Mark Scheme (Results) Summer 04 Pearson Edecel GCE in Core Mathematics 3 (6665_0)
Edecel and BTEC Qualifications Edecel 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 www.edecel.com or www.btec.co.uk. Alternatively, you can get in touch with us using the details on our contact us page at www.edecel.com/contactus. 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: www.pearson.com/uk Summer 04 Publications Code UA03846 All the material in this publication is copyright Pearson Education Ltd 04
General Marking Guidance All candidates must receive the same treatment. Eaminers must mark the first candidate in eactly 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. Eaminers 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. Eaminers should always award full marks if deserved, i.e. if the answer matches the mark scheme. Eaminers 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 eemplification may be limited. When eaminers 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.
PEARSON EDEXCEL GCE MATHEMATICS General Instructions for Marking. The total number of marks for the paper is 75. The Edecel 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. 3. 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 4. All A marks are correct answer only (cao.), unless shown, for eample, 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. 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.
General Principles for Core Mathematics Marking (But note that specific mark schemes may sometimes override these general principles). Method mark for solving 3 term quadratic:. Factorisation ( + b + c) ( + p)( + q), where pq c, leading to ( a + b + c) ( m + p)( n + q), where pq c and mn a, leading to. Formula Attempt to use the correct formula (with values for a, b and c). 3. Completing the square b Solving + b + c 0 : ± ± q ± c 0, q 0, leading to Method marks for differentiation and integration:. Differentiation Power of at least one term decreased by. ( n n ). Integration Power of at least one term increased by. ( n n+ )
Use of a formula Where a method involves using a formula that has been learnt, the advice given in recent eaminers 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 small errors 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. Eact answers Eaminers reports have emphasised that where, for eample, an eact answer is asked for, or working with surds is clearly required, marks will normally be lost if the candidate resorts to using rounded decimals.
Question Number Scheme Marks.(a) 4 + f( ), > vu ' uv' ( ) 4 (4+ ) Applies to get v ( ) MA 9 ( ) A* (b) 9 ( ).. M (5,7) A,A (3) (3) 6 marks Alt.(a) 4 + 9 f( ) 4+ Applies chain rule to get M f ' ( ) A ( ) 9( ) 9 ( ) A, A* (3)
(a) M A 4 + Applies the quotient rule to f( ) with u 4 + and v. If the rule is quoted it must be vu ' uv' correct. It may be implied by their u 4+, v, u'.., v'.. followed by. v ( ) A (4+ ) B If neither quoted nor implied only accept epressions of the form, 0 AB > ( ) allowing for a sign slip inside the brackets. Condone missing brackets for the method mark but not the final answer mark. Alternatively they could apply the product rule with u 4+ and v ( ). If the rule is quoted it must be correct. It may be implied by their ( ) u 4+, v, u'.., v'.. followed by vu ' + uv'. If it is neither quoted nor implied only accept epressions of the form/ or equivalent to the form ( ) C+ (4+ ) D( ) A third alternative is to use the Chain rule. For this to score there must have been some attempt to 4 +.. divide first to achieve f ( ).. + before applying the chain rule to get f ' ( ) A ( ) A correct and unsimplified form of the answer. ( ) 4 (4+ ) Accept from the quotient rule ( ) 4 8 4 Accept from the quotient rule even if the brackets were missing in line ( ) Accept ( ) 4 + (4 + ) ( ) or equivalent from the product rule Accept 9 ( ) from the chain rule A* 9 Proceeds to achieve the given answer. Accept 9( ) ( ) All aspects must be correct including the bracketing. If they differentiated using the product rule the intermediate lines must be seen. 4 4 4( ) (4 ) 9 Eg. ( ) 4 + (4 + ) ( ) + + ( ) ( ) ( ) ( ) (b) M 9 Sets ( ) and proceeds to. The minimum epectation is that they multiply by ( ) and then either, divide by - before square rooting or multiply out before solving a 3TQ equation. 9 A correct answer of 5 would also score this mark following as long as no incorrect ( ) work is seen. A 5 A (5, 7) or 5, y 7. Ignore any reference to - (and y ). Do not accept /3 for 7 If there is an etra solution,, then withhold this final mark.
