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CAMBRIDGE INTERNATIONAL EXAMINATIONS

Cambridge International Advanced Subsidiary and Advanced Level

MARK SCHEME for the March 2016 series

9709 MATHEMATICS

9709/12 Paper 1 (Pure Mathematics), maximum raw mark 75

This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes 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 March 2016 series for most Cambridge IGCSE

®

and Cambridge International A and AS Level components.

Mark Scheme Syllabus Paper

Cambridge International AS/A Level – March 2016 9709 12

© Cambridge International Examinations 2016

Mark Scheme Notes Marks are of the following three types:

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

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

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

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

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

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

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

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

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

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

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




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

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

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

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

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

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

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

Penalties

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

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

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

PA−1 penalty is usually discussed at the meeting.




1 (i) 480( )x , 532( )− x

B1B1 [2]

Fully simplified

(ii) ( )( )532 80 0− + =p x

2 / 5=p or 32/80 oe

M1

A1 [2]

Attempt to mult. relevant terms & put = 0

2 3 2

3 2

3 2

−

= −

−

x xy (+c)

3 1 1= − + + c 3 2

3−

= + +y x x

B1B1

M1

A1 [4]

Sub 1, 3= − =x y . c must be present Accept 3=c www

3 11 17+ =a d

( )31

2 30 10232

+ =a d

Solve simultaneous equations 4, 27= = −d a

31st term = 93

B1

B1

M1 A1 A1

[5]

At least one correct

4 (a) 3 3 / 2= −x

3

6

−

=x oe

M1

A1 [2]

Accept −0.866 at this stage

Or 3

6 3

−

or 1

2 3

−

(b) (2cos 1)(sin 1) 0θ θ− − = cos 1 / 2 or sin 1θ θ= =

/ 3 / 2θ π π= or

M1

A1 A1A1

[4]

Reasonable attempt to factorise and solve

Award B1B1 www Allow 1.05, 1.57. SCA1for both 60°, 90°

5 (i) Mid-point of AB = (7, 3) soi Grad. of AB = −2 →grad of perp. bisector = 1/2 soi

Eqn of perp. bisector is ( )1

3 72

− = −y x

B1 M1

A1 [3]

Use of

1 2m m = –1

(ii) Eqn of CX is ( )2 2 1− = − −y x

1 1

2 2−x = −2x + 4

x = 9/5, y = 2/5 2 2 2

7.2 1.4= +BX soi

BX = 7.33

M1

DM1

A1

M1 A1

[5]

Using their original gradient and (1,2)

Solve simultaneously dependent on both previous M’s




6 (i) 22 2π π= +A r rh

2

2

10001000 π

π

= → =r h hr

Sub for h into A → 2 20002π= +A r

r AG

B1

M1

A1 [3]

(ii) 2

d 20000 4 0

dπ= ⇒ − =

Ar

r r

=r = 5.4 2

2 3

d 40004

dπ= +

A

r r

0 > hence MIN hence MOST EFFICIENT AG

M1A1

DM1 A1

B1 [5]

Attempt differentiation & set = 0

Reasonable attempt to solve to 3r =

Or convincing alternative method

7 (i) 3

5CP CA= soi

3

5CP = (4i – 3k) = 2.4i – 1.8k AG

M1

A1 [2]

(ii) OP = 2.4i + 1.2k BP = 2.4i −2.4j + 1.2k

B1 B1

[2]

(iii)

BP.CP = 5.76 – 2.16 = 3.6

│BP││CP│= 2 2 2 2 22.4 2.4 1.2 2.4 1.8+ + +

3.6cos

12.96 9

=BPC 1

3

=

Angle BPC = 70.5° (or 1.23 rads) cao

M1

M1

M1

A1 [4]

Use of 1 2 1 2 1 2

+ +x x y y z z

Product of moduli

All linked correctly

8 (i) 2 4 8+ =a b 2

2 3 4 14+ + =a a b ( ) ( )( )2

2 3 8 2 14 2 2 3 0+ + − = → + − =a a a a a

2 or 3 / 2= −a 3 or 5 / 4=b

M1

A1

M1

A1 A1

[5]

Substitute in –2 and –3

Sub linear into quadratic & attempt solution

If A0A0 scored allow SCA1 for either ( )2, 3− or (3/2, 5/4)

(ii) 2

1 13

2 4y x

= − −

Attempt completing of square

( )1 13

2 4− = ± +x y oe

( )1 1 13f

2 4

−

= − +x x oe

Domain of 1f− is ( ) 13 / 4−x �

M1A1

DM1

A1 B1

[5]

Allow with x/y transposed

Allow with x/y transposed

Allow y =..... Must be a function of x

Allow > , 13 / 4− ∞x� � , 13

,4

− ∞

etc




9 (a) (i) BAO = OBA = 2

π

α−

AOB = 2 2 2

π π

π α α α

− − − − =

AG

M1A1

[2]

Allow use of 90º or 180º

Or other valid reasoning

(ii) ( )2 21 12 sin 2

2 2α α−r r oe

B2,1,0 [2]

SCB1 for reversed subtraction

(b) Use of , 46

π

α = =r

1 segment 2 21 1 4 4 sin

2 3 2 3

π π = −

S

84 3

3

π = −

Area ABC 21 4 sin

2 3

π =

T ( )4 3=

213 4 sin

2 3

π − =

T S – 3

2 21 14 4 sin

2 3 2 3

π π −

16√3 −8π cao

B1B1

M1

B1

M1

A1

[6]

