2024 IB DP Mathematics - Applications and Interpretation Practice Paper with Answers

(a) The value of the machinery follows a geometric sequence with \(u_1 = 120\,000\) and \(r = 0.88\). After 5 years, the value is given by \(u_6 = 120\,000 \times 0.88^5 = 63\,327.83\) dollars. To the nearest dollar, this is $63,328. (b) We set up the inequality \(120\,000 \times 0.88^n < 30\,000\). Dividing by 120,000 gives \(0.88^n < 0.25\). Taking logarithms on both sides: \(n \log(0.88) < \log(0.25) \Rightarrow n > \frac{\log(0.25)}{\log(0.88)} \approx 10.84\). Since \(n\) must be an integer for complete years, the number of complete years is 11.

(a) [M1] for setting up the geometric sequence term: \(120\,000 \times 0.88^5\). [A1.23] for the correct unrounded value 63327.83. [A1] for rounding to the nearest dollar: $63,328. (b) [M1] for setting up the inequality or equation: \(120\,000 \times 0.88^n < 30\,000\). [M1] for attempting to solve using logarithms or GDC. [A1.24] for the correct final answer of 11 years.

(a) The daily profit function \(P(x) = -2x^2 + 15x - 10\) is a quadratic function opening downwards. The maximum occurs at the vertex where \(x = -\frac{b}{2a} = -\frac{15}{2(-2)} = 3.75\) dollars. (b) Substituting \(x = 3.75\) into \(P(x)\) gives \(P(3.75) = -2(3.75)^2 + 15(3.75) - 10 = 18.125\) hundred dollars. Since the profit is in hundreds of dollars, the maximum daily profit is \(18.125 \times 100 = 1812.50\) dollars.

(a) [M1] for identifying the vertex formula or setting \(P'(x) = -4x + 15 = 0\). [A2.23] for finding \(x = 3.75\). (b) [M1] for substituting 3.75 into the profit equation. [A1] for 18.125 hundred dollars. [A1.24] for converting to actual dollars: $1,812.50.

(a) Since \(OABC\) is a rectangle, the triangle \(OAB\) is a right-angled triangle. By Pythagoras' theorem, \(OB = \sqrt{OA^2 + AB^2} = \sqrt{40^2 + 30^2} = 50\text{ m}\). (b) The triangle \(OPB\) is right-angled at \(O\). Therefore, \(\tan(15^\circ) = \frac{h}{OB} = \frac{h}{50}\). Thus, \(h = 50 \tan(15^\circ) \approx 13.397\text{ m} \approx 13.4\text{ m}\) (to 3 s.f.). (c) The triangle \(OPA\) is right-angled at \(O\). The angle of elevation \(\theta\) of \(P\) from \(A\) satisfies \(\tan(\theta) = \frac{h}{OA} = \frac{13.397}{40} \approx 0.3349\). Therefore, \(\theta = \arctan(0.3349) \approx 18.5^\circ\) (to 3 s.f.).

(a) [M1] for applying Pythagoras' theorem: \(\sqrt{40^2 + 30^2}\). [A1.15] for \(OB = 50\text{ m}\). (b) [M1] for setting up the trigonometric ratio: \(\tan(15^\circ) = \frac{h}{50}\). [A1.16] for \(h = 13.4\text{ m}\). (c) [M1] for setting up the trigonometric ratio: \(\tan(\theta) = \frac{13.397}{40}\). [A1.16] for \(\theta = 18.5^\circ\) (accept 18.5 or 0.323 radians).

(a) First, find the midpoint of \(AB\): \(M_{AB} = \left(\frac{2+8}{2}, \frac{8+10}{2}\right) = (5, 9)\). The gradient of the line segment \(AB\) is \(m_{AB} = \frac{10-8}{8-2} = \frac{2}{6} = \frac{1}{3}\). The gradient of the perpendicular bisector is \(m_{\perp} = -\frac{1}{m_{AB}} = -3\). The equation of the perpendicular bisector is \(y - 9 = -3(x - 5) \Rightarrow y = -3x + 24\). (b) The vertex is the intersection point of the perpendicular bisectors. Solve the system of equations: \(y = -3x + 24\) and \(y = -0.25x + 7.75\). Equating them gives \(-3x + 24 = -0.25x + 7.75 \Rightarrow 16.25 = 2.75x \Rightarrow x = \frac{65}{11} \approx 5.91\). Substituting back: \(y = -3\left(\frac{65}{11}\right) + 24 = \frac{69}{11} \approx 6.27\). The coordinates of the vertex are \((5.91, 6.27)\) (to 3 s.f.) or \((\frac{65}{11}, \frac{69}{11})\).

(a) [M1] for finding the midpoint (5, 9). [M1] for finding the gradient of AB as \(1/3\). [M1] for using the negative reciprocal gradient -3. [A0.23] for the correct equation \(y = -3x + 24\). (b) [M1] for equating the two equations: \(-3x + 24 = -\frac{1}{4}x + \frac{31}{4}\). [M1] for solving for \(x\). [A1.24] for both coordinates correct: \((5.91, 6.27)\).

(a) Let \(X\) represent the mass of an apple. \(X \sim N(150, 12^2)\). We need to find \(P(140 < X < 165)\). Using a graphic display calculator: \(P(140 < X < 165) \approx 0.6912\). (b) We need to find \(k\) such that \(P(X < k) = 0.05\). Using the inverse normal function on a graphic display calculator: \(k \approx 130.26\text{ grams}\). To 3 s.f., the maximum mass is 130 grams.

(a) [M1] for setting up the normal distribution parameters \(X \sim N(150, 12^2)\) and writing the probability statement \(P(140 < X < 165)\). [A2.23] for the correct probability 0.691 (accept 0.6912). (b) [M1] for writing the probability statement \(P(X < k) = 0.05\). [A2.24] for the correct value of \(k = 130\text{ grams}\) (accept 130.26).

(a) The null hypothesis \(H_0\) is that preferred music genre and age group are independent. (b) Expected frequency = \(\frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}} = \frac{90 \times 50}{200} = 22.5\). (c) Entering the observed frequencies into a \(2 \times 3\) contingency matrix on the GDC: Row 1 (Under 30): \([45, 35, 10]\), Row 2 (30 and over): \([25, 45, 40]\). Performing a \(\chi^2\) test of independence yields \(\chi^2 \approx 23.196\) and a p-value of approximately \(9.18 \times 10^{-6}\). (d) Since the p-value \(\approx 9.18 \times 10^{-6} < 0.05\), we reject the null hypothesis \(H_0\) at the 5% significance level.

(a) [A1.61] for stating that preferred music genre and age group are independent. (b) [M1] for using the formula \(\frac{90 \times 50}{200}\). [A0.62] for 22.5. (c) [M1] for entering data into the GDC matrix. [A1.62] for the correct p-value \(9.18 \times 10^{-6}\). (d) [R1] for comparing the p-value with 0.05. [A0.62] for concluding that \(H_0\) is rejected.

