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

Let \(PV = -15000\), \(N = 36\), \(I\% = 4.8\), \(P/Y = 12\), \(C/Y = 12\), \(FV = 0\).

(a) Using a GDC TVM Solver:
\(PMT = 447.82\text{ USD}\).

(b) Total amount paid = \(447.82 \times 36 = 16121.52\text{ USD}\).
Total interest paid = \(16121.52 - 15000 = 1121.52\text{ USD}\) (accept $1121.45 if the unrounded monthly payment of \(447.818...\) is used).

(c) To find the outstanding balance after 18 payments, we find \(FV\) at \(N = 18\):
Using a GDC TVM Solver with \(PV = -15000\), \(I\% = 4.8\), \(PMT = 447.82\), \(N = 18\):
\(FV = 7775.53\text{ USD}\) (accept $7775.55 if the unrounded PMT is used).

(a)
(M1) for substituting correct values into GDC TVM Solver or equivalent formula.
(A1) for \(447.82\).

(b)
(M1) for multiplying the monthly payment by 36 and subtracting 15000.
(A1) for \(1121.52\) (or \(1121.45\) from unrounded values).

(c)
(M1) for setting \(N = 18\) in GDC TVM Solver or equivalent formula.
(A1) for \(7775.53\) (or \(7775.55\)).

(a) The reservoirs \(A(2, 8)\) and \(B(8, 8)\) lie on a horizontal line.
The midpoint of \(AB\) is \(\left(\frac{2+8}{2}, 8\right) = (5, 8)\).
The perpendicular bisector is the vertical line \(x = 5\).

(b) The midpoint of \(AC\) is \(\left(\frac{2+1}{2}, \frac{8+1}{2}\right) = (1.5, 4.5)\).
The gradient of the line segment \(AC\) is \(m_{AC} = \frac{1-8}{1-2} = 7\).
The gradient of the perpendicular bisector is \(m_{\perp} = -\frac{1}{7}\).
Using the point-slope formula with \((1.5, 4.5)\):
\(y - 4.5 = -\frac{1}{7}(x - 1.5)\)
\(y = -\frac{1}{7}x + \frac{1.5}{7} + 4.5 = -\frac{1}{7}x + \frac{3}{14} + \frac{63}{14} = -\frac{1}{7}x + \frac{66}{14}\)
\(y = -\frac{1}{7}x + \frac{33}{7}\).

(c) Substitute \(x = 5\) into the equation of the perpendicular bisector of \(AC\):
\(y = -\frac{1}{7}(5) + \frac{33}{7} = \frac{28}{7} = 4\).
So the Voronoi vertex is \(V(5, 4)\).

(a)
(M1) for finding the midpoint \((5,8)\) or recognizing it is vertical.
(A1) for \(x = 5\).

(b)
(M1) for finding the midpoint \((1.5, 4.5)\) and the gradient of \(AC\) as \(7\).
(M1) for finding the perpendicular gradient \(-\frac{1}{7}\) and substituting into the line equation.
(A1) for \(y = -\frac{1}{7}x + \frac{33}{7}\) (or equivalent).

(c)
(M1) for attempting to solve the system of equations.
(A1) for \((5, 4)\).

(a) The area of the rectangular region is \(xy = 242\).
Thus, \(y = \frac{242}{x}\).

(b) The total length of fencing is the sum of the three sides:
\(L = x + 2y\)
Substituting \(y = \frac{242}{x}\) into the formula:
\(L = x + 2\left(\frac{242}{x}\right) = x + \frac{484}{x}\).

(c) \(\frac{\text{d}L}{\text{d}x} = \frac{\text{d}}{\text{d}x}\left(x + 484x^{-1}\right) = 1 - 484x^{-2} = 1 - \frac{484}{x^2}\).

(d) To find the minimum total length, set \(\frac{\text{d}L}{\text{d}x} = 0\):
\(1 - \frac{484}{x^2} = 0 \Rightarrow x^2 = 484 \Rightarrow x = 22\) (since \(x > 0\)).
Substituting \(x = 22\) back into the expression for \(L\):
\(L = 22 + \frac{484}{22} = 22 + 22 = 44\text{ m}\).

(a)
(A1) for \(y = \frac{242}{x}\).

(b)
(M1) for expressing \(L = x + 2y\) and substituting \(y\).
(AG) for showing \(L = x + \frac{484}{x}\) clearly with no missing steps.

(c)
(M1) for taking derivative.
(A1) for \(1 - \frac{484}{x^2}\).

(d)
(M1) for setting \(\frac{\text{d}L}{\text{d}x} = 0\) and solving for \(x\).
(A1) for \(44\text{ m}\).

Let \(X \sim N(1.02, 0.015^2)\).

(a) Using GDC normalcdf with lower bound \(-\infty\), upper bound \(1.00\), \(\mu = 1.02\), \(\sigma = 0.015\):
\(P(X < 1.00) \approx 0.0912\) (to 3 s.f.).

(b) Expected value \(E(Y) = n \times p = 500 \times 0.091211 \approx 45.6\).

(c) This is a conditional probability:
\(P(1.01 \le X \le 1.03 \mid X \ge 1.00) = \frac{P(1.01 \le X \le 1.03)}{P(X \ge 1.00)}\).
Using GDC:
\(P(1.01 \le X \le 1.03) \approx 0.495015\)
\(P(X \ge 1.00) = 1 - P(X < 1.00) \approx 1 - 0.091211 = 0.908789\).
Conditional probability = \(\frac{0.495015}{0.908789} \approx 0.545\) (to 3 s.f.).

(a)
(M1) for setting up the normal probability on GDC.
(A1) for \(0.0912\).

(b)
(M1) for multiplying their probability from (a) by 500.
(A1) for \(45.6\).

(c)
(M1) for setting up the conditional probability.
(M1) for calculating the numerator \(0.495015\) or the denominator \(0.908789\).
(A1) for \(0.545\) (accept \(0.544\) depending on intermediate rounding).

