2023 IB DP Mathematics - Applications and Interpretation Practice Paper with Answers

(a) The null hypothesis \(H_0\) is that music preference is independent of age group.

(b) Expected frequency for (Under 30, Classical) = \(\frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}} = \frac{110 \times 45}{200} = 24.75\).

(c) Using a GDC for a \(\chi^2\) test of independence on the 3x2 matrix of observed values: \(p\)-value \(= 0.0032205... \approx 0.00322\) (to 3 significant figures).

(a) [1 mark] for a correct statement of independence. (b) [2 marks]: [1 mark] for correct substitution into expected value formula, [1 mark] for 24.75. (c) [2.875 marks]: [1.875 marks] for correct GDC input and test setup, [1 mark] for 0.00322.

(a) Using the compound interest formula: \(FV = 12000 \times \left(1 + \frac{0.045}{12}\right)^{12 \times 5} = 12000 \times (1.00375)^{60} \approx 15021.55\) USD.

(b) Interest earned = \(15021.55 - 12000 = 3021.55\) USD.

(c) For simple interest: \(I = P \times \frac{R}{100} \times t\). Here, \(I = 17000 - 15021.55 = 1978.45\) and \(P = 15021.55\), so \(1978.45 = 15021.55 \times \frac{R}{100} \times 3 \Rightarrow 1978.45 = 450.6465 \times R \Rightarrow R \approx 4.39\).

(a) [2.875 marks]: [1.875 marks] for correct compound interest formula setup, [1 mark] for 15021.55. (b) [1 mark] for subtracting 12000 to get 3021.55. (c) [3 marks]: [1 mark] for finding the interest amount of 1978.45, [1 mark] for setting up the simple interest equation, [1 mark] for 4.39.

(a) Using Pythagoras\' theorem on the base: \(AC = \sqrt{8^2 + 6^2} = \sqrt{100} = 10\text{ cm}\).

(b) Let \(O\) be the center of the base, so \(AO = \frac{AC}{2} = 5\text{ cm}\). The vertical height \(VO = 12\text{ cm}\). Using Pythagoras on right-angled triangle \(AOV\): \(AV = \sqrt{AO^2 + VO^2} = \sqrt{5^2 + 12^2} = \sqrt{169} = 13\text{ cm}\).

(c) The angle \(\theta\) is angle \(VAO\). Using trigonometry: \(\tan(\theta) = \frac{VO}{AO} = \frac{12}{5} = 2.4 \Rightarrow \theta = \arctan(2.4) \approx 67.4^{\circ}\) (or \(1.18\) radians).

(a) [2 marks]: [1 mark] for Pythagoras setup, [1 mark] for 10. (b) [2.875 marks]: [1 mark] for finding AO = 5, [1 mark] for 3D Pythagoras setup, [0.875 marks] for 13. (c) [2 marks]: [1 mark] for trig ratio setup, [1 mark] for 67.4 degrees.

(a) The vertical translation (mean temperature) is \(d = \frac{22 + 8}{2} = 15\). The amplitude is \(a = \frac{22 - 8}{2} = 7\).

(b) The period of the temperature cycle is 24 hours. Therefore, \(b = \frac{2\pi}{\text{period}} = \frac{2\pi}{24} = \frac{\pi}{12} \approx 0.262\).

(c) At \(10:00\), \(t = 10\). Substituting the parameters into the model: \(T(10) = -7 \cos\left(\frac{\pi}{12}(10 - 4)\right) + 15 = -7 \cos\left(\frac{\pi}{2}\right) + 15 = -7(0) + 15 = 15^{\circ}\text{C}\).

(a) [2 marks]: [1 mark] for a = 7, [1 mark] for d = 15. (b) [2 marks]: [1 mark] for period calculation, [1 mark] for b = pi/12 or 0.262. (c) [2.875 marks]: [1.875 marks] for substituting t = 10 into the model, [1 mark] for 15.

(a) The volume \(V = \pi r^2 h = 500 \Rightarrow h = \frac{500}{\pi r^2}\). The surface area \(A = 2\pi r^2 + 2\pi r h\). Substituting \(h\): \(A = 2\pi r^2 + 2\pi r \left(\frac{500}{\pi r^2}\right) = 2\pi r^2 + \frac{1000}{r}\).

(b) Differentiating with respect to \(r\): \(\frac{dA}{dr} = 4\pi r - 1000r^{-2} = 4\pi r - \frac{1000}{r^2}\).

(c) To minimize the surface area, set \(\frac{dA}{dr} = 0\): \(4\pi r - \frac{1000}{r^2} = 0 \Rightarrow 4\pi r^3 = 1000 \Rightarrow r^3 = \frac{250}{\pi} \Rightarrow r = \left(\frac{250}{\pi}\right)^{1/3} \approx 4.30\text{ cm}\).

(a) [2 marks]: [1 mark] for expressing h in terms of r, [1 mark] for substituting into surface area formula and showing intermediate steps. (b) [2 marks]: [1 mark] for 4*pi*r, [1 mark] for -1000/r^2. (c) [2.875 marks]: [1.875 marks] for setting the derivative to 0, [1 mark] for 4.30.

(a) Let \(X\) be the weight of an apple, where \(X \sim N(150, 12^2)\). Using a GDC to find \(P(X > 165)\) gives \(0.105649... \approx 0.106\).

(b) We are given \(P(X < w) = 0.15\). Using the inverse normal function on a GDC: \(w \approx 137.56\text{ g}\). To three significant figures, \(w = 138\text{ g}\).

(a) [2.875 marks]: [1.875 marks] for setting up normal distribution parameters on GDC, [1 mark] for 0.106. (b) [4 marks]: [2 marks] for probability equation setup, [1 mark] for GDC inverse normal process, [1 mark] for 138.

(a) Substitute \(t = 5\) into the rate function: \(R(5) = 15 - 15e^{-0.2(5)} = 15 - 15e^{-1} \approx 9.48\text{ liters/min}\).

(b) The total volume \(V\) is given by the integral of the rate function: \(V = \int_{0}^{5} (15 - 15e^{-0.2t}) dt\). Evaluating this using a GDC: \(V = [15t + 75e^{-0.2t}]_{0}^{5} = (75 + 75e^{-1}) - (0 + 75) = 75e^{-1} \approx 27.6\text{ liters}\).

(a) [2 marks]: [1 mark] for substitution of t = 5, [1 mark] for 9.48. (b) [4.875 marks]: [2 marks] for correct integral limits and integrand, [1.875 marks] for GDC integration or analytical integration steps, [1 mark] for 27.6.

(a) \(u_{10} = u_1 + 9d = 12 + 9(4) = 48\).

(b) Using the geometric sum formula: \(S_8 = \frac{v_1(r^8 - 1)}{r - 1} = \frac{3(1.5^8 - 1)}{1.5 - 1} \approx 147.8 \approx 148\) (to 3 significant figures).

