Wednesday 18 June 2014 Afternoon

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1 Wednesday 8 June 04 Afternoon A GCE MATHEMATICS (MEI) 4754/0A Applications of Advanced Mathematics (C4) Paper A QUESTION PAPER * * Candidates answer on the Printed Answer Book. OCR supplied materials: Printed Answer Book 4754/0A MEI Examination Formulae and Tables (MF) Other materials required: Scientific or graphical calculator Duration: hour 30 minutes INSTRUCTIONS TO CANDIDATES These instructions are the same on the Printed Answer Book and the Question Paper. The Question Paper will be found inside the Printed Answer Book. Write your name, centre number and candidate number in the spaces provided on the Printed Answer Book. Please write clearly and in capital letters. Write your answer to each question in the space provided in the Printed Answer Book. Additional paper may be used if necessary but you must clearly show your candidate number, centre number and question number(s). Use black ink. HB 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 This information is the same on the Printed Answer Book and the Question Paper. The number of marks is given in brackets [ ] at the end of each question or part question on the Question Paper. 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. The Printed Answer Book consists of 6 pages. The Question Paper consists of 8 pages. Any blank pages are indicated. This paper will be followed by Paper B: Comprehension. INSTRUCTION TO EXAMS OFFICER / INVIGILATOR Do not send this Question Paper for marking; it should be retained in the centre or recycled. Please contact OCR Copyright should you wish to re-use this document. OCR 04 [T/0/653] DC (KN/SW) 7565/5 OCR is an exempt Charity Turn over

2 Section A (36 marks) 3x Express in partial fractions. [5] ^ - xh^4 + x h 3 4 Find the first three terms in the binomial expansion of ^ + xh. State the set of values of x for which the expansion is valid. [5] 3 Fig. 3 shows the curve y = x + ^sin xh for 0 G x G r. 4 3 y O r 4 x Fig. 3 (i) Use the trapezium rule with 4 strips to estimate the area of the region bounded by the curve, the x-axis r and the line x =, giving your answer to 3 decimal places. [4] 4 (ii) Suppose the number of strips in the trapezium rule is increased. Without doing further calculations, state, with a reason, whether the area estimate increases, decreases, or it is not possible to say. [] tan a tan b 4 (i) Show that cos^a + bh = -. [3] sec a sec b - tan a (ii) Hence show that cos a =. [] + tan a (iii) Hence or otherwise solve the equation - tan + tan i = for 0c G i G 80c. [3] i 3t t 5 A curve has parametric equations x = e, y = te. dy (i) Find in terms of t. Hence find the exact gradient of the curve at the point with parameter t =. [4] dx (ii) Find the cartesian equation of the curve in the form determined. b y = ax ln x, where a and b are constants to be [3] OCR /0A Jun4

3 6 Fig. 6 shows the region enclosed by the curve y x 3 = ^ + h and the line y =. y 3 O x Fig. 6 This region is rotated about the y-axis. Find the volume of revolution formed, giving your answer as a multiple of r. [6] Question 7 begins on page 4. OCR /0A Jun4 Turn over

4 4 Section B (36 marks) 7 Fig. 7 shows a tetrahedron ABCD. The coordinates of the vertices, with respect to axes Oxyz, are A( 3, 0, 0), B(, 0, ), C(0, 4, 0) and D(0, 4, 5). z D(0, 4, 5) A( 3, 0, 0) C(0, 4, 0) y O B(, 0, ) x Fig. 7 (i) Find the lengths of the edges AB and AC, and the size of the angle CAB. Hence calculate the area of triangle ABC. [7] (ii) (A) Verify that 4i 3j + 0k is normal to the plane ABC. [] (B) Hence find the equation of this plane. [] (iii) Write down a vector equation for the line through D perpendicular to the plane ABC. Hence find the point of intersection of this line with the plane ABC. [5] The volume of a tetrahedron is 3 area of base height. (iv) Find the volume of the tetrahedron ABCD. [] OCR /0A Jun4

