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(a) Express \( \mathrm{f}(x) \) in completed square form: \( \mathrm{f}(x) = 2(x^2 - 6x) + 13 = 2[(x-3)^2 - 9] + 13 = 2(x-3)^2 - 5 \). The vertex of the quadratic curve is at \( (3, -5) \). For a quadratic function to be one-to-one (and thus have an inverse) for \( x \ge k \), \( k \) must be at least the x-coordinate of the vertex. Thus, the smallest value of \( k \) is \( 3 \). (b) Let \( y = 2(x-3)^2 - 5 \). Rearranging to make \( x \) the subject: \( y + 5 = 2(x-3)^2 \implies \frac{y+5}{2} = (x-3)^2 \). Since \( x \ge 3 \), we take the positive square root: \( x - 3 = \sqrt{\frac{y+5}{2}} \implies x = 3 + \sqrt{\frac{y+5}{2}} \). Thus, \( \mathrm{f}^{-1}(x) = 3 + \sqrt{\frac{x+5}{2}} \). The domain of \( \mathrm{f}^{-1} \) is the range of \( \mathrm{f} \). Since \( x \ge 3 \), the minimum value of \( \mathrm{f}(x) \) is \( -5 \), so the domain of \( \mathrm{f}^{-1} \) is \( x \ge -5 \).

M1: Attempt to complete the square on f(x). A1: Identify correct vertex or write in form 2(x-3)^2 - 5. A1: Conclude k = 3. M1: Rearrange y = 2(x-3)^2 - 5 to make x the subject. A1: Obtain correct inverse function expression. A1: State correct domain x >= -5.

(a) The first term is \( a = 10 \). The common ratio is \( r = \frac{10 \sin^2 \theta}{10} = \sin^2 \theta \). The sum to infinity is \( S_{\infty} = \frac{a}{1-r} = \frac{10}{1 - \sin^2 \theta} = \frac{40}{3} \). This gives \( 1 - \sin^2 \theta = \frac{3}{4} \implies \sin^2 \theta = \frac{1}{4} \). Since \( 0 < \theta < \frac{\pi}{2} \), \( \sin \theta \) must be positive, so \( \sin \theta = \frac{1}{2} \). (b) With \( r = \sin^2 \theta = \frac{1}{4} \), the sum of the first 4 terms is \( S_4 = \frac{a(1 - r^4)}{1 - r} = \frac{10(1 - (1/4)^4)}{1 - 1/4} = \frac{10(1 - 1/256)}{3/4} = \frac{40}{3} \times \frac{255}{256} = \frac{425}{32} \approx 13.3 \) to 3 significant figures.

M1: Formulate the equation for sum to infinity in terms of sin^2(theta). A1: Correct equation for sin^2(theta). A1: Obtain sin(theta) = 1/2 (rejecting negative value). M1: Identify r = 1/4 and apply the geometric sum formula for n = 4. A1: Obtain exact sum of 425/32. A1: Provide correct decimal answer of 13.3.

(a) Integrating \( \frac{\mathrm{d}y}{\mathrm{d}x} = 6(3x+4)^{-1/2} \) gives \( y = \frac{6(3x+4)^{1/2}}{(1/2)(3)} + C = 4\sqrt{3x+4} + C \). Substituting \( (4, 11) \) to find \( C \): \( 11 = 4\sqrt{3(4)+4} + C \implies 11 = 16 + C \implies C = -5 \). Thus the equation of the curve is \( y = 4\sqrt{3x+4} - 5 \). (b) Set the gradient equal to 2: \( \frac{6}{\sqrt{3x+4}} = 2 \implies \sqrt{3x+4} = 3 \implies 3x + 4 = 9 \implies x = \frac{5}{3} \). Substitute \( x = \frac{5}{3} \) into the equation of the curve: \( y = 4\sqrt{3(5/3)+4} - 5 = 4\sqrt{9} - 5 = 7 \). The coordinates are \( (\frac{5}{3}, 7) \).

M1: Attempt integration of (3x+4)^(-1/2). A1: Correct term 4(3x+4)^(1/2). M1: Substitute x = 4, y = 11 to solve for C. A1: Correct constant C = -5. M1: Set gradient function equal to 2 and solve for x. A1: Correct coordinates (5/3, 7).

The area of the sector \( OAB \) is \( \frac{1}{2} r^2 \theta = \frac{1}{2} (8^2) \theta = 32\theta \). In the right-angled triangle \( OAP \), \( AP = 8 \sin \theta \) and \( OP = 8 \cos \theta \). The area of triangle \( OAP \) is \( \frac{1}{2} \times OP \times AP = \frac{1}{2} (8 \cos \theta) (8 \sin \theta) = 32 \sin \theta \cos \theta \). The area of the shaded region is the area of the sector minus the area of the triangle: \( 32\theta - 32\sin\theta\cos\theta = \frac{16}{3}\pi - 8\sqrt{3} \). Comparing the terms or solving, if we let \( \theta = \frac{\pi}{6} \), then \( 32(\frac{\pi}{6}) - 32\sin(\frac{\pi}{6})\cos(\frac{\pi}{6}) = \frac{16}{3}\pi - 32(\frac{1}{2})(\frac{\sqrt{3}}{2}) = \frac{16}{3}\pi - 8\sqrt{3} \). This matches the given area perfectly, so \( \theta = \frac{\pi}{6} \).

M1: State sector area as 32*theta. M1: Express AP and OP in terms of sin(theta) and cos(theta). A1: State triangle area as 32*sin(theta)*cos(theta). M1: Formulate the shaded area equation. A1: Substitute theta = pi/6 and verify it matches the given value. A1: Correct final value of theta = pi/6.

For the line and curve not to intersect, the equation \( x^2 - 4x + m = mx - 5 \) has no real solutions. Rearranging into standard quadratic form gives \( x^2 - (m+4)x + (m+5) = 0 \). The discriminant must be negative for no intersection: \( \Delta = [-(m+4)]^2 - 4(1)(m+5) < 0 \implies m^2 + 8m + 16 - 4m - 20 < 0 \implies m^2 + 4m - 4 < 0 \). Solving the critical equation \( m^2 + 4m - 4 = 0 \) using the quadratic formula gives \( m = \frac{-4 \pm \sqrt{16 - 4(1)(-4)}}{2} = -2 \pm 2\sqrt{2} \). Since we require the quadratic in m to be less than 0, the set of values is \( -2 - 2\sqrt{2} < m < -2 + 2\sqrt{2} \).

M1: Form a quadratic equation by equating the line and curve. A1: Correct quadratic in standard form. M1: Use discriminant b^2 - 4ac < 0. A1: Obtain the simplified inequality m^2 + 4m - 4 < 0. M1: Find critical values using the quadratic formula. A1: Correct final range of m.

Substitute \( \tan\theta = \frac{\sin\theta}{\cos\theta} \) into the equation: \( 3\sin\theta \left(\frac{\sin\theta}{\cos\theta}\right) = 8 - 5\cos\theta \implies 3\sin^2\theta = 8\cos\theta - 5\cos^2\theta \). Using \( \sin^2\theta = 1 - \cos^2\theta \), we get \( 3(1 - \cos^2\theta) = 8\cos\theta - 5\cos^2\theta \implies 3 - 3\cos^2\theta = 8\cos\theta - 5\cos^2\theta \implies 2\cos^2\theta - 8\cos\theta + 3 = 0 \). Let \( u = \cos\theta \). Using the quadratic formula: \( u = \frac{8 \pm \sqrt{64 - 24}}{4} = 2 \pm \frac{\sqrt{10}}{2} \). Since \( 2 + \frac{\sqrt{10}}{2} \approx 3.58 > 1 \), there is no solution. For \( \cos\theta = 2 - \frac{\sqrt{10}}{2} \approx 0.4189 \), we find the principal angle \( \theta = \cos^{-1}(0.4189) \approx 65.23^\circ \). The other angle in the range is \( 360^\circ - 65.23^\circ = 294.77^\circ \). Rounding to 1 decimal place gives \( \theta = 65.2^\circ \) and \( \theta = 294.8^\circ \).

M1: Substitute tan(theta) = sin(theta)/cos(theta). M1: Substitute sin^2(theta) = 1 - cos^2(theta). A1: Obtain the correct quadratic equation in cos(theta). M1: Solve the quadratic equation. A1: Identify cos(theta) = 2 - sqrt(10)/2 as the only valid value. A1: Correct angle 65.2 degrees. A1: Correct angle 294.8 degrees.

(a) The centre of the circle \( M \) is the midpoint of \( AB \): \( M = \left(\frac{-2+4}{2}, \frac{3+1}{2}\right) = (1, 2) \). The length of the diameter is \( AB = \sqrt{(4 - (-2))^2 + (1 - 3)^2} = \sqrt{36 + 4} = \sqrt{40} \). The radius \( r \) is \( \frac{\sqrt{40}}{2} = \sqrt{10} \). Thus, the equation of the circle is \( (x - 1)^2 + (y - 2)^2 = 10 \). (b) The gradient of the radius \( MB \) is \( m = \frac{1 - 2}{4 - 1} = -\frac{1}{3} \). The tangent at \( B \) is perpendicular to \( MB \), so its gradient is \( m_t = 3 \). The equation of the tangent is \( y - 1 = 3(x - 4) \implies y - 1 = 3x - 12 \implies 3x - y = 11 \).

M1: Find the midpoint of AB. A1: Correct centre (1, 2). M1: Find the distance between AB or from centre to A/B. A1: Correct equation of the circle. M1: Find the gradient of MB and its negative reciprocal. A1: Obtain correct final form of tangent 3x - y = 11.

(a) Write the equation as \( y = 16x^{-1} + x^2 \). Differentiating gives \( \frac{\mathrm{d}y}{\mathrm{d}x} = -16x^{-2} + 2x = -\frac{16}{x^2} + 2x \). Setting \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \) gives \( 2x = \frac{16}{x^2} \implies 2x^3 = 16 \implies x^3 = 8 \implies x = 2 \). Substituting \( x = 2 \) into the curve equation gives \( y = \frac{16}{2} + 2^2 = 12 \). So the stationary point is \( (2, 12) \). (b) Differentiating again gives \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 32x^{-3} + 2 = \frac{32}{x^3} + 2 \). At \( x = 2 \), \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{32}{8} + 2 = 6 \). Since this is greater than 0, the stationary point is a minimum.

M1: Attempt to differentiate y. A1: Correct derivative -16x^-2 + 2x. M1: Set derivative to 0 and solve for x. A1: Obtain x = 2 and y = 12. M1: Differentiate again to find d^2y/dx^2. A1: Correctly evaluate d^2y/dx^2 at x = 2 and conclude minimum.

(a) Expanding \( (2x - p)^2 + q \) gives \( 4x^2 - 4px + p^2 + q \).
Comparing with \( 4x^2 - 12x + 11 \):
- From the \( x \) term: \( -4p = -12 \implies p = 3 \).
- From the constant term: \( p^2 + q = 11 \implies 9 + q = 11 \implies q = 2 \).
Thus, \( f(x) = (2x - 3)^2 + 2 \).

