MetadataPastPaper.sampleTitle

First, factorise the numerator by taking out the common factor of 2, giving 2(x^2 - 4), which is a difference of two squares and factorises to 2(x - 2)(x + 2). Next, factorise the quadratic in the denominator, which has a product of 6 and sum of -5, giving (x - 2)(x - 3). Write the expression as the fraction of these factored forms: \(\frac{2(x-2)(x+2)}{(x-2)(x-3)}\). Cancel the common factor of (x - 2) from the numerator and denominator to get the final simplified expression \(\frac{2x+4}{x-3}\).

M1 for factorising the numerator to 2(x-2)(x+2). M1 for factorising the denominator to (x-2)(x-3). A0.5 for the final simplified fraction \(\frac{2x+4}{x-3}\) or \(\frac{2(x+2)}{x-3}\).

First, calculate the gradient m of the line using the formula m = (y2 - y1) / (x2 - x1). This gives m = (-5 - 7) / (4 - (-2)) = -12 / 6 = -2. Next, use the equation y = mx + c with the gradient m = -2 and substitute the coordinates of one point, for example (-2, 7): 7 = -2(-2) + c. This simplifies to 7 = 4 + c, which gives c = 3. Therefore, the equation of the line is y = -2x + 3.

M1 for finding the gradient m = -2. M1 for substituting one of the points into y = -2x + c to solve for c. A0.5 for the correct equation y = -2x + 3.

The perimeter of a sector consists of the arc length plus two radii. The arc length is calculated using the formula: Arc Length = (theta / 360) * 2 * pi * r. Substituting theta = 140 and r = 9 gives Arc Length = (140 / 360) * 2 * pi * 9 = (7 / 18) * 18 * pi = 7*pi. Adding the two straight edges (each equal to the radius of 9 cm): Perimeter = 7*pi + 9 + 9 = 7*pi + 18 cm.

M1 for correct method to find the arc length: (140 / 360) * 2 * pi * 9. M1 for adding 2 * 9 to their arc length. A0.5 for the exact answer 7*pi + 18.

First, find the midpoint of the line segment PQ: Midpoint = ((1 + 5)/2, (3 + 11)/2) = (3, 7). Next, find the gradient of the line PQ: Gradient of PQ = (11 - 3)/(5 - 1) = 8/4 = 2. The gradient of the perpendicular line is the negative reciprocal: m = -1/2 = -0.5. Now, find the equation of the line passing through the midpoint (3, 7) with gradient -0.5: y - 7 = -0.5(x - 3) which simplifies to y = -0.5x + 1.5 + 7, so y = -0.5x + 8.5.

M1 for finding the midpoint (3, 7). M1 for finding the perpendicular gradient of -0.5. A0.5 for the correct equation y = -0.5x + 8.5 or y = -1/2 x + 17/2.

To find the coordinates of the turning point, we can complete the square for the expression x^2 - 6x + 14. This gives (x - 3)^2 - 9 + 14 = (x - 3)^2 + 5. The minimum value of the expression is 5, which occurs when x - 3 = 0, so when x = 3. Therefore, the turning point has coordinates (3, 5).

M1 for completing the square to get (x - 3)^2 + 5 or using x = -b/(2a) to find x = 3. M1 for substituting x = 3 to find y = 5. A0.5 for the coordinates (3, 5).

Express both 27 and 9 as powers of 3: 27 = 3^3 and 9 = 3^2. Substitute these into the equation: (3^3)^(2x - 1) = (3^2)^(x + 4). Using the laws of indices, multiply the powers: 3^(3(2x - 1)) = 3^(2(x + 4)) which gives 3^(6x - 3) = 3^(2x + 8). Since the bases are the same, equate the indices: 6x - 3 = 2x + 8. Solve for x: 4x = 11, which gives x = 11/4 = 2.75.

M1 for writing both sides with base 3: 3^(6x - 3) = 3^(2x + 8). M1 for equating the exponents and forming a correct linear equation. A0.5 for x = 2.75 or 11/4.

The formula for the area of a trapezium is Area = 0.5 * (a + b) * h. Substitute the given values: 34 = 0.5 * ((x + 3) + (2x - 1)) * 4. Simplify the equation: 34 = 2 * (3x + 2). Divide both sides by 2: 17 = 3x + 2. Subtract 2 from both sides: 15 = 3x. Dividing by 3 gives x = 5.

M1 for setting up the equation 0.5 * (x + 3 + 2x - 1) * 4 = 34. M1 for simplifying to 3x + 2 = 17 or equivalent. A0.5 for x = 5.

Multiply both sides by (2 - u) to clear the fraction: w(2 - u) = 3u + 5. Expand the left side: 2w - uw = 3u + 5. Rearrange to group all terms containing u on one side: 2w - 5 = 3u + uw. Factorise the right side to isolate u: 2w - 5 = u(3 + w). Divide both sides by (3 + w) to find u: u = (2w - 5) / (w + 3).

M1 for multiplying both sides by (2 - u) and expanding. M1 for collecting terms in u on one side and factorising. A0.5 for the final correct expression u = (2w - 5)/(w + 3) or equivalent.

First, factor out the common factor of 3 to get \(3(4x^2 - y^2)\). Then, factorise the difference of two squares within the parentheses: \(4x^2 - y^2 = (2x - y)(2x + y)\). Combining these, the completely factorised expression is \(3(2x - y)(2x + y)\).

M1 for factorising out 3 to get \(3(4x^2 - y^2)\) (or equivalent partial factorisation), A1.5 for the fully factorised form: \(3(2x - y)(2x + y)\) (or equivalent).

Factorise the numerator: \(2a^2 - 8a = 2a(a - 4)\). Factorise the denominator using the difference of two squares: \(a^2 - 16 = (a - 4)(a + 4)\). Substituting these back into the fraction gives \(\frac{2a(a - 4)}{(a - 4)(a + 4)}\). Cancelling the common factor of \((a - 4)\) from the numerator and denominator leaves the simplified fraction \(\frac{2a}{a + 4}\).

M1 for factorising the numerator to \(2a(a - 4)\), M1 for factorising the denominator to \((a - 4)(a + 4)\), A0.5 for the fully simplified answer.

The gradient of the given line is \(4\). The gradient of the perpendicular line, \(m\), is the negative reciprocal: \(m = -\frac{1}{4}\). Using the point-slope formula with the point \((8, 2)\): \(y - 2 = -\frac{1}{4}(x - 8)\). Simplifying this gives \(y - 2 = -\frac{1}{4}x + 2\), which simplifies to \(y = -\frac{1}{4}x + 4\) or \(y = -0.25x + 4\).

