2023 Edexcel IGCSE Mathematics (Specification A) Practice Paper with Answers

In the right-angled triangle \(ABC\):
\(\sin(35^\circ) = \frac{AB}{AC} = \frac{5}{AC}\)
\(AC = \frac{5}{\sin(35^\circ)} \approx 8.7172\text{ cm}\)

In the right-angled triangle \(ACD\):
\(AD^2 = AC^2 + CD^2\)
\(AD^2 = (8.7172)^2 + 6^2\)
\(AD^2 \approx 75.990 + 36 = 111.990\)
\(AD = \sqrt{111.990} \approx 10.582\text{ cm}\)

To 3 significant figures, \(AD = 10.6\text{ cm}\).

M1 for a correct trigonometric ratio to find the length of \(AC\), e.g., \(\sin(35) = \frac{5}{AC}\) or \(AC = \frac{5}{\sin(35)}\)
M1 for a correct application of Pythagoras' theorem to find \(AD\), e.g., \(AD = \sqrt{\text{their } AC^2 + 6^2}\)
A1 for 10.6 (accept answers in the range 10.5 to 10.6)

The total age of the first group is \(15 \times 24 = 360\) years.
The total age of the second group is \(40 \times n = 40n\) years.
The total number of people is \(15 + n\).

The combined mean age is given by:
\(\frac{360 + 40n}{15 + n} = 30\)

Multiply both sides by \(15 + n\):
\(360 + 40n = 30(15 + n)\)
\(360 + 40n = 450 + 30n\)
\(10n = 90\)
\(n = 9\)

M1 for finding the total age of the first group, i.e., \(15 \times 24 = 360\)
M1 for setting up a correct equation for the combined mean, e.g., \(\frac{360 + 40n}{15 + n} = 30\) or \(360 + 40n = 450 + 30n\)
A1 for 9

Let \(V\) be the original value of the car.
After the first year, the value is \(V \times (1 - 0.12) = 0.88V\).
After the second year, the value is \(0.88V \times (1 - 0.08) = 0.88V \times 0.92 = 0.8096V\).

We are given:
\(0.8096V = 14168\)
\(V = \frac{14168}{0.8096} = 17500\)

The original value of the car was £17,500.

M1 for finding the combined depreciation multiplier, e.g., \(0.88 \times 0.92\) (or 0.8096)
M1 for a complete method to find the original value, e.g., \(\frac{14168}{0.88 \times 0.92}\)
A1 for 17500 (or 17,500)

First, factorise the numerator:
\(2x^2 + 5x - 3 = 2x^2 + 6x - x - 3 = 2x(x + 3) - 1(x + 3) = (2x - 1)(x + 3)\)

Next, factorise the denominator using the difference of two squares:
\(4x^2 - 1 = (2x - 1)(2x + 1)\)

Now, substitute these back into the fraction and simplify by cancelling common terms:
\(\frac{(2x - 1)(x + 3)}{(2x - 1)(2x + 1)} = \frac{x + 3}{2x + 1}\)

M1 for factorising the numerator correctly as \((2x - 1)(x + 3)\)
M1 for factorising the denominator correctly as \((2x - 1)(2x + 1)\)
A1 for \(\frac{x + 3}{2x + 1}\)

Rearrange the equation of \(\mathbf{L}_1\) into the form \(y = mx + c\):
\(2y = 3x - 8 \implies y = \frac{3}{2}x - 4\)
The gradient of \(\mathbf{L}_1\) is \(\frac{3}{2}\).

Since \(\mathbf{L}_2\) is perpendicular to \(\mathbf{L}_1\), its gradient is the negative reciprocal:
\(m_2 = -\frac{2}{3}\)

Using the point-slope form with the point \((6, -1)\):
\(y - (-1) = -\frac{2}{3}(x - 6)\)
\(y + 1 = -\frac{2}{3}x + 4\)

Multiply the entire equation by 3 to eliminate the fraction:
\(3y + 3 = -2x + 12\)
\(3y + 2x = 9\)

This is in the form \(ay + bx = c\) with integer coefficients.

M1 for finding the gradient of \(\mathbf{L}_1\) is \(\frac{3}{2}\) or stating that the gradient of \(\mathbf{L}_2\) is \(-\frac{2}{3}\)
M1 for substituting \((6, -1)\) and their perpendicular gradient into a linear equation formula, e.g., \(y - (-1) = -\frac{2}{3}(x - 6)\)
A1 for \(3y + 2x = 9\) (or any integer multiple equivalent such as \(2x + 3y = 9\))

The volume of a hemisphere is given by \(V = \frac{2}{3}\pi r^3\).
\(\frac{2}{3}\pi r^3 = 250\)
\(r^3 = \frac{375}{\pi} \approx 119.366\)
\(r = (119.366)^{1/3} \approx 4.9237\text{ cm}\)

The total surface area of a solid hemisphere consists of its curved surface area and its circular base:
\(A = 2\pi r^2 + \pi r^2 = 3\pi r^2\)
\(A = 3\pi (4.9237)^2 \approx 3\pi (24.243) \approx 228.48\text{ cm}^2\)

To 3 significant figures, the total surface area is \(228\text{ cm}^2\).

M1 for equating \(\frac{2}{3}\pi r^3 = 250\) and finding a value for \(r\) or \(r^3\) (e.g. \(r \approx 4.92\) or \(r^3 \approx 119.4\))
M1 for substituting their value of \(r\) into \(3\pi r^2\)
A1 for an answer in the range 228 to 229

Express all terms with a base of 3:
\(9 = 3^2\)
\(27 = 3^3\)

Substitute these into the equation:
\(3^{x+1} \times (3^2)^{x-2} = (3^3)^2\)
\(3^{x+1} \times 3^{2(x-2)} = 3^6\)
\(3^{x+1 + 2x - 4} = 3^6\)
\(3^{3x-3} = 3^6\)

Equating the exponents:
\(3x - 3 = 6\)
\(3x = 9\)
\(x = 3\)

M1 for expressing 9 and 27 as powers of 3, e.g., \((3^2)^{x-2}\) and \((3^3)^2\) seen
M1 for using laws of indices to form the linear equation \(x + 1 + 2(x - 2) = 6\) (or equivalent)
A1 for 3

First, find the derivative \(\frac{dy}{dx}\) to represent the gradient function:
\(\frac{dy}{dx} = 3x^2 - 12x + 15\)

Set the gradient function equal to 3:
\(3x^2 - 12x + 15 = 3\)
\(3x^2 - 12x + 12 = 0\)

Divide the entire equation by 3:
\(x^2 - 4x + 4 = 0\)
\((x - 2)^2 = 0 \implies x = 2\)

Find the corresponding \(y\)-coordinate by substituting \(x = 2\) back into the equation of the curve:
\(y = (2)^3 - 6(2)^2 + 15(2) - 3\)
\(y = 8 - 24 + 30 - 3 = 11\)

Thus, the coordinates of the point are \((2, 11)\).

M1 for differentiating the curve equation correctly to obtain \(3x^2 - 12x + 15\) (at least two terms correct)
M1 for setting their derivative equal to 3 and solving the quadratic equation to find \(x = 2\)
A1 for \((2, 11)\)

Rewrite the equation of the curve in index form: \(y = 3x^2 - 16x^{-1}\). Differentiate \(y\) with respect to \(x\) to find the gradient function: \(\frac{dy}{dx} = 6x - 16(-1)x^{-2} = 6x + \frac{16}{x^2}\). Substitute \(x = 2\) into this gradient function: \(\frac{dy}{dx} = 6(2) + \frac{16}{2^2} = 12 + \frac{16}{4} = 12 + 4 = 16\).

M1: for differentiating at least one term correctly to obtain either \(6x\) or \(16x^{-2}\) (or \(\frac{16}{x^2}\)). M1: for substituting \(x = 2\) into their derivative of the form \(ax + bx^{-2}\) where \(ab \ne 0\). A1: for the correct final answer of 16.

