2023 Cambridge IGCSE International Mathematics (0607) 模擬試題連答案詳解

First differences: \(8-3=5\), \(15-8=7\), \(24-15=9\), \(35-24=11\). Second differences: \(7-5=2\), \(9-7=2\), \(11-9=2\). Since the second difference is constant and equal to \(2\), the sequence is quadratic with leading term \(an^2\) where \(2a = 2 \implies a = 1\). Comparing the sequence with \(n^2 = 1, 4, 9, 16, 25, \dots\), the difference is \(3-1=2\), \(8-4=4\), \(15-9=6\), \(24-16=8\), \(35-25=10\). This difference sequence \(2, 4, 6, 8, 10, \dots\) has \(n\)-th term \(2n\). Therefore, the \(n\)-th term of the original sequence is \(n^2 + 2n\).

M1 for finding second differences are constant and equal to 2, leading to \(n^2\) coefficient of 1. M1 for identifying the linear part as \(2n\). A0.5 for the complete correct expression \(n^2 + 2n\).

The total surface area of a cone is given by \(A = \pi r^2 + \pi r l\), where \(r\) is the radius and \(l\) is the slant height. Substituting \(r = 3\) and \(l = 5\): \(A = \pi (3)^2 + \pi (3)(5) = 9\pi + 15\pi = 24\pi\).

M1 for base area \(9\pi\). M1 for curved surface area \(15\pi\). A0.5 for correct total of \(24\pi\).

Rearrange the given line equation to gradient-intercept form: \(2y = 3x + 8 \implies y = \frac{3}{2}x + 4\). The gradient of this line is \(m_1 = \frac{3}{2}\). The gradient \(m_2\) of a perpendicular line satisfies \(m_1 \times m_2 = -1\), so \(m_2 = -\frac{2}{3}\). Using the point \((6, -1)\) in the equation \(y = mx + c\): \(-1 = -\frac{2}{3}(6) + c \implies -1 = -4 + c \implies c = 3\). Thus, the equation is \(y = -\frac{2}{3}x + 3\).

M1 for finding perpendicular gradient \(-\frac{2}{3}\). M1 for substituting point \((6, -1)\) to find \(c = 3\). A0.5 for writing the final equation \(y = -\frac{2}{3}x + 3\).

Using the Cosine Rule: \(BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos(BAC)\). Substituting the given values: \(BC^2 = 5^2 + 8^2 - 2(5)(8)\cos(60^\circ) = 25 + 64 - 80(0.5) = 89 - 40 = 49\). Therefore, \(BC = \sqrt{49} = 7\text{ cm}\).

M1 for substituting correctly into the Cosine Rule. M1 for simplifying to \(BC^2 = 49\) using \(\cos(60^\circ) = 0.5\). A0.5 for the final answer of 7.

By completing the square: \(y = (x - 3)^2 - 3^2 + 14 = (x - 3)^2 - 9 + 14 = (x - 3)^2 + 5\). The minimum point of the curve occurs when the squared term is zero, i.e., when \(x = 3\), giving \(y = 5\). Thus, the coordinates of the turning point are \((3, 5)\). Alternatively, using \(x = -\frac{b}{2a} = -\frac{-6}{2(1)} = 3\), and \(y = 3^2 - 6(3) + 14 = 9 - 18 + 14 = 5\).

M1 for attempting to complete the square or using \(x = -b/(2a)\). M1 for finding \(x = 3\) and calculating \(y\). A0.5 for writing the coordinates as \((3, 5)\).

The total number of beads is \(4 + 3 = 7\). The probability of choosing a Red then a Blue bead is: \(P(R, B) = \frac{4}{7} \times \frac{3}{6} = \frac{12}{42} = \frac{2}{7}\). The probability of choosing a Blue then a Red bead is: \(P(B, R) = \frac{3}{7} \times \frac{4}{6} = \frac{12}{42} = \frac{2}{7}\). Since these are mutually exclusive events, the total probability is \(P(R, B) + P(B, R) = \frac{2}{7} + \frac{2}{7} = \frac{4}{7}\).

M1 for calculating the probability of one combination (e.g., Red then Blue) as \(\frac{2}{7}\) or \(\frac{12}{42}\). M1 for accounting for both orders of different colours. A0.5 for the final simplified fraction \(\frac{4}{7}\).

Multiply both sides by \((x - 5)\) to clear the fraction: \(y(x - 5) = 3x + 2\). Expand the left side: \(xy - 5y = 3x + 2\). Move all terms involving \(x\) to one side and other terms to the other side: \(xy - 3x = 5y + 2\). Factor out \(x\) on the left side: \(x(y - 3) = 5y + 2\). Divide both sides by \((y - 3)\) to isolate \(x\): \(x = \frac{5y + 2}{y - 3}\).

M1 for expanding to \(xy - 5y = 3x + 2\). M1 for gathering \(x\) terms and factoring to \(x(y-3) = 5y+2\). A0.5 for the correct final formula \(x = \frac{5y + 2}{y - 3}\).

An increase of \(20\%\) corresponds to a multiplier of \(1.2\) (or \(\frac{6}{5}\)). After 1 hour, the population is \(2500 \times 1.2 = 3000\). After 2 hours, the population is \(3000 \times 1.2 = 3600\). Alternatively, the formula is \(P = 2500 \times (1.2)^2 = 2500 \times 1.44 = 3600\).

M1 for identifying the growth multiplier \(1.2\) or showing \(2500 \times 1.2\). M1 for calculating population after 2 hours (e.g., \(2500 \times 1.2^2\)). A0.5 for the correct final population \(3600\).

The terms of the sequence can be rewritten as: \(\frac{3}{2}, \frac{5}{5}, \frac{7}{10}, \frac{9}{17}, \dots\)

1. Analyze the numerators: \(3, 5, 7, 9, \dots\). This is an arithmetic sequence with first term \(a = 3\) and common difference \(d = 2\). The \(n\)-th term of the numerator is:
\(3 + (n - 1) \times 2 = 2n + 1\).

2. Analyze the denominators: \(2, 5, 10, 17, \dots\). This sequence represents squares of consecutive integers plus 1:
\(1^2 + 1, 2^2 + 1, 3^2 + 1, 4^2 + 1, \dots\). The \(n\)-th term of the denominator is:
\(n^2 + 1\).

Combining these, the \(n\)-th term of the sequence is \(\frac{2n+1}{n^2+1}\).