5 5 8 Fig. 8. shows an upright cylindrical barrel containing water. The water is leaking out of a hole in the side of the barrel. h Fig. 8. The height of the water surface above the hole t seconds after opening the hole is h metres, where dh = A h dt and where A is a positive constant. Initially the water surface is metre above the hole. (i) Verify that the solution to this differential equation is The water stops leaking when h = 0. This occurs after 0 seconds. h = a Atk. [3] (ii) Find the value of A, and the time when the height of the water surface above the hole is 0.5 m. [4] Fig. 8. shows a similar situation with a different barrel; h is in metres. h For this barrel, where B is a positive constant. When t = 0, h =. Fig. 8. dh h = B dt ^ + hh, (iii) Solve this differential equation, and hence show that (iv) Given that h = 0 when t = 0, find B. h ^30 + 0h + 6h h = 56-5Bt. [7] Find also the time when the height of the water surface above the hole is 0.5 m. [4] END OF QUESTION PAPER OCR /0A Jun4

6 4754A Mark Scheme June 04 3x A Bx C ( x)(4 x ) x 4 x correct form of partial fractions ( condone additional coeffs eg A B BUT x 4 x ** is M0 ) Ax B Cx D x 4 x * for 3x = A(4 + x ) + (Bx + C)( x) Multiplying through oe and substituting values or equating coeffs at LEAST AS FAR AS FINDING A VALUE for one of their unknowns (even if incorrect) Can award in cases * and ** above Condone a sign error or single computational error for but not a conceptual error Eg 3x = A( x) + (Bx + C)(4 + x²) is M0 3x( x)(4 + x²) = A(4 + x²) + (Bx + C)( x) is M0 Do not condone missing brackets unless it is clear from subsequent work that they were implied. Eg 3x = A(4 + x²) + Bx + C( x) = 4A + Ax² + Bx + C Cx is M0 = 4A + Ax² + Bx Bx² + C Cx is x = 6 = 8A, A = ¾ oe www [SC A = 3/4 from cover up rule can be applied, then the applies to the other coefficients] A B NB A = 3/4 is A0 ww (wrong working) x 4 x x coeffs: 0 = A B B = ¾ oe www constants: 0 = 4A + C C = ½ oe www [In the case of * above, all 4 constants are needed for the final ] Ignore subsequent errors when recompiling the final solution provided that the coeffs were all correct [5]

7 4754A Mark Scheme June (4 x) 4 ( x ) 4 dealing with the 4 to obtain 3/ x 3/ 4 ( ) 4 3/ 3 / 3 / x 4 4 x 4... (or expanding as! and having all the powers of 4 correct) 3 3 4! 4 8( ( x).. ( x ) x +3/6 x² correct binomial coeffs for n = 3/ ie, 3/, 3/././! Not ncr form Indep of coeff of x Indep of first 8 + 3x www + 3/6 x² www Ignore subsequent terms Valid for 4 < x < 4 or x < 4 accept s or a combination of < and, but not 4 > x > 4, x > 4, or say 4 < x condone 4 < x < 4 Indep of all other marks [5] Allow MR throughout this question for n = m/ where m N, and m odd and then MR provided it is at least as difficult as the original.

8 4754A Mark Scheme June 04 3 (i) x y B,,0 For values ,0.679, (4dp or better soi) [accept truncated to 4 figs after dec point] A = ( /3) [( ) + ( )] = (ii) Not possible to say, eg some trapezia are above and some below curve oe. [4] [cannot assume values of form (π/6)³ + (sin π/6) are correct unless followed by correct total at some later stage as some will be in degree mode] Use of the trapezium rule. Trapezium rule formula for 4 strips must be seen, with or without substitution seen. Correct h must be soi. [accept separate trapezia added] www 3dp only (NB using.35 is ww) SC B without any working as no indication of strips or method used SC with some indication of 4 strips but no values seen Correct values followed by scores B B0 Correct values followed by correct formula for 4 strips, with or without substitution seen, then A= scores 4/4. Correct formula for 4 strips and values of form ((π/6)³ + (sinπ/6) followed by correct answer scores 4/4 (or ¾ with wrong dp) NB Values given in the table to only 3dp give apparently the correct answer, but scores B0,A0 ww Need a reason. Must be without further calculation. [] 3

