2023 Edexcel IGCSE Mathematics (Specification B) 模擬試題連答案詳解

We need to find two numbers that multiply to \(3 \times (-20) = -60\) and add to \(11\). These numbers are \(15\) and \(-4\). We can split the middle term: \(3x^2 + 15x - 4x - 20 = 3x(x + 5) - 4(x + 5) = (3x - 4)(x + 5)\).

M1 for any correct partial factorisation or finding the correct pair of numbers 15 and -4, or for a product of two linear brackets that when expanded gives two of the three terms of the quadratic correct. A1 for \((3x - 4)(x + 5)\).

Let \(x = 0.2777...\). Then \(10x = 2.777...\) and \(100x = 27.777...\). Subtracting the two equations gives: \(100x - 10x = 27.777... - 2.777...\) which simplifies to \(90x = 25\). Solving for \(x\) gives \(x = \frac{25}{90} = \frac{5}{18}\).

M1 for setting up two appropriate equations and attempting to subtract them, or for writing \(\frac{25}{90}\) (or equivalent). A1 for \(\frac{5}{18}\).

The determinant of \(\mathbf{M}\) is given by \(y(y+9) - (3)(12) = y^2 + 9y - 36\). Setting the determinant to 0 gives: \(y^2 + 9y - 36 = 0\). Factoring the quadratic: \((y+12)(y-3) = 0\), which gives \(y = -12\) or \(y = 3\). Since \(y > 0\), the value of \(y\) must be \(3\).

M1 for setting up the equation \(y(y+9) - 36 = 0\) (or equivalent). A1 for \(3\) (the negative solution must be rejected for this mark).

First find \(n(A \cup B)\): \(n(A \cup B) = n(\mathcal{E}) - n(A \cup B)' = 50 - 8 = 42\). Using the set identity \(n(A \cup B) = n(A) + n(B) - n(A \cap B)\), we get: \(42 = 28 + 21 - n(A \cap B) \implies 42 = 49 - n(A \cap B) \implies n(A \cap B) = 7\).

M1 for finding \(n(A \cup B) = 42\) or for writing a correct equation such as \(28 + 21 - x + 8 = 50\). A1 for \(7\).

The magnitude of \(\mathbf{v}\) is given by \(|\mathbf{v}| = \sqrt{k^2 + (k-2)^2} = 10\). Squaring both sides: \(k^2 + (k^2 - 4k + 4) = 100 \implies 2k^2 - 4k - 96 = 0 \implies k^2 - 2k - 48 = 0\). Factoring the quadratic: \((k-8)(k+6) = 0\). Since \(k > 0\), we have \(k = 8\).

M1 for setting up the equation \(k^2 + (k-2)^2 = 100\). A1 for \(8\) (the negative root must be rejected).

The reference angle for \(\cos \theta = \frac{\sqrt{3}}{2}\) is \(30^\circ\). Since \(\cos \theta\) is negative, \(\theta\) must lie in the second or third quadrant. The solutions are \(\theta = 180^\circ - 30^\circ = 150^\circ\) (second quadrant) and \(\theta = 180^\circ + 30^\circ = 210^\circ\) (third quadrant). In the given range \(180^\circ \le \theta \le 360^\circ\), the only solution is \(\theta = 210^\circ\).

M1 for identifying the reference angle of \(30^\circ\) or finding one correct solution such as \(150^\circ\) or \(210^\circ\). A1 for \(210^\circ\) (and no other values in the range).

Let \(y = \frac{4}{x-3}\). Rearranging to make \(x\) the subject: \(y(x-3) = 4 \implies xy - 3y = 4 \implies xy = 3y + 4 \implies x = \frac{3y+4}{y}\). Thus, the inverse function is \(\text{f}^{-1}(x) = \frac{3x+4}{x}\) (or \(3 + \frac{4}{x}\)).

M1 for a correct first step in making \(x\) the subject, such as \(y(x-3) = 4\) or \(x(y-3) = 4\) (if variables are swapped first). A1 for \(\frac{3x+4}{x}\) or equivalent.

The perimeter of a sector is given by \(2r + L\), where \(r\) is the radius and \(L\) is the arc length. Here, \(r = 12\text{ cm}\). The arc length is \(L = \frac{150}{360} \times 2\pi(12) = \frac{5}{12} \times 24\pi = 10\pi\text{ cm}\). Therefore, the perimeter is \(12 + 12 + 10\pi = 24 + 10\pi\text{ cm}\).

M1 for finding the arc length as \(10\pi\) (or showing a correct substitution into the arc length formula). A1 for \(24 + 10\pi\).

The value of the car after 1 year is \(18000 \times 0.85 = 15300\). The value after 2 years is \(15300 \times 0.90 = 13770\). Alternatively, we can calculate this in one step: \(18000 \times 0.85 \times 0.90 = 13770\).

M1 for a complete method to find the depreciated value, such as \(18000 \times 0.85 \times 0.90\) or finding \(15300\) and then multiplying by \(0.90\). A1 for 13770.

Expanding the brackets on the right-hand side gives \(5x - 3 > 3x + 15\). Subtracting \(3x\) from both sides gives \(2x - 3 > 15\). Adding 3 to both sides gives \(2x > 18\). Finally, dividing by 2 gives \(x > 9\).

M1 for expanding the brackets correctly and attempting to collect like terms, arriving at \(2x > a\) or \(bx > 18\). A1 for \(x > 9\).

The determinant of matrix \(\mathbf{A}\) is given by \(k(k-1) - (4 \times 3) = 8\). This simplifies to \(k^2 - k - 12 = 8\), which rearranges to \(k^2 - k - 20 = 0\). Factorising the quadratic yields \((k - 5)(k + 4) = 0\), which gives solutions \(k = 5\) or \(k = -4\). Since we require the positive value, we have \(k = 5\).

