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UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS
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Pre-U Certificate
MARK SCHEME for the May/June 2012 question paper
for the guidance of teachers
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 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,
Pre-U, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level
syllabuses.
Page 2
Mark Scheme: Teachers’ version
Pre-U – May/June 2012
n
∑ (r
1
n
)
n
n
r =1
r =1
− r + 1 = ∑ r − ∑ r + ∑1
2
r =1
2
r =1
=
1
6
n( n + 1)(2n + 1) − 12 n( n + 1) + n
=
1
3
n ( n 2 + 2)
=
=
1
2
1
2
Splitting summation and use of given results
1st for Σr2; 2nd for Σr & Σ1 = n
Paper
01
M1
B1 B1
A1
legitimately
∫
including squaring attempt; ignore limits and k ≠
∫ (1 + sin 2θ ) dθ
for use of the double-angle formula
1

π / 2
θ
−
cos
2
θ


2

 0
ft (constants only) in the integration;
A = k (sin θ + cos θ ) 2 dθ
2
Syllabus
1348
1
2
[4]
M1
B1
OR integration of sinθ cosθ as k sin2θ or k cos2θ
A1
MUST be 2 separate terms
= 14 π + 12
A1
1
3
(i)
=
(ii)
1
y = (sinh x) 2 ⇒
∫
−
dy 1
= 2 (sinh x) 2 . cosh x OR
dx
y 2 = sinh x ⇒ 2 y
dy
= cosh x
dx
1+ y4
2y
2y
1+ y
⇒x=
∫
4
[4]
M1 A1
A1
[3]
∫
dy = 1.dx
2y
1+ y4
By Sepg. Vars. in (i)’s answer
dy
M1
A1
But x = sinh–1 y2 so
∫
2t
1+ t
4
dx = sinh–1 (t 2) + C
condone missing “ + C ”
A1
ALT.1 Set t2 = sinhθ , 2t dt = coshθ dθ M1 Full substn.
∫
2t
1+ t
4
dt =
∫
cosh θ
2
1 + sinh θ
∫
dθ A1 = 1 .dθ = θ = sinh–1 (t2) A1
ALT.2 Set t2 = tanθ , 2t dt = sec2θ dθ M1 Full substn.
∫
2t
1+ t
4
dt =
∫
sec 2 θ
2
1 + tan θ
∫
dθ A1 = secθ .dθ
= ln|secθ + tanθ | = ln | t 2 + 1 + t 4 | A1
© University of Cambridge International Examinations 2012
[3]
Page 3
4
(i)
Mark Scheme: Teachers’ version
Pre-U – May/June 2012
x +1
⇒ y.x2 – x + (3y – 1) = 0
x2 + 3
Syllabus
1348
Creating a quadratic in x
M1
For real x, 1 – 4y(3y – 1) ≥ 0
Considering the discriminant
M1
12y2 – 4y – 1 ≤ 0
Creating a quadratic inequality
M1
For real x, (6y + 1)(2y – 1) ≤ 0
Factorising/solving a 3-term quadratic
M1
y=
− 16 ≤ y ≤
1
2
A1
cso
NB lack of inequality earlier with unjustified correct answer loses only the 3
M mark
(ii)
Paper
01
y=
y
−
1
2
1
6
substd. back ⇒
1
2
= substd. back ⇒
(x2 – 2x + 1) = 0 ⇒ x = 1
−
1
6
rd
[y=1]
(x2 + 6x + 9) = 0 ⇒ x = –3
2
[y = − 16 ]
[5]
M1 A1
M1 A1
Allow alternative approach via calculus:
dy − x 2 − 2 x + 3
M1
=
2
dx
x2 + 3
(
)
Solving quadratic to find 2 values of x M1
Then A1 A1 each pair of correct (x, y) coordinates
5
(i)
(ii)
 cos α − sin α 

