9709 s10_ms31

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS

GCE Advanced Subsidiary Level and GCE Advanced Level

MARK SCHEME for the May/June 2010 question paper

for the guidance of teachers

9709 MATHEMATICS

9709/11 Paper 11, maximum raw mark 75

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 May/June 2010 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

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Mark Scheme: Teachers’ version Syllabus Paper

GCE AS/A LEVEL – May/June 2010 9709 11

© 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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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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1 tan x = k

(i) )tan( xπ = −k B1

[1] co. www Mark final answers

(ii)

x

2tan

π

= k

1

B1

[1] co. www

(iii) sin x = 2

1 k

k

+

from 90° triangle.

M1 A1 [2]

Any valid method – 90° triangle or formulae.

2 5

32

x

x

(i) xxx 7202403235+

3 × B1

[3]

co. SC B2 for other 3 terms (i.e. ascending)

(ii)

+

2

21

x

( xxx 7202403235+ )

Coeff of x (1 × 720) + (2 × –240) → 240

M1 A1√

[2]

Looks at exactly 2 terms. co from his answer to (i).

3 9th term = 22, S4 = 49 (i) 228+ da

49)32(2 + da Soln of sim eqns

→ d = 1.5, a = 10 (ii) 46)1(+ dna Substitutes for a and d → n = 25

B1

B1 M1 A1

[4] M1 A1

[2]

co

co Solution of two linear sim eqns. co Correct formula needed and attempt to solve. co.

4 26 xxy

Meets y = 5 when x = 1 or x = 5. B1

co

Integral = 3

3

123 xx M1 A1 attempt to integrate. co.

Their limits (1 to 5) used → 30⅔ Area of rectangle = 20 Shaded area = 10⅔ (integral of 56

2xx B1, M1, A1

DM1 as above, then “−5x” B1√ A1)

DM1 B1√ A1

[6]

value at top limit − value at lower co to his x values co

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5 xxx22

cos3sin2a

(i) xx22 cos3)cos1(2 M1 Uses 1

22+ cs

→ x2

cos52 (a = 2, b = −5) A1 co [2] (ii) Values are −3 and 2 B1√ B1√

[2]

(iii) x2

cos52 = −1

→ cos2 x = 0.6 x = 0.685, 2.46 (accept 0.684)

B1√ B1 B1√

[3]

co √ for π − (first answer) SC B1 for both 39.2 and 140.8

6 x

y

d

d = 63 x (9, 2)

(i) )( 63

2

3

2

3

cxx

y +

(9, 2) 2 = 54 − 54 + c

→ c = 2.

B2,1

M1 A1

[4]

Loses 1 for each error – ignore +c

Uses (9, 2) with integration to find c. co.

(ii) x

y

d

d = 0 → x = 4

2

3

d

d 2

1

2

2x

x

y

→ +ve (or ¾) Minimum

B1

M1 A1

[3]

Ignore any y-value

Any valid method. co.

7 32

182

+xy

(i) A is (3, 0) B1 Anywhere – but not from given answer

x

y

d

d = 2)32(18 +x × 2 B1 B1 B1 for 2)32(18 +x , B1 for ×2

If x = 3, m = 9

4 .

m of normal = − 4

9 M1 Use of m1m2 = −1 with m from dy/dx

Equation of normal )3(4

9xy M1 Correct method for normal

→ 2794 + xy A1 [6]

co (answer was given)

(ii) Normal meets y-axis at (0, 6¾) Curve meets y-axis at (0, −4) → BC = 10¾

M1 A1

[2]

Needs to put x = 0 in both normal and curve. co

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8 (i) y-step ÷ x-step = 2 → m = 1 (ii) Eqn of AC )3(22+ xy Eqn of BC )15(22 xy Sim eqns 42+ xy , 7+xy

→ C (–1, 6) (iii) M is (9, 10) Perp gradient is −½ → ,292 + xy 7+xy

Sim eqns → D (5, 12)

M1 A1

[2] M1 A1√

A1√

A1

[4] B1 M1 M1 A1

[4]

Gradient = y-step ÷ x step used co Correct form of one of lines. √ to his m

√ to his m

co co Use of m1m2 = −1 Solve sim eqns for their BC & perp. bis co

9 (i) 71222

+xx = 11)3(2 2x 3 × B1

[3] B1 for each value – accept if a, b, c not specifically quoted.

(ii) Range of f [ − 11 B1√ [1]

√ to his “c”. allow > or [.

(iii) 71222

+xx < 21

→ 141222

xx or

32)3(2 2<x

M1

3-term quadratic to 0 or 32)3(2 2

<x

→ end-values of 7 or −1 → −1 < x < 7

A1 A1

[3]

Correct end-values co

(iv) gf(x) = kxx ++ )7122(2 2 = 0 Use of b2 − 4ac → 242 − 16(14 + k) → k = 22

M1 A1

M1 A1

[4]

Puts f into g. co.

Used correctly with quadratic co.

10 OA i + 3j + 3k, OC = 3i − j + k.

(i) OB OA + OC = 4i + 2j + 4k B1 co

Unit vector = 6

1 (4i + 2j + 4k) M1 A1√ Divides by the modulus. √ on OB .

[3]

(ii) AC = OC − OA = 2i − 4j − 2k B1 co

AC .OB 8 − 8 − 8 = −8 M1 Use of x1 x2 + y1 y2 + z1 z2

OB = 6; 24AC M1 Correct method for a modulus.

−8 = 6 × 24 × cos θ M1 Connected correctly provided ,OB AC used

θ = 105.8° → 74.2° A1 co (accept acute or obtuse) [5]

(iii) OA = 19 or OC = 11 Perimeter = 2(√ + √) → 15.4

B1 M1 A1

[3]

Used as a length. co (accept 15.3)

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Penalties





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1 (i) )cos3(sin2)cossin2(3 xxxx

→ cscs 6236 → cs 34

→ 43

tan x

M1

A1 [2]

Expanding, collecting, use of t = s ÷ c

Answer given. All correct.

(ii) x = 180 – 36.9 = 143.1° or x = 360 – 36.9 = 323.1°

B1 B1√

[2]

co For 180 + first answer.

2 x

ay

Volume = ∫

x

x

a

d2

2

π = ( )

x

a2

π

M1 B1

For using correct formula with π. For correct integration of x 2 only

Use of limits 1 to 3 → 3

22

aπ

M1 Must be using y2 or πy2.

Equates to 24π → a = 6 A1 [4]

Co, allow ±6.

3 f : 224 xxx a ,

g : 35 +xx a . (i) Turning point at x = 1. Range is Y 2. (ii) gf(x) = 3)24(5

2+xx

= k and use of acb 42

→ k = 13

M1 A1

[2]

B1 M1 A1

[3]

Calculus or completing the square etc. Condone < instead of Y.

For putting f into g. Setting to k, using acb 4

2 co

4 Gradient of L1 is 3

1 .

Equation of L1 is )1(33

1+xy M1 A1 M1 for equation for his m. A1 co.

Gradient of AB is 2

1 . Perp = 2. M1 Use of 121

mm

Equation of L2 is )3(21 xy . A1 co

Sim eqns 103 +xy , 52xy .

→ (5, 5)

M1 A1

[6]

Method of solution co

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5 (i) −8 + 3 + p = 0 → p = 5.

(ii) Vector AB = b − a

= 6i − 2j + (p − 1) k

36 + 4 + (p – 1)² = 49 → p = 4 or p = −2

M1 A1

[2] M1 M1 A1 A1

[4]

Must be scalar. co. Must be b – a or a − b

Must be sum of 3 squares. A1 √ lost. co.

6 (i) 221051 xaax ++

(ii) × (1 − 2x) → xax 25 → a =

5

2

B2,1 [2]

M1 A1

[2]

Loses 1 mark for each incorrect term. Needs to consider exactly 2 terms. co

(iii) Coeff of x² is 21010 aa +

→ −4 + 1.6 = −2.4

M1 A1√ A1

[3]

Needs to consider exactly 2 terms. co

7 (a) a = 100, d = 5, n = 41 → S = 8200

B1 M1 A1

[3]

co Use of correct sum formula. co

(b) (i) 2arara ++ or

r

r

a

1

)1(3

B1 co

= 35 → a = 45 M1 A1 [3]

Solution of equation. co

(ii) r

aS∞

1 = 27

M1 A1√

[2] Correct use of formula. √ for his a.

8 (i) 96242

+ xxh

→ 2

24 x

xh

hxV2 →

224

3x

xV .

(ii) x

V

d

d=

2

324

2x

= 0 when x = 4 → V = 64.

M1

A1

M1

[3]

B1

M1 A1

[3]

Needs to consider at least 5 areas.

co

for hxV2 with h as f(x)

co

Sets differential to 0 and solves. co

(iii) x

x

V3

d

d

2

2

→ Max. M1 A1√

[2] Any valid method. co.

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9 2)2(xy and 72+ xy

Elimination of y → 0322

xx

→ A (−1, 9) and B (3, 1) Area under line = ½ × 4 × 10 or [ ]27 xx from −1 to 3.

M1 DM1A1 M1

y (or x) removed completely. Soln of quadratic. Both points correct. Uses any valid method – integration or area of trapezium etc.

Area under curve =

3

)2(3

x

M1

A1

Any attempt at integration.

Correct integration in either form.

or

+ xx

x

423

2

3

from −1 to 3

→ 10⅔.

[ok to use ( )∫ xyy d 21

– marks the same]

M1

A1 [8]

Correct use of limits in an integral.

co

10 ( ) xxy 4323

6

1

(i) x

y

d

d= 2

6

1 )32(3×× x × 2 − 4

B2,1

B1 [3]

Everything but the “×2”

For the “×2”, even if B0 given above.

(ii) x = 0, y = 6

27 ,

xy 56

27+ → xy 1092 +

B1

M1 A1 [3]

For correct y value

Must be using calculus for m. co. (ok unsimplified)

(iii) 4)32(2

x ( > 0)

→ x = 2½ or ½ → x > 2½, x < ½.

M1

DM1 A1

[3]

Links x

y

d

d with 0

Method for quadratic – lead to 2 answers Correct set of values.

11 f : xx sin34a (i) 2sin34 x → sin x = ⅔ → x = 0.730 or 2.41 (ii)

(iii) k < 1, k > 7.

(iv) A = 2

3π.

(v) sin x = ⅓ – or using inverse g 1(3) = 2.80

M1 A1 A1√

[3] B1 B1

[2] B1 B1

[2] B1

[1] M1A1

[2]

Makes sin x the subject + solution. co. √ for π − first answer. Must be 1 complete oscillation. Shape and position correct, in 1st quadrant, curve not lines. B1 for k = 1, 7, B1 for answer Or B1 for k < 1, B1 for k > 7 co M1 for soln of 3 = 4 − 3sin x or inverse.

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Penalties





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1 (i) a = 12, ar = –6 → r = –½

ar9 =128

3

(ii) S∞ = r

a

1

used → 8

M1

M1 A1

[3]

M1 A1

[2]

Attempt at r from “ar”

ar9 must be correct. co

Correct formula used. M1 needs r < 1

2 (i) 6

2

x

x = x6 – 12x4 + 60x2

(ii) × (1 + x2) → 60 – 12 = 48

B1 ×3 [3]

M1 A1√ [2]

co

Must be exactly 2 terms. √ from his (i).

3 f : xbax cos+a (i) f(0) = 10, a + b = 10

f(2/3π) = 1, 1

2

ba

→ a = 4, b = 6 (ii) Range is –2 to 10.

(iii) 2

3

6

1cos

6

5cos

ππ

→ 334

B1

B1 [2]

B1√ [1]

B1

B1

[2]

EITHER OF THESE

both co √ for his “a – b” to “a + b”

For 330cos2

1° used somewhere.

co

4 (i) 03tansin2 +xx

03cos

sinsin2 +

x

x

x M1 For using tan sin ÷ cos

( )

03cos

cos12

2

+

x

x

M1 For using sin2 + cos2 = 1 and everything correct

→ 02cos3cos22

xx [2] Answer given check.

(ii) 02cos3cos22

xx → xcos –½ or 2 x = 120° or 240°

M1 A1 B1√

[3]

Solution of quadratic.

co. √ for 360 his answer.

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5 23

6

d

d

xx

y

(i) x = 2, tangent has gradient 3

→ normal has gradient 3

1

→ ( )23

111 xy

M1

M1 A1

[3]

Use of mlm2 = –1 with dy/dx

Correct form of line eqn. for normal

(ii) Integrate → 323

6

2

1÷

x

B1 B1

Without the ÷3 For ÷3, even if B0 above

→ cxy +234 through (2,11)

→ 3234 +xy

M1

A1 [4]

Using (2, 11) for c

co

6 OA = i – 2j + 4k, OB = 3i + 2j + 8k,

OC = –i – 2j + 10k (i) (±) 2i + 4j + 4k

(±) 4i + 4j – 2k

16.CBAB

θcos3636.CBAB °6.63θ

(ii) Perimeter 4066 ++ or 28.31sin666 ×°++ → 18.32

B1 B1 M1

M1

M1 A1 [6]

M1

A1 [2]

co co

Needs to be scalar.

