FP1 Mar Schemes from old P4, P5, P6 and FP1, 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 FP1 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. 1
[P4 January 00 Qn 1]. [P4 January 00 Qn ]
. [P4 January 00 Qn 4] 4. [*P4 January 00 Qn 5] 5. Σ6r Σ6 = n(n + 1)(n + 1), 6n, A1 = n(n + n 5) = n(n 1)(n 5) (*) A1 (4 mars) [P4 June 00 Qn 1] 6. (a) w = 50 (or equivalent) B1 (1)
(b) B A B1 (1) (c) OA = 5 4 BA = BA = 5, isosceles B1, A1 5 + 5 = ( 50), right-angled (or gradient method), A1 (5) (d) arg z w = arg z arg w = ( ) AOB = 4 π, A1 () (10 mars) [P4 June 00 Qn 5] 7. dy 4 dy 1 = ; at x = p = dx x dx p, A1 Equation of tangent at P, y 1 = ( x p) p p ( y = 1 4 x +, p y + p p x = 4 p etc) At Q q y + x + 4q Two correct equations in any form A1 (p q )y = 4(p q) y = 4 p + q ( ) A1 x = 4 p 4 p 4 pq = p + q p + q ( ), A1 (8) [*P5 June 00 Qn 7] 4
[P6 June 00 Qn 6] 8. (a) For n = 1 5 + 5 = 57, which is divisible by, A1 Assume true for n = ( + 1)th term is +5 + 5 + B1 ( + 1)th term ± th term = +5 + 5 + ± + + 5 +1 = + ( ± 1) + 5 +1 (5 ± 1), A1 = 6( + + 5 +1 ) +. + or = 4( + + 5 +1 ) +. + which is divisible by ( + 1)th term is divisible by A1 Thus by induction true for all n cso B1 (9) (b) For n = 1 RHS = 4 9 1 B1 Assume true for n = + + + = + = + 4 1 9 9 9 1 1 9 1 4 9 1 4 9 1 1 A//1/0 (-1 each error) + + + + + = 1 1) ( 1) 9( 1) ( 1) ( 1 B1 If true for then true for + 1 by induction true for all n B1 (7) (16 mars) 5
9. f() = 1.514 B1 f(π) = 1.14 α B1 π 1.14 1.514 π α 1.14 α = 1.514 π 1.514 + 1.14 = (1.14 + 1.514)α α =.65 A1 (4) [*P4 January 00 Qn 4] 10. (a) z = ( i)( i) = 18i A1 () (b) 1 ( + i) + i 1 + i = = = z ( i)( + i) 18 6 A1 () (c) z = (9 + 9) = 18 = z = 18 two correct 1 1 1 = = = z 18 6 all three correct A1 () (d) C D O two correct B1 A four correct B1 () (e) B. OB OA = 18, = = 18 OD OC / 6 A1 AOB = COD = 45 similar B1 () (11 mars) [P4 January 00 Qn 6] 11. (a) x 7 = (x )(x + x + 9) (x = is one root). Others satisfy (x + x + 9) = 0 (*) A1 () (b) Roots are x = B1 ± 9 6 and x = 6
(c) = + i, i A1 () and one other root in B1 correct quad B1 ft () Root in complex conjugate posn. (7 mars) [#P4 June 00 Qn ] 1. (a) f(1) = 4 f() = 1 Change of sign (and continuity) implies α (1, ) A1 () (b) f(1.5) =. 1.5 < α < B1 f(1.75) = 0.9 1.75 < α < B1 () f(1.875) = 0.0 1.875 < α < Alt to (a) NB Exact answer is 1.8789 y y = x 9 8 y = x + 6 7 Two graphs with single point of intersection (x > 0) Two calculations at both x = 1 and x = A1 O 1 x [*P4 June 00 Qn 4] 1. (a) 4 + + i 4i z = ( 4 + i)( 4i) = = 0 1 5 1 + ( ), = 0 10i 0 ( = 1 1 i ) awrt 1.1, accept exact equivalents, A1 () (b) ( a + i)( ( + ai)( π (tan 4 = ) 1 = ai) ai) = 6 a 5a 5a + (6 a 4 + a )i accept in (a) if clearly applied to (b) obtaining quadratic or equivalent A1 7
