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7/23/2019 STEP 2015 Mark Scheme
1/71
STEP Mark Scheme 2015
Mathematics
STEP 9465/9470/9475
October 2015
7/23/2019 STEP 2015 Mark Scheme
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This mark scheme is published as an aid to teachers and students, to indicate therequirements of the examination. It shows the basis on which marks were awardedby the Examiners and shows the main valid approaches to each question. It isrecognised that there may be other approaches and if a different approach wastaken in the exam these were marked accordingly after discussion by the marking
team. These adaptations are not recorded here.
All Examiners are instructed that alternative correct answers and unexpectedapproaches in candidates scripts must be given marks that fairly reflect the relevantknowledge and skills demonstrated.
Mark schemes should be read in conjunction with the published question papers andthe Report on the Examination.
The Admissions Testing Service will not enter into any discussion or correspondencein connection with this mark scheme.
UCLES 2015
More information about STEP can be found at:www.stepmathematics.org.uk
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Contents
STEP Mathematics (9465, 9470, 9475)
Mark Scheme PageSTEP Mathematics I 4STEP Mathematics II 18STEP Mathematics III 45
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SI-2015/Q1
(i) y= ex(2x 1)(x 2) B1 Correct factorisation of quadratic term (or formula, etc.)
(21 , 0) & (2, 0) B1 Noted or shown on sketch
y
d
d= ex(2x2x 3) M1 Derivative attempted and equated to zero for TPs
= ex(2x 3)(x+ 1)
( 23 , e1.5) & (1, 9e 1) A1 A1Noted or shown on sketch
(ify-coords. missing, allow one A1 for 2 correctx-coords.)
G1 Generally correct shape
G1 for (0, 2) noted or shown on sketch
G1 for negative-x-axis asymptote
(penalise curves that clearly turn up away from axis
or that do not actually seem to approach it)
Give M1for either 0, 1, 2 or 3 solutions ORclear indication they know these arise from where a
horizontal line meets the curve (e.g. by a line on their diagram) implied by any correct answer(s)
Then y = k has NO solutions for k < e1.5 A1
ONE solution for k= e1.5 and k > 9e 1 A1 A1
TWO solutions for e1.5
< k0 and k= 9e 1
A1 A1
THREE solutions for 0< k < 9e 1 A1
FTfrom theiry-coords.of the Max. &Min. points.
(ii) G1 Any curve clearly symmetric iny-axis
G1 Shape correct
G1 A Max. TP at (0, 2) FT
G1 Min. TPs at (23 , e1.5) FT
G1 Zeroes at x=21 , 2 FT
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SI-2015/Q2
(i) M1 Use of cos(A B) formula withA= 60o,B= 45o OR A= 45o,B= 30o
or 2 cos215o 1 etc.
A1 Exact trig.values used (visibly) to gain cos 15o=22
13
legitimately(Given Answer)
M1 Similar method OR sin = 2cos1 (as 15ois acute, no requirement to justify +vesq.rt.)
A1 sin 15o=22
13(however legitimately obtained)
(ii) M1 Use of cos(A + B) formula anddouble-angle formulae ORde MoivresThm. (etc.)
A1 cos34cos3 3cos
A1 Justifying/noting that x= cos is thus a root of 03cos34 3 xx
M1 For serious attempt to factorise cxcx 34 33 as linear quadratic factorsor via Vietas Theorem(roots/coefficients)
A1 34)( 22 ccxxcx M1 Solving 03444 22 ccxx FTtheir quadratic factor
Remaining roots arex= 34 2221 ccc
M1 Use of 21 cs to simplify sq.rt. term
A1 x= sin3cos21
(iii) M1 022
23
21 3 yy
A1 02
2
2
13
2
14
3
yy
M1 cos3=2
2= cos 45o
A1 = 15o
M1 cos2
1y , sin3cos
21 , sin3cos
21 with their
A1 y= 2 cos 15o=2
13
A1 oo 15cos15sin3 =2
13
A1 oo 15cos15sin3 = 2
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B1 For correct lengths in smaller
M1 By similar s (ORtrig.ORcoord.geom.)
A1)(
)(
21
21
ab
ab
b
PQ
PQ=ab
abb
)(
M1 so a guard at a corner can see 2(b+PQ)
A1 =ab
b
24(might be given as all but
ab
ba
4or as a
fraction of the perimeter)
Lengths a21 and )(
21 ab in smaller
M1 By similar s (ORtrig.ORcoord.geom.)
A1)(
21
21
21
ab
a
PQ
b
PQ=a
abb
2
)(
M1 so a guard at a midpoint can see b + 2PQ
A1 =a
b2
(might be given as all buta
bab )4( or as a
fraction of the perimeter)
Lengths a21 and )(
21 ab in smaller
M1 By similar s (ORtrig. OR coord.geom.)
A1
)(21
21
ab
a
b
PQ
PQ=
ab
ba
M1 so a guard at a midpoint can see 4b 2PQ
A1 =ab
abb
)32(2(might be given as all but
ab
ba
2or as a
fraction of the perimeter)
SI-2015/Q3
)(21 ab
)(21 ab
b a Q
b21
M
P
M Q
B1 Recognition that b= 3a is the case when guard atM/ Cequally preferable
(Pat corner in the twoMcases)
M1A1 Relevant algebra for comparison of one case
a
b
ab
b 224
baaba
b
3
)(
2
A1 Correct conclusion: Guard stands at C for b< 3aand atMfor b> 3a
M1A1 Relevant algebraab
abb
ab
b
)32(24 2 ba
abab
ba
3
))((
2
A1 Correct conclusion: Guard stands at C for b< 3aand atMfor b> 3a
C P
Q P
Overall, I am anticipating that most attempts will do the Corner scenario and o of the Middle scenarios. This
will allow for a maximum of 12 = 5 (for the Corner work) 4(for the Middle work) 3(for the comparison).
In this circumstance, it wont generally be suitable to give the B1for the b= 3aobservation.
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SI-2015/Q4
M1 WhenPis at (x, 241x ) ... and makes an angle of with the positivex-axis
A1 ... the lower end, Q, is at sin,cos 241 bxbx
M1 Also, 241xy x
y21
d
d = tan
A1 x= 2 tan i.e.P=
2 tan,tan2
A1A1 so thatQ= sintan,costan2 2 bb obtained legitimately(Given Answer)
M1A1 When x= 0, costan2 b
cos
tan2b
M1A1 Substg. intoy-coordinate yA=
22 tan
cos
sintan2tan
M1A1 Eqn. of lineAPis 2tantan xy
M1A1 Area between curve and line is tantan22
41 xx dx
B1 Correct limits (0, 2tan )
A1A1 = 22213
121 tantan xxx (Any 2 correct terms; all 3)
A1A1 = 333
3
2 tan2tan2tan (Any 2 correct terms; all 3 FT)
A1 = 332 tan obtained legitimately(Given Answer)
ALTERNATIVE
O B C
A
M1 A1 for obtaining the conversion factor bcos = 2 tan or tan2= sin2
1 b
M1 A1 for distances OB=BC(= cos21 b ) and so PC= OA= tan
2
M1 A1 giving OABCPB
A1 Area is 2
41x dx
B1 Correct limits (0, 2 tan) used
A1 A1 Correct integration; correct Given Answer
ALTERNATIVE Translate whole thing up by tan2 and calculate
cos
0
22
41 tan
b
x dx
P
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SI-2015/Q5
(i) M1A1 f(x) =1
3)1(
x
tx
A1 =x
x2
M1 Differentiating by use of Quotient RuleORtaking logs.anddiffg. implicitly)
B1 for 2ln.22d
d xx
x seen at any stage
A1y
d
d=
2
22ln.2.
