SOTA 2023 Prelim MAA HL Paper 2 Solutions

Yr6/5/MATAA/HP2/Aug2023 14 pages © School of the Arts, Singapore Year 6 Mathematics: analysis and approaches Higher level MARK SCHEME Paper 2 Preliminary Examinations Monday 28 August 2023 2 hours Instructions to candidates • Do not open this examination paper until instructed to do so. • A graphic display calculator is required for this paper. • Section A: answer all questions. Answers must be written within the answer boxes provided. • Section B: answer all questions on the answer sheets provided. Write your name and class on each answer sheet. • Unless otherwise stated in the question, all numerical answers should be given exactly or correct to three significant figures. • A clean copy of the mathematics: analysis and approaches formula booklet is required for this paper • The maximum mark for this examination paper is [110 marks]. Section A Section B Question Marks Question Marks 1 10 2 11 3 12 4 5 6 7 8 9 Name: 110 Class: Index:

- 2 - Yr6/5/MATAA/HP2/Aug2023 Full marks are not necessarily awarded for a correct answer with no working. Answers must be supported by working and/or explanations. Solutions found from a graphic display calculator should be supported by suitable working. For example, if graphs are used to find a solution, you should sketch these as part of your answer. Where an answer is incorrect, some marks may be given for a correct method, provided this is shown by written working. You are therefore advised to show all working. Section A No Solutions Mark scheme 1(a) BC = 3.42166 BC = 3.42 Attempt to use cosine rule (M1) Correct substitution (A1) BC = 3.42 (3s.f.) A1 1(b) Attempt to use sine rule (M1) Correct substitution (A1) A1 2(a) a = 30.5 (3 s.f.) b = – 419.462 b = – 419 (3 s.f.) a = 30.5 (3 s.f.) A1 b = – 419 (3 s.f.) A1 2(b) A1 2(c) y = 30.4932(29) – 419.462 y = 464 841 $465 000 Correct substitution of 29 into their regression equation (M1) $465 000 or 466 000 A1 2(d) EITHER correlation does not imply causation (there is an association but not causation) OR there could be another factor involved. EITHER correlation does not imply causation OR there could be another factor involved (eg other factors may affect profits or advertising not the sole factor affecting profits.) R1 NOT More money spent on ads will affect profits. 3 -------------- (1) ---------- (2) Attempt to use or (M1) Attempt to solve simultaneous equations (M1) A1 A1

- 3 - Yr6/5/MATAA/HP2/Aug2023 Turn Over Turn over [exact value] (M1) [must be exact value] (A1) 4(a) Both graphs are decreasing functions. hours hours < 23.105 hours Since Betacold took a shorter time for concentration to reach 1 unit, Betacold becomes ineffective before Anticold. Attempt to use a graph or mention that functions are decreasing (M1) or find Time for Anticold hours. Time for Betacold hours. A1 AG 4(b)(i) 24 hours after 1st dose units g(24) = 0.586 units Attempt to find g(24) M1 g(24)=0.586 units A1 4(b)(ii) Total concentration g(24) + g(9) = = 2.53344 = 2.53 units 24 – 15 = 9 hours or 12 – 3 = 9 hours for 2nd dose A1 g(24) + g(9) M1 2.53 units A1

- 4 - Yr6/5/MATAA/HP2/Aug2023 5 = k = 3.5 Valid approach for expansion for (k+x)2 (M1) Valid approach for expansion for (3 – 2x)7 (M1) Eg Recognizing that the term in x8 is needed (M1) Correct term or coefficient in binomial expansion (seen anywhere) (A1) Eg (either seen) Add their term in x8 or coefficient of x8 equal to 448x8 or 448. (M1) Eg k = 3.5 A1

- 5 - Yr6/5/MATAA/HP2/Aug2023 Turn Over Turn over 6(a) A1 6(b) Options Total M1 – recognizing the options. A1 – At least one correct option from Section A, B and C (M1) finding all are the same, times by 3 A1 – Final Answer 18 8 43758C= 666 233 666 323 666 332 | 15 20 20 6000 | 20 15 20 6000 | 20 20 15 6000 AABC BBCC CCC AABC ABCC CCC AABC ABBC CCC ´´ = ´ ´ = ´´= ´ ´ = ´´ = ´ ´ = 6000 3 18000´=

