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CAMBRIDGE INTERNATIONAL EXAMINATIONS
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Pre-U Certificate
MARK SCHEME for the May/June 2013 series
1348 FURTHER MATHEMATICS
1348/01
Paper 1 (Further Pure Mathematics),
maximum raw mark 120
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 May/June 2013 series for most IGCSE, Pre-U,
GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level
components.
Page 2
Mark Scheme
Pre-U – May/June 2013
Syllabus
1348
x2 – 6x + 12 ≡ (x – 3)2 + 3
1
6
B1
6
 1
−1  x − 3  

tan 
dx = 
2
 3  2
 3
3 + ( x − 3) 2
∫2 ( )
1
=
Paper
01
M1 tan – 1
 x −3
1st A1 

 3 
1  π  π 
π
 −  −   =
3  3  6  2 3
M1
A1
A1
[4]
ln(1 + x) = x − 12 x 2 + 13 x 3 − 14 x 4 + 15 x 5 + ... from the Formula Book
2
ln(1 – x) = − x − 12 x 2 − 13 x 3 − 14 x 4 − 15 x 5 − ...
1
2
1+ x 
ln 
=
1− x 
1
2
=
{ln(1 + x) − ln(1 − x)}
1
2
B1
M1
{
}
× 2 x + 13 x 3 + 15 x 5 + ... = tanh – 1x from the Formula Book
ln(1 + x) valid for − 1 < x ≤ 1 and so ln(1 – x) is valid for − 1 ≤ x < 1
so LHS valid for − 1 < x < 1 , which matches the range for RHS
A1
B1
[4]
3
(i)
(
)
(
dy
x 2 − 4 .1 − ( x + 1).2 x
Use of quotient rule; correct unsimplified
=
2
dx
x2 − 4
= −
)
(x + 2 x + 4) = − ( x + 1) + 3 or clear explanation this is < 0
(x − 4)
(x − 4)
ALT: y =
M1 A1
2
2
2
2
3
4
x−2
+
2
1
4
x+2
⇒
2
E1
− 34
− 14
dy
=
+
<0
dx ( x − 2 )2 ( x + 2 )2
[3]
© Cambridge International Examinations 2013
Page 3
3
(ii)
Mark Scheme
Pre-U – May/June 2013
Asymptotes y = 0
x=±2
Syllabus
1348
Paper
01
Stated or clear from graph
Stated or clear from graph
Crossing-points (0, – 14 ) and (–1, 0)
B1
B1
Noted or clearly shown on graph
B1 B1
3 regions
M1
All correct (incl. no TPs)
A1
[6]
4
(i)
d1 × d2 attempted
= 14i + 35j – 21k
(ALT: Use of 2 scalar prods. & attempt to get 2 components in terms of the 3rd)
M1
A1
[2]
(ii)
)
Sh. Dist. = | (b – a). n |
=
1
38
(b – a) = ± (i + 3 j − 7k )
(2i + 5 j − 3k ) • (i + 3j − 7k ) =
1
38
)
n=
1
38
(2i + 5 j − 3k ) ft
M1
B1 B1
2 + 15 + 21 ft scalar prod.
B1
=
A1
38 cao
 2 1  1   9   2 
         
ALT: Solving  4  + λ  2  −  1  − µ  − 3  = k  5  to find closest points on line,
 3   4  10   1   − 3 
         
(3, 6, 7) from λ = 1 and (1, 1, 10) from µ = 0 giving k = 1 and Sh.D. =
38
[5]
© Cambridge International Examinations 2013
Page 4
5
(i)
Mark Scheme
Pre-U – May/June 2013
zn −
Syllabus
1348
1
= (cos nθ + i.sin nθ ) – (cos[− nθ ] − i.sin[− nθ ])
zn
Paper
01
M1
De Moivre’s Thm. used for at least zn
= cos nθ + i.sin nθ − (cos nθ − i.sin nθ ) = 2i sin nθ
Given answer obtained from 2 correct uses of de Moivre’s Thm. and correct trig.
A1
[2]
(ii)
5
1

