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Part (a): Use the compound interest formula: \(FV = PV \times (1 + \frac{r}{100k})^{kt}\). Here, \(PV = 8000\), \(r = 4.5\), \(k = 12\), and \(t = 5\). \(FV = 8000 \times (1 + \frac{4.5}{1200})^{60} = 8000 \times (1.00375)^{60} \approx 10014.37\) USD. Part (b): To double, \(FV = 16000\). This gives \(16000 = 8000 \times (1.00375)^{12t} \Rightarrow 2 = (1.00375)^{12t}\). Taking the natural logarithm of both sides: \(\ln(2) = 12t \ln(1.00375) \Rightarrow t = \frac{\ln(2)}{12 \ln(1.00375)} \approx 15.432\) years. To 3 significant figures, \(t = 15.4\) years.

(a) M1 for substituting values into the compound interest formula. A1 for 10014.37. (b) M1 for setting up the equation to double: \(2 = (1.00375)^{12t}\) or equivalent. A1 for 15.4.

Part (a): The maximum of a quadratic function \(P(x) = ax^2 + bx + c\) occurs at the vertex \(x = -\frac{b}{2a}\). Here, \(x = -\frac{28}{2(-2)} = 7\). Part (b): Substitute \(x = 7\) into the profit equation: \(P(7) = -2(7)^2 + 28(7) - 40 = -98 + 196 - 40 = 58\) thousand dollars. Part (c): To find the break-even points, set \(P(x) = 0\): \(-2x^2 + 28x - 40 = 0 \Rightarrow x^2 - 14x + 20 = 0\). Solving using the quadratic formula: \(x = \frac{14 \pm \sqrt{(-14)^2 - 4(1)(20)}}{2} = \frac{14 \pm \sqrt{116}}{2} = 7 \pm \sqrt{29}\). The smaller value of \(x\) is \(7 - \sqrt{29} \approx 1.6148\). To 3 significant figures, \(x = 1.61\).

(a) M1 for finding the vertex of the parabola (e.g., using \(-b/(2a)\) or derivative). A1 for 7. (b) A1 for 58. (c) M1 for setting \(P(x) = 0\). A1 for 1.61.

Part (a): The midpoint of \(AB\) is \((\frac{1+5}{2}, \frac{4+8}{2}) = (3, 6)\). The gradient of \(AB\) is \(\frac{8 - 4}{5 - 1} = 1\). The gradient of the perpendicular bisector is the negative reciprocal, which is \(-1\). Using the point-slope form: \(y - 6 = -1(x - 3) \Rightarrow y = -x + 9\). Part (b): Find the equation of the perpendicular bisector of \(BC\). Midpoint of \(BC\) is \((\frac{5+9}{2}, \frac{8+2}{2}) = (7, 5)\). Gradient of \(BC\) is \(\frac{2 - 8}{9 - 5} = -1.5\). Gradient of the perpendicular bisector of \(BC\) is \(\frac{2}{3}\). The equation is \(y - 5 = \frac{2}{3}(x - 7) \Rightarrow y = \frac{2}{3}x + \frac{1}{3}\). Set the two bisector equations equal: \(-x + 9 = \frac{2}{3}x + \frac{1}{3} \Rightarrow \frac{26}{3} = \frac{5}{3}x \Rightarrow x = 5.2\). Substituting back: \(y = -5.2 + 9 = 3.8\). The coordinates are \((5.2, 3.8)\).

(a) M1 for finding midpoint \((3, 6)\) and gradient 1 of \(AB\). A1 for showing the step-by-step simplification to \(y = -x + 9\). (b) M1 for finding the equation of another perpendicular bisector (e.g., of \(BC\)). M1 for setting up and solving a system of two linear equations. A1 for \(x = 5.2\). A1 for \(y = 3.8\).

Part (a): In triangle \(APB\), angle \(PAB = 22^{\circ}\). Angle \(PBA = 180^{\circ} - 35^{\circ} = 145^{\circ}\). Therefore, angle \(APB = 180^{\circ} - (22^{\circ} + 145^{\circ}) = 13^{\circ}\). Using the sine rule: \(\frac{PB}{\sin(22^{\circ})} = \frac{500}{\sin(13^{\circ})} \Rightarrow PB = \frac{500 \sin(22^{\circ})}{\sin(13^{\circ})} \approx 832.61\) meters. To 3 significant figures, the distance is 833 meters. Part (b): Let \(H\) be the point directly below \(P\) on the horizontal path. In right-angled triangle \(PBH\): \(\sin(35^{\circ}) = \frac{\text{height}}{PB} \Rightarrow \text{height} = 832.61 \times \sin(35^{\circ}) \approx 477.56\) meters. To 3 significant figures, the height is 478 meters.

(a) M1 for finding angle \(APB = 13^{\circ}\). M1 for correctly applying the sine rule. A1 for 833 (or 832.61). (b) M1 for using basic right-angled trigonometry with their \(PB\) value. A1 for 478 (or 477.56).

Part (a): \(H_0\): Choice of major and preferred study environment are independent. Part (b): The total number of Science students is 80, the total number of students preferring the Library is 90, and the grand total is 240. Expected frequency = \(\frac{\text{row total} \times \text{column total}}{\text{grand total}} = \frac{80 \times 90}{240} = 30\). Part (c): The degrees of freedom are \((3 - 1) \times (3 - 1) = 4\). Using a GDC with \(\chi^2 = 16.6\) and \(df = 4\), the p-value is approximately 0.00231. Since \(0.00231 < 0.05\), we reject the null hypothesis. There is sufficient evidence at the 5% significance level to suggest that choice of major and preferred study environment are associated.

(a) A1 for stating \(H_0\) clearly. (b) M1 for substituting correctly into the expected frequency formula. A1 for showing the result is 30. (c) A1 for \(df = 4\). A1 for p-value \(\approx 0.00231\) (accept 0.0023 to 0.0024). R1 for rejecting \(H_0\) and stating the conclusion in context.

Part (a): Let \(X\) be the weight of a box, \(X \sim N(505, 4^2)\). Using GDC, \(P(X < 500) \approx 0.10565\). To 3 significant figures, the probability is 0.106. Part (b): We want to find \(w\) such that \(P(X < w) = 0.025\). Using the inverse normal cumulative function on a GDC, we get \(w \approx 497.16\) grams. To 3 significant figures, the minimum weight is 497 grams. Part (c): Let \(Y\) be the number of boxes that weigh less than 500 grams. \(Y\) follows a binomial distribution \(Y \sim B(10, 0.10565)\). Using binomial probability: \(P(Y = 1) = \binom{10}{1} (0.10565)^1 (1 - 0.10565)^9 \approx 0.3865\). To 3 significant figures, the probability is 0.387.

