An original Thinka practice paper modelled on the structure and difficulty of the Nov 2023 SL (TZ2) IB Diploma Programme Mathematics - Applications and Interpretation paper. Not affiliated with or reproduced from IB.
卷一 (Standard Level)
Answer all twelve questions. Answers must be written within the answer boxes provided. Working must be shown.
12 題目 · 77.36000000000001 分
題目 1 · Short-Response
6.67 分
A company invests $10,000 in a savings account that pays a nominal annual interest rate of 4.5%, compounded monthly.
(a) Find the value of the investment after 5 years, to the nearest dollar.
(b) Find the minimum number of complete years required for the investment to double in value.
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解題
(a) Using the compound interest formula: \(FV = PV \times \left(1 + \frac{r}{100k}\right)^{kn}\) where \(PV = 10000\), \(r = 4.5\), \(k = 12\), and \(n = 5\). \(FV = 10000 \times \left(1 + \frac{4.5}{1200}\right)^{60}\) \(FV = 10000 \times (1.00375)^{60} \approx 12517.96\). To the nearest dollar, the value is $12,518.
(b) To double in value, we require \(FV = 20000\). \(20000 = 10000 \times (1.00375)^{12t}\) \(2 = (1.00375)^{12t}\) Taking the natural logarithm of both sides: \(\ln(2) = 12t \ln(1.00375)\) \(12t = \frac{\ln(2)}{\ln(1.00375)} \approx 185.16\text{ months}\) To find the number of years: \(t = \frac{185.16}{12} \approx 15.43\text{ years}\). Since we need the minimum number of complete years, we round up to the next integer. Therefore, it takes 16 complete years.
評分準則
(a) [3 marks] M1 for substituting correct values into the compound interest formula. A1 for a correct intermediate value (e.g. 12517.96). A1 for the final answer rounded to the nearest dollar (12518). (b) [3.67 marks] M1 for setting up the equation with 20000 or 2. M1 for solving the exponential equation (using logs or GDC). A1 for finding the time in years (approx 15.4). A1 for rounding up to the nearest complete year (16).
題目 2 · Short-Response
6.67 分
A biologist models the population, \(P\), of a bacteria culture after \(t\) hours using the function \(P(t) = a \cdot b^t + 150\), for \(t \ge 0\).
(a) Given that \(P(0) = 450\), show that \(a = 300\).
(b) Given that \(P(2) = 1350\), find the value of \(b\).
(c) Find the population of the bacteria culture after 5 hours.
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解題
(a) Substitute \(t = 0\) and \(P(0) = 450\) into the model: \(450 = a \cdot b^0 + 150\) Since \(b^0 = 1\): \(450 = a + 150 \Rightarrow a = 300\).
(b) Substitute \(t = 2\), \(P(2) = 1350\), and \(a = 300\) into the model: \(1350 = 300 \cdot b^2 + 150\) \(1200 = 300 \cdot b^2\) \(b^2 = 4\) Since population growth base \(b > 0\), \(b = 2\).
(c) To find the population after 5 hours, substitute \(t = 5\) into the function: \(P(5) = 300 \cdot 2^5 + 150\) \(P(5) = 300 \cdot 32 + 150 = 9600 + 150 = 9750\).
評分準則
(a) [2 marks] M1 for substituting t=0 and P=450 into the equation. A1 for showing a=300 clearly. (b) [2.67 marks] M1 for substituting P(2)=1350 and a=300 into the equation. A1 for obtaining b^2 = 4. A1 for selecting the positive root b=2. (c) [2 marks] M1 for substituting t=5 into the completed formula. A1 for 9750.
題目 3 · Short-Response
6.67 分
Three food trucks are located at coordinates \(A(1, 2)\), \(B(5, 2)\), and \(C(3, 6)\) on a coordinate grid.
(a) Find the equation of the perpendicular bisector of the line segment \(AB\).
(b) Find the equation of the perpendicular bisector of the line segment \(BC\).
(c) Hence, find the coordinates of the intersection of these two perpendicular bisectors.
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解題
(a) The midpoint of \(AB\) is \(\left(\frac{1+5}{2}, \frac{2+2}{2}\right) = (3, 2)\). Since the line segment \(AB\) is horizontal (its gradient is 0), its perpendicular bisector is a vertical line passing through the midpoint. Thus, the equation is \(x = 3\).