Ft their (ii), , α r

OR AXB 4 tan3 6

π

= =

T or

21 4 2 4 3( ) sin

2 3 33

π =

OR 4 3 8

3 3 4 33 3 3

π − = − −

TS

10 (i) 1 / 3=x B1

[1]

(ii) ( ) [ ]d 2

3 1 3d 16

= −

yx

x

When x = 3 d

d

y

x = 3 soi

Equation of QR is ( )4 3 3− = −y x When 0 5 / 3= =y x

B1B1

M1

M1

A1

[5]

(iii) Area under curve ( )31 1

3 1 16 3 3

= − × ×

x

318 0

16 9

− ×

32

9=

Area of 8 / 3∆=

Shaded area 32 8 8

9 3 9= − = (or 0.889)

B1B1

M1A1

B1

A1

[6]

Apply limits: their 1

3 and 3





9709 MATHEMATICS



®





Mark Scheme Notes

Marks are of the following three types:

M Method mark, awarded for a valid method applied to the problem. Method marks are

not lost for numerical errors, algebraic slips or errors in units. However, it is not usually

sufficient for a candidate just to indicate an intention of using some method or just to

quote a formula; the formula or idea must be applied to the specific problem in hand,

e.g. by substituting the relevant quantities into the formula. Correct application of a

formula without the formula being quoted obviously earns the M mark and in some

cases an M mark can be implied from a correct answer.


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

implied).


• When a part of a question has two or more "method" steps, the M marks are generally

independent unless the scheme specifically says otherwise; and similarly when there are

several B marks allocated. The notation DM or DB (or dep*) is used to indicate that a

particular M or B mark is dependent on an earlier M or B (asterisked) mark in the scheme.

When two or more steps are run together by the candidate, the earlier marks are implied and

full credit is given.

• The symbol implies that the A or B mark indicated is allowed for work correctly following

on from previously incorrect results. Otherwise, A or B marks are given for correct work only.

A and B marks are not given for fortuitously "correct" answers or results obtained from

incorrect working.

• Note: B2 or A2 means that the candidate can earn 2 or 0.

B2/1/0 means that the candidate can earn anything from 0 to 2.

The marks indicated in the scheme may not be subdivided. If there is genuine doubt whether

a candidate has earned a mark, allow the candidate the benefit of the doubt. Unless

otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working

following a correct form of answer is ignored.


scheme specifically indicates otherwise.

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

or which would be correct to 3 s.f. if rounded (1 d.p. in the case of an angle). As stated

above, an A or B mark is not given if a correct numerical answer arises fortuitously from

incorrect working. For Mechanics questions, allow A or B marks for correct answers which

arise from taking g equal to 9.8 or 9.81 instead of 10.





AEF Any Equivalent Form (of answer is equally acceptable)

AG Answer Given on the question paper (so extra checking is needed to ensure that

the detailed working leading to the result is valid)

BOD Benefit of Doubt (allowed when the validity of a solution may not be absolutely

clear)

CAO Correct Answer Only (emphasising that no "follow through" from a previous error

is allowed)

CWO Correct Working Only – often written by a ‘fortuitous' answer

ISW Ignore Subsequent Working

MR Misread

PA Premature Approximation (resulting in basically correct work that is insufficiently

accurate)

SOS See Other Solution (the candidate makes a better attempt at the same question)

SR Special Ruling (detailing the mark to be given for a specific wrong solution, or a

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

particular circumstance)

Penalties

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

part question are genuinely misread and the object and difficulty of the question

remain unaltered. In this case all A and B marks then become "follow through "

marks. MR is not applied when the candidate misreads his own figures – this is

regarded as an error in accuracy. An MR–2 penalty may be applied in particular

cases if agreed at the coordination meeting.

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

PA –1 penalty is usually discussed at the meeting.




1 Attempt division at least as far as quotient 22 +x kx M1

Obtain quotient 22 2− +x x A1

Obtain remainder 6 A1 [3]

Special case: Use of Remainder Theorem to give 6 B1

2 Either State or imply non-modular inequality ( ) ( )

2 2

5 2 3− < +x x or corresponding pair of linear equations B1 Attempt solution of 3-term quadratic equation or of 2 linear equations M1

Obtain critical values 8− and 2

3 A1

State answer 8,< −x 2

3>x A1

Or Obtain critical value 8− from graphical method, inspection, equation B1

Obtain critical value 2

3 similarly B2

State answer 8,< −x 2

3>x B1 [4]

3 Use 22ln ln=x x B1

Use law for addition or subtraction of logarithms M1 Obtain 2 (3 )(2 )= + −x x x or equivalent with no logarithms A1 Solve 3-term quadratic equation M1 Obtain 3

2=x and no other solutions A1 [5]

4 (i) Use the iterative formula correctly at least once M1

Obtain final answer 1.516 A1 Show sufficient iterations to justify accuracy to 3 dp or show sign change in interval (1.5155,1.5165) B1 [3]

(ii) State equation 2 31

24

−

= +x x x or equivalent B1

Obtain exact value 58 or 0.2

8 B1 [2] 5 Obtain integral of form 2 1

e+x

k M1

Obtain correct 2 13e

+x A1

Apply both limits correctly and rearrange at least to 2 1e ...

