MARK SCHEME for the October/November 2010 question paper ...maxpapers.com/.../2012/11/9231_Oct-Nov-2010-All-Mark-Schemes.pdf · 9231/01 Paper 1, maximum raw mark 100 ... Page 2 Mark

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS

GCE Advanced Level

MARK SCHEME for the October/November 2010 question paper

for the guidance of teachers

9231 FURTHER MATHEMATICS

9231/01 Paper 1, maximum raw mark 100

This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers.

Mark schemes must be read in conjunction with the question papers and the report on the examination.

• CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the October/November 2010 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

www.maxpapers.com

Page 2 Mark Scheme: Teachers’ version Syllabus Paper

GCE A LEVEL – October/November 2010 9231 01

© UCLES 2010

Mark Scheme Notes

Marks are of the following three types:

M Method mark, awarded for a valid method applied to the problem. Method marks are not lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. Correct application of a formula without the formula being quoted obviously earns the M mark and in some cases an M mark can be implied from a correct answer.

A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained.

Accuracy marks cannot be given unless the associated method mark is earned (or implied).

B Mark for a correct result or statement independent of method marks.

• When a part of a question has two or more “method” steps, the M marks are generally independent unless the scheme specifically says otherwise; and similarly when there are several B marks allocated. The notation DM or DB (or dep*) is used to indicate that a particular M or B mark is dependent on an earlier M or B (asterisked) mark in the scheme. When two or more steps are run together by the candidate, the earlier marks are implied and full credit is given.

• The symbol √ implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A or B marks are given for correct work only. A and B marks are not given for fortuitously “correct” answers or results obtained from incorrect working.

• Note: B2 or A2 means that the candidate can earn 2 or 0. B2/1/0 means that the candidate can earn anything from 0 to 2.

The marks indicated in the scheme may not be subdivided. If there is genuine doubt whether a candidate has earned a mark, allow the candidate the benefit of the doubt. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored.

• Wrong or missing units in an answer should not lead to the loss of a mark unless the

scheme specifically indicates otherwise. • For a numerical answer, allow the A or B mark if a value is obtained which is correct to 3 s.f.,

or which would be correct to 3 s.f. if rounded (1 d.p. in the case of an angle). As stated above, an A or B mark is not given if a correct numerical answer arises fortuitously from incorrect working. For Mechanics questions, allow A or B marks for correct answers which arise from taking g equal to 9.8 or 9.81 instead of 10.

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© UCLES 2010

The following abbreviations may be used in a mark scheme or used on the scripts:

AEF Any Equivalent Form (of answer is equally acceptable) AG Answer Given on the question paper (so extra checking is needed to ensure that

the detailed working leading to the result is valid) BOD Benefit of Doubt (allowed when the validity of a solution may not be absolutely

clear) CAO Correct Answer Only (emphasising that no “follow through” from a previous error

is allowed) CWO Correct Working Only – often written by a ‘fortuitous' answer ISW Ignore Subsequent Working MR Misread PA Premature Approximation (resulting in basically correct work that is insufficiently

accurate) SOS See Other Solution (the candidate makes a better attempt at the same question) SR Special Ruling (detailing the mark to be given for a specific wrong solution, or a

case where some standard marking practice is to be varied in the light of a particular circumstance)

Penalties

MR –1 A penalty of MR –1 is deducted from A or B marks when the data of a question or

part question are genuinely misread and the object and difficulty of the question remain unaltered. In this case all A and B marks then become “follow through √” marks. MR is not applied when the candidate misreads his own figures – this is regarded as an error in accuracy. An MR–2 penalty may be applied in particular cases if agreed at the coordination meeting.

PA –1 This is deducted from A or B marks in the case of premature approximation. The

PA –1 penalty is usually discussed at the meeting.

