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

(a) Value of the machine after 5 years is given by: \(V_M(5) = 12000 \times (1 - 0.085)^5 = 12000 \times 0.915^5 \approx $7697.30\).

(b) Value of the investment after \(t\) years is: \(V_I(t) = 8000 \times \left(1 + \frac{0.042}{12}\right)^{12t} = 8000 \times (1.0035)^{12t}\). We need to find the smallest integer \(t\) such that \(V_I(t) > V_M(t)\). Using a GDC to compare values:
At \(t = 3\): \(V_M(3) \approx $9193.31\) and \(V_I(3) \approx $9072.93\) (investment is less).
At \(t = 4\): \(V_M(4) \approx $8411.88\) and \(V_I(4) \approx $9462.43\) (investment is greater).
Hence, it takes 4 complete years.

(a) [3 marks]
M1 for setting up the depreciation equation: \(12000(0.915)^5\).
A1 for \(0.915^5 \approx 0.6414\).
A1 for final answer \($7697.30\) (accept \(7697\) or \(7700\)).

(b) [4 marks]
M1 for setting up the investment equation with correct compounding: \(8000(1.0035)^{12t}\).
M1 for setting up inequality or equation: \(8000(1.0035)^{12t} = 12000(0.915)^t\).
A1 for finding the intersection point \(t \approx 3.10\) years.
A1 for stating 4 complete years (accept 4).

(a) The maximum of a quadratic function \(y = at^2 + bt + c\) occurs at the axis of symmetry \(t = -\frac{b}{2a}\). Here, \(6 = -\frac{k}{2(-0.4)} \implies 6 = \frac{k}{0.8} \implies k = 4.8\).

(b) The maximum height is at \(t = 6\): \(h(6) = -0.4(6)^2 + 4.8(6) + 1.8 = -14.4 + 28.8 + 1.8 = 16.2\) meters.

(c) The drone hits the ground when \(h(t) = 0\): \(-0.4t^2 + 4.8t + 1.8 = 0\). Using the quadratic formula or GDC: \(t \approx 12.369\) or \(t \approx -0.364\). Since \(t \ge 0\), we select \(t \approx 12.4\) seconds (to the nearest tenth of a second).

(a) [2 marks]
M1 for utilizing the axis of symmetry formula \(t = -\frac{b}{2a}\) or setting the derivative \(h'(t) = -0.8t + k = 0\) at \(t=6\).
A1 for showing clear algebraic steps leading to \(k = 4.8\).

(b) [2 marks]
M1 for substituting \(t=6\) and \(k=4.8\) into the equation.
A1 for \(16.2\) (m).

(c) [3 marks]
M1 for setting \(h(t) = 0\).
A1 for finding both roots or identifying the positive root: \(t \approx 12.369\).
A1 for rounding to the nearest tenth: \(12.4\) seconds.

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

(b) Set the two equations equal to find their intersection: \(-3x + 24 = \frac{1}{4}x + \frac{19}{4}\). Multiply by 4: \(-12x + 96 = x + 19 \implies 13x = 77 \implies x = \frac{77}{13} \approx 5.92\). Substitute \(x\) back to find \(y\): \(y = -3\left(\frac{77}{13}\right) + 24 = \frac{81}{13} \approx 6.23\). The coordinates of the circumcenter are \(\left(\frac{77}{13}, \frac{81}{13}\right)\) or approximately \((5.92, 6.23)\).

(a) [3 marks]
A1 for midpoint \((5, 9)\).
M1 for perpendicular gradient \(m_{\perp} = -3\) (derived from gradient of \(AB = \frac{1}{3}\)).
A1 for \(y = -3x + 24\).

(b) [4 marks]
M1 for equating the two perpendicular bisector equations: \(-3x + 24 = 0.25x + 4.75\).
M1 for attempting to solve the resulting equation for \(x\).
A1 for \(x = \frac{77}{13}\) (approx. \(5.92\)).
A1 for \(y = \frac{81}{13}\) (approx. \(6.23\)). Accept coordinates as \((5.92, 6.23)\).

Let \(X\) be the weight of a bag of coffee beans, where \(X \sim N(505, 4^2)\).

(a) Using GDC: \(P(X > 500) \approx 0.89435\). To 3 s.f., this is \(0.894\).

(b) Let \(Y\) be the number of bags weighing more than 500 grams. \(Y\) follows a binomial distribution: \(Y \sim B(12, 0.89435)\). We want: \(P(Y = 10) = \binom{12}{10} (0.89435)^{10} (1 - 0.89435)^2 \approx 0.24118\). To 3 s.f., this is \(0.241\).

(c) We want to find \(w\) such that \(P(X < w) = 0.025\). Using the inverse normal function on GDC: \(w = \text{invNorm}(0.025, 505, 4) \approx 497.16\) grams. To 3 s.f., this is \(497\) grams.

(a) [2 marks]
M1 for normal distribution setup or standardizing: \(P\left(Z > \frac{500-505}{4}\right)\).
A1 for \(0.894\) (accept \(0.89435\)).

(b) [3 marks]
M1 for recognizing binomial distribution with parameters \(n=12\) and \(p\) from part (a).
M1 for binomial probability calculation: \(P(Y=10)\).
A1 for \(0.241\) (accept \(0.24118\)).

(c) [2 marks]
M1 for setup: \(P(X < w) = 0.025\) or standardizing \(z = -1.96\).
A1 for \(497\) (grams) (accept \(497.16\)).

(a) Let the radius be \(r\) and the height be \(h\). The volume is \(V = \pi r^2 h = 500\), which gives \(h = \frac{500}{\pi r^2}\). The total surface area of a closed cylinder is: \(A = 2\pi r^2 + 2\pi r h\). Substituting \(h\) into the area formula: \(A = 2\pi r^2 + 2\pi r \left(\frac{500}{\pi r^2}\right) = 2\pi r^2 + \frac{1000}{r}\).

(b) Differentiate \(A\) with respect to \(r\): \(\frac{dA}{dr} = 4\pi r - \frac{1000}{r^2}\). Set the derivative equal to zero to find the minimum: \(4\pi r - \frac{1000}{r^2} = 0 \implies 4\pi r^3 = 1000 \implies r^3 = \frac{250}{\pi}\). Taking the cube root: \(r = \sqrt[3]{\frac{250}{\pi}} \approx 4.301\) cm. To 3 s.f., this is \(4.30\) cm.

