ACSI 2016 Promo Paper 1 Ans

ACS (Independent) Mathematics Department / Mathematics HL / Year 5 / 2016 Final Examinations / Paper 1 / Solutions 1 FINAL EXAMINATION 2016 YEAR 5 IB DIPLOMA PROGRAMME PAPER 1 MATHEMATICS HIGHER LEVEL – SOLUTIONS Qn Solution 1. General Term for  10 1 x :   10 210 1011 r rrr x xrr          101 11x x  r = 2:  22 1210 10.914 52 2.1x xx  r = 0:  00 210 110 x  1 114 5 . . 4 514 4xx      Comment: Mostly well done although a small number do not know how to evaluate 10C2 and a handful do not remember how to do Binomial Expansion. 2. (i)   18 27 ... 99 10 18 992 5 117 585      Comment: A number of students did by pure arithmetic calculation and some missed out on 99 as the last term. (ii) 10 1 99 r r   (divisible by 9) Total sum – sum of all divisible by 9 =   90 10 11 99 9 rr rr    Comment: Well done.

ACS (Independent) Mathematics Department / Mathematics HL / Year 5 / 2016 Final Examinations / Paper 1 / Solutions 2 Qn Solution 3.  22 2 22 2 22 2 22 A : B 11 1 : 22 2 : 3 4 20 Area R rr Rr r Rr r Rr Rr R        Comment: Very well done except some forgot the formula for area of sector. 4. (i) 22 12 2 1 21 12 1 21 21 21 21 21 x x x x x S           Comment: Some did not manipulate answer to the required form. (ii)    23 1 1 123. .1 1 11 11 12 0.5 66 33 2 2 2 .... 2 2 2 2 1 2 x xx x x x                Comment: A number forgot to substitute the value of x in their final answer.

ACS (Independent) Mathematics Department / Mathematics HL / Year 5 / 2016 Final Examinations / Paper 1 / Solutions 3 Qn Solution 5.      1 2 3 1 1 1 1 Let P be statement ! 2 3, 3! 6 24 is true. Assume P is true, ! 2 To show that P is true: 1 ! 2 !2 1! 2 1 11! 2 2 1 2, 1, 2 11! 2 2 n n k k k k k k k k n nL H S RHS LHS P k k k kk kk kif k kk                         31 21 1! 2 Since P and P is true, by Mathematical Induction, P is true for all 2, k k kn k nn        Comment: Some vague argument on the validity of P(k+1). 6.  32 32 32 16 28 2 2 81 68 1 6 022 2 81 6 4 1 6 084 4 4 16 0 or equivalent let y x yx yy y yy y yyy                        Comment: A number of the students did by finding out the coefficients of the function in relation to the sum and product of roots. A small group of students did by solving for the original roots.

ACS (Independent) Mathematics Department / Mathematics HL / Year 5 / 2016 Final Examinations / Paper 1 / Solutions 4 Qn Solution 7.  1sin arctan44 6 3 sin 46 sin cos cos sin46 4 6 13 1 1 2222 31 22 arc k k             