(a) M1 for setting up standard normal parameters. A1 for 0.106. (b) M1 for setting up the inverse normal equation. A1 for 497 (or 497.2). (c) M1 for identifying the binomial parameters. A1 for 0.387.

Part (a): The step size is \(h = \frac{10 - 0}{5} = 2\). Using the trapezoidal rule: \(\text{Area} \approx \frac{h}{2} [d(0) + d(10) + 2(d(2) + d(4) + d(6) + d(8))]\) \(\text{Area} \approx \frac{2}{2} [0 + 0 + 2(1.2 + 1.8 + 2.1 + 1.5)] = 2 \times 6.6 = 13.2\) square meters. Part (b): Volume flow rate \(= \text{Area} \times \text{velocity} = 13.2 \times 0.8 = 10.56\) cubic meters per second. To 3 significant figures, the flow rate is 10.6 cubic meters per second.

(a) M1 for finding step size \(h = 2\). M1 for substituting values into the trapezoidal rule. A1 for 13.2. (b) M1 for multiplying the estimated area by the water speed. A1 for 10.6 (or 10.56).

Part (a): Acceleration \(a(t)\) is the derivative of velocity: \(a(t) = v'(t) = 6t - 12\). At \(t = 3\), \(a(3) = 6(3) - 12 = 6\) meters per second squared. Part (b): Total distance is the integral of the absolute value of velocity: \(\text{Distance} = \int_{0}^{4} |3t^2 - 12t + 9| dt\). Set \(v(t) = 0\) to find when the particle changes direction: \(3(t^2 - 4t + 3) = 0 \Rightarrow 3(t-1)(t-3) = 0\), so the direction changes at \(t = 1\) and \(t = 3\). Evaluating the integrals: \(\int_{0}^{1} (3t^2 - 12t + 9) dt = [t^3 - 6t^2 + 9t]_0^1 = 4\). \(\int_{1}^{3} (3t^2 - 12t + 9) dt = [t^3 - 6t^2 + 9t]_1^3 = (27-54+27) - 4 = -4\). \(\int_{3}^{4} (3t^2 - 12t + 9) dt = [t^3 - 6t^2 + 9t]_3^4 = (64-96+36) - 0 = 4\). Total distance \(= |4| + |-4| + |4| = 12\) meters.

(a) M1 for differentiating the velocity expression. A1 for 6. (b) M1 for setting up the integral of absolute velocity (or identifying critical points at 1 and 3). M1 for splitting the integral or using a GDC to compute the total area. A1 for 12.

(a) We use the financial solver or annuity formula with:
\(FV = 30000\)
\(N = 5 \times 12 = 60\)
\(I\% = 4.2\)
\(PV = 0\)
\(P/Y = 12\)
\(C/Y = 12\)

Solving for \(PMT\):
\(PMT = \frac{30000}{\frac{(1 + 0.042/12)^{60} - 1}{0.042/12}} \approx 450.35\text{ USD}

(b) Total amount deposited = \)450.35 \times 60 = 27021\text{ USD}\) (or \(450.345 \times 60 = 27020.70\text{ USD}\)).

Total interest earned = \(30000 - 27021 = 2979.00\text{ USD}\) (or \(30000 - 27020.70 = 2979.30\text{ USD}\)).

(a)
(M1) for setting up the financial solver parameters or the correct formula.
(A1) for the correct substitution.
(A1) for the correct monthly payment: $450.35 (accept $450.34 from unrounded values).

(b)
(M1) for calculating total deposits: \(PMT \times 60\).
(A1) for finding the interest by subtracting total deposits from $30,000.
(A1) for $2979.00 (or $2979.30).

(a) At \(t=0\), \(T(0) = 85\).
\(22 + a e^0 = 85 \implies 22 + a = 85 \implies a = 63\).

(b) At \(t=10\), \(T(10) = 45\).
\(22 + 63 e^{-10k} = 45\)
\(63 e^{-10k} = 23\)
\(e^{-10k} = \frac{23}{63}\)
\(-10k = \ln\left(\frac{23}{63}\right) \approx -1.0076\)
\(k \approx 0.10076 \approx 0.101\) (to 3 significant figures).

(c) We find \(t\) when \(T(t) = 30\).
\(22 + 63 e^{-0.10076 t} = 30\)
\(63 e^{-0.10076 t} = 8\)
\(e^{-0.10076 t} = \frac{8}{63}\)
\(-0.10076 t = \ln\left(\frac{8}{63}\right) \approx -2.0637\)
\(t \approx 20.48\) minutes.
To the nearest tenth, \(t = 20.5\) minutes.

(a)
(M1) for substituting \(t=0\) and \(T=85\) into the model.
(A1) for \(a = 63\).

(b)
(M1) for substituting \(t=10\), \(T=45\), and their value of \(a\).
(M1) for using logarithms to solve for \(k\).
(AG) for showing clearly that \(k \approx 0.101\).

(c)
(M1) for setting up the equation \(T(t) = 30\).
(A1) for \(20.5\) (accept \(20.4\) if the rounded value of \(k=0.101\) is used).

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

(b) Find the midpoint of \(BC\):
\(M_{BC} = \left(\frac{8+8}{2}, \frac{10+2}{2}\right) = (8, 6)\).
Since \(B\) and \(C\) both have the \(x\)-coordinate of \(8\), the line \(BC\) is vertical (\(x = 8\)).
Thus, the perpendicular bisector of \(BC\) is a horizontal line passing through the midpoint \((8, 6)\).
Equation: \(y = 6\).

(c) The circumcentre is the intersection of the two perpendicular bisectors.
Substitute \(y = 6\) into \(y = -3x + 24\):
\(6 = -3x + 24 \implies 3x = 18 \implies x = 6\).
Thus, the circumcentre coordinates are \((6, 6)\).

(a)
(M1) for finding the midpoint \((5, 9)\) or the gradient \(1/3\).
(M1) for using the perpendicular gradient \(-3\).
(A1) for \(y = -3x + 24\) (or any equivalent linear form).

(b)
(M1) for finding the midpoint of \(BC\) as \((8, 6)\).
(A1) for \(y = 6\).

(c)
(M1) for setting up a system of equations with their bisectors.
(A1) for \((6, 6)\).

(a) The angle \(QPR\) is the difference between the two bearings:
\(\angle QPR = 115^\circ - 45^\circ = 70^\circ\).
We use the Cosine Rule to find the distance \(QR\):
\(QR^2 = 120^2 + 150^2 - 2(120)(150)\cos(70^\circ)\)
\(QR^2 = 14400 + 22500 - 36000\cos(70^\circ) \approx 36900 - 12312.72 = 24587.28\)
\(QR \approx 156.80\text{ m}\).
To the nearest metre, \(QR = 157\text{ m}\).

