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

Transcription

1 FP Mar Schemes from old P4, P5, P6 and FP, FP, FP papers (bac to June 00) Please note that the following pages contain mar schemes for questions from past papers which were not written at an AS standard and may be less accessible than those you will find on future AS FP papers from Edexcel. Some questions would certainly worth more mars at AS level. The standard of the mar 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. FP question mar schemes from old P4, P5, P6 and FP, FP, FP papers Version March 009

2 [P4 January 00 Qn ]. [P4 January 00 Qn ] FP question mar schemes from old P4, P5, P6 and FP, FP, FP papers Version March 009

3 . [P4 January 00 Qn 4] 4. [*P4 January 00 Qn 5] 5. Σ6r Σ6 = n(n + )(n + ), 6n, A = n(n + n 5) = n(n )(n 5) (*) A (4 mars) [P4 June 00 Qn ] FP question mar schemes from old P4, P5, P6 and FP, FP, FP papers Version March 009

4 6. (a) w = 50 (or equivalent) B () (b) B A B () (c) OA = 5 B 4 BA = BA = 5, isosceles, A = ( 50), right-angled (or gradient method), A (5) (d) arg z = arg z arg w w = ( ) AOB = 4 π, A () (0 mars) [P4 June 00 Qn 5] dy 4 dy 7. = ; at x = p =, A dx x dx p Equation of tangent at P, y = ( x p) p p 4 ( y = x +, p y + x = 4 p etc) p p At Q q y + x + 4q Two correct equations in any form A (p q )y = 4(p q) y = 4 p + q ( ) A x = 4 p 4 p = p + q 4 pq p + q ( ), A (8) [*P5 June 00 Qn 7] FP question mar schemes from old P4, P5, P6 and FP, FP, FP papers Version March 009 4

5 FP question mar schemes from old P4, P5, P6 and FP, FP, FP papers Version March [P6 June 00 Qn 6] 8. (a) For n = = 57, which is divisible by, A Assume true for n = ( + )th term is B ( + )th term ± th term = ± = + ( ± ) (5 ± ), A = 6( ) +. + or = 4( ) +. + which is divisible by ( + )th term is divisible by A Thus by induction true for all n cso B (9) (b) For n = RHS = 4 9 B Assume true for n = = + = A///0 (- each error) = ) ( ) 9( ) ( ) ( B If true for then true for + by induction true for all n B (7) (6 mars)

6 9. f() =.54 B α.4 f(π) =.4 B π.54 π α α =.4.54 π = ( )α α =.65 A (4) [*P4 January 00 Qn 4] 0. (a) z = ( i)( i) = 8i A () (b) ( + i) = = z ( i)( + i) + i 8 (c) z = (9 + 9) = 8 = = + i 6 z = 8 two correct A () = = z 8 = 6 all three correct A () (d) O C D E de xc A two correct four correct B B () B (e) OB OD OA =8, = OC / 6 = 8 A AOB = COD = 45 similar B () ( mars) [P4 January 00 Qn 6] FP question mar schemes from old P4, P5, P6 and FP, FP, FP papers Version March 009 6

7 . (a) x 7 = (x )(x + x + 9) (x = is one root). Others satisfy (x + x + 9) = 0 (*) A () (b) Roots are x = B and x = ± 9 6 (c) = + i, i A () and one other root in B correct quad B ft () Root in complex conjugate posn. (7 mars) [#P4 June 00 Qn ]. (a) f() = 4 f() = Alt to (a) Change of sign (and continuity) implies α (, ) A () (b) f(.5) =..5 < α < B f(.75) = < α < f(.875) = < α < NB Exact answer is.8789 y y = x 9 8 y = x Two graphs with single point of intersection (x > 0) Two calculations at both x = and x = B () A O x [*P4 June 00 Qn 4] FP question mar schemes from old P4, P5, P6 and FP, FP, FP papers Version March 009 7

