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Advanced Subsidiary GCE Further Mathematics (MEI) (3896). Advanced Subsidiary GCE Further ... 4751 (C1) Introduction to Advanced Mathematics. 3. 4752 (C2) Concepts for Advanced Mathematics. 8. 4753 (C3) Methods for Advanced ...
GCE Mathematics (MEI) Advanced GCE A2 7895-8 Advanced Subsidiary GCE AS 3895-8

Mark Schemes for the Units June 2009

3895-8/7895-8/MS/R/09 1

OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of pupils of all ages and abilities. OCR qualifications include AS/A Levels, GCSEs, OCR Nationals, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching/training, administration and secretarial skills. It is also responsible for developing new syllabuses to meet national requirements and the needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support which keep pace with the changing needs of today’s society. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by Examiners. It does not indicate the details of the discussions which took place at an Examiners’ meeting before marking commenced. 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 should be read in conjunction with the published question papers and the Report on the Examination. OCR will not enter into any discussion or correspondence in connection with this mark scheme. © OCR 2009 Any enquiries about publications should be addressed to: OCR Publications PO Box 5050 Annesley NOTTINGHAM NG15 0DL Telephone: Facsimile: E-mail:

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4751

Mark Scheme

June 2009

CONTENTS Advanced GCE Mathematics (MEI) (7895) Advanced GCE Further Mathematics (MEI) (7896) Advanced GCE Further Mathematics (Additional) (MEI) (7897) Advanced GCE Pure Mathematics (MEI) (7898) Advanced Subsidiary GCE Mathematics (MEI) (3895) Advanced Subsidiary GCE Further Mathematics (MEI) (3896) Advanced Subsidiary GCE Further Mathematics (Additional) (MEI) (3897) Advanced Subsidiary GCE Pure Mathematics (MEI) (3898) MARK SCHEME FOR THE UNITS Unit/Content

Page

4751 (C1) Introduction to Advanced Mathematics

3

4752 (C2) Concepts for Advanced Mathematics

8

4753 (C3) Methods for Advanced Mathematics

11

4754 (C4) Applications of Advanced Mathematics

17

4755 (FP1) Further Concepts for Advanced Mathematics

23

4756 (FP2) Further Methods for Advanced Mathematics

28

4757 Further Pure 3

34

4758 Differential Equations

42

4761 Mechanics 1

46

4762 Mechanics 2

52

4763 Mechanics 3

56

4764 MEI Mechanics 4

60

4766 Statistics 1

63

4767 Statistics 2

67

4768 Statistics 3

71

4769 Statistics 4

76

4771 Decision Mathematics 1

80

4772 Decision Mathematics 2

85

4773 Decision Mathematics Computation

90

1

4751

Mark Scheme

June 2009

4776 Numerical Methods

95

4777 MEI Numerical Computation

97

Grade Thresholds

103

2

4751

Mark Scheme

June 2009

4751 (C1) Introduction to Advanced Mathematics Section A 1

2

3

(0, 14) and (14/4, 0) o.e. isw

2 ( s − ut ) o.e. as final answer t2 ( s − ut ) ] [condone [a =] 0.5t 2 [ a =]

10 www

4

M2 for evidence of correct use of gradient with (2, 6) eg sketch with ‘stepping’ or y − 6 = −4(x − 2) seen or y = −4x + 14 o.e. or M1 for y = −4x + c [accept any letter or number] and M1 for 6 = −4 × 2 + c; A1 for (0, 14) [c = 14 is not sufficient for A1] and A1 for (14/4, 0) o.e.; allow when x = 0, y = 14 etc isw

3

4

M1 for each of 3 complete correct steps, ft from previous error if equivalent difficulty [eg dividing by t does not count as step – needs to be by t2]

[ a =]

( s − ut ) 1 2 2

t

gets M2 only (similarly

other triple-deckers) M1 for f(3) = 1 soi and A1 for 31 − 3k = 1 or 27 − 3k = −3 o.e. [a correct 3-term or 2-term equation]

3

3

long division used: M1 for reaching (9 − k)x + 4 in working and A1 for 4 + 3(9 − k) = 1 o.e.

4

5

x < 0 or x > 6 (both required) (i) 10 www

2 2

equating coeffts method: M2 for (x − 3)(x2 + 3x −1) [+ 1] o.e. (from inspection or division)

3

B1 each; if B0 then M1 for 0 and 6 identified;

2

M1 for

5× 4× 3 5× 4 or or for 3 × 2 (×1) 2 (×1)

1 5 10 (ii) 80 www or ft 8 × their (i)

2

10

5

1 seen

B2 for 80x3; M1 for 23 or (2x)3 seen

3

4

16

4751 6

Mark Scheme

June 2009

any general attempt at n being odd and n being even

M1

M0 for just trying numbers, even if some odd, some even

n odd implies n3 odd and odd − odd = even

A1

n even implies n3 even and even − even = even

A1

or n(n2 − 1) used with n odd implies n2 − 1 even and odd × even = even etc [allow even × odd = even] or A2 for n(n − 1)(n + 1) = product of 3 consecutive integers; at least one even so product even; odd3 − odd = odd etc is not sufft for A1 SC1 for complete general method for only one of odd or even eg n = 2m leading to 2(4m3 − m)

7

8

(i) 1

2

(ii) 1000

1

(i) 2/3 www

2

B1 for 50 or for 25 × 1/25 o.e. 3 M1 for 4/6 or for

48 = 2 12 or 4 3 or

27 = 3 3 or 108 = 3 12 or for (ii) 43 − 30 2 www as final answer 9

10

3

(i) (x + 3)2 − 4

3

(ii) ft their (−a, b); if error in (i), accept (−3, −4) if evidence of being independently obtained

2

3

4 9

M2 for 3 terms correct of 25 − 15 2 − 15 2 + 18 soi, M1 for 2 terms correct B1 for a = 3, B2 for b = − 4 or M1 for 5 − 32 soi B1 each coord.; allow x = −3, y = −4; or M1

5

 −3  o.e. oe for sketch with −3 and −4  −4 

for 

marked on axes but no coords given

(x2 − 9)(x2 + 4)

M2

or correct use of quad formula or comp sq reaching 9 and −4; allow M1 for attempt with correct eqn at factorising with factors giving two terms correct, or sign error, or attempt at formula or comp sq [no more than two errors in formula/substn]; for this first M2 or M1 allow use of y etc or of x instead of x2

x2 = 9 [or −4] or ft for integers /fractions if first M1 earned x = ±3 cao

M1

must have x2; or M1 for (x + 3)(x − 3); this M1 may be implied by x = ±3 A0 if extra roots if M0 then allow SC1 for use of factor theorem to obtain both 3 and −3 as roots or (x + 3) and (x − 3) found as factors and SC2 for x2 + 4 found as other factor using factor theorem [ie max SC3]

A1

4

5

4

20

4751

Mark Scheme

June 2009

Section B 11

i

ii

y = 3x eqn AB is y = -1/3 x + 3 o.e. or ft

3x = -1/3x + 3 or ft x = 9/10 or 0.9 o.e. cao

2

M1 for grad AB =

1− 3 or − 1/3 o.e. 6

M1 A1

need not be simplified; no ft from midpt used in (i); may be seen in (i) but do not give mark unless used in (ii) eliminating x or y , ft their eqns if find y first, cao for y then ft for x

A1

ft dep on both Ms earned

M1

4

y = 27/10 oe ft their 3 × their x iii

2

9 2   (1 + 3 ) o.e and completion  10 

2

2

or square root of this; M1 for 2

2

 9   27    +   or 0.81 + 7.29 soi or ft  10   10 

to given answer

their coords (inc midpt) or M1 for distance = 3 cos θ and tan θ = 3 and M1 for showing sin θ =

3 and 10

completion iv

v

2 10

2

9 www or ft their a 10

2

M1 for 62 + 22 or 40 or square roots of these

2

M1 for ½ × 3 × 6 or

1 9 × their2 10 × 10 2 10

5

2

2

12

4751

12

Mark Scheme

iA

expansion of one pair of brackets

M1

correct 6 term expansion

M1

June 2009

eg [(x + 1)](x2 − 6x + 8); need not be simplified eg x3 − 6x2 + 8x + x2 − 6x + 8; or M2 for correct 8 term expansion: x3 − 4x2 + x2 − 2x2 + 8x − 4x − 2x + 8, M1 if one error allow equivalent marks working backwards to factorisation, by long division or factor theorem etc or M1 for all three roots checked by factor theorem and M1 for comparing coeffts of x3

iB

cubic the correct way up x-axis: −1, 2, 4 shown y-axis 8 shown

G1 G1 G1

with two tps and extending beyond the axes at ‘ends’ ignore a second graph which is a translation of the correct graph

iC

ii

[y=](x − 2)(x − 5)(x − 7) isw or (x − 3)3 − 5(x − 3)2 + 2(x − 3) + 8 isw or x3 − 14x2 + 59x − 70

2

M1 if one slip or for [y =] f(x − 3) or for roots identified at 2, 5, 7 or for translation 3 to the left allow M1 for complete attempt: (x + 4)( x + 1)(x − 1) isw or (x + 3)3 − 5(x + 3)2 + 2(x + 3) + 8 isw

(0, −70) or y = −70

1

allow 1 for (0, −4) or y = −4 after f(x + 3) used

27 − 45 + 6 + 8 = −4 or 27 − 45 + 6 + 12 = 0

B1

or correct long division of x3 − 5x2 + 2x + 12 by (x − 3) with no remainder or of x3 − 5x2 + 2x + 8 with rem −4

long division of f(x) or their f(x) + 4 by (x − 3) attempted as far as x3 − 3x2 in working

M1

or inspection with two terms correct eg (x − 3)(x2 ………..− 4)

x2 − 2x − 4 obtained

A1

[ x =]



(x − 1)2 = 5

( −2 )

2

− 4 × ( −4 )

2

2 ± 20 o.e. isw or 1 ± 5 2

or

M1

2

3

3

dep on previous M1 earned; for attempt at formula or comp square on their other ‘factor’

A1 5

6

13

4751

13

Mark Scheme

i

1 1

(5, 2)

20 or 2 5 ii

iii

iv

no, since 20 < 5 or showing roots of y2 − 4y + 9 = 0 o.e. are not real

y = 2x − 8 or simplified alternative

2

June 2009

0 for ± 20 etc or ft from their centre and radius M1 for attempt (no and mentioning

20 or 5) or sketch or solving by formula or comp sq (−5)2 + (y − 2)2 = 20 [condone one error]

2

or SC1 for fully comparing distance from x axis with radius and saying yes

2

M1 for y − 2 = 2(x − 5) or ft from (i) or M1 for y = 2x + c and subst their (i) or M1 for ans y = 2x + k, k ≠ 0 or −8

2

(x − 5)2 + (2x)2 = 20 o.e.

M1

subst 2x + 2 for y [oe for x]

5x2 − 10x + 5[ = 0] or better equiv.

M1

expanding brackets and rearranging to 0; condone one error; dep on first M1

obtaining x = 1 (with no other roots) or showing roots equal

M1

one intersection [so tangent]

A1

o.e.; must be explicit; or showing line joining (1,4) to centre is perp to y = 2x +2

(1, 4) cao

A1

allow y = 4

alt method y − 2 = −½ (x − 5) o.e. 2x + 2 − 2 = −½ (x − 5) o.e. x=1 y = 4 cao showing (1, 4) is on circle

M1 M1 A1 A1 B1

line through centre perp to y = 2x + 2 dep; subst to find intn with y = 2x + 2

alt method perp dist between y = 2x − 8 and y = 2x + 2 = 10 cos θ where tan θ = 2 showing this is

20 so tgt

x = 5 − 20 sin θ x=1 (1, 4) cao

2

by subst in circle eqn or finding dist from centre = 20 [a similar method earns first M1 for eqn of diameter, 2nd M1 for intn of diameter and circle A1 each for x and y coords and last B1 for showing (1, 4) on line – award onlyA1 if (1, 4) and (9, 0) found without (1, 4) being identified as the soln]

M1 M1 M1 A1 A1

7

or other valid method for obtaining x allow y = 4

5

11

4752

Mark Scheme

June 2009

4752 (C2) Concepts for Advanced Mathematics Section A 1

2

3

4

using Pythagoras to show that hyp. of right angled isos. triangle with sides a and a is √2a completion using definition of cosine

M1

www

A1

a any letter or a number NB answer given

2 x6 + 5x

value at 2 − value at 1 131

M2 M1 A1

M1 if one error ft attempt at integration only

(i) 193

2

M1 for 8 + 15 + …+ 63

(ii) divergent + difference between terms increasing o.e.

1

(i) 2.4

2

(ii) 138

2

4

3 M1 for 43.2 ÷ 18 M1 for their (i) ×

180 or π

43.2 × 360 o.e. or for other rot 36π versions of 137.50…

θ=

(i)sketch of cosx ; one cycle, sketch of cos2x; two cycles, both axes scaled correctly

1 1 D1

(ii) (1-way) stretch parallel to y-axis sf 3

1 D1

6

y′ = 3x2 − 12x − 15 use of y′ = 0, s.o.i. ft x = 5, – 1 c.a.o. x < −1 or x > 5 ft

7

use of cos2 θ = 1 − sin2 θ at least one correct interim step in obtaining 4 sin2 θ − sin θ = 0.

M1 M1 A1 A1 A1 M1 M1

5

8

B1 B1 B1

attempt to integrate 3 x − 5

M1

[y=] 2 x − 5 x + c subst of (4, 6) in their integrated eqn 3

c = 10 or [y=] 2 x 2 − 5 x + 10

4

5 for two terms correct

5

NB answer given

θ = 0 and 180, 14.(47…) 165 − 166

3 2

2

r.o.t to nearest degree or better -1 for extras in range

A2 M1

5

A1 for two terms correct

A1 5

8

4751 9

Mark Scheme

(i) 7

1

(ii) 5.5 o.e.

2

June 2009

M1 for at least one of 5 log10a or ½ log10a or log10a5.5 o.e.

3

Section B 10

0.6(0...), 0.8(45...), [1], 1.1(76...) 1.3(0...), 1.6(0...) points plotted correctly ft ruled line of best fit

T1

Correct to 2 d.p. Allow 0.6, 1.3 and 1.6

P1 L1

tol. 1 mm

b = their intercept a = their gradient -11 ≤ b ≤ -8 and 21 ≤ a ≤ 23.5

M1 M1 A1

3

iii

34 to 35 m

1

1

iv

29 = ‘22’logt – ‘9’

M1

t = 10‘1.727...’

M1

55 [years] approx

A1

For small t the model predicts a negative height (or h = 0 at approx 2.75) Hence model is unsuitable

1

iA

10 + 20 + 30 + 40 + 50 + 60

B1

iB

correct use of AP formula with a = 10 and d = 10

M1

n(5 + 5n) or 5n(n + 1) or 5(n2 + n) or (5n2 + 5n)

A1

10n2 + 10n − 20700 = 0 45 c.a.o.

M1 A1

iiA

4

1

iiB

£2555

2

iiC

correct use of GP formula with a =5, r = 2

M1

5(2n − 1) o.e. = 2621435

D M1

i

ii

v

11

2n = 524288 www

3

accept 53 to 59

3

D1

2 or

6 2

(2 × 10 + 5 × 10) or 62 (10 + 60)

1

Or better 4 1 M1 for 5(1 + 2 + …28) or 5(29 − 1) o.e.

2

‘S’ need not be simplified

M1 19 c.a.o. 12

i

6.1

4 A1 2

9

(3.1

2

M1 for

) (

)

− 7 − 32 − 7 o.e. 3 .1 − 3

2

4751

Mark Scheme ii

iii

iv

v

((3 + h)

2

) (

− 7 − 32 − 7 h numerator = 6h + h2 6+h

)

M1

June 2009

s.o.i.

M1 A1

3

as h tends to 0, grad. tends to 6 o.e. f.t.from “6”+h

M1 A1

2

y − 2 = ‘6’ (x − 3) o.e. y = 6x − 16

M1 A1

At P, x = 16/6 o.e. or ft

M1 M1 A1

At Q, x = 7 0.021 c.a.o.

10

6 may be obtained from

dy dx

2

3

4753

Mark Scheme

June 2009

4753 (C3) Methods for Advanced Mathematics Section A 1



π 6

0

sin 3x dx = [− 13 cos 3 x ]06 π

= − 13 cos π2 + cos 0 =

B1 M1

1 3

A1cao [3]

2(i)

100 = Ae0 = A  A = 100

  

50 = 100 e–1500k e–1500k = 0.5 –1500k = ln 0.5 k = –ln 0.5 ÷ 1500 = 4.62 × 10–4

(ii) 1 = 100e–kt  -kt = ln 0.01  t = -ln 0.01 ÷ k = 9966 years

-1

g(x) = 2 x − 1



b = 2 0 − 1 = 2 or (0, 2) 2 x −1 = 0



x = 1, so a = 1 or (1, 0)

0.33 or better.

M1

50 = A e–1500k ft their ‘A’ if used

M1 A1 [5]

taking lns correctly 0.00046 or better

M1

ft their A and k

M1 A1 [3]

taking lns correctly art 9970

M1

Can use degrees or radians reasonable shape (condone extra range)

B1

passes through (-1, 2π), (0, π) and (1, 0)

A1

good sketches – look for curve reasonably vertical at (–1, 2π) and (1, 0), negative gradient at (0, π). Domain and range must be clearly marked and correct.

1

[3]

4

substituting correct limits in ±k cos …

M1A1

3 2π

[− 13 cos 3x] or [− 13 cos u ]

B1 M1 A1 [3]

11

Allow unsupported answers. www x = 1 is A0

www

4753

Mark Scheme

5(i) e2y = 1 + sin x dy = cos x  2e2y dx cos x dy =  dx 2e 2 y

(ii) 2y = ln(1 + sin x)  y = ½ ln(1 + sin x) dy = 1 cos x  dx 2 1 + sin x cos x = as before 2e 2 y

6

x +1 +1 f f( x ) = x − 1 x +1 −1 x −1 x +1 + x −1 = x +1− x +1

=

2x = x* 2

f–1(x) = f(x) Symmetrical about y = x. 7(i) (A) (x – y)(x2 + xy + y2) = x3 + x2y + xy2 – yx2 – xy2 – y3 = x3 – y3 *

(B) (x + ½ y)2 + ¾ y2 = x2 + xy + ¼ y2 + ¾ y2 = x2 + xy + y2

(ii) x3 – y3 = (x – y)[(x + ½ y)2 + ¾ y2]

(x + ½ y)2 + ¾ y2 > 0 [as squares ≥ 0]  

if x – y > 0 then x3 – y3 > 0 if x > y then x3 > y3 *

June 2009

M1

Their 2e2y ×

B1

2e2y

A1 [3]

o.e. cao

B1 M1 B1 E1 [4]

dy dx

chain rule (can be within ‘correct’ quotient rule with dv/dx = 0) 1/u or 1/(1 + sin x) soi www

M1

correct expression

M1

without subsidiary denominators e.g. = x + 1 + x − 1 × x −1

x −1 x +1− x +1

E1 stated, or shown by inverting B1 B1 [5] M1

expanding - allow tabulation

E1

www

M1

(x + ½ y)2 = x2 + ½ xy + ½ xy + ¼ y2 o.e.