(b) The midpoint of \(BC\) is \(\left(\frac{5+3}{2}, \frac{2+6}{2}\right) = (4, 4)\). The gradient of \(BC\) is \(m_{BC} = \frac{6-2}{3-5} = \frac{4}{-2} = -2\). The gradient of the perpendicular bisector is the negative reciprocal: \(m_{\perp} = -\frac{1}{-2} = 0.5\). Using the point-slope formula with midpoint \((4, 4)\): \(y - 4 = 0.5(x - 4)\) \(y = 0.5x + 2\).
(c) Substitute \(x = 3\) from part (a) into the equation from part (b): \(y = 0.5(3) + 2 = 1.5 + 2 = 3.5\). Thus, the coordinates of the intersection are \((3, 3.5)\).
評分準則
(a) [2 marks] M1 for finding the midpoint of AB as (3, 2). A1 for x = 3. (b) [3 marks] M1 for finding the gradient of BC as -2. M1 for finding the midpoint of BC as (4, 4) and using the perpendicular gradient 0.5. A1 for y = 0.5x + 2 (or equivalent format). (c) [1.67 marks] M1 for solving the simultaneous equations. A1 for (3, 3.5) (or x=3, y=3.5).
題目 4 · Short-Response
6.67 分
A right cone has a base radius of \(r\text{ cm}\) and a slant height of \(l\text{ cm}\). The total surface area of the cone is \(90\pi\text{ cm}^2\).
Given that the radius is \(r = 5\text{ cm}\):
(a) Show that the slant height \(l = 13\text{ cm}\).
(b) Find the vertical height of the cone.
(c) Find the volume of the cone, giving your answer in terms of \(\pi\).
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解題
(a) The total surface area \(A\) of a cone is given by \(A = \pi r^2 + \pi r l\). Given \(A = 90\pi\) and \(r = 5\): \(90\pi = \pi(5)^2 + \pi(5)(l)\) \(90\pi = 25\pi + 5\pi l\) Subtract \(25\pi\) from both sides: \(65\pi = 5\pi l\) Divide by \(5\pi\): \(l = 13\text{ cm}\).
(b) The radius, vertical height \(h\), and slant height \(l\) form a right-angled triangle. By Pythagoras' theorem: \(h = \sqrt{l^2 - r^2}\) \(h = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\text{ cm}\).
(c) The volume \(V\) of a cone is given by \(V = \frac{1}{3}\pi r^2 h\). Substituting the known values: \(V = \frac{1}{3}\pi (5)^2 (12) = 4 \times 25 \times \pi = 100\pi\text{ cm}^3\).
評分準則
(a) [2.67 marks] M1 for using the formula for total surface area of a cone. M1 for substituting r=5 and A=90pi. A1 for showing l=13 clearly. (b) [2 marks] M1 for applying Pythagoras' theorem. A1 for h = 12 cm. (c) [2 marks] M1 for using the volume formula with r=5 and h=12. A1 for 100pi (accept approx 314).
題目 5 · Short-Response
6.67 分
A school offers three sports: Football, Basketball, and Tennis. A surveyor asks a random sample of 120 students about their favorite sport. The results are classified by gender in the table below:
A \(\chi^2\) test for independence is conducted at the 5% significance level.
(a) State the null hypothesis for this test.
(b) Calculate the expected frequency of male students who prefer Tennis.
(c) Find the \(p\)-value for this test.
(d) State, with a reason, whether the null hypothesis should be rejected.
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解題
(a) The null hypothesis \(H_0\) is: Favorite sport is independent of gender (or there is no association between favorite sport and gender).
(b) The expected frequency is calculated as: \(E = \frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}} = \frac{60 \times 30}{120} = 15\).
(c) Using a GDC to perform a \(\chi^2\) two-way test on the \(2 \times 3\) contingency table: \(\chi^2 \approx 16.933\) with degrees of freedom \(df = (2-1)(3-1) = 2\). The corresponding \(p\)-value is \(p \approx 0.000210\) (or \(2.10 \times 10^{-4}\)).
(d) Since the \(p\)-value (\(0.000210\)) is less than the significance level (\(0.05\)), we reject the null hypothesis.
評分準則
(a) [1 mark] A1 for stating favorite sport and gender are independent. (b) [1.67 marks] M1 for calculating row total times column total divided by grand total. A1 for 15. (c) [2 marks] M1 for setting up the matrix in GDC. A1 for p-value = 0.000210 (accept 0.00021 or 2.10 x 10^-4). (d) [2 marks] R1 for comparing p-value with 0.05. A1 for correct decision (reject H0).
