PhysicsAndMathsTutor.com

Transcription

1

2 GCE Edecel GCE Core Mathematics C (6665) January 006 Mark (Results) Edecel GCE Core Mathematics C (6665)

3 January Core Mathematics C Mark. (a) y Shape unchanged B (, 7) Point B () O (b) y (, 4) Shape B Point B () O (c) (,4) y (,4) Shape B (, 4) B (,4) B () [7] O

4 . ( )( ) + ( + ) ( )( ) ( )( ) + 6 ( + ) 6 ( + )( ) ( )( + ) Alternative method + At any stage B ( )( + ) ( + )( ) ( )( + ) B A A + + A (7) [7] ( )( + ) At any stage B ( + ) appearing as a factor of the numerator at any stage B + 6 ( + )( + ) 6( + ) ( + )( ) ( )( + ) ( + )( )( + ) ( )( ) ( )( )( ) ( + )( )( + ) ( )( ) ( )( )( ) can be implied ( )( ) or or ( )( )( ) Any one linear factor quadratic + + Complete factors + + ( )( )( ) ( )( )( ) A + A (7) + A

5 . dy accept A d At, d y m Use of mm d ( ) y ln y + 9 Accept y 9 A (5) [5] dy leading to y 9+ 7 is a maimum of A0 A0 /5 d (a) (i) ( ) ( (ii) ( ) d e e or e e ) d At any stage B dy + + e + e d Or equivalent A+A (4) d ( cos ) 6 sin ( ) d At any stage A ( ) ( ) dy 8 sin cos d 9 A (4) Alternatively using the product rule for second A ( ) cos( ) y y d ( ) ( ) cos 6 ( ) sin ( ) d Accept equivalent unsimplified forms d y d d dy dy d 8cos y+ 6 (b) 8cos( y + 6) or 8cos( + 6) ( ) dy ( ± ) d 6 8cos arcsin 4 y ( ) A A (5) []

6 5. (a) 0 Dividing equation by 4 + Obtaining + cso A () 4.4,.9,.9 B, B, B If answers given to more than dp, penalise first time then accept awrt above. () (b) (c) Choosing (.95,.95) or a tighter interval ( ) ( ) Change of sign (and continuity) α (.95,.95) f.95 0, f Both, awrt A α.9 to decimal places cso A () [9] 6. (a) Rcosα, Rsinα 4 ( ) R Accept if just written down, awrt.6 A 4 tan α, α 8.4 awrt 8. 4, A(4) (b) 7 cos ( + their α ) ( 0.554) their R + their α 56.4 awrt 56 A, their principal value 8.0, 85. Ignore solutions out of range A, A (5) If answers given to more than dp, penalise first time then accept awrt above. (c)(i) minimum value is 60 ft their R Bft (ii) ( α ) cos + their 6.57 cao A () []

7 7. (a) (i) Use of cos cos sin in an attempt to prove the identity. cos cos sin ( cos sin )( cos + sin ) cos sin cso A () cos + sin cos + sin cos + sin (ii) Use of cos cos in an attempt to prove the identity. Use of si n sin cos in an attempt to prove the identity. ( cos sin ) ( cos sin cos ) cos cos sin cso A () (b) cosθ( cosθ sinθ) Using (a)(i) cos θ cosθsinθ 0 ( θ sin θ) 0 Using (a)(ii) cos θ sin θ A () (c) tan θ π 5π 9π π θ,,, any one correct value of θ A θ π,,, Obtaining at least solutions in range The 4 correct solutions A (4) If decimals ( 0.9,.96,.54,5.05 )or degrees (.5,.5,0.5,9.5 ) are [] given, but all 4 solutions are found, penalise one A mark only. Ignore solutions out of range.

8 8. (a) gf ( ) ( + ) ln e 4 ln e e 4 ln4 e e 4 4e Give mark at this point, cso A (4) 4 Hence gf : a 4e, ( ) (b) y 4 Shape and point B () O + ( ) (c) Range is Accept d gf 6e 4 d (d) ( ) 4 gf > 0, y > 0 B () e 6 A 4 ln A (4) [0]

9 Mark (Results) January 007 GCE GCE Mathematics Core Mathematics C (6665) Edecel Limited. Registered in England and Wales No Registered Office: One90 High Holborn, London WCV 7BH

10 January Core Mathematics C Mark sin θ sin θ + θ sin θ cosθ + cos θsinθ B. (a) ( ) ( ) sinθ cos θ + sin θ sinθ B B sinθ sin θ + sinθ sin θ sinθ 4sin θ cso A (5) 9 (b) sin θ equivalent or eact A () [7]. (a) f ( ) ( ) ( ) ( + ) +, ( + ) ( + ) A, A cso A (4) (b) , > 0 for all values of. A, A () (c) f ( ) ( + ) Numerator is positive from (b) ( ) + > 0 (Denominator is positive) Hence f ( ) > 0 B () [8] Alternative to (b) d ( + + ) A d 4 A parabola with positive coefficient of Accept equivalent arguments has a minimum + + > 0 A ()

11 π π. (a) y sin P C 4 4 Accept equivalent (reversed) arguments. In any method it must be clear π that sin or eact equivalent is used. 4 B () (b) d cos y dy dy cosy d A dy d cosy May be awarded after substitution π dy y 4 d cso A (4) (c) m B π y ( ) 4 A π y A (4) [9] 4. (i) ( 9+ ) ( ) ( + ) ( + ) dy 9 A d 9 9 dy d ± A,,, Final two A marks depend on second M only A, A (6) 6 6 (ii) ( ) dy e e d + A A dy ln ln ln ( + e ) e 4 8 d A (5) []

12 5. (a) R ( ) + R A π tanα α accept awrt.05 A (4) (b) sin ( + their α ) π 5π π + their α, A π π, 6 accept awrt.57, 5.76 A (4) [8] The use of degrees loses only one mark in this question. Penalise the first time it occurs in an answer and then ignore.

13 6. (a) y ln ( 4 ) y e 4 leading to y e Domain of y e Changing subject and removing ln A f a e cso A f is B (4) (b) Range of f is ( ) f < (and f - ( ) ) B () (c) y ln 4 Shape B.5 B ln 4 B B (4) (d) 0.704, 0.45 cao B, B () If more than 4 dp given in this part a maimum on one mark is lost. Penalise on the first occasion. (e) Calculating to at least 6 to at least four dp k 0.5 cao A () [] Alternative to (e) k 0.5 Found in any way Let g( ) + e g ( 0.55) , g( 0.55) 0.00 Change of sign (and continuity) k ( 0.55, 0.55 ) k 0.5 (to dp) A () 4

14 7. (a) f ( ) ( 6) > 0 B f ( ) + 4 8( ) < 0 B Change of sign (and continuity) root in interval (, ) Bft () ft their calculation as long as there is a sign change (b) dy d A Turning point is (, ) A () (c) a, b 4, c 4 B B B () (d) Shape B ft their turning point in correct quadrant only B ft and 8 B () (e) Shape B () [] 5

15 8. (i) sec cosec ( tan ) ( cot ) + + A tan cot cso A () (ii)(a) y arccos cos y B π π sin y arcsin y B () Accept arcsin arcsin cos y π π (b) arccos + arcsin y+ y B () [6] Alternatives for (i) Rearranging cso sec tan cosec cot A sec cosec tan cot A () sin cos LHS cos sin cos sin RHS ( sin cos )( sin + cos ) 4 4 sin cos sin cos cos sin cos sin cos sin sin cos A cos sin LHS or equivalent A () 6

16 Mark (Results) January 008 GCE GCE Mathematics (6665/0) Edecel Limited. Registered in England and Wales No Registered Office: One90 High Holborn, London WCV 7BH

17 January Core Mathematics C Mark a stated or implied A c stated or implied A a, b 0, c, d, e 0 d and b 0, e 0 stated or implied A [4]. (a) dy e tan e sec d A+A dy 0 e tan e sec 0 d tan tan A ( tan + ) 0 tan cso A (6) (b) dy d 0 Equation of tangent at ( 0, 0 ) is y A () [8]

18 . (a) f ( ) 0.8 f ( ) 0.9 Change of sign (and continuity) root in (, ) cso A () (b) ln A.5058 A () (c) Selecting [.5045,.5055 ], or appropriate tighter range, and evaluating at both ends. f ( ) ( ) 0 f.5055 Change of sign (and continuity) root (.5045,.5055 ) root.505 to dp cso A () [7] Note: The root, correct to 5 dp, is.5054

19 4. (a) y ( 5, 4) ( 5, 4) O Shape B 5, 4 B ( ) ( 5, 4) B () (b) For the purpose of marking this paper, the graph is identical to (a) Shape B 5, 4 B (c) ( ) ( 5, 4) B () y ( 4, 8) O ( 6, 8) General shape unchanged B Translation to left B 4, 8 B In all parts of this question ignore any drawing outside the domains shown in the diagrams above. ( ) ( 6, 8) B (4) [0]

20 5. (a) 000 B () (b) 570c 000 e c e A 570c ln c cao A (4) (c) R 90c 000 e 6.5 Accept 6-6 A () (d) R 000 O t Shape B 000 B () [9]

21 cos + cos cos sin sin 6. (a) ( ) ( ) ( ) ( ) cos cos sin cos sin cos cos ( cos )cos any correct epression A 4cos cos A (4) (b)(i) + ( + ) ( ) cos + sin cos sin + + sin cos + sin cos cos + + sin + sin sin cos ( + ) ( + sin ) ( + sin) cos A sec cso A (4) cos 7. (a) (c) sec or cos π 5π, accept awrt.05, 5.4 A, A () [] dy 6cos 8sin d A dy 6 d 0 B y 4 6 or equivalent A (5) (b) R ( ) A 4 tan α, α 0.97 awrt 0.97 A (4) (c) ( α ) sin + their 0.0, 0.46,.,.68 A A A (4) First A any correct solution; second A a second correct solution; third A all four correct and to the specified accuracy or better. Ignore the y-coordinate. []

22 8. (a) y y or f : a Ignore domain A () gf 4 A 4( ) 8 cso A (4) 8 gf : a Ignore domain (b) ( ) (c) (d) Attempting solution of numerator 0 Correct answer and no additional answers A () 8 0 ( ) ( ) ( ) dy d 8 ( ) A A Solving their numerator 0 and substituting to find y. 0, y A (5) []

23 Mark (Final) January 009 GCE GCE Core Mathematics C (6665/0) Edecel Limited. Registered in England and Wales No Registered Office: One90 High Holborn, London WCV 7BH

24 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.

25 Version for Online Standardisation January Core C Mark d d. (a) ( ) d d ( ) ( 5 ) ( 5 ) ( ) 5 5 A dy 5 ( ) ( ) Aft d At, dy d 9 46 Accept awrt 5. A (6) (b) d sin cos sin 4 d A+A A (4) [0] Alternative to (b) d ( sin ) cos sin ( + ) A + A d cos sin cos sin A (4)

26 . (a) ( )( ) + ( + )( + ) ( )( + ) ( + )( ) ( )( + ) Accept, A A (4) (b) ( )( ) ( ) ( ) d d + + ( ) ( ) A cso A () [7] Alternative to (a) + ( + ) ( )( + ) A + ( + ) A (4) Alternatives to (b) ( ) f ( ) f A ( ) ( )( )( ) ( ) ( ) ( )( ) f cso A () ( ) ( )( ) ( )( )( ) ( ) ( ) ( ) ( ) f + ( ) A () A

27 . (a) y O (, 6) ( 7, 0) Shape B, 6 B ( ) ( 7, 0 ) B () (b) y (, 5) ( 7, ) Shape B, 5 B ( ) ( 7, ) B () [6] O

28 4. cos( y+ π ) At π y, 4 d sin( y + π ) dy A dy Follow through their d d sin y+ π dy Aft ( ) dy d π sin before or after substitution B π y 4 π y + 4 A (6) [6]

29 5. (a) g( ) B () (b) ( ) ( ) fg f e e + lne + e A () ( fg : a + e ) (c) fg ( ) B () d (d) ( ) e 0, d + e + 6e A + 6e e + ( ) e A 0, 6 A A (6) [0]

30 6. (a)(i) sin θ sin ( θ + θ) sin θ cosθ + cos θsin θ sinθ cos θ.cosθ + sin θ sinθ A ( ) ( ) sinθ sin θ + sinθ sin θ sinθ 4sin θ cso A (4) (ii) 8sin θ 6sinθ + 0 sinθ + 0 A sin θ π 5π θ, 6 6 π 5π θ, 8 8 A A (5) (b) sin5 sin ( ) sin 60 cos 45 cos 60 sin 45 A 6 ( 6 ) cso A (4) [] Alternatives to (b) sin5 sin 45 0 sin 45 cos0 cos 45 sin 0 ( ) Using cos θ sin θ, A 6 ( 6 ) cso A (4) cos 0 sin 5 sin 5 cos 0 sin 5 A 4 ( 6 ) ( 6+ ) sin5 6 cso A (4) Hence ( ) 4

31 7. (a) ( ) f e + e A ( ) e + e e + 0 A f e B (5) ( ) (b) B 0.57 B B () (c) Choosing ( , ) or an appropriate tighter interval. f ( ) f ( ) A Change of sign (and continuity) root ( , ) cso A ( 0.576, is correct to 4 decimal places) () [] Note: is accurate

32 8. (a) 4 R 5 A 4 tanα α 5... awrt 5 A (4) R + (b) Maimum value is 5 ft their R B ft At the maimum, ( θ α) cos or θ α 0 θ α 5... ft their α A ft () (c) f () t 0+ 5cos( 5t α ) Minimum occurs when cos( 5t α ) The minimum temperature is ( 0 5) 5 A ft () (d) 5t α 80 t 5.5 awrt 5.5 A () []

33 Mark (Results) January 00 GCE Core Mathematics C (6665) Edecel Limited. Registered in England and Wales No Registered Office: One90 High Holborn, London WCV 7BH

34 Edecel is one of the leading eamining and awarding bodies in the UK and throughout the world. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. Through a network of UK and overseas offices, Edecel s centres receive the support they need to help them deliver their education and training programmes to learners. For further information, please call our GCE line on , our GCSE team on , or visit our website at If you have any subject specific questions about the content of this Mark that require the help of a subject specialist, you may find our Ask The Epert service helpful. Ask The Epert can be accessed online at the following link: January 00 Publications Code US070 All the material in this publication is copyright Edecel Ltd 00

35 + Q + January Core Mathematics C Mark + + ( ) + ( + )( ) + ( ) + ( + )( ) or ( + )( ) or ( + )( ) seen or implied anywhere in candidate s working. Award below + ( ) ( ) ( + ) + ( ) or ( ) ( + ) ( ) ( + ) Attempt to combine. Correct result. A Decide to award here!! 4 ( ) ( + ) or 4 Either ( )( + ) 4 4 or ( )( + ) ( )( + ) 4 or 9 6 A aef [4] GCE Core Mathematics C (6665) January 00