9 4754A Mark Scheme June 04 4 (i) EITHER Use of cos = /sec (or sin= /cosec) From RHS tan tan sec sec sin / cos.sin / cos / cos./ cos sin sin cos cos ( ) cos cos cos cos sin sin Must be used Substituting and simplifying as far as having no fractions within a fraction tt [need more than secsec cc ss ie an intermediate step that can lead to cc ss] cos( ) Convincing simplification and correct use of cos(α + β) Answer given OR From LHS, cos = /sec or sin = /cosec used cos( ) cos cos sin sin sec sec sin sin sec sin sec sin sec sec Correct angle formula and substitution and simplification to one term OR eg cosαcosβ sinαsinβ cos cos ( tan tan ) tan tan sec sec [3] Simplifying to final answer www Answer given Or any equivalent work but must have more than cc ss = answer. 4

10 4754A Mark Scheme June 04 4 (ii) cos tan tan sec tan. β = α used, Need to see sec²α Use of sec²α = + tan²α to give required result Answer Given OR, without Hence, cos cos ( ) sec tan tan ( tan ) sin cos Use of cosα = cos²α sin²α soi Simplifying and using sec²α = + tan²α to final answer Answer Given Accept working in reverse to show RHS=LHS, or showing equivalent 4 (iii) cos = ½ Soi or from tan²θ = /3 oe from sin²θ or cos²θ = 60, 300 = 30, 50 [] [3] First correct solution Second correct solution and no others in the range SC for π/6and 5π/6 and no others in the range 5

11 4754A Mark Scheme June 04 5 (i) EITHER x = e 3t, y = te t soi d / d e e y t x t in terms of t t t y t t Their d / d d / d dy/dx = (te t + e t )/3e 3t oe cao allow for unsimplified form even if subsequently cancelled incorrectly ie can isw when t =, dy/dx = 3e /3e 3 = /e cao www must be simplified to /e oe OR 3t = ln x, y ln e 3 3 /3 x /3ln x x ln x d y / dx x ln x x 3 x 9 /3 /3 = t oe cao t t 3e 3e Any equivalent form of y in terms of x only Differentiating their y provided not eased ie need a product including ln kx and x p and subst x 3 e t to obtain dy/dx in terms of t dy/dx = /3e + /3e =/e www cao exact only must be simplified to /e or e 5 (ii) 3t = ln x t = (ln x)/3 Finding t correctly in terms of x (ln x)/3 y = (ln x ) / 3e Subst in y using their t y = 3 ln 3 x x b Required form ax ln [4] [3] x only NB If this work was already done in 5(i), marks can only be scored in 5(ii) if candidate specifically refers in this part to their part (i). 6

12 4754A Mark Scheme June y ( x ) y x x y 3 finding x (or x) correctly in terms of y ( ) 3 V πx d y π ( y )d y 4 π y y π( ) π 8 [6] For need πx dy with substitution for their x² (in terms of y only) Condone absence of dy throughout if intentions clear. (need π) www For it must be correct with correct limits and, but they may appear later /[y 4 /4 y] independent of π and limits substituting both their limits in correct order in correct expression, condone a minor slip for (if using y = 0 as lower limit then -0 is enough) condone absence of π for oe exact only www (⅜ π or.375π) 7

