2025 Cambridge IAL Mathematics (9709) 模擬試題連答案詳解

To find the points of intersection, we equate the equations of the line and the curve:
\( x^2 + 2x + 2 = mx - 7 \)

Rearranging into a standard quadratic equation form:
\( x^2 + (2-m)x + 9 = 0 \)

Since the line is a tangent to the curve, there is only one point of contact. Therefore, the discriminant of this quadratic equation must be equal to zero:
\( b^2 - 4ac = 0 \)
\( (2-m)^2 - 4(1)(9) = 0 \)
\( (2-m)^2 - 36 = 0 \)
\( (2-m)^2 = 36 \)

Taking square roots:
\( 2-m = 6 \) or \( 2-m = -6 \)

This gives:
\( m = -4 \) or \( m = 8 \)

Since we are given that \( m \) is a positive constant, we reject \( m = -4 \) and choose:
\( m = 8 \)

To find the coordinates of the point of contact, we substitute \( m = 8 \) back into the quadratic equation:
\( x^2 + (2-8)x + 9 = 0 \)
\( x^2 - 6x + 9 = 0 \)
\( (x-3)^2 = 0 \implies x = 3 \)

Substituting \( x = 3 \) into the equation of the line to find the \( y \)-coordinate:
\( y = 8(3) - 7 = 17 \)

Thus, the point of contact is \( (3, 17) \).

M1: Equate the line and curve equations and form a quadratic equation in \( x \).
M1: Use the discriminant \( b^2 - 4ac = 0 \) to set up an equation in \( m \).
A1: Correctly solve for \( m \) and select the positive value \( m = 8 \).
M1: Substitute \( m = 8 \) back into the quadratic equation and solve for \( x \).
A1: Find the correct \( x \)-coordinate, \( x = 3 \).
A1: Find the correct \( y \)-coordinate, \( y = 17 \), and state the coordinates of the point of contact as \( (3, 17) \).

(i) Completing the square for \( f(x) = 3x^2 - 12x + 7 \):
\( f(x) = 3(x^2 - 4x) + 7 \)
\( f(x) = 3\left[(x-2)^2 - 4\right] + 7 \)
\( f(x) = 3(x-2)^2 - 12 + 7 \)
\( f(x) = 3(x-2)^2 - 5 \)

(ii) The quadratic expression has a minimum point at \( (2, -5) \). Since the domain is restricted to \( x \le 1 \), the vertex is outside the domain, and the function is strictly decreasing on the interval \( (-\infty, 1] \).
At the boundary \( x = 1 \):
\( f(1) = 3(1-2)^2 - 5 = -2 \)
As \( x \) decreases, \( f(x) \) increases. Thus, the range of \( f \) is:
\( f(x) \ge -2 \)

(iii) To find the inverse function, we let \( y = f(x) \) and solve for \( x \):
\( y = 3(x-2)^2 - 5 \)
\( y + 5 = 3(x-2)^2 \)
\( (x-2)^2 = \frac{y+5}{3} \)
Since \( x \le 1 \), we have \( x-2 \le -1 \), which means \( x-2 \) is negative. Taking the negative square root:
\( x - 2 = -\sqrt{\frac{y+5}{3}} \)
\( x = 2 - \sqrt{\frac{y+5}{3}} \)

Therefore, the inverse function is:
\( f^{-1}(x) = 2 - \sqrt{\frac{x+5}{3}} \)

The domain of \( f^{-1} \) is the range of \( f \), which is:
\( x \ge -2 \)

(i)
M1: For attempting to complete the square to obtain \( 3(x-b)^2 + c \).
A1: For the correct expression \( 3(x-2)^2 - 5 \).

(ii)
B1: For stating the correct range \( f(x) \ge -2 \) (or \( y \ge -2 \)).

(iii)
M1: For making \( x \) the subject of \( y = 3(x-2)^2 - 5 \).
A1: For identifying and justifying the choice of the negative square root due to the domain restriction.
A1: For the correct expression \( f^{-1}(x) = 2 - \sqrt{\frac{x+5}{3}} \).
B1: For stating the correct domain of \( f^{-1} \) as \( x \ge -2 \).

(i) First, we find the midpoint \( M \) of \( AB \):
\( M = \left(\frac{1+5}{2}, \frac{1+9}{2}\right) = (3, 5) \)

Next, we find the gradient \( m \) of the line segment \( AB \):
\( m = \frac{9-1}{5-1} = \frac{8}{4} = 2 \)

The gradient of the perpendicular bisector is the negative reciprocal of \( m \):
\( m_{\perp} = -\frac{1}{2} \)

The equation of the perpendicular bisector is:
\( y - 5 = -\frac{1}{2}(x - 3) \)
\( 2y - 10 = -x + 3 \)
\( x + 2y = 13 \)

(ii) The centre of the circle must lie on the perpendicular bisector of the chord \( AB \). Thus, the coordinates of the centre satisfy both the equation of the perpendicular bisector and the given line equation:
\( x + 2y = 13 \)
\( y = x + 2 \)

Substituting the second equation into the first:
\( x + 2(x + 2) = 13 \)
\( x + 2x + 4 = 13 \)
\( 3x = 9 \implies x = 3 \)

Using \( y = x + 2 \) to find the \( y \)-coordinate:
\( y = 3 + 2 = 5 \)

So, the coordinates of the centre of the circle are \( (3, 5) \).

(i)
B1: Find the correct midpoint \( (3, 5) \).
M1: Calculate the gradient of \( AB \) and find the perpendicular gradient.
A1: Obtain the correct equation of the perpendicular bisector, e.g., \( x + 2y = 13 \) or \( y = -0.5x + 6.5 \).

(ii)
M1: Set up simultaneous equations using the perpendicular bisector and the line \( y = x + 2 \).
A1: Solve the simultaneous equations to find \( x = 3 \).
A1: Find \( y = 5 \) and state the centre coordinates as \( (3, 5) \).

(i) The area \( A \) of a sector is given by the formula:
\( A = \frac{1}{2} r^2 \theta \)

Given \( A = 24\pi \) and \( \theta = \frac{3}{4}\pi \):
\( 24\pi = \frac{1}{2} r^2 \left(\frac{3}{4}\pi\right) \)
\( 24 = \frac{3}{8} r^2 \)
\( r^2 = 24 \times \frac{8}{3} = 64 \)
Since \( r > 0 \):
\( r = 8 \)

(ii) The perimeter of the region consists of three parts:
1. The arc length \( AB \):
\( \text{Arc } AB = r\theta = 8 \times \frac{3}{4}\pi = 6\pi \approx 18.850\text{ cm} \)

2. The line segment \( CB \):
Since \( OB = r = 8\text{ cm} \) and \( OC = 6\text{ cm} \):
\( CB = OB - OC = 8 - 6 = 2\text{ cm} \)

3. The line segment \( AC \):
We use the cosine rule on triangle \( OAC \) where \( OA = r = 8 \), \( OC = 6 \), and \( \angle AOC = \frac{3}{4}\pi \):
\( AC^2 = OA^2 + OC^2 - 2(OA)(OC)\cos\left(\frac{3}{4}\pi\right) \)
\( AC^2 = 8^2 + 6^2 - 2(8)(6)\left(-\frac{\sqrt{2}}{2}\right) \)
\( AC^2 = 64 + 36 + 48\sqrt{2} = 100 + 48\sqrt{2} \approx 167.882 \)
\( AC = \sqrt{100 + 48\sqrt{2}} \approx 12.957\text{ cm} \)

Total Perimeter:
\( P = \text{Arc } AB + CB + AC \approx 18.850 + 2 + 12.957 = 33.807\text{ cm} \)

To 3 significant figures, the perimeter is \( 33.8\text{ cm} \).

(i)
M1: Use \( \frac{1}{2} r^2 \theta = 24\pi \) with \( \theta = \frac{3}{4}\pi \) to form an equation for \( r \).
A1: Solve to find \( r = 8 \).

(ii)
B1: Find the correct arc length \( AB = 6\pi \) (or decimal equivalent \( 18.85 \)).
M1: Use the cosine rule in triangle \( OAC \) with correct side lengths and angle.
A1: Correctly calculate the length of \( AC \approx 12.96 \).
A1: Sum all three components to get \( 33.8 \) (accept \( 33.8 \) to \( 33.9 \)).

We rewrite \( \tan \theta \) in terms of sine and cosine:
\( 4\sin \theta \left(\frac{\sin \theta}{\cos \theta}\right) = 15 - 13\cos \theta \)
\( 4\frac{\sin^2 \theta}{\cos \theta} = 15 - 13\cos \theta \)

Multiplying both sides by \( \cos \theta \) (assuming \( \cos \theta \ne 0 \)):
\( 4\sin^2 \theta = \cos \theta(15 - 13\cos \theta) \)
\( 4\sin^2 \theta = 15\cos \theta - 13\cos^2 \theta \)

Using the Pythagorean identity \( \sin^2 \theta = 1 - \cos^2 \theta \):
\( 4(1 - \cos^2 \theta) = 15\cos \theta - 13\cos^2 \theta \)
\( 4 - 4\cos^2 \theta = 15\cos \theta - 13\cos^2 \theta \)

Rearranging to form a quadratic equation in \( \cos \theta \):
\( 9\cos^2 \theta - 15\cos \theta + 4 = 0 \)

Let \( u = \cos \theta \):
\( 9u^2 - 15u + 4 = 0 \)

We factorise this quadratic:
\( (3u - 1)(3u - 4) = 0 \)

This gives:
\( u = \frac{1}{3} \) or \( u = \frac{4}{3} \)

Since \( |\cos \theta| \le 1 \), the equation \( \cos \theta = \frac{4}{3} \) has no real solutions.

Thus, we solve:
\( \cos \theta = \frac{1}{3} \)

In the range \( 0^\circ \le \theta \le 360^\circ \):
\( \theta = \cos^{-1}\left(\frac{1}{3}\right) \approx 70.53^\circ \approx 70.5^\circ \) (to 1 d.p.)

The second solution is in the fourth quadrant:
\( \theta = 360^\circ - 70.53^\circ \approx 289.47^\circ \approx 289.5^\circ \) (to 1 d.p.)

M1: Substitute \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) and eliminate the fraction.
M1: Substitute \( \sin^2 \theta = 1 - \cos^2 \theta \) to obtain a quadratic in \( \cos \theta \).
A1: Obtain the correct quadratic equation \( 9\cos^2 \theta - 15\cos \theta + 4 = 0 \).
M1: Attempt to solve the quadratic equation to get values of \( \cos \theta \).
A1: Obtain \( \theta = 70.5^\circ \) (or 1 d.p. equivalent).
A1: Obtain \( \theta = 289.5^\circ \) (or 1 d.p. equivalent) and no extra solutions in range.

(i) The 1st, 3rd, and 4th terms of the arithmetic progression are:
\( T_1 = a \)
\( T_3 = a + 2d \)
\( T_4 = a + 3d \)

Since these three terms form a geometric progression, the common ratio is constant:
\( \frac{a+2d}{a} = \frac{a+3d}{a+2d} \)

Cross-multiplying:
\( (a+2d)^2 = a(a+3d) \)
\( a^2 + 4ad + 4d^2 = a^2 + 3ad \)
\( 4ad - 3ad + 4d^2 = 0 \)
\( ad + 4d^2 = 0 \)
\( d(a + 4d) = 0 \)

Since \( d \ne 0 \), we must have:
\( a + 4d = 0 \implies d = -\frac{1}{4}a \) (as required).