36 Q f( ) + (a) f( ) ( ) 0 Sets f( ) 0(can be implied) and takes out a factor of from +, or from + (slip). ( + ) then rearranges to give the quoted result on the question paper. A AG () n + (b) Iterative formula: n +, 0 n + (0) + (0) + An attempt to substitute 0 into the iterative formula. Can be implied by 5.5 or.5 or awrt Both awrt.45 and awrt.07 4 awrt.059 A A () (c) Let f( ) + 0 f (.0565) f (.0575) Sign change (and f ( ) is continuous) therefore a root α is α α.057 ( dp) such that (.0565,.0575) Choose suitable interval for, e.g. [.0565,.0575] or tighter any one value awrt sf d both values correct awrt sf, sign change and conclusion A () As a minimum, both values must be correct to sf, candidate states change of sign, hence root. [8] Core Mathematics C (6665) January 00 4

37 π 5cos sin Rcos + α, R > 0, 0 < < Q (a) ( ) 5cos sin Rcoscosα Rsinsinα Equate cos : 5 R cosα Equate sin : Rsinα R + { } 5 ; R 5 + ; 4 or awrt 5.8 A tanα α c 5 Hence, 5cos sin 4 cos( ) (b) 5cos sin 4 5 tanα ± or tanα ± or 5 5 sinα ± or cosα ± their R their R α awrt 0.54 or π α awrt 0.7π or α A awrt 5.8 (4) 4 cos( ) 4 4 cos ( + ) { } cos( their α ) c ( ) For applying 4 ± their R cos 4 their R c awrt 0.7 c A ( ) c c π { } + π their dd c c awrt 4.9 A Hence, { 0.7, 4.9} (5) [9] Part (b): If there are any EXTRA solutions inside the range 0 < π, then withhold the final accuracy mark if the candidate would otherwise score all 5 marks. Also ignore EXTRA solutions outside the range 0 < π. Core Mathematics C (6665) January 00 5

38 Q4 (i) y ln( ) + u du ln( + ) something + + ln( + ) A + ln( ) d + Apply quotient rule: u ln( + ) v du dv d + d + dy d ( ) ln( + ) u ln( + ) v Applying correctly. Correct differentiation with correct bracketing but allow recovery. A (4) d ( + ) dy ln( + ) {Ignore subsequent working.} (ii) tan y tan y differentiate sin y using either the d sec y cos y dy quotient rule or product rule. d sec y dy dy d { cos y} Finding d sec y d sec y or an attempt to y by reciprocating d. dy * A d* dy d + tan y For writing down or applying the identity sec y + tan y, which must be applied/stated completely in y. d* dy Hence, d +, (as required) For the correct proof, leading on from the previous line of working. A AG (5) [9] Core Mathematics C (6665) January 00 6

39 Q5 y ln y Right-hand branch in quadrants 4 and. Correct shape. B (,0) O ( ),0 Left-hand branch in quadrants and. Correct shape. Completely correct sketch and both, 0 and, 0 ( { }) ( { }) B B () [] Core Mathematics C (6665) January 00 7

40 Q6 (i) y f( ) + A' (, 4) y ( ) { 0, } Shape of and must have a maimum in quadrant and a minimum in quadrant or on the positive y-ais. Either ({ } ) Both ({ } ) 0, or '(, 4) B A B 0, and '(, 4) A B () (ii) y f( + ) + y A '({} 0, 6) Any translation of the original curve. The translated maimum has either -coordinate of 0 (can be implied) or y-coordinate of 6. The translated curve has maimum ({ 0,6 } ) and is in the correct position on the Cartesian aes. B B B O () (iii) y f ( ) y ( 0, ) ( ) A ', 6 Shape of with a minimum in quadrant and a maimum in quadrant. Either ({ } ) Both ({ } ) 0, or '(, 6) B A B 0, and '(, 6) A B O () [9] Core Mathematics C (6665) January 00 8

41 Q7 (a) y sec (cos ) cos dy d (cos ) ( sin ) dy ± ((cos ) (sin ) ) d (cos ) ( sin ) or (cos ) (sin ) A dy sin sin d cos cos cos sec tan Convincing proof. Must see both underlined steps. A AG () (b) y e sec du dv d d u e v sec e sec tan Either e e or Seen sec sec tan or implied Both e e and sec sec tan A Applies vu + uv correctly for their dy e sec e sectan u, u, v, v d + e sec + e sec tan A isw (4) dy (c) Turning point 0 d Hence, e sec ( + tan ) 0 {Note e 0, sec 0, so + tan 0, } Sets their d y 0 and factorises (or cancels) d out at least e from at least two terms. giving tan tan ± k ; k 0 { } a Either y b e sec( 0.96) ( 0.96) Hence, { } c awrt 0.96 or awrt. A (sf ) 0.8 A cao (4) π Part (c): If there are any EXTRA solutions for (or a) inside the range π < <, ie < < 0.54 or ANY 6 6 EXTRA solutions for y (or b), (for these values of ) then withhold the final accuracy mark. π Also ignore EXTRA solutions outside the range π < <, ie < < [] Core Mathematics C (6665) January 00 9

42 Q8 cosec cot, ( eqn *) 0 80 Using cosec + cot gives + cot cot o Writing down or using cosec ± ± cot or cosec θ ± ± cot θ. cot cot 0 or cot cot For either cot cot { 0} or cot cot A cot ( cot ) 0 or cot Attempt to factorise or solve a quadratic (See rules for factorising quadratics) or cancelling out cot from both sides. d cot 0 or cot Both cot 0 and cot. A cot 0 (tan ) 90, 70 45, 5 cot tan 45, 5.5,.5 Candidate attempts to divide at least one of their principal angles by. This will be usually implied by seeing.5 resulting from cot. dd Overall, {.5, 45,.5, 5} Both.5 and.5 A Both 45 and 5 B [7] If there are any EXTRA solutions inside the range 0 80 and the candidate would otherwise score FULL MARKS then withhold the final accuracy mark (the sith mark in this question). Also ignore EXTRA solutions outside the range Core Mathematics C (6665) January 00 0

43 Q9 (i)(a) ln( 7) 5 e e ln( 7) 5 Takes e of both sides of the equation. 5 This can be implied by 7 e. 5 5 e + 7 { } 7 e Then rearranges to make the subject. Eact answer of 5 e + 7. d A () (b) 7 + e ( ) ln e ln5 Takes ln (or logs) of both sides of the equation. 7+ ln ln e ln5 + Applies the addition law of logarithms. ln ln5 ln + 7+ ln5 A oe (ln + 7) + ln5 Factorising out at least two terms on one side and collecting number terms on the other side. dd (ii) (a) + ln5 7 + ln f( ) e, + { } Eact answer of + ln5 7 + ln A oe (5) y e + y e Attempt to make (or swapped y) the subject ln ( y ) ln ( y ) Makes e the subject and takes ln of both sides Hence f ( ) f ( ) ln( ) or ln( ) or ln ( ) f ( y) ln( y ) (see appendi) : Domain: > or (, ) Either > or (, ) or Domain >. B (b) g( ) ln( ),, > A cao (4) { } ln( ) fg( ) e + ( ) + An attempt to put function g into function f. ln( ) e + or ( ) + or + 4. A isw fg( ) : Range: y > or (, ) Either y > or (, ) or Range > or fg( ) >. B () [5] Core Mathematics C (6665) January 00

44 Further copies of this publication are available from Edecel Publications, Adamsway, Mansfield, Notts, NG8 4FN Telephone Fa Order Code US070 January 00 For more information on Edecel qualifications, please visit Edecel Limited. Registered in England and Wales no Registered Office: One90 High Holborn, London, WCV 7BH

45 Mark (Results) January 0 GCE GCE Core Mathematics C (6665) Paper Edecel Limited. Registered in England and Wales No Registered Office: One90 High Holborn, London WCV 7BH

46 Edecel is one of the leading eamining and awarding bodies in the UK and throughout the world. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. Through a network of UK and overseas offices, Edecel s centres receive the support they need to help them deliver their education and training programmes to learners. For further information, please call our GCE line on , our GCSE team on , or visit our website at If you have any subject specific questions about the content of this Mark that require the help of a subject specialist, you may find our Ask The Epert service helpful. Ask The Epert can be accessed online at the following link: January 0 Publications Code US068 All the material in this publication is copyright Edecel Ltd 0

47 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) 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) dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper The second mark is dependent on gaining the first mark

48 January 0 Core Mathematics C 6665 Mark. (a) 7cos 4sin Rcos( + α ) 7 cos 4sin Rcos cosα Rsin sinα Equate cos : 7 Rcosα Equate sin : 4 Rsinα R ; 5 R 5 B tanα α c tanα 7 or tanα 4 awrt.87 A Hence, 7cos 4sin 5cos( +.87) (b) Minimum value 5 5 or R Bft () () (c) 7cos 4sin 0 5cos( +.87) ( + ) cos ( their α ) cos.87 0 ± ( their R) c PV or For applying 0 cos their R c c So, +.87 { , , c } either π + or their PV c or 60 + or their PV gives, { , } awrt.84 OR 6.6 awrt.84 AND 6.6 A A (5) [9] GCE Core Mathematics C (6665) January 0

49 . (a) 4 ( ) ( )( ) (4 )( ) ( )( ) 8 6 { ( )( ) } An attempt to form a single fraction Simplifies to give a correct quadratic numerator over a correct quadratic denominator A aef ( )(4 + ) { ( )( ) } 4 + (b) 4 f( ), > ( ) ( )( ) An attempt to factorise a term quadratic numerator A (4) (4 + ) f( ) ( ) (c) (4 + ) ( ) ( ) ( ) ( ) f( ) ( ) ( ) An attempt to form a single fraction Correct result A () f( ) ( )( ) () k ± ( ) 6 f() Either or A aef A () [9] GCE Core Mathematics C (6665) January 0

50 . cosθ sinθ ( θ ) sin sinθ Substitutes either sin θ or cos θ or cos θ sin θ for cos θ. θ 4sin sin θ θ θ Forms a quadratic in sine 0 (*) 4sin sin 0 sinθ ± 4 4(4)( ) 8 Applies the quadratic formula See notes for alternative methods. PVs: α 54 or α 8 θ { 54, 6, 98, 4} Any one correct answer 80-their pv All four solutions correct. A d(*) A [6] GCE Core Mathematics C (6665) January 0

51 4. (a) θ 0 + Ae kt (eqn *) (0) { t 0, θ 90 } 90 0 Ae k + Substitutes t 0 and θ 90 into eqn * A A 70 A 70 (b) θ e kt A () (5) { t 5, θ 55 } e k 5 e 70 5 ln ( 70 ) 5k 5k ln 5k ( ) Substitutes t 5 and θ 55 into eqn * and rearranges eqn * to make e ±5k the subject. Takes lns and proceeds to make ±5k the subject. d 5k ln ln 5k ln k ln 5 Convincing proof that k ln 5 A () (c) 5 ln θ e t ± α e kt dθ where k ln t 5 ln.(70)e 5 ln dt 5 5 4ln e t ln A oe When t 0, dθ 4ln e dt ln dθ 7 ln dt Rate of decrease of θ.46 C / min ( dp.) awrt ±.46 A () [8] GCE Core Mathematics C (6665) January 0 4

52 5. (a) Crosses -ais f( ) 0 (8 )ln 0 Either (8 ) 0 or ln 0 8, Either one of {} OR {8} B Coordinates are A (, 0) and B (8, 0). Both A (, { 0} ) and B( 8, { 0} ) B () (b) Apply product rule: u (8 ) v ln du dv d d vu + uv f( ) 8 ln + Any one term correct A Both terms correct A () (c) f (.5) f (.6) Sign change (and as f( ) is continuous) therefore the -coordinate of Q lies between.5 and.6. Attempts to evaluate both f(.5) and f(.6) both values correct to at least sf, sign change and conclusion A () (d) At Q, 8 f( ) 0 ln + 0 Setting f( ) 0. 8 ln ln + 8 (ln + ) Splitting up the numerator and proceeding to 8 ln + (as required) For correct proof. No errors seen in working. A () GCE Core Mathematics C (6665) January 0 5

53 (e) Iterative formula: n + 8 ln + n 8 ln(.55) An attempt to substitute 0.55 into the iterative formula. Can be implied by.58(97)... Both awrt.59 and awrt.58 A.59,.58,.54, to dp.,, all stated correctly to dp A () [] GCE Core Mathematics C (6665) January 0 6

54 6. (a) y y( 5) 5 Attempt to make (or swapped y) the subject y 5y y y ( y + ) + 5y Collect terms together and factorise. + 5y y + 5 f ( ) A oe () (b) Range of g is -9 g() 4 or -9 y 4 Correct Range B () (c) Deduces that g() is 0. Seen or implied. g g() g (0) 6, from sketch. -6 A () (d) fg(8) f (4) Correct order g followed by f 4() A () GCE Core Mathematics C (6665) January 0 7

55 (e)(ii) y Correct shape B -6 Graph goes through ({ } ) ( 6, { 0} ) 0,and which are marked. B (4) (f) Domain of g is -9 4 Either correct answer or a follow through from part (b) answer B () [] GCE Core Mathematics C (6665) January 0 8

56 7 (a) y + sin + cos Apply quotient rule: u + sin v + cos du dv cos sin d d dy cos ( + cos ) sin ( + sin ) d cos ( + ) Applying Any one term correct on the numerator Fully correct (unsimplified). A A 4cos + cos + 6sin + sin ( + cos) 4cos + 6sin + (cos + sin ) ( + cos) ( + cos) 4cos + 6sin + (as required) For correct proof with an understanding that cos + sin. No errors seen in working. A* (4) π (b) When, + sinπ y + cosπ y B At π 6sinπ + 4cosπ + 4+ (, ),m( T ) ( + cos π ) π Either T: y ( ) or y + c and π + c c + π ; ( ) m( T) π y y m( ) with their TANGENT gradient and their y ; or uses y m+ c with their TANGENT gradient ; B T: y ( π ) + + y + π + A (4) [8] GCE Core Mathematics C (6665) January 0 9

57 8. (a) y sec (cos ) cos dy d (cos ) ( sin ) Writes sec as (cos ) and gives dy ± ((cos ) (sin ) ) d (cos ) ( sin ) or (cos ) (sin ) A dy sin sin d cos cos cos sec tan Convincing proof. Must see both underlined steps. A AG () (b) sec y, y (n+ ), n. π 4 d Ksecytan y sec ytan y dy secytan y A () (c) dy Applies d secytany dy d d ( dy ) dy Substitutes for sec y. d tany A A y y + tan sec tan sec Attempts to use the identity + tan A sec A So y tan dy d ( ) dy d ( ) A (4) [9] GCE Core Mathematics C (6665) January 0 0

58

59 Further copies of this publication are available from Edecel Publications, Adamsway, Mansfield, Notts, NG8 4FN Telephone Fa Order Code US068 January 0 For more information on Edecel qualifications, please visit Edecel Limited. Registered in England and Wales no Registered Office: One90 High Holborn, London, WCV 7BH

60 Mark (Results) January 0 GCE Core Mathematics C (6665) Paper

61 Edecel is one of the leading eamining and awarding bodies in the UK and throughout the world. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. Through a network of UK and overseas offices, Edecel s centres receive the support they need to help them deliver their education and training programmes to learners. For further information, please call our GCE line on , our GCSE team on , or visit our website at If you have any subject specific questions about the content of this Mark that require the help of a subject specialist, you may find our Ask The Epert service helpful. Ask The Epert can be accessed online at the following link: January 0 Publications Code US000 All the material in this publication is copyright Pearson Education Ltd 0

62 General Marking Guidance PhysicsAndMathsTutor.com 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.

63 EDEXCEL GCE MATHEMATICS PhysicsAndMathsTutor.com 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) should not be subdivided.. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes and can be used if you are using the annotation facility on epen. 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) dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper 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.