M1 for finding the numerator term \(2n+1\) (or equivalent clear working of numerator sequence).
M1 for finding the denominator term \(n^2+1\) (or identifying the sequence of squares + 1).
A0.5 for the fully correct fraction \(\frac{2n+1}{n^2+1}\).

1. Use the formula for the volume of a cone, \(V = \frac{1}{3}\pi r^2 h\), to find the radius \(r\):
\(12\pi = \frac{1}{3}\pi r^2 (4)\)
Multiply both sides by 3 and divide by \(\pi\):
\(36 = 4r^2\)
\(r^2 = 9 \implies r = 3\text{ cm}\).

2. Use Pythagoras' theorem to find the slant height \(l\) of the cone:
\(l = \sqrt{r^2 + h^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5\text{ cm}\).

3. Calculate the total surface area \(A\) of the cone:
\(A = \pi r^2 + \pi r l = \pi (3)^2 + \pi (3)(5) = 9\pi + 15\pi = 24\pi\text{ cm}^2\).

M1 for setting up the volume equation and correctly solving to find \(r = 3\).
M1 for calculating the slant height \(l = 5\).
A0.5 for obtaining the final total surface area of \(24\pi\).

1. Find the midpoint, \(M\), of the line segment \(AB\):
\(M = \left(\frac{1+5}{2}, \frac{-3+5}{2}\right) = (3, 1)\).

2. Find the gradient, \(m\), of \(AB\):
\(m = \frac{5 - (-3)}{5 - 1} = \frac{8}{4} = 2\).

3. The gradient of the perpendicular bisector, \(m_{\perp}\), is the negative reciprocal of \(m\):
\(m_{\perp} = -\frac{1}{2}\).

4. Use the point-slope formula with midpoint \((3, 1)\) and gradient \(-\frac{1}{2}\):
\(y - 1 = -\frac{1}{2}(x - 3)\)
\(2(y - 1) = -(x - 3)\)
\(2y - 2 = -x + 3\)
\(x + 2y - 5 = 0\).

M1 for finding the midpoint \((3, 1)\) and the gradient of line \(AB\) as \(2\).
M1 for using the perpendicular gradient relation \(m_{\perp} = -\frac{1}{2}\) to write an equation.
A0.5 for the final equation in the required form \(x + 2y - 5 = 0\) (or any integer multiple such as \(2x + 4y - 10 = 0\)).

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

Substitute the given values into the equation:
\(AC^2 = 5^2 + 6^2 - 2(5)(6)\left(\frac{1}{5}\right)\)
\(AC^2 = 25 + 36 - 12\)
\(AC^2 = 49\)

Since length must be positive, we take the positive square root:
\(AC = \sqrt{49} = 7\text{ cm}\).

M1 for correctly writing down the cosine rule or substituting the given values.
M1 for calculating \(AC^2 = 49\).
A0.5 for \(7\).

1. First, apply the translation by vector \(\begin{pmatrix} -3 \\ 4 \end{pmatrix}\). This shifts the graph of \(y = f(x)\) horizontally by \(-3\) (to the left) and vertically by \(4\) (upwards), resulting in:
\(y = f(x - (-3)) + 4 = f(x + 3) + 4\).

2. Next, apply the vertical stretch of scale factor \(2\). This multiplies the entire function representing the vertical coordinates by \(2\):
\(g(x) = 2(f(x + 3) + 4) = 2f(x + 3) + 8\).

M1 for applying the translation correctly to get \(f(x+3) + 4\).
M1 for applying the vertical stretch of scale factor 2 to the entire translated function.
A0.5 for the simplified expression \(2f(x+3) + 8\) (also accept \(2(f(x+3) + 4)\)).

The total number of counters in the box is \(4 + 6 = 10\).

To find the probability of getting at least one red counter, we can subtract the probability of drawing no red counters (i.e., drawing two blue counters) from 1.

1. Calculate the probability of choosing two blue counters without replacement:
\(P(\text{Blue, Blue}) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3}\).

2. Calculate the probability of drawing at least one red counter:
\(P(\text{at least one red}) = 1 - P(\text{Blue, Blue}) = 1 - \frac{1}{3} = \frac{2}{3}\).

M1 for finding the probability of drawing two blue counters, \(\frac{6}{10} \times \frac{5}{9}\) (or alternative correct method summation).
M1 for subtracting this value from 1, \(1 - \frac{1}{3}\).
A0.5 for the simplified fraction \(\frac{2}{3}\).

Let \(y = f(x)\):
\(y = \frac{2x + 1}{x - 3}\)

Multiply both sides by \(x - 3\):
\(y(x - 3) = 2x + 1\)
\(xy - 3y = 2x + 1\)

Rearrange to gather terms with \(x\) on one side:
\(xy - 2x = 3y + 1\)

Factor out \(x\):
\(x(y - 2) = 3y + 1\)

Divide by \(y - 2\):
\(x = \frac{3y + 1}{y - 2}\)

Therefore, the inverse function is:
\(f^{-1}(x) = \frac{3x + 1}{x - 2}\).

M1 for multiplying both sides to get \(y(x - 3) = 2x + 1\).
M1 for isolating \(x\) correctly to get \(x(y - 2) = 3y + 1\).
A0.5 for \(\frac{3x + 1}{x - 2}\) (must be written in terms of \(x\)).

Substitute \(t = 6\) and \(P = 200\) into the given formula:
\(200 = 1600 \times \left(\frac{1}{2}\right)^{6k}\)

Divide both sides by 1600:
\(\frac{200}{1600} = \left(\frac{1}{2}\right)^{6k}\)
\(\frac{1}{8} = \left(\frac{1}{2}\right)^{6k}\)

Express \(\frac{1}{8}\) as a power of \(\frac{1}{2}\):
\(\left(\frac{1}{2}\right)^3 = \left(\frac{1}{2}\right)^{6k}\)

Equate the exponents:
\(3 = 6k\)
\(k = \frac{3}{6} = \frac{1}{2}\) (or \(0.5\)).

M1 for substituting the given values into the equation to obtain \(200 = 1600 \times \left(\frac{1}{2}\right)^{6k}\).
M1 for simplifying to \(\frac{1}{8} = \left(\frac{1}{2}\right)^{6k}\) and recognizing that \(\frac{1}{8} = \left(\frac{1}{2}\right)^3\).
A0.5 for \(k = \frac{1}{2}\) (or \(0.5\)).

