a level math testttt

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS
GCE Advanced Subsidiary Level and GCE Advanced Level



MARK SCHEME for the May/June 2010 question paper

for the guidance of teachers



9709 MATHEMATICS
9709/11
Paper 11, maximum raw mark 75


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

Mark schemes must be read in conjunction with the question papers and the report on the
examination.



*
CIE will not enter into discussions or correspondence in connection with these mark schemes.



CIE is publishing the mark schemes for the May/June 2010 question papers for most IGCSE, GCE
Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.


www.XtremePapers.net

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Page 2
Mark Scheme: Teachers' version
Syllabus
Paper

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

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.

(c) UCLES 2010
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Page 3
Mark Scheme: Teachers' version
Syllabus
Paper

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



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.
(c) UCLES 2010
www.XtremePapers.net

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Page 4
Mark Scheme: Teachers' version
Syllabus
Paper

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


1 tan
x = k



(i) tan( - x) = -k
B1
co. www Mark final answers
[1]


1


(ii) tan
- x =
B1
co. www
2

k
[1]
k



(iii) sin x =
from 90 triangle.
M1 A1
Any valid method - 90 triangle or
2
1+ k
[2] formulae.
5


2

3

2x -



x



(i) 32x5 - 240x3 + 720x
3 x B1
co.
[3] SC B2 for other 3 terms (i.e. ascending)

2


(ii) 1+
( 32x5 - 240x3 + 720x )
2

x
Coeff
of
x (1 x 720) + (2 x -240)
M1
Looks at exactly 2 terms.


240
A1
co from his answer to (i).
[2]
3 9th term = 22, S4 = 49






(i) a + 8d = 22
B1
co


(
2 2a + 3d) = 49
B1
co


Soln of sim eqns
M1 A1
Solution of two linear sim eqns. co
d = 1.5, a = 10
[4]




(ii) a + (n - )
1 d = 46


Substitutes
for
a and d
M1
Correct formula needed and attempt to


n = 25
A1
solve. co.
[2]
4
2
y = 6x - x


Meets
y = 5 when x = 1 or x = 5.
B1
co
Integral
=
2
1 3
3x - x
3
M1 A1
attempt to integrate. co.

Their limits (1 to 5) used 30
DM1
value at top limit - value at lower

Area of rectangle = 20
B1
co to his x values

Shaded area = 10
A1
co

[6]
(integral
of
6
2
x - x - 5 B1, M1, A1
DM1 as above, then "-5x" B1 A1)
(c) UCLES 2010
www.XtremePapers.net

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Page 5
Mark Scheme: Teachers' version
Syllabus
Paper

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

5
x
2
a 2sin x
2
- 3cos x


(i)
2
1
(
2 - cos x
2
) - 3cos x
M1
Uses 2
2
s + c = 1



2
2 - 5cos x (a = 2, b = -5)
A1
co
[2]


(ii) Values are -3 and 2
B1 B1

[2]

(iii)
2
2 - 5cos x = -1



cos2 x = 0.6
B1



x = 0.685, 2.46 (accept 0.684)
B1 B1
co for - (first answer)
[3] SC B1 for both 39.2 and 140.8


6
y
d

= 3 x - 6 (9, 2)
x
d
3
3 2
x

(i) y =
- 6x ( +c)
B2,1
Loses 1 for each error - ignore +c
3
2


(9, 2) 2 = 54 - 54 + c
M1
Uses (9, 2) with integration to find c.


c = 2.
A1
co.
[4]
y
d

(ii)
= 0 x = 4
B1
Ignore any
x
y-value
d
1
d2
-
y 3 2


= x
M1 A1
Any valid method. co.
d2x
2


+ve (or 3/4) Minimum
[3]

7
18

y = 2 -

2x + 3


(i) A is (3, 0)
B1
Anywhere - but not from given answer
y
d


=
2
(
18 2x
)
3 -
+
x 2
B1 B1
B1 for
2
(
18 2x
)
3 -
+
, B1 for x2
x
d
If
x = 3, m = 49 .




m of normal = - 94
M1 Use
of
m1m2 = -1 with m from dy/dx


Equation of normal
9
y = - (x - )
3
4
M1
Correct method for normal


4y + 9x = 27
A1
co (answer was given)
[6]

(ii) Normal meets y-axis at (0, 63/4)


Curve
meets
y-axis at (0, -4)
M1
Needs to put x = 0 in both normal and


BC = 103/4
A1
curve. co
[2]
(c) UCLES 2010
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Page 6
Mark Scheme: Teachers' version
Syllabus
Paper

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

8 (i) y-step / x-step = 2
M1
Gradient = y-step / x step used


m = 1
A1
co

[2]




(ii) Eqn of AC y + 2 = - (
2 x - )
3
M1 A1
Correct form of one of lines. to his m
Eqn
of
BC y - 22 = (x - )
15
A1
to his m


Sim eqns y + 2x = 4 , y = x + 7




C (-1, 6)
A1
co

[4]
(iii) M is (9, 10)
B1
co


Perp gradient is -1/2
M1
Use of m1m2 = -1


2y + x = ,
29 y = x + 7
M1
Solve sim eqns for their BC & perp. bis


Sim eqns D (5, 12)
A1
co
[4]
9 (i) 2 2
x -12x + 7 = (
2 x - )
3 2 -11
3 x B1
B1 for each value - accept if a, b, c not
[3] specifically quoted.

(ii) Range of f [ - 11
B1
to his "c". allow > or [.
[1]

(iii) 2 2
x -12x + 7 < 21




2 2
x -12x -14 or
M1
3-term quadratic to 0 or (
2 x - )
3 2 < 32
(
2 x - )
3 2 < 32




end-values of 7 or -1
A1
Correct end-values


-1 < x < 7
A1
co
[3]

(iv) gf(x) = (
2 2x2 -12x + 7) + k = 0
M1 A1
Puts f into g. co.
Use
of
b2 - 4ac
M1
Used correctly with quadratic


242 - 16(14 + k)




k = 22
A1
co.
[4]
10 OA = i + 3j + 3k, OC = 3i - j + k.




(i) OB = OA + OC = 4i + 2j + 4k
B1 co
Unit
vector
=
1
6 (4i + 2j + 4k)
M1 A1
Divides by the modulus. on OB .
[3]

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


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


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


-8 = 6 x 24 x cos
M1
Connected correctly provided OB, AC used


= 105.8 74.2
A1
co (accept acute or obtuse)

[5]

(iii) OA = 19 or OC = 11
B1
Used as a length.


Perimeter = 2( + )
M1



15.4
A1
co (accept 15.3)
[3]

(c) UCLES 2010
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