Coimisiún na Scrúduithe Stáit State Examinations Commission Leaving Certificate 016 Marking Scheme Mathematics Higher Level
Note to teachers and students on the use of published marking schemes Marking schemes published by the State Examinations Commission are not intended to be standalone documents. They are an essential resource for examiners who receive training in the correct interpretation and application of the scheme. This training involves, among other things, marking samples of student work and discussing the marks awarded, so as to clarify the correct application of the scheme. The work of examiners is subsequently monitored by Advising Examiners to ensure consistent and accurate application of the marking scheme. This process is overseen by the Chief Examiner, usually assisted by a Chief Advising Examiner. The Chief Examiner is the final authority regarding whether or not the marking scheme has been correctly applied to any piece of candidate work. Marking schemes are working documents. While a draft marking scheme is prepared in advance of the examination, the scheme is not finalised until examiners have applied it to candidates work and the feedback from all examiners has been collated and considered in light of the full range of responses of candidates, the overall level of difficulty of the examination and the need to maintain consistency in standards from year to year. This published document contains the finalised scheme, as it was applied to all candidates work. In the case of marking schemes that include model solutions or answers, it should be noted that these are not intended to be exhaustive. Variations and alternatives may also be acceptable. Examiners must consider all answers on their merits, and will have consulted with their Advising Examiners when in doubt. Future Marking Schemes Assumptions about future marking schemes on the basis of past schemes should be avoided. While the underlying assessment principles remain the same, the details of the marking of a particular type of question may change in the context of the contribution of that question to the overall examination in a given year. The Chief Examiner in any given year has the responsibility to determine how best to ensure the fair and accurate assessment of candidates work and to ensure consistency in the standard of the assessment from year to year. Accordingly, aspects of the structure, detail and application of the marking scheme for a particular examination are subject to change from one year to the next without notice.
Contents Page Paper 1 Solutions and marking scheme... Structure of the marking scheme... 3 Summary of mark allocations and scales to be applied... 4 Model solutions and detailed marking notes... 5 Paper Solutions and marking scheme... 3 Structure of the marking scheme... 4 Summary of mark allocations and scales to be applied... 5 Model solutions and detailed marking notes... 6 Marcanna breise as ucht freagairt trí Ghaeilge... 44 [1]
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Coimisiún na Scrúduithe Stáit State Examinations Commission Leaving Certificate 016 Model Solutions and Marking Scheme Mathematics Higher Level Paper 1 []
