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Coimisiún na ScrúduitheScrúduithe StáitStáitState ExaminationsExaminations CommissionStateCommissionLeaving CertificateCertificate 20122015Marking SchemeSchemeMathematics GraphicsDesign and CommunicationHigher Level

Note to teachers and students on the use of published marking schemesMarking schemes published by the State Examinations Commission are not intended to bestandalone documents. They are an essential resource for examiners who receive training inthe correct interpretation and application of the scheme. This training involves, among otherthings, marking samples of student work and discussing the marks awarded, so as to clarifythe correct application of the scheme. The work of examiners is subsequently monitored byAdvising Examiners to ensure consistent and accurate application of the marking scheme.This process is overseen by the Chief Examiner, usually assisted by a Chief AdvisingExaminer. The Chief Examiner is the final authority regarding whether or not the markingscheme has been correctly applied to any piece of candidate work.Marking schemes are working documents. While a draft marking scheme is prepared inadvance of the examination, the scheme is not finalised until examiners have applied it tocandidates’ work and the feedback from all examiners has been collated and considered inlight of the full range of responses of candidates, the overall level of difficulty of theexamination and the need to maintain consistency in standards from year to year. Thispublished 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 notedthat these are not intended to be exhaustive. Variations and alternatives may also beacceptable. Examiners must consider all answers on their merits, and will have consultedwith their Advising Examiners when in doubt.Future Marking SchemesAssumptions 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 aparticular type of question may change in the context of the contribution of that question tothe overall examination in a given year. The Chief Examiner in any given year has theresponsibility 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 aparticular examination are subject to change from one year to the next without notice.

ContentsPagePaper 1Model Solutions .3Marking Scheme . 21Structure of the marking scheme . 21Summary of mark allocations and scales to be applied . 22Detailed marking notes . 23Paper 2Model Solutions . 33Marking Scheme . 52Structure of the marking scheme . 52Summary of mark allocations and scales to be applied . 53Detailed marking notes . 54Marcanna breise as ucht freagairt trí Ghaeilge . 63[1]

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2015. M29Coimisiún na Scrúduithe StáitState Examinations CommissionLeaving Certi cate Examination 2015MathematicsPaper 1Higher LevelFriday 5 JuneAfternoon 2:00 – 4:30300 marksModel Solutions – Paper 1Note: The model solutions for each question are not intended to be exhaustive – there may be othercorrect solutions. Any examiner unsure of the validity of the approach adopted by a particularcandidate to a particular question should contact his / her advising examiner.[3]

InstructionsThere are two sections in this examination paper.Section AConcepts and Skills150 marks6 questionsSection BContexts and Applications150 marks3 questionsAnswer all nine questions.Write your answers in the spaces provided in this booklet. You may lose marks if you do not do so.You may ask the superintendent for more paper. Label any extra work clearly with the questionnumber and part.The superintendent will give you a copy of the Formulae and Tables booklet. You must return it atthe end of the examination. You are not allowed to bring your own copy into the examination.You will lose marks if all necessary work is not clearly shown.You may lose marks if the appropriate units of measurement are not included, where relevant.You may lose marks if your answers are not given in simplest form, where relevant.Write the make and model of your calculator(s) here:[4]

Section AConcepts and Skills150 marksAnswer all six questions from this section.Question 1(25 marks)Mary threw a ball onto level ground from a height of 2 m. Each time the ball hit the ground itHeightbounced back up to34of the height of the previous bounce, as shown.2m0(a)(b)BounceComplete the table below to show the maximum height, in fraction form, reached by the ballon each of the first four bounces.Bounce0Height (m)2113223498273281128Find, in metres, the total vertical distance (up and down) the ball had travelled when it hit theground for the 5th time. Give your answer in fraction form. 3 9 27 81 525 653 10 132 2 2 2 64 m283212812864 or 3 9 27 81 2 2 2 2S4 2 8 32 128 32 (1 ( 34 )4 2 2 1 34 525 653 2 10 1364 m6464[5]

(c)If the ball were to continue to bounce indefinitely, find, in metres, the total vertical distance itwould travel. 3 9 a 2 2 . 2 2 2 8 1 r 3 2 2 2 3 1 4 2 12 14 m[6]

