Mark Scheme (Results) January Pearson Edexcel International A Level in Core Mathematics 34 (WMA02/01) January 2015 (IAL)

Transcription

1 Mark Scheme (Results) January 05 Pearson Edexcel International A Level in Core Mathematics (WMA0/0)

2 Edexcel and BTEC Qualifications Edexcel and BTEC qualifications are awarded by Pearson, the UK s largest awarding body. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information visit our qualifications websites at or Alternatively, you can get in touch with us using the details on our contact us page at Pearson: helping people progress, everywhere Pearson aspires to be the world s leading learning company. Our aim is to help everyone progress in their lives through education. We believe in every kind of learning, for all kinds of people, wherever they are in the world. We ve been involved in education for over 50 years, and by working across 70 countries, in 00 languages, we have built an international reputation for our commitment to high standards and raising achievement through innovation in education. Find out more about how we can help you and your students at: January 05 Publications Code IA0087 All the material in this publication is copyright Pearson Education Ltd 05

3 General Marking Guidance All candidates must receive the same treatment. Examiners must mark the first candidate in exactly the same way as they mark the last. Mark schemes should be applied positively. Candidates must be rewarded for what they have shown they can do rather than penalised for omissions. Examiners should mark according to the mark scheme not according to their perception of where the grade boundaries may lie. There is no ceiling on achievement. All marks on the mark scheme should be used appropriately. All the marks on the mark scheme are designed to be awarded. Examiners should always award full marks if deserved, i.e. if the answer matches the mark scheme. Examiners should also be prepared to award zero marks if the candidate s response is not worthy of credit according to the mark scheme. Where some judgement is required, mark schemes will provide the principles by which marks will be awarded and exemplification may be limited. When examiners are in doubt regarding the application of the mark scheme to a candidate s response, the team leader must be consulted. Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response.

4 EDEXCEL GCE MATHEMATICS General Instructions for Marking. The total number of marks for the paper is 5.. 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: Accuracy marks can only be awarded if the relevant method (M) marks have been earned. B marks are unconditional accuracy marks (independent of M marks) Marks should not be subdivided.. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes. bod benefit of doubt ft follow through the symbol will be used for correct ft cao correct answer only cso - correct solution only. There must be no errors in this part of the question to obtain this mark isw ignore subsequent working awrt answers which round to SC: special case oe or equivalent (and appropriate) d or dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper or ag- answer given or d The second mark is dependent on gaining the first mark. All A marks are correct answer only (cao.), unless shown, for example, as A ft to indicate that previous wrong working is to be followed through. After a misread however, the subsequent A marks affected are treated as A ft, but manifestly absurd answers should never be awarded A marks.

5 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 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 the attempts and score the highest single attempt. 7. Ignore wrong working or incorrect statements following a correct answer.

6 General Principles for Core Mathematics Marking (But note that specific mark schemes may sometimes override these general principles). Method mark for solving term quadratic:. Factorisation ( x bx c) ( x p)( x q), where pq c, leading to x = ( ax bx c) ( mx p)( nx q), where pq c and mn a, leading to x =. Formula Attempt to use correct formula (with values for a, b and c).. Completing the square Solving x bx c 0 : b ( x ) q c, q 0, leading to x = Method marks for differentiation and integration:. Differentiation Power of at least one term decreased by. ( x n x n ). Integration Power of at least one term increased by. ( x n x n ) 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 quoting 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 reports 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 will 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.

7 Question Number Scheme y 7 at point P B x d y ( x) (x) ( x) y d ( x) x ( x) MA Marks dy Sub x = into M y 7 xy 7 0 x cso MA cso (6 marks) B For seeing y 7 when x. This may be awarded if embedded within an equation. M Application of Quotient rule. If the rule is quoted it must be correct. It may be implied by their, '.., vu ' uv ' u x u v x, v'.. followed by their v If the rule is neither stated nor implied only accept expressions of the form A ( x) A(x) B( x) AB, 0 ( x ) condoning missing brackets Alternatively applies the Product rule to x x It may be implied by their If the rule is quoted it must be correct. uor v x, u', vor u x, v' followed by their vu ' uv ' If the rule is neither stated nor implied only accept expressions of the form Ax Bxx If they use partial fractions expect to see x P Q dy R S y y ( P, Q ) ( ) d ( x) x ( x) x ( x) ( x) You may also see implicit differentiation etc where the scheme is easily applied. A correct (unsimplified) form of the derivative. Accept from the quotient rule versions equivalent to d y ( x) (x) ( x) ( x ) dy Accept from the product rule versions equivalent to x xx dy Accept from partial fractions x 8x or dy ( x) y( x) from implicit differentiation FYI: Correct simplified expressions are dy x x x or d x ( x) ( x) M Sub x = into what they believe is their derivative to find a numerical value of d y. M Uses x = and their numerical value of y with their numerical at x = to form an equation of dy a normal. If the form ymxcis used then it must be a full method reaching a value for c. A Correct solution only Accept A( x y 7) 0where A N. from correct working. Watch for correct answers coming from incorrect versions of d y with eg. ( x ) on the denominator