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© UCLES 2010

1 222222

2

)ee(4

1))ee(

2

1(1

d

d1

xxxx

x

y++

+ M1A1 expression simplified

Length = ( ) ( )[ ]21

0

2222

2

1

0

ee4

1d ee

2

1xxxx

x+∫ M1 integrate

= ( ) ( )e4

1eee

4

1ee

4

12

0011 AG A1 cao [4]

2 nth term is

+ 2

11

2

1

nn

M1A1

+

+

+

++

+

3

1

1

1

4

1

2

1

...1

2

1

1

1

1

1

2

11

2

1 NNNNNNSN

M1 sum of terms

=

++ 1

1

2

1

2

3

2

1

NN A1 after cancellation [4]

Limit = ¾ B1√ [1]

3 Area = 3/823

2d

4

1

4

1

2

1

2

3

2

1

2

1

∫ xxxxx B1

A

xxx

A

xx

x

y

4

1

2

4

1

ln222

1d)

12(

2

1

++∫

M1 use of A

xy∫ d2

1 2

M1 integrate A1 correct Final answer:

+4

32ln

8

3 or

+2

34ln

16

3 or

32

92ln

8

3+ etc (ACF) A1 [5]

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© UCLES 2010

4 n = 0: 71 + 53 = 132 which is divisible by 44 B1 Assume 72k + 1 + 5k + 3 is divisible by 44 B1 Consider 72(k + 1) + 1 + 5(k + 1) + 3 = 7272k + 1 + 5.5k + 3 M1 (k + 1) th term = 49(72k + 1 + 5k + 3) – 44.5k + 3 M1 in appropriate form which is divisible by 44 A1 convincing argument [5] Alternative solution for final three marks: Consider (72k + 3 + 5k + 4) – (72k + 1 + 5k + 3) M1 = 48(72k + 1 + 5k + 3) – 44.5k + 3 M1 in appropriate form which is divisible by 44 A1 convincing argument

5 In + 2 = [–(1–x)n + 2 cosx] – ∫+

+ xxxnn

dcos)1)(2(1 M1A1

]dsin)1()sin)1)[((2())2(1( 1 ∫++++++

xxxxxnnnn

M1 integrate by parts again

In + 2 = 1 – (n + 1)(n + 2) In AG A1 [4] I6 = 1 – 5 × 6I4; I4 = 1 – 4 × 3I2; I2 = 1 – 1 × 2I0 M1

∫1

0

01cos1dsin xxI B1

I6 1 30(1 12(1 2I0)) 0.0177 M1A1 [4]

OR I0 = 1 – cos1 B1 I2 = 2cos1 – 1 M1 (use of RF) I4 = 13 – 24cos1 A1 I6 = 0.0177 A1 cao Accept decimal versions

6

2310

22430

2110

121

α

α

α

→

66000

24100

2110

121

α

α

α

α

M1A1

Dim = 4 ⇒ α ≠ 1 AG A1 [3] a + 2b – c = 0 2a + 3b – c = 0 Show a = b = c = 0 M1 attempt to solve 2a + b + 2c = 0 b – 3c = 0 Linearly independent and dim R(T) not 4: basis A1 [2] a + 2b – c = p 2a + 3b – c = 1 Attempt to find a, b, c in terms of q or p 2a + b + 2c = 1 b – 3c = q M1A1 6p + q = 3 A1 [3] Alternative solution: Use row operations as in (i) M1

Final column

+ 36

24

21

qp

p

p

p

A1

6p + q = 3 A1

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© UCLES 2010

7 1

1

+xy

y

yx∴

1 M1 use in given cubic equation

Gives 6y3 – 7y2 + 3y – 1 = 0 AG A1 [2] n = 1: given expression = sum of roots = 7/6 B1

n = 2:( ) ( ) ∑∑ ∑

++ 36

13""2

1

1

1

12

2αβ

αα

B1 [2]

From cubic in y,

036

73

36

13.7

1

16

3

+

+∑α

M1

216/731

13

+∑

α

A1 [2]

LHS = ∑

+

+++3)1(

1)(1)(1(

α

αγβ M1

216

73

6

11

×

M1 recognise product of roots

= 73/36 AG A1 [3]