(c) We want to find the smallest integer \(n\) such that \(3 \times 1.5^{n-1} > 12 + 4(n-1)\). Creating a table on GDC or testing values:
For \(n = 7\): \(u_7 = 12 + 24 = 36\) and \(v_7 = 3 \times 1.5^6 = 34.17\) (so \(v_7 < u_7\)).
For \(n = 8\): \(u_8 = 12 + 28 = 40\) and \(v_8 = 3 \times 1.5^7 = 51.26\) (so \(v_8 > u_8\)).
Thus, the smallest value is \(n = 8\).

(a) [1 mark] for 48. (b) [2.875 marks]: [1.875 marks] for correct geometric sum setup, [1 mark] for 148. (c) [3 marks]: [1 mark] for establishing the inequality, [1 mark] for testing values or GDC table, [1 mark] for concluding n = 8.

(a) The compound interest formula is given by:
\(FV = PV \left(1 + \frac{r}{100k}\right)^{kn}\)
Here, \(FV = 12\,500\), \(r = 4.5\), \(k = 12\) (compounded monthly), and \(n = 5\) years.
\(12\,500 = P \left(1 + \frac{4.5}{12 \times 100}\right)^{12 \times 5}\)
\(12\,500 = P (1.00375)^{60}\)
\(P = \frac{12\,500}{(1.00375)^{60}} \approx 9985.6534...\)
So, \(P \approx 9985.65\) USD.

(b) For simple interest, the formula for the future value is:
\(FV = P \left(1 + \frac{r \cdot t}{100}\right)\)
Here, \(FV = 12\,500\), \(P = 9985.65\), and \(t = 5\).
\(12\,500 = 9985.65 \left(1 + \frac{5r}{100}\right)\)
\(1 + 0.05r = \frac{12\,500}{9985.65} \approx 1.2517958\)
\(0.05r \approx 0.2517958\)
\(r \approx 5.035916...\)
To three significant figures, \(r = 5.04\%\).

(a) (M1) for substituting correct values into the compound interest formula.
(A1) for \(P (1.00375)^{60} = 12\,500\) or equivalent.
(A1) for \(P = 9985.65\) (accept \(9985.65\) or \(9990\) to 3 s.f.). [3 marks]

(b) (M1) for setting up the simple interest formula.
(M1) for substituting their \(P\) and \(FV = 12\,500\) into the formula.
(A1) for \(1 + 0.05r \approx 1.2518\) or equivalent.
(A1) for \(r = 5.04\) (accept \(5.03\) if \(9990\) is used). [3.875 marks]

(a) At \(t = 0\), the temperature is \(85^\circ\text{C}\).
\(T(0) = a \text{e}^{0} + 20 = 85 \implies a + 20 = 85 \implies a = 65\).

(b) At \(t = 10\), the temperature is \(50^\circ\text{C}\).
\(T(10) = 65 \text{e}^{-10k} + 20 = 50\)
\(65 \text{e}^{-10k} = 30 \implies \text{e}^{-10k} = \frac{6}{13}\)
\(-10k = \ln\left(\frac{6}{13}\right) \implies k = -\frac{1}{10} \ln\left(\frac{6}{13}\right) \approx 0.077319...\)
To three significant figures, \(k = 0.0773\).

(c) We want to find \(t\) such that \(T(t) = 30\).
\(65 \text{e}^{-0.077319 t} + 20 = 30\)
\(65 \text{e}^{-0.077319 t} = 10 \implies \text{e}^{-0.077319 t} = \frac{2}{13}\)
\(-0.077319 t = \ln\left(\frac{2}{13}\right) \implies t = \frac{\ln(2/13)}{-0.077319} \approx 24.212...\)
To three significant figures, \(t = 24.2\) minutes.

(a) (M1) for setting \(T(0) = 85\).
(A1) for \(a = 65\). [2 marks]

(b) (M1) for substituting \(t=10\), \(T(10)=50\) and their \(a=65\) into the equation.
(A1) for \(\text{e}^{-10k} = \frac{30}{65}\).
(A1) for \(k = 0.0773\). [2.875 marks]

(c) (M1) for setting \(T(t) = 30\).
(A1) for solving to find \(t = 24.2\). [2 marks]

(a) The base is an \(8 \times 8\) square. The diagonal \(AC\) is given by:
\(AC = \sqrt{8^2 + 8^2} = \sqrt{128} = 8\sqrt{2} \approx 11.3137...\text{ cm}\).
So \(AC \approx 11.3\text{ cm}\).

(b) The point \(O\) is the midpoint of \(AC\), so:
\(AO = \frac{1}{2} AC = 4\sqrt{2} \approx 5.65685...\text{ cm}\).
In the right-angled triangle \(AOV\), the slant edge \(AV\) is:
\(AV = \sqrt{AO^2 + VO^2} = \sqrt{(4\sqrt{2})^2 + 12^2} = \sqrt{32 + 144} = \sqrt{176} = 4\sqrt{11} \approx 13.2665...\text{ cm}\).
So \(AV \approx 13.3\text{ cm}\).

(c) The angle \(VAO\) is the angle \(\theta\) between the slant edge \(AV\) and the base.
\(\tan \theta = \frac{VO}{AO} = \frac{12}{4\sqrt{2}} = \frac{3}{\sqrt{2}}\)
\(\theta = \arctan\left(\frac{3}{\sqrt{2}}\right) \approx 64.7605...^\circ\)
So \(\theta \approx 64.8^\circ\).

(a) (M1) for applying Pythagoras' theorem to the base.
(A1) for \(AC \approx 11.3\) (or \(8\sqrt{2}\)). [2 marks]

(b) (M1) for finding \(AO = 4\sqrt{2}\) (or \(5.66\)).
(M1) for applying Pythagoras' theorem to triangle \(AOV\).
(A1) for \(AV \approx 13.3\) (or \(4\sqrt{11}\)). [2.875 marks]

(c) (M1) for identifying the angle \(\angle VAO\) and using an appropriate trigonometric ratio.
(A1) for \(64.8^\circ\). [2 marks]

(a) Find the midpoint \(M\) of \(AB\):
\(M = \left(\frac{2+8}{2}, \frac{5+3}{2}\right) = (5, 4)\).
Find the gradient of \(AB\):
\(m_{AB} = \frac{3 - 5}{8 - 2} = \frac{-2}{6} = -\frac{1}{3}\).
The gradient of the perpendicular bisector is:
\(m_{\perp} = -\frac{1}{m_{AB}} = 3\).
The equation of the perpendicular bisector is:
\(y - 4 = 3(x - 5) \implies y = 3x - 11\).

