MATHEMATICS - M1-M3 & S1-S3

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1 GCE MARKING SCHEME MATHEMATICS - -M3 & S1-S3 AS/Advanced SUMMER 015

2 INTRODUCTION The marking schemes which follow were those used by WJEC for the Summer 015 examination in GCE MATHEMATICS -M3 & S1-S3. They were finalised after detailed discussion at examiners' conferences by all the examiners involved in the assessment. The conferences were held shortly after the papers were taken so that reference could be made to the full range of candidates' responses, with photocopied scripts forming the basis of discussion. The aim of the conferences was to ensure that the marking schemes were interpreted and applied in the same way by all examiners. It is hoped that this information will be of assistance to centres but it is recognised at the same time that, without the benefit of participation in the examiners' conferences, teachers may have different views on certain matters of detail or interpretation. WJEC regrets that it cannot enter into any discussion or correspondence about these marking schemes. Page 1 M 10 M3 18 S1 5 S 8 S3 31

3 Q Solution Mark Notes 1. R a Mg NL applied to man R and Mg opposing. dim correct R Mg = Ma 680 = M( ) M = 68 cao NL applied to Lift and Man T and weight opposing. dim correct. T 1868g = 1868a ft M T = (N) ft M 1

4 . Apply NL to B dim correct, all forces. allow 5a RHS 5g T = 0 5g and T opposing. Resolve perpendicular to plane for A allow sin R = 4gcos Apply NL to A Friction opposes motion. Allow 4a RHS and/or cos T 4gsin - F = 0 At limiting equilibrium F = R used F 45g 15 = convincing R 48g 16 T = 5g = 49 45g 441 F = T 4gsin = = = g 35 R = 4g = =

5 3(a) Conservation of momentum attempted, equation, dim correct = 3v A + 5 v B 3v A + 5 v B = 34 Restitution v B v A = ( 8) v B v A = 3v A + 5 v B = 34-3v A + 3v B = 6 Adding m1 dep on both M s 8v B = 40 v B = 5 (ms -1 ) cao v A = 3(ms -1 ) cao 3(b) Impulse = change of momentum used I = = 15 (Ns) ft v A or v B 3

6 4 Moments about x-axis =5(-1) + (3) y = 16 si y = 1 cao Moments about y-axis = (-) + 6(-3) 16x = 0 si x = 0 cao 4

7 5(a) R A R B A 0.9 C B 15g 80g Moments about A 3 terms, dim correct Equation required.8r B = 80g g1.4 correct equation any correct moment R B = 35.5 (N) cao Vertical forces in equilibrium all forces, no extra R A + R B = 80g + 15g R A = (N) cao 5(b) 1.5R R A 0.9 C x B 15g 80g Resolve vertically 1.5R + R =95g R = 38g Moments about A 3 terms, dim correct.8r = 80g gx oe 17 x = =.3 (m) 75 cao 5

8 6(a) v ms -1 0 O T t s labels, units and shape (0, 0) to (10, 0) (10, 0) to (10+T, 0) 6(b) v = u + at, v=0, u=0, t=10 0 = a a = (ms - ) 6(c) Total distance = area under graph attempted D = T one correct area D = T (m) cao 6(d) s = ut + 0.5at, u=0, t=5+t, a= s = 0.5 (5+T) D = T + T T + T = T Ft exp for D in (d) and (c) T 10T 75 = 0 (T + 5)(T 15) = 0 T = 15 cao D = 400 (m) cao 6

9 7 Resolve in 80 N direction Equation required 80 = Pcos60º + Qcos45º Resolve in 5 N direction Equation required 5 = Psin60º - Qsin45º 160 = P + Q 50 = P3- Q Adding m1 dep on both M s (1 + 3)P = 10 P = 76.9 cao Q = 58.8 cao penalise once if not 1 d.p. 7

10 8(a) Use of v =u +as with u=(±).1,a=(±)9.8, s=(±)4. v = v = 9.1 allow - 4 speed of rebound = m1 = 5. (ms -1 ) convincing 8(b) n 4 We require smallest n st < 1 oe, si trial & error 4 bounces 8

11 9 BCD (5) for 19 ABDE (5) Circle 9 7 (5) both 8 and 7 required Lamina 05-9 x (y) expressions for areas, oe Moments about AE (05-9)x = signs correct. Ft table if at least one for c of m gained. x = cao y = 5 9

