MetadataPastPaper.sampleTitle

(a) Using Pythagoras' theorem on the square base: \(AC = \sqrt{12^2 + 12^2} = \sqrt{288} \approx 16.97 \approx 17.0\text{ cm}\). (b) Let \(O\) be the center of the base. The distance from the vertex \(A\) to the center \(O\) is \(AO = \frac{1}{2} AC = \frac{\sqrt{288}}{2} \approx 8.485\text{ cm}\). The angle \(\theta\) that \(VA\) makes with the base is given by \(\tan\theta = \frac{VO}{AO} = \frac{10}{8.485}\). Thus, \(\theta = \arctan(1.1785) \approx 49.7^\circ\).

[M1] for applying Pythagoras to find AC. [A1] for \(17.0\text{ cm}\). [M1] for finding AO. [M1] for setting up trig ratio \(\tan\theta = \frac{VO}{AO}\). [A1] for \(49.7^\circ\).

(a) Midpoint of \(AB\) is \(M = \left(\frac{2+8}{2}, \frac{8+10}{2}\right) = (5, 9)\). (b) Gradient of \(AB\) is \(m = \frac{10-8}{8-2} = \frac{2}{6} = \frac{1}{3}\). The gradient of the perpendicular line is \(m_{\perp} = -3\). Using the point-slope formula with midpoint \((5, 9)\): \(y - 9 = -3(x - 5) \implies y = -3x + 15 + 9 \implies y = -3x + 24\).

[M1] for calculating midpoint coordinates. [A1] for \((5, 9)\). [M1] for calculating gradient of AB. [M1] for finding perpendicular gradient. [A1] for \(y = -3x + 24\).

(a) Let \(X\) be the wingspan, \(X \sim N(32, 4.5^2)\). Using a GDC, \(P(X > 35) \approx 0.25249 \approx 0.252\). (b) We want to find \(w\) such that \(P(X < w) = 0.15\). Using the inverse normal function on a GDC: \(w \approx 27.336 \approx 27.3\text{ cm}\).

[M1] for setting up the normal probability \(P(X > 35)\). [A1] for \(0.252\). [M1] for using the inverse normal distribution. [A1] for \(27.3\).

(a) Total number of Science majors is \(35+45+20 = 100\). Total number of Humanities majors is \(20+40+40 = 100\). Total who prefer Evening is \(20+40 = 60\). Grand total is 200. Expected frequency for Humanities and Evening is \(\frac{100 \times 60}{200} = 30\). (b) Performing a \(\chi^2\) test of independence on a GDC with a \(2 \times 3\) matrix gives \(\chi^2 \approx 11.054\) and a \(p\)-value \(\approx 0.003977 \approx 0.00398\). (c) Since the \(p\)-value (\(0.00398\)) is less than the significance level (\(0.05\)), we reject the null hypothesis.

[M1] for calculating expected frequency formula. [A1] for \(30\). [M1] for entering data in GDC. [A1] for \(0.00398\). [R1] for comparing \(p\)-value to \(0.05\). [A1] for concluding to reject \(H_0\).

(a) Set the equations equal to find intersection points: \(3x^2 = 12x - 3x^2 \implies 6x^2 - 12x = 0 \implies 6x(x - 2) = 0\). Thus, \(x = 0\) and \(x = 2\). (b) The area is given by the integral \(\int_{0}^{2} ((12x - 3x^2) - 3x^2) dx = \int_{0}^{2} (12x - 6x^2) dx\). Integrating gives \([6x^2 - 2x^3]_{0}^{2} = (6(2)^2 - 2(2)^3) - 0 = (24 - 16) = 8\).

[M1] for setting equations equal. [A1] for \(x = 0\) and \(x = 2\). [M1] for setting up the correct definite integral. [M1] for correct integration. [A1] for area of \(8\).

(a) If the side perpendicular to the wall has length \(x\), the side parallel to the wall has length \(80 - 2x\). The area \(A\) is length times width: \(A = x(80 - 2x) = 80x - 2x^2\). (b) To maximize the area, find the derivative and set it to zero: \(\frac{dA}{dx} = 80 - 4x = 0 \implies 4x = 80 \implies x = 20\text{ m}\). (c) The maximum area is \(A(20) = 80(20) - 2(20)^2 = 1600 - 800 = 800\text{ m}^2\).

[M1] for establishing the side parallel to the wall is \(80 - 2x\). [A1] for showing \(A = 80x - 2x^2\). [M1] for finding derivative \(\frac{dA}{dx} = 80 - 4x\). [A1] for \(x = 20\). [A1] for \(800\text{ m}^2\).

(a) Initially, at \(t = 0\), \(P(0) = P_0 e^0 = P_0 = 500\). (b) When \(t = 3\), \(P(3) = 1200 \implies 500 e^{3k} = 1200 \implies e^{3k} = 2.4 \implies 3k = \ln(2.4) \approx 0.875468 \implies k \approx 0.29182 \approx 0.292\). (c) When \(t = 8\), \(P(8) = 500 e^{8 \times 0.29182} \approx 5162.75\). To the nearest hundred, this is \(5200\).

[A1] for \(P_0 = 500\). [M1] for setting up equation \(500 e^{3k} = 1200\). [A1] for \(k \approx 0.292\). [M1] for substituting \(t = 8\) into the formula. [A1] for \(5200\) (rounded to the nearest hundred).

(a) Using the compound interest formula: \(A = P\left(1 + \frac{r}{100k}\right)^{kn}\), where \(P = 4000\), \(r = 3.5\), \(k = 12\), and \(n = 5\). \(A = 4000\left(1 + \frac{3.5}{1200}\right)^{60} \approx 4763.76\). To the nearest dollar, the amount is \(\$4764\). (b) To double the investment, we need \(A = 8000\). So \(4000\left(1 + \frac{3.5}{1200}\right)^{12n} = 8000 \implies \left(1 + \frac{3.5}{1200}\right)^{12n} = 2\). Using GDC or logarithms: \(12n = \frac{\ln(2)}{\ln(1 + 3.5/1200)} \approx 238.01\text{ months}\). Thus, \(n \approx 19.83\text{ years}\). The minimum number of complete years for the money to double is \(20\text{ years}\).

[M1] for substituting correct values into the compound interest formula. [A1] for \(\$4764\). [M1] for setting up the inequality or equation for doubling. [M1] for solving for \(n\). [A1] for \(20\).