M1 for finding the perpendicular gradient of \(-\frac{1}{4}\) (or \(-0.25\)), M1 for substituting \((8, 2)\) into \(y = mx + c\) with their gradient to find \(c\), A0.5 for the correct final equation.

First, find the midpoint of \(PQ\): \(M = \left(\frac{-3 + 5}{2}, \frac{7 + (-1)}{2}\right) = (1, 3)\). Next, find the gradient of \(PQ\): \(m_{PQ} = \frac{-1 - 7}{5 - (-3)} = \frac{-8}{8} = -1\). The gradient of the perpendicular bisector is the negative reciprocal of \(-1\), which is \(m = 1\). Using the midpoint \((1, 3)\) and gradient \(1\): \(y - 3 = 1(x - 1)\), which simplifies to \(y = x + 2\).

M1 for finding the correct midpoint \((1, 3)\), M1 for finding the perpendicular gradient of \(1\), A0.5 for the correct final equation \(y = x + 2\).

The formula for the arc length of a sector is \(L = \frac{\theta}{360} \times 2\pi r\). Substituting the given values: \(L = \frac{150}{360} \times 2\pi(12) = \frac{5}{12} \times 24\pi = 10\pi\text{ cm}\). The perimeter of the sector includes the arc length plus two radii: \(P = L + 2r = 10\pi + 2(12) = 10\pi + 24\text{ cm}\).

M1 for calculating the correct arc length of \(10\pi\), M1 for adding two radii (i.e., \(24\)) to their arc length, A0.5 for the correct final expression \(10\pi + 24\).

The area of a sector is given by \(A = \frac{\theta}{360} \times \pi r^2\). Substituting the known values gives: \(24\pi = \frac{60}{360} \times \pi r^2\). Simplify the fraction: \(24\pi = \frac{1}{6}\pi r^2\). Divide both sides by \(\pi\): \(24 = \frac{1}{6}r^2\). Multiply both sides by \(6\): \(r^2 = 144\). Taking the positive square root for the radius gives \(r = 12\).

M1 for setting up the correct equation \(24\pi = \frac{60}{360} \pi r^2\), M1 for simplifying to \(r^2 = 144\), A0.5 for obtaining \(r = 12\).

First, substitute the point \((0, 5)\) into the function: \(5 = a \cdot 2^0 + b\). Since \(2^0 = 1\), this simplifies to \(a + b = 5\) (Equation 1). Next, substitute the point \((2, 17)\) into the function: \(17 = a \cdot 2^2 + b\), which simplifies to \(4a + b = 17\) (Equation 2). Subtract Equation 1 from Equation 2: \((4a + b) - (a + b) = 17 - 5 \implies 3a = 12 \implies a = 4\). Substitute \(a = 4\) back into Equation 1 to find \(b\): \(4 + b = 5 \implies b = 1\). Therefore, \(a = 4\) and \(b = 1\).

M1 for forming the equation \(a + b = 5\), M1 for forming the equation \(4a + b = 17\), A0.5 for solving the simultaneous equations to get \(a = 4\) and \(b = 1\).

(a) First, we factor out the common factor of 3:
\(12x^2 - 3y^2 = 3(4x^2 - y^2)\).
Next, recognise that \(4x^2 - y^2\) is a difference of two squares, which factorises to \((2x - y)(2x + y)\).
Thus, the fully factorised expression is:
\(3(2x - y)(2x + y)\).

(b) First, factorise the numerator \(2x^2 - 5x - 3\):
Find two numbers that multiply to \(2 \times (-3) = -6\) and add to \(-5\). These are \(-6\) and \(1\).
\(2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3)\).
Next, factorise the denominator \(4x^2 - 1\) as a difference of two squares:
\(4x^2 - 1 = (2x - 1)(2x + 1)\).
Now, simplify the fraction by cancelling the common factor \((2x + 1)\):
\(\frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)} = \frac{x - 3}{2x - 1}\).

(a) [2 marks]
- M1 for writing \(3(4x^2 - y^2)\) or for partial factorisation such as \((6x - 3y)(2x + y)\).
- A1 for \(3(2x - y)(2x + y)\) as the final answer.

(b) [3 marks]
- M1 for factorising the numerator correctly to \((2x + 1)(x - 3)\).
- M1 for factorising the denominator correctly to \((2x - 1)(2x + 1)\).
- A1 for \(\frac{x - 3}{2x - 1}\) as the final answer.

(a) First, find the gradient (\(m_1\)) of line \(L_1\):
\(m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - 5}{6 - (-2)} = \frac{-6}{8} = -\frac{3}{4}\).
Now, use the line equation \(y - y_1 = m(x - x_1)\) with the point \((-2, 5)\):
\(y - 5 = -\frac{3}{4}(x - (-2))\)
\(y - 5 = -\frac{3}{4}x - \frac{3}{2}\)
\(y = -\frac{3}{4}x + \frac{7}{2}\) (or \(y = -0.75x + 3.5\)).

(b) The gradient of the perpendicular line \(L_2\) is the negative reciprocal of \(m_1\):
\(m_2 = -\frac{1}{m_1} = -\frac{1}{-\frac{3}{4}} = \frac{4}{3}\).
Using the point \((3, 7)\) and gradient \(m_2 = \frac{4}{3}\):
\(y - 7 = \frac{4}{3}(x - 3)\)
\(y - 7 = \frac{4}{3}x - 4\)
\(y = \frac{4}{3}x + 3\).

(a) [2 marks]
- M1 for finding the gradient of \(L_1\): \(m = \frac{-1 - 5}{6 - (-2)}\) (or equivalent).
- A1 for \(y = -\frac{3}{4}x + \frac{7}{2}\) or any equivalent form with \(y\) as the subject, e.g., \(y = -0.75x + 3.5\).

(b) [2 marks]
- M1 for using the perpendicular gradient rule \(m_2 = -\frac{1}{m_1}\) to find \(m_2 = \frac{4}{3}\) (or follow through their gradient from part (a)).
- A1 for \(y = \frac{4}{3}x + 3\) (or equivalent).

(a) The formula for arc length \(s\) is:
\(s = \frac{\theta}{360} \times 2\pi r\)
Substituting \(\theta = 105^\circ\) and \(r = 12\text{ cm}\):
\(s = \frac{105}{360} \times 2\pi \times 12\)
\(s = \frac{105}{30} \times \pi = \frac{7}{2} \times 2\pi = 7\pi\text{ cm}\).