E1 [4]

cao www

M1

substituting results of (i)

M1 E1 [3]

12

4753

Mark Scheme

8(i) A: 1 + ln x = 0  ln x = –1 so A is (e–1, 0)  x = e–1 B: x = 0, y = e0 – 1 = e–1 so B is (0, e–1)

(ii) Either by invertion: e.g. y = ex – 1 x ↔ y x = ey – 1  ln x = y – 1  1 + ln x = y or by composing e.g. fg(x) = f(1 + ln x) = e1 + ln x – 1 = eln x = x

1



0

0

d

1 dx x = x ln x – x + c







e −1

g ( x ) dx =



e −1

M1

e1 + ln x – 1 or 1 + ln(ex – 1)

M1

= x ln x − x. 1

taking lns or exps

M1 A1cao [3]

 ln x dx =  ln x dx ( x) dx

1

SC1 if obtained using symmetry condone use of symmetry Penalise A = e–1, B = e–1, or co-ords wrong way round, but condone labelling errors.

M1 E1

M1

1

= e0 – e–1 = 1 – e–1

(iv)

A1 B1

E1 [2]

[ ]

e x −1dx = e x −1

M1

E1 E1 [5]

C: f(1) = e1–1 = e0 = 1 g(1) = 1 + ln 1 = 1

(iii)

June 2009

[e ] o.e. or u = x – 1  [e ] x −1

substituting correct limits for x or u o.e. not e0 , must be exact.

parts: u = ln x, du/dx = 1/x, v = x, dv/dx = 1

A1 A1cao

(1 + ln x) dx

condone no ‘c’

B1ft

ft their ‘x ln x – x’ (provided ‘algebraic’)

DM1

substituting limits dep B1

= [ x + x ln x − x ]e−1 1

= [ x ln x)]e−1 1

–1

–1

= 1ln 1 – e ln(e ) = e–1 *

E1 [6]

www

M1

Must have correct limits

(v) Area =



1 0

1

f( x) d x −  −1 g( x) d x e

1



f ( x ) dx −

0

= (1 – e–1) – e–1 2 =1– e

u

1



e −1

g ( x ) dx

A1cao

13

0.264 or better.

4753

Mark Scheme

June 2009

or Area OCB = area under curve – triangle = 1 – e–1 – ½ × 1 × 1 = ½ – e–1

M1

Area OAC = triangle – area under curve = ½ × 1 × 1 – e–1 = ½ – e–1 2 Total area = 2(½ – e–1) = 1 – e

A1cao

OCA or OCB = ½ – e–1

or

[2]

14

0.264 or better

4753

Mark Scheme

June 2009

Section B 9(i)

1 3

a=

2 (ii) d y = (3x − 1)2 x −2 x .3 dx (3x − 1) 2 6 x − 2 x − 3x 2 = (3x − 1) 2 3x 2 − 2 x = (3x − 1) 2 x(3x − 2) * = (3x − 1)2

dy = 0 when x(3x – 2) = 0 dx

(iii) 

2 3

x = 0 or x = , so at P, x =

2 3

2 (2 / 3)2 4 when x = , y = = 3 3 × (2 / 3) − 1 9

dy = –0.1875 dx dy = 0.1633 when x = 0.8, dx Gradient increasing  minimum when x = 0.6,

B1 [1]

or 0.33 or better

M1 A1

quotient rule

E1 [3]

www – must show both steps; penalise missing brackets.

M1

if denom = 0 also then M0

A1

o.e e.g. 0.6 , but must be exact

M1 A1cao

o.e e.g. 0.4 , but must be exact

.

.

B1

-3/16, or -0.19 or better

B1 E1 [7]

8/49 or 0.16 or better o.e. e.g. ‘from negative to positive’. Allow ft on their gradients, provided –ve and +ve respectively. Accept table with indications of signs of gradient.



B1

(u + 1)2 9 o.e. u



M1

× 13 (du)

M1

expanding

E1

condone missing du’s

B1

[u

M1

substituting correct limits, dep integration

x2 dx u = 3x − 1  du = 3dx (iv) 3x − 1 (u + 1) 2 1 9 du = u 3 1 (u + 1) 2 1 u 2 + 2u + 1 du = du = 27 u 27 u 1 1 = (u + 2 + ) du * 27 u







x2 dx 2 3 3x − 1 When x = 23 , u =1, when x = 1, u = 2 Area =

1



=

1 27

=

1 1 27 2

=

1 27

1

2

 (u + 2 + u ) du 1

[u

2

]

2

+ 2u + ln u 1

[(2 + 4 + ln 2) − ( 12 + 2 + ln1)]

15

1 2

2

+ 2u + ln u

]

4753 =

Mark Scheme 1 27

 7 + 2 ln 2  (3 12 + ln 2) = 54  

A1cao [7]

16

June 2009

o.e., but must evaluate ln 1 = 0 and collect terms.

4754

Mark Scheme

June 2009

4754 (C4) Applications of Advanced Mathematics Section A 1

4 cos θ − sin θ = R cos(θ + α ) = R cos θ cos α − R sin θ sin α  R cos α = 4, R sin α = 1



 R2 = 12 + 42 = 17, R = 17 = 4.123 tan θ = ¼  θ = 0.245 17 cos(θ + 0.245) = 3 cos(θ + 0.245) = 317

 

θ + 0.245 = 0.756, 5.527 θ = 0.511, 5.282

2



 





dy = 3x 2 y dx dy = 3 x 2 dx y ln y = x3 + c when x = 1, y = 1,  ln 1 = 1 + c  c = -1 ln y = x3 – 1



y = ex

4

When x = 0, y = 4

3

 







3

M1

θ + 0.245 = arcos 3/√17 ft their R, α for method (penalise extra solutions in the range (-1))

M1

correct partial fractions

M1 A1 A1

substituting, equating coeffts or cover-up A=1 B = -1

B1

ln(x + 1) ft their A

B1

-½ ln(2x + 1) ft their B

A1 [7]

cao – must have c

M1

separating variables

2

0

= π 4 y − 12 y 2

condone absence of c c = -1 o.e. o.e.

B1

0 4

[

R = 17 = 4.123 tan θ = ¼ o.e. θ = 0.245

A1 [4]

 x dy = π  ( 4 − y ) dy

V =π

B1 M1 A1

A1 B1

−1

4

correct pairs

A1A1 [7]

x A B = − ( x + 1)(2 x + 1) x + 1 (2 x + 1) x = A(2x + 1) + B(x + 1) x = -1  -1 = -A A=1 x = -½  -½ = ½ B  B = -1 1 1 x = − ( x + 1)(2 x + 1) x + 1 (2 x + 1) x 1 1 dx = − dx x + 1 (2 x + 1) ( x + 1)(2 x + 1) = ln(x + 1) – ½ ln(2x + 1) + c



M1

]

4

0

= π(16 – 8) = 8π

M1

must have integral, π, x² and dy s.o.i.

M1

must have π ,their (4−y), their numerical y limits

B1

1 2   4 y − 2 y 

A1 [5]

17

4754

5



Mark Scheme dy = − a(1 + t 2 ) −2 .2t dt dx = 3at 2 dt dy dy / dt − 2at = = 2 dx dx / dt 3at (1 + t 2 ) 2

M1 A1

6    

(1+t²)-2 × kt for method

B1

−2 * 3t (1 + t 2 ) 2 At (a, ½ a) , t = 1 1 gradient = −2 2 = − 6 3× 2

M1

M1 A1 [7]

finding t

cosec2 θ = 1 + cot2 θ 1 + cot2 θ – cot θ = 3 * cot2 θ – cot θ – 2 = 0 (cot θ – 2)(cot θ + 1) = 0 cot θ = 2, tan θ = ½ , θ = 26.57° cot θ = -1, tan θ = -1, θ = 135°

E1

clear use of 1 + cot² θ = cosec² θ

M1 A1 M1 A1 A1 [6]

factorising or formula roots 2, -1 cot = 1/tan used θ = 26.57° θ = 135° (penalise extra solutions in the range (-1))

=



June 2009

ft

E1

18

4754

Mark Scheme

June 2009

Section B 7(i)

 −1  →   AB =  −2  0    0 1      r = 0 + λ  2  2 0    

B1

B1 [2]

or equivalent alternative

1 

(ii) n =  0   

B1

1   

1  1      0 . 2  1   0  1 cos θ =     = 2 5 10



(iii)



B1 M1 M1

θ = 71.57°

A1 [5]

 −1  −2      0  .  −2   −1  −1  2 + 1 1 cos φ =     = = 2 9 3 2 2

θ = 45° *

(iv) sin 71.57° = k sin 45°  k = sin 71.57° / sin 45° = 1.34

0

 −2 

 2  

 −1   

correct vectors (any multiples) scalar product used finding invcos of scalar product divided by two modulae 72° or better

M1 A1 E1 [3]

ft their n for method ±1/√2 o.e. exact

M1 A1 [2]

ft on their 71.57° o.e.

M1

s.o.i.

M1 A1 A1

subst in x + z = -1

B1 [5]

www dep on μ = 1

(v) r =  0  + μ  −2     

  

x = -2μ, z = 2 − μ x + z = -1 -2θ + 2 – θ = -1 3θ = 3, θ = 1 point of intersection is (-2, -2, 1)

distance travelled through glass = distance between (0, 0, 2) and (-2, -2, 1) = (2 2 + 2 2 + 12 ) = 3 cm

19

4754

8(i)

Mark Scheme

(A)  

(B) 

360° ÷ 24 = 15° CB/OB = sin 15° CB = 1 sin 15° AB = 2CB = 2 sin 15° *

M1 E1 [2]

cos 30° = 1 – 2 sin2 15° cos 30° = 32

B1

= 1 – 2 sin2 15°

B1

3 2



2 sin2 15° = 1 –



sin215° =



sin 15° =

3 2

M1

= (2 – 3 )/2

June 2009

AB = 2AC or 2CB ∠AOC = 15° o.e.

simplifying

2− 3 4 2− 3 = 4

1 2

2− 3 *

(C) Perimeter = 12 × AB = 24 ×

1 2

E1 [4]

M1

2− 3

= 12 2 − 3 circumference of circle > perimeter of polygon



2π > 12 2 − 3



π > 6 2− 3

E1 [2] M1

(ii) (A) tan 15° = FE ÷ OF  FE = tan 15°  DE = 2FE = 2tan 15°

(B) tan 30 = tan 30 =

2 tan 15 2t = 2 1 − tan 15 1 − t 2 1

2t 1 =  2 3t = 1 − t 2 2 1− t 3



t2 + 2√3 t – 1 = 0 *

 

B1 M1

3



(C)

E1 [2]

E1 [3]

−2 3 ± 12 + 4 = 2− 3 2 circumference < perimeter 2π < 24( 2 − 3 ) t=

M1 A1 M1 using positive root E1 [4]

π < 12( 2 − 3 ) *

20

from exact working

4754

Mark Scheme

(iii) 6 2 − 3 < π < 12( 2 − 3 )  3.106 < π < 3.215

B1 B1 [2]

21

June 2009

3.106, 3.215

4754

Mark Scheme

June 2009

Comprehension

1. 2.

1 4

× [3 + 1 + ( −1) + ( −2)] = 0.25 *

M1, E1

(i) b is the benefit of shooting some soldiers from the other side while none of yours are shot. w is the benefit of having some of your own soldiers shot while not shooting any from the other side. Since it is more beneficial to shoot some of the soldiers on the other side than it is to have your E1 own soldiers shot, b > w . (ii) c is the benefit from mutual co-operation (i.e. no shooting). d is the benefit from mutual defection (soldiers on both sides are shot). With mutual co-operation people don’t get shot, while they do with mutual defection. So c > d . E1

3.

4.

5.

1 × 2 + (−2) × (n − 2) = −1.999 or equivalent (allow n, n+2) n n = 6000 so you have played 6000 rounds.

M1, A1 A1

No. The inequality on line 132, b + w < 2c, would not be satisfied since 6 + (-3) > 2 × 1. b + w < 2c and subst No, 3 > 2 o.e.

M1 A1

(i) Round

You

Opponent

1 2 3 4 5 6 7 8 …

C D C D C D C D …

D C D C D C D C …

Your score -2 3 -2 3 -2 3 -2 3 …

Opponent’s score 3 -2 3 -2 3 -2 3 -2 …

M1 Cs and Ds in correct places, A1 C=-2, A1 D=3

(ii) 6.

1 2

× [3 + (−2)] = 0.5

DM1 A1ft their 3, -2

(i) All scores are increased by two points per round

B1

(ii) The same player wins. No difference/change. The rank order of the players remains the same. B1 7.

(i) They would agree to co-operate by spending less on advertising or by sharing equally. (ii) Increased market share (or more money or more customers).

22

B1 DB1

4755

Mark Scheme

June 2009

4755 (FP1) Further Concepts for Advanced Mathematics Section A 1(i)

(ii)

2

M −1 =

1  2 1   11  −3 4 

1  2 1  49   x  1  198    = =   11  −3 4 100   y  11  253   x = 18, y = 23

M1 A1 [2]

Dividing by determinant

M1

Pre-multiplying by their inverse

A1(ft) A1(ft) [3]

z 3 + z 2 − 7 z − 15 = ( z − 3) ( z 2 + 4 z + 5 )

B1 M1 A1 M1

−4 ± 16 − 20 2  z = −2 + j and z = −2 − j

z2 + 4z + 5 = 0  z =

A1 [5]

Show z = 3 is a root; may be implied Attempt to find quadratic factor Correct quadratic factor Use of quadratic formula or other valid method Both solutions

3(i)

(ii)

B1 B1 [2]

Asymptote at x = -4 Both branches correct

2 = x + 3  x 2 + 7 x + 10 = 0 x+4

M1

 x = −2 or x = −5

A1

Attempt to find where graphs cross or valid attempt at solution using inequalities Correct intersections (both), or -2 and -5 identified as critical values

x ≥ −2 or − 4 > x ≥ −5

A1 A2

[5] 4

2w − 6w + 3w = − 12 w=

1 2

 roots are 1, -3, −

3 2

9 q = αβγ = −  q = 9 2 2

p = αβ + αγ + βγ = −6  p = −12 2

M1 A1

Use of sum of roots – can be implied

A1 M1

Correct roots seen Attempt to use relationships between roots s.c. M1 for other valid method One mark each for p = -12 and q = 9

A2(ft) [6] 23

x ≥ −2 −4 > x ≥ −5 s.c. A1 for −4 ≥ x ≥ −5 or − 4 > x > −5

4755

5(i)

Mark Scheme

1 1 5r + 3 − 5r + 2 − ≡ 5r − 2 5r + 3 ( 5r + 3)( 5r − 2 ) ≡

(ii)

M1

Attempt to form common denominator

A1 [2]

Correct cancelling

B1

First two terms in full

B1

Last term in full

M1

Attempt to cancel terms

5

( 5r + 3)( 5r − 2 )

  r =1  ( 5r − 2 ) ( 5r + 3 )    1 1   1 1   1 1   −  +  −  +  −  + ...   1  3 8   8 13   13 18   =  5  1 1   1 1   +  5n − 7 − 5n − 2  +  5n − 2 − 5n + 3  1

30

 ( 5r − 2 )( 5r + 3)

=

June 2009

=

1

30



 5

1



1

r =1

1 1

1  n − =   5  3 5 n + 3  3 ( 5n + 3 )

A1

[4] 6

When n = 1,

1 2

n ( 7 n − 1) = 3, so true for n = 1

Assume true for n = k

= = =

= =

1 2 1 2 1

2

Assume true for n = k

M1

Add ( k + 1) th term to both sides

M1

Valid attempt to factorise

A1

c.a.o. with correct simplification

E1

Dependent on previous E1 and immediately previous A1 Dependent on B1 and both previous E marks

k ( 7 k − 1) + (7 ( k + 1) − 4)

[ k ( 7 k − 1) + (14 ( k + 1) − 8)]  7 k 2 + 13k + 6 

2 1 2 1

E1 1

k ( 7 k − 1) 2  3 + 10 + 17 + ..... + (7 ( k + 1) − 4)

3 + 10 + 17 + ..... + (7 k − 4) =

B1

( k + 1)( 7 k + 6 ) ( k + 1) ( 7 ( k + 1) − 1)

But this is the given result with k + 1 replacing k. Therefore if it is true for k it is true for k + 1. Since it is true for n = 1, it is true for n = 1, 2, 3 and so true for all positive integers.

E1

[7] Section A Total: 36

24

4755

Mark Scheme

Section B 7(i) 5 ( 0 , 10 ) , ( −2 , 0 ) ,  , 0  3 

(ii) x=

−1 2

, x =1, y =

June 2009

B1 B1 B1 [3]

3

B1 B1 B1 [3]

2

+

(iii)

(iv)

3 2 (e.g. consider x = 100 ) − 3 Large negative x, y → 2 (e.g. consider x = −100 ) Large positive x, y →

M1 B1 B1 [3]

Curve 3 branches of correct shape Asymptotes correct and labelled Intercepts correct and labelled

B1 B1 B1 [3]

25

Clear evidence of method required for full marks

4755

8 (i)

Mark Scheme

z − ( 4 + 2 j) = 2

June 2009

B1 B1 B1

Radius = 2 z − ( 4 + 2 j) or z − 4 − 2 j All correct

[3] (ii)

arg ( z − ( 4 + 2 j) ) = 0

B1 B1 B1

Equation involving the argument of a complex variable Argument = 0 All correct

[3] (iii)

a = 4 − 2cos b = 2 + 2sin

π 4

π 4

(

M1

=4− 2

Valid attempt to use trigonometry involving

=2+ 2

)

P =4− 2 + 2+ 2 j

A2

π 4

, or coordinate geometry

1 mark for each of a and b s.c. A1 only for a = 2.59, b = 3.41

[3] (iv)

3 π > arg ( z − ( 4 + 2 j) ) > 0 4 and z − ( 4 + 2 j) < 2

B1 B1 B1

[3]

26

arg ( z − ( 4 + 2 j) ) > 0

3 arg ( z − ( 4 + 2 j) ) < π 4 z − ( 4 + 2 j) < 2 Deduct one mark if only error is use of inclusive inequalities

4755

9(i)

Mark Scheme

Matrix multiplication is associative

B1 [1]

 3 0  0 1  MN =     0 2  1 0   0 3  MN =    2 0  −2 0  QMN =    0 3

(ii)

June 2009

M1

Attempt to find MN or QM

A1

 0 −2  or QM =   3 0 

A1(ft) [3]

M is a stretch, factor 3 in the x direction, factor 2 in the y direction.

