ACSI 2021 Promo Paper 2 ans

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2021 Final Exam / Paper 2 / Solutions 1 ` FINAL EXAMINATION 2021 YEAR 5 IB DIPLOMA PROGRAMME MATHEMATICS HIGHER LEVEL PAPER 2 SECTION A Qn Solution Marks 1. [6 marks] (a) Given that 𝑎𝑎 = 1 and 𝑎𝑎 + 𝑎𝑎𝑎𝑎 + 𝑎𝑎𝑎𝑎2 = 2, we have 𝑎𝑎 + 𝑎𝑎2 = 1 𝑎𝑎2 + 𝑎𝑎 − 1 = 0 Solving the equation, we get 𝑎𝑎 = −1±√5 2 Since all terms are positive, common ratio is positive. So 𝑎𝑎 = −1+√5 2 (≈ 0.618) Many were unable to solve 3 2 10rr− += when they use sum to 3 terms. (b) 𝑆𝑆∞ = 𝑎𝑎 1−𝑟𝑟 = 1 1−0.618 = 2.618 … ≈ 2.62 (3 𝑆𝑆. 𝐹𝐹. ) Mostly ok except some wrote formula as 1 a r− .

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2021 Final Exam / Paper 2 / Solutions 2 Qn Solution Marks 2. [6 marks] (a) ( )( ) 5 13 12 13 12 5 (1 2 ) (1 3 ) Comparing: 3, 2 AB xx x x Ax Bx AB = ++− +− =−++ = = Alternative: By cover-up method, ( )( ) 5 32 13 12 13 12xx x x = ++− +− Mostly well done except some who used 1 2x=− for substitution. (b) ( ) ( ) 11 3 1 3 + 2 1 2xx −− +− ( ) 1 233 1 3 3 1 3 9 27 ...x xx x − + = −+ − + ( ) 1 232 1 2 2 1 2 4 8 ...x xx x − − = ++ + + ( )( ) 2 3 23 23 5 3 1 3 9 27 ... 2 1 2 4 8 ...13 12 5 5 35 65 xx x xx xxx xx x   = −+ − ++ ++ + +  +− ≈− + − Some expanded ( ) 1 12 x − + instead when they used 2x instead of 2x− while others messed up in evaluating the binomial coefficients. Qn Solution Marks 3. [4 marks] (a) For Aceh earthquake, 2 3 log10(𝐼𝐼) − 10.7 = 8.6 log10 𝐼𝐼 = 3 2 (19.3) 𝐼𝐼 = 8.9125 × 1028 Many left answer as 28.9510 . (b) For Indian Ocean earthquake, 𝑀𝑀 = 2 3 log10 �8 × 8.9125 × 1028� − 10.7 = 2 3 log10 �7.1300 × 1029� − 10.7 = 2 3 (29.853) − 10.7 = 9.202 … ≈ 9.2 (1.d.p.)

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2021 Final Exam / Paper 2 / Solutions 3 Qn Solution Marks Very well done except a few who were careless and left out 10.7− . Qn Solution Marks 4. [5 marks] (a) 8 3 7! 1693440 P × = Badly done. Every kind of erroneous thinking. (b) Number of ways to select 3 black pencils 3 3C= Number of ways to select 2 black and 1 non-black pencils 36 21CC= × Number of ways to select 1 black and 2 non-black pencils 36 12CC= × Total number of ways = 1+18+45=64 Alternative: Complement Method Number of ways without restrictions 9 3 84C= = Number of ways with no black pencils 6 3 20C= = Required number of ways 96 33 64CC=−= Many were able to think about cases of no black, one black and two black but made mistakes including using permutation instead of combination. Qn Solution Marks 5. [8 marks] (a) ( 14) ( 14) ( 15) ( 16) 321 3 64 32 PS PS PS PS ≥ == += += ++= = Many under counted the cases.

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2021 Final Exam