qualifications for schools and colleges in the United Kingdom, including ...
Advanced Subsidiary GCE Further Mathematics (MEI) (3896) ..... ft only if integral
attempted, not for x4 .... [4]. Implicit differentiation – allow one slip (but with dy/dx
both.
GCE Mathematics (MEI) Advanced GCE A2 7895-8 Advanced Subsidiary GCE AS 3895-8
Mark Schemes for the Units January 2008
3895-8/7895-8/MS/R/08J
Oxford Cambridge and RSA Examinations
OCR (Oxford, Cambridge and RSA Examinations) is a unitary awarding body, established by the University of Cambridge Local Examinations Syndicate and the RSA Examinations Board in January 1998. OCR provides a full range of GCSE, A level, GNVQ, Key Skills and other qualifications for schools and colleges in the United Kingdom, including those previously provided by MEG and OCEAC. It is also responsible for developing new syllabuses to meet national requirements and the needs of students and teachers. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by Examiners. It does not indicate the details of the discussions which took place at an Examiners’ meeting before marking commenced. All Examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the Report on the Examination. OCR will not enter into any discussion or correspondence in connection with this mark scheme. © OCR 2008 Any enquiries about publications should be addressed to: OCR Publications PO Box 5050 Annesley NOTTINGHAM NG15 0DL Telephone: Facsimile: E-mail:
0870 770 6622 01223 552310
[email protected]
2
CONTENTS Advanced GCE Further Mathematics (MEI) (7896) Advanced GCE Further Mathematics (Additional) (MEI) (7897) Advanced GCE Mathematics (MEI) (7895) Advanced GCE Pure Mathematics (MEI) (7898) Advanced Subsidiary GCE Further Mathematics (MEI) (3896) Advanced Subsidiary GCE Further Mathematics (Additional) (MEI) (3897) Advanced Subsidiary GCE Mathematics (MEI) (3895) Advanced Subsidiary GCE Pure Mathematics (MEI) (3898)
MARK SCHEME FOR THE UNITS Unit/Content
Page
4751 (C1)
Introduction to Advanced Mathematics
1
4752 (C2)
Concepts for Advanced Mathematics
5
4753 (C3)
Methods for Advanced Mathematics
7
4754 (C4)
Applications of Advanced Mathematics
11
4755 (FP1)
Further Concepts for Advanced Mathematics
16
4756 (FP2)
Further Methods for Advanced Mathematics
20
4758
Differential Equations
27
4761
Mechanics 1
31
4762
Mechanics 2
37
4763
Mechanics 3
41
4766
Statistics 1
46
4767
Statistics 2
50
4768
Statistics 3
54
4771
Decision Mathematics 1
58
4776
Numerical Methods
64
Grade Thresholds
66
3
4751
Mark Scheme 4751 (C1)
January 2008
Introduction to Advanced Mathematics
Section A 1
[v = ] [ ± ]
2E www m
3
M2 for v 2 =
2E or for [v =] [ ± ] m
E or m
1 2
M1 for a correct constructive first step and M1 for v = [ ± ] k ft their v2 = k;
2
3x − 4 7 or 3 − www as final x +1 x +1
3
(i) 1
1
(ii) 1/64 www
3
if M0 then SC1 for √E/ ½ m or √2E/m etc M1 for (3x − 4)(x − 1) and M1 for (x + 1)(x − 1)
3
answer 3
4
5
3
6x + 2(2x − 5) = 7
M1
10x = 17
M1
x = 1.7 o.e. isw y = −1.6 o.e .isw
A1 A1
M1 for dealing correctly with each of reciprocal, square root and cubing (allow 3 only for 1/64) eg M2 for 64 or −64 or 1/√4096 or ¼3 or M1 for 1/163/2 or 43 or −43 or 4-3 etc
4
for subst or multn of eqns so one pair of coeffts equal (condone one error) simplification (condone one error) or appropriate addn/subtn to eliminate variable allow as separate or coordinates as requested graphical soln: M0
4
(i) −4/5 or −0.8 o.e.
2
M1 for 4/5 or 4/−5 or 0.8 or −4.8/6 or correct method using two points on the line (at least one correct) (may be graphical) or for −0.8x o.e.
(ii) (15, 0) or 15 found www
3
M1 for y = their (i) x + 12 o.e. or 4x + 5y = k and (0, 12) subst and M1 for using y = 0 eg −12 = −0.8x or ft their eqn or M1 for given line goes through (0, 4.8) and (6, 0) and M1 for 6 × 12/4.8 graphical soln: allow M1 for correct required line drawn and M1 for answer within 2mm of (15, 0)
1
5
19
4751
6
Mark Scheme
f(2) used
M1
23 + 2k + 7 = 3
M1
k = −6 7
8
9
January 2008
or division by x − 2 as far as x2 + 2x obtained correctly or remainder 3 = 2(4 + k) + 7 o.e. 2nd M1 dep on first
A1
(i) 56
2
(ii) −7 or ft from −their (i)/8
2
3 M1 for
8× 7 × 6 or more simplified 3 × 2 ×1
M1 for 7 or ft their (i)/8 or for 56 × (−1/2)3 o.e. or ft; condone x3 in answer or in M1 expression; 0 in qn for just Pascal’s triangle seen
(i) 5√3
2
M1 for √48 = 4√3
(ii) common denominator = (5 − √2)(5 + √2) =23 numerator = 10
M1 A1 B1
allow M1A1 for
5− 2 5+ 2 + 23 23
allow 3 only for 10/23
(i) n = 2m
M1
or any attempt at generalising; M0 for just trying numbers
3n2 + 6n = 12m2 + 12m or = 12m(m + 1)
M2
or M1 for 3n2 + 6n = 3n (n + 2) = 3 × even × even and M1 for explaining that 4 is a factor of even × even or M1 for 12 is a factor of 6n when n is even and M1 for 4 is a factor of n2 so 12 is a factor of 3n2
(ii) showing false when n is odd e.g. 3n2 + 6n = odd + even = odd
B2
or 3n (n + 2) =3 × odd × odd = odd or counterexample showing not always true; M1 for false with partial explanation or incorrect calculation
2
4
5
5
17
4751
Mark Scheme
January 2008
Section B 10
i
ii iii
correct graph with clear asymptote x = 2 (though need not be marked)
(0, − ½ ) shown 11/5 or 2.2 o.e. isw
x=
1 x−2
x(x − 2) = 1 o.e. x2 − 2x − 1 [= 0]; ft their equiv eqn attempt at quadratic formula 1 ±√2 cao position of points shown
11
i
(x − 2.5)2 o.e. − 2.52 + 8 (x − 2.5)2 + 7/4 o.e. min y = 7/4 o.e. [so above x axis] or commenting (x − 2.5)2 ≥ 0
ii
iii
iv
G2
G1 for one branch correct; condone (0, − ½ ) not shown SC1 for both sections of graph shifted two to left
G1
allow seen calculated
3
2
M1 for correct first step
2
M1
or equivs with ys
M1 M1 M1 A1 B1
M1 M1 A1 B1
or (x − 1)2 − 1 = 1 o.e. or (x − 1) = ±√2 (condone one error) on their curve with y = x (line drawn or y = x indicated by both coords); condone intent of diagonal line with gradient approx 1through origin as y = x if unlabelled
11
for clear attempt at −2.52 allow M2A0 for (x − 2.5) + 7/4 o.e. with no (x − 2.5)2 seen ft, dep on (x − a)2 + b with b positive; condone starting again, showing b2 − 4ac < 0 or using calculus
correct symmetrical quadratic shape 8 marked as intercept on y axis tp (5/2, 7/4) o.e. or ft from (i)
G1 G1 G1
or (0, 8) seen in table
x2 − 5x − 6 seen or used −1 and 6 obtained x < − 1 and x > 6 isw or ft their solns
M1 M1 M1
or (x − 2.5)2 [> or =] 12.25 or ft 14 − b also implies first M1 if M0, allow B1 for one of x < − 1 and x>6
min = (2.5, − 8.25) or ft from (i) so yes, crosses
M1 A1
3
6
4
3
or M1 for other clear comment re translated 10 down and A1 for referring to min in (i) or graph in (ii); or M1 for correct method for solving x2 −5x −2 = 0 or using b2 − 4ac with this and A1 for showing real solns eg b2 − 4ac = 33; allow M1A0 for valid comment but error in −8.25 ft; allow M1 for showing y can be neg eg (0, −2) found and A1 for correct conclusion
3
2
12
4751
12
Mark Scheme
i
(x − 4)2 − 16 + (y − 2)2 − 4 = 9 o.e.
M2
B1 rad = √29
ii
iii
iv
42 + 22 o.e = 20 which is less than 29
M1 A1
showing midpt of AB = (4, 2) and showing AB = 2√29 or showing AC or BC = √29 or that A or B lie on circle
2
or showing both A and B lie on circle (or AC = BC = √29), and showing AB = 2√29 or that C is midpt of AB or that C is on AB or that gradients of AB and AC are the same or equiv.
2
or showing C is on AB and showing both A and B are on circle or AC = BC = √29
2
grad AC or AB or BC = −5/2 o.e.
M1
grad tgt = −1/their grad AC tgt is y − 7 = their m (x − 2) o.e.
M1 M1
y = 2/5 x + 31/5 o.e.
A1
2
2
January 2008
M1 for one completing square or for (x − 4)2 or (y − 2)2 expanded correctly or starting with (x − 4)2 + (y − 2)2 = r2: M1 for correct expn of at least one bracket and M1 for 9 + 20 = r2 o.e. or using x2 −2gx + y2 − 2fy+ c = 0 M1 for using centre is (g, f) [must be quoted] and M1 for r2 = g2 + f2 − c
3
allow 2 for showing circle crosses x axis at −1 and 9 or equiv for y (or showing one positive; one negative); 0 for graphical solutions (often using A and B from (iii) to draw circle)
2
in each method, two things need to be established. Allow M1 for the concept of what should be shown and A1 for correct completion with method shown allow M1A0 for AB just shown as √116 not 2√29 allow M1A0 for stating mid point of AB = (4,2) without working/method shown NB showing AB = 2√29 and C lies on AB is not sufficient – earns 2 marks only
2
4
if M0, allow SC2 for accurate graph of circle drawn with compasses and AB joined with ruled line through C. may be seen in (iii) but only allow this M1 if they go on to use in this part allow for m1m2=−1 used eg y = their mx + c then (2, 7) subst; M0 if grad AC used condone y = 2/5x + c and c = 31/5 o.e.
4
4
13
4752
Mark Scheme 4752 (C2)
January 2008
Concepts for Advanced Mathematics
Section A 1
40x3
2
2
(i) 3
1
(ii) 141
2
M1 for 9 × (1 + 2 + 3 + 4 + 5) + 1 + 2 +3
right angled triangle with 1 and 2 on correct sides Pythagoras used to obtain hyp = √5
M1
or M1 for sinθ = ½cosθ and M1 for substituting in sin2 θ + cos2θ = 1 E1 for sufficient working
3
a 2 cos θ = = h 5 4
5
6
7
-1 if extra term
M1 A1 2
1 for two points correct
(ii) line along y = 3 with V (−2,3), (−1,1), (0,3)
2
1 for two points correct 4
3 2 x6 + x + 7 x + c 4
5
(i) correct sine shape through O amplitude of 1 and period 2π shown
1 1
(ii) 7π/6 and 11π/6
3
B2 for one of these; 1 for −π/6 found
(i) 60
2
M1 for 22 + 23 + 24 + 25 o.e.
(ii) −6 (iii)
1
4 3
1 for 2x6; 2 for
4 3
5
5
y
1 1
Correct in both quadrants Through (0, 1) shown dep.
2 -
9
4 3
3 x or 1 for other kx ; 1 for 7x; 4
1 for +c
4
0
5 2
r = 1/3 s.o.i. a = 54 or ft 18 ÷ their r S=
3
3
(i)line along y = 6 with V (1, 6), (2, 2), (3, 6)
6
8
2
x
a used with -1 < r < 1 1− r
1 mark for ar = 18 and ar3 = 2 s.o.i.
2 M1 M1 A1
5
S = 81 c.a.o. (i) 0.23 c.a.o.
1
(ii) 0.1 or 1/10
1
(iii) 4(3x + 2) or 12x + 8
1
(iv) [y = ] 103x + 2 o.e.
1 5
10-1 not sufficient 4
5
4752
Mark Scheme
January 2008
Section B 10
i
ii
iii
11
iA
iB
ii
12
iA
iB
h = 120/x2 A = 2x2 + 4xh o.e.
completion to given answer
B1 M1 A1
A′ = 4x − 480/x2 o.e. A′′ = 4 + 960 / x3
2 2
use of A′ = 0 x = 3 120 or 4.9(3..) Test using A′ or A′′ to confirm minimum Substitution of their x in A A = 145.9 to 146
M1 A1
BC2 = 3482 + 3022 − 2 × 348 × 302 × cos 72° BC = 383.86… 1033.86…[m] or ft 650 + their BC
M2
sin B sin 72 = 302 their BC B = 48.4.. 355 − their B o.e. answer in range 306 to307
M1
224 × 2π × 120 360 o.e. or 469.1... to 3 sf or more QP = 222.5…to 3 sf or more answer in range 690 to 692 [m]
Arc length PQ =
A1 1
A1 M1 A1 M2
3
1 for kx-2 o.e. included ft their A′ only if kx-2 seen ; 1 if one error
4
Dependent on previous M1 M1 for recognisable attempt at Cosine Rule to 3 sf or more accept to 3 sf or more
5
4
Cosine Rule acceptable or Sine Rule to find C or 247 + their C 4 M1 for
136 × 2π × 120 360
B1 A1 4
x4 = 8x (2, 16) c.a.o. PQ = 16 and completion to show ½ × 2 × 16 = 16
M1 A1
x5/5
M1 M1
A1
NB answer 16 given
3
3
9.6 o.e.
