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Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level

MATHEMATICS 9709/11 Paper 1 May/June 2017

MARK SCHEME

Maximum Mark: 75

Published

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 should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2017 series for most Cambridge IGCSE®, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

9709/11 Cambridge International AS/A Level – Mark Scheme PUBLISHED

May/June 2017

© UCLES 2017 Page 2 of 12

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 FT 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.


May/June 2017


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

AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent

AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid)

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

SOI Seen or implied

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.


May/June 2017


Question Answer Marks Guidance

1 (3–2x)6

Coeff of x² = 34 × (−2)² × 6C2 = a B3,2,1 Mark unsimplified forms. –1 each independent error but powers

Coeff of x³ = 3³ × (−2)³ × 6C3 = b must be correct. Ignore any ‘x’ present.

98

ab=−

B1 OE. Negative sign must appear before or in the numerator

Total: 4

2 36OAp

= −

uuur and

267

OB = − −

uuur

2(i) Angle AOB = 90° → 6 + 36 −7p = 0 M1 Use of x1x2 + y1y2 + z1z2 = 0 or Pythagoras

→ p = 6 A1

Total: 2


May/June 2017



2(ii) 32 63

OCp

= −

uuur =

24

4

−

B1 FT CAO FT on their value of p

BCuuur

= c – b = 02

11

; magnitude = √125

M1 M1 Use of c – b. Allow magnitude of b + c or b – c Allow first M1 in terms of p

Unit vector = 0

1 2125 11

A1 OE Allow ± and decimal equivalent

3(i) 1 cos sin 2 sin 1 cos sin

θ θθ θ θ

++ ≡

+.

( )( )

21 ²1c s

s c+ +

+ =

( )1 2 ² ²

1c c s

s c+ + +

+

M1 Correct use of fractions

=

( )( )( )

2 12 21 1

ccs c s c

++=

+ + →

2s

M1 A1 Use of trig identity, A1 needs evidence of cancelling

Total: 3

3(ii) 2 3=s c

→ 23

t = M1 Use part (i) and t = s ÷ c, may restart from given equation

→ θ = 33.7° or 213.7° A1 A1FT FT for 180° + 1st answer. 2nd A1 lost for extra solns in range

Total: 3


May/June 2017



4(a) a = 32, a + 4d = 22, → d = −2.5 B1

a + (n – 1)d = −28 → n = 25 B1

S25 = ( )25 64 2.5 24

2− × = 50

M1 A1 M1 for correct formula with n = 24 or n = 25

Total: 4

4(b) a = 2000, r = 1.025 B1 1 2.5%r = + ok if used correctly in Sn formula

S10 = 2000(

101 .025 11.025 1

−−

) = 22400 or a value which rounds to this M1 A1 M1 for correct formula with n = 9 or n = 10 and their a and r

SR: correct answer only for n = 10 B3, for n = 9, B1 (£19 900)

Total: 3


May/June 2017



5 y = 2cosx

5(i) B1 One whole cycle – starts and finishes at –ve value

DB1 Smooth curve, flattens at ends and middle. Shows (0, 2).

Total: 2

5(ii) P(

3π

, 1) Q(π, −2)

→ PQ² =

22 3²3π +

→ PQ = 3.7

M1 A1 Pythagoras (on their coordinates) must be correct, OE.

Total: 2


May/June 2017



5(iii) Eqn of PQ

912 3

y x ππ − = − −

M1 Correct form of line equation or sim equations from their P & Q

If y = 0 →

59

h π=

A1 AG, condone

59

x π=

If x = 0 → k =

52

, A1 SR: non-exact solutions A1 for both

Total: 3

6(i) Volume =

1 3²2 2

x h

= 2000 → h = 8000

3 ²x√

M1 Use of (area of triangle, with attempt at ht) × h =2000, ( )fh x=

A = ( ) 21 33 2

2 2xh x + × × ×

M1 Uses 3 rectangles and at least one triangle

Sub for h → 2 13 24 000

2 3A x x−√= +

A1 AG

Total: 3

6(ii) 2d 3 240002

d 2 3A x xx

−= − B1 CAO, allow decimal equivalent

= 0 when x³ = 8000 → x = 20 M1 A1 Sets their

ddAx

to 0 and attempt to solve for x

Total: 3


May/June 2017



6(iii) 3d² 3 480002

d ² 2 3A x

x−= + > 0

M1 Any valid method, ignore value of

d²d ²

Ax

providing it is positive

→ Minimum A1 FT FT on their x providing it is positive

Total: 2

7 d 7 ² 6 dy x xx= − −

7(i) ³ 6 ²73 2x xy x= − − (+ c)

B1 CAO

Uses (3, − 10) → c = 5 M1 A1 Uses the given point to find c

Total: 3

7(ii) 7 ² 6 x x− − = ( )16 3 ²x− + B1 B1 B1 a = 16, B1 b = 3.

Total: 2

7(iii) ( )16 3 ² 0x− + > → (x + 3)² < 16, and solve M1 or factors (x + 7)(x – 1)

End-points x = 1 or −7 A1

→ −7 < x < 1 A1 needs <, not ⩽. (SR x < 1 only, or x > −7 only B1 i.e. 1/3)

Total: 3


May/June 2017



8(i) Letting M be midpoint of AB

OM = 8 (Pythagoras) → XM = 2 B1 (could find √40 and use sin−1or cos−1)

tan AXM =

62

AXB = 2tan−13 = 2.498 M1 A1 AG Needs × 2 and correct trig for M1

(Alternative 1:

6sin , 0.6435, 0.643510

AOM AOM AXB π= = = − ) (Alternative 1: Use of isosceles triangles, B1 for AOM, M1,A1 for

completion) (Alternative 2: Use of circle theorem, B1 for AOB, M1,A1for completion)

Total: 3

8(ii) AX = √(6² + 2²) = √40 B1 CAO, could be gained in part (i) or part (iii)

Arc AYB = rθ = √40 × 2.498 M1 Allow for incorrect √40 (not 6 1 2 1 0r or or= )

Perimeter = 12 + arc = 27.8 cm A1

Total: 3

8(iii) area of sector AXBY = ½ × (√40)² × 2.498 M1 Use of ½r²θ with their r , (not 6 10r or r= = )

Area of triangle AXB = ½ × 12 × 2, Subtract these → 38.0 cm² M1 A1 Use of ½bh and subtraction. Could gain M1 with 10r = .

Total: 3


May/June 2017



9 f : x ⟼ 2

3 2x− g : x ⟼ 4x + a,

9(i) 23 2

yx

=−

→ ( ) 23 2 2 3 2y x xy

− = → − = M1 Correct first 2 steps

→

22 3xy

= − → f−1(x) = 3 12 x−

M1 A1 Correct order of operations, any correct form with ( )f x or y =

Total: 3

9(ii) gf(−1) = 3 f(−1) =

25

M1 Correct first step

8 73 5 5

a a+ = → = M1 A1 Forms an equation in a and finds a , OE

(or

8 33 2

ax+ =

−, M1 Sub and solves M1, A1)

Total: 3

9(iii) g−1(x) =

4x a−

= f−1(x) M1 Finding ( )1g x− and equating to their ( )1f x− even if 7 / 5a =

→ ( ) ( )2 6 4 0x x a− + + = M1 Use of b² − 4ac on a quadratic with a in a coefficient

Solving ( ) ( )2 26 16 12 20 0a or a a+ = + + = M1 Solution of a 3 term quadratic

→ a = −2 or −10 A1

Total: 4


May/June 2017



10(i)

( )d 4d 5 3 ²yx x

−=

− × (−3)

B1 B1 B1 without ×(−3) B1 For ×(−3)

Gradient of tangent = 3, Gradient of normal – ⅓ *M1 Use of m1m2 = −1 after calculus

→ eqn: ( )12 1

3y x− = − −

DM1 Correct form of equation, with (1, their y), not (1,0)

→

1 73 3

y x= − + A1 This mark needs to have come from y = 2, y must be subject

Total: 5

10(ii) Vol = π

( )1

0

16 d5 3 ²

xx−∫

M1 Use of ²dV y xπ= ∫ with an attempt at integration

π( )

16 35 3x

−÷ −

−

A1 A1 A1 without( ÷ −3), A1 for (÷ −3)

= ( π

16 166 15

−

) = 85π

(if limits switched must show – to +) M1 A1 Use of both correct limits M1

Total: 5






MARK SCHEME

Maximum Mark: 75

Published



May/June 2017
















May/June 2017








SOI Seen or implied


Penalties




May/June 2017



1(i) Coefficient of x = 80(x) B2 Correct value must be selected for both marks. SR +80 seen in an expansion gets B1 or −80 gets B1 if selected.

Total: 2

1(ii) Coefficient of 1

x = −40 1

x

B2 Correct value soi in (ii), if powers unsimplified only allow if selected. SR

+40 soi in (ii) gets B1.

Coefficient of x = (1 × their 80) + (3 ×their − 40) = −40(x) M1 A1 Links the appropriate 2 terms only for M1.

Total: 4

2(i) Gradient = 1.5 Gradient of perpendicular = −⅔ B1

Equation of AB is ( )6 2x− = − +y ⅔ Or 3 2 14y x+ = oe

M1 A1 Correct use of straight line equation with a changed gradient and (− 2, 6), the (− (− 2)) must be resolved for the A1 ISW.

Using = +y mx c gets A1 as soon as c is evaluated.

Total: 3

2(ii) Simultaneous equations → Midpoint (1, 4) M1 Attempt at solution of simultaneous equations as far as x =, or y =.

Use of midpoint or vectors → B (4, 2) M1A1 Any valid method leading to x, or to y.

Total: 3


May/June 2017



3(i) LHS =

21 sc

− c

M1

Eliminates tan by replacing with sincos

leading to a function of sin and/or cos

only.

= ( )21

1 ²−−

ss

M1 Uses s² + c² = 1 leading to a function of sin only.

= ( )( )( )( )1 11 1− −− +

s ss s

= 1 sin1 sin

θθ

−+

A1 AG. Must show use of factors for A1.

Total: 3

3(ii) Uses part (i) → 2 – 2s = 1 + s

→ s = ⅓ M1 Uses part (i) to obtain s = k

θ = 19.5º or 160.5° A1A1 FT FT from error in 19.5°Allow 0.340ᶜ (0.3398ᶜ) & 2.80(2) or 0.108πᶜ & 0.892πᶜ for A1 only. Extra answers in the range lose the second A1 if gained for 160.5°.

Total: 3

4(i) (AB) = 2rsinθ (or 2 2 2θ−r cos or

2

sin 2

θπ θ −

rsin)

B1Allow unsimplifed throughout eg r + r, 2

2θ etc

(Arc AB) = 2rθ B1

(P =) 2r + 2rθ + 2rsinθ (or

22 2 2 or sin

2

θθπ θ

− −

rsinr cos ) B1

Total: 3


May/June 2017



4(ii) Area sector AOB = ( ½ r² 2θ) 25 or1 3.1

6π

B1 Use of segment formula gives 2.26 B1B1

Area triangle AOB = (½×2rsinθ×rcosθ or ½ × r2 sin2θ) 25 3 or1 0.8

4

B1

Area rectangle ABCD = (r × 2rsinθ) 25 B1

(Area =) Either 25 – (25π/6 – 25√3/4) or 22.7 B1 Correct final answer gets B4.

Total: 4

5(i) Crosses x-axis at (6, 0) B1 6=x is sufficient.

dd

yx

= (0 +) −12 (2 – x)−2 × (−1) B2,1,0 −1 for each incorrect term of the three or addition of + C.

Tangent ( )6y x= −¾ or 4 3 18y x= − M1 A1 Must use dy/dx, x= their 6 but not x = 0 (which gives m = 3), and correct form of line equation.

Using = +y mx c gets A1 as soon as c is evaluated.

Total: 5

5(ii) If x = 4, dy/dx = 3

d 3 0.04 0.12

dyt= × =

M1 A1FTM1 for (“their m” from d

dyx

and x = 4) × 0.04.

Be aware: use of x = 0 gives the correct answer but gets M0.

Total: 2


May/June 2017



6 Vol = π ( )2 165 d d

²π− −∫ ∫x x x

x

M1* Use of volume formula at least once, condone omission of π and limits and dx .

DM1 Subtracting volumes somewhere must be after squaring.

( )5 ²d∫ − x x =

( )5 – ³3

x ÷ −1

B1 B1 B1 Without ÷ (−1). B1 for ÷( −1)

(or 25x – 10x²/2 +⅓x³) (B2,1,0) −1 for each incorrect term

16 d²

∫ xx

= −16x

B1

Use of limits 1 and 4 in an integrated expression and subtracted. DM1 Must have used“y²” at least once. Need to see values substituted.

→ 9π or 28.3 A1

Total: 7

7(a) (Sn =) ( )32 1 8

2 + −

n n and 20000 M1 M1 correct formula used with d from 16 24+ =d

A1 A1 for correct expression linked to 20000.

→ n² + 3n – 5000 (<,=,> 0) DM1 Simplification to a three term quadratic.

→ (n = 69.2) → 70 terms needed. A1 Condone use of 20001 throughout. Correct answer from trial and improvement gets 4/4.

Total: 4


May/June 2017



7(b) a = 6, 18

1=

−a

r → r = ⅔

M1A1 Correct S∞ formula used to find r.

New progression a = 36, r = 4

9 oe

M1 Obtain new values for a and r by any valid method.

New S∞ = 36

419

− → 64.8 or 324

5 oe

A1 (Be aware that r =−⅔ leads to 64.8 but can only score M marks)

Total: 4

8(i) Uses scalar product correctly: 3 × 6 + 2 × 6 + (−4) × 3 = 18

M1 Use of dot product with oruuur uuurOA AO & or

uuur uuurOB BO only.

uuuurOA = 29 ,

uuuurOB = 9 M1 Correct method for any one of

uuurOA ,

uuurAO ,

uuurOB or .

uuurBO

29 × 9 × cos AOB = 18 M1 All linked correctly.

→ AOB = 68.2° or 1.19ᶜ A1 Multiples of π are acceptable (e.g. 0.379πᶜ)

Total: 4

8(ii) =uuurAB 3i + 4j +(3+2p)k *M1 For use of −

uuur uuurOB OA , allow with p = 2

Comparing “j” DM1 For comparing, uuurOC must contain p & q.

Can be implied by =uuurAB 2

uuurOC .

→ p = 2½ and q = 4 A1 A1 Accuracy marks only available if uuurAB is correct.

Total: 4


May/June 2017



9(i) ½d 4 2

d−= −

y xx

B1 Accept unsimplified.

= 0 when x = 2

x = 4, y = 8 B1B1

Total: 3

9(ii) 32d² 2

d ²−

= −y x

x

B1FT FT providing –ve power of x

d² 1 d ² 4

= −

yx

→ Maximum B1

Correct d²d ²

yx

and x=4 in (i) are required.

Followed by“< 0 or negative” is sufficient” but d²d ²

yx

must be correct if

evaluated.

Total: 2

9(iii) EITHER: Recognises a quadratic in x

(M1 Eg x =u → 2 ² 8 6 0− + =u u

1 and 3 as solutions to this equation A1

→ x = 9, x = 1. A1)


May/June 2017



OR: Rearranges then squares

(M1 x needs to be isolated before squaring both sides.

→ 2 10 9 0− + =x x oe A1

→ x = 9, x = 1. A1) Both correct by trial and improvement gets 3/3

Total: 3

9(iv) k > 8 B1

Total: 1

10(i) 3tan 1

2

x = −2 → tan 12

x = −⅔ M1

Attempt to obtain tan 12

=

x k from 3tan 12

x + 2 = 0

½x = −0.6 (‒ 0.588) → x = ‒1.2 M1 A1 1tan− k . Seeing ½x = ‒ 33.69° or x= ‒ 67.4° implies M1M1.