a + 5a 6 = (a + 6)(a 1) a = 6, 1 A1 π Reject a = 6, wrong quadrant/ 4, one value A1 (6) (9 mars) Alt. (a) 4 + i = 5, + 4i = 0 z = 5 5 A1 ( = ) 0 () (b) arg z = arg (a + i) arg ( + ai) π a arctan arctan 4 a a a A1 1 = 1 + () leading to a + 5a 6 = 0, then as before [P4 June 00 Qn 5] 8
14. [P6 June 00 Qn ] 15. (a) z = w =, A1 wz = ( ) = 16, A1 π arg z = 4 5π arg w = 6 ; arg wz π = 4 π 4 5π + 6 π =, 60, A1 (6) ALT z = 8i; z w = 8 + 8 i, A1 z w = 8 + 8 = 16 A1 arg z w = tan 1 π = A1 (b) C Points A and B B1 Point C B1ft B 5π angle BOC = 6 = π A π, 90 A1 (4) (10 mars) [P4 January 004 Qn ] 9
16. [P4 June 004 Qn 1] 17. f ( 1) = 1 and ( ) α = α = 1 1 α 1 f = B1 1 B1 () [*P4 June 004 Qn ] 18. [P4 June 004 Qn ] 10
19. [P4 June 004 Qn 5] 11
0. dy c dy 1 = 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 A1 B1 A1 (5) [*P5 June 004 Qn 8] 1. (a) z = ( ) + ( ) = 16, w ( ) z w = 1 + = 4 z 4 = = = A1 () w z + i + i 1 + i (b) = = w 1 i 1 i 1 + i = ( 1 i ( 1) ) 4 1 arctan = 15 + 1 z In correct quadrant arg = 165 w (c) 11π 1 A1 () A B1 B B1 C ft B1ft () (d) DOB = 60 B1 Correct method for AOC AOC = 45 + 15 = 60 A1 () (e) 1! AOC = = awrt.46 A1 () 4 sin 60 [#FP1/P4 January 005 Qn 8] 1
. A (a) z + 6z + 0 = ( z + ) ( z z + 10) Long division or any complete method ± ( 4 40) z = = 1± i C B A1 () (b) B A C In correct quadrants and on negative B1ft (1) x - axis only required (c) mab = = 1, mac = 1 Full method mabm AC = 1 triangle is right angled A1 () (6) Alternative to (c) AB + AC = 18 = 18 = 6 = BC Result follows by (converse of) Pythagoras, or any complete method. A1 [#FP1/P4 June 005 Qn ]. f ( 1.) = 0.97 ( ) f ( 1.1) = 0.4..., f ( 1.15) =.05 Attempt at f ( 1.1 ), f ( 1.15) f 1. to 1sf or better α 1. A1 () B1 [*FP1/P4 June 005 Qn 4] 1
4. [FP1/P4 June 005 Qn 5] 5. [*FP/P5 June 005 Qn 5] 14
6. [FP/P6 June 005 Qn 1] 7. [FP1/P4 January 006 Qn 1] 15
8. [FP1/P4 January 006 Qn ] 9. (a) f ( 1.8) = 19.6686... 0 = 0.1 awrt ± 0. B1 f ( ) = 0.644... = 0.644 awrt ± 0.64 B1 α 1.8 α = "0." "0.64" or equivalent α 1.87 cao A1 (4) (b) 11 (min) (1 hr 5 min) B1 (1) (5) [#*FP1/P4 January 006 Qn 5] 16
0. [*FP/P5 Jan 006 Qn 9] 1. [*FP/P6 Jan 006 Qn ] 17
