2023 Edexcel IGCSE Mathematics (Specification A) 模擬試題連答案詳解

First, factorise the numerator: \(3x^2 - 14x - 5 = (3x + 1)(x - 5)\). Next, factorise the denominator: \(2x^2 - 10x = 2x(x - 5)\). Write as a single fraction: \(\frac{(3x+1)(x-5)}{2x(x-5)}\). Cancel the common factor of \((x-5)\) to get \(\frac{3x+1}{2x}\).

M1 for factorising the numerator into two linear brackets, one of which is \((x - 5)\), e.g., \((3x+1)(x-5)\). M1 for factorising the denominator, e.g., \(2x(x-5)\). A1 for the correct simplified fraction \(\frac{3x+1}{2x}\).

The area of a sector is given by \(\frac{\theta}{360} \times \pi r^2\). Here, \(\frac{120}{360} \times \pi r^2 = 48\pi \implies \frac{1}{3}r^2 = 48 \implies r^2 = 144 \implies r = 12\text{ cm}\). The arc length is \(\frac{120}{360} \times 2\pi r = \frac{1}{3} \times 24\pi = 8\pi\text{ cm}\). The perimeter of the sector is the sum of two radii and the arc length: \(\text{Perimeter} = 2r + \text{arc length} = 2(12) + 8\pi = 24 + 8\pi\text{ cm}\).

M1 for setting up the equation for the area of the sector to find the radius: \(\frac{120}{360} \pi r^2 = 48\pi\). M1 for finding \(r = 12\) and calculating the arc length: \(\frac{120}{360} \times 2 \times \pi \times 12 = 8\pi\). A1 for the correct perimeter \(24 + 8\pi\).

First, calculate the diagonal of the base using Pythagoras' theorem: \(d^2 = 8^2 + 6^2 = 64 + 36 = 100 \implies d = 10\text{ cm}\). Let \(\theta\) be the angle between the space diagonal and the base. In the right-angled triangle formed by the base diagonal, the vertical height of the cuboid, and the space diagonal, we have: \(\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{5}{10} = 0.5\). Therefore, \(\theta = \tan^{-1}(0.5) \approx 26.565^\circ\). To 1 decimal place, the angle is \(26.6^\circ\).

M1 for finding the base diagonal \(d = 10\text{ cm}\). M1 for setting up a correct trigonometric ratio, e.g., \(\tan(\theta) = \frac{5}{10}\). A1 for \(26.6\) (or \(26.6^\circ\)).

First, find the gradient of the line \(3x - 4y = 12\). Rearranging gives \(4y = 3x - 12 \implies y = \frac{3}{4}x - 3\), so the gradient is \(\frac{3}{4}\). The perpendicular gradient \(m\) satisfies \(m \times \frac{3}{4} = -1 \implies m = -\frac{4}{3}\). Using the point \((6, -1)\) in the line equation \(y - y_1 = m(x - x_1)\), we get: \(y - (-1) = -\frac{4}{3}(x - 6) \implies y + 1 = -\frac{4}{3}x + 8 \implies y = -\frac{4}{3}x + 7\). Multiplying by 3 to eliminate the fraction gives \(3y = -4x + 21 \implies 3y + 4x = 21\).

M1 for finding the gradient of the original line as \(\frac{3}{4}\) or the perpendicular line as \(-\frac{4}{3}\). M1 for a method to find the equation of the line using their perpendicular gradient and \((6, -1)\), e.g., \(y - (-1) = -\frac{4}{3}(x - 6)\). A1 for \(3y + 4x = 21\) or any equivalent integer equation (e.g. \(4x + 3y = 21\)).

Let \(c\) be the cost of a coffee and \(t\) be the cost of a tea. We write the simultaneous equations: \(3c + 2t = 11.55\) (Equation 1) and \(5c + 4t = 20.25\) (Equation 2). Multiply Equation 1 by 2 to align the \(t\) coefficients: \(6c + 4t = 23.10\) (Equation 3). Subtract Equation 2 from Equation 3: \((6c + 4t) - (5c + 4t) = 23.10 - 20.25 \implies c = 2.85\). Substitute \(c = 2.85\) back into Equation 1: \(3(2.85) + 2t = 11.55 \implies 8.55 + 2t = 11.55 \implies 2t = 3 \implies t = 1.50\). Thus, a coffee costs £2.85 and a tea costs £1.50.

M1 for setting up two correct simultaneous equations, e.g., \(3c + 2t = 11.55\) and \(5c + 4t = 20.25\). M1 for a complete method to eliminate one variable to find either \(c\) or \(t\). A1 for coffee = £2.85 and tea = £1.50 (both required).

Let the original value of the car be \(V\). After a \(12\%\) depreciation in the first year, its value is \(0.88V\). After an \(8\%\) depreciation in the second year, its value is \(0.88 \times 0.92 \times V = 0.8096V\). We are given that this final value is £16,192: \(0.8096V = 16192 \implies V = \frac{16192}{0.8096} = 20000\). Therefore, the original value of the car was £20,000.

M1 for identifying the correct multipliers, e.g., \(0.88\) and \(0.92\) (or equivalent percentage calculations). M1 for setting up the equation \(V \times 0.88 \times 0.92 = 16192\) or equivalent. A1 for \(20000\) (or £20,000).

The ratio of shares is \(3 : 5 : 8\). The difference between Beatrice's share (5 parts) and Amanda's share (3 parts) is \(5 - 3 = 2\) parts. We are given that Beatrice receives £2,400 more than Amanda, so \(2\text{ parts} = 2400 \implies 1\text{ part} = 1200\). The total number of parts is \(3 + 5 + 8 = 16\) parts. The total inheritance is \(16 \times 1200 = 19200\).

M1 for finding the difference in ratio parts, e.g., \(5 - 3 = 2\) parts. M1 for calculating the value of 1 part (\(1200\)) or finding the total number of parts (\(16\)). A1 for \(19200\) (or £19,200).

Express both 27 and 9 as powers of 3: \(27 = 3^3\) and \(9 = 3^2\). Substitute these into the expression: \(\frac{27^{x+1}}{9^{x-1}} = \frac{(3^3)^{x+1}}{(3^2)^{x-1}} = \frac{3^{3(x+1)}}{3^{2(x-1)}} = \frac{3^{3x+3}}{3^{2x-2}}\). Apply the division rule of indices, \(3^a / 3^b = 3^{a-b}\): \(\frac{3^{3x+3}}{3^{2x-2}} = 3^{(3x+3) - (2x-2)} = 3^{3x+3-2x+2} = 3^{x+5}\). Thus, the expression in the form \(3^y\) is \(3^{x+5}\).

M1 for writing \(27^{x+1}\) as \(3^{3(x+1)}\) or \(9^{x-1}\) as \(3^{2(x-1)}\) (or equivalent). M1 for applying index division laws to subtract the exponents, e.g., \(3^{(3x+3) - (2x-2)}\). A1 for \(3^{x+5}\) (or \(y = x + 5\)).

To solve \(\frac{2}{x-1} - \frac{3}{x+1} = \frac{1}{4}\), first find a common denominator for the left-hand side: \(\frac{2(x+1) - 3(x-1)}{(x-1)(x+1)} = \frac{1}{4}\). Simplify the numerator: \(\frac{2x + 2 - 3x + 3}{x^2 - 1} = \frac{1}{4}\), which gives \(\frac{-x + 5}{x^2 - 1} = \frac{1}{4}\). Multiply both sides by \(4(x^2 - 1)\) to clear the fractions: \(4(-x + 5) = x^2 - 1\). Expand the left side: \(-4x + 20 = x^2 - 1\). Rearrange the equation into standard quadratic form: \(x^2 + 4x - 21 = 0\). Factorise the quadratic expression: \((x + 7)(x - 3) = 0\). Therefore, the solutions are \(x = 3\) or \(x = -7\).

M1 for expressing LHS as a single fraction with a correct common denominator. M1 for correctly expanding and simplifying to get \(\frac{-x + 5}{x^2 - 1} = \frac{1}{4}\) or equivalent. M1 for clearing fractions and forming a three-term quadratic equation, e.g., \(x^2 + 4x - 21 = 0\). M1 for factorising the quadratic to \((x+7)(x-3) = 0\) or using the quadratic formula correctly. A1 for both correct solutions \(x = 3\) and \(x = -7\).

