4755
Mark Scheme
June 2005
Section A
1(i)
1 ⎛ 2 −3 ⎞
A −1 = ⎜
⎟
5 ⎝ −1 4 ⎠
1(ii)
1⎛ 2
−3 ⎞⎛ 5 ⎞
⎛ x ⎞ 1 ⎛ 22 ⎞
=⎜ ⎟= ⎜
⎟⎜
⎟
⎟
4 ⎠⎝ −4 ⎠ ⎝ y ⎠ 5 ⎝ −21 ⎠
⎜
5 ⎝ −1
⇒x=
2
3
M1
A1
22
5
, y=
M1
Dividing by determinant
Pre-multiplying by their inverse
−21
A1(ft)
Follow through use of their
,
inverse
A1(ft)
No marks for solving without
[5]
using inverse matrix
5
4 − j, 4 + j
M1
A1
[2]
Use of quadratic formula
Both roots correct
17 ( cos 0.245 + jsin 0.245 )
M1
17 ( cos 0.245 − jsin 0.245 )
F1,
F1
[3]
Attempt to find modulus and
argument
One mark for each root
Accept ( r , θ ) form
Allow any correct arguments in
radians or degrees, including
negatives: 6.04, 14.0° , 346° .
Accuracy at least 2s.f.
S.C. F1 for consistent use of their
incorrect modulus or argument
(not both, F0)
⎛ 3 −1 ⎞ ⎛ x ⎞ ⎛ x ⎞
M1 for ⎜
⎟ ⎜ ⎟ = ⎜ ⎟ (allow if
⎝ 2 0 ⎠⎝ y⎠ ⎝ y⎠
implied)
⎛ 3 −1⎞ ⎛ k ⎞ ⎛ K ⎞
⎜ 2 0 ⎟ ⎜ mk ⎟ = ⎜ mK ⎟ can lead to
⎝
⎠⎝ ⎠ ⎝
⎠
full marks if correctly used. Lose
second A1 if answer includes two
lines
⎛ 3 −1⎞ ⎛ x ⎞ ⎛ x ⎞
⎜ 2 0 ⎟⎜ y ⎟ = ⎜ y ⎟ ⇒ x = 3x − y, y = 2 x
⎝
⎠⎝ ⎠ ⎝ ⎠
M1
A1
⇒ y = 2x
A1
[3]
B1
4(i)
α + β = 2, αβ = 4
4(ii)
α + β = (α + β ) − 2αβ = 4 − 8 = −4
4(iii)
Sum of roots = 2α + 2 β = 2 (α + β ) = 4
2
2
2
M1A1
(ft)
M1
Both
Accept method involving
calculation of roots
Or substitution method, or method
4755
Mark Scheme
June 2005
Product of roots = 2α × 2 β = 4αβ = 16
x 2 − 4 x + 16 = 0
5(i)
5(ii)
involving calculation of roots
A1(ft)
[5]
The = 0, or equivalent, is
necessary for final A1
Sketch of Argand diagram with:
Point 3 + 4 j .
Circle, radius 2.
B1
B1
[2]
Half-line:
Starting from (4, 0)
Vertically upwards
B1
B1
[2]
Points where line crosses circle clearly
indicated.
B1
[1]
Circle must not touch either axis.
B1 max if no labelling or scales.
Award even if centre incorrect.
5(iii)
Identifying 2 points where their
line cuts the circle
4755
Mark Scheme
Qu
Answer
June 2005
Mark
Comment
Section A (continued)
6
For k = 1, 13 = 1 and
1 2
1
4
(1+1)2 = 1 , so true for
B1
k =1
Assume true for n = k
B1
Next term is ( k + 1)
Add to both sides
2
3
RHS = 14 k 2 ( k + 1) + ( k + 1)
3
=
1
4
( k + 1) ⎡⎣ k
=
1
2
( k + 1) ( k + 2 )
=
1
4
4
2
2
B1
M1
+ 4 ( k + 1) ⎤⎦
2
( k + 1) ( ( k + 1) + 1)
2
Add to both sides
M1
A1
E1
[7]
7
3∑ r 2 − 3∑ r
= 3 × 16 n ( n + 1)( 2n + 1) − 3 × 12 n ( n + 1)
= 12 n ( n + 1) [( 2n + 1) − 3]
= n ( n + 1)( 2n − 2 )
1
2
= n ( n + 1)( n − 1)
Factor of ( k + 1)
2
Allow alternative correct methods
2
But this is the given result with ( k + 1)
replacing k.
Therefore if it is true for k it is true for
( k + 1) . Since it is true for k = 1 it is true for
k = 1, 2, 3, … .
Assuming true for k, (k +1)th term
for alternative statement, give this
mark if whole argument logically
correct
M1,A
1
For fully convincing algebra
leading to true for k ⇒ true for k
+1
Accept ‘Therefore true by
induction’ only if previous A1
awarded
S.C. Give E1 if convincing
explanation of induction
following acknowledgement of
earlier error
Separate sums
Use of formulae
M1,A
1
Attempt to factorise, only if
earlier M marks awarded
M1
Must be fully factorised
A1
c.a.o.
[6]
Section A Total: 36
4755
8(i)
8(ii)
8(iii)
Mark Scheme
x = 23 and y = 19
B1,
B1
+
Large positive x, y → 19
(e.g. consider x = 100 )
−
Large negative x, y → 19
(e.g. consider x = −100 )
Curve
y=
shown with correct approaches
(from below on left, above on right).
1
9
(2, 0), (-2, 0) and (0, -1) shown
−1 =
x2 − 4
( 3x − 2 )
2
-1 if any others given. Accept
min of 2s.f. accuracy
[2]
M1
Approaches horizontal asymptote,
not inconsistent with their (i)
Correct approaches
A1
E1
x = 23 shown with correct approaches
8(iv)
June 2005
Reasonable attempt to justify
approaches
[3]
B1(ft)
B1(ft)
B1(ft)
1 for each branch, consistent with
horizontal asymptote in (i) or (ii)
B1
B1
[5]
Both x intercepts
y intercept
(give these marks if coordinates
shown in workings, even if not
shown on graph)
M1
Reasonable attempt at solving
inequality
⇒ −9 x 2 + 12 x − 4 = x 2 − 4
⇒ 10 x 2 − 12 x = 0
⇒ 2x (5x − 6) = 0
⇒ x = 0 or x =
6
5
From sketch,
y ≥ −1 for x ≤ 0
and x ≥
6
5
Both values – give for seeing 0
A1
B1
F1
[4]
and
6
5
, even if inequalities are
wrong
For x ≤ 0
Lose only one mark if any strict
inequalities given
4755
9(i)
Mark Scheme
2-j
B1
B1
[2]
2j
9(iii)
( x − 2 − j)( x − 2 + j)( x + 2 j)( x − 2 j)
(
June 2005
= x − 4x + 5
2
)( x
2
+4
)
M1,
M1
A1,A1
= x 4 − 4 x 3 + 9 x 2 − 16 x + 20
So A = -4, B = 9, C = -16 and D = 20
− A = ∑ α = 4 ⇒ A = −4
B = ∑ αβ = 9 ⇒ B = 9
−C = ∑ αβγ = 16 ⇒ C = −16
D = ∑ αβγδ = 20 ⇒ D = 20
A1 for each quadratic - follow
through sign errors
A4
[8]
OR
M1 for each attempted factor pair
M1,
A1
M1,
A1
M1,
A1
M1,
A1
Minus 1 each error – follow
through sign errors only
M1s for reasonable attempt to find
sums
S.C. If one sign incorrect, give
total of A3 for A, B, C, D values
If more than one sign incorrect,
give total of A2 for A, B, C, D
values
[8]
OR
Attempt to substitute two correct roots into
x 4 + Ax 3 + Bx 2 + Cx + D = 0
M1
M1
One for each root
Produce 2 correct equations in two
unknowns
A2
One for each equation
A4
A = -4, B = 9, C = -16, D = 20
One mark for each correct.
S.C. If one sign incorrect, give
total of A3 for A, B, C, D values
If more than one sign incorrect, give
total of A2 for A, B, C, D values
4755
Mark Scheme
June 2005
10(i)
n
∑
r =1
n
2
r ( r + 1)( r + 2 )
⎛1
=
⎡1
2
1
⎤
∑ ⎢ r − ( r + 1) + ( r + 2 ) ⎥
r =1
⎣
⎦
1⎞ ⎛1 2 1⎞ ⎛1 2 1⎞
= ⎜ − + ⎟+⎜ − + ⎟+⎜ − + ⎟+
⎝1 2 3⎠ ⎝ 2 3 4 ⎠ ⎝ 3 4 5 ⎠
Give if implied by later working
M1
2
2
1 ⎞ ⎛1
2
1 ⎞
⎛ 1
− +
+
⎟+⎜ −
⎟
⎝ n −1 n n +1 ⎠ ⎝ n n +1 n + 2 ⎠
........ + ⎜
1 2 1
1
2
1
= − + +
−
+
1 2 2 n +1 n +1 n + 2
1
1
1
= −
+
2 n +1 n + 2
1
1
= −
2 ( n + 1)( n + 2 )
M1
Writing out terms in full, at least three
terms
All terms correct. A1 for at least two
correct
A2
M1
A3
Attempt at cancelling terms
Correct terms retained (minus 1 each
error)
Attempt at single fraction leading to
given answer.
M1
10(ii)
1
1× 2 × 3
=
1
n
2
∑
r =1
+
1
2 × 3× 4
+
1
3× 4 × 5
[9]
+ ...
⎞
1⎛1
1
= ⎜ −
⎟
2 ⎝ 2 ( n + 1)( n + 2 ) ⎠
r ( r + 1)( r + 2 )
2
M1
M1
⇒
1
1× 2 × 3
+
1
2 × 3× 4
+
1
3× 4 × 5
+ ... =
1
4
A1
[3]
M1 relating to previous sum, M1 for
recognising that
1
→ 0 as n → ∞
( n + 1)( n + 2 )
(could be implied)
Section A
1(i)
1(ii)
−6 ⎞
⎟ , A + C is impossible,
8⎠
⎛4
2B = ⎜
⎝2
⎛3
CA = ⎜⎜ 2
⎜1
⎝
1⎞
4 ⎟⎟ , A - B =
2 ⎟⎠
⎛2 6 ⎞
⎜
⎟
⎝ 0 −2 ⎠
B1
B1
M1, A1
B1
CA 3 × 2 matrix M1
[5]
⎛ 4 3 ⎞⎛ 2 −3 ⎞ ⎛ 11
AB = ⎜
⎟⎜
⎟=⎜
⎝ 1 2 ⎠⎝ 1 4 ⎠ ⎝ 4
⎛ 2 −3 ⎞⎛ 4 3 ⎞ ⎛ 5
BA = ⎜
⎟⎜
⎟=⎜
⎝ 1 4 ⎠⎝ 1 2 ⎠ ⎝ 8
AB ≠ BA
0⎞
⎟
5⎠
0⎞
⎟
11⎠
M1
Or AC impossible, or student’s
own correct example. Allow M1
even if slip in multiplication
E1
Meaning of commutative
[2]
2(i)
(a
z =
2
)
+ b 2 , z * = a − bj
B1
B1
[2]
2(ii)
zz * = ( a + bj )( a − bj ) = a 2 + b 2
⇒ zz * − z = a 2 + b2 − ( a 2 + b2 ) = 0
2
M1
Serious attempt to find zz * ,
consistent with their z *
M1
A1
ft their z in subtraction
[3]
All correct
3
∑ ( r + 1)( r − 1) = ∑ ( r
n
n
r =1
r =1
=
1
6
=
1
6
=
=
1
6
1
6
2
n ( n + 1)( 2n + 1) − n
)
−1
M1
M1,
A1, A1
Condone missing brackets
Attempt to use standard results
Each part correct
n [( n + 1)( 2n + 1) − 6 ]
(
n 2n 2 + 3n − 5
)
M1
Attempt to collect terms with
common denominator
A1
c.a.o.
n ( 2n + 5 )( n − 1)
[6]
6x − 2 y = a
−3 x + y = b
B1
B1
Determinant = 0
B1
The equations have no solutions or infinitely
many solutions.
E1
E1
5(i)
α + β + γ = −3 , αβ + βγ + γα = −7 , αβγ = −1
Minus 1 each error to minimum
B2
[2] of 0
5(ii)
Coefficients A, B and C
4(i)
4(ii)
[2]
−B
2α + 2 β + 2γ = 2 × −3 = −6 =
A
2α × 2 β + 2 β × 2γ + 2γ × 2α = 4 × −7 = −28 =
2α × 2 β × 2γ = 8 × −1 = −8 =
No solution
or infinitely many solutions
Give E2 for ‘no unique solution’
s.c. 1: Determinant = 12, allow
‘unique solution’ B0 E1 E0
1
s.c. 2: Determinant =
give
0
[3]
maximum of B0 E1
M1
Attempt to use sums and
products of roots
A3
ft their coefficients, minus one
each error (including ‘= 0’
missing), to minimum of 0
C
A
−D
A
⇒ x + 6 x − 28 x + 8 = 0
3
2
[4]
OR
ω = 2x ⇒ x =
ω
2
⎛ω ⎞
⎛ω ⎞
⎛ω⎞
⎜ ⎟ + 3⎜ ⎟ − 7 ⎜ ⎟ + 1 = 0
⎝2⎠
⎝2⎠
⎝2⎠
3
2
ω 3ω 7ω
⇒
+
−
+1 = 0
8
4
2
⇒ ω 3 + 6ω 2 − 28ω + 8 = 0
3
2
M1
A1
A1
Attempt at substitution
Correct substitution
Substitute into cubic (ft)
A1
[4]
c.a.o.
6
n
1
n
∑ r ( r + 1) = n + 1
r =1
n = 1, LHS = RHS =
1
2
Assume true for n = k
Next term is
1
1
k
+
( )( k + 2 )
B1
E1
B1
Add to both sides
1
k
RHS =
+
k + 1 ( k + 1)( k + 2 )
=
=
Assuming true for k (must be
explicit)
(k + 1)th term seen c.a.o.
Add to
M1
k
(ft)
k +1
k ( k + 2) + 1
( k + 1)( k + 2 )
k 2 + 2k + 1
( k + 1)( k + 2 )
( k + 1)
=
( k + 1)( k + 2 )
2
=
k +1
k+2
But this is the given result with k + 1
replacing k. Therefore if it is true for k it is
true for k + 1. Since it is true for k = 1, it is
true for k = 1, 2, 3
c.a.o. with correct working
A1
E1
True for k, therefore true for k + 1
k +1
(dependent on
seen)
k +2
Complee argument
E1
[7]
Section A Total: 36
7(i)
7(ii)
Section B
3 + x2 ≠ 0 for any real x.
y = -1, x = 2, x = −2
E1
[1]
B1, B1
B1
[3]
7(iii)
Large positive x, y → −1−
(e.g. consider x = 100 )
Large negative x, y → −1−
(e.g. consider x = −100 )
M1
B1
Evidence of method required
From below on each side c.a.o.
[2]
7(iv)
Curve
3 branches correct
Asymptotes labelled
Intercept labelled
B1
B1
Consistent with (i) and their (ii),
(iii)
Consistent with (i) and their (ii),
(iii)
B1
Labels may be on axes
Lose 1 mark if graph not
[3] symmetrical
May be written in script
7(v)
3 + x2
= −2 ⇒ 3 + x 2 = −8 + 2 x 2
4 − x2
⇒ 11 = x 2
M1
⇒ x = ( ± ) 11
From graph, − 11 ≤ x < −2 or 2 < x ≤ 11
A1
B1
A1
Reasonable attempt to solve
[4]
Accept 11
x < −2 and 2 < x seen
c.a.o.
8(i)
α 2 = (1 + j) = 2 j
2
α 3 = (1 + j) 2 j = −2 + 2 j
z 3 + 3z 2 + pz + q = 0
⇒ 2 j − 2 + 3 × 2 j + p (1 + j) + q = 0
⇒ (8 + p ) j + p + q − 2 = 0
p = −8 and p + q − 2 = 0 ⇒ q = 10
8(ii)
1 − j must also be a root.
The roots must sum to –3, so the other root
is
z = -5
M1, A1
A1
M1
M1
Substitute their α 2 and α 3 into
cubic
Equate real and imaginary parts to
A1
0
[6]
Results obtained correctly
B1
M1
A1
[3] Any valid method
c.a.o.
8(iii)
B2
Argand diagram with all three
roots clearly shown; minus 1 for
[2] each error to minimum of 0
ft their real root
Section B (continued)
9(i)
( 25,50 )
9(ii)
⎛1
⎞
⎜ y, y ⎟
⎝2
⎠
B1
[1]
B1,
B1
[2]
9(iii)
y=6
B1
[1]
9(iv)
All such lines are parallel to the x-axis.
B1
[1]
9(v)
9(vi)
All such lines are parallel to y = 2 x .
B1
[1]
⎛
⎜0
⎜⎜
⎝0
1⎞
2⎟
⎟
1 ⎟⎠
B3
9(vii)
⎛
0
det ⎜
⎜⎜
⎝0
1⎞
1
2 ⎟ = 0 ×1 − 0 × = 0
⎟⎟
2
1⎠
Or equivalent
[3]
M1
Transformation many to one.
Or equivalent
Minus 1 each error
s.c. Allow 1 for reasonable
attempt but incorrect working
Attempt to show determinant = 0
or other valid argument
E2
May be awarded without previous
M1
Allow E1 for ‘transformation has
[3] no inverse’ or other partial
explanation
Section B Total: 36
Total: 72
4755
Qu
Mark Scheme
June 2006
Mark
Answer
Comment
Section A
1 (i)
1(ii)
1(iii)
2
B1
[1]
Reflection in the x-axis.
⎛ 0 −1⎞
⎜1 0 ⎟
⎝
⎠
⎛ 1 0 ⎞ ⎛ 0 −1⎞ ⎛ 0 −1⎞
⎜ 0 −1⎟ ⎜ 1 0 ⎟ = ⎜ −1 0 ⎟
⎝
⎠⎝
⎠ ⎝
⎠
B1
[1]
M1
A1
c.a.o.
[2]
Multiplication of their matrices in
the correct order
or B2 for correct matrix without
working
( x + 2 ) ( Ax 2 + Bx + C ) + D
= Ax 3 + Bx 2 + Cx + 2 Ax 2 + 2 Bx + 2C + D
= Ax3 + ( 2 A + B ) x 2 + ( 2 B + C ) x + 2C + D
M1
⇒ A = 2, B = −7, C = 15, D = −32
B1
B1
F1
F1
OR
B5
Valid method to find all
coefficients
For A = 2
For D = -32
F1 for each of B and C
For all correct
[5]
3(i)
3(ii)
α + β + γ = −4
B1
αβ + βγ + αγ = −3
B1
αβγ = −1
B1
[3]
α + β + γ = (α + β + γ ) − 2 (αβ + βγ + αγ )
2
2
2
2
= 16 + 6 = 22
4 (i)
4(ii)
Argand diagram with solid circle, centre 3 – j,
radius 3, with values of z on and within the
circle clearly indicated as satisfying the
inequality.
M1
A1
E1
[3]
Attempt to use (α + β + γ )
Correct
Result shown
B1
B1
B1