A1
ft only if integral attempted, not for x4 or differentiation c.a.o.
6x2h2 + 4xh3 + h4
2
B1 for two terms correct.
evaluating their integral at their co-ord of P and zero [or 32/5 o.e.] iiA
T1 M1 A1
at least one interim step shown
2 iiB
4x3 + 6x2h + 4xh2 + h3
2
iiC
4x3
1
1
iiD
gradient of [tangent to] curve
1
1
2
6
B1 for three terms correct
2
4753
Mark Scheme 4753 (C3)
January 2008
Methods for Advanced Mathematics
Section A y = (1 + 6 x 2 )1/ 3 dy 1 = (1 + 6 x 2 ) −2 / 3 .12 x dx 3 = 4 x(1 + 6 x 2 ) −2 / 3
M1
chain rule used
B1
1 −2 / 3 u 3
A1 A1 [4]
×12x cao (must resolve 1/3 × 12) Mark final answer
2 (i) fg(x) = f(x – 2) = (x – 2)2 gf(x) = g(x2) = x2 – 2.
M1 A1 A1 [3]
forming a composite function mark final answer If fg and gf the wrong way round, M1A0A0
B1ft
fg – must have (2, 0)labelled (or inferable from scale). Condone no y-intercept, unless wrong
1 ⇒
(ii)
fg(x)
B1ft gf – must have (0, –2) labelled (or inferable from scale) Condone no x-intercepts, unless wrong
2 gf(x) -2
3 (i) When n = 1, 10 000 = A eb when n = 2,2b 16 000 = A e2b 16000 Ae = = eb ⇒ 10000 Aeb ⇒ ⇒
eb = 1.6 b = ln 1.6 = 0.470 A = 10000/1.6 = 6250.
(ii) When n = 20, P = 6250×e0.470×20 = £75,550,000 4 (i) 5 = k/100 ⇒ k = 500*
(ii)
dP 500 = −500V −2 = − 2 dV V
(iii) dP = dP . dV dt dV dt When V = 100, dP/dV = –500/10000 = –0.05 dV/dt = 10 ⇒ dP/dt = –0.05 × 10 = –0.5 So P is decreasing at 0.5 Atm/s
[2]
Allow ft only if fg and gf are correct but wrong way round.
B1 B1 M1
soi soi eliminating A (do not allow verification)
E1 B1 B1 [6]
SCB2 if initial ‘B’s are missing, and ratio of years = 1.6 b =e ln 1.6 or 0.47 or better (mark final answer) cao – allow recovery from inexact b’s
M1 A1 [2]
substituting n = 20 into their equation with their A and b Allow answers from £75 000 000 to £76 000 000.
E1 [1]
NB answer given
M1 A1 [2]
(–1)V–2 o.e. – allow –k/V2
M1
chain rule (any correct version)
B1ft B1 A1 [4]
(soi) (soi) –0.5 cao
7
4753
Mark Scheme
January 2008
5(i) p = 2, 2p – 1 = 3, prime p = 3, 2p – 1 = 7, prime p = 5, 2p – 1 = 31, prime p = 7, 2p – 1 = 127, prime
M1 E1 [2]
Testing at least one prime testing all 4 primes (correctly) Must comment on answers being prime (allow ticks) Testing p = 1 is E0
(ii) 23×89 = 2047 = 211 – 1 11 is prime, 2047 is not So statement is false.
M1 E1 [2]
211 – 1 must state or imply that 11 is prime (p = 11 is sufficient)
M1
Implicit differentiation – allow one slip (but with dy/dx both sides)
6 (i) e2y = x2 + y dy dy ⇒ 2e 2 y = 2x + dx dx dy 2y ⇒ (2e − 1) = 2 x dx dy 2x * ⇒ = dx 2e2 y − 1 (ii) Gradient is infinite when 2e2y – 1 = 0 ⇒ ⇒ ⇒ ⇒
e2y = ½ 2y = ln ½ y = ½ ln ½ = –0.347 (3 s.f.) x2 = e2y – y = ½ – (–0.347) = 0.8465 x = 0.920
A1 M1
collecting terms
E1 [4] M1 A1 M1
must be to 3 s.f. substituting their y and solving for x
A1 [4]
cao – must be to 3 s.f., but penalise accuracy once only.
8
4753
Mark Scheme
January 2008
Section B y = 2x ln(1 + x) dy 2x ⇒ = + 2 ln(1 + x) dx 1 + x When x = 0, dy/dx = 0 + 2 ln 1 = 0 ⇒ origin is a stationary point.
M1 B1 A1
product rule d/dx(ln(1+x)) = 1/(1+x) soi
E1 [4]
www (i.e. from correct derivative)
d 2 y (1 + x).2 − 2 x.1 2 = + dx 2 (1 + x) 2 1+ x 2 2 = + 2 (1 + x) 1 + x When x = 0, d2y/dx2 = 2 + 2 = 4 > 0 ⇒ (0, 0) is a min point
M1 A1ft
Quotient or product rule on their 2x/(1 + x) correctly applied to their 2x/(1+x)
7(i)
(ii)
(iii) ⇒
⇒
Let u = 1 + x ⇒ du = dx x2 (u − 1)2 dx = ∫ 1 + x ∫ u du 2 = (u − 2u + 1) du ∫ u = (u − 2 + 1 )du * ∫ u 2 1 x 2 1 ∫0 1 + x dx = ∫1 (u − 2 + u )du 2 ⎡1 ⎤ = ⎢ u 2 − 2u + ln u ⎥ ⎣2 ⎦1 = 2 – 4 + ln 2 – ( ½ – 2 + ln 1) = ln 2 – ½
A1 M1 E1 [5]
o.e., e.g. 4 + 2 x cao (1 + x) 2 substituting x = 0 into their d2y/dx2 www – dep previous A1
M1
(u − 1) 2 u
E1
www (but condone du omitted except in final answer)
B1
changing limits (or substituting back for x and using 0 and 1)
B1
⎡1 2 ⎤ ⎢⎣ 2 u − 2u + ln u ⎥⎦
M1 A1 [6]
substituting limits (consistent with u or x) cao
M1
soi
A1 M1
substituting their ln 2 – ½ for
1
(iv) A = 2 x ln(1 + x)dx ∫ 0
Parts: u = ln(1 + x), du/dx = 1/(1 + x) dv/dx = 2x ⇒ v = x2 2 1 x 1 = ⎡⎣ x 2 ln(1 + x) ⎤⎦ − ∫ dx 0 0 1+ x = ln 2 – ln 2 + ½ = ½
A1 [4]
cao
9
x2 ∫0 1+ x dx 1
4753
Mark Scheme
8 (i) Stretch in x-direction s.f. ½ translation in y-direction 1 unit up
(ii)
A=∫
π /4 −π / 4
(1 + sin 2 x ) dx
M1 A1 M1
(in either order) – allow ‘contraction’ dep ‘stretch’ allow ‘move’, ‘shift’, etc – direction can be inferred from
A1 [4]
⎛0⎞ ⎜1 ⎟ ⎝ ⎠ or ⎛ 0 ⎞ dep ‘translation’. ⎜ ⎟ ⎝1 ⎠
M1
π /4
1 ⎡ ⎤ = ⎢ x − cos 2 x ⎥ 2 ⎣ ⎦ −π / 4
M1
= π/2
A1 [4] M1 A1
(iii) y = 1 + sin 2x ⇒ dy/dx = 2cos 2x When x = 0, dy/dx = 2 So gradient at (0, 1) on f(x) is 2 ⇒ gradient at (1, 0) on f–1(x) = ½
(iv) Domain is 0 ≤ x ≤ 2. y 2
O
π/4
correct integral and limits. Condone dx missing; limits may be implied from subsequent working.
2
substituting their limits (if zero lower limit used, must show evidence of substitution) or 1.57 or better – cao (www)
differentiating – allow 1 error (but not x + 2cos 2x)
A1ft B1ft [4]
If 1, then must show evidence of using reciprocal, e.g. 1/1
B1
Allow 0 to 2, but not 0 < x < 2 or y instead of x
M1 A1
clear attempt to reflect in y = x correct domain indicated (0 to 2), and reasonable shape
x [3]
(v) ⇒ ⇒ ⇒
y = 1 + sin 2x x ↔y x = 1 + sin 2y sin 2y = x – 1 2y = arcsin(x – 1) y = ½ arcsin(x – 1)
⎛ 0 ⎞ alone is M1 A0 ⎜ ⎟ ⎝1 ⎠
B1
= π/4 – ½ cos π/2 + π/4 + ½ cos (–π/2)
–π/4
January 2008
M1
or sin 2x = y – 1
A1 [2]
cao
10
4754
Mark Scheme 4754 (C4)
January 2008
Applications of Advanced Mathematics
Section A 1 3 cos θ + 4sin θ = R cos(θ − α) = R(cos θ cos α + sin θ sin α) ⇒ R cos α = 3 , R sin α = 4 ⇒ R2 = 32 + 42 = 25 ⇒ R = 5 tan α = 4/3 ⇒ α = 0.9273
M1 B1 M1A1
5 cos(θ − 0.9273)= 2 ⇒ cos(θ − 0.9273) = 2/5 θ − 0.9273 = 1.1593, –1.1593 ⇒ θ = 2.087, –0.232
M1 A1 A1 [7]
and no others in the range
M1 A1
binomial expansion with p = -½ correct expression
A1 M1 A1 [5]
cao
M1
substitituting their 1 + x + 3 x 2 + ... and 2 expanding
2(i) (1 − 2 x)
−
1 2
1 = 1 − (−2 x) + 2
1 3 (− ).(− ) 2 2 (−2 x) 2 + ... 2!
3 = 1 + x + x 2 + ... 2 Valid for −1 < –2x < 1 ⇒ - ½ < x < ½
(ii) 1 + 2 x = (1 + 2 x)(1 + x + 3 x 2 + ...) 2 1− 2x 3 2 2 = 1 + x + x + 2 x + 2 x + ... 2 7 = 1 + 3 x + x 2 + ... 2
A1ft A1 [3]
R=5 cwo
cao
2
3 V = π x 2 dy ∫ 1
y = 1 + x2 ⇒ x2 = y– 1
B1
2
⇒ V = π ( y − 1)dy ∫ 1
M1
2
⎡1 ⎤ = π ⎢ y2 − y ⎥ 2 ⎣ ⎦1
= π(2 – 2 – ½ +1) =½π
B1 M1 A1 [5]
11
⎡1 2 ⎤ ⎢⎣ 2 y − y ⎥⎦
substituting limits into integrand
4754
4(i) ⇒ ⇒ ⇒ ⇒ ⇒
Mark Scheme
sin(θ + 45°) = cos θ sin θ cos 45 + cos θ sin 45 = cos θ (1/√2) sin θ + (1/√2) cos θ = cos θ sin θ + cos θ = √2 cos θ sin θ = (√2 – 1) cos θ sin θ = tan θ = √2 – 1 * cos θ
(ii) ⇒
tan θ = √2 – 1 θ = 22.5°, 202.5°
4 A Bx + C = + x( x 2 + 4) x x 2 + 4 A( x 2 + 4) + ( Bx + C ) x = x( x 2 + 4) 2 ⇒ 4 = A( x + 4) + ( Bx + C ) x x = 0 ⇒ 4 = 4A ⇒ A = 1 coefft of x2: 0 = A + B ⇒ B = –1 coeffts of x: 0 = C 4 1 x ⇒ = − x( x 2 + 4) x x 2 + 4 5
6
cosec θ = 3 ⇒ sin θ = 1/3 ⇒ θ = 19.47°, 160.53°
M1 B1 A1 M1
January 2008
compound angle formula sin 45 = 1/√2, cos 45 = 1/√2 collecting terms
E1 [5]
B1 B1 [2]
and no others in the range
M1
correct partial fractions
M1 B1 DM1 A1 A1 [6]
A=1 Substitution or equating coeffts B= –1 C= 0
M1 A1 A1 [3]
12
and no others in the range
4754
Mark Scheme
January 2008
Section B
⎛0 ⎞ ⎛ −6 ⎞ 7(i) uuur ⎜ ⎟ uuur ⎜ ⎟ . CD = ⎜ 6 ⎟ C B = ⎜ 20 ⎟ ⎜ 24 ⎟ ⎝ ⎠
(ii)
⎜0 ⎟ ⎝ ⎠
( −6) 2 + 62 + 242
= 25.46 cm
−6 ⎞ ⎛ 4 ⎞ ⎟⎜ ⎟ ⎟ . ⎜ 0 ⎟ = −24 + 0 + 24 = 0 ⎜1 ⎟ ⎜ 24 ⎟ ⎜1 ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎛ 4⎞ ⎛0 ⎞ ⎛ 4⎞ uuur ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ C B. ⎜ 0 ⎟ = ⎜ 20 ⎟ . ⎜ 0 ⎟ = 0 + 0 + 0 = 0 ⎜1 ⎟ ⎜ 0 ⎟ ⎜1 ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠
⎛ ⎞ ⎛ (iii) uuur ⎜ ⎟ ⎜ CD. ⎜ 0 ⎟ = ⎜ 6
B1 B1 [2] M1 A1 [2]
4
⇒ plane BCDE is 4x + z = c At C (say) 4×15 + 0 = c ⇒ c = 60 ⇒ plane BCDE is 4x + z = 60 ⎛0⎞ ⎛3 ⎞ ⎜ ⎟ ⎜ ⎟ r = ⎜0⎟ + λ ⎜6 ⎟ ⎜0⎟ ⎜ 24 ⎟ ⎝ ⎠ ⎝ ⎠ ⎛0 ⎞ ⎛3 ⎞ AF: ⎜ ⎟ ⎜ ⎟ r = ⎜ 20 ⎟ + μ ⎜ −6 ⎟ ⎜0 ⎟ ⎜ 24 ⎟ ⎝ ⎠ ⎝ ⎠
(iv) OG:
At (5, 10, 40) , 3λ = 5 ⇒ λ = 5/3 ⇒ 6λ = 10 , 24λ = 40, so consistent. At (5, 10, 40) , 3μ = 5 ⇒ μ = 5/3 ⇒ 20 − 6μ = 10, 24μ = 40, so consistent. So lines meet at (5, 10, 40)*
h=40 POABC: V = 1/3 × 20 × 15 × 40 = 4000 cm3. PDEFG: V = 1/3 × 8 × 6 × (40−24) = 256 cm3 ⇒ vol of ornament = 4000 − 256 = 3744 cm3 (v)
M1 A1
using scalar product
B1 M1 A1 [5]
or other equivalent methods
B1 B1
M1 E1 E1 [5]
evaluating parameter and checking consistency. [or other methods, e.g. solving]
B1 M1
soi 1/3 x w x d x h used for either –condone one error
A1 A1
both volumes correct cao
[4]
13
4754
8(i)
⇒ ⇒ ⇒
Mark Scheme
x 2y ,sin θ = k k 2 2 cos θ + sin θ = 1 cos θ = 2
2
⎛ x ⎞ ⎛ 2y ⎞ ⎜ ⎟ +⎜ ⎟ =1 ⎝k⎠ ⎝ k ⎠ x2 4 y 2 + =1 k2 k2 x2 + 4y2 = k2 *
M1
Used
M1
substitution
E1 [3]
dx dy 1 = −k sin θ , = k cos θ dθ dθ 2 1 k cos θ dy dy / dθ = =−2 dx dx / dθ k sin θ = – ½ cot θ x 2k cos θ 1 dy − =− = − cot θ = 4y 4k sin θ 2 dx or, by differentiating implicitly 2x + 8y dy/dx = 0 ⇒ dy/dx = –2x/8y = –x/4y*
(ii)
M1 A1
E1
oe
M1 A1 E1 [3]
B1 [1]
(iii) k = 2
(iv)
January 2008
y
B1 B1
k=1 k=2
B1 1
2 3
4
1 correct curve –shape and position 2 or more curves correct shape- in concentric form all 3 curves correct
x
k=3 k=4
[3]
(v) grad of stream path = –1/grad of contour dy 1 4y * ⇒ =− = dx (− x / 4 y ) x
M1
(vi) dy = 4 y ⇒ dy = 4dx ∫y ∫ x dx x ⇒ ln y = 4 ln x + c = ln ecx4 ⇒ y = Ax4 where A = ec.
