ACSI 2021 Promo Paper 1 ans

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2021 Final Exam / Paper 1 / Solutions 1 FINAL EXAMINATION 2021 YEAR 5 IB DIPLOMA PROGRAMME MATHEMATICS HIGHER LEVEL Paper 1 – SOLUTIONS SECTION A Qn Solution Comments 1 1 10 32 5 x x x + += 2 2 5 10 3x x x  = + 2 10 10 3xx = + 24 10 10 3xx = + Let 10xy= 24 3 0yy− − = (4 3)( 1) 0yy+ − = 3 14y or=− 0x= This question was less well - done than expected. Even though most candidates were able to break down to bases 2 and 5, many were not able to use substitution . A handful were also unable to correctly factorise the resultant quadratic equation. Finally, a few candidates were penalized for not knowing that . 2(a) 33log log ( 4) 2 0xx− + + = Generally well done, though there were many who made careless mistake on the last

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2021 Final Exam / Paper 1 / Solutions 2 Qn Solution Comments 3 9log 0 4 x x =+ 9 14 x x =+ 94xx=+ 1 2x= step and erroneously wrote down instead of . 2(b) 5log 5 log 2x x+= 1log 5 2 log 5 x x += 2(log 5) 2log 5 1 0xx − + = 2(log 5 1) 0x −= log 5 1x = 5x= Mostly well done. Most candidates were able to perform change of base for log but a few had difficulty solving the resultant quadratic equation. 3(a) If ( ) ( ) 1P X a P X b +  = ( ) 1 ( )P X b P X a = −  ( ) ( )P X b P X a =  . A very simple 2 -liner proof which took some candidates many lines. Candidates should not assume upper/lower tail regions to be 0. 3(b) 2 ab += 100 2 = 50 = Well done.

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2021 Final Exam / Paper 1 / Solutions 3 Qn Solution Comments 4(a) Disappointingly, many candidates were not able to expand in descending powers. Some who were able to made careless calculation errors. Candidates who did not simplify coefficients or powers of for final answer were penalized. 4(b) Mark was awarded as long as was seen as most students had difficulty manipulating to the correct final form. 5(a)(i) Solving (1) and (2) yields: Generally well done. 5(a)(ii) Generally well done. 5(b) The general term is given by Poorly done. Many candidates mistook to be the expression for the common difference . Instead of considering , some went to

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2021 Final Exam / Paper 1 / Solutions 4 Qn Solution Comments Since , then we must have , so we have , which is a constant. Thus the series is an AP. calculate specific terms instead. Candidates wasted time finding individual expressions for and instead of using one to obtain the other. 6(a) When , there will be a unique solution. Most candidates were able to perform manual row operations to obt