Advanced Subsidiary GCE Further Mathematics (MEI) (3896). Advanced
Subsidiary GCE Further ... 4751 (C1) Introduction to Advanced Mathematics. 3.
4752 (C2) Concepts for Advanced Mathematics. 8. 4753 (C3) Methods for
Advanced ...
GCE Mathematics (MEI) Advanced GCE A2 7895-8 Advanced Subsidiary GCE AS 3895-8
Mark Schemes for the Units June 2009
3895-8/7895-8/MS/R/09 1
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4751
Mark Scheme
June 2009
CONTENTS Advanced GCE Mathematics (MEI) (7895) Advanced GCE Further Mathematics (MEI) (7896) Advanced GCE Further Mathematics (Additional) (MEI) (7897) Advanced GCE Pure Mathematics (MEI) (7898) Advanced Subsidiary GCE Mathematics (MEI) (3895) Advanced Subsidiary GCE Further Mathematics (MEI) (3896) Advanced Subsidiary GCE Further Mathematics (Additional) (MEI) (3897) Advanced Subsidiary GCE Pure Mathematics (MEI) (3898) MARK SCHEME FOR THE UNITS Unit/Content
Page
4751 (C1) Introduction to Advanced Mathematics
3
4752 (C2) Concepts for Advanced Mathematics
8
4753 (C3) Methods for Advanced Mathematics
11
4754 (C4) Applications of Advanced Mathematics
17
4755 (FP1) Further Concepts for Advanced Mathematics
23
4756 (FP2) Further Methods for Advanced Mathematics
28
4757 Further Pure 3
34
4758 Differential Equations
42
4761 Mechanics 1
46
4762 Mechanics 2
52
4763 Mechanics 3
56
4764 MEI Mechanics 4
60
4766 Statistics 1
63
4767 Statistics 2
67
4768 Statistics 3
71
4769 Statistics 4
76
4771 Decision Mathematics 1
80
4772 Decision Mathematics 2
85
4773 Decision Mathematics Computation
90
1
4751
Mark Scheme
June 2009
4776 Numerical Methods
95
4777 MEI Numerical Computation
97
Grade Thresholds
103
2
4751
Mark Scheme
June 2009
4751 (C1) Introduction to Advanced Mathematics Section A 1
2
3
(0, 14) and (14/4, 0) o.e. isw
2 ( s − ut ) o.e. as final answer t2 ( s − ut ) ] [condone [a =] 0.5t 2 [ a =]
10 www
4
M2 for evidence of correct use of gradient with (2, 6) eg sketch with ‘stepping’ or y − 6 = −4(x − 2) seen or y = −4x + 14 o.e. or M1 for y = −4x + c [accept any letter or number] and M1 for 6 = −4 × 2 + c; A1 for (0, 14) [c = 14 is not sufficient for A1] and A1 for (14/4, 0) o.e.; allow when x = 0, y = 14 etc isw
3
4
M1 for each of 3 complete correct steps, ft from previous error if equivalent difficulty [eg dividing by t does not count as step – needs to be by t2]
[ a =]
( s − ut ) 1 2 2
t
gets M2 only (similarly
other triple-deckers) M1 for f(3) = 1 soi and A1 for 31 − 3k = 1 or 27 − 3k = −3 o.e. [a correct 3-term or 2-term equation]
3
3
long division used: M1 for reaching (9 − k)x + 4 in working and A1 for 4 + 3(9 − k) = 1 o.e.
4
5
x < 0 or x > 6 (both required) (i) 10 www
2 2
equating coeffts method: M2 for (x − 3)(x2 + 3x −1) [+ 1] o.e. (from inspection or division)
3
B1 each; if B0 then M1 for 0 and 6 identified;
2
M1 for
5× 4× 3 5× 4 or or for 3 × 2 (×1) 2 (×1)
1 5 10 (ii) 80 www or ft 8 × their (i)
2
10
5
1 seen
B2 for 80x3; M1 for 23 or (2x)3 seen
3
4
16
4751 6
Mark Scheme
June 2009
any general attempt at n being odd and n being even
M1
M0 for just trying numbers, even if some odd, some even
n odd implies n3 odd and odd − odd = even
A1
n even implies n3 even and even − even = even
A1
or n(n2 − 1) used with n odd implies n2 − 1 even and odd × even = even etc [allow even × odd = even] or A2 for n(n − 1)(n + 1) = product of 3 consecutive integers; at least one even so product even; odd3 − odd = odd etc is not sufft for A1 SC1 for complete general method for only one of odd or even eg n = 2m leading to 2(4m3 − m)
7
8
(i) 1
2
(ii) 1000
1
(i) 2/3 www
2
B1 for 50 or for 25 × 1/25 o.e. 3 M1 for 4/6 or for
48 = 2 12 or 4 3 or
27 = 3 3 or 108 = 3 12 or for (ii) 43 − 30 2 www as final answer 9
10
3
(i) (x + 3)2 − 4
3
(ii) ft their (−a, b); if error in (i), accept (−3, −4) if evidence of being independently obtained
2
3
4 9
M2 for 3 terms correct of 25 − 15 2 − 15 2 + 18 soi, M1 for 2 terms correct B1 for a = 3, B2 for b = − 4 or M1 for 5 − 32 soi B1 each coord.; allow x = −3, y = −4; or M1
5
−3 o.e. oe for sketch with −3 and −4 −4
for
marked on axes but no coords given
(x2 − 9)(x2 + 4)
M2
or correct use of quad formula or comp sq reaching 9 and −4; allow M1 for attempt with correct eqn at factorising with factors giving two terms correct, or sign error, or attempt at formula or comp sq [no more than two errors in formula/substn]; for this first M2 or M1 allow use of y etc or of x instead of x2
x2 = 9 [or −4] or ft for integers /fractions if first M1 earned x = ±3 cao
M1
must have x2; or M1 for (x + 3)(x − 3); this M1 may be implied by x = ±3 A0 if extra roots if M0 then allow SC1 for use of factor theorem to obtain both 3 and −3 as roots or (x + 3) and (x − 3) found as factors and SC2 for x2 + 4 found as other factor using factor theorem [ie max SC3]
A1
4
5
4
20
4751
Mark Scheme
June 2009
Section B 11
i
ii
y = 3x eqn AB is y = -1/3 x + 3 o.e. or ft
3x = -1/3x + 3 or ft x = 9/10 or 0.9 o.e. cao
2
M1 for grad AB =
1− 3 or − 1/3 o.e. 6
M1 A1
need not be simplified; no ft from midpt used in (i); may be seen in (i) but do not give mark unless used in (ii) eliminating x or y , ft their eqns if find y first, cao for y then ft for x
A1
ft dep on both Ms earned
M1
4
y = 27/10 oe ft their 3 × their x iii
2
9 2 (1 + 3 ) o.e and completion 10
2
2
or square root of this; M1 for 2
2
9 27 + or 0.81 + 7.29 soi or ft 10 10
to given answer
their coords (inc midpt) or M1 for distance = 3 cos θ and tan θ = 3 and M1 for showing sin θ =
3 and 10
completion iv
v
2 10
2
9 www or ft their a 10
2
M1 for 62 + 22 or 40 or square roots of these
2
M1 for ½ × 3 × 6 or
1 9 × their2 10 × 10 2 10
5
2
2
12
4751
12
Mark Scheme
iA
expansion of one pair of brackets
M1
correct 6 term expansion
M1
June 2009
eg [(x + 1)](x2 − 6x + 8); need not be simplified eg x3 − 6x2 + 8x + x2 − 6x + 8; or M2 for correct 8 term expansion: x3 − 4x2 + x2 − 2x2 + 8x − 4x − 2x + 8, M1 if one error allow equivalent marks working backwards to factorisation, by long division or factor theorem etc or M1 for all three roots checked by factor theorem and M1 for comparing coeffts of x3
iB
cubic the correct way up x-axis: −1, 2, 4 shown y-axis 8 shown
G1 G1 G1
with two tps and extending beyond the axes at ‘ends’ ignore a second graph which is a translation of the correct graph
iC
ii
[y=](x − 2)(x − 5)(x − 7) isw or (x − 3)3 − 5(x − 3)2 + 2(x − 3) + 8 isw or x3 − 14x2 + 59x − 70
2
M1 if one slip or for [y =] f(x − 3) or for roots identified at 2, 5, 7 or for translation 3 to the left allow M1 for complete attempt: (x + 4)( x + 1)(x − 1) isw or (x + 3)3 − 5(x + 3)2 + 2(x + 3) + 8 isw
(0, −70) or y = −70
1
allow 1 for (0, −4) or y = −4 after f(x + 3) used
27 − 45 + 6 + 8 = −4 or 27 − 45 + 6 + 12 = 0
B1
or correct long division of x3 − 5x2 + 2x + 12 by (x − 3) with no remainder or of x3 − 5x2 + 2x + 8 with rem −4
long division of f(x) or their f(x) + 4 by (x − 3) attempted as far as x3 − 3x2 in working
M1
or inspection with two terms correct eg (x − 3)(x2 ………..− 4)
x2 − 2x − 4 obtained
A1
[ x =]
2±
(x − 1)2 = 5
( −2 )
2
− 4 × ( −4 )
2
2 ± 20 o.e. isw or 1 ± 5 2
or
M1
2
3
3
dep on previous M1 earned; for attempt at formula or comp square on their other ‘factor’
A1 5
6
13
4751
13
Mark Scheme
i
1 1
(5, 2)
20 or 2 5 ii
iii
iv
no, since 20 < 5 or showing roots of y2 − 4y + 9 = 0 o.e. are not real
y = 2x − 8 or simplified alternative
2
June 2009
0 for ± 20 etc or ft from their centre and radius M1 for attempt (no and mentioning
20 or 5) or sketch or solving by formula or comp sq (−5)2 + (y − 2)2 = 20 [condone one error]
2
or SC1 for fully comparing distance from x axis with radius and saying yes
2
M1 for y − 2 = 2(x − 5) or ft from (i) or M1 for y = 2x + c and subst their (i) or M1 for ans y = 2x + k, k ≠ 0 or −8
2
(x − 5)2 + (2x)2 = 20 o.e.
M1
subst 2x + 2 for y [oe for x]
5x2 − 10x + 5[ = 0] or better equiv.
M1
expanding brackets and rearranging to 0; condone one error; dep on first M1
obtaining x = 1 (with no other roots) or showing roots equal
M1
one intersection [so tangent]
A1
o.e.; must be explicit; or showing line joining (1,4) to centre is perp to y = 2x +2
(1, 4) cao
A1
allow y = 4
alt method y − 2 = −½ (x − 5) o.e. 2x + 2 − 2 = −½ (x − 5) o.e. x=1 y = 4 cao showing (1, 4) is on circle
M1 M1 A1 A1 B1
line through centre perp to y = 2x + 2 dep; subst to find intn with y = 2x + 2
alt method perp dist between y = 2x − 8 and y = 2x + 2 = 10 cos θ where tan θ = 2 showing this is
20 so tgt
x = 5 − 20 sin θ x=1 (1, 4) cao
2
by subst in circle eqn or finding dist from centre = 20 [a similar method earns first M1 for eqn of diameter, 2nd M1 for intn of diameter and circle A1 each for x and y coords and last B1 for showing (1, 4) on line – award onlyA1 if (1, 4) and (9, 0) found without (1, 4) being identified as the soln]
M1 M1 M1 A1 A1
7
or other valid method for obtaining x allow y = 4
5
11
4752
Mark Scheme
June 2009
4752 (C2) Concepts for Advanced Mathematics Section A 1
2
3
4
using Pythagoras to show that hyp. of right angled isos. triangle with sides a and a is √2a completion using definition of cosine
M1
www
A1
a any letter or a number NB answer given
2 x6 + 5x
value at 2 − value at 1 131
M2 M1 A1
M1 if one error ft attempt at integration only
(i) 193
2
M1 for 8 + 15 + …+ 63
(ii) divergent + difference between terms increasing o.e.
1
(i) 2.4
2
(ii) 138
2
4
3 M1 for 43.2 ÷ 18 M1 for their (i) ×
180 or π
43.2 × 360 o.e. or for other rot 36π versions of 137.50…
θ=
(i)sketch of cosx ; one cycle, sketch of cos2x; two cycles, both axes scaled correctly
1 1 D1
(ii) (1-way) stretch parallel to y-axis sf 3
1 D1
6
y′ = 3x2 − 12x − 15 use of y′ = 0, s.o.i. ft x = 5, – 1 c.a.o. x < −1 or x > 5 ft
7
use of cos2 θ = 1 − sin2 θ at least one correct interim step in obtaining 4 sin2 θ − sin θ = 0.
M1 M1 A1 A1 A1 M1 M1
5
8
B1 B1 B1
attempt to integrate 3 x − 5
M1
[y=] 2 x − 5 x + c subst of (4, 6) in their integrated eqn 3
c = 10 or [y=] 2 x 2 − 5 x + 10
4
5 for two terms correct
5
NB answer given
θ = 0 and 180, 14.(47…) 165 − 166
3 2
2
r.o.t to nearest degree or better -1 for extras in range
A2 M1
5
A1 for two terms correct
A1 5
8
4751 9
Mark Scheme
(i) 7
1
(ii) 5.5 o.e.
2
June 2009
M1 for at least one of 5 log10a or ½ log10a or log10a5.5 o.e.
3
Section B 10
0.6(0...), 0.8(45...), [1], 1.1(76...) 1.3(0...), 1.6(0...) points plotted correctly ft ruled line of best fit
T1
Correct to 2 d.p. Allow 0.6, 1.3 and 1.6
P1 L1
tol. 1 mm
b = their intercept a = their gradient -11 ≤ b ≤ -8 and 21 ≤ a ≤ 23.5
M1 M1 A1
3
iii
34 to 35 m
1
1
iv
29 = ‘22’logt – ‘9’
M1
t = 10‘1.727...’
M1
55 [years] approx
A1
For small t the model predicts a negative height (or h = 0 at approx 2.75) Hence model is unsuitable
1
iA
10 + 20 + 30 + 40 + 50 + 60
B1
iB
correct use of AP formula with a = 10 and d = 10
M1
n(5 + 5n) or 5n(n + 1) or 5(n2 + n) or (5n2 + 5n)
A1
10n2 + 10n − 20700 = 0 45 c.a.o.
M1 A1
iiA
4
1
iiB
£2555
2
iiC
correct use of GP formula with a =5, r = 2
M1
5(2n − 1) o.e. = 2621435
D M1
i
ii
v
11
2n = 524288 www
3
accept 53 to 59
3
D1
2 or
6 2
(2 × 10 + 5 × 10) or 62 (10 + 60)
1
Or better 4 1 M1 for 5(1 + 2 + …28) or 5(29 − 1) o.e.
2
‘S’ need not be simplified
M1 19 c.a.o. 12
i
6.1
4 A1 2
9
(3.1
2
M1 for
) (
)
− 7 − 32 − 7 o.e. 3 .1 − 3
2
4751
Mark Scheme ii
iii
iv
v
((3 + h)
2
) (
− 7 − 32 − 7 h numerator = 6h + h2 6+h
)
M1
June 2009
s.o.i.
M1 A1
3
as h tends to 0, grad. tends to 6 o.e. f.t.from “6”+h
M1 A1
2
y − 2 = ‘6’ (x − 3) o.e. y = 6x − 16
M1 A1
At P, x = 16/6 o.e. or ft
M1 M1 A1
At Q, x = 7 0.021 c.a.o.
10
6 may be obtained from
dy dx
2
3
4753
Mark Scheme
June 2009
4753 (C3) Methods for Advanced Mathematics Section A 1
π 6
0
sin 3x dx = [− 13 cos 3 x ]06 π
= − 13 cos π2 + cos 0 =
B1 M1
1 3
A1cao [3]
2(i)
100 = Ae0 = A A = 100
50 = 100 e–1500k e–1500k = 0.5 –1500k = ln 0.5 k = –ln 0.5 ÷ 1500 = 4.62 × 10–4
(ii) 1 = 100e–kt -kt = ln 0.01 t = -ln 0.01 ÷ k = 9966 years
-1
g(x) = 2 x − 1
b = 2 0 − 1 = 2 or (0, 2) 2 x −1 = 0
x = 1, so a = 1 or (1, 0)
0.33 or better.
M1
50 = A e–1500k ft their ‘A’ if used
M1 A1 [5]
taking lns correctly 0.00046 or better
M1
ft their A and k
M1 A1 [3]
taking lns correctly art 9970
M1
Can use degrees or radians reasonable shape (condone extra range)
B1
passes through (-1, 2π), (0, π) and (1, 0)
A1
good sketches – look for curve reasonably vertical at (–1, 2π) and (1, 0), negative gradient at (0, π). Domain and range must be clearly marked and correct.
1
[3]
4
substituting correct limits in ±k cos …
M1A1
3 2π
[− 13 cos 3x] or [− 13 cos u ]
B1 M1 A1 [3]
11
Allow unsupported answers. www x = 1 is A0
www
4753
Mark Scheme
5(i) e2y = 1 + sin x dy = cos x 2e2y dx cos x dy = dx 2e 2 y
(ii) 2y = ln(1 + sin x) y = ½ ln(1 + sin x) dy = 1 cos x dx 2 1 + sin x cos x = as before 2e 2 y
6
x +1 +1 f f( x ) = x − 1 x +1 −1 x −1 x +1 + x −1 = x +1− x +1
=
2x = x* 2
f–1(x) = f(x) Symmetrical about y = x. 7(i) (A) (x – y)(x2 + xy + y2) = x3 + x2y + xy2 – yx2 – xy2 – y3 = x3 – y3 *
(B) (x + ½ y)2 + ¾ y2 = x2 + xy + ¼ y2 + ¾ y2 = x2 + xy + y2
(ii) x3 – y3 = (x – y)[(x + ½ y)2 + ¾ y2]
(x + ½ y)2 + ¾ y2 > 0 [as squares ≥ 0]
if x – y > 0 then x3 – y3 > 0 if x > y then x3 > y3 *
June 2009
M1
Their 2e2y ×
B1
2e2y
A1 [3]
o.e. cao
B1 M1 B1 E1 [4]
dy dx
chain rule (can be within ‘correct’ quotient rule with dv/dx = 0) 1/u or 1/(1 + sin x) soi www
M1
correct expression
M1
without subsidiary denominators e.g. = x + 1 + x − 1 × x −1
x −1 x +1− x +1
E1 stated, or shown by inverting B1 B1 [5] M1
expanding - allow tabulation
E1
www
M1
(x + ½ y)2 = x2 + ½ xy + ½ xy + ¼ y2 o.e.
