Paper Reference(s) 6663/01 Edexcel GCE
Paper Reference(s)6663/01Edexcel GCECore Mathematics C1Gold Level G4Time: 1 hour 30 minutesMaterials required for examinationpapersMathematical Formulae (Green)Items included with questionNilCandidates may use any calculator allowed by the regulations of the JointCouncil for Qualifications. Calculators must not have the facility for symbolicalgebra manipulation, differentiation and integration, or have retrievablemathematical formulas stored in them.Instructions to CandidatesWrite the name of the examining body (Edexcel), your centre number, candidate number,the unit title (Core Mathematics C1), the paper reference (6663), your surname, initialsand signature.Information for CandidatesA booklet ‘Mathematical Formulae and Statistical Tables’ is provided.Full marks may be obtained for answers to ALL questions.There are 11 questions in this question paper. The total mark for this paper is 75.Advice to CandidatesYou must ensure that your answers to parts of questions are clearly labelled.You must show sufficient working to make your methods clear to the Examiner. Answerswithout working may gain no credit.Suggested grade boundaries for this paper:Gold 4A*ABCDE605243342516This publication may only be reproduced in accordance with Edexcel Limited copyright policy. 2007–2013 Edexcel Limited.
1. 1(a) Find the value of 16 4 .(2)4 14 (b) Simplify x 2 x . (2)January 20112.Given that 32 2 2a, find the value of a.(3)June 20093.Show that2can be written in the form a b, where a and b are integers. 12 8(5)May 20121 dy 5 x 2 x x,dx4.x 0.Given that y 35 at x 4, find y in terms of x, giving each term in its simplest form.(7)January 2010Gold 4: 12/122
y5.xOFigure 1Figure 1 shows a sketch of the curve with equation y 3, x 0.x3, x 2, showing thex 2coordinates of any point at which the curve crosses a coordinate axis.(3)(a) On a separate diagram, sketch the curve with equation y (b) Write down the equations of the asymptotes of the curve in part (a).(2)May 20076.The equation x2 3px p 0, where p is a non-zero constant, has equal roots.Find the value of p.(4)June 2009Gold 4: 12/123
f(x) x2 (k 3)x k,7.where k is a real constant.(a) Find the discriminant of f(x) in terms of k.(2)(b) Show that the discriminant of f(x) can be expressed in the form (k a)2 b, where a andb are integers to be found.(2)(c) Show that, for all values of k, the equation f(x) 0 has real roots.(2)May 20114x – 5 – x2 q – (x p)2,8.where p and q are integers.(a) Find the value of p and the value of q.(3)(b) Calculate the discriminant of 4x – 5 – x2.(2)(c) Sketch the curve with equation y 4x – 5 – x2, showing clearly the coordinates of anypoints where the curve crosses the coordinate axes.(3)May 20129.The first term of an arithmetic series is a and the common difference is d.The 18th term of the series is 25 and the 21st term of the series is 32 12 .(a) Use this information to write down two equations for a and d.(2)(b) Show that a –17.5 and find the value of d.(2)The sum of the first n terms of the series is 2750.(c) Show that n is given byn2 – 15n 55 40.(4)(d) Hence find the value of n.(3)January 2009Gold 4: 12/124
10.Figure 2Figure 2 shows a sketch of the curve C with equationy 2 1,xx 0.The curve crosses the x-axis at the point A.(a) Find the coordinates of A.(1)(b) Show that the equation of the normal to C at A can be written as2x 8y 1 0.(6)The normal to C at A meets C again at the point B, as shown in Figure 2.(c) Find the coordinates of B.(4)January 2012Gold 4: 12/125
11.The curve C has equationy x3 – 2x2 – x 9,x 0.The point P has coordinates (2, 7).(a) Show that P lies on C.(1)(b) Find the equation of the tangent to C at P, giving your answer in the form y mx c,where m and c are constants.(5)The point Q also lies on C.Given that the tangent to C at Q is perpendicular to the tangent to C at P,(c) show that the x-coordinate of Q is1(2 6).3(5)June 2009TOTAL FOR PAPER: 75 MARKSENDGold 4: 12/126
QuestionNumberScheme1. (a) 16 4 2 or111Marksor betterM116 4 16 12 or 0.5 14A1(2)(b) 2x 144 24 x 44or24or equivalent4M1x4 x 2x 14 4 24 or 16A1 cao(2)[4]2.32 2 5 or 2048 211 ,a 3.1122 12 12or12048 2048 2B1, B11 or 5 or 5.5 2 8 Gold 4: 12/122 2B1 12 8 2 12 8 2 12 12 M1B1 B112 83 8A112 82 2 3 2 2 8 [3]A1 cso[5]27