(a) Let \(h\) be the height of the box. The volume is \(V = x^2 h = 1080\), which gives \(h = \frac{1080}{x^2}\). The surface area of the open box is \(A = x^2 + 4xh\). Substituting \(h\): \(A = x^2 + 4x\left(\frac{1080}{x^2}\right) = x^2 + \frac{4320}{x}\). (b) Differentiating \(A\) with respect to \(x\): \(\frac{\text{d}A}{\text{d}x} = 2x - 4320x^{-2} = 2x - \frac{4320}{x^2}\). (c) To minimize surface area, set \(\frac{\text{d}A}{\text{d}x} = 0\): \(2x - \frac{4320}{x^2} = 0 \Rightarrow 2x^3 = 4320 \Rightarrow x^3 = 2160 \Rightarrow x = \sqrt[3]{2160} \approx 12.926 \approx 12.9\text{ cm}\) (to 3 s.f.).

(a) [M1] for expressing \(h\) in terms of \(x\): \(h = \frac{1080}{x^2}\). [M1] for writing the surface area formula \(A = x^2 + 4xh\). [A0.15] for substituting \(h\) to obtain the given expression clearly. (b) [M1] for attempting to differentiate, with at least one term correct. [A1.16] for \(2x - \frac{4320}{x^2}\). (c) [M1] for setting their derivative to 0. [A1.16] for \(12.9\text{ cm}\) (accept 12.926 or \(\sqrt[3]{2160}\)).

(a) Acceleration is the derivative of speed with respect to time: \(a(t) = v'(t) = -0.45t^2 + 1.8t + 1.2\). At \(t = 2\): \(a(2) = -0.45(2)^2 + 1.8(2) + 1.2 = -1.8 + 3.6 + 1.2 = 3\text{ m s}^{-2}\). (b) The total distance traveled in the first 6 seconds is the integral of the speed function from \(t = 0\) to \(t = 6\): \(s = \int_{0}^{6} (-0.15t^3 + 0.9t^2 + 1.2t) \text{d}t\). Evaluating the definite integral: \(s = \left[ -0.0375t^4 + 0.3t^3 + 0.6t^2 \right]_0^6 = -0.0375(6)^4 + 0.3(6)^3 + 0.6(6)^2 = -48.6 + 64.8 + 21.6 = 37.8\text{ m}\).

(a) [M1] for differentiating \(v(t)\) to find \(a(t)\). [A1] for \(a(t) = -0.45t^2 + 1.8t + 1.2\). [A1.23] for evaluating \(a(2) = 3\text{ m s}^{-2}\). (b) [M1] for setting up the definite integral: \(\int_{0}^{6} v(t) \text{d}t\). [A2.24] for the correct integrated value 37.8 m.

The value after \(t\) years is modeled by \(V(t) = 45000 \times (0.88)^t\). (a) For \(t=6\), \(V(6) = 45000 \times (0.88)^6 \approx 20898.18\). To the nearest dollar, the value is $20,898. (b) We solve \(45000 \times (0.88)^t < 15000\). This simplifies to \((0.88)^t < \frac{1}{3}\). Taking the natural logarithm of both sides: \(t \ln(0.88) < \ln(1/3) \implies t > 8.59\). Thus, during the 9th year (at \(t = 9\)), the value first falls below $15,000.

M1 for setting up the equation \(V(6) = 45000 \times (0.88)^6\). A1 for \(20898\). M1 for setting up the inequality \(45000 \times (0.88)^t < 15000\). A1 for finding the boundary \(t \approx 8.59\). A1 for concluding 9 years.

(a) The profit function reaches its maximum at the vertex where \(x = -\frac{b}{2a} = -\frac{28}{2(-2)} = 7\). Since \(x\) is in hundreds of items, the manufacturer should produce \(7 \times 100 = 700\) items. (b) The maximum profit is \(P(7) = -2(7)^2 + 28(7) - 50 = -98 + 196 - 50 = 48\) thousand dollars, which is $48,000. (c) For no loss, \(P(x) \ge 0\). Solving \(-2x^2 + 28x - 50 = 0\) yields \(x^2 - 14x + 25 = 0\). Using the quadratic formula, \(x = \frac{14 \pm \sqrt{(-14)^2 - 4(1)(25)}}{2} = \frac{14 \pm \sqrt{96}}{2}\). This gives \(x \approx 2.10\) and \(x \approx 11.90\). In terms of items, the range is between 210 and 1190 items.

M1 for finding the vertex or taking the derivative and setting to zero. A1 for 700 items. A1 for $48,000. M1 for setting the inequality \(P(x) \ge 0\). A1 for finding roots 2.10 and 11.90. A1 for specifying the interval in terms of items (210 to 1190).

(a) First, find the distance from the center of the base \(O\) to a corner \(A\). The diagonal of the square base is \(AC = \sqrt{8^2 + 8^2} = \sqrt{128} = 8\sqrt{2} \approx 11.31\) cm. The distance \(AO = \frac{1}{2} AC = 4\sqrt{2} \approx 5.657\) cm. In the right-angled triangle \(VOA\), the slant edge \(VA = \sqrt{AO^2 + VO^2} = \sqrt{(4\sqrt{2})^2 + 12^2} = \sqrt{32 + 144} = \sqrt{176} \approx 13.27\) cm. (b) The angle \(\theta\) between the slant edge and the base is given by \(\tan(\theta) = \frac{VO}{AO} = \frac{12}{4\sqrt{2}} \approx 2.1213\). Thus, \(\theta = \arctan(2.1213) \approx 64.76^\circ\) (or 1.13 radians).

M1 for calculating the diagonal \(AC\) or distance \(AO\). A1 for \(AO = 5.66\) cm. M1 for applying Pythagoras' theorem to find \(VA\). A1 for \(VA \approx 13.3\) cm. M1 for setting up the trigonometric ratio for the angle. A1 for \(64.8^\circ\) (or \(1.13\) radians).

Using a graphic display calculator: (a) Enter the independent variable \(T\) in List 1 and dependent variable \(D\) in List 2. From the linear regression analysis, the Pearson's product-moment correlation coefficient is \(r \approx 0.998\). (b) The regression line equation is of the form \(D = aT + b\), where \(a \approx 3.88\) and \(b \approx -14.8\). So, the equation is \(D = 3.88T - 14.8\). (c) Substituting \(T = 30\) into the regression equation: \(D = 3.8846(30) - 14.8077 = 116.538 - 14.8077 \approx 101.73\). Rounding to the nearest integer, the estimated number of drinks sold is 102.

M1 for entering the data into GDC. A1 for \(r = 0.998\). A2 for the correct regression equation (A1 for gradient 3.88, A1 for y-intercept -14.8). M1 for substituting \(T=30\). A1 for 102.

