Mark scheme

GCE Mathematics Advanced GCE A2 7890 - 2 Advanced Subsidiary GCE AS 3890 - 2

Mark Schemes for the Units June 2007

3890-2/7890-2/MS/R/07

OCR (Oxford, Cambridge and RSA Examinations) is a unitary awarding body, established by the University of Cambridge Local Examinations Syndicate and the RSA Examinations Board in January 1998. OCR provides a full range of GCSE, A level, GNVQ, Key Skills and other qualifications for schools and colleges in the United Kingdom, including those previously provided by MEG and OCEAC. It is also responsible for developing new syllabuses to meet national requirements and the needs of students and teachers. 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 2007 Any enquiries about publications should be addressed to: OCR Publications PO Box 5050 Annesley NOTTINGHAM NG15 0DL Telephone: Facsimile: E-mail:

0870 870 6622 0870 870 6621 [email protected]

2

CONTENTS Advanced GCE Mathematics (7890) Advanced GCE Pure Mathematics (7891) Advanced GCE Further Mathematics (7892) Advanced Subsidiary GCE Mathematics (3890) Advanced Subsidiary GCE Pure Mathematics (3891) Advanced Subsidiary GCE Further Mathematics (3892)

MARK SCHEME ON THE UNITS Unit

Content

Page

4721

Core Mathematics 1

1

4722

Core Mathematics 2

7

4723

Core Mathematics 3

11

4724

Core Mathematics 4

15

4725

Further Pure Mathematics 1

19

4726


25

4727


31

4728

Mechanics 1

37

4729

Mechanics 2

43

4730

Mechanics 3

47

4731

Mechanics 4

53

4732

Probability & Statistics 1

59

4733


65

4734


69

4735


75

4736

Decision Mathematics 1

79

4737

Decision Mathematics 2

85

*

Grade Thresholds

91

Mark Scheme 4721 June 2007

1

4721

Mark Scheme

(4x2 + 20x + 25) – (x2 - 6x + 9) = 3x2 + 26x + 16

1

June 2007

A1

Square one bracket to give an expression of the form ax2 + bx + c (a ≠ 0, b ≠ 0, c ≠ 0) One squared bracket fully correct

A1 3

All 3 terms of final answer correct

M1

Alternative method using difference of two squares: (2x + 5 + (x – 3))(2x + 5 – (x – 3))

M1 2 brackets with same terms but different signs A1 One bracket correctly simplified A1 All 3 terms of final answer correct

= (3x + 2)(x + 8) = 3x2 + 26x + 16 3 2 (a)(i)

Excellent curve for

B1

1 in either x

quadrant B1 2

Excellent curve for

1 in other quadrant x

SR B1 Reasonably correct curves in 1st and 3rd quadrants (ii) B1 1 Correct graph, minimum point at origin, symmetrical

(b)

B1 B1 2

Stretch Scale factor 8 in y direction or scale factor ½ in x direction

5 M1 3 (i)

(ii)

3 20 or 3 2 or 90 × 2

5 × 2 or 180

=6 5

A1 2

Correctly simplified answer

10 5 + 5 5

M1 B1

Attempt to change both surds to 5 One part correct and fully simplified

= 15 5

A1 3

cao

5

2

4721

4 (i) (ii)

Mark Scheme

June 2007

(-4)2 – 4 x k x k = 16 – 4k2

M1 A1 2

Uses b 2 − 4ac (involving k) 16 – 4k2

16 – 4k2 = 0

M1

k2 = 4 k=2 or k = -2

B1 B1 3

Attempts b 2 − 4ac = 0 (involving k) or attempts to complete square (involving k)

Length = 20 – 2x

M1

5 5 (i)

A1 2 Area = x(20 – 2x) = 20x – 2x2 (ii)

dA = 20 – 4x dx For max, 20 – 4x = 0

M1

Differentiates area expression

x = 5 only Area = 50

M1 A1 A1 4

Uses

6

2

6

Expression for length of enclosure in terms of x Correctly shows that area = 20x – 2x2 AG

dy =0 dx

Let y = (x + 2) y2 + 5y – 6 = 0

B1

Substitute for (x + 2) 2 to get y2 + 5y – 6 (= 0)

(y + 6)(y - 1) = 0

M1 A1

Correct method to find roots Both values for y correct

M1 A1 A1 6

Attempt to work out x One correct value Second correct value and no extra real values

y = -6 or y = 1 (x + 2)2 = 1 x = -1 or x = -3 7 (a)

(b)

6

f(x) = x + 3x-1

M1

Attempt to differentiate

f′ (x) = 1 – 3x-2

A1

First term correct

A1

x-2 soi www

A1 4

Fully correct answer

M1

Use of differentiation to find gradient

B1

5 c x 2

3

dy 5 2 = x dx 2

3

B1 When x = 4,

dy 5 = dx 2

kx 2

4 3 soi

M1

43

A1 5

= 20

SR If 0 scored for first 3 marks, award

9 B1 if

3

4 n correctly evaluated.

4721

Mark Scheme

June 2007

8 (i)

(x + 4)2 – 16 + 15 = (x + 4)2 – 1

B1 M1 A1 3

a=4 15 – their a2 cao in required form

(ii)

( -4, -1 )

B1 ft B1 ft 2

Correct x coordinate Correct y coordinate

M1 A1

Correct method to find roots -5, -3

x2 + 8x + 15 > 0 (x + 5)(x + 3) > 0

M1

Correct method to solve quadratic inequality eg +ve quadratic graph

x < -5, x > -3

A1 4

(x - 3)2 – 9 + y2 - k = 0 (x - 3)2 + y2 = 9 + k Centre (3, 0) 9 + k = 42 k=7

B1

(iii)

9 (i)

x < -5, x > -3 (not wrapped, strict inequalities, no 9 ‘and’) ( x − 3) 2 soi Correct centre

B1

M1 A1 4

Correct value for k (may be embedded) Alternative method using expanded form: Centre (-g, -f) M1 Centre (3, 0) A1 4= f k=7

(ii)

(3 - 3)2 + y2 = 16 y2 = 16 y = 4 Length of AB = =

M1

(iii)

M1

32

A1 ft

y – 0 = m(x + 1) (x – 3)

A1 5

a 4

M1 A1

Correct method to find line length using Pythagoras’ theorem

(−1 − 3) 2 + (0 − 4) 2

Gradient of AB = 1 or

+ g 2 − (− k )

Attempt to substitute x = 3 into original equation or their equation y = 4 (do not allow ± 4)

A1

= 4 2

2

32 or

16 + a 2

cao

B1 ft

or y – 4 = m

M1 A1 3

y=x+1 12

4

Attempts equation of straight line through their A or B with their gradient Correct equation in any form with simplified constants

4721

10 (i)

Mark Scheme

( 3x + 1)( x – 5) = 0 x=

M1 A1 A1 3

−1 or x = 5 3

June 2007

Correct method to find roots Correct brackets or formula Both values correct SR B1 for x = 5 spotted www

B1

Positive quadratic (must be reasonably symmetrical)

B1

y intercept correct

B1 ft 3

both x intercepts correct

M1*

Use of differentiation to find gradient of curve Equating their gradient expression to 4

On curve, when x = 3, y = -20

M1* A1 A1 ft

-20 = (4 x 3) + c c = -32

M1dep A1 6

N.B. dependent on both previous M marks

Alternative method: 3x2 – 14x – 5 = 4x + c

M1

Equate curve and line (or substitute for x)

3x2 – 18x – 5 – c = 0 has one solution

B1

b2 – 4ac = 0

M1

Statement that only one solution for a tangent (may be implied by next line) Use of discriminant = 0

(-18)2 – (4 x 3 x (-5 –c)) = 0

M1

Attempt to use a, b, c from their equation

c = -32

A1

Correct equation

(ii)

(iii)

dy = 6x – 14 dx 6x – 14 = 4 x=3

A1

c = -32 12

5

Finding y co ordinate for their x value

4721

Mark Scheme

6

June 2007


7

4722 1

Mark Scheme

(i)

(ii)

u2 = 12 u3 = 9.6 , u4 = 7.68 (or any exact equivs)

B1 B1√

20 S 20 = 15 (11−−00..88 )

June 2007

2

M1

= 74.1

A1 A1

State u2 = 12 Correct u3 and u4 from their u2 Attempt use of

3

n S n = a (11−−rr ) , with n = 20 or 19

Obtain correct unsimplified expression Obtain 74.1 or better

OR M1

List all 20 terms of GP

A2

Obtain 74.1 5

2

(x + 2x )4 = x 4 + 4 x 3 ( 2x ) + 6 x 2 ( 2x )2 + 4 x( 2x )3 + ( 2x )4

= x 4 + 8 x 2 + 24 +

32 x2

+ 16 (or equiv) x4

M1*

Attempt expansion, using powers of x and 2/x (or

M1* A1dep* A1

the two terms in their bracket), to get at least 4 terms Use binomial coefficients of 1, 4, 6, 4, 1 Obtain two correct, simplified, terms Obtain a further one correct, simplified, term

A1

Obtain a fully correct, simplified, expansion

5

OR M1* M1*

Attempt expansion using all four brackets Obtain expansion containing the correct 5 powers only (could be unsimplified powers eg x3. x-1 )

A1dep* A1 A1

Obtain two correct, simplified, terms Obtain a further one correct, simplified, term Obtain a fully correct, simplified, expansion

5 3

OR

log 3 (2 x +1) = log 5 200

(2 x + 1) log 3 = 200 log 5

M1 M1 A1

Introduce logarithms throughout Drop power on at least one side Obtain correct linear equation (now containing no powers)

2x + 1 =

M1

Attempt solution of linear equation

200 log 5 log 3

x = 146

A1

(2x + 1) = log35200 2x + 1 = 200log35

M1 M1 A1 M1 A1

5

Obtain x = 146, or better Intoduce log3 on right-hand side Drop power of 200 Obtain correct equation Attempt solution of linear equation Obtain x = 146, or better

5 4

(i)

area ≈ 12 × 12 ×

{ 5 + 2(

)

7 + 9 + 11 + 13

≈ 0.25 × 23.766... ≈ 5.94

(ii)

This is an underestimate…… …as the tops of the trapezia are below the curve

}

M1

Attempt y-values for at least 4 of the x = 1, 1.5, 2,

M1 A1

2.5, 3 only Attempt to use correct trapezium rule Obtain 12 × 12 × 5 + 2 7 + 9 + 11 + 13 , or decimal equiv

A1

{

4

(

8

}

Obtain 5.94 or better (answer only is 0/4)

*B1 State underestimate B1dep*B1 Correct statement or sketch 2 6

)

4722 5

Mark Scheme (i) 3(1 − sin 2 θ ) = sin θ + 1 3 − 3 sin 2 θ = sin θ + 1 3 sin 2 θ + sin θ − 2 = 0 (ii) (3 sin θ − 2)(sin θ + 1) = 0

sin θ =

2

3

June 2007 Use cos 2 θ = 1 − sin 2 θ

M1 A1 M1

or -1

2

A1

θ = 42 , 138o, 270o o

A1 A1 A1√

5

Show given equation correctly Attempt to solve quadratic equation in sin θ Both values of sinθ correct Correct answer of 270o Correct answer of 42o For correct non-principal value answer, following their first value of θ in the required range (any extra values for θ in required range is max 4/5) (radians is max 4/5) SR: answer only (or no supporting method) is B1 for 42o, B1√ for 138o, B1 for 270o

7 6

∫x

(a) (i)

[

(ii)

1 4

3

− 4 x = 14 x 4 − 2 x 2 + c

x 4 − 2x 2

]

6

M1

Expand and attempt integration

A1 B1

Obtain ¼ x4– 2x2 (A0 if ∫ or dx still present) + c (mark can be given in (b) if not gained here)

3

M1

1

Use limits correctly in integration attempt (ie F(6) – F(1))

=(324 – 72) – ( ¼ – 2) = 253¾ (b)

∫ 6x

−3

A1

d x = − 3 x −2 + c

2

Obtain 253¾ (answer only is M0A0) 1 x3

= x −3

B1

Use of

M1 A1

Obtain integral of the form kx-2 Obtain correct -3x-2 (+ c) (A0 if ∫ or dx still present, but only penalise once in question)

3

8 7

(a) S70 =

70 2

{(2 × 12) + (70 − 1)d }

M1

Attempt S70

35(24 + 69d) = 12915

A1 M1

d=5

A1

Obtain correct unsimplified expression Equate attempt at S70 to 12915, and attempt to find d Obtain d = 5

OR 70 2

{12 + l} = 12915

M1

Attempt to find d by first equating n/2 (a + l) to

A1 M1 A1

12915 Obtain l = 357 Equate u70 to l Obtain d = 5

B1 B1

Correct statement for second term Correct statement for sum to infinity

or a = 9 − (9 × −a4 )

M1

Attempt to eliminate either a or r

a − 9a − 36 = 0

A1

(a + 3)(a − 12) = 0

M1

Obtain correct equation (no algebraic denominators/brackets) Attempt solution of three term quadratic equation