(a) Substitute \(t = 1\) into the model:
\(C(1) = \frac{12(1)}{1^2 + 4} = 2.4 \text{ mg L}^{-1}\).

(b) Using the quotient rule to find the derivative:
\(C'(t) = \frac{12(t^2 + 4) - 12t(2t)}{(t^2 + 4)^2} = \frac{48 - 12t^2}{(t^2 + 4)^2}\).
Set \(C'(t) = 0 \Rightarrow 48 - 12t^2 = 0 \Rightarrow t = 2\) (since \(t \ge 0\)).
At \(t = 2\), the maximum concentration is \(C(2) = \frac{12(2)}{2^2 + 4} = 3 \text{ mg L}^{-1}\).
(Alternatively, this can be solved using the GDC maximum feature).

(c) Set \(C(t) \ge 2.5\):
\(\frac{12t}{t^2 + 4} \ge 2.5 \Rightarrow 12t \ge 2.5t^2 + 10 \Rightarrow 2.5t^2 - 12t + 10 \le 0\).
Using GDC to solve the quadratic equation \(2.5t^2 - 12t + 10 = 0\):
\(t_1 \approx 1.07\) hours, \(t_2 \approx 3.73\) hours.
Thus, the concentration is at least \(2.5 \text{ mg L}^{-1}\) for \(1.07 \le t \le 3.73\) hours.

(a)
(A1) for \(2.4\).

(b)
(M1) for attempting to find the derivative or using GDC graph maximum finder.
(A1) for \(t = 2\).
(A1) for maximum concentration \(3\).

(c)
(M1) for setting up the inequality \(C(t) \ge 2.5\).
(M1) for finding the critical values \(t \approx 1.07\) and \(t \approx 3.73\).
(A1) for \(1.07 \le t \le 3.73\) hours.

(a) The interval width is \(h = 3\).
Using the trapezoidal rule:
\(\text{Distance} \approx \frac{3}{2} [v(0) + v(12) + 2(v(3) + v(6) + v(9))]\)
\(\text{Distance} \approx 1.5 [0 + 9.8 + 2(4.2 + 7.5 + 9.1)] = 1.5 [9.8 + 41.6] = 77.1\text{ m}\).

(b) (i) Acceleration is the derivative of velocity:
\(a(t) = u'(t) = 10 \times (0.25 e^{-0.25t}) = 2.5 e^{-0.25t}\).
At \(t = 3\):
\(a(3) = 2.5 e^{-0.75} \approx 1.18\text{ m s}^{-2}\) (to 3 s.f.).
(Alternatively, using GDC numerical derivative at \(t = 3\)).

(b) (ii) Distance is the integral of velocity:
\(\text{Distance} = \int_{0}^{12} 10(1 - e^{-0.25t}) \text{ d}t\).
Using GDC to integrate:
\(\text{Distance} \approx 82.0\text{ m}\) (to 3 s.f.).

(a)
(M1) for correct substitution into the trapezoidal rule formula.
(A1) for \(77.1\text{ m}\).

(b)(i)
(M1) for finding the derivative function \(u'(t)\) or using GDC numerical derivative.
(A1) for \(1.18\text{ m s}^{-2}\).

(b)(ii)
(M1) for setting up the definite integral expression.
(A1) for \(82.0\text{ m}\).

(a) \(H_0\): The type of exercise preferred and age group are independent (or there is no association between exercise preferred and age group).

(b) Expected value = \(\frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}} = \frac{120 \times 65}{250} = 31.2\).

(c) (i) Inputting the \(2 \times 3\) table into GDC and performing a \(\chi^2\) test of independence:
\(\chi^2 \approx 6.235\), degrees of freedom = 2.
\(p\text{-value} \approx 0.0443\).

(c) (ii) Since the \(p\text{-value} \approx 0.0443 < 0.05\) (the significance level), we reject the null hypothesis. There is sufficient evidence at the \(5\%\) significance level to suggest that preferred exercise type and age group are associated.

(a)
(A1) for stating the null hypothesis clearly.

(b)
(M1) for identifying row, column and grand totals.
(A1) for showing the calculation resulting in \(31.2\).

(c)(i)
(M1) for putting values into GDC matrix.
(A1) for \(0.0443\).

(c)(ii)
(R1) for comparing their \(p\)-value with \(0.05\).
(A1) for a consistent conclusion of rejecting the null hypothesis.

(a) \(r = \frac{1.2}{1.5} = 0.8\).

(b) Growth in the 5th year:
\(G_5 = 1.5 \times 0.8^4 = 0.6144 \approx 0.614\text{ m}\) (to 3 s.f.).

(c) (i) The total height of the tree is the sum of an infinite geometric series:
\(S_{\infty} = \frac{1.5}{1 - 0.8} = \frac{1.5}{0.2} = 7.5\text{ m}\).

(c) (ii) We want to find the smallest \(n\) such that \(S_n > 0.9 \times 7.5 = 6.75\text{ m}\).
\(S_n = \frac{1.5(1 - 0.8^n)}{1 - 0.8} = 7.5(1 - 0.8^n)\).
Set up inequality:
\(7.5(1 - 0.8^n) > 6.75 \Rightarrow 1 - 0.8^n > 0.9 \Rightarrow 0.8^n < 0.1\).
Solving this using logarithms:
\(n > \frac{\ln(0.1)}{\ln(0.8)} \approx 10.3\).
Since \(n\) must be an integer, \(n = 11\).
Therefore, after 11 years, the tree's height will exceed \(90\%\) of its maximum possible height.

(a)
(A1) for \(0.8\).

(b)
(M1) for \(1.5 \times 0.8^4\).
(A1) for \(0.614\text{ m}\) (or \(0.6144\)).

(c)(i)
(M1) for using the infinite sum formula.
(A1) for \(7.5\text{ m}\).

(c)(ii)
(M1) for setting up the inequality \(S_n > 6.75\).
(M1) for using GDC table, solver or logarithms to solve the inequality.
(A1) for 11 years.

