2025 IB DP Mathematics - Applications and Interpretation 模擬試題連答案詳解

(a) Using Pythagoras' theorem on the base: \(AC = \sqrt{10^2 + 10^2} = \sqrt{200} \approx 14.1\text{ cm}\). (b) Let \(O\) be the center of the base. The length \(AO = \frac{AC}{2} = \frac{\sqrt{200}}{2} = \sqrt{50} \approx 7.07\text{ cm}\). In the right-angled triangle \(AOV\), \(AV^2 = AO^2 + VO^2 = 50 + 12^2 = 194\). Therefore, \(AV = \sqrt{194} \approx 13.9\text{ cm}\). (c) The angle \(\theta\) between \(AV\) and the base is angle \(VAO\). \(\tan(\theta) = \frac{VO}{AO} = \frac{12}{\sqrt{50}}\). Thus, \(\theta = \arctan\left(\frac{12}{\sqrt{50}}\right) \approx 59.5^\circ\) (or \(1.04\text{ radians}\)).

(M1) for attempting Pythagoras to find AC. (A1) for \(14.1\text{ cm}\). (M1) for finding half the diagonal \(AO = 7.07\text{ cm}\). (A1) for finding \(AV = 13.9\text{ cm}\). (M1) for using a correct trigonometric ratio to find the angle. (A1) for \(59.5^\circ\). (A0.33) for correct units throughout.

(a) The angle \(QPR\) is the difference between the two bearings: \(\angle QPR = 130^\circ - 60^\circ = 70^\circ\). Using the Cosine Rule to find \(QR\): \(QR^2 = 45^2 + 70^2 - 2(45)(70)\cos(70^\circ) = 2025 + 4900 - 6300\cos(70^\circ) \approx 4770.27\). Taking the square root gives \(QR \approx 69.1\text{ km}\). (b) Using the Sine Rule to find angle \(PQR\): \(\frac{\sin(\angle PQR)}{70} = \frac{\sin(70^\circ)}{69.067}\), which gives \(\sin(\angle PQR) \approx 0.9523\). Thus, \(\angle PQR \approx 72.2^\circ\). The bearing from \(Q\) to \(P\) is the back-bearing of \(060^\circ\), which is \(60^\circ + 180^\circ = 240^\circ\). The bearing of \(R\) from \(Q\) is therefore \(240^\circ - 72.2^\circ = 167.8^\circ \approx 168^\circ\).

(M1) for finding the angle \(\angle QPR = 70^\circ\). (M1) for applying the Cosine Rule. (A1) for \(QR = 69.1\text{ km}\). (M1) for applying the Sine Rule to find angle \(PQR\). (A1) for \(\angle PQR \approx 72.2^\circ\). (M1) for finding the back-bearing or setting up the bearing calculation. (A1.33) for the final bearing \(168^\circ\).

(a) The dimensions of the base of the box will be \(24 - 2x\) and \(15 - 2x\), and the height will be \(x\). The volume is \(V(x) = x(24 - 2x)(15 - 2x) = x(360 - 48x - 30x + 4x^2) = 4x^3 - 78x^2 + 360x\). (b) Differentiating \(V(x)\) with respect to \(x\) gives \(\frac{\text{d}V}{\text{d}x} = 12x^2 - 156x + 360\). (c) Setting \(\frac{\text{d}V}{\text{d}x} = 0\) for maximum volume: \(12x^2 - 156x + 360 = 0 \Rightarrow x^2 - 13x + 30 = 0 \Rightarrow (x-10)(x-3) = 0\). Since \(x\) must be less than \(7.5\) (half of \(15\)), we choose \(x = 3\text{ cm}\). The maximum volume is \(V(3) = 4(3)^3 - 78(3)^2 + 360(3) = 108 - 702 + 1080 = 486\text{ cm}^3\).

(M1) for expressing volume as \(x(24-2x)(15-2x)\). (A1) for showing the steps leading to the given cubic expression. (A1) for \(\frac{\text{d}V}{\text{d}x} = 12x^2 - 156x + 360\). (M1) for setting their derivative to zero. (A1) for finding \(x = 3\) (and rejecting \(x = 10\)). (M1) for substituting \(x = 3\) back into the volume formula. (A1.33) for finding the maximum volume of \(486\text{ cm}^3\).

(a) Using the trapezoidal rule with interval width \(h = 2\): \(\text{Volume} \approx \frac{2}{2} [ R(0) + 2R(2) + 2R(4) + 2R(6) + R(8) ] = 1 [ 12 + 2(18) + 2(25) + 2(21) + 15 ] = 12 + 36 + 50 + 42 + 15 = 155\text{ m}^3\). (b) The total volume of water released over 8 hours is \(17 \times 8 = 136\text{ m}^3\). The net change is the water in minus the water out: \(\text{Net Change} \approx 155 - 136 = 19\text{ m}^3\) (which represents an increase).

(M1) for identifying interval width \(h=2\). (M2) for substituting correctly into the trapezoidal rule formula. (A1.33) for \(155\text{ m}^3\). (M1) for calculating total water released \(17 \times 8 = 136\text{ m}^3\). (M1) for subtracting the total water released from the total water entered. (A1) for finding the correct net increase of \(19\text{ m}^3\).

(a) We use the financial solver on a GDC with the following parameters: \(N = 60\) (since \(5 \times 12 = 60\)), \(I\% = 6.5\), \(PV = 28000\), \(FV = 0\), \(P/Y = 12\), \(C/Y = 12\). Solving for \(PMT\) yields \(PMT \approx -547.78\). Thus, the monthly payment is \(\$547.78\). (b) The total amount paid over the 5 years is \(60 \times 547.78 = \$32\,866.80\). The total interest paid is the difference between total payments and the original loan amount: \(\text{Total Interest} = 32\,866.80 - 28\,000 = \$4\,866.80\).

(M2) for correctly identifying GDC parameters (such as \(N = 60\), \(I\% = 6.5\), \(PV = 28000\)). (A1.33) for finding the correct monthly payment of \(\$547.78\). (M2) for calculating total payments \(60 \times 547.78\). (A2) for finding the total interest paid of \(\$4\,866.80\) (accept answers close due to rounding).

(a) Let \(X\) be the mass of an apple, where \(X \sim N(150, 18^2)\). Using a GDC: \(P(130 < X < 165) = \text{normalcdf}(130, 165, 150, 18) \approx 0.663\). (b) We need to find \(k\) such that \(P(X \le k) = 0.10\). Using the inverse normal function on a GDC: \(k = \text{invNorm}(0.10, 150, 18) \approx 127\text{ g}\). (c) The probability that an apple is suitable for local sales is \(1 - 0.10 = 0.90\). The expected number of apples is \(250 \times 0.90 = 225\).

(M1) for writing down the correct normal distribution setup. (A1.33) for finding the probability \(0.663\). (M1) for setting up the inverse normal equation \(P(X \le k) = 0.10\). (A1) for finding \(127\text{ g}\). (M1) for identifying the probability of being suitable is \(0.90\). (M1) for multiplying sample size by probability. (A1) for finding 225 apples.