A1

Obtain at least

l = 357 12 + 69d = 357 d=5 (b) ar = -4 a 1− r −4 r

=9 = 9 − 9r

9r − 9r − 4 = 0 2

(3r − 4)(3r + 1) = 0 r = 43 , r = − 13 Hence r = − 13

4

2

a = -3, a = 12

A1

7

Obtain

r = − 13 (from correct working only)

r = − 13 only (from correct working only)

SR: answer only / T&I is B2 only 11

9

4722 8

Mark Scheme

(i)

1 2

M1

× AB 2 × 0.9 = 16.2 AB 2 = 36 ⇒ AB = 6

(ii)

1 2

June 2007

A1

× 6 × AC × sin 0.9 = 32.4

Use 2 16.2)

M1* M1dep* A1 3

AC = 13.8 cm (iii) BC 2 = 6 2 + 13.8 2 − 2 × 6 × 13.8 × cos 0.9

Hence BC = 11.1 cm

M1 A1√ A1

BD = 6 × 0.9 = 5.4 cm Hence perimeter = 11.1 + 5.4 + (13.8 – 6) = 24.3 cm

B1 M1 A1

6

( 12 )r 2θ = 16.2

Confirm AB = 6 cm (or verify ½ x 62 x 0.9 = Use Δ = 12 bc sin A , or equiv Equate attempt at area to 32.4 Obtain AC = 13.8 cm, or better Attempt use of correct cosine formula in ΔABC Correct unsimplified equation, from their AC Obtain BC = 11.1 cm, or anything that rounds to this State BD = 5.4 cm (seen anywhere in question) Attempt perimeter of region BCD Obtain 24.3 cm, or anything that rounds to this

11 9

(i) (a) (b)

(ii) (a)

f (-1) = – 1 + 6 – 1 – 4 = 0

B1

x = -1 f ( x ) = ( x + 1) x 2 + 5 x − 4

B1 M1 A1 A1

State x = -1 at any point Attempt complete division by (x + 1), or equiv Obtain x2 + 5x + k Obtain completely correct quotient

M1

Attempt use of quadratic formula, or equiv, find

(

x=

− 5 ± 25 + 16 2

x=

1 2

(− 5 ±

41

)

)

log 2 (x + 3) + log 2 x − log 2 (4 x + 2 ) = 1 2

log 2

(

( x + 3 )2 x 4 x+2

)= 1

A1

1

6

Confirm f(-1) = 0, through any method

roots Obtain

1 2

(− 5 ±

41

)

B1 M1

State or imply that 2log (x + 3) = log (x + 3)2 Add or subtract two, or more, of their algebraic logs correctly

A1

Obtain correct equation (or any equivalent, with single term on each side)

( x + 3 )2 x 4 x+2

(x

2

Use log 2 a = 1 ⇒ a = 2 at any point

B1

)

+ 6 x + 9 x = 8x + 4

x + 6x + x − 4 = 0

A1

x > 0, otherwise log 2 x is undefined x = 12 − 5 + 41

B1* B1√dep*

3

(b)

=2 2

(

)

5

2 14

10

Confirm given equation correctly State or imply that log x only defined for x > 0 State x = 12 − 5 + 41 (or x = 0.7) only, following

(

)

their single positive root in (i)(b)


11

4723 1 (i)

(ii)

Mark Scheme

June 2007

Attempt use of product rule M1 A1 2 or equiv Obtain 3x 2 ( x + 1)5 + 5 x 3 ( x + 1) 4 [Or: (following complete expansion and differentiation term by term) B2 allow B1 if one term incorrect] Obtain 8 x 7 + 35 x 6 + 60 x5 + 50 x 4 + 20 x3 + 3x 2 3 4 n Obtain derivative of form kx (3 x + 1) M1 any constants k and n Obtain derivative of form kx3 (3x 4 + 1)

− 12

M1

− 12

Obtain correct 6 x3 (3x 4 + 1) A1 3 or (unsimplified) equiv ____________________________________________________________________________ 2

Identify critical value x = 2 Attempt process for determining both critical values Obtain 13 and 2

B1

Attempt process for solving inequality

M1

Obtain

1 3

M1 A1

0 . 3 Dep on previous 4 marks. 5

2

2

Use parts with u = x , dv = e 2

∫

x

Obtain x e − 2 xe

x

x

(dx )

(

)

x

Final = x − 2 x + 2 e AEF incl brackets Use limits correctly throughout

e

(1)

− 2 ISW

∫

A1

Attempt parts again with u = (− )(2 )x , dv = e x 2

obtaining a result f (x ) + / − g(x )(dx )

*M1

Exact answer only

M1 s.o.i. eg e + (− 2 x + 2) e x Tolerate (their value for x = 1 )

A1 dep*M1 A1

(−0)

6 Allow 0.718 → M1 6

π

3

(k )∫ sin 2 x (dx )

B1

where k = π ,2π or 1 ; limits necessary

Suitable method for integrating sin 2 x

*M1

eg

Volume =

0

∫ + / − 1 + / − cos 2x (dx ) or single integ by parts & connect to ∫ sin x (dx ) or − sin x cos x + ∫ cos x (dx ) or − sin x cos x + ∫ 1 − sin x (dx ) 2

∫ sin x (dx ) = ∫ 1 − cos 2x (dx ) ∫ cos 2x (dx ) = sin 2x 2

1 2

1 2

Volume =

2

A1

Use limits correctly 1 π2 2

2

A1

dep*M1

WWW

Exact answer

A1

6 Beware: wrong working leading to

1 π2 2

6

4

(i)

(1 + 2x )−2

= 1 + (− 2)( 2x ) + = 1− x + 34 x 2 − 21 x 3

(2 + x )−2

=

1 4

()

− 2.−3 x 2 2 2

+

()

− 2.− 3.− 4 x 3 3! 2

(their exp of (1 + ax ) ) mult out −2

x < 2 or − 2 < x < 2 (but not 2

3

1 2

x < 1)

(ii) If (i) is a + bx + cx + dx evaluate b + d

−

3 8

(x ) 3

M1

Clear indication of method of ≥ 3 terms

B1

First two terms, not dependent on M1

A1

For both third and fourth terms

√B1 B1

Correct:

1 1 3 2 1 3 − x+ x − x 4 4 16 8

5

M1 √A1

2 Follow-through from b + d 7

16

4724 5(i)

Mark Scheme

dy = dx

dy dt dx dt

June 2007

M1

− 4 sin 2t − sin t = 8 cos t ≤8 AG 2 (ii) Use cos 2t = 2 cos t + / − 1 or 1 − 2 cos 2 t =

A1

Accept

4 sin 2t WWW sin t

A1 A1

4 with brief explanation eg cos t ≤ 1

M1

If starting with y = 4 x 2 + 1, then

Use correct version cos 2t = 2 cos 2 t − 1

A1

Subst x = cos t , y = 3 + 2 cos 2t

Produce WWW y = 4 x 2 + 1 AG

A1

3 Either substitute a formula for cos 2t M1

B1 B1

Obtain 0=0 or 4cos 2 t + 1 = 4cos 2 t + 1 A1 Or Manip to give formula for cos 2t M1 Obtain corr formula & say it’s correct A1 Any labelling must be correct 2 either x = ±1 or y = 5 must be marked

(iii) U-shaped parabola abve x-axis, sym abt y-axis Portion between (− 1, 5 ) and (1, 5 )

N.B. If (ii) answered or quoted before (i) attempted, allow in part (i) B2 for 6

( )

d 2 dy y = 2y dx dx Using d(uv ) = u dv + v du for the (3 )xy term dy dy d 2 + 3y + 8y x + 3 xy + 4 y 2 = 2 x + 3 x dx dx dx dy Solve for & subst ( x, y ) = (2,3) dx

(i)

(

)

dy dx

M1

= 8 x +B1,B1 if earned.

9

B1 M1 A1 M1

or v.v. Subst now or at normal eqn stage; ( M1 dep on either/both B1 M1 earned)

dy 13 = − dx 30 Grad normal =

A1

30 13

follow-through

√B1

Find equ any line thro (2,3) with any num grad M1 30 x − 13 y − 21 = 0 AEF A1

7

(i) Leading term in quotient = 2 x Suff evidence of division or identity process Quotient = 2 x + 3

B1 M1 A1

Remainder = x

A1

(ii) their quotient +

their remainder x2 + 4

√B1

Implied if grad normal =

30 13

This f.t. mark awarded only if numerical 8 No fractions in final answer

8

Stated or in relevant position in division

x as remainder x +4 x 1 2x + 3 + 2 x +4 4 Accept

2

(iii) Working with their expression in part (ii) their Ax + B integrated as

1 2

Ax 2 + Bx

(

Cx integrated as k ln x 2 + 4 x +4 1 k = 2C their

2

Limits used correctly throughout

14 + 21 ln 13 5

√B1

)

M1

Ignore any integration of

D x +4 2

√A1 M1 A1

5 logs need not be combined. 10

17

4724 8

Mark Scheme (i) Sep variables eg LHS = − ln(6 − h )

1

1

∫ 6 − h (dh) = ∫ 20 (dt )

dt 20 = → M1 dh 6 − h & then t = −20 ln(6 − h ) (+c) → A1+A1

*M1

s.o.i. Or

A1

1 t (+c) 20 Subst t = 0, h = 1 into equation containing ‘c’ Correct value of their c = − (20 ) ln 5 WWW 5 Produce t = 20 ln WWW AG 6−h RHS =

(ii) When h = 2, t = 20 ln (iii) Solve 10 = 20 ln

June 2007

5 4

=4.46(2871)

5 5 = e 0. 5 to 6−h 6−h

A1 dep*M1 A1

or (20 ) ln 5 if on LHS

A1

6 Must see ln 5 − ln(6 − h )

B1

1

Accept 4.5, 4 21

6−h = e −0.5 or suitable 21 -way stage 5 6 − 5 e −0 ,5 or 6 − e1.109

M1

or

h = 2.97(2.9673467…) A1 2 [In (ii),(iii) accept non-decimal (exact) answers but − 1 once.] Accept truncated values in (ii),(iii). (iv) Any indication of (approximately) 6 (m)

B1

1 10

9

(i) Use − 6 i + 8j − 2 k and i + 3j + 2k only Correct method for scalar product

M1 M1

of any two vectors ( − 6 + 24 − 4 = 14 )

Correct method for magnitude

M1

of any vector ( 36 + 64 + 4 = 104 or

1 + 9 + 4 = 14 ) 68 or 68.5 (68.47546); 1.2(0) (1.1951222) rad A1 4 [N.B. 61 (60.562) will probably have been generated by 5i – j -2k and 3i – 8j] M1

− 6 i + 8j − 2 k & 3i + cj + k with some

c= −4

A1

indic of method of attack eg − 6 i + 8j − 2 k =λ(3i + cj + k) c = −4 WW → B2

(iii) Produce 2/3 equations containing t,u (& c)

M1

(ii) Indication that relevant vectors are parallel

Solve the 2 equations not containing ‘c’ M1 t = 2, u = 1 A1 Subst their (t,u) into equation containing c M1 c = −3 A1 Alternative method for final 4 marks Solve two equations, one with ‘c’, for t and u in terms of c, and substitute into third equation (M2) c = −3 (A2)

18

2

eg 3 + t = 2 + 3u ,−8 + 3t = 1 + cu and 2t = 3 + u

5

11


19

4725

1

Mark Scheme

EITHER a=2

b = 2 3, OR

a=2 2

3

M1 A1 M1 A1 M1 M1 A1 A1

Use trig to find an expression for a (or b) Obtain correct answer Attempt to find other value Obtain correct answer a.e.f. (Allow 3.46 ) State 2 equations for a and b 4

b=2 3

(1 = )

2 2 1 ×1 ×2 4

2 3 1 2 n (n + 1) + (n + 1) 4 2 2 1 (n + 1) (n + 2) 4

June 2007

4

Attempt to solve these equations Obtain correct answers a.e.f. SR ± scores A1 only

B1

Show result true for n = 1

M1 M1(indep) A1 A1

Add next term to given sum formula Attempt to factorise and simplify Correct expression obtained convincingly

5

Specific statement of induction conclusion 5

3

3Σr – 3Σr + Σ 1

M1

1 3Σr = n(n + 1)(2n + 1) 2 3 3Σr = n(n + 1) 2

A1

2

2

Σ1 = n

n3

Consider the sum of three separate terms Correct formula stated

A1

Correct formula stated

A1 M1 A1

Correct term seen Attempt to simplify Obtain given answer correctly

6 6

B1

4 (i)

1 2

⎛ 5 –1⎞ ⎟ ⎜ ⎝ –3 1 ⎠

(ii)

B1

2

M1 1 2

⎛ 2 0 ⎞ ⎜ ⎟ ⎝ 23 – 5 ⎠

M1(indep)

4 6

A1ft A1ft

20

Transpose leading diagonal and negate other diagonal or solve sim. eqns. to get 1st column Divide by the determinant or solve 2nd pair to get 2nd column Attempt to use B-1A-1 or find B Attempt at matrix multiplication One element correct, a.e.f, All elements correct, a.e.f. NB ft consistent with their (i)

4725

5

Mark Scheme

(i)

1 r(r + 1)

B1

(ii)

June 2007

1

M1 M1 A1

1 1 – n+1 (iii)

B1ft M1 A1 c.a.o.