8 Question Number Scheme Marks cos5 sin sin sin 0 MA (sin )(sin ) 0 sin M awrt 0.5,.889 (dp) A,A cso (5 marks) M A M Uses cos sin to get a quadratic equation in just sin. If candidate uses cos cos sin or cos they must use cos sin to form a quadratic equation in just sin before scoring the M. (sin sin) 0. The = 0 may be implied by subsequent working Solves their TQ in sin with usual rules by factorisation, formula or completing the square. They must proceed as far as sin.. Accept an answer from a calculator. You may have to pick up a calculator to check their values. A Either of awrt 0.5,.89 (dp) in radians or either of awrt.5,65.5 (dp) in degrees Accept either of awrt 0.08,0.9 A Correct solution with only two solutions awrt 0.5,.889 (dp) within the given range. Accept equivalents such as awrt 0.080,0.996 Ignore any extra answers outside the range. Note that incorrect factorisation (sin )(sin ) 0 would lead to correct answers. As this mark is cso, it would be withheld in such circumstances.

9 Question Number (a) Scheme V shape just in Quad and correct position B Marks Meets/cuts y axis at (0,8) B Meets x axis at (,0) B () (b) x B x 5 (8 x) x MA () (c) fg(5) f () MA () (d) 5 f'( x) x minatx min MA Maximum value = 5 B 5 f( x) 5 A () ( marks)

10 (a) B Accept a V shape just in quadrant one with the left hand end meeting the y - axis, the minimum point on the x - axis and the right hand section being at least as high as the left hand section. Look for either or shape just in quadrant one. Don t accept a curved base. B The graph meets or cuts the y axis at (0, 8) only. Allow just 8 and condone (8,0) written on the correct axis. There needs to be a graph for this to be awarded B The graph meets the x axis at (, 0) only. Allow and condone (0,) written on the correct axis There needs to be a graph for this to be awarded. (b) B For stating that x M For an attempt at the second solution. Accept x 5 (8 x) x... or ( x 5) 8 x x... or equivalent Do NOT condone invisible brackets in this case Accept x 5 (8 x) x.. A x and no other solutions (apart from x = ). Accept this for both marks as long as no incorrect working is seen. Eg x58x xis M0 A0 (c) M A (d) M A Scored for a full method to find fg(5). Accept x 5 being substituted into 8 x and the result being substituted into x Accept an attempt to substitute x 5 into 8 x 8 x Accept for an attempt at f()but not f( ) - only. Accept this for both marks as long as no incorrect working is seen. x An acceptable method of finding a turning point. A full method using calculus or a full method by completion of the square is acceptable. The y value must be attempted. Using calculus look for f'( x) axb 0 x.. followed by an attempt to find y. Using completing the square look for followed by a statement that For achieving the minimum value of y x 5 y. Award for y.5 following the M mark B For achieving the maximum value of y =5. This may be scored from an inequality. Accept... f 5 and even... f<5 A CSO Allow.5,5 and y.5- and y 5 Do not allow y.5- or y 5or [.5,5) or 5 f( x ) 5 Special case: Allow the answer from a graphical calculator as long as it is given with the evidence of a correct sketch. Allow from a table as long as the value at.5 is calculated. Score / Just the (correct) answer, no working, special case award,,0,0

11 Question Number Scheme x sin cos B d M cos (d ) x sin Marks M sec (d ) OR (d ) cos tan dma Uses limits 0 and in their integrated expression tan 0 MA (7 marks) B States either d x cos or d cos d d x. Condone x ' cos M Attempt to produce integral in just θ by substituting x sin and using Acos (d ) You may condone a missing d M Uses sin cos and simplifies integral to C sec ( d ) C or (d ) cos Again you may condone a missing d dm Dependent upon previous M for sec A tan c. No requirement for the +c M tan Changes limits in x to limits in of 0 and, then subtracts their integrated expression either way around. The subtraction of 0 can be implied if f (0) =0. If the candidate changes the limits to 0 and 60 (degrees) it scores M0, A0. Alternatively they could attempt to change their integrated expression in back to a function in x and use the original limits. Such a method would require x x seeing either cos or tan x A.