(b) We use the Sine Rule to find the angle \(PQR\):
\(\frac{\sin(\angle PQR)}{150} = \frac{\sin(70^\circ)}{156.80}\)
\(\sin(\angle PQR) = \frac{150 \sin(70^\circ)}{156.80} \approx 0.8989\)
\(\angle PQR \approx \arcsin(0.8989) \approx 64.01^\circ\).

To find the bearing of \(R\) from \(Q\):
Draw parallel North lines at \(P\) and \(Q\).
Since the bearing of \(Q\) from \(P\) is \(045^\circ\), the back-bearing (direction from \(Q\) to \(P\)) is \(45^\circ + 180^\circ = 225^\circ\).
Because \(R\) is situated east of \(PQ\), the bearing of \(R\) from \(Q\) is:
\(225^\circ - 64.01^\circ = 160.99^circ \approx 161^\circ\).

(a)
(M1) for finding \(\angle QPR = 70^\circ\).
(M1) for substituting correctly into the Cosine Rule.
(A1) for \(157\text{ m}\).

(b)
(M1) for substituting correctly into the Sine Rule to find \(\angle PQR\).
(A1) for \(\angle PQR \approx 64.0^\circ\).
(M1) for finding the bearing using back-bearing concepts.
(A1) for \(161^\circ\).

(a) Let \(X\) be the weight of an apple: \(X \sim N(150, 12^2)\).
Using GDC: \(P(X > 165) \approx 0.1056\).
To 3 significant figures, this is \(0.106\).

(b) We are given \(P(X < w) = 0.08\).
Using the inverse normal function on a GDC:
\(w = \text{invNorm}(0.08, 150, 12) \approx 133.14\text{ g}\).
To the nearest gram, \(w = 133\).

(c) Let \(Y\) be the number of small apples in a box of 12. \(Y\) follows a binomial distribution:
\(Y \sim B(12, 0.08)\).
We want to find \(P(Y = 2)\).
Using GDC binomial probability density function:
\(P(Y = 2) = \binom{12}{2} (0.08)^2 (0.92)^{10} \approx 0.183\).

(a)
(M1) for representing the normal distribution probability mathematically.
(A1) for \(0.106\) (accept \(0.1056\)).

(b)
(M1) for setting up the inverse normal equation.
(A1) for \(133\).

(c)
(M1) for identifying the binomial distribution model with parameters \(n=12\) and \(p=0.08\).
(A1) for \(0.183\) (accept \(0.1835\)).

(a) \(H_0\): Preferred exercise type and age group are independent.

(b) Expected frequency \(= \frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}} = \frac{100 \times 75}{200} = 37.5\).

(c) We input the contingency table into our GDC matrix and run a \(\chi^2\) two-way test.
Degrees of freedom \(= (2-1) \times (3-1) = 2\).
\(\chi^2\) statistic \(\approx 5.333\).
\(p\)-value \(\approx 0.0695\) (to 3 significant figures).

(d) Since the \(p\)-value \(0.0695\) is greater than the significance level \(0.05\), we do not reject the null hypothesis. There is insufficient evidence to conclude that preferred exercise type is associated with age group.

(a)
(A1) for stating that the preferred exercise type and age group are independent.

(b)
(M1) for correct substitution into the expected value formula.
(A1) for \(37.5\).

(c)
(M1) for setting up the test on the GDC.
(A1) for \(0.0695\).

(d)
(R1) for comparing the \(p\)-value with \(0.05\) and stating the correct conclusion (do not reject).

(a) Let \(y\) be the length of the side parallel to the brick wall.
The perimeter consisting of three sides is:
\(2x + y = 80 \implies y = 80 - 2x\).

The area \(A\) is given by:
\(A = x \times y = x(80 - 2x) = 80x - 2x^2\). [Shown]

(b) Differentiating \(A\) with respect to \(x\):
\(\frac{\text{d}A}{\text{d}x} = 80 - 4x\).

(c) To find the maximum area, set the derivative to zero:
\(80 - 4x = 0 \implies 4x = 80 \implies x = 20\text{ m}\).

(d) Substitute \(x = 20\) into the area function:
\(A(20) = 80(20) - 2(20)^2 = 1600 - 800 = 800\text{ m}^2\).

(a)
(M1) for expressing the third side in terms of \(x\): \(y = 80 - 2x\).
(AG) for multiplying by \(x\) to obtain the given expression.

(b)
(A1) for \(80 - 4x\).

(c)
(M1) for setting their derivative equal to zero.
(A1) for \(x = 20\).

(d)
(A1) for \(800\text{ m}^2\).

(a) The width of each interval is \(h = \frac{4-0}{4} = 1\).
The values of \(y\) at each interval boundary are:
\(y_0 = 0.1(0)^2 + 1 = 1.0\)
\(y_1 = 0.1(1)^2 + 1 = 1.1\)
\(y_2 = 0.1(2)^2 + 1 = 1.6\)
\(y_3 = 0.1(3)^2 + 1 = 1.9\)
\(y_4 = 0.1(4)^2 + 1 = 2.6\)

Applying the trapezoidal rule:
\(\text{Area} \approx \frac{h}{2} [y_0 + y_4 + 2(y_1 + y_2 + y_3)]\)
\(\text{Area} \approx \frac{1}{2} [1.0 + 2.6 + 2(1.1 + 1.6 + 1.9)]\)
\(\text{Area} \approx 0.5 [3.6 + 2(4.6)] = 0.5 [3.6 + 9.2] = 6.4\text{ m}^2\).

(b) Exact area \(= \int_{0}^{4} (0.1x^2 + 1) \text{ d}x\)
\(= \left[ \frac{0.1x^3}{3} + x \right]_{0}^{4}\n= \left( \frac{0.1(64)}{3} + 4 \right) - 0 = \frac{6.4}{3} + 4 = \frac{92}{15} \approx 6.13\text{ m}^2\) (to 3 significant figures).

(c) Percentage error \(= \left| \frac{\text{Estimate} - \text{Exact}}{\text{Exact}} \right| \times 100\%
= \left| \frac{6.4 - 6.1333}{6.1333} \right| \times 100\% \approx 4.35\%\).

(a)
(M1) for calculating the correct \(y\) values.
(M1) for substituting correctly into the trapezoidal rule.
(A1) for \(6.4\).

(b)
(M1) for the integration process (correct antiderivative).
(A1) for \(6.13\) (or exact fraction \(92/15\)).

(c)
(A1) for \(4.35\%\) (accept answers based on slightly different rounding paths, e.g., \(4.4\%\)).

