2025 Cambridge IGCSE Mathematics (0580) Practice Paper with Answers

To simplify the algebraic fraction, we factorise the numerator and the denominator separately.

1. **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 \(1\).
\(2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3)\).

2. **Factorise the denominator:** \(4x^2 - 1\).
This is a difference of two squares: \((2x)^2 - 1^2 = (2x - 1)(2x + 1)\).

3. **Simplify:**
\(\frac{(2x+1)(x-3)}{(2x-1)(2x+1)} = \frac{x-3}{2x-1}\).

M1 for correctly factorising the numerator as \((2x+1)(x-3)\).
M1 for correctly factorising the denominator as \((2x-1)(2x+1)\).
A0.5 for the fully simplified final fraction: \(\frac{x-3}{2x-1}\).

We can express \(\overrightarrow{OP}\) by choosing a path from \(O\) to \(P\):
\(\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP}\)

First, find the vector \(\overrightarrow{AB}\):
\(\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}\).

Since the ratio \(AP : PB = 3 : 2\), the point \(P\) lies \(\frac{3}{5}\) of the way along the vector \(\overrightarrow{AB}\):
\(\overrightarrow{AP} = \frac{3}{5}\overrightarrow{AB} = \frac{3}{5}(\mathbf{b} - \mathbf{a})\).

Now substitute this back to find \(\overrightarrow{OP}\):
\(\overrightarrow{OP} = \mathbf{a} + \frac{3}{5}(\mathbf{b} - \mathbf{a})\)
\(\overrightarrow{OP} = \mathbf{a} + \frac{3}{5}\mathbf{b} - \frac{3}{5}\mathbf{a}\)
\(\overrightarrow{OP} = \frac{2}{5}\mathbf{a} + \frac{3}{5}\mathbf{b}\).

M1 for finding \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\) (or equivalent).
M1 for writing a correct path equation for \(\overrightarrow{OP}\), e.g., \(\overrightarrow{OA} + \frac{3}{5}\overrightarrow{AB}\).
A0.5 for the simplified final vector: \(\frac{2}{5}\mathbf{a} + \frac{3}{5}\mathbf{b}\) (or equivalent).

1. **Find the gradient of line \(L\):**
\(m_1 = \frac{7 - (-1)}{6 - 2} = \frac{8}{4} = 2\).

2. **Find the gradient of the perpendicular line:**
Since the line is perpendicular, its gradient \(m_2\) satisfies \(m_1 \times m_2 = -1\).
\(m_2 = -\frac{1}{2}\).

3. **Find the midpoint of the segment:**
Midpoint \(M = \left(\frac{2 + 6}{2}, \frac{-1 + 7}{2}\right) = (4, 3)\).

4. **Find the equation of the perpendicular line:**
Using the formula \(y - y_1 = m_2(x - x_1)\) with point \((4, 3)\) and gradient \(-\frac{1}{2}\):
\(y - 3 = -\frac{1}{2}(x - 4)\)
\(y - 3 = -\frac{1}{2}x + 2\)
\(y = -\frac{1}{2}x + 5\).

M1 for finding the gradient of \(L\) is \(2\) and showing the perpendicular gradient is \(-\frac{1}{2}\).
M1 for finding the midpoint \((4, 3)\).
A0.5 for the correct equation: \(y = -\frac{1}{2}x + 5\).

The total surface area of a solid cone consists of the base area plus the curved surface area.

1. **Base Area:**
\(\text{Area} = \pi r^2 = \pi \times 6^2 = 36\pi\text{ cm}^2\).

2. **Curved Surface Area:**
\(\text{Area} = \pi r l = \pi \times 6 \times 10 = 60\pi\text{ cm}^2\).

3. **Total Surface Area:**
\(\text{Total} = 36\pi + 60\pi = 96\pi\text{ cm}^2\).

M1 for calculating the base area as \(36\pi\) or the curved surface area as \(60\pi\).
M1 for finding the sum of both the base area and curved surface area.
A0.5 for the correct final answer: \(96\pi\).

We want the probability of drawing two counters of different colours. Since the selection is without replacement, the total number of counters decreases from 9 to 8 on the second draw.

There are two ways to get different colours:
1. **Red first, then Blue (R, B):**
\(\text{P}(R, B) = \frac{5}{9} \times \frac{4}{8} = \frac{20}{72}\).