x
x xx
A1 TP at
2lne,
2ln
1
(y-coordinate not required)
B1 Jusitfying that the TP is a minimum
G1 Generally correct -shape
G1 Asymptotic toy-axis
and TP in FTcorrect position
(ii) M1 Let xtxu 21 22
A1 2udu= 2xdt
B1 t:(1, 1) u:( | 1 +x|, | 1 x| ) Correct limits seen at any stage
M1A1 Full substn. attempt; correct g(x) =
ud.11
A1 g(x) = xx 111 n. may be done directly, but be strict on the limits
org(x) =
12
112
12
xx
x
xx
(Must have completely correct three intervals: x< 1, 11 x , x> 1)
M1 Graph split into two or three regions
A1 A1 Reciprocal graphs on LHS & RHS (must be asymptotic tox-axis)
(Allow even if they approachy-axis also)
A1 Horizontal line for middle segment
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SI-2015/Q6
LetP, Q,Rand Sbe the midpoints of sides (as shown)
Then P
M1A1 p= ba21
21 , q= ab
21
21 ,
r= ba 21
21 , s= ab
21
21 Q
and
M1A1 aa 21SRPQ S
A1 bb 21PSQR
A1 so thatPQSR is a //gm. R
(opposite sides // and equal)
M1 bbaa 21
21QSPQQRPQ for use of the scalar product
A1 = babababa 41 Do not accept ba etc.
M1 Use of perpendicularity of OA, OB and OA, OB
= baba 41
M1 AOB= AOB = 180o ; and cos(180o ) = cos
A1 = 0 since cosba ba and cosba ba
and we are given that a = b and a= b
A1 so that PQRS is a rectangle (adjacent sides perpendicular)
B1 PQPQSRPQ 22 = aa 2)( 2241 aa
B1 22 PSQR = bb 2)( 2241 bb
M1 Since a = b, a= b and o90cosaaaa , o90cosbbbab
A1 it follows thatPQRSis a square (adjacent sides equal)
M1A1 AreaPQRS= o2241 90cos2)( aaaa
M1 ... which is maximal when 190cos o
A1 i.e. when o90
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SI-2015/Q7
M1 f (x) = 6ax 18x2
= 6x(a 3x)
A1A1 = 0 for x= 0 and x=31 a
A1A1 f(0) = 0 f(31 a) = 9
1 a3
A1 (Min. TP) (Max. TP) since f(x) is a negative cubic
(f(0) = 0 and the TPs may be shown on a sketch award the marks here if necessary)
M1 Evaluating at the endpoints
A1A1 f(31 ) = 9
1 (3a+ 2); f(1) = 3a 6
M191 (3a+ 2) 9
1 a3 a3 3a 2 0
M1 (a+ 1)2(a 2) 0
A1 and since a 0, a 2
M1 91 a3 3a 6 a3 27a+ 54 0
M1 (a 3)2(a+ 6) 0
A1 which holds for all a 0
M191 (3a+ 2) 3a 6 3a+ 2 27a 54
8(3a 7) 0
A1 a 37
(which, actually, affects nothing, but working should appear)
Thus
B1B1B1 M(a) =
363
32
20)23(3
91
91
aa
aa
aa
(Ignore non-unique allocation of endpoints)
(Do not award marks for correct answers unsupported or from incorrect working)
7/23/2019 STEP 2015 Mark Scheme
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SI-2015/Q8
(i) S = 1 + 2 + 3 + + (n 2) + (n 1) + n
M1 S= n+ (n 1) + (n 2) + + 3 + 2 + 1 Method
M1 2S= n(n+ 1) Adding
A1 S=21 n(n+ 1) obtained legitimately(Given Answer)
(Allow alternatives using induction or theMethod of Differences, for instance, but NOTby stating
that it is an AP and just quoting a formula; ditto -number formula)
(ii) (N m)k+ mk (kodd)
M1A1 =kkkkkk
mmNmk
kNm
kmN
kN
1221
1...
21
E1 which is clearly divisible byN (since each term has a factor ofN)
(Allow alternatives using induction, for instance)
Let S= 1k+ 2k+ + nk an odd no. of terms
M1 = 0k+ 1k+ 2k+ + nk an even no. of terms
M1 = kkkkkk nnnn )()(...1)1(0)0(21
21
21
21
(no need to demonstrate final pairing but must explain fully the pairing up orthe single extra nkterm)
E1 and, by (ii), each term is divisible by n.
For S= 1k+ 2k+ + nk an even no. of terms
M1 = 0k+ 1k+ 2k+ + nk an odd no. of terms
M1 = kkkkkkk nnnnn21
21
21 )1()1(...1)1(0)0(
(no need to demonstrate final pairing but must explain the pairing and note the separate, single term)
and, by (ii), each paired term is divisible by n
E1 and the final single term is divisible by21 n required result
M1 By the above result for neven, so that (n+ 1) is oddA1 (n+ 1) |1k+ 2k+ + nk + (n+ 1)k
E1 (n+ 1)|S+ (n+ 1)k (n+ 1) |S
M1 By the above result for nodd, so that (n+ 1) is even
A121 (n+ 1) |1k+ 2k+ + nk + (n+ 1)k
E121 (n+ 1)| S+ (n+ 1)k
21 (n+ 1) |S (as
21 (n+ 1) is an integer)
E1 Since hcf(n, n+ 1) = 1 hcf( 21 n, n+ 1) = 1 for neven
E1 and hcf(n,21 (n+ 1)) = 1 for nodd
So it follows that21 n(n+ 1) | S for all positive integers n
7/23/2019 STEP 2015 Mark Scheme
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SI/15/Q9
M1 Time taken to land (at the level of the projection) (fromy = utsin21 gt2, y= 0, t 0)
A1 isg
ut
sin2 (may be implicit)
M1 Bullet fired at time t
60 t lands at time
A1
t
g
utTL
3sin
2
M1A1
t
g
u
t
TL
3cos
21
d
d=
tk
k
3cos
1
A1 = 0 when
tk
3cos
M1A1 Horizontal range isg
uR
cossin2 2
(fromy = utsin 21 gt
2 with above time)
A1 22
12
kkg
uRL obtained legitimately(Given Answer)
M1A1 03
sin2
d
d 2
2
2
t
g
u
t
TL
maximum distance
M1A1
60 t in
tk
3cos
2
3
2
1 k
M1 If k 0, rain hitting top of bus is the same, and rain hits back of bus
as before, but with vcos u instead of vcos
E1 When vcos u < 0, rain hitting top of bus is the same, and rain hits front of bus
as before, but with u vcos instead of vcos
A1 Together, A h |vcos u|+ avsin Fully justified (Given Answer)
M1 Journey time u
1so we need to minimise
A1u
uvh
u
avJ
|cos|sin
(Ignore additional constant-of-proportionality factors)
M1 For vcos u > 0,
if w vcos, we minimise hu
hv
u
avJ
cossin
E1 and this decreases as uincreases
E1 and this is done by choosing uas large as possible; i.e. u = w
M1 For u vcos> 0,
we minimise hu
hv
u
avJ
cossin
E1 and this decreases as uincreases if asin> hcos
E1 so we again choose uas large as possible; i.e. u = w
[Note: minimisation may be justified by calculus in either case or both.]
M1 If asin< hcos, thenJincreases with uwhen uexceeds vcos
A1 so we choose u= vcos in this case
M1A1 If asin = hcos then Jis independent of u, so we may as well take u = w
M1 Replacing by 180o gives hu
hv
u
avJ
cossin
A1 Which always decreases as uincreases, so take u = w again
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SI-2015/Q11
(i) B1 O1: F=F1
O2: F=F2 (Both, with reason)
(ii) B1 Resg. ||plane (for C1): sin11 WRF
B1 Resg.r. plane (for C1): cos11 WFR
B1 Resg.||plane (for C2): sin22 WRF
B1 Resg
.r
. plane (for C2): cos22 WFR Max 4 marks to be given for four independent statements (though only 3 are required).