M1 A1 M1 M1 A1
Separating variables ln y = 4 ln x (+c) antilogging correctly (at any stage) substituting x = 2, y = 1 evaluating a correct constant
E1 [6]
www
When x = 2, y = 1 ⇒ 1 = 16A ⇒ A = 1/16 ⇒ y = x4/16 *
E1 [2]
14
4754
Mark Scheme
Paper B
Comprehension 4754 (C4)
1
4, 1, 5, 6, 11, 17
2
Even, odd, odd, even, odd, odd recurs 100th term is therefore even
3 4
B1 B1 M1 A1 B1
for 11 and 17 for 1 and 4 for reason www
1 − EH = 1 − CG = 1 − (φ − 1)
M1
oe
⎛ 1+ 5 ⎞ = 2 − φ = 2 − ⎜⎜ ⎟⎟ ⎝ 2 ⎠
A1
φ 6 = ( 3φ + 2 ) + ( 5φ + 3) = 8φ + 5
= 5
A1
3− 5 2
(i)Gradients −
1
φ
and
B1 B1
1 φ −1
(ii) Product of gradients: −
6
1
1 1 =− 2 φ φ −1 φ −φ 1 = − = −1 1 ×
1+ 5 +1 φ +1 = 2 2φ − 1 1 + 5 − 1 =
M1 E1
M1 A1
3+ 5 2 5
(3 + 5 ) =
E1
5
3 5 +5 = 10 2 5× 5
7
January 2008
a + ( a + d ) = a + 2d
M1
⇒ a=d
( a + d ) + ( a + 2d ) = a + 3d
⇒ a=0
a=d =0 *
15
M1 E1 [18]
4755
Mark Scheme 4755 (FP1)
Further Concepts for Advanced Mathematics
Qu Answer Section A ⎛ 3 1 ⎞⎛ 2 −1⎞ ⎛ 6 0 ⎞ 1(i) BA = ⎜ ⎟⎜ ⎟=⎜ ⎟ ⎝ −2 4 ⎠⎝ 0 3 ⎠ ⎝ −4 14 ⎠
1(ii)
Mark
Comment
Attempt to multiply M1 c.a.o. A1 [2]
det BA = ( 6 × 14 ) − ( −4 × 0 ) = 84
M1 A1 A1(ft) [3] M1
3 × 84 = 252 square units 2(i)
January 2008
α 2 = ( −3 + 4 j)( −3 + 4 j) = ( −7 − 24 j)
Attempt to calculate any determinant c.a.o. Correct area Attempt to multiply with use of j2 = −1
c.a.o. A1 [2]
2(ii)
α =5
B1 B1
arg α = π − arctan 43 = 2.21 (2d.p.) (or
Accept 2.2 or 127o
126.87o ) B1(ft)
α = 5 ( cos 2.21 + jsin 2.21)
3(i)
3
s.c. lose 1 mark only if α 2 used throughout (ii)
[3] Showing 3 satisfies the equation B1 (may be implied) Valid attempt to factorise M1 Correct quadratic factor A1
2
3 + 3 − 7 × 3 − 15 = 0
(
z 3 + z 2 − 7 z − 15 = ( z − 3) z 2 + 4 z + 5
z=
−4 ± 16 − 20 2
Accept degrees and ( r , θ ) form
)
= −2 ± j
So other roots are −2 + j and − 2 − j
M1
Use of quadratic formula, or other valid method
A1
One mark for both c.a.o.
[5] 3(ii)
B2
[2]
16
Minus 1 for each error ft provided conjugate imaginary roots
4755
4
Mark Scheme
n
n
r =1
r =1
n
∑ [( r + 1)( r − 2 )] =∑ r − ∑ r − 2n 2
5(i)
)
)
)
M1
Attempt to split sum up
A2
Minus one each error
M1
Attempt to factorise
M1
Collecting terms
r =1
= 1 n ( n + 1)( 2n + 1) − 1 n ( n + 1) − 2n 6 2 1 = n ⎡⎣( n + 1)( 2n + 1) − 3 ( n + 1) − 12 ⎤⎦ 6 1 = n 2n 2 + 3n + 1 − 3n − 3 − 12 6 = 1 n 2n 2 − 14 6 = 1 n n2 − 7 3
( ( (
January 2008
All correct A1 [6]
p = −3, r = 7
One mark for each B2 [2] s.c. B1 if b and d used instead of p and r B1
5(ii) q = αβ + αγ + βγ
α 2 + β 2 + γ 2 = (α + β + γ ) − 2 (αβ + αγ + βγ ) 2
M1
= (α + β + γ ) − 2 q 2
⇒ 13 = 32 − 2q ⇒ q = −2 6(i)
c.a.o. A1 [3] Use of inductive definition M1 c.a.o. A1
a2 = 7 × 7 − 3 = 46 a3 = 7 × 46 − 3 = 319
6(ii) When n = 1,
13 × 7 0 + 1
[2]
= 7 , so true for n = 1
2 Assume true for n = k 13 × 7 k −1 + 1 ak = 2 13 × 7 k −1 + 1 ⇒ ak +1 = 7 × −3 2 = =
13 × 7 k + 7 2
B1 E1
Correct use of part (i) (may be implied) Assuming true for k
M1
Attempt to use ak +1 = 7 ak − 3
A1
Correct simplification
E1
Dependent on A1 and previous E1
−3
13 × 7 k + 7 − 6 2 k
=
Attempt to find q using α 2 + β 2 + γ 2 and α + β + γ , but not αβγ
13 × 7 + 1 2
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 k = 1, it is true for k = 1, 2, 3 and so true for all positive integers.
17
E1 [6] Dependent on B1 and previous E1 Section A Total: 36
4755
Mark Scheme
Section B 7(i) (1, 0 ) and ( 0,
7(ii)
1 18
x = 2, x = −3, x =
)
January 2008
B1 B1 [2]
−3 2
Minus 1 for each error B4 [4]
,y=0
7(iii)
1 18
B1
Correct approaches to vertical asymptotes
B1
Through clearly marked (1, 0 )
and ( 0,
1
1 18
)
[2]
7(iv)
B1
x < −3 , x > 2 −3 2
B2
< x ≤1
B1 for
−3 2
< x < 1 , or
−3 2
≤ x ≤1
[3]
B3
Circle, B1; radius 2, B1; centre 3j, B1
B3
Half line, B1; from -1, B1; π 4 to x-axis, B1
8(i)
[6] B2(ft)
8(ii)
Correct region between their circle and half line indicated s.c. B1 for interior of circle
[2]
M1 Sketch should clearly show the radius and centre of the circle and the starting point and angle of the half-line.
8(iii)
arg z =
π 2
A1(ft)
M1
− arcsin 23 = 0.84 (2d.p.)
A1 [4]
18
Tangent from origin to circle Correct point placed by eye where tangent from origin meets circle Attempt to use right angled triangle c.a.o. Accept 48.20o (2d.p.)
4755
Mark Scheme
January 2008
9(i)
( −3, − 3)
9(ii)
( x, x )
B1 [1] B1 B1 [2]
9(iii)
⎛1 0 ⎞ ⎜1 0 ⎟ ⎝ ⎠
Minus 1 each error to min of 0 B3 [3]
9(iv)
Rotation through
π 2
anticlockwise about the
origin
9(v)
⎛ 0 −1 ⎞ ⎛ 1 0 ⎞ ⎛ −1 0 ⎞ ⎜ 1 0 ⎟ × ⎜1 0 ⎟ = ⎜ 1 0 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
Rotation and angle (accept 90o ) B1 Centre and sense B1 [2] M1 A1
Attempt to multiply using their T in correct order c.a.o.
[2]
9(vi)
⎛ −1 0 ⎞⎛ x ⎞ ⎛ − x ⎞ ⎜ 1 0 ⎟⎜ y ⎟ = ⎜ x ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠
M1 A1(ft)
May be implied
So (-x, x) A1
Line y = -x
[3]
19
c.a.o. from correct matrix
4756
Mark Scheme 4756 (FP2)
January 2008
Further Methods for Advanced Mathematics M1
1(a) π
⌠ 1 2 a (1 − cos 2θ ) 2 dθ 2 ⌡0
Area is ⎮
For ∫ (1 − cos 2θ ) 2 dθ
π
⌠ = ⎮ 12 a 2 1 − 2 cos 2θ + 12 (1 + cos 4θ ) dθ ⌡0
(
= 12 a 2 = πa 3 4
Correct integral expression including limits (may be implied by later work)
A1
)
B1
[ 32 θ − sin 2θ + 18 sin 4θ ] 0
π
B1B1B1 ft
2
A1
For cos 2 2θ = 12 (1 + cos 4θ ) Integrating a + b cos 2θ + c cos 4θ [ Max B2 if answer incorrect and no mark has previously been 7 lost ] d 1 arctan u = du 1+ u2 dy 1 = or dx sec2 y
(b)(i)
Applying
M1 1
f ′ ( x) =
A1
1 + ( 3 + x) 2
f ′′ ( x) =
− 2( 3 + x )
(1 + (
3 + x) 2
)
2
4
(ii) f (0) = 13 π f ′ (0) =
1 4
Applying chain (or quotient) rule
M1 A1
, f ′′ (0) = − 18 3
arctan( 3 + x) = 13 π + 14 x − 161 3 x 2 + ...