QuestionNumber4.Schemex x x 123x2B1 kx y 5x1112or x32 kx5M125x 2( C). 52225 435 12C Marks1A1 A1524 2 C52115M11or equivalent 2 , 2.25A15y 10 x122 x 2 11 55(or equivalent simplified)A1ft[7]y5. (a)x(b)6.Translation parallel to x-axisM1Top branch intersects ve y-axisLower branch has no intersectionsNo obvious overlapA13 3 marked on y- axis 0, or2 2 B1b 2 4ac attempted, in terms of p.(3)B1 B1(2)[5]M1(3 p) 2 4 p 0A1x 2, y 0or equivalentAttempt to solve for p e.g. p 9 p 4 0Must potentially lead to p k, k 0p 49(Ignore p 0, if seen)M1A1 cso[4]Gold 4: 12/128
QuestionNumberSchemeMarks7. (a) Discriminant: b 2 4ac (k 3) 2 4k or equivalent(b)M1 A1(2)M1 A1(2)M1(k 3) 2 4k k 2 2k 9 (k 1) 2 8(c) For real roots, b 2 4ac 0 or b 2 4ac 0 or (k 1)2 8 0(k 1) 2 0 for all k, so b 2 4ac 0 , so roots are real for all k8. (a)(oe)4 x 5 x 2 q ( x p) 2 , p, q are integers. 4 x 5 x 2 x 2 4 x 5 ( x 2) 2 4 5 ( x 2) 2 1 1 ( x 2) 2(b) "b2 4ac " 42 4( 1)( 5)yOxM1Correct shapeA1 A1(3)M1A1(2)M1Maximum within the 4th quadrantA1Curve cuts through -5 or(0, 5) marked on the y-axisB1 16 20 4(c)A1 cso(2)[6]-5(3)[8]Gold 4: 12/129
QuestionNumber9. (a)Schemea 17d 25 o.e.,(b)a 20d 32.5 o.e.3d 7.52750 B1 B1(2)M1so d 2.5a 32.5 20 2.5(c)Marksso a -17.5(*)A1 cso(2)n 35 52 n 1 2M1 A1ft{ 4 2750 n(5n 75) }4 550 n n 15 M1n2 15n 55 40A1 cso(4)M1(*)n 2 15n 55 40 0 or n 2 15n 2200 0(d) n 55 n 40 0n 55n M1(ignore - 40)A1(3)[11]10. (a)(b) 1 , 0 2 dy x 2dxB1M1 A1 2dy 1 4dx 2 1Gradient of normal m1At x ,2( m)A11 4 M11 1 Equation of normal: y 0 x 4 2 M12x 8y – 1 0(c)2 111 x x48(*)A1cso(6)M1 2 x 2 15x 8 0 ](2 x 1)( x 8) 0(1)or[ 8 y 2 17 y 0 ]leading to x 1 x or 8 2 17y (or exact equivalent)8M1A1A1ft (4)[11]Gold 4: 12/1210
QuestionNumber11. (a)SchemeMarksy 8 8 2 9 7 (*)x 2:B1(1)(b)dy 3x 2 4 x 1dxdyx 2: 12 8 1( 3)dxy 7 3( x 2) ,M1 A1A1fty 3x 1M1, A1(5)(c)1with their m)m1423x 2 4 x 1 , 9 x 2 12 x 2 0 or x 2 x 0 (o.e.)339m 13(for 12 144 72 x 18 3x 2 2x B1ftM1, A1216 36 6 6 6 orM1 6 3x 2 612 63 (*)A1cso(5)[11]Gold 4: 12/1211
Examiner reportsQuestion 1Overall this question was done poorly, with very few candidates scoring full marks. Therewere, however, many correct answers to part (a). Candidates were usually able to deal witheither the negative part of the power or the fractional part, but some had problems in dealingwith both. There were also some who did not understand the significance of the negative signin front of the 14 and the fact that it implied a reciprocal. Some assumed it implied a negativefinal answer, e.g. –2.Part (b) was successfully completed by very few candidates, with the vast majority of errorsbeing caused by the failure to raise 2 to the power 4. The power of the x inside the bracketwas also often incorrectly calculated. It was common for candidates to multiply by x beforeraising to the power and so ending up with either 2x3 or 16x3. Others added the powers – 14and 4. Even when the bracket was correctly expanded, the extra x was often omitted or notcombined with the other term.Question 2Many candidates could not deal with this test of indices. Two simple properties of indiceswere required: that a square root leads to a power of 12 and the rule for adding the powerswhen multiplying. Those who identified these usually made good progress but the remainderstruggled. Some re-wrote 32 2 as 64 and then obtained a 3 whilst others did obtain2048 but usually failed to identify 2048 as 211 . A number of candidates tried to use logs butthis approach was rarely successful.Question 3This question proved discriminating with just under a half of the candidature gaining all 5marks. About a third of the candidature gained only two marks by either correctlyrationalising the denominator or by correctly simplifying 12 and 8.A significant number of candidates answered this question by multiplying 12 12 212 8by to obtain 1 12 8 and proceeded no further. A small number of candidates28 8multiplied the denominator incorrectly to give either 12 20 20 8 or12 72 72 8.Those candidates who realised that 12 2 3 andanswer ofas8 2 2 usually obtained the correct13 2 , although a few proceeded to the correct answer by writing 12 821 12 41 8.4A number of candidates started this question by firstly simplifying 12 and 8 to give2, but many did not take the easy route of cancelling this to2 3 2 2Gold 4: 12/12121before3 2