S∞ = 1 1

Express terms as differences using (i) Show that terms cancel Obtain correct answer, must be n not any other letter 3 State correct value of sum to infinity Ft their (ii) Use sum to infinity – their (ii)

n+1

6

Show correct process to obtain given result

(i) (a)

α + β + γ = 3, αβ + βγ + γα = 2

B1 B1

3 7

Obtain correct answer a.e.f.

2

State correct values

(b)

α 2 + β 2 + γ 2 = (α + β + γ ) 2 − 2(αβ + βγ + γα ) M1 =9–4=5 A1 ft 3 9 6

– 2 + +2 = 0 3 u u u (ii) (a) 3 2 2u + 6u − 9u + 3 = 0

State or imply the result and use their values 2 Obtain correct answer

M1 2

1 (b)

α

+

1 β

+

1 γ

= –3

A1

Use given substitution to obtain an equation

M1 Obtain correct answer A1ft

21

2 8

Required expression is related to new cubic stated or implied -(their “b” / their “a”)

4725

7

Mark Scheme

(i)

a(a – 12) + 32 (ii) det M = 12 non-singular (iii) EITHER

M1 M1 A1 M1 A1ft B1

June 2007

3 2

Substitute a = 2 in their determinant

M1 OR

A1

Obtain correct answer and state a consistent conclusion 3

M1 A1 A1

det M = 0 so non-unique solutions Attempt to solve and obtain 2 inconsistent equations Deduce that there are no solutions

8 8

(i) Circle, centre (3, 0), y-axis a tangent at origin Straight line, through (1, 0) with +ve slope In 1st quadrant only (ii) Inside circle, below line, above x-axis

B1B1 B1 B1 B1 B1 B2ft

22

Show correct expansion process Show evaluation of a 2 x 2 determinant Obtain correct answer a.e.f.

6 2 8

Substitute a = 4 and attempt to solve Obtain 2 correct inconsistent equations Deduce no solutions Sketch showing correct features N.B. treat 2 diagrams asa MR

Sketch showing correct region SR: B1ft for any 2 correct features

4725

Mark Scheme

9 (i)

⎛ 2 ⎜ ⎜ 0 ⎝

0 ⎞ ⎟ 2 ⎟⎠

(ii) Rotation (centre O), 450, clockwise

June 2007

B1

1

Correct matrix

B1B1B1

3

Sensible alternatives OK, must be a single transformation

B1

1

Matrix multiplication or combination of transformations

M1 A1

2

For at least two correct images For correct diagram

(iii)

⎛0⎞ (iv) ⎜ ⎟ ⎝0⎠

⎛1⎞ ⎜ ⎟ ⎝1⎠

⎛ 1 ⎞ ⎜ ⎟ ⎝ –1⎠

⎛2⎞ ⎜ ⎟ ⎝0⎠

(v) det C = 2 area of square has been doubled

State correct value

B1 B1

2

State correct relation a.e.f.

9 10

(i) 2

M1

Attempt to equate real and imaginary parts of (x + iy)2 and 16+30i

A1A1

Obtain each result

M1

Eliminate to obtain a quadratic in x2 or y2

M1

Solve to obtain

2

x – y = 16 and xy = 15

±( 5 + 3i ) (ii)

x = (±) 5 or y = (±) 3 A1

6

z = 1± 16 + 30i M1* 6 + 3i,

-4 - 3i

A1 *M1dep A1 A1ft

Use quadratic formula or complete the square 5 11

23

Obtain correct answers as complex numbers

Simplify to this stage Use answers from (i) Obtain correct answers

4725

Mark Scheme

24

June 2007


25

4726

Mark Scheme

June 2007

Allow r2 = 2 sin2 3θ a, b ≠ 0 From a + bcos 6θ cao

1

Correct formula with correct r Rewrite as a + bcos 6θ Integrate their expression correctly Get ⅓π

2

(i)

Expand to sin2x cos¼π + cos2x sin¼π Clearly replace cos¼π, sin¼πto A.G.

B1 B1

(ii)

Attempt to expand cos2x Attempt to expand sin2x Get ½√2 ( 1 + 2x – 2x2 – 4x3/3)

M1 Allow 1 – 2x2/2 M1 Allow 2x – 2x3/3 A1 Four correct unsimplified terms in any order; allow bracket; AEEF SR Reasonable attempt at f n(0) for n= 0 to 3 M1 Attempt to replace their values in Maclaurin M1 Get correct answer only A1

(i)

Express as A/(x-1) + (Bx+C)/(x2+9) Equate (x2+9x) to A(x2+9) + (Bx+C)(x-1) Sub. for x or equate coeff.

M1 Allow C=0 here M1√ May imply above line; on their P.F. M1 Must lead to at least 3 coeff.; allow cover-up method for A A1 cao from correct method

3

M1 M1 A1√ A1

Get A=1, B=0,C=9 (ii)

4

Get Aln(x-1) Get C/3 tan-1(x/3)

B1√ On their A B1√ On their C; condone no constant; ignore any B ≠ 0

(i)

Reasonable attempt at product rule Derive or quote diff. of cos –1x Get -x2(1 - x2)-½ + (1 - x2)½ + (1 - x2)-½ Tidy to 2(1 - x2)½

M1 Two terms seen M1 Allow + A1 A1 cao

(ii)

Write down integral from (i) Use limits correctly Tidy to ½π

B1 On any k√(1-x2) M1 In any reasonable integral A1 SR Reasonable sub. B1 Replace for new variable and attempt to integrate (ignore limits) M1 Clearly get ½π A1

26

4726

5

6

Mark Scheme

June 2007

(i)

Attempt at parts on ∫ 1 (ln x)n dx Get x (ln x)n - ∫ n (ln x)n-1 dx Put in limits correctly in line above Clearly get A.G.

M1 Two terms seen A1 M1 A1 ln e =1, ln1 =0 seen or implied

(ii)

Attempt Ι3 to Ι2 as Ι3 = e – 3Ι2 Continue sequence in terms of In Attempt Ι0 or Ι1 Get 6 – 2e

M1 A1 Ι2 = e-2Ι1 and/or Ι1 = e–Ι0 M1 (Ι0 =e-1, Ι1=1) A1 cao

(i)

Area under graph (= ∫ 1/x2 dx, 1 to n+1) < Sum of rectangles (from 1 to n)

B1

Sum (total) seen or implied eg diagram; accept areas (of rectangles)

Area of each rectangle = Width x Height = 1 x 1/x2

B1

Some evidence of area worked out – seen or implied

(ii)

Indication of new set of rectangles Similarly, area under graph from 1 to n > sum of areas of rectangles from 2 to n Clear explanation of A.G.

B1 B1 B1

(iii) Show complete integrations of RHS, using correct, different limits Correct answer, using limits, to one integral Add 1 to their second integral to get complete series Clearly arrive at A.G.

M1 A1

(iv) Get one limit Get both 1 and 2

B1 B1

27

Sum (total) seen or implied Diagram; use of left-shift of previous areas

M1 Reasonable attempt at ∫ x-2 dx A1

Quotable Quotable; limits only required

4726

7

Mark Scheme

(i)

Use correct definition of cosh or sinh x Attempt to mult. their cosh/sinh Correctly mult. out and tidy Clearly arrive at A.G.

B1 Seen anywhere in (i) M1 A1√ A1 Accept ex-y and ey-x

(ii)

Get cosh(x – y) = 1 Get or imply (x – y) = 0 to A.G.

M1 A1

(iii) Use cosh2 x = 9 or sinh2 x = 8 Attempt to solve cosh x = 3 (not –3) or sinh x = ±√8 (allow +√8 or -√8 only) Get at least one x solution correct Get both solutions correct, x and y

8

June 2007

(i)

(ii)

B1 M1 x = ln( 3 + √8) from formulae book or from basic cosh definition A1 A1 x, y = ln(3 ±2√2); AEEF SR Attempt tanh = sinh/cosh B1 Get tanh x = ±√8/3 (+ or -) M1 Get at least one sol. correct A1 Get both solutions correct A1 SR Use exponential definition B1 M1 Get quadratic in ex or e2x A1 Solve for one correct x A1 Get both solutions, x and y

x2 = 0.1890 x3 = 0.2087 x4 = 0.2050 x5 = 0.2057 x6 = 0.2055 x7 (= x8) = 0.2056 (to x7 minimum) α = 0.2056

B1 B1√ From their x1 (or any other correct) B1√ Get at least two others correct, all to a minimum of 4 d.p.

Attempt to diff. f(x) Use α to show f ′(α) ≠ 0

M1 k/(2+x)3 A1√ Clearly seen, or explain k/(2+x)3 ≠ 0 as k ≠ 0; allow ± 0.1864 M1 SR Translate y=1/x2 State/show y=1/x2 has no TP A1

B1

(iii) δ3 = -0.0037 (allow –0.004)

cao; answer may be retrieved despite some errors

B1√ Allow ±, from their x4 and x3

(iv) Develop from δ10= f ′(α) δ9 etc. to get δi or quote δ10= δ3 f ′(α)7 Use their δi and f ′(α) Get 0.000000028

28

M1 Or any δi eg use δ9 = x10 – x9 M1 A1 Or answer that rounds to ± 0.00000003

4726

9

Mark Scheme

(i)

(ii)

Quote x = a Attempt to divide out

June 2007

Get y = x – a

B1 M1 Allow M1 for y=x here; allow A1 (x-a) + k/(x-a) seen or implied A1 Must be equations

Attempt at quad. in x (=0) Use b2 – 4ac ≥ 0 for real x Get y2 + 4a2 ≥ 0 State/show their quad. is always >0

M1 M1 Allow > A1 B1 Allow ≥

(iii)

B1√ Two asymptotes from (i) (need not be labelled) B1

Both crossing points

B1√ Approaches – correct shape SR Attempt diff. by quotient/product rule M1 x for d y /d x = 0 Get quadratic in A1 and note b2 – 4ac < 0 Consider horizontal asymptotes B1 Fully justify answer B1

29

4726

Mark Scheme

30

June 2007


31

4727 1 (i) z z* = r eiθ . r e−iθ = r 2 = z

Mark Scheme 2

(ii) Circle Centre 0 ( + 0i ) OR (0, 0) OR O, radius 3

B1

1

For verifying result AG

2

For stating circle For stating correct centre and radius

B1 B1

June 2007

3

2 EITHER: (r =) [3 + t , 1 + 4t , − 2 + 2t ] 8(3 + t ) − 7(1 + 4t ) + 10(−2 + 2t ) = 7 ⇒ (0 t ) + (−3) = 7 ⇒ contradiction

M1 M1 A1 A1

l is parallel to Π, no intersection OR: [1, 4, 2] . [8, − 7, 10] = 0 ⇒ l is parallel to Π (3, 1, − 2) into Π

B1 5 M1 A1 M1

For parametric form of l seen or implied For substituting into plane equation For obtaining a contradiction For conclusion from correct working For finding scalar product of direction vectors For correct conclusion For substituting point into plane equation

⇒ 24 − 7 − 20 ≠ 7 l is parallel to Π, no intersection

A1 B1

For obtaining a contradiction For conclusion from correct working

M1 A1 M1 A1 B1

For eliminating one variable For eliminating another variable For obtaining a contradiction For conclusion from correct working

x − 3 y −1 z + 2 = = and 8 x − 7 y + 10 z = 7 1 4 2 eg y − 2 z = 3 , 2 y − 2 = 4 z + 8

OR:Solve

eg 4 z + 4 = 4 z + 8 l is parallel to Π, no intersection

5

3 Aux. equation m2 − 6m + 8 (= 0) m = 2, 4

M1 A1

For auxiliary equation seen For correct roots

CF ( y =) Ae 2 x + Be 4 x

A1√

For correct CF. f.t. from their m

PI ( y =) C e3 x 9C − 18C + 8C = 1 ⇒ C = −1

M1 A1

For stating and substituting PI of correct form For correct value of C

GS y = Ae2 x + Be4 x − e3 x

B1√ 6

For GS. f.t. from their CF + PI with 2 arbitrary constants in CF and none in PI

6

32

4727

Mark Scheme

4 (i) q( st ) = qp = s (qs )t = tt = s

B1 B1

(ii) METHOD 1 Closed: see table Identity = r

Inverses: p −1 = s, q −1 = t , ( r −1 = r ), s

−1

= p, t

−1

=q

2

For obtaining s For obtaining s

B1 B1

For stating closure with reason For stating identity r

M1

For checking for inverses For stating inverses OR For giving sufficient explanation to justify each element has an inverse eg r occurs once in each row and/or column

A1

METHOD 2 Identity = r

June 2007

4

B1 M1

For stating identity r For attempting to establish a generator ≠ r

eg p 2 = t , p3 = q , p 4 = s

A1

For showing powers of p (OR q, s or t) are different elements of the set

⇒ p5 = r , so p is a generator

A1

For concluding p5 (OR q5 , s 5 or t 5 ) = r

(iii) e, d , d 2 , d 3 , d 4

B2

2

For stating all elements AEF eg d −1 , d −2 , dd

8

5 (i) (cos 6θ =) Re ( c + i s )