B1
Stated; or appearing in series Accept 1.05
M1
Evaluating f ′ (0) or f ′′ (0)
A1A1 ft
For 4
1 4
1 x and − 16 3 x2
ft provided coefficients are non-zero
(iii) ⌠ h
3 1 1 2 1 ⎮ ( 3 π x + 4 x − 16 3 x + ...) dx ⌡− h
⎡ ⎤ 1 3 1 x − 64 3 x 4 + ... ⎥ = ⎢ 16 π x 2 + 12 ⎣ ⎦ ≈
( 16
2
πh +
1 12
3
h −
1 64
M1
h
A1 ft
−h
4
Integrating (award if x is missed) for
3h )
1 12
x3
2
1 3 1 h − 64 − ( 16 π h − 12 3 h4 )
= 16 h 3
A1 ag
2 1 3 Allow ft from a + 4 x + cx provided that a ≠ 0 Condone a proof which neglects
h4
20
4756
2(a)
Mark Scheme
1π j
are r e jθ where
4th roots of 16 j = 16 e 2
1
r=2 θ = 18 π
2kπ + 8 4 7 θ = − 8 π , − 83 π ,
θ=
January 2008
π
5 π 8
B1 B1
Accept 16 4
M1
Implied by at least two correct (ft) further values or stating k = −2 , − 1 , (0) , 1
A1
Points at vertices of a square centre O or 3 correct points (ft) or 1 point in each quadrant
M1 A1 6
(b)(i)
jθ
(1 − 2 e )(1 − 2 e
− jθ
) = 1− 2 e
jθ
−2e
= 5 − 2(e
jθ
− jθ
+e
M1 A1
+4
− jθ
For e jθ e − jθ = 1
)
= 5 − 4 cos θ
A1 ag 3
OR (1 − 2 cos θ − 2 j sin θ )(1 − 2 cos θ + 2 j sin θ ) 2
M1
2
= (1 − 2 cos θ ) + 4 sin θ 2
A1 2
= 1 − 4 cos θ + 4(cos θ + sin θ )
= 5 − 4 cos θ
A1
(ii) C + jS = 2 e jθ + 4 e 2 jθ + 8 e 3 jθ + ... + 2 n e njθ = = =
(
2 e jθ 1 − ( 2 e jθ ) n 1 − 2 e jθ
)
Obtaining a geometric series
M1 A1
Summing (M0 for sum to infinity)
2 e jθ (1 − 2 n e njθ )(1 − 2 e − jθ ) (1 − 2 e jθ )(1 − 2 e − jθ ) 2e
jθ
n +1 ( n +1) jθ
−4−2 e +2 5 − 4 cos θ
M1 n+2
e
njθ
A2
2 cos θ − 4 − 2 n +1 cos(n + 1)θ + 2 n + 2 cos nθ C= 5 − 4 cos θ
S=
M1
Give A1 for two correct terms in numerator
M1 A1 ag
2 sin θ − 2 n +1 sin( n + 1)θ + 2 n + 2 sin nθ 5 − 4 cos θ
Equating real (or imaginary) parts
A1 9
21
4756
3 (i)
Mark Scheme
Characteristic equation is (7 − λ )(−1 − λ ) + 12 = 0
January 2008
M1
2
λ − 6λ + 5 = 0 λ = 1, 5 3 ⎞⎛ x⎞ ⎛ x⎞ ⎛ 7 ⎟⎟ ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ When λ = 1 , ⎜⎜ ⎝ − 4 − 1⎠ ⎝ y ⎠ ⎝ y ⎠
A1A1
7x + 3y = x −4 x − y = y
⎛ 1 ⎞ y = −2 x , eigenvector is ⎜⎜ ⎟⎟ ⎝ − 2⎠ 3 ⎞⎛ x⎞ ⎛ 7 ⎛ x⎞ ⎟⎟ ⎜⎜ ⎟⎟ = 5 ⎜⎜ ⎟⎟ When λ = 5 , ⎜⎜ ⎝ − 4 − 1⎠ ⎝ y ⎠ ⎝ y⎠
3 ⎞ ⎛ x ⎞ ⎛0⎞ ⎛ 6 ⎟⎜ ⎟ = ⎜ ⎟ − 4 − 2 ⎟⎠ ⎜⎝ y ⎟⎠ ⎜⎝ 0 ⎟⎠ ⎝
M1
or ⎜⎜
M1
can be awarded for either eigenvalue Equation relating x and y
A1
7 x + 3 y = 5x −4 x − y = 5 y
or any (non-zero) multiple
M1
⎛ 3 ⎞ y = − 23 x , eigenvector is ⎜⎜ ⎟⎟ ⎝ − 2⎠
A1 8
(ii)
3 ⎞ ⎛ 1 ⎟⎟ P = ⎜⎜ ⎝ − 2 − 2⎠ ⎛1 0⎞ ⎟⎟ D = ⎜⎜ ⎝ 0 5⎠
B1 ft B1 ft
22
SR (M − λ I )x = λ x can earn M1A1A1M0M1A0M1A0 B0 if P is singular
For B2, the order must be 2 consistent
4756 (iii)
Mark Scheme
n
M =PD P
−1
M1
⎛ 1 0 ⎞ −1 ⎟ = P ⎜⎜ n ⎟P ⎝0 5 ⎠ 3 ⎞ ⎛ 1 0 ⎞ 1 ⎛ − 2 − 3⎞ ⎛ 1 ⎟⎟ ⎜⎜ ⎟ ⎜ ⎟ = ⎜⎜ n⎟ ⎜ 1 ⎟⎠ ⎝ − 2 − 2⎠ ⎝0 5 ⎠ 4 ⎝ 2 ⎛ 1 3 × 5 n ⎞⎟ 1 ⎛ − 2 − 3 ⎞ ⎜ ⎟ = ⎜⎜ n⎟ ⎜ 1 ⎟⎠ ⎝− 2 − 2 × 5 ⎠ 4 ⎝ 2 =
1 ⎛⎜ − 2 + 6 × 5 n 4 ⎜⎝ 4 − 4 × 5 n
B1 ft
For P −1 − 3⎞ ⎟ 5 n ⎟⎠
Obtaining at least one element in a product of three matrices
M1 A1 ag
b = − 34 + 34 × 5 n 3 2
Dependent on M1M1
3 ⎞1⎛ −2 ⎛ 1 ⎟⎟ ⎜⎜ n ⎝ − 2 − 2⎠ 4 ⎝ 2 × 5
− 3 + 3 × 5 n ⎞⎟ 6 − 2 × 5 n ⎟⎠
d=
A1 ft
or ⎜⎜
a = − 12 + 32 × 5 n c = 1 − 5n
May be implied
M1
M = P D P −1 n
January 2008
− 12 × 5 n
A2 8
Give A1 for one of b, c, d correct SR If M n = P −1 D n P is used, max marks are M0M1A0B1M1A0A1 (d should be correct) SR If their P is singular, max marks are M1M1A1B0M0
23
4756
Mark Scheme
(e x + e − x ) = k
M1
or cosh x + sinh x = e x
e 2 x − 2k e x + 1 = 0
M1
or k ± k 2 − 1 = e x
A1
One value sufficient
M1
or cosh x is an even function (or equivalent)
1 2
4 (i)
January 2008
ex =
2k ± 4k 2 − 4 = k ± k 2 −1 2
x = ln(k + k 2 − 1) or ln(k − k 2 − 1) 2
2
2
2
(k + k − 1)(k − k − 1) = k − (k − 1) = 1 1
2
ln(k − k − 1) = ln(
2
k + k 2 −1
) = − ln(k + k − 1)
x = ± ln(k + k 2 − 1)
A1 ag 5 For arcosh or
M1
(ii)
ln(λx + λ2 x 2 − ... )
or any cosh substitution
A1 2
⌠ ⎮ ⌡1
For arcosh 2 x or 2 x = cosh u or
2
⎡ 1 ⎤ dx = ⎢ arcosh 2 x ⎥ 2 ⎣ ⎦1 4x −1 1
= 12 ( arcosh 4 − arcosh 2 ) =
1 2
( ln(4 +
15 ) − ln(2 + 3 )
ln(2 x + 4 x 2 − 1) or ln( x + x 2 − 14 )
A1
2
For
M1
)
1 2
or
∫ 12 du
Exact numerical logarithmic form
A1 5 M1 M1
(iii) 6 sinh x − 2 sinh x cosh x = 0 cosh x = 3 (or sinh x = 0 ) x=0
B1
x = ± ln(3 + 8 )
A1
Obtaining a value for cosh x or x = ln(3 ± 8 ) 4
OR e 4 x − 6 e 3 x + 6 e x − 1 = 0 (e 2 x − 1)(e 2 x − 6 e x + 1) = 0
or (e x − e − x )(e x + e − x − 6) = 0
M2 B1 A1
x=0 x = ln(3 ± 8 )
(iv) dy
= 6 cosh x − 2 cosh 2 x dx dy = 5 then 6 cosh x − 2(2 cosh 2 x − 1) = 5 If dx
B1 Using cosh 2 x = 2 cosh 2 x − 1
M1
4 cosh 2 x − 6 cosh x + 3 = 0
Discriminant D = 6 2 − 4 × 4 × 3 = −12
Considering D, or completing square, or considering turning point
M1
Since D < 0 there are no solutions
A1 4
24
4756
Mark Scheme OR Gradient g = 6 cosh x − 2 cosh 2 x
January 2008
B1
g ′ = 6 sinh x − 4 sinh 2 x = 2 sinh x(3 − 4 cosh x) = 0 when x = 0 (only)
M1 g ′′ = 6 cosh x − 8 cosh 2 x = −2 when x = 0 M1 Max value g = 4 when x = 0 So g is never equal to 5 A1
25
Final A1 requires a complete proof showing this is the only turning point
4756
5 (i)
Mark Scheme
λ = −1
λ =0
January 2008
λ =1
B1B1B1
cusp
(ii)
B1B1
loop
x =1
Two different features (cusp, 5 loop, asymptote) correctly identified
B1 1
(iii) Intersects itself when y = 0
(iv)
M1
t = (±) λ
A1
⎞ ⎛ λ , 0⎟ ⎜ 1 + λ ⎝ ⎠
A1 3
dy = 3t 2 − λ = 0 dt
λ
t=± x=
M1
3 λ
3
1+ λ
= 3
λ 3+ λ
A1 ag
λ ⎛ λ y = ± ⎜ ( ) −λ ( ) 3 ⎝ 3 3 2
⎛ 1 1 = ± λ ⎜⎜ − 3 ⎝ 3 3 3 2
=±
1 2
⎞ ⎟ ⎠
3 ⎞ ⎟ = ± λ2 ⎟ ⎠
⎛ 2 ⎜ − ⎜ ⎝ 3 3
One value sufficient
M1
⎞ ⎟ ⎟ ⎠
4λ 3 27
A1 ag 4
(v)
From asymptote, a = 8 From intersection point, λ=
B1 aλ =2 1+ λ
M1
1 3
From maximum point, b
A1 4λ 3 =2 27
M1
b = 27
A1 5
26
4758
Mark Scheme 4758
1(i )
January 2008
Differential Equations
α 2 + 2α + 1 = 0
M1
Auxiliary equation
α = −1 (repeated)
A1
−t
F1
CF for their roots
PI y = a in DE ⇒ y = 2
B1 B1
y = 2 + ( A + Bt ) e −t
F1
t = 0, y = 0 ⇒ 0 = 2 + A ⇒ A = −2
M1
Constant PI PI correct Their PI + CF (with two arbitrary constants) Condition on y
y& = ( B − A − Bt ) e
M1
CF y = ( A + Bt ) e
−t
t = 0, y& = 0 ⇒ 0 = B − A ⇒ B = −2
M1
y = 2 − ( 2 + 2t ) e
A1
−t
Differentiate (product rule) Condition on y& 10
(ii)
Both terms in CF hence will give zero if substituted in LHS PI y = bt 2 e −t
(
y& = 2bt − bt
(
2
)e
−t
(
)e + 2 ( 2bt − bt ) + bt ) e
, && y = 2b − 4bt + bt
2
E1 B1
−t
M1
Differentiate twice and substitute PI correct Their PI + CF (with two arbitrary constants) Condition on y
t = 0, y& = 0 ⇒ 0 = D − C ⇒ D = 0
M1
Condition on y&
y = 12 t 2 e −t
A1
t > 0 ⇒ 12 t 2 > 0 and e −t > 0 ⇒ y > 0
E1
y& = t − 12 t 2 e −t so y& = 0 ⇔ t − 12 t 2 = 0 ⇔ t = 0 or 2
M1
Solve y& = 0
Maximum at t = 2, y = 2 e −2
A1 B1 B1 B1
Maximum value of y Starts at origin Maximum at their value of y y>0
in DE ⇒ 2b − 4bt + bt 2 ⇒b=
(
1 2
2
2
−t
= e−t
M1 A1
)
y = C + Dt + 12 t 2 e −t t = 0, y = 0 ⇒ 0 = C
(
y& = D + t − C − Dt −
1 2 t 2
F1
)e
−t
8 (iii)
(
)
6
27
4758
2(i )
Mark Scheme
dv 3 3 + v=g− dt 1 + t 1+ t I = exp
(1 + t )3
(∫
3 1+ t
)
3 3ln 1+ t dt = e ( ) = (1 + t )
dv 2 3 2 + 3 (1 + t ) v = g (1 + t ) − 3 (1 + t ) dt
(
) (1 + t ) v = ∫ ( g (1 + t )
January 2008
M1
Rearrange
M1 A1 A1
Attempt integrating factor Correct Simplified
F1
Multiply DE by their I
M1
Integrate
A1
RHS
F1
Divide by their I (must also divide constant)
M1
Use condition
E1
Convincingly shown
d (1 + t )3 v = g (1 + t )3 − 3 (1 + t )2 dt 3
3
4
2
)
− 3 (1 + t ) dx
3
= 14 g (1 + t ) − (1 + t ) + A
v = 14 g (1 + t ) − 1 + A (1 + t )
−3
t = 0, v = 0 ⇒ 0 = g − 1 + A 1 4
(
)
v = 14 g (1 + t ) − 1 + 1 − 14 g (1 + t )
−3
10 (ii)
dv (1 + t ) + 5v = (1 + t ) g dt dv 5 + v=g dt 1 + t
M1
Rearrange
M1 A1
Attempt integrating factor Simplified
F1
Multiply DE by their I
(1 + t )5 v = ∫ g (1 + t )5 dx
M1
Integrate
6
A1
RHS
F1
Divide by their I (must also divide constant)
M1
Use condition
F1
Follow a non-trivial GS
I = exp
(1 + t )5
(∫
5 1+ t
)
5 5ln 1+ t dt = e ( ) = (1 + t )
dv 4 5 + 5 (1 + t ) v = g (1 + t ) dt
(
)
d (1 + t )5 v = g (1 + t )5 dt = 16 g (1 + t ) + B v = 16 g (1 + t ) + B (1 + t )
−5
t = 0, v = 0 ⇒ 0 = g + B 1 6
(
v = 16 g 1 + t − (1 + t )
−5
)
9 (iii)
First model:
dv 1 −4 = g − 3 1 − 14 g (1 + t ) dt 4
As t → ∞, (1 + t )
(
−4
)
→0
Hence acceleration tends to
(
1 4
g
dv 1 −6 = g 1 + 5 (1 + t ) dt 6 Hence acceleration tends to 16 g
Second model
M1
Find acceleration
B1
Identify term(s) → 0 in their solution for either model
A1
)
M1
Find acceleration
A1 5
28
4758
3(i)
Mark Scheme
P = A e0.5t t = 0, P = 2000 ⇒ A = 2000
January 2008
M1 M1 A1
Any valid method Use condition
CF P = A e0.5t PI P = a cos 2t + b sin 2t P& = −2a sin 2t + 2b cos 2t −2a sin 2t + 2b cos 2t = 0.5 ( a cos 2t + b sin 2t ) + 170sin 2t
F1 B1 M1
Correct or follows (i) Differentiate
M1
Substitute
−2a = 0.5b + 170 2b = 0.5a solving ⇒ a = −80, b = −20
M1 M1 A1
Compare coefficients Solve
GS P = A e0.5t − 80 cos 2t − 20sin 2t
F1
Their PI + CF (with one arbitrary constant)
M1 F1
Use condition Follow a non-trivial GS
P = 2000 e0.5t
3 (ii)
(iii)
8
t = 0, P = 2000 ⇒ A = 2080 P = 2080 e0.5t − 80 cos 2t − 20sin 2t
2 (iv)
t 0 0.1 0.2
P& 1000 1082.58
P 2000 2100 2208
M1 A1 A1 A1
Use of algorithm 2100 1082.5… 2208
M1
Set P& = 0
M1
Solve
4 (v)
(A) Limiting value ⇒ P& = 0 1 2
⎛ P ⎞ ⇒ P ⎜1 − ⎟ =0 ⎝ 12 000 ⎠ (as limit non-zero) limiting value = 12000
A1 3
(B) Growth rate max when 1
⎛ P ⎞2 f ( P ) = P ⎜1 − ⎟ max ⎝ 12 000 ⎠ 1
⎛ ⎛ P ⎞2 P ⎞ 1 P ⎜1 − f ′ ( P ) = ⎜1 − ⎟ − ⎟ × 12 000 2 12000 12 000 ⎠ ⎝ ⎠ ⎝ ⎛ P ⎞ 1 P=0 f ′ ( P ) = 0 ⇔ ⎜1 − ⎟− × 12 000 2 12000 ⎝ ⎠ ⇔ P = 8000
− 12
M1
Recognise expression to maximise
M1
Reasonable attempt at derivative
M1
Set derivative to zero
A1 4
29
4758
4(i)
Mark Scheme
&& x = −3x& + y&
M1 M1 M1
Differentiate first equation Substitute for y& y in terms of x, x&
M1 E1
Substitute for y
λ + 2λ + 2 = 0 λ = −1 ± j
M1 A1
Auxiliary equation
CF x = e −t ( A cos t + B sin t )
M1 F1 B1 B1
CF for complex roots CF for their roots Constant PI PI correct Their CF + PI (with two arbitrary constants)
= −3 x& + (−5 x + y + 15) y = 3 x − 9 + x&
&& x = −3x& − 5 x + ( 3 x − 9 + x& ) + 15 && x + 2 x& + 2 x = 6
(ii)
January 2008
5
2
PI x = a 2a = 6 ⇒ a = 3 GS x = 3 + e −t ( A cos t + B sin t )
F1
y = 3 x − 9 + x&
M1
y in terms of x, x&
M1
Differentiate x and substitute
A1
Constants must correspond with those in x
0 = 3 + A ⇒ A = −3 0 = 2A + B ⇒ B = 6
x = 3 + 3e −t ( 2sin t − cos t )
M1 M1 F1
Condition on x Condition on y Follow their GS
y = 15e −t sin t
F1
Follow their GS
B1 B1
Sketch of x starts at origin Asymptote x = 3
B1
Sketch of y starts at origin Decaying oscillations (may decay rapidly) Asymptote y = 0
7 (iii)
= 9 + 3e−t ( A cos t + B sin t ) − 9 − e−t ( A cos t + B sin t ) + e−t ( − A sin t + B cos t ) y = e −t ( ( 2 A + B ) cos t + ( 2 B − A ) sin t )
(iv)
3
4 (v)
B1 B1
5
30
4761
Mark Scheme 4761
Q1 (i)
January 2008
Mechanics 1
v 15 m s -1
Mark
Comment
B1
Acc and dec shown as straight lines
B1
Horizontal straight section All correct with v and times marked and at least one axis labelled. Accept (t, v) or (v, t) used.