(b) The total perimeter of the sector consists of the arc length plus the two straight edges (radii):
\(P = \text{arc length} + 2r\)
\(P = 7\pi + 2(12) = 7\pi + 24\)
Using \(\pi \approx 3.14159265\):
\(P = 7(3.14159265) + 24 = 21.991 + 24 = 45.991\text{ cm}\).
Correct to 1 decimal place, the perimeter is \(46.0\text{ cm}\).

(a) [2 marks]
- M1 for \(\frac{105}{360} \times 2\pi \times 12\) (or equivalent expression).
- A1 for \(7\pi\).

(b) [2 marks]
- M1 for adding \(2r\) (which is \(24\)) to their arc length.
- A1 for \(46.0\) (must be correct to 1 decimal place; do not accept \(46\) without the decimal place).

(a) To find the points of intersection, set the equation of the curve equal to the equation of the line:
\(x^2 - 4x - 5 = 2x + 11\)
Subtract \(2x\) and \(11\) from both sides to form a quadratic equation equal to zero:
\(x^2 - 4x - 2x - 5 - 11 = 0\)
\(x^2 - 6x - 16 = 0\) (as required).

(b) Solve the quadratic equation \(x^2 - 6x - 16 = 0\) by factorising:
Look for two numbers that multiply to \(-16\) and add to \(-6\). These are \(-8\) and \(2\).
\((x - 8)(x + 2) = 0\)
So, \(x = 8\) or \(x = -2\).
Now substitute these values back into the equation of the line \(y = 2x + 11\) (or the curve) to find the corresponding \(y\)-coordinates:
For \(x = -2\):
\(y = 2(-2) + 11 = -4 + 11 = 7\).
So, one point is \((-2, 7)\).
For \(x = 8\):
\(y = 2(8) + 11 = 16 + 11 = 27\).
So, the other point is \((8, 27)\).
Therefore, the coordinates of \(A\) and \(B\) are \((-2, 7)\) and \((8, 27)\).

(a) [2 marks]
- M1 for setting the expressions equal: \(x^2 - 4x - 5 = 2x + 11\).
- A1 for correctly simplifying and rearranging to \(x^2 - 6x - 16 = 0\) with no errors shown.

(b) [3 marks]
- M1 for factorising correctly to \((x - 8)(x + 2) = 0\) or correct use of the quadratic formula.
- A1 for \(x = 8\) and \(x = -2\).
- A1 for both coordinate points: \((-2, 7)\) and \((8, 27)\).

(a) Start by multiplying both sides of the equation by the denominator \(t + 2\):
\(w(t + 2) = 3t - 5\)
Expand the left-hand side:
\(wt + 2w = 3t - 5\)
Rearrange the equation to group all terms containing \(t\) on one side and the rest on the other:
\(2w + 5 = 3t - wt\)
Factor out \(t\) on the right-hand side:
\(2w + 5 = t(3 - w)\)
Divide both sides by \(3 - w\) to isolate \(t\):
\(t = \frac{2w + 5}{3 - w}\) (or equivalently, \(t = \frac{-2w - 5}{w - 3}\)).

(b) Substitute \(w = 4\) into the rearranged formula:
\(t = \frac{2(4) + 5}{3 - 4} = \frac{8 + 5}{-1} = \frac{13}{-1} = -13\).

(a) [3 marks]
- M1 for clearing the fraction to obtain \(w(t + 2) = 3t - 5\) or \(wt + 2w = 3t - 5\).
- M1 for separating the \(t\) terms on one side: e.g., \(3t - wt = 2w + 5\) or equivalent.
- A1 for the final answer \(t = \frac{2w + 5}{3 - w}\) (or equivalent form like \(t = \frac{-2w - 5}{w - 3}\)).

(b) [1 mark]
- B1 for \(-13\) (or correct follow through from their equation in part (a)).

(a) The midpoint formula is \(\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\).
Using \(P(1, 3)\) and \(Q(5, 11)\):
\(M = \left(\frac{1 + 5}{2}, \frac{3 + 11}{2}\right) = \left(\frac{6}{2}, \frac{14}{2}\right) = (3, 7)\).

(b) First, find the gradient of the line segment \(PQ\):
\(m_{PQ} = \frac{11 - 3}{5 - 1} = \frac{8}{4} = 2\).
The perpendicular bisector will have a gradient that is the negative reciprocal of \(2\):
\(m_{\perp} = -\frac{1}{2}\).
The perpendicular bisector passes through the midpoint \(M(3, 7)\). Using the line equation formula \(y - y_1 = m(x - x_1)\):
\(y - 7 = -\frac{1}{2}(x - 3)\)
\(y - 7 = -\frac{1}{2}x + \frac{3}{2}\)
Add 7 to both sides:
\(y = -\frac{1}{2}x + \frac{17}{2}\) (or \(y = -0.5x + 8.5\)).

(a) [1 mark]
- B1 for \((3, 7)\).

(b) [4 marks]
- M1 for finding the gradient of \(PQ = 2\).
- M1 for finding the gradient of the perpendicular line \(m = -\frac{1}{2}\) (or negative reciprocal of their gradient of \(PQ\)).
- M1 for substituting their midpoint and their perpendicular gradient into a straight line equation, e.g., \(y - 7 = -\frac{1}{2}(x - 3)\).
- A1 for \(y = -\frac{1}{2}x + \frac{17}{2}\) (or equivalent form like \(y = -0.5x + 8.5\) or \(x + 2y = 17\)).

(a) The area of a complete circle of radius \(r\) is given by the formula \(\pi r^2\).
A sector with a central angle of \(\theta^\circ\) represents a fraction of \(\frac{\theta}{360}\) of the complete circle.
Therefore, the area \(A\) of the sector is:
\(A = \frac{\theta}{360} \times \pi r^2 = \frac{\theta \pi r^2}{360}\) (as required).

(b) We are given \(A = 15\pi\) and \(r = 6\). Substitute these values into the formula from part (a):
\(15\pi = \frac{\theta \pi (6)^2}{360}\)
\(15\pi = \frac{36\theta \pi}{360}\)
Divide both sides by \(\pi\):
\(15 = \frac{36\theta}{360}\)
Simplify the fraction \(\frac{36}{360}\) to \(\frac{1}{10}\):
\(15 = \frac{\theta}{10}\)
Multiply both sides by 10:
\(theta = 150\).

(a) [1 mark]
- B1 for explaining or showing that the sector is the fraction \(\frac{\theta}{360}\) of the full area of a circle \(\pi r^2\).

