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M11/5/MATSD/SP2/ENG/TZ1/XX/M

MARKSCHEME
May 2011

MATHEMATICAL STUDIES

Standard Level

Paper 2

26 pages

–2–

M11/5/MATSD/SP2/ENG/TZ1/XX/M

This markscheme is confidential and for the exclusive use of
examiners in this examination session
...


–3–

M11/5/MATSD/SP2/ENG/TZ1/XX/M

Paper 2 Markscheme
Instructions to Examiners
Notes:

If in doubt about these instructions or any other marking issues, contact your team leader for
clarification
...


1

Abbreviations
The markscheme may make use of the following abbreviations:
M

Marks awarded for Method

A

Marks awarded for an Answer or for Accuracy

G

Marks awarded for correct solutions obtained from a Graphic Display Calculator, irrespective of
working shown
...


ft

Marks that can be awarded as follow through from previous results in the question
In paper 2 candidates are expected to demonstrate their ability to communicate mathematics using
appropriate working
...
Marks to be awarded for unsupported answers are designated G in the
mark scheme as such answers will usually arise from working performed on a graphic display
calculator
...


(b)

Marks must be noted on candidates’ scripts as in the markscheme and show the breakdown of
individual marks using the abbreviations (M1), (A2) etc;

(c)

Working crossed out by the candidate should not be awarded any marks
...


(e)

If correct working results in a correct answer but then further working is developed, full marks are
not always awarded
...
Full
marks can be awarded if the candidate demonstrates clear understanding of the task and the result
...


(f)

Candidate drawn graphs will have a single (A1) available for scales and labels combined
...

In papers which have two candidate drawn graphs, consistent errors in showing labels or scales can
follow through on the second graph, though not if the error is complete omission of these features
...
Marks for actual examination questions will not necessarily follow the same pattern
...
61 3s
...
)

OR

Case (i) 13 or 3
...
50

Marking
(G2)

(M1)
(A0)

Question: Calculate the gradient of the line passing through the points (5,3) and (0,9)
...
)
(M1)
(A0)
(There is
confusion about
what is
required
...

Markscheme

Candidate’s Script

Marking

OR

answer only

sin( A) sin(32)
=
5
61

(M1)(A1)

Case (i)

OR

3

(A2)

A = 55
...
9

61

A = 55
...
To limit the severity of the
penalty, follow through (ft) marks can be awarded
...


•

If an answer resulting from follow through is extremely unrealistic (e
...
negative distances or wrong
by a large order of magnitude) then the final A mark should not be awarded
...


•

If a question is transformed by an error into a different, much simpler question then follow through
might not apply or might be reduced
...


•

To award follow through marks for a question part, there must be working present for that part
and not just an answer based on the follow through
...


•

Inadvertent use of radians will be penalised the first time it occurs
...
Cases of
this kind will be addressed on an individual basis
...

Question: An investment problem with two different rates of interest and a total amount of $600 split
across the rates in consecutive periods:

(a)

Markscheme
$ 600 × 1
...
02) + (
× 1
...
36
(A1)(ft)

$(

answer only
(G1)
OR
Note: The (M1) is for splitting the value
from (a) and forming a sum of products
...


Candidate’s Script
Case (i)
(a) Final amount after 1st period
= $ 600 × 1
...
02 + 301 × 1
...
06

but note
Case (ii)
an (M0) almost always prohibits
the associated (ft) so
(a) $ 600 × 1
...
04 = $626
...
02 = $ 602
(b)

No working
...
06 given
as answer
...
36

–7–

M11/5/MATSD/SP2/ENG/TZ1/XX/M

Question: Using trigonometry to calculate angles and sides of triangles
...
0
(A1)
answer only

Candidate’s Script
sin A sin 30
=
4
3

Marking
(M1) (A0) (use of sine
rule but with wrong
values)

A = 41
...
)
(b)

OR

x = 7 tan A
= 2
...
83 only

case (i)

(G1)

x = 7 tan A
= 6
...
26

4

(M1) (A1)(ft)

(G0)

Using the Markscheme
This markscheme presents a particular way in which each question might be worked and how it should be
marked
...
Once an (M0) has been awarded, all subsequent A marks are lost in that part
of the question, even if calculations are performed correctly, until the next M mark, unless
otherwise instructed in the markscheme
...

Similarly (A1)(R0) cannot be awarded for an answer which is accidentally correct for the wrong
reasons given
...
(Dependence of A and R marks
...
92

(d)

χ crit = 9
...
92
(a)

(A1)

(b)

2

Marking

n=4

(A1)

(c)

Don’t know?

