4754 Mark Scheme June Mark Scheme 4754 June 2005

Transcription

1 4754 Mark Scheme June 005 Mark Scheme 4754 June 005

2 4754 Mark Scheme June 005. (a) Please mark in red and award part marks on the right side of the script, level with the work that has earned them. (b) If a part of a question is completely correct, or only one accuracy mark has been lost, the total mark or slightly reduced mark should be put in the margin at the end of the section, shown as, for example, 7 or 7, without any ringing. Otherwise, part marks should be shown as in the mark scheme, as,,, etc. (c) The total mark for the question should be put in the right hand margin at the end of each question, and ringed.. Every page of the script should show evidence that it has been assessed, even if the work has scored no marks. 3. Do not assume that, because an answer is correct, so is the intermediate working; nor that, because an answer is wrong, no marks have been earned. 4. Errors, slips, etc. should be marked clearly where they first occur by underlining or ringing. Missing work should be indicated by a caret ( ). For correct work, use, For incorrect work, use X, For correct work after and error, use For error in follow through work, use 5. An M mark is earned for a correct method (or equivalent method) for that part of the question. A method may contain incorrect working, but there must be sufficient evidence that, if correct, it would have given the correct answer. An A mark is earned for accuracy, but cannot be awarded if the corresponding M mark has not been earned. An A mark shown as f.t. or shows that the mark has been awarded following through on a previous error. A B mark is an accuracy mark awarded independently of any M mark. E marks are accuracy marks dependent on an M mark, used as a reminder that the answer has been given in the question and must be fully justified. 6. If a question is misread or misunderstood in such a way that the nature and difficulty of the question is unaltered, follow the work through, awarding all marks earned, but deducting one mark once, shown as MR, from any accuracy or independent marks earned in the affected work. If the question is made easier by the misread, then deduct more marks appropriately. 7. Mark deleted work if it has not been replaced. If it has been replaced, ignore the deleted work and mark the replacement.

3 4754 Mark Scheme June Other abbreviations: c.a.o. b.o.d. X X s.o.i. s.c. w.w.w : correct answer only : benefit of doubt (where full work is not shown) : work of no mark value between crosses : seen or implied : special case (as defined in the mark scheme) : without wrong working Procedure. Before the Examiners Meeting, mark at least 0 scripts of different standards and bring them with you to the meeting. List any problems which have occurred or that you can foresee.. After the meeting, mark 7 scripts and the 3 photocopied scripts provided and send these to your team leader. Keep a record of the marks, and enclose with your scripts a stamped addressed envelope for their return. Your team leader will contact you by telephone or as soon as possible with any comments. You must ensure that the corrected marks are entered on to the mark sheet. 3. By a date agreed at the standardisation meeting prior to the batch date, send a further sample of about 40 scripts, from complete centres. You should record the marks for these scripts on your marksheets. They will not be returned to you, but you will receive feedback on them. If all is well, you will then be given clearance to send your batch scripts and marksheets to Cambridge. 4. Towards the end of the marking period, your team leader will request a final sample of about 60 scripts. This sample will consist of complete centres and will not be returned to you. The marks must be entered on the mark sheets before sending the scripts, and should be sent, with the remainder of your marksheets, to the office by the final deadline. 5. Please contact your team leader by telephone or in case of difficulty. Contact addresses and telephone numbers will be found in your examiner packs.

4 4754 Mark Scheme June 005 SECTION A 3cos θ + 4sin θ = R cos(θ α) = R(cos θ cos α + sin θ sin α) R cos α = 3, R sin α = 4 R = = 5, R = 5 tan α = 4/3 α = 0.97 f(θ) = 7 + 5cos(θ 0.97) Range is to R=5 tan α=4/3 oe ft their R 0.93 or 53. or better their cos (θ ) = or - used (condone use of graphical calculator) and seen cao Greatest value of 7+ 3cosθ+ 4sinθ is ½. ft [6] simplified 4+ x = ( + x) 3.( ).( ).( ) 3 = { +.( x) + ( x) + ( x) +...}! 3! Taking out 4 oe correct binomial coefficients 3 = k + x x + x A,,0 4 3 x, x, + 8 x 3 3 = + x x + x cao Valid for < x <. cao [6] 3 sec θ = 4 4 cos θ = cos θ = ¼ cos θ = ½ or ½ θ = π/3, π/3 OR sec θ = + tan θ tan θ = 3 tan θ = 3 or 3 θ = π/3, π/3 [4] sec θ = /cos θ used ± ½ allow unsupported answers ± 3 allow unsupported answers

5 4754 Mark Scheme June V = πy dx x = π( + e ) dx 0 = x π x e = π( ½ e + ½ ) = π( ½ ½ e ) 0 [5] Correct formula k 0 x ( + e )dx x x e substituting limits. Must see 0 used. Condone omission of π o.e. but must be exact 5 cos x = (cos x ) = 4cos x 4cos x = + cos x 4cos x cos x 3 = 0 (4cosx + 3)(cos x ) = 0 cos x = 3/4 or x = 38.6 or.4 or 0 6 (i) y x = (t + /t) (t /t) = t + + /t t + /t = 4 (ii) EITHER dx/dt = + /t, dy/dt = /t dy dy / dt = dx dx / dt / t = + / t t ( t )( t+ ) = = * t + t + dy OR y x = 0 dx dy x t / t = = dx y t + / t t ( t )( t + ) = = t + t + OR y= (4+x ), dep [7] E [] E E Any double angle formula used getting a quadratic in cos x attempt to solve for -3/4 and 39, or better www - extra solutions in range Substituting for x and y in terms of t oe For both results dy = dx = = t = t x 4 + x t / t 4 + t + / t t / t t / t = ( t + / t ) ( t + / t) = + ( t )( t + ) t + E dy/dx = 0 when t = or t =, (0, ) t = (0, ) [6]

6 4754 Mark Scheme June 005 SECTION B 7 (i) OR t + t t dt = ½ ln( + t ) + c + t dt let u = + t, du = tdt / = du u = ½ ln u + c = ½ ln( + t ) + c A [3] k ln( + t ) ½ ln( + t ) [+c] substituting u = + t condone no c (ii) + = A + Bt C t( + t ) t + t = A(+ t ) + (Bt + C)t t = 0 = A coeff t of t 0 = A + B B = coeff t of t 0 = C t = t( + t ) t + t [5] Equating numerators substituting or equating coeff t s dep st A = B = C = 0 (iii) d M M = d t t(+ t ) t dm = dt = [ ] ( + ) dt M t t t + t ln M = ln t ½ ln( + t ) + c c et = ln( ) + t Kt M = * where K = e c + t ft E [6] Separating variables and substituting their partial fractions ln M = ln t ½ ln( + t ) + c combining ln t and ½ ln( + t ) K = e c o.e. (iv) t =, M = 5 5 = K/ K = 5 = As t, M K So long term value of M is grams ft [4] 5 or 35 or better soi ft their K. 8 (i) P is (0, 0, 30) Q is (0, 0, 5) R is ( 5, 0, 30) * PQ = 0 0 = * P R = 0 0 = B,,0 E E [4]

7 4754 Mark Scheme June 005 (ii) = = = = 0 0 is normal to the plane 3 E Scalar product with vector in the plane OR vector x product oe equation of plane is x + 3y + z = c At P (say), x = 0, y = 0, z = 30 c = = 90 equation of plane is x + 3y + z = 90 dep cao [5] x + 3y + z = c or an appropriate vector form substituting to find c or completely eliminating parameters (iii) S is ( 7,0, ) OT = OP+ PS = = So T is ( 5,6,5)* 3 ft E [4] Or Or 3 (OP + OR + OQ) oe ft their S 7 0 ft their S (iv) 6 r = + λ At C ( 30, 0, 0): 5 + λ = 30, λ =, 5 + λ = 0 st and 3 rd eqns give λ = ½, not compatible with nd. So line does not pass through C., E [5] λ Substituting coordinates of C into vector equation At least relevant correct equations for λ oe www

8 4754 Mark Scheme June 005 COMPREHENSION. The masses are measured in units. The ratio is dimensionless. Converting from base 5, = = [] [] 3. n x n Condone variations in last digits 4. φ φ = φ = φ φ φ = 0 Using the quadratic formula gives ± 5 φ = 5. φ = + 5 = = 5+ 5 E Or complete verification B Must discount ± ( 5 ) ( 5 ) = = 4 ( 5) = 5 E OR = φ φ = = E [3] Must discount ± Substituting for φ and simplifying

9 4754 Mark Scheme June Let r a a n+ = and = n r a a n n For either ratio used dividing through by a = a + 3a n n + n 3 a n r = + r r r 3 = 0 ( r 3)( r + ) = 0 3 r = + r r = 3 (discounting -) E [4] SC B if dividing terms as far as a9 a = 0 =3.00 a a The length of the next interval = l, where = l l = So next bifurcation at D [4]

10 Mark Scheme 4754 January 006

11 Section A x 4x = 3 x x + x(x + ) 4x(x ) = 3(x )(x + ) x + x 4x + 8x = 3x 3x 6 0 = 5x 3x 6 = (5x + )(x 3) x = /5 or 3. cao [5] Clearing fractions expanding brackets oe factorising or formula dx/dt = /t dy/dt = + /t dy dy / dt = dx dx / dt = + t t When t =, dy/dx = + = 3 Either dx/dt or dy/dt soi www [5] 4 3 BA =, BC = BA. BC =. 5 = ( 4) ( 3) ( ) 3 = = 0 angle ABC = 90 Area of triangle = ½ BA BC ( 4) 3 5 ( ) = ½ 6 30 = 3.96 sq units = [6] soi, condone wrong sense scalar product = 0 area of triangle formula oe length formula accept 4.0 and 95

12 4(i) sin θ + cos θ = 4sin θ cos θ + sin θ = sin θ (cos θ sin θ) = 0 or 4 tanθ- tan²θ=0 sin θ = 0 or tan θ =0, θ = 0, 80 or cos θ sin θ = 0 tan θ = θ = 63.43, OR Using Rsin(θ+α) R= 5 and α=6.57 θ +6.57=arcsin /R θ=0,80 θ=63.43, (i) Plane has equation x y + z = c At (,, 4), = c c =. x 7 + λ (ii) y = + 3λ z 9+ λ 7 + λ ( + 3 λ) + (9 + λ) = λ = λ = Coordinates are (6, 9, 7), [6], [6] [7] Using double angle formulae Correct simplification to factorisable or other form that leads to solutions 0 and 80 tan θ = (- for extra solutions in range) (- for extra solutions in range) x y + z = c finding c ft their equation from (i) ft their x-y+z=c cao 6 (i) = 4 ( x ) 4 x 4 3 ( )( ) = ! 4 [ ( )( x ) ( x )...] = + x + x +... Binomial coeffs correct Complete correct expression inside bracket cao (ii) 3 dx x x dx 4 x ( ) = x + x + x = = 0.53 (to 4 s.f.) 0 ft (iii) x dx = arcsin 0 4 x = π/6 = [7]

13 Section B 7(i) A O P = 80 β = 80 α θ β = α + θ θ = β α tan θ = tan (β α) tan β tan α = + tanβtanα y y = 0 6 y y y 0y = 60 + y 6y = * 60 + y When y = 6, tan θ = 36/96 θ = 0.4 E E cao [8] Use of sum of angles in triangle OPT and AOP oe SC for β=α+θ, θ= β-α no justification Use of Compound angle formula Substituting values for tan α and tan β www accept radians (ii) sec θ = dy dθ (60 + y )6 6 y. y = dθ = dy (60 + y ) (60 + y ) 6(60 + y y ) (60 + y ) 6(60 y ) cos θ * E dθ sec θ... dy = quotient rule correct expression simplifying numerator www [5] (iii) dθ/dy = 0 when 60 y = 0 y = 60 y =.65 When y =.65, tan θ = 0.37 θ = 3.3 cao [4] oe accept radians

14 8 (i) x = a( + kt) dx/dt = ka( + kt) = ka(x/a) = kx /a * OR kt=a/x, t= a/kx /k dt/dx= -a/kx² dx/dt= -kx²/a (ii) When t = 0, x = a a =.5 When t =, x =.6.6 =.5/( + k) + k =.565 k = E [3] E [3] [3] Chain rule (or quotient rule) Substitution for x a =.5 (iii) In the long term, x 0 (iv) A B = = + y y y( y) y y = A( y) + By y = 0 A = A = ½ y = = B B = ½ = + y y y ( y) [] [4] or, for example, they die out. partial fractions evaluating constants by substituting values, equating coefficients or cover-up (v) dy = dt y y [ + ] dy = dt y ( y) ½ ln y ½ ln( y) = t + c When t = 0, y = 0 0 = 0 + c c = 0 ln y ln( y) = t y ln t y = * y t y = e y = e t ye t y + ye t = e t y( + e t ) = e t t e y = = * t t + e + e (vi) As t e t 0 y So long term population is 000 ft E D E [7] [] Separating variables ½ ln y ½ ln( y) ft their A,B evaluating the constant Anti-logging Isolating y or y =

15 Comprehension. It is the largest number in the Residual column in Table 5.. (i) Acceptance percentage, a% 0% 4% % % 0.5% Party Votes (%) Seats Seats Seats Seats Seats P Q R.4 S 4.8 T U Total seats Seat Allocation P Q R S T U 0 (ii) 0% & 4% Trial 0.5% (0.3<x 0.9) Allocation Round Party Residual P Q R S T U Seat allocated to P R P S Q R T Seat Allocation P Q R S T U 0

16 General method Round correct Round 5 correct (condone minor arithmetic error) Residuals www Allocation cso 3... < +, 5.6 <. for either or both only for 5.6<a. 4. (i) The end-points of the intervals are the largest values in successive columns of Table 5.( or two largest within a column) So in 6.6 < a.. is the largest number in Round. 6.6 is the largest number in Round 3. (ii) Seats a Seats a. < a < a. 6.6 < a < a < a < a < a (i) means, means < (greater or less than) (ii) Vk < a N + V an a Vk a N k k k < k + ank Vk k k < 0 Vk ank 0 Vk ank < a V an a (iii) The unused votes may be zero but must be less than a.

17 4754 Mark Scheme June 006 Mark Scheme 4754 June 006

18 4754 Mark Scheme June 006 sin x 3 cos x = R sin(x α) = R(sin x cos α cos x sin α) R cos α =, R sin α = 3 R = + ( 3) = 4, R = tan α = 3/ = 3 α = π/3 sin x 3 cos x = sin(x π/3) x coordinate of P is when x π/3 = π/ x = 5π/6 y = So coordinates are (5π/6, ) ft ft [6] R = tan α = 3 or sinα= 3/their R or cosα=/their R α = π/3, 60 or.05 (or better) radians www Using x-their α=π/or 90 α 0 exact radians only (not π/) their R (exact only) (i) 3+ x A B C = + + ( + x) ( 4 x) + x ( + x) 4x 3+ x = A( + x)( 4 x) + B( 4 x) + C( + x) x = 5 = 5B B = x = ¼ 5 3 = C C = 8 6 coeff t of x : = 4A + C A = 0. E [4] Clearing fractions (or any correct equations) B = www C = www A = 0 needs justification (ii) ( + x) = + ( )x + ( )( 3)x /! + = x + 3x + ( 4x) = + ( )( 4x)+( )( )( 4x) /!+ = + 4x + 6x + 3+ x = ( + x) + ( 4 x) ( + x) ( 4 x) x + 3x + ( + 4x + 6x ) = 3 + 6x + 35x ft [4] Binomial series (coefficients unsimplified - for either) or (3+x²)(+x) (-4x) expanded theira,b,c and their expansions 3 sin(θ + α) = sin θ sin θ cos α + cos θ sin α = sin θ tan θ cos α + sin α = tan θ sin α = tan θ tan θ cos α = tan θ ( cos α) tan θ = sinα cosα * sin(θ + 40 ) = sin θ tan θ = sin 40 cos40 = θ = 7.5, 07.5 E [7] Using correct Compound angle formula in a valid equation dividing by cos θ collecting terms in tan θ or sin θ or dividing by tan θ oe www (can be all achieved for the method in reverse) tan θ = sin 40 cos40 - if given in radians - extra solutions in the range

19 4754 Mark Scheme June (a) dx k dt = x [] dx... dt = k x (b) dy 0000 = dt y ydy = 0000dt y = t+ c When t = 0, y = = c 3 3 y = [ (0000t+ 8000)] =(500(0t+8)) When t = 0, y = 35 3 [6] separating variables condone omission of c evaluating constant for their integral any correct expression for y = for method allow substituting t=0 in their expression cao 5 (i) x xe dx let u = x, dv/dx = e x v = ½ e x x x = xe e dx + x x = xe e + c 4 x = e (+ x) + c * 4 d [ xe e + c] = e + xe + e dx 4 = xe x or x x x x x E E [3] Integration by parts with u = x, dv/dx = e x = + x x xe e dx condone omission of c product rule (ii) V = π 0 0 y dx = π ( ) = π 0 / x x e dx xe x dx x = π e (+ x) 4 0 = π( ¼ e ¼ ) 5 = π ( ) * 4 4 e D E [4] Using formula condone omission of limits y²=xe x condone omission of limits and π condone omission of π (need limits)

20 4754 Mark Scheme June 006 Section B 6 (i) At E, θ = π x = a(π sin π) = aπ So OE = aπ. Max height is when θ = π y = a( cos π) = a [4] θ=π, 80,cosθ=- (ii) dy dy / dθ = dx dx / dθ a sinθ = a( cos θ ) sinθ = ( cos θ ) [3] dy dy / dθ = for theirs dx dx / dθ d d (sin θ) = cos θ, (cos θ) = sin θ both dθ dθ or equivalent www condone uncancelled a (iii) tan 30 = / 3 sinθ = ( cos θ ) 3 sin θ = ( cos θ ) * 3 When θ = π/3, sin θ = 3/ ( cos θ)/ 3 = ( + ½)/ 3 = = BF = a( + ½) = 3a/* OF = a( π/3 3/) E E E [6] Or gradient=/ 3 sin θ = 3/, cos θ = ½ or equiv. (iv) BC = aπ a(π/3 3/) = a(π/3 + 3) AF = 3 3a/ = 3 3a/ AD = BC + AF = a( π/ ) = a(π/ ) = 0 a =. m ft [5] their OE -their OF

21 4754 Mark Scheme June (i) AE = ( ) = 5 [] (ii) 5 3 uuur AE = 0 = Equation of BD is r = 7 + λ 4 0 BD = 5 λ = 3 D is (8, 9, ) cao [4] Any correct form 5 or r = 7 + λ 0 0 λ = 3 or 3/5 as appropriate (iii) At A: = 30 At B: 3 ( ) + 4 ( 7) + 5 = 30 At C: 3 ( 8) + 4 ( 6) = 30 3 Normal is 4 5 A,,0 [4] One verification (OR Normal, scalar product with vector in the plane, two correct, verification with a point OR vector form of equation of plane eg r=0i+0j+6k+μ(i+7j-5k)+ν(8i+6j+0k) elimination of both parameters equation of plane Normal * ) (iv) AE = 0 = = AB = 7 = = is normal to plane 3 5 Equation is 4x + 3y + 5z = 30. (v) Angle between planes is angle between 4 3 normals and ( 3) cosθ = = θ = 60 E [4] [4] scalar product with one vector in plane = 0 scalar product with another vector in plane = 0 4x + 3y + 5z = 30 OR as * above OR for subst point in 4x+3y+5z=, for subst further points =30 correct equation, Normal Correct method for any vectors their normals only ( rearranged) or 0 cao

22 4754 Mark Scheme June 006 Comprehension Paper Qu Answer Mark Comment minutes 760 hour 44 minutes 5.5 seconds Accept all equivalent forms, with units. Allow. R= ( T 900) ( T ) = 0 T = R will become negative in and 53 seconds. R=0 and attempting to solve. T=563,564,563.9 any correct cao 3. The value of L is 0.5 and this is over hours or (0 minutes) 4.(i) Substituting t = 0 in R = L+ ( U L)e kt gives R = L+ ( U L) = U 4.(ii) kt As t, e 0 and so R L 5.(i) E E E or R>0.5minutes or showing there is no solution for 0= e... 0 e = Increasing curve Asymptote A and B marked correctly 5.(ii) Any field event: long jump, high jump, triple jump, pole vault, javelin, shot, discus, hammer, etc. 6.(i) t = t R = e 6.(ii) ( ) R = e.89 R = e R = hours 4 minutes 3 seconds Substituting their t 4, 4.05, etc.

23 7 Mark Scheme 4754 January 007

24 4754 Mark Scheme Jan 007 Paper A Section A x + = x x + x+ + x = x( x+ ) = x + x x = D [4] Clearing fractions solving cao (i) x y A 0.5[(+3)/ ] = 3.8 (3 s.f.) [3] At least one value calculated correctly or 3.3 or seen (ii) 3.5 (or Chris) area should decrease with the number of strips used. [] ft (i) or area should decrease as concave upwards 3(i) sin 60 = 3/, cos 60 = /, sin 45 = /, cos 45 = / sin(05 ) = sin( ) = sin60 cos45 + cos60 sin45 = = 3+ * (ii) Angle B = 05 By the sine rule: AC sin B = sin 30 AC = sin =. sin 30 = 3+ * 4 sin θ + + tan θ = cos θ tan θ sin θ cos θ = cos θ + sin θ cos θ sin θ = cos θ = secθ secθ = cosθ = ½ θ = 60, 300 θ = 30, 50 E [4] E [3] E [7] splitting into 60 and 45, and using the compound angle formulae Sine rule with exact values www sinθ tanθ = or +tan²θ=sec²θ used cosθ simplifying to a simple fraction in terms of sinθ and/or cos θ only cos θ sin θ = cos θ oe used or +tan θ = ( tan θ) tan θ = ±/ 3oe and no others in range 8

25 4754 Mark Scheme Jan (+ 3 x) 3 = 5.( ) ( )( ) = (3 x) + (3 x) + (3 x) ! 3! 5 3 = + x x + x Valid for < 3x < -/3 < x < /3 A,,0 [5] binomial expansion (at least 3 terms) correct binomial coefficients (all) x, x, 5x 3 /3 6(i) A B = + (x + )( x+ ) x+ x+ = A(x + ) + B(x + ) x = : = B B = x = ½ : = ½ A A = [3] or cover up rule for either value (ii) dy y = dx (x + )( x + ) dy = dx y (x+ )( x+ ) = ( ) x + x + dx ln y = ln(x + ) ln(x + ) + c When x = 0, y = ln = ln ln + c c = ln ln y = ln(x + ) ln(x + ) + ln (x + ) = ln x + 4x + y = x + * ft E [5] separating variables correctly condone omission of c. ft A,B from (i) calculating c, no incorrect log rules combining lns www 9

26 4754 Mark Scheme Jan 007 Section B 7(i) At A, cos θ = θ = 0 At B, cos θ = θ = π At C x = 0, cos θ = 0 θ = π/ y = π sin sin π = 8 [4] or subst in both x and y allow 80 (ii) dy dy / dθ = dx dx / dθ cosθ cos θ = 4 sin θ = cos θ 4cosθ 4sinθ dy/dx = 0 when cos θ 4cosθ = 0 cos θ 4cosθ = 0 cos θ 4cosθ = 0* E [5] finding dy/dθ and dx/dθ correct numerator correct denominator =0 or their num=0 (iii) 4± 6+ 8 cosθ = = ± 6 4 ( + ½ 6 > so no solution) θ =.7975 y = sin θ sin θ =.09 8 ft cao cao [5] ± 6 or (.47,-.47) both or ve their quadratic equation.80 or 03 their angle.03 or better (iv) V = π ydx = π (6 8 x x )( x ) dx 6 + = 3 4 π (6 8x x 6x 8 x x ) dx = 3 4 π (6 8x 5x + 8 x x ) dx* 6 = π 6x 4x 5x x x = π (3 0 ) 6 5 =.35π = 4.4 E cao [6] correct integral and limits expanding brackets correctly integrated substituting limits 0

27 4754 Mark Scheme Jan (i) (40 0) + (0+ 40) + ( 0 0) = 60 m [] 40 (ii) BA = 40 = cosθ = = θ = 48 [4] or AB 3 oe eg oe eg 60 6 cao (or radians) 40 3 (iii) r = 0 + λ 4 0 At C, z = 0 λ = 0 a = = 00 b = = 80 [5] λ a 40 or.+λ b 0 6 (iv) 5. = + 0+ = = 8 0+ = is perpendicular to plane. Equation of plane is 6x 5y + z = c At B (say) = c c = 00 so 6x 5y + z = 00 [5] ( alt. method finding vector equation of plane eliminating both parameters D correct equation stating Normal hence perpendicular B)

28 4754 Mark Scheme Jan 007 Paper B Comprehension (i) Leading digit Frequency 6 4 Table (ii) (iii) (iv) Leading digit Frequency Leading digit Frequency Any sensible comment such as: The general pattern of the frequencies/results is the same for all three tables. Due to the small number of data items we cannot expect the pattern to follow Benford s Law very closely. Table any 4 correct other 4 correct E Evidence of from Table 4 frequencies is the same as 5 in Table p = p + p + p : on multiplication by 3, numbers with a leading digit of will be mapped to numbers with a leading digit of 3, 4 or 5 and no other numbers have this property. n+ n log0 ( n+ ) log0 n= log0 = log0 + = log0 + n n n n Multiplication by 3 E 5 Substitute L( 4) = L( ) and L( 6) = L( 3) + L( ) in L8 ( ) L6 ( ) = L4 ( ) L3 ( ) : this gives L( 8) = L( 6) L( 3) + L( 4) = L( ) + L( ) = 3 L( ) 6 a = 8. All entries with leading digit or 3 will, on multiplying by 5, have leading digit. None of the other original daily wages would have this property. b = 9. Similarly, all entries with leading digit 8 or 9 will, on multiplying by 5, have leading digit 4. None of the other original daily wages would have this property. subst E (or alt for or more Ls used use of at least given results oe E) Total 8

29 Mark Scheme 4754 June 007

30 4754 Mark Scheme June 007 Section A sin θ 3 cos θ = R sin(θ α) = R(sin θ cos α cos θ sin α) R cos α =, R sin α = 3 R = + 3 = 0 R = 0 tan α = 3 α = sin(θ 7.57 ) = θ 7.57 = sin (/ 0) θ 7.57 = 8.43, 6.57 θ = 90, 33. Normal vectors are and = 6+ 4= 0 4 planes are perpendicular. 3 (i) y = ln x x = e y V (ii) = π x dy 0 = y y π( e ) dy = πe dy * 0 0 y y πe dy= π e 0 0 = ½ π(e 4 ) 4 x = x t t = + t = x + + x x+ + x+ 3 y = + = = + x+ + x+ x + [7] E [4] E [3] [3] E equating correct pairs oe ft www cao (7.6 or better) oe ft R, α www and no others in range (MR- for radians) ½ e y substituting limits in kπe or equivalent, but must be exact and evaluate e 0 as. y Solving for t in terms of x or y Subst their t which must include a fraction, clearing subsidiary fractions/ changing the subject oe www or t x = t + x t + t = 3 t+ t t+ t = t + = y t + E [4] substituting for x or y in terms of t clearing subsidiary fractions/changing the subject 4

31 4754 Mark Scheme June 007 x λ 5 r = + λ y = + λ 3 z + 3λ When x =, λ =, λ = y = + λ = 6, z = + 3λ = 5 point lies on first line 0 x μ r = 6 + μ 0 y = 6 3 z 3 μ When x =, μ =, y = 6, z = 3 μ = 5 point lies on second line Angle between and is θ, where cosθ = 4. 5 = 7 70 θ = 46.8 acute angle is 33. E E cao [7] Finding λ or μ checking other two coordinates checking other two co-ordinates Finding angle between correct vectors use of formula 7 ± 70 Final answer must be acute angle 6(i) ( A 0.5[ +.060] =. (3 s.f.) ( + e ). = + e + ( e! ) +... x x + e e * 8 (ii) x / x x (iii) I = 8 ( + x e x e ) dx = x x x e + e 6 = 4 ( e e ) ( e + + e ) 6 6 = =. (3 s.f.) cao [] E [3] [3] Correct expression for trapezium rule Binomial expansion with p = ½ Correct coeffs integration substituting limits into correct expression 5

32 4754 Mark Scheme June 007 Section B 7 (a) (i) P max = = P min = = /3. + (ii) P = = ( sin t) sint dp = ( sin t). cost dt = cost ( sin t) 4 P cost = cost ( sin t) = cost = dp ( sin t) dt (b)(i) A B = + P(P ) P P = A(P ) + BP P(P ) = A(P ) + BP P = 0 = A A = P = ½ = A.0 + ½ B B = So = + P(P ) P P (ii) dp ( P P )cos t dt = dp = cos tdt P P ( ) dp = costdt P P ln(p ) ln P = ½ sin t + c When t = 0, P = ln ln = ½ sin 0 + c c = 0 P ln( ) = sin t * P [] D E [5] [4] E [5] chain rule ( ) soi (or quotient rule,numerator,denominator ) attempt to verify or by integration as in (b)(ii) correct partial fractions substituting values, equating coeffs or cover up rule A = B = separating variables ln(p ) ln P ft their A,B from (i) ½ sin t finding constant = 0 (iii) P max = =.847 / e P min = = 0.78 / e [4] www www 6

33 4754 Mark Scheme June 007 dy 0cosθ + 0cos θ 8 (i) = dx 0sinθ 0sin θ cosθ + cos θ = * sinθ + sin θ When θ = π/3, dy = cos π / 3 + cos π / 3 dx sin π / 3+ sin π / 3 = 0 as cos π/3 = ½, cos π/3 = ½ At A x = 0 cos π/3 + 5 cos π/3 = ½ y = 0 sin π/3 + 5 sin π/3 = 5 3/ E [6] dy/dθ dx/dθ or solving cosθ+cosθ=0 substituting π/3 into x or y ½ 5 3/ (condone 3 or better) (ii) x + y = (0 cosθ + 5cos θ) + (0sinθ + 5sin θ) = 00cos θ + 00cosθcos θ + 5cos θ + 00sin θ + 00sin θsin θ + 5sin θ = cos(θ θ) + 5 = cos θ * D E [4] expanding cos θ cos θ + sin θ sin θ = cos(θ θ) or substituting for sin θ and cos θ (iii) Max = 5 min 5 00 = 5 [] (iv) cos θ +cos θ = 0 cos θ = ± ± 3 = 4 4 At B, cos θ = + 3 OB = ( + 3) = = 6.6 OB = 6.6 =.7 (m) [4] quadratic formula or θ=68.53 or.0radians, correct root selected or OB=0sinθ+5sinθ ft their θ/cosθ oe cao 7

34 4754 Mark Scheme June 007 Paper B Comprehension ) M (aπ, a), θ=π N (4aπ, 0), θ=4π ) Compare the equations with equations given in text, x = aθ - bsinθ, y = bcosθ Seeing a=7, b=0.5 Wavelength = πa = 4π ( 44) Height = b = 0.5 3i) Wavelength = 0 a = π 0 (=3.8 ) Height = b = ii) In this case, the ratio is observed to be :8 Trough length : Peak length = πa + b : πa b and this is (0 + ) : (0 ) So the curve is consistent with the parametric equations substituting 4i) x = aθ, y = bcosθ is the sine curve V and x = aθ - bsinθ, y = bcosθ is the curtate cycloid U. The sine curve is above mid-height for half its wavelength (or equivalent) ii) d = aθ-(aθ-bsinθ) πa πa θ=π/, d= b = b iii) Because b is small compared to a, the two curves are close together. 5) Measurements on the diagram give Wavelength 3.5cm, Height 0.8cm Wavelength 3.5 = Height 0. 8 Since < 7, the wave will have become unstable and broken. E E E Subtraction Using θ=π/ Comparison attempted Conclusion measurements/reading ratio [8] 8

35 ADVANCED GCE UNIT MATHEMATICS (MEI) Applications of Advanced Mathematics (C4) Paper B: Comprehension INSERT THURSDAY 4 JUNE (B)/0 Afternoon Time: Up to hour INSTRUCTIONS TO CANDIDATES This insert contains the text for use with the questions. This document consists of 8 printed pages. HN/5 OCR 007 [T/0/653] OCR is an exempt Charity [Turn over

36 Introduction Modelling sea waves There are many situations in which waves and oscillations occur in nature and often they are accurately modelled by the sine curve. However, this is not the case for sea waves as these come in a variety of shapes. The photograph in Fig. shows an extreme form of sea wave being ridden by a surfer. 5 Fig. At any time many parts of the world s oceans are experiencing storms. The strong winds create irregular wind waves. However, once a storm has passed, the waves form into a regular pattern, called swell. Swell waves are very stable; those resulting from a big storm would travel several times round the earth if they were not stopped by the land. Fig. illustrates a typical swell wave, but with the vertical scale exaggerated. The horizontal distance between successive peaks is the wavelength; the vertical distance from the lowest point in a trough to a peak is called the height. The height is twice the amplitude which is measured from the horizontal line of mid-height. The upper part is the crest. These terms are illustrated in Fig OCR B/0 Insert June 07

37 3 Peak Crest Height Crest Peak Crest Line of mid-height Trough Trough Amplitude Trough Wavelength Fig. The speed of a wave depends on the depth of the water; the deeper the water, the faster the wave. (This is, however, not true for very deep water, where the wave speed is independent of the depth.) This has a number of consequences for waves as they come into shallow water. Their speed decreases. Their wavelength shortens. Their height increases. 0 Observations show that, as their height increases, the waves become less symmetrical. The troughs become relatively long and the crests short and more pointed. The profile of a wave approaching land is illustrated in Fig. 3. Eventually the top curls over and the wave breaks. 5 Fig. 3 If you stand at the edge of the sea you will see the water from each wave running up the shore towards you. You might think that this water had just travelled across the ocean. That would be wrong. When a wave travels across deep water, it is the shape that moves across the surface and not the water itself. It is only when the wave finally reaches land that the actual water moves any significant distance. 30 Experiments in wave tanks have shown that, except near the shore, each drop of water near the surface undergoes circular motion (or very nearly so). This has led people to investigate the possiblility that a form of cycloid would provide a better model than a sine curve for a sea wave. Cycloids There are several types of cycloid. In this article, the name cycloid refers to one of the family of curves which form the locus of a point on a circle rolling along a straight horizontal path. 35 Fig. 4 illustrates the basic cycloid; in this case the point is on the circumference of the circle. OCR B/0 Insert June 07 [Turn over

38 4 Fig. 4 Two variations on this basic cycloid are the prolate cycloid, illustrated in Fig. 5, and the curtate cycloid illustrated in Fig. 6. The prolate cycloid is the locus of a point attached to the circle but outside the circumference (like a point on the flange of a railway train s wheel); the curtate cycloid is the locus of a point inside the circumference of the circle. 40 Fig. 5 Fig. 6 When several cycles of the curtate cycloid are drawn upside down, as in Fig. 7, the curve does indeed look like the profile of a wave in shallow water. Fig. 7 The equation of a cycloid The equation of a cycloid is usually given in parametric form. 45 Fig. 8. and Fig. 8. illustrate a circle rolling along the x-axis. The circle has centre Q and radius a. Pand R are points on its circumference and angle PQR = q, measured in radians. Fig. 8. shows the initial position of the circle with P at its lowest point; this is the same point as the origin, O. Some time later the circle has rolled to the position shown in Fig. 8. with R at its lowest point. y y Q q P O a R x P O Q q a R x Fig. 8. Fig. 8. OCR B/0 Insert June 07

39 5 In travelling to its new position, the circle has rolled the distance OR in Fig. 8.. Since it has rolled along its circumference, this distance is the same as the arc length PR, and so is aq. Thus the coordinates of the centre, Q, in Fig. 8. are (aq, a). To find the coordinates of the point P in Fig. 8., look at triangle QPZ in Fig. 9. Q q a acosq 50 P asinq Fig. 9 Z R You can see that PZ a sin q and QZ a cos q. 55 (aq a sin q, a a cos q), Hence the coordinates of P are described by the curve with parametric equations and so the locus of the point P is x aq a sin q, y a a cos q. This is the basic cycloid. These parametric equations can be generalised to 60 x aq b sin q, y a b cos q, where b is the distance of the moving point from the centre of the circle. For b a b a b a the curve is a curtate cycloid, the curve is a basic cycloid, the curve is a prolate cycloid. 65 The equivalent equations with the curve turned upside down, and with the mid-height of the curve now on the x-axis, are x aq b sin q, y b cos q. (Notice that positive values of y are still measured vertically upwards.) Modelling a particular wave 70 A question that now arises is how to fit an equation to a particular wave profile. If you assume that the wave is a cycloid, there are two parameters to be found, a and b. OCR B/0 Insert June 07 [Turn over

40 4754 Mark Scheme June (C4) Applications of Advanced Mathematics Section A x x + = + x 4 x+ ( x )( x+ ) x+ x+ ( x ) = ( x+ )( x ) 3x 4 = ( x+ )( x ) [3] combining fractions correctly factorising and cancelling (may be 3x²+x-8) V = π ydx= 0 = x π x+ e 0 = π ( + e ) = ( π + e ) * 3 cos θ = sin θ sin θ = sin θ sin θ sin θ = 0 x π ( + e ) dx 0 ( sin θ)( + sin θ) = 0 sin θ = ½ or θ = π/6, 5π/6, 3π/ 4 sec θ = x/, tan θ = y/3 sec θ = + tan θ x /4 = + y /9 x /4 y /9 = * OR x²/4-y²/9= 4sec²θ/4-9tan²θ/9 =sec²θ-tan²θ = E [4] A,,0 [7] E [3] must be π x their y² in terms of x x + e x only substituting both x limits in a function of x www cosθ = sin θ oe substituted forming quadratic( in one variable) =0 correct quadratic www factorising or solving quadratic ½, oe www cao penalise extra solutions in the range sec θ = + tan θ used (oe, e.g. converting to sines and cosines and using cos θ + sin θ = ) eliminating θ (or x and y) www 5(i) dx/du = u, dy/du = 6u dy / 6 = dy du = u dx dx / du u = 3u OR y=(x-) 3/, dy/dx=3(x-) / =3u [3] both u and 6u (y=f(x)), differentiation, (ii) At (5, 6), u = dy/dx = 6 [] cao

41 4754 Mark Scheme June ( ).( ) 6(i) (+ 4 x ) =.4 x + (4 x ) +...! 4 = x + 6 x +... Valid for < 4x < - ½ < x < ½ [5] binomial expansion with p= -/ x + 6x 4 (ii) x = x x + x + + 4x 4 4 = x + 6x x + x +... = 3x + 8x ( )( 6...) [3] 4 substitituting their x + 6 x +... and expanding ft their expansion (of three terms) cao 7 3 sin x cos x = Rsin(x α) = R(sin xcos α cos x sin α) 3 = Rcos α, = R sin α R = 3 + =4 R = tan α = / 3 α = π/6 y = sin(x π/6) Max when x π/6 = π/ x = π/6 + π/ = π/3 max value y = So maximum is (π/3, ) [6] correct pairs soi R = ft cao www cao ft their R SC (, π/3) no working 3

42 4754 Mark Scheme June 008 Section B 8(i) At A: ( 5) = 0 At B: ( 30) = 0 At C: ( 5) = 0 So ABC has equation 3x + y + 0 z = (ii) DE = DF 0 = = = = = = = Equation of plane is x y + 0z = c At D (say) c = 0 40 = 800 So equation is x y + 0z = 0 A,,0 [3] [6] substituting co-ords into equation of plane for ABC OR using two vectors in the plane form vector product then 3x +y +0z = c = -300 OR using vector equation of plane,elim both parameters, need evaluation need evaluation (iii) Angle is θ, where cosθ = = + ( ) θ = 8.95 cao [4] formula with correct vectors top bottom (or 0.56 radians) 5 3 (iv) RS: r = 34 + λ λ = 34 + λ 0λ 3(5+3λ) + (34+λ) +0.0λ+300 = λ λ λ = λ = 0 λ = so S is (, 3, 0) [5] λ 0 solving with plane λ = cao 4

43 4754 Mark Scheme June 008 9(i) v = 0e t t dt = 0e + c when t = 0, v = 0 0 = 0 + c c = 0 so v= 0 0 t e [4] separate variables and intend to integrate t e 0 finding c cao (ii) As t e / t 0 v 0 So long term speed is 0 m s [] ft (for their c>0, found) (iii) A B = + ( w 4)( w+ 5) w 4 w+ 5 Aw ( + 5) + Bw ( 4) = ( w 4)( w+ 5) A(w + 5) + B(w 4) w = 4: = 9A A = /9 w = 5: = 9B B = /9 /9 /9 = ( w 4)( w+ 5) w 4 w+ 5 = 9( w 4) 9( w+ 5) [4] cover up, substitution or equating coeffs /9 /9 (iv) dw ( w 4)( w 5) dt = + dw = dt ( w 4)( w+ 5) [ ] dw = dt 9( w 4) 9( w+ 5) ln( w 4) ln( w+ 5) = t+ c 9 9 w ln 4 = t + c 9 w + 5 When t = 0, w = 0 6 c = ln = ln w 4 9 ln = t + ln w ln w 4 t+ t = 5 4.5t e = e = 0.4e * w ft E [6] separating variables substituting their partial fractions integrating correctly (condone absence of c) correctly evaluating c (at any stage) combining lns (at any stage) www (v) As t e 4.5 t 0 w 4 0 So long term speed is 4 m s. [] 5

44 4754 Mark Scheme June 008 Comprehension. (i) 3 3 cao (ii) cao 3. Dividing the grid up into four x blocks gives Lines drawn on diagram or reference to x blocks. One (or more) block does not contain all 4 of the symbols,, 3 and 4. oe. E Many possible answers Row correct Rest correct 6

45 4754 Mark Scheme June Either Or B In the top row there are 9 ways of allocating a symbol to the left cell, then 8 for the next, 7 for the next and so on down to for the right cell, giving = 9! ways. So there must be 9! the number of ways of completing the rest of the puzzle. E 6. (i) Block side length, b Sudoku, s s M , 5 and 6 (ii) 4 M = b 4 4 b -4 7

46 4754 Mark Scheme June (i) (ii) There are neither 3s nor 5s among the givens. So they are interchangeable and therefore there is no unique solution The missing symbols form a 3 3 embedded Latin square. There is not a unique arrangement of the numbers, and 3 in this square. E E [8] 8

47 4754 Mark Scheme January (C4) Applications of Advanced Mathematics Section A 3x + A Bx+ C = + xx ( + ) x ( x + ) 3x + = A(x + ) + (Bx + C)x x = 0 = A coefft of x : 0 = A + B B = coefft of x: 3 = C correct partial fractions equating coefficients at least one of B,C correct 3x + 3 x = + xx x x ( + ) ( + ) [6].( ) (+ x) = +. x+ 3 3 ( x) ! = + x 4 x = + x x +... * ( )( ) Next term ( ) 3 = x 3! = x (i) /3 E binomial expansion correct unsimplified expression simplification www Valid for < x < ½ < x < ½ [6] 3 4j 3k = λ a + μ b = λ(i + j k) + μ(4i j + k) 0 = λ + 4μ 4 = λ μ 3 = λ + μ λ = μ, λ = 4 λ =, μ =, [5] equating components at least two correct equations 4 LHS = cot β cot α cos β cosα = sin β sinα sinα cos β cosαsin β = sinαsin β sin( α β) = sinα sin β OR RHS= sin( α β) sin α cos β cos α sin β = sin α sin β sin α sin β sin α sin β =cot β cot α E E [3] cot = cos / sin combining fractions www using compound angle formula splitting fractions using cot=cos/sin 3

48 4754 Mark Scheme January 009 5(i) Normal vectors and 0 Angle between planes is θ, where cosθ = + ( ) 0+ ( ) + ( ) ( ) = / θ = 73. or.8 rads [4] scalar product finding invcos of scalar product divided by two modulae (ii) r = 0 + λ + λ = λ + λ ( + λ) ( λ) + ( + λ) = 5 + 6λ = λ = ½ So point of intersection is (, ½, ½ ) [4] 6(i) cosθ + 3 sinθ = r cos( θ α) = Rcosθ cosα + Rsinθsinα Rcosα =, Rsinα = 3 R = + ( 3) = 4, R = tan α = 3 α = π/3 [4] R = equating correct pairs tan α = 3 o.e. (ii) derivative of tan θ is sec θ π π sec ( ) 0 (cos 3 sin ) d π θ = θ 4 3 d θ θ + θ π = tan( θ ) 4 3 = ¼ (0 ( 3)) = 3/4 * π 3 0 E [4] ft their α R www [tan (θ π/3] ft their R,α(in radians) 4

49 4754 Mark Scheme January 009 Section B 7(i) (A) 9 /.5 = 6 hours (B) 8/.5 = hours [] (ii) dθ = k( θ θ0) dt dθ kdt θ θ = 0 θ θ0 ln( ) = kt + c separating variables ln( θ θ0) kt + c θ θ = e 0 kt+ c anti-logging correctly(with c) θ kt = θ0 + Ae * E [5] A=e c (iii) 98 = 50 + Ae 0 A = 48 Initially d θ = k(98 50) = 48k =.5 dt k = 0.035* E [4] 0.035t (iv) (A) 89 = e 39/48 = e 0.035t t = ln(39/48)/( 0.035) = 6.64 hours 0.035t (B) 80 = e 30/48 = e 0.035t t = ln(30/48)/( 0.035) = 5 hours [5] equating taking lns correctly for either (v) Models disagree more for greater temperature loss [] 5

50 4754 Mark Scheme January 009 8(i) dy cosθ sinθ dθ =, dx cosθ dθ = dy dy / dθ = dx dx / dθ dy cosθ sinθ cos θ sinθ = = dx cosθ cosθ, [4] substituting for theirs oe (ii) When θ = π/6, dy cos π / 3 sin π / 6 = dx cos π / 6 / / = = 0 3/ Coords of B: x = + sin(π/6) = 3 y = cos(π/6) + sin(π/3) = 3 3/ BC = 3 3/ = 3 3 E, ft [5] for either exact (iii) (A) y = cosθ + sinθ = cos θ + sin θ cos θ = cos θ ( + sin θ) = xcos θ * (B) sinθ = ½ (x ) cos θ = sin θ = ¼ (x ) = ¼ x + x = (x ¼ x ) * (C) Cartesian equation is y = x cos θ = x (x ¼ x ) = x 3 ¼ x 4 * E E E [7] sinθ = sin θ cos θ squaring and substituting for x (iv) 4 V = π y dx = π ( x x ) dx = π x x = π(64 5.) =.8π = 40. (m 3 ) [3] need limits 4 5 x x 4 0.8π or 40 or better. 6

51 4754 Mark Scheme January 009 Comprehension 400π d = d = = 7.96 π V= π0 h+ ( π0 H π0 h) = ( π0 H + π0 h) cm 3 = 00π ( H + h) cm 3 = π ( H + h) litres 5 3 H = 5+ 40tan30 or H=h+ 40tanθ V= ( ) π H + h = π ( tan 30 ) =0.8 litres V= 80 ( 40 5 ) + 30 cm 3 = cm 3 = 54 litres 5 (i) Accurate algebraic simplification to give y 60y+ 400 = 0 E E divide by 000 or evaluated including substitution values 30 of (ii) Use of quadratic formula (or other method) to find other root: d = 57.5 cm. This is greater than the height of the tank so not possible 6 y=0 Substitute for y in( 4 ): 00 V = 375dx V = = 37.5 * 000 E E [8] 7

52 4754 Mark Scheme June (C4) Applications of Advanced Mathematics Section A 4cosθ sinθ = R cos( θ + α) = Rcosθ cosα Rsinθsinα Rcosα = 4, Rsinα = R = + 4 = 7, R = 7 = 4.3 tan α = ¼ α = cos(θ ) = 3 cos(θ ) = 3/ 7 θ = 0.756, 5.57 θ = 0.5, 5.8 [7] correct pairs R = 7 = 4.3 tan α = ¼ o.e. α = 0.45 θ = arcos 3/ 7 ft their R, α for method (penalise extra solutions in the range (-)) x = A + B ( x+ )(x+ ) x+ (x+ ) x = A(x + ) + B(x + ) x = = A A = x = ½ ½ = ½ B B = x = ( x+ )(x+ ) x+ (x+ ) x dx = dx ( x+ )(x+ ) x+ (x+ ) = ln(x + ) ½ ln(x + ) + c [7] correct partial fractions substituting, equating coeffts or cover-up A = B = ln(x + ) ft their A -½ ln(x + ) ft their B cao must have c 3 d y 3x y d x = d y 3x dx y = ln y = x 3 + c When x =, y =, ln = + c c = ln y = x 3 x 3 y = e [4] separating variables condone absence of c c = oe o.e. 4 When x = 0, y = 4 4 V = π x dy 0 4 = π (4 ydy ) 0 = π 4y y 0 = π(6 8) = 8π 4 [5] must have integral, π, x² and dy soi must have π,their (4-y), their numerical y limits 4y y 3

53 4754 Mark Scheme June dy = a( + t ).t dt dx 3at dt = dy dy / dt at = = dx dx / dt 3 at ( + t ) = 3( t + t ) * At (a, ½ a), t = gradient = = /6 3 E [7] (+t²) kt for method ft finding t 6 cosec θ = + cot θ + cot θ cot θ = 3 * cot θ cot θ = 0 (cot θ )(cot θ + ) = 0 cot θ =, tan θ = ½, θ = 6.57 cot θ =, tan θ =, θ = 35 E [6] clear use of +cot²θ =cosec²θ factorising or formula roots, cot = /tan used θ = 6.57 θ = 35 (penalise extra solutions in the range (-)) 4

54 4754 Mark Scheme June 009 Section B 7(i) AB = 0 0 r = 0 + λ 0 [] or equivalent alternative (ii) n = cosθ = = 5 0 θ = 7.57 [5] correct vectors (any multiples) scalar product used finding invcos of scalar product divided by two modulae 7 or better (iii) 0. + cosφ = = = 9 3 ft their n for method ±/ oe exact φ = 45 * E [3] (iv) sin 7.57 = k sin 45 k = sin 7.57 / sin 45 =.34 [] ft on their 7.57 oe 0 (v) r = 0 + μ x= -μ, z=-μ x + z = μ + μ = 3 μ = 3, μ = point of intersection is (,, ) distance travelled through glass = distance between (0, 0, ) and (,, ) = ( + + ) = 3 cm [5] soi subst in x+z = - www dep on μ= 5

55 4754 Mark Scheme June 009 8(i) (A) 360 4=5 CB/OB = sin 5 CB = sin 5 AB = CB = sin 5 * E [] AB=AC or CB AOC= 5 oe (B) cos 30 = sin 5 cos 30 = 3/ 3/ = sin 5 sin 5 = 3/ = ( 3)/ sin 5 = ( 3)/4 sin 5 = 3 = 4 3 * E [4] simplifying (C) Perimeter = AB = 4 ½ ( 3) = ( 3) circumference of circle > perimeter of polygon π > ( 3) π > 6 ( 3) E [] (ii) (A) tan 5 = FE/OF FE = tan 5 DE = FE = tan 5 E [] (B) tan5 t tan 30 = = tan 5 t tan 30 = / 3 t = 3t = t t 3 t + 3 t = 0 * E [3] (C) 3± + 4 t = = 3 circumference < perimeter π < 4( 3) π < ( 3) * E [4] using positive root from exact working (iii) 6 ( 3) < π < ( 3) 3.06 < π < 3.5 [] 3.06, 3.5 6

56 4754 Mark Scheme June 009 Comprehension. [3 + + ( ) + ( )] = 0.5 *, E 4. (i) b is the benefit of shooting some soldiers from the other side while none of yours are shot. w is the benefit of having some of your own soldiers shot while not shooting any from the other side. (ii) Since it is more beneficial to shoot some of the soldiers on the other side than it is to have your own soldiers shot, b> w. E c is the benefit from mutual co-operation (i.e. no shooting). d is the benefit from mutual defection (soldiers on both sides are shot). With mutual co-operation people don t get shot, while they do with mutual defection. So c> d. E 3. + ( ) ( n ) =.999 or equivalent (allow n,n+), n n = 6000 so you have played 6000 rounds. 4. No. The inequality on line 3, b+ w< c, would not be satisfied since 6 + ( 3) >. b+w<c and subst No,3>oe 5. (i) Round You Opponent Your score Opponent s score C D - 3 D C 3-3 C D D C 3-5 C D D C 3-7 C D D C 3 - Cs and Ds in correct places, C=-, D=3 (ii) [3 + ( )] = 0.5 D ft their 3,- 6. (i) All scores are increased by two points per round (ii) The same player wins. No difference/change. The rank order of the players remains the same. 7. (i) They would agree to co-operate by spending less on advertising or by sharing equally. (ii) Increased market share (or more money or more customers). D 7

57 4754 A Mark Scheme January (C4) Applications of Advanced Mathematics + x = (+ x)( x) ( x) ( )( 3) = = ( + x)[ + 4 x+ x +...] = + 4x+ x + x+ 8 x +... = + 6x+ 0 x +... ( x)[ ( )( x) ( x)...] Valid for < x < ½ < x < ½ [7] binomial expansion power - unsimplified,correct sufficient terms tanθ tan θ = tan θ tan θ cot θ = = * tan θ tanθ cot θ = + tan θ tan θ = + tanθ tanθ tan θ = tanθ + tan θ 3tan θ + tanθ = 0 (3tanθ )(tanθ + ) = 0 tan θ = /3, θ = 8.43, or tan θ =, θ = 35, 35 E A3,,, 0 [7] oe eg converting either side into a one line fraction(s) involving sin θ and cos θ. quadratic = 0 factorising or solving 8.43, 98.43, 35, 35 extra solutions in the range 3 (i) d y ( + t). t. = = d t ( + t) ( + t) d x t e d t = d y d /d = y t dx d x/dt d y ( + t) = = t t dx e e ( + t) t = 0 dy/dx = ft [6] 4

58 4754 A Mark Scheme January 00 (ii) t = ln x t = ½ ln x ln x ln x y = = + lnx + lnx [] or t in terms of y 4 (i) AB =, A C = 3 [] (ii) n.ab =. = = 0 3 n.a C =. = + 9 = plane is x y 3z = d x =, y = 3, z = d = = 5 plane is x y 3z = 5 E E [5] scalar product 5 (i) x = 5 + 3λ = λ = y = = 3 z = 4 =, so (, 3, ) lies on st line. x = + μ = μ = y = 4 = 3 z = + 0 =, so (, 3, ) lies on nd line. E E [3] finding λ or μ verifying two other coordinates for line verifying two other coordinates for line (ii) Angle between 3 0 and 0 direction vectors only ( ) + ( ) 0 cosθ = 0 5 = θ = 3.9 [4] allow for any vectors or radians 5

59 4754 A Mark Scheme January 00 6 (i) BAC = 0 90 (90 θ) = θ 60 BC = b sin(θ 60) CD = AE = a sin θ h = BC + CD = a sin θ + b sin (θ 60 ) * E [3] (ii) h = a sin θ + b sin (θ 60 ) = a sin θ + b (sinθ cos 60 cosθ sin 60) = a sin θ + ½ b sin θ 3/ b cos θ 3 = ( a+ b)sinθ bcosθ * E [3] corr compound angle formula sin 60 = 3/, cos 60 = ½ used (iii) OB horizontal when h = 0 3 ( a+ b)sinθ bcosθ = 0 3 ( a+ b)sinθ = bcosθ 3 b sinθ cosθ = a+ b 3b tanθ = a+ b * E [3] sinθ tanθ cosθ = (iv) sinθ 3 cosθ = R sin( θ α) = R(sinθ cosα cosθ sin α) R cos α =, R sin α = 3 R = + ( 3) = 7, R = 7 =.646 m tan α = 3/, α = 40.9 So h = 7 sin(θ 40.9 ) h max = 7 =.646 m when θ 40.9 = 90 θ = 30.9 ft [7] 6

60 4754 A Mark Scheme January 00 7 (i) d x t t = ( + e ). e dt t e = t (+ e ) x = + e t t t + e e x = = t t + e + e t t e e x( x) = = t t t + e + e (+ e ) dx x( x) dt = When t = 0, x = 0 + e = 0.5 E [6] chain rule substituting for x( x) t t + e e x = = t t + e + e [OR, for solving differential equation for t, use of initial condition, making x the subject, E required form] (ii) 3 = t (+ e ) 4 e t = /3 t = ln /3 =.0 years [3] correct log rules (iii) = A + B + C x ( x) x x x = A( x) + Bx( x) + Cx x = 0 A = x = C = coefft of x : 0 = B + C B = B(,,0) [4] clearing fractions substituting or equating coeffs for A,B or C A =, B =, C = www (iv) d x dx = dt x ( x) t = ( + + )dx x x x = /x + ln x ln( x) + c When t = 0, x = ½ 0 = + ln ½ ln ½ + c separating variables -/x + ln x ln( x) ft their A,B,C substituting initial conditions c =. t = /x + ln x ln( x) + x = + ln * x x E [5] (v) 3/4 t = + ln = ln3+ =.77yrs 3/4 3/4 3 [] 7

61 4754 B Mark Scheme January 00 5 THE MATHEMATICIAN 3 M H X I Q 3 or 4 correct award mark B 4 Two from Ciphertext N has high frequency E would then correspond to ciphertext R which also has high frequency T would then correspond to ciphertext G which also has high frequency A is preceded by a string of six letters displaying low frequency oe oe 5 The length of the keyword is a factor of both 84 and 40. The only common factors of 84 and 40 are, and 4 (and a keyword of length can be dismissed in this context) E 6 Longer strings to analyse so letter frequency more transparent. Or there are fewer -letter keywords to check B 7 OQH DRR EBG One or two accurate award mark B 8 (i) (ii) (iii) Evidence of intermediate H FACE Evidence of intermediate HCEG award marks Evidence of accurate application of one of the two decoding ciphers - award mark 800 = ( 3 66) + ; the second row gives T so plaintext is R B3 Use of second row 8

62 GCE Mathematics (MEI) Advanced GCE 4754A Applications of Advanced Mathematics (C4) Paper A Mark Scheme for June 00 Oxford Cambridge and RSA Examinations

63 4754A Mark Scheme June 00 Section A x x x x ( x)( x) x x( x) = ( x)( x) (3x ) = ( x)( x) x = (x + )(x ) correct method for addition of fractions (3x ) or do not isw for incorrect x subsequent cancelling or x x( x) ( x ) x x ( x )( x) = 3x x ( x )( x) (3x)( x) = ( x )( x) (3x ) ( x ) [3] correct method for addition of fractions (3x )(x + ) accept denominator as x²- or(x-)(x+) do not isw for incorrect subsequent cancelling (i) When x = 0.5, y =.80 A 0.5/{+.44+( )} = =.5475 =.5 (3 d.p.)* (ii) Explain that the area is an over-estimate. or The curve is below the trapezia, so the area is an over- estimate. This becomes less with more strips. or Greater number of strips improves accuracy so becomes less E [3] [] 4dp (0.5 x 9.8) need evidence or use a diagram to show why (iii) V = = y dx 0 0 ( x ) dx 3 = ( xx /3) = 3 0 [3] allow limits later x + x 3 /3 exact

64 4754A Mark Scheme June 00 3 y = sin cos = ½ sin x = cos sin + cos = x + (y) = x + 4y = * or x² +4y²= (cos θ)² +4 (sinθcosθ)² = cos ²θ +sin²θ = * or cos θ=cos²θ- cos²θ= (x+)/ cos θ =-sin ²θ sin²θ= (-x)/ y²=sin²θcos²θ= x x y²= (-x²)/4 x²+ 4y²=* or x=cos θ=cos²θ sin²θ x²=cos 4 θ -sin²θcos²θ+sin 4 θ y²=sin²θcos²θ x²+4y²=cos 4 θ-sin²θcos²θ+sin 4 θ+4sin²θcos²θ =(cos²θ+sin²θ)² =* E E E E use of sin θ substitution use of sinθ for both correct use of double angle formulae correct squaring and use of sin²θ+cos²θ = ½ ellipse correct intercepts [5]

65 4754A Mark Scheme June 00 4 x 4x ( ) 4. x (. x ( )...) 4 4 ( x x...) 8 8 x x = = = dealing with 4 (or terms in 4, 4 correct binomial coefficients correct unsimplified expression for (+x/4) or (4+x) cao, etc) Valid for < x/4 < 4 < x < 4 [5] 5(i) 3 A B ( y)( y) y y = Ay ( ) By ( ) ( y)( y) 3 = A(y + ) + B(y ) y = 3 = 3A A = y = 3 = 3B B = (ii) dy x ( y )( y ) dx 3d y 3x dx ( y)( y) ( )dy 3x dx ( y) y ln(y ) ln(y + ) = x 3 + c y 3 ln x c y y x c x c x e e.e Ae y * [3] ft E [5] substituting, equating coeffs or cover up separating variables ln(y ) ln(y + ) ft their A,B x 3 + c anti-logging including c www tan tan 45 6 tan( 45) tan tan45 = tan tan tan tan tan + tan = ( tan )( tan ) = 3 tan + tan 0 = tan 4 tan = tan (tan ) tan = 0 or = 0 or [7] oe using sin/cos multiplying up and expanding any correct one line equation solving quadratic for tan θ oe www - extra solutions in the range 3

66 4754A Mark Scheme June 00 Section B 7(i) 00 ( 00) 300 * AB AB = ( ) = 33 m E [3] accept surds (ii) r Angle is between and cos = 7.45 [6] oe 0 and 0 complete scalar product method(including cosine) for correct vectors 7.5º or better, accept.6 radians (iii) Meets plane of layer when ( ) + (00+00) = = 30 = / r so meets layer at ( 80, 40, 40) [4] (iv) Normal to plane is 3 3 Angle is between and cos = angle with layer = 40. [5] complete method ft 90-theirθ accept radians 4

67 4754A Mark Scheme June 00 8(i) At A, y = 0 4cos = 0, = / At B, cos =, = x-coord of A = / sin / = x-coord of B = sin = OA =, AC = + = + ratio is ( ):( + ) * E [5] for either A or B/C for both A and B/C (ii) d y 4sin d d x cos d dy d y/d dx d x /d = 4sin cos At A, gradient = 4sin( /) = cos( /) [4] either dx/dor dy/d www (iii) d y 4sin dx cos 4sin = cos cos 4sin = * (iv) cos 4sin = Rcos( + = R(cos cos sin sin ) Rcos Rsin R = + 4 = 7, R = 7 tan = 4, =.36 7 cos( +.36) = cos( +.36) = / =.064, 5.9, = ( 0.6), 3.89, 6.0 E [] [7] their dy/dx = corr pairs accept 76.0º,.33 radians inv cos (/ 7) ft their R for method - extra solutions in the range 5

68 GCE Mathematics (MEI) Advanced GCE 4754B Applications of Advanced Mathematics (C4) Paper B: Comprehension Greener Travel Mark Scheme for June 00 Oxford Cambridge and RSA Examinations

69 4754B Mark Scheme June 00. Rail: = = 8.5 kg (3 sf) Road: = = 39.8 kg (3 sf) Reduction =.3 kg for either. 3 4 y = ( x 00 x 0000 x+ 0000).. 0 dy = 4 (3 x 00 x 0 000) dx 0 dy 0 3x 00x dx = = ( 3x+ 00)( x 00) = 0 00 x= 00 (or x= ) 3 The graph shows the minimum emission occurs at speed of 00 km hour - Substituting x = 00 gives y =0.0 Minimum rate of emission is 0 grams per km. 3. (i) Substituting p= 50, d = 79, s= 4 in E = ( pd ) + 00s E = (in kg) So emissions are 3.7 tonnes to s.f. * (ii) Emission rate =.5 g km - Distance = 79 km Emissions =.5 79 =48.5 g = 0.4 kg ( s.f.), and so is less than ½ kg. solving quadratic d y or dx justify min subst E E or p=5 in formula gives E= , difference=0.485kg<0.5kg 4. (i) 5 n 0 5 (ii) P (millions) 3 Approximate Step graph Correct with scales shown There is a basic service of 0 trains a day for up to million passengers per year. For every half million extra passengers above million, an extra daily train is provided.

70 4754B Mark Scheme June miles = km = 60 km 934 m 40 cm So it appears to give the answer to the nearest 0 cm (option B). [8]

71 4754A Mark Scheme January 0 Section A (i) x y A ½ { ( )} = B,,0 [4] table values formula 6.5 or better www (ii) Smaller, as the trapezium rule is an over-estimate in this case and the error is less with more strips [] x t t x t x t y x t x x x x x [5] attempt to solve for t oe substituting for t in terms of x clearing subsidiary fractions (3 x) 3 ( x) 3 ( 3)( 4) ( ( 3)( x) ( x)...) ( x x...) x x Valid for 3 x 3 3 x B,,0 [7] dealing with the 3 correct binomial coeffs,, 8/3 oe cao

72 4754A Mark Scheme January 0 5 4(i) AB 3, BC AB.B C ( 5) 0 5 AB is perpendicular to BC. E [4] (ii) AB = ( ( 5) )= 38 BC = ( )= 9 Area = ½ 38 9 = ½ 0 or 6.6 units [3] complete method ft lengths of both AB, BC oe www 5 LHS = sin cos cos sin cos cos sin tan RHS cos E [3] one correct double angle formula used cancelling cos s x 83 6(i) y z 6 Substituting into plane equation: ( 8 3) 3( ) = = 5 = 5, = 3 So point of intersection is (,, 3) ft [4] 3 (ii) Angle between and 3 0 ( 3) ( 3) 0 cos 4 0 = ( )0.43 acute angle = 65 [4] allow for a complete method only for any vectors

73 4754A Mark Scheme January 0 Section B 7(i) When t = 0, v = 5( e 0 ) = 0 As t, e t 0, v 5 When t = 0.5, v = 3.6 m s E E [3] d v t t (ii) 5 ( )e 0e d t 0 v = 0 0( e t ) = 0e t d v 0 v d t E [3] (iii) d v 0 0.4v d t 0 d v 00 4v d t 0 d v 4 5 v dt 0 d v 4 (5 v)(5 v) d t * 0 A B (5 v)(5 v) 5 v 5 v 0 = A(5 + v) + B(5 v) v = 5 0 = 0A A = v = 5 0 = 0B B = 0 (5 v)(5 v) 5 v 5 v ( )dv 4 t 5v 5v d ln(5 v) ln(5 v) 4tc when t = 0, v = 0, 0 = c c = 0 5 v ln 4t 5 v 5 v t ln * 4 5 v E E [8] for both A=,B= separating variables correctly and indicating integration ft their A,B, condone absence of c ft finding c from an expression of correct form (iv) When t, e 4t 0, v 5/ = 5 5( e ) when t = 0.5, t 3.8ms e E [3] (v) The first model E [] www 3

74 4754A Mark Scheme January 0 8(i) AC = 5sec oe CF = AC tan = 5sec tan GF = CF = 0sec tan * E [3] ACtanβ (ii) CE = BE BC = 5 tan( + ) 5 tan = 5(tan( + ) tan ) tan tan 5 tan tan tan tan tan tan tan tan 5 tan tan 5( tan ) tan tan tan 5tan sec * tan tan E D E [5] compound angle formula combining fractions sec = + tan (iii) sec 45 =, tan 45 = 5t 0t CE t t 0t CD t 0t 0t DE 0 t( ) t t t t tt 0t 0 t * ( )( ) t t t E [5] used substitution for both in CE or CD oe for both adding their CE and CD (iv) cos 45 = / sec = GF = 0 tan = 0 t E [] (v) DE = GF 0t 0 t t t = / t = / * t = 0.54 = 8.4 E [3] invtan t 4

75 4754B Mark Scheme January 0 Qn Answer Marks (i) 6 correct marks (ii) Either state both m and n odd or give a diagram (doorways between rooms not necessary) justification ft (i) 9 4 B 4 ( for LHS correct) (ii) x B,,0 x 3 3. If each of A, B and C appeared at least four times then the total number of vertices would have to be at least 34= E 4(i) A C B A A allow if one error C A C B 4(ii) Two points labelled B above clearly marked (or f.t. from (i)) 5(i) True. Two cameras at the vertices labelled A or at the vertices labelled B would cover the entire gallery 5(ii) False. One camera at either vertex labelled A would be sufficient (or C on RHS) 6 Anywhere in shaded region for either correct construction correct shading

76 4754A Mark Scheme June 0 A Bx C (x )( x ) x x = A(x + ) + (Bx + C) (x + ) x = ½ : = ¼ A A = 4/5 coeff of x : 0 = A + B B = /5 constants: = A + C C = /5 correct form of partial fractions mult up and equating or substituting oe soi www www www for omission of B or C on numerator, M0,, then (x= -/, A= 4/5), B0, B0 is possible. A Dx Bx C for x x,, then for both A=4/5 and D=0,, is possible. isw for incorrect assembly of final partial fractions following correct A,B & C..( ) 3 3 3! = + x x + 3 (3 x) (3 x) (3 x)... Valid for 3x /3 x /3 3 sin 3cos R sin( ) Rsin cos Rcos sin Rcosα =, Rsinα=3 R = + 3 = 3, R = 3 tan = 3/, = [5] [5] correct binomial coefficients + x x or 3x oe or x /3 (correct final answer scores ) correct pairs R = 3 or 3.6 or better 0.98 or better condone omission of brackets for second only if the brackets are implied by subsequent working. ie, /3, (/3)(-/3)/ not ncr form simplified www in this part simplified www in this part, ignore subsequent terms using (3x)² as 3x² can score B0 condone omission of brackets if 3x² is used as 9x² do not allow MR for power 3 or -/3 or similar condone inequality signs throughout or say < at one end and at the other condone -/3 x /3, x /3 is M0A0 the last two marks are not dependent on the first three condone wrong sign at this stage correct division, ft from first radians only accept multiples of π that round to 0.98 minimum 3, maximum + 3 or.6, 4.6 or better allow, ft for - R and + R for their R to dp or better [6] 4

77 4754A Mark Scheme June 0 4(i) x = sin, y = cos When = /3, x = sin /3 = 3 y = cos /3 = ½ x = 3 y = ½ exact only (isw all dec answers following exact ans ) EITHER dx/d = cos, dy/d = sin d y sin d x cos sin / 3 3 / cos / 3 / 3 dy/dx = (dy/d / (dx/dused any correct equivalent form exact www ft their derivatives if right way up (condone one further minor slip if intention clear) condone poor notation can isw if incorrect simplification..... OR expressing y in terms of x, y=-x²/ dy =-x or -sinθ dx = - 3 exact www (ii) y = sin = (x/) = ½ x [] [5] or reference to (i) if used there for, need correct trig identity and attempt to substitute for x allow SC for y=cos arcsin(x/) or equivalent 5

78 4754A Mark Scheme June 0 5 cosec = + cot + cot = + cot cot cot = 0 cot (cot ) = 0 cot = 0, and cot =, tan = ½ = 6.6, 53.4,-90,90.. OR cos sin cos sin sin sin sin + sin cos = 0 sin cos cos = 0 cos (sin cos ) = 0 cos = 0, and tan = ½ = 6.6, 53.4, 90, 90 6 Vol = vol of rev of curve + vol of rev of line vol of rev of curve = x d y 0 y d y 0 y 4 = 0 B3,,,0 B3,,,0 [6] correct trig identity used factorising oe both needed and cot =/tan soi 90, 90, 7, 53 or better www correct trig equivalents and a one line equation (or common denominator) formed use of Pythagoras and factorising both needed and tan = sin /cos oe soi accept 7º, -53º as above. answers, no working, award B3,,,0 (it is possible to score say then B3 ow) (soi) at any stage substituting x = y/ y 4 (use of -cot²θ could lead to M0 ) allow if cot θ =0 not seen (ie quadratic equation followed by cot θ-=0 or cot θ= ) (omission of cot θ=0 could gain,, M0, )... as above allow if cos θ=0 not seen (as above) in both cases, - if extra solutions in the range are given ( dependent on at least being scored)-not their incorrect solutions eg 6.6,-53.4, 0,80,-80 would obtain - MR if answers given in radians (-π/,π/,0.464, -.68 ( ) or multiples of π that round to these, or better) (dependent on at least being scored) to lose both of these, at least B would need to be scored. for need π, substitution for x², (dy soi), intention to integrate and correct limits even if π missing or limits incorrect or missing cao height of cone = 3 = so vol of cone = /3 x = /3 so total vol = 4/3 [7] h= soi www cao 3 OR (3 y) dy (even if expanded incorrectly) =π/3 www 6

79 4754A Mark Scheme June 0 7(i) 4 AB 0, A C ( 4).( ) 0.4 ( ). cos BAC AB.AC 0 = 0.93 Section B dot product evaluated cos BAC= dot product / AB. AC 0.93 or cos ABC=correct numerical expression as RHS above, or better condone rows substituted, ft their vectors AB, AC for method only need to see method for modulae as far as use of vectors BA and CA could obtain B0 B0 BAC = 73.0 (ii) A: x + y z + d = 6 + d = 0 d = 4 B: = 0 C: = 0 Normal n = 0 n. 0 cos 6 = 44.7 acute angle = 35.3 (iii) At D, + 4 k + 4 = 0 k = 6, k = 3 * CD 0 AB CD is parallel to AB CD : AB = : [6] D [7] [5] or rounds to 73.0º (accept 73º www) substituting one point evaluating for other two points d = 4 www stated or used as normal anywhere in part (ii) finding angle between normal vector and k allow ±/ 6 or 44.7º for or rounds to 35.3º substituting into plane equation AG CD 0 mark final answer www allow CD:AB=/, 5:0 oe, AB is twice CD oe (or.7 radians) alternatively, finding the equation of the plane using any valid method (eg from vector equation, for using valid equation and eliminating both parameters, for required form, or using vector cross product to get x+y-z =c oe,finding c and required form,, or showing that two vectors in the plane are perpendicular to normal vector and finding d, ) oe (may have deliberately made +ve to find acute angle) do not need to find 44.7º explicitly (or 0.65 radians) finding vector CD (or vector DC) or DC parallel to AB or BA oe (or hence two parallel sides, if clear which) but A0 if their vector CD is vector DC for allow vector CD used as vector DC 7

80 4754A Mark Scheme June 0 dv 8(i) kx d t V = /3 x 3 dv/dx = x d V d d d V. x x x dt dx dt dt d x x kx d t d x x k d t * (ii) d x x k d t x dx kdt ½ x = kt + c When t = 0, x = 0 50 = c ½ x = 50 kt x = (00 kt) * [3] [4] oe eg dx/dt=dx /dv. dv/dt =/x². kx =-k/x AG separating variables and intention to integrate condone absence of c finding c correctly ft their integral of form ax²= bt+c where a,b non zero constants AG (iii) When t = 50, x = 0 0 = k k = (iv) dv/dt = kx = x x dx/dt = x dx x * dt x (v) ( x) x( x) x x x x x x x * x x x dx dt x ( x ) dx t c x ln( x) ½ x x = t + c When t = 0, x = 0 c = ln 0 0 = 0 ln t x x * x (vi) understanding that ln (/0) or /0 is undefined oe [] [] [6] [] for dv/dt= -kx or better AG combining to single fraction AG separating variables & subst for x²/(-x) and intending to integrate condone absence of c finding c for equation of correct form eg c = 0, or ln (allow c=0 without evaluation here) cao AG www or long division or cross multiplying check signs need both sides of integral accept ln (/(-x)) as ln(-x) www ie aln(-x)+bx²+dx=et +c a,b,d,e non zero constants do not allow if c=0 without evaluation ln (/0) = ln 0, /0 = and ln (/0) = are all B0 8

81 4754B Mark Scheme June 0 Question Answer Marks Guidance % * or x 00 = 6.4* or 50-(64+70) =6.4% need evaluation 50 oe [] (i) The smallest possible PIN that does not begin with zero is 000 and the largest is 9999, giving However the 9 numbers,, 9999 are disallowed. The other disallowed numbers are 34, 345, 6789 (6 numbers) And 9876, 8765, 30 (7 numbers). from a correct starting point (eg 0,000 or 9000), clear attempt to eliminate (or not include) numbers starting with 0 clear attempt to eliminate all three of these categories (with approx correct values in each category) Alt) for (no 0 start), nos starting with,,7,8,9 give 000-, nos starting with 3,4,5,6 give =5(000-)+4(000-3)=8978, or) eg starting with,,not,any,any+,,not3,any +,,3,not4 = =999- (term)=998 can lead to 5( )+4( )=8978 oe if unclear, M0 M marks not dependent So, in all, there are 9000 ( ) = 8978 possible PINs SC 8978 www B3 [3] (ii) ft from (i) The average is about ft accept sf (or sf) only for [] 3 People with no breaches of security People with breaches of security numbers total 8-4 all correct [] 4

82 4754B Mark Scheme June 0 Question Answer Marks Guidance transactions from 80 people over 3½ years with 365 days per allow approximate number of year days in a year eg 360 for ( ) Approximately transaction per person per day cao [] 5 Allow any one of the following for mark 6 (i) An attack can happen without a breach of the card s security. The probabilities that a successful attack followed or did not follow a breach of card security are so close that a court would look for other evidence before reaching a decision. In many cases of unauthorised withdrawals the banks refund the money. The banks software does not detect all the attacks that occur. [] only accept versions of these statements Transactions Authorised Unauthorised Total for top row 480, 0, 500 Queried Not queried Total B all five other entries correct ( is given) allow for three or four correct from ,40,499500,499940,60 [3] 5

83 4754B Mark Scheme June 0 Question Answer Marks Guidance 6 (ii) 480 ft from (i) their 480: their 40 isw or to accept unsimplified answers 40 [] 6 (iii) Transactions Authorised Unauthorised Total Queried Not queried NB they are not required to complete the table. Total D ft from (i) {500or 5xtheir 500}-(their 60-5) [=their 445] their 445 ft from (i) : or 489 to 5 cao [3] 6

84 4754A Mark Scheme June 0 Question Answer Marks Guidance 4x 3 x x 4 x(x) 3( x) ( x)(x) Multiplying throughout by (x + )(x + ) or combining fractions and multiplying up oe (eg can retain denominator throughout) Condone a single numerical error, sign error or slip provided that there is no conceptual error in the process involved Do not condone omission of brackets unless it is clear from subsequent work that they were assumed eg 4x(x + ) 3(x + ) = (x + )(x ) gets 4x(x + ) 3(x + ) = gets M0 4x(x + )(x + ) 3(x + )(x + ) = (x + )(x + ) gets M0 4x(x + ) 3(x + ) = (x + ) gets, just, for slip in omission of (x + ) 8x + 4x 3x 3 = x + 3x + D Multiplying out, collecting like terms and forming quadratic = 0. 6x x 4 = 0 Follow through from their equation provided the algebra is not significantly eased and it is a quadratic. Condone a further sign or numerical error or minor slip when rearranging. 3x x = 0 or 6x² x 4 = 0 oe www, (not fortuitously obtained - check for double errors) (3x + )(x ) = 0 Solving their three term quadratic provided b² 4ac 0. Use of correct quadratic equation formula (can be an error when substituting into correct formula) or factorising (giving their correct x² and constant terms when factors multiplied out) or comp the square oe. soi x = /3 or cao for both obtained www (accept 4/6 oe, or exact decimal equivalent (condone 0.667or better)) [5] SC x = with or without any working 5

85 4754A Mark Scheme June 0 Question Answer Marks Guidance 3.( ).( )( )! 3! / 3 ( x) ( x) ( x) ( x)... Do not MR for n / All four correct binomial coeffs (not ncr form) soi Accept unsimplified coefficients if a subsequent error when simplifying. = + x ½ x + ½ x 3 + ½ x² www Condone absence of brackets only if followed by correct work eg x² = 4x² must be soi for second B mark. + x www + ½ x³ www If there is an error in say the third coeff of the expansion, M0,, B0, can be scored Valid for x < / or / < x < / Independent of expansion x / and / x / are actually correct in this case so we will accept them. Condone a combination of inequalities. Condone also, say / < x < / but not x < / or < x < or / > x > / [5] 6

86 4754A Mark Scheme June 0 Question Answer Marks Guidance 3 (i) dv/dt = kv cao condone different k (allow MR for = kv²) V = (½ kt + c) dv/dt = (½ kt + c).½ k (/ kt + c) constant multiple of k (or from multiplying out oe; or implicit differentiation) = k(½ kt + c) cao www any equivalent form (including unsimplified) = kv Allow SCB if V=(/ kt + c)² fully obtained by integration including convincing change of constant if used Can score M0 SCB [4] (ii) (½ k + c) = ½ k + c = 00 substituting any one from t =, V = 0,000 or t = 0, V = 0 or t =, V = 40,000 into squared form or rooted form of equation (Allow /00 or /00) (k + c) = k + c = 00 substituting any other from above ½ k = 00 k = 00, c = 0 Solving correct equations for both www (possible solutions are (00,0), ( 00,0), (600, 400), ( 600,400) (some from ve root)) V = (00t) = 0000t either form www SC B for V = (00t)² oe stated without justification SCB4 if justification eg showing substitution SC those working with (k + c)² = 30,000 can score a maximum of B0 A0 (leads to k 46, c 6.8) [4] 7

87 4754A Mark Scheme June 0 Question Answer Marks Guidance 4 LHS = sec θ + cosec θ Use of secθ = /cosθ and cosecθ = /sinθ not just stating cos θ sin θ sin θ cos θ adding cos θsin θ use of cos²θ + sin²θ = soi cos θsin θ =sec²θ cosec²θ AG OR sec²θ+cosec²θ=tan²θ++cot²θ+=sin²θ/cos²θ+cos²θ/sin²θ+ correct formulae oe 4 4 = cos sin sin cos sin cos (cos sin ) = sin cos sin cos sec²θ cosec²θ adding use of Pythagoras AG OR working with both sides Eg LHS sec²θ +cosec²θ =tan²θ++cot²θ+=tan²θ+cot²θ + RHS =(+tan²θ)(+cot²θ)= + tan²θ+ cot²θ+tan²θcot²θ =tan²θ +cot²θ +=LHS [4] Correct formulae used on one side Use of same formulae on other side Use of tanθcotθ = oe, dependent on both method marks Showing equal 8

88 4754A Mark Scheme June 0 Question Answer Marks Guidance 5 sin(x + 45) = sin x cos 45 + cos x sin 45 Use of correct compound angle formula = sin x./ + cos x. / = (/)(sin x + cos x) = cos x sin x + cos x = cos x * Since AG, sin x cos 45º + cos x sin 45º = cos x sin x + cos x = cos x only gets need the second line or statement of cos 45º = sin 45º = / oe as an intermediate step to get sin x = ( ) cos x tan x = terms collected and tan x = sin x /cos x used x = 6.3, 4.3 for first correct solution for second correct solution and no others in the range dp but allow overspecification ignore solutions outside the range [6] SC for both 6.3º and 4.3º SC for both.07 and 4. radians (or better) SC for incorrect answers that round to 6.3 and 80 + their ans eg 6.33 and 4.33 Do not award SC marks if there are extra solutions in the range. 9

89 4754A Mark Scheme June 0 Question Answer Marks Guidance 6 d y y d x x( x) correctly separating variables and intending to integrate (ie need to dy dx y x( x) see attempt at integration or integral signs) A B partial fractions soi x( x) x x = A(x + ) + B x x = 0 A = A = www x = = B B = B = www ft their A,B ln y ( )d x ln xln( x) c x x condone absence of c or lnc x =, y = 0 = 0 ln + c c = ln evaluating their c at any stage dependent on x and y terms all being logs of correct form but do not award following incorrect log rules, ft their A,B. c could be say a decimal. (eg y = x/(x + ) + c then c being found is B0) ln y = ln x ln(x + ) + ln = ln (x/(x + )) correctly combining lns and antilogging throughout (must have included the constant term). Apply this strictly. Do not allow if c is included as an afterthought unless completely convinced. ft A,B Logs must be of correct form ie not following say dx ln( x x) unless ft from partial fractions and xx ( ) B = y = x/(x + ) cao www (y=.693 x e loses final ) x NB evaluating c and log work can be in either order. eg y = cx/(x + ), at x =, y =, c = [8] 0

90 4754A Mark Scheme June 0 Question Answer Marks Guidance 7 (i) θ = /: O (0, 0) Origin or O, condone omission of (0, 0) or O θ = 0: P (, 0) Or, say at P x =, y = 0, need P stated θ = /: O (0, 0) Origin or O, condone omission of (0,0) or O [3] 7 (ii) d y d y /d θ their dy/dθ /dx/dθ dx d x /dθ cosθ cosθ sinθ sinθ any equivalent form www (not from cosθ/sinθ) When θ = / dy/dx = cos / sin/ = subst θ= π/ in their equation When θ = / dy/dx = cos ()/ sin(/) = Obtaining dy/dx =,and dy/dx = shown (or explaining using symmetry of curve) www Either = so perpendicular Or gradient tangent = meets axis at 45º, similarly, gradient = meets axis at 45º oe justification that tangents are perpendicular www dependent on previous [5] 7 (iii) At Q, sin θ = θ = /, θ = /4 or, using the derivative, cos θ = 0 soi or their dy/dx = 0 to find θ. If the only error is in the sign or the coeff of the derivative in (ii), allow full marks in this part (condone θ = 45) coordinates of Q are (cos /4, sin /) = (, ) www (exact only) accept / [3] 7 (iv) sin θ = ( cos θ) = ¼ x oe, eg may be x² =.. y = sin θ = sin θ cos θ Use of sinθ = sinθcosθ = () x( ¼ x ) subst for x or y² = 4sin²θcos²θ (squaring) either order oe y = x ( ¼ x )* squaring or subst for x either order oe AG [4]

91 4754A Mark Scheme June 0 Question Answer Marks Guidance 7 (v) integral including correct limits but ft their from (i) V πx ( x ) d x 0 4 (limits may appear later) condone omission of dx if intention clear 4 ( πx πx )dx π x x 3 0 x x 3 0 ie allow if no π and/or incorrect/no limits 0 (or equivalent by parts) 8 3 π substituting limits into correct expression (including π) ft their 3 0 = 6/5 cao oe, 3.35 or better (any multiple of π must round to 3.35 or better) [4] 8 (i) AA' 4 finding AA or AA by subtraction, subtraction must be seen 4 3 B0 if AA, AA confused Assume they have found AA if no label This vector is normal to x + y 3z = 0 reference to normal or n, or perpendicular to x + y 3z = 0, or statement that vector matches coefficients of plane and is therefore perpendicular, or showing AA is perpendicular to two vectors in the plane M is (½, 3, ½) for finding M correctly (can be implied by two correct coordinates) x + y 3z = ½ + 6 7½ = 0 M lies in plane showing numerical subst of M in plane = 0 [4]

92 4754A Mark Scheme June 0 Question Answer Marks Guidance 8 (ii) λ r λ λ meets plane when 4 4 λ + λ + ( λ ) 3(4 + λ ) = 0 subst of AB in the plane 7 7 λ = 0, λ = So B is (0, 3, ) cao 0 or BA, ft only on their B (condone AB used as BA or no label) A'B 3 4 (can be implied by two correct coordinates) Eqn of line A B is r 4 λ ft 4 or their B + ft λ their AB (or BA ) ft only their B correctly 8 (iii) Angle between and [6] correct vectors but ft their A B.Allow say, condone a minor slip if intention is clear and/or.( ) ( ).( ). cosθ 6. 6 correct formula (including cosθ) for their direction vectors from (ii) condone a minor slip if intention is clear = /6 ±/6 or 99.6º from appropriate vectors only soi Do not allow either A mark if the correct B was found fortuitously in (ii) θ = 80.4 cao or better [4] 3

93 4754A Mark Scheme June 0 Question Answer Marks Guidance 8 (iv) λ Equation of BC is r 4 λ 4λ λ Crosses Oxz plane when y = 0 NB this is not unique 0 eg 3 For putting y = 0 in their line BC and solving for λ λ = 4 6 r 0 so ( 6,0,5) 5 cao [3] Do not allow either A mark if B was found fortuitously in (ii) for A marks need fully correct work only NB this is not unique 0 eg 3 leads to μ = 3 4

94 4754A Mark Scheme January 03 Question Answer Marks Guidance x x x xx ( ) ( x) ( x)( x ) mult throughout by (x + )(x ) or combining fractions and mult up oe (can retain denominator throughout). Condone a single computational error provided that there is no conceptual error. Do not condone omission of brackets unless it is clear from subsequent work that they were assumed. x 3x x any fully correct multiplied out form (including say, if s correctly cancelled) soi x 3x = 0 = x(x 3) D dependent on first.for any method leading to both solutions. Collecting like terms and forming quadratic (= 0) and attempting to solve *(provided that it is a quadratic and b² 4ac 0).Using either correct quadratic equation formula (can be error when substituting), factorising (giving correct x² and constant terms when factors multiplied out), completing the square oe soi.* x = 0 or 3 for both solutions www. [4] SC for x = 0 (or x = 3) without any working SC B for x = 0 (or x = 3) without above algebra but showing that they satisfy equation SC M0 SC for first two stages followed by stating x = 0 SC M0 SC for first two stages and cancelling x to obtain x = 3 only. 5

95 4754A Mark Scheme January 03 Question Answer Marks Guidance 3 x ( x) /3 n = /3 only. Do not MR for n /3 5 ( ) ( )( ) ! 3! 3 ( x) ( x) ( x)... all four correct unsimplified binomial coeffs (not ncr) soi condone absence of brackets only if it is clear from subsequent work that they were assumed x x x... Valid for ½ < x < ½ or x <½ [6] x www in this term x www in this term (not if used x for ( x) throughout) x www in this term If there is an error in say the third coeff of the expansion then M0 B0 is possible. Independent of expansion Allow s (valid in this case) or a combination. Condone also, say, ½ < x < ½ but not x < ½ or < x < or ½ > x > ½ 6

96 4754A Mark Scheme January 03 Question Answer Marks Guidance d y d /d cos y θ θ their dy/dθ / their dx/dθ 3 (i) dx d x /dθ cosθ www correct (can isw) When = /6 = dy cos( π /3) d x cos( π / 6) 3/ 3 D subst θ = π/6 in theirs oe exact only, www (but not / 3/).. OR y x ( x ) dy x ( x ) ( x ) dx / / at /6,sin /6/ dy / 3 / ( ) ( ) dx [4] 3 (ii) y = sin = sin cos using sin θ = sin θ cos θ D full method for differentiation including product rule and function of a function oe oe cao (condone lack of consideration of sign) subst sin π/6 = ½ in theirs oe,exact only, www (but not / 3/) y = 4 sin cos = 4x ( x ) using cos = sin to eliminate cos θ = 4x 4x 4 * AG need to see sufficient working or A0. [3] 7

97 4754A Mark Scheme January 03 Question Answer Marks Guidance 4 (a) V = x πy d x π(e )dx 0 0 π x e x 0 x π( e )dx limits must appear but may be later 0 condone omission of dx if intention clear e x x independent of π and limits = ( + ½ e 4 ½ ) D dependent on first.need both limits substituted in their integral of the form ax + b e x, where a,b non-zero constants. Accept answers including e 0 for. Condone absence of π for at this stage 4 (b) (i) = ½ (3 + e 4 ) cao exact only [4] x = 0, y =.44; x =, y = , or better A = 0.5/{( ) correct formula seen (can be implied by correct intermediate step eg /4) + ( )} = or 6.93 (do not allow more dp) [3] 4 (b) (ii) 8 strips: 6.83, 6 strips: Trapezium rule overestimates this area, but the overestimate gets less as the no of strips increases. [] oe 8

98 4754A Mark Scheme January 03 Question Answer Marks Guidance 5 sec = 5 tan ( + tan ) = 5 tan sec²θ = + tan² θ used tan 5 tan + = 0 correct quadratic oe (tan )(tan ) = 0 solving their quadratic for tan θ (follow rules for solving as in Question [*,*] tan = ½ or www = 0.464,.07 first correct solution (or better) second correct solution (or better) and no others in the range Ignore solutions outside the range. SC for both 0.46 and. SC for both 6.6 and 63.4 (or better) Do not award SCs if there are extra solutions in range.. OR /cos²θ=5sinθ/cosθ cos 5sin cos, cos 0 cos ( 5sincos ) 0 cos 0, or sin 0.8 sin or ,.07 [6] using both sec = /cos and tan = sin/cos correct one line equation 5 sin θ cos θ = 0 or cos θ = 5 sin θ cos² θ oe (or common denominator). Do not need cosθ 0 at this stage. using sin θ = sin θ cos θ oe eg = 5 sin θ ( sin² θ) and squaring sinθ = 0.8 or, say, 5sin 4 θ 5 sin² θ + 4 = 0 first correct solution (or better) second correct solution (or better) and no others in range Ignore solutions outside the range SCs as above 9

99 4754A Mark Scheme January 03 Question Answer Marks Guidance 6 (i) AC = cosec or /sin AD = cosec sec φ oe but not if a fraction within a fraction [] 6 (ii) DE = AD sin ( + φ) = cosec θ sec φ sin (θ+ φ) = cosec sec φ(sin cos φ + cos sin φ) AD sin(θ + φ) with substitution for their AD correct compound angle formula used sin θcosφ cosθsin φ Do not award both M marks unless they are part of the same method. (They may sin θcos φ appear in either order.) cosθ sin φ sin θ cos φ = + tan φ / tan * simplifying using tan = sin/cos. A0 if no intermediate step as AG.. OR equivalent, eg from DE = CB + CD cos θ = + CD cos θ = + AD sin φ cos θ = + cosec θ sec φ sin φ cos θ.. from triangle formed by using X on DE where CX is parallel to BE to get DX = CD cos θ and CB = (oe trigonometry) substituting for both CD = AD sinφ and their AD oe to reach an expression for DE in terms of θ and φ only (M marks must be part of same method) = + tanφ /tanθ* AG simplifying to required form [3] 7 (i) DE = [(5) ] = 6 oe 6 oe using scalar products eg 5i + k with i oe cos = 5/6 oe 5 =.3 or better (or 68.7). Allow radians. [4] 0

100 4754A Mark Scheme January 03 Question Answer Marks Guidance 5 7 (ii) AE 4, ED 0 two relevant direction vectors (or 6i + 4j + k oe) 3 scalar product with a direction vector in the plane (including evaluation and = 0) (OR forms vector cross product with at least two correct terms in solution) scalar product with second direction vector, with evaluation. 5 (following OR above, all correct ie a multiple of i 4j + 5k) i 4j + 5k is normal to AED x 0 eqn of AED is y z (NB finding only one direction vector and its scalar product is only.) for x 4y + 5z = c oe x 4y + 5z = 6 A0 for i 4j + 5k = 6 B lies in plane if 8 4(a) = 6 allow for subst in their plane or if found from say scalar product of normal a = with vector EB can also get For first five marks above SC, if states, if i 4j + 5k is normal then of form x 4y + 5z = c and substitutes one coordinate gets, then substitutes other two coordinates A (not,).then states so is normal can get provided that there is a clear 4 [7] 5 argument ie A. Without a clear argument this is B0. SC, if finds two relevant vectors, and then finds equation of the plane from vector form, r = a + μb + λc gets. Eliminating parameters cao. If then states so is normal can get (4/5). 4 5

101 4754A Mark Scheme January 03 Question Answer Marks Guidance 7 (iii) D: 6 + = 8 B: = 8 C: = 8 plane BCD is x + z = 8 B,,0 or any valid method for finding x + z = 8 gets Angle between i 4j + 5k and i + k is between two correct relevant vectors cos = (+(4)0 + 5 )/4 = 6/84 complete method (including cosine) (for ft their normal(s) to their plane(s)) allow correct substitution or ±6/ 84, correct only = 49. or radians (or better) acute only [6] 8 (i) h = 0, stops growing AG need interpretation (ii) h = 0 0e t/0 dh/dt = e t /0 0 e t /0 = 0 0( e t /0 =0dh/dt when t = 0, h = 0( ) = 0 ) = 0 h OR verifying by integration dh dt 0 h 0 ln(0 h) 0.tc h 0, t 0, cln 0 ln(0 h) 0.tln 0 0 h ln( ) 0.t 0 0.t 0 h0e 0.t h0( e ) [] [5] differentiation (for need ke t /0, k const) oe eg 0 h = 0 0( e t/0 ) = 0e t/0 = 0dh/dt (showing sides equivalent) initial conditions.. sep correctly and intending to integrate correct result (condone omission of c, although no further marks are possible) condone ln (h 0) as part of the solution at this stage constant found from expression of correct form (at any stage) but B0 if say c = ln ( 0) (found using ln (h 0)) combining logs and anti-logging (correct rules) correct form (do not award if B0 above)

102 4754A Mark Scheme January 03 Question Answer Marks Guidance 8 (iii) 00 A B (0 h)(0 h) 0 h 0 h 00 = A(0 h) + B(0 + h) cover up, substitution or equating coeffs h = 0 00 = 40 B, B = 5 h = 0 00 = 40A, A = 5 00 dh/dt = 400 h 00 dh dt 400 h separating variables and intending to integrate (condone sign error) 5 5 ( ) dh dt 0 h 0 h substituting partial fractions 5ln(0 + h) 5ln(0 h) = t + c ft their A,B, condone absence of c, Do not allow ln(h-0) for. When t = 0, h = 0 0 = 0 + c c = 0 cao need to show this. c can be found at any stage. NB c = ln( ) (from ln(h 0)) or similar scores B0. 0 h 5ln t 0 h anti-logging an equation of the correct form. Allow if c = 0 clearly stated 0 h t/5 e (provided that c = 0) even if B mark is not awarded, but do not allow if c omitted. 0 h Can ft their c. 0 + h = (0 h)e t/5 = 0 e t/5 h e t/5 h + h e t/5 = 0 e t/5 0 h(e t/5 + ) = 0(e t/5 ) D making h the subject, dependent on previous mark t/5 0(e ) NB method marks can be in either order, in which case the dependence is the other h t/5 e way around.(in which case, 0 + h is divided by 0 h first to isolate h). t/5 0( e ) h * t/5 e AG must have obtained (for c) in order to obtain final. [9] 8 (iv) As t, h 0. So long-term height is 0m. www [] 8 (v) st model h = 0( e 0. ) =.90.. Or st model h = gives t =.05.. nd model h = 0(e /5 )/(e /5 + )=.99.. nd model h = gives t = so nd model fits data better dep dep previous s correct [3] 3

103 4754B Mark Scheme January 03 Question Answer Marks Guidance (4,0) and (5,) and (,4) B3 All three points correct (and no additional points) SC any correct point SC B any two correct points (and no more than two incorrect additional points) B0 if points unclear Accept a list of coordinates. [3] B3 B3: Exactly as shown (with line extended beyond (4,) and before (0,3)) 3 (i) tp,a 3 (ii) tp,a 3 Award for a locus including one partially correct line segment or at least two correct discrete points in that segment Award B for a locus that contains parts of all three correct line segments (or at least two discrete points within each of them) and no additional incorrect points. [3] t P,B (not t(p,a)=t(p,b)=3) Accept words, such as, [distance] t(p,a) is equal to [distance] t(p,b). [] Accept words such as, all the points with [distance] t(p,a) equal to 3. [] 4 Evidence of 35 + n(3,4) www gets B. [] 5

104 4754B Mark Scheme January 03 Question Answer Marks Guidance 5 The 0 routes shown from (0,0) to (3,) each continue in one way via (4,) to (4,3) and each continue in one way via (3,3) to (4,3) Or, the 35 all pass through either 0(3,3) or 5 (4,). 0 of the 0 at (3,3) come from (3,) [and the rest from (,3)] 0 of the 5 at (4,) come from (3,). [and the rest from (4,)] So = 0 are from (3,) oe Hence 0 B www Need explanation 6 (i) 0.7, ,0.7 [] SC without explanation SC for (y = ) 5.3 and 0.7. [] 6 (ii) A square with straight edges as shown with a vertex at (,). (and nothing more) [] 7 (i) B must lie on one of the two axes oe, say, horizontal and vertical lines from (0,0). [] Do not accept a list of points unless an overall statement that includes all points is given (not just integers). 6

105 4754B Mark Scheme January 03 Question Answer Marks Guidance 7 (ii) B lies on line y x oe As (i) or on line y x [] 7

106 4754A Mark Scheme June 03 Question Answer Marks Guidance (i) x A B ( x)( x) x x x = A( x) + B( + x) expressing in partial fractions of correct form (at any stage) and attempting to use cover up, substitution or equating coefficients Condone a single sign error for only. x = ½ ½ = B( + ½ ) B = /3 www cao x = = 3A A = /3 www cao [3] (accept A/(+x) +B/(-x), A = /3, B = /3 as sufficient for full marks without needing to reassemble fractions with numerical numerators) 4

107 4754A Mark Scheme June 03 Question Answer Marks Guidance (ii) x /3 /3 ( x)( x) x x ( ) ( ) x x 3 correct binomial coefficients throughout for first three terms of ( )( ) ( )( ) 3 [ ( )( x) ( x)... ( ( ) x x...)] [ 4... ( x x x x...)] 3 either (x) or (+x) oe ie,( ),( )( )/, not ncr form. Or correct simplified coefficients seen. + x + 4x x + x (or /3/-/3 of each expression, ft their A/B) (3 3...)... 3 x x xx so a = and b = OR x ( x x ²) = x ( ( x + x ²)) = x ( + x + x ² + (-)(-)( x + x ²)²/ + ) If k(-x+x²) () not clearly stated separately, condone absence of inner brackets (ie +x+4x²x + x²) only if subsequently it is clear that brackets were assumed, otherwise A0. [ie x+x² is A0 unless it is followed by the correct answer] Ignore any subsequent incorrect terms or from expansion of www cao = x ( + x + x ² + x ² ) A x(+x) www = x + x².. so a = and b = www cao x( x) ( x) correct binomial coefficients throughout for (-(x+x²)) oe (ie,-), at least as far as necessary terms (+x) (NB third term of expansion unnecessary and can be ignored) Valid for ½ < x < ½ or x < ½ independent of expansion. Must combine as one overall range. condone s (although incorrect) or a combination. Condone also, say ½ < x < ½ but not x < ½ or < x < or ½ > x > ½ [5] 5

108 4754A Mark Scheme June 03 Question Answer Marks Guidance cosec x + 5 cot x = 3 sin x 5cosx 3sin x using cosec x = /sin x and cot x = cos x / sin x sin x sin x + 5 cos x = 3 sin x = 3( cos x) cos x + sin x = used (both M marks must be part of same solution in order to score both marks) 3 cos x + 5 cos x = 0 * AG (Accept working backwards, with same stages needed) (3 cos x )(cos x + ) = 0 use of correct quadratic equation formula (can be an error when substituting into correct formula) or factorising (giving correct coeffs 3 and - when multiplied out) or comp square oe cos x = /3, cos x = /3 www x = 70.5, 89.5 [7] for 70.5 or first correct solution, condone over-specification (ie 70.5 or better eg 70.53, etc), for 89.5 or second correct solution (condone over-specification) and no others in the range Ignore solutions outside the range SCA0 for incorrect answers that round to 70.5 and 360-their ans, eg 70.5 and SC Award A0 for.,5. radians (or better) Do not award SC marks if there are extra solutions in the range 6

109 4754A Mark Scheme June 03 3 Question Answer Marks Guidance 45 tan 45 = / = * tan 30 = /3* 3 30 For both B marks AG so need to be convinced and need triangles but further explanation need not be on their diagram. Any given lengths must be consistent. Need or indication that triangle is isosceles oe Need all three sides oe tan 75 = tan (45+30) use of correct compound angle formula with 45º,30º soi tan 45 tan 30 / 3 tan45tan30 / ( 3) 3 (3 3 ) 3* (3 3) (oe eg ) substitution in terms of 3 in any correct form eliminating fractions within a fraction (or rationalising, whichever comes first) provided compound angle formula is used as tan(a+b)= tan(a±b)/(±tanatanb). rationalising denominator (or eliminating fractions whichever comes second) correct only, AG so need to see working [7] 7

110 4754A Mark Scheme June 03 Question Answer Marks Guidance 4 (i) 0 x r... 3 y need r (or another letter) = or z for first 4 (ii) x + 3y + z = 4... λ [] NB answer is not unique eg r 5 Accept i/j/k form and condone row vectors. + 3(+ ) + (3+ ) = 4 substituting their line in plane equation (condone a slip if intention clear) 5 = 5, = www cao NB λ is not unique as depends on choice of line in (i) so point of intersection is (, 0, ) www cao 4 (iii) Angle between i + j + k and i + 3j +k is where Angle between i+3j+k and their direction from (i) ft condone a single sign slip if intention clear 3 5 cosθ [3] correct formula (including cosine), with substitution, for these vectors condone a single numerical or sign slip if intention is clear = 63.5 www cao (63.5 in degrees (or better) or.09 radians or better) [3] 8

111 4754A Mark Scheme June 03 Question Answer Marks Guidance λ μ 5 required form, can be soi from two or more correct equations 3 = 0 forming at least two equations and attempting to solve oe + = 5 5 = 5, = 3 www 9 = 0, = www 5 = +, 5 = 3 + true verifying third equation, do not give BOD 0 3 accept a statement such as 5 3 as 5 verification Must clearly show that the solutions satisfy all the equations. coplanar oe independent of all above marks [6] 9

112 4754A Mark Scheme June 03 Question Answer Marks Guidance 6 (i) vdv/dx + 4x = 0 vdv = 4x dx separating variables and intending to integrate ½ v = x + c oe condone absence of c. [Not immediate v = -4x (+c)] When x =, v = 4, so c = 0 finding c, must be convinced as AG, need to see at least the statement given here oe (condone change of c) so v = 0 4x * AG following finding c convincingly 6 (ii) x = cos t + sint when t = 0, x = cos 0 + sin 0 = * AG need some justification v = dx/dt = sin t + 4 cos t v = 4 cos 0 sin 0 = 4* www AG [4] [4] Alternatively, SC v²=0-4x², by differentiation, v dv/dx= -8x vdv/dx + 4x =0 scores B if, in addition, they check the initial conditions a further is scored (ie 6=0-4). Total possible 3/4. differentiating, accept ±,±4 as coefficients but not ±,± and not ±/,± from integrating cao 0

113 4754A Mark Scheme June 03 Question Answer Marks Guidance 6 (iii) cos t + sin t = Rcos(t α) = R(cos t cos α + sin t sin α ) SEE APPENDIX for further guidance R = 5 or.4 or better (not ± unless negative rejected) R cos α =, R sin α = correct pairs soi tan α =, correct method α =.07 cao radians only,. or better (or multiples of π that round to.) x = 5cos(t.07) v = 5sin(t.07) differentiating or otherwise, ft their numerical R, α (not degrees) required form SC for v = 0 cos(t ) oe EITHER v = 0sin (t α) 0 4x = 0 0cos (t α) squaring their v (if of required form with same α as x), and x, and attempting to show v² = 0 4x² ft their R, α (incl. degrees) [α may not be specified]. = 0( cos (t α)) = 0sin (t α) cao www (condone the use of over-rounded α (radians) or so v = 0 4x degrees) OR multiplying out v² = (sin t + 4cos t) = 4 sin²t6sintcost +6cos²t and 4x² = 4(cos²t + 4sintcost + 4sin²t) = 4cos²t +6sintcost +6sin²t (need middle term) and attempting to show that v² + 4x² = 4 (sin²t +cos²t) + 6(cos²t +sin²t) = 4+6 = 0 (or 0 4x² = v²) oe so v² = 0 4x² cao www [7] differentiating to find v (condone coefficient errors), squaring v and x and multiplying out (need attempt at middle terms) and attempting to show v² = 0 4x²

114 4754A Mark Scheme June 03 Question Answer Marks Guidance 6 (iv) x = 5cos(t α) or otherwise x max = 5 when cos(t α) =, t -.07=0, t =.07 t = 0.55 [3] ft their R oe (say by differentiation) ft their α in radians or degrees for method only cao (or answers that round to 0.554) 7 (i) u = 0, x = 5 ln 0 =.5 Using u 0 to find OA so OA = 5ln0 accept.5 or better when u =, Using u to find OB or u 0 to find AC y = + = so OB = When u = 0, y = 0 + /0 = 0. So AC = 0. [5] In the case where values are given in coordinates instead of OA=,OB=,AC=, then give A0 on the first occasion this happens but allow subsequent As. Where coordinates are followed by length eg B(0, ), length= then allow.

115 4754A Mark Scheme June 03 Question Answer Marks Guidance 7 (ii) dy d y/du / u dx d x/ du 5/ u u 5u their dy/du /dx/du Award if any correct form is seen at any stage including unsimplified (can isw) EITHER When u = 0, dy/dx = 99/50 =.98 substituting u =0 in their expression tan (90 )=.98 = or by geometry, say using a triangle and the gradient of the line OR When u = 0, dy/dx = 99/50 =.98 tan(90 θ ) = 99/50 tanθ = 50/99 θ = 6.8º = 6.8 A 6.8º, or radians (or better) cao SC M0A0 for 63.º (or better) or.03 radians(or better) 7 (iii) x = 5 ln u x/5 = ln u, u = e x/5 Need some working A [6] allow use of their expression for M marks 6.8º, or radians (or better) cao y = u + /u = e x/5 + e x/5 Need some working as AG [] 3

116 4754A Mark Scheme June 03 Question Answer Marks Guidance 7 (iv) Vol of rev = 5ln0 5ln0 x /5 x/5 0 0 πy d x π( e e ) dx 5ln0 x/5 x/5 π(e e )dx 0 5ln0 5 x/5 5 x/5 ( e x e ) 0 π = (50 + 0ln ) x/5 x/5 need π ( e e ) and dx soi. Condone wrong limits or omission of limits for. /5 /5 Allow if y prematurely squared as eg ( e x e x ) including correct limits at some stage (condone.5 for this mark) 5 x/5 5 x/5 [ e x e ] allow if no π and/or no limits or incorrect limits substituting both limits (their OA and 0) in an expression of correct form ie x /5 x/5 ae be cx, a,b,c 0 and subtracting in correct order (- 0 is sufficient for lower limit) Condone absence of π for = 858 accept 73π and answers rounding to 73π or 858 [5] NB The integral can be evaluated using a change of variable to u. This involves changing dx to (dx/du)x du. For completely correct work from this method award full marks. Partially correct solutions must include the change in dx. If in doubt consult your TL. Remember to indicate second box has been seen even if it has not been used. 4

117 4754B Mark Scheme June 03 Question Answer Marks Guidance Point R marked correctly at (-60, 45) half way up the relevant square Need labelling with letters or co-ordinates oe Point M marked correctly at (0, 0) Generously applied if above half way If unclear, B0B0. For 30 o S, 0 hours daylight cao soi 0 only For 60 o S, 5.5 hours daylight. allow 5< t <6 soi not 5 or 6 [] (condone one not labelled if other is labelled (and no others marked)) SC BOD both marked (and no others) and no labelling. Difference 4.5 hours ft 0- t dependent on both previous B marks soi SC() allow B3 for 4<difference in length<5 (without wrong working) SC()allow B if uses t=5 or t=6 eg, 0,6,4 www SC(3) If calculates (using the equation (4)) can obtain all three marks. Approximate values are 0.07, 5.5, ( May also work with 5 and.5<t<3 if doubled at end [3] eg 5 dependent on later doubling.6 dependent on later doubling (5-.6) x=4.8 ) IT IS ESSENTIAL TO CHECK & ANNOTATE PAGES ATTACHED TO QUESTIONS 0, 0, 03 7

118 4754B Mark Scheme June 03 Question Answer Marks Guidance 3 0 o north is B o north is A B All four answers correct 5 o south is D 5 o south is C SC Any two answers correct 4 tan y cos(5 t) tan 3.44, y 60, t is to be found [] cos(5 t) substitute in formula and attempt to 5t solve (as far as 5t=invcos.)oe t accept 9. or better Daylight hours are D doubling, dependent on first So 8.5 hours (to 3s.f.) or approx 8 hours www (accept ,8.489,8.49,8.5 or 8.4 (from x9.),8.48) OR Using t=9 α=3.44, t=9, y is to be found tan y = y=58.485º so approx 60º D A [4] t=8/=9 and substituted in formula and attempt to solve (as far as y= inv tan constant) oe or 58.49º/58.5º/approx 60º www any reasonable accuracy or stating error is approx 0.49 oe any reasonable accuracy 8

119 4754B Mark Scheme June 03 Question Answer Marks Guidance 5 (i) cos n (ii) On February nd, n 3 33 calculate n= 3+ =33 (days in Jan + Feb) soi cos substitution of their n+0 in equation (3) and attempt to evaluate a 7.3 or or rounds to -7.3 tan y cos5t tan tan53 cos(5 t) tan( 7.306) t arccos( tan53 tan( 7.306)) 5 [3] use of their α in equation (4) D making t the subject SC condone 30+ Where n=3,3,33,34 only NB n= 3+0 gives gaining A0 t or better SC ft from Obtains ft for (or 4.34) Sunset is at hrs + 4 hours minutes, and so 6: hrs cao And then ft for 6: (or 6:0) [4] 9

120 4754A Mark Scheme June 04. Annotations and abbreviations Annotation in scoris Meaning Blank Page this annotation must be used on all blank pages within an answer booklet (structured or unstructured) and on each page of an additional object where there is no candidate response. and BOD Benefit of doubt FT Follow through ISW Ignore subsequent working M0, Method mark awarded 0, A0, Accuracy mark awarded 0, B0, Independent mark awarded 0, SC Special case ^ Omission sign MR Misread Highlighting Other abbreviations Meaning in mark scheme E Mark for explaining U Mark for correct units G Mark for a correct feature on a graph dep* Method mark dependent on a previous mark, indicated by * cao Correct answer only oe Or equivalent rot Rounded or truncated soi Seen or implied www Without wrong working

121 4754A Mark Scheme June 04. Subject-specific Marking Instructions for GCE Mathematics (MEI) Pure strand a Annotations should be used whenever appropriate during your marking. The A, M and B annotations must be used on your standardisation scripts for responses that are not awarded either 0 or full marks. It is vital that you annotate standardisation scripts fully to show how the marks have been awarded. For subsequent marking you must make it clear how you have arrived at the mark you have awarded. b An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed. When a candidate adopts a method which does not correspond to the mark scheme, award marks according to the spirit of the basic scheme; if you are in any doubt whatsoever (especially if several marks or candidates are involved) you should contact your Team Leader. c The following types of marks are available. M A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, eg by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 cannot ever be awarded. B Mark for a correct result or statement independent of Method marks. 4

122 4754A Mark Scheme June 04 E A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, eg wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument. d e When a part of a question has two or more method steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation dep * is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given. The abbreviation ft implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, exactly what is acceptable will be detailed in the mark scheme rationale. If this is not the case please consult your Team Leader. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be follow through. In such cases you must ensure that you refer back to the answer of the previous part question even if this is not shown within the image zone. You may find it easier to mark follow through questions candidate-by-candidate rather than question-by-question. f g Wrong or missing units in an answer should not lead to the loss of a mark unless the scheme specifically indicates otherwise. Candidates are expected to give numerical answers to an appropriate degree of accuracy, with 3 significant figures often being the norm. Small variations in the degree of accuracy to which an answer is given (e.g. or 4 significant figures where 3 is expected) should not normally be penalised, while answers which are grossly over- or under-specified should normally result in the loss of a mark. The situation regarding any particular cases where the accuracy of the answer may be a marking issue should be detailed in the mark scheme rationale. If in doubt, contact your Team Leader. Rules for replaced work If a candidate attempts a question more than once, and indicates which attempt he/she wishes to be marked, then examiners should do as the candidate requests. 5

123 4754A Mark Scheme June 04 If there are two or more attempts at a question which have not been crossed out, examiners should mark what appears to be the last (complete) attempt and ignore the others. NB Follow these maths-specific instructions rather than those in the assessor handbook. h For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate s data. A penalty is then applied; mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A mark in the question. Note that a miscopy of the candidate s own working is not a misread but an accuracy error. 6

124 4754A Mark Scheme June 04 Question Answer Marks Guidance 3x A Bx C ( x)(4 x ) x 4 x correct form of partial fractions ( condone additional coeffs eg A B BUT x 4 x ** is M0 ) Ax B Cx D x 4 x * for 3x = A(4 + x ) + (Bx + C)( x) Multiplying through oe and substituting values or equating coeffs at LEAST AS FAR AS FINDING A VALUE for one of their unknowns (even if incorrect) Can award in cases * and ** above Condone a sign error or single computational error for but not a conceptual error Eg 3x = A( x) + (Bx + C)(4 + x²) is M0 3x( x)(4 + x²) = A(4 + x²) + (Bx + C)( x) is M0 Do not condone missing brackets unless it is clear from subsequent work that they were implied. Eg 3x = A(4 + x²) + Bx + C( x) = 4A + Ax² + Bx + C Cx is M0 = 4A + Ax² + Bx Bx² + C Cx is x = 6 = 8A, A = ¾ oe www [SC A = 3/4 from cover up rule can be applied, then the applies to the other coefficients] A B NB A = 3/4 is A0 ww (wrong working) x 4 x x coeffs: 0 = A B B = ¾ oe www constants: 0 = 4A + C C = ½ oe www [In the case of * above, all 4 constants are needed for the final ] Ignore subsequent errors when recompiling the final solution provided that the coeffs were all correct [5]

125 4754A Mark Scheme June 04 Question Answer Marks Guidance (4 x) 4 ( x ) 4 dealing with the 4 to obtain 3/ x 3/ 4 ( ) 4 3/ 3 / 3 / x 4 4 x 4... (or expanding as! and having all the powers of 4 correct) 3 3 4! 4 8( ( x).. ( x ) x +3/6 x² correct binomial coeffs for n = 3/ ie, 3/, 3/././! Not ncr form Indep of coeff of x Indep of first 8 + 3x www + 3/6 x² www Ignore subsequent terms Valid for 4 < x < 4 or x < 4 accept s or a combination of < and, but not 4 > x > 4, x > 4, or say 4 < x condone 4 < x < 4 Indep of all other marks [5] Allow MR throughout this question for n = m/ where m N, and m odd and then MR provided it is at least as difficult as the original.

126 4754A Mark Scheme June 04 Question Answer Marks Guidance 3 (i) x y B,,0 For values ,0.679, (4dp or better soi) [accept truncated to 4 figs after dec point] A = ( /3) [( ) + ( )] = (ii) Not possible to say, eg some trapezia are above and some below curve oe. [4] [cannot assume values of form (π/6)³ + (sin π/6) are correct unless followed by correct total at some later stage as some will be in degree mode] Use of the trapezium rule. Trapezium rule formula for 4 strips must be seen, with or without substitution seen. Correct h must be soi. [accept separate trapezia added] www 3dp only (NB using.35 is ww) SC B without any working as no indication of strips or method used SC with some indication of 4 strips but no values seen Correct values followed by scores B B0 Correct values followed by correct formula for 4 strips, with or without substitution seen, then A= scores 4/4. Correct formula for 4 strips and values of form ((π/6)³ + (sinπ/6) followed by correct answer scores 4/4 (or ¾ with wrong dp) NB Values given in the table to only 3dp give apparently the correct answer, but scores B0,A0 ww Need a reason. Must be without further calculation. [] 3

127 4754A Mark Scheme June 04 Question Answer Marks Guidance 4 (i) EITHER Use of cos = /sec (or sin= /cosec) From RHS tan tan sec sec sin / cos.sin / cos / cos./ cos sin sin cos cos ( ) cos cos cos cos sin sin Must be used Substituting and simplifying as far as having no fractions within a fraction tt [need more than secsec cc ss ie an intermediate step that can lead to cc ss] cos( ) Convincing simplification and correct use of cos(α + β) Answer given OR From LHS, cos = /sec or sin = /cosec used cos( ) cos cos sin sin sec sec sin sin sec sin sec sin sec sec Correct angle formula and substitution and simplification to one term OR eg cosαcosβ sinαsinβ cos cos ( tan tan ) tan tan sec sec [3] Simplifying to final answer www Answer given Or any equivalent work but must have more than cc ss = answer. 4

128 4754A Mark Scheme June 04 Question Answer Marks Guidance 4 (ii) cos tan tan sec tan. β = α used, Need to see sec²α Use of sec²α = + tan²α to give required result Answer Given OR, without Hence, cos cos ( ) sec tan tan ( tan ) sin cos Use of cosα = cos²α sin²α soi Simplifying and using sec²α = + tan²α to final answer Answer Given Accept working in reverse to show RHS=LHS, or showing equivalent 4 (iii) cos = ½ Soi or from tan²θ = /3 oe from sin²θ or cos²θ = 60, 300 = 30, 50 [] [3] First correct solution Second correct solution and no others in the range SC for π/6and 5π/6 and no others in the range 5

129 4754A Mark Scheme June 04 Question Answer Marks Guidance 5 (i) EITHER x = e 3t, y = te t soi d / d e e y t x t in terms of t t t y t t Their d / d d / d dy/dx = (te t + e t )/3e 3t oe cao allow for unsimplified form even if subsequently cancelled incorrectly ie can isw when t =, dy/dx = 3e /3e 3 = /e cao www must be simplified to /e oe OR 3t = ln x, y ln e 3 3 /3 x /3ln x x ln x d y / dx x ln x x 3 x 9 /3 /3 = t oe cao t t 3e 3e Any equivalent form of y in terms of x only Differentiating their y provided not eased ie need a product including ln kx and x p and subst x 3 e t to obtain dy/dx in terms of t dy/dx = /3e + /3e =/e www cao exact only must be simplified to /e or e 5 (ii) 3t = ln x t = (ln x)/3 Finding t correctly in terms of x (ln x)/3 y = (ln x ) / 3e Subst in y using their t y = 3 ln 3 x x b Required form ax ln [4] [3] x only NB If this work was already done in 5(i), marks can only be scored in 5(ii) if candidate specifically refers in this part to their part (i). 6

130 4754A Mark Scheme June 04 Question Answer Marks Guidance y ( x ) y x x y 3 finding x (or x) correctly in terms of y ( ) 3 V πx d y π ( y )d y 4 π y y π( ) π 8 [6] For need πx dy with substitution for their x² (in terms of y only) Condone absence of dy throughout if intentions clear. (need π) www For it must be correct with correct limits and, but they may appear later /[y 4 /4 y] independent of π and limits substituting both their limits in correct order in correct expression, condone a minor slip for (if using y = 0 as lower limit then -0 is enough) condone absence of π for oe exact only www (⅜ π or.375π) 7

131 4754A Mark Scheme June 04 Question Answer Marks Guidance 7 (i) AB = 5 ( ) or better (condone sign error in vector for ) AC = Accept 5 (condone sign error in vector for ) cos cosθ = scalar product of AB with AC (accept BA/CA) AB. AC with substitution condone a single numerical error provided method is clearly understood [OR Cosine Rule, as far as cos θ= correct numerical expression ] www ± 0.557, 0.557,5/5 9,5/ 5 9 oe or better soi ( ± for method only) = 56.5 www Accept answers that round to 56.º or 56.º or 0.98 radians (or better) NB vector 5i+0j+k leads to apparently correct answer but loses all A marks in part(i) Area = ½ 5 9 sin Using their AB,AC, CAB. Accept any valid method using trigonometry =.8 Accept 5 5 and answers that round to.8 or.9 (dp) www or SC for accurate work soi rounded at the last stage to. (but not from an incorrect answer, say from an incorrect angle or from say.7 or. stated and rounded to.) We will not accept inaccurate work from over rounded answers for the final mark. [7] 8

132 4754A Mark Scheme June 04 Question Answer Marks Guidance 7 (ii) (A) AB ( 3) ( ) Scalar product with one vector in the plane with numerical expansion shown AC ( 3) Scalar product, as above, with evaluation, with a second vector. NB vectors are not unique SCB finding the equation of plane first by any valid method (or using vector product) and then clearly stating that the normal is proportional to the coefficients. SC For candidates who substitute all three points in the plane 4x-3y+0z = c and show that they give the same result, award If they include a statement explaining why this means that 4i-3j+0k is normal they can gain. 7 (ii) (B) 4x 3y+0z=c Required form and substituting the co-ordinates of a point on the plane [] 4x 3y + 0z + = 0 oe If found in (A) it must be clearly referred to in (B) to gain the marks. Do not accept vector equation of the plane, as Hence. [] 4i-3j+0k = is A0 9

133 4754A Mark Scheme June 04 Question Answer Marks Guidance 7 (iii) 0 r Meets 4x 3y + 0z + = 0 when 6 3(4 3 ) + 0(5 + 0 ) + = 0 5 = 50, = 0.4 Need r = (or oe x y z ) Subst their 4λ, 4 3λ, 5+0λ in equation of their plane from (ii) λ= 0.4 (NB not unique) So meets plane ABC at (.6, 5., ) cao www (condone vector) 7 (iv) height = (.6 + (.) + 4 )= 0 ft ft their (iii) [5] volume =.8 0 / 3 = 6.7 cao 50/3 or answers that round to 6.7 www and not from incorrect answers from (iii) ie not from say (.6,.8,9) [] 0

134 4754A Mark Scheme June 04 Question Answer Marks Guidance 8 (i) Either h = ( ½ At) dh/dt = A ( ½ At) Including function of a function, need to see middle step when t = 0, h = ( 0) = as required OR dh h At d = A h AG Separating variables correctly and integrating / h At c Including c. [Condone change of c.] At c h at t =0, h =, = (c/)² c =, h = ( At/)² Using initial conditions AG 8 (ii) When t = 0, h = 0 Subst and solve for A 0 A= 0, A = 0. cao When the depth is 0.5 m, 0.5 = ( 0.05t) substitute h= 0.5 and their A and solve for t 0.05t = 0.5, t = ( 0.5)/0.05 = 5.86s www cao accept 5.9 [3] [4]

135 4754A Mark Scheme June 04 Question Answer Marks Guidance 8 (iii) dh h B d t ( h) ( h) d d h EITHER, LHS h h h h B t dh separating variables correctly and intend to integrate both sides (may appear later) [NB reading (+h)²as +h² eases the question. Do not mark as a MR] In cases where (+h)² is MR as +h²or incorrectly expanded, as say +h+h² or +h², allow first for correct separation and attempt to integrate and can then score a max of M0A0A0 (for Bt+c) A0A0, max /7. expanding (+ h) and dividing by h to form a one line function of h (indep of first ) with each term expressed as a single power of h eg must simplify say / h+h/ h +h² h,condone a single error for (do not need to see integral signs) / ( h / h 3/ h )dh h / + h / + h 3/ cao dep on second M only -do not need integral signs OR,LHS, EITHER / / ( h h )h h ( h)dh using udv uv vdu correct formula used correctly, indep of first OR 5/ 3 / 3/ h 3/ h h h h ( h h )dh 3 3 / 4h h h 3 5 3/ 5/ h / + 4h 3/ /3 + h 5/ /5 = Bt + c condone a single error for if intention clear cao oe cao oe, both sides dependent on first mark = Bt +c cao need Bt and c for second but the constant may be on either side When t = 0, h = c = 56/5 from correct work only (accept 3.73 or rounded answers here but not for final ) or c= 56/5 if constant on opposite side. h / (30 + 0h + 6h ) = 56 5Bt * NB AG must be from all correct exact work including exact c. [7]

136 4754A Mark Scheme June 04 Question Answer Marks Guidance 8 (iv) h = 0 when t = 0 Substituting h = 0,t = 0 B = 56/300 = 0.87 Accept 0.87 When h = t = Subst their h = 0.5, ft their B and attempt to solve t = 9.5s Accept answers that round to 9.5s www. [4] 3

137 47543 Mark Scheme June 04. Annotations and abbreviations Annotation in scoris Meaning Blank Page this annotation must be used on all blank pages within an answer booklet (structured or unstructured) and on each page of an additional object where there is no candidate response. and BOD Benefit of doubt FT Follow through ISW Ignore subsequent working M0, Method mark awarded 0, A0, Accuracy mark awarded 0, B0, Independent mark awarded 0, SC Special case ^ Omission sign MR Misread Highlighting Other abbreviations Meaning in mark scheme E Mark for explaining U Mark for correct units G Mark for a correct feature on a graph dep* Method mark dependent on a previous mark, indicated by * cao Correct answer only oe Or equivalent rot Rounded or truncated soi Seen or implied www Without wrong working

138 4754B Mark Scheme June 04. Subject-specific Marking Instructions for GCE Mathematics (MEI) Pure strand a Annotations should be used whenever appropriate during your marking. The A, M and B annotations must be used on your standardisation scripts for responses that are not awarded either 0 or full marks. It is vital that you annotate standardisation scripts fully to show how the marks have been awarded. For subsequent marking you must make it clear how you have arrived at the mark you have awarded. b An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed. When a candidate adopts a method which does not correspond to the mark scheme, award marks according to the spirit of the basic scheme; if you are in any doubt whatsoever (especially if several marks or candidates are involved) you should contact your Team Leader. c The following types of marks are available. M A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, eg by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 cannot ever be awarded. B Mark for a correct result or statement independent of Method marks.

139 4754B Mark Scheme June 04 E A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, eg wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument. d e When a part of a question has two or more method steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation dep * is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given. The abbreviation ft implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, exactly what is acceptable will be detailed in the mark scheme rationale. If this is not the case please consult your Team Leader. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be follow through. In such cases you must ensure that you refer back to the answer of the previous part question even if this is not shown within the image zone. You may find it easier to mark follow through questions candidate-by-candidate rather than question-by-question. f Wrong or missing units in an answer should not lead to the loss of a mark unless the scheme specifically indicates otherwise. Candidates are expected to give numerical answers to an appropriate degree of accuracy, with 3 significant figures often being the norm. Small variations in the degree of accuracy to which an answer is given (e.g. or 4 significant figures where 3 is expected) should not normally be penalised, while answers which are grossly over- or under-specified should normally result in the loss of a mark. The situation regarding any particular cases where the accuracy of the answer may be a marking issue should be detailed in the mark scheme rationale. If in doubt, contact your Team Leader. 3

140 4754B Mark Scheme June 04 g Rules for replaced work If a candidate attempts a question more than once, and indicates which attempt he/she wishes to be marked, then examiners should do as the candidate requests. If there are two or more attempts at a question which have not been crossed out, examiners should mark what appears to be the last (complete) attempt and ignore the others. NB Follow these maths-specific instructions rather than those in the assessor handbook. h For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate s data. A penalty is then applied; mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A mark in the question. Note that a miscopy of the candidate s own working is not a misread but an accuracy error. 4

141 4754B Mark Scheme June 04 Question Answer Marks Guidance (i) for all three entries 30,60,45 correct Group P Q R S Number of people for all three entries.5, 3, 9 correct Average number of accidents in a year for 3000,4500 both correct Average cost of accidents per year SC B for all entries in any three columns correct (ii) Adding their bottom row ( = 4000 and dividing by 50 soi (and not divided or multiplied by any additional values) = 60 ft their (corr to dp if inexact) (iii) The 45 members of Group S pay So each pays [3] [] their oe soi as part of solution = 50. cao www (must be from correct final column) []

142 4754B Mark Scheme June 04 Question Answer Marks Guidance Basic premium = 35% of driver s premium use of 35% or 0.35 oe [not 65, 0.65 unless 0.65] Drivers premium = Basic premium 0.35 =.86 Basic premium k.86 accept.86 or better (k=.8574 or 6 oe) 7 3 Year % discount Total 30 [] [ k =.9, k = /0.35 scores A0 ] obtaining 30 or 3. oe [from adding all relevant terms ie =30 or = 3. soi (with or without first zero term) or from oe] 30% of the premium is 3875 The premium is = cao 3875 their 3. even if one term was missing from addition but must come from attempt at the appropriate addition [ie an error in adding to 30 or an omission of one term, an inclusion of say an extra 65,or an addition of 00,70,60, etc with subsequent subtraction from 700 ] [3]

143 4754B Mark Scheme June 04 Question Answer Marks Guidance 4 (i) Percentage Age A sketch. Must be at least for values of x between 8 and 45. Must be correct shape ie descending curve, does not curve up at end, does not cross the horizontal axis (even if extended) ie must. Does not need to go through points exactly as a sketch. 4 (ii) k( x 7) For large values of x, b e 0, and so y awhich is in this case. Need x becoming large or, exponential term 0 oe and y a, a = or y =,a = oe [] [] NOT just at x = 45, y = 4 (iii) Substituting x 3 y 6.6 subst x = 3 and finding y = 6.6 (6.605 ) (or subst y = 6 and finding x = 3.3 ( ) Only accept comparing x or y, not coefficients This is quite close to the observed value of y = 6. (or close to x = 3) Accept say, y = But not say at x = 3, y 6 with no evaluation seen Accept if states, say, y = 6.6 which is not consistent with y = 6 oe [] 3

144 4754B Mark Scheme June 04 Question Answer Marks Guidance 5 (A) With no more than 3 points on his licence the driver s premium is not altered: 50 (B) The driver now has = 9 points. The new premium is or 47 [but not ] [] 4

145 GCE Mathematics (MEI) Unit 4754B: Applications of Advanced Mathematics: Paper B Advanced GCE Mark Scheme for June 05 Oxford Cambridge and RSA Examinations

146 OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of candidates of all ages and abilities. OCR qualifications include AS/A Levels, Diplomas, GCSEs, Cambridge Nationals, Cambridge Technicals, Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching/training, administration and secretarial skills. It is also responsible for developing new specifications to meet national requirements and the needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support, which keep pace with the changing needs of today s society. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by examiners. It does not indicate the details of the discussions which took place at an examiners meeting before marking commenced. All examiners are instructed that alternative correct answers and unexpected approaches in candidates scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the report on the examination. OCR will not enter into any discussion or correspondence in connection with this mark scheme. OCR 05

147 4754B Mark Scheme June 05. Annotations and abbreviations Annotation in scoris Meaning and BOD Benefit of doubt FT Follow through ISW Ignore subsequent working M0, Method mark awarded 0, A0, Accuracy mark awarded 0, B0, Independent mark awarded 0, SC Special case ^ Omission sign MR Misread Highlighting Other abbreviations Meaning in mark scheme E Mark for explaining U Mark for correct units G Mark for a correct feature on a graph dep* Method mark dependent on a previous mark, indicated by * cao Correct answer only oe Or equivalent rot Rounded or truncated soi Seen or implied www Without wrong working 3

148 4754B Mark Scheme June 05. Subject-specific Marking Instructions for GCE Mathematics (MEI) Pure strand a Annotations should be used whenever appropriate during your marking. The A, M and B annotations must be used on your standardisation scripts for responses that are not awarded either 0 or full marks. It is vital that you annotate standardisation scripts fully to show how the marks have been awarded. For subsequent marking you must make it clear how you have arrived at the mark you have awarded. b An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed. When a candidate adopts a method which does not correspond to the mark scheme, award marks according to the spirit of the basic scheme; if you are in any doubt whatsoever (especially if several marks or candidates are involved) you should contact your Team Leader. c The following types of marks are available. M A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, eg by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 cannot ever be awarded. B Mark for a correct result or statement independent of Method marks. 4

149 4754B Mark Scheme June 05 E A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, eg wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument. d e When a part of a question has two or more method steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation dep * is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given. The abbreviation ft implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, exactly what is acceptable will be detailed in the mark scheme rationale. If this is not the case please consult your Team Leader. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be follow through. In such cases you must ensure that you refer back to the answer of the previous part question even if this is not shown within the image zone. You may find it easier to mark follow through questions candidate-by-candidate rather than question-by-question. f g Wrong or missing units in an answer should not lead to the loss of a mark unless the scheme specifically indicates otherwise. Candidates are expected to give numerical answers to an appropriate degree of accuracy, with 3 significant figures often being the norm. Small variations in the degree of accuracy to which an answer is given (e.g. or 4 significant figures where 3 is expected) should not normally be penalised, while answers which are grossly over- or under-specified should normally result in the loss of a mark. The situation regarding any particular cases where the accuracy of the answer may be a marking issue should be detailed in the mark scheme rationale. If in doubt, contact your Team Leader. Rules for replaced work If a candidate attempts a question more than once, and indicates which attempt he/she wishes to be marked, then examiners should do as the candidate requests. 5

150 4754B Mark Scheme June 05 If there are two or more attempts at a question which have not been crossed out, examiners should mark what appears to be the last (complete) attempt and ignore the others. NB Follow these maths-specific instructions rather than those in the assessor handbook. h For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate s data. A penalty is then applied; mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A mark in the question. Note that a miscopy of the candidate s own working is not a misread but an accuracy error. 6

151 4754B Mark Scheme June 05 Question Answer Marks Guidance (i) E. Pick up Q [] (ii) E5. Drop off Q [] (iii) E4. Drop off P and R [] 0 00% 75.3% 46 allow (90 or 95 or 5) / or better (eg etc), 75 is A0 B 75.3 with no working, (oe), [] [] * Substitution of the correct values into the correct formula 5. (must be dp), B 5. www (isw after correct answer seen) Correct calculation eg or 30 (30 + ) (oe) but not 00 Lift calls at floors to 8 (only), leaving floor 8 at 64 seconds. Descent takes 4 seconds Edep* [] Correct justification of the correct calculation (dependent on ). If the departure time of 64 is used then their explanation must imply the use of the first 8 floors (not just 8 floors ) 7

152 4754B Mark Scheme June 05 Question Answer Marks Guidance 5 (i) Car C Ground floor Arrival time (seconds) Departure time (seconds) 0 0 Floor Floor , 74 or their , their Return to ground 98 floor 98 or their [3] 5 (ii) 498s 39s (6 mins 3s) cao [] [] 7 n5, f 6 gives n5, f 4 gives Substitution of the correct values into the correct formula Accept to 4 dp so usually 6 and 4 stops respectively E Dependent on additional values of f considered E0 [3] 8

153 OCR (Oxford Cambridge and RSA Examinations) Hills Road Cambridge C EU OCR Customer Contact Centre Education and Learning Telephone: Facsimile: general.qualifications@ocr.org.uk For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee Registered in England Registered Office; Hills Road, Cambridge, C EU Registered Company Number: OCR is an exempt Charity OCR (Oxford Cambridge and RSA Examinations) Head office Telephone: Facsimile: OCR 05

154 GCE Mathematics (MEI) Unit 4754A: Applications of Advanced Mathematics: Paper A Advanced GCE Mark Scheme for June 05 Oxford Cambridge and RSA Examinations

155 OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of candidates of all ages and abilities. OCR qualifications include AS/A Levels, Diplomas, GCSEs, Cambridge Nationals, Cambridge Technicals, Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching/training, administration and secretarial skills. It is also responsible for developing new specifications to meet national requirements and the needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support, which keep pace with the changing needs of today s society. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by examiners. It does not indicate the details of the discussions which took place at an examiners meeting before marking commenced. All examiners are instructed that alternative correct answers and unexpected approaches in candidates scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the report on the examination. OCR will not enter into any discussion or correspondence in connection with this mark scheme. OCR 05

156 4754A Mark Scheme June 05. Annotations and abbreviations Annotation in scoris Meaning and BOD Benefit of doubt FT Follow through ISW Ignore subsequent working M0, Method mark awarded 0, A0, Accuracy mark awarded 0, B0, Independent mark awarded 0, SC Special case ^ Omission sign MR Misread Highlighting Other abbreviations Meaning in mark scheme E Mark for explaining U Mark for correct units G Mark for a correct feature on a graph dep* Method mark dependent on a previous mark, indicated by * cao Correct answer only oe Or equivalent rot Rounded or truncated soi Seen or implied www Without wrong working 3

157 4754A Mark Scheme June 05. Subject-specific Marking Instructions for GCE Mathematics (MEI) Pure strand a Annotations should be used whenever appropriate during your marking. The A, M and B annotations must be used on your standardisation scripts for responses that are not awarded either 0 or full marks. It is vital that you annotate standardisation scripts fully to show how the marks have been awarded. For subsequent marking you must make it clear how you have arrived at the mark you have awarded. b An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed. When a candidate adopts a method which does not correspond to the mark scheme, award marks according to the spirit of the basic scheme; if you are in any doubt whatsoever (especially if several marks or candidates are involved) you should contact your Team Leader. c The following types of marks are available. M A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, eg by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 cannot ever be awarded. B Mark for a correct result or statement independent of Method marks. 4

158 4754A Mark Scheme June 05 E A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, eg wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument. d e When a part of a question has two or more method steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation dep * is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given. The abbreviation ft implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, exactly what is acceptable will be detailed in the mark scheme rationale. If this is not the case please consult your Team Leader. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be follow through. In such cases you must ensure that you refer back to the answer of the previous part question even if this is not shown within the image zone. You may find it easier to mark follow through questions candidate-by-candidate rather than question-by-question. f g Wrong or missing units in an answer should not lead to the loss of a mark unless the scheme specifically indicates otherwise. Candidates are expected to give numerical answers to an appropriate degree of accuracy, with 3 significant figures often being the norm. Small variations in the degree of accuracy to which an answer is given (e.g. or 4 significant figures where 3 is expected) should not normally be penalised, while answers which are grossly over- or under-specified should normally result in the loss of a mark. The situation regarding any particular cases where the accuracy of the answer may be a marking issue should be detailed in the mark scheme rationale. If in doubt, contact your Team Leader. Rules for replaced work If a candidate attempts a question more than once, and indicates which attempt he/she wishes to be marked, then examiners should do as the candidate requests. 5

159 4754A Mark Scheme June 05 If there are two or more attempts at a question which have not been crossed out, examiners should mark what appears to be the last (complete) attempt and ignore the others. NB Follow these maths-specific instructions rather than those in the assessor handbook. h For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate s data. A penalty is then applied; mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A mark in the question. Note that a miscopy of the candidate s own working is not a misread but an accuracy error. 6

160 4754A Mark Scheme June 05 Question Answer Marks Guidance 5x(x + ) 3(x + ) = (x + )(x + ) * Multiplying throughout by (x + )(x + ) or combining fractions and multiplying up oe (eg can retain denominator throughout) Condone a single numerical error, sign error or slip provided that there is no conceptual error in the process involved Do not condone omission of brackets unless it is clear from subsequent work that they were assumed eg 5x(x + ) 3(x + ) = (x + )(x ) gets 5x(x + ) 3(x + ) = gets M0 5x(x + )(x + ) 3(x + )(x + ) = (x + )(x + ) gets M0 5x(x + ) 3(x + ) = (x + ) gets, just, for slip in omission of (x + ) dep* Multiplying out, collecting like terms and forming quadratic (= 0). Follow through from their equation provided the algebra is not significantly eased and it is a quadratic. Condone a further sign or numerical error or a minor slip when rearranging 3x 4x 4 = 0 oe www (not fortuitously obtained check for double errors) (3x + )(x ) = 0 Solving their three term quadratic (= 0) providedb 4ac 0. Use of correct quadratic equation formula (if formula is quoted correctly then only one sign slip is permitted, if the formula is quoted incorrectly M0, if not quoted at all substitution must be completely correct to earn the ) or factorising (giving their x term and one other term when factors multiplied out) or comp. the square (must get to the square root stage involving and arithmetical errors may be condoned provided their 3x / 3 seen or implied) x = /3 or cao for both obtained www (condone or better) (If no factorisation (oe) seen for each answer stated following correct quadratic) [5] 7

161 4754A Mark Scheme June 05 Question Answer Marks Guidance cos sin * cos sin (maybe implied in substitution) (6cos sin ) 6 sin sin 6cos + sin = 0 sin sin 6 = 0 (4sin 3)(3sin + ) = 0 dep* Use of correct quadratic equation formula or factorising or comp. the square on their three term quadratic in sin (see guidance in question for awarding this method mark) providedb 4ac 0 sin = 3/4 or /3 www First correct solution to dp or better (eg etc) Three correct solutions sin = 3/4, = 48.6, 3.4 sin = /3, =.8, 38. All four correct solutions and no others in the range Ignore solutions outside the range SC Award max B0 for answers in radians (0.85,.9, 3.87, 5.55 or better so one correct, three correct ). Award max if there are extra solutions in the range with radians [7] SC If awarded and both values of sin but B0B0B0 then award only for evidence of usingsin sin 80 8

162 4754A Mark Scheme June 05 Question Answer Marks Guidance 3 (i) /3 ( x) 3 x 4 ( )( ) 3 3 3! ( )( x) ( x)... n = /3. See below SC for those with n = /3 All three correct unsimplified binomial coefficients (not ncr) soi condone absence of brackets only if it is clear from subsequent work that they were assumed ( / 3) x... www x x... Valid for x or x (8 / 9)x www in this term If there is an error, in say, the third coefficient of the expansion then M0B0 is possible SC For n = /3 award for ( / 3)x and for (4 / 9)x (so max out of the first 4 marks) Independent of expansion. Accept, say, / x / [5] or / x / (must be strict inequality for +/) 3 (ii) 3x 8 ( 3 x)( x x...) 3 x x x 3x x... x x... Use of ( 3 x) their ( / 3) x (8 / 9) x ft [3] and attempt at removal of brackets (condone absence of brackets but must have two terms in x and two terms in x ) Correct simplified expansion following their expansion in (i). This mark is dependent on scoring both M marks in (i) and (ii) cao or B3 www in either part SC following either M0 or, for either a or b correct 9

163 4754A Mark Scheme June 05 Question Answer Marks Guidance 4 (i) cos x + sin x = R cos(x α) = R cos x cos α + R sin x sin α R cos α =, R sin α = Correct pairs. Condone sign error (so accept Rsin ) R = +, R = ( + ) Positive square root only - isw. Accept R / cos(arctan ) or R / sin(arctan ) tan α = (oe) Follow through their pairs. tan α = with no working implies both M marks. However, cos α =, sin α = tan α = scores M0. First two M marks may be implied by combining one of the pairs with R, eg, cos or sin ( ) α = arctan (oe) arccos, arcsin [4] Accept embedded answers, eg, co sx arct a n marks for full 4 (ii) max is R so R = + = 4 = 3 for using their ( ) Rmax, A0 for 3as final answer α = arctan 3 = /3 www (eg = and cos ( ) / 3 is B0) Exact answers only for final A and B marks [4] 0

164 4754A Mark Scheme June 05 Question Answer Marks Guidance 5 (i) x = sec, y = tan dy d y/ d sec dx d x/ d sec tan sec cos. cosec * tan cos sin sin for their (d y / d ) sec tan in terms of [3] cao (oe) allow for unsimplified form even if subsequently cancelled incorrectly ie can isw cao www (NB AG) must be at least one intermediate step between sec tan (oe) and either or cosec sin 5 (ii) x = sec = + tan = + ¼ y sec = + tan (oe) used y = 4(x ) = 4x 4 * www NB AG [] OR 4tan 4sec 4 * Correct substitution of x and y into the given answer tan sec which is true dep* Dependent on previous mark must simplify/remove the factor of 4 5 (iii) V y d x (4x 4) d x x x 3 x 4x from each term and state that the correctly derived trig identity is true k x x k 4 4 (d ) with or /, allow correct limits later condone lack of dx (4 / 3) x 3 4 x or (/3) x 3 x exact mark final answer [3]

165 4754A Mark Scheme June 05 Question Answer Marks Guidance 6 (i) A: ( ) + = 0 B: (.5) + = 0 E: ( ) + = 0 B,,0 for two points verified (must see as a minimum + = 0, = 0, + = 0) or any valid complete method for either finding or verifying that x 6y 0 gets At F, + 6a + = 0 Substitution of F into x 6y 0 6a = 4, a = 4/6 = 7/3 * www NB AG [4] 6 (ii) (A) 0 DH ( 6) scalar product with a direction vector in the plane (including evaluation and = 0) (OR forms a vector product with at least two correct terms in solution) 6 (ii) (B) 3 D C ( 6) r.(i 6j) = j.(i 6j ) [] scalar product with second direction vector, with evaluation. (following OR above, correct ie a multiple of i 6j) (NB finding only one direction vector and its scalar product is only) rn an with n 6 and a or or.5 or substituting H(0,, 3) or D(0,, 0) or C(3,.5, 0) into x 6y d x 6y + 6 = 0 oe (isw if d found correctly and x 6y = d stated) B www correct equation stated [] 6 (ii) (C) 6b + 6 = 0 b = 4/3 oe exact answer FG [] oe exact answer

166 4754A Mark Scheme June 05 Question Answer Marks Guidance 6 (iii) ( FE =) i + (/3)j + k, ( FB=) i (/6)j k or (EF ) i ( / 3) jk or (BF ) i (/ 6) j k OR () (/ 3)( / 6) ( ) cos 4 / 9 / 36 4 arccos q = 43 EF = 46 / 3, FB = 8 / 6, EB = 73 / cos FE FB / FE FB (oe) follow through their FE and FB (allow any combination of FE, EF with FB, BF) allow one sign slip only / arccos arccos [5] 3sf or better (or.5(0) radians or better). Allow candidates who find the acute angle using either EF with FBor FEwith BF and then state the obtuse angle. Do not isw those who find the obtuse angle and then state the acute angle. Note: 90 arctan(/ ) is 0/5 B3,,,0 One mark for each (.6,.4, 4.7) arccos 46 / 3 8 / 6 73 / ( 46 / 3)( 8 / 6) cosine rule correct with their EF, FB, EB q = 43 6 (iv) z coordinate of P is 5/ stating the correct z-coordinate of P; ignore incorrect x and y coordinates (or stated in a position vector) 0 OQ OP PQ 3 / 6 3 / / 3 5 / so height of Q is 8/3 (metres above ground).67 or better [3] Complete method for finding the z-coordinate of Q oroq (OH) ( / 3)(HP) or OQ ( / 3)(OP) (/ 3)(OH) 3

167 4754A Mark Scheme June 05 Question Answer Marks Guidance 7 (i) A B A( x) B( x) ( x)( x) x x Cover up, substitution or equating coefficients x = 3B =, B = /3 x = ½ = 3A/, A = /3 isw after correct A and B stated 7 (ii) x x ( x)( x) May be seen in separation of variables (may be implied by later working) implied by the use of factors ( + x) and ( x) [ ]d d 3 x k t ( x) x ln x ln( x) kt( c) Any non-zero constant, ln( + x) ln( x) = 3kt (+ c) www oe (condone absence of c) [3] Separating variables and substituting partial fractions. If no subsequent work integral signs needed, but allow omission of dx or dt, but must be correctly placed if present When t = 0, x = 0 c = 0 cao (must follow previous ) need to show (at some stage) that c = 0. As a minimum t = 0, x = 0, c = 0. Note that c = ln(-) (usually from incorrect integration of ( x)) or similar scores B0 x ln 3kt x Combining both their log terms correctly. Follow through their c. Allow if c = 0 clearly stated (provided that c = 0) even if B mark is not awarded, but do not allow if c omitted x x 3kt e * [7] AG www must have obtained all previous marks in this part 4

168 4754A Mark Scheme June 05 Question Answer Marks Guidance 7 (iii) ( + (0.75)) / ( 0.75)=e 3k substituting t =, x = 0.75 at any stage k = (/3)ln0 (= (3 s.f.)) 3sf or better t = ln(.8/0.)/3k =.45 hours.45 (or better) or hr 7 mins [3] 7 (iv) + x = e 3kt xe 3kt x + xe 3kt = e 3kt * Multiplying out and collecting x terms (condone one error) x( + e 3kt ) = e 3kt dep* Factorising their x terms correctly x = (e 3kt )/ ( + e 3kt ) = ( e 3kt )/( + e 3kt ) * www (AG) as AG must be an indication of how previous line leads when t e 3kt 0 x = ( e 3kt )/( + e 3kt ) / = OR x 3 e x kt x e xe 3kt 3kt x( e ) e [5] to the required result (eg stating or showing multiplying by 3 e kt ) clear indication that e 3kt 0 so, for example, 0 accept as a minimum ( x ) or e 3kt 0 ( x ) (NB 0 substitution of large values of t with no further explanation is B0) * Multiplying up and expanding (condone one error) 3kt 3kt dep* Factorising their x terms correctly x = ( e 3kt )/( + e 3kt ) * www (AG) final B mark as in scheme above 5

169 OCR (Oxford Cambridge and RSA Examinations) Hills Road Cambridge C EU OCR Customer Contact Centre Education and Learning Telephone: Facsimile: general.qualifications@ocr.org.uk For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee Registered in England Registered Office; Hills Road, Cambridge, C EU Registered Company Number: OCR is an exempt Charity OCR (Oxford Cambridge and RSA Examinations) Head office Telephone: Facsimile: OCR 05

170 GCE Mathematics (MEI) Unit 4754A: Applications of Advanced Mathematics: Paper A Advanced GCE Mark Scheme for June 06 Oxford Cambridge and RSA Examinations

171 OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of candidates of all ages and abilities. OCR qualifications include AS/A Levels, Diplomas, GCSEs, Cambridge Nationals, Cambridge Technicals, Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching/training, administration and secretarial skills. It is also responsible for developing new specifications to meet national requirements and the needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support, which keep pace with the changing needs of today s society. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by examiners. It does not indicate the details of the discussions which took place at an examiners meeting before marking commenced. All examiners are instructed that alternative correct answers and unexpected approaches in candidates scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the report on the examination. OCR will not enter into any discussion or correspondence in connection with this mark scheme. OCR 06

172 4754A Mark Scheme June 06 Annotations and abbreviations Annotation in scoris and Meaning Benefit of doubt Follow through Ignore subsequent working,,, Method mark awarded 0, Accuracy mark awarded 0, Independent mark awarded 0, Special case Omission sign Highlighting Misread Other abbreviations in Meaning mark scheme E Mark for explaining U Mark for correct units G Mark for a correct feature on a graph dep* Method mark dependent on a previous mark, indicated by * cao Correct answer only oe Or equivalent rot Rounded or truncated soi Seen or implied www Without wrong working 3

173 4754A Mark Scheme June 06 Subject-specific Marking Instructions for GCE Mathematics (MEI) Pure strand a Annotations should be used whenever appropriate during your marking. The A, M and B annotations must be used on your standardisation scripts for responses that are not awarded either 0 or full marks. It is vital that you annotate standardisation scripts fully to show how the marks have been awarded. For subsequent marking you must make it clear how you have arrived at the mark you have awarded. b An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed. When a candidate adopts a method which does not correspond to the mark scheme, award marks according to the spirit of the basic scheme; if you are in any doubt whatsoever (especially if several marks or candidates are involved) you should contact your Team Leader. c The following types of marks are available. M A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, eg by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 cannot ever be awarded. B Mark for a correct result or statement independent of Method marks. E A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. 4

174 4754A Mark Scheme June 06 Unless otherwise indicated, marks once gained cannot subsequently be lost, eg wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument. d e When a part of a question has two or more method steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation dep * is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given. The abbreviation ft implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, exactly what is acceptable will be detailed in the mark scheme rationale. If this is not the case please consult your Team Leader. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be follow through. In such cases you must ensure that you refer back to the answer of the previous part question even if this is not shown within the image zone. You may find it easier to mark follow through questions candidate-by-candidate rather than question-by-question. f g Wrong or missing units in an answer should not lead to the loss of a mark unless the scheme specifically indicates otherwise. Candidates are expected to give numerical answers to an appropriate degree of accuracy, with 3 significant figures often being the norm. Small variations in the degree of accuracy to which an answer is given (e.g. or 4 significant figures where 3 is expected) should not normally be penalised, while answers which are grossly over- or under-specified should normally result in the loss of a mark. The situation regarding any particular cases where the accuracy of the answer may be a marking issue should be detailed in the mark scheme rationale. If in doubt, contact your Team Leader. Rules for replaced work If a candidate attempts a question more than once, and indicates which attempt he/she wishes to be marked, then examiners should do as the candidate requests. If there are two or more attempts at a question which have not been crossed out, examiners should mark what appears to be the last (complete) attempt and ignore the others. NB Follow these maths-specific instructions rather than those in the assessor handbook. 5

175 4754A Mark Scheme June 06 h For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate s data. A penalty is then applied; mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A mark in the question. Note that a miscopy of the candidate s own working is not a misread but an accuracy error. 6

176 4754A Mark Scheme June 06 Question Answer Marks Guidance cos 3sin R cos cos sin sin Rcos,3 Rsin Correct pairs. Condone sign errors for the M mark (so accept Rsin 3) R 3 0 R 0 Or 3. or better, not 0 unless 0 chosen tan 3.49 ft their pairs (condone sign errors but division must be the correct way round), for.49 or better (accept.5), with no errors seen in method for angle Maximum value of cos 3sin is 0 4 Or equivalent convincing numerical statement that no solutions exist e.g. 4. Maybe embedded in an attempt at a solution. Do not accept general 0 statements e.g. doesn t work must be clear why no solutions exist dependent on first SC: If candidates state that cos,sin 3 tan 3 this could score M0A0 (so max 4/6) [6] Note that those candidates who state R 0 and tan 3 with no (wrong) working seen could go on to score full marks 7

177 4754A Mark Scheme June 06 Question Answer Marks Guidance q x ( ) x q q x q p p! p * One of q x p or q( q ) x! p q p qq ( ) 3 Allow x s on both sides of equations (if correct) p 4 q (soi), for example, scores p p( p) 3 qq ( ) 3 q p or dep* Eliminating p (or q ) from simultaneous equations (not involving x) involving p 4 q 4 both variables oe if awarded followed by either p or q correct (www) this implies this M mark p p www (or q ) q ft q (or p ) for second value, ft their p or q (e.g. the negative of their p or q) provided first 4 marks awarded and only a single computational error in the method so must be a correct method for solving their equation in p or q (ignore mention of p and/or q 0 ) x Valid for x or x www, allow x but not say, x SC If M0 M0 awarded and no wrong working seen then for p = and q = -, for - < x < (oe) so max marks [7] Guidance for solving quadratics on this paper: use of correct quadratic equation formula (if formula is quoted correctly then only one sign slip is permitted, if the formula is quoted incorrectly M0, if not quoted at all substitution must be completely correct to earn the ) or factorising (giving their x term and one other term when factors multiplied out) or completing the square (must get to the square root stage involving and arithmetical errors may be condoned provided that perfect square term was correct) 8

178 4754A Mark Scheme June 06 Question Answer Marks Guidance 3 4 V x dy for ( ) (d ) k x y with correct limits and k or, allow correct limits 0 seen or implied later, if formula not stated then must substitute for their correctly to imply this formula condone lack of for the M mark and dy throughout (condone incorrect use of dx too) V y dy Correct (or with a /) limits may be seen or implied through later working 3 y 3 3 y [4] 3 3 y or 3 3 y (but only if k ), condone Exact mark final answer (so no isw if correct answer is subsequently halved) but if exact value seen and is then followed by then isw y.5.5 (oe) x 9

179 4754A Mark Scheme June 06 Question Answer Marks Guidance 4sin cos cos * Use of correct double angle formulae: sin sin cos and any one of cos cos sin or sin or cos 4 cos sin cos 0 Correct equation in solvable form e.g. sincos 0(oe) or 4 4 5sin 6sin 0 or 5cos 4cos 0 but not 4sin cos cos tan dep* Use of sin tan on their sin cos 0 cos or correct method for solving quadratic in eithersin or cos (See guidance in question for solving quadratics) 6.6 www (6.6 or better) 90 Not from incorrect working Ignore additional solutions outside the range. If any additional solutions given inside the range 0 80 and full marks would have been awarded then remove last mark (so 4/5) Both answers in radians: (or better) and / scores [5] Answers with no working scores (so max /5) 0

180 4754A Mark Scheme June 06 Question Answer Marks Guidance 5 (i) AC xsec Accept any equivalent forme.g. ACcos x. If AC not seen then there must be a diagram as evidence of correct sides - xsec with no AC is B0 AD sec and AE sec 3 x x Accept 3 x xsec (as AE = x) or any equivalent form. Otherwise there must be a corresponding diagram as evidence of correct sides. Accept 3 cos x/ AC AC / AD AD / xfor the first two marks 3 xsec x This line (oe) must be seen before the x s cancelled 3 sec * NB AG dependent on all previous marks [3] 5 (ii) OR AD xcos Same principles as above for each corresponding mark 3 AC xcos and AB xcos 3 or x xcos (as AB = x) 3 3 xcos x sec * 3 Must see xcos x(oe) before given answer ED xsin oe e.g. ED 4x AD or ED ADtan with AD correctly expressed in terms of x and (or using 37.5 or better) - see (i) for alternatives for AD. Allow ED =.x (or better) but B0 if ED = missing CB xtan oe e.g. CB AC x or CB ACsin with AC correctly expressed in terms of x and (or using 37.5 or better) - see (i) for alternatives for AC. Allow CB = 0.77x (or better) but B0 if CB = missing ED xsin cos CB x tan www must come from exact working (so not using oe) - accept ED ED or sec 3 (oe) (as from (i): sec ) CB sec CB 3 3 / * NB AG dependent on all previous marks in (ii) must be one step of intermediate working from cos to given answer [4]

181 4754A Mark Scheme June 06 Question Answer Marks Guidance dy d y / dt / t * for their (d y/ d t ) / theird x/ dt in terms of t with at least one term correct dx d x / dt t 6 cao (oe) allow unsimplified even if subsequently cancelled incorrectly i.e. can isw with any non-zero gradient expressed as a function of t - or t any equivalent form (e.g. y = mx + c) but must have used the correct point, t t - if using y = mx + c must explicitly have c = before can be awarded dep* y f ( t) x t y x t t t 4 When x = 0, y t ft Must be a function of t When y = 0, x = 4t ft Must be a function of t oe need not be simplified So area of triangle = 4 4t 8 (which is t independent of t) OR (for the first two marks) 4 dy 4 y x x dx x dy 4 d x ( t) [7] No ft on this mark an answer of 8 (www) with no additional comment is sufficient to award this mark * Attempt to eliminate t and correctly differentiates their Cartesian equation

182 4754A Mark Scheme June 06 Question Answer Marks Guidance 7 (i) AG Condone 4 etc. if recovered = 50 5 Correct answer implies both marks accept 7. or better [] 7 (ii) DP = 4i + j 5k or PD= 4i j + 5k One correct direction vector in plane DPF (oe e.g. expressed as a column vector) DF= 4i + 3j (or PF = j + 5k) Any other correct direction vector in plane DPF n DP = ( 5) = 0 Scalar product with a direction vector in the plane (including evaluation and = 0) (OR forms vector cross product with at least two correct terms in solution) n DF = = 0 (or n P F ) Scalar product with a second direction vector in the plane (including evaluation and = 0) (following OR above, all correct ie a multiple of 5i 0j 4k ) (NB finding only one direction vector and its scalar product is B0 B0) 5x 0y 4z c oe (accept any non-cartesian form for only) rn an 5x 0 y + 4z = 0 A0 for 5i 0j 4k 0 or 5x 0 y + 4z 0 SC: if states if 5i 0j 4k is normal then of the form5x 0y 4z c and substitutes one coordinate gets, then substitutes the other two coordinates A3 (not,, ). Then states so 5i 0j 4k is normal and states the correct equation of the plane this can get provided that there is a clear argument ie A3. Without a clear argument this is B0 [6] SC: if finds two relevant direction vectors and then finds equation of the plane from vector form, r a b c gets. Eliminating parameters cao. If then states so 5i 0j 4k is normal can get, and then a valid reason why 5i 0j 4k is normal scores the final B mark (each B mark is dependent on the previous one) 3

183 4754A Mark Scheme June 06 Question Answer Marks Guidance 7 (iii) x r = 4i + Need r (or another single letter) = or y for first z r 3j 5k 4i 3j5k - accept column.+ ( 4i + 3j + 5k) NB answer is not unique e.g. vector form and condone row vectors (non-vector form scores only) 5(4 4) 0(3) + 4(5) = 0 Substituting their line in their plane equation from (ii) (condone a slip if intention clear) their line and plane must be of the correct form (e.g. the line must be of the form r a tb ) 40 = 00, = 0.4 www cao NB is not unique as depends on choice of line seen in this part Q is (.4,., ) www - condone answer given as a position vector 7 (iv) AQ : QG = :3 oe www [6] * Selecting 5i 0j 4k and their direction vector from (iii) Angle between 4i 3j 5k and (5i 0j4 k) is where 45) (30) (5 4 cos dep* Correct formula (including cosine), with correct substitution, using their direction vector from (iii) and the correct normal vector - condone either a single numerical slip or a single sign slip (but not one of each) if intention is clear. So it must be clear where the slip comes from e.g. if the magnitude of one vector is stated incorrectly with no working then this is M or 4.0 www cao (accept 56 or better, 4 or better,.6 radians or better, radians or better) angle between line and plane = 34.0 www cao (accept 34 or better, radians or better) [4] 4

184 4754A Mark Scheme June 06 Question Answer Marks Guidance 8 (i) x x 4 * NB AG must be at least one intermediate step before given answer correct x x ( x)( x) ( x)( x) application of partial fractions is fine [] 8 (ii) x x ln 0 x x 0 or e or ln( x) ln( x) x x x 0 If only this line seen then award B0 8 (iii) x t ln ln( x) ln( x) x dt dx x x [] SC: Allow only for verifying that when x = 0, t = 0 Correct differentiation of their t OR for first two marks - If no subtraction law of logs seen e.g. dt ( x)() ( x)( ) d x x award for correct first bracket ( x) x (reciprocal expression) and for second correct bracket (quotient/chain rule)(oe) if additional constant(s) added (e.g. t = k ln( )) then award only for a constant times a fully correct derivative dt 4 dx ( x )( x ) d x ( x)( x) dt 4 dt dx and d x must be correctly attributed to the correct expression for this mark dt k 4 Explicitly stating (that the constant of proportionality is) 4 possible to score A0in this part See next page for an alternative solution [4] - therefore it is 5

185 4754A Mark Scheme June 06 Question Answer Marks Guidance OR dx dx k( x)( x) k dt d t ( x)( x) Allow omission of dx and/or dt ln( x) ln( x) kt( c) Any non-zero constant, - further guidance in (v) for this and the next mark www oe (condone absence of c) x ) ln( x ) kt ( c ) 4 x 0, t 0 c 0 x ln 4kt k www must include c and show that c = 0. Must explicitly state that k x (iv) t x e x t ( x)e x t t e xe x t t x( e ) e Multiplying out, collecting x terms (condone sign slips and numerical errors (eg loss of a ) only but M0 if e t incorrectly replaced with e t ) and factorising their x terms correctly t t t t (e ) (e )e ( e ) x t t t t * e ( e )e e ( x ) (as t ) [4] OR (for first three marks) t x e x t ( x)e x t t e xe x t x( e ) e t ( e ) x t * e t www NB AG as AG must be an indication of how previous line leads to the required result (eg stating or showing multiplying by e t ) Multiplying out, collecting x terms (condone sign slips as above) and factorising their x terms correctly www NB AG 6

186 4754A Mark Scheme June 06 Question Answer Marks Guidance 8 (v) Separating variables - condone sign slips and issues with placement of k but M0 t dx k e dt xx * for ( x)( x)d x... or equivalent algebraic error in separating variables unless recovered. If no subsequent work integral signs needed, but allow omission of dx and/or dt but must be correctly placed if present t ln( x) ln( x) e ( c) Any non-zero constants,, - this line must be seen and cannot be implied by later working (as this is an AG) condone absence of c or if a constant present condone the use of k for their constant. Do not condone invisible brackets e.g. ln + x unless recovered before subtraction law of logs applied all of these points apply to the next A mark too t ln( x) ln( x) 4ke ( c) www oe When t 0, x 0 c 4k dep* Substituting x0, t 0 into each term in an attempt to find their c (must get c = ) if they integrate and use k as their constant they must use x0, t 0 to find this single k term only x t ln 4 k( e )* x www NB AG must have obtained all previous marks in this part [5] OR (for first 3 marks) final as above t dx k e dt xx * Separating variables. If no subsequent work integral signs needed, but allow omission of dx or dt, but must be correctly placed if present x t t dx k e dt ln ke ( c) 4 x 4 x A Must see / 4 x on lhs please note that one A mark cannot be awarded 8 (vi) as t, x.85 ln 3.85/0.5 = 4k Sets e t to 0 and substitutes x =.85 condone substitution of a large value of t only if it leads to the correct value of k k = 0.8 k 0.5ln 77 / 3 or 0.8 or better [] 7

187 OCR (Oxford Cambridge and RSA Examinations) Hills Road Cambridge C EU OCR Customer Contact Centre Education and Learning Telephone: Facsimile: general.qualifications@ocr.org.uk For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee Registered in England Registered Office; Hills Road, Cambridge, C EU Registered Company Number: OCR is an exempt Charity OCR (Oxford Cambridge and RSA Examinations) Head office Telephone: Facsimile: OCR 06

188 GCE Mathematics (MEI) Unit 4754B: Applications of Advanced Mathematics: Paper B Advanced GCE Mark Scheme for June 06 Oxford Cambridge and RSA Examinations

189 OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of candidates of all ages and abilities. OCR qualifications include AS/A Levels, Diplomas, GCSEs, Cambridge Nationals, Cambridge Technicals, Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching/training, administration and secretarial skills. It is also responsible for developing new specifications to meet national requirements and the needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support, which keep pace with the changing needs of today s society. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by examiners. It does not indicate the details of the discussions which took place at an examiners meeting before marking commenced. All examiners are instructed that alternative correct answers and unexpected approaches in candidates scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the report on the examination. OCR will not enter into any discussion or correspondence in connection with this mark scheme. OCR 06

190 4754B Mark Scheme June 06 Annotations and abbreviations Annotation in scoris and Meaning Benefit of doubt Follow through Ignore subsequent working,,, Method mark awarded 0, Accuracy mark awarded 0, Independent mark awarded 0, Special case Omission sign Highlighting Misread Other abbreviations in Meaning mark scheme E Mark for explaining U Mark for correct units G Mark for a correct feature on a graph dep* Method mark dependent on a previous mark, indicated by * cao Correct answer only oe Or equivalent rot Rounded or truncated soi Seen or implied www Without wrong working 3

191 4754B Mark Scheme June 06 Subject-specific Marking Instructions for GCE Mathematics (MEI) Pure strand a Annotations should be used whenever appropriate during your marking. The A, M and B annotations must be used on your standardisation scripts for responses that are not awarded either 0 or full marks. It is vital that you annotate standardisation scripts fully to show how the marks have been awarded. For subsequent marking you must make it clear how you have arrived at the mark you have awarded. b An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed. When a candidate adopts a method which does not correspond to the mark scheme, award marks according to the spirit of the basic scheme; if you are in any doubt whatsoever (especially if several marks or candidates are involved) you should contact your Team Leader. c The following types of marks are available. M A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, eg by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 cannot ever be awarded. B Mark for a correct result or statement independent of Method marks. E A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. 4

192 4754B Mark Scheme June 06 Unless otherwise indicated, marks once gained cannot subsequently be lost, eg wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument. d e When a part of a question has two or more method steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation dep * is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given. The abbreviation ft implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, exactly what is acceptable will be detailed in the mark scheme rationale. If this is not the case please consult your Team Leader. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be follow through. In such cases you must ensure that you refer back to the answer of the previous part question even if this is not shown within the image zone. You may find it easier to mark follow through questions candidate-by-candidate rather than question-by-question. f g Wrong or missing units in an answer should not lead to the loss of a mark unless the scheme specifically indicates otherwise. Candidates are expected to give numerical answers to an appropriate degree of accuracy, with 3 significant figures often being the norm. Small variations in the degree of accuracy to which an answer is given (e.g. or 4 significant figures where 3 is expected) should not normally be penalised, while answers which are grossly over- or under-specified should normally result in the loss of a mark. The situation regarding any particular cases where the accuracy of the answer may be a marking issue should be detailed in the mark scheme rationale. If in doubt, contact your Team Leader. Rules for replaced work If a candidate attempts a question more than once, and indicates which attempt he/she wishes to be marked, then examiners should do as the candidate requests. If there are two or more attempts at a question which have not been crossed out, examiners should mark what appears to be the last (complete) attempt and ignore the others. NB Follow these maths-specific instructions rather than those in the assessor handbook. 5

193 4754B Mark Scheme June 06 h For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate s data. A penalty is then applied; mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A mark in the question. Note that a miscopy of the candidate s own working is not a misread but an accuracy error. 6

194 4754B Mark Scheme June 06 Question Answer Marks Guidance Note that throughout units are not required. Blade radius = 45 m Hub height = = 04.5 m. Scale Photomontage: Real size =.35 cm : 35 m values.35, 35 and 0.8 or 6.7, 99.5 and 0.8 or 99.5, 35,.35 and 5.9 seen in a calculation [] 0.8 cm corresponds to rounding to m E.35 [] or in both cases must see.9 (or better) followed by or leading to so.35 must see.6 (or better) followed by (or equivalent correct calculation) 3. (i) 5 tan.3.3 or better - allow.35 rounded to.4 but B0 if only.4 seen 63 (ii) (iii) Height cm 0.3 or better or their(0.3).4.4 or better 97 e.g., or or 7.3 '.4' tan [4] Must see 0.3 (or better) or 7.3 (or better) 0.3 with no working scores M0A0.3 or.4 7

195 4754B Mark Scheme June 06 Question Answer Marks Guidance 4. Height above A of lowest visible point is All three values 99.5, 0, 0 seen in a calculation 99.5 with no working scores = 99.5 m Using similar triangles or trigonometry x x 40 or tan and 40tan where79.5 x (must be a complete method for finding the height) Height of C m 87.3 (or better) 30 [4] 5. 7 arctan ( ) 800 Beware: QB = arctan arctan ( ) Or cosine rule e.g. cos (oe) The angles are both 5. o to significant figures. E Must see 5.4 (or better), 5.3 (or better) and 5. [3] 8

196 4754B Mark Scheme June 06 Question Answer Marks Guidance 6. Those thinking they appear about right might be those who opted for 70 mm and 80 mm, so = 70 people Assumption stated explicitly and must mention the bolded values/words (for 70 allow or use of 85 twice may be seen in % calculation) any reasonable assumptions must be justified (as in the case above) it must be clear where the numbers are coming from (e.g. 0 is about right which comes from 60, 70 and 80 would score ) Percentage so 47% For 47% (or better) or 3% (or better accept 3.5%) (comes from 36 using just the value of 85) or a percentage which comes from a justified assumption (so if the % is not 47 or 3 then they must have been awarded the first B mark) Those saying too large would be people who opted for 50 mm or 60 mm, so = 66 people Those saying too small would be people who opted for 90 mm, 00mm or 0 mm, so = 6 people Assumptions stated explicitly and must mention the bolded values/words (allow values less than or equal to 60 for 50 and 60, allow values greater than or equal to 90 for 90,00 and 0) see first B mark for further details. Note that for this mark the too large and too small must be attributed to the correct values Percentages are 8% (too large) and 35% (too small) For 8% (or better) and 35% (or better) or a percentage which comes from a justified assumption (so if the % are not 8 and 35 then they must have been awarded the third B mark). Note that for this mark the values of 8 and 35 do not need to be attributed correctly but any other values must correspond correctly to large and small [4] For guidance: Focal length Number of %

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