(a) [3 marks]
A1 for correct volume formula in terms of \(r\) and \(h\): \(\pi r^2 h = 500\).
M1 for isolating \(h\) and substituting into the surface area formula \(A = 2\pi r^2 + 2\pi r h\).
A1 for showing clear algebraic steps leading to the given expression.

(b) [4 marks]
M1 for finding the derivative: \(4\pi r - \frac{1000}{r^2}\).
M1 for equating the derivative to zero.
A1 for solving for \(r^3\) or \(r\): \(r^3 = \frac{250}{\pi}\).
A1 for final value \(r \approx 4.30\) (cm).

(a) The given information translates to the following system of linear equations:
1) \(50x + 30y + 20z = 1140\)
2) \(40x + 45y + 25z = 1170\)
3) \(z - y = 2\) (or \(z = y + 2\))

(b) We can solve this system using a GDC matrix solver or substitution. Substituting \(z = y+2\) into the first two equations (after simplifying by dividing by 10 and 5 respectively):
\(5x + 3y + 2(y+2) = 114 \implies 5x + 5y = 110 \implies x + y = 22\).
\(8x + 9y + 5(y+2) = 234 \implies 8x + 14y = 224 \implies 4x + 7y = 112\).
From \(x = 22 - y\), substitute into the second: \(4(22-y) + 7y = 112 \implies 88 + 3y = 112 \implies 3y = 24 \implies y = 8\).
Then: \(x = 22 - 8 = 14\) and \(z = 8 + 2 = 10\).
So an Adult ticket costs $14, a Child ticket costs $8, and a Senior ticket costs $10.

(a) [3 marks]
A1 for \(50x + 30y + 20z = 1140\) (or equivalent).
A1 for \(40x + 45y + 25z = 1170\) (or equivalent).
A1 for \(z - y = 2\) (or equivalent).

(b) [4 marks]
M1 for a valid method to solve the system (e.g., substitution, elimination, or matrix row reduction on a GDC).
A1 for \(x = 14\) (Adult).
A1 for \(y = 8\) (Child).
A1 for \(z = 10\) (Senior).

(a) Since the ground is horizontal and the mast is vertical, \(\triangle OAP\) and \(\triangle OBP\) are right-angled triangles at \(O\).
In \(\triangle OAP\): \(\tan(32^\circ) = \frac{h}{OA} \implies OA = \frac{h}{\tan(32^\circ)}\).
In \(\triangle OBP\): \(\tan(23^\circ) = \frac{h}{OB} \implies OB = \frac{h}{\tan(23^\circ)}\).

(b) Since \(A\) is due South of \(O\) and \(B\) is due East of \(O\), \(\angle AOB = 90^\circ\). In the right-angled triangle \(OAB\):
\(OA^2 + OB^2 = AB^2\)
\(\left(\frac{h}{\tan(32^\circ)}\right)^2 + \left(\frac{h}{\tan(23^\circ)}\right)^2 = 85^2\)
\(h^2 \left[ \frac{1}{\tan^2(32^\circ)} + \frac{1}{\tan^2(23^\circ)} \right] = 7225\)
Compute the terms: \(\frac{1}{\tan^2(32^\circ)} \approx 2.5610\) and \(\frac{1}{\tan^2(23^\circ)} \approx 5.5501\).
\(h^2 (2.5610 + 5.5501) = 7225 \implies 8.1111 h^2 = 7225\)
\(h^2 \approx 890.75 \implies h \approx 29.845\) meters.
The height of the mast is \(29.8\) meters (to 3 s.f.).

(a) [2 marks]
A1 for \(OA = \frac{h}{\tan(32^\circ)}\) (accept \(1.60h\)).
A1 for \(OB = \frac{h}{\tan(23^\circ)}\) (accept \(2.36h\)).

(b) [5 marks]
M1 for recognizing \(\angle AOB = 90^\circ\).
M1 for applying Pythagoras' theorem: \(OA^2 + OB^2 = 85^2\).
M1 for substituting expressions in terms of \(h\).
A1 for finding \(h^2 \approx 891\).
A1 for final answer \(29.8\) (m).

(a) \(H_0\): Student's grade level and preferred method of transport to school are independent.

(b) Expected Value = \(\frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}} = \frac{65 \times 60}{180} = \frac{3900}{180} = 21.6667 \approx 21.67\).

(c) Entering the contingency table into a GDC to perform a \(\chi^2\) test of independence yields: \(p\text{-value} \approx 0.00292\).

(d) Since the \(p\)-value (\(0.00292\)) is less than the significance level \(0.05\), we reject the null hypothesis. There is sufficient evidence to suggest that grade level and preferred method of transport are not independent.

(a) [1 mark]
A1 for stating that the grade level and preferred method of transport are independent (or 'no association').

(b) [2 marks]
M1 for using the formula \(\frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}}\).
A1 for showing \(\frac{65 \times 60}{180} = 21.67\).

(c) [2 marks]
M1 for correctly entering the table in the GDC and choosing the \(\chi^2\) test.
A1 for \(p \approx 0.00292\) (accept 0.0029).

(d) [2 marks]
R1 for comparing the \(p\)-value to \(0.05\) (e.g. \(0.00292 < 0.05\)).
A1 for concluding to reject \(H_0\) (or reject the claim of independence).

(a) The value of the van after 6 years under a constant depreciation of 15% per year is given by:
\(V = 45000 \times (1 - 0.15)^6 = 45000 \times (0.85)^6\)
\(V \approx 16971.73\)
To the nearest dollar, the value is $16,972.

(b) Under the continuous model, we have:
\(V(3) = 45000 e^{-3k} = 28000\)
Dividing both sides by 45,000:
\(e^{-3k} = \frac{28}{45}\)
Taking the natural logarithm of both sides:
\(-3k = \ln\left(\frac{28}{45}\right)\)
\(-3k \approx -0.475046\)
\(k \approx 0.158348\)
To 4 decimal places, \(k = 0.1583\).

(a)
[M1] for a correct expression for constant depreciation: \(45000 \times (0.85)^6\).
[A1] for the unrounded value \(16971.73\).
[A1] for rounding to the nearest dollar: \(16972\).