Using Pythagoras' theorem: \((x + 1)^2 + (x + 3)^2 = 10^2\). Expanding the brackets gives: \((x^2 + 2x + 1) + (x^2 + 6x + 9) = 100\). Simplifying and collecting terms on one side: \(2x^2 + 8x + 10 = 100 \implies 2x^2 + 8x - 90 = 0\). Divide the entire equation by 2: \(x^2 + 4x - 45 = 0\). Factorise the quadratic equation: \((x + 9)(x - 5) = 0\). This gives \(x = -9\) or \(x = 5\). Since length must be a positive quantity, \(x\) must be positive, so we reject \(x = -9\) and obtain \(x = 5\).

M1: for establishing the initial equation using Pythagoras' theorem: \((x + 1)^2 + (x + 3)^2 = 10^2\). M1: for expanding brackets and simplifying to a three-term quadratic equation equal to zero, e.g., \(2x^2 + 8x - 90 = 0\) or \(x^2 + 4x - 45 = 0\). A1: for \(x = 5\) (accepting both solutions only if \(x = -9\) is clearly identified as rejected or inappropriate for side lengths).

(a) Differentiating \(y\) with respect to \(x\):
\(\frac{dy}{dx} = 6x^2 - 18x - 24\)

(b) At stationary points, \(\frac{dy}{dx} = 0\):
\(6x^2 - 18x - 24 = 0\)
Divide by 6:
\(x^2 - 3x - 4 = 0\)
\((x - 4)(x + 1) = 0\)
So, \(x = 4\) or \(x = -1\).

To determine the nature of the stationary points, find the second derivative:
\(\frac{d^2y}{dx^2} = 12x - 18\)

At \(x = 4\):
\(\frac{d^2y}{dx^2} = 12(4) - 18 = 30 > 0\) (local minimum)

At \(x = -1\):
\(\frac{d^2y}{dx^2} = 12(-1) - 18 = -30 < 0\) (local maximum)

Thus, the minimum point occurs at \(x = 4\).
Substitute \(x = 4\) into the original equation to find the y-coordinate:
\(y = 2(4)^3 - 9(4)^2 - 24(4) + 10 = 128 - 144 - 96 + 10 = -102\)

The coordinates of the local minimum point are \((4, -102)\).

(a)
M1: For differentiating at least one non-zero term correctly.
A1: Correct derivative: \(6x^2 - 18x - 24\).

(b)
M1: Equates their derivative to 0 and attempts to solve the quadratic equation to find \(x\).
M1: Identifies \(x = 4\) as the minimum point (e.g., by testing the second derivative or surrounding points) and substitutes \(x = 4\) into the original curve's equation.
A1: \((4, -102)\).

(a) In the right-angled triangle \(ABD\):
\(\tan(38^\circ) = \frac{AB}{BD} = \frac{7.5}{BD}\)
\(BD = \frac{7.5}{\tan(38^\circ)} \approx 9.5995\text{ cm}\)
To 3 significant figures, \(BD = 9.60\text{ cm}\).

(b) In triangle \(BCD\), using the Cosine Rule to find \(CD\):
\(CD^2 = BD^2 + BC^2 - 2 \times BD \times BC \times \cos(\angle CBD)\)
Using the unrounded value for \(BD = 9.5995\):
\(CD^2 = (9.5995)^2 + 12.4^2 - 2 \times 9.5995 \times 12.4 \times \cos(65^\circ)\)
\(CD^2 = 92.150 + 153.76 - 238.068 \times 0.422618\)
\(CD^2 = 245.910 - 100.612 = 145.298\)
\(CD = \sqrt{145.298} \approx 12.0539\text{ cm}\)
To 3 significant figures, \(CD = 12.1\text{ cm}\).

(a)
M1: Formulates a correct trigonometric ratio, e.g. \(\tan(38^\circ) = \frac{7.5}{BD}\).
A1: Correct value \(9.60\) (or \(9.6\)) (accept \(9.59 - 9.61\)).

(b)
M1: Substitutes their value of \(BD\), \(12.4\) and \(65^\circ\) correctly into the Cosine Rule: \(CD^2 = BD^2 + 12.4^2 - 2 \times BD \times 12.4 \times \cos(65^\circ)\).
M1: Correct order of operations leading to \(CD^2 \approx 145.3\).
A1: \(12.1\) (accept \(12.0 - 12.2\)).

(a) Let the normal price of the television be \(P\).
Since the price is reduced by \(15\%\), the multiplier is \(0.85\).
\(P \times 0.85 = 544\)
\(P = \frac{544}{0.85} = 640\)
So, the normal price is £\(640\).

(b) Using the compound interest formula:
\(4000 \times \left(1 + \frac{x}{100}\right)^2 = 4326.40\)
Divide both sides by \(4000\):
\(\left(1 + \frac{x}{100}\right)^2 = \frac{4326.40}{4000} = 1.0816\)
Take the square root of both sides:
\(1 + \frac{x}{100} = \sqrt{1.0816} = 1.04\)
\(\frac{x}{100} = 0.04\)
\(x = 4\)

(a)
M1: Formulates a correct reverse percentage equation, e.g., \(544 \div 0.85\).
A1: \(640\).

(b)
M1: Sets up the compound interest equation, e.g., \(4000 \times r^2 = 4326.40\).
M1: Solves for the multiplier \(r = 1.04\).
A1: \(4\).

(a) First, factorize the quadratic numerator:
\(2x^2 + 5x - 3 = (2x - 1)(x + 3)\)

Next, factorize the denominator using the difference of two squares:
\(x^2 - 9 = (x - 3)(x + 3)\)

Simplify the fraction by cancelling the common factor \((x + 3)\):
\(\frac{(2x - 1)(x + 3)}{(x - 3)(x + 3)} = \frac{2x - 1}{x - 3}\)

(b) To make \(t\) the subject:
Multiply both sides by \((t - 1)\):
\(p(t - 1) = 2t + 3\)
\(pt - p = 2t + 3\)

Collect the terms with \(t\) on one side and the other terms on the opposite side:
\(pt - 2t = p + 3\)

Factorize out \(t\):
\(t(p - 2) = p + 3\)

Divide by \((p - 2)\):
\(t = \frac{p + 3}{p - 2}\)

(a)
M1: Factorizes the numerator into two binomial brackets: \((2x - 1)(x + 3)\).
M1: Factorizes the denominator using the difference of two squares: \((x - 3)(x + 3)\).
A1: Correct simplified fraction: \(\frac{2x - 1}{x - 3}\).

(b)
M1: Eliminates the fraction by multiplying and expands: \(pt - p = 2t + 3\).
A1: Correct final expression for \(t\): \(t = \frac{p + 3}{p - 2}\) (or equivalent e.g. \(t = \frac{-p - 3}{2 - p}\)).

(a) From the first equation, make \(y\) the subject:
\(y = 2x + 3\)

Substitute this expression for \(y\) into the second equation:
\(x^2 + (2x + 3)^2 = 26\)
\(x^2 + (4x^2 + 12x + 9) = 26\)
\(5x^2 + 12x + 9 - 26 = 0\)
\(5x^2 + 12x - 17 = 0\) (as required).

(b) Factorize the quadratic equation:
\((5x + 17)(x - 1) = 0\)

This gives the two solutions for \(x\):
\(x = 1\) or \(x = -\frac{17}{5} = -3.4\)

Now, find the corresponding \(y\) values using \(y = 2x + 3\):
When \(x = 1\):
\(y = 2(1) + 3 = 5\)

When \(x = -3.4\):
\(y = 2(-3.4) + 3 = -3.8\)

So the solutions are:
\(x = 1, y = 5\) and \(x = -3.4, y = -3.8\).

(a)
M1: Expresses \(y = 2x + 3\) and substitutes into the quadratic equation.
A1: Expands and groups terms correctly to establish \(5x^2 + 12x - 17 = 0\) with no algebraic errors.

(b)
M1: Factorizes the quadratic or uses the quadratic formula correctly to find at least one value of \(x\).
M1: Finds both values of \(x\) (\(1\) and \(-3.4\)) and attempts to substitute back to find \(y\).
A1: Correct pairs: \(x = 1, y = 5\) and \(x = -3.4, y = -3.8\).

(a) The graph of \(y = f(x) + 3\) is a vertical translation of \(y = f(x)\) upwards by 3 units.
Therefore, the x-coordinate of the turning point remains \(2\), and the y-coordinate becomes \(-6 + 3 = -3\).
The coordinates are \((2, -3)\).