Extra answers between 1.57 &1 .57− lose the A1. Multiples of π are acceptable ( eg – 0.374π)

Total: 3

10(ii) 2

3y + = tan 1

2

x M1 Attempt at isolating tan(½x)

→ f−1(x) = 1 22tan

3x− +

M1 A1 Inverse tan followed by × 2. Must be function of x for A1.

−5,1 B1 B1 Values stated B1 for -5, B1 for 1.

Total: 5


May/June 2017



10(iii) B1 B1 B1 A tan graph through the first, third and fourth quadrants. (B1) An invtan graph through the first, second and third quadrants.(B1) Two curves clearly symmetrical about y = x either by sight or by exact end points. Line not required. Approximately in correct domain and range. (Not intersecting.) (B1) Labels on axes not required.

Total: 3






MARK SCHEME

Maximum Mark: 75

Published



May/June 2017




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 FT 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.




• 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.


May/June 2017





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

allowed)



SOI Seen or implied


Penalties


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.


May/June 2017



1 7C1 ( )6 2 a x× × , 7C2 ( ) 252 a x× × B1 B1 SOI Can be part of expansion. Condone ax2 only if followed by a2. ALT [ ] ( ) ( ) 2772 1 / 2 7 1 / 2 7 2 / 2ax C a x C a x+ → =

6

57 2 2

321 2a ×= =

×

B1 Ignore extra soln a = 0. Allow a = 0.667. Do not allow an extra x in the answer

Total: 3


2(i) 2 3 21

r rSr

− +=

−

M1

( )( ) ( )( )1 2 1 2 2

1 1r r r r

S rr r

− − − − −= = = −

− −OR

( )( )1 22

1r r

rr

− −= −

− OE

A1 AG Factors must be shown. Expressions requiring minus sign taken out must be shown

Total: 2

2(ii) Single range 1 3S< < or (1, 3) B2 Accept 1 2 3r< − < . Correct range but with S = 2 omitted scores SR B1 1 3S scores SR B1. [S > 1 and S < 3] scores SR B1.

Total: 2


May/June 2017



3 EITHER Elim y to form 3-term quad eqn in 1/3x (or u or y or even x) (M1

Expect ( )2/3 1/3 2 0x x− − = or ( )2 2 0u u− − = etc.

1/3x (or u or y or x) = 2, 1− *A1 Both required. But x = 2,‒1 and not then cubed or cube rooted scores A0

Cube solution(s) DM1 Expect x = 8, ‒ 1. Both required

(8, 3), (‒1,0) A1)

OR Elim x to form quadratic equation in y (M1

Expect ( )21 1y y+ = −

2 3 0y y− = *A1

Attempt solution DM1 Expect y = 3, 0

(8, 3), (‒1,0) A1)

Total: 4


May/June 2017



4(i)

( )5 5 0

4 1 33 3 6

OB OA AB − = = − = − −

uuur uuur uuur

B1

5 0 511 3 23

3 6 1OP

= + = −

uuur

M1 A1 If OPuuur

not scored in (i) can score SR B1 if seen correct in (ii). Other

equivalent methods possible

Total: 3

4(ii) Distance OP = 2 2 25 2 1+ + = 30 or 5.48 B1 FT FT on their OPuuur

from (i)

Total: 1

4(iii) Attempt . .AB OPuuur uuur

Can score as part of ( )( ). cosAB OP AB OP θ=uuur uuur

Rare ALT: Pythagoras 2 2 2

5 30OP AP OA+ = + =uuur uuur uuur

M1 Allow any combination of . AB POuuur uuur

etc. and also if APuuur

or used PBuuur

instead of ABuuur

giving 2‒2 = 0 & 4‒4 = 0 respectively. Allow notation × instead of . .

(0 + 6 ‒ 6) = 0 hence perpendicular. (Accept 90º)

A1 FT If result not zero then 'Not perpendicular' can score A1FT if value is 'correct' for their values of ,AB OP

uuur uuur etc. from (i).

Total: 2


May/June 2017



5(i) 2sin cos 2sinsin cos cos

θ θ θθ θ θ+

=+

M1 Replace tan by sin / cosθ θ θ

2 2 2 22sin cos cos 2sin 2sin cos 2c sθ θ θ θ θ θ+ = + ⇒ = M1 A1 Mult by c(s + c) or making this a common denom.. For A1 simplification to AG without error or omission must be seen.

Total: 3

5(ii) 2 2 2tan 1 / 2 or cos 2 / 3 or sin 1 / 3θ θ θ= = = B1 Use 2 2tan / c or c s 1sθ = + = and simplify to one of these results

35.3 or 1 44.7θ = ° ° B1 B1 FT FT for 180 ‒ other solution. SR B1 for radians 0.615, 2.53 (0.196π, 0.804π) Extra solutions in range amongst solutions of which 2 are correct gets B1B0

Total: 3


May/June 2017



6 Gradient of normal is – 1/3 → gradient of tangent is 3 SOI B1 B1 FT FT from their gradient of normal.

dy/dx = 2x – 5 = 3 M1 Differentiate and set = their 3 (numerical).

x = 4 *A1

Sub x = 4 into line → y = 7 & sub their (4, 7) into curve DM1 OR sub x = 4 into curve → y = k ‒ 4 and sub their(4, k ‒ 4) into line OR other valid methods deriving a linear equation in k (e.g. equating curve with either normal or tangent and sub x = 4).

k = 11 A1

Total: 6


May/June 2017



7(i) ( )sin 8 /10 0.927 3ABC ABC= → = B1 Or cos = 6/10 or tan = 8/6. Accept 0.295π.

Total: 1

7(ii) ( )6 Pythagoras 8 6 48.0AB BCD= → ∆ = × = M1A1 OR 8×10sin0.6435 or ½×10×10sin((2)×0.927)=48. 24or 40or80 gets M1A0

Area sector ( )2½ 10 2 0.9273BCD their= × × × *M1 Expect 92.7(3). 46.4 gets M1

Area segment = 92.7(3) – 48 *A1 Expect 44.7(3). Might not appear until final calculation.

Area semi-circle ‒ segment = ( )2½ 8 92.7 48theirπ× × − − DM1 Dep. on previous M1A1 OR ( )2 28 ½ 8 44.7 .theirπ π× − × × +

Shaded area = 55.8 – 56.0 A1

Total: 6


May/June 2017



8(i) ( ) ( )1 / 1 2 b a− + = M1 OR Equation of AP is ( )1 2 1 2 3y x y x− = + → = +

2 3b a= + CAO A1 Sub , 2 3x a y b b a= = → = +

Total: 2

8(ii) 2 2 211 2 125AB = + = oe B1 Accept AB = √125

( ) ( )2 21 1 125a b+ + − = B1 FT FT on their 125.

( ) ( )2 21 2 2 125a a+ + + = M1 Sub from part (i) → quadratic eqn in a (or possibly in b → 2 2 99 0b b− − = )

( )( ) ( )( )25 2 24 0 eg 4 6 0a a a a+ − = → − + = M1 Simplify and attempt to solve

a = 4 or −6 A1

b = 11 or −9 A1 OR (4, 11), (‒6, ‒9) If A0A0, SR1 for either (4, 11) or (‒6, ‒9)

Total: 6


May/June 2017



9(i) ( )23 1 5x − + B1B1B1 First 2 marks dependent on correct ( )2ax b+ form. OR 3, 1, 5a b c= = − = e.g. from equating coefs

Total: 3

9(ii) Smallest value of p is 1/3 seen. (Independent of (i)) B1 Allow 1 / 3 or 1 / 3 p p = or 1/3 seen. But not in terms of x.

Total: 1

9(iii) ( ) ( )23 1 5 3 1 5y x x y= − + ⇒ − = ± − B1 FT OR ( )

2 21 1 9 5 5 / 93 3

y x y x = − + ⇒ − = −

(Fresh start)

( ) 5x y= ± − +⅓ ⅓ OE B1 FT Both starts require 2 operations for each mark. FT for their values from part (i)

( )1f 5x x− = − +⅓ ⅓ OE domain is 5x their≥ B1B1 FT Must be a function of x and ± removed. Domain must be in terms of x. Note: 5y − expressed as 5y − scores Max B0B0B0B1 [See below for general instructions for different starts]

Total: 4

9(iv) 5q < CAO B1

Total: 1

Alt 9(iii) For start ( − ) + or ( − ) + (a≠ 0) ft for their a, b, c For start ( − ) + ft but award only B1 for 3 correct operations For start ( − ) + ft but award B1 for first2 operations correct and B1 for the next 3 operations correct


May/June 2017



10(a)(i) ( ) ( )Attempt to integrate 1 d V y yπ= ∫ + M1 Use of h in integral e.g. ( ) 21 ½h h h∫ + = + is M0. Use of 2dy x∫ is M0

( )

2

2y yπ

= +

A1

2

2h hπ

= +

A1 AG. Must be from clear use of limits 0→ h somewhere.

Total: 3

10(ii) ( )1/21 d y y∫ + ALT 6 ‒ ( )2 1 dx x∫ − M1 Correct variable and attempt to integrate

( )3/21y +⅔ oe ALT 6 ‒ ( 3x x−⅓ ) CAO *A1 Result of integration must be shown

[ ]8 1−⅔ ALT 8 16 [ 1 1

3 3 − − − −

] DM1 Calculation seen with limits 0→3 for y. For ALT, limits are 1→2 and

rectangle.

14/3 ALT 6 ‒ 4/3 = 14/3 A1 16/3 from 8×⅔ gets DM1A0 provided work is correct up to applying limits.

Total: 4


May/June 2017



10(b) Clear attempt to differentiate wrt h M1 Expect ( )d 1

dV hh

π= + . Allow h + 1. Allow h.

Derivative = 4π SOI *A1

2derivativetheir

. Can be in terms of h DM1

2 1 or or 0.1594 2 π π

A1

Total: 4


May/June 2017



11(i) ( ) ( ) ] [ ( )1/2f ' [ 4 1 ½ 4] x x c= + ÷ ÷ + B1 B1 Expect ( ) ( )1/2½ 4 1 x c+ +

( ) 3 3f ' 2 0 0

2 2c c= ⇒ + = ⇒ = − (Sufficient)

B1 FT Expect ( )1/2 3½ 4 1

2x + − . FT on their ( ) ( )1/2f 4 1x k x c= + +′ . (i.e. c = 3 )k−

Total: 3

11(ii) ( )f 0 1″ = SOI B1

( )f ' 0 1 / 2 1½ 1= − = − SOI B1 FT Substitute x = 0 into their f ′(x) but must not involve c otherwise B0B0

f(0) = – 3 B1 FT FT for 3 terms in AP. FT for 3rd B1 dep on 1st B1. Award marks for the AP method only.

Total: 3

11(iii) ( ) ( ) ] [ ( )3/2f ½ 4 1 3 / 2 4 1½ x x x k = + ÷ ÷ − + B1 FT B1 FT

Expect (1/12) ( ) ( )3/24 1 1½ x x k+ − + . FT from their f ′(x) but c numerical.

3 1 /12 0 37 /12k k− = − + ⇒ = − CAO M1A1 Sub ( )0, f 0x y their= = into their f(x). Dep on cx & k present (c numerical)

Minimum value = f(2) = 27 37 23 3 or 3.83

12 12 6− − = − −

A1

Total: 5




Cambridge International Examinations Cambridge International Advanced Subsidiary Level


MARK SCHEME

Maximum Mark: 50

Published


9709/21 Cambridge International AS Level – Mark Scheme PUBLISHED

May/June 2017

















May/June 2017






allowed)



SOI Seen or implied


Penalties





May/June 2017



1 Take logarithms of both sides and apply power law to both sides

M1Allow log5

4log3=y for M1 A1

Rearrange to the form ln5

4ln3=y x or equivalent

A1

Obtain 0.366=m A1

Total: 3

2 State or imply non-modulus inequality ( )24 x− ⩽ ( )23 2x− or corresponding equation, pair of linear equations or linear inequalities

M1

Attempt solution of 3-term quadratic equation, of two linear equations or of two linear inequalities

M1

Obtain critical values 1− and 73 A1

SR Allow B1 for x ⩽ –1 only or x ⩾ 73

only if

first M1 is not given

State answer x ⩽ –1, x ⩾ 73 A1

Do not accept 73

⩽ x ⩽–1 or –1⩾ x ⩾ 73

for A1

Total: 4


May/June 2017



3 Integrate to obtain form 12 3+xke where k is constant not equal to 4 M1

Obtain correct 12 38 +xe A1 Allow unsimplified for A1

Obtain 12 3 38 8 835+ − =ae e or equivalent A1

Carry out correct process to find a from equation of form 12 3+ =ake c M1

Obtain 3.65 A1 If 3.65 seen with no actual attempt at integration, award B1 if it is thought that trial and improvement with calculator has been used.

Total: 5

4(i) Use iteration correctly at least once M1

Obtain final answer 2.08 A1

Show sufficient iterations to 4 dp to justify answer or show sign change in interval ( )2.075, 2.085

A1

Total: 3

4(ii) State or clearly imply equation

( )

2

22 9

1+ +

=+

x xxx

or same equation using α B1

Carry out relevant simplification M1

Obtain 3 9 A1

3


May/June 2017



5(i) State 3=R B1 Allow marks for (i) if seen in (ii)

Use appropriate trigonometric formula to find α M1

Obtain 48.19 with no errors seen A1

Total: 3

5(ii) Carry out evaluation of ( )1 13cos 70.528...− = M1

M1 for 1 1cos− R

Obtain correct answer 118.7 A1

Carry out correct method to find second answer M1

Obtain 337.7 and no others between 0 and 360 A1

Total: 4

6(i) State or imply correct y-values 0, 1 26 6tan , tanπ π B1 Some candidates have their calculator in

degree mode when working out πtan6

etc. this

gives 0.00915 and 0.0183. Allow B1.