2. **Blue first, then Red (B, R):**
\(\text{P}(B, R) = \frac{4}{9} \times \frac{5}{8} = \frac{20}{72}\).

Add the two probabilities together:
\(\text{P}(\text{different}) = \frac{20}{72} + \frac{20}{72} = \frac{40}{72}\).

Simplify the fraction:
\(\frac{40}{72} = \frac{5}{9}\).

M1 for calculating a correct probability for one pathway, e.g., \(\frac{5}{9} \times \frac{4}{8} = \frac{20}{72}\).
M1 for summing both pathways: \(\frac{20}{72} + \frac{20}{72}\).
A0.5 for simplifying the final fraction to \(\frac{5}{9}\).

Let's find the differences between successive terms of the sequence:
- Terms: \(3,\ 10,\ 21,\ 36,\ 55\)
- First differences: \(7,\ 11,\ 15,\ 19\)
- Second differences: \(4,\ 4,\ 4\)

Since the second differences are constant and equal to \(4\), the sequence is quadratic and the coefficient of \(n^2\) is \(a = \frac{4}{2} = 2\).

Now, subtract \(2n^2\) from the original terms to find the linear part:
- For \(n=1\): \(3 - 2(1)^2 = 1\)
- For \(n=2\): \(10 - 2(2)^2 = 2\)
- For \(n=3\): \(21 - 2(3)^2 = 3\)
- For \(n=4\): \(36 - 2(4)^2 = 4\)

The resulting sequence is \(1,\ 2,\ 3,\ 4,\ \dots\), which is simply \(n\).

Therefore, the \(n\)-th term of the sequence is \(2n^2 + n\).

M1 for identifying the second difference is \(4\), which gives the \(2n^2\) term.
M1 for finding the linear part by subtraction or solving simultaneous equations.
A0.5 for the correct expression: \(2n^2 + n\).

1. **Multiply both sides by \(2 - x\) to clear the fraction:**
\(y(2 - x) = 3x - 5\)
\(2y - xy = 3x - 5\).

2. **Group all terms containing \(x\) on one side and constant/y terms on the other:**
\(2y + 5 = 3x + xy\).

3. **Factorise \(x\) out of the right-hand side:**
\(2y + 5 = x(3 + y)\).

4. **Divide by \(3 + y\) to isolate \(x\):**
\(x = \frac{2y + 5}{3 + y}\) (or equivalent).

M1 for clearing the fraction to get \(2y - xy = 3x - 5\).
M1 for gathering \(x\) terms on one side and factorising to get \(x(3 + y) = 2y + 5\).
A0.5 for the correct final answer: \(x = \frac{2y + 5}{3 + y}\) (or equivalent).

We begin by rearranging the equation to isolate \(2\mathbf{w}\):
\(3\mathbf{u} - \mathbf{v} = 2\mathbf{w}\)

Next, compute \(3\mathbf{u}\):
\(3\mathbf{u} = 3\begin{pmatrix} 3 \\ -2 \end{pmatrix} = \begin{pmatrix} 9 \\ -6 \end{pmatrix}\).

Now, subtract \(\mathbf{v}\) from \(3\mathbf{u}\):
\(3\mathbf{u} - \mathbf{v} = \begin{pmatrix} 9 \\ -6 \end{pmatrix} - \begin{pmatrix} -1 \\ 4 \end{pmatrix} = \begin{pmatrix} 9 - (-1) \\ -6 - 4 \end{pmatrix} = \begin{pmatrix} 10 \\ -10 \end{pmatrix}\).

Since \(2\mathbf{w} = \begin{pmatrix} 10 \\ -10 \end{pmatrix}\), divide by \(2\) to find \(\mathbf{w}\):
\(\mathbf{w} = \frac{1}{2}\begin{pmatrix} 10 \\ -10 \end{pmatrix} = \begin{pmatrix} 5 \\ -5 \end{pmatrix}\).

M1 for correctly evaluating \(3\mathbf{u} = \begin{pmatrix} 9 \\ -6 \end{pmatrix}\).
M1 for calculating \(3\mathbf{u} - \mathbf{v} = \begin{pmatrix} 10 \\ -10 \end{pmatrix}\).
A0.5 for the correct column vector: \(\begin{pmatrix} 5 \\ -5 \end{pmatrix}\).

Factorise the numerator: \(2x^2 - 5x - 3 = 2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3)\). Factorise the denominator: \(4x^2 - 1 = (2x - 1)(2x + 1)\). Divide the numerator and denominator by the common factor \((2x + 1)\): \(\frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)} = \frac{x - 3}{2x - 1}\).

