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Transcription:

Cambridge Assessment International Education Cambridge International General Certificate of Secondary Education MATHEMATICS 0580/43 Paper 4 (Extended) 017 MARK SCHEME Maximum Mark: 130 Published This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It ds not indicate the details of the discussions that took place at an Examiners meeting befe marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Rept f Teachers. Cambridge International will not enter into discussions about these mark schemes. Cambridge International is publishing the mark schemes f the 017 series f most Cambridge IGCSE, Cambridge International A and AS Level components and some Cambridge O Level components. IGCSE is a registered trademark. This document consists of 7 printed pages. UCLES 017 [Turn over

017 Abbreviations cao dep FT isw SC nfww soi crect answer only dependent follow through after err igne subsequent wking equivalent Special Case not from wrong wking seen implied 1(a)(i) 180 ( + 3 + 5) 5 [= 90] 1 with no errs seen 1(a)(ii) 7.05 7.053. 3 x M f = sin36 better 1 B1 f 36 54 seen 1(b)(i) 13 M1 f 7.8 3 soi 1(b)(ii) 36.9 36.86 to 36.87 3 B1 f smallest angle identified (a) 343 1 (b)(i) 1 1 (b)(ii) x 10 final answer 1 M1 f sin[ ] = 3 5 7.8 sin[ ] = their ( b(i) ) If zero sced, SC1 f calculation of 53.1 (b)(iii) 9x 16 final answer B1 f x 1 x 16 (3x 8 ) seen (c)(i) (x 3)(x + 3) final answer M1 f (x + 6)(x 3) (x 6)(x + 3) (x 3)(x + 3) (c)(ii) ( x + 3) x + 6 x + 10 x + 10 final answer nfww 3 M f (x + 10)(x 3) M1 f (x + a)(x + b) where ab = 30 a + b = 7 UCLES 017 Page of 7

017 3(a)(i) 1890 M1 f 16 4 [ 60] If zero sced, SC1 f answer 31.5 3(a)(ii) 103.95 4 44 55 M3 f 0.5 + 16 60 60 SC3 f figs 10395 figs 104 3(b)(i) 16 1000 (60 60) 1 M f two crect area methods f a full method without minutes to hours conversion M1 f one crect area with without minutes to hours conversion 3(b)(ii) 46.3 46.8 to 46.9 3 M f (1400 + 0) 35 M1 f distance speed 1400 + 0 3(c) 180 nfww 4 B3 f final answer 3 OR M3 f 17.5 60 7.5 M f 17.5 7.5 10 to 0 60 7.5 17.5 7 to 74 60 4(a) 80 < t 100 1 M1 f 17.5 7.5 seen 15 60 73 4(b) 86 nfww 4 M1 f midpoints soi M1 f use of Σfx with x in crect interval including both boundaries M1 (dep on nd M1) f Σfx 150 4(c)(i) Reference to not knowing the individual values so we do not know the highest the lowest values 1 4(c)(ii) 6.4 M1 f 6 150 360 150 soi 4(d) 150 1 UCLES 017 Page 3 of 7

017 4(e)(i) 90 350 10 9 M1 f 150 149 After zero sced, SC1 f answer 100 500 4(e)(ii) 440 350 3 10 10 M f + 150 149 150 149 10 10 M1 f 150 149 150 149 440 After zero sced, SC1 f answer 500 4(f) 13, 8.5, 7.5, 1.1 3 B f 3 crect B1 f 1 crect f 3 crect FD.s 5., 3.4,.9, 0.44 5(a)(i) Image at (0, 1), (0, ), ( 3, 1) B1 f reflection in y = 0 x = k 5(a)(ii) Image at (0, 0), (0, ), (6, ) B1 f crect size and crect ientation wrong position f crect vertices plotted 5(a)(iii) Image at ( 5, 4), ( 5, 5), (, 4) 5 B1 f translation by k k 3 5(b) Rotation 90 clockwise (4, 1) 3 B1 f each 5(c)(i) (4, 1) 0 1 1 M1 f 1 0 4 5(c)(ii) (8, 1) 0 1 3 1 1 M1 f 1 0 0 4 0 1 3 1 4 0 1 1 1 0 8 5(c)(iii) Rotation 90 anti-clockwise Origin 3 B1 f each UCLES 017 Page 4 of 7

017 6(a)(i) 5.5 5.46 M1 f π 5 h = 000 6(a)(ii) 9.85 9.847 3 M f [r 3 =] 000 π 3 M1 f 3 πr3 = 000 6(a)(iii) 95 95.4. 3 M f [6 ] 000 M1 f 3 000 6 times their area of one face 6(b)(i).5.49 1 M1 f 7 10 sin40 6(b)(ii) (10 + 7 10 7 cos40) + 7 + 10 M3 M f 10 + 7 10 7 cos40 M1 f crect implicit cosine rule 3.46 A A1 f 6.46 41.7 to 41.8 6(c) 64.9 64.9 to 64.94 3 c M f 8. 9 = 360 π 9 c M1 f 360 π 9 soi 7(a) 9, 6, 9 3 B1 f each 7(b) Crect graph 4 B3FT f 6 7 crect points BFT f 4 5 crect points B1FT f 3 crect points 7(c) 3.5 to 3.35 and 0.8 to 0.9.. FT FT their graph B1FT f either 7(d) a = 5 4 1 1 1.5 4 49 1 b = 6 6.15 8 8 3 B f either crect 5 M1 f [] x + seen isw 4 f x + 4ax + a + b 8(a)(i) 5 1 8(a)(ii) 3 1 8(b) 4,0 5 M1 f 5x 4 = 0 soi UCLES 017 Page 5 of 7

017 8(c) y = 0.x + 11 final answer 4 M f y = 0.x + c (any fm) FT their (a) 1 B1FT f grad = their (a)(i) soi and M1 f substitution of (10, 9) into their equation 8(d) (, 6) 3 M1 f elimination of one variable A1 f x = y = 6 8(e) 13 3 M f (4 + 9) their B1 f 9 4 4 seen 9(a) 10 x 0.5 final answer 1 0 Accept x 1 9(b)(i) 10 10 x 0.5 x = 0.5 M1 FT their (a) 10x 10(x 0.5) = 0.5x (x 0.5) 10x 10x + 5 = 0.5x 0.15x better M1 B1 Clears algebraic denominats collects as a single fraction FT their algebraic fractions dep on two fractions with algebraic denominats Expands brackets x x 40 = 0 A1 Dep on M1M1B1 and no errs seen 9(b)(ii) 1 ± ( 1) 4 40 B B1 f ( 1) 4()( 40) better 1+ q 1 q B1 f both 4.3 and 4.73 final answers B1 B1 SC1 f 4.9 and 4.79 f 4.3 and 4.73 seen in wking f 4.73 and 4.3 as final answer f 4. 4. and 4.7 4.7 as final answer 9(b)(iii) [hours] 7 [minutes] 3 B f.11.114 to.115 16.8 to 16.9 17 M1 f 10 their positive root from (b)(ii) 10(a)(i) 3 5 M1 f 3 crect prime facts in a tree table seen befe the first err f, 3, 5 identified 10(a)(ii) 540 M1 f 3 3 5 3 3 shown answer 540k UCLES 017 Page 6 of 7

017 10(b) X = 8575 Y = 615 4 B3 f X = 8575 Y = 615 B f a = 5 b = 1 soi B1 f 15 = 5 7 4 875 = 5 3 7 3 M1 f a² 7² [= 15] a 3 7 b + [= 4 875] UCLES 017 Page 7 of 7