(a) Using the geometric decay formula: \(V = P(1 - r)^t\). Here, \(P = 45000\), \(r = 0.12\), and \(t = 5\). \(V = 45000 \times (0.88)^5 \approx 23747.98\). To the nearest dollar, the value is $23748. (b) We need to find the smallest integer \(n\) such that \(45000 \times (0.88)^n < 15000\). This simplifies to \((0.88)^n < \frac{1}{3}\). Taking the natural logarithm on both sides: \(n \ln(0.88) < \ln(1/3)\). Since \(\ln(0.88)\) is negative, reversing the inequality gives \(n > \frac{\ln(1/3)}{\ln(0.88)} \approx 8.59\). Thus, it takes 9 complete years.

Part (a): [M1] for setting up the correct depreciation equation: \(45000 \times 0.88^5\). [A1] for the correct final answer of 23748 (accept 23748 or $23748). Part (b): [M1] for setting up the inequality or equation \(45000 \times 0.88^n < 15000\). [A1] for finding the decimal value \(n \approx 8.59\). [A1] for the final answer of 9 (years).

(a) Let \(x\) be the width perpendicular to the wall and \(y\) be the length parallel to the wall. The total length of the fence is \(2x + y = 80\), which gives \(y = 80 - 2x\). The area \(A\) of the rectangle is \(A = x \times y = x(80 - 2x) = 80x - 2x^2\). (b) To find the maximum area, differentiate \(A\) with respect to \(x\): \(\frac{dA}{dx} = 80 - 4x\). Setting the derivative to zero for a maximum: \(80 - 4x = 0 \implies x = 20\). Substitute \(x = 20\) back into the area formula: \(A = 80(20) - 2(20)^2 = 1600 - 800 = 800 \text{ m}^2\).

Part (a): [M1] for expressing the length \(y\) in terms of \(x\) as \(y = 80 - 2x\). [A1] for multiplying by \(x\) to obtain the area formula \(A = 80x - 2x^2\) clearly. Part (b): [M1] for differentiating \(A\) to get \(80 - 4x\). [M1] for setting their derivative to zero. [A1] for finding \(x = 20\). [A1] for calculating the correct maximum area of 800 (accept \(800 \text{ m}^2\)).

(a) Let \(X\) be the weight of an apple, where \(X \sim N(150, 12^2)\). Using a GDC to find \(P(X > 165)\) yields \(P(X > 165) \approx 0.1056\). To three significant figures, this is 0.106. (b) We are given \(P(X < w) = 0.15\). Using the inverse normal function on a GDC with mean 150 and standard deviation 12, we find \(w \approx 137.56\). To three significant figures, \(w = 138\).

Part (a): [M1] for writing a correct probability statement \(P(X > 165)\). [A1] for 0.106 (accept 0.1056). Part (b): [M1] for setting up the equation \(P(X < w) = 0.15\) or showing an inverse normal setup. [A1] for 137.56. [A1] for rounding to 138 (accept 138g).

(a) The midpoint \(M\) of \(AB\) is given by \(\left(\frac{2+8}{2}, \frac{5 + (-1)}{2}\right) = (5, 2)\). (b) The gradient of the line segment \(AB\) is \(m_{AB} = \frac{-1 - 5}{8 - 2} = \frac{-6}{6} = -1\). The gradient of the perpendicular bisector is the negative reciprocal of \(-1\), which is \(m_{\perp} = 1\). Using the midpoint \((5, 2)\) in the point-slope form: \(y - 2 = 1(x - 5) \implies y = x - 3\).

Part (a): [M1] for the midpoint formula setup. [A1] for the correct midpoint coordinates (5, 2). Part (b): [M1] for finding the gradient of \(AB\) as \(-1\). [M1] for finding the perpendicular gradient as \(1\). [A1] for substituting the midpoint and perpendicular gradient into a line equation. [A1] for the final simplified equation \(y = x - 3\).

(a) At \(t=0\), \(T(0) = 22 + a e^0 = 22 + a = 85\). Solving for \(a\) gives \(a = 63\). (b) At \(t=5\), \(T(5) = 22 + 63 e^{-5k} = 60\). This simplifies to \(63 e^{-5k} = 38 \implies e^{-5k} = \frac{38}{63}\). Taking the natural logarithm: \(-5k = \ln\left(\frac{38}{63}\right) \implies k = -\frac{1}{5}\ln\left(\frac{38}{63}\right) \approx 0.1011\). To three significant figures, \(k = 0.101\). (c) When \(t = 15\), \(T(15) = 22 + 63 e^{-15(0.1011...)} \approx 22 + 13.87 = 35.87\). To three significant figures, the temperature is 35.9 degrees Celsius.

Part (a): [M1] for setting up the equation \(22 + a = 85\). [A1] for \(a = 63\). Part (b): [M1] for substituting \(T=60\), \(t=5\), and their \(a\) value into the model. [M1] for algebraic steps or GDC usage to solve for \(k\). [A1] for \(k = 0.101\) (accept 0.1011). Part (c): [M1] for substituting \(t=15\) and their parameters into the model. [A1] for 35.9.

(a) From the statements, we write: \(2A + 3C + S = 51\), \(3A + C + 2S = 57\), and \(A + 4C + S = 45\). (b) This system can be written in matrix form: \(\begin{pmatrix} 2 & 3 & 1 \\ 3 & 1 & 2 \\ 1 & 4 & 1 \end{pmatrix} \begin{pmatrix} A \\ C \\ S \end{pmatrix} = \begin{pmatrix} 51 \\ 57 \\ 45 \end{pmatrix}\). Solving using a graphic display calculator yields: \(A = 12.50\), \(C = 6.50\), \(S = 6.50\). Thus, an Adult ticket costs $12.50, a Child ticket costs $6.50, and a Senior ticket costs $6.50.

