GCE Edexcel GCE. Core Mathematics C4 (6666) June Edexcel GCE. Mark Scheme (Final) Core Mathematics C4 (6666)

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1 GCE Edecel GCE Core Mathematics C4 (6666) June 006 Mark (Final) Edecel GCE Core Mathematics C4 (6666)

2 June Pure Mathematics C4 Mark Question. dy dy dy = 6 4y + = 0 d d d Differentiates implicitly to include either dy dy ± ky or ±. (Ignore dy = d d d.) Correct equation. M A dy 6 + = d 4y + not necessarily required. At (0, ), dy + = 0 = d 4 + Substituting = 0 & y = into an equation involving dy d ; to give or dm; A cso Hence m(n) = or Uses m(t) to correctly find m(n). Can be ft from their tangent gradient. A oe. Either N: y = ( 0) or N: y = + y = m( 0) with their tangent or normal gradient ; or uses y = m + with their tangent or normal gradient ; M; N: + y = 0 Correct equation in the form 'a + by + c = 0', where a, b and c are integers. A oe cso marks [] Beware: dy = does not necessarily imply the award of all the first four marks in this question. d So please ensure that you check candidates initial differentiation before awarding the first A mark. Beware: The final accuracy mark is for completely correct solutions. If a candidate flukes the final line then they must be awarded A0. Beware: A candidate finding an m(t) = 0 can obtain Aft for m(n) =, but obtains M0 if they write y = ( 0). If they write, however, N: = 0, then can score M. Beware: A candidate finding an m(t) = y = 0( 0) or y =. can obtain Aft for m(n) = 0, and also obtains M if they write Beware: The final cso refers to the whole question. 6666/0 Core Maths C4 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

3 . d d d = 6 4y + = 0 dy dy dy Way d 4y + = dy 6 + Differentiates implicitly to include either d d d ± k or ±. (Ignore = dy dy.) dy Correct equation. not necessarily required. M A At (0, ), d 4 + = = dy 0 + Substituting = 0 & y = into an equation involving d dy ; to give dm; A cso Hence m(n) = or Uses m(t) or d dy to correctly find m(n). d Can be ft using.. dy A oe. Either N: y = ( 0) or N: y = + y = m( 0) with their tangent, d or normal gradient ; dy or uses y = m + with their tangent, d or normal gradient ; dy M; N: + y = 0 Correct equation in the form 'a + by + c = 0', where a, b and c are integers. A oe cso marks 6666/0 Core Maths C4 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

4 . y + y 5 = 0 Way 9 5 y+ = y = ( 6) ( ) dy = d Differentiates using the chain rule; Correct epression for dy d. M; A oe At (0, ), dy 49 4 = = = d 6 Substituting = 0 into an equation involving dy d ; to give or dm A cso Hence m(n) = Uses m(t) to correctly find m(n). Can be ft from their tangent gradient. A Either N: y = ( 0) or N: y = + y = m( 0) with their tangent or normal gradient ; or uses y = m + with their tangent or normal gradient M N: + y = 0 Correct equation in the form 'a + by + c = 0', where a, b and c are integers. A oe [] marks 6666/0 Core Maths C4 4 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

5 . (a) A( ) + B Let = ; = B B = Considers this identity and either substitutes =, equates coefficients or solves simultaneous equations complete M Equate terms; = A A = A = ;B = A;A (No working seen, but A and B correctly stated award all three marks. If one of A or B correctly stated give two out of the three marks available for this part.) [] (b) f() = + Moving powers to top on any one of the two epressions M ( )( ) ( )( )( ) = ;...!! Either ± or ± 4 from either first or second epansions respectively dm; ( )( ) ( )( )( 4) ;...!! Ignoring and, any one correct {...} epansion. Both {...} correct. A A { } { 4...} = = ; ; (0 ) 4 A; A 9 marks Beware: In part (a) take care to spot that A = and B = are the right way around. Beware: In epen, make sure you aware the marks correctly in part (a). The first A is for A = second A is for B =. Beware: If a candidate uses a method of long division please escalate this to you team leader. and the 6666/0 Core Maths C4 5 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

