9231_w08_ms_2

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the October/November 2008 question paper 9231 FURTHER MATHEMATICS 9231/02 Paper 2, 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. All Examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes must be read in conjunction with the question papers and the report on the examination. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the May/June 2008 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses. www.XtremePapers.com Page 2 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2008 9231 02 © UCLES 2008 Mark Scheme Notes 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. A 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). B Mark for a correct result or statement independent of method marks. • 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. 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. Page 3 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2008 9231 02 © UCLES 2008 The following abbreviations may be used in a mark scheme or used on the scripts: AEF Any Equivalent Form (of answer is equally acceptable) AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid) BOD Benefit of Doubt (allowed when the validity of a solution may not be absolutely clear) CAO Correct Answer Only (emphasising that no "follow through" from a previous error is allowed) CWO Correct Working Only – often written by a ‘fortuitous' answer ISW Ignore Subsequent Working MR Misread PA Premature Approximation (resulting in basically correct work that is insufficiently accurate) SOS See Other Solution (the candidate makes a better attempt at the same question) SR 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. PA –1 This is deducted from A or B marks in the case of premature approximation. The PA –1 penalty is usually discussed at the meeting. Page 4 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2008 9231 02 © UCLES 2008 Qu No Mark Scheme Details Part Mark Total 1 Find MI about A of AB, AC (M1 for either): IAB = (6m/24)(⅓3 2 + 32)a2 = 3ma2 M1 IAC = (10m/24)(⅓5 2 + 52)a2 = (125/9) ma2 A1 Find MI about A of BC: IBC = (8m/24)(⅓4 2 + 62 + 42)a2 = (172/9)ma2 M1 A1 Sum to find MI about A of wire: I = (324/9)ma2 = 36ma2 A1 5 [5] 2 Find speeds from cons. of energy (M1 for either): ½mv1 2 = mga (1 – cos θ) M1 A1 ½mv2 2 = mga (1 + cos θ) A1 Find R1, R2 by radial resolution (M1 for either): R1 = mg cos θ – mv1 2 /a M1 A1 R2 = mg cos θ + mv2 2 /a A1 EITHER: Substitute in R1, R2 and combine: R1 = 3mg cos θ – 2mg R2 = 3mg cos θ + 2mg R2 – R1 = 4mg A.G. M1 A1 OR: Combine R1, R2 and substitute: R2 – R1 = m(v2 2 + v1 2)/a = 4mg A.G. (M1 A1) 8 [8] 3 EITHER: Relate angular acceln. to tension for block: 2ma d2θ/dt2 = 2mg – T – mg/10 M1 A1 Relate angular acceln. to tension for disc: I d2θ/dt2 = aT, I = ½ ma2 M1 A1 Eliminate tension T: (½ + 2)ma2 