Part (a): [M1] for attempting to formulate three linear equations. [A1] for all three correct equations. Part (b): [M1] for setting up the matrix system or displaying an algebraic elimination method. [A1] for Adult = $12.50. [A1] for Child = $6.50. [A1] for Senior = $6.50. (Award at most [A2] if currency/decimal format is totally omitted, but accept correct numerical answers).

(a) The null hypothesis \(H_0\): A person's favorite music genre is independent of their gender. (b) Total females = 100. Total who prefer Rock = 45 + 30 = 75. Expected frequency \(= \frac{100 \times 75}{200} = 37.5\). (c) Using the GDC's chi-squared two-way test with observed matrix: \(\begin{pmatrix} 35 & 45 & 20 \\ 45 & 30 & 25 \end{pmatrix}\), we obtain \(\chi^2 \approx 4.806\) and \(p\text{-value} \approx 0.0905\). Since the \(p\text{-value} = 0.0905 > 0.05\), we fail to reject the null hypothesis at the 5% significance level.

Part (a): [A1] for stating a correct null hypothesis mentioning 'independent'. Part (b): [M1] for the formula \((100 \times 75) / 200\). [A1] for 37.5. Part (c): [M1] for entering data correctly into GDC and finding the \(p\)-value. [A1] for \(p \approx 0.0905\). [R1] for comparing \(p\)-value to 0.05 and concluding correctly to not reject \(H_0\).

(a) The initial velocity is when \(t = 0\): \(v(0) = 3(0)^2 - 4(0) + 5 = 5 \text{ m s}^{-1}\). (b) Acceleration is the derivative of velocity: \(a(t) = \frac{dv}{dt} = 6t - 4\). At \(t = 3\), \(a(3) = 6(3) - 4 = 14 \text{ m s}^{-2}\). (c) The total distance is given by \(\int_0^4 |v(t)| dt\). Since \(v(t) = 3t^2 - 4t + 5\) has no real roots (the discriminant is negative), \(v(t) > 0\) for all \(t\). Therefore, distance \(= \int_0^4 (3t^2 - 4t + 5) dt = [t^3 - 2t^2 + 5t]_0^4 = (4^3 - 2(4)^2 + 5(4)) - 0 = 64 - 32 + 20 = 52 \text{ meters}\).

Part (a): [A1] for 5. Part (b): [M1] for differentiating to find \(a(t) = 6t - 4\). [A1] for substituting \(t=3\) to get 14. Part (c): [M1] for setting up the integral \(\int_0^4 (3t^2 - 4t + 5) dt\). [M1] for correct integration to obtain \(t^3 - 2t^2 + 5t\). [A1] for substituting limits and obtaining 52.

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

(b) The total cost consists of the cost of the base and top, and the cost of the curved side walls.
Area of base and top = \(2 \pi r^2\).
Cost of base and top = \(15 \times 2 \pi r^2 = 30 \pi r^2\).
Area of curved side walls = \(2 \pi r h\).
Cost of curved side walls = \(10 \times 2 \pi r h = 20 \pi r h\).
Substitute \(h = \frac{2}{\pi r^2}\) into the curved side walls cost:
Cost of side walls = \(20 \pi r \left(\frac{2}{\pi r^2}\right) = \frac{40}{r}\).
Thus, the total cost \(C(r)\) is:
\(C(r) = 30 \pi r^2 + \frac{40}{r}\).

(c) Differentiating \(C(r)\) with respect to \(r\):
\(\frac{\text{d}C}{\text{d}r} = 60\pi r - 40r^{-2} = 60\pi r - \frac{40}{r^2}\).

(d) To find the minimum cost, set \(\frac{\text{d}C}{\text{d}r} = 0\):
\(60\pi r - \frac{40}{r^2} = 0 \implies 60\pi r^3 = 40 \implies r^3 = \frac{40}{60\pi} = \frac{2}{3\pi}\).
\(r = \left(\frac{2}{3\pi}\right)^{1/3} \approx 0.596\text{ m}\) (3 s.f.).

(e) Substitute \(r \approx 0.5963\) back into the cost function:
\(C(0.5963) = 30\pi(0.5963)^2 + \frac{40}{0.5963} \approx 33.51 + 67.08 = 100.59\text{ dollars}\).

(f) Since \(r\) represents a physical radius, it must be strictly positive. Thus, the restriction is \(r > 0\).

(a)
M1: Attempt to use cylinder volume formula.
A1: Correct expression for \(h\) in terms of \(r\).

(b)
M1: Attempt to write cost expression for base and top.
M1: Attempt to write cost expression for side walls.
M1: Substituting \(h\) into the total cost equation.
A1: Correctly showing the final given expression.

(c)
M1: Attempt to differentiate at least one term.
A1: Correct derivative of the quadratic term.
A1: Correct derivative of the reciprocal term.

(d)
M1: Setting derivative equal to zero.
M1: Rearranging to solve for \(r^3\).
A1: Correct value of \(r\) (approx. 0.596).

(e)
M1: Substituting their \(r\) into the cost function.
A1: Correct minimum cost to 2 decimal places.

(f)
R1: Reasoning that radius must be positive.
A1: Correct inequality or interval notation.

(a) (i) Using a GDC on the given data:
\(r \approx 0.993\) (0.99270...).
(ii) The regression line of \(y\) on \(x\) is \(y = ax + b\), where:
\(a \approx 8.295\), \(b \approx 14.49\).
Thus, \(y = 8.30x + 14.5\) (3 s.f.).

(b) The gradient (8.30) represents that for each additional hour of sleep a student gets, their exam score is predicted to increase by 8.30 marks.

(c) Substituting \(x = 7.5\) into the regression equation:
\(y = 8.295(7.5) + 14.49 \approx 76.7\) (3 s.f.).

(d) The estimate is highly reliable because \(x = 7.5\) lies within the range of the original data (interpolation) and there is a very strong positive correlation between hours of sleep and exam scores (\(r \approx 0.993\)).