- 6 - Yr6/5/MATAA/HP2/Aug2023 7 Finding the intercept P. (A1) – or seen as a limit on the integral. Region 1 Either Method 1 – Cone OR Method 2 – Integration Region 2 Integration Total Volume M1 Finding the volume of the cone A1 M1 - Using the volume equation on either part (condone the lack of limits) M1 – arranging either equation into A1 A1 A1 ( )( ) 1 22e e 2.2916 3.14496 2.29,3.14 3sf x x x y P -= = = 2 1 2 1 1 1 1 3 1(2.2916) (2e 3.145)3 12.60196.. 12.602 Vr h V V V p p = =- = = ( ) 1 2e 2 1 1 1 2e d 12.6016 12.602 y Vy y V V p=- = = ò ( ) 1 2 2 1 2 2 ln( ) d 15.2281 y Vy y V p= = ò 12 3 12.6 15.2281 27.8303 27.8units T T T VV V V V =+ =+ = = xf ( y )=

- 7 - Yr6/5/MATAA/HP2/Aug2023 Turn Over Turn over 8(a) If using DEG on GDC M1 for integral within limits =1 A1 A1 8(b) The data is NOT symmetrical For GDC in DEGREE mode mark correct with values below. The data is symmetrical M1 Attempt to find 1 of the values required. A1 A1 A1 A1 5 sin 0 e1x kx dx =ò 0.073643k=0.0755k=5 sin 0 e mean 2.53189 xx kx dx´= = ò sin 0 e 0.5 median 2.39507 m xkx dx = = ò ( ) 5 2 sin 0 2 e 7.65031 SD 7.65031 2.53189 SD 1.11348 xxk x d x´= =- = ò ( )3 2.53189-2.39507 1.11348 0.368623 g g = = ( )2 mean 3.3589 median 3.56452 12.6504 sd 1.1697 EX = = = = 0.513g=-

- 8 - Yr6/5/MATAA/HP2/Aug2023 9 Solving for the roots. A1 – product of roots A1 – sum of roots M1 – attempting to solve for the roots A1 one of the roots correct. M1- using their roots to find the equation. A1 ( ) ( ) ( ) 2 2 2sin 24cos sin 0 sec sec xx g gg gg+- = 2 22 224 cos sin 24 cos sin 6sin 2sec 2sin 2 cos sin sin 2sec ggab g g g g gab g g g g -== - = - += - = - = - ( ) ( ) 2 22 sin 2 sin 2 6sin 2 sin 2 6sin 2 0 2sin 2 ( 3sin 2 ) 0 2sin 2 3sin 2 (rej) sin 2 2sin 2 3sin 2 ag b gb b g bg b g bg bg bg bg ag g ag =- - -- = - +- = -+ = = =- =- - =- ( ) 2 2 6sin2 32sin2 2sin2 1 2 6sin2 3 1 1031 33yx x x x ag bg bg ag -== - == -- æö=+ + =+ +ç÷èø

- 9 - Yr6/5/MATAA/HP2/Aug2023 Turn Over Turn over Section B 10(a) Recognise P(X > 2300) (M1) A1 10b) 39 days Attempt to multiply P(X > 2300) and 365 (M1) Accept 39 or 39.4 or 40 days A1 10(c) Valid approach (M1) Eg 1 – 0.4, 0.6 z = 0.253347 A1 attempt to standardize (M1) Eg Correct substitution with their z (do not accept a probability) A1 Eg A1 10(d)(i) P(X<2500 | X>2300) Recognition of conditional probability (M1) P(X<2500 | X>2300) (accept P(X>2300|X<2500) (do not accept P(A| B)) (A1) (A1) (A1) 10(d)(ii) Let Y be the number of special days that the daily revenue is less than $2500. = 0.022408 = 0.0224 Recognizing binomial probability (M1) n = 10, p =0.743438… (A1) (using their p) P(Y<5) A1 = 0.0224 (3s.f.) A1 (0.0226 using p = 0.743)

- 10 - Yr6/5/MATAA/HP2/Aug2023 11(a) (i) A1 condone lack of rejection or 0 11(a) (ii) M1 - product rule on first term M1 - Implicit on the y^2 A1 – differentiating RHS AG 11(a)(iii) M1 input x=0 and y=1 Or M1 attempt to find as a A1 For the rest of the question only penalise once for answers not to 6sf. 11(b) (M1) – Attempt at Euler’s evidence of the first term (A1) – y=1.1 (A1) – any other correct intermediary values A