(b) Since \( (2x - 3)^2 \ge 0 \) for all real \( x \), the minimum value of \( f(x) \) is \( 2 \).
Therefore, the range of \( f \) is \( f(x) \ge 2 \) (or \( y \ge 2 \)).

(c) A quadratic function has an inverse if and only if its domain is restricted to a region where it is one-to-one. For a parabola opening upwards, this corresponds to either side of its vertex.
The vertex is at \( x = \frac{p}{2} = \frac{3}{2} = 1.5 \).
Since the domain is restricted to \( x \le k \), the largest value of \( k \) is the \( x \)-coordinate of the vertex.
Thus, the largest value of \( k \) is \( 1.5 \).

To find the inverse, we set \( y = (2x - 3)^2 + 2 \):
\( (2x - 3)^2 = y - 2 \)

Taking the square root, and since \( x \le 1.5 \), we have \( 2x - 3 \le 0 \). We must choose the negative square root:
\( 2x - 3 = -\sqrt{y - 2} \)
\( 2x = 3 - \sqrt{y - 2} \)
\( x = \frac{3 - \sqrt{y - 2}}{2} \)

Replacing \( y \) with \( x \), we obtain:
\( g^{-1}(x) = \frac{3 - \sqrt{x - 2}}{2} \).

(a)
M1: Attempt to complete the square, e.g., identifying \( p = 3 \).
A1: Correct expression \( (2x - 3)^2 + 2 \) (or stating \( p=3, q=2 \)).

(b)
B1: Correct range, e.g., \( f(x) \ge 2 \) or \( y \ge 2 \) (do not accept \( x \ge 2 \)).

(c)
B1: State \( k = 1.5 \) (or \( \frac{3}{2} \)).
M1: Attempt to make \( x \) the subject of \( y = (2x - p)^2 + q \).
M1: Correctly choose the negative square root because of the domain restriction \( x \le k \).
A1: Correct expression for \( g^{-1}(x) \), i.e., \( \frac{3 - \sqrt{x - 2}}{2} \) or \( 1.5 - 0.5\sqrt{x-2} \).

To find the points of intersection, substitute \( y = 2x + k \) into the equation of the circle:
\( x^2 + (2x + k)^2 - 6x + 4(2x + k) - 12 = 0 \)

Expand the terms:
\( x^2 + (4x^2 + 4kx + k^2) - 6x + 8x + 4k - 12 = 0 \)

Group the terms by powers of \( x \):
\( 5x^2 + (4k + 2)x + (k^2 + 4k - 12) = 0 \)

Since the line is a tangent to the circle, there is exactly one point of intersection. Therefore, the discriminant of this quadratic equation in \( x \) must equal zero:
\( \Delta = b^2 - 4ac = 0 \)
\( (4k + 2)^2 - 4(5)(k^2 + 4k - 12) = 0 \)

Expand and simplify:
\( 16k^2 + 16k + 4 - 20(k^2 + 4k - 12) = 0 \)

\( 16k^2 + 16k + 4 - 20k^2 - 80k + 240 = 0 \)

\( -4k^2 - 64k + 244 = 0 \)

Divide the entire equation by \( -4 \):
\( k^2 + 16k - 61 = 0 \)

Use the quadratic formula to solve for \( k \):
\( k = \frac{-16 \pm \sqrt{16^2 - 4(1)(-61)}}{2} \)

\( k = \frac{-16 \pm \sqrt{256 + 244}}{2} \)

\( k = \frac{-16 \pm \sqrt{500}}{2} \)

Since \( \sqrt{500} = \sqrt{100 \times 5} = 10\sqrt{5} \):
\( k = \frac{-16 \pm 10\sqrt{5}}{2} \)

\( k = -8 \pm 5\sqrt{5} \).

M1: Substitute \( y = 2x + k \) into the circle equation.
A1: Obtain a correct simplified quadratic equation in \( x \), i.e., \( 5x^2 + (4k + 2)x + (k^2 + 4k - 12) = 0 \).
M1: Apply the discriminant condition \( b^2 - 4ac = 0 \) for tangency.
A1: Obtain a correct quadratic equation in \( k \), e.g., \( -4k^2 - 64k + 244 = 0 \) or \( k^2 + 16k - 61 = 0 \).
M1: Solve their three-term quadratic in \( k \) using the quadratic formula or by completing the square.
A1: Obtain the correct exact values \( k = -8 \pm 5\sqrt{5} \) (or equivalent exact forms; do not accept decimal approximations).

(a) Integrate \( \frac{\text{d}y}{\text{d}x} \) with respect to \( x \) to find the equation of the curve:
\( y = \int \left( 3x^{-1/2} - 1 \right) \text{d}x \)
\( y = \frac{3x^{1/2}}{1/2} - x + c \)
\( y = 6\sqrt{x} - x + c \)

Substitute \( x = 4 \) and \( y = 5 \) to find the constant of integration \( c \):
\( 5 = 6\sqrt{4} - 4 + c \)
\( 5 = 12 - 4 + c \)
\( 5 = 8 + c \implies c = -3 \)

So, the equation of the curve is:
\( y = 6\sqrt{x} - x - 3 \).

(b) At a stationary point, \( \frac{\text{d}y}{\text{d}x} = 0 \):
\( \frac{3}{\sqrt{x}} - 1 = 0 \implies \sqrt{x} = 3 \implies x = 9 \)

Substitute \( x = 9 \) back into the curve equation to find the \( y \)-coordinate:
\( y = 6\sqrt{9} - 9 - 3 = 18 - 12 = 6 \)

Hence, the coordinates of the stationary point are \( (9, 6) \).

To determine its nature, find the second derivative:
\( \frac{\text{d}^2y}{\text{d}x^2} = \frac{\text{d}}{\text{d}x} \left( 3x^{-1/2} - 1 \right) = -\frac{3}{2}x^{-3/2} \)

Substitute \( x = 9 \):
\( \frac{\text{d}^2y}{\text{d}x^2} = -\frac{3}{2}(9)^{-3/2} = -\frac{3}{2} \times \frac{1}{27} = -\frac{1}{18} \)

Since \( \frac{\text{d}^2y}{\text{d}x^2} < 0 \), the stationary point \( (9, 6) \) is a maximum.

(a)
M1: Attempt to integrate \( 3x^{-1/2} - 1 \), increasing at least one power by 1.
A1: Correct integration: \( 6\sqrt{x} - x \) (constant \( c \) not required yet).
M1: Substitute the coordinates \( (4, 5) \) to find \( c \).
A1: Obtain the correct equation of the curve \( y = 6\sqrt{x} - x - 3 \).

(b)
M1: Set \( \frac{\text{d}y}{\text{d}x} = 0 \) and solve for \( x \).
A1: Obtain \( x = 9 \) and \( y = 6 \).
M1: Find the second derivative and substitute their value of \( x \) (or use a sign change method on the first derivative).
A1: Obtain \( \frac{\text{d}^2y}{\text{d}x^2} = -\frac{1}{18} < 0 \) and state that the stationary point is a maximum.

(i) Since \((x-1)\) is a factor, \(p(1) = 0\):
\(a(1)^3 + 5(1)^2 + b(1) - 6 = 0 \implies a + b = 1\).

Since the remainder when divided by \((x+2)\) is \(-12\), \(p(-2) = -12\):
\(a(-2)^3 + 5(-2)^2 + b(-2) - 6 = -12 \implies -8a + 20 - 2b - 6 = -12 \implies 4a + b = 13\).

We have a system of linear equations:
1) \(a + b = 1\)
2) \(4a + b = 13\)

Subtracting (1) from (2) gives:
\(3a = 12 \implies a = 4\).

Substituting \(a=4\) into (1) gives:
\(4 + b = 1 \implies b = -3\).

So, \(a = 4\) and \(b = -3\).

(ii) With \(a = 4\) and \(b = -3\), \(p(x) = 4x^3 + 5x^2 - 3x - 6\).

Since \((x-1)\) is a factor, we can express \(p(x)\) as:
\(p(x) = (x-1)(4x^2 + kx + 6)\).

Equating the coefficient of the \(x^2\) term:
\(k - 4 = 5 \implies k = 9\).

So, \(p(x) = (x-1)(4x^2 + 9x + 6)\).

For the equation \(p(x) = 0\), we either have:
\(x - 1 = 0 \implies x = 1\), or \(4x^2 + 9x + 6 = 0\).

We examine the discriminant of the quadratic factor \(4x^2 + 9x + 6 = 0\):
\(\Delta = b^2 - 4ac = 9^2 - 4(4)(6) = 81 - 96 = -15\).

Since the discriminant is negative (\(\Delta < 0\)), the quadratic equation has no real roots.

Therefore, the equation \(p(x) = 0\) has only one real root, which is \(x = 1\).

(i)
M1: Attempt to apply the factor theorem \(p(1) = 0\) to obtain an equation in \(a\) and \(b\).
A1: Obtain the correct equation \(a + b = 1\).
M1: Attempt to apply the remainder theorem \(p(-2) = -12\) and solve the resulting simultaneous equations.
A1: Obtain \(a = 4\) and \(b = -3\).

(ii)
M1: Attempt to factorise \(p(x)\) into the form \((x-1)(4x^2 + kx + 6)\) and find \(k = 9\).
M1: Attempt to calculate the discriminant of the quadratic factor.
A1: Obtain discriminant \(-15 < 0\) and conclude that \(x = 1\) is the only real root.

We use the laws of logarithms to simplify the right-hand side of the equation:
\(2\ln(x) = \ln(x^2)\)
\(\ln(x^2) + \ln(3) = \ln(3x^2)\)

The equation becomes:
\(\ln(2x + 3) = \ln(3x^2)\)

Equating the arguments:
\(2x + 3 = 3x^2\)

Rearranging into a quadratic equation:
\(3x^2 - 2x - 3 = 0\)

Using the quadratic formula:
\(x = \frac{2 \pm \sqrt{(-2)^2 - 4(3)(-3)}}{2(3)} = \frac{2 \pm \sqrt{40}}{6} = \frac{1 \pm \sqrt{10}}{3}\)

This gives two values:
\(x_1 = \frac{1 + \sqrt{10}}{3} \approx 1.387\)
\(x_2 = \frac{1 - \sqrt{10}}{3} \approx -0.721\)

Since the domain of \(\ln(x)\) requires \(x > 0\), the negative root is rejected.

Therefore, the only valid solution is \(x \approx 1.39\) (correct to 3 significant figures).

M1: Use the power law of logarithms to write \(2\ln(x) = \ln(x^2)\).
M1: Use the product law of logarithms to combine terms to \(\ln(3x^2)\).
A1: Eliminate logarithms to obtain the correct quadratic equation \(3x^2 - 2x - 3 = 0\).
M1: Apply the quadratic formula correctly to find roots of their quadratic.
A1: Obtain the roots \(x = \frac{1 \pm \sqrt{10}}{3}\) (or equivalent decimals).
M1: Identify and state why the negative root must be rejected.
A1: Obtain the final answer \(x = 1.39\).