Part (a): For Sequence A, the common difference is \(3\), so the next term is \(17 + 3 = 20\). For Sequence B, the successive differences are \(4, 6, 8, 10\), so the next difference is \(12\), giving \(30 + 12 = 42\). Part (b)(i): Sequence A is a linear sequence with a common difference of \(3\). The first term is \(5\), so the \(n\)-th term is \(3n + 2\). Part (b)(ii): Sequence B has second differences equal to \(2\), indicating a quadratic sequence of the form \(n^2 + bn + c\). Let us find \(b\) and \(c\): for \(n=1\), \(1 + b + c = 2\); for \(n=2\), \(4 + 2b + c = 6\). Solving this system yields \(b = 1\) and \(c = 0\), so the \(n\)-th term is \(n^2 + n\). Part (b)(iii): Sequence C is geometric with first term \(3\) and common ratio \(2\), so the \(n\)-th term is \(3 \times 2^{n-1}\). Part (c): Set \(n^2 + n = 500\), which gives the quadratic equation \(n^2 + n - 500 = 0\). Using the quadratic formula, \(n = \frac{-1 \pm \sqrt{1 - 4(1)(-500)}}{2} = \frac{-1 \pm \sqrt{2001}}{2}\). This yields \(n \approx 21.87\) or \(n \approx -22.87\). Since the position \(n\) must be a positive integer, there is no term in Sequence B equal to 500.

Part (a): 2 marks (1 mark for each correct next term). Part (b)(i): 2 marks (1 mark for \(3n\), 1 mark for \(+ 2\)). Part (b)(ii): 3 marks (1 mark for setting up \(n^2\), 1 mark for the method to find coefficients, 1 mark for the final expression \(n^2 + n\)). Part (b)(iii): 2 marks (1 mark for \(2^{n-1}\), 1 mark for coefficient \(3\)). Part (c): 2 marks (1 mark for setting up the quadratic equation, 1 mark for finding a non-integer value for \(n\) and concluding).

Part (a): The radius \(r = 3\text{ cm}\). The total height is \(11\text{ cm}\), so the height of the cylinder is \(h = 11 - r = 8\text{ cm}\). Volume of the cylinder = \(\pi r^2 h = \pi (3^2)(8) = 72\pi\). Volume of the hemisphere = \(\frac{2}{3}\pi r^3 = \frac{2}{3}\pi (3^3) = 18\pi\). Total volume = \(72\pi + 18\pi = 90\pi\text{ cm}^3\). Part (b): Total surface area includes the base of the cylinder, the curved surface of the cylinder, and the curved surface of the hemisphere. Base area = \(\pi r^2 = 9\pi\). Cylinder curved area = \(2\pi r h = 2\pi (3)(8) = 48\pi\). Hemisphere curved area = \(2\pi r^2 = 18\pi\). Total surface area = \(9\pi + 48\pi + 18\pi = 75\pi \approx 235.62\text{ cm}^2\), which is \(236\text{ cm}^2\) to 3 s.f. Part (c): Mass = density \(\times\) volume = \(0.75 \times 90\pi \approx 212.06\text{ g}\), which is \(212\text{ g}\) to 3 s.f. Part (d): The scale factor of heights is \(22 / 11 = 2\). Since volume is proportional to the cube of the scale factor, the volume of the larger toy is \(2^3 \times 90\pi = 720\pi \approx 2261.95\text{ cm}^3\), which is \(2260\text{ cm}^3\) to 3 s.f.

Part (a): 3 marks (1 mark for cylinder volume \(72\pi\), 1 mark for hemisphere volume \(18\pi\), 1 mark for correct total). Part (b): 4 marks (1 mark for base area, 1 mark for cylinder curved area, 1 mark for hemisphere curved area, 1 mark for correct sum rounded to 3 s.f.). Part (c): 2 marks (1 mark for formula mass = density \(\times\) volume, 1 mark for correct mass). Part (d): 2 marks (1 mark for volume scale factor \(2^3 = 8\), 1 mark for final volume).

Part (a): Gradient \(m = \frac{-3 - 5}{4 - (-2)} = \frac{-8}{6} = -\frac{4}{3}\). Part (b): Using \(y - y_1 = m(x - x_1)\), we get \(y - 5 = -\frac{4}{3}(x + 2) \implies y = -\frac{4}{3}x - \frac{8}{3} + 5 \implies y = -\frac{4}{3}x + \frac{7}{3}\). Part (c): Midpoint \(M = \left(\frac{-2 + 4}{2}, \frac{5 - 3}{2}\right) = (1, 1)\). Part (d): The gradient of the perpendicular line is \(-\frac{1}{m} = \frac{3}{4}\). It passes through the midpoint \((1, 1)\). The equation is \(y - 1 = \frac{3}{4}(x - 1) \implies y = \frac{3}{4}x - \frac{3}{4} + 1 \implies y = \frac{3}{4}x + \frac{1}{4}\). Part (e): The perpendicular bisector crosses the x-axis when \(y = 0\). Substituting \(y = 0\) gives \(0 = \frac{3}{4}x + \frac{1}{4} \implies \frac{3}{4}x = -\frac{1}{4} \implies x = -\frac{1}{3}\). The coordinates of P are \(\left(-\frac{1}{3}, 0\right)\) or \((-0.333, 0)\).

Part (a): 2 marks (1 mark for substitution into gradient formula, 1 mark for gradient). Part (b): 2 marks (1 mark for substituting their gradient correctly, 1 mark for equation in correct form). Part (c): 2 marks (1 mark for each coordinate). Part (d): 3 marks (1 mark for perpendicular gradient \(3/4\), 1 mark for using their midpoint, 1 mark for correct equation). Part (e): 2 marks (1 mark for setting \(y = 0\), 1 mark for coordinates).

Part (a): Using the Cosine Rule: \(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(ABC) = 7.2^2 + 9.5^2 - 2(7.2)(9.5)\cos(64^\circ) = 51.84 + 90.25 - 136.8 \times 0.43837 = 142.09 - 59.97 = 82.12\). So, \(AC = \sqrt{82.12} \approx 9.06\text{ cm}\). Part (b): Using the Sine Rule: \(\frac{\sin(BAC)}{BC} = \frac{\sin(ABC)}{AC} \implies \frac{\sin(BAC)}{9.5} = \frac{\sin(64^\circ)}{9.06} \implies \sin(BAC) = \frac{9.5 \times \sin(64^\circ)}{9.06} \approx 0.9424\). Therefore, angle \(BAC = \sin^{-1}(0.9424) \approx 70.5^\circ\). Part (c): \(\text{Area} = \frac{1}{2} \times AB \times BC \times \sin(ABC) = \frac{1}{2} \times 7.2 \times 9.5 \times \sin(64^\circ) = 34.2 \times 0.89879 \approx 30.7\text{ cm}^2\). Part (d): In right-angled triangle ABD, \(\sin(64^\circ) = \frac{AD}{AB} \implies AD = 7.2 \times \sin(64^\circ) \approx 6.47\text{ cm}\).