(b) The quadratic equation \(f(x) = k\) has exactly one real solution when the line \(y = k\) is tangent to the vertex (turning point) of the curve.
Since the minimum value of the curve is \(y = -6\), we must have \(k = -6\).

(c) A translation by \(\begin{pmatrix} 0 \\ -5 \end{pmatrix}\) shifts the graph vertically downwards by 5 units.
\(g(x) = f(x) - 5\)
\(g(x) = x^2 - 4x - 2 - 5\)
\(g(x) = x^2 - 4x - 7\)
So, the equation is \(y = x^2 - 4x - 7\).

(a)
M1: Identifies the x-coordinate remains \(2\) or the y-coordinate becomes \(-3\).
A1: \((2, -3)\).

(b)
B1: \(-6\).

(c)
M1: Translates the equation down by 5 units, e.g., \(x^2 - 4x - 2 - 5\).
A1: Correct equation: \(y = x^2 - 4x - 7\) (must include '\(y =\)').

(a) The total surface area \(A\) of a cone is:
\(A = \pi r^2 + \pi r l\)
Substitute \(l = 3r\):
\(A = \pi r^2 + \pi r (3r) = \pi r^2 + 3\pi r^2 = 4\pi r^2\)
Given that \(A = 48\pi\):
\(4\pi r^2 = 48\pi\)
\(r^2 = 12\)
\(r = \sqrt{12} = 2\sqrt{3}\)

(b) By Pythagoras' theorem, the vertical height \(h\) of the cone is:
\(h = \sqrt{l^2 - r^2}\)
Substitute \(l = 3r\):
\(h = \sqrt{(3r)^2 - r^2} = \sqrt{9r^2 - r^2} = \sqrt{8r^2} = r\sqrt{8}\)
Using \(r = \sqrt{12}\):
\(h = \sqrt{12} \times \sqrt{8} = \sqrt{96} = 4\sqrt{6}\text{ cm}\)

Now, calculate the volume \(V\):
\(V = \frac{1}{3}\pi r^2 h\)
\(V = \frac{1}{3}\pi (12) (4\sqrt{6}) = 16\sqrt{6}\pi\)

(a)
M1: Expresses the surface area of the cone in terms of \(r\), e.g., \(\pi r^2 + \pi r(3r) = 48\pi\).
M1: Simplifies to \(4r^2 = 48\) or \(r^2 = 12\).
A1: \(2\sqrt{3}\) (or \(\sqrt{12}\)).

(b)
M1: Uses Pythagoras to find the height: \(h = \sqrt{(3r)^2 - r^2} = \sqrt{96}\) or \(4\sqrt{6}\).
A1: \(16\sqrt{6}\pi\).

(a) The angles \(\angle OAT = 90^\circ\) and \(\angle OCT = 90^\circ\) because a tangent to a circle is perpendicular to the radius at the point of contact.
In the quadrilateral \(OATC\), the angles add up to \(360^\circ\):
\(\angle AOC = 360^\circ - 90^\circ - 90^\circ - 52^\circ = 128^\circ\)

The angle at the circumference is half the angle at the center subtended by the same arc:
\(\angle ADC = 128^\circ \div 2 = 64^\circ\).

(b) \(ABCD\) is a cyclic quadrilateral because all of its vertices lie on the circle.
The opposite angles of a cyclic quadrilateral sum to \(180^\circ\):
\(\angle ABC = 180^\circ - 64^\circ = 116^\circ\).

(a)
M1: Calculates \(\angle AOC = 128^\circ\) with at least one correct reason mentioned (e.g., "tangent is perpendicular to radius" or "angles in quadrilateral sum to \(360^\circ\)").
A1: \(\angle ADC = 64^\circ\).
B1: For providing both correct circle theorems in part (a):
1) "The tangent to a circle is perpendicular to the radius"
2) "The angle subtended by an arc at the center is twice the angle subtended at the circumference".

(b)
M1: \(\angle ABC = 180^\circ - 64^\circ = 116^\circ\).
B1: For the reason "Opposite angles of a cyclic quadrilateral sum to \(180^\circ\)".

(a) Differentiating the equation term by term gives:
\(\frac{dy}{dx} = 3(2x^2) - 2(9x) - 24 = 6x^2 - 18x - 24\).

(b) At stationary points, \(\frac{dy}{dx} = 0\):
\( 6x^2 - 18x - 24 = 0 \)

Dividing the equation by 6 gives:
\( x^2 - 3x - 4 = 0 \)

Factoring the quadratic:
\( (x - 4)(x + 1) = 0 \)

So the \( x \)-coordinates of the stationary points are \( x = 4 \) and \( x = -1 \).

Substitute these \( x \)-values into the original curve equation to find the corresponding \( y \)-coordinates:
For \( x = 4 \):
\( y = 2(4)^3 - 9(4)^2 - 24(4) + 7 = 128 - 144 - 96 + 7 = -105 \)

For \( x = -1 \):
\( y = 2(-1)^3 - 9(-1)^2 - 24(-1) + 7 = -2 - 9 + 24 + 7 = 20 \)

Therefore, the coordinates of the two stationary points are \( (4, -105) \) and \( (-1, 20) \).

(a)
M1: For differentiating at least one term correctly (e.g. \(6x^2\) or \(-18x\)).
A1: For the complete correct derivative: \(6x^2 - 18x - 24\).

(b)
M1: For setting their derivative to 0 and attempting to solve the resulting quadratic equation (e.g., by factoring, completing the square, or using the quadratic formula).
M1: For finding both correct \(x\)-values (\(x = 4\) and \(x = -1\)) and attempting to find at least one corresponding \(y\)-value by substituting into the original equation.
A1: For both correct coordinates: \( (4, -105) \) and \( (-1, 20) \).

(a) Applying the Cosine Rule to find the length of \( BC \):
\( BC^2 = AB^2 + AC^2 - 2 \times AB \times AC \times \cos(BAC) \)
\( BC^2 = 8^2 + 11^2 - 2 \times 8 \times 11 \times \cos(112^\circ) \)
\( BC^2 = 64 + 121 - 176 \times (-0.374606...) \)
\( BC^2 = 185 + 65.9307... = 250.9307... \)
\( BC = \sqrt{250.9307...} \approx 15.8407... \text{ cm} \)

To 3 significant figures, \( BC = 15.8 \text{ cm} \).

(b) Applying the trigonometric area formula for a triangle:
\( \text{Area} = \frac{1}{2} \times AB \times AC \times \sin(BAC) \)
\( \text{Area} = \frac{1}{2} \times 8 \times 11 \times \sin(112^\circ) \)
\( \text{Area} = 44 \times 0.92718... = 40.796... \text{ cm}^2 \)

To 3 significant figures, \( \text{Area} = 40.8 \text{ cm}^2 \).

(a)
M1: For a correct substitution into the Cosine Rule: \( 8^2 + 11^2 - 2(8)(11)\cos(112^\circ) \).
M1: For correctly evaluating to find \( BC^2 \approx 250.9... \) or finding \( BC = \sqrt{250.9...} \).
A1: For the correct answer in the range \( 15.8 \) to \( 15.84 \).

(b)
M1: For a correct substitution into the triangle area formula: \( \frac{1}{2} \times 8 \times 11 \times \sin(112^\circ) \).
A1: For the correct area in the range \( 40.7 \) to \( 40.8 \).

First, factorise each term in the product: \(2x^2 + 5x - 3 = (2x - 1)(x + 3)\). \(x^2 + 5x + 6 = (x + 2)(x + 3)\). \(x^2 - 4 = (x - 2)(x + 2)\). \(2x^2 - x = x(2x - 1)\). Now, substitute these back into the expression: \(\frac{(2x - 1)(x + 3)}{(x + 2)(x + 3)} \times \frac{(x - 2)(x + 2)}{x(2x - 1)}\). Next, cancel the common terms \((2x - 1)\), \((x + 3)\), and \((x + 2)\) from the numerator and denominator. This leaves \(\frac{x - 2}{x}\). Finally, divide each term in the numerator by the denominator: \(\frac{x}{x} - \frac{2}{x} = 1 - \frac{2}{x}\).

