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2 Edecel GCE Core Mathematics C4 (6666) PhysicsAndMathsTutor.com GCSE Edecel GCE Core Mathematics C4 (6666) Summer 005 Mark (Results)

3 Final Version June Core C4 Mark. ( ) = 4 B 9 ( ) 9 ( )( ) 9 = M = =,, A, A, A [5] Note The M is gained for ( ) (... ). or ( )( ) (... ).. Special Case If the candidate reaches = and goes no further allow A A0 A Core C4 June 005 Advanced Subsidiary/Advanced Level in GCE Mathematics

4 . + + y 6y = 0 M (A) A 0 y 0 = + = or equivalent M Eliminating either variable and solving for at least one value of or y. M y y y + 6= 0 or the same equation in y = ± or =± A Note: d y + y = y (, ), (, ) A [7] Alternative ( ) ( ) y y + 6 = 0 ± y = 6 8 = ± ( ) 8 = 0 =± 6 9 ( + ) 64 = M A± A M = ± M A (, ), (, ) A [7] 6666 Core C4 June 005 Advanced Subsidiary/Advanced Level in GCE Mathematics

5 . (a) ( )( ) 5 + A B = ( ) ( ) 5+ = A + + B Substituting = or = and obtaining A or B; or equating coefficients and solving a pair of simultaneous equations to obtain A or B. M A=, B= A, A If the cover-up rule is used, give M A for the first of A or B found, A for the second. () (b) 5 + = ln( ) + ln( + ) M Aft ( )( + ) 6... ln 9 ln = M A = ln 54 cao A (5) [8] 6666 Core C4 June 005 Advanced Subsidiary/Advanced Level in GCE Mathematics

6 4. = cosθ dθ ( ) ( sin θ ) Use of = sinθ and d cosθ dθ = M = dθ cos θ M A = sec θ dθ = tanθ M A Using the limits 0 and 6 to evaluate integral [ tanθ ] 6 0 = = cao M A [7] Alternative for final M A Returning to the variable and using the limits 0 and to evaluate integral M = = ( ) 0 cao A 6666 Core C4 June 005 Advanced Subsidiary/Advanced Level in GCE Mathematics

7 5. (a) e d = e e d Attempting parts in the right direction M A 4 = e e A e e e 4 = + M A (5) (b) = 0.4 y = 0.8 y.96 4 Both are required to 5 d.p B. () I B (c) 0. [... ] Note ( ) M Aft ft their answers to (b) cao A (4) [0] 4 4 e Core C4 June 005 Advanced Subsidiary/Advanced Level in GCE Mathematics

8 6. (a) = cosec t, = 4sin tcost both M A sintcost = ( = sin tcost) cosec t M A (4) (b) At t = 4, =, y = both and y B Substitutes t = 4 into an attempt at d y M Equation of tangent is y ( ) = M A Accept + y = 4 or any correct equivalent (4) (c) Uses + cot t = cosec t, or equivalent, to eliminate t M + = y correctly eliminates t A 8 y 4 cao A The domain is 0 B (4) [] An alternative in (c) y y sin t = ; cost = sin t = y y sin t+ cos t = + = 4 M A 8 Leading to y 4 A 6666 Core C4 June 005 Advanced Subsidiary/Advanced Level in GCE Mathematics

9 7. (a) k component + 4λ = λ = M A Note µ = Substituting their λ (or µ ) into equation of line and obtaining B M (b) B: (,, ) Accept vector forms A = 8; = 4 0 both B (4) (c) = + + 0( = ) 4 0 B cosθ = = 8 cao M A (4) AB = i+ j 4k AB = 8 or AB = 8 ignore direction of vector M BC = i j BC = 8 or BC = 8 ignore direction of vector M Hence AB = BC A () (d) OD = 6i j+ k Allow first B for any two correct B B Accept column form or coordinates () [] 6666 Core C4 June 005 Advanced Subsidiary/Advanced Level in GCE Mathematics

10 8. (a) (b) dv is the rate of increase of volume (with respect to time) B kv : k is constant of proportionality and the negative shows decrease (or loss) giving d V 0 kv = These Bs are to be awarded independently B dv = 0 kv separating variables M ln ( 0 kv ) = t ( + C ) M A k Using V = 0, t = 0 to evaluate the constant of integration M () c = ln 0 k 0 t = ln k 0 kv Obtaining answer in the form V = A+ Be kt M 0 0 e kt 0 V = Accept ( e kt ) k k k A (6) (c) dv kt = 0e Can be implied M dv kt = 0, t = 5 0 = 0 e k = ln M A At t = 0, 75 V = ln awrt 08 M A (5) [] Alternative to (b) Using printed answer and differentiating dv kt = kb e M Substituting into differential equation kt kt kb e = 0 ka kb e M 0 A = k M A Using V = 0, t = 0 in printed answer to obtain A+ B = 0 M 0 B = k A (6) 6666 Core C4 June 005 Advanced Subsidiary/Advanced Level in GCE Mathematics

11 GCE Edecel GCE Core Mathematics C4 (6666) January 006 Mark (Results) Edecel GCE Core Mathematics C4 (6666)

12 January Core Mathematics C4 Mark. Differentiates 6+ 8y, to obtain :... + (6 + 6 y) = 0 6 6y = 6 + 8y M A, +(B) Substitutes =, y = into epression involving, to give = 8 0 M, A Uses line equation with numerical gradient y ( ) = (their gradient)( ) or finds c and uses y = ( their gradient) + " c" To give 5y =0 (or equivalent = 0) M A [7]. (a) y M A M for one correct, A for all correct () (b) Integral = { ( ) 6 = = } M A A cao () (c) Percentage error = M gained for ( ± ) appro = 0.5 % (allow 0.5% to 0.54% for A) appro ln ( + ln( + ) ) M A () [7]