12 Question Number 5 (a) Scheme x! x x... x x x MA Marks (b) Sub x ( x)( x x... x M 5x 6x A* x into both sides of x 5x 6x M 0 x 0 5 oe 5 dm Accept 00 A 0 59 () () (7 marks)

13 (a) M Uses the correct form of the binomial expansion with n and ' x ' x to achieve x x You may condone missing/invisible brackets. Candidates cannot just write down the answer x x. There must be an intermediate line showing some working for at least the x term. A Correct (unsimplified expression) x x Condone poor notation such as x for x if it is subsequently corrected Evidence of this could be 0.5x x x x... M An attempt to multiply their 'quadratic' binomial expansion by ( x ). Look for at least terms. If they have simplified their binomial expansion to x x. then it is possible to write out the final answer of 5x 6x from ( x)( x x ) This is acceptable for the final MAonly if the quadratic expansion x x has been simplified from an intermediate line. A* 5x 6x Correct solution only. This is a given answer and all aspects must be correct including bracketing. (b) M Sub x into both sides of the given expression. Condone missing brackets. 0 Accept for this any equivalent to dm For an attempt to simplify both sides of the expression resulting in an expression involving 0 a 0 c Look for an equation of the from or equivalent where a, b, c and d are integers b d Sight of 0, on the left hand side and 5 on the right hand side an example of correct work An alternative would be 5 Accept mixed numbers for fractions such as A Accept 0 or by using the rationalised form

14 Question Number 6(i) (ii) Scheme Marks x tan y 8tanysec y oe MA dy dy, d 8tan sec 8tan ( tan M,MA x y y y y) 8 x( x) 8x 0.5 x.5 (5) dv B,B Uses d d d dv, V x x V V x M cm s MA x x (5) (0 marks) (i) M Ctanysec y Differentiates tan yto get an expression equivalent to the form You may see tan y Asec y tan y B sec y from the product rule or versions appearing 0.5 x tan yax... Bsec y or 0.5 from Ax Bsec y... sin y cos y Asin ycos y sin ybcos ysin y x cos y dy cos y from the quotient rule A Any fully correct answer, or equivalent, including the left hand side. tany sec y dy dy Also accept the equivalent by implicit differentiation 8tanysec y M Uses d y d x d x Follow through on their d x d y dy. Condone issues with reciprocating the 8 but not the trigonometrical terms. If implicit differentiation is used it is scored for writing d y as the subject. M Uses sec ytan ywhere x tan y to get their expression for d y or d x in terms of just x. dy x If they use other functions it is for using sin y and cos y where x tan y to x x get their expression for d y or d x in terms of just x. dy A Correct answer and solution. Accept, or A=8, p =0.5and q = x x 8 x x Candidates do not have to explicitly state the values of A, p and q. Remember to isw after the sight of an acceptable answer.

15 dy x x x Alt (i) using y arctan M Changes the subject of the formula to get arctan A Achieves y and proceeds to M M A y D x and proceeds to dy arctan x... x dy Correctly proceeds to E x x dy Writes x and multiplies out bracket to get E x x x Correct answer and solution. Accept, x x 8 x x dy x... (ii) B States or uses d V. It may be awarded if embedded within the chain rule and assigned to d V dv B States or uses x. It may be awarded if embedded within the chain rule and assigned to d V You may also see x V V dv Accept any variable, for example s, l, a in place of x. M Uses a correct chain rule, eg. d V d V d x with d V and their value of d V OR d x dv You may see different versions of this. Eg dv dv M Substitutes x= into their chain rule to find a numerical value for d x... Accept a substitution of V 6 V to find a numerical value for d x.. dv Condone poor notation for d x and the appearance of an answer from the substitution of x = into an incorrect chain rule expression will be sufficient to award this mark. A cso cm s. Accept awrt 0.07 x

16 Question Number Scheme Marks 7(a) cos( x 0) sin( x 0) (cos xcos0 sin xsin 0 ) sin xcos0 cos xsin 0 MA cos0tanxsin 0 tan xcos0sin0 sin 0, cos0 B tan x tan x dma* (b) tan( +0) M +0=50., 0... dm 0., 0. A,A () (9 marks) (5)