M1 for factorising \(2x^2 + 5x - 3\) as \((2x - 1)(x + 3)\). M1 for factorising \(x^2 + 5x + 6\) as \((x + 2)(x + 3)\). M1 for factorising \(x^2 - 4\) as \((x - 2)(x + 2)\) and \(2x^2 - x\) as \(x(2x - 1)\). M1 for cancelling common factors to obtain \(\frac{x - 2}{x}\). A1 for completing the division to show \(1 - \frac{2}{x}\) with all steps shown clearly.

The total surface area \(A\) of a closed cylinder is given by \(A = 2\pi r^2 + 2\pi r h\). We are given that \(2\pi r^2 + 2\pi r h = 96\pi\). Divide the entire equation by \(2\pi\) to simplify: \(r^2 + rh = 48\). Rearrange this equation to express \(h\) in terms of \(r\): \(rh = 48 - r^2\), which gives \(h = \frac{48 - r^2}{r}\). The volume \(V\) of a cylinder is given by \(V = \pi r^2 h\). Substitute our expression for \(h\) into the volume formula: \(V = \pi r^2 \left(\frac{48 - r^2}{r}\right)\). Simplify the expression: \(V = \pi r (48 - r^2) = 48\pi r - \pi r^3\).

M1 for writing down the correct surface area formula equated to the given value: \(2\pi r^2 + 2\pi r h = 96\pi\). M1 for simplifying to \(r^2 + rh = 48\). M1 for rearranging to make \(h\) the subject: \(h = \frac{48 - r^2}{r}\). M1 for substituting this expression for \(h\) into the volume formula \(V = \pi r^2 h\). A1 for fully simplifying to obtain the required formula \(V = 48\pi r - \pi r^3\).

First, we find the position vectors of the relevant points. Since \(OP : PA = 2 : 1\), \(\vec{OP} = \frac{2}{3}\mathbf{a}\). Since \(Q\) is the midpoint of \(AB\), we have \(\vec{OQ} = \vec{OA} + \frac{1}{2}\vec{AB} = \mathbf{a} + \frac{1}{2}(\mathbf{b} - \mathbf{a}) = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\). Since \(R\) lies on \(OB\) produced with \(OB = BR\), we have \(\vec{OR} = 2\mathbf{b}\). Now, find vector \(\vec{PQ}\): \(\vec{PQ} = \vec{OQ} - \vec{OP} = \left(\frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\right) - \frac{2}{3}\mathbf{a} = -\frac{1}{6}\mathbf{a} + \frac{1}{2}\mathbf{b}\). Next, find vector \(\vec{QR}\): \(\vec{QR} = \vec{OR} - \vec{OQ} = 2\mathbf{b} - \left(\frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\right) = -\frac{1}{2}\mathbf{a} + \frac{3}{2}\mathbf{b}\). Notice that \(\vec{QR} = 3\left(-\frac{1}{6}\mathbf{a} + \frac{1}{2}\mathbf{b}\right) = 3\vec{PQ}\). Since \(\vec{QR}\) is a scalar multiple of \(\vec{PQ}\) and they share the common point \(Q\), the points \(P\), \(Q\), and \(R\) lie on a straight line.

M1 for writing \(\vec{OP} = \frac{2}{3}\mathbf{a}\) or \(\vec{OR} = 2\mathbf{b}\). M1 for finding \(\vec{OQ} = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\). M1 for finding \(\vec{PQ} = -\frac{1}{6}\mathbf{a} + \frac{1}{2}\mathbf{b}\) or equivalent vector. M1 for finding \(\vec{QR} = -\frac{1}{2}\mathbf{a} + \frac{3}{2}\mathbf{b}\) or \(\vec{PR} = -\frac{2}{3}\mathbf{a} + 2\mathbf{b}\). A1 for showing \(\vec{QR} = 3\vec{PQ}\) (or equivalent relationship) and concluding that they share a common point \(Q\), hence they are collinear.

Let \(O\) be the center of the circle, and let \(A\) and \(B\) be the endpoints of the chord. Since \(OA = r\), \(OB = r\), and \(AB = r\), triangle \(OAB\) is equilateral. Therefore, the angle subtended at the center is \(\theta = 60^\circ = \frac{\pi}{3}\) radians. The area of the sector \(OAB\) is given by \(\frac{60}{360} \times \pi r^2 = \frac{1}{6}\pi r^2\). The area of the equilateral triangle \(OAB\) is given by \(\frac{1}{2} \times r \times r \times \sin(60^\circ) = \frac{1}{2} r^2 \left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{4}r^2\). The area of the smaller segment is the area of the sector minus the area of the triangle: \(\text{Area} = \frac{1}{6}\pi r^2 - \frac{\sqrt{3}}{4}r^2\). Finding a common denominator of 12: \(\text{Area} = \frac{2\pi r^2}{12} - \frac{3\sqrt{3}r^2}{12} = \frac{r^2}{12}(2\pi - 3\sqrt{3})\).

M1 for identifying that the triangle is equilateral and finding the angle at the center is \(60^\circ\) (or \(\frac{\pi}{3}\) radians). M1 for finding the area of the sector: \(\frac{1}{6}\pi r^2\). M1 for finding the area of the triangle: \(\frac{\sqrt{3}}{4}r^2\). M1 for subtracting the area of the triangle from the area of the sector: \(\frac{1}{6}\pi r^2 - \frac{\sqrt{3}}{4}r^2\). A1 for factorising out \(\frac{r^2}{12}\) to arrive at the correct final expression \(\frac{r^2}{12}(2\pi - 3\sqrt{3})\).

We use the area formula for a triangle:
\(\text{Area} = \frac{1}{2} a c \sin B\)

Substitute the given values into the formula:
\(45 = \frac{1}{2} \times 12.5 \times 8.4 \times \sin(ABC)\)
\(45 = 52.5 \sin(ABC)\)

Solve for \(\sin(ABC)\):
\(\sin(ABC) = \frac{45}{52.5} = \frac{6}{7}\)

Find the acute angle:
\(ABC = \sin^{-1}\left(\frac{6}{7}\right) \approx 58.997^{\circ}\)

Since angle \(ABC\) is obtuse, we find the corresponding obtuse angle:
\(ABC = 180^{\circ} - 58.997^{\circ} = 121.003^{\circ}\)

Correct to 1 decimal place, the angle is \(121.0^{\circ}\).

M1: For a correct substitution into the area formula, e.g., \(45 = \frac{1}{2} \times 12.5 \times 8.4 \times \sin(ABC)\)
M1: For finding the acute angle \(\sin^{-1}(6/7) \approx 59.0\) or showing \(180 - \sin^{-1}(6/7)\)
A1: For \(121.0\) (accept \(121\))

Let \(x\) be the original price of the coat.

After a 15% increase, the price is:
\(x \times 1.15 = 1.15x\)

After a 20% reduction during the sale, the price is:
\(1.15x \times 0.80 = 0.92x\)

We are given that the sale price is £64.40, so:
\(0.92x = 64.40\)

Solve for \(x\):
\(x = \frac{64.40}{0.92} = 70\)

The original price of the coat was £70.

M1: For writing a correct expression/multiplier representing both percentage changes, e.g., \(1.15 \times 0.8\) or \(0.92\)
M1: For setting up a correct equation, e.g., \(0.92x = 64.40\) or calculation \(64.40 \div 0.92\)
A1: For 70

First, simplify the fraction inside the parentheses by dealing with the negative exponent of \(y\):
\(\frac{64x^6}{27y^{-3}} = \frac{64x^6y^3}{27}\)

Now apply the exponent of \(-\frac{2}{3}\). The negative power reciprocates the fraction:
\(\left(\frac{64x^6y^3}{27}\right)^{-\frac{2}{3}} = \left(\frac{27}{64x^6y^3}\right)^{\frac{2}{3}}\)

Now apply the fractional power of \(\frac{2}{3}\) to each part of the fraction (cube root, then square):
\(\frac{27^{\frac{2}{3}}}{(64x^6y^3)^{\frac{2}{3}}} = \frac{(\sqrt[3]{27})^2}{(\sqrt[3]{64})^2 \cdot (x^6)^{\frac{2}{3}} \cdot (y^3)^{\frac{2}{3}}} = \frac{3^2}{4^2 \cdot x^4 \cdot y^2} = \frac{9}{16x^4y^2}\)