Sub
t 0
10
30 35
B1
3
(ii)
Distance is found from the area area is (or
1 2
1 2
× 10 × 15 + 20 × 15 + 12 × 5 × 15
M1 A1
× ( 20 + 35) × 15 )
= 412.5 so distance is 412.5 m
A1
At least one area attempted or equivalent uvast attempted over one appropriate interval. Award for at least two areas (or equivalent) correct Allow if a trapezium used and only 1 substitution error. FT their diagram. cao (Accept 410 or better accuracy) 3
6 2 (i)
⎛6⎞ ⎛ 4⎞ ⎛ 4⎞ -2 ⎜ 9 ⎟ = 1.5a giving a = ⎜ 6 ⎟ so ⎜ 6 ⎟ m s ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
M1
Use of N2L with an attempt to find a. Condone spurious notation.
A1
Must be a vector in proper form. Penalise only once in paper. 2
(ii)
Angle is arctan ( 64 )
M1
= 56.309… so 56.3° (3 s. f.)
F1
Use of arctan with their 64 or 46 or equiv. May use F. FT their a provided both cpts are +ve and non-zero. 2
Using s = tu + 0.5t 2a we have
M1
Appropriate single uvast (or equivalent sequence of uvast). If integration used twice condone omission of r(0) but not v(0).
⎛ −2 ⎞ ⎛4⎞ s = 2 ⎜ ⎟ + 0.5 × 4 ⎜ ⎟ ⎝ 3⎠ ⎝6⎠
A1
FT their a only
⎛4⎞ so ⎜ ⎟ m ⎝ 18 ⎠
A1
cao. isw for magnitude subsequently found.
(iii)
Vector must be in proper form (penalise only once in paper). 3 7
31
4761
Mark Scheme
Q3 (i)
m × 9.8 = 58.8 so m = 6
Mark
Comment
M1 A1
T = mg. Condone sign error. cao. CWO.
January 2008
Sub
2
(ii)
Resolve → 58.8cos 40 − F = 0
M1
F = 45.043… so 45.0 N (3 s. f.)
B1 A1
Resolving their tension. Accept s ↔ c . Condone sign errors but not extra forces. (their T) × cos 40 (or equivalent) seen Accept ± 45 only. 3
(iii)
Resolve ↑
R + 58.8sin 40 − 15 × 9.8 = 0
R = 109.204… so 109 N (3 s. f.)
M1 A1 A1
Resolving their tension. All forces present. No extra forces. Accept s ↔ c . Condone errors in sign. All correct cao 3
8
Q4 (i)
⎛ 4 ⎞ ⎛ −6 ⎞ ⎛ −2 ⎞ Resultant is ⎜⎜ 1 ⎟⎟ + ⎜⎜ 2 ⎟⎟ = ⎜⎜ 3 ⎟⎟ ⎜2⎟ ⎜ 4 ⎟ ⎜ 6 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
Mark
Comment
M1
Adding the vectors. Condone spurious notation.
A1 Magnitude is
( −2) 2 + 32 + 62 = 49 = 7 N
M1 F1
Sub
Vector must be in proper form (penalise only once in the paper). Accept clear components. Pythagoras on their 3 component vector. Allow e.g. – 2² for ( – 2)² even if evaluated as – 4. FT their resultant. 4
(ii)
F + 2G + H = 0 ⎛ −12 ⎞ ⎛ 4 ⎞ So H = - 2G - F = − ⎜⎜ 4 ⎟⎟ − ⎜⎜ 1 ⎟⎟ ⎜ 8 ⎟ ⎜2⎟ ⎝ ⎠ ⎝ ⎠ ⎛ 8 ⎞ = ⎜⎜ −5 ⎟⎟ ⎜ −10 ⎟ ⎝ ⎠
M1
Either F + 2G + H = 0 or F + 2G = H
A1
Must see attempt at H = – 2 G – F
A1
cao. Vector must be in proper form (penalise only once in the paper). 3
7
32
4761
Mark Scheme
Q5
January 2008
Mark
Comment Differentiation, at least one term correct.
a = 0 gives t = 2
M1 A1 F1
x = ∫ (2 + 12t − 3t 2 )dx
M1
2t + 6t 2 − t 3 + C
A1
a = 12 − 6t
Sub
Follow their a Integration indefinite or definite, at least one term correct. Correct. Need not be simplified. Allow as definite integral. Ignore C or limits 2
x = 3 when t = 0
M1
Allow x = ± 3 or argue it is
∫
from A
0
then ± 3
so 3 = C and x = 2t + 6t 2 − t 3 + 3
A1
x (2) = 4 + 24 − 8 + 3 = 23 m
B1
Award if seen WWW or x = 2t + 6t 2 − t 3 seen with +3 added later. FT their t and their x if obtained by integration but not if -3 obtained instead of +3. [If 20 m seen WWW for displacement award SC6] [Award SC1 for position if constant acceleration used for displacement and then +3 applied] 8
8
33
4761 Q6 (i)
Mark Scheme
3.5 = 0.5 + 1.5T so T = 2 so 2 s 3.5 + 0.5 ×2 2 so s = 4 so 4 m s=
January 2008
Mark
Comment
M1 A1
Suitable uvast, condone sign errors. cao
M1
Suitable uvast, condone sign errors.
F1
FT their T. [If s found first then it is cao. In this case when finding T, FT their s, if used.]
Sub
4
(ii) (A)
(B)
N2L ↓ : 80 × 9.8 − T = 80 × 1.5
M1
T = 664 so 664 N
B1 A1
N2L ↓ : 80 × 9.8 − T = 80 × ( −1.5)
M1
T = 904 so 904 N
A1
Use of N2L. Allow weight omitted and use of F = mga Condone errors in sign but do not allow extra forces. weight correct (seen in (A) or (B)) cao N2L with all forces and using F = ma. Condone errors in sign but do not allow extra forces. cao [Accept 904 N seen for M1 A1] 5
(iii)
M1
N2L ↑ : 2500 − 80 × 9.8 − 116 = 80a
Use of N2L with F = ma. Allow 1 force missing. No extra forces. Condone errors in sign.
A1 a = 20 so 20 m s
-2
upwards.
A1 A1
±20 , accept direction wrong or omitted upwards made clear (accept diagram)
4
(iv)
N2L ↑ on equipment: 80 − 10 × 9.8 = 10a
M1
a = −1.8
A1
N2L ↑
M1
Use of N2L on equipment. All forces. F = ma. No extra forces. Allow sign errors. Allow ±1.8 N2L for system or for man alone. Forces correct (with no extras); accept sign errors; their ±1.8 used
either all: T − (80 + 10) × 9.8 − 116 = 90 × ( −1.8) or on man: T – (80×9.8) – 116 – 80 = 80×( – 1.8) T = 836 so 836 N
A1
cao [NB The answer 836 N is independent of the value taken for g and hence may be obtained if all weights are omitted.] 4
17
34
4761
Mark Scheme
Q7 (i)
Horiz 21t = 60
Mark
Comment
M1
Use of horizontal components and a = 0 or s = vt − 0.5at 2 with v = 0. Any form acceptable. Allow M1 A1 for answer seen WW. [If s = ut + 0.5at 2 and u = 0 used without justification award M1 A0] [If u = 28 assumed to find time then award SC1] Use of v = u + at (or v 2 = u 2 + 2as ) with v = 0. or Use of v = u + at with v = – u and appropriate t. or Use of s = ut + 0.5at 2 with s = 40 and appropriate t Condone sign errors and, where appropriate, u↔v. Accept signs not clear but not errors. Enough working must be given for 28 to be properly shown. [NB u = 28 may be found first and used to find time]
s (2.8571…)
A1
either 0 = u − 9.8 × 207
M1
so
20 7
or −u = u − 9.8 × ( 407 ) or 40 = u × 207 − 4.9 ( 207 )
2
so u = 28 so 28 m s -1
January 2008
E1
Sub
4 (ii)
y = 28t − 0.5 × 9.8t 2
E1
Clear & convincing use of g = – 9.8 in s = ut + 0.5at 2 or s = vt − 0.5at 2 NB: AG
(iii)
Start from same height with same (zero) vertical speed at same time, same acceleration Distance apart is 0.75 × 21t = 15.75t
E1
M1 A1
1
For two of these reasons
0.75×21t seen or 21t and 5.25t both seen with intention to subtract. Need simplification - LHS alone insufficient. CWO. 3
(iv) (A)
either or
Time is 207 s by symmetry so 15.75 × 207 = 45 so 45 m Hit ground at same time. By symmetry one travels 60 m so the other travels 15 m in
B1 B1
Symmetry or uvast FT their (iii) with t =
20 7
B1
this time ( 14 speed) so 45 m.
B1 [SC1 if 90 m seen] 2
(B)
see next page
35
4761
Q7
Mark Scheme
January 2008
continued
(B)
[SC1 if either and or methods mixed to give ±30 = 28t − 4.9t 2 or ±10 = 4.9t 2 ] either Time to fall is 40 − 10 = 0.5 × 9.8 × t 2
t = 2.47435… need 15.75 × 2.47435.. = 38.971.. so 39.0 (3sf) or Need time so 10 = 28t − 4.9t 2 4.9t 2 − 28t + 10 = 0
M1 A1 A1 A1
Considering time from explosion with u = 0. Condone sign errors. LHS. Allow ±30 All correct cao
F1
FT their (iii) only.
M1
Equating 28t − 4.9t 2 = ±10 Dep. Attempt to solve quadratic by a method that could give two roots.