(b) [3 marks]
- M1 for substituting the given values into the formula: \(15\pi = \frac{\theta \pi (6)^2}{360}\).
- M1 for isolating \(\theta\) correctly, e.g., \(15 = 0.1\theta\) or \(36\theta = 5400\).
- A1 for \(150\).

(a) Area = Length \(\times\) Width
\[\text{Area} = \frac{2x^2 + 5x - 3}{x^2 - 9} \times \frac{x^2 - 3x}{4x^2 - 1}\]
Factorise each expression:
- \(2x^2 + 5x - 3 = (2x - 1)(x + 3)\)
- \(x^2 - 9 = (x - 3)(x + 3)\)
- \(x^2 - 3x = x(x - 3)\)
- \(4x^2 - 1 = (2x - 1)(2x + 1)\)

Substitute back into the product:
\[\text{Area} = \frac{(2x - 1)(x + 3)}{(x - 3)(x + 3)} \times \frac{x(x - 3)}{(2x - 1)(2x + 1)}\]
Cancel common terms \((x + 3)\), \((x - 3)\), and \((2x - 1)\):
\[\text{Area} = \frac{x}{2x + 1}\]

(b)(i)
\[P = \frac{3a - 2}{b} + \frac{5b}{b} = \frac{3a - 2 + 5b}{b}\]

(b)(ii)
\[P - 5 = \frac{3a - 2}{b}\]
\[b(P - 5) = 3a - 2\]
\[b = \frac{3a - 2}{P - 5}\]

(c)
\[\frac{2}{y - 3} - \frac{1}{y} = \frac{1}{2}\]
Multiply the entire equation by the common denominator \(2y(y - 3)\):
\[2(2)(y) - 2(y - 3) = y(y - 3)\]
\[4y - 2y + 6 = y^2 - 3y\]
\[2y + 6 = y^2 - 3y\]
\[y^2 - 5y - 6 = 0\]
Factorise:
\[(y - 6)(y + 1) = 0\]
So, \(y = 6\) or \(y = -1\).

(a) [4 marks]:
M1 for factorising \(2x^2 + 5x - 3\) to \((2x-1)(x+3)\)
M1 for factorising \(x^2-9\) to \((x-3)(x+3)\)
M1 for factorising \(x^2-3x\) and \(4x^2-1\) to \(x(x-3)\) and \((2x-1)(2x+1)\)
A1 for fully correct simplification showing all cancellations.

(b)(i) [2 marks]:
M1 for finding a common denominator of \(b\).
A1 for \(\frac{3a - 2 + 5b}{b}\).

(b)(ii) [3 marks]:
M1 for isolating the term with \(b\), i.e., \(P - 5 = \frac{3a - 2}{b}\)
M1 for multiplying by \(b\) to get \(b(P - 5) = 3a - 2\)
A1 for \(b = \frac{3a - 2}{P - 5}\).

(c) [3 marks]:
M1 for multiplying by \(2y(y - 3)\) to eliminate denominators.
M1 for forming the quadratic equation \(y^2 - 5y - 6 = 0\).
A1 for both correct solutions \(y = 6\) and \(y = -1\).

(a) Gradient of \(AB\), \(m = \frac{17 - 5}{4 - (-2)} = \frac{12}{6} = 2\).
Using the point-slope formula with \(A(-2, 5)\):
\[y - 5 = 2(x - (-2))\]
\[y - 5 = 2x + 4 \implies y = 2x + 9\]

(b)(i) Midpoint of \(AB\) is:
\[M = \left(\frac{-2 + 4}{2}, \frac{5 + 17}{2}\right) = (1, 11)\]

(b)(ii) Since \(L\) is perpendicular to \(AB\), its gradient is the negative reciprocal of \(2\), which is \(-\frac{1}{2} = -0.5\).
Using the midpoint \((1, 11)\):
\[y - 11 = -0.5(x - 1)\]
\[y - 11 = -0.5x + 0.5 \implies y = -0.5x + 11.5\]

(c) The line parallel to \(AB\) has gradient \(m = 2\).
Since it passes through \((0, -3)\), its y-intercept is \(-3\). The equation of this parallel line is:
\[y = 2x - 3\]
Substitute the coordinates of \(C(6, k)\) into this equation:
\[k = 2(6) - 3\]
\[k = 12 - 3 = 9\]

(a) [3 marks]:
M1 for calculating the gradient \(m = 2\).
M1 for substituting \(m\) and a point into \(y = mx + c\) or equivalent.
A1 for \(y = 2x + 9\).

(b)(i) [2 marks]:
M1 for attempting the midpoint formula with correct coordinates.
A1 for \((1, 11)\).

(b)(ii) [3 marks]:
M1 for stating perpendicular gradient is \(-0.5\) (or negative reciprocal of their gradient from (a)).
M1 for substituting their midpoint and perpendicular gradient into line equation.
A1 for \(y = -0.5x + 11.5\) or \(y = -\frac{1}{2}x + \frac{23}{2}\).

(c) [4 marks]:
M1 for identifying the gradient of the parallel line is 2.
M1 for writing the equation of the parallel line as \(y = 2x - 3\).
M1 for substituting \(x = 6\) and \(y = k\) into their parallel line equation.
A1 for \(k = 9\).

(a) Area of sector \(OAB\):
\[\text{Area} = \frac{120}{360} \times \pi \times r^2 = \frac{1}{3} \times \pi \times 8^2 = \frac{64\pi}{3} \approx 67.02 \approx 67.0\text{ cm}^2\]

(b) In right-angled triangle \(OBC\), \(OB\) is the adjacent side to angle \(BOC\), and \(OC\) is the hypotenuse.
\[\cos(\angle BOC) = \frac{OB}{OC} = \frac{8}{15}\]
\[\angle BOC = \cos^{-1}\left(\frac{8}{15}\right) \approx 57.769^\circ \approx 57.8^\circ\]

(c) The outer boundary consists of:
- Arc \(AB = \frac{120}{360} \times 2 \times \pi \times 8 = \frac{16\pi}{3} \approx 16.755\text{ cm}\).
- Radius \(OA = 8\text{ cm}\).
- Hypotenuse \(OC = 15\text{ cm}\).
- Side \(BC\): using Pythagoras' theorem,
\[BC = \sqrt{OC^2 - OB^2} = \sqrt{15^2 - 8^2} = \sqrt{225 - 64} = \sqrt{161} \approx 12.689\text{ cm}\]
\[\text{Total Perimeter} = \text{Arc } AB + OA + OC + BC = 16.755 + 8 + 15 + 12.689 = 52.444 \approx 52.4\text{ cm}\]

(d) Area of the segment \(AB\) is the area of the sector \(OAB\) minus the area of the triangle \(OAB\).
\[\text{Area of Triangle } OAB = \frac{1}{2} a b \sin\theta = \frac{1}{2} \times 8 \times 8 \times \sin(120^\circ) = 32 \times \frac{\sqrt{3}}{2} = 16\sqrt{3} \approx 27.713\text{ cm}^2\]
\[\text{Area of Segment} = 67.021 - 27.713 = 39.308 \approx 39.3\text{ cm}^2\]

(a) [2 marks]:
M1 for \(\frac{120}{360} \times \pi \times 8^2\).
A1 for \(67.0\) or \(67.02\).