(b)
[M1] for substituting into the continuous formula: \(45000 e^{-3k} = 28000\).
[M1] for correctly rearranging and using logarithms: \(-3k = \ln\left(\frac{28}{45}\right)\).
[A1] for \(0.1583\) (accept 0.158).

(a) First, find the midpoint \(M\) of \(AB\):
\(M = \left(\frac{2+8}{2}, \frac{9+11}{2}\right) = (5, 10)\).

Next, find the gradient of \(AB\):
\(m_{AB} = \frac{11-9}{8-2} = \frac{2}{6} = \frac{1}{3}\).

The gradient of the perpendicular bisector is the negative reciprocal:
\(m_{\perp} = -3\).

The equation of the perpendicular bisector is:
\(y - 10 = -3(x - 5)\)
\(y - 10 = -3x + 15\)
\(3x + y = 25\).

(b) The control station \(P\) is located at the intersection of the two perpendicular bisectors. We solve the simultaneous equations:
1) \(3x + y = 25 \implies y = 25 - 3x\)
2) \(x - 3y = -15\)

Substituting (1) into (2):
\(x - 3(25 - 3x) = -15\)
\(x - 75 + 9x = -15\)
\(10x = 60 \implies x = 6\).

Substituting \(x = 6\) back into (1):
\(y = 25 - 3(6) = 7\).

So the coordinates of \(P\) are \((6, 7)\).

(a)
[M1] for finding the correct midpoint of \(AB\): \((5, 10)\).
[M1] for calculating the perpendicular gradient \(-3\).
[A1] for the correct final equation in the required form: \(3x + y = 25\) (or equivalent integer form).

(b)
[M1] for setting up a correct system of simultaneous equations.
[M1] for solving for one coordinate (e.g. \(x=6\) or \(y=7\)).
[A1] for the correct coordinates: \((6, 7)\).

(a) The null hypothesis \(H_0\): Preferred movie genre and age group are independent.

(b) The total number of people under 30 (Row 1 Total) is \(45 + 35 + 20 = 100\).
The total number of people who prefer Drama (Column 3 Total) is \(20 + 35 = 55\).
The grand total of participants is \(100 + 100 = 200\).

The expected frequency is:
\(E = \frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}} = \frac{100 \times 55}{200} = 27.5\).

(c) Using a GDC to run the \(\chi^2\) two-way test on the matrix, the test statistic is \(\chi^2 \approx 10.138\) with \(df = 2\).
The corresponding \(p\)-value is approximately \(0.00629\) (or \(0.0063\) to 3 significant figures).

(d) Since the \(p\)-value (\(0.00629\)) is less than the significance level of \(0.05\), we reject the null hypothesis \(H_0\).

(a)
[A1] for stating a correct null hypothesis referencing independence between movie genre and age group.

(b)
[M1] for finding correct row, column, and grand totals.
[A1] for \(27.5\).

(c)
[A1] for \(0.00629\) (accept range \([0.0062, 0.0063]\)).

(d)
[R1] for comparing their \(p\)-value with \(0.05\).
[A1] for the correct conclusion (Reject \(H_0\)).

(a) The vertical translation \(d\) is the midline of the function, which is the average of the maximum and minimum heights:
\(d = \frac{8.2 + 2.4}{2} = 5.3\).

The amplitude \(a\) is half the difference between the maximum and minimum heights:
\(a = \frac{8.2 - 2.4}{2} = 2.9\).

Since the maximum occurs at \(t=0\), \(H(0) = 2.9\cos(0) + 5.3 = 2.9 + 5.3 = 8.2\), which is correct.

(b) The time between a maximum and a minimum is half the period. Thus, the period is:
\(T = 2 \times 6 = 12\) hours.

Using the formula for period:
\(12 = \frac{2\pi}{b} \implies b = \frac{\pi}{6} \approx 0.524\).

(c) We set \(H(t) = 4.0\):
\(2.9 \cos\left(\frac{\pi}{6}t\right) + 5.3 = 4.0\)
\(2.9 \cos\left(\frac{\pi}{6}t\right) = -1.3\)
\(\cos\left(\frac{\pi}{6}t\right) = -\frac{1.3}{2.9} \approx -0.448276\)

Solving for the first positive value:
\(\frac{\pi}{6}t = \arccos(-0.448276) \approx 2.0354\) radians
\(t \approx 2.0354 \times \frac{6}{\pi} \approx 3.887\) hours.

To 3 significant figures, the first time is \(3.89\) hours after midnight.

(a)
[A1] for \(d = 5.3\).
[A1] for \(a = 2.9\).

(b)
[M1] for finding the period to be 12 hours.
[A1] for \(b = \frac{\pi}{6}\) (or \(0.524\)).

(c)
[M1] for setting up the equation \(2.9 \cos(bt) + 5.3 = 4.0\).
[M1] for solving for \(t\).
[A1] for \(3.89\).

(a) Rewrite the function as \(A(r) = 2\pi r^2 + 1000 r^{-1}\).
Differentiating with respect to \(r\):
\(A'(r) = 4\pi r - 1000 r^{-2} = 4\pi r - \frac{1000}{r^2}\).

(b) Set the derivative equal to 0 to find the minimum:
\(4\pi r - \frac{1000}{r^2} = 0\)
\(4\pi r^3 = 1000\)
\(r^3 = \frac{250}{\pi}\)
\(r = \sqrt[3]{\frac{250}{\pi}} \approx 4.30127\text{ cm}\).
To 3 significant figures, \(r = 4.30\text{ cm}\).

(c) Substitute the value of \(r\) into the original equation:
\(A(4.30127) = 2\pi(4.30127)^2 + \frac{1000}{4.30127}\)
\(A \approx 116.22 + 232.49 = 348.71\text{ cm}^2\).
To the nearest square centimeter, the minimum surface area is \(349\text{ cm}^2\).

(a)
[M1] for attempting to differentiate with power rule.
[A1] for \(4\pi r\).
[A1] for \(-\frac{1000}{r^2}\).

(b)
[M1] for setting their derivative equal to 0.
[A1] for \(4.30\).

(c)
[M1] for substituting their value of \(r\) into the surface area equation.
[A1] for \(349\).

Let \(X\) be the weight of an apple, where \(X \sim N(150, 12^2)\).

(a) We want to find \(P(135 \le X \le 165)\).
Using GDC normal cumulative distribution function:
\(P(135 \le X \le 165) \approx 0.7887\).
To 3 significant figures, this is \(0.789\).

