ACSI 2020 Promo Paper 2 ans

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2020 Final Exam / Paper 2 / Solutions 1 FINAL EXAM 2020 YEAR 5 IB DIPLOMA PROGRAMME MATHEMATICS HIGHER LEVEL Paper 2 MARKING SCHEME SECTION A Qn Solution 1. Eqn 1 : 3 8xy = 3 2 22log log log 8xy⇒ += 223log log 3xy⇒ += Eqn 2 : 24log log 3xy+= 22 1log log 32xy⇒ += Using GDC, letting 2logvx= and 2logwy= 3v=− and 12w= So 3 2 1log 3 2 8xx −= −⇒ = = 12 2log 12 2 4096yy= ⇒= = Comments : Generally well attempted.

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2020 Final Exam / Paper 2 / Solutions 2 Qn Solution 2. Let the two numbers be 122 1, 2 1kk++ ( )( ) ( ) 1 2 12 1 2 12 1 2 2 1 21 4 221 22 1 k k kk k k kk k k + += + + + = ++ + Comments : generally poorly attempted. Common mistakes included : 1. Letting the two odd numbers be (2𝑘𝑘 + 1)(2𝑘𝑘 + 1) 2. Letting the two odd numbers be consecutive (2𝑘𝑘 − 1)(2𝑘𝑘 + 1) 3. Letting 𝑛𝑛 be an even number and then (𝑛𝑛 − 1) is an odd number As they do not fully prove the result on its own. A few students attempted this question by substituting values. This reveals the students’ incorrect concepts as this technique does not prove the result for a high generality. A few students attempted to prove this question using contrapositive. While this attempt is commendable, they proved the result that “the product of two even numbers is even” instead of the correct contrapositive statement “suppose the product of two numbers is even, prove that the two numbers are both even”. Qn Solution 3. (a) From GDC, The product moment correlation coefficient is -0.606 (3 sf) There is moderate negative correlation (b) 0.0831 27.5404yx= −+ (c) (i) ( )0.0831 168 27.5404 13.6y= − += (ii) We find the regression line of x on y From GDC, The regression line is 229.219 4.4202xy= − So the estimated height when 100y= is ( ) ( )4.4202 100 229.219 177 3 sf− += Comment : a surprising number of students used the regression line of y on x to answer part (c), when the regression line of x on y ought to be used. For (a), the key word markers were looking out for was “moderate”.

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2020 Final Exam / Paper 2 / Solutions 3 Qn Solution 4. (a) There are 2! ways of arranging the man and woman. So there are 2! 13! 12454041600×= ways (b) Fix the position of the women. There are 8! ways of doing so ___ W ___ W ___ W ___ W ___ W ___ W ___ W ___ W ___ Now fix the position of the men in the gaps, for doing so, no two men will be standing next to each other. There are 9 6P ways of doing so The number of ways is 9 6 8! 2438553600P ×= Comment : generally well attempted. Some students rounded the answer off to 3 significant figures when there is no need to as the answer is exact. Students must be reminded that they only need to round their answer off to 3 significant figures when their answers are not exact. Qn Solution 5. (a) Draw a probability tree diagram: ( )Ryan is full 0.64P = So ( ) ( )( )0.6 1 0.7 0.64pp +− = 0.6 0.7 0.7 0.64 0.1 0.06 0.6 pp p p +− = = = (b) ( ) ( ) ( ) ( ) ( ) ( )( ) Ryan eats chicken rice|he is not full Ryan eats chicken rice and he is not full Ryan is not full 0.6 0.4 0.6 0.4 0.4 0.3 20.667 3 P P P= = + = = Comment : generally well done. Some students were unable to identify that part (b) is a conditional probability, instead finding the probability that Ryan eats chicken rice and he is not full. rice noodles full full not full not full p 1 – p 0.6 0.4 0.7 0.3