(a) Differentiating \(V(t) = -t^3 + 9t^2 + 48t + 200\) with respect to \(t\) gives \(\frac{\text{d}V}{\text{d}t} = -3t^2 + 18t + 48\). (b) At \(t=5\), the rate of change is \(V'(5) = -3(5)^2 + 18(5) + 48 = -75 + 90 + 48 = 63\, \text{m}^3/\text{week}\). (c) To find the maximum volume, we set the derivative to zero: \(-3t^2 + 18t + 48 = 0 \implies t^2 - 6t - 16 = 0 \implies (t-8)(t+2) = 0\). Since \(0 \le t \le 12\), we reject \(t = -2\). Thus, the maximum volume occurs at \(t = 8\) weeks.

M1 for attempting to differentiate. A1 for \(-3t^2 + 18t + 48\). A1 for substituting \(t=5\) and getting 63. M1 for setting derivative to 0. A1 for solving the quadratic. A1 for rejecting negative root and stating \(t=8\).

(a) Substitute \(x = 10\) into the function: \(y = 12 \sin(0.05 \times 10) = 12 \sin(0.5) \approx 5.75\,\text{m}\). (b) The cross-sectional area is the integral of the curve from \(x = 0\) to \(x = 20\pi\): \(\text{Area} = \int_{0}^{20\pi} 12 \sin(0.05x) \, \text{d}x\). Integrating gives: \([-12 \times \frac{1}{0.05} \cos(0.05x)]_{0}^{20\pi} = [-240 \cos(0.05x)]_{0}^{20\pi}\). Evaluating this at the limits: \(-240 \cos(\pi) - (-240 \cos(0)) = -240(-1) + 240(1) = 240 + 240 = 480\,\text{m}^2\).

M1 for substituting \(x = 10\) in the equation. A1 for \(5.75\). M1 for setting up the definite integral with correct limits. M1 for integrating \(\sin(0.05x)\) to get \(-20\cos(0.05x)\). A1 for substituting limits. A1 for \(480\).

Let \(X\) be the mass of a peach, so \(X \sim N(135, 15^2)\). (a) We seek \(P(120 < X < 150)\). Using the normal cumulative distribution function on a GDC, we find \(P(120 < X < 150) \approx 0.6827\) (or \(0.683\) to 3 sig figs). (b) We seek \(M\) such that \(P(X > M) = 0.10\), which is equivalent to \(P(X \le M) = 0.90\). Using the inverse normal function on a GDC, we find \(M \approx 154.2\,\text{g}\). To the nearest gram, the minimum mass is 154 grams.

M1 for identifying the normal distribution and setting up the probability expression \(P(120 < X < 150)\). A1 for \(0.683\). M1 for writing the equation \(P(X \le M) = 0.90\) or setting up the GDC inverse normal command. A1 for \(154.2\). A1 for rounding to 154.

(a) Let \(S\), \(R\), and \(C\) be the prices of Sourdough, Rye, and Ciabatta loaves respectively. The system of equations is: \(20S + 15R + 10C = 255\), \(15S + 25R + 12C = 287\), \(30S + 10R + 8C = 262\). (b) Solving this system using a GDC or algebraic substitution: From the system matrix, we find the unique solution: \(S = 4.60\), \(R = 3.20\), \(C = 11.50\). Thus, the price of a Sourdough loaf is $4.60, a Rye loaf is $3.20, and a Ciabatta loaf is $11.50.

M2 for writing down all three equations correctly (1 mark for any two correct, 2 marks for all three). M2 for attempting to solve using matrix methods on a GDC or algebraic substitution. A2 for the correct values (A1 for any two correct, A2 for all three).

(a) The width of each interval is \(h = \frac{12 - 0}{4} = 3\).

Using the trapezoidal rule:
\(\text{Area} \approx \frac{h}{2} \left[ y_0 + y_4 + 2(y_1 + y_2 + y_3) \right]\)

\(\text{Area} \approx \frac{3}{2} \left[ 0 + 0 + 2(1.8 + 2.5 + 1.4) \right]\)

\(\text{Area} \approx 1.5 \times [ 2(5.7) ] = 1.5 \times 11.4 = 17.1\text{ m}^2\).

(b) The volume of water flowing past the cross-section per second is given by:
\(\text{Rate of flow} = \text{Area} \times \text{speed} = 17.1 \times 0.5 = 8.55\text{ m}^3\text{ s}^{-1}\).

In one minute (60 seconds), the volume of water is:
\(\text{Volume} = 8.55 \times 60 = 513\text{ m}^3\).

(a)
(M1) for attempting to use the trapezoidal rule with \(h = 3\).
(A1) for a correct substitution into the formula: \(\frac{3}{2}[0 + 0 + 2(1.8 + 2.5 + 1.4)]\).
(A1) for correct area \(17.1\text{ (m}^2)\).

(b)
(M1) for multiplying their area by \(0.5\).
(M1) for multiplying by \(60\) to convert to minutes.
(A1) for \(513\text{ (m}^3)\) (accept follow-through from their part a).

(a) Deposit paid is \(0.10 \times 35000 = 3500\).
Loan amount is \(35000 - 3500 = 31500\). [1 mark]

(b) Using a financial calculator (TVM Solver):
\(N = 60\)
\(I\% = 4.8\)
\(PV = 31500\)
\(FV = 0\)
\(P/Y = 12\)
\(C/Y = 12\)
This yields \(PMT = -591.43\).
So, the monthly repayment is $591.43. [3 marks]

(c) Total payments = \(591.43 \times 60 = 35485.80\).
Total interest paid = \(35485.80 - 31500 = 3985.80\) USD. [2 marks]

(d) To find the remaining balance after 36 months, use TVM Solver with:
\(N = 36\)
\(I\% = 4.8\)
\(PV = 31500\)
\(PMT = -591.43\)
\(P/Y = 12\)
\(C/Y = 12\)
This yields \(FV = -13634.54\).
So, the remaining balance is $13,634.54. [3 marks]

(e) Using compound interest formula or TVM Solver:
\(PV = -35000\)
\(I\% = 6\)
\(N = 5 \times 4 = 20\)
\(P/Y = 4\)
\(C/Y = 4\)
\(FV = 35000 \times (1 + 0.06/4)^{20} = 47139.93\) USD. [3 marks]

(f) Real value adjusted for 2.5% inflation over 5 years:
\(\text{Real Value} = \frac{47139.93}{(1.025)^5} = 41665.17\) USD. [4 marks]

(a) M1 for calculating 90% of $35,000 or subtracting 10% deposit.

(b) M1 for setting up TVM parameters. A1 for correct inputs. A1 for correct final answer to 2 decimal places.

(c) M1 for multiplying monthly payment by 60. A1 for subtracting loan amount to find interest.

(d) M1 for setting N = 36 or remaining N = 24. A1 for correct TVM setup. A1 for correct remaining balance.

(e) M1 for recognizing quarterly compounding. A1 for substituting values into correct formula. A1 for correct final amount.