M1 for factorising the numerator to \((2x + 1)(x - 3)\). M1 for factorising the denominator to \((2x - 1)(2x + 1)\). A0.5 for the simplified fraction \(\frac{x - 3}{2x - 1}\).

First find \(\overrightarrow{AB}\): \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (5\mathbf{a} + 3\mathbf{b}) - (3\mathbf{a} - \mathbf{b}) = 2\mathbf{a} + 4\mathbf{b}\). Since \(AP : PB = 3 : 1\), the point \(P\) is \(\frac{3}{4}\) of the way along \(AB\) from \(A\): \(\overrightarrow{AP} = \frac{3}{4}\overrightarrow{AB} = \frac{3}{4}(2\mathbf{a} + 4\mathbf{b}) = 1.5\mathbf{a} + 3\mathbf{b}\). Now, find \(\overrightarrow{OP}\): \(\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP} = (3\mathbf{a} - \mathbf{b}) + (1.5\mathbf{a} + 3\mathbf{b}) = 4.5\mathbf{a} + 2\mathbf{b}\).

M1 for finding \(\overrightarrow{AB} = 2\mathbf{a} + 4\mathbf{b}\) or using the formula \(\overrightarrow{OP} = \frac{\overrightarrow{OA} + 3\overrightarrow{OB}}{4}\). M1 for \(\overrightarrow{OP} = \overrightarrow{OA} + \frac{3}{4}\overrightarrow{AB}\) or correct substitution into the formula. A0.5 for the final answer \(4.5\mathbf{a} + 2\mathbf{b}\) (or \(\frac{9}{2}\mathbf{a} + 2\mathbf{b}\)).

1. Find the midpoint of \(AB\): Midpoint \(M = \left(\frac{1 + 5}{2}, \frac{-3 + 5}{2}\right) = (3, 1)\). 2. Find the gradient of \(AB\): \(m_{AB} = \frac{5 - (-3)}{5 - 1} = \frac{8}{4} = 2\). 3. Find the gradient of the perpendicular line: \(m = -\frac{1}{2}\). 4. Find the equation of the line passing through \((3, 1)\) with gradient \(-0.5\): \(y - 1 = -0.5(x - 3) \implies y = -0.5x + 2.5\).

M1 for finding the midpoint \((3, 1)\). M1 for finding the perpendicular gradient \(-0.5\) from the gradient of \(AB = 2\). A0.5 for the final equation \(y = -0.5x + 2.5\) (or \(y = -\frac{1}{2}x + \frac{5}{2}\)).

Let \(h\) be the height of the cone. The volume of the sphere is \(V = \frac{4}{3} \pi r^3\). The volume of the cone is \(V = \frac{1}{3} \pi (2r)^2 h = \frac{4}{3} \pi r^2 h\). Equating the volumes: \(\frac{4}{3} \pi r^2 h = \frac{4}{3} \pi r^3\). Dividing both sides by \(\frac{4}{3} \pi r^2\) gives \(h = r\).

M1 for equating volume of sphere to volume of cone: \(\frac{1}{3} \pi (2r)^2 h = \frac{4}{3} \pi r^3\). M1 for simplifying the cone volume to \(\frac{4}{3} \pi r^2 h\). A0.5 for the final answer \(h = r\) (or \(r\)).

The total number of counters is \(5 + 3 = 8\). The probability of choosing at least one blue counter is \(1 - \text{P}(\text{both red})\). First find \(\text{P}(\text{both red}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}\). Thus, \(\text{P}(\text{at least one blue}) = 1 - \frac{5}{14} = \frac{9}{14}\).

M1 for calculating the probability of two red counters: \(\frac{5}{8} \times \frac{4}{7}\) (or \(\frac{20}{56}\)). M1 for subtracting this from 1, or summing \(\left(\frac{15}{56} + \frac{15}{56} + \frac{6}{56}\right)\). A0.5 for the simplified final fraction \(\frac{9}{14}\).

The terms are \(u_1 = 3, u_2 = 8, u_3 = 15, u_4 = 24\). First differences: \(5, 7, 9\). Second differences: \(2, 2\). Since the second difference is constant and equal to 2, the sequence is quadratic of the form \(an^2 + bn + c\) with \(2a = 2 \implies a = 1\). Subtracting \(n^2\) from each term gives the sequence: \(2, 4, 6, 8, \dots\) which has \(n\)-th term \(2n\). Thus, the \(n\)-th term is \(n^2 + 2n\).