Part (a): 3 marks (1 mark for correct Cosine Rule formula, 1 mark for correct substitution, 1 mark for final value of \(9.06\)). Part (b): 3 marks (1 mark for correct Sine Rule formula, 1 mark for correct substitution, 1 mark for final angle \(70.5^\circ\)). Part (c): 2 marks (1 mark for area formula, 1 mark for final value \(30.7\)). Part (d): 3 marks (1 mark for setting up trigonometric ratio in ABD, 1 mark for substitution, 1 mark for final value \(6.47\)).

Part (a): Set \(f(x) = 0 \implies x^3 - 3x^2 - 4x + 12 = 0\). Grouping terms: \(x^2(x - 3) - 4(x - 3) = 0 \implies (x^2 - 4)(x - 3) = 0 \implies (x - 2)(x + 2)(x - 3) = 0\). The x-intercepts are \(x = -2\), \(x = 2\), and \(x = 3\). Part (b): Differentiate: \(f'(x) = 3x^2 - 6x - 4\). Set \(f'(x) = 0 \implies 3x^2 - 6x - 4 = 0\). Using the quadratic formula, \(x = \frac{6 \pm \sqrt{36 - 4(3)(-4)}}{6} = \frac{6 \pm \sqrt{84}}{6}\). This gives \(x \approx 2.53\) (local minimum) and \(x \approx -0.528\) (local maximum). Substituting these back into \(f(x)\): \(f(-0.528) \approx 13.1\) and \(f(2.53) \approx -1.13\). The local maximum is \((-0.528, 13.1)\) and the local minimum is \((2.53, -1.13)\). Part (c): Set \(x^3 - 3x^2 - 4x + 12 = 5 \implies x^3 - 3x^2 - 4x + 7 = 0\). Solving this equation using a graphic display calculator yields three roots: \(x \approx -1.63\), \(x \approx 1.25\), and \(x \approx 3.38\).

Part (a): 3 marks (1 mark for factorization method, 2 marks for finding all three correct intercepts). Part (b): 4 marks (1 mark for differentiating, 1 mark for finding both x-values, 2 marks for finding the corresponding y-values and pairing correctly). Part (c): 4 marks (1 mark for writing the equation in standard form, 1 mark for each of the three correct roots).

Part (a): Total balls = \(6 + 4 + 2 = 12\). The probability that the first ball is red is \(\frac{6}{12} = \frac{1}{2}\). Part (b): Since there are only 2 green balls, we cannot select 3 green balls. Thus, the three balls can only be all red or all blue. \(P(\text{all same}) = P(RRR) + P(BBB) = \left(\frac{6}{12} \times \frac{5}{11} \times \frac{4}{10}\right) + \left(\frac{4}{12} \times \frac{3}{11} \times \frac{2}{10}\right) = \frac{120}{1320} + \frac{24}{1320} = \frac{144}{1320} = \frac{6}{55} \approx 0.109\). Part (c): Use the complement: \(P(\text{at least 1 green}) = 1 - P(\text{no green})\). The number of non-green balls is \(6 + 4 = 10\). \(P(\text{no green}) = \frac{10}{12} \times \frac{9}{11} \times \frac{8}{10} = \frac{720}{1320} = \frac{6}{11}\). Thus, \(P(\text{at least 1 green}) = 1 - \frac{6}{11} = \frac{5}{11} \approx 0.455\). Part (d): The possible arrangements of exactly two blue balls are \((B, B, B')\), \((B, B', B)\), and \((B', B, B)\). The probability for each arrangement is identical: \(\frac{4}{12} \times \frac{3}{11} \times \frac{8}{10} = \frac{96}{1320}\). Total probability is \(3 \times \frac{96}{1320} = \frac{288}{1320} = \frac{12}{55} \approx 0.218\).

Part (a): 1 mark. Part (b): 3 marks (1 mark for \(P(RRR)\), 1 mark for \(P(BBB)\), 1 mark for final sum \(6/55\)). Part (c): 3 marks (1 mark for recognizing complementary event method, 1 mark for calculating \(P(\text{no green}) = 6/11\), 1 mark for final answer \(5/11\)). Part (d): 4 marks (1 mark for calculating a single order probability, 1 mark for multiplying by 3, 2 marks for simplifying to \(12/55\)).

Part (a): The cumulative frequencies are: \(m \le 120\) is 14; \(m \le 140\) is \(14 + 28 = 42\); \(m \le 160\) is \(42 + 42 = 84\); \(m \le 180\) is \(84 + 24 = 108\); \(m \le 200\) is \(108 + 12 = 120\). Part (b)(i): The median is the 60th value, which lies in the \(140 < m \le 160\) class. Cumulative frequency up to 140 is 42. We need \(60 - 42 = 18\) more values from the 42 in this class of width 20. Estimated median = \(140 + \frac{18}{42} \times 20 \approx 148.57\text{ g}\), which is \(149\text{ g}\) to 3 s.f. Part (b)(ii): The lower quartile (LQ) is the 30th value, in the \(120 < m \le 140\) class. \(\text{LQ} = 120 + \frac{30 - 14}{28} \times 20 = 120 + 11.43 = 131.43\text{ g}\). The upper quartile (UQ) is the 90th value, in the \(160 < m \le 180\) class. \(\text{UQ} = 160 + \frac{90 - 84}{24} \times 20 = 160 + 5 = 165\text{ g}\). Therefore, \(\text{IQR} = \text{UQ} - \text{LQ} = 165 - 131.43 = 33.57\text{ g}\), which is \(33.6\text{ g}\) to 3 s.f. Part (c): The class \(160 < m \le 180\) has a frequency of 24. The proportion of this class above 175 g is \(\frac{180 - 175}{20} = \frac{5}{20} = 0.25\). Number of apples in this range is \(0.25 \times 24 = 6\). Adding the 12 apples in the \(180 < m \le 200\) class, the total estimated number of Grade A apples is \(6 + 12 = 18\).