M1*
2
− 4×4.9×10 so t = 28± 28 9.8 so 0.382784… or 5.33150…
Time required is 5.33150… − 207 = 2.47435.. need 15.75 × 2.47435.. = 38.971.. so 39.0 (3sf)
A1
Larger root correct to at least 2 s. f. Both method marks may be implied from two correct roots alone (to at least 1 s. f.). [SC1 for either root seen WW]
M1 F1
FT their (iii) only. 5
(v)
Horiz ( x = ) 21t Elim t between x = 21t and y = 28t − 4.9t 2
B1
so y = 28 ( 21x ) − 4.9 ( 21x )
A1
Intention must be clear, with some attempt made. t completely and correctly eliminated from their expression for x and correct y. Only accept wrong notation if subsequently explicitly given correct value 2 x2 . e.g. x21 seen as 441
E1
Some simplification must be shown.
so y =
4x 3
2
− 0.19x =
2
1 (120 x − x 2 ) 90
M1
[SC2 for 3 points shown to be on the curve. Award more only if it is made clear that (a) trajectory is a parabola (b) 3 points define a parabola] 4 19
36
4762
Mark Scheme 4762
Q1 (a) (i)
either In direction of the force I = Ft = mv so 1500 × 8 = 4000v giving v = 3 so 3 m s -1 or 1500 N2L gives a = 4000 1500 ×8 v=0+ 4000 giving v = 3 so 3 m s -1
January 2008
Mechanics 2 Mark
Comment
M1 A1 A1
Use of Ft = mv
M1
Appropriate use of N2L and uvast
Sub
A1 A1 3
(ii)
0
3 m s -1
before 500
4000
500
4000
after
V S m s -1
V Rm s -1
PCLM 12000 = 4000VR + 500VS so 24 = 8VR + VS VS − VR = −0.2 0−3 so VS − VR = 0.6 Solving VR = 2.6, VS = 3.2 so ram 2.6 m s -1 and stone 3.2 m s -1
NEL
M1 A1
Appropriate use of PCLM Any form
M1
Appropriate use of NEL
A1
Any form
A1 F1
Either value 6
(iii)
0.5 × 4000 × 32 − 0.5 × 4000 × 2.62 − 0.5 × 500 × 3.22
M1 B1
= 1920 J
A1
Change in KE. Accept two terms Any relevant KE term correct (FT their speeds) cao 3
(b)
see over
37
4762
Mark Scheme
1 (b) (i)
72i N s 8(9 cos 60i + 9 sin 60 j) = ( 36i + 36 3j ) N s
January 2008
Mark
Comment
B1
Neglect units but must include direction
E1
Evidence of use of 8 kg , 9 m s -1 and 60°
Sub
2 (ii)
(iii)
72i + (36i + 36 3j) = 12(ui + vj) Equating components 72 + 36 = 12u so u = 9 36 3 = 12v so v = 3 3
either 4 × 18 = 8 × 9 so equal momenta so 60/2 = 30° or arctan ( 3
3
9
) = arctan ( ) = 30° 1
3
M1 M1
PCLM. Must be momenta both sides
A1
Both 3
M1
Must be clear statements
A1
cao
M1
FT their u and v.
A1
cao 2
19
Q2 (i) (A)
0.5 × 80 × 32 = 360 J
Mark
Comment
M1 A1
Use of KE
Sub
2 (B)
360 = F × 12 so F = 30 so 30 N
M1 F1
W = Fd attempted FT their WD 2
(ii)
Using the WE equation
M1
Attempt to use the WE equation. Condone one missing term
0.5 × 80 × 102 − 0.5 × 80 × 4 2
M1
Δ KE attempted
= 80 × 9.8 × h − 1600
B1 A1 A1
1600 with correct sign All terms present and correct (neglect signs) cao
h = 6.32653… so 6.33 (3 s. f.)
5 (iii) (A)
We have driving force F = 40 so 200 = 40v and v = 5 so 5 m s -1
B1 M1 A1
May be implied Use of P = Fv 3
(B)
From N2L, force required to give accn is F − 40 = 80 × 2 so F = 200 P = 200 × 0.5 = 100 so 100 W
M1 A1 A1 M1 A1
Use of N2L with all terms present (neglect signs) All terms correct correct use of P = Fv cao 5
17
38
4762
Mark Scheme
Q3 (i)
January 2008
Mark
Comment
For z ( 2 × 20 × 100 + 2 × 50 × 120 ) z = 2 × 2000 × 50 + 2 × 6000 × 60 so z = 57.5
M1 B1 B1 A1
and y = 0
B1
Method for c.m. Total mass of 16000 (or equivalent) At least one term correct NB This result is given below. NB This result is given below. Statement (or proof) required. N.B. If incorrect axes specified, award max 4/5
Sub
5 (ii)
y and z are not changed with the folding For x 100 × 120 × 0 + 2 × 20 × 100 × 10 = 16000x
so x =
40000 = 2.5 16000
E1
A statement, calculation or diagram required.
M1 B1
Method for the c.m. with the folding Use of the 10
E1
Clearly shown 4
(iii)
Moments about AH. Normal reaction acts through this line c.w. P × 120 − 72 × ( 20 − 2.5 ) = 0 so P = 10.5
M1 B1
May be implied by diagram or statement
B1 A1 A1
20 − 2.5 or equivalent All correct cao
5 (iv)
Fmax = μ R
so Fmax = 72 μ For slipping before tipping we require 72 μ < 10.5 so μ < 0.1458333...
( 48 ) 7
M1 A1
Allow F = μ R
M1
Accept ≤ . Accept their Fmax and R.
A1
cao
Must have clear indication that this is max F
4 18
39
4762
Mark Scheme
Q4
January 2008
Mark
Comment
B1 M1 E1
Used below correctly Use of their 0.5 0.5 must be clearly established.
M1 B1 A1
Use of moments about D in an equation Use of 1100 and 2.75 or equiv
Sub
(i) Centre of CE is 0.5 m from D a.c. moment about D 2200 × 0.5 = 1100 so 1100 N m c.w moments about D R × 2.75 − 1100 = 0 R = 400 so 400 N
6 (ii)
c.w moments about D W × 1.5 − 1100 − 440 × 2.75 = 0 so W = 1540
M1 A1 E1
Moments of all relevant forces attempted All correct Some working shown 3
(iii) (A)
c.w. moments about D 1.5 × 1540 cos 20 − 1.75T −1100 cos 20 − 400 × 2.75cos 20 = 0
M1 M1
A1
T = 59.0663… so 59.1 N (3 s. f.)
B1 A1 A1
Moments equation. Allow one missing term; there must be some attempt at resolution. At least one res attempt with correct length Allow sin ↔ cos Any two of the terms have cos 20 correctly used (or equiv) 1.75 T All correct cao Accept no direction given 6
(iii) (B)
either Angle required is at 70° to the normal to CE so T1 cos 70 = 59.0663...
M1
so T1 = 172.698... so 173 N (3 s.f.)
A1
or 400 cos 20 × 2.75 + 1100 cos 20 = 1540 cos 20 × 1.5 − T sin 20 × 1.75
M1
Moments attempted with all terms present
A1 A1
All correct (neglect signs) FT(iii)(A)
T = 172.698… so 173 N (3s.f.)
B1 FT (iii) (A)
3 18
40
4763
Mark Scheme 4763
January 2008
Mechanics 3
1(a)(i) [ Force ] = M L T −2
B1
[ Density ] = M L−3
B1 2
(ii)
[E]=
[ F ] [ l0 ] (M L T −2 )(L) = [ A ] [ l − l0 ] ( L2 )(L)
= M L−1 T − 2
B1
for [ A ] = L2
M1 A1
Obtaining the dimensions of E 3
(iii)
T = Lα ( M L−3 ) β (M L−1 T −2 )γ −2γ = 1 , β + γ = 0
γ = − 12 β=
B1 cao
1 2
F1 M1 A1 A1
α − 3β − γ = 0 α =1
Obtaining equation involving α, β,γ 5
(b)
AP = 1.7 m F = T cos θ R + T sin θ = 5 × 9.8
B1 M1 M1
T cos θ = 0.4(49 − T sin θ )
M1
8 T 17
= 0.4(49 −
15 T) 17
Resolving in any direction Resolving in another direction (M1 for resolving requires no force omitted, with attempt to resolve all appropriate forces) Using F = 0.4 R to obtain an equation involving just one force (or k) Correct equation Allow T cos 61.9 etc
A1
T = 23.8
A1
T = k (1.7 − 1.5)
M1
Stiffness is 119 N m −1
or R = 28 or F = 11.2 May be implied λ Allow M1 for T = × 0.2
A1
1.5
8
41
If R = 49 is assumed, max marks are B1M1M0M0A0A0M1A0
4763
2(a)(i)
Mark Scheme
0.1 + 0.01 × 9.8 = 0.01 ×
January 2008
M1 A1
u2 0.55
Speed is 3.3 m s −1
Using acceleration u 2 / 0.55
A1 3
(ii)
1 2
m(v 2 − u 2 ) = mg (2 × 0.55 − 0.15)
1 2
(v 2 − 3.32 ) = 9.8 × 0.95
M1 A1
Using conservation of energy
v 2 = 29.51
( ft is v 2 = u 2 + 18.62 )
R − mg cos θ = m R − 0.01 × 9.8 ×
v2 a
Forces and acceleration towards centre
M1
0.4 29.51 = 0.01 × 0.55 0.55
A1
Normal reaction is 0.608 N
A1
( ft is
5 (b)(i)
u 2 + 22.54 ) 55
B1
T = 0.8 r ω 2 160 ( r − 2) 2 80(r − 2) 100(r − 2) ω2 = = 0.8r r 200 ω 2 = 100 − < 100 , so ω < 10 r T=
B1 E1 E1 4
(ii)
EE =
1 160 × × (r − 2) 2 = 40(r − 2) 2 2 2
B1
KE = 12 m(rω ) 2
Use of
M1
= 12 × 0.8 × r 2 ×
1 2
mv 2 with v = rω
100(r − 2) r
= 40r (r − 2)
A1
40r (r − 2) > 40(r − 2) 2
Since r > r − 2 ,
i.e. KE > EE
From fully correct working only
E1 4
(iii)
100(r − 2) r r = 3.125
When ω = 6 , 36 =
M1
T = 80(r − 2) = 80(3.125 − 2)
M1
Tension is 90 N
A1 cao
Obtaining r
3
42
4763
3 (i)
Mark Scheme
dx = Aω cos ωt − Bω sin ωt dt 2
d x dt 2
= − Aω 2 sin ωt − Bω 2 cos ωt
B1
B1
Aω = −1.44
M1 A1 cao
Period is
2π
ω
=
Using d2x
M1 A1 cao
or
dt 2
dx = −1.44 when t = 0 dt
= −0.18 when t = 0 (or x = 2 )
A1 cao
ω = 0.3 , A = −4.8
(iii)
Fully correct completion 3 SR For x& = − Aω cos ωt + Bω sin ωt && x = − Aω 2 sin ωt − Bω 2 cos ωt award B0B1E0
B=2
− Bω 2 = −0.18 − 0.18 = −ω 2 (2)
Must follow from their x&
B1 ft
= −ω 2 ( A sin ωt + B cos ωt ) = −ω 2 x E1
(ii)
January 2008
6
2π = 20.94 = 20.9 s 0.3
E1
(3 sf) Amplitude is
or 1.442 = 0.32 (a 2 − 22 )
A2 + B 2 = 4.82 + 2 2 = 5.2 m
M1 A1 3
(iv)
x = −4.8 sin 0.3t + 2 cos 0.3t v = −1.44 cos 0.3t − 0.6 sin 0.3t
M1
Finding x when t = 12 and t = 24
A1
Both displacements correct
M1 M1
Considering change of direction Correct method for distance
When t = 12 , x = 0.3306
( v = 1.56 )
When t = 24 , x = −2.5929
( v = −1.35 )
Distance travelled is (5.2 − 0.3306) + 5.2 + 2.5929 = 12.7 m
A1
5 ft from their A , B , ω and amplitude: Third M1 requires the method to be comparable to the correct one A1A1 both require ω ≈ 0.3, A ≠ 0, B ≠ 0 Note ft from A = +4.8 is x12 = −3.92 (v < 0) x24 = 5.03 (v > 0)
Distance is (5.2 − 3.92) + 5.2 + 5.03 = 11.5
43
4763
4 (i)
Mark Scheme
January 2008
π may be omitted throughout
8
−1 ⌠ V = ⎮ π ( x 3 ) 2 dx ⌡1
M1
8
⎡ 1 ⎤ = π ⎢ 3 x 3 ⎥ = 3π ⎣ ⎦1
A1
8
−1 ⌠ V x = ⎮ π x ( x 3 ) 2 dx ⌡1
⎡ =π ⎢ ⎣
x=
M1
8
3 4
4 45 ⎤ x3 ⎥ = π 4 ⎦1
A1
45 π 4
3π 15 = = 3.75 4
Dependent on previous M1M1
M1 A1 6
(ii)
8
⌠ −1 A = ⎮ x 3 dx ⌡1 ⎡ =⎢ ⎣
M1
8
3 2
2 9 ⎤ x 3 ⎥ = = 4.5 ⎦1 2
A1
8
−1 ⌠ A x = ⎮ x ( x 3 ) dx ⌡1
⎡ =⎢ ⎣
3 5
x=
x
5 3
93 ⎤ ⎥ = 5 = 18.6 ⎦1
⎡ =⎢ ⎣
3 2
y=
A1
18.6 62 = ( ≈ 4.13 ) 4.5 15
8
⌠ Ay=⎮ ⌡1
M1
8
1 2
x
) 2 dx
M1
3 ⎤ ⎥ = 2 = 1.5 ⎦1
A1
(x 1 3
−1
A1
3
If
8
1.5 1 = 4.5 3
A1 8
44
1 2
omitted, award M1A0A0
4763 (iii)
Mark Scheme M1
⎛ 62 ⎞ ⎛18.6 ⎞ ⎛x⎞ ⎛ 4.5 ⎞ ⎟⎟ = (4.5) ⎜⎜ 15 ⎟⎟ = ⎜⎜ ⎟⎟ (1) ⎜⎜ ⎟⎟ + (3.5) ⎜⎜ 1 ⎝ y⎠ ⎝ 0.25 ⎠ ⎝ 3 ⎠ ⎝ 1.5 ⎠
M1
A1 A1
x = 2.85 y = 0.625
45
January 2008 Attempt formula for CM of composite body (one coordinate sufficient) Formulae for both coordinates; signs must now be correct, but areas (1 and 3.5) may be wrong. ft only if 1 < x < 8 4 ft only if 0.5 < y < 1 Other methods: M1A1 for x M1A1 for y (In each case, M1 requires a complete and correct method leading to a numerical value)
4766
Mark Scheme 4766
January 2008
Statistics 1
Q1 (i)
Mode = 7 Median = 12.5
B1 cao B1 cao
(ii)
Positive or positively skewed (A) Median
E1 E1 cao
1
(iii)
(B) There is a large outlier or possible outlier of 58 / figure of 58.