(e) First, rank the data for \(x\) and \(y\):
Ranks of \(x\): [2.5, 5.5, 4, 7, 1, 8, 5.5, 2.5]
Ranks of \(y\): [2, 5, 4, 7, 1, 8, 6, 3]
Using the GDC to calculate the Pearson correlation on these ranks (Spearman's rank correlation coefficient):
\(r_s \approx 0.982\) (or \(r_s \approx 0.988\) if using the formula \(1 - \frac{6\sum d^2}{n(n^2-1)}\)).

(f) Spearman's rank correlation is less sensitive to extreme outliers and does not assume a strictly linear relationship, only requiring the relationship to be monotonic.

(a) (i)
M1: Enter data in GDC list.
A1: Correct value of \(r\) to 3 s.f.
(ii)
A1: Correct gradient (8.30).
A1: Correct y-intercept (14.5).

(b)
R1: Mentioning the rate of change (increase of score).
R1: Linking the change to one additional hour of sleep.

(c)
M1: Substituting 7.5 into their regression equation.
A1: Correct score of 76.7.

(d)
R1: Stating the estimate is reliable.
R1: Providing both valid reasons (strong correlation and interpolation).

(e)
M1: Attempt to rank the \(x\) values.
M1: Attempt to rank the \(y\) values.
M1: Using ranks in GDC calculation or formula.
A1: Correct value of \(r_s\) (0.982 or 0.988).

(f)
R1: Mentioning robustness to outliers.
R1: Mentioning non-linearity / monotonicity.

(a) Loan amount = \(350000 \times (1 - 0.15) = 297500\text{ dollars}\).

(b) Using the TVM Solver on a GDC:
\(N = 300\) (25 years \(\times\) 12 months)
\(I\% = 4.2\)
\(PV = 297500\)
\(FV = 0\)
\(P/Y = 12\)
\(C/Y = 12\)
Solving for \(PMT\) yields \(-1604.03\).
So the monthly repayment is $1604.03.

(c) Total amount paid = \(1604.03 \times 300 = 481209\text{ dollars}\).
Total interest paid = \(481209 - 297500 = 183709\text{ dollars}\).

(d) To find the remaining balance after 120 payments, we evaluate the remaining 180 payments using the TVM Solver:
\(N = 180\)
\(I\% = 4.2\)
\(PMT = -1604.03\)
\(FV = 0\)
\(P/Y = 12\)
\(C/Y = 12\)
Solving for \(PV\) yields \(213925.44\).
The remaining balance is $213,925.44.

(e) Using compound interest formula for the investment account:
\(A = P\left(1 + \frac{r}{k}\right)^{kt}\)
\(P = 120000\), \(r = 0.055\), \(k = 4\), \(t = 10\).
\(A = 120000\left(1 + \frac{0.055}{4}\right)^{40} \approx 120000(1.01375)^{40} \approx 207212.43\text{ dollars}\).
Since $207,212.43 < $213,925.44, Elena does not have enough in her savings to pay off the remaining balance of the mortgage.

(a)
M1: Calculating the 85% remaining portion.
A1: Correct loan amount ($297,500).

(b)
M1: Attempt to use TVM solver with correct parameters (N, I, PV).
M1: Setting P/Y and C/Y to 12.
A1: Correct monthly repayment ($1604.03).

(c)
M1: Attempt to find total repayments made.
M1: Subtracting original loan amount from total repayments.
A1: Correct total interest ($183,709).

(d)
M1: Set up TVM solver for remaining 180 periods (or equivalent method).
M1: Correct use of monthly repayment value.
A2: Correct remaining balance ($213,925.44).

(e)
M1: Correct compound interest formula setup with \(k=4\) and \(n=40\).
A1: Correct calculation of future value ($207,212.43).
R1: Comparison of the two values.
A1: Clear final statement that she does not have enough.

(a) Midpoint of \(AB = \left(\frac{2+8}{2}, \frac{9+7}{2}\right) = (5, 8)\).
Gradient of \(AB = m_{AB} = \frac{7-9}{8-2} = -\frac{1}{3}\).
Gradient of the perpendicular bisector = \(-\frac{1}{m_{AB}} = 3\).
Equation of the perpendicular bisector:
\(y - 8 = 3(x - 5) \implies y = 3x - 7\).

(b) Midpoint of \(AC = \left(\frac{2+4}{2}, \frac{9+3}{2}\right) = (3, 6)\).
Gradient of \(AC = m_{AC} = \frac{3-9}{4-2} = -3\).
Gradient of the perpendicular bisector = \(-\frac{1}{m_{AC}} = \frac{1}{3}\).
Equation of the perpendicular bisector:
\(y - 6 = \frac{1}{3}(x - 3) \implies y = \frac{1}{3}x + 5\).

(c) Set the two equations equal to find the intersection point:
\(3x - 7 = \frac{1}{3}x + 5 \implies \frac{8}{3}x = 12 \implies x = 4.5\).
Substitute \(x = 4.5\) into \(y = 3x - 7\):
\(y = 3(4.5) - 7 = 6.5\).
The intersection point is \((4.5, 6.5)\).

(d) Distance from \((4.5, 6.5)\) to \(A(2, 9)\):
\(d = \sqrt{(4.5 - 2)^2 + (6.5 - 9)^2} = \sqrt{2.5^2 + (-2.5)^2} = \sqrt{12.5} \approx 3.54\text{ km}\) (3 s.f.).

(e) The intersection point is the circumcenter of the triangle \(ABC\). In a Voronoi diagram, this point represents the Voronoi vertex where the cell boundaries of sites \(A\), \(B\), and \(C\) meet. It is the point that is equidistant to all three sites.

(a)
A1: Correct midpoint of \(AB\).
M1: Correct gradient of \(AB\).
M1: Taking negative reciprocal of gradient.
A1: Correct equation in requested form.

(b)
A1: Correct midpoint of \(AC\).
M1: Taking negative reciprocal of gradient.
A1: Correct equation in requested form.

(c)
M1: Setting equations equal to solve.
A1: Correct \(x\)-coordinate.
A1: Correct \(y\)-coordinate.

