9231 FURTHER MATHEMATICS

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1 CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the May/June 201 series 921 FURTHER MATHEMATICS 921/21 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, 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 201 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.

2 Page 2 Mark Scheme Syllabus Paper Question Mark Scheme Details Part Mark Total Number 1 Equate impulse to momentum to find initial speed v and Newton s law of restitution to find new speed: v = u, v = ev = [-] u A Find v 2 at both A and B: Find amplitude a m from given K.E. ratio: v A 2 = ω 2 (a ) and v B 2 = ω 2 (a ) ½ mv 2 2 A = (12 / 11) ½ mv B 11 (a ) = 12 (a ) a 2 = ¼ (27 11) =, a = 2 A1 Find ω from v max = aω: Find time ( on a ) at A or at B, e.g.: Combine correctly to find time from A to B: Evaluate to d.p.: 0 6 = 2ω, ω = 0 ω -1 sin -1 (0 5 / 2) or ω -1 cos -1 (0 5 / 2) ω -1 sin -1 (0 75 / 2) or ω -1 cos -1 (0 75 / 2) ω -1 sin -1 (0 75 / 2) ω -1 sin -1 (0 5 / 2) or ω -1 cos -1 (0 5 / 2) ω -1 cos -1 (0 75 / 2) = ω -1 ( ) or ω -1 ( ) = A = 0 9 [s] A1 5 8

3 Page Mark Scheme Syllabus Paper Use conservation of momentum, e.g.: Use Newton s law of restitution (consistent signs): mv A + 9mv B = mu v B v A = eu Relate v A to v B using K.E. (A.E.F.): Combine two eqns to find v A and v B e.g.: ½ mv A 2 + ½ 9mv B 2 = ¼ mu 2 v A = (1 9e) u /10, v B = (1 + e) u /10 or v A, v B = u /2, u /6 [or 7 u/10, u /0] A1 Use in rd eqn to find e, e.g.: (A0 if finally ±⅔) (1 9 e) 2 + 9(1 + e) 2 = e 2 = 0, e = ⅔ A1 7 Use Newton s law of restitution with Use conservation of momentum to find k: v C = 2v B, e.g.: v C v B = ev B, v B = ⅔v B [v B = u/6, v B = u/9, v C = 2u/9] 9mv B + kmv C = 9mv B 9v B + 2kv B = 1 5v B, k = 9/ A1 10

4 Page Mark Scheme Syllabus Paper (i) Use conservation of energy at lowest point: Use F = ma radially at lowest point: Eliminate v 2 to find R [v 2 = 2 ga]: ½ mv 2 = ½ mu 2 + mga R mg = mv 2 / a R = mu 2 / a + mg = mg (ii) Use conservation of energy at B to find V B : (A.E.F.) ½ mv 2 B = ½ mu 2 + mga sin θ A1 V 2 B = ( ) ga, V B = (0 8 ga) or 2 (ga / 5) or 0 89 (ga) A1 (iii) Use vertical component v B of speed V B at B: Find height h reached above B: Find height h reached above level of O: v B = V B cos θ [= ¼ 15 V B = (¾ ga)] h = v B 2 /2g = a / 8 A1 h a sin θ = a / 8 ¼ a = a / 8 A.G. A Find MI of components about A: ( for BC or CD) Find total MI about A: (OR can first find total MI about centre of mass) State or imply total mass acts at mid-point of AC Use eqn of circular motion to find d 2 θ / dt 2 : Approximate sin θ by θ and substitute for I: Glass (M / 5) {⅓ (5 a) a 2 } = 20 Ma 2 A1 AB M{⅓ ( a) 2 + ( a) 2 } = 6 Ma 2 / AD ⅓ M{⅓ (a) 2 + ( a) 2 } = Ma 2 BC ⅓ M{⅓ (a) a 2 } = 76 Ma 2/ A1 CD M{⅓ ( a) a 2 } = 172 Ma 2/ A1 I = 128 Ma 2 A.G. A1 I d 2 θ/dt 2 = [ ] (9Mg/15) 5a sin θ A1 d 2 θ/dt 2 = (9g/8a) θ A1 8 Find period T = 2 π / ω with ω = (9 g / 8 a): T = 2π (8a/9g) or (16π/7) (6a/g) or 17 6 (a/g) (A.E.F.) 5 1

5 Page 5 Mark Scheme Syllabus Paper 6 State or find the expected value of X: using p = ¼: E(X) = 1 / p = 1 /¼ = 1 (i) Find P (X = ): P (X = ) = (¾) ¼ = 27 / 256 or A1 2 (ii) Find P (XI6): P (X < 6) = 1 (¾) 5 or {1 + ¾ + (¾) 2 + (¾) + (¾) }¼ S.R. Using p = ½ can earn B0 A0 M0 A0 = 781 / 102 or 0 76 A (i) State probability density function of T: f (t) = exp ( t) (t 0) [ = 0 (otherwise or t < 0)] 1 (ii) Find P (T > 2000): S.R. 1 e -2 = earns only (max 1/) State inequality for t (lose A1 if = or ): Solve for t max: (Omitting power 10 earns 0/; using 1 (exp ( t )) 10 can earn A0 A0 only) P (t > 2000) = 1 F (2000) = 1 (1 e -2 ) = e -2 or 0 15 A1 (exp ( t)) 10 [or >] 0 9 A1 t max = (ln 0 9) / (-0 01) = 10 5 A1 8