64 General Principals for Core Mathematics Marking (But note that specific mark schemes may sometimes override these general principles). Method mark for solving 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 correct formula (with values for a, b and c), leading to. Completing the square Solving + b + c 0 b : ( ) ± ± q ± c, 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. ( Use of a formula + n n ) 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 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.

65 No PhysicsAndMathsTutor.com (a) for any constant B Applying vu +uv, ln, A A (b) Applying A+A A (4) A (9 MARKS) (5) (a) Differentiates the ln term to. Note that is fine for this mark. Applies the product rule to ln. If the rule is quoted it must be correct. There must have been some attempt to differentiate both terms. If the rule is not quoted (or implied by their working) only accept answers of the form ln where A and B are non- zero constants A A One term correct and simplified, either or. ln and ln are acceptable forms Both terms correct and simplified on the same line., ln, oe (b) A A A A Applies the quotient rule. A version of this appears in the formula booklet. If the formula is quoted it must be correct. There must have been some attempt to differentiate both terms. If the formula is not quoted (nor implied by their working) only accept answers of the form 0 Correct first term on numerator 4 4 Correct second term on numerator sin 4 Correct denominator, the needs to be simplified Fully correct simplified epression Accept 4 cos 4 sin 4 oe Alternative method using the product rule.,a,. Writes as sin 4 and applies the product rule. They will score both of these marks or neither of them. If the formula is quoted it must be correct. There must have been some attempt to differentiate both terms. If the formula is not quoted (nor implied by their working) only accept answers of the form 4 sin 4 A One term correct, either 4 4 or sin 4 A Both terms correct,eg. 4 4 sin 4. A Fully correct epression. 4 4 sin 4 or 4 4 sin 4 oe The negative must have been dealt with for the final mark.

66 No (a) Shape coordinates correct y coordinates correct B B B () (b) Shape Ma at (,4) Min at (-,0) B B B () (a) B B B (b) B B B 6 marks Shape unchanged. The positioning of the curve is not significant for this mark. The right hand section of the curve does not have to cross ais. The - coordinates of P and Q are -5 and 0 respectively. This is for translating the curve units left. The minimum point Q must be on the y ais. Accept if -5 is marked on the ais for P with Q on the y ais (marked -). The y- coordinates of P and Q are 0 and - respectively. This is for the stretch parallel to the y ais. The maimumm P must be on the ais. Accept if - is marked on the y ais for Q with P on the ais (marked -5) The curve below the ais reflected in the ais and the curve above the ais is unchanged. Do not accept if the curve is clearly rounded off with a zero gradient at the ais but allow small curvature issues. Use the same principles on the lhs- do not accept if this is a cusp. Both the - and y- coordinates of Q, (,4) given correctly and associated with the maimum point in the first quadrant. To gain this mark there must be a graph and it must only have one maimum. Accept as and 4 marked on the correct aes or in the script as long as there is no ambiguity. Both the - and y- coordinates of P, (-,0) given correctly and associated with the minimum point in the second quadrant. To gain this mark there must be a graph. Toleratee two cusps if this mark has been lost earlier in the question. Accept (0, -) marked on the correct ais. No ) (a) 0 (mm B ()

67 (b) A Correct order A A.8 or 8 (minutes} (5) (6 marks) (a) B (b) A A A Sight of 0 relating to the value of A at t0. Do not worry about (incorrect) units. Accept its sight in (b) Substitutes A40 or twice their answer to (a) and proceeds to.. Accept non numerical answers.. Correct ln work to find t. Eg..5 ln The order must be correct. Accept non numerical answers. See below for correct alternatives Achieves either or awrt 0.46 sf. Either :8 or 8 (minutes). Cao Alternatives in (b) Alt - taking ln s of both sides on line A Substitutes A40, or twice (a) takes ln s of both sides and proceeds to ln ln Make t the subject with correct ln work. ln ln.5 A,A are the same Alt - trial and improvement-hopefully seen rarely A A A Substitutes t 0.46 and t0.47 into 0. to obtain A at both values. Must be to at least dp but you may accept tighter interval but the interval must span the correct value of Obtains A(0.46)9.87 AND A(0.47)40.47 or equivalent Substitutes t0.46 and t0.465 into 40. Obtains A(0.46)9.99 AND A(0.465)40.0 or equivalent and states t0.46 (sf) AS ABOVE No working leading to fully correct accurate answer (sf or better) send/escalate to team leader

68 No 4,A substitute PhysicsAndMathsTutor.com into their 8, A When y. awrt.46 B 4 8 oe A (7 marks) For differentiation of tan. There is no need to identify this with A For correctly writing or Substitute into their. Accept if substituted into A 8 or oe B Obtains the value of corresponding to y. Accept awrt.46 This mark requires all of the necessary elements for finding a numerical equation of the normal. Either Invert their value of, to find m - to find the numerical gradient of the normal Or use their numerical value of - Having done this then use The could appear as awrt.46, the as awrt 0.79, This cannot be awarded for finding the equation of a tangent. Watch for candidates who correctly use If they use ym+c it must be a full method to find c. A Any correct form of the answer. It does not need to be simplified and the question does not ask for an eact answer. 8, 8, 8 6, y-8+ (awrt) 8.5

69 Alternatives using arctan (first marks) Differentiates arctan to get. Don t worry about the lhs A Achieves PhysicsAndMathsTutor.com This method mark requires to be found, which then needs to be substituted into The rest of the marks are then the same. Or implicitly (first marks) A Rearranges to get or The rest of the marks are the same Differentiates implicitly to get in terms of y Or by compound angle identities tan oe Differentiates using quotient rule-see question in applying this. Additionally the tany must have been differentiated to. There is no need to assign to A The correct answer for The rest of the marks are as the main scheme No 5. Uses the identity 7 5

70 7 0 0 A d A, dd, A value,5.5 Correct nd dd,a Correct rd value dd All 4 correct answers awrt 6.5⁰,5.5⁰,6.5 ⁰or 7.5⁰ A (0 marks) Uses the substitution to produce a quadratic equation in cosec( ) Accept invisible brackets in which is replaced by A (longer) but acceptable alternative is to convert everything to sin( ). For this to be scored must be replaced by, cosec( ) must be replaced by. An attempt must be made to multiply by and finally replaced by A A correct equation (0) written or implied by working is obtained. Terms must be collected together on one side of the equation. The usual alternatives are 7 0 or 7 0 d Either an attempt to factorise a term quadratic in cosec( ) or sin( ) with the usual rules Or use of a correct formula to produce a solution in cosec( ) or sin( ) A Obtaining the correct value of or sin. Ignore other values dd Correct method to produce the principal value of θ. It is dependent upon the two M s being scored. Look for A Awrt 6.5 dd Correct method to produce a secondary value. This is dependent upon the candidate having scored the first... M s. Usually you look for or Note 80-their 6.5 must be marked correct BUT 60+their 6.5 is incorrect A Any other correct answer. Awrt 6.5,5.5,6.5 or 7.5 dd Correct method to produce a THIRD value. This is dependent upon the candidate having scored the first M s. See above for alternatives A All 4 correct answers awrt 6.5,5.5,6.5 or 7.5 and no etras inside the range. Ignore any answers outside the range. Radian answers: awrt 0., 0.9,.,.0. Accuracy must be to dp. Lose the first mark that could have been scored. Fully correct radian answer scores,,,,,0,,,,9 marks Candidates cannot mi degrees and radians for method marks. Special case: Some candidates solve the equation in, to produce.

71 No PhysicsAndMathsTutor.com 6 (a) f(0.8) 0.08, f(0.9) Change of sign root (0.8,0.9) A (b) sin A () Sets 0 A* ( c) Sub 0 into awrt.9, awrt.9(0) and awrt.908 A,A (4) (d) [.90775,.90785] f (.90775) AND f (.90785) Change of sign.9078 A () () ( marks) (a) A (b) A Calculates both f(0.8) and f(0.9). Evidence of this mark could be, either, seeing both substitutions written out in the epression, or, one value correct to sig fig, or the appearance of incorrect values of f(0.8)awrt 0. or f(0.9)awrt 0. from use of degrees This requires both values to be correct as well as a reason and a conclusion. Accept f(0.8) 0.08 truncated or rounded (dp) or 0. rounded (dp) and f(0.9)-0.08 truncated or rounded as (dp) or -0.(dp) Acceptable reasons are change of sign, <0 >0, +ve ve, f(0.8)f(0.9)<0. Acceptable conclusion is hence root or Attempts to differentiate f(). Seeing any of, or ±Asin(½) is sufficient evidence. f () correct. Accept sin Sets their f ()0 and proceeds to.. You must be sure that they are setting what they think is f ()0. Accept sin going to..only if f () 0 is stated first A *. This is a given answer so don t accept just the sight of this answer. It is cso (c ) Substitutes 0 into. Evidence of this mark could be awrt.9 or.5 (from degrees) A awrt.9 A awrt.9(0) and awrt.908 (d) Continued iteration is not acceptable for this part. states By choosing a suitable interval Chooses the interval [.90775,.90785] or tighter containing the root Calculates f (.90775) and f (.90785) or tighter with at least one correct, rounded or truncated f (.90775) truncated or awrt rounded f (.90785) truncated or awrt rounded Accept versions of g()- where g(). When.90775, 8 0 rounded and truncated When.90785, 0 truncated or 4 0 rounded A Both values correct, rounded or truncated, a valid reason (see part a) and a minimal conclusion (see part a). Saying hence root is acceptable. There is no need to refer to the turning point.

72 No 7 (a) B (b) A* (4) (c ) >0 B (d) A () () ln 4 A A (4) (a) B Factorises the epression This may not be on line

73 PhysicsAndMathsTutor.com Combines the two fractions to form a single fraction with a common denominator. Cubic denominators are fine for this mark. Allow slips on the numerator but one must have been adapted. Allow invisible brackets. Accept two separate fractions with the same denominator. Amongst many possible options are Correct,Invisible bracket, Cubic and separate Simplifies the (now) single fraction to one with a linear numerator divided by a quadratic factorised denominator. Any cubic denominator must have been fully factorised (check first and last terms) and cancelled with terms on a fully factorised numerator (check first and last terms). A* Cso. This is a given solution and it must be fully correct. All bracketing/algebra must have been correct. You can however accept going to without the need for seeing the cancelling For eample scores B,,,A0. Incorrect line leading to solution. (b) Whereas scores B,,,A A ( c) B (d) This is awarded for an attempt to make or a swapped y the subject of the formula. The minimum criteria is that they start by multiplying by (-) and finish with or swapped y. Allow invisible brackets. For applying the order of operations correctly. Allow maimum of one slip. Eamples of this are (allow slip on sign) Must be written in terms of but can be (allow slip on sign) (allow slip on ) or equivalent inc,, Accept >0, (0,, domain is all values more than 0. Do not accept 0, y>0, [0,, 0 Attempt to write down fg() and set it equal to /7. The order must be correct but accept incorrect or lack of bracketing. Eg A Achieving correctly the line ln 4. Accept also 8 A Moving from ln 0 The ln work must be correct Alternatively moving from to Full solutions to calculate leading from, that is ln can score this mark. Correct answer only. Accept No 8 (a) tan A (

74 A * (b) tan (4) A * (c ) tan tan. d θ dd A () tan tan ddd θ A ( MARKS) (6) (a) A Uses the identity { tan. Accept incorrect signs for this. Just the right hand side is acceptable. Fully correct statement in terms of cos and sin { tan

75 Divide both numerator and denominator by cosacosb. This can be stated or implied by working. If implied you must have seen at least one term modified on both the numerator and denominator. A* This is a given solution. The last two principal s reports have highlighted lack of evidence in such questions. Both sides of the identity must be seen or implied. Eg lhs The minimum epectation for full marks is tan PhysicsAndMathsTutor.com sin cos The solution tan scores AM0A0 (b) The solution tan scores AA0 An attempt to use part (a) with Aθ and B. Seeing is enough evidence. Accept sign slips Uses the identity tan in the rhs of the identity on both numerator and denominator A* cso. This is a given solution. Both sides of the identity must be seen. All steps must be correct with no unreasonable jumps. Accept tan However the following is only worth out of as the last step is an unreasonable jump without further eplanation. (c ) tan Use the given identity in (b) to obtain tan tan. Accept sign slips d Writes down an equation that will give one value of θ, usually. This is dependent upon the first M mark. Follow through on slips dd Attempts to solve their equation in θ. It must end θ and the first two marks must have been scored. A Cso θ or ddd Writes down an equation that would produce a second value of θ. Usually A cso θ and with no etra solutions in the range. Ignore etra solutions outside the range.. Note that under this method one correct solution would score 4 marks. A small number of candidates find the second solution only. They would score,,,,0,0 Alternative to (a) starting from rhs A Uses correct identities for both tana and tanb in the rhs epression. Accept only errors in signs Multiplies both numerator and denominator by cosacosb. This can be stated or implied by working. If implied you must have seen at least one term modified on both the numerator and denominator

76 A This is a given answer. Correctly completes proof. All three epressions must be seen or implied. tan Alternative to (a) starting from both sides The usual method can be marked like this A A Uses correct identities for both tana and tanb in the rhs epression. Accept only errors in signs Multiplies both numerator and denominator by cosacosb. This can be stated or implied by working. If implied you must have seen at least one term modified on both the numerator and denominator Completes proof. Starting now from the lhs writes tan And then states that the lhs is equal to the rhs Or hence proven. There must be a statement of closure Alternative to (b) from sin and cos Writes tan A Uses the identities sin and oe in the rhs of the identity on both numerator and denominator and divides both numerator and denominator by cos to produce an identity in tan As in original scheme Alternative solution for c. Starting with tan Let tanθ t t ( )( ) + t t t t 0 ± ( + 4) t ± Must find an eact surd

77 5π π θ, Accept the use of a calculator for the A marks as long as there is an eact surd for the solution of the quadratic and eact answers are given. Starting with tan pand tan by the correct compound angle identity (or otherwise) and substitute tan π0 to produce an equation in tanθ d Collect terms and produce a term quadratic in tanθ dd Correct use of quadratic formula to produce eact solutions to tanθ. All previous marks must have been scored. ddd All previous marks must have been scored. This is for producing two eact values for θ A A One solution or. Both solutions the range.. PhysicsAndMathsTutor.com and and no etra solutions inside the range. Ignore etra solutions outside Special case: Watch for candidates who write and proceed correctly. They will lose the first mark but potentially can score the others. Solutions in degrees Apply as before. Lose the first correct mark that would have been scored-usually 75⁰

78

79 Further copies of this publication are available from Edecel Publications, Adamsway, Mansfield, Notts, NG8 4FN Telephone Fa Order Code US000 January 0 For more information on Edecel qualifications, please visit Pearson Education Limited. Registered company number 8788 with its registered office at Edinburgh Gate, Harlow, Esse CM0 JE

80 Mark (Results) January 0 GCE Core Mathematics C (6665/0)

81 Edecel and BTEC Qualifications Edecel and BTEC qualifications come from Pearson, the world s leading learning company. 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 for our BTEC qualifications. Alternatively, you can get in touch with us using the details on our contact us page at If you have any subject specific questions about this specification that require the help of a subject specialist, you can speak directly to the subject team at Pearson. Their contact details can be found on this link: You can also use our online Ask the Epert service at You will need an Edecel username and password to access this service. Pearson: helping people progress, everywhere 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 0 Publications Code US0465 All the material in this publication is copyright Pearson Education Ltd 0

82 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. Unless indicated in the mark scheme a correct answer with no working should gain full marks for that part of the question.