E1 [4]
cao www
M1
substituting results of (i)
M1 E1 [3]
12
4753
Mark Scheme
8(i) A: 1 + ln x = 0 ln x = –1 so A is (e–1, 0) x = e–1 B: x = 0, y = e0 – 1 = e–1 so B is (0, e–1)
(ii) Either by invertion: e.g. y = ex – 1 x ↔ y x = ey – 1 ln x = y – 1 1 + ln x = y or by composing e.g. fg(x) = f(1 + ln x) = e1 + ln x – 1 = eln x = x
1
0
0
d
1 dx x = x ln x – x + c
e −1
g ( x ) dx =
e −1
M1
e1 + ln x – 1 or 1 + ln(ex – 1)
M1
= x ln x − x. 1
taking lns or exps
M1 A1cao [3]
ln x dx = ln x dx ( x) dx
1
SC1 if obtained using symmetry condone use of symmetry Penalise A = e–1, B = e–1, or co-ords wrong way round, but condone labelling errors.
M1 E1
M1
1
= e0 – e–1 = 1 – e–1
(iv)
A1 B1
E1 [2]
[ ]
e x −1dx = e x −1
M1
E1 E1 [5]
C: f(1) = e1–1 = e0 = 1 g(1) = 1 + ln 1 = 1
(iii)
June 2009
[e ] o.e. or u = x – 1 [e ] x −1
substituting correct limits for x or u o.e. not e0 , must be exact.
parts: u = ln x, du/dx = 1/x, v = x, dv/dx = 1
A1 A1cao
(1 + ln x) dx
condone no ‘c’
B1ft
ft their ‘x ln x – x’ (provided ‘algebraic’)
DM1
substituting limits dep B1
= [ x + x ln x − x ]e−1 1
= [ x ln x)]e−1 1
–1
–1
= 1ln 1 – e ln(e ) = e–1 *
E1 [6]
www
M1
Must have correct limits
(v) Area =
1 0
1
f( x) d x − −1 g( x) d x e
1
f ( x ) dx −
0
= (1 – e–1) – e–1 2 =1– e
u
1
e −1
g ( x ) dx
A1cao
13
0.264 or better.
4753
Mark Scheme
June 2009
or Area OCB = area under curve – triangle = 1 – e–1 – ½ × 1 × 1 = ½ – e–1
M1
Area OAC = triangle – area under curve = ½ × 1 × 1 – e–1 = ½ – e–1 2 Total area = 2(½ – e–1) = 1 – e
A1cao
OCA or OCB = ½ – e–1
or
[2]
14
0.264 or better
4753
Mark Scheme
June 2009
Section B 9(i)
1 3
a=
2 (ii) d y = (3x − 1)2 x −2 x .3 dx (3x − 1) 2 6 x − 2 x − 3x 2 = (3x − 1) 2 3x 2 − 2 x = (3x − 1) 2 x(3x − 2) * = (3x − 1)2
dy = 0 when x(3x – 2) = 0 dx
(iii)
2 3
x = 0 or x = , so at P, x =
2 3
2 (2 / 3)2 4 when x = , y = = 3 3 × (2 / 3) − 1 9
dy = –0.1875 dx dy = 0.1633 when x = 0.8, dx Gradient increasing minimum when x = 0.6,
B1 [1]
or 0.33 or better
M1 A1
quotient rule
E1 [3]
www – must show both steps; penalise missing brackets.
M1
if denom = 0 also then M0
A1
o.e e.g. 0.6 , but must be exact
M1 A1cao
o.e e.g. 0.4 , but must be exact
.
.
B1
-3/16, or -0.19 or better
B1 E1 [7]
8/49 or 0.16 or better o.e. e.g. ‘from negative to positive’. Allow ft on their gradients, provided –ve and +ve respectively. Accept table with indications of signs of gradient.
B1
(u + 1)2 9 o.e. u
M1
× 13 (du)
M1
expanding
E1
condone missing du’s
B1
[u
M1
substituting correct limits, dep integration
x2 dx u = 3x − 1 du = 3dx (iv) 3x − 1 (u + 1) 2 1 9 du = u 3 1 (u + 1) 2 1 u 2 + 2u + 1 du = du = 27 u 27 u 1 1 = (u + 2 + ) du * 27 u
x2 dx 2 3 3x − 1 When x = 23 , u =1, when x = 1, u = 2 Area =
1
=
1 27
=
1 1 27 2
=
1 27
1
2
(u + 2 + u ) du 1
[u
2
]
2
+ 2u + ln u 1
[(2 + 4 + ln 2) − ( 12 + 2 + ln1)]
15
1 2
2
+ 2u + ln u
]
4753 =
Mark Scheme 1 27
7 + 2 ln 2 (3 12 + ln 2) = 54
A1cao [7]
16
June 2009
o.e., but must evaluate ln 1 = 0 and collect terms.
4754
Mark Scheme
June 2009
4754 (C4) Applications of Advanced Mathematics Section A 1
4 cos θ − sin θ = R cos(θ + α ) = R cos θ cos α − R sin θ sin α R cos α = 4, R sin α = 1
R2 = 12 + 42 = 17, R = 17 = 4.123 tan θ = ¼ θ = 0.245 17 cos(θ + 0.245) = 3 cos(θ + 0.245) = 317
θ + 0.245 = 0.756, 5.527 θ = 0.511, 5.282
2
dy = 3x 2 y dx dy = 3 x 2 dx y ln y = x3 + c when x = 1, y = 1, ln 1 = 1 + c c = -1 ln y = x3 – 1
y = ex
4
When x = 0, y = 4
3
3
M1
θ + 0.245 = arcos 3/√17 ft their R, α for method (penalise extra solutions in the range (-1))
M1
correct partial fractions
M1 A1 A1
substituting, equating coeffts or cover-up A=1 B = -1
B1
ln(x + 1) ft their A
B1
-½ ln(2x + 1) ft their B
A1 [7]
cao – must have c
M1
separating variables
2
0
= π 4 y − 12 y 2
condone absence of c c = -1 o.e. o.e.
B1
0 4
[
R = 17 = 4.123 tan θ = ¼ o.e. θ = 0.245
A1 [4]
x dy = π ( 4 − y ) dy
V =π
B1 M1 A1
A1 B1
−1
4
correct pairs
A1A1 [7]
x A B = − ( x + 1)(2 x + 1) x + 1 (2 x + 1) x = A(2x + 1) + B(x + 1) x = -1 -1 = -A A=1 x = -½ -½ = ½ B B = -1 1 1 x = − ( x + 1)(2 x + 1) x + 1 (2 x + 1) x 1 1 dx = − dx x + 1 (2 x + 1) ( x + 1)(2 x + 1) = ln(x + 1) – ½ ln(2x + 1) + c
M1
]
4
0
= π(16 – 8) = 8π
M1
must have integral, π, x² and dy s.o.i.
M1
must have π ,their (4−y), their numerical y limits
B1
1 2 4 y − 2 y
A1 [5]
17
4754
5
Mark Scheme dy = − a(1 + t 2 ) −2 .2t dt dx = 3at 2 dt dy dy / dt − 2at = = 2 dx dx / dt 3at (1 + t 2 ) 2
M1 A1
6
(1+t²)-2 × kt for method
B1
−2 * 3t (1 + t 2 ) 2 At (a, ½ a) , t = 1 1 gradient = −2 2 = − 6 3× 2
M1
M1 A1 [7]
finding t
cosec2 θ = 1 + cot2 θ 1 + cot2 θ – cot θ = 3 * cot2 θ – cot θ – 2 = 0 (cot θ – 2)(cot θ + 1) = 0 cot θ = 2, tan θ = ½ , θ = 26.57° cot θ = -1, tan θ = -1, θ = 135°
E1
clear use of 1 + cot² θ = cosec² θ
M1 A1 M1 A1 A1 [6]
factorising or formula roots 2, -1 cot = 1/tan used θ = 26.57° θ = 135° (penalise extra solutions in the range (-1))
=
June 2009
ft
E1
18
4754
Mark Scheme
June 2009
Section B 7(i)
−1 → AB = −2 0 0 1 r = 0 + λ 2 2 0
B1
B1 [2]
or equivalent alternative
1
(ii) n = 0
B1
1
1 1 0 . 2 1 0 1 cos θ = = 2 5 10
(iii)
B1 M1 M1
θ = 71.57°
A1 [5]
−1 −2 0 . −2 −1 −1 2 + 1 1 cos φ = = = 2 9 3 2 2
θ = 45° *
(iv) sin 71.57° = k sin 45° k = sin 71.57° / sin 45° = 1.34
0
−2
2
−1
correct vectors (any multiples) scalar product used finding invcos of scalar product divided by two modulae 72° or better
M1 A1 E1 [3]
ft their n for method ±1/√2 o.e. exact
M1 A1 [2]
ft on their 71.57° o.e.
M1
s.o.i.
M1 A1 A1
subst in x + z = -1
B1 [5]
www dep on μ = 1
(v) r = 0 + μ −2
x = -2μ, z = 2 − μ x + z = -1 -2θ + 2 – θ = -1 3θ = 3, θ = 1 point of intersection is (-2, -2, 1)
distance travelled through glass = distance between (0, 0, 2) and (-2, -2, 1) = (2 2 + 2 2 + 12 ) = 3 cm
19
4754
8(i)
Mark Scheme
(A)
(B)
360° ÷ 24 = 15° CB/OB = sin 15° CB = 1 sin 15° AB = 2CB = 2 sin 15° *
M1 E1 [2]
cos 30° = 1 – 2 sin2 15° cos 30° = 32
B1
= 1 – 2 sin2 15°
B1
3 2
2 sin2 15° = 1 –
sin215° =
sin 15° =
3 2
M1
= (2 – 3 )/2
June 2009
AB = 2AC or 2CB ∠AOC = 15° o.e.
simplifying
2− 3 4 2− 3 = 4
1 2
2− 3 *
(C) Perimeter = 12 × AB = 24 ×
1 2
E1 [4]
M1
2− 3
= 12 2 − 3 circumference of circle > perimeter of polygon
2π > 12 2 − 3
π > 6 2− 3
E1 [2] M1
(ii) (A) tan 15° = FE ÷ OF FE = tan 15° DE = 2FE = 2tan 15°
(B) tan 30 = tan 30 =
2 tan 15 2t = 2 1 − tan 15 1 − t 2 1
2t 1 = 2 3t = 1 − t 2 2 1− t 3
t2 + 2√3 t – 1 = 0 *
B1 M1
3
(C)
E1 [2]
E1 [3]
−2 3 ± 12 + 4 = 2− 3 2 circumference < perimeter 2π < 24( 2 − 3 ) t=
M1 A1 M1 using positive root E1 [4]
π < 12( 2 − 3 ) *
20
from exact working
4754
Mark Scheme
(iii) 6 2 − 3 < π < 12( 2 − 3 ) 3.106 < π < 3.215
B1 B1 [2]
21
June 2009
3.106, 3.215
4754
Mark Scheme
June 2009
Comprehension
1. 2.
1 4
× [3 + 1 + ( −1) + ( −2)] = 0.25 *
M1, E1
(i) b is the benefit of shooting some soldiers from the other side while none of yours are shot. w is the benefit of having some of your own soldiers shot while not shooting any from the other side. Since it is more beneficial to shoot some of the soldiers on the other side than it is to have your E1 own soldiers shot, b > w . (ii) c is the benefit from mutual co-operation (i.e. no shooting). d is the benefit from mutual defection (soldiers on both sides are shot). With mutual co-operation people don’t get shot, while they do with mutual defection. So c > d . E1
3.
4.
5.
1 × 2 + (−2) × (n − 2) = −1.999 or equivalent (allow n, n+2) n n = 6000 so you have played 6000 rounds.
M1, A1 A1
No. The inequality on line 132, b + w < 2c, would not be satisfied since 6 + (-3) > 2 × 1. b + w < 2c and subst No, 3 > 2 o.e.
M1 A1
(i) Round
You
Opponent
1 2 3 4 5 6 7 8 …
C D C D C D C D …
D C D C D C D C …
Your score -2 3 -2 3 -2 3 -2 3 …
Opponent’s score 3 -2 3 -2 3 -2 3 -2 …
M1 Cs and Ds in correct places, A1 C=-2, A1 D=3
(ii) 6.
1 2
× [3 + (−2)] = 0.5
DM1 A1ft their 3, -2
(i) All scores are increased by two points per round
B1
(ii) The same player wins. No difference/change. The rank order of the players remains the same. B1 7.
(i) They would agree to co-operate by spending less on advertising or by sharing equally. (ii) Increased market share (or more money or more customers).
22
B1 DB1
4755
Mark Scheme
June 2009
4755 (FP1) Further Concepts for Advanced Mathematics Section A 1(i)
(ii)
2
M −1 =
1 2 1 11 −3 4
1 2 1 49 x 1 198 = = 11 −3 4 100 y 11 253 x = 18, y = 23
M1 A1 [2]
Dividing by determinant
M1
Pre-multiplying by their inverse
A1(ft) A1(ft) [3]
z 3 + z 2 − 7 z − 15 = ( z − 3) ( z 2 + 4 z + 5 )
B1 M1 A1 M1
−4 ± 16 − 20 2 z = −2 + j and z = −2 − j
z2 + 4z + 5 = 0 z =
A1 [5]
Show z = 3 is a root; may be implied Attempt to find quadratic factor Correct quadratic factor Use of quadratic formula or other valid method Both solutions
3(i)
(ii)
B1 B1 [2]
Asymptote at x = -4 Both branches correct
2 = x + 3 x 2 + 7 x + 10 = 0 x+4
M1
x = −2 or x = −5
A1
Attempt to find where graphs cross or valid attempt at solution using inequalities Correct intersections (both), or -2 and -5 identified as critical values
x ≥ −2 or − 4 > x ≥ −5
A1 A2
[5] 4
2w − 6w + 3w = − 12 w=
1 2
roots are 1, -3, −
3 2
9 q = αβγ = − q = 9 2 2
p = αβ + αγ + βγ = −6 p = −12 2
M1 A1
Use of sum of roots – can be implied
A1 M1
Correct roots seen Attempt to use relationships between roots s.c. M1 for other valid method One mark each for p = -12 and q = 9
A2(ft) [6] 23
x ≥ −2 −4 > x ≥ −5 s.c. A1 for −4 ≥ x ≥ −5 or − 4 > x > −5
4755
5(i)
Mark Scheme
1 1 5r + 3 − 5r + 2 − ≡ 5r − 2 5r + 3 ( 5r + 3)( 5r − 2 ) ≡
(ii)
M1
Attempt to form common denominator
A1 [2]
Correct cancelling
B1
First two terms in full
B1
Last term in full
M1
Attempt to cancel terms
5
( 5r + 3)( 5r − 2 )
r =1 ( 5r − 2 ) ( 5r + 3 ) 1 1 1 1 1 1 − + − + − + ... 1 3 8 8 13 13 18 = 5 1 1 1 1 + 5n − 7 − 5n − 2 + 5n − 2 − 5n + 3 1
30
( 5r − 2 )( 5r + 3)
=
June 2009
=
1
30
5
1
−
1
r =1
1 1
1 n − = 5 3 5 n + 3 3 ( 5n + 3 )
A1
[4] 6
When n = 1,
1 2
n ( 7 n − 1) = 3, so true for n = 1
Assume true for n = k
= = =
= =
1 2 1 2 1
2
Assume true for n = k
M1
Add ( k + 1) th term to both sides
M1
Valid attempt to factorise
A1
c.a.o. with correct simplification
E1
Dependent on previous E1 and immediately previous A1 Dependent on B1 and both previous E marks
k ( 7 k − 1) + (7 ( k + 1) − 4)
[ k ( 7 k − 1) + (14 ( k + 1) − 8)] 7 k 2 + 13k + 6
2 1 2 1
E1 1
k ( 7 k − 1) 2 3 + 10 + 17 + ..... + (7 ( k + 1) − 4)
3 + 10 + 17 + ..... + (7 k − 4) =
B1
( k + 1)( 7 k + 6 ) ( k + 1) ( 7 ( k + 1) − 1)
But this is the given result with k + 1 replacing k. Therefore if it is true for k it is true for k + 1. Since it is true for n = 1, it is true for n = 1, 2, 3 and so true for all positive integers.
E1
[7] Section A Total: 36
24
4755
Mark Scheme
Section B 7(i) 5 ( 0 , 10 ) , ( −2 , 0 ) , , 0 3
(ii) x=
−1 2
, x =1, y =
June 2009
B1 B1 B1 [3]
3
B1 B1 B1 [3]
2
+
(iii)
(iv)
3 2 (e.g. consider x = 100 ) − 3 Large negative x, y → 2 (e.g. consider x = −100 ) Large positive x, y →
M1 B1 B1 [3]
Curve 3 branches of correct shape Asymptotes correct and labelled Intercepts correct and labelled
B1 B1 B1 [3]
25
Clear evidence of method required for full marks
4755
8 (i)
Mark Scheme
z − ( 4 + 2 j) = 2
June 2009
B1 B1 B1
Radius = 2 z − ( 4 + 2 j) or z − 4 − 2 j All correct
[3] (ii)
arg ( z − ( 4 + 2 j) ) = 0
B1 B1 B1
Equation involving the argument of a complex variable Argument = 0 All correct
[3] (iii)
a = 4 − 2cos b = 2 + 2sin
π 4
π 4
(
M1
=4− 2
Valid attempt to use trigonometry involving
=2+ 2
)
P =4− 2 + 2+ 2 j
A2
π 4
, or coordinate geometry
1 mark for each of a and b s.c. A1 only for a = 2.59, b = 3.41
[3] (iv)
3 π > arg ( z − ( 4 + 2 j) ) > 0 4 and z − ( 4 + 2 j) < 2
B1 B1 B1
[3]
26
arg ( z − ( 4 + 2 j) ) > 0
3 arg ( z − ( 4 + 2 j) ) < π 4 z − ( 4 + 2 j) < 2 Deduct one mark if only error is use of inclusive inequalities
4755
9(i)
Mark Scheme
Matrix multiplication is associative
B1 [1]
3 0 0 1 MN = 0 2 1 0 0 3 MN = 2 0 −2 0 QMN = 0 3
(ii)
June 2009
M1
Attempt to find MN or QM
A1
0 −2 or QM = 3 0
A1(ft) [3]
M is a stretch, factor 3 in the x direction, factor 2 in the y direction.