(d)
M1: Correct use of distance formula.
A2: Correct distance of \(3.54\text{ km}\) (accept \(\sqrt{12.5}\)).

(e)
R1: Identify as circumcenter.
R1: Identify as Voronoi vertex.
R1: Identify that it is equidistant from all three sites.

(a) As \(t \to \infty\), \(e^{-0.045t} \to 0\). Therefore, \(T(t) \to a\), which must represent the constant room temperature. Hence, \(a = 21\).

(b) (i) Substitute \(t = 0\) and \(T = 85\):
\(85 = 21 + b \cdot e^{0} \implies 85 = 21 + b \implies b = 64\).
(ii) Complete function is \(T(t) = 21 + 64 e^{-0.045t}\).

(c) Substitute \(t = 15\):
\(T(15) = 21 + 64 e^{-0.045(15)} = 21 + 64 e^{-0.675} \approx 21 + 32.59 = 53.6^\circ\text{C}\).

(d) Set \(T(t) = 50\):
\(21 + 64 e^{-0.045t} = 50 \implies 64 e^{-0.045t} = 29\).
\(e^{-0.045t} = \frac{29}{64} \implies -0.045t = \ln\left(\frac{29}{64}\right)\).
\(t = \frac{\ln(29/64)}{-0.045} \approx 17.6\text{ minutes}\).

(e) Rearranging \(T = 21 + 64 e^{-0.045t}\) to make \(t\) the subject:
\(T - 21 = 64 e^{-0.045t}\)
\(\frac{T - 21}{64} = e^{-0.045t}\)
\(\ln\left(\frac{T - 21}{64}\right) = -0.045t\)
\(t = -\frac{1}{0.045} \ln\left(\frac{T - 21}{64}\right)\) (or equivalent form).

(f) Graph requirements:
- Starts at \((0, 85)\) on the vertical axis.
- Smooth decreasing exponential curve.
- Horizontal asymptote indicated clearly at the line \(T = 21\) (or marked on the axis).

(a)
A1: Correct value of \(a = 21\).

(b) (i)
M1: Setting up equation with initial condition.
A1: Correct value of \(b = 64\).
(ii)
A1: Correct full function.

(c)
M1: Substituting \(t = 15\) into the model.
A1: Correct temperature \(53.6\).

(d)
M1: Equating function to 50.
M1: Use of logs to solve for \(t\).
A1: Correct time \(17.6\).

(e)
M1: Subtracting 21 and dividing by 64.
M1: Applying natural logarithm to both sides.
A1: Correct final expression.

(f)
A1: Curve with correct general shape.
A1: Correct y-intercept labeled at 85.
A1: Asymptote line sketched or clearly indicated at \(T = 21\).
A1: Domain restricted correctly (starts at \(t=0\) and goes to \(t=60\)).

(a) Initial rate at \(t = 0\):
\(f(0) = 0^2 e^{0} + 5 = 5\text{ litres per minute}\).

(b) To find maximum, solve \(f'(t) = 0\):
\(f'(t) = 2t e^{-0.2t} - 0.2 t^2 e^{-0.2t} = t e^{-0.2t}(2 - 0.2t)\).
Set \(f'(t) = 0 \implies t = 10\text{ minutes}\) (since \(t > 0\)).
Maximum rate \(f(10) = 10^2 e^{-2} + 5 \approx 18.5\text{ litres per minute}\).

(c) The definite integral is:
\(\int_{0}^{10} (t^2 e^{-0.2t} + 5) \text{ d}t\).

(d) Width of interval \(h = \frac{10 - 0}{5} = 2\).
The points are: \(t_0=0, t_1=2, t_2=4, t_3=6, t_4=8, t_5=10\).
Evaluate \(f(t)\) at each point:
\(f(0) = 5\)
\(f(2) = 2^2 e^{-0.4} + 5 \approx 7.6812\)
\(f(4) = 4^2 e^{-0.8} + 5 \approx 12.1891\)
\(f(6) = 6^2 e^{-1.2} + 5 \approx 15.8430\)
\(f(8) = 8^2 e^{-1.6} + 5 \approx 17.9211\)
\(f(10) = 10^2 e^{-2} + 5 \approx 18.5335\)
Applying trapezoidal rule:
\(\text{Volume} \approx \frac{2}{2} [f(0) + f(10) + 2(f(2) + f(4) + f(6) + f(8))]\)
\(\text{Volume} \approx 23.5335 + 2(7.6812 + 12.1891 + 15.8430 + 17.9211) = 130.802\text{ litres}\) (or 131 litres to 3 s.f.).

(e) Using GDC numerical integration:
Exact value \(\approx 130.831\text{ litres}\).
Percentage error:
\(\text{Percentage Error} = \left| \frac{\text{Estimate} - \text{Exact}}{\text{Exact}} \right| \times 100\%\)
\(\text{Percentage Error} = \left| \frac{130.802 - 130.831}{130.831} \right| \times 100\% \approx 0.021\%\).

(a)
A1: Correct initial rate of 5.

(b)
M1: Attempting to differentiate or find maximum using GDC.
A1: Correct time of 10 minutes.
A1: Correct maximum flow rate of 18.5.

(c)
M1: Attempt at integration notation.
A1: Correct limits and integrand.

(d)
M1: Determining correct step size \(h=2\).
M1: Evaluating function at the correct coordinates.
M1: Correct substitution into trapezoidal formula.
A1: Correct values inside parentheses.
A1: Correct estimated volume of 130.80 or 131.

(e)
A2: Correct exact GDC integral value of 130.83.
M1: Correct percentage error formula structure.
A2: Correct percentage error of approx. 0.021% (accept 0.022% depending on rounding).

(a) \(P(W < 150)\) is found using GDC normal cumulative distribution function:
\(P(W < 150) \approx 0.1587\) (or 0.159 to 3 s.f.).