6

(cos 6θ =) c 6 − 15c 4 s 2 + 15c 2 s 4 − s 6 (cos 6θ =)

(

)

(

c 6 − 15c 4 1 − c 2 + 15c 2 1 − c

) − (1 − c )

2 2

2 3

(cos 6θ =) 32c6 − 48c 4 + 18c 2 − 1

(ii) 64 x6 − 96 x 4 + 36 x 2 − 3 = 0 ⇒ cos 6θ = 1 π, 5 π, 7 π etc. ⇒ (θ =) 18 18 18

cos 6θ =

1 2

has multiple roots

largest x requires smallest θ 1 π ⇒ largest positive root is cos 18

For expanding (real part of) ( c + i s )

A1

at least 4 terms and 1 evaluated binomial coefficient needed For correct expansion

M1

For using s 2 = 1 − c 2

A1 1 2

6

M1

4

For correct result AG

M1

For obtaining a numerical value of cos 6θ

A1

For any correct solution of cos 6θ =

M1

For stating or implying at least 2 values of θ

A1

4

1 2

1 π AEF as the largest positive root For identifying cos 18

from a list of 3 positive roots OR from general solution OR from consideration of the cosine function 8

33

4727

Mark Scheme

6 (i) n = l1 × l2

B1

n = [2, − 1, 1] × [4, 3, 2]

M1* A1 M1 (*dep) A1 5 M1

n = k [−1, 0, 2]

[3, 4, − 1] . k [−1, 0, 2] = −5k r . [−1, 0, 2] = −5

(ii) [5, 1, 1] . k [−1, 0, 2] = −3k r . [−1, 0, 2] = −3 (iii) d =

−5 + 3

[2, − 3, 2] . [−1, 0, 2]

OR d =

5

OR d from (5, 1, 1) to Π1 = OR d from (3, 4, − 1) to Π 2 =

For substituting a point of l1 into r .n For obtaining correct p. AEF in this form

A1√ 2 M1

For using a distance formula from their equations Allow omission of | |

5 5(−1) + 1(0) + 1(2) + 5 5 3(−1) + 4(0) − 1(2) + 3 5

OR [5 − t , 1, 1 + 2t ] . [−1, 0, 2] = −5 ⇒ t = 2

For stating or implying in (i) or (ii) that n is perpendicular to l1 and l2 For finding vector product of direction vectors For correct vector (any k)

For using same n and substituting a point of l2 For obtaining correct p. AEF in this form f.t. on incorrect n

OR For finding intersection of n1 and Π 2 or n 2 and Π1

2 5 − 52

OR [3 − t , 4, − 1 + 2t ] . [−1, 0, 2] = −3 ⇒ t =

2 5 = 0.894427... 5 5 (iv) d is the shortest OR perpendicular distance between l1 and l2 d=

June 2007

=

A1√ 2

For correct distance AEF f.t. on incorrect n

B1

1

For correct statement

1

For correct justification AG

10 (eiφ + e−iφ ) +1 (2) ≡ z 2 − (2 cos φ) z + 1

7 (i ) ( z − eiφ )( z − e−iφ ) ≡ z 2 − (2) z

2 k πi

B1

B1

(ii) z = e 7

for k = 0, 1, 2, 3, 4, 5, 6 OR 0, ± 1, ± 2, ± 3

For general form OR any one non-real root For other roots specified

B1

(k=0 may be seen in any form, eg 1, e0 , e 2πi ) For answers in form cos θ + i sin θ allow maximum B1 B0

1 For any 7 points equally spaced round unit circle (circumference need not be shown)

B1 B1

(

)

2 πi

(z − e = (z − e

2 πi 7

)( z − e

)( z − e

(z − e

−2 πi 7

−2 πi 7

6 πi 7

)( z − e

) × (z − e

)( z − e

−6 πi 7

For 1 point on + ve real axis, and other points in correct quadrants

4 πi

(iii ) z 7 − 1 = ( z − 1)( z − e 7 )( z − e 7 ) 6 πi 7

4

−4 πi 7

4 πi 7

)( z − e

)( z − e

−6 πi 7

−4 πi 7

M1

For using linear factors from (ii), seen or implied

M1

For identifying at least one pair of complex conjugate factors

B1

For linear factor seen

A1

For any one quadratic factor seen

)

)

)× × ( z − 1)

2

= ( z − (2 cos 72 π) z + 1) × ( z 2 − (2 cos 74 π) z + 1) × ( z 2 − (2 cos 76 π) z + 1) ×

A1

× ( z − 1) 10

34

5

For the other 2 quadratic factors and expression written as product of 4 factors

4727

Mark Scheme

8 (i) Integrating factor e ∫

=e

tan x (dx )

− ln cos x

= (cos x)

−1

OR sec x

(

)

For correct IF

M1

For integrating to ln form

A1

For correct simplified IF AEF

M1 M1

y (cos x )−1 = 12 x + 14 sin 2 x (+c)

A1

For correct integration both sides AEF

y=

A1

B1√

y (cos x) −1 = ∫ 12 (1 + cos 2 x) (dx)

(ii)

B1

d ( y.their IF ) = cos3 x . their IF dx For integrating LHS For attempting to use cos 2x formula OR parts

⇒

d y (cos x)−1 = cos2 x dx

June 2007

( 12 x + 14 sin 2 x + c ) cos x 2 = ( 12 π + c ) . − 1 ⇒ c = −2 − 12 π y=

For

for ∫ cos 2 x dx 8

For correct general solution AEF

2

For substituting (π, 2) into their GS and solve for c For correct solution AEF

M1

( 12 x + 14 sin 2 x − 2 − 12 π ) cos x

A1 10

9 (i) 3n × 3m = 3n + m , n + m ∈ Z

(3 p × 3q ) × 3r = (3 p+q ) × 3r = 3 p+q+r = 3 p × ( 3q + r ) = 3 p × ( 3q × 3r ) ⇒ associativity

Identity is 30 −n

Inverse is 3

3n × 3m = 3n + m = 3m + n = 3m × 3n ⇒ commutativity

(

(ii) (a) 32n × 32 m = 32n + 2 m = 32( n + m)

)

Identity, inverse OK (b) For 3− n , −n ∉ subset 2

2

2

≠ 3r ⇒ not a subgroup 2

OR: 3n × 3m = 3n r2

2

A1

For showing closure For considering 3 distinct elements, seen bracketed 2+1 or 1+2 For correct justification of associativity

B1

For stating identity. Allow 1

B1

For stating inverse

M1

B1

6

For showing commutativity

B1*

For showing closure

B1 (*dep) 2 M1

For stating other two properties satisfied and hence a subgroup For considering inverse

A1

For justification of not being a subgroup

2

3− n must be seen here or in (i)

(c) EITHER: eg 31 × 32 = 35

2

B1

2

2

+ m2

≠ 3 eg 1 + 2 = 5 ⇒ not a subgroup

M1 A1

2

For attempting to find a specific counter-example of closure For a correct counter-example and statement that it is not a subgroup

M1

For considering closure in general

A1

For explaining why n 2 + m 2 ≠ r 2 in general and statement that it is not a subgroup

12

35

4727

Mark Scheme

36

June 2007


37

4728

Mark Scheme

June 2007

1(i)

X=5 Y = 12

B1 B1

(ii)

R2 = 52 + 122 Magnitude is 13 N tan θ = 12/5 Angle is 67.4o

[2] M1 A1 M1 A1 [4]

X=-5 B0. Both may be seen/implied in (ii) No evidence for which value is X or Y available from (ii) award B1 for the pair of values 5 and 12 irrespective of order For using R2 = X2 + Y2 Allow 13 from X=-5 For using correct angle in a trig expression SR: p=14.9 and Q=11.4 giving R=13+/-0.1 B2, Angle = 67.5+/-0.5 B2

2(i)

250 + ½ (290 – 250)

M1

Use of the ratio 12:12 (may be implied), or v = u+at

t = 270

A1 [2] M1 M1

(ii) ½ x40x12+210x12+½x20x12– ½x20x12 or ½ x40x12+210x12 or ½ x(210+250)x12etc Displacement is 2760m (iii)

3(i) (ii)

(iii)

4(i)

(iia)

(iib)

appropriate structure, ie triangle + rectangle + triangle + |triangle|, triangle + rectangle + 2triangle, etc Distance is 3000m

R + Tsin72o = 50g T = 50g/sin72o T = 515 T = mg m = 52.6

(AG)

X = Tcos72o

A1 [3] M1

All terms positive

A1 [2]

Treat candidate doing (ii) in (iii) and (iii) in (ii) as a mis-read.

M1 A1 [2] M1 A1 B1 B1 [4] B1

An equation with R, T and 50 in linear combination. R + 0.951T = 50g

X = 159

B1 [2]

In Q4 right to left may be used as the positive sense throughout. 0.18 x 2 – 3m = 0 m = 0.12

M1

Momentum after = -0.18 x 1.5 + 1.5m 0.18 x 2 – 3m = -0.18 x 1.5 + 1.5m m = 0.14 0.18 x 2 – 3m = (0.18 + m)1.5 m = 0.02 0.18 x 2 – 3m= - (0.18 + m)1.5 m = 0.42

The idea that area represents displacement Correct structure, ie triangle1 + rectangle2 + triangle3 |triangle4| with triangle3 = |triangle4|, triangle1 + rectangle2, trapezium1&2, etc

A1 A1 [3] B1 M1 A1 [3] B1ft B1 B1ft B1 [4]

Using R = 0 (may be implied) and Tsin72o = 50(g) Or better Accept 52.5 Implied by correct answer Or better For using Momentum ‘before’ is zero

3 marks possible if g included consistently For using conservation of momentum 3 marks possible if g included consistently ft wrong momentum ‘before’

0 marks if g included

38

4728

5(i)

Mark Scheme

2

8.4 – 2gsmax = 0 Height is 3.6m

(AG)

(ii) u = 5.6 (iii)

EITHER (time when at same height) s+/-2 = 8.4t – ½ gt2 and (s+/-2) = 5.6t – ½ gt2 t = 5/7 (0.714) vP =8.4 -0.714g and vQ =5.6-0.714g

M1 A1 A1 [3] M1 A1 [2] M1 A1 A1 M1 A1 A1 [6]

vP = 1.4 and vQ = -1.4 OR (time when at same speed in opposite directions) v = 8.4 -gt and -v =5.6-gt v = 1.4 {or t = 5/7 (0.714)}

June 2007

Using v2 = u2 +/– 2gs with v = 0 or u = 0

Using u2 =+/- 2g(ans(i) – 2) Using s = ut + ½ at2 for P and for Q, a = +/-g, expressions for s terms must differ Or 8.4t (– ½ gt2 )=5.6t (– ½ gt2 )+/- 2 Correct sign for g, cv(5.6), +/-2 in only one equation cao Using v = u +at for P and for Q, a = +/-g, cv(t) Correct sign for g, cv(5.6), candidates answer for t (including sign) cao

M1 A1 A1

Using v = u+at for P and for Q, a = +/-g Correct sign for g, cv(5.6) Only one correct answer is needed

(with v = 1.4) 1.42 = 8.42 - 2gsP and (-1.4)2 = 5.62 - 2gsQ

M1

Using v2 = u2 + 2as for P and for Q, a = +/-g, cv(v)

A1

sP = 3.5 and sQ = 1.5 {(with t=5/7)

A1

Correct sign for g, cv(5.6), candidate's answer for v (including - for Q) cao

M1

Using s = ut + ½ at2 for P and for Q, a = +/-g, cv(t)

s = 8.4x0.714 – ½ gx0.714 and s = 5.6x0.714 – ½ gx0.7142

A1

sP = 3.5 and sQ = 1.5

A1

Correct sign for g, cv(5.6), candidate's answer for t (including sign of t if negative) cao}

M1

Using v = u+at t for P and for Q, a = +/-g

A1

Both values correct mid-interval t (6/7+4/7)/2 = 0.714 {Or semi-interval = 6/7-4/7)/2=1/7} cao s = ut + ½ at2 for P and for Q, correct sign for g, cv(5.6) and cv(t) {s = vt - ½ at2 for P and s = ut + ½ at2 for Q}

2

OR (motion related to greatest height and verification) 0 = 8.4 -gt and 0 =5.6-gt t = 6/7 and t = 4/7 vP =8.4 -0.714g and vQ =5.6-0.714g {0 = vP - g/7 and vQ = 0 +g/7} vP = 1.4 and vQ = -1.4 sP = 8.4x0.714 – ½ gx0.7142 and sQ = 5.6x0.714 – ½ gx0.7142 { sP = 0/7 – ½(- g)x(1/7)2 and sQ = 0/7 + ½ gx(1/7)2} sP = 3.5 sQ = 1.5 { sP = 0.1 sQ = 0.1}