Question Number Scheme 5.(a) ln(+ ) 0 0 ln(+ ) 5 + e.. M (b) Alt (b) Alt (b) Using logs Alt 3 (b) Using 3 5 e A 3 e e ln(3 e ) ln e 4 7 4 7 Marks 4 7 ln 3 + ln e ln e ln 3 + 4ln e 7ln e M,M (ln3 + 4) 7... dm 7 (ln 3 + 4) oe A (4) 6 marks 7 4 7 e 3e e 3 4 e 7 4 3 e ln 3 7 4 ln e M,M ( ) (ln3 + 4) 7... dm 7 (ln 3 + 4) A 3 e e log(3 e ) log e 4 7 4 7 4 7 3e e 3 + + M, M 4 7 log3 log e log e log3 4 log e 7log e (log3+ 4log e) 7log e... dm 7log e (log 3 + 4log e) A e e 7 4 log ( ) Alt 4 (b) Using ln 3 3 e e e e e 4 7 ln3 4 7 7 4 3 e 7 4 log e M,M 3 3 3 ( + 4log e) 7log e... dm 7 log3 e (+ 4 log e) 3 ln3+ 4 7 e e, ln 3 + 4 7 M,M 3 e (ln3 + 4) 7... 7 (ln 3 + 4) () (4) (4) A (4) dm A (4)
(a) M Proceeds from ln(+ ) 0 0 to ln(+ ) 5 before taking ep's to achieve in terms of A Accept for M 5 ln(+ ) 0 0 ln(+ ) 5 f (e ) Alternatively they could use the power law before taking ep's to achieve in terms of ( + ) ( + ) ( + ) 0 ( 0 ) ln 0 ln 0 e g e 5 5 e e cso. Accept or other eact simplified alternatives such as. Remember to isw. The decimal answer of 73.7 will score MA0 unless the eact answer has also been given. The answer 0 e ± e 5 does not score this mark unless simplified. is MA0 (b) M Takes ln s or logs of both sides and applies the addition law. 4 4 4 ln(3 e ) ln 3 + ln e or ln(3 e ) ln 3 + 4 is evidence for the addition law 4 If the e 7 4 was moved over to the right hand side score for either e or the subtraction law. 7 e 7 4 ln ln e ln e 4 e or 7 4 7 e 7 4 3e e 3 3 e is evidence of the subtraction law 4 e 4 7 M Uses the power law of logs (seen at least once in a term with as the inde Eg,e or e 4 3 ). 4 7 ln 3 +lne ln e ln 3 + 4ln e 7 ln e is an eample after the addition law 7 4 3 e log 3 (7 4 ) log e is an eample after the subtraction law. It is possible to score M0M by applying the power law after an incorrect addition/subtraction law 4 7 4 7 For eample 3 e e ln(3 ) ln(e ) ln e ln 3 4ln e 7 ln e dm This is dependent upon both previous M s. Collects/factorises out term in and proceeds to. Condone sign slips for this mark. An unsimplified answer can score this mark. 7 A If the candidate has taken ln s then they must use ln e and achieve or equivalent. (ln 3 + 4) If the candidate has taken log s they must be writing log as oppose to ln and achieve 7log e 7 log e or other eact equivalents such as. 4 (log 3 + 4log e) log 3e 0 e 5 e
Question Number Scheme Marks 3.(a) 8 tan 8 8 B* (b) d 8tan y 6ysec ( y) dy MAA () d dy 8 8 8 At P 8 tan + 6 sec { 8 + 4 } M y 8, 8+ 4 8 accept y 0.049( ) MA (8 + 4 ) y + A (7) (8 marks) (a) B* Either sub y into 8ytan ( y) 8 tan 8 8 8 Or sub, y into 8ytan ( y) 8 tan 8 8 8 This is a proof and therefore an epectation that at least one intermediate line must be seen, including a term in tangent. Accept as a minimum y 8 tan 4 Or tan 4 This is a given answer however, and as such there can be no errors. (b)
