Mark Scheme (Results) June GCE Further Pure Mathematics FP2 (6668/01) Original Paper

Transcription

1 Mark (Results) June 03 GCE Further Pure Mathematics FP (6668/0) Original Paper

2 Edexcel and BTEC Qualifications Edexcel and BTEC qualifications come from Pearson, the world s leading learning company. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information, please visit our website at Our website subject pages hold useful resources, support material and live feeds from our subject advisors giving you access to a portal of information. If you have any subject specific questions about this specification that require the help of a subject specialist, you may find our Ask The Expert service helpful. Pearson: helping people progress, everywhere Our aim is to help everyone progress in their lives through education. We believe in every kind of learning, for all kinds of people, wherever they are in the world. We ve been involved in education for over 50 years, and by working across 70 countries, in 00 languages, we have built an international reputation for our commitment to high standards and raising achievement through innovation in education. Find out more about how we can help you and your students at: Summer 03 Publications Code N/A All the material in this publication is copyright Pearson Education Ltd 0

3 General Marking Guidance All candidates must receive the same treatment. Examiners must mark the first candidate in exactly the same way as they mark the last. Mark schemes should be applied positively. Candidates must be rewarded for what they have shown they can do rather than penalised for omissions. Examiners should mark according to the mark scheme not according to their perception of where the grade boundaries may lie. There is no ceiling on achievement. All marks on the mark scheme should be used appropriately. All the marks on the mark scheme are designed to be awarded. Examiners should always award full marks if deserved, i.e. if the answer matches the mark scheme. Examiners should also be prepared to award zero marks if the candidate s response is not worthy of credit according to the mark scheme. Where some judgement is required, mark schemes will provide the principles by which marks will be awarded and exemplification may be limited. When examiners are in doubt regarding the application of the mark scheme to a candidate s response, the team leader must be consulted. Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response. 3

4 June 03 Further Pure Mathematics FP 6668 Mark Question (a) y''' + xy'' + y' sin x so y''' xy'' y' sin x M, A, A (b) y ''' so y ''' 3 B d y (c) Substitute to give d Use Taylor Expansion to give x B 3 x y + 3 x+ x... M A () (7). (a) B (V shape) B (Parabola) B (positions correct) (b) Put 4 x x 3 or 4 x x+ 3 M Solve x + x 7 0, to give x + M A Solve x x 0, to give 4+ 4 x M A So < x < B (6) (9) 4

5 Question 3. (a) kcos kx M A f( x) + sinkx () (b) ( sin ) sin cos ( cos ) + kx k kx k kx k kx f ( x) (+ sin kx) M k sin kx k (sin kx+ cos kx) so f ( x) (+ sin kx) f() x k ( sin kx) (+ sin kx) + k * + sinkx and use sin kx + cos kx M Acao 3 k cos kx (c) f B ( x) (+ sin kx) Finds f(0), f (0), f (0) and f (0) M 3 k k 3 Uses MacLaurin Expansion to obtain f(x) 0 + kx x + x... 6 M A (4) Alternative method for (c) Uses ln ( + y) Use sinkx Obtains 3 y y 3 y + + with y sinkx B 3 3 k x kx in ln expansion M k k 3 M A 0 + kx x + x... 6 (4) 5

6 Question 4. + x cotx B IF d x x Obtains ln x + ln sin x M A {So IF xsinx} d sin x (IF y) IF dx x Aft M cosx x M A yxsinx sin xdx dx sin x ( + c ) 4 So y cosec c x cos x+ M A x xsinx (9) 6

7 Question 5 (a) + r ( r+ ) ( r+ ) M A A (b) LHS ( n ) ( n ) + n M + + ( n ) n ( n+ ) + + n ( n+ ) ( n+ ) A + n + n + n n n n ( n+ )( n+ ) M dep n + 3n ( n+ )( n+ ) nn ( + 3) ( n+ )( n+ ) A (4) 7

8 Question 6 Modulus 3 B Argument 5π arctan( ) 3 6 MA 5π 5π π π 5 5 z 3 (cos( ) + isin( )) (cos + isin ) M A OR π nπ π nπ (cos + + isin + ) M 7π 7π (cos + isin ) 30 30, 9π 9π (cos + isin ) 30 30, A 7π 7π (cos + isin ) 30 30, 9π 9π (cos + isin ) A (8) 7(a) Differentiate twice and obtaining dy x x d y x x x λxe + λe and 4λxe + λe + λe dx dx M A 3 Substitute to give λ M A (4) (b) Complementary function is y A B x x e + e M A So general solution is y A B x x e + e + 3 e x x A (7) 8

9 Question 8. (a) B () (b) w(z-i) z+7i so zw ( ) 7i + iw and z M 7i + iw A w So 7i+ iw w M Using w u + iv, ( ) ( 7) ( ) v + u+ u + v M So 3u + 3v + 30u+ 48 0, which is a circle equation A As ( u+ 5) + v 3 So centre is 5 and radius is 3 M A (5) () (8) 9

10 Question Q9 (a) sinθ sinθ 0.5 π 5π θ or 6 6, M A, π 5π Points are (, 6 ) and (, ) A 6 (b) Uses Area θ M ( sin θ ) d (4 8sin θ + ( cos θ))dθ M [(6 θ 8cos θ sin θ ))] + A Uses limits π and 6 π to give 7 3 π or π M A Finds area of sector of circle π 3 B So required area is 8 π M A (8) 0

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