rationalising. Although some candidates wrote1as3 23 2, those candidates whorationalised this usually obtained the correct answer.A small minority of candidates attempted to rationalise the denominator incorrectly bymultiplying212 8by either 12 12 or 8 8 .8 12 812 Question 4In this question the main problem for candidates was the integration of x x , for which a33x 2 2x 2common result was. Those who replaced x x by x 2 generally made good 23progress, although the fractional indices tended to cause problems. Some differentiatedinstead of integrating. Most candidates used the given point (4, 35) in an attempt to find thevalue of the integration constant, but mistakes in calculation were very common. A significantminority of candidates failed to include the integration constant or failed to use the value of yin their working, and for those the last three marks in the question were unavailable.Question 5There were very mixed responses to this question; some answered it very well with neatsketches and clearly identified asymptotes but others appeared to have little idea. In part (a)most candidates knew that a translation was required and the majority knew it was horizontaland to the left. A few moved their graph up and a number of candidates ended up with graphscutting both axes. Many identified the intersection with the y-axis as (0, 1.5) but some ignoredthis part of the question. It was clear in part (b) that a number of candidates still do not knowthe meaning of the term asymptote but many did give x –2 and slightly fewer gave y 0.However a number of candidates gave answers of y –2 and x 0, often despite having a3correct dotted line on their sketch, and some gave other non-linear equations such as y .xSome candidates gave no answer to part (b) despite having the lines x 2 and y 0 clearlyidentified on their diagram.Question 6Some candidates opted out of this question but most realised that they needed to use thediscriminant and made some progress. The most common error was to simplify 3 p 2 to 3 p 2but there were many correct equations seen and usually these led to a correct answer. Manycandidates chose to solve their equation by cancelling a p. In this case that was fine, sincethey were told that p was non-zero, but this is not good practice in general and factorising9 p 2 4 p to p 9 p 4 is recommended.Gold 4: 12/1213
Question 7Candidates were required to give (k 3)2 4k as their answer to part (a). Any x termsincluded resulted in zero marks. Some candidates tried to solve (k 3)2 4k 0 and this wasalso not given any credit in this part.Most candidates managed to complete the square correctly in part (b) and those starting withk 2 2k 9 usually arrived at the correct answer for this part. However, several left theirsolution as (k 3)2 4k , thus gaining no credit.Part (c) was poorly done, with a substantial minority of the candidates not understanding whatthe question required. Quite a few realised that the determinant had to be greater than zero,but didn’t know how to show this. M1 A0 was a common mark for those who tried a numberof values for k. Candidates were expected to use their completion of the square and to arguethat (k 1) 2 0 for all values of k.A large minority were intent on trying to solve k 2 2k 9 0, and concluded that there wereno real roots. They demonstrated some confusion between the values of k and the informationprovided by the discriminant. Good candidates scored full marks on this question.Question 8This question was poorly answered with about 10% of the candidature gaining all 8 marks.In part (a), the x 2 term and the order of the terms in 4 x 5 x 2 created problems for themajority of candidates. The most popular (and most successful) method was to complete thesquare. A large number of candidates struggled to deal with the x 2 term with bracketingerrors leading to incorrect answers such as ( x 2)2 9 or ( x 