2023 Edexcel IGCSE Mathematics (Specification B) Practice Paper with Answers

Multiply both sides of the equation by the common denominator \(4x(x-2)\):
\(12(x-2) + 8x = 7x(x-2)\)
\(12x - 24 + 8x = 7x^2 - 14x\)
\(20x - 24 = 7x^2 - 14x\)
\(7x^2 - 34x + 24 = 0\)

Factorise the quadratic equation:
\(7x^2 - 28x - 6x + 24 = 0\)
\(7x(x-4) - 6(x-4) = 0\)
\((7x-6)(x-4) = 0\)

This gives the solutions:
\(x = 4\) or \(x = \frac{6}{7}\).

M1: For multiplying by the common denominator and expanding terms correctly to obtain a quadratic equation.
M1: For solving the quadratic equation by factorisation, formula, or completing the square.
A1: For both correct values: \(x = 4\) and \(x = \frac{6}{7}\) (or equivalents).

To rationalise the denominator, multiply the numerator and denominator by \(3 - 2\sqrt{5}\):

\(\frac{(4 - \sqrt{5})(3 - 2\sqrt{5})}{(3 + 2\sqrt{5})(3 - 2\sqrt{5})}\)

Expand the denominator:
\((3 + 2\sqrt{5})(3 - 2\sqrt{5}) = 3^2 - (2\sqrt{5})^2 = 9 - 20 = -11\)

Expand the numerator:
\((4 - \sqrt{5})(3 - 2\sqrt{5}) = 12 - 8\sqrt{5} - 3\sqrt{5} + 2(5) = 22 - 11\sqrt{5}\)

Divide the terms:

\(\frac{22 - 11\sqrt{5}}{-11} = -2 + \sqrt{5}\)

Therefore, \(a = -2\) and \(b = 1\).

M1: For multiplying numerator and denominator by \(3 - 2\sqrt{5}\).
M1: For correctly expanding the numerator to \(22 - 11\sqrt{5}\) or the denominator to \(-11\).
A1: For both correct values of \(a = -2\) and \(b = 1\).

The determinant of a \(2 \times 2\) matrix \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\) is \(ad - bc\).

\(\det \mathbf{A} = k(k+1) - (3)(-2) = k^2 + k + 6\)

We are given that \(\det \mathbf{A} = 12\):
\(k^2 + k + 6 = 12\)
\(k^2 + k - 6 = 0\)

Factorise the quadratic equation:
\((k+3)(k-2) = 0\)

So the possible values are:
\(k = -3\) or \(k = 2\).

M1: For setting up the determinant equation: \(k(k+1) - (3)(-2) = 12\).
M1: For simplifying to a standard 3-term quadratic equation: \(k^2 + k - 6 = 0\).
A1: For the correct solutions: \(k = -3\) and \(k = 2\).

First list the elements of each set within \(\mathcal{E}\):
\(\mathcal{E} = \{1, 2, 3, \dots, 20\}\)
\(A = \{3, 6, 9, 12, 15, 18\}\)
\(B = \{4, 8, 12, 16, 20\}\)
\(C = \{1, 2, 3, 4, 6, 8, 12\}\)

Find the union of \(A\) and \(B\):
\(A \cup B = \{3, 4, 6, 8, 9, 12, 15, 16, 18, 20\}\)

Find the complement of this union:
\((A \cup B)' = \{1, 2, 5, 7, 10, 11, 13, 14, 17, 19\}\)

Now find the intersection with \(C\):
\((A \cup B)' \cap C = \{1, 2\}\)

Thus, the number of elements is:
\(n((A \cup B)' \cap C) = 2\).

M1: For listing the elements of at least two of the sets \(A\), \(B\), or \(C\) correctly within the universal set.
M1: For identifying the elements of \((A \cup B)'\) or for showing the correct intersection process.
A1: For the correct answer 2.

First, find the vector sum \(\mathbf{p} + \mathbf{q}\):
\(\mathbf{p} + \mathbf{q} = \begin{pmatrix} 3 \\ -4 \end{pmatrix} + \begin{pmatrix} 12 \\ k \end{pmatrix} = \begin{pmatrix} 15 \\ k-4 \end{pmatrix}\)

Since \(\mathbf{p} + \mathbf{q}\) is parallel to \(\mathbf{r} = \begin{pmatrix} 5 \\ -1 \end{pmatrix}\), there exists a scalar \(\lambda\) such that:
\(\begin{pmatrix} 15 \\ k-4 \end{pmatrix} = \lambda \begin{pmatrix} 5 \\ -1 \end{pmatrix}\)

Comparing the top components:
\(15 = 5\lambda \implies \lambda = 3\)

Comparing the bottom components:
\(k-4 = -\lambda\)
\(k-4 = -3 \implies k = 1\).

M1: For writing the vector sum \(\mathbf{p} + \mathbf{q} = \begin{pmatrix} 15 \\ k-4 \end{pmatrix}\).
M1: For setting up an equation using the parallel condition, e.g., \(\frac{15}{5} = \frac{k-4}{-1}\).
A1: For the correct value \(k = 1\).

Since \(AC\) is a diameter of the circle, \(\angle ADC = 90^\circ\).

By the Alternate Segment Theorem, the angle between the tangent \(TD\) and the chord \(CD\) is equal to the angle in the alternate segment:
\(\angle CAD = \angle CDT = 28^\circ\).

In triangle \(ADC\), the angles sum to \(180^\circ\):
\(\angle ACD = 180^\circ - 90^\circ - 28^\circ = 62^\circ\).

Since \(ACT\) is a straight line, the angle \(DCT\) is:
\(\angle DCT = 180^\circ - 62^\circ = 118^\circ\).

In triangle \(CDT\), the angles sum to \(180^\circ\):
\(\angle ATC = 180^\circ - \angle DCT - \angle CDT = 180^\circ - 118^\circ - 28^\circ = 34^\circ\).

M1: For applying the Alternate Segment Theorem to find \(\angle CAD = 28^\circ\) or using the radius-tangent theorem to get \(\angle ODT = 90^\circ\).
M1: For finding \(\angle ACD = 62^\circ\) or \(\angle DCT = 118^\circ\) or \(\angle COD = 56^\circ\).
A1: For \(\angle ATC = 34^\circ\).

First, use the Cosine Rule to find the length of side \(AC\):
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)\)

\(AC^2 = 8^2 + 3^2 - 2(8)(3)\cos(60^\circ)\)

\(AC^2 = 64 + 9 - 48(0.5) = 73 - 24 = 49\)

So, \(AC = 7\text{ cm}\).

Next, apply the Sine Rule to find \(\sin(\angle BAC)\):

\(\frac{\sin(\angle BAC)}{BC} = \frac{\sin(\angle ABC)}{AC}\)

\(\frac{\sin(\angle BAC)}{3} = \frac{\sin(60^\circ)}{7}\)

Since \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\):

\(\sin(\angle BAC) = 3 \times \frac{\sqrt{3}}{14} = \frac{3\sqrt{3}}{14}\).