12 M Q Solution Mark Notes 1. x.y = 0 used (sin i + cos j).(i j) (= 0) correct method for dot product, no i, j s sin -cos (= 0) sin - (1 sin ) = 0 m1 cos = 1 sin depends on both M s sin + sin -1 = 0 (sin - 1)(sin + 1) = 0 sin = =, 6 6 both values sin = -1 3 = 10

13 (a) Apply NL to object 1600 R = 50a 1600 kt = 50a R = kt When t =, a= -4 m1 used 1600 k = 50 (-4) k = 900 dv t = 50 dt d v = 3 18t dt convincing (b) dv 3 18t dt increase in power at least once v = 3t 9t (+ C) When t =, v = 41 m1 used C = C = 13 cao v = 9t + 3t + 13 When v = 8, 8 = 9t + 3t + 13 m1 substitution of v=8 in c s expression for v(t). 9t 3t + 15 = 0 (9t 5)(t 3) = 0 t = 9 5, 3 cao 11

14 3. NL dim correct, all forces T mgsin - R = ma correct equation P T = v used si 5P R = 6000 correct equation in P & R 5P - R = NL with a = 0 dim correct, all forces T mgsin - R = 0 correct equation 3P R = 0 correct equation in P & R 3P - R = Solving simultaneously m1 eliminating one variable, depends on both M s P 16 = 1000 P = 96000; R = both answers cao 1

15 4(a) NL (4t - 3)i + (3t 5t)j = 0.5a use of F = ma a = (8t - 6)i + (6t 10t)j cao v = a dt attempted, i, j retained, power of t increased once v = (4t 6t)i + (t 3 5t )j + (c) ft a of same diff, not F When t = 0, v = 8i 7j c = 8i 7j v = (4t 6t)i + (t 3 5t )j + 8i 7j v = (4t 6t + 8)i + (t 3 5t -7)j 4(b) Impulse = change in momentum attempted, vector form required When t = 3, v = 6i + j si ft c s v 0.5(xi + yj) 0.5(6i + j) = i 9j (xi + yj) = 30i 16j cao ft c s x, y Speed = 34 ms -1 cao Speed =

16 5(a) T = 15g si Hooke s Law x 1470 x T l x 1470 x = 0.04 (m) cao 5(b) Let PE be zero at the natural length level. PE = mgh used Initial PE = (-0.16) Initial PE = -3.5 J 1 Initial EE = used l Initial EE = 0 4 Initial EE = J x Final KE = 0.5mv Final KE = 7.5v Final PE = Final PE = J Final EE = 0 4 Final EE = J Conservation of energy equation, all 3 types 7.5v = all correct, any form v = v = = 1.87 (ms -1 )(to d.p.) 14

17 6(a) Initial u H = 35cos = ( = 1) (ms -1 ) si Initial u V = 35sin = ( = 8) (ms -1 ) si use of s= ut+0.5at with s=0,u=8(c),a=(±)9.8 complete method 0 = 8t+0.5(-9.8)t ft u V t(8 4.9t) = 0 40 t = (0), 7 Total distance travelled by ball = Ball will not fall into lake. = 10 (m) (b) time to tree = 1 6 Use v=u+at with u=8(c),a=(±)9.8,t=5/6(c) 5 v = oe complete method v = 119 (= ) speed = 1 m1 6 speed = 8.89 (ms -1 ) cao = tan m1 = 43.36º cao 15

18 7 a R 100g Resolve vertically equation, dim correct No extra force Rcos1º = 100g R = (N) NL towards the centre of motion dim correct, no extra force 100 v Rsin1º = 80 v = 1.91 cao 16

19 8(a)(i) Conservation of energy KE and PE = (1 cos ) v 5 = v cos v = cos cao 8(a)(ii) NL towards centre of motion dim correct, 3 terms T, 3gcos opposing 3v T 3gcos = 0 8 T = 3gcos ( cos ) m1 ft v of form a±bsin/cos T = cos cao 8(b) Greatest value of occurs when T= cos = 0 ft Tof form a±bsin/cos cos = - 88 = º ft a+bcos Motion stops being circular when = º as string cannot support negative tension. P moves under the action of gravity only. E1 ft >90º 17

20 M3 Q Solution Mark Notes 1(a). Use of NL F = 400a x dv dv v use of a= v v dx d x dv 5x 4vv d x 1(b)(i) 5xdx 4v vdv 5 x 3 v 4 3 sep variables v C When x = 0, v = 0, hence C = v x = 5 3 v m1 any correct form 1(b)(ii)When v = 3.5x = 4(9 + 9) m1 1 x = m = 5.37 m 5 cao a = a = 5 1 m1 substitution of x and v= = 1.34 (ms - ) cao 5 18