2)2 9. A successfulstrategy for some candidates involved negating the quadratic to get x 2 4 x 5 which wasusually manipulated correctly to ( x 2) 2 1. Whilst a good number negated this correctly to ( x 2)2 1, some candidates wrote down incorrect results such as ( x 2)2 1,( x 2)2 1 or ( x 2) 2 1.Whilst a number of candidates stopped after multiplying out q ( x p) 2 , those whoattempted to use the method of equating coefficients were less successful.In part (b), most candidates wrote down b 2 4ac for the discriminant and the majorityachieved the correct answer of 4, although some incorrectly evaluated 42 4( 1)( 5) as36. The most common error was for candidates to substitute the incorrect values ofa 4 , b 5 and c 1 in b 2 4ac. Those candidates who applied the quadratic formulagained no credit unless they could identify the discriminant part of the formula.In part (c), a majority of candidates were able to draw the correct shape of the graph and anumber correctly identified the y-intercept as 5. Only a small minority were able tocorrectly position the maximum turning point in the fourth quadrant and some labelled it as(2 , 1). A number of correct sketches followed from either candidates using differentiationor from candidates plotting points from a table of values. A number of candidates aftercorrectly identifying the discriminant as 4, (including some who stated “that this meant noroots”) could not relate this information to their sketch with some drawing graphs crossing thex-axis at either one or two points. Common errors included drawing graphs of cubics orpositive quadratics, drawing negative quadratics with a maximum at (5, 0) or with amaximum at (0 , 5).Gold 4: 12/1214
Question 9Although most candidates made a reasonable attempt at this question, only those whodemonstrated good skills in algebra managed to score full marks.The structure of parts (a) and (b) was intended to help candidates, but when the initial strategywas to write down (correctly) 3d 32.5 25 , there was sometimes confusion over what wasrequired for the two equations in part (a). Even when correct formulae such as u18 a 17dwere written down, the substitution of u18 25 did not always follow. The work seen in thesefirst two parts was often poorly presented and confused, but credit was given for any validmethod of obtaining the values of d and a without assuming the value of a.In part (c), many candidates managed to set up the correct sum equation but weresubsequently let down by poor arithmetic or algebra, so were unable to proceed to the givenquadratic equation. Being given 55 40 (to help with the factorisation in the last part of thequestion) rather than 2200 sometimes seemed to be a distraction.Despite being given the 55 40 , many candidates insisted on using the quadratic formula inpart (d). This led to the problem of having to find the square root of 9025 without a calculator,at which point most attempts were abandoned.Question 10This was a substantial question to end the paper and a number of candidates made littleattempt beyond part (a). Part (c) proved quite challenging but there were some clear andsuccinct solutions seen.Some stumbled at the first stage obtaining x 2 or even 1 instead of12 0 but most scored the mark for part (a).x12to the solution ofThe key to part (b) was to differentiate to find the gradient of the curve and most attempts didtry this but a number had x 2 . Some however tried to establish the result withoutdifferentiation and this invariably involved inappropriate use of the printed answer. Those 2who did differentiate correctly sometimes struggled to evaluate 12 correctly. A correct“show that” then required clear use of the perpendicular gradient rule and the use of theiranswer to part (a) to form the equation of the normal. There were a good number of fullycorrect solutions to this part but plenty of cases where multiple slips were made to arrive atthe correct equation.Most candidates set up a correct equation at the start of part (c) but simplifying this to acorrect quadratic equation proved too challenging for many. Those who did arrive at2x2 15x – 8 0 or 8y2 –17y 0 were usually able to proceed to find the correct coordinates1of B, but there were sometimes slips here in evaluating 2 for example. 8There were a few candidates who used novel alternative approaches to part (c) such assubstituting xy for 2x – 1 from the equation of the curve into the equation of the normal toobtain the simple equation 8y xy 0 from whence the two intersections y 0 and x –8were obtained.Gold 4: 12/1215