One or other of
Resg.||plane (for system) : sin2121 WWFF
Resg.r. plane (for system) : cos2121 WWRR
may also appear instead of one or more of the above.
(F1andF2may or may not appear in these statements asF, but should do so below)
M1A1 Equating for sin:21 W
RF
W
RF
usingand
M1A1 Re-arranging forFin terms ofR: RWW
WWF
21
21
M1 Use of the Friction Law, RF
A1
21
21
WW
WW obtained legitimately(Given Answer)
O1
O2
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M1A1 (e.g.) FR
RF
1
tan
M1A1 Substg. forR =FR
WW
WWFF
1
21
21
using FWW
WWR
21
21
A1 =11
21
1
2
FR
WW
WF
M1A1 Substg. forR1 (correct inequality)using Friction Law 111 RF 1
11
FR
1
1
1
21
12
FF
WW
WF
M1 Tidying-up algebra =
1
1
21
1
1
2
F
WW
WF
A1 tan 211
11
1
2
WW
W
obtained legitimately(Given Answer)
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SI-2015/Q12
(i) M1A1 P(exactly rout of nneed surgery) =
rnr
r
n
4
3
4
1
(A binomial prob. term; correct)
(ii) M1 P(S= r) = !)(!
!
!
8e
8
rnr
n
nrn
n
rnr
4
34
1
Attempt at sum of appropriate product terms
B1B1A1 Limits All internal terms correct; allow nCrfor the A mark
M1 =
rnr
rn
n
rnr
8
4
3
4
1
!)(
8
!
e
Factoring out these two terms
M1 =n
rn
rn
n
rnr 4
3
!)(
8
!
e 8
Attempting to deal with the powers of 3 and 4
A1 =
rn
rnn
rnr
8
!)(
32
!
e
Correctly
M1 =
rn
rnr
rnr
8
!)(
6
!
2eSplitting off the extra powers of 2 ready to
M1 =
0
8
!
6
!
2e
m
mr
mr adjust the lower limit (i.e. using m= n r)
A1 = 68
e!
2e
r
r
i.e.!
2e 2
r
r
A1 which is Poisson with mean 2 (Give B1for noting this without the working)
(iii) M1 P(M = 8 |M + T= 12) Identifying correct conditional probability outcome
A1A1A1 =
!12
4e
!4
2e
!8
2e
124
4282
One A mark for each correct term (& no extras for 3 rdA mark)
A1A1 =!4!84
!12212
12
Powers of e cancelled; factorials in correct part of the fraction
(unsimplified is okay at this stage)
A1 =4096
495
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SI-2015/Q13
Reminder
A: the 1st6 arises on the n
ththrow
B: at least one 5 arises before the 1st6
C: at least one 4 arises before the 1st6
D: exactly one 5 arises before the 1st6
E: exactly one 4 arises before the 1st6
(i) M1A1 P(A) = 611
65
n
(ii) M1A1 By symmetry (either a 5 or a 6 arises before the other), P(B) =21
(iii) M1 The first 4s, 5s, 6s can arise in the orders 456, 465, 546, 564, 645, 654
A1 P(BC) = 31 (i.e. by symmetry but with three pairs)
(iv) M1A1A1 P(D) = ...1
3
1
2612
64
61
61
64
61
61
61
M1 for infinite series with 1stterm; A1 for 2
ndterm; A1 for 3
rdterm and following pattern
M1 = ...321 232
32
361 For factorisation and an infinite series
M1 =
2
3
2
36
1
1
Use of the given series result
A1 = 41
(v) M1 P(DE) = P(D) + P(E) P(DE) Stated or used
B1 P(E) = P(D) = answer to (iv) Stated or used anywhere
M1A1A1 P(DE) = ...24
1
36
1
6
1
6
22
6
3
6
1
6
1
6
2
6
3
6
1
6
1
6
2
M1 for infinite series with 1stterm; A1 for 2ndterm; A1 for 3rdterm and following pattern
M1 = ...631 221
21
1081 For factorisation and an infinite series
M1 = 321
1081 1
Use of the given series result
A1 P(DE) = 5423
272
21
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Question
1
(i) ln1 1
11 B1
For 0, 1 M1Therefore
ln1 0for 0 A1If 0, ln1 0Therefore ln1 ispositiveforallpositive . B1Therefore
ln 1 0forallpositive .
So,
1
ln 1 1
B1
ln 1 1 ln
1 ln 1 l n
M1
So,
ln 1 1
ln 1 l n
ln 1 l n 1 M1
Therefore,
1
ln 1 A1
(ii)
ln1 1 1
1 B1
For0 1, 1 M1
Therefore
ln1 0for0 1. A1If 0, ln1 0Therefore ln1 isnegativefor0 1. B1Therefore
ln 1 0forall 1.
So,
1
ln1 1
B1
ln1 1 ln 1 ln 1 2l n ln 1 M1M1A1So,
ln1 1
ln 2 ln ln 1 M1A1
As ,ln ln 1 0 B1Therefore,
1
11
1
1 ln 2 A1
7/23/2019 STEP 2015 Mark Scheme
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Question
1
Notethatthestatementofthequestionrequirestheuseofaparticularmethodinbothparts.
(i)
B1 Correctdifferentiationoftheexpression.
M1
Considerationof
the
sign
of
the
derivative
for
positive
values
of
.A1 Deductionthatthederivativeispositiveforallpositivevaluesof.B1 Clearexplanationthat ln1 ispositiveforallpositive .
Notethatanswerisgiveninthequestion.
B1 Useof andsummation.M1 Manipulationoflogarithmicexpressiontoformdifference.
M1 Attempttosimplifythesum(somepairscancelledoutwithinsum).
A1 Clearexplanationofresult.
Note
that
answer
is
given
in
the
question.
(ii)
B1 Correctdifferentiationoftheexpression.
M1 Considerationofthesignofthederivativefor0 1.A1 Deductionthatthederivativeisnegativeforthisrangeofvalues.
B1 Deductionthat ln1 isnegativeforthisrangeofvalues.B1 Useof andsummation.M1 Expressionwithinlogarithmasasinglefractionandnumeratorsimplified.
M1 Logarithmsplittochangeatleastoneproducttoasumoflogarithmsoronequotientasa
differenceoflogarithms.
A1 Completesplitoflogarithmtorequiredform.
M1 Useofdifferencestosimplifysum.
A1 ln 2correct.B1 Correctlydealingwithlimitas .
Notethatanswerswhichuseastheupperlimitonthesumfromthebeginningmusthave
clear
justification
of
the
limit.
Those
beginning
with
as
the
upper
limit
must
haveln ln 1correctinsimplifiedsum.
A1 Inclusionof 1 tothesumtoreachthefinalanswer.Notethatanswerisgiveninthequestion.
7/23/2019 STEP 2015 Mark Scheme
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Question
2
3 B1
sin 3 sin M1A1
sin 3 sin3 B1
sin 3sin sin cos2 cos sin2sin M1sin 1 2 s i n cos 2 sin cos sin M1M1 3 4 s i n A1 (or ) B1 2 3 4 s i n cos2 B1B1 2 3 4 s i n 21 2 s i n
2 M1M1A1
sin ,so B1M1A1
Therefore 3 2 M1M1 13 3
13
So
trisectstheangle
A1
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Question
2
B1 Expressionfor(maybeimpliedbylaterworking).M1 Applicationofthesinerule.
A1 Correctstatement.
B1
Does
not
need
to
be
stated
as
long
as
implied
in
working.
M1 Useofsin formula.M1 Useofdoubleangleformulaforsin.
M1 Useofdoubleangleformulaforcos.
A1 Simplificationofexpression.
