9231 FURTHER MATHEMATICS

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1 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 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. 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 October/November 015 series for most Cambridge IGCSE, Cambridge International A and AS Level components and some Cambridge O Level components. IGCSE is the registered trademark of Cambridge International Examinations.

2 Page Scheme Syllabus Paper Scheme Details 1 Find 3 independent equations for T, R A, R B : Resolve horizontally: R B = T cos α M1 A1 Resolve vertically: R A = W + T sin α M1 A1 Take moments about A: R B 3a sin θ = W (3a/) cos θ (a may be omitted from moment eqns) + T a(sin α cos θ + cos α sin θ ) or + T a sin (α + θ ) or + T 3a cos θ sin α M1 A1 Take moments about B: Take moments about C: Take moments about D: R A 3a cos θ = W (3a/) cos θ + T a(sin α cos θ + cos α sin θ ) or + T a sin (α + θ ) or + T 3a sin θ cos α (M1 A1) R A a cos θ + W (a/) cos θ = R B a sin θ (M1 A1) R A 3a cos θ W (3a/) cos θ = R B 3a sin θ (M1 A1) Solve for T, R A, R B (AEF in W and α): T = W / sin α or ½W cosec α B1 R A = 3W / B1 R B = W / tan α or ½W cot α B1 9 9 For A & B use conservation of momentum, e.g.:mv A + mv B = mu (allow v A + v B = u) Use Newton s law of restitution (consistent signs): v B v A = eu M1 M1 Combine to find v A and v B : v A = ( e) u/3, v B = (1 + e) u/3 A1, A1 Find e from v A = v B with v B = [ ] 0 4 v B : ( e) = 0 8 (1 + e), e = /3 M1 A1 4 EITHER: Equate times in terms of reqd. distance x: (d x)/ v A = d/v B + x/v B (AEF) M1 A1 [speeds need not be found: v A = v B = 4u/9, v B = 10u/9] Use v A = v B = 0 4 v B to solve for x: d x = 0 4 d + x, x = 0 3 d M1 A1 OR: Find dist. moved by A when B reaches wall: d A = (d/v B ) v A = 0 4 d (M1 A1) Find reqd. distance x: x = ½ (d d A ) = 0 3 d (M1 A1) 4 10

3 Page 3 Scheme Syllabus Paper Scheme Details 3 Find k by equating equilibrium tensions: mg (a/)/a = mg (3a/ ka) / ka M1 A1 (vertical motion can earn M1 only) ½ = 3/k, k = 6/5 or 1 A1 3 Apply Newton s law at general point, e.g.: m d x/dt = mg (a/ + x)/a (lose A1 for each incorrect term) + mg (3a/ ka x) / ka or m d y/dt = + mg (a/ y)/a mg (3a/ ka + y) / ka M1 A Simplify to give standard SHM eqn, e.g.: d x/dt = (1 + /k)gx / a S.R.: B1 if no derivation (max /5) = 8gx/3a A1 State or find period using π/ω with ω = (8g/3a): T = π (3a/8g) or π (3a/g) ( on ω) or 3 85 (a/g) or 1 a [s] B1 5 Substitute values in v = ω (x 0 x ): 0 7 = (8g/3a){(0 a) (0 05a) } M1 A1 Solve to find numerical value of a: 0 49 = (8g/3) a, a = 0 49 A1 3 11