13 u +. Uses substitution to obtain = f(u) du and to obtain u = const. or equiv., M M Reaches ( u + ) udu or equivalent u A Simplifies integrand to Integrates to u + u u + du or equiv. M M A A dependent on all previous Ms Uses new limits and substituting and subtracting (or returning to function of with old limits) M To give 6 cso A [8] By Parts Attempt at right direction by parts M [ ) { ( ) } ] M{MA}. ( ) MA Uses limits 5 and correctly; [4 6] 6 MA

14 4. Attempts V = e d M e = e (M needs parts in the correct direction) M A e e e = [ ] (M needs second application of parts) M A MA refers to candidates e d, but dependent on prev. M e e e = [ ] 4 A cao Substitutes limits and and subtracts to give [dep. on second and third Ms] dm = e e or any correct eact equivalent. [Omission of loses first and last marks only] A [8]

15 5. (a) Considers + 6 = A( + ) + B( )( + ) + C( ) and substitutes =, or = /, M or compares coefficients and solves simultaneous equations To obtain A =, and C = 4 Compares coefficients or uses simultaneous equation to show B = 0. A, A B (4) (b) Writes ( ) + 4( + ) = ( +, ) + 4 ( ) ( )( ) ( )( )( 4) ( ) 4... M (M, A) ( M A ) = 4 + 8, A, A (7) Or uses + + ( 6)( ) ( ) M ( + 6) (, 9 7 ) = 4 + 8, ( ) ( )( ) ( )( )( 4) ¼( ) (MA) (MA) A, A (7) []

16 6. (a) (b) λ = 4 a= 8, µ = b= 9 8+ λ + λ =0 4 λ M A, A () M 8+ λ + + λ 4+ λ = Solves to obtain λ ( λ = ) 0 A dm Then substitutes value for λ to give P at the point (6, 0, 6) (any form) M, A (5) (c) OP = M (= 9 ) = 4 A cao () [0] 7. (a) dv dr = 4 r B () (b) dr dv dr Uses. = dv in any form, = r (t+ ) M,A () (c) V = 000(t+ ) and integrate to p (t + ), = 500(t+ ) ( + c) M, A Using V=0 when t=0 to find c, (c = 500, or equivalent) M V = 500( ) t + (any form) A (4) (d) (i) Substitute t = 5 to give V, V then use r = to give r, = M, M, A () (ii) Substitutes t = 5 and r = their value into their part (b) M d r = (.90 0 ) ( cm/s) AG A () []

17 8. (a) Solves y = 0 cost = to obtain 5 t = or (need both for A) M A Or substitutes both values of t and shows that y = 0 () (b) cost = M A Area= 5 y = ( cos t)( cos t) = 5 ( cos t) AG B () (c) Area = 4cost+ 4cos t terms = 4cost+ (cost+ ) (use of correct double angle formula) = 4cost+ cos t = [ t 4sint+ sint ] M M M A Substitutes the two correct limits 5 t = and and subtracts. M = 4 + AA (7) []

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

19 June Pure Mathematics C4 Mark. = 6 4y + = 0 Differentiates implicitly to include either ± ky or ±. (Ignore =.) Correct equation. M A 6 + = 4y + not necessarily required. At (0, ), + = 0 = Substituting = 0 & y = into an equation involving ; to give or 7 7 dm; A cso Hence m(n) = 7 or 7 Uses m(t) to correctly find m(n). Can be ft from their tangent gradient. A oe. 7 Either N: y = ( 0) 7 or N: y = + y = m( 0) with their tangent or normal gradient ; or uses y = m + with their tangent or normal gradient ; M; N: 7 + y = 0 Correct equation in the form 'a + by + c = 0', where a, b and c are integers. A oe cso 7 marks [7] Beware: = does not necessarily imply the award of all the first four marks in this question. 7 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

20 Aliter. = 6 4y + = 0 Way 4y + = 6 + Differentiates implicitly to include either ± k or ±. (Ignore =.) Correct equation. not necessarily required. M A At (0, ), 4 + = = Substituting = 0 & y = into an equation involving ; to give 7 dm; A cso Hence m(n) = 7 or 7 Uses m(t) or to correctly find m(n). Can be ft using.. A oe. 7 Either N: y = ( 0) 7 or N: y = + y = m( 0) with their tangent, or normal gradient ; or uses y = m + with their tangent, or normal gradient ; M; N: 7 + y = 0 Correct equation in the form 'a + by + c = 0', where a, b and c are integers. A oe cso 7 marks 6666/0 Core Maths C4 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

21 Aliter. y + y 5 = 0 Way ( ) 9 5 y+ = ( ) 6 49 y = ( 6) ( ) = Differentiates using the chain rule; Correct epression for. M; A oe At (0, ), 49 4 = = = Substituting = 0 into an equation involving ; to give or 7 7 dm A cso Hence m(n) = 7 Uses m(t) to correctly find m(n). Can be ft from their tangent gradient. A 7 Either N: y = ( 0) or N: y = + 7 y = m( 0) with their tangent or normal gradient ; or uses y = m + with their tangent or normal gradient M N: 7 + y = 0 Correct equation in the form 'a + by + c = 0', where a, b and c are integers. A oe [7] 7 marks 6666/0 Core Maths C4 4 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

22 . (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 [6] 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

