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® IGCSE is the registered trademark of Cambridge International Examinations.
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.
Page 2 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.
Page 3 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – March 2016 9709 12
© Cambridge International Examinations 2016
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.
Page 4 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – March 2016 9709 12
© Cambridge International Examinations 2016
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
Page 5 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – March 2016 9709 12
© Cambridge International Examinations 2016
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
Page 6 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – March 2016 9709 12
© Cambridge International Examinations 2016
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
® IGCSE is the registered trademark of Cambridge International Examinations.
CAMBRIDGE INTERNATIONAL EXAMINATIONS
Cambridge International Advanced Subsidiary and Advanced Level
MARK SCHEME for the March 2016 series
9709 MATHEMATICS
9709/22 Paper 2 (Pure Mathematics), maximum raw mark 50
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.
Page 2 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – March 2016 9709 22
© 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.
Page 3 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – March 2016 9709 22
© Cambridge International Examinations 2016
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.
Page 4 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – March 2016 9709 22
© Cambridge International Examinations 2016
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]
Page 5 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – March 2016 9709 22
© Cambridge International Examinations 2016
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]
® IGCSE is the registered trademark of Cambridge International Examinations.
CAMBRIDGE INTERNATIONAL EXAMINATIONS
Cambridge International Advanced Subsidiary and Advanced Level
MARK SCHEME for the March 2016 series
9709 MATHEMATICS
9709/32 Paper 3 (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.
Page 2 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – March 2016 9709 32
© 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.
Page 3 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – March 2016 9709 32
© Cambridge International Examinations 2016
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.
Page 4 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – March 2016 9709 32
© Cambridge International Examinations 2016
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]
Page 5 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – March 2016 9709 32
© Cambridge International Examinations 2016
(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]
Page 6 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – March 2016 9709 32
© Cambridge International Examinations 2016
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]
Page 7 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – March 2016 9709 32
© Cambridge International Examinations 2016
(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]
® IGCSE is the registered trademark of Cambridge International Examinations.
CAMBRIDGE INTERNATIONAL EXAMINATIONS
Cambridge International Advanced Subsidiary and Advanced Level
MARK SCHEME for the March 2016 series
9709 MATHEMATICS
9709/42 Paper 4 (Mechanics), maximum raw mark 50
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.
Page 2 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – March 2016 9709 42
© 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.
Page 3 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – March 2016 9709 42
© Cambridge International Examinations 2016
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.
Page 4 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – March 2016 9709 42
© Cambridge International Examinations 2016
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
Page 5 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – March 2016 9709 42
© Cambridge International Examinations 2016
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
Page 6 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – March 2016 9709 42
© Cambridge International Examinations 2016
(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 =
Page 7 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – March 2016 9709 42
© Cambridge International Examinations 2016
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
Page 8 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – March 2016 9709 42
© Cambridge International Examinations 2016
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
® IGCSE is the registered trademark of Cambridge International Examinations.
CAMBRIDGE INTERNATIONAL EXAMINATIONS
Cambridge International Advanced Subsidiary and Advanced Level
MARK SCHEME for the March 2016 series
9709 MATHEMATICS
9709/52 Paper 5 (Mechanics), maximum raw mark 50
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.
Page 2 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – March 2016 9709 52
© 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.
Page 3 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – March 2016 9709 52
© Cambridge International Examinations 2016
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.
Page 4 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – March 2016 9709 52
© Cambridge International Examinations 2016
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
Page 5 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – March 2016 9709 52
© Cambridge International Examinations 2016
(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
® IGCSE is the registered trademark of Cambridge International Examinations.
CAMBRIDGE INTERNATIONAL EXAMINATIONS
Cambridge International Advanced Subsidiary and Advanced Level
MARK SCHEME for the March 2016 series
9709 MATHEMATICS
9709/62 Paper 6 (Probability and Statistics), maximum raw mark 50
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.
Page 2 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – March 2016 9709 62
© 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.
Page 3 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – March 2016 9709 62
© Cambridge International Examinations 2016
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.
Page 4 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – March 2016 9709 62
© Cambridge International Examinations 2016
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
Page 5 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – March 2016 9709 62
© Cambridge International Examinations 2016
(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
Page 6 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – March 2016 9709 62
© Cambridge International Examinations 2016
(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
® IGCSE is the registered trademark of Cambridge International Examinations.
CAMBRIDGE INTERNATIONAL EXAMINATIONS
Cambridge International Advanced Subsidiary and Advanced Level
MARK SCHEME for the March 2016 series
9709 MATHEMATICS
9709/72 Paper 7 (Probability and Statistics), maximum raw mark 50
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.
Page 2 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – March 2016 9709 72
© 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.
Page 3 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – March 2016 9709 72
© Cambridge International Examinations 2016
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.
Page 4 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – March 2016 9709 72
© Cambridge International Examinations 2016
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
Page 5 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – March 2016 9709 72
© Cambridge International Examinations 2016
(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