(f) M1 for using inflation formula. A1 for substituting 2.5%. A1 for correct denominator. A1 for correct real value.

(a) (i) \(a = \frac{14.6 - 8.2}{2} = 3.2\). [2 marks]
(ii) \(d_0 = \frac{14.6 + 8.2}{2} = 11.4\). [2 marks]

(b) The time between a high tide (03:00) and the next low tide (09:15) is 6 hours and 15 minutes, which is 6.25 hours.
This represents half of the tidal cycle (period).
\(\text{Period} = 2 \times 6.25 = 12.5\) hours.
\(b = \frac{2\pi}{\text{Period}} = \frac{2\pi}{12.5} \approx 0.50265\).
Thus, \(b \approx 0.503\) to three significant figures. [3 marks]

(c) Since high tide (maximum depth) occurs at \(t = 3\) and we are using a cosine function, the horizontal shift \(c = 3\). [2 marks]

(d) At 14:00, \(t = 14\).
\(d(14) = 3.2 \cos(0.50265(14 - 3)) + 11.4\)
\(d(14) = 3.2 \cos(5.52915) + 11.4 \approx 3.2(0.72897) + 11.4 \approx 13.7\) meters. [3 marks]

(e) We require \(d(t) \ge 12\) for \(12 \le t \le 24\).
Set \(3.2 \cos(0.50265(t - 3)) + 11.4 = 12\).
\(\cos(0.50265(t - 3)) = \frac{0.6}{3.2} = 0.1875\).
\(0.50265(t - 3) = \arccos(0.1875) \approx 1.3817\).
Evaluating for the cycle within the range \([12, 24]\), we find the boundaries around the high tide at \(t = 15.5\):
\(t_1 = 15.5 - \frac{1.3817}{0.50265} \approx 12.75\) (which is 12:45).
\(t_2 = 15.5 + \frac{1.3817}{0.50265} \approx 18.25\) (which is 18:15).
So, the ship can enter safely during the interval \([12.75, 18.25]\), corresponding to 12:45 to 18:15. [4 marks]

(a) (i) M1 for correct formula for amplitude. A1 for 3.2.
(ii) M1 for correct formula for vertical shift. A1 for 11.4.

(b) M1 for finding half period as 6.25 hours. A1 for period = 12.5. A1 for showing b = 2pi / 12.5 = 0.503.

(c) M1 for relating high tide to cosine maximum. A1 for c = 3.

(d) M1 for substituting t = 14. M1 for evaluating in radians. A1 for 13.7 m.

(e) M1 for setting equation equal to 12. M1 for finding critical values. A1 for correct boundaries 12.75 and 18.25. A1 for expressing as a time interval.

(a) Midpoint of \(AB = \left(\frac{2+8}{2}, \frac{9+7}{2}\right) = (5, 8)\). [2 marks]

(b) Gradient of \(AB = m_{AB} = \frac{7-9}{8-2} = -\frac{2}{6} = -\frac{1}{3}\).
Gradient of perpendicular bisector \(m_{\perp} = -\frac{1}{m_{AB}} = 3\).
Equation: \(y - 8 = 3(x - 5) \Rightarrow y - 8 = 3x - 15 \Rightarrow 3x - y = 7\). (Shown) [3 marks]

(c) Midpoint of \(AC = \left(\frac{2+4}{2}, \frac{9+3}{2}\right) = (3, 6)\).
Gradient of \(AC = m_{AC} = \frac{3-9}{4-2} = -3\).
Gradient of perpendicular bisector \(m_{\perp} = \frac{1}{3}\).
Equation: \(y - 6 = \frac{1}{3}(x - 3) \Rightarrow 3y - 18 = x - 3 \Rightarrow x - 3y = -15\). [3 marks]

(d) Solve the system of equations:
1) \(3x - y = 7 \Rightarrow y = 3x - 7\)
2) \(x - 3y = -15\)
Substitute (1) into (2):
\(x - 3(3x - 7) = -15 \Rightarrow x - 9x + 21 = -15 \Rightarrow -8x = -36 \Rightarrow x = 4.5\).
Substitute \(x = 4.5\) into (1):
\(y = 3(4.5) - 7 = 6.5\).
The Voronoi vertex is \(D(4.5, 6.5)\). [3 marks]

(e) Since \(D\) is the Voronoi vertex, it is equidistant to \(A\), \(B\), and \(C\).
Distance \(DA = \sqrt{(4.5 - 2)^2 + (6.5 - 9)^2} = \sqrt{2.5^2 + (-2.5)^2} = \sqrt{12.5} \approx 3.54\) m. [2 marks]

(f) Adding a new site \(D\) at the Voronoi vertex will create a new Voronoi cell around \(D\). Because \(D\) is closer to the surrounding points than \(A\), \(B\), and \(C\) were previously, the new cell will consume part of the territory of all three existing regions. Thus, the regions of \(A\), \(B\), and \(C\) will all decrease in size. [3 marks]

(a) M1 for midpoint formula. A1 for (5, 8).

(b) M1 for gradient of AB. M1 for perpendicular gradient. A1 for showing the final equation.

(c) M1 for midpoint of AC. M1 for gradient of perpendicular bisector. A1 for correct equation.

(d) M1 for setting up system of equations. M1 for solving. A1 for D(4.5, 6.5).

(e) M1 for distance formula. A1 for 3.54 m.

(f) R1 for mentioning that a new cell is formed around D. R1 for stating D is closer to points previously in cells A, B, and C. A1 for concluding that all three regions decrease in size.

(a) \(H_0\): Favorite subject and preferred study time are independent (not associated).
\(H_1\): Favorite subject and preferred study time are not independent (associated). [2 marks]

(b) Expected frequency \(= \frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}} = \frac{80 \times 65}{200} = 26\). (Shown) [2 marks]

(c) Using a GDC on the contingency table:
(i) Degrees of freedom \(= (3-1) \times (3-1) = 4\).
(ii) \(\chi^2 \approx 11.6\) (11.596).
(iii) \(p\)-value \\approx 0.0206\). [4 marks]

(d) Since the \(p\)-value \(0.0206 < 0.05\) (or \(\chi^2 = 11.6 > 9.488\)), we reject the null hypothesis \(H_0\).
There is sufficient evidence at the 5% significance level to suggest that favorite subject and preferred study time are associated. [2 marks]

(e) Let \(H \sim N(168, 8^2)\).
Using a GDC (normalCDF):
\(P(H > 175) \approx 0.191\) (0.19078). [2 marks]

(f) This is a binomial situation: \(X \sim B(3, 0.19078)\).
We want \(P(X \ge 2) = P(X = 2) + P(X = 3)\).
\(P(X = 2) = \binom{3}{2} (0.19078)^2 (1 - 0.19078) \approx 0.08837\).
\(P(X = 3) = (0.19078)^3 \approx 0.00694\).
\(P(X \ge 2) \approx 0.08837 + 0.00694 = 0.0953\). [4 marks]

(a) A1 for correct null hypothesis. A1 for correct alternative hypothesis.