Question 2(25 marks)32Solve the equation x 3x 9 x 11 0.Write any irrational solution in the form a b c, where a , b , c .f ( x ) x 3 3 x 2 9 x 11f (1) 13 3(1) 2 9 11 0 x 1 is a solution.(x ‒ 1) is a factorx2 2xx 13 112x 3 x 9 x 11x3 x2 2 x 2 9 x 11 2x 2 2x( x 1) ( x 2 Ax 11) x 3 3 x 2 9 x 11or 11x 11 11x 11 x 3 Ax 2 x x 2 Ax 1 x 3 3 x 2 9 x 11 A 1 3 A 2orx 1x2x3 x2 2x 2x 22x 11 11 x11Hence, other factor is x 2 2 x 11x 2 ( 2) 2 4(1)( 11) 2 48 2 4 3 1 2 32(1)22Solutions: {1, 1 2 3 , 1 23 }[7]

Question 3Let(a)(25 marks)f ( x ) x 2 12 x 27 , x ℝ.(i)Complete Table 1 below.Table 1(ii)x3456789f ( x)0589850Use Table 1 and the trapezoidal rule to find the approximate area of the region boundedby the graph of f and the x-axis.h y 1 yn 2 ( y2 y3 yn 1 ) 2 1 0 0 2 ( 5 8 9 8 5 ) 2 35 square unitsA (b)(i)Find9 f ( x ) dx .39 ( x32 12 x 27 ) dx9 x 3 12 x 2 27 x 2 3 3 ( 243 486 243 ) ( 9 54 81) 36(ii)Use your answers above to find the percentage error in your approximation of the area,correct to one decimal place.1 100 2 8 %36[8]

Question 4(a)(25 marks)The complex numbers z 1 , z 2 a n d z 3 are such that211 , ݖ ଶ ʹ ݅ andz1 z 2 z 3z3 3 2i, where i 2 1. Write ݖ ଵ in the form a b i , where a , b .2 1 111 z1 z 2 z3 2 3i 3 2i3 2i 2 3i5 i (2 3i )(3 2i) 12 5iz 12 5i 1 25 i12 5i 5 i 5 i 5 i65 13i 26 z1 5 ior11 2 3i 2 3i 2 3i 2 3i 2 3i 2 3i 4 91311 3 2i 3 2i 3 2i 3 2i 3 2i 3 2i 4 913112 3i 3 2i 5 i 2 3i 3 2i1313132 5 i 13z1Let z1 a bi25 i a bi 1326 (5 i)(a bi)26 (0)i 5a 5bi ai b26 (0)i (5a b) ( a 5b)i[9]

5a b 26 (i) and a 5b 0 (ii)5a b 26(i):(ii): 5a 25b 026b 26b 1From (ii): 5b a a 5z1 5 i(b)Let ω be a complex number such that ω 1, ω 1, and S 1 ω ω 2 ω n 1 . Usethe formula for the sum of a finite geometric series to write the value of S in its simplest form.nS 1 ω ω 2 ω n 1a 1, r ω1(1 ω n ) 1(1 1) 0S 1 ω1 ω[10]

Question 5(a)(25 marks)Solve the equation x x 6, x .x x 6 x2 x 6 x2 x 6 0 ( x 2)( x 3) 0 x 2, x 3x 2 : 2 2 6 4 2x 3: 3 3 6 9 3(b)Differentiate x x 6 with respect to x.f ( x) x x 6 x ( x 6) 21 1f ′( x) 1 12 (x 6) 2 1 2 x 61(c)Find the co-ordinates of the turning point of the function y x x 6, x 6.f ′(x) 0 1 12 x 6 0 2 x 6 1 x 6 14 x 5 34f ( 5 34 ) 5 34 14 6 14( 5 34 , 6 14)[11]

Question 6(a)(25 marks)Donagh is arranging a loan and is examining two different repayment options.(i) Bank A will charge him a monthly interest rate of 0·35%. Find, correct to threesignificant figures, the annual percentage rate (APR) that is equivalent to a monthlyinterest rate of 0·35%.F P (1 i ) 1(1 0 0035t)12 1 042818 i 4 28 %(ii)Bank B will charge him a rate that is equivalent to an APR of 4·5%. Find, correct tothree significant figures, the monthly interest rate that is equivalent to an APR of 4·5%.F P (1 i )t1 045 1(1 i )12121 i 1 045 1 0036748 i 0 367 %[12]