Marking Scheme Paper 1, Section A and Section B Structure of the marking scheme Candidate responses are marked according to different scales, depending on the types of response anticipated. Scales labelled A divide candidate responses into two categories (correct and incorrect). Scales labelled B divide responses into three categories (correct, partially correct, and incorrect), and so on. The scales and the marks that they generate are summarised in this table: Scale label A B C D E No of categories 3 4 5 6 5 mark scales 0, 5 0,, 5 0,, 4, 5 0,, 3, 4, 5 10 mark scales 0, 10 0, 5, 10 0, 3, 7, 10 0,, 5, 8, 10 15 mark scales 0, 15 0, 7, 15 0, 5, 10, 15 0, 4, 7, 11, 15 0 mark scales 0, 0 0, 10, 0 0, 7, 13, 0 0, 5, 10, 15, 0 5 mark scales 0, 5 0, 1, 5 0, 8, 17, 5 0, 6, 1, 19, 5 0, 5, 10, 15, 0, 5 A general descriptor of each point on each scale is given below. More specific directions in relation to interpreting the scales in the context of each question are given in the scheme, where necessary. Marking scales level descriptors A-scales (two categories) incorrect response correct response B-scales (three categories) response of no substantial merit partially correct response correct response C-scales (four categories) response of no substantial merit response with some merit almost correct response correct response D-scales (five categories) response of no substantial merit response with some merit response about half-right almost correct response correct response E-scales (six categories) response of no substantial merit response with some merit response almost half-right response more than half-right almost correct response correct response In certain cases, typically involving incorrect rounding, omission of units, a misreading that does not oversimplify the work or an arithmetical error that does not oversimplify the work, a mark that is one mark below the full-credit mark may also be awarded. Thus, for example, in scale 10C, 9 marks may be awarded. Throughout the scheme indicate by use of * where an arithmetic error occurs. [3]
Summary of mark allocations and scales to be applied Section A Section B Question 1 (c) Question Question 3 (i) (ii) (iii) 5B 10C 10C 10C 15C 5C 5C 5B 10C Question 7 Question 8 (i) (ii) (i) (ii) (i) (ii) (iii) (iv) (i) (ii) (iii) 10C 10C 10C 10C 10C 5B 5B 10D 10D 5B 10C Question 4 (i) (ii) Question 5 (i) (ii) (i) (ii) 15D 5C 5D 10D 5B 5B 5B Question 9 (i) (ii) (iii) (i) (ii) (iii) 10C 10C 15D 5B 10C 5B Question 6 (i)+(ii) 10D 15D [4]
Model Solutions & Marking Notes Note: The model solutions for each question are not intended to be exhaustive there may be other correct solutions. Any Examiner unsure of the validity of the approach adopted by a particular candidate to a particular question should contact his / her Advising Examiner. Q1 Model Solution 5 Marks Marking Notes 4 3 Scale 5B (0,, 5) Partial Credit: real or imaginary part correct = 1 +1 = = (1+ ) = cos 4 + sin 4 (1+ ) = 16(cos + sin ) (1+ ) = 16(1) = 16 : correct answer without use of De Moivre s modulus or argument correct formula statement of De Moivre s : 16(cos + sin ) Note: not De Moivre and incorrect answer merits 0 marks (c) = ( ) ± ( + ) 4(3 ) ( ) ± 4 4 1 1+4 = = ± 9 = ±3 =1 1+ Or + + =0 + =0 Sum of roots = 1+ + = =1 : root formula with some substitution formula fully substituted Or : equation rearranged correct substitution [5]
Or Or ( 1 )( ) =. + + = (1+ + ) + (1 + ) = + ( + ) +(3 ) (1+ ) =3 = 3 1+. 1 1 =1 : correct factor(s) identification of equal terms z 1 i z z Or z 1+ i z + iz + 3 i z iz z + iz + 3 i z + 1 + i iz + i iz + i Or : long division formulated correctly two correct lines in division z 1+ i = 0 z = 1 i Or Or (1+ )( + ) =3 ( ) + ( + ) =3+( 1) = 3 + = 1 Solving = 1 = : correct multiplication substitution of ( + ) into quadratic and stops identification of equal terms Note: substitution of (1+ ) merits 0 marks [6]