(ii) The sum of the first \( n \) terms of an arithmetic progression is:
\( S_n = \frac{n}{2}[2a + (n-1)d] \)

For \( n = 20 \) and \( S_{20} = -220 \):
\( -220 = \frac{20}{2}[2a + 19d] \)
\( -220 = 10[2a + 19d] \)
\( 2a + 19d = -22 \)

We substitute \( a = -4d \) into this equation:
\( 2(-4d) + 19d = -22 \)
\( -8d + 19d = -22 \)
\( 11d = -22 \implies d = -2 \)

Then:
\( a = -4(-2) = 8 \)

The terms of the geometric progression are:
1st term: \( A = T_1 = a = 8 \)
2nd term: \( T_3 = a + 2d = 8 + 2(-2) = 4 \)

Therefore, the common ratio of the geometric progression is:
\( r = \frac{4}{8} = \frac{1}{2} \)

Since \( |r| < 1 \), the sum to infinity of the geometric progression exists and is given by:
\( S_{\infty} = \frac{A}{1-r} = \frac{8}{1 - 1/2} = 16 \)

(i)
M1: Express terms in terms of \( a \) and \( d \) and set up the geometric progression condition.
M1: Expand and simplify to obtain \( ad + 4d^2 = 0 \).
A1: Obtain the given relation \( d = -\frac{1}{4}a \) clearly showing the division by \( d \, (d \ne 0) \).

(ii)
M1: Set up the equation for the sum of 20 terms of the AP.
A1: Correctly solve for \( a = 8 \) and \( d = -2 \).
M1: Find the first term \( A = 8 \) and common ratio \( r = 0.5 \) of the GP, and apply the sum to infinity formula.
A1: Find the correct sum to infinity, \( 16 \).

(i) Let the radius of the water surface at depth \( h \) be \( r \). By similar triangles, the ratio of the radius of the water surface to the depth of the water is equal to the ratio of the base radius of the cone to its height:
\( \frac{r}{h} = \frac{4}{12} = \frac{1}{3} \implies r = \frac{1}{3}h \)

The volume \( V \) of a cone is given by:
\( V = \frac{1}{3}\pi r^2 h \)

Substituting \( r = \frac{1}{3}h \):
\( V = \frac{1}{3}\pi \left(\frac{1}{3}h\right)^2 h = \frac{1}{27}\pi h^3 \) (as required).

(ii) We are given the rate of change of volume with respect to time:
\( \frac{dV}{dt} = 3 \)

Differentiating \( V = \frac{1}{27}\pi h^3 \) with respect to \( h \):
\( \frac{dV}{dh} = \frac{3}{27}\pi h^2 = \frac{1}{9}\pi h^2 \)

Using the chain rule:
\( \frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt} \)
\( 3 = \left(\frac{1}{9}\pi h^2\right) \times \frac{dh}{dt} \)
\( \frac{dh}{dt} = \frac{27}{\pi h^2} \)

When \( h = 6 \):
\( \frac{dh}{dt} = \frac{27}{\pi (6)^2} = \frac{27}{36\pi} = \frac{3}{4\pi}\text{ cm s}^{-1} \)

(i)
M1: Use similar triangles to relate \( r \) and \( h \).
A1: Correctly substitute \( r = \frac{1}{3}h \) into the cone volume formula and obtain the given relation.

(ii)
M1: Differentiate \( V \) with respect to \( h \) to obtain \( \frac{dV}{dh} = k h^2 \).
A1: Obtain the correct derivative \( \frac{dV}{dh} = \frac{1}{9}\pi h^2 \).
M1: Use the chain rule connecting \( \frac{dh}{dt} \), \( \frac{dV}{dt} \), and \( \frac{dV}{dh} \).
A1: Correctly solve to find the rate of change is \( \frac{3}{4\pi} \).

(i) At \( x = 0 \), the \( y \)-coordinate is:
\( y = 3(2(0)+1)^{1/4} = 3(1)^{1/4} = 3 \)
So the point of contact is \( (0, 3) \).

To find the gradient of the tangent, we differentiate \( y = 3(2x+1)^{1/4} \) using the chain rule:
\( \frac{dy}{dx} = 3 \times \frac{1}{4}(2x+1)^{-3/4} \times 2 = \frac{3}{2}(2x+1)^{-3/4} \)

At \( x = 0 \):
\( \frac{dy}{dx} = \frac{3}{2}(1)^{-3/4} = \frac{3}{2} \)

Since the gradient of the tangent is \( \frac{3}{2} \), the gradient of the normal is:
\( m_{\text{normal}} = -\frac{2}{3} \)

The equation of the normal at \( (0, 3) \) is:
\( y - 3 = -\frac{2}{3}(x - 0) \implies y = -\frac{2}{3}x + 3 \)

(ii) The volume \( V \) of the solid of revolution about the \( x \)-axis is:
\( V = \pi \int_{0}^{12} y^2 \, dx \)

First, find \( y^2 \):
\( y^2 = \left[ 3(2x+1)^{1/4} \right]^2 = 9(2x+1)^{1/2} \)

Now set up the integral:
\( V = \pi \int_{0}^{12} 9(2x+1)^{1/2} \, dx \)

Using the integration formula \( \int (ax+b)^n \, dx = \frac{(ax+b)^{n+1}}{a(n+1)} \):
\( \int 9(2x+1)^{1/2} \, dx = 9 \left[ \frac{(2x+1)^{3/2}}{2 \times \frac{3}{2}} \right] = 3(2x+1)^{3/2} \)

Evaluating this from \( 0 \) to \( 12 \):
\( V = \pi \left[ 3(2(12)+1)^{3/2} - 3(2(0)+1)^{3/2} \right] \)
\( V = \pi \left[ 3(25)^{3/2} - 3(1)^{3/2} \right] \)
\( V = \pi [3(125) - 3] = \pi [375 - 3] = 372\pi \)

(i)
B1: Find the coordinates of the point where \( x = 0 \) as \( (0, 3) \).
M1: Differentiate the curve equation using the chain rule.
A1: Correctly evaluate the gradient of the normal as \( -\frac{2}{3} \) and write down the normal equation.

(ii)
M1: Use the volume formula \( V = \pi \int y^2 \, dx \) with correct limits.
A1: Correctly integrate \( 9(2x+1)^{1/2} \) to obtain \( 3(2x+1)^{3/2} \).
M1: Substitute the limits \( 12 \) and \( 0 \) into the integrated expression.
A1: Obtain the correct final volume of \( 372\pi \).

(i) First, find the \( y \)-coordinate when \( x = 1 \):
\( y = \frac{8}{\sqrt{3(1)+1}} = \frac{8}{2} = 4 \).

Now, differentiate \( y = 8(3x+1)^{-\frac{1}{2}} \) with respect to \( x \) using the chain rule:
\( \frac{dy}{dx} = 8 \left(-\frac{1}{2}\right) (3x+1)^{-\frac{3}{2}} \cdot 3 = -12(3x+1)^{-\frac{3}{2}} \).

At \( x = 1 \):
\( \frac{dy}{dx} = -12(4)^{-\frac{3}{2}} = -12 \cdot \frac{1}{8} = -\frac{3}{2} \).

The equation of the tangent is:
\( y - 4 = -\frac{3}{2}(x - 1) \)
\( 2y - 8 = -3x + 3 \implies 3x + 2y - 11 = 0 \).

(ii) At \( x = 5 \), find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = -12(3(5)+1)^{-\frac{3}{2}} = -12(16)^{-\frac{3}{2}} = -12 \cdot \frac{1}{64} = -\frac{3}{16} \).

Using the chain rule:
\( \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} \)

Given that \( \frac{dy}{dt} = -0.15 \):
\( -0.15 = -\frac{3}{16} \cdot \frac{dx}{dt} \)
\( \frac{dx}{dt} = -0.15 \cdot \left(-\frac{16}{3}\right) = 0.8 \) units per second.

(i)
* **M1**: Attempt to differentiate \( y = 8(3x+1)^{-1/2} \) using chain rule, obtaining \( k(3x+1)^{-3/2} \).
* **A1**: Correct derivative \( \frac{dy}{dx} = -12(3x+1)^{-3/2} \).
* **M1**: Substitute \( x = 1 \) to find gradient and attempt equation of tangent.
* **A1**: Correct equation \( 3x + 2y - 11 = 0 \) or equivalent (e.g. \( y = -1.5x + 5.5 \)).

(ii)
* **M1**: Substitute \( x = 5 \) into their \( \frac{dy}{dx} \).
* **M1**: Use chain rule \( \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} \) with \( \frac{dy}{dt} = -0.15 \).
* **A1**: Correct value of \( \frac{dx}{dt} = 0.8 \) (or \( \frac{4}{5} \)).

(i) To find the points of intersection, set the curve equation equal to the line equation:
\( 2x^2 + kx + 5 = 3x + 3 \)
\( 2x^2 + (k-3)x + 2 = 0 \)

For the line and the curve not to intersect, this quadratic equation must have no real roots. Therefore, the discriminant must be negative:
\( b^2 - 4ac < 0 \)
\( (k-3)^2 - 4(2)(2) < 0 \)
\( (k-3)^2 - 16 < 0 \)
\( (k-3)^2 < 16 \)

Taking the square root:
\( -4 < k - 3 < 4 \)
\( -1 < k < 7 \).

(ii) When \( k = 7 \), the quadratic equation becomes:
\( 2x^2 + (7-3)x + 2 = 0 \)
\( 2x^2 + 4x + 2 = 0 \)
\( 2(x^2 + 2x + 1) = 0 \)
\( 2(x+1)^2 = 0 \)

Solving for \( x \):
\( x = -1 \)

Substitute \( x = -1 \) into the line equation to find \( y \):
\( y = 3(-1) + 3 = 0 \).

So the coordinates of the point of intersection are \( (-1, 0) \).

(i)
* **M1**: Equate the curve and the line to form a single quadratic equation.
* **A1**: Obtain correct quadratic equation \( 2x^2 + (k-3)x + 2 = 0 \) or equivalent.
* **M1**: Set the discriminant of their quadratic equation to be less than 0.
* **A1**: Correct final range \( -1 < k < 7 \) (or interval notation \( (-1, 7) \)).

(ii)
* **M1**: Substitute \( k = 7 \) into the quadratic equation and attempt to solve for \( x \).
* **A1**: Correct \( x \)-value of \( -1 \).
* **A1**: Correct coordinates \( (-1, 0) \).

(i) The area of \( R \) is given by the integral:
\( \text{Area} = \int_{2}^{6} \sqrt{2x-3} \, dx = \int_{2}^{6} (2x-3)^{\frac{1}{2}} \, dx \)

Integrate using the reverse chain rule:
\( \int (2x-3)^{\frac{1}{2}} \, dx = \left[ \frac{(2x-3)^{\frac{3}{2}}}{\frac{3}{2} \cdot 2} \right]_{2}^{6} = \left[ \frac{1}{3}(2x-3)^{\frac{3}{2}} \right]_{2}^{6} \)

Substitute the limits:
\( \text{Area} = \frac{1}{3}(2(6)-3)^{\frac{3}{2}} - \frac{1}{3}(2(2)-3)^{\frac{3}{2}} \)
\( \text{Area} = \frac{1}{3}(9)^{\frac{3}{2}} - \frac{1}{3}(1)^{\frac{3}{2}} \)
\( \text{Area} = \frac{1}{3}(27) - \frac{1}{3}(1) = 9 - \frac{1}{3} = \frac{26}{3} \) (or \( 8.67 \)).

