2024 Cambridge IGCSE Mathematics (0580) 模拟试题及答案详解

Using the quadratic formula:
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
Here, \(a = 2\), \(b = 5\), and \(c = -11\).
\(x = \frac{-5 \pm \sqrt{5^2 - 4(2)(-11)}}{2(2)}\)
\(x = \frac{-5 \pm \sqrt{25 + 88}}{4}\)
\(x = \frac{-5 \pm \sqrt{113}}{4}\)
\(x_1 = \frac{-5 + 10.6301...}{4} \approx 1.41\)
\(x_2 = \frac{-5 - 10.6301...}{4} \approx -3.91\)

M1 for \(\frac{-5 \pm \sqrt{5^2 - 4(2)(-11)}}{2(2)}\) oe
B1 for 1.41 or 1.407 to 1.408
B1 for -3.91 or -3.907 to -3.908

Factorise the numerator:
\(3x^2 - 12 = 3(x^2 - 4) = 3(x - 2)(x + 2)\)
Factorise the denominator:
\(x^2 + 5x + 6 = (x + 2)(x + 3)\)
Divide both by the common factor \((x+2)\):
\(\frac{3(x-2)(x+2)}{(x+2)(x+3)} = \frac{3(x-2)}{x+3}\)

M1 for factorising the numerator to \(3(x-2)(x+2)\) oe
M1 for factorising the denominator to \((x+2)(x+3)\)
A1 for \(\frac{3(x-2)}{x+3}\) or \(\frac{3x-6}{x+3}\) as final answer

Complete the square for \(y = x^2 - 8x + 22\):
\(y = (x - 4)^2 - 4^2 + 22\)
\(y = (x - 4)^2 - 16 + 22\)
\(y = (x - 4)^2 + 6\)
From \(y = (x - h)^2 + k\), the turning point is \((h, k)\), which is \((4, 6)\).

M1 for \((x - 4)^2\) seen
A1 for completing the square to \((x - 4)^2 + 6\)
A1 for coordinates \((4, 6)\)

Using the cosine rule:
\(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(ABC)\)
\(AC^2 = 7.2^2 + 9.5^2 - 2 \cdot 7.2 \cdot 9.5 \cdot \cos(112^\circ)\)
\(AC^2 = 51.84 + 90.25 - 136.8 \cdot (-0.374606...)\)
\(AC^2 = 142.09 + 51.246\)
\(AC^2 = 193.336\)
\(AC = \sqrt{193.336} \approx 13.904\text{ cm}\)
To 3 significant figures, \(AC = 13.9\text{ cm}\).

M2 for \(\sqrt{7.2^2 + 9.5^2 - 2 \cdot 7.2 \cdot 9.5 \cdot \cos(112)}\)
or M1 for \(7.2^2 + 9.5^2 - 2 \cdot 7.2 \cdot 9.5 \cdot \cos(112)\)
A1 for 13.9 or 13.90 to 13.91

Find the differences between consecutive terms:
First differences: \(8, 12, 16, 20\)
Second differences: \(4, 4, 4\)
Since the second difference is constant, the sequence is quadratic of the form \(an^2 + bn + c\).
Here, \(2a = 4 \implies a = 2\).
Subtract \(2n^2\) from the terms of the sequence:
For \(n=1\): \(3 - 2(1)^2 = 1\)
For \(n=2\): \(11 - 2(2)^2 = 3\)
For \(n=3\): \(23 - 2(3)^2 = 5\)
This leaves the linear sequence \(1, 3, 5, \dots\), which has \(n\)th term \(2n - 1\).
Thus, the general \(n\)th term is \(2n^2 + 2n - 1\).

M1 for finding the second difference is 4 (or indicating \(2n^2\))
M1 for subtracting \(2n^2\) to find the remaining linear sequence \(1, 3, 5, \dots\) or setting up simultaneous equations
A1 for \(2n^2 + 2n - 1\) (or equivalent)

Let \(W_A\) and \(B_A\) be the events of selecting a white or black button from Box A respectively.
\(P(W_A) = \frac{4}{10}\) and \(P(B_A) = \frac{6}{10}\).
If a white button is moved from A to B, Box B now contains 6 white and 3 black buttons. The probability of choosing white from B is \(\frac{6}{9}\).
If a black button is moved from A to B, Box B now contains 5 white and 4 black buttons. The probability of choosing white from B is \(\frac{5}{9}\).
Using the law of total probability:
\(P(W_B) = P(W_A) \cdot P(W_B | W_A) + P(B_A) \cdot P(W_B | B_A)\)
\(P(W_B) = \frac{4}{10} \cdot \frac{6}{9} + \frac{6}{10} \cdot \frac{5}{9} = \frac{24}{90} + \frac{30}{90} = \frac{54}{90} = \frac{3}{5} = 0.6\)

M1 for \(\frac{4}{10} \cdot \frac{6}{9}\) or \(\frac{6}{10} \cdot \frac{5}{9}\)
M1 for \(\frac{4}{10} \cdot \frac{6}{9} + \frac{6}{10} \cdot \frac{5}{9}\)
A1 for \(\frac{3}{5}\) or \(0.6\) (or equivalent fraction)

Using the compound interest formula:
\(4500 \cdot \left(1 + \frac{r}{100}\right)^4 = 4985\)
\(\left(1 + \frac{r}{100}\right)^4 = \frac{4985}{4500}\)
\(1 + \frac{r}{100} = \left(\frac{4985}{4500}\right)^{0.25}\)
\(1 + \frac{r}{100} \approx 1.025916\)
\(\frac{r}{100} \approx 0.025916\)
\(r \approx 2.59\)

M1 for \(4500 \cdot \left(1 + \frac{r}{100}\right)^4 = 4985\) oe
M1 for \(\left(\frac{4985}{4500}\right)^{0.25} - 1\) oe
A1 for 2.59 or 2.591 to 2.592

Since ABCD is a cyclic quadrilateral, opposite angles add up to \(180^\circ\):
\(\text{Angle } ABC = 180^\circ - 115^\circ = 65^\circ\).
Since AB is a diameter, the angle in the semicircle is a right angle:
\(\text{Angle } ACB = 90^\circ\).
In triangle ABC, angles add up to \(180^\circ\):
\(\text{Angle } CAB = 180^\circ - 90^\circ - 65^\circ = 25^\circ\).