1. Since the radius \(OA\) is perpendicular to the tangent \(PAT\) at the point of contact, angle \(OAP = 90^\circ\). Therefore, angle \(OAD = 90^\circ - 54^\circ = 36^\circ\). 2. Since \(ABCD\) is a cyclic quadrilateral, the opposite angles sum to \(180^\circ\). Therefore, angle \(ABC = 180^\circ - 115^\circ = 65^\circ\). 3. By the alternate segment theorem, the angle between the tangent \(AP\) and chord \(AD\) is equal to the angle subtended by \(AD\) in the alternate segment. Thus, angle \(ABD = 54^\circ\). 4. Angle \(DBC = \angle ABC - \angle ABD = 65^\circ - 54^\circ = 11^\circ\). 5. Since angles subtended by the same arc \(CD\) at the circumference are equal, angle \(DAC = \angle DBC = 11^\circ\). 6. Finally, angle \(OAC = \angle OAD - \angle DAC = 36^\circ - 11^\circ = 25^\circ\).

M1 for finding \(\angle OAD = 36^\circ\) with correct reason (angle between tangent and radius is \(90^\circ\)). M1 for finding \(\angle ABC = 65^\circ\) with correct reason (opposite angles of cyclic quad sum to \(180^\circ\)). M1 for finding \(\angle ABD = 54^\circ\) with correct reason (alternate segment theorem). M1 for finding \(\angle DAC = 11^\circ\) with correct reason (angles subtended by the same arc are equal). A1 for final correct answer of \(25^\circ\) with all reasons correctly stated.

First, apply the Cosine Rule to find the length of \(BC\) in triangle \(ABC\): \(BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos(72^\circ)\). Substituting the values: \(BC^2 = 8.4^2 + 12.6^2 - 2(8.4)(12.6)\cos(72^\circ) = 70.56 + 158.76 - 211.68(0.309017) = 229.32 - 65.4127 = 163.9073\). This gives \(BC = \sqrt{163.9073} \approx 12.8026\text{ cm}\). Since \(BD:DC = 2:1\), the length of \(BD\) is \(\frac{2}{3}\) of \(BC\): \(BD = \frac{2}{3} \times 12.8026 \approx 8.5351\text{ cm}\). Now, find angle \(B\) using the Cosine Rule on triangle \(ABC\): \(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(B)\). Substituting the values: \(12.6^2 = 8.4^2 + 12.8026^2 - 2(8.4)(12.8026)\cos(B) \implies 158.76 = 70.56 + 163.9073 - 215.0837\cos(B) \implies 158.76 = 234.4673 - 215.0837\cos(B)\). This gives \(\cos(B) = \frac{234.4673 - 158.76}{215.0837} \approx 0.351988\). Finally, apply the Cosine Rule in triangle \(ABD\) to find \(AD\): \(AD^2 = AB^2 + BD^2 - 2 \cdot AB \cdot BD \cdot \cos(B)\). Substituting the values: \(AD^2 = 8.4^2 + 8.5351^2 - 2(8.4)(8.5351)(0.351988) = 70.56 + 72.8479 - 50.4715 = 92.9364\). This gives \(AD = \sqrt{92.9364} \approx 9.64\text{ cm}\) (to 3 significant figures).

M1 for correct application of Cosine Rule to find \(BC^2\) or \(BC\). A1 for \(BC \approx 12.8\text{ cm}\) (or more accurate). M1 for finding \(BD = \frac{2}{3} BC \approx 8.54\text{ cm}\) and calculating \(\cos(B)\) using Cosine Rule or Sine Rule. M1 for correct application of Cosine Rule in triangle \(ABD\) to find \(AD^2\). A1 for \(9.64\) (accept 9.63 - 9.65).

First, find the midpoint \(M\) of the line segment \(AB\): \(M = \left(\frac{2+6}{2}, \frac{-3+5}{2}\right) = (4, 1)\). Next, find the gradient of \(AB\): \(m_{AB} = \frac{5 - (-3)}{6 - 2} = \frac{8}{4} = 2\). Since line \(L\) is perpendicular to \(AB\), its gradient \(m_L\) is the negative reciprocal: \(m_L = -\frac{1}{2}\). Use the midpoint and perpendicular gradient to find the equation of line \(L\): \(y - 1 = -\frac{1}{2}(x - 4)\), which simplifies to \(y = -\frac{1}{2}x + 3\). To find coordinates of \(P\) (the \(y\)-intercept), set \(x = 0\): \(y = 3\), so \(P = (0, 3)\). To find coordinates of \(Q\) (the \(x\)-intercept), set \(y = 0\): \(0 = -\frac{1}{2}x + 3 \implies x = 6\), so \(Q = (6, 0)\). The triangle \(OPQ\) is right-angled at the origin \(O\), with perpendicular side lengths \(OP = 3\) and \(OQ = 6\). Thus, the area is \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 3 = 9\).

M1 for finding the midpoint \((4, 1)\) and gradient of \(AB = 2\). M1 for using the negative reciprocal to find the perpendicular gradient \(-1/2\). M1 for forming the correct equation of line \(L\): \(y = -1/2x + 3\). M1 for finding both intercepts \(P(0, 3)\) and \(Q(6, 0)\). A1 for the correct area of \(9\).

Rearrange the linear equation to express \(y\) in terms of \(x\): \(y = 3 - 2x\). Substitute this expression into the quadratic equation: \(x^2 - 3x(3 - 2x) + 2(3 - 2x)^2 = 12\). Expand the terms: \(x^2 - (9x - 6x^2) + 2(9 - 12x + 4x^2) = 12\), which simplifies to \(x^2 - 9x + 6x^2 + 18 - 24x + 8x^2 = 12\). Combine like terms: \(15x^2 - 33x + 18 = 12\). Subtract 12 from both sides to form a quadratic equation equal to zero: \(15x^2 - 33x + 6 = 0\). Divide the entire equation by 3: \(5x^2 - 11x + 2 = 0\). Factorise the quadratic equation: \((5x - 1)(x - 2) = 0\). This gives two possible solutions for \(x\): \(x = 0.2\) or \(x = 2\). Now, substitute these values back into the linear equation to find the corresponding \(y\) values: For \(x = 0.2\): \(y = 3 - 2(0.2) = 2.6\). For \(x = 2\): \(y = 3 - 2(2) = -1\). Thus, the solutions are \(x = 0.2, y = 2.6\) and \(x = 2, y = -1\).

M1 for expressing \(y\) in terms of \(x\) (e.g. \(y = 3 - 2x\)) or vice versa and substituting into the quadratic equation. M1 for expanding correctly to obtain a quadratic expression in one variable. M1 for reducing to a three-term quadratic equation, e.g., \(5x^2 - 11x + 2 = 0\) (or equivalent). M1 for solving the quadratic equation to find two values of \(x\) (or \(y\)). A1 for both correct pairs of solutions: \(x = 0.2, y = 2.6\) and \(x = 2, y = -1\).

According to the intersecting chords theorem, when two chords intersect inside a circle, the products of their segments are equal: \(AP \times PB = CP \times PD\). Substituting the given values: \(x(x + 5) = 6 \times 8 \implies x^2 + 5x = 48\). Rearranging gives the quadratic equation: \(x^2 + 5x - 48 = 0\). Apply the quadratic formula to solve for \(x\): \(x = \frac{-5 \pm \sqrt{5^2 - 4(1)(-48)}}{2} = \frac{-5 \pm \sqrt{25 + 192}}{2} = \frac{-5 \pm \sqrt{217}}{2}\). Since length must be positive, discard the negative root: \(x = \frac{-5 + \sqrt{217}}{2} \approx \frac{-5 + 14.73092}{2} \approx 4.86546\text{ cm}\). The length of chord \(AB\) is the sum of \(AP\) and \(PB\): \(AB = x + (x + 5) = 2x + 5\). Substitute the value of \(x\): \(AB = 2(4.86546) + 5 \approx 14.73092\text{ cm}\). Rounding to 3 significant figures gives \(14.7\text{ cm}\).

M1 for setting up the equation \(x(x + 5) = 6 \times 8\) using the intersecting chords theorem. M1 for forming the standard quadratic equation \(x^2 + 5x - 48 = 0\). M1 for solving the quadratic equation using a correct method (formula or completing the square) to find the positive root \(x \approx 4.87\). M1 for writing an expression for the total length \(AB = 2x + 5\). A1 for \(14.7\) (accept 14.7 - 14.8).