(b) [3 marks]:
M1 for identifying the correct trigonometric ratio: \(\cos(\theta) = \frac{8}{15}\).
M1 for \(\theta = \cos^{-1}\left(\frac{8}{15}\right)\).
A1 for \(57.8^\circ\) or \(57.769^\circ\).

(c) [4 marks]:
M1 for calculating arc length \(AB \approx 16.76\).
M1 for calculating \(BC = \sqrt{15^2 - 8^2} \approx 12.69\).
M1 for adding \(\text{Arc } AB + OA + OC + BC\).
A1 for \(52.4\) or \(52.44\).

(d) [3 marks]:
M1 for \(\frac{1}{2} \times 8 \times 8 \times \sin(120^\circ)\) or equivalent to find triangle area.
M1 for subtracting triangle area from sector area.
A1 for \(39.3\) or \(39.31\).

(a) Let us calculate each value:
- For \(x = -4\): \(f(-4) = \frac{12}{-4} + (-4)^2 - 6 = -3 + 16 - 6 = 7\), so \(a = 7\).
- For \(x = -3\): \(f(-3) = \frac{12}{-3} + (-3)^2 - 6 = -4 + 9 - 6 = -1\), so \(b = -1\).
- For \(x = -2\): \(f(-2) = \frac{12}{-2} + (-2)^2 - 6 = -6 + 4 - 6 = -8\), so \(c = -8\).
- For \(x = 4\): \(f(4) = \frac{12}{4} + 4^2 - 6 = 3 + 16 - 6 = 13\), so \(d = 13\).

(b) As \(x\) approaches 0 from the positive side, the term \(\frac{12}{x}\) becomes extremely large and positive, while \(x^2 - 6\) approaches \(-6\). Thus, \(f(x) \to \infty\) (or the graph rises infinitely up along the y-axis).

(c)(i) We want to solve:
\[\frac{12}{x} + x^2 - x - 8 = 0\]
Rewrite this to include the expression for \(f(x)\):
\[\left(\frac{12}{x} + x^2 - 6\right) - x - 2 = 0\]
\[y - x - 2 = 0 \implies y = x + 2\]
Thus, the equation of the line is \(y = x + 2\).

(c)(ii) Starting with:
\[\frac{12}{x} + x^2 - x - 8 = 0\]
Multiply the entire equation by \(x\) (since \(x \neq 0\)):
\[12 + x^3 - x^2 - 8x = 0\]
Rearranging the terms gives:
\[x^3 - x^2 - 8x + 12 = 0\]

(c)(iii) To show \(x = 2\) is a solution:
\[2^3 - 2^2 - 8(2) + 12 = 8 - 4 - 16 + 12 = 0\]
This is correct, so \(x = 2\) is a solution.
To find the other solution, we factorise the cubic expression using polynomial division by \(x - 2\):
\[(x^3 - x^2 - 8x + 12) \div (x - 2) = x^2 + x - 6\]
Factorise \(x^2 + x - 6\):
\[x^2 + x - 6 = (x - 2)(x + 3)\]
Thus, the cubic equation is:
\[(x - 2)^2(x + 3) = 0\]
The other distinct solution is \(x = -3\).

(a) [3 marks]:
B2 for 3 correct values, B1 for 2 correct values.
B3 for all 4 values correct (a = 7, b = -1, c = -8, d = 13).

(b) [1 mark]:
B1 for stating that the graph approaches infinity / grows very large / vertical asymptote.

(c)(i) [3 marks]:
M1 for attempting to write the equation in terms of \(y\).
M1 for isolating \(y\) to get \(y = x + c\).
A1 for \(y = x + 2\).

(c)(ii) [2 marks]:
M1 for multiplying the algebraic equation by \(x\).
A1 for obtaining the correct cubic expression shown.

(c)(iii) [3 marks]:
B1 for substituting \(x = 2\) into the equation and showing it equals 0.
M1 for factoring out \((x - 2)\) to get \(x^2 + x - 6\) or showing division.
A1 for finding the other solution \(x = -3\).

(a)(i) Time taken by the express train is \(\frac{180}{x}\) hours.

(a)(ii) Speed of regional train is \(x - 30\) km/h.
Time taken by the regional train is \(\frac{180}{x - 30}\) hours.

(b)(i) 1 hour and 12 minutes \(= 1\frac{12}{60} = 1.2\) hours.
The regional train is slower, so it takes longer:
\[\frac{180}{x - 30} - \frac{180}{x} = 1.2\]
Divide the entire equation by \(1.2\):
\[\frac{150}{x - 30} - \frac{150}{x} = 1\]
Multiply by \(x(x - 30)\) to eliminate the denominators:
\[150x - 150(x - 30) = x(x - 30)\]
\[150x - 150x + 4500 = x^2 - 30x\]
\[4500 = x^2 - 30x \implies x^2 - 30x - 4500 = 0\]

(b)(ii) Solve by factorisation:
\[(x - 90)(x + 50) = 0\]
Thus, \(x = 90\) or \(x = -50\).

(b)(iii) Since speed must be positive, \(x = 90\) km/h (speed of the express train).
The speed of the regional train is \(90 - 30 = 60\) km/h.
Time taken by the regional train is:
\[t = \frac{180}{60} = 3\text{ hours}\]
This is exactly 3 hours 0 minutes.

(a)(i) [1 mark]:
B1 for \(\frac{180}{x}\).

(a)(ii) [1 mark]:
B1 for \(\frac{180}{x - 30}\).

(b)(i) [4 marks]:
M1 for converting 1h 12m to 1.2h (or \(\frac{6}{5}\)h).
M1 for establishing the correct difference equation: \(\frac{180}{x-30} - \frac{180}{x} = 1.2\).
M1 for multiplying through to remove algebraic denominators.
A1 for correctly simplifying to the given equation \(x^2 - 30x - 4500 = 0\).