23 Aliter. (b) Way f() = ( )( ) Moving power to top M ( )( ) + ( )( ); + ( ) +! = ( ) ( )( )( 4) ( )... +! ± 4; Ignoring ( ), correct (...) epansion dm; A = ( )( ) = Correct epansion A = ; Aliter. (b) Maclaurin epansion Way f() = ( ) + ( ) f () = ( ) + ( ) + ; (0 ) 4 Bringing both powers to top Differentiates to give a( ) ± b( ) ; ( ) + ( ) A; A [6] M M; A oe f () = ( ) + ( ) = + Correct f()andf () f () 7( ) 96( ) A f(0) =, f (0) =, f (0) = 0 and f (0) = 4 gives f() = ; ; (0 ) 4 A; A [6] 6666/0 Core Maths C4 6 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

24 Aliter. (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 [6] 6666/0 Core Maths C4 7 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

25 . (a) Area Shaded = sin( ) 0 ( ) 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 9 sin 0 0 Use of V = y. 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( ) 0 ( ) 9 = ( cos ) 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 [6] 9 marks 6666/0 Core Maths C4 8 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

26 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( ) in part (a). 0 Volume = ( sin( )) 0 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

27 4. (a) sint, = y = sin( t+ ) 6 = cos t, = cos( t + 6 ) Attempt to differentiate both and y wrt t to give two terms in cos Correct and M A When t =, 6 cos( + 6 6) = = = = awrt 0.58 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(,awrt0.87 ) 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 = + [6] 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

28 Aliter 4. (a) sint, Way = ( ) y sin t sint cos costsin (Do not give this for part (b)) = + = = cos t, = cos t cos sin t sin 6 6 Attempt to differentiate and y wrt t to give in terms of cos and in the form ± acos t ± bsin t Correct and M A When t =, 6 When t =, 6 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 (,awrt0.87 ) 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 = + [6] 6666/0 Core Maths C4 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

29 Aliter 4. (a) y = + ( ) Way = + ( ) ( ) Attempt to differentiate two terms using the chain rule for the second term. Correct M A = + ( (0.5) ) ( (0.5) ) = When t =, 6, y = = The point ( ) Correct substitution of = into a correct A, or(,awrt0.87 ) 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: Aliter 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 [6] ( ) 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

30 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. OP = 6 +λ i λ j + λ k (b) ( ) ( ) ( ) [] 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 + 7k 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 + 7k M i + j + 7k or P (,, 7) A i j k and d = i + 4j k [6] 6666/0 Core Maths C4 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

31 Aliter (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 + 76λ + 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 + 7k i + j + 7k or P(,, 7) A [6] 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 + 7k λ + λ+ = ( 5) ( ) ( ) ( )k λ + λ+ = λ = λ = 6666/0 Core Maths C4 4 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

32 5. (c) OP = i + j + 7k 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] Aliter 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).

33 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.7 B (or miture of decimals and ln s) [] { (b)(i) I 0 + ( ln) + ln} For structure of trapezium rule{...} ; M; = = =.79 (4sf).79 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

34 du u = ln = 6. (d) dv = v = Use of integration by parts formula in the correct direction M I = ln Correct epression A = ln 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] Aliter 6. (d) ( )ln = ln ln Way ln = ln. Correct application of by parts M = ln (+ c) Correct integration A 4 ln = ln. Correct application of by parts M = ln (+ c) Correct integration A 9 ( ) ln ( ln ) ( ln ) = Substitutes limits of and = ln AG into both integrands and subtracts. ln ddm A cso [6] 6666/0 Core Maths C4 7 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

35 Aliter 6. (d) Way u = ln = = ( ) v = du ( ) dv ( ) ( ) Use of integration by parts formula in the correct direction I = ln Correct epression A M ( ) = ln + ( ) = ln + 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 [6] Beware: 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

36 Aliter 6. (d) Way 4 By substitution u = ln = du u u ( ) u Correct epression I = e.ue d u u ( ) = ue e du = u u u u u e e e e 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 [6] marks 6666/0 Core Maths C4 9 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics

37 7. (a) From question, ds = 8 ds = 8 B ds S = 6 = ds = B ds ds 8 ; = = = ( k ) = Candidate s ds ds ; 8 M; Aoe [4] (b) dv V = = dv = B dv dv = = =. ; Candidate s dv ; λ M; A As = V, then dv = V AG Use of = V, to give dv = V A [4] dv (c) = V V dv = Separates the variables with dv or V dvon one side and V 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 [7] 6666/0 Core Maths C4 0 June 006 Advanced Subsidiary/Advanced Level in GCE Mathematics 5 marks

38 Aliter 7. (b) Way = V & S = 6 S = 6V ds dv = 4V or dv V = ds 4 ds dv = 4V or S = 6V B dv V ds 4 = B dv ds dv = = 8. ; V ds = = 4V V AG Candidate s ds dv ; ds V M; A In epen, award for Way in the order they appear on this mark scheme. [4] Aliter dv 7. (c) V = Way V dv = ( )( ) 4 Separates the variables with dv or V dv 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 [7] 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

39 Aliter similar to way. (b) dv V = = Way dv dv ds = =.8. ; ds = Candidate s dv = B dv ds ; λ M; ds A Aliter As dv (c) = V Way = V, then V dv = 4 V = t (+c) dv dv = V AG Use of = V, to give = V A Separates the variables with dv or V dvon one side and V 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 4 Hence: V = t + 4 Having found their c candidate 4 4 ( 6 ) = t = t + 4 substitutes V = 6 into an equation involving V, t and c. depm giving t =. t = A cao [7] Beware when marking question 7(c). There are a variety of valid ways that a candidate can use to find the constant c. In questions 7(b) and 7(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 7(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

40 Mark (Results) January 007 GCE GCE Mathematics Core Mathematics C4 (6666) Edecel Limited. Registered in England and Wales No Registered Office: One90 High Holborn, London WCV 7BH