For product of 2 moduli and cosine

All correct.

Correct overall method for perimeter.

co

7 (i) 10

6sin

21 θ

Angle DOE = 1.287 radians. (ii) P = 12 + 12 + 2 × 10 × angle BOD

Angle BOD = (π – 1.287)

→ 61.1 (iii) Sector DOE = ½ × 102 × 1.287 Triangle DOE = ½ × 102 × sin 1.287 Area = π × 102 – (2 sectors – 2 triangles) (or 48 + 48 + 2×½×102×(π – 1.287) M1 M1 → 281 or 282

M1

A1 [2]

M1 M1 A1

[3] M1 M1 A1

[3]

Use of trig with/without radians

co answer given. Use of s = rθ for arc length. Correct angle co Correct formula used with radians. Correct formula used with radians. co

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8 (i) Mid-point of AC = (2, 3) Gradient of AC = 1/3 Gradient of BD = –3 Equation y – 3 = –3(x – 2) (ii) If x = 0, y = 9, B (0, 9) Vector move D (4, –3)

(iii) AC = 40

BD = 160 Area = 40 (or by matrix method M2 A1)

B1 M1 A1

[3] B1√ M1 A1

[3]

M1 M1 A1

[3]

Co Use of mlm2 = –1 Co √ on his equation. Valid method. co.

Correct use on either AC or BD, Full and correct method. co

9 x

xy4

+

(i) 54

+

x

x → A (1, 5), B(4, 5)

2

41

d

d

xx

y

= 0 when x = 2, M (2, 4).

B1 B1

M1

DM1 A1 [5]

co. co.

Differentiates.

Setting to 0. co.

(ii) Vol of cylinder = π52.3

Vol under curve = xy d 2∫π

B1 M1

Any valid method. Attempt at integrating y2

Integral = x

x

x

816

3

3

+ A2, 1, 0 Allow if no π present.

Uses his limits “1 to 4” → πππ 185775

DM1 A1

[6]

Using his limits. co.

10 f : 14822

+xxx a (i) y + kx = 12, Sim Eqns. → 2x2 – 8x + kx + 2 = 0 Use of b2 – 4ac

→ (k – 8)2 =16 → k = 12 or 4. (ii) 2x2 – 8x + 14 = 2(x – 2)2 + 6 (iii) Range of f [ 6. (iv) Smallest A = 2 (v) Makes x the subject Order of operations correct.

M1 A1 M1 A1

[4] B1×3

[3] B1√

[1] B1√

[1] M1 M1

Complete elimination of y (or x) Uses b2 – 4ac on eqn = 0, no “x” in a, b, c. co.co √ for c or from calculus. √ to answer to (ii). Could interchange x, y first. Order must be correct.

( ) 22

6g

1+

x

x

A1

[3]

co

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• 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.






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© UCLES 2010




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



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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1 EITHER: State or imply non-modular inequality (2x – 3)2 > 52 , or corresponding equation

or pair of linear equations M1

Obtain critical values –1 and 4 A1

State correct answer x < –1, x > 4 A1

OR: State one critical value, e.g. x = 4, having solved a linear equation (or inequality)

or from a graphical method or by inspection B1

State the other critical value correctly B1

State correct answer x < –1, x > 4 B1 [3]

2 Obtain integral ln(x + 2) B1

Substitute correct limits correctly M1

Use law for the logarithm of a product, a quotient or a power M1

Obtain given answer following full and correct working A1 [4]

3 (i) Use tan(A ± B) formula to obtain an equation in tan x M1

Use tan 45° = 1 and obtain a correct equation in any form A1

Obtain the given equation correctly A1 [3]

(ii) Solve the given quadratic in tan x and evaluate an inverse tangent M1

Obtain a correct answer, e.g. 18.4° A1

Obtain second answer, e.g. 26.6°, and no others in the given interval A1 [3]

[Treat the giving of answers in radians as a misread. Ignore answers outside the given interval.]

4 (i) Commence division by x2 + x – 1 obtaining quotient of the form x + k M1

Obtain quotient x + 2 A1

Obtain remainder 3x + 4 A1

Identify the quotient and remainder correctly A1√ [4]

(ii) Substitute x = –1 and evaluate expression M1

Obtain answer 0 A1 [2]

5 (i) State or imply 2 x = y

1, or 2 x = y 1 B1

Substitute and obtain a 3-term quadratic in y M1

Obtain the given answer correctly A1 [3]

(ii) Solve the given quadratic and carry out correct method for solving an equation of the form

2x = a, where a > 0 M1

Obtain answer x = 1.58 or 1.585 A1

Obtain answer x = 0 B1 [3]

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6 (i) State 2xy + x2 x

y

d

d as derivative of x2y B1

State 2y x

y

d

d as derivative of y2 B1

Equate derivatives of LHS and RHS, and solve for x

y

d

d M1

Obtain given answer A1 [4]

(ii) Substitute and obtain gradient 5

2 , or equivalent B1

Form equation of tangent at the given point (1, 2) M1

Obtain answer 2x – 5y + 8 = 0, or equivalent A1 [3]

[The M1 is dependent on at least one of the B marks being obtained.]

7 (i) Make a recognisable sketch of a relevant graph, e.g. y = 2 – x B1

Sketch an appropriate second graph, e.g. y = e2x, and justify the given statement B1 [2]

(ii) Consider sign of e2x – (2 – x) at x = 0 and x = 0.5, or equivalent M1

Complete the argument correctly with correct calculations A1 [2]

(iii) Show that e2x = 2 – x is equivalent to x = 2

1ln(2 – x), or vice versa B1 [1]

(iv) Use the iterative formula correctly at least once M1

Obtain final answer 0.27 A1

Show sufficient iterations to justify its accuracy to 2 d.p., or show there is a sign change

in the interval (0.265, 0.275) A1 [3]

8 (i) Use quotient rule M1

Obtain correct derivative in any form A1

Obtain given result correctly A1 [3]

(ii) State cot2 x ≡ –1 + cos ec2x , or equivalent B1

Obtain integral –x – cotx (f.t. on signs in the identity) B1√

Substitute correct limits correctly M1

Obtain given answer A1 [4]

(iii) Use trig formulae to convert integrand to xk

2

sin

1 where k = ±2, or ±1 M1

Obtain given answer 2

1 cos ec2x correctly A1

Obtain answer –2

1cot x + c, or equivalent B1 [3]

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Penalties





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1 State or imply y log 2.8 = x log 13 B1

Rearrange into form xy8.2log

13log or equivalent B1

Obtain answer k = 2.49 B1 [3]

2 (i) State or imply correct ordinates 0.27067..., 0.20521..., 0.14936... B1

Use correct formula, or equivalent, correctly with h = 0.5 and three ordinates M1

Obtain answer 0.21 with no errors seen A1 [3]

(ii) Justify statement that the trapezium rule gives an over-estimate B1 [1]

3 EITHER State or imply non-modular inequality (2x –1)2 < (x + 4)2, or corresponding equation


Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations M1


State correct answer –1 < x < 5 A1 [4]

OR Obtain one critical value, e.g. x = 5, by solving a linear equation (or inequality) or

from a graphical method or by inspection B1

Obtain the other critical value similarly B2

State correct answer –1 < x < 5 B1

4 (a) Obtain integral a sin 2x with a = ±

2

1or 2,1 M1

Use limits and obtain 2

1 (AG) A1 [2]

(b) Use tan2 x = sec2 x – 1 and attempt to integrate both terms M1

Obtain 3tan x – 3x A1

Attempt to substitute limits, using exact values M1

Obtain answer 2

32π

A1 [4]

5 (i) Use product rule M1


Show that derivative is equal to zero when x = 3 A1 [3]

(ii) Substitute x = 1 into gradient function, obtaining 2e l or equivalent M1

State or imply required y-coordinate is e 1 B1

Form equation of line through (l, e 1) with gradient found (NOT the normal) M1

Obtain equation in any correct form A1 [4]

6 (i) Make a recognisable sketch of a relevant graph, e.g. y = ln x or y = 2 – x2 B1

Sketch a second relevant graph and justify the given statement B1 [2]

(ii) Consider sign of In x – (2 – x2) at x = 1.3 and x = 1.4, or equivalent M1

Complete the argument correctly with appropriate calculations A1 [2]

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(iii) Show that given equation is equivalent to )ln2( xx or vice versa B1 [1]



Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change

in the interval (1.305, 1.315) B1 [3]

7 (i) Substitute x = 3 and equate to 30 M1

Substitute x = –1 and equate to 18 M1

Obtain a correct equation in any form A1

Solve a relevant pair of equations for a or for b M1

Obtain a = 1 and b = –13 A1 [5]

(ii) Either show that f(2) = 0 or divide by (x – 2), obtaining a remainder of zero B1

Obtain quadratic factor 2x2 + 5x – 3 B1

Obtain linear factor 2x – 1 B1

Obtain linear factor x + 3 B1

[Condone omission of repetition that x – 2 is a factor.]

[If linear factors 2x – 1, x + 3 obtained by remainder theorem or inspection, award B2 + B1.] [4]

8 (i) Use correct sin(A – B) and cos(A – B) formulae M1

Substitute exact values for sin 30° etc. M1

Obtain given answer correctly A1 [3]

(ii) State xx sec2

1sin3 B1

Rearrange to sin 2x = k, where k is a non-zero constant M1

Carry out evaluation of

3

1sin

2

1 1 M1

Obtain answer 17.6° A1

Carry out correct method for second answer M1

Obtain remaining 3 answers from 17.6°, 72.4°, 197.6°, 252.4° and no others in the

range A1 [6]

[Ignore answers outside the given range]

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Penalties





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1 State or imply y log 2.8 = x log 13 B1

Rearrange into form xy8.2log

13log or equivalent B1

Obtain answer k = 2.49 B1 [3]

2 (i) State or imply correct ordinates 0.27067..., 0.20521..., 0.14936... B1

Use correct formula, or equivalent, correctly with h = 0.5 and three ordinates M1

Obtain answer 0.21 with no errors seen A1 [3]

(ii) Justify statement that the trapezium rule gives an over-estimate B1 [1]

3 EITHER State or imply non-modular inequality (2x –1)2 < (x + 4)2, or corresponding equation




State correct answer –1 < x < 5 A1 [4]

OR Obtain one critical value, e.g. x = 5, by solving a linear equation (or inequality) or

from a graphical method or by inspection B1

Obtain the other critical value similarly B2

State correct answer –1 < x < 5 B1

4 (a) Obtain integral a sin 2x with a = ±

2

1or 2,1 M1

Use limits and obtain 2

1 (AG) A1 [2]

(b) Use tan2 x = sec2 x – 1 and attempt to integrate both terms M1

Obtain 3tan x – 3x A1

Attempt to substitute limits, using exact values M1

Obtain answer 2

32π

A1 [4]

5 (i) Use product rule M1


Show that derivative is equal to zero when x = 3 A1 [3]

(ii) Substitute x = 1 into gradient function, obtaining 2e l or equivalent M1

State or imply required y-coordinate is e 1 B1

Form equation of line through (l, e 1) with gradient found (NOT the normal) M1

Obtain equation in any correct form A1 [4]

6 (i) Make a recognisable sketch of a relevant graph, e.g. y = ln x or y = 2 – x2 B1

Sketch a second relevant graph and justify the given statement B1 [2]

(ii) Consider sign of In x – (2 – x2) at x = 1.3 and x = 1.4, or equivalent M1

Complete the argument correctly with appropriate calculations A1 [2]

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(iii) Show that given equation is equivalent to )ln2( xx or vice versa B1 [1]



Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change

in the interval (1.305, 1.315) B1 [3]

7 (i) Substitute x = 3 and equate to 30 M1

Substitute x = –1 and equate to 18 M1

Obtain a correct equation in any form A1

Solve a relevant pair of equations for a or for b M1

Obtain a = 1 and b = –13 A1 [5]

(ii) Either show that f(2) = 0 or divide by (x – 2), obtaining a remainder of zero B1

Obtain quadratic factor 2x2 + 5x – 3 B1

Obtain linear factor 2x – 1 B1

Obtain linear factor x + 3 B1

[Condone omission of repetition that x – 2 is a factor.]