8 . (a) 4 + i + 4i z = ( 4 + i)( 4i) = = 0 + ( ), = 5 0 0i 0 ( = i ) awrt., accept exact equivalents, A () (b) ( a + i)( ai) ( + ai)( ai) = 5a + (6 a 4 + a )i accept in (a) if clearly applied to (b) π (tan 4 =) = 6 a 5a a + 5a 6 = (a + 6)(a ) a = 6, obtaining quadratic or equivalent A A π Reject a = 6, wrong quadrant/ 4, one value A (6) Alt. (a) 4 + i = 5, + 4i = 0 z = 5 0 ( = (b) arg z = arg (a + i) arg ( + ai) π a arctan arctan 4 = a = a a + (9 mars) 5 A ) () leading to a + 5a 6 = 0, then as before A () [P4 June 00 Qn 5] FP question mar schemes from old P4, P5, P6 and FP, FP, FP papers Version March 009 8

9 4. [P6 June 00 Qn ] 5. (a) z = w =, A wz = ( ) = 6, A π 5π arg z = 4 arg w = 6 ; arg wz 5π = 4π 4π + 6 = π, 60, A (6) ALT z = 8i; z w = i, A z w = = 6 A arg z w = tan = π A (b) B C Points A and B Point C B Bft A 5π π angle BOC = 6 = π, 90 A (4) (0 mars) [P4 January 004 Qn ] FP question mar schemes from old P4, P5, P6 and FP, FP, FP papers Version March 009 9

10 6. [P4 June 004 Qn ] 7. f( ) = and ( ) f = B α = α = B α () [*P4 June 004 Qn ] 8. [P4 June 004 Qn ] FP question mar schemes from old P4, P5, P6 and FP, FP, FP papers Version March 009 0

11 9. [P4 June 004 Qn 5] FP question mar schemes from old P4, P5, P6 and FP, FP, FP papers Version March 009

12 0. dy c dy = dt = t = dx dx c t dt The normal to the curve has gradient t. c The equation of the normal is y = t ( x ct) t c The equation may be written y = t x + ct t A B A (5) [*P5 June 004 Qn 8]. (a) z = ( ) + ( ) =, w ( ) z w = + = 4 z 4 = = = A () w z + i + i + i (b) = = w i i + i = ( i( ) ) 4 arctan = 5 + z In correct quadrant arg = 65 w π A () (c) A A B B B C ft Bft () (d) DOB C = 60 Correct method for B AOC B AOC = = 60 A () (e)! AOC = 4 sin 60 = awrt.46 A () [#FP/P4 January 005 Qn 8] FP question mar schemes from old P4, P5, P6 and FP, FP, FP papers Version March 009

13 . (a) z + 6z+ 0 = ( z+ )( z z+ 0) Long division or any complete method ± ( 4 40) z = = ± i A () (b) A B In correct quadrants and on negative Bft () x - axis only required C (c) mab = =, mac = Full method mabm AC = triangle is right angled A () (6) Alternative to (c) AB + AC = 8 = 8 = 6 = BC Result follows by (converse of) Pythagoras, or any complete method. A [#FP/P4 June 005 Qn ]. f (.) = 0.97 ( ) f (.) = , f (.5) =.05 Attempt at f (. ), f (.5 ) f. to sf or better α. A () B [*FP/P4 June 005 Qn 4] FP question mar schemes from old P4, P5, P6 and FP, FP, FP papers Version March 009

14 4. [FP/P4 June 005 Qn 5] 5. [*FP/P5 June 005 Qn 5] FP question mar schemes from old P4, P5, P6 and FP, FP, FP papers Version March 009 4

15 6. [FP/P6 June 005 Qn ] 7. [FP/P4 January 006 Qn ] FP question mar schemes from old P4, P5, P6 and FP, FP, FP papers Version March 009 5

16 8. [FP/P4 January 006 Qn ] 9. (a) f (.8) = = 0. awrt ± 0. B f ( ) = = awrt ± 0.64 B α.8 α = "0." "0.64" or equivalent α.87 cao A (4) (b) (min) ( hr 5 min) B () (5) [#*FP/P4 January 006 Qn 5] FP question mar schemes from old P4, P5, P6 and FP, FP, FP papers Version March 009 6

17 0. [*FP/P5 Jan 006 Qn 9]. [*FP/P6 Jan 006 Qn ] FP question mar schemes from old P4, P5, P6 and FP, FP, FP papers Version March 009 7

18 . [*FP/P6 January 006 Qn 5]. (a) z+ iw= i z iw= i Adding z+ i z = 4+ i Eliminating either variable 4+ i z = + i 4+ i i z = + i i i+ 6i = 5 A = + i A (4) (b) arg z = π arctan.0 cao A () (6 mars) [FP June 006 Qn ] FP question mar schemes from old P4, P5, P6 and FP, FP, FP papers Version March 009 8