Note
that
answer
is
given
in
the
question.
B1 Identificationofthisrelationshipbetweendistances.(just issufficient)B1 Correctexpressionsubstitutedforthelengthof.B1 Correctexpressionsubstitutedforthelengthof.M1 Useofdoubleangleformulaforcos.
M1
Simplificationof
expression
obtained.
A1 Correctexpressionindependentof.B1 Identificationofarightangledtriangletocalculatesin.M1 Deductionthatoneofthelengthsinsineofthisangleisequalto.A1 Valueoftheangle(Degreesorradiansarebothacceptable).
M1 Obtaining 2M1 Useof .A1 Expressiontoshowthat andconclusionstated.
7/23/2019 STEP 2015 Mark Scheme
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Question
3
isthenumberoftrianglesthatcanbemadeusingarodoflength8andtwoother,shorterrods.
M1
Ifthemiddlelengthrodhaslength7thentheotherrodcanbe1,2,3,4,5or6. M1
Ifthemiddlelengthrodhaslength6thentheotherrodcanbe2,3,4or5.
Ifthemiddlelengthrodhaslength5thentheotherrodcanbe3or4. M1 2 4 6. A1Assumethatthelongestofthethreerodshaslength7: M1
Ifthemiddlelengthrodhaslength6thentheotherrodcanbe1,2,3,4or5. M1
Ifthemiddlelengthrodhaslength5thentheotherrodcanbe2,3or4.
Ifthemiddlelengthrodhaslength4thentheotherrodmustbe3. M1
Therefore 1 3 5. A1 1 2 3 4 5 6. A1
2 4 2 1 B1 1 2 3 2 1 B1 3.(Thepossibilitiesare1,2,3, 1,3,4and2,3,4.) B1Substituting 2intotheequationgives 22 14 2 1 3.Thereforetheformulaiscorrectinthecase 2. B1Assumethattheformulaiscorrectinthecase :
M1
14 1
2 1
M1
4 3 1 12 6 4 1 1,whichisastatementoftheformulawhere 1. M1Therefore,byinduction, 14 1 A1 2 4 2 1 1. M1A1Therefore 14 1 1. 14 5.(Or 14 1) A1
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Question
3
M1 Appreciationofthemeaningof .M1 Identifythenumberofpossibilitiesforthelengthofthethirdrodinonecase.
M1 Identifythesetofpossiblecasesandfindnumbersofpossibilitiesforeach.
A1
Clear
explanation
of
the
result.
Notethatanswerisgiveninthequestion.
M1 Anattempttoworkout .M1 Correctcalculationforanyonedefinedcase.
M1 Identificationofacompletesetofcases.
A1 Correctvaluefor .A1 Correctdeductionofexpressionfor .B1 Correctexpression.Nojustificationisneededforthismark.
B1 Correctexpression.Nojustificationisneededforthismark.
B1 Correctjustificationthat 3.Requiressightofpossibilitiesorotherjustification.B1 Evidenceofcheckingabasecase.(Acceptconfirmationthat 1 gives 0here.)M1
Application
of
the
previously
deduced
result.
M1 Substitutionofformulafor andtheformulaforthesum.M1 Takingcommonfactortogiveasingleproduct.
A1 Rearrangementtoshowthatitisastatementoftherequiredformulawhen 1andconclusionstated.
M1 Useofresultfromstartofquestion.
A1 Correctsummationof2 4 2 1.A1 Correctformulareached(anyequivalentexpressionisacceptable).
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Question
4
(i)
B1
B1
(ii)
,sostationarypointswhen 1.
B1
B1
M1
A1
A1
B1
B1
B1B1
(iii)
B1
B1
B1
B1
B1
B1
B1
B1
B1
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Question4
Penaliseadditionalsectionstographs(verticaltranslationsby)onlyonthefirstoccasionprovidingthatthecorrectsectionispresentinlaterparts.
B1 Correctshape.
B1 Asymptotes and shown.B1 Rotationalsymmetryaboutthepoint0,0.B1 Correctshape.
M1 Differentiationtofindstationarypoints.
A1 Correctstationarypoints 1,.(coordinates)A1 Correctcoordinates.
B1
Rotationalsymmetry
about
the
point
0, .B1 Correctshape.B1 Stationarypointshavesamecoordinatesaspreviousgraph.(Followthroughincorrect
stationarypointsinpreviousgraphifconsistenthere).
B1 Correctcoordinatesforstationarypoints 1, arctan B1 Correctasymptotes 1.B1 axisasanasymptote.B1 Middlesectioncorrectshape.
B1 Outsidesectionscorrectshape.
B1
Sectionfor
1 1correctshape.B1 1 .B1 1 .B1 Sectionfor 1 correctwithasymptote 2.B1 Sectionfor 1correctwithasymptote 0orarotationof 1sectionabout0, .
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Question
5
(i) tan tan arctan ,sotheformulaiscorrectfor 1. B1Assumethattan :
arctan .
M1
tan
M1
tan ,whichsimplifiestotan .
M1
A1
Hence,byinductiontan . A1Clearly, arctan . B1Suppose
that
it
is
not
true
that
arctan forallvaluesof.Thenthereisasmallestpositivevalue,suchthat arctan .Since , arctan andtan ,but arctan . M1M1However, arctan ,sothisisnotpossible. A1Therefore arctan . A1
(ii) tan2 . M1A1So,
B1Whichsimplifiesto2 tan 1 4 tan 2 0 M1A1tan 22 tan 1 0 A1Sincemustbeacute,tancannotequal2. B1Therefore arctan .
lim arctan1
4. M1A1
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Question
5
B1 Confirmationthattheformulaiscorrectfor 1.M1 Expressionofintermsof.M1 Useoftan formula.M1
Simplification
of
fraction.
A1 Expressionoftoshowthatitmatchesresult.A1 Conclusionstated.
B1 Confirmationfor 1.M1 Observationthat 0M1 Evidenceofunderstandingthatsuccessivevaluesofwiththesamevalueoftanmust
differby.A1 Evidenceofunderstandingthat cannotbesufficientlylargefortobeofthe
formarctanifis.A1 Clearjustification.
M1 Identificationoftherelevantsidesofthetriangle(diagramissufficient).
A1 Correctexpressionfortan2.B1 Useofdoubleangleformula.
M1 Rearrangementtoremovefractions.
A1 Correctquadraticreached.
A1 Quadraticfactorised.
B1 Irrelevantcaseeliminated(mustbejustified).
M1 SumexpressedaslimitofA1 Correctvaluejustified.
Notethatanswerisgiveninthequestion.
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Question
6
(i) sec 1cos B1
sec
1cos cos sin sin
B1
sec 2cos sin
cos sin cos 2 sin cos sin 1 sin M1Therefore,sec M1A1Therefore, tan . M1A1
(ii)Limits:
0 0
1 B1sin sin B1Therefore, sin sin 1 sin sinSo,2 sin sin M1 sin sin A1 ,andapplyingtheresultfrompart(i): tan tantan 2. B1 2 B1
(iii) Consider sin .Makingthesubstitution : sin sin 1 M1A1So, sin sinTherefore, 2 3sin 3 sin B1 sec
sec ta n sec
M1
2tan tan A1
Andso, B1Therefore, 3 . B1
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Question
6
B1 Expressionofsec intermsofanyothertrigonometricfunctions.B1 Correctuseofaformulasuchasthatforcos toobtainexpressionwith
trigonometricfunctionsof.
M1
Expanding
the
squared
brackets.
M1 Useofsin 2 sin cos andsin cos 1A1 Fullyjustifiedanswer.
Note
that
answer
is
given
in
the
question.