(a) \(H_0\): Preferred music genre is independent of age group. (b) Total 'Under 30' = \(45 + 25 + 10 = 80\). Total 'Rock' = \(25 + 45 = 70\). Grand Total = 200. Expected frequency = \(\frac{80 \times 70}{200} = 28\). (c) Entering the observed matrix into the GDC and performing a \(\chi^2\) two-way test yields a \(p\)-value of \(\approx 0.000145\) (or \(1.45 \times 10^{-4}\)). (d) Since \(p\text{-value} = 0.000145 < 0.05\), we reject the null hypothesis. There is significant evidence to suggest that preferred music genre is associated with age group.

(A1) for a correctly written null hypothesis. (M1) for calculating row/column/grand totals. (A1) for expected frequency \(28\). (A2) for \(p\)-value \(\approx 0.000145\). (R1) for comparing their \(p\)-value to \(0.05\). (A1.33) for rejecting \(H_0\) with a consistent conclusion.

(a) As \(t \to \infty\), \(T(t) \to a\), which represents the room temperature. Thus, \(a = 20\). (b) At \(t = 0\), the initial temperature is \(85^\circ\text{C}\). Therefore, \(T(0) = 20 + b e^0 = 85 \Rightarrow 20 + b = 85 \Rightarrow b = 65\). (c) Since \(T(5) = 60\), we have \(20 + 65 e^{-5k} = 60 \Rightarrow 65 e^{-5k} = 40 \Rightarrow e^{-5k} = \frac{40}{65} \Rightarrow -5k = \ln\left(\frac{8}{13}\right) \Rightarrow k \approx 0.0971\). (d) To find the time to reach \(40^\circ\text{C}\): \(20 + 65 e^{-0.0971 t} = 40 \Rightarrow 65 e^{-0.0971 t} = 20 \Rightarrow e^{-0.0971 t} = \frac{4}{13} \Rightarrow t = \frac{\ln(4/13)}{-0.0971} \approx 12.1\text{ minutes}\).

(A1) for \(a = 20\). (M1) for substituting \(t = 0\) and \(T(0) = 85\) into the model. (A1) for showing \(b = 65\). (M1) for setting up the equation for \(T(5) = 60\). (A1) for \(k \approx 0.0971\). (M1) for setting up the equation for \(T(t) = 40\). (A1.33) for \(12.1\text{ minutes}\).

(a) The distance from \(A(0,0,0)\) to \(B(120, 80, 50)\) is given by: \(AB = \sqrt{(120-0)^2 + (80-0)^2 + (50-0)^2} = \sqrt{120^2 + 80^2 + 50^2} = \sqrt{14400 + 6400 + 2500} = \sqrt{23300} \approx 152.64\text{ m}\). To 3 significant figures, this is \(153\text{ m}\). (b) The horizontal distance on the \(xy\)-plane from \(A\) to \(B\) is: \(d_{xy} = \sqrt{120^2 + 80^2} = \sqrt{14400 + 6400} = \sqrt{20800} \approx 144.22\text{ m}\). The vertical height is \(z = 50\text{ m}\). The angle of elevation \(\theta\) is: \(\theta = \arctan\left(\frac{50}{144.22}\right) \approx 19.1^\circ\) (or \(0.334\text{ rad}\)). (c) The distance from \(B(120, 80, 50)\) to \(C(200, -30, 110)\) is: \(BC = \sqrt{(200-120)^2 + (-30-80)^2 + (110-50)^2} = \sqrt{80^2 + (-110)^2 + 60^2} = \sqrt{6400 + 12100 + 3600} = \sqrt{22100} \approx 148.66\text{ m}\). Total distance is \(AB + BC \approx 152.64 + 148.66 = 301.30\text{ m}\). To 3 significant figures, this is \(301\text{ m}\).

(a) M1 for applying the 3D distance formula, A1 for \(153\text{ m}\) (or \(152.64\dots\)). (b) M1 for finding horizontal distance \(\approx 144\text{ m}\), M1 for using tangent ratio, A1 for \(19.1^\circ\) (or \(0.334\text{ rad}\)). (c) M1 for calculating \(BC \approx 149\text{ m}\), A1.33 for final answer \(301\text{ m}\) (accept \(301.30\)).

(a) When squares of side \(x\) are cut from each corner of the 30 cm square, the remaining base of the box is a square with side length \(30 - 2x\). The height of the box is \(x\). Therefore, the volume \(V\) is: \(V(x) = x(30 - 2x)^2 = x(900 - 120x + 4x^2) = 4x^3 - 120x^2 + 900x\). (b) Differentiating \(V(x)\) with respect to \(x\): \(\frac{dV}{dx} = 12x^2 - 240x + 900\). (c) Setting the derivative to zero for maximum volume: \(12x^2 - 240x + 900 = 0 \implies 12(x^2 - 20x + 75) = 0 \implies 12(x - 5)(x - 15) = 0\). Since \(x\) must be less than 15 (otherwise the sides would meet or exceed the length), we have \(x = 5\text{ cm}\). The maximum volume is: \(V(5) = 4(5)^3 - 120(5)^2 + 900(5) = 500 - 3000 + 4500 = 2000\text{ cm}^3\).

(a) M1 for establishing volume expression \(x(30-2x)^2\), A1 for correct expansion to show the given equation. (b) M1 for power rule application, A1 for \(12x^2 - 240x + 900\). (c) M1 for setting their derivative to zero, A1 for choosing the correct root \(x = 5\), A1.33 for the maximum volume of \(2000\text{ cm}^3\).

(a) Using the compound interest formula \(A = P\left(1 + \frac{r}{100k}\right)^{kn}\), where \(P = 12000\), \(r = 4.5\), \(k = 12\), and \(n = 5\): \(A = 12000\left(1 + \frac{4.5}{1200}\right)^{12 \times 5} = 12000(1.00375)^{60} \approx 15021.55\text{ USD}\). (b) Interest earned is: \(15021.55 - 12000 = 3021.55\text{ USD}\). (c) We solve for the number of years \(n\): \(12000(1.00375)^{12n} > 20000 \implies (1.00375)^{12n} > \frac{5}{3}\). Taking the natural logarithm on both sides: \(12n \ln(1.00375) > \ln(5/3) \implies 12n > \frac{\ln(1.6667)}{\ln(1.00375)} \approx 136.48\text{ months}\). Dividing by 12: \(n > 11.37\text{ years}\). Thus, the minimum number of complete years required is \(12\) years.

(a) M1 for compound interest formula with correct substitution, A1 for \(\$15,021.55\). (b) A1 for \(\$3,021.55\). (c) M1 for establishing the inequality or equation, A1 for obtaining \(n \approx 11.37\) years, A2.33 for rounding up to \(12\) complete years.

(a) The null hypothesis \(H_0\): A student's preferred sport is independent of their grade level. (b) The total number of Grade 12 students is 50. The total number of students preferring Tennis across all grades is \(10 + 8 + 17 = 35\). The grand total is 150. Expected value: \(E = \frac{50 \times 35}{150} = \frac{35}{3} \approx 11.7\) (or 11.67). (c) Entering the observed matrix into a graphic display calculator: Row 1: \([25, 15, 10]\), Row 2: \([20, 22, 8]\), Row 3: \([15, 18, 17]\). Performing a \(\chi^2\) two-way test yields a test statistic of \(\chi^2 \approx 7.673\) with \(df = 4\), giving a \(p\)-value of \(0.104\). (d) Since \(p\)-value \(= 0.104 > 0.05\), we do not reject the null hypothesis. There is insufficient evidence to suggest that preferred sport and grade level are associated.