+

=a M1

Use logarithms correctly to find a M1

Obtain 1.097 A1 [5]




6 (i) Use product rule to obtain expression of form 1 2e sin 2 e cos2− −

+x x

k x k x M1

Obtain correct 3e sin 2 6e cos2− −

− +x x

x x A1

Substitute 0=x in first derivative to obtain equation of form =y mx M1

Obtain 6=y x or equivalent with no errors in solution A1 [4] (ii) Equate first derivative to zero and obtain tan 2 =x k M1*

Carry out correct process to find value of x dep M1* Obtain 0.554=x A1 Obtain 1.543=y A1 [4]

7 (i) State 2 d3

d

yy

x as derivative of 3

y B1

Equate derivative of left-hand side to zero and solve for d

d

y

x M1

Obtain 2

2

d 6

d 3= −

y x

x y

or equivalent A1

Observe 2x and 2

y never negative and conclude appropriately A1 [4] (ii) Equate first derivative to 2− and rearrange to 2 2

=y x or equivalent B1

Substitute in original equation to obtain at least one equation in 3x or 3

y M1

Obtain 33 24=x or 3

24=x or 33 24=y or 3

24− =y A1 Obtain (2, 2) A1

Obtain 3 3( 24, 24)− or (2.88, 2.88)− and no others A1 [5]

8 (i) State cos

2sin cos .sin

x

x x

x

B1

Simplify to confirm 22cos x B1 [2]

(ii) (a) Use 2

cos2 2cos 1= −x x B1 Express in terms of cos x M1 Obtain 2

16cos 3+x or equivalent A1

State 3, following their expression of form 2cos +a x b A1 [4]

(b) Obtain integrand as 21sec 2

2x B1

Integrate to obtain form tan 2k x M1*

Obtain correct 1tan 2

4x A1

Apply limits correctly dep M1*

Obtain 1 1

34 4

− or exact equivalent A1 [5]





9709 MATHEMATICS



®





Mark Scheme Notes


M Method mark, awarded for a valid method applied to the problem. Method marks are not

lost for numerical errors, algebraic slips or errors in units. However, it is not usually

sufficient for a candidate just to indicate an intention of using some method or just to quote

a formula; the formula or idea must be applied to the specific problem in hand, e.g. by

substituting the relevant quantities into the formula. Correct application of a formula without

the formula being quoted obviously earns the M mark and in some cases an M mark can be

implied from a correct answer.


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


• When a part of a question has two or more “method” steps, the M marks are generally

independent unless the scheme specifically says otherwise; and similarly when there are several

B marks allocated. The notation DM or DB (or dep*) is used to indicate that a particular M or B

mark is dependent on an earlier M or B (asterisked) mark in the scheme. When two or more

steps are run together by the candidate, the earlier marks are implied and full credit is given.

• The symbol implies that the A or B mark indicated is allowed for work correctly following on

from previously incorrect results. Otherwise, A or B marks are given for correct work only. A and

B marks are not given for fortuitously “correct” answers or results obtained from incorrect

working.



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.






AG Answer Given on the question paper (so extra checking is needed to ensure that the

detailed working leading to the result is valid)

BOD Benefit of Doubt (allowed when the validity of a solution may not be absolutely clear)

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

allowed)

CWO Correct Working Only – often written by a ‘fortuitous’ answer


MR Misread


accurate)


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.




1 Use law of the logarithm of a power, quotient or product M1 Remove logarithms and obtain a correct equation in x, e.g. 2 2

4 4x x+ = A1 Obtain final answer 2 / 3x = , or exact equivalent A1 [3]

2 Use tan(A ± B) formula and obtain an equation in tan θ M1

Using tan45 1° = , obtain a horizontal equation in tan θ in any correct form A1 Reduce the equation to 2

7 tan 2 tan 1 0θ θ− − = , or equivalent A1

Solve a 3-term quadratic for tan θ M1 Obtain a correct answer, e.g. 28.7θ = ° A1

Obtain a second answer, e.g. 165.4θ = ° , and no others A1 [6] [Ignore answers outside the given interval. Treat answers in radians as a misread (0.500, 2.89).]

3 (i) Consider sign of 5 3 2

3 4x x x− + − at x = 1 and x = 2, or equivalent M1 Complete the argument correctly with correct calculated values A1 [2]

(ii) Rearrange the given quintic equation in the given form, or work vice versa B1 [1] (iii) Use the iterative formula correctly at least once M1

Obtain final answer 1.78 A1

Show sufficient iterations to 4 d.p. to justify 1.78 to 2 d.p., or show there is a sign change in the interval (1.775, 1.785) A1 [3]

4 (i) Substitute 1

2x = − and equate to zero, or divide by (2 1)x + and equate constant remainder

to zero M1 Obtain a = 3 A1 [2]

(ii) (a) Commence division by (2x + 1) reaching a partial quotient of 2

2x kx+ M1

Obtain factorisation 2(2 1)(2 2)x x x+ − + A1 [2]

[The M1 is earned if inspection reaches an unknown factor 22x Bx C+ + and an

equation in B and/or C, or an unknown factor 22Ax Bx+ + and an equation in

A and/or B.]

(b) State or imply critical value 1

2x = − B1

Show that 22 2x x− + is always positive, or that the gradient of 3

4 3 2x x+ + is always positive B1* Justify final answer 1

2x > − B1(dep*) [3]

5 (i) State or imply 2

d 3 sec dx θ θ= B1 Substitute for x and dx throughout M1

Obtain the given answer correctly A1 [3]




(ii) Replace integrand by 1 1

2 2cos2θ + B1

Obtain integral 1 1

4 2sin2θ θ+ B1

Substitute limits correctly in an integral of the form sin2c bθ θ+ , where cb ≠ 0 M1 Obtain answer 31

12 83 π + , or exact equivalent A1 [4]

[The f.t. is on integrands of the form cos2a bθ + , where ab ≠ 0.] 6 (i) EITHER: State correct derivative of sin y with respect to x B1

Use product rule to differentiate the LHS M1 Obtain correct derivative of the LHS A1 Obtain a complete and correct derived equation in any form A1