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2020 Final Exam / Paper 2 / Solutions 4 Qn Solution 6. (a) Let X be the number of times the target is hit in a training set. ( )~ 6,0.68XB ( ) ( ) ( )3 4 6,0.68,4,6 0.706441 0.706P X P X binomcdf>= ≥= = = Comment : Some students incorrectly found ( )3PX ≥ costing them precious marks. (b) Let N be the number of training sets in a day where the archer will hit the target more than thrice. Then ( )~ 8,0.706NB ( ) ( )8 0.706 5.65EN = = Comment : generally well done. (c) Let Y be the number of times the target is hit out of n times From GDC, When 7n= , ( )4 0.8465PY ≥= When 8n= , ( )4 0.925PY ≥= Therefore number of trials required is 8. Comment : generally well attempted if students knew how to use their GDC. Students may have lost marks by getting engaged in repetitive calculations when the answer could have been obtained by a few steps on GDC. Students must realize that since GDCs are allowed f or P2, they must know how the GDC can be used to make their lives easier. Qn Solution 7. (a) ( ) ( ) ( ) 61 0.25 61 0.75 61 0.67449 1 PX invNormµ σ µσ >= −⇒= ⇒−= ( ) ( ) ( ) 41 0.21 41 0.21 41 0.80642 2 PX invNormµ σ µσ <= −⇒= ⇒−= − Solve equation (1) and (2) 51.8909, 13.5052µσ= = (b) ( )0.99,51.8909,13.5052 83.3invNorm = Comment : generally well done

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2020 Final Exam / Paper 2 / Solutions 5 Qn Solution 8. (a) ( ) ( )( ) ( ) 22 22 22 22 2 22 * 2 2 4 zi z i i i ii αβ αβ αβ α β αβ α β α β αβ α β αβ −= + −= −+ − −− = −+ Since it is purely imaginary, 2 *Re 0z z  = So 32 2 20α αβ αβ−− = ( ) 22 30αα β−= So either 0α = (NA) or 22 1 3βα= Therefore 3 zi αα= + or 3 i αα − (b) If ( )arg 0z > , z is in the first or second quadrant If z is in the first quadrant, ( ) 3arg arctan 6z α π α  = = If z is in the second quadrant, ( ) 5arg 66z πππ=−= Comment : generally poorly done by students who are weak in algebraic manipulation. Qn Solution 9. (a) Project a line that is perpendicular to both BC and AD from B to AD. By Pythagoras’ theorem, 222BF AF AB+= 2 22 14 2 192BF⇒ = −= Therefore 192CD BF= = (b) In ,AFB∆

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2020 Final Exam / Paper 2 / Solutions 6 Qn Solution 21sin 14 7θ = = 1arcsin27 1.714 1.71 CBA CBF ABF π ∠ =∠ +∠ = +  = = By symmetry, 1.71CBA ABE∠= ∠= Therefore, obtuse ( )2 2 1.714 2.855CBE π∠= − = (c) By symmetry, EF = CD = 192 Length of minor arc ( )6 2.855 17.13CE = = Also, 1arccos 7DAB ∠=  , so reflex 12 2arccos 7DAF π ∠= −  Length of major arc DF = 18 2 2arccos 27.437π −= Therefore, the length of string is 2 192 27.43 17.13 72.27 72.3++== Comment : A surprising number of students gave a reflex angle for the answer in (b). Students who skipped (a) in order to proceed with the question tended to get (b) and (c) correct. Some students computed the area of the sector instead of the arc length. Some students wrongly assumed that ∠𝐶𝐶𝐶𝐶𝐶𝐶 = ∠𝐷𝐷𝐶𝐶𝐶𝐶 = 90∘. Solution 10. (a) (b) 0.838 9.16x≤≤

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2020 Final Exam / Paper 2 / Solutions 7 (c) range of : 10 6fy −≤≤ (d) 10 9 k− < <− (e) 0.702,9.61x= Comments : generally well done. Students lost marks for not putting down the maximum point at (10,6) on the graph. A surprising number of students were careless in the direction of the inequality for (d) – giving their answer as −9 < 𝑘𝑘 < −10. Solution 11. (a) 5 0 ( 5) 1 ( 5) 1 50 xP X F X dx≥= − <= − ∫ 3 4=

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2020 Final Exam / Paper 2 / Solutions 8 Comment : Some students have mistakenly find the probability using discrete method, i.e. summation. (b) 10 0 ( ) () 50 xE X xf x dx x dx = = ∫∫ 20 3= Comment : Well attempted, take note that GDC can be used to evaluate