(a) A1 for stating independence correctly in context. (b) M1 for formula \((\text{row total} \times \text{col total}) / \text{grand total}\), A1 for \(11.7\) (or \(11.67\)). (c) A2 for \(0.104\) (accept \(0.103\) to \(0.105\)). (d) R1 for comparing \(p\)-value with \(0.05\), A1.33 for correctly stating we fail to reject the null hypothesis.

(a) Using \(T(0) = 85\): \(22 + A e^{0} = 85 \implies 22 + A = 85 \implies A = 63\). (b) Using \(T(10) = 50\): \(22 + 63 e^{-10k} = 50 \implies 63 e^{-10k} = 28 \implies e^{-10k} = \frac{28}{63} = \frac{4}{9}\). Taking natural logarithm: \(-10k = \ln(4/9) \implies k = -0.1 \ln(4/9) \approx 0.0811\). (c) Using \(t = 25\): \(T(25) = 22 + 63 e^{-25 \times 0.081093} = 22 + 63(e^{-10k})^{2.5} = 22 + 63\left(\frac{4}{9}\right)^{2.5} = 22 + 63\left(\frac{32}{243}\right) \approx 30.3^\circ\text{C}\).

(a) M1 for setting \(t=0\) and \(T=85\), A1 for \(A=63\). (b) M1 for substituting \(A=63\), \(t=10\), and \(T=50\), A1 for \(k \approx 0.0811\) (or exact \(-0.1\ln(4/9)\)). (c) M1 for substituting \(t=25\) into formula, A2.33 for \(30.3^\circ\text{C}\) (accept \(30.3\)).

(a) Midpoint of \(P_1 P_2\) is \(M_{12} = \left(\frac{2+8}{2}, \frac{8+10}{2}\right) = (5, 9)\). Gradient of \(P_1 P_2\) is \(m_{12} = \frac{10-8}{8-2} = \frac{1}{3}\). The perpendicular gradient is \(m_{\perp} = -3\). Equation: \(y - 9 = -3(x - 5) \implies y = -3x + 24\). (b) Midpoint of \(P_1 P_3\) is \(M_{13} = \left(\frac{2+6}{2}, \frac{8+2}{2}\right) = (4, 5)\). Gradient of \(P_1 P_3\) is \(m_{13} = \frac{2-8}{6-2} = -1.5 = -\frac{3}{2}\). The perpendicular gradient is \(m_{\perp} = \frac{2}{3}\). Equation: \(y - 5 = \frac{2}{3}(x - 4) \implies y = \frac{2}{3}x + \frac{7}{3}\). (c) Set the two equations equal: \(-3x + 24 = \frac{2}{3}x + \frac{7}{3} \implies -9x + 72 = 2x + 7 \implies 11x = 65 \implies x = \frac{65}{11} \approx 5.91\). Then \(y = -3\left(\frac{65}{11}\right) + 24 = \frac{69}{11} \approx 6.27\). The coordinates of the vertex are \(\left(\frac{65}{11}, \frac{69}{11}\right) \approx (5.91, 6.27)\).

(a) M1 for finding midpoint and negative reciprocal gradient, A1 for \(y = -3x + 24\) (or equivalent). (b) M1 for finding midpoint and negative reciprocal gradient, A1 for \(y = \frac{2}{3}x + \frac{7}{3}\) (or equivalent). (c) M1 for equating both linear equations to find intersection, A2.33 for \((5.91, 6.27)\) (accept exact values).

(a) The particle is at rest when \(v(t) = 0\): \(3t^2 - 12t + 9 = 0 \implies 3(t^2 - 4t + 3) = 0 \implies 3(t-1)(t-3) = 0\). Thus, \(t = 1\text{ s}\) and \(t = 3\text{ s}\). (b) Acceleration \(a(t)\) is the derivative of velocity: \(a(t) = \frac{dv}{dt} = 6t - 12\). At \(t = 3\): \(a(3) = 6(3) - 12 = 6\text{ m s}^{-2}\). (c) The total distance traveled is given by \(\int_{0}^{4} |v(t)| dt\). We split the integral at the stationary points: \(d = \int_{0}^{1} (3t^2-12t+9) dt + \int_{1}^{3} -(3t^2-12t+9) dt + \int_{3}^{4} (3t^2-12t+9) dt\). Let \(s(t) = t^3 - 6t^2 + 9t\). Then \(s(0) = 0\), \(s(1) = 4\), \(s(3) = 0\), and \(s(4) = 4\). Distance \(= |4 - 0| + |0 - 4| + |4 - 0| = 4 + 4 + 4 = 12\text{ m}\).

(a) M1 for setting \(v(t) = 0\), A1 for both \(t = 1\) and \(t = 3\). (b) M1 for differentiation to find \(a(t) = 6t - 12\), A1 for \(6\text{ m s}^{-2}\). (c) M1 for setting up the integral of absolute value, A2.33 for \(12\text{ m}\) (accept GDC solution directly).

(a) Find the midpoint of \(AB\):
\(M_{AB} = \left(\frac{0+6}{2}, \frac{0+8}{2}\right) = (3, 4)\)
The gradient of \(AB\) is:
\(m_{AB} = \frac{8 - 0}{6 - 0} = \frac{4}{3}\)
The perpendicular gradient is:
\(m_{\perp} = -\frac{3}{4} = -0.75\)
The equation of the perpendicular bisector is:
\(y - 4 = -0.75(x - 3) \implies y = -0.75x + 6.25\) (or \(3x + 4y = 25\))

(b) Find the midpoint of \(BC\):
\(M_{BC} = \left(\frac{6+14}{2}, \frac{8+2}{2}\right) = (10, 5)\)
The gradient of \(BC\) is:
\(m_{BC} = \frac{2 - 8}{14 - 6} = \frac{-6}{8} = -\frac{3}{4}\)
The perpendicular gradient is:
\(m_{\perp} = \frac{4}{3}\)
The equation of the perpendicular bisector is:
\(y - 5 = \frac{4}{3}(x - 10) \implies y = \frac{4}{3}x - \frac{25}{3}\) (or \(4x - 3y = 25\))

(c) Solve the system of two equations to find the intersection point \(P\):
\(3x + 4y = 25\)
\(4x - 3y = 25\)
Multiplying the first by 3 and the second by 4:
\(9x + 12y = 75\)
\(16x - 12y = 100\)
Adding them:
\(25x = 175 \implies x = 7\)
Substituting back:
\(3(7) + 4y = 25 \implies 21 + 4y = 25 \implies 4y = 4 \implies y = 1\)
So, the coordinates of \(P\) are \((7, 1)\).

(d) Distance from \(P(7, 1)\) to \(A(0, 0)\):
\(d = \sqrt{(7-0)^2 + (1-0)^2} = \sqrt{49 + 1} = \sqrt{50} \approx 7.07\text{ dam}\).

