2014年Alevel 數(shù)學試卷解析(精排版)

1、edexcel gce mathematics general instructions for marking 1. the total number of marks for the paper is 75.2. the edexcel mathematics mark schemes use the following types of marks:m marks: method marks are awarded for knowing a method and attempting to apply it,unless otherwise indicated.a marks: acc

2、uracy marks can only be awarded if the relevant method (m) marks havebeen earned.b marks are unconditional accuracy marks (independent of m marks)marks should not be subdivided.3. abbreviationsthese are some of the traditional marking abbreviations that will appear in the mark schemes. bod benefit o

3、f doubtft follow throughthe symbol will be used for correct ftcao correct answer onlycso - correct solution only. there must be no errors in this part of the question toobtain this markisw ignore subsequent workingawrt answers which round tosc: special caseoe or equivalent (and appropriate)dep depen

4、dentindep independentdp decimal placessf significant figures? the answer is printed on the paperthe second mark is dependent on gaining the first mark4. all a marks are correct answer only (cao.), unless shown, for example, as a1 ft toindicate that previous wrong working is to be followed through. a

5、fter a misread however,the subsequent a marks affected are treated as a ft, but manifestly absurd answersshould never be awarded a marks.5. for misreading which does not alter the character of a question or materially simplify it,deduct two from any a or b marks gained, in that part of the question

6、affected.6. if a candidate makes more than one attempt at any question:if all but one attempt is crossed out, mark the attempt which is not crossed out.if either all attempts are crossed out or none are crossed out, mark all theattempts and score the highest single attempt.7. ignore wrong working or

7、 incorrect statements following a correct answer.pmtgeneral principles for core mathematics marking (but note that specific mark schemes may sometimes override these general principles). method mark for solving 3 term quadratic: 1. factorisationcpqqxpxcbxx=+=+where),)()(2, leading to x = amncpqqnxpm

8、xcbxax=+=+andwhere),)()(2, leading to x = 2. formulaattempt to use the correct formula (with values for a, b and c). 3. completing the squaresolving 02=+cbxx:0, 022=qcqbx, leading to x = method marks for differentiation and integration: 1. differentiationpower of at least one term decreased by 1. (1

9、nnxx)2. integrationpower of at least one term increased by 1. (1+nnxx)use of a formula where a method involves using a formula that has been learnt, the advice given in recent examiners reports is that the formula should be quoted first. normal marking procedure is as follows: method mark for quotin

10、g a correct formula and attempting to use it, even if there are mistakes in the substitution of values. where the formula is not quoted, the method mark can be gained by implication from correct working with values, but may be lost if there is any mistake in the working. exact answers examiners repo

11、rts have emphasised that where, for example, an exact answer is asked for, or working with surds is clearly required, marks will normally be lost if the candidate resorts to using rounded decimals. answers without working the rubric says that these may not gain full credit. individual mark schemes w

12、ill give details of what happens in particular cases. general policy is that if it could be done “in your head”, detailed working would not be required. pmtquestion number scheme marks 1. 2665466222 .2 .12222xxx =+ m1 264, 96 , 60.xx=+ b1, a1, a1 4264, 192, 240.22xx=+special casethis is correct but

13、unsimplified m1b1a1a0 4 marks alternative method 2666662 12 1.12444xxx =+ m1 264, 96 , 60.xx=+ b1, a1, a1 notesm1: the method mark is awarded for an attempt at binomial to get the second and/or third term need correct binomial coefficient combined with correct power of x. ignore bracket errors or er

14、rors (or omissions) in powers of 2 or sign or bracket errors. accept any notation for 61c and 62c, e.g. 61 and 62 (unsimplified) or 6 and 15 from pascals triangle this mark may be given if no working is shown, but either or both of the terms including x is correct. b1: must be simplified to 64 (writ