[If linear factors 2x – 1, x + 3 obtained by remainder theorem or inspection, award B2 + B1.] [4]

8 (i) Use correct sin(A – B) and cos(A – B) formulae M1

Substitute exact values for sin 30° etc. M1

Obtain given answer correctly A1 [3]

(ii) State xx sec2

1sin3 B1

Rearrange to sin 2x = k, where k is a non-zero constant M1

Carry out evaluation of

3

1sin

2

1 1 M1


Carry out correct method for second answer M1

Obtain remaining 3 answers from 17.6°, 72.4°, 197.6°, 252.4° and no others in the

range A1 [6]

[Ignore answers outside the given range]

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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



Penalties


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.



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1 EITHER: State or imply non-modular inequality 22 ))2(2()3( axax >+ , or corresponding

quadratic equation, or pair of linear equations )2(2 )3( axax ±+ B1

Make reasonable solution attempt at a 3-term quadratic, or solve two linear

equations M1

Obtain critical values ax3

1 and x = 7a A1

State answer axa 73

1<< A1

OR: Obtain the critical value x = 7a from a graphical method, or by inspection, or by

solving a linear equation or inequality B1

Obtain the critical value ax3

1 similarly B2

State answer axa 73

1<< B1 [4]

[Do not condone Y for <; accept 0.33 for 3

1 .]

2 Use correct cos 2A formula and obtain an equation in sin θ M1

Obtain 03sinsin42

+ θθ , or equivalent A1

Make reasonable attempt to solve a 3-term quadratic in sin θ M1


Obtain answer 131.4° and no others in the given range A1 √

Obtain answer 270° and no others in the given range A1 [6]

[Treat the giving of answers in radians as a misread. Ignore answers outside the given range.]

3 (i) EITHER: State or imply Cyxn lnlnln + B1

Substitute x- and y-values and solve for n M1

Obtain n = 1.50 A1

Solve for C M1

Obtain C = 6.00 A1

OR: Obtain two correct equations by substituting x- and y-values in Cyxn B1

Solve for n M1

Obtain n = 1.50 A1

Solve for C M1

Obtain C = 6.00 A1 [5]

(ii) State that the graph of ln y against ln x has equation nln x + ln y = ln C which is

linear in ln y and ln x, or has equation of the form nX + Y = ln C, where X = ln x and

Y = ln y, and is thus a straight line B1 [1]

4 (i) State correct expansion of cos(3x – x) or cos(3x + x) B1

Substitute expansions in )4cos2(cos2

1xx , or equivalent M1

Simplify and obtain the given identity correctly A1 [3]

(ii) Obtain integral xx 4sin2sin8

1

4

1 B1

Substitute limits correctly in an integral of the form xbxa 4sin2sin + M1

Obtain given answer following full, correct and exact working A1 [3]

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5 Separate variables correctly B1

Integrate and obtain term ln x B1

Integrate and obtain term )4ln( 2

2

1+y B1

Evaluate a constant or use limits y = 0, x = 1 in a solution containing aln x and bln(y2 + 4) M1

Obtain correct solution in any form, e.g. 4lnln)4ln(2

12

2

1++ xy A1

Rearrange as )1(4 22xy , or equivalent A1 [6]

6 (i) Using the formulae θ2

2

1r and θsin

2

2

1r , or equivalent, form an equation M1

Obtain a correct equation in r and x and/or x/2 in any form A1

Obtain the given equation correctly A1 [3]

(ii) Consider the sign of )sin(4

3xx π at x = 1.3 and x = 1.5, or equivalent M1

Complete the argument with correct calculations A1 [2]

(iii) Use the iterative formula correctly at least once M1


Show sufficient iterations to at least 4 d.p. to justify its accuracy to 2 d.p., or show

there is a sign change in the interval (1.375, 1.385) A1 [3]

7 (i) Obtain modulus 8 B1

Obtain argument π4

1 or 45° B1 [2]

(ii) Show 1, i and u in relatively correct positions on an Argand diagram B1

Show the perpendicular bisector of the line joining 1 and i B1

Show a circle with centre u and radius 1 B1

Shade the correct region B1 [4]

(iii) State or imply relevance of the appropriate tangent from O to the circle B1 √

Carry out complete strategy for finding z for the critical point M1

Obtain answer 7 A1 [3]

8 (i) State or imply the form 31 +

+

+ x

B

x

Aand use a relevant method to find A or B M1

Obtain A = 1, B = 1 A1 [2]

(ii) Square the result of part (i) and substitute the fractions of part (i) M1


(iii) Integrate and obtain ( )3

1)3ln(1ln

1

1

+

+++

+ x

xx

x

B3

Substitute limits correctly in an integral containing at least two terms of the correct

form M1

Obtain given answer following full and exact working A1 [5]

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9 (i) Use quotient or product rule to differentiate (1 – x)/(1 + x) M1


Use chain rule to find x

y

d

d M1

Obtain a correct expression in any form A1

Obtain the gradient of the normal in the given form correctly A1 [5]

(ii) Use product rule M1


Equate derivative to zero and solve for x M1

Obtain x = 2

1 A1 [4]

10 (i) Express general point of l or m in component form, e.g. (1 + s, 1 – s, 1 + 2s) or

(4 + 2t, 6 + 2t, 1 + t) B1

Equate at least two corresponding pairs of components and solve for s or t M1

Obtain s = 1 or t = 2 A1

Verify that all three component equations are satisfied A1 [4]

(ii) Carry out correct process for evaluating the scalar product of the direction vectors of

l and m M1

Using the correct process for the moduli, divide the scalar product by the product of

the moduli and evaluate the inverse cosine of the result M1

Obtain answer 74.2° (or 1.30 radians) A1 [3]

(iii) EITHER: Use scalar product to obtain a – b + 2c = 0 and 2a + 2b + c = 0 B1

Solve and obtain one ratio, e.g. a : b M1

Obtain a : b : c = 5 : 3 : 4, or equivalent A1

Substitute coordinates of a relevant point and values for a, b and c in

general equation of plane and evaluate d M1

Obtain answer 5x – 3y – 4z = 2, or equivalent A1

OR 1: Using two points on l and one on m, or vice versa, state three equations in

a, b, c and d B1

Solve and obtain one ratio, e.g. a : b M1

Obtain a ratio of three of the unknowns, e.g. a : b : c = 5 : 3 : 4 A1

Use coordinates of a relevant point and found ratio to find the fourth

unknown, e.g. d M1

Obtain answer –5x + 3y + 4z = 2, or equivalent A1

OR 2: Form a correct 2-parameter equation for the plane,

e.g. r = i + j + k + λ(i – j + 2k) + µ(2i + 2j + k) B1

State three equations in x, y, z, λ and µ M1

State three correct equations A1

Eliminate λ and µ M1

Obtain answer 5x – 3y – 4z = 2, or equivalent A1

OR 3: Attempt to calculate vector product of direction vectors of l and m M1

Obtain two correct components of the product A1

Obtain correct product, e.g. –5i + 3j + 4k A1

Form a plane equation and use coordinates of a relevant point to

calculate d M1

Obtain answer –5x + 3y + 4z = 2, or equivalent A1 [5]

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1 EITHER: Attempt to solve for x

2 M1

Obtain 2x = 6/4, or equivalent A1

Use correct method for solving an equation of the form 2x = a, where a > 0 M1 Obtain answer x = 0.585 A1

OR: State an appropriate iterative formula, e.g. xn + 1 = ln((2 nx + 6) / 5) / ln 2 B1

Use the iterative formula correctly at least once M1 Obtain answer x = 0.585 A1 Show that the equation has no other root but 0.585 A1 [4] [For the solution 0.585 with no relevant working, award B1 and a further B1 if 0.585 is shown to

be the only root.]

2 Integrate by parts and reach ∫±± xxxxx dcos2cos2

M1

Obtain ∫+ xxxxx dcos2cos2

, or equivalent A1

Complete the integration, obtaining xxxxx cos2sin2cos2

++ , or equivalent A1

Substitute limits correctly, having integrated twice M1 Obtain the given answer correctly A1 [5] 3 (i) State or imply sin a = 4/5 B1 Use sin(A – B) formula and substitute for cos a and sin a M1

Obtain answer )334(10

1, or exact eqivalent A1 [3]

(ii) Use tan 2A formula and substitute for tan a, or use sin 2A and cos 2A formulae, substitute sin a and cos a, and divide M1

Obtain 7

242tan a , or equivalent A1

Use tan(A + B) formula with A = 2a , B = a and substitute for tan 2a and tan a M1

Obtain 117

443tan a A1 [4]

4 (i) Use correct quotient or product rule M1 Obtain correct derivative in any form A1 Equate derivative to zero and solve for x M1


(ii) Use the iterative formula correctly at least once M1 Obtain final answer 4.49 A1

Show sufficient iterations to at least 4 d.p. to justify its accuracy to 2 d.p., or show that there is a sign change in the interval (4.485, 4.495) A1 [3] 5 (i) Substitute

2

1x , equate to zero and obtain a correct equation, e.g.

02

1

4

5

4

1++ ba B1

Substitute x = 2 and equate to 9 M1 Obtain a correct equation, e.g. 922016 ++ ba A1 Solve for a or for b M1 Obtain a = 4 and b = 3 A1 [5]

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(ii) Attempt division by 2x + 1 reaching a partial quotient of kxx +2 M1

Obtain quadratic factor 322+ xx A1

Obtain factorisation )1)(3)(12( ++ xxx A1 [3] [The M1 is earned if inspection has an unknown factor of fexx ++

2 and an equation in e and/or

f, or if two coefficients with the correct moduli are stated without working.] [If linear factors are found by the factor theorem, give B1 + B1 for (x 1) and (x + 3), and then B1

for the complete factorisation.]

6 (i) EITHER: State or imply x

y

y d

d1 as derivative of ln y B1

State correct derivative of LHS, e.g. x

y

y

xy

d

dln + B1

Differentiate RHS and obtain an expression for x

y

d

d M1

Obtain given answer A1

OR 1: State x

xy

12ln

+, or equivalent, and differentiate both sides M1

State correct derivative of LHS, e.g. x

y

y d

d1 B1

State correct derivative of RHS, e.g. 2/1 x B1

Rearrange and obtain given answer A1 OR 2: State )/12exp( xy + , or equivalent, and attempt differentiation by chain rule M1

State correct derivative of RHS, e.g. 2/)/12exp( xx+ B1 + B1 Obtain given answer A1 [4] [The B marks are for the exponential term and its multiplier.] (ii) State or imply

2

1x when y = 1 B1

Substitute and obtain gradient of 4 B1√ Correctly form equation of tangent M1 Obtain final answer y + 4x + 1 = 0, or equivalent A1 [4] 7 (i) Separate variables correctly and attempt integration of both sides B1 Obtain term tan x B1 Obtain term t2

2

1e B1

Evaluate a constant or use limits x = 0, t = 0 in a solution containing terms a tan x and be 2t M1 Obtain correct solution in any form, e.g. t

x2

2

1

2

1etan A1

Rearrange as )e(tan 2

2

1

2

11 tx , or equivalent A1 [6]

(ii) State that x approaches )(tan

2

11 B1 [1]

(iii) State that t2

e1 increases and so does the inverse tangent, or state that xt 22cose is

positive B1 [1]

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8 (i) EITHER: State a correct expression for zor 2

z , e.g. 22 )2(sin)2cos1( θθ ++ B1

Use double angle formulae throughout or Pythagoras M1 Obtain given answer 2cos θ correctly A1 State a correct expression for tangent of argument, e.g. )2cos1/(2(sin θθ + B1

Use double angle formulae to express it in terms of cos θ and sin θ M1 Obtain tan θ and state that the argument is θ A1 OR: Use double angle formulae to express z in terms of cos θ and sin θ M1 Obtain a correct expression, e.g. θθθθ cossin2sincos1

22i++ A1

Convert the expression to polar form M1 Obtain )sin(coscos2 θθθ i+ A1

State that the modulus is 2 cosθ A1 State that the argument is θ A1 [6] (ii) Substitute for z and multiply numerator and denominator by the conjugate of z, or equivalent M1 Obtain correct real denominator in any form A1 Identify and obtain real part equal to

2

1 A1 [3]

9 (i) State or imply a correct normal vector to either plane, e.g. 3i + 2j + 4k or ai + j + k B1 Equate scalar product of normals to zero and obtain an equation in a, e.g. 3a + 2 + 4 = 0 M1 Obtain a = 2 A1 [3] (ii) Express general point of the line in component form, e.g. (λ , 1 + 2λ , 1 + 2λ) B1 Either substitute components in the equation of p and solve for λ , or substitute components and the value of a in the equation of q and solve for λ M1* Obtain λ = 1 for point A A1 Obtain λ = 2 for point B A1 Carry out correct process for finding the length of AB M1(dep*) Obtain answer AB = 3 A1 [6] [The second M mark is dependent on both values of λ being found by correct methods.]