To find the angle between the line \(EC\) and the horizontal base \(ABCD\), we need to identify the angle \(ECA\), because the projection of \(EC\) onto the base is the diagonal \(AC\). First, calculate the length of the diagonal \(AC\) of the horizontal rectangle \(ABCD\) using Pythagoras' Theorem: \(AC^2 = AB^2 + BC^2 = 15^2 + 8^2 = 225 + 64 = 289 \implies AC = \sqrt{289} = 17\text{ cm}\). Since the vertex \(E\) is vertically above \(A\), the line \(EA\) is perpendicular to the base, making triangle \(EAC\) a right-angled triangle with the right angle at \(A\). Use trigonometry on triangle \(EAC\) to find angle \(ECA\): \(AE = 6\text{ cm}\) (opposite) and \(AC = 17\text{ cm}\) (adjacent). Therefore, \(\tan(\angle ECA) = \frac{AE}{AC} = \frac{6}{17}\). Thus, \(\angle ECA = \tan^{-1}\left(\frac{6}{17}\right) \approx 19.44^\circ\). Correct to 1 decimal place, the angle is \(19.4^\circ\).

M1 for identifying that the required angle is \(\angle ECA\). M1 for applying Pythagoras' Theorem on the base to find \(AC^2 = 15^2 + 8^2\). A1 for \(AC = 17\). M1 for using the tangent ratio on right-angled triangle \(EAC\): \(\tan(\theta) = 6/17\). A1 for \(19.4\) (accept 19.4 - 19.5).

To find the coordinates of the turning points, first differentiate \(y\) with respect to \(x\): \(\frac{dy}{dx} = 6x^2 - 18x - 24\). Set the derivative equal to zero to find the stationary points: \(6x^2 - 18x - 24 = 0\). Divide the entire equation by 6: \(x^2 - 3x - 4 = 0\). Factorise the quadratic equation: \((x - 4)(x + 1) = 0\), which gives critical values at \(x = 4\) and \(x = -1\). To determine which of these points is the local maximum, differentiate again to find the second derivative: \(\frac{d^2y}{dx^2} = 12x - 18\). Test both values: 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). Therefore, the local maximum occurs at \(x = -1\). Now, substitute \(x = -1\) into the original equation of \(C\) to find the corresponding \(y\)-coordinate: \(y = 2(-1)^3 - 9(-1)^2 - 24(-1) + 7 = -2 - 9 + 24 + 7 = 20\). The coordinates of the local maximum point are \((-1, 20)\).

M1 for differentiating correctly to get \(\frac{dy}{dx} = 6x^2 - 18x - 24\). M1 for setting their derivative to 0 and solving to find \(x = -1\) and \(x = 4\). M1 for using a correct method (e.g. second derivative test or sign of first derivative) to identify \(x = -1\) as the local maximum. M1 for substituting \(x = -1\) into the original equation of the curve. A1 for the correct coordinates \((-1, 20)\).

\(a)\) The area of a triangle is given by \(\frac{1}{2} a b \sin(C)\).
Here, \(\text{Area} = \frac{1}{2} (x+2)(2x-1) \sin(60^\circ)\).
Since \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\), we have:
\(\frac{1}{2} (2x^2 + 3x - 2) \left(\frac{\sqrt{3}}{2}\right) = 10\sqrt{3}\)
\(\frac{\sqrt{3}}{4} (2x^2 + 3x - 2) = 10\sqrt{3}\)
Multiply both sides by \(\frac{4}{\sqrt{3}}\):
\(2x^2 + 3x - 2 = 40\)
\(2x^2 + 3x - 42 = 0\) (as required).

\(b)\) Solve \(2x^2 + 3x - 42 = 0\) using the quadratic formula:
\(x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-42)}}{2(2)}
\)x = \frac{-3 \pm \sqrt{9 + 336}}{4} = \frac{-3 \pm \sqrt{345}}{4}\)
Since length must be positive, \(x = \frac{-3 + 18.574}{4} \approx 3.8936\).
Thus, \(AB = x + 2 = 5.8936\text{ cm}\) and \(BC = 2x - 1 = 6.7871\text{ cm}\).
Using the cosine rule to find \(AC\):
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(60^\circ)\)
\(AC^2 = (5.8936)^2 + (6.7871)^2 - 2(5.8936)(6.7871)(0.5)\)
\(AC^2 = 34.734 + 46.065 - 39.999 = 40.800\)
\(AC = \sqrt{40.800} \approx 6.39\text{ cm}\).

- M1: For setting up the area equation: \(\frac{1}{2}(x+2)(2x-1)\sin(60^\circ) = 10\sqrt{3}\)
- A1: For correctly simplifying to show \(2x^2 + 3x - 42 = 0\)
- M1: For solving the quadratic equation to find a valid positive value of \(x \approx 3.89\)
- M1: For substituting \(AB\), \(BC\), and \(\cos(60^\circ)\) into the cosine rule formula
- A1: For \(6.39\) (accept \(6.38 - 6.40\))

\(a)\) Let \(y = \frac{3}{x-2}\).
Rearrange to make \(x\) the subject:
\(y(x-2) = 3\)
\(xy - 2y = 3\)
\(xy = 3 + 2y\)
\(x = \frac{3+2y}{y}\)
So, \(f^{-1}(x) = \frac{2x+3}{x}\).

\(b)\) Find \(gf(x)\):
\(gf(x) = g\left(\frac{3}{x-2}\right) = 2\left(\frac{3}{x-2}\right) + 1 = \frac{6}{x-2} + 1 = \frac{6 + (x-2)}{x-2} = \frac{x+4}{x-2}\).
Set \(gf(x) = x\):
\(\frac{x+4}{x-2} = x\)
\(x+4 = x(x-2)\)
\(x+4 = x^2 - 2x\)
\(x^2 - 3x - 4 = 0\)
\((x-4)(x+1) = 0\)
So, \(x = 4\) or \(x = -1\).

- M1: For setting \(y = \frac{3}{x-2}\) and attempting to isolate \(x\) (e.g., \(xy - 2y = 3\))
- A1: For \(f^{-1}(x) = \frac{2x+3}{x}\) (or equivalent)
- M1: For correctly setting up the composite function equation \(\frac{6}{x-2} + 1 = x\)
- M1: For forming the quadratic equation \(x^2 - 3x - 4 = 0\)
- A1: For the solutions \(x = 4\) and \(x = -1\)

\(a)\) The area of a sector is given by \(\frac{\theta}{360} \pi R^2\).
\(\frac{120}{360} \pi R^2 = 75\pi\)
\(\frac{1}{3} R^2 = 75\)
\(R^2 = 225\)
\(R = 15\) (since \(R > 0\)).

\(b)\) Let \(C\) be the centre of the inscribed circle of radius \(r\). The line \(OC\) bisects the angle \(AOB\), so angle \(AOC = 60^\circ\).
Let \(D\) be the point of contact where the inner circle touches \(OA\). Triangle \(ODC\) is right-angled at \(D\) with angle \(DOC = 60^\circ\).
Thus, \(\sin(60^\circ) = \frac{CD}{OC} = \frac{r}{OC}\).
\(OC = \frac{r}{\sin(60^\circ)} = \frac{r}{\sqrt{3}/2} = \frac{2r}{\sqrt{3}}\).
Also, the distance from \(O\) to the outer arc through \(C\) is \(R\), so \(OC + r = R\).
\(\frac{2r}{\sqrt{3}} + r = 15\)
\(r\left(\frac{2}{\sqrt{3}} + 1\right) = 15\)
\(r(1.1547 + 1) = 15 \implies 2.1547r = 15 \implies r \approx 6.9615\text{ cm}\).
Area of the inscribed circle \(= \pi r^2 = \pi (6.9615)^2 \approx 152.25\text{ cm}^2\).
To 3 significant figures, the area is \(152\text{ cm}^2\).

