Advanced Programme Maths P2 Memo 2019 hlayiso.com

Downloaded from hlayiso.com GRADE 12 EXAMINATION NOVEMBER 2019 ADVANCED PROGRAMME MATHEMATICS: PAPER II MARKING GUIDELINES Time: 1 hour 100 marks These marking guidelines are prepared for use by examiners and sub-examiners, all of whom are required to attend a standardisation meeting to ensure that the guidelines are consistently interpreted and applied in the marking of candidates' scripts. The IEB will not enter into any discussions or correspondence about any marking guidelines. It is acknowledged that there may be different views about some matters of emphasis or detail in the guidelines. It is also recognised that, without the benefit of attendance at a standardisation meeting, there may be different interpretations of the application of the marking guidelines. IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER II – MARKING GUIDELINES Page 2 of 11 MODULE 2 STATISTICS QUESTION 1  47    1.1 (a)  1   2  = 28 =0,5091  11 55   3 (b)  4  3  7   4  7  6   7  4  6   7  6  5  7     +    +    +    =  11   10   9   11   10   9   11   10   9   11   10   9  11 1.2 (a) 20 ( 0,1) = 2 5 5 P ( X ≤ 3 ) =1−    ( 0,3 ) ( 0,7 ) + ( 0,3 )  4 (b) 4  = 0,9692 (c) X ~ B ( 200;0,6 ) since np > 5 and nq > 5 ( X ~ N 120; 48 2 ) P ( X >125 ) → P ( X >125,5 )  125,5 − 120  = P Z >   48  = P ( Z > 0,79 ) = 0,5 − 0,2852 = 0,2148 IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER II – MARKING GUIDELINES Page 3 of 11 QUESTION 2  1  1 2  1 2.1 (a) E [ X ] = 1  + 2   + 3   + 4   6 2 9 9 = 2,28  1  1 2  1 Var ( X ) = 1  + 4   + 9   +16   − ( 2,28 ) 2 6 2 9 9 = 0,746 σ x = 0,86 (b) The mean would decrease and the standard deviation would increase. 4 k 2.2 (a) ∫ 0 x + 1 dx = 1 k ln ( x + 1)  0 = 1 4 k ( ln 5 − ln 1) 1 = k ln 5 = 1 1 ∴k = ln5 1 1 ln ( x +1)  0 = m (b) ln5 2 1 ln ( m +1) − ln1 = ln5 2 ln ( m +1) = ln 5 m +1= 5 ∴ m = 5 − 1 or (1,2361) IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER II – MARKING GUIDELINES Page 4 of 11 QUESTION 3 3.1 (a) P(R) = P(Z > 1,1) = 0,5 – 0,3643 = 0,1357 (b) P�R ∪ Q� = P(R) + P(Q) – P�R ∩ Q� = 0,1357 + 0,9282 – P(1,1 < Z < 1,8) = 0,1357 + 0,9282 – (0,4641 – 0,3643) = 0,9641 OR P�R ∪ Q� = P(Z > –1,8) = 0,5 + 0,4641 = 0,9641 3.2 X ~ N(200; 502) P ( X > c) P ( X > c|X > = 280 ) = 0, 625 P ( X > 280 )  280 − 200  P ( X > 280 ) = P  Z >   50  = P(Z > 1,6) = 0,5 – 0,4452 = 0,0548 P ( X > c) ∴ = 0, 625 0, 0548 P(X > c) = 0,0343 c − 200 ∴1, 82 = 50 c = 291 IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER II – MARKING GUIDELINES Page 5 of 11 QUESTION 4 4.1 (a) A 98% CI for p is: 1 ± 2, 33 ( 0, 2 )( 0,8 ) 5 300 (0,1462; 0,2538) (b) Since 15% is in the interval there is no evidence to suggest that the percentage of residents have approved the revised plan. 4.2 (a) H0 : µx = µy H1 : µx > µy Reject H0 if z > 2,05 Test Statistic: 30,06 − 29,84 z= = 3,22 0,0784 0,168 + 60 50 Conclusion: Since z > 2,05 reject H0 and suggest sufficient evidence to support the claim that the mean volume from the first machine is greater than the mean volume of the second machine. 30,06 − 29,84 − 0,1 =(b) z = 1,76 0,0784 0,168 + 60 50 P ( z > 1,76 ) =0,5 − 0,4608 = 0,0392 ∴ α > 3,9% QUESTION 5 9! 5.1 = 10080 3!3! 5.2 An example of such an arrangement: * C * A * L (EE) S * S * S * 6 places for other E 7! 8!  7!  ∴ × 6 = 5040 or − 2   = 5040 3! 3!  3!  