Use correct formula, or equivalent, with 112h π= and y–values M1 Must be convinced they have considered 3

values for y for M1

Obtain 0.378 A1

Total: 3


May/June 2017



6(ii) State or imply ( )2sec 2 1 dx xπ −∫ B1

Integrate to obtain 1 2tan 2k x k x+ , any non-zero constants including π or not M1

Obtain 12 tan 2x x− or ( )1

2 tan 2x xπ − A1

Obtain ( )1 12 63π π− or equivalent A1

Total: 4

7(i) Differentiate x and y and form ddyx M1

Obtain

3 2

24 6 8 12

3 6t t t

t− + −

+

A1 First 2 marks may be implied by an attempt at division

Carry out division at least as far as kt or equivalent M1 For M1, it must be division by a quadratic factor. Allow attempt at factorisation with same conditions as for division

Obtain 43 t A1

Obtain 43 2t − with complete division shown and no errors seen A1

Total: 5


May/June 2017



7(ii) State or imply gradient of straight line is 12 B1

Allow B1 if 1 92 2

y x= + is seen

Attempt value of t from their ddyx = their negative reciprocal of gradient of line M1

Obtain 0t = and hence ( )1,5 A1

Total: 3

8(i) Apply product rule to find first derivative *M1

Obtain ( )166 1n 3x x x+ or equivalent A1 Allow unsimplified for A1

Identify 6x = at P B1

Substitute their value of x at P into attempt at first derivative DM1 dep *M

Obtain 18 A1

Total: 5


May/June 2017



8(ii) Equate their first derivative to zero and attempt solution of equation of form ( )1

61n 0k x m+ = *M1

Obtain x–coordinate of form 21

aa e DM1 dep *M

Obtain 126ex −= or exact equivalent A1

Substitute exact x–value in the form 21

aa e and attempt simplification to remove ln M1

Obtain 154e−− or exact equivalent A1

Total: 5






MARK SCHEME

Maximum Mark: 50

Published



May/June 2017

















May/June 2017






allowed)



SOI Seen or implied


Penalties





May/June 2017



1 State or imply non-modulus equation ( ) ( )2 22 5x a x a+ = − or pair of linear equations B1 SR B1 for 6x a=

Attempt solution of quadratic equation or of pair of linear equations M1Allow M1 if 4

3 and 6 seen

Obtain, as final answers, 6a and 43 a A1

Total: 3

2 Apply logarithms to both sides and apply power law *M1

Obtain ( )4 log3 2 log5x x+ = or equivalent A1

Solve linear equation for x DM1 dep *M

Obtain 2.07 A1 Allow greater accuracy

Total: 4

3(i) Draw sketch of 3y x= *B1 May be implied by part graph in first quadrant

Draw straight line with negative gradient crossing positive y-axis and indicate one intersection

DB1 dep *B

Total: 2

3(ii) Use iterative formula correctly at least once M1


Show sufficient iterations to justify 4 sf or show sign change in interval ( )1.9255,1.9265

A1

Total: 3


May/June 2017



4 Use quotient rule (or product rule) to find first derivative M1

Obtain

( )

4 4

28 e 10e

2 3

x xxx+

+ or equivalent

A1

Substitute 0x = to obtain gradient 109 A1

Form equation of tangent through ( )130, with numerical gradient M1

Obtain 10 9 3 0x y− + = A1

Total: 5


May/June 2017



5 State or imply ln ln 2 ln y K x a= − B1

EITHER:

Obtain 0.525− as gradient of line (M1

Equate their 2ln a− to their gradient and solve for a M1 Allow 2ln a = their gradient for M1

Obtain 1.3a = A1

Substitute to find value of K M1

Obtain 8.4K = A1)

OR:

Obtain two equations using coordinates correctly (M1

Solve these equations to obtain 2 ln a or equivalent M1

Obtain 1.3a = A1


Obtain 8.4K = A1)

Total: 6


May/June 2017



6(i) Evaluate expression when 2x = − M1

Obtain 0 with all necessary detail present A1 Use of ( ) ( )( )2f 2x x ax bx c= + + + to find a, b

and c, allow M1 A0 Use of ( ) ( )( )2f 2x x ax bx c d= + + + + to find a, b and c, and show d = 0, allow M1 A1

Carry out division, or equivalent, at least as far as 2x and x terms in quotient M1

Obtain 26 35x x+ − A1

Obtain factorised expression ( )( )( )2 2 5 3 7x x x+ + − A1

Total: 5

6(ii) State or imply substitution 1x

y= or equivalent

M1

Obtain 31 22 5 7, , − − A1

Total: 2

7(a) Obtain ( )22cos cos 3 dθ θ θ− −∫ B1

Attempt use of identity to obtain integrand involving cos 2θ and cosθ M1

Integrate to obtain form 1 2 3sin 2 sink k kθ θ θ+ + for non-zero constants M1

Obtain 12 sin 2 sin 2 cθ θ θ− − + A1

Total: 4


May/June 2017



7(b)(i) Integrate to obtain form ( ) ( )1 2ln 2 1 lnk x k x+ + or ( ) ( )1 2ln 2 1 ln 2k x k x+ + M1

Obtain ( ) 122ln 2 1 lnx x+ + or ( ) ( )1

22ln 2 1 ln 2x x+ + A1

Total: 2

7(b)(ii) Use relevant logarithm power law for expression obtained from application of limits M1

Use relevant logarithm addition / subtraction laws M1

Obtain ln18 A1

Total: 3

8(i) Obtain dd 2sin 2xt t= B1

Obtain d 2 2d 6sin cos 9cos siny

t t t t t= − B1

Use d d d/

d d dy y xx t t= for their first derivatives

M1

Use identity sin 2 2sin cost t t= B1

Simplify to obtain 3 92 4sin cost t− with necessary detail present A1

Total: 5


May/June 2017



8(ii) Equate ddyx to zero and obtain tan t k= M1

Obtain 32tan t = or equivalent A1

Substitute value of t to obtain coordinates ( )2.38, 2.66 A1

Total: 3

8(iii) Identify 14t π= B1

Substitute to obtain exact value for gradient of the normal M1

Obtain gradient 4

3 2 , 83 2

or similarly simplified exact equivalent A1

Total: 3






MARK SCHEME

Maximum Mark: 50

Published



May/June 2017

















May/June 2017






allowed)



SOI Seen or implied


Penalties





May/June 2017



1 State or imply non-modulus equation ( ) ( )2 22 5x a x a+ = − or pair of linear equations B1 SR B1 for 6x a=

Attempt solution of quadratic equation or of pair of linear equations M1Allow M1 if 4

3 and 6 seen

Obtain, as final answers, 6a and 43 a A1

Total: 3

2 Apply logarithms to both sides and apply power law *M1

Obtain ( )4 log3 2 log5x x+ = or equivalent A1

Solve linear equation for x DM1 dep *M

Obtain 2.07 A1 Allow greater accuracy

Total: 4

3(i) Draw sketch of 3y x= *B1 May be implied by part graph in first quadrant

Draw straight line with negative gradient crossing positive y-axis and indicate one intersection

DB1 dep *B

Total: 2

3(ii) Use iterative formula correctly at least once M1


Show sufficient iterations to justify 4 sf or show sign change in interval ( )1.9255,1.9265

A1

Total: 3


May/June 2017



4 Use quotient rule (or product rule) to find first derivative M1

Obtain

( )

4 4

28 e 10e

2 3

x xxx+

+ or equivalent

A1

Substitute 0x = to obtain gradient 109 A1

Form equation of tangent through ( )130, with numerical gradient M1

Obtain 10 9 3 0x y− + = A1

Total: 5


May/June 2017



5 State or imply ln ln 2 ln y K x a= − B1

EITHER:

Obtain 0.525− as gradient of line (M1

Equate their 2ln a− to their gradient and solve for a M1 Allow 2ln a = their gradient for M1

Obtain 1.3a = A1


Obtain 8.4K = A1)

OR:

Obtain two equations using coordinates correctly (M1

Solve these equations to obtain 2 ln a or equivalent M1

Obtain 1.3a = A1


Obtain 8.4K = A1)

Total: 6


May/June 2017



6(i) Evaluate expression when 2x = − M1

Obtain 0 with all necessary detail present A1 Use of ( ) ( )( )2f 2x x ax bx c= + + + to find a, b

and c, allow M1 A0 Use of ( ) ( )( )2f 2x x ax bx c d= + + + + to find a, b and c, and show d = 0, allow M1 A1

Carry out division, or equivalent, at least as far as 2x and x terms in quotient M1

Obtain 26 35x x+ − A1

Obtain factorised expression ( )( )( )2 2 5 3 7x x x+ + − A1

Total: 5

6(ii) State or imply substitution 1x

y= or equivalent

M1

Obtain 31 22 5 7, , − − A1

Total: 2

7(a) Obtain ( )22cos cos 3 dθ θ θ− −∫ B1

Attempt use of identity to obtain integrand involving cos 2θ and cosθ M1

Integrate to obtain form 1 2 3sin 2 sink k kθ θ θ+ + for non-zero constants M1

Obtain 12 sin 2 sin 2 cθ θ θ− − + A1

Total: 4


May/June 2017



7(b)(i) Integrate to obtain form ( ) ( )1 2ln 2 1 lnk x k x+ + or ( ) ( )1 2ln 2 1 ln 2k x k x+ + M1

Obtain ( ) 122ln 2 1 lnx x+ + or ( ) ( )1

22ln 2 1 ln 2x x+ + A1

Total: 2

7(b)(ii) Use relevant logarithm power law for expression obtained from application of limits M1

Use relevant logarithm addition / subtraction laws M1

Obtain ln18 A1

Total: 3

8(i) Obtain dd 2sin 2xt t= B1

Obtain d 2 2d 6sin cos 9cos siny

t t t t t= − B1

Use d d d/

d d dy y xx t t= for their first derivatives

M1

Use identity sin 2 2sin cost t t= B1

Simplify to obtain 3 92 4sin cost t− with necessary detail present A1

Total: 5


May/June 2017



8(ii) Equate ddyx to zero and obtain tan t k= M1

Obtain 32tan t = or equivalent A1

Substitute value of t to obtain coordinates ( )2.38, 2.66 A1

Total: 3

8(iii) Identify 14t π= B1

Substitute to obtain exact value for gradient of the normal M1

Obtain gradient 4

3 2 , 83 2

or similarly simplified exact equivalent A1

Total: 3




Cambridge International Examinations Cambridge International Advanced Level


MARK SCHEME

Maximum Mark: 75

Published


9709/31 Cambridge International A Level – Mark Scheme PUBLISHED

May/June 2017

















May/June 2017






allowed)



SOI Seen or implied


Penalties





May/June 2017


Question Answer Marks

1 EITHER: State or imply non-modular inequality 2 2(2 1) (3( 2))x x+ < − , or corresponding quadratic equation, or pair of linear equations (2 1) 3( 2)x x+ = ± −

(B1

Make reasonable solution attempt at a 3-term quadratic e.g. 25 40 35 0x x− + = or solve two linear equations for x

M1

Obtain critical values x = 1 and x = 7 A1

State final answer x < 1 and x > 7 A1)

OR: Obtain critical value x = 7 from a graphical method, or by inspection, or by solving a linear equation or inequality

(B1

Obtain critical value x = 1 similarly B2

State final answer x < 1 and x > 7 B1)

Total: 4

2 EITHER: State a correct unsimplified version of the x or 2x or 3x term in the expansion of

13(1 6 )x −+

(M1

State correct first two terms 1 2x− A1

Obtain term 28x A1

Obtain term 3112

3 x− 3137

3x

in final answer A1)

OR: Differentiate expression and evaluate f (0) and f (0)′ , where

43f ( ) (1 6 )x k x −′ = + (M1

Obtain correct first two terms 1 2x− A1

Obtain term 28x A1

Obtain term 31123 x− in final answer A1)

Total: 4


May/June 2017



3(i) Remove logarithms correctly and obtain 1ex y

y−

= B1

Obtain the given answer e

1 e

x

xy

−

−=

+ following full working

B1

Total: 2

3(ii) State integral ln(1 e )xk −+ where k = ± 1 *M1

State correct integral ln(1 e )x−− + A1

Use limits correctly DM1

Obtain the given answer 2e

lne 1+

following full working A1

Total: 4

4(i) Use chain rule to differentiate x d sin

d cos

x θ

θ θ= −

M1

State 2d

3 secd

yθ

θ= −

B1

Use d d d

d d dy y xx θ θ

= ÷ M1

Obtain correct d

d

y

x in any form e.g.

23 sec

tan

θ

θ

−

−

A1

Obtain

2d tan 2

d tan

y

x

θ

θ

−= , or equivalent

A1

Total: 5

4(ii) Equate gradient to −1 and obtain an equation in tanθ M1

Solve a 3 term quadratic ( )2tan tan 2 0θ θ+ − = in tanθ M1

Obtain

4

πθ = and 3

14

yπ

= − only A1

Total: 3


May/June 2017



5(i) Use correct sector formula at least once and form an equation in r and x M1

Obtain a correct equation in any form A1

Rearrange in the given form A1

Total: 3

5(ii) Calculate values of a relevant expression or expressions at x = 1 and x = 1.5 M1

Complete the argument correctly with correct calculated values A1

Total: 2

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


Show sufficient iterations to 5 d.p. to justify 1.374 to 3 d.p., or show there is a sign change in the interval (1.3745, 1.3755)

A1

Total: 3

6(i) State or obtain coordinates (1, 2, 1) for the mid-point of AB B1

Verify that the midpoint lies on m B1

State or imply a correct normal vector to the plane, e.g. 2 2+ −i j k B1

State or imply a direction vector for the segment AB, e.g. 4 4 2− − +i j k B1

Confirm that m is perpendicular to AB B1

Total: 5

6(ii) State or imply that the perpendicular distance of m from the origin is 53 , or

unsimplified equivalent

B1

State or imply that n has an equation of the form 2 2x y z k+ − = B1

Obtain answer 2 2 2x y z+ − = B1

Total: 3


May/June 2017



7(i) State that u – 2w = – 7 – i B1

EITHER:

Multiply numerator and denominator of uw

by 3 – 4i, or equivalent (M1

Simplify the numerator to 25 + 25i or denominator to 25 A1

Obtain final answer 1 + i A1)

OR: Obtain two equations in x and y and solve for x or for y (M1

Obtain x = 1 or y = 1 A1

Obtain final answer 1 + i A1)

Total: 4

7(ii) Find the argument of u

w

M1

Obtain the given answer A1

Total: 2

7(iii) State that OB and CA are parallel B1

State that CA = 2OB, or equivalent B1

Total: 2

8(i) Use sin( )A B− formula and obtain an expression in terms of sin x and cos x M1

Collect terms and reach 3sin 2cosx x− , or equivalent A1

Obtain 7R = A1

Use trig formula to find α M1

Obtain α = 49.11° with no errors seen A1

Total: 5


May/June 2017



8(ii) Evaluate 1sin (1/ 7)− to at least 1 d.p. (22.21° to 2 d.p.) B1 FT

Use a correct method to find a value of x in the interval 0° < x < 180° M1

Obtain answer 71.3° A1

[ignore answers outside given range.]

Total: 3

9(i) Carry out a relevant method to obtain A and B such that 1

(2 3) 2 3A B

x x x x≡ +

+ +, or

equivalent

M1

Obtain 13A = and 2

3B = − , or equivalent A1

Total: 2

9(ii) Separate variables and integrate one side B1

Obtain term ln y B1

Integrate and obtain terms 1 13 3ln ln(2 3)x x− + , or equivalent B2 FT

Use x = 1 and y = 1 to evaluate a constant, or as limits, in a solution containing ln , ln , ln(2 3)a y b x c x +

M1

Obtain correct solution in any form, e.g. 1 1 13 3 3ln ln ln(2 3) ln5y x x= − + + A1

Obtain answer y = 1.29 (3s.f. only) A1

Total: 7

10(i) State or imply d sin du x x= − B1

Using correct double angle formula, express the integral in terms of u and du M1

Obtain integrand 2 2(2 1)u± − A1

Change limits and obtain correct integral

12

12 2(2 1) du u−∫ with no errors seen

A1

Substitute limits in an integral of the form 5 3au bu cu+ + M1

Obtain answer 115 (7 4 2)− , or exact simplified equivalent A1

Total: 6


May/June 2017



10(ii) Use product rule and chain rule at least once M1

Obtain correct derivative in any form A1

Equate derivative to zero and use trig formulae to obtain an equation incos and sinx x

M1

Use correct methods to obtain an equation in cos x or sin x only M1

Obtain 210cos 9x = or 210sin 1x = , or equivalent A1

Obtain answer 0.32 A1

Total: 6






MARK SCHEME

Maximum Mark: 75

Published



May/June 2017


Marks are of the following three types:















May/June 2017






allowed)



SOI Seen or implied


Penalties





May/June 2017



1 Use law of the logarithm of a power or a quotient M1

Remove logarithms and obtain a correct equation in x. e.g. 2 21 e+ =x x A1

Obtain answer 0.763 and no other A1

Total: 3

2 EITHER:

State or imply non-modular inequality 2 2( 3) (3 4)− < −x x , or corresponding equation

(B1

Make reasonable attempt at solving a three term quadratic M1

Obtain critical value 74=x A1

State final answer 74>x only A1)

OR1: State the relevant critical inequality 3 3 4− < −x x , or corresponding equation