(b) The Voronoi vertex is the intersection point of the perpendicular bisectors of \(AB\) and \(BC\).
Set the equations equal:
\(3x - 11 = -x + 7 \implies 4x = 18 \implies x = 4.5\).
Substitute \(x = 4.5\) into \(y = -x + 7\):
\(y = -4.5 + 7 = 2.5\).
Thus, the coordinates of the Voronoi vertex are \((4.5, 2.5)\).

(a) (M1) for finding the midpoint of \(AB\).
(M1) for finding the gradient of \(AB\) and its perpendicular gradient.
(A1) for \(y = 3x - 11\). [3 marks]

(b) (M1) for setting up a system of equations using their perpendicular bisector from (a) and \(y = -x + 7\).
(M1) for solving for \(x\).
(A1) for \(x = 4.5\).
(A1) for \(y = 2.5\) (or final coordinate representation \((4.5, 2.5)\)). [3.875 marks]

(a) The null hypothesis \(H_0\) is that juice preference and age group are independent (or there is no association between them).

(b) Expected frequency of Adults who prefer Orange juice is:
\(\text{Expected} = \frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}} = \frac{100 \times 75}{200} = 37.5\).

(c) Using GDC for a \(\chi^2\) test of independence on the \(2 \times 3\) contingency table, we find:
\(\chi^2 \approx 4.8056\)
\(df = (2-1) \times (3-1) = 2\)
\(p\)-value \(\approx 0.090466...\)
To three significant figures, the \(p\)-value is \(0.0905\).

(d) Since the \(p\)-value is \(0.0905\), which is greater than the significance level \(0.05\) (\(0.0905 > 0.05\)), we fail to reject (or we do not reject) the null hypothesis. There is insufficient evidence at the \(5\%\) level to suggest that juice preference is dependent on the age group.

(a) (A1) for stating the correct null hypothesis (must mention independence/no association between preference and age group). [1 mark]

(b) (M1) for correct formula used with correct values.
(A1) for \(37.5\). [1.875 marks]

(c) (M1) for entering data into GDC matrix.
(A1) for \(0.0905\). [2 marks]

(d) (R1) for comparing their \(p\)-value with \(0.05\).
(A1) for concluding "do not reject \(H_0\)" (or equivalent). [2 marks]

(a) Let \(X\) be the weight of a bag of apples, where \(X \sim \text{N}(1.05, 0.03^2)\).
Using GDC (normalcdf):
\(\text{P}(X < 1.00) \approx 0.0477903...\)
To three significant figures, the probability is \(0.0478\).

(b) The expected number of bags that cannot be sold in a batch of 500 is:
\(\text{E}(N) = 500 \times \text{P}(X < 1.00) = 500 \times 0.0477903... \approx 23.895...\)
So the expected number of bags is \(23.9\) (or \(24\)).

(c) Let \(w\) be the minimum weight for a bag to be labeled as "Premium".
\(\text{P}(X > w) = 0.10 \implies \text{P}(X \le w) = 0.90\)
Using GDC (inverse normal function):
\(w \approx 1.08845...\text{ kg}\)
To three significant figures, the minimum weight is \(1.09\text{ kg}\).

(a) (M1) for setting up the normal distribution parameters.
(A1) for \(0.0478\). [2 marks]

(b) (M1) for multiplying \(500\) by their probability from part (a).
(A1) for \(23.9\) (or \(24\)). [1.875 marks]

(c) (M1) for writing down \(\text{P}(X > w) = 0.10\) or equivalent.
(M1) for using inverse normal function on GDC.
(A1) for \(1.09\) (or \(1.088\)). [3 marks]

(a) The total surface area of a closed rectangular box with a square base of side length \(x\) and height \(h\) is:
\(A = 2x^2 + 4xh = 600\)
Expressing \(h\) in terms of \(x\):
\(4xh = 600 - 2x^2 \implies h = \frac{600-2x^2}{4x} = \frac{150}{x} - \frac{x}{2}\)
The volume of the box is:
\(V = x^2 h = x^2 \left(\frac{150}{x} - \frac{x}{2}\right) = 150x - \frac{1}{2}x^3\).

(b) Differentiating \(V\) with respect to \(x\):
\(\frac{\text{d}V}{\text{d}x} = 150 - \frac{3}{2}x^2\).

(c) Setting \(\frac{\text{d}V}{\text{d}x} = 0\):
\(150 - \frac{3}{2}x^2 = 0 \implies x^2 = 100 \implies x = 10\) (since \(x > 0\)).
Substitute \(x = 10\) back into the volume formula:
\(V_{\text{max}} = 150(10) - \frac{1}{2}(10)^3 = 1500 - 500 = 1000\text{ cm}^3\).

(a) (M1) for setting up the correct surface area formula: \(2x^2 + 4xh = 600\).
(M1) for expressing \(h\) in terms of \(x\).
(A1) for substituting \(h\) into \(V = x^2 h\) and simplifying to get the required equation. [2.875 marks]

(b) (A1) for \(150 - \frac{3}{2}x^2\). [1 mark]

(c) (M1) for setting their derivative equal to 0 and solving for \(x\).
(A1) for \(x = 10\).
(A1) for \(1000\) (or \(1000\text{ cm}^3\)). [3 marks]

(a) Substitute \(t = 3\) into the rate function:
\(R(3) = 3(3^2) - 12(3) + 15 = 27 - 36 + 15 = 6\text{ liters per minute}\).

(b) The total volume of water is given by the definite integral of the rate function:
\(V = \int_{1}^{4} (3t^2 - 12t + 15) \text{d}t\)
Using the antiderivative:
\(F(t) = t^3 - 6t^2 + 15t\)
Applying the fundamental theorem of calculus:
\(V = [t^3 - 6t^2 + 15t]_1^4\)
\(V = \left(4^3 - 6(4^2) + 15(4)\right) - \left(1^3 - 6(1^2) + 15(1)\right)\)
\(V = (64 - 96 + 60) - (1 - 6 + 15)\)
\(V = 28 - 10 = 18\text{ liters}\).
(Note: This calculation can also be completed directly on the GDC).

(a) (M1) for substituting \(t = 3\) into \(R(t)\).
(A1) for \(6\). [1.875 marks]

(b) (M1) for setting up the definite integral: \(\int_{1}^{4} (3t^2 - 12t + 15) \text{d}t\).
(M1) for finding the correct antiderivative \(t^3 - 6t^2 + 15t\) (or setting up GDC integral correctly).
(M1) for evaluating the integral using limits of integration.
(A1) for \(18\). [5 marks]

(a) \(H_0\): The choice of physical exercise is independent of the age group of the gym members.

(b) Expected value = \(\frac{\text{Total Over 50} \times \text{Total Running}}{\text{Grand Total}} = \frac{60 \times 90}{250} = 21.6\).

(c) Using the \(\chi^2\) test for independence on the \(3 \times 3\) table on a GDC:
\(\chi^2 \approx 9.72\)
Degrees of freedom \( = (3-1)(3-1) = 4\)
\(p\)-value \(\approx 0.0454\) (to 3 s.f.).