Q Model Solution 5 Marks Marking Notes 8 +16 4 8 +1 0 ( )( 6) 0 = = 6 6 Or 4 4 6 : either side squared one correct linear inequality written stating range of natural numbers only : correct solutions to quadratic Full Credit: correct answer without work Note: use of natural numbers in range merits at most Or Graphical method (must indicate range on X-axis somehow) 6 Or Or : any one straight line : three straight lines [7]
= 3 1 3 1 + 3 1 ( ) + =4 11 +4 15=0 (11 + 15)( 1) =0 = 15 = 1 11 3 15 1 = 11 = = 17 = 11 or = 1 3 + 1 + 1 3 3 11 +5 34=0 (11 17)( +) =0 = 17 11 = 15 11 or = or = 1 3(1) 1 =4 Scale 15C (0, 5, 10,15) : effort to isolate (or ) : fully correct substitution into quadratic [8]
Q3 Model Solution 5 Marks Marking Notes (i) x 0 0 5 1 ln( 4) f ( x) = 1 1 0 74 0 5 x e g ( x) = e x 1 0 0 65 1 7 3 Scale 5C (0,, 4, 5) one entry correct 5 entries correct (ii) g(x) Scale 5C (0,, 4, 5) one plot correct 5 plots correct one correct graph no labelling f(x) Notes: - straight lines NOT acceptable - one clear label merits full credit - one ambiguous label merits High Partial Credit at most (iii) f ( x) = g( x) when x 0 7 Scale 5B (0,, 5) Partial Credit point of intersection clearly indicated on graph, but value of not stated [9]
Q3 Model Solution Continued Marking Notes 1 = 1 = ( ) =0 ( )( +1) =0 = = 1 = ln or = 0 693 substitution correct correct factors of quadratic root formula correctly substituted = ( 1) ± ( 1) 4(1)( ) (1) Or ( ) =0 Let = =0 = ( 1) ± ( 1) 4(1)( ) (1) = 1± 1+8 = 1±3 = or = 1 (not possible) = = = ln or = 0 693 Note: oversimplification of equation (i.e. not treating as quadratic) merits Low Partial Credit at most Or substitution correct root formula correctly substituted = ( 1) ± ( 1) 4(1)( ) (1) Note: oversimplification of equation (i.e. not treating as quadratic) merits Low Partial Credit at most [10]
Q4 Model Solution 5 Marks Marking Notes : 8 1 = 7 (divisible by 7) : Assume 8 1 is divisible by 7 8 1=7 8 =7 +1 : 8 1=8 8 1 =8(7 + 1) 1 = 56 + 7 =7(8 +1) is divisible by 7 Scale 15D (0, 4, 7, 11, 15) step Mid Partial Credit step step use of step to prove step is true true is true So, true whenever true. Since true, then, by induction, is true for all natural numbers 1 Note: accept step, step and step in any order Or = 8 1 = 8. 8 1 = (7+1).8 1 = 7 8 +(8 1) Obviously divisible by 7 From So, true whenever true. Since true, then, by induction, is true for all natural numbers 1 [11]
(i) =log, = log 3 8 log 3 =log 8 log 3 =log () log 3 =3log log 3 = 3 Scale 5C (0,, 4, 5) low Partial Credit log 8 log 3 log 8=3log (and/or = 3p) (ii) 9 log 16 =log (3 ) log () =log 3+log 4log = +(1) 4 = + 4 Scale 5D (0,, 3, 4, 5) log 9 log 16 Mid Partial Credit log 3 log 4log 4 or or (log 3+log ) 4 log or equivalent [1]
Q5 Model Solution 5 Marks Marking Notes (i) (5 9) = ( 1) + (4 ) 8 88 + 80 = 0 11 + 10 = 0 ( 1)( 10) =0 = 1 or = 10 = 10 Scale 10D (0,, 5, 8, 10) any use of Pythagoras Mid Partial Credit fully correct substitution both roots correct (ii) Sides=9, 40, 41 9 +40 =41 81 + 1600 = 1681 1681 = 1681 Scale 5B (0,, 5) Partial Credit 9 or 40 or 41 using 1 or 10 from candidates work (i) Function is bijective if inverse exists ( ) = + 3 Function is injective. or Horizontal line test. or ( ) = ( ) 3 = 3 = or,, ( ) = ( ) = Scale 5B (0,, 5) Partial Credit ( ) written ( ) drawn ( ) = ( ) (ii) ( ) =3 ( ) = + 3 Scale 5B (0,, 5) Partial Credit any relevant transpose [13]
