C4 June 2008 mark scheme.pdf

EDEXCEL CORE MATHEMATICS C4 (6666) – JUNE 2008PROVISIONAL MARK SCHEMEQuestionNumberSchemeMarksx 0 0.4 0.8 1.2 1.6 2y00.080.320.721.2821. (a) e e e eeeB1 (1)or y 1 1.08329… 1.37713… 2.05443… 3.59664… 7.38906…(b)1Area 0.4 ; e 2 e e e e e20 0.08 0.32 0.72 1.28 2B1; M10.2 24.61203164... 4.922406... 4.922 (4sf) A1 (3)(4 marks)2. (a)udvdxxexdudxve1xx x xxe dx xe e .1 dx M1 A1xexxe dxx xxee c A1 (3)(b)u x 2xdvdxe2 dudxxvex2 x 2 x xx e dx x e e .2x dx M1 A12 e xxx 2 xe dx2 x x xx e 2 xe e c A1 (3)(6 marks)1

EDEXCEL CORE MATHEMATICS C4 (6666) – JUNE 2008PROVISIONAL MARK SCHEMEQuestionNumber3. (a) From question, d A0.032dtSchemeMarksB12 dAA x 2 xdxB1dx dA dA1 0.0160.032 ;dt dt dx 2 x xM1Whenx 2cm , d x 0.016dt2(b)Hence, d x0.002546479... (cm s -1 ) A1 cso (4)dt2 3V x (5 x) 5 x B1dVdx15x2B1 ft4. (a)dV dV dx0.016dt dx dt xWhenx 2cm ,215 x . ; 0.24dVdt2 23x y xy 4 ( eqn )3 10.24(2) 0.48 (cm s )dy dy dy6x 2y y x 0dx dx dxdy 8 6x y 8dx 3 x 2y3giving 18x 3y 8x 16yxM1A1 (4)(8 marks)M1 B1 A1M1giving 13y 26x M1Hence, 2 2 0y x y x A1 cso (6)(b) At P & Q, y 2x. Substituting into eqngives2 23 x (2 x) x(2 x ) 4M1Simplifying gives,2x 4 x 2A1y 2x y 4, hence coordinates are (2,4) and ( 2, 4) A1 (3)(9 marks)2

EDEXCEL CORE MATHEMATICS C4 (6666) – JUNE 2008PROVISIONAL MARK SCHEMEQuestionNumber5. (a)Scheme1 12 21 11 3x1 3x22(4 3 x) 4 1 1(4 3 x)4 2 4MarksB1( )( )1 31 12 221 ( )(** x); (** x ) ...2 2with ** 12!M1; A1 ft1 31 (3 2)( 12)1 (3 22 )(x4 ) (x4 ) ...2 2!1 3 27 21 ; ...2 8 128x x A1 A1 (5)(b)1 3 27x x x M12 16 2562( 8) ...x1 32 16x3 27 24 x x .....2 322.....M133x x A1; A1 (4)3224 2 ; ...(9 marks)3

EDEXCEL CORE MATHEMATICS C4 (6666) – JUNE 2008PROVISIONAL MARK SCHEMEQuestionNumber6. (a) Lines meet where:Scheme9 2 3 30 1 1 110 1 17 5MarksAny two ofi : 9 2 3 3 (1)j: 1 (2)k : 10 17 5 (3)M1(1) – 2(2) gives: 9 1 5 2 M1(2) gives: 1 2 3 A19 2 3 3r 0 3 1 or r 1 2 1M110 1 17 5Intersect at3r 3 or r 3i 3j 7k A17Either check k:3 : LHS 10 10 3 72 : RHS 17 5 17 10 7B1 (6)(b) d12i j k , d23i j 5k2As d 1d 1 1 (2 3) (1 1) ( 1 5) 02 31 5M1 A1 (2)Then l 1 is perpendicular to l 2 .(c) Equating i ; 9 2 5 79 2 5r 0 7 1 7 ( OA.Hence the point A lies on l 1 .) B1 (1)10 1 3(d) Let OX 3i 3j 7k be point of intersection3 5 8AX OX OA 3 7 4M1 ft7 3 4OB OA AB OA 2AX4

EDEXCEL CORE MATHEMATICS C4 (6666) – JUNE 2008PROVISIONAL MARK SCHEMEQuestionNumberSchemeMarks5 8OB 7 2 4M1 ft3 4Hence,11OB 1 or OB 11i j 11k A111(12 marks)2 2A B7. (a) 24 y (2 y)(2 y) (2 y) (2 y)so 2 (2 ) (2 )A y B y M1Let 2,2 B 4 B , Let 2,1y22 A 4 A M11y2giving1 12 2(2 y) (2 y)A1 cao (3)(b) 2 dy1 dx24 y cot xB11 12 2 d y(2 ) (2 )tan x d xy yln(2 y) ln(2 y) ln(sec x)c B1 M1 A1 ft1 12 2y 0, x3ln 2 ln 2 ln c M11 1 12 2 cos30 ln 2 c c ln 2ln(2 y) ln(2 y) ln(sec x ) ln 21 12 21 2 y sec xln ln2 2 y 2M1ln2 y sec x2ln2 y 2ln2 y sec xln2 y 22M12 y2sec x2 y 4Hence,2 8 4ysec x2 yA1 (8)(11 marks)5