41 ** represents a constant. ( ) January Core Mathematics C4 Mark 5 5 f() = ( 5) = = 4 Takes outside the bracket to give any of () - or 4. B ( )( ) ( )( )( 4) = + ( )(* * ); + (* * ) + (* * ) !! Epands ( + * * ) to give an unsimplified + ( )(* * ) ; A correct unsimplified {... epansion } with candidate s (**) M A ( )( ) ( )( )( 4) ( )( ); ( ) ( ) = !! 75 5 = 5; = + ; Anything that cancels to 5 + ; Simplified A; A 5 = + ; [5] 5 marks

42 Aliter. Way f() = ( 5) ( )( ) 4 () + ( )() (* * ); + () (* * )! = ( )( )( 4) 5 + () (* * ) +...! or () 4 Epands ( 5) to give an unsimplifed () + ( )() (* * ) ; A correct unsimplified {...} epansion with candidate s (**) B M A ( )( ) 4 () + ( )() ( 5); + () ( 5)! = ( )( )( 4) 5 + () ( 5) +...! + ( )( )( 5); + ()( )(5 ) = + ( 4)( )( 5 ) = + ; Anything that cancels to 5 + ; Simplified A; A 5 = + ; [5] 5 marks Attempts using Maclaurin epansions need to be referred to your team leader.

43 . (a) Volume = = 9 ( + ) ( + ) 4 4 Use of V = y. Can be implied. Ignore limits. B 9 = ( + ) 4 Moving their power to the top. (Do not allow power of -.) Can be implied. Ignore limits and 9 M (+ ) = 9 ( )() 4 Integrating to give ± p( + ) ( ) M + A = + 9 ( ) 4 = 9 () ( ) = ( ) 4 9 Use of limits to give eact = values of or or or aef (b) From Fig., AB = ( ) = units A aef [5] As 4 units cm then scale factor k = = 4. ( ) 4 Hence Volume of paperweight = ( 4) ( 4 ) (their answer to part (a)) M V = 6 cm = cm 6 or awrt or or aef A [] Note: 9 (or implied) is not needed for the middle three marks of question (a). 7 marks

44 Aliter. (a) Volu me Way = = 6 ( + ) ( + ) 4 4 = ( ) ( + ) 4 6 Use of V y d =. Can be implied. Ignore limits. Moving their power to the top. (Do not allow power of -.) Can be implied. Ignore limits and B M ( + 6) = ( ) ( )(6) 4 Integrating to give ± p( + 6) M ( + 6) 6 A ( ) = + ( 6 6) 4 = ( ) 6(6) 6( ) ( ) = ( ) 6 9 Use of limits to give eact = values of or or aef or 6 4 A aef [5] Note: is not needed for the middle three marks of question (a). 4

45 . (a) = 7cos t cos7t, y = 7 sin t sin7t, = 7sint + 7sin7t, = 7cost 7cos7t Attempt to differentiate and y with respect to t to give in the form ± A sin t ± B sin7t in the form ± Ccost ± Dcos7t Correct and M A 7cos t 7cos7t = 7 sin t + 7 sin7t Candidate s B [] (b) 7 7cos 7cos W hen t =, m(t) = = sin 7sin ; o Substitutes t = or 0 into their 6 epression; M ( ) = = = = awrt to give any of the four underlined epressions oe (must be correct solution only) A cso Hence m(n) = or = awrt 0.58 Uses m(t) to correctly find m(n). Can be ft from their tangent gradient. A oe. W hen t, = 6 ( ) = 7co co 6 6 s s = = = 4 ( ) y 7sin sin = 4 7 = = 7 8 = 6 6 N : y 4 = ( ) 4 The point ( 4, 4 ) or ( awrt 6.9, 4) Finding an equation of a normal with their point and their normal gradient or finds c by using y = (their gradient) + "c ". B M N: y = or y = or y = Correct simplified EXACT equation of normal. This is dependent on candidate using correct ( 4, 4 ) A oe or ( ) 4 = 4 + c c = 4 4 = 0 Hence N: y = or y = or y = [6] 9 marks 5

46 Aliter. (a) = 7cos t cos7t, y = 7 sin t sin7t, Way = 7sint + 7sin7t, = 7cost 7cos t 7 cos7t 7( sin 4t sint) = = 7 sin t + 7 sin7t 7(cos 4t sint) 7cos7t Attempt to differentiate and y with respect to t to give in the form ± A sin t ± B sin7t in theform ± Ccost ± Dcos7t Correct and = tan4t Candidate s M A B [] (b) 4 When t =, m(t) = = tan ; 6 6 o Substitutes t = or 0 into their 6 epression; M ( )( ) ( ) = = = awrt.7 () to give any of the three underlined epressions oe (must be correct solution only) A cso Hence m(n) = or = awrt 0.58 Uses m(t) to correctly find m(n). Can be ft from their tangent gradient. A oe. When t = 6, 7 = 7cos cos = 7 = 8 = 4 ( ) = = 7 ( 8 ) = = 6 6 y 7sin sin 4 N: y 4 = ( ) 4 The point ( 4, 4 ) or ( awrt 6.9, 4) Finding an equation of a normal with their point and their normal gradient or finds c by using y = (their gradient) + "c ". B M N: y or y = or y = = Correct simplified EXACT equation of normal. This is dependent on candidate using correct ( 4, 4 ) A oe or ( ) 4 = 4 + c c = 4 4 = 0 Hence N: y = or y = or y = [6] 9 marks 6