10 (i) EITHER: Divide by denominator and obtain quadratic remainder M1 Obtain A = 1 A1 Use any relevant method to obtain B, C or D M1 Obtain one correct answer A1 Obtain B = 2, C = 1 and D = 3 A1 OR: Reduce RHS to a single fraction and equate numerators, or equivalent M1 Obtain A = 1 A1 Use any relevant method to obtain B, C or D M1

Obtain one correct answer A1 Obtain B = 2, C = 1 and D = 3 A1 [5] [SR: If A = 1 stated without working give B1.]

(ii) Integrate and obtain )12ln(2

31ln2+ x

x

xx , or equivalent B3√

(The f.t. is on A, B, C, D. Give B2√ if only one error in integration; B1√ if two.) Substitute limits correctly in the complete integral M1 Obtain given answer correctly following full and exact working A1 [5]

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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.


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

implied).


• 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.


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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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


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.



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1 EITHER: State or imply non-modular inequality (x – 3)2 > (2(x + 1))2 , or corresponding quadratic

equation, or pair of linear equations (x – 3) = ± 2(x + 1) B1


Obtain critical values 5 and 3

1 A1

State answer 31

5 << x A1

OR: Obtain the critical value x = 5 from a graphical method, or by inspection,

or by solving a linear equation or inequality B1

Obtain the critical value 3

1x similarly B2

State answer 31

5 << x B1 [4]

[Do not condone ≤ for <; accept 0.33 for 3

1 .]

2 (i) State or imply xAy 2lnln3 + at any stage B1

State gradient is 3

2, or equivalent B1 [2]

(ii) Substitute x = 0, ln y = 0.5 and solve for A M1

Obtain A = 4.48 A1 [2]

3 Attempt to use tan(A ± B) formula and obtain an equation in tan x M1

Obtain 3-term quadratic 2 tan2 x + 3 tan x – 1 = 0, or equivalent A1

Solve a 3-term quadratic and find a numerical value of x M1


Obtain answer 119.3° and no others in the given interval A1 [5]

[Ignore answers outside the given interval. Treat answers in radians, 0.274 and 2.08, as a misread.]

4 Separate variables correctly B1

Obtain term k ln(4 – x2), or terms k1 ln(2 – x) + k2 ln(2 + x) B1

Obtain term –2 ln(4 – x2), or –2 ln(2 – x) –2 ln(2 + x), or equivalent B1

Obtain term t, or equivalent B1

Evaluate a constant or use limits x = 1, t = 0 in a solution containing terms a ln(4 – x2) and bt

or terms c ln(2 – x), d ln(2 + x) and bt M1

Obtain correct solution in any form, e.g. –2 ln(4 – x2) = t – 2 ln3 A1

Rearrange and obtain )(3exp42

12

tx , or equivalent (allow use of 2 ln 3 = 2.20) A1 [7]

5 (i) State derivative xx 2

e)2(e , or equivalent B1 + B1

Equate derivative to zero and solve for x M1

Obtain p = ln 2, or exact equivalent A1 [4]

(ii) State indefinite integral xx 2

2

1e)(e , or equivalent B1 + B1

Substitute limits x = 0 and x = p correctly M1

Obtain given answer following full and correct working A1 [4]

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6 (i) Use correct quotient or product rule M1

Obtain correct derivative in any form, e.g. 2

)1(

ln

)1(

1

++x

x

xx

A1

Equate derivative to zero and obtain the given equation correctly A1

Consider the sign of x

xx

ln

)1( +at x = 3 and x = 4, or equivalent M1

Complete the argument with correct calculated values A1 [5]

(ii) Use the iterative formula correctly at least once, using or reaching a value in the interval (3, 4) M1


Show sufficient iterations to at least 4 d.p. to justify its accuracy to 2 d.p.,

or show there is a sign change in the interval (3.585, 3.595) A1 [3]

7 (i) Use correct cos(A + B) formula to express cos 3θ in terms of trig functions of 2θ and θ M1

Use correct trig formulae and Pythagoras to express cos 3θ in terms of cosθ M1

Obtain a correct expression in terms of cosθ in any form A1

Obtain the given identity correctly A1 [4]

[SR: Give M1 for using correct formulae to express RHS in terms of cos θ and cos 2θ ,

then M1A1 for expressing in terms of either only cos 3θ and cos θ , or only cos 2θ , sin 2θ ,

cos θ , and sin θ , and A1 for obtaining the given identity correctly.]

(ii) Use identity and integrate, obtaining terms )3sin3

1(

4

1θ and )sin3(

4

1θ , or equivalent B1 + B1

Use limits correctly in an integral of the form ksin 3θ + lsin θ M1

Obtain answer 38

3

3

2, or any exact equivalent A1 [4]

8 (a) EITHER: Substitute 3i1+ , attempt complete expansions of the x3 and x2 terms M1

Use i2 = –1 correctly at least once B1

Complete the verification correctly A1

State that the other root is 3i1 B1

OR1: State that the other root is 3i1 B1

State quadratic factor 422

+xx B1

Divide cubic by 3-term quadratic reaching partial quotient 2x + k M1

Complete the division obtaining zero remainder A1

OR2: State factorisation )42)(32(

2

++ xxx , or equivalent B1

Make reasonable solution attempt at a 3-term quadratic and use i2 = –1 M1

Obtain the root 3i1+ A1

State that the other root is 3i1 B1 [4]

(b) Show point representing 3i1+ in relatively correct position on an Argand diagram B1

Show circle with centre at 3i1+ and radius 1 B1√

Show line for arg z = π3

1 making π3

1 with the real axis B1

Show line from origin passing through centre of circle, or the diameter which would contain

the origin if produced B1

Shade the relevant region B1√ [5]

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9 (i) State or imply partial fractions of the form 2

)2(221 x

C

x

B

x

A

+

+

+

+ B1

Use any relevant method to determine a constant M1

Obtain one of the values A = 1, B = 1, C = 2 A1

Obtain a second value A1

Obtain the third value A1 [5]

[The form 2

)2(21x

EDx

x

A

+

++ , where A = 1, D = 1, E = 0, is acceptable

scoring B1M1A1A1A1 as above.]

(ii) Use correct method to obtain the first two terms of the expansion of 1

)21( x ,1

)2( + x ,

2

)2( + x , 1

2

1 )1( + x , or 2

2

1 )1( + x M1

Obtain correct unsimplified expansions up to the term in x2 of each partial fraction A1√ + A1√ + A1√

Obtain answer 2

4

15

4

9

1 xx ++ , or equivalent A1 [5]

[Symbolic binomial coefficients, e.g.

1

1

, are not sufficient for the M1. The f.t. is on A, B, C.]

[For the A, D, E form of partial fractions, give M1A1√A1√ for the expansions then, if D ≠ 0, M1 for

multiplying out fully and A1 for the final answer.]

[In the case of an attempt to expand212

)2()21)(54( ++ xxxx , give M1A1A1 for the expansions,

M1 for multiplying out fully, and A1 for the final answer.]

[SR: If B or C omitted from the form of fractions, give B0M1A0A0A0 in (i); M1A1√A1√ in (ii).]

[SR: If D or E omitted from the form of fractions, give B0M1A0A0A0 in (i); M1A1√A1√ in (ii).]

10 (i) Express general point of the line in component form, e.g. (2 + λ , 1 + 2λ , 4 + 2λ) B1

Substitute in plane equation and solve for λ M1

Obtain position vector 4i + 3j, or equivalent A1 [3]

(ii) State or imply a correct vector normal to the plane, e.g. 3i – j + 2k B1

Using the correct process, evaluate the scalar product of a direction vector for l and a normal for p M1

Using the correct process for the moduli, divide the scalar product by the product of the moduli

and evaluate the inverse cosine or inverse sine of the result M1

Obtain answer 26.5° (or 0.462 radians) A1 [4]

(iii) EITHER: State a + 2b + 2c = 0 or 3a – b + 2c = 0 B1

Obtain two relevant equations and solve for one ratio, e.g. a : b M1

Obtain a : b : c = 6 : 4 : 7, or equivalent A1

Substitute coordinates of a relevant point in 6x + 4y – 7z = d and evaluate d M1

Obtain answer 6x + 4y – 7z = 36, or equivalent A1

OR1: Attempt to calculate vector product of relevant vectors,

e.g. (i + 2j + 2k) × (3i – j + 2k) M1

Obtain two correct components of the product A1

Obtain correct product, e.g. 6i + 4j – 7k A1

Substitute coordinates of a relevant point in 6x + 4y – 7z = d and evaluate d M1

Obtain answer 6x + 4y – 7z = 36, or equivalent A1

OR2: Attempt to form 2-parameter equation with relevant vectors M1

State a correct equation, e.g. r = 2i – j – 4k + λ(i + 2j + 2k) +µ(3i j + 2k) A1

State three equations in x, y, z, λ, µ A1

Eliminate λ and µ M1

Obtain answer 6x + 4y – 7z = 36, or equivalent A1 [5]

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1 DF = 35000/16 B1

M1 For using Newton’s second law

DF – 1150g sin1.2o – 975 = 1150a A1

Acceleration is 0.845 ms-2 A1

[4]

2 (i) Acceleration is 0.09 ms 2 B1

[1]

(ii) [D = ½ (8 + 4)0.18 or

D = (0 + ½ 0.09 × 22) + (0.18 × 4 + ½ 0 × 42)

+ (0.18 × 2 – ½ 0.09 × 22)]

M1 For using the idea that area represents

distance or for repeated use of

s = ut + ½ at2

Distance is 1.08 m A1

[2]

(iii) [½ 3V = 1.08] M1 For using area of triangle

= area of trapezium

Greatest speed is 0.72 ms 1 A1

[2]

SR (max 1 out of 2) for candidates who

assume (implicitly) that speed is greatest

at a specific time

(t = 11 or t = 9.5) 0.72 ms 1 B1

from ½ (0 + V) × 3 = 1.08 or

from ½ (0 + V) × 1.5 = ½ 1.08

3 (i) [R + 7sin45o = 0.8g] M1 For resolving forces vertically (needs 3

terms)

Normal component is 3.05 N A1

[2]

AG

(ii) F = 7cos45o B1

M1 For using µ = F/3.05

Coefficient is 1.62 A1

[3]

4 M1 For resolving forces in the x-direction or

in the y-direction

X = 160 + 250cosα A1

Y = 370 – 250sinα A1

M1 For using R2 = X2 + Y2

Magnitude is 500 N A1ft ft 264 N for consistent sin/cos mix

M1 For using tan θ = Y/X

Required angle is 36.9o (or 0.644 rads) A1ft

[7]

ft 29.5o for consistent sin/cos mix

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Alternative for 4 M1 For finding the resultant in magnitude and

direction of two forces and obtaining a

triangle enabling the calculation of the

resultant of the three forces

Triangle has sides 403, 250 and R A1 or equivalent for different choice of two

forces*

Triangle has angle opposite R equal to 97.1o A1 As *

[R2 = 4032 + 2502 – 2 × 403 × 250cos97.1o] M1 For using cosine rule to find R

Magnitude is 500 N A1

[sin(66.6o – z) ÷ 250 = sin97.1o ÷ R] M1 For using sine rule to find z

Required angle is 36.9o A1

5 (i) M1 For using KE loss = PE gain or

02 = u2 – 2(g sinα)(0.45/sinα)

½ (m)u2 = (m)g(0.45) A1

Speed is 3 ms 1 A1

[3]

(ii) [PE gain = ½ 0.3 × 32 – 0.39] M1 For using PE gain = KE lost – WD

PE gain is 0.96 J A1ft ft incorrect u

[0.3gh = 0.96] DM1 For using PE = mgh; dependent on the

given WD being reflected in the value for

PE used

R is 0.32 m higher than the level of P A1

[4]

6 (i) M1 For applying Newton’s second law to A or

to B or using (M + m)a = Mg – F

0.45a = 0.45g – T and 0.2a = T – F or

(0.45 + 0.2)a = 0.45g – F

A1

F = 0.3 × 0.2g B1

M1 For substituting for F and solving for a

Acceleration is 6 ms 2 A1

[v2 = 2 × 6 × [2 – (2.8 – 2.1)] M1 For using v2 = (02) + 2as

(s must be less than 2)

Speed is 3.95 ms 1 A1

[7]

AG

(ii) 0.2a2 = –0.06g B1ft ft incorrect F

M1 For using v2 = 3.952 +

2a2[2.1 – distance moved by B]

v2 = 15.6 + 2(–3)(0.8) A1


[4]

Alternative for 6(ii)

WD against friction = 0.06g × [2.1 – (2 – 0.7)] B1

M1 For using KE loss = WD against friction

½ 0.2 × 3.952 – ½ 0.2v2 = 0.48 A1


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7 (i) M1 For integrating v1 to find s1