M1 Anymultipleoftan .A1 Correctanswer
B1 Dealswithchangeoflimitscorrectly.AND
Correctlydealswithchangetointegralwithrespectto.Note
that
both
these
steps
need
to
be
seen
the
correct
result
reached
without
evidence
of
these
steps
should
not
score
this
mark.
B1
Useof
sin sin (maybejustseenwithinworking)M1 Groupingsimilarintegrals.
A1 Fullyjustifiedanswer.
Note
that
answer
is
given
in
the
question.
B1 Evaluationoftheintegralfrom(i)withtheappropriatelimits.
B1 Useofresultfrom(ii)toevaluaterequiredintegral.
M1 Attempttomakethesubstitution.
A1 Substitutionallcompletedcorrectly.
B1 Rearrangetogivesomethingthatcanrepresenttherequiredintegralononeside.
M1
Use
of
sec 1 tan
within
integral.
A1 Correctevaluationofthisintegral.
B1 Correctuseofresultfrompart(i).
B1 Correctapplicationofresultdeducedearliertoreachfinalanswer.
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Question
7
(i) Mostlikelyexamples:
and M1
M1
A1
If
thentherecannotbetwopointsonthecirclethatareadistanceof
2
apartand
any
two
diametrically
opposite
points
on
mustbeadistanceof2apart. B1If thenthecirclemustbethesameas,sothereisnotexactly2pointsofintersection.
B1
(ii) Thedistancesofthecentreoffromthecentresofandare and .
M1A1
B1
Ifthecoordinateofthecentreofis,thenthecoordinateisgivenby and B1B1Therefore, M14 andso
. M1A1
Therefore,thecoordinateofthecentreofsatisfies
and
B1
So2 2
2 2 2 2
So,
Therefore, 0 B116 8 8 16 0 M1M116 16 8 8 164 M1164 4 M1164 24 2 164 4 A1
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Question
7
M1 Calculationthatthedistancebetweenthecentresofthecirclesmustbe .M1 Anexamplewhichshowsthatitispossibleforatleastonevalueof.A1 Exampleshowingthatitispossibleforall .B1
Statement
that
the
two
intersection
points
must
be
adistance
2apart.B1 Explanationthatinthecase itwouldhavetobethesamecircle.M1 Thelinejoiningthecentreof(or)andtheradiitoapointofintersectionformaright
angledtriangleineachcase.(onecase)
A1 Useofthistofindthedistancebetweencentresofcircles.
B1 Applyingthesametotheothercircle.
B1 Expressionrelatingthecoordinatesandradiiobtainedfromconsidering.B1 Expressionrelatingthecoordinatesandradiiobtainedfromconsidering.M1 Eliminationoffromtheequations.M1 Eitherexpansionofsquaredtermsorrearrangementtoapplydifferenceoftwosquares.
A1
Expressionfor
reached.Note
that
answer
is
given
in
the
question.
B1 Substitutiontofindexpressionforcoordinate.Note
that
any
expression
for
in
terms
of
,
,
andissufficient,butitmustbeexpressed
as
,
not
.B1 Observationthatmustbepositive.
Alternativemarkschemeforthismayberequiredoncesomesolutionsseen.
M1 Attempttorearrangetheinequalitytoget16ontheleft.M1 Reachapointsymmetricinand.M1 Reachacombinationofsquaredterms.
M1
Applydifference
of
two
squares
to
simplify.
A1 Reachtherequiredinequality.
Notethatanswerisgiveninthequestion.
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Question
8
(i) Letbethevectorfromthecentreofto.Usingsimilartriangles,thevectorfromthecentreoftois . M1A1Therefore
,sincethesearebothexpressionsforthevector
fromthecentreoftothecentreof. M1So A1Thepositionvectorofis M1A1
(ii) Thepositionvectorsofandwillbe and . B1Therefore,
M1A1 M1A1Similarly,
M1A1M1A1Sincetheyaremultiplesofeachotherthepoints,andmustlieonthesamestraightline.
B1
(iii) lieshalfwaybetweenandif B1Therefore
M1So, Whichsimplifiesto 2 M1A1
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Question
8
M1 Identificationofsimilartriangleswithinthediagram.
A1 Relationshipbetweenthetwovectorsto.M1 Equatingtwoexpressionsforthevectorbetweenthecentresofthecircles.
A1
Correct
simplified
expression.
M1 Calculationofvectorfromcentreofonecircleto.A1 Correctpositionvectorfor.
Note
that
answer
is
given
in
the
question.
B1 Identifyingthecorrectvectorsforthefocioftheotherpairsofcircles.
M1 Expressionforvectorbetweenanytwoofthefoci.
A1 Termsgroupedbyvector.
M1 Simplificationofgroupedterms.
A1 Extractionofcommonfactor.
M1
A1M1
A1
Expressionforavectorbetweenadifferentpairoffoci.
Award
marks
as
same
scheme
for
previous
example,
but
award
all
four
marks
for
thecorrect
answer
written
down
as
it
can
be
obtained
by
rotating
1,
2
and
3
in
the
previous
answer.
B1 Statementthattheylieonastraightline.
B1 Statementthatthetwovectorsmustbeequal.
M1 Reductiontostatementinvolvingonlyterms.M1 Attempttosimplifyexpressionobtained(ifnecessary).
A1 Anysimplifiedform.
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Question
9
(i) Takingmomentsabout: 3 sin3 0 M1A1 5 sin3 0 M1A1
B1
5cos 30 sin cos sin 30 3cos 30 sin cos sin 30 M15 32 sin
12 cos 3
32 sin
12 cos A1
Therefore
43 sin cos A1Either
Usesin cos 1andjustifychoiceofpositivesquareroot.Or
Drawrightangledtrianglesuchthattan andcalculatethelengthofthehypotenuse.
M1
sin A1(ii) Letbetheverticaldistanceofbelow.
Letbetheverticaldistanceofbelow. sin M1M1A1
sin M1A1Ifisthecentreofmassofthetriangle:
M1A1
Conservationof
energy:
4 8.2forcompleterevolutions. M1A1Therefore . A1
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Question
9
M1 Attempttofindthemomentofabout.A1 Correctexpressionformoment(sin3 0 maybereplacedbycos6 0 ).M1 Attempttofindthemomentofabout.A1
Correct
expression
for
moment
(sin3 0
may
be
replaced
by
cos6 0 ).
B1 Correctstatementthatthesemustbeequal.
M1 Useofsin orcos formulae.A1 Correctvaluesusedforsin30andcos30.A1 Correctlysimplified.
M1 Useofacorrectmethodtofindthevalueofsin.A1 Fullyjustifiedsolution.Ifusingrightangledtrianglemethodthenchoiceofpositiverootnot
needed,ifchoiceofpositiverootnotgivenwhenapplyingsin cos 1methodthenM1A0shouldbeawarded.
Note
that
answer
is
given
in
the
question.
M1
Attemptto
find
.M1 Correctlydealwithsineorcosineterm.A1 Correctvalue.
M1 Attempttofind.A1 Correctvalue.
M1 Combinetwovaluestoobtaindistanceofcentreofmassfrom.A1 Correctvalue
M1 Applyconservationofenergy.
A1 Correctinequality.
A1 Correctminimumvalue.
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Question
9
Alternative
part
(i)
(i) Letbethecentreofmassofthetriangleandletthedistancebe.Takingmomentsabout:5cos 3 cos M1A1Therefore
5 3 ,so
.
A1
mustlieonand 30 . B1sin3 0 cos M1sin 30 cos cos 30 sin cos M1
. A1Thereforecos 43sinandsocos 48sin M1sin ,andso(sinceisacute)sin . M1A1
M1
Takingmoments.
A1 Correctequation.
A1 Correctrelationshipbetweenand.B1 Identificationthatliesonandcalculationof.M1 Useofsineofidentifiedangle.
M1 Useofsin formula.A1 Directrelationshipbetweensinandcos.M1 Rearrangementandsquaringbothsides.