(ii) The volume of revolution about the \( x \)-axis is given by:
\( V = \pi \int_{2}^{6} y^2 \, dx = \pi \int_{2}^{6} (2x-3) \, dx \)

Integrate \( 2x-3 \):
\( \int (2x-3) \, dx = [x^2 - 3x]_{2}^{6} \)

Substitute the limits:
At \( x = 6 \): \( 6^2 - 3(6) = 36 - 18 = 18 \).
At \( x = 2 \): \( 2^2 - 3(2) = 4 - 6 = -2 \).

\( V = \pi [18 - (-2)] = 20\pi \).

(i)
* **M1**: Attempt to integrate \( (2x-3)^{1/2} \) to obtain \( a(2x-3)^{3/2} \).
* **A1**: Correct integration of \( \frac{1}{3}(2x-3)^{3/2} \).
* **A1**: Correct final area of \( \frac{26}{3} \) or equivalent exact fraction/decimal to 3sf.

(ii)
* **M1**: State or use the formula \( V = \pi \int y^2 \, dx \) with correct limits.
* **A1**: Show correct integrand \( (2x-3) \).
* **M1**: Integrate \( 2x-3 \) to obtain \( x^2 - 3x \) and substitute limits \( 2 \) and \( 6 \).
* **A1**: Obtain \( 20\pi \) (accept \( 62.8 \) from correct working).

To solve \(|2x - 5| < |x + 1|\), we can square both sides since both sides are non-negative:
\[(2x - 5)^2 < (x + 1)^2\]

Expanding both sides:
\[4x^2 - 20x + 25 < x^2 + 2x + 1\]

Rearranging into a quadratic inequality:
\[3x^2 - 22x + 24 < 0\]

Factorising the quadratic expression:
\[(3x - 4)(x - 6) < 0\]

Thus, the critical values are:
\[x = \frac{4}{3} \quad \text{and} \quad x = 6\]

Since the inequality is less than zero, the solution lies between these critical values:
\[\frac{4}{3} < x < 6\]

M1: For squaring both sides or equating both sides to find critical values.
A1: For obtaining a correct 3-term quadratic inequality, e.g., \(3x^2 - 22x + 24 < 0\).
M1: For attempting to factorise or solve their 3-term quadratic equation.
A1.25: For finding the correct critical values \(x = \frac{4}{3}\) and \(x = 6\).
A2: For obtaining the correct final range \(\frac{4}{3} < x < 6\) (or equivalent interval notation).

We can rewrite the equation using indices:
\[3 \cdot (3^x)^2 - 10 \cdot 3^x + 3 = 0\]

Let \(y = 3^x\). The equation becomes:
\[3y^2 - 10y + 3 = 0\]

Factorising this quadratic equation:
\[(3y - 1)(y - 3) = 0\]

This gives two possible values for \(y\):
\[y = \frac{1}{3} \quad \text{or} \quad y = 3\]

Now, substitute back \(y = 3^x\):
1) \(3^x = \frac{1}{3} = 3^{-1} \implies x = -1\)
2) \(3^x = 3^1 \implies x = 1\)

Therefore, the exact solutions are \(x = -1\) and \(x = 1\).

M1: For writing the equation as a quadratic in \(3^x\) or using a suitable substitution like \(y = 3^x\).
A1.25: For obtaining the correct quadratic equation \(3y^2 - 10y + 3 = 0\).
M1: For attempting to factorise or solve their quadratic equation.
A1: For obtaining both correct values of \(3^x\), which are \(\frac{1}{3}\) and \(3\).
M1: For using indices or logarithms to solve for \(x\).
A1: For obtaining the final exact answers \(x = -1\) and \(x = 1\).

Using the trigonometric identity \(\sec^2 \theta = 1 + \tan^2 \theta\), we substitute this into the equation:
\[2(1 + \tan^2 \theta) + 3\tan \theta = 5\]
\[2 + 2\tan^2 \theta + 3\tan \theta = 5\]
\[2\tan^2 \theta + 3\tan \theta - 3 = 0\]

Let \(t = \tan \theta\). Then:
\[2t^2 + 3t - 3 = 0\]

Using the quadratic formula:
\[t = \frac{-3 \pm \sqrt{3^2 - 4(2)(-3)}}{2(2)} = \frac{-3 \pm \sqrt{33}}{4}\]

This gives two values for \(\tan \theta\):
\[\tan \theta \approx 0.68614 \quad \text{or} \quad \tan \theta \approx -2.18614\]

Now we find the values of \(\theta\) in the interval \(0^\circ \le \theta \le 180^\circ\):
1) For \(\tan \theta \approx 0.68614\):
\[\theta = \tan^{-1}(0.68614) \approx 34.5^\circ\]

2) For \(\tan \theta \approx -2.18614\):
\[\theta = 180^\circ - \tan^{-1}(2.18614) \approx 180^\circ - 65.4^\circ = 114.6^\circ\]

Thus, the solutions in the given interval are \(\theta = 34.5^\circ\) and \(\theta = 114.6^\circ\).

M1: For applying the identity \(\sec^2 \theta = 1 + \tan^2 \theta\) to form an equation in \(\tan \theta\).
A1: For the correct quadratic equation \(2\tan^2 \theta + 3\tan \theta - 3 = 0\).
M1: For using a valid method to solve the quadratic equation.
A1.25: For obtaining correct decimal values for \(\tan \theta\) (approx. \(0.686\) and \(-2.186\)).
M1: For applying the inverse tangent function correctly to obtain at least one angle in the given range.
A1: For both correct solutions \(\theta = 34.5^\circ\) and \(\theta = 114.6^\circ\).

To find the stationary points, we need to calculate the derivative \(\frac{dy}{dx}\) and set it to 0.
Using the product rule on \(y = x^2 e^{-3x}\):
Let \(u = x^2 \implies \frac{du}{dx} = 2x\)
Let \(v = e^{-3x} \implies \frac{dv}{dx} = -3e^{-3x}\)

Applying the formula \(\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}\):
\[\frac{dy}{dx} = x^2(-3e^{-3x}) + e^{-3x}(2x) = e^{-3x}(2x - 3x^2)\]

Set \(\frac{dy}{dx} = 0\):
\[e^{-3x}(2x - 3x^2) = 0\]

Since \(e^{-3x} \neq 0\) for all real \(x\), we solve:
\[2x - 3x^2 = 0 \implies x(2 - 3x) = 0\]

This gives \(x = 0\) or \(x = \frac{2}{3}\).

Now we find the corresponding \(y\)-coordinates:
- For \(x = 0\):
\[y = (0)^2 e^{-3(0)} = 0\]
- For \(x = \frac{2}{3}\):
\[y = \left(\frac{2}{3}\right)^2 e^{-3(2/3)} = \frac{4}{9} e^{-2}\]

Thus, the stationary points are \((0, 0)\) and \(\left(\frac{2}{3}, \frac{4}{9}e^{-2}\right)\).

M1: For applying the product rule to differentiate \(x^2 e^{-3x}\).
A1.25: For obtaining the correct derivative \(\frac{dy}{dx} = e^{-3x}(2x - 3x^2)\) (or equivalent).
M1: For setting their \(\frac{dy}{dx}\) to 0 and attempting to solve for \(x\).
A1: For finding the correct x-coordinates: \(x = 0\) and \(x = \frac{2}{3}\).
M1: For substituting at least one of their x-values into the original curve equation to find a y-coordinate.
A1: For obtaining both correct exact coordinates: \((0, 0)\) and \(\left(\frac{2}{3}, \frac{4}{9}e^{-2}\right)\).

We use the double-angle identity for cosine to rewrite \(\cos^2(2x)\):
\[\cos(4x) = 2\cos^2(2x) - 1 \implies \cos^2(2x) = \frac{1 + \cos(4x)}{2}\]

Substituting this into the integral:
\[\int_{0}^{\frac{\pi}{4}} \cos^2(2x) \, dx = \int_{0}^{\frac{\pi}{4}} \frac{1 + \cos(4x)}{2} \, dx\]
\[= \frac{1}{2} \left[ x + \frac{1}{4} \sin(4x) \right]_{0}^{\frac{\pi}{4}}\]

Now, evaluating with the limits:
- At the upper limit \(x = \frac{\pi}{4}\):
\[\frac{1}{2} \left( \frac{\pi}{4} + \frac{1}{4} \sin\left(4 \cdot \frac{\pi}{4}\right) \right) = \frac{1}{2} \left( \frac{\pi}{4} + \frac{1}{4} \sin(\pi) \right) = \frac{1}{2} \left( \frac{\pi}{4} + 0 \right) = \frac{\pi}{8}\]
- At the lower limit \(x = 0\):
\[\frac{1}{2} \left( 0 + \frac{1}{4} \sin(0) \right) = 0\]

Subtracting the lower limit value from the upper limit value gives the exact value:
\[\frac{\pi}{8}\]

M1.25: For stating and applying the double-angle identity \(\cos^2(2x) = \frac{1}{2}(1 + \cos(4x))\).
M1: For attempting to integrate a term of the form \(a + b \cos(4x)\) to obtain \(ax + c \sin(4x)\).
A1: For obtaining the correct integration result \(\frac{1}{2}x + \frac{1}{8}\sin(4x)\) (or equivalent).
M1: For substituting limits \(0\) and \(\frac{\pi}{4}\) correctly into their integrated expression.
A2: For obtaining the correct exact value \(\frac{\pi}{8}\).

(a) Let \(f(x) = x^3 - 5x + 1\).
We evaluate \(f(x)\) at the interval endpoints:
\[f(2) = (2)^3 - 5(2) + 1 = 8 - 10 + 1 = -1\]
\[f(2.5) = (2.5)^3 - 5(2.5) + 1 = 15.625 - 12.5 + 1 = 4.125\]
Since there is a change of sign between \(f(2) < 0\) and \(f(2.5) > 0\) and \(f(x)\) is a continuous function, a root must lie in the interval \(2 < x < 2.5\).

(b) Using the formula \(x_{n+1} = \sqrt[3]{5x_n - 1}\) with \(x_1 = 2\):
\[x_2 = \sqrt[3]{5(2) - 1} = \sqrt[3]{9} \approx 2.08008\]
\[x_3 = \sqrt[3]{5(2.08008) - 1} \approx 2.11046\]
\[x_4 = \sqrt[3]{5(2.11046) - 1} \approx 2.12182\]
\[x_5 = \sqrt[3]{5(2.12182) - 1} \approx 2.12602\]
\[x_6 = \sqrt[3]{5(2.12602) - 1} \approx 2.12757\]
\[x_7 = \sqrt[3]{5(2.12757) - 1} \approx 2.12814\]
\[x_8 = \sqrt[3]{5(2.12814) - 1} \approx 2.12835\]
\[x_9 = \sqrt[3]{5(2.12835) - 1} \approx 2.12843\]

Since successive iterations round to \(2.128\), the root is \(2.128\) correct to 3 decimal places.

Part (a):
M1: For evaluating \(f(x) = x^3 - 5x + 1\) at \(x = 2\) and \(x = 2.5\).
A1.25: For showing \(f(2) = -1\) and \(f(2.5) = 4.125\), noting the sign change and concluding that a root exists.