M1 for \(\text{Angle } ABC = 180 - 115 = 65^\circ\)
M1 for \(\text{Angle } ACB = 90^\circ\)
A1 for 25

Multiply all terms by the common denominator \((x-1)(x-2)\):
\(12(x-2) - 6(x-1) = (x-1)(x-2)\)
Expand the brackets:
\(12x - 24 - 6x + 6 = x^2 - 3x + 2\)
Simplify the equation:
\(6x - 18 = x^2 - 3x + 2\)
Rearrange into a quadratic equation set to 0:
\(x^2 - 9x + 20 = 0\)
Factorise the quadratic expression:
\((x-4)(x-5) = 0\)
Solve for \(x\):
\(x = 4\) or \(x = 5\).

M1 for correctly eliminating denominators: \(12(x-2) - 6(x-1) = (x-1)(x-2)\) oe
M1 for reducing to quadratic form: \(x^2 - 9x + 20 = 0\) oe
A1 for \(x = 4\) and \(x = 5\) both correct

Substitute the coordinates of both points into the equation:
For \((1, 6)\): \(6 = a \cdot b^1 \implies ab = 6\)
For \((3, 24)\): \(24 = a \cdot b^3\)
Divide the second equation by the first:
\(\frac{ab^3}{ab} = \frac{24}{6}\)
\(b^2 = 4\)
Since \(b > 0\), we have \(b = 2\).
Substitute \(b = 2\) back into the first equation:
\(a \cdot 2 = 6 \implies a = 3\).

M1 for setting up two equations: \(ab = 6\) and \(ab^3 = 24\) oe
M1 for dividing to find \(b^2 = 4\) or \(b = 2\)
A1 for both \(a = 3\) and \(b = 2\) correct

Factorise the numerator:
\(2x^2 - 5x - 3 = (2x + 1)(x - 3)\)
Factorise the denominator as a difference of two squares:
\(4x^2 - 1 = (2x + 1)(2x - 1)\)
Substitute the factored forms into the fraction:
\(\frac{(2x + 1)(x - 3)}{(2x + 1)(2x - 1)}\)
Cancel the common factor of \((2x + 1)\):
\(\frac{x - 3}{2x - 1}\).

B1 for factorising the numerator: \((2x + 1)(x - 3)\)
B1 for factorising the denominator: \((2x + 1)(2x - 1)\)
B1 for the final simplified fraction \(\frac{x - 3}{2x - 1}\) (dependent on previous factorisation marks)

Express both sides with base 3:
\(27 = 3^3\)
So the equation becomes:
\(3^{2x-1} = (3^3)^{x+2}\)
\(3^{2x-1} = 3^{3(x+2)}\)
Equate the exponents:
\(2x - 1 = 3(x + 2)\)
Expand and solve for \(x\):
\(2x - 1 = 3x + 6\)
\(-1 - 6 = 3x - 2x\)
\(x = -7\).

M1 for writing \(27\) as \(3^3\) or converting equation to a common base oe
M1 for equating exponents: \(2x - 1 = 3(x + 2)\) oe
A1 for \(x = -7\) correct

Let \(y = f(x)\):
\(y = \frac{3x}{x-2}\)
Multiply by \(x-2\):
\(y(x-2) = 3x\)
\(xy - 2y = 3x\)
Rearrange to group all \(x\) terms together:
\(xy - 3x = 2y\)
Factorise \(x\):
\(x(y - 3) = 2y\)
Divide by \(y-3\):
\(x = \frac{2y}{y-3}\)
Swap \(x\) and \(y\) to find the inverse function:
\(f^{-1}(x) = \frac{2x}{x-3}\).

M1 for a correct first step to rearrange: e.g. \(y(x-2) = 3x\) or \(x(y-2) = 3y\)
M1 for grouping terms in \(x\) (or \(y\)) and factorising: e.g. \(x(y-3) = 2y\)
A1 for \(\frac{2x}{x-3}\) oe

Multiply both sides by \(3-t\):
\(S(3 - t) = 5t + 2\)
\(3S - St = 5t + 2\)
Rearrange to group all terms with \(t\) on one side:
\(3S - 2 = 5t + St\)
Factorise \(t\):
\(3S - 2 = t(5 + S)\)
Divide both sides by \(5+S\):
\(t = \frac{3S - 2}{5 + S}\).

M1 for clearing the fraction: \(S(3-t) = 5t + 2\) oe
M1 for grouping terms with \(t\) and factorising: \(t(5+S) = 3S - 2\) oe
A1 for \(t = \frac{3S - 2}{5 + S}\) oe

Let \(\theta\) be the angle of the sector in degrees.
\(\text{Area of sector} = \frac{\theta}{360} \times \pi r^2\)
Substitute \(r = 8\) and \(\text{Area} = 20\pi\):
\(20\pi = \frac{\theta}{360} \times \pi \times 8^2\)
\(20 = \frac{\theta}{360} \times 64 \implies \frac{\theta}{360} = \frac{20}{64} = \frac{5}{16}\)
Now calculate the arc length:
\(\text{Arc length} = \frac{\theta}{360} \times 2\pi r = \frac{5}{16} \times 2\pi \times 8 = 5\pi\text{ cm}\)
The perimeter of the sector is the arc length plus two radii:
\(\text{Perimeter} = 5\pi + 2(8) = 5\pi + 16 \approx 15.708 + 16 = 31.708...\text{ cm}\)
To 3 significant figures, the perimeter is \(31.7\text{ cm}\).