(B1

Solve for x M1

Obtain critical value 74=x A1


OR2: Make recognizable sketches of 3= −y x and 3 4= −y x on a single diagram

(B1

Find x-coordinate of the intersection M1

Obtain 74=x A1


Total: 4


May/June 2017



3(i) Use correct formulae to express the equation in terms of cos θ and sin θ M1

Use Pythagoras and express the equation in terms of cos θ only M1

Obtain correct 3-term equation, e.g. 4 22cos cos 2 0θ θ+ − = A1

Total: 3

3(ii) Solve a 3-term quadratic in 2cos θ for cos θ M1

Obtain answer θ = 152.1°only A1

Total: 2

4(i) State d 24

d 2 1= +

−yt t

B1

Use d d d

d d dy y xx t t= ÷

M1

Obtain answer d 8 2

d 2 (2 1)−

=−

y tx t t

, or equivalent e.g. 2 224 2

+−t t t

A1

Total: 3

4(ii) Use correct method to find the gradient of the normal at t = 1 M1

Use a correct method to form an equation for the normal at t = 1 M1

Obtain final answer 3 14 0+ − =x y , or horizontal equivalent A1

Total: 3


May/June 2017



5(i) State 2

d 2d (1 )y yt t= −

+, or equivalent

B1

Separate variables correctly and attempt integration of one side M1

Obtain term ln y , or equivalent A1

Obtain term 2

(1 )t+, or equivalent

A1

Use y = 100 and t = 0 to evaluate a constant, or as limits in an expression containing terms of

the form lna y and 1

b

t+

M1

Obtain correct solution in any form, e.g. 2ln 2 ln100

1y

t= − +

+

A1

Total: 6

5(ii) State that the mass of B approaches 2

100e

, or exact equivalent B1

State or imply that the mass of A tends to zero B1

Total: 2


May/June 2017



6(i) EITHER:

Substitute x = 2 – i (or 2 i= +x ) in the equation and attempt expansions of 2x and

3x

(M1

Equate real and/or imaginary parts to zero M1

Obtain a = – 2 A1

Obtain b = 10 A1)

OR1:

Substitute x = 2 – i in the equation and attempt expansions of 2x and

3x

(M1

Substitute 2 i= +x in the equation and add/subtract the two equations M1

Obtain a = – 2 A1

Obtain b = 10 A1)

OR2:

Factorise to obtain ( )( )( )2 i 2 i− + − − −x x x p ( )( )2 4 5 = − + −

x x x p

(M1

Compare coefficients M1

Obtain a = – 2 A1

Obtain b = 10 A1)

OR3:

Obtain the quadratic factor ( 2 4 5− +x x )

(M1

Use algebraic division to obtain a real linear factor of the form −x p and set the remainder equal to zero

M1

Obtain a = – 2 A1

Obtain b = 10 A1)

OR4: Use 5αβ = and 4α β+ = in 3αβ βγ γα+ + = −

(M1

Solve for γ and use in αβγ = −b and/or α β γ+ + = −a M1

Obtain a = – 2 A1

Obtain b = 10 A1)


May/June 2017



OR5: Factorise as (x–- (2-i))(x2 + ex + g) and compare coefficients to form an equation in a and b

(M1

Equate real and/or imaginary parts to zero M1

Obtain a = – 2 A1

Obtain b = 10 A1)

Total: 4

6(ii) Show a circle with centre 2 i− in a relatively correct position B1

Show a circle with radius 1 and centre not at the origin B1

Show the perpendicular bisector of the line segment joining 0 to – i B1

Shade the correct region B1

Total: 4

7(i) Use quotient or chain rule M1

Obtain given answer correctly A1

Total: 2

7(ii) EITHER: Multiply numerator and denominator of LHS by 1 sinθ+

(M1

Use Pythagoras and express LHS in terms of sec θ and tanθ M1

Complete the proof A1)

OR1: Express RHS in terms of cos θ and sin θ

(M1

Use Pythagoras and express RHS in terms of sin θ M1


OR2: Express LHS in terms of secθ and tanθ

(M1

Multiply numerator and denominator by secθ + tanθ and use Pythagoras M1


Total: 3


May/June 2017



7(iii) Use the identity and obtain integral 2 tan 2secθ θ θ+ − B2

Use correct limits correctly in an integral containing terms a tanθ and b secθ M1

Obtain answer 12 2 4π− A1

Total: 4

8(i) State or imply the form 23 2 5

A Bx Cx x

++

+ +

B1

Use a relevant method to determine a constant M1

Obtain one of the values A = 2, B = 1, C = −3 A1

Obtain a second value A1

Obtain the third value A1

Total: 5

8(ii) Use correct method to find the first two terms of the expansion of 1(3 2)−+x , 13(1 )2−+ x ,

2 1(5 )−+ x or 2 11(1 )5−+ x

[Symbolic coefficients, e.g.1

2−

are not sufficient]

M1

Obtain correct unsimplified expansions up to the term in2x of each partial fraction.

The FT is on A, B, C. from part (i)

A1FT + A1FT

Multiply out up to the term in 2x by Bx +C, where BC ≠ 0 M1

Obtain final answer 213 23725 10 100− +x x , or equivalent A1

Total: 5

9(i) EITHER: Find

uuurAP for a general point P on l with parameter λ , e.g.(8 + 3λ, – 3 − λ, 4 + 2λ) (B1

Equate scalar product of uuurAP and direction vector of l to zero and solve for λ M1

Obtain 52λ = − and foot of perpendicular 3 3 32 2+ +i j k A1

Carry out a complete method for finding the position vector of the reflection of A in l M1

Obtain answer 2 2+ +i j k A1)


May/June 2017



OR: Find

uuurAP for a general point P on l with parameter λ , e.g.(8 + 3λ, – 3 − λ, 4 + 2λ)

(B1

Differentiate

2AP and solve for λ at minimum M1

Obtain 52λ = − and foot of perpendicular 3 3 32 2+ +i j k A1

Carry out a complete method for finding the position vector of the reflection of A in l M1

Obtain answer 2 2+ +i j k A1)

Total: 5

9(ii) EITHER: Use scalar product to obtain an equation in a, b and c, e.g. 3a − b + 2 c = 0

(B1

Form a second relevant equation, e.g. 9a – b + 8c = 0 and solve for one ratio, e.g. a : b M1

Obtain final answer a : b : c = 1 : 1 : – 1 and state plane equation x + y – z = 0 A1)

OR1: Attempt to calculate vector product of two relevant vectors, e.g. (3 2 ) (9 8 )− + × − +i j k i j k

(M1

Obtain two correct components A1

Obtain correct answer, e.g. 6 6 6− − +i j k , and state plane equation 0− − + =x y z A1)

OR2: Using a relevant point and relevant vectors, attempt to form a 2-parameter equation for the plane, e.g. 6 6 (3 2 ) (9 8 )= + + − + + − +s tr i k i j k i j k

(M1

State 3 correct equations in x, y, z, s and t A1

Eliminate s and t and state plane equation 0+ − =x y z , or equivalent A1)

OR3: Using a relevant point and relevant vectors, attempt to form a determinant equation for the

plane, e.g. 3 1 4

3 1 2 09 1 8

− − −− =−

x y z

(M1

Expand a correct determinant and obtain two correct cofactors A1

Obtain answer 6 6 6 0x y z− − + = , or equivalent A1)

Total: 3


May/June 2017



9(iii) EITHER: Using the correct processes, divide the scalar product of

uuurOA and a normal to the plane by the

modulus of the normal or make a recognisable attempt to apply the perpendicular formula

(M1

Obtain a correct expression in any form, e.g. 1 2 4

2 2 2(1 1 ( 1) )

+ −

+ + −, or equivalent

A1 FT

Obtain answer 1 3 , or exact equivalent A1)

OR1: Obtain equation of the parallel plane through A, e.g. x + y – z = – 1 [The f.t. is on the plane found in part (ii).]

(B1 FT

Use correct method to find its distance from the origin M1

Obtain answer 1 3 , or exact equivalent

A1)

OR2: Form equation for the intersection of the perpendicular through A and the plane [FT on their n]

(B1 FT

Solve for λ M1

13

λ =n A1)

Total: 3

10(i) Use correct product rule M1

Obtain correct derivative in any form ( )22 cos 2 2 sin 2′ = −y x x x x A1

Equate to zero and derive the given equation A1

Total: 3

10(ii) Use the iterative formula correctly at least once e.g. 0.5 0.55357 0.53261 0.54070 0.53755→ → → →

M1



A1

Total: 3


May/June 2017



10(iii) Integrate by parts and reach 2 sin 2 sin 2 d+ ∫ax x b x x x *M1

Obtain 21 1sin 2 2 . sin 2 d2 2− ∫x x x x x A1

Complete integration and obtain 21 1 1sin 2 cos2 sin 22 2 4+ −x x x x x , or equivalent A1

Substitute limits x = 0, 14π=x , having integrated twice DM1

Obtain answer 21 ( 8)32 π − , or exact equivalent A1

Total: 5






MARK SCHEME

Maximum Mark: 75

Published



May/June 2017

















May/June 2017






allowed)



SOI Seen or implied


Penalties





May/June 2017



1 Express the LHS in terms of either cos x and sin x or in terms of tan x B1

Use Pythagoras M1

Obtain the given answer A1

Total: 3

2 EITHER: State a correct unsimplified version of the x or 2x term in the expansion of

( ) 3231

−+ x or ( ) 33 2 −+ x

[Symbolic binomial coefficients, e.g. 3

2−

, are not sufficient for M1.]

(M1

State correct first term 127 B1

Obtain term 227− x A1

Obtain term 2881 x A1)

OR: Differentiate expression and evaluate ( ) ( )f 0 and f ' 0 , where ( ) ( ) 4f ' 3 2 −= +x k x

(M1

State correct first term 127 B1

Obtain term 227− x A1

Obtain term 2881 x A1)

Total: 4

3 Rearrange as 23 4 4 0+ − =u u , or 23e 4e 4 0+ − =x x , or equivalent B1

Solve a 3-term quadratic for ex or for u M1

Obtain 23e =x or 2

3=u A1

Obtain answer x = –0.405 and no other A1

Total: 4


May/June 2017



4 Integrate by parts and reach 1 12 2cos cos da bθ θ θ θ+ ∫ *M1

Complete integration and obtain indefinite integral 1 12 22 cos 4sinθ θ θ− + A1

Substitute limits correctly, having integrated twice DM1

Obtain final answer ( )4 / 2π− , or exact equivalent A1

Total: 4

5(i) Use the chain rule M1


Use correct trigonometry to express derivative in terms of tan x M1

Obtain 2

d 4 tand 4 tan

= −+

y xx x

, or equivalent A1

Total: 4

5(ii) Equate derivative to –1 and solve a 3–term quadratic for tan x M1

Obtain answer x=1.11 and no other in the given interval A1

Total: 2

6(i) Calculate the value of a relevant expression or expressions at x = 2.5 and at another relevant value, e.g. x = 3

M1

Complete the argument correctly with correct calculated values A1

Total: 2

6(ii) State a suitable equation, e.g. ( )( )1tan 1 / 1π −= + −x x without suffices B1

Rearrange this as cot 1= −x x , or commence working vice versa B1

Total: 2

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

Obtain final answer 2.576 only A1


A1

Total: 3


May/June 2017



7(i) Use correct quotient rule or product rule M1


Equate derivative to zero and solve for x M1

Obtain x = 2 A1

Total: 4

7(ii) State or imply ordinates 1.6487…, 1.3591…, 1.4938… B1

Use correct formula, or equivalent, with h = 1 and three ordinates M1

Obtain answer 2.93 only A1

Total: 3

7(iii) Explain why the estimate would be less than E B1

Total: 1

8(i) Justify the given differential equation B1

Total: 1

8(ii) Separate variables correctly and attempt to integrate one side B1

Obtain term kt, or equivalent B1

Obtain term ( )ln 50− − x , or equivalent B1

Evaluate a constant, or use limits x = 0, t = 0 in a solution containing terms ( )ln 50−a x and bt

M1*

Obtain solution ( )ln 50 ln50− − = −x kt , or equivalent A1

Use x = 25, t = 10 to determine k DM1

Obtain correct solution in any form, e.g. ( ) ( )110ln50 ln 50 ln 2− − =x t A1

Obtain answer ( )( )50 1 exp 0.0693= − −x t , or equivalent A1

Total: 8


May/June 2017



9(i) State or imply the form 2 3 2

+ ++

A B Cx xx

B1

Use a relevant method to determine a constant M1

Obtain one of the values A = 3, B = –2, C = –6 A1

Obtain a second value A1

Obtain the third value

[Mark the form 2 3 2+

++

Ax B Cxx

using same pattern of marks.]

A1

Total: 5

9(ii) Integrate and obtain terms ( )23ln 2ln 3 2= − +x x

x

[The FT is on A, B and C]

Note: Candidates who integrate the partial fraction 23 2−x

x by parts should obtain

23ln 3+ −xx

or equivalent

B3 FT

Use limits correctly, having integrated all the partial fractions, in a solution containing terms ( )ln ln 3 2+ + +b

xa x c x M1

Obtain the given answer following full and exact working A1

Total: 5

10(i) Carry out a correct method for finding a vector equation for AB M1

Obtain ( )2 2 2 3λ= − + + + −r i j k i j k , or equivalent A1

Equate two pairs of components of general points on AB and l and solve for λ or for µ

M1

Obtain correct answer for λ or µ, e.g. 57λ = or 3

7µ = A1

Obtain m = 3 A1

Total: 5


May/June 2017



10(ii) EITHER: Use scalar product to obtain an equation in a, b and c, e.g. 2 4 0− − =a b c

(B1

Form a second relevant equation, e.g. 2 3 0+ − =a b c and solve for one ratio, e.g. a : b

M1

Obtain final answer a : b : c = 14 : – 7 : 7 A1

Use coordinates of a relevant point and values of a, b and c and find d M1

Obtain answer 14 7 7 42− + =x y z , or equivalent A1)

OR 1: Attempt to calculate the vector product of relevant vectors, e.g. ( ) ( )2 4 2 3− − × + −i j k i j k

(M1

Obtain two correct components A1

Obtain correct answer, e.g. 14 7 7− +i j k A1

Substitute coordinates of a relevant point in 14 7 7− + =x y z d , or equivalent, and find d

M1


OR 2: Using a relevant point and relevant vectors, form a 2–parameter equation for the plane

(M1

State a correct equation, e.g. ( ) ( )2 2 2 4 2 3= − + + − − + + −s tr i j k i j k i j k A1

State 3 correct equations in x, y, z, s and t A1

Eliminate s and t M1

Obtain answer 2x – y + z = 6, or equivalent A1)

OR 3: Using a relevant point and relevant vectors, form a determinant equation for the plane

(M1

State a correct equation, e.g.