15、ing just 62 is b0 ). this must be the only constant term (do not isw here) a1: is cao and is for 96 x. the x is required for this mark. allow +(-96x) a1: is cao and is for 260 x (can follow omission of negative sign in working) any extra terms in higher powers of x should be ignored isw if this is f

16、ollowed by 216, 24 , 15.xx=+ allow terms separated by commas and given as list alternative method m1: does not require power of 2 to be accurate b1: if answer is left as 26664 1.1244xx + allow m1 b1 a0 a0 pmtquestion number scheme marks 2.(a) 1f ( ) = 16xx3 22x+33or f ( x) =162+3xxm1 a1 a1 3 (b) 322

17、1324382f(x)d = xx+xx+x( ) cm1 a1 a1 3222183883f(x)d83232xx xxxx x+c orx x+cx= +a1 47 marks notes(a) m1: attempt to differentiate power reduced 1nnxxor 3x becomes 3 a1: two correct terms ( of the three shown). they may be unsimplified a1: fully correct and simplified then isw (any equivalent simplifi

18、ed form acceptable) (b) m1: attempt to integrate original f(x) one power increased 1nnxx+ a1: two of the four terms in x correct unsimplified (ignore lack of constant here) a1: three terms correct unsimplified (ignore lack of constant here) a1: all correct simplified with constant allow -1x for x n.

19、b integrating answer to part (a) is m0 pmtquestion number scheme marks 3. (a) f (x) = 1027xx32+13x12 attempts f( 2) or f(3) or uses long division as far as a remainder m1 (i) a1 (ii) f(2) = 150 f(-3)= 0 a1 3(b) 1027xx32+13x 12(= x3+ )(10 x2.+ m1 1027xx32+13x 12(= x3+ )(10 x23 x4)a1 dm1 (10 x23x4)= (

20、ax+b)(cx+d) where ac =10 and bd = 4 (=+3)(5xx4)(21x+ ) a1 47 marks notes(a) (b) m1: as on scheme a1: for 150, next a1: for 0 both cao (if division has been used it should be clear that they know these values are the remainders) m1: recognises (x+3) is factor and obtains correct first term of quadrat

21、ic factor by division or any other method a1: correct quadratic may have been done in part (a) dm1: attempt to factorise their quadratic 4152a1: need all three factors together, accept any correct equivalent e.g. 10( +3)(xx)(+x) if the three roots of f(x) = 0 are given after correct factorisation th

22、en isw special case. just writes down the three factors = (3)(5xx+4)(21x)+with no working : full marks allow trial and error or use of calculator for completely correct answer so 4 marks or 0 marks if “hence” is not used. question number scheme marks 4.(i)()()()4 2 262 262 26+m1 ()()2 262 26862+=b1

23、62 3= used in numerator - may be implied by a correct factorisation of numeratorb1 concludes ()()4 2 262 2 232+=+* a1 * 4(ii) 1st two terms 273 3=and2177 3=b1 3rd term6 3see 2 3or3b1 3 37 32 38 3+= or 6 33 37 38 33+= * b1 * 3alternative for (i) assume result and multiply both sides by ()2 26m1 ()()2

24、 264 22 6+ = 16 12 = 4b1 b1 so lhs = rhs and result is true a1 4alternative for (ii) 8121 7363+ 8121 7368 3 3or+ =b1 92163+ 9216+=b1 2438 333= 921624+= so equation is true b1 3(7 marks)notes (i) m1: multiplies numerator and denominator by ()2 26+ b1: correct treatment of denominator to give 2 (may b

25、e implied by answer obtained with no errors seen) b1: splits 62 3= - may be implied , but b0 for 2 62 2(2 3.)= a1 cao reaches result and no errors should be seen n.b.()()4 2 262 2 232+=+may be awarded b1 a1 as there is an implication that 62 3=(ii) b1: expresses both of first two terms as multiple o