M1 for identifying the second difference is 2. M1 for finding the linear sequence of differences \(2, 4, 6, 8, \dots\) or solving for coefficients. A0.5 for \(n^2 + 2n\) (or \(n(n+2)\)).

Multiply both sides by \((5 - 2x)\): \(y(5 - 2x) = 3x + 2 \implies 5y - 2xy = 3x + 2\). Rearrange to group the \(x\) terms on one side: \(5y - 2 = 3x + 2xy\). Factorise \(x\): \(5y - 2 = x(3 + 2y)\). Divide by \((3 + 2y)\) to isolate \(x\): \(x = \frac{5y - 2}{2y + 3}\).

M1 for multiplying by the denominator and expanding: \(5y - 2xy = 3x + 2\). M1 for grouping \(x\) terms and factorising: \(x(3 + 2y) = 5y - 2\). A0.5 for the final answer \(x = \frac{5y - 2}{2y + 3}\) (or equivalent, such as \(x = \frac{5y - 2}{3 + 2y}\)).

(a) Group the terms:
\( (6x^2 - 15xy) - (4x - 10y) \)
Factorise out the common terms from each group:
\( 3x(2x - 5y) - 2(2x - 5y) \)
Factorise out the common binomial factor \( (2x - 5y) \):
\( (3x - 2)(2x - 5y) \)

(b) Factorise the numerator and the denominator:
Numerator: \( 2x^2 - 5x - 3 = (2x + 1)(x - 3) \)
Denominator: \( 4x^2 - 1 = (2x - 1)(2x + 1) \)
Simplify by dividing out the common factor \( 2x + 1 \):
\( \frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)} = \frac{x - 3}{2x - 1} \)

(a) M1 for finding a partial factorization such as \( 3x(2x - 5y) \) or \( -2(2x - 5y) \)
A1 for final answer \( (3x - 2)(2x - 5y) \)

(b) M1 for factorising numerator into \( (2x + 1)(x - 3) \)
M1 for factorising denominator into \( (2x - 1)(2x + 1) \)
A1 for final simplified answer \( \frac{x - 3}{2x - 1} \)

(a) \( \vec{AB} = \vec{AO} + \vec{OB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a} \)

(b) Since \( AP : PB = 3 : 1 \), we have \( \vec{AP} = \frac{3}{4}\vec{AB} \).
\( \vec{OP} = \vec{OA} + \vec{AP} = \mathbf{a} + \frac{3}{4}(\mathbf{b} - \mathbf{a}) = \mathbf{a} - \frac{3}{4}\mathbf{a} + \frac{3}{4}\mathbf{b} = \frac{1}{4}\mathbf{a} + \frac{3}{4}\mathbf{b} \)

(c) First, find \( \vec{OQ} \):
\( \vec{OQ} = 2\vec{OP} = 2\left(\frac{1}{4}\mathbf{a} + \frac{3}{4}\mathbf{b}\right) = \frac{1}{2}\mathbf{a} + \frac{3}{2}\mathbf{b} \)
Now find \( \vec{BQ} \):
\( \vec{BQ} = \vec{BO} + \vec{OQ} = -\mathbf{b} + \frac{1}{2}\mathbf{a} + \frac{3}{2}\mathbf{b} = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b} \)

(a) B1 for \( \mathbf{b} - \mathbf{a} \)
(b) M1 for writing \( \vec{OP} = \mathbf{a} + \frac{3}{4}\vec{AB} \) or equivalent
A1 for \( \frac{1}{4}\mathbf{a} + \frac{3}{4}\mathbf{b} \)
(c) M1 for finding \( \vec{OQ} = \frac{1}{2}\mathbf{a} + \frac{3}{2}\mathbf{b} \) or writing \( \vec{BQ} = \vec{BO} + \vec{OQ} \)
A1 for \( \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b} \) or equivalent factored form

(a) Calculate the gradient of \( L_1 \):
\( m_1 = \frac{-7 - 5}{4 - (-2)} = \frac{-12}{6} = -2 \)
Using the line equation formula with point \( A(-2, 5) \):
\( y - 5 = -2(x - (-2)) \)
\( y - 5 = -2x - 4 \)
\( y = -2x + 1 \)

