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UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS

GCE Advanced Level

MARK SCHEME for the May/June 2010 question paper for the guidance of teachers

9231 FURTHER MATHEMATICS

9231/23 Paper 23, maximum raw mark 100

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 must be read in conjunction with the question papers and the report on the examination.

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CIE will not enter into discussions or correspondence in connection with these mark schemes.

CIE is publishing the mark schemes for the May/June 2010 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

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Page 2 Mark Scheme Notes

Mark Scheme: Teachers’ version GCE A LEVEL – May/June 2010

Syllabus 9231

Paper 23

Marks are of the following three types: M Method mark, awarded for a valid method applied to the problem. Method marks are not lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. Correct application of a formula without the formula being quoted obviously earns the M mark and in some cases an M mark can be implied from a correct answer. Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated method mark is earned (or implied). Mark for a correct result or statement independent of method marks.

A B ?

When a part of a question has two or more "method" steps, the M marks are generally independent unless the scheme specifically says otherwise; and similarly when there are several B marks allocated. The notation DM or DB (or dep*) is used to indicate that a particular M or B mark is dependent on an earlier M or B (asterisked) mark in the scheme. When two or more steps are run together by the candidate, the earlier marks are implied and full credit is given. The symbol √ implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A or B marks are given for correct work only. A and B marks are not given for fortuitously "correct" answers or results obtained from incorrect working. Note: B2 or A2 means that the candidate can earn 2 or 0. B2/1/0 means that the candidate can earn anything from 0 to 2.

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The marks indicated in the scheme may not be subdivided. If there is genuine doubt whether a candidate has earned a mark, allow the candidate the benefit of the doubt. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. ? Wrong or missing units in an answer should not lead to the loss of a mark unless the scheme specifically indicates otherwise. For a numerical answer, allow the A or B mark if a value is obtained which is correct to 3 s.f., or which would be correct to 3 s.f. if rounded (1 d.p. in the case of an angle). As stated above, an A or B mark is not given if a correct numerical answer arises fortuitously from incorrect working. For Mechanics questions, allow A or B marks for correct answers which arise from taking g equal to 9.8 or 9.81 instead of 10.

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Mark Scheme: Teachers’ version GCE A LEVEL – May/June 2010

Syllabus 9231

Paper 23

The following abbreviations may be used in a mark scheme or used on the scripts: AEF AG BOD CAO CWO ISW MR PA SOS SR Any Equivalent Form (of answer is equally acceptable) Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid) Benefit of Doubt (allowed when the validity of a solution may not be absolutely clear) Correct Answer Only (emphasising that no "follow through" from a previous error is allowed) Correct Working Only – often written by a ‘fortuitous' answer Ignore Subsequent Working Misread Premature Approximation (resulting in basically correct work that is insufficiently accurate) See Other Solution (the candidate makes a better attempt at the same question) Special Ruling (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the light of a particular circumstance)

Penalties MR –1 A penalty of MR –1 is deducted from A or B marks when the data of a question or part question are genuinely misread and the object and difficulty of the question remain unaltered. In this case all A and B marks then become "follow through √" marks. MR is not applied when the candidate misreads his own figures – this is regarded as an error in accuracy. An MR–2 penalty may be applied in particular cases if agreed at the coordination meeting. This is deducted from A or B marks in the case of premature approximation. The PA –1 penalty is usually discussed at the meeting.

PA –1

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Mark Scheme: Teachers’ version GCE A LEVEL – May/June 2010

Syllabus 9231

Paper 23 Part Mark M1 M1 A1 M1 A1 Total

Question Number 1

Mark Scheme Details Find eqn of motion for P, denoting P’s accel. by A: 2mg – T = 2mA Find eqn of motion for disc: Ta = IdiscA/a (2mga = IdiscA/a can earn this M1; usually 1/5) Substitute ?ma2 for MI Idisc of disc: T = ?mA Eliminate T to find accel. A: 2mg = (5/2)mA A = 4g/5 or 8 EITHER: Resolve vertically:

5

[5]