[3]
Circle, radius 3, shown on
diagram
Circle centred on 3 - j
Solution set indicated (solid circle
with region inside)
B1
B1
[2]
Hole, radius 1, shown on diagram
Boundaries dealt with correctly
2
4755
Qu
Mark Scheme
June 2006
Section A (continued)
4(iii)
Im
B1
B1
B1
π
4
Line through their 3 – j
Half line
π
4
to real axis
R
3–j
5(i)
Comment
Mark
Answer
[3]
⎛ −1 2 ⎞⎛ 1⎞ ⎛ 1⎞
⎜
⎟⎜ ⎟ = ⎜ ⎟
⎝ −3 4 ⎠⎝ 1⎠ ⎝ 1⎠
B1
1 ⎛ 4 −2 ⎞
S −1 = ⎜
⎟
2 ⎝ 3 −1 ⎠
M1,
1 ⎛ 4 −2 ⎞⎛1⎞ ⎛ 1⎞
⎜
⎟⎜ ⎟ = ⎜ ⎟
2 ⎝ 3 −1 ⎠⎝1⎠ ⎝ 1⎠
A1
Attempt to divide by determinant
and manipulate contents
Correct
E1
[4]
5(ii)
⎛ x⎞ ⎛ x⎞
T⎜ ⎟ = ⎜ ⎟
⎝ y⎠ ⎝ y⎠
⎛ x⎞
⎛ x⎞
⇒ T −1T ⎜ ⎟ = T−1 ⎜ ⎟
⎝ y⎠
⎝ y⎠
⎛ x⎞
⎛ x⎞
⇒ ⎜ ⎟ = T −1 ⎜ ⎟
⎝ y⎠
⎝ y⎠
6
3 + 6 + 12 + ........ + 3 × 2n −1 = 3 ( 2 n − 1)
M1
Pre-multiply by T−1
A1
[2]
Invariance shown
n = 1, LHS = 3, RHS = 3
B1
Assume true for n = k
Next term is 3 × 2k +1−1 = 3 × 2k
Add to both sides
RHS = 3 ( 2k − 1) + 3 × 2k
E1
B1
Assuming true for k
(k + 1)th term.
M1
Add to both sides
A1
Working must be valid
E1
Dependent on previous A1and E1
E1
[7]
Dependent on B1 and previous E1
= 3 ( 2 k − 1 + 2k )
= 3 ( 2 × 2k − 1)
= 3 ( 2k +1 − 1)
But this is the given result with k + 1
replacing k. Therefore if it is true for k it is
true for k + 1. Since it is true for k = 1, it is
true for all positive integers n.
Section A Total: 36
4755
Mark Scheme
Section B
x = 2 , x = −1 and y = 1
7(i)
7(ii)
(A)
(B)
7(iii)
B1
B1B1
[3]
Large positive x, y → 1+ (from above)
(e.g. consider x = 100 )
Large negative x, y → 1− (from below)
(e.g. consider x = −100 )
M1
B1
B1
[3]
Curve
3 branches
B1
Correct approaches to horizontal
asymptote
Asymptotes marked
Through origin
7(iv)
June 2006
x < −1, x > 2
B1
B1
B1
[4]
B1B1,
B1,
[3]
One mark for each
Evidence of method needed for
M1
With correct approaches to
vertical asymptotes
Consistent with their (i) and (ii)
Equations or values at axes clear
s.c. 1 for inclusive inequalities
Final B1 for all correct with no
other solutions
4755
8(i)
Mark Scheme
( 2 + j) = 3 + 4 j
3
( 2 + j) = 2 + 11j
June 2006
B1
2
Substituting into 2 x 3 − 11x 2 + 22 x − 15 :
2 ( 2 + 11j) − 11( 3 + 4 j) + 22 ( 2 + j) − 15
= 4 + 22 j − 33 − 44 j + 44 + 22 j − 15
=0
So 2 + j is a root.
B1
M1
Attempt at substitution
A1
Correctly substituted
A1
Correctly cancelled
(Or other valid methods)
[5]
8(ii)
2-j
B1
[1]
8(iii)
( x − ( 2 + j ) ) ( x − ( 2 − j) )
M1
Use of factor theorem
= ( x − 2 − j)( x − 2 + j)
= x 2 − 2 x + j x − 2 x + 4 − 2 j − jx + 2 j + 1
A1
= x2 − 4x + 5
(x
(x
2
2
)
− 4 x + 5 ) ( 2 x − 3) = 2 x
− 4 x + 5 ( ax + b ) = 2 x 3 − 11x 2 + 22 x − 15
3
Comparing coefficients or long
division
A1
[4]
Correct third root
− 11x 2 + 22 x − 15
3
( 2 x − 3) = 0 ⇒ x =
M1
2
OR
Sum of roots =
11
2
or product of roots =
leading to
α + 2+ j+ 2− j =
⇒α =
11
15
2
M1
A1
M1
2
3
A1
[4]
2
or
α ( 2 + j)( 2 − j) =
⇒ 5α =
15
2
15
⇒α =
2
3
2
M1
A1
M1
A1
[4]
(Or other valid methods)
4755
9(i)
Mark Scheme
r ( r + 1)( r + 2 ) − ( r − 1) r ( r + 1)
(
≡ r2 + r
) ( r + 2) − r
3
−r
June 2006
M1
Accept ‘=’ in place of ‘ ≡ ’
throughout working
E1
[2]
Clearly shown
M1
Using identity from (i)
M1
A2
Writing out terms in full
At least 3 terms correct (minus 1
each error to minimum of 0)
M1
Attempt at eliminating terms
(telescoping)
Correct result
≡ r 3 + 2r 2 + r 2 + 2r − r 3 + r
≡ 3r 2 + 3r ≡ 3r ( r + 1)
9(ii)
n
∑ r ( r + 1)
r =1
=
1
n
∑ [ r ( r + 1)( r + 2 ) − ( r − 1) r ( r + 1)]
3
r =1
1
= [(1 × 2 × 3 − 0 × 1 × 2 ) + ( 2 × 3 × 4 − 1× 2 × 3) +
3
3
( × 4 × 5 − 2 × 3 × 4 ) + ......
+ ( n ( n + 1)( n + 2 ) − ( n − 1) n ( n + 1) )]
=
1
3
n ( n + 1)( n + 2 ) or equivalent
A1
[6]
9(iii)
n
n
n
∑ r ( r + 1) = ∑ r + ∑ r
r =1
=
=
=
=
1
6
1
6
1
6
1
3
2
r =1
r =1
n ( n + 1)( 2n + 1) +
1
2
n ( n + 1)
n ( n + 1) [( 2n + 1) + 3]
n ( n + 1)( 2n + 4 )
n ( n + 1)( n + 2 ) or equivalent
B1
B1
M1
Use of standard sums (1 mark
each)
Attempt to combine
A1
E1
[5]
Correctly simplified to match
result from (ii)
Section B Total: 36
Total: 72
4755
Qu
Mark Scheme
Jan 2007
Mark
Answer
Comment
Section A
1
2(i)
The statement is false. The ‘if’ part is true, but
the ‘only if’ is false since x = −2 also satisfies
the equation.
M1
A1
4 ± 16 − 28
[2]
M1
2
A1
=
4 ± 12
2
j = 2 ± 3j
A1
A1
[4]
2(ii)
‘False’, with attempted
justification (may be implied)
Correct justification
Attempt to use quadratic formula
or other valid method
Correct
Unsimplified form.
Fully simplified form.
[2]
One correct, with correct labelling
Other in correct relative position
s.c. give B1 if both points
consistent with (i) but no/incorrect
labelling
B3
B1
Points correctly plotted
Points correctly labelled
ELSE
M1
A1
[4]
Applying matrix to points
Minus 1 each error
B1(ft)
B1(ft)
3(i)
⎛ 2 0⎞⎛1 1 2⎞ ⎛ 2 2 4⎞
⎜ 0 1 ⎟⎜ 2 0 2⎟ = ⎜ 1 0 1⎟
⎠ ⎝
⎠
⎝
2 ⎠⎝
3(ii)
Stretch, factor 2 in x-direction, stretch factor
half in y-direction.
B1
B1
B1
[3]
24
1 mark for stretch (withhold if
rotation, reflection or translation
mentioned incorrectly)
1 mark for each factor and
direction
4755
4
Mark Scheme
n
∑ r (r
2
r =1
=
=
=
1
4
1
4
1
4
n
)
n
+1 = ∑ r + ∑ r
3
r =1
n 2 ( n + 1) +
2
2
n ( n + 1) [ n ( n + 1) + 2]
(
n ( n + 1) n + n + 2
A1
A1
c.a.o.
M1
A1
M1
n ( n + 1)
2
Separate into two sums (may be
implied by later working)
Use of standard results
Correct
Attempt to factorise (dependent
on previous M marks)
Factor of n(n + 1)
M1
r =1
1
Jan 2007
)
[6]
5
ω = 2x + 1 ⇒ x =
ω −1
3
M1
A1
M1
2
2
⎛ ω −1⎞
⎛ ω −1⎞ ⎛ ω −1⎞
⎟ − 3⎜
⎟ +⎜
⎟−4= 0
⎝ 2 ⎠
⎝ 2 ⎠ ⎝ 2 ⎠
2⎜
⇒
+
1
4
1
2
(ω
3
3
)
− 3ω + 3ω − 1 −
2
4
(ω
2
A1(ft)
A1(ft)
)
− 2ω + 1
Attempt to give substitution
Correct
Substitute into cubic
Cubic term
Quadratic term
(ω − 1) − 4 = 0
⇒ ω 3 − 6ω 2 + 11ω − 22 = 0
A2
Minus 1 each error (missing ‘= 0’
is an error)
[7]
5
OR
α + β +γ =
3
2
αβ + αγ + βγ =
M1
Attempt to find sums and
products of roots
A1
All correct
M1
Use of sum of roots
M1
Use of sum of product of roots in
pairs
M1
Use of product of roots
A2
Minus 1 each error (missing ‘= 0’
is an error)
1
2
αβγ = 2
Let new roots be k, l, m then
k + l + m = 2 (α + β + γ ) + 3 = 6 =
kl + km + lm = 4 (αβ + αγ + βγ ) +
−B
A
C
4 (α + β + γ ) + 3 = 11 =
A
klm = 8αβγ + 4 (αβ + βγ + βγ )
+2 (α + β + γ ) + 1 = 22 =
−D
A
⇒ ω − 6ω + 11ω − 22 = 0
3
2
[7]
25
4755
6
Mark Scheme
n
∑r
2
Jan 2007
= 16 n ( n + 1)( 2n + 1)
r =1
n = 1, LHS = RHS = 1
Assume true for n = k
2
Next term is ( k + 1)
Add to both sides
2
RHS = 16 k ( k + 1)( 2k + 1) + ( k + 1)
B1
M1
B1
Assuming true for k.
(k + 1)th term.
M1
Add to both sides
= 16 ( k + 1) [k ( 2k + 1) + 6 ( k + 1)]
M1
Attempt to factorise
A1
Correct brackets required – also
allow correct unfactorised form
Showing this is the expression with
n=k+1
( k + 1) [2k + 7k + 6]
= ( k + 1)( k + 2 )( 2k + 3)
= 16 ( k + 1) ( ( k + 1) + 1) ( 2 ( k + 1) + 1)
=
1
6
2
1
6
E1
But this is the given result with k + 1
replacing k. Therefore if it is true for k it is
true for k + 1. Since it is true for k = 1, it is
true for k = 1, 2, 3
and so true for all positive integers.
E1
Only if both previous E marks
awarded
[8]
Section A Total: 36
26
4755
Mark Scheme
Jan 2007
Section B
7(i)
7(ii)
7(iii)
y=
B1
[1]
5
8
B1,
B1
B1
[3]
x = −2, x = 4, y = 0
3 correct branches
Correct, labelled asymptotes
y-intercept labelled
B1
B1
B1
Ft from (ii)
Ft from (i)
[3]
5
7(iv)
=1
( x + 2 )( 4 − x )
⇒ 5 = ( x + 2 )( 4 − x )
Or evidence of other valid method
M1
⇒ 5 = −x + 2x + 8
2
⇒ x2 − 2 x − 3 = 0
⇒ ( x − 3)( x + 1) = 0
⇒ x = 3 or x = −1
Both values
A1
From graph:
x < −2 or
−1 < x < 3 or
x>4
B1
B1
B1
[5]
27
Ft from previous A1
Penalise inclusive inequalities
only once
4755
8(i)
Mark Scheme
1
m
=
8(ii)
=
−1
5
m =
1
−4 + 2 j
−
1
10
=
−4 − 2 j
2
M1
Attempt to multiply top and
bottom by conjugate
A1
[2]
Or equivalent
( −4 + 2 j)( −4 − 2 j)
j
( −4 )
Jan 2007
+ 22 = 20
B1
⎛1⎞
arg m = π − arctan ⎜ ⎟ = 2.68
⎝ 2⎠
M1
A1
So m = 20 ( cos 2.68 + jsin 2.68 )
A1(ft)
[4]
Attempt to calculate angle
Accept any correct expression for
angle, including 153.4 degrees, –
206 degrees and –3.61 (must be
at least 3s.f.)
Also accept ( r , θ ) form
8(iii)
(A)
B1
B1
[2]
π
4
8(iii)
(B)
Shaded region,
excluding
boundaries
B1(ft)
π
4
B1(ft)
B1(ft)
[3]
28
Correct initial point
Half-line at correct angle
Correct horizontal half-line from
starting point
Correct region indicated
Boundaries excluded (accept
dotted lines)
4755
Qu
Mark Scheme
Jan 2007
Mark
Answer
Comment
Section B (continued)
9(i)
N −1 =
9(ii)
M1
A1
A1
[3]
Dividing by determinant
One for each inverse c.a.o.
M1
A1
Must multiply in correct order
1 ⎛ 4 1⎞
21 ⎜⎝ −1 5 ⎟⎠
A1
Ft from MN
3 ⎞ 1 ⎛ 1 −2 ⎞
⎜
⎟× ⎜
⎟
7 ⎝ −1 1 ⎠ 3 ⎝ 0 3 ⎠
M1
A1
Multiplication in correct order
Ft from (i)
A1
Statement of equivalence to
−1
( MN )
1 ⎛1
M −1 = ⎜
3⎝0
−2 ⎞
3 ⎟⎠
1⎛ 4
3⎞
⎜
7 ⎝ −1 1 ⎟⎠
⎛ 3 2 ⎞⎛ 1 −3 ⎞ ⎛ 5 −1⎞
MN = ⎜
⎟⎜
⎟=⎜
⎟
⎝ 0 1 ⎠⎝ 1 4 ⎠ ⎝ 1 4 ⎠
( MN )
−1
=
N −1M −1 =
=
1⎛ 4
1 ⎛4
1⎞
⎜
⎟
21 ⎝ −1 5 ⎠
= ( MN )
−1
[6]
⇒ ( PQ ) PQQ −1 = IQ −1
−1
9(iii)
⇒ ( PQ ) PI = Q −1
−1
⇒ ( PQ ) P = Q
−1
M1
M1
−1
M1
⇒ ( PQ ) PP −1 = Q −1P −1
−1
⇒ ( PQ ) I = Q −1P −1
−1
⇒ ( PQ )
−1
= Q −1P −1
A1
−1
=I
Correctly eliminate I from LHS
QQ
Post-multiply both sides by P −1 at
an appropriate point
Correct and complete argument
[4]
Section B Total: 36
Total: 72
29
4755
Mark Scheme
June 2007
Section A
1 ⎛ 3 1⎞
⎜
⎟
10 ⎝ −4 2 ⎠
1(i)
M −1 =
1(ii)
20 square units
2
z − (3 − 2 j) = 2
3
x 3 − 4 = ( x − 1) Ax 2 + Bx + C + D
(
)
⇒ x − 4 = Ax + ( B − A ) x + ( C − B ) x − C + D
3
3
2
⇒ A = 1, B = 1, C = 1, D = −3
M1
A1
[2]
Attempt to find determinant
B1
[1]
2 × their determinant
B1
B1
B1
[3]
z ± ( 3 − 2 j) seen
radius = 2 seen
Correct use of modulus
M1
Attempt at equating coefficients or
long division (may be implied)
For A = 1
B1
B1
B1
B1
[5]
B1 for each of B, C and D
B1
B1
One for each correctly shown.
s.c. B1 if not labelled correctly but
position correct
4(i)
[2]
4(ii)
4(iii)
αβ = (1 − 2 j)( −2 − j) = −4 + 3j
α + β (α + β ) β
αβ * + ββ * 5j + 5
=
=
=
= j+1
5
β
ββ *
ββ *
*
M1
A1
[2]
Attempt to multiply
M1
Appropriate attempt to use
conjugate, or other valid method
5 in denominator or correct working
consistent with their method
All correct
A1
A1
[3]
20
4755
5
Mark Scheme
June 2007
Scheme A
w = 3x ⇒ x =
w
3
3
B1
Substitution. For substitution x = 3w
give B0 but then follow through for a
maximum of 3 marks
M1
Substitute into cubic
A3
Correct coefficients consistent with
x 3 coefficient, minus 1 each error
A1
[6]
Correct cubic equation c.a.o.
M1
Attempt to find sums and products of
roots (at least two of three)
M1
Attempt to use sums and products of
roots of original equation to find
sums and products of roots in related
equation
A3
Correct coefficients consistent with
x 3 coefficient, minus 1 each error
A1
[6]
M1
A1
Correct cubic equation c.a.o.
2
⎛ w ⎞ + 3⎛ w ⎞ − 7 ⎛ w ⎞ +1 = 0
⎟
⎜ ⎟
⎜ ⎟
⎝3⎠
⎝3⎠
⎝3⎠
3
2
⇒ w + 9 w − 63w + 27 = 0
⇒⎜
OR
Scheme B
α + β + γ = −3
αβ + αγ + βγ = −7
αβγ = −1
Let new roots be k, l, m then
k + l + m = 3 ( α + β + γ ) = −9 =
−B
A
kl + km + lm = 9 (αβ + αγ + βγ ) = −63 =
klm = 27αβγ = −27 =
C
A
−D
A
⇒ ω + 9ω − 63ω + 27 = 0
3
1
6(i)
r+2
6(ii)
2
−
1
r +3
=
1
50
r + 3 − ( r + 2)
1
=
( r + 2 )( r + 3) ( r + 2 )( r + 3)
50
⎡ 1
1 ⎤
∑ ( r + 2 )( r + 3) = ∑ ⎢⎣ r + 2 − r + 3 ⎥⎦
[2]
M1
Correct use of part (i) (may be
implied)
M1,
First two terms in full
⎛1 1⎞ ⎛ 1 1⎞
− ⎟+⎜ − ⎟
⎝ 51 52 ⎠ ⎝ 52 53 ⎠
M1
Last two terms in full (allow in terms
of n)
1
A1
Give B4 for correct without working
Allow 0.314 (3s.f.)
r =1
r =1
⎛1 1⎞ ⎛1 1⎞ ⎛1 1⎞
= ⎜ − ⎟ + ⎜ − ⎟ + ⎜ − ⎟ + .....
⎝3 4⎠ ⎝ 4 5⎠ ⎝ 5 6⎠
+⎜
=
Attempt at common denominator
3
−
1
53
=
50
159
[4]
21
4755
7
Mark Scheme
n
3n − 1
r =1
2
∑ 3r −1 =
n = 1, LHS = RHS = 1
Assume true for n = k
Next term is 3k
Add to both sides
3k − 1 k
RHS =
+3
2
3k − 1 + 2 × 3k
=
2
k
3× 3 −1
=
2
k +1
3 −1
=
2
But this is the given result with k + 1 replacing
k. Therefore if it is true for k it is true for k + 1.