Since 3 and 14 are coprime integers, \(p = 3\) and \(q = 14\).

M1: For applying the Cosine Rule to find \(AC = 7\text{ cm}\).
M1: For applying the Sine Rule to set up \(\frac{\sin(\angle BAC)}{3} = \frac{\sin(60^\circ)}{7}\).
A1: For the correct values of \(p = 3\) and \(q = 14\).

The probability that the first chocolate selected is dark is \(\frac{5}{n}\).

Since the selection is without replacement, there are now \(n-1\) chocolates remaining in the box, of which 4 are dark. The probability that the second chocolate selected is dark is \(\frac{4}{n-1}\).

The probability that both chocolates chosen are dark is:
\(\frac{5}{n} \times \frac{4}{n-1} = \frac{20}{n(n-1)}\)

We are given this probability is \(\frac{5}{33}\):
\(\frac{20}{n(n-1)} = \frac{5}{33}\)

Divide both sides by 5:
\(\frac{4}{n(n-1)} = \frac{1}{33}\)

\(n(n-1) = 132\)

\(n^2 - n - 132 = 0\)

Factorise the quadratic:
\((n-12)(n+11) = 0\)

Since \(n\) must be a positive integer, \(n = 12\).

M1: For expressing the probability of selecting two dark chocolates in terms of \(n\): \(\frac{5}{n} \times \frac{4}{n-1}\).
M1: For setting up the equation \(\frac{20}{n(n-1)} = \frac{5}{33}\) and deriving the quadratic equation \(n^2 - n - 132 = 0\).
A1: For finding the correct positive integer solution \(n = 12\).

To subtract the fractions, we first find a common denominator.
Factorize the denominator of the second fraction:
\(x^2 - 4 = (x-2)(x+2)\)

Now rewrite the first fraction with the common denominator \((x-2)(x+2)\):
\(\frac{3}{x-2} = \frac{3(x+2)}{(x-2)(x+2)}\)

Subtract the second fraction from the first:
\(\frac{3(x+2) - (2x-3)}{(x-2)(x+2)} = \frac{3x + 6 - 2x + 3}{(x-2)(x+2)}\)
\(= \frac{x+9}{(x-2)(x+2)}\) or \(\frac{x+9}{x^2-4}\)

M1: For factorizing \(x^2-4\) to \((x-2)(x+2)\) and writing the first term with the common denominator, e.g., \(\frac{3(x+2)}{(x-2)(x+2)}\).
M1: For attempting to subtract the numerators, with at least one sign correct in expanding \(3(x+2) - (2x-3)\) to obtain \(3x+6-2x+3\).
A1: For the correct simplified fraction \(\frac{x+9}{x^2-4}\) or \(\frac{x+9}{(x-2)(x+2)}\).

A matrix is singular if its determinant is zero.
\(\det(\mathbf{A}) = (k-2)(k) - (5)(3) = 0\)

Expand the terms:
\(k^2 - 2k - 15 = 0\)

Factorize the quadratic equation:
\((k - 5)(k + 3) = 0\)

Thus, \(k = 5\) or \(k = -3\).

M1: For setting up the determinant equation \((k-2)k - (5)(3) = 0\).
M1: For expanding and forming a three-term quadratic equation \(k^2 - 2k - 15 = 0\).
A1: For finding both correct values of \(k\) (\(k = 5\) and \(k = -3\)).

Let the universal set \(\mathcal{E}\) represent the group of students, so \(n(\mathcal{E}) = 50\).
We are given:
\(n(A) = 28\)
\(n(B) = 24\)
\(n(A' \cap B') = 10\)

The number of students who study Art or Biology or both is:
\(n(A \cup B) = n(\mathcal{E}) - n(A' \cap B') = 50 - 10 = 40\)

Using the set formula:
\(n(A \cup B) = n(A) + n(B) - n(A \cap B)\)
\(40 = 28 + 24 - n(A \cap B)\)
\(40 = 52 - n(A \cap B)\)
\(n(A \cap B) = 52 - 40 = 12\)

M1: For calculating \(n(A \cup B) = 50 - 10 = 40\).
M1: For setting up a correct equation using the values, such as \((28-x) + x + (24-x) = 40\) or \(28 + 24 - x = 40\).
A1: For the correct answer 12.

Using the cosine rule in triangle \(ABC\):
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)\)

Substitute the given values:
\(AC^2 = 7^2 + 9^2 - 2(7)(9)\cos(60^\circ)\)
\(AC^2 = 49 + 81 - 126 \times 0.5\)
\(AC^2 = 130 - 63\)
\(AC^2 = 67\)

Calculate the square root:
\(AC = \sqrt{67} \approx 8.18535...\text{ cm}\)

To 3 significant figures, \(AC = 8.19\text{ cm}\).

M1: For substituting the values correctly into the cosine rule: \(7^2 + 9^2 - 2(7)(9)\cos(60^\circ)\).
M1: For simplifying the expression to find \(AC^2 = 67\).
A1: For \(8.19\) (accept answers in the range \(8.18\) to \(8.20\)).

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

Since \(P\) divides \(AB\) in the ratio \(2 : 3\), we have:
\(\overrightarrow{AP} = \frac{2}{5}\overrightarrow{AB} = \frac{2}{5}(-\mathbf{a} + \mathbf{b})\)

Now, find \(\overrightarrow{OP}\):
\(\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP}\)
\(\overrightarrow{OP} = \mathbf{a} + \frac{2}{5}(-\mathbf{a} + \mathbf{b})\)
\(\overrightarrow{OP} = \mathbf{a} - \frac{2}{5}\mathbf{a} + \frac{2}{5}\mathbf{b}\)
\(\overrightarrow{OP} = \frac{3}{5}\mathbf{a} + \frac{2}{5}\mathbf{b}\)

M1: For finding \(\overrightarrow{AB} = -\mathbf{a} + \mathbf{b}\) or \(\overrightarrow{BA} = \mathbf{a} - \mathbf{b}\).
M1: For writing a correct vector path for \(\overrightarrow{OP}\), e.g., \(\overrightarrow{OP} = \overrightarrow{OA} + \frac{2}{5}\overrightarrow{AB}\).
A1: For the simplified answer \(\frac{3}{5}\mathbf{a} + \frac{2}{5}\mathbf{b}\) or \(\frac{1}{5}(3\mathbf{a} + 2\mathbf{b})\).

Let \(y = \text{h}(x)\):
\(y = \frac{2x + 3}{x - 5}\)

Multiply both sides by \(x - 5\):
\(y(x - 5) = 2x + 3\)
\(xy - 5y = 2x + 3\)

Rearrange to isolate \(x\) terms on one side:
\(xy - 2x = 5y + 3\)

Factorize \(x\):
\(x(y - 2) = 5y + 3\)

Divide by \(y - 2\):
\(x = \frac{5y + 3}{y - 2}\)

Replace \(y\) with \(x\) to find the inverse function:
\(\text{h}^{-1}(x) = \frac{5x + 3}{x - 2}\)

M1: For writing \(y(x-5) = 2x+3\) or equivalent with switched variables.
M1: For isolating terms in \(x\) and factorizing, resulting in \(x(y-2) = 5y + 3\) or equivalent.
A1: For the final correct expression \(\frac{5x + 3}{x - 2}\) (must be in terms of \(x\)).