M1: For simplifying the fraction inside the parentheses or applying the negative index, e.g., \(\frac{64x^6y^3}{27}\) or \(\left(\frac{27y^{-3}}{64x^6}\right)^{\frac{2}{3}}\)
M1: For correctly evaluating the power \(\frac{2}{3}\) on at least one coefficient or variable, e.g., finding \(27^{\frac{2}{3}} = 9\) or \(64^{\frac{2}{3}} = 16\) or \(x^4\) or \(y^2\)
A1: For \(\frac{9}{16x^4y^2}\) or \(\frac{9}{16}x^{-4}y^{-2}\) or equivalent fully simplified form

Multiply the first equation by 6 to clear the denominators:
\(3x + 2y = 22\) --- (Equation 1)

Multiply the second equation by 6 to clear the denominators:
\(4x - 3y = 1\) --- (Equation 2)

To eliminate \(y\), multiply Equation 1 by 3 and Equation 2 by 2:
\(9x + 6y = 66\)
\(8x - 6y = 2\)

Add the two equations together:
\(17x = 68\)
\(x = 4\)

Substitute \(x = 4\) back into Equation 1:
\(3(4) + 2y = 22\)
\(12 + 2y = 22\)
\(2y = 10\)
\(y = 5\)

M1: For clearing the fractions to obtain linear equations, e.g., \(3x + 2y = 22\) and \(4x - 3y = 1\) (or an equivalent correct substitution setup)
M1: For a complete and correct algebraic method to eliminate one variable to find the value of either \(x\) or \(y\)
A1: For both \(x = 4\) and \(y = 5\)

First, find the velocity \(v\) by differentiating displacement \(s\) with respect to time \(t\):
\(v = \frac{ds}{dt} = 3t^2 - 12t + 9\)

Find the times when the velocity is \(0\) by solving \(v = 0\):
\(3t^2 - 12t + 9 = 0\)
\(t^2 - 4t + 3 = 0\)
\((t - 1)(t - 3) = 0\)

So the velocity is \(0\) at \(t = 1\) and \(t = 3\). The velocity is first \(0\) when \(t = 1\).

Next, find the acceleration \(a\) by differentiating velocity \(v\) with respect to time \(t\):
\(a = \frac{dv}{dt} = 6t - 12\)

Substitute \(t = 1\) into the acceleration formula:
\(a = 6(1) - 12 = -6\text{ m/s}^2\)

M1: For differentiating the displacement formula to get a correct expression for velocity, \(v = 3t^2 - 12t + 9\) (allow at least two correct terms)
M1: For setting their velocity expression to \(0\) to find the first time \(t = 1\), and differentiating the velocity to find acceleration \(a = 6t - 12\)
A1: For \(-6\)

First, calculate the total sum of marks for the first class:
\(\text{Total}_1 = 20 \times 64 = 1280\)

Next, calculate the total sum of marks for the second class:
\(\text{Total}_2 = 30 \times 72 = 2160\)

Calculate the sum of marks for the four absent students:
\(\text{Total}_{\text{absent}} = 68 + 71 + 72 + 75 = 286\)

Calculate the overall total sum of marks for all 54 students:
\(\text{Overall Total} = 1280 + 2160 + 286 = 3726\)

Calculate the overall mean:
\(\text{Overall Mean} = \frac{3726}{54} = 69\)

M1: For finding the total marks of at least one of the classes (e.g., \(20 \times 64 = 1280\) or \(30 \times 72 = 2160\))
M1: For a complete method to find the sum of all 54 marks divided by 54, i.e., \(\frac{1280 + 2160 + 286}{54}\)
A1: For 69

Let \(\theta\) be the angle of the sector in degrees.

Using the area of a sector formula:
\(\text{Area} = \frac{\theta}{360} \times \pi r^2\)
\(48\pi = \frac{\theta}{360} \times \pi \times 12^2\)
\(48\pi = \frac{\theta}{360} \times 144\pi\)

Divide both sides by \(\pi\):
\(48 = \frac{144\theta}{360}\)
\(\frac{\theta}{360} = \frac{48}{144} = \frac{1}{3}\)

So the sector represents \(\frac{1}{3}\) of a full circle.

Now, calculate the arc length of the sector:
\(\text{Arc Length} = \frac{1}{3} \times 2\pi r = \frac{1}{3} \times 2\pi(12) = 8\pi\text{ cm}\)

The perimeter of the sector consists of the arc length plus two radii:
\(\text{Perimeter} = \text{Arc Length} + 2r = 8\pi + 2(12) = 8\pi + 24\text{ cm}\)

M1: For setting up an equation to find the fraction of the circle or the sector angle, e.g., \(\frac{\theta}{360} \times \pi \times 12^2 = 48\pi\) or stating the fraction is \(\frac{1}{3}\)
M1: For a correct method to find the arc length, e.g., \(\frac{1}{3} \times 2 \times \pi \times 12\) resulting in \(8\pi\)
A1: For \(8\pi + 24\) (or \(24 + 8\pi\))

First, find the gradient of the line \(3x - 5y = 12\) by rearranging it into the form \(y = mx + c\):
\(5y = 3x - 12\)
\(y = \frac{3}{5}x - \frac{12}{5}\)

So, the gradient of this line is \(\frac{3}{5}\).

The gradient \(m\) of the perpendicular line is the negative reciprocal:
\(m = -\frac{1}{\frac{3}{5}} = -\frac{5}{3}\)

Now, use the point-slope formula with the point \((6, -2)\) and gradient \(-\frac{5}{3}\):
\(y - y_1 = m(x - x_1)\)
\(y - (-2) = -\frac{5}{3}(x - 6)\)
\(y + 2 = -\frac{5}{3}x + 10\)
\(y = -\frac{5}{3}x + 8\)

M1: For finding the gradient of the given line is \(\frac{3}{5}\) or determining the perpendicular gradient is \(-\frac{5}{3}\)
M1: For substituting the point \((6, -2)\) and their perpendicular gradient into a linear equation form, e.g., \(y - (-2) = -\frac{5}{3}(x - 6)\) or \(-2 = -\frac{5}{3}(6) + c\)
A1: For \(y = -\frac{5}{3}x + 8\) (or equivalent form in \(y = mx + c\), such as \(y = -1.\dot{6}x + 8\))

Let \(V\) be the value of the company at the start of 2021.

Value at the end of 2021 after a 12% increase:
\(V \times 1.12\)

Value at the end of 2022 after an 8% decrease:
\((V \times 1.12) \times (1 - 0.08) = V \times 1.12 \times 0.92 = 1.0304V\)

We are given that this final value is £2,471,040:
\(1.0304V = 2,471,040\)

Solving for \(V\):
\(V = \frac{2,471,040}{1.0304} = 2,400,000\)

M1 for expressing the final value in terms of the initial value, e.g., \(V \times 1.12 \times 0.92 = 2,471,040\) or finding the combined multiplier \(1.0304\)
M1 for a complete method to solve for \(V\), e.g., \(\frac{2,471,040}{1.0304}\)
A1 for 2,400,000 (or 2400000)

The volume of a cylinder is given by \(V = \pi r^2 h\).
Here, \(h = 9r\), so:
\(\text{Volume of cylinder} = \pi r^2 (9r) = 9\pi r^3\)

The volume of a sphere is given by \(V = \frac{4}{3}\pi R^3\).
Since there are 2 identical spheres:
\(\text{Total volume of 2 spheres} = 2 \times \frac{4}{3}\pi R^3 = \frac{8}{3}\pi R^3\)

Since the cylinder is recast into the 2 spheres, their volumes are equal:
\(9\pi r^3 = \frac{8}{3}\pi R^3\)

Divide both sides by \(\pi\):
\(9r^3 = \frac{8}{3}R^3\)

Multiply by 3:
\(27r^3 = 8R^3\)

Take the cube root of both sides:
\(3r = 2R\)
\(R = 1.5r\) (or \(R = \frac{3}{2}r\))

M1 for writing a correct expression for the volume of the cylinder (\(9\pi r^3\)) or the total volume of the 2 spheres (\(\frac{8}{3}\pi R^3\))
M1 for equating the two volumes and attempting to simplify, e.g., \(27r^3 = 8R^3\) or \(R^3 = 3.375r^3\)
A1 for \(R = 1.5r\) or \(R = \frac{3}{2}r\)