Since it is true for k = 1, it is true for k = 1, 2, 3
and so true for all positive integers.
June 2007
B1
E1
M1
Assuming true for k
Attempt to add 3k to RHS
A1
c.a.o. with correct simplification
E1
Dependent on previous E1 and
immediately previous A1
E1
Dependent on B1 and both previous
E marks
[6]
Section A Total: 36
22
4755
Mark Scheme
June 2007
Section B
8(i)
( 2, 0 ) , ( −2, 0 ) , ⎛⎜ 0,
⎝
−4 ⎞
⎟
3 ⎠
8(ii)
x = 3, x = −1, x = 1, y = 0
B1
B1
B1
[3]
B4
[4]
1 mark for each
B1
Direction of approach must be clear
for each B mark
B1
M1
[3]
Evidence of method required
s.c. B2 for 2, − 2,
−4
3
Minus 1 for each error
8(iii)
Large positive x, y → 0+ , approach from above
(e.g. consider x = 100 )
Large negative x, y → 0− , approach from below
(e.g. consider x = −100 )
8(iv)
Curve
4 branches correct
Asymptotes correct and labelled
Intercepts labelled
B2
B1
B1
[4]
23
Minus 1 each error, min 0
4755
9(i)
9(ii)
Mark Scheme
x = 1− 2j
June 2007
B1
[1]
Complex roots occur in conjugate pairs. A cubic
has three roots, so one must be real. Or, valid
argument involving graph of a cubic or
behaviour for large positive and large negative
x.
E1
[1]
9(iii)
Scheme A
( x − 1 − 2 j)( x − 1 + 2 j) = x 2 − 2 x + 5
( x − α ) ( x 2 − 2 x + 5) = x 3 + Ax 2 + Bx + 15
M1
A1
A1(ft)
comparing constant term:
−5α = 15 ⇒ α = −3
M1
M1
So real root is x = −3
A1(ft)
( x + 3) ( x 2 − 2 x + 5) = x3 + Ax 2 + Bx + 15
M1
M1
A1
[9]
⇒ x 3 + x 2 − x + 15 = x 3 + Ax 2 + Bx + 15
⇒ A = 1, B = −1
OR
Scheme B
Product of roots = −15
M1
A1
M1
A1
(1 + 2 j)(1 − 2 j) = 5
Attempt to use factor theorem
Correct factors
Correct quadratic(using their factors)
Use of factor involving real root
Comparing constant term
From their quadratic
Expand LHS
Compare coefficients
1 mark for both values
Attempt to use product of roots
Product is –15
Multiplying complex roots
⇒ 5α = −15
A1
⇒ α = −3
Sum of roots = -A
⇒ − A = 1 + 2 j + 1 − 2 j − 3 = −1 ⇒ A = 1
A1
c.a.o.
M1
Attempt to use sum of roots
Substitute root x = –3 into cubic
M1
Attempt to substitute, or to use sum
A1
[9]
c.a.o.
( −3) + ( −3)
3
2
− 3B + 15 = 0 ⇒ B = −1
A = 1 and B = -1
OR
Scheme C
α = −3
(1 + 2 j)
6
3
+ A (1 + 2 j) + B (1 + 2 j) + 15 = 0
2
⇒ A( −3 + 4 j) + B (1 + 2 j) + 4 − 2 j = 0
⇒ −3 A + B + 4 = 0 and 4 A + 2 B − 2 = 0
⇒ A = 1 and B = −1
24
As scheme A, or other valid method
M1
Attempt to substitute root
M1
Attempt to equate real and imaginary
parts, or equivalent.
A1
[9]
c.a.o.
4755
Mark Scheme
Section B (continued)
10(i)
⎛ 1 −2
⎜
⎜
⎜3
⎝
AB = 2
1
2
k ⎞⎛ −5
−2 + 2k
−4 − k ⎞
2
−1 − 3k
−2 + 2k
−8
5
⎟⎜ 8
⎟⎜
−1⎟⎜
⎠⎝ 1
June 2007
M1
⎟
⎟
⎟
⎠
Attempt to multiply matrices
(can be implied)
0
0 ⎞
⎛ k − 21
⎜
⎟
0
k − 21
=
⎜ 0
⎟
⎜ 0
0
k − 21⎟⎠
⎝
n = 21
10(ii)
A1
[2]
⎛ −5 −2 + 2k
⎜ 8 −1 − 3k
A =
k − 21 ⎜⎜
−8
⎝1
−1
1
−4 − k ⎞
⎟
−2 + 2k
⎟
5 ⎠⎟
k ≠ 21
M1
M1
A1
Use of B
Attempt to use their answer to (i)
Correct inverse
A1
Accept n in place of 21 for full
marks
[4]
10
(iii)
Scheme A
⎛ −5 0
1 ⎜
8 −4
−20 ⎜⎜
⎝ 1 −8
−5 ⎞ ⎛ 1 ⎞
⎛ −20 ⎞ ⎛ 1 ⎞
1 ⎜
⎟
⎜
⎟
⎟ ⎜ ⎟
0 12 =
⎟ ⎜ ⎟ −20 ⎜ −40 ⎟ = ⎜ 2 ⎟
⎜ −80 ⎟ ⎜ 4 ⎟
5 ⎟⎠ ⎜⎝ 3 ⎟⎠
⎝
⎠ ⎝ ⎠
x = 1, y = 2, z = 4
M1
M1
Attempt to use inverse
Their inverse with k = 1
A3
[5]
One for each correct (ft)
OR
Scheme B
Attempt to eliminate 2 variables
Substitute in their value to attempt to find others
x = 1, y = 2, z = 4
M1
M1
A3
[5]
s.c. award 2 marks only for
x = 1, y = 2, z = 4 with no working.
Section B Total: 36
Total: 72
25
4755
Mark Scheme
4755 (FP1)
Further Concepts for Advanced Mathematics
Qu
Answer
Section A
⎛ 3 1 ⎞⎛ 2 −1⎞ ⎛ 6 0 ⎞
1(i)
BA = ⎜
⎟⎜
⎟=⎜
⎟
⎝ −2 4 ⎠⎝ 0 3 ⎠ ⎝ −4 14 ⎠
1(ii)
Mark
Comment
Attempt to multiply
M1
c.a.o.
A1
[2]
det BA = ( 6 × 14 ) − ( −4 × 0 ) = 84
M1
A1
A1(ft)
[3]
M1
3 × 84 = 252 square units
2(i)
January 2008
α 2 = ( −3 + 4 j)( −3 + 4 j) = ( −7 − 24 j)
Attempt to calculate any
determinant
c.a.o.
Correct area
Attempt to multiply with use of
j2 = −1
c.a.o.
A1
[2]
2(ii)
α =5
B1
B1
arg α = π − arctan 43 = 2.21 (2d.p.) (or
Accept 2.2 or 127D
126.87D )
α = 5 ( cos 2.21 + jsin 2.21)
3(i)
B1(ft)
s.c. lose 1 mark only if α 2 used
throughout (ii)
[3]
Showing 3 satisfies the equation
B1
(may be implied)
Valid attempt to factorise
M1
Correct quadratic factor
A1
3 + 3 − 7 × 3 − 15 = 0
3
2
(
z 3 + z 2 − 7 z − 15 = ( z − 3) z 2 + 4 z + 5
z=
−4 ± 16 − 20
2
Accept degrees and ( r , θ ) form
)
= −2 ± j
So other roots are −2 + j and − 2 − j
M1
Use of quadratic formula, or other
valid method
A1
One mark for both c.a.o.
[5]
3(ii)
B2
[2]
16
Minus 1 for each error
ft provided conjugate imaginary
roots
4755
4
Mark Scheme
n
n
r =1
r =1
n
∑ [( r + 1)( r − 2 )] =∑ r − ∑ r − 2n
2
r =1
= 1 n ( n + 1)( 2n + 1) − 1 n ( n + 1) − 2n
6
2
1
= n ⎡⎣( n + 1)( 2n + 1) − 3 ( n + 1) − 12 ⎤⎦
6
1
= n 2n 2 + 3n + 1 − 3n − 3 − 12
6
= 1 n 2n 2 − 14
6
= 1 n n2 − 7
3
(
(
(
5(i)
5(ii)
January 2008
)
)
)
M1
Attempt to split sum up
A2
Minus one each error
M1
Attempt to factorise
M1
Collecting terms
All correct
A1
[6]
p = −3, r = 7
One mark for each
B2
[2] s.c. B1 if b and d used instead of
p and r
B1
q = αβ + αγ + βγ
α 2 + β 2 + γ 2 = (α + β + γ ) − 2 (αβ + αγ + βγ )
2
M1
= (α + β + γ ) − 2 q
2
⇒ 13 = 32 − 2q
⇒ q = −2
6(i)
c.a.o.
A1
[3]
Use of inductive definition
M1
c.a.o.
A1
a2 = 7 × 7 − 3 = 46
a3 = 7 × 46 − 3 = 319
6(ii)
When n = 1,
13 × 7 0 + 1
[2]
= 7 , so true for n = 1
2
Assume true for n = k
13 × 7 k −1 + 1
ak =
2
13 × 7 k −1 + 1
⇒ ak +1 = 7 ×
−3
2
=
=
13 × 7 k + 7
2
B1
E1
Correct use of part (i) (may be
implied)
Assuming true for k
M1
Attempt to use ak +1 = 7 ak − 3
A1
Correct simplification
E1
Dependent on A1 and previous
E1
−3
13 × 7 k + 7 − 6
2
k
=
Attempt to find q using α 2 + β 2 + γ 2
and α + β + γ , but not αβγ
13 × 7 + 1
2
But this is the given result with k + 1
replacing k. Therefore if it is true for k it is
true for k + 1. Since it is true for k = 1, it is
true for k = 1, 2, 3 and so true for all positive
integers.
17
E1
[6] Dependent on B1 and previous
E1
Section A Total: 36
4755
Mark Scheme
Section B
7(i)
(1, 0 ) and ( 0,
7(ii)
1
18
x = 2, x = −3, x =
)
January 2008
B1
B1
[2]
−3
2
Minus 1 for each error
B4
[4]
,y=0
7(iii)
1
18
B1
Correct approaches to vertical
asymptotes
B1
Through clearly marked (1, 0 )
and ( 0,
1
1
18
)
[2]
B1
x < −3 , x > 2
7(iv)
−3
2
B2
< x ≤1
B1 for
−3
2
< x < 1 , or
−3
2
≤ x ≤1
[3]
B3
Circle, B1; radius 2, B1;
centre 3j, B1
B3
Half line, B1; from -1, B1;
π
4 to x-axis, B1
8(i)
[6]
B2(ft)
8(ii)
Correct region between their
circle and half line indicated
s.c. B1 for interior of circle
[2]
M1
Sketch should clearly show the radius and
centre of the circle and the starting point
and angle of the half-line.
8(iii)
arg z =
π
2
A1(ft)
M1
− arcsin 23 = 0.84 (2d.p.)
A1
[4]
18
Tangent from origin to circle
Correct point placed by eye
where tangent from origin meets
circle
Attempt to use right angled
triangle
c.a.o. Accept 48.20D (2d.p.)
4755
Mark Scheme
January 2008
9(i)
( −3, − 3)
9(ii)
( x, x )
B1
[1]
B1
B1
[2]
9(iii)
⎛1 0 ⎞
⎜1 0 ⎟
⎝
⎠
Minus 1 each error to min of 0
B3
[3]
9(iv)
Rotation through
π
2
anticlockwise about the
origin
9(v)
⎛ 0 −1 ⎞ ⎛ 1 0 ⎞ ⎛ −1 0 ⎞
⎜ 1 0 ⎟ × ⎜1 0 ⎟ = ⎜ 1 0 ⎟
⎝
⎠ ⎝
⎠ ⎝
⎠
Rotation and angle (accept 90D )
B1
Centre and sense
B1
[2]
M1
A1
Attempt to multiply using their T in
correct order
c.a.o.
[2]
9(vi)
⎛ −1 0 ⎞⎛ x ⎞ ⎛ − x ⎞
⎜ 1 0 ⎟⎜ y ⎟ = ⎜ x ⎟
⎝
⎠⎝ ⎠ ⎝ ⎠
M1
A1(ft)
May be implied
So (-x, x)
A1
Line y = -x
[3]
19
c.a.o. from correct matrix
4755
Mark Scheme
June 2008
4755 (FP1) Further Concepts for Advanced Mathematics
Qu
Mark
Answer
Section A
⎛ −1 0 ⎞
1(i)
⎜
⎟
⎝ 0 1⎠
1(ii)
1(iii)
Comment
B1
⎛ 3 0⎞
⎜
⎟
⎝ 0 3⎠
B1
M1
Multiplication, or other valid
A1
method (may be implied)
[4] c.a.o.
⎛ 3 0 ⎞ ⎛ −1 0 ⎞ ⎛ −3 0 ⎞
⎜
⎟⎜
⎟ =⎜
⎟
⎝ 0 3⎠ ⎝ 0 1⎠ ⎝ 0 3⎠
2
B3
Circle, B1; centre −3 + 2 j , B1;
radius = 2, B1
B3
Line parallel to real axis, B1;
through (0, 2), B1;
correct half line, B1
B1
Points −1 + 2 j and −5 + 2 j
indicated
c.a.o.
-3
[7]
3
⎛ −1 −1⎞⎛ x ⎞ ⎛ x ⎞
⎜
⎟⎜ ⎟ = ⎜ ⎟
⎝ 2 2 ⎠⎝ y ⎠ ⎝ y ⎠
⇒ − x − y = x, 2 x + 2 y = y
⇒ y = −2 x
4
M1
⎛ −1 −1⎞⎛ x ⎞ ⎛ x ⎞
For ⎜
⎟⎜ ⎟ = ⎜ ⎟
⎝ 2 2 ⎠⎝ y ⎠ ⎝ y ⎠
M1
B1
[3]
3 x3 − x 2 + 2 ≡ A ( x − 1) + ( x3 + Bx 2 + Cx + D )
3
≡ Ax3 − 3 Ax 2 + 3 Ax − A + x3 + Bx 2 + Cx + D
M1
Attempt to compare coefficients
B4
One for each correct value
≡ ( A + 1) x 3 + ( B − 3 A) x 2 + ( 3 A + C ) x + ( D − A )
⇒ A = 2, B = 5, C = −6, D = 4
[5]
19
4755
5(i)
Mark Scheme
⎛7 0 0⎞
⎜
⎟
AB = 0 7 0
⎜
⎟
⎜0 0 7⎟
⎝
⎠
5(ii)
⎛ −1
−1
A =
6
B3
2⎞
M1
⎟
14 −14 7
⎟
7 ⎜⎜
⎟
⎝ −5 7 −4 ⎠
w = 2x ⇒ x =
Use of B
A1
[2] c.a.o.
w
B1
2
3
Minus 1 each error to minimum of
0
[3]
0
1⎜
June 2008
2
⎛ w ⎞ + ⎛ w ⎞ − 3⎛ w ⎞ +1 = 0
⎟ ⎜ ⎟
⎜ ⎟
⎝2⎠ ⎝2⎠
⎝2⎠
⇒ 2⎜
M1
A1
⇒ w3 + w2 − 6 w + 4 = 0
A2
Substitution. For substitution
x = 2 w give B0 but then follow
through for a maximum of 3
marks
Substitute into cubic
Correct substitution
Minus 1 for each error (including
‘= 0’ missing), to a minimum of 0
Give full credit for integer multiple
of equation
[5]
6
OR
α + β + γ = − 12
αβ + αγ + βγ = −
B1
All three
M1
Attempt to use sums and
products of roots of original
equation to find sums and
products of roots in related
equation
Sums and products all correct
3
2
αβγ = − 12
Let new roots be k, l, m then
k + l + m = 2 ( α + β + γ ) = −1 =
−B
A
kl + km + lm = 4 (αβ + αγ + βγ ) = −6 =
klm = 8αβγ = −4 =
C
A
A1
−D
A
A2
ft their coefficients; minus one for
each error (including ‘= 0’
missing), to minimum of 0
Give full credit for integer multiple
[5] of equation
⇒ ω + ω − 6ω + 4 = 0
3
2
20
4755
Mark Scheme
7(i)
1
−
1
3r − 1 3r + 2
≡
≡
3r + 2 − ( 3r − 1)
( 3r − 1)( 3r + 2 )
3
( 3r − 1)( 3r + 2 )
June 2008
M1
Attempt at correct method
A1
Correct, without fudging
[2]
n
7(ii)
1
1
n
⎡ 1
1
⎤
∑ ( 3r − 1)( 3r + 2 ) = 3 ∑ ⎢⎣ 3r − 1 − 3r + 2 ⎥⎦
M1
Attempt to use identity
1 ⎡⎛ 1 1 ⎞ ⎛ 1 1 ⎞
1 ⎞⎤
⎛ 1
= ⎢⎜ − ⎟ + ⎜ − ⎟ + ....... + ⎜
−
⎟
3 ⎣⎝ 2 5 ⎠ ⎝ 5 8 ⎠
⎝ 3n − 1 3n + 2 ⎠ ⎥⎦
1 ⎡1
1 ⎤
= ⎢ −
3 ⎣ 2 3n + 2 ⎥⎦
A1
M1
Terms in full (at least two)
Attempt at cancelling
r =1
r =1
21
1
missing,
3
A1 max if answer not in terms
[5]
of n
Section A Total: 36
A2
A1 if factor of
4755
Mark Scheme
June 2008
Section B
8(i)
x = 3 , x = −2 , y = 2
B1
B1
B1
[3]
8(ii)
Large positive x, y → 2+
(e.g. consider x = 100 )
Large negative x, y → 2−
(e.g. consider x = −100 )
M1
Evidence of method required
B1
B1
[3]
8(iii)
Curve
Central and RH branches correct
Asymptotes correct and labelled
LH branch correct, with clear minimum
B1
B1
B1
[3]
y=2
8(iv)
B2
−2 < x < 3
B1
[3]
x≠0
22
B2 max if any inclusive
inequalities appear
B3 for −2 < x < 0 and 0 < x < 3 ,
4755
9(i)
Mark Scheme
2 + 2 j and − 1 − j
June 2008
B2
1 mark for each
[2]
9(ii)
B2
1 mark for each correct pair
[2]
9(iii)
( x − 2 − 2 j)( x − 2 + 2 j)( x + 1 + j)( x + 1 − j)
M1
B2
= ( x 2 − 4 x + 8 )( x 2 + 2 x + 2 )
A1
= x 4 + 2 x 3 + 2 x 2 − 4 x3 − 8 x 2 − 8 x + 8 x 2 + 16 x + 16
M1
Expanding
A2
Minus 1 each error, A1 if only
errors are sign errors
4
3
Attempt to use factor theorem
Correct factors, minus 1 each
error
B1 if only errors are sign errors
One correct quadratic with real
coefficients (may be implied)
2
= x − 2 x + 2 x + 8 x + 16
⇒ A = −2, B = 2, C = 8, D = 16
OR
[7]
∑α = 2
αβγδ = 16
∑αβ = αα * + αβ + αβ * + ββ * + βα * + β *α *
∑αβγ = αα β + αα β
*
*
*
+ αββ * + α * ββ *
B1
B1
M1
M1
∑αβ = 2 , ∑αβγ = −8
A1
A = −2, B = 2, C = 8, D = 16
A2
Both correct
Minus 1 each error, A1 if only
[7] errors are sign errors
OR
M1
M1
A1
Attempt to substitute in one root
Attempt to substitute in a second root
Equating real and imaginary parts to 0
Attempt to solve simultaneous equations
A = −2, B = 2, C = 8, D = 16
M1
M1
A2
[7]
Both correct
Minus 1 each error, A1 if only
errors are sign errors
23
4755
Qu
Mark Scheme
June 2008
Mark
Answer
Section B (continued)
n
n
n
10(i)
∑ r 2 ( r + 1) = ∑ r 3 +∑ r 2
r =1
r =1
A1
Separation of sums (may be
implied)
One mark for both parts