Express both bases as powers of 2:
\(8 = 2^3\)
\(32 = 2^5\)

Substitute into the equation:
\((2^3)^{2x-1} = (2^5)^{x+2}\)

Use index laws to multiply the exponents:
\(2^{3(2x-1)} = 2^{5(x+2)}
\)2^{6x-3} = 2^{5x+10}\)

Since the bases are identical, equate the exponents:
\(6x - 3 = 5x + 10\)

Solve for \(x\):
\(6x - 5x = 10 + 3\)
\(x = 13\)

M1: For expressing 8 and 32 as powers of the same base (e.g., \(2^3\) and \(2^5\)).
M1: For equating the power expressions correctly to form a linear equation: \(3(2x-1) = 5(x+2)\) or \(6x-3 = 5x+10\).
A1: For the correct solution \(x = 13\).

The total number of counters is \(5 + 7 = 12\).

There are two mutually exclusive outcomes where the counters have different colours:
1) Choosing a Red counter first, then a Blue counter (\(RB\)):
\(P(RB) = \frac{5}{12} \times \frac{7}{11} = \frac{35}{132}\)

2) Choosing a Blue counter first, then a Red counter (\(BR\)):
\(P(BR) = \frac{7}{12} \times \frac{5}{11} = \frac{35}{132}\)

Total probability of choosing different colours is the sum of these two probabilities:
\(P(\text{different}) = \frac{35}{132} + \frac{35}{132} = \frac{70}{132} = \frac{35}{66}\)

M1: For calculating the probability of a single valid path (e.g., \(\frac{5}{12} \times \frac{7}{11} = \frac{35}{132}\)).
M1: For combining both paths by addition, i.e., \(\frac{35}{132} + \frac{35}{132}\).
A1: For the correct simplified fraction \(\frac{35}{66}\).

To solve the equation:
\(\frac{4}{x - 1} - \frac{3}{x} = 1\)

Multiply all terms by the common denominator \(x(x - 1)\) to eliminate the fractions:
\(4x - 3(x - 1) = 1 \times x(x - 1)\)

Expand both sides of the equation:
\(4x - 3x + 3 = x^2 - x\)
\(x + 3 = x^2 - x\)

Rearrange the terms to form a standard quadratic equation:
\(x^2 - 2x - 3 = 0\)

Factorise the quadratic expression:
\((x - 3)(x + 1) = 0\)

This gives two solutions:
\(x = 3\) or \(x = -1\)

M1: For removing fractions by multiplying by the common denominator \(x(x-1)\) to obtain \(4x - 3(x - 1) = x(x - 1)\) or equivalent.
M1: For expanding and simplifying to a three-term quadratic equation, e.g., \(x^2 - 2x - 3 = 0\).
A1.57: For both correct solutions \(x = 3\) and \(x = -1\).

First, simplify the numerator by expressing both terms with the same power of 10:
\(4.5 \times 10^7 + 1.2 \times 10^8 = 0.45 \times 10^8 + 1.2 \times 10^8 = 1.65 \times 10^8\)

Next, divide this result by the denominator:
\(\frac{1.65 \times 10^8}{3.0 \times 10^{-3}} = \left(\frac{1.65}{3.0}\right) \times 10^{8 - (-3)}\)
\(= 0.55 \times 10^{11}\)

To write the final answer in standard form:
\(5.5 \times 10^{10}\)

M1: For adding the numerator correctly to obtain \(1.65 \times 10^8\) (or \(165,000,000\)).
M1: For dividing their numerator by \(3.0 \times 10^{-3}\) to obtain \(0.55 \times 10^{11}\) or for demonstrating the correct subtraction of indices \(8 - (-3) = 11\).
A1.57: For the correct standard form answer: \(5.5 \times 10^{10}\).

First, express the vector \(\mathbf{u} + k\mathbf{v}\) in terms of \(\mathbf{i}\) and \(\mathbf{j}\):
\(\mathbf{u} + k\mathbf{v} = (\mathbf{i} + 4\mathbf{j}) + k(2\mathbf{i} - \mathbf{j})\)
\(= (1 + 2k)\mathbf{i} + (4 - k)\mathbf{j}\)

Since this vector is parallel to \(5\mathbf{i} + 2\mathbf{j}\), the ratio of the \(\mathbf{i}\)-component to the \(\mathbf{j}\)-component must equal the ratio for \(5\mathbf{i} + 2\mathbf{j}\):
\(\frac{1 + 2k}{5} = \frac{4 - k}{2}\)

Cross-multiply to solve for \(k\):
\(2(1 + 2k) = 5(4 - k)\)
\(2 + 4k = 20 - 5k\)
\(9k = 18\)
\(k = 2\)

M1: For finding \(\mathbf{u} + k\mathbf{v}\) in terms of \(k\) as \((1 + 2k)\mathbf{i} + (4 - k)\mathbf{j}\) or equivalent component-wise representation.
M1: For setting up a correct equation using the parallel condition, e.g., \(\frac{1+2k}{5} = \frac{4-k}{2}\) or \(1+2k = 2.5(4-k)\).
A1.57: For the correct value \(k = 2\).

Let the exterior angle of the regular polygon be \(x^\circ\).
Then, the interior angle of the regular polygon is \(11x^\circ\).

Since the interior and exterior angles at any vertex of a polygon lie on a straight line, they sum to \(180^\circ\):
\(11x + x = 180\)
\(12x = 180\)
\(x = 15\)

So each exterior angle of the polygon is \(15^\circ\).
The formula for the exterior angle of a regular polygon is \(\frac{360^\circ}{n}\), where \(n\) is the number of sides:
\(\frac{360}{n} = 15\)
\(n = \frac{360}{15} = 24\)

M1: For establishing an equation summing the interior and exterior angles, e.g., \(11x + x = 180\) or setting up \(\frac{(n-2) \times 180}{n} = 11 \times \frac{360}{n}\).
M1: For solving to find the exterior angle \(x = 15^\circ\) or simplifying the equation in terms of \(n\) to \(180(n-2) = 3960\).
A1.57: For the correct value \(n = 24\).

The total number of balls in the bag is \(5 + n\).
The probability of selecting a red ball first is:
\(P(R_1) = \frac{5}{5 + n}\)

Since the selection is made without replacement, there are now 4 red balls remaining, and a total of \(4 + n\) balls remaining.
The probability of selecting a red ball second is:
\(P(R_2) = \frac{4}{4 + n}\)

The probability of selecting two red balls is:
\(P(R_1 \text{ and } R_2) = \frac{5}{5 + n} \times \frac{4}{4 + n} = \frac{20}{(5+n)(4+n)}\)

We are given that this probability is \(\frac{2}{9}\):
\(\frac{20}{(5+n)(4+n)} = \frac{2}{9}\)

Cross-multiply to clear the fractions:
\(180 = 2(5+n)(4+n)\)
\(90 = (5+n)(4+n)\)
\(90 = n^2 + 9n + 20\)
\(n^2 + 9n - 70 = 0\)

Factorise the quadratic equation:
\((n + 14)(n - 5) = 0\)

Since the number of green balls \(n\) must be a positive integer, we discard \(n = -14\).
Thus, \(n = 5\).