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

(c) Let \(Y\) be the number of small apples in a box of 20.
Then \(Y \sim B(20, 0.08)\).
We want to find \(P(Y \ge 2) = 1 - P(Y \le 1)\).
Using binomial cumulative distribution:
\(P(Y \le 1) \approx 0.5169\).
Thus, \(P(Y \ge 2) = 1 - 0.5169 = 0.4831\).
To 3 significant figures, the probability is \(0.483\).

(a)
[M1] for setting up correct normal interval: \(P(135 \le X \le 165)\).
[A1] for \(0.789\).

(b)
[M1] for setting up the inverse normal equation \(P(X < w) = 0.08\).
[A1] for \(133\).

(c)
[M1] for modeling the situation using Binomial distribution \(B(20, 0.08)\).
[M1] for writing \(1 - P(Y \le 1)\).
[A1] for \(0.483\).

(a) In the right-angled triangle \(ABT\):
\(\tan(28^\circ) = \frac{BT}{AB} = \frac{BT}{60}\)
\(BT = 60 \tan(28^\circ) \approx 31.902\text{ m}\).
To 3 significant figures, the height of the tower is \(31.9\text{ m}\).

(b) (i) In triangle \(ABC\), using the Cosine Rule to find \(AC\):
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(110^\circ)\)
\(AC^2 = 60^2 + 45^2 - 2(60)(45)\cos(110^\circ)\)
\(AC^2 = 3600 + 2025 - 5400(-0.342020)\)
\(AC^2 = 5625 + 1846.9 = 7471.9\)
\(AC \approx \sqrt{7471.9} \approx 86.44\text{ m}\).
To 3 significant figures, \(AC = 86.4\text{ m}\).

(ii) In the right-angled triangle \(CBT\), the angle of elevation of \(T\) from \(C\) is \(\angle BCT\):
\(\tan(\angle BCT) = \frac{BT}{BC} = \frac{31.902}{45} \approx 0.70893\)
\(\angle BCT = \arctan(0.70893) \approx 35.33^\circ\).
To 3 significant figures, the angle of elevation is \(35.3^\circ\).

(a)
[M1] for using tangent ratio: \(\tan(28^\circ) = \frac{BT}{60}\).
[A1] for \(31.9\).

(b)
(i)
[M1] for substituting correctly into the Cosine Rule.
[A1] for \(86.4\).

(ii)
[M1] for setting up the tangent ratio in triangle \(CBT\): \(\tan(\theta) = \frac{31.902}{45}\).
[A1] for \(35.3^\circ\) (or \(35.3\)).

(a) Integrating the function:
\(\int_{0}^{12} (12x - x^2) \mathrm{d}x = \left[ 6x^2 - \frac{x^3}{3} \right]_{0}^{12} = \left(6(144) - \frac{1728}{3}\right) - 0 = 864 - 576 = 288\).
The exact area is 288 square decameters.

(b) (i) Since \(n=4\) and the interval is from \(0\) to \(12\), the width of each interval is:
\(h = \frac{12 - 0}{4} = 3\) decameters.

(ii) The x-coordinates are \(x_0 = 0\), \(x_1 = 3\), \(x_2 = 6\), \(x_3 = 9\), \(x_4 = 12\).
The corresponding y-values are:
\(y_0 = 12(0) - 0^2 = 0\)
\(y_1 = 12(3) - 3^2 = 27\)
\(y_2 = 12(6) - 6^2 = 36\)
\(y_3 = 12(9) - 9^2 = 27\)
\(y_4 = 12(12) - 12^2 = 0\).

Using the trapezoidal rule formula:
\(\text{Area} \approx \frac{h}{2} [y_0 + y_4 + 2(y_1 + y_2 + y_3)]\)
\(\text{Area} \approx \frac{3}{2} [0 + 0 + 2(27 + 36 + 27)] = \frac{3}{2} [2(90)] = 270\).

(c) The percentage error is:
\(\text{Percentage Error} = \left| \frac{\text{Approximate} - \text{Exact}}{\text{Exact}} \right| \times 100\%\)
\(\text{Percentage Error} = \left| \frac{270 - 288}{288} \right| \times 100\% = \frac{18}{288} \times 100\% = 6.25\%\).

(a)
[M1] for a correct integration attempt resulting in \(6x^2 - \frac{x^3}{3}\).
[A1] for \(288\).

(b)
(i)
[A1] for showing the calculation \(\frac{12-0}{4} = 3\).

(ii)
[M1] for finding the correct mid-values of \(y\): \(27, 36, 27\).
[M1] for substituting values correctly into the trapezoidal rule formula.
[A1] for \(270\).

(c)
[M1] for setting up the correct percentage error formula.
[A1] for \(6.25\%\) (or \(6.25\)).

(a) Let the width be \(w\) and the length be \(2w\). The volume is given by \(V = \text{length} \times \text{width} \times \text{height} = 2w \times w \times h = 2w^2 h\). Since \(V = 12\), we have \(2w^2 h = 12 \implies h = \frac{12}{2w^2} = \frac{6}{w^2}\).
(b) The container has an open top. The total surface area \(A\) is the sum of the base and the four vertical sides: \(A = \text{Base} + 2(\text{Front}) + 2(\text{Side}) = (2w \times w) + 2(2w \times h) + 2(w \times h) = 2w^2 + 4wh + 2wh = 2w^2 + 6wh\). Substituting \(h = \frac{6}{w^2}\) gives: \(A = 2w^2 + 6w\left(\frac{6}{w^2}\right) = 2w^2 + \frac{36}{w}\).
(c) Differentiating \(A\) with respect to \(w\): \(\frac{\text{d}A}{\text{d}w} = 4w - \frac{36}{w^2}\).
(d) To minimize the surface area, set \(\frac{\text{d}A}{\text{d}w} = 0 \implies 4w - \frac{36}{w^2} = 0 \implies 4w^3 = 36 \implies w^3 = 9 \implies w = \sqrt[3]{9} \approx 2.08\text{ m}\).
(e) Substitute \(w = 2.08008\) back into the surface area equation: \(A = 2(2.08008)^2 + \frac{36}{2.08008} \approx 25.96 \approx 26.0\text{ m}^2\).