Attempt to factorise (at least two
linear algebraic factors)
Correct
E1
Complete, convincing argument
M1
r =1
= 14 n 2 ( n + 1) + 16 n ( n + 1)( 2n + 1)
2
B1
M1
= 121 n ( n + 1) ⎡⎣3n ( n + 1) + 2 ( 2n + 1)⎤⎦
(
= 121 n ( n + 1) 3n 2 + 7 n + 2
)
= 121 n ( n + 1)( n + 2 )( 3n + 1)
Comment
[5]
10(ii)
n
∑ r 2 ( r + 1) =
1
12
r =1
n ( n + 1)( n + 2 )( 3n + 1)
B1
n = 1, LHS = RHS = 2
E1
Assume true for n = k
k
∑ r ( r + 1) =
2
r =1
k +1
1
12
2 must be seen
Assuming true for k
k ( k + 1)( k + 2 )( 3k + 1)
∑ r ( r + 1)
2
r =1
= 121 k ( k + 1)( k + 2 )( 3k + 1) + ( k + 1) ( k + 2 )
2
= 121 ( k + 1)( k + 2 ) [ k ( 3k + 1) + 12 ( k + 1)]
(
= 121 ( k + 1)( k + 2 ) 3k 2 + 13k + 12
)
B1
M1
A1
A1
(k + 1)th term
Attempt to factorise
Correct
Complete convincing argument
= 121 ( k + 1)( k + 2 )( k + 3 )( 3k + 4 )
= 121 ( k + 1) ( ( k + 1) + 1) ( ( k + 1) + 2 ) ( 3 ( k + 1) + 1)
But this is the given result with k + 1 replacing
k. Therefore if it is true for k it is true for k +
1. Since it is true for k = 1, it is true for k = 1,
2, 3 and so true for all positive integers.
E1
E1
[8]
Dependent on previous A1 and
previous E1
Dependent on first B1 and
previous E1
Section B Total: 36
Total: 72
24
4755
Mark Scheme
January 2009
4755 (FP1) Further Concepts for Advanced
Mathematics
Section A
1(i)
1(ii)
6 ± 36 − 40
2
⇒ z = 3 + j or z = 3 − j
M1
3 + j = 10 = 3.16 (3s.f.)
M1
⎛1⎞
arg ( 3 + j) = arctan ⎜ ⎟ = 0.322 (3s.f.)
⎝3⎠
M1
z=
A1
[2]
Method for modulus
Method for argument
(both methods must be seen
following A0)
One mark for both roots in modulusA1
argument form – accept surd and
decimal equivalents and ( r , θ )
[3]
form. Allow +18.4o for θ.
⇒ roots are 10 ( cos 0.322 + jsin 0.322 )
and 10 ( cos 0.322 − jsin 0.322 )
or √10(cos(-0.322)+jsin(-0.322))
2
Use of quadratic formula/completing
the square
For both roots
2 x 2 − 13 x + 25 = A ( x − 3) − B ( x − 2 ) + C
2
⇒ 2 x 2 − 13 x + 25
B1
M1
For A=2
Attempt to compare coefficients of
x1or x0, or other valid method.
A1
A1
For B and C,
cao.
= Ax 2 − ( 6 A + B ) x + ( 2 B + C ) + 9 A
A=2
B=1
C=5
[4]
3(i)
⎛2 1⎞
⎜0 2⎟
⎝
⎠
B1
[1]
3(ii)
⎛ 2 −1 ⎞ ⎛ 0 2 3 1 ⎞ ⎛ 0 4 4 0 ⎞
⎜
⎟⎜
⎟=⎜
⎟
⎝ 0 3 ⎠⎝0 0 2 2⎠ ⎝ 0 0 6 6⎠
⇒ A′′= ( 4,
3(iii)
0 ) , B′′ = ( 4,
6 ) , C′′= ( 0,
M1
6)
Applying matrix to column vectors,
with a result.
All correct
A1
[2]
Both factor and direction for each
B1
mark. SC1 for “enlargement”, not
B1
[2] stretch.
Stretch factor 4 in x-direction.
Stretch factor 6 in y-direction
18
4755
4
Mark Scheme
arg ( z − ( 2 − 2 j) ) =
π
January 2009
B1
4
B1
Equation involving arg(complex
variable).
Argument (complex expression) =
π
B1
[3]
5
Sum of roots = α + ( −3α ) + α + 3 = 3 − α = 5
M1
⇒ α = −2
A1
Product of roots
M1
M1
= − 2 × 6 × 1 = −12
4
All correct
Use of sum of roots
Attempt to use product of roots
Attempt to use sum of products of
roots in pairs
Product of roots in pairs
= −2 × 6 + ( −2 ) × 1 + 6 × 1 = −8
A1
One mark for each, ft if α incorrect
A1
[6]
⇒ p = −8 and q = 12
Alternative solution
(x-α)(x+3α)(x-α-3)
=x3+(α-3)x2+(-5α2-6α)x+3α3+9α2
=> α = -2,
M1
2
M1A1 Matching coefficient of x ,cao.
Matching other coefficients
M1
A1A1 One mark for each, ft incorrect α.
[6]
p = -8 and q = 12
6
Attempt to multiply factors
n
n
n
M1
r =1
r =1
r =1
M1
∑ ⎡⎣r ( r 2 − 3)⎤⎦ = ∑ r 3 − 3∑ r
1
3
2
= n 2 ( n + 1) − n ( n + 1)
4
2
1
= n ( n + 1) ( n ( n + 1) − 6 )
4
1
1
= n ( n + 1) ( n 2 + n − 6 ) = n ( n + 1)( n + 3)( n − 2 )
4
4
19
A2
Separate into separate sums.
(may be implied)
Substitution of standard result in
terms of n.
For two correct terms (indivisible)
M1
Attempt to factorise with n(n+1).
A1
Correctly factorised to give fully
[6] factorised form
4755
7
Mark Scheme
When n = 1, 6 ( 3n − 1) = 12 , so true for n = 1
January 2009
B1
E1
Assume true for k
M1
Add correct next term to both sides
= 6 ⎢ 3k − 1 +
M1
Attempt to factorise with a factor 6
= 6 ( 3k +1 − 1)
A1
c.a.o. with correct simplification
But this is the given result with k + 1 replacing
k. Therefore if it is true for n = k, it is true for
n = k + 1.
E1
Dependent on A1 and first E1
E1
Dependent on B1 and second E1
Assume true for n = k
(
k
k
)
12 + 36 + 108 + ..... + (4 × 3 ) = 6 3 − 1
⇒ 12 + 36 + 108 + ..... + (4 × 3
(
)
k +1
= 6 3 − 1 + (4 × 3
k
k +1
)
)
2
⎡
( ) 3 × 3k +1 ⎤⎥
⎣
⎦
k
k
= 6 ⎡⎣3 − 1 + 2 × 3 ⎤⎦
Since it is true for n = 1, it is true for n = 1, 2,
3... and so true for all positive integers.
[7]
Section A Total: 36
20
4755
Mark Scheme
January 2009
Section B
8(i)
8(ii)
(
)(
)
⎛
⎝
3, 0 , − 3, 0 ⎜ 0,
B1
B1
Intercepts with x axis (both)
Intercept with y axis
SC1 if seen on graph or if x = +√3,
y = 3/8 seen without y = 0, x = 0
[2] specified.
3⎞
⎟
8⎠
B3
x = 4, x = −2, y = 1
Minus 1 for each error. Accept
equations written on the graph.
[3]
8(iii)
B1
B1B1
B1
Correct approaches to vertical
asymptotes, LH and RH branches
LH and RH branches approaching
horizontal asymptote
On LH branch 0<y<1 as x→-∞.
[4]
8(iv)
−2 < x ≤ − 3 and 4 > x ≥ 3
B1
B2
[3]
21
LH interval and RH interval correct
(Award this mark even if errors in
inclusive/exclusive inequality signs)
All inequality signs correct, minus 1
each error
4755
9(i)
Mark Scheme
α +β =3
B1
αα * = (1 + j)(1 − j) = 2
M1 Attempt to multiply (1 + j)(1 − j)
A1
M1 Multiply top and bottom by 1 − j
A1
[5]
3 (1 − j)
3
3 3
α +β
=
=
= − j
1 + j (1 + j)(1 − j) 2 2
α
9(ii)
( z − (1 + j) ) ( z − (1 − j) )
M1 Or alternative valid methods
A1
(Condone no “=0” here)
[2]
= z − 2z + 2
2
9(iii)
January 2009
1 − j and 2 + j
Either
( z − ( 2 − j) ) ( z − ( 2 + j) )
B1
For both
M1
For attempt to obtain an equation
using the product of linear factors
involving complex conjugates
M1
Using the correct four factors
= z2 − 4z + 5
( z − 2 z + 2 )( z − 4 z + 5)
2
2
= z 4 − 6 z 3 + 15 z 2 − 18 z + 10
So equation is z 4 − 6 z 3 + 15 z 2 − 18 z + 10 = 0
A2
[5]
Or alternative solution
Use of ∑α = 6, ∑αβ = 15,
∑αβγ = 18 and αβγδ = 10
All correct, -1 each error (including
omission of “=0”) to min of 0
Use of relationships between roots
M1 and coefficients.
A3 All correct, -1 each error, to min of 0
to obtain the above equation.
[5]
22
4755
10(i)
Mark Scheme
α = 3 × −5 + 4 × 11 + −1 × 29 = 0
January 2009
B1
M1
A1
β = −2 × −7 + 7 × ( 5 + k ) + −3 × 7 = 28 + 7 k
Attempt at row 3 x column 3
[3]
10(ii)
⎛ 42 0 0 ⎞
AB = ⎜⎜ 0 42 0 ⎟⎟
⎜ 0 0 42 ⎟
⎝
⎠
10(iii)
A
−1
⎛ 11
1 ⎜
= ⎜1
42 ⎜
⎝ −5
B2
[2]
−5
−7 ⎞
11
7
29
Minus 1 each error to min of 0
⎟
⎟
7 ⎟⎠
M1
Use of B
1
B1
42
A1
Correct inverse, allow decimals to 3
sf
[3]
10(iv)
⎛ x⎞
⎜ ⎟ ⎜ ⎟
⎟
11 7 ⎜ −9 ⎟ = ⎜ y ⎟
⎟
29 7 ⎟⎠ ⎜⎝ 26 ⎟⎠ ⎜⎝ z ⎟⎠
⎛ −126 ⎞ ⎛ −3 ⎞
1 ⎜
= ⎜ 84 ⎟⎟ = ⎜⎜ 2 ⎟⎟
42 ⎜
⎟ ⎜ ⎟
⎝ −84 ⎠ ⎝ −2 ⎠
⎛ 11
1 ⎜
1
42 ⎜⎜
⎝ −5
−5
−7 ⎞ ⎛ 1 ⎞
M1
Attempt to pre-multiply by A −1
SC B2 for Gaussian elimination with
3 correct solutions, -1 each error to
min of 0
x = −3, y = 2, z = −2
A3
[4] Minus 1 each error
Section B Total: 36
Total: 72
23
4755
Mark Scheme
June 2009
4755 (FP1) Further Concepts for Advanced
Mathematics
Section A
1(i)
1(ii)
2
M −1 =
1 ⎛ 2 1⎞
⎜
⎟
11 ⎝ −3 4 ⎠
M1
Dividing by determinant
A1
[2]
1 ⎛ 2 1 ⎞⎛ 49 ⎞ ⎛ x ⎞ 1 ⎛ 198 ⎞
⎜
⎟⎜
⎟=⎜ ⎟= ⎜
⎟
11 ⎝ −3 4 ⎠⎝100 ⎠ ⎝ y ⎠ 11 ⎝ 253 ⎠
⇒ x = 18, y = 23
M1
Pre-multiplying by their inverse
A1(ft)
A1(ft)
[3]
B1
M1
A1
M1
z + z − 7 z − 15 = ( z − 3) ( z + 4 z + 5 )
3
2
2
−4 ± 16 − 20
2
⇒ z = −2 + j and z = −2 − j
z2 + 4z + 5 = 0 ⇒ z =
A1
Show z = 3 is a root; may be
implied
Attempt to find quadratic factor
Correct quadratic factor
Use of quadratic formula or other
valid method
Both solutions
[5]
3(i)
B1
Asymptote at x = -4
B1
Both branches correct
[2]
2
= x + 3 ⇒ x 2 + 7 x + 10 = 0
x+4
3(ii)
M1
A1
⇒ x = −2 or x = −5
x ≥ −2 or − 4 > x ≥ −5
A1
A2
[5]
18
Attempt to find where graphs
cross or valid attempt at solution
using inequalities
Correct intersections (both), or -2
and -5 identified as critical values
x ≥ −2
−4 > x ≥ −5
s.c.
A1 for −4 ≥ x ≥ −5 or − 4 > x > −5
4755
4
Mark Scheme
2 w − 6 w + 3w =
⇒w=
−1
2
p
2
M1
A1
2
1
Use of sum of roots – can be
implied
2
⇒ roots are 1, −3,
−q
June 2009
= αβγ =
−9
2
3
A1
M1
2
⇒q=9
= αβ + αγ + βγ = −6 ⇒ p = −12
Correct roots seen
Attempt to use relationships
between roots
s.c. M1 for other valid method
A2(ft)
[6] One mark each for p = -12 and q
=9
19
4755
5(i)
Mark Scheme
1
1
5r + 3 − 5r + 2
−
≡
5r − 2 5r + 3 ( 5r + 3)( 5r − 2 )
≡
5(ii)
June 2009
M1
5
( 5r + 3)( 5r − 2 )
1
30
Attempt to form common
denominator
Correct cancelling
A1
[2]
1
30
⎡
1
1
⎤
∑ ( 5r − 2 )( 5r + 3) = 5 ∑ ⎢ ( 5r − 2 ) − ( 5r + 3) ⎥
⎣
⎦
⎡⎛ 1 1 ⎞ ⎛ 1 1 ⎞ ⎛ 1 1 ⎞
⎤
⎜ − ⎟ + ⎜ − ⎟ + ⎜ − ⎟ + ... ⎥
1 ⎢⎝ 3 8 ⎠ ⎝ 8 13 ⎠ ⎝ 13 18 ⎠
⎥
= ⎢
5⎢ ⎛ 1
1 ⎞ ⎛ 1
1 ⎞⎥
+
−
+
−
⎢⎣ ⎜⎝ 5n − 7 5n − 2 ⎟⎠ ⎜⎝ 5n − 2 5n + 3 ⎟⎠ ⎥⎦
r =1
r =1
1 ⎡1
n
⎤
= ⎢ −
=
5 ⎣ 3 5n + 3 ⎥⎦ 3 ( 5n + 3)
1
B1
First two terms in full
B1
Last term in full
M1
Attempt to cancel terms
A1
[4]
6
When n = 1,
1
2
n ( 7 n − 1) = 3 , so true for n =
1
B1
E1
Assume true for n = k
M1
Add ( k + 1) th term to both sides
M1
Valid attempt to factorise
A1
c.a.o. with correct simplification
E1
Dependent on previous E1 and
immediately previous A1
Dependent on B1 and both previous
E marks
Assume true for n = k
1
k ( 7 k − 1)
2
⇒ 3 + 10 + 17 + ..... + (7 ( k + 1) − 4)
3 + 10 + 17 + ..... + (7 k − 4) =
=
=
=
=
=
1
2
1
2
1
2
1
2
1
2
k ( 7 k − 1) + (7 ( k + 1) − 4)
[ k ( 7 k − 1) + (14 ( k + 1) − 8) ]
2
⎣⎡ 7 k + 13k + 6 ⎦⎤
( k + 1)( 7k + 6 )
( k + 1) ( 7 ( k + 1) − 1)
But this is the given result with k + 1
replacing k. Therefore if it is true for k it is
true for k + 1.
Since it is true for n = 1, it is true for n = 1,
2, 3 and so true for all positive integers.
E1
[7]
Section A Total: 36
20
4755
Mark Scheme
Section B
7(i)
5
( 0 , 10 ) , ( −2 , 0 ) , ⎛⎜ , 0 ⎞⎟
⎝3
⎠
June 2009
B1
B1
B1
[3]
7(ii)
x=
−1
2
, x =1, y =
3
B1
B1
B1
[3]
2
+
7(iii)
7(iv)
3
2
(e.g. consider x = 100 )
−
3
Large negative x, y →
2
(e.g. consider x = −100 )
Large positive x, y →
M1
B1
B1
[3]
Curve
3 branches of correct shape
Asymptotes correct and labelled
Intercepts correct and labelled
B1
B1
B1
[3]
21
Clear evidence of method
required for full marks
4755
8 (i)
Mark Scheme
z − ( 4 + 2 j) = 2
June 2009
B1
B1
B1
Radius = 2
z − ( 4 + 2 j) or z − 4 − 2 j
All correct
[3]
8(ii)
arg ( z − ( 4 + 2 j) ) = 0
B1
B1
B1
8(iii)
a = 4 − 2cos
b = 2 + 2sin
π
4
π
4
(
Equation involving the argument
of a complex variable
Argument = 0
All correct
[3]
=4− 2
M1
Valid attempt to use trigonometry
involving
= 2+ 2
)
P = 4− 2 + 2+ 2 j
8(iv)
3
π > arg ( z − ( 4 + 2 j) ) > 0
4
and z − ( 4 + 2 j) < 2
4
, or coordinate
A2
geometry
[3]
1 mark for each of a and b
s.c. A1 only for a = 2.59, b = 3.41
B1
B1
22
π
arg ( z − ( 4 + 2 j) ) > 0
B1
3
arg ( z − ( 4 + 2 j) ) < π
4
z − ( 4 + 2 j) < 2
[3]
Deduct one mark if only error is
use of inclusive inequalities
4755
Mark Scheme
Section B (continued)
9(i) Matrix multiplication is associative
B1
[1]
⎛ 3 0 ⎞⎛ 0 1 ⎞
MN = ⎜
⎟⎜
⎟
⎝ 0 2 ⎠⎝ 1 0 ⎠
⎛ 0 3⎞
⇒ MN = ⎜
⎟
⎝ 2 0⎠
⎛ −2 0 ⎞
QMN = ⎜
⎟
⎝ 0 3⎠
9(ii)
9(iii)
June 2009
M1
Attempt to find MN or QM
A1
⎛ 0 −2 ⎞
or QM = ⎜
⎟
⎝3 0 ⎠
A1(ft)
[3]
M is a stretch, factor 3 in the x direction,
factor 2 in the y direction.
B1
B1
N is a reflection in the line y = x.
B1
Q is an anticlockwise rotation through 90D
about the origin.
B1
⎛ −2 0 ⎞⎛ 1 1 2 ⎞ ⎛ −2 −2 −4 ⎞
⎜
⎟⎜
⎟=⎜
⎟
⎝ 0 3 ⎠⎝ 2 0 2 ⎠ ⎝ 6 0 6 ⎠
Stretch factor 3 in the x direction
Stretch factor 2 in the y direction
[4]
M1
A1(ft)
Applying their QMN to points.
Minus 1 each error to a minimum
of 0.
B2
Correct, labelled image points,
minus 1 each error to a minimum
of 0. Give B4 for correct diagram
with no workings.
[4]
Section B Total: 36
Total: 72
23
4755
Mark Scheme
January 2010
4755 (FP1) Further Concepts for Advanced
Mathematics
αβ = ( −3 + j)( 5 − 2 j) = −13 + 11j
1
Use of j2 = −1
M1
A1
[2]
α −3 + j ( −3 + j)( 5 + 2 j) −17 1
=
=
=
−
j
β 5 − 2j
29
29 29
2
(i)
AB is impossible
B1
CA = ( 50 )
B1
3
B+D=
1 
B1