(e) Distance from \(P(7, 1)\) to \(B(6, 8)\):
\(PB = \sqrt{(7-6)^2 + (1-8)^2} = \sqrt{1 + 49} = \sqrt{50} \approx 7.07\text{ dam}\).
Distance from \(P(7, 1)\) to \(D(12, 10)\):
\(PD = \sqrt{(12-7)^2 + (10-1)^2} = \sqrt{5^2 + 9^2} = \sqrt{25 + 81} = \sqrt{106} \approx 10.3\text{ dam}\).
Since \(PB = \sqrt{50} < PD = \sqrt{106}\), the restroom \(P\) is closer to shelter \(B\) than to shelter \(D\).

(a)
- \(M1\) for finding midpoint \((3, 4)\).
- \(M1\) for finding gradient of \(AB\) as \(\frac{4}{3}\).
- \(M1\) for perpendicular gradient of \(-\frac{3}{4}\).
- \(A1\) for the correct equation in any linear form (e.g. \(3x + 4y = 25\)).

(b)
- \(M1\) for finding midpoint \((10, 5)\).
- \(M1\) for finding gradient of \(BC\) as \(-\frac{3}{4}\).
- \(M1\) for perpendicular gradient of \(\frac{4}{3}\).
- \(A1\) for the correct equation in any linear form (e.g. \(4x - 3y = 25\)).

(c)
- \(M1\) for setting up a valid system of equations.
- \(A1\) for \(x = 7\).
- \(A1\) for \(y = 1\).

(d)
- \(M1\) for using distance formula.
- \(A1\) for \(7.07\text{ dam}\) (accept \(\sqrt{50}\)).

(e)
- \(A1\) for stating distance to \(B\) is \(\sqrt{50}\) or \(7.07\).
- \(M1\) for calculating distance to \(D\) as \(\sqrt{106}\) or \(10.3\).
- \(R1\) for comparing values and concluding that \(P\) is closer to \(B\).

(a) The volume of a cylinder is given by:
\(V = \pi r^2 h = 150\pi \implies h = \frac{150}{r^2}\)
The total surface area of a closed cylinder is:
\(A = 2\pi r^2 + 2\pi r h\)
Substitute \(h\):
\(A(r) = 2\pi r^2 + 2\pi r \left(\frac{150}{r^2}\right) = 2\pi r^2 + \frac{300\pi}{r}\).

(b) Find the derivative with respect to \(r\):
\(A'(r) = \frac{\text{d}}{\text{d}r}\left(2\pi r^2 + 300\pi r^{-1}\right) = 4\pi r - 300\pi r^{-2} = 4\pi r - \frac{300\pi}{r^2}\).

(c) Set \(A'(r) = 0\) to find the minimum:
\(4\pi r - \frac{300\pi}{r^2} = 0 \implies 4\pi r = \frac{300\pi}{r^2}\)
\(4r^3 = 300 \implies r^3 = 75\)
\(r = \sqrt[3]{75} \approx 4.21716 \approx 4.22\text{ m}\).

(d) Calculate \(A(4.21716)\):
\(A = 2\pi (4.21716)^2 + \frac{300\pi}{4.21716} \approx 111.733 + 223.466 = 335.199 \approx 335\text{ m}^2\).

(e) The height is given by:
\(h = \frac{150}{r^2}\)
Since at the minimum \(r^3 = 75 \implies r^2 = \frac{75}{r}\):
\(h = \frac{150}{\frac{75}{r}} = \frac{150r}{75} = 2r\)
Since the diameter of the cylinder is \(d = 2r\), we have shown that \(h = d\).

(f) Accept any reasonable practical reason. For example: sheet metal is typically manufactured in rectangular sizes, so cutting circular bases would cause significant material wastage; or space/height constraints on-site where the tank is to be installed limit the maximum diameter or height.

(a)
- \(M1\) for expressing \(h\) in terms of \(r\) using the volume formula.
- \(M1\) for writing down the correct surface area formula.
- \(A1\) for substituting and showing the algebraic steps clearly to arrive at the given formula.

(b)
- \(M1\) for attempting to differentiate \(r^2\) or \(r^{-1}\).
- \(A1\) for \(4\pi r\).
- \(A1\) for \(-\frac{300\pi}{r^2}\).

(c)
- \(M1\) for setting \(A'(r) = 0\).
- \(A1\) for \(r^3 = 75\).
- \(A1\) for \(r \approx 4.22\text{ m}\).

(d)
- \(M1\) for substituting their \(r\) back into \(A(r)\).
- \(A1\) for \(335\text{ m}^2\).

(e)
- \(M1\) for using \(h = \frac{150}{r^2}\).
- \(M1\) for substituting \(r = 75^{1/3}\) or relating \(r^3 = 75\).
- \(A1\) for completing the proof to show \(h = 2r\).

(f)
- \(R1\) for naming a realistic physical limitation (e.g., standard material sizes, shipping/transport limits).
- \(R1\) for linking this clearly to why the chosen radius would differ.

(a) For Plan A, \(P = x\), \(r = 0.045\), \(k = 12\), \(t = 5\). Total periods \(n = 60\).
Value is:
\(V_A = x \left(1 + \frac{0.045}{12}\right)^{60} = x (1.00375)^{60}\).

(b) For Plan B, \(P = y\), \(r = 0.068\), \(k = 4\), \(t = 5\). Total periods \(n = 20\).
Value is:
\(V_B = y \left(1 + \frac{0.068}{4}\right)^{20} = y (1.017)^{20}\).

(c) The total value after 5 years is:
\(x (1.00375)^{60} + y (1.017)^{20} = 67\,810\)
Substitute \(y = 50\,000 - x\):
\(x (1.251796) + (50\,000 - x)(1.400949) = 67\,810\)
\(1.251796 x + 70\,047.47 - 1.400949 x = 67\,810\)
\(-0.149153 x = 67\,810 - 70\,047.47 = -2237.47\)
\(x = \frac{-2237.47}{-0.149153} \approx 15\,001.14\)
To the nearest dollar, Mia invested \(\$15\,001\) (accept \(\$15\,000\) if using unrounded steps/financial solver).

(d) (i) Using TVM solver or amortization formulas:
\(PV = 15\,000\)
\(I\% = 5.2\)
\(PMT = -350\)
\(FV = 0\)
\(P/Y = 12\), \(C/Y = 12\)
Solving for \(N\):
\(N \approx 47.51\) payments.
Since payments are monthly, it will take 48 months to completely pay off the loan.
(Accept either 47.5 or 48 months depending on interpretation of final payment period).

(ii) Total payments made:
\(47.5115 \times 350 = \$16\,629.03\)
Total interest paid:
\(16\,629.03 - 15\,000 = \$1\,629.03\) (or \(\$1\,628.50\) if using precise amortization schedules).

(a)
- \(M1\) for substituting compound interest formula values.
- \(A1\) for \(x (1.00375)^{60}\) or equivalent.

(b)
- \(M1\) for substituting compound interest formula values.
- \(A1\) for \(y (1.017)^{20}\) or equivalent.