6 . (b) Way f() = ( )( ) Moving power to top M ( )( ) + ( )( ); + ( ) +! = ( ) ( )( )( 4)... +! ± 4; Ignoring ( ), correct (...) epansion dm; A = ( )( ) = Correct epansion A = ; (b) Maclaurin epansion Way f() = + f () = ( ) + ( ) + ; (0 ) 4 Bringing both powers to top Differentiates to give a ± b( ) ; ( ) + ( ) A; A M M; A oe f () = ( ) + ( ) = + Correct f()andf () f () 96 A f(0) =, f (0) =, f (0) = 0 and f (0) = 4 gives f() = ; ; (0 ) 4 A; A 6666/0 Core Maths C4 6 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

7 . (b) Way 4 f() = ( 4) + ( ) = ( )( ) + +! () () ( 4); () ( 4) ( )( )( ) () 4 ( 4) ! Moving powers to top on any one of the two epressions Either ± or ± 4 from either first or second epansions respectively M dm; ( )( ) ( )( )( 4) ;...!! Ignoring and, any one correct {...} epansion. Both {...} correct. A A { } { } = = ; ; (0 ) 4 A; A 6666/0 Core Maths C4 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

8 . (a) Area Shaded = sin 0 d cos = 0 sin to give Integrating ( ) kcos with k. Ignore limits. M 6cos 6cos ( ) or cos( ) = 0 A oe. = [ 6( ) ] [ 6() ] = = A cao [] (Answer of with no working scores M0A0A0.) (b) Volume = ( ) = sin d 9 sin d 0 0 Use of V = y d. Can be implied. Ignore limits. M NB : cos =± ± sin gives sin = NB : cos =± ± sin gives sin = cos cos Consideration of the Half Angle sin or the Formula for Double Angle Formula for sin M cos Volume = 9 d 0 9 = ( cos ) d 0 Correct epression for Volume Ignore limits and. A [ ] 9 = sin 9 = ( 0) (0 0) [ ] 0 Integrating to give ± a ± bsin ; Correct integration k kcos k ksin depm ; A 9 = ( ) = 9 or Use of limits to give either 9 or awrt 88.8 Solution must be completely correct. No flukes allowed. A cso 9 marks 6666/0 Core Maths C4 8 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

9 Note: is not needed for the middle four marks of question (b). Beware: Owing to the symmetry of the curve between = 0 and = candidates can find: Area = sin d in part (a). 0 Volume = ( sin) 0 d Beware: If a candidate gives the correct answer to part (b) with no working please escalate this response up to your team leader. 6666/0 Core Maths C4 9 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

10 4. (a) sint, = y = sin( t+ ) 6 d dy = cos t, = cos( t + 6 ) Attempt to differentiate both and y wrt t to give two terms in cos Correct d and dy M A When t =, 6 dy cos( + 6 6) = = = = awrt 0.58 d cos When t =, 6 6, y = = The point ( ) Divides in correct way and substitutes for t to give any of the four underlined oe: Ignore the double negative if candidate has differentiated sin cos A, or(,awrt0.8 ) B T: y = ( ) or = + c c = = 6 Finding an equation of a tangent with their point and their tangent gradient or finds c and uses y = (their gradient) + "c ". Correct EXACT equation of tangent oe. dm A oe or T: y = + y sin t sint cos costsin = + = (b) Use of compound angle formula for sine. M Nb : sin t + cos t cos t sin t = sint gives cos t = ( ) Use of trig identity to find cos t in terms of or cos t in terms of. M y = sint + cost gives y ( ) = + AG Substitutes for sin t, cos, cos t and sin to 6 6 give y in terms of. A cso [] 9 marks 6666/0 Core Maths C4 0 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

11 4. (a) sint, Way d = y sin t sint cos costsin (Do not give this for part (b)) = + = dy = cos t, = cos t cos sin t sin 6 6 Attempt to differentiate and y wrt t to give d in terms of cos and dy in the form ± acos t ± bsin t Correct d and dy M A When t =, 6 When t =, 6 dy d cos cos sin sin = cos = = = = awrt 0.58 =, y = 6 Divides in correct way and substitutes for t to give any of the four underlined oe: The point (, ) A B or (,awrt0.8 ) T: y = ( ) or = + c c = = 6 Finding an equation of a tangent with their point and their tangent gradient or finds c and uses y = (their gradient) + "c ". Correct EXACT equation of tangent oe. dm A oe or T: y = /0 Core Maths C4 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