 6 −2 
 20 4 32 


AC = −10 −2 −16


 20 4 32 


(ii)
Use of conjugate
M1
29 in denominator
A1
All correct
A1
[3]
B2
[5]
 −2 0   5 1   −10 −2 

=

 4 1   2 −3   22 1 
DB = 
M1
A1
[2]
3
α + β +γ = a−d + a+ a+d =
(a − d ) a (a + d ) =
So the roots are
αβ + αγ + βγ =
1
2
k
4
3
4
d =±
, 1 and
=
11
4
–1 each error
12
4
 a =1
1
2
3
2
 k = 11
Attempt to multiply in correct
order
c.a.o.
M1
A1
Valid attempt to use sum of roots
a = 1 , c.a.o.
M1
Valid attempt to use product of
roots
A1
All three roots
M1
Valid attempt to use
αβ + αγ + βγ , or to multiply out
factors, or to substitute a root
A1
k = 11 c.a.o.
[6]
19
4755
Mark Scheme
4
4
MM
−1
=
5
=
1
k 
0
0
1   −3



−6 1 1 −35 −15 10 


k 

  17
8
−4 
5
2
5


.
0 0

5 0 k =5

0 5 
0
1
−2
1
 x
 −3 −2 1   9 
  1
 
 y  = 5  −35 −15 10   32 


 
8
−4  81 
 17
z
 −10 
 9 
1




15
−35 −15 10 32 =






5
 81  5  85 
17
8
−
4

 


