2024 Edexcel IGCSE Mathematics (Specification B) Practice Paper with Answers

The determinant of matrix \(\mathbf{M}\) is given by \(\det(\mathbf{M}) = k(k-2) - (4 \times 2) = k^2 - 2k - 8\). Setting this equal to 16 gives \(k^2 - 2k - 8 = 16\), which simplifies to \(k^2 - 2k - 24 = 0\). Factoring the quadratic expression, we get \((k-6)(k+4) = 0\). This yields two solutions, \(k = 6\) and \(k = -4\). Since we are looking for the positive value of \(k\), we have \(k = 6\).

M1: Setting up the determinant equation \(k(k-2) - 8 = 16\) and simplifying to a 3-term quadratic equation, e.g., \(k^2 - 2k - 24 = 0\). A1: Correctly solving the quadratic equation and selecting the positive solution \(k = 6\).

Factorise the numerator: \(3x^2 - 12 = 3(x^2 - 4) = 3(x-2)(x+2)\). Factorise the denominator: \(x^2 - x - 6 = (x-3)(x+2)\). Divide out the common factor of \((x+2)\) from both the numerator and the denominator: \(\frac{3(x-2)(x+2)}{(x-3)(x+2)} = \frac{3(x-2)}{x-3} = \frac{3x-6}{x-3}\).

M1: Factorising either the numerator to \(3(x-2)(x+2)\) or the denominator to \((x-3)(x+2)\). A1: Completely simplified correct answer, either in the form \(\frac{3(x-2)}{x-3}\) or \(\frac{3x-6}{x-3}\).

First list the elements of each set: \(\mathcal{U} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}\). The prime numbers in this range are \(A = \{2, 3, 5, 7, 11\}\). The multiples of 3 in this range are \(B = \{3, 6, 9, 12\}\). Find the union \(A \cup B = \{2, 3, 5, 6, 7, 9, 11, 12\}\). The complement of the union is \((A \cup B)' = A' \cap B' = \{1, 4, 8, 10\}\). Therefore, the number of elements in \(A' \cap B'\) is 4.

M1: Finding the union set \(A \cup B\) or listing elements of the complement set \(A' \cap B'\). A1: Correct answer of 4.

Find the upper bound of the length: Since it is to the nearest metre, the upper bound is \(80.5\text{ m}\). Find the upper bound of the width: Since it is to the nearest \(5\text{ m}\), the upper bound is \(45 + 2.5 = 47.5\text{ m}\). The upper bound for the area is the product of these two upper bounds: \(80.5 \times 47.5 = 3823.75\text{ m}^2\).

M1: Finding the correct upper bounds for length (\(80.5\)) and width (\(47.5\)) and multiplying them. A1: Correct calculation giving \(3823.75\).

The probability of choosing either a blue or a yellow counter is \(1 - 0.3 = 0.7\). Since the ratio of blue to yellow is \(2:5\), the fraction of the non-red counters that are blue is \(\frac{2}{2+5} = \frac{2}{7}\). Therefore, the probability of choosing a blue counter is \(\frac{2}{7} \times 0.7 = 0.2\).

M1: Subtracting the probability of red from 1 to get \(0.7\), and establishing that the probability of blue is \(\frac{2}{7}\) of this value. A1: Correct probability of \(0.2\) (or equivalent fraction).

First evaluate the inner function \(\text{g}(2) = \frac{3}{2+1} = \frac{3}{3} = 1\). Next, substitute this output into the outer function \(\text{f}\) to find \(\text{fg}(2) = \text{f}(1) = 2(1) - 5 = -3\).

M1: Evaluating \(\text{g}(2) = 1\) or expressing the composite function \(\text{fg}(x) = 2\left(\frac{3}{x+1}\right) - 5\). A1: Correct final answer of \(-3\).

The volume of the sphere is \(V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (3)^3 = 36\pi\text{ cm}^3\). The volume of a cylinder is given by \(V = \pi R^2 h\). Equating the volumes, we have \(\pi (2)^2 h = 36\pi\), which simplifies to \(4\pi h = 36\pi\). Dividing both sides by \(4\pi\) gives \(h = 9\text{ cm}\).

M1: Writing down a correct volume equation for the sphere or the cylinder with substituted values, such as \(\frac{4}{3}\pi (3)^3 = \pi (2)^2 h\). A1: Correct height of \(9\text{ cm}\).

Find the sum of the vectors: \(\mathbf{a} + \mathbf{b} = \begin{pmatrix} 5+3 \\ -12+k \end{pmatrix} = \begin{pmatrix} 8 \\ k-12 \end{pmatrix}\). The magnitude is \(|\mathbf{a} + \mathbf{b}| = \sqrt{8^2 + (k-12)^2} = 17\). Squaring both sides yields \(64 + (k-12)^2 = 289\). Solving for the term in brackets: \((k-12)^2 = 225\), which gives \(k-12 = 15\) or \(k-12 = -15\). This gives \(k = 27\) or \(k = -3\). Since \(k > 0\), the correct value of \(k\) is 27.

M1: Correctly adding the two vectors to get \(\begin{pmatrix} 8 \\ k-12 \end{pmatrix}\) and setting up the magnitude equation \(\sqrt{8^2 + (k-12)^2} = 17\). A1: Correct value of \(k = 27\) (ignoring negative solution).

To write \(\frac{3}{2x - 1} - \frac{2}{x + 3}\) as a single fraction, first find a common denominator: \((2x - 1)(x + 3)\). Express each fraction with the common denominator: \(\frac{3(x + 3)}{(2x - 1)(x + 3)} - \frac{2(2x - 1)}{(2x - 1)(x + 3)}\). Combine the numerators: \(\frac{3(x + 3) - 2(2x - 1)}{(2x - 1)(x + 3)}\). Expand the numerator: \(\frac{3x + 9 - 4x + 2}{(2x - 1)(x + 3)}\). Simplify the terms: \(\frac{11 - x}{(2x - 1)(x + 3)}\).

M1 for obtaining a common denominator of \((2x - 1)(x + 3)\) and expressing at least one numerator correctly. M1 for expanding and simplifying the numerator to \(11 - x\). A0.67 for the final correct simplified fraction: \(\frac{11 - x}{(2x - 1)(x + 3)}\) or \(\frac{11 - x}{2x^2 + 5x - 3}\).

The determinant of a matrix \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\) is given by \(ad - bc\). For \(\mathbf{A}\): \(\det(\mathbf{A}) = y(y - 1) - (4 \times 3)\), so \(\det(\mathbf{A}) = y^2 - y - 12\). We are given that the determinant is 8: \(y^2 - y - 12 = 8 \implies y^2 - y - 20 = 0\). Factorise the quadratic equation: \((y - 5)(y + 4) = 0\). Therefore, \(y = 5\) or \(y = -4\).

M1 for setting up the determinant equation: \(y(y - 1) - 12 = 8\). M1 for forming the standard quadratic equation: \(y^2 - y - 20 = 0\) and attempting to solve by factorisation or formula. A0.67 for both correct solutions: \(y = 5\) and \(y = -4\).

Let \(C\) be the set of students studying Chemistry, and \(P\) be the set of students studying Physics. The total number of students is 30, and 5 study neither, so the number of students studying Chemistry or Physics (or both) is: \(n(C \cup P) = 30 - 5 = 25\). Using the set union formula: \(n(C \cup P) = n(C) + n(P) - n(C \cap P)\). Substitute the known values: \(25 = 18 + 15 - n(C \cap P) \implies 25 = 33 - n(C \cap P) \implies n(C \cap P) = 33 - 25 = 8\). So, 8 students study both.

