Mark Scheme (Results) June GCE Core Mathematics C4 (6666) Paper 1

Transcription

1 Mark Scheme (Results) June 0 GCE Core Mathematics C4 (6666) Paper

2 Edexcel is one of the leading examining and awarding bodies in the UK and throughout the world. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. Through a network of UK and overseas offices, Edexcel s centres receive the support they need to help them deliver their education and training programmes to learners. For further information, please call our GCE line on or visit our website at If you have any subject specific questions about the content of this Mark Scheme that require the help of a subject specialist, you may find our Ask The Expert service helpful. Ask The Expert can be accessed online at the following link: June 0 Publications Code UA0766 All the material in this publication is copyright Edexcel Ltd 0

3 EDEXCEL GCE MATHEMATICS General Instructions for Marking. The total number of marks for the paper is 75.. The Edexcel Mathematics mark schemes use the following types of marks: M marks: method marks are awarded for knowing a method and attempting to apply it, unless otherwise indicated. A marks: Accuracy marks can only be awarded if the relevant method (M) marks have been earned. B marks are unconditional accuracy marks (independent of M marks) Marks should not be subdivided.. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes and can be used if you are using the annotation facility on epen. bod benefit of doubt ft follow through the symbol will be used for correct ft cao correct answer only cso - correct solution only. There must be no errors in this part of the question to obtain this mark isw ignore subsequent working awrt answers which round to SC: special case oe or equivalent (and appropriate) dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper The second mark is dependent on gaining the first mark

4 Number June 0 FINAL Core Mathematics C Mark Scheme Scheme Marks. 9x A( x )( x ) B( x ) C( x ) = B x 9= B B = M x 9 = C C = 4 Any two of A, B, C A x terms 9= A+ C A= 4 All three correct A (4) [4] Alternatives for finding A. x terms 0= A+ B C A= 4 Constant terms 0= A+ B+ C A= 4 GCE Core Mathematics C4 (6666) June 0 4

5 Number Scheme Marks. f x = M ( kx )... = n nkx..., or + = + + n not a natural number, k ( kx ) ( kx ) + =... + ft their k 9 B M A ft x = x + x A f = A (6) [6] 4 ( x) x x GCE Core Mathematics C4 (6666) June 0 5

6 Number Scheme Marks. (a) dv dh h h = π π or equivalent M A h =, At 0. d V = π 0. π 0. = 0.04π dh π 5 M A (4) dh dv dv π (b) = = dt dt dh 800 π h π h or π their (a) 800 M At h = 0., h π = 5 = dt 800 π awrt 0.0 A () [6] GCE Core Mathematics C4 (6666) June 0 6

7 Number 4. Scheme (a) 0.0,.596 awrt 0.0,.596 Marks B B () (b) Area ( R) [...]. (c) B ( ) +.90 M.0 Accept A () du u = x + = x B Area ( R) = x ln ( x + ) 0 B x ln ( + ) d = x ln ( + ) d = ( )( ln ) d M Area R ln d u u u 4 Hence = ( ) cso (d) A (4) u u u lnudu = u lnu u du M A u u u = u lnu du u u = u lnu u + C 4 M A u u Area ( R) = u ln u u 4 = ( 8 8) ln ( ( 4) ln + 4) 4 M = ln+ ln + A (6) [5] GCE Core Mathematics C4 (6666) June 0 7

8 Number Scheme Marks 5. dy y = B = ln x+ x M A x At x =, ln y = ()ln M leading to y = 6 Accept 4ln y = e A At (,6 ) Alternative dy ln 6 M dy 6( ln ) A (7) [7] ln y = e x x B d ( xlnx) lnx x = + M A x dy = lnx x e + x xlnx M A x =, 4ln At dy ln e = + M = 6( + ln ) A (7) GCE Core Mathematics C4 (6666) June 0 8

9 Number Scheme Marks 6. (a) i: 6 λ = 5+ µ j: + λ = 5 µ Any two equations M leading to λ =, µ = 4 M A 6 5 r = + = or r = = 7 7 M A RHS = + 4 = 7 B (6) k: LHS 7 = + =, (As LHS = RHS, lines intersect) Alternatively for B, showing that λ = and µ = 4 both give 7 (b). 6 4 = + = 4 cosθ ( θ 0.9 ) M A Acute angle is 69. awrt 69. A () (c) 6 5 = + = r ( B lies on l) (d) Let d be shortest distance from B to l 5 uuur AB = = uuur AB d 56 ( ( 4) ( 6) ) = + + B () M = 56 awrt 7.5 A = sinθ M A d = 56sin awrt 6.99 A (4) [4] θ B d l l GCE Core Mathematics C4 (6666) June 0 9

10 Number Scheme Marks 7. (a) tanθ = or sinθ = M π θ = awrt.05 A () (b) sec θ dθ =, d y cosθ dθ = dy cosθ = ( = cos θ ) M A sec θ At P, π m = cos = 8 Can be implied A Using mm =, m = 8 M For normal y = 8( x ) M At Q, 0 y = = 8( x ) 7 x = leading to 7 6 k =.065 A (6) 6 (c) dθ y = y dθ = sin θsec θdθ M A = tan θ dθ A ( sec θ ) = dθ M = tanθ θ + C A π π [ ] π d tan M = π π ( p, q ) V = π y x= θ θ = π = = A (7) [5] GCE Core Mathematics C4 (6666) June 0 0

11 Number Scheme Marks 8. (a) ( 4y ) ( 4)( ) ( = 4 y + + C) + 4y+ = + C M A () (b) dy = ( 4y+ ) x + = 4y dy x 4y+ = ( + C) M x B Using (,.5) C leading to C = 4y + = + x 4y M A + = M x y = 4 x 4 or equivalent A (6) [8] GCE Core Mathematics C4 (6666) June 0

12 Further copies of this publication are available from Edexcel Publications, Adamsway, Mansfield, Notts, NG8 4FN Telephone Fax Order Code UA0766 June 0 For more information on Edexcel qualifications, please visit Pearson Education Limited. Registered company number 8788 with its registered office at Edinburgh Gate, Harlow, Essex CM0 JE