(d) Since the \(p\)-value \((0.0454) < 0.05\), we reject the null hypothesis. There is significant evidence at the \(5\%\) level to suggest that exercise choice is associated with age group.

(e) Let \(X \sim N(45, 8^2)\).
\(P(35 \le X \le 50) = \text{normalCDF}(35, 50, 45, 8) \approx 0.628\) (to 3 s.f.).

(f) \(P(X > t) = 0.15 \Rightarrow P(X \le t) = 0.85\).
Using inverse normal function on GDC:
\(t \approx 53.3\) minutes (to 3 s.f.).

(g) Since 45 minutes is the mean, the probability that any individual runner takes longer than 45 minutes is \(0.5\).
Let \(Y\) be the number of runners taking longer than 45 minutes. Then \(Y \sim B(5, 0.5)\).
We want \(P(Y \ge 3) = P(Y = 3) + P(Y = 4) + P(Y = 5) = 0.5\).

a) **A1** for stating that choice of exercise and age group are independent (or not associated).

b) **M1** for setting up the calculation \(\frac{60
times 90}{250}\).
**A1** for \(21.6\).

c) **M1** for entering data into GDC matrix.
**A1** for \(p\)-value \(\approx 0.0454\) (accept \(0.04537...\)).

d) **R1** for comparing \(p\)-value to \(0.05\) (\(0.0454 < 0.05\)).
**A1** for concluding to reject \(H_0\).

e) **M1** for standardizing or setting up normal boundaries.
**M1** for correct parameters in GDC.
**A1** for \(0.628\) (accept \(0.6282...\)).

f) **M1** for setting up equation \(P(X > t) = 0.15\) or \(P(X < t) = 0.85\).
**M1** for using inverse normal.
**A1** for \(53.3\) (accept \(53.28...\)).

g) **A1** for recognizing \(p = 0.5\).
**M1** for setting up binomial probability \(P(Y \ge 3)\) with \(n = 5\).
**A1** for \(0.5\) (accept \(0.500\)).

(a) \(AB = \sqrt{(50-10)^2 + (80-20)^2} = \sqrt{40^2 + 60^2} = \sqrt{5200} \approx 72.1\text{ m}\).

(b) Using vectors \(\vec{CA}\) and \(\vec{CB}\):
\(\vec{CA} = A - C = (10-90, 20-10, 0) = (-80, 10, 0)\)
\(\vec{CB} = B - C = (50-90, 80-10, 0) = (-40, 70, 0)\)
Lengths of these vectors:
\(|\vec{CA}| = \sqrt{(-80)^2 + 10^2} = \sqrt{6500} \approx 80.62\text{ m}\)
\(|\vec{CB}| = \sqrt{(-40)^2 + 70^2} = \sqrt{6500} \approx 80.62\text{ m}\)
Dot product:
\(\vec{CA} \cdot \vec{CB} = (-80)(-40) + (10)(70) + 0 = 3200 + 700 = 3900\)
Using cosine formula:
\(\cos(\angle ACB) = \frac{\vec{CA} \cdot \vec{CB}}{|\vec{CA}||\vec{CB}|} = \frac{3900}{6500} = 0.6\)
\(\angle ACB = \arccos(0.6) \approx 53.1^\circ\) (to 3 s.f.).

(c) The drone is at \(P(10, 20, 100)\). The horizontal distance \(AC = \sqrt{(90-10)^2 + (10-20)^2} = \sqrt{80^2 + (-10)^2} = \sqrt{6500} \approx 80.62\text{ m}\).
Let \(\theta\) be the angle of depression (which is equal to the angle of elevation from \(C\) to \(P\)):
\(\tan \theta = \frac{100}{AC} = \frac{100}{\sqrt{6500}}\)
\(\theta = \arctan\left(\frac{100}{80.62}\right) \approx 51.1^\circ\).

(d) For \(ABCD\) to be a parallelogram with \(AB\) parallel and equal in length to \(CD\), the vector \(\vec{DC}\) must equal \(\vec{AB}\):
\(\vec{AB} = B - A = (40, 60, 0)\)
\(\vec{DC} = C - D \Rightarrow D = C - \vec{AB} = (90 - 40, 10 - 60, 0 - 0) = (50, -50, 0)\).

(e) The area of the parallelogram is twice the area of the triangle \(ABC\):
\(\text{Area of } \triangle ABC = \frac{1}{2} \cdot CA \cdot CB \cdot \sin(\angle ACB)\)
Since \(\cos(\angle ACB) = 0.6\), \(\sin(\angle ACB) = 0.8\).
\(\text{Area} = 2 \times \left( \frac{1}{2} \times \sqrt{6500} \times \sqrt{6500} \times 0.8 \right) = 6500 \times 0.8 = 5200\text{ m}^2\).

a) **M1** for attempting to use the 3D/2D distance formula.
**A1** for \(72.1\) (accept \(\sqrt{5200}\)).

b) **M1** for finding vectors \(\vec{CA}\) and \(\vec{CB}\).
**M1** for calculating lengths of \(\vec{CA}\) and \(\vec{CB}\).
**M1** for applying the dot product and cosine formula.
**A1** for \(53.1^\circ\) (accept \(0.927\) radians).

c) **M1** for recognizing height is \(100\) and finding horizontal distance \(AC\).
**M1** for setting up \(\tan \theta = \frac{100}{AC}\).
**A1** for \(51.1^\circ\) (accept \(0.892\) radians).

d) **M1** for setting up vector equivalence \(\vec{AB} = \vec{DC}\) (or equivalent).
**M1** for finding components of \(\vec{AB}\).
**A1** for \(x = 50\), **A1** for \(y = -50\) (coordinates \(D(50, -50, 0)\)).

e) **M1** for attempting to find the area of \(\triangle ABC\) or using cross product.
**M1** for correct substitution into an area formula.
**A1** for \(5200\).

(a) Using TVM solver on a GDC:
\(N = 300\)
\(I\% = 4.8\)
\(PV = 350000\)
\(FV = 0\)
\(P/Y = 12\)
\(C/Y = 12\)
We get \(PMT \approx -2005.14\).
So the monthly repayment is $2005.14 USD.

(b) Total payments to Bank A = \(2005.1447 \times 300 \approx 601,543.41\text{ USD}\).
Total interest = \(601,543.41 - 350,000 = 251,543.41\text{ USD}\) (or $251,542 if using 3 s.f.).

(c) For Bank B:
\(PV = 350000\)
\(PMT = -2000\)
\(I\% = 4.5\)
\(FV = 0\)
\(P/Y = 12\)
\(C/Y = 12\)
Solving for \(N\) gives \(N \approx 285.29\) months.
Therefore, she will make 285 full monthly payments.