1 2 1 1 2 4 0

2 3 1

− + −− − =

−

x y z

A1

Attempt to expand the determinant M1

Obtain or imply two correct cofactors A1


Total: 5


May/June 2017



11(a) Solve for z or for w M1

Use 2i 1= − M1

Obtain i

2 i=

−w or 2 i

2 i+

=−

z A1

Multiply numerator and denominator by the conjugate of the denominator M1

Obtain 1 25 5 i= − +w A1

Obtain 3 45 5 i= +z A1

Total: 6

11(b) EITHER: Find ( )2 2 2 3 i ± + −

(B1

Multiply by 2i (or –2i) M1*

Add result to v DM1

Obtain answer 4 3 1 6i− + A1)

OR:

State i−=

−z v kv u

, or equivalent

(M1

State k = 2 A1

Substitute and solve for z even if i omitted M1

Obtain answer 4 3 1 6i− + A1)

Total: 4






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Maximum Mark: 50

Published



May/June 2017
















May/June 2017








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Penalties




May/June 2017


Question Answer Mark Guidance

1 PE loss = 0.6 × 10 × 8 [= 48] B1

KE gain = ½ (0.6) 10 2 [= 30] B1

WD against Res = 48 – 30 = 18 J B1

Total: 3

2(i) R = 0.8g cos 10 [= 7.88] B1

F = 0.4 × 8 cos 10 [= 3.15] M1 Use F = µR

–8 sin 10 – 3.2 cos 10 = 0.8a M1 Newton 2 along the plane

a = –5.68 ms–2 A1

Total: 4

2(ii) 0 = 12 2 – 2 × 5.68 × s M1 Using v2 = u2 + 2as

s = 144/(2 × 5.68) = 12.7 m A1

Total: 2


May/June 2017



3 EITHER: (M1

Resolve horizontally and/or vertically at the 25 N weight

A cos 30 + B cos 40 = 25 A1

A sin 30 = B sin 40 A1

M1 Solve for A and/or B

A = 17.1 A1

B = 13.3 A1)

OR: (M1

Attempt Lami’s theorem

25

sin 70 sin140 sin150= =

A B A1 One correct equation

A1 A second correct equation

M1 Solve for A and/or B

A = 17.1 A1

B = 13.3 A1)

Total: 6


May/June 2017



4(i) M1 Attempt KE and/or PE with correct dimensions

KE gain = ½ × 800 × (142 – 82) = 52800 J

A1

PE gain = 800 × 10 × 120 × 0.15 = 144000 J

A1

Total: 3

4(ii) WD by engine = 32000 × 12 B1

32000 × 12 = 144000 + 52800 + WD against F

M1 Work/Energy equation 4 terms

WD against F = 187200 J A1 WD = 187000 to 3sf

Total: 3

5(i) [12 2 = 20 2 – 2a × AB 6 2 = 12 2 – 2a × BC]

M1 Use v 2 = u 2 + 2(–a)s for AB or BC where a is the deceleration

AB = 128/a A1

BC = 54/a A1

AB : BC = 64:27 A1 Allow equivalent unsimplified ratio

Total: 4


May/June 2017



5(ii) 0 = 20 2 – 2a × 80 → a = 2.5 M1 Use v 2 = u 2 + 2(–a)AD to find a

BC = 54/2.5 M1 Use a to find BC

BC = 21.6 m A1

Total: 3

6(i) [q + r = 4 and 2q + 4r = 4] M1 Use v = 4 at t = 1 and t = 2

q = 6 and r = –2 so v = 6t – 2t 2 A1

a = 6 – 4t M1 Differentiation used for a

At t = 0.5, a = 4 A1 AG

Total: 4

6(ii) v = 6t – 2t 2 = 0 M1 Set v = 0 and solve for t

t = 0 and t = 3 A1

Total: 2


May/June 2017



6(iii) EITHER: s = ∫(6t – 2t 2) dt (M1

Attempt to integrate v to find s

s = 3t 2 – ⅔t 3 + C A1

0 = 3 × 32 – ⅔ × 33 + C M1 Use s = 0 when t = 3 to find C

C = –9 so distance = 9 m A1) Valid argument

OR:

( )

32

0

6 2 d= −∫s t t t(M1

Attempt integration with limits

32 3

0

233

− t t

A1 Correct integration and correct limits but no evaluation

[27 – 18 = 9] M1 Evaluation of integral between limits

Distance from O at t = 0 is 9 m A1) With explanation

Total: 4


May/June 2017



7(i) [T – 0.8g sin 30 = 0.8a 1.2g sin 60 – T = 1.2a 1.2g sin 60 – 0.8g sin 30 = 2a]

M1 Resolve along the plane for either A or for B or for the system

For A 4 0.8− =T a A1

For B

6 3 10.4 1.2− = − =T T aA1 System equation is

6 3 4 6.4 2− = = a

M1 Solve for a or T

3 3 2 3.20= − =a ms–2 A1

( )12 1 3 6.56 N5

T = + = A1

Total: 6


May/June 2017



7(ii) RA = 0.8 cos30 4 3=g RB = 1.2 cos60 6=g

B1 For either RA or RB

FA = 4 3 µ and FB = 6µ M1 Either FA or FB used

M1 Resolve parallel to the plane for both particles A and B or system

12 sin 60 – 6µ – T = 0 or T – 8 sin 30 – 4√3 µ = 0

A1 System equation is 12 sin 60 – 8 sin 30 – 6µ – 4√3 µ = 0

M1 Eliminate T and/or find µ

( ) ( )µ 6 3 4 / 6 4 3 = √ − + √

= 0.494

A1

Total: 6






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Published



May/June 2017
















May/June 2017








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Penalties




May/June 2017



1 EITHER: WD = 20 cos θ × 1.5 × 12 (J) (B1

Using WD = Fd cos θ

[cos θ = 50/360] θ = .... M1 Use WD = 50 and solve for θ

82(.0)θ = A1)

OR: Power P = 50/12 = 4.1666… (B1

Using Power = WD/time

[50/12 = 20 cos θ × 1.5] θ = .... M1 Use P = Fv and solve for θ

( )82 .0θ = A1)

Total: 3

2(i) 2 2.5 5 = × ×v (ms–1) B1 AG Using 2 2 2= +v u as

Total: 1

2(ii)(a) M1 Attempting PE loss or KE gain

PE loss = 0.2 × 10 × 6 sin 30 [= 6] and KE gain = 0.5 × 0.2 × (v 2 – 5 2)

A1 Both PE and KE correct both unsimplified

[6 = 0.1(v 2 – 52)] M1 PE loss = KE gain (3 terms)

v 2 = 85 → v = 9.22 ms–1 A1

Total: 4


May/June 2017



2(ii)(b) Max velocity at lowest point [0.2 × 10 × 6 = 0.5 × 0.2 × (v 2 – 5 2)]

M1 PE loss = KE gain

v 2 = 145 → v = 12(.0) ms–1 A1

Total: 2

3(i) M1 Attempt sA as sA = k + 10t (any k)

sA = 20 + 10t A1

sB = 16t + ½(–2)t 2 [= 16t – t 2] B1 FT Allow FT only if sA = 10t and sB = 16(t – 2) + ½(–2)(t – 2) 2

i.e. t measured from when A passes O

Total: 3

3(ii) vB = 16 – 2t → vB = 0, t = 8 B1

s = sA – sB [= 20 + 10t + t2 – 16t = t 2 – 6t + 20]

M1 Finding distance between A and B at time t = T ( T > 0 ) found from a valid method for vB = 0

t = 8, s = 36 (m) A1

Total: 3


May/June 2017



3(iii) d 2 6

d= −

s tt

or s = t 2 – 6t + 20 = (t – 3) 2 + 11

M1 Either use differentiation or complete the square, or state value of t when speeds are the same

[t = 3] M1 Solve for t and evaluate sA – sB at this value of t

s = sA – sB = 11 m A1

Total: 3

4(i)(a) [P = 850 × 42] M1 Using P = Fv

P = 35700 W = 35.7 kW A1 Must be in kW to 3sf

Total: 2

4(i)(b) P = 41700 → [DF = 41700/42]

M1 Find new power and new DF based on power found in 4(i)(a)

[(993 – 850) = 1200a] M1 Apply Newton 2, three terms

a = 5/42 = 0.119 ms–2 A1

Total: 3


May/June 2017



4(ii) DF = 80000/24 B1 DF = P/v

[DF – 850 – mg sin θ = 0] M1 Newton 2 along the hill, 3 terms

[12000 sin θ = 80000/24 – 850] θ = …..

M1 Solve for θ, from a three term equation

θ = 11.9 A1

Total: 4

5 M1 Resolve perpendicular to the plane, three terms

R + P sin 30 = 0.12g cos 40 A1 R does not need to be the subject

F = 0.32R M1 Use F = µR

[Pmin cos 30 + F = 0.12g sin 40] M1 About to slip down, 3 terms

[Pmax cos 30 – F = 0.12g sin 40] M1 About to slip up, 3 terms

[P cos 30 = 0.12g sin 40 ±0.32 (0.12g cos 40 – P sin 30)] OR [P cos 30 ± 0.32R = 0.12g sin 40 R + P sin 30 = 0.12g cos 40] Must reach P =… in either method

M1 Substitute for F and solve for P in either case, 4 terms OR solve a pair of simultaneous equations (each with 3 terms) in R and P for P in one of the cases

Pmax = 1.04 Pmin = 0.676 A1 For either correct

0.676 ⩽ P ⩽ 1.04 A1

Total: 8


May/June 2017



6(i) A [T = 0.3a] B [1.5g sin θ – T = 1.5a] System [1.5g sin θ = 1.8a]

M1 Apply Newton’s second law to A or to B or to the system

A1 Any two correct equations

M1 Solve 2 simultaneous equations for a and/or T or use the system equation.

a = 9/1.8 = 5 ms–2 A1

T = 1.5 N A1

Total: 5


May/June 2017



6(ii) [5 = 3a] M1 v = u + at used with t = 3, u = 0, v = 5

a = 5/3 = 1.67 A1

RA = 3 RB = 15 cos 36.9 = 12 B1 For either reaction

[FA = 3µ FB = 12µ] M1 Use F = µR for either term

EITHER: A [T – FA = 0.3a] B [15 sin 36.9 – T – FB = 1.5a] System equation is [1.5g sin 36.9 – FA – FB = 1.8a]

(M1 Apply Newton’s second law to A or to B or to the system

A2/1/0 A1 Correct equation for A or B A2 Correct equations for A and B OR A2 Correct system equation

[9 – 15µ = 3] M1 Solve for µ from equations with correct number of terms

µ = 0.4 = 2/5 A1)

OR: s = ½ (5/3) × 32 = 7.5 (B1

Find distance travelled in 3 secs

PE loss = 1.5 × 10 × 7.5 × (3/5) = 67.5

B1

KE gain = ½ (1.8) × 52 = 22.5 B1

[67.5 = 22.5 + 3µ × 7.5 + 12µ × 7.5] M1 Use Work/Energy equation

µ = 2/5 = 0.4 A1)

Total: 9






MARK SCHEME

Maximum Mark: 50

Published



May/June 2017

















May/June 2017






allowed)



SOI Seen or implied


Penalties




9709/43 Cambridge International AS/A Level – Mark SchemePUBLISHED

May/June 2017



1(i) WD = 35 cos 20 × 12 M1 Uses WD = Fd cos θ

395 J A1

Total: 2

1(ii) EITHER:

WD against resistance = 15 × 12 (B1

35cos20 × 12 = 15 × 12 + ½ (25v2) M1 Uses WDman = WDresistance + KE gain

v = 4.14 ms–1 A1)

OR: 35 cos 20 – 15 = 25 a [a = 0.716] (B1

Applies Newton’s Second Law

v2 = 2 × 0.7155. × 12 M1 Uses v2 = u2 + 2as

4.14v = ms–1 A1)

Total: 3


May/June 2017



2 EITHER: 3P sin 55 + P sin θ = 20 + P sin θ or 3P sin 55 = 20

(M1 Resolves forces vertically

P = 8.14 A1

3P cos 55 = 2P cos θ M1 Resolves forces horizontally

cos θ = 1.5 cos 55 → θ = … M1 Attempt to solve for θ

θ = 30.6 A1)

OR: 3

sin90P = 20

sin125

(M1 Uses Lami’s Theorem (forces 3P and 20)

P = 8.14 A1

3sin90

P = 2 cos sin145P θ

M1 Uses Lami’s Theorem (forces 3P and 2P cos θ)

cos θ = 1.5 sin 145→ θ = … M1 Attempt to solve for θ

θ = 30.6 A1)

Total: 5


May/June 2017



3(i) Trapezium, right-hand steeper than left-hand slope B1

Total: 1

3(ii) Deceleration 0.5 T B1 May be implied

Constant speed 180 – 1.5 T B1

Total: 2

3(iii) 0.5[180 + (180 – 1.5T)] × 25 = 3300 M1 Uses area property

T = 64 A1

Distance decelerating = [0.5 × 32 × 25 =] 400 m B1

Total: 3

4(i) a = 3 × 2 × (2t – 5)2 [= 54] *M1 Uses a = dv/dt

6(2t – 5)2 = 54 → t = ... DM1 Solves for t

t = 1, 4 A1

Total: 3


May/June 2017



4(ii) ( ) 42 5

4 2t

s−

=×

(+ C) *M1 Uses s = ∫ v dt

C = − 625

8

DM1 Uses s = 0 at t = 0

( ) 42 5 625 8 8

ts −

−=

A1

Total: 3

Alternative method for Question 4

4(i) v = 8t3 – 60t2 + 150t – 125 → a = 24t2 – 120t + 150

*M1 Uses a = dv/dt

24t2 – 120t + 150 = 54 → t = ... DM1 Solves for t

t = 1, 4 A1

Total: 3

4(ii) s = ∫ 8t 3 – 60t 2 + 150t – 125 dt

→ s = 84

t 4 – 603

t 3 + 1502

t 2 – 125t (+ C)

*M1 Uses s = ∫ v dt

C = 0 DM1 Uses s = 0 at t = 0 (may be implied)

s = 2t 4 – 20t 3 + 75t 2 – 125t A1

Total: 3


May/June 2017



5(i) s2 = 20t – 0.5gt2 B1 Second particle

M1 Uses s = ut + ½ at2 for first particle

s1 = 12(t + 2) – 0.5g(t + 2)2 *A1

12(t + 2) – 0.5g(t + 2)2 = 20t – 0.5gt2 → t=…

DM1 Solves s1 = s2

t = 1

7 = 0.143

A1

Total: 5

5(ii) [s = 20 × 1

7 – 5 × 21( )

7 = 2.755…]

Height is 2.76 m

B1

Total: 1


May/June 2017



6(i)(a) 16 000 = F × 40 M1 Using P = Fv

Resistance is 400 N A1

Total: 2

6(i)(b) 22 500 = F × 45 F = 500

B1

500 – 400 = 1200a M1 Applying Newton’s Second Law

a = 1

12 = 0.0833 (ms–2)

A1

Total: 3

6(ii) 16 000 = (590 + 2v)v M1 Using P = Fv

[2v2 + 590 v – 16 000 = 0] → v = … M1 Solving for v

v = 25 (ms–1) A1

Total: 3


May/June 2017



7(i) R = mg cos30 B1 Resolves normally

F = 2m cos 30 [= m√3] M1 Uses F = µR

T = 4g [= 40] B1 Particle B

T = mgsin30 + F M1 Resolves parallel to plane for particle A

40 = 5m + m√3 A1 Equation in m

m = 40

5 3+√ = 5.94

A1 AG All correct and no errors seen

Total: 6


May/June 2017



7(ii) EITHER: [R = 3g cos30] F = 0.2 × 3g cos30 (3√3 = 5.196)

(B1

4g – T = 4a or T – 3gsin30 – F = 3a or 4g – 3gsin30 –F =7a

M1 Applies Newton’s Second Law to one of the particles or forms system equation in a (mBg – mAgsin30 – F = (mA + mB)a)

T – 3gsin30 – 3√3 = 3a or 40 – T = 4a or 4g – 3gsin30 – 3√3 = 7a →a = …

M1 Applies Newton’s Second Law to form second equation in T and a and solves for a or solves system equation for a

25 3 3 7

a − √=

= 2.83.

