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CAMBRIDGE INTERNATIONAL EXAMINATIONS
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GCE Advanced Level
MARK SCHEME for the October/November 2012 series
9231 FURTHER MATHEMATICS
9231/11
Paper 1, maximum raw mark 100
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 October/November 2012 series for most IGCSE,
GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level
components.
Page 2
Mark Scheme
GCE A LEVEL – October/November 2012
Syllabus
9231
Paper
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.
© Cambridge International Examinations 2012
Page 3
Mark Scheme
GCE A LEVEL – October/November 2012
Syllabus
9231
Paper
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.
© Cambridge International Examinations 2012
Page 4
Qu
No
1
Mark Scheme
GCE A LEVEL – October/November 2012
Commentary
Solution
Changes to cartesian.
⇒ r cos θ + r sin θ = 2 ⇒ x + y = 2 or y = 2 − x .
A1
Sketches graph.
π
to the initial line.
4
Point (2,0) clearly indicated.
B1
Uses formula for mean
value.
Integrates
Uses formula for ycoordinate.
Integrates.
3
Marks
and uses compound
angle formula.
Straight line at –
∫
2 x 2 dx
0
∫
=
∫
M1
A1
B1
4
4
0
4 xdx
M1A1
[5]
3
M1
[x ]
2 4
0
4

 3
3
x2
4


 0
=
16 × 3 3
=
32
2
M1A1
M1
Form for PI and
differentiates.
PI: x = pt 2 + qt + r ⇒ x& = 2 pt + q ⇒ &x& = 2 p
M1
Compares cefficients
and solves.
13 p = 26 ⇒ p = 2 8 p + 13q = 3 ⇒ q = −1
2 p + 4q + 13r = 13 ⇒ r = 1
Finds CF.
2
1
4
2 x 2 dx
0
m 2 + 4m + 13 = 0 ⇒ m = −2 ± 3 i
CF: e −2t ( A cos 3t + B sin 3t )
Solves AQE.
3
M1
1 3 
8
=  x2  =
 3  0 3
1
2
Part Total
Mark
1
4
4
(ii)
Paper
11
π

r cosθ −  = 2
4

 cos θ sin θ 
 = 2
r 
+
2 
 2
Re-writes equation
2 (i)
Syllabus
9231
GS:
x = e −2t ( A cos 3t + B sin 3t ) + 2t 2 − t + 1
© Cambridge International Examinations 2012
3
[6]
6
[6]
A1
M1A1
A1
Page 5
Qu
No
4
Mark Scheme
GCE A LEVEL – October/November 2012
Commentary
Verifies given result.
Uses method of
differences to
sum first series.
n
Subtracts
∑r
r =1
Solution
Applies sum of squares
formula to obtain result.
5
Integrates by parts to
obtain reduction formula.
B1
∑ r (r + 1) =
Part Total
Mark
1
r =1
1
{[f (n) − f (n − 1)] + [f (n − 1) − f (n − 2)] + ... + [f (1) − f (0]}
3
1
= n(n + 1)(n + 2) (AG) (Award B1 if ‘not
3
hence’.)
n
n
2
r =1
n(n + 1)(n + 2) n(n + 1)
−
3
2
r =1
r =1
1
1
= n(n + 1)(2n + 4 − 3) = n( n + 1)(2n + 1) (AG)
6
6
(12 + 2 2 + ... + n 2 ) + (2 2 + 4 2 + ... + (n − 1) 2 ) =
 n − 1  n + 1 
4

n
1
n(n + 1)(2n + 1)
2  2 
=…= n 2 (n + 1)
+ 
2
6
6
∫
∞
0
M1
A1
2
n
∑ r = ∑ r (r + 1) − ∑ r =
M1
A1
2
M1
M1A1
3
[8]
∞
n −2 x
x e
- 2x
∞
 − e −2 x 
n −1 e
dx =  x n
+
nx
dx