(a) Since \(ABCD\) is a rectangle with \(AB = 8\text{ cm}\) and \(BC = 6\text{ cm}\), we can find the diagonal \(AC\) using Pythagoras' Theorem:
\(AC = \sqrt{AB^2 + BC^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10\text{ cm}\).

(b) The center of the base \(O\) is the midpoint of the diagonal \(AC\). Therefore, \(AO = \frac{AC}{2} = \frac{10}{2} = 5\text{ cm}\).
Consider the right-angled triangle \(VOA\), where \(VO\) is the vertical height of \(12\text{ cm}\) and \(AO\) is \(5\text{ cm}\). The angle between the edge \(VA\) and the base is the angle \(\angle VAO\).
\(\tan(\angle VAO) = \frac{VO}{AO} = \frac{12}{5} = 2.4\)
\(\angle VAO = \arctan(2.4) \approx 67.3801^\circ\).
To 3 significant figures, the angle is \(67.4^\circ\) (or \(1.18\text{ radians}\)).

(a)
[M1] for attempting to use Pythagoras' Theorem to find \(AC\).
[A1] for \(AC = 10\text{ cm}\).

(b)
[M1] for finding \(AO = 5\text{ cm}\).
[M1] for setting up a correct trigonometric ratio (e.g. \(\tan\theta = \frac{12}{5}\)).
[A1] for correct angle: \(67.4^\circ\) (accept \(67.38^\circ\) or \(1.18\text{ rad}\)).

(a) The diameter of the wheel is \(60\text{ m}\), so the radius is \(r = 30\text{ m}\). This gives the amplitude \(a = 30\).
The lowest point is \(5\text{ m}\) and the highest point is \(5 + 60 = 65\text{ m}\). The vertical shift (midline) is \(d = \frac{65 + 5}{2} = 35\text{ m}\).
The period is \(12\text{ minutes}\), so \(b = \frac{2\pi}{\text{Period}} = \frac{2\pi}{12} = \frac{\pi}{6}\).
Thus, the model is \(h(t) = -30\cos\left(\frac{\pi}{6}t\right) + 35\).

(b) Substitute \(t = 5\) into the model:
\(h(5) = -30\cos\left(\frac{5\pi}{6}\right) + 35\)
Using a GDC (set in radians mode):
\(h(5) = -30\left(-\frac{\sqrt{3}}{2}\right) + 35 = 15\sqrt{3} + 35 \approx 60.9807\text{ m}\).
To 3 significant figures, the height is \(61.0\text{ m}\).

(a)
[A1] for \(a = 30\).
[M1] for attempting to find \(b\) using \(\frac{2\pi}{12}\).
[A1] for \(b = \frac{\pi}{6}\) (or \(0.524\)).
[A1] for \(d = 35\).

(b)
[M1] for substituting \(t = 5\) into their model.
[A1] for \(61.0\text{ m}\) (accept answers rounding to \(61\) or \(61.0\)).

(a) Find the midpoint of \(AB\):
\(M_{AB} = \left(\frac{1+5}{2}, \frac{2+8}{2}\right) = (3, 5)\).
Find the gradient of \(AB\):
\(m_{AB} = \frac{8 - 2}{5 - 1} = \frac{6}{4} = 1.5\).
The gradient of the perpendicular bisector is \(m = -\frac{1}{1.5} = -\frac{2}{3}\).
Using the point-slope formula with \((3, 5)\):
\(y - 5 = -\frac{2}{3}(x - 3) \Rightarrow y - 5 = -\frac{2}{3}x + 2 \Rightarrow y = -\frac{2}{3}x + 7\).

(b) To find the circumcenter, we can find the intersection of the perpendicular bisector of \(AB\) with another perpendicular bisector, for example, the perpendicular bisector of \(AC\).
Since \(A(1, 2)\) and \(C(9, 2)\) lie on the horizontal line \(y = 2\), the perpendicular bisector of \(AC\) is a vertical line passing through their midpoint.
Midpoint of \(AC = \left(\frac{1+9}{2}, 2\right) = (5, 2)\).
So the perpendicular bisector of \(AC\) is the vertical line \(x = 5\).
Substitute \(x = 5\) into the equation of the perpendicular bisector of \(AB\):
\(y = -\frac{2}{3}(5) + 7 = -\frac{10}{3} + \frac{21}{3} = \frac{11}{3} \approx 3.67\).
Thus, the coordinates of the vertex are \((5, 3.67)\) (or \((5, \frac{11}{3})\)).

(a)
[M1] for finding the midpoint \((3, 5)\) of \(AB\).
[M1] for finding the gradient of \(AB\) and taking the negative reciprocal to find the perpendicular gradient \(-\frac{2}{3}\).
[A1] for the equation \(y = -\frac{2}{3}x + 7\) (or equivalent).

(b)
[M1] for identifying the perpendicular bisector of \(AC\) as \(x = 5\) (or constructing the perpendicular bisector of \(BC\) as \(y = \frac{2}{3}x + 0.333\)).
[M1] for equating the two lines to find the intersection.
[A1] for the coordinates \((5, 3.67)\) or \((5, \frac{11}{3})\).

(a) The null hypothesis \(H_0\) states that the two categorical variables are independent.
\(H_0\): Preference for a hot drink is independent of gender.

(b) To find the expected frequency, we use:
\(E = \frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}}\).

Row total for Males: \(45 + 30 + 15 = 90\).
Column total for Coffee: \(45 + 35 = 80\).
Grand total: \(200\).

\(E = \frac{90 \times 80}{200} = 36\).

(c) Using a GDC, input the \(2 \times 3\) matrix of observed frequencies and run a \(\chi^2\) two-way test:
Matrix:
\(\begin{pmatrix} 45 & 30 & 15 \\ 35 & 45 & 30 \\ \end{pmatrix}\)

The GDC output gives:
\(\chi^2 \approx 7.3232\) with degrees of freedom \(df = (2-1)(3-1) = 2\).
\(p\)-value \(\approx 0.0257\) (or \(2.57\%\)).

(a)
[A1] for stating the null hypothesis clearly, mentioning independence and both variables (drink preference and gender).

(b)
[M1] for calculating Row, Column and Grand totals (e.g. \(90\) and \(80\)).
[A1] for \(36\).

(c)
[M1] for entering correct observed frequencies into a matrix on GDC.
[A1] for \(p \approx 0.0257\) (accept \(0.026\) or \(0.02569\)).

Let \(X\) be the weight of a pear, so \(X \sim N(150, 15^2)\).