Part (a): 2 marks (1 mark for first 3 values, 1 mark for last 2 values). Part (b)(i): 3 marks (1 mark for identifying correct class, 1 mark for correct interpolation calculation, 1 mark for correct value 149). Part (b)(ii): 4 marks (1 mark for LQ calculation, 1 mark for UQ calculation, 1 mark for subtracting UQ - LQ, 1 mark for correct IQR 33.6). Part (c): 2 marks (1 mark for interpolating 160-180 class to find 6 apples, 1 mark for correct total 18).

Part (a): Savings = \(5000 \times (1 + 0.032)^5 = 5000 \times 1.032^5 \approx \$5852.86 \approx \$5853\). Part (b): Car value = \(18000 \times (1 - 0.12)^5 = 18000 \times 0.88^5 \approx \$9499.17 \approx \$9499\). Part (c): Let \(t\) be the number of years. We want to find the smallest integer \(t\) such that \(18000 \times 0.88^t < 5000 \times 1.032^t \implies \frac{18000}{5000} < \left(\frac{1.032}{0.88}\right)^t \implies 3.6 < (1.172727...)^t\). Taking logs: \(\ln(3.6) < t \ln(1.172727...) \implies t > \frac{\ln(3.6)}{\ln(1.172727...)} \approx 8.04\text{ years}\). Thus, it takes 9 complete years. Part (d): Monthly option after 5 years = \(5000 \times \left(1 + \frac{0.0315}{12}\right)^{60} = 5000 \times (1.002625)^{60} \approx \$5851.46\). The original option yields \(\$5852.86\), which is more money. The difference is \(5852.86 - 5851.46 = \$1.40\).

Part (a): 2 marks (1 mark for correct formula, 1 mark for correct value \(\$5853\)). Part (b): 2 marks (1 mark for correct formula, 1 mark for correct value \(\$9499\)). Part (c): 4 marks (1 mark for setting up inequality, 1 mark for simplifying base and exponent, 1 mark for using logarithms, 1 mark for rounding up to 9 years). Part (d): 3 marks (1 mark for calculating monthly compounding total \(\$5851.46\), 1 mark for identifying original option as better, 1 mark for correct difference \(\$1.40\)).

**(a)(i)**
Find the next terms by continuing the pattern of differences:
Differences: 7, 11, 15, 19, 23, 27, 31...
6th term: \(57 + 23 = 80\)
7th term: \(80 + 27 = 107\)
8th term: \(107 + 31 = 138\).

**(a)(ii)**
First differences: 7, 11, 15, 19
Second differences: 4, 4, 4
Since the second differences are constant and equal to 4, the sequence is quadratic with a leading term of \(\frac{4}{2}n^2 = 2n^2\).
Subtracting \(2n^2\) from each term:
- \(n=1\): \(5 - 2(1)^2 = 3\)
- \(n=2\): \(12 - 2(2)^2 = 4\)
- \(n=3\): \(23 - 2(3)^2 = 5\)
The sequence of remainders is \(3, 4, 5, \dots\), which is a linear sequence with formula \(n + 2\).
Therefore, the \(n\)-th term of Sequence \(A\) is \(2n^2 + n + 2\).

**(b)(i)**
Sequence \(B\) is a geometric sequence with first term \(a = 3\) and common ratio \(r = 2\).
Therefore, the \(n\)-th term is \(3 \times 2^{n-1}\).

**(b)(ii)**
For the 12th term, substitute \(n = 12\):
\(3 \times 2^{12-1} = 3 \times 2^{11} = 3 \times 2048 = 6144\).

**(c)**
We want to find the smallest \(n\) such that \(3 \times 2^{n-1} > 2n^2 + n + 2\).
Let us test values of \(n\):
- For \(n = 5\):
\(3 \times 2^4 = 48\)
\(2(5)^2 + 5 + 2 = 57\)
\(48 < 57\)
- For \(n = 6\):
\(3 \times 2^5 = 96\)
\(2(6)^2 + 6 + 2 = 80\)
\(96 > 80\)
Thus, the smallest value is \(n = 6\).

**(a)(i)**
[1] for listing or finding 6th and 7th terms (80 and 107).
[1] for final answer 138.

**(a)(ii)**
[1] for establishing second difference of 4 or showing \(2n^2\) is part of the formula.
[1] for finding the linear remainder sequence \(n+2\).
[1] for correct final formula \(2n^2 + n + 2\).

**(b)(i)**
[1] for identifying common ratio \(r = 2\).
[1] for correct final formula \(3 \times 2^{n-1}\) (or equivalent).

**(b)(ii)**
[1] for setting up calculation \(3 \times 2^{11}\).
[1] for final answer 6144.

**(c)**
[1] for showing comparisons for \(n=5\) and \(n=6\), or writing inequality \(3 \times 2^{n-1} > 2n^2 + n + 2\).
[1] for concluding \(n = 6\).

**(a)**
Volume of the cylinder:
\(V_{\text{cyl}} = \pi r^2 H = \pi \times 3.5^2 \times 9 = 110.25\pi \approx 346.36\text{ m}^3\)

Volume of the cone:
\(V_{\text{cone}} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \times 3.5^2 \times 3 = 12.25\pi \approx 38.48\text{ m}^3\)

Total Volume:
\(V = 110.25\pi + 12.25\pi = 122.5\pi \approx 384.85\text{ m}^3\)

To 3 significant figures, the volume is \(385\text{ m}^3\).

**(b)**
First, find the slant height \(l\) of the cone:
\(l = \sqrt{r^2 + h^2} = \sqrt{3.5^2 + 3^2} = \sqrt{12.25 + 9} = \sqrt{21.25} \approx 4.6098\text{ m}\)

Curved surface area of the cylinder:
\(A_{\text{cyl}} = 2 \pi r H = 2 \pi \times 3.5 \times 9 = 63\pi \approx 197.92\text{ m}^2\)

Curved surface area of the cone:
\(A_{\text{cone}} = \pi r l = \pi \times 3.5 \times \sqrt{21.25} \approx 50.69\text{ m}^2\)

Total Surface Area:
\(A = 63\pi + 3.5\sqrt{21.25}\pi \approx 197.92 + 50.69 = 248.61\text{ m}^2\)

To 3 significant figures, the total surface area is \(249\text{ m}^2\).

**(c)**
Number of tins needed:
\(\frac{248.61}{15} \approx 16.57\)

Since paint must be purchased in complete tins, we round up to the next integer:
17 tins.

