - 2\cos\theta = 2(1 - \cos\theta) \]

Substitute this back into the fraction:
\[ \frac{2(1 - \cos\theta)}{\sin\theta(1 - \cos\theta)} = \frac{2}{\sin\theta} \quad \text{(as required)} \]

(b) Using the identity from part (a) with \(\theta = 2x\), the equation becomes:
\[ \frac{2}{\sin 2x} = \frac{4}{3}\tan 2x \]

Since \(\tan 2x = \frac{\sin 2x}{\cos 2x}\):
\[ \frac{2}{\sin 2x} = \frac{4\sin 2x}{3\cos 2x} \]

Cross-multiplying gives:
\[ 6\cos 2x = 4\sin^2 2x \]

Using \(\sin^2 2x = 1 - \cos^2 2x\):
\[ 6\cos 2x = 4(1 - \cos^2 2x) \]
\[ 4\cos^2 2x + 6\cos 2x - 4 = 0 \]
\[ 2\cos^2 2x + 3\cos 2x - 2 = 0 \]

Factorising the quadratic in \(\cos 2x\):
\[ (2\cos 2x - 1)(\cos 2x + 2) = 0 \]

Since \(\cos 2x = -2\) has no real solutions, we have:
\[ \cos 2x = \frac{1}{2} \]

Given \(0^\circ < x < 180^\circ\), the range for \(2x\) is \(0^\circ < 2x < 360^\circ\):
\[ 2x = 60^\circ \quad \text{or} \quad 2x = 300^\circ \]
\[ x = 30^\circ \quad \text{or} \quad x = 150^\circ \]

(a)
M1: Express LHS over a common denominator.
A1: Correctly expand \((1-\cos\theta)^2\) and use the identity \(\sin^2\theta + \cos^2\theta = 1\).
A0.8: Factorise out the common factor of 2 and complete the proof.

(b)
M1: Substitute the identity into the given equation and express \(\tan 2x\) as \(\frac{\sin 2x}{\cos 2x}\).
A1: Obtain a quadratic equation in \(\cos 2x\): \(2\cos^2 2x + 3\cos 2x - 2 = 0\).
M1: Solve the quadratic equation to get \(\cos 2x = \frac{1}{2}\).
A1: Obtain both correct solutions: \(x = 30^\circ\) and \(x = 150^\circ\).

(a) The terms of the arithmetic progression are:
- 15th term: \(T_{15} = a + 14d\)
- 5th term: \(T_{5} = a + 4d\)
- 1st term: \(T_{1} = a\)

These are the first three terms of a geometric progression, so:
\[ G_1 = a + 14d, \quad G_2 = a + 4d, \quad G_3 = a \]

Since they form a geometric progression, the common ratio is constant:
\[ G_2^2 = G_1 G_3 \]
\[ (a + 4d)^2 = a(a + 14d) \]
\[ a^2 + 8ad + 16d^2 = a^2 + 14ad \]
\[ 16d^2 = 6ad \]

Since \(d \neq 0\), we can divide both sides by \(2d\):
\[ 8d = 3a \implies 3a = 8d \quad \text{(as required)} \]

(b) The common ratio \(r\) of the geometric progression is given by:
\[ r = \frac{G_2}{G_1} = \frac{a + 4d}{a + 14d} \]

Substitute \(d = \frac{3}{8}a\):
\[ r = \frac{a + 4(\frac{3}{8}a)}{a + 14(\frac{3}{8}a)} = \frac{a + 1.5a}{a + 5.25a} = \frac{2.5a}{6.25a} = \frac{2.5}{6.25} = 0.4 \]

(c) The sum to infinity is:
\[ S_\infty = \frac{G_1}{1 - r} = 50 \]

With \(r = 0.4\) and \(G_1 = a + 14d = a + 14(\frac{3}{8}a) = 6.25a\):
\[ \frac{6.25a}{1 - 0.4} = 50 \]

\[ \frac{6.25a}{0.6} = 50 \]
\[ 6.25a = 30 \]
\[ a = \frac{30}{6.25} = 4.8 \quad \left(\text{or } \frac{24}{5}\right) \]

(a)
M1: Formulate the geometric relationship: \((a+4d)^2 = a(a+14d)\).
A1: Correctly expand and simplify to \(16d^2 = 6ad\).
A0.8: Complete the proof to show \(3a = 8d\).

(b)
M1: Express \(r = \frac{a+4d}{a+14d}\) and substitute \(d = \frac{3}{8}a\).
A1: Correctly obtain \(r = 0.4\) (or \(\frac{2}{5}\)).

(c)
M1: Substitute \(r = 0.4\) and \(G_1 = 6.25a\) into the sum to infinity formula.
A1: Correctly obtain \(a = 4.8\) (or \(\frac{24}{5}\)).

(a) Differentiating \(y = \frac{1}{3}(2x-1)^3 - 8x\) using the chain rule:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{3} \cdot 3(2x-1)^2 \cdot 2 - 8 = 2(2x-1)^2 - 8 \]

Differentiating again to find the second derivative:
\[ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2 \cdot 2(2x-1) \cdot 2 = 8(2x-1) \]

(b) To find the stationary points, set \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\[ 2(2x-1)^2 - 8 = 0 \]
\[ 2(2x-1)^2 = 8 \]
\[ (2x-1)^2 = 4 \]

Taking the square root of both sides:
\[ 2x - 1 = 2 \implies 2x = 3 \implies x = 1.5 \]
\[ 2x - 1 = -2 \implies 2x = -1 \implies x = -0.5 \]

So the \(x\)-coordinates of the stationary points are \(x = 1.5\) and \(x = -0.5\).

To determine the nature of each stationary point, substitute these values into the second derivative:
- At \(x = 1.5\):
\[ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 8(2(1.5)-1) = 8(2) = 16 > 0 \]
Since the second derivative is positive, the point at \(x = 1.5\) is a **minimum**.

- At \(x = -0.5\):
\[ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 8(2(-0.5)-1) = 8(-2) = -16 < 0 \]
Since the second derivative is negative, the point at \(x = -0.5\) is a **maximum**.

(a)
M1: Correctly apply the chain rule to differentiate the first term.
A1: Obtain \(\frac{\mathrm{d}y}{\mathrm{d}x} = 2(2x-1)^2 - 8\).
A0.8: Differentiate again to get \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 8(2x-1)\).

(b)
M1: Set \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) and solve the quadratic equation.
A1: Obtain both \(x = 1.5\) and \(x = -0.5\).
M1: Substitute the \(x\)-values into \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\).
A1: Correctly identify \(x = 1.5\) as a minimum and \(x = -0.5\) as a maximum.

(a) Expand \(y^2 = (x^{1/4} + 2x^{-1/4})^2\):
\[ y^2 = (x^{1/4})^2 + 2(x^{1/4})(2x^{-1/4}) + (2x^{-1/4})^2 \]
\[ y^2 = x^{1/2} + 4 + 4x^{-1/2} \quad \text{(as required)} \]

(b) The formula for the volume of revolution about the \(x\)-axis is:
\[ V = \pi \int_{a}^{b} y^2\ \mathrm{d}x \]

Substitute the limits \(x=1\) to \(x=4\) and the expression for \(y^2\):
\[ V = \pi \int_{1}^{4} (x^{1/2} + 4 + 4x^{-1/2})\ \mathrm{d}x \]

Integrate term by term:
\[ \int (x^{1/2} + 4 + 4x^{-1/2})\ \mathrm{d}x = \frac{2}{3}x^{3/2} + 4x + 8x^{1/2} \]

Evaluate the integral between the limits 1 and 4:
\[ \left[ \frac{2}{3}x^{3/2} + 4x + 8x^{1/2} \right]_{1}^{4} \]

At the upper limit \(x = 4\):
\[ \frac{2}{3}(4)^{3/2} + 4(4) + 8(4)^{1/2} = \frac{2}{3}(8) + 16 + 8(2) = \frac{16}{3} + 32 = \frac{112}{3} \]

At the lower limit \(x = 1\):
\[ \frac{2}{3}(1)^{3/2} + 4(1) + 8(1)^{1/2} = \frac{2}{3} + 4 + 8 = \frac{38}{3} \]

Subtract the lower limit evaluation from the upper limit evaluation:
\[ \frac{112}{3} - \frac{38}{3} = \frac{74}{3} \]

Thus, the volume is:
\[ V = \frac{74}{3}\pi \]

(a)
M1: Attempt to square the expression for \(y\).
A0.8: Correctly obtain the expanded form \(y^2 = x^{1/2} + 4 + 4x^{-1/2}\).

(b)
M1: State or use the correct volume of revolution formula \(V = \pi \int y^2\ \mathrm{d}x\).
M1: Integrate the expression \(x^{1/2} + 4 + 4x^{-1/2}\) correctly term by term.
A1: Obtain the correct integrated expression \(\frac{2}{3}x^{3/2} + 4x + 8x^{1/2}\).
M1: Correctly substitute limits 1 and 4 into their integrated expression.
A1: Obtain the final exact value \(\frac{74}{3}\pi\) (or \(24\frac{2}{3}\pi\)).

**(a)**
The first, third, and eleventh terms of the arithmetic progression are:
\(u_1 = a\)
\(u_3 = a + 2d\)
\(u_{11} = a + 10d\)

Since these terms form a geometric progression with common ratio \(r\), we have:
\(a + 2d = ar\) --- (1)
\(a + 10d = ar^2\) --- (2)

From (1), we can express \(d\) in terms of \(a\) and \(r\):
\(2d = ar - a = a(r - 1) \implies d = \frac{a(r - 1)}{2}\)

Substitute \(d\) into (2):
\(a + 10\left(\frac{a(r - 1)}{2}\right) = ar^2\)
\(a + 5a(r - 1) = ar^2\)

Since \(d \neq 0\), we must have \(a \neq 0\) and \(r \neq 1\). Dividing both sides by \(a\):
\(1 + 5(r - 1) = r^2\)
\(1 + 5r - 5 = r^2\)
\(r^2 - 5r + 4 = 0\)

Factorising the quadratic equation:
\((r - 1)(r - 4) = 0\)

Since \(r \neq 1\) (as \(d \neq 0\)), we have:
\(r = 4\)

**(b)**
Using \(r = 4\) in the expression for \(d\):
\(2d = 3a \implies a = \frac{2}{3}d\)

The sum of the first 20 terms of the arithmetic progression is given by:
\(S_{20} = \frac{20}{2}[2a + 19d] = 10[2a + 19d]\)

Given that \(S_{20} = 610\):
\(10[2a + 19d] = 610 \implies 2a + 19d = 61\)

Substitute \(a = \frac{2}{3}d\) into this equation:
\(2\left(\frac{2}{3}d\right) + 19d = 61\)
\(\frac{4}{3}d + 19d = 61\)
\(\frac{61}{3}d = 61\)
\(d = 3\)

Now, find \(a\):
\(a = \frac{2}{3}(3) = 2\)

So, \(a = 2\) and \(d = 3\).