d2θ/dt2 = (2 – 0⋅1)mga M1 Find d2θ/dt2 : 19g/25a or 0⋅76g/a or 7⋅6/a A1 Use (dθ/dt)2 = 2 d2θ/dt2 2π (√ on d2θ/dt2): (dθ/dt)2 = 76πg/25a A.E.F. M1 A1√ Find dθ/dt (A.E.F.): dθ/dt = 3⋅09√(g/a) or 9⋅77/√a A1 OR: Use conservation of energy for system: ½I (dθ/dt)2 + ½ 2m (a dθ/dt)2 (M1 A1) = 2mgaθ – 0⋅1 mgaθ (M1 A1) Put θ = 2π and find dθ/dt (A.E.F.): dθ/dt = 3⋅09√(g/a) or 9⋅77/√a (M1 A1) Differentiate energy eqn w.r.t. t: (5ma2/4) 2 d2θ/dt2 = 1⋅9 mga (M1 A1) Find d2θ/dt2 : 19g/25a or 0⋅76g/a or 7⋅6/a (A1) 9 (9) [9] Page 5 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2008 9231 02 © UCLES 2008 Qu No Mark Scheme Details Part Mark Total 4 (i) Use conservation of momentum: 0⋅1vA + mvB = 0⋅1 × 5 – m × 2 M1 Find m: m = (0⋅5 – 0⋅1 vA) / (2 + vB) A1 Use vA > 0 to find lower bound on m: vB > 0, m < 0⋅5/2 = 0⋅25 A.G. M1 A1 (ii) Use Newton’s law of restitution: vB – vA = ½( 2 + 5) = 7/2 M1 A1 Put m = 0⋅2 and find one of vA, vB: 2 vB + vA = 1, vA = –2 or vB = 1⋅5 M1 A1 Find magnitude of impulse [N s]: 0⋅1 (5 – vA) or 0⋅2 (1⋅5 + 2) = 0⋅7 M1 A1 4 6 [10] 5 Find equation of motion at general point: m d²x/dt² = mg ((a – x)/a)½ – mg ((a + x)/a)–¼ M1 Expand terms and approximate: ≈ mg ((1 – x/2a) – (1 – x/4a)) M1 A1 Simplify to give SHM eqn: d²x/dt² = – gx/4a A1 Use SHM eqn to find speed when x = 0: vmax² = (g/4a) (0⋅04a)² M1 A1 Simplify (A.E.F.): vmax = 0⋅02 √(ag) or 0⋅0632 √a A1 Use SHM eqn to find time when v = ½vmax: ½aω = aω sin ωt (A.E.F.) M1 A1 Substitute ω = √(g/4a) and simplify: t = √(4a/g) sin–1 ½ M1 = (π/3) √(a/g) (A.E.F.) A1 4 7 [11] 6 Use standard formula for pooled estimate, e.g.: ((128 – 15²/5) + (980 – 36²/10))/13 M1 Award A1 for one term in numerator, e.g.: 5 × 16⋅6 or 10 × 85⋅04 or 83 or 850⋅4 or 4 × 20⋅75 or 9 × 94⋅5 A1 Calculate value of pooled estimate: 71⋅8 A1 3 [3] 7 (i) Find sample mean: x = ½(61⋅21 + 64⋅39) = 62⋅8 M1 A1 Use confidence interval formula: x ± ts/√n for any t M1 Use correct tabular t: t24,099 = 2⋅492 A1 Calculate standard deviation: s = 1⋅59 × 5 / 2⋅492 = 3⋅19 A1 (ii) State assumption (A.E.F.): Population has normal distribution B1 (iii) State valid reason (A.E.F.): 72 exceeds upper limit of interval *B1 State conclusion (A.E.F., dep *B1): Yes, it does reduce pulse rate B1 5 1 2 [8] Page 6 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2008 9231 02 © UCLES 2008 Qu No Mark Scheme Details Part Mark Total 8 (i) Formulate two eqns for means: y + 0⋅425 x = 1⋅28 and x + 0⋅516y = 1⋅05 M1 Solve for means: x = 0⋅499, y = 1⋅068 or 1⋅07 M1 A1 (ii) Find correlation coefficient for sample: r² = 0⋅425 × 0⋅516; r = – 0⋅468 M1; *A1 (iii) State hypotheses: H0: ρ = 0, H1: ρ ≠ 0 B1 Valid method for reaching conclusion: ρ ≠ 0 if |r| > tabular value M1 Use of correct tabular value: ρ25,2⋅5% = 0⋅396 *B1 Correct conclusion (A.E.F., dep *A1, *B1): Coefficient does differ from zero A1 3 2 4 [9] 9 Integrate