83 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) should not be subdivided. In some instances, the mark distributions (e.g., B and A) printed on the candidate s response may differ from the final mark scheme.. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes and can be used. 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) dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper 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 incorrect 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.

84 7. Ignore wrong working or incorrect statements following a correct answer. 8. The maimum mark allocation for each question/part question(item) is set out in the marking grid and you should allocate a score of 0 or for each mark, or trait, as shown: 0 am aa b ba bb bm ba

85 January Core Mathematics C Mark. (a) (b) When (w ) w oe A dy d 5 5 ( ) or 0( ) 4 4 A, Gradient 60 () Equation of tangent is '60' y ( ) ' ' oe d y 60 cso A (5) (7 marks) (a) 5 Substitute y- into y (w ) and proceed to w.. [Accept positive sign used of y, ie y+] A Obtains wor oe with no incorrect working seen. Accept alternatives such as 0.5. Sight of just the answer would score both marks as long as no incorrect working is seen. (b) 5 Attempts to differentiate y ( ) using the chain rule. 4 Sight of ± A( ) where A is a non- zero constant is sufficient for the method mark. A A correct (un simplified) form of the differential. dy 4 Accept 5 ( ) d dy 4 0( ) d d This is awarded for an attempt to find the gradient of the tangent to the curve at P Award for substituting their numerical value to part (a) into their differential to find the numerical gradient of the tangent Award for a correct method to find an equation of the tangent to the curve at P. It is dependent upon the previous M mark being awarded. y ( ) Award for ' their 60' their ' ' If they use y m+ cit must be a full method, using m their 60, their and -. An attempt must be seen to find c

86 A cso y 60. The question is specific and requires the answer in this form. You may isw in this question after a correct answer.

87 . (a) 0 e 6 ln(6 ) + + A* (b) Sub 0 into n+ ln(6 n) +.86, A AWRT 4 dp A (c ) Chooses interval [.065,.075] () () g(.065)-0.000(7), g(.075)0.004(4) d Sign change, hence root (correct to dp) (a) Sets g()0,and using correct ln work, makes the of the e Look for e e ± 6± ln( ± 6 ± ) ± A () (8 marks) term the subject of the formula. Do not accept e 6 without firstly seeing e + 6 0or a statement that g()0 A* cso. ln(6 ) + Note that this is a given answer (and a proof). Invisible brackets are allowed for the M but not the A Do not accept recovery from earlier errors for the A mark. The solution below scores 0 marks. 0 e ln( 6) ln(6 ) + (b) Sub 0 into n+ ln(6 n) + to produce a numerical value for. Evidence for the award could be any of ln(6 ) +, ln 4 +,... or awrt.4 A Answer correct to 4 dp.86. The subscript is not important. Mark as the first value given/found. A Awrt 4 dp..847 and.5 The subscripts are not important. Mark as the second and third values given/found (c ) Chooses the interval [.065,.075] or smaller containing the root d Calculates g(.065) and g(.075) with at least one of these correct to sf. The answers can be rounded or truncated g(.065) rounded, g(.065) truncated g(.075) (+) rounded and truncated A Both values correct (rounded or truncated), A reason which could include change of sign, >0 <0, g(.065) g(.075) <0 AND a minimal conclusion such as hence root, α.07 or Do not accept continued iteration as question demands an interval to be chosen. Alternative solution to (a ) working backwards Proceeds from ln(6 ) + using correct ep work to.0 A Arrives correctly at e and makes a statement to the effect that this is g()0 Alternative solution to (c ) using f( ) ln(6 ) + {Similarly h( ) ln(6 ) } Chooses the interval [.065,.075] or smaller containing the root d Calculates f(.065) and f(.075) with at least correct rounded or truncated f(.065) Accept rounded or truncated. Also accept 0.000

88 f(.075) Accept rounded or truncated

89 PhysicsAndMathsTutor.com. (a) ff(-) f(0),,a M () (b) y f-( ) y Shape B B (0,-) annd (,0) B B (,0)) (0,-) (c ) () y yf(( )- Shape B B (0,0) B B () (0,0) (d) y Shape B B (-6,0)) or (0,4) B B (0,4) (-6,0) aand (0,4) B B (-6,00) () (9 marrks) (a) A full methhod of findinng ff(-). f(0) is accepttable but f(-)0 is not.. + as the line passes through botth points. Do D not allow w for y ln( + 4), whicch only passses through one of the points. p Cao ff(-). Writing down on its own is enough for both b marks pprovided no o incorrect working is seen. Accept a soolution obtaained from tw wo substitu utions into thhe equationn y A (b) B B For the corrrect shape. Award F A this mark for an n increasingg function inn quadrants, 4 and only. o Do not awaard if the cuurve bends back b on itseelf or has a clear c minim mum This is indeependent to the first maark and for the t graph paassing throuugh (0,-) and a (, 0)

90 Accept - and marked on the correct aes. Accept (-,0) and (0,) instead of (0,-) and (,0) as long as they are on the correct aes Accept P (0,-), Q (,0) stated elsewhere as long as P and Q are marked in the correct place on the graph There must be a graph for this to be awarded (c ) B Award for a correct shape roughly symmetrical about the y- ais. It must have a cusp and a gradient that decreases either side of the cusp. Do not award if the graph has a clear maimum B (0,0) lies on their graph. Accept the graph passing through the origin without seeing (0, 0) marked (d) B Shape. The position is not important. The gradient should be always positive but decreasing There should not be a clear maimum point. B The graph passes through (0,4) or (-6,0). See part (b) for allowed variations B The graph passes through (0,4) and (-6,0). See part (b) for allowed variations

91 4. (a) R R 0 A 8 tanα α awrt 0.97 A 6 (4) (b)(i) 4 p( ) + 0cos( θ 0.97) 4 p( ) 0 Maimum A (b)(ii) θ ' theirα' π θ awrt 4.07 A () () (8 marks) (a) Using Pythagoras Theorem with 6 and 8 to find R. Accept R If α has been found first accept R ± or R ± sin' α ' cos' α ' A R 0. Many candidates will just write this down which is fine for the marks. Accept ± 0 but not For tanα ± or tanα ± If R is used then only accept sinα ± or cosα ± R R A α awrt Note that 5.⁰ is A0 (b) Note that (b)(i) and (b)(ii) can be marked together 4 (i) Award for p( ). ' R ' A Cao p( ) ma. The answer is acceptable for both marks as long as no incorrect working is seen (ii) For setting θ ' theirα' π and proceeding to θ.. If working eclusively in degrees accept θ ' theirα ' 80 Do not accept mied units A θ awrt If the final A mark in part (a) is lost for 5., then accept awrt.

92 5. (i)(a) (i)(b) (ii) Uses dy ln+ d AA ln+ dy ( cos ) d +sin ) + BA d cosec y dy A dy d cosec y cosec y + cot yand cot y in d y d or d to get an epression in dy () () dy d cosec y + cot y + cso, A* (5) ( marks) (i)(a) Applies the product rule vu +uv to ln. If the rule is quoted it must be correct. There must have been some attempt to differentiate both terms. If the rule is not quoted (nor implied by their working, with terms written out u,u.,v.,v.followed by their vu +uv ) then only accept answers of the form B A ln + where A, B are constants 0 A One term correct, either ln or A dy Cao. ln+. The answer does not need to be simplified. d dy For reference the simplified answer is ln (ln ) d (i)(b) B Sight of ( + sin ) For applying the chain rule to ( + sin ). If the rule is quoted it must be correct. If it is not quoted possible forms of evidence could be sight of C ( + sin ) (± Dcos ) where C and D are non- zero constants. Alternatively accept u + sin, u followed by Cu their u ' Do not accept C( + sin ) cosunless you have evidence that this is their u Allow invisible brackets for this mark, ie. C ( + sin ) ± Dcos A dy Cao ( + sin ) ( + cos ). There is no requirement to simplify this. d You may ignore subsequent working (isw) after a correct answer in part (i)(a) and (b)

93 (ii) Writing the derivative of coty as -cosec y. It must be in terms of y A d cosec y dy dy cosec y. Both lhs and rhs must be correct. d Using d y d d dy Using cosec y + cot y and cot y to get d y d just in terms of. dy dy A cso d + Alternative to (a)(i) when ln() is written ln+ln Writes ln as ln + ln. Achieves A for differential of ln and applies the product rule vu +uv to ln. A A Either ln+ lnor A correct (un simplified) answer. Eg ln+ ln+ Alternative to 5(ii) using quotient rule Writes cot y as cos y and applies the quotient rule, a form of which appears in the sin y formula book. If the rule is quoted it must be correct. There must have been some attempt to differentiate both terms. If the rule is not quoted (nor implied by their working, vu ' uv' meaning terms are written out u,u.,v.,v.followed by their ) v sin y ± sin y cos y ± cos y only accept answers of the form (sin y) A Correct un simplified answer with both lhs and rhs correct. d sin y sin y cosy cosy { cot y} d y (sin y) Using d y d d dy Using sin y+ cos y, cosec y sin y and cosec y + cot yto get d y d or d dy in dy A cso d + Alternative to 5(ii) using the chain rule, first two marks

94 Writes cot y as (tan ) y and applies the chain rule (or quotient rule). Accept answers of the form (tan y) sec y A Correct un simplified answer with both lhs and rhs correct. d (tan y) sec y dy cot y tan y Alternative to 5(ii) using a triangle last Uses triangle with tan y to find siny sin y and get d y d or d + just in terms of dy

95 6. (i) (sin.5 + cos.5) sin.5 + cos sin.5 + cos.5 + sin.5cos.5 States or uses sin.5 + cos.5 B Uses sin cos sin sin.5cos.5 sin 45 (sin.5 + cos.5) + sin 45 A + or+ cso A (5) (ii) (a) cos θ + sinθ sin θ + sinθ sinθ sin θ 0 sin θ sinθ 0or k A* () (b) sin θ (sinθ ) 0 sinθ 0, sinθ A Any two of 0,0,50,80 B All four answers 0,0,50,80 A (4) ( marks) (i) Attempts to epand (sin.5 + cos.5). Award if you see sin.5 + cos There must be > two terms. Condone missing brackets ie sin.5 + cos B Stating or using sin.5 + cos.5. Accept sin.5 + cos.5 as the intention is clear. Note that this may also come from using the double angle formula cos 45 + cos 45 sin.5 + cos.5 ( ) + ( ) Uses sin cos sin to write sin.5cos.5 as sin 45 or sin(.5) A Reaching the intermediate answer + sin45 A Cso+ or +. Be aware that both.707 and + can be found by using a calculator for +sin45. Neither can be accepted on their own without firstly seeing one of the two answers given above. Each stage should be shown as required by the mark scheme. Note that if the candidates use (sinθ + cos θ ) they can pick up the first M and B marks, but no others until they use θ.5. All other marks then become available. (iia) Substitutes cos θ sin θ in cos θ + sinθ to produce an equation in sinθ only. It is acceptable to use cos θ cos θ or cos θ sin θ as long as the cos θ is subsequently replaced by sin θ A* Obtains the correct simplified equation in sinθ. sinθ sin θ 0 or sinθ sin θ must be written in the form sin θ sinθ 0as required by the question. Also accept k as long as no incorrect working is seen. (iib) Factorises or divides by sinθ. For this mark ' k 'sinθ is acceptable. If they have a TQ in sinθ this can be scored for correct factorisation A Both sinθ 0, and sinθ B Any two answers from 0, 0, 50, 80. A All four answers 0, 0, 50, 80 with no etra solutions inside the range. Ignore solutions outside the range.