We use the identity \(\sec^2 \theta = 1 + \tan^2 \theta\) to rewrite the equation:
\((1 + \tan^2 \theta) + 3\tan \theta = 5\)
\(\tan^2 \theta + 3\tan \theta - 4 = 0\)

Factorising the quadratic in terms of \(\tan \theta\):
\((\tan \theta + 4)(\tan \theta - 1) = 0\)

This yields two possible cases:

Case 1: \(\tan \theta = 1\)
The basic angle is \(45^\circ\).
Since tangent is positive, \(\theta\) is in the first or third quadrant:
\(\theta = 45^\circ\) or \(\theta = 180^\circ + 45^\circ = 225^\circ\)

Case 2: \(\tan \theta = -4\)
The basic angle is \(\tan^{-1}(4) \approx 76.0^\circ\).
Since tangent is negative, \(\theta\) is in the second or fourth quadrant:
\(\theta = 180^\circ - 76.0^\circ = 104.0^\circ\)
\(\theta = 360^\circ - 76.0^\circ = 284.0^\circ\)

Thus, the solutions in the given range are \(\theta = 45^\circ\), \(104.0^\circ\), \(225^\circ\), and \(284.0^\circ\).

M1: Substitute \(1 + \tan^2 \theta\) for \(\sec^2 \theta\).
A1: Obtain the correct simplified quadratic equation \(\tan^2 \theta + 3\tan \theta - 4 = 0\).
M1: Solve the quadratic to find values for \(\tan \theta\).
A1: Obtain both \(\tan \theta = 1\) and \(\tan \theta = -4\).
A1: Obtain solutions \(\theta = 45^\circ\) and \(225^\circ\).
M1: Find the basic angle for \(\tan \theta = -4\) and compute quadrant angles.
A1: Obtain \(\theta = 104.0^\circ\) and \(284.0^\circ\).

(i) We use the quotient rule with \(u = e^{2x}\) and \(v = 3x + 1\):
\(\frac{du}{dx} = 2e^{2x}\)
\(\frac{dv}{dx} = 3\)

\(\frac{dy}{dx} = \frac{2e^{2x}(3x+1) - 3e^{2x}}{(3x+1)^2} = \frac{e^{2x}(6x - 1)}{(3x+1)^2}\).

The curve crosses the \(y\)-axis at \(x = 0\).
Substituting \(x = 0\) into the derivative:
\(\frac{dy}{dx} = \frac{e^0(6(0) - 1)}{(3(0)+1)^2} = -1\).

Thus, the gradient of the curve at this point is \(-1\).

(ii) To find the stationary point, we set \(\frac{dy}{dx} = 0\):
\(\frac{e^{2x}(6x - 1)}{(3x+1)^2} = 0 \implies e^{2x}(6x - 1) = 0\).

Since \(e^{2x} > 0\) for all real \(x\):
\(6x - 1 = 0 \implies x = \frac{1}{6}\).

Substituting \(x = \frac{1}{6}\) into the original equation of the curve:
\(y = \frac{e^{2(1/6)}}{3(1/6) + 1} = \frac{e^{1/3}}{\frac{1}{2} + 1} = \frac{2}{3}e^{1/3}\).

Therefore, the exact coordinates of the stationary point are \(\left(\frac{1}{6}, \frac{2}{3}e^{1/3}\right)\).

(i)
M1: Apply the quotient rule to differentiate \(y\).
A1: Obtain the correct simplified expression \(\frac{dy}{dx} = \frac{e^{2x}(6x - 1)}{(3x+1)^2}\).
A1: Substitute \(x = 0\) to obtain the gradient of \(-1\).

(ii)
M1: Set their numerator equal to zero to find the stationary x-coordinate.
A1: Solve to find the correct value \(x = \frac{1}{6}\).
M1: Substitute \(x = \frac{1}{6}\) back into the equation for \(y\).
A1: Obtain the exact coordinates \(\left(\frac{1}{6}, \frac{2}{3}e^{1/3}\right)\).

We use the double-angle identity \(2\cos^2 x = 1 + \cos 2x\) to rewrite the integrand:
\(\int_{0}^{\frac{\pi}{4}} (2\cos^2 x + \sin 2x) \, dx = \int_{0}^{\frac{\pi}{4}} (1 + \cos 2x + \sin 2x) \, dx\).

Now we integrate term by term:
\(\int (1 + \cos 2x + \sin 2x) \, dx = \left[ x + \frac{1}{2}\sin 2x - \frac{1}{2}\cos 2x \right]_{0}^{\frac{\pi}{4}}\).

Substituting the limits:
At the upper limit \(x = \frac{\pi}{4}\):
\(\left(\frac{\pi}{4} + \frac{1}{2}\sin\left(\frac{\pi}{2}\right) - \frac{1}{2}\cos\left(\frac{\pi}{2}\right)\right) = \frac{\pi}{4} + \frac{1}{2}(1) - 0 = \frac{\pi}{4} + \frac{1}{2}\).

At the lower limit \(x = 0\):
\(\left(0 + \frac{1}{2}\sin(0) - \frac{1}{2}\cos(0)\right) = 0 + 0 - \frac{1}{2} = -\frac{1}{2}\).

Subtracting the lower limit value from the upper limit value:
\(\left(\frac{\pi}{4} + \frac{1}{2}\right) - \left(-\frac{1}{2}\right) = \frac{\pi}{4} + 1\).

M1: Apply the double-angle identity to express \(2\cos^2 x\) as \(1 + \cos 2x\).
A1: Obtain correct integrand \(1 + \cos 2x + \sin 2x\).
M1: Integrate \(\cos 2x\) to get \(\frac{1}{2}\sin 2x\).
M1: Integrate \(\sin 2x\) to get \(-\frac{1}{2}\cos 2x\).
A1: State the correct overall integrated expression.
M1: Substitute the limits of integration \(0\) and \(\frac{\pi}{4}\) correctly.
A1: Obtain the correct exact answer \(\frac{\pi}{4} + 1\).

(i) Let \(f(x) = x^3 + 2x - 5\).

Evaluate \(f(x)\) at \(x = 1.3\):
\(f(1.3) = (1.3)^3 + 2(1.3) - 5 = 2.197 + 2.6 - 5 = -0.203\).

Evaluate \(f(x)\) at \(x = 1.4\):
\(f(1.4) = (1.4)^3 + 2(1.4) - 5 = 2.744 + 2.8 - 5 = 0.544\).

Since \(f(x)\) is a continuous function and there is a sign change between \(x = 1.3\) and \(x = 1.4\), there must be a real root \(\alpha\) in this interval.

(ii) Applying the iterative formula \(x_{n+1} = \sqrt[3]{5 - 2x_n}\) starting with \(x_1 = 1.3\):
\(x_1 = 1.3\)
\(x_2 = \sqrt[3]{5 - 2(1.3)} = \sqrt[3]{2.4} \approx 1.3389\)
\(x_3 = \sqrt[3]{5 - 2(1.338865)} \approx 1.3243\)
\(x_4 = \sqrt[3]{5 - 2(1.324263)} \approx 1.3298\)
\(x_5 = \sqrt[3]{5 - 2(1.329780)} \approx 1.3277\)
\(x_6 = \sqrt[3]{5 - 2(1.327697)} \approx 1.3285\)

We observe that the values are converging towards \(1.33\).

Therefore, \(\alpha = 1.33\) correct to 2 decimal places.

(i)
M1: Evaluate \(f(1.3)\) and \(f(1.4)\) for a continuous function.
A1: Obtain correct values of \(-0.203\) and \(0.544\), and draw the conclusion of a root.

(ii)
M1: Perform the first iteration successfully to find \(x_2\).
A1: Obtain \(x_2 = 1.3389\).
A1: Obtain \(x_3 = 1.3243\).
A1: Perform further iterations up to at least \(x_5\) showing convergence.
A1: Round correctly to conclude \(\alpha = 1.33\).

(i) We differentiate \(x\) and \(y\) with respect to \(t\):
\(\frac{dx}{dt} = 2 + \frac{1}{t+1}\)
\(\frac{dy}{dt} = 2t + e^{-t}\)

Using the chain rule, we obtain:
\(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2t + e^{-t}}{2 + \frac{1}{t+1}}\).

(ii) To find the equation of the tangent at \(t = 0\):
First, compute the coordinates of the point on the curve:
At \(t = 0\):
\(x = 2(0) + \ln(0 + 1) = 0\)
\(y = 0^2 - e^0 = -1\)
So, the coordinates are \((0, -1)\).

Next, compute the gradient of the curve at this point:
At \(t = 0\):
\(m = \frac{2(0) + e^0}{2 + \frac{1}{0+1}} = \frac{1}{2 + 1} = \frac{1}{3}\).

The equation of the tangent line is:
\(y - y_1 = m(x - x_1) \implies y - (-1) = \frac{1}{3}(x - 0)\)
\(y + 1 = \frac{1}{3}x\)

Multiplying by 3 and rearranging to the required integer form:
\(x - 3y - 3 = 0\).

(i)
M1: Differentiate \(x\) correctly with respect to \(t\).
M1: Differentiate \(y\) correctly with respect to \(t\).
M1: Apply the parametric chain rule.
A1: Obtain the correct derivative \(\frac{2t + e^{-t}}{2 + \frac{1}{t+1}}\).

(ii)
M1: Find the coordinates of the point at \(t = 0\) as \((0, -1)\).
M1: Substitute \(t = 0\) to find the gradient \(m = \frac{1}{3}\).
A1: Express the tangent line in the correct integer form \(x - 3y - 3 = 0\) (or any integer multiple).

Let \( f(x) = \frac{A}{1 - 2x} + \frac{B}{1 + x} + \frac{C}{(1 + x)^2} \).

Multiplying by the denominator, we get:
\[ 7x - 5 = A(1 + x)^2 + B(1 - 2x)(1 + x) + C(1 - 2x) \]

Let \( x = -1 \):
\[ 7(-1) - 5 = C(1 - 2(-1)) \implies -12 = 3C \implies C = -4 \]

Let \( x = \frac{1}{2} \):
\[ 7\left(\frac{1}{2}\right) - 5 = A\left(1 + \frac{1}{2}\right)^2 \implies -\frac{3}{2} = \frac{9}{4}A \implies A = -\frac{2}{3} \]

Let \( x = 0 \):
\[ -5 = A + B + C \implies -5 = -\frac{2}{3} + B - 4 \implies B = -\frac{1}{3} \]

So the partial fractions are:
\[ f(x) = -\frac{2}{3}(1 - 2x)^{-1} - \frac{1}{3}(1 + x)^{-1} - 4(1 + x)^{-2} \]

Using the binomial expansion:
\[ (1 - 2x)^{-1} = 1 + 2x + 4x^2 + \dots \]
\[ (1 + x)^{-1} = 1 - x + x^2 - \dots \]
\[ (1 + x)^{-2} = 1 - 2x + 3x^2 - \dots \]

Substituting these expansions back into the partial fraction decomposition:
\[ f(x) = -\frac{2}{3}\left(1 + 2x + 4x^2\right) - \frac{1}{3}\left(1 - x + x^2\right) - 4\left(1 - 2x + 3x^2\right) \]

Now, collect coefficients:
- Constant term: \( -\frac{2}{3} - \frac{1}{3} - 4 = -5 \)
- Coefficient of \( x \): \( -\frac{4}{3} + \frac{1}{3} + 8 = 7 \)
- Coefficient of \( x^2 \): \( -\frac{8}{3} - \frac{1}{3} - 12 = -15 \)

Thus, the expansion is \( -5 + 7x - 15x^2 \).