(b)Donagh borrowed 80 000 at a monthly interest rate of 0·35%, fixed for the term of the loan,from Bank A. The loan is to be repaid in equal monthly repayments over ten years. The firstrepayment is due one month after the loan is issued. Calculate, correct to the nearest euro, theamount of each monthly repayment. i (1 i )t A P t (1 i ) 1 0 0035 (1 0035 )120 80000 120 (1 0035 ) 1 0 00532296 80000 0 520846 817 59 818 or80000 AAA . 21 0035 1 00351 0035 120 1 11 A . 2120 1 0035 1 0035 1 0035 1 1 120 1 1 0035 1 0035 A 1 1 1 0035 0 342471198 A 0.0035 A [ 97 8489137 ] A 817 58 818[13]

Section BContexts and Applications150 marksAnswer all three questions from this section.Question 7(50 marks)A plane is flying horizontally at P at a height of 150 m above level ground when it begins its descent.P is 5 km, horizontally, from the point of touchdown O. The plane lands horizontally at O.P0·15 kmO5 kmTaking O as the origin, (x, f ( x ) ) approximately describes the path of the plane’s descent wheref (x) 0 0024x3 0 018x2 cx d,(a)(i) 5 x 0, and both x and f ( x ) are measured in km.Show that d 0.f ( x ) 0 0024 x 3 0 018 x 2 cx df (0) 0 0 0 d 0(ii) d 0Using the fact that P is the point ( 5, 0 15), or otherwise, show that c 0.f ( x ) 0 0024 x 3 0 018 x 2 cxf ( 5) 0 0024( 5)3 0 018( 5) 2 c( 5) 0 15 0 15 5 c 0 15 c 0orThe plane lands horizontally at O f ′( x ) 0when x 0f ′( x ) 0 0072 x 0 036 x cf ′(0) 0 0 c 0 c 02(b)(i)Find the value of f ′ ( x ), the derivative of f ( x ), when x 4.f ( x) 0 0024x 3 0 018x 2 cx df ′( x) 0 0072x 2 0 036xf ′( 4 ) 0 0072 ( 4 ) 2 0 036 ( 4 ) 0 0288[14]

(ii)Use your answer to part (b) (i) above to find the angle at which the plane is descendingwhen it is 4 km from touchdown. Give your answer correct to the nearest degree.tan θ f ′( x ) 0 0288 θ 178 3503 Angle of descent α 1 6497 2 (c)Show that ( 2 5, 0 075) is the point of inflection of the curve y f ( x ).f ′( x ) 0 0072 x 2 0 036 xf ′′ ( x ) 0 0144 x 0 036 0 x 2 5f ( x) 0 0024x 3 0 018x 2f ( 2 5 ) 0 0024 ( 2 5 ) 3 0 018 ( 2 5 ) 2 0 0375 0 1125 0 075( 2 5, 0 075 )(d)(i)If (x, y) is a point on the curve y f ( x ), verify that ( x 5, y 0 15) is also apoint on y f ( x ).f ( x) 0 0024x 3 0 018x 2f ( x 5) 0 0024 ( x 5)3 0 018( x 5) 2 0 0024 ( x3 15 x 2 75 x 125 ) 0 018 ( x 2 10 x 25 ) 0 0024 x3 0 018 x 2 0 x 0 15 y 0 15(ii)Find the image of ( x 5, y 0 15) under symmetry in the point of inflection.Point: ( x 5, y 0 15)Point of inflection: ( 2 5, 0 075)Change in x value: ( 2 5) ( x 5) x 2 5Change in y value: 0 075 ( y 0 15) y 0 075Image of point of inflection:x value: 2 5 ( x 2 5) xy value: 0 075 ( y 0 075) y (x, y) is image[15]

orLet ( x , y ) be the image. x 5 x y 0 15 y , ( 2 5, 0 075) , the point of inflection22 [16]