M1 for finding the number of students studying at least one subject: \(30 - 5 = 25\). M1 for setting up the equation \(25 = 18 + 15 - x\) (or showing equivalent work on a Venn diagram). A0.67 for the correct final answer of 8.

First, simplify \(\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}\). So the expression becomes \(\frac{6}{2\sqrt{3} - 3}\). To rationalise the denominator, multiply the numerator and the denominator by the conjugate \(2\sqrt{3} + 3\): \(\frac{6(2\sqrt{3} + 3)}{(2\sqrt{3} - 3)(2\sqrt{3} + 3)}\). Expand the numerator and the denominator: Numerator: \(6(2\sqrt{3} + 3) = 12\sqrt{3} + 18\). Denominator: \((2\sqrt{3})^2 - 3^2 = 12 - 9 = 3\). Combine and divide: \(\frac{12\sqrt{3} + 18}{3} = 4\sqrt{3} + 6\). Therefore, \(a = 4\) and \(b = 6\).

M1 for writing \(\sqrt{12}\) as \(2\sqrt{3}\) or multiplying numerator/denominator directly by \(\sqrt{12} + 3\). M1 for rationalising the denominator: multiplying by the conjugate and correctly expanding the denominator to get 3 (or 3 for the unsimplified equivalent). A0.67 for the final answer in the form \(4\sqrt{3} + 6\).

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

M1 for setting up the equation for the area of the sector to find the fraction of the circle (\(\frac{1}{3}\) or \(\theta = 120^\circ\)). M1 for calculating the arc length as \(8\pi\). A0.67 for the total perimeter: \(8\pi + 24\) (or \(24 + 8\pi\)).

The probability that Sarah wins tennis is 0.6, so the probability that she loses tennis is \(1 - 0.6 = 0.4\). The probability that Sarah wins chess is 0.35, so the probability that she loses chess is \(1 - 0.35 = 0.65\). There are two mutually exclusive ways she can win exactly one game: 1) Wins tennis and loses chess: \(P(\text{Wins T and Loses C}) = 0.6 \times 0.65 = 0.39\). 2) Loses tennis and wins chess: \(P(\text{Loses T and Wins C}) = 0.4 \times 0.35 = 0.14\). Total probability of winning exactly one game: \(0.39 + 0.14 = 0.53\).

M1 for calculating the complementary probabilities (0.4 and 0.65) or setting up the correct calculation: \(0.6(1-0.35) + (1-0.6)0.35\). M1 for summing the two correct probabilities: \(0.39 + 0.14\). A0.67 for the correct probability of 0.53.

First, find the vector \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \mathbf{b} - \mathbf{a}\). Since \(AP:PB = 3:2\), the point \(P\) divides the line \(AB\) in the ratio 3 to 2, so: \(\overrightarrow{AP} = \frac{3}{5}\overrightarrow{AB} = \frac{3}{5}(\mathbf{b} - \mathbf{a})\). Now, express \(\overrightarrow{OP}\): \(\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP} = \mathbf{a} + \frac{3}{5}(\mathbf{b} - \mathbf{a})\). Expand and simplify: \(\overrightarrow{OP} = \mathbf{a} + \frac{3}{5}\mathbf{b} - \frac{3}{5}\mathbf{a} = \frac{2}{5}\mathbf{a} + \frac{3}{5}\mathbf{b}\).

M1 for writing \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\) or \(\overrightarrow{AP} = \frac{3}{5}(\mathbf{b} - \mathbf{a})\). M1 for setting up the route \(\overrightarrow{OP} = \overrightarrow{OA} + \frac{3}{5}\overrightarrow{AB}\). A0.67 for the final simplified vector: \(\frac{2}{5}\mathbf{a} + \frac{3}{5}\mathbf{b}\) or equivalent fraction/decimal form.

Multiply both sides of the equation by the common denominator \(x(x-1)\):

\(4x - 3(x-1) = x(x-1)\)

Expand the brackets on both sides:

\(4x - 3x + 3 = x^2 - x\)

Simplify the left-hand side:

\(x + 3 = x^2 - x\)

Rearrange the terms to form a quadratic equation set to zero:

\(x^2 - 2x - 3 = 0\)

Factorise the quadratic expression:

\((x-3)(x+1) = 0\)

Therefore, the solutions are \(x = 3\) or \(x = -1\).

M1: For multiplying by the common denominator \(x(x-1)\) to get a correct unsimplified equation, e.g., \(4x - 3(x-1) = x(x-1)\).
M1: For expanding the brackets correctly on both sides, e.g., \(4x - 3x + 3 = x^2 - x\).
M1: For rearranging into the standard quadratic form \(x^2 - 2x - 3 = 0\).
M1: For correctly factorising their quadratic equation, e.g., \((x-3)(x+1) = 0\) (or using the quadratic formula).
A1: For both correct solutions: \(x = 3\) and \(x = -1\).

First, write down the product of the matrices \(\mathbf{A}\) and \(\mathbf{B}\):

\(\mathbf{A}\mathbf{B} = \begin{pmatrix} 2 & p \\ 1 & -3 \end{pmatrix} \begin{pmatrix} q & -2 \\ 1 & r \end{pmatrix} = \begin{pmatrix} 2q + p & -4 + pr \\ q - 3 & -2 - 3r \end{pmatrix}\)

We are given that \(\mathbf{A}\mathbf{B} = \begin{pmatrix} 11 & 10 \\ -1 & -8 \end{pmatrix}\). By equating corresponding elements, we get:

1) \(2q + p = 11\)
2) \(-4 + pr = 10\)
3) \(q - 3 = -1\)
4) \(-2 - 3r = -8\)

From equation (3):
\(q = -1 + 3 = 2\)

From equation (4):
\(-3r = -6 \Rightarrow r = 2\)

Now, substitute \(q = 2\) into equation (1):
\(2(2) + p = 11 \Rightarrow 4 + p = 11 \Rightarrow p = 7\)

Verify with equation (2):
\(-4 + (7)(2) = -4 + 14 = 10\), which is consistent.

Thus, \(p = 7\), \(q = 2\), and \(r = 2\).

M1: For expressing \(\mathbf{A}\mathbf{B}\) with at least two correct algebraic elements, e.g., \(2q+p\) and \(q-3\).
M1: For setting up the equation \(q - 3 = -1\) or \(-2 - 3r = -8\).
M1: For setting up the equation \(2q + p = 11\) or \(-4 + pr = 10\).
A1: For finding \(q = 2\) and \(r = 2\).
A1: For finding \(p = 7\).

Let \(H\) be the set of students studying History and \(G\) be the set of students studying Geography.