M1 for setting up the determinant equation \(k(k-1) - 12 = 8\) and reducing to a quadratic equation, e.g., \(k^2 - k - 20 = 0\). A1 for \(k = 5\).

Let \(x\) be the number of students who study both. The number of students who study at least one of the subjects is \(40 - 5 = 35\), so \(n(H \cup G) = 35\). Using the set formula \(n(H \cup G) = n(H) + n(G) - n(H \cap G)\), we have \(35 = 23 + 18 - x\). This simplifies to \(35 = 41 - x\), which gives \(x = 6\).

M1 for \(23 + 18 - (40 - 5)\) or for setting up a correct equation such as \((23 - x) + x + (18 - x) + 5 = 40\). A1 for 6.

The size of each exterior angle of the regular polygon is \(180^\circ - 162^\circ = 18^\circ\). Since the sum of the exterior angles of any polygon is \(360^\circ\), the number of sides is \(360^\circ / 18^\circ = 20\). Alternatively, using the interior angle formula: \((n-2) \times 180 / n = 162\), which gives \(180n - 360 = 162n\), leading to \(18n = 360\) and \(n = 20\).

M1 for finding the exterior angle \(180 - 162 = 18\) or for setting up the equation \((n-2) \times 180 / n = 162\). A1 for 20.

We first find the product matrix \(\mathbf{AB}\): \(\mathbf{AB} = \begin{pmatrix} 2 & x \\ -1 & 3 \end{pmatrix} \begin{pmatrix} y & 4 \\ 2 & -1 \end{pmatrix} = \begin{pmatrix} 2y + 2x & 8 - x \\ -y + 6 & -4 - 3 \end{pmatrix} = \begin{pmatrix} 2y + 2x & 8 - x \\ -y + 6 & -7 \end{pmatrix}\). Given that \(\mathbf{AB} = \begin{pmatrix} 14 & 5 \\ 2 & -7 \end{pmatrix}\), we can equate the corresponding elements: From the top-right element: \(8 - x = 5 \implies x = 3\). From the bottom-left element: \(-y + 6 = 2 \implies y = 4\). We can check consistency with the top-left element: \(2y + 2x = 2(4) + 2(3) = 8 + 6 = 14\), which is correct.

M1: For an attempt to multiply the two matrices, obtaining at least two correct algebraic elements in terms of \(x\) and \(y\) (e.g. \(8 - x\) or \(-y + 6\)). A1: For establishing at least two correct equations, e.g., \(8 - x = 5\) and \(-y + 6 = 2\). A1: For finding \(x = 3\). A1: For finding \(y = 4\).

Combine the fractions on the left-hand side of the equation using a common denominator of \((x - 1)(x + 2)\): \(\frac{2(x + 2) - 1(x - 1)}{(x - 1)(x + 2)} = \frac{1}{2}\) which simplifies to \(\frac{2x + 4 - x + 1}{x^2 + x - 2} = \frac{1}{2}\) and then \(\frac{x + 5}{x^2 + x - 2} = \frac{1}{2}\). Cross-multiply to eliminate the fractions: \(2(x + 5) = x^2 + x - 2\) which gives \(2x + 10 = x^2 + x - 2\). Rearrange the terms to form a quadratic equation equal to zero: \(x^2 - x - 12 = 0\). Factorise the quadratic equation: \((x - 4)(x + 3) = 0\). Thus, the solutions are \(x = 4\) or \(x = -3\).

M1: For writing the left-hand side as a single fraction with denominator \((x-1)(x+2)\) or equivalent, with a correct attempt at expanding the numerator, e.g., \(2(x+2) - (x-1)\). M1: For removing the fractions to obtain a 3-term quadratic equation, e.g., \(x^2 - x - 12 = 0\). M1: For a correct method to solve their 3-term quadratic equation (factorisation or quadratic formula). A1: For both correct solutions: \(x = 4\) and \(x = -3\).

(a) Equating the expressions for \(y\):
\[ \frac{x^2 - 5x + 6}{x - 1} = 2x - 4 \]
Multiply both sides by \((x - 1)\):
\[ x^2 - 5x + 6 = (2x - 4)(x - 1) \]
\[ x^2 - 5x + 6 = 2x^2 - 2x - 4x + 4 \]
\[ x^2 - 5x + 6 = 2x^2 - 6x + 4 \]
Rearrange to form a quadratic equal to zero:
\[ 2x^2 - x^2 - 6x + 5x + 4 - 6 = 0 \]
\[ x^2 - x - 2 = 0 \]

(b) Solve the quadratic equation \(x^2 - x - 2 = 0\):
\[ (x - 2)(x + 1) = 0 \]
So \(x = 2\) or \(x = -1\).
Find the corresponding \(y\)-values using \(y = 2x - 4\):
For \(x = 2\):
\[ y = 2(2) - 4 = 0 \]
For \(x = -1\):
\[ y = 2(-1) - 4 = -6 \]
Thus, the coordinates of the two points of intersection are \((2, 0)\) and \((-1, -6)\).

(a)
M1: For equating the two expressions for \(y\) and multiplying both sides by \((x - 1)\).
M1: For expanding \((2x - 4)(x - 1)\) correctly to get \(2x^2 - 6x + 4\).
A1: For correct simplification leading to the given equation \(x^2 - x - 2 = 0\).

(b)
M1: For factoring or solving the quadratic to find \(x = 2\) and \(x = -1\).
M1: For substituting at least one of their \(x\)-values into \(y = 2x - 4\) to find a corresponding \(y\)-value.
A1: Both correct coordinates \((2, 0)\) and \((-1, -6)\).