Q6 Model Solution 5 Marks Marking Notes ( +h) ( ) = ( + h + 4) ( +4) ( +h) ( ) lim = h ( + h + 4) ( +4) lim h +8h +4h + 16 + 16h + 16) (4 (4 = lim + 16 + 16) h =lim 8h + 4h + 16h h =8 +16 or Scale 10D (0,, 5, 8, 10) any ( +h) Mid Partial Credit limit of ( ) ( ) limit of ( ) ( ) Notes: - omission of limit sign penalised once only - answer not from 1 st Principles merits 0 marks ( ) = ( + 4) =4 + 16 + 16 ( +h) =4( +h) +16( +h) +16 =4 +8h +4h + 16 + 16h + 16 ( +h) ( ) lim h 8h + 4h + 16h lim h =8 +16 [14]
(i)+ (ii) =.sin 1 =sin1 + cos 1 1 =sin1 1 cos 1 =sin 4 4 cos 4 =0 15 Scale 15D (0, 4, 7, 11, 15) any correct differentiation Mid Partial Credit product rule applied correct differentiation Note: one penalty for calculator in wrong mode [15]
Q7 Model Solution 40 Marks Marking Notes (i) = 4 3 =4 = 50 cm /s =. = 1. 50 4 = 50 4 400 = 5 3 cm/s work towards or correct expression for or (ii) =4 =8 =. =8. 5 3 = 5(0) 4 =5 cm /s work towards or correct expression for (i) + 10 = 0 ( + 10) =0 = 0 or = 10 quadratic equation formed gets =0 only quadratic factorised Note: ( ) =0 10=0 =5 merits 0 marks (ii) 1 10 0 ( + 10 ) = 1 10 3 +5 = 1 10 1000 + 500 0 3 = 100 3 +50= 50 3 m integration set up correct integration with some substitution [16]
Q8 Model Solution 55 Marks Marking Notes (i) ( ) = 0 74 + 1 193 + 3 3 ( ) = 0 548 + 1 193 = 0 = 177 m ( 177) = 0 74( 177) + 1 193( 177) + 3 3 = 1 986 + 597 + 3 3 = 4 59 m 0 74( 1193 74 1615 137 ) 0 74( 1193 548 ) + 4.585 Max Height = 4 59 m any correct differentiation effort made at completing square trial and error with more than one value of tested value correct Note: if correct answer by trial and error, must show points on each side of max point to be lower to earn full credit (ii) tan = 0 548(4 5)+1 193 tan = 1 73 = 51 8 = 5 Scale 5B (0,, 5) Partial Credit tan Note: right angled triangles may appear in diagram given in equation (iii) Map C ( 0 5, 565) (0, ) 177 ( 0 5) = 677 4 59 0 565 = 3 964 ( 177, 4 59) ( 677, 3 964) Scale 5B (0,, 5) Partial Credit ( 0 5, 565) (0, ) [17]
(iv) ( ) = + + (0, ) ( ) => = (4 5, 3 05) ( ) 3 05=a(4 5) + b(4 5)+ 0 5 + 4 5 = 1 05 ( ) = + =0 ( 677) + =0 (i) Scale 10D (0,, 5, 8, 10) c value found relevant equation in a, b and/or c Mid Partial Credit formulated correctly any two equations formulated correctly any three equations 5 354 + = 0 (ii) Note: + + not in an equation merits 0 marks From (i) and (ii) = 0 73 = 1 46 ( ) = 0 73 + 1 46 + [Note: a third equation that could be used is 3 964 = ( 677) + ( 677) + (iii)] Or Or Equation of parabola with vertex (h, ): ( ) = ( h) + (0, ) on curve: (h, ) = ( 677, 3.964) = ( 677) +3 964 1 964 = (7 16639) = 0 7405 = 0 74 Parabola: ( ) = 0 74 ( 677) +3 964 or ( ) = ( 0 5) 0 565 ( ) = 0 74( 0 5) +1 193( 0 5) + 3 3 0 565 ( ) = 0 74 + 1 467 + Scale 10D (0,, 5, 8, 10) equation of curve use of C Mid Partial Credit using peak value value of a found [18]
(i) 00 m Race: = ( ) = 4 99087(4 5 3 8) 1 = 1000 Javelin: = ( ) = 15 9803(58 3 8) 1 = 100 Scale 10D (0,, 5, 8, 10) some relevant substitution into one formula Mid Partial Credit one value of found some relevant substitution into both formulas one value correct and some relevant substitution into second formula uses incorrect formula (once only) (ii) = ( ) 195 = 15 9803( 3 8) 81 0373=( 3 8) = log 81 0373 log = 1 04 = 68 4343 = ( 3 8) = 7 343 = 7 3 m Scale 5B (0,, 5) Partial Credit some relevant substitution into formula (iii) = ( ) 1087 = 0 11193(54 11 84) c 1087 0 11193 = (13 16) log 9711 46 = c log 13 16 log 9711 46 = log 13 16 =1 88 some relevant substitution into formula fully correct substitution into formula [19]