4 Page 4 Scheme Syllabus Paper Scheme Details 4 EITHER: Find tension at top from F = ma vertically: T = mu /a mg B1 OR: Use energy at e.g. θ to upward vertical: ½ mv = ½ mu + mga (1 cos θ ) Find tension T by using F = ma radially: T = mv /a mg cos θ Eliminate v : = mu /a + mg ( 3 cos θ ) Find T at top by taking θ = 0: T = mu /a mg (B1) Find u min by requiring T 0 at top [or T > 0]: u /a g 0 so u min = ag A.G. B1 Find v at bottom from conservation of energy: ½mv = ½mu + mg a M1 v = ag + 4ag, v = (5ag) A1 Find new speed V from conservation of momentum: m V = mv with m = m + ¼m M1 V = 4v/5 = 4 (ag/5) or (4/5) (5ag) AEF A1 4 Find w at angle θ from conservation of energy: ½ m w = ½ m V (condone m instead of m here since cancels out) m ga (1 + cos θ ) M1 A1 [ w = ag (6/5 cos θ )] S.R. Invalid energy method (max /5): ½ m w = ½ mu [gives T = (5mg/4)( 3 cos θ ) ] + mga (1 cos θ ) ¼mga (1 + cos θ ) (B1) Find tension T there by using F = ma radially: T = m w /a m g cos θ B1 Eliminate w : = m V /a m g ( + 3 cos θ ) A1 Substitute for m and V: = (5mg/4)(6/5 3 cos θ ) AEF or 3mg/ (15/4) mg cos θ A1 Find cos θ when string becomes slack from T = 0: cos θ = ⅓ 6/5 = /5 or 0 4 M1 A1 S.R. Allow if found from T = mg (6/5 3 cos θ ) Find or use sample mean x = 8 / 10 = 8 and estimate population variance: s = 4 1 / 9 (allow biased here: 0 41 or 0 64 ) = or 103/5 or M1 Find confidence interval (allow z in place of t) e.g.: 8 ± t (0 458 / 10) M1 A1 Use of correct tabular value: t 9, = 6[] A1 Evaluate C.I. correct to 3 s.f.: 3 ± 0 48[4] or [1 8, 8] A1 5 5

5 Page 5 Scheme Syllabus Paper Scheme Details 6 Find prob. p of head from mean = variance: 1/p = (1 p)/p, p = ⅔ A.G. M1 A1 (i) Find P(X = 4) (denoting 1 p by q [= ⅓]): P(X = 4) = q 3 p = /81 or B1 1 (ii) (iii) Find or state P(X > 4): P(X > 4) [= 1 (1 + q + q + q 3 ) p = 1 (1 q 4 ) ] = q 4 = 1/81 or M1 A1 Formulate condition for N: 1 q N > 0 999, [ (⅓) N < ] M1 Take logs (any base) to give bound for N: N > log / log ⅓ M1 Find N min : N > 6 9, N min = 7 A1 (N < 6 9 or N = 6 9 earns M A0) Find F(x) for 1 x 4: F(x) = (x 3 1)/63 B1 Find G(y) from Y = X for 1 x 4: G(y) = P(Y < y) = P(X < y) = P(X < y 1/ ) = F(y 1/ ) (result may be stated) = (y 3/ 1)/63 M1 A1 Find g(y) for corresponding range of y: g(y) = y 1/ /4 A.G. A1 (i) (ii) Find or state corresponding range of y : 1 y 16 A.G. B1 Find median value m of Y: (m 3/ 1)/63 = ½ m = 3 5 /3 = 10 M1 A1 Find E(Y) [or equivalently E(X )]: E(Y) = y g(y) dy = y 3/ dy /4 = [y 5/ ] 1 16 /105 = 103/105 = 341/35 or 9 74 M1 A1 5 9

6 Page 6 Scheme Syllabus Paper Scheme Details 8 Find mean of sample data [for use in Poisson distn.]: λ = 0/100 = B1 State (at least) null hypothesis (AEF): H 0 : Poisson distn. fits data or λ = B1 Find expected values 100λ r e -λ /r! (to 1 d.p.): (ignore incorrect final value here for M1) M1 A1 Combine last two cells so that exp. value 5: O i : 3 E i : 7 5 M1* Calculate value of χ (to d.p.; A1 dep M1*): χ = (allow 7 95 if 1 d.p. exp.values used) = 7 99 M1 A1 State or use consistent tabular value (to 3 s.f.): 5 cells: χ 3,0.95 = cells: χ 4, 0.95 = (correct) 7 cells: χ 5, 0.95 = B1 State or imply valid method for conclusion e.g.: Accept H 0 if χ < tabular value M1 Conclusion (AEF, requires both values correct): Distn fits or λ = A1 Not combining cells [so χ = 8 64] can earn B1 B1 M1 A1 M0 M1 B1 M1 (max 7) Calculate gradient b 1 in y y = b 1 (x x) : S xy = /8 = 1 79 S xx = /8 = 10 b 1 = S xy / S xx = (3 s.f.) M1 A1 Find regression line of y on x: y = 400/8 + b 1 (x 47/8) M1 A1 = (x 59) = x Find y when x = 7: = 57 9 or 58 Allow use of regression line of x on y (since neither variable clearly independent): S yy = /8 = 1 6 b = S xy / S yy = (M1 A1) x = 47/8 + b (y 400/8) (M1 A1) = y Y = 6 5 or 6 (A1) 5