(b) \(P(145 < W < 185)\) is found using normal cumulative distribution function:
\(P(145 < W < 185) \approx 0.81757\).
Therefore, the percentage of standard apples is 81.8% (3 s.f.).

(c) Let \(X\) be the number of Standard apples in a sample of 12. \(X \sim \text{B}(12, 0.81757)\).
(i) \(P(X = 9) = \binom{12}{9} (0.81757)^9 (1 - 0.81757)^3 \approx 0.218\) (3 s.f.).
(ii) \(P(X \ge 10) = 1 - P(X \le 9)\).
Using GDC binomial cumulative distribution:
\(P(X \le 9) \approx 0.38072\).
\(P(X \ge 10) = 1 - 0.38072 \approx 0.619\) (3 s.f.).

(e) Let the premium threshold be \(w\). We want \(P(W > w) = 0.05 \implies P(W \le w) = 0.95\).
Using GDC inverse normal function:
\(w = \text{invNorm}(0.95, 165, 15) \approx 189.69\text{g}\) (or 190g to 3 s.f.).

(f) To find IQR, find the 25th percentile (\(Q_1\)) and 75th percentile (\(Q_3\)):
\(Q_1 = \text{invNorm}(0.25, 165, 15) \approx 154.88\text{g}\).
\(Q_3 = \text{invNorm}(0.75, 165, 15) \approx 175.12\text{g}\).
\(\text{IQR} = Q_3 - Q_1 = 175.12 - 154.88 = 20.24\text{g}\) (or 20.2g to 3 s.f.).

(a)
M1: Attempt to write correct normal probability syntax.
A1: Correct probability of 0.159.

(b)
M1: Set up of normal cumulative distribution with correct limits.
A1: Correct probability 0.81757.
A1: Expressing as a percentage 81.8%.

(c) (i)
M1: Identifying binomial distribution parameter \(n=12\) and \(p\) from (b).
A1: Correct probability of 0.218.
(ii)
M1: Recognizing need to sum terms or subtract from 1.
A1: Correct binomial cumulative distribution input.
A1: Correct probability of 0.619.

(d)
M1: Equating probability to 0.95 (or 0.05).
M1: Attempt to use Inverse Normal on GDC.
A1: Correct weight of 190g.

(e)
M1: Finding \(Q_1\) or \(Q_3\).
M1: Calculation of \(Q_3 - Q_1\).
A1: Correct IQR of 20.2g.

Part (a): Set the derivatives to zero: \(x(4 - x - 2y) = 0\) and \(y(3 - y - x) = 0\). Case 1: \(x = 0, y = 0 \implies (0,0)\). Case 2: \(x = 0, 3 - y - x = 0 \implies y = 3 \implies (0,3)\). Case 3: \(y = 0, 4 - x - 2y = 0 \implies x = 4 \implies (4,0)\). Case 4: \(4 - x - 2y = 0\) and \(3 - y - x = 0\). Subtracting the second equation from the first yields \(1 - y = 0 \implies y = 1\). Substituting back gives \(x = 2\). Thus, the fourth point is \((2,1)\). Part (b): Let \(f(x,y) = 4x - x^2 - 2xy\) and \(g(x,y) = 3y - y^2 - xy\). The partial derivatives are: \(\frac{\partial f}{\partial x} = 4 - 2x - 2y\), \(\frac{\partial f}{\partial y} = -2x\), \(\frac{\partial g}{\partial x} = -y\), \(\frac{\partial g}{\partial y} = 3 - 2y - x\). This yields \(J(x,y) = \begin{pmatrix} 4 - 2x - 2y & -2x \\ -y & 3 - x - 2y \end{pmatrix}\). Part (c)(i): At \((2,1)\), the Jacobian is \(J(2,1) = \begin{pmatrix} -2 & -4 \\ -1 & -1 \end{pmatrix}\). The characteristic equation is \((-2 - \lambda)(-1 - \lambda) - 4 = 0 \implies \lambda^2 + 3\lambda - 2 = 0\). Solving gives \(\lambda = \frac{-3 \pm \sqrt{17}}{2}\), so \(\lambda_1 \approx 0.562\) and \(\lambda_2 \approx -3.56\). Part (c)(ii): Since the eigenvalues are real and have opposite signs, the equilibrium point is a saddle point, which is unstable. Part (d)(i): Let \(x_0 = 1.5\), \(y_0 = 1.2\), \(h = 0.1\). At \(t=0\): \(f(1.5, 1.2) = 1.5(4 - 1.5 - 2.4) = 0.15\) and \(g(1.5, 1.2) = 1.2(3 - 1.2 - 1.5) = 0.36\). Next values: \(x_1 = 1.5 + 0.1(0.15) = 1.515\), \(y_1 = 1.2 + 0.1(0.36) = 1.236\). At \(t=0.1\): \(f(1.515, 1.236) = 1.515(4 - 1.515 - 2.472) = 0.019695\) and \(g(1.515, 1.236) = 1.236(3 - 1.236 - 1.515) = 0.307764\). Next values: \(x_2 = 1.515 + 0.1(0.019695) \approx 1.51697 \approx 1.52\), \(y_2 = 1.236 + 0.1(0.307764) \approx 1.26678 \approx 1.27\). Part (d)(ii): The system has stable equilibria at \((0,3)\) (Species B only) and \((4,0)\) (Species A only), separated by the stable manifold of the saddle point \((2,1)\). Since the initial state \((1.5, 1.2)\) is on the side of the boundary where \(x\) is low and \(y\) is high, the trajectory will converge to the stable node at \((0,3)\). Thus, Species B will dominate and Species A will become extinct. Part (e): The modified equations are \(\frac{dx}{dt} = x(4 - x - 2y) - 0.5\) and \(\frac{dy}{dt} = y(3 - y - x)\). For a coexistence equilibrium, setting both derivatives to zero with \(y > 0\) requires \(y = 3 - x\). Substituting this into the first equation: \(x(4 - x - 2(3 - x)) - 0.5 = 0 \implies x(x - 2) - 0.5 = 0 \implies x^2 - 2x - 0.5 = 0\). Solving for \(x > 0\): \(x = 1 + \frac{\sqrt{6}}{2} \approx 2.22\). Substituting back: \(y = 3 - (1 + \frac{\sqrt{6}}{2}) = 2 - \frac{\sqrt{6}}{2} \approx 0.775\). The new equilibrium is \((2.22, 0.775)\).