(a) 
 sin α cos α 
 cos β sin β 

(b) 
 sin β − cos β 
 cos φ

 sin φ
sin φ  cos θ

− cos φ  sin θ
B1
B1
[2]
sin θ 
 Multn. of 2 reflection matrices. Correct order.
− cos θ 
 cos φ cosθ + sin φ sin θ
 sin φ cosθ − cos φ sin θ
= 
[4]
M1 M1
cos φ sin θ − sin φ cosθ 

cos φ cosθ + sin φ sin θ 
 cos(φ − θ ) − sin(φ − θ ) 

 sin(φ − θ ) cos(φ − θ ) 
= 
… giving a Rotation (about O)
Use of the addition formulae; correctly done
through (φ – θ ) acw [or (θ – φ ) cw]
Those who get the initial matrices in the wrong order, can get 5/6, losing only that
M mark
© University of Cambridge International Examinations 2012
M1 A1
M1 A1
[6]
Page 4
6
(i)
(ii)
Mark Scheme: Teachers’ version
Pre-U – May/June 2012
Syllabus
1348
Possible orders are 1, 2, 3, 4, 6 & 12
B1
By Lagrange’s Theorem, the order of an element divides the order of the group
(since the order of an element ≡ the order of the subgroup generated by that element)
B1
E.g. y = xyx ⇒ y . x2y = xyx . x2y
by M1
by M1
= xy . x3 . y = xy . y2 . y
= x . (x3)2 = x . e
by 2 M’s for first, correct uses of 2 different conditions; the A for the 3rd condition used
to clinch the result.
SPECIAL CASE Allow 2/3 for those who correctly argue the converse
7
(i)
Proving G not abelian: [e.g. by xyx = y but x2 ≠ e] ⇒ G not cyclic
OR establishing a contradiction
cos4θ + i sin4θ = (c + is)4
Use of de Moivre’s Theorem
A1
[3]
B1 B1
[2]
M1
= c4 + 4c3.is + 6c2.i2s2 + 4c.i3s3 + i4s4 Binomial expansion attempted
M1
cos4θ = c4 – 6c2s2 + s4 and sin4θ = 4c3s – 4cs3
M1
Equating Re & Im parts
4c 3 s − 4cs 3
tan4θ =
= 4
c − 6c 2 s 2 + s 4
cos 4θ
sin 4θ
M1
4t − 4t 3
Dividing throughout by c to get
legitimately
1 − 6t 2 + t 4
4
(ii)
[2]
[by ]
= x . y4 = x . (y2)2
(iii)
Paper
01
t=
1
5
(
⇒ tan4θ =
120
119
)
1
tan 14 π + tan −1 239
=
A1
[5]
B1
1
1 + 239
= 120
119
1
1 − 239
Noting that this is tan(4 tan–1 1 ) so that 4tan–1 1 =
5
5
M1 A1
1
4
1
π + tan–1 239
© University of Cambridge International Examinations 2012
A1
[4]
Page 5
8
Mark Scheme: Teachers’ version
Pre-U – May/June 2012
Syllabus
1348
(i)
Substg. x = 1, f(1) = 2 and f ′(1) = 3 into (*) ⇒ f ″(1) = 5
(ii)
{x f ′′′( x) + 2 xf ′′( x)} + {(2 x − 1)f ′′( x) + 2f ′( x)} − 2f ′( x) = 3e
2
M1 A1
x −1
Substg. x = 1, f′(1) = 3 and f ″(1) = 5 into this ⇒ f ″′(1) = –12 ft their f ″(1)
f(x) = f(1) + f ′(1)(x – 1) +
1
2
f ″(1)(x – 1)2 +
1
6
[2]
M1
A1
Product Rule used twice; at least one bracket correct
(iii)
Paper
01
f ″′(1)(x – 1)3 + …
M1 A1
[4]
M1
Use of the Taylor series
= 2 + 3(x – 1) +
5
2
(x – 1)2 – 2(x – 1)3 + … 1st two terms cao;
A1 A1
nd
2 two terms ft (i) & (ii)’s answers
9
⇒ f(1.1) ≈ 2.323 to 3d.p. cso (i.e. exactly this answer)
(iv)
Substg. x = 1.1
(i)
dy
+ y = 3x y4 is a Bernouilli (differential) equation
dx
u=
(ii)
M1 A1
1
du
3 dy
⇒
=−
×
3
dx
y
y 4 dx
Then
[3]
[2]
B1
du
3 dy 3
dy
+ y = 3x y4 becomes −
× −
= −9 x ⇒
− 3u = −9 x
dx
dx
y 4 dx y 3
AG
M1 A1
[3]
Method 1
IF is e–3x
M1 A1
∫
⇒ ue −3 x = − 9 xe −3 x dx
∫
= 3 xe −3 x − 3e −3 x dx
M1
Use of “parts”
= (3x + 1)e–3x + C
Gen. Soln. is u = 3x + 1 + Ce3x
⇒ y3 =
Using x = 0, y =
1
2
1
3 x + 1 + Ce 3 x
to find C
M1
A1
ft
B1
ft
B1
C = 7 or y 3 =
1
3 x + 1 + 7e 3 x
© University of Cambridge International Examinations 2012
M1
A1
[9]
Page 6
Mark Scheme: Teachers’ version
Pre-U – May/June 2012
Syllabus
1348
Paper
01
Method 2
Aux. Eqn. m – 3 = 0 ⇒ uC = Ae3x is the Comp. Fn.
For Part. Intgl. try uP = ax + b , uP′ = a
M1
Substg. uP = ax + b and uP′ = a into the d.e. and comparing terms
M1
a – 3ax – 3b = –9x ⇒ a = 3, b = 1
A1
i.e. uP = 3x + 1
Gen. Soln. is u = 3x + 1 Ae3x
⇒ y3 =
Using x = 0, y =
(i)
ft PI + CF provided PI has no
arbitrary constants and CF has one
1
3 x + 1 + Ae 3 x
1
to find A
2
ft
A = 7 or y 3 =
 1 + 3λ 