(a) We need to find \(P(X > 165)\).
Using a GDC (Normal Cumulative Distribution):
\(P(X > 165) \approx 0.158655\).
To 3 significant figures, \(P(X > 165) = 0.159\).

(b) We want to find the value \(w\) such that \(P(X > w) = 0.10\), which is equivalent to \(P(X \le w) = 0.90\).
Using a GDC (Inverse Normal Distribution) with area \(0.90\), mean \(150\), and standard deviation \(15\):
\(w \approx 169.224\text{ grams}\).
To the nearest gram, the minimum weight is \(169\text{ grams}\) (or \(169\text{ grams}\) to 3 significant figures).

(a)
[M1] for setting up the normal distribution calculation, e.g., writing \(P(X > 165)\) or using the standard Z-value \(Z = \frac{165-150}{15} = 1\).
[A1] for \(0.159\) (accept \(0.1587\)).

(b)
[M1] for setting up the equation \(P(X > w) = 0.10\) or \(P(X \le w) = 0.90\).
[M1] for applying inverse normal on GDC.
[A1] for \(169\text{ g}\) (accept \(169.2\) or \(169.22\)).

(a) The volume of a cylinder is given by \(V = \pi r^2 h\). Since \(V = 500\), we have:
\(\pi r^2 h = 500 \Rightarrow h = \frac{500}{\pi r^2}\).

The total surface area of a closed cylinder is given by:
\(A = 2\pi r^2 + 2\pi r h\).

Substitute the expression for \(h\) into the surface area formula:
\(A = 2\pi r^2 + 2\pi r \left(\frac{500}{\pi r^2}\right)\).

Simplify the second term:
\(A = 2\pi r^2 + \frac{1000}{r}\) (as required).

(b) To minimize the surface area, find the derivative \(A'(r)\) and set it to \(0\):
\(A'(r) = 4\pi r - \frac{1000}{r^2}\).

Set \(A'(r) = 0\):
\(4\pi r - \frac{1000}{r^2} = 0 \Rightarrow 4\pi r^3 = 1000 \Rightarrow r^3 = \frac{250}{\pi}\).

Using a GDC:
\(r = \sqrt[3]{\frac{250}{\pi}} \approx 4.30127\text{ cm}\).

To 3 significant figures, the optimal radius is \(4.30\text{ cm}\).

(a)
[M1] for using the volume formula to express \(h\) in terms of \(r\).
[M1] for substituting the expression for \(h\) into the total surface area formula.
[A1] for clear algebraic steps leading to \(A(r) = 2\pi r^2 + \frac{1000}{r}\).

(b)
[M1] for differentiating \(A(r)\) to find \(A'(r) = 4\pi r - 1000 r^{-2}\).
[M1] for setting \(A'(r) = 0\) (or using GDC solver/graphical minimum).
[A1] for \(r \approx 4.30\text{ cm}\) (accept \(4.301\)).

(a) The step size is \(h = 2\).
Using the trapezoidal rule:
\(\text{Volume} \approx \frac{h}{2} \left[ R(0) + 2R(2) + 2R(4) + 2R(6) + R(8) \right]\)
\(\text{Volume} \approx \frac{2}{2} \left[ 12 + 2(15) + 2(17) + 2(14) + 10 \right]\)
\(\text{Volume} \approx 1 \cdot \left[ 12 + 30 + 34 + 28 + 10 \right]\)
\(\text{Volume} \approx 114\text{ liters}\).

(b) Since the actual rate function \(R(t)\) is concave down, any straight-line chord connecting two points on the curve lies entirely below the curve itself. Therefore, the area of each trapezoid is smaller than the actual area under the curve, making the trapezoidal rule approximation an underestimate.

(a)
[M1] for identifying the interval width \(h = 2\).
[M1] for substituting values correctly into the trapezoidal rule formula.
[A1] for \(114\text{ liters}\).

(b)
[R1] for stating "underestimate".
[R1] for a valid geometric explanation mentioning that the chords/trapezoid tops lie below a concave down curve.

(a) At \(t = 0\), the temperature is \(85^\circ\text{C}\).
\(T(0) = 22 + A e^0 = 22 + A = 85\)
\(A = 85 - 22 = 63\).

(b) At \(t = 5\), the temperature is \(60^\circ\text{C}\).
\(60 = 22 + 63 e^{-5k}\)
\(38 = 63 e^{-5k}\)
\(e^{-5k} = \frac{38}{63}\)
\(-5k = \ln\left(\frac{38}{63}\right)\)
\(k = -\frac{1}{5}\ln\left(\frac{38}{63}\right) \approx 0.10110\).
To 3 significant figures, \(k \approx 0.101\).

(c) Substitute \(t = 15\) into the model:
\(T(15) = 22 + 63 e^{-15(0.10110)}\)
Using GDC:
\(T(15) = 22 + 63 (0.21935) \approx 35.819^\circ\text{C}\).
To 3 significant figures, the temperature is \(35.8^\circ\text{C}\).

(a)
[A1] for \(A = 63\).

(b)
[M1] for setting up the equation \(60 = 22 + 63e^{-5k}\).
[M1] for taking logarithms or using GDC solver.
[A1] for \(k \approx 0.101\).

(c)
[M1] for substituting \(t = 15\) and their \(k\) into the model.
[A1] for \(35.8^\circ\text{C}\) (accept \(35.8\)).

(a) Midpoint of \(AB\) is \(M_{AB} = \left(\frac{2+8}{2}, \frac{8+10}{2}\right) = (5, 9)\).
Gradient of \(AB\) is \(m_{AB} = \frac{10-8}{8-2} = \frac{2}{6} = \frac{1}{3}\).
Gradient of the perpendicular bisector is \(m_{\perp AB} = -3\).
Equation: \(y - 9 = -3(x - 5) \implies y = -3x + 24\).

(b) Midpoint of \(BC\) is \(M_{BC} = \left(\frac{8+6}{2}, \frac{10+2}{2}\right) = (7, 6)\).
Gradient of \(BC\) is \(m_{BC} = \frac{2-10}{6-8} = \frac{-8}{-2} = 4\).
Gradient of the perpendicular bisector is \(m_{\perp BC} = -0.25\).
Equation: \(y - 6 = -0.25(x - 7) \implies y = -0.25x + 7.75\).

(c) Equating the two perpendicular bisectors:
\(-3x + 24 = -0.25x + 7.75\)
\(16.25 = 2.75x \implies x = \frac{65}{11} \approx 5.91\).
Substituting back:
\(y = -3(5.909) + 24 \approx 6.27\).
Circumcenter \(V\) is \((5.91, 6.27)\).

