Cambridge Assessment International Education Cambridge Ordinary Level. Published

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1 Cambridge Assessment Internatinal Educatin Cambridge Ordinary Level ADDITIONAL MATHEMATICS 4037/1 Paper 1 Octber/Nvember 017 MARK SCHEME Maximum Mark: 80 Published This mark scheme is published as an aid t teachers and candidates, t indicate the requirements f the examinatin. It shws the basis n which Examiners were instructed t award marks. It des nt indicate the details f the discussins that tk place at an Examiners meeting befre marking began, which wuld have cnsidered the acceptability f alternative answers. Mark schemes shuld be read in cnjunctin with the questin paper and the Principal Examiner Reprt fr Teachers. Cambridge Internatinal will nt enter int discussins abut these mark schemes. Cambridge Internatinal is publishing the mark schemes fr the Octber/Nvember 017 series fr mst Cambridge IGCSE, Cambridge Internatinal A and AS Level cmpnents and sme Cambridge O Level cmpnents. IGCSE is a registered trademark. This dcument cnsists f 10 printed pages. UCLES 017 [Turn ver

2 4037/1 Cambridge O Level Mark Scheme Octber/Nvember 017 MARK SCHEME NOTES The fllwing ntes are intended t aid interpretatin f mark schemes in general, but individual mark schemes may include marks awarded fr specific reasns utside the scpe f these ntes. Types f mark M A B Methd marks, awarded fr a valid methd applied t the prblem. Accuracy mark, awarded fr a crrect answer r intermediate step crrectly btained. Fr accuracy marks t be given, the assciated Methd mark must be earned r implied. Mark fr a crrect result r statement independent f Methd marks. When a part f a questin has tw r mre methd steps, the M marks are in principle independent unless the scheme specifically says therwise; and similarly where there are several B marks allcated. The ntatin dep is used t indicate that a particular M r B mark is dependent n an earlier mark in the scheme. Abbreviatins awrt ca dep FT isw nfww e rt SC si answers which rund t crrect answer nly dependent fllw thrugh after errr ignre subsequent wrking nt frm wrng wrking r equivalent runded r truncated Special Case seen r implied UCLES 017 Page f 10

3 4037/1 Cambridge O Level Mark Scheme Octber/Nvember 017 Questin Answer Marks Guidance 1(i) 1 X Y 1(ii) Either fr C with n intersectin with either A r B (allw if C is nt A B represented by a circle) C fr all crrect, C must be represented by a circle Or A B C a = 4 b = 6 c =, π fr use f, t btain c, 1 using their values f a and f b 3(i) 4 3 0x + 5x B3 fr each crrect term 3(ii) 4 ( x x ) x x 1 x 9 and 4 x Independent f x: attempt t deal with terms independent f x, must be lking 1 at terms in x and x and terms in 4 1 x and 4 x = 5 FT their answers frm (i) (their 0 1) + (their 5 9) UCLES 017 Page 3 f 10

4 4037/1 Cambridge O Level Mark Scheme Octber/Nvember 017 Questin Answer Marks Guidance 4 crrect differentiatin f ln ( 3 + ) x attempt t differentiate a qutient r a prduct 6x ( x + 1) ln x ( 3x + ) dy 3x + = dx 1 ( x + ) all ther terms crrect. When 1 5 4ln14 dy 14 x =, = d x 5 dep fr substitutin and attempt t simplify 6 4 = ln A fr each crrect term, must be in simplest frm 5(i) Either Gradient = 0. lg y = 0.x+ c lg y= mx+ c si crrect attempt t find c must have previus lg y= x r Or 1 lg y = x 50 5 line in either frm, allw equivalent fractins 0.3 = 0.6m+ c 0. = 1.1m+ c attempt t slve fr bth m and c must have at least ne f the previus B marks Leading t lg y= x r 1 lg y = x 50 5 line in either frm, allw equivalent fractins UCLES 017 Page 4 f 10

5 4037/1 Cambridge O Level Mark Scheme Octber/Nvember 017 Questin Answer Marks Guidance 5(ii) Either ( x) y = 10 dealing with the index, using their answer t (i) =10 ( 10 x y ) y =.63( 10 0.x ) Or A fr each bx ( 10 ) y = A leads t lg y = lg A+ bx Cmpare this frm with their equatin frm (i) cmparing their answer t (i) with lg y = lg A+ bx may be implied by ne crrect term frm crrect wrk lg A = 0.4 s A =.63 b = 0. fr each 6(i) y e Must have crrect ntatin i.e. n use f x 6(ii) y > 3 e Must have crrect ntatin i.e. n use f x 6(iii) 1 f ( x ) = e x r ( ) ( ) g 4 = 35 First may be implied by crrect answer r by use f f g 4 = e 6(iv) y 3 = x r x 3 = y valid attempt t btain the inverse g 1 3 ( x ) = x 1 crrect frm, must be ( ) y = g x = r Dmain x > 3 Must have crrect ntatin 7(i) 1 a b p : = substitutin f x = and equating t zer (allw unsimplified) ( ) p : 8a+ 3 b+ 5= 5 substitutin f x = and equating t 5 (allw unsimplified) leading t a+ 4b + 56= 0 4a+ b 31= 0 e dep fr slutin f simultaneus equatins t btain a and b a = 1, b = 17 A fr each UCLES 017 Page 5 f 10