6 Page 6 Mark Scheme Syllabus Paper 8 State hypotheses (B0 for χ ): Estimate both popln. variances using two samples: (allow use of biased: σ X, 60 2 = 26 or ) (allow use of biased: σ Y, 50 2 = 265 or ) Estimate population variance for combined sample: (allow σ X, 60 2 /60 + σ Y, 50 2 /50: 9 2 or 09 2 ) Calculate value of z (to 2 d.p., either sign): State or use correct tabular z value (to 2 d.p.): (or can compare 6 with e.g s = 7 1 or 7 07) Correct conclusion (A.E.F, on z values): S.R. Assuming equal population variances: Find pooled estimate of common variance s 2 : Calculate value of z (to 2 d.p., either sign): Tabular value; conclusion H 0: µ X = µ Y, H 1: µ X µ Y S 2 x = ( / 60) / 59 [= 20 or ] And s 2 Y = ( / 50) / 9 [= 270 or 16 2 ] A1 s = s X / 60 + s Y / 50 = 9 08 or A1 z = (101 95) / s = 6/ 067 = 1 96 (or 1 97) A1 z 0.99 = 2 26 or 2 (allow 2 6) [Accept H 0 ] Claims are the same Hypotheses; Explicit assumption (; ) : s 2 = ( / /50) / 108 ( A1) z = 6 / s (1/60+1/50) = 1 97 ( A1) = 25 8 or (A1) As above (; ) 9

7 Page 7 Mark Scheme Syllabus Paper 9 Find expected frequency p: Find q by similar method or by using total of 200: State (at least) null hypothesis: Calculate χ 2 (to s.f.): State or use correct tabular χ 2 value (to s.f.):: Valid method for reaching conclusion: Conclusion consistent with correct values (A.E.F): p = (1 / x ln 8) dx = (200 / ln 8) [ln x] 2 = = 9 00 A.G. A1 q = 21 6 or 21 5 A1 H 0 : f (x) fits data (A.E.F.) χ 2 = = 11 A1 χ 2 6, 0.95 = Accept H 0 if χ 2 Y tabular value Distribution fits observations A Find correlation coefficient r: r = ( / 10) / {( / 10) ( / 10)} A1 (A.E.F.; A0 if only s.f. clearly used) = / ( ) A1 = *A1 State both hypotheses (B0 for r ): H 0 : ρ = 0, H 1 : ρ 0 State or use correct tabular two-tail r-value: r 10, 5% = 0 62 * Valid method for reaching conclusion: Reject H 0 if r > tabular value Correct conclusion (A.E.F, dep *A1, *): There is non-zero correlation A1 Calculate gradient p in x χ = p ( y γ ) : p = / = 1 20 Find regression line of x on y: x = ( y 6 9 ) = 1 20 y A1 11

8 Page 8 Mark Scheme Syllabus Paper 11 A (i) (ii) (iii) Use Pythagoras to find AB: AB = (a a 2 ) = a A.G. A1 Find SAB: CAB = sin -1 2a / a or cos -1 2a / a or tan -1 2a / 2a = 60 so SAB = 0 A.G. A1 EITHER Resolve vertically and horizontally, e.g.: ½ N A + ½ N B + ½ F A = W (F A may be in either direction) and ½ N A = ½ N B + ½ F A A1 Eliminate N B + F A to find N A: N A = ½ W A.G. A1 OR Resolve in dirn. PQ to find N A: N A = ½ W A.G. ( A1) Second resolution, e.g. in dirn. PS: N B + F A = ½ W (A1) Take moments, e.g. about A: ½ W a / 2 + ½ W (2 )a (A1 for each side of eqn) = N B 2 a A1 A1 Solve to find N B: N B = {(7 6)/8} W A1 Use N B to find F A: F A = N A N B or ½ W N B = {(2 ) / 8}W (A.E.F.) A1 7 1

9 Page 9 Mark Scheme Syllabus Paper B Estimate population variance: s 2 P = ( / 8) / 7 (allow biased here: or ) = 51 / 50 or 1 00 or Find confidence interval (allow z in place of t) e.g.: 2 8 / 8 ± t (s 2 P / 8) Use correct tabular t-value: t 7, = 2 65 A1 Evaluate C.I. correct to 2 d.p.: 5 5 ± 0 8 or [ 51, 6 19] A1 Formulate inequality for k (or equality for k max ): (5 5 k) / (s 2 P / 8) [ [or K] t Use correct tabular t-value: t 7, 0.9 = 1 15 A1 Solve for k max (A0 if = or Y=was used for k above): 5 5 k [ 0 50, k max = 85 A1 State hypotheses (B0 for x ), e.g.: H 0 : µ P = µ Q, H 1 : µ P K=µ Q State assumption (A.E.F.): Normal distns. for [P and] Q Estimate (pooled) common variance: s 2 = ( ) / 18 and equal variances = or A1 Calculate value of t (to s.f.): t = (5 5 60) / (s (1 / / 12)) = 1 0 A1 Correct conclusion (A.E.F., on t): t I t 18, 0.9 = 1 so Q s mean is not less than P s 7 1

9231 FURTHER MATHEMATICS

9231 FURTHER MATHEMATICS CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International Advanced Level MARK SCHEME for the October/November 015 series 931 FURTHER MATHEMATICS 931/1 Paper, maximum raw mark 100 This mark scheme is