step 2015 mark scheme

7/23/2019 STEP 2015 Mark Scheme

1/71

STEP Mark Scheme 2015

Mathematics

STEP 9465/9470/9475

October 2015


2/71

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


3/71

Contents

STEP Mathematics (9465, 9470, 9475)

Mark Scheme PageSTEP Mathematics I 4STEP Mathematics II 18STEP Mathematics III 45


4/71

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


5/71

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


6/71

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


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


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.


7/71

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


8/71

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


9/71

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


10/71

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)


11/71

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


12/71

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


14/71

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


15/71

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)


16/71

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


17/71

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


18/71

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


19/71

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.


20/71

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


21/71

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.


22/71

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


23/71

Question

3

M1 Appreciationofthemeaningof .M1 Identifythenumberofpossibilitiesforthelengthofthethirdrodinonecase.

M1 Identifythesetofpossiblecasesandfindnumbersofpossibilitiesforeach.

A1

Clear

explanation

of

the

result.


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).


24/71

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


25/71

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, .


26/71

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


27/71

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.



28/71

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


29/71

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.


30/71

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


31/71

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.



32/71

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


33/71

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.


34/71

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


35/71

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.


36/71

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.


37/71

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


38/71

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.


39/71

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


40/71

Question

11

B1 Horizontalcomponent.

B1 Verticalcomponent.

M1 Differentiation.

A1

Complete

justification,

including

clear

explanation

that

.


M1 Statementthatmomentumwillbeconserved.

M1 Identificationthathorizontalmomentumofandwillbeequal.A1 Correctequationreached.


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.


41/71

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


42/71

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.


43/71

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


44/71

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.


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.


45/71

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 ,


46/71

1 M1A1*(6)(iii)

M1A1

M1So M1 !! !! A1 (5)


47/71

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)


48/71

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


49/71

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)


50/71

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)


51/71

(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)


52/71


53/71

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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