Applications of Advanced Mathematics (C4) Paper A TUESDAY 22 JANUARY 2008

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1 ADVANCED GCE 4754/0A MATHEMATICS (MEI) Applications of Advanced Mathematics (C4) Paper A TUESDAY JANUARY 008 Additional materials: Answer Booklet (8 pages) Graph paper MEI Eamination Formulae and Tables (MF) Afternoon Time: hour 30 minutes INSTRUCTIONS TO CANDIDATES Write our name in capital letters, our Centre Number and Candidate Number in the spaces provided on the Answer Booklet. Read each question carefull and make sure ou know what ou have to do before starting our answer. Answer all the questions. You are permitted to use a graphical calculator in this paper. Final answers should be given to a degree of accurac appropriate to the contet. INFORMATION FOR CANDIDATES The number of marks is given in brackets [ ] at the end of each question or part question. The total number of marks for this paper is 7. You are advised that an answer ma receive no marks unless ou show sufficient detail of the working to indicate that a correct method is being used. NOTE This paper will be followed b Paper B: Comprehension. This document consists of 4 printed pages. OCR 008 [T/0/653] OCR is an eempt Charit [Turn over

2 Section A (36 marks) Epress 3 cos θ + 4sinθ in the form R cos(θ α),wherer > 0and0< α < π. Hence solve the equation 3 cos θ + 4sinθ = for π θ π. [7] (i) Find the first three terms in the binomial epansion of. State the set of values of for which the epansion is valid. (ii) Hence find the first three terms in the series epansion of +. 3 Fig. 3 shows part of the curve = +, together with the line =. O Fig. 3 Theregionenclosedbthecurve,the-ais and the line = is rotated through 360 about the -ais. Find the volume of the solid generated, giving our answer in terms of π. 4 The angle θ satisfies the equation sin(θ + 45 )=cos θ. (i) Using the eact values of sin 45 and cos 45,showthattanθ =. (ii) Find the values of θ for 0 < θ < 360. [] 5 Epress 4 ( in partial fractions. [6] + 4) 6 Solve the equation cosec θ = 3, for 0 < θ < 360. OCR /0A Jan08

3 3 Section B (36 marks) 7 A glass ornament OABCDEFG is a truncated pramid on a rectangular base (see Fig. 7). All dimensions are in centimetres. G (3, 6, 4) F (3, 4, 4) D (9, 6, 4) z E (9, 4, 4) O A (0, 0, 0) C (5, 0, 0) B (5, 0, 0) Fig. 7 (i) Write down the vectors CD and CB. [] (ii) Find the length of the edge CD. [] (iii) Show that the vector 4i + k is perpendicular to the vectors CD and CB. Hence find the cartesian equation of the plane BCDE. (iv) Write down vector equations for the lines OG and AF. Show that the meet at the point P with coordinates (5, 0, 40). You ma assume that the lines CD and BE also meet at the point P. The volume of a pramid is area of base height. 3 (v) Find the volumes of the pramids POABC and PDEFG. Hence find the volume of the ornament. [4] OCR /0A Jan08 [Turn over

4 4 8 A curve has equation where k is a positive constant. (i) Verif that + 4 = k, = k cos θ, = k sin θ, are parametric equations for the curve. (ii) Hence or otherwise show that d d = 4. (iii) Fig. 8 illustrates the curve for a particular value of k. Write down this value of k. [] O Fig. 8 (iv) Cop Fig. 8 and on the same aes sketch the curves for k =, k = 3andk = 4. On a map, the curves represent the contours of a mountain. A stream flows down the mountain. Its path on the map is alwas at right angles to the contour it is crossing. (v) Eplain wh the path of the stream is modelled b the differential equation d d = 4. [] (vi) Solve this differential equation. Given that the path of the stream passes through the point (, ), show that its equation is = 4 6. [6] Permission to reproduce items where third-part owned material protected b copright is included has been sought and cleared where possible. Ever reasonable effort has been made b the publisher (OCR) to trace copright holders, but if an items requiring clearance have unwittingl been included, the publisher will be pleased to make amends at the earliest possible opportunit. OCR is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of Universit of Cambridge Local Eaminations Sndicate (UCLES), which is itself a department of the Universit of Cambridge. OCR /0A Jan08