M Applies the product rule to 8ytan yachieving Atan y + Bysec ( y) A One term correct. Either 8tan y or + 6ysec ( y). There is no requirement for d dy d A Both lhs and rhs correct. 8tan y 6ysec ( y) dy + It is an intermediate line and the epression does not need to be simplified. d Accept tan y 8 8y sec ( y) dy + or dy or using implicit d tan y 8 + 8y sec ( y) dy dy differentiation tan y 8 + 8y sec ( y) d d M For fully substituting y 8 into their d d y or d y to find a 'numerical' value d Accept d awrt 0.6 dy or d y awrt 0.05 d as evidence M For a correct attempt at an equation of the tangent at the point (, 8 ). The gradient must be an inverted numerical value of their d dy y Look for 8, d numerical d y Watch for negative reciprocals which is M0 If the form y m + c is used it must be a full method to find a numerical value to c. A A correct equation of the tangent. y Accept 8 or if y m + c is used accept m 8+ 4 8 + 4 and Watch for answers like this which are correct ( 8 4) y 8 c 8 8 + 4 A y 0.39 Accept the decimal answers awrt sf y 0.049+ 0.4, awrt sf y + 4.9, 0.049 3. Accept a miture of decimals and s ' for eample 0. 6 y 8 Correct answer and solution only. (8 + 4 ) y + Accept eact alternatives such as 4( + ) y + 0.5 and because the question does not ask for a and b to be simplified in the form ay + b, accept versions like (8+ 4 ) y + ( 8+ 4) and (8+ 4 ) y + ( 8+ 4) 8 8 8+ 4
Question Number Scheme Marks 4.(a) W Shape (0, ) and (6, ) B B () (b) V shape (-6,) (0,5) B B B (3) (c ) One of a or b 6 B a and b 6 B (7 marks) ()
(a) B A W shape in any position. The arms of the W do not need to be symmetrical but the two bottom points must appear to be at the same height. Do not accept rounded W s. A correct sketch of y f ( ) would score this mark. B A W shape in quadrants and sitting on the ais with P ' (0,) and Q ' (6,). It is not necessary to see them labelled. Accept being marked on the y ais for P'. Condone P ' (,0) marked on the correct ais, but Q ' (, 6) is B0 (b) B Score for a V shape in any position on the grid. The arms of the V do not need to be symmetrical. Do not accept rounded or upside down V s for this mark. B Q ' ( 6,). It does not need to be labelled but it must correspond to the minimum point on the curve and be in the correct quadrant. B P ' (0,5). It does not need to be labelled but it must correspond to the y intercept and the line must cross the ais. Accept 5 marked on the correct ais. Condone P ' (5,0) marked on the positive y ais. Special case: A candidate who mistakenly sketches y f ( ) + 3 or y f ( ) + 3 will arrive at one of the following. They can be awarded SC BB0B0 (6,5) ( 6,5) (0, 9) (0, 9) (c) B Either states a or b 6. This can be implied (if there are no stated answers given) by the candidate writing that y.. 6 or y... If they are both stated and written, the stated answer takes precedence. B States both a and b 6 This can be implied by the candidate stating that y 6 If they are both stated and written, the stated answer takes precedence.