Q9 Model Solution 55 Marks Marking Notes (i) 4,, 1, 1, 1 4, 1 8, 1 16 = (1 ) 1 4 1 1 = 1 1 = 7 9375 1 1 = 18 =7 some listing of terms formula listing of exactly 7 correct terms formula fully substituted (ii) = 1 = 4 1 1 =8 formula formula fully substituted [0]
1 3 4 5 6 7 8 9 Chg x +4 0-1 0 1 4 0 1 16 0 1 64 Chg y 0 0 1 (iii) = 4 1 1 =3 = 16 4 5 = 1 1 =1 6= 8 4 5 16 5, 8 or (3, 1 6) 5 0 1 8 Scale 15D (0, 4, 7, 11, 15) extra entries correct in either row Mid Partial Credit either row fully correct 0 1 3 one co-ordinate correct Notes: - need to see correctly used to move beyond Mid Partial Credit - no merits Mid Partial Credit at most 0 (i) =Female,Male,Female,Female,Male Scale 5B (0,, 5) Partial Credit one correct entry (ii) = + =5+3=8 = + =8+5=13 = + = + or correct 8 and/or 13 without work correct substitution in both [1]
(iii) = (1 + 5) (1 5) 5 = (1 + 5) = 1+3 5 +3 5 + 5 = 16 + 8 5 (1 5) = 1 3 5 +3 5 5 = 16 8 5 6 5 + 5 = 8 5 = 6+ 5 8 = 16 8 = Q. E. D. Scale 5B (0,, 5) Partial Credit some correct substitution using approximate value for 5 = some effort at cubing Note: use of 5 as approximation, even if rounded off to at end of work merits at most Partial Credit []
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Coimisiún na Scrúduithe Stáit State Examinations Commission Leaving Certificate 016 Model Solutions and Marking Scheme Mathematics Higher Level Paper [3]
Marking Scheme Paper 1, Section A and Section B Structure of the marking scheme Candidate responses are marked according to different scales, depending on the types of response anticipated. Scales labelled A divide candidate responses into two categories (correct and incorrect). Scales labelled B divide responses into three categories (correct, partially correct, and incorrect), and so on. The scales and the marks that they generate are summarised in this table: Scale label A B C D E No of categories 3 4 5 6 5 mark scales 0,, 5 0,, 4, 5 10 mark scales 0, 5, 10 0, 3, 7, 10 0, 3, 5, 8, 10 15 mark scales 0, 5, 10, 15 0, 4, 7, 11, 15 0 mark scales 5 mark scales A general descriptor of each point on each scale is given below. More specific directions in relation to interpreting the scales in the context of each question are given in the scheme, where necessary. Marking scales level descriptors A-scales (two categories) incorrect response correct response B-scales (three categories) response of no substantial merit partially correct response correct response C-scales (four categories) response of no substantial merit response with some merit almost correct response correct response D-scales (five categories) response of no substantial merit response with some merit response about half-right almost correct response correct response E-scales (six categories) response of no substantial merit response with some merit response almost half-right response more than half-right almost correct response correct response In certain cases, typically involving incorrect rounding, omission of units, a misreading that does not oversimplify the work or an arithmetical error that does not oversimplify the work, a mark that is one mark below the full-credit mark may also be awarded. Thus, for example, in scale 10C, 9 marks may be awarded. Throughout the scheme indicate by use of * where an arithmetic error occurs. [4]
Summary of mark allocations and scales to be applied Section A Section B Question 1 Question Question 3 Question 4 (i) (ii) Question 5 (i) (ii) Question 6 (c) 10C 15D 10C 15D 15C 10D 15C 5C 5C 5B 10C 10C 10C 10C 5C Question 7 (i) (ii) (iii) (iv) (v) Question 8 (c) (d)(i) (d)(ii) (e) (f) Question 9 (i) (ii) (iii) (c) (d) 10C 10B 10C 10C 10D 5C 5C 5B 5C 10C 10C 5B 5B 10D 5C 15D 10C 5B 5C [5]