M1 for finding the fraction of the circle: \(\frac{\theta}{360} = \frac{20}{64}\) or sector angle \(\theta = 112.5^\circ\) oe
M1 for calculating arc length: \(5\pi\) or \(15.7\) or \(15.707...\) oe
A1 for \(31.7\) or \(31.71\) to \(31.72\)

The total number of balls in the bag is \(5 + 3 = 8\).
We want the probability of getting either (Red, Blue) or (Blue, Red).
Probability of first Red and second Blue:
\(P(\text{Red, Blue}) = \frac{5}{8} \times \frac{3}{7} = \frac{15}{56}\)
Probability of first Blue and second Red:
\(P(\text{Blue, Red}) = \frac{3}{8} \times \frac{5}{7} = \frac{15}{56}\)
Add these two mutually exclusive probabilities:
\(P(\text{different colours}) = \frac{15}{56} + \frac{15}{56} = \frac{30}{56} = \frac{15}{28}\).
In decimals, this is approximately \(0.536\).

M1 for \(\frac{5}{8} \times \frac{3}{7}\) or \(\frac{3}{8} \times \frac{5}{7}\) oe
M1 for addition of two different cases: e.g. \(2 \times \left(\frac{5}{8} \times \frac{3}{7}\right)\) oe
A1 for \(\frac{15}{28}\) or equivalent fraction (e.g., \(\frac{30}{56}\)) or \(0.536\) or \(0.5357...\)

Multiply the first equation by 3 and the second equation by 4 to get: \(9x + 12y = 9\) and \(20x + 12y = 64\). Subtracting the first of these new equations from the second gives: \(11x = 55\), which simplifies to \(x = 5\). Substitute \(x = 5\) into the first equation to get: \(3(5) + 4y = 3\), which simplifies to \(15 + 4y = 3\), then \(4y = -12\), so \(y = -3\). Therefore, the solution is \(x = 5\) and \(y = -3\).

M1 for multiplying to find common coefficients (e.g. \(9x + 12y = 9\) and \(20x + 12y = 64\)) or substituting expression for one variable. M1 for adding/subtracting to eliminate one variable. A1 for \(x = 5\) and \(y = -3\).

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 the fraction by cancelling the common factor \((x + 2)\): \(\frac{2(x - 2)(x + 2)}{(x + 2)(x + 3)} = \frac{2(x - 2)}{x + 3} = \frac{2x - 4}{x + 3}\).

B1 for numerator factorised to \(2(x - 2)(x + 2)\) or \(2(x^2 - 4)\). B1 for denominator factorised to \((x + 2)(x + 3)\). B1 for final answer \(\frac{2x - 4}{x + 3}\) or \(\frac{2(x - 2)}{x + 3}\).

By completing the square: \(y = 3(x^2 - 4x) + 7 = 3[(x - 2)^2 - 4] + 7 = 3(x - 2)^2 - 12 + 7 = 3(x - 2)^2 - 5\). The turning point is therefore \((2, -5)\).

M1 for attempting to complete the square or using \(x = -\frac{b}{2a}\) to find the x-coordinate. M1 for substituting \(x = 2\) to find the y-coordinate. A1 for \((2, -5)\).

Since \(y\) is inversely proportional to the square root of \(x\), we can write \(y = \frac{k}{\sqrt{x}}\). Substituting the given values \(x = 16\) and \(y = 3\) allows us to find \(k\): \(3 = \frac{k}{\sqrt{16}}\), so \(3 = \frac{k}{4}\), which gives \(k = 12\). Now, substituting \(x = 36\) into our equation \(y = \frac{12}{\sqrt{x}}\) yields: \(y = \frac{12}{\sqrt{36}} = \frac{12}{6} = 2\).

M1 for \(y = \frac{k}{\sqrt{x}}\). M1 for finding \(k = 12\). A1 for 2.

First, convert the volume of the block from \(\text{m}^3\) to \(\text{cm}^3\): \(V = 0.045 \times 1,000,000 = 45,000\text{ cm}^3\). Next, calculate the mass in grams using \(\text{mass} = \text{density} \times \text{volume}\): \(\text{mass} = 7.8 \times 45,000 = 351,000\text{ g}\). Finally, convert this mass from grams to kilograms: \(\text{mass} = \frac{351,000}{1000} = 351\text{ kg}\).

M1 for converting \(0.045\text{ m}^3\) to \(45,000\text{ cm}^3\) or converting density to \(7800\text{ kg/m}^3\). M1 for mass = density \(\times\) volume. A1 for 351.

The probability of picking two black pens is \(P(\text{Black, Black}) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90}\). The probability of picking two red pens is \(P(\text{Red, Red}) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90}\). The probability that both pens are the same colour is the sum of these probabilities: \(P(\text{Same Colour}) = \frac{30}{90} + \frac{12}{90} = \frac{42}{90} = \frac{7}{15}\).

M1 for \(P(\text{BB}) = \frac{6}{10} \times \frac{5}{9}\) or \(P(\text{RR}) = \frac{4}{10} \times \frac{3}{9}\). M1 for adding their two probabilities with a denominator of 90 (or equivalent). A1 for \(\frac{7}{15}\) or equivalent.