21 (a)(i). NL 0.5a = -6.5x v dimensionally correct 1 d x 13 dx d x d x x a= dt dt dt, v=. d t d x d x x = 0 dt d t (a)(ii) Axilliary equation m + 4m + 13 = 0 m = - ± 3i C. F. is x = e -t (Asin3t + Bcos3t) ft m if complex d x When t=0, x=6, =3 dt m1 used B = 6 d x = -e -t (Asin3t + Bcos3t) dt + e -t (3Acos3t - 3Bsin3t) ft e kt (Asinpt + Bcospt) -B + 3A = 3 A = 5 Solution is x = e -t (5sin3t + 6cos3t) cao When t is large, x 0 (b) Try PI x = at + b 4a + 13(at + b) = 91t a = 91 m1 equating coefficients a = 7 4a + 13b = 15 b = -1 cao both G.S. is x = e -t (Asin3t + Bcos3t) +7t 1 19

22 3(a) NL 50a = 50g 50v dimensionally correct d v 5 d t = 5g v convincing 3(b) 5dv 5g v -5ln dt separation of variables 5 g v = t (+C) correct integration When t = 0, v = 0 m1 used -5ln 5 g = C 5 t t 5g v ln 5g 5 5ge 5g v m1 correct inversion t v 5g 1 e 5 cao When t = 5, v = 5g(1 e -1 ) = (ms -1 ) cao numerical answer. 3(c) t d x = 5g - 5 5ge v = dt t 5 x = 5gt + 5ge (+C) correct integration ft similar expression dx dt When t = 0, x = 0 m1 used C = -5g t 5 x = 5gt + 5ge -5g When t = 5, x = 5ge -1 = (m) cao 0

23 4(a) A y C 1.4-y B Tension of spring at A = Tension of spring at B = 15( y 03) 0.3 0(1 4 y 06) 0.6 When in equilibrium T A = T B 15( y 03) 0(1 4 y 06) = all correct 30y 9 = 16 0y 50y = 5 y = 0.5 (m) convincing 1

24 4(b)(i) A x 0.5 C 0.9-x B 15(0 x) T A = 0.3 0(0 3 x) T B = 0.6 either 0(0 3 x) 15(0 x) Force to right = allow =/- 50x = - 3 d 50x Apply NL to P, 7.5 = - dtx 3 d x 100 = x dt si or =100/9 Period = Therefore motion is SHM with = 3 3 = 5 convincing 4(b)(ii) Amplitude = 0.5 (m) 4(b)(iii)Use v = (a -x 10 ), =,a=0.5,x=0. ft a and. oe 3 v 10 =( ) ( ) 3 v = 0.5 (ms -1 ) cao 4(b)(iv)x = acos(t) oe allow sin/cos, c s a, = 0.5cos( t) 3 3 t = cos -1 0 ( ) t = (s) cao

25 5. A v J l 3l 60º J u B 8 ms -1 v 5(a) Sine rule sin sin10 l l 3 sin = 0.5= 30º = 60º 30º = 30º 5(b) Impulse = change in momentum used. Allow sin/cos. Apply to B J = 5 8cos30º - 5v Apply to A J = 3v Solving simultaneously m v = 3v Speed of A = v = u = 8sin30º = 4 (ms -1 ) 5 3 = 4.33 (ms -1 ) cao 5 3 Speed of B = 4 m1 = 5.9 (ms -1 ) cao J = 3v = 1.99 (Ns) ft c s 3v 3

26 6 A S 80g 0g R 0.6R B Resolve vertically equation, no missing and no extra force. R = 80g + 0g (= 100g) Resolve horizontally equation, no missing and no extra force. S = 0.6R = 60g = 588 (N) Moments about B equation, no missing and no extra force. Dimensionally correct. 80g 5cos + 0g 3cos = S 6sin A -1 each error 360sin = 460cos 460 = tan = 51.95º cao The ladder is modelled as a rigid rod. 4

