4754A * * A A. MATHEMATICS (MEI) Applications of Advanced Mathematics (C4) Paper A ADVANCED GCE. Friday 14 January 2011 Afternoon


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1 ADVANCED GCE MATHEMATICS (MEI) Applications of Advanced Mathematics (C4) Paper A 4754A Candidates answer on the answer booklet. OCR supplied materials: 8 page answer booklet (sent with general stationery) MEI Examination Formulae and Tables (MF) Other materials required: Scientific or graphical calculator Friday 4 January 0 Afternoon Duration: hour 30 minutes * * A A * * INSTRUCTIONS TO CANDIDATES Write your name, centre number and candidate number in the spaces provided on the answer booklet. Please write clearly and in capital letters. Use black ink. Pencil may be used for graphs and diagrams only. Read each question carefully. Make sure you know what you have to do before starting your answer. Answer all the questions. Do not write in the bar codes. You are permitted to use a scientific or graphical calculator in this paper. Final answers should be given to a degree of accuracy appropriate to the context. INFORMATION FOR CANDIDATES The number of marks is given in brackets [ ] at the end of each question or part question. You are advised that an answer may receive no marks unless you show sufficient detail of the working to indicate that a correct method is being used. The total number of marks for this paper is 7. This document consists of 4 pages. Any blank pages are indicated. NOTE This paper will be followed by Paper B: Comprehension. OCR 0 [T/0/653] OCR is an exempt Charity RP 0F0 Turn over
2 Section A (36 marks) (i) Use the trapezium rule with four strips to estimate + e x dx, showing your working. Fig. shows a sketch of y = + e x. y x Fig. (ii) Suppose that the trapezium rule is used with more strips than in part (i) to estimate + e x dx. State, with a reason but no further calculation, whether this would give a larger or smaller estimate. [] A curve is defined parametrically by the equations x = + t, y = t + t. Find t in terms of x. Hence find the cartesian equation of the curve, giving your answer as simply as possible. [5] 3 Find the first three terms in the binomial expansion of in ascending powers of x. State the 3 (3 x) set of values of x for which the expansion is valid. [7] 4 The points A, B and C have coordinates (, 0, ), (4, 3, 6) and (9, 3, 4) respectively. (i) Show that AB is perpendicular to BC. (ii) Find the area of triangle ABC. 5 Show that sin θ = tan θ. + cos θ OCR A Jan
3 3 6 8 (i) Find the point of intersection of the line r = ( ) + λ ( 3 0) and the plane x 3y + =. 6 (ii) Find the acute angle between the line and the normal to the plane. Section B (36 marks) 7 A particle is moving vertically downwards in a liquid. Initially its velocity is zero, and after t seconds it is v m s. Its terminal (longterm) velocity is 5 m s. A model of the particle s motion is proposed. In this model, v = 5( e t ). (i) Show that this equation is consistent with the initial and terminal velocities. Calculate the velocity after 0.5 seconds as given by this model. (ii) Verify that v satisfies the differential equation dv dt In a second model, v satisfies the differential equation As before, when t = 0, v = 0. dv dt = 0 0.4v. (iii) Show that this differential equation may be written as 0 dv (5 v)(5 + v) dt = 4. Using partial fractions, solve this differential equation to show that = 0 v. t = 4 ln(5 + v ). [8] 5 v This can be rearranged to give v = 5( e 4t ) + e 4t. [You are not required to show this result.] (iv) Verify that this model also gives a terminal velocity of 5 m s. Calculate the velocity after 0.5 seconds as given by this model. The velocity of the particle after 0.5 seconds is measured as 3 m s. (v) Which of the two models fits the data better? [] OCR A Jan Turn over