(a) M1 for volume formula \(V = 2w^2 h\), A1 for setting equal to 12, A1 for rearranging to \(h = \frac{6}{w^2}\).
(b) M1 for area formula \(A = 2w^2 + 6wh\), M1 for substitution of \(h\), A1 for showing the final expression.
(c) M1 for attempting to differentiate, A1 for \(4w\), A1 for \(-\frac{36}{w^2}\).
(d) M1 for setting derivative to 0, M1 for algebraic step \(4w^3 = 36\), A1 for \(w^3 = 9\), A1 for \(w \approx 2.08\).
(e) M1 for substituting their \(w\) into the area formula, A1.71 for the final answer \(26.0\text{ m}^2\) (accept 26 or 25.96).

(a) The midpoint of \(AB\) is \(M = \left(\frac{2+8}{2}, \frac{8+10}{2}\right) = (5, 9)\). The gradient of \(AB\) is \(m = \frac{10-8}{8-2} = \frac{2}{6} = \frac{1}{3}\). The gradient of the perpendicular bisector is \(m_{\perp} = -3\). The equation of the perpendicular bisector is \(y - 9 = -3(x - 5) \implies y = -3x + 24\).
(b) Set the two perpendicular bisectors equal to find the circumcenter: \(-3x + 24 = -0.5x + 9 \implies 2.5x = 15 \implies x = 6\). Substituting \(x = 6\) into \(y = -3x + 24\) gives \(y = -3(6) + 24 = 6\). The circumcenter is at \((6, 6)\).
(c) Find the squared distance from \(H(5, 6)\) to each tower: \(HA^2 = (5-2)^2 + (6-8)^2 = 9 + 4 = 13 \implies HA = \sqrt{13} \approx 3.61\text{ km}\); \(HB^2 = (5-8)^2 + (6-10)^2 = 9 + 16 = 25 \implies HB = 5\text{ km}\); \(HC^2 = (5-4)^2 + (6-2)^2 = 1 + 16 = 17 \implies HC = \sqrt{17} \approx 4.12\text{ km}\). Since \(HA\) is the shortest distance, Tower \(A\) is the closest.
(d) The distance is \(HA = \sqrt{13} \approx 3.61\text{ km}\).

(a) M1 for finding midpoint (5, 9), M1 for finding gradient 1/3, M1 for perpendicular gradient -3, A1 for final equation \(y = -3x + 24\).
(b) M1 for setting the two linear equations equal, M1 for solving for \(x\), A1 for \(x = 6\), A1 for \(y = 6\).
(c) M1 for attempting to use the distance formula, A1 for calculating distances (e.g., \(\sqrt{13}\), \(5\), \(\sqrt{17}\)), A1 for concluding Tower A is closest.
(d) M2 for identifying the distance as \(\sqrt{13}\), A2.71 for the final value 3.61.

(a) \(\bar{T} = \frac{4+6+8+10+12+14+16+18}{8} = 11\). \(\bar{C} = \frac{110+95+80+75+55+45+30+20}{8} = 63.75 \approx 63.8\).
(b) Using a GDC, the product-moment correlation coefficient is \(r \approx -0.997\) (or exact \(-0.996839\)).
(c) There is a very strong, negative linear correlation.
(d) Using a GDC, the linear regression line is \(C = -6.43T + 134.46\) (or \(C = -6.43T + 134\) to 3 significant figures).
(e) When \(T = 11\), \(C = -6.42857(11) + 134.464 = 63.75 \approx 64\) cups.
(f) The estimate is highly reliable because \(11^{\circ}\text{C}\) lies within the range of the original data (interpolation) and the correlation coefficient is extremely high in magnitude.
(g) Using \(T = 25^{\circ}\text{C}\) involves extrapolation (predicting outside the range of given data), which is unreliable because the linear trend may not continue (for instance, the model would predict negative cups of hot chocolate sold, which is physically impossible).

(a) A1 for \(\bar{T} = 11\), A1 for \(\bar{C} = 63.8\) (or 63.75).
(b) A2 for \(r \approx -0.997\) (accept -0.996 to -1.00).
(c) A1 for 'strong negative' (both words required).
(d) M1 for recognizing linear regression, A1 for slope \(a \approx -6.43\), A1 for intercept \(b \approx 134\).
(e) M1 for substituting 11 into their equation, A1 for \(63.8\) or 64.
(f) R1 for mentioning interpolation / within data range, R1 for pointing out the very strong correlation.
(g) R2 for identifying this as extrapolation / outside range, R1.71 for discussing the physical/logical limitation (e.g., sales cannot be negative).

(a) Using the TVM Solver on a GDC: \(N = 60\), \(I\% = 6.5\), \(PV = 25000\), \(FV = 0\), \(P/Y = 12\), \(C/Y = 12\). Solving for \(PMT\) gives \(PMT = -489.15\). Thus, the monthly payment is $489.15.
(b) Total amount paid: \(489.15 \times 60 = \$29,349.00\).
(c) Total interest: $29,349.00 - $25,000.00 = $4,349.00.
(d) To find the remaining balance after 36 payments, use TVM Solver with \(N = 36\), \(I\% = 6.5\), \(PV = 25000\), \(PMT = -489.15\), \(P/Y = 12\), \(C/Y = 12\). Solving for \(FV\) gives \(FV = -10981.18\). Thus, the remaining balance is $10,981.18.
(e) For the savings account: \(PV = 20000\), \(I\% = 4.8\), \(N = 5 \times 4 = 20\) quarters, \(P/Y = 4\), \(C/Y = 4\). Using GDC TVM Solver or formula \(FV = 20000(1 + 0.048/4)^{20} \approx 25387.74\). The value of the investment is $25,387.74.

(a) M2 for setting up TVM Solver parameters (e.g. \(N=60\), \(I=6.5\), etc.), A2 for \(PMT = \$489.15\) (accept $489.15 or $489.14 depending on rounding).
(b) M1 for \(489.15 \times 60\), A1 for $29,349.00.
(c) M1 for subtracting 25000 from their part (b), A1 for $4,349.00.
(d) M2 for updating TVM parameters for \(N = 36\) (or using remaining \(N = 24\)), A2 for $10,981.18 (accept $10,981.02 depending on exact PMT rounding).
(e) M1 for compounding quarterly setup, A1 for \(N = 20\), A1.71 for $25,387.74.