題目 6 · Short-Response
6.67 分
The weight of apples in an orchard is normally distributed with a mean of 150 grams and a standard deviation of 12 grams.
(a) Find the probability that a randomly selected apple weighs more than 165 grams.
(b) Apples that weigh less than \(w\) grams are classified as \("small"\) and are sold at a discount. It is known that 10% of the apples are classified as small. Find the value of \(w\).
(c) In a box of 50 randomly selected apples, find the expected number of apples that weigh more than 165 grams.
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解題
(a) Let \(X\) be the weight of an apple, where \(X \sim N(150, 12^2)\). Using a GDC to find \(P(X > 165)\): \(P(X > 165) \approx 0.10565 \approx 0.106\) (to 3 significant figures).
(b) We are given \(P(X < w) = 0.10\). Using the inverse normal function on a GDC with area = 0.10, \(\mu = 150\), and \(\sigma = 12\): \(w \approx 134.62 \approx 135\text{ grams}\) (to 3 significant figures).
(c) The probability of an apple weighing more than 165 grams is \(p = 0.10565\). For a sample size of \(n = 50\), the expected number is given by: \(E = n \times p = 50 \times 0.10565 = 5.2825 \approx 5.28\) apples (to 3 significant figures).
評分準則
(a) [2 marks] M1 for setting up normal CDF expression. A1 for 0.106 (accept 0.10565). (b) [2.67 marks] M1 for setting up inverse normal equation. A1 for finding raw z-score of approx -1.28. A1 for 135 (accept 134.62). (c) [2 marks] M1 for multiplying probability from (a) by 50. A1 for 5.28 (accept 5.2825 or answer consistent with their part a).
題目 7 · Short-Response
6.67 分
A study is conducted on the number of hours students spent studying for a math exam (\(x\)) and their resulting exam score (\(y\)). The study hours for the 8 students in the study ranged from 2 to 10 hours.
The data is summarized with the following mean values: \(\bar{x} = 8\) \(\bar{y} = 72\)
The equation of the regression line of \(y\) on \(x\) is \(y = 3.5x + c\).
(a) Find the value of \(c\).
(b) Interpret the meaning of the gradient of the regression line in the context of this study.
(c) Estimate the exam score of a student who studied for 12 hours.
(d) State whether this estimate is reliable, giving a reason.
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解題
(a) The regression line of \(y\) on \(x\) must pass through the mean point \((\bar{x}, \bar{y})\), which is \((8, 72)\). Substitute \(x = 8\) and \(y = 72\) into the regression equation: \(72 = 3.5(8) + c\) \(72 = 28 + c \Rightarrow c = 44\).
(b) The gradient of 3.5 means that for each additional hour a student spends studying, their exam score is expected to increase by 3.5 marks.
(c) Substitute \(x = 12\) into the regression equation: \(y = 3.5(12) + 44 = 42 + 44 = 86\).
(d) The estimate is not reliable because \(x = 12\) lies outside the range of the studied data (which is 2 to 10 hours). This is extrapolation.
評分準則
(a) [2 marks] M1 for substituting the mean point into the line equation. A1 for c = 44. (b) [1.67 marks] A1 for explaining 'each additional hour of study' leads to an expected increase of 3.5 marks. (c) [1.5 marks] M1 for substituting x = 12 into their equation. A1 for 86. (d) [1.5 marks] R1 for stating 'not reliable' with the reason 'extrapolation' or 'outside the range [2, 10]'.
題目 8 · Short-Response
6.67 分
A closed rectangular box has a square base of side length \(x\text{ cm}\) and a height of \(h\text{ cm}\). The volume of the box is \(500\text{ cm}^3\).
(a) Show that the total surface area, \(A(x)\text{ cm}^2\), of the box is given by \(A(x) = 2x^2 + \frac{2000}{x}\).
(b) Find \(A'(x)\).
(c) Find the value of \(x\) that minimizes the total surface area.
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解題
(a) The volume of the rectangular box with a square base is \(V = x^2 h = 500\). Expressing \(h\) in terms of \(x\): \(h = \frac{500}{x^2}\). The total surface area of a closed box consists of 2 square bases and 4 rectangular sides: \(A(x) = 2x^2 + 4xh\) Substitute \(h = \frac{500}{x^2}\) into the surface area equation: \(A(x) = 2x^2 + 4x\left(\frac{500}{x^2}\right) = 2x^2 + \frac{2000}{x}\).
