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. Mathematics ...
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:
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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
Further Pure Mathematics 2
25
4727
Further Pure Mathematics 3
31
4728
Mechanics 1
37
4729
Mechanics 2
43
4730
Mechanics 3
47
4731
Mechanics 4
53
4732
Probability & Statistics 1
59
4733
Probability & Statistics 2
65
4734
Probability & Statistics 3
69
4735
Probability & Statistics 4
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
Mark Scheme 4722 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)
Mark Scheme 4723 June 2007
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
Mark Scheme 4725 June 2007
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
Mark Scheme 4726 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
Mark Scheme 4727 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
Mark Scheme 4728 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
Mark Scheme 4729 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
Mark Scheme 4730 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
Mark Scheme 4731 June 2007
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
Mark Scheme 4732 June 2007
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
Mark Scheme 4733 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
Mark Scheme 4734 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
Mark Scheme 4736 June 2007
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
Mark Scheme 4737 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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