Total for Module 2: 100 marks IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER II – MARKING GUIDELINES Page 6 of 11 MODULE 3 FINANCE AND MODELLING QUESTION 1 1.1 B 1.2 C 1.3 A 1.4 B QUESTION 2 2.1 920 000 = 1 850 000 (1 – i)4 ∴ i = 16,02%  0, 042  46   0, 042 3 x  1 +  − 1  1 +   12    12  2.2 2 680 000 − 920 000 = 0, 042 12 ∴ x = 34 961,87 QUESTION 3 0,082 4 3.1 x + 1 000 = x�1 + 4 � x = 11 826,46 3.2 2 600(x + 0,025) + 1 800(x) = 274 4 400x = 209 x = 0,0475 x = 4,75% + 2,5% = 7,25% 0,072 n 0,064 n 3.3 10 000 �1 + 12 � =12 000 �1 + 12 � n  377  5  375   1 508  n = =    6  503   1 509   500  n = 275 months IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER II – MARKING GUIDELINES Page 7 of 11 QUESTION 4 4.1 Logistics: carrying capacity present 4.2 (a) S-shaped (b) Linear 4 4.3 0, 65 × 0, 82 – = 0, 453 50  Rn  4.4 =+ Rn +1 Rn a.Rn  1 −  − 4 000 with Rn +1 = Rn  40000   18 000  a. (18 000 )  1 − =  4 000 a = 0,404  40 000  4.5 1 + 0,4 = (1 + a)4 a = 0,087 757 per annum  Rn  = Rn +1 Rn + 0, 087.Rn  1 −  = with R0 18 000  40 000  R6 = 23 235/6 OR a 4 1 + 0,4 = �1 + � 4 a = 0,351 092 four-yearly cycle, compounded per annum 0, 351092  Rn  Rn +1 Rn + = .Rn  1 −  = with R0 18 000 4  40 000  R6 = 23 235/6 IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER II – MARKING GUIDELINES Page 8 of 11 QUESTION 5 PREDATOR-PREY LOTKA-VOLTERRA MODEL B C D Predator populations Q A A P Prey populations 5.1 (a) A on phase plot (b) B on phase plot 5.2 (a) C on phase plot (b) D on phase plot 5.3 pair of axes accuracy (passing through equilibrium pt) accuracy (passing through max/min values of prey) accuracy (passing through max/min values of predator) QUESTION 6  0, 048  6.1 = T1 20 000  1 +  + 400 = 20 480  12   0, 048  T2= 20 480  1 +  + 400 (1, 005 )= 20 963,92  12   0, 048   + 400 (1, 005 ) 21 451, 48 2 = T3 20 963,92  1 + =  12  6.2 Tn = 1,004. Tn–1 + 400 (1,005)n–1, T0 = 20 000 Total for Module 3: 100 marks IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER II – MARKING GUIDELINES Page 9 of 11 MODULE 4 MATRICES AND GRAPH THEORY QUESTION 1 1  −1 −4  1  1 4 A−1 = − =  7  −3  1.1 −5  7  3 5 1.2 3 – 3z = 12 z=–3 y–3=0 y=3 1 – 3(1) = x x=–2 1.3 (a) k (b) –k (c) – 3k (d) k QUESTION 2 2.1 (a) translation 2 units right (b) factor = – 3  cos 2 A sin 2 A   3   3, 232  2.2    =    sin 2 A −cos 2 A   −2   −1, 598  3cos2A – 2sin2A = 3,232 and 3sin2A + 2cos2A = – 1,598 cos2A = 0,5 and sin2A = – 0,866 2A = 360o – 60o A = 150o 1 k t t   t + kv t + kr  2.3 (a)   1   v  =  0 r  v r  v −r v −r 1 (b) m= = = ( t + kv ) − ( t + kr ) k (v − r ) k IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER II – MARKING GUIDELINES Page 10 of 11 QUESTION 3 3.1 More zeroes, hence easier multiplications. 2 1 0 2 2 0 3.2 det =− ( −1) . 9 3 1 + 0 − 3. 4 9 1 + 0 = –248 −1 5 7 0 −1 7 OR 2 2 1 2 2 1 Det = -0 + 0 – 1. -1 0 3 +7. -1 0 3 = –248 0 -1 5 4 9 3  −192 32 42 −6    1  100 4 −49 7  3.3 −248  −64 −72 14 −2     60 52 −17 −33  QUESTION 4 4.1 (a) n–1 (b) n/2(n – 1) (c) n(n – 1) 4.2 (a) A, B, C, D, E, B, D or its reverse or many other options Start at A or D, end at D or A, use all edges once only. (b) 5 vertices, 4 edges, connectivity IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER II – MARKING GUIDELINES Page 11 of 11 QUESTION 5 5.1 all vertices do not have the same degrees 5.2 no graphs have HCs 5.3 A, C, D QUESTION 6 6.1 DF 10 HC 9 HA, DE 8 JD, GF 7 AJ 6 AB 5 max spanning tree = 60 6.2 A B C D E F G H J E 8E 6E J 12J 9J 11J 13 J D 18D B 14B C 20C A 17A 16A 20A F 23F H 24F E J A F G = 23 OR E J A F G = 23 Total for Module 4: 100 marks IEB Copyright © 2019