M1: For a correct expression of the probability of obtaining two red balls without replacement in terms of \(n\), e.g., \(\frac{5}{5+n} \times \frac{4}{4+n}\).
M1: For setting the expression equal to \(\frac{2}{9}\) and simplifying to a three-term quadratic equation, e.g., \(n^2 + 9n - 70 = 0\).
A1.57: For the correct positive integer solution \(n = 5\).

Let \(a = BC = 8\text{ cm}\), \(b = AC = 5\text{ cm}\), and \(c = AB = 7\text{ cm}\).
We can find the angle \(BAC\) (which is angle \(A\)) using the Cosine Rule:
\(a^2 = b^2 + c^2 - 2bc \cos(A)\)

Substitute the given lengths into the formula:
\(8^2 = 5^2 + 7^2 - 2(5)(7) \cos(BAC)\)
\(64 = 25 + 49 - 70 \cos(BAC)\)
\(64 = 74 - 70 \cos(BAC)\)

Rearrange to make \(\cos(BAC)\) the subject:
\(70 \cos(BAC) = 74 - 64\)
\(70 \cos(BAC) = 10\)
\[\cos(BAC) = \frac{10}{70} = \frac{1}{7}\]

Find the angle:
\(BAC = \arccos\left(\frac{1}{7}\right) \approx 81.78678...^\circ\)

To 3 significant figures, the angle is \(81.8^\circ\).

M1: For substituting the values correctly into the Cosine Rule: \(8^2 = 5^2 + 7^2 - 2 \times 5 \times 7 \cos(BAC)\).
M1: For rearranging the equation to make \(\cos(BAC)\) the subject, yielding \(\cos(BAC) = \frac{1}{7}\) or \(0.143\) (or better).
A1.57: For the angle \(81.8\) (accept answers in the range \(81.7\) to \(81.8\)).

First, find an expression for the composite function \(\mathrm{fg}(x)\):
\(\mathrm{fg}(x) = \mathrm{f}(\mathrm{g}(x))\)
\(\mathrm{fg}(x) = \mathrm{f}(2x + 5)\)

Substitute \(2x + 5\) in place of \(x\) in the expression for \(\mathrm{f}(x)\):
\(\mathrm{fg}(x) = \frac{3}{(2x + 5) - 2} = \frac{3}{2x + 3}\)

We need to find \(x\) when \(\mathrm{fg}(x) = \frac{1}{3}\):
\(\frac{3}{2x + 3} = \frac{1}{3}\)

Cross-multiply to solve for \(x\):
\(3 \times 3 = 1 \times (2x + 3)\)
\(9 = 2x + 3\)
\(2x = 6\)
\(x = 3\)

M1: For a correct attempt to find the composite function \(\mathrm{fg}(x)\) to obtain \(\frac{3}{2x + 3}\), or for solving \(\mathrm{f}(y) = \frac{1}{3}\) to find \(y = 11\).
M1: For setting up the linear equation \(\frac{3}{2x + 3} = \frac{1}{3}\) or \(2x + 5 = 11\).
A1.57: For the final answer \(x = 3\).

We can solve this using properties of determinants: \(\det(\mathbf{AB}) = \det(\mathbf{A}) \times \det(\mathbf{B})\).

First, find the determinant of \(\mathbf{A}\):
\(\det(\mathbf{A}) = (3)(4) - (-2)(1) = 12 + 2 = 14\)

Next, find the determinant of \(\mathbf{B}\):
\(\det(\mathbf{B}) = (2)(a) - (1)(5) = 2a - 5\)

Using the determinant product property:
\(\det(\mathbf{AB}) = \det(\mathbf{A}) \times \det(\mathbf{B})\)
\(42 = 14(2a - 5)\)

Divide both sides by 14:
\(3 = 2a - 5\)
\(2a = 8\)
\(a = 4\)

*Alternative Method (by multiplying the matrices first):*
\(\mathbf{AB} = \begin{pmatrix} 3 & -2 \\ 1 & 4 \end{pmatrix} \begin{pmatrix} 2 & 1 \\ 5 & a \end{pmatrix} = \begin{pmatrix} 3(2) + (-2)(5) & 3(1) + (-2)a \\ 1(2) + 4(5) & 1(1) + 4a \end{pmatrix} = \begin{pmatrix} -4 & 3 - 2a \\ 22 & 1 + 4a \end{pmatrix}\)

Find the determinant of this product matrix:
\(\det(\mathbf{AB}) = (-4)(1 + 4a) - (22)(3 - 2a)\)
\(42 = -4 - 16a - 66 + 44a\)
\(42 = 28a - 70\)
\(112 = 28a\)
\(a = 4\)

M1: For a correct method to find the determinant of \(\mathbf{A}\) as 14, OR for multiplying the two matrices with at least two correct elements in \(\mathbf{AB}\).
M1: For setting up the determinant equation using the product property, i.e., \(14(2a - 5) = 42\), OR for expanding the determinant of their product matrix to get \(28a - 70 = 42\).
A1.57: For the correct value \(a = 4\).

First, solve the quadratic inequality: \(2x^2 - 7x - 15 \le 0\). Factoring the quadratic expression gives \((2x + 3)(x - 5) \le 0\). The critical values are \(x = -1.5\) and \(x = 5\). Thus, the solution to the quadratic inequality is \(-1.5 \le x \le 5\). Next, solve the linear inequality: \(4 - 3x < -2 \implies -3x < -6 \implies x > 2\). To find the range of values of \(x\) that satisfy both inequalities, we find the intersection of \(-1.5 \le x \le 5\) and \(x > 2\), which is \(2 < x \le 5\).

M1: For factoring the quadratic to find critical values \(x = -1.5\) and \(x = 5\), or for finding the interval \(-1.5 \le x \le 5\\. M1: For solving the linear inequality to get \)x > 2\\. A1: For the correct combined range \(2 < x \le 5\).

We find the product \(\mathbf{A}\mathbf{B}\): \(\mathbf{A}\mathbf{B} = \begin{pmatrix} 2 & p \\\\ -3 & 4 \end{pmatrix} \begin{pmatrix} q & -2 \\\\ 1 & 3 \end{pmatrix} = \begin{pmatrix} 2q + p & -4 + 3p \\\\ -3q + 4 & 6 + 12 \end{pmatrix} = \begin{pmatrix} 2q + p & -4 + 3p \\\\ -3q + 4 & 18 \end{pmatrix}\\. Equating the elements of \)\\mathbf{A}\\mathbf{B}\) to the given matrix: From the top-right element: \(-4 + 3p = -13 \implies 3p = -9 \implies p = -3\\. From the bottom-left element: \)-3q + 4 = -11 \\implies -3q = -15 \\implies q = 5\\. We can verify with the top-left element: \(2q + p = 2(5) + (-3) = 10 - 3 = 7\), which matches.