A1

v2 = 2 × 2.83 × 0.5 v = 1.68…

B1 FT v as T becomes zero FT on a

–3gsin30 – 0.2(3gcos30) = 3a –15 – 3√3 = 3a →a = …(–5–√3 = –6.73)

M1 Applies Newton’s Second Law and solves for a

0 = 1.682 – 2 × 6.73s s = …(0.210)

M1 Uses v2 = u2 + 2as and solves for s

Total distance = 0.710 m A1)

OR: [R = 3g cos30] F = 0.2 × 3g cos30 (3√3 = 5.196)

(B1


May/June 2017



M1 For 4kg mass, uses PE loss – WDT = KE gain

M1 For 3kg mass, uses WDT = KE gain + PE gain + WDFr

4g(0.5) – 0.5T = ½ (4v2) and 0.5T = ½ (3v2) + 3g(0.5sin30) + 3√3(0.5)

A1

v2 = (25 – 3√3)/7 or v = 1.68 B1

½ (3)(1.68)2 = 3g(s sin30) + 3√3s M1 For 3kg mass, uses KE loss = PE gain + WDFr

s = …(0.210) M1 Solves for s

Total distance = 0.710 m A1)

Total: 8






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May/June 2017
















May/June 2017








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May/June 2017



1 tan40 = v / 20cos60 M1

v = 10tan40 ( = 8.3909...) A1

–10tan40 = 20sin60 – gt M1 Uses v = u + at vertically

t = 1.27 s A1

Total: 4

2(i) 7 = 0.35λ / 0.25 M1 Uses T = λx / L

λ = 5 A1

Total: 2

2(ii) EE = 20.35 × 5 / (2 × 0.25) or 20.05 × 5 / (2 × 0.05) B1 Uses EE = λ 2x / 2L

PE = mg × 0.3sin30 B1

mg × 0.3sin30 = 20.35 × 5 / (2 × 0.25) − 20.05 × 5 / (2 × 0.25) M1 Sets up a 3 term energy equation involving EE, KE and PE

m = 0.8 A1

Total: 4


May/June 2017



3(i) CofM of hemisphere = 3

8 × 0.56 or 3

8 × 0.28

B1

[ 23π × 30.56 – 2

3π × 30.28 ]X = 2

3π × 30.56 × 3

8 × 0.56 – 2

3π × 30.28 ×

38

× 0.28

M1A1

Take moments about O

X = 0.225 m A1

Total: 4

3(ii) 24 × 0.225 + W(3 × 0.28 / 8) = (24 + W) × 0.15 M1A1 Attempts to take moments about O W = weight of uniform hemi-sphere

W = 40 N A1

Total: 3

4(i)

x = 10t or y = 2gt / 2 B1

y = 15x / 10 – g(x / 10 2) / 2 M1A1 Attempts to eliminate t

y = 1.5x – 0.05 2x

A1

Total: 4


May/June 2017



4(ii) 0 = 1.5x – 0.05 2x M1 Substitute y = 0 into the trajectory equation

x = 30 A1

Total: 2

4(iii) –14 = 1.5x – 20.05x M1 Sets up a quadratic equation and attempts to solve it

x = 37.5 A1

Total: 2

5(i) OG = 2 × 0.7sin(π / 2) / (3π / 2) (= 0.297) B1

0.9R = 14(0.7cos30 – 0.297sin30) M1A1 Attempts to take moments about A

R = 7.12 N A1

Total: 4

5(ii) H = 7.12sin30 and V = 14 − Rcos30 M1 Resolves horizontally and vertically

tanθ = (14 – 7.12cos30) / (7.12sin30) M1 Uses tanθ = V / H, where θ is the required angle

θ = 65.6 A1

Total: 3


May/June 2017



6(i) T = 12 × 0.1 / 0.4 ( = 3 N) B1 Uses T = λx / L

3sinθ = 0.15 2ω (0.5sinθ) M1 Uses Newton's Second Law horizontally

ω = 6.32 rad 1s− A1

Tcosθ = 0.15g (cosθ = 0.5) M1 Resolves vertically

θ = 60 A1

Total: 5

6(ii) v = 6.32 × 0.5sin60 B1 FT Uses v = rω and r = 0.5sin60

KE = 0.15(6.32 × 0.5sin60 2) / 2 (=0.5625J) B1

Difference = 0.5625 – 12 × 0. 21 / (2 × 0.4) M1 Uses EE = λ 2x / (2L)

Difference = 0.4125 J A1

Total: 4

7(i) µ = 0.6 × 0. 25 / (0.5 g) ( = 0.03) B1 Uses F = µ R

0.5dv / dt = 0.6 2t – 0.03 × 0.5g M1 Uses Newton's Second Law horizontally

dv / dt = 1.2 2t – 0.3 A1

Total: 3


May/June 2017



7(ii) dv∫ = (∫ 1.2 2t – 0.3) dt

v = 0.4 3t – 0.3t ( + c)

M1 Separates the variables and attempts to integrate

t = 0.5, v = 0 hence c = 0.1 M1 Attempts to find c

v = 0.4 3t – 0.3t + 0.1 A1

Total: 3

7(iii) dx∫ = (∫ 0.4 3t – 0.3t + 0.1) dt

x = 0.1 4t – 0.15 2t + 0.1t ( + c)

M1 Attempts to integrate

t = 0.5, x = 0 hence c = –0.01875 M1 Finds c or substitutes the limits

x(1.2) = 0.0926(1) A1

Total: 3






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May/June 2017
















May/June 2017








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May/June 2017


Question Answer Marks Notes

1(i) (4 = 5r ) r = 0.8 m B1 Uses v = rω

Total: 1

1(ii) T = 0.2 × 25 × 0.8 M1 Uses Newton’s Second Law horizontally

T = 4 N A1 FT FT with their radius from part (i)

4 = λ(0.8 – 0.6) / 0.6 M1 Uses T = λx / L

λ = 12 A1

Total: 4

2(i) 6cos60 = 4cos60 + mg M1 Resolve vertically

m = 0.1 kg A1

Total: 2

2(ii) radius = 0.7sin60 B1

6sin60 + 4sin60 = 0.1 2v / (0.7sin60) M1 Uses Newton's Second Law horizontally with 3 terms

v = 7.25 m 1s− A1

Total: 3

3(i) Height of C of M of each vertical face above the base = 0.1 m B1

5 × 3y = 4 × 3 × 0.1 M1 Takes moments about the base. y is the height of the C of M above the base

y = 0.08 m A1

Total: 3


May/June 2017



3(ii) Moment of lid about the base = 3 × (0.2 + 0.1sinθ)

B1 θ is the angle the lid makes with the horizontal

(6 × 3 + 2) × 0.12 = 5 × 3 × 0.08 + 2 × 0.2 + 3 × (0.2 + 0.1sinθ) M1 Take moments about the base

A1

θ = 41.8° A1

Total: 4

4(i) 0.4a = 0.4g – 0.2 2v M1 Uses Newton’s Second Law vertically

vdv / dx = 10 – 0.5 2v A1 AG

Total: 2

4(ii) 2 d / (10 0. )5v v v∫ − = dx∫ M1 Separates the variables and attempts to integrate

– ln(10 – 0.5 2v ) = x ( + c ) A1

x = 0, v = 0 hence c = –ln10 M1 Attempts to find c using x = 0, v = 0

v = (20 20 )e x−− A1 10–0.5 2v = 10e x ln− + = 10 e x−

Total: 4

4(iii) Increase= 8(20 20e−− − 4(20 20e−− M1 M1 if x values are substituted into their value for part (ii)

Increase = 0.0404 m 1s− A1 Allow 0.04

Total: 2


May/June 2017



5(i) 0.3g = 6 e / 0.8 M1 Uses T = λx / L

e = 0.4 m A1

EE = 6 × 20.4 /(2 × 0.8) B1 FT FT for their e

0.3 2v / 2 – 0.3 × 22 / 2 = 0.3 g(0.8 + 0.4) – 6 × 20.4 / (2 × 0.8) M1 Sets up a 4 term energy equation involving EE, KE and PE

v = 4.9(0) m 1s− or 2 6 A1

Total: 5

5(ii) 0.3 × 22 /2 + 0.3 gL = 6(L− 0.8 2) / (2 × 0.8) M1 Sets up a 3 term energy equation involving EE, KE and PE

A1

L = 2.18 m A1 Ignore answers less than 0.8

Total: 3

6(i) 3 × 0.6 = 8cos60 x M1 Takes moments about A

x = 0.45 m A1

Total: 2

6(ii) Pcos60 × 0.6 = 8 × 0.45cos60 M1 Takes moments about A

P = 6 N A1

Total: 2


May/June 2017



6(iii) µ = 3cos30 / (8 – 3sin30) M1 Uses F = µR used

µ = 6cos30 / (8 + 6sin30) M1

µ = 0.4 or 0.472 A1

µ = 0.472 accept 0.47 A1

Total: 4

7(i) tanθ = 2 B1 Note θ = 63.4349..°

Total: 1


May/June 2017



7(ii) EITHER: a = 2a – 25 2a / 2V (25a = 2V )

(B1

Substitutes x = y = a into the trajectory equation

a = Vcos63.4349.. × 4 B1 Horizontal motion

2V = 25 × 4 × Vcos63.4349.. M1 Attempts to eliminate a

V = 44.7(213..) or 20 5 A1

a = 80 A1)

OR: a = Vsin63.4349..× 4 – g 24 /2

(B1

Uses s = ut + a 2t / 2 vertically

a = Vcos63.4349..× 4 B1 Horizontal motion

Vsin63.4349..× 4 − g 24 / 2 = Vcos63.4349..× 4 M1 Attempts to solve the 2 equations

V = 44.7(213..)or 20 5 A1

a = 80 A1)

Total: 5

7(iii) vv = 44.7213..sin63.4349.. – 4g ( = 0) M1 vv = vertical component of the velocity

α = 1tan− +/− 0 / (44.7213..cos63.4349..) M1 tanα = vv / hv where hv = horizontal velocity

α = 0° A1

Total: 3






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May/June 2017



1 tan40 = v / 20cos60 M1

v = 10tan40 ( = 8.3909...) A1

–10tan40 = 20sin60 – gt M1 Uses v = u + at vertically

t = 1.27 s A1

Total: 4

2(i) 7 = 0.35λ / 0.25 M1 Uses T = λx / L

λ = 5 A1

Total: 2

2(ii) EE = 20.35 × 5 / (2 × 0.25) or 20.05 × 5 / (2 × 0.05) B1 Uses EE = λ 2x / 2L

PE = mg × 0.3sin30 B1

mg × 0.3sin30 = 20.35 × 5 / (2 × 0.25) − 20.05 × 5 / (2 × 0.25) M1 Sets up a 3 term energy equation involving EE, KE and PE

m = 0.8 A1

Total: 4


May/June 2017



3(i) CofM of hemisphere = 3

8 × 0.56 or 3

8 × 0.28

B1

[ 23π × 30.56 – 2

3π × 30.28 ]X = 2

3π × 30.56 × 3

8 × 0.56 – 2

3π × 30.28 ×

38

× 0.28

M1A1

Take moments about O

X = 0.225 m A1

Total: 4

3(ii) 24 × 0.225 + W(3 × 0.28 / 8) = (24 + W) × 0.15 M1A1 Attempts to take moments about O W = weight of uniform hemi-sphere

W = 40 N A1

Total: 3

4(i)

x = 10t or y = 2gt / 2 B1

y = 15x / 10 – g(x / 10 2) / 2 M1A1 Attempts to eliminate t

y = 1.5x – 0.05 2x

A1

Total: 4


May/June 2017



4(ii) 0 = 1.5x – 0.05 2x M1 Substitute y = 0 into the trajectory equation

x = 30 A1

Total: 2

4(iii) –14 = 1.5x – 20.05x M1 Sets up a quadratic equation and attempts to solve it

x = 37.5 A1

Total: 2

5(i) OG = 2 × 0.7sin(π / 2) / (3π / 2) (= 0.297) B1

0.9R = 14(0.7cos30 – 0.297sin30) M1A1 Attempts to take moments about A

R = 7.12 N A1

Total: 4

5(ii) H = 7.12sin30 and V = 14 − Rcos30 M1 Resolves horizontally and vertically

tanθ = (14 – 7.12cos30) / (7.12sin30) M1 Uses tanθ = V / H, where θ is the required angle

θ = 65.6 A1

Total: 3


May/June 2017



6(i) T = 12 × 0.1 / 0.4 ( = 3 N) B1 Uses T = λx / L

3sinθ = 0.15 2ω (0.5sinθ) M1 Uses Newton's Second Law horizontally

ω = 6.32 rad 1s− A1

Tcosθ = 0.15g (cosθ = 0.5) M1 Resolves vertically

θ = 60 A1

Total: 5

6(ii) v = 6.32 × 0.5sin60 B1 FT Uses v = rω and r = 0.5sin60

KE = 0.15(6.32 × 0.5sin60 2) / 2 (=0.5625J) B1

Difference = 0.5625 – 12 × 0. 21 / (2 × 0.4) M1 Uses EE = λ 2x / (2L)

Difference = 0.4125 J A1

Total: 4

7(i) µ = 0.6 × 0. 25 / (0.5 g) ( = 0.03) B1 Uses F = µ R

0.5dv / dt = 0.6 2t – 0.03 × 0.5g M1 Uses Newton's Second Law horizontally

dv / dt = 1.2 2t – 0.3 A1

Total: 3


May/June 2017



7(ii) dv∫ = (∫ 1.2 2t – 0.3) dt

v = 0.4 3t – 0.3t ( + c)

M1 Separates the variables and attempts to integrate

t = 0.5, v = 0 hence c = 0.1 M1 Attempts to find c

v = 0.4 3t – 0.3t + 0.1 A1

Total: 3

7(iii) dx∫ = (∫ 0.4 3t – 0.3t + 0.1) dt

x = 0.1 4t – 0.15 2t + 0.1t ( + c)

M1 Attempts to integrate

t = 0.5, x = 0 hence c = –0.01875 M1 Finds c or substitutes the limits

x(1.2) = 0.0926(1) A1

Total: 3






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May/June 2017
















May/June 2017








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May/June 2017



1(i) EITHER: 315 10.5

30 30− = =∑ x

k

(M1 Dividing 315 by ±30 and + or – from 50.5 need both and no more

5.5 10.5 40= − =k A1) Correct answer from correct working

OR: 50.5 30 1515= × =∑ x , 1515 30 315− =k

(M1 Mult by 50.5 by 30 and + or – 315 and dividing by ±30 need all these

k = 40 A1) Correct answer from correct working. 1200 gets M0

Total: 2

1(ii) EITHER: var = 4022/30–10.52(=23.817)

(M1 Subst in correct coded variance formula

sd = 4.88 A1)

OR: ( ) ( )22 2 40 30 40 4022− + =∑ ∑x x , 2 77222=∑ x

Var = 77222/30 – 50.52 (= 23.817)

(M1 Expanding with ± 40Σx and ± 30(40)2 seen

sd = 4.88 A1)

Total: 2


May/June 2017



2 P(R) = 4/36 = 1/9 M1 Attempt at P(R) by probability space diag or listing more than half the options, must see a prob, just a list is not enough

( ) ( ) ( )P P O, E P E, O 1/ 4 1/ 4 1/ 2T = + = + = OR ( )P | 1/ 9R T = M1 Attempt at P(T) or P(R|T) involving more than half the options

( ) ( ) ( )P P 3, 4 P 4, 3 2 / 36 1/18= + = =IR T OR ( )P | 1/ 9R T = B1 Value stated, not from P(R) × P(T) e.g. from probability space diagram

( ) ( ) ( )As P P P× = IR T R T OR ( ) ( )as P | PR T R= M1 Comparing product values with ( )P IR T , or comparing P (R|T) with P(R)

The events are independent. A1 Correct conclusion must have all probs correct

Total: 5


May/June 2017



3(i) M1 Correct shape i.e. 3 branches then 3 by 3 branches, labelled and clear annotation Condone omission of lines for first match result providing the probabilities are there.

A1 All correct probs with fully correct shape and probs either fractions or decimals not 1.5/5 etc.

Total: 2

W7/10

W 2/10 D3/5 1/10 L

1/3 W

1/5 D 1/3 D 1/3 L

1/5 3/10 WL

1/20 D13/20

L


May/June 2017



3(ii) ( ) ( )

( )1 2

1 22

PP given

P∩

=L W

L WW

M1 Attempt at P(L1∩W2) as a two-factor prod only as

num or denom of a fraction

1 / 5 3 /103 / 5 7 /10 1 / 5 1 / 3 1 / 5 3 /10

×=

× + × + ×

M1 Attempt at P(W2) as sum of appropriate 3 two-factor probs OE seen anywhere

A1 Unsimplified correct P(W2) num or denom of a fraction

( )3 / 50 9 / 82 0.11041 / 75

= = A1

Total: 4


May/June 2017



4(i) fd 16, 14, 11, 505, 2.5 M1 Attempt at fd (must be at least 3 freq/cw) – may be implied by graph

A1 Correct heights seen on graph i.e. must see a gap for fd = 2.5 etc.

B1 Correct end points of bars and correct widths

B1 labels fd, sec. Time can be optional. Linear axes, condone 0 ⩽ t < 20 etc.

