9794 MATHEMATICS. 9794/03 Paper 3 (Applications of Mathematics), maximum raw mark 80

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1 CAMBRIDGE INTERNATIONAL EXAMINATIONS Pre-U Certificate MARK SCHEME for the May/June 013 series 9794 MATHEMATICS 9794/03 Paper 3 (Applications of Mathematics), maximum raw mark 80 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 013 series for most IGCSE, Pre-U, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.

2 Page Mark Scheme Syllabus Paper Pre-U May/June Where appropriate, accept answers to 3 sf or better, then, except in Q4 (iii), ISW if rounded to sf or fewer. Answers given to sf or fewer without an unrounded answer score A x = = Use of correct formula for mean; may be implied. c.a.o (5...) s = = = 100 [4] [4] Use of correct formula for standard deviation; may be implied. c.a.o. Accept unbiased estimate 1.09(80 ) If no working shown, answer must be correct to 3 sf (or better) to score. (i) P( A B) = P( A) P( B A) = = 4 8 [] Conditional probability rule applied, s.o.i. c.a.o. Accept solutions based on Venn diagrams. (ii) P( B) = P( A B) P( A) + P( A B) = + = [] [4] Probability rule applied, s.o.i. Ft (i) provided both P( A B) and P(B) lie between 0 and 1. 3 (i) S xy = 7753 = Use of formula for numerator. 100 S xx = = Use of formula for either term in denominator S yy = = r = = 0.799(36...) [4] Use of formula for r. c.a.o. (ii) Form y = ax + b Sxy 78.8 a = = = 0.83(1...) S xx Use of formula for a. S xy and S xx from above. AG. b = y ax b = = = 41.8(46...) [4] Use of formula for b. AG. Must be convincing. Allow for use of a = 0.83 to find b (= ), or b = 41.8 to find a (= ), but not both, but do not award the corresponding A mark. Cambridge International Examinations 013

3 Page 3 Mark Scheme Syllabus Paper Pre-U May/June (iii) When x = 50, y = Accept a.w.r.t. 8.8 This is ok; it is within the range of the data. At least one of the comments must refer to within/beyond the range of the data. (o.e.) When x = 65, y = Accept a.w.r.t. 95. This is not ok; it is beyond the range of the data. 4 (i) X ~ N( 85.1, 3.4 ) [4] [1] P Z < 3.4 Standardising. =Φ( 1.5) = 1 Φ (1.5) = to deal with negative z value. = [3] (ii) P(B(6, ) <1) Recognise need for B(6, p).possibly implied by partially correct terms in the next line. = Either term correct. Sum of two correct terms. = = 0.944(1 ) [4] Ft their p from (i). (iii) 50 ( ) 50 = [3] [10] (1 (ii)). Must be at least 1 dp. Do not allow answer rounded to the nearest integer, even following an answer to 3sf or better. Cambridge International Examinations 013

4 Page 4 Mark Scheme Syllabus Paper Pre-U May/June (i) 7! 5040 = = 50! [3] 7!! c.a.o. (ii) 6 C Consider selections when all digits are 5 different. 6 C P or 6 P 5 = 70 Arrangements when all digits different. C Consider selections of the form 11xxx. 5! (10) = 600! Arrangements of 11xxx Adding two (or more) relevant cases. = 130 [7] Fully correct. OR: (e.g.) Using no 1 s + one 1 + two 1 s = 5 P P P 3 = = 130 [10] 6 (i) v= t( t )( t 4) Set v = 0 and attempt to solve. t 0 so t = and 4. Fully correct. SC: for both t = and t = 4 found by substitution or stated without working, and if shows/explains there are no other values. Cubic graph crossing the t axis at 0 & other places. Fully correct curve, axes and intercepts labelled and curve only between t = 0 and 4. [4] (ii) a = 3t 1t+ 8 Differentiate v. All terms correct. Allow if found in (i) and used here. = = 4 (ms - ) [3] Substitute t =. c.a.o Cambridge International Examinations 013

5 Page 5 Mark Scheme Syllabus Paper Pre-U May/June (iii) 4 t 3 x = t + 4t + c 4 Integrate v. All terms correct; condone omission of + c. Allow definite integral as alternative. x = 0 when t = 0 therefore c = 0 Deal with c correctly or consider lower limit of definite integral. When t =, x = = 4 Indep of previous. So average speed = 4 / Use formula for average speed. = (ms -1 ) [6] [13] Ft their x when t =. 7 (i) Let the velocities of A and B after the collision be v and w. 4mu = 4mv + mw u = v+ w Use of conservation of momentum: a correct equation, consistent with a diagram, if present. eu = w v Use of N.E.L.: a correct equation, consistent with a diagram, if present v= ( e) u and w= ( + ) e u [4] Solve simultaneous equations. Both correct. Accept w unsimplified. (ii) 1 1 If e = then v = u and w = u [1] Ft their v and w in (i). (iii) After A collides with B velocities are: u/, u (and 0) respectively. Apply the result from (i) at least once. Or a complete correct method for the BC collision. After B collides with C velocities are: u/, u/ and u respectively. [] All correct, including A. (iv) A and B have the same velocity and C is moving away from them so there can be no further collisions. [1] [8] Ft (iii). Must consider all 3 particles. 8 (i) x = Utcosθ 1 y = Utsinθ gt Allow g = 9.8. t = U x cos θ Make t the subject of x equation and substitute. x 1 x y = U sinθ g Ucosθ Ucosθ gx = x tanθ U cos θ [4] Accept any correct form/unsimplified. Cambridge International Examinations 013

6 Page 6 Mark Scheme Syllabus Paper Pre-U May/June (ii) y =0 and x 0 gives U x = sin θ Set y = 0 and attempt to make x or sinθ the g subject. Allow other equivalent methods e.g by solving a quadratic (t 4t + 1 = 0) in tan θ (= ± 3). gx sin θ = = = 0.5 Substitute and obtain 0.5 (or tan θ) U 30 correctly. This has solutions so there are trajectories. Require an explicit statement to this effect. θ = 15 or 75 [4] Both correct. (iii) θ = 15 is fast (and low). Advantage of one. (ft (ii)) θ = 75 is high (more likely to clear any obstacles). [] Advantage of the other. (ft (ii)) [10] SC only for just high and low. Allow other reasonable advantages. 9 (i) Diagram with weight, normal contact and friction forces added. [1] Do not accept both T and the components of T shown. (ii) F = Tcosθ Resolve horizontally. mg = R + Tsinθ Resolve vertically. F = µr Limiting friction Tcosθ = µ(mg Tsinθ) µ mg T = cosθ + µ sinθ [4] Eliminate F and R and rearrange to given answer. Must be convincing require at least one intermediate line. (iii) With µ = 0.75, min T occurs at max (cosθ sinθ). Allow substitution for µ at any stage. EITHER sinθ cosθ = 0 Differentiate and set = 0. tanθ = 0.75 θ = invtan(0.75) = 36.9 [4] OR Use of Rcos(θ α) or Rsin(θ + α). And set cos( ) or sin( ) = 1. α = 36.9 or 53.1 As appropriate. θ = 36.9 [9] Cambridge International Examinations 013

MARK SCHEME for the May/June 2011 question paper for the guidance of teachers 0580 MATHEMATICS

MARK SCHEME for the May/June 2011 question paper for the guidance of teachers 0580 MATHEMATICS UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MARK SCHEME for the May/June 0 question paper for the guidance of teachers 0580 MATHEMATICS 0580/4