M1: For a correct expression of at least one of the non-constant elements in the product matrix \(\mathbf{A}\mathbf{B}\) (e.g., \(-4 + 3p\) or \(-3q + 4\)). A1: For finding the correct value of \(p = -3\). A1: For finding the correct value of \(q = 5\).

We are given \(n(\mathcal{E}) = 40\). The number of students studying neither Music nor Art is \(n(M' \cap A') = 6\). Thus, the number of students studying Music, Art, or both is \(n(M \cup A) = 40 - 6 = 34\). Using the formula \(n(M \cup A) = n(M) + n(A) - n(M \cap A)\), we have: \(34 = 23 + 18 - n(M \cap A) \implies 34 = 41 - n(M \cap A) \implies n(M \cap A) = 7\\. The number of students who study Music but do not study Art is given by \)n(M \\cap A') = n(M) - n(M \\cap A) = 23 - 7 = 16\).

M1: For finding the union \(n(M \cup A) = 34\) or representing this in a Venn diagram. M1: For calculating the intersection \(n(M \\cap A) = 7\\. A1: For the final correct answer of 16.

First, find the vector \(\overrightarrow{AB}\): \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} -1 \\\\ 6 \end{pmatrix} - \begin{pmatrix} 3 \\\\ -2 \end{pmatrix} = \begin{pmatrix} -4 \\\\ 8 \end{pmatrix}\\. Since the ratio \)AC : CB = 3 : 1\), the point \(C\) divides \(AB\) in the ratio \(3:1\). Thus, \(\overrightarrow{AC} = \frac{3}{4} \overrightarrow{AB} = \frac{3}{4} \begin{pmatrix} -4 \\\\ 8 \end{pmatrix} = \begin{pmatrix} -3 \\\\ 6 \end{pmatrix}\\. The position vector of \)C\) is: \(\\overrightarrow{OC} = \\overrightarrow{OA} + \\overrightarrow{AC} = \\begin{pmatrix} 3 \\\\ -2 \\end{pmatrix} + \\begin{pmatrix} -3 \\\\ 6 \\end{pmatrix} = \\begin{pmatrix} 0 \\\\ 4 \\end{pmatrix}\\.

M1: For finding the vector \(\overrightarrow{AB} = \begin{pmatrix} -4 \\\\ 8 \end{pmatrix}\\. M1: For a correct vector equation to find \)\\overrightarrow{OC}\), such as \(\overrightarrow{OC} = \mathbf{a} + \frac{3}{4}(\mathbf{b} - \mathbf{a})\) or equivalent. A1: For the correct column vector \(\begin{pmatrix} 0 \\\\ 4 \end{pmatrix}\).

(a) The total number of students is the sum of all regions in the Venn diagram:
\(\text{Total} = 18 + 24 + (3x - 1) + 2x + (x + 3) + 5 + x + 15\)
\(\text{Total} = (18 + 24 - 1 + 3 + 5 + 15) + (3x + 2x + x + x) = 7x + 64\)

(b) We are given that the total number of students is 120.
\(7x + 64 = 120\)
\(7x = 56\)
\(x = 8\)

(c) The number of students who study at least two subjects is represented by the regions with intersections of two or more sets:
\(\text{At least two} = \text{n}(P \cap C \cap B') + \text{n}(P \cap B \cap C') + \text{n}(C \cap B \cap P') + \text{n}(P \cap C \cap B)\)
\(\text{At least two} = 2x + (x + 3) + 5 + x = 4x + 8\)
Substitute \(x = 8\):
\(4(8) + 8 = 32 + 8 = 40\)

(a) M1: Summing all eight regions in terms of x.
A1: Correct simplified expression: \(7x + 64\).

(b) M1: Setting their expression equal to 120 and attempting to solve for x.
A1: \(x = 8\).

(c) M1: Identifying the correct regions for 'at least two' as \(2x + (x + 3) + 5 + x\) or \(4x + 8\).
A1: Correctly substituting their value of \(x\) into their expression.
A1: 40.

(a) Time = Distance / Speed.
Time taken by the cyclist = \(\frac{36}{x}\) hours.
Time taken by the runner = \(\frac{18}{x-5}\) hours.
Since 45 minutes is \(\frac{45}{60} = \frac{3}{4}\) hours:
\(\frac{18}{x-5} - \frac{36}{x} = \frac{3}{4}\)

(b) Multiply the entire equation by \(4x(x - 5)\) to clear denominators:
\(4x(18) - 4(36)(x - 5) = 3x(x - 5)\)
\(72x - 144(x - 5) = 3x^2 - 15x\)
\(72x - 144x + 720 = 3x^2 - 15x\)
\(-72x + 720 = 3x^2 - 15x\)
Rearrange to form a quadratic equation:
\(3x^2 + 57x - 720 = 0\)
Divide the entire equation by 3:
\(x^2 + 19x - 240 = 0\) (as required).

(c) Factorise the quadratic equation:
\((x + 24)(x - 10) = 0\)
So, \(x = -24\) or \(x = 10\).
Since speed \(x\) must be positive, we reject \(x = -24\).
Therefore, \(x = 10\).

(a) M1: Correct expressions for both times in hours: \(\frac{36}{x}\) and \(\frac{18}{x-5}\).
A1: Correct equation setup: \(\frac{18}{x-5} - \frac{36}{x} = \frac{3}{4}\) (or equivalent with consistent units).

(b) M1: Multiplying by the common denominator \(4x(x-5)\) or \(x(x-5)\).
M1: Expanding brackets and grouping terms correctly.
A1: Fully correct algebraic derivation leading to \(x^2 + 19x - 240 = 0\).

(c) M1: Solving the quadratic equation by factoring, formula, or completing the square.
A1: Finding solutions \(x = 10\) and \(x = -24\).
A1: Choosing \(x = 10\) and stating that speed must be positive (rejecting the negative solution).

(a) Let \(y = \frac{2x+4}{x-1}\).
Multiply by \(x-1\):
\(y(x - 1) = 2x + 4\)
\(yx - y = 2x + 4\)
Rearrange to make \(x\) the subject:
\(yx - 2x = y + 4\)
\(x(y - 2) = y + 4\)
\(x = \frac{y+4}{y-2}\)
So, \(\text{f}^{-1}(x) = \frac{x+4}{x-2}\), where \(x \neq 2\).

(b) Find \(\text{g}(3)\):
\(\text{g}(3) = 3(3) + a = 9 + a\)
Now, substitute into \(\text{f}\):
\(\text{fg}(3) = \text{f}(9+a) = \frac{2(9+a) + 4}{(9+a) - 1} = 3\)
\(\frac{18 + 2a + 4}{8 + a} = 3\)
\(\frac{2a + 22}{a + 8} = 3\)
\(2a + 22 = 3(a + 8)\)
\(2a + 22 = 3a + 24\)
\(a = -2\).