Question Number Scheme Marks 5.(a) + 6 ( + 3)( ) B 3( + ) ( ) + 3( + ) + M + 3 ( + 3)( ) ( + 3)( ) + 4+ 3 ( + 3)( ) ( + 3)( + ) ( + 3)( ) ( + ) ( ) A cso A* (b) One end either ( y),( y) or ( y) 4,( y) 4 B < y < 4 B (c ) Attempt to set Either g ( ) or ( + ) ( ) g ( ) g ( ) or + + g ( ) or + g ( ) + + + M (4) () 3 0... A, dm 3+ 3 a oe (.5 + 3.5 ) cso A (4) (0 marks)
(a) B M A + 6 ( + 3)( ) This can occur anywhere in the solution. For combining the two fractions with a common denominator. The denominator must be correct for their fractions and at least one numerator must have been adapted. Accept as separate fractions. Condone missing brackets. 3(+ ) ( + 6) + 3(+ )( + 3) Accept + + 3 + 6 ( + 3)( + 6) 3( + ) 3( + ) Condone + + + 3 ( + 3)( ) ( + 3)( ) ( + 3)( ) simplified quadratic A correct intermediate form of simplified quadratic Accept 3 + +, + +, OR + + + + + + ( + 3)( ) + 6 ( + 3)( + 6) ( + 3)( + 6) 4 3 4 3 7 5 9 ( )( 3)( 3) As in question one they can score this mark having 'invisible' brackets on line. A* Further factorises and cancels (which may be implied) to complete the proof to reach the given ( + ) answer. All aspects including bracketing must be correct. If a cubic is formed then it needs ( ) to be correct. (b) B States either end of the range. Accept either y< 4, y 4 or y >, y with or without the y's. B Correct range. Accept < y < 4, < g < 4, y and y 4, <, Do not accept < < 4, y 4, 4 etc. Special case, allow BB0 for < < 4 (c) M Attempting to set g ( ), g ( ) or A dm A > < g ( ) g ( ) or 4,, < Range < 4, even < f < 4, g ( ). If g ( ) has been used then a full attempt must have been made to make the subject of the formula. A full attempt would involve cross multiplying, collecting terms, factorising and ending with division. As a result, it must be in the form g ( ) ± ± ± ± + Accept as evidence ( + ) + ± ± OR ( ) ± ± OR ± ± + OR ± ± + 3 0or eact equivalent. The 0 may be implied by subsequent work. For solving a 3TQ0. It is dependent upon the first M being scored. Do not accept a method using factors unless it clearly factorises. Allow the answer written down awrt 3.30 (from a graphical calculator). 3 + 3 aor. Ignore any reference to 3 3 Withhold this mark if additional values are given for, > 3
Question Number Scheme Marks 6.(a) y. 0.4, y. ( + )0.546 M (b) At R Change of sign Q lies between dy + d sin 3 3 A MA () sin + 3 3 0 + sin 3 cso MA* (4) (c ) +.3sin.3 3 M awrt.84 awrt.76 A () (8 marks)
(a) M Sub both. and. into y and achieve at least one correct to sig fig In radians y. awrt 0. y. awrt/truncating to0.5 In degrees y. awrt 3 y. awrt 4 A Both values correct to sf with a reason and a minimal conclusion. y. awrt 0. y. awrt/truncating to0.5 Accept change of sign, positive and negative, y. y. as reasons and hence root, Q lies between. and., QED as a minimal conclussion. Accept a smaller interval spanning the root of.358, say.3 and.4, but the A can only be scored when the candidate refers back to the question, stating that as root lies between.3 and.4 it lies between. and. (b) dy M Differentiating to get...sin + 3 3where... is a constant, or a d linear function in. dy A sin + 3 3 d M Sets their d y 0 d and proceeds to make the of their 3 the subject of the formula Alternatively they could state d y 0 and write a line such as d sin 3 3, before making the of 3 the subject of the formula A* Correct given solution. + sin 3 Watch for missing 's in their formula (c) M Subs.3 into the iterative formula to find at least. This can be implied by awrt.3(not just.3) or +.3sin.3 3 or awrt.006 (degrees) A Both answers correct (awrt 3 decimal places). The subscripts are not important. Mark as the first and second values seen. awrt.84 awrt.76