26、f root 3 correctly b1: rationalises denominator in second term -may not see working b1: has used 3 37 32 38 3+= n.b. 63 37 38 33+= is b1b0b0 (i) alternative m1: assume result and multiply both sides by ()2 262nd b1: uses2 36= 1st b1: multiplies out these two brackets to give 4 a1: conclusion (ii)alt

27、ernatives b1: uses common denominator or multiplies both sides by root 3 and obtains correct unsimplified equationb1: lhs numerator correctly simplified or just see 9 + 21 6 b1: in the first alternative must see multiplication of numerator and denominator by 3 to give 8 3 in the second need statemen

28、t lhs = rhs and so truepmtquestion number scheme marks 5. (a) 2342342442,24,23334uuu= = m1 a1 a1 3(b) 613u=. b1 1(c) 992223331(34)(34)(34).iiu=+m1 99133 (. . .) ,11iiu=+= a1, a1 3(c) alternative method for part (c) adds 233 4 nnn+ + m1 uses n = 33 a1 -11 a1 37 marks notes(a) (b) (c) m1: attempt to u

29、se formula correctly (implied by first term correct, or given as 0.67, or third term following through from their second etc) a1: two correct answers a1: 3 correct answers (allow 0.6 recurring but not 0.667) look for the values. ignore the labelru b1: cao (nb use of ap is b0) m1: uses sum of at leas

30、t 3 terms found from part (a) (may be implied by correct answer). attempt to sum an ap here is m0. a1: obtains 33 (sum of three adjacent terms)or11 (sum of nine adjacent terms) a1: - 11 cao ( -11 implies both a marks) n.b. use of n = 99 is m1a0a0 pmtquestion number scheme marks 6. 4b4444253aloga= 3

31、or loglogab+= log 25 or loga= or254 logbb=3 (if this is preceded by wrong algebra (e.g. b = 25 a) m1 can still be given if their b is used m1 34log 643 or 4= 64 (may be implied by the use of 64) 12(log25+3)12or see log = (log253) becomeaa4=+12(log253)12or see log = (log253) becomebb4= (these latter

32、two statements will be implied by correct answers) b1 correct algebraic elimination of a variable to obtain expression in a or b without logs dm1 5840 orab=a1 substitutes to give second variable or solves again from startdm1 5840 andab= and no other answers. a1 66 marks notesm1: uses addition or sub

33、traction law correctly for logs (n.b. 44loglog25ab+= is m0) b1: see number 64 used (independent of m mark) or 12(log25 3)12or see log = (log253) become4aa+=12(log25 3)12or see log = (log253) become4bb=dm1: dependent on first m mark. eliminates a or b (with appropriate algebra) and eliminates logs a1

34、: either a or b correct dm1: dependent on first m mark . attempts to find second variable a1: both a and b correct allow b = 0.625 if a = -40 and b = -5/8 are also given as answers lose the last a mark. nb log a + log b = 2.3219.will not yield exact answers if they round their answers to 40 and 0.62

35、5 after decimal work, do not give final a mark. nb: some will change the base of the log and use loga log b = 3log4 pmtquestion number scheme marks (7.12sin2cosxx110= 12(1 cos2)cosxx 110 =and so 12cos2+cosxx1 = 0 * b1 * 1 (b) solve quadratic to obtain (cos x) =or 1134m1 a1 m1 a1cao x = 75.5, 109.5,

36、250.5, 284.5 answers in radians (see notes) 45 marks notes(a) (b) b1: replaces sin x by (1cos22x)- or replace 11 by 11(sin x+cos22x)and no errors seen to give printed answer including = 0 m1: solving the correct quadratic equation (allow sign errors), by the usual methods (see notes) implied by corr

37、ect answers a1: both answers needed allow 0.25 and awrt 0.33 m1 uses inverse cosine to obtain two correct values for x for their values of cosx e.g. (75.5 and 109.4 or 109.5) or (75.5 and 284.5) or (109.5 and 250.5) allow truncated answers or awrt here. a1: all four correct allow awrt. ignore extra