M1: State or imply the form of partial fractions with three terms.
A1: Obtain any one of the constants \( A = -\frac{2}{3} \), \( B = -\frac{1}{3} \), or \( C = -\frac{4}{3} \).
A1: Obtain all three constants.
M1: Use binomial expansions for \( (1-2x)^{-1} \), \( (1+x)^{-1} \), or \( (1+x)^{-2} \).
A1: Obtain correct expansions of the terms.
A1: Obtain final answer \( -5 + 7x - 15x^2 \).

Let \( u = e^x \). Since \( e^x > 0 \), we must have \( u > 0 \).

The given equation is:
\[ u^2 - 4u - 12 + \frac{27}{u} = 0 \]

Multiply by \( u \) to get a cubic equation:
\[ u^3 - 4u^2 - 12u + 27 = 0 \]

We test possible integer roots. Let \( P(u) = u^3 - 4u^2 - 12u + 27 \):
\[ P(-3) = (-3)^3 - 4(-3)^2 - 12(-3) + 27 = -27 - 36 + 36 + 27 = 0 \]

Thus, \( (u + 3) \) is a factor.

Perform algebraic division or factorisation:
\[ u^3 - 4u^2 - 12u + 27 = (u + 3)(u^2 - 7u + 9) = 0 \]

Since \( u = e^x > 0 \), the root \( u = -3 \) yields no real solution.

Now solve the quadratic equation \( u^2 - 7u + 9 = 0 \):
\[ u = \frac{7 \pm \sqrt{(-7)^2 - 4(1)(9)}}{2} = \frac{7 \pm \sqrt{13}}{2} \]

Since \( \sqrt{13} \approx 3.61 \), both roots are positive:
\[ u_1 = \frac{7 + \sqrt{13}}{2} > 0 \quad \text{and} \quad u_2 = \frac{7 - \sqrt{13}}{2} > 0 \]

Taking natural logarithms:
\[ x = \ln\left(\frac{7 + \sqrt{13}}{2}\right) \quad \text{and} \quad x = \ln\left(\frac{7 - \sqrt{13}}{2}\right) \]

M1: Substitute \( u = e^x \) and form a cubic equation in \( u \).
A1: Obtain the correct cubic equation \( u^3 - 4u^2 - 12u + 27 = 0 \).
M1: Find a real root of the cubic equation (e.g. \( u = -3 \)).
A1: Factorise the cubic to obtain the quadratic factor \( u^2 - 7u + 9 \).
M1: Solve the quadratic equation to find the positive values of \( u \).
A1: Obtain both final answers in exact logarithmic form.

Let \( w_1 = 1 + i\sqrt{3} \).
\[ |w_1| = \sqrt{1^2 + (\sqrt{3})^2} = 2 \]
\[ \arg(w_1) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \]
Thus,
\[ |w_1^4| = 2^4 = 16 \]
\[ \arg(w_1^4) = 4 \times \frac{\pi}{3} = \frac{4\pi}{3} \]

Let \( w_2 = 1 - i \).
\[ |w_2| = \sqrt{1^2 + (-1)^2} = \sqrt{2} \]
\[ \arg(w_2) = \tan^{-1}(-1) = -\frac{\pi}{4} \]
Thus,
\[ |w_2^6| = (\sqrt{2})^6 = 8 \]
\[ \arg(w_2^6) = 6 \times \left(-\frac{\pi}{4}\right) = -\frac{3\pi}{2} \]

Now find the modulus and argument of \( z = \frac{w_1^4}{w_2^6} \):
\[ |z| = \frac{|w_1^4|}{|w_2^6|} = \frac{16}{8} = 2 \]
\[ \arg(z) = \arg(w_1^4) - \arg(w_2^6) = \frac{4\pi}{3} - \left(-\frac{3\pi}{2}\right) = \frac{17\pi}{6} \]

Subtract \( 2\pi \) to get the principal argument:
\[ \arg(z) = \frac{17\pi}{6} - 2\pi = \frac{5\pi}{6} \]

Express \( z \) in exponential/polar form to find \( x + iy \):
\[ z = 2\left( \cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6} \right) \]
\[ z = 2\left( -\frac{\sqrt{3}}{2} + i\frac{1}{2} \right) = -\sqrt{3} + i \]

M1: Obtain modulus and argument of \( 1 + i\sqrt{3} \) and \( 1 - i \).
A1: Obtain modulus \( 16 \) and argument \( \frac{4\pi}{3} \) for the numerator (or equivalent polar form).
A1: Obtain modulus \( 8 \) and argument \( -\frac{3\pi}{2} \) (or equivalent polar form).
M1: Apply rules for modulus and argument of division.
A1: State correct modulus \( |z| = 2 \) and principal argument \( \arg(z) = \frac{5\pi}{6} \).
A1: Deduce the exact Cartesian form \( z = -\sqrt{3} + i \).

(i) We expand \( R \cos(2\theta - \alpha) \):
\[ R \cos(2\theta - \alpha) = R \cos 2\theta \cos \alpha + R \sin 2\theta \sin \alpha \]

Comparing with \( 3\sin 2\theta + 4\cos 2\theta \), we get:
\[ R \cos \alpha = 4 \]
\[ R \sin \alpha = 3 \]

Thus:
\[ R = \sqrt{3^2 + 4^2} = 5 \]
\[ \tan \alpha = \frac{3}{4} \implies \alpha = 36.87^\circ \]

So the expression is \( 5 \cos(2\theta - 36.87^\circ) \).

(ii) We now solve:
\[ 5 \cos(2\theta - 36.87^\circ) = 2.5 \implies \cos(2\theta - 36.87^\circ) = 0.5 \]

Since \( 0^\circ < \theta < 180^\circ \), we have:
\[ 0^\circ < 2\theta < 360^\circ \implies -36.87^\circ < 2\theta - 36.87^\circ < 323.13^\circ \]

Let \( \phi = 2\theta - 36.87^\circ \).
The solutions for \( \cos \phi = 0.5 \) in this interval are \( \phi = 60^\circ \) and \( \phi = 300^\circ \).

Case 1:
\[ 2\theta - 36.87^\circ = 60^\circ \implies 2\theta = 96.87^\circ \implies \theta = 48.435^\circ \approx 48.4^\circ \]

Case 2:
\[ 2\theta - 36.87^\circ = 300^\circ \implies 2\theta = 336.87^\circ \implies \theta = 168.435^\circ \approx 168.4^\circ \]

M1: Use correct identities to find \( R \) and \( \alpha \).
A1: Obtain \( R = 5 \).
A1: Obtain \( \alpha = 36.87^\circ \).
M1: Set up equation \( \cos(2\theta - 36.87^\circ) = 0.5 \) and find one correct value of \( 2\theta - 36.87^\circ \).
A1: Obtain \( \theta = 48.4^\circ \).
A1: Obtain \( \theta = 168.4^\circ \) and no other values in the range.

(i) Let \( f(x) = x^3 - 5x - 3 \).
Evaluate \( f(x) \) at the boundaries of the interval:
\[ f(2.4) = (2.4)^3 - 5(2.4) - 3 = 13.824 - 12 - 3 = -1.176 \]
\[ f(2.5) = (2.5)^3 - 5(2.5) - 3 = 15.625 - 12.5 - 3 = 0.125 \]

Since \( f(x) \) is continuous and there is a change of sign between \( x = 2.4 \) and \( x = 2.5 \), the equation has a real root in this interval.

(ii) Applying the iterative formula with initial value \( x_1 = 2.5 \):
\[ x_2 = \sqrt[3]{5(2.5) + 3} = \sqrt[3]{15.5} \approx 2.49330 \]
\[ x_3 = \sqrt[3]{5(2.49330) + 3} = \sqrt[3]{15.46650} \approx 2.49150 \]
\[ x_4 = \sqrt[3]{5(2.49150) + 3} = \sqrt[3]{15.45750} \approx 2.49102 \]
\[ x_5 = \sqrt[3]{5(2.49102) + 3} = \sqrt[3]{15.45510} \approx 2.49089 \]
\[ x_6 = \sqrt[3]{5(2.49089) + 3} = \sqrt[3]{15.45445} \approx 2.49085 \]

Since the successive values converge to \( 2.491 \) to 3 decimal places, the root is \( 2.491 \).

M1: Evaluate the function at \( x=2.4 \) and \( x=2.5 \).
A1: State correct values (or signs) and conclude there is a root.
M1: Use the iterative formula correctly at least once.
A1: Obtain \( x_2 = 2.49330 \).
A1: Obtain \( x_3 = 2.49150 \) and continue to \( x_5 \) or \( x_6 \).
A1: State final root as \( 2.491 \) with sufficient evidence of convergence.

(i) Differentiating \( x \) with respect to \( t \):
\[ \frac{dx}{dt} = \frac{2}{2t + 1} \]

Differentiating \( y \) with respect to \( t \) using the quotient rule:
\[ \frac{dy}{dt} = \frac{2t(2t + 1) - 2(t^2)}{(2t + 1)^2} = \frac{4t^2 + 2t - 2t^2}{(2t + 1)^2} = \frac{2t^2 + 2t}{(2t + 1)^2} = \frac{2t(t+1)}{(2t + 1)^2} \]

Using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t(t+1)}{(2t+1)^2} \times \frac{2t+1}{2} = \frac{t(t+1)}{2t+1} \]

(ii) At \( t = 1 \):
\[ x = \ln(3) \]
\[ y = \frac{1^2}{2(1)+1} = \frac{1}{3} \]
\[ m = \frac{1(1+1)}{2(1)+1} = \frac{2}{3} \]

The equation of the tangent is:
\[ y - \frac{1}{3} = \frac{2}{3}(x - \ln 3) \]
\[ 3y - 1 = 2x - 2\ln 3 \implies 2x - 3y = 2\ln 3 - 1 \]

M1: Differentiate \( x \) to get \( \frac{dx}{dt} \).
M1: Differentiate \( y \) using the quotient rule.
A1: Obtain correct simplified \( \frac{dy}{dt} \).
M1: Apply the chain rule \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \).
A1: Obtain \( \frac{dy}{dx} = \frac{t(t+1)}{2t+1} \).
M1: Substitute \( t = 1 \) to find coordinates and gradient, and form the equation of the tangent.
A1: Obtain \( 2x - 3y = 2\ln 3 - 1 \) (or equivalent).

We use integration by parts, \( \int u \, dv = uv - \int v \, du \).

Let \( u = x \implies du = dx \).
Let \( dv = \sec^2 x \, dx \implies v = \tan x \).

