AP Maths P2 MG 2017 hlayiso.com

Downloaded from hlayiso.com GRADE 12 EXAMINATION NOVEMBER 2017 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 © 2017 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 (53)(72) = 0.2652 (125) 1.1 1.2 ( ) P ( X = 7 ) = 10 ( 0.7 ) ( 0.3 ) = 0.2668 7 7 3 5! 5! 1.3 (a) + 2 × = 50 2!2! 2!3! (b) 5 × 2 + 2 = 12 {(CHLHL) x 5 + (HLCLH) x 2 = 12} QUESTION 2 2.1 (a) 0,3 × (0,7) + (0,3)(0,7)2 + (0,3)(0,7)3 + (0,3)(0,7)4 + C = 1 C = 0,4681 (b) P(X > 3) = P(X = 4) + P(X = 5) = 0,3 (0,7)4 + 0,4681 = 0,5401 2.2 (a) A n 0,2278 + 0,2922 (b) (i) = 500 2 ∴ n = 130 (ii) 0,26 + Z ( 0,26 )( 0,74 ) = 0,2922 500 Z = 1,64 ∴ α = 90 IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER II – MARKING GUIDELINES Page 3 of 11 QUESTION 3 3.1 X ~ N(9, 0,12)  8,9 − 9  P ( X > 8,9 ) = P  Z >  0,1  = P(Z > –1) = 0,5 + 0,3413 = 0,8413 3.2 X ~ B(6, 0,8413) () () P ( X ≥ 2 ) = 1 −  6 ( 0,8413 ) ( 0,1587 ) + 6 ( 0,8413 ) ( 0,1587 )   0 0 6 1 1 5  = 0,9995 3.3 P(X < a) = 0,04 a−9 –1,75 = 0,1 ∴ a = 8,825 cm QUESTION 4 4.1 (a) 1 + m2 + (m + 1)2 + 42 + 52 = 55 2 m2 + 2 m – 12 = 0 m2 + m – 6 = 0 (m + 3)(m – 2) = 0 m ≠ -3 or m = 2 5 + t − 1+ 4 + 3 + t =3 5 2t + 11 = 15 2t = 4 t =2 (b) r = –0,4 (c) (i) y = 4,2 – 0,4x (ii) y = 4,2 – 0,4(6) y = 1,8 This is an unreliable estimation as the correlation is weak. IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER II – MARKING GUIDELINES Page 4 of 11 4.2 H0 : μ = 49,5 H1 : μ < 49,5 Rejection Region Reject H0 if Z < –1,48 Test Statistic: 48 − 49,5 Z= = − 1,98 4,8 40 Conclusion: since Z < –1,48, we reject H0 at the 7% l.o.s. and suggest sufficient evidence to support the claim that Basi's working hours per week have decreased significantly. QUESTION 5 5.1 0,8 + 0,15(0,9) = 0,935 5.2 (0.8)(0.8) + (0.8)(0.15)(0.9)+(0.15)(0.9)(0.8) + (0.15)(0.9)(0.15)(0.7) = 0.8702 Total for Module 2: 100 marks IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER II – MARKING GUIDELINES Page 5 of 11 MODULE 3 FINANCE AND MODELLING QUESTION 1 1.1 R1 200 000 1.2 50 1.3 ± R1 100 000 method: 2,3 – 1,2 = 1,1 million 1.4 ± R900 000 1.5 (a) straight line gradient unaffected (b) interest has been increased OR Withdrawal from loan QUESTION 2   0,0568  − 34  5154,26 1−  1+     12  2  0,0568   2.1 P  1+  =  12  0,0568 12 P = R160 000   0,0568  − 12  5154,261−  1+ 12   2.2 OB =     0,0568 12 OB = R59 989,47 2.3 Jude paid: 5 154,26 × 12 = 61 851,12 Balance decreased: 116 674,09 – 59 989,47 = 56 684,62 Interest paid: 61 851,12 – 56 684,62 = R5 166,50 IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER II – MARKING GUIDELINES Page 6 of 11 QUESTION 3 3.1 2 000 000 = A(1 + 0,068)8 A = 1 181 571,41 3.2 1 181 571,41 = 3 400 000(1 – i)8 i = 12,38% 12 2  0,0764   i 3.3  1+  =  1+  i = 7,7626%  12   2 6  0,077626  5 500 000 + 300 000  1+  = 5 877 003,11  2   0,077626 9   0,077626 4  1+  − 1  1+   2    2  x = 12,264 x 0,07626 2 5 877 003,611 = 12,264x x = 479 200,70 QUESTION 4 1 396 − 1 300 4.1 = 7,4% 1 300 4.2 Qn + 1 = 1,05. Qn – 50, Q0 = 6 500 4.3 A = 8 020 B = 8 371 C = 8 739 D = 9 126 E = 9 532 F = 9 959 4.4 9 126 / 4 = 2 281 < 2 301 during 8th year 2 655 − 2 472 4.5 = 7,4% constant exponential growth;/thus 2 472 Malthusian IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER II – MARKING GUIDELINES Page 7 of 11 QUESTION 5 5.1 (a) prey ≈ 526 000 predator ≈ 4 500 (b) ± 4 180 – 4190 (c) B 5.2 Sn + 1 = 4 000 + 760 – 0,2 × 4 000 = 3 960 5.3 2/3 × 3 × 8 × 0,05 = 0,8  500 000  5.4 533 300 = 500 000 + 0,8 (500 000)  1 −  – 0,4 (500 000)  K  K = 1 200 000 QUESTION 6 6.1 (a) 8th (b) A 2a + 3b a + 2b + a + b a+ b 1 1 6.2 = =1+ =1+ =1+ a + 2b a + 2b a + 2b a + b + b b 1+ a+ b a+ b 1 1 1 =1+ 1 =1+ 1 =1+ 1+ a + b 1+ a 1+ 1+1Tn 1+ b b OR 2a + 4b − b b 1 =2− = 2 − a + 2b a + 2b a + 2b b 1 =2− a 2b 1 =2− 2 + Tn Total for Module 3: 100 marks IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER II – MARKING GUIDELINES Page 8 of 11 MODULE 4 MATRICES AND GRAPH THEORY QUESTION 1 1.1 (a) 1×1 k  (b) (k + 3 k +2 3 )  3  = ( k 2 + 3k ) + ( 3k + 6 ) + 3 1    OR  k+3  ( k 1 1)  3k + 2  = ( k 2 + 3k ) + ( 3k + 2 ) + 7  7  2 k + 6k + 9 = 0 k=–3 1.2 (a) 1 (b) pr (c) 27p (d) pr/q QUESTION 2 1 −3 6 4  1 −3 6 4 2.1 0 3 −1 1  ⇒ 0 3 −1 1  0 −18   0 −16   2 2 8 0 y = – 2; z=–7 x – 3(– 2) + 6(– 7) = 4 x = 40 2.2 L: 0x + 0y + 0z ≠ 1 (equation is an inconsistency) 2.3 D: 0x + 0y + 0z = 0 IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER II – MARKING GUIDELINES Page 9 of 11 QUESTION 3 3.1 (a) Reflection; direction y = –x; coordinates (b) Enlargement; scale k = 3; coordinates (c) Stretch; invariant line y = 0; scale k = 2; coordinates (d) Shear, invariant line y = 0; scale k = 2, coordinates 3.2 (cos sinθ θ )( ) ( −sinθ 6 = −5,55 + 10 = 4,45 cosθ 1 5,15 − 1 ) ( ) 4,15 6cos Ɵ – sin Ɵ = 4,45 AND cos Ɵ + 6sin Ɵ = 4,15 cos Ɵ = 0,83 378 378 … OR sin Ɵ = 0,55 270 270 … Ɵ = 33,5o Ɵ = 33,6o QUESTION 4 4.1 6 edges 4.2 not symmetrical 4.3 all vertices loop to themselves 4.4 five vertices, four edges correct edges, disconnect IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER II – MARKING GUIDELINES Page 10 of 11 QUESTION 5 5.1 NA 17 AD 15 DF 11 DB 21 FX 24 XG 8 XH 18 NC 25 EF 13 EH 18 = 170 min 5.2 ∴ N A D F E G X = 85 min OR A B C D E F G H X N N17 N23 N25 A N17 N23 N25 A32  A48 B N17 N23 N25 B44 B58 A48 C N17 N23 N25 A32 C58 A48  C53  D N17 N23 N25 A32 B/C58 D43  C53  F N17 N23 N25 A32 F56 D43  C53 F67 But half caves not visited H N17 N23 N25 A32 H71 D43 H77 C53 H71 But half caves not visited E N17 N23 N25 A32 F56 D43 E77 C53 H71 G N17 N23 N25 A32 F56 D43 H/E77 C53 G85 G N17 N23 N25 A32 F56 D43 H77 C53 G85 But half caves not visited ∴N AD F E G × = 85 min IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER II – MARKING GUIDELINES Page 11 of 11 QUESTION 6 6.1 (a) n–1 (b) n(n – 1) 6.2 (a) ABDCA ACDBA ACBDA ADBCA (b) (n – 1)! Total for Module 4: 100 marks Total: 100 marks IEB Copyright © 2017