B1 B1
N is a reflection in the line y = x.
B1
Q is an anticlockwise rotation through 90° about the origin.
B1
Stretch factor 3 in the x direction Stretch factor 2 in the y direction
[4]
(iii)
−2 0 1 1 2 −2 −2 −4 = 0 3 2 0 2 6 0 6
M1 A1(ft)
Applying their QMN to points. Minus 1 each error to a minimum of 0.
B2
Correct, labelled image points, minus 1 each error to a minimum of 0. Give B4 for correct diagram with no workings.
[4]
Section B Total: 36 Total: 72
27
4756
Mark Scheme
June 2009
4756 (FP2) Further Methods for Advanced Mathematics x2 x3 x4 x5 + − + ... 2 3 4 5 2 3 4 x x x x5 − − − ... ln(1 – x) = − x − 2 3 4 5 1+ x ln = ln(1 + x) – ln(1 – x) 1− x 2x3 2x5 + ... = 2x + 3 5 Valid for -1 < x < 1
1(a)(i) ln(1 + x) = x −
(ii)
Series for ln(1 – x) as far as x5 s.o.i.
M1
Seeing series subtracted
A1
1+ x =3 1− x 1 + x = 3(1 – x) 1 + x = 3 – 3x 4x = 2 x = 12
ln 3 ≈ 2 × 23 × ( 12 ) + 52 × ( 12 ) 3
B1
5
B1 4
Inequalities must be strict
M1
Correct method of solution
A1
B2 for x =
M1
Substituting their x into their series in (a)(i), even if outside range of validity. Series must have at least two terms SR: if >3 correct terms seen in (i), allow a better answer to 3 d.p. Must be 3 decimal places
= 1 + 121 + 801 = 1.096 (3 d.p.) (b)(i)
A1 4
1 2
stated
y 1
G1 G1 G1
0.5
–1
–0.5
0.5
1
3 M1
(ii) r + y = r + r sin θ a = r(1 + sin θ) = × (1 + sin θ) 1 + sin θ =a r=a–y
r(0) = a, r(π/2) = a/2 indicated Symmetry in θ = π/2 Correct basic shape: flat at θ = π/2, not vertical or horizontal at ends, no dimple Ignore beyond 0 ≤ θ ≤ π Using y = r sin θ
A1 (AG)
x2 + y2 = (a – y)2 x2 + y2 = a2 – 2ay + y2 2ay = a2 – x2 a2 − x2 y= 2a
M1 A1
Using r2 = x2 + y2 in r + y = a Unsimplified
A1
A correct final answer, not spoiled
5
28
16
4756
Mark Scheme
3− λ 2 (i) M − λ I = 0 2
−2 −1 − λ 0 0 1 − λ 1
det(M – λI) = (3 – λ)[(–1 – λ)(1 – λ)] + 2[2(–1 – λ)] = (3 – λ)(λ2 – 1) + 4(–1 – λ) λ3 – 3λ2 + 3λ + 7 = 0 det M = -7
M1 A1
Attempt at det(M – λI) with all elements present. Allow sign errors Unsimplified. Allow signs reversed. Condone omission of = 0
B1 3
(ii) f (λ) = λ3 – 3λ2 + 3λ + 7 f (-1) = -1 – 3 – 3 + 7 = 0 -1 eigenvalue
B1
f (λ) = (λ + 1)(λ2 – 4λ + 7) λ2 – 4λ + 7 = (λ – 2)2 + 3 ≥ 3 so no real roots (M – λI)s = 0, λ = -1
June 2009
M1 A1
Showing -1 satisfies a correct characteristic equation Obtaining quadratic factor www (M – λI)s = (λ)s M0 below
4 1 −2 x 0 0 0 0 y = 0 2 0 2 z 0
4x + y – 2z = 0 2 x + 2z = 0 x = -z y = 2z – 4 x = 2z + 4z = 6z
M1 M1
−1 s = 6 1 3 1 −2 x −0.1 0 −1 0 y = 0.6 2 0 1 z 0.1
x = 0.1, y = -0.6, z = -0.1
A1
Or any non-zero multiple
M1
Solution by any method, e.g. use of multiple of s, but M0 if s itself quoted without further work Give A1 if any two correct
A2 9 B1
(iii) C-H: a matrix satisfies its own characteristic equation M3 – 3M2 + 3M + 7I = 0 M3 = 3M2 – 3M – 7I
Obtaining equations relating x, y and z Obtaining equations relating two variables to a third. Dep. on first M1
Idea of λ ↔ M
B1 (AG) Must be derived www. Condone omitted I M1 Multiplying by M–1 A1 o.e.
M2 = 3M – 3I – 7M–1 M–1 = − 17 M 2 + 73 M − 73 I
4 3 1 −2 3 1 −2 5 2 (iv) M = 0 −1 0 0 −1 0 = 0 2 0 1 2 0 1 8 5 2 −8 3 1 −2 1 1 3 3 − 0 1 0 + 0 −1 0 − 0 7 7 2 0 1 7 0 8 2 −3 1 1 2 7 1 1 2 7 7 1 = 0 −1 0 or 0 −7 0 7 − 2 − 2 3 −2 −2 3 7 7 7
2 −8 1 0 2 −3 0 0 1 0 0 1
29
M1
Correct attempt to find M2
M1
Using their (iii)
A1
SC1 for answer without working
4756
Mark Scheme
OR Matrix of cofactors: −1 −1 Adjugate matrix 0 7 2 2
−1 0 2 −1 7 2 −2 0 −3 −2 0 : det M = –7 −3
June 2009
M1
Finding at least four cofactors
M1
Transposing and dividing by determinant. Dep. on M1 above
3 G1
3(a)(i) 1
19 Correct basic shape (positive gradient, through (0, 0))
–1
1 M1
y = arcsin x sin y = x dy = cos y dx dy 1 1 = = dx cos y 1− x2
A1 A1
Positive square root because gradient positive
1
(ii)
0
=
4 M1
1
x dx = arcsin 2 2 0 2− x 1
A1
π
A1
4 jθ
3 jθ
3 M1 M1
5 jθ
(b) C + jS = e + e + e + ... This is a geometric series 1 3
1 9
with first term a = e jθ , common ratio r = 13 e 2 jθ Sum to infinity = =
= =
B1
jθ
sin y = and attempt to diff. both sides dy Or cos y =1 dx www. SC1 if quoted without working Dep. on graph of an increasing function arcsin function alone, or any sine substitution x 2
, or 1 dθ www without limits
Evaluated in terms of π
A1
Forming C + jS as a series of powers Identifying geometric series and attempting sum to infinity or to n terms Correct a and r
A1
Sum to infinity
jθ
a e 3e = (= ) 1 − r 1 − 13 e 2 jθ 3 − e 2 jθ
3e jθ 3 − e −2 jθ × 3 − e 2 jθ 3 − e −2 jθ
M1*
9e jθ − 3e − jθ 9 − 3e −2 jθ − 3e 2 jθ + 1 9 ( cos θ + j sin θ ) − 3 ( cos θ − j sin θ )
M1 M1
10 − 3 ( cos 2θ − j sin 2θ ) − 3 ( cos 2θ + j sin 2θ )
30
Multiplying numerator and denominator by 1 − 13 e −2 jθ o.e. Or writing in terms of trig functions and realising the denominator Multiplying out numerator and denominator. Dep. on M1* Valid attempt to express in terms of trig functions. If trig functions used from start, M1 for using the compound angle formulae and
4756
Mark Scheme
June 2009
Pythagoras Dep. on M1* =
6 cos θ + 12 j sin θ 10 − 6 cos 2θ
C=
A1
6 cos θ 10 − 6 cos 2θ
M1
3cos θ 5 − 3cos 2θ 6sin θ S= 5 − 3cos 2θ
=
Equating real and imaginary parts. Dep. on M1*
A1 (AG) A1
o.e.
11 4 (i) cosh u =
19
eu + e − u 2
e 2u + 2 + e −2u 2 e 2u + e −2u 2 2 cosh u – 1 = 2
2 cosh2 u =
B1
(e
B1
cosh 2u =
= cosh 2u
u
+ e − u ) = e 2 u + 2 + e −2 u 2
e 2u + e −2u 2
B1 (AG) Completion www 3
(ii) x = arcsinh y sinh x = y e x − e− x 2
M1
Expressing y in exponential form ( 12 , – must be correct)
ex = y ± y 2 + 1
M1
Take + because ex > 0
B1
Reaching ex by quadratic formula or completing the square. Condone no ± Or argument of ln must be positive
y=
e2x – 2yex – 1 = 0 (ex – y)2 – y2 – 1 = 0 (ex – y)2 = y2 + 1 ex – y = ± y 2 + 1
x = ln(y + y + 1 ) 2
dx = 2 cosh u du
(iii) x = 2 sinh u
x 2 + 4 dx =
A1 (AG) Completion www but independent of B1 4 dx and substituting for all elements M1 du
4 sinh 2 u + 4 × 2 cosh u du
A1
Substituting for all elements correctly
4 cosh u du = 2 cosh 2u + 2 du
M1
Simplifying to an integrable form
= sinh 2u + 2u + c
A1
Any form, e.g. 12 e2u − 12 e−2u + 2u Condone omission of + c throughout
M1
Using double ‘angle’ formula and attempt to express cosh u in terms of x
2
=
= 2 sinh u cosh u + 2u + c = x 1+
=
1 2
x2 x + 2 arcsinh + c 4 2
x x 4 + x 2 + 2 arcsinh + c 2
A1 (AG) Completion www 31
4756
Mark Scheme
June 2009
6 (iv) t2 + 2t + 5 = (t + 1)2 + 4 1
B1
Completing the square
M1
Simplifying to an integrable form, by substituting x = t + 1 s.o.i. or complete alternative method Correct limits consistent with their method seen anywhere
1
t 2 + 2t + 5 dt =
−1
(t + 1) 2 + 4 dt
−1 2
=
x 2 + 4 dx
0
A1 2
x = 12 x 4 + x 2 + 2 arcsin h 20
M1
(v) = 8 + 2 arcsinh 1 = 2 2 + 2 ln(1 + 2 ) = 2(ln(1 + 2 ) + 2 )
A1 (AG) Completion www. Condone 8 etc. 5 18
5 (i) If a = 1, angle OCP = 45° so P is (1 – cos 45°, sin 45°) 1
P (1 –
,
2
1
M1
)
2
A1 (AG) Completion www
OR Circle (x – 1)2 + y2 = 1, line y = -x + 1 (x – 1)2 + (-x + 1)2 = 1 1
x=1±
2 1
Q (1 +
2
M1
and hence P ,-
1 2
Using (iii) or otherwise reaching the result of integration, and using limits
Complete algebraic method to find x
A1
)
B1
3 a
(ii) cos OCP =
a2 + 1 1
sin OCP =
a2 + 1
P is (a – a cos OCP, a sin OCP) a2 a Pa − , a2 +1 a2 +1 OR Circle (x – a)2 + y2 = a2, line y = −
M1
Attempt to find cos OCP and sin OCP in terms of a
A1
Both correct
A1 (AG) Completion www 1 x +1 a
2
1 (x – a) + − x + 1 = a2 a 2
2
x=
2a +
2 2 ± 2a + a a 1 2 1 + 2 a
1 − 4 1 + 2 a
32
M1
Complete algebraic method to find x
A1
Unsimplified
4756
Mark Scheme
x =a±
a2 a2 +1
and hence P
June 2009
A1
a2 a Qa + ,− a2 +1 a2 +1
B1 4
(iii)
1
0.5
–5
G1 G1 G1 G1ft
5
–0.5
–1
As a → ∞, P → (0, 1) As a → –∞, y-coordinate of P → –1 a 2
a +1
→
B1 B1 M1 A1 8
a = −1 as a → –∞ −a
(iv) POQ = 90° Angle in semicircle Loci cross at 90°
B1 B1 B1 3
33
Locus of P (1st & 3rd quadrants) through (0, 0) Locus of P terminates at (0, 1) Locus of P: fully correct shape Locus of Q (2nd & 4th quadrants: dotted) reflection of locus of P in yaxis Stated separately Stated Attempt to consider y as a → –∞ Completion www
o.e. 18
4757
Mark Scheme
June 2009
4757 Further Pure 3 1 (i) Putting x = 0, -3y + 10z = 6, -4y −2z = 8 y = -2, z = 0 8 3 Direction is given by − 3 × − 4 10 − 2
46 = 46 − 23
=
or finding a second point
5 M1
− 1 2 14 ⋅ 5 5 14 2
2
2 + 5 + 14
2
138 46 = = 9.2 15 5
6 AB × d = 15 = 6 2 + 15 2 + 42 2 42
Distance is
M1
A1 ft
2 = 3 5 14
7 2 6 AB × d = -14 × 2 = 15 4 − 1 42
− 1 0 Distance is 12 − − 2 ⋅ nˆ = 5 0
(iii)
Finding coords of a point on the line or (2, 0, -1), (1, -1, - 12 ) etc
A1
0 2 Equation of L is r = − 2 + λ 2 0 − 1 (ii)
M1 A1
AB × d d
=
=
6 2 + 15 2 + 42 2 2
2
2
2 + 2 +1 45 = 15 3
Evaluating AB × d
A2 ft
Give A1 ft if just one error
M1
Appropriate scalar product
A1 ft
Fully correct expression
A1 6 M1
For AB × d
M1
Evaluating magnitude
M1 A1ft
In this part, M marks are dependent on previous M marks
A1 5
34
Dependent on M1M1 Accept any form Condone omission of ‘r =’
4757
(iv)
Mark Scheme −1 k + 1 6 2 At D, 12 + λ −12 = −2 + μ 2 5 −3 9 −1
M1
12 − 12λ = −2 + 2 μ 5 − 3λ = 9 − μ
A1 ft M1 M1
λ = 13 , μ = 5
Condone use of same parameter on both sides Two equations for λ and μ Obtaining λ and μ (numerically) Give M1 for λ and μ in terms of k Equation for k
M1
−1 + 13 (k + 1) = 6 + 10 k = 50
A1
D is (6 + 2μ, -2 + 2μ, 9 − μ) i.e. (16, 8, 4)
M1 A1 8
Alternative solutions for Q1 1 (i) e.g. 23x – 23y = 46 x = t, y = t − 2 3t − 4(t − 2) − 2z = 8 x = t, y = t − 2, z = - 12 t
(ii)
June 2009
Obtaining coordinates of D
M1A1 Eliminating one of x, y, z M1 A1 ft A1 5
− 1 + 7μ 2 λ PQ = 12 − 14μ − − 2 + 2 λ 5 + 4μ -λ
M1 A1 ft
PQ ⋅ d = PQ ⋅ AB = 0 A1 ft 2( −1 + 7μ − 2 λ) + 2(14 − 14μ − 2 λ) − (5 + 4μ + λ) =0 7(−1 + 7μ − 2 λ) − 14(14 − 14μ − 2 λ) + 4(5 + 4μ + λ) M1 =0 27 47 λ= , μ= 25 75 2
2
A1 6
2
92 230 644 PQ = + + = 9.2 75 75 75
(iii)
A1 ft
6 + 2λ + 1 2 AX ⋅ d = − 2 + 2 λ − 12 ⋅ 2 = 0 9 − λ − 5 − 1
M1
2(7 + 2 λ) + 2(2 λ − 14) − (4 − λ) = 0 λ=2 11 AX = − 10 2
A1ft
AX = 112 + 10 2 + 2 2 = 15
M1 A1 5
M1
35
Two equations for λ and μ Expression for shortest distance
4757
Mark Scheme
June 2009
(iv) 7 k + 1 2 −14 ⋅ 4
−12 × 2 −3 −1
=0
7 18 −14 ⋅ k − 5 = 0 4 2k + 26 126 − 14k + 70 + 8k + 104 = 0 k = 50 −1 51 6 2 At D, 12 + λ −12 = −2 + μ 2 5 −3 9 −1
M1
Appropriate scalar triple product equated to zero
M1 A1
Equation for k
M1
Condone use of same parameter on both sides
−1 + 51λ = 6 + 2μ 12 − 12λ = −2 + 2μ 5 − 3λ = 9 − μ
A1 ft M1
Two equations for λ and μ Obtaining λ and μ
M1 A1 8
Obtaining coordinates of D
M1 A2
Partial differentiation Give A1 if just one minor error
λ = 13 , μ = 5
D is (6 + 2 μ , − 2 + 2μ , 9 − μ ) i.e. (16, 8, 4)
2(i) ∂z
= 3( x + y )3 + 9 x( x + y ) 2 − 6 x 2 + 24
∂x ∂z = 9 x( x + y)2 ∂y
A1 4
(ii) At stationary points,
∂z ∂z = 0 and =0 ∂x ∂y
M1
9 x ( x + y ) 2 = 0 x = 0 or y = − x
If x = 0 then 3 y 3 + 24 = 0 y = −2 ; one stationary point is (0, − 2, 0) If y = − x then −6 x 2 + 24 = 0 x = ±2 ; stationary points are (2, − 2, 32) and (−2, 2, − 32)
(iii)
At P (1, − 2, 19) ,
∂z ∂z = 24 , =9 ∂x ∂y
B1
1 24 Normal line is r = −2 + λ 9 19 −1
(iv)
M1 A1A1 M1 A1 A1 If A0A0, give A1 for x = ±2 7
M1 A1 ft 3
∂z ∂z δx+ δ y ∂x ∂y = 24 δ x + 9 δ y
δz ≈
M1 A1 ft
3h ≈ 24k + 9h
M1
36
For normal vector (allow sign error) Condone omission of ‘r =’
4757
Mark Scheme
A1 4
k ≈ − 14 h
M2 A1 ft A1
OR Tangent plane is 24 x + 9 y − z = −13 24(1 + k ) + 9(−2 + h) − (19 + 3h) ≈ −13 k ≈ − 14 h
(v) ∂z ∂x
June 2009
= 27 and
∂z =0 ∂y
M1
(Allow M1 for
∂z = −27 ) ∂x
9 x ( x + y ) 2 = 0 x = 0 or y = − x
If x = 0 then 3 y 3 + 24 = 27 y = 1, z = 0 ; point is (0 , 1, 0)
M1 A1
d =0
A1 M1
If y = − x then −6 x 2 + 24 = 27 x 2 = − 12 ; there are no other points
A1 6
3(i) dx 2 dy 2 2 2 dθ + dθ = [a(1 + cos θ )] + (a sin θ )
M1 A1
2
Forming dx + dy dθ dθ
2
= a 2 (2 + 2 cos θ ) = 4a 2 cos 2 12 θ
s = 2a cos
1 θ 2
M1 M1 A1 A1(AG) 6
dθ
= 4a sin 12 θ + C s = 0 when θ = 0 C = 0
(ii) dy dx
= =
a sin θ a (1 + cos θ )
ρ=
Correctly obtained (+C not needed) Dependent on all previous marks
M1
2sin 12 θ cos 12 θ 2a cos 2 12 θ
M1 A1
= tan 12 θ
ψ = 12 θ , and so s = 4a sinψ
(iii)
Using half-angle formula Integrating to obtain k sin 12 θ
Using half-angle formulae
A1 4 M1 A1 ft
ds = 4a cosψ dψ
= 4a cos 12 θ
Differentiating intrinsic equation
A1(AG) 3
OR ρ= =
( 4a
2
cos 2 12 θ
)
3
2
M1 A1 ft
a (1 + cos θ )(a cos θ ) − (−a sin θ )(a sin θ ) 8a 3 cos3 12 θ a 2 (1 + cos θ )
=
8a 3 cos3 12 θ 2a 2 cos 2 12 θ
= 4a cos 12 θ
A1(AG)
37
Correct expression for ρ or κ
4757
Mark Scheme
(iv) When θ = 2 π , ψ = 1 π , x = a( 2 π + 1 3), y = 3 a 2 3 3 3 2 ρ = 2a − sinψ − 12 3 nˆ = = 1 cosψ 2 a ( 2 π + 1 3) − 1 3 2 + 2a 21 c= 3 3 a 2 2
Centre of curvature is ( a( 23 π − 12 3) ,
B1 M1 A1
5 2
a)
A1A1 Accept (1.23a , 2.5a) 6 M1
π
M1
Correct integral expression in any form (including limits; may be implied by later working) Obtaining an integrable form
M1
Obtaining k sin 3 12 θ
A1 ft
= 2π a (1 − cos θ ) 2a cos 12 θ dθ 0
π
= 8π a 2 sin 2 12 θ cos 12 θ dθ 0
=
16 π a2 3
Obtaining a normal vector Correct unit normal (possibly in terms of θ )
M1
(v) Curved surface area is 2π y ds
June 2009
π
sin 3 12 θ 0
= 163 π a 2
or equivalent
A1 5
4 (i) In G, 32 = 2, 33 = 6, 34 = 4, 35 = 5, 36 = 1
M1
All powers of an element of order 6
A1
All powers correct in both groups
[ or 52 = 4, 53 = 6, 54 = 2, 55 = 3, 56 = 1 ] In H, 52 = 7 , 53 = 17 , 54 = 13, 55 = 11, 56 = 1 [ or 112 = 13, 113 = 17 , 114 = 7 , 115 = 5, 116 = 1 ] B1 B1 4
G has an element 3 (or 5) of order 6 H has an element 5 (or 11) of order 6 {1, 6} (ii) {1, 2, 4}
(iii) G 1 2 3 4 5 6
H ↔ 1 ↔ 7 ↔ 5 ↔ 13 ↔ 11 ↔ 17
G
OR
1 2 3 4 5 6
B1 B2 3
Ignore {1} and G Deduct 1 mark (from B1B2) for each proper subgroup in excess of two
B4 4
Give B3 for 4 correct, B2 for 3 correct, B1 for 2 correct
M1
Evaluating e.g. ad(1) (one case sufficient; intermediate value must be shown) For ad = c correctly shown Evaluating e.g. da(1) (one case sufficient; no need for any working)
H ↔ 1 ↔ 13 ↔ 11 ↔ 7 ↔ 5 ↔ 17
ad(1) = a(3) = 1 (iv) ad(2) = a(2) = 3 ad(3) = a(1) = 2, so ad = c
A1 da(1) = d(2) = 2 da(2) = d(3) = 1
M1
38
4757
Mark Scheme
da(3) = d(1) = 3, so da = f
June 2009
A1 4
(v) S is not abelian; G is abelian
B1 1
or S has 3 elements of order 2; G has 1 element of order 2 or S is not cyclic etc
(vi)
Element Order
a 3
b 3
c 2
d 2
e 1
f 2
B4
4 (vii) {e, c} {e, d} {e, f} {e, a, b}
Give B3 for 5 correct, B2 for 3 correct, B1 for 1 correct
B1 B1 B1 B1 4
Ignore { e } and S If more than 4 proper subgroups are given, deduct 1 mark for each proper subgroup in excess of 4
B2 2
Give B1 for two columns correct
M1 A2 3
Using P13 (or P14 ) Give A1 for 2 probabilities correct (Max A1 if not at least 3dp) Tolerance ±0.0001
Pre-multiplication by transition matrix 5 (i) 0 0.1 0 0.3 0.7 0.8 0 0.6 P= 0.1 0 1 0.1 0.2 0.1 0 0
(ii)
0.6 0.0810 0.4 0.5684 P13 = 0 0.2760 0 0.0746
(iii) 0.5684 × 0.8 + 0.2760
M1M1 For 0.5684 × 0.8 and 0.2760 A1 ft Accept 0.73 to 0.7312 3
= 0.731
(iv)
0.6 . . 30 0.4 , = P 0 0.4996 0 .
