Paper Reference(s) Edexcel GCE
Paper Reference(s)6664/01Edexcel GCECore Mathematics C2Advanced SubsidiaryMonday 14 January 2013 MorningTime: 1 hour 30 minutesMaterials required for examinationMathematical Formulae (Pink)Items included with question papersNilCandidates may use any calculator allowed by the regulations of the Joint Councilfor Qualifications. Calculators must not have the facility for symbolic algebramanipulation, differentiation or integration, or have retrievable mathematicalformulae stored in them.Instructions to CandidatesWrite the name of the examining body (Edexcel), your centre number, candidate number, theunit title (Core Mathematics C2), the paper reference (6664), your surname, initials andsignature.Information for CandidatesA booklet ‘Mathematical Formulae and Statistical Tables’ is provided.Full marks may be obtained for answers to ALL questions.The marks for the parts of questions are shown in round brackets, e.g. (2).There are 9 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.Answers without working may not gain full credit.P41487AThis publication may only be reproduced in accordance with Edexcel Limited copyright policy. 2013 Edexcel Limited.
1.Find the first 3 terms, in ascending powers of x, in the binomial expansion of(2 5x)6.Give each term in its simplest form.(4)f(x) ax3 bx2 4x 3, where a and b are constants.2.Given that (x – 1) is a factor of f(x),(a) show that a b 7.(2)Given also that, when f(x) is divided by (x 2), the remainder is 9,(b) find the value of a and the value of b, showing each step in your working.(4)3.A company predicts a yearly profit of 120 000 in the year 2013. The company predicts that theyearly profit will rise each year by 5%. The predicted yearly profit forms a geometric sequencewith common ratio 1.05.(a) Show that the predicted profit in the year 2016 is 138 915.(1)(b) Find the first year in which the yearly predicted profit exceeds 200 000.(5)(c) Find the total predicted profit for the years 2013 to 2023 inclusive, giving your answer to thenearest pound.(3)4.Solve, for 0 x 180 ,cos (3x 10 ) 0.4,giving your answers to 1 decimal place. You should show each step in your working.(7)P41487A2
5.The circle C has equationx2 y2 20x 24y 195 0.The centre of C is at the point M.(a) Find(i) the coordinates of the point M,(ii) the radius of the circle C.(5)N is the point with coordinates (25, 32).(b) Find the length of the line MN.(2)The tangent to C at a point P on the circle passes through point N.(c) Find the length of the line NP.(2)6.Given that 2 log2 (x 15) log2 x 6,(a) show that x2 34x 225 0.(5)(b) Hence, or otherwise, solve the equation 2 log2 (x 15) log2 x 6.(2)P41487A3Turn over
7.Figure 2The triangle XYZ in Figure 1 has XY 6 cm, YZ 9 cm, ZX 4 cm and angle ZXY .The point W lies on the line XY.The circular arc ZW, in Figure 1 is a major arc of the circle with centre X and radius 4 cm.(a) Show that, to 3 significant figures, 2.22 radians.(2)(b) Find the area, in cm2, of the major sector XZWX.(3)The region enclosed by the major arc ZW of the circle and the lines WY and YZ is shown shadedin Figure 1.Calculate(c) the area of this shaded region,(3)(d) the perimeter ZWYZ of this shaded region.(4)P41487A4
8.The curve C has equation y 6 3x 4, x 0.x3(a) Use calculus to show that the curve has a turning point P when x 2.(4)(b) Find the x-coordinate of the other turning point Q on the curve.(1)d2 y(c) Find.dx 2(1)(d) Hence or otherwise, state with justification, the nature of each of these turning pointsP and Q.(3)P41487A5