A1 M1 A1 A1

cao continued

39

4728

5(iii) cont

Mark Scheme

OR (without finding exactly where or when) vP2 = 8.42 -2g(s+/-2) and vQ2 = 5.62 -2g[(s+/-2)] = for all values of s so that the speeds are always the same at the same heights. vP2

vQ2

0 = 8.4 -gt and 0 =5.6-gt t P = 6/7 and tQ = 4/7 means there is a time interval when Q has started to descend but P is still rising, and there will be a position where they have the same height but are moving in opposite directions. 6(i)

(ii)

(iii)

M1

v = 0.004t3 – 0.12t2 + 1.2t v(10) = 4 – 12 + 12 = 4ms-1

(AG)

v = 0.8t – 0.04t2 (+ C) 8-4+C= 4 v = 0.8x20 – 0.04x202 (+ C) v(20) = 16 – 16 = 0 (AG) S = 0.4t2 – 0.04t3/3

(+K)

s(10) = 10 – 40 + 60 = 30 40 – 40/3 + K = 30 Î K = 10/3 S(20) = 160 – 320/3 + 10/3 = 56.7m OR s(10) = 10 – 40 + 60 = 30

A1

Using v2 = u2 + 2as for P and for Q, a = +/-g, cv(5.6), different expressions for s. Correct sign for g, cv(5.6), (s+/-2) used only once cao. Verbal explanation essential Using v = u+at t for P and for Q, a = +/-g Correct sign for g, correct choice for velocity of zero, cv(5.6)

A1 M1 A1 cao. Verbal explanation essential

A1 M1 A1 A1 [3] M1 A1 M1* M1 DA1 [5] M1 A1 B1 M1 A1

S(20) - S(10) = 26.6, 26.7

B1 [6] B1 M1 A1 M1 A1

displacement is 56.7m

B1

S = 0.4t2 – 0.04t3/3

June 2007

For differentiating s Condone the inclusion of +c Correct formula for v (no +c) and t=10 stated sufficient For integrating a Only for using v(10) = 4 to find C Dependant on M1* For integrating v Accept 0.4t2 – 0.013t3 (+ ct +K, must be linear) For using S(10) = 30 to find K Not if S includes ct term Accept 56.6 to 56.7, Adding 30 subsequently is not isw, hence B0 For integrating v Accept 0.4t2 – 0.013t3 (+ ct +K, must be linear) Using limits of 10 and 20 (limits 0, 10 M0A0B0) For 53.3 - 26.7 or better (Note S(10) = 26.7 is fortuitously correct M0A0B0) Accept 56.6 to 56.7

40

4728

7(i)

Mark Scheme R = 1.5gcos21o Frictional force is 10.98N (AG)

(ii)

(iii)

(iva)

B1 M1 A1 [3] M1

T + 1.5gsin21o – 10.98 = 1.5a 1.2g – T = 1.2a

A2 A2 [5]

T – 1.5a = 5.71 and 1.2a + T = 11.76 a = 2.24 (AG)

M1

v2 = 2 x 2.24 x 2 Speed of the block is 2.99ms-1

(ivb) a = -3.81 v2 = 2.992 + 2 x (-3.81) x 0.8 Speed of the block is 1.69ms-1

A1 [2] M1 A1 [2] M1 A1 M1 A1 [4]

June 2007

For using F = μ R Note 1.2gcos21=10.98 fortuitously, B0M0A0 For obtaining an N2L equation relating to the block in which F, T, m and a are in linear combination or For obtaining an N2L equation relating to the object in which T, m and a are in linear combination -A1 for each error to zero -A1 for each error to zero Error is a wrong/omitted term, failure to substitute a numerical value for a letter (excluding g), excess terms. Minimise error count. For solving the simultaneous equations in T and a for a. Evidence of solving needed For using v2 = 2as with cv (a) or 2.24 Accept 3 For using T = 0 to find a For using v2 = u2 + 2as with cv(2.99) and s = 2.8 - 2 and any value for a Accept art 1.7 from correct work

41

4728

Mark Scheme

42

June 2007


43

4729

Mark Scheme

June 2007

40 cos35°

B1

WD = 40cos35° x 100 3280 J

M1 A1 3

ignore units

0 = 12sin27°t – 4.9t2 any correct. t = 1.11 …..method for total time R = 12cos27° x t 11.9

M1

or R = u2sin2θ/g (B2)

A1 M1 A1 4

correct formula only 122 x sin54° / 9.8 sub in values 11.9

WD = ½x250x1502 –½x250x1002 1 560 000 450 000 = 1 560 000/t 3.47 F = 450 000/120 3750 3750 = 250a 15 ms-2

M1 A1 M1 A1 4 M1 A1 M1 A1 4

x = 7t y = 21t – 4.9t2 y = 21.x/7 – 4.9 x2/49 y = 3x – x2 /10 –25 = 3x – x2 /10 (must be -25) solving quadratic 36.8 m

B1 M1 A1 M1 A1 5 M1 M1 A1 3

5(i)

½ . 70 .42

M1

(ii)

560 J 70 x 9.8 x 6

A1 2 M1

4120 60d 8000 = 560 + 4120 + 60d

A1 2 B1 M1 A1 A1 4

1

2

3 (i)

(ii)

4 (i)

(ii)

(iii)

55.4 m

44

3

4

1 562 500

8

or – g/2

AG or method for total time (5.26) or 7 x total time 8

4116 4 terms their KE and PE 8

4729

Mark Scheme

6 (i)

5cos30° = 0.3x9.8 + Scos60°

(ii)

2.78 N r = 0.4sin30° = 0.2

M1 A1 A1 3 B1

res. vertically (3 parts with comps)

5sin30° + Ssin60° =0.3 x 0.2 x ω 9.04 rads-1 v = 0.2 x 9.04 KE = ½ x 0.3 x (0.2 x 9.04)2 0.491 J or 0.49

M1 A1 3 M1 M1 A1 3

res. horizontally (3 parts with comps)

1.8 = –0.3 + 3m

M1

m = 0.7 e = 4/6 2/3 ± 3f 1/3 f ( : 1 ) I = 3f x 0.7 – – 3 x 0.7 I = 2.1 (f + 1) 0.3 + 6.3/4 = 0.3a + 0.7b 3a + 7b = 18.75 2/3 = (a – b)/ 5/4 3a – 3b = 5/2 solve a = 2.5 b = 1.6

A1 M1 A1 B1 B1 M1 A1 A1 M1 A1 M1 A1 M1 A1 A1

com of hemisphere 0.3 from O com of cylinder h/2 from O 0.6x45 = 40x0.5 + (0.8+h/2) x 5 or 45(h+0.2) = 5h/2 +40(h+0.3) 27 = 20 + (0.8+h/2) x 5 h = 1.2 1.2 T 0.8 F 0.8F = 1.2T F = 3T/2 F + Tcos30° 45sin30° must be involved in res. resolving parallel to the slope F + Tcos30° = 45sin30° aef T = 9.51 F = 14.3

B1 B1 M1 A1 M1 A1 6 B1 B1 M1 A1 4 B1 B1 M1 A1 A1 A1 6

T + Fcos30° = Rsin30° Rcos30° + Fsin30° = 45 tan30°=(T+Fcos30°)/(45-Fsin30°)

B1 B1 M1

2

(iii)

7 (i) (ii) (iii) (iv)

(v)

8 (i)

(ii)

(iii)

or (iii)

June 2007

45

2 2

may be on diagram

or previous v via mv2/r their ω2 x 0.006

9

AG accept 2/6 for M1 accept 0.67

2 ok for only one minus sign for M1

*

aef 2 marks only for -2.1(f + 1) can be – 0.7b aef allow e=3/4 or their e for M1 aef * means dependent.

7

(2.46) allow ± (59/24) (1.625) allow ± (13/8)

3 *

16

or 0.5 from base or 40x0.3 – 5xh/2 = 45 x 0.2 or 5(0.2 + h/2) = 40x0.1 solving AG

aef or 45 x 0.8 sin30° T x (1.2 + 0.8cos30°) mom. about point of contact 45.0.8sin30°=T(1.2+0.8cos30°) 16 res. horizontally res. vertically eliminating R

4729

Mark Scheme

46

June 2007


47

4730

1

Mark Scheme [ ω = 2 π /6.1 = 1.03]

(i)

M1 M1 A1 M1

Speed is 3.09ms-1 (ii)

2

2.52 = 1.032(32 – x2) or x = 3sin(1.03x0.60996..) Distance is 1.76m

A1ft

[Magnitudes 0.6, 0.057 x 7, 0.057 x 10]

M1

For magnitudes of 2 sides correctly marked For magnitudes of all 3 sides correctly marked

A1 A1 M1

A1

June 2007 For using T = 2 π / ω For using vmax = a ω 3

3 For triangle with magnitudes shown For attempting to find angle ( α ) opposite to the side of magnitude 0.057 x 7 For correct use of the cosine rule or equivalent

M1 0.3992 = 0.572 + 0.62 – 2 x 0.57 x 0.6cos α Angle is 140o 2

A1ft A1

7

ALTERNATIVE METHOD

0.6sin α = 0.057 x 7sin β or 0.057x10sin α = 0.057x 7sin γ

For using Ι= Δ mv perpendicular to the initial direction of motion or perpendicular to the impulse

A1

0.399 = (0.57 – 0.6cos α ) + (0.6sin α ) or 0.3992 = (0.6 – 0.57cos α )2 + (0.057sin α )2 Angle is 140o 2

– 39.8)o

A1 M1

2

(180

For using Ι= Δ mv parallel to the initial direction of motion or parallel to the impulse

M1 -0.6cos α = 0.057 x 7cos β - 0.057 x 10 or 0.6 = 0.057x10cos α +0.057x7cos γ

For using v2 = ω 2(A2 – x2) or for using v = A ω cos ω t and x = Asin ω t ft incorrect ω

2

48

For eliminating β *or γ

M1 A1ft A1

7

(180 – 39.8)o

4730

3

Mark Scheme

[0.2v dv/dx = -0.4v2]

(i)

M1

(1/v) dv/dx = -2 (ii) [ (1 / v)dv = − 2dx ]

A1 M1

ln v = -2x (+C) [ln v = -2x + ln u] v = ue-2x (iii) [ e 2 x dx = udt ]

A1 M1 A1 M1

∫

∫

∫

June 2007

∫

2x

e /2 = ut (+C) [e2x/2 = ut + ½ ] u = 6.70

A1 M1 A1

2

4

4

For using Newton’s second law with a = v dv/dx AG For separating variables and attempting to integrate For using v(0) = u AG For using v = dx/dt and separating variables For using x(0) = 0 Accept (e4 – 1)/8

ALTERNATIVE METHOD FOR PART (iii) M1

[ ∫ 12 dv = − 2 ∫ dt Î-1/v = -2t + A, and

For using a = dv/dt, separating variables, attempting to integrate and using v(0) = u For substituting v = ue-2x

v

A = -1/u]

4

-e2x/u = -2t – 1/u u = 6.70

M1 A1 A1

y= 15sin α ( =12) [4(15cos α ) – 3 x 12 = 4a + 3b]

B1 M1

Equation complete with not more than one error 4a + 3b = 0

A1 A1 M1

0.5(15cos α + 12) = b - a [a = -4.5, b = 6] [Speed =

2

Direction tan-1(12/(-4.50)] Speed of A is 12.8ms-1 and direction is 111o anticlockwise from ‘i’ direction

A1

Speed of B is 6ms-1 to the right

A1

49

Accept (e4 – 1)/8

For using principle of conservation of momentum in the direction of l.o.c. For using NEL in the direction of l.o.c.