The first differences of the sequence are 7, 9, 11, and the second differences are constant and equal to 2. This indicates that the \(n\)th term formula has an \(n^2\) term with coefficient \(\frac{2}{2} = 1\). Subtracting \(n^2\) from the terms: \(5 - 1^2 = 4\), \(12 - 2^2 = 8\), \(21 - 3^2 = 12\), \(32 - 4^2 = 16\). The remaining sequence is 4, 8, 12, 16, which is linear with formula \(4n\). Therefore, the \(n\)th term is \(n^2 + 4n\).

M1 for finding second difference is 2. M1 for subtracting \(n^2\) from terms to find the linear part \(4n\). A1 for \(n^2 + 4n\).

(a) Differentiate to find \(\frac{dy}{dx}\):
\(\frac{dy}{dx} = 6x^2 - 6x - 12\)
Set \(\frac{dy}{dx} = 0\):
\(6x^2 - 6x - 12 = 0 \implies x^2 - x - 2 = 0\)
\((x - 2)(x + 1) = 0 \implies x = 2\) or \(x = -1\)

Substitute the \(x\)-values back into the original equation to find the \(y\)-coordinates:
For \(x = 2\):
\(y = 2(2)^3 - 3(2)^2 - 12(2) + 5 = 16 - 12 - 24 + 5 = -15\) \(\implies (2, -15)\)
For \(x = -1\):
\(y = 2(-1)^3 - 3(-1)^2 - 12(-1) + 5 = -2 - 3 + 12 + 5 = 12\) \(\implies (-1, 12)\)

(b) (i) The curve crosses the \(y\)-axis when \(x = 0\):
\(y = 2(0)^3 - 3(0)^2 - 12(0) + 5 = 5 \implies (0, 5)\)

(ii) Let \(f(x) = 2x^3 - 3x^2 - 12x + 5\).
Evaluate \(f(0.3)\) and \(f(0.5)\):
\(f(0.3) = 2(0.3)^3 - 3(0.3)^2 - 12(0.3) + 5 = 0.054 - 0.27 - 3.6 + 5 = 1.184\)
\(f(0.5) = 2(0.5)^3 - 3(0.5)^2 - 12(0.5) + 5 = 0.25 - 0.75 - 6 + 5 = -1.5\)
Since there is a sign change between \(f(0.3) > 0\) and \(f(0.5) < 0\), a root must exist between \(x = 0.3\) and \(x = 0.5\).

(c) Equate the curve equation and the equation to be solved:
\(2x^3 - 3x^2 - 12x + 5 = y\)
We want to solve:
\(2x^3 - 3x^2 - 14x + 1 = 0\)
Rewrite this to incorporate the curve's expression:
\((2x^3 - 3x^2 - 12x + 5) - 2x - 4 = 0\)
\(y - 2x - 4 = 0 \implies y = 2x + 4\)

(d) At \(x = 1\), the gradient of the tangent is \(m = \frac{dy}{dx}\) evaluated at \(x = 1\):
\(m = 6(1)^2 - 6(1) - 12 = -12\)
The \(y\)-coordinate at \(x = 1\) is:
\(y = 2(1)^3 - 3(1)^2 - 12(1) + 5 = -8\)
Using the equation of a straight line \(y - y_1 = m(x - x_1)\):
\(y - (-8) = -12(x - 1) \implies y + 8 = -12x + 12 \implies y = -12x + 4\)

(a)
M1 for differentiating: at least two terms correct in \(6x^2 - 6x - 12\)
A1 for \(\frac{dy}{dx} = 6x^2 - 6x - 12\)
M1 for setting their derivative to 0 and solving to find \(x = 2\) and \(x = -1\)
A1 for \((2, -15)\)
A1 for \((-1, 12)\)

(b) (i)
B1 for \((0, 5)\)

(ii)
M1 for evaluating \(f(0.3)\) and \(f(0.5)\)
A1 for correct values of \(1.184\) and \(-1.5\) (or equivalent) and stating there is a sign change

(c)
M1 for attempting to write \(2x^3 - 3x^2 - 14x + 1 = 0\) in terms of \(y\)
A1 for \(y = 2x + 4\)

(d)
M1 for substituting \(x = 1\) into \(\frac{dy}{dx}\) to find gradient \(= -12\)
M1 for substituting \(x = 1\) into the curve equation to find \(y = -8\)
A1 for \(y = -12x + 4\)

(a) The path has a width of 1.5 m all around the outside.
Total length of the garden and path is \(x + 2(1.5) = x + 3\) metres.
Total width of the garden and path is \((x - 4) + 2(1.5) = x - 1\) metres.
Total Area \(= (x + 3)(x - 1) = x^2 + 3x - x - 3 = x^2 + 2x - 3\) square metres.

(b) (i) \(x^2 + 2x - 3 = 140 \implies x^2 + 2x - 143 = 0\).

(ii) Factorise \(x^2 + 2x - 143 = 0\):
We look for two numbers that multiply to \(-143\) and add to \(2\). These are \(13\) and \(-11\).
\((x + 13)(x - 11) = 0\)
So, \(x = 11\) or \(x = -13\).

(iii) Since length cannot be negative, we reject \(x = -13\), so \(x = 11\).
Length of the garden \(= x = 11\text{ m}\).
Width of the garden \(= x - 4 = 11 - 4 = 7\text{ m}\).

(c) Using Pythagoras' theorem:
\(y^2 + (2y - 1)^2 = (2y + 1)^2\)
\(y^2 + (4y^2 - 4y + 1) = 4y^2 + 4y + 1\)
\(y^2 - 8y = 0\)
\(y(y - 8) = 0\)
Since \(y\) must be greater than 0, \(y = 8\).