Obtain a correct expression ford

d

y

x in any form A1

OR: State correct derivative of sin y with respect to x B1 Rearrange the given equation as sin / (ln 2)y x x= + and attempt to differentiate both sides B1 Use quotient or product rule to differentiate the RHS M1 Obtain correct derivative of the RHS A1

Obtain a correct expression ford

d

y

xin any form A1 [5]

(ii) Equated

d

y

xto zero and obtain a horizontal equation in ln x or sin y M1

Solve for ln x M1 Obtain final answer 1/ ex = , or exact equivalent A1 [3]

7 (i) Separate variables and attempt integration of one side M1 Obtain term e

y−− A1

Integrate ex

x by parts reaching e e dx x

x x± ∫ M1

Obtain integral e ex x

x − A1

Evaluate a constant, or use limits x = 0, y = 0 M1

Obtain correct solution in any form A1

Obtain final answer ln(e (1 ))x

y x= − − , or equivalent A1 [7] (ii) Justify the given statement B1 [1]




8 (i) EITHER: Substitute for r in the given equation of p and expand scalar product M1

Obtain equation in λ in any correct form A1

Verify this is not satisfied for any value of λ A1 OR1: Substitute coordinates of a general point of l in the Cartesian equation of plane p M1

Obtain equation in λ in any correct form A1

Verify this is not satisfied for any value of λ A1

OR2: Expand scalar product of the normal to p and the direction vector of l M1

Verify scalar product is zero A1

Verify that one point of l does not lie in the plane A1

OR3: Use correct method to find the perpendicular distance of a general point of l from p M1

Obtain a correct unsimplified expression in terms of λ A1 Show that the perpendicular distance is 5 / 6 , or equivalent, for all λ A1

OR4: Use correct method to find the perpendicular distance of a particular point of l from p M1

Show that the perpendicular distance is 5 / 6 , or equivalent A1

Show that the perpendicular distance of a second point is also5 / 6 , or equivalent A1 [3]

(ii) EITHER: Calling the unknown direction vector a b c+ +i j k state equation 2 3 0a b c+ + = B1

State equation 2 0a b c− − = B1 Solve for one ratio, e.g. a : b M1

Obtain ratio a : b : c = 1 : 4 : − 2, or equivalent A1

OR: Attempt to calculate the vector product of the direction vector of l and the normal vector of the plane p, e.g. (2 3 ) (2 )+ + × − −i j k i j k M2

Obtain two correct components of the product A1

Obtain answer 2 8 4+ −i j k , or equivalent A1

Form line equation with relevant vectors M1

Obtain answer 5 3 ( 4 2 )µ= + + + + −r i j k i j k , or equivalent A1 [6]

9 (i) State or obtain A = 3 B1

Use a relevant method to find a constant M1

Obtain one of B = −4, C = 4 and D = 0 A1

Obtain a second value A1

Obtain the third value A1 [5]

(ii) Integrate and obtain 3 4lnx x− B1

Integrate and obtain term of the form 2ln( 2)k x + M1

Obtain term 22ln( 2)x + A1

Substitute limits in an integral of the form 2ln ln( 2)ax b x c x+ + + , where abc ≠ 0 M1

Obtain given answer 3 ln 4− after full and correct working A1 [5]

10 (a) Substitute and obtain a correct equation in x and y B1

Use 2i 1= − and equate real and imaginary parts M1

Obtain two correct equations, e.g. x + 2y +1 = 0 and y + 2x = 0 A1

Solve for x or for y M1

Obtain answer 1 2

3 3iz = − A1 [5]




(b) (i) Show a circle with centre 1 3 i− + B1 Show a circle with radius 1 B1 Show the line Im z = 3 B1 Shade the correct region B1 [4]

(ii) Carry out a complete method to calculate the relevant angle M1

Obtain answer 0.588 radians (accept 33.7°) A1 [2]





9709 MATHEMATICS

9709/42 Paper 4 (Mechanics), maximum raw mark 50


®





Mark Scheme Notes











implied).










A and B marks are not given for fortuitously “correct” answers or results obtained from

incorrect working.






















clear)

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

is allowed)



MR Misread


accurate)





Penalties



remain unaltered. In this case all A and B marks then become “follow through ”


regarded as an error in accuracy. An MR –2 penalty may be applied in particular







1 M1 Attempt KE gain or WD against Res

KE gain = ½ × 105 × (102 – 52) WD against Resistance = 50 × 40

A1

Both correct (unsimplified) KE gain = 3937.5 J WD = 2000 J

Total WD = 5937.5 J B1 3 WD = KE gain + WD against Res

Alternative method

102 = 52 + 2 × 50 × a [a = 0.75] DF – 40 = 105a

M1

Using v2 = u2 + 2as and applying Newton’s 2nd law to the system

DF = 40 + 105 × 0.75 = 118.75 A1

Total WD = 118.75 × 50 = 5937.5 J B1 3 WD = DF × 50

2 (i) DF = 1350 B1

P = 1350 × 32 = 43.2 kW B1 2

(ii) DF – 1350 – 1200g × 0.1 = 0 [DF = 2550]

M1

For using Newton’s 2nd law applied to the car up the hill (3 terms) Allow use of θ = 5.7o

DF = 76500/v M1 For using DF = P/v

v = 30 ms–1 A1 3

3 (i) M1 For resolving forces horizontally

Rx = 40 × (24/25) – 30 × (7/25) [= 30]

A1

Allow Rx = 40 cos 16.3 – 30 sin 16.3

M1 For resolving forces vertically

Ry = 50 – 40 × (7/25) – 30 × (24/25) [= 10]