(a) The volume of the cylinder is given by \(V = \pi r^2 h\).
Since the volume is \(350\text{ cm}^3\), we have:
\(\pi r^2 h = 350 \implies h = \frac{350}{\pi r^2}\)

The area of the top and bottom circles is \(2\pi r^2\).
The cost of the top and bottom is:
\(C_{\text{base}} = 0.05 \times 2\pi r^2 = 0.1\pi r^2\)

The area of the curved surface is \(2\pi r h\).
The cost of the curved surface is:
\(C_{\text{side}} = 0.03 \times 2\pi r h = 0.06\pi r h\)

Substituting \(h = \frac{350}{\pi r^2}\) into the side cost expression:
\(C_{\text{side}} = 0.06\pi r \left(\frac{350}{\pi r^2}\right) = \frac{21}{r}\)

Therefore, the total cost \(C\) is:
\(C(r) = 0.1\pi r^2 + \frac{21}{r}\)

(b) To find the value of \(r\) that minimizes the cost, we find the minimum of the function \(C(r)\) for \(r > 0\) using a graphic display calculator, or by setting the derivative equal to zero:
\(C'(r) = 0.2\pi r - \frac{21}{r^2} = 0\)
\(0.2\pi r^3 = 21 \implies r^3 = \frac{105}{\pi} \approx 33.4225\)
\(r \approx 3.22\text{ cm}\) (to 3 significant figures)

(c) Substituting \(r \approx 3.2212\) into the cost function:
\(C(3.2212) = 0.1\pi (3.2212)^2 + \frac{21}{3.2212} \approx 9.779\text{ cents}\)

Thus, the minimum cost is \(9.78\) cents (correct to 2 decimal places).

**(a)**
\(V = \pi r^2 h = 350 \implies h = \frac{350}{\pi r^2}\) *(M1)*

Attempt to write an expression for the total cost of the top/bottom or the side: *(M1)*
\(C(r) = 0.05(2\pi r^2) + 0.03(2\pi r h)\)

Substituting \(h\) to obtain the final expression: *(A1)*
\(C(r) = 0.1\pi r^2 + 0.03 \times 2\pi r \left(\frac{350}{\pi r^2}\right) = 0.1\pi r^2 + \frac{21}{r}\)

**Note:** The *(A1)* mark should only be awarded if the intermediate step showing the multiplication/simplification is clear.

**(b)**
An attempt to find the minimum of \(C(r)\) either by sketching the graph on GDC, finding the minimum point, or setting \(C'(r) = 0\). *(M1)*
\(r \approx 3.22\text{ cm}\) (accept \(3.22124...\)) *(A1)*

**(c)**
Substituting their value of \(r\) into \(C(r)\) to find the minimum cost: *(A1)*
\(C = 9.78\text{ cents}\)

**Note:** Accept \(9.78\) cents. Do not penalize if units are omitted.

(a) \(H_0\): Weekly exercise hours and sleep quality rating are independent. \(H_1\): Weekly exercise hours and sleep quality rating are not independent.
(b) Total for High Exercise = \(5 + 15 + 40 = 60\). Total for Good Sleep = \(10 + 25 + 40 = 75\). Grand total \(N = 180\). Expected frequency = \(\frac{60 \times 75}{180} = 25\).
(c) Degrees of freedom \(= (\text{rows} - 1)(\text{columns} - 1) = (3 - 1)(3 - 1) = 4\).
(d) Using a GDC or manual calculation of \(\sum \frac{(O - E)^2}{E}\): \(\chi^2 \approx 38.1\) (3 significant figures, or 38.075), \(p\text{-value} \approx 1.11 \times 10^{-7}\).
(e) Since the \(p\text{-value} < 0.05\), we reject the null hypothesis. There is significant evidence to conclude that weekly exercise hours and sleep quality rating are not independent.
(f) Total number of participants with Good sleep is 75. Out of these, 40 do High exercise. Probability \(= \frac{40}{75} = \frac{8}{15} \approx 0.533\).

(a) M1 for stating H0 and M1 for stating H1. Accept alternative equivalent wording.
(b) M1 for calculating High Exercise total (60) and Good Sleep total (75), M1 for using the formula \(\frac{\text{Row} \times \text{Col}}{\text{Total}}\), A1 for showing the final answer is 25.
(c) A1 for 4.
(d) M2 for setup of the chi-square sum (or indicating GDC use), A1 for \(\chi^2 \approx 38.1\), A1 for \(p \approx 1.11 \times 10^{-7}\).
(e) R1 for comparing p-value to 0.05 (or critical value), A1 for the correct conclusion of rejecting H0 in context.
(f) M1 for identifying denominator of 75, M1 for numerator of 40, A1.7 for final answer 0.533 (or \(\frac{8}{15}\)).

(a) The volume of a cylinder is \(V = \pi r^2 h = 1000\). Rearranging gives \(h = \frac{1000}{\pi r^2}\).
(b) The surface area consists of the base (area \(\pi r^2\)), top (area \(\pi r^2\)), and curved side (area \(2\pi r h\)).
Total cost \(C = 0.05(\text{base area}) + 0.03(\text{top area}) + 0.03(\text{side area})\)
\(C = 0.05(\pi r^2) + 0.03(\pi r^2) + 0.03(2\pi r h)\)
\(C = 0.08\pi r^2 + 0.06\pi r h\)
Substituting \(h = \frac{1000}{\pi r^2}\):
\(C(r) = 0.08\pi r^2 + 0.06\pi r \left(\frac{1000}{\pi r^2}\right) = 0.08\pi r^2 + \frac{60}{r}\).
(c) Differentiating with respect to \(r\): \(C'(r) = 0.16\pi r - \frac{60}{r^2}\).
(d) To minimize cost, set \(C'(r) = 0\):
\(0.16\pi r - \frac{60}{r^2} = 0 \implies 0.16\pi r^3 = 60 \implies r^3 = \frac{60}{0.16\pi} \approx 119.366\).
\(r = \sqrt[3]{119.366} \approx 4.92\text{ cm}\).
(e) Substitute \(r \approx 4.9239\) back into the cost function:
\(C(4.9239) = 0.08\pi (4.9239)^2 + \frac{60}{4.9239} \approx 6.091 + 12.185 = 18.28\text{ dollars}\).

(a) M1 for using volume formula \(\pi r^2 h = 1000\), A1 for isolating \(h\).
(b) M1 for setting up individual cost components, M1 for adding top, bottom, and side costs properly, M1 for substituting \(h\), A2 for clear algebra steps leading to the given expression.
(c) M1 for differentiating power terms, A1 for \(0.16\pi r - \frac{60}{r^2}\).
(d) M1 for setting derivative equal to 0, M1 for solving for \(r^3\), A1.7 for \(r \approx 4.92\text{ cm}\).
(e) M1 for substituting their \(r\) value into the cost formula, A2 for getting \(\$18.28\) (accept 18.3).