9.Figure 2The finite region R, as shown in Figure 2, is bounded by the x-axis and the curve with equationy 27 2x 9 x 16,x2x 0.The curve crosses the x-axis at the points (1, 0) and (4, 0).(a) Copy and complete the table below, by giving your values of y to 3 decimal places.x11.5y05.86622.55.21033.541.8560(2)(b) Use the trapezium rule with all the values in the completed table to find an approximatevalue for the area of R, giving your answer to 2 decimal places.(4)(c) Use integration to find the exact value for the area of R.(6)TOTAL FOR PAPER: 75 MARKSENDP41487A6
January 20136664 Core Mathematics C2Mark SchemeQuestionNumberScheme1.( 2 5x )( 2 )6Marks6Award this when first seen (not 64x0)64B1Attempt binomial expansion with correctstructure for at least one of these terms. E.g. aterm of the form: 6 ( 2 ) ( 5 x ) 56 54( 2 ) ( 5 x ) 22 6 6 pp ( 2 ) ( 5 x) with p 1 or p 2 p consistently. Condone sign errors. Condonemissing brackets if later work implies correctstructure and allow alternative forms forbinomial coefficients e.g. 6 6 C1 or or even 1 1 Do not allow 960xM16 960x( )6000x2Allow this to come from (5 x )A1 (first)2A1 (Second)Ignore any extra terms and isw e.g. divides all terms by 2The terms do not have to form a sum i.e. they can be listed with commas or given onseparate lines.Special Case - decreasing powers can score M1 with the conditions as above for thesecond and third terms.( 2 5x )6() 6 6 2 64 ( 25 5 x ) 24 ( 5 x ) scores B1 only as the 1 2 powers of 2 and (-5x) are being added not multiplied.Fully correct answer with no working can score full marks. If either the second or thirdterm is correct, the M1 can be implied and the A1 scored for that term.(4)Way 264 (1 .)64 and (1 . – Award when first seen.B1Correct structure for at least one of theunderlined terms. E.g. a term of the form: 6 p (kx) with p 1or p 262 5x 6 5 5x p 5x 1 1 6 2 22 2 consistently and k 5 M1Condone sign errors. Condoned missingbrackets if later work implies correctstructure but it must be an expansion of(1 kx ) 960x( ) 6000x26where k 5Do not allow 960xA1 5x 2 Allow this to come from 2A1(4)
QuestionNumber2.(a)Schemef (1) a b 4 3 0 or a b – 7 0a b 7*MarksAttempt f( 1)Must be f(1) and 0 needs to beseenM1A1(2)(b)f ( 2) a ( 2 ) b ( 2 ) 4 ( 2 ) 3 9Attempt f( 2) and uses f( 2) 9M1 8a 4b 8 3 9Correct equation with exponentsof (-2) removedA132(–8a 4b 4)Solves the given equation from part (a)and their equation in a and b from partM1(b) as far as a . or b .a 2 and b 5Both correctA1Attempts at trial and improvement in (b): Allow the first M1 if they tryvalues for a and b where a b 7 and substitute their values into the cubicalong with x 2 and sets 9. For completion to a 2 and b 5 fully shownto be correct allow 4/4. For incomplete or incorrect solutions allow the firstM1 only. If in doubt consult your team leader.(4)[6]((a)Long Divisionax bx 4 x 3 ( x 1) ax 2 px q32)where p and q arein terms of a or b or bothand sets their remainder 0NB Quotient ax 2 ( a b ) x ( a b 4 )a b 7*( ax(b)3M1A1(2) bx 4 x 3) ( x 2 ) ax px q22where p and q arein terms of a or b or bothand sets their remainder 9NB Quotient ax 2 ( b 2a ) x ( 4a 4 2b )4b – 8a 5 9Follow scheme for final 2 marksM1A1