To find the stationary points, we first differentiate the curve equation to find the gradient function \(\frac{dy}{dx}\):
\(y = 2x^3 - 9x^2 - 24x + 7\)

\(\frac{dy}{dx} = 6x^2 - 18x - 24\)

At stationary points, the gradient is zero, so we set \(\frac{dy}{dx} = 0\):
\(6x^2 - 18x - 24 = 0\)

Divide the entire equation by 6 to simplify:
\(x^2 - 3x - 4 = 0\)

Factorise the quadratic expression:
\((x - 4)(x + 1) = 0\)

This gives the solutions:
\(x = 4\) or \(x = -1\)

M1 for differentiating the equation correctly with at least two terms correct (e.g., \(6x^2 - 18x\))
M1 for setting their derivative to 0 and attempting to solve the quadratic equation by factorisation or formula
A1 for both \(x = 4\) and \(x = -1\)

Let the given equations be:
(1) \(3x + 2y = 4.5\)
(2) \(4x - 5y = 17.5\)

Multiply equation (1) by 5 and equation (2) by 2 to make the coefficients of \(y\) equal in magnitude:
(3) \(15x + 10y = 22.5\)
(4) \(8x - 10y = 35\)

Add equations (3) and (4):
\((15x + 8x) + (10y - 10y) = 22.5 + 35\)
\(23x = 57.5\)
\(x = \frac{57.5}{23} = 2.5\)

Substitute \(x = 2.5\) back into equation (1):
\(3(2.5) + 2y = 4.5\)
\(7.5 + 2y = 4.5\)
\(2y = 4.5 - 7.5\)
\(2y = -3\)
\(y = -1.5\)

Thus, the solution is \(x = 2.5\) and \(y = -1.5\).

M1 for a correct method to eliminate one variable, e.g. multiplying equations to align coefficients of \(x\) or \(y\) (allowing 1 arithmetic error)
M1 for a correct method to find the second variable using their value for the first variable
A1 for \(x = 2.5\) and \(y = -1.5\) (or equivalent fractions)

For (a): In the right-angled triangle \(ABC\), \(BC = \sqrt{AB^2 + AC^2} = \sqrt{6^2 + 8^2} = \sqrt{100} = 10\text{ cm}\). For (b): The projection of \(BF\) onto the base is \(AF\). In the rectangular base, \(AF = \sqrt{AC^2 + CF^2} = \sqrt{8^2 + 15^2} = \sqrt{289} = 17\text{ cm}\). The angle \(\theta\) that \(BF\) makes with the base is \(\angle BFA\). Since \(AB\) is vertical, \(\tan \theta = \frac{AB}{AF} = \frac{6}{17}\). Thus, \(\theta = \tan^{-1}\left(\frac{6}{17}\right) \approx 19.44^{\circ}\), which is \(19.4^{\circ}\) to 1 decimal place.

(a) M1 for using Pythagoras' theorem to find BC. A1 for 10. (b) M1 for finding the diagonal AF of the base: \(AF = 17\). M1 for \(\tan(\theta) = \frac{6}{17}\) or equivalent trigonometric equation. A1 for 19.4 (accept 19.4 - 19.5).

(a) The midpoints of the intervals are 5, 15, 25, and 35. The sum of the products of midpoints and frequencies is \(\sum fx = 4(5) + 15a + 12(25) + 35b = 20 + 15a + 300 + 35b = 320 + 15a + 35b\). Since the estimated mean is 21 and the total frequency is 30, we have \(\frac{320 + 15a + 35b}{30} = 21 \implies 320 + 15a + 35b = 630 \implies 15a + 35b = 310\). Dividing by 5 gives \(3a + 7b = 62\). (b) Since the total number of students is 30, we have \(4 + a + 12 + b = 30 \implies a + b = 14\). Multiplying this by 3 gives \(3a + 3b = 42\). Subtracting this from \(3a + 7b = 62\) gives \(4b = 20 \implies b = 5\). Substituting back gives \(a = 14 - 5 = 9\).

(a) M1 for identifying correct midpoints 5, 15, 25, 35 and writing an expression for \(\sum fx\). A1 for correctly deriving \(3a + 7b = 62\) with intermediate steps shown. (b) M1 for writing the equation \(a + b = 14\) (or equivalent) and attempting to solve simultaneously. A1 for both \(a = 9\) and \(b = 5\).

(a) Let \(P\) be the original price of the laptop. After a 20% reduction, the price is \(0.80P\). After a further 15% reduction, the price is \(0.80P \times 0.85 = 0.68P\). We are given \(0.68P = 374 \implies P = \frac{374}{0.68} = 550\). So the original price was £550. (b) After an increase of \(x\%\) followed by a decrease of \(x\%\), the price is multiplied by \((1 + \frac{x}{100})(1 - \frac{x}{100}) = 1 - \frac{x^2}{10000}\). Thus, \(250 \left(1 - \frac{x^2}{10000}\right) = 246.40 \implies 1 - \frac{x^2}{10000} = 0.9856 \implies \frac{x^2}{10000} = 0.0144 \implies x^2 = 144 \implies x = 12\).

(a) M1 for setting up the equation \(0.80 \times 0.85 \times P = 374\) or equivalent. A1 for 550. (b) M1 for setting up the equation \(250 \times (1 - \frac{x^2}{10000}) = 246.40\) or equivalent. A1 for 12.

(a) Factorize the numerator: \(2x^2 + 5x - 3 = (2x - 1)(x + 3)\). Factorize the denominator as a difference of two squares: \(4x^2 - 1 = (2x - 1)(2x + 1)\). Divide both by the common factor \((2x - 1)\) to get \(\frac{x + 3}{2x + 1}\). (b) Multiply both sides by \((3w + 1)\): \(T(3w + 1) = 5 - 2w \implies 3Tw + T = 5 - 2w\). Rearrange to collect terms with \(w\) on one side: \(3Tw + 2w = 5 - T \implies w(3T + 2) = 5 - T\). Divide by \((3T + 2)\) to make \(w\) the subject: \(w = \frac{5 - T}{3T + 2}\).

(a) M1 for factorizing either the numerator to \((2x-1)(x+3)\) or the denominator to \((2x-1)(2x+1)\). A1 for \(\frac{x+3}{2x+1}\). (b) M1 for multiplying by the denominator and rearranging to collect \(w\) terms. A1 for \(w = \frac{5-T}{3T+2}\) (or equivalent).

(a) Set the line equation equal to the curve equation: \(x^2 - 4x + 6 = kx + 2 \implies x^2 - (4+k)x + 4 = 0\). For the line to be a tangent, this quadratic must have exactly one real solution, so its discriminant must be zero: \(B^2 - 4AC = 0 \implies (-(4+k))^2 - 4(1)(4) = 0 \implies (4+k)^2 - 16 = 0 \implies (4+k)^2 = 16\). Thus, \(4+k = 4 \implies k = 0\), or \(4+k = -4 \implies k = -8\). (b) Under a translation by vector \(\begin{pmatrix} -3 \\ 1 \end{pmatrix}\), the function \(y = f(x)\) becomes \(y = f(x + 3) + 1\). Substituting into the original equation: \(y = (x + 3)^2 - 4(x + 3) + 6 + 1 = (x^2 + 6x + 9) - (4x + 12) + 7 = x^2 + 2x + 4\). Thus, \(a = 2\) and \(b = 4\).

(a) M1 for equating and setting discriminant of the resulting quadratic to zero. A1 for \(k = 0\) and \(k = -8\). (b) M1 for substituting \(x + 3\) for \(x\) and adding 1 to the function (or translating the vertex \((2, 2) \to (-1, 3)\)). A1 for \(a = 2\) and \(b = 4\).

(a) By the alternate segment theorem, the angle between the tangent \(TA\) and the chord \(AD\) is equal to the angle subtended by \(AD\) in the alternate segment, which is angle \(ABD\). Therefore, angle \(ABD = \text{angle } DAT = 35^\circ\). (b) Since \(A\), \(B\), \(C\) and \(D\) are points on the circumference of the circle, \(ABCD\) is a cyclic quadrilateral. The opposite angles of a cyclic quadrilateral add up to \(180^\circ\). Therefore, angle \(ABC + \text{angle } ADC = 180^\circ \implies \text{angle } ABC = 180^\circ - 112^\circ = 68^\circ\).