0.6 . . 31 0.4 P = 0 0.5103 0 .
M1 A1
Level 32
Finding P(C) for some powers of P For identifying P31
A1 3
(v) Expected number of levels including the next 1 change of location is =5 0.2
Expected number of further levels in B is 4
39
M1
For 1/(1 − 0.8) or 0.8 / (1 − 0.8)
A1
For 5 or 4
A1 3
For 4 as final answer
4757
(vi)
Mark Scheme 0 0.1 0 0.3 0.7 0.8 0 0.6 Q= 0.1 0 0.9 0.1 0.2 0.1 0.1 0 0.0916 0.0916 0.0916 0.6183 0.6183 0.6183 n Q → 0.1908 0.1908 0.1908 0.0992 0.0992 0.0992
0.0916 0.6183 0.1908 0.0992
June 2009
B1
Can be implied
M1
Evaluating powers of Q or Obtaining (at least) 3 equations from Qp = p Limiting matrix with equal columns or Solving to obtain one equilib prob or M2 for other complete method Give A1 for two correct (Max A1 if not at least 3dp) Tolerance ±0.0001
M1
A: 0.0916 B: 0.6183 C: 0.1908 D: 0.0992
0.1 a 0.3 0.11 0.11 0.7 0.8 b 0.6 0.75 = 0.75 0.1 0 c 0.1 0.04 0.04 0.2 0.1 d 0 0.1 0.1
A2 5
(vii) 0
M1 A1
0.075 + 0.04a + 0.03 = 0.11 0.077 + 0.6 + 0.04b + 0.06 = 0.75 0.011 + 0.04c + 0.01 = 0.04 0.022 + 0.075 + 0.04d = 0.1
M1
0.11 0.75 Transition matrix and 0.04 0.1
Forming at least one equation
or a + b + c + d = 1
a = 0.125, b = 0.325, c = 0.475, d = 0.075
A2 5
Give A1 for two correct
B2 2
Give B1 for two rows correct
M1 A2 3
Using P13 (or P14 ) Give A1 for 2 probabilities correct (Max A1 if not at least 3dp) Tolerance ±0.0001
Post-multiplication by transition matrix 5 (i) 0 0.7 0.1 0.2 0.1 0.8 0 0.1 P= 0 0 1 0 0.3 0.6 0.1 0
(ii)
( 0.6
0.4 0 0 ) P13
= ( 0.0810 0.5684 0.2760 0.0746 )
(iii) 0.5684 × 0.8 + 0.2760
M1M1 For 0.5684 × 0.8 and 0.2760 A1 ft Accept 0.73 to 0.7312 3
= 0.731
(iv)
( 0.6 ( 0.6
0.4 0 0 ) P 30 = (. . 0.4996 .)
M1
0.4 0 0 ) P
A1
31
= (. . 0.5103 .)
A1 3
Level 32
(v) Expected number of levels including the next change of location is
Finding P(C) for some powers of P For identifying P 31
1 =5 0.2
Expected number of further levels in B is 4
40
M1
For 1/(1 − 0.8) or 0.8 / (1 − 0.8)
A1
For 5 or 4
A1 3
For 4 as final answer
4757
(vi)
(vii)
Mark Scheme 0 0.7 0.1 0.2 0.1 0.8 0 0.1 Q= 0 0 0.9 0.1 0.3 0.6 0.1 0 0.0916 0.6183 0.1908 0.0916 0.6183 0.1908 n Q → 0.0916 0.6183 0.1908 0.0916 0.6183 0.1908
June 2009
B1
Can be implied
0.0992 0.0992 0.0992 0.0992
M1
Evaluating powers of Q or Obtaining (at least) 3 equations from pQ = p Limiting matrix with equal rows or Solving to obtain one equilib prob or M2 for other complete method
A: 0.0916 B: 0.6183 C: 0.1908 D: 0.0992
A2 5
Give A1 for two correct (Max A1 if not at least 3dp) Tolerance ±0.0001
M1
Transition matrix and
0 0.7 0.1 0.2 0.1 0.8 0 0.1 0.11 0.75 0.04 0.1 ( ) a b c d 0.3 0.6 0.1 0 = ( 0.11 0.75 0.04 0.1)
M1
( 0.11
0.75 0.04 0.1)
A1
0.075 + 0.04a + 0.03 = 0.11 0.077 + 0.6 + 0.04b + 0.06 = 0.75 0.011 + 0.04c + 0.01 = 0.04 0.022 + 0.075 + 0.04d = 0.1
M1
A2 5
a = 0.125, b = 0.325, c = 0.475, d = 0.075
41
Forming at least one equation
or a + b + c + d = 1 Give A1 for two correct
4758
Mark Scheme
June 2009
4758 Differential Equations 1(i)
α 2 + 25 = 0 α = ±5 CF y = A cos 5t + B sin 5t PI y = at cos 5t + bt sin 5t y = a cos 5t − 5at sin 5t + b sin 5t + 5bt cos 5t y = −10a sin 5t − 25at cos 5t + 10b cos 5t − 25bt sin 5t In DE 10b cos 5t − 10a sin 5t = 20 cos 5t b = 2, a = 0 PI y = 2t sin 5t GS y = 2t sin 5t + A cos 5t + B sin 5t
M1 A1 F1 B1
Auxiliary equation CF for their roots
M1 M1
Differentiate twice Substitute and compare coefficients
A1 F1
8 (ii)
t = 0, y = 1 A = 1 y = 2 sin 5t + 10t cos 5t − 5 A sin 5t + 5 B cos 5t t = 0, y = 0 B = 0 y = 2t sin 5t + cos 5t
B1 M1 M1 A1
From correct GS Differentiate Use condition on y
4 (iii)
Curve through (0, 1) Curve with zero gradient at (0, 1) Oscillations Oscillations with increasing amplitude
B1 B1 B1 B1
4 (iv)
y = 2 sin 5t , y = 10 cos 5t , y = −50 sin 5t y + 2 y + 25 y = −50 sin 5t + 20 cos 5t + 50 sin 5t = 20 cos 5t 2 α + 2α + 25 = 0 α = −1 ± i 24 CF e −t C cos 24t + D sin 24t
(
GS y = 2 sin 5t + e
−t
(C cos
)
24t + D sin 24t
)
M1 E1 M1 A1 F1
Substitute into DE
F1
Their PI + their CF with two arbitrary constants
Auxiliary equation CF for their complex roots
6
(v)
Oscillations of amplitude 2
B1
Compared to unbounded oscillations in first model
B1
or bounded oscillations; or both oscillate o.e. or one bounded, one unbounded 2
2(i)
dy 3 sin x + y= 2 dx x x
M1
Rearrange
M1 A1
Attempting integrating factor
3
dx I =e x = e 3 ln x 42
4758
Mark Scheme
= x3 d 3 x y = x sin x dx x 3 y = x sin x dx = − x cos x + cos x dx
( )
= − cos x + sin x + A − x cos x + sin x + A y= x3
June 2009
A1
Correct and simplified
M1
Multiply and recognise derivative
M1
Integrate
A1 A1
All correct
F1
Must include constant 9
(ii)
y≈
(
)
− x 1 − 12 x 2 + x − 16 x 3 + A
M1
x3
Substitute given approximations
F1 1 A = + 3 3 x A=0 sin x − x cos x y= x3 lim y = 13
M1
Use finite limit to deduce A
A1
x →0
B1
Correct particular solution
B1
Correct limit 6
(iii)
y = 0 sin x − x cos x = 0
M1
tan x = x
E1
Equate to zero and attempt to get tanx Convincingly shown 2
(iv)
dy 3 1 1 + y = − x , multiply by I = x 3 x 6 dx x
( )
d 3 1 x y = x2 − x4 dx 6
M1
Rearrange and multiply by IF
B1
Same IF as in (i) or correct IF
A1
Recognise derivative and RHS correct
1 1 5 x y = x3 − x +B 3 30 1 1 2 B y= − x + 3 3 30 x Finite limit B = 0 lim y = 13 3
x →0
M1
Integrate
A1
c.a.o
M1 E1
Use condition to find constant Show correct limit (or same limit as (ii)) 7
3(a)(i) 2α + 4 = 0 α = −2 CF Ae −2t PI I = a cos 2t + b sin 2t I = −2a sin 2t + 2b cos 2t − 4a sin 2t + 4b cos 2t + 4a cos 2t + 4b sin 2t = 3 cos 2t − 4a + 4b = 0, 4b + 4a = 3 a = b = 83
43
M1
Find root of auxiliary equation
A1 B1 M1 M1 M1
Differentiate Substitute Compare coefficients and solve
4758
Mark Scheme
June 2009
PI I = 83 (cos 2t + sin 2t )
A1
GS I = Ae −2t + 83 (cos 2t + sin 2t )
F1
Their PI + their CF with one arbitrary constant 8
(ii)
t = 0, l = 0 0 = A + 83
A = − 83
I = 83 (cos 2t + sin 2t − e −2t )
M1
Use condition
A1
c.a.o 2
(iii)
For large t, I ≈ 83 (cos 2t + sin 2t )
M1
Amplitude = 83 12 + 12 = 83 2 Curve with oscillations with constant amplitude Their amplitude clearly indicated
A1 B1 B1
Consider behaviour for large t (may be implied)
4
(b)(i)
(A) t = 0, y = 0
dy = 2 − 2(0) + e 0 dt
M1
Gradient =3
Substitute into DE
A1
(B) At stationary point,
dy = 0, y = dt e − t = 14
9 8
0 = 2 − 2(98 ) + e −t t = ln 4 dy → 0, e −t → 0 (C) dt Giving 0 = 2 − 2y + 0, so y → 1
M1
Substitute into DE
M1 A1
Solve for t
M1
Substitute into DE
A1 7
(ii)
Curve through origin with positive gradient With maximum at (ln4, 98 ) With y → 1 as x → ∞
B1 B1 B1
Follow their ln4 Follow their (C) 3
4(i)
x = 7 x + 6 y − 6e −3t
= 7 x + 6(−12 x − 10 y + 5 sin t ) − 6e y=
1 6
−3t
(x − 7 x − 2e ) −3t
x = 7 x − 72 x − 10( x − 7 x − 2e −3t ) + 30 sin t − 6e −3t
x + 3 x + 2 x = 14e
−3t
+ 30 sin t
M1
Differentiate
M1
Substitute for y
M1
y in terms of x, x , t
M1
Substitute for y
E1
Complete argument 5
(ii)
x = ae −3t − 9 cos t + 3 sin t x = −3ae −3t + 9 sin t + 3 cos t x = 9ae −3t + 9 cos t − 3 sin t In DE gives 9ae −3t + 9 cos t − 3 sin t + 3(−3ae −3t + 9 sin t + 3 cos t )
44
M1 M1
Differentiate twice Substitute
4758
Mark Scheme
June 2009
+ 2(ae −3t − 9 cos t + 3 sin t )
= 2ae −3t + 30 sin t So PI with 2a = 14 a=7
E1 A1
Correct form shown
AE α 2 + 3α + 2 = 0 α = −1,−2
M1 A1
Auxiliary equation
CF Ae −t + Be −2t GS x = Ae − t + Be −2t + 7e −3t − 9 cos t + 3 sin t
F1 F1
CF for their roots Their PI + their CF with two arbitrary constants 8
(iii)
x = 16 ( x − 7 x − 2e −3t )
M1
y in terms of x, x , t
x = − Ae − t − 2 Be −2t − 21e −3t + 9 sin t + 3 cos t
M1 F1 A1
Differentiate GS for x Follow their GS c.a.o
y = − 43 Ae −t − 32 Be −2t − 12e −3t + 11cos t − 2 sin t
4
(iv)
x ≈ 3 sin t − 9 cos t y ≈ 11 cos t − 2 sin t x = y 11 cos t − 2 sin t ≈ 3 sin t − 9 cos t 20 cos t ≈ 5 sin t tan t ≈ 4
B1 B1 M1 A1
Follow their x Follow their y Equate Complete argument 4
(v)
Amplitude of x ≈ 3 2 + 9 2 = 3 10 2
2
Amplitude of y ≈ 11 + 2 = 5 5 Ratio is
5 6
2
M1
Attempt both amplitudes
A1
One correct
A1
c.a.o (accept reciprocal) 3
45
4761
Mark Scheme
June 2009
4761 Mechanics 1 Q1 (i)
0.5 × 8 × 10 = 40 m
Mark M1
A1
Comment Attempt to find whole area or … If suvat used in 2 parts, accept any t value 0 ≤ t ≤ 8 for max. c.a.o.
Sub
2 (ii)
0.5 × (T − 8) = 10
M1
T = 12
B1 A1
0.5 × 5 × k = 10 seen. Accept ±5 and ±10 only. If suvat used need whole area; if in 2 parts, accept any t value 8 ≤ t ≤ T for min. Attempt to use k = T – 8. c.a.o. [Award 3 if T = 12 seen] 3
(iii)
40 – 10 = 30 m
B1
ft their 40. 1
6 Q2 (i)
10 + 24 = 26 so 26 N 2
2
Mark B1
arctan 10 24
M1
= 22.619… so 22.6° (3 s.f.)