**(a)**
- **M1**: Write down expressions for \(u_1, u_3, u_{11}\) and use the GP property to set up an equation, e.g., \((a+2d)^2 = a(a+10d)\) or equivalent.
- **M1**: Eliminate \(a\) or \(d\) to obtain a quadratic equation in terms of \(r\).
- **A1**: Solve the quadratic and explain why \(r = 4\) (rejecting \(r = 1\)).

**(b)**
- **M1**: Use the AP sum formula \(S_{20} = 10(2a+19d)\) and equate to 610.
- **M1**: Use the relation from part (a) (e.g., \(2d = 3a\)) to form a single-variable equation.
- **A1**: Find \(d = 3\) (or \(a = 2\)).
- **A1**: Find the other value, \(a = 2\) (or \(d = 3\)).

**(a)**
To find the gradient of the tangent, we differentiate \(y = (2x + 1)^{\frac{1}{2}}\):
\(\frac{dy}{dx} = \frac{1}{2}(2x + 1)^{-\frac{1}{2}} \cdot 2 = (2x + 1)^{-\frac{1}{2}}\)

At the point \(P(4, 3)\), where \(x = 4\):
\(\frac{dy}{dx} = (2(4) + 1)^{-\frac{1}{2}} = 9^{-\frac{1}{2}} = \frac{1}{3}\)

So the gradient of the tangent is \(\frac{1}{3}\).
The gradient of the normal, \(m_n\), is:
\(m_n = -\frac{1}{1/3} = -3\)

The equation of the normal at \(P(4, 3)\) is:
\(y - 3 = -3(x - 4)\)
\(y - 3 = -3x + 12\)
\(y = -3x + 15\)

**(b)**
The volume of revolution \(V\) about the x-axis is given by:
\(V = \pi \int_{a}^{b} y^2 dx\)

Here, \(y^2 = 2x + 1\), and the limits are from \(x = 0\) to \(x = 4\):
\(V = \pi \int_{0}^{4} (2x + 1) dx\)

Integrate with respect to \(x\):
\(V = \pi \left[ x^2 + x \right]_0^4\)

Substitute the limits:
\(V = \pi \left[ (4^2 + 4) - (0^2 + 0) \right]\)
\(V = \pi [ 16 + 4 - 0 ] = 20\pi\)

Thus, the volume of the solid is \(20\pi\).

**(a)**
- **M1**: Differentiate \(y\) correctly using the chain rule to find \(\frac{dy}{dx}\).
- **A1**: Substitute \(x=4\) to find the gradient of the normal is \(-3\).
- **A1**: Obtain the correct equation of the normal in the form \(y = mx + c\), which is \(y = -3x + 15\).

**(b)**
- **M1**: Set up the volume integral \(\pi \int y^2 dx\) with limits \(0\) and \(4\).
- **M1**: Integrate \(2x + 1\) correctly to obtain \(x^2 + x\) (condone omission of \(\pi\) at this stage).
- **A1**: Apply the limits \(0\) and \(4\) correctly to find the numerical value \(20\).
- **A1**: State the final volume as \(20\pi\).

**(a)**
To prove the identity, we start with the left-hand side (LHS):
\(\text{LHS} = \frac{\sin \theta}{1 + \cos \theta} + \frac{1 + \cos \theta}{\sin \theta}\)

Find a common denominator:
\(\text{LHS} = \frac{\sin^2 \theta + (1 + \cos \theta)^2}{\sin \theta(1 + \cos \theta)}\)

Expand the numerator:
\(\text{LHS} = \frac{\sin^2 \theta + 1 + 2\cos \theta + \cos^2 \theta}{\sin \theta(1 + \cos \theta)}\)

Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\):
\(\text{LHS} = \frac{1 + 1 + 2\cos \theta}{\sin \theta(1 + \cos \theta)}\)
\(\text{LHS} = \frac{2 + 2\cos \theta}{\sin \theta(1 + \cos \theta)}\)

Factor out 2 from the numerator:
\(\text{LHS} = \frac{2(1 + \cos \theta)}{\sin \theta(1 + \cos \theta)}\)

Cancel the common factor \((1 + \cos \theta)\):
\(\text{LHS} = \frac{2}{\sin \theta} = \text{RHS}\) (Proved)

**(b)**
Using the identity proven in part (a), the equation becomes:
\(\frac{2}{\sin \theta} = \frac{3\tan \theta}{\cos \theta}\)

Substitute \(\tan \theta = \frac{\sin \theta}{\cos \theta}\):
\(\frac{2}{\sin \theta} = \frac{3\sin \theta}{\cos^2 \theta}\)

Multiply both sides by \(\sin \theta \cos^2 \theta\) (noting \(\sin \theta \neq 0\) and \(\cos \theta \neq 0\)):
\(2\cos^2 \theta = 3\sin^2 \theta\)

Use \(\cos^2 \theta = 1 - \sin^2 \theta\):
\(2(1 - \sin^2 \theta) = 3\sin^2 \theta\)
\(2 - 2\sin^2 \theta = 3\sin^2 \theta\)
\(5\sin^2 \theta = 2\)
\(\sin^2 \theta = 0.4\)

Taking the square root:
\(\sin \theta = \pm \sqrt{0.4}\)

Since \(0^\circ \le \theta \le 180^\circ\), \(\sin \theta\) must be positive, so:
\(\sin \theta = \sqrt{0.4} \approx 0.63246\)

Find the principal angle:
\(\theta = \sin^{-1}(0.63246) \approx 39.2^\circ\) (to 1 d.p.)

The second angle in the given range is:
\(\theta = 180^\circ - 39.2^\circ = 140.8^\circ\)

Thus, the solutions are \(\theta = 39.2^\circ\) and \(\theta = 140.8^\circ\).

**(a)**
- **M1**: Put LHS over a single common denominator \(\sin \theta(1+\cos \theta)\).
- **M1**: Use the fundamental identity \(\sin^2 \theta + \cos^2 \theta = 1\) to simplify the numerator to \(2 + 2\cos \theta\).
- **A1**: Factorise numerator and cancel \((1+\cos \theta)\) correctly to obtain \(\frac{2}{\sin \theta}\).

**(b)**
- **M1**: Equate \(\frac{2}{\sin \theta}\) to \(\frac{3\tan \theta}{\cos \theta}\) and substitute \(\tan \theta = \frac{\sin \theta}{\cos \theta}\).
- **M1**: Form a quadratic equation in either \(\sin \theta\) or \(\cos \theta\) (e.g., \(5\sin^2 \theta = 2\)).
- **A1**: Solve to find \(\theta = 39.2^\circ\) (accept 39.2).
- **A1**: Find the second correct angle \(\theta = 140.8^\circ\) (accept 140.8), and show no other solutions in range.

To solve \(|3x - 5| > |2x + 1|\), we square both sides:
\((3x - 5)^2 > (2x + 1)^2\)
\(9x^2 - 30x + 25 > 4x^2 + 4x + 1\)
\(5x^2 - 34x + 24 > 0\)

We find the critical values by solving \(5x^2 - 34x + 24 = 0\):
\((5x - 4)(x - 6) = 0\)
Thus, the critical values are \(x = 0.8\) and \(x = 6\).

Since the inequality is \(> 0\), the solution set represents the regions outside these critical values:
\(x < 0.8\) or \(x > 6\).

- **M1**: For attempting to square both sides or solve the two linear equations \(3x - 5 = \pm(2x + 1)\).
- **A1**: For obtaining the correct quadratic inequality \(5x^2 - 34x + 24 > 0\) or the correct boundary equations.
- **M1**: For solving the quadratic equation to find the critical values \(0.8\) and \(6\).
- **A1**: For identifying both correct critical values.
- **A1.1**: For the final correct inequality \(x < 0.8\) or \(x > 6\) (or equivalent interval notation).

Let \(u = 2^x\).
Then \(2^{2x+1} = 2 \cdot (2^x)^2 = 2u^2\).
Substitute \(u\) into the equation:
\(3(2u^2) - 17u + 10 = 0\)
\(6u^2 - 17u + 10 = 0\)

Factor the quadratic equation:
\((6u - 5)(u - 2) = 0\)
So, \(u = \frac{5}{6}\) or \(u = 2\).

Now solve for \(x\):

Case 1: \(2^x = \frac{5}{6}\)
\(x \ln 2 = \ln\left(\frac{5}{6}\right)\)
\(x = \frac{\ln(5/6)}{\ln 2} \approx -0.263\) (to 3 s.f.)

Case 2: \(2^x = 2\)
\(x = 1\)

- **M1**: For using a substitution such as \(u = 2^x\) to obtain a quadratic equation in \(u\).
- **A1**: For obtaining the correct quadratic equation \(6u^2 - 17u + 10 = 0\) (or equivalent).
- **M1**: For attempting to solve the quadratic equation to find values of \(u\).
- **A1**: For obtaining \(u = 2\) and \(u = \frac{5}{6}\).
- **M1**: For using logarithms to solve \(2^x = k\) for \(x\).
- **A1**: For \(x = 1\) (exact).
- **A1.1**: For \(x = -0.263\) (correct to 3 s.f.).

Expand the left-hand side using the compound angle identity \(\cos(A + B) = \cos A \cos B - \sin A \sin B\):
\(\cos 2\theta \cos 45^\circ - \sin 2\theta \sin 45^\circ = 2 \sin 2\theta\)

Since \(\cos 45^\circ = \sin 45^\circ = \frac{1}{\sqrt{2}}\):
\(\frac{1}{\sqrt{2}}\cos 2\theta - \frac{1}{\sqrt{2}}\sin 2\theta = 2 \sin 2\theta\)

Multiply the entire equation by \(\sqrt{2}\):
\(\cos 2\theta - \sin 2\theta = 2\sqrt{2} \sin 2\theta\)
\(\cos 2\theta = (2\sqrt{2} + 1) \sin 2\theta\)

Divide by \(\cos 2\theta\) (since \(\cos 2\theta \neq 0\)) to express in terms of tangent:
\(\tan 2\theta = \frac{1}{2\sqrt{2} + 1} \approx 0.26120\)

Since \(0^\circ < \theta < 180^\circ\), we have \(0^\circ < 2\theta < 360^\circ\):
\(2\theta = \tan^{-1}(0.26120) \approx 14.64^\circ\)
or \(2\theta = 180^\circ + 14.64^\circ = 194.64^\circ\)

Solving for \(\theta\):
\(\theta \approx 7.3^\circ\) or \(\theta \approx 97.3^\circ\) (to 1 decimal place).

- **M1**: For using the compound angle formula to expand \(\cos(2\theta + 45^\circ)\).
- **A1**: For substituting the exact values of \(\sin 45^\circ\) and \(\cos 45^\circ\).
- **M1**: For collecting terms to obtain an equation of the form \(\tan 2\theta = k\).
- **A1**: For obtaining \(\tan 2\theta = \frac{1}{2\sqrt{2} + 1}\) or equivalent.
- **M1**: For finding one correct value of \(2\theta\) in the range.
- **A1**: For \(\theta = 7.3^\circ\).
- **A1.1**: For \(\theta = 97.3^\circ\). Deduct 1 mark for any extra angles in the range.