0
2 0
2

n
= I n −1
(AG)
2
Pn : I n =
n!
2 n +1
Proves base case.
n=1
I0 =
States conclusion.
Marks
n
(States proposition.)
Shows Pk ⇒ Pk +1 .
Paper
11
r (r + 1)(r + 2) − (r − 1)r (r + 1) = r (r + 1)(r + 2 − r + 1)
= 3r (r + 1) (AG)
to obtain sum of second
series.
Splits series into two
series.
Syllabus
9231
∫
∞
1
 1

e -2x dx = − e -2x  =
0
2

0 2
1 1 1 1!
I1 = × = = 2
∴ P1 true.
2 2 4 2
∫
∞
k!
for some integer k.
2 k +1
(k + 1)!
k + 1 k!
∴ I k +1 =
× k +1 = k + 2
2
2
2
∴ Pk ⇒ Pk +1
Pk : I k =
Hence by PMI Pn is true for all positive integers n.
© Cambridge International Examinations 2012
M1
A1
2
B1
B1
B1
M1A1
A1
6
[8]
Page 6
Qu
No
6
Mark Scheme
GCE A LEVEL – October/November 2012
Commentary
Proves initial result.
Syllabus
9231
Solution
Marks
cos 4θ = c 4 − 6c 2 (1 − c 2 ) + (1 − c 2 ) 2
4
2
= 8 cos θ − 8 cos θ + 1
Verifies remaining two
cases (B1)
Shows roots of
equation to be as
given.
States sum of roots of
equation.
7
Finds asymptotes to C.
Differentiates and
equates to 0.
Sketch of graph.
Deduct 1 mark
for poor forms at
infinity.
Deduct 1 mark if
intersections
with axes not shown.
Part Total
Mark
(cos θ + i sin θ ) 4 = cos 4θ + i sin 4θ
= c 4 + 4c 3 (is) + 6c 2 (is) 2 + 4c(is) 3 + (is) 4
Verifies any two cases
(B1)
Paper
11
M1
M1
(AG)
A1
4
4 
3

cos π = − cos π − π  = − cos π
7
7 
7

12
2 
9
 2 

cos π = cos − π  = − cos π + π  = − cos π
7
7 
7
 7 

20
6
1
15
cos π = cos π = − cos π = − cos π
7
7
7
7
cos 4π = 1 = −(−1) = − cos 3π
8 cos 4 θ − 8 cos 2 θ + 1 = – ( 4 cos 3 θ − 3 cos θ )
⇒ 8c 4 + 4c 3 − 8c 2 − 3c + 1 = 0 (*)
1
3
5
⇒ cos π , cos π , cos π , - 1 are the roots. (AG)
7
7
7
−4
1
=−
⇒ Result (AG)
Sum of roots of (*) are
8
2
B1
B1
A1
2
M1A1
2
M1A1
2
> 0 ⇒ no turning points if λ < 0 .
( x − 2) 2
A1
[9]
B1
3
M1A1
Or y ′ = 0 ⇒ λx 2 − 4λx + 4λ − 2 = 0
Uses discriminant to show 8λ < 0 ⇒ no T.P.s.
(M1A1)
(A1)
Axes and asymptotes.
LH branch. RH branch. (Indicating intersections with
axes at (0,0) and (3,0).)
B1
B1B1
© Cambridge International Examinations 2012
2
M1
Vertical asymptote x = 2.
2
y = λx + 1 +
⇒ y = λx + 1 is oblique asymptote.
x−2
y ′ = λ − 2( x − 2) −2 = 0 for turning points.
λ=
3
3
3
[9]
Page 7
Qu
No
8
Mark Scheme
GCE A LEVEL – October/November 2012
Commentary
Differentiates.
Solution
1
x& = t −
t
2
y& =
Uses arc length formula.
Paper
11
Marks
1
2t 2
Part Total
Mark
B1
2
Squares and adds.
Syllabus
9231
ds
1
1


=  t 2 −  + 4t =  t 2 + 
dt
t
t


3
1
s =  t 2 + dt
1 
t
2
M1A1
∫
M1
3
Integrates.
Obtains result.
Uses surface area formula.
Integrates.
Inserts limits.
Obtains result.
t 3

=  + ln t 
3
1
1 26
= 9 + ln 3 − =
+ ln 3
3 3
S = 2π
∫
8π
=
3
3
1
A1
(= 9.77)
3

 4 t 2  t 2 + 1  dt = 8π
3 
t  
3

∫
3
1
7
1

 t 2 + t 2 dt




8π
3
6
M1
3
2 9 2 3 
 t2 + t2
3 
 9
1
[
A1
]