(b) The gradient of \( L_2 \) is perpendicular to \( L_1 \):
\( m_2 = -\frac{1}{m_1} = -\frac{1}{-2} = \frac{1}{2} \)
Find the midpoint of \( AB \):
\( M = \left(\frac{-2 + 4}{2}, \frac{5 - 7}{2}\right) = (1, -1) \)
Using the gradient-point form for \( L_2 \) through \( (1, -1) \):
\( y - (-1) = \frac{1}{2}(x - 1) \)
\( y + 1 = \frac{1}{2}x - \frac{1}{2} \)
\( y = \frac{1}{2}x - \frac{3}{2} \)

(a) M1 for finding the gradient of \( L_1 \) as \( -2 \)
M1 for substituting a point into \( y = mx + c \) to find \( c \)
A1 for \( y = -2x + 1 \)

(b) M1 for finding the midpoint \( (1, -1) \) or stating gradient is \( \frac{1}{2} \)
A1 for \( y = \frac{1}{2}x - \frac{3}{2} \)

(a) Total surface area of a cylinder is given by:
\( A = 2\pi r^2 + 2\pi r h \)
Substitute \( h = 3r \):
\( A = 2\pi r^2 + 2\pi r (3r) = 2\pi r^2 + 6\pi r^2 = 8\pi r^2 \)

(b) The volume of the cylinder is:
\( V_{\text{cylinder}} = \pi r^2 h = \pi r^2 (3r) = 3\pi r^3 \)
The volume of a sphere of radius \( R \) is:
\( V_{\text{sphere}} = \frac{4}{3}\pi R^3 \)
Since the cylinder is melted to make the sphere, their volumes are equal:
\( \frac{4}{3}\pi R^3 = 3\pi r^3 \)
Divide both sides by \( \pi \):
\( \frac{4}{3}R^3 = 3r^3 \)
Multiply by \( \frac{3}{4} \):
\( R^3 = \frac{9}{4}r^3 \)
Take the cube root of both sides:
\( R = \sqrt[3]{\frac{9}{4}r^3} = r\sqrt[3]{\frac{9}{4}} \)

(a) M1 for applying the cylinder surface area formula with \( h = 3r \)
A1 for final answer \( 8\pi r^2 \)

(b) M1 for expressing volume of cylinder as \( 3\pi r^3 \)
M1 for equating \( \frac{4}{3}\pi R^3 = 3\pi r^3 \)
A1 for final answer \( R = r\sqrt[3]{\frac{9}{4}} \) or equivalent, e.g. \( R = r\left(\frac{9}{4}\right)^{1/3} \)

(a) The total number of balls is \( 4 + 6 = 10 \).
To get the same colour, we can either draw two red balls or two blue balls:
\( P(\text{Red, Red}) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} \)
\( P(\text{Blue, Blue}) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} \)
\( P(\text{Same Colour}) = P(\text{Red, Red}) + P(\text{Blue, Blue}) = \frac{12}{90} + \frac{30}{90} = \frac{42}{90} \)
Simplifying the fraction:
\( \frac{42}{90} = \frac{7}{15} \)

(b) The probability of drawing at least one red ball is equal to \( 1 - P(\text{No Red Ball}) \).
No red ball means both balls are blue:
\( P(\text{Blue, Blue}) = \frac{30}{90} = \frac{1}{3} \)
Therefore, \( P(\text{At least one Red}) = 1 - \frac{1}{3} = \frac{2}{3} \).

(a) M1 for drawing with replacement/probabilities with denominator 9
M1 for adding \( P(\text{R, R}) \) and \( P(\text{B, B}) \) together
A1 for simplified final answer \( \frac{7}{15} \)

(b) M1 for \( 1 - P(\text{B, B}) \) or summing \( P(\text{R, R}) + P(\text{R, B}) + P(\text{B, R}) \)
A1 for final answer \( \frac{2}{3} \)

(a) For \( n = 1 \):
\( u_1 = a(1)^2 + b(1) + 3 = 9 \Rightarrow a + b + 3 = 9 \Rightarrow a + b = 6 \) (Equation 1)
For \( n = 2 \):
\( u_2 = a(2)^2 + b(2) + 3 = 21 \Rightarrow 4a + 2b + 3 = 21 \Rightarrow 4a + 2b = 18 \Rightarrow 2a + b = 9 \) (Equation 2)
Subtracting Equation 1 from Equation 2:
\( (2a + b) - (a + b) = 9 - 6 \Rightarrow a = 3 \)
Substitute \( a = 3 \) into Equation 1:
\( 3 + b = 6 \Rightarrow b = 3 \).