(i) Differentiate \(x\) and \(y\) with respect to \(t\):
\(\frac{dx}{dt} = \frac{2}{2t + 1}\)
\(\frac{dy}{dt} = 2t - 3\)

Using the chain rule:
\(\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t - 3}{\frac{2}{2t + 1}} = \frac{(2t - 3)(2t + 1)}{2} = 2t^2 - 2t - 1.5\)

(ii) Set \(\frac{dy}{dx} = 2.5\):
\(2t^2 - 2t - 1.5 = 2.5\)
\(2t^2 - 2t - 4 = 0\)
\(t^2 - t - 2 = 0\)
\((t - 2)(t + 1) = 0\)

Since \(t > -0.5\), we discard \(t = -1\). Thus, \(t = 2\).

Find the coordinates at \(t = 2\):
\(x = \ln(2(2) + 1) = \ln 5\)
\(y = 2^2 - 3(2) = 4 - 6 = -2\)

The coordinates of the point are \((\ln 5, -2)\).

**(i)**
- **M1**: For differentiating \(x\) correctly to get \(\frac{2}{2t+1}\).
- **M1**: For differentiating \(y\) correctly to get \(2t - 3\).
- **A1.1**: For obtaining \(\frac{dy}{dx} = \frac{(2t - 3)(2t + 1)}{2}\) (or equivalent form).

**(ii)**
- **M1**: For setting their expression for \(\frac{dy}{dx}\) equal to \(2.5\) and attempting to solve the quadratic equation.
- **A1**: For obtaining \(t = 2\) and explaining why \(t = -1\) is rejected.
- **M1**: For substituting \(t = 2\) back into both parametric equations.
- **A1**: For finding the exact coordinates \((\ln 5, -2)\).

The area \(A\) is given by:
\(A = \int_{0}^{\pi/6} \left( 2x + 1 + \cos^2(3x) \right) dx\)

First, use the double-angle identity for \(\cos^2(3x)\):
\(\cos^2(3x) = \frac{1}{2}(1 + \cos 6x)\)

Substitute this into the integrand:
\(y = 2x + 1 + \frac{1}{2} + \frac{1}{2}\cos 6x = 2x + \frac{3}{2} + \frac{1}{2}\cos 6x\)

Now integrate term by term:
\(\int_{0}^{\pi/6} \left( 2x + \frac{3}{2} + \frac{1}{2}\cos 6x \right) dx = \left[ x^2 + \frac{3}{2}x + \frac{1}{12}\sin 6x \right]_{0}^{\pi/6}\)

Evaluate at the limits:
At \(x = \frac{\pi}{6}\):
\(\left(\frac{\pi}{6}\right)^2 + \frac{3}{2}\left(\frac{\pi}{6}\right) + \frac{1}{12}\sin\left(6 \cdot \frac{\pi}{6}\right) = \frac{\pi^2}{36} + \frac{\pi}{4} + \frac{1}{12}\sin\pi = \frac{\pi^2}{36} + \frac{\pi}{4}\)

At \(x = 0\):
\(0^2 + 0 + \frac{1}{12}\sin 0 = 0\)

Thus, the exact area is:
\(\frac{\pi^2}{36} + \frac{\pi}{4}\)

- **M1**: For using the double-angle identity to express \(\cos^2(3x)\) in terms of \(\cos 6x\).
- **A1**: For obtaining the correct simplified integrand: \(2x + \frac{3}{2} + \frac{1}{2}\cos 6x\).
- **M1**: For integrating the linear terms to get \(x^2 + \frac{3}{2}x\).
- **M1**: For integrating \(\cos 6x\) to get \(k \sin 6x\).
- **A1**: For the correct integrated expression: \(x^2 + \frac{3}{2}x + \frac{1}{12}\sin 6x\).
- **M1**: For substituting limits of \(\frac{\pi}{6}\) and \(0\) into their integrated expression.
- **A1.1**: For the final exact answer \(\frac{\pi^2}{36} + \frac{\pi}{4}\).

(i) Let \(f(x) = x^3 - 5x - 3\).
\(f(2.4) = (2.4)^3 - 5(2.4) - 3 = 13.824 - 12 - 3 = -1.176 < 0\)
\(f(2.5) = (2.5)^3 - 5(2.5) - 3 = 15.625 - 12.5 - 3 = 0.125 > 0\)

Since there is a change of sign and \(f(x)\) is continuous, \(\alpha\) lies between \(2.4\) and \(2.5\).

(ii) From \(x^3 - 5x - 3 = 0\):
\(x^3 = 5x + 3\)
Divide by \(x\) (since \(x > 0\)):
\(x^2 = 5 + \frac{3}{x}\)
Since \(x > 0\), we take the positive square root:
\(x = \sqrt{5 + \frac{3}{x}}\)

(iii) Applying the iterative formula:
\(x_1 = 2.4\)
\(x_2 = \sqrt{5 + \frac{3}{2.4}} = 2.50000\)
\(x_3 = \sqrt{5 + \frac{3}{2.5}} \approx 2.48998\)
\(x_4 = \sqrt{5 + \frac{3}{2.48998}} \approx 2.49095\)
\(x_5 = \sqrt{5 + \frac{3}{2.49095}} \approx 2.49086\)
\(x_6 = \sqrt{5 + \frac{3}{2.49086}} \approx 2.49087\)

Thus, \(\alpha \approx 2.491\) correct to 3 decimal places.

**(i)**
- **M1**: For evaluating \(f(2.4)\) and \(f(2.5)\) with at least one correct calculation.
- **A1.1**: For obtaining correct values \(-1.176\) and \(0.125\) and concluding with a valid statement about the sign change.

**(ii)**
- **B1**: For showing clear algebraic steps to rewrite the cubic equation into the required form.

**(iii)**
- **M1**: For calculating \(x_2\) and \(x_3\) correctly.
- **A1**: For obtaining \(x_2 = 2.50000\) and \(x_3 = 2.48998\).
- **A1**: For obtaining subsequent iterations to 5 decimal places: \(x_4 \approx 2.49095\), \(x_5 \approx 2.49086\), and \(x_6 \approx 2.49087\).
- **A1**: For concluding that \(\alpha = 2.491\) and showing sufficient convergence to justify this rounding.

(i) Perform polynomial division or equate coefficients:
\(4x^2 + 8x + 5 \equiv (2x + 1)(ax + b) + c\)
\(4x^2 + 8x + 5 \equiv 2ax^2 + (a + 2b)x + (b + c)\)

Equating coefficients:
For \(x^2\): \(2a = 4 \Rightarrow a = 2\)
For \(x\): \(a + 2b = 8 \Rightarrow 2 + 2b = 8 \Rightarrow b = 3\)
For the constant term: \(b + c = 5 \Rightarrow 3 + c = 5 \Rightarrow c = 2\)

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

(ii) Integrate \(f(x)\) from \(0\) to \(2\):
\(\int_{0}^{2} \left( 2x + 3 + \frac{2}{2x+1} \right) dx\)
\(= \left[ x^2 + 3x + \ln|2x + 1| \right]_{0}^{2}\)

Substitute the upper limit \(x = 2\):
\(2^2 + 3(2) + \ln|2(2) + 1| = 4 + 6 + \ln 5 = 10 + \ln 5\)

Substitute the lower limit \(x = 0\):
\(0^2 + 3(0) + \ln|1| = 0\)

Subtract the lower limit:
\((10 + \ln 5) - 0 = 10 + \ln 5\)

**(i)**
- **M1**: For an appropriate method to divide polynomials or equate coefficients.
- **A1**: For obtaining \(a = 2\) and \(b = 3\).
- **A1.1**: For obtaining \(c = 2\).

**(ii)**
- **M1**: For integrating \(2x + 3\) to obtain \(x^2 + 3x\).
- **M1**: For integrating \(\frac{2}{2x+1}\) to obtain \(\ln|2x+1|\).
- **M1**: For substituting limits of \(2\) and \(0\) into their integrated expression.
- **A1**: For obtaining the final exact answer \(10 + \ln 5\) (with no decimals).

Let \(z = x + \mathrm{i}y\) be a square root of \(21 - 20\mathrm{i}\), where \(x\) and \(y\) are real numbers.

Squaring both sides:
\((x + \mathrm{i}y)^2 = 21 - 20\mathrm{i} \Rightarrow (x^2 - y^2) + 2xy\mathrm{i} = 21 - 20\mathrm{i}\)

Equating real and imaginary parts:
1) \(x^2 - y^2 = 21\)
2) \(2xy = -20 \Rightarrow y = -\frac{10}{x}\)

Substitute \(y = -\frac{10}{x}\) into equation 1:
\(x^2 - \left(-\frac{10}{x}\right)^2 = 21\)
\(x^2 - \frac{100}{x^2} = 21\)
\(x^4 - 21x^2 - 100 = 0\)

Factorise the quadratic in \(x^2\):
\((x^2 - 25)(x^2 + 4) = 0\)

Since \(x\) is real, \(x^2 = 25\), which gives \(x = \pm 5\).

If \(x = 5\), then \(y = -\frac{10}{5} = -2\).
If \(x = -5\), then \(y = -\frac{10}{-5} = 2\).

Thus, the square roots are \(5 - 2\mathrm{i}\) and \(-5 + 2\mathrm{i}\).

M1: Attempt to square \(x + \mathrm{i}y\) and equate real and imaginary parts.
A1: Obtain \(x^2 - y^2 = 21\) and \(2xy = -20\).
M1: Eliminate one variable to obtain a 3-term quartic in \(x\) (or \(y\)).
A1: Obtain \(x^4 - 21x^2 - 100 = 0\) (or equivalent for \(y\)).
M1: Solve the quartic to find real values for \(x\) (or \(y\)).
A1.8: Obtain both correct complex numbers: \(5 - 2\mathrm{i}\) and \(-5 + 2\mathrm{i}\).

Let \(\frac{5x^2 + x + 8}{(x+1)(x^2+3)} = \frac{A}{x+1} + \frac{Bx+C}{x^2+3}\).

Multiply both sides by the denominator:
\(5x^2 + x + 8 = A(x^2+3) + (Bx+C)(x+1)\)

Let \(x = -1\):
\(5(-1)^2 + (-1) + 8 = A((-1)^2+3) + 0\)
\(12 = 4A \Rightarrow A = 3\)

Equating coefficients of \(x^2\):
\(5 = A + B \Rightarrow 5 = 3 + B \Rightarrow B = 2\)

Equating constant terms:
\(8 = 3A + C \Rightarrow 8 = 3(3) + C \Rightarrow 8 = 9 + C \Rightarrow C = -1\)

Thus, \(\frac{5x^2 + x + 8}{(x+1)(x^2+3)} = \frac{3}{x+1} + \frac{2x-1}{x^2+3}\).