96 6.alt (i) (sin.5 + cos.5) sin.5 + cos Uses sin.5 + cos.5 + sin.5cos.5 States or uses sin.5 + cos.5 B cos cos + sin cos cos 45 + cos 45 Hence cos 45 A (sin.5 + cos.5) + or + A (5) 6.alt (i) Uses Factor Formula (sin.5 sin 67.5) ( sin 45cos.5) Uses the double angle formula +,A Reaching the stage cos.5 B cos.5 + cos 45 + or+ A (5) 6.alt (i) Uses Factor Formula (cos 67.5 cos.5) ( cos 45cos.5) Uses the double angle formula +,A Reaching the stage cos.5 B cos.5 + cos 45 + or+ A (5)

97 7. (a) 4 8 ( + 5) + 4( + ) ( + )( + 5) ( + )( + 5) A ( + ) + + ( )( 5) (b) h'( ) + ( 5) ( + 5) ( + 5) A* A (4) 0 h'( ) ( + 5) cso A () (c ) Maimum occurs when h'( ) A 5 When 5 h( ),A 5 5 Range of h() is 0 h ( ) Aft 5 (5) ( marks) (a) Combines the three fractions to form a single fraction with a common denominator. Allow errors on the numerator but at least one must have been adapted. Condone invisible brackets for this mark. Accept three separate fractions with the same denominator. Amongst possible options allowed for this method are Eg An eample of invisible brackets ( + )( + 5) ( + 5) Eg An eample of an error (on middle term), st term has been adapted ( + )( + 5) ( + )( + 5) ( + )( + 5) ( + 5) ( + ) + 4( + ) ( + 5) 8( + 5)( + ) ( + ) ( + 5) Eg An eample of a correct fraction with a different denominator A Award for a correct un simplified fraction with the correct (lowest) common denominator. ( + 5) + 4( + ) 8 ( + )( + 5) Accept if there are three separate fractions with the correct (lowest) common denominator. Eg ( + 5) 4( + ) ( )( 5) ( )( 5) ( )( 5)

98 Note, Eample would score A0 as it does not have the correct lowest common denominator There must be a single denominator. Terms must be collected on the numerator. A factor of (+) must be taken out of the numerator and then cancelled with one in the denominator. The cancelling may be assumed if the term disappears A* Cso This is a given solution and this mark should be withheld if there are any errors ( + 5) (b) Applies the quotient rule to, a form of which appears in the formula book. ( + 5) If the rule is quoted it must be correct. There must have been some attempt to differentiate both terms. If the rule is not quoted (nor implied by their working, meaning terms are written out vu ' uv' u,u.,v.,v.followed by their ) then only accept answers of the form v ( + 5) A B where AB, > 0 ( + 5) A Correct unsimplified answer ( + 5) h'( ) ( + 5) A 0 The correct simplified answer. Accept (5 ) ( 5), 0 h'( ) ( + 5) ( + 5) ( + 5) DO NOT ISW FOR PART (b). INCORRECT SIMPLIFICATION IS A0 4 ( ) (c ) Sets their h ()0 and proceeds with a correct method to find. There must have been an attempt to differentiate. Allow numerical errors but do not allow solutions from unsolvable equations. A Finds the correct value of the maimum point 5. Ignore the solution - 5 but withhold this mark if other positive values found. Substitutes their answer into their h ()0 in h() to determine the maimum value A Cso-the maimum value of h() 5 5. Accept equivalents such as 5 0 but not Aft Range of h() is 0 h( ). Follow through on their maimum value if the M s have been scored. Allow 0 y, 0 Range, 0, but not 0, 0, If a candidate attempts to work out h ( ) in (b) and does all that is required for (b) in (c), then allow. Do not allow h ( ) to be used for h () in part (c ). For this question (b) and (c) can be scored together. Alternative to (b) using the product rule Sets h( ) ( + 5) and applies the product rule vu +uv with terms being and ( +5) - If the rule is quoted it must be correct. There must have been some attempt to differentiate both terms. If the rule is not quoted (nor implied by their working, meaning terms are written out u,u.,v.,v.followed by their vu +uv ) then only accept answers of the form ( 5) + A+ ± B( + 5) A Correct un simplified answer ( 5) + + ( + 5) A The question asks for h () to be put in its simplest form. Hence in this method the terms need to be combined to form a single correct epression.

99 For a correct simplified answer accept 0 (5 ) ( 5) h'( ) (0 )( + 5) ( + 5) ( + 5) ( + 5)

100 8. (a) ( ) 9500 B (b) (c ) 0.5t 0.5t e + 000e t 0.5t 7e + e 9 ( e ) 7e + 9e 0.5 t 0.5 t 0.5 t dv ( ) 450 dt 0.5t 0.5t 0 9e 7e 0.5t 0.5t 0 (9e + )( e ) 0.5t e A t 4ln() oe A 0.5t 0.5t e 000e A () (4) When t8 Decrease 59 ( /year) A (4) (9 marks) (a) B The sign is not important for this mark (b) Substitute V9500, collect terms and set on side of an equation 0. Indices must be correct 0.5t 0.5t Accept 7000e + 000e and where e 0.5t 0.5t Factorise the quadratic in e or e 0.5t 0.5t For your information the factorised quadratic in e 0.5t 0.5t is (e )( e + 9) 0 0.5t Alternatively let ' ' e or otherwise and factorise a quadratic equation in A Correct solution of the quadratic. Either 0.5t e or e oe. 0.5t A Correct eact value of t. Accept variations of 4ln(), such as ln(6), ln( ), ln() , 4ln( ) 0.5t 0.5t.(c ) Differentiates V 7000e + 000e by the chain rule. dv 0.5t 0.5t Accept answers of the form ( ) ± Ae ± Be A, B are constants 0 dt dv 0.5t 0.5t A Correct derivative ( ) 450e 000e. dt There is no need for it to be simplified so accept dv 0.5t 0.5t ( ) e e oe dt Substitute t8 into their d V dt. This is not dependent upon the first but there must have been some attempt to differentiate. Do not accept t8 in V

101 A ±59. Ignore the sign and the units. If the candidate then divides by 8, withhold this mark. This would not be isw. Be aware that sub t8 into V first and then differentiating can achieve 59. This is M0A0M0A0.

102 Further copies of this publication are available from Edecel Publications, Adamsway, Mansfield, Notts, NG8 4FN Telephone Fa Order Code US0465 January 0 For more information on Edecel qualifications, please visit our website Pearson Education Limited. Registered company number 8788 with its registered office at Edinburgh Gate, Harlow, Esse CM0 JE

103 GCE Edecel GCE Core Mathematics C (6665) Summer 005 Mark (Results) Edecel GCE Core Mathematics C (6665)

104 June Core C Mark. (a) Dividing by cos θ : sin θ cos θ + cos θ cos θ cos θ Completion : + tan θ sec θ (no errors seen) A () (b) Use of + tan θ sec θ : (sec θ ) + secθ [ sec θ + secθ 0] Factorising or solving: ( secθ + )(secθ ) 0 [ sec θ or sec θ ] θ 0 B cos θ ; θ. 8 θ 8. A A (6) [Aft for θ 60 θ ] [ 8 ]

105 . (a) (b) (i) 6 sin cos + sec tan or sin + sec tan [ for 6 sin] (ii) ( + ln ) ( + ) [ B for ( + ln ) ] Differentiating numerator to obtain 0 0 Differentiating denominator to obtain (-) AA () BA () B B Using quotient rule formula correctly: dy ( ) (0 0) (5 0 To obtain 4 d ( ) + 9)( ) A ( )[5( ) (5 Simplifying to form 4 ( ) 8 * (c.s.o.) ( ) 0 + 9) A (6) [] Alternatives for (b) Either Using product rule formula correctly: Obtaining 0 0 Obtaining ( ) dy To obtain (5 0+ 9){ ( ) } + (0 0)( ) d 0( ) (5 0+ 9) Simplifying to form ( ) 8 * (c.s.o.) ( ) 4 Or Splitting fraction to give 5 + ( ) Then differentiating to give answer B B A cao A (6) B B A A (6)

106 (a) ( )( ) + B 5 + ( ) ( + )( ) for combining fractions even if the denominator is not lowest common (b) (c) y + 4 ( + ) ( + )( ) ( + )( ) must have linear numerator y y y + y f + () fg() (attempt) [ ] + 4 " g" Setting and finding ; ± o.e. * A cso (4) A A () ; A () [0]

107 4 (a) f () e AA () (b) e 0 6 α e α α e 6 α (*) A cso () (c) , , , 4 [ at least correct, A all correct to 4 d.p.] A () (d) Using f () e with suitable interval e.g. f (0.445) f (0.445) () Accuracy (change of sign and correct values) A () [9]

108 5. (a) cos A cos A sin A ( + use of cos A + sin A ) ( sin A); sin A sin A (*) A () (b) sin θ cos θ sinθ + 4sinθ cos θ; ( sin θ) sinθ + B; 4sinθ cosθ + 6sin θ sinθ sinθ (4cosθ + 6sinθ ) (*) A (4) (c) 4cosθ + 6sinθ R sinθ cosα + R cosθ sinα Complete method for R (may be implied by correct answer) [ R 4 + 6, R sin α 4, R cos α 6] R 5 or 7. Complete method for α ; α (allow.7º) A A (4) (d) sinθ ( 4 cosθ + 6sinθ ) 0 θ 0 sin( θ ) (4.6º) 5 θ (0.49),.75 [or θ +. 7 (4.6º), 55.4 ] θ. cao B d A (5) [5]

109 6. (a) y Translation by Intercepts correct 0 a A () (b) y 0 b 0, correct shape provided graph is not original graph Reflection in y-ais Intercepts correct B B B () (c) a, b BB () (d) Intersection of y 5 with y Solving to give 6 A A (4) [Notes: (i) If both values found for 5 and 5, or solved algebraically, can score out of 4 for and ¾; 6 required to eliminate ¾ for final mark. (ii) Squaring approach: correct method, ( correct term quadratic, any form) A Solving, Final correct answer A.] []

110 7. (a) Setting p 00 at t a + a (00 500a); a 0. (c.s.o) * da () (b) 800(0.)e t + 0.e 0.t ; e 0. t 6... A Correctly taking logs to 0. t ln k t 4 (.9..) A (4) (c) Correct derivation: (Showing division of num. and den. by 0.t e ; using a) B () (d) 0.t Using t, e 0, 6 p A () [0]

111 GCE Edecel GCE Mathematics Core Mathematics C (6665) June 006 Mark (Results) Edecel GCE Mathematics

112 EDEXCEL CORE MATHEMATICS C (6665) JUNE 006 MARK SCHEME. (a) ( + )( ) +, ( + )( ) +. (a) B, A () Notes attempt to factorise numerator, usual rules B factorising denominator seen anywhere in (a), A given answer If factorisation of denom. not seen, correct answer implies B (b) Epressing over common denominator + + ( + ) ( + ) ( + ) ( ) ( ) [Or Otherwise : ] ( ) Multiplying out numerator and attempt to factorise dy d [ ( )( ) ] + + Notes B e : a b (b) ( ) e + Answer: A: e B A () (6 marks) BA() ( ) 5 +. ( 5 ) + for k (5 ) m + A () (6 marks)

113 EDEXCEL CORE MATHEMATICS C (6665) JUNE 006 MARK SCHEME. (a) Mod graph, reflect for y < 0 (0, ), (, 0) or marked on aes A Correct shape, including cusp A () (b) Attempt at reflection in y Curvature correct A (c) (, 0), (0, ) or equiv. Attempt at stretches B () (0, ) or equiv. B (, 0) B () (9 marks) 4. (a) 45 ºC B () t (b) e e t 75 sub. T 00 and attempt to rearrange to e 0.05t a, where a Q e A t 75 correct application of logs t 7.49 A (4) dt (c) 0 dt t t e ( for k e ) A At t 50, rate of decrease ( ± ).64 ºC/min A () (d) T > 5, (since e t 0 as t ) B () (9 marks)

114 EDEXCEL CORE MATHEMATICS C (6665) JUNE 006 MARK SCHEME 5. (a) Using product rule: d y tan + ( )sec d A A sin Use of tan " and sec " cos cos sin [ + ( ) ] cos cos d y Setting 0 and multiplying through to eliminate fractions d [ sin cos + ( ) 0] Completion: producing 4 k + sin 4k 0 with no wrong working seen and at least previous line seen. AG A* (6) (b).670, 0.809, 0.746, 0.774, A A () 0 4 Note: for first correct application, first A for two correct, second A for all four correct Ma deduction, if ALL correct to > 4 d.p. A0 A SC: degree mode: , A for 0.494, then A0; ma (c) Choose suitable interval for k: e.g. [0.765, 0.775] and evaluate f() at these values Show that 4k + sin 4k changes sign and deduction A () [f(0.765) , f(0.775) ] Note: Continued iteration: (no marks in degree mode) Some evidence of further iterations leading to or better ; Deduction A ( marks) 4

115 EDEXCEL CORE MATHEMATICS C (6665) JUNE 006 MARK SCHEME 6. (a) Dividing sin θ + cos θ by sin θ to give sin θ cos θ + sin θ sin θ sin θ Completion: + cot θ cosec θ cosec θ cot θ AG A* () 4 4 (b) cos ec θ cot θ ( cosec θ cot θ )( cosec θ + cot θ ) ( cos ec θ + cot θ ) using (a) AG A* () Notes: (i) Using LHS ( + cot θ ) cot 4 θ, using (a) & elim. cot 4 θ, conclusion {using (a) again} A* ( cos θ )(+ cos θ ) (ii) Conversion to sines and cosines: needs for 4 sin θ (c) Using (b) to form cosec θ + cot θ cotθ Forming quadratic in cotθ + cot θ + cot θ cotθ {using (a)} cot θ + cotθ 0 A Solving: ( cot )( cotθ + ) 0 θ to cot θ cotθ or cotθ A θ 5 º (or correct value(s) for candidate dep. on Ms) A (6) Note: Ignore solutions outside range Etra solutions in range loses A, but candidate may possibly have more than one correct solution. (0 marks) 5

116 EDEXCEL CORE MATHEMATICS C (6665) JUNE 006 MARK SCHEME 7. (a) Log graph: Shape B Intersection with ve -ais db (0, ln k), ( k, 0) Mod graph :V shape, verte on +ve -ais B B k (0, k) and, 0 B (5) (b) f() R, < f ( ) <, < y < B () (c) fg k 4 k k ln{k + k } or f 4 k ln ( ) A () (d) dy d + k B Equating (with ) to grad. of line; + k 9 ; A k ½ A (4) ( marks) 6

117 EDEXCEL CORE MATHEMATICS C (6665) JUNE 006 MARK SCHEME 8. (a) Method for finding sin A sin A 7 4 A A Note: Use of 7 First A for, eact. 4 Second A for sign (even if dec. answer given) sin A sin Acos A (b)(i) 7 sin A or equivalent eact A (5) 8 Note: ± f.t. Requires eact value, dependent on nd M π π π π π π cos + + cos cos cos sin sin + cos cos + sin sin π cos cos A [This can be just written down (using factor formulae) for A] cos AG A* () (b)(ii) Note: π A earned, if cos cos just written down, using factor theorem Final A* requires some working after first result. dy 6sin cos sin B B d or 6 sin cos sin + π sin π Note: First B for sin sin sin AG A* (4) 6 sin cos ; second B for remaining term(s) ( marks) 7

118 Mark (Results) Summer 007 GCE GCE Mathematics Core Mathematics C (6665) Edecel Limited. Registered in England and Wales No Registered Office: One90 High Holborn, London WCV 7BH

119 June Core Mathematics C Mark. (a) 6 ln ln 6 or ln ln [implied by 0.69 ] or ln 0 6 (only this answer) A (cso) () (b) (e ) 4e + 0 (any term form) (e )(e ) 0 e or e Solving quadratic dep ln, 0 (or ln ) A (4) (6 marks) Notes: (a) Answer with no working or no incorrect working seen: A ln 6 Beware from ln ln M0A0 ln (ln 6 ln ) ln ln 6 ln e allow, (no wrong working) A (b) st for attempting to multiply through by e : Allow y, X, even, for e Be generous for e.g e + 4, e + 4e, y + y (from e - ), e + 4e e nd is for solving quadratic (may be by formula or completing the square) as far as getting two values for e or y or X etc rd is for converting their answer(s) of the form e k to lnk (must be eact) A is for ln and ln or 0 (Both required and no further solutions)

120 . (a) + ( )( + ) at any stage B ( + )( ) (9 + ) f() f.t. on error in denominator factors ( )( + ), A (need not be single fraction) Simplifying numerator to quadratic form [ ] ( )( + ) 4 + Correct numerator A [( )( + )] ( )( + ) Factorising numerator, with a denominator o.e. ( )( + ) ( ) [ ] 4 6 ( ) A cso (7) Alt.(a) + ( )( + ) at any stage B (+ )( + ) (9 + )( + ) f() A f.t. ( + )( + ) ( + )( + 4 ) ( + )( + 6) ( + )( + ) or ( )( ) ( + )( + + ) Any one linear factor quadratic factor in numerator ( + )( + )( ) ( + )( + ) o.e. ( ) 4 6 ( ) o.e., A A ( ) 4 (4 6) (b) Complete method for f ( ) ; e.g f( ) o.e A ( ) 8 or 8( ) A () ( ) Not treating f (for f ) as misread (0 marks) Notes: (a) st in either version is for correct method st + ( ) (9 + ) ( + )( ) ( ) 9+ A Allow or or ( fractions) ( )( + ) ( )( + ) ( )( + ) nd in (main a) is for forming term quadratic in numerator rd is for factorising their quadratic (usual rules) ; factor of need not be etracted ( ) A is given answer so is cso Alt (a) rd is for factorising resulting quadratic Notice that B likely to be scored very late but on epen scored first (b) SC: For M allow ± given epression or one error in product rule Alt: Attempt at f() 4 ( ) and diff. ; k( ) A; A as above Accept 8 ( ). Differentiating original function mark as scheme.

