C4 June 2008 mark scheme.pdf

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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
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C4 June 2008 mark scheme.pdf

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