f(t) to give F(t): F(t) = – 9/8t2 M1 Apply limits: F(2⋅5) – F(2) = – (9/8) (2⋅5–2 – 2–2) A1 Evaluate and multiply by 100: 10⋅125 A.G. A1 State hypotheses (A.E.F.): H0: f(t) fits data, H1: doesn’t fit B1 Find χ2 (A1 if at least 3 terms correct): χ2 = 1⋅52/62⋅5 + 4⋅8752/21⋅875 + 5⋅8752/10⋅125 + 2⋅52/5⋅5 M1 A1 Evaluate χ2: = 0⋅036 + 1⋅086 + 3⋅409 + 1⋅136 = 5⋅67 [± 0⋅01] *A1 Compare with consistent tab. value (to 2 dp): χ3, 0.9 2 = 6⋅251 *B1√ (χ2, 0.9 2 = 4⋅605, χ1, 0.9 2 = 2⋅706) Consistent conclusion (A1 dep *A1, *B1): Distribution fits data (A.E.F.) M1√ A1 3 7 [10] 10 Replace 2x by ekx to find k: f(x) = ae–kx; k = ln 2 M1; A1 Show a = k by e.g. ∫ ∞ = 0 1 )f(x : [ – (a/k) e–kx] 0 ∞ = 1, a = k or ln 2 M1 A1 State value of E(X): 1 / ln 2 or 1⋅44 B1 Find distribution fn G of Y: G(y) = P(Y ≤ y) = P(X ≤ k–1 ln y) M1 A1 = F(k–1ln y) = (a/k)(1 – e–ln y ) M1 A1 = 1 – 1/y (CAO) A1 Find probability density function g of Y: g(y) = 1/y2 (CAO) M1 A1 State interval for either G or g: y ≥ 1 B1 4 1 8 [13] Page 7 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2008 9231 02 © UCLES 2008 Qu No Mark Scheme Details Part Mark Total 11 a EITHER: Observe or deduce when RB is maximised: RB, max occurs when dog at B M1 Moments for ladder about A: 6a RB, max = 4aW + 8a × ¼W M1 OR: Moments when dog is x hor. from wall: 6a RB = 4aW + (8a – x) ¼W (M1) Deduce limit on RB : RB, max occurs when dog at B (M1) Find max. value of RB: RB, max = W A1 Resolve horizontally for ladder AB: FA = RB B1 Resolve vertically for ladder AB: RA = W + ¼W [= 5W/4] B1 Find bound on µ from FA ≤ µRA : µ ≥ FA / RA ≥ RB, max / (5W/4) M1 µ ≥ 4/5 A.G. A1 Find friction Fcube along DE by hor. resolution: Fcube = FA or RB B1 Find reaction Rcube from floor by vert. resolution: Rcube = W + ¼W + kW B1 Show that Fcube ≤ µ Rcube : µ Rcube ≥ W + 4kW/5 ≥ W = RB, max ≥ Fcube M1 A1 Find moments about D opposing effect of Rcube : 2akW + a5W/4 – 4a FA M1 Find smallest value of k for which moments ≥ 0: (4W – 5W/4) / 2W = 11/8 M1 A1 7 4 3 [14] 11 b State hypotheses (A.E.F.): H0: µ 2 = µ 1, H1: µ 2 > µ 1 B1 Calculate Σ (xi – x ) 2 (M1 for either) 8⋅24, 4⋅62[4] or estimate variances: 0⋅168 or 0⋅165, 0⋅0784 or 0⋅0771 M1 A1 A1 Find s2 (M0 if inconsistent denominators used): s2 = 0⋅168/50 + 0⋅0784/60 or 0⋅165/49 + 0⋅0771/59 [= 0⋅00467] *M1 Find value of z (dep *M1): z = (1803⋅6/60 – 1492⋅0/50) / s M1 = (30⋅06 – 29⋅84)/0⋅0683 = 3⋅22 *A1 S.R. Using pooled estimate of variance: s2 = (8⋅24 + 4⋅62)/108 = 0⋅119 (M0) z = 0⋅22 / s√ (1/50 + 1/60) = 3⋅33 (B1) Find tabular. value (to 2 dp): Φ–1(0⋅98) = 2⋅05[4] *B1 Compare values for conclusion (A1 dep *A1, *B1): µ 2 > µ 1 (A.E.F., M1√ on values) M1√ A1 Find limiting value of z (to 2 dp): z = (0⋅22 – 0⋅1)/s = 1⋅756 M1 A1 Find Φ(z) and hence values of α (to 1 dp): Φ(z) = 0⋅9604, α ≥ 3⋅9 or 4⋅0 M1 A1 s2 = 0⋅119 gives: z = 1⋅816, α ≥ 3⋅47 (M1 M1) 10 4 [14]