(a)
M1 for writing total length as \(x + 3\) or total width as \(x - 1\)
M1 for multiplying \((x+3)(x-1)\)
A1 for completing the show clearly to obtain \(x^2 + 2x - 3\)

(b) (i)
B1 for showing \(x^2 + 2x - 3 = 140 \implies x^2 + 2x - 143 = 0\)

(ii)
M1 for attempting to factorise (e.g. finding two numbers multiplying to \(-143\) and adding to \(2\))
A1 for \((x - 11)(x + 13) = 0\)
A1 for \(x = 11\) and \(x = -13\)

(iii)
B1 for choosing \(x = 11\) and rejecting \(-13\)
B1 for dimensions: Length \(= 11\text{ m}\), Width \(= 7\text{ m}\)

(c)
M1 for writing \(y^2 + (2y-1)^2 = (2y+1)^2\)
M1 for expanding correctly: \(4y^2 - 4y + 1\) and \(4y^2 + 4y + 1\)
M1 for simplifying to \(y^2 - 8y = 0\) or equivalent
A1 for \(y = 8\)

(a) Find a common denominator:
\(\frac{3}{2x - 1} - \frac{2}{x + 3} = \frac{3(x + 3) - 2(2x - 1)}{(2x - 1)(x + 3)}\)
\(= \frac{3x + 9 - 4x + 2}{(2x - 1)(x + 3)}\)
\(= \frac{11 - x}{(2x - 1)(x + 3)}\)

(b) Multiply both sides by \(5 - t\):
\(w(5 - t) = 3t + 2\)
\(5w - wt = 3t + 2\)
Rearrange to group all terms containing \(t\) on one side:
\(5w - 2 = 3t + wt\)
Factorise \(t\):
\(5w - 2 = t(3 + w)\)
Divide by \(3 + w\):
\(t = \frac{5w - 2}{w + 3}\)

(c) (i) Identify the common factor \(3ab\):
\(6a^2b - 15ab^2 = 3ab(2a - 5b)\)

(ii) Use the difference of two squares:
\(4x^2 - 25y^2 = (2x)^2 - (5y)^2 = (2x - 5y)(2x + 5y)\)

(iii) Group the terms:
\(2px - 6py + qx - 3qy = 2p(x - 3y) + q(x - 3y)\)
Factorise out the common bracket \((x - 3y)\):
\(= (2p + q)(x - 3y)\)

(a)
M1 for finding common denominator \((2x-1)(x+3)\)
M1 for \(3(x+3) - 2(2x-1)\) as the numerator
A1 for expanding and simplifying numerator to \(11 - x\) (or \(-x + 11\))
A1 for final answer \(\frac{11-x}{(2x-1)(x+3)}\) (or equivalent with expanded denominator)

(b)
M1 for multiplying to clear the fraction: \(w(5-t) = 3t+2\)
M1 for expanding and grouping \(t\) terms on one side: \(5w - 2 = t(3+w)\) (or equivalent)
M1 for factorising \(t\) correctly
A1 for \(t = \frac{5w-2}{w+3}\) (or equivalent)

(c) (i)
B1 for taking out common factor \(3ab\) or partial factorisation like \(ab(6a-15b)\)
B1 for \(3ab(2a - 5b)\)

(ii)
B1 for \((2x - 5y)(2x + 5y)\)

(iii)
M1 for grouping: e.g. \(2p(x - 3y) + q(x - 3y)\)
A1 for \((2p + q)(x - 3y)\)

(a) Use the Cosine Rule to find \(AC\):
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(ABC)\)
\(AC^2 = 65^2 + 80^2 - 2(65)(80)\cos(72^\circ)\)
\(AC^2 = 4225 + 6400 - 10400(0.309017)\)
\(AC^2 = 10625 - 3213.78 = 7411.22\)
\(AC = \sqrt{7411.22} \approx 86.088\text{ m} \approx 86.1\text{ m}\)

(b) Use the Sine Rule to find angle \(ACB\):
\(\frac{\sin(ACB)}{AB} = \frac{\sin(ABC)}{AC}\)
\(\frac{\sin(ACB)}{65} = \frac{\sin(72^\circ)}{86.088}\)
\(\sin(ACB) = \frac{65 \times \sin(72^\circ)}{86.088} \approx 0.71809\)
Angle \(ACB = \arcsin(0.71809) \approx 45.89^\circ \approx 45.9^\circ\)

(c) The mast \(TB\) is vertical and stands at \(B\). Triangle \(TBA\) is right-angled at \(B\).
We have \(TB = 15\text{ m}\) and \(AB = 65\text{ m}\).
Let \(\theta\) be the angle of elevation of \(T\) from \(A\):
\(\tan(\theta) = \frac{TB}{AB} = \frac{15}{65}\)
\(\theta = \arctan\left(\frac{15}{65}\right) \approx 12.99^\circ \approx 13.0^\circ\)

(d) Area of the triangular field \(ABC\):
\(\text{Area} = \frac{1}{2}(AB)(BC)\sin(ABC) = \frac{1}{2}(65)(80)\sin(72^\circ) = 2600 \times 0.951056 \approx 2472.7\text{ m}^2 \approx 2470\text{ m}^2\)

(e) The shortest path from \(B\) to \(AC\) is the perpendicular distance \(h\) from \(B\) to \(AC\):
Using the area formula: \(\text{Area} = \frac{1}{2}(AC)(h)\)
\(2472.7 = \frac{1}{2}(86.088)(h)\)
\(h = \frac{2 \times 2472.7}{86.088} \approx 57.44\text{ m} \approx 57.4\text{ m}\)
Alternatively: \(h = BC \times \sin(ACB) = 80 \times \sin(45.89^\circ) \approx 57.4\text{ m}\).