225dtv

15

0 1∫ �

A[(152/2 – 0.05 × 153/3) – (0 – 0)] = 225

A1

A = 4 A1

[4(15 – 0.05 × 152) = B/152] M1 For using v1(15) = v2(15)

B = 3375 A1

[5]

AG

(ii) s2(t) = Bt 1/(–1) (+ C) B1

[–3375/15 + C = 225] M1 For using s2(15) = 225 to find C

Distance travelled is [450 – 3375/t] m

(for t [ 15)

A1

[3]

(iii) [450 – 3375/t = 315] M1 For attempting to solve s2(t) = 315

[v = 3375/252] M1 For substituting into v = 3375/t2


[3]


s = )

15

1

t

1(3375dt3375

15

2−−=∫

t

t

= 225 – 3375/t

B1

Distance travelled = 225 + (225 – 3375/t) M1


(for t [ 15)

A1

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1 DF = 35000/16 B1

M1 For using Newton’s second law

DF – 1150g sin1.2o – 975 = 1150a A1

Acceleration is 0.845 ms-2 A1

[4]

2 (i) Acceleration is 0.09 ms 2 B1

[1]

(ii) [D = ½ (8 + 4)0.18 or

D = (0 + ½ 0.09 × 22) + (0.18 × 4 + ½ 0 × 42)

+ (0.18 × 2 – ½ 0.09 × 22)]

M1 For using the idea that area represents

distance or for repeated use of

s = ut + ½ at2

Distance is 1.08 m A1

[2]

(iii) [½ 3V = 1.08] M1 For using area of triangle

= area of trapezium

Greatest speed is 0.72 ms 1 A1

[2]

SR (max 1 out of 2) for candidates who

assume (implicitly) that speed is greatest

at a specific time

(t = 11 or t = 9.5) 0.72 ms 1 B1

from ½ (0 + V) × 3 = 1.08 or

from ½ (0 + V) × 1.5 = ½ 1.08

3 (i) [R + 7sin45o = 0.8g] M1 For resolving forces vertically (needs 3

terms)

Normal component is 3.05 N A1

[2]

AG

(ii) F = 7cos45o B1

M1 For using µ = F/3.05

Coefficient is 1.62 A1

[3]

4 M1 For resolving forces in the x-direction or

in the y-direction

X = 160 + 250cosα A1

Y = 370 – 250sinα A1

M1 For using R2 = X2 + Y2

Magnitude is 500 N A1ft ft 264 N for consistent sin/cos mix

M1 For using tan θ = Y/X

Required angle is 36.9o (or 0.644 rads) A1ft

[7]

ft 29.5o for consistent sin/cos mix

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Alternative for 4 M1 For finding the resultant in magnitude and

direction of two forces and obtaining a

triangle enabling the calculation of the

resultant of the three forces

Triangle has sides 403, 250 and R A1 or equivalent for different choice of two

forces*

Triangle has angle opposite R equal to 97.1o A1 As *

[R2 = 4032 + 2502 – 2 × 403 × 250cos97.1o] M1 For using cosine rule to find R

Magnitude is 500 N A1

[sin(66.6o – z) ÷ 250 = sin97.1o ÷ R] M1 For using sine rule to find z

Required angle is 36.9o A1

5 (i) M1 For using KE loss = PE gain or

02 = u2 – 2(g sinα)(0.45/sinα)

½ (m)u2 = (m)g(0.45) A1

Speed is 3 ms 1 A1

[3]

(ii) [PE gain = ½ 0.3 × 32 – 0.39] M1 For using PE gain = KE lost – WD

PE gain is 0.96 J A1ft ft incorrect u

[0.3gh = 0.96] DM1 For using PE = mgh; dependent on the

given WD being reflected in the value for

PE used

R is 0.32 m higher than the level of P A1

[4]

6 (i) M1 For applying Newton’s second law to A or

to B or using (M + m)a = Mg – F

0.45a = 0.45g – T and 0.2a = T – F or

(0.45 + 0.2)a = 0.45g – F

A1

F = 0.3 × 0.2g B1

M1 For substituting for F and solving for a

Acceleration is 6 ms 2 A1

[v2 = 2 × 6 × [2 – (2.8 – 2.1)] M1 For using v2 = (02) + 2as

(s must be less than 2)


[7]

AG

(ii) 0.2a2 = –0.06g B1ft ft incorrect F

M1 For using v2 = 3.952 +

2a2[2.1 – distance moved by B]

v2 = 15.6 + 2(–3)(0.8) A1


[4]


WD against friction = 0.06g × [2.1 – (2 – 0.7)] B1

M1 For using KE loss = WD against friction

½ 0.2 × 3.952 – ½ 0.2v2 = 0.48 A1


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7 (i) M1 For integrating v1 to find s1

225dtv

15

0 1∫ �

A[(152/2 – 0.05 × 153/3) – (0 – 0)] = 225

A1

A = 4 A1

[4(15 – 0.05 × 152) = B/152] M1 For using v1(15) = v2(15)

B = 3375 A1

[5]

AG

(ii) s2(t) = Bt 1/(–1) (+ C) B1

[–3375/15 + C = 225] M1 For using s2(15) = 225 to find C


(for t [ 15)

A1

[3]

(iii) [450 – 3375/t = 315] M1 For attempting to solve s2(t) = 315

[v = 3375/252] M1 For substituting into v = 3375/t2


[3]


s = )

15

1

t

1(3375dt3375

15

2−−=∫

t

t

= 225 – 3375/t

B1

Distance travelled = 225 + (225 – 3375/t) M1


(for t [ 15)

A1

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1 [7.3 sinα = 5.5]

α = 48.9

[R = 6.8 – 7.3 cos48.9°]

Magnitude of resultant is 2 N

M1

A1

M1

A1

[4]

For using Ry = 0

For using R = Rx

2 [1.2 – 0.24t = 0.6]

t = 2.5

[s = 0.6t2 – 0.04t3]

s = (0.6 × 2.52 – 0.04 × 2.53) – (0 – 0)

Displacement is 3.125 m

M1

A1

M1

DM1

A1

[5]

For using a = dv/dt and attempting to

solve a = 0.6

For using ∫ vdts

For using limits 0 to 2.5 or equivalent

(dependent on integration)

Accept 3.12 or 3.13

3 (i) [WD = 25 × 40 cos30°]

Work done is 866 J

M1

A1

[2]

For using WD = Fdcosθ

(ii) [50 × 40 cos30° = 866 + KE gain]

KE gain is 866 J

½ 35(v2 – 1.22) = 866

Speed is 7.14 ms-1

M1

A1ft

M1

A1ft

A1

[5]

For using WD by P = WD against

resistance + KE gain

ft incorrect ans (i)

For using KE gain = ½ m(v2 – u2)

ft incorrect KE

SR (max 2/3 for the last three marks) for

using Newton’s second law and constant

acceleration formula

50 cos30° – 25 cos30° = 35a and

v2 = 1.22 + 2 × 40a M1

� speed is 7.14 ms-1 A1

4 (i) 0.36g sin60° – T = 0.36 × 0.25

Tension is 3.03 N

B1

B1

[2]

AG

(ii)

T ± F – 0.24g sin60° = 0.24 × 0.25

F = 3.03 – 0.24g sin60° – 0.24 × 0.25

(F = 0.889)

R = 0.24g cos60° (R = 1.2)

Coefficient is 0.74

M1

A1

A1

B1

M1

A1

[6]

For applying Newton’s second law to B.

For using µ = F/R

5 (i)

[s = ½ (1.4 + 1.1) × 1.2; 1.1 = 1.4 + (–d) × 1.2]

AB = 1.5 m or d = 0.25

d = 0.25 or AB = 1.5 m

M1

A1

B1ft

[3]

For using s = ½ (u + v)t to find AB

or v = u + at to find d

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(ii) [0 = u2 + 2(–0.25)2;

2 = 0 – ½ (–0.25)t2 ]

Speed is 1 ms-1 or time is 4 s

Time is 4 s or speed is 1 ms-1

M1

A1

B1ft

[3]

For using 0 = u2 + 2(–d)s to find u or

s = 0 – ½ (–d)t2 to find t

(iii) For line joining (0, 1.4) and (1.2, 1.1)

For line joining (1.2, –1) and (5.2, 0)

B1

B1ft

[2]

ft wrong answer(s) in (ii)

SR (max 1/2)

For two correct lines and values missing

B1ft

6 (i) [2a = 3.5]

Acceleration is 1.75 ms-2

[1.75 = gsin α] or

[0.5 × 3.52 = gh; s = 0.5 × 3.5 × 2 and

sinα = h/s]

Angle is 10.1o or 0.176c

M1

A1

M1

A1

[4]

For using v = 0 + at

For using a = gsin α

or for using ½mv2 = mgh, s = ½vt and

sinα = h/s

(ii) [sP = ½ a22 + {(a2)t + ½ at2}]

or [sP = ½ a (t + 2)2]

[sP – sQ = ½ a22 + (a2)t + ½ at2 – ½ at2]

2 × 1.75 + 2 × 1.75t

[4.9 = 2a + 2at]

t = 0.4

M1

M1

A1

M1

A1

[5]

For constructing an expression in t for sp

For constructing an expression in t for

sP – sQ

Correct expression for sP – sQ

For using sP – sQ = 4.9 to construct an

equation in t

7 (i) R = 4500 N

3150 = µ4500

Coefficient is 0.7

B1

M1

A1

[3]

For using limiting equilibrium of boxes

� P = µR

(ii)

0.2 × 200g = 200a

No sliding � a Y 2

M1

A1

A1

[3]

For resolving forces horizontally on A

when A is about to slide

AG

(iii) [P – F = 450a; P – F – F2 = 250a]

Pmax = 3150 + 450 × 2 or

Pmax = 3150 + 0.2 × 2000 + 250 × 2

Pmax = 4050 N

M1

A1

A1

[3]

For applying Newton’s second law to A

and B combined or to B

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1 c of m of arc = 20sin(π/2)/(π/2)

(2 + 0.9) x = 2×20sin(π/2)/(π/2)

Distance is 8.78cm

B1

M1

A1

A1

[4]

For attempting to take moments about the

diameter

2 (i)

tan35° = r/7.5

r = 5.25

M1

A1ft

A1

[3]

For using the idea that the c.m. is

vertically above the lowest point of

contact

ft using their c of m from the base

(ii) [µmgcos35° > mgsin35°]

µ > tan35° → Coefficient is greater than 0.7

M1

A1

[2]

For using ‘no sliding → µR > weight

component’

Do not allow µ [ 0.7

AG

3 (i) mg = Tcosθ

ma = Tsinθ

tanθ = a/g = 0.75

T = 0.24 × 10/cosθ = 3

B1

B1

B1

B1

[4]

SR B1 not B2 for tanθ = v2/gr or a/g used

AG

For using Tcosθ = mg to find T

(ii) [v2 = 7.5 × 2sinθ]

Speed is 3ms 1

M1

A1

[2]

For using v2 = ar to find v

4 Weight split is 9N:6N

For lamina 9 × 0.75 + 6 × 0.5

= T × 1.5sin30°

Tension is 13N

Alternatively

B1

M1

A1ft

A1

A1

[5]

For taking moments about A

[(1.52+2

1 1.5×2) x = 1.52×0.75+2

1 1.5×2×0.5] M1 For using A x = A1x1 + A2x2

x = 0.65

15 × 0.65 = T × 1.5sin30°

Tension is 13N

A1

M1

A1ft

A1

[5]


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5 (i) 7 = 16tanθ – 10×162/(2×202)cos2θ

[7 = 16T – 3.2(1 + T2)]

3.2T2 – 16T + 10.2 = 0

T = ¾, 17/4

B1

M1

A1

A1

[4]

For using cosθ = 1/secθ and the given

identity to obtain a quadratic in T(tanθ)

AEF

AG

(ii) [x = tanθ cos2θ/0.0125 or x = 202sin2θ/g]

For tanθ = 0.75, distance is 38.4 m


M1

A1

A1

[3]

For solving y = 0 for x or for using

R = V2sin2θ/g

(iii) For sketching two parabolic arcs which intersect

once, both starting at the origin, each with y [ 0

throughout, and each returning to the x-axis, the arc

for which the angle of projection is smaller having

the greater range.

The ranges appear significantly greater than x at the

intersection, and slightly greater, respectively.