Model Solutions & Detailed Marking Notes Note: The model solutions for each question are not intended to be exhaustive there may be other correct solutions. Any Examiner unsure of the validity of the approach adopted by a particular candidate to a particular question should contact his/her Advising Examiner. Q1 Model Solution 5 Marks Marking Notes Slope = 3 perp. slope = 3 3= 3 ( 5) 3 = 9 slope formula with some relevant substitution 3 = 5m+c = ( ) with or or both substituted perpendicular slope equation of line through B parallel to AC Point of intersection of the altitudes Slope = 3+ 5 6 = 5 1 perp. slope = 1 5 4= ( + 3) 5 +3=0 If BC chosen: Orthocentre: 3 = 9 5 = 3 = 6 = 7 (7, 6) or Slope = 3 4 5+3 = 1 8 Scale 15D (0, 4, 7,11,1 5) demonstration of understanding of orthocentre ( e.g. mentions altitude) slope formula with some relevant substitution altitude from part Mid Partial Credit equation of an altitude other than some relevant substitution towards finding a second altitude and altitude from correct construction two correct altitudes correct construction with orthocentre (7, 6) perp. slope = 8 Equation of altitude: + = 8( 6) Equation: 8 = 50 Orthocentre: 3 = 9 8 = 50 = 6 = 7 (7, 6) [6]
Q Model Solution 5 Marks Marking Notes 6= 1 7 ( +1) 7 +43=0 : equation of line formula with some relevant substitution : equation of line not in required form = + + + 3( ) +4( ) 1 = 3 +4 5 = 3 4 1 3 4 1 = ±5 3 + 4 = 46 (i) and 3 + 4 = 4 (ii) But (, ) 7 +43 = 0 + 7 + 43 = 0... (iii) = 7 +43 Scale 15D (0, 4, 7,11,1 5) some correct substitution into relevant formula (line, circle, perpendicular distance). Mid Partial Credit one relevant equation in g and f ( either(i) or (ii) or (iii)) two relevant equations ( either (i) and (iii) or (ii) and (iii)) Solving : = 7 +43 and 3 + 4 = 46 = 7 and = 6 Centre (6, 7) ( 6) + ( 7) =5 or Solving: = 7 +43 and 3 + 4 = 4 = 5 and = 8 Centre (- 8, 5) ( + 8) +( 5) =5 [7]
Q3 Model Solution 5 Marks Marking Notes 7 + cos cos 7 + sin 7 cos cos4 cos3 cos4 sin3 = cos 3 sin 3 =cot 3 7 Scale 15C (0, 5, 10, 15) sum to product formula with some substitution sum to product formula fully substituted Method 1: Method : cos = 1 (1+cos ) = 1 1+1 9 = 5 9 cos = ± 5 3 or cosθ =1 sin = 9 18 sin =1 sin 4 = sin =± 9 cos =± Method 3: or Scale 10D (0, 3, 5, 8, 10) Use of a relevant formula in cos cos =83 6 =41 8 Mid Partial Credit correct substitution (method 1) expression in sin (method ) expression in tan (method 3) expression in cos (method 4) =41 8 and = 13 or = 1 8 one value only (e.g.+ 5 ) 3 values found for cos 41 8 and cos 138 or cos 1 8 cosθ = 1 tan 1+ tan θ = θ 1 9 9 9 tan =1+tan tan = 4 5 tan =± 5 cos =± 5 3 [8]
Method 4: sin θ = 1 or ( 1 cos θ ) 1 1 cos θ = cos θ = 1 cos θ ( 1 cos θ ) 1 1+ 1+ cos θ cos θ = = 9 5 cos θ = 9 5 cosθ = ± 3 [9]
Q4 Model Solution 5 Marks Marking Notes (i) = = 90.(i) + = 90 angles in triangle sum to 180 + = 90. angle in semicircle + = + =..(ii) ( ) = = 90.(i) = = ( ) h h (ii) Scale 15C (0, 5, 10, 15) identifies one angle of same size in each triangle identifies second angle of same size in each triangle implies triangles are similar without justifying (ii) in model solution or equivalent (ii) 1 = = = or + = = + = +1 + + +1=( +1) = = Or Scale 5C (0,, 4, 5) one set of corresponding sides identified indicates relevant use of Pythagoras corresponding sides fully substituted expression in or, i.e. fails to finish x y + y + 1 y = x 1 + y = y ( y 4 y = x y = x y = + 1) x [30]