5
 z −  = 32i sin θ
z


M1
5
3
= z − 5 z + 10 z −


=  z5 −
10 5 1
+ −
z z3 z5
1  3 1
1

− 5 z − 3  + 10 z − 
5 
z
z  
z 

= 2i sin 5θ – 10i sin 3θ + 20i sinθ
⇒ sin5θ =
1
16
5
16
sin 5θ –
sin 3θ +
5
8
Use of binomial expansion
M1
Pairing up terms
M1
Use of (i)’s result (×3)
M1
A1
sinθ
[5]
6
(i)
r = 1 + r sinθ ⇒
M1 M1
x2 + y2 = 1+ y
Squaring and cancelling: x 2 + y 2 = y 2 + 2 y + 1 ⇒ y =
1
2
(x
2
)
−1
A1
[3]
(ii)
Parabola
All correct: Crossing-points at (± 1, 0) and (0, – 12 )
M1 A1
[2]
(iii)
2π
2π
1
∫π (1 − sin θ )2 dθ = 2 ×
∫
r 2 dθ
Recognition that this is related to area
M1
Matching up with parabola-related
M1
π
1
= –2
1
2
∫ (x
1
2
−1
2
)
− 1 dx
region
1
 x3
4
− x =
−1 3
3
=– 
Ignore –ve answer
A1
[3]
© Cambridge International Examinations 2013
Page 5
7
(i)
Mark Scheme
Pre-U – May/June 2013
Syllabus
1348
x 3 + y 3 = ( x + y ) 3 − 3xy( x + y ) or equivalent
Paper
01
M1 A1
[2]
(ii)
(a) α + β (= 3) and αβ (=
8
9
) substd. into (i)’s result ft ⇒ α 3 + β 3 = 19
M1 A1
[2]
(b) 9t 2 − 27t + 8 = 0 ⇒ (3t – 1)(3t – 8) = 0 ⇒ α , β =
Then α 3 + β 3 = 19 =
(13 )3 + (83 )3
1
3
,
8
3
Explicit statement required
M1 A1
A1
[3]
8
(i)
(a) x ∈ G ⇒ ∃ x –1 ∈ G and pre-multiplying by this (or x in the ⇐ case) gives the
result
(NB Both directions must be dealt with)
B1
B1
[2]
(b) Since each xgi is distinct, and there are n of them, the set xG is just a
permutation of the elements of G OR mention that it is just a row of the group table
and hence contains a permutation of the elements of G
B1
[1]
(ii)
Multiply all elements together:
(Since G is abelian)
⇒
xg1 xg2 xg3 … xgn = g1 g2 g3 … gn
xn.(g1 g2 g3 … gn) = (g1 g2 g3 … gn)
Since g1 g2 g3 … gn is an element of G, it has an inverse;
Pre/post-multg. by this inverse then gives xn = e
E1
E1
E1
[3]
(iii)
(a) Elements may have an order which divides into (is a factor of) n
B1
[1]
(b) Because the change of the order of multns. in
g.g1 g.g2 g.g3 … g.gn = gn.(g1 g2 g3 … gn)
is only valid in an abelian group
B1
[1]
© Cambridge International Examinations 2013
Page 6
9
Mark Scheme
Pre-U – May/June 2013
Syllabus
1348
Reflection in y = x tan 18 π :




Shear // y-axis, mapping (1, 0) to (1, 2):
1 0


2
1


Rotation through
1
4
π clockwise about O:
Shear // x-axis, mapping (0, 1) to (–2, 1):
1
2
1
2
−
1
2
1
2