38、answers outside range but lose last a mark for extra answers inside range answers in radians are 1.3, 5.0, 1.9 and 4.4 allow m1a0 for two or more correct asnwers question number scheme marks 8. kx2x+82 k +()7 = 0 uses 2 4bac with ak, b= 8 and attempt at c =2(k + 7) m1 ba=c46456k 228kor 6456k=+8k2o.e

39、. a1 attempts to solve 27kk+8=0to give k =dm1 critical values, k1,=8.a1cso m1 uses4bac0 or22b402kk+78 0 gives 1(or)kk 8 m1 a1 77 marks notesm1: attempts 24bac for ,8ak b= and c = 2(k+7) or attempt at c from quadratic = 0 (may omit bracket or make sign slip or lose the 2, so 2k +7 or k + 7 for exampl

40、e) or uses quadratic formula to solve equation or uses on two sides of an equation or inequation a1: correct three term quadratic expression for 24bac - (may be under root sign) dm1: uses factorisation, formula, or completion of square method to find two values for k , or finds two correct answers w

41、ith no obvious method for their three term quadratic a1: obtains 1 and -8 m1: states 22404bacor bac anywhere (may be implied by the following work) m1: chooses outside region ( k ) for appropriate 3 term quadratic inequality . do not award simply for diagram or table. a1: 1or8kk 1 and k 1, x -8 is a

42、0 ( only lose 1 mark for using x instead of k ) and 1(or)8kk is a0 also 1 k -8 is m1 a0 n.b. lack of working: if there is no mention of 2240 or4bacbac3000or 300(1.05)n1=3000 m1 1.05or (n1)=log10 or correct equivalent log work ftm1 (n 1)log1.05log10or (n 1)log1.05=log10 n 48.19 n = 49 a1 37 marks not

43、es(a) (b) (c) m1: for correct statement of formula with correct a, r and n a1: cao (this answer implies the m1) m1: correct formula with r = 1.05 and n = 24 ft their a (if they list all the terms correct answer implies method mark) a1: answers which round to 13400 are acceptable m1: correct inequali

44、ty or uses equality and interprets correctly later (ft their a ) m1: correct algebra then correct use of logs on their previous line (may follow use of =, or use of n instead of n -1) can get m0m1a0 a1: need to see 49 or 49th month special case: uses sum formula: if they reach12 (1.05)n 1 and then u

45、se logs correctly to give 12nlog(1.05)log1then give m0m1a0 if trial and error is used then the correct answer implies the method. so 49 is m1m1a1 and 48 scores m1m0a0. similar marks follow answer only with no working. question number scheme marks 10. (a)m1 a1 2(b) (0 , ) ; 5, 06 or (150, 0) and11, 0

46、6 or (330,0)b1; b1 b1 3(c) or344x= m1 7or1212x=m1 a1 a1 49 marks notes(a) m1: could be part of a cycle, or several cycles but needs at least one max and one min a1: needs to be only one cycle. needs to be positive y intercept and positive gradient at start and finish (not zero gradient). needs to be

47、 solely 0 x and to finish at the same y value as it started. (b) each answer is cao need coordinates with zeroes (as given) unless points are indicated correctly on the graph e.g. or (1/2 , 0) on the y axis may be given credit etc allow degrees on x axis. extra in range lose last b1. if answers are

48、given in text and on diagram, text takes precedence. (c) m1: uses inverse cos correctly to obtain at least one correct answer (may be in degrees) m1: adds 3 to their previous answer, which must have been in radians but may add 60 to answer in degrees a1: one correct answer a1: both answers correct e

49、xtra answers in range lose final a1 extra answers outside range isw special case: all answers given as decimals (b) b1 b0 b0 (c) m1 m1 a0 a0 all answers in degrees: (b) 150 and 330 then (c) 15 and 105 (just lose final two a marks) so b1, b1 in (b) then m1m1a0a0 method mark for harmonic curve i.e. an