(A0)

(d)

Do not reject null hypothesis
because χ calc 2 > 0

(A0)(ft)

(A1)
(A1)(ft)

Do not reject null hypothesis (A1)(ft)
because χ calc 2 < χ crit 2
(R1)(ft)

(R0)(ft)
((A0) was awarded
here because the

–8–

M11/5/MATSD/SP2/ENG/TZ1/XX/M
reason is wrong
...
92
(a)
(b)

n=4

(c)

(A1)

χ crit 2 = 4
...
92
(a)
(b)

n=1

(c)

(A1)

χ crit 2 = 3
...
Thus, if an answer is wrong then the
working must be carefully analysed in order that marks are awarded for a different method in a
manner that is consistent with the markscheme
...
This includes alternatives obtained with a graphic display calculator
...


Example: Question to find the coordinates of a vertex of a given quadratic
...


or (G1)

Coordinates are (–7/4, –73/8)

(A1)(ft)

OR

OR

–9–

M11/5/MATSD/SP2/ENG/TZ1/XX/M

(–7/4, –73/8) (with no working at all)

(G2)(G1)

OR

OR

f '( x) = 4 x + 7 ,

4x+7 = 0

(M1)

so x = –7/4
(M1) for attempting to take a derivative and setting it to 0
(A1) for answer
146
73
f (− 7 ) = −
=−
4
16
8
(M1) for using f(-7/4), (A1) for answer
...
For example:

sin θ
for tan θ
...

(d)

As this is an international examination, all valid alternative forms of notation should be accepted
...
7; 1’7; 1 ⋅ 7 ; 1,7
...

Different forms of notation for set properties (e
...
complement): A′ ; A ; Ac ; U − A ; (A
Different forms of logic notation:

¬ p ; p′ ; p ; p ; ~ p
...


(e)

Discretionary (d) marks: There will be rare occasions where the markscheme does not cover the
work seen
...
It
must be accompanied by a brief note to explain the decision made
...

A penalty known as an ACCURACY PENALTY (AP) is applied if an answer is either
(i)

rounded incorrectly to 3 significant figures or

(ii)

rounded correctly or incorrectly to some other level of accuracy
...
It applies also when an exact answer
is incorrectly rounded
...
Please see section G in the guidance
document which clearly explains, with the use of screenshots how this works in scoris
...
This is different
to what we have done previously awarding A0AP
...


If the level of accuracy is specified in the question, a mark will be allocated for giving the answer to
the required accuracy
...
This is NOT an accuracy penalty
...

Rounding of an exact answer to 3 significant figures should be accepted if performed correctly
...
Exact answers such as
1
can be written as decimals to less than three significant figures if the result is still exact
...

Ratios of π and answers taking the form of square roots of integers (even if exact squares) or any
2

rational power of an integer (e
...
13, 2 3 , 4 5 , 9 ) may be accepted as exact answers
...
g
...

Answers with no supporting working which are written correct to more than 3 significant figures should
be marked according to the scheme for correct answers with no working, but with an (AP) then applied
...
g
...
If there
is no working shown, and answers are given to the correct two significant figures, apply the (AP)
...

An accuracy penalty should not be applied to an answer that is already incorrect for some other reason
...

For judging equivalence between 3 significant figures and use of minutes and seconds for angles,
guidelines have been issued to paper setters
...


– 11 –

M11/5/MATSD/SP2/ENG/TZ1/XX/M

Examples: The Pythagoras example used before:
Markscheme

Candidates’ Scripts

Marking

OR

answer only

(i)

4

(G0)

(ii)

3
...
6

(G1)(G1)(AP)

(iii)

9 + 4 = 13 (M1)(A1)
(3
...
f
...
6
(iv)

(M1)

9 + 4 = 13

(A1)(AP)

= 3
...
60

(M1)(A1)(AP)

(vi)

9 + 4 = 14 = 3
...
g
...
p
...
605551 = 3
...
p
...
606

(A0)

(iii)

3
...
6056

(G2)

= 3
...
p
...
606 or
3
...
7417

(vi)

9−4 = 5
= 2
...
606

(M1)(A0)
(A1)(ft)
(M0)(A0)
(A1)(ft)
(Note: this is a special case,
where the initial (M0) does not
determine the final (A0)
because the correction to 4dp is
an entirely new task
...
There are two situations
...
If
the first stage of the answer is correct but rounded further on, then it should be penalised at an appropriate
place close to where it is rounded
...