(a)
M1 for substituting correctly into Cosine Rule: \(65^2 + 80^2 - 2 \times 65 \times 80 \times \cos(72)\)
A1 for \(AC^2 = 7411.22\) (or better)
A1 for \(86.1\) (or 86.08 to 86.09)

(b)
M1 for correct substitution into Sine Rule: \(\frac{\sin(ACB)}{65} = \frac{\sin(72)}{86.1}\)
M1 for rearranging to make \(\sin(ACB)\) the subject
A1 for \(45.9^\circ\) (or 45.88 to 45.90)

(c)
M1 for identifying a right-angled triangle with sides 15 and 65
M1 for \(\tan(\theta) = \frac{15}{65}\) (or equivalent)
A1 for \(13.0^\circ\) (or 12.99 to 13.01)

(d)
M1 for \(\frac{1}{2} \times 65 \times 80 \times \sin(72)\)
A1 for \(2470\) (or 2472 to 2473)

(e)
M1 for equating \(\frac{1}{2} \times 86.1 \times h = \text{their Area}\) or using \(80 \sin(\text{their } ACB)\)
A1 for \(57.4\) (or 57.40 to 57.45)

(a) The perimeter consists of two radii of length \(r\) and the arc length of the sector.
Arc length \(= \frac{120}{360} \times 2\pi r = \frac{2}{3}\pi r\).
Perimeter \(= 2r + \frac{2}{3}\pi r = r\left(2 + \frac{2}{3}\pi\right)\text{ cm}\).

(b) Set the perimeter equal to 45:
\(r\left(2 + \frac{2}{3}\pi\right) = 45\)
\(r(2 + 2.0944) = 45\)
\(4.0944 r = 45 \implies r = \frac{45}{4.0944} \approx 10.99\text{ cm} \approx 11.0\text{ cm}\).

(c) (i) When the sector is folded to form a cone, the arc length of the sector becomes the circumference of the circular base of the cone.
Circumference of cone base \(= 2\pi R\).
\(2\pi R = \frac{2}{3}\pi r \implies R = \frac{1}{3}r\).

(ii) The slant height, \(L\), of the cone is equal to the radius of the sector, \(r = 10.99\text{ cm}\).
The radius of the base of the cone is \(R = \frac{1}{3}r = \frac{10.99}{3} \approx 3.663\text{ cm}\).
Using Pythagoras' theorem to find the height, \(h\), of the cone:
\(h = \sqrt{L^2 - R^2} = \sqrt{10.99^2 - 3.663^2} = \sqrt{120.78 - 13.417} = \sqrt{107.36} \approx 10.36\text{ cm} \approx 10.4\text{ cm}\).

(iii) Volume of the cone:
\(V = \frac{1}{3}\pi R^2 h = \frac{1}{3}\pi (3.663)^2 (10.36) \approx \frac{1}{3}\pi (13.418)(10.36) \approx 145.58\text{ cm}^3 \approx 146\text{ cm}^3\).

(a)
M1 for arc length \(= \frac{120}{360} \times 2\pi r\) (or equivalent)
A1 for adding \(2r\) and factorising to show \(r\left(2 + \frac{2}{3}\pi\right)\)

(b)
M1 for setting up the equation \(r\left(2 + \frac{2}{3}\pi\right) = 45\)
M1 for rearranging to make \(r\) the subject: \(r = \frac{45}{2 + \frac{2}{3}\pi}\)
A1 for \(11.0\) (or 10.99 to 11.00)

(c) (i)
M1 for equating the base circumference \(2\pi R\) to the arc length of the sector \(\frac{2}{3}\pi r\)
A1 for completing the division to obtain \(R = \frac{1}{3}r\)

(ii)
M1 for identifying that slant height \(L = r\) and using their value of \(r\)
M1 for using Pythagoras: \(h = \sqrt{L^2 - R^2}\) with their values
A1 for \(10.4\) (or 10.35 to 10.37)

(iii)
M1 for substituting their values of \(R\) and \(h\) into the formula \(V = \frac{1}{3}\pi R^2 h\)
M1 for correct evaluation steps
A1 for \(146\) (or 145 to 146)

(a) First draw probabilities: Red: \(\frac{5}{8}\), Blue: \(\frac{3}{8}\).
Second draw probabilities (no replacement):
- If first was Red: Red remaining \(= 4\), Blue remaining \(= 3\), Total \(= 7\). Probabilities: Red: \(\frac{4}{7}\), Blue: \(\frac{3}{7}\).
- If first was Blue: Red remaining \(= 5\), Blue remaining \(= 2\), Total \(= 7\). Probabilities: Red: \(\frac{5}{7}\), Blue: \(\frac{2}{7}\).

(b) (i) \(P(RR) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14} \approx 0.357\).

(ii) Different colours can occur as Red then Blue (RB) or Blue then Red (BR):
\(P(\text{different}) = P(RB) + P(BR) = \left(\frac{5}{8} \times \frac{3}{7}\right) + \left(\frac{3}{8} \times \frac{5}{7}\right) = \frac{15}{56} + \frac{15}{56} = \frac{30}{56} = \frac{15}{28} \approx 0.536\).

(iii) \(P(\text{at least one blue}) = 1 - P(RR) = 1 - \frac{5}{14} = \frac{9}{14} \approx 0.643\).