(b) M1 for formula. A1 for calculation showing 26.

(c) A1 for df = 4. A2 for chi2 = 11.6. A1 for p-value = 0.0206.

(d) R1 for comparing p-value to 0.05. A1 for rejecting H0 and stating correct conclusion.

(e) M1 for standardizing or GDC setup. A1 for 0.191.

(f) M1 for identifying Binomial distribution. M1 for correct parameters. A1 for setup of P(X >= 2). A1 for 0.0953.

(a) Initial velocity is \(v(0) = 32\) m s\(^{-1}\). [1 mark]

(b) Acceleration \(a(t) = v'(t) = -3t^2 + 24t - 36\). [2 marks]

(c) The particle is at rest when \(v(t) = 0\).
\(-t^3 + 12t^2 - 36t + 32 = 0\).
Factorizing gives \(-(t-2)^2(t-8) = 0\).
Thus, \(t = 2\) s and \(t = 8\) s. [3 marks]

(d) To find maximum velocity, set \(a(t) = 0\):
\(-3t^2 + 24t - 36 = 0 \Rightarrow t^2 - 8t + 12 = 0 \Rightarrow (t-2)(t-6) = 0\).
Critical points are \(t = 2\) and \(t = 6\).
Evaluate \(v(t)\) at critical points and boundaries:
\(v(0) = 32\)
\(v(2) = 0\)
\(v(6) = -(6)^3 + 12(6)^2 - 36(6) + 32 = -216 + 432 - 216 + 32 = 32\)
\(v(10) = -1000 + 1200 - 360 + 32 = -128\).
Comparing values, the maximum velocity is \(32\) m s\(^{-1}\) (occurring at \(t = 0\) and \(t = 6\)). [4 marks]

(e) The particle is at rest at \(t = 2\) and \(t = 8\). However, since \(v(t) = -(t-2)^2(t-8)\), \(v(t) \ge 0\) for all \(t \in [0, 6]\) (the factor \((t-2)^2\) is non-negative, and \(t-8\) is negative on this interval, so the product with the leading negative is positive).
Since there is no change of direction in \([0, 6]\), the total distance is simply the integral of \(v(t)\):
\(\text{Distance} = \int_0^6 (-t^3 + 12t^2 - 36t + 32) \, dt\)
\(\text{Distance} = \left[ -\frac{t^4}{4} + 4t^3 - 18t^2 + 32t \right]_0^6\)
\(\text{Distance} = \left(-\frac{1296}{4} + 4(216) - 18(36) + 32(6)\right) - 0\)
\(\text{Distance} = -324 + 864 - 648 + 192 = 84\) meters. [6 marks]

(a) A1 for 32.

(b) M1 for attempting to differentiate v(t). A1 for correct expression.

(c) M1 for setting v(t) = 0. A1 for recognizing root t=2. A1 for root t=8.

(d) M1 for setting derivative equal to 0. A1 for identifying critical points t=2 and t=6. M1 for evaluating at boundaries and critical points. A1 for concluding max velocity is 32.

(e) M1 for recognizing that v(t) >= 0 on [0, 6]. M1 for setting up the integral. A2 for correct integration. M1 for substituting boundaries. A1 for 84.

(a) \(V = \pi r^2 h = 40\pi \Rightarrow r^2 h = 40 \Rightarrow h = \frac{40}{r^2}\). (Shown) [2 marks]

(b) The cost of the two circular bases is \(2 \times (\pi r^2) \times 15 = 30\pi r^2\).
The cost of the curved side is \((2\pi r h) \times 10 = 20\pi r h\).
Substituting \(h = \frac{40}{r^2}\) into the cost of the curved side:
\(\text{Curved cost} = 20\pi r \left(\frac{40}{r^2}\right) = \frac{800\pi}{r}\).
Thus, \(C(r) = 30\pi r^2 + \frac{800\pi}{r}\). (Shown) [4 marks]

(c) \(\frac{dC}{dr} = \frac{d}{dr}\left(30\pi r^2 + 800\pi r^{-1}\right) = 60\pi r - 800\pi r^{-2} = 60\pi r - \frac{800\pi}{r^2}\). [3 marks]

(d) To find the minimum cost, set \(\frac{dC}{dr} = 0\):
\(60\pi r - \frac{800\pi}{r^2} = 0 \Rightarrow 60\pi r^3 = 800\pi \Rightarrow r^3 = \frac{800}{60} = \frac{40}{3}\).
\(r = \sqrt[3]{\frac{40}{3}} \approx 2.37\) m. [4 marks]

(e) Substitute \(r \approx 2.3712\) into the cost function:
\(C(2.3712) = 30\pi (2.3712)^2 + \frac{800\pi}{2.3712} \approx 529.91 + 1059.83 \approx 1589.74\).
To the nearest dollar, the minimum cost is $1,590. [3 marks]

(a) M1 for writing volume formula. A1 for showing h = 40 / r^2.

(b) M1 for cost of bases. M1 for cost of curved side. M1 for substituting h. A1 for obtaining the final cost expression.

(c) M1 for differentiating first term. M1 for differentiating second term. A1 for correct final derivative.

(d) M1 for setting derivative equal to 0. M1 for simplifying to r^3. A1 for exact value. A1 for 2.37 m.

(e) M1 for substituting r back into cost formula. A1 for accurate calculation. A1 for rounding to nearest dollar.

(a) The diagram should show point \(P\) with a north arrow. \(Q\) is at distance 12 km at an angle of \(75^\circ\) clockwise from North. \(R\) is at distance 18 km at an angle of \(130^\circ\) clockwise from North. [2 marks]

(b) Angle \(\widehat{QPR}\) is the difference between the two bearings:
\(\widehat{QPR} = 130^\circ - 75^\circ = 55^\circ\). (Shown) [2 marks]

(c) Using the cosine rule on \(\triangle PQR\):
\(QR^2 = PQ^2 + PR^2 - 2(PQ)(PR)\cos(\widehat{QPR})\)
\(QR^2 = 12^2 + 18^2 - 2(12)(18)\cos(55^\circ)\)
\(QR^2 = 144 + 324 - 432(0.573576) = 468 - 247.78 = 220.22\)
\(QR = \sqrt{220.22} \approx 14.8\) km. [3 marks]

(d) First, find the angle \(\widehat{PQR}\) using the sine rule:
\(\frac{\sin(\widehat{PQR})}{PR} = \frac{\sin(\widehat{QPR})}{QR} \Rightarrow \frac{\sin(\widehat{PQR})}{18} = \frac{\sin(55^\circ)}{14.84}\)
\(\sin(\widehat{PQR}) = \frac{18 \times \sin(55^\circ)}{14.84} \approx 0.99355\)
Since \(12^2 + 14.84^2 > 18^2\), \(\widehat{PQR}\) is an acute angle:
\(\widehat{PQR} \approx 83.50^\circ\).