(a) The determinant of matrix \(\mathbf{A}\) is:
\[ \det(\mathbf{A}) = (k)(k - 1) - (2)(3) \]
\[ \det(\mathbf{A}) = k^2 - k - 6 \]
Since the determinant of \(\mathbf{A}\) is 8:
\[ k^2 - k - 6 = 8 \]
Subtract 8 from both sides:
\[ k^2 - k - 14 = 0 \]

(b) Solve the quadratic equation \(k^2 - k - 14 = 0\) using the quadratic formula:
\[ k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 1\), \(b = -1\), and \(c = -14\):
\[ k = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-14)}}{2(1)} \]
\[ k = \frac{1 \pm \sqrt{1 + 56}}{2} \]
\[ k = \frac{1 \pm \sqrt{57}}{2} \]

(a)
M1: For writing a correct expression for the determinant of matrix \(\mathbf{A}\) in terms of \(k\), i.e., \(k(k - 1) - 6\).
M1: For setting their determinant expression equal to 8.
A1: For correct simplification leading to the given equation \(k^2 - k - 14 = 0\).

(b)
M1: For applying the quadratic formula with \(a = 1, b = -1, c = -14\) (allow one sign error in substitution).
M1: For correctly evaluating the discriminant to get \(1 - 4(1)(-14) = 57\).
A1: For the final correct expression \(k = \frac{1 \pm \sqrt{57}}{2}\).

(a) The probability that the first counter drawn is green is \(\frac{6}{n}\).
Since the counter is not replaced, there are now \(n - 1\) counters remaining in the bag, of which 5 are green.
The probability that the second counter drawn is green is \(\frac{5}{n - 1}\).
Thus, the probability that both counters are green is:
\[ P(\text{both green}) = \frac{6}{n} \times \frac{5}{n - 1} = \frac{30}{n(n - 1)} \]

(b) (i) We are given that \(P(\text{both green}) = \frac{1}{3}\):
\[ \frac{30}{n(n - 1)} = \frac{1}{3} \]
Multiply both sides by \(3n(n - 1)\):
\[ 90 = n(n - 1) \]
\[ n^2 - n - 90 = 0 \]

(ii) Solve the quadratic equation \(n^2 - n - 90 = 0\):
\[ (n - 10)(n + 9) = 0 \]
This gives \(n = 10\) or \(n = -9\).
Since \(n\) represents the total number of counters, \(n\) must be positive.
Therefore, \(n = 10\).

(a)
M1: For representing the probability of the first green counter as \(\frac{6}{n}\) and the second green counter as \(\frac{5}{n-1}\) and multiplying them.
A1: For completing the multiplication to show \(\frac{30}{n(n-1)}\) clearly.

(b) (i)
M1: For setting their probability expression from (a) equal to \(\frac{1}{3}\) and cross-multiplying.
A1: For expanding and simplifying to the given quadratic equation \(n^2 - n - 90 = 0\).

(b) (ii)
M1: For factoring the quadratic expression to \((n - 10)(n + 9) = 0\) (or using the quadratic formula).
A1: For choosing \(n = 10\) and rejecting the negative root with explanation (e.g., total counters must be positive).

(a) First, find the vector \(\overrightarrow{AB}\):
\[ \overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -\mathbf{a} + \mathbf{b} \]
Since \(AP : PB = 3 : 1\), the point \(P\) lies \(\frac{3}{4}\) of the way along \(AB\) from \(A\).
Thus, \(\overrightarrow{AP} = \frac{3}{4}\overrightarrow{AB} = \frac{3}{4}(\mathbf{b} - \mathbf{a})\).
Now, find \(\overrightarrow{OP}\):
\[ \overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP} \]
\[ \overrightarrow{OP} = \mathbf{a} + \frac{3}{4}(\mathbf{b} - \mathbf{a}) = \mathbf{a} + \frac{3}{4}\mathbf{b} - \frac{3}{4}\mathbf{a} \]
\[ \overrightarrow{OP} = \frac{1}{4}\mathbf{a} + \frac{3}{4}\mathbf{b} \]

(b) The point \(Q\) lies on \(OB\) extended such that \(OB : BQ = 1 : 2\).
This means \(\overrightarrow{BQ} = 2\overrightarrow{OB} = 2\mathbf{b}\).
So, \(\overrightarrow{OQ} = \overrightarrow{OB} + \overrightarrow{BQ} = \mathbf{b} + 2\mathbf{b} = 3\mathbf{b}\).
Now find \(\overrightarrow{PQ}\):
\[ \overrightarrow{PQ} = \overrightarrow{PO} + \overrightarrow{OQ} = -\overrightarrow{OP} + \overrightarrow{OQ} \]
\[ \overrightarrow{PQ} = -\left(\frac{1}{4}\mathbf{a} + \frac{3}{4}\mathbf{b}\right) + 3\mathbf{b} \]
\[ \overrightarrow{PQ} = -\frac{1}{4}\mathbf{a} - \frac{3}{4}\mathbf{b} + 3\mathbf{b} \]
\[ \overrightarrow{PQ} = -\frac{1}{4}\mathbf{a} + \frac{9}{4}\mathbf{b} \]

(a)
M1: For expressing \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\) or \(\overrightarrow{AP} = \frac{3}{4}(\mathbf{b} - \mathbf{a})\).
M1: For the vector addition path \(\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP}\).
A1: For the correct simplified vector \(\frac{1}{4}\mathbf{a} + \frac{3}{4}\mathbf{b}\).

(b)
M1: For finding \(\overrightarrow{OQ} = 3\mathbf{b}\) or expressing \(\overrightarrow{PQ}\) as a correct vector path, e.g., \(\overrightarrow{PB} + \overrightarrow{BQ}\).
M1: For substituting their vector expressions into a correct path equation.
A1: For the correct simplified vector \(-\frac{1}{4}\mathbf{a} + \frac{9}{4}\mathbf{b}\).