Applying the formula:
\[ \int x \sec^2 x \, dx = x \tan x - \int \tan x \, dx \]

Since \( \int \tan x \, dx = \ln|\sec x| \) (or \( -\ln|\cos x| \)):
\[ \int x \sec^2 x \, dx = \left[ x \tan x - \ln|\sec x| \right]_{0}^{\frac{\pi}{3}} \]

Now evaluate the limits:
Upper limit \( x = \frac{\pi}{3} \):
\[ \frac{\pi}{3} \tan\left(\frac{\pi}{3}\right) - \ln\left|\sec\left(\frac{\pi}{3}\right)\right| = \frac{\pi}{3}(\sqrt{3}) - \ln(2) = \frac{\pi\sqrt{3}}{3} - \ln 2 \]

Lower limit \( x = 0 \):
\[ 0 \cdot \tan(0) - \ln|\sec(0)| = 0 - \ln(1) = 0 \]

Therefore, the exact value of the integral is:
\[ \frac{\pi\sqrt{3}}{3} - \ln 2 \]

M1: State or imply use of integration by parts with \( u = x \) and \( v' = \sec^2 x \).
A1: Obtain \( x \tan x \).
M1: Integrate \( \tan x \) to obtain \( \ln(\sec x) \) or \( -\ln(\cos x) \).
A1: State the correct indefinite integral \( x \tan x - \ln(\sec x) \).
M1: Substitute the limits of \( 0 \) and \( \frac{\pi}{3} \) correctly.
A1: Obtain the exact final answer \( \frac{\pi\sqrt{3}}{3} - \ln 2 \).

Separate variables to get:
\[ \frac{1}{y^2} \, dy = \frac{\ln x}{x} \, dx \]

Integrate both sides:
\[ \int y^{-2} \, dy = \int \frac{\ln x}{x} \, dx \]

The left side integrates to:
\[ -y^{-1} = -\frac{1}{y} \]

For the right side, let \( u = \ln x \implies du = \frac{1}{x} \, dx \):
\[ \int u \, du = \frac{1}{2} u^2 + C = \frac{1}{2}(\ln x)^2 + C \]

So we have:
\[ -\frac{1}{y} = \frac{1}{2}(\ln x)^2 + C \]

Using the initial condition \( y = 1 \) when \( x = 1 \):
\[ -\frac{1}{1} = \frac{1}{2}(\ln 1)^2 + C \implies -1 = 0 + C \implies C = -1 \]

Thus:
\[ -\frac{1}{y} = \frac{1}{2}(\ln x)^2 - 1 \]

Multiply by \( -1 \):
\[ \frac{1}{y} = 1 - \frac{1}{2}(\ln x)^2 = \frac{2 - (\ln x)^2}{2} \]

Inverting both sides, we get the explicit solution:
\[ y = \frac{2}{2 - (\ln x)^2} \]

M1: Separate variables correctly.
A1: Integrate \( y^{-2} \) to get \( -\frac{1}{y} \).
M1: Integrate \( \frac{\ln x}{x} \) (by substitution or inspection).
A1: Obtain \( \frac{1}{2}(\ln x)^2 \).
M1: Evaluate constant of integration \( C \) using given boundary conditions.
A1: Obtain the correct explicit solution \( y = \frac{2}{2 - (\ln x)^2} \).

To solve the quadratic equation \( z^2 - (4 - 2\mathrm{i})z + (6 - 8\mathrm{i}) = 0 \), we use the quadratic formula:
\[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -(4 - 2\mathrm{i}) \), and \( c = 6 - 8\mathrm{i} \).

First, we find the discriminant \( \Delta \):
\[ \Delta = b^2 - 4ac = [-(4 - 2\mathrm{i})]^2 - 4(1)(6 - 8\mathrm{i}) \]
\[ \Delta = (16 - 4 - 16\mathrm{i}) - (24 - 32\mathrm{i}) \]
\[ \Delta = 12 - 16\mathrm{i} - 24 + 32\mathrm{i} = -12 + 16\mathrm{i} \]

Now, we find the complex square roots of \( -12 + 16\mathrm{i} \). Let:
\[ (x + \mathrm{i}y)^2 = -12 + 16\mathrm{i} \]
Expanding and equating real and imaginary parts:
1) \( x^2 - y^2 = -12 \)
2) \( 2xy = 16 \implies xy = 8 \implies y = \frac{8}{x} \)

Substitute \( y = \frac{8}{x} \) into equation (1):
\[ x^2 - \frac{64}{x^2} = -12 \]
\[ x^4 + 12x^2 - 64 = 0 \]
\[ (x^2 + 16)(x^2 - 4) = 0 \]
Since \( x \) must be real, \( x^2 = 4 \implies x = \pm 2 \).
- If \( x = 2 \), then \( y = 4 \).
- If \( x = -2 \), then \( y = -4 \).

Thus, the square roots of the discriminant are \( \pm(2 + 4\mathrm{i}) \).

Now, substitute back into the quadratic formula:
\[ z = \frac{(4 - 2\mathrm{i}) \pm (2 + 4\mathrm{i})}{2} \]

This gives two solutions:
\[ z_1 = \frac{(4 - 2\mathrm{i}) + (2 + 4\mathrm{i})}{2} = \frac{6 + 2\mathrm{i}}{2} = 3 + \mathrm{i} \]
\[ z_2 = \frac{(4 - 2\mathrm{i}) - (2 + 4\mathrm{i})}{2} = \frac{2 - 6\mathrm{i}}{2} = 1 - 3\mathrm{i} \]

**M1**: Attempt to use the quadratic formula with correct coefficients.
**A1**: Obtain correct discriminant \( \Delta = -12 + 16\mathrm{i} \).
**M1**: Setup equations \( x^2 - y^2 = -12 \) and \( 2xy = 16 \) to find the square root of the discriminant.
**A1**: Solve to find the square roots \( \pm(2 + 4\mathrm{i}) \).
**M1**: Substitute the roots of the discriminant back into the quadratic formula expression.
**A1**: Obtain the final root \( z = 3 + \mathrm{i} \).
**A1**: Obtain the final root \( z = 1 - 3\mathrm{i} \).

To evaluate the integral, we can use the substitution method.
Let \( u = 1 + x^2 \). Then:
\[ \mathrm{d}u = 2x \, \mathrm{d}x \implies x \, \mathrm{d}x = \frac{1}{2} \, \mathrm{d}u \]
Also, \( x^2 = u - 1 \).

Now find the new limits of integration:
- When \( x = 0 \), \( u = 1 + 0^2 = 1 \).
- When \( x = 1 \), \( u = 1 + 1^2 = 2 \).

We rewrite the integral in terms of \( u \):
\[ \int_{0}^{1} \frac{x^3}{\sqrt{1+x^2}} \, \mathrm{d}x = \int_{0}^{1} \frac{x^2 \cdot x}{\sqrt{1+x^2}} \, \mathrm{d}x \]
\[ = \int_{1}^{2} \frac{u - 1}{\sqrt{u}} \cdot \frac{1}{2} \, \mathrm{d}u \]
\[ = \frac{1}{2} \int_{1}^{2} \left( u^{1/2} - u^{-1/2} \right) \, \mathrm{d}u \]

Now integrate term by term:
\[ = \frac{1}{2} \left[ \frac{2}{3}u^{3/2} - 2u^{1/2} \right]_{1}^{2} \]
\[ = \left[ \frac{1}{3}u^{3/2} - u^{1/2} \right]_{1}^{2} \]

Evaluate at the limits:
- At \( u = 2 \):
\[ \frac{1}{3}(2^{3/2}) - 2^{1/2} = \frac{2\sqrt{2}}{3} - \sqrt{2} = -\frac{1}{3}\sqrt{2} \]
- At \( u = 1 \):
\[ \frac{1}{3}(1^{3/2}) - 1^{1/2} = \frac{1}{3} - 1 = -\frac{2}{3} \]

Subtracting the lower limit from the upper limit:
\[ \left( -\frac{1}{3}\sqrt{2} \right) - \left( -\frac{2}{3} \right) = \frac{2 - \sqrt{2}}{3} \]

**M1**: Initiate a suitable substitution, e.g., \( u = 1+x^2 \) or \( u = \sqrt{1+x^2} \), finding the differential relation.
**A1**: Obtain the fully correct transformed integral in terms of \( u \) with correct limits.
**M1**: Integrate the simplified terms correctly, finding an expression of the form \( a u^{3/2} - b u^{1/2} \).
**A1**: Obtain the correct integrated expression \( \frac{1}{3}u^{3/2} - u^{1/2} \) (or equivalent in \( x \)).
**M1**: Apply limits correctly to their integrated expression.
**A1**: Obtain the final exact answer \( \frac{2 - \sqrt{2}}{3} \) (or equivalent simplified form).

For the lines \( l_1 \) and \( l_2 \) to intersect, there must exist values of the parameters \( \lambda \) and \( \mu \) such that their coordinates are identical. Equating the components from both vector equations:
1) \( 1 + 2\lambda = k + \mu \)
2) \( 2 + \lambda = -3 - 2\mu \)
3) \( -1 + 3\lambda = 8 + 2\mu \)

We first solve equations (2) and (3) simultaneously, as they do not contain the unknown constant \( k \).
From equation (2):
\[ \lambda = -5 - 2\mu \]

Substitute this expression for \( \lambda \) into equation (3):
\[ -1 + 3(-5 - 2\mu) = 8 + 2\mu \]
\[ -1 - 15 - 6\mu = 8 + 2\mu \]
\[ -16 - 6\mu = 8 + 2\mu \]
\[ -24 = 8\mu \implies \mu = -3 \]

Now find \( \lambda \) using equation (2):
\[ \lambda = -5 - 2(-3) = 1 \]

Substitute \( \lambda = 1 \) and \( \mu = -3 \) into equation (1) to determine the value of \( k \):
\[ 1 + 2(1) = k + (-3) \]
\[ 3 = k - 3 \implies k = 6 \]

To find the position vector of the point of intersection, substitute \( \lambda = 1 \) into the equation of \( l_1 \):
\[ \mathbf{r} = \begin{pmatrix} 1 + 2(1) \\ 2 + 1 \\ -1 + 3(1) \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \\ 2 \end{pmatrix} \]

Hence, \( k = 6 \) and the point of intersection is \( \begin{pmatrix} 3 \\ 3 \\ 2 \end{pmatrix} \).

**M1**: Equate components of \( l_1 \) and \( l_2 \) to obtain three equations involving \( \lambda \), \( \mu \), and \( k \).
**A1**: Obtain the correct set of simultaneous equations.
**M1**: Solve the two equations in \( \lambda \) and \( \mu \) simultaneously.
**A1**: Obtain the correct parameter values \( \lambda = 1 \) and \( \mu = -3 \).
**M1**: Substitute \( \lambda \) and \( \mu \) into the third equation to find \( k \).
**A1**: Obtain the correct value \( k = 6 \).
**A1**: Substitute the parameter back to obtain the correct position vector \( \begin{pmatrix} 3 \\ 3 \\ 2 \end{pmatrix} \) (or equivalent coordinate form).

The weight of the block is \(W = mg = 5 \times 10 = 50\text{ N}\).
Given \(\sin \theta = 0.6\), we have \(\cos \theta = 0.8\).

Resolving forces perpendicular to the inclined plane:
\(R = W \cos \theta = 50 \times 0.8 = 40\text{ N}\).