(b) Differentiate \(A(x)\) with respect to \(x\): \(A'(x) = \frac{d}{dx}\left(2x^2 + 2000x^{-1}\right) = 4x - 2000x^{-2} = 4x - \frac{2000}{x^2}\).
(a) [2.67 marks] M1 for writing volume equation in terms of x and h. M1 for writing total surface area equation. A1 for combining them to show the desired expression. (b) [2 marks] M1 for differentiating 2x^2 to get 4x. M1 for differentiating 2000/x to get -2000/x^2. (c) [2 marks] M1 for setting their derivative equal to 0. A1 for 7.94 (accept cuberoot(500)).
題目 9 · Short-Response
6 分
Elena invests $5000 in a savings account that pays a nominal annual interest rate of 3.6\% compounded monthly.
(a) Find the value of Elena's savings account after 3 years.
At the same time, Elena buys a laptop for $1500, which depreciates at a rate of 15\% per annum.
(b) Find the number of complete years it will take for the value of the laptop to fall below $500.
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解題
(a) Using the compound interest formula: \(A = P\left(1 + \frac{r}{100k}\right)^{kn}\) where \(P = 5000\), \(r = 3.6\), \(k = 12\), and \(n = 3\). \(A = 5000\left(1 + \frac{3.6}{1200}\right)^{36} = 5000(1.003)^{36} \approx 5569.34\). So the value of the savings account is $5569.34 (or $5570 to 3 s.f.).
(b) Using the depreciation formula: \(V = V_0(1 - r)^t\) where \(V_0 = 1500\), \(r = 0.15\). We want \(1500(0.85)^t < 500\), so \((0.85)^t < \frac{1}{3}\). Taking natural logarithms on both sides: \(t \ln(0.85) < \ln\left(\frac{1}{3}\right)\) Since \(\ln(0.85)\) is negative, the inequality sign reverses: \(t > \frac{\ln(1/3)}{\ln(0.85)} \approx 6.76\) years. Therefore, the number of complete years is 7.
評分準則
(a) M1 for substituting into the compound interest formula. A1 for correct values substituted: \(5000(1.003)^{36}\). A1 for correct final answer: $5569.34 (or $5570).
(b) M1 for setting up the inequality or equation: \(1500(0.85)^t < 500\). A1 for finding the decimal value \(t \approx 6.76\). A1 for rounding up to the next complete year: 7.
題目 10 · Short-Response
6 分
The height, \(h\) metres, of a drone \(t\) seconds after it is launched is modelled by the quadratic function \(h(t) = -2t^2 + bt + c\) for \(0 \le t \le T\), where \(T\) is the time the drone lands back on the ground. The drone is launched from a platform at a height of 4 metres, and it reaches a maximum height of 16.5 metres.
(b) The maximum height of a quadratic function occurs at the vertex. The time \(t\) at the vertex is given by \(t = -\frac{b}{2a} = -\frac{b}{2(-2)} = \frac{b}{4}\). The maximum height is \(h\left(\frac{b}{4}\right) = 16.5\). Substituting this into the function: \(-2\left(\frac{b}{4}\right)^2 + b\left(\frac{b}{4}\right) + 4 = 16.5\), which simplifies to \(-\frac{b^2}{8} + \frac{b^2}{4} = 12.5 \implies \frac{b^2}{8} = 12.5 \implies b^2 = 100\). Since \(t \ge 0\), we have \(b > 0\), hence \(b = 10\).
(c) The drone lands when \(h(T) = 0\). This gives \(-2T^2 + 10T + 4 = 0\). Dividing by \(-2\) gives \(T^2 - 5T - 2 = 0\). Using the quadratic formula: \(T = \frac{5 + \sqrt{(-5)^2 - 4(1)(-2)}}{2} = \frac{5 + \sqrt{33}}{2} \approx 5.37\) seconds.
評分準則
(a) A1 for \(c = 4\).
(b) M1 for expressing the axis of symmetry in terms of \(b\): \(t = \frac{b}{4}\). M1 for substituting \(t = \frac{b}{4}\) into the function and setting it equal to 16.5. AG for obtaining \(b = 10\) through correct algebraic steps.
(c) M1 for setting \(h(T) = 0\). M1 for using the quadratic formula or a GDC to solve the equation. A1 for \(5.37\) (accept 5.37228...).
題目 11 · Short-Response
6 分
A triangular garden plot \(ABC\) has sides of length \(AB = 12\text{ m}\) and \(BC = 15\text{ m}\), and the angle \(\widehat{ABC} = 74^\circ\).