12 4. (a) y = + ( ) Way dy = + d Attempt to differentiate two terms using the chain rule for the second term. Correct dy d M A dy = + ( (0.5) ) ( (0.5) ) d = When t =, 6, y = = The point ( ) Correct substitution of = into a correct dy d A, or(,awrt0.8 ) B T: y = ( ) Finding an equation of a tangent with their point and their tangent gradient or finds c and uses y = (their gradient) + "c ". Correct EXACT equation of tangent oe. dm A oe or = + c c = = 6 or T: 4. (b) sint Way y = + = gives y = sint + ( sin t ) Nb : sin t + cos t cos t sin t Substitutes = sintinto the equation give in y. M cos t = sin t Use of trig identity to deduce that cos t = sin t. M gives y = sin t + cos t = + = sin( t + ) Hence y sint cos costsin Using the compound angle formula to prove y = sin( t + 6 ) A cso [] 9 marks 6666/0 Core Maths C4 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

13 5. (a) Equating i ; 0= 6 +λ λ = 6 λ= 6 B d Can be implied Using λ= 6 and equating j ; a = 9 + 4( 6) = 5 For inserting their stated λ into either a correct j or k component Can be implied. M d equating k ; b = ( 6) = a = 5 and b = A With no working only one of a or b stated correctly gains the first marks. both a and b stated correctly gains marks. (b) OP = ( 6 +λ ) i + ( 9 + 4λ ) j + ( λ) k [] direction vector or l = d = i + 4j k OP l OP d = 0 Allow this statement for M if OP and d are defined as above. ie. 6+λ 9 + 4λ 4 = 0 λ ( or + 4y z = 0) Allow either of these two underlined statements M 6+λ + 4(9+ 4 λ) ( λ ) = 0 Correct equation A oe 6 +λ λ + + 4λ = 0 Attempt to solve the equation in λ dm λ+ 84 = 0 λ= 4 λ= 4 A OP = ( 4) + ( 4) k i j OP = i + j + k Substitutes their λ into an epression for OP Note: A similar method may be used by using OP = ( 0 +λ ) + ( 5 + 4λ ) + ( λ) OP d = 0 yields 6 +λ + 4( λ) ( λ ) = 0 This simplifies to λ 4 = 0 λ=. OP = ( 0 + ) i + ( 5 + 4() ) j + ( () ) k OP = i + j + k M i + j + k or P (,, ) A i j k and d = i + 4j k 6666/0 Core Maths C4 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

14 (b) OP = ( 6 +λ ) i + ( 9 + 4λ ) j + ( λ) Way AP = 6 +λ λ+ 5 + λ k i j k direction vector or l = d = i + 4j k Allow this statement AP OP AP OP = 0 for M if AP and OP are defined as above. ie. 6+λ 6+λ 4 + 4λ 9 + 4λ = 0 λ λ underlined statement M ( 6 +λ )(6 +λ ) + (4 + 4 λ )(9 + 4 λ ) + ( λ)( λ ) = 0 Correct equation A oe 6 + λ +λ λ + 6λ + 6λ + + 4λ + λ + 4λ = 0 Attempt to solve the equation in λ dm λ + 0λ+ 504 = 0 λ + 0λ+ 4 = 0 λ = 6 λ = 4 λ= 4 A OP = ( 4) + ( 4) k i j Substitutes their λ into an epression for OP M OP = i + j + k i + j + k or P(,, ) A i j k Note: A similar method to way may be used by using OP = ( 5 +λ ) + ( 5 + 4λ ) + ( λ) and AP = ( 5 +λ 0) i + ( 5 + 4λ+ 5) j + ( λ )k AP OP = 0 yields ( 5 +λ )(5 +λ ) + (0 + 4 λ )(5 + 4 λ ) + ( 0 λ)( λ ) = 0 This simplifies to OP = 5 i ( ) j + ( ) OP = i + j + k λ + λ+ = ( 5) k λ + λ+ = λ = λ = 6666/0 Core Maths C4 4 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