B1 B1

N is a reflection in the line y = x.

B1

Q is an anticlockwise rotation through 90° about the origin.

B1

Stretch factor 3 in the x direction Stretch factor 2 in the y direction

[4]

(iii)

 −2 0  1 1 2   −2 −2 −4    =   0 3  2 0 2   6 0 6 

M1 A1(ft)

Applying their QMN to points. Minus 1 each error to a minimum of 0.

B2

Correct, labelled image points, minus 1 each error to a minimum of 0. Give B4 for correct diagram with no workings.

[4]

Section B Total: 36 Total: 72

27

4756

Mark Scheme

June 2009

4756 (FP2) Further Methods for Advanced Mathematics x2 x3 x4 x5 + − + ... 2 3 4 5 2 3 4 x x x x5 − − − ... ln(1 – x) = − x − 2 3 4 5 1+ x  ln  = ln(1 + x) – ln(1 – x) 1− x  2x3 2x5 + ... = 2x + 3 5 Valid for -1 < x < 1

1(a)(i) ln(1 + x) = x −

(ii)

Series for ln(1 – x) as far as x5 s.o.i.

M1

Seeing series subtracted

A1

1+ x =3 1− x  1 + x = 3(1 – x)  1 + x = 3 – 3x  4x = 2  x = 12

ln 3 ≈ 2 × 23 × ( 12 ) + 52 × ( 12 ) 3

B1

5

B1 4

Inequalities must be strict

M1

Correct method of solution

A1

B2 for x =

M1

Substituting their x into their series in (a)(i), even if outside range of validity. Series must have at least two terms SR: if >3 correct terms seen in (i), allow a better answer to 3 d.p. Must be 3 decimal places

= 1 + 121 + 801 = 1.096 (3 d.p.) (b)(i)

A1 4

1 2

stated

y 1

G1 G1 G1

0.5

–1

–0.5

0.5

1

3 M1

(ii) r + y = r + r sin θ a = r(1 + sin θ) = × (1 + sin θ) 1 + sin θ =a  r=a–y

r(0) = a, r(π/2) = a/2 indicated Symmetry in θ = π/2 Correct basic shape: flat at θ = π/2, not vertical or horizontal at ends, no dimple Ignore beyond 0 ≤ θ ≤ π Using y = r sin θ

A1 (AG)

 x2 + y2 = (a – y)2  x2 + y2 = a2 – 2ay + y2  2ay = a2 – x2 a2 − x2  y= 2a

M1 A1

Using r2 = x2 + y2 in r + y = a Unsimplified

A1

A correct final answer, not spoiled

5

28

16

4756

Mark Scheme

3− λ  2 (i) M − λ I =  0  2 

−2   −1 − λ 0  0 1 − λ  1

det(M – λI) = (3 – λ)[(–1 – λ)(1 – λ)] + 2[2(–1 – λ)] = (3 – λ)(λ2 – 1) + 4(–1 – λ)  λ3 – 3λ2 + 3λ + 7 = 0 det M = -7

M1 A1

Attempt at det(M – λI) with all elements present. Allow sign errors Unsimplified. Allow signs reversed. Condone omission of = 0

B1 3

(ii) f (λ) = λ3 – 3λ2 + 3λ + 7 f (-1) = -1 – 3 – 3 + 7 = 0  -1 eigenvalue

B1

f (λ) = (λ + 1)(λ2 – 4λ + 7) λ2 – 4λ + 7 = (λ – 2)2 + 3 ≥ 3 so no real roots (M – λI)s = 0, λ = -1 

June 2009

M1 A1

Showing -1 satisfies a correct characteristic equation Obtaining quadratic factor www (M – λI)s = (λ)s M0 below

 4 1 −2   x   0       0 0 0  y  = 0  2 0 2  z  0     

 4x + y – 2z = 0 2 x + 2z = 0  x = -z y = 2z – 4 x = 2z + 4z = 6z

M1 M1

 −1  s =  6  1    3 1 −2   x   −0.1       0 −1 0   y  =  0.6   2 0 1   z   0.1      

 x = 0.1, y = -0.6, z = -0.1

A1

Or any non-zero multiple

M1

Solution by any method, e.g. use of multiple of s, but M0 if s itself quoted without further work Give A1 if any two correct

A2 9 B1

(iii) C-H: a matrix satisfies its own characteristic equation  M3 – 3M2 + 3M + 7I = 0  M3 = 3M2 – 3M – 7I

Obtaining equations relating x, y and z Obtaining equations relating two variables to a third. Dep. on first M1

Idea of λ ↔ M

B1 (AG) Must be derived www. Condone omitted I M1 Multiplying by M–1 A1 o.e.

 M2 = 3M – 3I – 7M–1  M–1 = − 17 M 2 + 73 M − 73 I

4  3 1 −2   3 1 −2   5     2 (iv) M =  0 −1 0   0 −1 0  =  0  2 0 1  2 0 1  8      5 2 −8   3 1 −2  1 1  3  3 −  0 1 0  +  0 −1 0  −  0 7  7 2 0 1  7 0  8 2 −3     1 1 2  7  1 1 2 7 7 1   =  0 −1 0  or  0 −7 0  7  − 2 − 2 3   −2 −2 3  7 7  7

2 −8   1 0 2 −3  0 0  1 0 0 1 

29

M1

Correct attempt to find M2

M1

Using their (iii)

A1

SC1 for answer without working

4756

Mark Scheme

OR Matrix of cofactors:  −1 −1 Adjugate matrix  0 7 2 2 

 −1 0 2     −1 7 2   −2 0 −3    −2   0  : det M = –7 −3 

June 2009

M1

Finding at least four cofactors

M1

Transposing and dividing by determinant. Dep. on M1 above

3 G1

3(a)(i) 1

19 Correct basic shape (positive gradient, through (0, 0))

–1

1 M1

y = arcsin x  sin y = x dy = cos y dx dy 1 1 = = dx cos y 1− x2

 

A1 A1

Positive square root because gradient positive

1

(ii)

 0

=

4 M1

1

 x  dx = arcsin  2 2 0  2− x 1

A1

π

A1

4 jθ

3 jθ

3 M1 M1

5 jθ

(b) C + jS = e + e + e + ... This is a geometric series 1 3

1 9

with first term a = e jθ , common ratio r = 13 e 2 jθ Sum to infinity = =

= =

B1



sin y = and attempt to diff. both sides dy Or cos y =1 dx www. SC1 if quoted without working Dep. on graph of an increasing function arcsin function alone, or any sine substitution x 2



, or 1 dθ www without limits

Evaluated in terms of π

A1

Forming C + jS as a series of powers Identifying geometric series and attempting sum to infinity or to n terms Correct a and r

A1

Sum to infinity



a e 3e = (= ) 1 − r 1 − 13 e 2 jθ 3 − e 2 jθ

3e jθ 3 − e −2 jθ × 3 − e 2 jθ 3 − e −2 jθ

M1*

9e jθ − 3e − jθ 9 − 3e −2 jθ − 3e 2 jθ + 1 9 ( cos θ + j sin θ ) − 3 ( cos θ − j sin θ )

M1 M1

10 − 3 ( cos 2θ − j sin 2θ ) − 3 ( cos 2θ + j sin 2θ )

30

Multiplying numerator and denominator by 1 − 13 e −2 jθ o.e. Or writing in terms of trig functions and realising the denominator Multiplying out numerator and denominator. Dep. on M1* Valid attempt to express in terms of trig functions. If trig functions used from start, M1 for using the compound angle formulae and

4756

Mark Scheme

June 2009

Pythagoras Dep. on M1* =

6 cos θ + 12 j sin θ 10 − 6 cos 2θ

 C=

A1

6 cos θ 10 − 6 cos 2θ

M1

3cos θ 5 − 3cos 2θ 6sin θ S= 5 − 3cos 2θ

=

Equating real and imaginary parts. Dep. on M1*

A1 (AG) A1

o.e.

11 4 (i) cosh u =

19

eu + e − u 2

e 2u + 2 + e −2u 2 e 2u + e −2u 2 2 cosh u – 1 = 2

 2 cosh2 u =

B1

(e



B1

cosh 2u =

= cosh 2u

u

+ e − u ) = e 2 u + 2 + e −2 u 2

e 2u + e −2u 2

B1 (AG) Completion www 3

(ii) x = arcsinh y  sinh x = y e x − e− x 2

M1

Expressing y in exponential form ( 12 , – must be correct)

 ex = y ± y 2 + 1

M1

Take + because ex > 0

B1

Reaching ex by quadratic formula or completing the square. Condone no ± Or argument of ln must be positive

 y=

 e2x – 2yex – 1 = 0  (ex – y)2 – y2 – 1 = 0  (ex – y)2 = y2 + 1  ex – y = ± y 2 + 1

 x = ln(y + y + 1 ) 2

dx = 2 cosh u du

(iii) x = 2 sinh u 



x 2 + 4 dx =

A1 (AG) Completion www but independent of B1 4 dx and substituting for all elements M1 du



4 sinh 2 u + 4 × 2 cosh u du

A1

Substituting for all elements correctly

 4 cosh u du =  2 cosh 2u + 2 du

M1

Simplifying to an integrable form

= sinh 2u + 2u + c

A1

Any form, e.g. 12 e2u − 12 e−2u + 2u Condone omission of + c throughout

M1

Using double ‘angle’ formula and attempt to express cosh u in terms of x

2

=

= 2 sinh u cosh u + 2u + c = x 1+

=

1 2

x2 x + 2 arcsinh + c 4 2

x x 4 + x 2 + 2 arcsinh + c 2

A1 (AG) Completion www 31

4756

Mark Scheme

June 2009

6 (iv) t2 + 2t + 5 = (t + 1)2 + 4 1

B1

Completing the square

M1

Simplifying to an integrable form, by substituting x = t + 1 s.o.i. or complete alternative method Correct limits consistent with their method seen anywhere

1



t 2 + 2t + 5 dt =

−1



(t + 1) 2 + 4 dt

−1 2

=



x 2 + 4 dx

0

A1 2

x  =  12 x 4 + x 2 + 2 arcsin h  20 

M1

(v) = 8 + 2 arcsinh 1 = 2 2 + 2 ln(1 + 2 ) = 2(ln(1 + 2 ) + 2 )

A1 (AG) Completion www. Condone 8 etc. 5 18

5 (i) If a = 1, angle OCP = 45° so P is (1 – cos 45°, sin 45°) 1

 P (1 –

,

2

1

M1

)

2

A1 (AG) Completion www

OR Circle (x – 1)2 + y2 = 1, line y = -x + 1 (x – 1)2 + (-x + 1)2 = 1 1

 x=1±

2 1

Q (1 +

2

M1

and hence P ,-

1 2

Using (iii) or otherwise reaching the result of integration, and using limits

Complete algebraic method to find x

A1

)

B1

3 a

(ii) cos OCP =

a2 + 1 1

sin OCP =

a2 + 1

P is (a – a cos OCP, a sin OCP)  a2 a   Pa − ,   a2 +1 a2 +1   OR Circle (x – a)2 + y2 = a2, line y = −

M1

Attempt to find cos OCP and sin OCP in terms of a

A1

Both correct

A1 (AG) Completion www 1 x +1 a

2

 1  (x – a) +  − x + 1 = a2  a  2

2

 x=

2a +

2 2  ±  2a +  a a  1  2 1 + 2  a

1   − 4 1 + 2   a    

32

M1

Complete algebraic method to find x

A1

Unsimplified

4756

Mark Scheme

 x =a±

a2 a2 +1

and hence P

June 2009

A1

 a2 a  Qa + ,−   a2 +1 a2 +1  

B1 4

(iii)

1

0.5

–5

G1 G1 G1 G1ft

5

–0.5

–1

As a → ∞, P → (0, 1) As a → –∞, y-coordinate of P → –1 a 2

a +1



B1 B1 M1 A1 8

a = −1 as a → –∞ −a

(iv) POQ = 90° Angle in semicircle Loci cross at 90°

B1 B1 B1 3

33

Locus of P (1st & 3rd quadrants) through (0, 0) Locus of P terminates at (0, 1) Locus of P: fully correct shape Locus of Q (2nd & 4th quadrants: dotted) reflection of locus of P in yaxis Stated separately Stated Attempt to consider y as a → –∞ Completion www

o.e. 18

4757

Mark Scheme

June 2009

4757 Further Pure 3 1 (i) Putting x = 0, -3y + 10z = 6, -4y −2z = 8 y = -2, z = 0  8   3      Direction is given by  − 3  ×  − 4   10   − 2     

 46    =  46   − 23   

=

or finding a second point

5 M1

 − 1  2       14  ⋅  5   5  14      2

2

2 + 5 + 14

2

138 46 = = 9.2 15 5

6   AB × d =  15  = 6 2 + 15 2 + 42 2  42   

Distance is

M1

A1 ft

  2     = 3 5   14   

 7  2 6       AB × d =  -14  ×  2  =  15   4   − 1  42       

 − 1  0      Distance is  12  −  − 2  ⋅ nˆ =  5   0   

(iii)

Finding coords of a point on the line or (2, 0, -1), (1, -1, - 12 ) etc

A1

 0  2     Equation of L is r =  − 2  + λ 2   0   − 1     (ii)

M1 A1

AB × d d

=

=

6 2 + 15 2 + 42 2 2

2

2

2 + 2 +1 45 = 15 3

Evaluating AB × d

A2 ft

Give A1 ft if just one error

M1

Appropriate scalar product

A1 ft

Fully correct expression

A1 6 M1

For AB × d

M1

Evaluating magnitude

M1 A1ft

In this part, M marks are dependent on previous M marks

A1 5

34

Dependent on M1M1 Accept any form Condone omission of ‘r =’

4757

(iv)

Mark Scheme  −1  k + 1  6  2       At D,  12  + λ  −12  =  −2  + μ  2  5  −3   9   −1        

M1

12 − 12λ = −2 + 2 μ 5 − 3λ = 9 − μ

A1 ft M1 M1

λ = 13 , μ = 5

Condone use of same parameter on both sides Two equations for λ and μ Obtaining λ and μ (numerically) Give M1 for λ and μ in terms of k Equation for k

M1

−1 + 13 (k + 1) = 6 + 10 k = 50

A1

D is (6 + 2μ, -2 + 2μ, 9 − μ) i.e. (16, 8, 4)

M1 A1 8

Alternative solutions for Q1 1 (i) e.g. 23x – 23y = 46 x = t, y = t − 2 3t − 4(t − 2) − 2z = 8 x = t, y = t − 2, z = - 12 t

(ii)

June 2009

Obtaining coordinates of D

M1A1 Eliminating one of x, y, z M1 A1 ft A1 5

 − 1 + 7μ   2 λ      PQ = 12 − 14μ  −  − 2 + 2 λ   5 + 4μ   -λ     

M1 A1 ft

PQ ⋅ d = PQ ⋅ AB = 0 A1 ft 2( −1 + 7μ − 2 λ) + 2(14 − 14μ − 2 λ) − (5 + 4μ + λ) =0 7(−1 + 7μ − 2 λ) − 14(14 − 14μ − 2 λ) + 4(5 + 4μ + λ) M1 =0 27 47 λ= , μ= 25 75 2

2

A1 6

2

 92   230   644  PQ =   +   +  = 9.2  75   75   75 

(iii)

A1 ft

 6 + 2λ + 1   2      AX ⋅ d =  − 2 + 2 λ − 12  ⋅  2  = 0  9 − λ − 5   − 1    

M1

2(7 + 2 λ) + 2(2 λ − 14) − (4 − λ) = 0 λ=2  11    AX =  − 10   2   

A1ft

AX = 112 + 10 2 + 2 2 = 15

M1 A1 5

M1

35

Two equations for λ and μ Expression for shortest distance

4757

Mark Scheme

June 2009

(iv)  7    k + 1  2       −14  ⋅   4     

     −12  ×  2   −3   −1    

 =0  

 7   18       −14  ⋅  k − 5  = 0  4   2k + 26      126 − 14k + 70 + 8k + 104 = 0 k = 50  −1  51   6  2       At D,  12  + λ  −12  =  −2  + μ  2          5  −3   9   −1

M1

Appropriate scalar triple product equated to zero

M1 A1

Equation for k

M1

Condone use of same parameter on both sides

−1 + 51λ = 6 + 2μ 12 − 12λ = −2 + 2μ 5 − 3λ = 9 − μ

A1 ft M1

Two equations for λ and μ Obtaining λ and μ

M1 A1 8

Obtaining coordinates of D

M1 A2

Partial differentiation Give A1 if just one minor error

λ = 13 , μ = 5

D is (6 + 2 μ , − 2 + 2μ , 9 − μ ) i.e. (16, 8, 4)

2(i) ∂z

= 3( x + y )3 + 9 x( x + y ) 2 − 6 x 2 + 24

∂x ∂z = 9 x( x + y)2 ∂y

A1 4

(ii) At stationary points,

∂z ∂z = 0 and =0 ∂x ∂y

M1

9 x ( x + y ) 2 = 0  x = 0 or y = − x

If x = 0 then 3 y 3 + 24 = 0 y = −2 ; one stationary point is (0, − 2, 0) If y = − x then −6 x 2 + 24 = 0 x = ±2 ; stationary points are (2, − 2, 32) and (−2, 2, − 32)

(iii)

At P (1, − 2, 19) ,

∂z ∂z = 24 , =9 ∂x ∂y

B1

1  24    Normal line is r =  −2  + λ  9   19   −1     

(iv)

M1 A1A1 M1 A1 A1 If A0A0, give A1 for x = ±2 7

M1 A1 ft 3

∂z ∂z δx+ δ y ∂x ∂y = 24 δ x + 9 δ y

δz ≈

M1 A1 ft

3h ≈ 24k + 9h

M1

36

For normal vector (allow sign error) Condone omission of ‘r =’