(c) Apply the new rules:
- Case 1: First ball is Red (probability \(\frac{5}{8}\)). Since it is replaced, the bag returns to 5 Red and 3 Blue (total 8). The probability of drawing a Red on the second draw is \(\frac{5}{8}\).
- Case 2: First ball is Blue (probability \(\frac{3}{8}\)). It is not replaced, and 1 more Blue is added. The bag now has 5 Red and 3 Blue (total 8). The probability of drawing a Red on the second draw is \(\frac{5}{8}\).

Total probability of second ball being Red:
\(P(\text{Red}_2) = \left(\frac{5}{8} \times \frac{5}{8}\right) + \left(\frac{3}{8} \times \frac{5}{8}\right) = \frac{25}{64} + \frac{15}{64} = \frac{40}{64} = \frac{5}{8} = 0.625\).

(a)
B1 for first draw branches with correct probabilities (\(\frac{5}{8}\), \(\frac{3}{8}\))
B2 for second draw branches with correct probabilities (\(\frac{4}{7}\), \(\frac{3}{7}\) and \(\frac{5}{7}\), \(\frac{2}{7}\))

(b) (i)
M1 for \(\frac{5}{8} \times \frac{4}{7}\)
A1 for \(\frac{5}{14}\) (or equivalent fraction, or 0.357)

(ii)
M1 for \(\frac{5}{8} \times \frac{3}{7}\) or \(\frac{3}{8} \times \frac{5}{7}\)
M1 for adding both products
A1 for \(\frac{15}{28}\) (or equivalent fraction, or 0.536)

(iii)
M1 for \(1 - P(RR)\) or \(P(RB) + P(BR) + P(BB)\)
A1 for \(\frac{9}{14}\) (or equivalent fraction, or 0.643)

(c)
M1 for calculating \(P(\text{Red}_2 \mid \text{Red}_1) = \frac{5}{8}\) or \(P(\text{Red}_2 \mid \text{Blue}_1) = \frac{5}{8}\)
M1 for \(\left(\frac{5}{8} \times \frac{5}{8}\right) + \left(\frac{3}{8} \times \frac{5}{8}\right)\)
A1 for \(0.625\) (or \(\frac{5}{8}\))

(a) First, find the midpoints of each interval:
- \(0 < h \le 10\): midpoint \(= 5\)
- \(10 < h \le 15\): midpoint \(= 12.5\)
- \(15 < h \le 25\): midpoint \(= 20\)
- \(25 < h \le 40\): midpoint \(= 32.5\)

Now, calculate \(\sum f \cdot x\):
\(\sum f \cdot x = 18(5) + 24(12.5) + 38(20) + 20(32.5) = 90 + 300 + 760 + 650 = 1800\).
Calculate the mean:
\(\text{Mean} = \frac{1800}{100} = 18\text{ cm}\).

(b) (i) Cumulative frequencies are:
- Up to 10: 18
- Up to 15: 42
- Up to 25: 80
The median is the 50th/51st value, which lies in the interval \(15 < h \le 25\).

(ii) Cumulative frequency reaches 42 by the end of \(10 < h \le 15\), and 80 by the end of \(15 < h \le 25\). Therefore, the 50th and 51st values must lie in the interval \(15 < h \le 25\).

(c) Frequency Density \(= \frac{\text{Frequency}}{\text{Class Width}}\).
(i) For \(10 < h \le 15\), class width is 5:
\(\text{FD} = \frac{24}{5} = 4.8\).
(ii) For \(25 < h \le 40\), class width is 15:
\(\text{FD} = \frac{20}{15} = \frac{4}{3} \approx 1.33\).

(d) The seedlings with height greater than 20 cm are:
- Part of the \(15 < h \le 25\) interval. Since 20 is exactly halfway, half of the 38 seedlings in this interval are estimated to have height greater than 20 cm:
\(38 \times 0.5 = 19\).
- All of the 20 seedlings in the \(25 < h \le 40\) interval.

Total seedlings with height > 20 cm \(= 19 + 20 = 39\).
Probability \(= \frac{39}{100} = 0.39\).

(a)
M1 for identifying the correct midpoints: \(5\), \(12.5\), \(20\), \(32.5\)
M1 for calculating \(\sum f \cdot x = 1800\) (allow one error in midpoints)
M1 for dividing their sum by 100
A1 for \(18\)

(b) (i)
B1 for \(15 < h \le 25\)

(ii)
B1 for explaining using cumulative frequencies (e.g. cumulative frequency up to 15 is 42, and up to 25 is 80)

(c) (i)
M1 for \(\frac{24}{5}\)
A1 for \(4.8\)

(ii)
M1 for \(\frac{20}{15}\)
A1 for \(1.33\) (or \(1.333\dots\) or \(1\frac{1}{3}\))

(d)
M1 for calculating \(\frac{25 - 20}{25 - 15} \times 38 = 19\)
M1 for adding 20 to their 19
A1 for \(0.39\) (or \(\frac{39}{100}\))

(a) (i) Sequence A has a common difference of 7.
Next term \(= 33 + 7 = 40\).
(ii) Sequence B terms are \(n^3 + 1\):
Next term \(= 6^3 + 1 = 216 + 1 = 217\).

(b) (i) Sequence A is linear with a common difference of 7:
\(n\)-th term \(= 7n + c\).
When \(n = 1\), \(7(1) + c = 5 \implies c = -2\).
\(n\)-th term \(= 7n - 2\).
(ii) Sequence B consists of perfect cubes plus 1:
\(n\)-th term \(= n^3 + 1\).
(iii) Sequence C is a geometric sequence with a common ratio of 4:
\(n\)-th term \(= a \cdot r^{n-1} = 1 \cdot 4^{n-1} = 4^{n-1}\).

(c) Set the \(n\)-th term of Sequence C equal to 4096:
\(4^{k-1} = 4096\)
Since \(4^6 = 4096\):
\(k - 1 = 6 \implies k = 7\).