Example: Question: sine rule used to find angle A, with angle B and side b known but side a is first
calculated using Pythagoras in an adjoining triangle
...
9

OR

Candidate’s Script

(M1)(A1)

a = 25 + 36 = 61
= 7
...
8
5

(M1)(A1)(ft)

A = 55
...
8
5
A = 55
...
8
5

(M1)(A0)

A = sin −1 (0
...
1

(i)

(A0)

a = 25 + 36 = 61 = 8

(M1)(A0)

sin( A) sin(32)
=
8
5

(M1)(A1)(ft)

(G2)
(M1)(A1)(ft)

(A1)(ft)
answer only

Marking

(G2)

(ii)

(iii)

(iv)

(A1)(ft)

– 13 –

M11/5/MATSD/SP2/ENG/TZ1/XX/M

A = 58
...
8
A = 55
...

(G0)
(G0)(ft)
(there is no working
to justify the follow
through
...
This will
usually be either whole units or two decimal places, but could differ in rare instances depending on the
currency in question
...
This penalty is applied to the final answer of a question part
only
...


THE FINANCIAL ACCURACY PENALTY IS APPLIED AT MOST ONCE PER PAPER!
Subsequent financial accuracy errors can be ignored and full marks awarded, if all else is correct
...
This is
different to what we have done previously awarding (A0)(FP)
...

The financial accuracy penalty is imposed only for rounding to the wrong level of accuracy and NOT for
incorrect rounding to the required number of places
...

No single answer can receive two penalties
...

Please see the examples below
...
If this
instruction is not present, then do not apply the penalty
...


– 14 –

M11/5/MATSD/SP2/ENG/TZ1/XX/M

Example: A financial question demands accuracy correct to 2dp
...
6189
Markscheme
Financial accuracy penalty (FP) applies
in this question
$231
...
62 or 231
...
)

231
...
)

231
...
)

231

(A1)(FP)
(Both types of
error occurred but
(FP) takes
priority
...
00

(A1)(AP)
(It’s not clear
whether nearest
dollar or 2dp was
really intended but
we interpret as 2dp
rounded
incorrectly
...

This applies both to missing units and to incorrect units
...


THE UNIT PENALTY IS APPLIED AT MOST ONCE PER PAPER! Subsequent unit errors can be
ignored and full marks awarded if all else is correct
...

THE UNIT PENALTY IS APPLIED AT MOST ONCE PER PAPER! Subsequent unit errors can be
ignored and full marks awarded if all else is correct
...
This is different to
what we have done previously awarding (A0)(UP)
...

NOTE: The unit penalty will be flagged in the markscheme at the start of each answer where it could
apply, with the words “Unit penalty (UP) applies in parts (a)…”
...
A (UP) will also be present in the left hand column next to where it applies
...
Candidates
are encouraged to include them but should not be penalised if they are missing
...


No single answer can receive two penalties
...
If the (AP) has already been used, such an
answer is eligible for the unit penalty
...
2 cm
...
213 cm
...
Assume that the (UP) has not been used previously
...


Candidate’s Script

Marking

(i)

66
...
2cm

(A1)

(ii)

1
...
10 cms-1

(A1)

(i)

66
...
10

(A1)

(i)

66
...
10

(A1)(UP)

(i)

66

(A1)(AP) if (AP)
not used previously
but (A1)(UP)
otherwise
...
1

(i)

(A1)(UP) if (AP)
used in part (i) but
(A1)(ft) for correct
follow through to
exact answer if
(UP) used in part (i)
...

(ii)

1
...
They must use mathematical
notation, not calculator notation
...
The comment ‘I used my GDC’ cannot receive a method mark
...

(b)

x = 42

(A1)

(ii)
(c)

(i)

y = 64

(A1)

[2 marks]

(A1)(ft)(A1)

[2 marks]

(G2)

[2 marks]

(A1)(ft)(A1)

[2 marks]

( x , y ) plotted on graph and labelled, M

Note: Award (A1)(ft) for position, (A1) for label
...
998

Note: Award (G1) for correct sign, (G1) for correct absolute value
...
It is not necessary that
the line is seen to intersect the y-axis
...

(f)

y = −0
...
7

(M1)

Note: Award (M1) for substitution into formula or some indication of
method on their graph
...
470(0
...
7 is incorrect
...
0 (accept 71
...

Accept 72 ± 0
...

(g)

Yes since 25 % lies within the data set and r is close to -1

Note: Accept Yes, since r is close to –1
Note: Do not award (R0)(A1)
...

=0
number of elements in A = 36

(A1)
(A1)(ft)(G3)

[3 marks]

(A1)(ft)

[1 mark]

(M1)(A1)(ft)(G2)

[2 marks]

Note: Follow through from (b)
...