A1 M1 M1

(−4.5) + 12 , 2

4

For solving for a and b For correct method for speed or direction of A

10

Direction may be stated in any form , including θ = 69o with θ clearly and appropriately indicated Depends on first three M marks

4730

5

Mark Scheme

June 2007

(i)

M1

For taking moments of forces on BC about B

80 x 0.7cos60o = 1.4T Tension is 20N [X = 20cos30o] Horizontal component is 17.3N [Y = 80 – 20sin30o] Vertical component is 70N (ii)

A1 A1 M1 A1ft M1 A1ft 7 M1

17.3 x 1.4sin α = (80 x 0.7 + 70 x1.4)cos α or 80x0.7cos α + 80(1.4cos α + 0.7cos60o) = 20cos60o(1.4cos α +1.4cos60o) + 20sin60o(1.4sin α +14sin60o) [tan α = ( ½ 80 + 70)/17.3= 11/ 3 ]

A1ft

M1

α = 81.1o

A1

For resolving forces horizontally ft X = Tcos30o For resolving forces vertically ft Y = 80 – Tsin30o For taking moments of forces on AB, or on ABC, about A

For obtaining a numerical expression for tan α 4

ALTERNATIVE METHOD FOR PART (i) M1 Hx1.4sin60o + Vx1.4cos60o = 80x0.7cos60o

A1 M1

Tension is 20N Horizontal component is 17.3N [Y = 80 – V] Vertical component is 70N

A1 B1ft M1 A1ft 7

50

For taking moments of forces on BC about B Where H and V are components of T For using H = V 3 and solving simultaneous equations ft value of H used to find T For resolving forces vertically ft value of V used to find T

4730

6

Mark Scheme

June 2007 For using T = λ x/L

(i) [T = 2058x/5.25] 2058x/5.25 = 80 x 9.8 (x = 2) OP = 7.25m (ii) Initial PE = (80 + 80)g(5) (= 7840) or (80 + 80)gX used in energy equation Initial KE = ½ (80 + 80)3.52 (= 980) [Initial EE = 2058x22/(2x5.25) ( = 784), ( = 9604), or Final EE = 2058x72/(2x5.25) 2058(X + 2)2/(2x5.25)] [Initial energy = 7840 + 980 + 784, final energy = 9604 or 1568X + 980 + 784 = 196(X2 + 4X + 4) Î 196X2 – 784X – 980 = 0]

M1 A1 A1 B1

Initial energy = final energy or X = 5 ÎP&Q just reach the net (iii) [PE gain = 80g(7.25 + 5)]

A1

PE gain = 9604 PE gain = EE at net level Î P just reaches O (iv) For any one of ‘light rope’, ‘no air resistance’, ‘no energy lost in rope’ For any other of the above

A1 A1 B1

3

B1

2

FIRST ALTERNATIVE METHOD FOR PART (ii) [160g – 2058x/5.25 = 160v dv/dx]

B1 M1 M1

5

M1

C = -8.575 [v(7) 2]/2 = 68.6 – 60.025 – 8.575 = 0 ÎP&Q just reach the net SECOND ALTERNATIVE METHOD FOR PART (ii) &x& = g − 2 .45 x (= -2.45(x – 4)) (A = 3)

distance travelled downwards by P and Q = 5 ÎP&Q just reach the net

51

A1 M1 A1 A1

For attempting to verify compatibility with the principle of conservation of energy, or using the principle and solving for X AG For finding PE gain from net level to O AG

For using Newton’s second law with a = v dv/dx, separating the variables and attempting to integrate Any correct form For using v(2) = 3.5 5

B1 M1

AG

For using n2 = 2.45 and v2 = n2(A2 – (x – 4)2)

A1 M1

A1

AG From 5.25 + 2

For using EE = λ x2/2L

M1

v2/2 = gx – 1.225x2 (+ C)

3.52 = 2.45(A2 – (-2)2) [(4 – 2) + 3]

3

5

For using ‘distance travelled downwards by P and Q = distance to new equilibrium position + A AG

4730

7

Mark Scheme

June 2007

(i) [a = 0.72/0.4] For not more than one error in T – 0.8gcos60o = 0.8x0.72/0.4 Above equation complete and correct Tension is 4.9N (ii)

M1 A1

½ 0.8v2 = ½ 0.8(0.7)2 + 0.8g0.4 – 0.8g0.4 cos60o (2.1 – 0)/7 = 2u Q’s initial speed is 0.15ms-1 (iii)

A1 M1 A1 M1

(m)0.4 θ&& = -(m)g sin θ

A1

[0.4 θ&& ≈ -g θ ] [ ½ m0.152 = mg0.4(1 – cos θ max) Î θ max = 4.34o (0.0758rad)]

M1 M1

θ max small justifies 0.4 θ&& ≈ -g θ , and this implies

A1

SHM (iv)

M1

A1 A1 M1

For using a = v2/r

4 For using the principle of conservation of energy (v = 2.1) 4

For using NEL AG For using Newton’s second law transversely *Allow m = 0.8 (or any other numerical value) For using sin θ ≈ θ For using the principle of conservation of energy to find

θ max

[T = 2 π / 24 . 5 = 1.269..] [ 24 . 5 t = π ]

Time interval is 0.635s

A1ft

52

5

2

For using T = 2 π /n or for solving either sin nt = 0 (non-zero t) (considering displacement) or cos nt = -1 (considering velocity) From t = ½ T


53

4731 1 (i)

Mark Scheme Using θ = ω0t + 12 αt 2 , 56 = 0 + 12 α × 82

M1

α = 1.75 rad s −2

A1

June 2007

2 (ii)

M1

Using ω12 = ω0 2 + 2αθ , 36 2 = 20 2 + 2 × 1.75θ θ = 256 rad

ft is 448 ÷ α

A1 ft 2

2

[

a

⌠ ⌡0

Volume is ⎮ π (4a 2 − x 2 ) dx = π 4a 2 x − 13 x 3 = 11 πa 3

]

a 0

3

π may be omitted throughout (Limits not required)

M1 A1

a

⌠ 2 2 ⎮ π x(4a − x ) dx ⌡0

[

= π 2a 2 x 2 − =

x=

7 π 4

a

1 4

x4

]

M1 a 0

A1

4

(Limits not required)

A1

7 π a4 4 11 π a3 3

∫ x y dx 2 ∫ y dx 2

for

M1

21 = a 44

A1 7

3 (i) (ii)

I = 6.2 + 2.8 = 9.0 kg m 2

B1

WD against frictional couple is L × 12 π

B1

Loss of PE is 6 × 9.8 ×1.3 ( = 76.44 )

B1

1

2

Gain of KE is × 9.0 × 2.4 ( = 25.92 ) By work-energy principle, L × 12 π = 76.44 − 25.92 1 2

B1 ft Equation involving WD, KE and PE

M1

L = 32.2 N m

A1

5 Accept 32.1 to 32.2 (iii)

Moments equation

M1 A1 ft

6 × 9.8 × 0.8 − L = I α

α = 1.65 rad s − 2

A1 3

54

4731

Mark Scheme

June 2007

4 (i)

M may be ρ π a 2 throughout 3a (condone use of ρ = 1 )

MI of elemental disc about a diameter is 1⎛M ⎞ 2 ⎜ δx ⎟a 4 ⎝ 3a ⎠

B1

MI of elemental disc about AB is

A1

3a

I=

M⌠ 1 2 ( 4 a + x 2 ) dx 3a ⎮ ⌡0

[

M 1 2 a x + 13 x 3 3a 4 M 3 3 = ( 4 a + 9a 3 ) 3a = M ( 14 a 2 + 3a 2 ) =

]

Using parallel axes rule (can award A1 for 14 ma 2 + mx 2 )

M1

1⎛M ⎞ 2 ⎛M ⎞ 2 ⎜ δx ⎟a + ⎜ δx ⎟ x 4 ⎝ 3a ⎠ ⎝ 3a ⎠

Integrating MI of disc about AB Correct integral expression for I

M1 A1

3a 0

Obtaining an expression for I in terms of M and a Dependent on previous M1

M1

= 134 M a 2

A1 (ag) 7 (ii)

Period is 2π

I Mgh = 2π = 2π

M1 13 4

or − Mgh sin θ = I θ&&

Ma 2

A1

Mg 32 a 13a 6g

A1 3

55

4731

Mark Scheme

M1 A1

5 (i)

sin θ sin 115 = 12 16 θ = 42.8° Bearing of v B is 007.2°

M1

June 2007

Relative velocity on bearing 050 Correct velocity diagram; or ⎛ u sin 50 ⎞ ⎛ 16sin α ⎞ ⎛ 12sin 345 ⎞ ⎜ ⎟=⎜ ⎟−⎜ ⎟ ⎝ u cos 50 ⎠ ⎝16 cos α ⎠ ⎝12 cos 345 ⎠ or eliminating u (or α )

A1

u 16 = sin 22.2 sin 115 u = 6.66 2400 Time taken is = 360 s 6.664

M1

or obtaining equation for u (or α )

A1

For equations in α and t M1*M1A1 for equations M1*A1 ft M1 for eliminating t (or α ) 8 A1 for α = 7.2 M1A1 ft for equation for t (or α ) A1 cao for t = 360 M1 A1

(ii) 10 12 φ = 33.6° Bearing of v B is 018.6° cos φ =

M1 A1

56

Relative velocity perpendicular to v B Correct velocity diagram For alternative methods: M2 for a completely correct method A2 for 018.6 (give A1 for a correct relevant angle) 4

4731

6 (i)

Mark Scheme

= 94 ma 2

A1

a cos θ ) = I α 1 3

α=

(ii)

M1

mga cos θ 4 9

ma

2

=

3g cos θ 4a

4 M1 A1 ft

ma 2ω 2 = 13 mga sin θ

ω=

3 g sin θ 2a

dω 3g cos θ = dθ 4a g cos θ 3 1 2 ω =⌠ dθ ⎮ 2 ⌡ 4a 3g sin θ = (+C) 4a

OR ω

ω=

(iii)

A1 (ag)

By conservation of energy, 1 I ω 2 = mg ( 13 a sin θ ) 2 2 9

Using parallel axes rule

M1

I = 13 ma 2 + m( 13 a) 2

mg ( 13

June 2007

3g sin θ 2a

Condone ω 2 =

A1 3

3 g sin θ 2a

M1

A1 A1

Acceleration parallel to rod is ( 13 a)ω 2

B1

F − mg sin θ = m( 13 a )ω 2

M1

Radial equation with 3 terms

F − mg sin θ = mg sin θ 1 2 3 2

F = mg sin θ

A1

Acceleration perpendicular to rod is ( 13 a )α

B1 ft

ft is r α with r the same as before

mg cos θ − R = m( 13 a )α

M1

Transverse equation with 3 terms

mg cos θ − R = mg cos θ 1 4 3 4

R = mg cos θ

A1 6

OR R( 13 a) = I G α

(iv)

Must use I G

M1

⎛ 3g cos θ ⎞ R( 13 a ) = ( 13 ma 2 ) ⎜ ⎟ ⎝ 4a ⎠

A1

R = 34 mg cos θ

A1

On the point of slipping, F = μR 3 2

mg sin θ = μ ( 34 mg cos θ )

M1

tan θ = 12 μ

A1 (ag)

57

Correctly obtained 2 Dependent on 6 marks earned in (iii)

4731

7 (i)

Mark Scheme

GPE = (−) mg (2a cos θ ) cos θ EPE = =

1 2

mg

2a 1 mg 2

2a

June 2007

or (−) mg (a + a cos 2θ )

B1

( AR − a) 2

M1

(2a cos θ − a ) 2

A1

2

2

V = mga(2 cos θ − 1) − 2mga cos θ 1 4

= mga(cos 2 θ − cos θ +

1 4

− 2 cos 2 θ )

= mga( 14 − cos θ − cos 2 θ )

A1 (ag) 4

(ii)

dV = mga(sin θ + 2 cos θ sin θ ) dθ = mga sin θ (1 + 2 cos θ )

Equilibrium when

B1

dV =0 dθ

M1

ie when θ = 0

A1 (ag) 3

(iii)

1 2 2 &2

KE is

B1

m(2aθ&) 2

2ma θ + V = constant Differentiating with respect to t,

M1

dV & θ =0 dθ 4ma 2θ& θ&& + mga sin θ (1 + 2 cos θ )θ& = 0

M1

4ma 2θ& θ&& +

θ&& = −

(iv)

A1 ft

g sin θ (1 + 2 cos θ ) 4a

A1 (ag)

When θ is small, sin θ ≈ θ , cos θ ≈ 1 θ&& ≈ −

SR B2 (replacing the last 3 marks) for the given result correctly 5 obtained by differentiating w.r.t. θ

M1

g 3g θ (1 + 2) = − θ 4a 4a

Period is 2π

(can award this M1 if no KE term)

A1

4a 3g

A1 3

58


59

4732

Mark Scheme

June 2007

Note: “3 sfs” means an answer which is equal to, or rounds to, the given answer. If such an answer is seen and then later rounded, apply ISW.

1

(0×0.1) + 1×0.2 + 2×0.3 + 3×0.4 = 2(.0) (02×0.1) +1×0.2 + 22×0.3 + 32×0.4 (= 5) - 22 =1

Total 2

Total 3i ii

Total 4ia b

c

÷ 4: M0 > 2 non-zero terms correct Indep, ft their μ. Dep +ve result (-2)2×0.1+(-1)2×0.2 +02×0.3+ 12×0.4:M2 ÷ 4: M0 > 2 non-0 correct: M1 5

UK Fr Ru Po Ca 1 2 3 4 5 or 5 4 3 2 1 4 3 1 5 2 2 3 5 1 4 2 Σd (= 24) rs = 1 – _6 × “24” 5 × (52–1) 1 = – /5 or – 0.2

15

C7 or 15!/7!8! 6435

6

C3 × 9C4 or 6!/3!3! × 9!/4!5!