(d) Calculate the 10th term of Sequence B:
\(10^3 + 1 = 1000 + 1 = 1001\).
Calculate the 10th term of Sequence A:
\(7(10) - 2 = 70 - 2 = 68\).
Difference \(= 1001 - 68 = 933\).

(a) (i)
B1 for \(40\)
(ii)
B1 for \(217\)

(b) (i)
M1 for \(7n + c\) (or any linear term with gradient 7)
A1 for \(7n - 2\)
(ii)
M1 for recognizing \(n^3\)
A1 for \(n^3 + 1\)
(iii)
M1 for powers of 4 (e.g. \(4^n\))
A1 for \(4^{n-1}\) (or equivalent)

(c)
M1 for setting up \(4^{k-1} = 4096\)
A1 for \(k = 7\)

(d)
M1 for finding 10th term of B \(= 1001\)
M1 for finding 10th term of A \(= 68\)
A1 for \(933\)

(a) Substituting the x-values into \(y = x^3 - 3x^2 + 2\):
When \(x = -1\), \(y = (-1)^3 - 3(-1)^2 + 2 = -1 - 3 + 2 = -2\).
When \(x = 1\), \(y = (1)^3 - 3(1)^2 + 2 = 1 - 3 + 2 = 0\).
When \(x = 3\), \(y = (3)^3 - 3(3)^2 + 2 = 27 - 27 + 2 = 2\).

(b) Plotting the points from the table and connecting them with a smooth curve.

(c) Draw the horizontal line \(y = 3\). The intersection with the curve occurs at \(x \approx 3.1\) (accept values in the range \(3.0 \le x \le 3.2\)).

(d) Differentiating term by term:
\(\frac{dy}{dx} = 3x^2 - 6x\).

(e) At \(x = 3\), the y-coordinate is \(y = 2\).
Substituting \(x = 3\) into the derivative to find the gradient \(m\):
\(m = 3(3)^2 - 6(3) = 27 - 18 = 9\).
Using \(y = mx + c\) with point \((3, 2)\):
\(2 = 9(3) + c \implies 2 = 27 + c \implies c = -25\).
So the equation of the tangent is \(y = 9x - 25\).

(a) B2 for all three correct values (-2, 0, 2)
(B1 for two correct values)

(b) M1 for plotting at least 8 points correctly
A1 for a smooth curve through the points
A1 for correct shape with turning points in correct quadrants

(c) M1 for drawing the line \(y = 3\) or showing clear method of finding intersection
A1 for \(x \approx 3.1\) (accept 3.0 to 3.2)

(d) B2 for \(3x^2 - 6x\)
(B1 for \(3x^2\) or \(-6x\))

(e) M1 for substituting \(x = 3\) into their derivative
A1 for gradient = 9
M1 for substituting \((3, 2)\) into \(y = mx + c\) (or equivalent) to find \(c\)
A1 for \(y = 9x - 25\) (or equivalent)

(a)(i) Time taken for the first part: \(\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{36}{x}\) hours.

(a)(ii) Time taken for the second part: \(\text{Time} = \frac{45}{x-3}\) hours.

(b) Since the total time is 6 hours:
\(\frac{36}{x} + \frac{45}{x-3} = 6\)
Divide the entire equation by 3:
\(\frac{12}{x} + \frac{15}{x-3} = 2\)
Multiply by \(x(x-3)\):
\(12(x-3) + 15x = 2x(x-3)\)
\(12x - 36 + 15x = 2x^2 - 6x\)
\(27x - 36 = 2x^2 - 6x\)
Rearranging gives:
\(2x^2 - 33x + 36 = 0\).

(c) Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\(x = \frac{33 \pm \sqrt{(-33)^2 - 4(2)(36)}}{2(2)}\)
\(x = \frac{33 \pm \sqrt{1089 - 288}}{4}\)
\(x = \frac{33 \pm \sqrt{801}}{4}\)
\(x = \frac{33 \pm 28.3019}{4}\)
\(x = 15.325...\) or \(x = 1.174...\)
Correct to 2 decimal places:
\(x = 15.33\) or \(x = 1.17\).

(d) Since the speed of the second part must be positive, \(x - 3 > 0\), so we select \(x \approx 15.325\) km/h (rejecting \(x = 1.17\)).
Time taken for the second part is:
\(\text{Time} = \frac{45}{15.325 - 3} = \frac{45}{12.325} \approx 3.651\) hours.
Converting to hours and minutes:
\(0.651 \times 60 = 39.06\) minutes.
This is 3 hours 39 minutes (to the nearest minute).

(a)(i) B1 for \(\frac{36}{x}\)

(a)(ii) B1 for \(\frac{45}{x-3}\)

(b) M1 for their (a)(i) + their (a)(ii) = 6
M1 for common denominator method: \(36(x-3) + 45x = 6x(x-3)\) or equivalent
M1 for expansion: \(36x - 108 + 45x = 6x^2 - 18x\)
A1 for final convincing step leading to \(2x^2 - 33x + 36 = 0\) with no algebraic errors

(c) B1 for \(\sqrt{(-33)^2 - 4(2)(36)}\) or \(\sqrt{801}\) or \(28.30...\) seen
M1 for \(\frac{-(-33) \pm \sqrt{\text{their } 801}}{2(2)}\)
A1 for 15.33
A1 for 1.17

(d) M1 for selecting the correct positive speed root \(x \approx 15.33\) and calculating \(\frac{45}{x - 3}\)
A1 for \(3.65\) or \(3.6509...\) hours
A1 for 3 hours 39 minutes (accept 3h 39m, or 219 minutes)