 12
 1
−
 2
Allow cos ( 14 π ) ' s , etc.
B1
B1
1
2
1
2




B1
1 − 2


0 1 
B1
0 1

1 0
Multiplying them together in this order (from right-to-left) = 
Reflection
Paper
01
in y = x
M1
A1
M1 A1
[8]
NB 1 Multiplying the matrices in the reverse order scores max. 4 × B1 + M0 ; then
1 0 
 and M1 for “Reflection” and A1 for “in x-axis”
 0 −1
B1 for correct 
NB 2 Incorrect final matrices automatically lose the last 2 marks
10
(a)
y = k x cos x ⇒
d2 y
dy
= –k x sin x + k cos x and
= –k x cos x – 2k sin x
dx
dx 2
Attempt at 1st and 2nd derivatives using the Product Rule
Substituting both of these into the given DE
M1
M1
–k x cos x – 2k sin x + k x cos x = 4 sin x
Comparing terms to evaluate k:
k = –2
Aux. Eqn. m2 + 1 = 0 solved ⇒ m = ± i
Comp. Fn. is yC = A cos x + B sin x ft
M1 A1
M1 A1
Accept yC = Aeix + Be – ix here
G. S. is y = A cos x + B sin x – 2x cos x ft provided yP has no arb. consts. & yC
has 2
B1
B1
Do not accept final answer involving complex numbers
[8]
© Cambridge International Examinations 2013
Page 7
10
(b)(i)
Mark Scheme
Pre-U – May/June 2013
x = 1, y = 2 &
Differentiating
dy
=1 ⇒
dx
d2 y
+ y2
2
dx
d2 y 
dx
B1
+ xy = 5x – 19
M1
Use of Product Rule and implicit differentiation (at least once)
⇒
3
d y
dx 3


2
d y
+ y2
dx 2
 dy 

 dx 
+ 2 y
Paper
01
= –20

dx2  x = 1
dy
Syllabus
1348
2
  dy

d2 y 
= 78
 + x + y = 5 ⇒

  dx

dx 2  x = 1
M1
A1 A1
A1
2
 dy 
FT “78” from “–20” and also from
instead of   (both = 1)
dx
 dx 
dy
(b)(ii)
Use of y = y(1) + (x – 1).y′ (1) +
1
2
(x – 1)2.y″ (1) +
1
6
[6]
(x – 1)3.y′″ (1) + …
= 2 + (x – 1) – 10(x – 1)2 + 13(x – 1)3 + … ft
M1
A1
Substituting x = 1.1 into this series ⇒ y(1.1) ≈ 2.013 ft
M1 A1
[4]
11
(i)
( p + iq )2 =
(p
2
)
− q 2 + i.2pq
B1
Comparing real and imaginary parts: p 2 − q 2 = 63 and 2pq = –16
Solving simultaneously :
p = ± 8, q = m 1
M1
i.e. ± (8 − i ) = 63 − 16i
2
M1 A1
[4]
(ii)
(a) Use of z 3 − (α + β + γ ) z 2 + (αβ + βγ + γα ) z − (αβγ ) = 0
M1
A = 4 – 4i, B = 21 – 16i, C = 84
i.e. f(z) = z3 – (4 – 4i)z2 + (21 – 16i)z – 84 = 0
A1 A1
A1
[4]
(b) Differentiating to get f′ (z) = 3z2 – 8(1 – i)z + (21 – 16i) OR
3z2 – 2Az + B = 0 ft
Solving z =
z=
B1
8 − 8i ± 64( 1 − 2i − 1 ) − 12( 21 − 16i )
using the quadratic formula
6
1
3
(4 − 4i ±
) (4 − 4i ± i
16i − 63 =
1
3
Use of (i)’s result (on the right thing): z =
1
3
63 − 16i
)
(4 − 4i ± i(8 − i)) =
M1
A1
5
3
+ 43 i or 1− 4i
M1 A1
[5]
© Cambridge International Examinations 2013
Page 8
12
(i)
Mark Scheme
Pre-U – May/June 2013
y ′(x) = (2x + 1) e 2x ,
Syllabus
1348
B1 B1
y ′′(x) = (4x + 4) e 2x ,
2x
y
y′′′ (x) = (8x + 12) e ,
(4)
Paper
01
(x) = (16x + 32) e
B1 B1
2x
[4]
(ii)
Conjecture
dn y
dx
n
= (2n x + n.2n – 1) e 2x
One mark each: coefft. of x, constant
B1 B1
[2]
(iii)
Diffferentiating their conjectured expression (must be linear × e2x)
d n +1 y
dx n +1
= 2 × (2n x + n.2n – 1) e 2x + 2n × e 2x
FT max 1/2
= (2n + 1 x + (n + 1).2(n+1) – 1) e 2x
Shown of correct form
Usual induction round-up/explanation of proof, including clear demonstration that
(n+1)th formula is in the right form.
M1
A1 A1
A1
E1
[5]
13
(i)
(a) 1 – sech2θ ≡
(b)
(e
θ
(e
+ e −θ
θ
+e
(
)
2
−4
)
−θ 2
)(
≡
(e
(e
θ
θ
− e −θ
+e
) (
)
)
2
≡ tanh2θ shown legitimately
−θ 2
)(
)
θ
−θ
θ
−θ
θ
−θ
θ
−θ
d
(tanh θ ) = e + e e + e θ − −eθ 2− e e − e ≡ sech2θ from (a)
dθ
e +e
(
)
M1
A1
M1
A1
[4]
α
(ii)
(a) I n =
∫ tanh
2n − 2
α
∫ tanh
2
θ . tanh θ dθ =
2n−2
(
)
θ 1 − sech 2θ dθ
M1 M1
0
0
 tanh 2 n −1 θ  α
(tanh α ) 2 n −1
⇒
I
–
I
=
= In – 1 – 
n
–
1
n