8 (i) 2

1sinsin3sin1 ⇒+ θθθ M1

6,

2

3 π

and

6

5,

2

3 π

A1 (both) [2]

(ii)

B1 circle B1 cardioid behaviour at origin B1 cardioid closed and symmetry [3] (iii) Subtract integrands M1

θθθ

π

π

d)sin22cos43(22

6

21 ∫× M1

[ ] 2

6

cos22sin23π

πθθθ + M1A1

= π AG A1 [5] Alternative: Area inside C1:

2

6

2

6

22sin

2

1

2

9d sin9

2

12

π

π

π

π

θθθθ∫

× M1

+

4

3

32

9 π

A1

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© UCLES 2010

Area inside C2:

θθθ

π

π

d)2cos1(2

1sin21

2

12

2

6

∫ ++×

2

6

2sin4

1cos2

2

3π

π

θθθ

M1

+

8

39

2

π

(A1 if not earned earlier)

Subtraction M1 Required area = π AG A1 [5] 9 [ ] 0))3((11)3)(2()3( + λλλλ M1 characteristic equation 0)4)(1)(3( λλλ M1 factorise λ = 1, 3, 4 A1

0

0

0

310

121

013

z

y

x

λ

λ

λ

Solve for λ = 1: (1, 2, 1) M1A1 Solve for λ = 3: (1, 0, –1) A1 Solve for λ = 4: (1, –1, 1) A1 [7]

M =

111

102

111

B1√ eigenvectors as columns

(except

0

0

0

)

D =

800

010

001

M1A1√ ft on eigenvalues [3]

10 4235

5105cos csscc +θ M1A1 use of de Moivre for (c + is)5

5324

1055sin sscsc +θ A1

42

35

5101

5105tan

tt

ttt

+

+θ AG M1A1 intermediate step needed [5]

5

05tanπ

θθn

⇒ M1

Solutions 5

tanπn

for n = 1, 2, 3, 4 A1 justify values of n [2]

Roots ±5

tanπ

, ±5

2tan

π

B1

Product of these roots = 5 M1

55

2tan

5tan

ππ

A1 [3]

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© UCLES 2010

11 z′ = y + xy′ B1 z′′ = 2y′ + xy′′ B1 Obtain result B1 [3] Auxiliary equation: m2 + 4 = 0 : m = ±2i M1 CF: xBxA 2sin2cos + A1 PI: z = ax2 + bx + c

Differentiate twice and substitute M1 a = 2, b = 0, c = 3 A1

GS: 322sin2cos2

+++ xxBxAz A1√ their CF + their PI

y = 0, x = 2

1π : (z = 0) gives 3

2

2

+π

A B1

z′ = –2Asin2x + 2Bcos2x + 4x M1

y′ = –2, x = 2

π

: (z′ = –π) gives 2

3πB A1

+++

+ 322sin

2

32cos3

2

1 2

2

xxxx

yππ

A1 [9]

12 EITHER (i) y′ = 0 ⇒ (x2 – 2x + λ)(2x + 2λ) – (x2 + 2λx)(2x – 2) = 0 M1 ⇒ ... ⇒ (λ + 1)x2 – λx – λ2 = 0 A1 Hence at most 2 values of x and at most 2 stationary points A1 [3] (ii) For 2 real distinct roots, λ2 > 4(λ + 1)(– λ2) M1 use of discriminant

4

50)45(

2

>∴>+ λλλ AG A1 [2]

(iii) Vert. asymptotes when x2 – 2x + λ = 0 M1 b2 – 4ac > 0 ⇒ 4 – 4 λ > 0 For two vert. asymp. λ < 1 A1 [2] (iv) (a) y = 0 ⇒ x2 + 2λx = 0 M1 ⇒ x = 0 or –2λ A1 (both)

(b) y = 1: x = 22 +λ

λ B1 [3]

(v) (a) λ < –2: no stat points: 2 vert. asymp B1 3 branches B1 completely correct shape (b) λ < 2: 2 stats points: no vert. asymp B1 max, min, horiz asymp B1 correct shape [4]