Part (b):
M1: For performing at least two iterations using the given formula correctly.
A1: For obtaining correct values \(x_2 = 2.08008\) and \(x_3 = 2.11046\).
A1: For obtaining correct further iterations reaching at least \(x_6 = 2.12757\) or showing convergence.
A1: For stating the root as \(2.128\) and justifying it with sufficient iterations.

(a) Since \(x - 2\) is a factor of \(p(x)\), by the Factor Theorem:
\[p(2) = 0\]
\[2(2)^3 - (2)^2 + a(2) + 6 = 0\]
\[16 - 4 + 2a + 6 = 0\]
\[18 + 2a = 0 \implies a = -9\]

(b) With \(a = -9\), the polynomial is \(p(x) = 2x^3 - x^2 - 9x + 6\).
We can factorise \(p(x)\) by dividing by \(x - 2\):
\[2x^3 - x^2 - 9x + 6 = (x - 2)(2x^2 + kx - 3)\]
Matching coefficients of \(x^2\):
\[-4 + k = -1 \implies k = 3\]
So, the quadratic factor is \(2x^2 + 3x - 3\).

To find the remaining roots, we solve \(2x^2 + 3x - 3 = 0\) using the quadratic formula:
\[x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-3)}}{2(2)} = \frac{-3 \pm \sqrt{9 + 24}}{4} = \frac{-3 \pm \sqrt{33}}{4}\]

Therefore, the exact roots of \(p(x) = 0\) are \(x = 2\), \(x = \frac{-3 + \sqrt{33}}{4}\), and \(x = \frac{-3 - \sqrt{33}}{4}\).

Part (a):
M1: For setting \(p(2) = 0\) and attempting to solve for \(a\).
A1: For obtaining \(a = -9\).

Part (b):
M1: For attempting division of \(2x^3 - x^2 - 9x + 6\) by \(x - 2\) or identifying coefficients to find the quadratic factor.
A1: For finding the correct quadratic factor \(2x^2 + 3x - 3\).
M1: For attempting to solve their quadratic factor using the quadratic formula.
A1.25: For obtaining all three correct exact roots: \(x = 2\), \(x = \frac{-3 + \sqrt{33}}{4}\), and \(x = \frac{-3 - \sqrt{33}}{4}\).

The area \(A\) of the bounded region is given by the definite integral:
\[A = \int_{1}^{4} \frac{4}{2x+1} \, dx\]

To integrate \(\frac{4}{2x+1}\), we use the log integration rule \(\int \frac{1}{ax+b} \, dx = \frac{1}{a} \ln|ax+b|\):
\[\int \frac{4}{2x+1} \, dx = 4 \cdot \frac{1}{2} \ln|2x+1| = 2 \ln|2x+1|\]

Now, evaluating this integral from \(x = 1\) to \(x = 4\):
\[A = \left[ 2 \ln(2x+1) \right]_{1}^{4}\]
\[A = 2 \ln(2(4)+1) - 2 \ln(2(1)+1)\]
\[A = 2 \ln(9) - 2 \ln(3)\]

Using laws of logarithms, we simplify the expression:
\[A = 2 (\ln 9 - \ln 3) = 2 \ln\left(\frac{9}{3}\right) = 2 \ln 3\]

Thus, the exact area is \(2 \ln 3\) (or \(\ln 9\)).

M1.25: For setting up the correct definite integral for the area: \(\int_{1}^{4} \frac{4}{2x+1} \, dx\).
M1: For attempting to integrate to get an expression of the form \(k \ln(2x+1)\).
A1: For the correct integrated expression \(2 \ln(2x+1)\).
M1: For substituting the limits 4 and 1 correctly into their integrated expression.
A2: For simplifying to the correct exact single term \(2 \ln 3\) (or equivalent like \(\ln 9\)).

We write the algebraic fraction in the form of partial fractions with a repeated linear factor: \(\frac{3x^2 + 6x}{(x-1)^2(2x+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{2x+1}\). Multiplying both sides by the denominator gives: \(3x^2 + 6x = A(x-1)(2x+1) + B(2x+1) + C(x-1)^2\). To find the constants: Substitute \(x = 1\): \(3(1)^2 + 6(1) = B(2(1)+1) \implies 9 = 3B \implies B = 3\). Substitute \(x = -\frac{1}{2}\): \(3(-\frac{1}{2})^2 + 6(-\frac{1}{2}) = C(-\frac{1}{2}-1)^2 \implies \frac{3}{4} - 3 = C(-\frac{3}{2})^2 \implies -\frac{9}{4} = \frac{9}{4}C \implies C = -1\). To find \(A\), we can equate the coefficients of \(x^2\): \(3 = 2A + C \implies 3 = 2A - 1 \implies 2A = 4 \implies A = 2\). Thus, the partial fractions are \(\frac{2}{x-1} + \frac{3}{(x-1)^2} - \frac{1}{2x+1}\).

B1: State or imply the correct form of partial fractions with three terms. M1: Substitute values of x or equate coefficients to find at least one constant. A1: Find B = 3 correctly. A1: Find C = -1 correctly. M1: Use a valid method to find the remaining constant A. A1.8: Obtain the correct final expression: 2/(x-1) + 3/(x-1)^2 - 1/(2x+1).

Take the natural logarithm of both sides: \(\ln(5^{2x-1}) = \ln(3^{x+2})\). Apply the power rule of logarithms: \((2x-1)\ln 5 = (x+2)\ln 3\). Expand the terms: \(2x\ln 5 - \ln 5 = x\ln 3 + 2\ln 3\). Collect the terms involving \(x\) on one side: \(2x\ln 5 - x\ln 3 = 2\ln 3 + \ln 5\). Factor out \(x\): \(x(2\ln 5 - \ln 3) = 2\ln 3 + \ln 5\). Solve for \(x\): \(x = \frac{2\ln 3 + \ln 5}{2\ln 5 - \ln 3}\). Using numerical approximations (\(\ln 3 \approx 1.0986\) and \(\ln 5 \approx 1.6094\)): \(x = \frac{2(1.0986) + 1.6094}{2(1.6094) - 1.0986} \approx \frac{3.8066}{2.1202} \approx 1.79537\). Thus, to 3 decimal places, \(x = 1.795\).

M1: Take natural or base-10 logarithms of both sides and apply the power rule correctly. A1: Obtain a correct linear equation in x. M1: Group x terms and factorise to make x the subject. A1: Obtain the correct exact expression for x. A2.8: Calculate the correct numerical value to 3 decimal places (1.795).

Use the double angle identity \(\cos 2\theta = 1 - 2\sin^2 \theta\) to rewrite the equation in terms of \(\sin \theta\) only: \(3(1 - 2\sin^2 \theta) + 8\sin \theta = 5\). Expand and rearrange: \(3 - 6\sin^2 \theta + 8\sin \theta = 5 \implies 6\sin^2 \theta - 8\sin \theta + 2 = 0\). Divide the equation by 2: \(3\sin^2 \theta - 4\sin \theta + 1 = 0\). Factorise the quadratic equation: \((3\sin \theta - 1)(\sin \theta - 1) = 0\). This yields two cases: Case 1: \(\sin \theta = 1 \implies \theta = 90^\circ\). Case 2: \(\sin \theta = \frac{1}{3} \implies \theta \approx 19.47^\circ\) or \(\theta \approx 180^\circ - 19.47^\circ = 160.53^\circ\). Rounded to 1 decimal place, the solutions are \(\theta = 19.5^\circ, 90.0^\circ, 160.5^\circ\).

M1: Use the correct identity for cos 2θ to obtain a quadratic equation in sin θ. A1: Obtain the correct quadratic equation 3sin^2 θ - 4sin θ + 1 = 0 (or equivalent). M1: Solve the quadratic equation to find values for sin θ. A1: Obtain θ = 90°. A1: Obtain θ = 19.5° (or 19.4°). A1.8: Obtain θ = 160.5° (or 160.6°) and no other solutions within the range.

First, we find the derivatives of \(x\) and \(y\) with respect to the parameter \(t\): \(\frac{dx}{dt} = 2 e^{2t} - 3\) and \(\frac{dy}{dt} = 2 e^t + 2t\). At \(t = 0\), these derivatives evaluate to: \(\frac{dx}{dt} = 2(1) - 3 = -1\) and \(\frac{dy}{dt} = 2(1) + 2(0) = 2\). Using the chain rule, the gradient of the curve is: \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2}{-1} = -2\). Next, we find the coordinates of the point at \(t = 0\): \(x = e^0 - 3(0) = 1\) and \(y = 2 e^0 + 0^2 = 2\). The point is \((1, 2)\). The equation of the tangent line is: \(y - 2 = -2(x - 1) \implies y - 2 = -2x + 2 \implies y = -2x + 4\).

M1: Differentiate x and y with respect to t correctly. A1: Obtain both dx/dt = 2e^(2t) - 3 and dy/dt = 2e^t + 2t. M1: Evaluate dy/dx at t = 0 using the chain rule. A1: Find the gradient of the tangent is -2. B1: Find the coordinates of the contact point (1, 2). A1.8: State the correct equation of the tangent in the required form: y = -2x + 4.

We use integration by parts: \(\int u \, dv = uv - \int v \, du\). Let \(u = \ln x\) and \(dv = x^2 \, dx\). Then, \(du = \frac{1}{x} \, dx\) and \(v = \frac{x^3}{3}\). Applying the integration by parts formula: \(\int x^2 \ln x \, dx = \left[ \frac{x^3}{3} \ln x \right] - \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx = \frac{x^3}{3} \ln x - \frac{1}{3} \int x^2 \, dx = \frac{x^3}{3} \ln x - \frac{x^3}{9}\). Now we evaluate this between the limits \(1\) and \(e\): At \(x = e\): \(\frac{e^3}{3} \ln e - \frac{e^3}{9} = \frac{e^3}{3} - \frac{e^3}{9} = \frac{2e^3}{9}\). At \(x = 1\): \(\frac{1^3}{3} \ln 1 - \frac{1^3}{9} = 0 - \frac{1}{9} = -\frac{1}{9}\). Subtracting the lower limit value: \(\frac{2e^3}{9} - \left(-\frac{1}{9}\right) = \frac{2e^3 + 1}{9}\).

M1: Apply integration by parts with correct assignment of u and dv. A1: Obtain correct first stage: (x^3 / 3) * ln x - Integral of (x^2 / 3) dx. A1: Complete integration to obtain (x^3 / 3) * ln x - (x^3 / 9). M1: Substitute the limits 1 and e correctly. A2.8: Obtain the correct exact final answer (2e^3 + 1) / 9.

Let \(f(x) = x^3 - 5x - 3\). Evaluating at \(x = 2\) and \(x = 3\): \(f(2) = 2^3 - 5(2) - 3 = 8 - 10 - 3 = -5 < 0\), and \(f(3) = 3^3 - 5(3) - 3 = 27 - 15 - 3 = 9 > 0\). Since there is a change of sign and the function is continuous, there is a root in the interval \(2 < x < 3\). Now we apply the iterative formula starting with \(x_1 = 2.5\): \(x_2 = \sqrt[3]{5(2.5) + 3} = \sqrt[3]{15.5} \approx 2.4933\), \(x_3 = \sqrt[3]{5(2.4933) + 3} = \sqrt[3]{15.4665} \approx 2.4915\), \(x_4 = \sqrt[3]{5(2.4915) + 3} = \sqrt[3]{15.4575} \approx 2.4910\), \(x_5 = \sqrt[3]{5(2.4910) + 3} = \sqrt[3]{15.4550} \approx 2.4909\). The values converge to \(2.49\) correct to 2 decimal places.