(d) Distances from \(P(5,5)\) to each center:
\(PA = \sqrt{(5-2)^2 + (5-8)^2} = \sqrt{18} \approx 4.24\text{ km}\)
\(PB = \sqrt{(5-8)^2 + (5-10)^2} = \sqrt{34} \approx 5.83\text{ km}\)
\(PC = \sqrt{(5-6)^2 + (5-2)^2} = \sqrt{10} \approx 3.16\text{ km}\)
\(PD = \sqrt{(5-4)^2 + (5-4)^2} = \sqrt{2} \approx 1.41\text{ km}\)
Since \(PD\) is the shortest distance, \(P\) belongs to the Voronoi cell of center \(D\).

(e) To be as far away as possible from \(A\), \(B\), and \(C\), the park should be placed at the circumcenter \(V\), which is \((5.91, 6.27)\).
The distance is \(VA = \sqrt{(5.91-2)^2 + (6.27-8)^2} \approx 4.28\text{ km}\).

(a) M1 for finding the midpoint of \(AB\), M1 for finding the perpendicular gradient, A1 for the correct equation. (b) M1 for finding the midpoint of \(BC\), M1 for finding the perpendicular gradient, A1 for the correct equation. (c) M1 for equating bisectors, A1 for \(x\)-coordinate, A1 for \(y\)-coordinate. (d) M2 for calculating distances (at least two correctly), A1 for identifying \(PD\) is shortest, A1 for stating \(P\) is in cell \(D\). (e) R1 for identifying circumcenter, A1 for correct coordinates, A1 for correct distance.

(a) The diagram should show the horizontal triangle \(ABF\) with North arrows at \(A\) and \(B\). Bearings: \(B\) from \(A\) at \(060^\circ\), \(F\) from \(A\) at \(020^\circ\), and \(F\) from \(B\) at \(310^\circ\).

(b) In triangle \(ABF\):
\(\angle FAB = 060^\circ - 020^\circ = 40^\circ\).
Bearing of \(A\) from \(B\) is \(060^\circ + 180^\circ = 240^\circ\).
Bearing of \(F\) from \(B\) is \(310^\circ\).
Thus, \(\angle ABF = 310^\circ - 240^\circ = 70^\circ\).
So, \(\angle AFB = 180^\circ - (40^\circ + 70^\circ) = 70^\circ\).

(c) Using the sine rule in triangle \(ABF\):
\(\frac{AF}{\sin(\angle ABF)} = \frac{AB}{\sin(\angle AFB)}\)
\(\frac{AF}{\sin(70^\circ)} = \frac{150}{\sin(70^\circ)} \implies AF = 150\text{ m}\).

(d) In the vertical right-angled triangle \(TAF\):
\(\tan(25^\circ) = \frac{TF}{AF} = \frac{h}{150}\)
\(h = 150 \tan(25^\circ) \approx 69.9\text{ m}\).

(e) In triangle \(ABF\), since \(\angle ABF = \angle AFB = 70^\circ\), the triangle is isosceles, meaning \(AF = AB = 150\text{ m}\).
Using the sine rule or properties of isosceles triangles to find \(BF\):
\(\frac{BF}{\sin(40^\circ)} = \frac{150}{\sin(70^\circ)} \implies BF = 150 \times \frac{\sin(40^\circ)}{\sin(70^\circ)} \approx 102.6\text{ m}\).
Now, in the vertical right-angled triangle \(TBF\):
\(\tan(\theta) = \frac{TF}{BF} = \frac{69.946}{102.606} \approx 0.6817\)
\(\theta = \arctan(0.6817) \approx 34.3^\circ\).

(a) A1 for relative positions, A1 for angles/bearings correctly labeled. (b) M1 for \(\angle FAB = 40^\circ\), M1 for \(\angle ABF = 70^\circ\), A1 for \(\angle AFB = 70^\circ\). (c) M1 for sine rule setup, A1 for substituting values, M1 for calculation, A1 for \(AF = 150\text{ m}\). (d) M1 for using tangent ratio, A1 for setting up formula, A1 for \(69.9\text{ m}\). (e) M1 for identifying need for \(BF\), A1 for \(BF \approx 102.6\text{ m}\), M1 for tangent setup, A1 for \(34.3^\circ\).

(a) \(H_0\): Sleep duration and preferred revision method are independent.
\(H_1\): Sleep duration and preferred revision method are dependent.

(b) Expected frequency = \(\frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}} = \frac{120 \times 80}{250} = 38.4\).

(c) Degrees of freedom = \((\text{rows} - 1) \times (\text{columns} - 1) = (3 - 1) \times (3 - 1) = 4\).

(d) Using a GDC on the contingency table:
\(\chi^2\) test statistic \(\approx 7.10\) (3 s.f.)
\(p\)-value \(\approx 0.131\) (3 s.f.)

(e) Since \(p\text{-value} = 0.131 > 0.05\), we fail to reject the null hypothesis. There is insufficient evidence to conclude that sleep duration and revision methods are dependent.

(f) (i) \(P(\text{less than 6 hours}) = \frac{90}{250} = 0.36\).

(ii) \(P(\text{Active Recall} \mid \text{more than 8 hours}) = \frac{15}{40} = 0.375\).

(iii) Let \(A\) be "sleeping less than 6 hours" and \(B\) be "preferring Group Discussion".
\(P(A) = \frac{90}{250} = 0.36\)
\(P(B) = \frac{60}{250} = 0.24\)
\(P(A \cap B) = \frac{15}{250} = 0.06\)
\(P(A) \times P(B) = 0.36 \times 0.24 = 0.0864\).
Since \(P(A \cap B) \neq P(A) \times P(B)\), the events are not independent.

(a) A1 for correct null hypothesis, A1 for correct alternative hypothesis. (b) M1 for correct formula setup, A1 for 38.4. (c) A1 for 4. (d) A2 for \(\chi^2 \approx 7.10\) (award A1 for 7.1), A1 for \(p \approx 0.131\). (e) R1 for comparing p-value to 0.05, A1 for correct conclusion. (f)(i) A1 for fraction, A1 for 0.36. (f)(ii) A1 for fraction, A1 for 0.375. (f)(iii) M1 for computing products of individual probabilities, A1 for comparison and clear conclusion.

(a) Independent variable: Temperature (\(T\)). Dependent variable: Sales (\(S\)).