(a) The volume of a cylinder is given by \(V = \pi r^2 h\).
Since the cylinder has height \(h = 4r\):
\[ V_{\text{cylinder}} = \pi r^2 (4r) = 4\pi r^3 \]
The volume of a sphere is given by \(V = \frac{4}{3}\pi R^3\).
Since the cylinder is melted down and recast into the sphere, their volumes are equal:
\[ 4\pi r^3 = \frac{4}{3}\pi R^3 \]
Divide both sides by \(4\pi\):
\[ r^3 = \frac{1}{3}R^3 \]
Multiply by 3:
\[ R^3 = 3r^3 \]
Taking the cube root of both sides:
\[ R = 3^{\frac{1}{3}} r \]

(b) The total surface area of the cylinder is:
\[ A_{\text{cylinder}} = 2\pi r^2 + 2\pi r h = 2\pi r^2 + 2\pi r (4r) = 10\pi r^2 \]
The surface area of the sphere is:
\[ A_{\text{sphere}} = 4\pi R^2 \]
Substitute \(R = 3^{\frac{1}{3}} r\):
\[ A_{\text{sphere}} = 4\pi (3^{\frac{1}{3}} r)^2 = 4 \times 3^{\frac{2}{3}} \pi r^2 \]
Now, find the ratio of the total surface area of the cylinder to the surface area of the sphere:
\[ \text{Ratio} = 10\pi r^2 : 4 \times 3^{\frac{2}{3}} \pi r^2 \]
Divide both terms of the ratio by the common factor of \(2\pi r^2\):
\[ \text{Ratio} = 5 : 2 \times 3^{\frac{2}{3}} \]

(a)
M1: For writing a correct expression for the volume of the cylinder in terms of \(r\), i.e., \(4\pi r^3\).
M1: For setting their cylinder volume equal to the volume of a sphere, \(\frac{4}{3}\pi R^3\).
A1: For correct algebra leading to \(R = 3^{\frac{1}{3}} r\).

(b)
M1: For writing a correct expression for the total surface area of the cylinder, \(10\pi r^2\).
M1: For substituting \(R = 3^{\frac{1}{3}} r\) into \(4\pi R^2\) to obtain the sphere's surface area in terms of \(r\), \(4 \times 3^{\frac{2}{3}} \pi r^2\).
A1: For dividing both parts of the ratio by \(2\pi r^2\) to reach the required form \(5 : 2 \times 3^{\frac{2}{3}}\).

(a) Time taken for the outward run is \(\frac{36}{x}\) hours. (b) Time taken for the return run is \(\frac{36}{x-3}\) hours. (c) Since the return run takes 1 hour longer than the outward run: \(\frac{36}{x-3} - \frac{36}{x} = 1\). Multiplying both sides by the common denominator \(x(x-3)\) gives: \(36x - 36(x-3) = x(x-3)\) which simplifies to \(36x - 36x + 108 = x^2 - 3x\), so \(108 = x^2 - 3x\), which rearranges to \(x^2 - 3x - 108 = 0\). (d) To solve \(x^2 - 3x - 108 = 0\), we factorise to get \((x-12)(x+9) = 0\). Since speed must be positive, \(x = 12\) km/h (rejecting \(x = -9\)). Thus, the time taken for the return run is \(\frac{36}{12-3} = \frac{36}{9} = 4\) hours.

(a) B1: for \(\frac{36}{x}\) or equivalent. (b) B1: for \(\frac{36}{x-3}\) or equivalent. (c) M1: for setting up the equation \(\frac{36}{x-3} - \frac{36}{x} = 1\). M1: for multiplying by the common denominator to obtain a quadratic. A1: for showing the target equation \(x^2 - 3x - 108 = 0\) with no errors. (d) M1: for factorising or using the quadratic formula to solve \(x^2 - 3x - 108 = 0\). A1: for finding \(x = 12\) (optionally rejecting \(x = -9\)). B1: for calculating the return time of 4 hours.

(a) \(\vec{AB} = \vec{AO} + \vec{OB} = -\mathbf{a} + \mathbf{b}\). (b) Since \(OP:PA = 2:1\), \(\vec{OP} = \frac{2}{3}\mathbf{a}\). Thus \(\vec{BP} = \vec{BO} + \vec{OP} = -\mathbf{b} + \frac{2}{3}\mathbf{a} = \frac{2}{3}\mathbf{a} - \mathbf{b}\). (c) \(\vec{OR} = \vec{OB} + \vec{BR} = \mathbf{b} + \lambda \vec{BP} =
\mathbf{b} + \lambda\left(\frac{2}{3}\mathbf{a} - \mathbf{b}\right) = \frac{2}{3}\lambda\mathbf{a} + (1-\lambda)\mathbf{b}\). (d) We also know \(\vec{OR} = \mu \vec{OQ}\). Since \(Q\) is the midpoint of \(AB\), \(\vec{OQ} = \frac{1}{2}(\mathbf{a} + \mathbf{b})\). So \(\vec{OR} = \frac{1}{2}\mu\mathbf{a} + \frac{1}{2}\mu\mathbf{b}\). Equating coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) gives: \(\frac{2}{3}\lambda = \frac{1}{2}\mu \implies \mu = \frac{4}{3}\lambda\) and \(1-\lambda = \frac{1}{2}\mu\). Substituting \(\mu\) gives \(1-\lambda = \frac{2}{3}\lambda \implies 1 = \frac{5}{3}\lambda \implies \lambda = \frac{3}{5}\). Using this, \mu = \frac{4}{3} \times \frac{3}{5} = \frac{4}{5}.

(a) B1: for \(-\mathbf{a} + \mathbf{b}\) or equivalent. (b) M1: for identifying \(\vec{OP} = \frac{2}{3}\mathbf{a}\). A1: for \(\frac{2}{3}\mathbf{a} - \mathbf{b}\) or equivalent. (c) M1: for expressing \(\vec{OR}\) as \(\vec{OB} + \lambda\vec{BP}\) or equivalent. A1: for \(\frac{2}{3}\lambda\mathbf{a} + (1-\lambda)\mathbf{b}\). (d) M1: for equating coefficients using \(\vec{OQ} = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\). A1: for \(\lambda = \frac{3}{5}\). A1: for \(\mu = \frac{4}{5}\).