B1: Calculate f(2) and f(3) with correct signs, and state a valid conclusion. M1: Attempt to use the iterative formula with x1 = 2.5. A1: Obtain x2 = 2.4933 correctly. A1: Obtain x3 = 2.4915 correctly. A1: Obtain x4 = 2.4910 and x5 = 2.4909. A1.8: Conclude that the root is 2.49 and provide evidence of sufficient convergence.

The angle between two lines is the angle between their direction vectors: \(\mathbf{d}_1 = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}\) and \(\mathbf{d}_2 = \begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix}\). First, compute the scalar product: \(\mathbf{d}_1 \cdot \mathbf{d}_2 = (2)(1) + (-1)(1) + (3)(-2) = 2 - 1 - 6 = -5\). Next, calculate the magnitudes of both vectors: \(|\mathbf{d}_1| = \sqrt{2^2 + (-1)^2 + 3^2} = \sqrt{14}\) and \(|\mathbf{d}_2| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{6}\). The cosine of the angle \(\theta\) is: \(\cos \theta = \frac{|\mathbf{d}_1 \cdot \mathbf{d}_2|}{|\mathbf{d}_1| |\mathbf{d}_2|} = \frac{|-5|}{\sqrt{14}\sqrt{6}} = \frac{5}{\sqrt{84}} \approx 0.54554\). Thus, \(\theta = \cos^{-1}(0.54554) \approx 56.94^circ\). To the nearest degree, the angle is \(57^\circ\).

B1: Identify the correct direction vectors. M1: Calculate the scalar product of the two direction vectors. M1: Calculate the magnitudes of both direction vectors. A1: Obtain correct scalar product (-5) and magnitudes (sqrt(14) and sqrt(6)). M1: Substitute values into the cosine formula. A1.8: Obtain 57° (or 123°) correct to the nearest degree.

To express \(u\) in the form \(a + b\mathrm{i}\), we multiply the numerator and the denominator by the complex conjugate of the denominator, which is \(3 + \mathrm{i}\): \(u = \frac{5 + 5\mathrm{i}}{3 - \mathrm{i}} \times \frac{3 + \mathrm{i}}{3 + \mathrm{i}} = \frac{(5 + 5\mathrm{i})(3 + \mathrm{i})}{3^2 + (-1)^2} = \frac{15 + 5\mathrm{i} + 15\mathrm{i} + 5\mathrm{i}^2}{9 + 1} = \frac{15 + 20\mathrm{i} - 5}{10} = \frac{10 + 20\mathrm{i}}{10} = 1 + 2\mathrm{i}\). The modulus of \(u\) is: \(|u| = \sqrt{1^2 + 2^2} = \sqrt{5}\). The argument of \(u\) is: \(\arg(u) = \tan^{-1}\left(\frac{2}{1}\right) \approx 1.107\) radians. Correct to 2 decimal places, \(\arg(u) = 1.11\) radians.

M1: Multiply numerator and denominator by the conjugate of the denominator (3 + i). A1: Obtain the correct real part (1). A1: Obtain the correct imaginary part (2). M1: Use a correct method to find the modulus of u. A1: Obtain exact modulus as sqrt(5). A1.8: Obtain the correct argument as 1.11 radians.

For \(u = -\sqrt{3} + \mathrm{i}\):

(a) The modulus of \(u\) is:
\(r = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{3+1} = 2\).
Since \(u\) lies in the second quadrant, the argument is:
\(\theta = \pi - \arctan\left(\frac{1}{\sqrt{3}}\right) = \pi - \frac{\pi}{6} = \frac{5}{6}\pi\).
Thus, \(u = 2\mathrm{e}^{\frac{5}{6}\pi\mathrm{i}}\).

(b) Let \(w = x + \mathrm{i}y\) be a square root of \(u\), so \(w^2 = u\).
\((x + \mathrm{i}y)^2 = -\sqrt{3} + \mathrm{i}\)
\(x^2 - y^2 + 2x y \mathrm{i} = -\sqrt{3} + \mathrm{i}\)

Equating real and imaginary parts:
1) \(x^2 - y^2 = -\sqrt{3}\)
2) \(2xy = 1\)

Also, using the modulus relation:
\(x^2 + y^2 = |u| = 2\)

Adding equation (1) and the modulus equation:
\(2x^2 = 2 - \sqrt{3} \implies x^2 = \frac{2-\sqrt{3}}{2} = \frac{4-2\sqrt{3}}{4} = \frac{(\sqrt{3}-1)^2}{4}\)
Since \(x\) is real, \(x = \pm \frac{\sqrt{3}-1}{2}\).

Subtracting equation (1) from the modulus equation:
\(2y^2 = 2 + \sqrt{3} \implies y^2 = \frac{2+\sqrt{3}}{2} = \frac{4+2\sqrt{3}}{4} = \frac{(\sqrt{3}+1)^2}{4}\)
Since \(y\) is real, \(y = \pm \frac{\sqrt{3}+1}{2}\).

From equation (2), \(2xy = 1 > 0\), so \(x\) and \(y\) must have the same sign.
Thus, the square roots of \(u\) are:
\(\pm \left( \frac{\sqrt{3}-1}{2} + \frac{\sqrt{3}+1}{2}\mathrm{i} \right)\).

(a)
M1: For attempting to find the modulus or argument of \(u\).
A1: For obtaining \(r = 2\) and \(\theta = \frac{5}{6}\pi\) (or state \(2\mathrm{e}^{\frac{5}{6}\pi\mathrm{i}}\)).

(b)
M1: For equating real and imaginary parts of \((x+\mathrm{i}y)^2 = -\sqrt{3}+\mathrm{i}\) to form two equations.
M1: For using the modulus relation \(x^2+y^2=2\) (or substituting \(y=1/(2x)\) to form a quartic in \(x\)).
A1: For solving for \(x^2\) (or \(y^2\)) and obtaining a correct exact expression, e.g., \(x^2 = \frac{2-\sqrt{3}}{2}\) or \(y^2 = \frac{2+\sqrt{3}}{2}\).
M1: For taking the square root to find exact values of \(x\) and \(y\), resolving the sign from \(2xy>0\).
A1: For both correct square roots: \(\pm \left( \frac{\sqrt{3}-1}{2} + \frac{\sqrt{3}+1}{2}\mathrm{i} \right)\).

Let \(u = \mathrm{e}^x\), then \(\mathrm{d}u = \mathrm{e}^x \, \mathrm{d}x\).

When \(x = 0\), \(u = \mathrm{e}^0 = 1\).
When \(x = \ln 2\), \(u = \mathrm{e}^{\ln 2} = 2\).

Substituting these into the integral:
\(\int_{0}^{\ln 2} \frac{\mathrm{e}^{2x}}{\mathrm{e}^x + 2} \, \mathrm{d}x = \int_{0}^{\ln 2} \frac{\mathrm{e}^x}{\mathrm{e}^x + 2} \cdot \mathrm{e}^x \, \mathrm{d}x = \int_{1}^{2} \frac{u}{u+2} \, \mathrm{d}u\).

Rewrite the integrand:
\(\frac{u}{u+2} = \frac{u+2-2}{u+2} = 1 - \frac{2}{u+2}\).

Now perform the integration:
\(\int_{1}^{2} \left(1 - \frac{2}{u+2}\right) \, \mathrm{d}u = [u - 2\ln(u+2)]_{1}^{2}\)

Substitute the limits:
\(= (2 - 2\ln 4) - (1 - 2\ln 3)\)

Simplify the expression:
\(= 2 - 2\ln 4 - 1 + 2\ln 3\)
\(= 1 - 2(\ln 4 - \ln 3)\)
\(= 1 - 2\ln\left(\frac{4}{3}\right)\)
\(= 1 - \ln\left(\frac{16}{9}\right)\).

Hence, the result is shown.

M1: For using an appropriate substitution, e.g., \(u = \mathrm{e}^x\) or \(u = \mathrm{e}^x + 2\), and finding the corresponding expression for \(\mathrm{d}u\).
A1: For obtaining a completely correct integral in terms of \(u\), including correct limits (e.g., limits 1 and 2 for \(u = \mathrm{e}^x\)).
M1: For expressing the integrand in a form that can be integrated, e.g., by division or splitting: \(1 - \frac{2}{u+2}\).
A1: For correct integration to obtain \(u - 2\ln(u+2)\) (or \(u - 2\ln u\) if \(u = \mathrm{e}^x+2\) is used).
M1: For substituting the correct limits into their integrated expression.
M1: For applying laws of logarithms correctly to combine terms of the form \(a\ln b - c\ln d\).
A1: For obtaining the final given result \(1 - \ln\frac{16}{9}\) with no errors seen.

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

Integrate both sides:
LHS: \(\int y^{-2} \, \mathrm{d}y = -y^{-1} = -\frac{1}{y}\)
RHS: \(\int \sin^2 x \cos x \, \mathrm{d}x = \frac{1}{3}\sin^3 x + C\)

Combining these gives:
\(-\frac{1}{y} = \frac{1}{3}\sin^3 x + C\)

Apply the initial condition \(y = 1\) when \(x = 0\):
\(-\frac{1}{1} = \frac{1}{3}\sin^3(0) + C \implies -1 = C\)

So the equation is:
\(-\frac{1}{y} = \frac{1}{3}\sin^3 x - 1\)

Multiply by \(-1\):
\(\frac{1}{y} = 1 - \frac{1}{3}\sin^3 x = \frac{3 - \sin^3 x}{3}\)

Invert to find \(y\):
\(y = \frac{3}{3 - \sin^3 x}\).

M1: For separating variables correctly, obtaining \(\int \frac{1}{y^2} \, \mathrm{d}y = \int \sin^2 x \cos x \, \mathrm{d}x\) (or equivalent).
A1: For integrating the LHS correctly to get \(-\frac{1}{y}\).
M1: For attempting to integrate the RHS using substitution or inspection.
A1: For obtaining \(\frac{1}{3}\sin^3 x\).
M1: For introducing a constant of integration and using the boundary condition \(y = 1\), \(x = 0\) to find its value.
A1: For finding the correct value of the constant (e.g., \(C = -1\) if written as \(-\frac{1}{y} = \frac{1}{3}\sin^3 x + C\)).
A1: For expressing \(y\) explicitly as \(y = \frac{3}{3 - \sin^3 x}\) (or equivalent form).

To find the total distance travelled, we must identify the times at which the particle changes direction. We do this by setting the velocity to zero:

\( 3t^2 - 12t + 9 = 0 \)

Dividing by 3:

\( t^2 - 4t + 3 = 0 \)

\( (t - 1)(t - 3) = 0 \)

This gives critical times at \( t = 1 \) and \( t = 3 \). Since both values lie within the interval \( [0, 4] \), the particle reverses direction at these times.

