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 / Paper 2 / Solutions 4 Qn Solution Marks (b) ( ) ( ) ( ) 17 7 72 8 than two '7 ' P one '7' + 8 88 14 49 63 64 6 P no 4 '7' P less   = +     += = = Alternative: ( ) 2 than two 1 ( '7') 11 8 '7' 63 64 Pt Pl w e o ss −   = = −  = All kinds of errors. (c) (less than 2 dice shows 7| 14) (one 7 14) ( 7 14) ( 14) 3264 64 664 5 6 PS P S P no S PS ≥ ∩≥ + ∩≥= ≥ + = = Badly done with many students not treating it as conditional probability.

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2021 Final Exam / Paper 2 / Solutions 5 Qn Solution Marks 6. [8 marks] (a) 0.8992.... 0.899 (3 S.F.) r =− =− Strong negative linear correlation. Well done although no table was drawn and some gave the value of r2 instead of r. (b) 11.5 0.738yx= − Mostly correct although several gave the wrong regression line. (c) 11.5 0.738(6.5) 6.703.... 6.7 (1 d.p.) y= − = = Many did not correct to one decimal place. (d) The line x on y should be used instead of y on x to estimate this value. Extrapolation is always unreliable. Most could only give one reason. Qn Solution Marks 7. [5 marks] (a) 3 0 13 2 01 13 2 01 Since ( ) is p.d.f, () 1 21 1 2 1 43 3 4 fx f x dx k x dx x dx k x dx x dx = ∴ +− = = +− = = ∫ ∫∫ ∫∫ Many did not realise they can use the GDC to evaluate 3 1 2 x dx−∫ and got stuck.

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2021 Final Exam / Paper 2 / Solutions 6 Qn Solution Marks Testing the median of X: 1 2 0 Median cannot be in first piece 3 0.254 Since 0.25 .0.5, x dx = < ∫ Assume median in second piece: 1 2 01 33 2 0.544 m x dx x dx+ −=∫∫ Median = 1.42265… (Median > 1) = 1.42 (3 S.F.) Poorly done with many trying to solve manually. Yet others did not understand the concept of median. Qn Solution Marks 8. [5 marks] (a) ( )( ) ( )( ) ( ) 2 2 2 2 22 2 2 32 1 21 12 24 1 5 5 5 rejected 5 k iki k kiz kiki k k k kk k k k ++ +−= =−+ + − =+ −= + = = ±− =  Quite well done although some wrote 2 1k − or 1k+ as the denominator for the LHS.

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2021 Final Exam / Paper 2 / Solutions 7 Qn Solution Marks (b) ( ) ( ) 1 155 35 2 51 2 5arg tan 1 1.15026 1.15 (3 S.F.) iiz z − ++−= =+ =  = = Those who got part (a) correct mostly got this part right as well. Many got the concept of arctan of ( imaginary part/real part). Qn Solution Marks 9. [8 marks] a ( ) 2( )( )( ) ( ) ( )xxx x x xα β γ α β αβ γ− − −= −+ + − = 32 2 ( )( )xx x x xγ α β αγ βγ αβ αβγ− −+ − + + − 32 ( )( ) .xx x α β γ αβ βγ αγ αβγ=− ++ + + + − Hence q = αβ βγ αγ++ Well done although some made careless mistake in expanding. b 2 2 22 ()αβγ α αβ αγ αβ β βγ αγ βγ γ ++ =++++++++ 2 22 2( ).α β γ αβ βγ αγ=+++ ++ Careless mistakes and some could not write the symbols properly and misread their own handwriting. c 2 22 2( ) 2( ).α β γ α β γ αβ βγ αγ+ + = ++ − + + 2(2) 2(3)= − 2=− Mostly correct with several making careless substitution.

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2021 Final Exam / Paper 2 / Solutions 8 Qn Solution Marks d Recognize di and di− are both roots of the equation ( )2 2di diαα+ +− = ⇒ = 22 2 () ( ) 2di diα + +− = − 222 2 22 26 3 dd d d −−= − = =± The roots are 2, 3 , 3 .ii− Alternative: 2 2 3 2( ) 2( ) 3 3 3 q di di d d d αβ βγ γα=++= +− + = = =± The conjugate roots are 3, 3ii− Many could not do this part or have no time for this part.

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2021 Final Exam / Paper 2 / Solutions 9 Mark Scheme for 2021 EOY P2 Section B Qn Solution Marks 10(a) [16 marks] (a) i ~ (50,0.3)XB or X follows a binomial distribution. Several students used weird notations like X ~(50, 0.3) or P ~ X(50,0.3) or are unsure how to state the distribution. (a) ii ( 20) (50,0.3, 21,50) 0.04776... 0.0478 (3 S.F.) PX binomCDF > = = = Significant numbers of students use (50,0.3, 20,50)binomCDF including the 20 by mistake. (a) iii ~ (30,0.0478) ( 6) (30,0.0478,6) 0.002178... 0.00218 (3 S.F.) YB PY binomPdf = = = = Generally not well done if part (ii) is wrong. Most could see that it is binomial here but may have made careless mistakes in keying in numbers. (a) iv ( ) 0.12 ( ) (50,0.3, ) PX n f x binomPdf x = > = N = 15 Alternative:

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2021 Final Exam / Paper 2 / Solutions 10 50 50(0.3 )(0.7 ) 0.12nn nC − > Using GDC, n = 15. Badly done. Some tried using inverse binomial gdc function which is for questions with the form ( ) 0.12PX n≤> . Hence tend to get wrong answers. A number use brute force to list out all the possibilities. Students should use the graph method, the