M1 Applyingsin cos 1.A1 Finalanswer(choiceofpositiverootmustbeexplained).
Note
that
answer
is
given
in
the
question.
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Question
10
Ifthelengthofstringfromtheholeatanymomentis,then . B1Thedistance,,fromthepointbeneaththeholesatisfies, . B1Therefore
2. M1A1
,and cosec M1Therefore,thespeedoftheparticleiscosec. A1Acceleration:
cosec cosec cot M1A1
sin ,socos
M1Therefore
cot A1
Theaccelerationis
cot M1Since
sec ,theaccelerationcanbewrittenas
cot
. M1A1
Horizontally:
sin cot ,so
cot cosecM1M1
A1
Theparticlewillleavethefloorwhen cos M1A1
cot andsotan
M1A1
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Question
10
B1 Aninterpretationofintermsofothervariables(includinganynewlydefinedones).B1 Anyvalidrelationshipbetweenthevariables.
M1 Differentiationtofindhorizontalvelocity.
A1
Correct
differentiation.
M1 Attempttoeliminateanyintroducedvariables.
A1 Correctresult.
Answers
which
make
clear
reference
to
the
speed
of
the
particle
in
the
direction
of
the
string
being
V.
M1 Differentiationofspeedfoundinfirstpart.
A1 Correctanswer.
M1 Attempttodifferentiatetofindanexpressionfor .
A1 Correctanswer.
M1 Substitutiontofindexpressionforacceleration.
M1
Relationshipbetween
required
variables
and
any
extra
variables
identified.
A1 Substitutiontogiveanswerintermsofcorrectvariables.
M1 Horizontalcomponentoftension.
M1 ApplicationofNewtonssecondlaw.
A1 Correctanswer.
M1 Verticalcomponentoftensionfound.
A1 Identificationthatparticleleavesgroundwhentensionisequaltothemass.
M1 Substitutionoftheirvaluefor.A1
Rearrangement
to
give
required
result.
Note
that
answer
is
given
in
the
question.
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Question
11
(i) cos , sin B1B1Differentiating: sin , cos M1Sinceismovingwithvelocityandisatthepoint, 0attime, :Velocity
of
is sin , cos . A1(ii) Initialmomentumwas(horizontally). M1
Horizontalvelocityofwillbethesameasthatof,sohorizontallythetotalmomentumisgivenby 2 sin M1Therefore3 2sin . A1Initialenergywas
M1
Totalenergyis 2 sin cos M1A1
Therefore
2 2sin sin cos M1
So 3 4sin 2 Substituting gives3 2sin 4sin 2sin 6 M16 3 4sin 4 sin 4sin 8 sin 6 4 sin 2So, . A1
(iii) 0,sotherecanonlybeaninstantaneouschangeofdirectioninwhichvariesatacollision.Sincethefirstcollisionwillbewhen
0,thesecond
collisionmust
be
when
.B1
B1
(iv) Sincehorizontalmomentummustbe, 0 2sin . B1TheKEofmustbe ,so B1 sin sin ,so or M1A1isonly0whentakesthesevaluesand ispositiveaswouldneedanonzerovaluetosatisfy3 2sin if isnegative.(Therelationshipisstilltruesincecollisionsareelastic).
B1
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Question
11
B1 Horizontalcomponent.
B1 Verticalcomponent.
M1 Differentiation.
A1
Complete
justification,
including
clear
explanation
that
.
Notethatanswerisgiveninthequestion.
M1 Statementthatmomentumwillbeconserved.
M1 Identificationthathorizontalmomentumofandwillbeequal.A1 Correctequationreached.
Notethatanswerisgiveninthequestion.
M1 Statementthatenergywillbeconserved.
M1 Useofsymmetrytoobtainenergyof(acceptanswerswhichsimplydoubletheenergyofratherthanstatingtheverticalvelocityinoppositedirection).A1 Correctrelationship.
M1
Useof
sin cos 1.M1 Substitutingtheotherrelationshiptoeliminate.A1 Correctequationreached.
Note
that
answer
is
given
in
the
question.
B1 Correctvalueof.B1 Answerjustified.
B1 Firstequationidentified.
B1 Secondequationidentified.
M1 Solvingsimultaneouslytofind
.
A1
Correctvalues
for
.B1 Justifiedanswerthatisnotalways0whentakesthesevalues.
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Question
12
(i) Ifatailoccursthenplayermustalwayswinbeforecanachievethesequencerequired.Thereforetheonlywayfortowinisifbothofthefirsttwotossesareheads.
B1
Afterthefirsttwotossesareheadsitdoesnotmatterifmoretossesresultin
headsasthefirsttimetailsoccurswillwin. B1Theprobabilitythatwinsistherefore B1
(ii) Asbefore,after,onlycanwin. B1Similarly,after,onlycanwin. B1Inallothercasesforthefirsttwotossesonlyandwillbeabletowin. M1Theprobabilitiesforandtowinmustbeequal. M1Theprobabilityofwinningis
foralloftheplayers. A1
(iii)
If
the
first
two
tosses
are
then
must
win
(as
soon
as
a
occurs),
so
the
probabilityis1. B1After:mustwinifthenexttossisaasneedstwostowin,butwillwinthenexttimeanoccurs. M1Ifthenexttossis,thenthepositionisasifthefirsttwotosseshadbeen,andsotheprobabilitythatwinsfromthispointis. M1Therefore, 1 A1After:Ifthenexttossis
then
willwinwithprobability
.
Ifthe
next
toss
is
thenwillwinwithprobability. M1Therefore ,andso . A1After:Ifthenexttossisthenplayerwinsimmediately.Ifthenexttossisthenwillwinwithprobability. M1Therefore . A1Solvingthetwoequationsinand,gives , From
the
third
equation
M1
A1
Theprobabilitythatwinsis 1 M1A1
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Question
12
B1 Identifyingthatcannotwinonceatailhasbeentossed.B1 Identifyingthatmustwinoncethefirsttwotosseshavebeenheads.B1 Showingthecalculationtoreachtheanswer.
Note
that
answer
is
given
in
the
question.
B1 Recognisingthatthesituationisunchangedforplayer.B1 Recognisingthatthesamelogicappliestoplayer.M1 Allothercasesleadtowinsforoneoftheremainingplayers.
M1 Recognisingthattheprobabilitiesmustbeequal.
A1 Correctstatementoftheprobabilities.
If
no
marks
possible
by
this
scheme
award
one
mark
for
each
probability
correctly
calculated
with
supporting
working.
All
four
calculated
scores
5
marks.
B1 Explanationthatprobabilitymustbe1.M1
Explanation
of
the
case
that
the
next
toss
is
.Thismarkandthenextcouldbeawardedforanappropriatetreediagram.M1 Explanationofthecasethatthenexttossis.A1 Justificationoftherelationshipbetweenand.
Note
that
answer
is
given
in
the
question.
M1 Considerationofonecasefollowing.This
mark
could
be
awarded
for
an
appropriate
tree
diagram.
A1 Establishmentoftherelationship.
M1 Considerationofonecasefollowing
.
This
mark
could
be
awarded
for
an
appropriate
tree
diagram.
A1 Establishmentoftherelationship.
M1 Attempttosolvethesimultaneousequations.
A1 Correctvaluesfor,and.M1 Attempttocombineprobabilitiestoobtainoverallprobabilityofwin.
A1 Correctprobability.
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Question
13
(i) for for B1
M1
M1
A1
Usethesubstitution intheintegral:
B1
M1Therefore . A1
,sothestationarypointoccurswhen ln . M1A1If
1
thenchoose
ln
asitispositive.