(c) Solve \(\text{f}(x) = x\):
\(\frac{2x + 4}{x - 1} = x\)
\(2x + 4 = x(x - 1)\)
\(2x + 4 = x^2 - x\)
\(x^2 - 3x - 4 = 0\)
\((x - 4)(x + 1) = 0\)
\(x = 4\) or \(x = -1\).

(a) M1: Setting \(y = \text{f}(x)\) and multiplying by \(x-1\).
M1: Collecting terms in \(x\) on one side and factorising.
A1: Correct inverse function \(\text{f}^{-1}(x) = \frac{x+4}{x-2}\).

(b) M1: Writing down \(\text{g}(3) = 9 + a\).
M1: Substituting \(9+a\) into \(\text{f}(x)\) and setting equal to 3.
A1: \(a = -2\).

(c) M1: Setting \(\frac{2x+4}{x-1} = x\) and multiplying to obtain a quadratic.
A1: Formulating \(x^2 - 3x - 4 = 0\).
A1: \(x = 4\) and \(x = -1\).

(a)(i) \(\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}\).

(a)(ii) Since \(OP : PA = 2 : 1\), \(\overrightarrow{OP} = \frac{2}{3}\mathbf{a}\).
Since \(Q\) is the midpoint of \(AB\), \(\overrightarrow{OQ} = \overrightarrow{OA} + \frac{1}{2}\overrightarrow{AB} = \mathbf{a} + \frac{1}{2}(\mathbf{b} - \mathbf{a}) = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\).
\(\overrightarrow{PQ} = \overrightarrow{PO} + \overrightarrow{OQ} = -\frac{2}{3}\mathbf{a} + \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b} = -\frac{1}{6}\mathbf{a} + \frac{1}{2}\mathbf{b}\).

(b) Since \(OB : BR = 1 : k\), \(\overrightarrow{BR} = k\mathbf{b}\), which means \(\overrightarrow{OR} = (1 + k)\mathbf{b}\).
\(\overrightarrow{PR} = \overrightarrow{PO} + \overrightarrow{OR} = -\frac{2}{3}\mathbf{a} + (1 + k)\mathbf{b}\).

(c) Given \(\overrightarrow{PR} = m\overrightarrow{PQ}\):
\(-\frac{2}{3}\mathbf{a} + (1 + k)\mathbf{b} = m \left( -\frac{1}{6}\mathbf{a} + \frac{1}{2}\mathbf{b} \right)\)
Comparing coefficients of \(\mathbf{a}\):
\(-\frac{2}{3} = -\frac{1}{6}m \implies m = 4\)

Comparing coefficients of \(\mathbf{b}\):
\(1 + k = \frac{1}{2}m\)
Substitute \(m = 4\):
\(1 + k = \frac{1}{2}(4) = 2 \implies k = 1\).

(a)(i) B1: \(\mathbf{b} - \mathbf{a}\).
(a)(ii) M1: Writing a correct vector path for \(\overrightarrow{PQ}\), e.g., \(\overrightarrow{PO} + \overrightarrow{OQ}\), and identifying \(\overrightarrow{OP} = \frac{2}{3}\mathbf{a}\).
A1: Correct simplified vector: \(-\frac{1}{6}\mathbf{a} + \frac{1}{2}\mathbf{b}\).

(b) M1: Writing a correct path for \(\overrightarrow{PR}\) and using \(\overrightarrow{OR} = (1+k)\mathbf{b}\).
A1: \(-\frac{2}{3}\mathbf{a} + (1 + k)\mathbf{b}\).

(c) M1: Equating their \(\overrightarrow{PR}\) with \(m\) times their \(\overrightarrow{PQ}\).
M1: Equating the coefficients of \(\mathbf{a}\) to solve for \(m\).
A1: \(m = 4\).
A1: \(k = 1\).

(a) The angle \(BAC = 115^\circ - 60^\circ = 55^\circ\).
Using the Cosine Rule on triangle \(ABC\):
\(BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos(BAC)\)
\(BC^2 = 80^2 + 120^2 - 2(80)(120)\cos(55^\circ)\)
\(BC^2 = 6400 + 14400 - 19200(0.573576)\)
\(BC^2 = 20800 - 11012.67 = 9787.33\)
\(BC = \sqrt{9787.33} \approx 98.93\text{ m} \approx 98.9\text{ m}\) (to 3 s.f.).

(b) To find the bearing of \(C\) from \(B\), we first need the angle \(ABC\).
Using the Sine Rule:
\(\frac{\sin(ABC)}{AC} = \frac{\sin(BAC)}{BC}\)
\(\frac{\sin(ABC)}{120} = \frac{\sin(55^\circ)}{98.93}\)
\(\sin(ABC) = \frac{120 \sin(55^\circ)}{98.93} \approx 0.9936\)
\(ABC \approx \arcsin(0.9936) \approx 83.54^\circ\).
Since \(AB^2 + BC^2 = 80^2 + 98.93^2 = 16187 > 120^2 = 14400\), angle \(ABC\) is acute, so \(ABC = 83.5^\circ\).

Now, the bearing of \(A\) from \(B\) is \(60^\circ + 180^\circ = 240^\circ\).
The line \(BC\) lies to the east of the line \(BA\), so we subtract \(ABC\) from the back bearing:
\(\text{Bearing of } C \text{ from } B = 240^\circ - 83.54^\circ = 156.46^\circ \approx 156^\circ\) or \(157^\circ\) (depending on intermediate rounding; using 98.93 gives 156.5 degrees, rounding to 157).

(c) The tower is vertical at \(A\), so triangle \(TAB\) is right-angled at \(A\) with base \(AB = 80\) m and angle of elevation \(15^\circ\).
\(\tan(15^\circ) = \frac{TX}{AB} = \frac{TX}{80}\)
\(TX = 80 \tan(15^\circ) \approx 80(0.26795) \approx 21.43\text{ m} \approx 21.4\text{ m}\).

(a) M1: Finding the angle \(BAC = 55^\circ\).
M1: Applying the Cosine Rule with their values: \(80^2 + 120^2 - 2(80)(120)\cos(55^\circ)\).
A1: \(BC \approx 98.9\text{ m}\) (accept 98.9 - 99.0).

(b) M1: Correct application of Sine Rule (or Cosine Rule) to find angle \(ABC\).
A1: Finding angle \(ABC \approx 83.5^\circ\).
M1: Method for calculating the bearing, e.g., using back-bearing of \(240^\circ\) and subtracting angle \(ABC\).
A1: \(157^\circ\) (accept \(156^\circ\) to \(157^\circ\)).

(c) M1: Correct trigonometric ratio setup: \(TX = 80 \tan(15^\circ)\).
A1: \(21.4\text{ m}\) (accept 21.4 - 21.5).

(a) The total number of balls in the bag initially is \(r + 6\).
Since the balls are drawn without replacement:
- The probability of drawing the first red ball is \(\frac{r}{r+6}\).
- The probability of drawing the second red ball is \(\frac{r-1}{r+5}\).