We are given:
\(n(U) = 80\)
\(n(H) = 45\)
\(n(G) = 40\)
\(n(H \cup G)' = 12\)

Therefore, the number of students who study History or Geography (or both) is:
\(n(H \cup G) = 80 - 12 = 68\)

Using the formula for union of sets:
\(n(H \cup G) = n(H) + n(G) - n(H \cap G)\)
\(68 = 45 + 40 - n(H \cap G)\)
\(68 = 85 - n(H \cap G)\)
\(n(H \cap G) = 17\)

(a) The probability that the student studies both subjects is:
\(\mathrm{P}(H \cap G) = \frac{17}{80}\)

(b) The number of students who study History but not Geography is:
\(n(H \cap G') = n(H) - n(H \cap G) = 45 - 17 = 28\)

The probability is:
\(\mathrm{P}(H \cap G') = \frac{28}{80} = \frac{7}{20}\) (or \(0.35\))

(c) The probability that the student studies Geography given that they study History is:
\(\mathrm{P}(G | H) = \frac{n(H \cap G)}{n(H)} = \frac{17}{45}\)

M1: For finding the number of students studying at least one of the subjects, \(n(H \cup G) = 80 - 12 = 68\).
A1: For finding the number of students studying both, \(n(H \cap G) = 17\), and the probability \(\frac{17}{80}\).
A1: For calculating the probability of History only as \(\frac{28}{80}\) or \(\frac{7}{20}\).
M1: For using the conditional probability formula \(\mathrm{P}(G|H) = \frac{n(H \cap G)}{n(H)}\) with their values.
A1: For the correct probability of \(\frac{17}{45}\).

The total volume \(V\) of the ornament is the sum of the volume of the hemisphere and the volume of the cone:

\(V = V_{\text{hemisphere}} + V_{\text{cone}}\)

\(V = \frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2 h\)

We are given that \(h = \frac{4}{3}r\). Substitute this into the volume equation:

\(V = \frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2 \left(\frac{4}{3}r\right) = \frac{2}{3}\pi r^3 + \frac{4}{9}\pi r^3\)

Combine the terms:

\(V = \left(\frac{6}{9} + \frac{4}{9}\right)\pi r^3 = \frac{10}{9}\pi r^3\)

We are given \(V = 30\pi\), so:

\(\frac{10}{9}\pi r^3 = 30\pi \Rightarrow \frac{10}{9}r^3 = 30\)

\(r^3 = 30 \times \frac{9}{10} = 27 \Rightarrow r = 3\ \mathrm{cm}\)

Now, calculate the height of the cone:

\(h = \frac{4}{3}(3) = 4\ \mathrm{cm}\)

Find the slant height \(l\) of the cone using Pythagoras' theorem:

\(l = \sqrt{r^2 + h^2} = \sqrt{3^2 + 4^2} = \sqrt{25} = 5\ \mathrm{cm}\)

The total surface area \(A\) of the ornament consists of the curved surface area of the hemisphere and the curved surface area of the cone (the flat bases are joined inside):

\(A = 2\pi r^2 + \pi r l\)

\(A = 2\pi (3^2) + \pi (3)(5) = 18\pi + 15\pi = 33\pi\ \mathrm{cm}^2\)

Thus, the total surface area is \(33\pi\).

M1: For expressing the volume of the ornament in terms of \(r\) only by substituting \(h = \frac{4}{3}r\), e.g., \(\frac{2}{3}\pi r^3 + \frac{4}{9}\pi r^3 = 30\pi\).
M1: For solving to find \(r = 3\).
M1: For using Pythagoras' theorem to find the slant height \(l = 5\).
M1: For stating the formula for total surface area of this combined solid, \(2\pi r^2 + \pi r l\).
A1: For obtaining \(33\pi\).

Let \(x = 0.2\dot{3}\dot{6}\), which means:

\(x = 0.2363636...\)

Multiply \(x\) by 10 to bring the non-repeating part to the left of the decimal point:

\(10x = 2.363636...\) (Equation 1)

Multiply \(x\) by 1000 to bring one full cycle of the repeating part to the left of the decimal point:

\(1000x = 236.363636...\) (Equation 2)

Subtract Equation 1 from Equation 2:

\(1000x - 10x = 236.3636... - 2.3636...\)

\(990x = 234\)

Solve for \(x\):

\(x = \frac{234}{990}\)

To simplify the fraction, divide both the numerator and the denominator by their greatest common divisor, which is 18:

\(234 \div 18 = 13\)

\(990 \div 18 = 55\)

So,

\(x = \frac{13}{55}\) (as required).

M1: For setting up the definition, e.g., let \(x = 0.23636...\).
M1: For multiplying by appropriate powers of 10 to align the recurring decimals, e.g., showing both \(10x = 2.3636...\) and \(1000x = 236.3636...\) (or equivalent equations).
M1: For subtracting the two equations to eliminate the recurring decimal part, e.g., obtaining \(990x = 234\).
M1: For writing \(x = \frac{234}{990}\).
A1: For clearly showing the simplification step to arrive at \(\frac{13}{55}\).

First, let us find expressions for \(\overrightarrow{OQ}\) and \(\overrightarrow{BP}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).

Since \(Q\) is the midpoint of \(AB\):
\(\overrightarrow{OQ} = \frac{1}{2}(\mathbf{a} + \mathbf{b})\)

Since \(OP : PA = 2 : 1\), we have \(\overrightarrow{OP} = \frac{2}{3}\mathbf{a}\).
\(\overrightarrow{BP} = \overrightarrow{BO} + \overrightarrow{OP} = -\mathbf{b} + \frac{2}{3}\mathbf{a} = \frac{2}{3}\mathbf{a} - \mathbf{b}\)

Now, express \(\overrightarrow{OX}\) in two different ways using the parameters \(\mu\) and \(\lambda\):

First way:
\(\overrightarrow{OX} = \mu \overrightarrow{OQ} = \mu \left(\frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\right) = \frac{1}{2}\mu\mathbf{a} + \frac{1}{2}\mu\mathbf{b}\) (Equation 1)

Second way:
\(\overrightarrow{OX} = \overrightarrow{OB} + \overrightarrow{BX} = \mathbf{b} + \lambda\overrightarrow{BP} = \mathbf{b} + \lambda\left(\frac{2}{3}\mathbf{a} - \mathbf{b}\right) = \frac{2}{3}\lambda\mathbf{a} + (1 - \lambda)\mathbf{b}\) (Equation 2)

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

For \(\mathbf{a}\):
\(\frac{1}{2}\mu = \frac{2}{3}\lambda\) (Equation 3)

For \(\mathbf{b}\):
\(\frac{1}{2}\mu = 1 - \lambda\) (Equation 4)

Equating Equation 3 and Equation 4:
\(\frac{2}{3}\lambda = 1 - \lambda\)

Multiply by 3:
\(2\lambda = 3 - 3\lambda \Rightarrow 5\lambda = 3 \Rightarrow \lambda = \frac{3}{5}\)

Now substitute \(\lambda = \frac{3}{5}\) into Equation 4:
\(\frac{1}{2}\mu = 1 - \frac{3}{5} = \frac{2}{5} \Rightarrow \mu = \frac{4}{5}\).

Thus, \(\mu = \frac{4}{5}\) and \(\lambda = \frac{3}{5}\).

M1: For finding \(\overrightarrow{OQ} = \frac{1}{2}(\mathbf{a} + \mathbf{b})\) or \(\overrightarrow{BP} = \frac{2}{3}\mathbf{a} - \mathbf{b}\).
M1: For setting up two expressions for \(\overrightarrow{OX}\), one using \(\mu\) and one using \(\lambda\).
M1: For equating coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) to obtain two simultaneous equations, e.g., \(\frac{1}{2}\mu = \frac{2}{3}\lambda\) and \(\frac{1}{2}\mu = 1 - \lambda\).
A1: For solving to find \(\lambda = \frac{3}{5}\) (or 0.6) or \(\mu = \frac{4}{5}\) (or 0.8).
A1: For finding both correct values: \(\mu = \frac{4}{5}\, \lambda = \frac{3}{5}\).