(c)
- \(M1\) for setting up the equation \(V_A + V_B = 67\,810\).
- \(M1\) for substituting \(y = 50\,000 - x\).
- \(M1\) for expanding and gathering terms.
- \(A1\) for obtaining a linear equation in \(x\).
- \(A1\) for \(\$15\,001\) (or \(\$15\,000\)).

(d) (i)
- \(M1\) for identifying correct financial variables (\(PV=15\,000\), \(I=5.2\), \(PMT=-350\)).
- \(M1\) for attempting to solve using TVM solver or GP formula.
- \(A1\) for \(47.5\).
- \(A1\) for rounding to \(48\) months for the final payment.

(ii)
- \(M1\) for calculating total payments.
- \(M1\) for subtracting initial loan principal.
- \(A1\) for \(\$1\,629\) (accept values between \(\$1\,628\) and \(\$1\,630\)).

(a) Let \(X \sim N(50, 1.5^2)\).
(i) \(P(X < 48.5) = P\left(Z < \frac{48.5 - 50}{1.5}\right) = P(Z < -1) \approx 0.1587 \approx 0.159\).
(ii) \(P(49 < X < 52) = P\left(\frac{49 - 50}{1.5} < Z < \frac{52 - 50}{1.5}\right) = P(-0.6667 < Z < 1.3333) \approx 0.656\).

(b) Let \(Y\) be the number of bars weighing less than \(48.5\) grams.
\(Y \sim B(6, p)\) where \(p = 0.158655\).
We want \(P(Y \ge 2) = 1 - P(Y \le 1)\):
\(P(Y = 0) = (1 - p)^6 = (0.841345)^6 \approx 0.3547\)
\(P(Y = 1) = 6 \times (0.158655) \times (0.841345)^5 \approx 0.4012\)
\(P(Y \ge 2) = 1 - (0.3547 + 0.4012) = 0.2441 \approx 0.244\).

(c) (i) \(H_0: \mu = 50\)
\(H_1: \mu < 50\)

(ii) The test statistic is:
\(z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} = \frac{49.2 - 50}{1.5 / \sqrt{20}} = \frac{-0.8}{0.33541} \approx -2.385\)
The \(p\)-value for a one-tailed test is:
\(p\text{-value} = P(Z < -2.385) \approx 0.00853\).

(iii) Since the \(p\)-value \((0.00853) < 0.05\), we reject the null hypothesis \(H_0\).
There is sufficient evidence at the \(5\%\) significance level to conclude that the mean weight of the chocolate bars is less than \(50\) grams.

(a) (i)
- \(M1\) for attempting normal CDF.
- \(A1\) for \(0.159\).
(ii)
- \(M1\) for normal boundaries.
- \(A1\) for \(0.656\).

(b)
- \(M1\) for recognizing Binomial distribution.
- \(M1\) for identifying correct parameters \(n=6, p=0.159\).
- \(M1\) for attempting to calculate \(1 - P(Y \le 1)\).
- \(A1\) for \(0.244\).

(c) (i)
- \(A1\) for correct \(H_0\).
- \(A1\) for correct \(H_1\).
(ii)
- \(M1\) for substituting into standard error formula \(\frac{\sigma}{\sqrt{n}}\).
- \(M1\) for finding \(z\)-score.
- \(A1\) for \(p\text{-value} \approx 0.00853\).
(iii)
- \(R1\) for comparing \(p\)-value with \(0.05\).
- \(A1\) for a correct contextual conclusion.

(a) (i) The vertical shift \(d\) represents the average value of the maximum and minimum temperatures:
\(d = \frac{16 + (-4)}{2} = 6\).
(ii) The amplitude is \(10\). Since the minimum occurs at \(t=c\) and a negative cosine starts at its minimum value:
\(a = -10\).
(Alternatively: \(T(4) = a\cos(0) + 6 = -4 \implies a = -10\)).

(b) Since the period is 24 hours:
\(b = \frac{2\pi}{24} = \frac{\pi}{12} \approx 0.262\).

(c) The minimum temperature occurs at 04:00, which corresponds to the start of the negative cosine wave (where phase shift is zero for the argument). Hence, \(c = 4\).

(d) Substituting \(t = 10\):
\(T(10) = -10 \cos\left(\frac{\pi}{12}(10 - 4)\right) + 6 = -10 \cos\left(\frac{\pi}{2}\right) + 6 = -10(0) + 6 = 6^\circ\text{C}\).

(e) (i) Find the times when \(T(t) = 5\):
\(-10 \cos\left(\frac{\pi}{12}(t - 4)\right) + 6 = 5\)
\(\cos\left(\frac{\pi}{12}(t - 4)\right) = 0.1\)
Let \(\theta = \frac{\pi}{12}(t-4)\).
\(\theta = \arccos(0.1) \approx 1.4706\) or \(2\pi - 1.4706 \approx 4.8126\)
Solving for \(t\):
\(t_1 - 4 = 1.4706 \times \frac{12}{\pi} \approx 5.617 \implies t_1 \approx 9.617\) (which is 09:37)
\(t_2 - 4 = 4.8126 \times \frac{12}{\pi} \approx 18.383 \implies t_2 \approx 22.383\) (which is 22:23)
So, the rink is closed between 09:37 (or 9.62) and 22:23 (or 22.4).

(ii) Total hours closed:
\(22.383 - 9.617 = 12.766 \approx 12.8\) hours.

(a) (i)
- \(M1\) for attempting to find the mean value.
- \(A1\) for \(6\).
(ii)
- \(M1\) for using amplitude definition or equation substitution.
- \(A1\) for showing \(a = -10\).

(b)
- \(M1\) for using \(\frac{2\pi}{\text{Period}}\).
- \(A1\) for \(\frac{\pi}{12}\) (or \(0.262\)).

(c)
- \(A1\) for \(c = 4\).

(d)
- \(M1\) for substituting \(t = 10\).
- \(A1\) for \(6^\circ\text{C}\).

(e) (i)
- \(M1\) for setting \(T(t) = 5\).
- \(M1\) for finding first angle/time \(t_1 \approx 9.62\).
- \(M1\) for finding second angle/time \(t_2 \approx 22.4\).
- \(A1\) for both times correct (accept decimals or HH:MM format).
(ii)
- \(M1\) for subtracting the two times.
- \(A1\) for \(12.8\) hours (or 12 hours 46 minutes).

(a) First, find the interior angle \(\angle PAB\).
Looking at the parallel north-south lines at \(P\) and \(A\):
- The line from \(A\) back to \(P\) makes an angle of \(60^\circ\) with the South direction.
- The bearing of \(B\) from \(A\) is \(140^\circ\), which means the angle of \(AB\) with the North direction is \(140^\circ\), and with the South direction is \(180^\circ - 140^\circ = 40^\circ\).
Therefore, \(\angle PAB = 60^\circ + 40^\circ = 100^\circ\).
Using the Cosine Rule in \(\triangle PAB\):
\(PB^2 = PA^2 + AB^2 - 2(PA)(AB)\cos(\angle PAB)\)
\(PB^2 = 12^2 + 8^2 - 2(12)(8)\cos(100^\circ)\)
\(PB^2 = 144 + 64 - 192(-0.173648) = 208 + 33.34 = 241.34\)
\(PB = \sqrt{241.34} \approx 15.535 \approx 15.5\text{ km}\).