B1

B1

[2]

6 (i) [0.35g = 2T{0.7/ (2.42 + 0.72)1/2}]

Tension is 6.25N

[6.25 = λ × ¼]

Modulus is 25N

M1

A1

M1

A1

[4]

For resolving forces on P vertically

For using T = λx/L

AG

(ii)

EE on release = 25×22/(2×4)

EE when P is at M = 25×0.82/(2×4)

M1

A1

A1

M1

For using EE = λx2/2L

For using EE on release = mgh + EE

when P is at M + 2

1 mv2

25×22/(2×4) = 0.35g×1.8+25×0.82/(2×4) + 21 0.35v2 A1

Speed is 4.90ms 1 A1

[6]

7 (i) [0.25v(dv/dx) = –(5 – x)] B1 For using Newton’s second law and

a = v(dv/dx)

]5)dx(x4vdv[∫ ∫

M1

For separating variables and attempting

to integrate

v2/2 = 4(x – 5)2/2 (+ A)

v2 = 4(x – 5) 2

Selects correct square root to obtain v = 10 – 2x

A1

M1

A1

A1

[6]

For using v(0) = 10

Any correct expression in x

AG

(ii) [ ∫x210

dx= ∫dt ] M1

For using v = dx/dt and separating

variables

–2

1 ln(10 – 2x) = t(–2

1 lnB) A1

B = 10 (or equivalent)

x = 5(1 – e 2t)

0 < e 2t < 1 for all t → x < 5 for all t

A1

B1ft

B1

[5]

ft x = (B/2)(1 – e 2t)

AG

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1 c of m of arc = 20sin(π/2)/(π/2)

(2 + 0.9) x = 2×20sin(π/2)/(π/2)

Distance is 8.78cm

B1

M1

A1

A1

[4]

For attempting to take moments about the

diameter

2 (i)

tan35° = r/7.5

r = 5.25

M1

A1ft

A1

[3]

For using the idea that the c.m. is

vertically above the lowest point of

contact

ft using their c of m from the base

(ii) [µmgcos35° > mgsin35°]

µ > tan35° → Coefficient is greater than 0.7

M1

A1

[2]

For using ‘no sliding → µR > weight

component’

Do not allow µ [ 0.7

AG

3 (i) mg = Tcosθ

ma = Tsinθ

tanθ = a/g = 0.75

T = 0.24 × 10/cosθ = 3

B1

B1

B1

B1

[4]

SR B1 not B2 for tanθ = v2/gr or a/g used

AG

For using Tcosθ = mg to find T

(ii) [v2 = 7.5 × 2sinθ]

Speed is 3ms 1

M1

A1

[2]

For using v2 = ar to find v

4 Weight split is 9N:6N

For lamina 9 × 0.75 + 6 × 0.5

= T × 1.5sin30°

Tension is 13N

Alternatively

B1

M1

A1ft

A1

A1

[5]


[(1.52+2

1 1.5×2) x = 1.52×0.75+2

1 1.5×2×0.5] M1 For using A x = A1x1 + A2x2

x = 0.65

15 × 0.65 = T × 1.5sin30°

Tension is 13N

A1

M1

A1ft

A1

[5]


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5 (i) 7 = 16tanθ – 10×162/(2×202)cos2θ

[7 = 16T – 3.2(1 + T2)]

3.2T2 – 16T + 10.2 = 0

T = ¾, 17/4

B1

M1

A1

A1

[4]

For using cosθ = 1/secθ and the given

identity to obtain a quadratic in T(tanθ)

AEF

AG

(ii) [x = tanθ cos2θ/0.0125 or x = 202sin2θ/g]



M1

A1

A1

[3]

For solving y = 0 for x or for using

R = V2sin2θ/g

(iii) For sketching two parabolic arcs which intersect

once, both starting at the origin, each with y [ 0

throughout, and each returning to the x-axis, the arc

for which the angle of projection is smaller having

the greater range.

The ranges appear significantly greater than x at the

intersection, and slightly greater, respectively.

B1

B1

[2]

6 (i) [0.35g = 2T{0.7/ (2.42 + 0.72)1/2}]

Tension is 6.25N

[6.25 = λ × ¼]

Modulus is 25N

M1

A1

M1

A1

[4]

For resolving forces on P vertically

For using T = λx/L

AG

(ii)

EE on release = 25×22/(2×4)

EE when P is at M = 25×0.82/(2×4)

M1

A1

A1

M1

For using EE = λx2/2L

For using EE on release = mgh + EE

when P is at M + 2

1 mv2

25×22/(2×4) = 0.35g×1.8+25×0.82/(2×4) + 21 0.35v2 A1

Speed is 4.90ms 1 A1

[6]

7 (i) [0.25v(dv/dx) = –(5 – x)] B1 For using Newton’s second law and

a = v(dv/dx)

]5)dx(x4vdv[∫ ∫

M1

For separating variables and attempting

to integrate

v2/2 = 4(x – 5)2/2 (+ A)

v2 = 4(x – 5) 2

Selects correct square root to obtain v = 10 – 2x

A1

M1

A1

A1

[6]

For using v(0) = 10

Any correct expression in x

AG

(ii) [ ∫x210

dx= ∫dt ] M1

For using v = dx/dt and separating

variables

–2

1 ln(10 – 2x) = t(–2

1 lnB) A1

B = 10 (or equivalent)

x = 5(1 – e 2t)

0 < e 2t < 1 for all t → x < 5 for all t

A1

B1ft

B1

[5]

ft x = (B/2)(1 – e 2t)

AG

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1 vdown = 2g

tanθ = 2g/12

θ = 59.0°

B1

M1

A1

[3]

tanα =12/2g

2 XG = 20/4

8 × 3 = (20/4)mg

m = 0.48 kg

B1

M1

A1

A1

[4]

5

Attempt at moments about P

3 (i) 15 = 40tanθ – g402/(2×402cos2θ)

15 = 40tanθ – 5sec2θ

tan2θ – 8tanθ + 4 = 0 AG

M1

M1

A1

[3]

Substitutes in projectile equation

Uses sec2θ = 1 + tan2θ

(ii) θ = tan 1 (4 +/– 2 3 ) M1 Solves quadratic equation for θ

θ = 28.2° or 82.4°

R = 402sin(2×28.2°)/g or

R = 402sin(2×82.4°)/g

R = 133 or R = 41.9 (or 42.0)

Difference = 91.1 m

A1

M1

A1

A1

[5]

Valid formula for one range

0 = Rtan28.2° – gR2/(2×402cos28.2°)

or 0 = rtan82.4° –gr2/(2×402cos82.4°)

Using exact angles. Allow +/– 0.2

4 (i) d = 2×0.3sin(π/2)/(3π/2)

T(0.6cos30) =

0.4g(0.3sin30° + 0.1273cos30°)

T = 2 N AG

B1

M1

A1

A1

[4]

d = 0.1273

2.003…

(ii) R = 22 )g4.0(2( + ) or tanθ = 2/(0.4g) M1 Either (or tanα = 0.4g/2 with horizontal)

R = 4.47 N

θ = 26.6° (with vertical)

A1

A1

[3]

α = 63.4° (with horizontal)

5 (i) 3Tcos30° – Tcos30° = 0.4g

T = 2.31

0.4×62/r = 4Tsin30°

r = 3.12

M1

A1

M1

A1

[4]

Resolves vertically, 3 terms

Newton’s 2nd Law horizontally

(ii) TPB = 0

Tcos30° = 0.4g (T = 4.62)

0.4v2/3.12 = Tsin30°

v = 4.24 ms 1

B1

M1

M1

A1

[4]

Resolves vertically, 2 terms

Newton’s 2nd Law horizontally

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6 (i) 0.5vdv/dx = –3v1/2 M1 Newton’s 2nd Law with a = vdv/dx

∫2/1

v dv = – ∫6dx M1 Separates variables and integrates

v3/2/(3/2) = –6x (+ c)

x = 0, v = 9 hence c = 18

v3/2 = 3(18 – 6x)/2

v = (27 – 9x)2/3 AG

A1

M1

A1

[5]

Or uses limits

(ii) dx/dt = (27 – 9x)2/3 M1 0.5dv/dt = –3v1/2

∫ dx 9x)(27 32 / = ∫dt ∫2/1

v dv = – ∫ dt6

(27 – 9x)1/3/–3 = t (+ c)

t = 0, x = 0 hence c = –1

t = 0.5, x = 2.625

A1ft

M1

A1

[4]

v1/2 = –3t + c

t = 0, v = 9 hence c = 3 and t = 0.5,

giving v = 2.25

v = 2.25, x = 2.625

7 (i) 0.4v2/2 + 24x2/(2×3)

0.4g(3 + x) + 0.4×22/2

v2 = 64 + 20x – 20x2 AG

M1

A2

A1

[4]

PE, EE, KE terms

–1 each error to zero

(ii) 2vdv/dx = 20 – 40x = 0

x = 0.5

v = 8.31

M1

A1ft

A1

[3]

0.4g = 24x/3

(iii) 20x2 – 20x – 64 = 0

x = 2.357

T = 24×2.357/3

T = 18.9

M1

A1

M1

A1

[4]

And attempts to solve

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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



Penalties


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.



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1 a + b = 0.45

–3a – b + 1.6 = 0.75

a = 0.2 b = 0.25

B1

M1

A1

A1

[4]

Correct sum probs = 1 o.e.

Attempt at Σxp = 0.75

Correct a

Correct b

2 (i)

Key

1 2 represents 0

1

2

3

2 5 6 8 8

2 4 6 7 7 9

1 2 3 3 3 5 6 7

1 5

12 people

B1

B1

B1

[3]

Correct stem

Correct leaves must be sorted and

accurate

Key; must have people o.e

(ii) median = 19 people

LQ = 10, UQ = 24

IQ range = 24 – 10 = 14 people

B1

B1

B1ft

[3]

Correct median

Correct quartiles

Ft their quartiles

(iii) median because mode could be any number

which is duplicated more than twice

B1

[1]

Correct answer must say something about

the mode being not much use or another

sensible reason

3 (+/–) 1.045, (+/–) 0.313

20.9 – µ = –0.313 σ

30 – µ = 1.045 σ

σ = 6.70

µ = 23.0

B1, B1

M1

A1

A1

[5]

1 correct z-value, the other correct

z-value.

Valid attempt to solve 2 equations

relating to µ, σ, 30, 20.9. No σ , σ2

correct answer

correct answer

4 (i) sd = 0

so all rides must cost the same i.e. the mean.

B1*

B1 dep

[2]

Must see this and some relevant

comment, e.g. no change

o.e.

(ii) 1 × 2.5 + 3 × 2.5 + 6 × x = 3.76 × 10

6x = 37.6 – 10

x = 4.6 for revolving drum

σ2 = (2.52 × 1 + 2.52 × 3 + 4.62 × 6)/10 – 3.76 2

σ = 1.03

M1

A1

A1

M1

A1

[5]

attempt to find cost of revolving drum

ride

correct equation

correct x

substituting in correct variance formula

correct answer

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5 (i) P(X = 2)) = (0.25)2 × (0.75)6 × 8C2

= 0.311

M1

A1

[2]

3 term binomial expression involving 8C

something, powers summing to 8

correct answer

(ii) 12 × 0.25 = 3, < 5 so not possible B1

[1]

(iii) mean = 40 × 0.25 (= 10)

variance = 40 × 0.25 × 0.75 ( = 7.5)

B1

40 × 0.25 and 40 × 0.25 × 0.75 seen, o.e.

P(X at least 13) = P

>

5.7

105.12z M1

standardising, ±, with or without cc, must

have sq rt

= P(z > 0.913)

= 1 – Φ(0.913)

= 1 – 0.8194

= 0.181

M1

M1

A1

[5]

continuity correction 12.5 or 13.5

correct area, i.e. < 0.5 legit

correct answer

6 (i) 10C1+ 10C3 + 10C5 + 10C7 + 10C9

= 512

M1

A1

A1

[3]

Summing some 10C combinations with

odd numbers, all different

At least 3 correct unsimplified

expressions

Correct answer

(ii) 6! × 7 × 6 × 5

= 151200

B1

M1

A1

[3]

6! seen

multiplying by 7P3 o.e.

correct answer

(iii) 12! / (4! × 7!)

= 3960

B1

M1

A1

[3]

12! Seen

dividing by 4!7!

correct answer

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7 (i) P(1st correct) = 0.7 + 0.2 × 0.95

= 0.89 AG

B1

(ii)

C

0.7

0.7 C 0.1 C A

0.2 0.95

0.1 C H 0.7 C

A

0.2 H 0.95 0.1 C

0.2 0.65 P

H

M1

M1

M1

Considering any 2 of CC, CHA, HAC or

HAHP [where C = Peter correct, H = ask

for help, A = audience correct, P = phone

correct] or tree diagram with ‘top half’

labels and probs shown

Considering other 2

Summing 4 probabilities

P(CC) = 0.7 × 0.7 (= 0.49)

P(CHA) = 0.7 × 0.2 × 0.95 (= 0.133)

P(HAC) = 0.2 × 0.95 × 0.7 (= 0.133)

P(HAHP) = 0.2 × 0.95 × 0.2 × 0.65 (= 0.0247)

P(both correctly answered) = 0.781

B1

B1

A1

[6]

Two correct probabilities

Three correct probabilities

Correct

(iii) P(audience | both correct)

=(ii)

)()()(

ans

HAHPPHACPCHAP ++ M1*

Summing two or three 3-factor terms in

numerator of a fraction

=

7807.0

65.02.095.02.07.095.02.095.02.07.0 ×××+××+××

= 0.2907/0.7807

= 0.372

M1dep

A1

[3]

Dividing by their (ii)

Correct answer

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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



Penalties


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.