Part (a) [4 marks]: M1 for setting derivatives to 0, A1 for each correct non-zero equilibrium point: (0,0), (0,3), (4,0), (2,1). Part (b) [3 marks]: M1 for differentiating f with respect to x and y, A1 for differentiating g with respect to x and y, A1 for showing the final Jacobian matrix. Part (c)(i) [4 marks]: M1 for substituting (2,1) into the Jacobian, M1 for setting up the characteristic equation, A1 for the correct quadratic equation in lambda, A1 for correct eigenvalues. Part (c)(ii) [2 marks]: R1 for recognizing opposite signs of real eigenvalues, A1 for concluding it is a saddle point / unstable. Part (d)(i) [5 marks]: M1 for calculating f(x0, y0) and g(x0, y0), A1 for correct x1, y1 values, M1 for calculating f(x1, y1) and g(x1, y1), A2 for correct x2, y2 values. Part (d)(ii) [3 marks]: R1 for identifying the stable nodes and the role of the saddle point boundary, A1 for predicting Species B survives, A1 for predicting Species A goes extinct. Part (e) [6 marks]: M1 for writing the modified x differential equation, M1 for writing the modified y differential equation, M1 for setting dy/dt = 0 to get y = 3 - x, M1 for substituting to get a quadratic in x, A1 for the correct value of x, A1 for the correct value of y.

Part (a): A diagram with 5 states (T, B, P, C, L) with directed edges labeled with transition probabilities. Part (b): Ordering transient states as T, B, P and absorbing states as C, L: \(Q = \begin{pmatrix} 0.25 & 0.20 & 0.10 \\ 0 & 0.60 & 0.15 \\ 0 & 0.05 & 0.70 \end{pmatrix}\) and \(R = \begin{pmatrix} 0.40 & 0.05 \\ 0.20 & 0.05 \\ 0.15 & 0.10 \end{pmatrix}\). Part (c): \(I - Q = \begin{pmatrix} 0.75 & -0.20 & -0.10 \\ 0 & 0.40 & -0.15 \\ 0 & -0.05 & 0.30 \end{pmatrix}\). Invertibility is guaranteed because the probability of remaining in transient states approaches zero as time approaches infinity (since every state has a path to absorption). Part (d): Using GDC to invert \(I-Q\): \(F = \begin{pmatrix} 4/3 & 104/135 & 112/135 \\ 0 & 8/3 & 4/3 \\ 0 & 4/9 & 32/9 \end{pmatrix} \approx \begin{pmatrix} 1.333 & 0.770 & 0.830 \\ 0 & 2.667 & 1.333 \\ 0 & 0.444 & 3.556 \end{pmatrix}\). Part (e)(i): Expected months in Trial: \(F_{11} = 4/3 \approx 1.33\) months. Total expected months: \(F_{11} + F_{12} + F_{13} = 4/3 + 104/135 + 112/135 = 396/135 \approx 2.93\) months. Part (e)(ii): For Basic: \(F_{21} + F_{22} + F_{23} = 0 + 8/3 + 4/3 = 4\) months. For Premium: \(F_{31} + F_{32} + F_{33} = 0 + 4/9 + 32/9 = 4\) months. Both groups have the same expected lifetime in transient states (4 months). Part (f): \(B = F R = \begin{pmatrix} 4/3 & 104/135 & 112/135 \\ 0 & 8/3 & 4/3 \\ 0 & 4/9 & 32/9 \end{pmatrix} \begin{pmatrix} 0.40 & 0.05 \\ 0.20 & 0.05 \\ 0.15 & 0.10 \end{pmatrix} = \begin{pmatrix} 548/675 & 127/675 \\ 11/15 & 4/15 \\ 28/45 & 17/45 \end{pmatrix} \approx \begin{pmatrix} 0.812 & 0.188 \\ 0.733 & 0.267 \\ 0.622 & 0.378 \end{pmatrix}\). Part (g)(i): Expected lifetime members: \(5000 \times B_{12} = 5000 \times (127/675) \approx 941\) users. Part (g)(ii): Monthly revenue from one Trial user: \(0 \times F_{11} + 15 \times F_{12} + 30 \times F_{13} = 15(104/135) + 30(112/135) = 328/9 \approx \$36.44\). One-off lifetime membership payment: \(200 \times B_{12} = 200(127/675) \approx \$37.63\). Total expected revenue per trial user: \(328/9 + 25400/675 = 50000/675 = 2000/27 \approx \$74.07\). For 5,000 users, total revenue: \(5000 \times (2000/27) \approx \$370,370\).

Part (a) [3 marks]: M1 for drawing a diagram with five states, A2 for labeling transitions correctly. Part (b) [4 marks]: A2 for the correct Q matrix, A2 for the correct R matrix. Part (c) [3 marks]: A2 for calculating I - Q, R1 for the explanation of invertibility based on eventual absorption. Part (d) [5 marks]: M1 for setup of the inverse matrix, A4 for the correct entries (or fractions). Part (e)(i) [4 marks]: A1 for expected trial months, M1 for setting up the sum of row 1, A2 for the correct sum of transient time. Part (e)(ii) [2 marks]: A1 for Basic expected time of 4 months, A1 for Premium expected time of 4 months. Part (f) [4 marks]: M1 for multiplying F and R, A3 for the correct matrix values. Part (g)(i) [2 marks]: M1 for multiplying 5000 by B12, A1 for 941. Part (g)(ii) [3 marks]: M1 for setting up subscription revenue calculation, M1 for adding lifetime membership fee, A1 for the correct total of $370,370 (accept $370,300 to $370,400 due to rounding).