6 4037/1 Cambridge O Level Mark Scheme Octber/Nvember 017 Questin Answer Marks Guidance 7(ii) 3 1x + 8x 17x = 0 fr x = 0 x = x = ± e A 5 V C 4 B 8(i) ABC = = sin BAC sin 67.4 attempt at the sine rule, using 4 and 5 (r e.g. use f csine rule fllwed by sine rule n triangle shwn) BAC = 47.6 may be implied by later wrk Angle required = (ii) = ( 5 4 cs65 ) = 65 Answer Given V attempt at the csine rule r sine rule t btain V allw if seen in (i) V = r = V sin BAC sin 65 Distance t travel: 10 sin 67.4 distance t travel allw if seen in (i) 130 r Time taken: dep fr crrect methd t find the time, must have bth f the previus M marks 6.5 UCLES 017 Page 6 f 10

7 4037/1 Cambridge O Level Mark Scheme Octber/Nvember 017 Questin Answer Marks Guidance Alternative methd 10 AC = e cs 5 crrect attempt at AC = 13.4 Allw 13 Speed fr this distance = 5 dep fr speed, it must be 5 exactly fr, must have first M mark Time taken = dep fr a crrect methd t find the time, must have bth f the previus M marks = 6.5 9(a) B3 fr line jining ( 0,5 ) and ( 10,5 ) fr a line jining ( 10,0.5 ) and ( 30,0.5 ) all crrect with n slid line 10,0.5 jining ( 10,5 ) t ( ) 9(b)(i) 3 9(b)(ii) dv 5 3 = 15e t + dt attempt t differentiate, must be in 5 the frm ae t + b When dv 5t = 0, e = 0.1 dt dep fr equating t zer and attempt t slve, must be f the 5t frm ae = b, b > 0 t btain an equatin in the frm 5t = k where k is a lgarithm r < 0 t = UCLES 017 Page 7 f 10

8 4037/1 Cambridge O Level Mark Scheme Octber/Nvember 017 Questin Answer Marks Guidance 9(b)(iii) Either attempt t integrate, must be in the frm 3 5t 3 s = e + t + c 5 4 ( ) When t = 0, s = 0 s 3 c = 5 5 ce t + dt dep fr attempt t find c and substitute t = 0.5 s = Or attempt t integrate, must be in the frm 3 5t 3 e + t ce t + dt crrect use f limits dep leading t s = (i) 5 BAC = 6., BAC = (ii) sin 0.6 = BD, BD =.905,.91 5 valid methd t find BD Arc BFC: π ( = 9.13) BD attempt t find arc length BFC, using their BD Perimeter: = (iii) Area: 1 π sin1.4 B3 fr area f semi circle (= 13.3) fr area f sectr (= 15.5) fr area f triangle (= 11.8) 9.58 Area 9.6 final answer UCLES 017 Page 8 f 10

9 4037/1 Cambridge O Level Mark Scheme Octber/Nvember 017 Questin Answer Marks Guidance 11(a) ( ) tan φ + 35 = 5 dealing crrectly with ct and an attempt at slutin f tan φ + 35 =c, rder must be ( ) crrect, t btain a value fr φ + 35 φ + 35 = 1.8, 01.8, dep fr an attempt at a secnd slutin in the range, ( their first slutin in the range e) φ = 166.8, A fr each 11(b)(i) Either 1 csθ csθ sinθ + sinθ csθ 1 sinθcsθ = csθ cs θ + sin θ sinθ = Or () 1 secθ 1 tanθ tanθ + secθ = 1+ tan θ tanθ expressing all terms in terms f sinθ and csθ where necessary dealing with the fractins crrectly sin θ + cs θ t get in sinθ csθ denminatr r as in left hand clumn use f identity, tgether with a cmplete and crrect slutin, withhld fr incrrect use f brackets dealing with fractins in the denminatr crrectly t get 1+ tan θ in the denminatr, tanθ allw tanθ taken t the numeratr secθ tanθ = sec θ use f the identity t get sec θ = tan θ secθ = sin θ csθ csθ = sinθ expressing all terms in terms f sinθ and csθ and simplificatin t the given answer, withhld fr incrrect use f brackets UCLES 017 Page 9 f 10

10 4037/1 Cambridge O Level Mark Scheme Octber/Nvember 017 Questin Answer Marks Guidance 11(b)(ii) 3 sin 3θ = crrect attempt t slve fr θ, rder must be crrect, may be implied by ne crrect slutin π π 4π 3 θ =,, π π 4π θ =,, A3 fr each UCLES 017 Page 10 f 10

0606 ADDITIONAL MATHEMATICS

PAPA CAMBRIDGE CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge Internatinal General Certificate f Secndary Educatin MARK SCHEME fr the Octber/Nvember 0 series 0606 ADDITIONAL MATHEMATICS 0606/ Paper, maimum