M1: State or imply the correct form of partial fractions \(\frac{A}{x+1} + \frac{Bx+C}{x^2+3}\).
A1: Correctly form the identity \(5x^2 + x + 8 = A(x^2+3) + (Bx+C)(x+1)\).
M1: Apply a correct method to find at least one constant (e.g., substituting \(x = -1\) or equating coefficients).
A1: Obtain \(A = 3\).
A1: Obtain \(B = 2\).
A1.8: Obtain \(C = -1\) and write final answer clearly.

Using the laws of logarithms:
\(\ln(x+6) - \ln(x^2) = \ln(2)\)
\(\ln\left(\frac{x+6}{x^2}\right) = \ln(2)\)

Taking exponentials of both sides:
\(\frac{x+6}{x^2} = 2\)
\(x+6 = 2x^2\)
\(2x^2 - x - 6 = 0\)

Factorise the quadratic equation:
\((2x+3)(x-2) = 0\)

This gives two potential solutions:
\(x = -\frac{3}{2}\) or \(x = 2\).

However, for the term \(\ln(x)\) to be defined, we must have \(x > 0\).
Therefore, we reject \(x = -\frac{3}{2}\).

The only valid solution is \(x = 2\).

M1: Apply logarithm law \(\ln(a) - \ln(b) = \ln(a/b)\) or \(2\ln(x) = \ln(x^2)\).
A1: Obtain correct equation \(\frac{x+6}{x^2} = 2\).
M1: Solve the resulting quadratic equation to find two roots.
A1: Obtain \(x = 2\) and \(x = -1.5\).
M1: Identify that \(x\) must be positive for \(\ln(x)\) to be defined.
A1.8: State \(x = 2\) as the unique solution, clearly rejecting \(x = -1.5\).

Express \(12\cos(2\theta) + 5\sin(2\theta)\) in the form \(R\cos(2\theta - \alpha)\).
\(R\cos(2\theta - \alpha) = R\cos(2\theta)\cos\alpha + R\sin(2\theta)\sin\alpha\)

Comparing coefficients:
\(R\cos\alpha = 12\)
\(R\sin\alpha = 5\)

\(R = \sqrt{12^2 + 5^2} = 13\)

\(\tan\alpha = \frac{5}{12} \Rightarrow \alpha \approx 22.62^\circ\)

So the equation becomes:
\(13\cos(2\theta - 22.62^\circ) = 6.5\)
\(\cos(2\theta - 22.62^\circ) = 0.5\)

Since \(0^\circ \le \theta \le 180^\circ\), the interval for \((2\theta - 22.62^\circ)\) is:
\(-22.62^\circ \le 2\theta - 22.62^\circ \le 337.38^\circ\)

Solving within this interval:
\(2\theta - 22.62^\circ = 60^\circ \Rightarrow 2\theta = 82.62^\circ \Rightarrow \theta \approx 41.3^\circ\)

\(2\theta - 22.62^\circ = 300^\circ \Rightarrow 2\theta = 322.62^\circ \Rightarrow \theta \approx 161.3^\circ\)

The solutions are \(\theta = 41.3^\circ\) and \(\theta = 161.3^\circ\).

M1: State correct equations for \(R\) and \(\alpha\) or use a valid identity method.
A1: Obtain \(R = 13\) and \(\alpha = 22.62^\circ\) (or 22.6°).
M1: Formulate the equation \(\cos(2\theta - 22.62^\circ) = 0.5\).
A1: Obtain one correct value for \(2\theta - \alpha\) (e.g. \(60^\circ\) or \(300^\circ\)).
M1: Carry out a correct process to find at least one value of \(\theta\).
A1.8: Obtain both correct answers \(\theta = 41.3^\circ\) and \(\theta = 161.3^\circ\), and no others in the range.

First, differentiate \(x\) and \(y\) with respect to \(t\):
\(\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{2}{2t+1}\)

Using the quotient rule for \(y\):
\(\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{(2t)(t+2) - (t^2)(1)}{(t+2)^2} = \frac{2t^2 + 4t - t^2}{(t+2)^2} = \frac{t^2 + 4t}{(t+2)^2}\)

Using the chain rule:
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y}{\mathrm{d}t} \div \frac{\mathrm{d}x}{\mathrm{d}t} = \frac{t^2+4t}{(t+2)^2} \times \frac{2t+1}{2} = \frac{(t^2+4t)(2t+1)}{2(t+2)^2}\)

At \(t = 1\):
\(\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{2}{3}\)

\(\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{1+4}{3^2} = \frac{5}{9}\)

So, the gradient of the tangent is:
\(m = \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{5/9}{2/3} = \frac{5}{6}\)

Find the coordinates of the point at \(t = 1\):
\(x = \ln(2(1)+1) = \ln(3)\)

\(y = \frac{1^2}{1+2} = \frac{1}{3}\)

The equation of the tangent is:
\(y - \frac{1}{3} = \frac{5}{6}(x - \ln(3))\)

\(y = \frac{5}{6}x - \frac{5}{6}\ln(3) + \frac{1}{3}\)

\(y = \frac{5}{6}x + \frac{2 - 5\ln(3)}{6}\)

M1: Differentiate \(x\) with respect to \(t\) to obtain \(\frac{2}{2t+1}\).
M1: Use quotient rule to differentiate \(y\) with respect to \(t\).
A1: Obtain correct derivative \(\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{t^2+4t}{(t+2)^2}\).
M1: Calculate numerical value of \(\frac{\mathrm{d}y}{\mathrm{d}x}\) at \(t=1\).
A1: Obtain gradient \(m = \frac{5}{6}\) and points \(x = \ln(3)\), \(y = \frac{1}{3}\).
A1.8: Obtain correct equation \(y = \frac{5}{6}x + \frac{2 - 5\ln(3)}{6}\) (or equivalent form).

We use integration by parts, \(\int u \, \mathrm{d}v = uv - \int v \, \mathrm{d}u\).

Let \(u = \ln(x)\) and \(\mathrm{d}v = x \, \mathrm{d}x\).
Then \(\mathrm{d}u = \frac{1}{x} \, \mathrm{d}x\) and \(v = \frac{1}{2}x^2\).

Applying integration by parts:
\(\int x \ln(x) \, \mathrm{d}x = \left[ \frac{1}{2}x^2 \ln(x) \right] - \int \frac{1}{2}x^2 \cdot \frac{1}{x} \, \mathrm{d}x\)

\(= \left[ \frac{1}{2}x^2 \ln(x) \right] - \int \frac{1}{2}x \, \mathrm{d}x\)

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

Now, apply the limits from \(1\) to \(e\):
\(\left( \frac{1}{2}e^2 \ln(e) - \frac{1}{4}e^2 \right) - \left( \frac{1}{2}(1)^2 \ln(1) - \frac{1}{4}(1)^2 \right)\)

\(= \left( \frac{1}{2}e^2 - \frac{1}{4}e^2 \right) - \left( 0 - \frac{1}{4} \right)\)

\(= \frac{1}{4}e^2 + \frac{1}{4}\)

\(= \frac{1}{4}(e^2 + 1)\)

M1: Identify integration by parts as the method and choose correct \(u = \ln(x)\), \(\mathrm{d}v = x \, \mathrm{d}x\).
A1: Obtain \(\mathrm{d}u = \frac{1}{x} \, \mathrm{d}x\) and \(v = \frac{1}{2}x^2\).
M1: Substitute into integration by parts formula to get \(\frac{1}{2}x^2 \ln(x) - \int \frac{1}{2}x \, \mathrm{d}x\).
A1: Complete the integration to get \(\frac{1}{2}x^2 \ln(x) - \frac{1}{4}x^2\).
M1: Correctly substitute limits \(1\) and \(e\) into their expression.
A1.8: Obtain final exact answer \(\frac{1}{4}(e^2 + 1)\) (or equivalent).

To find the intersection, equate the components of \(r_1\) and \(r_2\):
1) \(1 + 2\lambda = 4 - \mu\)
2) \(-2 + \lambda = -3 + 2\mu\)
3) \(4 - \lambda = 1 + 2\mu\)

Rearranging equations (1) and (2):
1) \(2\lambda + \mu = 3 \Rightarrow \mu = 3 - 2\lambda\)
2) \(\lambda - 2\mu = -1\)

Substitute (1) into (2):
\(\lambda - 2(3 - 2\lambda) = -1\)
\(\lambda - 6 + 4\lambda = -1\)
\(5\lambda = 5 \Rightarrow \lambda = 1\)

Substitute \(\lambda = 1\) back into (1):
\(\mu = 3 - 2(1) = 1\)

Now, verify that these values of \(\lambda\) and \(\mu\) satisfy the third component equation:
LHS of (3): \(4 - 1 = 3\)
RHS of (3): \(1 + 2(1) = 3\)
Since LHS = RHS, the lines intersect.

To find the point of intersection, substitute \(\lambda = 1\) into \(r_1\):
\(r_1 = \begin{pmatrix} 1 \\ -2 \\ 4 \end{pmatrix} + 1 \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 3 \\ -1 \\ 3 \end{pmatrix}\).

M1: Equate components of \(r_1\) and \(r_2\) to set up three equations in \(\lambda\) and \(\mu\).
A1: Correctly solve a pair of equations to find values for \(\lambda\) and \(\mu\).
A1: Obtain \(\lambda = 1\) and \(\mu = 1\).
M1: Verify the third equation is satisfied using these values to show intersection.
M1: Substitute \(\lambda\) or \(\mu\) back to find the position vector.
A1.8: Obtain the correct position vector \(\begin{pmatrix} 3 \\ -1 \\ 3 \end{pmatrix}\).

Separate variables:
\(\int \frac{1}{y^2} \, \mathrm{d}y = \int \frac{\cos x}{1 + 2\sin x} \, \mathrm{d}x\)

Integrate both sides:
\(\int y^{-2} \, \mathrm{d}y = -\frac{1}{y}\)

For the RHS, let \(u = 1 + 2\sin x\), so \(\mathrm{d}u = 2\cos x \, \mathrm{d}x\).
\(\int \frac{\cos x}{1 + 2\sin x} \, \mathrm{d}x = \frac{1}{2} \ln|1 + 2\sin x|\)

So, the general solution is:
\(-\frac{1}{y} = \frac{1}{2} \ln(1 + 2\sin x) + C\)

Use the boundary condition \(y = 1\) when \(x = 0\):
\(-\frac{1}{1} = \frac{1}{2} \ln(1 + 0) + C \Rightarrow -1 = C\)

Thus, the particular solution is:
\(-\frac{1}{y} = \frac{1}{2} \ln(1 + 2\sin x) - 1\)

\(\frac{1}{y} = 1 - \frac{1}{2} \ln(1 + 2\sin x)\)

\(\frac{1}{y} = \frac{2 - \ln(1 + 2\sin x)}{2}\)

\(y = \frac{2}{2 - \ln(1 + 2\sin x)}\)

M1: Separate variables correctly to obtain \(\int \frac{1}{y^2} \, \mathrm{d}y = \int \frac{\cos x}{1 + 2\sin x} \, \mathrm{d}x\).
A1: Integrate LHS to get \(-\frac{1}{y}\).
M1: Integrate RHS to obtain \(k \ln|1 + 2\sin x|\) form.
A1: Obtain correct RHS integration \(\frac{1}{2} \ln(1 + 2\sin x) + C\).
M1: Apply boundary condition \(y = 1\) when \(x = 0\) to find \(C\).
A1.8: Correctly solve for \(y\) to obtain \(y = \frac{2}{2 - \ln(1 + 2\sin x)}\).