QuestionNumber3.Scheme(a)120000 (1.05) 138915 *(b)120000 (1.05)MarksOr 120000 1.05 1.05 1.05 138915Or 120000, 126000, 132000, 1389153Or a 120000 and a (1.05 ) 138915B13(1)n 1 200000 5 3 log1.05n 1 log ( n 1 ) 5 log 3 log1.05or equivalent 7 log 4 e.g ( n )Allow n or n – 1 and “ ”, “ ”, or “ ” etc.M1Takes logs correctlyAllow n or n – 1 and “ ”, “ ”, or “ ” etc.M1Allow n or n – 1 and “ ”, “ ”, or “ ” etc. Allow1.6 or awrt 1.67 for 5/3.A1log1.05M1: Identifies a calendar year using their valueof n or n - 12024M1A1A1: 2024 only cso2024 with no working no marksSee appendix for alternative taking logs base 1.05 and mis-read as total profita(1 r n ) 120000 (1 1.05 1 r1 1.0511(c)1704814)(5)M1: Correct sum formula with n 10, 11 or 12A1: Correct numerical expression withn 11M1 A1Cao (Allow 1704814.00)A1(3)[9]Listing or trial/improvement in (b)U10 186 159.39, U11 195 467.36, U12 205 240.72Attempt to find at least the 10th or 11th or 12th terms correctly using a common ratio of 1.05(all the terms need not be listed)Forms the geometric progression correctly to reach a term 200 000(May be implied e.g. reaches 195 467.36 – Hence the next year)Obtains an “11th” term of awrt 195 500 and a “12th” term of awrt 205 200Uses their number of terms to identify a calendar year2024If you are not sure how to award the marks please consult your Team LeaderM1M1A1M1A1(5)
QuestionNumber4.Schemecos 1 ( 0.4 ) 113.58 (α )3x 10 α x Note: If x α 103α 103MarksAwrt 114B1Uses their α to find x.α 10αnot 10Allow x 33M1is not clearly applied from their first angle it may be recovered ifapplied to their second or third angle.x 41.2( 3x 10 ) 360 α(246.4.)x 85.5( 3x 10 ) 360 α ( 473.57. )x 161.2AwrtA1360 α (can be implied by 246.4.)Awrt360 α (Can be implied by 473.57.)AwrtM1A1M1A1Note 1: Do not penalise incorrect accuracy more than once and penalise it the first time itoccurs. E.g if answers are only given to the nearest integer (41, 85, 161) only the first A markthat would otherwise be scored is lost.Note 2: Ignore any answers outside the range. For extra answers in range in an otherwisefully correct solution lose final A1Note 3: Lack of working means that it is sometimes not clear where their intermediate anglesare coming from. In these cases, if the final answers are incorrect score M0.Note 4: Candidates are unlikely to be working in radians deliberately but may have theircalculator in radian mode ( gives α 1.98). In such cases the main scheme should be appliedand the method marks are available. If you suspect that the candidate is working in radianscorrectly then please use the review mechanism and/or consult your team leader.Way 2cos 1 ( 0.4 ) 66.42 (α )180 – 66.42 113.583x 10 113.58 x 113.58 103Awrt 114B1Uses their 113.58 to find xM1x 41.2Awrt3x 10 180 α (246.4.)180 αto give x 85.53x 10 540 α (473.57.)540 - αto give x 161.2Special case - takes 0.4 as -0.4 1cos ( 0.4 ) 66.42 (α )3x 10 66.4 x 66.4 103x 41.23x 10 360 α (293.6.)x 101.23x 10 360 α (426.4.)x 145.5A1M1A1M1A1B0M1A0M1A0M1A0(3/7)
QuestionNumber5.(a)(i)SchemeMarksParts (i) and (ii) are likely to be solved together so mark as one partB1: x 10The centre is at (10, 12)B1: y 12Uses ( x 10) ( y 12) 195 100 144 r .Completes the square for both x and y in an attempt to find r.