- M1: For setting up the sector area equation: \(\frac{120}{360}\pi R^2 = 75\pi\)
- A1: For showing \(R = 15\)
- M1: For using trigonometry to express \(OC\) in terms of \(r\), e.g., \(OC = \frac{r}{\sin(60^\circ)}\)
- M1: For setting up the equation \(\frac{2r}{\sqrt{3}} + r = 15\) and solving for \(r\)
- A1: For \(152\) (accept \(152 - 153\))

Let \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). Since \(M\) is the midpoint of \(OA\), we have \(\overrightarrow{OM} = \frac{1}{2}\mathbf{a}\). Since \(N\) lies on \(OB\) with \(ON : NB = 2 : 1\), we have \(\overrightarrow{ON} = \frac{2}{3}\mathbf{b}\). The point \(P\) lies on the line segment \(AN\), so we can express \(\overrightarrow{AP}\) as a fraction \(\lambda\) of \(\overrightarrow{AN}\): \(\overrightarrow{AP} = \lambda \overrightarrow{AN}\). Thus, \(\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP} = \mathbf{a} + \lambda(\overrightarrow{ON} - \overrightarrow{OA}) = \mathbf{a} + \lambda(\frac{2}{3}\mathbf{b} - \mathbf{a}) = (1 - \lambda)\mathbf{a} + \frac{2}{3}\lambda \mathbf{b}\). Similarly, the point \(P\) lies on the line segment \(BM\), so we can express \(\overrightarrow{BP}\) as a fraction \(\mu\) of \(\overrightarrow{BM}\): \(\overrightarrow{BP} = \mu \overrightarrow{BM}\). Thus, \(\overrightarrow{OP} = \overrightarrow{OB} + \overrightarrow{BP} = \mathbf{b} + \mu(\overrightarrow{OM} - \overrightarrow{OB}) = \mathbf{b} + \mu(\frac{1}{2}\mathbf{a} - \mathbf{b}) = \frac{1}{2}\mu \mathbf{a} + (1 - \mu)\mathbf{b}\). Since \(\mathbf{a}\) and \(\mathbf{b}\) are non-parallel vectors, their coefficients in both expressions for \(\overrightarrow{OP}\) must be equal: 1) \(1 - \lambda = \frac{1}{2}\mu\) and 2) \(\frac{2}{3}\lambda = 1 - \mu\). From equation (2), we get \(\mu = 1 - \frac{2}{3}\lambda\). Substituting this into equation (1) gives: \(1 - \lambda = \frac{1}{2}(1 - \frac{2}{3}\lambda) \implies 1 - \lambda = \frac{1}{2} - \frac{1}{3}\lambda \implies \frac{1}{2} = \frac{2}{3}\lambda \implies \lambda = \frac{3}{4}\). Since \(\lambda = \frac{3}{4}\), we have \(\overrightarrow{AP} = \frac{3}{4}\overrightarrow{AN}\). This means that \(P\) is three-quarters of the way along the line segment \(AN\), which gives the ratio \(AP : PN = 3 : 1\). This completes the proof.

M1: States correct vector expressions for \(\overrightarrow{OM} = \frac{1}{2}\mathbf{a}\) and \(\overrightarrow{ON} = \frac{2}{3}\mathbf{b}\). M1: Writes a valid vector expression for \(\overrightarrow{OP}\) in terms of \(\mathbf{a}\), \(\mathbf{b}\), and a parameter \(\lambda\) along the line \(AN\). M1: Writes a valid vector expression for \(\overrightarrow{OP}\) in terms of \(\mathbf{a}\), \(\mathbf{b}\), and a parameter \(\mu\) along the line \(BM\). M1: Equates the coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) from both expressions to form a pair of simultaneous equations. A1: Correctly solves the simultaneous equations to find \(\lambda = \frac{3}{4}\) (or \(\mu = \frac{1}{2}\)). A1: Concludes with a clear explanation that \(\lambda = \frac{3}{4}\) means \(AP\) is \(\frac{3}{4}\) of \(AN\), hence \(AP : PN = 3 : 1\).

To find the points of intersection of the line and the circle, we substitute the equation of the line \(y = mx + c\) into the equation of the circle \(x^2 + y^2 = r^2\): \(x^2 + (mx + c)^2 = r^2\). Expanding the squared bracket gives: \(x^2 + m^2x^2 + 2mcx + c^2 = r^2\). Grouping the terms by powers of \(x\), we get a quadratic equation in \(x\): \((1 + m^2)x^2 + (2mc)x + (c^2 - r^2) = 0\). For the line to be a tangent to the circle, there must be exactly one point of intersection. This means that the quadratic equation must have exactly one real root, which requires its discriminant \(\Delta = b^2 - 4ac\) to be equal to zero. Here, \(A = 1 + m^2\), \(B = 2mc\), and \(C = c^2 - r^2\). Setting the discriminant to zero: \(\Delta = (2mc)^2 - 4(1 + m^2)(c^2 - r^2) = 0\). Expanding this expression: \(4m^2c^2 - 4(c^2 - r^2 + m^2c^2 - m^2r^2) = 0 \implies 4m^2c^2 - 4c^2 + 4r^2 - 4m^2c^2 + 4m^2r^2 = 0\). Simplifying the equation by cancelling out the term \(4m^2c^2\): \(-4c^2 + 4r^2 + 4m^2r^2 = 0 \implies -4c^2 + 4r^2(1 + m^2) = 0\). Dividing by 4 and rearranging gives: \(c^2 = r^2(1 + m^2)\). This completes the proof.

M1: Substitutes \(y = mx + c\) into \(x^2 + y^2 = r^2\). M1: Correctly expands and groups terms to form a quadratic equation in standard form \(Ax^2 + Bx + C = 0\). M1: States that for the line to be a tangent, the quadratic must have equal roots, meaning the discriminant \(\Delta = 0\). M1: Sets up the discriminant equation \((2mc)^2 - 4(1 + m^2)(c^2 - r^2) = 0\). A1: Correctly expands and simplifies the discriminant to eliminate the \(m^2c^2\) term. A1: Correctly isolates \(c^2\) to obtain the final required result \(c^2 = r^2(1 + m^2)\).

First, we draw the line segment \(PQ\) connecting the two points of intersection of the circles. Consider the quadrilateral \(APQC\). Since all four of its vertices \(A\), \(P\), \(Q\), and \(C\) lie on the circumference of the first circle, \(APQC\) is a cyclic quadrilateral. The opposite angles of a cyclic quadrilateral sum to \(180^\circ\). Let \(\angle PQC = x\). Therefore, \(\angle PAC = 180^\circ - x\). Since \(CQD\) is a straight line, the angles on a straight line sum to \(180^\circ\). Therefore, \(\angle PQD = 180^\circ - \angle PQC = 180^\circ - x\). Next, consider the quadrilateral \(PQDB\). Since all four of its vertices lie on the circumference of the second circle, \(PQDB\) is also a cyclic quadrilateral. The opposite angles of a cyclic quadrilateral sum to \(180^\circ\), which means \(\angle PBD + \angle PQD = 180^\circ\). Substituting \(\angle PQD = 180^\circ - x\) into this equation, we get: \(\angle PBD + (180^\circ - x) = 180^\circ \implies \angle PBD = x\). Now consider the line segments \(AC\) and \(BD\) intersected by the transversal line \(AB\) (since \(APB\) is a straight line). The interior angles on the same side of the transversal are \(\angle PAC\) and \(\angle PBD\). Their sum is: \(\angle PAC + \angle PBD = (180^\circ - x) + x = 180^\circ\). Since the co-interior angles sum to \(180^\circ\), the lines \(AC\) and \(BD\) must be parallel. This completes the proof.

M1: Draws or defines the line segment \(PQ\) to divide the shape into two cyclic quadrilaterals. M1: Identifies that \(APQC\) is a cyclic quadrilateral and uses the theorem that opposite angles of a cyclic quadrilateral sum to \(180^\circ\) to write \(\angle PAC = 180^\circ - \angle PQC\). M1: Uses the straight line property on \(CQD\) to state that \(\angle PQD = 180^\circ - \angle PQC\) with appropriate reason. M1: Identifies that \(PQDB\) is a cyclic quadrilateral and uses the opposite angles theorem to find \(\angle PBD = \angle PQC\). A1: Correctly establishes that \(\angle PAC + \angle PBD = 180^\circ\) with fully coherent algebraic/geometric working. A1: Correctly concludes that because the co-interior angles sum to \(180^\circ\), the line segment \(AC\) is parallel to \(BD\) (all steps must be justified with geometric reasons).