Question 8(50 marks)An oil-spill occurs off-shore in an area of calm water with no currents. The oil is spilling at a rateof 4 106 cm3 per minute. The oil floats on top of the water.(a)(i)Complete the table below to show the total volume of oil on the water after each of thefirst 6 minutes of the oil-spill.Time (minutes)123456Volume ( 10 6 cm 3 )4812162024(ii)Draw a graph to show the total volume of oil on the water over the first 6 minutes.24Volume (106 cm3)20161284123456Time (minutes)(iii) Write an equation for V(t), the volume of oil on the water, in cm3, after t minutes.Line, slope 4 106, passing through (0, 0).V(t) ( 4 10 6 ) t(b)The spilled oil forms a circular oil slick 1 millimetre thick.(i) Write an equation for the volume of oil in the slick, in cm3, when the radius is r cm.V π r 2h π r 2 (0 1) 0 1π r 2 cm 3[17]

(ii)Find the rate, in cm per minute, at which the radius of the oil slick is increasing whenthe radius is 50 m.dV 4 10 6 cm3 per minutedtV π r 2 h whereh 0 1 cmdV 2πrhdrdV 0 2π rdrdr dr dV1 4 106dt dV dt 0 2π r (c)4 10 6 1273 3 cm per minute0 2π (5000)Show that the area of water covered by the oil slick is increasing at a constant rateof 4 10 7 cm2 per minute.A π r2 dA 2π rdrdA dA dr4 106 2π r 4 107 cm 2 per minutedt dr dt0 2π ror(0 1)π r 2 (4 106 )t A π r 2 (4 107 )tdA 4 107dt(d)The nearest land is 1 km from the point at which the oil-spill began. Find how long it willtake for the oil slick to reach land. Give your answer correct to the nearest hour.A π r 2 π (105 )2 π 1010 cm2t π10104 107 π1034 785 398 minutes 13 09 13 hours[18]

Question 9(50 marks)The approximate length of the day in Galway, measured in hours from sunrise to sunset, may becalculated using the function 2π f (t ) 12 25 4 75sin t , 365 2π t is expressed in radians.where t is the number of days after March 21st and 365 (a)Find the length of the day in Galway on June 5th (76 days after March 21st). Give youranswer in hours and minutes, correct to the nearest minute. 2π f (t ) 12 25 4 75 sin t 365 2π f (76) 12 25 4 75sin 76 365 12 25 4 587 16 837 16 hours 50 minutes(b)Find a date on which the length of the day in Galway is approximately 15 hours. 2π f (t ) 12 25 4 75 sin t 15 365 2π t 0 578947 sin 365 2πt 0 6174371 365 t 35 8736 days after March 21 is April 26.(c)Find f ′(t ), the derivative of f ( t ). 2π f (t ) 12 25 4 75 sin t 365 2π 2π f ′(t ) 0 4 75 t cos 365 365 9 5π 2π cos t365 365 [19]

(d)Hence, or otherwise, find the length of the longest day in Galway. 2π t is a maximum of 1. 365 f (t ) is a maximum when sin t 12 25 4 75 17 hoursor9 5π 2π cos t 0365 365 2π cos t 0 365 π2π t 3652365 t 91 254 2πf ′(t ) 0 91 25 f (91 25) 12 25 4 75sin 365 12 25 4 75sin 17 hours(e)π2Use integration to find the average length of the day in Galway over the six months fromMarch 21st to September 21st (184 days). Give your answer in hours and minutes, correct tothe nearest minute.b11f ( x ) dx b a a184184 2π 12 25 4 75 sin 365 t dt01841 365 2π t 12 25t 4 75 cos 184 2π 365 01 ( 2254 275 843 ) ( 0 275 934 ) 184 1 [ 2805 777 ]184 15 24879 15 hours 15 minutes [20]

