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Cambridge International Examinations
Cambridge Pre-U Certificate
FURTHER MATHEMATICS (SHORT COURSE)
Paper 1 Further Pure Mathematics
1348/01
For Examination from 2016
SPECIMEN MARK SCHEME
3 hours
MAXIMUM MARK: 120
The syllabus is approved for use in England, Wales and Northern Ireland as a Cambridge International Level 3 Pre-U Certificate.
This document consists of 10 printed pages.
© UCLES 2013
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2
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.
The following abbreviations may be used in a mark scheme:
AG
Answer Given on the question paper (so extra checking is needed to ensure that the detailed working
leading to the result is valid)
CAO Correct Answer Only (emphasising that no “follow through” from a previous error is allowed)
aef Any equivalent form
art
Answers rounding to
cwo Correct working only (emphasising that there must be no incorrect working in the solution)
ft
Follow through from previous error is allowed
o.e. Or equivalent
© UCLES 2013
1348/01/SM/16
3
n
∑ (r
1
)
n
n
n
r =1
r =1
r =1
− r + 1 = ∑ r 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
1st for Σr2; 2nd for Σr & Σ1 = n
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
Splitting summation and use of given results
1
2
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+ y4
⇒x=
∫
M1 A1
A1
∫
dy = 1.dx
2y
1+ y4
By separating variables 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 substitution
∫
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 substitution
∫
2t
1+ t4
dt =
∫
sec 2 θ
1 + tan 2 θ
∫
dθ A1 = secθ .dθ
= ln|secθ + tanθ | = ln | t 2 + 1 + t 4 | A1
© UCLES 2013
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4
4
y=
(i)
x +1
⇒ y.x2 – x + (3y – 1) = 0
x2 + 3
Creating a quadratic in x
M1
For real x, 1 – 4y(3y – 1) [ 0
Considering the discriminant
M1
12y2 – 4y – 1 Y 0
Creating a quadratic inequality
M1
For real x, (6y + 1)(2y – 1) Y 0
Factorising/solving a 3-term quadratic
M1
− 16 Y
≤ yY
≤ 12
(ii)
y=
y
5
(i) (a)
(b)
(ii)
−
1
2
1
6
CAO
A1
substituted back ⇒
1
2
= substituted back ⇒
 cos α

 sin α
 cos β

 sin β
− sin α 

cos α 
sin β 

− cos β 
 cos φ

 sin φ
sin φ  cos θ

− cos φ  sin θ
(x2 – 2x + 1) = 0 ⇒ x = 1
−
1
6
[y=1]
(x2 + 6x + 9) = 0 ⇒ x = –3
M1 A1
2
[y = − 16 ]
M1 A1
B1
B1
sin θ 
 multiplication of 2 reflection matrices. Correct
− cos θ 
M1 M1
order.
 cos φ cos θ + sin φ sin θ
=
 sin φ cos θ − cos φ sin θ
 cos(φ − θ )
=
 sin(φ − θ )
cos φ sin θ − sin φ cos θ 

cos φ cos θ + sin φ sin θ 
− sin(φ − θ ) 
 Use of the addition formulae; correctly done
cos(φ − θ ) 
… giving a Rotation (about O)
6
(i)
(ii)
through (φ – θ )
M1 A1
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
[by ]
= x . y4 = x . (y2)2
= 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.
(iii)
© UCLES 2013
M1 A1
Proving G not abelian: [e.g. by xyx = y but x2 ≠ e] ⇒ G not cyclic
OR establishing a contradiction
1348/01/SM/16
A1
B1 B1
5
7
(i)
cos4θ + i sin4θ = (c + is)4
Use of de Moivre’s Theorem
= 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
cos 4θ
c − 6c 2 s 2 + s 4
sin 4θ
M1
4t − 4t 3
legitimately
Dividing throughout by c to get
1 − 6t 2 + t 4
4
(ii)
t=
1
5
⇒ tan4θ =
A1
120
119
(
)
1
tan 14 π + tan −1 239
=
B1
1
1 + 239
= 120
1
1 − 239 119
M1 A1
Noting that this is tan(4tan–1 1 ) so that 4tan–1 1 =
5
5
8
1
4
1
π + tan–1 239
(i)
Substituting 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
A1
M1 A1
x −1
M1
A1
Product Rule used twice; at least one bracket correct
Substituting x = 1, f′(1) = 3 and f ″(1) = 5 into this ⇒ f ″′(1) = –12 ft their f ″(1)
(iii)
M1
f(x) = f(1) + f ′(1)(x – 1) +
1
2
f ″(1)(x – 1)2 +
1
6
f ″′(1)(x – 1)3 + …
M1 A1
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. CAO
(iv)
Substituting x = 1.1
(i)
dy
+ y = 3x y4 is a Bernouilli (differential) equation
dx
u=
du
1
3 dy
⇒
=−
×
3
dx
y
y 4 dx
Then
© UCLES 2013
M1 A1
B1
du
3 dy 3
dy
+ y = 3x y4 becomes −
− 3u = −9 x
× −
= −9 x ⇒
dx
dx
y 4 dx y 3
1348/01/SM/16
AG
M1 A1
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6
(ii)
Method 1
IF is e–3x
M1 A1
∫
⇒ ue −3 x = − 9 xe −3 x dx
M1
∫
= 3 xe −3 x − 3e −3 x dx
Use of “parts”
= (3x + 1)e–3x + C
A1
General Solution is u = 3x + 1 + Ce3x
⇒ y3 =
Using x = 0, y =
1
2
M1
1
3 x + 1 + Ce 3 x
ft
B1
ft
B1
C = 7 or y 3 =
to find C
1
3 x + 1 + 7e 3 x
M1
A1
Method 2
Auxiliary equation m – 3 = 0 ⇒ uC = Ae3x is the complementary function
For particular integral try uP = ax + b , uP′ = a
M1
Substituting 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
General solution is u = 3x + 1 Ae3x ft particular integral + complementary function
provided particular integral has no arbitrary
constants and complementary function has
one
⇒ y3 =
Using x = 0, y =
10
(i)
M1 A1
1
3 x + 1 + Ae 3 x
ft
1
to find A
2
B1
A = 7 or y 3 =
 1 + 3λ 