( x "10") 2 a and ( y "12")2 b and 195 0, ( a, b 0 )2(ii)2Allow slips in obtaining their r2 but must find square rootA correct numerical expression for r22including the square root and canr 10 12 195implied by a correct value for rNot r 7 unless 7 is rejectedr 7B1 B1M1A1A1(5)(a)Way 2Compares the given equation withx 2 y 2 2 gx 2 fy c 0 to writedown centre ( g , f ) i.e. (10, 12)B1: x 10B1B1B1: y 12Uses r ( "10") 2 ( "12") c2r 10 2 12 2 195r 7M1A correct numerical expression for rA1A1(5)Note that although the marks for the centre are B marks, they do need to22come from correct work. E.g. ( x 10) , ( y 12) giving a centre of(10, 12) scores B0 B0 but could score the M1A1ftA1ft for the radius as a22special case. Similarly ( x 10) , ( y 12) giving a centre of (-10, 12)scores B0 B1, ( x 10) , ( y 12) giving a centre of (10, -12) scores B1 B0but both could score M1A1ftA1ft for the radius as a special case also.2(b)2MN (25 "10") 2 (32 "12") 2(Correct use of Pythagoras)MN 625 25M1A1(2)(c)NP ("25"2 "7"2 )()NP ( MN 2 r 2 )M1NP (252 7 2 ) is M0 (Quite common)NP 576 24A1(2)(c)Way 2cos ( NMP ) NP 247 NP "25"sin ( NMP )"25"Correct strategy for finding NPM1A1(2)[9]
QuestionNumber6.(a)SchemeMarks2log( x 15) log( x 15)2log( x 15)2 log x logB1( x 15)2xCorrect use of log a log b log ( x 15 )22 log( x 15) log x 6 log x abM1 6 with no incorrect work scores B1M1together2 log 2 ( x 15) log 2 x 2 log 226 64 or log 2 64 6( x 15)( x 15) 6 64xx2log2( x 15)is M0x64 used in the correct contextB1Removes logs correctlyM122log( x 15) log x 6 log( x 15) log x 6 2( x 15)2 64xIs acceptable for the first 4 marksThis method mark should only be awarded for the removal of logs in an appropriateway. Some examples are below,log( x 15) 2log xloglog( x 15)2x( x 15)2x 6 6 6 ( x 15) 2x( x 15)2x( x 15)2x 6 M0log( x 15)2 6 x( x 15)2x 6 M0 log 2 6 M0( x 15) ( x 15) 64 M1 6 M0 log 6 x x 222 x 30 x 225 64 x2or x 30 225 x 1 64 x2 34x 225 0 *Must see expansion of ( x 15 ) toscore the final mark.Correct completion to printed answerwith no errors but allow recovery from‘invisible’ brackets e.g.2A1x 15 x 30 x 22522(5)(b)( x 25)( x 9) 0 x 25 or x 9M1: Correct attempt to solve the givenquadratic as far as x .It must be an attempt at solving thegiven quadratic but allow mis-copye.g. 255 for 225A1: Both 25 and 9M1 A1(2)[7]See appendix for some alternative correct and incorrect methods for (a)
QuestionNumber7.(a)SchemeMarks92 42 62 2 4 6cos α cos α .Correct use of cosine ruleleading to a value for cos αM1Cso (2.22 must be seen here)A142 62 92 29 cos α 0.604. 2 4 6 48 α 2.22 (NB α 2.219516005)(a) Way 2(2)XY 2 42 62 2 4 6cos 2.22 XY 2 .Correct use of cosine ruleleading to a value for XY2M12XY 81.01.XY 9.00.(b)2π 2.22( 4.06366.)12A12π 2.22 or awrt 4.06or 2π 2.2 or awrt 4.08(May be implied)Correct method for majorsector area. Allow π 2.22 forthe major sector angle. 42 "4.06"Awrt 32.532.5(2)B1M1A1(3)Finding the minor sector area here (17.8) is 0/3Circle – Minor sector(b) Way2π 4212π 4 2 4 2 2.22 32.5 32.5Correct expression for circle areaCorrect method forcircle - minor sector areaAwrt 32.5B1M1A1(3)(c)Area of triangle Correct expression for the area oftriangle XYZ (allow 2.2 or awrt 2.22)B1So area required “9.56” “32.5”Their Triangle XYZ (Not triangleZXW) (part (b) answer or correctattempt at major sector)M1 42.1 cm2 or 42.0 cm2Awrt 42.1 or 42.0 (Or just 42)A112 4 6 sin 2.22 ( 9.56 )(3)Note: The minor sector area (17.76) the triangle (9.56) 27.32 which looks like theanswer to (d) – beware!(d)Arc length 4 4.06 ( 16.24 )Or 8π 4 2.22Perimeter ZY WY Arc LengthM1: 4 their ( 2π 2.22 )Or circumference – minor arcA1: Correct ft expressionM1A1ft9 2 Any ArcM1Perimeter 27.2 or 27.3Awrt 27.2 or awrt 27.3Note the order of marks on Epen is M1M1A1A1 – the M’s and A’s mustcorrespond so that the second mark on Epen is the second M1 on the schemeA1(4)(Generally do not apply isw in this question and mark their final answer unless a correct answer issubsequently rounded incorrectly)In this question we will need to be careful with labelling as each part has clear demands and mustbe marked as labelled by the candidate.[12]