(b) Substitute \( n = 10 \) into \( u_n = 3n^2 + 3n + 3 \):
\( u_{10} = 3(10)^2 + 3(10) + 3 = 300 + 30 + 3 = 333 \)

(c) Let \( 3n^2 + 3n + 3 = 500 \):
\( 3(n^2 + n + 1) = 500 \)
\( n^2 + n + 1 = \frac{500}{3} \)
Since 500 is not divisible by 3 (the sum of its digits is 5, which is not a multiple of 3), \( \frac{500}{3} \) is not an integer. For any integer term number \( n \), \( n^2 + n + 1 \) must be an integer. Therefore, 500 cannot be a term in this sequence.

(a) M1 for writing \( a + b = 6 \) or \( 4a + 2b = 18 \)
M1 for solving the simultaneous equations to find \( a \) or \( b \)
A1 for completing the proof clearly showing both values

(b) B1 for \( 333 \)

(c) B1 for a correct mathematical reason showing why 500 cannot be a term, e.g., showing 500 is not divisible by 3 or solving the quadratic to get a non-integer value of \( n \).

(a) Find a common denominator, which is \( (x-2)(x+3) \):
\( \frac{3(x+3) - 2(x-2)}{(x-2)(x+3)} \)
Expand the numerator:
\( \frac{3x + 9 - 2x + 4}{(x-2)(x+3)} = \frac{x + 13}{(x-2)(x+3)} \)

(b) Factorise the quadratic equation \( 2x^2 - 7x - 4 = 0 \):
Find two numbers that multiply to \( 2 \times (-4) = -8 \) and add to \( -7 \). These are \( -8 \) and \( 1 \).
\( 2x^2 - 8x + x - 4 = 0 \)
\( 2x(x - 4) + 1(x - 4) = 0 \)
\( (2x + 1)(x - 4) = 0 \)
So, \( 2x + 1 = 0 \Rightarrow x = -\frac{1}{2} \)
or \( x - 4 = 0 \Rightarrow x = 4 \)

(a) M1 for writing with a common denominator, e.g., \( 3(x+3) - 2(x-2) \) over the common denominator
A1 for \( \frac{x+13}{(x-2)(x+3)} \) or equivalent

(b) M1 for factorising into \( (2x+a)(x+b) \) where \( ab=-4 \) or \( a+2b=-7 \)
A1 for \( (2x + 1)(x - 4) = 0 \)
A1 for solutions \( x = -\frac{1}{2} \) and \( x = 4 \)

(a) The curved surface area of a sphere of radius \( R \) is \( 4\pi R^2 \).
Therefore, the curved surface area of a hemisphere is \( 2\pi R^2 \):
\( A = 2\pi (6)^2 = 2\pi (36) = 72\pi \text{ cm}^2 \)

(b) Substitute \( d = 3 \) into the volume formula:
\( V = \frac{1}{3}\pi (3)^2 (18 - 3) \)
\( V = \frac{1}{3}\pi (9)(15) \)
\( V = 3\pi (15) = 45\pi \text{ cm}^3 \)

(a) M1 for using \( 2\pi r^2 \) with \( r = 6 \)
A1 for \( 72\pi \)

(b) M1 for substituting \( d = 3 \) into the volume formula
M1 for performing correct operations, e.g., calculating \( 3^2 = 9 \) and \( 18 - 3 = 15 \)
A1 for final answer \( 45\pi \)

For (a), the intersection with the \(y\)-axis occurs where \(x = 0\):
\(y = (0 - 1)(0 + 2)(3 - 0) = (-1)(2)(3) = -6\).
So the coordinates are \((0, -6)\).

For (b), the intersections with the \(x\)-axis occur where \(y = 0\):
\((x - 1)(x + 2)(3 - x) = 0\)
This yields \(x = 1\), \(x = -2\), and \(x = 3\).
So the coordinates are \((1, 0)\), \((-2, 0)\), and \((3, 0)\).

For (c), expanding the expression for \(y\) gives:
\(y = (x^2 + x - 2)(3 - x) = -x^3 + 2x^2 + 5x - 6\).
As \(x \to \infty\), the term with the highest power, \(-x^3\), dominates. Since the coefficient of \(x^3\) is negative, the value of \(y\) goes to \(-\infty\).