Subst .  − 3 + 4λ  into plane equation; i.e.
 2 + 6λ 


g
10
M1 A1
B1
B1
1
3 x + 1 + 7e 3 x
 1 + 3λ   2 

  
 − 3 + 4λ  •  − 6  = k
 2 + 6λ   3 

  
M1
A1
[9]
M1
OR any point on line (since “given”)
(ii)
k = 2 + 6λ + 18 – 24λ + 6 + 18λ = 26
A1
10 + 2m 


Working with vector  2 − 6m  .
 3m − 43 


B1
10 + 2m   2 

  
Subst . into the plane equation:  2 + 6m  •  − 6  = k
 3m − 43   3 

  
M1
Solving a linear equation in m: 20 + 4m – 12 + 36m + 9m – 129 = 26
M1
m = 3 ⇒ Q = (16, –16, –34)
A1
g
 2 
 
Sh. Dist. is |m|  − 6  = 21 or PQ =
 3 
 
6 2 + 182 + 9 2 = 21
Alternate methods that find only Sh. Dist. but not Q can score M1 A1 only
© University of Cambridge International Examinations 2012
[2]
A1
[5]
Page 7
(iii)
Mark Scheme: Teachers’ version
Pre-U – May/June 2012
Syllabus
1348
Paper
01
Finding 3 points in the plane: e.g. A(1, –3, 2), B(4, 1, 8), C(10, 2, –43)
M1
 3
 9 
 6 
 




Then 2 vectors in (// to) plane: e.g. AB =  4 , AC =  5 , BC =  1 
6
 − 45 
 − 51
 




M1
OR B1 B1 for any two vectors in the plane
 10 
 
Vector product of any two of these to get normal to plane:  − 9 
 1 
 
M1
A1
(any non-zero multiple)
 10 
 
d =  − 9  • (any position vector) =
 1 
 
 10   1 
   
 − 9  •  − 3  e.g. = 39
 1   2 
   
M1
A1
⇒ 10x – 9y + z = 39 cao (or any correct equivalent form)
[6]
ALTERNATE SOLUTION
 1 + 3λ 
 10 




ax + by + cz = d contains  − 3 + 4λ  and  2 
 2 + 6λ 
 − 43 




... so a + 3aλ + 4bλ – 3b + 2c + 6cλ = d
Then a – 3b + 2c = d
and
and
10a + 2b – 43c = d
3a + 4b + 6c = 0 (λ terms)
i.e. equating terms
Eliminating (e.g.) c from 1st two eqns. ⇒ 9a + 10b = 0
Choosing a = 10, b = –9 ⇒ c = 1 and d = 39 i.e. 10x – 9y + z = 39 cao
© University of Cambridge International Examinations 2012
M1 B1
M1
M1
M1 A1
[6]
Page 8
11
(i)
Mark Scheme: Teachers’ version
Pre-U – May/June 2012
(
|w|=
) (
2
3 −1 +
Syllabus
1348
)
3 + 1 = 4 − 2 3 + 4 + 2 3 = 8 or 2 2
M1
A1
 3 +1
3 + 1
 = tan −1 2 + 3 = 125 π
×
 3 −1
3 + 1 