Next, we find the displacement \( s(t) \) from the initial position by integrating the velocity, setting the constant of integration to 0:

\( s(t) = \int (3t^2 - 12t + 9) \, dt = t^3 - 6t^2 + 9t \)

Now we evaluate the displacement at the boundaries and critical points:

- At \( t = 0 \): \( s(0) = 0 \)
- At \( t = 1 \): \( s(1) = 1^3 - 6(1)^2 + 9(1) = 4 \)
- At \( t = 3 \): \( s(3) = 3^3 - 6(3)^2 + 9(3) = 27 - 54 + 27 = 0 \)
- At \( t = 4 \): \( s(4) = 4^3 - 6(4)^2 + 9(4) = 64 - 96 + 36 = 4 \)

The total distance travelled is the sum of the absolute differences in displacement over each sub-interval:

\( d = |s(1) - s(0)| + |s(3) - s(1)| + |s(4) - s(3)| \)

\( d = |4 - 0| + |0 - 4| + |4 - 0| = 4 + 4 + 4 = 12 \text{ m} \).

M1: Set velocity equal to zero to find the stationary times.
A1: Correct values of \( t = 1 \) and \( t = 3 \).
M1: Integrate velocity to find the displacement function \( s(t) \).
A1: Correct integrated expression \( s(t) = t^3 - 6t^2 + 9t \).
M1: Evaluate displacement at the critical points and endpoints.
A1: Find individual segment distances (4, 4, 4).
A1: State final correct total distance of 12.

Let \( a \) be the acceleration of the system while both blocks are moving.
Using Newton's second law for both blocks:
- For block \( B \): \( 5g - T = 5a \)
- For block \( A \): \( T - 3g = 3a \)

Adding these two equations:
\( 2g = 8a \Rightarrow a = 0.25g = 2.5\text{ m s}^{-2} \).

When block \( B \) hits the ground, it has travelled \( 1.6\text{ m} \). The velocity \( v \) of both blocks at this instant is given by:
\( v^2 = u^2 + 2as = 0 + 2(2.5)(1.6) = 8 \).

At this point, block \( A \) has risen through a height of \( 1.6\text{ m} \) and the string becomes slack. Block \( A \) then continues to move upwards freely under gravity with deceleration \( g = 10\text{ m s}^{-2} \).

The extra height \( h \) reached by \( A \) while moving freely is:
\( v^2 = u^2 - 2gh \Rightarrow 0 = 8 - 2(10)h \Rightarrow 20h = 8 \Rightarrow h = 0.4\text{ m} \).

Therefore, the maximum height reached by block \( A \) above its initial position is:
\( 1.6\text{ m} + 0.4\text{ m} = 2.0\text{ m} \).

M1: Set up equations of motion for both blocks.
A1: Correct acceleration of the system, \( a = 2.5\text{ m s}^{-2} \).
M1: Use equations of motion to find the velocity \( v \) (or \( v^2 \)) of \( A \) when \( B \) hits the ground.
A1: Correct velocity term, \( v^2 = 8 \).
M1: Treat the subsequent motion of \( A \) as a free projectile under gravity to find the extra height \( h \).
A1: Obtain the extra height \( h = 0.4\text{ m} \).
A1: Calculate the correct total maximum height above the starting position, \( 2\text{ m} \).

Let \( F \) be the driving force provided by the car's engine. The equation of motion along the plane up the hill is:

\( F - R - mg\sin\theta = ma \)

We are given:
- Mass, \( m = 1200\text{ kg} \)
- Slope angle component, \( \sin\theta = 0.05 \)
- Resistance force, \( R = 400\text{ N} \)
- Acceleration, \( a = 0.5\text{ m s}^{-2} \)
- Gravity, \( g = 10\text{ m s}^{-2} \)

Substitute these values into the equation:

\( F - 400 - 1200(10)(0.05) = 1200(0.5) \)

\( F - 400 - 600 = 600 \)

\( F = 1600\text{ N} \)

Power \( P \) in Watts is calculated using the relation \( P = Fv \), where \( v = 15\text{ m s}^{-1} \):

\( P_{\text{watts}} = 1600 \times 15 = 24000\text{ W} \)

Converting to kilowatts:

\( P = \frac{24000}{1000} = 24 \).

M1: Establish the equation of motion along the slope.
A1: Correct weight component term, \( mg\sin\theta = 600 \).
A1: Correct mass-acceleration term, \( ma = 600 \).
A1: Deduce the driving force \( F = 1600\text{ N} \).
M1: Apply the formula linking Power, Force, and Velocity (\( P = Fv \)).
A1: Correctly calculate the power in Watts (\( 24000\text{ W} \)).
A1: Provide final answer in kW, \( P = 24 \).

Let the direction of \( P \)'s initial motion be positive. This means:
- Initial velocity of \( P \), \( u_P = 4\text{ m s}^{-1} \)
- Initial velocity of \( Q \), \( u_Q = -3\text{ m s}^{-1} \) (since they are moving towards each other)
- Mass of \( P \), \( m_P = 0.2\text{ kg} \)
- Mass of \( Q \), \( m_Q = 0.3\text{ kg} \)

After the collision, the direction of \( P \) is reversed, so:
- Final velocity of \( P \), \( v_P = -2.6\text{ m s}^{-1} \)

Let \( v_Q \) be the final velocity of \( Q \). We apply the principle of conservation of momentum:

\( m_P u_P + m_Q u_Q = m_P v_P + m_Q v_Q \)

Substituting the values:

\( 0.2(4) + 0.3(-3) = 0.2(-2.6) + 0.3 v_Q \)

\( 0.8 - 0.9 = -0.52 + 0.3 v_Q \)

\( -0.1 = -0.52 + 0.3 v_Q \)

\( 0.3 v_Q = 0.42 \)

\( v_Q = 1.4\text{ m s}^{-1} \)

Since \( v_Q \) is positive, and the initial velocity of \( Q \) was negative, the direction of motion of \( Q \) is reversed.

Thus, the speed of \( Q \) is \( 1.4\text{ m s}^{-1} \) and its direction of motion is reversed.

M1: Formulate momentum conservation equation with correct signs representing opposite directions.
A1: Correct left-hand side initial momentum (\( -0.1\text{ kg m s}^{-1} \)).
A1: Correct right-hand side with reversed direction for P (\( 0.2(-2.6) + 0.3v_Q \)).
M1: Rearrange and solve for \( v_Q \).
A1: Obtain the correct velocity value of \( 1.4\text{ m s}^{-1} \).
A1: State the final speed of \( Q \) is \( 1.4\text{ m s}^{-1} \).
A1: Correctly state that its direction is reversed, with supporting explanation based on velocity sign change.

First, we calculate the normal reaction force \( R \) acting perpendicular to the inclined plane:

\( R = mg\cos 30^\circ = 8 \times 10 \times \frac{\sqrt{3}}{2} = 40\sqrt{3}\text{ N} \)

The limiting frictional force \( F_{\text{max}} \) is:

\( F_{\text{max}} = \mu R = \frac{\sqrt{3}}{5} \times 40\sqrt{3} = \frac{3 \times 40}{5} = 24\text{ N} \)

The component of gravity acting down the inclined plane is:

\( W_{\parallel} = mg\sin 30^\circ = 8 \times 10 \times 0.5 = 40\text{ N} \)

Now we consider the two extremes of limiting equilibrium:

**Case 1: Limiting equilibrium with the block on the point of sliding down the plane.**
In this case, the frictional force acts up the plane to oppose the motion:

\( P + F_{\text{max}} = W_{\parallel} \Rightarrow P + 24 = 40 \Rightarrow P = 16\text{ N} \)

**Case 2: Limiting equilibrium with the block on the point of sliding up the plane.**
In this case, the frictional force acts down the plane to oppose the potential motion:

\( P = W_{\parallel} + F_{\text{max}} \Rightarrow P = 40 + 24 = 64\text{ N} \)

For equilibrium to be maintained, the force \( P \) must satisfy:

\( 16 \le P \le 64 \).

M1: Correct resolution perpendicular to the plane to find \( R \).
A1: Obtain \( R = 40\sqrt{3}\text{ N} \).
M1: Calculate the maximum frictional force using \( F = \mu R \).
A1: Obtain \( F_{\text{max}} = 24\text{ N} \).
M1: Establish equilibrium equation for block on the point of slipping down the plane.
A1: Correctly calculate the lower limit \( P = 16\text{ N} \).
A1: Establish equilibrium equation for block on the point of slipping up the plane, giving upper limit \( P = 64\text{ N} \) and expressing the correct range.

To find the velocity, we integrate the acceleration with respect to time:

\( v = \int a \, dt = \int (6t - 18) \, dt = 3t^2 - 18t + C \)

Since the particle starts from rest at \( t = 0 \), \( v(0) = 0 \), which implies \( C = 0 \). Thus:

\( v = 3t^2 - 18t \)

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

\( 3t^2 - 18t = 0 \Rightarrow 3t(t - 6) = 0 \)

Since \( t = 0 \) is the start, the particle is next at instantaneous rest at \( t = 6\text{ s} \).

To find the displacement \( s \), we integrate the velocity function:

\( s = \int v \, dt = \int (3t^2 - 18t) \, dt = t^3 - 9t^2 + K \)

Since the displacement is measured from \( O \), at \( t = 0 \) we have \( s(0) = 0 \), which gives \( K = 0 \). Thus:

\( s = t^3 - 9t^2 \)

Evaluating the displacement at \( t = 6\text{ s} \):

\( s(6) = 6^3 - 9(6)^2 = 216 - 324 = -108\text{ m} \).

M1: Integrate acceleration to obtain a velocity expression.
A1: Obtain \( v = 3t^2 - 18t \), justifying that the constant of integration is zero.
M1: Set the velocity to zero to find the non-zero time of instantaneous rest.
A1: Correctly find \( t = 6\text{ s} \).
M1: Integrate the velocity expression to obtain a displacement function.
A1: Obtain \( s = t^3 - 9t^2 \).
A1: Substitute \( t = 6 \) to obtain the correct final displacement of \( -108 \).

We use the work-energy principle.

Let's calculate the initial kinetic energy of the bead at \( A \):
\( \text{KE}_A = \frac{1}{2} m u^2 = \frac{1}{2} \times 0.5 \times 4^2 = 4\text{ J} \).

The vertical height fallen by the bead from \( A \) to \( B \) is:
\( h = AB \times \sin\theta = 5 \times 0.6 = 3\text{ m} \).

The loss in gravitational potential energy is:
\( \text{PE loss} = mgh = 0.5 \times 10 \times 3 = 15\text{ J} \).

Since the bead comes to rest at \( B \), the final kinetic energy is 0. By conservation of energy, the total mechanical energy lost is equal to the work done against friction (\( W_f \)):
\( W_f = \text{KE}_A + \text{PE loss} = 4 + 15 = 19\text{ J} \).

The work done against friction is also defined as \( W_f = F \times d \), where \( F \) is the frictional force and \( d = 5\text{ m} \):
\( 5F = 19 \Rightarrow F = 3.8\text{ N} \).

The normal contact force \( R \) is:
\( R = mg\cos\theta \)
Since \( \sin\theta = 0.6 \), we have \( \cos\theta = 0.8 \). Hence:
\( R = 0.5 \times 10 \times 0.8 = 4\text{ N} \).

Finally, we find the coefficient of friction \( \mu \) using \( F = \mu R \):
\( \mu = \frac{F}{R} = \frac{3.8}{4} = 0.95 \).

M1: Correct calculation of initial kinetic energy.
A1: Obtain \( 4\text{ J} \).
M1: Correct calculation of loss in gravitational potential energy.
A1: Obtain \( 15\text{ J} \).
M1: Set up the work-energy relation to find work done against friction, and deduce friction force.
A1: Correct normal force \( R = 4\text{ N} \) and friction force \( F = 3.8\text{ N} \).
A1: Correct final coefficient of friction \( \mu = 0.95 \).