To find the bearing of \(R\) from \(Q\):
Let \(N_P\) and \(N_Q\) be the North directions at \(P\) and \(Q\).
Since the bearing of \(Q\) from \(P\) is \(075^\circ\), the back-bearing (direction of \(P\) from \(Q\)) is \(75^\circ + 180^\circ = 255^\circ\).
Turning clockwise from the line \(QP\) (which is at bearing \(255^\circ\)), we must subtract \(\widehat{PQR}\) because \(R\) lies inside the angle towards the south/east:
\(\text{Bearing of } R \text{ from } Q = 255^\circ - 83.50^\circ = 171.5^\circ\). [5 marks]

(e) \(\text{Area} = \frac{1}{2} \times PQ \times PR \times \sin(\widehat{QPR})\)
\(\text{Area} = \frac{1}{2} \times 12 \times 18 \times \sin(55^\circ) = 108 \times 0.81915 \approx 88.5\) km\(^2\). [4 marks]

(a) A1 for relative locations. A1 for correct angles/distances marked.

(b) M1 for subtracting bearings. A1 for 55.

(c) M1 for setting up cosine rule. A1 for correct substitutions. A1 for 14.8 km.

(d) M1 for setting up sine rule. A1 for finding angle PQR = 83.5. M1 for finding back-bearing of P from Q (255 degrees). M1 for subtracting angle PQR. A1 for 171.5 degrees (or 171 degrees).

(e) M1 for choosing correct area formula. A2 for substituting values. A1 for 88.5.

**(a)** For Tank A: Inflow rate = 10 (pure water) + 10 (from Tank B) = 20 L/min. Outflow rate = 20 L/min (to Tank B). Since inflow = outflow, volume is constant at 100 L.
For Tank B: Inflow rate = 20 (pure water) + 20 (from Tank A) = 40 L/min. Outflow rate = 10 (to Tank A) + 30 (to waste) = 40 L/min. Since inflow = outflow, volume is constant at 200 L.

**(b)** Rate of change of salt in Tank A:
\(\frac{dx}{dt} = \text{rate of salt in} - \text{rate of salt out} = 10 \times \frac{y}{200} - 20 \times \frac{x}{100} = 0.05y - 0.2x\).
Rate of change of salt in Tank B:
\(\frac{dy}{dt} = \text{rate of salt in} - \text{rate of salt out} = 20 \times \frac{x}{100} - (10+30) \times \frac{y}{200} = 0.2x - 0.2y\).
These match the given differential equations.

**(c)** The system in matrix form is \(\frac{d\mathbf{X}}{dt} = \begin{pmatrix} -0.2 & 0.05 \\ 0.2 & -0.2 \end{pmatrix} \mathbf{X}\).
To find the eigenvalues, solve \(\det(\mathbf{M} - \lambda \mathbf{I}) = 0\):
\((-0.2 - \lambda)^2 - (0.05)(0.2) = 0 \implies \lambda^2 + 0.4\lambda + 0.04 - 0.01 = 0 \implies \lambda^2 + 0.4\lambda + 0.03 = 0\).
Solving this quadratic equation gives \((\lambda + 0.1)(\lambda + 0.3) = 0\).
Thus, the eigenvalues are \(\lambda_1 = -0.1\) and \(\lambda_2 = -0.3\).

**(d)** For \(\lambda_1 = -0.1\):
\(\begin{pmatrix} -0.1 & 0.05 \\ 0.2 & -0.1 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies -0.1v_1 + 0.05v_2 = 0 \implies v_2 = 2v_1\).
The eigenvector is \(\mathbf{v}_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix}\).
For \(\lambda_2 = -0.3\):
\(\begin{pmatrix} 0.1 & 0.05 \\ 0.2 & 0.1 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies 0.1v_1 + 0.05v_2 = 0 \implies v_2 = -2v_1\).
The eigenvector is \(\mathbf{v}_2 = \begin{pmatrix} 1 \\ -2 \end{pmatrix}\).

**(e)** The general solution is:
\(\mathbf{X}(t) = C_1 e^{-0.1t} \begin{pmatrix} 1 \\ 2 \end{pmatrix} + C_2 e^{-0.3t} \begin{pmatrix} 1 \\ -2 \end{pmatrix}\).

**(f)** At \(t=0\), \(x(0) = 12\) and \(y(0) = 4\):
\(C_1 + C_2 = 12\)
\(2C_1 - 2C_2 = 4 \implies C_1 - C_2 = 2\).
Adding the two equations yields \(2C_1 = 14 \implies C_1 = 7\).
Subtracting the second from the first yields \(2C_2 = 10 \implies C_2 = 5\).
Thus, the particular solutions are:
\(x(t) = 7e^{-0.1t} + 5e^{-0.3t}\)
\(y(t) = 14e^{-0.1t} - 10e^{-0.3t}\).

**(g)** As \(t \to \infty\), both \(x(t) \to 0\) and \(y(t) \to 0\) because both exponential terms decay to zero.
For the sketch:
- \(x(t)\) starts at \(12\) and decreases monotonically to \(0\).
- \(y(t)\) starts at \(4\). Its initial derivative is \(y'(0) = 0.2(12) - 0.2(4) = 1.6 > 0\), so \(y(t)\) initially increases, reaches a maximum, and then decreases asymptotically to \(0\).

**(h)** Since both eigenvalues are real and negative (\(-0.1\) and \(-0.3\)), the equilibrium point at the origin is a stable node (or sink). Trajectories in the phase portrait converge to the origin, becoming tangent to the line \(y = 2x\) (the slow eigenspace associated with \(\lambda = -0.1\)) as \(t \to \infty\).

**(i)** Euler's method formulas with step size \(h = 1\):
\(x_{n+1} = x_n + h \frac{dx}{dt}\), \(y_{n+1} = y_n + h \frac{dy}{dt}\).
At \(t = 0\), \(x_0 = 12\) and \(y_0 = 4\).
\(\frac{dx}{dt} = -0.2(12) + 0.05(4) = -2.4 + 0.2 = -2.2\).
\(\frac{dy}{dt} = 0.2(12) - 0.2(4) = 2.4 - 0.8 = 1.6\).
Thus:
\(x(1) \approx 12 + 1 \times (-2.2) = 9.8\) kg.
\(y(1) \approx 4 + 1 \times (1.6) = 5.6\) kg.

**(j)** The exact value of salt in Tank A at \(t = 1\) is:
\(x(1) = 7e^{-0.1} + 5e^{-0.3} \approx 7(0.904837) + 5(0.740818) = 6.33386 + 3.70409 = 10.03795\) kg.
The percentage error is:
\(\frac{|10.03795 - 9.8|}{10.03795} \times 100\% \approx 2.37\%\).