 −3
−2
1
1
5
n
 ( r + 2 )( r − 3 ) =  ( r
r =1
=
=
6
6
1
6
1
6
M1
Attempt to multiply matrices
A1
Correct
n ( n + 1) − 6 n
A2
–1 each error
n [ ( n + 1)( 2n + 1) − 3 ( n + 1) − 36 ]
M1
Valid attempt to factorise (with n as
a factor)
Correct expression c.a.o.
Complete, convincing argument
n ( n + 1)( 2n + 1) −
(
2
)
n 2 n − 38 =
1
3
1
2
(
2
n n − 19
n ( n + 1)( n + 2 )
3
2 + 6 + ..... + k ( k + 1) =
)
A1
A1
[6]
=2,
B1
k ( k + 1)( k + 2 )
3
E1
Assume true for k
M1
Add ( k + 1) th term to both sides
A1
c.a.o. with correct simplification
E1
Dependent on A1 and previous E1
k ( k + 1)( k + 2 )
 2 + 6 + ..... + ( k + 1) ( k + 2 )
=
Attempt to pre-multiply by M −1
Separate into 3 sums
so true for n = 1
Assume true for n = k
=
M1
M1
r =1
When n = 1,
=
c.a.o.
A1
[2]
n
2
1
Attempt to consider MM −1 or
−1
M M (may be implied)
)
 r −  r − 6n
r =1
=
−r −6
r =1
n
=
2
M1
All 3 correct
A1
[4] s.c. B1 if matrices not used
 x = −2, y = 3, z = 17
n
January 2010
3
+ (k + 1) ( k + 2 )
1
( k + 1)( k + 2 )( k + 3)
3
( k + 1) ( ( k + 1) + 1) ( ( k + 1) + 2 )
3
But this is the given result with k + 1 replacing
k. Therefore if it is true for n = k it is true for
n = k + 1.
Since it is true for n = 1, it is true for n = 1, 2, 3
and so true for all positive integers.
20
E1
Dependent on B1 and previous E1
[6]
4755
7
(i)
Mark Scheme
x=
January 2010
B1
B1
B1
3
−7
, x= ,y=0
2
2
[3]
(ii)
Large positive x, y → 0+
(e.g. consider x = 100)
Large negative x, y → 0−
(e.g. consider x = –100)
(iii)
x=−
7
y
2
B1
B1
M1
Evidence of method
[3]
x=
3
2
3
B1
Intercepts correct and labelled
B1
LH and central branches correct
B1
RH branch correct, with clear
maximum
7
x
9
[3]
5
(iv)
x<−
or
7
2
B1
3
9
<x≤
2
5
B2
[3]
21
Award B1 if only error relates to
inclusive/exclusive inequalities
4755
8(a) (i)
Mark Scheme
z − ( 2 + 6 j) = 4
January 2010
B1
B1
B1
2 + 6j seen
(expression in z) = 4
Correct equation
[3]
(ii)
z − ( 2 + 6 j) < 4 and z − ( 3 + 7 j) > 1
B1
z − ( 2 + 6 j) < 4
B1
z − ( 3 + 7 j) > 1
(allow errors in inequality signs)
Both inequalities correct
B1
[3]
Im
(b) (i)
2+ j
Re
Any straight line through 2 + j
Both correct half lines
Region between their two half
lines indicated
B1
B1
B1
[3]
(ii)
43 + 47 j − ( 2 + j) = 41 + 46 j
M1
Attempt to calculate argument, or
other valid method such as
comparison with y = x – 1
A1
Correct
E1
Justified
 46 
 = 0.843
 41 
arg ( 41 + 46 j) = arctan 
π
< 0.843 <
3π
4
4
so 43 + 47 j does fall within the region
[3]
22
4755
9
(i)
Mark Scheme
2
r
=
−
3
r +1
+
1
r+2
2 ( r + 1)( r + 2 ) − 3r ( r + 2 ) + r ( r + 1)
(ii)
2
(iv)
2
2 r + 6 r + 4 − 3r − 6 r + r + r
r ( r + 1)( r + 2 )
=
4+r
r ( r + 1)( r + 2 )
Attempt a common denominator
Convincingly shown
A1
[2]
 r ( r + 1)( r + 2 ) =   r − r + 1 + r + 2 
M1
Use of the given result (may be
implied)
 2 3 1  2 3 1 2 3 1
=  − +  +  − +  +  − +  + ..
 1 2 3  2 3 4  3 4 5
3
1  2
3
1 
 2
.. + 
− +
+
+ −

 n −1 n n +1  n n +1 n + 2 
2 3 2
1
3
1
= − + +
−
+
1 2 2 n +1 n +1 n + 2
3
2
1
as required
= −
+
2 n +1 n + 2
M1
Terms in full (at least first and
one other)
A2
At least 3 consecutive terms
correct, -1 each error
M1
Attempt to cancel, including
algebraic terms
Convincingly shown
4+r
n
r =1
(iii)
M1
r ( r + 1)( r + 2 )
2
=
January 2010
n
2
3
1 
r =1
3
2
A1
[6]
B1
[1]
100
4+r
 r ( r + 1)( r + 2 )
r = 50
49
4+r
4+r
−
r =1 r ( r + 1)( r + 2 )
r =1 r ( r + 1)( r + 2 )
100
=
1  3 2
1
3 2
= −
+
− − + 
 2 101 102   2 50 51 
= 0.0104 (3s.f.)
M1
Splitting into two parts
M1
Use of result from (ii)
c.a.o.
A1
[3]
23
GCE
Mathematics (MEI)
Advanced Subsidiary GCE 4755
Further Concepts for Advanced Mathematics (FP1)
Mark Scheme for June 2010
Oxford Cambridge and RSA Examinations
4755
Mark Scheme
Qu
Section A
1
Answer
Mark
B1
A=4
⇔ 4 x 2 − 16 x + C ≡ Ax 2 + 2 ABx + AB 2 + 2
M1
Attempt to expand RHS or other
valid method (may be implied)
A2, 1 1 mark each for B and C, c.a.o.
[4]
2x − 5 y = 9
B1
B1
[2]
3 x + 7 y = −1
2(ii)
M -1 =
1  7 5


29  −3 2 
M1
Divide by determinant
A1
c.a.o.
[2]
1  7 5  9  1  58 

  = 

29  −3 2  −1 29  −29 
 x = 2, y = −1
3
Comment
4 x 2 − 16 x + C ≡ A ( x 2 + 2 Bx + B 2 ) + 2
⇔ A = 4, B = −2, C = 18
2(i)
June 2010
Pre-multiply by their inverse
M1
A1(ft) For both
[2]
z = 1− 2 j
B1
1
1+ 2 j +1− 2 j + α =
2
3
α = −
2
−k
3
15
= − (1 − 2 j)(1 + 2 j) = −
2
2
2
M1
Valid attempt to use sum of roots, or
other valid method
A1
c.a.o.
M1
Valid attempt to use product of roots,
or other valid method
Correct equation – can be implied
A1(ft)
k = 15
A1
c.a.o.
[6]
OR
( z − (1 + 2 j) ) ( z − (1 − 2 j) ) = z
2
− 2z + 5
2 z 3 − z 2 + 4 z + k = ( z 2 − 2 z + 5 ) ( 2 z + 3)
α=
−3
2
M1
A1
Multiplying correct factors
Correct quadratic, c.a.o.
M1
Attempt to find linear factor
A1(ft)
k = 15
A1
c.a.o.
[6]
1
4755
4
Mark Scheme
w = x +1  x = w −1
June 2010
B1
Substitution. For x = w + 1 give B0
but then follow for a maximum of 3
marks
M1
M1
A3
Attempt to substitute into cubic
Attempt to expand
-1 for each error
(including omission of = 0)
x 3 − 2 x 2 − 8 x + 11 = 0, w = x − 1
 ( w − 1) − 2 ( w − 1) − 8 ( w − 1) + 11 = 0
3
2
 w3 − 5w2 − w + 16 = 0
[6]
OR
α + β +γ = 2
αβ + αγ + βγ = −8
αβγ = −11
B1
All 3 correct
M1
M1
Valid attempt to use their sum of
roots in original equation to find sum
of roots in new equation
Valid attempt to use their product of
roots in original equation to find one
of  αβ or αβγ
A3
-1 each error
(including omission of = 0)
Let the new roots be k, l and m then
k +l + m =α + β +γ +3 = 2+3= 5
kl + km + lm = (αβ + αγ + βγ ) + 2 (α + β + γ ) + 3
= −8 + 4 + 3 = −1
klm = αβγ + (αβ + αγ + βγ ) + (α + β + γ ) + 1
= −11 − 8 + 2 + 1 = −16
 w3 − 5w2 − w + 16 = 0
[6]
5
n
1
n
 1

 ( 5r − 1)( 5r + 4 ) = 5   5r − 1 − 5r + 4 
1
r =1
=
1
r =1
1 1 1 
 1 − 1 
  −  +  −  + ..... + 

5   4 9   9 14 
 5n − 1 5n + 4  
M1
Attempt to use identity – may be
implied
A1
Terms in full (at least first and last)
M1
Attempt at cancelling
A1
1 
1
 −

 4 5n + 4 
A1
factor of
1  1
1 1
1 
=  −
=
5  4 5n + 4 
1  5n + 4 − 4 
n

=
5  4 ( 5n + 4 )  4 ( 5n + 4 )
A1
[6]
2
1
5
Correct answer as a single algebraic
fraction
4755
Mark Scheme
6(i)
2
u2 =
6(ii)
June 2010
2
1+ 2
2
=
3
When n = 1,
Assume uk =
3
, u3 =
1+
2
2 ×1 − 1
2
2
=
M1 Use of inductive definition
A1 c.a.o.
[2]
2
5
3
= 2 , so true for n = 1
2
B1
Showing use of un =
E1
Assuming true for k
M1
uk +1
A1
Correct simplification
E1
Dependent on A1 and previous E1
2n − 1
2k − 1
2
 u k +1 =
2k − 1
2
1+
2k − 1
2
=
2
2k − 1
=
2k − 1 + 2 2k + 1
2k − 1
=
2
2 ( k + 1) − 1
But this is the given result with k + 1 replacing k.
Therefore if it is true for k it is also true for k + 1.
Since it is true for k = 1, it is true for all positive
integers.
Dependent on B1 and previous E1
E1
[6]
Section A Total: 36
3
4755
Mark Scheme
June 2010
Section B
7(i)
 0, − 1 


2

B1
1
( −3, 0 ) ,  , 0 
2
7(ii)
7(iii)
B1
For both
[2]

x = 3, x = 2 and y = 2
B1
B1
B1
[3]
Large positive x, y → 2+
(e.g. substitute x = 100 to give 2.15…, or
convincing algebraic argument)
M1
Must show evidence of method
A1
A0 if no valid method
B1
Correct RH branch
[3]
7(iv)
( 2 x − 1)( x + 3)
=2
( x − 3)( x − 2 )
 ( 2 x − 1)( x + 3) = 2 ( x − 3)( x − 2 )
M1
 x =1
Or other valid method to find
intersection with horizontal
asymptote
A1
B1
For x < 1
B1
For 2 < x < 3
[4]
From graph x < 1 or 2 < x < 3
4
4755
8(i)
Mark Scheme
arg α =
arg β =
π
6
π
2
June 2010
, α =2
B1
B1
Modulus of α
Argument of α (allow 30 )
, β =3
B1
Both modulus and argument of β
[3]
8(ii)
αβ =
β
α
=
=
(
)
3 + j 3j = −3 + 3 3j
3j
3+j
3 + 3 3j
4
3j
=
(
=
3
4
(
3+ j
+
3−j
)(
)
3−j
(allow 90 )
M1
A1
Use of j2 = −1
Correct
M1
.
Correct use of conjugate of
denominator
)
Denominator = 4
A1
All correct
A1
[5]
3 3j
4
8(iii)
Argand diagram with at least one
correct point
A1(ft) Correct relative positions with
[2] appropriate labelling
M1
5
4755
Mark Scheme
Qu
Section B (continued)
9(i)
Answer
June 2010
Mark
Comment
Rotation about origin
90 degrees clockwise, or equivalent
P is a rotation through 90 degrees about the
origin in a clockwise direction.
B1
B1
Q is a stretch factor 2 parallel to the x-axis
Stretch factor 2
B1
Parallel to the x-axis
B1
[4]
9(ii)
 2 0 0 1  0 2

=

 0 1   −1 0   −1 0 
QP = 
Correct order
M1
c.a.o.
A1
[2]
9(iii)
 0 2  2 1 3   0 4 2 
 −1 0  0 2 1  =  −2 −1 −3 


 

M1
A′ = ( 0, − 2 ) , B′ = ( 4, − 1) , C ′ = ( 2, − 3)
9(iv)
0
−1 
 −1
0
R =
9(v)
A1(ft) For all three points
[2]
One for each correct column
B1
B1
[2]

0
RQP = 
 −1
Pre-multiply by their QP - may be
implied
−1  0
2 1 0 

=

0  −1 0   0 −2 
M1
Multiplication of their matrices in
correct order
A1(ft)
( RQP )
−1
=
−1  −2

2  0
0

1
M1
A1
[4]
Attempt to calculate inverse of their
RQP
c.a.o.
Section B Total: 36
Total: 72
6
GCE
Mathematics (MEI)
Advanced Subsidiary GCE
Unit 4755: Further Concepts for Advanced Mathematics
Mark Scheme for January 2011
Oxford Cambridge and RSA Examinations
4755
Mark Scheme
Qu
1
Answer
Mark
January 2011
Comment
3x 3  18 x 2  Px  31  Q  x  R   S
3
Q3
B1
Q = 3 anywhere
3x 3  18 x 2  Px  31  3x 3  9 Rx 2  9 R 2 x  3R 3  S
M1
Attempt to expand and compare at least another coefficient, or other valid method
A3
One mark for each correct constant cao
R  2, P  36, S  7
[5]
2(i)
2(ii)
det M  4  3   1  0
M1
Area = 12  3  36 square units
A1
[2]
M 1 
1 3 0
det M 1 
2(iii)

12  1
M1
A1

4
1
division by their det M
cao condone decimals 3sf or better
B1
cao condone decimal 3sf or better
[3]
12
det M  det M 1  12 
oe www
1
1
12
The inverse ‘undoes’ the transformation, so the
composite of M and its inverse must leave a shape
unchanged, meaning the area scale factor of the
composite transformation must be 1 and so the
determinant is 1.
B1
Seen or implied
E1
Any valid explanation involving transformations and unchanged area
[2]
6
4755
3
Mark Scheme
  2x 1  x 
 1
2
  1
  1
  1

  4
  8
3 0
 2 
 2 
 2 
3

1
8

3
M1
A1
Using a substitution
Correct
M1
Substitute into cubic
M1
Attempting to expand cubic and quadratic
2
 3 2  3  1    2  2  1
4    1  3  0
  3  5 2  19  49  0
3
A2
LHS oe, -1 each error
A1
Correct equation
[7]
OR
     4
      8
M1
Attempt to find   
  3
A1
All correct
M1
Attempt to use root relationships to find at least two of k kl klm
A1
A1
A1
A1
[7]
Quadratic coefficient
Linear coefficient
Constant term
Let new roots be k, l, m then
k  l  m  2        3  5 
kl  km  lm  4      
C
A
klm  8  4      
4        3  19 
2        1  49 
   5  19  49  0
3
2
B
A
D
A
Correct equation
7
January 2011
4755
Mark Scheme
4
B1
B1
B1
Circle
Centre 3 + 2j
Radius = 2 or 3, consistent with their centre
B1
Both circles correct cao
B1
Correct boundaries indicated, inner excluded, outer included
(f t concentric circles)
B1
Region between concentric circles indicated as solution
SC -1 if axes incorrect
[6]
5
n
n
n
r 1
r 1
r 1
 r 2  3  4 r   3 r 2  4 r 3

3
6

1
2
n  n  1 2n  1 
4
4
n
2
 n  1 2
n  n  1   2 n  1   2 n  n  1  
 n  n  1 1  2n 2 
M1
Separate into two sums involving r 2 and r 3 , may be implied
M1
A1
Appropriate use of at least one standard result
Both terms correct
M1
Attempt to factorise using both n and n + 1
1
2
A1
Complete, convincing argument
[5]
8
January 2011
4755
6
Mark Scheme
When n = 1, 211  1  5 , so true for n = 1
B1
Assume uk  2 k 1  1
E1
Assuming true for k
M1
Using this uk to find uk 1
A1
Correct simplification
E1
Dependent on A1 and previous E1
E1
[6]
Dependent on B1 and previous E1
 uk 1  2
 2 2
k 1
k 1
1 2
k 1
1
 2k 2  1
2
 k 11
1
But this is the given result with k + 1 replacing k.
Therefore if it is true for k it is also true for k + 1.
Since it is true for n = 1, it is true for all positive
integers.
9
January 2011
4755
7(i)
Mark Scheme
 1 
 0,
 ,  5, 0 
 8 
B1
B1
[2]
7(ii)
x
5
2
,x
8
3
,y0
7(iii) Large positive x, y  0
(e.g. consider x = 100)
Large negative x, y  0
(e.g. consider x = -100)
7(iv)
One mark for each point
SC1 for x = -5, y = -1/8
One mark for each equation
B1
B1
B1
[3]
B1
B1
Evidence of a valid method
M1
[3]
B1
RH branch correct
B1
LH branch correct
[2]
7(v)
x  5
8
5
or
x
3
2
B1
cao
B1
cao
[2]
10
January 2011
4755
8(i)
Mark Scheme
  1 j
B1
There must be a second real root because complex roots occur
in conjugate pairs.
E1
January 2011
[2]
8(ii)
     1
B1
        1  1  1  j     1  j   1
8(iii)
   2
M1
cao
A1
[3]
 z  1 z  2   z  1  j    z  1  j  
B1
Correct factors from their roots
  z 2  z  2  z 2  2 z  2 
M1
Attempt to expand using all 4 factors
 z  2z  2z  z  2z  2z  2z  4z  4
4
3
2
3
2
2
 z4  z3  2z2  6z  4
 a  2, b  6, c  4
One for each of a, b and c
A3
[5]
OR
            a  2
        b  6  b  6
  c  4
8(iv)
f ( z )  z 4  z 3  2 z 2  6 z  4
Roots of f ( z )  0 are 1, 2,  1  j and  1  j
M2
A1
A1
B1
[5]
Use of root relationships attempted, M2 evidence of all 3, M1 for
evidence of 2 OR substitution to get three equations and solving
a = -2 cao
b = 6 cao
c = -4 (SC f t on their 2nd real root)
f t on their a, b, c, simplified
B1
For all four roots, cao
B1
[2]
11
4755
9(i)
Mark Scheme
 2 1 5  2a  1 3 1  5a 