(b)(ii) [3 marks]:
M2 for factorising \((x-90)(x+50)\) or using the quadratic formula with correct substitution.
A1 for \(x = 90\) and \(x = -50\).

(b)(iii) [3 marks]:
M1 for selecting \(x = 90\) and calculating the regional speed \(90 - 30 = 60\).
M1 for dividing 180 by their regional speed.
A1 for 3 hours 0 minutes.

(a) Gradient of \(L_1\) is:
\[m = \frac{-6 - 10}{5 - (-3)} = \frac{-16}{8} = -2\]

(b) Using \(P(-3, 10)\) and gradient \(-2\):
\[y - 10 = -2(x - (-3))\]
\[y - 10 = -2x - 6 \implies y = -2x + 4\]

(c)(i) Midpoint of \(PQ\) is:
\[M = \left(\frac{-3 + 5}{2}, \frac{10 + (-6)}{2}\right) = (1, 2)\]

(c)(ii) Gradient of the perpendicular bisector \(L_2\) is the negative reciprocal of \(-2\), which is \(\frac{1}{2} = 0.5\).
Using the midpoint \((1, 2)\):
\[y - 2 = 0.5(x - 1)\]
\[y - 2 = 0.5x - 0.5 \implies y = 0.5x + 1.5\]

(d) The line \(L_2\) crosses the y-axis at point \(R\). Substituting \(x = 0\) into \(y = 0.5x + 1.5\) gives \(R(0, 1.5)\).
Since \(L_2\) is the perpendicular bisector of the base \(PQ\), the point \(R\) is equidistant from \(P\) and \(Q\), and the perpendicular distance from \(R\) to the base \(PQ\) is the distance between \(R\) and the midpoint \(M(1, 2)\).
Height of the triangle, \(h = RM\):
\[RM = \sqrt{(1 - 0)^2 + (2 - 1.5)^2} = \sqrt{1 + 0.25} = \sqrt{1.25} = \sqrt{\frac{5}{4}}\]
Length of the base, \(b = PQ\):
\[PQ = \sqrt{(5 - (-3))^2 + (-6 - 10)^2} = \sqrt{8^2 + (-16)^2} = \sqrt{64 + 256} = \sqrt{320} = \sqrt{64 \times 5} = 8\sqrt{5}\]
Area of the triangle \(PQR\):
\[\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8\sqrt{5} \times \sqrt{\frac{5}{4}} = 4\sqrt{5} \times \frac{\sqrt{5}}{2} = 2 \times 5 = 10\]

(a) [2 marks]:
M1 for attempt at rise over run: \(\frac{-6 - 10}{5 - (-3)}\).
A1 for \(-2\).

(b) [2 marks]:
M1 for substituting \(m = -2\) and point \(P\) or \(Q\) into equation of a line.
A1 for \(y = -2x + 4\).

(c)(i) [2 marks]:
M1 for using midpoint formula correctly.
A1 for \((1, 2)\).

(c)(ii) [3 marks]:
M1 for finding gradient of perpendicular line \(= 0.5\).
M1 for using their midpoint and perpendicular gradient to form line equation.
A1 for \(y = 0.5x + 1.5\) or \(y = \frac{1}{2}x + \frac{3}{2}\).

(d) [3 marks]:
M1 for finding the coordinates of \(R(0, 1.5)\).
M1 for calculating the length of \(PQ\) or finding perpendicular distance from \(R\) to \(PQ\), or using shoelace method.
A1 for Area \(= 10\).

(a) Arc length formula is \(s = \frac{\theta}{360} \times 2\pi r\).
Substitute \(s = 15\pi\) and \(r = 45\):
\[15\pi = \frac{\theta}{360} \times 2\pi(45)\]
\[15\pi = \frac{90\theta\pi}{360} = \frac{\theta\pi}{4}\]
Divide both sides by \(\pi\):
\[15 = \frac{\theta}{4} \implies \theta = 60\]

(b) Area of the sector:
\[\text{Area} = \frac{\theta}{360} \times \pi r^2 = \frac{60}{360} \times \pi \times 45^2 = \frac{1}{6} \times 2025\pi = 337.5\pi \approx 1060.29 \approx 1060\text{ cm}^2\]

(c)(i)
\[V = \frac{4}{3}\pi r^3\]
\[180 = \frac{4}{3}\pi r^3\]
\[r^3 = \frac{180 \times 3}{4\pi} = \frac{135}{\pi} \approx 42.9718\]
\[r = \sqrt[3]{42.9718} \approx 3.5025\text{ cm}\]
To 1 decimal place, the radius is \(3.5\) cm.

(c)(ii) \(\text{Mass} = \text{Volume} \times \text{Density} = 180 \times 8.5 = 1530\text{ g}\).
Convert grams to kilograms:
\[\text{Mass} = \frac{1530}{1000} = 1.53\text{ kg}\]

(d) At the maximum displacement of \(30^\circ\), the vertical distance of the bob from the pivot is:
\[d = 45 \cos(30^\circ) = 45 \times 0.8660 = 38.97\text{ cm}\]
The vertical height \(h\) it rises is the difference between the full pendulum length (at the lowest point) and this vertical distance:
\[h = 45 - 45 \cos(30^\circ) = 45(1 - \cos(30^\circ)) \approx 45(1 - 0.86603) = 45(0.13397) \approx 6.03\text{ cm}\]

(a) [3 marks]:
M1 for \(15\pi = \frac{\theta}{360} \times 2\pi \times 45\) or equivalent.
M1 for algebraic rearrangement to isolate \(\theta\).
A1 for reaching \(\theta = 60\) showing full steps.

(b) [2 marks]:
M1 for \(\frac{60}{360} \times \pi \times 45^2\).
A1 for \(1060\) or \(1060.3\).

(c)(i) [3 marks]:
M1 for setting up the volume equation \(180 = \frac{4}{3}\pi r^3\).
M1 for rearranging to \(r^3 = \frac{135}{\pi}\) or equivalent.
A1 for \(3.5\) (do not accept 3.50).

(c)(ii) [2 marks]:
M1 for calculating \(180 \times 8.5\).
A1 for \(1.53\) kg.

(d) [2 marks]:
M1 for \(45(1 - \cos 30^\circ)\).
A1 for \(6.03\) or \(6.029\).