4 4 8 Fig. 8 shows a searchlight, mounted at a point A, 5 metres above level ground. Its beam is in the shape of a cone with axis AC, where C is on the ground. AC is angled at α to the vertical. The beam produces an ovalshaped area of light on the ground, of length DE. The width of the oval at C is GF. Angles DAC, EAC, FAC and GAC are all β. A A side view 5 m 5 m B D C F B D C E G E Fig. 8 In the following, all lengths are in metres. (i) Find AC in terms of α, and hence show that GF = 0 sec α tan β. (ii) Show that CE = 5(tan(α + β) tan α). Hence show that CE = 5 tan β sec α tan α tan β. [5] Similarly, it can be shown that CD = 5 tan β sec α. [You are not required to derive this result.] + tan α tan β You are now given that α = 45 and that tan β = t. (iii) Find CE and CD in terms of t. Hence show that DE = (iv) Show that GF = 0 t. For a certain value of β, DE = GF. (v) Show that t =. 0t t. [5] [] Hence find this value of β. Copyright Information OCR is committed to seeking permission to reproduce all thirdparty content that it uses in its assessment materials. OCR has attempted to identify and contact all copyright holders whose work is used in this paper. To avoid the issue of disclosure of answerrelated information to candidates, all copyright acknowledgements are reproduced in the OCR Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download from our public website ( after the live examination series. If OCR has unwittingly failed to correctly acknowledge or clear any thirdparty content in this assessment material, OCR will be happy to correct its mistake at the earliest possible opportunity. For queries or further information please contact the Copyright Team, First Floor, 9 Hills Road, Cambridge CB GE. OCR is part of the Cambridge Assessment Group; Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. OCR A Jan
5 4754A Mark Scheme January 0 Section A (i) x y A ½ { ( )} = B,,0 table values formula 6.5 or better www (ii) Smaller, as the trapezium rule is an overestimate in this case and the error is less with more strips [] x t t x t x t y x t x x x x x [5] attempt to solve for t oe substituting for t in terms of x clearing subsidiary fractions (3 x) 3 ( x) 3 ( 3)( 4) ( ( 3)( x) ( x)...) ( x x...) x x Valid for 3 x 3 3 x B,,0 [7] dealing with the 3 correct binomial coeffs,, 8/3 oe cao
6 4754A Mark Scheme January 0 5 4(i) AB 3, BC AB.B C ( 5) 0 5 AB is perpendicular to BC. (ii) AB = ( ( 5) )= 38 BC = ( )= 9 Area = ½ 38 9 = ½ 0 or 6.6 units complete method ft lengths of both AB, BC oe www 5 LHS = sin cos cos sin cos cos sin tan RHS cos one correct double angle formula used cancelling cos s x 83 6(i) y z 6 Substituting into plane equation: ( 8 3) 3( ) = = 5 = 5, = 3 So point of intersection is (,, 3) ft 3 (ii) Angle between and 3 0 ( 3) ( 3) 0 cos 4 0 = ( )0.43 acute angle = 65 allow for a complete method only for any vectors
7 4754A Mark Scheme January 0 Section B 7(i) When t = 0, v = 5( e 0 ) = 0 As t, e t 0, v 5 When t = 0.5, v = 3.6 m s d v t t (ii) 5 ( )e 0e d t 0 v = 0 0( e t ) = 0e t d v 0 v d t (iii) d v 0 0.4v d t 0 d v 00 4v d t 0 d v 4 5 v dt 0 d v 4 (5 v)(5 v) d t * 0 A B (5 v)(5 v) 5 v 5 v 0 = A(5 + v) + B(5 v) v = 5 0 = 0A A = v = 5 0 = 0B B = 0 (5 v)(5 v) 5 v 5 v ( )dv 4 t 5v 5v d ln(5 v) ln(5 v) 4tc when t = 0, v = 0, 0 = c c = 0 5 v ln 4t 5 v 5 v t ln * 4 5 v [8] for both A=,B= separating variables correctly and indicating integration ft their A,B, condone absence of c ft finding c from an expression of correct form (iv) When t, e 4t 0, v 5/ = 5 5( e ) when t = 0.5, t 3.8ms e (v) The first model [] www 3
8 4754A Mark Scheme January 0 8(i) AC = 5sec oe CF = AC tan = 5sec tan GF = CF = 0sec tan * ACtanβ (ii) CE = BE BC = 5 tan( + ) 5 tan = 5(tan( + ) tan ) tan tan 5 tan tan tan tan tan tan tan tan 5 tan tan 5( tan ) tan tan tan 5tan sec * tan tan D [5] compound angle formula combining fractions sec = + tan (iii) sec 45 =, tan 45 = 5t 0t CE t t 0t CD t 0t 0t DE 0 t( ) t t t t tt 0t 0 t * ( )( ) t t t [5] used substitution for both in CE or CD oe for both adding their CE and CD (iv) cos 45 = / sec = GF = 0 tan = 0 t [] (v) DE = GF 0t 0 t t t = / t = / * t = 0.54 = 8.4 invtan t 4
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