First, find the expression for the composite function \(\mathrm{fg}(x)\):

\(\mathrm{fg}(x) = \mathrm{f}(\mathrm{g}(x)) = 3\left(\frac{2}{x-1}\right) - 4 = \frac{6}{x-1} - 4\)

Next, find the inverse function \(\mathrm{g}^{-1}(x)\):

Let \(y = \mathrm{g}(x) = \frac{2}{x-1}\)

Rearrange to make \(x\) the subject:

\(y(x - 1) = 2 \Rightarrow x - 1 = \frac{2}{y} \Rightarrow x = \frac{2}{y} + 1\)

So,

\(\mathrm{g}^{-1}(x) = \frac{2}{x} + 1\)

Now set \(\mathrm{fg}(x) = \mathrm{g}^{-1}(x)\):

\(\frac{6}{x-1} - 4 = \frac{2}{x} + 1\)

Add 4 to both sides:

\(\frac{6}{x-1} = \frac{2}{x} + 5\)

Subtract \(\frac{2}{x}\) from both sides:

\(\frac{6}{x-1} - \frac{2}{x} = 5\)

Multiply the entire equation by the common denominator \(x(x-1)\):

\(6x - 2(x-1) = 5x(x-1)\)

Expand the brackets:

\(6x - 2x + 2 = 5x^2 - 5x\)

\(4x + 2 = 5x^2 - 5x\)

Rearrange to form a quadratic equation:

\(5x^2 - 9x - 2 = 0\)

Factorise the quadratic equation:

\((5x + 1)(x - 2) = 0\)

Thus, \(x = 2\) or \(x = -\frac{1}{5}\) (or \(-0.2\)).

M1: For finding the composite function \(\mathrm{fg}(x) = \frac{6}{x-1} - 4\).
M1: For finding the inverse function \(\mathrm{g}^{-1}(x) = \frac{2}{x} + 1\).
M1: For equating and multiplying through by the common denominator \(x(x-1)\) to eliminate fractions.
M1: For obtaining a correct quadratic equation, e.g., \(5x^2 - 9x - 2 = 0\).
A1: For solving to find both correct solutions: \(x = 2\) and \(x = -0.2\) (or \(-\frac{1}{5}\)).

(a) The area of a triangle is given by the formula:

\(\text{Area} = \frac{1}{2} a c \sin B\)

Substitute the given values into the formula:

\(33 = \frac{1}{2} \times 11 \times 8 \times \sin(\angle ABC)\)

\(33 = 44 \sin(\angle ABC)\)

\(\sin(\angle ABC) = \frac{33}{44} = 0.75\)

Since angle \(ABC\) is obtuse (\(90^\circ < \angle ABC < 180^\circ\)):

\(\angle ABC = 180^\circ - \arcsin(0.75) \approx 180^\circ - 48.59^\circ = 131.41^\circ\)

To 1 decimal place, \(\angle ABC = 131.4^\circ\).

(b) To find the length of \(AC\), we use the Cosine Rule:

\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)\)

\(AC^2 = 8^2 + 11^2 - 2(8)(11)\cos(131.41^\circ)\)

\(AC^2 = 64 + 121 - 176(-0.66144)\)

\(AC^2 = 185 + 116.41 = 301.41\)

\(AC = \sqrt{301.41} \approx 17.36\ \mathrm{cm}\)

To 3 significant figures, \(AC = 17.4\ \mathrm{cm}\).

M1: For using the area formula, setting up the equation \(33 = \frac{1}{2} \times 8 \times 11 \times \sin(\angle ABC)\).
A1: For finding the correct obtuse angle \(\angle ABC = 131.4^\circ\) (accept \(131.4^\circ\) to \(131.41^\circ\)).
M1: For applying the Cosine Rule to find \(AC\): \(AC^2 = 8^2 + 11^2 - 2(8)(11)\cos(\text{their angle})\).
M1: For evaluating \(AC^2\) to get approximately \(301.4\) (or using their values correctly).
A1: For the correct length of \(AC = 17.4\ \mathrm{cm}\) (accept \(17.3 - 17.5\) depending on rounding).

(a) First, find the product \(\mathbf{A}\mathbf{B}\): \(\mathbf{A}\mathbf{B} = \begin{pmatrix} 3 & x \\ -1 & 2 \end{pmatrix} \begin{pmatrix} 1 & 3 \\ 2 & y \end{pmatrix} = \begin{pmatrix} 3(1) + x(2) & 3(3) + x(y) \\ -1(1) + 2(2) & -1(3) + 2(y) \end{pmatrix} = \begin{pmatrix} 3 + 2x & 9 + xy \\ 3 & -3 + 2y \end{pmatrix}\). We are given that \(\mathbf{A}\mathbf{B} = \begin{pmatrix} 11 & 17 \\ 3 & 1 \end{pmatrix}\). Comparing corresponding elements: From row 1, column 1: \(3 + 2x = 11 \implies 2x = 8 \implies x = 4\). From row 2, column 2: \(-3 + 2y = 1 \implies 2y = 4 \implies y = 2\). Checking row 1, column 2 with these values: \(9 + xy = 9 + (4)(2) = 17\), which is correct. (b) Since \(y = 2\), \(\mathbf{B} = \begin{pmatrix} 1 & 3 \\ 2 & 2 \end{pmatrix}\). The determinant of \(\mathbf{B}\) is: \(\det(\mathbf{B}) = (1)(2) - (3)(2) = 2 - 6 = -4\). The inverse matrix \(\mathbf{B}^{-1}\) is: \(\mathbf{B}^{-1} = \frac{1}{-4} \begin{pmatrix} 2 & -3 \\ -2 & 1 \end{pmatrix} = \begin{pmatrix} -0.5 & 0.75 \\ 0.5 & -0.25 \end{pmatrix}\).

M1: for attempting to multiply matrices A and B (at least two elements correct in terms of x and/or y). A1: for setting up correct equations and finding x = 4 and y = 2. M1: for a correct method to find the determinant of B using their value of y. M1: for a correct method to find the adjoint matrix of B. A1: for the correct inverse matrix B^{-1} = \begin{pmatrix} -0.5 & 0.75 \\ 0.5 & -0.25 \end{pmatrix} (or equivalent fraction form).

(a) Since \(OP : PA = 2 : 1\), we have \(\overrightarrow{OP} = \frac{2}{3}\overrightarrow{OA} = \frac{2}{3}\mathbf{a}\). Since \(Q\) is the midpoint of \(AB\), \(\overrightarrow{OQ} = \frac{1}{2}(\mathbf{a} + \mathbf{b}) = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\). Now, find \(\overrightarrow{PQ}\): \(\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP} = \left(\frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\right) - \frac{2}{3}\mathbf{a} = \left(\frac{1}{2} - \frac{2}{3}\right)\mathbf{a} + \frac{1}{2}\mathbf{b} = -\frac{1}{6}\mathbf{a} + \frac{1}{2}\mathbf{b}\). (b) Express \(\overrightarrow{OR}\) using \(\overrightarrow{PR} = k\overrightarrow{PQ}\): \(\overrightarrow{OR} = \overrightarrow{OP} + \overrightarrow{PR} = \frac{2}{3}\mathbf{a} + k\left(-\frac{1}{6}\mathbf{a} + \frac{1}{2}\mathbf{b}\right) = \left(\frac{2}{3} - \frac{1}{6}k\right)\mathbf{a} + \frac{1}{2}k\mathbf{b}\). We are given \(\overrightarrow{OR} = \frac{1}{3}\mathbf{a} + \mathbf{b}\). Comparing the coefficients of \(\mathbf{b}\): \(\frac{1}{2}k = 1 \implies k = 2\). Let us verify with the coefficient of \(\mathbf{a}\): \(\frac{2}{3} - \frac{1}{6}(2) = \frac{2}{3} - \frac{1}{3} = \frac{1}{3}\), which matches the coefficient of \(\mathbf{a}\) in \(\overrightarrow{OR}\). Thus, \(k = 2\).