**(a)**
M1: Attempt to balance inflow and outflow for both tanks.
A1: Shows constant volume of 100 L for Tank A and 200 L for Tank B with clear rates. [2 marks]

**(b)**
M1: Formulates rate in and rate out equations using concentration.
A1: Correctly derives \(\frac{dx}{dt} = 0.05y - 0.2x\).
A1: Correctly derives \(\frac{dy}{dt} = 0.2x - 0.2y\). [3 marks]

**(c)**
A1: Correctly identifies matrix \(\mathbf{M} = \begin{pmatrix} -0.2 & 0.05 \\ 0.2 & -0.2 \\ \end{pmatrix}\).
M1: Sets up determinant equation \(\det(\mathbf{M} - \lambda\mathbf{I}) = 0\).
A1: Obtains characteristic equation \(\lambda^2 + 0.4\lambda + 0.03 = 0\).
A1: Finds eigenvalues \(\lambda_1 = -0.1\) and \(\lambda_2 = -0.3\). [4 marks]

**(d)**
A1: Correctly finds eigenvector \(\begin{pmatrix} 1 \\ 2 \end{pmatrix}\) for \(\lambda_1 = -0.1\).
A1: Correctly finds eigenvector \(\begin{pmatrix} 1 \\ -2 \end{pmatrix}\) for \(\lambda_2 = -0.3\). [2 marks]

**(e)**
A2: Correct general solution expressing linear combination of eigenvectors and exponentials (award 1 mark if a minor notation error occurs). [2 marks]

**(f)**
M1: Substitutes \(t=0\) and sets up a system of equations for constants \(C_1\) and \(C_2\).
A1: Solves for \(C_1 = 7\) and \(C_2 = 5\).
A1: Writes correct particular solutions. [3 marks]

**(g)**
A1: Correct asymptotic limit as \(t \to \infty\) (both reach 0).
A1: Graph of \(x(t)\) is monotonically decreasing starting at 12.
A1: Graph of \(y(t)\) starts at 4, increases to a peak, then decays to 0. [3 marks]

**(h)**
A1: Correctly sketches phase portrait showing trajectories in the first quadrant moving towards the origin.
R1: Correctly identifies origin as a stable node (sink) based on both eigenvalues being real and negative.
A1: Shows trajectories approaching the origin along the slow eigenspace line \(y = 2x\). [3 marks]

**(i)**
M1: Correct use of Euler's formula for coupled systems.
A1: Correct calculation of rates at \(t=0\) (\(-2.2\) and \(1.6\)).
A1: Correct values: \(x(1) \approx 9.8\) and \(y(1) \approx 5.6\). [3 marks]

**(j)**
M1: Calculates exact value \(x(1) \approx 10.038\).
A1: Correct percentage error calculation, giving \(2.37\%\) (accept answers between \(2.3\%\) and \(2.4\%\)). [2 marks]

#### Part A: Regular Markov Chain
**(a)** The transition matrix \(\mathbf{T}\) with columns representing current state (U, S, R) and rows representing next state (U, S, R) is:
\(\mathbf{T} = \begin{pmatrix} 0.70 & 0.15 & 0.10 \\ 0.20 & 0.75 & 0.10 \\ 0.10 & 0.10 & 0.80 \end{pmatrix}\).

**(b)** Given initial distribution \(\mathbf{s}_0 = \begin{pmatrix} 0.40 \\ 0.40 \\ 0.20 \end{pmatrix}\):
\(\mathbf{s}_1 = \mathbf{T}\mathbf{s}_0 = \begin{pmatrix} 0.70 & 0.15 & 0.10 \\ 0.20 & 0.75 & 0.10 \\ 0.10 & 0.10 & 0.80 \end{pmatrix} \begin{pmatrix} 0.40 \\ 0.40 \\ 0.20 \end{pmatrix} = \begin{pmatrix} 0.28 + 0.06 + 0.02 \\ 0.08 + 0.30 + 0.02 \\ 0.04 + 0.04 + 0.16 \end{pmatrix} = \begin{pmatrix} 0.36 \\ 0.40 \\ 0.24 \end{pmatrix}\).
Next, calculate \(\mathbf{s}_2\):
\(\mathbf{s}_2 = \mathbf{T}\mathbf{s}_1 = \begin{pmatrix} 0.70 & 0.15 & 0.10 \\ 0.20 & 0.75 & 0.10 \\ 0.10 & 0.10 & 0.80 \end{pmatrix} \begin{pmatrix} 0.36 \\ 0.40 \\ 0.24 \end{pmatrix} = \begin{pmatrix} 0.7(0.36) + 0.15(0.40) + 0.1(0.24) \\ 0.2(0.36) + 0.75(0.40) + 0.1(0.24) \\ 0.1(0.36) + 0.1(0.40) + 0.8(0.24) \end{pmatrix} = \begin{pmatrix} 0.336 \\ 0.396 \\ 0.268 \end{pmatrix}\).
The population distribution in Year 2 is 33.6% Urban, 39.6% Suburban, and 26.8% Rural.

**(c)** To find the steady state, we solve \(\mathbf{T}\mathbf{s} = \mathbf{s}\) subject to \(u + s + r = 1\):
\(\begin{pmatrix} -0.30 & 0.15 & 0.10 \\ 0.20 & -0.25 & 0.10 \\ 0.10 & 0.10 & -0.20 \end{pmatrix} \begin{pmatrix} u \\ s \\ r \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\).
This gives the system of equations:
1) \(-0.3u + 0.15s + 0.1r = 0 \implies -6u + 3s + 2r = 0\)
2) \(0.2u - 0.25s + 0.1r = 0 \implies 4u - 5s + 2r = 0\)
3) \(u + s + r = 1\).

**(d)** Subtracting equation (2) from equation (1) yields:
\(-10u + 8s = 0 \implies s = 1.25u\).
Substituting \(s = 1.25u\) into the third equation derived from the third row (\(u + s - 2r = 0\)) gives:
\(u + 1.25u - 2r = 0 \implies 2r = 2.25u \implies r = 1.125u\).
Using the total population condition:
\(u + 1.25u + 1.125u = 1 \implies 3.375u = 1 \implies u = \frac{8}{27}\).
Thus:
\(s = 1.25 \times \frac{8}{27} = \frac{10}{27}\),
\(r = 1.125 \times \frac{8}{27} = \frac{9}{27} = \frac{1}{3}\).
The steady-state distribution is \(\mathbf{s} = \begin{pmatrix} 8/27 \\ 10/27 \\ 1/3 \end{pmatrix}\).

#### Part B: Introduction of Policy (Absorbing Markov Chain)
**(e)** This system is an absorbing Markov chain because:
1. The Urban state (U) is an absorbing state because once a resident enters, they cannot leave (the probability of staying is 1.0).
2. It is possible to reach the Urban state from any other state (S and R) in a finite number of steps. Hence, S and R are transient states.

