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 obtain final RREF matrix. Some candidates were not able to apply the condition for unique solutions to exist. 6(b) When , there are no solutions. Well done. 7(a) Area of sector = Area of triangle = Since ratio of area of sector: area of segment = 4 : 1, then area of sector: area of triangle = 4 : 3. Thus we have Thus the ratio is . Not as well done as expected. Surprisingly there were a handful candidates who did not know the formulae for the area of segment. Some also made careless mistakes in translating the ratio in question in terms of areas of the two regions. 7(a) Alternatively: Area of sector = Area of segment =

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2021 Final Exam / Paper 1 / Solutions 5 Qn Solution Comments Thus the ratio is . 7(b) Generally well done as long as candidates were able to use Addition Formula on and Double Angle Formula on . A few candidates went to use De Moivre’s Theorem to expand both numerator and denominator which was not necessary and took more time. 7(b) Alternative Solution: Only a handful candidates approached the question in this manner.

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2021 Final Exam / Paper 1 / Solutions 6 Qn Solution Comments 8(a) Most candidates were able to sketch the point B in the correct region though without the detail provided in the answer. 8(b) “Hence” by using the fact that triangle OAB is an isosceles triangle, and dropping a perpendicular bisector from A to OB: OB is diagonal of parallelogram formed. Poorly attempted. Many candidates skipped this part entirely. The few who were able to apply geometrical approaches to this question were able to obtain the answers swiftly. Candidates who wrote the correct argu ment down without any reason or explanation were penalized for their lack of working and brevity. 8(b) “Hence” by dropping a perpendicular from B to real axis to form a right-angled triangle: Some candidates wrote down the relations but were stuck in applying Half-Angle Formula.

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2021 Final Exam / Paper 1 / Solutions 7 Qn Solution Comments OB is diagonal of parallelogram formed. 8(b) “Hence” by using the fact that triangle OAB is an isosceles triangle, Sine Rule: Base angles of an isosceles triangle: 2 = Few candidates approached the question using Sine Rule. Out of those who did, most were stuck in their attempt at applying Half-Angle formula. 8(b) “Hence” by using the fact that triangle OAB is an isosceles triangle, Cosine Rule: Very few candidates approached the question using this method.

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2021 Final Exam / Paper 1 / Solutions 8 Qn Solution Comments Base angles of an isosceles triangle: 2 = 8(b) “Otherwise” method: Thus we have A handful of candidates used this method but were not able to read off the modulus and argument even after arrival at the . 8(b) “Brute force” method: Most common approach for modulus but most candidates were stuck when they reached , again hindered by inability to manipulate using Half Angle formula. Ditto for the argument calculation.

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2021 Final Exam / Paper 1 / Solutions 9 Qn Solution Comments 9 Let be the proposition that for all positive integers Since , thus is true. Assume is true for some positive integer i.e. To prove is true i.e. Not well attempted. Some candidates defined the proposition as the recurrence relation. Some candidates used as the first term when the small est positive integer is . Many candidates were penalized for not using induction hypothesis and starting from , or not showing explicitly that . Do not start by assuming . Candidates were also penalized when they did not write the final conclusion properly which showed a lack

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2021 Final Exam / Paper 1 / Solutions 10 Qn Solution Comments Thus, is true whenever is true. Since is true and is true whenever is true, hence by mathematical induction is true for all positive integers of understanding of the Principle of Mathematical Induction. There were some who were penalized for confusing the notation of with . Minor errors to note: -Defining “for all positive integers ” in the proposition -Assume is true for SOME (some candidates wrote “all” instead which is a grave error) -Spell out mathematical induction instead of writing MI – this is not a recognized acronym Section B 10. Total:19 (a) of 2 of 1 coeffi xy coeffi x== Generally well done. Some student went to rewrit