15 5. (c) OP = i + j + k OA = 0i 5 j + k and OB = 5i + 5 j + k AP =± AB =± ( i j k ), PB =± ( i + j 6k) ( i j k ) AP = + 6 = PB 5 5 AB 8 4 AP 5 5 AB = + 6 = PB PB 8 4 AP AP = = AB 5 5 PB = = AB etc As ( i j k) or = ( i + j k) = or ( i j k) or = ( i + j k) = or ( i j k) or ( i j k) 5 5 Subtracting vectors to find any two of AP, PB or AB ; and both are correctly ft using candidate s OA and OP found in parts (a) and (b) respectively. AP = PB 5 or AB = AP 5 or AB = PB or PB = AP or AP = AB 5 or PB = AB 5 M; A ± alternatively candidates could say for eample that AP = i + 4 j k PB = i + 4 j k then the points A, P and B are collinear. AP :PB = : A, P and B are collinear Completely correct proof. A : or : or 84 : 89 aef B oe allow SC [4] 5. (c) Way At B; 5= 6 +λ,5= 9+ 4λ or = λ or at B; λ= gives λ= for all three equations. or when λ=, this gives r = 5i + 5j + k Writing down any of the three underlined equations. λ = for all three equations or λ = gives r = 5i + 5j + k M A Hence B lies on l. As stated in the question both A and P lie on l. A, P and B are collinear. AP :PB = : Must state B lies on l A, P and B are collinear A : or aef B oe 6666/0 Core Maths C4 5 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics [4] marks Beware of candidates who will try to fudge that one vector is multiple of another for the final A mark in part (c).

16 6. (a).5.5 y ln.5 ln.5 ln.5 ln or y ln ln Either 0.5 ln.5 and.5 ln.5 or awrt 0.0 and. B (or miture of decimals and ln s) [] { (b)(i) I 0 + ( ln) + ln} For structure of trapezium rule{...} ; M; = = =.9 (4sf).9 A cao (ii) { ) I 0.5 ; ln.5 + ln +.5ln.5 + ln ( } Outside brackets 0.5 For structure of trapezium rule{...} ; B; M = = awrt.684 A 4 [5] (c) With increasing ordinates, the line segments at the top of the trapezia are closer to the curve. Reason or an appropriate diagram elaborating the correct reason. B [] Beware: In part (b) candidate can add up the individual trapezia: (b)(i) I ( 0 + ln) + ( ln + ln) (ii) I. ( ln.5 ) +. ( 0.5ln.5 + ln ) +. ( ln +.5 ln.5 ) +. (.5ln.5 + ln ) 6666/0 Core Maths C4 6 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

17 du u = ln = d 6. (d) dv = v = d Use of integration by parts formula in the correct direction M I = ln d Correct epression A = ln d An attempt to multiply at least one term through by and an attempt to... = ln 4 (+c) integrate; correct integration M; A I = ln + 4 ( ln 9 ) ( ln ) = Substitutes limits of and and subtracts. ddm = ln = ln AG ln 4 4 A cso 6. (d) ( )lnd = ln d ln d Way ln d = ln. d Correct application of by parts M = ln (+ c) Correct integration A 4 ln d = ln. d Correct application of by parts M = ln (+ c) Correct integration A 9 ( ) ln d ( ln ) ( ln ) = Substitutes limits of and = ln AG into both integrands and subtracts. ln ddm A cso 6666/0 Core Maths C4 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

18 6. (d) Way u = ln = = ( ) v = du d ( ) dv d ( ) ( ) Use of integration by parts formula in the correct direction I = ln d Correct epression A M = ln + d = ln + d Candidate multiplies out numerator to obtain three terms multiplies at least one term through by and then attempts to... = ln + ln 4 (+c) integrate the result; correct integration M; A I = ln + ln 4 ( ln 9 ln) ( 0 0) = Substitutes limits of and and subtracts. ddm = ln ln + + = ln AG ln 4 4 A cso Beware: d can also integrate to ln Beware: If you are marking using WAY please make sure that you allocate the marks in the order they appear on the mark scheme. For eample if a candidate only integrated ln correctly then they would be awarded M0A0MAM0A0 on epen. 6666/0 Core Maths C4 8 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