50、y sine or cosine curve accuracy for correct section and position relative to the pmtquestion number scheme marks 11. (a)uses (2p 6) 4p = 4p 60 or 60(26)42pp+=or 60+2(4p-60) =2p-6 or etc or two correct equations with d m1 so p = 9 * a1 * 2alternative to (a) use p=9 to give 60, 36 and 12 and deduce d

51、= -24 so conclude ap when p = 9 m1 a1 2(b) uses a + 19d with a = 60 m1 finds d = 36 60 = -24 b1 so obtains -396 a1 3(c) uses(260(1) )2nnd+m1 uses(26024(1)2nna1 = 12n (6-n) * a1* 38 marks notes(a) (b) m1: correct equation to enable p to be found or two correct equations if d introduced and solving si

52、multaneous equations to eliminate d and enable p to be found nb may add three terms and use sum formula giving e.g. 3260426(6026)ppp+=+ a1: cso (do not need intermediate step) m1: correct formula with their value for d b1: d = -24 seen in (a) or (b) a1: -396 if all terms are found and added 60 + 36

53、+ 12 + -12 + . need 20 terms for m1, need -24 implied by first 4 terms for b1 and correct answer for a1 m1:uses correct formula with their value for d a1: correct value for d a1: given answer must be no errors to award this mark special case: proves formula for sum of ap m1: correct method of proof

54、using their d a1: for d = -24 a1: given answer must be no errors to award this mark (c) pmtquestion number scheme marks 12. (a)2221510102 10 10cosboc=+ m1 22210101525cos0.1252 10 10200bocoror+=a1 1.696boc=(n.b. 97.2 degrees is a0) a1 3(b) uses22with theirs= from part (a) not (2 )m1 22 1.69637.3(15)r

55、= a1 perimeter15,3976.3 (m)rxxtheirarclength=+=+m1 a1ft 4(c) 212area of sector(22)= -not (2 ) b1 212area of triangle(10) sin= b1 area of paved area = 212(22)212(10) sin = 410.432 49.6 or 75 7410.4324= 360.8 or awrt 361 (m2) m1 a1 4(11 marks)notes(a) m1: uses cosine rule must be correct or other corr

56、ect trigonometry e.g. 7.52where sin10= a1: makes cos subject of formula correctly or uses 17.52 sin10a1: accept awrt 1.696 (answer in degrees is a0). if answer is given as 1.70 (3sf) then a0 but remaining as are available (special case below) m1: uses22with theirs=in radians, or correct formula for

57、degrees if working in degrees a1: accept awrt 37.3 (may be implied by their perimeter) m1: adds arc length to 15 to two further equal lengths for perimeter a1ft: accept awrt 76.3 do not need metres ft on their arc lengthso 39 + arc length b1: this formula used with their in radians or correct formul

58、a for degrees - allow miscopy of angle b1: correct formula for area may use half base times height m1: subtracts correct triangle ( two sides of length 10) from their sector a1: awrt 361 do not need units special case uses 3 sf instead of 3 dp in part (a) loses final a mark in part (a) but can have

59、a marks in part (b) for 37.4 and 76.4 and can have a mark in part (c) for awrt 362 (b) (c) pmtquestion number scheme marks 13. (a)75so334yxx=+b1 2d3750dyxx=+( x 0 ) accept 22d375dyxxx= or equivalentm1 a1 3(b) 2dput3750dyxx=m1 5x=a1 substitute to give y = - 4 m1 a1 4(c) 232dconsider1500dyxx=m1 so min

60、imum a1 2(d) when x = 2.5, y = 3.5b1 also gradient of curve found by substituting 2.5 into their ddyx (= -9)m1 191so gradient of normal is -()m=dm1 either : ()193.5 2.5yx= or: 19 yxc=+ and ()12993.5 2.53 cc=+= dm1 so 9290 xy+= or 9290yx= or any multiple of these answersa1 514 marks notes(a) b1: any