17 (a) M Uses identities for cos( A B) and sin( A B) with A x, B 0. Condone missing bracket and incorrect signs but the terms must be correct. A Fully correct equation in sinx and cosx B Replaces sin 0 by and cos0 by throughout their expanded equation. If candidate divides by cos0 it will be for tan 0 or equivalent dm Either for collecting terms in sinx and cosx to reach (...) sin x (...) cos x, before then dividing by cosx to reach sin x (...) cos x or (...) tan x. (...) (...) Alternatively, by dividing by cosx first, producing an equation in tanx, then collecting terms reaching (...) tan x (...) An intermediate line must be seen. ( )tan x tan x is dm0 Similarly ( )sin x ( ) cos x tan x is dm0 A* Reaches final answer by showing rationalisation with no errors. Accept as a minimium tan x tan x (b) M For using part (a) to produce (or imply) an equation tan( ) Condone 0 and being replaced by x. dm Dependent upon the previous M. Score for an attempt at the correct method to find one value of θ invtan Look for tan( ) A One correct answer awrt dp 0. or 0. A Both 0. and 0.awrt dp and no other solutions within the given range. Ignore extra solutions outside the given range. An otherwise case for students starting again in part (b). M Expands both sides (see part a) using correct identities, divides by cos and proceeds to an equation of the form tan... cos 0 sin 0 Note that the correct answer is tan 0.89 sin0 cos0 dm Uses correct order of operations from tan... arctan... to find at least one solution AA Follows Correct answers without working scores B, B

18 Question Number Scheme 8 (a) 000 V 000 B,B (b) dv dv 0.t 0.t e e M t e 0000.e awrt( )6 MA 0.t 0.t (c) e 000e t 0.t 0 9e e 7 0.t 0.t 0 (9e 7)(e ) MA 9e 7 t 0ln 7 0.t 9 Marks () () oe dma () (9 marks) (a) B Accept either boundary: V 000 or V 000 or Vmax 000 for the upper boundary and V 000 or V 000 or Vmin000 for the lower boundary. Answers like V 000 are B0 B Completely correct solution. Accept 000 V 000, 000 Range or y 000, (000,000], V 000 and V 000 (b) M dv 0.t 0.t Score for a Ae Be, where A 8000, B 000 M Sub t 0into a d V 0.t 0.t of the form Ae Be where A 8000, B 000 Condone substitution of t 0into a d V of the form A 0.t B 0.t A B dv 0.t 0.t A Correct solution and answer only. Accept 6 following correct 600e 00e Watch for students who subt 0into their V first and then differentiate. This is 0,0,0. dv 0.t 0.t Watch for students who achieve +6 following 600e 00e. This is,,0 A correct answer with no working can score all marks. (c) M Setting up TQ in 0. e t 0.t e t (9e 7)(e ) e e , 000 AND correct attempt to factorise or solve by the formula. For this to be scored the term must be the x term. t 0.t 0. t A Correct factors or (7e 9)(e ) 0.t 7 or a root e 9 dm Dependent upon the previous M. 0.t This is scored for setting the ae b 0 9 A t 0ln 7. Accept alternatives such as t ln, ln, 0ln If any extra solutions are given withhold this mark. and proceeding using correct ln work to t...

19 Question Number 9 (a) Scheme Marks, Area of R = yd x (d t) t t B, M ( t ) Correct proof with limits and no errors Area = d t ( t ) t A* (b) A B C t ( t) t t ( t) Att ( ) Bt ( ) Ct A B t ( t) t ( t) or Sub t0b Sub t C Compare t A+C=0 A MA B () d t ( t ) t = d ln ln( ) ( ) t t t t t t t MA ln ln5 ln ln 5 ln 9 dma (c) x Sub te into y t y, ( x ln) x e MA (7) () ( marks) (a) B States or implies d x. Accept t t You may award this if embedded within an integral before the final answer is given For example accept Area = y t t