The maximum possible frictional force is:
\(F_{\max} = \mu R = 0.25 \times 40 = 10\text{ N}\).

The component of the weight acting down the slope is:
\(W \sin \theta = 50 \times 0.6 = 30\text{ N}\).

There are two extreme cases for equilibrium:

1. **Block on the point of slipping down the plane:**
In this case, the frictional force acts up the plane.
\(P_{\min} + F_{\max} = W \sin \theta\)
\(P_{\min} + 10 = 30 \implies P_{\min} = 20\text{ N}\).

2. **Block on the point of slipping up the plane:**
In this case, the frictional force acts down the plane.
\(P_{\max} = W \sin \theta + F_{\max}\)
\(P_{\max} = 30 + 10 = 40\text{ N}\).

Thus, the range of values of \(P\) for which the block remains in equilibrium is \(20 \le P \le 40\).

M1: Resolves perpendicular to the plane to find the normal reaction \(R\).
A1: Obtains \(R = 40\text{ N}\).
B1: Calculates limiting friction \(F_{\max} = 10\text{ N}\).
M1: Resolves parallel to the plane for the case where the block is about to slide down to find \(P_{\min}\).
A1: Obtains \(P_{\min} = 20\).
M1: Resolves parallel to the plane for the case where the block is about to slide up to find \(P_{\max}\).
A1: Obtains \(P_{\max} = 40\) and states the final range: \(20 \le P \le 40\).

To find the total distance, we must identify if and when the particle changes its direction of motion. We find the times when \(v = 0\):
\(3t^2 - 12t + 9 = 0 \implies 3(t^2 - 4t + 3) = 0 \implies 3(t - 1)(t - 3) = 0\).
Thus, the particle is instantaneously at rest at \(t = 1\) and \(t = 3\).

We integrate the velocity function to obtain the displacement function \(s(t)\), assuming \(s(0) = 0\):
\(s(t) = \int (3t^2 - 12t + 9) \, dt = t^3 - 6t^2 + 9t\).

Now, we evaluate the displacement at the key times \(t = 0\), \(t = 1\), \(t = 3\), and \(t = 4\):
- \(s(0) = 0\)
- \(s(1) = 1^3 - 6(1)^2 + 9(1) = 4\)
- \(s(3) = 3^3 - 6(3)^2 + 9(3) = 27 - 54 + 27 = 0\)
- \(s(4) = 4^3 - 6(4)^2 + 9(4) = 64 - 96 + 36 = 4\)

We calculate the distance travelled in each distinct interval:
- From \(t = 0\) to \(t = 1\): \(|4 - 0| = 4\text{ m}\)
- From \(t = 1\) to \(t = 3\): \(|0 - 4| = 4\text{ m}\)
- From \(t = 3\) to \(t = 4\): \(|4 - 0| = 4\text{ m}\)

Total distance travelled = \(4 + 4 + 4 = 12\text{ m\}}.

M1: Factorises the velocity expression to find the times when the particle is at rest.
A1: Correctly identifies \(t = 1\) and \(t = 3\).
M1: Integrates the velocity function to find the displacement function.
A1: Obtains \(s(t) = t^3 - 6t^2 + 9t\).
M1: Evaluates the displacement at \(t = 0\), \(1\), \(3\), and \(4\).
M1: Sums the absolute values of the differences in displacement across the intervals.
A1: Obtains the correct total distance of \(12\text{ m}\).

First, we identify all the forces opposing the motion of the car as it moves up the incline:
1. The component of the weight down the slope:
\(W_{\parallel} = mg \sin \alpha = 1200 \times 10 \times 0.05 = 600\text{ N}\).
2. The constant resistance force:
\(R = 300\text{ N}\).

The total resistance to motion is:
\(F_{\text{total}} = 600 + 300 = 900\text{ N}\).

The engine works at a constant rate of \(P = 24\text{ kW} = 24000\text{ W}\).
Using the power formula \(P = DF \times v\), where \(DF\) is the driving force:
At \(v = 20\text{ m s}^{-1}\):
\(DF = \frac{24000}{20} = 1200\text{ N}\).

Now, applying Newton's second law in the direction of motion:
\(DF - F_{\text{total}} = ma\)
\(1200 - 900 = 1200 a\)
\(300 = 1200 a \implies a = 0.25\text{ m s}^{-2}\).

B1: Calculates the component of gravity down the slope as \(600\text{ N}\).
M1: Uses \(P = Fv\) to calculate the driving force \(DF\).
A1: Obtains driving force \(DF = 1200\text{ N\}}.
M1: Applies Newton's second law along the slope: \)DF - mg\sin\alpha - R = ma\).
A1: Substitutes values correctly into the equation: \(1200 - 600 - 300 = 1200a\).
A1: Obtains the correct acceleration of \(0.25\text{ m s}^{-2}\).

Let the initial direction of motion of particle \(A\) be positive.
- Initial velocity of \(A\), \(u_A = 6\text{ m s}^{-1}\).
- Initial velocity of \(B\), \(u_B = -4\text{ m s}^{-1}\) (since they move towards each other).
- Final velocity of \(A\), \(v_A = -1.2\text{ m s}^{-1}\) (since its direction is reversed).

Let \(v_B\) be the final velocity of \(B\) after the collision.
Applying the principle of conservation of linear momentum:
\(m_A u_A + m_B u_B = m_A v_A + m_B v_B\)
\(0.5(6) + 0.3(-4) = 0.5(-1.2) + 0.3 v_B\)
\(3.0 - 1.2 = -0.6 + 0.3 v_B\)
\(1.8 = -0.6 + 0.3 v_B\)
\(2.4 = 0.3 v_B \implies v_B = 8\text{ m s}^{-1}\).
Since \(v_B\) is positive, \(B\) now moves in the positive direction (its direction of motion was also reversed).
Thus, the speed of \(B\) after the collision is \(8\text{ m s}^{-1}\).

To find the magnitude of the impulse \(I\) exerted on \(A\):
\(I = |m_A v_A - m_A u_A|\)
\(I = |0.5(-1.2) - 0.5(6)| = |-0.6 - 3.0| = |-3.6| = 3.6\text{ N s}\).

M1: Sets up the conservation of momentum equation with correct initial velocities and signs.
A1: Substitutes the correct numerical values into the momentum equation.
A1: Correctly calculates the final speed of \(B\) as \(8\text{ m s}^{-1}\).
M1: Applies the impulse formula \(I = |m \Delta v|\) to particle \(A\) (or \(B\)).
A1: Correctly substitutes values for \(A\): \(0.5(-1.2) - 0.5(6)\).
A1: Obtains the correct magnitude of impulse as \(3.6\text{ N s}\).

Let \(a\) be the acceleration of the system and \(T\) be the tension in the string.

For block \(P\) (moving horizontally on the rough table):
- Normal reaction force: \(R = m_P g = 3 \times 10 = 30\text{ N}\).
- Frictional force: \(F = \mu R = 0.4 \times 30 = 12\text{ N}\).
Using Newton's second law:
\(T - F = m_P a \implies T - 12 = 3a\) --- (Equation 1)

For block \(Q\) (hanging vertically and moving downwards):
Using Newton's second law:
\(m_Q g - T = m_Q a \implies 20 - T = 2a\) --- (Equation 2)

Add Equation 1 and Equation 2 to eliminate \(T\):
\((T - 12) + (20 - T) = 3a + 2a\)
\(8 = 5a \implies a = 1.6\text{ m s}^{-2}\).

Substitute \(a = 1.6\) back into Equation 1 to find \(T\):
\(T - 12 = 3(1.6) \implies T = 12 + 4.8 = 16.8\text{ N}\).

B1: Calculates the friction force \(F = 12\text{ N}\) on block \(P\).
M1: Writes down the equation of motion for block \(P\).
M1: Writes down the equation of motion for block \(Q\).
A1: Obtains both correct equations: \(T - 12 = 3a\) and \(20 - T = 2a\).
M1: Solves the system of equations for either \(a\) or \(T\).
A1: Obtains the correct acceleration \(a = 1.6\text{ m s}^{-2}\).
A1: Obtains the correct tension \(T = 16.8\text{ N}\).

(i) At point \(A\), the kinetic energy of the bead is:
\(KE_A = \frac{1}{2} m v^2 = \frac{1}{2} \times 0.4 \times 6^2 = 7.2\text{ J}\).

As the bead moves to \(B\), the vertical height gained is:
\(h = AB \sin \theta = 2 \times 0.6 = 1.2\text{ m}\).

The gain in gravitational potential energy is:
\(GPE = mgh = 0.4 \times 10 \times 1.2 = 4.8\text{ J}\).

Since the bead is at rest at \(B\), its final kinetic energy is \(0\). Applying the work-energy principle:
\(KE_A = GPE + WD_{\text{friction}}\)
\(7.2 = 4.8 + WD_{\text{friction}} \implies WD_{\text{friction}} = 2.4\text{ J}\).

(ii) The work done against friction is given by:
\(WD_{\text{friction}} = F \times AB\), where \(F\) is the constant friction force.
\(2.4 = F \times 2 \implies F = 1.2\text{ N}\).

Resolving forces perpendicular to the wire:
\(R = mg \cos \theta\).
Since \(\sin \theta = 0.6\), then \(\cos \theta = 0.8\).
\(R = 0.4 \times 10 \times 0.8 = 3.2\text{ N}\).

Using the relation \(F = \mu R\):
\(\mu = \frac{F}{R} = \frac{1.2}{3.2} = 0.375\).

M1: Calculates initial kinetic energy at \(A\) and gain in potential energy at \(B\).
A1: Correctly determines \(KE_A = 7.2\text{ J}\) and \(GPE = 4.8\text{ J\}}.
A1: Applies work-energy relation to obtain \)WD_{\text{friction}} = 2.4\text{ J}\).
M1: Relates work done to friction force: \(WD_{\text{friction}} = F \times d\) to find \(F\).
A1: Obtains \(F = 1.2\text{ N}\).
M1: Resolves perpendicular to the wire to find \(R\) and uses \(\mu = F/R\).
A1: Obtains \(\mu = 0.375\).

Because the system of forces is in equilibrium, the sum of both the horizontal components and the vertical components must equal zero.

Resolving forces horizontally (along the \(x\)-axis):
\(15 - F \cos \alpha = 0 \implies F \cos \alpha = 15\) --- (Equation 1)

Resolving forces vertically (along the \(y\)-axis):
\(F \sin \alpha - 8 = 0 \implies F \sin \alpha = 8\) --- (Equation 2)

We square and add both equations to eliminate \(\alpha\):
\((F \cos \alpha)^2 + (F \sin \alpha)^2 = 15^2 + 8^2\)
\(F^2 (\cos^2 \alpha + \sin^2 \alpha) = 225 + 64\)
\(F^2 = 289 \implies F = 17\).

To find \(\alpha\), we divide Equation 2 by Equation 1:
\(\frac{F \sin \alpha}{F \cos \alpha} = \frac{8}{15}\)
\(\tan \alpha = \frac{8}{15}\)
\[\alpha = \arctan\left(\frac{8}{15}\right) \approx 28.07^\circ \approx 28.1^\circ\.\]

M1: Resolves forces horizontally to obtain an equation in terms of \(F\) and \(\alpha\).
M1: Resolves forces vertically to obtain a second equation.
A1: Correctly states \(F \cos \alpha = 15\) and \(F \sin \alpha = 8\).
M1: Employs a valid algebraic/trigonometric method to eliminate \(\alpha\).
A1: Obtains \(F = 17\).
M1: Employs a valid trigonometric method to calculate the angle \(\alpha\).
A1: Obtains \(\alpha = 28.1^\circ\) (or \(28.1\)).

Let \(y = w - 50\). Then \(\sum y = 480\) and \(\sum y^2 = 7920\) with \(n_1 = 120\).

(a) The mean of \(y\) is \(\bar{y} = \frac{480}{120} = 4\).
Therefore, the mean of \(w\) is \(\bar{w}_1 = \bar{y} + 50 = 4 + 50 = 54\).

The variance of \(w\) is equal to the variance of \(y\):
\(\text{Var}(w) = \frac{\sum y^2}{n_1} - \bar{y}^2 = \frac{7920}{120} - 4^2 = 66 - 16 = 50\).

(b) The second group has \(n_2 = 80\) and mean \(\bar{w}_2 = 53\).
Combined mean \(\bar{w}_C = \frac{n_1\bar{w}_1 + n_2\bar{w}_2}{n_1 + n_2} = \frac{120 \times 54 + 80 \times 53}{200} = \frac{6480 + 4240}{200} = \frac{10720}{200} = 53.6\).

(c) For the first group, \(\sum w_1^2 = n_1(\text{Var}_1 + \bar{w}_1^2) = 120(50 + 54^2) = 355920\).
For the second group, \(\text{Var}_2 = 4^2 = 16\).
\(\sum w_2^2 = n_2(\text{Var}_2 + \bar{w}_2^2) = 80(16 + 53^2) = 80 \times 2825 = 226000\).

For the combined group, total sum of squares is \(\sum w^2 = 355920 + 226000 = 581920\).
Combined variance \(\text{Var}_C = \frac{581920}{200} - 53.6^2 = 2909.6 - 2872.96 = 36.64\).
Combined standard deviation \(\sigma_C = \sqrt{36.64} \approx 6.05\) (to 3 s.f.).

(a) M1 for calculating \bar{y} = 4, A1 for \bar{w} = 54. M1 for variance formula, A1 for Variance = 50.
(b) M1 for combining total weights, A1 for 53.6.
(c) M1 for finding sum of squares for both groups, M1 for combined variance formula, A1 for 6.05.

The letters of PROPORTION are: O, O, O, P, P, R, R, T, I, N (total of 10 letters).

(a) The number of arrangements is \(\frac{10!}{3! \times 2! \times 2!} = \frac{3628800}{6 \times 2 \times 2} = 151200\).

(b) Let the block (OOO) be treated as one entity. We have 8 entities: (OOO), P, P, R, R, T, I, N.
To find arrangements where the two Ps are not adjacent, we first arrange the other 6 entities: (OOO), R, R, T, I, N.
Number of arrangements of these 6 entities is \(\frac{6!}{2!} = 360\).
These 6 entities create 7 possible spaces for the two Ps.
The number of ways to choose 2 spaces out of 7 for the identical Ps is \(\binom{7}{2} = 21\).
Total arrangements = \(360 \times 21 = 7560\).
Alternatively, total arrangements of the 8 entities is \(\frac{8!}{2! \times 2!} = 10080\).
Arrangements with adjacent Ps (treated as one entity (PP)) is \(\frac{7!}{2!} = 2520\).
Number of non-adjacent arrangements = \(10080 - 2520 = 7560\).

(c) We select 4 letters from O (3), P (2), R (2), T (1), I (1), N (1).
We require at least one O (1, 2, or 3) and at most one P (0 or 1).
We classify by the number of Os (\(o\)) and Ps (\(p\)) chosen:
- Case 1: \(o=1, p=0\). We need 3 other letters from {R, R, T, I, N}.
Ways = 3 (if two Rs, e.g. RRT, RRI, RRN) + 4 (if three distinct, \(\binom{4}{3}\)) = 7.
- Case 2: \(o=1, p=1\). We need 2 other letters from {R, R, T, I, N}.
Ways = 1 (RR) + \(\binom{4}{2}\) = 7.
- Case 3: \(o=2, p=0\). We need 2 other letters from {R, R, T, I, N}.
Ways = 7.
- Case 4: \(o=2, p=1\). We need 1 other letter from {R, R, T, I, N}.
Ways = \(\binom{4}{1}\) = 4.
- Case 5: \(o=3, p=0\). We need 1 other letter from {R, R, T, I, N}.
Ways = 4.
- Case 6: \(o=3, p=1\). We need 0 other letters.
Ways = 1.
Total number of selections = \(7 + 7 + 7 + 4 + 4 + 1 = 30\).

(a) M1 for \(\frac{10!}{3!2!2!}\), A1 for 151200.
(b) M1 for treating (OOO) as a single entity, M1 for subtracting adjacent Ps case or using the gap method, A1 for 7560.
(c) M1 for setting up a systematic approach based on counts of O and P, A1 for identifying correct cases, A1 for 30.

Let \(A, M, D\) represent the events that a student studies Art, Music, and Drama respectively.
Let \(X\) and \(Y\) represent male and female respectively.

(a) Using the law of total probability:
\(P(\text{Male}) = P(A \cap X) + P(M \cap X) + P(D \cap X)\)
\(P(\text{Male}) = 0.40 \times 0.30 + 0.35 \times 0.60 + 0.25 \times 0.50 = 0.12 + 0.21 + 0.125 = 0.455\).

(b) The probability that a student is female is \(P(\text{Female}) = 1 - 0.455 = 0.545\).
We want \(P(M|\text{Female}) = \frac{P(M \cap \text{Female})}{P(\text{Female})}\).
\(P(M \cap \text{Female}) = 0.35 \times (1 - 0.60) = 0.35 \times 0.40 = 0.14\).
So \(P(M|\text{Female}) = \frac{0.14}{0.545} = \frac{140}{545} \approx 0.257\) (to 3 s.f.).

(c) The probability that a student studies Art is \(0.40\).
Let \(N\) be the number of students who study Art out of 3 chosen at random.
\(N \sim \text{B}(3, 0.40)\).
\(P(N = 2) = \binom{3}{2} (0.40)^2 (0.60)^1 = 3 \times 0.16 \times 0.60 = 0.288\).

(a) M1 for sum of three correct products, A1 for 0.455.
(b) M1 for finding \(P(\text{Female})\), M1 for using conditional probability formula, A1 for 0.257.
(c) M1 for binomial formula with \(n=3, p=0.4\), A1 for 0.288.

Since the sum of the probabilities must be 1:
\(a + b + 0.15 + (b + 0.1) + 2a = 1 \implies 3a + 2b + 0.25 = 1 \implies 3a + 2b = 0.75\). (Equation 1)

Given that \(\text{E}(X) = 0.4\):
\(\text{E}(X) = -2(a) - 1(b) + 0(0.15) + 1(b + 0.1) + 2(2a) = 0.4\)
\(-2a - b + b + 0.1 + 4a = 0.4\)
\(2a + 0.1 = 0.4 \implies 2a = 0.3 \implies a = 0.15\).

Substitute \(a = 0.15\) into Equation 1:
\(3(0.15) + 2b = 0.75 \implies 0.45 + 2b = 0.75 \implies 2b = 0.30 \implies b = 0.15\).

(b) The probability distribution is now:
\(\text{P}(X = -2) = 0.15\)
\(\text{P}(X = -1) = 0.15\)
\(\text{P}(X = 0) = 0.15\)
\(\text{P}(X = 1) = 0.25\)
\(\text{P}(X = 2) = 0.30\)

First, find \(\text{E}(X^2)\):
\(\text{E}(X^2) = (-2)^2(0.15) + (-1)^2(0.15) + 0^2(0.15) + 1^2(0.25) + 2^2(0.30)\)
\(\text{E}(X^2) = 4(0.15) + 1(0.15) + 0.25 + 4(0.30) = 0.60 + 0.15 + 0.25 + 1.20 = 2.20\).

Now calculate the variance:
\(\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2 = 2.20 - (0.4)^2 = 2.20 - 0.16 = 2.04\).

(a) M1 for summing probabilities to 1, M1 for expectation formula in terms of \(a\) and \(b\), A1 for \(a = 0.15\), A1 for \(b = 0.15\).
(b) M1 for finding \(\text{E}(X^2)\), M1 for using the variance formula, A1 for 2.04.

(a) We are given \(P(T < 12) = 0.15\) and \(P(T < 24) = 0.70\).

For \(P(T < 12) = 0.15\):
\(\frac{12 - \mu}{\sigma} = \Phi^{-1}(0.15) = -1.036\)
\(12 - \mu = -1.036\sigma \implies \mu - 1.036\sigma = 12\) (Equation 1)

For \(P(T < 24) = 0.70\):
\(\frac{24 - \mu}{\sigma} = \Phi^{-1}(0.70) = 0.524\)
\(24 - \mu = 0.524\sigma \implies \mu + 0.524\sigma = 24\) (Equation 2)

Subtract Equation 1 from Equation 2:
\(1.56\sigma = 12 \implies \sigma \approx 7.69\) (to 3 s.f.)

Substitute \(\sigma = 7.6923\) into Equation 2:
\(\mu = 24 - 0.524(7.6923) \approx 20.0\) (to 3 s.f.)

(b) We want to find \(P(T > 28)\):
\(P\left(Z > \frac{28 - 19.97}{7.692}\right) = P(Z > 1.044) = 1 - \Phi(1.044) \approx 1 - 0.8517 = 0.148\).

(a) M1 for setting up standardisation equations with correct sign, A1 for correct z-values (-1.036 and 0.524), M1 for solving simultaneous equations, A1 for \(\sigma = 7.69\), A1 for \(\mu = 20.0\).
(b) M1 for standardising 28, M1 for calculating correct tail probability, A1 for 0.148.

Let \(X\) be the number of defective light bulbs in the sample of 150.
\(X \sim \text{B}(150, 0.08)\).

(a) We approximate \(X\) by a normal distribution \(Y \sim \text{N}(\mu, \sigma^2)\):
\(\mu = np = 150 \times 0.08 = 12\)
\(\sigma^2 = np(1-p) = 150 \times 0.08 \times 0.92 = 11.04\)

We want to find \(P(X \ge 15)\). Applying a continuity correction:
\(P(X \ge 15) \approx P(Y > 14.5)\)

Standardising:
\(Z = \frac{14.5 - 12}{\sqrt{11.04}} = \frac{2.5}{3.3226} \approx 0.7524\)

\(P(Z > 0.7524) = 1 - \Phi(0.7524) \approx 1 - 0.7741 = 0.226\) (to 3 s.f.).

(b) To justify the normal approximation, we must check that both \(np > 5\) and \(n(1-p) > 5\):
\(np = 150 \times 0.08 = 12 > 5\)
\(n(1-p) = 150 \times 0.92 = 138 > 5\)
Since both conditions are satisfied, the normal approximation is valid.

(a) B1 for finding mean 12 and variance 11.04, M1 for using continuity correction (14.5), M1 for standardising, M1 for correct tail probability calculation, A1 for 0.226.
(b) B1 for checking \(np = 12 > 5\), B1 for checking \(n(1-p) = 138 > 5\), B1 for conclusion.

(i) \(E(X) = \int_{1}^{3} x \cdot \frac{3}{8}(x-1)^2 \, dx = \frac{3}{8} \int_{1}^{3} (x^3 - 2x^2 + x) \, dx = \frac{3}{8} \left[ \frac{x^4}{4} - \frac{2x^3}{3} + \frac{x^2}{2} \right]_1^3\).
At \(x=3\): \(\frac{81}{4} - 18 + \frac{9}{2} = 6.75\).
At \(x=1\): \(\frac{1}{4} - \frac{2}{3} + \frac{1}{2} = \frac{1}{12}\).
Thus, \(E(X) = \frac{3}{8} \left( \frac{27}{4} - \frac{1}{12} \right) = \frac{3}{8} \times \frac{80}{12} = 2.5\).

(ii) For \(1 \le x \le 3\):
\(F(x) = \int_{1}^{x} \frac{3}{8}(t-1)^2 \, dt = \left[ \frac{1}{8}(t-1)^3 \right]_1^x = \frac{1}{8}(x-1)^3\).

(iii) The median \(m\) satisfies \(F(m) = 0.5\):
\(\frac{1}{8}(m-1)^3 = 0.5 \implies (m-1)^3 = 4 \implies m = 1 + 4^{1/3} \approx 2.59\).

(i)
M1: Attempt to integrate \(x f(x)\) with correct limits.
A1: Correct integration with values substituted.
A1: Correct answer of 2.5.

(ii)
M1: Attempt to integrate \(f(t)\) from 1 to \(x\).
A1: Obtain \(\frac{1}{8}(x-1)^3\).

(iii)
M1: Set \(F(m) = 0.5\) and attempt to solve for \(m\).
A1: Obtain 2.59 (or \(1 + 4^{1/3}\)).

Let \(\mu\) be the population mean recovery time under the new program.
\(H_0: \mu = 12.4\)
\(H_1: \mu < 12.4\)

Under \(H_0\), the sample mean \(\bar{X}\) is distributed as \(N\left(12.4, \frac{3.2^2}{40}\right)\).
Standard error \(= \frac{3.2}{\sqrt{40}} \approx 0.5060\).

Calculate the test statistic \(z\):
\(z = \frac{11.3 - 12.4}{0.5060} = \frac{-1.1}{0.5060} \approx -2.174\).

For a one-tailed test at the 5% level of significance, the critical region is \(z < -1.645\).
Since \(-2.174 < -1.645\), we reject the null hypothesis \(H_0\).

There is significant evidence at the 5% level of significance to support the claim that the new treatment program reduces the mean recovery time.

B1: State correct null and alternative hypotheses with \(\mu\).
M1: Calculate the standard error of the mean \(\frac{3.2}{\sqrt{40}}\).
M1: Calculate the correct test statistic \(z\).
A1: Obtain \(z = -2.17\) or \(-2.174\) (or p-value \(= 0.0149\)).
M1: Compare their \(z\)-value to \(-1.645\) (or p-value to \(0.05\)).
A1: Correct decision to reject \(H_0\).
A1: Contextual conclusion stating there is evidence that the treatment reduces recovery time.

(i) Let \(X\) be the number of defects in a 200-metre stretch.
The mean rate is \(\lambda = 1.2 \times 2 = 2.4\).
\(X \sim \text{Po}(2.4)\).
\(P(X = 3) = \frac{e^{-2.4} \times 2.4^3}{3!} \approx 0.209\).

(ii) Let \(Y\) be the number of defects in a 5 km (5000-metre) stretch.
The mean rate is \(\lambda = 1.2 \times 50 = 60\).
\(Y \sim \text{Po}(60)\).
Since \(\lambda > 15\), we can approximate \(Y\) by a normal distribution \(W \sim N(60, 60)\).
We want \(P(50 \le Y \le 65)\).
Applying a continuity correction, we calculate \(P(49.5 \le W \le 65.5)\):
\(z_1 = \frac{49.5 - 60}{\sqrt{60}} \approx -1.355\)
\(z_2 = \frac{65.5 - 60}{\sqrt{60}} \approx 0.710\)

\(P(-1.355 \le Z \le 0.710) = \Phi(0.710) - (1 - \Phi(1.355)) = 0.7611 - (1 - 0.9123) = 0.6734 \approx 0.673\).

(i)
M1: Calculate the mean for 200m as \(\lambda = 2.4\).
M1: Substitute into the Poisson formula for \(X = 3\).
A1: Obtain 0.209.

(ii)
B1: State the correct approximating normal distribution \(N(60, 60)\).
M1: Apply the continuity correction correctly to get limits \(49.5\) and \(65.5\).
M1: Standardize both values with their mean and standard deviation.
A1: Obtain 0.673 (or 0.674).

Let the mass of the 4 medium apples be \(M_1 + M_2 + M_3 + M_4\) and the mass of the 2 large apples be \(L_1 + L_2\).
We define a new random variable \(T = (M_1 + M_2 + M_3 + M_4) - (L_1 + L_2)\).

Since \(M\) and \(L\) are normally distributed, \(T\) is also normally distributed.
\(E(T) = 4 \times E(M) - 2 \times E(L) = 4(120) - 2(230) = 480 - 460 = 20\text{ g}\).

\(\text{Var}(T) = 4 \times \text{Var}(M) + 2 \times \text{Var}(L) = 4(8^2) + 2(15^2) = 4(64) + 2(225) = 256 + 450 = 706\text{ g}^2\).
So \(T \sim N(20, 706)\).

We want to find \(P(T > 0)\):
\(z = \frac{0 - 20}{\sqrt{706}} = \frac{-20}{26.5707} \approx -0.753\).

\(P(Z > -0.753) = \Phi(0.753) \approx 0.774\).

M1: State or imply expectation of the combined variable, \(E(T) = 20\).
M1: Identify that \(\text{Var}(T) = 4\text{Var}(M) + 2\text{Var}(L)\).
A1: Obtain \(\text{Var}(T) = 706\).
M1: Express the required probability as \(P(T > 0)\).
M1: Standardize 0 with their mean and standard deviation.
A1: Identify the correct \(z\)-value of \(-0.753\).
A1: Obtain 0.774.

(i) Unbiased estimate of the population mean \(\bar{x}\):
\(\bar{x} = \frac{\sum x}{n} = \frac{120500}{100} = 1205\text{ hours}\).

Unbiased estimate of the population variance \(s^2\):
\(s^2 = \frac{1}{n-1} \left( \sum x^2 - \frac{(\sum x)^2}{n} \right) = \frac{1}{99} \left( 145902500 - \frac{120500^2}{100} \right)\)
\(s^2 = \frac{1}{99} (145902500 - 145202500) = \frac{700000}{99} \approx 7070.7\text{ hours}^2\).

(ii) The standard error is \(\frac{s}{\sqrt{n}} = \sqrt{\frac{7070.7}{100}} \approx 8.4087\).
For a 95% confidence interval, the critical value \(z = 1.96\).

Confidence interval \(= \bar{x} \pm z \frac{s}{\sqrt{n}} = 1205 \pm 1.96(8.4087) = 1205 \pm 16.48\).
Lower limit: \(1205 - 16.48 = 1188.52 \approx 1189\).
Upper limit: \(1205 + 16.48 = 1221.48 \approx 1221\).
Thus, the 95% confidence interval is \((1189, 1221)\).

(i)
B1: State correct mean \(\bar{x} = 1205\).
M1: Apply the correct unbiased variance formula.
A1: Obtain \(7070.7\) (or \(7070\)).

(ii)
B1: Identify correct critical value \(z = 1.96\).
M1: Apply correct confidence interval formula \(\bar{x} \pm z \frac{s}{\sqrt{n}}\).
A1: Obtain the margin of error \(16.5\).
A1: Obtain correct interval \((1189, 1221)\) (or equivalent).

Let \(\lambda\) be the mean number of wildflowers per 3 square metres.
Under the null hypothesis \(H_0\), the density is unchanged:
\(H_0: \lambda = 7.5\) (since \(3 \times 2.5 = 7.5\))
\(H_1: \lambda < 7.5\)

Let \(X\) be the number of wildflowers in a 3 square metre plot. Under \(H_0\), \(X \sim \text{Po}(7.5)\).
We observe \(X = 3\). To test for a decrease, we find \(P(X \le 3)\):
\(P(X \le 3) = e^{-7.5} \left( 1 + 7.5 + \frac{7.5^2}{2} + \frac{7.5^3}{6} \right)\)
\(P(X \le 3) = e^{-7.5} (1 + 7.5 + 28.125 + 70.3125) = e^{-7.5} (106.9375)\)
\(P(X \le 3) \approx 0.00055308 \times 106.9375 \approx 0.0591\).

We compare this probability with the significance level \(\alpha = 0.10\).
Since \(0.0591 < 0.10\), the result is significant, and we reject \(H_0\).

There is significant evidence at the 10% level that the density of the wildflowers has decreased.

B1: State correct hypotheses in terms of \(\lambda\) or the population mean.
B1: Determine correct mean \(\lambda = 7.5\) for the 3 square metre plot.
M1: Establish that \(P(X \le 3)\) needs to be calculated.
M1: Write down the correct summation of Poisson terms for \(X = 0, 1, 2, 3\).
A1: Obtain the correct probability \(0.0591\).
M1: Compare their probability with \(0.10\).
A1: Correct conclusion in context.

Let \(T\) be the total mass of a filled bottle.
\(T = 1.2V + M\).

(i) Since \(V\) and \(M\) are independent:
\(E(T) = 1.2 E(V) + E(M) = 1.2(505) + 120 = 606 + 120 = 726\text{ g}\).

\(\text{Var}(T) = 1.2^2 \text{Var}(V) + \text{Var}(M) = 1.44(3^2) + 2^2 = 1.44(9) + 4 = 12.96 + 4 = 16.96\text{ g}^2\).

(ii) We want to find \(P(T > 735)\):
\(T \sim N(726, 16.96)\).

\(z = \frac{735 - 726}{\sqrt{16.96}} = \frac{9}{4.1183} \approx 2.185\).

\(P(T > 735) = P(Z > 2.185) = 1 - \Phi(2.185)\).
Using standard normal tables:
\(\Phi(2.185) \approx 0.9856\).
\(P(T > 735) = 1 - 0.9856 = 0.0144\).

(i)
M1: Attempt to find \(E(T)\) using linear combination of means.
A1: Obtain 726.
M1: Use \(\text{Var}(aX+bY) = a^2\text{Var}(X) + b^2\text{Var}(Y)\) to calculate variance.
A1: Obtain 16.96.

(ii)
M1: Standardize with their mean and standard deviation.
M1: Correct method to find the upper tail probability \(1 - \Phi(z)\).
A1: Obtain 0.0144 (accept 0.0143 to 0.0145).