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© UCLES 2010

1 (i) 9.18x (132/7)

sd = 12.3

B1

B1 [2]

(ii) median B1 [1]

(iii) mode inappropriate because it is 10 and this is

the lowest value.

mean inappropriate because it is affected by the

outlier (of 48).

B1

B1 [2]

Sensible reason allow if seen in (ii)

Sensible reason allow if seen in (ii) not

‘outliers’ in plural

2 (i) P(x > 10.9) = )095.0

119.10(P >z

= P(z > –1.0526)

= 0.8538 (0.854)

M1

A1 [2]

Standardising, no cc, no sq rt

Rounding to correct answer

(ii) P(at least 2 < 10.9) = 1 – P(0, 1)

= 1 – (0.8538)6 – 6C1(0.1462)(0.8538)5

= 0.215

M1

A1ft

A1 [3]

Bin expression with ∑ powers = 6, 6Cx,

p + q = 1.

Reasonably correct unsimplified

expression ft their (i)


3 median A = 2.0 – 2.1 or A

x = 2.0 – 2.1

median B = 3.8 – 3.9 or B

x = 3.4 – 3.5

Country B has heavier babies on average

IQ range A = 2.4 – 1.5 = 0.9 or sd = 0.5 – 0.7

IQ range B = 4.5 – 2.2 = 2.3 or sd = 1.2 – 1.4

Country B has greater spread of weights

M1

A1

B1

M1

A1

A1 [6]

For finding medians or using mid-pts and

freqs to find means, or seen on 2 box-

plots

Correct medians or means for A and B

Correct statement allow ’...higher

median…’ etc.

Finding spreads by IQ range or range or

sd or 2 box-plots

Correct IQ range or sd for A and B

(±0.1 kg) or correct IQR on box-plots

Correct statement

4 (i) P (X < 2 µ) = P

<

σ

µµ2z

= P(z < µ/σ) = P(z < 5/3)

= 0.952

M1

A1

A1 [3]

Standardising, and attempt to get

1 variable, no cc, no , no sq

±5/3 seen oe


(ii)

<

<

σ

µµ

3

2P

3P zX

047.13

2

σ

µ

µ = –1.57σ

M1

B1

A1 [3]

standardising attempt resulting in

z Y – some µ/σ

allow

±

σ

µµ 3/

±1.047 seen

correct single number, answer must have

a minus sign and µ = …..σ

5 (i) (2,12), (3, 8), (4, 6), (6, 4), (8,3), (12,2)

P(Q) = 6/144 (1/24) (0.0417)

M1

A1 [2]

Listing or picking out at least 3 different

options from a 12 by 12 (mult) table or

seeing 3, 4, 5 or 6/144

Correct answer

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(ii) P(both > 8) = 1/3 × 1/3

= 1/9 = P(R) (16/144)

M1

A1 [2]

Squaring a sensible prob or picking out

12 – 25 options

Correct answer

(iii) since P(R and Q) = 0

Yes, R and Q are exclusive

B1*

B1dep

[2]

o.e. in words

(iv) P(R and Q) = 0 ≠ P(R) × P(Q)

or P(R|Q) = 0, ≠ P(R)

No, not independent

B1*

B1dep

[2]

o.e. in words

6 (i)

x 0 1 2

P(X = x) 1/7 4/7 2/7

B1

B1

B1 [3]

0, 1, 2 only in table or listed with some

prob

3, 4… if in table must have blank or 0 for

prob

One correct probability

All correct

(ii) E(X) = 8/7 (1.14) AG

Var(X) = 12/7 – (8/7)2

= 20/49 (0.408)

B1

M1

A1 [3]

Legitimate correct given answer rounding

to 1.14

Correct method with mean2 subt

numerically no dividing by anything

Correct final answer

(iii) P(G | NA) = )(

)(

NAP

NAGP ∩

10/95/34/15/2

4/15/2

×+×

×

32

5 (0.156)

M1

M1

A1

A1 [4]

Attempt at P(G ∩ NA) or P(G ∩ A) as

numerator of a fraction

Attempt at P(NA) or P(A) in form of

summing two 2-factor products, seen

anywhere

Correct unsimplified denominator of a

fraction

Correct answer

7 (i) 362880 (363000) B1 [1]

(ii) PG or GP in 8! × 2 = 80640 or 7/9 of (i)

362880 – 80640 = 282240

M1

B1

A1ft

[3]

Considering together and also subtracting

from their (i) or using probabilities

8! × 2 or 80640 seen oe

correct answer ft 40320 only

(iii) 9P3 or 9C3 × 3! or 9!/6!

= 504

M1

A1 [2]

9P3 or 9C3 oe seen allow extra

multiplication

correct final answer

(iv) 8C2 × 3! or 504 – 8C3 × 3! or 8P2 × 3

= 168

M1

A1 [2]

8Cx or 8Px seen allow extra mult, or (iii)/9

or (iii)/3

correct final answer

(v) PG and x in 7 × 2 × 2 ways = 28

Answer 504 – 28 = 476

M1

A1 [2]

x × 2 × 2 seen or their (iii) – 7 or 7C1 or 7C2

correct answer

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1 7

52

4

39

3

13

C

CC ×

M1 M1

Using combinations with attempt to evaluate product of 2 in num and only 1 in denom Correct numerator or denominator

= 0.176 A1 Correct answer

OR P(RRR) =

3

7C

46

36

47

37

48

38

49

39

50

11

51

12

52

13×××××××

= 0.176

M1 M1 A1

[3]

OR Multiplying 3 unequal red probs with 4 unequal non-red probs Multiplying a probability by 7C3 Correct answer

2 (i) 82/287130x = 126.5 (126, 127) cm

M1 A1

[2]

287/82 seen added or subt to 130 OR 287 seen added or subt to 82 × 130 Correct answer

(ii) 22

2

9.6)5.3(82

)130(Σ x

Σ(x – 130)2 = 4908.5 cm (4910)

M1

A1

[2]

6.92 + (±their coded mean)2 seen or implied

correct answer

3 (i) P(> 5) = 7C6(0.6)6(0.4) + (0.6)7 = 0.1306 + 0.02799 = 0.159

M1 A1

[2]

Summing 2 or 3 binomial probs of the form 7Cr(0.6)r(0.4)7 r Correct answer

(ii) P(bark) = P(park, bark) + P(not park, bark) = 0.6 × 0.35 + 0.4 × 0.75 = 0.51

M1 A1

[2]

Summing two appropriate 2-factor probabilities Correct answer

(iii) Variance (number of times) = 7.2 B1 [1]

Correct final answer

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4 (i) ends cola, 5!/2!2! = 30 ends green tea, 5!/3!2! = 10 ends orange juice, 5!/3!2! = 10 total = 50 ways

OR P(ends same) 6

1

7

2

6

1

7

2

6

2

7

3×+×+×

21

5

×

!2!2!3

!7

21

5 50 ways

M1 A1 A1

M1

A1

A1

[3]

Considering all three options Any one option correct Correct answer

OR Considering all three options

Correct fraction

Correct answer

(ii) colas together, no restrictions, 5!/2!2! = 30 ways colas together and green tea together, 4!/2! = 12 ways 30 – 12 = 18 ways. OR1 Attempt to list

OR2 182

343

×

×

M1 A1 M1 A1 A1 M1A1 M1A1 A1

M1

A1 M1 A1 A1

[5]

Considering all colas together, or 5! seen Correct answer Considering all colas tog and all green tea tog, or 4! seen Correct answer Correct final answer OR1 10 or more, 12 or more correct 14 or more, 16 or more correct 18 correct

OR2 Considering all colas together, or 3! seen

3 ways for colas and orange juice Considering green teas not together 4 × 3 or (4 × 3)/2 Correct final answer

5 (i) P(2) = P(0,2) + P(2,0) = 6/10 × 3/7 + 3/10 × 4/7 = 30/70 = 3/7 AG

M1 A1

[2]

Summing two 2-factor probabilities Correct answer legit obtained

(ii)

x 0 2 4 6 P(X = x) 24/70 30/70 13/70 3/70

B1 B1

[2]

Correct values for rv X Correct probs

(iii) E(X) = 13/7 Var(X) = 120/70 + 208/70 + 108/70 – (13/7)2 = 2.78

B1ft M1 A1

[3]

Using variance formula correctly with mean2 subtracted numerically, no extra division Correct final answer

(iv) P(A2│Sum 2) = 70/30

7/410/3 ×

= 0.4

M1

A1 [2]

Correct numerator with a 0 < denom < 1

Correct answer

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6 (i) for X: Median = 0.825 cm IQ range = 0.019 cm (0.833 – 0.814)

B1 B1

[2]

Correct median Correct IQ range

(ii) q = 4 r = 2 SR q = 0.824 and r = 0.852

B1 B1

[2] B1

Must be 4 and 2 not 3 and 1

(iii)

Y

X

0.80 0.81 0.82 0.83 0.84 0.85 0.86 0.87 length in cm

B1 B1ft B1ft B1

[4]

Labels X, Y and length/cm, linear scale from 0.80 to 0.87 and both on one diagram Correct median and quartiles for X ft theirs must be a box Correct median and quartiles for Y ft theirs must be a box Whiskers correct no line through middle

(iv) Y has longer insects on average Y has larger range

B1 B1

[2]

Correct statement about lengths Correct statement about spreads

7 (i) 0.431 = σ

µ135

–0.842 = σ

µ127

σ = 6.29 µ = 132

B1

B1

M1

A1 A1

[5]

One ±z-value correct, accept 0.430

A second ±z-value correct

Solving two equations relating µ, σ, 135,

127 and their z-values (must be z-values) Correct answer accept 6.28 Correct answer

(ii) P(X < 145) =

<

284.6

3.132145P z

=P(z < 2.023) = 0.978

M1

M1 A1

[3]

Standardising no sq rt no cc

Correct use of normal tables Answer rounding to 0.978 or 0.979

(iii) p = 1/3 P(at least 2) = 1 – P(0, 1) = 1 – [ 71

1

88 )3/2()3/1(C)3/2( ×+ ] = 0.805

M1 A1 A1

[3]

Binomial expression with powers summing to 8 and 8Csomething. (any p) Correct unsimplified expression Answer rounding to 0.805

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1 (i) 1/12 B1

[1]

Accept 0.0833

(ii) trains arrive every 12 minutes B1

[1]

must have ‘every 12 minutes’

2 (i) 0.145

= 87 / n

n = 600

B1

M1

A1

[3]

correct mid-point

equating their mid-point with 87 / n

correct answer

(ii) 0.0321 = 600

)145.01(145.0×z B1 0.0321 seen or implied

M1 Equating half-width with

n

pqz×

z = 2.233 Φ(z) = 0.9872

width of CI is 1 – 2 × (1 – 0.9872)

α = 97.4%

M1

A1

[4]

Correct method to find width of CI

Correct answer

3 (i) z = 45/3.0

62.255.2= –1.565 M1 Standardising no cc

M1 Dividing 0.3 by 45 as denominator

P (z > –1.565) = 0.941 A1

[3]

Correct answer

(Accept equivalent method using totals)

(ii) rejection region is 1am < and

2am >

where 645.1

30/3.0

62.21a

B1 ±1.645 seen

and 645.1

30/3.0

62.22

a

M1 one correct unsimplified equation of correct form

m < 2.53 and m > 2.71

M1

A1

[4]

second unsimplified equation of correct form

(or clear use of 1-tail test and ±1.282 used)

correct answer

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4 (i) Mr – 5Mrs ~ N(512 – 5×89, 622 + 25×7.42)

~ N(67, 5213)

P(Mr > 5 Mrs) = P(Mr – 5 Mrs > 0)

= P

>

5213

670z

= P(z > –0.9280)

B1

B1

M1

M1

Correct unsimplified mean

Correct unsimplified variance

Using distribution Mr – 5 Mrs

Standardising and using tables

= 0.823 A1

[5]

Correct answer

(ii) Mr + Mrs ~ N(601, 622 + 7.42)

E[5/8(Mr + Mrs)] = 376 miles

Var[5/8(Mr + Mrs)] = 76.389864

25×

= 1520

B1

B1

Correct mean and variance

Correct answer

SR Two separate answers 320 and 55.6 B1

sd = 39.0 miles B1

[3]

Correct answer

5 (i) 1e

5

0

2.0∫ dtkt M1 Equating to 1 and attempting to integrate

1e2.0

e2.0

00.1

kk A1 Correct integrand and limits

( ) 11e2.0

k

)1e(5

1k AG A1 Correct answer legitimately obtained

[3]

(ii)

0 5

B1

B1

[2]

Correct curve shape

Correct horizontal lines (need to see a 5)

(iii) 2.0e

0

2.0∫T

tdtk M1 Equation relating T and 0.2 or 0.8

[ ] [ ] 2.05e52.0

kkT A1 Correct equation (can be in ‘k’)

344.115

2.0e

2.0+

k

T

T = 1.48 (seconds) A1

[3]

Correct answer

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6 (i) λA= np = 0.022 ×55 = 1.21

λB = 0.058 × 55 = 3.19

total λ = 4.4

P(more than 2) = 1 – P(0, 1, 2)

= 1 –

++

!2

4.44.41e

2

4.4

= 1 – 0.185

M1

A1

M1

Two different np (can be implied)

Correct total 4.4 (or alt method: 6 correct

combinations 0,0 1,0 etc stated and used)

Finding 1 – P(0, 1, 2), Poisson, any mean, allow

one end error.