Let \( u = t - 45 \).

(i) For the first 80 people:
\(\bar{u} = \frac{\sum u}{n} = \frac{240}{80} = 3\)
Therefore, the mean time \( \bar{t} \) is:
\(\bar{t} = \bar{u} + 45 = 3 + 45 = 48 \text{ minutes}\)

The variance of \( u \) is:
\(\text{Var}(u) = \frac{\sum u^2}{n} - \bar{u}^2 = \frac{18400}{80} - 3^2 = 230 - 9 = 221\)

Since the standard deviation of \( t \) is the same as the standard deviation of \( u \):
\(\text{SD}(t) = \sqrt{221} \approx 14.9 \text{ minutes}\) (to 3 s.f.)

(ii) For the remaining 20 people, let \( v = y - 45 \), so \(\sum v = 90\).
For all 100 people, the sum of the coded values is:
\(\sum (x - 45) = 240 + 90 = 330\)

The mean of the coded values for all 100 people is:
\(\frac{330}{100} = 3.3\)

Therefore, the combined mean time is:
\(3.3 + 45 = 48.3 \text{ minutes}\)

M1: For correct method to find the coded mean
A1: For correct mean of 48 minutes
M1: For correct formula for variance or standard deviation
A1: For SD = 14.9 minutes (accept 14.86)

M1: For adding the two sums to get 330
M1: For dividing by 100 and adding 45
A1: For 48.3 minutes

(i) The class width of the interval \( 40 < x \le 70 \) is \( 70 - 40 = 30 \).
Using the formula for frequency density:
\(\text{Frequency density} = \frac{\text{Frequency}}{\text{Class width}}\)
\(0.8 = \frac{f}{30} \implies f = 0.8 \times 30 = 24\)

(ii) We need to estimate the number of students in the interval \( 15 < x \le 55 \).
This range covers parts of three classes:
- From \( 15 < x \le 20 \): This is the upper half of the class \( 10 < x \le 20 \) (width 5, total width 10).
Estimated number of students = \(\frac{20 - 15}{20 - 10} \times 25 = 0.5 \times 25 = 12.5\)
- From \( 20 < x \le 40 \): This is the entire class of width 20.
Number of students = \(48\)
- From \( 40 < x \le 55 \): This is the lower half of the class \( 40 < x \le 70 \) (width 15, total width 30).
Estimated number of students = \(\frac{55 - 40}{70 - 40} \times 24 = 0.5 \times 24 = 12\)

Total estimated number of students = \(12.5 + 48 + 12 = 72.5\)

M1: For using frequency density formula with correct class width of 30
A1: For \( f = 24 \)

M1: For estimating frequency of first interval (\( 12.5 \))
M1: For using full frequency of second interval (\( 48 \))
M1: For estimating frequency of third interval (\( 12 \))
M1: For summing the three frequencies
A1: For \( 72.5 \) (accept 72 or 73 if clear working is shown)

(i) The word ELEVATE contains 7 letters: three E's, one L, one V, one A, and one T.
Number of arrangements = \(\frac{7!}{3!} = \frac{5040}{6} = 840\).

(ii) To find the number of arrangements where all three E's are consecutive, we treat the three E's as a single unit (EEE).
This gives 5 items to arrange: (EEE), L, V, A, T.
Since all these 5 items are distinct, they can be arranged in \( 5! = 120 \) ways.
Therefore, the number of arrangements in which the three E's are not all consecutive is:
\(840 - 120 = 720\).

(iii) The available letters are E (up to 3), and one each of L, V, A, T. We want to select 4 letters.
We consider cases based on the number of E's in the selection:
- Case 1: Three E's and 1 other letter from {L, V, A, T}.
Number of ways = \(\binom{4}{1} = 4\)
- Case 2: Two E's and 2 other letters from {L, V, A, T}.
Number of ways = \(\binom{4}{2} = 6\)
- Case 3: One E and 3 other letters from {L, V, A, T}.
Number of ways = \(\binom{4}{3} = 4\)
- Case 4: Zero E's and 4 other letters from {L, V, A, T}.
Number of ways = \(\binom{4}{4} = 1\)

Total number of selections = \(4 + 6 + 4 + 1 = 15\).

M1: For \(\frac{7!}{3!}\)
A1: For 840

M1: For treating three E's as a single unit to get \( 5! \) and subtracting from total
A1: For 720

M1: For identifying appropriate cases based on the number of E's
M1: For summing correct combinations for the cases
A1: For 15

(i) The committee must have more men than women. Since the total size is 6, the possible structures are:
- 6 men and 0 women: \(\binom{8}{6} \times \binom{6}{0} = 28 \times 1 = 28\)
- 5 men and 1 woman: \(\binom{8}{5} \times \binom{6}{1} = 56 \times 6 = 336\)
- 4 men and 2 women: \(\binom{8}{4} \times \binom{6}{2} = 70 \times 15 = 1050\)

Total number of ways = \(28 + 336 + 1050 = 1414\).

(ii) We need 3 men and 3 women, where Albert and Bernard cannot both be chosen.
Method 1 (Subtraction):
Total ways to choose 3 men and 3 women without restriction:
\(\binom{8}{3} \times \binom{6}{3} = 56 \times 20 = 1120\)

Ways where both Albert and Bernard are on the committee:
Albert and Bernard are chosen, so we choose 1 more man from the remaining 6 men, and 3 women from the 6 women:
\(\binom{6}{1} \times \binom{6}{3} = 6 \times 20 = 120\)

Ways where Albert and Bernard are not both on the committee:
\(1120 - 120 = 1000\).

Method 2 (Case-by-case):
- Neither Albert nor Bernard is chosen: Choose 3 men from 6, and 3 women from 6:
\(\binom{6}{3} \times \binom{6}{3} = 20 \times 20 = 400\)
- Exactly one of Albert or Bernard is chosen: Choose 1 of Albert or Bernard, then 2 men from 6, and 3 women from 6:
\(2 \times \binom{6}{2} \times \binom{6}{3} = 2 \times 15 \times 20 = 600\)

Total ways = \(400 + 600 = 1000\).

M1: For identifying the three valid scenarios (6M/0W, 5M/1W, 4M/2W)
M1: For calculating the combinations for at least two of the scenarios correctly
M1: For summing the combinations of the three scenarios
A1: For 1414

M1: For calculating total unrestricted combinations (\(1120\))
M1: For calculating combinations with both Albert and Bernard included (\(120\)) and subtracting, OR for summing the two valid cases (neither / exactly one)
A1: For 1000

(i) We are given:
\(P(X < 992) = 0.08 \implies P\left(Z < \frac{992 - \mu}{\sigma}\right) = 0.08\)
Since \( 0.08 < 0.5 \), the z-score is negative:
\(\frac{992 - \mu}{\sigma} = -1.405 \implies \mu - 1.405\sigma = 992\) (Eq 1)

We are also given:
\(P(X > 1012) = 0.15 \implies P\left(Z < \frac{1012 - \mu}{\sigma}\right) = 0.85\)
Since \( 0.85 > 0.5 \), the z-score is positive:
\(\frac{1012 - \mu}{\sigma} = 1.036 \implies \mu + 1.036\sigma = 1012\) (Eq 2)

Subtracting Eq 1 from Eq 2:
\((1.036 + 1.405)\sigma = 1012 - 992\)
\(2.441\sigma = 20 \implies \sigma \approx 8.193 \approx 8.19 \text{ g}\)

Substituting \( \sigma = 8.193 \) back into Eq 2:
\(\mu = 1012 - 1.036 \times 8.193 \approx 1003.51 \approx 1004 \text{ g}\).

(ii) We want to find \( P(995 < X < 1005) \) using \( \mu = 1003.51 \) and \( \sigma = 8.193 \):
\(P\left(\frac{995 - 1003.51}{8.193} < Z < \frac{1005 - 1003.51}{8.193}\right) = P(-1.039 < Z < 0.182)\)

Using standard normal tables:
\(\Phi(0.182) - \Phi(-1.039) = \Phi(0.182) - (1 - \Phi(1.039))\)
\(\approx 0.5722 - (1 - 0.8506) = 0.5722 - 0.1494 = 0.4228 \approx 0.423\).

M1: For standardizing and setting up an equation with a negative z-score
A1: For \(\mu - 1.405\sigma = 992\) (allow z-scores in range \([-1.41, -1.40]\))
M1: For standardizing and setting up an equation with a positive z-score
A1: For \(\mu + 1.036\sigma = 1012\) (allow z-scores in range \([1.03, 1.04]\))
A1: For \( \sigma = 8.19 \) and \( \mu = 1004 \) (accept \( \mu = 1000 \) if 2 d.p. z-scores were used)

M1: For standardizing both 995 and 1005 with their mean and standard deviation
A1: For \( 0.423 \) (accept range \( 0.420 - 0.425 \))

(i) Let \( X \) be the number of times a 6 is rolled.
\( X \sim \text{B}(180, \frac{1}{6}) \).

The mean is:
\(\mu = np = 180 \times \frac{1}{6} = 30\)

The variance is:
\(\sigma^2 = np(1 - p) = 180 \times \frac{1}{6} \times \frac{5}{6} = 25\)

So standard deviation \( \sigma = \sqrt{25} = 5 \).

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

Standardizing:
\(z_1 = \frac{24.5 - 30}{5} = -1.1\)
\(z_2 = \frac{35.5 - 30}{5} = 1.1\)

So we calculate:
\(P(-1.1 < Z < 1.1) = \Phi(1.1) - \Phi(-1.1) = 2\Phi(1.1) - 1\)
From normal distribution tables, \( \Phi(1.1) = 0.8643 \).
\(2 \times 0.8643 - 1 = 1.7286 - 1 = 0.7286 \approx 0.729\).

(ii) A normal approximation to the binomial distribution is appropriate because:
- \( np = 180 \times \frac{1}{6} = 30 \ge 5 \)
- \( n(1 - p) = 180 \times \frac{5}{6} = 150 \ge 5 \)
Since both \( np \ge 5 \) and \( nq \ge 5 \), the distribution is close enough to a normal distribution.

M1: For calculating the mean (\(30\)) and variance (\(25\))
M1: For using a continuity correction (bounds \(24.5\) and \(35.5\))
M1: For standardizing with their mean and standard deviation
M1: For calculating \( 2\Phi(z) - 1 \) from tables
A1: For \( 0.729 \) (accept range \( 0.728 - 0.730 \))

M1: For stating both conditions \( np \ge 5 \) and \( nq \ge 5 \)
A1: For calculating both values (\(30\) and \(150\)) and concluding they are greater than or equal to 5

Let \( H \) be the height of a tree. \( H \sim \text{N}(14.2, 3.6^2) \).

(i) \( P(H > 16.0) = P\left(Z > \frac{16.0 - 14.2}{3.6}\right) = P(Z > 0.5) \)
\( P(Z > 0.5) = 1 - \Phi(0.5) = 1 - 0.6915 = 0.3085 \approx 0.309 \).

(ii) \( P(H > h) = 0.75 \implies P\left(Z > \frac{h - 14.2}{3.6}\right) = 0.75 \)
Since the probability is greater than 0.5, the z-score must be negative:
\(\frac{h - 14.2}{3.6} = -0.674\)
\(h - 14.2 = -0.674 \times 3.6 = -2.4264 \implies h \approx 11.77 \approx 11.8 \text{ meters}\).

