4037 ADDITIONAL MATHEMATICS

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1 CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Ordinary Level MARK SCHEME for the October/November 0 series 40 ADDITIONAL MATHEMATICS 40/ Paper, maimum raw mark 80 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the eamination. It shows the basis on which Eaminers were instructed to award marks. It does not indicate the details of the discussions that took place at an Eaminers meeting before 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 Eaminer Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the October/November 0 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.

2 Page Mark Scheme Syllabus Paper GCE O LEVEL October/November 0 40 Mark Scheme Notes Marks are of the following three types: M A B Method mark, awarded for a valid method applied to the problem. Method marks are not lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. Correct application of a formula without the formula being quoted obviously earns the M mark and in some cases an M mark can be implied from a correct answer. Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated method mark is earned (or implied). Accuracy mark for a correct result or statement independent of method marks. When a part of a question has two or more method steps, the M marks are generally independent unless the scheme specifically says otherwise; and similarly when there are several B marks allocated. The notation DM or DB (or dep*) is used to indicate that a particular M or B mark is dependent on an earlier M or B (asterisked) mark in the scheme. When two or more steps are run together by the candidate, the earlier marks are implied and full credit is given. The symbol implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A or B marks are given for correct work only. A and B marks are not given for fortuitously correct answers or results obtained from incorrect working. Note: B or A means that the candidate can earn or 0. B,, 0 means that the candidate can earn anything from 0 to. Cambridge International Eaminations 0

3 Page Mark Scheme Syllabus Paper GCE O LEVEL October/November 0 40 ( + 6)( ) Critical values 6 and 6 < < ( 4 5 ) = Multiply top and bottom by OR ( 4 5 ) = ( 5 )( p 5 + q) = 5p q + 5( q p) Leading to 5p q = 84, q p = 6 p = q = [] [4] Attempt to solve a three term quadratic Allow > 6 AND < but not OR or a comma. Mark final answer. Attempt to epand, allow one error, must be in the form a + b 5. Must be attempt to epand top and bottom. Allow for c Must get to a pair of simultaneous equations for this mark (i) dy = k 5 dk 4 k = dy Use y = with = and d 56p = p [] [] on k needs both M marks only for 8kp and must be evaluated 4 (i) (iii) 0 5 log XY = log X + log Y = p p p [] [] [] Not log p 5 Or log XY p = log p XY Do not allow just log p X + log p Y = on log p XY Cambridge International Eaminations 0

4 Page 4 Mark Scheme Syllabus Paper GCE O LEVEL October/November y = 5 oe + y = 5 oe Solve their linear simultaneous equations = or y = 0. 5 OR from log y = y =.86 OR from ln y = y = 5.49 Final follows as before, [5] Each in two variables and not quadratic as far as = or y = 6 (a) (i) (b) 8 or 0 ( ) 60 isw 60( ) (i) + (their 60) 0 ( ) oe [] [] B,,,0 [] ± 40 implies ± 0 or +60 hence OK if seen in epansion Can be implied Terms must be evaluated (allow 4 0 ) B for 4 terms correct. for or terms correct. ISW once epansion is seen. (i) 500 l = L = l Substitute for l and correctly reach 000 L = 4 + Dag [] allow l = 500 RHS terms e.g. or better Dependent on both previous B marks dl 4000 = 4 d dl Equate to 0 and solve d = 0 L = 0 d y 4000 = 4 and minimum stated d D [5] either power reduced by one both terms correct n Must get = Both values Or use of gradient either side of turning point. Cambridge International Eaminations 0

5 Page 5 Mark Scheme Syllabus Paper GCE O LEVEL October/November (i) [] Implied by aes or values in a table. May be seen in y Plot against with linear scales Must be linear scales y [] At least correct points plotted and no incorrect points Line must be ruled and through at least correct points (iii) (iv) Finds gradient (0.4) a = 0.4 ± 0.0 b =. ± 0.4 y Read =. 5 [] Condone use of correct values from table/graph to find gradient and /or equation. Values read from graph must be correct. Obtaining ( ) = to 4 from graph or substitute in formula As far as = +ve constant 4.8 [] 4. to 4.9 ignore 4.8 or 0 9 Method A Takes components v sin α = 40 v cosα +.8 = v cosα = 48.4 Solve for v or α α = 9.6 v = 5. ( ) 0 Method B D [8] Allow 0.69 radians =.8 =.6 y = 0.6 = 48.4 D = D = 6.8 D V = V = tan α = 48.4 α = 9. 6 ( = 94.56) D [8] 5. or better Allow 0.69 radians Cambridge International Eaminations 0

6 Page 6 Mark Scheme Syllabus Paper GCE O LEVEL October/November 0 40 Method C z = ( = 80.6) v = ( = 6.) 4 tan δ = ( δ = 9.4) oe V = cos 9.4 V = 5. sin β sin =. 5 β = 9.8() or 9.8() α = β = 9.6 Method D [8] Or ( 90 δ ) tan = 4 Allow 0. radians Allow 0.69 radians ( = 80.6) z = =.8 =.6 4 tan δ = ( δ = 9.4) oe D = cos 9.4 V = 6.8 / = 5. ( ) sin β sin =.8 6 β = 9.8() or 9.8( ) α = β = 9.6 [8] This method has etra steps so note at this point the M mark is for an equation in D but the A mark is for a value of V. Allow 0. radians Allow 0.69 radians Cambridge International Eaminations 0

7 Page Mark Scheme Syllabus Paper GCE O LEVEL October/November (i) AB = to 5.5 θ = π.4( = 4.88) Use s = rθ ( = 58.6) 4. cos.4 [5] AB = sin 0. May be implied May be implied 4.9 or better oe π or π.4 = sin.4 = 0.9 or Area of major sector + Area of triangle 4 or 4 (Sector) (.4)( = 5) (Triangle) ( ) [4] May be implied. May be implied (i) y d = e d m = e y e = e ( 9) At Q y = 0, = 6 D [4] For insertion of = 9 into d y their. 6. or better if correct. d Using their evaluated m to find eqn y = or better if correct. Accept value that rounds to 6.0 to sf Area triangle.5e or 0. e d = e oe Uses limits of 0 and 9 in integrated function. e or 5. Area under curve subtract area of triangle.5e or. [6] ± must see both values inserted if incorrect answer Condone. if obtained from Cambridge International Eaminations 0

8 Page 8 Mark Scheme Syllabus Paper GCE O LEVEL October/November 0 40 (a) cosec = inserted into equation sin tan = D [4] One correct value. on 80 + ( 64.) Must come from tan = Condone64 and 44 Deduct mark for etras in range (b) (y ) = 0.9..or.4 Find y using radians (or 0.9 or 0.90).6, 4.04 and 4.8(45) [5] Allow 0.8,. or 45.6 Add then divide by on a correct angle One correct value Another correct value Final two values Deduct mark for etras in range Cambridge International Eaminations 0

4037 ADDITIONAL MATHEMATICS

4037 ADDITIONAL MATHEMATICS CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Ordinary Level MARK SCHEME for the October/November 202 series 07 ADDITIONAL MATHEMATICS 07/22 Paper 2, maimum raw mark 80 This mark scheme is published as an aid