We use the quadratic formula to solve \(z^2 - 2i\sqrt{3}z - (5 + 2i\sqrt{3}) = 0\):

\[z = \frac{2i\sqrt{3} \pm \sqrt{(-2i\sqrt{3})^2 - 4(1)(-(5 + 2i\sqrt{3}))}}{2}\]

First, simplify the discriminant \(\Delta\):
\[\Delta = (-2i\sqrt{3})^2 + 4(5 + 2i\sqrt{3}) = -12 + 20 + 8i\sqrt{3} = 8 + 8i\sqrt{3}\]

We need to find the square roots of \(8 + 8i\sqrt{3}\). Let \((u + iv)^2 = 8 + 8i\sqrt{3}\), where \(u\) and \(v\) are real.
Comparing real and imaginary parts:
1) \(u^2 - v^2 = 8\)
2) \(2uv = 8\sqrt{3} \implies uv = 4\sqrt{3}\)

Using the modulus equation:
\(u^2 + v^2 = \sqrt{8^2 + (8\sqrt{3})^2} = \sqrt{64 + 192} = \sqrt{256} = 16\)

Adding equation (1) and the modulus equation:
\[2u^2 = 24 \implies u^2 = 12 \implies u = \pm 2\sqrt{3}\]

Subtracting equation (1) from the modulus equation:
\[2v^2 = 8 \implies v^2 = 4 \implies v = \pm 2\]

Since \(uv > 0\), \(u\) and \(v\) must have the same sign. Thus, the square roots are \(\pm (2\sqrt{3} + 2i)\).

Now substitute this back into the expression for \(z\):
\[z = \frac{2i\sqrt{3} \pm (2\sqrt{3} + 2i)}{2} = i\sqrt{3} \pm (\sqrt{3} + i)\]

This yields two solutions:
\[z_1 = \sqrt{3} + i(\sqrt{3} + 1)\]
\[z_2 = -\sqrt{3} + i(\sqrt{3} - 1)\]

M1: Apply the quadratic formula correctly and identify the discriminant as \(8 + 8i\sqrt{3}\).
M1: Set up simultaneous equations to find the square root of the complex discriminant.
A1: Correctly find the square roots as \(\pm (2\sqrt{3} + 2i)\).
M1: Substitute the square roots back into the quadratic formula expression.
A1: Obtain the first root \(z = \sqrt{3} + i(\sqrt{3} + 1)\) or equivalent exact form.
A1: Obtain the second root \(z = -\sqrt{3} + i(\sqrt{3} - 1)\) or equivalent exact form.

We begin by simplifying the integrand using logarithm properties:
\[\ln(x^2) = 2\ln x\]

Thus, the integral becomes:
\[I = 2 \int_1^{\sqrt{e}} x^3 \ln x \, dx\]

We integrate \(\int x^3 \ln x \, dx\) using integration by parts:
Let \(u = \ln x \implies du = \frac{1}{x} \, dx\)
Let \(dv = x^3 \, dx \implies v = \frac{x^4}{4}\]

Using the formula \)\int u \, dv = uv - \int v \, du\):
\[\int x^3 \ln x \, dx = \frac{x^4}{4} \ln x - \int \frac{x^4}{4} \cdot \frac{1}{x} \, dx\]
\[= \frac{x^4}{4} \ln x - \frac{1}{4} \int x^3 \, dx = \frac{x^4}{4} \ln x - \frac{x^4}{16}\]

Multiplying by 2, we have:
\[I = 2 \left[ \frac{x^4}{4} \ln x - \frac{x^4}{16} \right]_1^{\sqrt{e}} = \left[ \frac{x^4}{2} \ln x - \frac{x^4}{8} \right]_1^{\sqrt{e}}\]

Now, substitute the upper and lower limits:
At \(x = \sqrt{e}\):
\[\frac{(\sqrt{e})^4}{2} \ln(\sqrt{e}) - \frac{(\sqrt{e})^4}{8} = \frac{e^2}{2} \left(\frac{1}{2}\right) - \frac{e^2}{8} = \frac{e^2}{4} - \frac{e^2}{8} = \frac{e^2}{8}\]

At \(x = 1\):
\[\frac{1^4}{2} \ln(1) - \frac{1^4}{8} = 0 - \frac{1}{8} = -\frac{1}{8}\]

Subtract the lower limit evaluation from the upper limit evaluation:
\[I = \frac{e^2}{8} - \left(-\frac{1}{8}\right) = \frac{e^2 + 1}{8}\]

M1: Use logarithmic identity \(\ln(x^2) = 2 \ln x\) to simplify the integrand (or apply integration by parts directly with \(u = \ln(x^2)\)).
M1: Correctly apply integration by parts, assigning \(u = \ln x\) and \(dv = x^3\).
A1: Obtain correct first-stage integration term, e.g., \(\frac{x^4}{4}\ln(x^2) - \int \frac{x^3}{2} \, dx\).
A1: Obtain the correct integrated expression: \(\frac{x^4}{2} \ln x - \frac{x^4}{8}\) (or equivalent).
M1: Substitute the limits \(\sqrt{e}\) and \(1\) into their integrated expression.
A1: Obtain the final exact value \(\frac{e^2 + 1}{8}\) (or equivalent single fraction).

We separate variables to solve the differential equation:
\[\frac{1}{y^2 + 1} \, dy = \frac{1}{x^{1.5}} \, dx\]
\[\int \frac{1}{y^2 + 1} \, dy = \int x^{-3/2} \, dx\]

Integrate both sides:
\[\arctan y = \frac{x^{-1/2}}{-1/2} + c\]
\[\arctan y = -\frac{2}{\sqrt{x}} + c\]

Using the given initial condition, \(y = 0\) when \(x = 1\):
\[\arctan(0) = -\frac{2}{\sqrt{1}} + c\]
\[0 = -2 + c \implies c = 2\]

So the equation is:
\[\arctan y = 2 - \frac{2}{\sqrt{x}}\]

Taking the tangent of both sides to express \(y\) explicitly in terms of \(x\):
\[y = \tan\left(2 - \frac{2}{\sqrt{x}}\right)\]

M1: Separate variables correctly, obtaining \(\int \frac{1}{y^2 + 1} \, dy = \int x^{-3/2} \, dx\).
A1: Obtain \(\arctan y\) on the left-hand side.
A1: Obtain \(-2x^{-1/2}\) (or equivalent) on the right-hand side.
M1: Substitute boundary conditions \(x = 1, y = 0\) into an integrated equation containing a constant \(c\).
A1: Obtain \(c = 2\).
A1: Rearrange and obtain \(y = \tan\left(2 - \frac{2}{\sqrt{x}}\right)\) as the final explicit function.

First, we find the driving force F of the engine using the formula P = F * v. Since P = 24000 W and v = 15 m/s, we have F = 24000 / 15 = 1600 N. Next, we resolve forces parallel to the slope. The component of the weight acting down the slope is m * g * \sin\theta = 1200 * 10 * 0.05 = 600 N. The resistance force is 300 N acting down the slope. Applying Newton's second law along the slope: F - R - m * g * \sin\theta = m * a. Substituting the known values: 1600 - 300 - 600 = 1200 * a, which simplifies to 700 = 1200 * a. Therefore, a = 700 / 1200 = 7/12 = 0.583 m/s^2.

M1: For using P = F * v to find the driving force. A1: For obtaining F = 1600 N. M1: For calculating the component of weight down the slope as 600 N. M1: For setting up the Newton's second law equation of motion. A1: For obtaining the correct equation 1600 - 300 - 600 = 1200 * a. A1.1: For the final answer 0.583 m/s^2 (or 7/12 m/s^2).

The normal contact force is R = m * g * \cos(30) = 8 * 10 * \cos(30) = 40 * \sqrt{3} N. The maximum frictional force is F_f = \mu * R = 0.4 * 40 * \sqrt{3} = 16 * \sqrt{3} N. The component of the weight down the slope is W_p = m * g * \sin(30) = 8 * 10 * 0.5 = 40 N. Case 1: The block is on the point of slipping down the plane. The frictional force acts up the plane. Resolving parallel to the plane: F + F_f = W_p => F = 40 - 16 * \sqrt{3} = 12.3 N. Case 2: The block is on the point of slipping up the plane. The frictional force acts down the plane. Resolving parallel to the plane: F = W_p + F_f => F = 40 + 16 * \sqrt{3} = 67.7 N. Thus, the range of possible values of F is 12.3 <= F <= 67.7.

M1: For resolving forces perpendicular to the plane to find R. A1: For obtaining R = 40*\sqrt{3} N or 69.3 N. M1: For calculating the maximum frictional force using F_f = \mu * R. M1: For considering the equilibrium equation when the block is about to slip down. A1: For finding the lower bound F = 12.3 N. M1: For considering the equilibrium equation when the block is about to slip up. A0.1: For obtaining the final range 12.3 <= F <= 67.7.

We first find the times when the particle is instantaneously at rest by setting v = 0. This gives 3*t^2 - 12*t + 9 = 0, which factorizes to 3*(t-1)*(t-3) = 0. Thus, the direction of motion changes at t = 1 and t = 3. To find the total distance, we integrate the velocity to find the displacement function s(t) = t^3 - 6*t^2 + 9*t + C. Taking s(0) = 0, we evaluate s(t) at the boundaries and turning points: s(0) = 0, s(1) = 4, s(3) = 0, s(4) = 4. The distance travelled in each interval is: from t=0 to t=1: |4 - 0| = 4 m; from t=1 to t=3: |0 - 4| = 4 m; from t=3 to t=4: |4 - 0| = 4 m. Total distance = 4 + 4 + 4 = 12 m.

M1: For setting v = 0 and solving for t. A1: For obtaining t = 1 and t = 3. M1: For integrating v with respect to t to obtain displacement s(t). A1: For obtaining s(t) = t^3 - 6*t^2 + 9*t. M1: For evaluating s(t) at t = 0, 1, 3, 4. M1: For summing the absolute differences of the displacements. A0.1: For the correct total distance of 12 m.