4757

Mark Scheme

A1 4

k ≈ − 14 h

M2 A1 ft A1

OR Tangent plane is 24 x + 9 y − z = −13 24(1 + k ) + 9(−2 + h) − (19 + 3h) ≈ −13 k ≈ − 14 h

(v) ∂z ∂x

June 2009

= 27 and

∂z =0 ∂y

M1

(Allow M1 for

∂z = −27 ) ∂x

9 x ( x + y ) 2 = 0  x = 0 or y = − x

If x = 0 then 3 y 3 + 24 = 27 y = 1, z = 0 ; point is (0 , 1, 0)

M1 A1

d =0

A1 M1

If y = − x then −6 x 2 + 24 = 27 x 2 = − 12 ; there are no other points

A1 6

3(i)  dx 2  dy 2 2 2  dθ  +  dθ  = [a(1 + cos θ )] + (a sin θ )    

M1 A1

2

Forming  dx  +  dy   dθ   dθ 

2

= a 2 (2 + 2 cos θ ) = 4a 2 cos 2 12 θ

s =  2a cos

1 θ 2

M1 M1 A1 A1(AG) 6



= 4a sin 12 θ + C s = 0 when θ = 0  C = 0

(ii) dy dx

= =

a sin θ a (1 + cos θ )

ρ=

Correctly obtained (+C not needed) Dependent on all previous marks

M1

2sin 12 θ cos 12 θ 2a cos 2 12 θ

M1 A1

= tan 12 θ

ψ = 12 θ , and so s = 4a sinψ

(iii)

Using half-angle formula Integrating to obtain k sin 12 θ

Using half-angle formulae

A1 4 M1 A1 ft

ds = 4a cosψ dψ

= 4a cos 12 θ

Differentiating intrinsic equation

A1(AG) 3

OR ρ= =

( 4a

2

cos 2 12 θ

)

3

2

M1 A1 ft

a (1 + cos θ )(a cos θ ) − (−a sin θ )(a sin θ ) 8a 3 cos3 12 θ a 2 (1 + cos θ )

=

8a 3 cos3 12 θ 2a 2 cos 2 12 θ

= 4a cos 12 θ

A1(AG)

37

Correct expression for ρ or κ

4757

Mark Scheme

(iv) When θ = 2 π , ψ = 1 π , x = a( 2 π + 1 3), y = 3 a 2 3 3 3 2 ρ = 2a  − sinψ   − 12 3  nˆ =  = 1   cosψ   2   a ( 2 π + 1 3)  − 1 3 2  + 2a  21  c= 3 3     a  2  2  

Centre of curvature is ( a( 23 π − 12 3) ,

B1 M1 A1

5 2

a)

A1A1 Accept (1.23a , 2.5a) 6 M1

π

M1

Correct integral expression in any form (including limits; may be implied by later working) Obtaining an integrable form

M1

Obtaining k sin 3 12 θ

A1 ft

=  2π a (1 − cos θ ) 2a cos 12 θ dθ 0 

π

=  8π a 2 sin 2 12 θ cos 12 θ dθ 0

= 

16 π a2 3

Obtaining a normal vector Correct unit normal (possibly in terms of θ )

M1

(v) Curved surface area is  2π y ds 

June 2009

π

sin 3 12 θ  0

= 163 π a 2

or equivalent

A1 5

4 (i) In G, 32 = 2, 33 = 6, 34 = 4, 35 = 5, 36 = 1

M1

All powers of an element of order 6

A1

All powers correct in both groups

[ or 52 = 4, 53 = 6, 54 = 2, 55 = 3, 56 = 1 ] In H, 52 = 7 , 53 = 17 , 54 = 13, 55 = 11, 56 = 1 [ or 112 = 13, 113 = 17 , 114 = 7 , 115 = 5, 116 = 1 ] B1 B1 4

G has an element 3 (or 5) of order 6 H has an element 5 (or 11) of order 6 {1, 6} (ii) {1, 2, 4}

(iii) G 1 2 3 4 5 6

H ↔ 1 ↔ 7 ↔ 5 ↔ 13 ↔ 11 ↔ 17

G

OR

1 2 3 4 5 6

B1 B2 3

Ignore {1} and G Deduct 1 mark (from B1B2) for each proper subgroup in excess of two

B4 4

Give B3 for 4 correct, B2 for 3 correct, B1 for 2 correct

M1

Evaluating e.g. ad(1) (one case sufficient; intermediate value must be shown) For ad = c correctly shown Evaluating e.g. da(1) (one case sufficient; no need for any working)

H ↔ 1 ↔ 13 ↔ 11 ↔ 7 ↔ 5 ↔ 17

ad(1) = a(3) = 1 (iv) ad(2) = a(2) = 3 ad(3) = a(1) = 2, so ad = c

A1 da(1) = d(2) = 2 da(2) = d(3) = 1

M1

38

4757

Mark Scheme

da(3) = d(1) = 3, so da = f

June 2009

A1 4

(v) S is not abelian; G is abelian

B1 1

or S has 3 elements of order 2; G has 1 element of order 2 or S is not cyclic etc

(vi)

Element Order

a 3

b 3

c 2

d 2

e 1

f 2

B4

4 (vii) {e, c} {e, d} {e, f} {e, a, b}

Give B3 for 5 correct, B2 for 3 correct, B1 for 1 correct

B1 B1 B1 B1 4

Ignore { e } and S If more than 4 proper subgroups are given, deduct 1 mark for each proper subgroup in excess of 4

B2 2

Give B1 for two columns correct

M1 A2 3

Using P13 (or P14 ) Give A1 for 2 probabilities correct (Max A1 if not at least 3dp) Tolerance ±0.0001

Pre-multiplication by transition matrix 5 (i)  0 0.1 0 0.3    0.7 0.8 0 0.6  P=  0.1 0 1 0.1     0.2 0.1 0 0 

(ii)

 0.6   0.0810      0.4 0.5684  P13   =   0   0.2760       0   0.0746 

(iii) 0.5684 × 0.8 + 0.2760

M1M1 For 0.5684 × 0.8 and 0.2760 A1 ft Accept 0.73 to 0.7312 3

= 0.731

(iv)

 0.6   .      .  30  0.4   , = P  0   0.4996       0   . 

 0.6   .      .  31  0.4   P =  0   0.5103       0   . 

M1 A1

Level 32

Finding P(C) for some powers of P For identifying P31

A1 3

(v) Expected number of levels including the next 1 change of location is =5 0.2

Expected number of further levels in B is 4

39

M1

For 1/(1 − 0.8) or 0.8 / (1 − 0.8)

A1

For 5 or 4

A1 3

For 4 as final answer

4757

(vi)

Mark Scheme  0 0.1 0 0.3    0.7 0.8 0 0.6  Q=  0.1 0 0.9 0.1     0.2 0.1 0.1 0   0.0916 0.0916 0.0916  0.6183 0.6183 0.6183 n Q →  0.1908 0.1908 0.1908   0.0992 0.0992 0.0992

0.0916   0.6183  0.1908   0.0992 

June 2009

B1

Can be implied

M1

Evaluating powers of Q or Obtaining (at least) 3 equations from Qp = p Limiting matrix with equal columns or Solving to obtain one equilib prob or M2 for other complete method Give A1 for two correct (Max A1 if not at least 3dp) Tolerance ±0.0001

M1

A: 0.0916 B: 0.6183 C: 0.1908 D: 0.0992

0.1 a 0.3   0.11   0.11        0.7 0.8 b 0.6   0.75  =  0.75   0.1 0 c 0.1   0.04   0.04        0.2 0.1 d 0   0.1   0.1 

A2 5

(vii)  0

M1 A1

0.075 + 0.04a + 0.03 = 0.11 0.077 + 0.6 + 0.04b + 0.06 = 0.75 0.011 + 0.04c + 0.01 = 0.04 0.022 + 0.075 + 0.04d = 0.1

M1

 0.11    0.75  Transition matrix and   0.04     0.1 

Forming at least one equation

or a + b + c + d = 1

a = 0.125, b = 0.325, c = 0.475, d = 0.075

A2 5

Give A1 for two correct

B2 2

Give B1 for two rows correct

M1 A2 3

Using P13 (or P14 ) Give A1 for 2 probabilities correct (Max A1 if not at least 3dp) Tolerance ±0.0001

Post-multiplication by transition matrix 5 (i)  0 0.7 0.1 0.2    0.1 0.8 0 0.1  P=  0 0 1 0     0.3 0.6 0.1 0 

(ii)

( 0.6

0.4 0 0 ) P13

= ( 0.0810 0.5684 0.2760 0.0746 )

(iii) 0.5684 × 0.8 + 0.2760

M1M1 For 0.5684 × 0.8 and 0.2760 A1 ft Accept 0.73 to 0.7312 3

= 0.731

(iv)

( 0.6 ( 0.6

0.4 0 0 ) P 30 = (. . 0.4996 .)

M1

0.4 0 0 ) P

A1

31

= (. . 0.5103 .)

A1 3

Level 32

(v) Expected number of levels including the next change of location is

Finding P(C) for some powers of P For identifying P 31

1 =5 0.2

Expected number of further levels in B is 4

40

M1

For 1/(1 − 0.8) or 0.8 / (1 − 0.8)

A1

For 5 or 4

A1 3

For 4 as final answer

4757

(vi)

(vii)

Mark Scheme  0 0.7 0.1 0.2    0.1 0.8 0 0.1  Q=  0 0 0.9 0.1     0.3 0.6 0.1 0   0.0916 0.6183 0.1908  0.0916 0.6183 0.1908 n Q →  0.0916 0.6183 0.1908   0.0916 0.6183 0.1908

June 2009

B1

Can be implied

0.0992   0.0992  0.0992   0.0992 

M1

Evaluating powers of Q or Obtaining (at least) 3 equations from pQ = p Limiting matrix with equal rows or Solving to obtain one equilib prob or M2 for other complete method

A: 0.0916 B: 0.6183 C: 0.1908 D: 0.0992

A2 5

Give A1 for two correct (Max A1 if not at least 3dp) Tolerance ±0.0001

M1

Transition matrix and

 0 0.7 0.1 0.2    0.1 0.8 0 0.1   0.11 0.75 0.04 0.1 ( ) a b c d     0.3 0.6 0.1 0  = ( 0.11 0.75 0.04 0.1)

M1

( 0.11

0.75 0.04 0.1)

A1

0.075 + 0.04a + 0.03 = 0.11 0.077 + 0.6 + 0.04b + 0.06 = 0.75 0.011 + 0.04c + 0.01 = 0.04 0.022 + 0.075 + 0.04d = 0.1

M1

A2 5

a = 0.125, b = 0.325, c = 0.475, d = 0.075

41

Forming at least one equation

or a + b + c + d = 1 Give A1 for two correct

4758

Mark Scheme

June 2009

4758 Differential Equations 1(i)

α 2 + 25 = 0 α = ±5 CF y = A cos 5t + B sin 5t PI y = at cos 5t + bt sin 5t y = a cos 5t − 5at sin 5t + b sin 5t + 5bt cos 5t y = −10a sin 5t − 25at cos 5t + 10b cos 5t − 25bt sin 5t In DE  10b cos 5t − 10a sin 5t = 20 cos 5t  b = 2, a = 0 PI y = 2t sin 5t GS y = 2t sin 5t + A cos 5t + B sin 5t

M1 A1 F1 B1

Auxiliary equation CF for their roots

M1 M1

Differentiate twice Substitute and compare coefficients

A1 F1

8 (ii)

t = 0, y = 1  A = 1 y = 2 sin 5t + 10t cos 5t − 5 A sin 5t + 5 B cos 5t t = 0, y = 0  B = 0 y = 2t sin 5t + cos 5t

B1 M1 M1 A1

From correct GS Differentiate Use condition on y

4 (iii)

Curve through (0, 1) Curve with zero gradient at (0, 1) Oscillations Oscillations with increasing amplitude

B1 B1 B1 B1

4 (iv)

y = 2 sin 5t , y = 10 cos 5t , y = −50 sin 5t y + 2 y + 25 y = −50 sin 5t + 20 cos 5t + 50 sin 5t = 20 cos 5t 2 α + 2α + 25 = 0 α = −1 ± i 24 CF e −t C cos 24t + D sin 24t

(

GS y = 2 sin 5t + e

−t

(C cos

)

24t + D sin 24t

)

M1 E1 M1 A1 F1

Substitute into DE

F1

Their PI + their CF with two arbitrary constants

Auxiliary equation CF for their complex roots

6

(v)

Oscillations of amplitude 2

B1

Compared to unbounded oscillations in first model

B1

or bounded oscillations; or both oscillate o.e. or one bounded, one unbounded 2

2(i)

dy 3 sin x + y= 2 dx x x

M1

Rearrange

M1 A1

Attempting integrating factor

3

 dx I =e x = e 3 ln x 42

4758

Mark Scheme

= x3 d 3 x y = x sin x dx x 3 y = x sin x dx = − x cos x + cos x dx

( ) 



= − cos x + sin x + A − x cos x + sin x + A y= x3

June 2009

A1

Correct and simplified

M1

Multiply and recognise derivative

M1

Integrate

A1 A1

All correct

F1

Must include constant 9

(ii)

y≈

(

)

− x 1 − 12 x 2 + x − 16 x 3 + A

M1

x3

Substitute given approximations

F1 1 A = + 3 3 x A=0 sin x − x cos x y= x3 lim y = 13

M1

Use finite limit to deduce A

A1

x →0

B1

Correct particular solution

B1

Correct limit 6

(iii)

y = 0  sin x − x cos x = 0

M1

 tan x = x

E1

Equate to zero and attempt to get tanx Convincingly shown 2

(iv)

dy 3 1 1 + y = − x , multiply by I = x 3 x 6 dx x

( )

d 3 1 x y = x2 − x4 dx 6

M1

Rearrange and multiply by IF

B1

Same IF as in (i) or correct IF

A1

Recognise derivative and RHS correct

1 1 5 x y = x3 − x +B 3 30 1 1 2 B y= − x + 3 3 30 x Finite limit  B = 0 lim y = 13 3

x →0

M1

Integrate

A1

c.a.o

M1 E1

Use condition to find constant Show correct limit (or same limit as (ii)) 7

3(a)(i) 2α + 4 = 0  α = −2 CF Ae −2t PI I = a cos 2t + b sin 2t I = −2a sin 2t + 2b cos 2t − 4a sin 2t + 4b cos 2t + 4a cos 2t + 4b sin 2t = 3 cos 2t − 4a + 4b = 0, 4b + 4a = 3  a = b = 83

43

M1

Find root of auxiliary equation

A1 B1 M1 M1 M1

Differentiate Substitute Compare coefficients and solve

4758

Mark Scheme

June 2009

PI I = 83 (cos 2t + sin 2t )

A1

GS I = Ae −2t + 83 (cos 2t + sin 2t )

F1

Their PI + their CF with one arbitrary constant 8

(ii)

t = 0, l = 0  0 = A + 83 

A = − 83

I = 83 (cos 2t + sin 2t − e −2t )

M1

Use condition

A1

c.a.o 2

(iii)

For large t, I ≈ 83 (cos 2t + sin 2t )

M1

Amplitude = 83 12 + 12 = 83 2 Curve with oscillations with constant amplitude Their amplitude clearly indicated

A1 B1 B1

Consider behaviour for large t (may be implied)

4

(b)(i)

(A) t = 0, y = 0 

dy = 2 − 2(0) + e 0 dt

M1

Gradient =3

Substitute into DE

A1

(B) At stationary point,

dy = 0, y = dt  e − t = 14

9 8

 0 = 2 − 2(98 ) + e −t  t = ln 4 dy → 0, e −t → 0 (C) dt Giving 0 = 2 − 2y + 0, so y → 1

M1

Substitute into DE

M1 A1

Solve for t

M1

Substitute into DE

A1 7

(ii)

Curve through origin with positive gradient With maximum at (ln4, 98 ) With y → 1 as x → ∞

B1 B1 B1

Follow their ln4 Follow their (C) 3

4(i)

x = 7 x + 6 y − 6e −3t

= 7 x + 6(−12 x − 10 y + 5 sin t ) − 6e y=

1 6

−3t

(x − 7 x − 2e ) −3t

x = 7 x − 72 x − 10( x − 7 x − 2e −3t ) + 30 sin t − 6e −3t

x + 3 x + 2 x = 14e

−3t

+ 30 sin t

M1

Differentiate

M1

Substitute for y

M1

y in terms of x, x , t

M1

Substitute for y

E1

Complete argument 5

(ii)

x = ae −3t − 9 cos t + 3 sin t x = −3ae −3t + 9 sin t + 3 cos t x = 9ae −3t + 9 cos t − 3 sin t In DE gives 9ae −3t + 9 cos t − 3 sin t + 3(−3ae −3t + 9 sin t + 3 cos t )

44

M1 M1

Differentiate twice Substitute

4758

Mark Scheme

June 2009

+ 2(ae −3t − 9 cos t + 3 sin t )

= 2ae −3t + 30 sin t So PI with 2a = 14  a=7

E1 A1

Correct form shown

AE α 2 + 3α + 2 = 0 α = −1,−2

M1 A1

Auxiliary equation

CF Ae −t + Be −2t GS x = Ae − t + Be −2t + 7e −3t − 9 cos t + 3 sin t

F1 F1

CF for their roots Their PI + their CF with two arbitrary constants 8

(iii)

x = 16 ( x − 7 x − 2e −3t )

M1

y in terms of x, x , t

x = − Ae − t − 2 Be −2t − 21e −3t + 9 sin t + 3 cos t

M1 F1 A1

Differentiate GS for x Follow their GS c.a.o

y = − 43 Ae −t − 32 Be −2t − 12e −3t + 11cos t − 2 sin t

4

(iv)

x ≈ 3 sin t − 9 cos t y ≈ 11 cos t − 2 sin t x = y  11 cos t − 2 sin t ≈ 3 sin t − 9 cos t  20 cos t ≈ 5 sin t  tan t ≈ 4

B1 B1 M1 A1

Follow their x Follow their y Equate Complete argument 4

(v)

Amplitude of x ≈ 3 2 + 9 2 = 3 10 2

2

Amplitude of y ≈ 11 + 2 = 5 5 Ratio is

5 6

2

M1

Attempt both amplitudes

A1

One correct

A1

c.a.o (accept reciprocal) 3

45

4761

Mark Scheme

June 2009

4761 Mechanics 1 Q1 (i)

0.5 × 8 × 10 = 40 m

Mark M1

A1

Comment Attempt to find whole area or … If suvat used in 2 parts, accept any t value 0 ≤ t ≤ 8 for max. c.a.o.