A1

Allow Ry = 50 – 40 sin16.3 – 30 cos16.3

2 2

x yR R R= +

and 1 y

x

R θ = tan

R

−

M1

For using Pythagoras to find the resultant force R and trigonometry to find the angle θ made by the resultant with the x-axis

R = 31.6 N and θ = 18.4o with the positive x-axis

A1

6




Alternative method for 3(i)

(i) M1 Resolve forces along 40 N direction

R1 = 40 – 50 × (7/25) [= 26] A1 Allow R1 = 40 – 50 sin 16.3

M1 Resolve forces along 30 N direction

R2 = 30 – 50 × (24/25) [= –18] A1 Allow R2 = 30 – 50 cos 16.3

R2 = R1

2 + R2

2 and arctan(–R2/R1) M1 Use Pythagoras and trigonometry

R = 31.6 N and direction is 34.7 – α = 18.4° with positive x–axis

A1

6

Using arctan(18/26) = 34.7° is the angle between R and the 40 N force

(ii) P = 40 B1 1

4 (i) 5cos α = F [F = 4] M1

For resolving forces horizontally Allow use of α = 36.9o throughout

R + 5sin α = 8 [R = 5] M1 For resolving forces vertically

4 = 5µ M1 For using F = µR

µ = 0.8 A1 4

(ii) R + 10sin α = 8 [R = 2] and

F = 0.8 × R [F =1.6]

B1

For resolving forces vertically to find the new value of R and using F = µR

10cos α – F = 0.8a M1 For resolving horizontally

a = 8 ms–2 A1 3

5 (i) [2500 – 2000g × 0.1 – 250 = 2000a]

M1

For using Newton’s 2nd law for the system or for applying Newton’s 2nd law to the car and to the trailer and for solving for a

Allow use of α = 5.7o throughout

a = 1/8 = 0.125 ms–2 A1

2500 – T – 100 – 1200g × 0.1 = 1200 × 0.125 or T – 150 – 800g × 0.1 = 800 × 0.125

M1

For applying Newton’s 2nd law either to the car or to the trailer to set up an equation for T

T = 1050 N A1 4




(ii) –2000g × 0.1 – 250 = 2000a [a = – 1.125]

M1

For applying Newton’s 2nd law to the system with no driving force to set up an equation for a

0 = 30 – 1.125t M1 For using v = u + at

t = 26.7 s A1 3 Allow t = 80/3 s

Alternative method for 5(ii)

(ii) [½ (2000) 302 = 250s + 2000 × g × 0.1s] → s = 400

M1

Apply work/energy equation to find s the distance travelled up the plane with no driving force (3 terms) as: KE loss = WD against F + PE gain

[400 = ½ (30 + 0)t] M1 For using x = ½(u + v)t

t = 26.7 s A1 3 Allow t = 80/3 s

6 (i) [T = 0.8a for A 2 – T = 0.2a for B

0.2g = (0.2 + 0.8)a system]

M1

For applying Newton’s 2nd law either to particle A or to particle B or to the system

M1

For applying N2 to a second particle (if needed) and solving for a

[a = 2] A1

[2.5 = ½ × 2 × t2]

M1

A complete method for finding t such as using s = ut + ½at2

t = 1.58 s

A1

5 Allow 1

102

t =

First Alternative Method for 6(i)

(i) [0.2 × g × 2.5 or ½(0.2 + 0.8)v2] M1 Finding PE loss or KE gain (system)

[0.2 × g × 2.5 = ½(0.2 + 0.8)v2] M1 Using PE loss = KE gain and find v

[v2 = 10] A1

[2.5 = ½ (0 + √10)t] M1 For using s = ½(u + v)t

t = 1.58 s

A1

5 Allow 1

102

t =




Second Alternative Method for 6(i)

(i) [T = 0.8a 2 – T = 0.2a

→ T = 1.6 N]

M1

Apply N2 to A and B and solve for T

[T × 2.5 = ½ (0.8) v2] M1 Use WD by T = KE gain by A, find v

[v2 = 10] A1

[2.5 = ½ (0 + √10)t] M1 Using s = ½(u + v)t

t = 1.58 s

A1

5 Allow1

102

t =

(ii) N = 8 and F = 0.1 × N = 0.8 B1

T – 0.8 = 0.8a and 2 – T = 0.2a

or 0.2g – 0.8 = (0.2 + 0.8)a

M1

For applying N2 to both particles or to the system and solving for a

a = 1.2 A1

v2 = 0 + 2 × 1.2 × 2.5 M1 For using v2 = u2 + 2as

v = √6 = 2.45 ms–1 A1 5

First Alternative Method for 6(ii)

(ii) N = 8 and F = 0.1 × N = 0.8 B1

[0.2 ×g × 2.5 = ½ (0.8 + 0.2) v2 + 0.8 × 2.5]

M1

Apply work/energy to the system as PE loss = KE gain + WD against resistance

A1 Correct Work/Energy equation

M1 For solving for v

v = √6 = 2.45 ms–1 A1 5

Second Alternative Method for 6(ii)

(ii) N = 8 and F = 0.1 × N = 0.8 B1

T – 0.8 = 0.8a and 2 – T = 0.2a M1 Use N2 for A and B and solve for T

T = 1.76 N A1

[T × 2.5 = 0.8 × 2.5 + ½ (0.8) v2] M1 Apply Work/Energy equation to A

v = √6 = 2.45 ms–1 A1 5




7 (i) k = 40 B1 1

(ii) Correct for 0 ⩽ t ⩽ 4

B1

Quadratic curve with minimum at t = 1 approximately, v = 0 at t = 2 and v = k at t = 4. ft on k