(d) First, we find the remaining balance after 285 full payments:
Set \(N = 285\), and solve for \(FV\) with \(PMT = -2000\):
\(FV \approx -575.14\) USD (meaning she still owes $575.14).
This balance accumulates interest for 1 month:
\(575.144 \times \left(1 + \frac{0.045}{12}\right) = 575.144 \times 1.00375 \approx 577.30\) USD.
So her final partial payment is $577.30 USD.

(e) Total amount paid to Bank B = \(285 \times 2000 + 577.30 = 570,577.30\) USD.
Total interest paid to Bank B = \(570,577.30 - 350,000 = 220,577.30\) USD.
Comparing the options:
Bank B interest = $220,577.30
Bank A interest = $251,543.41
Difference = \(251,543.41 - 220,577.30 = 30,966.11\) USD.
Thus, Bank B results in less interest, by $30,966 USD (accept values between $30,964 and $30,967 depending on rounding).

a) **M1** for choosing correct financial variables (\(N=300\), \(I=4.8\), etc.).
**M1** for attempting to use a financial calculator or TVM formula.
**A1** for $2005.14.

b) **M1** for calculating total payments (\(2005.14 \times 300\)).
**A1** for $251,543 (accept $251,542 from rounded PMT).

c) **M1** for setting up the TVM parameters with \(PMT = -2000\) and \(I = 4.5\).
**M1** for attempting to solve for \(N\).
**A1** for \(285.29...\).
**A1** for concluding 285 full months.

d) **M1** for calculating outstanding balance at \(N=285\).
**M1** for applying 1 month of interest (\(\times 1.00375\)).
**A1** for $577.30.

e) **M1** for finding total paid to Bank B (\(285 \times 2000 + 577.30\)).
**M1** for calculating Bank B interest.
**A1** for concluding Bank B is cheaper.
**A1** for finding the difference of approximately $30,966.

(a)
(i) \(d\) is the principal axis (mean value):
\(d = \frac{12.4 + 2.8}{2} = 7.6\).

(ii) \(a\) is the amplitude:
\(a = 12.4 - 7.6 = 4.8\).

(iii) The time between a high and low tide is half the period. Thus:
\(\text{Period} = 2 \times 6.2 = 12.4\) hours.
\(b = \frac{2\pi}{\text{Period}} = \frac{2\pi}{12.4} \approx 0.507\) rad/hour (or \(\frac{\pi}{6.2}\)).

(b) At \(t=8\):
\(h(8) = 4.8 \cos\left(\frac{8\pi}{6.2}\right) + 7.6 \approx 4.66\) meters.

(c) We want to solve \(h(t) \ge 10\) for \(0 \le t \le 12\):
\(4.8 \cos\left(\frac{\pi}{6.2} t\right) + 7.6 = 10 \Rightarrow \cos\left(\frac{\pi}{6.2} t\right) = 0.5\).
This yields:
\(\frac{\pi}{6.2} t = \frac{\pi}{3} \Rightarrow t = \frac{6.2}{3} \approx 2.067\) hours
and \(\frac{\pi}{6.2} t = \frac{5\pi}{3} \Rightarrow t = \frac{31}{3} \approx 10.33\) hours.
Between \(t=0\) and \(t=12\), the tide is above 10m on the intervals \([0, 2.067]\) and \([10.33, 12]\).
Total time = \(2.067 + (12 - 10.33) = 2.067 + 1.667 = 3.73\) hours (or 3 hours 44 minutes).

(d) At 04:00, \(t = 4\):
\(h(4) = 4.8 \cos\left(\frac{4\pi}{6.2}\right) + 7.6 \approx 5.50\) meters.
Since \(5.50 > 5.0\), the depth is sufficient and the boat can safely enter.

a) (i) **M1** for attempting to find the average. **A1** for \(d = 7.6\).
(ii) **A1** for \(a = 4.8\).
(iii) **M1** for finding period \(12.4\). **A1** for \(b = 0.507\) (or \(\frac{\pi}{6.2}\)).

b) **M1** for substituting \(t=8\) into their model. **A1** for \(4.66\) meters.

c) **M1** for setting up the equation \(h(t) = 10\).
**M1** for finding \(t_1 \approx 2.07\).
**M1** for finding \(t_2 \approx 10.33\).
**M1** for identifying the two valid time intervals \([0, 2.07]\) and \([10.33, 12]\).
**A1** for \(3.73\) hours.

d) **M1** for substituting \(t=4\) into the model.
**A1** for finding \(5.50\) meters.
**R1** for comparing \(5.50 > 5.0\).
**A1** for concluding that the boat can safely enter.

(a) Entering the data into a GDC, we find:
\(r \approx 0.756\).

(b) Crop yield increases with fertilizer up to a certain point (around \(x=120\)), but then starts to decline due to over-fertilization (toxicity). A linear model implies a constant rate of increase, which is physically unrealistic.

(c) Sorting \(y\) values:
\(3.2 \rightarrow 1\)
\(4.5 \rightarrow 2\)
\(5.1 \rightarrow 3\)
\(5.4 \rightarrow 4\)
\(5.8 \rightarrow 5\)
\(5.9 \rightarrow 6\)
\(6.0 \rightarrow 7\)
\(6.1 \rightarrow 8\)
So the ranks of \(y\) are:
Plot 1: 1, Plot 2: 2, Plot 3: 3, Plot 4: 5, Plot 5: 7, Plot 6: 8, Plot 7: 6, Plot 8: 4.

(d) Finding differences \(d_i = \text{Rank}(x_i) - \text{Rank}(y_i)\):
\(d = [0, 0, 0, -1, -2, -2, 1, 4]\)
\(d^2 = [0, 0, 0, 1, 4, 4, 1, 16]\)
\(\sum d^2 = 26\).
\(r_s = 1 - \frac{6 \sum d^2}{n(n^2 - 1)} = 1 - \frac{6 \times 26}{8(63)} = 1 - \frac{156}{504} \approx 0.690\) (to 3 s.f.).

(e) There is a moderate to strong positive monotonic relationship between fertilizer applied and crop yield, indicating that higher amounts of fertilizer are generally associated with higher yields, though the relationship is not strictly increasing.

(f) Fitting a quadratic regression \(y = a x^2 + b x + c\) on a GDC gives:
\(a \approx -0.000327\)
\(b \approx 0.0742\)
\(c \approx 1.91\)
So the equation is \(y = -0.000327 x^2 + 0.0742 x + 1.91\).

a) **M1** for attempting to enter data into GDC. **A1** for \(0.756\).

b) **R1** for recognizing that yield peaks and then decreases. **R1** for explaining that linear model implies constant increase, which is unrealistic.

c) **A3** for all 8 ranks of \(y\) correct (\(1, 2, 3, 5, 7, 8, 6, 4\)). Award **A2** if 6 or 7 are correct, **A1** if 4 or 5 are correct.

d) **M1** for attempting to calculate differences \(d\) and their squares. **M1** for substituting into formula. **A1** for \(0.690\).

e) **R1** for stating it shows a positive monotonic relationship. **R1** for describing it as moderate/strong.

f) **M1** for selecting quadratic regression on GDC. **A1** for \(a \approx -0.000327\), **A1** for \(b \approx 0.0742\), **A1** for \(c \approx 1.91\).