(b) To find the bearing, we first calculate \(\angle APB\) using the Sine Rule:
\(\frac{\sin(\angle APB)}{AB} = \frac{\sin(100^\circ)}{PB}\)
\(\sin(\angle APB) = \frac{8 \sin(100^\circ)}{15.535} \approx 0.5071\)
\(\angle APB = \arcsin(0.5071) \approx 30.47^\circ\).
Since the bearing of \(A\) from \(P\) is \(060^\circ\), the bearing of \(B\) from \(P\) is:
\(060^\circ + 30.47^\circ = 090.47^\circ \approx 090.5^\circ\) (or \(90.5^\circ\)).

(c) Time taken for return journey:
\(t = \frac{\text{distance}}{\text{speed}} = \frac{15.535\text{ km}}{6\text{ km h}^{-1}} \approx 2.589\text{ hours}\).
Converting the fractional part to minutes:
\(0.589 \times 60 = 35.34\text{ minutes}\).
To the nearest minute, this is \(2\text{ hours } 35\text{ minutes}\).

(d) Area of the triangle:
\(\text{Area} = \frac{1}{2} \times PA \times AB \times \sin(\angle PAB)\)
\(\text{Area} = \frac{1}{2} \times 12 \times 8 \times \sin(100^\circ) = 48 \sin(100^\circ) \approx 47.27 \approx 47.3\text{ km}^2\).

(a)
- \(M1\) for finding \(\angle PAB = 100^\circ\).
- \(M1\) for choosing the Cosine Rule.
- \(M1\) for substituting values correctly into the Cosine Rule.
- \(A1\) for \(PB^2 \approx 241.34\).
- \(A1\) for \(PB = 15.5\text{ km}\).

(b)
- \(M1\) for choosing the Sine Rule to find \(\angle APB\).
- \(A1\) for \(\angle APB \approx 30.5^\circ\).
- \(M1\) for adding their angle to \(60^\circ\).
- \(A1\) for bearing \(090.5^\circ\) (or \(90.5^\circ\)).

(c)
- \(M1\) for \(t = \frac{15.5}{6}\).
- \(M1\) for attempting to convert fractional hours to minutes.
- \(A1\) for \(2\text{ hours } 35\text{ minutes}\).

(d)
- \(M1\) for choosing the area formula \(\frac{1}{2}ab\sin C\).
- \(M1\) for substitution.
- \(A1\) for \(47.3\text{ km}^2\).

(a) Separate variables:
\(\frac{1}{T - 20} \text{d}T = -k \text{d}t\)
Integrate both sides:
\(\int \frac{1}{T - 20} \text{d}T = \int -k \text{d}t\)
\(\ln|T - 20| = -kt + C\)
\(T - 20 = A\text{e}^{-kt}\) (where \(A = \pm \text{e}^C\))
At \(t = 0\), \(T = 85\):
\(85 - 20 = A \implies A = 65\).
Thus, \(T(t) = 20 + 65\text{e}^{-kt}\).

(b) At \(t = 5\), \(T = 60\):
\(60 = 20 + 65\text{e}^{-5k}\)
\(40 = 65\text{e}^{-5k} \implies \text{e}^{-5k} = \frac{40}{65} = \frac{8}{13}\)
\(-5k = \ln\left(\frac{8}{13}\right) \implies k = -\frac{1}{5}\ln\left(\frac{8}{13}\right) = \frac{1}{5}\ln\left(\frac{13}{8}\right) \approx 0.0971\).

(c) At \(t = 15\):
\(T(15) = 20 + 65\text{e}^{-15(0.09708)} = 20 + 65\left(\frac{8}{13}\right)^3 \approx 35.15 \approx 35.1^\circ\text{C}\).

(d) (i)
For \(n=1\):
\(\left(\frac{\text{d}T}{\text{d}t}\right)_1 = -0.097(72.39 - 20) = -5.08183 \approx -5.082\)
\(T_2 = 72.39 + 2(-5.08183) = 62.226 \approx 62.23\)
For \(n=2\):
\(\left(\frac{\text{d}T}{\text{d}t}\right)_2 = -0.097(62.226 - 20) = -4.0959 \approx -4.096\)
\(T_3 = 62.226 + 2(-4.0959) = 54.034 \approx 54.03\)
Completed table values:
- At \(t=2\), derivative \(\approx -5.082\)
- At \(t=4\), \(T_2 \approx 62.23\), derivative \(\approx -4.096\)
- At \(t=6\), \(T_3 \approx 54.03\)

(ii) At \(t = 6\), the actual temperature with \(k = 0.097\) is:
\(T(6) = 20 + 65\text{e}^{-0.582} \approx 56.32^\circ\text{C\).
Since the approximation is \(54.03\), it is an underestimate.
This is because the second derivative is positive:
\(\frac{\text{d}^2T}{\text{d}t^2} = -k \frac{\text{d}T}{\text{d}t} = k^2(T-20) > 0\), indicating the curve is concave up. Tangent approximations lie below a concave up curve.

(a)
- \(M1\) for separation of variables.
- \(A1\) for correct integration of both sides with constant.
- \(M1\) for translating log equation to exponential form.
- \(M1\) for using initial condition \(T(0) = 85\).
- \(A1\) for showing the final expression.

(b)
- \(M1\) for setting up equation \(60 = 20 + 65\text{e}^{-5k}\).
- \(M1\) for isolating \(\text{e}^{-5k}\).
- \(A1\) for \(k = \frac{1}{5}\ln\left(\frac{13}{8}\right)\).

(c)
- \(M1\) for substituting \(t = 15\).
- \(A1\) for \(35.1^\circ\text{C}\).

(d) (i)
- \(A1\) for derivative at \(t=2\) as \(-5.082\).
- \(A1\) for \(T_2 \approx 62.23\).
- \(A1\) for derivative at \(t=4\) as \(-4.096\).
- \(A1\) for \(T_3 \approx 54.03\).
(ii)
- \(R1\) for indicating it is an underestimate.
- \(R1\) for justification based on the second derivative being positive (concave up).

**(a)**
Rate of change of mass in Pond A:
\(\frac{dx}{dt} = \text{rate in} - \text{rate out}\)
\(\text{rate in} = 20 \times 0.3 \text{ (external)} + 5 \times \frac{y}{100} \text{ (from B)}\)
\(\text{rate out} = 25 \times \frac{x}{166.7} \text{ (to B)} = 25 \times \frac{3}{500} x = 0.15x\)
Thus, \(\frac{dx}{dt} = 6 + 0.05y - 0.15x\) (as required).

Rate of change of mass in Pond B:
\(\frac{dy}{dt} = \text{rate in} - \text{rate out}\)
\(\text{rate in} = 25 \times \frac{x}{166.7} = 0.15x\)
\(\text{rate out} = 5 \times \frac{y}{100} \text{ (to A)} + 20 \times \frac{y}{100} \text{ (to stream)} = 0.25y\)
Thus, \(\frac{dy}{dt} = 0.15x - 0.25y\) (as required).