If 1thenchoose 0astheminimumoccursatanegativevalueof. B1(ii) 2
M1
A1
Usethesubstitution intheintegral: 2
B1
Applyingintegrationdonebefore:
2
2
Usingintegrationbyparts:
M1A1and,applyingtheintegrationalreadycompleted,
.Therefore
. A1Var M1
Var
.
Var
2
. A1
Var
2 1 M1
For 0, Var 0,sothevariancedecreasesasincreases. A1
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Question
13
B1 Statementofrandomvariable.
M1 Anycorrectterminexpectation(allowmultipliedbyanattemptattheprobabilityfornotneedinganyextracosts).
M1 Correctintegralstated(allow
missing).
A1 Fullycorrectstatement.
May
be
altered
to
accommodate
other
methods
once
solutions
seen.
B1 Substitutionperformedcorrectly.
M1 Integrationbypartsusedtocalculateintegral.
A1 Correctlyjustifiedsolution.
Notethatanswerisgiveninthequestion.
M1 Differentiationtofindminimumpoint.
A1 Correctidentificationofpoint.
B1 Bothcasesidentifiedwiththesolutionsstated.
M1
Attemptat
(atleasttwotermscorrect).A1 Correctstatementof.B1 Substitutionperformedcorrectly.
M1 Applyingintegrationbyparts.
A1 Correctintegration.
A1 Correctexpressionfor.M1 UseofVar A1 CorrectsimplifiedformforVarM1 DifferentiationofVar.A1 Correctinterpretation.
Note
that
answer
is
given
in
the
question.
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1. (i)
11 11
1 11
1 B1
1
121 121
integratingbyparts M1A1
0 12 11 12 A1*
(4) M1
M1
tan B1
!! !! M1Thus !! !! A1* (5)(ii)
.
usingthesubstitution , andthenthesubstitution , 1 M1A1*2
1
So 1 M1A1*Using
the
substitution
,
1 ,
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1 M1A1*(6)(iii)
M1A1
M1So M1 !! !! A1 (5)
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2. (i) True. B1
1000 B1If
1000,then
1000 ,so
1000 ,i.e.
1000
M1A1(4)
(ii) False. B1
E.G.Let 1 and 2 for odd,and 2 and 1 for even. B1Then forwhichfor , ,nor M1Soitisnotthecasethat ,butnorisitthecasethat A1(4)(iii)True. B1
meansthatthereexistsapositiveinteger,say
,forwhich for
,
.
E1 meansthatthereexistsapositiveinteger,say ,forwhich for , .
E1
Thenif max, , B1for , ,andso A1(5)(iv) True. B1
4 B1Assume 2 forsomevalue 4 . B1Then 1 1 1 2 2 2 2
M1A1
4 2 B1sobytheprincipleofmathematicalinduction,
2 for
4,andthus
2
A1(7)
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3. (i)
Symmetryaboutinitialline G1
Two
branches G1
Shapeandlabelling G1(3)
If | sec | ,then sec or sec So sec or sec M1A1If sec 0 , sec 0 as and sec 0 as and arebothpositive,andthusinbothcases, 0 whichisnotpermitted. B1If sec 0, sec 0 and sec 0 giving 0so
sec 0 asrequired. B1 (4)
So sec ,thuspointssatisfying(*)lieonacertainconchoidofNicomedeswithAbeingthepole(origin), B1
dbeingb, B1
andLbeingtheline sec . B1(3)(ii)
Symmetryaboutinitialline G1
Twobranches
G1
Loop,shapeandlabelling G1
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If ,thenthecurvehastwobranches, sec with sec 0 and sec with sec 0 ,theendpointsoftheloopcorrespondingto sec . B1(4)Inthecase 1 and 2 , sec 2 so Areaofloop
2 sec 2 M1A1 sec 4 sec 4 tan 4 ln|sec tan | 4 M1A1
4 3 4 ln2 3
3 4 l n 2 3
M1A1(6)
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4. (i) iscontinuous.For , andfor , . B1Sothesketchofthisgraphmustbeoneofthefollowing:
B1
Hence,itmustintersectthe axisatleastonce,andsothereisatleastonerealrootof 0 B1(3)(ii)
M1
Thus A1 A1
and,as 0, 0, 0addingthesethreeequationswehave,
3 0 M1(Alternatively,
3 6 )So
3 0 M1
Thus 6 3 2 A1*(6)
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(iii) Let cos sin for 1, 2,3 M1Then cos2 sin 2 and cos3 sin 3 bydeMoivre M1As
sin 0 sin2 0 sin3 0
0
0 0
andso , ,and arereal, M1andthereforesoare
,
,and
A1
Hence, as ,,and aretherootsof 0 with ,,and real,bypart(i),atleastoneof ,,and isreal. M1Soforatleastonevalueof ,cos sin isrealandthus, sin 0,andas , 0 asrequired. A1(6)If 0 then isreal. and aretherootsof 0whichis
0 (say
0)
and andsothequadraticofwhich and aretherootshasrealcoefficients. Thus, . ( 0 because 0) B1If 4 0, M1Thus cos cos,andso ,as .Butsin sin andso . M1A1If
4 0,then
and
arerealroots,so
sin sin 0,andthus
0,so
. B1(5)
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5.(i)Havingassumedthat 2 isrational(step1), 2 ,where , , 0 B1Thusfromthedefinitionof(step2),as and2 ,so provingstep3.B1
(2)
If ,then isanintegerand2isaninteger. B1So 2 1 2 isaninteger, B1and2 1 2 2 2 whichisanintegerandso2 1 provingstep5. B1(3)1 2 2 andso M10 2 1 1,andthus0 2 1 A1andthusthiscontradictsstep4that
isthesmallestpositiveintegerin
as
2 1 hasbeen
showntobeasmallerpositiveintegerandisin. A1(3)(ii) If 2 isrational,then2 ,where , , 0
So 2 ,thatis 2 ,whichcanbewritten 2 2 M1andhence2 2 provingthat 2 isrational. A1If
2 isrational,then
2 ,where
, , 0 M1
andso2 provingthat2 isrationalandthat 2 isrationalonlyif 2 isrational.A1
(4)
Assumethat2 isrational.Definetheset tobethesetofpositiveintegerswiththefollowingproperty: isinifandonlyif2 and 2 areintegers. B1The
set
containsatleastonepositiveintegerasif 2 ,where , , 0,then2 and 2 ,so . M1A1Define tobethesmallestpositiveintegerin . Then 2 and 2 areintegers. B1Consider 2 1. 2 1 2 whichisthedifferenceoftwointegersandsoisitselfaninteger. 2 1 2 2 2 whichisaninteger,and
2 1 2 2 2 2 2 2 whichisaninteger.
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Thus 2 1 isin. M1A11 2 2 andso 0 2 1 1,andthus0 2 1 ,andthusthiscontradictsthat
isthesmallestpositiveintegerin
as
2
1hasbeenshowntobeasmallerpositive
integerand
is
in
. M1A1(8)
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6. (i) , , B1For , ,werequiretosolve , M1
2 2 0 2 4 8 84 2 2 2 2
M1A1
Sofor
, ,as
mustbereal,
mustbereal,and
2 0
i.e. 2 B1*(5)(ii) 2 soif , then 2 so 3 1 M1A1
3 1M1A1
Thusif
, 1 1 M1A1Thus 1 1 0, M1 1 0 M1A1Thus 1 or Soas and 0 ,thevaluesof arereal. B1Therearethreedistinctvaluesof
unless
1 inwhichcase
1 4 3,i.e.
2
M1A1(12)
For , ,from(i)werequire whichitis, whichitis,and 2 inotherwords . M1So and neednotbereal. Acounterexamplewouldbe 1 B1forthen 1, ,so 1 , i.e. 2 2 0inwhichcasethe
discriminant
is
0
so
.