 2 2 
 18 3 + 2 3 −  +  
 9 3 

 160 3 64 
(= 283 or = 90.0π )
= π 
− 
27 
 3
=
A1
M1
A1
© Cambridge International Examinations 2012
4
[10]
Page 8
Qu
Commentary
No
Finds vector normal
9
to Π.
Mark Scheme
GCE A LEVEL – October/November 2012
Solution
i
Cartesian equation
of Π.
Substitutes general
point of l2.
Finds value of
parameter.
Finds p.v. of
intersection.
Finds distance along
l from known point
to foot of
perpendicular from
given point to l.
F.t. on nonhypotenuse side
(must be real).
Writes a set of three
equations in
three unknowns for
the intersection of
l with Π.
Paper
11
Marks
Part Total
Mark
k
2 = 2i + 8 j + 7k
1
M1A1
−2
 3 + 8t   2 
 8   2

  
   
 6 + 5t  ⋅  8  = 138 or  5  ⋅  8  = 0
12 − 8t   7 
 − 8  7 

  
   
Independent of t ⇒ parallel, or ⇒ parallel.
A1
Π: 2 x + 8 y + 7 z = 21
Sub. x = 5 + 2 s , y = −4 − s , z = 7 + s
⇒ s = −2
and line meets Π at point with p.v. i – 2j + 5k
A1
4
B1
M1
A1
A1
4
Take (9,11,2) as A, (3,6,12) as B and let C be foot of
perpendicular from A to l.
B1
AB = 62 + 52 +102 = 161
BC =
Finds distance from
point to known
point on l.
j
n= 1 −2
3
Dot product of this
with general point
on l1.
Deduces result.
Syllabus
9231
1
2
2
2
(6 5 + 8 )(6i + 5 j −10k )(8i + 5 j − 8k )
AC = 161 −153 = 8 or 2 2
=
153
= 153
153
(= 2.83)
M1
A1
A1
Alternatively:
5 + 2 s = 2 + λ + 3µ
− 4 − s = 3 − 2λ + µ
7 + s = −1 + 2λ − 2 µ
⇒ s = −2 , ⇒ λ = 2 , µ = −1
and line meets Π at point with p.v. i – 2j + 5k
Solves the set of
equations.
Finds p.v. of
intersection.
© Cambridge International Examinations 2012
(B1)
(M1A1)
(A1)
4
[12]
Page 9
Mark Scheme
GCE A LEVEL – October/November 2012
Qu
Commentary
No
9
Finds vector BA .
Syllabus
9231
Solution
Marks
Part
Mark
Total
B1
BA = 6i + 5j – 10j
i
1
j
k
6 5 − 10
82 + 52 + 8 8 5 − 8
M1A1
1224
= 8 or 2 2 ( = 2.83)
153
=
Paper
11
A1
(4)
Alternatively:
(A)
Take (9,11,2) as A, (3,6,12) as B and let C be
foot of perpendicular from A to l.
Finds distance from point to
known point on l.
(B1)
Finds distance along l from
known point to foot of
perpendicular from given point
to l.
F.t. on non-hypotenuse side
(must be real).
2
AB =
BC =
2
2
6 + 5 + 10 = 161
1
2
2
2
(6i + 5 j − 10k )(8i + 5 j − 8k )
8 +5 +8
153
=
= 153
153
AC = 161 − 153 = 8 or 2 2
(M1)
(A1)
(A1 )
(4)
( = 2.83)
(B)
 8t − 6 


AC =  5t − 5 
10 − 8t 


Finds vector . AC
Uses AC perpendicular to l to
find t.
 8t − 6   8 

  
 5t − 5  .  5  = 0 ⇒ t = 1
10 − 8t   − 8 

  
Finds length AC.
AC =
22 + 02 + 22 = 8
© Cambridge International Examinations 2012
(B1)
(M1A1)
(A1)
(4)
[12]
Page 10
Qu
No
10
Mark Scheme
GCE A LEVEL – October/November 2012
Commentary
States eigenvalues.
Finds eigenvectors.
States P and D.
Finds inverse of P.
Finds An.
Syllabus
9231
Solution
Marks
Part Total
Mark
1
B1
Eigenvalues are 1, 2, 3.
i
j
 28   1 
   