(iii) We want the conditional probability \( P(H < 18.5 \mid H > 16.0) \):
\(P(H < 18.5 \mid H > 16.0) = \frac{P(16.0 < H < 18.5)}{P(H > 16.0)}\)

We already have \( P(H > 16.0) = 0.3085 \).
Now we find \( P(H < 18.5) \):
\(P(H < 18.5) = P\left(Z < \frac{18.5 - 14.2}{3.6}\right) = P(Z < 1.194)\)
Using standard normal tables, \(\Phi(1.194) \approx 0.8838\).

So:
\(P(16.0 < H < 18.5) = P(H < 18.5) - P(H < 16.0) = 0.8838 - 0.6915 = 0.1923\).

Therefore, the conditional probability is:
\(\frac{0.1923}{0.3085} \approx 0.6233 \approx 0.623\).

M1: For standardizing 16.0
M1: For subtracting from 1
A1: For \( 0.309 \) (accept range \( 0.308 - 0.310 \))

M1: For equating the standardized term to a negative z-score (e.g. \(-0.674\) or \(-0.675\))
A1: For \( 11.8 \) (accept range \( 11.7 - 11.8 \))

M1: For standardizing 18.5 and setting up the conditional probability fraction
A1: For \( 0.623 \) (accept range \( 0.620 - 0.625 \))

Let \(X\) be the number of queries in 1 hour. \(X \sim \text{Po}(2.4)\).
(i) \(P(X = 3) = \frac{e^{-2.4} \cdot 2.4^3}{3!} \approx \frac{0.090718 \times 13.824}{6} \approx 0.209\) (3 s.f.).

Let \(Y\) be the number of queries in 2 hours. \(Y \sim \text{Po}(4.8)\).
(ii) \(P(Y > 4) = 1 - P(Y \le 4) = 1 - e^{-4.8}\left(1 + 4.8 + \frac{4.8^2}{2} + \frac{4.8^3}{6} + \frac{4.8^4}{24}\right) = 1 - e^{-4.8}(1 + 4.8 + 11.52 + 18.432 + 22.1184) = 1 - 57.8704 e^{-4.8} \approx 1 - 0.47626 = 0.524\) (3 s.f.).

M1 for using Poisson distribution formula with mean 2.4
A1 for 0.209 (3 s.f.)
M1 for calculating the correct mean for 2-hour period as 4.8
M1 for expressing \(P(Y > 4)\) as \(1 - P(Y \le 4)\) and setting up the sum
A1 for 0.524 (3 s.f.)

(i) Since \(f(x)\) is a probability density function, \(\int_{0}^{3} f(x) \, dx = 1\).
\(\int_{0}^{3} k(x^2 + 2x) \, dx = k \left[ \frac{x^3}{3} + x^2 \right]_0^3 = k(9 + 9) = 18k = 1 \implies k = \frac{1}{18}\).

(ii) \(P(1 \le X \le 2.5) = \int_{1}^{2.5} \frac{1}{18}(x^2 + 2x) \, dx = \frac{1}{18} \left[ \frac{x^3}{3} + x^2 \right]_1^{2.5}\).
Substituting \(x = 2.5\): \(\frac{2.5^3}{3} + 2.5^2 = \frac{15.625}{3} + 6.25 = 11.4583\).
Substituting \(x = 1\): \(\frac{1^3}{3} + 1^2 = 1.3333\).
\(P(1 \le X \le 2.5) = \frac{1}{18} (11.4583 - 1.3333) = \frac{10.125}{18} = 0.5625 \approx 0.563\) (3 s.f.).

M1 for integrating \(k(x^2+2x)\) from 0 to 3 and setting equal to 1
A1 for correctly showing \(k = \frac{1}{18}\)
M1 for integrating from 1 to 2.5
M1 for substituting limits 1 and 2.5 correctly into their integral
A1 for 0.563 (or exact 0.5625)

Let \(p\) be the germination rate.
\(H_0: p = 0.85\)
\(H_1: p < 0.85\)
Under \(H_0\), the number of germinated seeds is \(X \sim \text{B}(20, 0.85)\).
We find the probability of obtaining at most 14 successes under \(H_0\):
\(P(X \le 14)\).
Let \(Y = 20 - X\) be the number of failures. Under \(H_0\), \(Y \sim \text{B}(20, 0.15)\), and we want \(P(Y \ge 6) = 1 - P(Y \le 5)\).
\(P(Y = 0) = 0.85^{20} \approx 0.0388\)
\(P(Y = 1) = 20 \times 0.15 \times 0.85^{19} \approx 0.1368\)
\(P(Y = 2) = 190 \times 0.15^2 \times 0.85^{18} \approx 0.2293\)
\(P(Y = 3) = 1140 \times 0.15^3 \times 0.85^{17} \approx 0.2428\)
\(P(Y = 4) = 4845 \times 0.15^4 \times 0.85^{16} \approx 0.1821\)
\(P(Y = 5) = 15504 \times 0.15^5 \times 0.85^{15} \approx 0.1028\)
\(P(Y \le 5) \approx 0.0388 + 0.1368 + 0.2293 + 0.2428 + 0.1821 + 0.1028 = 0.9327\).
\(P(Y \ge 6) = 1 - 0.9327 = 0.0673\).
Since \(0.0673 > 0.05\), we do not reject \(H_0\).
There is insufficient evidence at the 5% significance level to support the gardener's belief that the germination rate is lower than 85%.

B1 for stating both hypotheses correctly
M1 for identifying Binomial distribution \(\text{B}(20, 0.85)\) and expressing cumulative probability \(P(X \le 14)\) or equivalent
M1 for calculating the sum of the individual Binomial terms
A1 for finding probability \(0.0673\) (or \(0.067\))
M1 for comparing their probability with the significance level of 0.05
A1 for concluding to not reject \(H_0\) and stating the conclusion in context

(i) A Poisson distribution is a suitable approximation because \(n = 500\) is large (i.e., \(n > 50\)), and \(p = 0.008\) is small (i.e., \(p < 0.1\)) such that the mean \(\lambda = np = 4\) is constant and moderate.
(ii) Let \(X\) be the number of defective ornaments in a box. \(X \approx \text{Po}(4)\).
(a) \(P(X = 3) = \frac{e^{-4} \cdot 4^3}{3!} \approx 0.195\) (3 s.f.).
(b) \(P(X \ge 2) = 1 - P(X \le 1) = 1 - e^{-4}(1 + 4) = 1 - 5e^{-4} \approx 1 - 0.091578 = 0.908\) (3 s.f.).

B1 for stating that \(n\) is large and \(p\) is small
B1 for finding the mean \(\lambda = 4\)
M1 for using Poisson probability formula for \(P(X = 3)\)
A1 for 0.195 (3 s.f.)
M1 for expressing \(P(X \ge 2)\) as \(1 - P(X \le 1)\) and evaluating
A1 for 0.908 (3 s.f.)

(i) Let \(\mu\) be the mean mass of the bags of flour.
\(H_0: \mu = 505\)
\(H_1: \mu < 505\)
Under \(H_0\), the sample mean mass is normally distributed: \(\bar{X} \sim \text{N}\left(505, \frac{12^2}{16}\right) \implies \bar{X} \sim \text{N}(505, 9)\).
We find the test statistic: \(z = \frac{499 - 505}{\sqrt{9}} = \frac{-6}{3} = -2\).
For a one-tailed test at the 2.5% significance level, the critical value of \(Z\) is \(-1.96\).
Since \(-2 < -1.96\), the test statistic lies in the critical region. Therefore, we reject \(H_0\).
There is sufficient evidence at the 2.5% level to support the manager's suspicion that the machine is under-filling the bags.

(ii) Since \(H_0\) was rejected, a Type I error (rejecting \(H_0\) when \(H_0\) is actually true) could have been made.

B1 for stating both hypotheses correctly
M1 for calculating standard error \(\sigma_{\bar{x}} = \frac{12}{\sqrt{16}} = 3\)
M1 for calculating the test statistic \(z = -2\) (or critical value \(499.12\))
A1 for comparing \(z\) with \(-1.96\) (or finding \(p\)-value 0.0228 and comparing with 0.025)
A1 for rejecting \(H_0\) and stating the correct contextual conclusion
B1 for stating Type I error with a valid reason

(i) \(E(X) = \int_{0}^{2} x f(x) \, dx = \int_{0}^{2} \frac{3}{16}(4x^2 - x^3) \, dx = \frac{3}{16} \left[ \frac{4x^3}{3} - \frac{x^4}{4} \right]_0^2 = \frac{3}{16} \left( \frac{32}{3} - 4 \right) = \frac{3}{16} \left( \frac{20}{3} \right) = 1.25\).

(ii) Let \(m\) be the median of \(X\). Then \(\int_{0}^{m} f(x) \, dx = 0.5\).
\(\int_{0}^{m} \frac{3}{16}(4x - x^2) \, dx = \frac{3}{16} \left[ 2x^2 - \frac{x^3}{3} \right]_0^m = 0.5\).
\(2m^2 - \frac{m^3}{3} = \frac{8}{3} \implies m^3 - 6m^2 + 8 = 0\).
Testing values in the range \(0 \le m \le 2\):
If \(m = 1.30\), \(1.3^3 - 6(1.3)^2 + 8 = 0.057 > 0\).
If \(m = 1.31\), \(1.31^3 - 6(1.31)^2 + 8 = -0.0486 < 0\).
Using numerical refinement, \(m \approx 1.31\) (3 s.f.).

M1 for setting up \(E(X) = \int x f(x) \, dx\)
M1 for integrating correctly
A1 for 1.25
M1 for integrating \(f(x)\) with limits \(0\) to \(m\) and equating to 0.5
A1 for establishing the cubic equation \(m^3 - 6m^2 + 8 = 0\)
M1 for showing a valid method of solving the cubic equation in the interval
A1 for 1.31 (3 s.f.)

(i) Let \(\lambda\) be the mean number of flaws per roll.
\(H_0: \lambda = 1.5\)
\(H_1: \lambda < 1.5\)

(ii) Under \(H_0\), the rate for 3 rolls is \(\lambda_3 = 3 \times 1.5 = 4.5\).
Let \(X\) be the number of flaws in 3 rolls. Under \(H_0\), \(X \sim \text{Po}(4.5)\).
Type I error = \(P(X \le 1 \mid \lambda_3 = 4.5) = e^{-4.5}(1 + 4.5) = 5.5 e^{-4.5} \approx 0.0611\) (3 s.f.).

(iii) Under the new process, \(\lambda = 0.4\), so for 3 rolls, \(\lambda_3' = 3 \times 0.4 = 1.2\).
Type II error = \(P(\text{Fail to reject } H_0 \mid H_1 \text{ is true}) = P(X \ge 2 \mid \lambda_3' = 1.2) = 1 - P(X \le 1 \mid \lambda_3' = 1.2)\).
\(1 - e^{-1.2}(1 + 1.2) = 1 - 2.2 e^{-1.2} \approx 1 - 0.66263 = 0.337\) (3 s.f.).

B1 for stating both hypotheses correctly
M1 for finding the mean for 3 rolls under \(H_0\) as 4.5
M1 for calculating \(P(X \le 1)\) using Poisson distribution
A1 for 0.0611 (3 s.f.)
M1 for finding the mean for 3 rolls under the new process as 1.2
M1 for calculating \(P(X \ge 2)\) with \(\lambda = 1.2\)
A1 for 0.337 (3 s.f.)