UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS

GCE Advanced Subsidiary Level and GCE Advanced Level

MARK SCHEME for the May/June 2012 question paper

for the guidance of teachers

9709 MATHEMATICS
9709/11 Paper 1, 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.

• Cambridge will not enter into discussions or correspondence in connection with these mark schemes.

Cambridge is publishing the mark schemes for the May/June 2012 question papers for most IGCSE,
GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level
syllabuses.
 Page 2 Mark Scheme: Teachers’ version Syllabus Paper
GCE AS/A LEVEL – May/June 2012 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.

© University of Cambridge International Examinations 2012

 Page 3 Mark Scheme: Teachers’ version Syllabus Paper
GCE AS/A LEVEL – May/June 2012 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.

© University of Cambridge International Examinations 2012

 Page 4 Mark Scheme: Teachers’ version Syllabus Paper
GCE AS/A LEVEL – May/June 2012 9709 11

1 tan 2x = 2 M1
2x = 63.4 or 243.4 A1 1 solution sufficient
x = 31.7 or 121.7 (allow 122) A1A1 For 2nd A1 allow 90 + 1st soln prov.
[4] only 2 solns in range. Alt methods
possible

2 [7C3] × [(2x3)4] × [(–1/x2)3] seen soi B1B1 2 elements correct, 3rd element correct
35 × 24 × (–1)3 leading to their answer soi B1 2 elements correct. Identifying reqd
‒560(x6) as answer B1 term
[4] SC B3 for [560(x)6] as answer

3 AQ (or r) = 3 B1 soi Allow 1.73

3 soi ft their 3 Allow 1.73
Area ∆ = 3 (or area ∆ AQC = ) B1
2 ft their 3 . Allow 1.57. SCA1 for π/4

Area sector APR =

1
2
( 3 ) × π3 = π2
2
M1A1 from
1
2
( 3 ) × π6
2
provided ∆ =
2
3

π
Shaded region = 3− oe cao A1
2 [5]

3.2 2
4 1000k = 3.2 ⇒ k = or or 0.0032 oe M1A1
1000 625
 dM  2
  = 3kr B1
 dr 
dM dM dr
= × used e.g. 3 × k × 10 2 × 0.1 M1 Must eventually make dM/dt subject
dt dr dt cao. Non-calculus methods (e.g. →
0.096 A1
0.09696) can score only 1st 2 marks
[5]

5 (i) ( ) 2
6x + 2 = 7 x ⇒ 6 x − 7 x + 2 = 0 M1 Expressing as a clear quadratic soi
(3 x − 2)(2 x − 1) = 0 M1 oe e.g. (3t − 2 )(2t − 1) = 0
2 1
x= or A1 1 solution sufficient. Accept e.g. t = 2/3
3 2
4 1
x = or (or 0.444, 0.25) A1 Both solutions required cao
9 4
OR (6 x + 2 ) = 49 x → 36 x 2 − 25 x + 4 = 0
2
M1A1 Attempt to square both sides
(9 x − 4)(4 x − 1) = 0 M1 Attempt to solve (or formula etc.)
4 1
x = or (or 0.444, 0.25) oe A1
9 4
[4]

(ii) 7 2 − 4 × 6 × k (= 0 ) M1 Apply b 2 − 4ac(= 0 )

49
k= or 2.04 A1 Attempt to equate derivatives
24
OR
d
dx
( ) 1 d
dx
7 −1
7 x 2 = (6 x + k ) → x 2 = 6
2
M1

49 49 49
x= , y= →k = or 2.04 A1
144 12 24 [2]

© University of Cambridge International Examinations 2012

 Page 5 Mark Scheme: Teachers’ version Syllabus Paper
GCE AS/A LEVEL – May/June 2012 9709 11

6 (i) 2 p 2 − 2 p + 2 + 12 p + 6 → 2 p 2 + 10 p + 8 M1 Correct method for scalar product

u.v = 0 B1 Scalar product = 0
( p + 1)( p + 4) = 0 → p = −1 or p = −4 A1 cao Both solutions required
[3]

(ii) u.v = 2 + 0 + 18 = 20 M1 Use of x1 x2 + y1 y2 + z1 z 2

│u│ = 41 or │v│ = 13 M1 Correct method for moduli
20 = 41 × 13 × cos θ oe M1 All connected correctly
θ = 30.0° or 0.523 rads A1 cao
[4]