(a) Find the distance from \(A\) to \(C\).
(b) Find the area of the garden plot.
(c) A gardener wants to put a fence along the side \(AC\). The fencing costs $18.50 per metre. Calculate the total cost of the fence.
(b) Using the area formula: \(\text{Area} = \frac{1}{2}ac\sin(B) = \frac{1}{2}(12)(15)\sin(74^\circ) = 90\sin(74^\circ) \approx 86.514\text{ m}^2 = 86.5\text{ m}^2\) (to 3 s.f.).
(c) Total cost = \(AC \times 18.50 = 16.4247 \times 18.50 \approx 303.86\text{ USD}\). (If using 3 s.f. value \(16.4\text{ m}\), the cost is \(16.4 \times 18.50 = 303.40\text{ USD}\). Accept both.)
評分準則
(a) M1 for substituting correct values into the cosine rule. A1 for \(16.4\text{ m}\) (accept 16.425... m).
(b) M1 for substituting correct values into the area of a triangle formula. A1 for \(86.5\text{ m}^2\) (accept 86.514... m^2).
(c) M1 for multiplying their side length \(AC\) by 18.50. A1 for $303.86 (accept $304 or $303.40 from the rounded \(16.4\)).
題目 12 · Short-Response
6 分
The weight of bags of apples packed in a farm is normally distributed with a mean of 2.05 kg and a standard deviation of 0.08 kg.
(a) Find the probability that a randomly chosen bag of apples weighs less than 2.00 kg.
(b) A supermarket will only accept bags of apples that weigh between 1.95 kg and 2.20 kg. Find the percentage of bags that are accepted by the supermarket.
(c) The farm packs 500 bags of apples in a day. Estimate the number of bags that are expected to be rejected because they weigh more than 2.20 kg.
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解題
Let \(W \sim N(2.05, 0.08^2)\).
(a) We want to find \(P(W < 2.00)\). Using a graphic display calculator: \(P(W < 2.00) \approx 0.26598... \approx 0.266\) (to 3 s.f.).
(b) We want to find \(P(1.95 < W < 2.20)\). Using a GDC: \(P(1.95 < W < 2.20) \approx 0.86395... \approx 86.4\%\) (to 3 s.f.).
(c) First, find the probability that a bag weighs more than 2.20 kg: \(P(W > 2.20) \approx 0.03039...\). Then, multiply by 500 to find the expected number: \(\text{Expected} = 500 \times 0.03039... \approx 15.199... \approx 15.2\) bags (or 15 bags to the nearest whole number).
評分準則
(a) M1 for writing a valid probability statement: \(P(W < 2.00)\). A1 for \(0.266\) (accept 0.26598...).
(b) M1 for finding the interval probability: \(P(1.95 < W < 2.20)\). A1 for \(86.4\%\) (accept 0.864 or 86.395...).
(c) M1 for calculating \(P(W > 2.20) \approx 0.0304\) and multiplying by 500. A1 for 15.2 (accept 15).
卷二 (Standard Level)
Answer all five structured long-response questions in the answer booklet. GDC required.
5 題目 · 80 分
題目 1 · Structured Long-Response
16 分
A fitness center is conducting a study on the preferences of its members regarding three types of exercise classes: Yoga, Spin, and HIIT. The members are categorized into three age groups: Under 30, 30-50, and Over 50. A random sample of 260 members is selected, and their preferences are recorded in the contingency table below:
A \(\chi^2\) test for independence is conducted at the \(5\%\) significance level.
(a) Write down the null hypothesis for this test. [1]
(b) Find the expected number of members under 30 who prefer HIIT. [2]
(c) Write down the number of degrees of freedom. [1]
(d) Find the \(p\)-value for this test. [2]
(e) State, with a reason, whether the null hypothesis should be rejected at the \(5\%\) significance level. [2]
A member is selected at random from the sample. Let \(Y\) be the event that the member prefers Yoga, and let \(O\) be the event that the member is over 50.