27 Ques Solution Mark Notes 1(a) (b) E(X) = 3,Var(X) =.1 si E(Y) = E(X) + 1 = 7 Var(Y) = 4Var(X) = 8.4 P(Y = 7) = P(X = 3) (a) S = = 0.67 P(AB) = P(A) + P(B) P(AB) oe P(AB) = P(AB) P(AB) = 0.3 Award just for this line Award M0A0 for no working Accept or Award for a valid verification (b) P( A B) P( A B) P( B) 0.3 = Accept the use of a Venn diagram in (b) and (c) (c) P( B A) P( B A ) ( = P( A) = = 3 1 (0.33) P( B) P( B A) ) 1 P( A) 3(a) P(A chooses G) = 0.3 (b) (c) 8 P(B chooses Y) = 10 = 0. 3 P(Diff colours) = = Accept 0. without working Accept C1 C1 C1 C1 C1 C 10 C 1 4(a)(i) (ii) e P(X = 9) = 9! = P(X < 1) = Accept or Award M0 if no working seen Award A0 if in adjacent row or column (b) Looking at the appropriate section of the table, n = 19 Award A0 for 18 or 0 5

28 5(a)(i) (ii) P(male and bike) = = P(owns a bike) = = 0.57 (b) 0.1 P(female bike) = 0.57 = 0.11 (4/19) cao num, denom FT denominator from (a) 6(a) (i) (ii) (b) Let X = no. of defective cups so X is B(50,0.05) P ( X ) = 0.61 P(3 X 8) = or = Let Y = no. of defective plates so Y is B(50,0.015) Po(3.75) si e 3.75 P( Y = 4) = 4! = si Accept or M0A0 if no working Award no marks if no working seen M0A0 if no working 7(a) (b) k k k E (X ) = 3. Or equivalent. Accept verification (c) The possible pairs are (3,4), (4,3), (,6),(6,) Prob = = 0.13 (16/75) for (3,4),(,6) A0A0 if factor missing 6

29 8(a) P(1 st hit with 3 rd throw) = = (b)(i) (ii) (iii) P(F wins 1 st throw) = P(G misses) P(F hits) = = 0.4 P(F wins with nd throw) = P(G miss) P(F miss) P(G miss) P(F hits) = = (a) P(F wins) = E = (4x x )dx X 9 x = 4 9 x x 3 1 = (0/7) Award this for realising that the probability is the sum of an infinite geometric series 1 for the integral of f ( x) x for completely correct although limits may be left until nd line Award M0 if no working (b)(i) 4 F( x) 9 x u 8x 9 (4u u 3 4 u 4 4 x 9 )du x Allow x as dummy variable Limits may be left until next line but must then be applied (ii) (iii) P(1.5 X 1.75) = F(1.75) F(1.5) = (9/16) The median m satisfies 4 8m m m 8m FT from (b)(i) if possible FT from (b)(i) if possible 8 m m Condone the absence of 7

30 Ques Solution Mark Notes 1(a) H 0 : 10; H1 : 10 (b) x x 10 = Test statistic = 1. /10 =.11 Value from tables = p-value = Strong evidence that the mean speed has changed. Award M0 if 10 omitted (a) (b) (c) S 95 th percentile = = 86.1 Let X=weight of a man, Y=weight of a woman z z 0.5 P( Y 1.5) or P( Y 0.5) P( Y 0.5) or P( Y 1.5) P(64 Y 68) Let U = 3 X 4 i i1 i1 E(U) = = 506 Var(U) = = z = Prob = Y i FT from line above Accept mean speed has decreased FT the p-value if less than 0.05 M0 if no working 3(a) Let X, Y = measured sugar contents of A,B ( x 161; y 1584 ) x 01.5; y 198 (b) SE of diff of means= (0.75) % confidence limits for the difference are [1.57, 5.43] z 0.75 z = 1.75 Confidence level = 9% m1 M0 if 8 omitted or only one term Award this for z if m1 given FT from (a) 8

31 4(a) (b) Under H 0, X is B(0,0.4) si P( X 13) P( X 14) X 14 has significance level closest to 1% Let Y = number of hits Under H 0, Y is B(10, 0.4) N(48, 8.8) si Test statistic = 8.8 = 1.1 p-value = Insufficient evidence to conclude that his shooting has improved Award for valid attempt at using tables Award A0 for 13 or 15 Award A0 for incorrect or no continuity correction but FT for following marks No cc gives z = 1.30, p = Wrong cc z = 1.40, p = FT the p-value 5 Let X = score on a single die. Then E(X) = 3.5 and Var(X) = Let Y = mean of scores on 100 dice. Then by the Central Limit Theorem, Y N(3.5, 35/100) z 35/100 = ()1.46 Prob = (a)(i) H 0 : 1.; H1 : 1. (ii) Under H 0, X is Po(1) si P ( X 9) = 0.44 Insufficient evidence to conclude that the (mean) number of breakdowns has decreased. (b) Under H 0, Y is Po(10) N(10,10) z p-value = Strong evidence to conclude that the (mean) number of breakdowns has decreased. m1 FT their mean and variance Use of continuity correction gives z = 1.43, p = Accept 1 in place of 1. FT the p-value Award A0 for incorrect or no continuity correction but FT for following marks No cc gives z = 1.73, p= Wrong cc, z = 1.78, p = FT the p-value if less than