For (d), to sketch the curve:
1. Plot the \(x\)-intercepts at \((-2, 0)\), \((1, 0)\), and \((3, 0)\).
2. Plot the \(y\)-intercept at \((0, -6)\).
3. Draw a smooth, continuous cubic curve with a negative \(x^3\) shape, starting in the upper-left (quadrant 2), passing through \((-2,0)\), dipping down to \((0,-6)\), rising to a local maximum between \(x=1\) and \(x=3\), and heading down to the lower-right (quadrant 4) after passing through \((3,0)\).

(a) M1 for substituting \(x = 0\) into the equation. A0.5 for \((0, -6)\) or \(y = -6\).
(b) B2 for all three points correct (deduct 0.5 marks for each incorrect or missing coordinate up to 2 marks).
(c) B1 for stating \(-\infty\) with a correct reason mentioning the negative coefficient of the \(x^3\) term.
(d) B1 for drawing/describing a negative cubic curve shape; B1 for correctly passing through \((-2, 0)\), \((1, 0)\), and \((3, 0)\); B1 for correctly passing through \((0, -6)\).

For (a), the reflection of a point \((x, y)\) in the line \(y = -x\) maps the point to \((-y, -x)\).
Applying this rule:
\(A(1, 1) \to A'(-1, -1)\)
\(B(3, 1) \to B'(-1, -3)\)
\(C(1, 4) \to C'(-4, -1)\)

For (b), a rotation of \(90^\circ\) clockwise about the origin maps \((x, y)\) to \((y, -x)\).
Applying this rule:
\(A(1, 1) \to A''(1, -1)\)
\(B(3, 1) \to B''(1, -3)\)
\(C(1, 4) \to C''(4, -1)\)

For (c), let us compare the coordinates of triangle \(U\) and triangle \(V\):
\(A'(-1, -1) \to A''(1, -1)\)
\(B'(-1, -3) \to B''(1, -3)\)
\(C'(-4, -1) \to C''(4, -1)\)
We can see that each point \((x, y)\) of triangle \(U\) maps to \((-x, y)\) in triangle \(V\). This is a reflection across the vertical axis, which is the \(y\)-axis or the line \(x = 0\).

(a) B3 for all three vertices correct (B1 for each correct vertex: \(A'(-1, -1)\), \(B'(-1, -3)\), \(C'(-4, -1)\)).
(b) B3 for all three vertices correct (B1 for each correct vertex: \(A''(1, -1)\), \(B''(1, -3)\), \(C''(4, -1)\)).
(c) B0.5 for identifying the transformation as a 'reflection'; B1 for giving the correct line of reflection 'y-axis' or 'x = 0' (Do not accept 'reflection in the x-axis').

For (a), on a cumulative frequency diagram, cumulative frequencies are plotted against the upper class boundary of each interval. Thus, the 6 coordinates to plot are:
\((120, 0)\), \((140, 12)\), \((160, 38)\), \((180, 65)\), \((200, 76)\), and \((220, 80)\).

For (b)(i), the total frequency is 80.
1. Calculate the position of the median: \(80 \div 2 = 40\).
2. Locate 40 on the vertical cumulative frequency axis.
3. Draw a horizontal line from 40 to intersect the cumulative frequency curve.
4. From the intersection point, draw a vertical line down to the horizontal axis and read the corresponding completion time.

For (b)(ii):
1. The lower quartile (LQ) corresponds to a cumulative frequency of \(0.25 \times 80 = 20\).
2. The upper quartile (UQ) corresponds to a cumulative frequency of \(0.75 \times 80 = 60\).
3. Use the curve to find the completion times corresponding to cumulative frequencies of 20 and 60 (let these times be \(T_{20}\) and \(T_{60}\)).
4. Calculate the IQR by subtracting the lower quartile time from the upper quartile time: \(\text{IQR} = T_{60} - T_{20}\).

(a) B3 for all 6 coordinates correctly identified (B2 for 5 correct, B1 for 3 or 4 correct, B0 if fewer than 3 are correct).
(b)(i) B0.5 for identifying cumulative frequency of 40; B1 for explaining how to read the corresponding time on the horizontal axis.
(b)(ii) B1 for identifying cumulative frequencies of 20 (lower quartile) and 60 (upper quartile); B1 for explaining how to find their corresponding times from the curve; B1 for stating that the IQR is the difference between these two times (Upper Quartile - Lower Quartile).