Question Number Scheme Marks 7.(a) cos cosec cot sin sin M + cos sin M + cos sin cos cos sin cos M A cos cot sin A* (b) cosec(4θ + 0 ) + cot(4θ + 0 ) 3 cot( θ ±.. ) 3 M θ ±... 30 θ.5 dm, A θ ±... 80 + PV θ.. dm θ 0.5 A (5) (5) (0 marks)
(a) M Writing cosec sin and cos cot or sin tan M Writing the lhs as a single fraction a + b. The denominator must be correct for c their terms. M Uses the appropriate double angle formulae/trig identities to produce a fraction in a form containing no addition or subtraction signs. A form p q s t or similar cos A A correct intermediate line. Accept sin cos or sin cos or similar sin cos tan This cannot be scored if errors have been made A* Completes the proof by cancelling and using either cos cot sin or cot tan The cancelling could be implied by seeing cos cos cot sin cos cos The proof cannot rely on epressions like cot (with missing s) for the sin final A (b) M Attempt to use the solution to part (a) with 4θ + 0 to write or imply cot( θ ±... ) 3 Watch for attempts which start cotα 3. The method mark here is not scored until the α has been replaced by θ ±... Accept a solution from cot( ±... ) 3 where θ has been replaced by another variable. dm Proceeds from the previous method and uses tan.. and cot.. arctan 30 to solve θ ±... 30 θ.. 3 A θ. 5 or eact equivalent. Condone answers such as. 5 dm This mark is for the correct method to find a second solution to θ. It is dependent upon the first M only. Accept θ ±... 80 + PV θ.. A θ 0.5 or eact equivalent. Condone answers such as 0. 5 Ignore any solutions outside the range. This mark is withheld for any etra solutions within the range. If radians appear they could just lose the answer marks. So for eample
Question Number 7.(a)Alt 7.(a)Alt θ ±... ( 0.54 ) θ.. is MdMA0 followed by 6 θ ±... + ' ' θ.. dma0 6 Special case : For candidates in (b) who solve cot(4 θ ±... ) 3 the mark scheme is severe, so we are awarding a special case solution, scoring 000. cot(4 θ + β ) 3 4θ + β 30 θ.. is M0M0A0 where β 5 or0 4θ + β 0 θ.. can score MA Special case. If β 5, θ 5.5 If β 0, θ 50 Special case : Just answers in (b) with no working scores 0 0 0 for.5 and 0.5 BUT cot(θ ± 5 ) 3 θ.5, 0.5 scores all available marks. Scheme Marks cosec cot sin tan ST M tan + sin cos tan tan + ( tan )sin cos tan + ( tan ) sin cos or sin cos tan 4sin cos tan nd M tan + sin cos tan sin cos sin cos tan tan + sin cos tan sin sin cos tan tan ( sin ) + sin cos sin cos tan tan cos + sin cos sin cos tan sin cos + sin cos sin cos tan sin cos oe sin cos tan 3 rd MA cot tan A* (5) Eample of how main scheme could work in a roundabout route cosec+ cot cot + sin tan tan st M tan tan + sin tan sin tan nd M tan sin tan tan + sin cos sin cos cos tan tan
Question Number Scheme Marks sin tan 4sin + tan tan ( tan ) tan + sin ( tan ) 4sin tan sin tan sin tan ( sin ) sin 3 rd M tan sin cos A As this is true, initial statement is true A* (5)
Question Number Scheme Marks 8.(a) 0 800e 800 0 + 3e + 3 P, 00 M,A () (b) 800e 50 + 3e 50( + 3e ) 800e 50e 50, e 5 M,A t ln ( 5 ) 0. M t 0 ln(5) A (4) (c ) 800e d P ( + 3e ) 800 0.e 800e 3 0.e P + 3e d t ( + 3e ) M,A At t0 d P ( + 3e) 80e 40e 80e d t (+ 3e) (+ 3e) M,A (4) (d) 800e 800 800 P ma P 66. Hence P cannot be 70 B + 3e e + 3 3 () ( marks) (a) M Sub t 0 into P and use 0 e in at least one of the two cases. Accept as evidence A 00. Accept this for both marks as long as no incorrect working is seen. (b) M 0. to Ae t A 800 P + 3 800e 0. Sub P50 into P, cross multiply, collect terms in e t and proceed + 3e B Condone bracketing issues and slips in arithmetic. 0. If they divide terms by e t you should epect to see Ce D e 5 or e 0.