Construction Scale 5C (0,, 4, 5) perpendicular line drawn at U or T relevant use of 1 cm length mid point of incorrect extended segment constructed correct mid-point constructed [31]
Q5 Model Solution 5 Marks Marking Notes (i) John David Mike Scale 5B (0,, 5) Partial Credit 1 correct column (ii) ( ) = 1 5 1 6 3 4 + 1 5 5 6 1 4 + 4 5 1 6 1 4 + 1 5 1 6 1 4 = 13 10 one correct triple (numerical or descriptive) probability of any one Miss 4 correct triples (numerical) ( ) = ( ) ( ) 0 1 = ( + 0 1) 0 4 0 4 = 0 06 = 0 15 or ( ) = ( ) = +0 1 = 0 15 formula written or implied writes P(A) = + 0 1 formula fully substituted [3]
Q6 Model Solution 5 Marks Marking Notes Event (, 3, 3) = 1 6 1 10 1 10 = 1 600 Payout Win 1000 letter 1 No. letter nd No letter only Fail to win 50 50 50 Prob (P(x)) 1 600 9 600 9 600 81 600 x.p(x) 1000 600 450 600 450 600 4050 600 0 0. ( ) = 5950 600 = 9 Club loses 9 cent per play any correct relevant probability High Partial credit correct probabilities but not expressed as single fraction or equivalent Note: Accept correct answer without supporting work 1 correct entry to table all entries correct but fails to finish or finishes incorrectly no conclusion Or Event Pay out Prob (P(x) x.p(x) Win 998 1 998 600 600 letter + 1 st No. Letter + nd No letter only Fail to Win 48 9 43 600 600 48 9 43 600 600 48 81 3888 600 600 + 500 5000 600 600. ( ) = = 9 cent [33]
(c) Profit = Revenue Pay-out 600 = 845( 9) = 600 + 845( 9) 845 =3 Scale 5C (0,, 4, 5) links profit, revenue and payout High partial Credit formula fully substituted or 600 845 =0 71 0 71 + 9 = 3 [34]
Q7 Model Solution 55 Marks Marking Notes (i) =3 + 5 = 15 5 = 15 5 =3 905 = 1 955 Pythagoras with relevant substitution correct = 15 5 =1 95 (ii) tan 50 = 1 95 =1 95(1 19175) = 339 = 3 Scale 10B (0, 5, 10) Partial Credit tan formulated correctly (iii) Also: sin 40 = 1 95 cos 50 = 1 95 =1 95 + 3 BC = 3 015377 =3 or cos 40 = 3 or sin 50 = 3 or Pythagoras with relevant substitution Pythagoras fully substituted 1 95 = ( i.e. isolated) sin 40 (iv) 3 =3 + 5 (3)( 5) cos 15 cos = 6 5 = 65 or cos = 1 5 3 = 65 cosine rule with some relevant substitution cosine ratio with some relevant substitutions identifies three sides of triangle BCD cosine rule with full relevant substitutions cosine ratio with full relevant substitutions [35]
(v) A = isosceles triangle + equilateral triangle = 1 ( 5)(3) sin 65 + Scale 10D (0,3,5,8,10) recognises area of 4 triangles 1 (3)(3) sin 60 = 14 59 A=15 Mid Partial Credit Area of 1 triangle correct area of isosceles triangle and equilateral triangle Note: Area = 4 isosceles or 4 equilateral triangles merit HPC at most tan 60 = 3 = 3 = 3 + = 3 = 6 Scale 5C (0,, 4, 5) effort at Pythagoras but without (or ) found = 3 [36]
Q8 Model Solution 45 Marks Marking Notes Period = = 1 hours Scale 5C (0,,4, 5) Range = 1 6 1 5, 1 6 + 1 5 = 0 1 m, 3 1 m some use of or range of cos function High partial credit period or range correct Note: Accept correct period and/or range without work Max = 1 6 + 1 5(1) =3 1 m. or 3 1 m from range Scale 5B (0,, 5) Partial Credit max occurs when cos A = 1 or t = 0 effort at h ( ) Note: Accept correct answer without work (c) h ( ) =1 5( sin 6 ) 6 h () =1 5( sin 6 ) 6 = 0 68017 = 0 68 m/h Tide is going out at a rate of 0 68 m per hour at am Scale 5C (0,, 4, 5) effort at differentiation correct numerical answer but not in context [37]
(d)(i) h( ) =1 6+1 5 cos 6 Time 1 am 3 am 6 am 9 am 1 pm 3 pm 6 pm 9 pm 1 am t 0 3 6 9 1 15 18 1 4 Height 3 1 1 6 1 1 6 3 1 1 6 1 1 6 3 1 (d) (i) one correct height five correct heights [38]