B1
(3)
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7. M1 M1A1(3)(i)
Suppose
isapolynomialofdegree i.e. forsomeinteger. B1Then 1 1 whichisapolynomialofdegree . M1A1Suppose ,then 1 sotheresultistruefor 1, M1A1andwehaveshownthatifitistruefor ,itistruefor 1. Hencebyinduction,itistrueforanypositiveinteger . B1(6)
(ii) Suppose 1 isdivisibleby1 i.e. 1 1 forsomeinteger,with 1. B1Then
1
1
1
1 1 1 1 whichisdivisibleby 1 . M1A11 1 1 soresultistruefor 1. M1A1Wehaveshownthatifitistruefor ,itistruefor 1. Hencebyinduction,itistrueforanypositiveinteger . B1(6)(iii)
1 1 M1So1 1 1 M1A1Butby(ii), 1 isdivisibleby1 andso 1 1 ,andthusif 1, 1 0, andhence
1 0
M1A1*
(5)
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8. (i) cos sin cos M1A1and sin cos sin M1A1Thus
becomes sin cos sin cos M1Thatis sin cos cos sinsin cos sin cosas 0, 0Multiplyingoutandcollectingliketermsgives
cos sin
sin cos 0whichis 0 . M1A1*(7)So 0 M1andthus , A1
A1
G1
(4)
(oralternatively M1so ln|| A1andhence A1)
(ii) becomes sin cos cos sin cos sinthatis
sin cos cos cos sin sin cos sin sin cos
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Multiplyingoutandcollectingliketermsgives
cos sin cos sin sin cos 0 M1whichis
0 . A1
M1So A1 ln A1So
1 with 0
11 thatis
A1*
G1
G1
G1
(9)
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9. Iftheinitialpositionofis ,thenattime, ,soconservingenergy,12 12 2
M1
A1
A1
Thus, M1
i.e. A1*
(5)
Thegreatestvalue,,attainedby,occurswhen 0. M1Thus So (negativerootdiscountedasallquantitiesarepositive)Thus
2
and
2 M1
A1
(3)
As
differentiatingwithrespectto2 2 12 2 M1A1
Thus
A1
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Sowhen ,theaccelerationofis
2
2 M1
A1
(5)
Thatis
andthus
1
where istheperiod. M1A1So
4 1
4 11
Let
B1
then
andso
2
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2 1 2 2 as Thus
1 2M1A1
andso
4 1 1
1 2 32 1 1
as
required.
M1
A1*
(7)
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10. Thepositionvectoroftheupperparticleis sin cosB1B1
sodifferentiatingwithrespecttotime,itsvelocityis
cos sinE1*
(3)
Itsacceleration,bydifferentiatingwithrespecttotime,isthus
cos sin sin cos
M1
A1
A1
sobyNewtonssecondlawresolvinghorizontallyandvertically
sin cos cos sin sin cosM1A1
Thatis
cos sin sin cos sincos 01Theotherparticlesequationis
cos sin sin cos sincos 01B1
(6)
Addingthesetwoequationswefind
2 2 01i.e. 0 and M1A1*Thus cos sin sin cos sincosi.e.
cos
sin sin and
sin
cos cos
Multiplyingthe
second
of
these
by
sinandthefirstby cosandsubtracting,
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0 andso 0. M1A1*(4)Thus andasinitially 2 , M1A1Thereforethetimetorotateby
isgivenby
,so
A1
As andinitially ,attime , ,andso asthecentreoftherodisinitially abovethetable. M1A1Hence,giventheconditionthattheparticleshitthetablesimultaneously,0 / 1/2 / .
Hence
0 2 2
,
or
2 2
as
required.
M1
A1*
(7)
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11. (i) Supposethattheforceexertedbyontherodhascomponentsperpendiculartotherodand paralleltotherod. Thentakingmomentsfortherodaboutthehinge, 0, M1whichas 0 yields 0 andhencetheforceexertedontherodby isparalleltotherod.A1*
(2)
Resolvingperpendiculartotherodfor , sin sin cos M1A1Dividingby sin, sin cotThatis cot cos orinotherwords cot cos asrequired. M1A1*(4)Theforceexertedbythehingeontherodisalongtherodtowards , B1andifthatforceis ,thenresolvingverticallyfor , cos M1A1so
sec
.
A1
(4)
(ii) Supposethattheforceexertedbyontherodhascomponentperpendiculartotherodtowardstheaxis,thattheforceexertedbyontherodhascomponentperpendiculartotherodtowardstheaxis, B1
thenresolvingperpendiculartotherodfor , sin sin cosM1A1
and
similarly
for
,
sin sin cos
M1A1
Takingmomentsfortherodaboutthehinge, 0 M1A1Somultiplyingthefirstequationby ,thesecondby andaddingwehave sin sin sin cos sin cosDividingby
sin,
cot
cos M1A1
Thatis cot cos ,where A1(10)
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12. (i) Theprobabilitydistributionfunctionof is 1 2 3 4 5 6
16
16
16
16
16
16
sotheprobabilitydistributionfunctionof is 0 1 2 3 4 5 16 16 16 16 16 16
andthus
1 . B1Theprobabilitydistributionfunctionof is 2 3 4 5 6 7 8 9 10 11 12 136 236 336 436 536 636 536 436 336 236 136
M1
sotheprobabilitydistributionfunctionof is 0 1 2 3 4 5 636 636 636 636 636 636
A1
whichisthesameasfor
andhenceitsprobabilitygeneratingfunctionisalso
. A1*
Therefore,theprobabilitygeneratingfunctionof isalso B1andthustheprobabilitythat isdivisibleby6is 16 . B1(6)(ii) Theprobabilitydistributionfunctionof is
0 1 2 3 4
16
26
16
16
16
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andthus 1 2 M1A1 wouldbe exceptthatthepowersmustbemultipliedcongruenttomodulus5. 1 2 1
B1
Thus wouldbe except 1 1 M1A1and 1 1 1 1 1 1 1 5 A1So
2
2 5
7
M1A1*(8)
16 16 7 16 7 7Thatis
16 7 7 16 7 35 16 43Wenoticethatthecoefficientof insidethebracketin is 1 6 6 6Thiscanbeshownsimplybyinduction. Itistruefor
1trivially.
Consider 1 6 6 6 1 6 6 6 1 6 6 6 1 6 6 6 1 6 6 6 1 6 6 6 51 6 6 6
5 1 6 6 6 6 1 1 6 6 6 6 1So
1 6 6
6
51 6 6
6
1 6 6
6
asrequired. M1
However,thiscoefficientisthesumofaGPandso where isanintegersuchthat 0 5 4. M1A1Soif isnotdivisibleby5,theprobabilitythat isdivisibleby5willbethecoefficientof whichinturnisthecoefficientof ,namely 1 asrequired. B1*If
isdivisibleby5,theprobabilitythat
isdivisibleby5willbe
1
as
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Thatis 1 M1A1(6)
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13. (i)
G1
if 0 1 B1
G1
and 1 2 if 1 2 B1 0if 0 and 1 if 2So
0 0 0 11 2 1 21 2
B1(5)
Thus
1
1 12 1 12 2 1 12 10 12
M1
A1
Soas
,
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0 121 2 1 12 1
1 1
asrequired. M1A1*(4)
1 2 . 2 l n
1 2 ln 12 2 1 2 ln 2M1
A1
(2)
(ii)
G1
12 0 11 12 1B1
(2)
Thus
1
1 1
1
So
1 12 1 1 12 112 1 1 0 12
i.e.
12 3
12 112 1 0 12
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M1A1
Soas ,
12 12 112 1 0 12M1A1
(4)
because,bysymmetry, and 1 1 B1
12 1 12
12 1 12 1
12 ln
12 12 ln1
12 ln 12
12 12 ln 12 12 ln 1212asrequired. M1A1(3)
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