20 M States and uses Area = y functions of t. with the y, the and the sign and replaces both y and by Alternatively states and uses Area = y with the y, the d x and the sign and replaces both y and d x by functions of t. There is no need for limits and you can award even if there is a lack of a A* Correct proof with no errors or omissions on any line for the integrand and there must be a in all integrals in t. The limits need only be correct on the final line and they may have just been written in. The two separate fractions must be combined into a single fraction (b) A B C B Scored for use of partial fractions. Accept the correct form but also t ( t ) t t ( t ) A B award for the form t ( t) t ( t) M A For Substitute values of t and/or use inspection to determine A, B and C from a form equivalent to ( ) ( ) =. t ( t ) t t ( t ) At t Bt Ct. The partial fraction must be of the correct form A C.. M d t..ln( t) t ( t ) t A B Note this can be scored from an incorrect assumption that t ( t) t ( t) A d t ( t ) t ln t ln( t) ( c) t There is no need to consider limits. dm Dependent upon previous M. Sub in limits, subtracts either way around and uses a correct log law at least once to get expression of the form a lnb. A 5 Correct solution only ln 9 (c) M x Rearranges x ln( t ) to reach t e and sub in y t to get y in terms of x Alternatively substitutes t or equivalent into x ln( t ) and attempts to rearrange to y =.. y A y. Remember to isw. x e You can ignore any reference to the domain, x ln, for this mark.

21 Question Number 0(a) Scheme ln d ln y x x, x y x x x x Marks MA, B xlnx x 0xlnx6 x.. dm x 6 x A* lnx (5) x (b) ln.7 6 awrt.7 MA x = awrt.7 and x =awrt.7 A () (c) A= (., 0.9) M A () (0 marks)

22 (a) M x ln x x ln x Applying the product rule to x ln x or multiples of it such as and even If the rule is quoted it must be correct. It may be implied by, for example, x u, v ln x, u'.., v'.. followed by their vu ' uv ' If it is not quoted nor implied only accept expressions of the form Axln x Bx x A ln ln A correct (unsimplified) derivative for x x x x B The derivative of the x term is seen or implied to be dm Dependent upon the previous M being scored. It is for setting their d y 0, taking out a common factor of x and proceeding to x.. Alternatively they could state that d y 0 and write out a line xlnx x from their derivative equivalent to x 6 A* Correct solution only x lnx. Note that this is a given answer. All aspects need to be correct. may just be stated ln x lnx dy Note: If the candidate multiplies by to get y x lnx6x xlnxx 6 before setting dy 0 they can score all marks if they proceed to the given answer. If they multiply by and leave the subject as y (or perhaps ignore the lhs) they can score a special case 0 0 for 6 6 x ln x x 6 0 x(ln x ) 6 x x ( ln x ) (ln x ) (b) M 6 Attempts ln.7. Awrt.7 implies this method A x awrt.7. The subscript is not important. Mark as the first value given. A x =awrt.7(dp) and x = awrt.7 (dp) (c) M Deduces the x coordinate of A is.. The sight of. is sufficient to award this as long as their values in (b) round to this. Alternatively uses their (rounded) answer from part (b) and substitutes it into equation for y to find the y coordinate of A. In a similar way to that of the x coordinate, the sight of 0.9 would be sufficient evidence for this award. A (., 0.9). Accept x =., y =0.9

23 Question Number (a) (b) Scheme Marks q. q 0 q MA* () Equate the y and z coordinates p M Full method to find either or dm () subbed into () 8 Sub into () Aeither Sub values back into x coordinates 5 p p 5 ddm A (c) (d ) ' 5' Point of intersection is 6 5 OR 7 6 AX OB OAAX 0 OB 0 OB 5 = (,,7) M,A M either AA (5) () () ( marks) (a) M Attempts to find a solution for q by setting the scalar product of the direction vectors = 0 Condone one sign error in q 0 leading to q.. Alternatively set q and attempt the scalar product A* q. This is a given answer. In the alternative, there must be a statement (=0) and a conclusion, (hence true/hence perpendicular) (b)

24 M Equate y and z coordinates. Condone sign errors dm Dependent upon the previous M. Scored for a full method to find either or A Either 5or ddm Dependent upon both previous M s. Either uses both of their values for and in the equation for the x coordinates in order to find a numerical value for p. Condone sign slips. Look for p with their and leading to a solution of p. Alternatively uses the value of one variable, expresses the other variable in terms of this and substitutes both in the equation for the x coordinates in order to find a numerical value for p. A p 5 (c) M Uses their value of in l, or their values of, p and qin l to find the coordinate of X A Coordinates of X (,,7). Accept in vector form OX 7 (d) 6 '' M Uses either OB OA AX AX ' ' with an attempt to find one possible value of '7' vector B Or uses mid points. Sight of ( abc,, ) appearing from solving 6 a b c,, 7 Or attempts to find the value of one value of and uses it correctly to find one position of B You may see B A A X or a version of ( 6) ( 6 ) ( ) (6 ' ') ( ' ) ( 7).. followed by being substituted into line l A more simple version of this could be using a diagram and deducing that by being substituted into line l or6 followed l. X (,,7) 5. A (6,,) l A 0 Gives one possible vector OB 0 ikor OB 0i j 5k 5 A Gives both possible position vectors or coordinates