Question 11A number of partial attempts at this question may suggest that some were short of timealthough the final part was quite challenging.Most secured the mark in part (a) although careless evaluation of 2 2 2 as 6 spoiled it forsome. Apart from the few who did not realise the need to differentiate to find the gradient ofthe curve, and hence the tangent, part (b) was answered well. Some candidates though thoughtthat the coefficient of x 2 (the leading term) in their derivative gave them the gradient. Therewas the usual confusion here between tangents and normals with some candidates thinkingthatdygave the gradient of the normal not the tangent. In part (c) many knew they needed todxuse the perpendicular gradient rule but many were not sure what to do. A common error wasto find the equation of a straight line (often the normal at P) and then attempt to find theintersection with the curve. Those who did embark on a correct approach usually solved theirquadratic equation successfully using the formula, completing the square often led todifficulties with the x 2 term, but a few provided a correct verification.Gold 4: 12/1216
Statistics for C1 Practice Paper Gold Level G4Mean score for students achieving grade:Qu1234567891011Maxscore4357546811111175Gold 4: 681.662.481.091.1513.60
2007–2013 Edexcel Limited. Paper Reference(s) 6663/01 Edexcel GCE Core Mathematics C1 Gold Level G4 Time: 1 hour 30 minutes Materials required for examination Items included with question papers Mathematical Formulae (Green) Nil Candidates may use any calculator allowed by the regulations of the Joint Council for Qualifications. Calculators must not have the facility for symbolic algebra .
Jun 07, 2019 · Edexcel GCE Mathematics PMT. 6663 Core Mathematics C1 June 2006 Advanced Subsidiary/Advanced Level in GCE Mathematics 2 June 2006 6663 Core Mathematics C1 Mark Scheme Question Scheme Marks number 1. 1 3 2 1 2 6 2 3 x x x ( c
Jun 18, 2005 · 6663 Core Mathematics C1 June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics Question Number Scheme Marks Shape Points -2 and 4 max B1 B1 (2) M1 A1 A1 (3) (5) 4. (a) (b) (a) (b) Marks for shape: graphs must have curved sides and round top. 1st
6663 Core Mathematics C1 June 2006 Advanced Subsidiary/Advanced Level in GCE Mathematics 6 Question Scheme Marks number 5. (a) (yx x y x x 433 4 6) 1122′ or x
6663-PAK, Islamabad, Pakistan) and RDF (Research and Development Foundation, an NGO, Hyderabad, Pakistan). The trainings were conducted in the study district of Gujranwala district and conducted crop modelling and integration. This was followed by training local government officials on the use of crop models, mainly DSSAT and geospatial .
CAPE Management of Business Specimen Papers: Unit 1 Paper 01 60 Unit 1 Paper 02 68 Unit 1 Paper 03/2 74 Unit 2 Paper 01 78 Unit 2 Paper 02 86 Unit 2 Paper 03/2 90 CAPE Management of Business Mark Schemes: Unit 1 Paper 01 93 Unit 1 Paper 02 95 Unit 1 Paper 03/2 110 Unit 2 Paper 01 117 Unit 2 Paper 02 119 Unit 2 Paper 03/2 134
Independent Personal Pronouns Personal Pronouns in Hebrew Person, Gender, Number Singular Person, Gender, Number Plural 3ms (he, it) א ִוה 3mp (they) Sֵה ,הַָּ֫ ֵה 3fs (she, it) א O ה 3fp (they) Uֵה , הַָּ֫ ֵה 2ms (you) הָּ תַא2mp (you all) Sֶּ תַא 2fs (you) ְ תַא 2fp (you
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Dr. Alfredo López Austin [National University of Mexico (UNAM)] Golden Eagle Ballroom 3:00 pm 3:15 pm BREAK 3:15 pm 4:00 pm BREAKING THROUGH MEXICO'S PAST: DIGGING THE AZTECS WITH EDUARDO MATOS MOCTEZUMA Dr. David Carrasco (Harvard University) Golden Eagle Ballroom 4:00 pm 4:30 pm 4:30 pm 5:00 pm TLAMATINI AWARD PRESENTATION to Dr. Eduardo Matos Moctezuma (Bestowed by Dr. Rennie Schoepflin .