Sub

2 (ii)

0.5 × (T − 8) = 10

M1

T = 12

B1 A1

0.5 × 5 × k = 10 seen. Accept ±5 and ±10 only. If suvat used need whole area; if in 2 parts, accept any t value 8 ≤ t ≤ T for min. Attempt to use k = T – 8. c.a.o. [Award 3 if T = 12 seen] 3

(iii)

40 – 10 = 30 m

B1

ft their 40. 1

6 Q2 (i)

10 + 24 = 26 so 26 N 2

2

Mark B1

arctan 10 24

M1

= 22.619… so 22.6° (3 s.f.)

A1

Comment

Sub

24 Using arctan or equiv. Accept arctan 10 or equiv. Accept 157.4°.

3 (ii)

W = −wj

B1

T1 + T2 + W = 0

M1

k = –10 w = 34

B1 B1

 0   0  Accept   and    −w   − wj 

1 (iii)

Accept in any form and recovery from W = wj. Award if not explicit and part (ii) and both k and w correct. Accept from wrong working. Accept from wrong working but not −34. [Accept –10i or 34j but not both] 3

7

46

4761

Q3 (i)

Mark Scheme

The line is not straight

Mark B1

June 2009

Comment Any valid comment

Sub

1 (ii)

a = 3−

6t 8

M1

a(4) = 0 The sprinter has reached a steady speed

Attempt to differentiate. Accept 1 term 3t correct but not 3 − . 8

F1 E1

Accept ‘stopped accelerating’ but not just a = 0. Do not ft a(4) ≠ 0. 3

(iii)

4



We require

 3t   3t −  dt 8   2

M1

Integrating. Neglect limits.

 3t 2 t 3  = −   2 8 1

A1

One term correct. Neglect limits.

 3 1 = (24 − 8) −  −   2 8

M1

= 14 85 m (14.625 m)

A1

Correct limits subst in integral. Subtraction seen. If arb constant used, evaluated to give s = 0 when t = 1 and then sub t = 4. c.a.o. Any form. [If trapezium rule used M1 use of rule (must be clear method and at least two regions) A1 correctly applied M1 At least 6 regions used A1 Answer correct to at least 2 s.f.)]

1

4

4 8

47

4761

Q4 (i)

Mark Scheme

Mark B1

32 cos α t

June 2009

Comment

Sub

1

(ii)

32 cos α × 5 = 44.8 so 160 cos α = 44.8 and cos α = 0.28

M1 E1

ft their x. Shown. Must see some working e.g cosθ = 44.8 ÷ 160 or 160 cosθ = 44.8. If 32 × 0.28 × 5 = 44.8 seen then this needs a statement that ‘hence cosθ = 0.28’.

(iii)

sin α = 0.96

B1

Need not be explicit e.g. accept sin(73.73…) seen.

M1

Allow use of ‘u’ = 32, g = ± (10, 9.8, 9.81). Correct substitution. c.a.o.

2

either 0 = (32 × 0.96) 2 − 2 × 9.8 × s

s = 48.1488… so 48.1 m (3 s. f.) or Time to max height is given by 32 × 0.96 − 9.8 T = 0 so T = 3.1349…. y = 32 × 0.96 t – 4.9 t²

A1 A1 B1 M1

Could use ½ total time of flight to the horizontal. Allow use of ‘u’ = 32, g = ± (10, 9.8, 9.81) May use s=

putting t = T, y = 48.1488 so 48.1 m (3 s. f.)

A1

(u + v ) t. 2

c.a.o. 4

7 Q5 (i)

Mark M1

v = i + (3 − 2t ) j

v(4) = i – 5j

A1 F1

a = – 2j

B1

Using N2L F = 1.5 × (–2j) so –3j N

M1 A1

Comment Differentiating r. Allow 1 error. Could use const accn.

Sub

Do not award if 26 is given as vel (accept if v given and v given as well called speed or magnitude). 3

(ii)

Diff v. ft their v. Award if – 2j seen & isw. Award for 1.5 × (± a or a) seen. c.a.o. Do not award if final answer is not correct. [Award M1 A1 for –3j WW] 3

(iii)

x = 2 + t and y = 3t − t

2

Substitute t = x − 2 so y = 3( x − 2) − ( x − 2) 2 [ = ( x − 2)(5 − x ) ]

B1

Must have both but may be implied.

B1

c.a.o. isw. Must see the form y = …. 2

8

48

4761

Q6 (i)

Mark Scheme

Up the plane T − 4 g sin 25 = 0

Mark M1

T = 16.5666… so 16.6 N (3 s. f.)

A1

Down the plane, (4 + m)g sin25 – 50 = 0

M1

June 2009

Comment Resolving parallel to the plane. If any other direction used, all forces must be present. Accept s ↔ c . Allow use of m. No extra forces.

Sub

2 (ii)

m = 8.0724… so 8.07 (3 s. f.)

A1 A1

Diagram

B1

No extra forces. Must attempt resolution in at least 1 term. Accept s ↔ c . Accept Mg sin25. Accept use of mass. Accept Mgsin25 3

(iii)

B1

Any 3 of weight, friction normal reaction and P present in approx correct directions with arrows. All forces present with suitable directions, labels and arrows. Accept W, mg, 4g and 39.2. 2

(iv)

Resolving up the plane

M1

P cos15 − 20 −4g sin25 = 0

B1 B1

P = 37.8565…. so 37.9 N (3 s. f.)

A1 A1

Resolving perpendicular to the plane

M1

R + P sin15 − 4g cos25 = 0

B1

Or resolving parallel to the plane. All forces must be present. Accept s ↔ c . Allow use of m. At least one resolution attempted and accept wrong angles. Allow sign errors. P cos15 term correct. Allow sign error. Both resolutions correct. Weight used. Allow sign errors. ft use of P sin 15. All correct but ft use of P sin 15. 5

(v)

F1 R = 25.729… so 25.7 N

A1

May use other directions. All forces present. No extras. Allow s ↔ c . Weight not mass used. Both resolutions attempted. Allow sign errors. Both resolutions correct. Allow sign errors. Allow use of Pcos15 if Psin15 used in (iv). All correct. Only ft their P and their use of Pcos15. c.a.o. 4

16 If there is a consistent s ↔ c error in the weight term throughout the question, penalise only two marks for this error. In the absence of other errors this gives (i) 35.52… (ii) 1.6294… (iv) 57.486… (v) 1.688… For use of mass instead of weight lose maximum of 2.

49

4761

Q7 (i)

Mark Scheme

Mark With the 11.2 N resistance acting to the left N2L F − 11.2 = 8 × 2 M1 F = 27.2 so 27.2 N

A1 A1

June 2009

Comment

Sub

Use of N2L (allow F = mga). Allow 11.2 omitted; no extra forces. All correct c.a.o. 3

(ii)

The string is inextensible

E1

Allow ‘light inextensible’ but not other irrelevant reasons given as well (e.g. smooth pulley).

B1

One diagram with all forces present; no extras; correct arrows and labels accept use of words. Both diagrams correct with a common label.

1 (iii)

B1

2 (iv)

Method (1)

M1

Box N2L → 105 − T − 11.2 = 8a Sphere N2L ↑ T − 58.8 = 6a

A1 A1

Adding 35 = 14a

M1

a = 2.5 so 2.5 ms−2 Substitute a = 2.5 giving T = 58.8 + 15 T = 73.8 so 73.8 N

E1 M1

Method (2) 105 – 11.2 – 58.8 = 14a a = 2.5

For either box or sphere, F = ma. Allow omitted force and sign errors but not extra forces. Need correct mass. Allow use of mass not weight. Correct and in any form. Correct and in any form. [box and sphere equns with consistent signs] Eliminating 1 variable from 2 equns in 2 variables. Attempt to substitute in either box or sphere equn.

A1 M1 A1 E1 M1

either: box N2L → 105 − T − 11.2 = 8a or: sphere N2L ↑ T − 58.8 = 6a Substitute a = 2.5 in either equn

A1 M1

T = 73.8 so 73.8 N

A1

For box and sphere, F = ma. Must be correct mass. Allow use of mass not weight. Method made clear. For either box or sphere, F = ma. Allow omitted force and sign errors but not extra forces. Need correct mass. Allow use of mass not weight. Correct and in any form. Attempt to substitute in either box or sphere equn. [If AG used in either equn award M1 A1 for that equn as above and M1 A1 for finding T. For full marks, both values must be shown to satisfy the second equation.] 7

50

4761

Mark Scheme

(v)(A) g downwards

June 2009

B1

Accept ±g, ±9.8, ±10, ±9.81

Taking ↑ + ve, s = −1.8, u = 3 and a = −9.8 so −1.8 = 3T − 4.9T 2

M1

and so 4.9T 2 − 3T − 1.8 = 0

E1

Some attempt to use s = ut + 0.5at2 with a = ±9.8 etc s = ±1.8 and u = ±3 . Award for a = g even if answer to (A) wrong. Clearly shown. No need to show +ve required.

1 (B)

(C)

2

Time to reach 3 ms−1 is given by 3 = 0 + 2.5t so t = 1.2 remaining time is root of quad

B1 M1

time is 0.98513… s

B1

Total 2.1851…so 2.19 s (3 s. f.)

A1

Quadratic solved and + ve root added to time to break. Allow 0.98. [Award for answer seen WW] c.a.o. 4

(i)

With the 11.2 N resistance acting to the right F + 11.2 = 8 × 2 so F = 4.8

The 11.2 N force may be in either direction, otherwise the same scheme

(iii) (iv)

The same scheme as above

The same scheme with + 11.2 N instead of – 11.2 N acting on the box Method (1) Box N2L → 105 − T + 11.2 = 8a Sphere as before Method (2) 105 + 11.2 – 58.8 = 14a These give a = 4.1 and T = 83.4 Allow 2.5 substituted in box equation to give T = 96.2 If the sign convention gives as positive the direction of the sphere descending, a = – 4.1. Allow substituting a = 2.5 in the equations to give T = 43.8 (sphere) or 136.2 (box). In (C) allow use of a = 4.1 to give time to break as 0.73117..s. and total time as 1.716…s

(v)

4 20

51

4762

Mark Scheme

June 2009

4762 Mechanics 2 Q1 (a)(i)

Mark before u P m kg after

Comment

Sub

u Q km kg

v

u

B1

3

1 (ii)

mu − kmu = mv + km

u 3

 4k  v = 1 −  u 3  

M1

PCLM applied

A1

Either side correct (or equiv)

E1

Must at least show terms grouped

E1

4k > 1 without reason Accept 3

3 (iii)

Need v < 0 so k >

3 4

B1 [SC1: v = 0 used and inequality stated without reason] 2

(iv)

1 u 3

−v

−u −u

M1

= − 12

Use of NEL

A1

2u so v = − 3 2u  4k  − = u 1 −  3 3  

so k = 1.25

E1 M1 A1

c.a.o. 5

(b)(i)

 1   3 9   + 5   = 8V  −2   2 

3 V=   −1 

M1

Use of PCLM

B1 M1

Use of mass 8 in coalescence Use of I = Ft

E1 4

(ii)

i cpt 3 → −3 × j cpt unchanged

1 2

 −1.5 

−1 New velocity   ms 1 −  

M1 B1

Allow wrong sign May be implied c.a.o. [Award 2/3 if barrier taken as

A1

1 0 ]  

3 18 Q2 (a)(i)(A) Yes. Only WD is against conservative forces.

Mark E1

52

Comment Accept only WD is against gravity or no work done against friction.

Sub

4762

Mark Scheme

June 2009

1 (B)

Block has no displacement in that direction

E1 2

(ii)

2

0.5 × 50 × 1.5 = 20 gx − 5 gx

x = 0.38265… so 0.383 m (3 s.f.)

M1 B1 M1 A1 A1

Use of WE with KE. Allow m = 25. Use of 50 At least 1 GPE term GPE terms correct signs c.a.o. 5

(iii)

2

0.5 × 50 × V − 0.5 × 50 × 1.5

2

= 2 × 20g − 2 × 5g − 180 V = 2.6095… so 2.61 ms−1

M1

WE equation with WD term. Allow GPE terms missing

B1 B1 B1 A1

Both KE terms. Accept use of 25. Either GPE term 180 with correct sign c.a.o.

M1

Both terms. Allow mass not weight

B1 M1 F1 A1

Weight term correct

5 (b)

Force down the slope is 2000 + 450g sin20 Using P = Fv P = (2000 + 450g sin20) × 2.5 P = 8770.77… so 8770 W (3 s.f.)

ft their weight term c.a.o. 5

17

53

4762

Q3 (i)

(ii)

Mark Scheme

c.w. moments about A 5RB − 3 × 85 so RB = 51 giving 51 N ↑ Either a.c. moments about B or resolve ↑ RA = 34 so 34 N ↑

Mark M1 A1

F1

Sub

Accept no direction given 4

M1

85 × 3cos α − 27.2 × 5sin α = 0

B1 B1

3 × 85 15 = 27.2 × 5 8

Comment Moments equation. Accept no direction given

M1

c.w. moments about A

so tan α =

June 2009

E1

Moments with attempt to resolve at least one force. Allow s ↔ c . Weight term Horiz force term Must see some arrangement of terms or equiv 4

(iii) A

S 34 N

B1

All forces present and labelled

a.c. moments about B

M1

85 × 2 × cos α + 34 × 2.5 − 5S × sin α = 0

Moments with attempt to resolve forces and all relevant forces present

B1 A1

34 × 2.5

S = 37.4 Resolving horizontally and vertically → S − F − 34 sin α = 0 so F = 7.4 ↑ R − 85 − 34 cos α = 0

A1 M1

Using F = μ R

M1

R F 85 N

μ=

α B

7.4 = 0.07326… 101

E1 A1

A1

All other terms correct. Allow sign errors. All correct Either attempted R = 101 need not be evaluated here [Allow A1 for the two expressions if correct other than s ↔ c ]

c.a.o.

so 0.0733(3 s.f.) 10 18

54

4762

Q4 (i)

Mark Scheme

Mark

Taking y-axis vert downwards from O 2πσ × 82 × 4 + 2πσ × 8 × k ×

k 2

= ( 2πσ × 82 + 2πσ × 8k ) y 2

64 + k 16 + 2k

so y =

June 2009

Comment Allow areas used as masses

M1

Method for c.m.

B1 B1 B1

‘4’ used 16π k k used 2

B1

Masses correct

E1

Must see some evidence of simplification Need no reference to axis of symmetry

B1

Allow for either. Allow σ = 1

M1 B1 B1

Method for c.m. combining with (i) or starting again One term correct Second term correct

E1

Some simplification shown

Sub

6 (ii)

k = 12 gives OG as 5.2 and mass as 320πσ 320πσ × 5.2 + πσ × 82 × 12

= ( 320πσ + 64πσ ) y

y = 6 13

5 (iii) O G

B1 B1 B1

G above edge of base 12 − 6 13 = 5 23 seen here or below 8 seen here or below

M1

Accept

θ = 54.6887… so 54.7° (3 s. f.)

A1

and 8. c.a.o.

Slips when μ = tan θ

M1

Or ….

8 = 1.4117... 5 23

B1

< 1.5 so does not slip

A1

12 − 6 13 = 5 23

12 8

θ

tan θ =

8 5 23

5 23 or attempts based on 6 13 8

5 (iv)

There must be a reason 3

19

55

4763

Mark Scheme

June 2009

4763 Mechanics 3 1 (i)

1 2

m(v 2 − 1.42 ) = m × 9.8(2.6 − 2.6 cos θ )

M1

2

v − 1.96 = 50.96 − 50.96 cos θ

Equation involving KE and PE

A1

2

v = 52.92 − 50.96 cos θ

E1 3

(ii)

v2 2.6 6.37 cos θ − R = 0.25(52.92 − 50.96 cos θ )

0.65 × 9.8cos θ − R = 0.65 ×

6.37 cos θ − R = 13.23 − 12.74 cos θ R = 19.11cos θ − 13.23

(iii)

Leaves surface when R = 0 13.23 9 (= ) ( θ = 46.19° ) 19.11 13 9 v 2 = 52.92 − 50.96 × 13

cos θ =

Radial equation involving

A1 M1 A1 4

Substituting for v 2 Dependent on previous M1 Special case: R = 13.23 − 19.11cos θ earns M1A0M1SC1

M1

( ft if R = a + bcosθ and 0 < −

A1

Dependent on previous M1

A1 4 M1 A1 M1 A1

T sin α + R cos α = 0.65 × 9.8 1.22 T cos α − R sin α = 0.65 × 2.4

 1.2 2 OR T − mg sin α = m  2.4

  cos α  

M1A1

 1.2 2 mg cos α − R = m  2.4

  sin α  

M1A1

sin α =

2.4 12 5 = , cos α = 2.6 13 13

a < 1) b

M1

Speed is 4.2 m s −1 (iv)

v2 r

M1

( α = 67.38° )

Tension is 6.03 N Normal reaction is 2.09 N

Horiz eqn involving

v2 or r ω 2 r

M1 M1

Solving to obtain a value of T or R

A1 A1

Dependent on necessary M1s ( Accept 6 , 2.1 ) Treat ω = 1.2 as a misread, leading to T = 6.744, R = 0.3764 for 7 / 8

8

56

Resolving vertically (3 terms)

4763

2 (i)

Mark Scheme

1 2

× 5000 x 2 = 12 × 400 × 32

M1 A1 A1

Compression is 0.849 m

June 2009

Equation involving EE and KE Accept

3 (ii)

Change in PE is 400 × 9.8 × (7.35 + 1.4) sin θ

M1

Or 400 × 9.8 × 1.4sin θ

A1

1 × 400 × 4.542 2 Or 784 + 4116

and

= 400 × 9.8 × 8.75 × 17 = 4900 J

Change in EE is

1 2

× 5000 × 1.4 = 4900 J

3 2 5

2

M1

Since Loss of PE = Gain of EE, car will be at rest

E1

M1M1A1 can also be given for a correct equation in x (compression): 2500 x 2 − 560 x − 4116 = 0

Conclusion required, or solving equation to obtain x = 1.4

4 (iii) WD against resistance is 7560(24 + x)

Change in EE is Change in KE is

2

1 × 5000x 2 1 × 400 × 302 2

Change in PE is 400 × 9.8 × (24 + x) ×

1 7

OR Speed 7.75 m s −1 when it hits buffer, then WD against resistance is 7560x Change in EE is 12 × 5000x 2 Change in KE is

1 2

B1 B1

( = 181440 + 7560x ) ( = 2500x 2 )

B1

( = 180000 )

B1

( = 13440 + 560x )