Total: 4

fd20

15

10

5

0 20 40 60 80 100 120 140 time sec


May/June 2017



4(ii) (10 × 320 + 30 × 280 + 50 × 220 + 80 × 220 + 120 × 100) / 1140 M1 using Σ fx / n with mid-point attempt ±0.5, not ends not class widths

= 45.8 A1

Total: 2

5(i) p = 0.07 B1

( ) ( ) ( )2 18202P 2 C 0.07 0.93= M1 Bin term ( )2020C 1 −− xx

x p p their p

= 0.252 A1

Total: 3

5(ii) P(at least 1 cracked egg)=1–(0.93)20=1–0.2342 M1 Attempt to find P(at least1 cracked egg) with their p from (i) allow 1 – P(0, 1) OE

= 0.766 A1 Rounding to 0.766

Total: 2

5(iii) (0.7658)n<0.01 M1 Eqn or inequal containing (their 0.766)n or (their 0.234)n, together with 0.01 or 0.99

n = 18 A1

Total: 2


May/June 2017



6(a)(i) z = 0.674 B1 rounding to ±0.674 or 0.675

6.80.6740.25

µµ−

= M1 standardising, no cc, no sq rt, no sq, σ may still be

present on RHS

M1 subst and sensible solving for µ must collect terms, no z-value needed can be 0.75 or 0.7734 need a value for µ

µ = 5.82 A1

Total: 4

6(a)(ii) ( ) 4.7 5.819P 4.7 P

1.4548− < = <

X z

M1 ± standardising no cc, no sq rt, no sq unless penalised in (a)(i)

= ɸ(–0.769) = 1 – 0.7791 M1 correct side for their mean i.e. 1–ɸ (final solution)

= 0.221 A1

Total: 3

6(b) ( ) ( )15.75 16P 15.75 P 1 P 1.25

0.2− < = < = − <

z z =1 – 0.8944 = 0.1056 and

P(>16.25) = 0.1056 by sym

*M1 Standardising for 15.75 or 16.25 no cc no sq no sq rt unless penalised in (a)(i) or (a)(ii)

P(usable) = 1 – 0.2112 = 0.7888 B1 2ɸ– 1 OE for required prob, (final solution)

Usable rods=1000 × 0.7888 = DM1 Mult their prob by 1000 dep on recognisable attempt to standardise

788 or 789 A1

Total: 4


May/June 2017



7(a) EITHER: e.g. xxxxx =5! for the other children

(B1 5! OE seen alone or mult by integer k ⩾ 1, no addition

Put y in 6 ways, then 5 then 4 for the youngest children B1 Mult by 6P3 OE

Answer 5! × 6P3 = 14400 B1) Correct answer

OR: total – 3 tog – 2 tog = 8! – 6!3! – 6! × 2 × 5 × 3 = 14400

(B1 8! – 6! × k ⩾ 1seen

B1 6!3! or 6! × 2 × 5 × 3 seen subtracted

B1) Correct answer

Total: 3

7(b) D W M 2 2 1 = 6C2 × 4C2 × 1 = 90

B1 One correct unsimplified option

3 1 1 = 6C3 × 4 × 1 = 80

M1 Summing 2 or more 3-factor options which can contain perms or 3 factors added. The 1 can be implied

1 3 1 = 6 × 4C3 × 1 = 24 M1 Summing the correct 3 unsimplified outcomes only

Total=194 ways A1

Total: 4


May/June 2017



7(c) C D S 2 1 1 = 26C2 × 9 × 5 × 4! = 351 000

M1 summing 2 or more options of the form (2 1 1), (1 2 1), (1 1 2), can have perms, can be added

1 2 1 = 26 × 9C2 × 5 × 4! = 112 320

M1 4 relevant products seen excluding 4! e.g. 26 × 9 × 8 × 5 or 26 × 9P2 × 5 for 2nd outcome, condone 26 × 9 × 5 × 37 as being relevant

1 1 2 = 26 × 9 × 5C2 × 4! = 56 160 M1 mult all terms by 4! or 4!/2!

Total = 519 480 A1

Total: 4






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May/June 2017






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May/June 2017



1(i) 4 × 5.5 + 3x + 90 = 8 × 29 M1 An expression to work out total cost of individual items = 8 × mean, x may be implied.

112 + 3x = 232 x = 40

A1 Correct complete unsimplified expression / calculation

(Cost = $)40 A1 Units not required

Total: 3

1(ii) sd = 0 so all cost the same M1 Must see comment interpreting sd = 0, OE

shirts cost 4 × $26 = $104 AG A1 See 4 × $26, $130 – $26 OE. Must have a final value of $104 stated

Total: 2

2(i) med = 3.2 B1 Accept 3.2 ± 0.05

UQ = 3.65 ⩽ uq ⩽ 3.7 LQ = 2.55⩽ lq ⩽ 2.6 M1 UQ – LQ, UQ greater than their ‘median’, LQ less than their ‘median’

IQR = 1.05 ⩽ iqr ⩽ 1.15 A1 Correct answer from both LQ and UQ in given ranges

Total: 3

2(ii) 134 – 24 = 110 B1 Accept 108 ⩽ n ⩽ 112, n an integer

Total: 1


May/June 2017



2(iii) 200 – 12 = 188 less than length l M1 188 seen, can be implied by answer in range, mark on graph.

l = 4.5 cm A1 Correct answer accept 4.4 ⩽ l ⩽ 4.5

Total: 2

3(i) k (–2)2 is the same as k (2)2 = 4k B1 need to see –22 k, 22k and 4k, algebraically correct expressions OE

Total: 1

3(ii) x –2 –1 2 4

Prob 4k k 4k 16k

B1 –2, -1, 2, 4 only seen in a table, together with at least one attempted probability involving k

4k + k + 4k + 16k = 1 M1 Summing 4 probs equating to 1. Must all be positive (table not required)

k = 1/25 (0.04) A1 CWO

Total: 3

3(iii) E(X) = –8k + –k + 8k + 64k = 63k M1 using Σpx unsimplified. FT their k substituted before this stage, no inappropriate dividing

= 63/25 (2.52) A1

Total: 2


May/June 2017



4 P(score is 6) = P(3, 3) M1 Realising that score 6 is only P(3, 3)

= r 2 = 1/36 r = 1/6

A1 Correct ans [SR B2 r = 1/6 without workings]

P(2, 3) + P(3, 2) = 1/9 qr +rq = 1/9

M1 Eqn involving qr (OE) equated to 1/9 (r may be replaced by their ‘r value’)

q/6 + q/6 = 1/9 M1 Correct equation with their ‘r value’ substituted

q = 1/3 A1 Correct answer seen, does not imply previous M’s

p = 1 – 1/6 – 1/3 = 1/2 B1 FT FT their p + their r + their q =1 , 0 < p < 1

Total: 6

5(i) 4.2 3.9( )zσ−

= M1 Standardising, not square root of σ, not σ2

z = 0.916 or 0.915 B1 Accept 0.915 ⩽ ± z ⩽ 0.916 seen

σ = 0.328 A1 Correct final answer (allow 20/61 or 75/229)

Total: 3


May/June 2017



5(ii) z = 4.4 – 3.9/their 0.328 or z = 3.4 – 3.9/their 0.328 = 1.5267 = –1.5267

M1 Standardising attempt with 3.4 or 4.4 only, allow square root of σ, or σ2

Φ = 0.9364 A1 0.936 ⩽ Φ ⩽ 0.937 or 0.063 ⩽ Φ ⩽ 0.064 seen

Prob = 2Φ – 1 = 2(0.9364) – 1 M1 Correct area 2Φ – 1OE i.e. Φ = – (1 – Φ), linked to final solution

= 0.873 A1 Correct final answer from 0.9363 ⩽ Φ ⩽0.9365

Total: 4

5(iii) dividing (0.5) by a larger number gives a smaller z-value or more spread out as sd larger or use of diagrams

*B1 No calculations or calculated values present e.g. (σ = )0.656 seen Reference to spread or z value required

Prob is less than that in (ii) DB1 Dependent upon first B1

Total: 2

6(i) EITHER: Route 1 A*********A in 9! / 2!2!5! = 756 ways (*M1

Considering AA and BB options with values

B*********B in 9! / 4!5! = 126 ways A1 Any one option correct

756 + 126 DM1 Summing their AA and BB outcomes only

Total = 882 ways A1)


May/June 2017



OR1: Route 2 A*********A in 9C5 × 4C2 = 756 ways (M1

Considering AA and BB options with values

B*********B in 9C4 × 5C5= 126 ways A1 Any one option correct

756 + 126 DM1 Summing their AA and BB outcomes only

Total = 882 A1)

Total: 4


May/June 2017



6(ii) EITHER: (The subtraction method) As together, no restrictions 8! / 2!5! = 168

(*M1 Considering all As together – 8! seen alone or as numerator – condone × 4! for thinking A’s not identical

As together and Bs together 7! / 5! = 42 M1 Considering all As together and all Bs together – 7! seen alone or numerator

M1 Removing repeated Bs or Cs – Dividing by 5! either expression or 2! 1st expression only – OE

Total 168 – 42 DM1 Subt their 42 from their 168 (dependent upon first M being awarded)

= 126 A1)

OR1: As together, no restrictions 8C5 x 3C1 = 168 (*M1 8C5 seen alone or multiplied

M1 7C5 seen alone or multiplied

As together and Bs together 7C5 x 2C1 = 42 M1 First expression x 3C1 or second expression x 2C1

Total 168 – 42 DM1 Subt their 42 from their 168 (dependent upon first M being awarded)

= 126 A1)

OR2: (The intersperse method ) (M1

Considering all “As together” with Cs – Mult by 6!

(AAAA)CCCCC then intersperse B and another B M1 Removing repeated Cs – Dividing by 5!– [Mult by 6 implies M2]

*M1 Considering positions for Bs – Mult by 7P2 oe –


May/June 2017



6!5!×7×6÷ 2

DM1 Dividing by 2! Oe – removing repeated Bs (dependent upon 3rd M being awarded)

= 126 A1)

Total: 5

7(i) P(H) = P(BH) + P(SH) = 0.6×0.05 + 0.4×0.75 M1 Summing two 2-factor probs using 0.6 with 0.05 or 0.95, and 0.4 with 0.75 or 0.25

= 0.330 or 33

100

A1 Correct final answer accept 0.33

Total: 2

7(ii) P ( )S H = ( )

( )P S H

P H∩ = 0.4 0.75

0.33× = 0.3

0.33

M1 FTTheir ( )

( )P S H

P H∩ unsimplified, FT from (i)

= 10

11 or 0.909

A1

Total: 2

7(iii) Var (B) = 45×0.6×0.4 Var (S)= 45×0.4×0.6

B1 One variance stated unsimplified

Variances same B1 Second variance stated unsimplified and at least one variance clearly identified, and both evaluated or showing equal or conclusion made SR B1 – Standard Deviation calculated Fulfil all the criteria for the variance method but calculated to Standard Deviation

Total: 2


May/June 2017



7(iv) 1 – P(0, 1) = 1 – [(0.6)10 + 10C1(0.4)(0.6)9] = 1 – 0.0464 OR P(2,3,4,5,6,7,8,9,10) = 10C2(0.4)2(0.6)8 + … + 10C9(0.4)9(0.6) + (0.4)10

M1M1

Bin term 10Cx px(1 – p)10 – x 0 < p < 1 Correct unsimplified answer

= 0.954 A1

Total: 3






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May/June 2017



1 P(6) = 0.3 B1 SOI

P(sum is 9) = P(3, 6) + P(4, 5) + P(5, 4) + P(6, 3) M1 Identifying the four ways of summing to 9 (3,6), (6,3) (4,5) and (5,4)

= (0.03 + 0.02) × 2 M1 Mult 2 probs together to find one correct prob of (3,6), (6,3) (4,5) or (5,4) unsimplified

= 0.1 A1 OE

Total: 4

2 np = 270 × 1/3 = 90, npq = 270 × 1/3 × 2/3 = 60 B1 Correct unsimplified np and npq, SOI

P ( )100>x = P 99.5 90

60− >

z = P(z > 1.2264)

M1M1

±Standardising using 100 need sq rt Continuity correction, 99.5 or 100.5 used

= 1 – 0.8899 M1 Correct area 1 – Φ implied by final prob. < 0.5

= 0.110 A1

Total: 5

3(i) P(S) = 0.65 × 0.6 + 0.35 × 0.75

M1 Summing two 2-factor probs or 1 – (sum of two 2-factor probs)

= 0.653 (261/400) A1

Total: 2


May/June 2017



3(ii) P ( )Std L = ( )

( )∩P Std L

P L= 0.35 0.25

1 0.6525×

− = 0.0875/0.3475

M1

M1

‘P(Std)’ × ‘P(L/Std)’as num of a fraction. Could be from tree diagram in 3(i). Denominator (1 - their (i)) or their (i) or 0.65 × 0.4(or 0.6) + 0.35 × 0.25(or 0.75) = 0.26+0.0875 or P(L) from their tree diagram

= 0.252 (35/139) A1

Total: 3

4(a) P(x > 0) = P 0 µ

σ− > ±

z

= P/1.5µ

µ −

>

z or P 1.5σσ

− >

z

M1 ±Standardising, in terms of µ and/or σ with 0 - …. in numerator, no continuity correction, no √

= P(z > -1.5) A1 Obtaining z value of ±1.5 by eliminating µ and σ, SOI

= 0.933 A1

Total: 3

4(b) z = –1.151 B1 ± z value rounding to 1.1 or 1.2

70 1201.151s−

− = M1 ± Standardising (using 70) equated to a z-value, no cc, no

squaring, no √

σ = 43.4 or 43.5 A1

Totals: 3


May/June 2017



5(i) constant probability (of completing) B1 Any one condition of these two

independent trials/events B1 The other condition

Totals: 2

5(ii) P(5, 6, 7) = 7C5(0.7)5(0.3)2 + 7C6(0.7)6(0.3)1 + (0.7)7 M1A1

Bin term 7Cx(0.7)x(0.3)7-x , x ≠ 0, 7 Correct unsimplified answer (sum) OE

= 0.647 A1

Total: 3

5(iii) P(0, 1, 2, 3, 4) = 1 – their ‘0.6471’ = 0.3529 M1 Find P( 4 ) either by subtracting their (ii) from 1 or from adding Probs of 0,1,2,3,4 with n=7 (or 10) and p = 0.7

P(3) = 10C3(0.3529)3(0.6471)7 M1 10C3 (their 0.353)3(1 – their 0.353)7 on its own

= 0.251 A1

6(a)(i) First digit in 2 ways. 2 × 4 × 3 × 2 or 2 × 4P3 M1 1, 2 or 3 × 4P3 OE as final answer

Total = 48 ways A1

Total: 2

6(a)(ii) 2 × 5 × 5 × 3 M1

M1

Seeing 52 mult; this mark is for correctly considering the middle two digits with replacement Mult by 6; this mark is for correctly considering the first and last digits

= 150 ways A1

Totals: 3


May/June 2017



6(b)(i) OO**** in 18C4 ways M1 18Cx or the sum of five 2-factor products with n = 14 and 4, may be × by 2C2: 4C0 × 14C4 + 4C1 × 14C3 + 4C2 × 14C2 + 4C3 × 14C1 + 4C4 (× 14C0)

= 3060 A1

Totals: 2


May/June 2017



6(b)(ii) Choc Not Choc 0 6= 1 × 16 C6 = 8008 0.2066 1 5= 4 C1 × 16 C5 = 17472 0.4508 2 4= 4 C2 × 16 C4 = 10920 0.2817

OR Choc Oats Ginger

0 0 6 0 1 5 0 2 4 1 0 5 1 1 4 1 2 3 2 0 4 2 1 3 2 2 2

B1 The correct number of ways with one of 0, 1 or 2 chocs , unsimplified or any three correct number of ways of combining choc/oat/ginger, unsimplified

Total = 36400 ways M1 sum the number of ways with 0, 1 and 2 chocs and two must be totally correct, unsimplified OR sum the nine combinations of choc, ginger, oats, six must be totally correct, unsimplified

Probability = 36400/ 20 C6

M1 dividing by 20C6 (38760) oe

= 0.939 (910/969) A1

Totals: 4

7(i) freq = fd × cw 10, 40, 120, 30 M1A1

Attempt to multiply at least 3 fds by their ‘class widths’

Totals: 2


May/June 2017



7(ii) length < 5 < 10 < 20 < 25

cf 10 50 170 200

B1

B1

M1

A1

3 or more correct cfs heights on graph 10, 50, 170, 200 Labels correct cf and length(cm), linear scales from zero (allow 0.5 on horizontal axis) Attempt (at least three) at plotting at upper end points (either 5 or 5.5, 10 or 10.5 etc.) Starting at (0, 0) polygon or smooth curve increasing with plotted points at lengths 5, 10, 20 and 25

Totals: 4

7(iii) median = 14.2 B1 Median (accept 13.2 – 15.2)

‘18.5’ – ‘10’ M1 Subt their LQ from their UQ if reasonable from their graph

IQ range = 8.5 A1FT Correct FT using LQ = 10 and UQ between 17.5 and 19.5

Totals: 3

7(iv) mean = (2.5×10 + 7.5×40 + 15×120 + 22.5×30) / 200 M1 Using mid points (± 0.5) and their frequencies from 7(i) in correct formula

= 14 A1

Totals: 2

cf200

150

100

50

0

5 10 15 20 25 length (cm)






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May/June 2017
















May/June 2017








SOI Seen or implied


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.