13 4754A Mark Scheme June 04 7 (i) AB = 5 ( ) or better (condone sign error in vector for ) AC = Accept 5 (condone sign error in vector for ) cos cosθ = scalar product of AB with AC (accept BA/CA) AB. AC with substitution condone a single numerical error provided method is clearly understood [OR Cosine Rule, as far as cos θ= correct numerical expression ] www ± 0.557, 0.557,5/5 9,5/ 5 9 oe or better soi ( ± for method only) = 56.5 www Accept answers that round to 56.º or 56.º or 0.98 radians (or better) NB vector 5i+0j+k leads to apparently correct answer but loses all A marks in part(i) Area = ½ 5 9 sin Using their AB,AC, CAB. Accept any valid method using trigonometry =.8 Accept 5 5 and answers that round to.8 or.9 (dp) www or SC for accurate work soi rounded at the last stage to. (but not from an incorrect answer, say from an incorrect angle or from say.7 or. stated and rounded to.) We will not accept inaccurate work from over rounded answers for the final mark. [7] 8

14 4754A Mark Scheme June 04 7 (ii) (A) AB ( 3) ( ) Scalar product with one vector in the plane with numerical expansion shown AC ( 3) Scalar product, as above, with evaluation, with a second vector. NB vectors are not unique SCB finding the equation of plane first by any valid method (or using vector product) and then clearly stating that the normal is proportional to the coefficients. SC For candidates who substitute all three points in the plane 4x-3y+0z = c and show that they give the same result, award If they include a statement explaining why this means that 4i-3j+0k is normal they can gain. 7 (ii) (B) 4x 3y+0z=c Required form and substituting the co-ordinates of a point on the plane [] 4x 3y + 0z + = 0 oe If found in (A) it must be clearly referred to in (B) to gain the marks. Do not accept vector equation of the plane, as Hence. [] 4i-3j+0k = is A0 9

15 4754A Mark Scheme June 04 7 (iii) 0 r Meets 4x 3y + 0z + = 0 when 6 3(4 3 ) + 0(5 + 0 ) + = 0 5 = 50, = 0.4 Need r = (or oe x y z ) Subst their 4λ, 4 3λ, 5+0λ in equation of their plane from (ii) λ= 0.4 (NB not unique) So meets plane ABC at (.6, 5., ) cao www (condone vector) 7 (iv) height = (.6 + (.) + 4 )= 0 ft ft their (iii) [5] volume =.8 0 / 3 = 6.7 cao 50/3 or answers that round to 6.7 www and not from incorrect answers from (iii) ie not from say (.6,.8,9) [] 0

16 4754A Mark Scheme June 04 8 (i) Either h = ( ½ At) dh/dt = A ( ½ At) Including function of a function, need to see middle step when t = 0, h = ( 0) = as required OR dh h At d = A h AG Separating variables correctly and integrating / h At c Including c. [Condone change of c.] At c h at t =0, h =, = (c/)² c =, h = ( At/)² Using initial conditions AG 8 (ii) When t = 0, h = 0 Subst and solve for A 0 A= 0, A = 0. cao When the depth is 0.5 m, 0.5 = ( 0.05t) substitute h= 0.5 and their A and solve for t 0.05t = 0.5, t = ( 0.5)/0.05 = 5.86s www cao accept 5.9 [3] [4]

17 4754A Mark Scheme June 04 8 (iii) dh h B d t ( h) ( h) d d h EITHER, LHS h h h h B t dh separating variables correctly and intend to integrate both sides (may appear later) [NB reading (+h)²as +h² eases the question. Do not mark as a MR] In cases where (+h)² is MR as +h²or incorrectly expanded, as say +h+h² or +h², allow first for correct separation and attempt to integrate and can then score a max of M0A0A0 (for Bt+c) A0A0, max /7. expanding (+ h) and dividing by h to form a one line function of h (indep of first ) with each term expressed as a single power of h eg must simplify say / h+h/ h +h² h,condone a single error for (do not need to see integral signs) / ( h / h 3/ h )dh h / + h / + h 3/ cao dep on second M only -do not need integral signs OR,LHS, EITHER / / ( h h )h h ( h)dh using udv uv vdu correct formula used correctly, indep of first OR 5/ 3 / 3/ h 3/ h h h h ( h h )dh 3 3 / 4h h h 3 5 3/ 5/ h / + 4h 3/ /3 + h 5/ /5 = Bt + c condone a single error for if intention clear cao oe cao oe, both sides dependent on first mark = Bt +c cao need Bt and c for second but the constant may be on either side When t = 0, h = c = 56/5 from correct work only (accept 3.73 or rounded answers here but not for final ) or c= 56/5 if constant on opposite side. h / (30 + 0h + 6h ) = 56 5Bt * NB AG must be from all correct exact work including exact c. [7]