**(f)** The new transition matrix \(\mathbf{P}\) is:
\(\mathbf{P} = \begin{pmatrix} 1 & 0.20 & 0.10 \\ 0 & 0.70 & 0.10 \\ 0 & 0.10 & 0.80 \end{pmatrix}\).
Partitioning as \(\mathbf{P} = \begin{pmatrix} \mathbf{I} & \mathbf{R} \\ \mathbf{0} & \mathbf{Q} \end{pmatrix}\), we have:
\(\mathbf{I} = [1]\) (dimension \(1 \times 1\))
\(\mathbf{R} = \begin{pmatrix} 0.20 & 0.10 \end{pmatrix}\) (dimension \(1 \times 2\))
\(\mathbf{0} = \begin{pmatrix} 0 \\ 0 \\ \\ \end{pmatrix}\) (dimension \(2 \times 1\))
\(\mathbf{Q} = \begin{pmatrix} 0.70 & 0.10 \\ 0.10 & 0.80 \end{pmatrix}\) (dimension \(2 \times 2\)).

**(g)** Calculate \(\mathbf{I}_2 - \mathbf{Q}\):
\(\mathbf{I}_2 - \mathbf{Q} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} - \begin{pmatrix} 0.70 & 0.10 \\ 0.10 & 0.80 \end{pmatrix} = \begin{pmatrix} 0.30 & -0.10 \\ -0.10 & 0.20 \end{pmatrix}\).
Find the determinant of this matrix:
\(\det(\mathbf{I}_2 - \mathbf{Q}) = (0.30)(0.20) - (-0.10)(-0.10) = 0.06 - 0.01 = 0.05\).
The inverse matrix \(\mathbf{F}\) is:
\(\mathbf{F} = \frac{1}{0.05} \begin{pmatrix} 0.20 & 0.10 \\ 0.10 & 0.30 \end{pmatrix} = 20 \begin{pmatrix} 0.20 & 0.10 \\ 0.10 & 0.30 \end{pmatrix} = \begin{pmatrix} 4 & 2 \\ 2 & 6 \end{pmatrix}\).

**(h)** The entries of the fundamental matrix \(\mathbf{F} = \begin{pmatrix} F_{SS} & F_{SR} \\ F_{RS} & F_{RR} \end{pmatrix} = \begin{pmatrix} 4 & 2 \\ 2 & 6 \end{pmatrix}\) represent:
- \(F_{SS} = 4\): A resident starting in S is expected to spend a total of 4 years in S before absorption.
- \(F_{SR} = 2\): A resident starting in R is expected to spend a total of 2 years in S before absorption.
- \(F_{RS} = 2\): A resident starting in S is expected to spend a total of 2 years in R before absorption.
- \(F_{RR} = 6\): A resident starting in R is expected to spend a total of 6 years in R before absorption.

**(i)** The expected number of years before absorption for a resident starting in R is the sum of the entries in the second column of \(\mathbf{F}\):
\(E(\text{time from R}) = F_{SR} + F_{RR} = 2 + 6 = 8\) years.

**(j)** The absorption probability matrix is given by \(\mathbf{B} = \mathbf{R}\mathbf{F}\):
\(\mathbf{B} = \begin{pmatrix} 0.20 & 0.10 \end{pmatrix} \begin{pmatrix} 4 & 2 \\ 2 & 6 \end{pmatrix} = \begin{pmatrix} 0.20(4) + 0.10(2) & 0.20(2) + 0.10(6) \end{pmatrix} = \begin{pmatrix} 0.80 + 0.20 & 0.40 + 0.60 \end{pmatrix} = \begin{pmatrix} 1.0 & 1.0 \end{pmatrix}\).
This shows that the probability of being absorbed into the Urban area is 1.0 for both states.

#### Part A: Regular Markov Chain [10 marks]
**(a)**
A1: Correct first column of \(\mathbf{T}\).
A1: Correct second and third columns of \(\mathbf{T}\). [2 marks]

**(b)**
M1: Attempts to calculate state vector \(\mathbf{s}_1 = \mathbf{T}\mathbf{s}_0\).
A1: Correctly finds \(\mathbf{s}_1 = \begin{pmatrix} 0.36 \\ 0.40 \\ 0.24 \end{pmatrix}\).
A1: Correctly calculates \(\mathbf{s}_2 = \begin{pmatrix} 0.336 \\ 0.396 \\ 0.268 \end{pmatrix}\). [3 marks]

**(c)**
M1: Sets up equation \(\mathbf{T}\mathbf{s} = \mathbf{s}\).
A1: Correctly writes the system of equations (e.g., \(-0.3u + 0.15s + 0.1r = 0\)).
A1: States the normalization condition \(u + s + r = 1\). [3 marks]

**(d)**
M1: Attempts to solve the system of equations simultaneously.
A1: Obtains exact fractions: \(u = \frac{8}{27}\), \(s = \frac{10}{27}\), \(r = \frac{1}{3}\). [2 marks]

#### Part B: Introduction of Policy (Absorbing Markov Chain) [18 marks]
**(e)**
R1: Explains that Urban (U) is an absorbing state because once entered, the probability of transitioning out of it is 0.
R1: Explains that all other states are transient as they can transition to the absorbing state. [2 marks]

**(f)**
A1: Correctly identifies \(\mathbf{I} = [1]\).
A1: Correctly identifies \(\mathbf{R} = \begin{pmatrix} 0.20 & 0.10 \end{pmatrix}\).
A1: Correctly identifies \(\mathbf{0} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\).
A1: Correctly identifies \(\mathbf{Q} = \begin{pmatrix} 0.70 & 0.10 \\ 0.10 & 0.80 \end{pmatrix}\). [4 marks]

**(g)**
A1: Obtains \(\mathbf{I}_2 - \mathbf{Q} = \begin{pmatrix} 0.30 & -0.10 \\ -0.10 & 0.20 \end{pmatrix}\).
M1: Attempts to calculate the determinant of \(\mathbf{I}_2 - \mathbf{Q}\).
A1: Determinant is \(0.05\).
M1: Applies the formula for the inverse of a \(2 \times 2\) matrix.
A1: Correctly simplifies to show \(\mathbf{F} = \begin{pmatrix} 4 & 2 \\ 2 & 6 \end{pmatrix}\). [5 marks]

**(h)**
A1: Correct interpretation of diagonal entries \(F_{SS} = 4\) and \(F_{RR} = 6\).
A1: Correct interpretation of off-diagonal entry \(F_{SR} = 2\).
A1: Correct interpretation of off-diagonal entry \(F_{RS} = 2\). [3 marks]

**(i)**
M1: Sums the column entries corresponding to Rural start (column 2 of \(\mathbf{F}\)).
A1: Obtains 8 years. [2 marks]

**(j)**
M1: Multiplies matrix \(\mathbf{R}\) by matrix \(\mathbf{F}\).
A1: Obtains \(\begin{pmatrix} 1.0 & 1.0 \end{pmatrix}\), proving absorption probability is 1.0 for both states. [2 marks]