(b) (i) Using GDC with the 8 pairs of data:
\(r \approx 0.997\) (0.99652...)

(ii) Ranking the data:
\(T\) ranks: 1, 2, 3, 4, 5, 6, 7, 8
\(S\) ranks: 1, 2, 3, 4, 5, 6, 7, 8
Since both lists of ranks are identical, \(r_s = 1.0\).

(c) Since \(r\) is extremely close to 1, it indicates an extremely strong positive linear correlation between daily average temperature and iced coffee sales.

(d) Using linear regression on the GDC:
Slope \(a \approx 15.5\) (15.5147...)
Intercept \(b \approx -111\) (-111.057...)
So, the regression equation is \(S = 15.5T - 111\) (or \(S = 15.5T - 111.1\)).

(e) (i) At \(T = 25\):
\(S = 15.5147(25) - 111.057 \approx 277\) USD (or \(276.81\) USD).

(ii) At \(T = 42\):
\(S = 15.5147(42) - 111.057 \approx 541\) USD (or \(540.56\) USD).

(f) The estimate at \(25^\circ\text{C}\) is highly reliable because \(25\) is within the range of original data points (interpolation) and \(r\) is close to 1.
The estimate at \(42^\circ\text{C}\) is unreliable because it lies outside the data range (extrapolation), and the linear relationship may not continue at extreme temperatures.

(a) A1 for correct identification of independent and dependent variables. (b)(i) A2 for \(r \approx 0.997\). (b)(ii) M1 for setting up the ranks, A2 for \(r_s = 1.0\). (c) A1 for 'strong', A1 for 'positive' and 'linear'. (d) A1 for \(a \approx 15.5\), A1 for \(b \approx -111\), A1 for final equation. (e)(i) M1 for substituting 25, A1 for correct value. (e)(ii) A1 for correct value. (f) A1 for evaluating \(25^\circ\text{C}\) as reliable (interpolation), A1 for evaluating \(42^\circ\text{C}\) as unreliable (extrapolation).

(a) The volume of a cylinder is \(V = \pi r^2 h\).
Given \(V = 64\pi\):
\(\pi r^2 h = 64\pi \implies r^2 h = 64 \implies h = \frac{64}{r^2}\).

(b) Since there is no lid, the surface area \(A\) is the sum of the circular base and the curved side:
\(A = \pi r^2 + 2\pi r h\).
Substituting \(h = \frac{64}{r^2}\):
\(A = \pi r^2 + 2\pi r \left(\frac{64}{r^2}\right) = \pi r^2 + \frac{128\pi}{r}\).

(c) Differentiating \(A\) with respect to \(r\):
\(\frac{\text{d}A}{\text{d}r} = 2\pi r - 128\pi r^{-2} = 2\pi r - \frac{128\pi}{r^2}\).

(d) For minimum surface area, set \(\frac{\text{d}A}{\text{d}r} = 0\):
\(2\pi r - \frac{128\pi}{r^2} = 0\)
\(2\pi r = \frac{128\pi}{r^2}\)
\(r^3 = 64 \implies r = 4\text{ dm}\).

(e) Substituting \(r = 4\) into the surface area equation:
\(A(4) = \pi (4)^2 + \frac{128\pi}{4} = 16\pi + 32\pi = 48\pi\text{ dm}^2\).

(f) The cost function \(C\) is given by:
\(C = 1.50 (\pi r^2) + 1.00 \left(\frac{128\pi}{r}\right) = 1.5\pi r^2 + \frac{128\pi}{r}\).
To minimize cost, find \(\frac{\text{d}C}{\text{d}r}\) and set it to 0:
\(\frac{\text{d}C}{\text{d}r} = 3\pi r - \frac{128\pi}{r^2} = 0\)
\(3\pi r = \frac{128\pi}{r^2} \implies r^3 = \frac{128}{3} \approx 42.667\)
\(r = \sqrt[3]{\frac{128}{3}} \approx 3.50\text{ dm}\).

(a) M1 for stating volume formula, A1 for rearranging and matching given result. (b) M1 for correct surface area formula, A1 for substitution, A1 for matching given result. (c) M2 for differentiating each term, A1 for correct final expression. (d) M1 for setting derivative to 0, A1 for solving equation, A1 for \(r = 4\text{ dm}\). (e) M1 for substituting \(r = 4\) back, A1 for \(48\pi\text{ dm}^2\). (f) M1 for establishing correct cost function, A1 for differentiating and setting to 0, A1 for \(r \approx 3.50\text{ dm}\).

(a) Using the trapezoidal rule with interval width \(h = 2\):
\(\text{Distance} \approx \frac{h}{2} \left[ v(0) + v(10) + 2(v(2) + v(4) + v(6) + v(8)) \right]\)
\(\text{Distance} \approx \frac{2}{2} \left[ 0 + 1.2 + 2(3.5 + 6.2 + 7.8 + 5.1) \right]\)
\(\text{Distance} \approx 1.2 + 2(22.6) = 1.2 + 45.2 = 46.4\text{ meters}\).

(b) According to the model:
\(f(6) = -0.02(6)^3 + 0.12(6)^2 + 1.2(6) = -4.32 + 4.32 + 7.2 = 7.2\text{ m s}^{-1}\).
Percentage error:
\(\text{Error} = \frac{|7.2 - 7.8|}{7.8} \times 100\% = \frac{0.6}{7.8} \times 100\% \approx 7.69\%\).

(c) The integral expression is:
\(\int_{0}^{10} (-0.02t^3 + 0.12t^2 + 1.2t)\text{ d}t\).

(d) Integrating term-by-term:
\(\int_{0}^{10} (-0.02t^3 + 0.12t^2 + 1.2t)\text{ d}t = \left[ -0.005t^4 + 0.04t^3 + 0.6t^2 \right]_{0}^{10}\)
\(= \left( -0.005(10000) + 0.04(1000) + 0.6(100) \right) - 0\)
\(= -50 + 40 + 60 = 50\text{ meters}\).

(e) To find maximum velocity, set \(f'(t) = 0\):
\(f'(t) = -0.06t^2 + 0.24t + 1.2 = 0\)
\(t^2 - 4t - 20 = 0\)
Using quadratic formula:
\(t = \frac{4 \pm \sqrt{16 - 4(1)(-20)}}{2} = \frac{4 \pm \sqrt{96}}{2} = 2 \pm \sqrt{24}\).
Since \(t \ge 0\), \(t = 2 + \sqrt{24} \approx 6.90\text{ seconds}\).
At \(t \approx 6.90\), the velocity is:
\(f(6.899) = -0.02(6.899)^3 + 0.12(6.899)^2 + 1.2(6.899) \approx 7.42\text{ m s}^{-1}\).