E1indep
2
M1 for 14.75 × 28 but 413
2
2
Just ‘outlier’ on its own without reference to either 58 or large scores E0 Accept the large outlier affects the mean (more) E1
(iv)
There are 14.75 × 28 = 413 messages So total cost = 413 × 10 pence = £41.30
can also imply the mark
A1cao TOTAL
Q2 (i)
⎛ 4⎞ 4 4 ⎜ ⎟ × 3! = 4 × 6 = 24 codes or P3 = 24 (M2 for P3 ) 3 ⎝ ⎠ 4 × 3 × 2 = 24
Or
(ii)
Probability = 0.3 × 0.8 = 0.24 Either: P(AUB) = P(A) + P(B) – P(A∩B)
(ii)
= 0.3 + 0.2 – 0.3 × 0.2 = 0.5 – 0.06 = 0.44 Or: P(AUB) = 0.7 × 0.2 + 0.3 × 0.8 + 0.3 × 0.2 = 0.14 + 0.24 + 0.06 = 0.44 Or: P(AUB) = 1 – P(A′ ∩B′ ) = 1 – 0.7 × 0.8 = 1 – 0.56 = 0.44
(iii)
P(A|B) =
3
M1 for 43 A1 cao
3
4 = 64 codes
Q3 (i)
M1 for 4 M1 for ×6 A1
P(A ∩ B) 0.06 6 = = = 0.136 P(B) 0.44 44
46
7
2 TOTAL
5
M1 for 0.8 from (1-0.2) A1 M1 for adding 0.3 and 0.2 M1 for subtraction of ( 0.3 × 0.2) A1 cao
2
M1 either of first terms M1 for last term A1 M1 for 0.7 × 0.8 or 0.56 M1 for complete method as seen A1 M1 for numerator of their 0.06 only M1 for ‘their 0.44’ in denominator A1 FT (must be valid p) TOTAL
3
3
8
4766
Q4 (i)
Mark Scheme
E(X) = 1 × 0.2 + 2 × 0.16 + 3 × 0.128 + 4 × 0.512 = 2.952
Division by 4 or other spurious value at end loses A mark
E(X2) = 1 × 0.2 + 4 × 0.16 + 9 × 0.128 + 16 × 0.512 = 10.184
(ii)
January 2008
M1 for Σ rp (at least 3 terms correct) A1 cao M1 for Σ x2p at least 3 terms correct
Var(X) = 10.184 – 2.9522 = 1.47 (to 3 s.f.)
M1 for E(X2) – E(X) 2 Provided ans > 0 A1 FT their E(X) but not a wrong E(X2)
Expected cost = 2.952 × £45000 = £133000 (3sf)
B1 FT ( no extra multiples / divisors introduced at this stage)
5
1
(iii) G1 labelled linear scales G1 height of lines
P(X = r )
0.6 0.4
2
0.2 0 1
2
3
4
r
TOTAL
Q5 (i) (ii) (iii)
Impossible because the competition would have finished as soon as Sophie had won the first 2 matches SS, JSS, JSJSS 2
2
2
0.7 + 0.3 × 0.7 + 0.7 × 0.3 × 0.7 = 0.7399 or 0.74(0) { 0.49 + 0.147 + 0.1029 = 0.7399}
47
E1 B1, B1, B1 (-1 each error or omission) M1 for any correct term M1 for any other correct term M1 for sum of all three correct terms A1 cao TOTAL
8 1 3 4
8
4766
Mark Scheme
January 2008
Section B Q6 (i)
180.6 = 15.05 or 15.1 12 180.62 = 3107.56 − or 3107.56 – 12(their 15.05)2 = 12
Mean =
B1 for mean
Sxx
M1 for attempt at Sxx
(389.53) s= (ii)
3
389.53 = 5.95 or better 11
A1 cao
NB Accept answers seen without working (from calculator) x + 2s = 15.05 + 2 × 5.95 = 26.95 x – 2s = 15.05 – 2 × 5.95 = 3.15 So no outliers
M1 for attempt at either M1 for both A1 for limits and conclusion FT their mean and sd
3
(iii)
New mean = 1.8 × 15.05 + 32 = 59.1
B1FT M1 A1FT E1FT using 0F ( x dep)
3
(iv)
New s = 1.8 × 5.95 = 10.7 New York has a higher mean or ‘ is on average’ higher (oe) New York has greater spread /range /variation or SD (oe)
E1FT using 0F ( σ dep)
2
(v)
Cumulative frequency
Upper bound (70) 100 110 120 150 170 190 Cumulative frequency (0) 6 14 24 35 45 48
G1 for linear scales
50
(linear from 70 to 190) ignore x < 70 vertical: 0 to 50 but not beyond 100 (no inequality scales)
40 30 20 10
G1 for labels
0 0
50
100
150
200
Hours
(vi)
B1 for all correct cumulative frequencies (may be implied from graph). Ignore cf of 0 at this stage
G1 for points plotted as (UCB, their cf). Ignore
5
(70,0) at this stage. No mid – point or LCB plots.
NB all G marks dep on attempt at cumulative frequencies.
NB All G marks dep on attempt at cumulative frequencies
G1 for joining all of ‘their points’(line or smooth curve) AND now including (70,0)
Line on graph at cf = 43.2(soi) or used 90th percentile = 166
M1 for use of 43.2 A1FT but dep on 3rd G mark earned
TOTAL
48
2
18
4766
Q7 (i)
Mark Scheme
X ~ B(12, 0.05)
⎛12 ⎞ (A) P(X = 1) = ⎜ ⎟ × 0.05 × 0.9511 = 0.3413 ⎝1⎠
OR from tables
(ii)
January 2008
0.8816 − 0.5404 = 0.3412
(B )
P(X ≥ 2) = 1 − 0.8816 = 0.1184
(C )
Expected number E(X ) = np = 12 × 0.05 = 0.6
Either: 1 – 0.95n ≤ ⅓ 0.95n ≥ ⅔ n ≤ log ⅔ /log0.95, so n ≤ 7.90 Maximum n = 7
M1 0.05 × 0.9511 M1
( ) × pq 12 1
11
(p+q) =
1 A1 cao OR: M1 for 0.8816 seen and M1 for subtraction of 0.5404 A1 cao M1 for 1 – P(X ≤ 1) A1 cao M1 for 12 × 0.05 A1 cao (= 0.6 seen) M1 for equation in n
3
2 2
M1 for use of logs A1 cao
Or: (using tables with p = 0.05): n = 7 leads to P(X ≥ 1) = 1 - P(X = 0) = 1 – 0.6983 = 0.3017 ( < ⅓ ) or 0.6983 ( > 2/3)
n = 8 leads to P(X ≥ 1) = 1 - P(X = 0) = 1 – 0.6634 = 0.3366 ( > ⅓ ) or 0.6634 ( < 2/3 )
Maximum n = 7 (total accuracy needed for tables)
M1indep M1indep A1 cao dep on both M’s 3
Or: (using trial and improvement): 1 – 0.957 = 0.3017 ( < ⅓) or 0.957 = 0.6983 ( > 2/3) 1 – 0.958 = 0.3366 ( > ⅓) or 0.968 = 0.6634 ( < 2/3) Maximum n = 7 (3 sf accuracy for calculations) NOTE: n = 7 unsupported scores SC1 only (iii)
Let X ~ B(60, p) Let p = probability of a bag being faulty H0: p = 0.05 H1: p < 0.05
M1indep (as above) M1indep (as above) A1 cao dep on both M’s
B1 for definition of p B1 for H0 B1 for H1
P(X ≤ 1) = 0.9560 + 60 × 0.05 × 0.9559 = 0.1916 > 10% So not enough evidence to reject H0 Conclude that there is not enough evidence to indicate that the new process reduces the failure rate or scientist incorrect/ wrong.
M1 A1 for probability M1 for comparison A1 E1
TOTAL
49
8
18
4767
Mark Scheme 4767
January 2008
Statistics 2
Question 1 (i)
(ii)
x is independent, y is dependent since the values of x are chosen by the student but the values of y are dependent on x x = 2.5, y = 80.63 b=
Sxy 2530.3 − 30 × 967.6 / 12 111.3 = = =7.42 Sxx 15 90 − 30 2 / 12
OR b =
2530.3 / 12 − 2.50 × 80.63 9.275 = =7.42 1.25 90 / 12 − 2.50 2
Hence least squares regression line is: y - y = b(x - x ) ⇒ y – 80.63 =7.42(x – 2.5) ⇒ y = 7.42x + 62.08
(v)
3
M1 for attempt at gradient (b) A1 for 7.42 cao M1 for equation of line A1 FT (b>0) for complete equation
5
M1 for at least one prediction attempted. A1 for both answers
(iii) (A) For x = 1.2, predicted growth = 7.42 × 1.2 + 62.08 = 71.0 (B) For x = 4.3, predicted growth = 7.42 × 4.3 + 62.08 = 94.0
(iv)
B1 E1 dep E1 dep B1 for x and y used (SOI)
(FT their equation if b>0)
Valid relevant comments relating to the predictions such as : Comment re interpolation/extrapolation Comment relating to the fact that x = 4.3 is only just beyond the existing data. Comment relating to size of residuals near each predicted value (need not use word ‘residual’) x=3⇒ predicted y = 7.42 × 3 + 62.08 = 84.3 Residual = 80 − 84.3 = −4.3 This point is a long way from the regression line. The line may be valid for the range used in the experiment but then the relationship may break down for higher concentrations, or the relationship may be non linear.
E1 (first comment)
4
E1 (second comment) M1 for prediction M1 for subtraction A1 FT (b>0) E1 E1 for valid in range E1 for either ‘may break down’ or ‘could be non linear’ or other relevant comment
3
3 18
50
4767
Mark Scheme
January 2008
Question 2 (i)
Binomial (94,0.1)
(ii)
n is large and p is small
(iii)
B1 for binomial B1 dep for parameters B1, B1 Allow appropriate numerical ranges B1 for mean
λ = 94 × 0.1 = 9.4 9.4 4 = 0.0269 (3 s.f.) 4! or from tables = 0.0429 − 0.0160 = 0.0269 cao (B) Using tables: P(X ≥ 4) = 1 − P(X ≤ 3)
(A) P(X = 4) = e−9.4
= 1 − 0.0160 = 0.9840 cao (iv) (v)
M1 for calculation or use of tables A1 M1 for attempt to find P(X ≥ 4) A1 cao
P(sufficient rooms throughout August) = 0.984031 = 0.6065 (A) 31 × 94 = 2914 Binomial (2914,0.1)
M1 A1 FT B1 for binomial B1 dep, for parameters
(B)Use Normal approx with µ = np = 2914 × 0.1 = 291.4 σ2 = npq = 2914 × 0.1 × 0.9 = 262.26
B1
⎛
P(X ≤ 300.5) = P ⎜⎜ Z ≤
⎝
2
2
5
2 2
B1 B1 for continuity corr. M1 for probability using correct tail A1 cao, (but FT wrong or omitted CC)
300.5 − 291.4 ⎞ ⎟⎟ 262.26 ⎠
= P(Z ≤ 0.5619) = Φ(0.5619) = 0.7130
5 18
51
4767
Mark Scheme
January 2008
Question 3 (i)
X ~ N(56,6.52) M1 for standardizing
57.5 − 56 ⎞ ⎛ 52.5 − 56 t)
M1
Use of 1 – P(…).
k ∴ F(t ) = 1 − 2 t
M1
F(1) = 0 k ∴1 − 2 = 0 1
∴k=1 (ii)
(iii)
d F( t ) dt 2 = 3 t
f (t ) =
∞
μ = ∫ tf (t )dt = ∫ 1
∞
1
M1
Attempt to differentiate c’s cdf.
A1
(For t ≥ 1, but condone absence of this.) Ft c’s cdf provided answer sensible. Correct form of integral for the mean, with correct limits. Ft c’s pdf. Correctly integrated. Ft c’s pdf.
2
Correct use of limits leading to correct value. Ft c’s pdf provided answer sensible. Both hypotheses. Hypotheses in words only must include “population”. For adequate verbal definition.
3
A1
∞
A1
H0: m = 5.4 H1: m ≠ 5.4 where m is the population median time for the task. Times
Beware: answer given.
M1
2 dt t2
⎡− 2⎤ =⎢ ⎥ ⎣ t ⎦1 = 0 − (−2) = 2
(b)
A1
Rank of |diff| 6.4 1.0 8 5.9 0.5 5 5.0 −0.4 4 6.2 0.8 7 6.8 1.4 10 6.0 0.6 6 5.2 −0.2 2 6.5 1.1 9 5.7 0.3 3 5.3 −0.1 1 W− = 1 +2 + 4 = 7 (or W+ = 3+5+6+7+8+9+10 = 48) Refer to tables of Wilcoxon single sample (/paired) statistic for n = 10. Lower (or upper if 48 used) double-tailed 5% point is 8 (or 47 if 48 used). Result is significant. Seems that the median time is no longer as previously thought.
B1 B1
3
− 5.4
54
M1
for subtracting 5.4.
M1 A1
for ranks. FT if ranks wrong.
B1 M1
No ft from here if wrong.
A1
i.e. a 2-tail test. No ft from here if wrong. ft only c’s test statistic. ft only c’s test statistic.