The total surface area \(S\) of a solid cylinder of radius \(r\) and height \(h\) is given by the formula: \(S = 2\pi r^2 + 2\pi r h\). Since the surface area \(S\) is a fixed constant, we can express \(h\) in terms of \(S\) and \(r\): \(2\pi r h = S - 2\pi r^2 \implies h = \frac{S - 2\pi r^2}{2\pi r}\). The volume \(V\) of a cylinder is given by: \(V = \pi r^2 h\). Substituting the expression for \(h\) into the volume formula gives: \(V = \pi r^2 \left(\frac{S - 2\pi r^2}{2\pi r}\right) = r \left(\frac{S - 2\pi r^2}{2}\right) = \frac{S}{2}r - \pi r^3\). To find the value of \(r\) that maximizes the volume, we differentiate \(V\) with respect to \(r\): \(\frac{dV}{dr} = \frac{S}{2} - 3\pi r^2\). Setting \(\frac{dV}{dr} = 0\) for a stationary point: \(\frac{S}{2} - 3\pi r^2 = 0 \implies S = 6\pi r^2\). Substituting this expression for \(S\) back into the formula for \(h\): \(h = \frac{6\pi r^2 - 2\pi r^2}{2\pi r} = \frac{4\pi r^2}{2\pi r} = 2r\). Since the diameter of the cylinder is \(d = 2r\), this shows that \(h = d\) at the stationary point. To prove that this stationary point corresponds to a maximum volume, we find the second derivative of \(V\) with respect to \(r\): \(\frac{d^2V}{dr^2} = \frac{d}{dr}\left(\frac{S}{2} - 3\pi r^2\right) = -6\pi r\). Since the radius \(r\) of a cylinder must be positive (\(r > 0\)), the second derivative is always negative (\(\frac{d^2V}{dr^2} < 0\)). A negative second derivative confirms that the volume is indeed a maximum when \(h = 2r\) (the height is equal to the diameter). This completes the proof.

M1: Recalls the correct formula for the total surface area of a cylinder \(S = 2\pi r^2 + 2\pi r h\) and rearranges to express \(h\) in terms of \(S\) and \(r\). M1: Substitutes \(h\) into the volume formula \(V = \pi r^2 h\) to obtain a volume formula in terms of \(S\) and \(r\) only. M1: Correctly differentiates \(V\) with respect to \(r\) to find \(\frac{dV}{dr} = \frac{S}{2} - 3\pi r^2\). M1: Sets \(\frac{dV}{dr} = 0\) and solves to find the relationship \(S = 6\pi r^2\) (or equivalent). A1: Substitutes \(S\) back into the expression for \(h\) to obtain \(h = 2r\), showing height equals diameter. A1: Correctly finds \(\frac{d^2V}{dr^2} = -6\pi r\) and states that because \(r > 0\), \(\frac{d^2V}{dr^2} < 0\), proving the stationary point is a maximum.

To factorise \(6x^2 - 11x - 10\), we look for two numbers that multiply to \(6 \times (-10) = -60\) and add to \(-11\). These numbers are \(-15\) and \(4\). We can split the middle term: \(6x^2 - 15x + 4x - 10\). Now, group the terms and factorise: \(3x(2x - 5) + 2(2x - 5) = (2x - 5)(3x + 2)\).

M1 for two numbers with product \(-60\) and sum \(-11\) (\(-15\) and \(4\)) or for \((2x \pm 5)(3x \pm 2)\) with incorrect signs. M1 for \(3x(2x-5) + 2(2x-5)\) or equivalent. A1 for \((2x - 5)(3x + 2)\) (or \((3x + 2)(2x - 5)\)).

Rearranging \(3x - 2y = 8\) into gradient-intercept form gives \(2y = 3x - 8\), so \(y = 1.5x - 4\). The gradient of this line is \(1.5\) (or \(\frac{3}{2}\)). The gradient of the perpendicular line is the negative reciprocal, which is \(-\frac{2}{3}\). Using the equation of a straight line with gradient \(m = -\frac{2}{3}\) and point \((6, -1)\): \(y - (-1) = -\frac{2}{3}(x - 6)\). This simplifies to \(y + 1 = -\frac{2}{3}x + 4\). Multiplying the entire equation by 3 to clear the fraction gives \(3y + 3 = -2x + 12\). Rearranging into the form \(ay + bx = c\) yields \(3y + 2x = 9\).

M1 for finding the gradient of the given line as \(1.5\) or establishing the perpendicular gradient is \(-\frac{2}{3}\). M1 for substituting \((6, -1)\) and gradient \(-\frac{2}{3}\) into \(y - y_1 = m(x - x_1)\) or \(y = mx + c\) to find \(c\). A1 for \(3y + 2x = 9\) (accept any integer equivalent, e.g., \(2x + 3y = 9\)).

Let \(d\) be the distance from the foot of the ladder to the wall. Using the cosine ratio in the right-angled triangle formed by the ladder, the ground, and the wall: \(\cos(68^\circ) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{d}{4.5}\). Solving for \(d\) gives \(d = 4.5 \times \cos(68^\circ) \approx 4.5 \times 0.374606 = 1.6857\text{ m}\). Rounding to 3 significant figures, we get \(1.69\text{ m}\).

M1 for setting up a correct trigonometric ratio, e.g., \(\cos(68^\circ) = \frac{d}{4.5}\) or \(\sin(22^\circ) = \frac{d}{4.5}\). M1 for \(4.5 \times \cos(68^\circ)\) or equivalent. A1 for \(1.69\) (accept answers in the range \(1.68\) to \(1.69\)).

Since \(TA\) and \(TB\) are tangents to the circle, the angles between the radius and the tangent at the points of contact are \(90^\circ\), so angle \(OAT = 90^\circ\) and angle \(OBT = 90^\circ\). The sum of angles in the quadrilateral \(AOBT\) is \(360^\circ\). Therefore, the obtuse angle \(AOB = 360^\circ - 90^\circ - 90^\circ - 44^\circ = 136^\circ\). The reflex angle \(AOB\) is \(360^\circ - 136^\circ = 224^\circ\).

M1 for identifying that angle \(OAT = 90^\circ\) or angle \(OBT = 90^\circ\). M1 for finding the obtuse angle \(AOB = 180^\circ - 44^\circ = 136^\circ\). A1 for \(224\).

To find \(fg(4)\), we first evaluate \(g(4)\). Substituting \(x = 4\) into \(g(x)\) gives \(g(4) = 2(4) - 3 = 8 - 3 = 5\). Next, we substitute this result into \(f(x)\) to find \(f(5)\): \(f(5) = \frac{4(5)+7}{5-2} = \frac{20+7}{3} = \frac{27}{3} = 9\). Thus, \(fg(4) = 9\).

M1 for evaluating \(g(4) = 2(4) - 3 = 5\). M1 for substituting their value of \(g(4)\) into the expression for \(f(x)\), i.e., \(\frac{4(5)+7}{5-2}\). A1 for 9.

First, express the terms in the denominator with the same power of 10 to add them: \(1.6 \times 10^{-3} + 0.4 \times 10^{-3} = 2.0 \times 10^{-3}\). Now perform the division: \(\frac{6.4 \times 10^7}{2.0 \times 10^{-3}} = \frac{6.4}{2.0} \times 10^{7 - (-3)} = 3.2 \times 10^{10}\).

M1 for simplifying the denominator to \(2.0 \times 10^{-3}\) or \(0.002\). M1 for \(\frac{6.4 \times 10^7}{\text{their } 2.0 \times 10^{-3}}\) or for showing subtraction of indices \(7 - (-3)\). A1 for \(3.2 \times 10^{10}\).

The sale price of $323 represents \(100\% - 15\% = 85\%\) of the normal price. Let \(x\) be the normal price. Then \(0.85x = 323\). Solving for \(x\) gives \(x = \frac{323}{0.85} = 380\). The normal price of the bicycle was $380.

M1 for identifying that $323 represents \(85\%\) of the normal price, or writing \(0.85\). M1 for \(323 \div 0.85\) or setting up \(0.85x = 323\). A1 for 380.

Multiply both sides of the equation by \(5 - w\) to remove the fraction: \(p(5 - w) = 3w + 2\). Expand the bracket on the left side: \(5p - pw = 3w + 2\). Collect all terms containing \(w\) on one side and other terms on the opposite side: \(5p - 2 = 3w + pw\). Factorise out \(w\) on the right side: \(5p - 2 = w(3 + p)\). Finally, divide by \(3 + p\) to isolate \(w\): \(w = \frac{5p - 2}{3 + p}\).

M1 for multiplying both sides by \(5 - w\) to obtain \(p(5 - w) = 3w + 2\). M1 for expanding brackets and gathering all terms in \(w\) on one side, e.g., \(5p - 2 = 3w + pw\). A1 for \(w = \frac{5p - 2}{3 + p}\) or equivalent.

We know that the arc length of a sector is given by:
\(L = \frac{\theta}{360} \times 2\pi r\)
where \(\theta\) is the angle subtended at the centre and \(r\) is the radius.