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T cos 30° + R cos 30° = mg M1 A1 [T + R = 2mg/√3] Resolve horizontally: T sin 30° – R sin 30° = mu2/a sin 60° M1 A1 [T – R = 4mu2/a√3] OR: Any 2 other independent resolutions, e.g. Along BV (gives T): T cos 30° – mg cos 60° = (mu2/a sin 60°) sin 60° Normal to BV: R + T cos 60° – mg cos 30° = – (mu2/a sin 60°) cos 60° Along BA: T + R cos 60° – mg cos 30° = (mu2/a sin 60°) cos 60° Normal to BA: R cos 30° – mg cos 60° = – (mu2/a sin 60°) sin 60° Solve for T: T = (m/√3) (g + 2u2/a) A.G. A1 Solve for R: R = (m/√3) (g – 2u2/a) A1* 2 R [ 0 (A.E.F.) B1 Valid reason for deducing u Y ?ga (dep A1*): 0?1g = λ 0?01/0?25 M1 λ = 25 A.G. A1 0?1 d2x/dt2 = 0?1g – λ(0?01 + x)/0?25 M1 A1 d2x/dt2 = – 1000x A.G. x = 0?02 cos (t√1000) [or sin] M1 A1 t = (1/√1000) cos-1(-0?01/0?02) 2π/4√1000 + (1/√1000) sin-1(1/2) M1 A1 t = (1/√1000) 2π/3 = 0?0662 [s] A1 uQ = √(10 g) or 10 [m s-1] vQ = ? uQ = 5 [m s-1] 0?04 (uQ + vQ) = 0?6 [N s] hQ = vQ t – ?g t2 = 5t – 5t2 uQ /g = 1 hP = ?g(1 + t)2 = 5 + 10t + 5t2 [P’s speed at B is 10] h′P = 10t + ?gt2 = 10t + 5t2 15t = 5, t = 1/3 hQ = 10/9 or 1?11 [m] M1 B1 M1 A1 M1 M1 M1 (M1) (M1) M1 A1 A1

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Resolve vertically at equilibrium: Evaluate λ: Use Newton’s Law at general point: Simplify: Use SHM formula for x: Find reqd. time t: or Evaluate:

2 2

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[9]

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Find speed uQ of Q when striking plane: Find speed vQ of Q when rebounding from plane: Find magnitude of impulse: Find height risen by Q to collision in time t: EITHER: Find time for Q to fall to plane: Find height fallen by P in time (1 + t): OR: State or imply P is at B when Q is at plane: Find height fallen by P in time t: Use hP + hQ = 10 or h′P + hQ = 5 to find t: [or relative motion can earn previous M1 M1 A1] Evaluate height above plane of collision:

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Mark Scheme: Teachers’ version GCE A LEVEL – May/June 2010

Syllabus 9231

Paper 23 Part Mark M1 M1 A1 M1 A1 M1 A1 M1 A1 M1 A1 (M1) (A1√) A1 M1 *A1 [4] Total

Question Number 5

Mark Scheme Details Consider speed components at Y normal to KN: Consider speed components at Y parallel to KN: Combine to eliminate u, v: Find XM in terms of a, θ: v sin φ = ? u sin θ v cos φ = u cos θ tan φ = ? tan θ A.G. XM = 2a – a cot θ – a cot φ = 2a – (7/3) a cot θ A.G. Find alternative form of XM in terms of a, θ: XM = a cot MXN = a (4/3)2 cot θ Equate two forms of XM: (37/9) cot θ = 2 [tan θ = 37/18] Evaluate θ: θ = 1?12 rad or 64?1° EITHER: Combine speed components at N: w = u √(cos2 θ + (9/16)2 sin2 θ) OR: Consider speed components along KN: w cos MXN = u cos θ and MXN = cot-1 ((16/9) cot θ) √ on θ = 0?858 rad or 49?1° Evaluate speed w at N: w = 0?669 u (allow 0?668 u) 110?4 ± t √(50?42/25)

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Find confidence interval (allow z in place of t): (using 24 in place of 25 loses A1) Use of correct tabular value: Evaluate C.I. correct to 3 sf (dep *A1, *B1): State (at least) null hypothesis (A.E.F.): Calculate expected values (to 1 dp): Calculate value of χ2 : Compare with consistent tabular value (to 2 dp): Valid method for reaching conclusion: Correct conclusion (A.E.F., requires correct values): (a) (i) Find prob. of 5 operating after 3 months:

*B1 t24, 0.95 = 1?71[1] 110?4 ± 2?4 or [108?0, 112?8] A1 H0: Data conforms to Law 30?10 17?61 12?49 9?69 7?92 22?18 χ2 = 4?23 ± 0?02 χ5, 0.9 2 = 9?236 Reject H0 if χ2 > tabular value Data conforms to Benford’s Law B1 M1 A1 M1 A1 B1 M1 A1

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[8]

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{exp (-3/2?5)}5 = {exp (-3 × 0?4)}5 = 0?30115 = 0?00248 (to 2 sf) M1 A1 B1 M1 A1

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(ii) Find prob. of 1 failing within one month: p1 = 1 – exp (-1/2?5) [= 0?3297] 5 Find prob. of 2 failing within one month: C2 p12 (1 – p1)3 = 0?327 Using 2?5 as parameter, not mean, can earn M1s only in (a) (b) Find prob. of n operating after c months: Show same as 1 operating after nc months: {exp (-cλ)}n = exp (-ncλ) [λ = 0?4]

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M1 A1 M1 A1

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Mark Scheme: Teachers’ version GCE A LEVEL – May/June 2010