(a) To find the stationary points, we find the first derivative \(g'(x)\) and set it to 0:
\[g'(x) = 3x^2 - 6x - 9\]
Set \(g'(x) = 0\):
\[3x^2 - 6x - 9 = 0\]
Divide by 3:
\[x^2 - 2x - 3 = 0\]
Factorise:
\[(x - 3)(x + 1) = 0 \implies x = 3\text{ or } x = -1\]
Now find the corresponding y-coordinates:
- For \(x = 3\): \(g(3) = 3^3 - 3(3)^2 - 9(3) + 15 = 27 - 27 - 27 + 15 = -12\). So the point is \((3, -12)\).
- For \(x = -1\): \(g(-1) = (-1)^3 - 3(-1)^2 - 9(-1) + 15 = -1 - 3 + 9 + 15 = 20\). So the point is \((-1, 20)\).

(b) Use the second derivative \(g''(x)\) to test the nature of the stationary points:
\[g''(x) = 6x - 6\]
- For \(x = -1\): \(g''(-1) = 6(-1) - 6 = -12 < 0\). Since the second derivative is negative, \((-1, 20)\) is a local maximum.
- For \(x = 3\): \(g''(3) = 6(3) - 6 = 12 > 0\). Since the second derivative is positive, \((3, -12)\) is a local minimum.

(c)(i) Since \(k = 25\) is strictly greater than the local maximum value of 20, the line \(y = 25\) only intersects the curve once. Therefore, there is 1 real solution.

(c)(ii) Since \(k = 0\) lies between the local minimum of -12 and the local maximum of 20 (i.e., \(-12 < 0 < 20\)), the line \(y = 0\) (the x-axis) intersects the curve three times. Therefore, there are 3 real solutions.

(d) When \(x = 2\), the y-coordinate is:
\[g(2) = 2^3 - 3(2)^2 - 9(2) + 15 = 8 - 12 - 18 + 15 = -7\]
The gradient of the tangent is the derivative evaluated at \(x = 2\):
\[g'(2) = 3(2)^2 - 6(2) - 9 = 12 - 12 - 9 = -9\]
Using the point-slope form with \((2, -7)\) and gradient \(-9\):
\[y - (-7) = -9(x - 2)\]
\[y + 7 = -9x + 18 \implies y = -9x + 11\]

(a) [5 marks]:
M1 for derivative \(g'(x) = 3x^2 - 6x - 9\).
M1 for setting \(g'(x) = 0\) and factorising or using formula to get \(x = 3\) and \(x = -1\).
M1 for substituting \(x = 3\) to get \(y = -12\).
M1 for substituting \(x = -1\) to get \(y = 20\).
A1 for both points correctly identified as \((3, -12)\) and \((-1, 20)\).

(b) [2 marks]:
M1 for finding the second derivative \(g''(x) = 6x - 6\) and substituting values (or testing gradients around points).
A1 for correctly identifying \((-1, 20)\) as the local maximum and \((3, -12)\) as the local minimum with valid mathematical justification.

(c)(i) [1 mark]:
B1 for 1 solution.

(c)(ii) [1 mark]:
B1 for 3 solutions.

(d) [3 marks]:
M1 for substituting \(x = 2\) to find \(y = -7\).
M1 for substituting \(x = 2\) into \(g'(x)\) to get gradient \(-9\).
A1 for \(y = -9x + 11\).

(a) Let the angle of the sector be \(\theta^\circ\). The perimeter of a sector is given by \(2r + \text{arc length}\).
\(2(8) + \text{arc length} = 22 \implies 16 + \text{arc length} = 22 \implies \text{arc length} = 6\text{ cm}\).
Using the formula for arc length: \(\frac{\theta}{360} \times 2\pi(8) = 6 \implies \frac{16\pi\theta}{360} = 6 \implies \theta = \frac{6 \times 360}{16\pi} = \frac{135}{\pi} \approx 42.9718^\circ\).
Correct to 1 decimal place, \(\theta = 43.0^\circ\).

(b)(i) The combined area of three identical sectors is \(150\text{ cm}^2\), so the area of one sector is \(\frac{150}{3} = 50\text{ cm}^2\).
Using the sector area formula: \(\frac{45}{360} \times ̀\pi r^2 = 50 \implies \frac{1}{8}\pi r^2 = 50 \implies r^2 = \frac{400}{\pi} \approx 127.324 \implies r = \sqrt{127.324} \approx 11.2837\text{ cm}\).
Rounding to 3 significant figures gives \(r = 11.3\text{ cm}\).

(b)(ii) The perimeter of one sector is \(2r + \text{arc length}\).
\(\text{Arc length} = \frac{45}{360} \times 2\pi r = \frac{\pi}{4}r\).
\(\text{Perimeter of one sector} = r\left(2 + \frac{\pi}{4}\right)\).
Using \(r = 11.28\): \(\text{Perimeter} = 11.28 \times \left(2 + \frac{\pi}{4}\right) \approx 31.419\text{ cm}\).
Total perimeter of three sectors = \(3 \times 31.419 = 94.257\text{ cm}\), which rounds to \(94.3\text{ cm}\).
(Using the unrounded \(r = 11.2837\) gives \(94.288\text{ cm} \approx 94.3\text{ cm}\)).

(c)(i) Area of the original sector = \(50\text{ cm}^2\). Area of the new similar sector = \(32\text{ cm}^2\).
Area scale factor = \​\frac{32}{50} = 0.64\.
Linear scale factor = \(\sqrt{0.64} = 0.8\).

(c)(ii) Radius of the new sector = \(0.8 \times \text{original radius}\).
Using \(r = 11.28\): \(0.8 \times 11.28 = 9.024\text{ cm} \approx 9.02\text{ cm}\).
Using the unrounded value of \(r = 11.2837\): \(0.8 \times 11.2837 = 9.027\text{ cm} \approx 9.03\text{ cm}\).

(a)
- M1 for writing 2(8) + arc length = 22 or arc length = 6
- M1 for (theta / 360) * 2 * pi * 8 = 6
- A1 for 43.0 or 43

(b)(i)
- M1 for (45/360) * pi * r^2 = 50 or 3 * (45/360) * pi * r^2 = 150
- M1 for r^2 = 400/pi or r = sqrt(127.3...)
- A1 for complete proof showing 11.28... before rounding to 11.3

(b)(ii)
- M1 for arc length = (45/360) * 2 * pi * r (or approx 8.86)
- M1 for total perimeter = 3 * (2r + arc length)
- A1 for 94.3 (accept 94.2 to 94.3)

(c)(i)
- B1 for 0.8 (or equivalent fraction 4/5)

(c)(ii)
- M1 for 0.8 * 11.28 or 0.8 * 11.3 or 0.8 * their r
- A1 for 9.02 or 9.03 (or 9.04 if using r=11.3)

(a) Gradient of \(AB\) is \(m = \frac{3 - 9}{10 - 2} = \frac{-6}{8} = -0.75\).
Substitute \(A(2, 9)\) into \(y = mx + c\):
\(9 = -0.75(2) + c \implies 9 = -1.5 + c \implies c = 10.5\).
So the equation of \(AB\) is \(y = -0.75x + 10.5\).