(a) Midpoint of \(AB\) is \(M_{AB} = \left(\frac{0+8}{2}, \frac{0+2}{2}\right) = (4, 1)\).
Gradient of \(AB\) is \(m_{AB} = \frac{2-0}{8-0} = \frac{1}{4}\).
Gradient of the perpendicular bisector is \(-4\).
Equation: \(y - 1 = -4(x - 4) \implies y = -4x + 17\).
(b) Midpoint of \(AC\) is \(M_{AC} = \left(\frac{0+3}{2}, \frac{0+7}{2}\right) = (1.5, 3.5)\).
Gradient of \(AC\) is \(m_{AC} = \frac{7-0}{3-0} = \frac{7}{3}\).
Gradient of the perpendicular bisector is \(-\frac{3}{7}\).
Equation: \(y - 3.5 = -\frac{3}{7}(x - 1.5) \implies y = -\frac{3}{7}x + \frac{29}{7}\) (or \(3x + 7y = 29\)).
(c) Solve the system of equations:
\(-4x + 17 = -\frac{3}{7}x + \frac{29}{7}\)
Multiply by 7: \(-28x + 119 = -3x + 29 \implies 25x = 90 \implies x = 3.6\).
Substitute \(x\) back: \(y = -4(3.6) + 17 = 2.6\).
Circumcenter coordinates are \((3.6, 2.6)\).
(d) Since the drone is equidistant from \(A\) and \(B\), the point must lie on the perpendicular bisector of \(AB\).
Set \(y = 3\) in \(y = -4x + 17\):
\(3 = -4x + 17 \implies 4x = 14 \implies x = 3.5\).
The coordinates of the point are \((3.5, 3)\).

(a) M1 for finding midpoint of AB, M1 for gradient of AB, M1 for perpendicular gradient, A1 for final equation.
(b) M1 for finding midpoint of AC, M1 for gradient of AC, M1 for perpendicular gradient, A1 for final equation.
(c) M2 for equating both perpendicular bisectors (or setting up system), A1.7 for \(x = 3.6\), A1 for \(y = 2.6\).
(d) M1 for recognizing the point lies on the perpendicular bisector, M1 for substituting \(y = 3\), A1 for \((3.5, 3)\).

(a) Using the product rule, the derivative is:
\(C'(t) = a e^{-b t} + a t (-b) e^{-b t} = a e^{-b t}(1 - b t)\).
Since the maximum is at \(t = 2\), we have \(C'(2) = 0 \implies a e^{-2b}(1 - 2b) = 0\). Since \(a > 0\) and \(e^{-2b} \ne 0\), then \(1 - 2b = 0 \implies b = 0.5\).
Using \(C(1) = 4\): \(a (1) e^{-0.5(1)} = 4 \implies a e^{-0.5} = 4 \implies a = 4 e^{0.5} \approx 6.59\).
(b) Maximum concentration is at \(t = 2\):
\(C(2) = 4 e^{0.5} (2) e^{-0.5(2)} = 8 e^{-0.5} \approx 4.85\text{ mg/L}\).
(c) The concentration is decreasing for \(t > 2\) (after the maximum).
We solve for when \(C(t) > 2 \implies 4 e^{0.5} t e^{-0.5 t} > 2\).
Using a GDC numerical solver, the values of \(t\) where \(C(t) = 2\) are \(t \approx 0.365\) and \(t \approx 5.95\).
Therefore, the interval where it is both decreasing and greater than 2 is \(2 < t < 5.95\).
(d) The sketch starts at \((0, 0)\), rises to a peak at \((2, 4.85)\), and then asymptotically approaches the horizontal axis as \(t \to \infty\).

(a) M1 for product rule setup, A1 for correct derivative, M1 for substituting \(t = 2\) and setting to 0, A1 for showing \(b = 0.5\). M1 for using \(C(1) = 4\), A1 for exact value \(a = 4 e^{0.5}\) (or accept 6.59).
(b) M1 for substituting \(t=2\) into function, A1 for \(8 e^{-0.5}\) or \(4.85\).
(c) M1 for noting decreasing when \(t > 2\), M2 for setting up inequality and finding the upper intersection boundary \(t \approx 5.95\) using GDC, A1.7 for the correct interval \(2 < t < 5.95\).
(d) A1 for starting at origin and positive domain, A1 for correct general shape, A1 for labeling max point \((2, 4.85)\).

(a) At \(t = 6\): \(R(6) = 15 + 10\sin\left(\frac{6\pi}{12}\right) = 15 + 10\sin\left(\frac{\pi}{2}\right) = 15 + 10 = 25\text{ m}^3/\text{h}\).
(b) For 4 sub-intervals on \([0, 12]\), the step size is \(\Delta t = \frac{12 - 0}{4} = 3\).
The boundary points are \(t = 0, 3, 6, 9, 12\).
\(R(0) = 15\)
\(R(3) = 15 + 10\sin\left(\frac{\pi}{4}\right) \approx 22.071\)
\(R(6) = 25\)
\(R(9) = 15 + 10\sin\left(\frac{3\pi}{4}\right) \approx 22.071\)
\(R(12) = 15\)
Applying the formula:
\(\text{Volume} \approx \frac{3}{2}\left[R(0) + 2(R(3) + R(6) + R(9)) + R(12)\right]\)
\(\approx 1.5\left[15 + 2(22.071 + 25 + 22.071) + 15\right] = 1.5\left[30 + 138.284\right] \approx 252.426\text{ m}^3\) (or \(252\text{ m}^3\) to 3 sf).
(c) Exact volume \(= \int_{0}^{12} \left(15 + 10\sin\left(\frac{\pi t}{12}\right)\right) dt\)
\(= \left[ 15t - \frac{120}{\pi}\cos\left(\frac{\pi t}{12}\right) \right]_{0}^{12}\)
\(= \left(180 - \frac{120}{\pi}\cos(\pi)\right) - \left(0 - \frac{120}{\pi}\cos(0)\right) = 180 + \frac{120}{\pi} + \frac{120}{\pi} = 180 + \frac{240}{\pi} \approx 256.394\text{ m}^3\).
(d) Percentage error \(= \frac{|252.426 - 256.394|}{256.394} \times 100\% \approx 1.55\%\).
(e) Underestimates because \(R''(t) = -10\left(\frac{\pi}{12}\right)^2\sin\left(\frac{\pi t}{12}\right) < 0\) on \((0, 12)\), meaning the curve is concave down and the trapezoidal lines lie below the curve.