32 7(a)(i) (ii) (b) P( Y y) P( X y) P( X y y a b a Attempting to differentiate, y giving b a y f ( y) for a y b b a = 0 otherwise ) We are given that a b 5.5 and Solving, a =.5, b = 8.5 ( b a)

33 Ques Solution Mark Notes 1 The sample space is as follows. EITHER S3 OR Samples R M 1,, 1 1,,4 3 1,,6 5 1,,6 5 1,,4 3 1,,6 5 1,,6 5 1,4, ,4, ,6,6 5 6,,4,,6 4,,6 4,4,6 4 4,4,6 4 4,6,6 4 6,4,6 4 4,4,6 4 4,6, ,6,6 6 Samples R M No. of ways 1,, 1 1 1,,4 3 1,, ,4, ,6, ,,4 1,,6 4,4, ,6, ,6,6 6 1 for the samples column for the R column for the M column Minus if 1 or rows omitted Minus A if 3 or 4 rows omitted for columns 1 and 4 for the R column for the M column Minus if 1 or rows omitted Minus A if 3 or 4 rows omitted 31

34 The probability distribution of R is therefore r P(R=r) 1/0 /0 /0 8/0 7/0 The probability distribution of M is therefore FT for both tables from (a) if sum of probabilities is 1 (a) (b) m 4 6 P(M=m) 10/0 6/0 4/0 x 19.9; x UE of = UE of = DF = 11 si Crit value = % confidence limits are giving [14.9,17.] Must be seen No working need be seen M0 division by 1 Answer only no marks FT their s and mean M0 use of z-values M0 if 1 omitted Answer only no marks 3(a) (b) H 0 : a b; H1 : SE = (= ) Test stat = = 1.68 Tabular value = p-value = Insufficient evidence to conclude that there is a difference in mean weight. a b Treat taking the variances as SDs as a misread, giving SE = 1.09, Test stat = 0.6, p-value = M0 if 100 omitted FT the p-value 4(a) 54 p ˆ 0.6 si ESE = = si 90 90% confidence limits are giving [0.515,0.685] 3

35 (b)(i) p ˆ Only award if used to find SE in (b)(i) (ii) n 1.96 n n = 34 cao Number of red squirrels = = 04 m1 Award this even if 0.6 used instead of FT the n from (b)(i) 5(a) (b)(i) x 100, x y , 50, xy S / si xy S /5 50 si xx 153 b = a = = SE of a = % confidence limits are [330., 331.9] ( ) B Minus 1 each error FT from (a) (ii) 0.5 SE of b = ( ) Test stat = Critical value = 1.96 or p-value = Reject H 0 FT from (a) 33

36 6(a)(i) (ii) (b)(i) (ii) (c) E( X ) 33 4(1 6 ) 4 10 Therefore E( X ) 4 10 si E( U ) a(4 10 ) b for all 1 4 a ; b U X Var( X ) (1 6 ) (4 10 ) 0 (1 5 ) Var( X ) Var( U ) a n (1 5 ) 5n Y is B(n,1 6) so E(Y) = n(1 6) Therefore E(V) = cn(1 6) + d = (for all ) 1 1 c ; d 6n V Y 6 6n Var( V ) c Var( Y) c npq (1 6 ) 6n Var( U) (1 5 ) 6n 6 30 Var( V ) 5n (1 6 ) 5 30 This ratio is greater than 1 so that V is the better estimator. Convincing 34

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MATHEMATICS - C1-C4 & FP1-FP3

MATHEMATICS - C1-C4 & FP1-FP3 GCE MARKING SCHEME MATHEMATICS - C-C4 & FP-FP AS/Advanced SUMMER 5 INTRODUCTION The marking schemes which follow were those used by WJEC for the Summer 5 eamination in GCE MATHEMATICS - C-C4 & FP-FP. They