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To find the points of intersection, we equate the equations of the line and the curve:

\( x^2 + kx - 1 = 2kx - 5 \)

Rearranging into a standard quadratic equation \( ax^2 + bx + c = 0 \):

\( x^2 - kx + 4 = 0 \)

For the line and the curve to intersect at two distinct points, the discriminant of this quadratic equation must be strictly greater than zero:

\( b^2 - 4ac > 0 \)

Substituting the coefficients \( a = 1 \), \( b = -k \), and \( c = 4 \):

\( (-k)^2 - 4(1)(4) > 0 \)
\( k^2 - 16 > 0 \)

Solving this inequality gives the critical values:

\( (k - 4)(k + 4) > 0 \)

Thus, the set of values of \( k \) is:

\( k < -4 \) or \( k > 4 \).

M1: Equating the line and curve equations and rearranging into standard quadratic form.
A1: Correct quadratic equation \( x^2 - kx + 4 = 0 \).
M1: Using the discriminant condition \( b^2 - 4ac > 0 \).
A1: Formulating the correct inequality \( k^2 - 16 > 0 \).
M1: Solving the quadratic inequality to find critical values and determining the outer regions.
A1.8: Correct final answer: \( k < -4 \) or \( k > 4 \).

To find the inverse function, we let \( y = f(x) \) and rearrange to make \( x \) the subject:

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

Multiply both sides by \( (x-1) \):

\( y(x-1) = 2x+3 \)
\( xy - y = 2x+3 \)

Rearrange to group all terms containing \( x \) on one side:

\( xy - 2x = y+3 \)
\( x(y-2) = y+3 \)
\( x = \frac{y+3}{y-2} \)

Thus, the inverse function is:

\( f^{-1}(x) = \frac{x+3}{x-2} \)

To find the domain of \( f^{-1} \), we determine the range of the original function \( f \) for \( x > 1 \):

\( f(x) = \frac{2(x-1)+5}{x-1} = 2 + \frac{5}{x-1} \)

Since \( x > 1 \), we have \( x-1 > 0 \), which implies \( \frac{5}{x-1} > 0 \). Therefore, \( f(x) > 2 \).
The range of \( f \) is \( f(x) > 2 \), so the domain of \( f^{-1} \) is \( x > 2 \).
The range of \( f^{-1} \) is the same as the domain of \( f \), which is \( f^{-1}(x) > 1 \).

M1: Attempting to rearrange and make \( x \) the subject of \( y = \frac{2x+3}{x-1} \) by multiplying out the fraction.
A1: Correctly factorizing to obtain \( x(y-2) = y+3 \).
A1: Obtaining the correct expression for \( f^{-1}(x) = \frac{x+3}{x-2} \).
M1: Attempting to find the range of \( f \) through algebraic manipulation or graphical analysis.
A1: Stating the correct domain of \( f^{-1} \) as \( x > 2 \).
A1.8: Stating the correct range of \( f^{-1} \) as \( f^{-1}(x) > 1 \).

First, find the center \( M \) of the circle, which is the midpoint of the diameter \( AB \):

\( M = \left( \frac{2+8}{2}, \frac{5 + (-3)}{2} \right) = (5, 1) \)

Next, find the gradient of the radius \( MA \):

\( m_{MA} = \frac{5 - 1}{2 - 5} = \frac{4}{-3} = -\frac{4}{3} \)

Since the tangent at \( A \) is perpendicular to the radius \( MA \), the gradient of the tangent, \( m \), is given by:

\( m = -\frac{1}{m_{MA}} = -\frac{1}{-\frac{4}{3}} = \frac{3}{4} \)

Now, use the point-slope formula to find the equation of the tangent passing through \( A(2, 5) \):

\( y - 5 = \frac{3}{4}(x - 2) \)

Multiply both sides by 4 to clear the fraction:

\( 4(y - 5) = 3(x - 2) \)
\( 4y - 20 = 3x - 6 \)

Rearranging into the form \( ax + by + c = 0 \):

\( 3x - 4y + 14 = 0 \)

M1: Attempting to find the coordinates of the center of the circle, \( M \), as the midpoint of \( AB \).
A1: Correct coordinates \( (5, 1) \).
M1: Finding the gradient of the line segment \( MA \) (or the diameter \( AB \)).
A1: Correct gradient of \( MA \) as \( -\frac{4}{3} \).
M1: Using the perpendicular relation \( m_1 m_2 = -1 \) to find the gradient of the tangent.
A1.8: Obtaining the correct equation of the tangent in the required integer form \( 3x - 4y + 14 = 0 \).

The perimeter \( P \) of the sector is given by:

\( P = 2r + r\theta = 30 \) --- (Equation 1)

The area \( A \) of the sector is given by:

\( A = \frac{1}{2}r^2\theta = 50 \) --- (Equation 2)

From Equation 1, we can express \( \theta \) in terms of \( r \):

\( r\theta = 30 - 2r \implies \theta = \frac{30 - 2r}{r} \)

Substitute this expression for \( \theta \) into Equation 2:

\( \frac{1}{2}r^2 \left( \frac{30 - 2r}{r} \right) = 50 \)
\( \frac{1}{2}r(30 - 2r) = 50 \)
\( r(15 - r) = 50 \)
\( 15r - r^2 = 50 \)
\( r^2 - 15r + 50 = 0 \)

Solve the quadratic equation by factoring:

\( (r - 5)(r - 10) = 0 \)

This gives two possible values of \( r \):

1) If \( r = 5 \):
\( \theta = \frac{30 - 2(5)}{5} = \frac{20}{5} = 4 \) radians.

2) If \( r = 10 \):
\( \theta = \frac{30 - 2(10)}{10} = \frac{10}{10} = 1 \) radian.

M1: Recalling and using correct formulas for the perimeter \( 2r + r\theta \) and area \( \frac{1}{2}r^2\theta \).
A1: Setting up both equations correctly as \( r(2+\theta) = 30 \) and \( \frac{1}{2}r^2\theta = 50 \).
M1: Eliminating one variable to produce a quadratic equation in terms of one variable.
A1: Correct quadratic equation \( r^2 - 15r + 50 = 0 \).
M1: Solving the quadratic equation to obtain the two values of \( r \).
A1.8: Correctly matching the values \( r = 5 \implies \theta = 4 \) and \( r = 10 \implies \theta = 1 \).

Use the identity \( \tan x = \frac{\sin x}{\cos x} \) to rewrite the equation:

\( \sin x \left( \frac{\sin x}{\cos x} \right) = 3\cos x - 1 \)
\( \frac{\sin^2 x}{\cos x} = 3\cos x - 1 \)

Multiply through by \( \cos x \) (where \( \cos x \neq 0 \)):

\( \sin^2 x = 3\cos^2 x - \cos x \)

Substitute the identity \( \sin^2 x = 1 - \cos^2 x \):

\( 1 - \cos^2 x = 3\cos^2 x - \cos x \)
\( 4\cos^2 x - \cos x - 1 = 0 \)

This is a quadratic equation in terms of \( \cos x \). Applying the quadratic formula:

\( \cos x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(4)(-1)}}{2(4)} \)
\( \cos x = \frac{1 \pm \sqrt{17}}{8} \)

Case 1: \( \cos x = \frac{1 + \sqrt{17}}{8} \approx 0.640388 \)
\( x = \cos^{-1}(0.640388) \approx 50.18^\circ \)
Since \( 0^\circ \le x \le 360^\circ \), the other solution is:
\( x = 360^\circ - 50.18^\circ \approx 309.82^\circ \)

Case 2: \( \cos x = \frac{1 - \sqrt{17}}{8} \approx -0.390388 \)
\( x = \cos^{-1}(-0.390388) \approx 112.98^\circ \)
The other solution in the range is:
\( x = 360^\circ - 112.98^\circ \approx 247.02^\circ \)

Rounding all solutions to 1 decimal place:
\( x = 50.2^\circ, 113.0^\circ, 247.0^\circ, 309.8^\circ \).

M1: Substituting \( \tan x = \frac{\sin x}{\cos x} \) and clearing the fraction.
M1: Applying the identity \( \sin^2 x = 1 - \cos^2 x \) to form a quadratic equation in terms of \( \cos x \).
A1: Obtaining the correct quadratic equation \( 4\cos^2 x -
\cos x - 1 = 0 \).
M1: Solving the quadratic equation to find two valid values for \( \cos x \).
A1: Finding the solutions in the first and second quadrants: \( 50.2^\circ \) and \( 113.0^\circ \).
A1.8: Finding the remaining solutions: \( 247.0^\circ \) and \( 309.8^\circ \).

Let the first term of the arithmetic progression be \( a = 8 \) and the common difference be \( d \).
The first, second, and fifth terms of the arithmetic progression are:

\( T_1 = 8 \)
\( T_2 = 8 + d \)
\( T_5 = 8 + 4d \)

These correspond to the first three terms of a geometric progression:

\( G_1 = 8 \)
\( G_2 = 8 + d \)
\( G_3 = 8 + 4d \)

Since these form a geometric progression, the ratio between consecutive terms is constant:

\( \frac{8 + d}{8} = \frac{8 + 4d}{8 + d} \)

Cross-multiplying gives:

\( (8 + d)^2 = 8(8 + 4d) \)
\( 64 + 16d + d^2 = 64 + 32d \)
\( d^2 - 16d = 0 \)

Since the common difference \( d \) is non-zero, we solve to find:

\( d = 16 \)

Now, calculate the common ratio \( r \) of the geometric progression:

\( r = \frac{8+d}{8} = \frac{8+16}{8} = 3 \)

Finally, calculate the sum of the first 12 terms of the arithmetic progression using the sum formula \( S_n = \frac{n}{2}[2a + (n-1)d] \):

\( S_{12} = \frac{12}{2}[2(8) + (12-1)(16)] \)
\( S_{12} = 6 [16 + 11(16)] \)
\( S_{12} = 6 [16 + 176] = 6 [192] = 1152 \).

M1: Expressing the 2nd and 5th terms of the AP in terms of \( d \).
M1: Setting up the equal-ratio equation \( \frac{8+d}{8} = \frac{8+4d}{8+d} \).
A1: Solving the quadratic equation to find \( d = 16 \).
A1: Finding the common ratio of the GP is \( r = 3 \).
M1: Applying the AP sum formula for \( n = 12 \).
A1.8: Correctly calculating the sum of the first 12 terms as \( 1152 \).

The curve is given by:

\( y = 2x + 18(x-1)^{-1} \)

Differentiating with respect to \( x \):

\( \frac{dy}{dx} = 2 - 18(x-1)^{-2} = 2 - \frac{18}{(x-1)^2} \)

Set \( \frac{dy}{dx} = 0 \) to find the stationary point:

\( 2 - \frac{18}{(x-1)^2} = 0 \)
\( 2 = \frac{18}{(x-1)^2} \)
\( (x-1)^2 = 9 \)

Since the domain is restricted to \( x > 1 \), we take the positive square root:

\( x - 1 = 3 \implies x = 4 \)

Substitute \( x = 4 \) back into the original curve equation to find the \( y \)-coordinate:

\( y = 2(4) + \frac{18}{4-1} = 8 + 6 = 14 \)

So the coordinates of the stationary point are \( (4, 14) \).