(a) Since the initial temperature is \(85^{\circ}\text{C}\), \(T(0) = 85 \implies A e^{0} + 20 = 85 \implies A + 20 = 85 \implies A = 65\).
(b) At \(t = 5\), \(T(5) = 60 \implies 65 e^{-5k} + 20 = 60 \implies 65 e^{-5k} = 40 \implies e^{-5k} = \frac{40}{65} = \frac{8}{13}\). Taking natural logs on both sides: \(-5k = \ln\left(\frac{8}{13}\right) \approx -0.485508 \implies k \approx 0.0971\) (to 4 decimal places).
(c) Substituting \(t = 15\) and \(k = 0.0971\): \(T(15) = 65 e^{-15(0.0971)} + 20 \approx 35.2^{\circ}\text{C}\).
(d) As \(t \to \infty\), \(e^{-kt} \to 0\). Thus, \(T(t) \to 20^{\circ}\text{C}\).
(e) The rate of change of temperature is given by the derivative \(T'(t) = -k A e^{-kt} = -0.0971 \times 65 e^{-0.0971 t}\). At \(t = 10\), \(T'(10) = -0.0971 \times 65 e^{-10(0.0971)} \approx -2.39^{\circ}\text{C/min}\). The rate of decrease is \(2.39^{\circ}\text{C/min}\).

(a) M1 for setting \(T(0) = 85\), A1 for showing \(A = 65\).
(b) M1 for setting up the equation \(65 e^{-5k} + 20 = 60\), M1 for simplifying to \(e^{-5k} = \frac{8}{13}\), M1 for taking natural logs, A1 for \(k \approx 0.0971\).
(c) M1 for substituting \(t=15\), A1 for \(35.2^{\circ}\text{C}\) (or \(35.1^{\circ}\text{C}\) using exact \(k\)).
(d) A2 for \(20^{\circ}\text{C}\).
(e) M2 for attempting to differentiate \(T(t)\), A2 for \(T'(t) = -6.3115 e^{-0.0971 t}\), M1 for substituting \(t = 10\), A0.71 for the rate of decrease of \(2.39^{\circ}\text{C/min}\).

(a) The particle is at rest when \(v(t) = 0 \implies 3t^2 - 12t + 9 = 0 \implies 3(t^2 - 4t + 3) = 0 \implies 3(t-1)(t-3) = 0\). Thus, \(t = 1\) s and \(t = 3\) s.
(b) Acceleration \(a(t) = v'(t) = 6t - 12\). The initial acceleration is at \(t = 0 \implies a(0) = 6(0) - 12 = -12\text{ m s}^{-2}\).
(c) The particle changes direction at \(t = 1\) and \(t = 3\). The displacement function is \(s(t) = \int (3t^2 - 12t + 9)\text{d}t = t^3 - 6t^2 + 9t + c\). Let \(s(0) = 0\). Then \(s(1) = 4\) and \(s(3) = 0\). The total distance is \(|s(1) - s(0)| + |s(3) - s(1)| = |4 - 0| + |0 - 4| = 4 + 4 = 8\text{ m}\).
(d) The displacement at \(t = 4\) is \(s(4) - s(0) = (4^3 - 6(4^2) + 9(4)) - 0 = 64 - 96 + 36 = 4\text{ m}\).

(a) M1 for setting \(v(t) = 0\), M1 for factoring or quadratic formula, A1 for \(t = 1\) and \(t = 3\) (both needed).
(b) M1 for differentiating \(v(t)\) to find \(a(t) = 6t - 12\), M1 for substituting \(t = 0\), A1 for \(-12\text{ m s}^{-2}\).
(c) M1 for realizing integration is needed, M1 for dividing interval at \(t = 1\), M1 for correct integration to \(t^3 - 6t^2 + 9t\), A1 for calculating intermediate positions 4 and 0, A1 for sum of absolute distances = 8 m.
(d) M2 for definite integral \(\int_{0}^{4} (3t^2 - 12t + 9)\text{d}t\), A2.71 for final answer 4 m.

(a) Let \(X \sim N(150, 20^2)\). We want to find \(P(130 < X < 170)\). Using GDC normalCDF(130, 170, 150, 20), we get \(0.682689 \approx 0.683\).
(b) We want to find \(P(X < 110)\). Using GDC normalCDF(-\infty, 110, 150, 20), we find \(0.02275\). Expressing this as a percentage: \(2.28\%\).
(c) We want to find the value \(x\) such that \(P(X > x) = 0.10 \implies P(X \le x) = 0.90\). Using inverse normal on GDC: \(x = \text{invNorm}(0.90, 150, 20) \approx 175.63\text{ g}\). Rounding to the nearest gram, the minimum mass is 176 g.
(d) Let \(Y\) be the number of premium apples in a sample of 5. This follows a binomial distribution \(Y \sim B(5, 0.10)\). We need to find \(P(Y = 2) = \binom{5}{2} (0.10)^2 (0.90)^3 = 10 \times 0.01 \times 0.729 = 0.0729\).

(a) M2 for normal distribution setup, A1 for 0.683 (or 68.3\%).
(b) M2 for finding \(P(X < 110)\), A1 for 2.28\% (accept 2.27\% to 2.28\%).
(c) M2 for writing the inverse normal statement \(P(X < x) = 0.90\), A1 for finding 175.63, A1 for rounding to 176 g.
(d) M2 for identifying Binomial distribution with \(n = 5\) and \(p = 0.10\), M2 for binomial formula or GDC binomialPDF setup, A1.71 for 0.0729.

(a) From the probabilities given, we write the transition matrix \(T\) by reading the destination stations down columns (origin stations):
\(T = \begin{pmatrix} 0.6 & 0.2 & 0.1 \\ 0.2 & 0.6 & 0.3 \\ 0.2 & 0.2 & 0.6 \end{pmatrix}\).

(b)
(i) Tuesday morning (Day 1):
\(\mathbf{x}_1 = T \mathbf{x}_0 = \begin{pmatrix} 0.6 & 0.2 & 0.1 \\ 0.2 & 0.6 & 0.3 \\ 0.2 & 0.2 & 0.6 \end{pmatrix} \begin{pmatrix} 160 \\ 100 \\ 100 \end{pmatrix} = \begin{pmatrix} 126 \\ 122 \\ 112 \end{pmatrix}\).
So there are 126 bikes at A, 122 at B, and 112 at C.
(ii) Wednesday morning (Day 2):
\(\mathbf{x}_2 = T \mathbf{x}_1 = \begin{pmatrix} 0.6 & 0.2 & 0.1 \\ 0.2 & 0.6 & 0.3 \\ 0.2 & 0.2 & 0.6 \end{pmatrix} \begin{pmatrix} 126 \\ 122 \\ 112 \end{pmatrix} = \begin{pmatrix} 111.2 \\ 132 \\ 116.8 \end{pmatrix}\).
So there are 111.2 (or 111) bikes at A, 132 at B, and 116.8 (or 117) at C.