19 6. (d) Way 4 By substitution u = ln = du d u u u Correct epression I = e.ue d u u = ue e du = u u u u u e e e e d Use of integration by parts formula in the correct direction Correct epression M A u u u u = u e e e e 4 (+c) Attempt to integrate; correct integration M; A u u u u I = ue ue e + e 4 ( 9 ln ln 9 ) ( 0 0 ) 4 ln = + + ln 4 Substitutes limits of ln and ln and subtracts. ddm = ln + + = ln AG ln 4 4 A cso marks 6666/0 Core Maths C4 9 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

20 . (a) From question, ds = 8 ds = 8 B ds S = 6 = d ds d = B d ds ds 8 ; d = = = ( k ) = Candidate s ds ds ; d 8 M; Aoe [4] (b) V = = d d = B d = = d =. ; Candidate s d d ; λ M; A As = V, then = V AG Use of = V, to give = V A [4] (c) = V V = V Separates the variables with or V on one side and on the other side. integral signs not necessary. Attempts to integrate and V = t (+c) must see V and t; Correct equation with/without + c. Use of V = 8 and t = 0 in a (8) = (0) + c c = 6 changed equation containing c ; c = 6 B M; A M ; A Hence: V = t + 6 Having found their c candidate 6 = t + 6 = t + 6 substitutes V = 6 into an equation involving V, t and c. depm giving t =. t = A cao [] 6666/0 Core Maths C4 0 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics 5 marks

21 . (b) Way = V & S = 6 S = 6V ds = 4V or V = ds 4 ds = 4V or S = 6V B V ds 4 = B ds = = 8. ; V ds = = 4V V AG Candidate s ds ; ds V M; A In epen, award for Way in the order they appear on this mark scheme. [4]. (c) V = Way V = 4 Separates the variables with or V oe on one V side and on the other side. integral signs not necessary. Attempts to integrate and V = t (+c) must see V and t; Correct equation with/without + c. Use of V = 8 and t = 0 in a (8) = (0) + c c = changed equation containing c ; c = B M; A M ; A Hence: V t 4 ( 6 4 ) t = + Having found their c candidate = + 6 = t + substitutes V = 6 into an equation involving V, t and c. depm giving t =. t = A cao [] Beware: On epen award the marks in part (c) in the order they appear on the mark scheme. 6666/0 Core Maths C4 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

22 similar to way. (b) V = = d Way ds d = =.8. ; d ds = Candidate s d = B ds d ; λ M; d ds A As (c) = V Way = V, then V = 4 V = t (+c) = V AG Use of = V, to give V = V A Separates the variables with or V on one side and on the other side. integral signs not necessary. Attempts to integrate and must see V and 4 t; Correct equation with/without + c. B M; A [4] 4 (8) = (0) + c c = 4 Use of V = 8 and t = 0 in a changed equation containing c ; c = 4 M ; A Hence: V 4 = t + 4 Having found their c candidate = t = t + 4 substitutes V = 6 into an equation involving V, t and c. depm giving t =. t = A cao [] Beware when marking question (c). There are a variety of valid ways that a candidate can use to find the constant c. In questions (b) and (c) there may be Ways that I have not listed. Please use the mark scheme as a guide of how the mark the students responses. In (c), if a candidate instead tries to solve the differential equation in part (a) escalate the response to your team leader. IF YOU ARE UNSURE ON HOW TO APPLY THE MARK SCHEME PLEASE ESCALATE THE RESPONSE UP TO YOUR TEAM LEADER VIA THE REVIEW SYSTEM. Note: dm denotes a method mark which is dependent upon the award of the previous method mark. ddm denotes a method mark which is dependent upon the award of the previous two method marks. depm denotes a method mark which is dependent upon the award of M. 6666/0 Core Maths C4 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics