FP3 mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers (back to June 2002)

Size: px

Start display at page:

Download "FP3 mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers (back to June 2002)"

Clare Lucas
5 years ago
Views:

1 FP mark schemes from old P, P5, P6 and FP, FP, FP papers (back to June ) Please note that the following pages contain mark schemes for questions from past papers. Where a question reference is marked with an asterisk (*), it is a partial version of the original. The standard of the mark schemes is variable, depending on what we still have many are scanned, some are handwritten and some are typed. The questions are available on a separate document, originally sent with this one. This document was circulated by in March 9; the mark schemes for questions and 7 have since been removed (8..9) since they are not on the specification. FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

2 . (a) y Closed shape, B () (b) b a ( e ) 9 6( e ) M 7 e oe awrt.66 A () (c) Foci are at (±ae, ) use of ae M ( 7, ) and ( 7, ) awrt.65, is required, ft their e A ft () (5 marks) [P5 June Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

3 e e e e. M 6e quadratic in e M, A ( e )(e ) e ½ and / ln ½ and ln / M A M, A (7 marks) Alt. cosh sinh R cosh( α) R 96 and tan α M, A 5 cosh( α) 96 α ln ± either A ln and 6 6 ln both as single ln A ln 6 ½ ln, ln 6 ½ ln 6 ln, ln combine either into single ln. 6 M Dependent on first two Ms ln ½ and ln / A, A [One answer by alt. method gains ma. MAMMMAA] ( marks) [P5 June Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

4 FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9. (a) n n n n d sin sin d cos M, A ] d cos ) ( cos [... n n n n n M Using limits ) ( n n n I n n I π ( ) cso M, A (5) (b) d cos π I [ ] sin π at any stage B 6 6 I I π M 6 6 I π π I π π π M Hence π π π I cao A () (9 marks) [P5 June Qn ]

5 5. (a) y arctan tan y M dy sec y d A dy d tan y 9 ( ) M, A () (b) 9 6 arctan d arctan d 9 M, A... 9 d 9 M... arctan A π π 9 (π ) ( ) cso A (6) 9 M ( marks) [P5 June Qn 6] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

6 6. (a) ½ y, d d y ½ M, A ½ d ½ π y d π d d M... π d ( ) A ()... () / (b) S d π ( ) π M, A... / ( ) 8 () π or any eact equivalent A () (c) dy d d d M d A (d) Using symmetry, s d ( ) A () d sinh θ, sinhθ coshθ oe B dθ sinh θ I. sinhθ coshθ dθ M sinh θ cosh θ dθ ( cosh θ ) dθ M θ sinh θ A Limits are and arsinh ( ln ( ) s θ sinhθ sinh θ arsinh ( ) ( ) arsinh ln M A (6) FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

7 6. (d) Alt The last four marks can be gained: θ θ e e θ θ I dθ (e e )dθ M θ θ e e θ A ( ) ( ) s arsinh.... ln( ) ½( ) [ ln( )] ( ) M, A 6. (d) Alt The last two marks may be gained by substituting back to the variable s... [ θ sinh θ ] [ θ sinhθ θ ] cosh... [ arsinh ] arsinh ln( ) [ ln( ] ( ) M, A [P5 June Qn 8] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

8 FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9 [P6 June Qn 5] 8. (a) 6 6 5, eigenvalue is MA, A () (b) Either 6 6 (6) 6) 8( 6 8 M A () Alt(b) or 5 λ λ λ ) 6(5 ) 6( ) )( )(5 ( λ λ λ λ λ ) 9)( )( ( λ λ λ λ is an eigenvalue M A z y eigenvector z 9, 5y z 9y, y z 9z At least two of these equations M Attempt to solve z, z y, y z A A () (c) Make e.vectors unit to obtain P columns in any order M, Aft D 9 λ, where λ, P and D consistent M, A, B (5) Alt P, D 7 8 7, P and D consistent MAft, MA, B

9 9. (a) AB 5i j AC i j k or BC i j k i j k AB AC 5 i 5j k M, A r i j k λ( i 5j k) B ft () (b) Volume AD.( AB AC) AD i j k B 6 / 6 (i j k).( i 5j k) M / 6 A () (c) r.( i 5j k) (i j).( i 5j k) M, A ft A () (d) [i.( λ) j( 5λ) k( λ)].( i 5j k) M, Aft 9λ 5λ λ 5λ λ 5 M E is 68, , 5 A () (e) Distance (f) 5 5 i 5 j k ( ) M A () 5 λ B 5 5 r D i j k 5 ( i 5j k) 8 D is,, M A () (8 marks) [P6 June Qn 7] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

10 . e e e 5e 6e e 8 M M A (5e )(e ) A e 5, (a) y artanh ln( 5 ), ln accept ln 5 M A (6) tanh y sech y d y M A d d y d tanh y cso (*) (b).artanh d artanh d (6 marks) [P5 June Qn ] A () M A artanh ln( ) ( c) M A () (7 marks) Alt. d ln( ) ln( ) This is acceptable (with the rest correct) for final M A [P5 June Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

11 . d 9 9 d M arsinh 5ln 9 M A [] 5 5 arsinh ft on 5 Area (m ) Using a substitution (i) 9 5 ln sinh θ; d cosh θ dθ d 9 then as before, cosh θ 5 dθ 5 arsinh cosh θ dθ complete subs. M 9 or changing limits to and arsinh (or ln ) can gain 9 this A (ii) tan θ; d sec θ dθ d 9 Limits are and arctan 9 tan θ 9 sec θ dθ M A ft M A M A M 5 sec θ dθ 5 ln (sec θ tan θ ) M arctan [] 5 ln etc M Aft 9 A (7 marks) [P5 June Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

12 . (a) d y d sinh a B dy d d sinh d a cosh d sinh a a M, A Length a sinh a (b) a sinh k 8a sinh k ka a sinh k (*) M A (5) B ka a arsinh a ln ( 7) B ka y a cosh a ( sinh k) a 7 a M A () (9 marks) [P5 June Qn 5] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

13 . (a) sec y e B d y sec y tan y e ( sec y) M A d d y e (*) cso M A (5) d sec y tan y sec y e (b) y π O Shape, curve (, ) Asymptote, (y ) π B B () (c) ( ln ) π y arcsec d y d B B π tangent is y ( ln ) M π, y ln eact answer only A () Alt to (a) cos y e B ( marks) sin y d y e d cos y M A e e e (*) cso M A (5) [P5 June Qn 6] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

14 5. (a) I n (b) J n n n e n e d e ni n (*) cso M A () n n e n e d M A e nj n A () (c) J e J J e J J and J M e e d e ( e ) ( e ) A J e ( e ) e 5 (*) A () (d) n cosh d n e e d (I n J n ) (*) B () (e) I e I e (e I ) I e e d e [e ] e ( e ) M A 5 5 (I J ) (e ) e (e ) M A () e ( marks) [P5 June Qn 7] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

15 6. (a) d d y a sec t tan t, b sec t M A dt dt d y b sec t b d a sec t tan t a sin t gradient of normal is a sin t b M A a sin t y b tan t ( a sec t) M b a sin t by (a b ) tan t (*) cso A (6) (b) y ( a b ) tan t a sin t a b a cos t B b a (e ) b OS ae and OA AS 5a M M a 5a cos t a cos t M A π 5π t, A By symmetry or (as OA a b a cos t a b ) a cost ae Alt. to (a) t π, π M A (8) ( marks) y d y M A a b d d y d b b a sect M A a y a b tan t then as before [P5 June Qn 8] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

16 7. [P6 June Qn ] 8. [P6 June Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

17 9. [P6 June Qn 6] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

18 .. [P6 June Qn 7]. (a) cosh sinh ( e e ) ( e e ) e e e e ) M A A () (b) cosh ) ( e e e e sinh sinh e Make the subject of the formula, ln( ) ln M A M, A () [P5 June Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

19 . (a) (b) a, b, c 6.5 d.5 ( ) 6 arc tan 8 π.5.5 B, B, B () M M A B () [P5 June Qn ]. (a) As 9( e ), e 5 9 Uses ae to obtain that the foci are at ( ± 5,) M, A M A () (b) PS PS e( PM PM ) M for single statement e.g, PS epm a e e M needs complete method a 6 M M A () [P5 June Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

20 . (a) (b) dy n n Using product rule ( n )sinh cosh sinh d Using cosh sinh in derived epression dy n n to obtain ( n )sinh ( sinh ) sinh d dy n n and ( n )sinh nsinh d ar sinh ar sinh n arsinh n n [sinh cosh ] ( n ) sinh d nsinh d So cosh( ar sinh) ( n ) I n ni n If sinh α then cosh α sinh α nin ( n ) In OR arsinh ar sinh n n arsinh n and use sinh sinh d sinh cosh ( n ) cosh sinh d cosh sinh collect I ( n) I to obtain nin ( n ) In n n M M A () M A () M A () (c) I ar sinh I I I I and use with previous results to obtain (ar sinh ).5 (either answer acceptable) 8 B M M A () [P5 June Qn 5] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

21 5. (a) d dy acos θ sinθ asin θ cosθ dθ dθ (9 ( )) s a c s s c dθ a c s dθ a cosθ sinθdθ a π Total length [sin θ ] 6a B M M A M M A (7) (b) A π asin θ acosθsinθdθ 6π a sin θ cosθdθ π 6π a 5 sin θ 5 π a 5 M A M M A (5) [P5 June Qn 7] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

22 FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9 6. (a) AB AC M AB AC k j i A: One value correct, A: All correct M A A () (b) 8.. r 7. r M Aft () (c) AD. AB AC (Attempt suitable triple scalar product) M or (if using AD) B Volume ( ) 6 6. M A(cso)() [P6 June Qn ]

23 FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9 7. k k k c p b a c b a p T MM (a) ( ) p p M A () (b) ( ) 8 k k (ft on their p, if used) M Aft () (c) equations: a b c a c M a and b in terms of c (or equiv.): a c b c (ft on their p) M Aft Using ) 8 ( 8 c b a c b a c b a. : a, b, c M A(,) (6) (d) det M ( ) ( ) ( ) 5 M A(cso) () Alternatives: (c) Require c b a parallel to, 6 M, M A (Then as in main scheme, scaling to give a, b and c.) M A(,) (6) (d) det ) 8 T (MM, det M det T M, det M 8 8 (5 ) M A () [P6 June Qn 5]

24 FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9 8. (a) det A ) ( λ M, ) )( ( 8 6 λ λ λ λ λ λ A Eigenvector,, : y y λ (or equiv.) M A Eigenvector,, : y y λ (or equiv.) A (5) (b) P k M: eigenvectors as columns, k M, A T P P, AP P D M, M A (5) (c). Rotation of π clockwise (about (, )).. Stretch, parallel to -ais, parallel to y-ais.. Rotation of π anticlockwise (about (, )).. and. both rotation, or both reflection. M Correct angles, opposite sense or correct lines (reflection). A Stretch. B All correct, including order. A () [P6 June Qn 6]

25 9. [FP/P5 June 5 Qn ].. [FP/P5 June 5 Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

26 . [FP/P5 June 5 Qn ]. [FP/P5 June 5 Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

27 . [FP/P5 June 5 Qn 6] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

28 . [FP/P5 June 5 Qn 8] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

29 5. [FP/P6 June 5 Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

30 6. [FP/P6 June 5 Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

31 7. [FP/P6 June 5 Qn 7] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

32 8. [FP/P5 January 6 Qn ] 9. [FP/P5 January 6 Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

33 . [FP/P5 January 6 Qn ]. [FP/P5 January 6 Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

34 . [FP/P5 January 6 Qn 7] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

35 . [FP/P5 January 6 Qn 8] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

36 . (a) (b) [*FP/P6 January 6 Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

37 5. B [FP/P6 January 6 Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

38 6. [FP/P6 January 6 Qn 7] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

39 e e e e 7. 5 B e e 7 M: Simplify to form quadratic in e M A ( e )(e 7) e, e 7 M: Solve term quadratic. M A ln ( or ln ) ln 7 A (6) 6 Marks [FP June 6 Qn ] 8. (a) Using b a ( e ) or equiv. to find e or ae: (a and b ) M A e Using y ( ae) y (M requires values for a and e) M A () (b) Bft () 5 Marks [FP June 6 Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

40 d y 9. (a) sech B d Put d y ( cosh cosh ) M d ln( ± ) or 8 ln(7 ± ) or e ± or A e 7 ± (± or ) ln( ) or ln(7 ) (or equiv.) A () 8 (b) y ln( ) tanh(...) (Substitute for ) M sech tanh, tanh M y ln( ) { ln( ) } (*) A () 7 Marks (a) Second solution, if seen, must be rejected to score the final mark. (b) nd M requires an epression in terms of without hyperbolics, eponentials and logarithms. [FP June 6 Qn 5] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

41 FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9 5. (a) d d d d t t y t t B B t t t t t, or t t M, A [ ] ( ) ln ln ln d Length t t t t (*) M M A (7) (b) Surface area t t t t t t t d 8 d π π M 6 6 ) (8 ) (8 π π π t t π 5 M M A () ( marks) [FP June 6 Qn 6] 5. d arsinh d arsinh M A A 9arsinh arsinh ( ) ( ) 9ln or B Let u u d d u u u d d M u u u u u u u u d 6 d 6 d. ( ) u u d M 6 u u u u M When, u and when, u [ ]... u 6 6 u u M A Area ( ) ( ) 7 9 9ln 6 arsinh 9 (*) Acso ()

42 Dependent M marks: du M: Choose an appropriate substitution & find or Set up integration by d parts. M: Get all in terms of u or Use integration by parts. M: Sound integration. M: Substitute both limits (for the correct variable) and subtract. 5. Alternative solution: Let sinh θ d coshθ dθ M ar sinh d θ sinh θ coshθ dθ M θsinh θ sinhθ ( cosh θ ) dθ M A A θ sinh θ sinhθ cosh arsinh 9arsinh cosh θ ( cosh θ ) dθ coshθ arsinh θ coshθ 7 Area 9 arsinh 9ln( ) ( 7 ) (*) 9 B M M A Acso () Marks FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

43 5. A few alternatives for: d. du (i) Let u d u u du du u u u No marks yet needs another substitution, or parts, or perhaps u u u u M u du du u M ( u ) ( u) M Limits ( to 9) M (ii) Let sinh θ d coshθ dθ M sinh θ coshθ dθ sinhθ ( cosh θ ) dθ coshθ M Then, as in the alternative solution, cosh θ sinhθ ( cosh θ ) dθ coshθ M Limits ( to arsinh) M FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

44 du 5. (iii) Let u tan θ sec θ M dθ tan θ sec θ dθ tanθ secθ ( sec θ ) dθ M secθ sec θ sec θ ( secθ tanθ ) dθ ( secθ tanθ ) dθ secθ M Limits ( sec θ to secθ ) M (iv) (By parts must be the right way round, not integrating ) du dv u,, v M d d d M ( ) Limits M (v) (By parts) du dv u, d d, v arsinh No progress M (vi) M ( ) ( ) M d d M ( ) ( ) M Limits M [FP June 6 Qn7] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

45 5. (a) n n n cosh d sinh n sinh d M A n n n sinh n cosh n( n ) In n n n sinh n cosh n( n ) cosh d M I (*) A () (b) I sinh cosh I M I sinh cosh sinh cosh I M ( ) (This M may also be scored by finding I by integration.) I cosh d sinh k B I ( ) sinh, ( ) cosh ( C) A, A (5) (c) [( ) sinh ( ) cosh ] 7sinh 8cosh M: substituted throughout (at some stage) M e e e e 7 8 M M: Use of ep. Definitions (can be in terms of ) ( 9e 65e ) A () Marks (b) Integration constant missing throughout loses the B mark [FP June 6 Qn 8] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

46 ( m c) b a ( m c) a b 5.(a) M a b ( b a m ) a mc a ( c b ) (*) A () (b) ( a mc) ( b a m ) a ( c b ) M a m c a ( b c b a m c a m b ) c b a m A () (*) (c) Find height and base of triangle (perhaps in terms of c). M c OB c ( b a m ) and AO b a m m m A c b a m Area of triangle OAB m m M: Find area and subs. for c. M A () (d) b a m b a Δ m m m d b m b a m dm m M A b a a b Δ ab b a (*) A () (e) a mc Root of quadratic: b a m (Should be correct if quoted M directly) b a Using m and c b a m : a M A () (The nd M is dependent on using the quadratic equation). Marks (d) Alternative: b a m bam (since ( b am) ) [M] b a m ab [A] m Conclusion [A] (e) Alternative: Begin with full eqn. ( b a m ) a mc a ( c b ). b In the eqn., use conditions m and c b a m ( b ) [M] a Simplify and solve eqn., e.g. a a a [FP June 6 Qn 9] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

47 5. ( ) A A B ( Hence true for n ) ( ) k k k. k. k k A A A M ( ) k k k k k ( ) ( ) ( ) k k k k 5k k k k M Dep (( k ) ( k )) A ( Hence, if result is true for n k, then it is true for n k ). By Mathematical Induction, above implies true for all positive integers. cso A cso (5) [5 marks] [FP June 6 Qn ] 55. (a) ( λ)( λ) M ( )( ) 5 6 M λ λ λ λ λ, λ both A () (b) M B B () 6 (c) M for either value ( hence is an eigenvalue of M ) M A FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

48 ( hence is an eigenvalue of ) A M () (d) Using eigenvalues. y y y y M A. y y y y M A () Alternative to (c), using characteristic polynomial of ( λ)( λ) 6 6 Leading to ( )( ) Alternative to (d) M M [] 6λ 5λ λ λ λ, A, A (). m m m, m m both M m m m Leading to ( )( ) m m m m m, M y y both A (), A [FP June 6 Qn 5] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

49 i j k b a c a M (a) ( ) ( ) 5i jk AAA () Allow M A for negative of above (b). ( ) ( ).( ) r i j k i j k i j k or equivalent M ( ) r.i j k 7 or multiple A () (c) Let λ, z λ, then y 7 ( ) y λ λ λ, y and z in terms of a single parameter M The direction of l is any multiple of ( ) i j k M ( ( )) ( ) r j k ij k or equivalent M A () Possible equivalents are ( ( )) ( ) The general form is r i k i j k and ( ( )) ( ) r i j i j k { c ( ) } c( ) r i k i j k i j k ( ) ( ) i j k. ij k M (d) λ ( λ) ( λ) λ λ 6 λ Alternative to (d) Leading to λ 9 M A ( ) P :,, A () [] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

50 ( ) ( ) ( ) ( 9 6 7) OP λ λ λ λ λ M d dλ ( OP ) λ M A 9 ( ) P :,, A () [FP June 6 Qn 7] ( ) 9 B d arcosh ( ) 9 M Aft ( ) ft their completing the square, requires arcosh 5 arcosh arcosh arcosh ( ) ln ln ln 9 M A (5) [5] Alternative 5 ( ) 9 B d Let secθ, secθ tanθ dθ secθ tanθ d dθ ( ) 9 M 9sec θ 9 ( ) ( ) secθ dθ Aft 5 arcsec ( ) 5 ln secθ tanθ ln ln arcsec M A (5) [FP June 7 Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

51 58. (a) y E D O 5 (b) Using b a ( e ) One ellipse centred at O B Another ellipse, centred at O, touching on y-ais B Intersections: At least, 5, and shown correctly B (), or equivalent, to find e or ae for D or E. M For S: a 5 and b, For T: a and b, e, ae ignore sign with ae A 5 5 e, ae 5 ignore sign with ae A ( ) ST 6 5 M A (5) [8] [FP June 7 Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

52 d 59. y d B dy d d M d d d d M A 6 ln A ln.5 ln 8 ln ln M A (7) 5 a, b 8 (7 marks) [FP June 7 Qn ] 6. (a) A A B B A A B B e e e e e e e e cosh Acosh B sinh Asinh B M A B A B AB A B A B A B AB A B ( e e e e e e e e ) AB ( AB) A B AB e e ( e e ) cosh( A B) cso M A () (b) cosh cosh sinh sinh sinh M cosh cosh sinh ( sinh) cosh tanh sinh M e e e e e tanh cso M A () e e e e e e [7] Alternative for (b) ( ) e e e e M FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

53 Leading to e e e M e tanh e e e e e (e ) e cso M A () e e e e e (e ) e e [FP June 7 Qn ] 6. (a) 8 n I n n (8 ) n (8 ) d M A n n (8 ) d ft numeric constants only Aft n n n n (8 )(8 ) d n 8(8 ) d n (8 ) d M A n In 6nIn nin In In cso A (6) n (b) ( ) d (8 ) 8 M A I 8 ( 5)(8 ) d 5 M I I I I I, I 8 I 8 I 576 I The previous line can be implied by I I 5I I I 5 (.) M A A (6) ( marks) [FP June 7 Qn 6] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

54 d arsinh M A dy d accept equivalents A () 5 6. (a) ( ) d ( ) ( ) At, (b) θ sinh, d sinhθ coshθ dθ arsinh d θ sinhθcoshθdθ M A θ cosh θ cosh θ θ sinh θdθ dθ sinh θ... M M A A arsinh θcosh θ sinh θ ( ) attempt at substitution M θ sinh θ sinhθcoshθ 5 arsinh ( 8 ) M A 9 ln ( 5 ) 5 A () [] Alternative for (a) dy sinh, sinh cosh d M dy d sinh ycosh y sinh y ( sinh y ) ( ) A At, dy d 5 accept equivalents A () An alternative for (b) is given on the net page FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

55 6. Alternative for (b) arsinh d arsinh d ( ) M A A arsinh d ( ) d Let sinh θ, sinhθ coshθ dθ sinhθ d sinhθ coshθdθ M A ( ) coshθ cosh θ sinh θ sinh θ dθ d θ, θ M, M arsinh arsinh sinh θ sinhθcoshθ 5 θ θ arsinh M A 5 9 arsinh d arsinh arsinh ln ( 5) 5 A () The last 7 marks of the alternative solution can be gained as follows d Let tan θ, tanθ sec θ dθ tanθ d tanθ sec θdθ dependent on first M M A ( ) secθ secθ tan θdθ sec θ tan θ tan θ d θ sec θ tan θ sec θ d θ ( ) M secθ tanθ secθ( tan θ) dθ secθ tan θdθ secθ tanθ secθdθ Hence secθ tanθ ln ( secθ tanθ) M [ arctan... ] 5 ln ( 5 ) M A 9 arsinh d arsinh 5 ln ( 5) ln ( 5) 5 A [FP June 7 Qn 7] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

56 p 6. (a) q. λ Third row λ λ M A () (b) p p q. q M A First row p p Method for either M Second row q q Both correct A ft () (c) l. m n l m n lm n l m n Obtaining linear equations M l m 6 l m 8 Reducing to a pair of equations and M solving for one variable l, m, n Solving for all three variables. M A () [] Alternative to (c) A 5 8 M M M A () [FP June 7 Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

57 6. (a) AB i j k, AC j k i j k AB AC 6i jk any two B M A A Give A for any two components correct or the negative of the correct answer. () (b) Cartesian equation has form y z p (,, ) 6 p or use of another point M y z 7 or any multiple A () 5 (c) Parametric form of line is r 5 λ or equivalent form M A Substituting into equation of plane 5 ( λ) ( 5 λ) ( λ) 7 M Leading to λ A T :,8,9 A (5) ( ) (d) AT i 9j 9 k, BT i 6j 6k both M These are parallel and hence A, B and T are collinear (by the aiom of parallels) M A () [] Alternative to (d) r i j μ i j k or equivalent The equation of AB: ( ) i: μ μ M μ OT i 8j 9k M Hence A, B and T are collinear cso A () Note: Column vectors or bold-faced vectors may be used at any stage. [FP June 7 Qn 7] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

58 65. [FP June 8 Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

59 66. [FP June 8 Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

60 67. [FP June 8 Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

61 68. [FP June 8 Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

62 69. [FP June 8 Qn 5] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

63 7. [FP June 8 Qn 6] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

64 7. [FP June 8 Qn 7] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

65 7. [FP June 8 Qn ] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

66 7. [FP June 8 Qn 7] FP question mark schemes from old P, P5, P6 and FP, FP, FP papers Version. March 9

FP2 Mark Schemes from old P4, P5, P6 and FP1, FP2, FP3 papers (back to June 2002)

FP2 Mark Schemes from old P4, P5, P6 and FP1, FP2, FP3 papers (back to June 2002) FP Mark Schemes from old P, P5, P6 and FP, FP, FP papers (back to June 00) Please note that the following pages contain mark schemes for questions from past papers. The standard of the mark schemes is