18 4754A Mark Scheme June 04 8 (iv) h = 0 when t = 0 Substituting h = 0,t = 0 B = 56/300 = 0.87 Accept 0.87 When h = t = Subst their h = 0.5, ft their B and attempt to solve t = 9.5s Accept answers that round to 9.5s www. [4] 3

19 For a description of how UMS marks are calculated see: Unit level raw mark and UMS grade boundaries June 04 series AS GCE / Advanced GCE / AS GCE Double Award / Advanced GCE Double Award GCE Mathematics (MEI) Max Mark a b c d e u 475/0 (C) MEI Introduction to Advanced Mathematics Raw /0 (C) MEI Concepts for Advanced Mathematics Raw /0 (C3) MEI Methods for Advanced Mathematics with Coursework: Written Paper Raw /0 (C3) MEI Methods for Advanced Mathematics with Coursework: Coursework Raw /8 (C3) MEI Methods for Advanced Mathematics with Coursework: Carried Forward Coursework Mark Raw (C3) MEI Methods for Advanced Mathematics with Coursework 4754/0 (C4) MEI Applications of Advanced Mathematics Raw /0 (FP) MEI Further Concepts for Advanced Mathematics Raw /0 (FP) MEI Further Methods for Advanced Mathematics Raw /0 (FP3) MEI Further Applications of Advanced Mathematics Raw /0 (DE) MEI Differential Equations with Coursework: Written Paper Raw /0 (DE) MEI Differential Equations with Coursework: Coursework Raw /8 (DE) MEI Differential Equations with Coursework: Carried Forward Coursework Mark Raw (DE) MEI Differential Equations with Coursework 476/0 () MEI Mechanics Raw /0 (M) MEI Mechanics Raw /0 (M3) MEI Mechanics 3 Raw /0 (M4) MEI Mechanics 4 Raw /0 (S) MEI Statistics Raw /0 (S) MEI Statistics Raw /0 (S3) MEI Statistics 3 Raw /0 (S4) MEI Statistics 4 Raw /0 (D) MEI Decision Mathematics Raw /0 (D) MEI Decision Mathematics Raw /0 (DC) MEI Decision Mathematics Computation Raw /0 (NM) MEI Numerical Methods with Coursework: Written Paper Raw /0 (NM) MEI Numerical Methods with Coursework: Coursework Raw /8 (NM) MEI Numerical Methods with Coursework: Carried Forward Coursework Mark Raw (NM) MEI Numerical Methods with Coursework 4777/0 (NC) MEI Numerical Computation Raw /0 (FPT) Further Pure Mathematics with Technology Raw GCE Statistics (MEI) Max Mark a b c d e u G4/0 (Z) Statistics Raw G4/0 (Z) Statistics Raw G43/0 (Z3) Statistics 3 Raw Unit level raw mark and UMS grade boundaries June 03 series: GCE

4754A Mark Scheme June 2014 BUT

4754A Mark Scheme June 2014 BUT 4754A Mark Scheme June 04 3x A Bx C ( x)(4 x ) x 4 x correct form of partial fractions ( condone additional coeffs eg A B BUT x 4 x ** is M0 ) Ax B Cx D x 4 x * for 3x = A(4 + x ) + (Bx + C)( x) Multiplying