(a) B1 for \(35^\circ\). B1 for stating 'alternate segment theorem' or equivalent. (b) B1 for \(68^\circ\). B1 for stating 'opposite angles of a cyclic quadrilateral sum to 180 degrees' or equivalent.

(a) Differentiating \(y\) with respect to \(x\): \(\frac{dy}{dx} = 6x^2 - 18x - 24\). (b) At the turning points, \(\frac{dy}{dx} = 0 \implies 6x^2 - 18x - 24 = 0\). Dividing by 6 gives \(x^2 - 3x - 4 = 0 \implies (x - 4)(x + 1) = 0\). So the \(x\)-coordinates of the turning points are \(x = 4\) and \(x = -1\). Substituting these into the original equation: For \(x = 4\), \(y = 2(4)^3 - 9(4)^2 - 24(4) + 11 = 128 - 144 - 96 + 11 = -101\). For \(x = -1\), \(y = 2(-1)^3 - 9(-1)^2 - 24(-1) + 11 = -2 - 9 + 24 + 11 = 24\). The coordinates of the turning points are \((4, -101)\) and \((-1, 24)\).

(a) B1 for \(6x^2 - 18x - 24\). (b) M1 for setting their derivative to 0 and finding both \(x = 4\) and \(x = -1\). M1 for substituting their \(x\)-values back into the original cubic equation to find \(y\). A1 for both correct points: \((4, -101)\) and \((-1, 24)\).

(a) Volume of sphere = \(\frac{4}{3}\pi r^3\) and Volume of cone = \(\frac{1}{3}\pi r^2 h\). Since the volume is conserved: \(\frac{4}{3}\pi r^3 = \frac{1}{3}\pi r^2 h\). Multiplying both sides by \(\frac{3}{\pi r^2}\) gives \(4r = h\). (b) We are given the volume of the sphere is \(288\pi\text{ cm}^3 \implies \frac{4}{3}\pi r^3 = 288\pi \implies r^3 = 216 \implies r = 6\text{ cm}\). Thus, the height of the cone is \(h = 4 \times 6 = 24\text{ cm}\). The slant height \(l\) of the cone is \(l = \sqrt{r^2 + h^2} = \sqrt{6^2 + 24^2} = \sqrt{612} = 6\sqrt{17}\text{ cm} \approx 24.7386\text{ cm}\). The total surface area of the cone is \(\pi r^2 + \pi r l = \pi (6^2) + \pi (6)(24.7386) = 36\pi + 148.432\pi = 184.432\pi \approx 579.41\text{ cm}^2\). To 3 significant figures, this is \(579\text{ cm}^2\).

(a) M1 for equating sphere volume formula to cone volume formula. A1 for correct algebraic simplification leading to \(h = 4r\). (b) M1 for finding \(r = 6\) and \(l = \sqrt{6^2 + 24^2}\) (or \(24.7\)). A1 for 579 (accept 579 - 580).

(a) The velocity \(v\) is the derivative of the displacement \(s\) with respect to time \(t\): \(v = \frac{ds}{dt} = 6t^2 - 30t + 24\). Substituting \(t = 3.5\) gives: \(v = 6(3.5)^2 - 30(3.5) + 24 = 73.5 - 105 + 24 = -7.5\) m/s. (b) The particle is instantaneously at rest when \(v = 0\): \(6t^2 - 30t + 24 = 0\) which simplifies to \(t^2 - 5t + 4 = 0\). Factoring gives \((t - 1)(t - 4) = 0\), so the times of rest are \(t = 1\) and \(t = 4\). The second time is \(t = 4\) seconds. The acceleration \(a\) is the derivative of the velocity \(v\) with respect to time \(t\): \(a = \frac{dv}{dt} = 12t - 30\). Substituting \(t = 4\) gives: \(a = 12(4) - 30 = 48 - 30 = 18\) m/s\(^2\).

Part (a): M1 for differentiating displacement to get velocity (at least two terms correct, e.g. \(6t^2 - 30t\)). A1 for \(-7.5\). Part (b): M1 for setting their velocity equal to 0, solving to find the second time \(t = 4\), and finding the acceleration expression \(12t - 30\). A1 for \(18\).

(a) First, find the diagonal length \(BF\) on the horizontal rectangular base \(BCFE\). Since angle \(BCF = 90^\circ\), by Pythagoras' theorem: \(BF^2 = BC^2 + CF^2 = 9^2 + 12^2 = 81 + 144 = 225\), so \(BF = \sqrt{225} = 15\text{ cm}\). Since \(AB\) is vertical and \(BF\) lies in the horizontal base, angle \(ABF = 90^\circ\). Applying Pythagoras' theorem in triangle \(ABF\): \(AF^2 = AB^2 + BF^2 = 8^2 + 15^2 = 64 + 225 = 289\), so \(AF = \sqrt{289} = 17\text{ cm}\). (b) The angle between the line \(AF\) and the base is angle \(AFB\). In the right-angled triangle \(ABF\): \(\tan(\angle AFB) = \frac{AB}{BF} = \frac{8}{15}\), so \(\angle AFB = \tan^{-1}\left(\frac{8}{15}\right) \approx 28.072\dots^\circ\). Rounding to 1 decimal place, we get \(28.1^\circ\).

Part (a): M1 for finding the base diagonal \(BF = 15\) using Pythagoras, or showing a complete method for \(AF = \sqrt{8^2 + 9^2 + 12^2}\). A1 for \(17\). Part (b): M1 for a correct trigonometric ratio to find angle \(AFB\), e.g., \(\tan(\theta) = \frac{8}{15}\) or \(\sin(\theta) = \frac{8}{17}\). A1 for \(28.1\) (accept answers in the range \(28.0\) to \(28.1\)).

Let the three consecutive odd integers be \(2n-1\), \(2n+1\), and \(2n+3\), where \(n\) is an integer.

The sum of their squares, \(S\), is given by:
\(S = (2n-1)^2 + (2n+1)^2 + (2n+3)^2\)

Expanding each of the squared terms:
\((2n-1)^2 = 4n^2 - 4n + 1\)
\((2n+1)^2 = 4n^2 + 4n + 1\)
\((2n+3)^2 = 4n^2 + 12n + 9\)

Summing these terms together:
\(S = (4n^2 - 4n + 1) + (4n^2 + 4n + 1) + (4n^2 + 12n + 9)\)
\(S = 12n^2 + 12n + 11\)

To show this is 1 less than a multiple of 12, we can rewrite the constant term 11 as \(12 - 1\):
\(S = 12n^2 + 12n + 12 - 1\)
\(S = 12(n^2 + n + 1) - 1\)

Since \(n\) is an integer, \(n^2 + n + 1\) must also be an integer (let's call it \(k\)).

Thus, \(S = 12k - 1\), which is exactly 1 less than a multiple of 12.

M1: Expresses three consecutive odd integers algebraically, e.g., \(2n-1\), \(2n+1\), and \(2n+3\) (or \(2n+1\), \(2n+3\), and \(2n+5\)).
M1: Attempts to expand all three squared terms (at least two expansions correct).
A1: Correctly expands all three terms to obtain \(4n^2 - 4n + 1\), \(4n^2 + 4n + 1\), and \(4n^2 + 12n + 9\) (or equivalent depending on the choice of odd integers).
M1: Simplifies the sum to a quadratic expression, e.g., \(12n^2 + 12n + 11\) (or \(12n^2 + 36n + 35\)).
M1: Rearranges the expression to factor out 12, showing the form \(12(\dots) - 1\), e.g., \(12(n^2 + n + 1) - 1\).
A1: Completes the proof with a clear concluding statement that \(n^2 + n + 1\) is an integer, so the sum is 1 less than a multiple of 12.

Draw a line segment from \(C\), passing through the centre \(O\), and extend it to a point \(D\) on the other side of the circle.

Consider triangle \(AOC\):
- \(OA = OC\) (both are radii of the circle).
- Therefore, triangle \(AOC\) is an isosceles triangle.
- This implies that the base angles are equal: \(\angle OCA = \angle OAC = x\).
- The exterior angle of a triangle is equal to the sum of its two opposite interior angles.
- Therefore, \(\angle AOD = \angle OCA + \angle OAC = x + x = 2x\).