Syllabus 9231

Paper 23 Part Mark B1 B1 3 Total

Question Number 9 (a)

Mark Scheme Details State valid comment on each diagram (A.E.F.): Str. line with negative gradient (i) (ii) (Almost) no linear relation

(iii)

Close to str. line with +ve gradient B1

(b) (i) Calculate correlation coefficient: r = (13040 – 700 × 275/7) / √{(149000 – 7002/7)(17351 – 2752/7)} = – 14460 /√(79000 × 6547?4285 = – 0?636 (ii) State hypotheses: Compare with consistent tabular value: Valid method for reaching conclusion: Correct conclusion (needs correct values): 10 H0: ρ = 0, H1: ρ < 0 r7, 5% = 0?669 Reject H0 if r < – tabular value No negative correlation (A.E.F.)

M1 A1 A1 B1 B1 M1 A1

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State valid assumption (A.E.F.): Popln. of diffs. has Normal distn. B1 State hypotheses: H0: ?2 – ?1 = 0, H1: ?2 – ?1 > 0 B1 Consider differences eg: 4?0 5?1 2?3 3?1 0 1?1 0?9 1?2 M1 x = 17?7/8 [= 2?2125] M1 Calculate sample mean: 2 2 Estimate population variance: s = (60?37 – 17?7 /8) / 7 (allow biased: 2?6511 or 1?6282) [= 3?0298 or 1?7412 ] M1 Calculate value of t (to 2 dp): t = x /(s/√8) = 3?60 [or 3?59] M1 *A1 *B1 Compare with correct tabular t value: t7, 0.99 = 2?998 (to 2 dp) Correct conclusion (AEF, dep *A1, *B1): There is an increase B1 Formulate inequality, with any tabular t value: ( x – w)/(s/√8) > t7, 0.9 M1 A1 Use correct tabular value (to 2 dp): t7, 0.9 = 1?415 Evaluate inequality for w: w < 1?34 [or ≤] A1 S.R. Allow M1 A1 A0 if = or < used in inequality EITHER: Find 2 indep. eqns for RB, FB only: M1 Moments for BA about A: FB 2a sin β – RB 2a cos β = Wa sin β M1 A1 Moments for system about C: FB 6a sin β + RB 2a cos β = 9Wa sin β M1 A1 Add eqns to find FB : FB = 5W/4 A.G. M1 A1 OR: If RC, FC introduced, resolve vertically: FB + FC = 3W (B1) Any 2 moment eqns indep. of above resln. e.g.: For BA about A: FB 2a sin β – RB 2a cos β = Wa sin β For CA about A: FC 4a sin β – RC 4a cos β = 4Wa sin β For system about B: FC 6a sin β – RC 2a cos β = 9Wa sin β (2 system eqns are equiv. to vert. resolution) (2 × M1 A1) Solve eqns using RB = RC to find FB : FB = 5W/4 A.G. (M1 A1) Find FC by eg vertical resolution for rods: FC = 7W/4 B1 Find RB [or RC] from a moment eqn: RB [= RC ] = ? W tan β B1 Find one of FB /RB , FC /RC eg: FB /RB = 5 / (3 tan β) M1 A1 Find other eg: FC /RC = 7 / (3 tan β) A1 Find set of possible values of ?: ? > 7 / (3 tan β) (allow [) M1 A1

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11 EITHER

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Mark Scheme: Teachers’ version GCE A LEVEL – May/June 2010

Syllabus 9231

Paper 23 Part Mark Total

Question Number 11 OR

Mark Scheme Details For time to pack 50 items, find E(T50) and Var(T50): State valid justification (A.E.F.) e.g.: Find Normal approximation to P(T50 < 70): For time Tn to pack n items, find E(Tn) and Var(Tn): Use Normal approximation for P(Tn < 70) > 0?9: Invert function: Rearrange as quadratic expression in √n: Find positive root when expression is zero: Find greatest integer value of n: For difference in times, find E(A–H ) and Var(A–H): Find Normal approximation to P(A–H > 0):

E(T50) = 50 × 1?5 [= 75] and Var(T50) = 50 × 0?42 [= 8] By central limit theorem or Since n [or 50] is large Φ((70 – 75)/√8) = 1 – Φ(1?768) = 0?0385 ± 0?0001 E(Tn) = 1?5n and Var(Tn) = 0?42 n Φ((70 – 1?5n) / (0?4√n)) [ 0?9 (70 – 1?5n)/ 0?4√n [ 1?282 (√n)2 + 0?3419√n – 140/3 [Y 0] 44?4 nmax = 44 E(A–H) = 75 – 65 = 10 and Var(A–H) = 8 + 12?5 = 20?5 Φ((10)/√20?5) = Φ(2?209) = 0?986

B1 B1 M1 A1 B1 M1 A1 M1 A1 A1

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M1 A1 M1 A1

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