QuestionNumberSchemey 6 3x 8.(a)1dx( 2)4or 3 120 4M1 A1A1: Correct derivativeSo x 4 4 and x 2 or12 3( x x or x x or 6 0)dy12 0 3 4 0 x . ordxxdy12 3 4dx2 3 4x3M1: x n x n 1dy12 3 4 or 3 12 x 4dxxdyMarks( 2) 4 0y ′ 0 and attempt to solve for xMay be implied bydy1212 3 4 0 4 3 x . ordxxxM1Substitutes x 2 into their y ′Correct completion to printed answer withno errors by solving their y ′ 0 orsubstituting x 2 into their y ′ 1 For solving, allow e.g. x x 4 4 41 A114 2The minimum for verification is as in the scheme which could be implied by -3 3 0Do not allow x 4 x 1.41. 4(b)x 2(c)d y 48 5 or 48x 52dxx2 for the final A1Awrt -1.41(4)B1(1)2Follow through their first derivativefrom part (a)B1ft(1)An appreciation that either(d)y′′ 0 a minimumor y′′ 0 a maximumMaximum at P as y ′′ 0A generous mark that is independent ofany previous workCsoNeed a fully correct solution for this mark. y ′′ need not be evaluated but must beB1B1correct and there must be reference to P or to 2 and negative or 0 and maximum.There must be no incorrect or contradictory statements (NB allow y ′′ awrt-8 or -9)Minimum at Q as y ′′ 0CsoNeed a fully correct solution for this mark. y ′′ need not be evaluated but must beB1correct and part (b) must be correct and there must be reference to P or to 2 and positive or 0 and minimum. There must be no incorrect or contradictorystatements (NB allow y ′′ awrt 8 or 9)(3)[9]Other methods for identifying the nature of the turning points are acceptable. The first B1 isfor finding values of y or dy/dx either side of 2 or their x at Q and the second and thirdB1’s for fully correct solutions to identify the maximum/minimum.
QuestionNumberSchemey 27 2 x 9 x 9.(a)6.272 , 3.634(b)1 11 or2 24Marks16x2Awrt in each caseSpecial case 6.27 and 3.63 scores B1B0B1, B1(2)B1.{(0 0) 2 ( 5.866 "6.272" 5.210 "3.634" 1.856 )}Need {} or impliedlater for A1ftM1A1ft(0 0) may be implied if omitted and follow through their f(2) and f(3) in anotherwise correct expression and allow one missing or mis-copied term in the2(.) bracket for the method mark1 0.5(0 0) 2 ( 5.866 "6.272" 5.210 "3.634" 1.856 )2Unless followed by an answer that implies correct (missing) brackets, scoresB1M1A0A0 (Usually implied by an answer of 45.676)1 0.5 {(0 0) 2 ( 5.866 "6.272" 5.210 "3.634" 1.856 )}21 45.6764 11.42caoA1Separate trapezia may be used : B1 for 0.25, M1 for 12 h(a b) used 5 or 6times (and A1ft all correct )1NB 0.5 {(0 0) 2 ( 0 5.866 "6.272" 5.210 "3.634" 1.856 0 )}2Scores B1M0A0A0Correct answer with no working scores 0/4(4)n 1M1: x x on any termA1: 27x x 2n y dx 27 x x(c)2 6 x 16 x 1 ( c )323A1: 6x 2A1: 16x 1Accept any correct and possibly unsimplified versions for the terms and markin this order on Epen( 27 ( 4) ( 4) 6 ( 4) 16 ( 4) ) ( 27 (1) (1) 6 (1) 16 (1) )221232 132 1Attempt to subtract either wayround using the limits 4 and 1.Dependent on the previous M1. Maybe implied by 48 – 36 but you mayneed to check both their values if theintegration has errors. (48 – 36)Cao (Penalise -12)M1A1A1A1dM1A1(6)[12]
Appendix3(b)Way 2Allow n or n – 1 and “ ”, “ ”, or “ ”etc.120000 (1.05)n 1 200000Takes logs correctlyAllow n or n – 1 and “ ”, “ ”, or “ ” etc. 5 n 1 log 1.05This may be implied by 3 5 log1.05 1.05n 1 log1.05 3 e.g. log1.05 (120000 (1.05)( n 1 ) log1.05"M1M1and effectively gets the next A1n 1) ( n 1) log (120000 (1.05) ) would be M05"320241.05Allow n or n – 1 and “ ”, “ ”, or “ ”etc.M1: Identifies a calendar year using theirvalue of n or n - 1A1M1A1A1: 2024 only cso(5)3(b)MR?120000 (1 1.05n ) 2000001 1.05131.05n 12 13 log1.05n log 12 13 log 12 n log1.052014M0Takes logs correctlyM1A0M1: Identifies a calendar year using their valueof n or n - 1M1A0A1: 2024 onlyTrial & Improvement for this MR is 0/5(2/5)