A1
Comment
Sub
24 Using arctan or equiv. Accept arctan 10 or equiv. Accept 157.4°.
3 (ii)
W = −wj
B1
T1 + T2 + W = 0
M1
k = –10 w = 34
B1 B1
0 0 Accept and −w − wj
1 (iii)
Accept in any form and recovery from W = wj. Award if not explicit and part (ii) and both k and w correct. Accept from wrong working. Accept from wrong working but not −34. [Accept –10i or 34j but not both] 3
7
46
4761
Q3 (i)
Mark Scheme
The line is not straight
Mark B1
June 2009
Comment Any valid comment
Sub
1 (ii)
a = 3−
6t 8
M1
a(4) = 0 The sprinter has reached a steady speed
Attempt to differentiate. Accept 1 term 3t correct but not 3 − . 8
F1 E1
Accept ‘stopped accelerating’ but not just a = 0. Do not ft a(4) ≠ 0. 3
(iii)
4
We require
3t 3t − dt 8 2
M1
Integrating. Neglect limits.
3t 2 t 3 = − 2 8 1
A1
One term correct. Neglect limits.
3 1 = (24 − 8) − − 2 8
M1
= 14 85 m (14.625 m)
A1
Correct limits subst in integral. Subtraction seen. If arb constant used, evaluated to give s = 0 when t = 1 and then sub t = 4. c.a.o. Any form. [If trapezium rule used M1 use of rule (must be clear method and at least two regions) A1 correctly applied M1 At least 6 regions used A1 Answer correct to at least 2 s.f.)]
1
4
4 8
47
4761
Q4 (i)
Mark Scheme
Mark B1
32 cos α t
June 2009
Comment
Sub
1
(ii)
32 cos α × 5 = 44.8 so 160 cos α = 44.8 and cos α = 0.28
M1 E1
ft their x. Shown. Must see some working e.g cosθ = 44.8 ÷ 160 or 160 cosθ = 44.8. If 32 × 0.28 × 5 = 44.8 seen then this needs a statement that ‘hence cosθ = 0.28’.
(iii)
sin α = 0.96
B1
Need not be explicit e.g. accept sin(73.73…) seen.
M1
Allow use of ‘u’ = 32, g = ± (10, 9.8, 9.81). Correct substitution. c.a.o.
2
either 0 = (32 × 0.96) 2 − 2 × 9.8 × s
s = 48.1488… so 48.1 m (3 s. f.) or Time to max height is given by 32 × 0.96 − 9.8 T = 0 so T = 3.1349…. y = 32 × 0.96 t – 4.9 t²
A1 A1 B1 M1
Could use ½ total time of flight to the horizontal. Allow use of ‘u’ = 32, g = ± (10, 9.8, 9.81) May use s=
putting t = T, y = 48.1488 so 48.1 m (3 s. f.)
A1
(u + v ) t. 2
c.a.o. 4
7 Q5 (i)
Mark M1
v = i + (3 − 2t ) j
v(4) = i – 5j
A1 F1
a = – 2j
B1
Using N2L F = 1.5 × (–2j) so –3j N
M1 A1
Comment Differentiating r. Allow 1 error. Could use const accn.
Sub
Do not award if 26 is given as vel (accept if v given and v given as well called speed or magnitude). 3
(ii)
Diff v. ft their v. Award if – 2j seen & isw. Award for 1.5 × (± a or a) seen. c.a.o. Do not award if final answer is not correct. [Award M1 A1 for –3j WW] 3
(iii)
x = 2 + t and y = 3t − t
2
Substitute t = x − 2 so y = 3( x − 2) − ( x − 2) 2 [ = ( x − 2)(5 − x ) ]
B1
Must have both but may be implied.
B1
c.a.o. isw. Must see the form y = …. 2
8
48
4761
Q6 (i)
Mark Scheme
Up the plane T − 4 g sin 25 = 0
Mark M1
T = 16.5666… so 16.6 N (3 s. f.)
A1
Down the plane, (4 + m)g sin25 – 50 = 0
M1
June 2009
Comment Resolving parallel to the plane. If any other direction used, all forces must be present. Accept s ↔ c . Allow use of m. No extra forces.
Sub
2 (ii)
m = 8.0724… so 8.07 (3 s. f.)
A1 A1
Diagram
B1
No extra forces. Must attempt resolution in at least 1 term. Accept s ↔ c . Accept Mg sin25. Accept use of mass. Accept Mgsin25 3
(iii)
B1
Any 3 of weight, friction normal reaction and P present in approx correct directions with arrows. All forces present with suitable directions, labels and arrows. Accept W, mg, 4g and 39.2. 2
(iv)
Resolving up the plane
M1
P cos15 − 20 −4g sin25 = 0
B1 B1
P = 37.8565…. so 37.9 N (3 s. f.)
A1 A1
Resolving perpendicular to the plane
M1
R + P sin15 − 4g cos25 = 0
B1
Or resolving parallel to the plane. All forces must be present. Accept s ↔ c . Allow use of m. At least one resolution attempted and accept wrong angles. Allow sign errors. P cos15 term correct. Allow sign error. Both resolutions correct. Weight used. Allow sign errors. ft use of P sin 15. All correct but ft use of P sin 15. 5
(v)
F1 R = 25.729… so 25.7 N
A1
May use other directions. All forces present. No extras. Allow s ↔ c . Weight not mass used. Both resolutions attempted. Allow sign errors. Both resolutions correct. Allow sign errors. Allow use of Pcos15 if Psin15 used in (iv). All correct. Only ft their P and their use of Pcos15. c.a.o. 4
16 If there is a consistent s ↔ c error in the weight term throughout the question, penalise only two marks for this error. In the absence of other errors this gives (i) 35.52… (ii) 1.6294… (iv) 57.486… (v) 1.688… For use of mass instead of weight lose maximum of 2.
49
4761
Q7 (i)
Mark Scheme
Mark With the 11.2 N resistance acting to the left N2L F − 11.2 = 8 × 2 M1 F = 27.2 so 27.2 N
A1 A1
June 2009
Comment
Sub
Use of N2L (allow F = mga). Allow 11.2 omitted; no extra forces. All correct c.a.o. 3
(ii)
The string is inextensible
E1
Allow ‘light inextensible’ but not other irrelevant reasons given as well (e.g. smooth pulley).
B1
One diagram with all forces present; no extras; correct arrows and labels accept use of words. Both diagrams correct with a common label.
1 (iii)
B1
2 (iv)
Method (1)
M1
Box N2L → 105 − T − 11.2 = 8a Sphere N2L ↑ T − 58.8 = 6a
A1 A1
Adding 35 = 14a
M1
a = 2.5 so 2.5 ms−2 Substitute a = 2.5 giving T = 58.8 + 15 T = 73.8 so 73.8 N
E1 M1
Method (2) 105 – 11.2 – 58.8 = 14a a = 2.5
For either box or sphere, F = ma. Allow omitted force and sign errors but not extra forces. Need correct mass. Allow use of mass not weight. Correct and in any form. Correct and in any form. [box and sphere equns with consistent signs] Eliminating 1 variable from 2 equns in 2 variables. Attempt to substitute in either box or sphere equn.
A1 M1 A1 E1 M1
either: box N2L → 105 − T − 11.2 = 8a or: sphere N2L ↑ T − 58.8 = 6a Substitute a = 2.5 in either equn
A1 M1
T = 73.8 so 73.8 N
A1
For box and sphere, F = ma. Must be correct mass. Allow use of mass not weight. Method made clear. For either box or sphere, F = ma. Allow omitted force and sign errors but not extra forces. Need correct mass. Allow use of mass not weight. Correct and in any form. Attempt to substitute in either box or sphere equn. [If AG used in either equn award M1 A1 for that equn as above and M1 A1 for finding T. For full marks, both values must be shown to satisfy the second equation.] 7
50
4761
Mark Scheme
(v)(A) g downwards
June 2009
B1
Accept ±g, ±9.8, ±10, ±9.81
Taking ↑ + ve, s = −1.8, u = 3 and a = −9.8 so −1.8 = 3T − 4.9T 2
M1
and so 4.9T 2 − 3T − 1.8 = 0
E1
Some attempt to use s = ut + 0.5at2 with a = ±9.8 etc s = ±1.8 and u = ±3 . Award for a = g even if answer to (A) wrong. Clearly shown. No need to show +ve required.
1 (B)
(C)
2
Time to reach 3 ms−1 is given by 3 = 0 + 2.5t so t = 1.2 remaining time is root of quad
B1 M1
time is 0.98513… s
B1
Total 2.1851…so 2.19 s (3 s. f.)
A1
Quadratic solved and + ve root added to time to break. Allow 0.98. [Award for answer seen WW] c.a.o. 4
(i)
With the 11.2 N resistance acting to the right F + 11.2 = 8 × 2 so F = 4.8
The 11.2 N force may be in either direction, otherwise the same scheme
(iii) (iv)
The same scheme as above
The same scheme with + 11.2 N instead of – 11.2 N acting on the box Method (1) Box N2L → 105 − T + 11.2 = 8a Sphere as before Method (2) 105 + 11.2 – 58.8 = 14a These give a = 4.1 and T = 83.4 Allow 2.5 substituted in box equation to give T = 96.2 If the sign convention gives as positive the direction of the sphere descending, a = – 4.1. Allow substituting a = 2.5 in the equations to give T = 43.8 (sphere) or 136.2 (box). In (C) allow use of a = 4.1 to give time to break as 0.73117..s. and total time as 1.716…s
(v)
4 20
51
4762
Mark Scheme
June 2009
4762 Mechanics 2 Q1 (a)(i)
Mark before u P m kg after
Comment
Sub
u Q km kg
v
u
B1
3
1 (ii)
mu − kmu = mv + km
u 3
4k v = 1 − u 3
M1
PCLM applied
A1
Either side correct (or equiv)
E1
Must at least show terms grouped
E1
4k > 1 without reason Accept 3
3 (iii)
Need v < 0 so k >
3 4
B1 [SC1: v = 0 used and inequality stated without reason] 2
(iv)
1 u 3
−v
−u −u
M1
= − 12
Use of NEL
A1
2u so v = − 3 2u 4k − = u 1 − 3 3
so k = 1.25
E1 M1 A1
c.a.o. 5
(b)(i)
1 3 9 + 5 = 8V −2 2
3 V= −1
M1
Use of PCLM
B1 M1
Use of mass 8 in coalescence Use of I = Ft
E1 4
(ii)
i cpt 3 → −3 × j cpt unchanged
1 2
−1.5
−1 New velocity ms 1 −
M1 B1
Allow wrong sign May be implied c.a.o. [Award 2/3 if barrier taken as
A1
1 0 ]
3 18 Q2 (a)(i)(A) Yes. Only WD is against conservative forces.
Mark E1
52
Comment Accept only WD is against gravity or no work done against friction.
Sub
4762
Mark Scheme
June 2009
1 (B)
Block has no displacement in that direction
E1 2
(ii)
2
0.5 × 50 × 1.5 = 20 gx − 5 gx
x = 0.38265… so 0.383 m (3 s.f.)
M1 B1 M1 A1 A1
Use of WE with KE. Allow m = 25. Use of 50 At least 1 GPE term GPE terms correct signs c.a.o. 5
(iii)
2
0.5 × 50 × V − 0.5 × 50 × 1.5
2
= 2 × 20g − 2 × 5g − 180 V = 2.6095… so 2.61 ms−1
M1
WE equation with WD term. Allow GPE terms missing
B1 B1 B1 A1
Both KE terms. Accept use of 25. Either GPE term 180 with correct sign c.a.o.
M1
Both terms. Allow mass not weight
B1 M1 F1 A1
Weight term correct
5 (b)
Force down the slope is 2000 + 450g sin20 Using P = Fv P = (2000 + 450g sin20) × 2.5 P = 8770.77… so 8770 W (3 s.f.)
ft their weight term c.a.o. 5
17
53
4762
Q3 (i)
(ii)
Mark Scheme
c.w. moments about A 5RB − 3 × 85 so RB = 51 giving 51 N ↑ Either a.c. moments about B or resolve ↑ RA = 34 so 34 N ↑
Mark M1 A1
F1
Sub
Accept no direction given 4
M1
85 × 3cos α − 27.2 × 5sin α = 0
B1 B1
3 × 85 15 = 27.2 × 5 8
Comment Moments equation. Accept no direction given
M1
c.w. moments about A
so tan α =
June 2009
E1
Moments with attempt to resolve at least one force. Allow s ↔ c . Weight term Horiz force term Must see some arrangement of terms or equiv 4
(iii) A
S 34 N
B1
All forces present and labelled
a.c. moments about B
M1
85 × 2 × cos α + 34 × 2.5 − 5S × sin α = 0
Moments with attempt to resolve forces and all relevant forces present
B1 A1
34 × 2.5
S = 37.4 Resolving horizontally and vertically → S − F − 34 sin α = 0 so F = 7.4 ↑ R − 85 − 34 cos α = 0
A1 M1
Using F = μ R
M1
R F 85 N
μ=
α B
7.4 = 0.07326… 101
E1 A1
A1
All other terms correct. Allow sign errors. All correct Either attempted R = 101 need not be evaluated here [Allow A1 for the two expressions if correct other than s ↔ c ]
c.a.o.
so 0.0733(3 s.f.) 10 18
54
4762
Q4 (i)
Mark Scheme
Mark
Taking y-axis vert downwards from O 2πσ × 82 × 4 + 2πσ × 8 × k ×
k 2
= ( 2πσ × 82 + 2πσ × 8k ) y 2
64 + k 16 + 2k
so y =
June 2009
Comment Allow areas used as masses
M1
Method for c.m.
B1 B1 B1
‘4’ used 16π k k used 2
B1
Masses correct
E1
Must see some evidence of simplification Need no reference to axis of symmetry
B1
Allow for either. Allow σ = 1
M1 B1 B1
Method for c.m. combining with (i) or starting again One term correct Second term correct
E1
Some simplification shown
Sub
6 (ii)
k = 12 gives OG as 5.2 and mass as 320πσ 320πσ × 5.2 + πσ × 82 × 12
= ( 320πσ + 64πσ ) y
y = 6 13
5 (iii) O G
B1 B1 B1
G above edge of base 12 − 6 13 = 5 23 seen here or below 8 seen here or below
M1
Accept
θ = 54.6887… so 54.7° (3 s. f.)
A1
and 8. c.a.o.
Slips when μ = tan θ
M1
Or ….