0. M Dependent upon gaining e t E, for taking ln s of both sides and proceeding to t Accept e E ln E t... It could be implied by t awrt 6. A t 0 ln(5) Accept eact equivalents of this as long as a and b are integers. Eg. t 5ln(5) is fine. (c) M Scored for a full application of the quotient rule and knowing that d e e 0. k and NOT kt e t dt If the rule is quoted it must be correct. It may be implied by their u 800e, v + 3e, u' pe, v' qe vu ' uv' followed by. v If it is neither quoted nor implied only accept epressions of the form ( + 3e ) pe 800e qe (+ 3e ) Condone missing brackets. You may see the chain or product rule applied to d 0. 0. For applying the product rule see question but still insist on e t e t k dt For the chain rule look for 800e 800 dp P 800 0. 0. ( e + 3) 0.e t t + 3e e + 3 dt A A correct unsimplified answer to d P ( + 3e ) 800 0.e 800e 3 0.e d t (+ 3e ) M For substituting t 0 into their d P dt, NOT P Accept numerical answers for this..59 is the numerical value if d P was correct dt dp 80e A d t dp 80e or equivalent such as 80e( + 3e), (+ 3e) dt + 6e + 9e Note that candidates who substitute t 0 before differentiation will score 0 marks (d) B Accept solutions from substituting P70 and showing that you get an unsolvable equation 800e Eg. 70 7 e ln( 7) which has no answers. + 3e 800e Eg. 70 7 e e /e is never negative + 3e
Accept solutions where it implies the ma value is 66.6 or 67. For eample accept sight of 800, with a comment so it cannot reach 70, or a large value 3 of t (t > 99) being substituted in to get 66.6 or 67 with a similar statement, or a graph drawn with an asymptote marked at 66.6 or 67 Do not accept ep's cannot be negative or you cannot ln a negative number without numerical evidence. Look for both a statement and a comment
Question Number Scheme Marks 9.(a) R 0 B 4 tanα α awrt.07 MA (3) (b)(i) '4 + 5 R ' 04 Bft (ii) 3 θ '.07' θ awrt 0.89 MA (3) (c )(i) 4 B (ii) 3 θ '.07' θ awrt.46 MA (3) ( 9 marks)
(a) B Accept R 0 or 5 or awrt 4.47 Do not accept R ± 0 This could be scored in parts (b) or (c) as long as you are certain it is R M 4 4 for sight of tan α ±, tanα ±. Condone sinα 4, cosα tanα 4 4 If R is found first only accept sin α ±, cosα ± R R A α awrt.07. The degrees equivalent 63.4 is A0. If a candidate does all the question in degrees they will lose just this mark. (b)(i) Bft Either 04 or if R was incorrect allow for the numerical value of their '4+ 5 R '. Allow a tolerance of dp on decimal R s. (b)(ii) M Using 3θ ± their '.07' θ.. Accept 3θ ± their '.07 ' (n + ) θ.. where n is an integer Allow slips on the lhs with an etra bracket such as 3( θ ± their '.07 ') θ.. The degree equivalent is acceptable 3θ their '63.4 ' 90 θ Do not allow mied units in this question A awrt 0.89 radians or 5.. Do not allow multiple solutions for this mark. (c)(i) B 4 (c)(ii) M Using 3θ ± their '.07' θ... Accept 3θ ± their '.07 ' n θ.. where n is an integer, including 0 Allow slips on the lhs with an etra bracket such as 3( θ ± their '.07 ') θ.. The degree equivalent is acceptable 3θ their '63.4 ' 360 θ but Do not allow mied units in this question A θ awrt.46 radians or 4. Do not allow multiple solutions for this mark.
Pearson Education Limited. Registered company number 8788 with its registered office at Edinburgh Gate, Harlow, Esse CM0 JE