(a) In the Venn diagram, the numbers in the regions are: \(H \cap G \cap F = x\); Only \(H \cap G = 12 - x\); Only \(G \cap F = 15 - x\); Only \(H \cap F = 10 - x\); Only \(H = 45 - (12-x + 10-x + x) = 23 + x\); Only \(G = 40 - (12-x + 15-x + x) = 13 + x\); Only \(F = 35 - (10-x + 15-x + x) = 10 + x\); Outside all three = 8. (b) The sum of all regions is 100: \((23+x) + (13+x) + (10+x) + (12-x) + (15-x) + (10-x) + x + 8 = 100 \implies 91 + x = 100 \implies x = 9\). (c) The number of students studying at least two subjects is \((12-x) + (15-x) + (10-x) + x = 37 - 2x = 37 - 18 = 19\). The probability is \(\frac{19}{100}\). (d) The number of students studying French is 35. The number of students studying both French and History is 10. The probability is \(\frac{10}{35} = \frac{2}{7}\).

(a) M1: for drawing a Venn diagram with three intersecting circles. A2: for all regions correctly written in terms of \(x\) (or as numeric values once \(x=9\) is shown). (b) M1: for setting up the sum of all regions plus 8 equal to 100. A1: for simplifying the expression to show \(x = 9\). (c) M1: for summing the correct four intersection regions. A1: for \(\frac{19}{100}\) or 0.19. (d) B1: for \(\frac{2}{7}\) (or \(\frac{10}{35}\), or 0.286).

(a) In the right-angled triangle \(TPQ\) (right-angled at \(P\)): \(\tan(25^\circ) = \frac{h}{15} \implies h = 15 \tan(25^\circ) \approx 6.9946 \approx 6.99\) m. (b) In the right-angled triangle \(PQR\) (right-angled at \(Q\)): \(PR = \sqrt{PQ^2 + QR^2} = \sqrt{15^2 + 20^2} = \sqrt{625} = 25\) m. (c) In the right-angled triangle \(TPR\) (right-angled at \(P\)), the angle of elevation of \(T\) from \(R\) is \(\angle TRP\): \(\tan(\angle TRP) = \frac{h}{PR} = \frac{6.9946}{25} \approx 0.27978 \implies \angle TRP \approx 15.63^\circ \approx 15.6^\circ\). (d) In the right-angled triangle \(TPR\): \(TR = \sqrt{TP^2 + PR^2} = \sqrt{6.9946^2 + 25^2} = \sqrt{673.925} \approx 25.96 \approx 26.0\) m.

(a) M1: for using \(\tan(25^\circ) = \frac{h}{15}\). A1: for 6.99 (accept answers in range 6.99 to 7.00). (b) M1: for \(PR = \sqrt{15^2 + 20^2}\). A1: for 25. (c) M1: for \(\tan(\theta) = \frac{\text{their } h}{\text{their } PR}\). A1: for 15.6 (accept answers in range 15.5 to 15.7). (d) M1: for applying Pythagoras' theorem or trigonometry to find \(TR\). A1: for 26.0 (accept answers in range 25.9 to 26.1).

(a) The total volume of the toy is the volume of the cylinder plus the volume of the hemisphere: \(V = \pi r^2 h + \frac{2}{3}\pi r^3 = 120\pi\). Dividing by \(\pi\) gives \(r^2 h + \frac{2}{3}r^3 = 120 \implies r^2 h = 120 - \frac{2}{3}r^3 \implies h = \frac{120}{r^2} - \frac{2}{3}r\). (b) The total surface area of the toy is the sum of the circular base area, the curved cylinder area, and the curved hemisphere area: \(A = \pi r^2 + 2\pi r h + 2\pi r^2 = 3\pi r^2 + 2\pi r h\). Substituting the expression for \(h\) from part (a): \(A = 3\pi r^2 + 2\pi r\left(\frac{120}{r^2} - \frac{2}{3}r\right) = 3\pi r^2 + \frac{240\pi}{r} - \frac{4}{3}\pi r^2 = \frac{5}{3}\pi r^2 + \frac{240\pi}{r}\). (c) When \(r = 6\): \(A = \frac{5}{3}\pi (6^2) + \frac{240\pi}{6} = 60\pi + 40\pi = 100\pi\).

(a) M1: for writing the volume of the hemisphere or cylinder correctly. M1: for setting up the equation \(\pi r^2 h + \frac{2}{3}\pi r^3 = 120\pi\). A1: for correctly rearranging to make \(h\) the subject. (b) M1: for writing the total surface area formula as \(A = 3\pi r^2 + 2\pi r h\). M1: for substituting \(h\) into their surface area expression. A1: for simplifying correctly to obtain the given expression. (c) M1: for substituting \(r=6\) into the given surface area formula. A1: for \(100\pi\).