To determine the nature of the stationary point, find the second derivative \( \frac{d^2y}{dx^2} \):

\( \frac{d^2y}{dx^2} = \frac{d}{dx} \left[ 2 - 18(x-1)^{-2} \right] = 36(x-1)^{-3} = \frac{36}{(x-1)^3} \)

At \( x = 4 \):

\( \frac{d^2y}{dx^2} = \frac{36}{(4-1)^3} = \frac{36}{27} = \frac{4}{3} \)

Since \( \frac{d^2y}{dx^2} > 0 \), the stationary point \( (4, 14) \) is a minimum.

M1: Differentiating the function to find \( \frac{dy}{dx} \).
A1: Correct first derivative \( \frac{dy}{dx} = 2 - \frac{18}{(x-1)^2} \).
M1: Setting the derivative to zero and solving for \( x \).
A1: Obtaining \( x = 4 \) (and rejecting the negative root because \( x > 1 \)).
A1: Finding the correct y-coordinate of \( 14 \).
M1: Finding the second derivative and substituting \( x = 4 \) to determine the sign.
A0.8: Correctly concluding that the stationary point \( (4, 14) \) is a minimum.

The volume \( V \) of a solid of revolution generated by rotating about the x-axis is given by:

\( V = \pi \int_{a}^{b} y^2 \, dx \)

First, express \( y^2 \):

\( y^2 = \left( \frac{1}{\sqrt[4]{2x+1}} \right)^2 = \left( (2x+1)^{-1/4} \right)^2 = (2x+1)^{-1/2} \)

Now, set up the integral with the limits \( x = 0 \) and \( x = 12 \):

\( V = \pi \int_{0}^{12} (2x+1)^{-1/2} \, dx \)

Integrate \( (2x+1)^{-1/2} \) using the rule \( \int (ax+b)^n \, dx = \frac{(ax+b)^{n+1}}{a(n+1)} \):

\( \int (2x+1)^{-1/2} \, dx = \left[ \frac{(2x+1)^{1/2}}{2 \cdot \frac{1}{2}} \right] = \left[ \sqrt{2x+1} \right] \)

Evaluating this definite integral from \( 0 \) to \( 12 \):

\( V = \pi \left[ \sqrt{2x+1} \right]_{0}^{12} \)
\( V = \pi \left( \sqrt{2(12)+1} - \sqrt{2(0)+1} \right) \)
\( V = \pi ( \sqrt{25} - \sqrt{1} ) \)
\( V = \pi (5 - 1) = 4\pi \)

M1: Recalling and setting up the volume formula \( V = \pi \int y^2 \, dx \).
A1: Correctly obtaining \( y^2 = (2x+1)^{-1/2} \).
M1: Attempting to integrate \( (2x+1)^{-1/2} \).
A1: Obtaining the correct integrated expression \( \sqrt{2x+1} \).
M1: Substituting the upper and lower limits \( 12 \) and \( 0 \) into the integrated function.
A1.8: Obtaining the correct final answer of \( 4\pi \).

Let the first term of the arithmetic progression (AP) be \( a \) and the common difference be \( d \). The 1st, 3rd, and 11th terms of the AP are given by:
\( T_1 = a \)
\( T_3 = a + 2d \)
\( T_{11} = a + 10d \)

Since these are the first three terms of a geometric progression (GP), their ratios must be equal:
\( \frac{a+2d}{a} = \frac{a+10d}{a+2d} \)
\( (a+2d)^2 = a(a+10d) \)
\( a^2 + 4ad + 4d^2 = a^2 + 10ad \)
\( 4d^2 = 6ad \)

Assuming a non-trivial progression where \( d \neq 0 \), we can divide by \( 2d \):
\( 2d = 3a \implies d = 1.5a \)

We are also given that the sum of the first five terms of the AP is 80:
\( S_5 = \frac{5}{2}(2a + 4d) = 80 \)
\( 5(a + 2d) = 80 \)
\( a + 2d = 16 \)

Substitute \( d = 1.5a \) into this equation:
\( a + 2(1.5a) = 16 \)
\( 4a = 16 \implies a = 4 \)

Using \( a = 4 \), we find the common difference:
\( d = 1.5(4) = 6 \)

Now we find the first term and common ratio of the GP:
\( G_1 = a = 4 \)
\( G_2 = a + 2d = 4 + 12 = 16 \)
\( r = \frac{16}{4} = 4 \)

The 5th term of the GP is:
\( G_5 = a r^4 = 4 \times 4^4 = 1024 \)

M1: Writes the geometric progression condition \( (a+2d)^2 = a(a+10d) \) or equivalent.
A1: Obtains the simplified relation \( 2d = 3a \) or equivalent.
M1: Uses the sum of AP formula \( S_5 = 80 \) to obtain \( a + 2d = 16 \).
M1: Solves the resulting simultaneous equations to find \( a \) and \( d \).
A1: Correctly finds the first term \( a = 4 \).
M1: Finds the common ratio \( r = 4 \) and applies the formula for the 5th term of a GP.
A1: Obtains the correct 5th term of 1024.

To find the stationary points, we first find the derivative of \( y = 4x + 9(x-1)^{-1} \) with respect to \( x \):
\( \frac{dy}{dx} = 4 - 9(x-1)^{-2} = 4 - \frac{9}{(x-1)^2} \)

At a stationary point, \( \frac{dy}{dx} = 0 \):
\( 4 - \frac{9}{(x-1)^2} = 0 \)

\( \frac{9}{(x-1)^2} = 4 \)

\( (x-1)^2 = \frac{9}{4} \)

\( x-1 = \pm 1.5 \)

Since the domain is restricted to \( x > 1 \), we only consider the positive root:
\( x - 1 = 1.5 \implies x = 2.5 \)

Substitute \( x = 2.5 \) back into the original curve equation to find the y-coordinate:
\( y = 4(2.5) + \frac{9}{2.5-1} = 10 + \frac{9}{1.5} = 10 + 6 = 16 \)

So the stationary point is \( (2.5, 16) \).

To determine the nature of this point, we find the second derivative:
\( \frac{d^2y}{dx^2} = \frac{d}{dx} \left[ 4 - 9(x-1)^{-2} \right] = 18(x-1)^{-3} = \frac{18}{(x-1)^3} \)

Evaluating this at \( x = 2.5 \):
\( \frac{d^2y}{dx^2} = \frac{18}{(2.5-1)^3} = \frac{18}{1.5^3} = \frac{18}{3.375} = \frac{16}{3} \)

Since \( \frac{d^2y}{dx^2} > 0 \), the stationary point \( (2.5, 16) \) is a local minimum.

M1: Attempts to differentiate the function, with at least one term correct.
A1: Obtains the correct derivative \( \frac{dy}{dx} = 4 - \frac{9}{(x-1)^2} \).
M1: Sets their derivative to 0 and attempts to solve for \( x \).
A1: Obtains \( x = 2.5 \) (and explicitly or implicitly rejects the root outside the domain \( x > 1 \)).
A1: Obtains the correct y-coordinate \( y = 16 \).
M1: Differentiates again to find the second derivative and evaluates it at their x-value.
A1: Correctly shows that \( \frac{d^2y}{dx^2} > 0 \) and concludes that the point is a minimum.

First, we find the x-coordinates of the points of intersection of the curve and the line by equating their expressions:
\( x^2 - 4x + 5 = x + 1 \)
\( x^2 - 5x + 4 = 0 \)
\( (x - 1)(x - 4) = 0 \)

This gives the limits of integration as \( x = 1 \) and \( x = 4 \).

The line lies above the curve in the interval \( 1 < x < 4 \). Thus, the area of the enclosed region is:
\( \text{Area} = \int_{1}^{4} [ (x + 1) - (x^2 - 4x + 5) ] \, dx \)
\( \text{Area} = \int_{1}^{4} ( -x^2 + 5x - 4 ) \, dx \)

Integrating each term with respect to \( x \):
\( \int ( -x^2 + 5x - 4 ) \, dx = \left[ -\frac{1}{3}x^3 + \frac{5}{2}x^2 - 4x \right]_{1}^{4} \)

Now, substitute the upper limit \( x = 4 \):
\( \left( -\frac{1}{3}(64) + \frac{5}{2}(16) - 4(4) \right) = -\frac{64}{3} + 40 - 16 = \frac{8}{3} \)

Substitute the lower limit \( x = 1 \):
\( \left( -\frac{1}{3}(1) + \frac{5}{2}(1) - 4(1) \right) = -\frac{1}{3} + \frac{5}{2} - 4 = -\frac{11}{6} \)

Subtract the lower limit value from the upper limit value:
\( \text{Area} = \frac{8}{3} - \left( -\frac{11}{6} \right) = \frac{16}{6} + \frac{11}{6} = \frac{27}{6} = 4.5 \)

M1: Equates the curve and line equations to find intersection points.
A1: Obtains the correct intersection limits \( x = 1 \) and \( x = 4 \).
M1: Sets up the definite integral of \( (line) - (curve) \) with their limits.
A1: Simplifies the integrand correctly to \( -x^2 + 5x - 4 \).
M1: Integrates the expression correctly, with at least two terms correct.
A1: Obtains the correct integrated expression \( -\frac{1}{3}x^3 + \frac{5}{2}x^2 - 4x \).
A1: Correctly substitutes the limits to obtain the final area as \( 4.5 \) (or \( \frac{9}{2} \)).

Using the factor theorem, since \(p(x)\) is divisible by \(x - 2\), we have \(p(2) = 0\). Substituting \(x = 2\) yields: \(2(2)^3 + a(2)^2 + b(2) - 10 = 0 \implies 16 + 4a + 2b - 10 = 0 \implies 4a + 2b = -6 \implies 2a + b = -3\) (Equation 1). Using the remainder theorem, when \(p(x)\) is divided by \(x + 1\), the remainder is \(-12\), so \(p(-1) = -12\). Substituting \(x = -1\) yields: \(2(-1)^3 + a(-1)^2 + b(-1) - 10 = -12 \implies -2 + a - b - 10 = -12 \implies a - b = 0 \implies a = b\) (Equation 2). Substituting Equation 2 into Equation 1 gives: \(2a + a = -3 \implies 3a = -3 \implies a = -1\). Since \(a = b\), we also have \(b = -1\). The polynomial is therefore \(p(x) = 2x^3 - x^2 - x - 10\). To find the quotient when \(p(x)\) is divided by \(x - 2\), we perform algebraic division: \(2x^3 - x^2 - x - 10 = (x - 2)(2x^2 + kx + 5)\). Comparing the coefficient of \(x^2\): \(-4 + k = -1 \implies k = 3\). Thus, the quotient is \(2x^2 + 3x + 5\).

M1: Set up the equation for \(p(2) = 0\) and simplify. A1: Correctly obtain \(2a + b = -3\). M1: Set up the equation for \(p(-1) = -12\) and simplify. A1: Correctly obtain \(a = -1\) and \(b = -1\). M1: Perform division or equate coefficients to find the quadratic quotient. A1: Correctly state the quotient as \(2x^2 + 3x + 5\).