47 Beware: A candidate finding an m(t) = 0 can obtain Aft for m(n), but obtains M0 if they write y 4 = ( 4 ). If they write, however, N: = 4, then they can score M. B eware: A candidate finding an m(t) = can obtain Aft for m(n) = 0, and also obtains M if they write y 4 = 0( 4 ) or y = 4. 7

48 4. (a) A B + ( )( ) ( ) ( ) A( ) + B( ) Let, ( ) = = B B = 4 Forming this identity. NB: A & B are not assigned in this question M Let =, = A( ) A = either one of A = or B = 4. A both correct for their A, B. A giving 4 + ( ) ( ) [] (b) & (c) = ( ) y ( )( ) Separates variables as shown Can be implied B 4 = + ( ) ( ) Replaces RHS with their partial fraction to be integrated. M ln y = ln( ) + ln( ) + c At least two terms in ln s At least two ln terms correct All three terms correct and + c M A A [5] y = 0, = gives c = ln0 c = ln0 B ln y = ln( ) + ln( ) + ln0 ln y = ln( ) + ln( ) + ln0 ( ) ln y = ln + ln0 or ( ) 0( ) ln y = ln ( ) Using the power law for logarithms Using the product and/or quotient laws for logarithms to obtain a single RHS logarithmic term with/without constant c. M M y 0( ) = ( ) y 0( ) = ( ) or aef. isw A aef [4] marks 8

49 Aliter 4. (b) & (c) Way = ( ) y ( )( ) 4 = + ( ) ( ) Separates variables as shown Can be implied Replaces RHS with their partial fraction to be integrated. B M ln y = ln( ) + ln( ) + c At least two terms in ln s At least two ln terms correct All three terms correct and + c M A A See below for the award of B decide to award B here!! B ln y = ln( ) + ln( ) + c ( ) ln y = ln + c Using the power law for logarithms Using the product and/or quotient laws for logarithms to obtain a single RHS logarithm ic term with/without constant c. M M ln y A( ) = ln where c = lna o r ( ) ( ) ln + c ln ln y e = e = e e c y = A( ) ( ) y = 0, = gives award A = 0 A = 0 for B above y 0( ) = ( ) y 0( ) = ( ) or aef & isw A aef [5] & [4] Note: The B mark (part (c)) should be awarded in the same place on epen as in the Way approach. 9

50 Aliter (b) & (c) Way = ( ) y ( )( ) Separates variables as shown Can be implied = + es RHS with their partial ( ) ( ) Replacfraction to be integrated. B M ln y = ln( ) + ln( ) + c At least two terms in ln s At least two ln terms correct All three terms correct and + c M A A [5] y = 0, = gives c ln0 ln( ) ln 40 = = c = ln0 ln( ) or c = ln40 B oe ln y = ln( ) + ln( ) + ln 40 ln ) + y = ln( ) + ln( ln0 Using the power law for logarithms M ( ) ln y = ln + ln 40 or ( ) 40( ) ln y = ln ( ) Using the product and/or quotient laws for logarithms to obtain a single RHS logarithmic term with/without constant c. M y = 40( ) ( ) y = 40( ) ( ) or aef. isw A aef [4] Note: Please mark parts (b) and (c) together for any of the three ways. 0

51 5. (a) sin + cos y = 0.5 ( eqn ) = cos sin y = 0 ( eqn # ) Differentiates implicitly to include ± sin y. (Ignore ( = ).) M = cos sin y cos sin y A cso [] (b) cos = 0 = 0 cos = 0 sin y Candidate realises that they need to solve their numerator = 0 or candidate sets = 0 in their (eqn #) and attempts to solve the resulting equation. M giving = or = both o =, or =± 90 or awrt = ±.57 required here A When When =, sin + cos y = 0.5 =, ( ) ( ) sin + cos y = 0.5 Substitutes either their = or = into eqn M cos y =.5 y has no solutions cos y = 0.5 y = or Only one of y o = or or 0 or 0 or awrt -.09 or awrt.09 A In ( ) (, and, ) A specified range (, y ) = (, ) and (, ) Only eact coordinates of Do not award this mark if candidate states other coordinates inside the required range. [5] 7 marks

52 6. (a) Way Aliter (a) Way ln y = = e ln.e = ln.e ln = ln Hence = ln.( ) = ln AG ln AG A cso ( ) ln y = ln leads to ln y = ln Takes logs of both sides, then uses the power law of logarithms and differentiates implicitly to ln y = give = ln y M M [] H ence = yln = ln AG ln AG A cso [] (b) ( ) y = ( ) =..ln ( ) A ( )..ln or.y.l n if y is define d M A When =, = 4 () ln Substitutes = into th eir which is of the form ± k or A ( ) ( ) M 64ln = = ln or awrt 44.4 A [4] 6 marks

53 Aliter 6. ) ln y = ln( ) (b leads to ln y Way.ln y = = ln A.ln y =.ln y = M A When =, = () 4 ln Substitutes = into their ( ) which is of the form ± k ( ) or A M 64ln = = ln or awrt 44.4 A [4]

54 7. a = OA = i + j + k OA = b = OB = i + j 4k OB = 8 BC =± ( i + j + k) BC = AC =± i + j 4k AC = 8 ( ) (a) c = OC = i + j k i + j k B cao [] (b) OA OB = = = 0 or B uu ur O BC = = = 0 or AC BC = = = 0 or AO AC = = = 0 An attempt to take the dot product between either OA and OB OA and AC, AC and BC or OB and BC Showing the result is equal to zero. M A and therefore OA is perpendicular to OB and hence OACB is a rectangle. perpendicular and OACB is a rectangle A cso Area = 8 = 8 = 9 Using distance formula to find either the correct height or wih. Multiplying the rectangle s height by its wih. eact value of 8, 9, 6 or aef (c) OD = d = ( i + j k ) ( + ) M M A i j k B [6] [] 4