M1: for expressing OP = 2/3 a and OQ = 1/2(a + b). M1: for the correct vector subtraction method to find PQ. A1: for the correct simplified expression PQ = -1/6 a + 1/2 b. M1: for expressing OR in terms of a, b, and k and equating coefficients. A1: for finding k = 2.

First, equate the volume of the cylinder to the volume of the sphere. The volume of the cylinder is: \(V_c = \pi r^2 h = \pi r^2 (3r) = 3\pi r^3\). The volume of the sphere is: \(V_s = \frac{4}{3}\pi R^3\). Since \(V_c = V_s\), we have: \(3\pi r^3 = \frac{4}{3}\pi R^3 \implies 9r^3 = 4R^3 \implies \frac{r^3}{R^3} = \frac{4}{9}\). Next, find the total surface area of the cylinder \(A_c\): \(A_c = 2\pi r^2 + 2\pi r h = 2\pi r^2 + 2\pi r(3r) = 8\pi r^2\). The surface area of the sphere \(A_s\) is: \(A_s = 4\pi R^2\). Now, express the ratio of the surface areas: \(\frac{A_c}{A_s} = \frac{8\pi r^2}{4\pi R^2} = 2\left(\frac{r}{R}\right)^2\). Cubing both sides of this ratio gives: \(\left(\frac{A_c}{A_s}\right)^3 = \left(2\left(\frac{r}{R}\right)^2\right)^3 = 8\left(\frac{r}{R}\right)^6 = 8\left(\frac{r^3}{R^3}\right)^2\). Substitute \(\frac{r^3}{R^3} = \frac{4}{9}\) into the expression: \(\left(\frac{A_c}{A_s}\right)^3 = 8\left(\frac{4}{9}\right)^2 = 8 \times \frac{16}{81} = \frac{128}{81}\). Since 128 and 81 have no common factors, they are coprime. Therefore, \(a = 128\) and \(b = 81\).

M1: for equating the volume of the cylinder to the volume of the sphere to find a relationship between r and R (e.g., r^3/R^3 = 4/9 or R = (9/4)^(1/3) r). A1: for the correct volume relationship. M1: for writing down correct expressions for total surface area of both the cylinder (Ac = 8*pi*r^2) and the sphere (As = 4*pi*R^2). M1: for cubing the ratio of the surface areas and substituting the volume relationship. A1: for finding a = 128 and b = 81.

(a) We are given \(\mathrm{f}^{-1}(5) = 4\), which is equivalent to \(\mathrm{f}(4) = 5\). Substitute \(x = 4\) into the definition of \(\mathrm{f}(x)\): \(\mathrm{f}(4) = \frac{2(4) + c}{4 - 3} = 5 \implies \frac{8 + c}{1} = 5 \implies 8 + c = 5 \implies c = -3\). (b) We want to solve \(\mathrm{g}\mathrm{f}(x) = 8\). Since \(\mathrm{g}(x) = 3x - 4\), the equation becomes: \(3\mathrm{f}(x) - 4 = 8 \implies 3\mathrm{f}(x) = 12 \implies \mathrm{f}(x) = 4\). Substitute the expression for \(\mathrm{f}(x)\) with \(c = -3\): \(\frac{2x - 3}{x - 3} = 4 \implies 2x - 3 = 4(x - 3) \implies 2x - 3 = 4x - 12 \implies 2x = 9 \implies x = 4.5\).

M1: for using the property f(4) = 5 to set up an equation in terms of c. A1: for finding c = -3. M1: for setting up the composite equation g(f(x)) = 8 or equivalent (e.g., f(x) = g^-1(8) = 4). M1: for resolving the equation 2x - 3 = 4(x - 3) to solve for x. A1: for finding x = 4.5 (or 9/2).

(a) Factorise the numerator and denominator of the first fraction: \( 3x^2 + 10x + 3 = (3x+1)(x+3) \) and \( x^2 + x - 6 = (x-2)(x+3) \). So, \( \frac{3x^2 + 10x + 3}{x^2 + x - 6} = \frac{(3x+1)(x+3)}{(x-2)(x+3)} = \frac{3x+1}{x-2} \). Now subtract the second fraction: \( f(x) = \frac{3x+1}{x-2} - \frac{x-1}{x-2} = \frac{(3x+1) - (x-1)}{x-2} = \frac{2x+2}{x-2} \). (b) Substitute the simplified expression into the given equation: \( \frac{2x+2}{x-2} + \frac{6}{x+3} = 2 \). Multiply the entire equation by \( (x-2)(x+3) \): \( (2x+2)(x+3) + 6(x-2) = 2(x-2)(x+3) \) which can be written as \( 2(x+1)(x+3) + 6(x-2) = 2(x^2 + x - 6) \). Dividing both sides by 2 gives: \( (x+1)(x+3) + 3(x-2) = x^2 + x - 6 \). Expand and simplify: \( x^2 + 4x + 3 + 3x - 6 = x^2 + x - 6 \) which simplifies to \( x^2 + 7x - 3 = x^2 + x - 6 \). Subtracting \( x^2 \) and \( x \) from both sides: \( 6x = -3 \implies x = -0.5 \).

(a) M1: Factorises \( 3x^2 + 10x + 3 \) correctly as \( (3x+1)(x+3) \). M1: Factorises \( x^2 + x - 6 \) correctly as \( (x-2)(x+3) \). M1: Cancels the common term \( (x+3) \) to get \( \frac{3x+1}{x-2} \). A1: Correctly subtracts and simplifies the numerator to obtain \( \frac{2x+2}{x-2} \). (b) M1: Sets up the equation \( \frac{2x+2}{x-2} + \frac{6}{x+3} = 2 \) and attempts to clear the denominators. M1: Expands brackets correctly: \( (2x+2)(x+3) = 2x^2 + 8x + 6 \) and \( 2(x-2)(x+3) = 2x^2 + 2x - 12 \). M1: Simplifies the equation to the linear form \( 6x = -3 \) or equivalent. A1: Obtains \( x = -0.5 \) or equivalent fraction.