7 Page 7 Scheme Syllabus Paper Scheme Details Find product moment correlation coefficient r: r = 1 79 / ( ) or ( ) = M1 A1* State both hypotheses (B0 for r ): H 0 : ρ = 0, H 1 : ρ 0 B1 State or use correct tabular two-tail r-value: r 8, 5% = B1* State or imply valid method for conclusion e.g.: Reject H 0 if r > tab. value (AEF) M1 Correct conclusion (AEF, dep A1*, B1*): There is non-zero correlation A A Find MI of lamina about Q: I lamina = ⅓m{(3a) + (3a/) } + m(9a/) M1 A1 [= (15/4 + 81/4) ma = 4 ma ] State or find MI of rod about Q: I rod = (⅓ + 1) M (3a/) [= 3Ma ] B1 Sum to find MI of object about Q: I 1 = 4 ma + 3 Ma = 3 (8m + M) a A.G. A1 Find MI of object about mid-point of PQ: I = (15/4 + 3 ) ma + ⅓ M (3a/) Use eqn of circular motion to find d θ/dt for axis l 1 : [ ]I 1 d θ/dt = (51/4) ma + ¾ Ma = ¾ (17m + M) a A.G. M1 A1 = mg (9a/) sin θ + Mg (3a/) sin θ M1 A1 [ = (9m/ + 3M/) ga sin θ ] [Approximate sin θ by θ and] find ω 1 in SHM eqn: ω 1 = (3m + M)g / (8m + M) a M1 4 Find period T 1 for axis l 1 from π/ω 1 : (AEF) T 1 = π {(8m + M) a / (3m + M)g} A1 Use eqn of circular motion to find d θ/dt for axis l : [ ]I d θ/dt = mg 3a sin θ M1 [Approximate sin θ by θ and] find ω in SHM eqn: ω = 4mg / (17m + M) a M1 Find period T for axis l from π/ω : (AEF) T = π {(17m + M) a / 4mg} A1 Verify that T 1 = T when m = M: (AEF) T 1 = π (18 a / 4g) = T B1 [Taking m = M throughout nd part can earn: M1 A1 M1 A0 M1 M1 A0 B1 (max 6/8)] 8 14

8 Page 8 Scheme Syllabus Paper Scheme Details 10B State hypotheses (B0 for x ), e.g.: H 0 : µ X = µ Y, H 1 : µ X µ Y B1 State assumption (AEF): Distributions have equal variances B1 Find sample means and estimate popln. variances: x = 4, y = 4 8 s X = (180 4 /10) / 9 (allow biased here: 0 36 or 0 6 ) = 0 4 or s Y = ( /1) / 11 (allow biased here: or ) = or 51/550 or M1 Estimate (pooled) common variance: s = (9 s X + 11 s Y ) / 0 (AEF) (note s X and s Y not needed explicitly) or (180 4 / /1) / 0 = or M1 A1 Calculate value of t (to 3 s.f.): [-] t = ( y x) / s (1/10 + 1/1) = 13 M1 A1 State or use correct tabular t value: t 0, = 086 [allow 09] B1* (or can compare y x = 0 6 with 0 586) Correct conclusion (AEF, on t, dep *B1): t > 09 so mean masses not same B1 S.R. Implicitly taking s X, s Y as popln. variances: z = ( y x ) / (s X /10 + s Y /1) (may also earn first B1 B1 M1) = 0 6 / (0 078 = 15 Comparison with z and conclusion: 15 > 1 96 (can earn at most 5/9) so mean masses not same (B1) 9 State hypotheses (B0 for x ), e.g.: H 0 : µ X = 3 8, H 1 : µ X > 3 8 or H 0 : µ X = µ Z, H 1 : µ X > µ Z B1 Calculate value of t using s X from above: t = (4 3 8) / (s X / 10) = 0 M1 A1 State or use correct tabular t value: t 9, 0.95 = [allow 1 83] B1* (or can compare 0 4 with 0 367) Correct conclusion (A.E.F., on t, dep *B1): t > 1 833, so claim is justified or mean mass of Royals > mean mass of Crowns B1 5 14

9231 FURTHER MATHEMATICS

9231 FURTHER MATHEMATICS 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