10
7 (a) S10 = M1 Correct formula with d = ± (cos 2 x − 1)
2[2 + 9(cos 2 x − 1)]
S10 = 5[2 − 9 sin 2 x] M1 Use of c 2 + s 2 = 1 in a correct S10
S10 = 10 − 45 sin 2 x A1 Or a = 10, b = 45
[3]

(b) (i) (0 < ) 1 tan 2 θ < 1 oe M1 Allow <

3
(0 < )θ < π A1 cao Allow <
3 [2]

1
(ii) S∞ = M1
1 π
1 − tan 2
3 6
9
S∞ = or 1.125 A1 cao
8 [2]

8 (i) (x − 2)2 − 4 + k B1B1 a = –2, b = –4

[2]

(ii) f(x) > k – 4 or [k – 4, ∞ ] or (k – 4, ∞ ) oe B1 ft their k – 4. Accept >

[1]

(iii) smallest value of p = 2 B1 ft their 2

[1]

(iv) x − 2 = (± ) y + 4 − k M1
x = 2+ y+4−k A1 ft from their part (i)
f −1 ( x ) = 2 + x + 4 − k A1 cao
Domain is x > k – 4 or [k – 4, ∞ ] B1 ft from their part (ii). Accept >
or (k – 4, ∞ ) oe [4]

© University of Cambridge International Examinations 2012

 Page 6 Mark Scheme: Teachers’ version Syllabus Paper
GCE AS/A LEVEL – May/June 2012 9709 11

1
9 (i) M = (1, 4) gradient = soi B1B1
2
grad of MB = –2 soi M1 Use of m1m2 = –1
Equation MB : y – 4 = –2(x – 1) A1 Or y = –2x + 6 ft on their ½ or M
When y = 0, x = 3 or B = (3, 0) A1 ft result of putting y = 0 into their eqn
[5]

2 6
(ii) grad of AB = − ; grad of BC = oe M1 At least one correct
6 2
m1m2 = −1(⇒ AB ⊥ AC ) A1 AG Allow omitted conclusion
[2]

(iii) D = (–1, 8) B1
AD = 40 or 6.32 B1
[2]

10 (i) 3 x 2 − 4 x + 1(< )5 M1 Attempt differentiate & put 5 on RHS

(3x + 2)(x − 2) < 0 M1 Attempt to factorise or solve
2  2   2  2
− < x < 2 or − , 2 or  − , 2  . A2 SC Allow A1 for − and 2 seen
3  3   3  [4] 3
Allow <

(ii) 3 x 2 − 4 x + 1 = 0 ⇒ (3 x − 1)(x − 1) = 0 M1 Derivative = 0 & any attempt to solve

1
x = or 1 A1 Both
3
4
y= or 0 A1 Both
27
1
f″(x) = 6x – 4 → f″   = –2 (< 0);
 3
f″(1) = 2 (>0) M1 Or other valid method
1 4 
max at  ,  ; min at (1, 0) cao A1 Allow just x values or just y values
 3 27  [5] given for identification

© University of Cambridge International Examinations 2012

 Page 7 Mark Scheme: Teachers’ version Syllabus Paper
GCE AS/A LEVEL – May/June 2012 9709 11

4 B1 AG At least 1 step of working needed

11 (i) x= −1 [1]
y2

 4   4  B1B1
(ii) ∫  y 2
− 1 dy = − − y 
  y 
4
Upper limit = 2 B1 For − , –y
y
 4  
 − 2 − 2  − (− 4 − 1) M1 Apply limits 1 and their 2 ‘correctly’
  
SC B2 for 2(x + 1) 2 dx − 3 → 1
−1
1 A1 ∫
[5]

 16 8 
(iii) (π ) x 2 dy = (π )  4 − 2 + 1 dy
∫ ∫ B1B1
y y 

 
(π ) − 163 + 8 + y  B1
 3y y 

(π ) − 16 + 4 + 2  −  − 16 + 8 + 1 M1 Apply limits 1 and their 2 ‘correctly’

 24   3 
5π
A1
3 [5]

© University of Cambridge International Examinations 2012