Correct for 4 ⩽ t ⩽ 14 B1 Horizontal line at v = k. ft on k

Correct 14 ⩽ t ⩽ 20

B1

3

Line with negative gradient from (14, k) to (20, 28). ft on k

(iii) For 0 ⩽ t ⩽ 4 a = 10t – 10 M1 Attempting to differentiate to find a

1 < t ⩽ 4 A1 2

(iv) 2(5 10 )t t dt∫ − =

3 255

3t t−

M1

For attempting to integrate the given quadratic expression and attempting to apply limits over the interval t = 0 to t = 4

2

3 2

0

55

3A t t

= − =

3 25

2 5 23

− ×

3 25

0 5 03

− − ×

4

3 2

2

55

3B t t

= − =

3 25

4 5 43

− ×

3 25

2 5 2 3

− − ×

A1

Use of limits to obtain A, the integral from t = 0 to t = 2 and B, the integral from t = 2 to t = 4

Full evaluation of A not necessary at this stage

20

3A

= −

Full evaluation of B not necessary at this stage

100

3B

=

C = (40 × 10) + 0.5 × (40 + 28) × 6

B1

For finding the distance travelled in the interval t = 4 to t = 20 using area properties or integration. ft on k

–A + B + C = [20/3 + 100/3 + 400 + 204]

M1

For attempting to evaluate the total distance travelled by P in the interval t = 0 to t = 20. The distance travelled in the first 4 seconds must have been found using integration methods.

Total distance travelled = 644 m A1 5





9709 MATHEMATICS

9709/52 Paper 5 (Mechanics), maximum raw mark 50


®





Mark Scheme Notes



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




implied).









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.






















clear)


is allowed)



MR Misread


accurate)





Penalties












1 Vsinθ = 2g ( = 20) B1 Using vertical motion to greatest height

Vcosθ = 30/2 ( = 15) B1 Using horizontal motion

2V = 2

15 + 220 or tanθ = 20/15

M1

Using Pythagoras or trigonometry

V = 25 ms–1 A1

θ = 53.1° A1 5

2 (i) 60(3 × 0.8/8) × 0.28 = P(0.8 − 0.8 × 0.28) M1

A1

An attempt at taking moments

P = 8.75 AG A1 3

(ii) µ = 8.75/60 M1

µ = 0.146 A1 2

3 h

V = 9cos60

B1

v

V = (±)4.5

B1

Or 9cos60

− 4.5 = 4.5 – gt M1

t = 0.9 A1

Distance = 4.05 A1 5 From 0.9 × 9cos60

4 (i) 2 × 0.56 × 0.28 + 21.2 (0.56 + 1.2/2) =

h(2 × 0.56 + 21.2 )

M1

Moments about BC

h = 0.775 A1

2 × 0.56 × 1 + 2 1.2 (1.2/2) = v(2 × 0.56 +

21.2 )

M1

Moments about BAG

v = 0.775 A1 4

(ii) 45° B1 1

(iii) tanθ=(0.56 + 1.2 – 0.775) / (1.2 – 0.775) M1

θ =66.7° A1 2

5 (i) 24e/0.8 = 0.2g M1

e = 0.2 A1 2

(ii) 24 × 20.2 / (2 × 0.8) (= 0.6)

B1

ft(cv0.2) Initial EE

0.6 × 24.5 / 2 + 0.6gd + 24 × 2

0.2 / (2 × 0.8) = 0.6 × 2

3.5 / 2 + 24 × (0.2 + d 2) / (2 × 0.8)

M1

A1

PE/EE/KE balance attempt

d = distance particle falls

d = 0.4 so AP ( = 0.8 + 0.2 + 0.4) = 1.4m A1 4

(iii) 24 × 20.2 / (2 × 0.8) + 0.6 × 2

4.5 / 2 = 0.6 2

v /2 + 0.6g × 0.5

M1

A1

PE/EE/KE balance, 4 terms. Award B1ft for initial KE if not already seen in part ii

v = 3.5 m 1−s

A1

3

6 (i) 0.2vdv/dx = 0.2gsin30 – 0.1 2x (0.2gcos30)

M1

N2L parallel to the slope

2vdv/dx=10 – ( 3 ) 2x AG

A1

2




(ii) 2 ∫vdv = 2(10 3∫ − x ) dx

M1

Integrates acceleration

2v = 10x – 3

3x /3

A1

No need to show c = 0

0 = 10 – 23x

M1*

Solves accn =0 (x = 2.4028..)

2v = 10 x 2.4 – 3 x 3

2.4 /3

dep

M1*

Puts solution of accn = 0 in 2v (x)

Max v = 4(.002) A1 5

(iii) 0 = 10x – 33x /3

M1

x = 4.16 A1 2

7 (i) Rcos60 + Tcos30 = 0.2 g M1 Resolving vertically

R + T 3 = 4 AG

A1

2

g = 10 must be used

(ii) Tsin30 – Rsin60 = 0.2 2ω × 0.6sin60

(T – R 3 = 0.12 23ω )

M1

A1

2 N2L horizontally with accn= 2ω r

Accept with trig ratios

(iii) (a) Rcos60sin30 + Rsin60cos30 = 2sin30–0.2 × 2

2 × 0.6sin60cos30 M1 Substitutes ω =2 and eliminates T

from (i) and (ii)

R = 0.64 N A1 2 Accept answers between 0.639 and 0.641 inclusive

OR

R + 3R = 4 – 0.12 × 22 × 3

M1

R = 0.64 A1

(b) Tcos30 = 2 M1

T = 2.31

A1 When R = 0, T = 4 3 /3 or 4/ 3

2.31sin30 = 0.2 2ω × 0.6sin60

AND v = ω × 0.6sin60

M1

v = 1.73 m 1−s

A1

4

OR

2.31sin30 = 0.2 2v /(0.6sin60)

M1

v = 1.73 m 1−s

A1

Final pair of marks





9709 MATHEMATICS

9709/62 Paper 6 (Probability and Statistics), maximum raw mark 50


®





Mark Scheme Notes











implied).