(c) At steady-state, \(T \mathbf{x} = \mathbf{x}\) where \(x_A + x_B + x_C = 360\). This yields:
\(-0.4 x_A + 0.2 x_B + 0.1 x_C = 0\)
\(0.2 x_A - 0.4 x_B + 0.3 x_C = 0\)
\(0.2 x_A + 0.2 x_B - 0.4 x_C = 0\)
From the third equation: \(x_C = 0.5(x_A + x_B)\).
Substituting into the first equation: \(-0.4 x_A + 0.2 x_B + 0.05 x_A + 0.05 x_B = 0 \implies 0.25 x_B = 0.35 x_A \implies x_B = 1.4 x_A\).
Then \(x_C = 0.5(x_A + 1.4 x_A) = 1.2 x_A\).
Since \(x_A + x_B + x_C = 360\):
\(x_A + 1.4 x_A + 1.2 x_A = 360 \implies 3.6 x_A = 360 \implies x_A = 100\).
Thus, \(x_B = 140\) and \(x_C = 120\).

(d)
(i) The redistribution equations are:
\(x_{A,\text{new}} = x_{A,\text{old}} + 0.10 x_{B,\text{old}}\) (100% of A's bikes stay, 10% of B's are added)
\(x_{B,\text{new}} = 0.85 x_{B,\text{old}}\) (85% of B's bikes stay)
\(x_{C,\text{new}} = 0.05 x_{B,\text{old}} + x_{C,\text{old}}\) (5% of B's bikes are added, 100% of C's bikes stay)
In matrix form \(\mathbf{x}_{\text{new}} = R \mathbf{x}_{\text{old}}\), this gives:
\(R = \begin{pmatrix} 1 & 0.10 & 0 \\ 0 & 0.85 & 0 \\ 0 & 0.05 & 1 \end{pmatrix}\).
(ii) \(M = R T = \begin{pmatrix} 1 & 0.10 & 0 \\ 0 & 0.85 & 0 \\ 0 & 0.05 & 1 \end{pmatrix} \begin{pmatrix} 0.6 & 0.2 & 0.1 \\ 0.2 & 0.6 & 0.3 \\ 0.2 & 0.2 & 0.6 \end{pmatrix} = \begin{pmatrix} 0.62 & 0.26 & 0.13 \\ 0.17 & 0.51 & 0.255 \\ 0.21 & 0.23 & 0.615 \end{pmatrix}\).

(e) To find the new steady-state, we solve \(M \mathbf{x} = \mathbf{x}\) subject to \(x_A + x_B + x_C = 360\). Using GDC or solving the systems of equations:
\(x_A \approx 119.7\)
\(x_B \approx 109.6\)
\(x_C \approx 130.7\)

(f)
(i) Let \(\mathbf{y}\) be the state of the bikes before redistribution. We have \(\mathbf{x}_{steady} = R \mathbf{y}\). Looking at the second row:
\(x_{B, steady} = 0.85 y_B \implies y_B = \frac{109.565}{0.85} \approx 128.9\) bikes.
(ii) Number of bikes moved to A: \(0.10 \times 128.9 = 12.89\).
Number of bikes moved to C: \(0.05 \times 128.9 = 6.445\).
Daily cost \(= (12.89 \times 3.50) + (6.445 \times 2.00) \approx 45.115 + 12.89 = \$58.01\).
(iii) Annual cost \(= 365 \times 58.01 \approx \$21,173.65\). Since \(\$21,173.65 < \$22,000\), the budget is sufficient.

(a)
M1: Attempt to write down transition matrix with entries in correct positions.
A1: Correct transition matrix \(T\). [2 marks]

(b)
M1: Method to find \(\mathbf{x}_1\).
A1: Correct results for Tuesday: 126 at A, 122 at B, 112 at C.
M1: Method to find \(\mathbf{x}_2\).
A1: Correct results for Wednesday: 111.2 at A, 132 at B, 116.8 at C. [4 marks]

(c)
M1: Setting up steady-state matrix equation \(T \mathbf{x} = \mathbf{x}\).
M1: Translating to a system of equations.
M1: Utilizing constraint \(x_A + x_B + x_C = 360\).
A1: Solving for ratios (e.g., \(x_B = 1.4 x_A\)).
A1: Correct steady-state distribution: 100 at A, 140 at B, 120 at C. [5 marks]

(d)
(i) R1: Explaining that since 15% leave B, 85% remain at B.
A1: Constructing equations for the transition \(\mathbf{x}_{\text{new}} = R \mathbf{x}_{\text{old}}\) to show the rows of \(R\).
(ii) M1: Setting up matrix multiplication \(R T\).
A2: Correctly calculated matrix \(M\) (award A1 for 1 or 2 correct rows). [5 marks]

(e)
M1: Attempting to solve \(M \mathbf{x} = \mathbf{x}\).
M1: Applying the constraint \(x_A + x_B + x_C = 360\).
A1: \(x_A \approx 119.7\).
A1: \(x_B \approx 109.6\).
A1: \(x_C \approx 130.7\). [5 marks]

(f)
(i) M1: Realizing relationship \(\mathbf{x}_{steady} = R \mathbf{y}\).
M1: Writing the equation \(x_{B, steady} = 0.85 y_B\).
A1: Finding \(y_B \approx 128.9\).
(ii) M1: Setting up the cost calculation \((0.10 y_B \times 3.50) + (0.05 y_B \times 2.00)\).
A1: Correct daily cost: \(\$58.01\) (accept \(\$58.00\)).
(iii) M1: Setting up the annual cost calculation \(365 \times 58.01\).
R1: Correctly comparing with budget and making a valid conclusion. [7 marks]

(a) Set \(\frac{dP}{dt} = 0 \implies 0.5 P \left(1 - \frac{P}{10}\right) = 0\). This gives two equilibrium solutions: \(P = 0\) and \(P = 10\).
- \(P = 0\) is unstable: if \(P > 0\), then \(\frac{dP}{dt} > 0\), and the population grows away from 0. Physically, it is the extinction state.
- \(P = 10\) is stable: if \(0 < P < 10\), then \(\frac{dP}{dt} > 0\) (population increases toward 10); if \(P > 10\), then \(\frac{dP}{dt} < 0\) (population decreases toward 10). Physically, this represents the carrying capacity of the lake (10,000 fish).