(a) For a matrix to be singular, its determinant must be 0. \(\det(\mathbf{A}) = k(k-1) - (2)(3) = k^2 - k - 6 = 0\). Factoring gives \((k-3)(k+2) = 0\), so \(k = 3\) or \(k = -2\). (b) When \(k = 4\), \(\mathbf{A} = \begin{pmatrix} 4 & 2 \\ 3 & 3 \end{pmatrix}\). The determinant is \(\det(\mathbf{A}) = 4(3) - 2(3) = 6\). The inverse is \(\mathbf{A}^{-1} = \frac{1}{6} \begin{pmatrix} 3 & -2 \\ -3 & 4 \end{pmatrix} = \begin{pmatrix} 0.5 & -1/3 \\ -0.5 & 2/3 \end{pmatrix}\). (c) The area of the image is given by \(|\det(\mathbf{A})| \times \text{Area of } T\). For \(k = 4\), the determinant is 6. Area of \(T'\) = \(6 \times 5 = 30\text{ cm}^2\).

(a) M1: Sets up determinant equation \(k(k-1) - 6 = 0\). M1: Solves quadratic equation by factoring or formula. A1: Obtains \(k = 3\) and \(k = -2\). (b) M1: Calculates determinant as 6. M1: Correctly swaps elements and negates off-diagonal elements. A1: Obtains correct inverse matrix. (c) M1: Knows that Area of \(T'\) = \(|\det(\mathbf{A})| \times \text{Area of } T\). M1: Multiplies determinant by 5. A1: Obtains \(30\text{ cm}^2\).

(a) The total number of counters is \(n+7\). The probability of drawing a red counter first is \(\frac{7}{n+7}\). The probability of drawing a red counter second is \(\frac{6}{n+6}\). Thus, the probability both are red is \(\frac{7}{n+7} \times \frac{6}{n+6} = \frac{42}{(n+7)(n+6)}\) or \(\frac{42}{n^2+13n+42}\). (b) Setting this expression equal to \(\frac{7}{22}\) gives \(\frac{42}{(n+7)(n+6)} = \frac{7}{22}\). Divide both sides by 7 to get \(\frac{6}{(n+7)(n+6)} = \frac{1}{22}\). Cross-multiplying gives \(132 = (n+7)(n+6)\). Expanding the brackets: \(132 = n^2 + 13n + 42\). Rearranging gives \(n^2 + 13n - 90 = 0\). (c) Factorise the quadratic: \((n+18)(n-5) = 0\). Since \(n\) must be positive, \(n = 5\). The total number of counters is 12. The probability of different colours is P(Red, Blue) + P(Blue, Red) = \(\frac{7}{12} \times \frac{5}{11} + \frac{5}{12} \times \frac{7}{11} = \frac{35}{132} + \frac{35}{132} = \frac{70}{132} = \frac{35}{66}\).

(a) M1: Correct fraction for first draw \(\frac{7}{n+7}\) or second draw \(\frac{6}{n+6}\). A1: Correct simplified product \(\frac{42}{(n+7)(n+6)}\). (b) M1: Equates expression to \(\frac{7}{22}\). M1: Multiplies out brackets to get \(n^2 + 13n + 42\). M1: Rearranges into quadratic equation form. A1: Correctly completes proof. (c) M1: Solves quadratic to find \(n = 5\). M1: Calculates sum of P(RB) and P(BR). A1: Correct probability of \(\frac{35}{66}\) or equivalent decimal (approx 0.530).

(a) Total Volume \(V = \text{Volume of cylinder} + \text{Volume of hemisphere} = \pi r^2 h + \frac{2}{3}\pi r^3\). Since \(V = 120\pi\), we have \(\pi r^2 h + \frac{2}{3}\pi r^3 = 120\pi\). Dividing by \(\pi\) gives \(r^2 h + \frac{2}{3}r^3 = 120\). Solving for \(h\): \(r^2 h = 120 - \frac{2}{3}r^3 \implies h = \frac{120}{r^2} - \frac{2}{3}r\). (b) Total surface area \(A\) is the sum of the curved area of the hemisphere, the curved area of the cylinder, and the bottom circular base of the cylinder: \(A = 2\pi r^2 + 2\pi r h + \pi r^2 = 3\pi r^2 + 2\pi r h\). Substitute \(h\): \(A = 3\pi r^2 + 2\pi r \left(\frac{120}{r^2} - \frac{2}{3}r\right) = 3\pi r^2 + \frac{240\pi}{r} - \frac{4}{3}\pi r^2 = \pi \left( \left(3 - \frac{4}{3}\right)r^2 + \frac{240}{r} \right) = \pi \left(\frac{5}{3}r^2 + \frac{240}{r}\right)\). (c) If \(A = 95\pi\), then \(\pi \left(\frac{5}{3}r^2 + \frac{240}{r}\right) = 95\pi \implies \frac{5}{3}r^2 + \frac{240}{r} = 95\). Multiply by \(3r\) to clear the denominators: \(5r^3 + 720 = 285r \implies 5r^3 - 285r + 720 = 0\). To verify \(r = 3\), substitute \(r = 3\) into the equation: \(5(3)^3 - 285(3) + 720 = 5(27) - 855 + 720 = 135 - 855 + 720 = 0\). Since the equation holds, \(r = 3\) is verified as a solution.

(a) M1: Recalls volume formulas and sets up equation \(\pi r^2 h + \frac{2}{3}\pi r^3 = 120\pi\). M1: Divides by \(\pi\) and attempts to isolate \(h\). A1: Obtains \(h = \frac{120}{r^2} - \frac{2}{3}r\) or equivalent. (b) M1: Recalls surface area components and sets up \(A = 3\pi r^2 + 2\pi r h\). M1: Substitutes expression for \(h\). A1: Correctly simplifies to given form \(A = \pi \left(\frac{5}{3}r^2 + \frac{240}{r}\right)\). (c) M1: Equates expression to \(95\pi\) and multiplies by \(3r\). A1: Shows derivation of \(5r^3 - 285r + 720 = 0\). B1: Substitutes \(r=3\) to show it equals 0.

(a) Since \(OP : PA = 2 : 1\), \(\overrightarrow{OP} = \frac{2}{3}\mathbf{a}\). (i) \(\overrightarrow{BP} = \overrightarrow{BO} + \overrightarrow{OP} = -\mathbf{b} + \frac{2}{3}\mathbf{a} = \frac{2}{3}\mathbf{a} - \mathbf{b}\). (ii) Since \(Q\) is the midpoint of \(AB\), \(\overrightarrow{OQ} = \overrightarrow{OA} + \frac{1}{2}\overrightarrow{AB} = \mathbf{a} + \frac{1}{2}(\mathbf{b} - \mathbf{a}) = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\). (b) First way: \(\overrightarrow{OX} = \lambda \overrightarrow{OQ} = \frac{1}{2}\lambda \mathbf{a} + \frac{1}{2}\lambda \mathbf{b}\). Second way: \(\overrightarrow{OX} = \overrightarrow{OB} + \overrightarrow{BX} = \mathbf{b} + \mu \overrightarrow{BP} = \mathbf{b} + \mu\left(\frac{2}{3}\mathbf{a} - \mathbf{b}\right) = \frac{2}{3}\mu \mathbf{a} + (1 - \mu)\mathbf{b}\). Equating the coefficients of \(\mathbf{a}\) and \(\mathbf{b}\): \(\frac{1}{2}\lambda = \frac{2}{3}\mu\) and \(\frac{1}{2}\lambda = 1 - \mu\). This implies \(\frac{2}{3}\mu = 1 - \mu \implies \frac{5}{3}\mu = 1 \implies \mu = \frac{3}{5}\). Then \(\lambda = \frac{4}{3}\mu = \frac{4}{3} \times \frac{3}{5} = \frac{4}{5}\). (c) Since \(\overrightarrow{BX} = \mu \overrightarrow{BP} = \frac{3}{5}\overrightarrow{BP}\), the point \(X\) lies \(\frac{3}{5}\) of the way along \(BP\). Therefore, \(BX : XP = 3 : 2\).