(d) (ii) (d) (ii) Graph Scale 10C (0, 3,7,10) one correct plot at least 7 correct plots plots correct but graph not sketched or sketched incorrectly [39]
(e) Low tide = 0 1 m High tide = 3 1 m Difference = 3 1 0 1 =3 m Scale 5B (0,, 5) Partial Credit height of Low tide or High tide correctly identified Notes: (i) candidates may show work for this section on graph (ii) accept values from candidate s graph (iii) accept correct answer from graph without work (f) Enter port at 9:30 approx Leave port before 15:15 approx Time = 15:15 9:30 = 5 hr 45 min approx. Scale 5B (0,, 5) Partial Credit time of entry to port or leave port correctly identified value(s) for h = and/or h = 1 5 on sketch time estimated using relevant values other than those required for the maximum time. Notes: (i) candidates may show relevant work for this section on graph (ii) accept values from candidate s graph [40]
Q9 Model Solution 50 Marks Marking Notes (i) = 39400, = 190 = 60000 39400 = 190 = 1 59 ( > 1 59) =1 ( < 1 59) = 1 0 9441 = 0 0559 =5 59% = 5 6% (ii) ( ) =0 9 =1 8 = 1 8 39400 = 1 8 190 = 86 40 = 86 Scale 10D (0, 3, 5, 8, 10) and identified Mid Partial Credit z = 1 59 identifies 0 9441 Scale 5C (0,, 4, 5) identifies 1 8 but fails to progress formula for fully substituted (iii) = 39400, = 190, = 3880, = 1000 = 39400 39400 3880 39400 = = 74 190 1000 74 < 1 96 Result is significant. There is evidence to reject the null hypothesis The mean income has changed. Scale 15D (0, 4, 7, 11,15) z formulated with some substitution states null and/or alternative hypothesis only reference to 1 96 Mid Partial Credit z fully substituted = 74 and stops fails to state the null and alternative hypothesis correctly fails to contextualise the answer [41]
or Confidence Interval: ±1 96 39400 ± 1 96 190 1000 38599, 4000 8 3880 outside range Result is significant. There is evidence to reject the null hypothesis The mean income has changed. or Confidence Interval: ±1 96 3880 ± 1 96 190 1000 3880 ± 1 96(408 57) 37479, 39080 8 39400 outside range Result is significant. There is evidence to reject the null hypothesis The mean income has changed. [4]
Q9 6974 1 96 510 400 6974 + 1 96 510 400 647 4 7475 76 Marking Notes interval formulated with some correct substitution interval formulated with fully correct substitution (c) The distribution of sample means will be normally distributed Scale 5B (0,, 5) Partial Credit mentions 30 (or more) but not contextualised (d) 1 =0 045 1 0 045 = 1 = 0 045 = 493 87 Scale 5C (0,, 4, 5) formulated with fully correct substitution Note: Accept 493 farmers or 494 farmers [43]
Marcanna breise as ucht freagairt trí Ghaeilge (Bonus marks for answering through Irish) Ba chóir marcanna de réir an ghnáthráta a bhronnadh ar iarrthóirí nach ngnóthaíonn níos mó ná 75% d iomlán na marcanna don pháipéar. Ba chóir freisin an marc bónais sin a shlánú síos. Déantar an cinneadh agus an ríomhaireacht faoin marc bónais i gcás gach páipéir ar leithligh. Is é 5% an gnáthráta agus is é 300 iomlán na marcanna don pháipéar. Mar sin, bain úsáid as an ngnáthráta 5% i gcás iarrthóirí a ghnóthaíonn 5 marc nó níos lú, e.g. 198 marc 5% = 9 9 bónas = 9 marc. Má ghnóthaíonn an t-iarrthóir níos mó ná 5 marc, ríomhtar an bónas de réir na foirmle [300 bunmharc] 15%, agus an marc bónais sin a shlánú síos. In ionad an ríomhaireacht sin a dhéanamh, is féidir úsáid a bhaint as an tábla thíos. Bunmharc Marc Bónais 6 11 7 33 10 34 40 9 41 46 8 47 53 7 54 60 6 61 66 5 67 73 4 74 80 3 81 86 87 93 1 94 300 0 [44]
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