(a) Since \(V = \pi r^2 h = 1500\), we have:
\(h = \frac{1500}{\pi r^2}\).

(b) The total cost is the sum of the cost of the circular ends and the curved surface:
\(\text{Area of ends} = 2 \pi r^2\)
\(\text{Cost of ends} = 0.05 \times 2 \pi r^2 = 0.1 \pi r^2\)
\(\text{Area of curved surface} = 2 \pi r h\)
\(\text{Cost of curved surface} = 0.03 \times 2 \pi r h = 0.06 \pi r h\)
Substitute \(h = \frac{1500}{\pi r^2}\):
\(\text{Cost of curved surface} = 0.06 \pi r \left(\frac{1500}{\pi r^2}\right) = \frac{90}{r}\)
Adding both components together:
\(C(r) = 0.1 \pi r^2 + \frac{90}{r}\) (as required).

(c) Differentiating with respect to \(r\):
\(\frac{\mathrm{d}C}{\mathrm{d}r} = \frac{\mathrm{d}}{\mathrm{d}r}(0.1 \pi r^2 + 90 r^{-1}) = 0.2 \pi r - 90 r^{-2} = 0.2 \pi r - \frac{90}{r^2}\).

(d) Setting \(\frac{\mathrm{d}C}{\mathrm{d}r} = 0\):
\(0.2 \pi r - \frac{90}{r^2} = 0 \Rightarrow 0.2 \pi r^3 = 90 \Rightarrow r^3 = \frac{450}{\pi} \approx 143.239\)
\(r = \sqrt[3]{143.239} \approx 5.23\text{ cm}\) (to 3 s.f.).

(e) Substituting \(r \approx 5.232\) into the cost function:
\(C(5.232) = 0.1 \pi (5.232)^2 + \frac{90}{5.232} \approx 8.60 + 17.20 = 25.80\) cents.
Rounding to the nearest cent, the minimum cost is \(26\) cents.

a) **M1** for setting up \(\pi r^2 h = 1500\). **A1** for \(h = \frac{1500}{\pi r^2}\).

b) **M1** for expressing cost of top/bottom as \(0.05
times 2 \pi r^2\).
**M1** for expressing cost of curved surface as \(0.03
times 2 \pi r h\).
**M1** for substituting \(h\) into the cost expression.
**A1** for obtaining the given expression clearly.

c) **M1** for differentiating \(0.1 \pi r^2\) to obtain \(0.2 \pi r\).
**M1** for differentiating \(\frac{90}{r}\) to obtain \(-\frac{90}{r^2}\).
**A1** for the complete derivative.

d) **M1** for setting their derivative to 0.
**M1** for rearranging to solve for \(r^3\).
**A1** for \(r \approx 5.23\text{ cm}\).

e) **M1** for substituting their value of \(r\) into \(C(r)\).
**A1** for obtaining \(25.80\) cents.
**A1** for rounding to \(26\) cents.

(a) The particle is at rest when \(v(t) = 0\):
\(-t^3 + 6t^2 - 9t + 4 = 0\).
Factoring this polynomial:
\((t-1)^2(t-4) = 0\).
Thus, the particle is at rest at \(t = 1\) second and \(t = 4\) seconds.

(b) Acceleration is the derivative of velocity:
\(a(t) = v'(t) = -3t^2 + 12t - 9\).
At \(t = 2\):
\(a(2) = -3(2)^2 + 12(2) - 9 = -12 + 24 - 9 = 3\text{ m s}^{-2}\).

(c) To find the maximum velocity, we examine the local extrema and boundary values on \(0 \le t \le 6\).
Setting \(v'(t) = 0 \Rightarrow -3t^2 + 12t - 9 = 0 \Rightarrow t^2 - 4t + 3 = 0 \Rightarrow (t-1)(t-3) = 0\).
So stationary points occur at \(t = 1\) and \(t = 3\).
Evaluating \(v(t)\) at these points and boundaries:
\(v(0) = 4\)
\(v(1) = 0\)
\(v(3) = -27 + 54 - 27 + 4 = 4\)
\(v(6) = -216 + 216 - 54 + 4 = -50\).
The maximum velocity is \(4\text{ m s}^{-1}\), which occurs at \(t = 0\) and \(t = 3\) seconds.

(d) The total distance travelled in the first 4 seconds is given by \(\int_{0}^{4} |v(t)| \mathrm{d}t\).
On the interval \([0, 4]\), \(v(t) \ge 0\) because \(v(t) = -(t-1)^2(t-4)\), where \((t-1)^2 \ge 0\) and \((t-4) \le 0\) (so the product is non-negative).
Therefore:
\(\text{Distance} = \int_{0}^{4} (-t^3 + 6t^2 - 9t + 4) \mathrm{d}t\)
\(\text{Distance} = \left[ -\frac{1}{4}t^4 + 2t^3 - \frac{9}{2}t^2 + 4t \right]_{0}^{4}\)
\(\text{Distance} = \left(-\frac{1}{4}(256) + 2(64) - 4.5(16) + 16\right) - 0\)
\(\text{Distance} = -64 + 128 - 72 + 16 = 8\text{ meters}\).

a) **M1** for setting \(v(t) = 0\). **A1** for \(t = 1\), **A1** for \(t = 4\).

b) **M1** for attempting to differentiate \(v(t)\). **A1** for \(a(t) = -3t^2 + 12t - 9\). **A1** for \(3\text{ m s}^{-2}\).

c) **M1** for setting \(v'(t) = 0\) to find critical points. **A1** for \(t = 1, 3\). **M1** for evaluating/comparing with boundary values. **A1** for max velocity of \(4\text{ m s}^{-1}\) at \(t=0, 3\).

d) **M1** for setting up integral \(\int_{0}^{4} |v(t)| \mathrm{d}t\).
**R1** for showing/explaining that \(v(t) \ge 0\) on \([0, 4]\).
**M1** for integrating terms correctly.
**A1** for finding correct anti-derivative.
**A1** for evaluating and finding \(8\) meters.