(a) M1 for substituting \(t = 6\), A1 for 25.
(b) M1 for calculating \(\Delta t = 3\), M1 for finding function values at intervals, M1 for substituting into the trapezoidal formula, A2 for \(252\text{ m}^3\) (accept 252.4).
(c) M1 for setting up the definite integral, M1 for integrating sine correctly, A1 for correct limits evaluation, A1 for \(256\text{ m}^3\) (accept \(180 + \frac{240}{\pi}\) or 256.4).
(d) M1 for the percentage error formula, A1.7 for \(1.55\%\) (accept 1.5%).
(e) R1 for stating 'underestimates', R1 for mentioning that the curve is concave down (or \(R''(t) < 0\)).

(a) The monthly interest rate is \(r = \frac{0.06}{12} = 0.005\).
The sum is given by: \(S_n = 500 + 500(1.005) + 500(1.005)^2 + \dots + 500(1.005)^{n-1}\).
This is a geometric series with first term \(a = 500\) and common ratio \(q = 1.005\).
\(S_n = 500 \times \frac{1.005^n - 1}{1.005 - 1} = 500 \times \frac{1.005^n - 1}{0.005} = 100\,000\left(1.005^n - 1\right)\).
(b) 15 years \(= 15 \times 12 = 180\) months. \(S_{180} = 100\,000\left(1.005^{180} - 1\right) \approx 100\,000(2.454094 - 1) = \$145\,409.36\).
(c) Total contribution over 15 years \(= 180 \times 500 = \$90\,000\).
Interest earned \(= \$145\,409.36 - \$90\,000 = \$55\,409.36\).
(d) Set \(100\,000\left(1.005^n - 1\right) = 250\,000 \implies 1.005^n - 1 = 2.5 \implies 1.005^n = 3.5\).
Taking the natural logarithm on both sides: \(n \ln(1.005) = \ln(3.5) \implies n = \frac{\ln(3.5)}{\ln(1.005)} \approx 251.18\text{ months}\).
Convert to years: \(\frac{251.18}{12} \approx 20.93\text{ years}\). To the nearest year, she needs 21 years.

(a) M1 for finding monthly rate \(0.005\), M1 for expressing sum of geometric series, M1 for identifying \(a = 500\) and \(r = 1.005\), A2 for showing the simplification step to reach the final formula.
(b) M1 for calculating 180 periods, M1 for substituting into the formula, A1 for \(\$145\,409.36\) (accept \(\$145\,409\) or \(\$145\,000\) to 3sf).
(c) M1 for calculating total contribution, M1 for subtraction logic, A1 for \(\$55\,409.36\).
(d) M1 for setting equation equal to \(250\,000\), M1 for algebraic manipulation to isolate power, M1 for using logarithms, A1.7 for 21 years (must round to nearest year).

(a) Let \(X\) be the mass of a bag of coffee. \(X \sim N(505, 4^2)\).
Using a GDC: \(P(X > 500) \approx 0.89435 \approx 0.894\).
(b) Let \(Y\) be the number of bags containing more than 500 grams out of 12. \(Y \sim B(12, 0.89435)\).
We want \(P(Y \ge 10) = 1 - P(Y \le 9)\).
Using binomial cumulative distribution on a GDC: \(P(Y \le 9) \approx 0.13437\).
So \(P(Y \ge 10) \approx 1 - 0.13437 \approx 0.86563 \approx 0.866\).
(c) We have \(P(X < k) = 0.025\).
Using inverse normal distribution: \(k \approx 497.16 \approx 497.2\text{ grams}\).
(d) Let the new mean mass be \(\mu_{\text{new}}\). We require \(P(X_{\text{new}} < 497.16) = 0.01\) with \(\sigma = 4\).
Using standard normal distribution: \(Z = \frac{497.16 - \mu_{\text{new}}}{4}\).
Since \(P(Z < z) = 0.01\), we have \(z \approx -2.32635\).
So: \(\frac{497.16 - \mu_{\text{new}}}{4} = -2.32635 \implies 497.16 - \mu_{\text{new}} = -9.3054 \implies \mu_{\text{new}} \approx 506.465 \approx 506.5\text{ grams}\).

(a) M1 for correct setup of normal CDF, A1 for \(0.894\).
(b) M1 for identifying binomial model, M1 for parameters \(n=12, p=0.894\), M1 for expressing \(P(Y \ge 10)\), A1 for \(0.866\).
(c) M1 for setting up inverse normal equation, A1 for inverse norm computation, A1 for \(k = 497.2\).
(d) M1 for setting up the standardized z-value equation, M2 for finding \(z = -2.32635\) (or using GDC solver directly), M2 for substituting values into the z-formula, A1.7 for \(506.5\text{ g}\) (accept 506).