25 Question Scheme Number (a) 0.9 exactly B (b) Strip wih =0.5 B 0.5 M =.9 A Area.958 (.0 '0.9' Marks () () (c) x ln x x x x = ln x d x, x x MA, B 9 9 x x x = ln 9 7 x x x A Area = x x x x x ln ln dm 6 ln 7 7 A (6) real approx (d) % error = 00 =Accept awrt ±.6% MA real (e) Increase the number of strips B () () ( marks) (a) B (b) B M A 0.9 exactly either in the table or within the trapezium rule in part (b) Uses a strip wih of 0.5 or equivalent. Uses the correct form of the trapezium rule, a form of which appears in the formula booklet.... Look for.958 (.0 their Accept for this the sum of four trapezia Awrt.9 (dp)

26 (c) M Uses integration by parts, the correct way around. ln x x x ln x Accept integration on either x ln x or multiples like or even qx Accept as evidence an expression of the form px ln x ( d) x x x x A Using integration by parts to arrive at an intermediate form ln x where k most k k x likely will be a multiple of. B Integrates the x to x x ( c). Ignore any constants. Watch for candidates who take out a common factor.. 6 x to... x x ( c ) ln x x x x A The correct integral for Accept equivalent expressions to ln x ( c) 9 7 This is independent of the integral for the x term. dm A Dependent upon the M mark it is for substituting in both x= and x= and subtracting (either way around). 6 Correct solution only ln 7. The answer must be in this form and ln ln7 7 (d) M Uses their answer obtained by integration in part c and their answer obtained by the trapezium rule in part b and calculates c b c A Accept awrt ±.6%. (e) B Makes a reference to increasing the number of strips. Accept decrease the wih of the strips, use more trapezia Be generous with statements like more values or strips as the intention is clear. Also accept more x s, more y s but don t accept use more decimal places.

27 Question Number Scheme (a) R 09 B tan awrt6.70 MA 0 (b)(i) Max height = + 09 =. m MA (ii) Occurs when 0 t'6.70' 80 t 5. MA Marks () (c) cos(0t 6.70) cos(0t 6.70) ( ) MA 09 () 6 0t6.70 arccos t.. 09 dm t awrt.6(dp) A () (d) Attempting 0t 60 t.. or 0t 70 t.. M revolutions in minutes A () ( marks)

28 (a) B Accept R 09. Remember to isw after a correct answer. Eg R M 0 0 For tan or tan. If R is used to find, only accept cos or sin 0 ' R ' ' R ' A awrt6.70(dp). Condone 6.7 Note that the answer of awrt 0.9radians scores A0. (b)(i) M For +their R A Awrt. m. Accept + 09 (b) (ii) M For arriving at a solution for t from 0 t '6.70' 80 t.. If radians were used in part a then accept 0 t '0.9' t.. A t awrt 5. only. If multiple solutions are found, 5. must be referred to as their chosen solution Answers from calculus will be rare. They can be scored as follows. From the original function: For (b)(ii) M d H 'arctan t ' t.. 0 A t awrt 5. only (b)(i) M Sub their t awrt 5. obtained from d H 0 A Awrt. m From the adapted function: For (b)(ii) M dh 0 0 t '6.70' 80 t.. A t awrt 5. only (b)(i) M Sub their t awrt 5. obtained from d H 0 A Awrt. m (c) M Attempts to substitute H 8 into H 0cos0tsin 0tand use their answer to part (a) to proceed to cos(0 ttheir'6.70')... 6 A cos(0 t their'6.70') or awrt It may be implied by 0 t'6.70' awrt5 09 dm Dependent upon previous M. Score for cos(0 t '6.70')... t.. The cos(..) must be negative, the order of operations must be seen to be correct with the invcos being attempted first and the second quadrant must be chosen for their calculation. A t.6. The answer with no incorrect working scores all marks. If multiple solutions are found,.6 must be referred to as their chosen solution (d) M Attempting 0t 60 t.. or 0t 70 t.. A minutes. Both and minutes are required.

29

30 Pearson Education Limited. Registered company number 8788 with its registered office at Edinburgh Gate, Harlow, Essex CM0 JE