Let a be the acceleration of the system and T be the tension in the string. For the heavier particle B, which moves downwards, the equation of motion is: 0.5 * g - T = 0.5 * a => 5 - T = 0.5 * a. For the lighter particle A, which moves upwards, the equation of motion is: T - 0.3 * g = 0.3 * a => T - 3 = 0.3 * a. Adding these two equations gives 2 = 0.8 * a, which leads to a = 2.5 m/s^2. Since B starts from rest, we use the equation of motion s = u * t + 0.5 * a * t^2. Substituting s = 1.6, u = 0, and a = 2.5 gives: 1.6 = 0.5 * 2.5 * t^2 => 1.6 = 1.25 * t^2 => t^2 = 1.28 => t = 1.13 s (to 3 s.f.).

M1: For applying Newton's second law to particle B. A1: For obtaining 5 - T = 0.5*a. M1: For applying Newton's second law to particle A. A1: For obtaining T - 3 = 0.3*a. M1: For solving the simultaneous equations to find acceleration a. A1: For obtaining a = 2.5 m/s^2. A0.1: For using s = 0.5*a*t^2 to find t = 1.13 s.

Let the direction of P's initial velocity be the positive direction. P's initial velocity is u_P = 4 m/s, and Q's initial velocity is u_Q = -v m/s since it is moving towards P. After the collision, P rebounds, so its velocity becomes v_P = -1 m/s. Q continues in its original direction, so its velocity is v_Q = -2 m/s. By conservation of linear momentum: m_P * u_P + m_Q * u_Q = m_P * v_P + m_Q * v_Q. Substituting the values: 2 * 4 + 3 * (-v) = 2 * (-1) + 3 * (-2) => 8 - 3 * v = -2 - 6 => 8 - 3 * v = -8 => 3 * v = 16 => v = 16/3 = 5.33 m/s (to 3 s.f.).

M1: For using the principle of conservation of momentum. A1: For establishing correct initial momentum terms with opposite signs, e.g., 2*4 - 3*v. A1: For establishing correct final momentum terms with correct signs, e.g., 2*(-1) - 3*2. M1: For equating initial and final momentum to form an equation in v. A1: For obtaining 8 - 3*v = -8. A1.1: For solving to find v = 5.33 m/s or 16/3 m/s.

We use the work-energy principle: Net Work Done = Change in Kinetic Energy. The work done by the pulling force is W_F = F * d = 120 * 8 = 960 J. The normal contact force is R = m * g * \cos(20) = 15 * 10 * \cos(20) = 140.95 N. The frictional force is F_f = \mu * R = 0.25 * 140.95 = 35.24 N. The work done against friction is W_f = F_f * d = 35.24 * 8 = 281.9 J. The gain in gravitational potential energy is \Delta PE = m * g * d * \sin(20) = 15 * 10 * 8 * \sin(20) = 410.42 J. The change in kinetic energy is \Delta KE = W_F - W_f - \Delta PE = 960 - 281.9 - 410.42 = 267.68 J. Since the block starts from rest, \Delta KE = 0.5 * m * v^2 = 7.5 * v^2 = 267.68. Solving for v, we get v^2 = 35.69, which gives v = 5.97 m/s.

M1: For calculating the work done by the pulling force as 960 J. M1: For calculating the normal force R = 150*\cos(20). M1: For calculating the work done against friction. A1: For obtaining work against friction as 281.9 J. M1: For calculating the change in potential energy as 410.4 J. M1: For using the work-energy relation to find the kinetic energy. A0.1: For solving 7.5*v^2 = 267.68 to obtain v = 5.97 m/s.

For the system to be in equilibrium, the sum of the horizontal and vertical components of all forces must be zero. Resolving forces horizontally: 10 + 8 * \cos(60) - P * \cos(\theta) = 0 => 10 + 4 = P * \cos(\theta) => P * \cos(\theta) = 14. Resolving forces vertically: 8 * \sin(60) - P * \sin(\theta) = 0 => 4 * \sqrt{3} = P * \sin(\theta) => P * \sin(\theta) = 6.928. Dividing the vertical equation by the horizontal equation: \tan(\theta) = (4 * \sqrt{3}) / 14 = 0.49487 => \theta = 26.3 degrees. Squaring and adding both components to find P: P^2 = 14^2 + (4 * \sqrt{3})^2 = 196 + 48 = 244 => P = \sqrt{244} = 15.6 N.

M1: For resolving forces horizontally. A1: For establishing P*\cos(\theta) = 14. M1: For resolving forces vertically. A1: For establishing P*\sin(\theta) = 4*\sqrt{3} (or 6.93). M1: For using division to find \tan(\theta) and solving for \theta. A1: For obtaining \theta = 26.3. A1.1: For obtaining P = 15.6.

**(i)**
We want to find the number of different committees of 5 that can be chosen from 6 men and 6 women if there must be more women than men. The possible cases for the composition of the committee (Women, Men) are:
- Case 1: 3 Women and 2 Men
Number of ways: \(^{6}C_{3} \times ^{6}C_{2} = 20 \times 15 = 300\)
- Case 2: 4 Women and 1 Man
Number of ways: \(^{6}C_{4} \times ^{6}C_{1} = 15 \times 6 = 90\)
- Case 3: 5 Women and 0 Men
Number of ways: \(^{6}C_{5} \times ^{6}C_{0} = 6 \times 1 = 6\)

Total number of ways = \(300 + 90 + 6 = 396\).

**(ii)**
The committee of 5 must contain at least one man and at least one woman, and Alice and Beatrice cannot both serve on the committee.
We consider two mutually exclusive cases:

- **Case 1: Neither Alice nor Beatrice is on the committee.**
We choose 5 people from the remaining 10 people (6 men, 4 other women).
Total ways to choose 5 from 10: \(^{10}C_{5} = 252\).
We must exclude committees with no men (all 5 women): \(^{4}C_{5} = 0\).
We must exclude committees with no women (all 5 men): \(^{6}C_{5} = 6\).
Valid ways for Case 1 = \(252 - 6 = 246\).

- **Case 2: Exactly one of Alice or Beatrice is on the committee.**
There are 2 choices (either Alice is on and Beatrice is off, or vice versa).
For each choice, we select the remaining 4 members from the 10 remaining people (6 men, 4 other women).
Total ways to choose 4 from 10: \(^{10}C_{4} = 210\).
Since we already have 1 woman (Alice or Beatrice), we already have at least one woman.
We only need to exclude committees with no men (all 4 chosen are women): \(^{4}C_{4} = 1\).
Valid ways for each choice = \(210 - 1 = 209\).
Total valid ways for Case 2 = \(2 \times 209 = 418\).

Summing the cases:
Total number of ways = \(246 + 418 = 664\).

**(i)**
* **M1**: For identifying the three valid cases of (W, M): (3,2), (4,1), (5,0).
* **M1**: For calculating the combinations for at least one case correctly.
* **A1**: For the correct final sum of 396.

**(ii)**
* **M1**: For a clear strategy identifying mutually exclusive cases or subtracting invalid combinations from total.
* **M1**: For correctly evaluating the case with neither Alice nor Beatrice.
* **M1**: For correctly evaluating the case with exactly one of Alice or Beatrice.
* **M1**: For adding the two correct cases together.
* **A1.3**: For the correct final answer of 664.

**(i)**
The sum of all probabilities in a probability distribution is 1:
\(a + 2b + c = 1 \implies a + c = 1 - 2b\) (1)

The expectation \(E(X) = 0.2\):
\(E(X) = -2a - b + b + 2c = -2a + 2c = 0.2 \implies -a + c = 0.1 \implies c = a + 0.1\) (2)

Substitute (2) into (1):
\(a + (a + 0.1) = 1 - 2b \implies 2a + 2b = 0.9 \implies b = 0.45 - a\) (3)

The variance is given by \(\text{Var}(X) = E(X^2) - [E(X)]^2 = 2.16\):
\(E(X^2) - (0.2)^2 = 2.16 \implies E(X^2) - 0.04 = 2.16 \implies E(X^2) = 2.2\)

Calculate \(E(X^2)\):
\(E(X^2) = (-2)^2 a + (-1)^2 b + 1^2 b + 2^2 c = 4a + 2b + 4c\)
So:
\(4a + 2b + 4c = 2.2 \implies 2a + b + 2c = 1.1\) (4)

Substitute (2) and (3) into (4):
\(2a + (0.45 - a) + 2(a + 0.1) = 1.1\)
\(3a + 0.65 = 1.1 \implies 3a = 0.45 \implies a = 0.15\)

Using \(a = 0.15\):
\(b = 0.45 - 0.15 = 0.3\)
\(c = 0.15 + 0.1 = 0.25\)

**(ii)**
We want to find \(P(X > E(X))\), which is \(P(X > 0.2)\).
The values of \(X\) greater than \(0.2\) are \(1\) and \(2\).
\(P(X > 0.2) = P(X = 1) + P(X = 2) = b + c = 0.3 + 0.25 = 0.55\).

**(i)**
* **M1**: For setting up the sum of probabilities equation \(a + 2b + c = 1\).
* **M1**: For setting up the expectation equation \(-2a + 2c = 0.2\).
* **M1**: For using \(\text{Var}(X) = E(X^2) - [E(X)]^2\) to find \(E(X^2) = 2.2\).
* **M1**: For setting up the equation \(4a + 2b + 4c = 2.2\) and solving the system.
* **A1**: For obtaining the correct values: \(a = 0.15, b = 0.3, c = 0.25\).

**(ii)**
* **M1**: For identifying that \(P(X > 0.2) = P(X=1) + P(X=2)\).
* **A1.3**: For the correct answer \(0.55\).

Let \(R_1\) and \(B_1\) be the events that the first ball drawn is red or blue respectively.
Let \(R_2\) be the event that the second ball drawn is red.

Initially: 4 Red, 6 Blue (10 total)
- \(P(R_1) = 0.4\)
- \(P(B_1) = 0.6\)

If \(R_1\) occurs, 2 red balls are added. The urn now has 6 Red, 6 Blue (12 total):
- \(P(R_2 | R_1) = \frac{6}{12} = 0.5\)

If \(B_1\) occurs, 2 blue balls are added. The urn now has 4 Red, 8 Blue (12 total):
- \(P(R_2 | B_1) = \frac{4}{12} = \frac{1}{3}\)

**(i)**
\(P(R_2) = P(R_1 \cap R_2) + P(B_1 \cap R_2) = P(R_1)P(R_2 | R_1) + P(B_1)P(R_2 | B_1)\)
\(P(R_2) = (0.4 \times 0.5) + (0.6 \times \frac{1}{3}) = 0.2 + 0.2 = 0.4\).

**(ii)**
Using Bayes' theorem:
\(P(B_1 | R_2) = \frac{P(B_1 \cap R_2)}{P(R_2)} = \frac{0.6 \times \frac{1}{3}}{0.4} = \frac{0.2}{0.4} = 0.5\).

