ACSI 2022 Promo paper 2 ans

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2022 Final Exam / Paper 2 / Solutions 1 FINAL EXAMINATION 2022 YEAR 5 IB DIPLOMA PROGRAMME MATHEMATICS HIGHER LEVEL PAPER 2 ANALYSIS AND APPROACHES SECTION A Qn Solution Marks 1*. In an infinite geometric sequence, 𝑢1 = 𝑎, 𝑢2 = 𝑎2 − 2. [6 marks] 1(a) Given that 𝑎 = 9 5 show that the sum to infinity of the sequence exists. Students found the sum to infinity, when what they are required to do is to find the common ratio r and show that -1<r<1. Common error to conclude that r < 1 for sum to infinity to exist. When 𝑎 = 9 5 , 𝑟 = 𝑎2−2 𝑎 = 31 45 Since −1 < 31 45 < 1, the sum to infinity exists 1(b) Find the least value of 𝑛 such that 𝑆∞ − 𝑆𝑛 < 1 100. There is mostly error in computing least value of n and handling of equations involving inequalities. 𝑆∞ − 𝑆𝑛 = 𝑎 1−𝑟 − 𝑎(1−𝑟𝑛) 1−𝑟 = 𝑎𝑟𝑛 1−𝑟 9 5(31 45) 𝑛 1−31 45 < 1 100 By GDC, 𝑛 > 17.1 The least value of 𝑛 is 18

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2022 Final Exam / Paper 2 / Solutions 2 Qn Solution Marks 2*. The time 𝑇 hours taken for a machine to cool down to temperature 𝑥℃ after it has been turned off can be modelled by 𝑇 = 5𝑙𝑛 ( 13 𝑥−𝑘), where 𝑇 ≥ 0 and 𝑘 is a positive constant. The machine has an initial temperature of 36℃ when it is turned off. [4 marks] 2(a) Show that 𝑘 = 23. Done correctly by most students 0 = 5𝑙𝑛 ( 13 36−𝑘) 𝑘 = 23 (Shown) 2(b) Sketch the graph of the machine’s temperature for 𝑇 ≥ 0, clearly stating the coordinates of any axial intercept and the equation of any asymptote. Sketching x against T which leads to wrong graph. Note that the equation given has T as the y- axis variable by convention from the way it is expressed. x-intercept, x = 36 Vertical asymptote, x = 23 Qn Solution Marks 3*. On 1st January 2022, Sam invested $600 in an account that pays a nominal annual interest rate of 1.2%, compounded monthly. The amount of money in Sam’s account at the end of each year follows a geometric sequence with common ratio, 𝑟. [6 marks] 3(a) Find the value of 𝑟, giving your answer correct to 4 significant figures. Generally well done 𝑟 = (1 + 0.012 12 ) 12 = 1.012 3(b) Sam makes no further deposits or withdrawals from the account. Find the year in which the amount of money in Sam’s account becomes more than $1000. There are several ways of doing it, by

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2022 Final Exam / Paper 2 / Solutions 3 Qn Solution Marks 600 × 1.012𝑛 > 1000 𝑛 > ln (1000 600 ) ln (1.012) = 42.8 Year 2064 letting the general term of a GP to be greater than 1000. Common error is to confuse the term with the sum of a GP. OR GDC Financial Solver I(%) = 1.2 PV = -600 FV = 1000 PpY = 12 CpY = 12 N = 511. 08 months (number of times investment is compounded) = 42.59 years Year = 2064 Qn Solution Marks 4*. The functions 𝑓 and 𝑔 are defined by 𝑓(𝑥) = 3−2𝑥 𝑥+1 , 𝑥 ∈ 𝑅, 𝑥 ≠ −1 and 𝑔(𝑥) = 𝑥 − 𝑎, where 𝑥 ∈ 𝑅, 𝑎 ∈ 𝑅. [5 marks] 4(a) Find the range of 𝑓. Generally well done. Common answer is to state the range as simply f(x) not equal to -2 From GDC or otherwise, range of f =] − ∞, −2[ ∪ [−2, ∞[ 4(b) Given that (𝑓 ∘ 𝑔)(𝑥) has an asymptote at 𝑥 = −4 for all 𝑥 ∈ 𝑅, determine the value of 𝑎. Generally well done. 𝑓 ∘ 𝑔(𝑥) = 3−2(𝑥−𝑎) (𝑥−𝑎)+1 [composite function] = 3+2𝑎−2𝑥 𝑥−𝑎+1 𝑥 − 𝑎 + 1 = 0 [equate denominator = 0] 𝑥 = 𝑎 − 1 = −4 𝑎 = −3 Qn Solution Marks 5. Consider the quartic equation 𝑧4 − 3𝑧3 + 𝑎𝑧2 + 𝑏𝑧 + 𝑐 = 0, 𝑧 ∈ 𝐶. Two of the roots of this equation are 2 and 1 + 2𝑖. Find the possible values of 𝑎, 𝑏 and 𝑐 where 𝑎, 𝑏, 𝑐 ∈ 𝑅. [7 marks]