Marking Scheme – Paper 1, Section A and Section BStructure of the marking schemeCandidate responses are marked according to different scales, depending on the types of responseanticipated. Scales labelled A divide candidate responses into two categories (correct andincorrect). Scales labelled B divide responses into three categories (correct, partially correct, andincorrect), and so on. The scales and the marks that they generate are summarised in this table:Scale labelABCDENo of categories234560, 50, 100, 150, 200, 250, 2, 50, 5, 100, 7, 150, 10, 200, 12, 250, 2, 4, 50, 4, 8, 100, 5, 10, 150, 7, 13, 200, 8, 17, 250, 2, 5, 8, 100, 4, 7, 11, 150, 5, 10, 15, 200, 6, 12, 19, 250, 5, 10, 15, 20, 255 mark scales10 mark scales15 mark scales20 mark scales25 mark scalesA general descriptor of each point on each scale is given below. More specific directions in relationto interpreting the scales in the context of each question are given in the scheme, where necessary.Marking scales – level descriptorsA-scales (two categories) incorrect responsecorrect response response of no substantial meritpartially correct responsecorrect response response of no substantial meritresponse with some meritalmost correct responsecorrect response response of no substantial meritresponse with some meritresponse about half-rightalmost correct responsecorrect response response of no substantial meritresponse with some meritresponse almost half-rightresponse more than half-rightalmost correct responsecorrect responseB-scales (three categories)C-scales (four categories)D-scales (five categories)E-scales (six categories)In certain cases, typically involving incorrect rounding, omission of units, a misreading that doesnot oversimplify the work or an arithmetical error that does not oversimplify the work, a mark thatis one mark below the full-credit mark may also be awarded. Thus, for example, in scale 10C, 9marks may be awarded.Throughout the scheme indicate by use of * where an arithmetic error occurs.[21]

Summary of mark allocations and scales to be appliedSection ASection BQuestion 1(a)(b)(c)Question 5B10D5C10CQuestion 0D10C10CQuestion 9(a)(b)(c)(d)(e)10C10C10B10D10Question 25C10C10C25EQuestion 3(a)(b)15D10CQuestion 4(a)(b)15D10CQuestion 5(a)(b)(c)10C5B10CQuestion 6(a)(i) (ii) 10C(b)(15C[22]

Detailed marking notesNOTE: In certain cases, typically involving incorrect rounding, omission of units, a misreadingthat does not oversimplify the work or an arithmetical error that does not oversimplify the work, amark that is one mark below the full-credit mark may also be awarded.Rounding and units penalty to be applied only once in each section (a), (b), (c) etc.Throughout the scheme indicate by use of * where an arithmetic error occurs.Section AQuestion 1(a)Note:(b)Scale 5C (0, 2, 4, 5)Low Partial Credit: Any term correctHigh Partial Credit: Any two terms correctଷDividing by gets high partial credit at most.ସCorrect decimal values high partial at most.Scale 10C (0, 4, 8, 10) – NOTE: two solutions1st solutionLow Partial Credit: Indicates addition of termsHigh Partial Credit: Recognises double distance after first hop Sum of all rises or dropsor2nd solutionLow Partial Credit: Indicates addition of terms Indicates Geometric ProgressionHigh Partial Credit: Correct Geometric Progression formula with correct substitution(c)Scale 10C (0, 4, 8, 10)Low Partial Credit: Recognition of sum to infinity ܵஶ formulaHigh Partial Credit Correct formula with correct substitution Sum of all rises or drops[23]

Question 2(a)Scale 25E (0, 5, 10, 15, 20, 25)Low Partial Credit: Effort at finding root, i.e. f (1) , f ( 1) , etc.Low Mid Partial Credit: Finds one root correctly after division by incorrect factor Correct answers in decimal form from calculator with or without workHigh Mid Partial Credit: Tries division and gets x 2 at very minimumNote:High Partial Credit: Having got a quadratic equation with no remainder, fills in quadratic formula 1 12If there is a remainder after division can only get maximum of 15 marks.[24]

Question 3(a)(i) and (ii) combinedScale 15D (0, 4, 7, 11, 15)Low Partial Credit: Any one correct value Writes formulaMid Partial Credit: Correct tableHigh Partial Credit: Correct formula for trapezoidal rule, and some correct substitution with ℎ 1 Completely incorrect table but applied correctly in a(ii) Correct table and 35 without workNote (1): Answers in terms of ℎ merit Mid Partial at most.Note (2): Correct formula and some substitution gets High Partial.1Note (3): No formula and [5 5 2 (8 9 8 )] 30 gets High Partial.2(b)(i) and (ii) combinedScale 10C (0, 4, 8, 10)Low Partial Credit: Any correct integration Correct substitution of f ( x ) Correct % error formula2 Correct substitution of f ( x ) i.e. ( x 12 x 27 )High Partial Credit: Correct integration with some correct substitution 97 2%Full Credit: Accept 2 8% without work for full credit.[25]