8 = 1.4117... 5 23
B1
< 1.5 so does not slip
A1
12 − 6 13 = 5 23
12 8
θ
tan θ =
8 5 23
5 23 or attempts based on 6 13 8
5 (iv)
There must be a reason 3
19
55
4763
Mark Scheme
June 2009
4763 Mechanics 3 1 (i)
1 2
m(v 2 − 1.42 ) = m × 9.8(2.6 − 2.6 cos θ )
M1
2
v − 1.96 = 50.96 − 50.96 cos θ
Equation involving KE and PE
A1
2
v = 52.92 − 50.96 cos θ
E1 3
(ii)
v2 2.6 6.37 cos θ − R = 0.25(52.92 − 50.96 cos θ )
0.65 × 9.8cos θ − R = 0.65 ×
6.37 cos θ − R = 13.23 − 12.74 cos θ R = 19.11cos θ − 13.23
(iii)
Leaves surface when R = 0 13.23 9 (= ) ( θ = 46.19° ) 19.11 13 9 v 2 = 52.92 − 50.96 × 13
cos θ =
Radial equation involving
A1 M1 A1 4
Substituting for v 2 Dependent on previous M1 Special case: R = 13.23 − 19.11cos θ earns M1A0M1SC1
M1
( ft if R = a + bcosθ and 0 < −
A1
Dependent on previous M1
A1 4 M1 A1 M1 A1
T sin α + R cos α = 0.65 × 9.8 1.22 T cos α − R sin α = 0.65 × 2.4
1.2 2 OR T − mg sin α = m 2.4
cos α
M1A1
1.2 2 mg cos α − R = m 2.4
sin α
M1A1
sin α =
2.4 12 5 = , cos α = 2.6 13 13
a < 1) b
M1
Speed is 4.2 m s −1 (iv)
v2 r
M1
( α = 67.38° )
Tension is 6.03 N Normal reaction is 2.09 N
Horiz eqn involving
v2 or r ω 2 r
M1 M1
Solving to obtain a value of T or R
A1 A1
Dependent on necessary M1s ( Accept 6 , 2.1 ) Treat ω = 1.2 as a misread, leading to T = 6.744, R = 0.3764 for 7 / 8
8
56
Resolving vertically (3 terms)
4763
2 (i)
Mark Scheme
1 2
× 5000 x 2 = 12 × 400 × 32
M1 A1 A1
Compression is 0.849 m
June 2009
Equation involving EE and KE Accept
3 (ii)
Change in PE is 400 × 9.8 × (7.35 + 1.4) sin θ
M1
Or 400 × 9.8 × 1.4sin θ
A1
1 × 400 × 4.542 2 Or 784 + 4116
and
= 400 × 9.8 × 8.75 × 17 = 4900 J
Change in EE is
1 2
× 5000 × 1.4 = 4900 J
3 2 5
2
M1
Since Loss of PE = Gain of EE, car will be at rest
E1
M1M1A1 can also be given for a correct equation in x (compression): 2500 x 2 − 560 x − 4116 = 0
Conclusion required, or solving equation to obtain x = 1.4
4 (iii) WD against resistance is 7560(24 + x)
Change in EE is Change in KE is
2
1 × 5000x 2 1 × 400 × 302 2
Change in PE is 400 × 9.8 × (24 + x) ×
1 7
OR Speed 7.75 m s −1 when it hits buffer, then WD against resistance is 7560x Change in EE is 12 × 5000x 2 Change in KE is
1 2
B1 B1
( = 181440 + 7560x ) ( = 2500x 2 )
B1
( = 180000 )
B1
( = 13440 + 560x )
B1 B1
( = 2500x 2 ) ( = 12000 ) ( = 560x )
× 400 × 7.752
B1
1 7
B1
Change in PE is 400 × 9.8 × x ×
M1
−7560(24 + x) = 12 × 5000 x 2 − 12 × 400 × 302
Equation involving WD, EE, KE, PE
−400 × 9.8 × (24 + x) × 17 −7560(24 + x) = 2500 x 2 − 180 000 − 560(24 + x) 2
−3.024(24 + x) = x − 72 − 0.224(24 + x)
F1 M1
2
x + 2.8 x − 4.8 = 0
A1
−2.8 + 2.82 + 19.2 2 = 1.2
M1
x=
A1 10
57
Simplification to three term quadratic
4763
Mark Scheme
3(a)(i) [ Velocity ] = L T −1
B1
[ Force ] = M L T −2
B1
−3
[ Density ] = M L
June 2009
Deduct 1 mark for ms−1 etc
B1 3
(ii) M L T −2 = (M L−3 )α (L T −1 ) β (L2 )γ α =1 β =2 −3α + β + 2γ = 1 γ =1
B1 B1 M1A1 ( ft if equation involves α , β and γ ) A1 5
(b)(i) 2π = 4.3 ω 2π ω= ( = 1.4612 )
M1 A1
4.3
θ 2 = 1.46122 (0.082 − 0.052 )
M1 F1
Angular speed is 0.0913 rad s −1
A1 5
OR θ = 0.08ω cos ωt
M1 F1 A1
= 0.08 ×1.4612 cos 0.6751 = 0.0913
(ii) θ = 0.08sin ωt When θ = 0.05, 0.08sin ωt = 0.05 ωt = 0.6751 t = 0.462
Time taken is 2 × 0.462 = 0.924 s
Using ω 2 ( A2 − θ 2 ) For RHS ( b.o.d. for v = 0.0913 m s −1 ) Or θ = (−) 0.08ω sin ωt = (−) 0.08 × 1.4612sin 0.8957
or θ = 0.08cos ωt Using θ = (±) 0.05 to obtain an equation for t A1 cao B1M1 above can be earned in (i) or t = 0.613 from θ = 0.08cos ωt or t = 1.537 from θ = 0.08cos ωt B1 M1
M1
Strategy for finding the required time ( 2 × 0.462 or 12 × 4.3 − 2 × 0.613 or 1.537 − 0.613 ) Dep on first M1 A1 cao For θ = 0.05sin ωt , max B0M1A0M0 ( for 0.05 = 0.05sin ωt ) 5
58
4763
4(a)
Mark Scheme
June 2009
ln 3
ln 3 e x dx = e x 0 0 =2
Area is
ln 3
x y dx = 0
M1 A1
x e x dx
M1 ln 3
= x ex − ex 0 = 3ln 3 − 2 x=
3ln 3 − 2 3 = 2 ln 3 − 1 2 ln 3
1 2
y 2 dx = 0 =
1 2
(e x ) 2 dx
1 2x e 4
Integration by parts For x e x − e x
A1
ww full marks (B4) Give B3 for 0.65
M1
For integral of (e x )2
A1
For
A1
If area wrong, SC1 for
ln 3
0
=2 y=
M1 A1
2 =1 2
1 2x e 4
x=
3ln 3 − 2 2 and y = area area
9 (b)(i)
a
36
2
x4
Volume is π y 2 dx = π
M1
dx
π may be omitted throughout
a
12 3 12 =π − 3 =π − 3 x 2 2 a a
36
2
x3
2 π xy dx = π
A1
dx
M1 a
18 9 18 =π − 2 =π − 2 x 2 2 a
A1
π xy dx 2 π y dx 2
x=
M1
9 18 2 a 2 3(a 3 − 4a ) = = 3 12 a3 − 8 π − 3 2 a
π −
E1 6
M1
(ii) Since a > 2, 4a > 8 3
3
so a − 4a < a − 8 Hence x =
3(a3 − 4a ) a3 − 8
A1
throughout
−3
→3
Fully acceptable explanation Dependent on M1A1
3
M1A1 Accept x ≈
Since x increases as a increases, x is less than 3
E1
3a 3
→ 3 , etc a3 ( M1 for x → 3 stated, but A1 requires
correct justification )
59
4764
Mark Scheme
June 2009
4764 MEI Mechanics 4 1(i)
d (mv ) = mg dt dm dv v+m = mg dt dt dv mg = mg v+m 2(v + 1) dt
v dv = = g 1 − ( dt 2 v + 1) v + 1 dv 1 = 2g v + 2 dt 1 dv 1 1 − =2g v + 2 dt
B1
Seen or implied
M1 Expand M1 Use
dm mg v= dt 2(v + 1)
v+2 g 2(v + 1)
M1 Separate variables (oe) E1 5
(ii)
1
1 − v + 2 dv =
1 2
g dt
M1 Integrate
v − ln v + 2 = 12 gt + c t = 0, v = 0 −ln2 = c v − ln v + 2 = 12 gt − ln 2
A1 LHS M1 Use condition
2 (v − ln v + 2 + ln 2) g v = 10 t ≈ 1.68
t=
A1 B1 5
(iii)
As t gets large, v gets large dv → 12 g (i.e. constant) So dt
M1 A1
Complete argument 2
2(i)
V = −mg ⋅ 2a sin θ +
1 2
mg
2a
(4a sin θ − a )2
B1
GPE
M1 Reasonable attempt at EPE A1 EPE correct dV mg (4a sin θ − a ) ⋅ 4a cos θ = −2mga cos θ + dθ 2a = −2mga cos θ + 2mga cos θ (4 sin θ − 1) = 4mga cos θ (2 sin θ − 1)
M1 Differentiate E1
Complete argument 5
(ii)
dV =0 dθ ⇔ cos θ = 0 or sin θ = ⇔ θ = 12 π or
1 6
M1 Set derivative to zero 1 2
M1 Solve
π
A1
Both
M1
Second derivative (or alternative method)
2
d V = 4mga cos θ (2 cos θ ) − 4mga sin θ (2 sin θ − 1) dθ 2 V ′′( 12 π )(= −4mga ) < 0 unstable
60
M1 Consider sign A1 One correct conclusion validly
4764
Mark Scheme
June 2009
shown
3 V ′′( 16 π ) = 4mga ⋅ 2
3 > 0 stable
( )
A1
Complete argument 7
3(i)
Mass of ‘ring’ ≈ 2πrδrρ
B1
May be implied
a
I C = r 2 ⋅ 2πρr dr
M1 Set up integral
0
[
= 2πρ ⋅
1 4
]
a r4 0
A1 4
= πa ρ 1 2
M1 Integrate
M = πa 2 ρ
M1 Use relationship between ρ and M 2
I C = 12 Ma (ii)
All correct
E1
6
I A = IC + M ( a)
2
1 10
2
Complete argument
M1 Use parallel axis theorem
2
1 = 12 Ma + 100 Ma = 0.51Ma
2
E1
Convincingly shown 2
(iii)
I A θ = − Mg
⋅ 101
a sin θ
B1 B1
g sin θ 5.1a θ small sin θ ≈ θ g θ =− θ , i.e. SHM 5.1a θ =−
M1 Expression for θ M1 Use small angle approximation
5.1a ≈ 4.53 a g
Period 2π
LHS RHS
E1
Complete argument and conclude SHM
F1
Follow their SHM equation 6
(iv)
e.g. C A
B1
P
M1 Moments or ( m )x = mx (oe)
mg ⋅ 109 a = Mg ⋅ 101 a m = 19 M
A1
I = 0.51Ma + m(109 a )
2
2
= 0.6 Ma (v)
Method may be implied
M1
2
E1
Convincingly shown 5
2
2
KE = Iω = (0.6Ma )ω 1 2
2
1 2
2
M1 Attempt to find KE
2
A1 M1 Work-energy equation A1 Correct equation
= 0.3Ma ω C ⋅ n ⋅ 2π = 0.3Ma 2 ω 2 C=
Show PAC in straight line (in any direction)
0.3Ma 2 ω 2 2 nπ
A1 5
61
4764
4(i)
Mark Scheme
June 2009
At terminal velocity, F = 0 k ⋅ 60 2 = 90 g
M1
Equilibrium of forces
k=
E1
Convincingly shown
1 40
g
2 (ii)
dv = 90 g − 401 gv 2 dx 90v dv = dx 90 g − 401 gv 2
90v
−
1800 ln 90 g − 401 gv 2 = x + c1 g
90 − 401 gv 2 = Ae v2 =
−
M1 A1
N2L
M1
Separate and integrate
A1
LHS
M1 M1
Rearrange, dealing properly with constant Use condition
E1
Complete argument
gx 1800
gx − 40 90 g − Ae 1800 g
x = 0, v = 0
A = 90 g
9x − v 2 = 36001 − e 1800
7 1800
kv dx 2
(iii) WD against R =
B1
0
1800
=
0
gx − 90 g 1 − e 1800 dx
1800
gx 1800 −1800 = 90 g x + e g 0
(
)
= 162000 g + e − g − 1 2
M1
(
x = 1800 v = 3600 1 − e
−g
Integrate
A1
)
(
Loss in energy = 90 g ⋅ 1800 − ⋅ 90 ⋅ 3600 1 − e 1 2
−g
= 162000( g + e − g − 1) = WD against R
)
B1 M1 M1
GPE KE
E1
Convincingly shown (including signs) 7
(iv)
v = 60 1 − e
−g
B1
≈ 59.9983
1 (v)
90
dv = 90 g − 90v dt
M1
N2L
A1 dv
g − v = dt
or
10 59.9983
t = dt 0 g −v
dv
M1
− ln g − v = t + c 2
A1
t = 0, v = 59.9983 c 2 = −3.91598 v = 10 t = − ln 0.2 + 3.91598 ≈ 5.53 s
M1 M1 A1
Separate and integrate
Use condition (or limits) Calculate t 7
62
4766
Mark Scheme
June 2009
4766 Statistics 1 Q1 (i)
Median = 2
B1 cao B1 cao
Mode = 1
2
(ii)
S1 labelled linear scales on both axes H1 heights
60 50 40 30
2
20 10 0 1
2
3
4
N umber o f P eo ple
(iii)
Positive
B1 TOTAL
Q2 (i)
25 different teams = 53130 5
25 M1 for 5 A1 cao
14 11 × = 364 × 55 = 20020 3 2
M1 for either combination M1 for product of both A1 cao
(ii)
1 5
2
TOTAL Q3 (i)
Mean =
126 = 10.5 12
Sxx = 1582 − s= (ii)
5
B1 for mean
126 2 = 259 12
M1 for attempt at Sxx 3
259 = 4.85 11
A1 cao
New mean = 500 + 100 × 10.5 = 1550
B1 ANSWER GIVEN 3
New s = 100 × 4.85 = 485
(iii)
3
M1A1 ft
On average Marlene sells more cars than Dwayne. Marlene has less variation in monthly sales than Dwayne.
E1 E1 ft
2 TOTAL
63
8
4766
Mark Scheme
June 2009
Q4 (i)
E(X) = 25 because the distribution is symmetrical.
E1 ANSWER GIVEN
1
M1 for Σr2p (at least 3 terms correct) M1dep for –25² A1 cao TOTAL
3
Allow correct calculation of Σrp (ii)
E(X2) = 10 2 × 0.2 + 20 2 × 0.3 + 30 2 × 0.3 + 40 2 × 0.2 = 730 Var(X) = 730 – 252 = 105
4
Q5 (i)
Distance 050100200-400
freq 360 400 307 133
width 50 50 100 200
f dens 7.200 8.000 3.070 0.665
M1 for fds A1 cao Accept any suitable unit for fd such as eg freq per 50 miles. L1 linear scales on both axes and label W1 width of bars
5
H1 height of bars
(ii)
Estimate = 50 +
Q6 (i)
(A) (B)
(ii)
B1 for 600th
Median = 600th distance 240
/400 × 50 = 50 + 30 = 80
M1 for attempt to interpolate A1 cao TOTAL
3 8
P(at most one) =
83 = 0.83 100
B1 aef
P(exactly two) =
13 10 + 2 + 1 = = 0.13 100 100
M1 for (10+2+1)/100 A1 aef
2
53 × 100 M1dep for product of next 2 correct fractions A1 cao TOTAL
3
P(all at least one) =
1
53 52 51 140556 × × = = 0.145 100 99 98 970200
64
M1 for
6
4766
Mark Scheme
June 2009
Q7 (i)
(ii)
P(Not delayed) = 0.8 × 0.85 × 0.9 = 0.612 P(Delayed) = 1 – 0.8 × 0.85 × 0.9 = 1 – 0.612 = 0.388
(iii)
= 0.153 + 0.108 + 0.068 = 0.329
P(Just one problem) 0.329 = 0.848 = 0.388 P(Delay)
B1 one product correct M1 three products M1 sum of 3 products A1 cao
M1 for denominator A1 ft
4
3
M1 for 0.15 + M1 for second term A1 cao
P(Delayed | No technical problems) Either = 0.15 + 0.85 × 0.1 = 0.235
M1 for product M1 for 1 – product A1 cao
Or = 1 – 0.9 × 0.85 = 1 – 0.765 = 0.235 Or = 0.15 × 0.1 + 0.15 × 0.9 + 0.85 × 0.1 = 0.235
M1 for all 3 products M1 for sum of all 3 products A1 cao
3
Or (using conditional probability formula) P(Delayed and no technical problems) P(No technical problems) 0.8 × 0.15 × 0.1 + 0.8 × 0.15 × 0.9 + 0.8 × 0.85 × 0.1 = 0 .8 0.188 = = 0.235 0.8
(vi)
4
M1 for numerator
P(Just one problem | delay) =
(v)
M1 for 1 – P(delayed) A1 ft
P(just one problem) = 0.2×0.85×0.9 + 0.8×0.15×0.9 + 0.8×0.85×0.1
(iv)
2
B1 for any one B1 for the other two M1 for product A1 cao
a = 0.8, b = 0.85, c = 0.9.
M1 for numerator M1 for denominator A1 cao
M1 for product A1 ft
Expected number = 110 × 0.388 = 42.7
2 TOTAL
65
18
4766
Q8 (i)
Mark Scheme
X ~ B(15, 0.2) (A)
15 P(V = 3) = × 0.2 3 × 0.812 = 0.2501 3
June 2009
M1 0.23 × 0.812 M1
( ) ×p q 15 3
3
12
A1 cao
3 Or from tables 0.6482 − 0.3980 = 0.2502
(B)
(C)
(ii)
(iii)
Or: M2 for 0.6482 – 0.3980 A1 cao M1 P(X≤2) M1 1-P(X≤2) A1 cao
P( X ≥ 3) = 1 − 0.3980 = 0.6020
E(X) = np = 15 × 0.2 = 3.0
(A) Let p = probability of a randomly selected child eating at least 5 a day H0: p = 0.2 H1: p > 0.2 (B) H1 has this form as the proportion who eat at least 5 a day is expected to increase. Let X ~ B(15, 0.2) P(X ≥ 5) = 1 – P(X ≤ 4) = 1 – 0.8358 = 0.1642 > 10% P(X ≥ 6) = 1 – P(X ≤ 5) = 1 – 0.9389 = 0.0611 < 10% So critical region is {6,7,8,9,10,11,12,13,14,15} 7 lies in the critical region, so we reject null hypothesis and we conclude that there is evidence to suggest that the proportion who eat at least five a day has increased.
66
M1 for product A1 cao B1 for definition of p in context B1 for H0 B1 for H1 E1
3
2
4
B1 for 0.1642 B1 for 0.0611 M1 for at least one comparison with 10% A1 cao for critical region dep on M1 and at least one B1 M1 dep for comparison A1 dep for decision and conclusion in context TOTAL
6
18
4767
Mark Scheme
June 2009
4767 Statistics 2 Question 1 (i)
EITHER: 1 1 Sxy = xy − x y = 316345 – × 2331.3 × 6724.3 n 50
M1 for method for Sxy
= 2817.8 Sxx = x 2 −
1 ( x )2 = 111984 − 1 × 2331.32 = 3284.8 n 50
M1 for method for at least one of Sxx or Syy
Syy = y 2 −
1 ( y )2 = 921361 − 1 × 6724.32 = 17036.8 n 50
A1 for at least one of Sxy, Sxx or Syy correct
r =
S xy S xx S yy
2817.8
=
3284.8 × 17036.8
= 0.377
M1 for structure of r A1 (AWRT 0.38)
OR: 316345 xy − xy = – 46.626 × 134.486 n 50 = 56.356 S xx 3284.8 = 65.696 = 8.105 = rmsd(x) = n 50
cov(x, y) =
M1 for method for at least one msd
17036.8 = 340.736 = 18.459 n 50 56.356 cov( x, y ) r = = = 0.377 8.105 × 18.459 rmsd( x)rmsd( y )
A1 for at least on of cov(x,y), rmsd(x) or rmsd(y) correct M1 for structure of r A1 (AWRT 0.38)
H0: ρ = 0 H1: ρ ≠ 0 (two-tailed test) where ρ is the population correlation coefficient For n = 50, 5% critical value = 0.2787
B1 for H0, H1 in symbols B1 for defining ρ
rmsd(y) =
(ii)
S yy
M1 for method for cov(x,y)
=
5
B1FT for critical value
M1 for sensible comparison leading to a conclusion There is sufficient evidence at the 5% level to suggest that A1 for result 1 B1 FT for conclusion in there is correlation between oil price and share cost context Since 0.377 > 0.2787 we can reject H0:
(iii)
(iv)
Population The scatter diagram has a roughly elliptical shape, hence the assumption is justified. Because the alternative hypothesis should be decided without referring to the sample data and there is no suggestion that the correlation should be positive rather than negative.
