MARK SCHEME for the May/June 2011 question paper for the guidance of teachers 9794 MATHEMATICS
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1 UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS Pre-U Certificate MARK SCHEME for the May/June 0 question paper for the guince of teachers 9794 MATHEMATICS 9794/0 Paper (Pure Mathematics and Mechanics), maximum raw mark 0 This mark scheme is published as an aid to teachers and candites, 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 must be read in conjunction with the question papers and the report on the examination. Cambridge will not enter into discussions or correspondence in connection with these mark schemes. Cambridge is publishing the mark schemes for the May/June 0 question papers for most IGCSE, Pre-U, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.
2 Page Mark Scheme: Teachers version Syllabus Paper Pre-U May/June (i) (i) Substitute x 4 into equation or attempt factorisation of (x 4) Verify y(4) 0 or that (x 4) is a factor [] May be seen in part (i) ( x 4)( x + 4x + 4) x x 6 ( 4 )( x + )( + ) x x [] Attempt to multiply out brackets Obtain 6 8. [] SC For answer given without working seen Multiply numerator and denominator by 5, and expand. and use of ( 5)( 5). + Obtain AG [4] 4 (i) dv sin x, u x du v cosx, Obtain an expression of the form f ( x ) ± g( x) Obtain x cos x + cos x x cosx + sin x + c 9 Shape of each graph (concavity). Asymptote at π CAO [5] Max/Min points clearly indicated at x 0 and π. [4] Evidence that sec x cos x Multiply by cos x, obtaining a quadratic. Solve quadratic. Solutions x π and x 0.84 [5] SC For either both in degrees or one in degress and one in radians A0 University of Cambridge International Examinations 0
3 Page Mark Scheme: Teachers version Syllabus Paper Pre-U May/June (i) 6 (i) 7 (i) Attempt to solve c ( or c < ) for at least one drug, and obtain a solution. Obtain 54.9 (hours) for Antiflu; Obtain.0 (hours) for Coldcure. [] Two decaying exponentials in the first quadrant showing correct intercepts on the c-axis and crossing for some t > 0. [] Assume additive nature of the concentrations: e + 5e.0. [] d u x or equivalent used Substitute to obtain u d e u u Obtain e Evaluate: 0.5 WWW [4] SC For 0.5 without working B x x e + x ( x) e Equate to zero and find at least one point Stationary points (, e 0.5 ); 0.5, e ). [4] (a) Not invertible Not or equivalent [] (b) (Minimum value of at x ) f ( x) [] [ for correct interval; for correct inequality] x (a) gh( ) sin. x Obtain ( cos x) with some working AG [] (b) Sine wave Period of π Completely correct [] University of Cambridge International Examinations 0
4 Page 4 Mark Scheme: Teachers version Syllabus Paper Pre-U May/June (i) (a) cosθ d y sin θ sinθ cosθ sin θ cos θ cot θ AG sin θ At least two of θ... π, 0, π... without any incorrect values [5] (b) Rearranging y x to give θ + sinθ cosθ ( θ α ) + A sin where A and π α [4] 4 π π (c) Consider sign of θ sin θ at θ, π 4 Change of sign implies root: π sinθ cosθ ( negative) and π ( positive) d y d or equivalent ( cosθ ) ( cosθ ) AEF, unsimplified [] d y 0 > y 4 [4] University of Cambridge International Examinations 0
5 Page 5 Mark Scheme: Teachers version Syllabus Paper Pre-U May/June (i) P has x-coordinate k. Region R has area k k ( k ) ( k ) k k x or 0 kx x 0 (iv) k AG [] 6 a kx x k or equivalent. 0 a kx x 0 > k 6ka + 4a 0 AG [] Differentiate the implicit equation wrt t: k a k 6a + a d a Make substitutions and obtain. OR: 0 Differentiate the implicit equation wrt a or k (< errors) CAO k ak 6a + a 0 ork a k 6a + a 0 Relate connected rates of change d a Make substitutions and obtain. k ( ) The formula may appear k Attempt to factorise 6k + 4 Obtain ( )( k + k ) k with linear factor ( ) [4] k k Solve quadratic factor and obtain either or both of k ± Correctly substitute into derivative formula and attempt to simplify Obtain either or both of ±. [5] University of Cambridge International Examinations 0
6 Page 6 Mark Scheme: Teachers version Syllabus Paper Pre-U May/June (i) (i) Any valid method, for example ( 4i + k )(. i 4k) AB. AC Hence result. [] 4 Resolving along AB: T AB 0 cos tan Obtain N. 4 Resolving along AC: T AC 0sin tan 6N [] SC Both answers either unassigned or swapped The vector tension is x unit vector in AB direction 9.6i + 7.j [] Or ai + bj where Use of Solving a 4 and a + b (their T AB ) b x Vt cosθ and y Vt sinθ gt y 0 for t and substitute in x formula V sinθ cosθ V sin θ R g g ( symmetry of the trajectory) implies 00m AG [4] R [] V 000 ( 0 0 ) ms ( g theirr ) Solving 0 0 0t sin π 4 Obtain t or 5 Obtain x t and substitute later V cosθ h m [5] University of Cambridge International Examinations 0
7 Page 7 Mark Scheme: Teachers version Syllabus Paper Pre-U May/June (i) All forces shown: Applied, weight and reaction. [] Net force up the slope 0 0 sin 0 0( N) Use ' Force mass acceleration' > a 5ms Applying suva with u 0 and a 5 v 5t. [4] Let U and V ( V > U ) be the speeds of the particles up the slope after the collision. An attempt at both of COM: 5 5 U + V NEL:. ( 5 ( 5) ) V U 0 Obtain - U 7ms suva gives v 7 5T, wheret is time after impact. [6] (i) As the system is in equilibrium, the tension in the string is T mg Resolving at right angles to the plane: R + T sin α mg cosα (iv) giving R mg( cosα sin α ). AG [] By implication α 45 (condone bounry case only) cosα ; sinα R mg AG [] Resolving up the slope ( sin α cosα ) F mg sin α T cosα mg For this to be positive and combined with first line of solution of 0.5 < tanα AG [] Using F µ R sin α cosα tanα µ cosα sin α tanα Max value of µ is when tanα. [] University of Cambridge International Examinations 0
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