Substituting  − 3 + 4λ  into plane equation; i.e.
 2 + 6λ 


B1
1
3 x + 1 + 7e 3 x
 1 + 3λ   2 

  
 − 3 + 4λ  •  − 6  = k
 2 + 6λ   3 

  
M1
A1
M1
OR any point on line (since “given”)
k = 2 + 6λ + 18 – 24λ + 6 + 18λ = 26
© UCLES 2013
1348/01/SM/16
A1
7
(ii)
10 + 2m 


Working with vector  2 − 6m  .
 3m − 43 


B1
10 + 2m   2 

  
Substituting 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
 2 
 
Shortest Distance is |m|  − 6  = 21 or PQ =
 3 
 
(iii)
6 2 + 182 + 9 2 = 21
A1
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 (aef)
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 equations ⇒ 9a + 10b = 0
Choosing a = 10, b = –9 ⇒ c = 1 and d = 39 i.e. 10x – 9y + z = 39 CAO
© UCLES 2013
1348/01/SM/16
M1 B1
M1
M1
M1 A1
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8
11
(
|w|=
(i)
) (
2
3 −1 +
)
3 + 1 = 4 − 2 3 + 4 + 2 3 = 8 or 2 2
M1
A1
 3 +1
3 + 1
−1
5

 3 − 1 × 3 + 1  = tan 2 + 3 = 12 π


(
arg(w) = tan −1 
(
(ii) (a) z3 = 2 2 ,
3
| w| ;
⇒ z=
(
5
12
) (
π , 2 2,
29
12
)
) (
π , 2 2 , − 19
12 π or
53
12
π
M1 A1
)
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
) (
) (
)
r e^(iθ) forms equally acceptable
(b) z1, z2, z3 the roots of z3 – 0.z2 + 0.z – w = 0
⇒ z1 z2 z3 = w = 3 − 1 + i 3 + 1
ALT. Multiplying the 3 roots together in any form
(
) (
M1 M1
A1 A1
M1
A1
)
(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
29
(d) k = exp {− i. 365 π } or exp {− i. 36
}
π } or exp {i. 19
36 π
© UCLES 2013
1348/01/SM/16
=
6
A1
M1
A1
B1
9
12
(i)
In =
∫
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
Correct splitting and use of parts
M1
(16 + x )
A1
2 3/ 2
3
0
(
dx
)
 n − 1  3 n−2
x 16 + x 2 16 + x 2 dx
= 3n − 2 ⋅ 125 − 

0
3


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 Ins
(n + 2) In = 125 × 3n – 1 – 16(n – 1) In – 2 AG
M1
M1
A1
(ii) (a)
Spiral (with r increasing)
B1
From O to just short of θ =π
B1
2
(b)
r= θ
1
4
L=
∫
4
dr
 dr 
8
6
= θ 3 and r 2 + 
⇒
 = 161 θ + θ
dθ
 dθ 
3
1 3
θ
04
16 + θ 2
(= 14 I 3 )
1
Now I 1 =  16 + x 2
3
(
)
3/ 2
M1 A1
3

 =
61
3
© UCLES 2013
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
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Use of given reduction formula
ft only from suitable k I3
M1
A1
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10
13
Base-line case: for n = 5, 13579 R5 = 1508 7 6269 contains a string of (5 – 4 = 1) 7s
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) 7s, as required. Proof follows by induction
(usual round-up).
© UCLES 2013
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A1
A1