**(iii)**
Two balls are drawn simultaneously from the original urn containing 4 red and 6 blue balls.
\(P(\text{different}) = \frac{^{4}C_{1} \times ^{6}C_{1}}{^{10}C_{2}} = \frac{4 \times 6}{45} = \frac{24}{45} = \frac{8}{15} \approx 0.533\).

**(i)**
* **M1**: For using the law of total probability with two distinct branches.
* **M1**: For calculating both branch probabilities correctly: \(0.4 \times 0.5\) and \(0.6 \times \frac{1}{3}\).
* **A1**: For the correct probability of \(0.4\).

**(ii)**
* **M1**: For using Bayes' theorem formula.
* **M1**: For substituting their values from part (i) correctly.
* **A1**: For the correct probability of \(0.5\).

**(iii)**
* **M1**: For a correct expression for choosing one red and one blue ball simultaneously.
* **A1.3**: For the correct answer of \(\frac{8}{15}\) or \(0.533\).

**(i)**
Given \(W \sim N(\mu, \sigma^2)\).
\(P(W < 100) = 0.1587 \implies P\left(Z < \frac{100 - \mu}{\sigma}\right) = 0.1587\).
Using standard normal tables, \(\Phi(-1.0) = 0.1587\).
So, \(\frac{100 - \mu}{\sigma} = -1.0 \implies \mu - \sigma = 100\) (1)

\(P(W > 150) = 0.0668 \implies P\left(Z > \frac{150 - \mu}{\sigma}\right) = 0.0668 \implies P\left(Z < \frac{150 - \mu}{\sigma}\right) = 0.9332\).
Using standard normal tables, \(\Phi(1.5) = 0.9332\).
So, \(\frac{150 - \mu}{\sigma} = 1.5 \implies \mu + 1.5\sigma = 150\) (2)

Subtracting (1) from (2):
\(2.5\sigma = 50 \implies \sigma = 20\).
Substituting \(\sigma = 20\) into (1):
\(\mu - 20 = 100 \implies \mu = 120\).

**(ii)**
For a single apple, let's find the probability that its mass is greater than 135 grams:
\(P(W > 135) = P\left(Z > \frac{135 - 120}{20}\right) = P(Z > 0.75) = 1 - \Phi(0.75)\).
From tables, \(\Phi(0.75) = 0.7734\).
So, \(P(W > 135) = 1 - 0.7734 = 0.2266\).

Let \(Y\) be the number of apples in a sample of 8 that have a mass greater than 135 grams.
\(Y \sim B(8, 0.2266)\).
We want to find \(P(Y \ge 2) = 1 - P(Y = 0) - P(Y = 1)\).
\(P(Y = 0) = (0.7734)^8 \approx 0.12922\)
\(P(Y = 1) = 8 \times 0.2266 \times (0.7734)^7 \approx 0.30268\)
\(P(Y \ge 2) = 1 - 0.12922 - 0.30268 = 0.5681 \approx 0.568\).

**(i)**
* **M1**: For standardizing 100 and equating to \(-1.0\).
* **M1**: For standardizing 150 and equating to \(1.5\).
* **M1**: For setting up two simultaneous equations in \(\mu\) and \(\sigma\).
* **A1**: For showing \(\sigma = 20\) clearly.
* **A1**: For showing \(\mu = 120\) clearly.

**(ii)**
* **M1**: For standardizing 135 and finding \(p = 0.2266\).
* **M1**: For using binomial distribution formula to find \(1 - P(0) - P(1)\).
* **A1.3**: For the correct final probability of \(0.568\).

**(i)**
For Group X (15 values):
The values in ascending order are:
142, 145, 148, 151, 154, 156, 157, 159, 160, 163, 165, 168, 172, 175, 181.
- **Median**: The 8th value, which is 159.
- **Lower Quartile (\(Q_1\))**: The 4th value, which is 151.
- **Upper Quartile (\(Q_3\))**: The 12th value, which is 168.
- **Interquartile Range (IQR)**: \(Q_3 - Q_1 = 168 - 151 = 17\).

**(ii)**
For Group Y (15 values):
The values in ascending order are:
143, 146, 151, 154, 155, 158, 162, 162, 166, 167, 169, 170, 173, 175, 184.
- **Median**: The 8th value, which is 162.
- **Lower Quartile (\(Q_1\))**: The 4th value, which is 154.
- **Upper Quartile (\(Q_3\))**: The 12th value, which is 170.
- **Interquartile Range (IQR)**: \(Q_3 - Q_1 = 170 - 154 = 16\).

- **Comparison**:
Group X was generally faster because its median time (159 minutes) is lower than the median time for Group Y (162 minutes).

**(i)**
* **M1**: For identifying the median as the 8th value of Group X.
* **A1**: For the correct median of 159.
* **M1**: For finding both quartiles correctly (151 and 168).
* **A1**: For the correct IQR of 17.

**(ii)**
* **M1**: For finding the correct median of Group Y (162).
* **M1**: For finding the correct quartiles and IQR of Group Y (154, 170, IQR = 16).
* **M1**: For making a comparison based on the median values.
* **A1.3**: For concluding Group X was faster because its median time (159) is lower than Group Y's median (162).

**(i)**
Let \(X\) be the number of times a 3 is rolled in 180 rolls.
\(X \sim B(180, \frac{1}{6})\).
Using a normal approximation \(N(\mu, \sigma^2)\):
- \(\mu = np = 180 \times \frac{1}{6} = 30\)
- \(\sigma^2 = np(1-p) = 180 \times \frac{1}{6} \times \frac{5}{6} = 25 \implies \sigma = 5\)

We want to find \(P(25 \le X \le 35)\).
Applying the continuity correction:
\(P(24.5 \le X_{normal} \le 35.5)\)

Standardizing:
\(P\left(\frac{24.5 - 30}{5} \le Z \le
\frac{35.5 - 30}{5}\right) = P(-1.1 \le Z
\le 1.1)\)
\(P(-1.1 \le Z \le 1.1) = 2\Phi(1.1) - 1\)
From normal tables, \(\Phi(1.1) = 0.8643\).
So, probability = \(2(0.8643) - 1 = 1.7286 - 1 = 0.7286 \approx 0.729\).

**(ii)**
The normal approximation is justified because both \(np = 30 > 5\) and \(n(1-p) = 150 > 5\).

**(iii)**
For 8 rolls, \(Y \sim B(8, \frac{1}{6})\).
\(P(Y = 2) = ^{8}C_{2} \times \left(\frac{1}{6}\right)^2 \times \left(\frac{5}{6}\right)^6 = 28 \times \frac{1}{36} \times \frac{15625}{46656} \approx 0.260\).

**(i)**
* **M1**: For calculating the correct mean (30) and variance (25).
* **M1**: For applying the continuity correction correctly (using 24.5 and 35.5).
* **M1**: For standardizing both values using their mean and standard deviation.
* **M1**: For calculating the probability from standard normal tables (e.g. \(2\Phi(1.1) - 1\)).
* **A1**: For the correct answer of 0.729.

**(ii)**
* **B1**: For showing \(np = 30 > 5\) and \(n(1-p) = 150 > 5\).

**(iii)**
* **M1**: For using the exact binomial distribution formula with \(n = 8\), \(p = \frac{1}{6}\), and \(x = 2\).
* **A1.3**: For the correct answer of 0.260.

(i) Let \(X\) be the number of flaws in a 2 m\(^2\) sheet.
Then \(X \sim \text{Po}(1.6)\).
\(P(X = 1) = e^{-1.6} \times 1.6^1 / 1! \approx 0.32304 = 0.323\) (to 3 s.f.).

(ii) Let \(Y\) be the number of flaws in an 80 m\(^2\) roll.
Then \(Y \sim \text{Po}(64)\).
Since \(\lambda = 64 > 15\), we can approximate \(Y\) by a normal distribution: \(Y_{normal} \sim N(64, 64)\).
We require \(P(55 \le Y \le 70)\).
Applying a continuity correction:
\(P(54.5 < Y_{normal} < 70.5)\)
Standardising:
\(z_1 = \frac{54.5 - 64}{\sqrt{64}} = -1.1875\)
\(z_2 = \frac{70.5 - 64}{\sqrt{64}} = 0.8125\)
\(P(-1.1875 < Z < 0.8125) = \Phi(0.8125) - (1 - \Phi(1.1875)) = 0.7917 - (1 - 0.8825) = 0.6742 = 0.674\) (to 3 s.f.).

Part (i):
M1: For using Poisson distribution with mean 1.6.
A1: Correct probability 0.323 (accept 0.323 or 0.32).

Part (ii):
B1: For stating normal approximation with mean 64 and variance 64.
M1: For applying continuity correction correctly (using 54.5 and 70.5).
M1: For standardising with their mean and standard deviation.
A1: For obtaining both correct z-values (-1.188 and 0.813).
A1: For final correct probability 0.674 (accept 0.673 - 0.675).

(i) Since \(f(x)\) is a PDF, the total area under the curve is 1:
\(\int_1^5 k(x - 1)(5 - x) \, dx = 1\)
\(k \int_1^5 (-x^2 + 6x - 5) \, dx = 1\)
\(k \left[ -\frac{x^3}{3} + 3x^2 - 5x \right]_1^5 = 1\)
At \(x = 5\): \(-\frac{125}{3} + 75 - 25 = \frac{25}{3}\)
At \(x = 1\): \(-\frac{1}{3} + 3 - 5 = -\frac{7}{3}\)
\(k \left( \frac{25}{3} - (-\frac{7}{3}) \right) = k \left( \frac{32}{3} \right) = 1 \implies k = \frac{3}{32}\).

(ii) By symmetry of the quadratic function about its vertex at \(x = 3\) on the domain \([1, 5]\), \(\text{E}(X) = 3\).
Alternatively, \(\text{E}(X) = \int_1^5 x f(x) \, dx = \frac{3}{32} \int_1^5 (-x^3 + 6x^2 - 5x) \, dx = \frac{3}{32} \left[ -\frac{x^4}{4} + 2x^3 - \frac{5x^2}{2} \right]_1^5\)
Evaluating this gives \(\frac{3}{32} \left( 31.25 - (-0.75) \right) = \frac{3}{32} \times 32 = 3\).

(iii) \(P(X > 4) = \frac{3}{32} \int_4^5 (-x^2 + 6x - 5) \, dx = \frac{3}{32} \left[ -\frac{x^3}{3} + 3x^2 - 5x \right]_4^5\)
At \(x = 5\): \(\frac{25}{3}\)
At \(x = 4\): \(-\frac{64}{3} + 48 - 20 = \frac{20}{3}\)
\(P(X > 4) = \frac{3}{32} \left( \frac{25}{3} - \frac{20}{3} \right) = \frac{3}{32} \times \frac{5}{3} = \frac{5}{32} = 0.15625 = 0.156\) (to 3 s.f.).

Part (i):
M1: Integrate the PDF between 1 and 5 and set equal to 1.
A1: Show correct integration and substitution to get \(k = 3/32\) (AG).