(a) \(\mathbf{AB} = \begin{pmatrix} 2 & k \\ -1 & 3 \end{pmatrix} \begin{pmatrix} 3 & 1 \\ 2 & 4 \end{pmatrix} = \begin{pmatrix} 2(3) + k(2) & 2(1) + k(4) \\ -1(3) + 3(2) & -1(1) + 3(4) \end{pmatrix} = \begin{pmatrix} 2k+6 & 4k+2 \\ 3 & 11
\end{pmatrix}\). (b) \(\det(\mathbf{AB}) = (2k+6)(11) - (4k+2)(3) = 22k + 66 - 12k - 6 = 10k + 60\). Since \(\det(\mathbf{AB}) = 80\), we have \(10k + 60 = 80 \implies 10k = 20 \implies k = 2\). (Alternatively, \(\det(\mathbf{B}) = 12-2 = 10\), and \(\det(\mathbf{A}) = 6+k\). Since \(\det(\mathbf{AB}) = \det(\mathbf{A})\det(\mathbf{B})\), \(10(6+k) = 80 \implies 6+k = 8 \implies k = 2\).) (c) For \(k = 2\), \(\mathbf{A} = \begin{pmatrix} 2 & 2 \\ -1 & 3 \end{pmatrix}\). The determinant of \(\mathbf{A}\) is \(\det(\mathbf{A}) = 2(3) - 2(-1) = 8\). The inverse matrix is \(\mathbf{A}^{-1} = \frac{1}{8} \begin{pmatrix} 3 & -2 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} \frac{3}{8} & -\frac{1}{4} \\ \frac{1}{8} & \frac{1}{4} \end{pmatrix}\).

(a) M1: for attempting matrix multiplication with at least two correct entries. A1: for \(\begin{pmatrix} 2k+6 & 4k+2 \\ 3 & 11 \end{pmatrix}\). (b) M1: for setting up the determinant of \(\mathbf{AB}\) or using \(\det(\mathbf{AB}) = \det(\mathbf{A})\det(\mathbf{B})\). M1: for forming the equation \(10k + 60 = 80\) or equivalent. A1: for \(k=2\). (c) M1: for substituting \(k=2\) and finding the determinant of \(\mathbf{A}\) is 8. M1: for swapping the leading diagonal entries and changing signs of the other entries. A1: for \(\begin{pmatrix} \frac{3}{8} & -\frac{1}{4} \\ \frac{1}{8} & \frac{1}{4} \end{pmatrix}\) or equivalent.

(a) The volume \(V\) of the box is:
\(V = 3x \times 2x \times h = 6x^2 h = 480\)
Thus, \(h = \frac{480}{6x^2} = \frac{80}{x^2}\).

The total surface area \(A\) of an open box (no top lid) is:
\(A = \text{Base Area} + 2 \times \text{Front Area} + 2 \times \text{Side Area}\)
\(A = (3x)(2x) + 2(3x)(h) + 2(2x)(h)\)
\(A = 6x^2 + 6xh + 4xh = 6x^2 + 10xh\)

Substitute \(h = \frac{80}{x^2}\):
\(A = 6x^2 + 10x \left(\frac{80}{x^2}\right) = 6x^2 + \frac{800}{x}\) (as required).

(b) Differentiating \(A\) with respect to \(x\):
\(\frac{\mathrm{d}A}{\mathrm{d}x} = \frac{\mathrm{d}}{\mathrm{d}x}(6x^2 + 800x^{-1}) = 12x - 800x^{-2} = 12x - \frac{800}{x^2}\).

(c) For a stationary value, \(\frac{\mathrm{d}A}{\mathrm{d}x} = 0\):
\(12x - \frac{800}{x^2} = 0\)
\(12x = \frac{800}{x^2}\)
\(12x^3 = 800\)
\(x^3 = \frac{200}{3} \approx 66.667\)
\(x = \sqrt[3]{\frac{200}{3}} \approx 4.0548 \approx 4.05\) cm (3 s.f.).

(d) Differentiating again:
\(\frac{\mathrm{d}^2A}{\mathrm{d}x^2} = 12 - 800(-2)x^{-3} = 12 + \frac{1600}{x^3}\).

At \(x^3 = \frac{200}{3}\):
\(\frac{\mathrm{d}^2A}{\mathrm{d}x^2} = 12 + \frac{1600}{\frac{200}{3}} = 12 + 24 = 36\).

Since \(\frac{\mathrm{d}^2A}{\mathrm{d}x^2} = 36 > 0\), the stationary value is a minimum.

(e) Substitute the exact value \(x = 4.0548\) into the expression for \(A\):
\(A = 6(4.0548)^2 + \frac{800}{4.0548}\)
\(A \approx 98.65 + 197.30 = 295.95 \approx 296\text{ cm}^2\) (to the nearest whole number).

(a)
M1: Writes an expression for volume \(V = 6x^2h = 480\) and makes \(h\) the subject.
M1: Writes the total surface area formula for an open box: \(A = 6x^2 + 10xh\).
A1: Substitutes \(h\) and simplifies to obtain the given formula.

(b)
M1: For differentiating at least one term correctly.
A1: Correct derivative \(12x - \frac{800}{x^2}\).

(c)
M1: Sets derivative equal to 0 and attempts to solve for \(x^3\).
M1: Finds \(x^3 = \frac{200}{3}\) or \(66.7\).
A1: \(4.05\) (3 s.f.).

(d)
M1: Correctly differentiates to find \(\frac{\mathrm{d}^2A}{\mathrm{d}x^2} = 12 + \frac{1600}{x^3}\).
A1: Evaluates derivative to find \(36\) (or shows it is positive) and concludes it is a minimum.

(e)
M1: Substitutes their value of \(x\) back into the formula for \(A\).
A1: Correct answer of \(296\) (nearest whole number).

(a) From triangle law:
\(\vec{AB} = \vec{AO} + \vec{OB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}\).
Since \(Q\) is the midpoint of \(AB\):
\(\vec{OQ} = \vec{OA} + \frac{1}{2}\vec{AB} = \mathbf{a} + \frac{1}{2}(\mathbf{b} - \mathbf{a}) = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\).

(b) \(P\) lies on \(OA\) such that \(OP : PA = 2 : 1\), which means \(\vec{OP} = \frac{2}{3}\mathbf{a}\).
\(\vec{BP} = \vec{BO} + \vec{OP} = -\mathbf{b} + \frac{2}{3}\mathbf{a} = \frac{2}{3}\mathbf{a} - \mathbf{b}\).