(b) For \(h = 0.8\), we solve \(0.5 P - 0.05 P^2 - 0.8 = 0 \implies 0.05 P^2 - 0.5 P + 0.8 = 0\).
Multiplying by 20: \(P^2 - 10P + 16 = 0 \implies (P - 2)(P - 8) = 0\).
Thus, the equilibrium solutions are \(P = 2\) (2,000 fish) and \(P = 8\) (8,000 fish).

(c)
(i) If \(P(0) = 1.5\), then since \(1.5 < 2\), \(\frac{dP}{dt} < 0\), meaning the population continuously decreases and goes extinct (approaches 0).
If \(P(0) = 5\), then since \(2 < 5 < 8\), \(\frac{dP}{dt} > 0\), meaning the population increases and asymptotically approaches the stable carrying capacity of \(P = 8\) (8,000 fish).
(ii) The threshold to survive is the unstable equilibrium; hence, the minimum initial population must be strictly greater than 2 (2,000 fish).

(d) For equilibria, \(0.05 P^2 - 0.5 P + h = 0\).
To prevent extinction for all starting populations above the threshold, we must have real roots. The maximum sustainable constant rate occurs when the quadratic has exactly one real root (the vertex).
Using the discriminant: \(\Delta = (-0.5)^2 - 4(0.05)h = 0 \implies 0.25 - 0.2h = 0 \implies h = 1.25\) (1,250 fish per year).
The corresponding equilibrium population is \(P = -\frac{-0.5}{2(0.05)} = 5\) (5,000 fish).

(e) Let \(f(t, P) = 0.5 P - 0.05 P^2 - 0.1 P \cos^2(\pi t)\). We use \(t_0 = 0\), \(P_0 = 4\), and \(\Delta t = 0.25\).
- Step 1: at \(t = 0\):
\(f(0, 4) = 0.5(4) - 0.05(16) - 0.1(4)\cos^2(0) = 2 - 0.8 - 0.4 = 0.8\).
\(P_1 = P_0 + \Delta t \times f(0, 4) = 4 + 0.25(0.8) = 4.2\).
- Step 2: at \(t = 0.25\):
\(f(0.25, 4.2) = 0.5(4.2) - 0.05(17.64) - 0.1(4.2)\cos^2(\pi/4) = 2.1 - 0.882 - 0.21 = 1.008\).
\(P_2 = P_1 + \Delta t \times f(0.25, 4.2) = 4.2 + 0.25(1.008) = 4.452\).
- Step 3: at \(t = 0.5\):
\(f(0.5, 4.452) = 0.5(4.452) - 0.05(4.452^2) - 0.1(4.452)\cos^2(\pi/2)\).
Since \(\cos(\pi/2) = 0\), the last term is 0.
\(f(0.5, 4.452) = 2.226 - 0.05(19.8203) = 1.23498\).
\(P_3 = P_2 + \Delta t \times f(0.5, 4.452) = 4.452 + 0.25(1.23498) \approx 4.761\).
Thus, the estimated population at \(t = 0.75\) is approximately 4.761 thousand fish (or 4,761 fish).

(f)
(i) Substituting \(P(t) \approx 4.5\):
\(H \approx \int_{0}^{2} 0.1(4.5) \cos^2(\pi t) \, dt = 0.45 \int_{0}^{2} \cos^2(\pi t) \, dt\).
(ii) Applying the double-angle identity:
\(\int_{0}^{2} \cos^2(\pi t) \, dt = \int_{0}^{2} \frac{1 + \cos(2\pi t)}{2} \, dt = \left[ \frac{t}{2} + \frac{\sin(2\pi t)}{4\pi} \right]_{0}^{2} = (1 + 0) - (0) = 1\).
Thus, \(H \approx 0.45 \times 1 = 0.45\) thousand fish (or 450 fish).
(iii) One limitation is that \(P(t)\) is not constant in reality; as seen in part (e), it changes continuously, which means this model ignores the correlation between fluctuations in fish population and seasonal changes (cosine term) over time.

(a)
M1: Setting \(\frac{dP}{dt} = 0\).
A1: Finding \(P = 0\) and \(P = 10\).
R1: Describing the stability and explaining physical meanings. [3 marks]

(b)
M1: Setting up equation with \(h=0.8\).
M1: Rearranging into standard quadratic form.
A2: Finding correct equilibrium populations: \(P=2\) and \(P=8\). [4 marks]

(c)
A1: Explaining extinction for \(P(0)=1.5\).
A1: Explaining asymptotic growth to 8 for \(P(0)=5\).
M1: Correct rationale for the threshold.
A1: Identifying \(P=2\) (2,000 fish) as the minimum initial population. [4 marks]

(d)
M1: Establishing condition for tangent/vertex of the parabola.
M1: Setting discriminant \(\Delta = 0\).
A1: Correct maximum harvest rate \(h_{\text{max}} = 1.25\) and population \(P = 5\). [3 marks]

(e)
M1: Correct use of Euler's formula.
M1: Step 1 calculations yielding \(P(0.25) = 4.2\).
M1: Step 2 calculations yielding \(f(0.25, 4.2) = 1.008\).
A1: Step 2 final value \(P(0.5) = 4.452\).
M1: Step 3 calculations yielding \(f(0.5, 4.452) = 1.23498\).
A1: Step 3 final value \(P(0.75) \approx 4.761\). [6 marks]

(f)
(i) A1: Substituting \(P(t) = 4.5\).
A1: Writing down correct definite integral. [2 marks]
(ii) M1: Applying cosine double-angle identity.
M1: Finding the antiderivative.
A1: Finding \(H = 0.45\) (450 fish). [3 marks]
(iii) R1: Identifying that the population \(P(t)\) is dynamic (or not constant).
R1: Describing the limitation (such as ignoring the interaction of population cycles with seasonal harvesting rates). [2 marks]