A and B marks are not given for fortuitously “correct” answers or results obtained from

incorrect working.






















clear)


is allowed)



MR Misread


accurate)





Penalties












1 (i) Σx = 862 B1 1 Must be stated or replaced in (ii) Can see (i) and (ii) in any order

(ii) 362/10 + a = 86.2 a = 50

M1

A1 2

86.2 ± 36.2 seen oe Correct answer, nfww

2

No of W 0 1 2

Prob 42/90 42/90 6/90 P(0) = 8/10 × 7/9 × 6/8 = 42/90 P(1W) = P(W,NW, NW) × 3 = 2/10 × 8/9 × 7/8 × 3 = 42/90 P(2W) = P(W, W, NW) × 3 = 2/10 × 1/9 × 8/8 × 3 = 6/90

B1

M1

M1

A1 4

0, 1, 2, seen in table with attempt at prob.

3-factor prob seen with different denoms. Mult by 3 All correct

3 (i) P(R) [ (1, 4),(2,5), (3,6),( 4,7),(5,8)] × 2/64 = 10/64

M1

A1 2

List of at least 4 different options or possibility space diagram Correct answer

(ii) P(S) = [(3,8)(3,7)(4,8)(4,7)(4,6)(4,5)(5,8) (5,7)(5,6)(6,8)(6,7)(7,8)] × 2 + (5,5)(6,6)(7,7)(8,8) = 28/64

M1

A1 2

List of at least 14 different options or ticks oe from possibility space Correct answer

(iii) P ( )R S∩ = 4/64 4/64 ≠ 10/64 × 28/64

Events are not independent

B1

M1

A1 3

Comparing their P(R∩S) with (i) ×(ii) with values Correct answer

4 (i) 32 B1 1

(ii) freqs 0 18 32 9 4 fd 0 1.2 1.6 0.6 0.2 cf 2 1 0 10 20 30 40 50 60 70 80

Time (mins)

M1

A1

B1

B1 4

attempt at fd or scaled freq (at least 3 f/cw attempt) correct heights seen on diagram Correct bar ends Labels fd and time (mins) and linear axes or squiggle




(iii) (17.5 × 18 + 35 × 32 + 52.5 × 9 + 70 × 4)/63 = 2187.5/63 = 34.7

M1

A1 2

Σfx/63 where x is midpoint attempt not end pt or cw Correct answer

5 (i) P(Abroad given camping)

= ( )

( ) ( )

P A C

P A C P H C

∩

∩ + ∩

= 0.35 0.15

0.35 0.15 0.65 0.4

×

× + ×

= 0.0525

0.3125

= 0.168

M1

A1

M1

A1

A1 5

Attempt at P(A∩C) seen alone anywhere

Correct answer seen as num or denom of a fraction Attempt at P(C) seen anywhere

Correct unsimplified answer seen as num or denom of a fraction

Correct answer

(ii) (0.65)n < 0.002

n > lg (0.002)/lg(0.65)

n = 15

M1

M1

A1 3

Eqn with 0.65 or 0.35, power n, 0.002 or 0.998 Attempt to solve their eqn by logs or trial and error need a power Correct answer

6 (i) 15P5 = 360360

M1

A1 2

oe, can be implied Not 15C5 Correct answer

(ii) 5 × 10 × 4 × 9 × 3 = 5400

M1

A1 2

Mult 5 numbers Correct answer

(iii) M(5) F(10) 3 2 = 5C3 × 10C2 = 450 ways 4 1 = 5C4 × 10C1 = 50 5 0 = 5C5 × 10C0 = 1 Total = 501 ways

M1

M1

A1 3

Mult 2 combs, 5Cx × 10Cy Summing 2 or 3 two-factor options, x + y =5 Correct answer

(iv) (Couple) M(4) F(9) ManWife + 3 0 = 4C3 × 9C0 = 4 ManWife + 2 1 = 4C2 × 9C1 = 54 Total = 58

M1

M1

A1 3

Mult 2 combs 4Cx and 9Cy

Summing both options x + y =3, gender correct Correct answer

7 (i) z = –1.645 0.9

1.6450.35

m−

− =

m = 1.48

B1

M1

A1 3

± 1.64 to 1.65 seen

Standardising with a z-value accept (0.35)2 Correct answer

(ii) P(< 2) = P2 1.476

0.35z

− <

= P(z < 1.50) = 0.933 Prob = (0.9332)4 = 0.758

M1

M1

A1

M1

A1 5

Standardising no sq , FT their m, no cc

Correct area i.e. F Accept correct to 2sf here Power of 4, from attempt at P(z) Correct answer




(iii) P(t > 0.6µ) = P0.6

/ 3z

µ µ

µ

−>

= P(z > –1.2) = 0.885

M1

M1

A1 3

Standardising attempt with 1 or 2 variables

Eliminating µ or σ

Correct final answer





9709 MATHEMATICS

9709/72 Paper 7 (Probability and Statistics), maximum raw mark 50


®





Mark Scheme Notes











implied).



independent unless the scheme specifically says otherwise; and similarly when there

are several B marks allocated. The notation DM or DB (or dep*) is used to indicate that

a particular M or B mark is dependent on an earlier M or B (asterisked) mark in the

scheme. When two or more steps are run together by the candidate, the earlier marks

are implied and full credit is given.