55 (d) Way using dot product formula 5 DA =± ( i + j + k ) DC =± i + j k BA =± i + j + 5k & OC =± i + j k or ( ) & ( ) ( ) cos D = ( ± ) = ( ± ) = ( ± ) Identifies a set of two relevant vectors Correct vectors ± Applies dot product formula on multiples of these vectors. Correct ft. application of dot product formula M A D = cos Attempts to find the correct angle D rather than 80 D. ddm o D = or awr c t09 or.9 A Aliter using dot product formula and direction vectors [6] (d) Identifies a set of two M d BA =± ( i + j + 5k) & d OC =± ( i + j k) direction vectors Correct vectors ± A Way Applies dot product formula on multiples dm of these vectors Correct ft. cos D = ( ± ) = ( ± ) = ( ± ) application of dot A product formula. D = cos Attempts to find the correct angle D rather than 80 D. ddm o D = or awrt09 or.9 c A [6] 5

56 Aliter using dot product formula a nd similar triangles doa = i + j + k OC = i + j k (d) ( ) Way cos ( D) & d ( ) + = = = D = cos Identifies a set of two direction vectors Correct vectors Applies dot product formula on multiples of these vectors. Correct ft. application of dot product formula. Attempts to find the correct angle D by doubling their angle for. D M A dm A ddm o D = or A awrt09 or.9 c [6] Aliter using cosine rule (d) D uu ur 5 A = i + j + k Way 4 DA = 7,, DC = + DC = 7 i j k, AC = 8 ( 8 ) cos D = = 7 7, AC = i + j 4k Attempts to find all the lengths of all three edges of ADC All Correct Using the cosine rule formula with correct subtraction. Correct ft application of the cosine rule formula M A dm A D = cos Attempts to find the correct angle D rather than 80 D. ddm o 09.5 or D = A awrt09 or.9 c [6] 6

57 Aliter using trigonometry on a right angled triangle 5 (d) DA = i + j + k OA = i + j + k AC = i + j 4k Way 5 Let X be the midpoint of AC 7 DA =, DX = OA =, AX = AC = 8 (hypotenuse), (adjacent), (opposite) Attempts to find two out of the three lengths in ADX Any two correct M A 8 7 sin( D) =, 7 cos( D) = or 8 tan( D) = Uses correct sohcahtoa to find D Correct ft application of sohcahtoa dm A eg. 8 D = tan Attempts to find the correct angle D by doubling their a ngle for. D ddm D = o Aliter using trigonometry on a right angled similar triangle OAC (d) OC = i + j k OA = i + j + k AC = i + j 4k Way 6 OC = 7, OA =, AC = 8 (hypotenuse), (adjacent), (opposite) 09.5 or awrt09 or.9 c A Attempts to find two out of the three lengths in OAC M Any two correct A [6] 8 sin( D) =, cos( D) = or tan( D) = Uses correct sohcahtoa to find dm Correct ft application of sohcahtoa A eg. 8 D = tan Attempts to find the correct angle D by doubling their angle for. D ddm o 09.5 or D = A awrt09 or.9 c [6] 7

58 Aliter 7. (b) (i) Way c = OC = ± i + j k AB =± i j 5k ( ) ( ) A complete method OC = () + () + ( ) = () + () + ( 5) = AB of proving that the diagonals are equal. A s OC = AB = 7 Correct result. M A then the diagonals are equal, and OACB is a rectangle. diagonals are equal and OACB is a rectangle A cso [] a = OA = i + j + k OA = b = OB = i + j 4k OB = 8 BC =± ( i + j + k) BC = AC =± ( i + j 4k) AC = 8 c = OC = ± ( i + j k) OC = 7 AB =± i j 5k AB = 7 ( ) Aliter 7. (b) (i) ( OA) + ( AC) = ( OC) or ( BC) + ( OB) = ( OC) or ( OA) + ( OB) = ( AB) Way or ( BC) + ( AC) = ( AB) + = ( ) () ( 8) 7 or equivalent A complete method of proving that Pythagoras holds using their values. Correct result M A and therefore OA is perpendicular to OB or AC is perpendicular to BC and hence OACB is a rectangle. perpendicular and OACB is a rectangle A cso [] 4marks 8

59 Sch eme 8. (a) y e e e 7 0 e e 4 e or y Either e, e and e or awrt 4.,.6 and 6.8 or e to the power awrt.65,.6,.6 (or miture of decimals and e s) At least two correct All three correct B B [] (b) { ( ) } I ; e + e + e + e + e + e Outside brackets For structure of trapezium rule{...} ; B; M = = = A 0.6 (4sf) 0.6 cao [] Beware: In part (b) candidates can add up the individual trapezia: (b) ( + ) + ( + ) + ( + ) + ( + ) + ( + ) I.e e.e e.e e.e e.e e 9

60 (c) A( ) t = ( + ) =..( + ) + or t = A or t = + t = ( + ) or t = M A so = =.( + ) t t = Candidate obt ains either or in terms of t (+ ) I e = = t e. = t t e.. and moves on to dm substitute this into I to convert an integral wrt to an integral wrt t. t I = te change limits: when = 0, t = & when = 5, t = 4 t te changes limits t so that 0 and 5 4 A B 4 t ence I = H te ; where a =, b = 4, k = (d) du u = t = dv t t = e v = e Let k be any constant for the first three marks of this part. [5] ( ) k te t = k te t e t. t t ( ) = k te e + c {( ) ( } 4 t = 4 4 te 4e e e e ) Use of integration by parts formula in the correct direction. Correct epression with a constant factor k. Correct integration with/without a constant factor k Substitutes their changed limits into the integrand and subtracts oe. M A A dm oe = = = 4 either e or awrt 09. A 4 4 (e ) e 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. [5] 5 marks 0