B1 B1

( = 2500x 2 ) ( = 12000 ) ( = 560x )

× 400 × 7.752

B1

1 7

B1

Change in PE is 400 × 9.8 × x ×

M1

−7560(24 + x) = 12 × 5000 x 2 − 12 × 400 × 302

Equation involving WD, EE, KE, PE

−400 × 9.8 × (24 + x) × 17 −7560(24 + x) = 2500 x 2 − 180 000 − 560(24 + x) 2

−3.024(24 + x) = x − 72 − 0.224(24 + x)

F1 M1

2

x + 2.8 x − 4.8 = 0

A1

−2.8 + 2.82 + 19.2 2 = 1.2

M1

x=

A1 10

57

Simplification to three term quadratic

4763

Mark Scheme

3(a)(i) [ Velocity ] = L T −1

B1

[ Force ] = M L T −2

B1

−3

[ Density ] = M L

June 2009

Deduct 1 mark for ms−1 etc

B1 3

(ii) M L T −2 = (M L−3 )α (L T −1 ) β (L2 )γ α =1 β =2 −3α + β + 2γ = 1 γ =1

B1 B1 M1A1 ( ft if equation involves α , β and γ ) A1 5

(b)(i) 2π = 4.3 ω 2π ω= ( = 1.4612 )

M1 A1

4.3

θ 2 = 1.46122 (0.082 − 0.052 )

M1 F1

Angular speed is 0.0913 rad s −1

A1 5

OR θ = 0.08ω cos ωt

M1 F1 A1

= 0.08 ×1.4612 cos 0.6751 = 0.0913

(ii) θ = 0.08sin ωt When θ = 0.05, 0.08sin ωt = 0.05 ωt = 0.6751 t = 0.462

Time taken is 2 × 0.462 = 0.924 s

Using ω 2 ( A2 − θ 2 ) For RHS ( b.o.d. for v = 0.0913 m s −1 ) Or θ = (−) 0.08ω sin ωt = (−) 0.08 × 1.4612sin 0.8957

or θ = 0.08cos ωt Using θ = (±) 0.05 to obtain an equation for t A1 cao B1M1 above can be earned in (i) or t = 0.613 from θ = 0.08cos ωt or t = 1.537 from θ = 0.08cos ωt B1 M1

M1

Strategy for finding the required time ( 2 × 0.462 or 12 × 4.3 − 2 × 0.613 or 1.537 − 0.613 ) Dep on first M1 A1 cao For θ = 0.05sin ωt , max B0M1A0M0 ( for 0.05 = 0.05sin ωt ) 5

58

4763

4(a)

Mark Scheme

June 2009

ln 3

ln 3  e x dx =  e x    0 0 =2

Area is 



ln 3

 x y dx = 0

M1 A1

x e x dx

M1 ln 3

=  x ex − ex   0 = 3ln 3 − 2 x=



3ln 3 − 2 3 = 2 ln 3 − 1 2 ln 3

1 2

 y 2 dx =  0 = 

1 2

(e x ) 2 dx

1 2x e 4

Integration by parts For x e x − e x

A1

ww full marks (B4) Give B3 for 0.65

M1

For integral of (e x )2

A1

For

A1

If area wrong, SC1 for

ln 3

 0

=2 y=

M1 A1

2 =1 2

1 2x e 4

x=

3ln 3 − 2 2 and y = area area

9 (b)(i)

a

36

2

x4

Volume is  π y 2 dx =   π

M1

dx

π may be omitted throughout

a

 12   3 12  =π  − 3  =π  − 3   x 2 2 a  a

36

2

x3

 2  π xy dx =  π

A1

dx

M1 a

 18   9 18  =π  − 2  =π  − 2   x 2 2 a 

A1

 π xy dx 2  π y dx 2

x=

M1

 9 18  2 a 2  3(a 3 − 4a ) =  =  3 12  a3 − 8 π − 3 2 a 

π −

E1 6

M1

(ii) Since a > 2, 4a > 8 3

3

so a − 4a < a − 8 Hence x =

3(a3 − 4a ) a3 − 8

A1

throughout

−3

→3

Fully acceptable explanation Dependent on M1A1

3

M1A1 Accept x ≈

Since x increases as a increases, x is less than 3

E1

3a 3

→ 3 , etc a3 ( M1 for x → 3 stated, but A1 requires

correct justification )

59

4764

Mark Scheme

June 2009

4764 MEI Mechanics 4 1(i)

d (mv ) = mg dt dm dv  v+m = mg dt dt dv mg  = mg v+m 2(v + 1) dt

 v  dv  = = g 1 − ( dt  2 v + 1)   v + 1  dv 1   = 2g   v + 2  dt 1  dv 1   1 − =2g   v + 2  dt 

B1

Seen or implied

M1 Expand M1 Use

dm mg v= dt 2(v + 1)

 v+2   g   2(v + 1) 

M1 Separate variables (oe) E1 5

(ii)



1 

 1 − v + 2 dv = 

1 2

g dt

M1 Integrate

v − ln v + 2 = 12 gt + c t = 0, v = 0  −ln2 = c v − ln v + 2 = 12 gt − ln 2

A1 LHS M1 Use condition

2 (v − ln v + 2 + ln 2) g v = 10  t ≈ 1.68

t=

A1 B1 5

(iii)

As t gets large, v gets large dv → 12 g (i.e. constant) So dt

M1 A1

Complete argument 2

2(i)

V = −mg ⋅ 2a sin θ +

1 2

mg

2a

(4a sin θ − a )2

B1

GPE

M1 Reasonable attempt at EPE A1 EPE correct dV mg (4a sin θ − a ) ⋅ 4a cos θ = −2mga cos θ + dθ 2a = −2mga cos θ + 2mga cos θ (4 sin θ − 1) = 4mga cos θ (2 sin θ − 1)

M1 Differentiate E1

Complete argument 5

(ii)

dV =0 dθ ⇔ cos θ = 0 or sin θ = ⇔ θ = 12 π or

1 6

M1 Set derivative to zero 1 2

M1 Solve

π

A1

Both

M1

Second derivative (or alternative method)

2

d V = 4mga cos θ (2 cos θ ) − 4mga sin θ (2 sin θ − 1) dθ 2 V ′′( 12 π )(= −4mga ) < 0  unstable

60

M1 Consider sign A1 One correct conclusion validly

4764

Mark Scheme

June 2009

shown

 3 V ′′( 16 π ) = 4mga ⋅ 2 

 3  > 0  stable 

( )

A1

Complete argument 7

3(i)

Mass of ‘ring’ ≈ 2πrδrρ

B1

May be implied

a



 I C = r 2 ⋅ 2πρr dr

M1 Set up integral

0

[

= 2πρ ⋅

1 4

]

a r4 0

A1 4

= πa ρ 1 2

M1 Integrate

M = πa 2 ρ

M1 Use relationship between ρ and M 2

 I C = 12 Ma (ii)

All correct

E1

6

I A = IC + M ( a)

2

1 10

2

Complete argument

M1 Use parallel axis theorem

2

1 = 12 Ma + 100 Ma = 0.51Ma

2

E1

Convincingly shown 2

(iii)

I A θ = − Mg

⋅ 101

a sin θ

B1 B1

g sin θ 5.1a θ small  sin θ ≈ θ g  θ =− θ , i.e. SHM 5.1a  θ =−

M1 Expression for θ M1 Use small angle approximation

5.1a ≈ 4.53 a g

Period 2π

LHS RHS

E1

Complete argument and conclude SHM

F1

Follow their SHM equation 6

(iv)

e.g. C A

B1

P

M1 Moments or ( m )x =  mx (oe)

mg ⋅ 109 a = Mg ⋅ 101 a  m = 19 M

A1

I = 0.51Ma + m(109 a )

2

2

= 0.6 Ma (v)

Method may be implied

M1

2

E1

Convincingly shown 5

2

2

KE = Iω = (0.6Ma )ω 1 2

2

1 2

2

M1 Attempt to find KE

2

A1 M1 Work-energy equation A1 Correct equation

= 0.3Ma ω C ⋅ n ⋅ 2π = 0.3Ma 2 ω 2  C=

Show PAC in straight line (in any direction)

0.3Ma 2 ω 2 2 nπ

A1 5

61

4764

4(i)

Mark Scheme

June 2009

At terminal velocity,  F = 0  k ⋅ 60 2 = 90 g

M1

Equilibrium of forces

 k=

E1

Convincingly shown

1 40

g

2 (ii)

dv = 90 g − 401 gv 2 dx 90v dv = dx 90 g − 401 gv 2

90v

 −



1800 ln 90 g − 401 gv 2 = x + c1 g

90 − 401 gv 2 = Ae v2 =



M1 A1

N2L

M1

Separate and integrate

A1

LHS

M1 M1

Rearrange, dealing properly with constant Use condition

E1

Complete argument

gx 1800

gx  − 40  90 g − Ae 1800   g  

x = 0, v = 0 

A = 90 g

9x   − v 2 = 36001 − e 1800     

7 1800

 kv dx 2

(iii) WD against R =

B1

0

1800

=

 0

gx   − 90 g 1 − e 1800 dx    

1800

gx   1800 −1800   = 90 g  x + e   g    0

(

)

= 162000 g + e − g − 1 2

M1

(

x = 1800  v = 3600 1 − e

−g

Integrate

A1

)

(

Loss in energy = 90 g ⋅ 1800 − ⋅ 90 ⋅ 3600 1 − e 1 2

−g

= 162000( g + e − g − 1) = WD against R

)

B1 M1 M1

GPE KE

E1

Convincingly shown (including signs) 7

(iv)

v = 60 1 − e

−g

B1

≈ 59.9983

1 (v)

90

dv = 90 g − 90v dt

M1

N2L

A1 dv

 g − v =  dt

 or 



10 59.9983

t  = dt  0 g −v 

dv



M1

− ln g − v = t + c 2

A1

t = 0, v = 59.9983  c 2 = −3.91598 v = 10  t = − ln 0.2 + 3.91598 ≈ 5.53 s

M1 M1 A1

Separate and integrate

Use condition (or limits) Calculate t 7

62

4766

Mark Scheme

June 2009

4766 Statistics 1 Q1 (i)

Median = 2

B1 cao B1 cao

Mode = 1

2

(ii)

S1 labelled linear scales on both axes H1 heights

60 50 40 30

2

20 10 0 1

2

3

4

N umber o f P eo ple

(iii)

Positive

B1 TOTAL

Q2 (i)

 25    different teams = 53130 5

 25  M1 for   5 A1 cao

14  11   ×   = 364 × 55 = 20020  3 2

M1 for either combination M1 for product of both A1 cao

(ii)

1 5

2

TOTAL Q3 (i)

Mean =

126 = 10.5 12

Sxx = 1582 − s= (ii)

5

B1 for mean

126 2 = 259 12

M1 for attempt at Sxx 3

259 = 4.85 11

A1 cao

New mean = 500 + 100 × 10.5 = 1550

B1 ANSWER GIVEN 3

New s = 100 × 4.85 = 485

(iii)

3

M1A1 ft

On average Marlene sells more cars than Dwayne. Marlene has less variation in monthly sales than Dwayne.

E1 E1 ft

2 TOTAL

63

8

4766

Mark Scheme

June 2009

Q4 (i)

E(X) = 25 because the distribution is symmetrical.

E1 ANSWER GIVEN

1

M1 for Σr2p (at least 3 terms correct) M1dep for –25² A1 cao TOTAL

3

Allow correct calculation of Σrp (ii)

E(X2) = 10 2 × 0.2 + 20 2 × 0.3 + 30 2 × 0.3 + 40 2 × 0.2 = 730 Var(X) = 730 – 252 = 105

4

Q5 (i)

Distance 050100200-400

freq 360 400 307 133

width 50 50 100 200

f dens 7.200 8.000 3.070 0.665

M1 for fds A1 cao Accept any suitable unit for fd such as eg freq per 50 miles. L1 linear scales on both axes and label W1 width of bars

5

H1 height of bars

(ii)

Estimate = 50 +

Q6 (i)

(A) (B)

(ii)

B1 for 600th

Median = 600th distance 240

/400 × 50 = 50 + 30 = 80

M1 for attempt to interpolate A1 cao TOTAL

3 8

P(at most one) =

83 = 0.83 100

B1 aef

P(exactly two) =

13 10 + 2 + 1 = = 0.13 100 100

M1 for (10+2+1)/100 A1 aef

2

53 × 100 M1dep for product of next 2 correct fractions A1 cao TOTAL

3

P(all at least one) =

1

53 52 51 140556 × × = = 0.145 100 99 98 970200

64

M1 for

6

4766

Mark Scheme

June 2009

Q7 (i)

(ii)

P(Not delayed) = 0.8 × 0.85 × 0.9 = 0.612 P(Delayed) = 1 – 0.8 × 0.85 × 0.9 = 1 – 0.612 = 0.388

(iii)

= 0.153 + 0.108 + 0.068 = 0.329

P(Just one problem) 0.329 = 0.848 = 0.388 P(Delay)

B1 one product correct M1 three products M1 sum of 3 products A1 cao

M1 for denominator A1 ft

4

3

M1 for 0.15 + M1 for second term A1 cao

P(Delayed | No technical problems) Either = 0.15 + 0.85 × 0.1 = 0.235

M1 for product M1 for 1 – product A1 cao

Or = 1 – 0.9 × 0.85 = 1 – 0.765 = 0.235 Or = 0.15 × 0.1 + 0.15 × 0.9 + 0.85 × 0.1 = 0.235

M1 for all 3 products M1 for sum of all 3 products A1 cao

3

Or (using conditional probability formula) P(Delayed and no technical problems) P(No technical problems) 0.8 × 0.15 × 0.1 + 0.8 × 0.15 × 0.9 + 0.8 × 0.85 × 0.1 = 0 .8 0.188 = = 0.235 0.8

(vi)

4

M1 for numerator

P(Just one problem | delay) =

(v)

M1 for 1 – P(delayed) A1 ft

P(just one problem) = 0.2×0.85×0.9 + 0.8×0.15×0.9 + 0.8×0.85×0.1

(iv)

2

B1 for any one B1 for the other two M1 for product A1 cao

a = 0.8, b = 0.85, c = 0.9.

M1 for numerator M1 for denominator A1 cao

M1 for product A1 ft

Expected number = 110 × 0.388 = 42.7

2 TOTAL

65

18

4766

Q8 (i)

Mark Scheme

X ~ B(15, 0.2) (A)

15  P(V = 3) =   × 0.2 3 × 0.812 = 0.2501 3

June 2009

M1 0.23 × 0.812 M1

( ) ×p q 15 3

3

12

A1 cao

3 Or from tables 0.6482 − 0.3980 = 0.2502

(B)

(C)

(ii)

(iii)

Or: M2 for 0.6482 – 0.3980 A1 cao M1 P(X≤2) M1 1-P(X≤2) A1 cao

P( X ≥ 3) = 1 − 0.3980 = 0.6020

E(X) = np = 15 × 0.2 = 3.0

(A) Let p = probability of a randomly selected child eating at least 5 a day H0: p = 0.2 H1: p > 0.2 (B) H1 has this form as the proportion who eat at least 5 a day is expected to increase. Let X ~ B(15, 0.2) P(X ≥ 5) = 1 – P(X ≤ 4) = 1 – 0.8358 = 0.1642 > 10% P(X ≥ 6) = 1 – P(X ≤ 5) = 1 – 0.9389 = 0.0611 < 10% So critical region is {6,7,8,9,10,11,12,13,14,15} 7 lies in the critical region, so we reject null hypothesis and we conclude that there is evidence to suggest that the proportion who eat at least five a day has increased.

66

M1 for product A1 cao B1 for definition of p in context B1 for H0 B1 for H1 E1

3

2

4

B1 for 0.1642 B1 for 0.0611 M1 for at least one comparison with 10% A1 cao for critical region dep on M1 and at least one B1 M1 dep for comparison A1 dep for decision and conclusion in context TOTAL

6

18

4767

Mark Scheme

June 2009

4767 Statistics 2 Question 1 (i)

EITHER: 1 1 Sxy =  xy −  x  y = 316345 – × 2331.3 × 6724.3 n 50

M1 for method for Sxy

= 2817.8 Sxx =  x 2 −

1 ( x )2 = 111984 − 1 × 2331.32 = 3284.8 n 50

M1 for method for at least one of Sxx or Syy

Syy =  y 2 −

1 ( y )2 = 921361 − 1 × 6724.32 = 17036.8 n 50

A1 for at least one of Sxy, Sxx or Syy correct

r =

S xy S xx S yy

2817.8

=

3284.8 × 17036.8

= 0.377

M1 for structure of r A1 (AWRT 0.38)

OR: 316345  xy − xy = – 46.626 × 134.486 n 50 = 56.356 S xx 3284.8 = 65.696 = 8.105 = rmsd(x) = n 50

cov(x, y) =

M1 for method for at least one msd

17036.8 = 340.736 = 18.459 n 50 56.356 cov( x, y ) r = = = 0.377 8.105 × 18.459 rmsd( x)rmsd( y )

A1 for at least on of cov(x,y), rmsd(x) or rmsd(y) correct M1 for structure of r A1 (AWRT 0.38)

H0: ρ = 0 H1: ρ ≠ 0 (two-tailed test) where ρ is the population correlation coefficient For n = 50, 5% critical value = 0.2787

B1 for H0, H1 in symbols B1 for defining ρ 

rmsd(y) =

(ii)

S yy

M1 for method for cov(x,y)

=

5

B1FT for critical value

M1 for sensible comparison leading to a conclusion There is sufficient evidence at the 5% level to suggest that A1 for result 1 B1 FT for conclusion in there is correlation between oil price and share cost context Since 0.377 > 0.2787 we can reject H0:

(iii)

(iv)

Population The scatter diagram has a roughly elliptical shape, hence the assumption is justified. Because the alternative hypothesis should be decided without referring to the sample data and there is no suggestion that the correlation should be positive rather than negative.