(a) - The value 0.8 (first row, second column) represents the fecundity of sub-adults: the average number of juvenile offspring produced per sub-adult bird per year. - The value 1.5 (first row, third column) represents the fecundity of adults: the average number of juvenile offspring produced per adult bird per year. - The value 0.5 (second row, first column) represents the survival probability of juveniles: the proportion of juveniles that survive to become sub-adults the following year. \\ (b) To find the characteristic equation, we set \(\det(L - \lambda I) = 0\): \\ \det\begin{pmatrix} -\lambda & 0.8 & 1.5 \\ 0.5 & -\lambda & 0 \\ 0 & 0.8 & -\lambda \end{pmatrix} = 0 \\ Expanding along the first row: \\ -\lambda\left((-\lambda)(-\lambda) - 0\right) - 0.8\left(0.5(-\lambda) - 0\right) + 1.5\left(0.5(0.8) - 0\right) = 0 \\ -\lambda^3 + 0.4\lambda + 0.6 = 0 \\ Multiplying by -1 gives the characteristic equation: \\ \lambda^3 - 0.4\lambda - 0.6 = 0 \\ Substituting \(\lambda = 1\) into the equation: \\ 1^3 - 0.4(1) - 0.6 = 1 - 0.4 - 0.6 = 0. Since the equation holds, \(\lambda = 1\) is verified as an eigenvalue. \\ (c) To find the eigenvector \(\mathbf{v} = \begin{pmatrix} j \\ s \\ a \end{pmatrix}\_1\) corresponding to \(\lambda = 1\): \\ \begin{pmatrix} -1 & 0.8 & 1.5 \\ 0.5 & -1 & 0 \\ 0 & 0.8 & -1 \end{pmatrix}\begin{pmatrix} j \\ s \\ a \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \\ From the second row: \\ 0.5j - s = 0 \implies s = 0.5j \\ From the third row: \\ 0.8s - a = 0 \implies a = 0.8(0.5j) = 0.4j \\ Setting \(j = 1\) gives the eigenvector \(\mathbf{v} = \begin{pmatrix} 1 \\ 0.5 \\ 0.4 \end{pmatrix}\). \\ The sum of the components is \(1 + 0.5 + 0.4 = 1.9\). \\ The long-term stable proportions are: \\ - Juveniles: \(\frac{1}{1.9} \approx 52.6\%\) \\ - Sub-adults: \(\frac{0.5}{1.9} \approx 26.3\%\) \\ - Adults: \(\frac{0.4}{1.9} \approx 21.1\%\) \\ (d) (i) \(\mathbf{x}_1 = L\mathbf{x}_0 = \begin{pmatrix} 0 & 0.8 & 1.5 \\ 0.5 & 0 & 0 \\ 0 & 0.8 & 0 \end{pmatrix}\begin{pmatrix} 100 \\ 40 \\ 10 \end{pmatrix} = \begin{pmatrix} 32+15 \\ 50 \\ 32 \end{pmatrix} = \begin{pmatrix} 47 \\ 50 \\ 32 \end{pmatrix}\). \\ \(\mathbf{x}_2 = L\mathbf{x}_1 = \begin{pmatrix} 0 & 0.8 & 1.5 \\ 0.5 & 0 & 0 \\ 0 & 0.8 & 0 \end{pmatrix}\begin{pmatrix} 47 \\ 50 \\ 32 \end{pmatrix} = \begin{pmatrix} 40+48 \\ 23.5 \\ 40 \end{pmatrix} = \begin{pmatrix} 88 \\ 23.5 \\ 40 \end{pmatrix}\). \\ (ii) Using a GDC to calculate \(\mathbf{x}_{20} = L^{20}\mathbf{x}_0\): \\ \mathbf{x}_{20} \approx \begin{pmatrix} 75.12 \\ 37.44 \\ 29.95 \end{pmatrix}\). \\ Total population after 20 years is \(75.12 + 37.44 + 29.95 = 142.51\), which rounds to 143 (or 142 depending on intermediate roundings). \\ (e) (i) \(L' = \begin{pmatrix} 0 & 0.8 & f \\ 0.6 & 0 & 0 \\ 0 & 0.8 & 0 \end{pmatrix}\). \\ (ii) For \(\lambda = 1\) to remain the eigenvalue, \(\det(L' - I) = 0\): \\ \det\begin{pmatrix} -1 & 0.8 & f \\ 0.6 & -1 & 0 \\ 0 & 0.8 & -1
\end{pmatrix} = -1(1) - 0.8(-0.6) + f(0.48) = -1 + 0.48 + 0.48f = 0 \\ -0.52 + 0.48f = 0 \implies f = \frac{0.52}{0.48} = \frac{13}{12} \approx 1.08\). \\ (iii) The new stable eigenvector is found from: \\ s = 0.6j \\ a = 0.8s = 0.48j \\ Giving \\mathbf{v}' = \begin{pmatrix} 1 \\ 0.6 \\ 0.48 \end{pmatrix}\). Total = 2.08. \\ New proportions: J: \(\frac{1}{2.08} \approx 48.1\%\), S: \(\frac{0.6}{2.08} \approx 28.8\%\), A: \(\frac{0.48}{2.08} \approx 23.1\%\). \\ Comment: The conservation program decreases the proportion of juveniles but increases the proportion of sub-adults and adults, creating a more mature population structure despite lower individual adult fecundity.

Part (a): [4 marks]
- M1 for identifying 0.8 as sub-adult fecundity.
- M1 for identifying 1.5 as adult fecundity.
- M1 for identifying 0.5 as juvenile survival rate.
- A1 for accurate description of the time step context (annual/per year).

Part (b): [5 marks]
- M1 for setting up the determinant equation \(\det(L - \lambda I) = 0\).
- A2 for correct algebraic expansion steps.
- A1 for obtaining the correct cubic characteristic polynomial.
- A1 for substituting \(\lambda = 1\) and demonstrating it yields 0.

Part (c): [5 marks]
- M1 for setting up the linear system \((L - I)\mathbf{v} = \mathbf{0}\).
- A1 for finding the relation \(s = 0.5j\).
- A1 for finding the relation \(a = 0.4j\).
- A1 for the correct stable eigenvector ratio.
- A1 for converting components to percentages (52.6%, 26.3%, 21.1%) rounding to 3 sig figs.

Part (d): [6 marks]
- A1.5 for calculating \(\mathbf{x}_1\) correctly.
- A1.5 for calculating \(\mathbf{x}_2\) correctly.
- M1 for expressing \(\mathbf{x}_{20} = L^{20}\mathbf{x}_0\).
- A1 for correct numeric calculation from GDC.
- A1 for correct rounding to 143 (accept 142).

Part (e): [7 marks]
- A1 for correct matrix \(L'\).
- M1 for setting up \(\det(L' - I) = 0\).
- A2 for solving the linear equation to get \(f = \frac{13}{12}\) (or 1.08).
- A1 for calculating new stable percentages (48.1%, 28.8%, 23.1%).
- A2 for a sensible comparison and comment on the shift in age structure.