**(a)**
[1] for cylinder volume calculation: \(\pi \times 3.5^2 \times 9\).
[1] for cone volume calculation: \(\frac{1}{3} \pi \times 3.5^2 \times 3\).
[1] for summing the two volumes.
[1] for \(385\) (accept answers in range \([384.8, 385.0]\)).

**(b)**
[2] for calculating slant height \(l = \sqrt{21.25} \approx 4.61\) (M1 for using Pythagoras, A1 for correct value).
[1] for curved surface area of cylinder: \(2 \pi \times 3.5 \times 9 = 63\pi \approx 198\).
[1] for curved surface area of cone: \(\pi \times 3.5 \times 4.61 \approx 50.7\).
[1] for final sum \(249\) (accept answers in range \([248.5, 249.0]\)).

**(c)**
[1] for dividing their calculated area by 15.
[1] for rounding up to the nearest integer (17 tins, or FT based on their part b).

**(a)**
Probability that the first ball is red is \(\frac{6}{12} = \frac{1}{2}\).
Probability that the second ball is red given the first was red is \(\frac{5}{11}\).
\(P(R_1 \cap R_2) = \frac{6}{12} \times \frac{5}{11} = \frac{30}{132} = \frac{5}{22} \approx 0.227\).

**(b)**
Same colour means both red, both blue, or both green:
\(P(R_1 \cap R_2) = \frac{6}{12} \times \frac{5}{11} = \frac{30}{132}\)
\(P(B_1 \cap B_2) = \frac{4}{12} \times \frac{3}{11} = \frac{12}{132}\)
\(P(G_1 \cap G_2) = \frac{2}{12} \times \frac{1}{11} = \frac{2}{132}\)

\(P(\text{same colour}) = \frac{30 + 12 + 2}{132} = \frac{44}{132} = \frac{1}{3} \approx 0.333\).

**(c)**
\(P(\text{at least one green}) = 1 - P(\text{no green})\).
The number of non-green balls is \(6 + 4 = 10\).
\(P(\text{no green}) = \frac{10}{12} \times \frac{9}{11} = \frac{90}{132} = \frac{15}{22}\).

\(P(\text{at least one green}) = 1 - \frac{15}{22} = \frac{7}{22} \approx 0.318\).

**(d)**
We want to find the conditional probability \(P(R_1 | B_2) = \frac{P(R_1 \cap B_2)}{P(B_2)}\).

First, find \(P(R_1 \cap B_2)\):
\(P(R_1 \cap B_2) = \frac{6}{12} \times \frac{4}{11} = \frac{24}{132}\).

Next, find \(P(B_2)\):
\(P(B_2) = P(R_1 \cap B_2) + P(B_1 \cap B_2) + P(G_1 \cap B_2)\)
\(P(B_2) = \frac{24}{132} + \frac{12}{132} + \left(\frac{2}{12} \times \frac{4}{11}\right)\)
\(P(B_2) = \frac{24}{132} + \frac{12}{132} + \frac{8}{132} = \frac{44}{132} = \frac{1}{3}\).

Now calculate the conditional probability:
\(P(R_1 | B_2) = \frac{24/132}{44/132} = \frac{24}{44} = \frac{6}{11} \approx 0.545\).

**(a)**
[1] for \(\frac{6}{12} \times \frac{5}{11}\) seen.
[1] for \(\frac{5}{22}\) or any equivalent fraction or decimal (e.g. \(0.227\) or \(22.7\%\)).

**(b)**
[1] for calculating individual same-colour probabilities (e.g., \(12/132\) and \(2/132\)).
[1] for adding three probabilities: \(\frac{30}{132} + \frac{12}{132} + \frac{2}{132}\).
[1] for \(\frac{1}{3}\) or equivalent decimal \(0.333\).

**(c)**
[1] for calculating \(P(\text{no green}) = \frac{10}{12} \times \frac{9}{11}\) or list of cases: \((G, \text{not } G)\), \((\text{not } G, G)\), \((G, G)\).
[1] for subtraction from 1 or correct summing expression: \(\frac{20}{132} + \frac{20}{132} + \frac{2}{132}\).
[1] for \(\frac{7}{22}\) or equivalent decimal \(0.318\).

**(d)**
[1] for finding \(P(R_1 \cap B_2) = \frac{24}{132}\).
[1] for finding \(P(B_2) = \frac{44}{132}\) (or \(\frac{1}{3}\)).
[1] for division of the two values giving \(\frac{6}{11}\) (or \(0.545\)).

### PART A: INVESTIGATION

**(a)**
* The number of hexagons in Ring 4 is \(18\).
* The total number of hexagons in Design 4 is \(1 + 6 + 12 + 18 = 37\).

**(b)**
For \(n \ge 2\), the number of hexagons in Ring \(n\) is \(R_n = 6(n - 1)\).

**(c)**
The total number of hexagons \(T_n\) is the sum of the hexagons in all rings from 1 to \(n\):
\[T_n = R_1 + R_2 + R_3 + \dots + R_n\]
\[T_n = 1 + 6(1) + 6(2) + \dots + 6(n - 1)\]
\[T_n = 1 + 6[1 + 2 + \dots + (n - 1)]\]
Using the formula for the sum of the first \(n-1\) integers, \(\sum_{i=1}^{n-1} i = \frac{(n-1)n}{2}\):
\[T_n = 1 + 6 \cdot \frac{n(n-1)}{2}\]
\[T_n = 1 + 3n(n-1)\]
\[T_n = 3n^2 - 3n + 1\]

**(d)(i)**
* In Design 1, the single hexagon has \(P_1 = 6\) cm, which fits \(12(1) - 6 = 6\).
* For \(n \ge 2\), the boundary of Design \(n\) is formed by the outer ring (Ring \(n\)), which has \(6(n-1)\) hexagons.
* There are 6 corner hexagons in this outer ring. Each corner hexagon shares 1 edge with Ring \(n-1\) and 2 edges with its adjacent neighbors in Ring \(n\). This leaves \(6 - 3 = 3\) exposed edges per corner hexagon. Total exposed edges for corners = \(6 \times 3 = 18\).
* There are \(6(n-1) - 6 = 6(n-2)\) non-corner (side) hexagons in the outer ring. Each side hexagon shares 2 edges with Ring \(n-1\) and 2 edges with adjacent neighbors in Ring \(n\). This leaves \(6 - 4 = 2\) exposed edges per side hexagon. Total exposed edges for side hexagons = \(6(n-2) \times 2 = 12n - 24\).
* Total perimeter \(P_n = 18 + 12n - 24 = 12n - 6\).
* Since this formula gives \(12(1) - 6 = 6\) for \(n = 1\), it holds for all \(n \ge 1\).