B1 B1 elliptical shape E1 conclusion E1 E1

TOTAL

67

6

3

2 16

4767

Mark Scheme

June 2009

Question 2 (i)

(ii)

Meteors are seen randomly and independently There is a uniform (mean) rate of occurrence of meteor sightings (A) Either P(X = 1) = 0.6268 – 0.2725 = 0.3543 1.31 = 0.3543 Or P(X = 1) = e −1.3 1! (B) Using tables: P(X ≥ 4) = 1 – P(X ≤ 3)

B1 B1 M1 for appropriate use of tables or calculation A1 M1 for appropriate probability

= 1 – 0.9569

calculation

= 0.0431

A1

(iii)

λ = 10 × 1.3 = 13

B1 for mean M1 for calculation A1 CAO

(iv)

131 = 0.0859 1! Mean no. per hour = 60 × 1.3 = 78 P(X = 10) = e −13

Normal approx. to the Poisson,

X ~ N(78, 78)

 99.5 − 78   P(X ≥ 100) = P  Z > 78  

4

3

B1 for Normal approx. B1 for correct parameters (SOI) B1 for continuity corr. M1 for correct Normal probability calculation using correct tail A1 CAO, (but FT wrong or omitted CC)

= P(Z >2.434) = 1 – Φ(2.434) = 1 – 0.9926 = 0.0074

(v)

2

5

Either P(At least one) = 1 − e

−λ

M1 formation of equation/inequality using P(X ≥ 1) = 1 – P(X = 0) with Poisson distribution. A1 for correct equation/inequality M1 for logs A1 for 3.54 A1 for t (correctly justified)

λ0 = 1 − e − λ ≥ 0.99 0!

e − λ ≤ 0.01 –λ ≤ ln 0.01, so λ ≥ 4.605 1.3 t ≥ 4.605, so t ≥ 3.54 Answer t = 4

Or t = 1, λ = 1.3, P(At least one) = 1 − e −1.3 = 0.7275 t = 2, λ = 2.6, P(At least one) = 1 − e −2.6 = 0.9257 t = 3, λ = 3.9, P(At least one) = 1 − e −3.9 = 0.9798 t = 4, λ = 5.2, P(At least one) = 1 − e −5.2 = 0.9944 Answer t = 4

M1 at least one trial with any value of t A1 correct probability. M1 trial with either t = 3 or t = 4 A1 correct probability of t = 3 and t = 4 A1 for t

TOTAL

68

5

19

4767

Mark Scheme

June 2009

Question 3 (i)

X ~ N(1720,902)

M1 for standardising

1700 − 1720   P(X – 1.645 so not significant. There is not sufficient evidence to reject H0 There is insufficient evidence to conclude that the mean intensity of bulbs made by this manufacturer is less than 1720

M1 must include √20

A1FT B1 for –1.645 No FT from here if wrong. Must be –1.645 unless it is clear that absolute values are being used. M1 for sensible comparison leading to a conclusion. FT only candidate’s test statistic A1 for conclusion in words in context

TOTAL

69

5 20

4767

Mark Scheme

June 2009

Question 4

(i)

H0: no association between type of car and sex; H1: some association between type of car and sex; EXPECTED Hatchback Saloon People carrier 4WD

Male 83.16 70.56 51.66 17.01

Female 48.84 41.44 30.34 9.99

Sports car

29.61

17.39

CONTRIBUTION Hatchback Saloon People carrier 4WD Sports car

Male 1.98 0.59 3.61 0.23 1.96

Female 3.38 1.00 6.15 0.40 3.33

B1 M1 A2 for expected values (to 2 dp) (allow A1 for at least one row or column correct)

M1 for valid attempt at (O − E)2/E A1 for all correct NB These M1A1 marks cannot be implied by a correct final value of X 2

X2 = 22.62

M1 for summation A1 for X2 CAO

Refer to Ξ42 Critical value at 5% level = 9.488

B1 for 4 deg of f

12

B1 CAO for cv 22.62 > 9.488 M1 sensible comparison Result is significant There is evidence to suggest that there is some association leading to a conclusion A1 between sex and type of car.

(ii)

NB if H0 H1 reversed, or ‘correlation’ mentioned, do not award first B1or final A1 • In hatchbacks, male drivers are more frequent than expected. • In saloons, male drivers are slightly more frequent than expected. • In people carriers, female drivers are much more frequent than expected. • In 4WDs the numbers are roughly as expected • In sports cars, female drivers are more frequent than expected.

E1 E1 E1 E1 E1

TOTAL

70

5

17

4768

Mark Scheme

June 2009

4768 Statistics 3 Q1

W ~ N(14, 0.552) G ~ N(144, 0.92)

(i)

145 − 144   P(G < 145) = P Z < = 1.1111  0 . 9   = 0·8667

M1 A1

For standardising. Award once, here or elsewhere.

A1

c.a.o.

W + G ~ N (14 + 144 = 158,

B1 B1

Mean. Variance. Accept sd (= 1.0547…).

A1

c.a.o.

B1 B1

Mean. Variance. Accept sd (= 2.6415).

(ii)

When a candidate’s answers suggest that (s)he appears to have neglected to use the difference columns of the Normal distribution tables penalise the first occurrence only.

σ 2 = 0.55 2 + 0.9 2 = 1.1125 ) P(this > 160) = 160 − 158   = 1 ⋅ 896  = 1 − 0 ⋅ 9710 = 0.0290 P Z > 1 . 054 7  

(iii)

(iv)

H = W1 + ... + W7 + G1 + ... + G6 ~ N (962, σ 2 = 0.55 2 + ... + 0.55 2 + 0.9 2 + ... + 0.9 2 = 6.9775) P(960 < this < 965) = 965 − 962   960 − 962 = 1.1357  P = −0.7571 < Z < 2 ⋅ 6415   2 ⋅ 6415 = 0·8720 – (1 – 0.7755) = 0.6475

M1

Two-sided requirement.

A1

c.a.o.

Now want P(B(4, 0.6475) ≥ 3)

M1

Evidence of attempt to use binomial. ft c’s p value. Correct terms attempted. ft c’s p value. Accept 1 – P(… ≤ 2) c.a.o.

= 4 × 0.64753 × 0.3525 + 0.64754

M1

= 0.38277 + 0.17577 = 0.5585

A1 B1 B1

D = H 1 − H 2 ~ N(0, 6.9775 + 6.9775 = 13.955) Want h s.t. P(−h < D < h) = 0.95 i.e. P(D < h) = 0 975 ∴ h = 13.955 × 1.96

M1 B1 A1

= 7.32

3

3

7

Mean. (May be implied.) Variance. Accept sd (= 3.7356). Ft 2 × c’s 6.9775 from (iii). Formulation of requirement as 2sided. For 1.96. c.a.o.

5 18

71

4768

Q2 (i)

Mark Scheme

H0: μ = 1 H1: μ < 1

B1

where μ is the mean weight of the cakes.

B1

x = 0.957375

B1

s n −1 = 0.07314(55)

M1

Test statistic is 0 . 957375 − 1 0 . 07314 8

June 2009

Both hypotheses. Hypotheses in words only must include “population”. For adequate verbal definition. Allow absence of “population” if correct notation μ is used, but do NOT allow “ X = ... ” or similar unless X is clearly and explicitly stated to be a population mean.

sn = 0.06842 but do NOT allow this here or in construction of test statistic, but FT from there. Allow c’s x and/or sn–1. Allow alternative: 1 + (c’s –1.895) × 0.07314 (= 0.950997) for 8

subsequent comparison with x . (Or x – (c’s –1.895) × 0.07314 8

= –1.648(24).

(ii)

(iii)

A1

Refer to t7.

M1

Single-tailed 5% point is –1.895.

A1

Not significant. Insufficient evidence to suggest that the cakes are underweight on average.

A1 A1

CI is given by 0.957375 ± 2·365 0.07314 × 8 = 0.957375 ± 0.061156= (0.896(2), 1.018(5))

M1 B1 M1

x ± 1.96 ×

0.006 n

72

(= 1.006377) for comparison with 1.) c.a.o. but ft from here in any case if wrong. Use of 1 – x scores M1A0, but ft. No ft from here if wrong. P(t < –1.648(24)) = 0.0716. Must be minus 1.895 unless absolute values are being compared. No ft from here if wrong. ft only c’s test statistic. ft only c’s test statistic.

A1

c.a.o. Must be expressed as an interval. ZERO/4 if not same distribution as test. Same wrong distribution scores maximum M1B0M1A0. Recovery to t7 is OK.

M1 B1 A1

Structure correct, incl. use of Normal. 1.96. All correct.

9

4

3

4768

(iv)

Mark Scheme

2 × 1.96 ×

0.006 < 0.025 n

June 2009

M1

Set up appropriate in equation. Condone an equation.

M1

Attempt to rearrange and solve.

A1

c.a.o. (expressed as an integer). S.C. Allow max M1A1(c.a.o.) when the factor “2” is missing. (n > 36.879)

2

 2 × 1.96  n>  × 0.006 = 147.517  0.025  So take n = 148

3

19

73

4768

Q3 (i)

(ii)

Mark Scheme

For a systematic sample • she needs a list of all staff • with no cycles in the list. All staff equally likely to be chosen if she • chooses a random start between 1 and 10 • then chooses every 10th. Not simple random sampling since not all samples are possible.

June 2009

E1 E1 E1 E1 E1

Nothing is known about the background population ...

E1

... of differences between the scores.

E1

H0: m = 0 H1: m ≠ 0

B1

where m is the population median difference for the scores.

B1

5 Any reference to unknown distribution or “non-parametric” situation. Any reference to pairing/differences. Both hypotheses. Hypotheses in words only must include “population”. For adequate verbal definition.

4

(iii) Diff Rank

−0.8 2

−2.6 5

8.6 12

6.2 10

6.0 9

−3.6 6

−2.4 4 M1 M1 A1 B1

W− = 1 + 2 + 4 + 5 + 6 + 7 = 25 Refer to tables of Wilcoxon paired (/single sample) statistic for n = 12. Lower (or upper if 53 used) 2½ % tail is 13 (or 65 if 53 used). Result is not significant. No evidence to suggest a preference for one of the uniforms.

M1 A1 A1 A1

−0.4 1

−4.0 7

5.6 8

6.6 11

2.2 3

For differences. ZERO in this section if differences not used. For ranks. ft from here if ranks wrong. (or W+ = 3 + 8 + 9 + 10 + 11 + 12 = 53) No ft from here if wrong. i.e. a 2-tail test. No ft from here if wrong. ft only c’s test statistic. ft only c’s test statistic.

8 17

74

4768

Mark Scheme

2x λ2

Q4

f ( x) =

(i)

f(x) > 0 for all x in the domain.

 (ii)

for 0 < x < λ , λ > 0

λ

x  λ dx =  2  = 2 = 1 2 λ λ  0 λ 2

λ2x 0

μ=



2

λ

 2x3 / 3 2λ d x =   = 2 2 3 λ  λ 0

λ 2x 2 0

P( X < μ) =

June 2009



μ 0

E1 M1

Correct integral with limits.

A1

Shown equal to 1.

M1

Correct integral with limits.

A1

c.a.o.

M1

Correct integral with limits.

A1

Answer plus comment. ft c’s μ provided the answer does not involve λ.

M1

Use of Var(X) = E(X2) – E(X)2.

A1

c.a.o.

μ

 x2  2x d x =  2 λ2 λ 0

μ 2 4 λ2 / 9 4 = = 9 λ2 λ2 which is independent of λ. =

(iii)

3

λ2 2 2 2 λ λ2 4λ − = σ2 = 2 9 18

4

Given E ( X 2 ) =

2

(iv) Probability Expected f

0.18573 9.2865

0.25871 12.9355

0.36983 18.4915 M1 A1 M1 A1 M1

X2 = 3.0094 + 0.2896 + 0.1231 + 3.5152 = 6.937(3) Refer to χ 32 .

Upper 5% point is 7.815. Not significant. Suggests model fits the data for these jars. But with a 10% significance level (cv = 6.251) a different conclusion would be reached.

A1 A1 A1 E1

0.18573 9.2865 Probs × 50 for expected frequencies. All correct. Calculation of X2. c.a.o. Allow correct df (= cells – 1) from wrongly grouped table and ft. Otherwise, no ft if wrong. P(X2 > 6.937) = 0.0739. No ft from here if wrong. ft only c’s test statistic. ft only c’s test statistic. Any valid comment which recognises that the test statistic is close to the critical values.

9

18

75

4769

Mark Scheme

June 2009

4769 Statistics 4 Q1 (i)

Follow-through all intermediate results in this question, unless obvious nonsense. P(X ≥ 2) = 1 – θ – θ (1 – θ) = (1 – θ)2 [o.e.] L = [θ] n0 [θ(1 – θ)] n1 [(1-θ)2] n−n0 −n1 = θ n0 + n1 (1 – θ) 2 n −2 n0 − n1

(ii)

ln L = (n0 + n1 ) ln θ + (2n − 2n0 − n1 ) ln (1 – θ) d ln L dθ n0 + n1 2n − 2n0 − n1 – = θ 1− θ =0  (1 - θˆ ) (n0 + n1 ) = θˆ (2n − 2n0 − n1 ) n +n  θˆ = 0 1 2 n − n0

(iii)

E(X) =



 xθ (1 − θ )

5

M1 A1

A1 M1 A1 6

x

M1

= θ {0 + (1 – θ) + 2(1 – θ) 2 + 3(1 − θ ) 3 + …} 1− θ = θ So could sensibly use (method of moments) ~ ~ 1− θ θ given by ~ = X θ ~ 1 θ = 1+ X

A2

Divisible, for algebra; e.g. by “GP of GPs” BEWARE PRINTED ANSWER

M1 A1

To use this, we need to know the exact numbers of faults for components with “two or more”. 14 x= = 0 ⋅14 100 ~ 1 θ = = 0·8772 1+ 0.14 Also, from expression given in question, ~ 0 ⋅ 8772 2 (1 − 0 ⋅ 8772) Var(θ ) = 100 = 0·000945 CI is given by 0·8772 ± 1·96 x (0·817, 0·937)

Product form Fully correct BEWARE PRINTED ANSWER

M1

x =0

(iv)

M1 A1 M1 A1 A1

0 ⋅ 000945 =

BEWARE PRINTED ANSWER + 6

E1 B1 B1

B1 M1 B1 M1 A1

For 0.8772 For 1.96 For 0 ⋅ 000945 7

76

4769

Mark Scheme

June 2009

Q2 (i)

tZ

Mgf of Z = E (e ) =



1

−∞





e

tz −

z2 2

M1 dz

Complete the square z2 1 1 tz − = − (z − t)2 + t 2 2 2 2 =

t2 e2





1

2π Pdf of N(t,1) ∴ =1 −∞

e



( z −t ) 2 2

M1 A1 A1 M1 M1 M1

t2 e2

dt =

A1



(ii)

Y has mgf M Y (t )

For final answer

t2 e2

= e bt E[e ( at )Y ] = e bt M Y (at )

For factor e bt For factor E [e ( at )Y ] For final answer

X −μ , so X = σZ + μ σ

M1 1

For factor e μt

Z= ∴M

(iv)

For taking out factor For use of pdf of N(t,1) For ∫ pdf = 1

M1 1 1 1

Mgf of aY + b is E [e t ( aY +b ) ]

(iii)

t2 e2

X

( t ) = e μ t .e

( σt )2 2

=e

μt +

σ 2t 2 2

1 1

W = eX E (W k ) = E[(e X ) k ] = E (e kX ) = M X (k )

M1 A1 A1

∴ E (W ) = M X (1) = e

μ+

σ2 2

E(W 2 ) = M X (2) = e 2 μ + 2σ 2

( σt ) 2 e 2

For factor For final answer

8

4

4

For E[(e X ) k ] For E (e kX ) For M X (k )

M1 A1 2

2

2

2

∴ Var(W ) = e 2 μ + 2σ − e 2 μ +σ [= e 2 μ +σ (e σ − 1)]

77

M1 A1 A1

8

4769

Mark Scheme

June 2009

Q3 (i)

A1

x = 126 ⋅ 2 s = 8·7002 s 2 = 75 ⋅ 693 y = 133⋅ 9 s = 10·4760 s 2 = 109 ⋅ 746 H 0 : μ A = μB H 0 : μ A ≠ μB Where μ A , μ B are the population means.

1 1

Pooled s 2 9 × 75 ⋅ 693 + 6 × 109 ⋅ 476 681 ⋅ 24 + 658 ⋅ 48 = = 15 15  = 89 ⋅ 3146 [√ = 9·4506] Test statistic is 126 ⋅ 2 − 133 ⋅ 9 1 1 89 ⋅ 3146 + 10 7

(ii)

=−

A1 if all correct. [No mark for use of sn , which are 8·2537 and 9·6989 respectively.] Do not accept X = Y or similar.

7⋅7 = − 1⋅ 653 4 ⋅ 6573

B1

M1 A1

Refer to t15

1

No FT if wrong

Double-tailed 10% point is 1·753 Not significant No evidence that population mean concentrations differ. There may be consistent differences between days (days of week, types of rubbish, ambient conditions, …) which should be allowed for.

1 1 1

No FT if wrong

Assumption: Normality of population of differences. Differences are 7·4 -1·2 11·1 5·5 6·2 3·7 −0·3 1·8 3·6 [ d = 4·2, s = 3·862 ( s 2 =14·915)] Use of s n (= 3 ⋅ 641) is not acceptable, even in a denominator of s n /

Test statistic is

10 E1 E1 1 M1 A1 Can be awarded here if NOT awarded in part (i)

n −1 ]

4⋅2 − 0 3 ⋅ 862 / 9

M1 A1

= 3·26

1 1 1 1

Refer to t 8 Double-tailed 5% point is 2·306 Significant Seems population means differ

No FT if wrong No FT if wrong 10

78

4769

(iii)

Mark Scheme

Wilcoxon rank sum test Wilcoxon signed rank test H0: medianA = medianB H1: medianA ≠ medianB

June 2009

B1 B1 1 1

[Or more formal statements]

4

Q4 (i)

Description must be in context. If no context given, mark according to scheme and then give half-marks, rounded down. Clear description of “rows”. And “columns” As extraneous factors to be taken account of in the design, with “treatments” to be compared. Need same numbers of each Clear contrast with situations for completely randomised design and randomised trends.

(ii)

(iii)

eij ~ ind N (0, σ2)

E1 E1 E1 E1 E1 E1 E1 E1 E1

αi is population mean effect by which ith treatment differs from overall mean

1 1 1 1 1

Source of Variation

1 1

SS

df

MS

MS ratio

Between Treatments

92·30

4

23·075

5·034

Residual

68·76

15

4·584

Total

161·06

19

9 Allow uncorrelated For 0 For σ2

5

1 1 1 1 1

Refer to F4,15 Upper 1% point is 4·89 Significant, seems treatments are not all the same

79

1

No FT if wrong

1 1

No FT if wrong 10

4771

Mark Scheme

June 2009

4771 Decision Mathematics 1 Question 1 (i)

1 and 6, 2 and 5, 3 and 4

B1

(ii) Vertex 6

Vertex 1

Vertex 5

M1 A4

10 to 14 edges (–1 each edge error)

B1 B1

identification sketch

Vertex 2

Vertex 4

Vertex 3

(iii)

Question 2. (i)

A's c takes 2, leaving 3. You have to take 1. A's c takes one and you lose.