Substituting the given values:
\(18 = \frac{\theta}{360} \times 2\pi \times 15\)
\(18 = \frac{\theta}{360} \times 30\pi\)

So we can express the fraction of the circle as:
\\frac{\\theta}{360} = \\frac{18}{30\\pi} = \\frac{3}{5\\pi}\

The area of the sector is given by:
\(A = \frac{\theta}{360} \times \pi r^2\)

Substituting the fraction and \(r = 15\):
\(A = \frac{3}{5\pi} \times \pi \times 15^2\)
\(A = \frac{3}{5} \times 225 = 135\)

Therefore, the area of the sector is \(135\text{ cm}^2\).

M1: For a correct expression for the arc length, e.g., \\frac{\\theta}{360} \\times 2 \\times \\pi \\times 15 = 18\\ or finding the angle \(\theta \approx 68.75^{\circ}\).
M1: For substituting their angle or fraction into the sector area formula, e.g., \\frac{18}{30\\pi} \\times \\pi \\times 15^2\\ or \\frac{68.75}{360} \\times \\pi \\times 15^2\\.
A1: For 135.

First, factorise the quadratic expression in the numerator by finding two numbers that multiply to \(2 \times (-3) = -6\) and add to \(-5\). These numbers are \(-6\) and \(1\):
\(2x^2 - 5x - 3 = 2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3)\)

Next, factorise the denominator, which is a difference of two squares:
\(x^2 - 9 = (x + 3)(x - 3)\)

Now write the original fraction with both numerator and denominator in their factorised forms:
\\frac{2x^2 - 5x - 3}{x^2 - 9} = \\frac{(2x + 1)(x - 3)}{(x + 3)(x - 3)}\

Cancel the common factor of \((x - 3)\) from both the numerator and the denominator to get the fully simplified expression:
\\frac{2x + 1}{x + 3}

M1: For factorising the numerator as \((2x + 1)(x - 3)\) (or correct factorisation of a 3-term quadratic).
M1: For factorising the denominator as \((x + 3)(x - 3)\).
A1: For \\frac{2x + 1}{x + 3} or equivalent fraction.

Multiply both sides by \((x-2)(x+1)\) to clear the fractions:
\(4(x+1) + 3(x-2) = (x-2)(x+1)\)

Expand both sides:
\(4x + 4 + 3x - 6 = x^2 - x - 2\)

Simplify:
\(7x - 2 = x^2 - x - 2\)

Rearrange into standard quadratic form:
\(x^2 - 8x = 0\)

Factorise the quadratic equation:
\(x(x-8) = 0\)

This gives two solutions:
\(x = 0\) or \(x = 8\).

- M1: For multiplying by the common denominator to get \(4(x+1) + 3(x-2) = (x-2)(x+1)\) or equivalent.
- M1: For expanding and simplifying to a quadratic equation of the form \(x^2 - 8x = 0\) or equivalent.
- M1: For factorising \(x(x-8) = 0\) or using the quadratic formula.
- A1: For both correct solutions \(x = 0\) and \(x = 8\).

First, find the intersection point of \(2x + 3y = 11\) and \(3x - y = 11\).
From the second equation, write \(y = 3x - 11\).

Substitute this into the first equation:
\(2x + 3(3x - 11) = 11\)
\(2x + 9x - 33 = 11\)
\(11x = 44 \Rightarrow x = 4\)

Substitute \(x = 4\) back into \(y = 3x - 11\):
\(y = 3(4) - 11 = 1\)
So the intersection point is \((4, 1)\).

Since the required line is parallel to \(y = 2x + 7\), its gradient is \(m = 2\).

Use the equation of a line formula \(y - y_1 = m(x - x_1)\):
\(y - 1 = 2(x - 4)\)
\(y - 1 = 2x - 8\)
\(y = 2x - 7\).

- M1: For a method to solve the simultaneous equations to find either \(x\) or \(y\).
- A1: For finding the correct intersection point \((4, 1)\).
- M1: For using the parallel gradient \(m = 2\) in \(y - y_1 = m(x - x_1)\) or \(y = mx + c\) with their intersection point.
- A1: For the correct final equation \(y = 2x - 7\).

Since \(A\), \(B\), \(C\), and \(D\) lie on the circumference in order, \(ABCD\) is a cyclic quadrilateral.

The opposite angles of a cyclic quadrilateral sum to \(180^\circ\).
Therefore, angle \(DAB = 180^\circ - 125^\circ = 55^\circ\).

Since \(OA\) and \(OD\) are both radii of the circle, triangle \(AOD\) is an isosceles triangle with \(OA = OD\).

The base angles of an isosceles triangle are equal, so angle \(ADO = \text{angle } DAB = 55^\circ\).

The angles in a triangle sum to \(180^\circ\).
Therefore, angle \(AOD = 180^\circ - 55^\circ - 55^\circ = 70^\circ\).

- M1: For calculating angle \(DAB = 55^\circ\) (or stating opposite angles of a cyclic quad sum to \(180^\circ\)).
- M1: For identifying that \(OA = OD\) (radii of the circle) and hence triangle \(AOD\) is isosceles.
- M1: For stating angle \(ADO = 55^\circ\) (or using base angles of an isosceles triangle are equal).
- A1: For \(70^\circ\).

First, find the length of the diagonal \(AC\) of the square base \(ABCD\).
Using Pythagoras' theorem on triangle \(ABC\):
\(AC^2 = 10^2 + 10^2 = 200 \Rightarrow AC = \sqrt{200} = 10\sqrt{2} \approx 14.142\text{ cm}\).

Since \(O\) is the center of the base, the distance \(AO\) is half the diagonal \(AC\):
\(AO = \frac{1}{2} \times 10\sqrt{2} = 5\sqrt{2} \approx 7.071\text{ cm}\).

In the right-angled triangle \(VOA\), the angle between the line \(VA\) and the base \(ABCD\) is angle \(VAO\).
Using trigonometry:
\(\tan(\text{angle } VAO) = \frac{VO}{AO} = \frac{12}{5\sqrt{2}}\)
\(\text{angle } VAO = \tan^{-1}\left(\frac{12}{5\sqrt{2}}\right) \approx \tan^{-1}(1.697) \approx 59.489^\circ\).

To 1 decimal place, the angle is \(59.5^\circ\).

- M1: For using Pythagoras' theorem to find \(AC\) or \(AO\) (e.g., \(AC = \sqrt{10^2+10^2}\)).
- A1: For \(AO = 5\sqrt{2}\) (or \(\sqrt{50}\) or \(7.07...\)).
- M1: For using a correct trigonometric ratio for angle \(VAO\), e.g., \(\tan(\theta) = \frac{12}{\text{their } AO}\).
- A1: For \(59.5\) (accept answers in range \(59.4\) to \(59.6\)).

First, factorise each of the quadratic expressions:
1) \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\)
2) \(2x^2 - 18 = 2(x^2 - 9) = 2(x - 3)(x + 3)\)
3) \(x^2 + 5x + 6 = (x + 2)(x + 3)\)

Now substitute these factorised forms back into the division expression and change the division to multiplication by the reciprocal:
\(\frac{(2x+1)(x-3)}{2(x-3)(x+3)} \times \frac{(x+2)(x+3)}{2x+1}\)

Cancel out common terms in the numerator and denominator:
- Cancel \((x-3)\) from top and bottom
- Cancel \((2x+1)\) from top and bottom
- Cancel \((x+3)\) from top and bottom

This leaves:
\(\frac{x+2}{2}\).

- M1: For factorising \(2x^2 - 5x - 3\) to \((2x+1)(x-3)\) or \(2x^2 - 18\) to \(2(x-3)(x+3)\).
- M1: For factorising \(x^2 + 5x + 6\) to \((x+2)(x+3)\).
- M1: For inverting the second fraction and multiplying, showing cancellation of at least two common factors.
- A1: For \(\frac{x+2}{2}\) or equivalent (e.g. \(0.5x + 1\)).

The total number of counters in the bag is \(n + 6\).

The probability of choosing a blue counter on the first draw is \(\frac{6}{n+6}\).
Since the first counter is not replaced, there are now 5 blue counters left and the total number of counters is \(n + 5\).

The probability of choosing a blue counter on the second draw is \(\frac{5}{n+5}\).

