ACSI 2023 Promo Paper 2 ans

ACS (Independent) / Mathematics Department / Mathematics AA HL / Year 5 / 2023 Final Exam / Paper 2 / Solutions 1 FINAL EXAMINATION 2023 YEAR 5 IB DIPLOMA PROGRAMME MATHEMATICS HIGHER LEVEL PAPER 2 ANALYSIS AND APPROACHES SECTION A Qn Solution Marks 1*. [7 marks] (a) 4 (b) 𝑢𝑢𝑛𝑛 = 2(4)𝑛𝑛−1 For 𝑢𝑢𝑛𝑛 < 10000 2(4)𝑛𝑛−1 < 10000 4𝑛𝑛−1 < 5000 𝑛𝑛 − 1 < lg(5000) lg(4) 𝑛𝑛 < lg(5000) lg 4 + 1 = 7.1438 The largest term is 𝑢𝑢7 = 2(46) = 8192 Alternative: From GDC, we generate a sequence with general term 𝑢𝑢𝑛𝑛 = 2(4)𝑛𝑛−1 The largest term smaller than 10000 is 𝑢𝑢7 = 8192 (as 𝑢𝑢8 = 32768 > 10000)

ACS (Independent) / Mathematics Department / Mathematics AA HL / Year 5 / 2023 Final Exam / Paper 2 / Solutions 2 Qn Solution Marks Generally well attempted. However many candidates did not find the term 7U but give the answer s n=7 . (c) 𝑆𝑆𝑛𝑛 = 2(4𝑛𝑛−1) 4−1 = 2 3 (4𝑛𝑛 − 1) 𝑆𝑆𝑛𝑛 < 100000 2 3 (4𝑛𝑛 − 1) < 100000 4𝑛𝑛 − 1 < 150000 4𝑛𝑛 < 150001 𝑛𝑛 < lg(150001) lg 4 = 8.59 The largest possible value of 𝑛𝑛 is 8. Alternative: 𝑆𝑆𝑛𝑛 = 2(4𝑛𝑛−1) 4−1 = 2 3 (4𝑛𝑛 − 1) From GDC, we generate a sequence with general term 𝑆𝑆𝑛𝑛 = 2 3 (4𝑛𝑛 − 1) The largest possible value of 𝑛𝑛 is 8 (as 𝑆𝑆9 = 174762 > 100000) Well attempted. Qn Solution Marks 2*. [7 marks]

ACS (Independent) / Mathematics Department / Mathematics AA HL / Year 5 / 2023 Final Exam / Paper 2 / Solutions 3 Qn Solution Marks (a) log√𝑎𝑎 𝑚𝑚 = log𝑎𝑎 𝑚𝑚 log𝑎𝑎 √𝑎𝑎 = log𝑎𝑎 𝑚𝑚 1 2 log𝑎𝑎 𝑎𝑎 = 2 log𝑎𝑎 𝑚𝑚 = log𝑎𝑎 𝑚𝑚2 Well attempted. However, a number of students just write 2 2 ()log loga amm= without showing the proof and no mark is awarded for this. (b) log√2 𝑥𝑥 + 3 log𝑥𝑥 2 = 5 log2 𝑥𝑥2 + 3 log2 𝑥𝑥 = 5 Let 𝑦𝑦 = log2 𝑥𝑥 2𝑦𝑦 + 3 𝑦𝑦 = 5 2𝑦𝑦2 − 5𝑦𝑦 + 3 = 0 (2𝑦𝑦 − 3)(𝑦𝑦 − 1) = 0 𝑦𝑦 = 3 2 or 𝑦𝑦 = 1 Therefore 𝑥𝑥 = 2 3 2 or 𝑥𝑥 = 2 Very well done Qn Solution Marks 3*. [8 marks] (a) 𝜆𝜆+2 2 = 4.5 𝜆𝜆 = 9 − 2 = 7 Generally well done (b) Let 𝑥𝑥 = 0, 𝑦𝑦 = −4 −4 = 𝑎𝑎(−7)(−2) 𝑎𝑎 = − 4 14 = − 2 7 Generally well done

ACS (Independent) / Mathematics Department / Mathematics AA HL / Year 5 / 2023 Final Exam / Paper 2 / Solutions 4 Qn Solution Marks (c) 𝑘𝑘𝑥𝑥 + 9 = − 2 7 (𝑥𝑥 − 7)(𝑥𝑥 − 2) −7𝑘𝑘𝑥𝑥 − 63 = 2(𝑥𝑥2 − 9𝑥𝑥 + 14) 2𝑥𝑥2 − 18𝑥𝑥 + 7𝑘𝑘𝑥𝑥 + 28 + 63 = 0 2𝑥𝑥2 + (7𝑘𝑘 − 18)𝑥𝑥 + 91 = 0 Since the line is a tangent to the curve, the discriminant is 0 (7𝑘𝑘 − 18)2 − 4(2)(91) = 0 (7𝑘𝑘 − 18)2 = 728 7𝑘𝑘 − 18 = ±√728 𝑘𝑘 = 18±√728 7 = 6.43 or −1.28 Generally well done Qn Solution Marks 4. [4 marks] 3 , 32α β αβ+= = − ( ) ( ) ( ) 22 2 3 2 2 3 22 4 43 12 pp p p qq q q α β αβ α β αβ + = −⇒ + = −  =− =− =⇒= −= =− Alternative: 2 2 2 2 2 3 5122 3 12 0 3, 12 yx ySubst x yy yy pq = =  − −=   −−= = −= −

ACS (Independent) / Mathematics Department / Mathematics AA HL / Year 5 / 2