5 4754 Mark Scheme Januar (C4) Applications of Advanced Mathematics Section A 3 cos θ + 4sin θ = R cos(θ α) = R(cos θ cos α + sin θ sin α) R cos α = 3, R sin α = 4 R = = 5 R = 5 tan α = 4/3 α = cos(θ 0.973)= cos(θ 0.973) = /5 θ =.593,.593 θ =.087, 0.3 [7] R = 5 cwo and no others in the range (i) 3 ( ) ( ).( ) = ( ) + ( ) +...! 3 = Valid for < < - ½ < < ½ binomial epansion with p = -½ correct epression cao (ii) + 3 = ( + )( ) 3 = = ft substitituting their and epanding cao 3 V = π d = + = V = π ( ) d = π = π( ½ +) = ½ π substituting limits into integrand

6 4754 Mark Scheme Januar 008 4(i) sin(θ + 45 ) = cos θ sin θ cos 45 + cos θ sin 45 = cos θ (/ ) sin θ + (/ ) cos θ = cos θ sin θ + cos θ = cos θ sin θ = ( ) cos θ sinθ = tan θ = * cosθ compound angle formula sin 45 = /, cos 45 = / collecting terms (ii) tan θ = θ =.5, 0.5 [] and no others in the range 5 4 A B + C = + ( + 4) + 4 A ( + 4) + ( B+ C ) = ( + 4) 4 = A( + 4) + ( B+ C) = 0 4 = 4A A = coefft of : 0 = A + B B = coeffts of : 0 = C 4 = ( + 4) + 4 D [6] correct partial fractions A= Substitution or equating coeffts B= C= 0 6 cosec θ = 3 sin θ = /3 θ = 9.47, and no others in the range

7 4754 Mark Scheme Januar 008 Section B 7(i) 6 CD = CB= 0 0 [] (ii) ( 6) = 5.46 cm [] (iii) CD. 0 = 6. 0 = = CB. 0 = 0. 0 = = 0 0 plane BCDE is 4 + z = c At C (sa) = c c = 60 plane BCDE is 4 + z = 60 using scalar product or other equivalent methods 0 3 (iv) OG: r = 0 + λ AF: r = 0 + μ At (5, 0, 40), 3λ = 5 λ = 5/3 6λ = 0, 4λ = 40, so consistent. At (5, 0, 40), 3μ = 5 μ = 5/3 0 6μ = 0, 4μ = 40, so consistent. So lines meet at (5, 0, 40)* evaluating parameter and checking consistenc. [or other methods, e.g. solving] (v) h=40 POABC: V = / = 4000 cm 3. PDEFG: V = /3 8 6 (40 4) = 56 cm 3 vol of ornament = = 3744 cm 3 [4] soi /3 w d h used for either condone one error both volumes correct cao 3

8 4754 Mark Scheme Januar 008 8(i) cos θ =,sinθ = k k cos θ + sin θ = + = k k 4 + = k k + 4 = k * Used substitution (ii) d d = ksin θ, = kcosθ dθ dθ k cos θ d d / dθ = = d d / dθ k sinθ = ½ cot θ kcosθ d = = cot θ = 4 4ksinθ d or, b differentiating implicitl + 8 d/d = 0 d/d = /8 = /4* oe (iii) k = [] (iv) k = k = 3 4 correct curve shape and position or more curves correct shape- in concentric form all 3 curves correct k = 3 k = 4 (v) grad of stream path = /grad of contour d 4 = = * d ( /4 ) [] (vi) d 4 d 4d = d = ln = 4 ln + c = ln e c 4 = A 4 where A = e c. When =, = = 6A A = /6 = 4 /6 * [6] Separating variables ln = 4 ln (+c) antilogging correctl (at an stage) substituting =, = evaluating a correct constant www 4

Tuesday 18 June 2013 Morning

Tuesday 18 June 2013 Morning Tuesda 8 June 0 Morning A GCE MATHEMATICS (MEI) 475/0 Methods for Advanced Mathematics (C) QUESTION PAPER * 4 7 5 6 6 0 6 * Candidates answer on the Printed Answer Book. OCR supplied materials: Printed