M1 A1 A1

(ii)

A's c takes 3 leaving 3. Then as above.

M1 A1

(iii)

A's c takes 3 leaving 4. You can then take 1, leading to a win.

M1 A1 A1

80

4771

Mark Scheme

June 2009

Question 3. (i) (0, 12) 48

(5, 10) 55

(11, 7) 61

(18, 0) 54

(ii)

B3 lines B1 shading

61 at (11, 7)

M1 optimisation A1

Intersection of 2x+5y=60 and x+y=18 is at (10,8)

M1

10 + 2×8 = 26

A1

81

4771

Mark Scheme

June 2009

Question 4. (i)

e.g. 0 – 4 exit

(ii)

e.g.

1 2 3 4 5 6 7 8 9 10

A A A A A A A A A A

E×A B E×A B B B B E×A B E×A

(iii)

e.g. 0–2 exit 6–8 other vertex

(iv)

e.g.

0.7, 0.1, 0.2

A

B

A

B

A E×B A A

B

A

E×A

B B

A E×B

B

E×B

A A A A A A A A A A

B C E×A B E×A C E×A B E×A E×A

M1 A1

process with exits

B1 M1 A1

probabilities duration

M1 DM1 A1 A1

ignore conditionality equal prob efficient

E×B

E×B

(Theoretical answers: 2/3, 1/3, 2) (Gambler's ruin)

0.5, 0.5, 1.9

1 2 3 4 5 6 7 8 9 10

B1 B1

5 – 9 other vertex

3–5 next vertex in cycle 9–ignore and re-draw

A A

B E×A

A

E×A

C

B

C

E×C

A

B

E B

C

E×C

(Theoretical probs are 0.5, 0.25, 0.25) (Markov chain)

82

M1 A2

M1 A1

4771

Mark Scheme

June 2009

Question 5. (i)

e.g. A

B

17 15

18

25 15

15 and/or 20 16 E (ii) A

connectivity lengths

M1 A2

connected tree (–1 each error)

C

20

D

M1 A1 A1

B

C D E

Order: AE; AD; DB; DC or AD; AE; DB; DC or AD; DB; AE; DC Length: 65 km OR B A

A1 B1

C D E

(iii)

Order: AD; DB; DE; DC

Length: 66 km B

A

X

C

M1 A2

D

E

Length: 53 km Advice: Close BC, AE and BD (iv)

B1 B3

facility (e.g. anglers) distances (e.g. B to C)

B1

83

connected tree (–1 each error)

4771

Mark Scheme

June 2009

Question 6. (i)&(ii) 60 200 D 60 0

A E B 160 160 0 100 100 100 40 60 C 150

200 200

F 2

220 220 H 10

G 30 160 190

230 230

sca (activity on arc) single start & end dummy rest

M1 A1 M1 A1

forward pass

B1 B1

time − 230 minutes critical − A; B; E; F; H (iii)

M1 A1 A1 A1

backward pass

cao

e.g.

H G F E D C B A

M1 A2

cascade

B1

Joan/Keith

Least time = 240 mins

B1

Minimum project completion times assumes no resource constraints.

B1

84

4772

Mark Scheme

June 2009

4772 Decision Mathematics 2 Question 1. (a)

e.g. "It is easy to overestimate the effect that your contribution will make."

M1 A1

(b)

remove double negatives same meaning

e.g.

A

B

combinatorial "ands" negations "ors" one for each alternative

M1 A1 A1 A1 A1

8 lines a, b, c negations "and"s "or"s

machine

C

(c)

M1 A1 A1 A1 A3

e.g. ((a 1 1 1 1 0 0 0 0 ~

∧ 1 1 0 0 0 0 0 0 ((~a 0 0 0 0 1 1 1 1

b) 1 1 0 0 1 1 0 0

∨ 1 1 0 0 1 0 1 0 ∧ 0 0 0 0 0 1 0 1

(~a 0 0 0 0 1 1 1 1 ~c) 0 1 0 1 0 1 0 1

∧ 0 0 0 0 1 0 1 0 ∨ 0 0 0 1 0 1 0 1

c)) 1 0 1 0 1 0 1 0 (~b 0 0 1 1 0 0 1 1

∨ 1 1 1 0 1 0 1 0 ∧ 0 0 0 1 0 0 0 1

(~b 0 0 1 1 0 0 1 1

∧ 0 0 1 0 0 0 1 0

~c)) 0 1 0 1 0 1 0 1

c) 1 0 1 0 1 0 1 0

M1 A1

85

4772

Mark Scheme

June 2009

Question 2. (i)

revised D D S S

60marks D S D S

score 48+20 = 68 32+30 = 62 36+28 = 64 24+42 = 66

M1 A1 A1 A1 A1

(ii) D worth 60 revise D

68

0.5 0.5 S worth 60 62

65

M1 A1 A1

65 revise S

D worth 60 65

64

M1 A1

0.5

0.5 S worth 60 66

65 S M predicts D worth 60

65.8

decision node

D worth 60

revise D do not ask M

chance node

68 0.7 66.2 0.3 S worth 60 62

66.2 revise S

D worth 60 64.6

64

0.7

0.3 S worth 60 66

0.5 ask M

M1 chances A1

65.8 D worth 60

0.5 M predicts S worth 60

revise D

68

0.3 0.7 S worth 60 62

63.8

65.4 revise S

D worth 60 65.4

64

0.3

0.7 S worth 60 66 B1

Worth of Michael's help = 0.8 marks

86

M1decisions A1 revise A1 ask/don't

4772

Mark Scheme

June 2009

Question 3. (i)

a is the number of acres of land put to crop A, etc a + b ≤ 20 is equivalent to a + b ≤ c + d Given that a + b + c + d ≤ 40, the maximisation will ensure that a + b + c + d = 40 (and it's easier to solve using simplex).

(ii) P

a 1 0 0 1 0 0 1 0 0

b –50 1 1 0 1 0 0 1 0

c –40 1 1 10 1 0 10 1 0

d –40 0 1 –40 0 1 0 0 1

s1 –30 0 1 –30 0 1 10 0 1

s2 0 1 0 50 1 –1 10 1 –1

M1 A1 A1 A1

RHS 0 20 40 1000 20 20 1800 20 20

0 0 1 0 0 1 40 0 1

M1 A1 M1 A1

20 acres to A and 20 acres to C, giving profit of £1800

B1 B1

(iii) Max 50a + 40b + 40c + 30d st a + b ≤ 20 a + b + c + d ≤ 40 a + b + c + d ≥ 40 A 1 0 0 0 0

P

P 0 1 0 0 0

B1 B1 B1

new obj surplus artificial

B1

3 constraints

Minimise A (to zero) then drop A row and art column and continue normally

B1 B1

OR

OR

0 0 0

b –50 –M 1 1 1

b 1 –40 1 1 1

c –40 –M 1 1 1

c 1 –40 0 1 1

B1 R 40 0 20 40 40

a 1

a 1 –50 1 1 1

B1 B1 B1

d –40 –M 0 1 1

–30 –M 0 1 1

d 1 –30 0 1 1

s1 0 0 1 0 0

s2 0 0 0 1 0

sur –1 0 0 0 –1

art 0 0 0 0 1

s1 s2 sur art R 0 0 M 0 –40M 1 0 0

0 1 0

0 0 –1

0 0 1

20 40 40

Proceed as per simplex, regarding M as a large fixed number.

87

M1 A1 B1 B1 B1 B1

surplus artificial

4772

Mark Scheme

June 2009

Question 4. (a) (i),(ii) and (iii)

1 2 3 4 5

1 ∞ 22 ∞ 15 15

2 22 ∞ 20 5 23

3 ∞ 20 ∞ 40 ∞

4 15 5 40 ∞ 16

5 15 23 ∞ 16 ∞

1 2 3 4 5

1 1 1 1 1 1

2 2 2 2 2 2

3 3 3 3 3 3

4 4 4 4 4 4

5 5 5 5 5 5

1 2 3 4 5

1 ∞ 22 ∞ 15 15

2 22 44 20 5 23

3 ∞ 20 ∞ 40 ∞

4 15 5 40 30 16

5 15 23 ∞ 16 30

1 2 3 4 5

1 1 1 1 1 1

2 2 1 2 2 2

3 3 3 3 3 3

4 4 4 4 1 4

5 5 5 5 5 1

1 2 3 4 5

1 44 22 42 15 15

2 22 44 20 5 23

3 42 20 40 25 43

4 15 5 25 10 16

5 15 23 43 16 30

1 2 3 4 5

1 2 1 2 1 1

2 2 1 2 2 2

3 2 3 2 2 2

4 4 4 2 2 4

5 5 5 2 5 1

1 2 3 4 5

1 44 22 42 15 15

2 22 44 20 5 23

3 42 20 40 25 43

4 15 5 25 10 16

5 15 23 43 16 30

1 2 3 4 5

1 2 1 2 1 1

2 2 1 2 2 2

3 2 3 2 2 2

4 4 4 2 2 4

5 5 5 2 5 1

1 2 3 4 5

1 30 20 40 15 15

2 20 10 20 5 21

3 40 20 40 25 41

4 15 5 25 10 16

5 15 21 41 16 30

1 2 3 4 5

1 4 4 2 1 1

2 4 4 2 2 4

3 4 3 2 2 4

4 4 4 2 2 4

5 5 4 2 5 1

1 2 3 4 5

1 30 20 40 15 15

2 20 10 20 5 21

3 40 20 40 25 41

4 15 5 25 10 16

5 15 21 41 16 30

1 2 3 4 5

1 4 4 2 1 1

2 4 4 2 2 4

3 4 3 2 2 4

4 4 4 2 2 4

5 5 4 2 5 1

Shortest distance from 3 to 1 is 40 (1st row and 3rd column of distance matrix)

88

M1 distance A1 1 to 5 etc A1 rest B1

route

Not part of the question

Not part of the question

Not part of the question

M1 A1 10 changed dists M1 2's in r3 of route A1 rest of route

B1 B1

4772

Mark Scheme

June 2009

B1

Shortest route is 3 2 4 1 3 followed by route matrix (3,1) = 2 followed by route matrix (2,1) = 4 followed by route matrix (4,1) = 1

M1 A1

(iv)

1

21

15

5

15

M1 A1

20 25

16

5

40

2

20

4

3

41

(v)

2 (5) 4 (15) 1 (15) 5 (41) 3 (20) 2

Total length = 96

B1 B1

2 4 1 5 (4 2) 3 2

M1 A1

Finds a (hopefully short) route visiting every vertex and returning to the start, or, upper bound to the TSP

B1

89

4773

Mark Scheme

June 2009

4773 Decision Mathematics Computation Question 1. (i)

Bn+2 = Bn+1 + (0 – Bn)

M1 A1

(ii)

Oscillation: 2, 4, 2, –2, –4, –2, 2, 4, …

M1 A1 B1

(iii)

Bn+2 – Bn+1 +

1 2

B1 B1 B1

Bn = 0

2, 4, 3, 1, –0.5, –1, …, 0.00391, –0.00195 Oscillatory convergence (iv)

2, 4, 3.5, 2.5, 1.625, 1, …, 0.00022, 0.00012 Faster and uniform convergence

(v)

Auxiliary eqn: x2 – x + x=

1 4

B1 B1 B1 B1 B1

=0

1 2

B1

Bn = A( ) + Bn( ) 2=A 4 = 1 + 12 B giving B = 6 1 n 2

Bn = ( 2 + 6n )

1 n 2

( ) 1 2

n

B1 B1

(12 )n

n −1 (2 + 6n ) or (1 + 3n ) 12

( )

or (1 + 3n )( 12 )

"the same"

90

n

B1 B1

4773

Mark Scheme

June 2009

Question 2. (i)

12

e.g.

2

10

20

20

10

20

10

10

0

10

2

18

18

8

10

28

10

2

8

18

20

0

2

12

10

20

10

2

8

18

0

18

36

10

0

10

20

10

20

10

0

2

12

10

20

10

2

8

10

18

36

reversal

B1

rest

0

0

20

8

0

M1 A1

10

10

18

2

18 10

0

0

10

8

18

M1 A1

10

10

2

10

10

10

10

2

20

0

12

20

10 10

18

12

0 0

10

20

10

10

10

12

10

M1 A1

10 10

91

0

4773

Mark Scheme

June 2009

(ii)

{S, A, C, D, E} / {B, T}

M1 A1

(iii)

e.g. Max SA + SE st SA + BA + CA – AB – AC = 0 AB + CB – BA – BC – BT = 0 AC + DC + BC – CA – CB – CD = 0 SE + DE – ED = 0 CD + ED – DC – DE – DT = 0 SA < 12 SE < 20 AB < 10 BA < 10 AC < 10 CA < 10 BC < 10 CB < 10 CD < 10 DC < 10 ED < 18 DE < 18 BT < 20 DT < 10 end

M1 A1

variables objective

M1 A1

balancing

M1 A1 A1

capacities forwards backwards

B1

running

(iv)

OBJECTIVE FUNCTION VALUE 1) 30.00000 VARIABLE VALUE REDUCED COST SA 12.000000 0.000000 SE 18.000000 0.000000 BA 0.000000 0.000000 CA 0.000000 0.000000 AB 0.000000 10.000000 0.000000 AC 2.000000 CB 10.000000 0.000000 0.000000 BC 0.000000 BT 0.000000 20.000000 DC 0.000000 8.000000 CD 0.000000 0.000000 0.000000 DE 1.000000 18.000000 ED 0.000000 10.000000 DT 0.000000

B1

Solution as per part (i)

92

4773

Mark Scheme

June 2009

Question 3. (i)

e.g. -1 1 1 1 1 1 -1 etc.

5 4 5 6 7 8 9 8

= if(rand() | b | + 2 will ensure convergence. ( > required as dominance has to be strict)

(ii) 4 1 2 1

1 4 1 2

2 1 4 1

1 2 1 4

0 0.25 0.321289 0.340733 0.344469 0.344515 0.344124 0.343886 0.343789 0.343758 0.34375 0.343749 0.34375 0.34375

0 -0.0625 -0.05103 -0.03941 -0.03388 -0.0319 -0.03134 -0.03123 -0.03123 -0.03124 -0.03125 -0.03125 -0.03125 -0.03125

0 -0.10938 -0.14691 -0.15599 -0.15715 -0.15681 -0.15648 -0.15633 -0.15627 -0.15625 -0.15625 -0.15625 -0.15625 -0.15625

0 -0.00391 -0.01808 -0.02648 -0.02989 -0.03098 -0.03124 -0.03127 -0.03127 -0.03126 -0.03125 -0.03125 -0.03125 -0.03125

2 1 4 1

1 2 1 4

4 1 2 1

1 4 1 2

0 0.5 2.03125 6.033203 12.69934 3.347054 -156.613 -1153.81 -5937.67

0 -0.25 -1.95313 -10.5127 -46.9236 -183.278 -632.515 -1881.51 -4365.6

0 -0.875 -3.42969 -9.11279 -13.2195 37.89147 456.5137 2690.835 12560.88

0 0.6875 4.605469 22.56519 94.10735 345.9377 1115.079 2994.509 5419.593

1 0 0 0

a 4

[E1] [E1E1] [subtotal 3] b 2

setup [M3A3]

values [A3]

1 0 0 0

a 2

b 4

values [A3] [subtotal 12]

(iii)

No convergence when a = 2, b = 0 Indicates that non-strict diagonal dominance is not sufficient 101

[M1A1] [E1E1]

4777

Mark Scheme

June 2009

[subtotal 4] (iv)

Use RHSs 1,0,0,0 0,1,0,0 to obtain inverse as 0.34375 -0.03125 -0.15625 -0.03125

-0.03125 0.34375 -0.03125 -0.15625

0,0,1,0 0,0,0,1 -0.15625 -0.03125 0.34375 -0.03125

-0.03125 -0.15625 -0.03125 0.34375

102

[M1]

[A1] [A1] [A1] [A1] [subtotal 5] [TOTAL 24]

Grade Thresholds Advanced GCE MEI Mathematics 7895-8 3895-8 June 2009 Examination Series Unit Threshold Marks Unit

Maximum Mark

A

B

C

D

E

U

All units

UMS

100

80

70

60

50

40

0

4751

Raw

72

59

52

45

39

33

0

4752

Raw

72

51

44

38

32

26

0

4753

Raw

72

57

52

47

42

37

0

4753/02

Raw

18

15

13

11

9

8

0

4754

Raw

90

67

59

51

43

35

0

4755

Raw

72

53

45

37

30

23

0

4756

Raw

72

51

45

39

33

27

0

4757

Raw

72

60

51

42

34

26

0

4758

Raw

72

61

55

49

43

36

0

4758/02

Raw

18

15

13

11

9

8

0

4761

Raw

72

57

48

39

30

21

0

4762

Raw

72

47

40

33

26

20

0

4763

Raw

72

55

46

38

30

22

0

4764

Raw

72

61

52

43

34

26

0

4766/G241

Raw

72

60

53

46

40

34

0

4767

Raw

72

57

50

44

38

32

0

4768

Raw

72

55

48

41

34

28

0

4769

Raw

72

56

49

42

35

28

0

4771

Raw

72

63

56

49

42

36

0

4772

Raw

72

57

51

45

39

33

0

4773

Raw

72

51

44

37

30

24

0

4776

Raw

72

62

53

45

37

28

0

4776/02

Raw

18

14

12

10

8

7

0

4777

Raw

72

55

47

39

32

25

0

103

Specification Aggregation Results

Overall threshold marks in UMS (ie after conversion of raw marks to uniform marks) A

B

C

D

E

U

7895-7898

Maximum Mark 600

480

420

360

300

240

0

3895-3898

300

240

210

180

150

120

0

The cumulative percentage of candidates awarded each grade was as follows: A

B

C

D

E

U

7895

44.1

65.4

81.4

92.1

97.9

100

Total Number of Candidates 10375

7896

57.2

78.0

88.9

95.4

98.9

100

1807

7897

87.1

93.55

100

100

100

100

31

7898

0

0

100

100

100

100

1

3895

35.3

52.9

67.4

79.1

88.1

100

16238

3896

52.1

70.2

82.4

90.4

95.7

100

2888

3897

80.4

88.2

91.2

96.1

97.1

100

102

3898

6.3

12.5

18.8

25.0

68.8

100

16

For a description of how UMS marks are calculated see: http://www.ocr.org.uk/learners/ums_results.html Statistics are correct at the time of publication.

104

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