Consistent

A1 5

2

Dep 2nd M1

5 M1 A1 2 M1

1

B1

31452 54321

All 5 d2 attempted & added. Dep ranks att’d

M1

A1 2

RCFUP 35214 12345

attempt rank other judge

M1 A1 M1

2520

______43 – 15 /5_____ 2 2 √((55-15 /5)(55-15 /5)) M1 Corr sub in > 2 S’s All correct: M1

Alone except allow ÷ 15C7 Or 6P3 × 9P4 or 6!/3! × 9!/5! Allow ÷ 15P7 NB not 6!/3!×9!/4! 362880

4 /3 oe

P(BB) + P(WB) attempted = 4/10 × 3/9 + 6/10 × 4/9 or 2/15 + 4/15 = 2/5 oe Denoms 9 & 8 seen or implied /9 × 2 /8 + 6 /9 × 3 /8

3

= 1 /3 ii

> 2 non-zero terms correct eg ÷ 4: M0

M1 A1 M1 M1 A1 5

oe

P(Blue) not constant or discs not indep, so no

1

Or 4/10 × 3/9 OR 6/10 × 4/9 correct

M1 M1 A1 3 B1 M1

NB 4/10 × 4/10 + 6/10 × 4/10 = 2/5: M1M0A0 Or 2/15 as numerator Or 2/15 Or ___4/10×6/9×3/8 + 4/10×3/9×2/8____ 4 6 5 4 6 4 3 above + /10× /9× /8 + /10× /9× /8 /10 May not see wking

A1 3 B1

Total

1

8

60

B↔W MR: max (a)B0(b)M1M1(c)B1M1

Prob changes as discs removed Limit to no. of discs. Fixed no. of discs Discs will run out Context essential: “disc” or “blue” NOT fixed no. of trials NOT because without repl Ignore extra

4732

5i

Mark Scheme

1991 100 000 to 110 000

iia

Median = 29 to 29.9 Quartiles 33 to 34, 24.5 to 26 = 7.5 to 9.5 140 to 155 23 to 26.3%

b

Older Median (or ave) greater } % older mothers greater oe} % younger mothers less oe}

June 2007

B1 ind B1 ind 2 B1

Or fewer in 2001 Allow digits100 to 110

M1 A1 M1 A1 5 B1

Or one correct quartile and subtr NOT from incorrect wking ×1000, but allow without Rnded to 1 dp or integer 73.7 to 77% : SC1

B1 B1

Total

3 10

61

Or 1991 younger Any two Or 1991 steeper so more younger: B2 NOT mean gter Ignore extra

4732

6ia

b

iia

Mark Scheme

Correct subst in > two S formulae 60 × 72 or ___227__ 767 − √698√162 8 2 2 60 72 (1148 − )(810 − ) 8 8 = 0.675 (3 sfs) 1 y always increases with x or ranks same oe

M1

Any version All correct. Or 767-8x7.5x9_____ 2 2 \/((1148-8x7.5 )(810-8x9 )) or correct substn in any correct formula for r

M1 A1 B1 B1

June 2007

3

2

+ve grad thro’out. Increase in steps. Same order. Both ascending order Perfect RANK corr’n Ignore extra NOT Increasing proportionately

Closer to 1, or increases because nearer to st line

B1 B1

2

Corr’n stronger. Fewer outliers. “They” are outliers Ignore extra

None, or remains at 1 Because y still increasing with x oe

B1 B1

2

Σd2 still 0. Still same order. Ignore extra NOT differences still the same. NOT ft (i)(b)

13.8 to 14.0 (iii) or graph or diag or my est

B1 B1

1

Takes account of curve

B1

2

P(contains voucher) constant oe Packets indep oe

12 B1 B1 2

ii

0.9857 or 0.986 (3 sfs)

B2

iii

(1 – 0.9857) = 0.014(3) (2 sfs)

b

iii iv

Total 7i

iv

B(11, 0.25) or 6 in 11 wks stated or impl 11 C6 × 0755 × 0.256 (= 0.0267663) P(6 from 11) × 0.25 = 0.00669 or 6.69 x 10-3 (3 sfs)

2

B1ft 1 B1 M1 M1 A1 4

Total

9

62

Must be clear which est. Can be implied. “This est” probably ⇒ using equn of line Straight line is not good fit. Not linear. Corr’n not strong. Context essential NOT vouchers indep B1 for 0.9456 or 0.946 or 0.997(2) or for 7 terms correct, allow one omit or extra NOT 1 – 0.9857 = 0.0143 (see (iii)) Allow 1- their (ii) correctly calc’d or 0.75a × 0.25b (a + b = 11) or 11C6 dep B1

4732

Mark Scheme

M1 M1 A1 3 B1 M1 M1

\/0.04

8i ii

(= 0.2) (1 – their \/0.04)2 = 0.64 1 – p seen M1 for either 2p(1 – p) = 0.42 or p(1 – p) = 0.21 oe 2p2 – 2p + 0.42(= 0) or p2- p + 0.21(= 0) 2±\/((-2)2– 4×0.42) or 1±\/((-1)2– 4×0.21) 2×2 2×1 or (p–0.7)(p–0.3)=0 or (10p–7)(10p–3)=0 p = 0.7 or 0.3

M1 A1 5

June 2007

2pq= 0.42 or pq =0.21 Allow pq=0.42 or opp signs, correct terms any order (= 0) oe Correct Dep B1M1M1 Any corr subst’n or fact’n Omit 2 in 2nd line: max B1M1M0M0A0 One corr ans with no or inadeq wking: SC1 eg 0.6 × 0.7 = 0.42 ⇒ p = 0.7 or 0.6

p2 + 2pq + q2 = 1 p2 + q2 = 0.58 } p = 0.21/q } p4- 0.58p2 + 0.0441 = 0 corr subst’n or fact’n

B1 M1 M1

B1 1 – p seen 2p(1 – p) = 0.42 or p(1 – p) = 0.21 M1 p2 – p = -0.21 p2 – p + 0.25 = -0.21 + 0.25 oe } M1 OR (p – 0.5)2 – 0.25 = -0.21 oe } M1 (p – 0.5)2 = 0.04 (p – 0.5) = ±0.02 p = 0.3 or 0.7 A1 Total 9ia 1 / 1/5 =5 b ( 4 / 5 )3 × 1 / 5 = 64/625 or 0.102 (3 sfs) c (4/5)4

iia

8 M1 A1 2 M1 A1 2 M1

= 256/625 or a.r.t 0.410 (3 sfs) or 0.41 P(Y=1) = p, P(Y=3) = q2p, P(Y=5) = q4p

A1 2

B1 1 b

Recog that c.r. = q2 or (1 – p)2 p p or S∞ = 2 1− q 1 − (1 − p) 2

M1

or 1- (1/5 + 4/5×1/5 + (4/5)2×1/5 + (4/5)3 ×1/5) NOT 1 - (4/5)4 P(Y =1)+P(Y =3)+P(Y =5)= p + q2p + q4p p, p(1 - p)2, p(1 – p)4 q1-1, q3-1, q5-1 or any of these with 1 – p instead of q “Always q to even power × p” Either associate each term with relevant prob Or give indication of how terms derived > two terms or eg r = q2p/p

M1

1− q 1 − q2 1− q = Must see this step for A1 (1 − q )(1 + q ) 1 AG) (= 1+ q P(odd) =

63

M1

( = __p__ ) = ___p__ p(2 – p) ( 2p – p2 )

A1 4

( = __1__ ) = ____1____ ( 2–p ) 2 – (1 – q)

4732

Mark Scheme

Total

11

64

June 2007


65

4733 1

(i)

(ii) 2

Mark Scheme

μˆ = 4830.0/100 = 48.3 249509.16/100 – (their x 2 ) × 100/99 = 163.84 No, Central Limit theorem applies, so can assume distribution is normal B(130, 1/40) ≈ Po(3.25) 4

B1 M1 M1 A1

4

B2

2

= 0.180

B1 M1 A1√ M1 A1

5

Binomial Each element equally likely Choices independent

B1 B1 B1

Two of: Distribution symmetric No substantial truncation Unimodal/Increasingly unlikely further from μ, etc Variance 82/20 47.0 − 50.0 = –1.677 z= 8 2 / 20 Φ(1.677) = 0.9532 H1: λ > 2.5 or 15 Use parameter 15 P(> 23)

B1 B1

2

M1 A1 A1 A1

4

B1 M1 M1

1

1 – 0.9805 = 0.0195 or 1.95%

A1

3

P(≤ 23 | λ = 17) = 0.9367 P(≤ 23 | λ = 18) = 0.8989 Parameter = 17

M1

λ = 17/6 or 2.83

M1

3

H0: p = 0.19, H1: p < 0.19 where p is population proportion 0.8120 + 20 × 0.8119 × 0.19 = 0.0841 Compare 0.1 Add binomial probs until ans > 0.1 Critical region ≤ 1 Reject H0 Significant evidence that proportion of e’s in language is less than 0.19 Letters not independent

B2 M1 A1 A1 B1 A1 B1 M1 A1√

8

B1

1

e− λ

3

4

(i) (ii) (i)

(ii)

5

(i) (ii)

(iii)

6

(i)

or

(ii)

λ

4!

1 2

A1

66

June 2007

48.3 seen Biased estimate: 162.2016: can get B1M1M0 Multiply by n/(n – 1) Answer, 164 or 163.8 or 163.84 “No” with statement showing CLT is understood (though CLT does not need to be mentioned) [SR: No with reason that is not wrong: B1] B(130, 1/40) stated or implied Poisson, or correct N on their B(n, p) Parameter their np, or correct parameter(s)√ Correct formula, or interpolation Answer, 0.18 or a.r.t. 0.180 [SR: N(3.25, 3.17) or N(3.25, 3.25): B1M1A1] Binomial stated or implied All elements, or selections, equally likely stated Choices independent [not just “independent”] [can get B2 even if (i) is wrong] One property Another definitely different property Don’t give both marks for just these two “Bell-shaped”: B1 only unless “no truncation” Standardise, allow cc, don’t need n Denominator (8 or 82 or √8) ÷ (20 or √20 or 202) z-value, a.r.t. –1.68 or +1.68 Answer, a.r.t. 0.953

λ > 2.5 or 15, allow μ, don’t need “H1” λ = 15 used [N(15, 15) gets this mark only] Find P(> 23 or ≥ 23), final answer < 0.5 eg 0.0327 or 0.0122 Answer, 1.95% or 2% or 0.0195 or 0.02 [SR: 2-tailed, 3.9% gets 3/3 here] One of these, or their complement: .9367, .8989, 0.9047, 0.8551, .9317, .8933, .9907, .9805 Parameter 17 [17.1076], needs P(≤ 23), cwo [SR: if insufficient evidence can give B1 for 17] [2.85] Their parameter ÷ 6 [SR: Solve (23.5 – λ)/√λ = 1.282 M1; 18.05 A0] Correct, B2. One error, B1, but x or x or r: B0 Binomial probabilities, allow 1 term only Correct expression [0.0148 + 0.0693] Probability, a.r.t. 0.084 Explicit comparison of “like with like” [P(≤ 2) = 0.239] Correct deduction and method [needs P(≤ 1)] Correct conclusion in context [SR: N(3.8, 3.078): B2M1A0B1M0] Correct modelling assumption, stated in context Allow “random”, “depends on message”, etc

4733

7

Mark Scheme

(i)

(ii) (iii)

S is equally likely to take any value in range, T is more likely at extremities

∫

1

3 t 2

(i)

(ii)

64.2 − 63 12.25 / 23

= 1.644

P(z > 1.644) = 0.05 3 .5 (a) 63 + 1.645 ×

50

(iii) 9

(a)

(b)

≥ 63.81 (b) P(< 63.8 | μ = 65) 63.8 − 65 = –2.3956 3.5 / 50 0.0083 B better: Type II error smaller (and same Type I error) np > 5 and nq > 5 0.75n > 5 is relevant n > 20 (i) 70.5 – μ = 1.75σ μ – 46.5 = 2.25σ

(ii)

3

B2

2

M1

1

⎡ x3 ⎤ x dx = ⎢ ⎥ ⎣ 2 ⎦t 2

½(1 – t3) = 0.2 or ½(t3 + 1) = 0.8 t3 = 0.6 t = 0.8434 8

B1 B1 B1

B1 M1 M1 5 A1 M1dep A1 dep M1 4 A1 M1 B1 3 A1 M1 M1 A1 A1

4

B2√

2

M2

Solve simultaneously μ = 60 σ=6

A1 M1 A1 B1 M1 A1√ A1√

np = 60, npq = 36 q = 36/60 = 0.6 p = 0.4 n = 150

M1dep depM1 A1√ 4 A1√

70.5

46.5

σ 6

71

46

6.25

71.5

46.5

6.25

70.5 71.5 70

45.5 45.5 46

6.25 6.5 6

3

6

μ 60 60.062 5 60.562 5 59.562 5 60.125 59.5

67

June 2007

Horizontal straight line Positive parabola, symmetric about 0 Completely correct, including correct relationship between two Don’t need vertical lines or horizontal lines outside range, but don’t give last B1 if horizontal line continues past “±1” Correct statement about distributions (not graphs) [Partial statement, or correct description for one only: B1] Integrate f(x) with limits (–1, t) or (t, 1) [recoverable if t used later] Correct indefinite integral Equate to 0.2, or 0.8 if [–1, t] used Solve cubic equation to find t Answer, in range [0.843, 0.844] Standardise 64.2 with √n z = 1.644 or 1.645, must be + Find Φ(z), answer < 0.5 Answer, a.r.t. 0.05 or 5.0% 63 + 3.5 × k / √50, k from Φ–1, not – k = 1.645 (allow 1.64, 1.65) Answer, a.r.t. 63.8, allow >, ≥, =, c.w.o. Use of correct meaning of Type II Standardise their c with √50 z = (±) 2.40 [or –2.424 or – 2.404 etc] Answer, a.r.t. 0.008 [eg, 0.00767] This answer: B2. “B because sample bigger”: B1. [SR: Partial answer: B1] Use either nq > 5 or npq > 5 [SR: If M0, use np > 5, or “n = 20” seen: M1] Final answer n > 20 or n ≥ 20 only Standardise once, and equate to Φ–1, ± cc Standardise twice, signs correct, cc correct Both 1.75 and 2.25 Correct solution method to get one variable μ, a.r.t. 60.0 or ± 154.5 σ, a.r.t. 6.00 [Wrong cc (below): A1 both] [SR: σ2: M1A0B1M1A1A0] np = 60 and npq = 62 or 6 Solve to get q or p or n p = 0.4 √ on wrong cc or z n = 150 √ on wrong cc or z q 0.6

p (±0.01) 0.4

n 150

0.6504

0.3496

171.8

0.6450

0.3550

170.6

0.6558 0.7027 0.6050

0.3442 0.2973 0.3950

173.0 202.2 150.6

4733

Mark Scheme

68

June 2007


69

4734

Mark Scheme

June 2007

______________________________________________________________________________ 1

∫

1

0

+

adx

[ ax ]10

∫

∞

1

a x2

dx = 1

M1

For sum of integrals =1

A1

For second integral.