**(d)(ii)**
Set \(P_n = 594\):
\[12n - 6 = 594\]
\[12n = 600\]
\[n = 50\]
So, **Design 50** has a perimeter of 594 cm.

---

### PART B: MODELLING

**(a)(i)**
Using \(T(t) = a - bt\):
* At \(t = 0\), \(180 = a - b(0) \implies a = 180\).
* At \(t = 5\), \(100 = 180 - b(5) \implies 5b = 80 \implies b = 16\).
Thus, \(a = 180\) and \(b = 16\).

**(a)(ii)**
Set \(T(t) = 20\):
\[180 - 16t = 20 \implies 16t = 160 \implies t = 10\text{ minutes}\]
For \(t > 10\), the model predicts that the temperature continues to decrease below \(20^\circ\text{C}\) (and eventually becomes negative), which is impossible because a cooling object cannot become colder than the room temperature. Thus, Model 1 is not valid for large values of \(t\).

**(b)(i)**
Using \(T(t) = 20 + c \cdot d^t\):
* At \(t = 0\), \(180 = 20 + c \cdot d^0 \implies 180 = 20 + c \implies c = 160\) (Shown).
* At \(t = 5\), \(100 = 20 + 160 \cdot d^5 \implies 160 \cdot d^5 = 80 \implies d^5 = 0.5\).
Thus, the exact value of \(d\) is:
\[d = 0.5^{1/5} \quad \text{or} \quad d = 2^{-1/5}\]

**(b)(ii)**
For \(t = 12\):
\[T(12) = 20 + 160 \cdot (0.5^{1/5})^{12} = 20 + 160 \cdot 0.5^{2.4}\]
\[T(12) = 20 + 160 \cdot 0.18946457... = 20 + 30.3143... \approx 50.3^\circ\text{C}\]

**(b)(iii)**
As \(t \to \infty\), \(d^t \to 0\) since \(0 < d < 1\).
Therefore, \(T(t) \to 20 + 0 = 20^\circ\text{C}\).
This is more realistic because it shows the temperature of the heating element asymptotically approaching the surrounding room temperature, rather than dropping below it indefinitely.

**(c)**
For equivalence between Model 2 and Model 3:
\[20 + 160 \cdot e^{-kt} = 20 + 160 \cdot d^t \implies e^{-kt} = d^t \implies (e^{-k})^t = d^t\]
This requires:
\[e^{-k} = d \implies -k = \ln(d) \implies k = -\ln(d)\]
Substitute \(d = 0.5^{1/5}\):
\[k = -\ln(0.5^{1/5}) = -\frac{1}{5}\ln(0.5) = \frac{\ln(2)}{5} \approx 0.138629...\]
Correct to 4 decimal places, \(k = 0.1386\).

### PART A: INVESTIGATION (15 Marks)

**(a) [2 Marks]**
* **1 Mark**: Correct number of hexagons in Ring 4 (18).
* **1 Mark**: Correct total number of hexagons in Design 4 (37).

**(b) [2 Marks]**
* **2 Marks**: \(R_n = 6(n - 1)\) or equivalent (e.g., \(6n - 6\)).
* Deduct 1 mark if condition \(n \ge 2\) is ignored but formula is otherwise correct.

**(c) [4 Marks]**
* **1 Mark**: Attempt to sum the terms, e.g., \(1 + 6 + 12 + \dots\) or writing \(1 + 6\sum(i-1)\).
* **1 Mark**: Use of sum of AP formula or identity \(1+2+\dots+n-1 = \frac{n(n-1)}{2}\).
* **1 Mark**: Correct algebraic expansion and simplification steps.
* **1 Mark**: Final correct expression \(3n^2 - 3n + 1\).

**(d)(i) [5 Marks]**
* **1 Mark**: Explicitly showing the base case \(P_1 = 6\) fits the formula.
* **1 Mark**: Identifying that the outer ring has \(6(n-1)\) hexagons.
* **1 Mark**: Identifying 6 corner hexagons, each contributing 3 exposed edges (total 18).
* **1 Mark**: Identifying \(6(n-2)\) non-corner hexagons, each contributing 2 exposed edges (total \(12n-24\)).
* **1 Mark**: Combining terms to show \(18 + 12n - 24 = 12n - 6\) clearly and logically.

**(d)(ii) [2 Marks]**
* **1 Mark**: Setting up the equation \(12n - 6 = 594\).
* **1 Mark**: Finding \(n = 50\).

---

### PART B: MODELLING (15 Marks)

**(a)(i) [2 Marks]**
* **1 Mark**: Correctly finding \(a = 180\).
* **1 Mark**: Correctly finding \(b = 16\).

**(a)(ii) [3 Marks]**
* **1 Mark**: Solving \(180 - 16t = 20\) to get \(t = 10\).
* **1 Mark**: Explaining that for \(t > 10\), the temperature falls below room temperature (or becomes negative).
* **1 Mark**: Concluding that this makes the model unrealistic/invalid for long-term predictions.

**(b)(i) [3 Marks]**
* **1 Mark**: Substituting \(t = 0, T = 180\) to show \(c = 160\).
* **1 Mark**: Setting up \(100 = 20 + 160 \cdot d^5\) and simplifying to \(d^5 = 0.5\).
* **1 Mark**: Expressing \(d\) exactly as \(0.5^{1/5}\), \(2^{-1/5}\), or \(\sqrt[5]{0.5}\).

**(b)(ii) [2 Marks]**
* **1 Mark**: Substituting \(t = 12\) into the model: \(20 + 160 \cdot (0.5)^{12/5}\).
* **1 Mark**: Finding the correct value \(50.3\) (accept answers rounding to 50.3).

**(b)(iii) [2 Marks]**
* **1 Mark**: Stating that as \(t \to \infty\), \(T(t) \to 20\).
* **1 Mark**: Explaining that this is realistic because the temperature of the element stabilizes at room temperature.

**(c) [3 Marks]**
* **1 Mark**: Equating the bases or functions: \(e^{-kt} = d^t\) or \(e^{-k} = d\).
* **1 Mark**: Showing \(k = -\ln(d)\) or \(k = \frac{\ln(2)}{5}\).
* **1 Mark**: Finding the correct rounded decimal value \(k = 0.1386\).