The probability that both counters are blue is:
\(\frac{6}{n+6} \times \frac{5}{n+5} = \frac{1}{3}\)
\(\frac{30}{(n+6)(n+5)} = \frac{1}{3}\)

Multiply both sides by \(3(n+6)(n+5)\):
\(90 = (n+6)(n+5)\)

Expand and simplify the equation:
\(90 = n^2 + 11n + 30\)
\(n^2 + 11n - 60 = 0\)

Factorise the quadratic equation:
\((n + 15)(n - 4) = 0\)

Since the number of counters \(n\) must be a positive integer, \(n = 4\).

- M1: For writing a correct expression for the probability of selecting two blue counters, e.g., \(\frac{6}{n+6} \times \frac{5}{n+5}\).
- M1: For setting up the equation \(\frac{30}{(n+6)(n+5)} = \frac{1}{3}\) and showing steps to remove fractions.
- M1: For forming a correct quadratic equation, e.g., \(n^2 + 11n - 60 = 0\).
- A1: For \(4\) (rejecting \(-15\)).

First, find the expression for the composite function \(fg(x)\) by substituting \(g(x)\) into \(f(x)\):
\(fg(x) = f(2x - 5) = \frac{3(2x - 5) + 2}{(2x - 5) - 1}\)

Simplify the numerator and denominator:
\(fg(x) = \frac{6x - 15 + 2}{2x - 6} = \frac{6x - 13}{2x - 6}\)

Now, set \(fg(x) = 4\) and solve for \(x\):
\(\frac{6x - 13}{2x - 6} = 4\)
\(6x - 13 = 4(2x - 6)\)
\(6x - 13 = 8x - 24\)
\(24 - 13 = 8x - 6x\)
\(11 = 2x\)
\(x = \frac{11}{2} = 5.5\).

- M1: For attempting to find \(fg(x)\) by substituting \(2x-5\) into \(f(x)\), or writing \(f(g(x)) = 4\).
- M1: For a correct equation \(\frac{3(2x-5) + 2}{(2x-5)-1} = 4\) or equivalent.
- M1: For expanding and rearranging to a linear equation, e.g., \(6x - 13 = 8x - 24\) or equivalent.
- A1: For \(5.5\) (or \(\frac{11}{2}\)).

First, find the velocity \(v\) by differentiating the displacement \(s\) with respect to \(t\):
\(v = \frac{ds}{dt} = 6t^2 - 30t + 24\)

Set the velocity to \(0\) to find the times when the particle is instantaneously at rest:
\(6t^2 - 30t + 24 = 0\)

Divide by 6:
\(t^2 - 5t + 4 = 0\)

Factorise the quadratic equation:
\((t - 1)(t - 4) = 0\)

So the velocity is \(0\) at \(t = 1\) second and \(t = 4\) seconds.
The velocity is *first* equal to \(0\) at \(t = 1\) second.

Next, find the acceleration \(a\) by differentiating the velocity \(v\) with respect to \(t\):
\(a = \frac{dv}{dt} = 12t - 30\)

Substitute \(t = 1\) into the acceleration formula:
\(a = 12(1) - 30 = -18\text{ m/s}^2\).

- M1: For differentiating \(s\) to find \(v = 6t^2 - 30t + 24\).
- M1: For setting their \(v = 0\) and solving to find \(t = 1\) (and \(t = 4\)).
- M1: For differentiating their \(v\) to find \(a = 12t - 30\) or equivalent.
- A1: For \(-18\).

First, find the midpoint \(M\) of the line segment \(AB\): \(M = \left(\frac{-2+4}{2}, \frac{5+17}{2}\right) = (1, 11)\). Next, calculate the gradient of \(L_1\): \(m_1 = \frac{17 - 5}{4 - (-2)} = \frac{12}{6} = 2\). Since \(L_2\) is perpendicular to \(L_1\), its gradient \(m_2\) is the negative reciprocal of \(m_1\): \(m_2 = -\frac{1}{2}\). Now, use the point-slope formula with the midpoint \((1, 11)\) and gradient \(-\frac{1}{2}\) to find the equation of \(L_2\): \(y - 11 = -\frac{1}{2}(x - 1)\). Multiply both sides by 2: \(2(y - 11) = -(x - 1)\) which gives \(2y - 22 = -x + 1\). Rearranging this into the required form: \(x + 2y = 23\).

M1: for a correct method to find the midpoint of \(AB\) (giving \((1, 11)\)) or the gradient of \(L_1\) (giving \(2\)).
M1: for a correct method to find the perpendicular gradient (giving \(-\frac{1}{2}\)).
M1: for substituting their midpoint and perpendicular gradient into a linear equation formula, e.g., \(y - 11 = -\frac{1}{2}(x - 1)\).
A1: for the correct final equation \(x + 2y = 23\) (or any equivalent integer form, e.g., \(2x + 4y = 46\)).

Multiply all terms by the common denominator \((x+2)(x-1)\) to eliminate the fractions: \(8(x-1) + 5(x+2) = 2(x+2)(x-1)\). Expand the terms on both sides: \(8x - 8 + 5x + 10 = 2(x^2 + x - 2)\). Simplify both sides: \(13x + 2 = 2x^2 + 2x - 4\). Rearrange into standard quadratic form: \(2x^2 - 11x - 6 = 0\). Factorise the quadratic equation: \((2x + 1)(x - 6) = 0\). Solve for \(x\): \(2x + 1 = 0 \implies x = -0.5\) and \(x - 6 = 0 \implies x = 6\).

M1: for a correct method to eliminate fractions, yielding \(8(x-1) + 5(x+2) = 2(x+2)(x-1)\) or equivalent.
M1: for expanding and simplifying to a correct 3-term quadratic equation, e.g., \(2x^2 - 11x - 6 = 0\).
M1: for a complete and correct method to solve their 3-term quadratic equation (by factorisation, completing the square, or quadratic formula).
A1: for both correct solutions: \(x = 6\) and \(x = -0.5\) (or \(x = -\frac{1}{2}\)).

The area \(A\) of a sector with radius \(12\) and angle \(\theta\) is: \(A = \frac{\theta}{360} \times \pi \times 12^2 = \frac{144\pi\theta}{360} = \frac{2\pi\theta}{5}\). The perimeter \(P\) of the sector consists of the arc length plus two radii: \(P = 2(12) + \frac{\theta}{360} \times 2\pi \times 12 = 24 + \frac{24\pi\theta}{360} = 24 + \frac{\pi\theta}{15}\). Since \(A = 1.5P\), we substitute the expressions into the equation: \(\frac{2\pi\theta}{5} = 1.5\left(24 + \frac{\pi\theta}{15}\right)\). Expand the right-hand side: \(\frac{2\pi\theta}{5} = 36 + \frac{1.5\pi\theta}{15} \implies \frac{2\pi\theta}{5} = 36 + \frac{\pi\theta}{10}\). Multiply the entire equation by 10 to clear fractions: \(4\pi\theta = 360 + \pi\theta\). Subtract \(\pi\theta\) from both sides: \(3\pi\theta = 360 \implies \pi\theta = 120\). Solve for \(\theta\): \(\theta = \frac{120}{\pi} \approx 38.197...\). Correct to 1 decimal place, \(\theta = 38.2\).

M1: for a correct expression for the area \(A\) or the arc length in terms of \(\theta\).
M1: for a correct expression for the perimeter \(P\), including both radii: \(24 + \frac{24\pi\theta}{360}\).
M1: for setting up the equation \(A = 1.5P\) and showing a complete algebraic method to isolate \(\theta\) (e.g., reaching \(3\pi\theta = 360\) or \(\theta = \frac{120}{\pi}\)).
A1: for \(38.2\) (accept answers in the range \(38.1\) to \(38.3\)).

Let \(\vec{OA} = \mathbf{a}\) and \(\vec{OB} = \mathbf{b}\).

Since \(M\) is the midpoint of \(OA\), \(\vec{OM} = \frac{1}{2}\mathbf{a}\).
Since \(ON : NB = 2 : 1\), \(\vec{ON} = \frac{2}{3}\mathbf{b}\).