61 Mark (Results) Summer 007 GCE GCE Mathematics Core Mathematics C4 (6666) Edecel Limited. Registered in England and Wales No Registered Office: One90 High Holborn, London WCV 7BH

62 June Core Mathematics C4 Mark ** represents a constant. (a) ( ) f( ) = ( + ) = + = + 7 ( )( 4) ( )( 4)( 5) = !! ( )(* * ); (* * ) (* * )... with ** Takes outside the bracket to give any of () or. 7 See note below. Epands ( + * * ) to give a simplified or an unsimplified + ( )(* * ) ; A correct simplified or an un-simplified {...} epansion with candidate s followed thro (**) B M; A ( )( 4) ( )( 4)( 5) = ( )( ) ( ) ( ) !! = = ; Anything that cancels to ; Simplified 8 79 A; A [5] 5 marks Note: You would award: BMA0 for ( )( 4) ( )( 4)( 5) = !! 7 ( )( ) () ()... because ** is not consistent. Special Case: If you see the constant in a candidate s final 7 binomial epression, then you can award B 6666/0 Core Maths C4 5 th June 007 June 007 Advanced Subsidiary/Advanced Level in GCE Mathematics Version 8: THE FINAL VERSION

63 Aliter. Way f() = ( + ) 4 ( )( 4) 5 () + ( )() (* * ); + () (* * )! = ( )( 4)( 5) () 6 (* * ) ! with ** or( ) (See note ) 7 Epands ( + ) to give an un-simplified or simplified 4 () + ( )() (* * ) ; A correct un-simplified or simplified {...} epansion with candidate s followed thro (**) B M A 4 ( )( 4) 5 () + ( )() (); + () ()! = ( )( 4)( 5) () 6 () ! + ( )( )(); + (6)( )(4 ) = + ( 0)( )(8 ) = ; Anything that cancels to ; Simplified 8 79 A; A [5] 5 marks Attempts using Maclaurin epansions need to be escalated up to your team leader. If you feel the mark scheme does not apply fairly to a candidate please escalate the response up to your team leader. Special Case: If you see the constant in a candidate s 7 final binomial epression, then you can award B 6666/0 Core Maths C4 5 th June 007 June 007 Advanced Subsidiary/Advanced Level in GCE Mathematics Version 8: THE FINAL VERSION

64 ., with substitution u = ( + ) 0 du.ln = = du.ln =.ln or du = u.ln d or ( ) d u = B ln u du d = du k ( u + ) ( + ) ln ( u + ) du where k is constant M = c ln + ( u + ) ( u + ) a( u + ) ( u ).( ) M + u + A change limits: when = 0 & = then u = & u = = ( + ) ln ( u ) + 0 = ln Correct use of limits u = and u = depm = 6ln Alternatively candidate can revert back to or 6ln ln4 ln8 or ln n A aef Eact value only! [6] = ln ( + ) ( + ) 0 0 = ln Correct use of limits = 0 and = depm = 6ln If you see this integration applied anywhere in a candidate s working then you can award M, A or 6ln ln4 ln8 There are other acceptable answers for A, eg: or ln8 ln64 NB: Use your calculator to check eg or ln Eact value only! ln A aef 6 marks 6666/0 Core Maths C4 4 5 th June 007 June 007 Advanced Subsidiary/Advanced Level in GCE Mathematics Version 8: THE FINAL VERSION

65 . (a) du u = = d v = cos v = sin (see note below) Int = cos = sin sin. ( cos ) = sin + c Use of integration by parts formula in the correct direction. Correct epression. sin cos or sink cos k with k, k > 0 k M A dm = sin + cos + c Correct epression with +c A 4 [4] cos + (b) = ( ) cos = cos + Substitutes correctly for cos in the given integral M = sin cos ; d their answer to (a) ; ( ) or underlined epression A; = c sin cos ( ) Completely correct epression with/without +c A [] 7 marks Notes: (b) This is acceptable for M M Int = cos d = sin ± sin.d du u = = d v = cos v = λ sin Int = cos = λsin ± λsin. This is also acceptable for M M 6666/0 Core Maths C4 5 5 th June 007 June 007 Advanced Subsidiary/Advanced Level in GCE Mathematics Version 8: THE FINAL VERSION

66 Aliter. (b) Way cos + ( ) cos d = d Substitutes correctly for cos in the given integral M du u = = = cos + v = sin + dv 4 or dv u = and = cos + ( ) = sin + sin = sin + + cos c their answer to (a) ; or underlined epression + ( ) A = c sin cos ( ) Completely correct epression with/without +c A [] Aliter (b) cos cos = ( ) Way Substitutes correctly for cos in cos M + 4 cos = sin + cos c cos = sin cos ; ( their answer to (a) ) ; or underlined epression A; = c sin cos ( ) Completely correct epression with/without +c A [] 7 marks 6666/0 Core Maths C4 6 5 th June 007 June 007 Advanced Subsidiary/Advanced Level in GCE Mathematics Version 8: THE FINAL VERSION