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© UCLES 2010

OR

Normal to plane: (2, 3, 4) × (–1, 0, 1) = (3, –6, 3) M1A1 r.(1, –2, 1) = d and point (2, 1, 4) M1 substitute point into plane eqn d = 4 x – 2y + z = 4 A1 [4] Alternative: x = 2 + 2λ – µ y = 1 + 2λ x + z = 6 + 6λ M1A1 z = 4 + 4λ + µ

)1(26 ++∴ yzx M1 42 +∴ zyx A1 x – 4y + 5z = 12 x – 2y + z = 4 Solve by eliminating one variable M1 Use parameter and express all 3 variables in terms of it M1 e.g. x = 3t – 4, y = 2t – 4, z = t r = (–4, –4, 0) + t (3, 2, 1) A1 or equivalent [3] Alternative:

Direction of line =

×

1

2

3

5

4

1

1

2

1

t M1A1

Find any point on line e.g.

0

4

4

,

2

0

2

etc.

∴ r =

+

1

2

3

0

4

4

t B1

Line l: r = (a, 2a + 1, –3) + α(3c, –3, c) Plane: x – 2y + z = 4 Distance A to plane:

6

15

6

43)12(2 +aa

M1

3a + 9 =15 M1 correct use of modulus sign a = 2 A1

22

996

63sin

cc

cc

++

++θ M1A1

6

2

1096

64

2

+

+∴

c

c

M1 solve for c

6c2 – 12c = 0: c = 2 A1 (Penalise only once for negative values.) [7]

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UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS

GCE Advanced Level

MARK SCHEME for the October/November 2010 question paper

for the guidance of teachers

9231 FURTHER MATHEMATICS

9231/02 Paper 2, maximum raw mark 100

This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers.

Mark schemes must be read in conjunction with the question papers and the report on the examination.

• CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the October/November 2010 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

www.maxpapers.com



© UCLES 2010

Mark Scheme Notes

Marks are of the following three types:

M Method mark, awarded for a valid method applied to the problem. Method marks are not lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. Correct application of a formula without the formula being quoted obviously earns the M mark and in some cases an M mark can be implied from a correct answer.

A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained.

Accuracy marks cannot be given unless the associated method mark is earned (or implied).

B Mark for a correct result or statement independent of method marks.

• When a part of a question has two or more “method” steps, the M marks are generally independent unless the scheme specifically says otherwise; and similarly when there are several B marks allocated. The notation DM or DB (or dep*) is used to indicate that a particular M or B mark is dependent on an earlier M or B (asterisked) mark in the scheme. When two or more steps are run together by the candidate, the earlier marks are implied and full credit is given.

• The symbol √ implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A or B marks are given for correct work only. A and B marks are not given for fortuitously “correct” answers or results obtained from incorrect working.

• Note: B2 or A2 means that the candidate can earn 2 or 0. B2/1/0 means that the candidate can earn anything from 0 to 2.

The marks indicated in the scheme may not be subdivided. If there is genuine doubt whether a candidate has earned a mark, allow the candidate the benefit of the doubt. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored.

• Wrong or missing units in an answer should not lead to the loss of a mark unless the

scheme specifically indicates otherwise. • For a numerical answer, allow the A or B mark if a value is obtained which is correct to 3 s.f.,

or which would be correct to 3 s.f. if rounded (1 d.p. in the case of an angle). As stated above, an A or B mark is not given if a correct numerical answer arises fortuitously from incorrect working. For Mechanics questions, allow A or B marks for correct answers which arise from taking g equal to 9.8 or 9.81 instead of 10.

www.maxpapers.com



© UCLES 2010

The following abbreviations may be used in a mark scheme or used on the scripts:

AEF Any Equivalent Form (of answer is equally acceptable) AG Answer Given on the question paper (so extra checking is needed to ensure that

the detailed working leading to the result is valid) BOD Benefit of Doubt (allowed when the validity of a solution may not be absolutely

clear) CAO Correct Answer Only (emphasising that no “follow through” from a previous error

is allowed) CWO Correct Working Only – often written by a ‘fortuitous' answer ISW Ignore Subsequent Working MR Misread PA Premature Approximation (resulting in basically correct work that is insufficiently

accurate) SOS See Other Solution (the candidate makes a better attempt at the same question) SR Special Ruling (detailing the mark to be given for a specific wrong solution, or a

case where some standard marking practice is to be varied in the light of a particular circumstance)

Penalties

MR –1 A penalty of MR –1 is deducted from A or B marks when the data of a question or

part question are genuinely misread and the object and difficulty of the question remain unaltered. In this case all A and B marks then become “follow through √” marks. MR is not applied when the candidate misreads his own figures – this is regarded as an error in accuracy. An MR–2 penalty may be applied in particular cases if agreed at the coordination meeting.

PA –1 This is deducted from A or B marks in the case of premature approximation. The

PA –1 penalty is usually discussed at the meeting.

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© UCLES 2010

Question Number

Mark Scheme Details Part Mark

Total

1 Find period T using v2 = ω2 (A2 – x2) and T = 2π/ω: ω = 6/4, T = 4π/3 or 4⋅19 [s] M1 A1(i) Find max speed using vmax = ωA: vmax = 15/2 or 7⋅5 [ms-1] M1 A1

(ii) Find mag. of max accel. using amax = ω2A: amax = 45/4 or 11⋅2[5] [ms-2] M1 A1

2 2 2

[6]

2 Apply conservation of energy: ½mv2 = ½mu2 – mga(1 – cos θ) M1Put v = ½u and simplify: 8ga(1 – cos θ) = 3u2 A.G. A1Equate radial forces to find contact force N: N = mg cos θ + m(½u)2/a M1 A1Replace cos θ by 1 – 3u2/8ga (A.E.F.): N = mg – mu2/8a M1 A1

2 4

[6]

3 Use conservation of momentum: mu + ¼α mu = α mvB M1 A1Use Newton’s law of restitution: – vB = – e(u – ¼u) [vB = ¾eu] M1 A1Eliminate vB to find e (A.E.F.): e = (1 + ¼α)/¾α or (4 + α)/3α M1 A1Use e ≤ 1 to find inequality for α: 4 + α ≤ 3α so α ≥ 2 A.G. M1 A1

6 2

[8]

4 Resolve in any two dirns. for rod, e.g. vertically: RA sin 2θ + RB cos θ = W or horizontally: RA cos 2θ – RB sin θ = 0 or parallel to rod: RA cos θ = W sin θ or normal to rod: RA sin θ + RB = W cos θ B1 B1

(i) Solve to find RA, e.g.: RA = W sin θ / (cos 2θ cos θ – sin 2θ sin θ) = W tan θ A.G. M1 A1

(ii) Solve to find RB, e.g.: RB = W tan θ cos 2θ / sin θ = W cos 2θ / cos θ A.G. M1 A1

(iii) Take moments for rod, e.g. about A: RB 2r cos θ = W a cos θ or about B: RA 2r cos θ sin θ = W (2r cos θ – a) cos θ M1 A1 Substitute and simplify: 2r cos 2θ = a cos θ A.G. A1

2 2 2 3

[9]

5 Find MI of disc about axis at A by par. axes thm: Idisc = ¼ 2ma2 + 2ma2 [= 5ma2 /2] M1 A1Find MI of particle about axis at A: Im = m(2a)2 [= 4ma2 ] B1Combine to find MI of system : I = 13ma2 /2 A.G. A1Use conservation of energy (lose A1 for one error): ½IΩ2 = 2mg × a + mg × 2a M1 A2Substitute for I to find angular speed Ω: Ω = √(16g/13a) A.E.F. M1 A1State eqn of motion (A.E.F.): I d2θ /dt2 = – 4mga sin θ M1 A1Approximate sin θ by θ (implied by use of SHM): I d2θ /dt2 = – 4mga θ M1Find approx. period T from SHM formula: T = 2π /√(8mga/13ma2) = 2π √(13a/8g) A.E.F. M1 A1

4 5

5

[14]

6 Use valid formula for C.I.: x B – x A ± zσ √(1/nA + 1/nB) M2 = 112 – 109 ± z 15 √(1/15 + 1/20) A1 = 3 ± 5⋅123 z A1Use of correct tabular value: z 0.995 = 1⋅64[5] *A1C.I. correct to 3 s.f. (dep *A1): 3 ± 8⋅43 or [-5⋅43, 11⋅4] A1

6

[6]

7 (i) Find or imply value of p: p = ¼ or 0⋅25 B1 Find P(X = 5): (1 – p)4 p or q4 p = 0⋅0791 M1 A1

(ii) Find P(X ≥ 5): 1 – (1 + q + q2 + q3) p or q4 or q4p + q5 = 0⋅316 M1 A1

(iii) Find least N with P(X ≤ N) > 0⋅9995: 1 – qN > 0⋅9995, qN < 0⋅0005 N > 26⋅4, Nmin = 27 M1 A1

3 2 2

[7]

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8 Find expected values to (at least) 1 dp: A B C (lose A1 if one or more errors Passes 18⋅33 14⋅67 22⋅00 or if rounded to integers) Failures 31⋅67 25⋅33 38⋅00 M1 A1State (at least) null hypothesis (A.E.F.): H0: Test result indep of school B1Calculate value of χ2 : χ2 = 3⋅7 ± 0⋅02 B1 S.R. If rounded to integers above allow: χ2 = 3⋅96 or 4⋅0 (earns max 6/7) (B1)Compare with tabular value (to 2 dp): χ2, 0.95

2 = 5⋅99 B1Valid method for reaching conclusion: Reject H0 if χ

2 > tabular value M1Correct conclusion (A.E.F., requires correct values): No association A1

7

[7]

9 State both hypotheses (A.E.F.): H0: µI – µO = 0⋅1, H1: µI – µO > 0⋅1 B1State valid assumption for paired-sample test: Popln. of diffs. has Normal distn. B1Consider differences eg: 0⋅4 0⋅1 0⋅2 0⋅3 0⋅1 0⋅4 0⋅1 0⋅2 M1Calculate sample mean: d = 1⋅8 / 8 [= 0⋅225] M1Estimate population variance: s2 = (0⋅52 – 1⋅82/8) / 7 (allow biased: 0⋅0144 or 0⋅1202) [= 0⋅0164 or 0⋅1282 ] M1Calculate value of t (to 2 dp): t = ( d – 0⋅1) / s√(1/8) = 2⋅76 M1 *A1Compare with correct tabular t value: t7, 0.975 = 2⋅36[5] *B1Valid method for reaching conclusion: Reject H0 if χ

2 > tabular value M1Correct conclusion (AEF, dep *A1, *B1): Coach’s suspicion is correct A1S.R.: State both hypotheses: H0: µI – µO = 0⋅1, H1: µI – µO > 0⋅1 (B1) State valid assumption for 2-sample test: Both poplns. have Normal distns. and a common variance (B1) Calculate sample means: 170⋅4/8, 168⋅6/8 [= 21⋅3, 21⋅075] and estimate population variance: s2 = (3630⋅1 – 170⋅42/8 + 3553⋅94 – 168⋅62/8)/14 [= 0⋅09107] (M1) Calculate value of t (to 2 dp): (0⋅225 – 0⋅1)/s√(1/8 + 1/8) = 0⋅828 (M1*A1) Compare with correct tabular t value: t14, 0.975 = 2⋅14[5] (*B1) Correct conclusion (AEF, dep *A1, *B1): Coach’s suspicion is not correct (B1 max 7)

10

[10]