Part (ii):
B1: State \(\text{E}(X) = 3\) by symmetry or correct integration.

Part (iii):
M1: Formulate integration for \(P(X > 4)\) from 4 to 5.
A1: Correct integration.
M1: Correct substitution of limits 4 and 5.
A1: Correct final answer \(\frac{5}{32}\) or 0.156 (3 s.f.).

(i) The null hypothesis represents no change from the historical value, and the alternative hypothesis represents the botanist's belief:
\(H_0: p = 0.7\)
\(H_1: p > 0.7\), where \(p\) is the germination probability of a seed.

(ii) A Type I error is rejecting the null hypothesis when it is true. This occurs if \(X \ge 17\) when \(p = 0.7\):
Under \(H_0\), \(X \sim \text{B}(20, 0.7)\).
\(P(X \ge 17) = P(X=17) + P(X=18) + P(X=19) + P(X=20)\)
\(P(X=17) = \binom{20}{17}(0.7)^{17}(0.3)^3 \approx 0.07160\)
\(P(X=18) = \binom{20}{18}(0.7)^{18}(0.3)^2 \approx 0.02785\)
\(P(X=19) = \binom{20}{19}(0.7)^{19}(0.3)^1 \approx 0.00684\)
\(P(X=20) = (0.7)^{20} \approx 0.00080\)
\(P(\text{Type I error}) = 0.07160 + 0.02785 + 0.00684 + 0.00080 = 0.10709 \approx 0.107\) (to 3 s.f.).

(iii) A Type II error is accepting the null hypothesis when it is false. This occurs if \(X \le 16\) when \(p = 0.85\):
Under \(H_1\) with \(p = 0.85\), \(X \sim \text{B}(20, 0.85)\).
\(P(X \le 16) = 1 - P(X \ge 17)\)
\(P(X=17) = \binom{20}{17}(0.85)^{17}(0.15)^3 \approx 0.2428\)
\(P(X=18) = \binom{20}{18}(0.85)^{18}(0.15)^2 \approx 0.2293\)
\(P(X=19) = \binom{20}{19}(0.85)^{19}(0.15)^1 \approx 0.1368\)
\(P(X=20) = (0.85)^{20} \approx 0.0388\)
\(P(X \ge 17) = 0.2428 + 0.2293 + 0.1368 + 0.0388 = 0.6477\)
\(P(\text{Type II error}) = 1 - 0.6477 = 0.3523 \approx 0.352\) (to 3 s.f.).

Part (i):
B1: Correct null and alternative hypotheses with parameter defined.

Part (ii):
M1: Correct binomial formulation for \(P(X \ge 17)\) using \(p=0.7\).
A1: Correct sum of at least three terms.
A1: Correct final probability 0.107.

Part (iii):
M1: Correct binomial formulation for \(1 - P(X \ge 17)\) using \(p=0.85\).
A1: Correct sum of terms for \(P(X \ge 17)\).
A1: Correct final probability 0.352.

(i) Unbiased estimate of population mean \(\mu\):
\(\bar{x} = \frac{\sum x}{n} = \frac{118.4}{80} = 1.48\) g.
Unbiased estimate of population variance \(s^2\):
\(s^2 = \frac{1}{n-1} \left( \sum x^2 - \frac{(\sum x)^2}{n} \right) = \frac{1}{79} \left( 181.76 - \frac{118.4^2}{80} \right)\)
\(s^2 = \frac{1}{79} \left( 181.76 - 175.232 \right) = \frac{6.528}{79} \approx 0.082633 = 0.0826\) (to 3 s.f.).

(ii) For a 98% confidence interval, the critical value \(z\) satisfies \(\Phi(z) = 0.99\), which gives \(z = 2.326\).
\(\text{CI} = \bar{x} \pm z \sqrt{\frac{s^2}{n}} = 1.48 \pm 2.326 \sqrt{\frac{0.082633}{80}}\)
\(\text{CI} = 1.48 \pm 2.326 \times 0.032139 = 1.48 \pm 0.07476\)
Interval is \([1.405, 1.555]\), which to 3 s.f. is \([1.41, 1.55]\).

(iii) Yes, it was necessary to use the Central Limit Theorem. The distribution of the masses of the population of beetles is unknown. Since the sample size \(n = 80\) is large (i.e., \(\ge 30\)), the Central Limit Theorem ensures that the sample mean \(\bar{X}\) is approximately normally distributed.

Part (i):
B1: Correct unbiased estimate of the mean (1.48).
M1: Correct formula used for the unbiased variance estimate.
A1: Correct variance estimate of 0.0826 (accept 0.0826 - 0.0827).

Part (ii):
B1: Correct critical value \(z = 2.326\) (or 2.33).
M1: Formulating confidence interval \(\bar{x} \pm z \sqrt{\frac{s^2}{n}}\).
A1: Correct interval \([1.41, 1.55]\) (accept \([1.40, 1.56]\) if \(z=2.33\) used).

Part (iii):
B1: Correctly state 'Yes' and give reason (population distribution is unknown and sample size is large).

Let \(B\) be the mass of an empty box, \(B \sim N(250, 12^2)\).
Let \(C\) be the mass of the contents, \(C \sim N(850, 35^2)\).
Let \(T = B + C\) be the total mass of a filled box.

(i) Since \(B\) and \(C\) are independent,
\(\text{E}(T) = \text{E}(B) + \text{E}(C) = 250 + 850 = 1100\) g.
\(\text{Var}(T) = \text{Var}(B) + \text{Var}(C) = 12^2 + 35^2 = 144 + 1225 = 1369\).
So, \(T \sim N(1100, 37^2)\) since \(\sqrt{1369} = 37\).
We want \(P(T > 1130) = P\left(Z > \frac{1130 - 1100}{37}\right) = P(Z > 0.811)\)
\(P(Z > 0.811) = 1 - \Phi(0.811) = 1 - 0.7913 = 0.2087 = 0.209\) (to 3 s.f.).

(ii) Let \(T_1\) and \(T_2\) be the masses of two randomly chosen filled boxes.
Let \(D = T_1 - T_2\).
Then \(D\) is normally distributed with:
\(\text{E}(D) = \text{E}(T_1) - \text{E}(T_2) = 1100 - 1100 = 0\).
\(\text{Var}(D) = \text{Var}(T_1) + \text{Var}(T_2) = 1369 + 1369 = 2738\).
So \(D \sim N(0, 2738)\).
We require \(P(|D| \ge 50) = 2 P(D \ge 50)\) due to symmetry.
\(P(D \ge 50) = P\left(Z \ge \frac{50 - 0}{\sqrt{2738}}\right) = P(Z \ge 0.9556)\)
\(P(Z \ge 0.9556) = 1 - \Phi(0.956) = 1 - 0.8304 = 0.1696\).
So \(P(|D| \ge 50) = 2 \times 0.1696 = 0.3392 = 0.339\) (to 3 s.f.).

Part (i):
M1: For finding the mean 1100 and variance 1369 of the total mass.
M1: Standardising with their mean and standard deviation.
A1: Correct final probability 0.209 (accept 0.208 - 0.210).

Part (ii):
M1: For finding the mean 0 and variance 2738 of the difference.
M1: For standardising 50 with their mean and standard deviation.
M1: For doubling the one-tailed probability.
A1: Correct final probability 0.339 (accept 0.338 - 0.340).

(i) Let \(\lambda\) be the mean weekly absences.
\(H_0: \lambda = 4.5\)
\(H_1: \lambda < 4.5\)
Under \(H_0\), the number of absences \(X \sim \text{Po}(4.5)\).
We test the probability of obtaining a result as extreme or more extreme than the observed 2 absences:
\(P(X \le 2) = e^{-4.5} \left(1 + 4.5 + \frac{4.5^2}{2}\right)\)
\(P(X \le 2) = e^{-4.5} (1 + 4.5 + 10.125) = e^{-4.5} (15.625) \approx 0.1736\).
Since \(0.1736 > 0.05\) (the 5% significance level), we do not reject \(H_0\).
There is insufficient evidence at the 5% level to suggest that the weekly absences have decreased.

(ii) We want to find the largest value \(c\) such that \(P(X \le c) \le 0.05\).
Let's test values:
For \(c = 0\): \(P(X = 0) = e^{-4.5} \approx 0.0111 \le 0.05\).
For \(c = 1\): \(P(X \le 1) = e^{-4.5}(1 + 4.5) = 5.5 e^{-4.5} \approx 0.0611 > 0.05\).
Thus, the largest number of weekly absences that would lead to rejecting \(H_0\) is 0.

Part (i):
B1: Correct null and alternative hypotheses.
M1: For calculating \(P(X \le 2)\) using Poisson distribution with mean 4.5.
A1: Correct probability of 0.174 (or 0.1736).
M1: For comparing their probability with 0.05.
A1: For drawing correct conclusion in context (do not reject \(H_0\)).

Part (ii):
M1: For calculating and comparing \(P(X \le 1)\) or \(P(X = 0)\) with 0.05.
A1: Correctly concluding that the largest value is 0.

(i) Let \(L\) be the weekly demand for LED bulbs and \(H\) be the weekly demand for Halogen bulbs.
The total weekly demand is \(T = L + H\).
Since \(L\) and \(H\) are independent Poisson variables, \(T \sim \text{Po}(3.5 + 1.8) = \text{Po}(5.3)\).
\(P(T = 4) = \frac{e^{-5.3} \times 5.3^4}{4!} = \frac{0.0049916 \times 789.0481}{24} = 0.1641 \approx 0.164\) (to 3 s.f.).

(ii) Let \(W\) be the total demand over a 30-week period.
The mean of \(W\) is \(\lambda = 30 \times 5.3 = 159\).
Since \(\lambda = 159 > 15\), we can approximate \(W\) by a normal distribution: \(W_{normal} \sim N(159, 159)\).
We require \(P(145 \le W \le 170)\).
Applying a continuity correction:
\(P(144.5 < W_{normal} < 170.5)\)
Standardising:
\(z_1 = \frac{144.5 - 159}{\sqrt{159}} = \frac{-14.5}{12.6095} \approx -1.150\)
\(z_2 = \frac{170.5 - 159}{\sqrt{159}} = \frac{11.5}{12.6095} \approx 0.912\)
\(P(-1.150 < Z < 0.912) = \Phi(0.912) - (1 - \Phi(1.150)) = 0.8191 - (1 - 0.8749) = 0.6940 = 0.694\) (to 3 s.f.).

Part (i):
M1: For stating or using Poisson distribution with mean 5.3.
A1: Correct probability 0.164.

Part (ii):
B1: For stating normal approximation with mean 159 and variance 159.
M1: For applying continuity correction correctly (using 144.5 and 170.5).
M1: For standardising with their mean and standard deviation.
A1: Correct z-values (-1.150 and 0.912).
A1: Correct final probability 0.694 (accept 0.693 - 0.695).