(Or combinations method – use at least 4 and

find 1 – P(Y2) )

= 0.815 A1

[4]

Correct answer

(ii) λ = 0.08n

P(at least 1 stained tablecloth ) = 1 – P(0)

1 – e 0.08n > 0.99

0.01 > e 0.08n

n > 57.6

least value of n = 58

B1

M1

M1

A1

[4]

Correct λ

Equation of correct form relating their λ and 0.99

Valid attempt to solve equation of correct form

by logs or trial and error

Correct answer

(SR Accept use of Binomial leading to n = 57)

7 (i) Type I error is made when we say the

number of white blood cells has decreased

when it hasn’t.

P(0) = e 5.2 = 0.005516

P(1) = e 5.2(5.2) = 0.02868 Σ < 0.10

P(2) = e 5.2(5.22/2) = 0.07458 Σ > 0.10

P(Type I error) = 0.0342

B1

M1

M1*

A1dep

[4]

Correct and relating to question

Evaluating at least 2 of P(X = 0, 1, 2)

Comparing their Σ 3 probs with 10% (must be Σ

probs)

Correct answer, dep on previous M

(ii) H0: λ = 5.2

H1: λ < 5.2

P(0+1+2) = 0.1087 > 10%

2 not in C Region.

Accept H0. Not enough evidence to say the

number of blood cells has decreased.

B1

M1

A1

[3]

Both hypotheses correct

Stating 2 is not in the critical region from above,

or evaluating P(0, 1, 2) and comparing with 10%

again

Correct conclusion no contradictions

(iii) P(Type II error) = 1 – P(0, 1)

= 1 – e 4.1(1 + 4.1)

= 0.915

B1

M1

A1

[3]

Identifying correct area

(indep) Some form of (Poisson) expression with

mean 4.1

Correct answer

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1 (i) 1/12 B1

[1]

Accept 0.0833

(ii) trains arrive every 12 minutes B1

[1]

must have ‘every 12 minutes’

2 (i) 0.145

= 87 / n

n = 600

B1

M1

A1

[3]

correct mid-point

equating their mid-point with 87 / n

correct answer

(ii) 0.0321 = 600

)145.01(145.0×z B1 0.0321 seen or implied

M1 Equating half-width with

n

pqz×

z = 2.233 Φ(z) = 0.9872

width of CI is 1 – 2 × (1 – 0.9872)

α = 97.4%

M1

A1

[4]

Correct method to find width of CI

Correct answer

3 (i) z = 45/3.0

62.255.2= –1.565 M1 Standardising no cc

M1 Dividing 0.3 by 45 as denominator

P (z > –1.565) = 0.941 A1

[3]

Correct answer

(Accept equivalent method using totals)

(ii) rejection region is 1am < and

2am >

where 645.1

30/3.0

62.21a

B1 ±1.645 seen

and 645.1

30/3.0

62.22

a

M1 one correct unsimplified equation of correct form

m < 2.53 and m > 2.71

M1

A1

[4]

second unsimplified equation of correct form

(or clear use of 1-tail test and ±1.282 used)

correct answer

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4 (i) Mr – 5Mrs ~ N(512 – 5×89, 622 + 25×7.42)

~ N(67, 5213)

P(Mr > 5 Mrs) = P(Mr – 5 Mrs > 0)

= P

>

5213

670z

= P(z > –0.9280)

B1

B1

M1

M1

Correct unsimplified mean

Correct unsimplified variance

Using distribution Mr – 5 Mrs


= 0.823 A1

[5]

Correct answer

(ii) Mr + Mrs ~ N(601, 622 + 7.42)

E[5/8(Mr + Mrs)] = 376 miles

Var[5/8(Mr + Mrs)] = 76.389864

25×

= 1520

B1

B1

Correct mean and variance

Correct answer

SR Two separate answers 320 and 55.6 B1

sd = 39.0 miles B1

[3]

Correct answer

5 (i) 1e

5

0

2.0∫ dtkt M1 Equating to 1 and attempting to integrate

1e2.0

e2.0

00.1

kk A1 Correct integrand and limits

( ) 11e2.0

k

)1e(5

1k AG A1 Correct answer legitimately obtained

[3]

(ii)

0 5

B1

B1

[2]

Correct curve shape

Correct horizontal lines (need to see a 5)

(iii) 2.0e

0

2.0∫T

tdtk M1 Equation relating T and 0.2 or 0.8

[ ] [ ] 2.05e52.0

kkT A1 Correct equation (can be in ‘k’)

344.115

2.0e

2.0+

k

T

T = 1.48 (seconds) A1

[3]

Correct answer

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6 (i) λA= np = 0.022 ×55 = 1.21

λB = 0.058 × 55 = 3.19

total λ = 4.4

P(more than 2) = 1 – P(0, 1, 2)

= 1 –

++

!2

4.44.41e

2

4.4

= 1 – 0.185

M1

A1

M1

Two different np (can be implied)

Correct total 4.4 (or alt method: 6 correct

combinations 0,0 1,0 etc stated and used)

Finding 1 – P(0, 1, 2), Poisson, any mean, allow

one end error.

(Or combinations method – use at least 4 and

find 1 – P(Y2) )

= 0.815 A1

[4]

Correct answer

(ii) λ = 0.08n

P(at least 1 stained tablecloth ) = 1 – P(0)

1 – e 0.08n > 0.99

0.01 > e 0.08n

n > 57.6

least value of n = 58

B1

M1

M1

A1

[4]

Correct λ

Equation of correct form relating their λ and 0.99

Valid attempt to solve equation of correct form

by logs or trial and error

Correct answer

(SR Accept use of Binomial leading to n = 57)

7 (i) Type I error is made when we say the

number of white blood cells has decreased

when it hasn’t.

P(0) = e 5.2 = 0.005516

P(1) = e 5.2(5.2) = 0.02868 Σ < 0.10

P(2) = e 5.2(5.22/2) = 0.07458 Σ > 0.10

P(Type I error) = 0.0342

B1

M1

M1*

A1dep

[4]

Correct and relating to question

Evaluating at least 2 of P(X = 0, 1, 2)

Comparing their Σ 3 probs with 10% (must be Σ

probs)

Correct answer, dep on previous M

(ii) H0: λ = 5.2

H1: λ < 5.2

P(0+1+2) = 0.1087 > 10%

2 not in C Region.

Accept H0. Not enough evidence to say the

number of blood cells has decreased.

B1

M1

A1

[3]

Both hypotheses correct

Stating 2 is not in the critical region from above,

or evaluating P(0, 1, 2) and comparing with 10%

again

Correct conclusion no contradictions

(iii) P(Type II error) = 1 – P(0, 1)

= 1 – e 4.1(1 + 4.1)

= 0.915

B1

M1

A1

[3]

Identifying correct area

(indep) Some form of (Poisson) expression with

mean 4.1

Correct answer

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1 H0: Pop prop = 1/3 (or unchanged) H1: Pop prop < 1/3 (or decreased)

(3

2 )20 + 30(3

2 )19(3

1 ) + 20C2(3

2 )18(3

1 )2

+ 20C3(3

2 )17(3

1 )3

= 0.0604/0.0605 comp “0.0604” with 0.025 No evidence that support decreased or support probably not decreased SC Use Of Normal Standardising with or without cc Obtains z = –1.502 Valid Comparison with z = –1.96 Correct conclusion

B1

M1

A1 M1 A1ft

[5] M1 A1 M1 A1ft

Accept p

Attempt Bin(20, ⅓) P(Y 3)

Allow one term omitted

For comparison of their 0.0604 Correct conclusion no contradictions

2 (i) 2-tail; H1: µ ¸ 35 B1 [1]

(ii) comp –1.75 with –1.645 (or 1.75 with 1.645) Evidence that µ is not 35 or reject µ = 35

M1 A1

[2]

Allow “Accept µ ¸ 35”. No contradictions

(iii) 8 B2 [2]

SR B1 for 4, 8.02, or 92%

3 (i) (Approx) normal mean 62

sd = 50

2.8 = 1.16 (3 sfs)

B1 B1

B1

[3]

or var = 50

2.82

= 1.34 (3 sfs)

(ii) "16.1"

6264 (= 1.725 or 1.724) M1 For standardising ÷ 50 essential (no CC)

1 – Φ(“1.725”) = (1 – 0.9577) = 0.0423 (3 sfs)

M1 A1

[3]

For correct area consistent with their mean

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4 (i) Mean is 4.8 but Y 1 breakdown B1 [1]

Accept reduction when none has occurred

(ii) e 4.8(1 + 4.8) = 0.0477

M1 A1

[2]

Poisson attempt at P(0) (+ P(1))

(iii) P(X > 1) = 1 – e 0.9(1 + 0.9) = 0.228 (3 sfs)

M1 M1 A1

[3]

Attempt correct probability for Type II error Allow any λ except 4.8; 1– (P(0)+(P(1))) using Poisson As final answer

5 (i) xx

kd

4

1 ∫∞

= 1 M1 Attempt integ f(x) & “= 1”; ignore limits

∞

13

3x

k = 1 oe

Correct integrand & limits leading to AG, no errors seen

(3

0k

+ = 1 ⇒ k = 3 AG)

A1 [2]

(ii) x

x

x d3

4

1

×∫∞

∞

12

2

3

x

= 2

3

x

x

x d3

4

2

1

×∫∞

∞

1

3

x

(= 3)

“3” – ( )2"2

3"

= 4

3

M1

A1

M1*

A1

M1*dep

A1

[6]

Attempt integ xf(x); ignore limits.

CWO

Attempt integ x2f(x); ignore limits.

Correct integrand; correct limits

dep 2nd M1 attempt E(X 2) – [E(X )]2

cwo

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6 (i) 5×(22.4 + 20.3) (= 213.5) & 5×(4.82 + 5.22) (= 250.4)

z = "4.250"

"5.213"180− (= –2.117)

1 – Φ(“–2.117”) = Φ(“2.117”) = 0.983 (3 sfs)

B1 B1

M1

A1

[4]

For correct expression for new mean For correct expression for new variance

Standardising and use of tables

(no sd/var mixes + no cc)

(ii) P(H – W > 0) 20.3 – 22.4 (= –2.1) & 4.82 + 5.22 (= 50.08)

z = "08.50"

)1.2(0 −−

(= 0.297)

1 – Φ(“0.297”) (= 1 – 0.6168) = 0.383 (3 sfs)

M1 B1 B1

M1

A1

[5]

Or P(W – H < 0) ± 2.1 Correct expression for new mean Correct expression for new variance


(no sd/var mixes + no cc)

7 (i) Patients arrive at constant mean rate Patients arrive at random Patients arrive independently Patients arrive singly

B1 B1

[2]

B1 For first correct B1 For second correct Must be in context SR B1 For two correct but not in context

(ii) (a) 1 – e–4.2 = 0.985

M1 A1

[2]

Correct expression

(b) 4.2 × 10/15 oe

e–2.8×(1 + 2.8 + !3

8.2

!2

8.232

+ )

= 0.692

B1

M1

A1 [3]

Allow extra term e–2.8×!4

8.24

Allow incorrect λ (not 4.2)

(iii) N(336, 336) stated or implied

336

3365.370 − (= 1.882)

1 – Φ(“1.882”) = 0.0300/0.0299

B1

M1

M1 A1

[4]

ft “336”Allow wrong or no cc or no √

Standardising with correct cc and no √ Allow 0.03

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9709 s10_ms31

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