10 (i) Find mean values to 3 s.f.: x = 2⋅024, y = 3⋅817 B1

(ii) Calculate gradient b in y – y = b(x – x): b = (88⋅415 – 24⋅29 × 45⋅8/12) / (50⋅146 – 24⋅292/12) M1 = – 4⋅292 / 0⋅979 or – 0⋅358 / 0⋅0816 = – 4⋅38[4] A1

(iii) Find regression line: y – 3⋅817 = – 4⋅384 (x – 2⋅024) M1 y = 12⋅7 – 4⋅38x A1

(iv) Find correlation coefficient r: r = (88⋅415 – 24⋅29 × 45⋅8/12) / √(50⋅146 – 24⋅292/12) (211⋅16 – 45⋅82/12) M1 = – 4⋅292 / √(0⋅979 × 36⋅36) A1 or – 0⋅358 / √(0⋅0816 × 3⋅03) = – 0⋅719 A1 State valid comment in context (A.E.F.): [Moderate] negative correlation between rainfall and sunshine A1

(v) State both hypotheses: H0: ρ = 0, H1: ρ < 0 B1 Use correct tabular r value: r12, 1% = 0⋅658 B1 Valid method for reaching conclusion: Reject H0 if |r| > tabular value M1 S.R. Calculate t-value: t = r√10 / √(1 – r2) = – 3⋅27 (B1) Use correct tabular t value: t10, 0.99 = 2⋅76[4] (B1) Correct conclusion (needs values correct): There is negative corrln. (A.E.F.) A1

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11 EITHER

Equate tensions at O (A1 for each): 60 (1 + MO)/2 = 20 (2 – MO)/1 M1 A1 A1Simplify to evaluate MO: MO = 10/50 = 0⋅2 A.G. A1Apply Newton’s law at general point: 0⋅1 d2y/dt2 = (lose A1 for each incorrect term) 20 (1⋅8 – y) – 60 (1⋅2 + y)/2 M1 A2Simplify: d2y/dt2 = – 500y A.G. A1State period (A.E.F.): T = 2π/√500 or π /5√5 or 0⋅281 [s] B1

(i) Find speed v when first at 0⋅3 from M: v2 = 500 (0⋅22 – 0⋅12) M1 v = √15 or 3⋅87 [m s-1] A1

(ii) Find time t to reach this point: t = (1/ω) cos-1 (-0⋅1/0⋅2) (A.E.F.) or ¼ T + (1/ω) sin-1 (0⋅1/0⋅2) M1 = (2π/3) /ω or (π/2 + π/6) /ω A1 = 2⋅094/√500 [or = 0⋅07025 + 0⋅02342] = 0⋅0937 [s] A1

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11 OR

Integrate f(t) to find F(t): F(t) = ∫0t λe-λx dx = [– e-λx] 0

t = 1 – e-λt A.G. M1 A1EITHER: Deduce λ directly from mean: λ = 1/20 or 0⋅05 OR: Deduce λ from a tabular value, e.g.: 1 – e-40λ = 0⋅8647, λ = 0⋅05 M1Substitute for λ and put t = 15 to give F(15) to 4 dp: 1 – e-15/20 = 0⋅5276 [or 0⋅5277] A1Calculate expected values to 2 dp (5 values earn A1): 22⋅12 17⋅23 13⋅41 10.45 8.14 6.34 4.93 3⋅85 13⋅53 M1 A2State (at least) null hypothesis: H0: 1 – e-t/20 fits data (A.E.F.) B1Combine two adjacent cells with exp. value < 5: O: . . . 8 6 17 E: . . . 6⋅34 8⋅78 13⋅53 M1Calculate value of χ2 (to 2 dp): χ2 = 3⋅58 M1 A1 (Cells not combined gives 4⋅81 earning M1 A0, max 4/7) Compare with consistent tabular value (to 2 dp): χ7, 0.95

2 = 14⋅07 (cells combined) χ8, 0.95

2 = 15⋅51 (not combined) B1Valid method for reaching conclusion: Reject H0 if χ

2 > tabular value M1Correct conclusion (A.E.F., requires correct values): 3⋅58 < 14⋅07 so suitable model A1

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