121 . (a) dy e + e d,a,a () d y (b) If 0, d e ( + ) 0 setting (a) 0 [e 0] ( + ) 0 ( 0 ) or A 0, y 0 and, y 4e ( 0.54 ) A () (c) d y d e e e e ( + 4+ )e, A () (d) d y d y 0, > 0 (), < 0 [ e ( 0.70 )] d d : Evaluate, or state sign of, candidate s (c) for at least one of candidate s value(s) from (b) minimum maimum A (cso) () Alt.(d) For : Evaluate, or state sign of, d y at two appropriate values on either side of at d least one of their answers from (b) or Evaluate y at two appropriate values on either side of at least one of their answers from (b) or Sketch curve (0 marks) Notes: (a) Generous M for attempt at f ( ) g ( ) + f ( ) g( ) st A for one correct, nd A for the other correct. Note that e on its own scores no marks (b) st A ( 0) may be omitted, but for nd A both sets of coordinates needed ; f.t only on candidate s (c) requires complete method for candidate s (a), result may be unsimplified for A (d) A is cso; 0, min, and, ma and no incorrect working seen., or (in alternative) sign of dy either side correct, or values of y appropriate to t.p. d Need only consider the quadratic, as may assume e > 0. If all marks gained in (a) and (c), and correct values, give A for correct statements with no working

122 4. (a) ( ) 0 o.e. (e.g. ( + ) ) ( ) A (cso) () Note( ), answer is given: need to see appropriate working and A is cso [Reverse process: Squaring and non-fractional equation, form f() A] Alt (i) (b) , 0.657, st B is for one correct, nd B for other two correct If all three are to greater accuracy, award B0 B Alt (ii) (c) Choose values in interval (0.655, 0.655) or tighter and evaluate both f(0.655) ( 7 f(0.655) 0.00 (0 At least one correct up to bracket, i.e or 0.00 Change of sign, 0.65 is a root (correct) to d.p. Requires both correct up to bracket and conclusion as above Continued iterations at least as far as , , 7 two correct to at least 4 s.f. A Conclusion : Two values correct to 4 d.p., so 0.65 is root to d.p. A If use g(0.655) >0.655 and g(0.655) <0.655 A Conclusion : Both results correct, so 0.65 is root to d.p. A B; B () A A () (7 marks) 5. 4 (a) Finding g(4) k and f(k). or fg() ln [ f() ln( ) fg(4) ln(4 )] ln A () y (b) y ln( ) e or e y, A (c) f () (e + ) Allow y (e + ) A Domain R [Allow R, all reals, (-, ) ] independent B (4) y Shape, and -ais should appear to be B O asymptote Equation needed, may see in diagram (ignore others) Intercept (0, ) no other; accept y ⅔ (0.67) or on graph B ind. B ind () (d) or eact equiv. B, or eact equiv., A () Note: ( + ) or ( ) o.e. is M0A0 Alt: Squaring to quadratic ( ) and solving ; BA ( marks)

123 6. (a) Complete method for R: e.g. Rcosα, Rsinα, R ( + ) R or.6 (or more accurate) A Complete method for tanα [Allow tan α ] α (Allow.7 ) A (4) (b) Greatest value ( ) 4 69, A () (c) sin( ) ( ) sin( + their α) their R ( ) 0.8( 0 or 6. A ( ) π 0.80 Must be π their 0.8 or 80 their 6. or ( ) π Must be π + their 0.8 or 60 + their 6..7 or (awrt) Both (radians only) A (5) If 0.8 or 6. not seen, correct answers imply this A mark ( marks) Notes: (a) st on Epen for correct method for R, even if found second nd for correct method for tanα No working at all: A for, A for or.7. N.B. Rcos α, Rsin α used, can still score A for R, but loses the A mark for α. cosα, sin α : apply the same marking. (b) for realising sin( + α ) ±, so finding R 4. (c) Working in mied degrees/rads : first two marks available Working consistently in degrees: Possible to score first 4 marks [Degree answers, just for reference, Only are 0. and 4.4 ] Third can be gained for candidate s 0.8 candidate s π or equiv. in degrees One of the answers correct in radians or degrees implies the corresponding M mark. Alt: (c) (i) Squaring to form quadratic in sin or cos [cos 4cos 8 0, sin 6sin 0] Correct values for cos 0.95, 0.646; or sin 0.767,.7 awrt A For any one value of cos or sin, correct method for two values of.7 or (awrt) Both seen anywhere A Checking other values (0.07, 4.0 or 0.869,.449) and discarding (ii) Squaring and forming equation of form a cos + bsin c 9sin + 4cos + sin sin + 5cos Setting up to solve using R formula e.g. cos(.76) (.76) cos 0.56(0... ( α) A (.76) π α, π + α,....7 or (awrt) Both seen anywhere A Checking other values and discarding

124 sinθ cosθ sin θ + cos θ + 7. (a) cosθ sinθ cosθ sinθ Use of common denominator to obtain single fraction cosθ sinθ Use of appropriate trig identity (in this case sin θ + cos θ ) Use of sin θ sinθcosθ sin θ Alt.(a) (b) cosecθ ( ) A cso (4) sinθ cosθ tan θ + + tanθ + cosθ sinθ tanθ tanθ sec θ tanθ cosθ sinθ sin θ cosecθ ( ) (cso) A If show two epressions are equal, need conclusion such as QED, tick, true. y Shape (May be translated but need to see 4 sections ) B O θ (c) cosecθ sin θ Allow sin θ T.P.s at y ±, asymptotic at correct - values (dotted lines not required) [ for equation in sin θ ] (θ ) [ 4.80, 8.89 ; 40.80, ] st for α, 80 α ; nd adding 60 to at least one of values θ 0.9, 69., 00.9, 49. ( d.p.) awrt B dep. (), A ; Note Alt.(c) st A for any two correct, nd A for other two Etra solutions in range lose final A only SC: Final 4 marks:θ 0.9, after M0M0 is B; record as M0M0AA0 tanθ + and form quadratic, tan θ tanθ + 0, A tanθ ( for attempt to multiply through by tanθ, A for correct equation above) ± 5 Solving quadratic [ ta nθ.68 or 0.89 ] θ 69., 49. θ 0.9, 00.9 ( d.p.), A, A ( is for one use of 80 + α, AA as for main scheme) A,A (6) ( marks)

125 8. (a) 8 D 0, t 5, 0e awrt A () (b) D 5 0 0e 8 +, t, e.549 ( ) A cso () Alt.(b) 0e e.549 ( ) A cso (c) T e T 8 e ln T T.06 or. or A () 5 Notes: (b) (main scheme) is for ( 0 0e ) e, or {0 + their(a)}e -(/8) N.B. The answer is given. There are many correct answers seen which deserve M0A0 or A0 (If adding two values, these should be 4.74 and 8.85) (7 marks) (c) st M is for ( 0 0e ) e 5 T T nd 8 M is for converting e k (k > 0) to T ln k 8. This is independent of st M. Trial and improvement: as scheme, correct process for their equation (two equal to s.f.) A as scheme

126 Mark (Results) Summer 008 GCE GCE Mathematics (6665/0) Edecel Limited. Registered in England and Wales No Registered Office: One90 High Holborn, London WCV 7BH

127 June Core Mathematics C Mark. (a) (b) e + + ln ( ln ) A () dy d dy ( ln ) 6 B d + 8e B y 8 6 ( ln ) y ln A (4) [6]

128 . (a) R 5 + R A tanα 5 α.76 cao A (4) (b) cos( α ) 6 6 α arccos.09 A awrt. A α.09 accept 5.9 for M awrt or A (5) (c)(i) R ma ft their R B ft (ii) At the maimum, ( α ) cos or α 0 α.76 awrt., ft their α Aft () []

129 . (a) y shape B Vertices correctly placed B () O (b) y O shape B Verte and intersections with aes correctly placed B () (c) P :(,) B Q :( 0,) B R :,0 ( ) B () (d) > ; A < ; Leading to A + + Leading to 6 A (5) []

130 + B 4. (a) ( )( ) ( ) ( ) ( ) f ( ) + + or ( )( + ) ( )( + ) ( )( + ) A + + cso A (4) ( )( ) (b) 0, 4 < < < < etc. B B () 4 4 Accept 0 y, 0 f( ) (c) Let y f ( ) (d) fg ( ) y + y + y + y or A f ( ) Domain of f is 0, 4 ft their part (b) B ft () A ± 5 both A () []

131 5. (a) sin θ Alternative for (a) sin θ + cos θ sin θ cos θ + sin θ sin θ sin θ + cot θ cosec θ cso A () cos θ sin θ + cos θ cot θ sin sin sin + + θ θ θ cosec θ cso A (b) ( θ ) cosec 9cosecθ cosec θ 9cosecθ 5 0 or 5sin θ + 9sinθ 0 ( cosecθ + )( cosecθ 5) 0 or ( 5sinθ )( sinθ + ) 0 cosecθ 5 or sinθ 5 A θ.5,68.5 A A (6) [8]

132 d ( e sin cos ) e ( sin cos ) e ( cos sin ) A A () d 6. (a)(i) ( ) (ii) ( ) ( e ( sin + 7cos) ) d 5 ( ln 5+ ) ln ( 5+ ) + d 5+ A A () (b) (c) ( + ) ( 6+ 6) ( + )( + 6 7) dy d ( + ) ( + )( ) 0 ( + ) ( + ) d y A A cso A (5) 4 ( + ) ( + ) 4 6, both A () [4] d 4 Note: The simplification in part (b) can be carried out as follows ( + ) ( 6+ 6) ( + )( + 6 7) ( + ) ( ) ( ) ( ) 4 ( + ) ( + ) ( ) ( ) A

133 7. (a) f (.4) < 0 f (.45) 0.45 > 0 Change of sign (and continuity) α (.4,.45) A () (b) A + cso A () (c).47 B.447 B.455 B () (d) Choosing the interval (.445,.455 ) or appropriate tighter interval. f (.445) f (.455) 0.00 Change of sign (and continuity) α (.445,.455) α.45, correct to decimal places cso A () [] Note: α

134 Mark (Results) Summer 009 GCE GCE Mathematics (6665/0)

135 June Core Mathematics C Mark Q (a) Iterative formula: n+ +, 0.5 ( ) n + (.5) An attempt to substitute 0.5 into the iterative formula. Can be implied by. or.0 Both.(0) and awrt.7 Both awrt.56 and 4 awrt.60 or.6 A A cso () (b) Let f ( ) Choose suitable interval for, f (.585) e.g. [.585,.595] or tighter f (.595) any one value awrt sf Sign change (and f ( ) is continuous) therefore a root or truncated sf α is such that α (.585,.595) α.59 ( dp) both values correct, sign change and conclusion At a minimum, both values must be correct to sf or truncated sf, candidate states change of sign, hence root. d A () [6] 6665/0 GCE Mathematics June 009

136 Q (a) cos θ + sin θ ( cos θ ) cos θ sin θ + cos θ cos θ cos θ Dividing cos θ + sin θ by cos θ to give underlined equation. + tan θ sec θ tan θ sec θ (as required) AG Complete proof. No errors seen. A cso () θ < 60 (b) tan θ 4secθ sec θ, ( eqn *) o (sec θ ) + 4secθ + sec θ Substituting tan θ sec θ into eqn * to get a quadratic in secθ only sec θ + 4secθ + sec θ sec θ + 4secθ 4 0 Forming a three term one sided quadratic epression in sec θ. ( θ )( θ ) sec + sec 0 Attempt to factorise or solve a quadratic. secθ or secθ or cosθ cosθ cos θ ; or cosθ cosθ A; o α 0 or α no solutions o θ 0 0 o A θ 40 o 40 o or θ 60 o θ when solving using cos θ... B θ { 0, 40 } o o Note the final A mark has been changed to a B mark. (6) [8] 6665/0 GCE Mathematics June 009

137 Q 5 P 80e t (a) t 0 0 P 80e 5 80() B () t 000 t 5 5 (b) P e e t 5ln 80 Substitutes P 000 and 5 rearranges equation to make e t the subject. t awrt.6 or years A Note t or t awrt.6 t will score A0 () dp t 5 (c) 6e dt (d) e t ke t 5 and k e t A () 50 t 5ln { } dp Using 50 and dt an attempt to solve t to find the value of t or ln 5 6 P 80e or ( ) P 80e 5 Substitutes their value of t back into the equation for P. d 80(50) P or awrt 50 A 6 () [8] 6665/0 GCE Mathematics June 009 4

138 Q4 (i)(a) y cos Apply product rule: u v cos du dv sin d d Applies vu + u v correctly for their u, u, v, v AND gives an epression of the form dy α cos ± β sin cos sin d Any one term correct A Both terms correct and no further simplification to terms in cosα or sin β. A () (b) y ln( + ) + u ln( ) du ln( + ) d + something + + ln( + ) A + Apply quotient rule: u ln( + ) v + du dv d + d ( ) ln( ) y + + d d + ( + ) vu u v Applying v Correct differentiation with correct bracketing but allow recovery. A (4) y d ( + ) d ln( + ) {Ignore subsequent working.} 6665/0 GCE Mathematics June 009 5