(a) M1 for stating formula with correct weights, A1 for substituting values, M1 for calculation, A1 for 46.4. (b) M1 for finding model value 7.2, M1 for percentage error formula, A1 for 7.69%. (c) A1 for limits, A1 for integrand. (d) M1 for correct integration antiderivative, A1 for substituting limits, A1 for 50. (e) M1 for differentiating, A1 for setting to 0, A1 for finding \(t \approx 6.90\), A1 for maximum velocity \(\approx 7.42\text{ m s}^{-1}\).

(a) Using the compound interest formula:
\(FV = PV \left(1 + \frac{r}{100k}\right)^{nk}\)
\(FV = 12000 \left(1 + \frac{4.5}{1200}\right)^{120} = 12000(1.00375)^{120} \approx \$18,803.90\).

(b) (i) Using TVM Solver parameters on a GDC:
\(N = 120\)
\(I\% = 3.6\)
\(PV = 0\)
\(PMT = -250\)
\(P/Y = 12\)
\(C/Y = 12\)
Solving for \(FV\) yields \(FV \approx \$36,053.69\).

(ii) Total amount of deposits = \(250 \times 120 = \$30,000\).
Interest earned = \(\$36,053.69 - \$30,000 = \$6,053.69\).

(c) (i) Total down payment available = \(\$18,803.90 + \$36,053.69 = \$54,857.59\).

(ii) Percentage = \(\frac{54857.59}{350000} \times 100\% \approx 15.7\%\) (or \(15.67\%\)).

(d) Remaining mortgage balance (loan amount) = \(\$350,000 - \$54,857.59 = \$295,142.41\).
Using TVM Solver for the mortgage payment:
\(N = 300\) (25 years \(\times\) 12 months)
\(I\% = 5.2\)
\(PV = 295,142.41\)
\(FV = 0\)
\(P/Y = 12\)
\(C/Y = 12\)
Solving for \(PMT\) yields \(PMT \approx -\$1,759.67\).
Thus, her monthly payment is \(\$1,759.67\).

(a) M1 for correct interest formula setup, A1 for substituting correct values, A1 for correct final answer. (b)(i) M2 for entering correct TVM values, A2 for \(\$36,053.69\) (accept slightly different values due to intermediate rounding). (b)(ii) M1 for calculating total deposits, A1 for final interest value. (c)(i) M1 for adding both values, A1 for correct sum. (c)(ii) M1 for percentage ratio setup, A1 for \(15.7\%\). (d) M1 for establishing correct loan amount, M1 for TVM solver setup, A1 for monthly payment of \(\$1,759.67\).

**(a)**
The transition matrix is formed by the given probabilities of moving from the state in each row to the state in each column:

\(\mathbf{T} = \begin{pmatrix} 0.5 & 0.3 & 0.2 \\ 0.3 & 0.6 & 0.1 \\ 0.4 & 0.2 & 0.4 \end{pmatrix}\)

**(b)**
With \(\mathbf{s}_0 = [120, 80, 50]\):

\(\mathbf{s}_1 = \mathbf{s}_0 \mathbf{T} = [120(0.5) + 80(0.3) + 50(0.4), \, 120(0.3) + 80(0.6) + 50(0.2), \, 120(0.2) + 80(0.1) + 50(0.4)] = [104, 94, 52]\)

\(\mathbf{s}_2 = \mathbf{s}_1 \mathbf{T} = [104(0.5) + 94(0.3) + 52(0.4), \, 104(0.3) + 94(0.6) + 52(0.2), \, 104(0.2) + 94(0.1) + 52(0.4)] = [101, 98, 51]\)

So there are expected to be 101 cars at Downtown, 98 cars at the Airport, and 51 cars in the Suburb on Wednesday morning.

**(c)**
To find the steady state vector \(\boldsymbol{\pi} = [x, y, z]\) where \(x+y+z=1\):

\([x, y, z] \begin{pmatrix} 0.5 & 0.3 & 0.2 \\ 0.3 & 0.6 & 0.1 \\ 0.4 & 0.2 & 0.4 \end{pmatrix} = [x, y, z]\)

This leads to the system of equations:

1) \(0.5x + 0.3y + 0.4z = x \implies -0.5x + 0.3y + 0.4z = 0\)
2) \(0.3x + 0.6y + 0.2z = y \implies 0.3x - 0.4y + 0.2z = 0\)
3) \(x + y + z = 1\)

Solving equations (1) and (2):
Multiply Eq (1) by 2: \(-x + 0.6y + 0.8z = 0\)
Multiply Eq (2) by 4: \(1.2x - 1.6y + 0.8z = 0\)

Subtracting the two:
\(2.2x - 2.2y = 0 \implies x = y\)

Substituting \(x = y\) into Eq (1):
\(-0.5x + 0.3x + 0.4z = 0 \implies -0.2x + 0.4z = 0 \implies x = 2z\)

Since \(x + y + z = 1 \implies 2z + 2z + z = 1 \implies 5z = 1\):
\(z = 0.2\), \(x = 0.4\), \(y = 0.4\)

Therefore, \(\boldsymbol{\pi} = [0.4, 0.4, 0.2]\).

**(d)**
With a fleet of 250 cars, the steady-state distribution is:
- Downtown: \(250 \times 0.4 = 100\) cars
- Airport: \(250 \times 0.4 = 100\) cars
- Suburb: \(250 \times 0.2 = 50\) cars

Expected daily profit:
\(\text{Profit} = 100(\$50) + 100(\$70) + 50(\$40) = 5000 + 7000 + 2000 = \$14000\)

**(e)**
The transition probabilities for the active states are scaled down by the retention probability \((1 - p_{\text{ret}})\).