. [*FP/P6 January 006 Qn 5]. (a) z + i w = 1 i z i w = i Adding z + i z = 4 + i Eliminating either variable z = 4 + i + i A1 z = 4 + i i + i i = 8 + + 4i+ 6i 5 = 1 + i A1 (4) (b) arg z = π arctan.0 cao A1 () (6 mars) [FP1 June 006 Qn 1] 18
4. (a) f ( 0.4) 0.058, f ( 0.8) = 0.089 accept 1sf Change of sign (and continuity) α ( 0.4, 0.8) A1 () (b) f ( 0.6) 0.017 ( α ( 0.4, 0.6) ) accept 1sf f ( 0.5) 0.00 ( α ( 0.5, 0.6) ) f ( 0.55) 0.001 α ( 0.55, 0.6) A1 () (5) [*FP1 June 006 Qn 6] 5. [FP1 Jan 007 Qn 1] 19
6. [FP1 Jan 007 Qn ] 0
7. [FP1 June 007 Qn 4] 1
8. [FP1 June 007 Qn 6]
ap aq 9. (a) Gradient of PQ = = ap aq Can be implied B1 p + q Use of any correct method or formula to obtain an equation of PQ in any form. p + q y = x + apq A1 () Leading to ( ) ( ) (b) Gradient of normal at P is p. Given or implied at any stage B1 Obtaining any correct form for normal at either point. A1 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 A1 Leading to x = a( p + q + pq + ) cso A1 y = apq( p + q) cso A1 (7) [*FP June 007 Qn8]
40. n = 1: + 1 1 = 1 1 B1 (Hence result is true for n = 1.) 1 ( r 1) = ( r 1) + ( + 1) r = 1 r = 1 = 1 ( 1) ( 1) ( 1) + + +, by induction hypothesis = 1 ( + 1 ) ( + 6 + ) = 1 ( + 1 ) ( + 5 + ) = 1 ( + 1 ) ( + ) ( + 1 ) A1 = 1 ( + 1)[( + 1) 1][( + 1) + 1] (Hence, if result is true for n =, then it is true for n = + 1. ) + By Mathematical Induction, above implies the result is true for all n. cso A1 (5) (5 mars) [FP June 007 Qn5] 4
41. (a) ( ) ( ) f + 1 f = + Note: f ( 1) f ( ) 4 + 4 4 + 6 4 4 + ( ) ( ) 4 4 4 + 4 = 1 + 1 4 4 + = 80 + 15 can be implied A1 4 1 4 4 1 4 40 + 15 15 16 + = + = + Hence 15 f ( 1) f ( ) ( ) + cso A1 (4) + is divisible by 40 and other appropriate multiples of 15 lead to the required result. (b) 1 = + = = B1 (Hence result is true for n = 1.) f 1 f 15 λ, f = 5 µ, say. 4 6 n = : f ( 1) 145 5 9 5 f ( 1) From (a) ( + ) ( ) = say. By induction hypothesis ( ) f ( 1) f ( ) 15λ 5 ( µ λ ) 5 f ( 1) + = + = + + (Hence, if result is true for n =, then it is true for n = + 1. ) + By Mathematical Induction, above implies the result is true for all n. Accept equivalent arguments cso A1 () (7 mars) [*FP June 007 Qn6] 5
4. [FP1 January 008 Qn ] 4. ( ) ( x ) 0.8 α f ( 0.8) Using = α 0.7 f ( 0.7) f 0.7 = 0.195 08 497, f 0.8 = 0.97 06 781 dp or better B1, B1 to obtain α = ( ) ( ) ( ) ( ) 0.8f 0.7 + 0.7 f 0.8 f 0.8 f 0.7 α = 0.79 60 991 dp or better A1 (4) [*FP1 January 008 Qn 4] 6
44. [FP1 January 008 Qn 6] 7
45. [FP1 June 008 Qn 1] 8
46. [FP1 June 008 Qn ] 9
47. [FP June 008 Qn 5] 0