 5

a
1
1
13
AB  3



 1 1 2  3  a 1 2a  3 



0
0

 4a  2  5  15  5a



0
9  a 1
0




4




0
0
1
5
a
13
a
6


0
0 
8  a


8 a
0
 0

 0
0
8  a 

 8  a  I
M1
Attempt to find AB with some justification of at least two leading
diagonal terms and any other
A1
Correct
B1
Relating correct diagonal matrix to I
[3]
9(ii)
A 1 does not exist for a = - 8
 2a  1
9(iii)
1 
3
1  5a 

A 
5
1
13

8  a 

 3  a 1 2a  3 
1
 9
A
1
3 21 

1 13 

 7 1 11 
1
  5
12 
 x
 9 3 21  55   2 
 y   1  5 1 13  9    6 
  12 

  
z
 7 1 11  26   9 
 


  
9(iv)
January 2011
There is no unique solution.
B1
M1
A1
kB, k not equal to 1
Correct A 1 as shown
[3]
B1
Correct use of their A 1
M1
x, y and z cao, -1 each error
A3
[5]
B1
[1]
12
GCE
Mathematics (MEI)
Advanced Subsidiary GCE
Unit 4755: Further Concepts for Advanced Mathematics
Mark Scheme for June 2011
Oxford Cambridge and RSA Examinations
4755
Mark Scheme
Qu
Answer
June 2011
Mark
Section A
 0 1
1(i)
1 0 


0
1

1(ii) 
1 0


B1
Comment
Accept expressions in sin and cos
B1
1(iii)
 0 1   0 1  1 0 
 1 0   1 0    0 1 


 

1(iv)
Reflection in the x axis
M1
A1ft
Ans (ii) x Ans (i) attempt evaluation
B1
[5]
zw
2(i)
w

2(ii)

3  5j
17
1  j 4  j

4  j 4  j

3
17

5
17
M1
Multiply top and bottom by -4 - j
A1
Denominator = 17
A1
Correct numerators
[3]
j
w  17
B1
1
 2.90
4
w  17  cos 2.90  jsin 2.90 
B1
Not degrees
B1
c.a.o. Accept
arg w    arctan
[3]
2(iii)
B1
B1

17, 2.90

Accept 166 degrees
Correct position
Mod w and Arg w correctly shown
[2]
3
     4  p
M1
A1
p  4
     
2
  2   2   2  2      
 16  6  2q
M1
May be implied
Attempt to use      
o.e. Correct
A1
c.a.o.
A1
[5]
q5
1
2
4755
Mark Scheme
4
5x
x 4
2
x
June 2011
M1*
 5 x  x3  4 x
Method attempted towards
factorisation to find critical values
 0  x3  x
A1
x=0
 0  x  x  1 x  1
A1
x = 1, x = -1
M1dep*
 x  1,  1  x  0
A1
A1
Valid method leading to required
intervals, graphical or algebraic
x 1
1  x  0
[6]
SC B2 No valid working seen
x 1
1  x  0
5
20
  3r  1 3r  2 
1
1
20
 1
1



3  3r  1 3r  2 
1  1
A1
Attempt to use identity – may be
implied
Correct use of 1/3 seen
1 1
A1
M1
Terms in full (at least first and last)
Attempt at cancelling
r 1


M1
r 1
1 1 1
 1 1 
....









  
3  2 5   5 8 
 59 62  
1  5
   
3  2 62  31
A1
c.a.o.
[5]
2
4755
6
Mark Scheme
When n = 1,
1
n 2  n  1  1 ,
June 2011
2
B1
4
so true for n = 1
Assume true for n = k
k
1
2
r 3  k 2  k  1
4
r 1
E1
Assume true for k
M1
Add  k  1 th term to both sides
M1
Factor of
A1
c.a.o. with correct simplification
E1
Dependent on A1 and previous E1
E1
Dependent on B1 and previous E1
and correct presentation

k 1
  r 3  k 2  k  1   k  1
1



1
4
1
4
1
4
1
3
4
r 1

2
 k  1
2
 k  1
2
[ k 2  4  k  1]
2
 k  1   k  1  1
2
4
 k  1
2
[ k 2  4k  4]
 k  1  k  2 
2
1
2
4
But this is the given result with k + 1 replacing
k. Therefore if it is true for k it is true for k + 1.
Since it is true for n = 1, it is true for n = 1, 2, 3
and so true for all positive integers.
[7]
Section A Total: 36
3
4755
Mark Scheme
June 2011
Section B
7(i)
 0, 18 
B1
8
 9, 0  ,  , 0 
3
B1
B1
[3]

7(ii)
x  2, x  2 and y  3
B1
B1
B1
[3]
7(iii)
Large positive x, y  3 from above
Large negative x, y  3 from below
B1
B1
(e.g. consider x  100 , or convincing algebraic
argument)
M1
Must show evidence of working
[3]
7(iv)
B1
B1
B1
-
[3]
4
3 branches correct
Asymptotes correct and labelled
Intercepts correct and labelled
4755
8(i)
Mark Scheme
Because a cubic can only have a maximum of
two complex roots, which must form a conjugate
pair.
June 2011
E1
[1] .
8(ii)
2  j,  1  2 j
P( z ) 

B1
B1
 z   2  j    z   2  j    z   1  2 j    z   1  2 j  
 z  2  1  z  1  4
2
2
M1
Use of factor theorem
M1
Attempt to multiply out factors
A4
-1 for each incorrect coefficient
M2
M1 for attempt to use all 4 root
relationships. M2 for all correct
a = -2
b, c, d correct -1 for each incorrect
  z 2  4 z  5  z 2  2 z  5 
 z 4  2 z 3  2 z 2  10 z  25
OR
        2  a  2
            2  b  2
        10  c  10
  25  d  25
B1
A3
 P( z )  z 4  2 z 3  2 z 2  10 z  25
-1 for P(z) not explicit, following A4
or B1A3
[8]
8(iii)
B1
z  5
B1
[2]
5
All correct with annotation on axes
or labels
4755
Mark Scheme
Qu
Answer
Mark
Section B (continued)
9(i)
 2 1
M

3 k 
9(ii)
M1
3
2
M 1 
Comment
- 1 each error
B2
[2]
M 1 does not exist for 2k  3  0
k
June 2011
May be implied
A1
1  k 1
2k  3  3 2 
1  5 1 1 
13  3 2   21
2
 
 3
 x  2, y  3
B1
Correct inverse
M1
A1ft
A1
Attempt to pre-multiply by their
inverse
Correct matrix multiplication
c.a.o.
A1ft
At least one correct
[7]
9(iii)
There are no unique solutions
B1
[1]
9(iv)
B1
B1
B1
[3]
(A) Lines intersect
(B) Lines parallel
(C) Lines coincident
Section B Total: 36
Total: 72
6
4755
Mark Scheme
Question
1
(i)
Answer
Transformation A is a reflection in the y-axis.
◦
Transformation B is a rotation through 90 clockwise about the
origin.
1
(ii)
0
1
(iii)
Reflection in the line y = x
2
(i)
2
(ii)
2
(iii)
3
Marks
B1
B1
[2]
M1
A1
1  1 0   0 1 
 1 0  0




1 1 0
z1  32  3 3


arg  z1   arctan
3 3 

3
3
z2 
2
[2]
B1
[1]
M1
A1
M1
A1
6
[4]
M1
A1
5 5 3

j
2
2
[2]
E1
[1]
M1
A1
Because z1 and z2 have the same argument


6
  7 
8
  2
3
Other roots are -5 and
Product of roots =
June 2012
Guidance
Attempt to multiply in correct order
cao
Use of Pythagoras
cao
cao
May be implied
cao
Consistent with (i)
Attempt to use sum of roots
Value of  (cao)
1
3
q 10

 q  10
3
3
Sum of products in pairs 
p
 11  p  33
3
5
M1
A1
Attempt to use product of roots
q  10 c.a.o.
M1
A1
Attempt to use sum of products of roots in pairs
p  33 cao
4755
Mark Scheme
Question
Marks
Answer
OR, for final four marks
 x  2  x  5 3x  1
 3x 3  8 x 2  33 x  10
 p  33 and q  10
4
3
 1  3 x  4   x  4
June 2012
2
M1
Express as product of factors
M1
A1
A1
[6]
M1*
Expanding
p  33 cao
q  10 cao
x4
 0  x 2  11x  28
 0   x  4  x  7 
4 x7
Guidance
M1dep*
B2
Multiply through by  x  4 
2
Factorise quadratic
One each for 4  x and x  7
OR
3
7x
0
1  0 
x4
x4
M1*
Consideration of graph sketch or table of values/signs
Obtain single fraction > 0
M1dep*
B2
One each for 4  x and x  7
4 x7
OR
3  x  4  x  7 (each side equal)
x  4 (asymptote)
Critical values at x  7 and x  4
Consideration of graph sketch or table of values/signs
4 x7
OR
Consider inequalities arising from both x  4 and x  4
Solving appropriate inequalities to their x > 7 and x  7
M1*
Identification of critical values at x  7 and x  4
M1dep*
B2
One each for 4  x and x  7
M1*
M1dep*
B2
One for each 4  x and x  7 , and no other solutions
[4]
4 x7
6
4755
5
Question
(i)
5
(ii)
Mark Scheme
Marks
M1
A1
Answer
2
r

3

2
r
  1
1
1
2



2r  1 2r  3  2r  1 2r  3  2r  1 2r  3
30
1
  2r  1 2r  3
1  1
1 


2 r 1  2r  1 2r  3 
30

r 1
1 1 1
 1 1   1 1 
        ...         

2  3 5   5 7 
 59 61   61 63  
1  1 1  10
   
2  3 63  63

6
(i)
6
(ii)
1  1
a2  3  2  6, a3  3  7  21
June 2012
Guidance
Attempt at common denominator
[2]
M1
Use of (i); do not penalise missing factor of
M1
Sufficient terms to show pattern
M1
A1
A1
[5]
B1
Cancelling terms
Factor ½ used
oe cao
1
2
cao
[1]
5 3  3
0
When n = 1,
Assume ak 
2
5  3k 1  3
2
k 1
 53  3
 a k 1  3 
2


5  3k  9
2
3
k

53  3
2

 1 , so true for n = 1

 1

5  3n1  3
B1
Showing use of an 
E1
Assuming true for n = k
M1
ak 1 , using ak and attempting to simplify
A1
Correct simplification to left hand expression.
2
5  3k  9  6
5 3
2
( k 1) 1
3
2
But this is the given result with k + 1 replacing k.
Therefore if it is true for n = k it is also true for n = k  1 .
Since it is true for n = 1, it is true for all positive integers.
E1
E1
[6]
7
May be identified with a ‘target’ expression using n  k  1
Dependent on A1 and previous E1
Dependent on B1 and previous E1
4755
7
Question
(i)
Mark Scheme
 5, 0  ,  5, 0  ,  0,

7
7
(ii)
(iii)
Marks
B1
B1
B1
[3]
B1
B1
B1
B1
[4]
M1
Answer
25 

24 
x  3 , x  4 , x  
2
3
and y  0
Some evidence of method needed e.g. substitute in ‘large’
values or argument involving signs
Large positive x, y  0
Guidance
-1 for each additional point
B1
Large negative x, y  0
7
June 2012
B1
[3]
B1*
B1dep*
B1
B1
(iv)
[4]
8
4 branches correct
Asymptotic approaches clearly shown
Vertical asymptotes correct and labelled
Intercepts correct and labelled
4755
8
Question
(i)
Mark Scheme
Answer
3 1  3j  2 1  3j  22 1  3j  40
3
2
 3  26  18j  2  8  6 j  22 1  3j  40
 (78  16  22  40)  (54  12  66) j
0
So z  1  3 j is a root
June 2012
Marks
M1
Guidance
Substitute z  1  3j into cubic
A1 A1
1  3j
A1
2
 8  6 j , 1  3j  26  18 j
3
Simplification (correct) to show that this comes to 0 and so
z  1  3j is a root
[4]
8
(ii)
All cubics have 3 roots. As the coefficients are real, the complex
conjugate is also a root. This leaves the third root, which must
therefore be real.
8
(iii)
1  3j must also be a root
Sum of roots = 
40
2 2
OR product of roots = 

3 3
3
OR
1  3j  1  3j   
  
OR
1  3j must also be a root
 2 z  10
3 z  2 z  22 z  40  ( z  2 z  10)(3 z  4)  0
4
is the real root
z
3
3
2
Attempt to use one of
  ,  ,  
Correct equation
22
3
4
is the real root
3
 z  1  3j z  1  3j  z
[1]
B1
M1
A2,1,0
40
2
OR (1  3 j )(1  3 j )  
3
3
2
Convincing explanation
22
3
OR (1  3 j )(1  3 j )  (1  3 j )  (1  3 j ) 
 
E1
2
A1
Cao
B1
M1
Use of factors
A1
A1
Correct quadratic factor
Correct linear factor (by inspection or division)
A1
Cao
[5]
9
4755
9
Question
(i)
Mark Scheme
Answer
p  7   4    1   19    1   9   0
Marks
E1
q  2  11  1   7   k   2  k 
 q  15  2k  k 2
9
(ii)
 79 0 0 


AB   0 79 0 
 0 0 79 


 4 5 11 
1 

A   19 4 7 
79 

 9 31 5 
June 2012
Guidance
AG must see correct working
M1
A1
[3]
B2
-1 each error
M1
Use of B
B1
1
79
Correct inverse
AG Correct working
1
9
(iii)
A1
[5]
M1
 x
 4 5 11  14   2 
  1 

  
 y   79  19 4 7  23    3 
 
 9 31 5  9   8 
z


  
 x  2, y  3, z  8
A1
A1
A1
[4]
10
Attempt to pre-multiply by their A 1
SC A2 for x, y, z unspecified
sSC B1 for A-1 not used or incorrectly placed.
GCE
Mathematics (MEI)
Advanced Subsidiary GCE
Unit 4755: Further Concepts for Advanced Mathematics
Mark Scheme for January 2013
Oxford Cambridge and RSA Examinations
4755
Question
1
1
2
(i)
(ii)
Mark Scheme
Answer
A is a reflection in the line y = x
B is a two way stretch, (scale) factor 2 in the x-direction and
(scale) factor 3 in the y-direction
 2 0  0 1   0 2 
BA  



 0 3  1 0   3 0 
[2]
M1
 a  bj
z a  bj


*
z
a  bj  a  bj a  bj
2
a 2  2abj  b 2
a 2  b2
2
2
2ab
 z  a b
 z 
 Re  *   2
and Im  *   2
2
2
 z  a b
 z  a b
Guidance
Stretch, with attempt at details.
Details correct.
Attempt to multiply in correct order
Multiply top and bottom by a + bj and
attempt to simplify
M1
Using j 2  1
A1
A1
cao correctly labelled
cao correctly labelled
[4]
B1
Stated, not just used.
M1
A1
Attempt to use roots in a relationship
Correct equation obtained for  .
Allow incorrect signs
M1
Attempt use of complex factors.
Allow incorrect signs (z+…)
  az  b  z 2  4 z  5  2 z 3  pz 2  22 z  15
A1
OR 2(2  11 j )  p(3  4 j )  22(2  j )  15  0
M1
Correct complex factors; one pair of
factors correctly multiplied
Substitution
correct equation
Allow an incorrect sign