• The symbol implies that the A or B mark indicated is allowed for work correctly

following on from previously incorrect results. Otherwise, A or B marks are given for

correct work only. A and B marks are not given for fortuitously “correct” answers or

results obtained from incorrect working.

Note: B2 or A2 means that the candidate can earn 2 or 0.


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.



• 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.









clear)


is allowed)



MR Misread


accurate)





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.




1 E(X) = 103

oe Var(X) = 25

9 oe

E(Y) = 10 Var(Y) = 5 E(X + Y) = 40

3 oe or 13.3 (3 sf)

Var (X + Y) = " 25

9" + "5"

sd = 70

3 oe or 2.79 (3 sf)

B1

B1

B1

M1

A1 [5]

For E(X) and Var(X)

For E(Y) and Var(Y) OR For E(X) and E(Y) For Var(X) and Var(Y)

For adding 2 (appropriate) variances

or sd = or 5

32 ×

2 H0: P(hit target) = 0.65 H1: P(hit target) > 0.65 20C2 × 0.352 × 0.6518 + 19 × 0.35 × 0.6519 + 0.6520 = 0.0121 (3 sf) Comp 0.01 There is no evidence (at the 1% level) that she has improved

B1

M1

A1

M1

A1 [5]

Allow p = 0.65 Allow p > 0.65 Allow one end error. Allow p/q mix. Allow (1– ) for M mark A mark recovered following valid comparison For valid comparison She has probably not improved. No contradictions. (SR Use of Normal M0, but M1A1 for valid comparison could be awarded)

3 (i) H0: pop mean journey time = 35.2 mins H1: pop mean journey time < 35.2 mins 34.7 35.2

5.6/ 25

− (= –0.446)

Φ(< "–0.446") = 1 – Φ("0.446") = 0.328 (3 sf)

B1

M1

M1

A1 [4]

Allow "µ". Not "mean journey time"

For standardising (√25 needed)

For correct area consistent with their working As final answer

(ii) H0 is rejected but Type II error can only be made if H0 is not rejected

B1 [1] Allow just "H0 is rejected." oe

4 X – 2Y ~ N(0.1, 0.22 + 4 × 0.12) soi (= N(0.1, 0.08)) 0 0.1

"0.08"

− (= –0.354)

Φ("–0.354 ") = 1 – Φ("0.354") = 0. 362 (3 sf)

B1 B1

M1

M1

A1 [5]

B1 for ± 0.1 B1 for 0.22 + 4 × 0.12

For standardising. Allow without √ sign

For correct area consistent with their working

5 (i) Est(µ) = 14 910

150 (= 99.4)

Est(σ2) = 150 1525000

149 150( – "99.4"2)

= 288.228 z = 2.576 "99.4" ± z × 288.228 150÷ CI = 95.8 to 103 (3 sf)

B1

M1

A1

B1

M1

A1 [6]

Allow M1 if 150149

omitted Accept 2.574–2.579

Any z

(NB Use of biased Var can score 5/6 max)

(ii) 100 lies within this CI Hence yes

B1 [1]

Both needed, ft their CI




(iii) To avoid bias or Necessary to enable statistical inference

B1 [1]

Or any equivalent

6 (i) λ = 3.3 × 25

30 = 2.75

e–2.75(1 + 2.75 + 2

2.75

2)

= 0.481 (3 sf)

B1

M1

A1 [3]

Allow any λ Allow one end error

As final answer. Accept 0.482

(ii) (a) λ (= 3.3 × 365

30) = 40.15

(X ~ Po(40.15) ⇒ X ~ N(40.15, 40.15)) 50.5 "40.15"

"40.15"

− (= 1.633)

1 – Φ("1.633") = 0.0513 (3 sf)

B1

M1

M1

A1 [4]

Accept 40.1 or 40.2

Allow with incorrect or no cc OR no √ sign

For correct area consistent with their working Accept 0.0512

(b) λ > 15 B1 [1] or similar

(iii)

λ = 73

30 oe or 1.1+1.33= 2.43 (3 sf)

1 – e–2.43(1 + 2.43 + 2 3

2.43 2.43

2 3!+ )

= 0.228 (3 sf)

B1

M1

A1 [3]

Allow any λ. Allow one end error

7 (a) (i) E(X) = 1.5 3

3 42

90

(3 )d−∫ x x x

= 4 5

32

9 4 5

3

0

− x x

= 243 2432

9 4 5 − (= 2.7)

Var(X) (= 2.7 – 1.52) = 0.45 oe

B1

M1

M1

A1 [4]

Attempt integ x2f(x) ignore limits

Sub correct limits into correct integral

Ft their E(X), but no ft for –ve Var.

(ii) 0.5 B1 [1]

(iii)

(1 – 13

27) ÷ 2

= 7

27 or 0.259

M1

A1 [2]

or 3

22

9

2

(3 ) dx x x−∫ oe

As final answer

(b) 1 1

2 22 2× × =a or

2

0

d∫ ax x = 1

2

a = 1

4

1 1

2 41× × =b b or

b1

40

d∫ x x = 1

or b = 2 × 2 b = 2 2

M1

A1

M1

A1 [4]

Attempt correct equation in ‘a’

or 1

21× × =b ab or

b

0

d∫ ax x = 1 attempt correct

equation in (a and) b

Allow b = 8 or 2.83 (3 sf) Ft incorrect a, both Ms needed

Total for

paper 50

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