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2022 Final Exam / Paper 2 / Solutions 4 Another root is 1 − 2𝑖. [since all real coefficients] (𝑧 − 1 + 2𝑖)(𝑧 − 1 − 2𝑖) = 𝑧2 − 2𝑧 + 5 [find quadratic root] Let 𝑑 ∈ 𝑅 be the last unknown root 𝑧4 − 3𝑧3 + 𝑎𝑧2 + 𝑏𝑧 + 𝑐 = (𝑧 − 2)(𝑧2 − 2𝑧 + 5)(𝑧 − 𝑑) [M1: equate coefficients] Equating coefficient of 𝑧3, −3 = −𝑑 − 2 − 2, 𝑑 = −1 Equating constant term, 𝑐 = 10𝑑 = 10 Equating coefficient of 𝑧2 𝑎 = 2𝑑 + 5 + 2𝑑 + 4 = 5 Equating coefficient of 𝑧 𝑏 = −10 − 5𝑑 − 4𝑑 = −1 Considerable effort in solving this part using factor/remainder theorem and leading to wrong answers because of careless mistakes. OR Another root is 1 − 2𝑖. [since all real coefficients] Let 𝑑 ∈ 𝑅 be the last unknown root Sum of roots = − −3 1 [sum/ product of roots] 2 + 1 + 2𝑖 + 1 − 2𝑖 + 𝑑 = 3 𝑑 = −1 Product of roots = (−1)3𝑐 2(1 + 2𝑖)(1 − 2𝑖)(−1) = −𝑐 𝑐 = 10 Hence, 𝑧4 − 3𝑧3 + 𝑎𝑧2 + 𝑏𝑧 + 10 = (𝑧 − 2)(𝑧2 − 2𝑧 + 5)(𝑧 + 1) [M1: equate coefficients] Equating coefficient of 𝑧2 𝑎 = −2 − 2 + 4 + 5 = 5 Equating coefficient of 𝑧 𝑏 = −10 + 5 + 4 = −1 Qn Solution Marks 6. Find the roots of the equation (𝑤 − 2𝑖)3 = 8𝑖, 𝑤 ∈ 𝐶. Give your answers in Cartesian form. [7 marks]

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2022 Final Exam / Paper 2 / Solutions 5 Qn Solution Marks (𝑤 − 2𝑖)3 = 8𝑖 (𝑤 − 2𝑖)3 = 8 (𝑒 𝜋 2+2𝑘𝜋) , 𝑘 ∈ 𝑍 𝑤 − 2𝑖 = 8 1 3 (𝑒 𝜋 6+2𝑘𝜋 3 ) [simplify, 2𝑘𝜋] 𝑤 − 2𝑖 = √3 + 𝑖, −√3 + 𝑖, −2𝑖 [Answers in cartesian form] 𝑤 = √3 + 3𝑖, −√3 + 3𝑖, 0 Possible to solve this using GDC using Cpolyroots as the answer required is not specified to be exact. Those who use analytic method and got it wrong failed to consider 2kpi to find other possible solutions. Qn Solution Marks 7. The following diagram shows two circles of radius 1 unit, with centres at O and A respectively. The circles intersects at B. [7 marks] 7(a) Find the value of angle AOB, giving your answers in radians. Generally well done. Some forget to express answer in radians. OB = OA = AB = 1 since they are the radius of the circle. Angle AOB = 𝜋 3 (since OAB is an equilateral triangle) [correct angle] 7(b) Find the area of the shaded region. Give your answer correct to 3 significant figures.

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2022 Final Exam / Paper 2 / Solutions 6 Qn Solution Marks Area of shaded region = area of square – 2(area of sector OCB) – area of equilateral triangle Angle COB = 𝜋 6 Area of square = 1 (sides of length = radius) Area of equilateral triangle = 1 2 sin(𝐵𝑂𝐴) = 1 2 sin ( 𝜋 3) = √3 4 Area of sector OCB = 1 2 × 12 × 𝜋 6 = 𝜋 12 Area of shaded region = area of square – 2(area of sector OCB) – area of equilateral triangle = 1 − 𝜋 6 − √3 4 𝑢𝑛𝑖𝑡𝑠2 = 0.0434 𝑢𝑛𝑖𝑡𝑠2 Students who added the lines to form the square were able to solve the question easily. Note: to use radians for angles in area of sector/ segment or arc length formulas. Qn Solution Marks 8. Consider the graph of the function 𝑓(𝑥) = cos(𝑝𝑥 − 𝑞) , 𝑥 ∈ 𝑅 where 𝑝 and 𝑞 are positive constants. The graph of 𝑓 intersects the x-axis at point A, point B and point C. This is shown in the following diagram. The coordinates of B are (1.1, 0) and t