For \(D\) (5% retired, 95% active):
- D to D: \(0.95 \times 0.5 = 0.475\)
- D to A: \(0.95 \times 0.3 = 0.285\)
- D to S: \(0.95 \times 0.2 = 0.190\)

For \(A\) (2% retired, 98% active):
- A to D: \(0.98 \times 0.3 = 0.294\)
- A to A: \(0.98 \times 0.6 = 0.588\)
- A to S: \(0.98 \times 0.1 = 0.098\)

For \(S\) (10% retired, 90% active):
- S to D: \(0.90 \times 0.4 = 0.360\)
- S to A: \(0.90 \times 0.2 = 0.180\)
- S to S: \(0.90 \times 0.4 = 0.360\)

The submatrix \(\mathbf{Q}\) of transient states is:

\(\mathbf{Q} = \begin{pmatrix} 0.475 & 0.285 & 0.190 \\ 0.294 & 0.588 & 0.098 \\ 0.360 & 0.180 & 0.360 \end{pmatrix}\)

The fundamental matrix is \(\mathbf{F} = (\mathbf{I} - \mathbf{Q})^{-1}\).

\(\mathbf{I} - \mathbf{Q} = \begin{pmatrix} 0.525 & -0.285 & -0.190 \\ -0.294 & 0.412 & -0.098 \\ -0.360 & -0.180 & 0.640 \end{pmatrix}\)

We solve \((\mathbf{I} - \mathbf{Q})\mathbf{t} = \mathbf{1}\) where \(\mathbf{t} = [t_D, t_A, t_S]^T\):
Using GDC to solve the system of linear equations:

\(t_D \approx 20.871\)
\(t_A \approx 21.954\)
\(t_S \approx 19.477\)

The expected number of days a car starting at Downtown remains in service is approximately 20.9 days.

**(f)**
We want the lifespan starting at Downtown, \(t_D\), to be at least 25 days. Set \(t_D = 25\).
The equations for transient states \(A\) and \(S\) do not contain \(p_D\):

2) \(-0.294(25) + 0.412 t_A - 0.098 t_S = 1 \implies 0.412 t_A - 0.098 t_S = 8.35\)
3) \(-0.360(25) - 0.180 t_A + 0.640 t_S = 1 \implies -0.180 t_A + 0.640 t_S = 10.0\)

Solving this \(2 \times 2\) linear system:
From (3): \(t_S = \frac{10 + 0.180 t_A}{0.640} = 15.625 + 0.28125 t_A\)
Substitute into (2):
\(0.412 t_A - 0.098(15.625 + 0.28125 t_A) = 8.35\)
\(0.3844375 t_A = 9.88125 \implies t_A \approx 25.7031\)
Then \(t_S \approx 22.8540\).

Now, substitute \(t_D = 25\), \(t_A \approx 25.7031\), and \(t_S \approx 22.8540\) into the first row equation of the system \((\mathbf{I} - \mathbf{Q})\mathbf{t} = \mathbf{1}\):

\((1 - 0.5(1-p_D))(25) - 0.3(1-p_D) t_A - 0.2(1-p_D) t_S = 1\)

Let \(k = 1 - p_D\):
\((1 - 0.5k)(25) - 0.3k(25.7031) - 0.2k(22.8540) = 1\)
\(25 - k [12.5 + 7.7109 + 4.5708] = 1\)
\(24.7817 k = 24 \implies k \approx 0.968456\)

Since \(p_D = 1 - k\):
\(p_D \approx 1 - 0.968456 = 0.031544\) or \(3.15\%\).

Thus, the maximum daily retirement rate at Downtown is 3.15%.

**(a)**
M1 for attempting to set up a \(3 \times 3\) matrix.
A1 for the correct transition matrix \(\mathbf{T}\). [2 marks]

**(b)**
M1 for finding \(\mathbf{s}_1 = \mathbf{s}_0 \mathbf{T}\) or evaluating \(\mathbf{s}_2 = \mathbf{s}_0 \mathbf{T}^2\).
A1 for calculating the correct values \([101, 98, 51]\).
A1 for communicating the final distribution clearly in context. [3 marks]

**(c)**
M1 for setting up the steady-state equation \(\boldsymbol{\pi} \mathbf{T} = \boldsymbol{\pi}\).
M1 for attempting to solve the resulting system of equations alongside \(x+y+z=1\).
A1 for obtaining \(x = y\) and \(x = 2z\) (or equivalent relationships).
A2 for the correct steady-state vector \([0.4, 0.4, 0.2]\). [5 marks]

**(d)**
M1 for finding the long-run fleet distribution \([100, 100, 50]\).
M1 for multiplying by corresponding profit values.
A2 for \(\$14000\). [4 marks]

**(e)**
M1 for calculating correct adjusted transition probabilities for each transient state (e.g. \(0.475, 0.285, 0.190\)).
A1 for the correct submatrix \(\mathbf{Q}\).
M1 for recognizing the fundamental matrix formula \(\mathbf{F} = (\mathbf{I} - \mathbf{Q})^{-1\).
M1 for setting up the system of linear equations \((\mathbf{I} - \mathbf{Q})\mathbf{t} = \mathbf{1}\).
M2 for solving the system of equations using a GDC.
A1 for \(t_D \approx 20.9\) days. [7 marks]

**(f)**
M1 for setting \(t_D = 25\) in the system of equations.
M1 for isolating equations 2 and 3 to solve for \(t_A\) and \(t_S\).
A1 for \(t_A \approx 25.7\) and \(t_S \approx 22.9\).
M1 for substituting these values into the first row equation using variable \(k = 1-p_D\).
A1 for obtaining \(k \approx 0.9685\).
A1 for the correct percentage: \(3.15\%\). [6 marks]

**(a)**
The coefficient matrix is:

\(\mathbf{A} = \begin{pmatrix} -2 & 1 \\ 2 & -3 \end{pmatrix}\)

To find the eigenvalues, solve \(\det(\mathbf{A} - \lambda \mathbf{I}) = 0\):

\(\det \begin{pmatrix} -2-\lambda & 1 \\ 2 & -3-\lambda \end{pmatrix} = (-2-\lambda)(-3-\lambda) - 2 = 0\)
\(\lambda^2 + 5\lambda + 6 - 2 = 0 \implies
\lambda^2 + 5\lambda + 4 = 0\)
\((\lambda+1)(\lambda+4) = 0\)

So the eigenvalues are \(\lambda_1 = -1\) and \(\lambda_2 = -4\).

Since both eigenvalues are real and negative, the origin \((0,0)\) is a stable equilibrium point (a sink).

**(b)**
Next, find the eigenvectors \(\mathbf{v}\) associated with each eigenvalue.

For \(\lambda_1 = -1\):
\(\begin{pmatrix} -1 & 1 \\ 2 & -2 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0
\end{pmatrix} \implies v_1 = v_2\)

So, an eigenvector is \(\mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}\).