B1 B1 elliptical shape E1 conclusion E1 E1
TOTAL
67
6
3
2 16
4767
Mark Scheme
June 2009
Question 2 (i)
(ii)
Meteors are seen randomly and independently There is a uniform (mean) rate of occurrence of meteor sightings (A) Either P(X = 1) = 0.6268 – 0.2725 = 0.3543 1.31 = 0.3543 Or P(X = 1) = e −1.3 1! (B) Using tables: P(X ≥ 4) = 1 – P(X ≤ 3)
B1 B1 M1 for appropriate use of tables or calculation A1 M1 for appropriate probability
= 1 – 0.9569
calculation
= 0.0431
A1
(iii)
λ = 10 × 1.3 = 13
B1 for mean M1 for calculation A1 CAO
(iv)
131 = 0.0859 1! Mean no. per hour = 60 × 1.3 = 78 P(X = 10) = e −13
Normal approx. to the Poisson,
X ~ N(78, 78)
99.5 − 78 P(X ≥ 100) = P Z > 78
4
3
B1 for Normal approx. B1 for correct parameters (SOI) B1 for continuity corr. M1 for correct Normal probability calculation using correct tail A1 CAO, (but FT wrong or omitted CC)
= P(Z >2.434) = 1 – Φ(2.434) = 1 – 0.9926 = 0.0074
(v)
2
5
Either P(At least one) = 1 − e
−λ
M1 formation of equation/inequality using P(X ≥ 1) = 1 – P(X = 0) with Poisson distribution. A1 for correct equation/inequality M1 for logs A1 for 3.54 A1 for t (correctly justified)
λ0 = 1 − e − λ ≥ 0.99 0!
e − λ ≤ 0.01 –λ ≤ ln 0.01, so λ ≥ 4.605 1.3 t ≥ 4.605, so t ≥ 3.54 Answer t = 4
Or t = 1, λ = 1.3, P(At least one) = 1 − e −1.3 = 0.7275 t = 2, λ = 2.6, P(At least one) = 1 − e −2.6 = 0.9257 t = 3, λ = 3.9, P(At least one) = 1 − e −3.9 = 0.9798 t = 4, λ = 5.2, P(At least one) = 1 − e −5.2 = 0.9944 Answer t = 4
M1 at least one trial with any value of t A1 correct probability. M1 trial with either t = 3 or t = 4 A1 correct probability of t = 3 and t = 4 A1 for t
TOTAL
68
5
19
4767
Mark Scheme
June 2009
Question 3 (i)
X ~ N(1720,902)
M1 for standardising
1700 − 1720 P(X – 1.645 so not significant. There is not sufficient evidence to reject H0 There is insufficient evidence to conclude that the mean intensity of bulbs made by this manufacturer is less than 1720
M1 must include √20
A1FT B1 for –1.645 No FT from here if wrong. Must be –1.645 unless it is clear that absolute values are being used. M1 for sensible comparison leading to a conclusion. FT only candidate’s test statistic A1 for conclusion in words in context
TOTAL
69
5 20
4767
Mark Scheme
June 2009
Question 4
(i)
H0: no association between type of car and sex; H1: some association between type of car and sex; EXPECTED Hatchback Saloon People carrier 4WD
Male 83.16 70.56 51.66 17.01
Female 48.84 41.44 30.34 9.99
Sports car
29.61
17.39
CONTRIBUTION Hatchback Saloon People carrier 4WD Sports car
Male 1.98 0.59 3.61 0.23 1.96
Female 3.38 1.00 6.15 0.40 3.33
B1 M1 A2 for expected values (to 2 dp) (allow A1 for at least one row or column correct)
M1 for valid attempt at (O − E)2/E A1 for all correct NB These M1A1 marks cannot be implied by a correct final value of X 2
X2 = 22.62
M1 for summation A1 for X2 CAO
Refer to Ξ42 Critical value at 5% level = 9.488
B1 for 4 deg of f
12
B1 CAO for cv 22.62 > 9.488 M1 sensible comparison Result is significant There is evidence to suggest that there is some association leading to a conclusion A1 between sex and type of car.
(ii)
NB if H0 H1 reversed, or ‘correlation’ mentioned, do not award first B1or final A1 • In hatchbacks, male drivers are more frequent than expected. • In saloons, male drivers are slightly more frequent than expected. • In people carriers, female drivers are much more frequent than expected. • In 4WDs the numbers are roughly as expected • In sports cars, female drivers are more frequent than expected.
E1 E1 E1 E1 E1
TOTAL
70
5
17
4768
Mark Scheme
June 2009
4768 Statistics 3 Q1
W ~ N(14, 0.552) G ~ N(144, 0.92)
(i)
145 − 144 P(G < 145) = P Z < = 1.1111 0 . 9 = 0·8667
M1 A1
For standardising. Award once, here or elsewhere.
A1
c.a.o.
W + G ~ N (14 + 144 = 158,
B1 B1
Mean. Variance. Accept sd (= 1.0547…).
A1
c.a.o.
B1 B1
Mean. Variance. Accept sd (= 2.6415).
(ii)
When a candidate’s answers suggest that (s)he appears to have neglected to use the difference columns of the Normal distribution tables penalise the first occurrence only.
σ 2 = 0.55 2 + 0.9 2 = 1.1125 ) P(this > 160) = 160 − 158 = 1 ⋅ 896 = 1 − 0 ⋅ 9710 = 0.0290 P Z > 1 . 054 7
(iii)
(iv)
H = W1 + ... + W7 + G1 + ... + G6 ~ N (962, σ 2 = 0.55 2 + ... + 0.55 2 + 0.9 2 + ... + 0.9 2 = 6.9775) P(960 < this < 965) = 965 − 962 960 − 962 = 1.1357 P = −0.7571 < Z < 2 ⋅ 6415 2 ⋅ 6415 = 0·8720 – (1 – 0.7755) = 0.6475
M1
Two-sided requirement.
A1
c.a.o.
Now want P(B(4, 0.6475) ≥ 3)
M1
Evidence of attempt to use binomial. ft c’s p value. Correct terms attempted. ft c’s p value. Accept 1 – P(… ≤ 2) c.a.o.
= 4 × 0.64753 × 0.3525 + 0.64754
M1
= 0.38277 + 0.17577 = 0.5585
A1 B1 B1
D = H 1 − H 2 ~ N(0, 6.9775 + 6.9775 = 13.955) Want h s.t. P(−h < D < h) = 0.95 i.e. P(D < h) = 0 975 ∴ h = 13.955 × 1.96
M1 B1 A1
= 7.32
3
3
7
Mean. (May be implied.) Variance. Accept sd (= 3.7356). Ft 2 × c’s 6.9775 from (iii). Formulation of requirement as 2sided. For 1.96. c.a.o.
5 18
71
4768
Q2 (i)
Mark Scheme
H0: μ = 1 H1: μ < 1
B1
where μ is the mean weight of the cakes.
B1
x = 0.957375
B1
s n −1 = 0.07314(55)
M1
Test statistic is 0 . 957375 − 1 0 . 07314 8
June 2009
Both hypotheses. Hypotheses in words only must include “population”. For adequate verbal definition. Allow absence of “population” if correct notation μ is used, but do NOT allow “ X = ... ” or similar unless X is clearly and explicitly stated to be a population mean.
sn = 0.06842 but do NOT allow this here or in construction of test statistic, but FT from there. Allow c’s x and/or sn–1. Allow alternative: 1 + (c’s –1.895) × 0.07314 (= 0.950997) for 8
subsequent comparison with x . (Or x – (c’s –1.895) × 0.07314 8
= –1.648(24).
(ii)
(iii)
A1
Refer to t7.
M1
Single-tailed 5% point is –1.895.
A1
Not significant. Insufficient evidence to suggest that the cakes are underweight on average.
A1 A1
CI is given by 0.957375 ± 2·365 0.07314 × 8 = 0.957375 ± 0.061156= (0.896(2), 1.018(5))
M1 B1 M1
x ± 1.96 ×
0.006 n
72
(= 1.006377) for comparison with 1.) c.a.o. but ft from here in any case if wrong. Use of 1 – x scores M1A0, but ft. No ft from here if wrong. P(t < –1.648(24)) = 0.0716. Must be minus 1.895 unless absolute values are being compared. No ft from here if wrong. ft only c’s test statistic. ft only c’s test statistic.
A1
c.a.o. Must be expressed as an interval. ZERO/4 if not same distribution as test. Same wrong distribution scores maximum M1B0M1A0. Recovery to t7 is OK.
M1 B1 A1
Structure correct, incl. use of Normal. 1.96. All correct.
9
4
3
4768
(iv)
Mark Scheme
2 × 1.96 ×
0.006 < 0.025 n
June 2009
M1
Set up appropriate in equation. Condone an equation.
M1
Attempt to rearrange and solve.
A1
c.a.o. (expressed as an integer). S.C. Allow max M1A1(c.a.o.) when the factor “2” is missing. (n > 36.879)
2
2 × 1.96 n> × 0.006 = 147.517 0.025 So take n = 148
3
19
73
4768
Q3 (i)
(ii)
Mark Scheme
For a systematic sample • she needs a list of all staff • with no cycles in the list. All staff equally likely to be chosen if she • chooses a random start between 1 and 10 • then chooses every 10th. Not simple random sampling since not all samples are possible.
June 2009
E1 E1 E1 E1 E1
Nothing is known about the background population ...
E1
... of differences between the scores.
E1
H0: m = 0 H1: m ≠ 0
B1
where m is the population median difference for the scores.
B1
5 Any reference to unknown distribution or “non-parametric” situation. Any reference to pairing/differences. Both hypotheses. Hypotheses in words only must include “population”. For adequate verbal definition.
4
(iii) Diff Rank
−0.8 2
−2.6 5
8.6 12
6.2 10
6.0 9
−3.6 6
−2.4 4 M1 M1 A1 B1
W− = 1 + 2 + 4 + 5 + 6 + 7 = 25 Refer to tables of Wilcoxon paired (/single sample) statistic for n = 12. Lower (or upper if 53 used) 2½ % tail is 13 (or 65 if 53 used). Result is not significant. No evidence to suggest a preference for one of the uniforms.
M1 A1 A1 A1
−0.4 1
−4.0 7
5.6 8
6.6 11
2.2 3
For differences. ZERO in this section if differences not used. For ranks. ft from here if ranks wrong. (or W+ = 3 + 8 + 9 + 10 + 11 + 12 = 53) No ft from here if wrong. i.e. a 2-tail test. No ft from here if wrong. ft only c’s test statistic. ft only c’s test statistic.
8 17
74
4768
Mark Scheme
2x λ2
Q4
f ( x) =
(i)
f(x) > 0 for all x in the domain.
(ii)
for 0 < x < λ , λ > 0
λ
x λ dx = 2 = 2 = 1 2 λ λ 0 λ 2
λ2x 0
μ=
2
λ
2x3 / 3 2λ d x = = 2 2 3 λ λ 0
λ 2x 2 0
P( X < μ) =
June 2009
μ 0
E1 M1
Correct integral with limits.
A1
Shown equal to 1.
M1
Correct integral with limits.
A1
c.a.o.
M1
Correct integral with limits.
A1
Answer plus comment. ft c’s μ provided the answer does not involve λ.
M1
Use of Var(X) = E(X2) – E(X)2.
A1
c.a.o.
μ
x2 2x d x = 2 λ2 λ 0
μ 2 4 λ2 / 9 4 = = 9 λ2 λ2 which is independent of λ. =
(iii)
3
λ2 2 2 2 λ λ2 4λ − = σ2 = 2 9 18
4
Given E ( X 2 ) =
2
(iv) Probability Expected f
0.18573 9.2865
0.25871 12.9355
0.36983 18.4915 M1 A1 M1 A1 M1
X2 = 3.0094 + 0.2896 + 0.1231 + 3.5152 = 6.937(3) Refer to χ 32 .
Upper 5% point is 7.815. Not significant. Suggests model fits the data for these jars. But with a 10% significance level (cv = 6.251) a different conclusion would be reached.
A1 A1 A1 E1
0.18573 9.2865 Probs × 50 for expected frequencies. All correct. Calculation of X2. c.a.o. Allow correct df (= cells – 1) from wrongly grouped table and ft. Otherwise, no ft if wrong. P(X2 > 6.937) = 0.0739. No ft from here if wrong. ft only c’s test statistic. ft only c’s test statistic. Any valid comment which recognises that the test statistic is close to the critical values.
9
18
75
4769
Mark Scheme
June 2009
4769 Statistics 4 Q1 (i)
Follow-through all intermediate results in this question, unless obvious nonsense. P(X ≥ 2) = 1 – θ – θ (1 – θ) = (1 – θ)2 [o.e.] L = [θ] n0 [θ(1 – θ)] n1 [(1-θ)2] n−n0 −n1 = θ n0 + n1 (1 – θ) 2 n −2 n0 − n1
(ii)
ln L = (n0 + n1 ) ln θ + (2n − 2n0 − n1 ) ln (1 – θ) d ln L dθ n0 + n1 2n − 2n0 − n1 – = θ 1− θ =0 (1 - θˆ ) (n0 + n1 ) = θˆ (2n − 2n0 − n1 ) n +n θˆ = 0 1 2 n − n0
(iii)
E(X) =
∞
xθ (1 − θ )
5
M1 A1
A1 M1 A1 6
x
M1
= θ {0 + (1 – θ) + 2(1 – θ) 2 + 3(1 − θ ) 3 + …} 1− θ = θ So could sensibly use (method of moments) ~ ~ 1− θ θ given by ~ = X θ ~ 1 θ = 1+ X
A2
Divisible, for algebra; e.g. by “GP of GPs” BEWARE PRINTED ANSWER
M1 A1
To use this, we need to know the exact numbers of faults for components with “two or more”. 14 x= = 0 ⋅14 100 ~ 1 θ = = 0·8772 1+ 0.14 Also, from expression given in question, ~ 0 ⋅ 8772 2 (1 − 0 ⋅ 8772) Var(θ ) = 100 = 0·000945 CI is given by 0·8772 ± 1·96 x (0·817, 0·937)
Product form Fully correct BEWARE PRINTED ANSWER
M1
x =0
(iv)
M1 A1 M1 A1 A1
0 ⋅ 000945 =
BEWARE PRINTED ANSWER + 6
E1 B1 B1
B1 M1 B1 M1 A1
For 0.8772 For 1.96 For 0 ⋅ 000945 7
76
4769
Mark Scheme
June 2009
Q2 (i)
tZ
Mgf of Z = E (e ) =
∞
1
−∞
2π
e
tz −
z2 2
M1 dz
Complete the square z2 1 1 tz − = − (z − t)2 + t 2 2 2 2 =
t2 e2
∞
1
2π Pdf of N(t,1) ∴ =1 −∞
e
−
( z −t ) 2 2
M1 A1 A1 M1 M1 M1
t2 e2
dt =
A1
(ii)
Y has mgf M Y (t )
For final answer
t2 e2
= e bt E[e ( at )Y ] = e bt M Y (at )
For factor e bt For factor E [e ( at )Y ] For final answer
X −μ , so X = σZ + μ σ
M1 1
For factor e μt
Z= ∴M
(iv)
For taking out factor For use of pdf of N(t,1) For ∫ pdf = 1
M1 1 1 1
Mgf of aY + b is E [e t ( aY +b ) ]
(iii)
t2 e2
X
( t ) = e μ t .e
( σt )2 2
=e
μt +
σ 2t 2 2
1 1
W = eX E (W k ) = E[(e X ) k ] = E (e kX ) = M X (k )
M1 A1 A1
∴ E (W ) = M X (1) = e
μ+
σ2 2
E(W 2 ) = M X (2) = e 2 μ + 2σ 2
( σt ) 2 e 2
For factor For final answer
8
4
4
For E[(e X ) k ] For E (e kX ) For M X (k )
M1 A1 2
2
2
2
∴ Var(W ) = e 2 μ + 2σ − e 2 μ +σ [= e 2 μ +σ (e σ − 1)]
77
M1 A1 A1
8
4769
Mark Scheme
June 2009
Q3 (i)
A1
x = 126 ⋅ 2 s = 8·7002 s 2 = 75 ⋅ 693 y = 133⋅ 9 s = 10·4760 s 2 = 109 ⋅ 746 H 0 : μ A = μB H 0 : μ A ≠ μB Where μ A , μ B are the population means.
1 1
Pooled s 2 9 × 75 ⋅ 693 + 6 × 109 ⋅ 476 681 ⋅ 24 + 658 ⋅ 48 = = 15 15 = 89 ⋅ 3146 [√ = 9·4506] Test statistic is 126 ⋅ 2 − 133 ⋅ 9 1 1 89 ⋅ 3146 + 10 7
(ii)
=−
A1 if all correct. [No mark for use of sn , which are 8·2537 and 9·6989 respectively.] Do not accept X = Y or similar.
7⋅7 = − 1⋅ 653 4 ⋅ 6573
B1
M1 A1
Refer to t15
1
No FT if wrong
Double-tailed 10% point is 1·753 Not significant No evidence that population mean concentrations differ. There may be consistent differences between days (days of week, types of rubbish, ambient conditions, …) which should be allowed for.
1 1 1
No FT if wrong
Assumption: Normality of population of differences. Differences are 7·4 -1·2 11·1 5·5 6·2 3·7 −0·3 1·8 3·6 [ d = 4·2, s = 3·862 ( s 2 =14·915)] Use of s n (= 3 ⋅ 641) is not acceptable, even in a denominator of s n /
Test statistic is
10 E1 E1 1 M1 A1 Can be awarded here if NOT awarded in part (i)
n −1 ]
4⋅2 − 0 3 ⋅ 862 / 9
M1 A1
= 3·26
1 1 1 1
Refer to t 8 Double-tailed 5% point is 2·306 Significant Seems population means differ
No FT if wrong No FT if wrong 10
78
4769
(iii)
Mark Scheme
Wilcoxon rank sum test Wilcoxon signed rank test H0: medianA = medianB H1: medianA ≠ medianB
June 2009
B1 B1 1 1
[Or more formal statements]
4
Q4 (i)
Description must be in context. If no context given, mark according to scheme and then give half-marks, rounded down. Clear description of “rows”. And “columns” As extraneous factors to be taken account of in the design, with “treatments” to be compared. Need same numbers of each Clear contrast with situations for completely randomised design and randomised trends.
(ii)
(iii)
eij ~ ind N (0, σ2)
E1 E1 E1 E1 E1 E1 E1 E1 E1
αi is population mean effect by which ith treatment differs from overall mean
1 1 1 1 1
Source of Variation
1 1
SS
df
MS
MS ratio
Between Treatments
92·30
4
23·075
5·034
Residual
68·76
15
4·584
Total
161·06
19
9 Allow uncorrelated For 0 For σ2
5
1 1 1 1 1
Refer to F4,15 Upper 1% point is 4·89 Significant, seems treatments are not all the same
79
1
No FT if wrong
1 1
No FT if wrong 10
4771
Mark Scheme
June 2009
4771 Decision Mathematics 1 Question 1 (i)
1 and 6, 2 and 5, 3 and 4
B1
(ii) Vertex 6
Vertex 1
Vertex 5
M1 A4
10 to 14 edges (–1 each edge error)
B1 B1
identification sketch
Vertex 2
Vertex 4
Vertex 3
(iii)
Question 2. (i)
A's c takes 2, leaving 3. You have to take 1. A's c takes one and you lose.