(a) The system can be written as: \\ \frac{d}{dt}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -0.4 & 0.1 \\ 0.3 & -0.2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} + \begin{pmatrix} 10 \\ 0 \end{pmatrix} \\ To find the equilibrium state, set \(\frac{dx}{dt} = 0\) and \(\frac{dy}{dt} = 0\): \\ 0.3x_s - 0.2y_s = 0 \implies y_s = 1.5x_s \\ Substitute into the first equation: \\ 10 - 0.4x_s + 0.1(1.5x_s) = 0 \\ 10 - 0.25x_s = 0 \implies x_s = 40 \text{ mg} \\ Then: \\ y_s = 1.5(40) = 60 \text{ mg} \\ So the equilibrium state is \(\mathbf{x}_s = \begin{pmatrix} 40 \\ 60 \end{pmatrix}\). \\ (b) Given \(X = x - 40\) and \(Y = y - 60\), we have \(\frac{dX}{dt} = \frac{dx}{dt}\) and \(\frac{dY}{dt} = \frac{dy}{dt}\). \\ Substituting \(x = X + 40\) and \(y = Y + 60\) into the original system: \\ \frac{dX}{dt} = 10 - 0.4(X + 40) + 0.1(Y + 60) = 10 - 0.4X - 16 + 0.1Y + 6 = -0.4X + 0.1Y \\ \frac{dY}{dt} = 0.3(X + 40) - 0.2(Y + 60) = 0.3X + 12 - 0.2Y - 12 = 0.3X - 0.2Y \\ In matrix form, this is: \\ \frac{d}{dt}\begin{pmatrix} X \\ Y \end{pmatrix} = \begin{pmatrix} -0.4 & 0.1 \\ 0.3 & -0.2 \end{pmatrix}\begin{pmatrix} X \\ Y \end{pmatrix} \\ which is indeed \(\frac{d\mathbf{X}}{dt} = M\mathbf{X}\). \\ (c) To find the eigenvalues of \(M\), we solve \(\det(M - \lambda I) = 0\): \\ \det\begin{pmatrix} -0.4 - \lambda & 0.1 \\ 0.3 & -0.2 - \lambda \end{pmatrix} = 0 \\ (-0.4 - \lambda)(-0.2 - \lambda) - 0.03 = 0 \\ \lambda^2 + 0.6\lambda + 0.08 - 0.03 = 0 \\ \lambda^2 + 0.6\lambda + 0.05 = 0 \\ (\lambda + 0.1)(\lambda + 0.5) = 0 \implies \lambda_1 = -0.1, \lambda_2 = -0.5 \\ Since both eigenvalues are real and negative, the deviations \(X(t)\) and \(Y(t)\) decay to zero as \(t \to \infty\) without oscillating. Therefore, the equilibrium point is a stable node. \\ (d) First, we find the eigenvectors corresponding to each eigenvalue. \\ For \(\lambda_1 = -0.1\): \\ \begin{pmatrix} -0.3 & 0.1 \\ 0.3 & -0.1 \end{pmatrix}\begin{pmatrix} v_1 \\ v_2
\end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies -0.3v_1 + 0.1v_2 = 0 \implies \mathbf{v}_1 = \begin{pmatrix} 1 \\ 3 \end{pmatrix} \\ For \(\lambda_2 = -0.5\): \\ \begin{pmatrix} 0.1 & 0.1 \\ 0.3 & 0.3
\end{pmatrix}\begin{pmatrix} v_1 \\ v_2
\end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies 0.1v_1 + 0.1v_2 = 0 \implies \mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix} \\ The general solution for the deviations is: \\ \begin{pmatrix} X(t) \\ Y(t) \end{pmatrix} = c_1 e^{-0.1t}\begin{pmatrix} 1 \\ 3 \end{pmatrix} + c_2 e^{-0.5t}\begin{pmatrix} 1 \\ -1 \end{pmatrix} \\ This yields the general solutions for \(x(t)\) and \(y(t)\): \\ x(t) = c_1 e^{-0.1t} + c_2 e^{-0.5t} + 40 \\ y(t) = 3c_1 e^{-0.1t} - c_2 e^{-0.5t} + 60 \\ Using the initial conditions \(x(0) = 0, y(0) = 0\): \\ c_1 + c_2 + 40 = 0 \implies c_1 + c_2 = -40 \\ 3c_1 - c_2 + 60 = 0 \implies 3c_1 - c_2 = -60 \\ Adding the equations yields: \\ 4c_1 = -100 \implies c_1 = -25 \\ Substituting back gives: \\ c_2 = -15 \\ The exact particular solutions are: \\ x(t) = -25e^{-0.1t} - 15e^{-0.5t} + 40 \\ y(t) = -75e^{-0.1t} + 15e^{-0.5t} + 60 \\ (e) (i) Sketch Description: - Axes labeled \(x\) (bloodstream concentration) and \(y\) (tissue concentration). - Equilibrium point at \((40, 60)\) marked. - The trajectory starts at the origin \((0, 0)\) and curves smoothly to end at the stable node \((40, 60)\), showing arrows pointing in the direction of increasing time. - Trajectories approach the node along the line corresponding to the dominant slow-decaying eigenvector \(\mathbf{v}_1\) which has slope 3 (i.e. \(y - 60 = 3(x - 40)\)). \\ (ii) Euler's method with \(h = 0.5\): \\ At \(t=0, x=0, y=0\): \\ \frac{dx}{dt} = 10 - 0.4(0) + 0.1(0) = 10 \\ \frac{dy}{dt} = 0.3(0) - 0.2(0) = 0 \\ x(0.5) \approx 0 + 0.5(10) = 5 \text{ mg} \\ y(0.5) \approx 0 + 0.5(0) = 0 \text{ mg} \\ At \(t=0.5, x=5, y=0\): \\ \frac{dx}{dt} = 10 - 0.4(5) + 0.1(0) = 8 \\ \frac{dy}{dt} = 0.3(5) - 0.2(0) = 1.5 \\ x(1.0) \approx 5 + 0.5(8) = 9 \text{ mg} \\ y(1.0) \approx 0 + 0.5(1.5) = 0.75 \text{ mg}.

Part (a): [4 marks]
- A2 for expressing the correct matrix equation (1 mark for the matrix M, 1 mark for the constant vector).
- M1 for setting up the equations for steady state.
- A1 for finding the correct values \(x_s = 40\) and \(y_s = 60\).

Part (b): [4 marks]
- M1 for identifying derivatives \(\frac{dX}{dt} = \frac{dx}{dt}\) and \(\frac{dY}{dt} = \frac{dy}{dt}\).
- M1 for substituting the deviation formulas into the system.
- A2 for simplifying correctly to eliminate constant terms and showing the final matrix representation matches \(M\mathbf{X}\).

Part (c): [7 marks]
- M1 for setting up \(\det(M - \lambda I) = 0\).
- A2 for expanding the determinant to get the quadratic equation \(\lambda^2 + 0.6\lambda + 0.05 = 0\).
- A2 for finding the correct eigenvalues \(\lambda = -0.1\) and \(\lambda = -0.5\).
- R1 for explaining that the negative eigenvalues ensure convergence as \(t \to \infty\).
- R1 for identifying the classification of the equilibrium point as a stable node.

Part (d): [7 marks]
- M1 for finding the eigenvectors associated with the eigenvalues.
- A1 for finding \(\mathbf{v}_1 = \begin{pmatrix} 1 \\ 3 \end{pmatrix}\) and \(\mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}\).
- A1 for setting up the general solution with constants \(c_1\) and \(c_2\).
- M1 for substituting \(t=0\) and equating to the initial conditions \(x(0)=0, y(0)=0\).
- A2 for solving the system of equations to find \(c_1 = -25\) and \(c_2 = -15\).
- A1 for the final complete particular equations for \(x(t)\) and \(y(t)\).

Part (e): [6 marks]
- A1 for sketching axes correctly labeled with \((40, 60)\) indicated.
- A1 for drawing the trajectory starting at \((0,0)\) and moving towards \((40,60)\) with direction arrows.
- A1 for demonstrating the correct shape tangent to the slow eigenvector direction as it approaches equilibrium.
- M1 for applying Euler's formula step 1 correctly to get \(x(0.5)=5, y(0.5)=0\).
- M1 for calculating correct derivative values at step 1.5.
- A1 for the correct final values \(x(1) = 9\) mg and \(y(1) = 0.75\) mg.