ACSI 2022 HL AA Prelim Paper 3 Solutions for students

ACS (Independent) / Mathematics Department / Mathematics HL / Year 6/ 2022 Prelim Exam / Paper 3 / Solutions 1 PRELIMINARY EXAMINATION 2022 YEAR 6 IB DIPLOMA PROGRAMME MATHEMATICS : ANALYSIS AND APPROACHES HIGHER LEVEL Paper 3 Qn Solution 1. (a) Hence is a root of the above equation Comments: Generally well done. (b) so is a root of Any root Any integer power of  can be reduced exclusively to integer power of k only. Since there are 10 values of k , this leads to 10 distinct complex number arguments. Comments: poorly done. Many did not comprehend the question.

ACS (Independent) / Mathematics Department / Mathematics HL / Year 6/ 2022 Prelim Exam / Paper 3 / Solutions 2 Qn Solution (c) Using the Argand diagram, it can be observed that there is a total of 6 equally spaced roots and thus 5  = , Comments: Some draw argand diagram with wrong angles and lack of symmetry. Some did not label correctly according to what is required in the question. (d) Let sin b = , sin 2 c = , cos d = and cos2 e = Considering the symmetry of the Argand Diagram about the Imaginary axes, it can be observed that sin 4 b = , sin 2 c = , cos4 d =− and cos3 e =− Therefore sin 4 sin= , cos4 cos=− and sin3 sin 2= . Hence ( ) 2 2 2 2 2 2 sin 3 sin 4tan 3 tan 4 cos3 cos 4 sin 2 sin cos 2 cos 2sin cos sin cos2cos 1 2sin 2cos 1 2 tan 2 sec 2 tan tan1 tan tan tan 2                = =−− = −−− = − = − = − =

ACS (Independent) / Mathematics Department / Mathematics HL / Year 6/ 2022 Prelim Exam / Paper 3 / Solutions 3 Qn Solution Alternative Solution for proving tan3 tan 4 tan tan 2   = Using the Argand Diagram Let gradient 1tan m = then by symmetry, 1tan 4 m =− Similarly, let gradient 2tan 2 m = then 2tan 3 m =− ( )( )12 12 tan 3 tan 4 tan tan 2 mm mm   = − − = = Comments: Surprising number did not utilize the argand diagram effectively and gave long winded workings to arrive at same result. Some misread the question and miss out the need to prove the trigo identities and lost marks then. Students generally quite weak in trigonometry. A few just used GDC to give approximate value and not the exact value needed for proof. (e) DE = 1− Or using 1 CD AB ED ED ED = = =  Since AB = 1 (alternative form of solution accepted) Comments: Generally well done.

ACS (Independent) / Mathematics Department / Mathematics HL / Year 6/ 2022 Prelim Exam / Paper 3 / Solutions 4 Qn Solution (f) 1 1=− (resulting from property of Golden Rectangle given in question or from the fact that DE can be expressed in the two forms above) 2 10 −− = Using quadratic formula, ( )1 1 4 1 2 15 2  − −= = Since  is the ratio of lengths, we reject the negative answer and therefore 15 2 += (p