(c) First expression for \(\vec{OX}\):
\(\vec{OX} = \mu \vec{OQ} = \mu \left(\frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\right) = \frac{1}{2}\mu \mathbf{a} + \frac{1}{2}\mu \mathbf{b}\).

Second expression for \(\vec{OX}\):
\(\vec{OX} = \vec{OB} + \vec{BX} = \mathbf{b} + \lambda \vec{BP} = \mathbf{b} + \lambda \left(\frac{2}{3}\mathbf{a} - \mathbf{b}\right) = \frac{2}{3}\lambda \mathbf{a} + (1 - \lambda)\mathbf{b}\).

(d) Equating the coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) from both expressions:
For \(\mathbf{a}\): \(\frac{1}{2}\mu = \frac{2}{3}\lambda\) (1)
For \(\mathbf{b}\): \(\frac{1}{2}\mu = 1 - \lambda\) (2)

From (1) and (2):
\(\frac{2}{3}\lambda = 1 - \lambda\)
\(\frac{5}{3}\lambda = 1 \implies \lambda = \frac{3}{5}\).

Substitute \(\lambda = \frac{3}{5}\) into (2):
\(\frac{1}{2}\mu = 1 - \frac{3}{5} = \frac{2}{5} \implies \mu = \frac{4}{5}\).

(e) Since \(\vec{OX} = \frac{4}{5}\vec{OQ}\), the point \(X\) lies \(\frac{4}{5}\) of the way along \(OQ\).
This means \(OX : XQ = 4 : 1\).

(a)
B1: For \(\vec{AB} = \mathbf{b} - \mathbf{a}\) (or equivalent).
B1: For \(\vec{OQ} = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\).

(b)
M1: Identifies \(\vec{OP} = \frac{2}{3}\mathbf{a}\) and writes a valid vector sum for \(\vec{BP}\).
A1: \(\vec{BP} = \frac{2}{3}\mathbf{a} - \mathbf{b}\).

(c)
M1: Substitutes \(\vec{OQ}\) into \(\mu \vec{OQ}\) to find the first expression.
A1: First expression correct.
M1: Expresses \(\vec{OX}\) as \(\vec{OB} + \lambda \vec{BP}\) and expands.
A1: Second expression correct.

(d)
M1: Equates coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) to set up two simultaneous equations.
M1: Solves the equations for either \(\lambda\) or \(\mu\).
A1: Both \(\lambda = \frac{3}{5}\) and \(\mu = \frac{4}{5}\).

(e)
B1: Ratio is \(4 : 1\).

(a) In the right-angled triangle \(ABC\):
\(AC^2 = AB^2 + BC^2 = 12^2 + 8^2 = 144 + 64 = 208\)
\(AC = \sqrt{208} = 4\sqrt{13} \approx 14.422\text{ cm} \approx 14.4\text{ cm}\) (3 s.f.).

(b) Since \(O\) is the center of the base, \(AO = \frac{1}{2}AC = \frac{\sqrt{208}}{2} = \sqrt{52} \approx 7.211\text{ cm}\).
The angle between the edge \(VA\) and the base \(ABCD\) is \(\angle VAO\) in the right-angled triangle \(VOA\):
\(\tan(\angle VAO) = \frac{VO}{AO} = \frac{15}{\sqrt{52}} \approx 2.0801\)
\(\angle VAO = \arctan(2.0801) \approx 64.318^\circ \approx 64.3^\circ\) (3 s.f.).

(c) Using Pythagoras' theorem in triangle \(VOA\):
\(VA^2 = AO^2 + VO^2 = 52 + 15^2 = 52 + 225 = 277\)
\(VA = \sqrt{277} \approx 16.643\text{ cm} \approx 16.6\text{ cm}\) (as required).

(d) Let \(M\) be the midpoint of \(AB\). The line \(OM\) is parallel to \(BC\), so \(OM = \frac{1}{2}BC = 4\text{ cm}\).
The angle between the face \(VAB\) and the base \(ABCD\) is \(\angle VMO\) in the right-angled triangle \(VOM\):
\(\tan(\angle VMO) = \frac{VO}{OM} = \frac{15}{4} = 3.75\)
\(\angle VMO = \arctan(3.75) \approx 75.068^\circ \approx 75.1^\circ\) (3 s.f.).

(e) The total surface area consists of the rectangular base \(ABCD\), two triangles \(VAB\) and \(VCD\), and two triangles \(VBC\) and \(VAD\).
Base area \(= 12 \times 8 = 96\text{ cm}^2\).
For triangle \(VAB\), height \(VM = \sqrt{VO^2 + OM^2} = \sqrt{15^2 + 4^2} = \sqrt{241} \approx 15.524\text{ cm}\).
Area of \(VAB = \frac{1}{2} \times AB \times VM = \frac{1}{2} \times 12 \times \sqrt{241} \approx 93.145\text{ cm}^2\).

For triangle \(VBC\), let \(N\) be the midpoint of \(BC\). \(ON = \frac{1}{2}AB = 6\text{ cm}\).
Height \(VN = \sqrt{VO^2 + ON^2} = \sqrt{15^2 + 6^2} = \sqrt{261} \approx 16.155\text{ cm}\).
Area of \(VBC = \frac{1}{2} \times BC \times VN = \frac{1}{2} \times 8 \times \sqrt{261} \approx 64.622\text{ cm}^2\).

Total Surface Area \(= 96 + 2(93.145) + 2(64.622) = 96 + 186.290 + 129.244 = 411.534 \approx 412\text{ cm}^2\) (3 s.f.).

(a)
M1: Uses Pythagoras' theorem on the base (e.g., \(12^2 + 8^2\)).
A1: \(14.4\) or \(\sqrt{208}\).