λ = 1 e1 = 0 4 − 16 =  0  ~  0 
 0  0
0 1
3
   
i j k
 4
 
λ = 2 e2 = − 1 4 − 16 =  1 
0
0 0 1
 
i
j
k
 − 4  − 2
   
λ = 3 e3 = − 2 4 − 16 =  6  ~  3 
 2   1 
0 −1 3
   
1 4 − 2


P = 0 1 3 
0 0 1 


k
1 0

D =  0 2n
0 0

Paper
11
0

0
3n 
 1 − 4 14 


Det P = 1 ⇒ Adj P = P =  0 1 − 3 
0 0
1 

-1
M1A1
A1
A1
B1
B1
4
2
M1A1
An = PDP-1
14 
1 − 4


= P 0 2 n − 3.2 n  or
0 0
3 n 

 1 4 . 2 n − 2. 3 n 


3 n +1 P
 0 2n
0
0
3 n 

M1A1
A1
5
 1 [ −4 + 4.2 n ] [14 − 12.2 n − 2.3 n ] 


2n
[ −3.2 n + 3 n +1 ] 
= 0
0

0
3n


States required limit.
 0 0 − 2


3 A →  0 0 3  as n → ∞ .
0 0 1 


−n
n
© Cambridge International Examinations 2012
B1
1
[13]
Page 11
Qu
No
11
(i)
Mark Scheme
GCE A LEVEL – October/November 2012
Commentary
∑α
2
=
(∑ α )
2
−2
Finds S 4 from formula.
(ii)
Solution
EITHER
Substitute α into equation.
Multiply by α n .
Obtain result.
Uses
S −1 =
∑ αβγ
αβγδ
Finds S 3 from formula.
Finds S 5 from formula.
Syllabus
9231
∑ αβ
Paper
11
Marks
Part Total
Mark
α is a root ⇒ α 4 − 3α 2 + 5α − 2 = 0
⇒ α n + 4 − 3α n + 2 + 5α n +1 − 2α n = 0
Repeat for β , γ , δ and sum
⇒ S n + 4 − 3S n + 2 + 5S n +1 − 2 S n = 0 (AG)
S 2 = 0 − 2 × ( −3) = 6
S 4 = 3 × 6 − 5 × 0 + 2 × 4 = 26
S −1 =
−5 5
=
−2 2
5
= −15
2
S 5 = 3 × (−15) − 5 × 6 + 2 × 0 = −75
S3 = 3 × 0 − 5 × 4 + 2 ×
∑α
2
β 3 = S 2 S3 − S5
= 6 × (−15) − (−75) = −15
© Cambridge International Examinations 2012
M1
A1
2
B1
M1A1
3
M1A1
M1A1
M1A1
6
M1
M1A1
3
[14]
Page 12
Qu
No
11
(i)
Mark Scheme
GCE A LEVEL – October/November 2012
Commentary
Solution
(iii)
Paper
11
Marks
OR
Reduces M to echelon form.
Uses dimension theorem.
(ii)
Syllabus
9231
 2 1 −1 4 
2 1 −1 4 




 3 4 6 1  → … →  0 1 3 - 2
−1 2 8 - 7
0 0 0 0 




States basis for R.
Basis for R is
Finds cartesian equation for R.
x = 2λ + µ
y = 3λ + 4 µ
z = −λ + 2 µ
⇒ 2x − y + z = 0
(OE)
2 x + y − z + 4t = 0
y + 3z − 2t = 0
t = λ and z = µ
Finds basis for null space.
⇒ y = 2λ − 3µ
2 × 8 − 7 + k = 0 ⇒ k = −9
Finds a particular solution.
Finds general solution.
M1A1
B1
M1A1
M1
and x = −3λ + 2 µ
 − 3   2 
   
 2   - 3 
⇒ Basis of null space is   ,   (OE)
 0   1 
 1   0 


Evaluates k.
Total
A1
Dim(M) = 4 – 2 = 2
 2   1 
   
 3  ,  4  .
 − 1  2 
   
Part
Mark
 2  1  8 
     
5 3  − 2 4  =  7  (OE) (via equations)
 − 1  2   − 9 
     
 2 
 5 
 − 3
 
 
 
 − 2
 2 
 − 3
x =   + λ  + µ  .
0
1
0
 
 
 
 0 
 1 
 0 
 
 
 
© Cambridge International Examinations 2012
A1A1
9
B1
M1A1
M1A1
5
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