(
arg(w) = tan −1 
(ii)
(a)
3
(
z3 = 2 2 ,
| w| ;
⇒ z=
(
5
12
) (
π , 2 2,
29
12
)
) (
π , 2 2 , − 19
12 π or
53
12
M1 A1
[4]
π
)
arg(w)
These method marks can be earned for just the first root
3
29
2 , 365 π ,
2 , 36
π ,
2 , − 19
A marks for the 2nd & 3rd roots:
36 π
) (
) (
(b) z1, z2, z3 the roots of z3 – 0.z2 + 0.z – w = 0
⇒ z 1 z 2 z3 = w = 3 − 1 + i 3 + 1
ALT. Multiplying the 3 roots together in any form
) (
M1M1
)
r e^(iθ) forms equally acceptable
(
Paper
01
A1 A1
[4]
M1
A1
)
[2]
(c)
Three points in approx. correct places
M1
All equally spaced around a circle,
centre O, radius 2
(Explained that ∆1 equilateral)
M1
l = 2 × 2 sin ( 12 × 23 π )
=
or by the Cosine Rule
6
A1
M1
A1
[5]
29
}
(d) k = exp {− i. 365 π } or exp {− i. 36
π } or exp {i. 19
36 π
© University of Cambridge International Examinations 2012
B1
[1]
Page 9
12
(i)
In =
Mark Scheme: Teachers’ version
Pre-U – May/June 2012
∫
3
0
x n −1  x 16 + x 2  dx


(

16 + x 2
=  x n −1.
3

)
3/ 2
3

 –

0
3
∫ (n − 1) x
n−2
M1
(16 + x )
A1
2 3/ 2
3
dx
)
 n − 1  3 n−2
x 16 + x 2 16 + x 2 dx
= 3n − 2 ⋅ 125 − 

 3  0
Method to get 2nd integral of correct form
∫
 n −1
 {16 I n − 2 + I n }
 3 
= 3 n − 2.125 − 
M1
[i.e. reverting to I’s in 2nd integral ft]
⇒ 3 In = 3n – 1.125 – 16(n – 1) In – 2– (n – 1) In
Collecting up In’s
(n + 2) In = 125 × 3n – 1 – 16(n – 1) In – 2 AG
(ii)
Paper
01
Correct splitting and use of parts
0
(
Syllabus
1348
M1
M1
A1
[6]
(a)
Spiral (with r increasing)
B1
From O to just short of θ =π
B1
[2]
2
(b) r = 14 θ 4 ⇒
L=
∫
3
1 3
θ
04
dr
 dr 
8
6
= θ 3 and r 2 + 
 = 161 θ + θ
dθ
 dθ 
16 + θ 2
(= 14 I 3 )
1
Now I 1 =  16 + x 2
3
(
)
3/ 2
M1 A1
3

 =
61
3
1
20
B1
0
and 5 I 3 = 125 × 9 − 16 × 2( 613 ) =
so that L =
M1 A1
1423
3
or 474 13
1423
43
or 23 60
or awrt 23.7
× 1423
3 = 60
Use of given reduction formula
ft only from suitable k I3
NB The last 3 marks can be earned by integrating in a variety ways
© University of Cambridge International Examinations 2012
M1
A1
[7]
Page 10
13
Mark Scheme: Teachers’ version
Pre-U – May/June 2012
Syllabus
1348
Paper
01
Base-line case: for n = 5, 13579 R5 = 1508 7 6269 contains a string of (5 – 4 = 1) 7’s
B1
13579 R6 = 1508776269, 13579 R7 = 15087776269, etc. or form of 1st & last 4 digits
B1
Assume that, for some k ≥ 5, 13579 Rk = 1508 77…7 6269. Induction hypothesis
M1
(k – 4) 7’s
Then, for n = k + 1,
13579 Rk + 1 = 13579(10Rk + 1)
M1
Give the M mark for the key observation that Rk + 1 = 10Rk + 1 or 10k + Rk , even if
not subsequently used.
= 1508 77…7 62690
(k – 4) 7’s
+ 13579
= 1508 77…7 76269
(k – 4 + 1) 7’s
which contains a string of (k – 4 + 1) 7’s, as required. Proof follows by induction
(usual round-up).
© University of Cambridge International Examinations 2012
A1
E1
[6]