To find the equilibrium state, we set \(\frac{dx}{dt} = 0\) and \(\frac{dy}{dt} = 0\):
\(0.15x - 0.25y = 0 \implies x = \frac{5}{3} y\)
Substitute into the first equation:
\(-0.15(\frac{5}{3}y) + 0.05y + 6 = 0\)
\(-0.25y + 0.05y + 6 = 0 \implies 0.20y = 6 \implies y_e = 30\text{ kg}\)
Thus, \(x_e = \frac{5}{3}(30) = 50\text{ kg}\).

**(b)**
Using Euler's formula: \(x_{n+1} = x_n + h f(x_n, y_n)\) and \(y_{n+1} = y_n + h g(x_n, y_n)\) with \(h = 2\).

Step 1: \(t = 0\), \(x_0 = 0\), \(y_0 = 0\)
\(f(0,0) = -0.15(0) + 0.05(0) + 6 = 6\)
\(g(0,0) = 0.15(0) - 0.25(0) = 0\)
\(x(2) \approx 0 + 2(6) = 12\)
\(y(2) \approx 0 + 2(0) = 0\)

Step 2: \(t = 2\), \(x_1 = 12\), \(y_1 = 0\)
\(f(12,0) = -0.15(12) + 0.05(0) + 6 = -1.8 + 6 = 4.2\)
\(g(12,0) = 0.15(12) - 0.25(0) = 1.8\)
\(x(4) \approx 12 + 2(4.2) = 20.4\)
\(y(4) \approx 0 + 2(1.8) = 3.6\)

**(c)**
(i) Since \(X = \begin{pmatrix} x - 50 \\ y - 30 \end{pmatrix}\), we have \(\frac{dX}{dt} = \begin{pmatrix} \frac{dx}{dt} \\ \frac{dy}{dt} \end{pmatrix}\).
Substitute \(x = X_1 + 50\) and \(y = X_2 + 30\):
\(\frac{dx}{dt} = -0.15(X_1 + 50) + 0.05(X_2 + 30) + 6 = -0.15X_1 + 0.05X_2 - 7.5 + 1.5 + 6 = -0.15X_1 + 0.05X_2\)
\(\frac{dy}{dt} = 0.15(X_1 + 50) - 0.25(X_2 + 30) = 0.15X_1 - 0.25X_2 + 7.5 - 7.5 = 0.15X_1 - 0.25X_2\)
Thus, \(\frac{dX}{dt} = M X\) where \(M = \begin{pmatrix} -0.15 & 0.05 \\ 0.15 & -0.25 \end{pmatrix}\).

(ii) Find the eigenvalues by solving \(\det(M - \lambda I) = 0\):
\((-0.15 - \lambda)(-0.25 - \lambda) - 0.0075 = 0\)
\(\lambda^2 + 0.4\lambda + 0.0375 - 0.0075 = 0\)
\(\lambda^2 + 0.4\lambda + 0.03 = 0\)
\((\lambda + 0.1)(\lambda + 0.3) = 0\)
Thus, \(\lambda_1 = -0.1\) and \(\lambda_2 = -0.3\).

(iii) For \(\lambda_1 = -0.1\):
\(\begin{pmatrix} -0.05 & 0.05 \\ 0.15 & -0.15 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies v_1 = v_2\)
An eigenvector is \(\vec{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}\).

For \(\lambda_2 = -0.3\):
\(\begin{pmatrix} 0.15 & 0.05 \\ 0.15 & 0.05 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies 3v_1 + v_2 = 0\)
An eigenvector is \(\vec{v}_2 = \begin{pmatrix} 1 \\ -3 \end{pmatrix}\).

(iv) The general solution is:
\(X(t) = c_1 e^{-0.1t} \begin{pmatrix} 1 \\ 1 \end{pmatrix} + c_2 e^{-0.3t} \begin{pmatrix} 1 \\ -3 \end{pmatrix}\)
Using initial conditions at \(t = 0\), \(X(0) = \begin{pmatrix} 0 - 50 \\ 0 - 30 \end{pmatrix} = \begin{pmatrix} -50 \\ -30 \end{pmatrix}\):
\(c_1 + c_2 = -50\)
\(c_1 - 3c_2 = -30\)
Subtracting the equations gives \(4c_2 = -20 \implies c_2 = -5\).
Then, \(c_1 = -45\).
Thus, \(X(t) = -45 e^{-0.1t} \begin{pmatrix} 1 \\ 1 \end{pmatrix} - 5 e^{-0.3t} \begin{pmatrix} 1 \\ -3 \end{pmatrix}\).
Using \(X(t) = \begin{pmatrix} x(t) - 50 \\ y(t) - 30 \end{pmatrix}\):
\(x(t) = 50 - 45e^{-0.1t} - 5e^{-0.3t}\)
\(y(t) = 30 - 45e^{-0.1t} + 15e^{-0.3t}\)

**(d)**
(i) At \(t = 4\):
\(x(4) = 50 - 45e^{-0.4} - 5e^{-1.2} \approx 50 - 30.164 - 1.506 = 18.33 \approx 18.3\text{ kg}\) (to 3 sf).
\(y(4) = 30 - 45e^{-0.4} + 15e^{-1.2} \approx 30 - 30.164 + 4.518 = 4.354 \approx 4.35\text{ kg}\) (to 3 sf).

(ii) Percentage errors:
\(x\) error: \(\left| \frac{20.4 - 18.33}{18.33} \right| \times 100\% \approx 11.3\%\).
\(y\) error: \(\left| \frac{3.6 - 4.354}{4.354} \right| \times 100\% \approx 17.3\%\).

(iii) The error can be reduced by using a smaller step size \(h\) (e.g., \(h = 1\) or \(h = 0.5\)).

**(a)**
* **M1** for attempting to set up rate in and rate out equations for Pond A.
* **A1** for obtaining correct DEs for both ponds.
* **M1** for setting both DEs to 0 and substituting to solve for the equilibrium state.
* **A1** for obtaining \(x_e = 50\) and \(y_e = 30\).

**(b)**
* **M1** for correct expression/substitution for step 1 of Euler's method.
* **A1** for \(x(2) \approx 12\) and \(y(2) \approx 0\).
* **M1** for calculating slopes at \(t = 2\) using \(x = 12, y = 0\).
* **A1** for \(dx/dt = 4.2\) and \(dy/dt = 1.8\).
* **A1** for final approximation of \(x(4) \approx 20.4\).
* **A1** for final approximation of \(y(4) \approx 3.6\).

**(c)**
* **M1** for substituting \(x = X_1 + 50\) and \(y = X_2 + 30\) into the DEs.
* **A1** for correctly simplifying to show \(\frac{dX}{dt} = MX\).
* **M1** for writing down the characteristic equation \(\det(M - \lambda I) = 0\).
* **A1** for simplifying to \(\lambda^2 + 0.4\lambda + 0.03 = 0\).
* **A1** for correct eigenvalues \(\lambda = -0.1\) and \(\lambda = -0.3\).
* **M1** for set-up to find eigenvectors.
* **A1** for correct eigenvector \(\begin{pmatrix} 1 \\ 1 \end{pmatrix}\) for \(\lambda = -0.1\).
* **A1** for correct eigenvector \(\begin{pmatrix} 1 \\ -3 \end{pmatrix}\) for \(\lambda = -0.3\).
* **M1** for set-up of particular solution with initial conditions.
* **M1** for solving the linear system for constants \(c_1, c_2\).
* **A1** for \(c_1 = -45, c_2 = -5\).
* **A1** for correct final equations for \(x(t)\) and \(y(t)\).