Using the laws of logarithms, we can combine the terms on the left-hand side: \(\ln\left(\frac{2x + 1}{x - 1}\right) = \ln(x + 3)\). Taking the exponential of both sides gives: \(\frac{2x + 1}{x - 1} = x + 3\). Multiplying both sides by \(x - 1\) (where \(x \ne 1\)): \(2x + 1 = (x + 3)(x - 1) \implies 2x + 1 = x^2 + 2x - 3\). Subtracting \(2x + 1\) from both sides: \(x^2 - 4 = 0\), which factorises to \((x - 2)(x + 2) = 0\). This gives two potential solutions: \(x = 2\) or \(x = -2\). However, we must check the validity of these solutions in the original logarithmic equation. For \(x = -2\), the terms \(\ln(2x + 1)\) and \(\ln(x - 1)\) would involve taking the logarithm of a negative number, which is undefined for real numbers. For \(x = 2\), all logarithmic terms are well-defined. Therefore, the only valid solution is \(x = 2\).

M1: Apply logarithm division law to combine LHS. M1: Remove logarithms correctly by equating arguments. A1: Obtain correct quadratic equation \(x^2 - 4 = 0\). A1: Find roots \(x = 2\) and \(x = -2\). B1: Correctly identify that \(x = -2\) is invalid and state the final answer \(x = 2\).

(a) Recall the double-angle identity \(\cos 2\theta = 1 - 2\sin^2\theta\). Substituting this into the equation: \(3(1 - 2\sin^2\theta) + 8\sin\theta + 1 = 0 \implies 3 - 6\sin^2\theta + 8\sin\theta + 1 = 0 \implies -6\sin^2\theta + 8\sin\theta + 4 = 0\). Multiplying the entire equation by \(-1\) yields: \(6\sin^2\theta - 8\sin\theta - 4 = 0\), as required.
(b) The equation simplifies to \(3\sin^2\theta - 4\sin\theta - 2 = 0\) by dividing by 2. Using the quadratic formula for \(\sin\theta\): \(\sin\theta = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(3)(-2)}}{2(3)} = \frac{4 \pm \sqrt{16 + 24}}{6} = \frac{4 \pm \sqrt{40}}{6}\). This gives: \(\sin\theta = \frac{4 + \sqrt{40}}{6} \approx 1.7208\) (which has no solution since \(\sin\theta \le 1\)), or \(\sin\theta = \frac{4 - \sqrt{40}}{6} \approx -0.3874\). For \(\sin\theta = -0.3874\), the basic angle is \(\alpha = \sin^{-1}(0.3874) \approx 22.79^\circ\). Since sine is negative, \(\theta\) lies in the third and fourth quadrants: \(\theta = 180^\circ + 22.79^\circ = 202.8^\circ\), or \(\theta = 360^\circ - 22.79^\circ = 337.2^\circ\).

Part (a) M1: Apply the double angle identity \(\cos 2\theta = 1 - 2\sin^2\theta\). A1: Complete showing the target equation with no algebraic errors.
Part (b) M1: Solve the quadratic equation in \(\sin\theta\). A1: Identify \(\sin\theta \approx -0.3874\) and state that the positive root is invalid. M1: Calculate the basic angle and find at least one correct quadrant angle. A1: State both solutions \(\theta = 202.8^\circ\) and \(\theta = 337.2^\circ\) correct to 1 decimal place.

To find the stationary points, we find \(\frac{\mathrm{d}y}{\mathrm{d}x}\) using the product rule: \(\frac{\mathrm{d}y}{\mathrm{d}x} = (2x)(\mathrm{e}^{-3x}) + (x^2)(-3\mathrm{e}^{-3x}) = x(2 - 3x)\mathrm{e}^{-3x}\). Setting \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\), and since \(\mathrm{e}^{-3x} \ne 0\) for all real \(x\), we get: \(x(2 - 3x) = 0 \implies x = 0\) or \(x = \frac{2}{3}\). When \(x = 0\), \(y = 0\). When \(x = \frac{2}{3}\), \(y = \left(\frac{2}{3}\right)^2 \mathrm{e}^{-3(2/3)} = \frac{4}{9}\mathrm{e}^{-2}\). To determine the nature of the stationary points, we find the second derivative: \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{\mathrm{d}}{\mathrm{d}x}[ (2x - 3x^2)\mathrm{e}^{-3x} ] = (2 - 6x)\mathrm{e}^{-3x} - 3(2x - 3x^2)\mathrm{e}^{-3x} = (2 - 12x + 9x^2)\mathrm{e}^{-3x}\). For \(x = 0\): \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2\mathrm{e}^{0} = 2 > 0 \implies\) local minimum at \((0, 0)\). For \(x = \frac{2}{3}\): \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \left(2 - 12\left(\frac{2}{3}\right) + 9\left(\frac{4}{9}\right)\right)\mathrm{e}^{-2} = (2 - 8 + 4)\mathrm{e}^{-2} = -2\mathrm{e}^{-2} < 0 \implies\) local maximum at \(\left(\frac{2}{3}, \frac{4}{9}\mathrm{e}^{-2}\right)\).

M1: Apply product rule correctly to find \(\frac{\mathrm{d}y}{\mathrm{d}x}\). A1: Obtain correct derivative \(x(2-3x)\mathrm{e}^{-3x}\). M1: Set derivative to 0 and find both \(x\) values. A1: Obtain both correct coordinate pairs: \((0,0)\) and \((\frac{2}{3}, \frac{4}{9}\mathrm{e}^{-2})\). M1: Use a valid method (such as the second derivative) to test the nature. A1: Correctly conclude that \((0,0)\) is a local minimum and \((\frac{2}{3}, \frac{4}{9}\mathrm{e}^{-2})\) is a local maximum.

First, we rewrite the integrand using algebraic division or inspection: \(\frac{4x + 3}{2x + 1} = \frac{2(2x + 1) + 1}{2x + 1} = 2 + \frac{1}{2x + 1}\). Now we integrate term by term: \(\int_{1}^{3} \left(2 + \frac{1}{2x + 1}\right) \mathrm{d}x = \left[ 2x + \frac{1}{2}\ln|2x + 1| \right]_{1}^{3}\). Evaluating this at the limits: At \(x = 3\): \(2(3) + \frac{1}{2}\ln|2(3) + 1| = 6 + \frac{1}{2}\ln 7\). At \(x = 1\): \(2(1) + \frac{1}{2}\ln|2(1) + 1| = 2 + \frac{1}{2}\ln 3\). Subtracting the lower limit value from the upper limit value: \(\left(6 + \frac{1}{2}\ln 7\right) - \left(2 + \frac{1}{2}\ln 3\right) = 4 + \frac{1}{2}(\ln 7 - \ln 3)\). Using logarithm laws: \(4 + \frac{1}{2}\ln\left(\frac{7}{3}\right) = 4 + \ln\sqrt{\frac{7}{3}}\). Thus, \(a = 4\) and \(b = \frac{7}{3}\).

M1: Perform algebraic division or express numerator in terms of the denominator. A1: Correctly express the integrand as \(2 + \frac{1}{2x+1}\). M1: Integrate to obtain terms of the form \(kx + c\ln(2x+1)\). A1: Correctly obtain \(2x + \frac{1}{2}\ln(2x+1)\). M1: Substitute limits correctly and apply logarithmic subtraction rules. A1: Express answer exactly in the form \(4 + \ln\sqrt{\frac{7}{3}}\).

(a) Let \(f(x) = \ln(x + 2) + x^2 - 3\). Evaluating \(f(x)\) at the boundaries of the interval: \(f(1.3) = \ln(1.3 + 2) + 1.3^2 - 3 = \ln(3.3) + 1.69 - 3 \approx 1.1939 + 1.69 - 3 = -0.1161 < 0\). \(f(1.4) = \ln(1.4 + 2) + 1.4^2 - 3 = \ln(3.4) + 1.96 - 3 \approx 1.2238 + 1.96 - 3 = 0.1838 > 0\). Since \(f(x)\) is continuous and there is a change of sign between \(x = 1.3\) and \(x = 1.4\), there is a root in this interval.
(b) Using \(x_1 = 1.3\): \(x_2 = \sqrt{3 - \ln(1.3 + 2)} = \sqrt{3 - 1.193922} \approx 1.3439\). \(x_3 = \sqrt{3 - \ln(1.3439 + 2)} = \sqrt{3 - 1.207137} \approx 1.3390\). \(x_4 = \sqrt{3 - \ln(1.3390 + 2)} = \sqrt{3 - 1.205670} \approx 1.3395\). \(x_5 = \sqrt{3 - \ln(1.3395 + 2)} = \sqrt{3 - 1.205820} \approx 1.3395\). Since successive values round to \(1.34\) to 2 decimal places, the root is \(1.34\).

Part (a) M1: Define a function \(f(x) = 0\) and evaluate at \(1.3\) and \(1.4\). A1: Show a change of sign and write a concluding sentence.
Part (b) M1: Calculate \(x_2\) correctly to 4 d.p. A1: Correctly calculate \(x_3\), \(x_4\), \(x_5\) to 4 d.p. A1: Correctly conclude the root is \(1.34\) to 2 d.p.

Since both sides of the inequality are non-negative, we can square both sides: \((3x - 5)^2 < (x + 3)^2 \implies 9x^2 - 30x + 25 < x^2 + 6x + 9\). Rearranging all terms to one side: \(8x^2 - 36x + 16 < 0\). Dividing the inequality by 4: \(2x^2 - 9x + 4 < 0\). Now, we factorise the quadratic expression: \((2x - 1)(x - 4) < 0\). The critical values are \(x = \frac{1}{2}\) and \(x = 4\). For the product to be negative, \(x\) must lie between the critical values: \(\frac{1}{2} < x < 4\).

M1: Attempt to square both sides to remove modulus or set up equations for boundary cases. A1: Expand correctly to get a quadratic in \(x\). A1: Reduce to \(2x^2 - 9x + 4 < 0\) (or equivalent). M1: Solve the quadratic to find critical values. A1: State the final inequality range correctly as \(\frac{1}{2} < x < 4\).

With 4 intervals over the domain \([0, 1]\), the width of each interval is \(h = \frac{1 - 0}{4} = 0.25\). The \(x\)-values are \(x_0 = 0\), \(x_1 = 0.25\), \(x_2 = 0.5\), \(x_3 = 0.75\), and \(x_4 = 1\). We calculate the corresponding \(y\)-values: \(y_0 = \frac{1}{1 + \mathrm{e}^0} = 0.5\); \(y_1 = \frac{1}{1 + \mathrm{e}^{0.25}} \approx 0.43782\); \(y_2 = \frac{1}{1 + \mathrm{e}^{0.5}} \approx 0.37754\); \(y_3 = \frac{1}{1 + \mathrm{e}^{0.75}} \approx 0.32082\); \(y_4 = \frac{1}{1 + \mathrm{e}^1} \approx 0.26894\). Applying the trapezium rule: \(I \approx \frac{h}{2} [ y_0 + y_4 + 2(y_1 + y_2 + y_3) ] = \frac{0.25}{2} [ 0.5 + 0.26894 + 2(0.43782 + 0.37754 + 0.32082) ] = 0.125 [ 0.76894 + 2(1.13618) ] = 0.125 [ 0.76894 + 2.27236 ] = 0.125 [ 3.04130 ] \approx 0.38016\). To 3 decimal places, the estimate is \(0.380\).