(a) From the given transition rules, the transition matrix \(T\) is:
\(T = \begin{pmatrix} 0 & 0.4 & 0.8 \\\\ 0.7 & 0 & 0.2 \\\\ 0.3 & 0.6 & 0 \end{pmatrix}\)

(b) Given \(X_0 = \begin{pmatrix} 1 \\\\ 0 \\\\ 0 \end{pmatrix}\):
\(X_1 = T X_0 = \begin{pmatrix} 0 & 0.4 & 0.8 \\\\ 0.7 & 0 & 0.2 \\\\ 0.3 & 0.6 & 0 \end{pmatrix} \begin{pmatrix} 1 \\\\ 0 \\\\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\\\ 0.7 \\\\ 0.3 \end{pmatrix}\)
\(X_2 = T X_1 = \begin{pmatrix} 0 & 0.4 & 0.8 \\\\ 0.7 & 0 & 0.2 \\\\ 0.3 & 0.6 & 0 \end{pmatrix} \begin{pmatrix} 0 \\\\ 0.7 \\\\ 0.3 \end{pmatrix} = \begin{pmatrix} 0(0) + 0.4(0.7) + 0.8(0.3) \\\\ 0.7(0) + 0(0.7) + 0.2(0.3) \\\\ 0.3(0) + 0.6(0.7) + 0(0.3) \end{pmatrix} = \begin{pmatrix} 0.52 \\\\ 0.06 \\\\ 0.42 \end{pmatrix}\)

(c) The equation \((T - I)q = 0\) can be expanded as:
\(\begin{pmatrix} -1 & 0.4 & 0.8 \\\\ 0.7 & -1 & 0.2 \\\\ 0.3 & 0.6 & -1 \end{pmatrix} \begin{pmatrix} x \\\\ y \\\\ z \end{pmatrix} = \begin{pmatrix} 0 \\\\ 0 \\\\ 0 \end{pmatrix}\)
Multiplying out the rows gives:
\(-x + 0.4y + 0.8z = 0\)
\(0.7x - y + 0.2z = 0\)
\(0.3x + 0.6y - z = 0\)

(d) From the third equation: \(z = 0.3x + 0.6y\).
Substitute this into the first equation:
\(-x + 0.4y + 0.8(0.3x + 0.6y) = 0\)
\(-x + 0.4y + 0.24x + 0.48y = 0\)
\(-0.76x + 0.88y = 0 \implies y = \frac{76}{88}x = \frac{19}{22}x\).
Substitute \(y\) back into the expression for \(z\):
\(z = 0.3x + 0.6\left(\frac{19}{22}x\right) = \frac{3}{10}x + \frac{57}{110}x = \frac{90}{110}x = \frac{9}{11}x = \frac{18}{22}x\).
Using the normalization constraint \(x + y + z = 1\):
\(x + \frac{19}{22}x + \frac{18}{22}x = 1 \implies \frac{59}{22}x = 1 \implies x = \frac{22}{59}\).
Then:
\(y = \frac{19}{22} \left(\frac{22}{59}\right) = \frac{19}{59}\)
\(z = \frac{18}{22} \left(\frac{22}{59}\right) = \frac{18}{59}\)
Thus, the steady-state probabilities are \(x = \frac{22}{59}\), \(y = \frac{19}{59}\), and \(z = \frac{18}{59}\).

(e) With \(d = 0.7\), we have \(1-d = 0.3\), so \(\frac{1-d}{3} J = 0.1 \begin{pmatrix} 1 & 1 & 1 \\\\ 1 & 1 & 1 \\\\ 1 & 1 & 1 \end{pmatrix} = \begin{pmatrix} 0.1 & 0.1 & 0.1 \\\\ 0.1 & 0.1 & 0.1 \\\\ 0.1 & 0.1 & 0.1 \end{pmatrix}\).
And \(d T = 0.7 \begin{pmatrix} 0 & 0.4 & 0.8 \\\\ 0.7 & 0 & 0.2 \\\\ 0.3 & 0.6 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0.28 & 0.56 \\\\ 0.49 & 0 & 0.14 \\\\ 0.21 & 0.42 & 0 \end{pmatrix}\).
Summing these two matrices:
\(G = \begin{pmatrix} 0.1 & 0.38 & 0.66 \\\\ 0.59 & 0.1 & 0.24 \\\\ 0.31 & 0.52 & 0.1 \end{pmatrix}\).

(f) Using a GDC to solve the system \((G - I)q_G = 0\) with \(x_G + y_G + z_G = 1\):
\(x_G \approx 0.3652\)
\(y_G \approx 0.3226\)
\(z_G \approx 0.3122\)
So \(q_G = \begin{pmatrix} 0.3652 \\\\ 0.3226 \\\\ 0.3122 \end{pmatrix}\).

(g) (i) If site \(D\) has no outgoing links, all transitions will eventually lead to \(D\). As \(n \to \infty\), the probability of being at site \(D\) approaches 1, and the probabilities for \(A\), \(B\), and \(C\) approach 0, rendering the relative rankings of \(A, B, C\) useless.
(ii) The damping factor introduces a non-zero probability \(1-d\) of jumping to any random page at each step, even from an absorbing site like \(D\). This ensures the transition matrix remains regular and yields a unique steady-state vector where all sites have non-zero ranking values.

(a) M1 for setting up a 3x3 matrix. A2 for all entries correct (A1 if 1 or 2 entries are incorrect).

(b) M1 for multiplying \(T X_0\). A1 for correct \(X_1\). M1 for multiplying \(T X_1\). A1 for correct \(X_2\).

(c) M1 for showing the setup of \(Tq - q = 0\) or \((T-I)q = 0\). A2 for showing all three equations derived correctly (A1 if only 1-2 are shown/correct).

(d) M1 for expressing \(z\) in terms of \(x\) and \(y\). M1 for finding the relation \(y = \frac{19}{22}x\). M1 for using the constraint \(x+y+z=1\). A1 for showing \(x = \frac{22}{59}\). A1 for \(y = \frac{19}{59}\). A1 for \(z = \frac{18}{59}\).

(e) M1 for calculating \(d T\). M1 for calculating \(\frac{1-d}{3} J\). A2 for adding both matrices correctly to obtain \(G\) (A1 if 1-2 elements are incorrect).

(f) M1 for utilizing GDC to set up the system. A2 for all three probabilities correct to 4 d.p. (A1 if only 1 or 2 are correct or if rounding is incorrect).

(g) (i) R1 for explaining that all probability flows to \(D\). R1 for concluding other sites end up with 0 ranking.
(ii) R1 for noting the random jump allows escape from \(D\). R1 for explaining this leads to a regular transition matrix with non-zero steady-state rankings for all sites.

(a) To find the equilibrium points, we set the rates of change to zero:
\(x(3 - y - x) = 0\)
\(y(2x - 3) = 0\)
Case 1: If \(x = 0\), then from the second equation: \(y(2(0) - 3) = -3y = 0 \implies y = 0\). This gives the equilibrium point \((0, 0)\).
Case 2: If \(y = 0\), then from the first equation: \(x(3 - 0 - x) = 0 \implies x(3 - x) = 0 \implies x = 3\) (since \(x \ge 0\)). This gives the equilibrium point \((3, 0)\).
Case 3: If \(x \neq 0\) and \(y \neq 0\), we solve:
\(3 - y - x = 0\)
\(2x - 3 = 0 \implies x = 1.5\)
Substituting \(x = 1.5\) into the first equation: \(3 - y - 1.5 = 0 \implies y = 1.5\).
This gives the equilibrium point \((1.5, 1.5)\).
Therefore, the three equilibrium points are \((0, 0)\), \((3, 0)\), and \((1.5, 1.5)\).