Consider triangle \(BOC\):
- \(OB = OC\) (both are radii of the circle).
- Therefore, triangle \(BOC\) is also an isosceles triangle.
- This implies that the base angles are equal: \(\angle OCB = \angle OBC = y\).
- Similarly, using the exterior angle theorem: \(\angle BOD = \angle OCB + \angle OBC = y + y = 2y\).

Now, combine these angles:
- The angle at the circumference is \(\angle ACB = \angle OCA + \angle OCB = x + y\).
- The angle at the centre is \(\angle AOB = \angle AOD + \angle BOD = 2x + 2y = 2(x + y)\).

Therefore, \(\angle AOB = 2\angle ACB\).

M1: States or shows the construction line \(CO\) extended to a point \(D\).
M1: Identifies that \(OA = OC\) (or \(OB = OC\)) because they are radii, and identifies triangle \(AOC\) (or \(BOC\)) as isosceles.
A1: Correctly writes \(\angle OCA = \angle OAC = x\) and \(\angle OCB = \angle OBC = y\) with the reason 'base angles of an isosceles triangle are equal'.
M1: Applies the exterior angle theorem to find \(\angle AOD = 2x\) and \(\angle BOD = 2y\) with the reason 'exterior angle of a triangle equals the sum of the two opposite interior angles'.
M1: Sets up the sum of angles: \(\angle ACB = x + y\) and \(\angle AOB = 2x + 2y\).
A1: Fully completes the algebraic steps to show \(\angle AOB = 2(x + y) = 2\angle ACB\) with all reasons clearly articulated.

The volume \(V\) of a cylinder is given by:
\(V = \pi r^2 h\)

We are given that \(V = 128\pi\), so:
\(\pi r^2 h = 128\pi \implies h = \frac{128}{r^2}\)

The total surface area \(S\) of a closed cylinder is given by:
\(S = 2\pi r^2 + 2\pi r h\)

Substitute \(h = \frac{128}{r^2}\) into the surface area equation:
\(S = 2\pi r^2 + 2\pi r \left(\frac{128}{r^2}\right)\)
\(S = 2\pi r^2 + \frac{256\pi}{r}\)

To find the value of \(r\) that minimizes the surface area, differentiate \(S\) with respect to \(r\):
\(\frac{dS}{dr} = 4\pi r - \frac{256\pi}{r^2}\)

Set the derivative equal to zero to find the stationary point:
\(4\pi r - \frac{256\pi}{r^2} = 0 \implies 4\pi r = \frac{256\pi}{r^2}\)
\(r^3 = 64 \implies r = 4\text{ cm}\)

To prove this is a minimum, we find the second derivative:
\(\frac{d^2S}{dr^2} = 4\pi + \frac{512\pi}{r^3}\)

Evaluate the second derivative at \(r = 4\):
\(\frac{d^2S}{dr^2} = 4\pi + \frac{512\pi}{64} = 4\pi + 8\pi = 12\pi > 0\)

Since the second derivative is positive, \(r = 4\) corresponds to a minimum surface area.

Now, calculate the minimum surface area by substituting \(r = 4\) back into the surface area formula:
\(S = 2\pi(4^2) + \frac{256\pi}{4}\)
\(S = 32\pi + 64\pi = 96\pi\text{ cm}^2\).

M1: Writes down the volume formula and expresses \(h\) in terms of \(r\) as \(h = \frac{128}{r^2}\).
M1: Substitutes the expression for \(h\) into the surface area formula to obtain \(S = 2\pi r^2 + \frac{256\pi}{r}\).
M1: Differentiates \(S\) with respect to \(r\) to obtain \(\frac{dS}{dr} = 4\pi r - \frac{256\pi}{r^2}\).
M1: Equates the first derivative to 0 and solves to find the critical value \(r = 4\).
A1: Finds the second derivative \(\frac{d^2S}{dr^2} = 4\pi + \frac{512\pi}{r^3}\) and evaluates it at \(r=4\) to show it is positive (confirming a minimum).
A1: Obtains the correct minimum surface area of \(96\pi\text{ cm}^2\) with clear final steps.

First, find vector expressions for \(\overrightarrow{OP}\) and \(\overrightarrow{OQ}\):
- Since \(OP : PA = 2 : 1\), \(P\) is \(\frac{2}{3}\) of the way along \(OA\), so \(\overrightarrow{OP} = \frac{2}{3}\mathbf{a}\).
- Since \(AQ : QB = 3 : 1\), \(Q\) is \(\frac{3}{4}\) of the way along \(AB\).
\( \overrightarrow{OQ} = \overrightarrow{OA} + \overrightarrow{AQ} = \mathbf{a} + \frac{3}{4}\overrightarrow{AB} = \mathbf{a} + \frac{3}{4}(\mathbf{b} - \mathbf{a}) = \frac{1}{4}\mathbf{a} + \frac{3}{4}\mathbf{b} \).

Next, express the position vector of \(X\), \(\overrightarrow{OX}\), in two different ways:

Way 1: Since \(X\) lies on the line segment \(OQ\), we can write:
\(\overrightarrow{OX} = \mu \overrightarrow{OQ} = \mu \left(\frac{1}{4}\mathbf{a} + \frac{3}{4}\mathbf{b}\right) = \frac{\mu}{4}\mathbf{a} + \frac{3\mu}{4}\mathbf{b}\), where \(\mu\) is a scalar.

Way 2: Since \(X\) lies on the line segment \(BP\), we can write:
\(\overrightarrow{OX} = \overrightarrow{OB} + \lambda \overrightarrow{BP}\), where \(\lambda\) is a scalar.
We have:
\(\overrightarrow{BP} = \overrightarrow{OP} - \overrightarrow{OB} = \frac{2}{3}\mathbf{a} - \mathbf{b}\).
Thus:
\(\overrightarrow{OX} = \mathbf{b} + \lambda \left(\frac{2}{3}\mathbf{a} - \mathbf{b}\right) = \frac{2\lambda}{3}\mathbf{a} + (1 - \lambda)\mathbf{b}\).

Now equate the coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) from the two expressions for \(\overrightarrow{OX}\):
1) For \(\mathbf{a}\): \(\frac{\mu}{4} = \frac{2\lambda}{3} \implies \mu = \frac{8\lambda}{3}\)
2) For \(\mathbf{b}\): \(\frac{3\mu}{4} = 1 - \lambda\)

Substitute the equation for \(\mu\) into the second equation:
\(\frac{3}{4}\left(\frac{8\lambda}{3}\right) = 1 - \lambda \implies 2\lambda = 1 - \lambda \implies 3\lambda = 1 \implies \lambda = \frac{1}{3}\)

Now substitute \(\lambda = \frac{1}{3}\) back to find \(\mu\):
\(\mu = \frac{8}{3}\left(\frac{1}{3}\right) = \frac{8}{9}\)

Since \(\overrightarrow{OX} = \frac{8}{9}\overrightarrow{OQ}\), the point \(X\) lies on the line segment \(OQ\) such that the ratio of \(OX : XQ\) is \(8 : 1\). This completes the proof.

M1: Finds \(\overrightarrow{OP} = \frac{2}{3}\mathbf{a}\) and expresses \(\overrightarrow{OQ} = \frac{1}{4}\mathbf{a} + \frac{3}{4}\mathbf{b}\).
M1: Formulates the first expression for \(\overrightarrow{OX}\) along the line \(OQ\): \(\overrightarrow{OX} = \mu\left(\frac{1}{4}\mathbf{a} + \frac{3}{4}\mathbf{b}\right)\).
M1: Formulates the second expression for \(\overrightarrow{OX}\) along the line \(BP\): \(\overrightarrow{OX} = \frac{2\lambda}{3}\mathbf{a} + (1-\lambda)\mathbf{b}\).
M1: Equates the coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) to create a system of simultaneous equations: \(\frac{\mu}{4} = \frac{2\lambda}{3}\) and \(\frac{3\mu}{4} = 1 - \lambda\).
A1: Solves the simultaneous equations to find \(\lambda = \frac{1}{3}\) and \(\mu = \frac{8}{9}\).
A1: Interprets \(\mu = \frac{8}{9}\) correctly to prove the ratio \(OX : XQ = 8 : 1\) with a complete and logical explanation.