(a) B1: Correct expression for \(\overrightarrow{BP}\). B1: Correct expression for \(\overrightarrow{OQ}\). (b) M1: Expresses \(\overrightarrow{OX}\) in terms of \(\lambda\). M1: Expresses \(\overrightarrow{OX}\) in terms of \(\mu\). M1: Equates coefficients to set up simultaneous equations. A1: Solves to find \(\mu = \frac{3}{5}\). A1: Solves to find \(\lambda = \frac{4}{5}\). (c) B1: Correct ratio \(3:2\).

(a) Let \(y = \frac{3x+2}{x-1}\). Multiply by \(x-1\): \(y(x-1) = 3x+2 \implies xy - y = 3x+2\). Rearrange to collect \(x\) terms: \(xy - 3x = y+2 \implies x(y-3) = y+2\). Thus, \(x = \frac{y+2}{y-3}\). Replacing \(y\) with \(x\), we get \(\mathrm{f}^{-1}(x) = \frac{x+2}{x-3}\), for \(x \neq 3\). (b) The equation \(\mathrm{fg}(x) = 4\) can be solved by applying \(\mathrm{f}^{-1}\) to both sides: \(\mathrm{g}(x) = \mathrm{f}^{-1}(4)\). Using the formula from part (a): \(\mathrm{f}^{-1}(4) = \frac{4+2}{4-3} = 6\). So, \(\mathrm{g}(x) = 6 \implies 2x^2 - 3 = 6 \implies 2x^2 = 9 \implies x^2 = 4.5\). Therefore, \(x = \pm \sqrt{4.5} = \pm \frac{3\sqrt{2}}{2} \approx \pm 2.12\) (to 3 sf). (c) Since \(2x^2 \geq 0\) for all real values of \(x\), the minimum value of \(\mathrm{g}(x) = 2x^2 - 3\) is \(-3\). Hence, the range is \(\mathrm{g}(x) \geq -3\).

(a) M1: Sets up equation \(y = \frac{3x+2}{x-1}\) and attempts to clear denominator. M1: Rearranges to isolate \(x\). A1: Correctly identifies \(\mathrm{f}^{-1}(x) = \frac{x+2}{x-3}\). (b) M1: Realises \(\mathrm{g}(x) = \mathrm{f}^{-1}(4)\). M1: Solves \(\mathrm{f}^{-1}(4) = 6\). M1: Sets \(2x^2 - 3 = 6\) and attempts to solve for \(x\). A1: Obtains \(x = \pm \frac{3\sqrt{2}}{2}\) or \(\pm 2.12\). (c) M1: Considers the minimum value of \(2x^2 - 3\) or draws graph. A1: Range is \(\mathrm{g}(x) \geq -3\) (accept \(y \geq -3\)).

(a) The Venn diagram consists of three intersecting circles \(P\), \(V\), and \(F\). The regions are defined as follows: The central intersection \(P \cap V \cap F\) is \(x\). The regions representing exactly two instruments are: \(P \cap V\) only is \(15-x\), \(P \cap F\) only is \(10-x\), \(V \cap F\) only is \(12-x\). The regions representing exactly one instrument are: \(P\) only is \(48 - (15-x) - (10-x) - x = 23 + x\). \(V\) only is \(42 - (15-x) - (12-x) - x = 15 + x\). \(F\) only is \(30 - (10-x) - (12-x) - x = 8 + x\). The region outside all three circles is 10. (b) The total number of students is 100, so we sum all eight regions: \((23+x) + (15+x) + (8+x) + (15-x) + (10-x) + (12-x) + x + 10 = 100 \implies 93 + x = 100 \implies x = 7\). (c) The number of students who play exactly two instruments is \((15-x) + (10-x) + (12-x) = (15-7) + (10-7) + (12-7) = 8 + 3 + 5 = 16\). The probability is \(\frac{16}{100} = 0.16\).

(a) M1: Represents intersections correctly as \(15-x\), \(10-x\), and \(12-x\). M1: Calculates singleton regions in terms of \(x\) correctly (e.g., \(23+x\)). A2: All regions correct (including the outside value of 10). (b) M1: Sums all regions and sets equal to 100. M1: Simplifies equation to \(93 + x = 100\). A1: Obtains \(x = 7\). (c) M1: Identifies regions representing exactly two instruments and adds them. A1: Obtains \(\frac{16}{100} = 0.16\) or equivalent fraction.

(a) Let Arthur's initial investment be \(P = 4x\) and Beatrice's initial investment be \(Q = 5x\). After 3 years, the value of Arthur's investment is \(A_3 = 4x \times (1.03)^3 = 4x \times 1.092727 = 4.370908x\). After 3 years, the value of Beatrice's investment is \(B_3 = 5x \times (1.025)^3 = 5x \times 1.076891 = 5.384453x\). The ratio \(A_3 : B_3\) is \(4.370908x : 5.384453x\). Dividing both parts by \(4.370908x\) gives \(1 : 1.23188... \approx 1 : 1.232\). (b) For \(P = 8000\), after 8 years, the total value of Arthur's investment is \(8000 \times (1.03)^8 = 8000 \times 1.266770 = 10134.16\) (to 2 decimal places). (c) Since \(P : Q = 4 : 5\), Beatrice's initial investment is \(Q = \frac{5}{4} \times 8000 = 10000\). After 8 years, the value of Beatrice's investment is \(10000 \times (1.025)^8 = 10000 \times 1.218403 = 12184.03\) (to 2 decimal places). Arthur's interest is \(I_A = 10134.16 - 8000 = 2134.16\). Beatrice's interest is \(I_B = 12184.03 - 10000 = 2184.03\). The difference is \(I_B - I_A = 2184.03 - 2134.16 = 49.87\).

(a) M1: Uses multipliers \(1.03\) and \(1.025\). M1: Calculates \(4 \times (1.03)^3\) and \(5 \times (1.025)^3\). M1: Divides to get the ratio in the form \(1 : r\). A1: Correctly rounds to \(1 : 1.232\). (b) M1: Sets up calculation \(8000 \times (1.03)^8\). A1: Obtains \(10134.16\) (allow 10134). (c) M1: Finds Beatrice's initial investment of \(10000\). M1: Calculates Beatrice's final value as \(12184.03\) or interest as \(2184.03\). A1: Finds correct difference of \(49.87\) (allow 49.86 to 49.88).

(a) Setting up the matrix multiplication for point \(P\):
\(\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 2a - b \\ 2c - d \end{pmatrix} = \begin{pmatrix} 7 \\ 4 \end{pmatrix}\)
This gives two equations:
1) \(2a - b = 7\)
2) \(2c - d = 4\)

Setting up the matrix multiplication for point \(Q\):
\(\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} -3 \\ 5 \end{pmatrix} = \begin{pmatrix} -3a + 5b \\ -3c + 5d \end{pmatrix} = \begin{pmatrix} -14 \\ -13 \end{pmatrix}\)
This gives two more equations:
3) \(-3a + 5b = -14\)
4) \(-3c + 5d = -13\)

From (1), \(b = 2a - 7\). Substituting into (3):
\(-3a + 5(2a - 7) = -14 \implies 7a - 35 = -14 \implies 7a = 21 \implies a = 3\).
Then \(b = 2(3) - 7 = -1\).