May/June 2017



1(i) Poisson with λ = 0.2 B1

1−e−0.2 (1 + 0.2 + 20.22 ) M1 1 – Poisson P(0, 1, 2, 3) attempted, any λ, allow one end

error

= 0.00115 (3 sf) A1 SR: using Bin, ans 0.00115: B1

Total: 3

1(ii) n large (n > 50) B1

np = 0.2 < 5 or p small B1

Total: 2

2 Assume sd still = 3.8 B1 or sd unchanged

H0: µ = 64.0 H1: µ < 64.0 B1

3.8100

63.3 64.0− M1 Standardising with their values (no sd / var mixes) Must have √100

= –1.842 A1

comp "1.842" with z-value "1.842" < 1.96

M1 comp +ve with +ve or –ve with –ve or comp Φ ("1.842") with 0.975 0.9672 < 0.975 OE

No evidence that heights are shorter A1FT OE FT their zcalc

Total: 6


May/June 2017



3(a) 7.1 ± z × 2.675 M1 Expression of correct form must be z (note MR var = 2.6² can

score M1) seen

z = 1.751 B1

6.77 to 7.43 (3 sfs) A1 Must be an interval

Total: 3

3(b) 0.043 M1 Allow 0.083 for M1

= 0.000064 A1

Total: 2

3(c) e.g. Particular day or time of day B1 Allow "Not random"

Total: 1

4(i) Greater area where x < 7.5 than x > 7.5 B1 Allow Graph higher for x < 7.5 than for x > 7.5 or Graph decreasing or equiv expl'n

Total: 1

4(ii) 10

25

kx∫ dx = 1

M1 Attempt Integ f(x) = 1 ignore limits

k 1 10

5x − = 1

k × 110 = 1

A1 Correct integration and limits

k = 10 AG A1 No errors seen

Total: 3


May/June 2017



4(iii) 10

101

5x∫ dx

M1 Attempt Integ xf(x) ignore limits

= 10 [ ]105ln x

= 10(ln10 – ln5)

M1 Correct integration and limits

= 10ln2 or 6.93 (3 sf) A1 OE

Total: 3

4(iv) 10

10

5

1∫ dx − "6.93"2 M1 Attempt (Integ x²f(x)) – (E(x))². No limits M0

= 1.95 (accept 1.96) A1 Use of 6.93 gives 1.97 A0

Total: 2

5(i) W ~ N(6210, 171.88) B2 seen or implied. B1 each parameter

6200 "6210""171.88"− (= – 0.763)

M1 Standardising with their values. No sd / var mix

1 – Φ(“0.763”) M1 For area consistent with their mean

= 0.223 (3 sfs) A1

Total: 5


May/June 2017



5(ii) E(C – 2B) = ̶ 50 M1 “6210”–2(3130) (or E(2B–C)=50

Var(C – 2B) = "171.88" + 22 × 12.12 (= 757.52)

M1

0 ( 50)"757.52"− − (= 1.817)

M1 Standardising with their values

Φ(“1.817”) M1 For area consistent with their mean

= 0.965 (3 sfs) A1

Total: 5

6(i) mean = 6.6 B1 B1 for 6.6 (could be scored in iii)

P(X ⩽ 1) = e–6.6 (1 + 6.6) = 0.0103 M1 Allow incorrect λ in both probs

P(X ⩽ 2) = e–6.6(1 + 6.6 + 26.62 )= 0.0400 M1A1 A1 for both values

CR is X ⩽ 1 DA1 Dep on at least one M

P(Type I error) = P(X ⩽ 1) = 0.0103 B1FT FT their P(X ⩽ 1)

Total: 6

6(ii) Wrongly concluding that (mean) no of (sports) injuries has decreased B1 Must be in context

Total: 1


May/June 2017



6(iii) H0: λ = 6.6 H1: λ < 6.6 B1 Can be scored in (i). Allow µ or λ / 1.1 or 6.6 or P(X ⩽ 2) = 0.0400 > 0.02

2 not in CR M1

No evidence mean no. of injuries has decreased A1FT

Total: 3

6(iv) N(39.6, 39.6) B1 May be implied

29.5 39.639.6− (= −1.605)

M1 Allow with wrong or no cc

Φ(“–1.605”) = 1 – Φ(“1.605”) M1 For area consistent with their mean

= 0.0543 (3 sfs) A1

Total: 4






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May/June 2017

















May/June 2017






allowed)



SOI Seen or implied


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May/June 2017



1 ( )0.801 1 0.8012000× −

(= 0.0000797) M1

0.801± z × "0.0000797" M1 Allow any z-value

z = 1.96 B1

0.784 to 0.818 (3 sf) A1 As final answer. Must be an interval Allow 0.783 to 0.819

Total: 4

2(i) E(X) = 4.197 B1

Var (X) = 4.196 B1 Both to 3dp or better

Total: 4

2(ii) E(X) ≈ Var(X) B1 Condone =

2(iii) ( )2 3 44.1968 4.1968 4.1968 4.19682 3! 4!1 4.1968e− + + + + M1 Any λ. Allow with one end error

= 0.59(0) (3 sfs) A1 Allow 0.591

Total: 2


May/June 2017



3(i) Est (µ) = 923/400 or 2.3075 or 2.31 (3 sf) B1

Est(σ2) = 2400 3170 "2.3075"

399 400 −

OE M1

= 2.60696 or 2.61 (3 sf) A1 (Note: Biased Var= 2.600 scores M0)

Total: 3

3(ii) H0: Pop mean (or µ) = "2.31" or "2310" H1: Pop mean (or µ) > "2.31" or "2310"

B1 FT

± 2.6 "2.310"2.60696 50−

÷= 1.27 M1 A1 Standardising using their values, Accept 1.28

Comp 1.645 (OE) M1 Valid comparison z values or areas

No evidence that incomes in the region greater A1 FT OE FT their z. No contradictions (No FT for 2 tail test – max score B0 M1 A1 M1 for comp 1.96 A0) Note: Accept alternative CV method

Total: 5


May/June 2017



4(i) 0.7520 + 20 × 0.7519 × 0.25 + 20C2× 0.7518× 0.252 M1 No end errors

= 0.0913 A1 As final answer

Total: 2

4(ii) H0: Pop proportion=0.25 H1: Pop proportion<0.25

B1 Allow p or π, not "proportion" (Accept anywhere in the question)

0.7525 + 25 × 0.7524 × 0.25 M1 Must be B(25,0,25) No end errors

= 0.00702 A1

comp 0.01 M1 Valid comparison

There is evidence that the claim is not justified A1 FT OE. No contradictions

Total: 5


May/June 2017



5(i) 0.5×1× h=0.25 h = 0.5 grad = 0.5

M1 P(X < 2) = 4 × P(X < 1) M1

f(x) = 0.5x A1 P(X < 2) = 1 A1 a = 2 A1

0.5 × a × 0.5a = 1 M1 0.5 × 2 × h/ = 1 M1 h/ = 1

a = 2 A1 grad = 0.5

P(X < 2) = 1 A1 f(x) = 0.5x A1

Total: 5

5(ii) 0 0.5 0.5m xdx =∫ M1 Attempt ( )f d 0.5x x =∫ Ignore limits

= 2

4 00.5

mx = A1FT Correct integration (ft f(x) ) & limits = 0.5

m = 2 or 1.41 (3 sf) A1or by similarity m = 1 2

2× M2

= 2 A1

Total: 3


May/June 2017



6(i) 22.4 2.42!e− × M1 Allow incorrect λ

= 0.261 (3 sfs) A1

Total: 2

6(ii) N(60, 60) B1 seen or implied

54.5 6060− (= ̶ 0.710) M1 allow with wrong or missing cc

1 ̶ φ(" ̶ 0.710") = φ("0.710") M1 For area consistent with their working

= 0.761 (3 sf) A1

Total: 4

6(iii) λ = 3.6 + 12 ÷ 7 (= 186/35) (= 5.314) M1

( )2 35.314 5.314 5.3142 3!1 5.314e− + + + M1 Allow incorrect λ. Allow one end error.

= 0.224 (3 sfs) A1

Total: 3


May/June 2017



7(a) E(X1+X2) = 2 × 4.2 = 8.4 Var(X1+X2) = 2 × 1.1² = 2.42

B1 Both. Seen or implied (or sd = 1.56)

10 8.42.42− (= 1.029) M1 Standardising with their mean and var (no sd / var mix)

1 ̶ φ("1.029") M1 For area consistent with their working

= 0.152 (3 sf) A1

Total: 4

7(b) E(X) = 20.5 B1

Var(X) = 105 + 0.52 × 15 (= 108.75) M1 correct expression oe

0 "20.5""108.75"− (= ̶ 1.966) M1 correct standardisation using their E & V (no sd/var mix)

ignore any attempted cc

φ(" ̶ 1.966") = 1 ̶ φ("1.966") (= (1 ̶ 0.9754))

M1 For area consistent with their working

= 0.0246 or 2.46% (3 sf) A1 Accept 0.0247

Total: 5






MARK SCHEME

Maximum Mark: 50

Published



May/June 2017
















May/June 2017








SOI Seen or implied


Penalties




May/June 2017



1 573, 43 (or 043), 289 B1B1B1 Ignore incorrect numbers. But allow other correct use of table (i.e. 573, 650, 431)

Total: 3

2(i) z = 1.751 B1

103200 ± z

103 103200 200(1 )

200× − oe

M1 all correct except for recognisable value of z, allow for one side only

= 0.453 to 0.577 (3 sf) as final answer A1 must be an interval

Total: 3

2(ii) 0.08 oe 8%, 8/100 B1

3 10 × 0.462 (= 2.116) or

0.4610

B1 SOI

Total mass of ore ~ N(70, 2.116) or

~N2

0.467,10

B1

71 "70""2.116"−

± or 7.1 "7.0"0.46 / 10

−± (= 0.687)

M1 correct, using their sd or √(their var) e.g. allow 71 "70"

4.6− for M1

1 – ɸ("0.687") M1 for correct area consistent with their working

= 0.246 (3 sf) A1

Total: 5


May/June 2017



4(i) x = 6.7/200 (= 67/2000 = 0.0335) B1

s2 = 2200 0.2312 "0.0335"

199 200 × −

M1

s2 = 20.2312 0.0335200

− M0

= 0.0000339(2) = 27/796000 A1 = 0.00003375 A0

Total: 3

4(ii) H0: Pop mean level = 0.034 H1: Pop mean level ≠ 0.034

B1 not just "mean", but allow just “µ”

"030335" 0.034"0.00003392"

200

− M1 must have 200

"0.00003375"200

0.0335 0.034− M1

= –1.21(4) (3 sfs) (–1.22↔–1.21) A1 = –1.217 (3 sfs) A1

Comp with z = −1.645 (or 0.1124>0.05) M1 0.112 > 0.05 valid comparison z or areas

No evidence that (mean) pollutant level has changed, accept H0 (if correctly defined)

A1FT correct conclusion no contradictions SR: One tail test: B0, M1A1 as normal, M1 (comparison with 1.282 consistent signs) A0

Total: 5


May/June 2017



5(i)(a) X ~ N(42, 42) B1 stated or implied

39.5 "42""42"− (= ̶ 0.386)

M1 allow with wrong or no cc

1 ̶ɸ (" ̶ 0.386") = ɸ ("0.386") M1 correct area consistent with their working

= 0.65(0) (3 sf) A1

Total: 4

5(i)(b) 42 > (e.g. 15) or mean is large B1 λ > 15 or higher, λ = large ignore subsequent work if not undermining what already written

Total: 1

5(ii)(a) Y ~ Po(1.2) B1 stated or implied

1 – e–1.2(1 + 1.2 + 21.22 ) M1 allow any λ allow one end error

= 0.121 (3 sf) A1 Using binomial: 0.119 SR B1

Total: 3

5(ii)(b) 60 × 0.02 = 1.2 < 5 or mean is small B1FT or large n small p FT Poisson only

Total: 1


May/June 2017



6(i) 12

0

( )−∫k x x dx = 1 M1 Attempt integ f(x) and "= 1", ignore limits

= 2 3

2 3

1

0 −

x xk = 1 A1 correct integration, limits 0 and 1

= 1 12 3 − k = 1 or 6

k = 1 A1 correctly obtained, no errors seen

Total: 3

6(ii) E(X) = 0.5 B1

13 4

0

6 ( )−∫ x x dx M1 Attempt integ x2f(x), limits 0 to 1

(= 1 14 56 − = 0.3)

"0.3" − "0.5"2

M1 their int x2f(x) – their (E(X))2 dep +ve result

= 0.05( = 1/20) A1

Total: 4

6(iii) 12

0.4

6 ( )−∫ x x dx M1

ignore limits, eg M1for 2

2

0.4

6 ( )−∫ x x dx

= 6{ 2 30.4 0.41 12 3 2 3( )− − − } A1FT subst correct limits into correct integration

= 0.648(= 81/125) A1 condone incorrect “k” for A1

Total: 3


May/June 2017



7(i) H0: Pop mean no. accidents = 5.64 H1: Pop mean no. accidents < 5.64

B1 or “= 0.47 (per month)” not just "mean", but allow just "λ" or “µ”

Use of λ = 5.64 B1 used in a Poisson calculation

= e−5.64 (1 + 5.64 + 25.642 ) M1 Allow incorrect λ in otherwise correct

= 0.08(0) A1

Comp with 0.05 M1 Valid comparison (Poisson only), no contradictions.

No evidence to believe mean no. of accidents has decreased; accept H0 (if correctly defined)

A1FT Normal distribution: M0M0

Total: 6

7(ii) Mean < 0.47 but conclude that this is not so B1 (Mean) no. of accidents reduced, but conclude not reduced. Must be in context.

Total: 1

7(iii) (Need greatest x such that P(X ⩽ x) < 0.05 ) P(X ⩽ 1) = e−5.64 (1 + 5.64) = 0.024 P(X ⩽ 2) = 0.08

B1 Both, could be seen in (i)

Hence rejection region is X ⩽ 1 B1 Can be implied

With λ = 12× 0.05 = 0.6, 1 ̶ P(X ⩽ 1) = 1 ̶ e−0.6(1+ 0.6)

M1 λ=0.6 and 1 ̶ P(X ⩽ 1)

= 0.122 (3 sf) A1 Normal scores 0

Total: 4

Documents

Cambridge International Examinations Cambridge ...maxpapers.com/wp-content/uploads/2012/11/9709_s17_ms_all.pdfMR –1 A penalty of MR –1 is deducted from A or B marks when the data