(f) Calculate the probability that the randomly chosen member prefers Yoga, given that they are over 50. [2]
(g) Determine if the events \(Y\) and \(O\) are independent. Justify your answer. [3]
(h) Two members who prefer Spin are chosen at random. Find the probability that both are under 30. [3]
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解題
(a) \(H_0\): The preferred type of exercise class is independent of the member's age group. [1 mark]
(d) Using a GDC to perform a \(\chi^2\) test for independence: \(\chi^2 \approx 36.8\) (or \(36.7728...\)) \(p\)-value \approx 2.01 \times 10^{-7} (or 0.000000201). [2 marks]
(e) Since the \(p\)-value \(\approx 2.01 \times 10^{-7} < 0.05\), we reject the null hypothesis. There is sufficient evidence at the \(5\%\) significance level to suggest that preferred exercise class is not independent of age group. [2 marks]
(g) First find \(P(Y) = \frac{80}{260} = \frac{4}{13} \approx 0.308\) (or \(0.307692...\)). We have \(P(Y | O) = \frac{4}{7} \approx 0.571\). Since \(P(Y | O) \neq P(Y)\) (or \(P(Y \cap O) = \frac{40}{260} \approx 0.154 \neq P(Y) \times P(O) = \frac{4}{13} \times \frac{70}{260} \approx 0.0828\)), the events \(Y\) and \(O\) are not independent. [3 marks]
(h) Total Spin members = 100. Spin members under 30 = 35. Probability that both are under 30 \(= \frac{35}{100} \times \frac{34}{99} = \frac{1190}{9900} = \frac{119}{990} \approx 0.120\) (or \(0.120202...\)). [3 marks]
評分準則
(a) M1 for stating independence between the two variables. (b) M1 for correct substituted fraction, A1 for correct value to 3 sf. (c) A1 for df = 4. (d) A2 for correct p-value (accept any equivalent scientific notation or decimal format). (e) R1 for comparing p-value with 0.05, A1 for correct conclusion. (f) M1 for correct fraction format, A1 for correct final answer. (g) M1 for finding P(Y), R1 for comparing P(Y) with P(Y|O), A1 for correct conclusion. (h) M1 for first probability, M1 for second probability without replacement, A1 for correct final answer.
題目 2 · Structured Long-Response
16 分
A weather station models the average daily temperature \(T\) (in \(^\circ\text{C}\)) in a city over a 12-month period. The temperature is modeled by the function:
$$T(t) = a \cos(b(t - c)) + d, \quad 0 \leq t \leq 12$$
where \(t\) is the number of months since the start of January (with \(t=0\) representing January 1st, and \(t=12\) representing January 1st of the following year).
It is given that the temperature cycle repeats exactly every 12 months, and that the constants satisfy \(a > 0\) and \(0 \leq c \leq 6\).
(a) Show that the period of the function is 12 months, hence find the exact value of \(b\). [2]
The maximum average daily temperature is \(26^\circ\text{C}\), which occurs at \(t = 6\) (July 1st). The minimum average daily temperature is \(8^\circ\text{C}\).
(b) Find the value of:
(i) \(d\); [2]
(ii) \(a\); [2]
(iii) \(c\). [2]
(c) Write down the complete model for \(T(t)\). [1]
(d) Find the temperature predicted by the model on April 1st (\(t = 3\)). [2]
(e) Find the total number of months during the 12-month period during which the average daily temperature is predicted to be strictly below \(12^\circ\text{C}\). [5]
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解題
(a) Since the seasonal cycle repeats annually, the period is 12. \(b = \frac{2\pi}{\text{Period}} = \frac{2\pi}{12} = \frac{\pi}{6} \approx 0.524\). [2 marks]
(b) (i) \(d\) is the mean/midline temperature: \(d = \frac{\text{Max} + \text{Min}}{2} = \frac{26 + 8}{2} = 17\). [2 marks]
(ii) \(a\) is the amplitude: \(a = \frac{\text{Max} - \text{Min}}{2} = \frac{26 - 8}{2} = 9\). [2 marks]
(iii) The maximum occurs when the cosine function takes its maximum value of 1. \(\cos(b(t - c)) = 1 \implies b(t - c) = 0 \implies t = c\). Since the maximum occurs at \(t = 6\), we must have \(c = 6\). [2 marks]
(e) We want to find when \(T(t) < 12\): \(9 \cos\left(\frac{\pi}{6}(t - 6)\right) + 17 < 12\) \(\cos\left(\frac{\pi}{6}(t - 6)\right) < -\frac{5}{9} \approx -0.5556\). Using a GDC to find the critical boundary values where \(T(t) = 12\): \(\frac{\pi}{6}(t - 6) = \pm \arccos\left(-\frac{5}{9}\right) \approx \pm 2.1598\) \(t - 6 = \pm 4.1250\) \(t_1 = 6 - 4.1250 = 1.875\) \(t_2 = 6 + 4.1250 = 10.125\) Since the cosine is less than \(-5/9\) when the angle is further from 0, this occurs for: \(0 \leq t < 1.875\) and \(10.125 < t \leq 12\). Total duration \(= (1.875 - 0) + (12 - 10.125) = 1.875 + 1.875 = 3.75\) months. [5 marks]
評分準則
(a) M1 for relating period to 12, A1 for exact value of b. (b) (i) M1 for average of max/min, A1 for 17. (ii) M1 for range half-width, A1 for 9. (iii) M1 for matching peak to t=6, A1 for c=6. (c) A1 for correct model. (d) M1 for substituting t=3, A1 for 17. (e) M1 for setting up inequality, M1 for solving for boundary values (accept 1.88 and 10.1), M1 for identifying the correct intervals, M1 for summing the interval lengths, A1 for 3.75 months.