Let \(\vec{AP} = \lambda \vec{AN}\). Since \(\vec{AN} = \vec{AO} + \vec{ON} = -\mathbf{a} + \frac{2}{3}\mathbf{b}\), we have:

\(\vec{OP} = \vec{OA} + \vec{AP} = \mathbf{a} + \lambda(-\mathbf{a} + \frac{2}{3}\mathbf{b}) = (1-\lambda)\mathbf{a} + \frac{2}{3}\lambda\mathbf{b}\) (Equation 1)

Now, let \(\vec{BP} = \mu \vec{BM}\). Since \(\vec{BM} = \vec{BO} + \vec{OM} = -\mathbf{b} + \frac{1}{2}\mathbf{a}\), we have:

\(\vec{OP} = \vec{OB} + \vec{BP} = \mathbf{b} + \mu(-\mathbf{b} + \frac{1}{2}\mathbf{a}) = \frac{1}{2}\mu\mathbf{a} + (1-\mu)\mathbf{b}\) (Equation 2)

Since \(\mathbf{a}\) and \(\mathbf{b}\) are non-parallel vectors, we can equate the coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) in Equation 1 and Equation 2:

1) \(1 - \lambda = \frac{1}{2}\mu\)
2) \(\frac{2}{3}\lambda = 1 - \mu\)

From (1), we get \(\mu = 2 - 2\lambda\).
Substituting this into (2):

\(\frac{2}{3}\lambda = 1 - (2 - 2\lambda)\)

\(\frac{2}{3}\lambda = 2\lambda - 1\)

\(1 = 2\lambda - \frac{2}{3}\lambda\)

\(1 = \frac{4}{3}\lambda \implies \lambda = \frac{3}{4}\)

Since \(\lambda = \frac{3}{4}\), we have \(\vec{AP} = \frac{3}{4}\vec{AN}\).
This means that \(P\) lies \(\frac{3}{4}\) of the way along the line segment \(AN\), leaving \(PN\) as the remaining \(\frac{1}{4}\) of \(AN\).
Therefore, the ratio is \(AP : PN = 3 : 1\).

M1: Finds expressions for \(\vec{OM}\) and \(\vec{ON}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).
M1: Formulates a vector expression for \(\vec{OP}\) along the line \(AN\) using a parameter (e.g., \(\lambda\)).
M1: Formulates a vector expression for \(\vec{OP}\) along the line \(BM\) using a second parameter (e.g., \(\mu\)).
M1: Equates coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) to set up a system of two simultaneous equations.
M1: Solves the equations to find the value of \(\lambda\) or \(\mu\) (e.g., \(\lambda = 3/4\)).
A1: Concludes logically with proof showing that \(AP : PN = 3 : 1\).

Consider triangles \(APB\) and \(DPC\).

1. \(\angle APB = \angle DPC\) because they are vertically opposite angles.
2. \(\angle ABP = \angle DCP\) because they are angles subtended at the circumference by the same arc, \(AD\).
3. \(\angle BAP = \angle CDP\) because they are angles subtended at the circumference by the same arc, \(BC\).

Since all three corresponding angles are equal, triangle \(APB\) is similar to triangle \(DPC\) (by Angle-Angle-Angle similarity).

Because triangle \(APB\) is similar to triangle \(DPC\), the ratios of their corresponding sides must be equal:

\(\frac{AP}{DP} = \frac{BP}{CP}\)

Multiplying both sides of the equation by \(DP \times CP\), we get:

\(AP \times CP = BP \times DP\)

Since the length of the line segments \(CP = PC\) and \(DP = PD\), we can rewrite this as:

\(AP \times PC = BP \times PD\)

This completes the proof.

M1: Identifies \(\angle APB = \angle DPC\) with the reason 'vertically opposite angles'.
M1: Identifies \(\angle ABP = \angle DCP\) or \(\angle BAP = \angle CDP\) with the reason 'angles subtended by the same arc/segment at the circumference are equal'.
M1: Identifies a second pair of equal angles subtended by the same arc with the correct reason.
A1: Concludes that triangle \(APB\) is similar to triangle \(DPC\) citing AAA (Angle-Angle-Angle).
M1: Uses similarity to write a ratio equation of corresponding sides, e.g., \(\frac{AP}{DP} = \frac{BP}{CP}\).
A1: Multiplies out and concludes with the required identity: \(AP \times PC = BP \times PD\).

Let the two consecutive positive integers be \(n\) and \(n+1\), where \(n\) is a positive integer.

The difference between their cubes is given by:
\((n+1)^3 - n^3\)

We expand \((n+1)^3\) using the binomial expansion or brackets multiplication:
\((n+1)^3 = n^3 + 3n^2 + 3n + 1\)

Now, subtract \(n^3\):
\((n^3 + 3n^2 + 3n + 1) - n^3 = 3n^2 + 3n + 1\)

Factorising \(3\) out of the first two terms gives:

\(3n(n+1) + 1\)

Now, let us consider the term \(n(n+1)\). Since \(n\) and \(n+1\) are consecutive integers, one of them must be an even number. Therefore, their product \(n(n+1)\) must always be an even number.

We can write \(n(n+1) = 2k\) for some integer \(k \ge 1\).

Substituting this back into our expression gives:

\(3(2k) + 1 = 6k + 1\)

Since \(k\) is an integer, \(6k\) is a multiple of 6. Therefore, \(6k + 1\) is always 1 more than a multiple of 6. This completes the proof.

M1: Defines consecutive integers algebraically (e.g., \(n\) and \(n+1\)).
M1: Formulates the difference of their cubes: \((n+1)^3 - n^3\).
M1: Expands \((n+1)^3\) correctly to get \(n^3 + 3n^2 + 3n + 1\).
A1: Simplifies the expression to \(3n^2 + 3n + 1\).
M1: Factorises to get \(3n(n+1) + 1\) and states that \(n(n+1)\) must be even because it is the product of consecutive integers.
A1: Expresses the final form as \(6k + 1\) (or equivalent explanation) and clearly concludes that it is 1 more than a multiple of 6.

Let the cylinder have radius \(r\) and height \(h\).

The formulas for total surface area \(S\) and volume \(V\) are:
\(S = 2\pi r^2 + 2\pi rh\) (1)
\(V =
\pi r^2 h\) (2)

From (1), express \(h\) in terms of \(S\) and \(r\):
\(2\pi rh = S - 2\pi r^2 \implies h = \frac{S - 2\pi r^2}{2\pi r}\) (3)

Substitute (3) into (2) to write \(V\) as a function of \(r\):
\(V = \pi r^2 \left(\frac{S - 2\pi r^2}{2\pi r}\right) = \frac{r(S - 2\pi r^2)}{2} = \frac{Sr}{2} - \pi r^3\)

Differentiate \(V\) with respect to \(r\):
\(\frac{dV}{dr} = \frac{S}{2} - 3\pi r^2\)

Set \(\frac{dV}{dr} = 0\) to find the stationary point for maximum volume:
\(\frac{S}{2} - 3\pi r^2 = 0 \implies 3\pi r^2 = \frac{S}{2} \implies S = 6\pi r^2\)

To verify it is a maximum, differentiate again:
\(\frac{d^2V}{dr^2} = -6\pi r\). Since \(r > 0\), \(\frac{d^2V}{dr^2} < 0\), which confirms a maximum.

Now, substitute \(S = 6\pi r^2\) back into the expression for \(h\) (Equation 3):
\(h = \frac{6\pi r^2 - 2\pi r^2}{2\pi r} = \frac{4\pi r^2}{2\pi r} = 2r\).
This proves that the height is equal to the diameter (\(h = 2r\)) at maximum volume.

To find the maximum volume \(V\), substitute \(r = \sqrt{\frac{S}{6\pi}}\) and \(h = 2r\) into the volume formula:
\(V = 2\pi r^3 = 2\pi \left(\sqrt{\frac{S}{6\pi}}\right)^3\)

Simplify the expression:
\(V = 2\pi \left(\frac{S}{6\pi}\right) \sqrt{\frac{S}{6\pi}} = \frac{S}{3} \sqrt{\frac{S}{6\pi}}\)

Bring \(\frac{S}{3}\) inside the square root:
\(V = \sqrt{\frac{S^2}{9} \times \frac{S}{6\pi}} = \sqrt{\frac{S^3}{54\pi}}\).

This completes the proof.

M1: States correct formulas for \(S\) and \(V\) in terms of \(r\) and \(h\).
M1: Rearranges surface area to find \(h\) in terms of \(S, r\) and substitutes to get \(V\) solely in terms of \(r\).
M1: Correctly differentiates \(V\) with respect to \(r\) to find \(\frac{dV}{dr}\).
M1: Sets \(\frac{dV}{dr} = 0\) and solves to get \(S = 6\pi r^2\) or \(r = \sqrt{\frac{S}{6\pi}}\) (and shows it is a maximum using the second derivative).
A1: Proves that \(h = 2r\) by substituting \(S = 6\pi r^2\) back into the equation for \(h\).
A1: Shows through rigorous algebraic steps that \(V\) simplifies to \(\sqrt{\frac{S^3}{54\pi}}\).