Question 4(a)Scale 15D (0, 4, 7, 11, 15) – NOTE: two solutionsLow Partial Credit: Some rationalisation Some relevant rearrangement.Mid Partial Credit:a bi2in the form of Gets z1 orc diz1High Partial Credit: Correct use of conjugate in12 5i5 iorLow Partial Credit: One complex number correctMid Partial Credit: Two complex numbers correctHigh Partial Credit: Correct use of conjugate in(b)25 i a bi13Scale 10C (0, 4, 8, 10)Low Partial Credit: Correct Geometric Progression formula Correct first term Correct ratioHigh Partial Credit: Values substituted in formula[26]

Question 5(a)Scale 10C (0, 4, 8, 10)Low Partial Credit: Indication of squaringHigh Partial Credit: Correct rootsNote: must indicate required root(b)Scale 5B (0, 2, 5)Partial Credit: Any correct differentiation (c) Indication of ( ) Scale 10C (0, 4, 8, 10)Low Partial Credit: Differentiation equals 0High Partial Credit: Finds x value.Note (1): A linear equation from ( ) gets low partial at most.Note (2): Must put ( ) 0 in (c) to get any marks.Note (3): ( ) only and ( ) only .no credit[27]

Question 6(a)(i) and (ii) combinedScale 10C (0, 4, 8, 10)Low Partial Credit: Correct formula in either part Correct substitution in incorrect formulaHigh Partial Credit: Any one section correctNote: Rate as 0 367% or 0 00367 gets High Partial.(b)Scale 15C (0, 5, 10, 15) – NOTE: two solutions1st solutionLow Partial Credit: Any correct step, i.e. correct formulaHigh Partial Credit: Substitution in correct formula.or2nd solutionLow Partial Credit: Correct equation. Listing some terms Some substitutionHigh Partial Credit: Complete substitution and effort at evaluation.Note: If A and 80 000 interchanged and remainder of work correct, may get High Partial credit.[28]

Section BQuestion 7(a)(i)Scale 5B (0, 2, 5)Partial Credit: Recognises x 0(a)(ii)Scale 5B (0, 2, 5) – NOTE: two solutions1st solutionPartial Credit: Uses x 5 or f ( x ) 0 15Full credit: Begins with 0 and shows ( 5) 0 · 15 or similaror2nd solutionPartial Credit: Uses x 5 Gets f ' ( x ) Uses f ' ( x ) 0 when x 0(b)(i)Scale 10C (0, 3, 7, 10)Low Partial Credit: Any term correctly differentiated.High Partial Credit: Correct differentiationFull credit: is a correct answer (b)(ii)(c)Scale 5B (0, 2, 5)Partial Credit: Recognition of connection between slope and tan θ Any right angled triangleScale 10D (0, 2, 5, 8, 10)Low Partial Credit: Some correct differentiation of f ' ( x ) Mention of f ' ( x )Mid Partial Credit: Correct f ''( x ) 0High Partial Credit: Value of x substituted[29]

(d)(i)Scale 5C (0, 2, 4, 5)Low Partial Credit: Some correct substitutionHigh Partial Credit: Correct expansions(d)(ii)Scale 10C (0, 4, 8, 10) – NOTE: two solutions1st solutionLow Partial Credit: Work leading to change in x -value or y -valueHigh Partial Credit: Correct change in x and y valuesor2nd solutionLow Partial Credit: Uses ( x, y ) as image, and no moreHigh Partial Credit: Effort at calculating mid-pointQuestion 8(a)(i)Scale 5B (0, 2, 5)Partial Credit: One correct box(a)(ii)Scale 5B (0, 2, 5)Partial Credit: At least two points plottedNo credit Bar chart(a)(iii) Scale 5B (0, 2, 5)Partial Credit: Incomplete equation for volume ܸ ൌ any function of t Attempt at finding slope(b)(i)Scale 5B (0, 2, 5)Partial Credit: Correct volume formula Converting mm to cm[30]

(b)(ii)Scale 10D (0, 2, 5, 8, 10)Low Partial Credit: Mentions a relevant rate of change.Mid Partial Credit:dVdrdVfromand Getsdtdrdt Writing down chain rule.High Partial Credit: Substitution of values(c)Scale 10C (0, 4, 8, 10) – NOTE: two solutions1st solutionLow Partial Credit: Mentions relevant rate of change.High Partial Credit: States chain rule i.e. or 2nd solutionLow Partial Credit: Effort to establish value of ANote:(d)High Partial Credit: A in terms of tMust use calculus to get any credit.Scale 10C (0, 4, 8, 10)Low Partial Credit: r in centimetres Effort at expression of areaHigh Partial Credit: Correct expression for time[31]