139 (ii) y 4 +, > 4 At P, y 4() + 9 At P, y 9 or B ± k (4 + ) * dy ( 4 ) + (4) d (4 + ) A aef dy d (4 + ) At P, dy d Substituting into an equation 4()+ involving d y ; d ( ) Hence m(t) Either T: y ( ) ; or y + c and 5 ( ) c c + ; 4 Either T: y 9 ( ) ; T: y 9 4 y y m( ) or y y m( their stated ) with their TANGENT gradient and their y ; or uses y m + c with their TANGENT gradient, their and their y. d*; T: y y Tangent must be stated in the form a + by + c 0, where a, b and c are integers. A 5 or T: y + (6) T: y + 5 T: y [] 6665/0 GCE Mathematics June 009 6

140 Q5 (a) y Curve retains shape when > ln k B ( 0, k ) Curve reflects through the -ais when < ln k B O ( ) ln k, 0 ( 0, k ) and ( ln k, 0) marked in the correct positions. B (b) ( k, 0) y ( 0, ln k ) Correct shape of curve. The curve should be contained in quadrants, and (Ignore asymptote) B () O ( k, 0) and ( 0, ln k ) B (c) Range of f: f ( ) > k or y > k or ( k, ) (d) y e k y + k e ( ) ( ) ln y + k Either f ( ) > k or y > k or + Makes e ln y k ( k, ) or f > k or Range > k. Attempt to make (or swapped y) the subject the subject and takes ln of both sides B () () Hence f ( ) ln( + k) ln( + k) or ln ( + k ) A cao () (e) f ( ) : Domain: > k or ( k, ) Either > k or ( k, ) or Domain > k or ft one sided inequality their part (c) RANGE answer B () [0] 6665/0 GCE Mathematics June 009 7

141 Q6 (a) ( ) A B cos A + A cos A cos A cos A sin Asin A Applies A B cos A + B to give the underlined equation or cos A cos A sin A to ( ) cos A cos A sin A and gives cos A + sin A cos A sin A sin A sin A (as required) Complete proof, with a link between LHS and RHS. No errors seen. A AG () (b) C C Eliminating y correctly. sin 4sin cos cos sin 4 cos ( ) sin cos cos sin cos cos Using result in part (a) to substitute for sin as ± ± cos or k sin as cos k ± ± to produce an equation in only double angles. sin + 4cos Rearranges to give correct result A AG () (c) sin + 4cos R cos( α ) sin + 4cos R cos cosα + R sin sinα Equate sin : Rsinα Equate cos : 4 R cosα R + 4 ; 5 5 R 5 B o tanα α tanα ± or tanα ± or 4 4 sinα ± or cosα ± their R their R awrt 6.87 A Hence, sin + 4cos 5cos( 6.87) () 6665/0 GCE Mathematics June 009 8

142 (d) sin + 4cos 5cos( 6.87) 5 ( ) cos( their α ) cos 6.87 ± their R ( ) o awrt 66 A ( ) o Hence, o, o One of either awrt 5.6 or awrt A 5.7 or awrt 65. or awrt 65. Both awrt 5.6 AND awrt 65. A If there are any EXTRA solutions inside the range 0 < 80 then withhold the final accuracy mark. Also ignore EXTRA solutions outside the range 0 < 80. (4) [] 6665/0 GCE Mathematics June 009 9

143 Q7 8 f ( ) + ( + 4) ( )( + 4) R, 4,. (a) f ( ) An attempt to combine to one ( )( + 4) ( ) + 8 fraction ( )( + 4) Correct result of combining all three fractions A ( )( + 4) + [( + 4)( ) ] Simplifies to give the correct numerator. Ignore omission of denominator A ( + 4)( ) [( + 4)( ) ] An attempt to factorise the numerator. d ( ) ( ) Correct result A cso AG (5) (b) g( ) e e R, ln. Apply quotient rule: u e v e du dv e e d d g ( ) e (e ) e (e ) (e ) vu u v Applying v Correct differentiation A e e e + e (e ) e (e ) Correct result A AG cso () 6665/0 GCE Mathematics June 009 0

144 (c) e g ( ) (e ) e (e ) e e e e + 4 Puts their differentiated numerator equal to their denominator. e 5e e 5e 4 + A (e 4)(e ) 0 e 4 or e Attempt to factorise or solve quadratic in e ln 4 or 0 both 0, ln 4 A (4) [] 6665/0 GCE Mathematics June 009

145 Q8 (a) sin sin cos sin cos B aef () (b) cosec 8cos 0, 0 < < π 8cos 0 sin Using cosec sin sin 8cos 8sin cos ( ) 4 sin cos 4sin sin 4 sin k, where < k < and k 0 sin 4 A { } { } Radians , Degrees , { } { } Radians , Degrees , Either arwt 7.4 or 8.76 or 0. or.44 or.45 or awrt 0.04π or awrt 0.46 π. A Both 0. and.44 A cao Solutions for the final two A marks must be given in only. If there are any EXTRA solutions inside the range 0 < < π then withhold the final accuracy mark. Also ignore EXTRA solutions outside the range 0 < < π. (5) [6] 6665/0 GCE Mathematics June 009

146 Mark (Results) Summer 00 GCE Core Mathematics C (6665) Edecel Limited. Registered in England and Wales No Registered Office: One90 High Holborn, London WCV 7BH

147 Edecel is one of the leading eamining and awarding bodies in the UK and throughout the world. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. Through a network of UK and overseas offices, Edecel s centres receive the support they need to help them deliver their education and training programmes to learners. For further information, please call our GCE line on , our GCSE team on , or visit our website at If you have any subject specific questions about the content of this Mark that require the help of a subject specialist, you may find our Ask The Epert service helpful. Ask The Epert can be accessed online at the following link: Summer 00 Publications Code UA070 All the material in this publication is copyright Edecel Ltd 00

148 June Core Mathematics C Mark sinθ cosθ. (a) + cos θ (b) sinθ cosθ cosθ cosθ tanθ (as required) AG A cso tanθ tanθ θ awrt 6.6 A θ awrt 5.4 A () () [5] (a) : Uses both a correct identity for sin θ and a correct identity for cos θ. Also allow a candidate writing + cosθ cos θ on the denominator. Also note that angles must be consistent in when candidates apply these identities. A: Correct proof. No errors seen. (b) st for either tanθ or tanθ, seen or implied. A: awrt 6.6 A : awrt 5.4 or θ 80 + θ Special Case: For candidate solving, tan θ k, where k, to give θ and θ 80 + θ, then award M0A0B in part (b). Special Case: Note that those candidates who writes tanθ, and gives ONLY two answers of 45 and 5 that are inside the range will be awarded SC M0A0B. GCE Core Mathematics C (6665) Summer 00

149 . At P, y B d y ( ) ( 5 ) 8 ( ) or A d (5 ) dy 8 { 8} d (5 ()) m(n) or 8 8 N: y ( ) 8 N: 8 y+ 5 0 A [7] st : ± k(5 ) can be implied. See appendi for application of the quotient rule. nd : Substituting into an equation involving their d y ; d rd : Uses m(n) their m( T ). 4 th : y y m( ) with their NORMAL gradient or a changed tangent gradient and their y. Or uses a complete method to epress the equation of the tangent in the form y m+ c with their NORMAL ( changed numerical) gradient, their y and. Note: To gain the final A mark all the previous 6 marks in this question need to be earned. Also there must be a completely correct solution given. GCE Core Mathematics C (6665) Summer 00

150 . (a) f (.) , f (.) Sign change (and as f ( ) is continuous) therefore a root α is such that [.,.] α A (b) 4cosec cosec + cosec sin 4 A (c) + sin(.5) , A A (d) f (.905) , f (.95) Sign change (and as f ( ) is continuous) therefore a root α is such that α α.9 ( dp) (.905,.95) (a) : Attempts to evaluate both f (.) and f (.) and evaluates at least one of them correctly to awrt (or truncated) sf. A: both values correct to awrt (or truncated) sf, sign change and conclusion. (b) : Attempt to make 4 or the subject of the equation. A: Candidate must then rearrange the equation to give the required result. It must be clear that candidate has made their initial f ( ) 0. (c) : An attempt to substitute 0.5 into the iterative formula Eg +. sin(.5) 4 Can be implied by awrt. or awrt 46. A: Both awrt.08 and awrt.867 A: awrt.97 (d) : Choose suitable interval for, e.g. [.905,.95] or tighter and at least one attempt to evaluate f(). A: both values correct to awrt (or truncated) sf, sign change and conclusion. A () () () () [9] GCE Core Mathematics C (6665) Summer 00

151 4. (a) y ( 0, 5 ) A (b) 0 B 0 5 (5 + ) ; ;A oe. (c) fg() f( ) ( ) 5; ;A (d) g( ) 4+ ( ) 4 + ( ). Hence gmin Either gmin or g( ) or g(5) B g( ) 6 or y 6 A (a) : V or or graph with verte on the -ais. 5 A: (, { 0 }) and ({ } ) 0, 5 seen and the graph appears in both the first and second quadrants. (b) : Either 5 (5 + ) or ( 5) 5+ (c) : Full method of inserting g() into f( ) 5 or for inserting into O 5 (,0 ) +. There must be evidence of the modulus being applied. ( 4 ) 5 (d) : Full method to establish the minimum of g. Eg: ( ± α ) + β leading to g min β. Or for candidate to differentiate the quadratic, set the result equal to zero, find and insert this value of back into f ( ) in order to find the minimum. B: For either finding the correct minimum value of g (can be implied by g( ) or g( ) > ) or for stating that g(5) 6. A: g( ) 6 or y 6 or g 6. Note that: 6 is A0. Note that: f( ) 6 is A0. Note that: g( ) 6 is A0. Note that: g( ) or g( ) > or or > with no working gains BA0. Note that for the final Accuracy Mark: If a candidate writes down < g( ) < 6 or < y < 6, then award BA0. If, however, a candidate writes down g( ), g( ) 6, then award A0. If a candidate writes down g( ) or g( ) 6, then award A0. () () () () [0] GCE Core Mathematics C (6665) Summer 00

152 5. (a) Either y or ( ) 0, B (b) When, y (8 0 + )e 0e 0 B ( 5+ ) 0 ( )( ) 0 Either (for possibly B above) or A (c) dy d. (4 5)e ( 5+ )e AA (d) (4 5)e ( 5+ )e ( 7)( ) 0 7 A When 7 7 y, 9e, when, y e dda (b) If the candidate believes that e 0 solves to 0 or gives an etra solution of 0, then withhold the final accuracy mark. (c) : (their u )e + ( 5+ )(their v') A: Any one term correct. A: Both terms correct. (d) st : For setting their d y d found in part (c) equal to 0. nd : Factorise or eliminate out e correctly and an attempt to factorise a -term quadratic or apply the formula to candidate s a + b + c. See rules for solving a three term quadratic equation on page of this Appendi. rd dd: An attempt to use at least one -coordinate on y ( 5+ )e. Note that this method mark is dependent on the award of the two previous method marks in this part. Some candidates write down corresponding y-coordinates without any working. It may be necessary on some occasions to use your calculator to check that at least one of the two y-coordinates found is correct to awrt sf. 7 7 Final A: Both{ }, y e and { }, y 9e. cao Note that both eact values of y are required. () () () (5) [] GCE Core Mathematics C (6665) Summer 00

153 6. (a) (i) (, 4 ) B B (ii) ( 6, 8) B B (b) y (4) 5 B B B O (, 4) (, 4) (c) f( ) ( ) 4 or f( ) 6 5 (d) Either: The function f is a many-one {mapping}. Or: The function f is not a one-one {mapping}. + A (b) B: Correct shape for 0, with the curve meeting the positive y-ais and the turning point is found below the -ais. (providing candidate does not copy the whole of the original curve and adds nothing else to their sketch.). B: Curve is symmetrical about the y-ais or correct shape of curve for < 0. Note: The first two BB can only be awarded if the curve has the correct shape, with a cusp on the positive y-ais and with both turning points located in the correct quadrants. Otherwise award BB0. B: Correct turning points of (, 4) and (, 4 ). Also, ({ 0,5 } ) is marked where the graph cuts through the y-ais. Allow (5, 0) rather than (0, 5) if marked in the correct place on the y-ais. (c) : Either states f ( ) in the form ( ± α) ± β ; α, β 0 Or uses a complete method on f( ) + a + b, with f (0) 5 and f() 4 to find both a and b. A: Either ( ) 4 or (d) B: Or: The inverse is a one-many {mapping and not a function}. Or: Because f (0) 5 and also f (6) 5. Or: One y-coordinate has corresponding -coordinates {and therefore cannot have an inverse}. B () () () [0] GCE Core Mathematics C (6665) Summer 00

154 7. (a) R 6.5 or.5 B tanα α awrt A.5 4 (b) (i) Ma Value.5 B π (ii) sin ( θ 0.645) or θ theirα ; θ awrt. ;A (c) H Ma 8.5 (m) B 4πt 4 sin or π t their (b) answer ; t awrt 4.4 ;A 5 5 (d) 4π t 4π t 6 +.5sin ; sin ; π t sin (0.4) or awrt 0.4 A 5 Either t awrt. or awrt 6.7 A c So, { π or } 4π t 5 dd Times { 4 : 06, 8: 4} A (6) [5] (a) B: R.5 or R 6.5. For R ±.5, award B0. :.5 tanα ± or tanα ± A: α awrt (b) B.5 :.5 or follow through the value of R in part (a). sin θ theirα : For ( ) π A : awrt. or + their α rounding correctly to sf. (c) B : 8.5 or 6 + their R found in part (a) as long as the answer is greater than 6. 4π t : sin ± their α or 4 π t their (b) answer 5 5 A: For sin (0.4) This can be implied by awrt 4.4 or awrt π t 6 + their sin ± their α 7, : 5 4π t sin ± their α 5 their R (d) : ( R) A: For sin (0.4). This can be implied by awrt 0.4 or awrt.7 or other values for different α ' s. Note this mark can be implied by seeing.055. A: Either t awrt. or t awrt 6.7 dd: either π their PV c. Note that this mark is dependent upon the two M marks. This mark will usually be awarded for seeing either.70 or.7 A: Both t 4 : 06 and t 8: 4 or both 6 (min) and 40 (min) or both hr 6 min and 6 hr 4 min. () () () GCE Core Mathematics C (6665) Summer 00

155 8. (a) (b) ( + 5)( ) ( + 5)( ) ( ) ( ) ln e + 5 e e (e ) e e (a) : An attempt to factorise the numerator. B: Correct factorisation of denominator to give ( + 5)( ). Can be seen anywhere. (b) : Uses a correct law of logarithms to combine at least two terms. This usually is achieved by the subtraction law of logarithms to give ln. + 5 The product law of logarithms can be used to achieve ln ln e + 5. ( ) ( ( )) The product and quotient law could also be used to achieve ln 0. e( + 5) d: Removing ln s correctly by the realisation that the anti-ln of is e. Note that this mark is dependent on the previous method mark being awarded. : Collect terms together and factorise. Note that this is not a dependent method mark. A: e e or or e. aef e e -e Note that the answer needs to be in terms of e. The decimal answer is Note that the solution must be correct in order for you to award this final accuracy mark. Note: See Appendi for an alternative method of long division. B A aef d A aef cso () (4) [7] GCE Core Mathematics C (6665) Summer 00