67 4. (a) A method of long division gives, Way (4 + ) 4 + ( + )( ) ( + )( ) A = B 4 B C + ( + )( ) ( + ) ( ) 4 B( ) + C( +) or their remainder, D + E B( ) + C ( + ) Forming any one of these two identities. Can be implied. M Let =, 4 = B B = Let =, 4 = C C = See note below either one of B = or C = A both B and C correct A [4] Aliter 4. (a) Way (4 + ) B C A + + ( + )( ) ( + ) ( ) See below for the award of B decide to award B here!! for A = B (4 ) A( )( ) B( ) C( ) Forming this identity. Can be implied. M Equate, 8 = 4 A A = Let =, 4 = B B = Let =, 4 = C C = See note below either one of B = or C = A both B and C correct A [4] If a candidate states one of either B or C correctly then the method mark M can be implied. 6666/0 Core Maths C4 7 5 th June 007 June 007 Advanced Subsidiary/Advanced Level in GCE Mathematics Version 8: THE FINAL VERSION

68 4. (b) (4 ) + = + ( + )( ) ( + ) ( ) = ln( + ) + ln( ) ( +c ) Either pln( + ) or qln( ) or either pln + or qln M A A B ln( + ) + ln( ) A or ln( + ) + ln( ) cso & aef See note below. (4 + ) ( + )( ) [ ] = ln( + ) + ln( ) = ( 4 ln5 + ln) ( ln + ln) = + ln + ln ln5 () = + ln 5 Substitutes limits of and and subtracts the correct way round. (Invisible brackets okay.) Use of correct product (or power) and/or quotient laws for logarithms to obtain a single logarithmic term for their numerical epression. depm M 9 = + ln ln 5 A 5 ln and k stated as 9. [6] 5 Or ( ) 9 0 marks Some candidates may find rational values for B and C. They may combine the denominator of their B or C with ( +) or ( ). Hence: a Either kln( b( )) or b( ) a kln( b( + )) is okay for M. b(+ ) To award this M mark, the candidate must use the appropriate law(s) of logarithms for their ln terms to give a one single logarithmic term. Any error in applying the laws of logarithms would then earn M0. Candidates are not allowed to fluke ln( + ) + ln( ) for A. Hence cso. If they do fluke this, however, they can gain the final A mark for this part of the question. Note: This is not a dependent method mark. 6666/0 Core Maths C4 8 5 th June 007 June 007 Advanced Subsidiary/Advanced Level in GCE Mathematics Version 8: THE FINAL VERSION

69 5. (a) If l and l intersect then: 0 +λ = +µ 0 6 Any two of i : +λ= + µ () j : λ= + µ () k : = 6 µ ( ) Writes down any two of these equations correctly. M () & () yields λ = 6, µ = Solves two of the above equations to find () & () yields λ = 4, µ = 7 either one of λ or µ correct A () & () yields λ = 0, µ = 7 both λ and µ correct A Either checking eqn (), - checking eqn (), 4 0 checking eqn (), 5 Complete method of putting their values of λ and µ into a third equation to show a contradiction. B or for eample: checking eqn (), LHS = -, RHS = Lines l and l do not intersect this type of eplanation is also allowed for B. [4] Aliter 5. (a) k : = 6 µ µ = 7 Way i : + λ = + µ + λ = + (7) j : λ= + µ λ= + (7) Uses the k component to find µ and substitutes their value of µ into either one of the i or j component. M i : λ= 4 either one of the λ s correct A j : λ= 0 both of the λ s correct A Either: These equations are then inconsistent Or: 4 0 Or: Lines l and l do not intersect Complete method giving rise to any one of these three eplanations. B [4] 6666/0 Core Maths C4 9 5 th June 007 June 007 Advanced Subsidiary/Advanced Level in GCE Mathematics Version 8: THE FINAL VERSION

70 Aliter 5. (a) Way If l and l intersect then: 0 +λ = +µ 0 6 Any two of i : +λ= + µ () j : λ= + µ () k : = 6 µ ( ) Writes down any two of these equations M () & () yields µ = either one of the µ s correct A () yields µ = 7 both of the µ s correct A Either: These equations are then inconsistent Or: 7 Or: Lines l and l do not intersect Complete method giving rise to any one of these three eplanations. B [4] Aliter 5. (a) Way 4 Any two of i : +λ= + µ () j : λ= + µ () k : = 6 µ ( ) Writes down any two of these equations M () & () yields µ = µ = A () RHS = 6 = RHS of () = A () yields Complete method giving rise to this eplanation. B [4] 6666/0 Core Maths C4 0 5 th June 007 June 007 Advanced Subsidiary/Advanced Level in GCE Mathematics Version 8: THE FINAL VERSION

71 5. (b) 5 λ = OA = & µ = OB = 5 4 Only one of either 5 OA = or OB = 5 or 4 A(,, ) or B (5,5,4). (can be implied) B 5 AB = OB OA = 5 = or BA = 4 5 Finding the difference between their OB and OA. (can be implied) M AB = i + 4j + 5k, d = i + j + 0k & θ is angle Applying the dot product formula between allowable vectors. See notes below. M AB d cos θ = = ± AB. d 50. Applies dot product formula between d and their ± AB. M Correct epression. A 7 cos θ = or 0.7 or 7 00 A cao but not 7 50 [6] 0 marks Candidates can score this mark if there is a complete method for finding the dot product between their vectors in the following cases: Case : their ft ± AB = ± ( i + 4j + 5k and d = i + j + 0 k cos θ =± 50. ) Case : d = i + j + 0k and d = i + j k cos θ = Case : d = i + j + 0k and d = (i + j k) cos θ = Case 4: their ft ± AB = ± ( i + 4j + 5k) Case 5: their ft OA = i + j k and d = i + j k and their ft OB = 5i + 5j + 4k cos θ =± cos θ =± Note: If candidate use cases,, 4 and 5 they cannot gain the final three marks for this part. Note: Candidate can only gain some/all of the final three marks if they use case. 6666/0 Core Maths C4 5 th June 007 June 007 Advanced Subsidiary/Advanced Level in GCE Mathematics Version 8: THE FINAL VERSION