M1: Calculate correct value of \(h = 0.25\). A1: Calculate all 5 ordinates correctly (at least to 4 d.p.). M1: Apply the trapezium rule formula with correct coefficients. A1: Correctly calculate the unrounded sum. A1: State final answer of \(0.380\) to 3 decimal places.

(i) Let \(\frac{7x-1}{(1+2x)(1-x)} = \frac{A}{1+2x} + \frac{B}{1-x}\). Multiplying by the denominator gives \(7x - 1 = A(1-x) + B(1+2x)\). Substituting \(x = 1\) gives \(6 = 3B \implies B = 2\). Substituting \(x = -1/2\) gives \(-4.5 = 1.5A \implies A = -3\). Thus, \(f(x) = \frac{-3}{1+2x} + \frac{2}{1-x}\). (ii) We rewrite \(f(x)\) as \(-3(1+2x)^{-1} + 2(1-x)^{-1}\). Using the binomial expansion: \((1+2x)^{-1} = 1 - 2x + \frac{(-1)(-2)}{2}(2x)^2 + \dots = 1 - 2x + 4x^2\). \((1-x)^{-1} = 1 + x + x^2\). Therefore, \(f(x) = -3(1 - 2x + 4x^2) + 2(1 + x + x^2) = -3 + 6x - 12x^2 + 2 + 2x + 2x^2 = -1 + 8x - 10x^2\).

M1: Set up partial fractions and attempt to find coefficients. A1: Correctly find \(A = -3\) and \(B = 2\). M1: Expand \((1+2x)^{-1}\) up to \(x^2\). A1: Correct expansion \(1 - 2x + 4x^2\). M1: Expand \((1-x)^{-1}\) up to \(x^2\). A1: Correct expansion \(1 + x + x^2\). A1: Obtain final simplified expression \(-1 + 8x - 10x^2\).

Using the laws of logarithms, we write \(2\ln(x) = \ln(x^2)\). The equation becomes \(\ln(2x+3) - \ln(x^2) = \ln(3)\), which simplifies to \(\ln\left(\frac{2x+3}{x^2}\right) = \ln(3)\). Taking exponentials of both sides, we get \(\frac{2x+3}{x^2} = 3\). Multiplying by \(x^2\) gives \(2x + 3 = 3x^2\), which rearranges to the quadratic equation \(3x^2 - 2x - 3 = 0\). Using the quadratic formula: \(x = \frac{2 \pm \sqrt{(-2)^2 - 4(3)(-3)}}{2(3)} = \frac{2 \pm \sqrt{40}}{6} = \frac{1 \pm \sqrt{10}}{3}\). Since \(\ln(x)\) is defined only for \(x > 0\), we reject the negative root. The only valid solution is \(x = \frac{1 + \sqrt{10}}{3} \approx 1.387\). Correct to 3 significant figures, \(x = 1.39\).

M1: Apply log laws to combine the LHS into a single logarithm. A1: Correct equation \(\ln\left(\frac{2x+3}{x^2}\right) = \ln(3)\). M1: Remove logarithms to form a quadratic equation. A1: Correct quadratic equation \(3x^2 - 2x - 3 = 0\). M1: Solve the quadratic equation. A1: Obtain \(x = 1.39\). A1: Explicitly reject the negative root \(\frac{1-\sqrt{10}}{3}\) with a valid reason.

(i) We have \(R = \sqrt{5^2 + 12^2} = \sqrt{169} = 13\). We also have \(\tan\alpha = \frac{12}{5} = 2.4\), which gives \(\alpha = \arctan(2.4) \approx 1.1760\) radians. Thus, \(5\sin\theta - 12\cos\theta = 13\sin(\theta - 1.1760)\). (ii) The equation \(5\sin(2x) - 12\cos(2x) = 4.5\) becomes \(13\sin(2x - 1.1760) = 4.5\), so \(\sin(2x - 1.1760) = \frac{4.5}{13} = \frac{9}{26}\). Let \(\phi = 2x - 1.1760\). Since \(0 < x < \pi\), we have \(-1.1760 < \phi < 2\pi - 1.1760 \approx 5.107\). The solutions for \(\phi\) are \(\phi = \arcsin(9/26) \approx 0.3535\) and \(\phi = \pi - 0.3535 \approx 2.7881\). For \(\phi = 0.3535\), \(2x = 0.3535 + 1.1760 = 1.5295 \implies x \approx 0.765\). For \(\phi = 2.7881\), \(2x = 2.7881 + 1.1760 = 3.9641 \implies x \approx 1.98\).

B1: Obtain \(R = 13\). M1: Use \(\tan\alpha = 12/5\) to find \(\alpha\). A1: Obtain \(\alpha = 1.1760\). M1: Set up the equation \(\sin(2x - 1.1760) = \frac{9}{26}\) and find one principal value of the angle. A1: Obtain \(x = 0.765\). M1: Find the second angle in the correct interval. A1: Obtain \(x = 1.98\).

To find the stationary points, we use parametric differentiation: \(\frac{\mathrm{dx}}{\mathrm{dt}} = 2\mathrm{e}^{2t - 2} - 2\) and \(\frac{\mathrm{dy}}{\mathrm{dt}} = 6t^2 - 18t + 12\). The gradient is \(\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}/\mathrm{dt}}{\mathrm{dx}/\mathrm{dt}} = \frac{6t^2 - 18t + 12}{2\mathrm{e}^{2t - 2} - 2}\). Setting \(\frac{\mathrm{dy}}{\mathrm{dx}} = 0\) gives \(\frac{\mathrm{dy}}{\mathrm{dt}} = 0\), which means \(6t^2 - 18t + 12 = 0 \implies 6(t^2 - 3t + 2) = 0 \implies 6(t - 1)(t - 2) = 0\). This gives \(t = 1\) or \(t = 2\). Since we are given \(t > 1\), we choose \(t = 2\). We must verify that \(\frac{\mathrm{dx}}{\mathrm{dt}} \neq 0\) at \(t = 2\): \(\frac{\mathrm{dx}}{\mathrm{dt}} = 2\mathrm{e}^2 - 2 \neq 0\). Substituting \(t = 2\) into the parametric equations: \(x = \mathrm{e}^{2(2) - 2} - 2(2) = \mathrm{e}^2 - 4\) and \(y = 2(2)^3 - 9(2)^2 + 12(2) = 16 - 36 + 24 = 4\). Thus, the exact coordinates of the stationary point are \((\mathrm{e}^2 - 4, 4)\).

M1: Differentiate \(x\) with respect to \(t\). A1: Correct \(\frac{\mathrm{dx}}{\mathrm{dt}} = 2\mathrm{e}^{2t - 2} - 2\). M1: Differentiate \(y\) with respect to \(t\). A1: Correct \(\frac{\mathrm{dy}}{\mathrm{dt}} = 6t^2 - 18t + 12\). M1: Set \(\frac{\mathrm{dy}}{\mathrm{dx}} = 0\) and select the correct parameter \(t = 2\). A1: Obtain the exact \(x\)-coordinate \(\mathrm{e}^2 - 4\). A1: Obtain the exact \(y\)-coordinate \(4\).

Let \(u = x^2 + 1\). Then \(\mathrm{du} = 2x \, \mathrm{dx}\), which means \(x \, \mathrm{dx} = \frac{1}{2} \mathrm{du}\). We also have \(x^2 = u - 1\). Changing the limits: when \(x = 0\), \(u = 0^2 + 1 = 1\); when \(x = \sqrt{3}\), \(u = (\sqrt{3})^2 + 1 = 4\). Substituting these into the integral: \(\int_{0}^{\sqrt{3}} \frac{x^3}{\sqrt{x^2+1}} \, \mathrm{dx} = \int_{0}^{\sqrt{3}} \frac{x^2 \cdot x \, \mathrm{dx}}{\sqrt{x^2+1}} = \int_{1}^{4} \frac{u-1}{\sqrt{u}} \cdot \frac{1}{2} \, \mathrm{du} = \frac{1}{2} \int_{1}^{4} \left(u^{1/2} - u^{-1/2}\right) \, \mathrm{du}\). Integrating term by term: \(\frac{1}{2} \left[ \frac{2}{3}u^{3/2} - 2u^{1/2} \right]_{1}^{4} = \left[ \frac{1}{3}u^{3/2} - u^{1/2} \right]_{1}^{4}\). Substituting the upper limit \(u = 4\): \(\frac{1}{3}(4)^{3/2} - (4)^{1/2} = \frac{8}{3} - 2 = \frac{2}{3}\). Substituting the lower limit \(u = 1\): \(\frac{1}{3}(1)^{3/2} - (1)^{1/2} = \frac{1}{3} - 1 = -\frac{2}{3}\). The exact value is \(\frac{2}{3} - \left(-\frac{2}{3}\right) = \frac{4}{3}\).

M1: Use the substitution \(u = x^2 + 1\) and find \(\mathrm{du} = 2x \, \mathrm{dx}\). A1: Correctly write the integrand in terms of \(u\). B1: Correctly transform the limits of integration to 1 and 4. M1: Integrate the terms of the form \(u^n\). A1: Correct antiderivative \(\frac{1}{3}u^{3/2} - u^{1/2}\). M1: Substitute limits 4 and 1 into the antiderivative. A1: Obtain the exact final value \(\frac{4}{3}\).

(i) Let \(f(x) = x^3 - 5x - 3\). We evaluate the function at the boundary points: \(f(2.4) = (2.4)^3 - 5(2.4) - 3 = 13.824 - 12 - 3 = -1.176\). \(f(2.5) = (2.5)^3 - 5(2.5) - 3 = 15.625 - 12.5 - 3 = 0.125\). Since \(f(x)\) is continuous on the interval \([2.4, 2.5]\) and there is a change of sign between \(f(2.4) < 0\) and \(f(2.5) > 0\), there must be a root of the equation \(f(x) = 0\) in this interval. (ii) Using the iterative formula \(x_{n+1} = \sqrt[3]{5x_n + 3}\): \(x_1 = 2.4\), \(x_2 = \sqrt[3]{5(2.4) + 3} = \sqrt[3]{15} \approx 2.46621\), \(x_3 = \sqrt[3]{5(2.46621) + 3} \approx 2.48424\), \(x_4 = \sqrt[3]{5(2.48424) + 3} \approx 2.48908\), \(x_5 = \sqrt[3]{5(2.48908) + 3} \approx 2.49038\), \(x_6 = \sqrt[3]{5(2.49038) + 3} \approx 2.49073\), \(x_7 = \sqrt[3]{5(2.49073) + 3} \approx 2.49082\), \(x_8 = \sqrt[3]{5(2.49082) + 3} \approx 2.49085\). Since successive iterations converge to \(2.491\) to 3 decimal places, the root is \(2.491\).

M1: Evaluate \(f(2.4)\) and \(f(2.5)\). A1: Obtain correct values \(-1.176\) and \(0.125\) and conclude a root exists due to the sign change. M1: Calculate \(x_2\) and \(x_3\) using the iterative formula. A1: Correct \(x_2 \approx 2.46621\) and \(x_3 \approx 2.48424\). M1: Continue iterations to find \(x_4, x_5, x_6, x_7\). A1: Correctly show convergence up to \(2.491\) (e.g. \(x_6 \approx 2.49073\)). A1: Conclude the root is \(2.491\) correct to 3 d.p.

(i) The position vector of any point \(F\) on \(l\) is given by \(\vec{OF} = (1 + 2\lambda)\mathbf{i} + (2 + \lambda)\mathbf{j} + (-1 - 2\lambda)\mathbf{k}\). The vector \(\vec{PF}\) is \(\vec{OF} - \vec{OP} = (1 + 2\lambda - 4)\mathbf{i} + (2 + \lambda - 2)\mathbf{j} + (-1 - 2\lambda - (-7))\mathbf{k} = (2\lambda - 3)\mathbf{i} + \lambda\mathbf{j} + (6 - 2\lambda)\mathbf{k}\). Since \(PF\) is perpendicular to \(l\), the scalar product of \(\vec{PF}\) and the direction vector of the line, \(\mathbf{d} = 2\mathbf{i} + \mathbf{j} - 2\mathbf{k}\), must be zero: \(\vec{PF} \cdot \mathbf{d} = 2(2\lambda - 3) + 1(\lambda) - 2(6 - 2\lambda) = 0 \implies 4\lambda - 6 + \lambda - 12 + 4\lambda = 0 \implies 9\lambda - 18 = 0 \implies \lambda = 2\). Substituting \(\lambda = 2\) into the equation of \(l\), we find the position vector of \(F\): \(\vec{OF} = (1 + 2(2))\mathbf{i} + (2 + 2)\mathbf{j} + (-1 - 2(2))\mathbf{k} = 5\mathbf{i} + 4\mathbf{j} - 5\mathbf{k}\). (ii) The perpendicular distance from \(P\) to \(l\) is the magnitude of the vector \(\vec{PF}\) when \(\lambda = 2\). Substituting \(\lambda = 2\) into \(\vec{PF}\) gives \(\vec{PF} = (2(2) - 3)\mathbf{i} + 2\mathbf{j} + (6 - 2(2))\mathbf{k} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k}\). The distance is \(|\vec{PF}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3\).

M1: Express \(\vec{PF}\) in terms of \(\lambda\). A1: Obtain correct vector \((2\lambda-3)\mathbf{i} + \lambda\mathbf{j} + (6-2\lambda)\mathbf{k}\). M1: Set up the equation \(\vec{PF} \cdot \mathbf{d} = 0\). A1: Solve for \(\lambda\) to find \(\lambda = 2\). A1: Find the position vector of \(F\) as \(5\mathbf{i} + 4\mathbf{j} - 5\mathbf{k}\). M1: Compute the magnitude of \(\vec{PF}\). A1: Obtain the distance \(3\).

(i) The modulus of \(w\) is \(r = |w| = \sqrt{(\sqrt{3})^2 + 1^2} = 2\). Since \(w\) is in the first quadrant, its argument is \(\theta = \arctan(1/\sqrt{3}) = \frac{\pi}{6}\). Thus, \(w = 2(\cos(\frac{\pi}{6}) + \mathrm{i}\sin(\frac{\pi}{6}))\). (ii) The region defined by \(|z - w| \le 1\) is a circle with center representing \(w\) and radius 1, including the boundary and its interior. In the Argand diagram, this is a circle centered at \((\sqrt{3}, 1)\) with radius 1. Since the center is at distance 1 from the real axis and the radius is 1, the circle is tangent to the real axis. (iii) The argument of \(z\) is the angle that the line from the origin to \(z\) makes with the positive real axis. Looking at the shaded region, the boundary of the region touches the positive real axis at \(z = \sqrt{3}\) (since the center is \(\sqrt{3} + \mathrm{i}\) and the radius is 1, the point directly below the center is \(\sqrt{3} + 0\mathrm{i}\)). Since the entire region lies on or above the real axis in the first quadrant, the least possible argument of any point in the region is at the point of tangency on the real axis, which is \(0\).

B1: Obtain \(r = 2\). B1: Obtain \(\theta = \frac{\pi}{6}\). B1: Draw a circle centered in the first quadrant. B1: Show the circle is tangent to the real axis and shade its interior. M1: Use a geometric argument or a diagram to locate the point with the minimum argument. A1: Explain that the circle touches the real axis because the distance of the center from the real axis is equal to the radius. A1: Obtain the minimum argument of \(0\).

Using the quadratic formula with coefficients \(a = 1\), \(b = -2(1+\text{i})\), and \(c = 3 - 2\text{i}\): \(\Delta = [-2(1+\text{i})]^2 - 4(1)(3-2\text{i}) = 4(1 + 2\text{i} - 1) - 12 + 8\text{i} = -12 + 16\text{i}\). We find the complex square roots of \(-12 + 16\text{i}\) by setting \((x + \text{i}y)^2 = -12 + 16\text{i}\), where \(x\) and \(y\) are real. This gives \(x^2 - y^2 = -12\) and \(2xy = 16\). Substituting \(y = \frac{8}{x}\) into the first equation yields \(x^2 - \frac{64}{x^2} = -12 \implies x^4 + 12x^2 - 64 = 0\). Factoring gives \((x^2 + 16)(x^2 - 4) = 0\). Since \(x\) is 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 \(\Delta\) are \(\pm(2 + 4\text{i})\). Substituting back into the quadratic formula: \(z = \frac{2(1+\text{i}) \pm (2 + 4\text{i})}{2} = (1+\text{i}) \pm (1 + 2\text{i})\). The roots are \(z = (1+\text{i}) + (1+2\text{i}) = 2 + 3\text{i}\) and \(z = (1+\text{i}) - (1+2\text{i}) = -\text{i}\).

M1: Attempt to apply the quadratic formula, finding the discriminant \(\Delta = -12 + 16\text{i}\). M1: Set up simultaneous equations for the square root of \(\Delta\) to get \(x^2 - y^2 = -12\) and \(2xy = 16\). A1: Correctly solve to find the square roots \(\pm(2 + 4\text{i})\). M1: Substitute the square roots back into the expression for \(z\). A1: Find one correct root, e.g., \(2 + 3\text{i}\). A1.8: Find the second correct root, e.g., \(-\text{i}\).

We use integration by parts \(\int u \, \text{d}v = uv - \int v \, \text{d}u\). Let \(u = x \implies \text{d}u = \text{d}x\), and \(\text{d}v = \sec^2 x \, \text{d}x \implies v = \tan x\). This gives \(\int x \sec^2 x \, \text{d}x = x \tan x - \int \tan x \, \text{d}x\). Since \(\int \tan x \, \text{d}x = -\ln|\cos x|\), we have \(\int x \sec^2 x \, \text{d}x = x \tan x + \ln|\cos x|\). Now we evaluate this between the limits \(0\) and \(\frac{\pi}{3}\): At \(x = \frac{\pi}{3}\), we have \(\frac{\pi}{3} \tan\left(\frac{\pi}{3}\right) + \ln\left|\cos\left(\frac{\pi}{3}\right)\right| = \frac{\pi}{3} \sqrt{3} + \ln\left(\frac{1}{2}\right) = \frac{\sqrt{3}}{3}\pi - \ln 2\). At \(x = 0\), we have \(0 \cdot \tan 0 + \ln|\cos 0| = 0\). Thus, the exact value of the integral is \(\frac{\sqrt{3}}{3}\pi - \ln 2\). Comparing with \(a\pi - \ln b\), we get \(a = \frac{\sqrt{3}}{3}\) and \(b = 2\).

M1: Use integration by parts with correct allocation of \(u = x\) and \(\text{d}v = \sec^2 x \, \text{d}x\). A1: Obtain the correct term \(x \tan x\). M1: Integrate \(\tan x\) correctly to obtain \(\ln|\sec x|\) or \(-\ln|\cos x|\). A1: Obtain the correct combined antiderivative \(x \tan x + \ln|\cos x|\). M1: Substitute limits \(0\) and \(\frac{\pi}{3}\) correctly into their antiderivative. A1.8: Obtain the correct exact value and state the values of \(a = \frac{\sqrt{3}}{3}\) and \(b = 2\).

We separate variables to obtain \(\int \frac{1}{y^2 + 1} \, \text{d}y = \int \frac{1}{x \ln x} \, \text{d}x\). Integrating the left-hand side gives \(\arctan y\). Integrating the right-hand side using the substitution \(u = \ln x\), where \(\text{d}u = \frac{1}{x} \, \text{d}x\), gives \(\int \frac{1}{u} \, \text{d}u = \ln|u| = \ln(\ln x)\) since \(x > 1\). Therefore, \(\arctan y = \ln(\ln x) + C\). We use the initial condition \(x = \text{e}\) and \(y = 1\) to find \(C\): \(\arctan 1 = \ln(\ln \text{e}) + C \implies \frac{\pi}{4} = \ln(1) + C \implies C = \frac{\pi}{4}\). The equation becomes \(\arctan y = \ln(\ln x) + \frac{\pi}{4}\). Taking the tangent of both sides, we get \(y = \tan\left(\ln(\ln x) + \frac{\pi}{4}\right)\).

M1: Separate variables correctly. A1: Integrate LHS correctly to obtain \(\arctan y\). M1: Integrate RHS correctly to obtain \(\ln(\ln x)\). A1: Use the boundary condition \(x = \text{e}, y = 1\) to find the constant of integration \(C\). A1: Correctly calculate \(C = \frac{\pi}{4}\). A1.8: Solve for \(y\) to find \(y = \tan\left(\ln(\ln x) + \frac{\pi}{4}\right)\).

First, calculate the normal reaction force \(R\):
\(R = m g \cos 30^\circ = 4 \times 10 \times \cos 30^\circ = 20\sqrt{3} \approx 34.64 \text{ N}\).

The maximum frictional force \(F_{\max}\) is given by:
\(F_{\max} = \mu R = 0.25 \times 20\sqrt{3} = 5\sqrt{3} \approx 8.66 \text{ N}\).

The component of the weight acting down the slope is:
\(W_{\parallel} = m g \sin 30^\circ = 4 \times 10 \times \sin 30^\circ = 20 \text{ N}\).

Case 1: The block is on the point of slipping down the plane. Here, friction acts up the plane.
\(P + F_{\max} = W_{\parallel} \implies P + 5\sqrt{3} = 20 \implies P = 20 - 5\sqrt{3} \approx 11.34 \text{ N}\).

Case 2: The block is on the point of slipping up the plane. Here, friction acts down the plane.
\(P - F_{\max} = W_{\parallel} \implies P - 5\sqrt{3} = 20 \implies P = 20 + 5\sqrt{3} \approx 28.66 \text{ N}\).

Therefore, the range of values for \(P\) to maintain equilibrium is:
\(20 - 5\sqrt{3} \le P \le 20 + 5\sqrt{3}\), or \(11.3 \le P \le 28.7\) (to 3 significant figures).

M1: Attempt to find the normal reaction \(R = 40 \cos 30^\circ\).
A1: Correct maximum friction \(F_{\max} = 5\sqrt{3}\) or \(8.66\).
M1: Set up the equilibrium equation for slipping down: \(P + F = W \sin 30^\circ\).
A1: Obtain lower limit \(P \ge 11.3\).
M1: Set up the equilibrium equation for slipping up: \(P - F = W \sin 30^\circ\).
A1: Obtain upper limit \(P \le 28.7\).

(i) Write the equations of motion for both particles:
For \(B\) (moving downwards): \(5g - T = 5a\)
For \(A\) (moving upwards): \(T - 3g = 3a\)

Adding these equations:
\(2g = 8a \implies a = 0.25g = 2.5 \text{ ms}^{-2}\).

Using the equation of motion for \(B\) to find its final speed \(v\) before hitting the floor:
\(v^2 = u^2 + 2as = 0 + 2(2.5)(2.4) = 12 \implies v = \sqrt{12} = 2\sqrt{3} \approx 3.46 \text{ ms}^{-1}\).

(ii) When \(B\) hits the floor, the string becomes slack. Particle \(A\) is now projected vertically upwards with an initial velocity of \(v = 2\sqrt{3} \approx 3.464 \text{ ms}^{-1}\) and moves under gravity alone (decelerating at \(g = 10 \text{ ms}^{-2}\)).

The string will become taut again when \(A\) returns to the level it was at when \(B\) hit the floor.
Using \(s = ut + \frac{1}{2}at^2\) with net displacement \(s = 0\):
\(0 = vt - \frac{1}{2}gt^2 \implies t = \frac{2v}{g} = \frac{2(3.464)}{10} \approx 0.693 \text{ s}\).

M1: Write equations of motion for both particles to find the acceleration.
A1: Correct acceleration \(a = 2.5 \text{ ms}^{-2}\).
M1: Use \(v^2 = 2as\) to find the speed of \(B\) before impact.
A1: Correct speed \(v = \sqrt{12} \approx 3.46 \text{ ms}^{-1}\).
M1: Realise that after impact, \(A\) moves under gravity with initial speed \(v\), and set up equation for time \(t = 2v/g\).
A1: Correct time interval \(t \approx 0.693 \text{ s}\).

(i) Find velocity by integrating acceleration:
\(v(t) = \int (6 - 2t) dt = 6t - t^2 + C\).

Using the initial condition \(v(0) = 16\):
\(C = 16 \implies v(t) = 16 + 6t - t^2\).

To find when \(P\) is at instantaneous rest, set \(v(t) = 0\):
\(16 + 6t - t^2 = 0 \implies t^2 - 6t - 16 = 0\)
\((t - 8)(t + 2) = 0\).

Since \(t \ge 0\), \(t = 8 \text{ s}\).

(ii) The total distance in the first 10 seconds is split into two intervals because the particle changes direction at \(t = 8\):

Interval 1: \(0 \le t \le 8\) (moving forward):
\(d_1 = \int_0^8 (16 + 6t - t^2) dt = \left[ 16t + 3t^2 - \frac{1}{3}t^3 \right]_0^8 = 16(8) + 3(64) - \frac{512}{3} = 128 + 192 - 170.67 = 149.33 \text{ m}\) (or \(\frac{448}{3}\) m).

Interval 2: \(8 \le t \le 10\) (moving backward):
\(d_2 = \left| \int_8^{10} (16 + 6t - t^2) dt \right| = \left| \left[ 16t + 3t^2 - \frac{1}{3}t^3 \right]_8^{10} \right|\).
At \(t = 10\), the displacement is \(16(10) + 3(100) - \frac{1000}{3} = 160 + 300 - 333.33 = 126.67 \text{ m}\) (or \(\frac{380}{3}\) m).

Thus, \(d_2 = \left| \frac{380}{3} - \frac{448}{3} \right| = \frac{68}{3} \approx 22.67 \text{ m}\).

Total distance = \(d_1 + d_2 = \frac{448}{3} + \frac{68}{3} = \frac{516}{3} = 172 \text{ m}\).

M1: Integrate acceleration to obtain velocity, including constant of integration.
A1: Correct expression \(v(t) = 16 + 6t - t^2\).
A1: Solve \(v(t) = 0\) to find \(t = 8 \text{ s}\).
M1: Set up two separate integrals for distance, taking account of the change of sign at \(t = 8\).
A1: Find correct distance for first part \(d_1 = 149.33\) or second part \(d_2 = 22.67\).
A1: Correct total distance of \(172 \text{ m}\).

(i) Use power to find the driving force \(F_d\) when speed \(v = 15 \text{ ms}^{-1}\):
\(P = F_d v \implies 36000 = F_d (15) \implies F_d = 2400 \text{ N}\).

Write down Newton's second law along the slope:
\(F_d - R - mg \sin \theta = ma\)
\(2400 - 400 - 1200(10)(0.05) = 1200 a\)
\(2400 - 400 - 600 = 1200 a\)
\(1400 = 1200 a \implies a = \frac{1400}{1200} = \frac{7}{6} \approx 1.17 \text{ ms}^{-2}\).

(ii) At the maximum steady speed, the acceleration is zero:
\(F_d - R - mg \sin \theta = 0\)
\(F_d = 400 + 600 = 1000 \text{ N}\).

Using the power equation again to find this maximum speed \(v_{\max}\):
\(v_{\max} = \frac{P}{F_d} = \frac{36000}{1000} = 36 \text{ ms}^{-1}\).

M1: Use \(P = F_d v\) to find driving force \(F_d = 2400 \text{ N}\).
M1: Formulate equation of motion \(F_d - R - mg \sin \theta = ma\).
A1: Correct acceleration \(a \approx 1.17 \text{ ms}^{-2}\) (accept \(7/6\)).
M1: Recognize steady speed means \(a = 0\) and find required driving force \(F_d = 1000 \text{ N}\).
A1: Correct maximum steady speed \(v_{\max} = 36 \text{ ms}^{-1}\).

(i) Define the positive direction to be the initial direction of motion of \(A\).
Initial velocities:
\(u_A = 5 \text{ ms}^{-1}\)
\(u_B = -2 \text{ ms}^{-1}\)

Final velocity of \(A\):
\(v_A = -1 \text{ ms}^{-1}\) (since it rebounds in the opposite direction).

Using the conservation of momentum:
\(m_A u_A + m_B u_B = m_A v_A + m_B v_B\)
\(0.4(5) + 0.6(-2) = 0.4(-1) + 0.6 v_B\)
\(2.0 - 1.2 = -0.4 + 0.6 v_B\)
\(0.8 = -0.4 + 0.6 v_B\)
\(1.2 = 0.6 v_B \implies v_B = 2 \text{ ms}^{-1}\).

Since \(v_B\) is positive, \(B\) moves in the direction of \(A\)'s initial motion, with a speed of \(2 \text{ ms}^{-1}\).

(ii) Calculate the total initial kinetic energy \(E_i\):
\(E_i = \frac{1}{2} m_A u_A^2 + \frac{1}{2} m_B u_B^2 = \frac{1}{2}(0.4)(25) + \frac{1}{2}(0.6)(4) = 5.0 + 1.2 = 6.2 \text{ J}\).

Calculate the total final kinetic energy \(E_f\):
\(E_f = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2 = \frac{1}{2}(0.4)(1) + \frac{1}{2}(0.6)(4) = 0.2 + 1.2 = 1.4 \text{ J}\).

The loss in kinetic energy during the collision is:
\(E_i - E_f = 6.2 - 1.4 = 4.8 \text{ J}\).

M1: Use the conservation of momentum with correct signs for velocities.
A1: Correct value \(v_B = 2 \text{ ms}^{-1}\).
A1: Correctly state direction of \(B\) (e.g., in direction of \(A\)'s initial motion).
M1: Calculate total initial KE and total final KE.
A1: Correct initial KE (\(6.2 \text{ J}\)) or final KE (\(1.4 \text{ J}\)).
A1: Correct loss in KE of \(4.8 \text{ J}\).

Let East be the positive \(x\)-direction and North be the positive \(y\)-direction.

Resolve each force into its \(x\)- and \(y\)-components:
- \(10\text{ N}\) force: \(F_{1x} = 10\), \(F_{1y} = 0\).
- \(18\text{ N}\) force: \(F_{2x} = 0\), \(F_{2y} = 18\).
- \(16\text{ N}\) force (acting \(30^\circ\) South of West):
- \(x\)-component: \(-16 \cos 30^\circ = -8\sqrt{3} \approx -13.856\text{ N}\).
- \(y\)-component: \(-16 \sin 30^\circ = -8\text{ N}\).

Sum the components:
- Total \(x\)-component \(R_x = 10 - 8\sqrt{3} \approx -3.856\text{ N}\).
- Total \(y\)-component \(R_y = 18 - 8 = 10\text{ N}\).

Find the magnitude of the resultant force \(R\):
\(R = \sqrt{R_x^2 + R_y^2} = \sqrt{(-3.856)^2 + 10^2} = \sqrt{14.87 + 100} = \sqrt{114.87} \approx 10.7\text{ N}\) (to 3 s.f.).

Find the angle \(\theta\) that the resultant makes with due North (positive \(y\)-axis):
\(\tan \theta = \frac{|R_x|}{R_y} = \frac{3.856}{10} = 0.3856 \implies \theta \approx 21.1^\circ\).

Since \(R_x\) is negative and \(R_y\) is positive, the resultant lies in the North-West quadrant. Thus, the angle is \(21.1^\circ\) West of North.

M1: Resolve forces horizontally and vertically.
A1: Correct horizontal sum \(R_x = 10 - 16 \cos 30^\circ\) and vertical sum \(R_y = 18 - 16 \sin 30^\circ\).
A1: Obtain accurate values: \(R_x \approx -3.86\) and \(R_y = 10\).
M1: Use Pythagoras' theorem to find magnitude \(R\).
A1: Correct magnitude \(R \approx 10.7\text{ N}\).
M1: Use trigonometry to find angle with North.
A1: Correct angle \(\theta \approx 21.1^\circ\) West of North.

(i) Find the normal reaction force \(R\):
\(R = mg \cos 15^\circ = 250 \cos 15^\circ \approx 241.48 \text{ N}\).

The friction force \(F\) opposing the motion up the slope is:
\(F = \mu R = 0.3 \times 241.48 = 72.44 \text{ N}\).

The component of weight down the slope is:
\(W_{\parallel} = mg \sin 15^\circ = 250 \sin 15^\circ \approx 64.70 \text{ N}\).

Apply Newton's second law along the slope:
\(P - F - W_{\parallel} = m a_1\)
\(150 - 72.44 - 64.70 = 25 a_1\)
\(12.86 = 25 a_1 \implies a_1 \approx 0.514 \text{ ms}^{-2}\).

(ii) Calculate the velocity squared \(v^2\) after travelling \(8 \text{ m}\) from rest:
\(v^2 = 2 a_1 s_1 = 2(0.514) \times 8 \approx 8.224 \text{ m}^2\text{s}^{-2}\).

Once the pulling force is removed, the deceleration \(a_2\) is given by:
\(-F - W_{\parallel} = m a_2\)
\(-72.44 - 64.70 = 25 a_2 \implies a_2 = -5.486 \text{ ms}^{-2}\).

Find the further distance \(s_2\) to rest using \(0 = v^2 + 2 a_2 s_2\):
\(s_2 = \frac{-v^2}{2 a_2} = \frac{-8.224}{2(-5.486)} \approx 0.750 \text{ m}\).

M1: Resolve forces perpendicular to the plane to find normal reaction \(R\).
A1: Correct acceleration \(a_1 \approx 0.514 \text{ ms}^{-2}\).
M1: Use kinematics to find the speed of the sledge when the force is removed.
A1: Correct value for \(v^2 \approx 8.22\).
M1: Set up the equation of motion without pulling force to find deceleration.
A1: Correct deceleration \(a_2 \approx -5.49 \text{ ms}^{-2}\).
A1: Correct further distance \(s_2 \approx 0.750 \text{ m}\) (accept \(0.75\)).

(i) Given \(s = 300\), \(t = 15\), \(v = 3u\).
Using the formula \(s = \frac{u+v}{2} t\):
\(300 = \frac{u + 3u}{2} \times 15\)
\(300 = 2u \times 15 = 30u \implies u = 10 \text{ ms}^{-1}\).

Therefore, the final velocity at \(B\) is \(v = 30 \text{ ms}^{-1}\).
Using \(v = u + at\) to find the acceleration:
\(30 = 10 + 15a \implies 20 = 15a \implies a = \frac{4}{3} \approx 1.33 \text{ ms}^{-2}\).

(ii) Find the distance \(x\) from \(A\) where the speed is \(2u = 20 \text{ ms}^{-1}\).
Using \(v^2 = u^2 + 2ax\):
\(20^2 = 10^2 + 2 \left(\frac{4}{3}\right) x\)
\(400 = 100 + \frac{8}{3} x\)
\(300 = \frac{8}{3} x \implies x = \frac{900}{8} = 112.5 \text{ m}\).

M1: Use \(s = \frac{u+v}{2} t\) to form an equation for \(u\).
A1: Correct initial velocity \(u = 10 \text{ ms}^{-1}\).
M1: Use a kinematic formula to solve for acceleration \(a\).
A1: Correct acceleration \(a = \frac{4}{3} \approx 1.33 \text{ ms}^{-2}\).
M1: Use \(v^2 = u^2 + 2as\) with speed \(2u = 20\) to find distance.
A1: Correct distance \(112.5 \text{ m}\).

(a) Let \(W_1\) and \(W_2\) represent winning the first and second games respectively, and \(L_1\), \(L_2\) represent losing them.
\(P(L_1) = 1 - 0.6 = 0.4\)
\(P(L_2 | L_1) = 1 - 0.45 = 0.55\)
The probability of winning at least one game is \(1 - P(L_1 \text{ and } L_2) = 1 - P(L_1) \times P(L_2 | L_1) = 1 - (0.4 \times 0.55) = 1 - 0.22 = 0.78\).

(b) The events where exactly one game is won are \((W_1 \text{ and } L_2)\) and \((L_1 \text{ and } W_2)\).
\(P(W_1 \text{ and } L_2) = 0.6 \times (1 - 0.7) = 0.6 \times 0.3 = 0.18\)
\(P(L_1 \text{ and } W_2) = 0.4 \times 0.45 = 0.18\)
Total probability of winning exactly one game = \(0.18 + 0.18 = 0.36\).
The probability that they won the first game given that they won exactly one game is \(\frac{P(W_1 \text{ and } L_2)}{P(\text{exactly one})} = \frac{0.18}{0.36} = 0.5\).

(a)
M1: Attempting to calculate \(P(L_1 \cap L_2)\) or summing the other three probabilities.
A1: Finding \(P(L_1 \cap L_2) = 0.22\) or listing all individual correct outcomes.
A1: Correct final answer of 0.78.

(b)
M1: Finding \(P(W_1 \cap L_2)\) and \(P(L_1 \cap W_2)\).
A1: Correctly calculating the sum as 0.36.
M1: Setting up conditional probability fraction with 0.18 in numerator and their 0.36 in denominator.
A1: Correct final answer of 0.5.

(a) The word PARAMETRIC has 10 letters, with 2 A's, 2 R's, and 1 each of P, M, E, T, I, C.
The number of different arrangements is \(\frac{10!}{2! 2!} = 907\\,200\).

(b) First, treat the two A's as a single block (AA). This leaves us with 9 entities to arrange: (AA), R, R, P, M, E, T, I, C.
Since there are 2 R's, the number of arrangements where the two A's are together is \(\frac{9!}{2!} = 181\\,440\).
Next, we find the number of arrangements where both the two A's are together and the two R's are together. Treating (AA) and (RR) as single blocks, we have 8 unique entities: (AA), (RR), P, M, E, T, I, C.
The number of arrangements for these is \(8! = 40\\,320\).
Therefore, the number of arrangements where the A's are together but the R's are not together is \(181\\,440 - 40\\,320 = 141\\,120\).

(a)
M1: Attempting division of 10! by 2! and 2!.
A1: Correct answer 907200.

(b)
M1: Treating AA as a single unit and attempting to arrange 9 units with 2 R's.
A1: Correctly calculating 181440.
M1: Treating AA and RR as single units and attempting to arrange 8 distinct units.
A1: Correctly calculating 40320.
A1: Subtracting to get 141120.

(a) Let \(X\) be the mass of an apple. We are given \(P(X > 180) = 0.15\) and \(P(X < 120) = 0.06\).
Standardising the first equation:
\(P\left(Z > \frac{180 - \mu}{\sigma}\right) = 0.15 \implies \frac{180 - \mu}{\sigma} = \Phi^{-1}(0.85)\)
From normal tables, \(\Phi^{-1}(0.85) = 1.036\).
So, \(\mu + 1.036\sigma = 180\) (Equation 1).

Standardising the second equation:
\(P\left(Z < \frac{120 - \mu}{\sigma}\right) = 0.06 \implies \frac{120 - \mu}{\sigma} = -\Phi^{-1}(0.94)\)
From normal tables, \(\Phi^{-1}(0.94) = 1.555\).
So, \(\frac{120 - \mu}{\sigma} = -1.555 \implies \mu - 1.555\sigma = 120\) (Equation 2).

(b) Subtracting Equation 2 from Equation 1:
\((\mu + 1.036\sigma) - (\mu - 1.555\sigma) = 180 - 120\)
\(2.591\sigma = 60 \implies \sigma = 23.157... \approx 23.2\) (3 s.f.).
Substituting \(\sigma\) back into Equation 2:
\(\mu = 120 + 1.555(23.157...) = 156.009... \approx 156\) (3 s.f.).

(a)
B1: Finding the correct z-value 1.036 (or 1.04).
M1: Standardising and setting equal to their first z-value to form a linear equation.
B1: Finding the correct z-value -1.555 (or -1.554).
M1: Standardising and setting equal to their second z-value to form a second linear equation.

(b)
M1: Solving the simultaneous equations to eliminate one variable.
A1: Correct value of \(\sigma = 23.2\) (accept 23.1 to 23.3).
A1: Correct value of \(\mu = 156\) (accept 155 to 157).

(a) Since the sum of all probabilities is 1:
\(a + b + 0.3 + 0.2 = 1 \implies a + b = 0.5\) (Equation 1).
Using the given expectation \(E(X) = -0.1\):
\(E(X) = -2(a) - 1(b) + 1(0.3) + 2(0.2) = -0.1\)
\(-2a - b + 0.7 = -0.1 \implies 2a + b = 0.8\) (Equation 2).
Subtracting Equation 1 from Equation 2:
\((2a + b) - (a + b) = 0.8 - 0.5 \implies a = 0.3\).
Then \(b = 0.5 - 0.3 = 0.2\).

(b) To find the variance, first calculate \(E(X^2)\):
\(E(X^2) = (-2)^2(a) + (-1)^2(b) + 1^2(0.3) + 2^2(0.2)\)
\(E(X^2) = 4(0.3) + 1(0.2) + 0.3 + 4(0.2) = 1.2 + 0.2 + 0.3 + 0.8 = 2.5\).
Using the formula \(Var(X) = E(X^2) - [E(X)]^2\):
\(Var(X) = 2.5 - (-0.1)^2 = 2.5 - 0.01 = 2.49\).

(a)
B1: Stating \(a + b = 0.5\) (or equivalent).
M1: Writing an expression for \(E(X)\) in terms of \(a\) and \(b\) and setting equal to -0.1.
A1: Correctly simplifying to \(2a + b = 0.8\) (or equivalent).
A1: Solving to get \(a = 0.3\) and \(b = 0.2\).

(b)
M1: Attempting to calculate \(E(X^2)\) using their values of \(a\) and \(b\).
A1: Correctly finding \(E(X^2) = 2.5\).
A1: Correctly calculating \(Var(X) = 2.49\).

(a) Let \(y = t - 30\).
The mean of \(y\) is \(\bar{y} = \frac{\sum y}{n} = \frac{240}{80} = 3\).
Since \(y = t - 30\), we have \(\bar{t} = \bar{y} + 30 = 3 + 30 = 33\) minutes.
The variance of \(y\) is \(Var(y) = \frac{\sum y^2}{n} - \bar{y}^2 = \frac{4160}{80} - 3^2 = 52 - 9 = 43\).
Since coding by subtraction does not affect variance, the variance of \(t\) is also 43.

(b) For the original group of 80 people, the sum of times is \(\sum t_1 = 80 \times 33 = 2640\) minutes.
For the additional 20 people, the mean is 28 minutes, so the sum of times is \(\sum t_2 = 20 \times 28 = 560\) minutes.
For the combined group of 100 people, the combined mean is:
\(\bar{t}_{\text{combined}} = \frac{\sum t_1 + \sum t_2}{80 + 20} = \frac{2640 + 560}{100} = \frac{3200}{100} = 32\) minutes.

(a)
M1: Finding the mean of \((t - 30)\) as 3.
A1: Obtaining the mean of \(t\) as 33.
M1: Attempting to calculate the variance of \((t - 30)\) using the correct variance formula.
A1: Correctly finding the variance of \(t\) as 43.

(b)
M1: Finding the sum of times for the first group (2640).
M1: Finding the sum of times for the second group (560) and adding to the first.
A1: Finding the correct combined mean of 32.

(a) Let \(X\) be the number of heads obtained, so \(X \sim B(120, 0.35)\).
Mean: \(\mu = np = 120 \times 0.35 = 42\).
Variance: \(\sigma^2 = np(1-p) = 42 \times 0.65 = 27.3\).
Standard deviation: \(\sigma = \sqrt{27.3} \approx 5.2249\).
We want to find \(P(38 \le X \le 46)\).
Applying continuity correction, this becomes \(P(37.5 < Y < 46.5)\), where \(Y \sim N(42, 27.3)\).
Standardising:
For \(37.5\): \(z_1 = \frac{37.5 - 42}{5.2249} = -0.861\).
For \(46.5\): \(z_2 = \frac{46.5 - 42}{5.2249} = 0.861\).
The probability is:
\(P(-0.861 < Z < 0.861) = \Phi(0.861) - \Phi(-0.861) = 2\Phi(0.861) - 1\).
From tables, \(\Phi(0.861) = 0.8054\).
So \(2(0.8054) - 1 = 1.6108 - 1 = 0.611\) (3 s.f.).