For \(\lambda_2 = -4\):
\(\begin{pmatrix} 2 & 1 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies 2u_1 + u_2 = 0 \implies u_2 = -2u_1\)

So, an eigenvector is \(\mathbf{v}_2 = \begin{pmatrix} 1 \\ -2 \end{pmatrix}\).

The general solution is:

\(\begin{pmatrix} x(t) \\ y(t)
\end{pmatrix} = C_1 e^{-t} \begin{pmatrix} 1 \\ 1 \end{pmatrix} + C_2 e^{-4t} \begin{pmatrix} 1 \\ -2 \end{pmatrix}\)

Using initial conditions \(x(0) = 6\) and \(y(0) = 0\):

1) \(C_1 + C_2 = 6\)
2) \(C_1 - 2C_2 = 0 \implies C_1 = 2C_2\)

Substitute (2) into (1):
\(2C_2 + C_2 = 6 \implies 3C_2 = 6 \implies C_2 = 2\)
Thus, \(C_1 = 4\).

The particular solutions are:
\(x(t) = 4e^{-t} + 2e^{-4t}\)
\(y(t) = 4e^{-t} - 4e^{-4t}\)

**(c)**
To find the maximum of \(y(t)\), set \(\frac{dy}{dt} = 0\):

\(\frac{dy}{dt} = -4e^{-t} + 16e^{-4t} = 0\)
\(16e^{-4t} = 4e^{-t} \implies e^{3t} = 4\)
\(3t = \ln(4) \implies t = \frac{1}{3}\ln(4) \approx 0.462\) hours.

At this time, the tissue concentration is:

\(y(0.462) = 4e^{-0.462} - 4e^{-4(0.462)} \approx 1.89\) mg/L.

**(d)**
The dominant eigenvalue is \(\lambda_1 = -1\) (closest to zero, decays slowest).
The trajectory corresponding to \(\mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}\) is the line \(y = x\).

In the first quadrant (\(x \ge 0, y \ge 0\)):
- Trajectories approach the line \(y = x\) as \(t\) increases, then converge to the origin along this line.
- Arrows on all trajectories should point towards the origin.
- A straight-line trajectory should be drawn along \(y = x\) starting in the first quadrant and ending at the origin.

**(e)**
(i) At steady-state, \(\frac{dx}{dt} = 0\) and \(\frac{dy}{dt} = 0\):

1) \(-2x + y + I_0 = 0\)
2) \(2x - 3y = 0 \implies 2x = 3y \implies x = 1.5y\)

Substitute (2) into (1):
\(-3y + y + I_0 = 0 \implies 2y = I_0 \implies y = 0.5 I_0\)
Then, \(x = 1.5(0.5 I_0) = 0.75 I_0\).

The new steady-state is at the coordinates \((0.75 I_0, 0.5 I_0)\).

(ii) Given \(I_0 = 4\), the coupled system is:

\(f(x, y) = \frac{dx}{dt} = -2x + y + 4\)
\(g(x, y) = \frac{dy}{dt} = 2x - 3y\)

Using Euler's method with \(h = 0.1\) and initial conditions \(x_0 = 2, y_0 = 1\):

**Step 1** (at \(t = 0\)):
\(f(2, 1) = -2(2) + 1 + 4 = 1\)
\(g(2, 1) = 2(2) - 3(1) = 1\)

\(x_1 = 2 + 0.1(1) = 2.1\)
\(y_1 = 1 + 0.1(1) = 1.1\)

**Step 2** (at \(t = 0.1\)):
\(f(2.1, 1.1) = -2(2.1) + 1.1 + 4 = 0.9\)
\(g(2.1, 1.1) = 2(2.1) - 3(1.1) = 0.9\)

\(x_2 = 2.1 + 0.1(0.9) = 2.19\)
\(y_2 = 1.1 + 0.1(0.9) = 1.19\)

So at \(t = 0.2\), the estimated concentrations are:
\(x(0.2) \approx 2.19\) mg/L, \(y(0.2) \approx 1.19\) mg/L.

**(a)**
M1 for setting up the characteristic equation \(\det(\mathbf{A}-\lambda\mathbf{I})=0\).
A1 for finding the correct eigenvalues: \(\lambda = -1\) and \(\lambda = -4\).
R1 for explaining that because both eigenvalues are real and negative, the origin is a stable equilibrium point (sink). [3 marks]

**(b)**
M1 for finding eigenvectors corresponding to both eigenvalues.
A1 for obtaining the correct eigenvectors (e.g., \(\begin{pmatrix} 1 \\ 1 \end{pmatrix}\) and \(\begin{pmatrix} 1 \\ -2 \end{pmatrix}\)).
M1 for constructing the general solution.
M1 for substituting \(t = 0\) with the given initial conditions.
A1 for finding \(C_1 = 4\) and \(C_2 = 2\).
A1 for correct final particular solutions: \(x(t) = 4e^{-t} + 2e^{-4t}\) and \(y(t) = 4e^{-t} - 4e^{-4t}\). [6 marks]

**(c)**
M1 for setting \(\frac{dy}{dt} = 0\).
A1 for obtaining \(e^{3t} = 4\) or \(t = \frac{1}{3}\ln(4)\).
A1 for \(t \approx 0.462\) hours.
M1 for substituting \(t\) back into the expression for \(y(t)\).
A1 for \(y_{\text{max}} \approx 1.89\) mg/L. [5 marks]

**(d)**
A1 for drawing the straight line trajectory \(y=x\) in the first quadrant.
A1 for labeling \(y=x\) as the dominant eigenvector trajectory.
A1 for drawing trajectories that approach the line \(y=x\) and flow towards the origin.
A2 for correct direction arrows pointing towards the origin on all drawn paths. [5 marks]

**(e)**
(i)
M1 for setting both derivatives to zero.
M1 for solving the resulting algebraic system.
A1 for the steady-state coordinate \((0.75 I_0, 0.5 I_0)\). [3 marks]

(ii)
M1 for writing down the formulas for \(x_{n+1}\) and \(y_{n+1}\).
M1 for calculating the slopes at the first step (\(f(2,1)=1\), \(g(2,1)=1\)).
A1 for the correct first step approximations: \(x(0.1) = 2.1\) and \(y(0.1) = 1.1\).
M1 for calculating the slopes at the second step (\(f(2.1,1.1)=0.9\), \(g(2.1,1.1)=0.9\)).
A2 for the correct second step approximations: \(x(0.2) \approx 2.19\) and \(y(0.2) \approx 1.19\). [6 marks]