Mark Scheme (Results) June GCE Further Pure FP2 (6668) Paper 1

Transcription

1 Mark (Results) June 0 GCE Further Pure FP (6668) 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 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 UA07968 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) 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 June 0 Further Pure Mathematics FP 6668 Mark. x= ( x 4)( x+ ) x 4x = 0 x=, x= 6 both Other critical values are x=, x= 0 B, B < x<, 0< x< 6 st for ±( x 4x ) =0 not required. B marks can be awarded for values appearing in solution e.g. on sketch of graph or in final answer. nd for attempt at method using graph sketch or +/- If cvs correct but correct inequalities are not strict award A0. (7) 7 GCE Further Pure Mathematics FP (6668) June 0

5 . (a) y dx x d y dy dy x dy = e y y e y y dx dx dx dx d y dx x d y dy dy = e y y y dx dx dx d (k = 4) () d y = = 6 dx 0 (b) e ( ) (a) (b) 0 d y 0 e = ( dx ) x 0x y = + x+ + = + x+ x + 5x 6 st for evidence of Product Rule st for completely correct expression or equivalent nd for correct expression or k = 4 stated nd require four terms and denominators of and 6 (might be implied) follow through from their values in the final answer. B B ft (4) 7 GCE Further Pure Mathematics FP (6668) June 0

6 . dy y ln x + 5 = dx x x 5 Integrating factor e x 5 x 5ln x 5 e = e = x 4 x ln x x x ln xdx = dx x ln x x = ( + C) x ln x x ln x C x y = + C y = x 6x x st for attempt at correct Integrating Factor st for simplified IF nd ln x for times their IF to give their x ln x x rd for attempt at correct Integration by Parts nd for both terms correct rd constant not required 4 th x 5 y = their answer + C (8) 8 GCE Further Pure Mathematics FP (6668) June 0

7 4. (a) ( r ) ( r) ( r) ( r) + = (b) ( r ) ( r) ( r) ( r) ( r ) ( r ) r (c) A= 8, B=, C = 6 = + + = 4 + (*) cso r = : = 4 + r = : 5 = 4 + : : ( ) ( ) r = n: n+ n = 4 n + () () (a) (b) (c) Summing: (n+ ) = 4 r + ( ) ( ) = n B n r = n n+ n+ cso 6 Proceeding to ( )( ) r= st require coefficients of,,, or equivalent st require,-,,- or equivalent st for attempt with at least, and n if summing expression incorrect. RHS of display not required at this stage. st for, and n correct. nd require cancelling and use of 4r + Award B for correct kn for their approach nd is for correct solution only (5) 9 GCE Further Pure Mathematics FP (6668) June 0 4

8 5. (a) x ( y ) + = 4 () (b) : Sketch of circle : Evidence of correct centre and radius (c) ( x+ i y) + i x+ i( y+ ) w = = + i( x + i y) ( y) + ix ( ) ( ) ( ) ( ) x + i y+ y ix = y + ix y ix On x-axis, so imaginary part = 0: ( y )( y) x + = 0 ( y + )( y) x = 0 x + ( y ) = 4, so Q is on C cso () (5) 9 Alt. (c) Let w= u+ iv: z + i u = + iz (since v = 0) u i z = ui d u i i u ( u i) z i = = ui ui u + z i = =, so Q is on C u + cso (a) Use of z = x + iy and find modulus (b) Award A0 if circle doesn t intersect x - axis twice (c) st M for subbing z = x + iy and collecting real and imaginary parts nd M for multiply numerator and denominator by their complex conjugate rd M for equating imaginary parts of numerator to 0 Award for equation matching part (a), statement not required. GCE Further Pure Mathematics FP (6668) June 0 5

9 6. 5 π + cosθ = θ = B ( ) cos ( 4 4 cos cos ) + θ d = d + θ + θ θ sinθ θ = 4θ 4sinθ Substituting limits 9π π = Area of triangle = ( )( ) cos sin 5 r θ r θ = = 4 π 7 5 π 9 Area of R = + = st for use of r dθ and correct attempt to expand nd for use of double angle formula - sin θ required in square brackets rd for substituting their limits 4 th for use of base x height 5 th area of sector area of triangle Please note there are no follow through marks on accuracy. (9) 9 GCE Further Pure Mathematics FP (6668) June 0 6

10 7. (a) sin 5θ Im(cosθ i sin θ) cos θ (i sin θ) + 0 cos θ(i sin θ) + i sin θ 4 5 = i(5cos θ sinθ 0cos θsin θ + sin θ) 5 Im(cosθ + i sin θ) 5 = 5sin θ( sin θ) 0sin θ( sin θ) + sin θ 5 = + B ( ) 5 sin 5 6sin 0sin 5sin θ = θ θ + θ (*) cso (5) (b) 6sin 5 θ 0sin θ 5sinθ 5( sinθ 4sin θ) + = 5 6sin θ 0sinθ = sin θ = 8 θ =.095 Inclusion of solutions from sinθ = Other solutions: θ =.046, 4.7, 5.88 sinθ = 0 θ = 0, θ = π (.4) B (a) Award B if solution considers Imaginary parts and equates to sin 5θ st for correct attempt at expansion and collection of imaginary parts nd for substitution powers of cos θ (b) st M for substituting correct expressions nd M for attempting to form equation Imply rd M if 4.7 or 5.88 seen. Award for their negative root. Ignore π but nd A0 if other extra solutions given. (6) GCE Further Pure Mathematics FP (6668) June 0 7

11 8. (a) m + 6m+ 9= 0 m= C.F. t x = ( A+ Bt) e P.I. x = Pcos t+ Q sin t B x = Psint+ Qcost x = 9Pcost 9Qsint ( P t Q t) ( P t Q t) ( P t Q t) 9 cos 9 sin + 6 sin + cos + 9 cos + sin = cos 9P+ 8Q+ 9P= and 9Q 8P+ 9Q= 0 P = 0 and Q = 8 x = ( A+ Bt) e + sint ft 8 t (b) t = 0: x = A = B t t x& = ( A + Bt)e + Be + cost 8 4 t = 0: x& = A + B + = 0 B = 6 4t t x = + e + sin t 8 (c) 59π t ( 0.9) 6 B x 8 Bft (a) (b) st Form auxiliary equation and correct attempt to solve. Can be implied from correct exponential. nd for attempt to differentiate PI twice rd for substituting their expression into differential equation 4 th for substitution of both boundary values st for correct attempt to differentiate their answer to part (a) nd for substituting boundary value (8) (5) () 5 GCE Further Pure Mathematics FP (6668) June 0 8

12 GCE Further Pure Mathematics FP (6668) June 0 9

13

14 Further copies of this publication are available from Edexcel Publications, Adamsway, Mansfield, Notts, NG8 4FN Telephone Fax Order Code UA07968 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