Question 9(a)Note:(b)Note:(c)Note:(d)Scale 10C (0, 4, 8, 10)Low Partial Credit: Uses t 76High Partial Credit: Correct substitutionUsing ߨ ൌ ͻͲι one error, but do not penalise again in (b)Scale 10C (0, 4, 8, 10)Low Partial Credit: Correct f (t ) ݂ሺͳͷሻ substituted.High Partial Credit: Correct equation with t onlyAccept 35 or 36 substituted correctly and tested.Scale 10B (0, 5, 10)Partial Credit: Any correct differentiation (note: ‘0’ could be correct differentiation here)Substituting 180 for ߨ one errorScale 10D (0, 2, 5, 8, 10) – both solutionsLow Partial Credit: f ' (t ) 0Mid Partial Credit: Value of tHigh Partial Credit: Value of t substituted into f (t ) f (t ) maximum when sin θ 1Note: Accept 91 or 92 substituted and evaluated correctly for full marks.(e)Scale 10D (0, 2, 5, 8, 10)Low Partial Credit: Correct expression in x or t Correct formula Correct limitsMid Partial Credit: Any correct integrationHigh Partial Credit: Correct integration and effort at substitutionNote: Integration with one error but finished correctly gets High Partial Credit.[32]

2015. M30Coimisiún na Scrúduithe StáitState Examinations CommissionLeaving Certi cate Examination 2015MathematicsPaper 2Higher LevelMonday 8 JuneMorning 9:30 – 12:00300 marksModel Solutions – Paper 2Note: The model solutions for each question are not intended to be exhaustive – there may be othercorrect solutions. Any examiner unsure of the validity of the approach adopted by a particularcandidate to a particular question should contact his / her advising examiner.[33]

InstructionsThere are two sections in this examination paper.Section AConcepts and Skills150 marks6 questionsSection BContexts and Applications150 marks3 questionsAnswer all nine questions.Write your answers in the spaces provided in this booklet. You may lose marks if you do not do so.There is space for extra work at the back of the booklet. You may also ask the superintendent formore paper. Label any extra work clearly with the question number and part.The superintendent will give you a copy of the Formulae and Tables booklet. You must return it atthe end of the examination. You are not allowed to bring your own copy into the examination.You will lose marks if all necessary work is not clearly shown.You may lose marks if the appropriate units of measurement are not included, where relevant.You may lose marks if your answers are not given in simplest form, where relevant.Write the make and model of your calculator(s) here:[34]

Section AConcepts and Skills150 marksAnswer all six questions from this section.Question 1(25 marks)An experiment consists of throwing two fair, standard, six-sided dice and noting the sum of the twonumbers thrown. If the sum is 9 or greater it is recorded as a “win” (W). If the sum is 8 or less it isrecorded as a “loss” (L).(a)Complete the table below to show all possible outcomes of the experiment.Die 1Die LWWWWFind the probability of a win on one throw of the two dice.P (W ) (ii)10 5 36 18Find the probability that each of 3 successive throws of the two dice results in a loss.Give your answer correct to four decimal places.3 13 P(L, L, L ) 0 3767 18 (c)The experiment is repeated until a total of 3 wins occur. Find the probability that the third winoccurs on the tenth throw of the two dice. Give your answer correct to four decimal places.2 9 5 13 P ( 2 wins in 9 ) 2 18 18 727 9 5 13 5 P ( 3 wins, 3rd on 10th throw ) 0 0791 2 18 18 18 [35]

Question 2(25 marks)A survey of 100 shoppers, randomly selected from a large number of Saturday supermarketshoppers, showed that the mean shopping spend was 90·45. The standard deviation of this samplewas 20·73.(a)Find a 95% confidence interval for the mean amount spent in a supermarket on that Saturday.σn 20 73 2 073100C. I. x 1 96σ 90 45 4 06nWe can be 95% confident that the mean amount spent was in the range 86·39 μ 94·51(b)A supermarket has claimed that the mean amount spent by shoppers on a Saturday is 94.Based on the survey, test the supermarket’s claim using a 5% level of significance. Clearlystate your null hypothesis, your

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

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