(b)
M1: Identifies \(AO = \frac{1}{2}AC\) and uses trigonometry involving \(VO\) and \(AO\).
M1: Sets up \(\tan \theta = \frac{15}{\text{their } AO}\).
A1: \(64.3^\circ\).

(c)
M1: Uses Pythagoras' theorem on triangle \(VOA\).
A1: Shows \(VA = \sqrt{277} \approx 16.6\) with sufficient working.

(d)
M1: Identifies the required angle \(\angle VMO\) and calculates \(OM = 4\).
M1: Sets up \(\tan \angle VMO = \frac{15}{4}\).
A1: \(75.1^\circ\).

(e)
M1: Calculates the areas of the two different triangular faces using calculated slant heights.
A1: Correct total surface area of \(412\) (3 s.f.).

(a) In 2010, \(t = 0\):
\(120,000 = A \times k^0 \implies A = 120,000\).

(b) In 2018, \(t = 8\) and \(P = 158,400\):
\(158,400 = 120,000 \times k^8\)
\(k^8 = \frac{158,400}{120,000} = 1.32\)
\(k = (1.32)^{1/8} \approx 1.035305\)
So, \(k \approx 1.0353\) (4 d.p.).

(c) In 2025, \(t = 15\):
Using \(k = 1.0353\):
\(P = 120,000 \times (1.0353)^{15} \approx 120,000 \times 1.68114 = 201,737\).
To the nearest hundred, \(P = 201,700\).
(If using exact \(k\), \(P \approx 201,800\); either value is acceptable depending on rounding step).

(d) We need \(P > 240,000\):
\(120,000 \times (1.0353)^t > 240,000\)
\(1.0353^t > 2\)
Taking logarithms on both sides:
\(t \ln(1.0353) > \ln(2)\)
\(t > \frac{\ln(2)}{\ln(1.0353)} \approx \frac{0.693147}{0.034692} \approx 19.98\) years.
Since \(t\) must be an integer for full calendar years of growth, \(t = 20\).
Year is \(2010 + 20 = 2030\).

(e) Increase in population \(= 201,737 - 120,000 = 81,737\).
Percentage increase \(= \frac{81,737}{120,000} \times 100\% \approx 68.114\% \approx 68.1\%\).
(Using exact \(k\), \(P = 201,754\), percentage increase \(= 68.128\% \approx 68.1\%\)).

(a)
B1: \(120,000\).

(b)
M1: Sets up the equation \(158,400 = 120,000 \times k^8\).
M1: Rearranges to make \(k^8\) the subject and attempts to find the eighth root.
A1: Shows clearly that \(k \approx 1.0353\).

(c)
M1: Substitutes \(t = 15\) into the model formula with \(k = 1.0353\).
A1: \(201,700\) (or \(201,800\) from exact value).

(d)
M1: Sets up the inequality \(1.0353^t > 2\).
M1: Uses logarithms to solve for \(t\).
A1: Obtains \(t \approx 19.98\) or \(20\).
A1: Correct calendar year of \(2030\).

(e)
M1: Calculates the difference in population and divides by \(120,000\).
A1: \(68.1\%\).

(a) The tree diagram should show three initial branches: \(A (0.40)\), \(B (0.35)\), and \(C (0.25)\).
From each of these branches, there should be two sub-branches for Defective (\(D\)) and Not Defective (\(ND\)):
- From \(A\): \(D (0.02)\), \(ND (0.98)\)
- From \(B\): \(D (0.03)\), \(ND (0.97)\)
- From \(C\): \(D (0.04)\), \(ND (0.96)\)

(b) Using the law of total probability:
\(P(D) = P(A \cap D) + P(B \cap D) + P(C \cap D)\)
\(P(D) = 0.40(0.02) + 0.35(0.03) + 0.25(0.04)\)
\(P(D) = 0.008 + 0.0105 + 0.01 = 0.0285\).

(c) Using Bayes' Theorem:
\(P(B | D) = \frac{P(B \cap D)}{P(D)} = \frac{0.35 \times 0.03}{0.0285}\)
\(P(B | D) = \frac{0.0105}{0.0285} = \frac{105}{285}\)
Dividing the numerator and denominator by 15 gives:
\(P(B | D) = \frac{7}{19}\).

(d) Out of 20 components, 4 are defective (\(D\)) and 16 are not defective (\(ND\)).
The probability of choosing exactly one defective component in two selections without replacement is:
\(P(\text{Exactly } 1) = P(D, ND) + P(ND, D)\)
\(P(D, ND) = \frac{4}{20} \times \frac{16}{19} = \frac{64}{380}\)
\(P(ND, D) = \frac{16}{20} \times \frac{4}{19} = \frac{64}{380}\)
\(P(\text{Exactly } 1) = \frac{64}{380} + \frac{64}{380} = \frac{128}{380} = \frac{32}{95} \approx 0.337\).

(a)
M1: Draws three main branches with correct labels and probabilities.
M1: Draws correct sub-branches representing defective and non-defective outcomes.
A1: Correctly labels all probabilities on the tree diagram.

(b)
M1: Identifies the three product pathways representing defective components.
M1: Computes the sum of products: \(0.40(0.02) + 0.35(0.03) + 0.25(0.04)\).
A1: \(0.0285\) (or \(\frac{57}{2000}\)).

(c)
M1: Recalls and writes down the formula for conditional probability: \(P(B|D) = \frac{P(B \cap D)}{P(D)}\).
M1: Substitutes \(0.0105\) and their \(P(D)\) into the fraction.
A1: \(\frac{7}{19}\) (must be in simplest form).

(d)
M1: Identifies the two possible scenarios: (D, ND) and (ND, D).
M1: Applies conditional probability without replacement (e.g., \(\frac{4}{20} \times \frac{16}{19}\)).
A1: \(\frac{32}{95}\) (or equivalent simplified fraction or decimal \(\approx 0.337\)).