M1 A1 A1
(ii)
A's c takes 3 leaving 3. Then as above.
M1 A1
(iii)
A's c takes 3 leaving 4. You can then take 1, leading to a win.
M1 A1 A1
80
4771
Mark Scheme
June 2009
Question 3. (i) (0, 12) 48
(5, 10) 55
(11, 7) 61
(18, 0) 54
(ii)
B3 lines B1 shading
61 at (11, 7)
M1 optimisation A1
Intersection of 2x+5y=60 and x+y=18 is at (10,8)
M1
10 + 2×8 = 26
A1
81
4771
Mark Scheme
June 2009
Question 4. (i)
e.g. 0 – 4 exit
(ii)
e.g.
1 2 3 4 5 6 7 8 9 10
A A A A A A A A A A
E×A B E×A B B B B E×A B E×A
(iii)
e.g. 0–2 exit 6–8 other vertex
(iv)
e.g.
0.7, 0.1, 0.2
A
B
A
B
A E×B A A
B
A
E×A
B B
A E×B
B
E×B
A A A A A A A A A A
B C E×A B E×A C E×A B E×A E×A
M1 A1
process with exits
B1 M1 A1
probabilities duration
M1 DM1 A1 A1
ignore conditionality equal prob efficient
E×B
E×B
(Theoretical answers: 2/3, 1/3, 2) (Gambler's ruin)
0.5, 0.5, 1.9
1 2 3 4 5 6 7 8 9 10
B1 B1
5 – 9 other vertex
3–5 next vertex in cycle 9–ignore and re-draw
A A
B E×A
A
E×A
C
B
C
E×C
A
B
E B
C
E×C
(Theoretical probs are 0.5, 0.25, 0.25) (Markov chain)
82
M1 A2
M1 A1
4771
Mark Scheme
June 2009
Question 5. (i)
e.g. A
B
17 15
18
25 15
15 and/or 20 16 E (ii) A
connectivity lengths
M1 A2
connected tree (–1 each error)
C
20
D
M1 A1 A1
B
C D E
Order: AE; AD; DB; DC or AD; AE; DB; DC or AD; DB; AE; DC Length: 65 km OR B A
A1 B1
C D E
(iii)
Order: AD; DB; DE; DC
Length: 66 km B
A
X
C
M1 A2
D
E
Length: 53 km Advice: Close BC, AE and BD (iv)
B1 B3
facility (e.g. anglers) distances (e.g. B to C)
B1
83
connected tree (–1 each error)
4771
Mark Scheme
June 2009
Question 6. (i)&(ii) 60 200 D 60 0
A E B 160 160 0 100 100 100 40 60 C 150
200 200
F 2
220 220 H 10
G 30 160 190
230 230
sca (activity on arc) single start & end dummy rest
M1 A1 M1 A1
forward pass
B1 B1
time − 230 minutes critical − A; B; E; F; H (iii)
M1 A1 A1 A1
backward pass
cao
e.g.
H G F E D C B A
M1 A2
cascade
B1
Joan/Keith
Least time = 240 mins
B1
Minimum project completion times assumes no resource constraints.
B1
84
4772
Mark Scheme
June 2009
4772 Decision Mathematics 2 Question 1. (a)
e.g. "It is easy to overestimate the effect that your contribution will make."
M1 A1
(b)
remove double negatives same meaning
e.g.
A
B
combinatorial "ands" negations "ors" one for each alternative
M1 A1 A1 A1 A1
8 lines a, b, c negations "and"s "or"s
machine
C
(c)
M1 A1 A1 A1 A3
e.g. ((a 1 1 1 1 0 0 0 0 ~
∧ 1 1 0 0 0 0 0 0 ((~a 0 0 0 0 1 1 1 1
b) 1 1 0 0 1 1 0 0
∨ 1 1 0 0 1 0 1 0 ∧ 0 0 0 0 0 1 0 1
(~a 0 0 0 0 1 1 1 1 ~c) 0 1 0 1 0 1 0 1
∧ 0 0 0 0 1 0 1 0 ∨ 0 0 0 1 0 1 0 1
c)) 1 0 1 0 1 0 1 0 (~b 0 0 1 1 0 0 1 1
∨ 1 1 1 0 1 0 1 0 ∧ 0 0 0 1 0 0 0 1
(~b 0 0 1 1 0 0 1 1
∧ 0 0 1 0 0 0 1 0
~c)) 0 1 0 1 0 1 0 1
c) 1 0 1 0 1 0 1 0
M1 A1
85
4772
Mark Scheme
June 2009
Question 2. (i)
revised D D S S
60marks D S D S
score 48+20 = 68 32+30 = 62 36+28 = 64 24+42 = 66
M1 A1 A1 A1 A1
(ii) D worth 60 revise D
68
0.5 0.5 S worth 60 62
65
M1 A1 A1
65 revise S
D worth 60 65
64
M1 A1
0.5
0.5 S worth 60 66
65 S M predicts D worth 60
65.8
decision node
D worth 60
revise D do not ask M
chance node
68 0.7 66.2 0.3 S worth 60 62
66.2 revise S
D worth 60 64.6
64
0.7
0.3 S worth 60 66
0.5 ask M
M1 chances A1
65.8 D worth 60
0.5 M predicts S worth 60
revise D
68
0.3 0.7 S worth 60 62
63.8
65.4 revise S
D worth 60 65.4
64
0.3
0.7 S worth 60 66 B1
Worth of Michael's help = 0.8 marks
86
M1decisions A1 revise A1 ask/don't
4772
Mark Scheme
June 2009
Question 3. (i)
a is the number of acres of land put to crop A, etc a + b ≤ 20 is equivalent to a + b ≤ c + d Given that a + b + c + d ≤ 40, the maximisation will ensure that a + b + c + d = 40 (and it's easier to solve using simplex).
(ii) P
a 1 0 0 1 0 0 1 0 0
b –50 1 1 0 1 0 0 1 0
c –40 1 1 10 1 0 10 1 0
d –40 0 1 –40 0 1 0 0 1
s1 –30 0 1 –30 0 1 10 0 1
s2 0 1 0 50 1 –1 10 1 –1
M1 A1 A1 A1
RHS 0 20 40 1000 20 20 1800 20 20
0 0 1 0 0 1 40 0 1
M1 A1 M1 A1
20 acres to A and 20 acres to C, giving profit of £1800
B1 B1
(iii) Max 50a + 40b + 40c + 30d st a + b ≤ 20 a + b + c + d ≤ 40 a + b + c + d ≥ 40 A 1 0 0 0 0
P
P 0 1 0 0 0
B1 B1 B1
new obj surplus artificial
B1
3 constraints
Minimise A (to zero) then drop A row and art column and continue normally
B1 B1
OR
OR
0 0 0
b –50 –M 1 1 1
b 1 –40 1 1 1
c –40 –M 1 1 1
c 1 –40 0 1 1
B1 R 40 0 20 40 40
a 1
a 1 –50 1 1 1
B1 B1 B1
d –40 –M 0 1 1
–30 –M 0 1 1
d 1 –30 0 1 1
s1 0 0 1 0 0
s2 0 0 0 1 0
sur –1 0 0 0 –1
art 0 0 0 0 1
s1 s2 sur art R 0 0 M 0 –40M 1 0 0
0 1 0
0 0 –1
0 0 1
20 40 40
Proceed as per simplex, regarding M as a large fixed number.
87
M1 A1 B1 B1 B1 B1
surplus artificial
4772
Mark Scheme
June 2009
Question 4. (a) (i),(ii) and (iii)
1 2 3 4 5
1 ∞ 22 ∞ 15 15
2 22 ∞ 20 5 23
3 ∞ 20 ∞ 40 ∞
4 15 5 40 ∞ 16
5 15 23 ∞ 16 ∞
1 2 3 4 5
1 1 1 1 1 1
2 2 2 2 2 2
3 3 3 3 3 3
4 4 4 4 4 4
5 5 5 5 5 5
1 2 3 4 5
1 ∞ 22 ∞ 15 15
2 22 44 20 5 23
3 ∞ 20 ∞ 40 ∞
4 15 5 40 30 16
5 15 23 ∞ 16 30
1 2 3 4 5
1 1 1 1 1 1
2 2 1 2 2 2
3 3 3 3 3 3
4 4 4 4 1 4
5 5 5 5 5 1
1 2 3 4 5
1 44 22 42 15 15
2 22 44 20 5 23
3 42 20 40 25 43
4 15 5 25 10 16
5 15 23 43 16 30
1 2 3 4 5
1 2 1 2 1 1
2 2 1 2 2 2
3 2 3 2 2 2
4 4 4 2 2 4
5 5 5 2 5 1
1 2 3 4 5
1 44 22 42 15 15
2 22 44 20 5 23
3 42 20 40 25 43
4 15 5 25 10 16
5 15 23 43 16 30
1 2 3 4 5
1 2 1 2 1 1
2 2 1 2 2 2
3 2 3 2 2 2
4 4 4 2 2 4
5 5 5 2 5 1
1 2 3 4 5
1 30 20 40 15 15
2 20 10 20 5 21
3 40 20 40 25 41
4 15 5 25 10 16
5 15 21 41 16 30
1 2 3 4 5
1 4 4 2 1 1
2 4 4 2 2 4
3 4 3 2 2 4
4 4 4 2 2 4
5 5 4 2 5 1
1 2 3 4 5
1 30 20 40 15 15
2 20 10 20 5 21
3 40 20 40 25 41
4 15 5 25 10 16
5 15 21 41 16 30
1 2 3 4 5
1 4 4 2 1 1
2 4 4 2 2 4
3 4 3 2 2 4
4 4 4 2 2 4
5 5 4 2 5 1
Shortest distance from 3 to 1 is 40 (1st row and 3rd column of distance matrix)
88
M1 distance A1 1 to 5 etc A1 rest B1
route
Not part of the question
Not part of the question
Not part of the question
M1 A1 10 changed dists M1 2's in r3 of route A1 rest of route
B1 B1
4772
Mark Scheme
June 2009
B1
Shortest route is 3 2 4 1 3 followed by route matrix (3,1) = 2 followed by route matrix (2,1) = 4 followed by route matrix (4,1) = 1
M1 A1
(iv)
1
21
15
5
15
M1 A1
20 25
16
5
40
2
20
4
3
41
(v)
2 (5) 4 (15) 1 (15) 5 (41) 3 (20) 2
Total length = 96
B1 B1
2 4 1 5 (4 2) 3 2
M1 A1
Finds a (hopefully short) route visiting every vertex and returning to the start, or, upper bound to the TSP
B1
89
4773
Mark Scheme
June 2009
4773 Decision Mathematics Computation Question 1. (i)
Bn+2 = Bn+1 + (0 – Bn)
M1 A1
(ii)
Oscillation: 2, 4, 2, –2, –4, –2, 2, 4, …
M1 A1 B1
(iii)
Bn+2 – Bn+1 +
1 2
B1 B1 B1
Bn = 0
2, 4, 3, 1, –0.5, –1, …, 0.00391, –0.00195 Oscillatory convergence (iv)
2, 4, 3.5, 2.5, 1.625, 1, …, 0.00022, 0.00012 Faster and uniform convergence
(v)
Auxiliary eqn: x2 – x + x=
1 4
B1 B1 B1 B1 B1
=0
1 2
B1
Bn = A( ) + Bn( ) 2=A 4 = 1 + 12 B giving B = 6 1 n 2
Bn = ( 2 + 6n )
1 n 2
( ) 1 2
n
B1 B1
(12 )n
n −1 (2 + 6n ) or (1 + 3n ) 12
( )
or (1 + 3n )( 12 )
"the same"
90
n
B1 B1
4773
Mark Scheme
June 2009
Question 2. (i)
12
e.g.
2
10
20
20
10
20
10
10
0
10
2
18
18
8
10
28
10
2
8
18
20
0
2
12
10
20
10
2
8
18
0
18
36
10
0
10
20
10
20
10
0
2
12
10
20
10
2
8
10
18
36
reversal
B1
rest
0
0
20
8
0
M1 A1
10
10
18
2
18 10
0
0
10
8
18
M1 A1
10
10
2
10
10
10
10
2
20
0
12
20
10 10
18
12
0 0
10
20
10
10
10
12
10
M1 A1
10 10
91
0
4773
Mark Scheme
June 2009
(ii)
{S, A, C, D, E} / {B, T}
M1 A1
(iii)
e.g. Max SA + SE st SA + BA + CA – AB – AC = 0 AB + CB – BA – BC – BT = 0 AC + DC + BC – CA – CB – CD = 0 SE + DE – ED = 0 CD + ED – DC – DE – DT = 0 SA < 12 SE < 20 AB < 10 BA < 10 AC < 10 CA < 10 BC < 10 CB < 10 CD < 10 DC < 10 ED < 18 DE < 18 BT < 20 DT < 10 end
M1 A1
variables objective
M1 A1
balancing
M1 A1 A1
capacities forwards backwards
B1
running
(iv)
OBJECTIVE FUNCTION VALUE 1) 30.00000 VARIABLE VALUE REDUCED COST SA 12.000000 0.000000 SE 18.000000 0.000000 BA 0.000000 0.000000 CA 0.000000 0.000000 AB 0.000000 10.000000 0.000000 AC 2.000000 CB 10.000000 0.000000 0.000000 BC 0.000000 BT 0.000000 20.000000 DC 0.000000 8.000000 CD 0.000000 0.000000 0.000000 DE 1.000000 18.000000 ED 0.000000 10.000000 DT 0.000000
B1
Solution as per part (i)
92
4773
Mark Scheme
June 2009
Question 3. (i)
e.g. -1 1 1 1 1 1 -1 etc.
5 4 5 6 7 8 9 8
= if(rand() | b | + 2 will ensure convergence. ( > required as dominance has to be strict)
(ii) 4 1 2 1
1 4 1 2
2 1 4 1
1 2 1 4
0 0.25 0.321289 0.340733 0.344469 0.344515 0.344124 0.343886 0.343789 0.343758 0.34375 0.343749 0.34375 0.34375
0 -0.0625 -0.05103 -0.03941 -0.03388 -0.0319 -0.03134 -0.03123 -0.03123 -0.03124 -0.03125 -0.03125 -0.03125 -0.03125
0 -0.10938 -0.14691 -0.15599 -0.15715 -0.15681 -0.15648 -0.15633 -0.15627 -0.15625 -0.15625 -0.15625 -0.15625 -0.15625
0 -0.00391 -0.01808 -0.02648 -0.02989 -0.03098 -0.03124 -0.03127 -0.03127 -0.03126 -0.03125 -0.03125 -0.03125 -0.03125
2 1 4 1
1 2 1 4
4 1 2 1
1 4 1 2
0 0.5 2.03125 6.033203 12.69934 3.347054 -156.613 -1153.81 -5937.67
0 -0.25 -1.95313 -10.5127 -46.9236 -183.278 -632.515 -1881.51 -4365.6
0 -0.875 -3.42969 -9.11279 -13.2195 37.89147 456.5137 2690.835 12560.88
0 0.6875 4.605469 22.56519 94.10735 345.9377 1115.079 2994.509 5419.593
1 0 0 0
a 4
[E1] [E1E1] [subtotal 3] b 2
setup [M3A3]
values [A3]
1 0 0 0
a 2
b 4
values [A3] [subtotal 12]
(iii)
No convergence when a = 2, b = 0 Indicates that non-strict diagonal dominance is not sufficient 101
[M1A1] [E1E1]
4777
Mark Scheme
June 2009
[subtotal 4] (iv)
Use RHSs 1,0,0,0 0,1,0,0 to obtain inverse as 0.34375 -0.03125 -0.15625 -0.03125
-0.03125 0.34375 -0.03125 -0.15625
0,0,1,0 0,0,0,1 -0.15625 -0.03125 0.34375 -0.03125
-0.03125 -0.15625 -0.03125 0.34375
102
[M1]
[A1] [A1] [A1] [A1] [subtotal 5] [TOTAL 24]
Grade Thresholds Advanced GCE MEI Mathematics 7895-8 3895-8 June 2009 Examination Series Unit Threshold Marks Unit
Maximum Mark
A
B
C
D
E
U
All units
UMS
100
80
70
60
50
40
0
4751
Raw
72
59
52
45
39
33
0
4752
Raw
72
51
44
38
32
26
0
4753
Raw
72
57
52
47
42
37
0
4753/02
Raw
18
15
13
11
9
8
0
4754
Raw
90
67
59
51
43
35
0
4755
Raw
72
53
45
37
30
23
0
4756
Raw
72
51
45
39
33
27
0
4757
Raw
72
60
51
42
34
26
0
4758
Raw
72
61
55
49
43
36
0
4758/02
Raw
18
15
13
11
9
8
0
4761
Raw
72
57
48
39
30
21
0
4762
Raw
72
47
40
33
26
20
0
4763
Raw
72
55
46
38
30
22
0
4764
Raw
72
61
52
43
34
26
0
4766/G241
Raw
72
60
53
46
40
34
0
4767
Raw
72
57
50
44
38
32
0
4768
Raw
72
55
48
41
34
28
0
4769
Raw
72
56
49
42
35
28
0
4771
Raw
72
63
56
49
42
36
0
4772
Raw
72
57
51
45
39
33
0
4773
Raw
72
51
44
37
30
24
0
4776
Raw
72
62
53
45
37
28
0
4776/02
Raw
18
14
12
10
8
7
0
4777
Raw
72
55
47
39
32
25
0
103
Specification Aggregation Results
Overall threshold marks in UMS (ie after conversion of raw marks to uniform marks) A
B
C
D
E
U
7895-7898
Maximum Mark 600
480
420
360
300
240
0
3895-3898
300
240
210
180
150
120
0
The cumulative percentage of candidates awarded each grade was as follows: A
B
C
D
E
U
7895
44.1
65.4
81.4
92.1
97.9
100
Total Number of Candidates 10375
7896
57.2
78.0
88.9
95.4
98.9
100
1807
7897
87.1
93.55
100
100
100
100
31
7898
0
0
100
100
100
100
1
3895
35.3
52.9
67.4
79.1
88.1
100
16238
3896
52.1
70.2
82.4
90.4
95.7
100
2888
3897
80.4
88.2
91.2
96.1
97.1
100
102
3898
6.3
12.5
18.8
25.0
68.8
100
16
For a description of how UMS marks are calculated see: http://www.ocr.org.uk/learners/ums_results.html Statistics are correct at the time of publication.
104
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