(e)

54

Note: Award (M1) for 17, 10, 27 seen
...


Continued…

– 20 –

M11/5/MATSD/SP2/ENG/TZ1/XX/M

Question 2 continued

Part B
(a)

40  1

 , 0
...
3 % 
120  3


(A1)(A1)(G2)

[2 marks]

(A1)(A1)(G2)

[2 marks]

(A1)(A1)(G2)

[2 marks]

(A1)

[1 mark]

Note: Award (A1) for numerator, (A1) for denominator
...
283, 28
...

(c)

8 2

 , 0
...
6 % 
28  7


Note: Award (A1) for numerator, (A1) for denominator
...

(e)

2

(A1)

[1 mark]

(f)

0
...
75 or for correct answer incorrectly
rounded to 3 s
...
or more, (G0) for 0
...

(g)

since χ 2calc < χ 2 crit (5
...

OR
Accept (or Do not reject) H 0 as p-value (0
...
05

(R1)(A1)(ft)

[2 marks]

Notes: Do not award (A1)(R0)
...


Total [23 marks]

– 21 –

M11/5/MATSD/SP2/ENG/TZ1/XX/M

QUESTION 3
(a)

f (2) = 23 +

48
2

= 32

(M1)
(A1)(G2)

[2 marks]

(A4)

[4 marks]

(A1)(A1)(A1)

[3 marks]

(b)

(A1) for labels and some indication of scale in an appropriate
window
(A1) for correct shape of the two unconnected and smooth
branches
(A1) for maximum and minimum in approximately correct positions
(A1) for asymptotic behaviour at y-axis
Notes: Please be rigorous
...

The branches must be smooth: a single continuous line that does
not deviate from its proper direction
...

The y-axis must be an asymptote for both branches
...

(c)

f ′( x ) = 3 x 2 −

48
x2

Notes: Award (A1) for 3 x 2 , (A1) for –48 , (A1) for x −2
...


Continued…

– 22 –

M11/5/MATSD/SP2/ENG/TZ1/XX/M

Question 3 continued

(d)

f ′(2) = 3(2)2 −

48
(2) 2

(M1)

Note: Award (M1) for substitution of x = 2 into their derivative
...

(f)

{ y ≥ 32 } ∪ { y ≤ −32}

Notes: Award (A1)(ft) y ≥ 32 or y > 32 seen, (A1)(ft) for y ≤ −32 or
y < −32 , (A1) for weak (non-strict) inequalities used in both of the
above
...
Accept alternative interval notation
...

If domain is given award (A0)(A0)(A0)
...

Award (A0)(A1)(ft)(A1)(ft) for ]–200 , –32] , [32 , 200[
...

Follow through from their derivative if working is seen
...

Total [20 marks]

– 23 –
QUESTION 4
(a)

I=

M11/5/MATSD/SP2/ENG/TZ1/XX/M

Financial Penalty applies in parts (b) and (e)
...

I = 4800

(A1)

OR
40 000 +

40 000(4)(3)
100

(M1)(M1)

Note: Award (M1) for substituted simple interest formula, (M1) for
addition of 40 000
...

(b)

(FP)

44 800 × 18
...

(c)

(d)

2
...
608

(M1)
(A1)(G2)

[2 marks]

(M1)(M1)

Note: Award (M1) for their 48750 seen or implied, (M1) for ×10
...


Continued…

– 24 –

M11/5/MATSD/SP2/ENG/TZ1/XX/M

Continued Question 4

(e)

517 000 × (1
...

(FP)

= 739 707 ZAR

(A1)(G2)

[3 marks]

Notes: Accept 739 908 if 517 140 used
...


BD2 = 1902 + 1202 − 2(190)(120)cos75

(M1)(A1)

Note: Award (M1) for substituted cosine formula, (A1) for correct substitution
...

(b)

(FP)

cost = 196
...

(c)

sin(ABD) sin(115 )
=
70
196
...

= 18
...
8

(A1)(ft)
(AG)

[3 marks]

Notes: Both the unrounded and rounded answers must be seen for the
final (A1) to be awarded
...
If 197
is used the unrounded answer is 18
...
2
Area =

70 × (196
...
2 )
2

(A1)
(M1)(A1)

Note: Award (M1) for substituted area formula, (A1) for correct
substitution
...

Notes: Follow through from (a) only
...
2 not seen
...
2 seen without subsequent
working, award (A1)(G2)
...
38…× 120
= 596 327 USD

(M1)
(A1)(ft)(G2)

[2 marks]

Notes: Follow through from their (d)
...

r = 4
...
73 with no working
...
7 with no working

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