(b) A normal approximation is appropriate because both \(np\) and \(nq\) are greater than 5.
Specifically:
\(np = 120 \times 0.35 = 42 > 5\)
\(nq = 120 \times 0.65 = 78 > 5\).

(a)
M1: Finding the mean \(\mu = 42\) and variance \(\sigma^2 = 27.3\).
M1: Applying continuity correction to 38 and 46, obtaining 37.5 and 46.5.
M1: Standardising both values with their mean and standard deviation.
A1: Correctly calculating \(z = \pm 0.861\) (or \(\pm 0.86\)).
A1: Correct final probability of 0.611 (accept 0.610 to 0.612).

(b)
B1: Calculating \(np = 42\) and \(n(1-p) = 78\).
B1: Stating that both values are greater than 5 (or both \(> 5\)), which justifies the normal approximation.

(a) To have more women than men in a team of 5, the possible compositions of the team are:
Case 1: 3 women and 2 men.
Number of ways: \(\binom{5}{3} \times \binom{6}{2} = 10 \times 15 = 150\).
Case 2: 4 women and 1 man.
Number of ways: \(\binom{5}{4} \times \binom{6}{1} = 5 \times 6 = 30\).
Case 3: 5 women and 0 men.
Number of ways: \(\binom{5}{5} \times \binom{6}{0} = 1 \times 1 = 1\).
Total number of ways = \(150 + 30 + 1 = 181\).

(b) First, find the total number of ways to choose any team of 5 from the 11 people:
\(\binom{11}{5} = 462\).
Next, find the number of teams that contain both Alan and Bella. If both are on the team, we must choose the remaining 3 team members from the remaining 9 people:
\(\binom{9}{3} = 84\).
Subtracting this from the total gives the number of teams where Alan and Bella are not both included:
\(462 - 84 = 378\).

Alternative method:
Case 1: Alan is included but Bella is not. Choose 4 members from the remaining 9: \(\binom{9}{4} = 126\).
Case 2: Bella is included but Alan is not. Choose 4 members from the remaining 9: \(\binom{9}{4} = 126\).
Case 3: Neither Alan nor Bella is included. Choose 5 members from the remaining 9: \(\binom{9}{5} = 126\).
Total = \(126 + 126 + 126 = 378\).

(a)
M1: Formulating and calculating one correct case (e.g., \(\binom{5}{3} \times \binom{6}{2}\) or \(\binom{5}{4} \times \binom{6}{1}\)).
M1: Correctly listing and summing all three possible cases (3W/2M, 4W/1M, 5W/0M).
A1: Correct answer of 181.

(b)
M1: Calculating total unrestricted combinations, \(\binom{11}{5} = 462\).
M1: Calculating combinations that contain both Alan and Bella, \(\binom{9}{3} = 84\).
A1: Subtracting these values (or setting up equivalent case-by-case calculations).
A1: Correct answer of 378.

Let \(X\) be the number of flaws in a 3-metre length of denim. The average rate for a 3-metre length is \(\lambda = 0.8 \times 3 = 2.4\). So, \(X \sim \text{Po}(2.4)\). We want to find \(P(2 \le X < 5) = P(X = 2) + P(X = 3) + P(X = 4)\). Using the Poisson probability formula \(P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}\): \(P(X = 2) = \frac{e^{-2.4} \times 2.4^2}{2!} = 2.88 e^{-2.4}\), \(P(X = 3) = \frac{e^{-2.4} \times 2.4^3}{3!} = 2.304 e^{-2.4}\), \(P(X = 4) = \frac{e^{-2.4} \times 2.4^4}{4!} = 1.3824 e^{-2.4}\). Summing these probabilities: \(P(2 \le X < 5) = (2.88 + 2.304 + 1.3824) e^{-2.4} = 6.5664 e^{-2.4}\). Using \(e^{-2.4} \approx 0.090718\): \(P(2 \le X < 5) \approx 6.5664 \times 0.090718 \approx 0.59569\). To 3 significant figures, the probability is 0.596.

M1: For calculating the correct parameter \(\lambda = 2.4\). M1: For writing down the expression for \(P(2 \le X < 5)\) as \(P(X=2) + P(X=3) + P(X=4)\). M1: For correctly evaluating the three Poisson terms. A1: For obtaining the correct final probability of 0.596 (accept 0.595 - 0.596).

(i) Since the total area under the probability density function is 1: \(\int_{0}^{4} k(4x^2 - x^3) \, dx = 1 \implies k \left[ \frac{4x^3}{3} - \frac{x^4}{4} \right]_{0}^{4} = 1 \implies k \left( \frac{256}{3} - 64 \right) = 1 \implies k \left( \frac{64}{3} \right) = 1 \implies k = \frac{3}{64}\). (ii) To find \(P(1 \le X \le 3)\): \(P(1 \le X \le 3) = \int_{1}^{3} \frac{3}{64} (4x^2 - x^3) \, dx = \frac{3}{64} \left[ \frac{4x^3}{3} - \frac{x^4}{4} \right]_{1}^{3} = \frac{3}{64} \left[ \left( 36 - \frac{81}{4} \right) - \left( \frac{4}{3} - \frac{1}{4} \right) \right] = \frac{3}{64} \left[ \frac{63}{4} - \frac{13}{12} \right] = \frac{3}{64} \times \frac{176}{12} = \frac{11}{16} = 0.6875\).

M1: For integrating \(x^2(4-x)\) and setting the integral from 0 to 4 equal to 1. A1: For correctly showing \(k = \frac{3}{64}\). M1: For setting up the integral for \(P(1 \le X \le 3)\) with limits 1 and 3. M1: For integrating and substituting the limits correctly. A1: For obtaining the final answer 0.6875 (or 11/16).

Let \(T = C_1 + C_2 + C_3 - S_1 - S_2 - S_3\). We need to find the expectation and variance of \(T\): \(E(T) = 3 E(C) - 3 E(S) = 3(120) - 3(90) = 90\). Since the variables are independent: \(\text{Var}(T) = 3 \text{Var}(C) + 3 \text{Var}(S) = 3(8^2) + 3(5^2) = 192 + 75 = 267\). Thus, \(T \sim N(90, 267)\). We want to find \(P(T \ge 80)\): \(P(T \ge 80) = P\left(Z \ge \frac{80 - 90}{\sqrt{267}}\right) = P\left(Z \ge -0.612\right) = \Phi(0.612) \approx 0.7297\). To 3 significant figures, the probability is 0.730.

M1: For calculating the correct expectation \(E(T) = 90\). M1: For calculating the correct variance \(\text{Var}(T) = 267\). M1: For standardizing to find the Z-score. A1: For obtaining \(z = -0.612\). A1: For the correct final probability of 0.730 (accept 0.729 - 0.731).

(i) Unbiased estimate of the population mean: \(\bar{x} = \frac{1440}{80} = 18\text{ mm}\). Unbiased estimate of the population variance: \(s^2 = \frac{1}{79} \left( 26360 - \frac{1440^2}{80} \right) = \frac{1}{79} \left( 26360 - 25920 \right) = \frac{440}{79} \approx 5.5696\text{ mm}^2\). (ii) Standard error of the mean is \(\sqrt{\frac{s^2}{n}} = \sqrt{\frac{5.5696}{80}} \approx 0.2639\). For a 95% confidence interval, the critical value of \(z\) is 1.960. The interval is \(18 \pm 1.960 \times 0.2639 = [17.48, 18.52]\). To 3 significant figures, this is \([17.5, 18.5]\).

M1: For calculating the unbiased estimate of the mean \(\bar{x} = 18\). M1: For using the correct formula for the unbiased estimate of variance. A1: For obtaining \(s^2 \approx 5.57\). M1: For using the formula \(\bar{x} \pm z \frac{s}{\sqrt{n}}\) with \(z = 1.96\). A1: For obtaining the correct interval [17.5, 18.5].

Let \(p\) be the population proportion of bulbs that last more than 15,000 hours. \(H_0: p = 0.90\), \(H_1: p < 0.90\). Under \(H_0\), the number of bulbs \(X\) that last more than 15,000 hours is \(X \sim B(20, 0.90)\). We find \(P(X \le 15) = 1 - P(X \ge 16)\). Calculating each term for \(X \ge 16\): \(P(X=20) = (0.90)^{20} \approx 0.12158\), \(P(X=19) = 20 \times (0.90)^{19} \times 0.10 \approx 0.27017\), \(P(X=18) = 190 \times (0.90)^{18} \times (0.10)^2 \approx 0.28518\), \(P(X=17) = 1140 \times (0.90)^{17} \times (0.10)^3 \approx 0.19012\), \(P(X=16) = 4845 \times (0.90)^{16} \times (0.10)^4 \approx 0.08978\). Summing these: \(P(X \ge 16) \approx 0.95683 \implies P(X \le 15) = 1 - 0.95683 = 0.04317\). Since \(0.04317 < 0.05\), we reject \(H_0\). There is sufficient evidence at the 5% level to suggest the proportion is less than 90%.

B1: For stating both hypotheses correctly. M1: For calculating the probability of the tail \(P(X \le 15)\). M1: For evaluating at least 3 terms of the binomial expansion. A1: For obtaining the probability \(P(X \le 15) \approx 0.0432\). A1: For a correct decision to reject \(H_0\) with a correct contextual conclusion.

Let \(X \sim B(600, 0.004)\). Since \(n = 600\) is large and \(p = 0.004\) is small, we approximate using a Poisson distribution \(Y \sim \text{Po}(\lambda)\) where \(\lambda = np = 600 \times 0.004 = 2.4\). We want to find \(P(Y \ge 4) = 1 - P(Y \le 3) = 1 - e^{-2.4} \left( 1 + 2.4 + \frac{2.4^2}{2!} + \frac{2.4^3}{3!} \right) = 1 - e^{-2.4} (1 + 2.4 + 2.88 + 2.304) = 1 - 8.584 e^{-2.4} \approx 1 - 8.584 \times 0.090718 \approx 1 - 0.77872 = 0.22128\). To 3 significant figures, this is 0.221.

M1: For identifying and justifying the Poisson approximation with \(\lambda = 2.4\). M1: For expressing the required probability as \(1 - P(Y \le 3)\). M1: For calculating the Poisson terms for \(k = 0, 1, 2, 3\). A1: For obtaining the correct final probability of 0.221.

Let \(\mu\) be the population mean score. \(H_0: \mu = 52.0\), \(H_1: \mu > 52.0\). Under \(H_0\), the sample mean is distributed as \(\bar{X} \sim N\left(52.0, \frac{10.5^2}{40}\right)\). The standard error of the mean is \(\sigma_{\bar{X}} = \frac{10.5}{\sqrt{40}} \approx 1.6602\). Calculating the test statistic: \(z = \frac{55.5 - 52.0}{1.6602} \approx 2.11\). For a one-tailed test at the 5% significance level, the critical value of \(z\) is 1.645. Since \(2.11 > 1.645\), we reject \(H_0\). There is sufficient evidence at the 5% level to conclude that the tutoring system has increased the mean score.

B1: For stating both hypotheses correctly. M1: For calculating the standard error. M1: For calculating the \(z\)-value. A1: For obtaining \(z \approx 2.11\). M1: For comparing their \(z\)-value with the critical value 1.645. A1: For concluding to reject \(H_0\) and stating the conclusion in context.