Part A:
1. (a) Since volume \(V = \pi r^2 h = 1000\), we rearrange to get: \(h = \frac{1000}{\pi r^2}\).
(b) The total surface area of a closed cylinder is \(A = 2\pi r^2 + 2\pi r h\). Substituting the expression for \(h\): \(A = 2\pi r^2 + 2\pi r \left(\frac{1000}{\pi r^2}\right) = 2\pi r^2 + \frac{2000}{r}\).
(c) To find the minimum surface area, differentiate with respect to \(r\): \(\frac{dA}{dr} = 4\pi r - \frac{2000}{r^2}\). Set \(\frac{dA}{dr} = 0\): \(4\pi r^3 = 2000 \implies r^3 = \frac{500}{\pi} \approx 159.155 \implies r = \sqrt[3]{\frac{500}{\pi}} \approx 5.42\text{ cm}\) (to 3 s.f.). Substitute \(r = 5.419\) back into \(A\): \(A \approx 2\pi (5.419)^2 + \frac{2000}{5.419} \approx 184.5 + 369.1 = 554\text{ cm}^2\).

2. (a) For a square prism, volume \(V = x^2 y = 1000 \implies y = \frac{1000}{x^2}\). The surface area is \(S = 2x^2 + 4xy\). Substituting \(y\): \(S = 2x^2 + 4x\left(\frac{1000}{x^2}\right) = 2x^2 + \frac{4000}{x}\).
(b) Differentiating \(S\): \(\frac{dS}{dx} = 4x - \frac{4000}{x^2}\). Set to 0: \(4x^3 = 4000 \implies x^3 = 1000 \implies x = 10\text{ cm}\). The minimum surface area is \(S = 2(10)^2 + \frac{4000}{10} = 200 + 400 = 600\text{ cm}^2\).

Part B:
3. A regular hexagon is made of 6 equilateral triangles of side length \(s\). The area of one equilateral triangle is \(\frac{1}{2} s^2 \sin(60^\circ) = \frac{\sqrt{3}}{4}s^2\). Thus, total area of the hexagon is \(6 \times \frac{\sqrt{3}}{4}s^2 = \frac{3\sqrt{3}}{2}s^2\).

4. Since \(V = \text{Base Area} \times H = 1000\), we have \(\frac{3\sqrt{3}}{2}s^2 H = 1000 \implies H = \frac{2000}{3\sqrt{3}s^2}\). The total surface area is \(A_{\text{hex}} = 2 \times \text{Base Area} + 6sH = 3\sqrt{3}s^2 + 6s\left(\frac{2000}{3\sqrt{3}s^2}\right)\). Simplifying: \(A_{\text{hex}} = 3\sqrt{3}s^2 + \frac{4000}{\sqrt{3}s} = 3\sqrt{3}s^2 + \frac{4000\sqrt{3}}{3s}\).

5. Differentiating \(A_{\text{hex}}\): \(\frac{dA_{\text{hex}}}{ds} = 6\sqrt{3}s - \frac{4000\sqrt{3}}{3s^2}\). Set to 0: \(6\sqrt{3}s = \frac{4000\sqrt{3}}{3s^2} \implies 18s^3 = 4000 \implies s^3 = \frac{2000}{9} \approx 222.22 \implies s = \sqrt[3]{\frac{2000}{9}} \approx 6.06\text{ cm}\) (to 3 s.f.).

6. Substituting \(s \approx 6.057\) into \(A_{\text{hex}}\): \(A_{\text{hex}} = 3\sqrt{3}(6.057)^2 + \frac{4000\sqrt{3}}{3(6.057)} \approx 190.6 + 381.3 = 572\text{ cm}^2\).

7. Comparing the minimum areas: Cylinder \(\approx 554\text{ cm}^2\), Hexagonal Prism \(\approx 572\text{ cm}^2\), Square Prism \(= 600\text{ cm}^2\). The cylinder is the most material-efficient. As the number of sides of the base polygon increases (from 4 for a square to 6 for a hexagon, and infinitely many for a circle), the minimum surface area needed to enclose a fixed volume decreases.

Part A (15 marks):
1. (a) [2 marks] M1 for volume formula \(\pi r^2 h = 1000\), A1 for correct rearrangement \(h = \frac{1000}{\pi r^2}\).
(b) [3 marks] M1 for total surface area formula \(A = 2\pi r^2 + 2\pi r h\), M1 for substitution of \(h\), A1 for clear algebraic reduction to the given formula.
(c) [4 marks] M1 for correct derivative \(4\pi r - \frac{2000}{r^2}\), M1 for setting to 0 and solving for \(r\), A1 for \(r \approx 5.42\text{ cm}\), A1 for \(A \approx 554\text{ cm}^2\).
2. (a) [3 marks] M1 for writing \(y = \frac{1000}{x^2}\), M1 for total surface area formula \(S = 2x^2 + 4xy\), A1 for showing the final expression.
(b) [3 marks] M1 for differentiating and setting to 0 (\(4x - \frac{4000}{x^2} = 0\)), A1 for \(x = 10\text{ cm}\), A1 for \(S = 600\text{ cm}^2\).

Part B (15 marks):
3. [3 marks] M1 for finding the area of one equilateral triangle of side \(s\) (\(\frac{1}{2}s^2\sin(60^\circ)\) or equivalent), M1 for multiplying by 6, A1 for establishing the given result.
4. [4 marks] M1 for volume expression \(\frac{3\sqrt{3}}{2}s^2 H = 1000\), M1 for expressing \(H\) in terms of \(s\), M1 for setting up total surface area \(A_{\text{hex}} = 2\text{Base} + 6sH\), A1 for showing the final expression.
5. [3 marks] M1 for differentiating and setting to 0 (\(6\sqrt{3}s - \frac{4000\sqrt{3}}{3s^2} = 0\)), M1 for simplifying to \(18s^3 = 4000\), A1 for \(s \approx 6.06\text{ cm}\).
6. [2 marks] M1 for substituting their value of \(s\) into the formula, A1 for \(A_{\text{hex}} \approx 572\text{ cm}^2\).
7. [3 marks] C1 for stating the correct order of surface areas (Cylinder < Hexagon < Square), C1 for concluding the cylinder is the most material-efficient, C1 for concluding that more sides leads to greater efficiency (approaching a circular cross-section).