**(d)**
* **A1** for \(x(4) \approx 18.3\text{ kg}\).
* **A1** for \(y(4) \approx 4.35\text{ kg}\).
* **A1** for percentage error of \(x\) being \(11.3\%\).
* **A1** for percentage error of \(y\) being \(17.3\%\).
* **R1** for explaining that a smaller step size improves accuracy.

**(a)**
(i) Perpendicular bisector of \(AB\):
- Midpoint \(M_{AB} = \left(\frac{2+8}{2}, \frac{6+8}{2}\right) = (5, 7)\).
- Gradient of \(AB\): \(m_{AB} = \frac{8-6}{8-2} = \frac{2}{6} = \frac{1}{3}\).
- Perpendicular gradient: \(m_{\perp} = -3\).
- Equation: \(y - 7 = -3(x - 5) \implies y = -3x + 22\).

(ii) Perpendicular bisector of \(AC\):
- Midpoint \(M_{AC} = \left(\frac{2+6}{2}, \frac{6+2}{2}\right) = (4, 4)\).
- Gradient of \(AC\): \(m_{AC} = \frac{2-6}{6-2} = -1\).
- Perpendicular gradient: \(m_{\perp} = 1\).
- Equation: \(y - 4 = 1(x - 4) \implies y = x\).

(iii) Circumcentre \(V\) is the intersection of these two lines:
\(x = -3x + 22 \implies 4x = 22 \implies x = 5.5\).
Since \(y = x\), \(y = 5.5\).
Thus, the coordinates of \(V\) are \((5.5, 5.5)\).

(iv) Sketch of Voronoi Diagram:
- Plot sites \(A(2,6)\), \(B(8,8)\), and \(C(6,2)\).
- Mark the vertex \(V(5.5, 5.5)\).
- Draw the boundary lines starting from \(V\):
- Boundary between cell A and B going up-left along \(y = -3x + 22\).
- Boundary between cell A and C going down-left along \(y = x\).
- Boundary between cell B and C going right along \(y = -\frac{1}{3}x + \frac{22}{3}\).

**(b)**
(i) Distances from \(P(5, 6.5)\):
- \(AP = \sqrt{(5-2)^2 + (6.5-6)^2} = \sqrt{9 + 0.25} = \sqrt{9.25} \approx 3.04\text{ km}\).
- \(BP = \sqrt{(5-8)^2 + (6.5-8)^2} = \sqrt{9 + 2.25} = \sqrt{11.25} \approx 3.35\text{ km}\).
- \(CP = \sqrt{(5-6)^2 + (6.5-2)^2} = \sqrt{1 + 20.25} = \sqrt{21.25} \approx 4.61\text{ km}\).

(ii) Service Station and Justification:
- Station \(A\) will service the customer because \(AP < BP < CP\) (the straight-line distance to \(A\) is the shortest).
- Geometrically, \(P(5, 6.5)\) lies in the Voronoi cell of \(A\) because:
- For \(x = 5\), the boundary between cell A and B is at \(y = -3(5) + 22 = 7\). Since \(6.5 < 7\), \(P\) is in cell A's region.
- The boundary between cell A and C is \(y = x = 5\). Since \(6.5 > 5\), \(P\) is in cell A's region.

**(c)**
(i) Perpendicular bisector of \(BD\) (with \(B(8,8)\), \(D(10,4)\)):
- Midpoint \(M_{BD} = (9, 6)\).
- Gradient of \(BD\): \(m_{BD} = \frac{4-8}{10-8} = -2\).
- Perpendicular gradient: \(m_{\perp} = 0.5\).
- Equation: \(y - 6 = 0.5(x - 9) \implies y = 0.5x + 1.5\).

(ii) Intersection of the perpendicular bisector of \(BC\) and \(BD\):
- Bisector of \(BC\) is \(y = -\frac{1}{3}x + \frac{22}{3}\).
- Bisector of \(BD\) is \(y = 0.5x + 1.5\).
- Equating them:
\(0.5x + 1.5 = -\frac{1}{3}x + \frac{22}{3} \implies 3(0.5x + 1.5) = -x + 22\)
\(1.5x + 4.5 = -x + 22 \implies 2.5x = 17.5 \implies x = 7\).
- Then \(y = 0.5(7) + 1.5 = 5\).
- The coordinates of the new vertex \(V_2\) are \((7, 5)\).

**(d)**
(i) Displacement vectors:
- \(\vec{u} = \vec{AH} = \begin{pmatrix} 5-2 \\ 7-6 \\ 2-0 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix}\).
- \(\vec{v} = \vec{HB} = \begin{pmatrix} 8-5 \\ 8-7 \\ 0-2 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix}\).

(ii) Angle between vectors:
- \(\vec{u} \cdot \vec{v} = 3(3) + 1(1) + 2(-2) = 9 + 1 - 4 = 6\).
- \(|\vec{u}| = \sqrt{3^2 + 1^2 + 2^2} = \sqrt{14}\).
- \(|\vec{v}| = \sqrt{3^2 + 1^2 + (-2)^2} = \sqrt{14}\).
- \(\cos \theta = \frac{6}{\sqrt{14}\sqrt{14}} = \frac{6}{14} = \frac{3}{7}\).
- \(\theta = \arccos\left(\frac{3}{7}\right) \approx 64.623^\circ \approx 64.6^\circ\).

**(a)**
* **M1** for finding midpoint and gradient of \(AB\).
* **A1** for perpendicular gradient \(-3\).
* **A1** for correct equation \(y = -3x + 22\).
* **A1** for midpoint and gradient of \(AC\).
* **A1** for correct equation \(y = x\).
* **M1** for attempting to solve the simultaneous equations for the intersection.
* **A1** for correct coordinates \((5.5, 5.5)\).
* **A2** for a neat sketch showing three cells, labeled vertices, and correct boundary directions.

**(b)**
* **M1** for correct substitution into distance formula.
* **A2** for all three distances correct to at least 3 significant figures.
* **A1** for concluding Station A services the customer based on shortest distance.
* **R2** for justifying using boundaries (showing \(P\) is within cell A using inequalities or boundary positions).

**(c)**
* **M1** for finding midpoint of \(BD\) and gradient of \(BD\).
* **A1** for perpendicular gradient \(0.5\).
* **A1** for correct equation \(y = 0.5x + 1.5\).
* **M1** for setting up simultaneous equations with the bisector of \(BC\).
* **A1** for finding \(x = 7\).
* **A1** for finding \(y = 5\).

**(d)**
* **A1** for correct vector \(\vec{u}\).
* **A1** for correct vector \(\vec{v}\).
* **M1** for calculating dot product \(\vec{u} \cdot \vec{v} = 6\).
* **M1** for calculating magnitudes of both vectors.
* **M1** for setting up cosine formula.
* **A1** for correct angle \(64.6^\circ\) (accept 1.13 radians).