(b) Let \(f(x, y) = x(3 - y - x)\) and \(g(x, y) = y(2x - 3)\).
Starting at \(t_0 = 0\) with \(x_0 = 1.4\), \(y_0 = 1.6\) and step size \(h = 0.1\):

Step 1 (to \(t_1 = 0.1\)):
\(f(1.4, 1.6) = 1.4(3 - 1.6 - 1.4) = 1.4(0) = 0\)
\(g(1.4, 1.6) = 1.6(2(1.4) - 3) = 1.6(2.8 - 3) = 1.6(-0.2) = -0.32\)
\(x_1 = x_0 + h \cdot f(x_0, y_0) = 1.4 + 0.1(0) = 1.4\)
\(y_1 = y_0 + h \cdot g(x_0, y_0) = 1.6 + 0.1(-0.32) = 1.568\)

Step 2 (to \(t_2 = 0.2\)):
\(f(1.4, 1.568) = 1.4(3 - 1.568 - 1.4) = 1.4(0.032) = 0.0448\)
\(g(1.4, 1.568) = 1.568(2(1.4) - 3) = 1.568(-0.2) = -0.3136\)
\(x_2 = x_1 + h \cdot f(x_1, y_1) = 1.4 + 0.1(0.0448) = 1.40448\)
\(y_2 = y_1 + h \cdot g(x_1, y_1) = 1.568 + 0.1(-0.3136) = 1.53664\)

Rounding to 3 decimal places:
\(x(0.2) \approx 1.404\)
\(y(0.2) \approx 1.537\).

(c) (i) For \(x > 0, y > 0\):
The \(x\)-nullcline is where \(\frac{dx}{dt} = 0 \implies 3 - y - x = 0 \implies y = 3 - x\).
The \(y\)-nullcline is where \(\frac{dy}{dt} = 0 \implies 2x - 3 = 0 \implies x = 1.5\).
(ii) In the region where \(x > 1.5\) and \(x + y < 3\):
- Since \(x + y < 3\), we have \(3 - y - x > 0\). Since \(x > 0\), this implies \(\frac{dx}{dt} = x(3 - y - x) > 0\), so the prey population is increasing (motion is to the right).
- Since \(x > 1.5\), we have \(2x - 3 > 0\). Since \(y > 0\), this implies \(\frac{dy}{dt} = y(2x - 3) > 0\), so the predator population is increasing (motion is upwards).
Thus, the population trajectories in this region move up and to the right (northeasterly direction).

(d) Substituting \(x = 1.5 + u\) and \(y = 1.5 + v\) into the differential equations:
\(\frac{du}{dt} = \frac{dx}{dt} = (1.5 + u)(3 - (1.5 + v) - (1.5 + u)) = (1.5 + u)( -u - v ) = -1.5u - 1.5v - u^2 - uv\)
Neglecting the second-order terms \(u^2\) and \(uv\) gives:
\(\frac{du}{dt} \approx -1.5u - 1.5v\)

For the predator equation:
\(\frac{dv}{dt} = \frac{dy}{dt} = (1.5 + v)(2(1.5 + u) - 3) = (1.5 + v)(3 + 2u - 3) = (1.5 + v)(2u) = 3u + 2uv\)
Neglecting the second-order term \(2uv\) gives:
\(\frac{dv}{dt} \approx 3u\).

(e) The matrix \(M\) is:
\(M = \begin{pmatrix} -1.5 & -1.5 \\\\ 3 & 0 \end{pmatrix}\)
To find the eigenvalues, solve \(\det(M - \lambda I) = 0\):
\(\det \begin{pmatrix} -1.5 - \lambda & -1.5 \\\\ 3 & -\lambda \end{pmatrix} = 0 \implies (-\lambda)(-1.5 - \lambda) - (-1.5)(3) = 0 \implies \lambda^2 + 1.5\lambda + 4.5 = 0\)
Using the quadratic formula:
\(\lambda = \frac{-1.5 \pm \sqrt{1.5^2 - 4(1)(4.5)}}{2} = \frac{-1.5 \pm \sqrt{2.25 - 18}}{2} = \frac{-1.5 \pm \sqrt{-15.75}}{2} = -0.75 \pm \frac{3\sqrt{7}}{4} i\)
Since the eigenvalues are complex with a negative real part (\(-0.75 < 0\)), the equilibrium point \(E(1.5, 1.5)\) is a stable spiral.
In the long term, starting close to this point, populations will exhibit damped oscillations and eventually converge to the equilibrium state of \(1500\) prey and \(1500\) predators.

(a) M1 for setting both derivatives to 0. A1 for \((0,0)\). A1 for \((3,0)\). M1 for substituting \(x = 1.5\) into \(3-y-x = 0\). A1 for \((1.5, 1.5)\).

(b) M1 for correctly evaluating the derivatives at \((1.4, 1.6)\). A1 for correct intermediate values \(x_1 = 1.4\) and \(y_1 = 1.568\). M1 for calculating derivatives at the second step. A1 for \(f(1.4, 1.568) = 0.0448\) and \(g(1.4, 1.568) = -0.3136\). A1 for final estimated \(x(0.2) \approx 1.404\). A1 for final estimated \(y(0.2) \approx 1.537\).

(c) (i) A1 for \(y = 3 - x\). A1 for \(x = 1.5\).
(ii) M1 for demonstrating \(\frac{dx}{dt} > 0\) because \(3 - y - x > 0\). A1 for stating prey population increases / moves right. M1 for demonstrating \(\frac{dy}{dt} > 0\) because \(x > 1.5\). A1 for stating predator population increases / moves up.

(d) M1 for substituting \(1.5 + u\) and \(1.5 + v\) into the equations. A1 for expanding the prey equation to obtain \(-1.5u - 1.5v - u^2 - uv\). M1 for expanding the predator equation to obtain \(3u + 2uv\). R1 for explicitly stating that higher-order terms like \(u^2\), \(uv\) are neglected. A1 for the correct final linearized forms.

(e) A1 for identifying \(M = \begin{pmatrix} -1.5 & -1.5 \\\\ 3 & 0 \end{pmatrix}\). M1 for setting up the determinant equation. A1 for the quadratic equation \(\lambda^2 + 1.5\lambda + 4.5 = 0\). A1 for the correct complex eigenvalues. A1 for classifying the equilibrium point as a stable spiral. R1 for describing the long-term behavior as damped oscillations converging to \(E\).