156

157 Further copies of this publication are available from Edecel Publications, Adamsway, Mansfield, Notts, NG8 4FN Telephone Fa Order Code UA070 Summer 00 For more information on Edecel qualifications, please visit Edecel Limited. Registered in England and Wales no Registered Office: One90 High Holborn, London, WCV 7BH

158 Mark (Results) June 0 GCE Core Mathematics C (6665) Paper

159 Edecel is one of the leading eamining and awarding bodies in the UK and throughout the world. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. Through a network of UK and overseas offices, Edecel s centres receive the support they need to help them deliver their education and training programmes to learners. For further information, please call our GCE line on or visit our website at If you have any subject specific questions about the content of this Mark that require the help of a subject specialist, you may find our Ask The Epert service helpful. Ask The Epert can be accessed online at the following link: June 0 Publications Code UA07660 All the material in this publication is copyright Edecel Ltd 0

160 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) should not be subdivided.. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes and can be used if you are using the annotation facility on epen. 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) dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper The second mark is dependent on gaining the first mark

161 (a),,a () (b) Applying, oe A,,0 () 5 (a) f(0.75) -0.8 f(0.85) Change of sign, hence root between 0.75 and 0.85 A () (b) Sub into to obtain Awrt and 0.80 A Awrt A (c ) f( ) f( ) A () Change of sign and conclusion A () See Notes for continued iteration method 8 GCE Core Mathematics C (6665) June 0 4

162 (a) y V shape verte on y ais &both branches of graph cross ais y co-ordinate of R is -6 B B B (0,-6) () (b) y (-4,) W shape vertices on the negative ais. W in both quad & quad. R (-4,) B Bdep B () 6 4 (a) oe A () (b) B () (c ) (d) da Bft () () 8 GCE Core Mathematics C (6665) June 0 5

163 5 (a) (b) p7.5 B () d A* See notes for additional correct solutions and the last A (4) (c ) ft on their p and k Aft A d t or awrt 4. A (6) GCE Core Mathematics C (6665) June 0 6

164 6 (a) A tan cso A* (4) (b)(i) cso d A* () (b)(ii) A A(any two) A (5) Alt for (b)(i) Rationalises to produce A* GCE Core Mathematics C (6665) June 0 7

165 7 (a) B A A* (5) (b) A A Uses m m - to give gradient of normal or any equivalent form A (8) GCE Core Mathematics C (6665) June 0 8

166 8 (a) A A (4) (b) AA A* cso (5) (c ) 0.96 awrt 0.0 A () Alternative to part (c ) 0.96 awrt 0.0 A () GCE Core Mathematics C (6665) June 0 9

167 Further copies of this publication are available from Edecel Publications, Adamsway, Mansfield, Notts, NG8 4FN Telephone Fa Order Code UA07660 June 0 For more information on Edecel qualifications, please visit Pearson Education Limited. Registered company number 8788 with its registered office at Edinburgh Gate, Harlow, Esse CM0 JE

168 Mark (Results) Summer 0 GCE Core Mathematics C (6665) Paper

169 Edecel and BTEC Qualifications Edecel and BTEC qualifications come from Pearson, the world s leading learning company. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information, please visit our website at Our website subject pages hold useful resources, support material and live feeds from our subject advisors giving you access to a portal of information. If you have any subject specific questions about this specification that require the help of a subject specialist, you may find our Ask The Epert service helpful. Pearson: helping people progress, everywhere 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: Summer 0 Publications Code UA0956 All the material in this publication is copyright Pearson Education Ltd 0

170 Summer Core Mathematics C Mark 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.

171 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) should not be subdivided.. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes and can be used if you are using the annotation facility on epen. 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) dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper 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.

172 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 ( + 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 correct formula (with values for a, b and c), leading to. Completing the square Solving + b + c 0 b : ( ) ± ± q ± c, 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 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. 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. 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.

173 . + At any stage B 9 4 ( )( ) Eliminating the common factor of (+) at any stage ( + ) ( )( + ) B Use of a common denominator + + ( )( ) (9 4) (9 4)( + ) (9 4)( + ) or ( + ) ( ) ( )( + ) ( + )( ) 6 6 or ( )( + ) 9 A (4 marks) Notes B For factorising 9 4 ( )( + ) using difference of two squares. It can be awarded at any stage of the answer but it must be scored on E pen as the first mark B For eliminating/cancelling out a factor of (+) at any stage of the answer. For combining two fractions to form a single fraction with a common denominator. Allow slips on the numerator but at least one must have been adapted. Condone invisible brackets. Accept two separate fractions with the same denominator as shown in the mark scheme. Amongst possible (incorrect) options scoring method marks are ( + ) (9 4) Only one numerator adapted, separate fractions (9 4)( + ) (9 4)( + ) + Invisible brackets, single fraction ( )( + ) 6 A ( )( + ) This is not a given answer so you can allow recovery from invisible brackets. Alternative method ( + ) ( + )( + ) (9 4) 8 + has scored 0,0,,0 so far (9 4) ( + ) (9 4)( + ) (9 4)( + ) 6( + ) is now,,,0 ( + )( )( + ) 6 and now,,, ( )( + )

174 . (a) (b) awrt ( + ) 4 4 4( ) ( + ) ( + ) da*.4,.0. A,A (c ) Choosing (.75,.75) or tighter containing root.7998 () () Notes (a) f (.75) ( + ) f (.75) Change of sign α.7 Moves from f()0, which may be implied by subsequent working, to ( ± ) ± ± 4 by separating terms and factorising in either order. No need to factorise rhs for this mark. d Divides by (+) term to make the subject, then takes square root. No need for rhs to be factorised at this stage A* CSO. This is a given solution. Do not allow sloppy algebra or notation with root on just numerator for instance. The -4 needs to have been factorised. A () (9 marks) (b) Note that this appears B,B,B on EPEN An attempt to substitute 0 into the iterative formula to calculate. This can be awarded for the sight of 4( ) 8,, and even.4 (+ ) 4 A.4. The subscript is not important. Mark as the first value found, is A0 A awrt.0 awrt.. Mark as the second and third values found. Condone. for (c ) Note that this appears AA on EPEN Choosing the interval (.75,.75) or tighter containing the root Continued iteration is not allowed for this question and is M0 Calculates f(.75) and f(.75), or the tighter interval with at least correct to sig fig rounded or truncated. Accept f(.75) sf rounded or truncated. Also accept f(.75) -0.0 dp Accept f(.75) (+)0.008 sf rounded or truncated. Also accept f(.75) (+)0.0 dp A Both values correct (see above), A valid reason; Accept change of sign, or >0 <0, or f(.75) f(.75)<0 And a (minimal) conclusion; Accept hence root or α.7 or QED or

175 Alternative to (a) working backwards (a) 4( ) 4( ) ( + ) ( + ) + ( ) 4( ) d States that this is f()0 A* Alternative starting with the given result and working backwards Square (both sides) and multiply by (+) d Epand brackets and collect terms on one side of the equation 0 A A statement to the effect that this is f()0 () An acceptable answer to (c ) with an eample of a tighter interval Choosing the interval (.75,.70). This contains the root.79(98) Calculates f(.75) and f(.70), with at least correct to sig fig rounded or truncated. Accept f(.75) sf rounded or truncated f(.75) -0.0 dp Accept f(.70) (+) sf rounded or f(.70) (+) truncated sf A Both values correct (see above), A valid reason; Accept change of sign, or >0 <0, or f(.75) f(.70)<0 And a (minimal) conclusion; Accept hence root or α.7 or QED or f() ( ) An acceptable answer to (c ) using g() where g() ( + ) nd Calculates g(.75) and g(.75), or the tighter interval with at least correct to sig fig rounded or truncated. g(.75) Accept g(.75) awrt (+) sf rounded or awrt truncated. g(.75) Accept g(.75) awrt sf rounded or awrt truncated.

176 . (a) dy e sin e cos d A dy 0 d e ( sin + cos ) 0 tan A π π A 9 (6) dy (b) At 0 d B y 0 Equation of normal is or any equivalent y A 0 () (9 marks) (a) Applies the product rule vu +uv to e sin. If the rule is quoted it must be correct and there must have been some attempt to differentiate both terms. If the rule is not quoted (nor implied by their working, ie. terms are written out u,u.,v.,v.followed by their dy vu +uv ) only accept answers of the form Ae sin e B cos d A dy Correct epression for e sin e cos d Sets their d y 0 d e producing an equation in sin and cos A Achieves either tan or tan Correct order of arctan, followed by. A 5π 5π π π Accept or but not arctan( ) 9 9 π CS0 Ignore etra solutions outside the range. Withhold mark for etra inside the range. 9 (b) B Sight of for the gradient A A full method for finding an equation of the normal. Their tangent gradient m must be modified to y 0 Eg or equivalent is acceptable their ' m' 0 y 0 y or any correct equivalent including. 0 and used together with (0, 0). m

177 Alternative in part (a) using the form Rsin( + α ) JUST LAST MARKS. (a) dy e sin e cos d A dy 0 d e ( sin + cos ) 0 π ( )sin( + ) 0 A π π A 9 (6) π π A Achieves either ( )sin( + ) 0 or ( ) cos( ) 0 6 Correct order of arcsin or arcos, etc to produce a value of π Eg accept + 0 or π or π... π A Cao Ignore etra solutions outside the range. Withhold mark for etra inside the range. 9 Alternative to part (a) squaring both sides JUST LAST MARKS. (a) dy e sin e cos d A dy 0 d e ( sin + cos ) 0 sin cos cos ( ) or sin ( ) 4 4 A arcos( ± ) 4 oe π 9 A

178 4.(a) Shape including cusp B (-.5, 0) and (0, 5) B () (b) Shape B (0,5) B () ( c) Shape B (0,0) B (-0.5, 0) B () (7 marks) (a) Note that this appears as A on EPEN B Shape (inc cusp) with graph in just quadrants and. Do not be overly concerned about relative gradients, but the left hand section of the curve should not bend back beyond the cusp B This is independent, and for the curve touching the -ais at (-.5, 0) and crossing the y-ais at (0,5) (b) Note that this appears as A on EPEN B For a U shaped curve symmetrical about the y- ais B (0,5) lies on the curve (c ) Note that this appears as BB on EPEN B Correct shape- do not be overly concerned about relative gradients. Look for a similar shape to f() B Curve crosses the y ais at (0, 0). The curve must appear in both quadrants and B Curve crosses the ais at (-0.5, 0). The curve must appear in quadrants and. In all parts accept the following for any co-ordinate. Using (0,) as an eample, accept both (,0) or written on the y ais (as long as the curve passes through the point) Special case with (a) and (b) completely correct but the wrong way around mark - SC(a) 0, SC(b) 0, Otherwise follow scheme

179 5. (a) 4 4cosec θ cosec θ sin θ sin θ 4 (sinθ cos θ ) sin θ B B 4 4 (sinθ cos θ ) sin θ 4sin θ cos θ sin θ (b) Using cos θ sin θ (c ) Note (a) and (b) can be scored together (a) B One term correct. Eg. writes cos θ sin θ cos θ sin θ cos θ sin θ sin θ cos θ sec θ cos θ A* sec θ 4 secθ ± cosθ ± π π θ, A,A 4 (sinθ cos θ ) 4cosec θ as or cosec θ as (9 marks) () (4) (). Accept terms like sin θ cos θ cosec θ + cot θ +. The question merely asks for an epression in sin θ and cos θ sin θ 4 B A fully correct epression in sin θ and cos θ. Eg. Accept equivalents (sinθ cos θ ) sin θ Allow a different variable say s instead of θ s but do not allow mied units. b) Attempts to combine their epression in sinθ and cosθ using a common denominator. The terms can be separate but the denominator must be correct and one of the numerators must have been adapted Attempts to form a single term on the numerator by using the identity cos θ sin θ Cancels correctly by sin θ terms and replaces with sec θ cos θ A* Cso. This is a given answer. All aspects must be correct IF IN ANY DOUBT SEND TO REVIEW OR CONSULT YOUR TEAM LEADER c ) A A For sec θ 4 leading to a solution of cosθ by taking the root and inverting in either order. Similarly accept tan θ, sin θ leading to solutions of tan θ, sin θ. Also accept cosθ 4 π Obtains one correct answer usuallyθ Do not accept decimal answers or degrees π π Obtains both correct answers. θ, Do not award if there are etra solutions inside the range. Ignore solutions outside the range.

180 6. (a) f()> B () (b) ln fg( ) e, + +,A () (c ) e + 6 e 4 A + ln 4 ln 4 or ln- A + + (d) Let y e + y e ln( y ) (4) f ( ) ln( ),. > A, Bft () (e) Shape for f() B (0, ) B Shape for f - () B (, 0) B (4) (4 marks) (a) B Range of f()>. Accept y>, (, ), f>, as well as range is the set of numbers bigger than but don t accept > (b) For applying the correct order of operations. Look for ln e +. Note that ln e + is M0 A ln Simplifies e + to +. Just the answer is acceptable for both marks (c ) Starts with e and proceeds to e +... A e + 4 Takes ln s both sides, + ln.. and proceeds to. ln 4 A oe. eg ln - Remember to isw any incorrect working after a correct answer

181 (d) Note that this is marked AA on EPEN y Starts with y e + or e + and attempts to change the subject. All ln work must be correct. The must be dealt with first. Eg. y e + ln y + ln ln y ln is M0 A There must be some form of bracket f ( ) ln( ) or y ln( ) or y ln Bft Either >, or follow through on their answer to part (a), provided that it wasn t y R Do not accept y> or f - ()>. (e) B B B B Shape for ye. The graph should only lie in quadrants and. It should start out with a gradient that is appro. 0 above the ais in quadrant and increase in gradient as it moves into quadrant. You should not see a minimum point on the graph. (0, ) lies on the curve. Accept written on the y ais as long as the point lies on the curve Shape for yln. The graph should only lie in quadrants 4 and. It should start out with gradient that is appro. infinite to the right of the y ais in quadrant 4 and decrease in gradient as it moves into quadrant. You should not see a maimum point. Also with hold this mark if it intersects ye (, 0) lies on the curve. Accept written on the ais as long as the point lies on the curve Condone lack of labels in this part Eamples Scores,0,,0. Both shapes are fine, do not be concerned about asymptotes appearing at, y. (See notes) Both co-ordinates are incorrect Scores 0,,, Shape for y e is incorrect, there is a minimum point on the graph. All other marks an be awarded