題目 3 · Structured Long-Response
16 分
Three tourist hubs are located at points \(A(2, 5)\), \(B(9, 12)\), and \(C(10, 5)\) on a regional map, where all coordinates are given in kilometers. A new security station, \(S\), is to be built such that it is equidistant from all three tourist hubs.
(a) Find the equation of the perpendicular bisector of the line segment \(AB\). [4]
(b) Find the equation of the perpendicular bisector of the line segment \(BC\). [4]
(c) Hence, determine the coordinates of the security station, \(S\). [3]
(d) Find the distance from the security station \(S\) to each of the three tourist hubs. [2]
(e) A straight pathway runs along the line segment \(AC\). Find the shortest distance from the security station \(S\) to this pathway. [3]
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解題
(a) Perpendicular bisector of \(AB\): Midpoint of \(AB\), \(M_{AB} = \left(\frac{2+9}{2}, \frac{5+12}{2}\right) = (5.5, 8.5)\). Gradient of \(AB\), \(m_{AB} = \frac{12 - 5}{9 - 2} = \frac{7}{7} = 1\). Gradient of the perpendicular line, \(m_{\perp AB} = -1\). Equation: \(y - 8.5 = -1(x - 5.5) \implies y = -x + 14\). [4 marks]
(b) Perpendicular bisector of \(BC\): Midpoint of \(BC\), \(M_{BC} = \left(\frac{9+10}{2}, \frac{12+5}{2}\right) = (9.5, 8.5)\). Gradient of \(BC\), \(m_{BC} = \frac{5 - 12}{10 - 9} = -7\). Gradient of the perpendicular line, \(m_{\perp BC} = \frac{1}{7}\). Equation: \(y - 8.5 = \frac{1}{7}(x - 9.5) \implies 7y - 59.5 = x - 9.5 \implies 7y = x + 50\) (or \(y = \frac{1}{7}x + \frac{50}{7}\)). [4 marks]
(c) Find the intersection \(S(x, y)\) of the two bisectors: \(-x + 14 = \frac{1}{7}x + \frac{50}{7}\) Multiply by 7: \(-7x + 98 = x + 50\) \(8x = 48 \implies x = 6\). Then \(y = -6 + 14 = 8\). So, the coordinates of \(S\) are \((6, 8)\). [3 marks]
(d) Distance from \(S(6, 8)\) to any hub, say \(A(2, 5)\): \(d = \sqrt{(6 - 2)^2 + (8 - 5)^2} = \sqrt{4^2 + 3^2} = \sqrt{25} = 5\text{ km}\). [2 marks]
(e) The pathway runs along \(AC\). Since the coordinates of \(A\) and \(C\) are \((2, 5)\) and \((10, 5)\) respectively, they both lie on the horizontal line \(y = 5\). The shortest distance from the point \(S(6, 8)\) to the horizontal line \(y = 5\) is the vertical distance: \(|8 - 5| = 3\text{ km}\). [3 marks]
評分準則
(a) M1 for midpoint, M1 for gradient of AB, M1 for perpendicular gradient, A1 for correct equation. (b) M1 for midpoint, M1 for gradient of BC, M1 for perpendicular gradient, A1 for correct equation. (c) M1 for equating both equations, A1 for finding x = 6, A1 for finding y = 8. (d) M1 for substituting coordinates of S and any hub into the distance formula, A1 for 5. (e) M1 for identifying the line equation of AC as y = 5 (or finding the line equation), M1 for setting up the distance calculation, A1 for 3 km.
題目 4 · Structured Long-Response
16 分
A logistics company is designing an open-topped storage box with a rectangular base.
The base of the box has a width of \(x\) meters and a length of \(2x\) meters. The total volume of the box is designed to be exactly \(36 \text{ m}^3\).