∞

+

a + a=½

⎡ a⎤ ⎢− 3 ⎥ = 1 ⎣ x ⎦1 a =1

For second a Or from F(x) M1A1 then a=1/2 A1 F(∞)=1 M1, ______________________________________________________________________________ 0.7 2 ) B1 If no parameters allow in (ii) 2 (i) X I N(5, 20 XE

N(4.5,

0.52 ) 25

A1 A1

4

B1

2

If 0.7/20, 0.5/25 then B1 for

both, with means in (ii) --------------------------------------------------------------------------------------------------------------------------------M1A1 OR X I − X E − 1 N(-0.5,σ 2 ) (ii) Use X I − X E N (0.5, σ 2 ) σ2 = 0.49/20 + 0.25/25 B1 cao 1- Φ([1-0.5]/ σ) M1 RH probability implied. If 0.7, 0.5 = 0.0036 or 0.0035 A1 5 in σ2, M1A1B0M1A1 for 0.165 ______________________________________________________________________________ 3 Assumes differences form a random sample from a normal distribution. B1 H0: μ = 0, H1: μ > 0 B1 Other letters if defined; or in words 2 B1B1 Or (12/11)(136.36/12-(17.2/12)2)aef x = 17.2 /12 ; s = 10.155 AEF ----------------------------------------------------------------------------------------------------------------------------------------------x M1 With 12 or 9.309/11 EITHER: t = (+ or - ) 2 s / 12 =1.558 A1 Must be positive. Accept 1.56 1.363 seen B1 Allow CV of 1.372 or 1.356 evidence 1.558 > 1.363, so reject H0 and accept that there that the readings from the aneroid Explicit comparison of CV(not B1√ with +) and conclusion in context. device overestimate blood pressure on average ------------------------------------------------------------------------------------------------------------------------------------------OR: For critical region or critical value of x 1.363√(s2/12) M1B1 B1 for correct t Giving 1.25(3) A1 Compare 1.43(3) with 1.25(3) 8 Conclusion in context B1√ _____________________________________________________________________________________

70

4734

Mark Scheme

June 2007

______________________________________________________________________________ 4 (i) Proper P F 11 42 B1 Two correct P 31 Trial 5 13 18 B1 Others correct F 24 60 2 36 ----------------------------------------------------------------------------------------------------------------------------------------------(ii) (H0: Trial results and Proper results are independent.) E-values: 25.2 16.8 M1 One correct. Ft marginals in (i) 10.8 7.2 A1 All correct χ2 = 5.32(25.2-1+10.8-1+16.8-1+7.2-1)

M1 A1 A1

= 9.289

Allow two errors With Yates’ correction art 9.29

Compare correctly with 7.8794 M1 Or 7.88 There is evidence that results are not 7 Ft χ2calc. independent. A1 √ _______________________________________________________________________________________ (i) e-μ = 0.45 M1 A1 2 0.799 or 0.798 or better seen μG = 0.799 ≈ 0.80 AG ----------------------------------------------------------------------------------------------------------------------------------------------(ii) μU ≈ 1.8 B1 Total, T ~ Po(2.6) M1 May be implied by answer 0.264 P(>3) = 0.264 A1 3 From table or otherwise ---------------------------------------------------------------------------------------------------------------------------------------------(iii) e-2.62.66/6! B1 Or 0.318 from table e-5.25.24/4! B1 Multiply two probabilities M1 Answers rounding to 0.0053 or 0.0054 A1 4 ______________________________________________________________________________

5

71

4734 6

Mark Scheme (i)

pˆ = 62/200=0.31

Use pˆα ± z

pˆα (1 − pˆα ) 200

June 2007

B1

aef

M1

With 200 or 199

z=1.96 B1 Seen Correct variance estimate A1√ ft pˆ 5 art (0.246,0.374) (0.2459,0.3741) A1 ----------------------------------------------------------------------------------------------------------------------------------------------(ii)EITHER: Sample proportion has an approximate normal distribution OR: Variance is an estimate B1 1 Not pˆ is an estimate, unless variance mentioned ----------------------------------------------------------------------------------------------------------------------------------------------(iii) H0: pα = pβ , H1: p α ≠ p β pˆ =(62+35)/(200+150) B1 aef ----------------------------------------------------------------------------------------------------------------------------------------------62 / 200 − 35 / 150 EITHER: z= (±) M1 s2 with, pˆ , 200,150 (or 199,149) −1 −1 ˆ ˆ (200 + 150 ) pq B1√

=1.586 (-1.96 1/√y)] = 1 – F(1/√y) y ≤ 0, ⎧0 ⎪ 2 = ⎨y 0 ≤ y ≤ 1, ⎪(1 y > 1.) ⎩

June 2007

M1 A1 A1

May be implied by following line Accept strict inequalities

Or F(x)=P(X≤x) = P(Y≥1/x2) M1 =1 – P(Y 12. M0 if W=27

A1 7 M1 A1A1 M1 A1 5

Conclusion in context.

M1

Correct method for any one

A1

76

2

All correct SR: B1 if no method indicated

4735

Mark Scheme

4 (i) H0: m = 2.70, H1: m > 2.7 Subtract 2.70 from each value and count the number of positive signs Obtain 13 Use B(20,1/2) to obtain P(X ≥ 13) = 0.1316 (0.132) Compare correctly with 0.05 Do not reject H0. Conclude that there is insufficient evidence to claim that median level of impurity is greater than 2.70

∞

1

∫ (α − 1)! x

In terms of medians Allow just ‘medians’ here

B1 M1 A1 M1 A1 M1 A1

(ii)Wilcoxon signed rank test Advantage: More powerful (uses more formation) Disadvantage: This test requires a symmetric population distribution, not required for sign test 5 (i)

June 2007

α −1 − x

e dx = 1, result follows

For finding tail probability Or CR: X ≥ 15 M1A1 Or: N(10, 5), p=0.132 7

B1 B1

Smaller P(Type II) Not ‘more time taken’

B1

3

B1

1

0

M1

∞

(ii) MX(t)=

1 α −1 − x xt ∫0 (α − 1)! x e e dx ∞

=

1

∫ (α − 1)! x

α −1 - x (1−t )

e

dx

0

x=u/(1 – t), dx=du/(1 – t) and limits unchanged ∞

=

1

u

α −1

e du 1− t

A1

1 uα −1e-u du α ∫ (α − 1)!(1 − t ) 0

A1

= (1 – t)-α

A1

∫ (α − 1)! (1 − t )α 0

Attempt to differentiate

M1

−u

−1

∞

=

AG

(iii) EITHER: M'(t)=α(1-t)-α -1 M"(t)= α(α + 1)(1 – t)-α – 2 Substitute t=0 E(X) = α Var(X) = α(α + 1) – α2 =α OR: (1 – t)-α = 1 + αt + ½ α(α+1)t2+… E(X) = α Var(X) = E(X2) – [E(X)]2 = α(α+1) – α2 ; α

5

B1 B1 M1 A1 M1 A1 M1A1 B1 M1 A1A1 6

77

With evidence AEF

M0 if t involved

4735

6

7

Mark Scheme

June 2007

Accept qt0+pt1

(i) q+ pt

B1

1

(ii) (q+ pt)n (= GS(t)) Binomial

B1 B1

2

(iii) E(S)=G'(1) = np(q+p) = np Var(S) = G"(1)+G'(1) – [G'(1)]2 = n(n – 1)p2(p + q) + np – n2p2 = npq

M1A1 A1 M1 A1 6 A1

AEF, properly obtained

(iv) (½ + ½t)10e-(1 – t) Find coefficient of t2 (1/210)(1 + 10t + ½×10×9t2) e-1(1 + t + ½t2) Required coefficient = e-12-10(1/2 + 10 + 45) = 0.0199

M1 M1 A1 A1

Seen May be implied OR: P(Y=0)P(Z=2)+….M1, Z is Po(1) M1 Ans:A1A1A1;A1

M1 A1

Not from e-(1-t)=1-(1-t)+(1-t)2/2 No more than one term missing--

1 2

(i) E(T1) = 2E(X ) = 2 × θ = θ

(ii) E(U) =

nu n

0

θn

∫

E(U2) = ∫

θ

0

θ

n

du

;

Var(U) = E(U2) – [E(U)]2 nθ 2 = AG

ϑ

x

0

θ

∫

dθ

2

⎡ nu n +1 ⎤ nθ du ; ⎢ n ⎥; ⎣ θ (n + 1) ⎦ n + 1

nu n +1

SR: B1 if X =

M1A1

(So T1 is an unbiased estimator of θ) θ

6

M1A1A1

n θ2 n+2

M1A1 A1

6

(n + 1) 2 (n + 2)

(iii) Var(T2) = θ2/[n(n+2)] Var(T1) = 4Var(X)/n ; θ2/3n Var(T2)/Var(T1) 3/(n+2) < 1 for n > 1 So T2 is more efficient than T1

B1 M1A1

For comparison of var. T1, T2

M1 M1A1

Idea used.

A1

78

7


79

4736

Mark Scheme

80

June 2007

4736

Mark Scheme

81

June 2007

4736

Mark Scheme

82

June 2007

4736

Mark Scheme

83

June 2007

4736

Mark Scheme

84

June 2007


85

4737

Mark Scheme

86

June 2007

4737

Mark Scheme

87

June 2007

4737

Mark Scheme

88

June 2007

4737

Mark Scheme

89

June 2007

4737

Mark Scheme

90

June 2007

Advanced GCE Mathematics (3892 – 2, 7890 - 2) June 2007 Assessment Series

Unit Threshold Marks Unit Maximum Mark Raw 72 4721

4722 4723 4724 4725 4726 4727 4728 4729 4730 4731 4732 4733

a

b

c

d

e

u

60

52

44

36

29

0

UMS

100

80

70

60

50

40

0

Raw

72

56

48

40

33

26

0

UMS

100

80

70

60

50

40

0

Raw

72

57

50

43

36

29

0

UMS

100

80

70

60

50

40

0

Raw

72

61

54

47

40

33

0

UMS

100

80

70

60

50

40

0

Raw

72

54

46

39

32

25

0

UMS

100

80

70

60

50

40

0

Raw

72

60

53

46

39

33

0

UMS

100

80

70

60

50

40

0

Raw

72

57

50

43

36

29

0

UMS

100

80

70

60

50

40

0

Raw

72

57

49

42

35

28

0

UMS

100

80

70

60

50

40

0

Raw

72

59

51

44

37

30

0

UMS

100

80

70

60

50

40

0

Raw

72

62

54

46

38

31

0

UMS

100

80

70

60

50

40

0

Raw

72

51

43

36

29

22

0

UMS

100

80

70

60

50

40

0

Raw

72

55

48

42

36

30

0

UMS

100

80

70

60

50

40

0

Raw

72

56

48

41

34

27

0

UMS

100

80

70

60

50

40

0

91

4734 4735 4736 4737

Raw

72

56

49

42

36

30

0

UMS

100

80

70

60

50

40

0

Raw

72

60

51

43

35

27

0

UMS

100

80

70

60

50

40

0

Raw

72

62

55

48

42

36

0

UMS

100

80

70

60

50

40

0

Raw

72

61

53

46

39

32

0

UMS

100

80

70

60

50

40

0

Specification Aggregation Results

Overall threshold marks in UMS (i.e. after conversion of raw marks to uniform marks) A

B

C

D

E

U

3890/3891/3892

Maximum Mark 300

240

210

180

150

120

0

7890/7891/7892

600

480

420

360

300

240

0

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

B

C

D

E

U

3890

31.2

47.9

62.0

74.4

84.9

100

Total Number of Candidates 13873

3891

20.0

20.0

20.0

20.0

20.0

100

10

3892

58.5

75.6

87.9

94.7

97.5

100

1384

7890

45.3

66.9

82.2

92.4

97.7

100

9663

7891

0

0

0

100

100

100

1

7892

58.2

78.1

89.1

96.0

98.8

100

1487

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

92

93

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