The combined probability of drawing two red balls is:
\(\text{P(Red, Red)} = \frac{r}{r+6} \times \frac{r-1}{r+5} = \frac{3}{8}\)
\(\frac{r(r-1)}{(r+6)(r+5)} = \frac{3}{8}\)
Multiply both sides by \(8(r+6)(r+5)\):
\(8r(r-1) = 3(r+6)(r+5)\)
\(8r^2 - 8r = 3(r^2 + 11r + 30)\)
\(8r^2 - 8r = 3r^2 + 33r + 90\)
Rearrange all terms to one side:
\(5r^2 - 41r - 90 = 0\) (as required).

(b) Solve the quadratic equation \(5r^2 - 41r - 90 = 0\):
We look for factors: \((5r + 9)(r - 10) = 0\).
This gives \(r = -1.8\) or \(r = 10\).
Since the number of balls \(r\) must be a positive integer, \(r = 10\).

(c) With \(r = 10\), the total number of balls is \(10 + 6 = 16\).
The two ways to get different colors are (Red, Blue) or (Blue, Red):
\(\text{P(Red, Blue)} = \frac{10}{16} \times \frac{6}{15} = \frac{5}{8} \times \frac{2}{5} = \frac{2}{8} = \frac{1}{4}\)
\(\text{P(Blue, Red)} = \frac{6}{16} \times \frac{10}{15} = \frac{3}{8} \times \frac{2}{3} = \frac{2}{8} = \frac{1}{4}\)

Therefore, the probability of different colors is:
\(\text{P(Different Colors)} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} = 0.5\).

(a) M1: Writing down the product of the probabilities for two red balls: \(\frac{r}{r+6} \times \frac{r-1}{r+5}\).
M1: Setting this product equal to \(\frac{3}{8}\) and clearing fractions.
A1: Correctly expanding and simplifying to obtain the given quadratic equation.

(b) M1: Solving the quadratic equation by factorisation, formula, or completing the square.
A1: Identifying \(r = 10\) and rejecting the negative/fractional solution.

(c) M1: Calculating one of the different-color combinations (e.g., Red-Blue or Blue-Red) using \(16\) as total.
M1: Adding the two combinations together: \(\text{P(R,B)} + \text{P(B,R)}\).
A1: \(0.5\) (or \(\frac{1}{2}\)).

(a) Work out \(2\mathbf{A} - \mathbf{B}\):
\(2\mathbf{A} = 2\begin{pmatrix} 3 & k \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 6 & 2k \\ -2 & 4 \end{pmatrix}\)
\(\mathbf{C} = \begin{pmatrix} 6 & 2k \\ -2 & 4 \end{pmatrix} - \begin{pmatrix} 2 & 1 \\ 4 & -3 \end{pmatrix} = \begin{pmatrix} 6 - 2 & 2k - 1 \\ -2 - 4 & 4 - (-3) \end{pmatrix} = \begin{pmatrix} 4 & 2k - 1 \\ -6 & 7 \end{pmatrix}\).

(b) Use the property of determinants: \(\det(\mathbf{AB}) = \det(\mathbf{A}) \times \det(\mathbf{B})\).
Calculate \(\det(\mathbf{A})\):
\(\det(\mathbf{A}) = 3(2) - k(-1) = 6 + k\).

Calculate \(\det(\mathbf{B})\):
\(\det(\mathbf{B}) = 2(-3) - 1(4) = -6 - 4 = -10\).

Set the product of determinants equal to \(-100\):
\((6 + k)(-10) = -100\)
\(6 + k = 10\)
\(k = 4\).

(c) For \(k = 4\), \(\mathbf{A} = \begin{pmatrix} 3 & 4 \\ -1 & 2 \end{pmatrix}\).
\(\det(\mathbf{A}) = 6 + 4 = 10\).

The inverse is given by:
\(\mathbf{A}^{-1} = \frac{1}{\det(\mathbf{A})} \begin{pmatrix} 2 & -4 \\ 1 & 3 \end{pmatrix} = \frac{1}{10} \begin{pmatrix} 2 & -4 \\ 1 & 3 \end{pmatrix} = \begin{pmatrix} 0.2 & -0.4 \\ 0.1 & 0.3 \end{pmatrix}\).

(a) M1: Multiplying \(\mathbf{A}\) by 2 correctly to obtain \(\begin{pmatrix} 6 & 2k \\ -2 & 4 \end{pmatrix}\).
A1: Correct matrix \(\begin{pmatrix} 4 & 2k-1 \\ -6 & 7 \end{pmatrix}\).

(b) M1: Correct expression for \(\det(\mathbf{A})\) in terms of \(k\).
M1: Calculating \(\det(\mathbf{B}) = -10\) and setting \(\det(\mathbf{A}) \times \det(\mathbf{B}) = -100\) (or multiplying out \(\mathbf{AB}\) first and then taking the determinant).
A1: \(k = 4\).

(c) M1: Swapping leading diagonal elements and negating other diagonal elements of \(\mathbf{A}\) with \(k=4\).
M1: Dividing by their determinant of \(\mathbf{A}\) (which is 10).
A1: Correct inverse matrix \(\begin{pmatrix} 0.2 & -0.4 \\ 0.1 & 0.3 \end{pmatrix}\) (or fractional equivalent).

(a) Since \(L\) and \(W\) are measured to the nearest 10 cm (0.1 m), the maximum error is 0.05 m.
- Upper bound for \(L\) is \(12.45\) m.
- Upper bound for \(W\) is \(8.55\) m.

Perimeter \(P = 2(L + W)\).
Upper bound for Perimeter \(P_{\max} = 2(12.45 + 8.55) = 2(21.00) = 42\) m.

(b) Lower bound for \(L\) is \(12.35\) m.
Lower bound for \(W\) is \(8.45\) m.

Lower bound for Area \(A_{\min} = L_{\min} \times W_{\min} = 12.35 \times 8.45 = 104.3575\text{ m}^2\).
Rounding to 3 significant figures, the lower bound is \(104\text{ m}^2\).

(c) The total cost including tax is \(\$1265\). The tax rate is \(15\%\).
Let the cost before tax be \(X\).
\(1.15 \times X = 1265\)
\(X = \frac{1265}{1.15} = 1100\).
So the cost of the fence before tax was \(\$1100\).

(a) M1: Identifying the upper bounds of length and width as \(12.45\) and \(8.55\).
A1: Correct perimeter calculation: \(42\) m.

(b) M1: Identifying the lower bounds of length and width as \(12.35\) and \(8.45\).
M1: Multiplying their lower bounds together.
A1: \(104\text{ m}^2\) (accept 104.3575).

(c) M1: Recognising that \(1265\) represents \(115\%\) of the pre-tax cost, setting up the division \(\frac{1265}{1.15}\).
A1: Correct pre-tax cost: \(\$1100\).