19 4. (a) f ( 0.4) 0.058, f ( 0.8) = accept sf Change of sign (and continuity) α ( 0.4, 0.8) A () ( ) (b) f ( 0.6) 0.07 α ( 0.4, 0.6) accept sf ( ) ( ) α ( ) f , 0.6 ( ) α ( ) f , 0.6 A () (5) [*FP June 006 Qn 6] 5. [FP Jan 007 Qn ] FP question mar schemes from old P4, P5, P6 and FP, FP, FP papers Version March 009 9

20 6. [FP Jan 007 Qn ] FP question mar schemes from old P4, P5, P6 and FP, FP, FP papers Version March 009 0

21 7. [FP June 007 Qn 4] FP question mar schemes from old P4, P5, P6 and FP, FP, FP papers Version March 009

22 8. [FP June 007 Qn 6] FP question mar schemes from old P4, P5, P6 and FP, FP, FP papers Version March 009

23 ap aq 9. (a) Gradient of PQ = = Can be implied ap aq B p + q Use of any correct method or formula to obtain an equation of PQ in any form. p + q y = x + apq A () Leading to ( ) ( ) (b) Gradient of normal at P is p. Given or implied at any stage B Obtaining any correct form for normal at either point. A Allow if just written down. y + px = ap + ap y + qx = aq + aq Using both normal equations and eliminating x or y. Allow in any unsimplified form. ( p q) x = a( p q) + a( p q ) Any correct form for x or y A Leading to x = a( p + q + pq + ) cso A y = apq( p + q) cso A (7) [*FP June 007 Qn8] FP question mar schemes from old P4, P5, P6 and FP, FP, FP papers Version March 009

24 40. n = : = B (Hence result is true for n =.) hypothesis + ( r ) = ( r ) + ( + ) r= r= = ( )( ) ( ) + + +, by induction = ( )( ) = ( )( ) = ( + )( + )( + ) A = ( + )[( + ) ][( + ) + ] (Hence, if result is true for n=, then it is true for n= +. ) + By Mathematical Induction, above implies the result is true for all n. cso A (5) (5 mars) [FP June 007 Qn5] FP question mar schemes from old P4, P5, P6 and FP, FP, FP papers Version March 009 4

25 f + f = (a) ( ) ( ) Note: f( ) f( ) ( ) ( ) = = can be implied A ( ) = + = + Hence 5 f ( ) f ( ) + cso A (4) + is divisible by 40 and other appropriate multiples of 5 lead to the required result. (b) 4 6 n = : f ( ) f ( ) = + = = B (Hence result is true for n =.) From (a) f ( + ) f ( ) = 5 λ, say. By induction hypothesis ( ) f( ) f( ) λ 5( µ λ) 5 f ( ) f = 5 µ, say. + = + = + + (Hence, if result is true for n=, then it is true for n= +. ) + By Mathematical Induction, above implies the result is true for all n. Accept equivalent arguments cso A () (7 mars) [*FP June 007 Qn6] FP question mar schemes from old P4, P5, P6 and FP, FP, FP papers Version March 009 5

26 4. [FP January 008 Qn ] 4. ( ) ( x) 0.8 α f ( 0.8) Using = α 0.7 f ( 0.7) f 0.7 = , f 0.8 = dp or better B, B ( ) ( ) ( ) ( ) 0.8f f 0.8 to obtain α = + f 0.8 f 0.7 α = dp or better A (4) [*FP January 008 Qn 4] FP question mar schemes from old P4, P5, P6 and FP, FP, FP papers Version March 009 6

27 44. [FP January 008 Qn 6] FP question mar schemes from old P4, P5, P6 and FP, FP, FP papers Version March 009 7

28 45. [FP June 008 Qn ] FP question mar schemes from old P4, P5, P6 and FP, FP, FP papers Version March 009 8

29 46. [FP June 008 Qn ] FP question mar schemes from old P4, P5, P6 and FP, FP, FP papers Version March 009 9

30 47. [FP June 008 Qn 5] FP question mar schemes from old P4, P5, P6 and FP, FP, FP papers Version March 009 0