(a) Find an expression for the height, \(h\), of the box in terms of \(x\). [2]
The box is constructed from a material that costs \(\$10\) per square meter for the base and \(\$2\) per square meter for the sides.
(b) Show that the cost, \(C(x)\) in dollars, of the material for the box is given by:
$$C(x) = 20x^2 + \frac{216}{x}$$
[4]
(c) Find \(C'(x)\). [2]
(d) Find the value of \(x\) that minimizes the cost of the box. [3]
(e) Calculate the minimum cost of the box, giving your answer to the nearest dollar. [2]
(f) Find the height of the box that minimizes the cost, and verify that this value indeed minimizes the cost. [3]
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解題
(a) Volume \(V = \text{length} \times \text{width} \times \text{height} = 2x \times x \times h = 2x^2 h\). Since \(V = 36\): \(2x^2 h = 36 \implies h = \frac{18}{x^2}\). [2 marks]
(b) Area of the base \(= x \times 2x = 2x^2 \text{ m}^2\). Cost of the base \(= 10 \times (2x^2) = 20x^2\). Area of the four side faces \(= 2(xh) + 2(2xh) = 6xh\). Cost of the sides \(= 2 \times 6xh = 12xh\). Substitute \(h = \frac{18}{x^2}\) into the cost of the sides: Cost of sides \(= 12x\left(\frac{18}{x^2}\right) = \frac{216}{x}\). Therefore, the total cost is: \(C(x) = 20x^2 + \frac{216}{x}\). [4 marks]
(e) Substituting the value of \(x\) into \(C(x)\): \(C(1.7544) = 20(1.7544)^2 + \frac{216}{1.7544} \approx 61.56 + 123.12 = 184.68\). To the nearest dollar, the minimum cost is \(\$185\). [2 marks]
(f) The height is: \(h = \frac{18}{(1.7544)^2} \approx 5.85\text{ m}\) (or \(5.84803...\)). To verify this is a minimum, we use the second derivative test: \(C''(x) = 40 + \frac{432}{x^3}\). At \(x = \sqrt[3]{5.4}\), \(C''(\sqrt[3]{5.4}) = 40 + \frac{432}{5.4} = 40 + 80 = 120 > 0\). Since the second derivative is positive, this value of \(x\) minimizes the cost. [3 marks]
評分準則
(a) M1 for volume formula set to 36, A1 for correct expression of h. (b) M1 for cost of base, M1 for sum of area of sides, M1 for cost of sides, A1 for substituting h and completing proof. (c) M1 for power rule derivative on first term, M1 for power rule on second term. (d) M1 for setting C'(x) = 0, M1 for rearranging to x^3 = 5.4, A1 for correct value. (e) M1 for substituting x back into C(x), A1 for correct cost rounded to nearest dollar. (f) M1 for calculating h, M1 for finding second derivative expression, A1 for evaluating and concluding a minimum.
題目 5 · Structured Long-Response
16 分
A real estate agency investigates the relationship between the distance from the city center, \(d\) (in kilometers), and the average price of a two-bedroom apartment, \(P\) (in thousands of USD). Data is collected for 8 apartments:
(c) (i) The gradient represents the average decrease in apartment price per additional kilometer from the city center. Specifically, for every additional kilometer, the price drops by \(21.6\) thousand USD (or \(\$21,587\)). [2 marks] (ii) The \(y\)-intercept represents the estimated average price of an apartment located at the city center (0 km), which is \(478.2\) thousand USD (or \(\$478,214\)). [2 marks]
(e) The estimate for 5.0 km is more reliable. [1 mark] This is because 5.0 km lies within the range of the sampled data (between 1.5 km and 12.0 km), which is an interpolation. [1 mark] The estimate for 15.0 km lies outside the range of the given data, which is an extrapolation, and the linear trend may not continue indefinitely. [1 mark]
評分準則
(a) (i) A2 for correct r to 3 sf. (ii) A1 for gradient, A1 for y-intercept. (b) A1 for writing down the equation correctly. (c) (i) R1 for stating the direction of change, R1 for quoting the price decrease with units. (ii) R1 for stating 'at the city center', R1 for quoting the estimated price with units. (d) (i) M1 for substitution of d=5, A1 for correct value. (ii) M1 for substitution of d=15, A1 for correct value. (e) A1 for selecting 5.0 km, R1 for explaining interpolation for 5.0 km, R1 for explaining extrapolation for 15.0 km.
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