A1 A1
10
4768
Mark Scheme
Q2
X ~ N(260, σ = 24)
(i)
P (X
(iii)
P ( this < 600) = P( Z
w ) = P⎜ Z > ⎟ 28.8 ⎠ ⎝ = P(Z > −1.96 ) ∴ w − 624 = 28.8 × −1.96 ⇒ w = 567.5(52)
(v)
On ~ N(150, σ = 18) X1 + X2 +X3 + On1 + On2 ~ N(1080, P ( this > 1000) = P( Z >
2376)
1000 − 1080 = −1.6412 ) 48.744
= 0.9496
(vi)
Given
x = 252 .4
CI is given by
3
3
3
3
s n −1 = 24 .6 252.4 ± 2.576 ×
24.6 100
= 252.4 ± 6.33(6) = (246.0(63), 258.7(36))
B1 A1
3 18
55
4768
Q3 (i)
(ii)
Mark Scheme
A t test should be used because the sample is small, the population variance is unknown, the background population is Normal H0: μ = 380 H1: μ < 380
E1 E1 E1 B1
where μ is the mean temperature in the chamber.
B1
x = 373.825
B1
s n −1 = 9.368
Test statistic is 373 .825 − 380
M1
9 . 368 √ 12
January 2008
3 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 = 8.969 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: 380 + (c’s – 1.796) × 9⋅368 (= 375.143) for 12
subsequent comparison with x . (Or x – (c’s –1.796) × 9⋅368 12
= –2.283(359).
(iii)
Refer to t11. Single-tailed 5% point is –1.796.
M1 A1
Significant. Seems mean temperature in the chamber has fallen. CI is given by 373.825 ± 2·201
A1 A1
×
(iv)
A1
9.368
(= 378.681) for comparison with 380.) c.a.o. but ft from here in any case if wrong. Use of 380 – x scores M1A0, but ft. No ft from here if wrong. Must be minus 1.796 unless absolute values are being compared. No ft from here if wrong. ft only c’s test statistic. ft only c’s test statistic.
9
M1 B1 M1
12
= 373.825 ± 5.952= (367.87(3), 379.77(7))
A1
Advantage: greater certainty. Disadvantage: less precision.
E1 E1
56
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 t11 is OK. Or equivalents.
4
2 18
4768
Mark Scheme
January 2008
Q4 (a) (i)
x=
1125 = 2.25 500
B1 M1
For binomial E(X) = n × p ∴ pˆ =
2.25 = 0.45 5
A1
Use of mean of binomial distribution. May be implicit. Beware: answer given.
3
(ii) fo fe (calc) fe (tables)
32 25.164 25.15
110 102.944 102.95
154 168.455 168.45
125 137.827 137.85 M1
X2 = 1.8571 + 0.4836 + 1.2404 + 1.1938 + 0.7763 + 4.9737 = 10.52(49)
(b)
A1 M1 A1
Refer to χ 42 .
M1
Upper 5% point is 9.488. Significant. Suggests binomial model does not fit.
A1 A1 A1
The model appears to overestimate in the middle and to underestimate at the tails. The biggest discrepancy is at X = 5.
E1
A binomial model assumes all trials are independent with a constant probability of “success”. It seems unlikely that there will be independence within families and/or that p will be the same for all families.
E2
She should try to choose a simple random sample which would involve establishing a sampling frame and using some form of random number generator.
E1
E1
E1 E1
63 56.384 56.35
16 9.226 9.25
Calculation of expected frequencies. All correct. Or using tables: 1.8657 + 0.4828 + 1.2396 + 1.1978 + 0.7848 + 4.9257 c.a.o. Or using tables: 10.49(64) Allow correct df (= cells – 2) from wrongly grouped or ungrouped table, and FT. Otherwise, no FT if wrong. No ft from here if wrong. ft only c’s test statistic. ft only c’s test statistic. Accept also any other sensible comment e.g. at 2.5% significance, the result would NOT have been significant. (E2, 1, 0) Any sensible comment which addresses independence and constant p.
Allow sensible discussion of practical limitations of choosing a random sample. Allow other sensible suggestions. E.g systematic sample - choosing every tenth family; stratified sample - by the number of girls in a family.
12
3
18
57
4771
Mark Scheme 4771
January 2008
Decision Mathematics 1
1 (i)
(ii)
(iii)
6 routes M→A→I→T→Pi→C M→A→I→T→Pi→R→C M→A→I→T→Pi→H→R→C M→V→I→T→Pi→C M→V→I→T→Pi→R→C M→V→I→T→Pi→H→R→C
B1
6 routes M→A→I→Pa→Pi→C M→A→I→Pa→Pi→R→C M→A→I→Pa→Pi→H→R→C M→V→I→Pa→Pi→C M→V→I→Pa→Pi→R→C M→V→I→Pa→Pi→H→R→C
B1
B1
B1
M→V→I→Pa→Pi→H→R→Pi→C B1 A
(iv)
R→H
e.g. P→T→I→V→M→A→I→Pa→P→H→R→C→P→R
58
M1 A2
ends at R (–1 each error/omission)
4771
Mark Scheme
January 2008
2. y
(i)
lines shading (3.6, 0.6) 25.8 at (3.6, 0.6) versus 21 and 24 (or profit line)
B2 x B1 B1 graph or sim. eqns M1 A1
(ii)
25 at (3, 1)
B1 B1
59
4771
Mark Scheme
January 2008
3. y = 2008 c = 2008/100 = 20 n = 2008 – 19 x (2008/19) = 2008 – 19 x (105) = 13 k = 3/25 = 0 i = 20 – 5 – 20 / 3 + 19 × 13 + 15 = 271 i=1 i=1–0=1 j = 2008 + 502 + 1 + 2 – 20 + 5 = 2498 j=6 p=–5 m=3 d = 23 So 23rd March
B1 B1 B1 B1 B1 B1 B1 B1
60
4771
Mark Scheme
January 2008
4. (i)
(ii)
(iii)
e.g.
e.g.
0–3→brown 4–7→blue 8–9→green
M1 A1 A1
0–1→brown 2–5→blue 6–7→green 8–9→reject
M1 A2
proportions OK efficient
A1
some rejected proportions OK (–1 each error) efficient
B1 B1 B1
br/br→br (4 times) br/gr→bl gr/gr→gr
M1 A1 A1
br/bl rule application application
brow n
B1
bl/bl application
green
M1 A1
gr/bl rule application
e.g. Eye colours Parent 1 Parent 2 Offspring brow n brow n brow n
brow n brow n brow n
gree n blue blue
brow n blue brow n
blue brow n brow n
gree n gree n gree n
brow n brow n brow n brow n brow n brow n
blue blue brow n
blue
61
4771
Mark Scheme
January 2008
5. (i)&(ii) e.g. 4
M1 A1
30
D
0
A
0 8
8
8
B 22
30 30
(v)
F
G 9
H 1 I
55 55
forward pass
M1 A1
backward pass
time − 55 weeks critical − A; B; F; G; J
B1 B1
cao cao
50 weeks (49 weeks if G crashed rather than H)
B1
11 30 30
(iv)
41 41
M1 A1
C 10
(iii)
E 9
A1 A1
sca (activity on arc) dummy activities + E and F A, B, C, D G, H, I, J
50 50
3 J 5
55 55
E − 1 week F − 3 weeks J − 2 weeks (G − 1 week, if crashed)
M1 A3
£115000 (£121000)
A1
62
4771
Mark Scheme
January 2008
6. (i)
e.g.
B
C 0.2 G
F
0.2 A
D
E H
0.3 0.1 0.1 Total length = 2.2 km
0.1
0.2
0.2 M1 A1
Prim: connect in nearest to connected set Kruskal: Shortest arc s.t. no cycles
(iii)
5
0.7
0.7
8
9
0.9
1.1
1.5 1.1
0.4
0.2
0.4
A
2
0.3
D
7
E
1.0
0.3
I
4
0.6
M1 A1
Dijkstra working values (see vertex G) order of labelling labels
0.2
G
F 3
name description
C
0.9
0
M1 A1
1.1
1.1
B
6
1
connecting tree DE FC, FG AD, DI, FH 2 of length 0.4
0.1
I
(ii)
M1 A1 A1 A1 A1
H
1.0
0.1
A1 A1
0.1
0.6
0.3 0.1 0.1 0.2 0.2 Arcs used: AD, DE, EF, FG, DI, IH, AB or DB, FC or BC Total length = 2.7 km (AB&FC) or 2.9 km (AB&BC) or 2.4 km (DB&FC) or 2.6 km (DB&BC)
63
M1 A1 A1
arcs counted only once
4776
Mark Scheme 4776
1
x f(x) Eg:
2 0.24
January 2008
Numerical Methods
root = (2 x 0.03 - 3 x 0.24) / (0.03 - 0.24) = 3.142857
3 0.03
graph showing turning point at x = 3 with root some way to the left or the right.
[M1A1] [A1] [G2] [TOTAL 5]
2
x 0 1 0.5
f(x) 1 0.333333 0.477592
T1 = M= hence and
0.666667 0.477592 T2 = (T1 + M)/2 = S = (T1 + 2*M)/3 =
0.572129 0.540617
[M1A1] [M1A1] [M1A1] [M1A1] [TOTAL 8]
3
x f(x)
f(2) = =
0 2
1 2.57
3 3.85
3 terms: form: use x=2: 2(2-1)(2-3)/(0-1)(0-3) + 2.57(2-0)(2-3)/(1-0)(1-3) + 3.85(2-0)(2-1)/(3-0)(3-1) 3.186667 (3.19)
[M1] [M1] [M1] [A1A1A1] [A1] [TOTAL 7]
4
x x3(2-x)-1
1.5 0.6875
2 -1
a 1.5 1.75 1.75
b 2 2 1.875
x 1.75 1.875 1.8125
change of sign, so root (may be implied) x3(2-x)-1 0.339844 -0.17603
mpe 0.25 0.125 0.0625
4 further iterations reqd: mpe 0.0325, 0.015625, 0.0078125, 0.00390625
[M1A1]
[M1A1] [A1] [A1] [M1A1] [TOTAL 8]
5
Sketch showing curve, tangent, chord, h. Makes clear that tangent and chord have substantially different gradients. h g(2 + h) est g '(2)
0 3.61
0.1 3.849 2.39
0.01 3.633 2.3
Clear loss of significant figures as h is reduced
0.001 3.612 2
[G3]
[M1A1A1A1] [E1] [TOTAL 8]
64
4776 6 (i)
Mark Scheme x 3 4 5 6 quadratic
f(x) 1 3 -1 -10
Δf
Δ2f
Δ3f
2 -4 -9
-6 -5
1
= 1 + 2(x-3) - 6(x-3)(x-4)/2 = 1 + 2x-6 - 3x2+21x-36 = -3x2 +23x -41
[M1A1A1] [M1A1] [A1] [A1]
q'(x) = -6x + 23 = 0 at x = 23/6 (= 3.833...)
[M1A1]
q(x) = 0 at x = 4.847(127); also at 2.81954 - not reqd.
[M1A1]
q(6) = -11 (or point out that the second differences not constant)
(ii)
January 2008
cubic est
= 1 + 2(4.5-3) - 6(4.5-3)(4.5-4)/2 + 1(4.5-3)(4.5-4)(4.5-5)/6 = 1.6875
S = 1.5/3 (1 + 4x1.6875 -10) =
-1.125
[A1] [subtotal 12] [M1A1A1] [A1] [M1A1] [subtotal 6] [TOTAL 18]
7
(i)
mpe 0.000 000 5 mpre 0.000 000 5 / 2.506 628 =
[B1] 1.99 x 10-7
[M1A1] [subtotal 3]
(ii)
mpe 1000 x 0.000 000 5 = 0.000 5 In practice the positive and negative errors will tend to cancel out
[M1A1] [E1] [subtotal 3]
(iii)
mpe 1000 x 0.000 001 = 0.001 In practice 1000 x 0.000 000 5 = 0.000 5 because average error in chopping will be 0.000 000 5
[M1A1] [M1A1] [E1] [subtotal 5]
(iv)
L to R: 1 (or 1.000 000) R to L: 1.000 001 L to R requires 8 sf, (R to L doesn't)
[B1] [B1] [E1] [subtotal 3]
(v)
Reverse order more accurate as that way allows the very small terms at the end of the series to contribute to the sum. The spreadsheet is likely to work to greater accuracy The spreadsheet works to more sf than are displayed
[E1] [E1] [E1] [E1] [subtotal 4] [TOTAL 18]
65
Grade Thresholds Advanced GCE (Subject) (Aggregation Code(s)) January 2008 Examination Series Unit Threshold Marks Unit All units
UMS
Maximum Mark 100
A
B
C
D
E
U
80
70
60
50
40
0
4751
Raw
72
54
46
38
31
24
0
4752
Raw
72
55
48
41
34
28
0
4753
Raw
72
57
50
43
36
28
0
4753/02
Raw
18
15
13
11
9
8
0
4754
Raw
90
77
68
59
50
41
0
4755
Raw
72
55
47
39
32
25
0
4756
Raw
72
59
51
44
37
30
0
4758
Raw
72
62
54
46
38
30
0
4758/02
Raw
18
15
13
11
9
8
0
4761
Raw
72
60
52
44
37
30
0
4762
Raw
72
61
53
45
37
30
0
4763
Raw
72
58
51
44
37
30
0
4766/ G241
Raw
72
56
49
42
35
28
0
4767
Raw
72
62
54
46
38
31
0
4768
Raw
72
54
47
40
33
27
0
4771
Raw
72
60
53
46
39
33
0
4776
Raw
72
58
50
42
35
27
0
4776/02
Raw
18
14
12
10
8
7
0
66
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
25.5
50.0
75.5
85.9
95.3
100
Total Number of Candidates 106
7896
42.9
85.7
85.7
85.7
85.7
100
7
7897
0
7898
0
3895
22.7
40.7
59.3
77.8
94.8
100
383
3896
80
80
95
95
100
100
20
3897
0
100
100
100
100
100
1
3898
56.4
76.9
87.2
97.4
97.4
100
39
556 candidates aggregated this series 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.
67
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