From (2), \(d = 2c - 4\). Substituting into (4):
\(-3c + 5(2c - 4) = -13 \implies 7c - 20 = -13 \implies 7c = 7 \implies c = 1\).
Then \(d = 2(1) - 4 = -2\).

So, \(\mathbf{A} = \begin{pmatrix} 3 & -1 \\ 1 & -2 \end{pmatrix}\).

(b) To find \(\mathbf{A}^{-1}\):
\(\det(\mathbf{A}) = 3(-2) - (-1)(1) = -6 + 1 = -5\).
\(\mathbf{A}^{-1} = -\frac{1}{5} \begin{pmatrix} -2 & 1 \\ -1 & 3 \end{pmatrix} = \begin{pmatrix} 0.4 & -0.2 \\ 0.2 & -0.6 \end{pmatrix}\).

(c) Let \(R\) have coordinates \((x, y)\). Since \(\mathbf{A}R = R'\), we have \(R = \mathbf{A}^{-1} R'\):
\(\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0.4 & -0.2 \\ 0.2 & -0.6 \end{pmatrix} \begin{pmatrix} 5 \\ 10 \end{pmatrix} = \begin{pmatrix} 0.4(5) - 0.2(10) \\ 0.2(5) - 0.6(10) \end{pmatrix} = \begin{pmatrix} 2 - 2 \\ 1 - 6 \end{pmatrix} = \begin{pmatrix} 0 \\ -5
\end{pmatrix}\).
So the coordinates of \(R\) are \((0, -5)\).

(a) M1: Setting up at least two simultaneous equations involving \(a, b\) or \(c, d\).
M1: Correctly solving for \(a, b\) or \(c, d\).
A1: Finding all four values \(a=3, b=-1, c=1, d=-2\) correctly.
A1: Writing down the correct matrix \(\mathbf{A}\).
(b) M1: For calculating the determinant \(-5\) or showing the adjoint matrix pattern.
A1: For correct inverse matrix \(\mathbf{A}^{-1}\) (either with fractions or decimals).
(c) M1: For attempting to multiply \(\mathbf{A}^{-1}\) by \(\begin{pmatrix} 5 \\ 10 \end{pmatrix}\) (order must be correct).
A1: For finding \(x = 0\) and \(y = -5\).
A1: Expressing the final answer as coordinates \((0, -5)\).

(a) The total surface area \(A\) of a closed cylinder is given by:
\(A = 2\pi r^2 + 2\pi r h\)
Given that \(A = 150\pi\):
\(2\pi r^2 + 2\pi r h = 150\pi\)
Divide the entire equation by \(2\pi\):
\(r^2 + rh = 75\)
Rearranging for \(h\):
\(rh = 75 - r^2 \implies h = \frac{75 - r^2}{r}\) (as required).

(b) The volume \(V\) of a cylinder is given by:
\(V =
\pi r^2 h\)
Substitute the expression for \(h\) from part (a):
\(V = \pi r^2 \left( \frac{75 - r^2}{r} \right)\)
\(V = \pi r (75 - r^2) = \pi (75r - r^3)\) (as required).

(c) To find the maximum volume, differentiate \(V\) with respect to \(r\):
\(\frac{dV}{dr} = \pi (75 - 3r^2)\)
Set \(\frac{dV}{dr} = 0\) for stationary points:
\(\pi (75 - 3r^2) = 0 \implies 75 - 3r^2 = 0 \implies r^2 = 25\)
Since radius must be positive, \(r = 5\) cm.

To confirm it is a maximum, check the second derivative:
\(\frac{d^2V}{dr^2} = \pi(-6r) = -6\pi r\)
When \(r = 5\), \(\frac{d^2V}{dr^2} = -30\pi < 0\), confirming a maximum.

Substitute \(r = 5\) back into the volume formula:
\(V = \pi(75(5) - 5^3) = \pi(375 - 125) = 250\pi\) cm\(^3\).

(a) M1: Writing down the correct total surface area formula \(2\pi r^2 + 2\pi r h = 150\pi\).
M1: Rearranging to make \(h\) the subject.
A1: Showing clearly that \(h = \frac{75 - r^2}{r}\).
(b) M1: Substituting the expression for \(h\) into the volume formula \(V = \pi r^2 h\).
A1: Simplifying algebra to obtain \(V = \pi(75r - r^3)\).
(c) M1: For differentiating to find \(\frac{dV}{dr} = \pi(75 - 3r^2)\).
M1: Setting \(\frac{dV}{dr} = 0\) and solving for \(r\).
A1: Finding \(r = 5\).
A1: Correctly substituting \(r = 5\) to find maximum volume \(250\pi\).

(a) The probability of choosing a red marble first is \(\frac{5}{n}\).
Since the marble is not replaced, the probability of choosing a red marble second is \(\frac{4}{n-1}\).
\(P(\text{both red}) = \frac{5}{n} \times \frac{4}{n-1} = \frac{20}{n(n-1)}\).

(b) The number of blue marbles is \(n - 5\).
Two marbles of different colours can be chosen in two ways: Red then Blue, or Blue then Red.
\(P(\text{different}) = P(R, B) + P(B, R)\)
\(P(\text{different}) = \left(\frac{5}{n} \times \frac{n-5}{n-1}\right) + \left(\frac{n-5}{n} \times \frac{5}{n-1}\right)\)
\(P(\text{different}) = \frac{5(n-5) + 5(n-5)}{n(n-1)} = \frac{10(n-5)}{n(n-1)}\).

(c) (i) The probability of choosing two blue marbles is:
\(P(\text{both blue}) = \frac{n-5}{n} \times \frac{n-6}{n-1} = \frac{(n-5)(n-6)}{n(n-1)}\)
We are given that this probability is \(\frac{7}{22}\):
\(\frac{n^2 - 11n + 30}{n^2 - n} = \frac{7}{22}\)
Cross-multiplying yields:
\(22(n^2 - 11n + 30) = 7(n^2 - n)\)
\(22n^2 - 242n + 660 = 7n^2 - 7n\)
\(15n^2 - 235n + 660 = 0\)
Dividing the entire equation by 5:
\(3n^2 - 47n + 132 = 0\) (as required).

(ii) Factorising the quadratic equation:
\((3n - 11)(n - 12) = 0\)
This gives \(n = \frac{11}{3}\) or \(n = 12\).
Since \(n\) must be a positive integer, we reject \(n = \frac{11}{3}\).
Thus, \(n = 12\).

(a) B1: Correct probability \(\frac{20}{n(n-1)}\).
(b) M1: Identifying the two scenarios \(P(R, B)\) and \(P(B, R)\) with correct denominators.
M1: Adding the two products together with correct numerators.
A1: Showing clear algebraic steps to simplify to \(\frac{10(n-5)}{n(n-1)}\).
(c)(i) M1: Writing down the probability of both blue as \(\frac{(n-5)(n-6)}{n(n-1)}\).
M1: Setting the expression equal to \(\frac{7}{22}\) and attempting to cross-multiply.
A1: Simplifying to \(3n^2 - 47n + 132 = 0\) with no errors shown.
(c)(ii) M1: Attempting to solve the quadratic equation (by factorisation or formula).
A1: Correctly obtaining \(n = 12\) and stating why \(n = 11/3\) is rejected.