(a) Differentiating \(y = 2x^3 - 9x^2 - 24x + 15\) with respect to \(x\) gives \(\frac{\mathrm{d}y}{\mathrm{d}x} = 6x^2 - 18x - 24\). (b) At stationary points, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\). So, \(6x^2 - 18x - 24 = 0\). Dividing by 6 gives \(x^2 - 3x - 4 = 0\). Factoring this quadratic equation gives \((x - 4)(x + 1) = 0\), which yields \(x = 4\) or \(x = -1\). When \(x = 4\), \(y = 2(4)^3 - 9(4)^2 - 24(4) + 15 = 128 - 144 - 96 + 15 = -97\). When \(x = -1\), \(y = 2(-1)^3 - 9(-1)^2 - 24(-1) + 15 = -2 - 9 + 24 + 15 = 28\). Thus, the stationary points are \((4, -97)\) and \((-1, 28)\). (c) Finding the second derivative: \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12x - 18\). For \(x = 4\): \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12(4) - 18 = 30 > 0\), so \((4, -97)\) is a local minimum. For \(x = -1\): \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12(-1) - 18 = -30 < 0\), so \((-1, 28)\) is a local maximum.

(a) M1 for differentiating at least one non-constant term correctly. A1 for correct derivative \(\frac{\mathrm{d}y}{\mathrm{d}x} = 6x^2 - 18x - 24\). (b) M1 for setting their derivative to 0. M1 for solving the resulting quadratic equation. A1 for one correct stationary point coordinate pair. A1 for both correct stationary points. (c) M1 for finding the second derivative and substituting at least one of their x values, or using a first-derivative sign test. A1 for correct classification of both stationary points with appropriate justification.

(a) Multiplying the matrices: \(\mathbf{AB} = \begin{pmatrix} 2 & p \\ -1 & 3 \end{pmatrix} \begin{pmatrix} q & -4 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} 2q + 2p & -8 + p \\ -q + 6 & 4 + 3 \end{pmatrix}\). Equating the top-left element of \(\mathbf{AB}\) to the top-left element of the given result gives \(2q + 2p = 16\). (b) Equating the top-right elements: \(-8 + p = -2\) which gives \(p = 6\). Substituting \(p = 6\) into \(2q + 2p = 16\) gives \(2q + 12 = 16\), so \(2q = 4\) and \(q = 2\). Checking the bottom-left elements: \(-q + 6 = -2 + 6 = 4\), which is consistent. Thus, \(p = 6\) and \(q = 2\). (c) With \(p = 6\), \(\mathbf{A} = \begin{pmatrix} 2 & 6 \\ -1 & 3 \end{pmatrix}\). The determinant of \(\mathbf{A}\) is \(\det(\mathbf{A}) = (2)(3) - (6)(-1) = 6 + 6 = 12\). The inverse is \(\mathbf{A}^{-1} = \frac{1}{12} \begin{pmatrix} 3 & -6 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} \frac{1}{4} & -\frac{1}{2} \\ \frac{1}{12} & \frac{1}{6} \end{pmatrix}\).

(a) M1 for a correct attempt to multiply row 1 of \(\mathbf{A}\) by column 1 of \(\mathbf{B}\). A1 for showing clearly that \(2q + 2p = 16\). (b) M1 for establishing another equation, e.g., \(-8 + p = -2\) or \(-q + 6 = 4\). A1 for \(p = 6\). A1 for \(q = 2\). (c) M1 for finding the determinant of \(\mathbf{A}\) as \(12\). M1 for the adjugate matrix swap and sign change. A1 for \(\begin{pmatrix} \frac{1}{4} & -\frac{1}{2} \\ \frac{1}{12} & \frac{1}{6} \end{pmatrix}\) or equivalent.

(a) Using the Cosine Rule: \(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(ABC)\). This gives \(AC^2 = 7^2 + 10^2 - 2(7)(10)\cos(60^\circ) = 49 + 100 - 140(0.5) = 149 - 70 = 79\). Thus, \(AC = \sqrt{79} \approx 8.89\text{ cm}\) (to 3 s.f.). (b) Area of \(\triangle ABC = \frac{1}{2}(AB)(BC)\sin(ABC) = \frac{1}{2}(7)(10)\sin(60^\circ) = 35 \times \frac{\sqrt{3}}{2} \approx 30.31 \approx 30.3\text{ cm}^2\) (to 3 s.f.). (c) The shortest distance \(h\) from \(B\) to \(AC\) is the height. Since \(\text{Area} = \frac{1}{2} \times AC \times h\), we have \(h = \frac{2 \times \text{Area}}{AC} = \frac{35\sqrt{3}}{\sqrt{79}} \approx 6.82\text{ cm}\) (to 3 s.f.).

(a) M1 for a correct substitution into the Cosine Rule. A1 for \(AC^2 = 79\). A1 for \(8.89\text{ cm}\) (accept \(8.88\) to \(8.90\)). (b) M1 for a correct substitution into the triangle area formula \(\frac{1}{2}ab\sin C\). A1 for \(30.3\text{ cm}^2\) (accept \(30.31\)). (c) M1 for writing a valid area equation \(\frac{1}{2} \times AC \times h = \text{Area}\). M1 for rearranging to solve for \(h\). A1 for \(6.82\text{ cm}\) (accept \(6.81\) to \(6.83\)).

(a) The probability that the first ball is red is \(\frac{5}{8}\). Since selection is without replacement, the probability that the second ball is red given the first was red is \(\frac{4}{7}\). Probability both are red = \(\frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}\). (b) The two balls have different colours if the first is red and second is blue, or first is blue and second is red. \(P(\text{Red, Blue}) = \frac{5}{8} \times \frac{3}{7} = \frac{15}{56}\). \(P(\text{Blue, Red}) = \frac{3}{8} \times \frac{5}{7} = \frac{15}{56}\). Total probability = \(\frac{15}{56} + \frac{15}{56} = \frac{30}{56} = \frac{15}{28}\). (c) Let \(E\) be the event that at least one ball is red. This is the complement of both being blue: \(P(E) = 1 - P(\text{both blue}) = 1 - \left(\frac{3}{8} \times \frac{2}{7}\right) = 1 - \frac{6}{56} = \frac{50}{56} = \frac{25}{28}\). The probability that both are red given at least one is red is: \(P(\text{both red} | E) = \frac{P(\text{both red})}{P(E)} = \frac{20/56}{50/56} = \frac{20}{50} = \frac{2}{5}\).

(a) M1 for multiplying two correct fractions \(\frac{5}{8} \times \frac{4}{7}\). A1 for \(\frac{5}{14}\) or equivalent fraction or decimal \(\approx 0.357\). (b) M1 for finding one correct combination (e.g. \(\frac{5}{8} \times \frac{3}{7} = \frac{15}{56}\)). M1 for adding both possible orderings: \(\frac{15}{56} + \frac{15}{56}\). A1 for \(\frac{15}{28}\) or equivalent fraction or decimal \(\approx 0.536\). (c) M1 for calculating the probability of getting at least one red ball: \(1 - \frac{6}{56} = \frac{50}{56}\). M1 for setting up the conditional probability division: \(\frac{P(\text{both red})}{P(\text{at least one red})}\). A1 for \(\frac{2}{5}\) or \(0.4\).