(b) The perpendicular bisector passes through the midpoint of \(AB\).
\(\text{Midpoint } M = \left(\frac{2+10}{2}, \frac{9+3}{2}\right) = (6, 6)\).
The gradient of the perpendicular line is \(m_{\perp} = -\frac{1}{m} = -\frac{1}{-0.75} = \frac{4}{3}\).
Equation of the perpendicular bisector:
\(y - 6 = \frac{4}{3}(x - 6)\)
Multiply by 3 to clear the fraction:
\(3(y - 6) = 4(x - 6) \implies 3y - 18 = 4x - 24\).
Rearranging into the form \(ay + bx = d\):
\(3y - 4x = -6\) (or \(4x - 3y = 6\)).

(c) Point \(C\) lies on the \(x\)-axis, so \(y = 0\):
\(0 = -0.75x + 10.5 \implies 0.75x = 10.5 \implies x = 14\).
So \(C\) has coordinates \((14, 0)\).
Point \(D\) lies on the \(y\)-axis, so \(x = 0\):
\(y = -0.75(0) + 10.5 = 10.5\).
So \(D\) has coordinates \((0, 10.5)\).

(d) Let \(P\) have coordinates \((k, y_P)\). Since \(P\) lies on the perpendicular bisector:
\(3y_P - 4k = -6 \implies y_P = \frac{4}{3}k - 2\).
The distance from \(A(2, 9)\) to \(P(k, y_P)\) is given by:
\(AP^2 = (k - 2)^2 + (y_P - 9)^2 = \left(5\sqrt{2}\right)^2 = 50\).
Substitute \(y_P = \frac{4}{3}k - 2\):
\((k - 2)^2 + \left(\frac{4}{3}k - 11\right)^2 = 50\)
\(k^2 - 4k + 4 + \frac{16}{9}k^2 - \frac{88}{3}k + 121 = 50\).
Multiply the entire equation by 9 to clear denominators:
\(9(k^2 - 4k + 4) + 16k^2 - 264k + 1089 = 450\)
\(9k^2 - 36k + 36 + 16k^2 - 264k + 1089 = 450\)
\(25k^2 - 300k + 1125 = 450 \implies 25k^2 - 300k + 675 = 0\).
Divide the entire equation by 25:
\(k^2 - 12k + 27 = 0\).
Factorise:
\((k - 3)(k - 9) = 0\).
So the two possible values of \(k\) are \(k = 3\) and \(k = 9\).

(a)
- M1 for gradient m = (3 - 9)/(10 - 2) = -0.75
- A1 for y = -0.75x + 10.5

(b)
- B1 for Midpoint M(6, 6)
- M1 for perpendicular gradient m_perp = -1 / (their m) = 4/3
- A1 for 3y - 4x = -6 (or 4x - 3y = 6, or equivalent form with integer coefficients)

(c)
- B1 for C(14, 0)
- B1 for D(0, 10.5)
- B1 for correct coordinates shown clearly

(d)
- M1 for distance equation (k - 2)^2 + (y_P - 9)^2 = 50
- M1 for substituting y_P = (4/3)k - 2
- M1 for simplifying to a quadratic equation of the form 25k^2 - 300k + 675 = 0 (or k^2 - 12k + 27 = 0)
- A1 for k = 3 and k = 9

(a) Factorise the numerator: \(2x^2 - 8 = 2(x^2 - 4) = 2(x - 2)(x + 2)\).
Factorise the denominator: \(x^2 - 5x + 6 = (x - 2)(x - 3)\).
Simplify by cancelling the common factor \((x - 2)\):
\(\frac{2(x - 2)(x + 2)}{(x - 2)(x - 3)} = \frac{2(x + 2)}{x - 3} = \frac{2x + 4}{x - 3}\).

(b) Multiply both sides by \((5 - p)\):
\(w(5 - p) = 3p + 2 \implies 5w - wp = 3p + 2\).
Collect terms containing \(p\) on one side:
\(5w - 2 = 3p + wp\).
Factorise out \(p\):
\(5w - 2 = p(3 + w)\).
Divide by \((w + 3)\):
\(p = \frac{5w - 2}{w + 3}\).

(c) Multiply both sides by the common denominator \((2x - 1)(x + 2)\):
\(4(x + 2) - 3(2x - 1) = (2x - 1)(x + 2)\)
\(4x + 8 - 6x + 3 = 2x^2 + 4x - x - 2\)
\(-2x + 11 = 2x^2 + 3x - 2\).
Rearrange into standard quadratic form:
\(2x^2 + 5x - 13 = 0\).
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\(x = \frac{-5 \pm \sqrt{5^2 - 4(2)(-13)}}{2(2)} = \frac{-5 \pm \sqrt{25 + 104}}{4} = \frac{-5 \pm \sqrt{129}}{4}\).
Calculating the roots:
\(x_1 = \frac{-5 + 11.3578}{4} \approx 1.59\)
\(x_2 = \frac{-5 - 11.3578}{4} \approx -4.09\).

(a)
- M1 for factorising the numerator: 2(x - 2)(x + 2)
- M1 for factorising the denominator: (x - 2)(x - 3)
- A1 for (2x + 4)/(x - 3) (or 2(x + 2)/(x - 3))

(b)
- M1 for multiplying by (5 - p): w(5 - p) = 3p + 2
- M1 for expanding and separating p terms: 5w - 2 = p(w + 3)
- A1 for p = (5w - 2)/(w + 3) (or equivalent)

(c)
- M1 for multiplying by (2x - 1)(x + 2) or finding common denominator
- M1 for expanding: 4x + 8 - 6x + 3 = 2x^2 + 3x - 2
- A1 for correct quadratic: 2x^2 + 5x - 13 = 0
- M1 for using the quadratic formula: x = (-5 +/- sqrt(5^2 - 4*2*(-13))) / 4
- A1 for 1.59
- A1 for -4.09