2n − 1
 2n − 1  0
α
ALT: In – 1 – I n =
∫
(
)
tanh 2 n −2 θ 1 − tanh 2 θ dθ =
0
α
∫ tanh
2n−2
θ . sech 2θ dθ
M1 A1
M1 M1
0
 tanh 2 n −1 θ  α (tanh α ) 2 n −1
= 
 =
2n − 1
 2n − 1  0
M1 A1
[4]
© Cambridge International Examinations 2013
Page 9
Mark Scheme
Pre-U – May/June 2013
α
13
(ii)
∫
(b) I0 = 1 dθ = α =
1
2
Syllabus
1348
ln3
Paper
01
B1
0
[1]
(c) (I n−1 − I n ) + (I n−2 − I n−1 ) + (I n−3 − I n−2 ) + ... + (I 2 − I 3 ) + (I1 − I 2 ) + (I 0 − I1 )
Use of the method of differences
n
=
∑
r =1
(tanhα )2 r −1
2r − 1
⇒ I0 – In =
n
=
∑ 2r − 1
when α =
r =1
( 12 )2r−1
n
( 12 )2r −1
∑ 2r − 1
1
2
ln3
Cancellation of terms in the summation
r =1
⇒ I n = 12 ln3 −
n
( 12 )2r −1
∑ 2r − 1
AG
M1
A1
M1
A1
r =1
Ignoring “method of differences”, but opting for a direct iterative approach scores
max 3/4 … M0 M1 A1 A1
As n → ∞, In → 0 since | tanh | < 1
⇒
⇒
1
2
ln 3 =
E1
(12 )2r −1 = 12 + (12 )3 + (12 )5 + (12 )7 + ...
n
∑ 2r − 1
r =1
ln 3 = 1 +
1
3
5
7
∞
1
1
1
1
+
+
+
...
=
2
3
r
3.4 5.4
7.4
r = 0 ( 2r + 1 )4
∑
Ignoring “method of differences”, but opting for a direct iterative approach scores
max 3/4 … M0 M1 A1 A1
© Cambridge International Examinations 2013
M1
A1
[[7]