4.Way 3General Solutioncos 1 ( 0.4 ) 113.58 (α )Awrt 114B13x 10 360n 113.58360n αM13x 10 360n 113.583 x 10 α 3 x α 10360n αM1360n 123.58 360 n 103.58or33x 41.2x 85.5x 161.2360 n 113.58 103AwrtAwrtAwrtx x M1A1A1A1(7)4.Special Case 1cos ( 3x 10 ) cos ( 3x ) cos (10 )cos ( 3x ) 0.4 cos (10 )cos ( 3x ) 0.5848.3x 54.2 αx 18.1B0M0A0 so far360 αAwrt360 αAwrt3x 360 αx 101.93x 360 αx 138.1M1A0M1A0(2/7)4.cos 1( 0.4 )Special Case 2 – Quite common 113.58 (α )Awrt 1143x 10 α x α 103x 41.2Uses their α to find x.α 10αnot 10Allow x 33AwrtB1M1A13 x 10 α 3 x α 103x 360 (α
2013 Edexcel Limited. Paper Reference(s) 6664/01 Edexcel GCE Core Mathematics C2 Advanced Subsidiary Monday 14 January 2013 Morning Time: 1 hour 30 minutes Materials required for examination Items included with question papers Mathematical Formulae (Pink) Nil Candidates may use any calculator allowed by the regulations of the Joint Council for Qualifications. Calculators must not have the .
University of Cambridge International Examinations London GCE AS/A-Level / IGCSE / GCSE Edexcel International. 6 Examination Date in 2011 Cambridge IGCSE Oct/Nov X 9 Cambridge GCE / May/Jun 9 9 London GCE London GCSE May/Jun 9 X Chinese London IGCSE Jan X 9 Cambridge IGCSE / May/Jun 9 9 London IGCSE London GCE Jan 9 9 Cambridge GCE Oct/Nov X 9 Private Candidates School Candidates Exam Date. 7 .
GCE AS / A Level M A Y / J U N E 2 0 1 5 EDEXCEL INTERNATIONAL EXAMINATIONS GCE AS / A Level MAY/JUNE 2016 UAE Edexcel International Examinations GCE AS / A Level May/June 2014 If there are any subjects not listed on the form that you would like to take please c
This publication may only be reproduced in accordance with Edexcel Limited copyright policy. 2007–2013 Edexcel Limited. Paper Reference(s) 6666/01 . Edexcel GCE . Core Mathematics C4 . Gold Level (Hard
Silver 1 This publication may only be reproduced in accordance with Edexcel Limited copyright policy. 2007–2013 Edexcel Limited. Paper Reference(s) 6683/01 . Edexcel GCE . Statistics S1 . Silver Level S1 . Time: 1 hour 30 minutes . Materials req
The AS GCE is both a 'stand-alone' qualification and also the first half of the corresponding Advanced GCE. The AS GCE is assessed at a standard appropriate for candidates who have completed the first year of study (both in terms of teaching time and content) of the corresponding two-year Advanced GCE course, ie between GCSE and Advanced GCE.
GCE Geography Edexcel Advanced Subsidiary GCE in Geography (8GE01) First examination 2009 Edexcel Advanced GCE in
GCE Economics Edexcel Advanced Subsidiary GCE in Economics (8EC01) First examination 2009 Edexcel Advanced GCE in Economics (9EC01) First examination 2010 Issue 4 . Level Assessment information Number of marks allocated in
.3 ISA / ANSI, ANSI-A300, Standards for Tree Care Operations. 2.2 Planting Layout, Massing and Plant Selection.1 Consider the limits and frequencies of institutional maintenance practices at UBC, and design accordingly for efficiency, servicing accessibility, low maintenance, weed control, pest, disease and drought tolerance. .1 Regardless of whether irrigation will be installed on site, the .