3
Marks
B1
B1
B1
[3]
M1
A1
January 2013
z  2  j is also a root
  152 ,
or
     
  (2  j )(2  j )  5 used.
22
2
, with
OR  az  b  z  2  j  z  2  j   2 z  pz  22 z  15
3

2

A1
Complete valid method for then obtaining the other unknown.
3
real root = , p  11
2
M1
A1 A1
[6]
5
Root relation, obtaining linear factor,
equating real and imaginary parts
FT one value
Signs correct
4755
Question
4 (i)
Mark Scheme
Answer
x  x  2 has discriminant -7, so x  x  2  0 and when
2
e.g. x  0 , x  x  2  0 so positive for all x
2
2
January 2013
Marks
E2,1,0
Guidance
Allow complex roots found, with
Discriminant < 0 shown and sign of
x 2  x  2 or curve position discussed. discussion
E2,1,0
Completing square and minimum value
discussed
E2,1,0
Calculus, showing minimum value>0.
OR
x 2  x  2 = ( x  12 ) 2  74  74  0 for all x.
OR using y 
x2  x  2
dy
 2 x  1  0 when x 
dx
1
2
and y 
Hence y has minimum value, and y 
7
4
7
4
;
d2 y
20
dx 2
> 0 for all x.
[2]
4
(ii)
2x
x
x x2
2
 2 x  x3  x 2  2 x
 0  x 3  x 2  0  x 2  x  1
M1
M1
Valid attempt to eliminate fraction
Simplification and factors
0, 1 critical values
A1
A1
Both, no other values given.
A1
[5]
cao
x 1
 x  0 or 0  x  1 or x  1, x  0
OR
Graphical approach by sketching
y
2x
2x
and y  x or y  2
x
2
x x2
x x2
M2,1,0
Critical values 0 and 1
A1
x 1
A1
 x  0 or 0  x  1 or x  1, x  0
A1
[5]
6
Accuracy of sketch
Both
cao
Or combine to one fraction > or<0
In numerator
4755
Question
5 (i)
Mark Scheme
Answer
Marks
M1
  5  3r  2  3r   k   2  3r  5  3r 
100
100
1
r 1
1
1
January 2013
Guidance
r 1
1 1 1 1 
 k[        ...
 5 8   8 11 
1 
 1


]
 302 305 
M1
1 
1
 k 

 5 305 
20
4

 , oe
305 61
M1
Cancelling inner terms
A1
cao
Write out terms (at least first and last
terms in full)
A1
[5]
(ii)
1
15
B1
[1]
6
When n = 1,  1
0
1 2
1
2
B1
and 1  1 , so true for n = 1
Assume true for n = k
2
 1  2  3  .....   1
2
2
2
k 1
 1  2  3  .....   1
2
  1
2
k 1
2
k  k  1
  1
2

k
  k  1
k   1
k 1
k 11
2
k   1
2
 k  1
2
  k  1 
2

k
 k

  1  k  1 
 k  1
 2

  1 

k
k 1
k  k  1
E1
Assuming true result for some n .
Condone series shown incomplete
2
k 11
 k  1
2
M1*
Adding  k  1 th term to both sides.
M1
Dep*
Attempt to factorise (at least one valid
factor)
2
A1
7
Correct factorisation Accept ( 1)
provided expression correct.
k m
4755
Mark Scheme
Question
Answer
k
k 2
  1  k  1 

 2 
 n 1 n  n  1
, n  k 1
  1
2
Therefore if true for n = k it is also true for n = k  1
Since it is true for n = 1, it is true for all positive
integers, n.
Marks
January 2013
Guidance
A1
Valid simplification with (-1)k
E1
Or target seen
E1
Dependent on A1 and previous E1
Dependent on B1 and previous E1
[8]
7
(i)
Asymptotes
y = 0,
x = 5, x = 8
Crosses axes at  4, 0  ,  0,  101 
B1
B1
B1 B1
x4
 0  x  8 or 4  x  5
 x  5 x  8
B1 B1
both
[6]
7
(ii)
x4
 k  x  4  kx 2  13kx  40k


x
5
x
8



Attempt to remove fraction and simplify
M1
 kx 2  13k  1 x  40k  4  0
A1
3 term quadratic (= 0)
b 2  4ac  13k  1  4k (40k  4)
M1
Attempt to use discriminant
 9k 2  10k  1
A1
A1
Correct 3-term quadratic
Roots found or factors shown
2
Critical values –1, –1/9
For no solutions to exist, 9k 2  10k  1  0
 1  k   19
E1
No point on the graph has a y coordinate in the range
E1
 1  y  
1
9
[7]
8
Accept equivalent statement
4755
Mark Scheme
Question
8 (i)
Answer
Marks
January 2013
Guidance
Im
The circle should be reasonably
circular.
A
10
The radius should be shown to be 10
by annotation as in the diagram or by
other positions marked.
 8, 15 
The centre point should be indicated
and correct.
B
p
The region should be shown by a key
or by description. Accept a “dotty”
outline to a shaded interior.
Re
The set of points for which z   8  15j  10 is all points
B4
inside the circle, radius 10, centre  8, 15  , excluding the
Circle, B1; radius 10, B1;
centre  8, 15  , B1;all points inside but
not on circumference of the correctly
placed circle, B1
points on the circumference.
[4]
9
Correctly placed: the “circle” must lie
above the Re axis and intersect the Im
axis twice as in the diagram.
4755
Question
8 (ii)
Mark Scheme
Answer
Origin to centre of circle =
 8
2
 152  17 .
Origin to centre of the circle  10
Point A is the point on the circle furthest from the origin. Since
the radius of the circle is 10, OA = 27. Point B is the point on
the circle closest to the origin. Since the radius of the circle is
10, OB=7. Hence for z in the circle
January 2013
Marks
M1
Guidance
Allow centre at 8  15 j and FT
M1
E1
Use of radius of circle
Correct explanation for both
[3]
B1
Correctly positioned on circle
B1
Accept
M1
Attempt to calculate the correct angle.
A1
cao Accept 154
[4]
M1
Any valid method soi
7  z  27
8
(iii)
P is the point where a line from the origin is a tangent to the
circle giving the greatest
argument  ,      
p  17 2  102  189  13.7 (3 s.f.)
arg p 

2
 arcsin
 2.69 (3 s.f.)
9
(i)
(ii)
17
 arcsin
10
1
15
 x
 4 2 3   14   1 
1
 
  

 y    15  5 4 0   25    2 


 
  
 1 1 2   3   3 
z
x = –1, y = 2, z = –3
9
(iii)
189 or 3 21 or 13.7
Correct circle only
17
8  4    7  5  12  1  15
k 
9
8
Allow circles centred as in (ii)
1 a    8  4    21 2   0  a  10
 7  5    5  1  15  b   0  b  2
A1
[2]
B1
No working or wrong working
SC B1
1
Use of A in correct position(s)
M1
Attempt to multiply matrices to obtain
column vector
A2
[4]
M1
–1 each error
A1
[2]
10
Attempt to multiply BB 1 matrices to
find a or b soi
For both
Condone missing k
4755
Question
9 (iv)
Mark Scheme
Answer
 AB 
1
 B 1 A 1
1 0
1
  4 3
3
2 1
 9
1 
   30
45 
 15
 4 2 3
1


1   5 4 0
 15


2 
 1 1 2 
3 13 

21 10 
6
10 
5
Marks
B1
January 2013
Guidance
By notation or explicitly
M1
Attempt to multiply in correct sequence,
may be implied by the answer (at least 7
elements correct)
A2
–1 each error FT their value of b.
[4]
11
Must include k
4755
Question
1
Mark Scheme
Answer
2x  x  5 
 x  2 Ax
2
2
Marks
 Bx  C
 D
Comparing coefficients of x , A  2
3
B1
Comparing coefficients of x , B  2 A  0  B  4
Comparing coefficients of x, C  2 B   1 0  C   2
Comparing constants, D  2 C  0  D   4
B1
2
2
M1
3
z 
2
is a root   2 z  3  is a factor.
June 2013
Guidance
Evidence of comparing coefficients, or multiplying out
the RHS, or substituting. May be implied by A = 2 or
D=-4
B1
B1
[5]
Unidentified, max 4 marks.
M1
Use of factor theorem, accept 2 z  3 , z 
M1
Attempt to factorise cubic to linear x quadratic
3
2

2z
 3  z  bz  c    2 z  9 z  2 z  30 
2
3
2
Other roots when z 2  6 z  1 0  0
M1
A1
6 
z 
36  40
M1
2
 3  j or  3  j
3
A1
9 3
3
3
,    1 5, o r
      1
2 2
2
2
Compare coefficients to find quadratic (or other valid
complete method leading to a quadratic)
Correct quadratic
Use of quadratic formula (or other valid method) in
their quadratic
oe for both complex roots FT their 3-term quadratic
provided roots are complex.
M1
Two root relations (may use  )
    6,    10
M1
leading to sum and product of unknown roots
z  6 z  10  0 ,
M1
A1
and quadratic equation
M1
Use of quadratic formula (or other valid method) in
their quadratic
oe For both complex roots FT their 3-term quadratic
provided roots are complex.
OR
    
2
2
z 
6 
36  40
2
 3  j or  3  j
or roots must be complex, so a  b j , 2 a   6 , 9  b 2  1 0
z  3  j, z  3  j
A1
M1
A1
[6]
which is correct
SCM0B1 if conjugates not justified
4755
Question
3 (i)
Mark Scheme
Answer
2  4 p  0
Marks
M1
1
 p  
June 2013
Guidance
Any valid row x column leading to p
B1
2
[2]
3
(ii)
 x
 
y  N
 
 z
 
 1

 2

 7
 2
1





39 

5

2 2 
M1
Attempt to use N
1
2   39 
0

5


6   22
1
3
1
2
 5 


 7


 2 






M1
Attempt to multiply matrices (implied by 3x1 result)
A1
One element correct
A1
All 3 correct. FT their p
[4]
4
(i)

 

z2  5  cos
 j s in

4
4 


5
2
2

5
2
j
M1
May be implied
A1
oe (exact numerical form)
2
[2]
Correct solution by means
of simultaneous equations
can earn full marks.
M1 elimination of one
unknown,
M1 solution for one
unknown
A1 one correct,
A1 all correct
4755
Mark Scheme
Question
4 (ii)
Answer
z1  z 2  3 
z1  z 2  3 
5
2
2
5
2
2
Marks

5 2
  2 

2


 j  6 .5 4  1 .5 4 j


5 2
  2 

2


 j   0 .5 4  5 .5 4 j

June 2013
Guidance
M1
Attempt to add and subtract z 1 and their z 2 - may be
implied by Argand diagram
B3
For points cao, -1 each error – dotted lines not needed.
z2
z1  z 2
z1
z1  z 2
[4]
5
n
1

r 1



4r
 3   4 r  1

n
1

4
r 1
1 
 1



4r  1
 4r  3
1  1 1   1
1
1
1




         ...  
4  1 5   5
9
4n  1 
 4n  3
1 
1

1


4 
4n  1
1  4n  1  1
n



4  4n  1 
4n  1
M1
For splitting summation into two. Allow missing 1/4
M1
Write out terms (at least first and last terms in full)
A1
Allow missing 1/4
M1
Cancelling inner terms; SC insufficient working shown
above,M1M0M1A1 (allow missing 1/4 )
A1
Inclusion of 1/4 justified
A1
Honestly obtained (AG)
[6]
4755
Question
6
Mark Scheme
Answer
w 
x
Marks
 1  3  w  1  x
3
June 2013
Guidance
M1
x  5x  3x  6  0
3
2

3 w
 1   5  3  w  1   3  3  w  1   6  0
3
2
M1
Substituting
A1
Correct
 27 w  126 w  180 w  87  0
A3
 9w  42w  60 w  29  0
A1
FT x  3 w  3, 3 w  1 , -1 each error
cao

3
 27 w  3w
2


 3w  1  45 w
3
2

 2 w  1  9 w  15  0
2
3
2
OR
In original equation

 5 ,     3,     6
M1A1
all correct for A1
New roots  ,  , 

 
 

 3,
3

27


 
9
Fully correct equation
 
 
9


3
1

2
3

 3
M1
A3
At least two relations attempted
Correct -1 each error FT their 5,3,6
A1
[7]
Cao, accept rational coefficients here
4755
Question
7 (i)
Mark Scheme
Answer
Vertical asymptotes at
x  2
and
x 
Marks
1
occur when
M1
2
 b x  1  x  a  
3
so when x gets very large,
2
7
2
 1  x  2 
Some evidence of valid reasoning – may be implied
A1 A1
Horizontal asymptote at y 
2x
Guidance
0
 a  2 and b  2
cx
June 2013

3
A1
 c  3
2
[4]
(ii)
Valid reasoning seen
Large positive x, y 
3
Some evidence of method needed e.g. substitute in
‘large’ values with result
A1
Both approaches correct (correct b,c)
B1
B1
LH branch correct
RH branch correct
Each one carefully drawn.
fro m b e lo w
2
Large negative x, y 
M1
3
fro m a b o v e
2
x  2
x 
1
2
y 
3
2
[4]
4755
Mark Scheme
Question
7 (iii)
Answer
3x
2x
Marks
Guidance
2
 1  3x
 1  x  2 
 0 
June 2013
x
2
  2 x  1  x  2 
M1
 2   x  1
Or other valid method, to values of x (allow valid solution of
inequality)
Explicit values of x
 x  1 or x  2
A1
(1, 1)
From the graph
for  2  x 
1
3x
2x
 1  x  2 
or 1  x  2
1
B1 B1
2
[4]
FT their x=1,2 provided >1/2.
(2, 1)
y 1
4755
Question
8 (i)
Mark Scheme
Answer
n
n
  r  r  1   1  
r 1
1

2


1
Guidance
r  n
M1
Split into separate sums
n  n  1  n
M1
Use of at least one standard result (ignore 3rd term)
A1
Correct
M1
Attempt to factorise. If more than two errors, M0
A1
Correct with factor
r 1
1
2
n[ n  1   2 n  1   3  n  1   6 ]
6

r
n  n  1  2 n  1 
6
Marks
n
r 1
1

June 2013
n[ 2 n  8 ]
2
6
1

n[ n  4 ]
2
3
1

3
1
n oe
3
n n  2n  2
Answer given
[5]
8
(ii)
When n = 1,
n
  r  r  1   1  1  0   1 
1
r 1
1
and
n n  2n  2 
1
1 3  1  1
3
3
So true for n = 1
Assume true for n = k
k
1
r 1
3
  r  r  1   1 
k 1

1
k  k
3
2

3

1
3
4
1
3
 3k
2
k  k  2   k  2    k  1 k  1
3
k  k 1
3
k
Or “if true for n=k, then…”
k k  2k  2
  r  r  1   1 
r 1

B1
E1
 k  3
M1*
Add (k + 1)th term to both sides
4755
Mark Scheme
Question
Answer

1
3

1
3

1
3
k
 1  k
k
 1  k  3   k  1
k
 1   k  1  2    k  1  2 
2
 2k  3
But this is the given result with n = k + 1 replacing n = k.
Therefore if the result is true for n = k, it is also true for n =
k+1.
Since it is true for n = 1, it is true for all positive integers, n.
9
(i)
Q represents a rotation
90 degrees clockwise about the origin
Marks
M1dep
*
June 2013
Guidance
Attempt to factorise a cubic with 4 terms
A1
Or = 13 n ( n  2 )( n  2 ) where n  k  1 ;
or target seen
E1
Depends on A1 and first E1
E1
[7]
B1
B1
Depends on B1 and second E1
Angle, direction and centre
[2]
9
(ii)
0

0
P 
 2 
   

1 2
 2 
1   x 
 2,
2
M1
A1
Allow both marks for P(-2, 2) www
[2]
9
(iii)
0

0
1  x    y 
   

1  y   y 
l is th e lin e y   x
9
(iv)
0

0
1   x 

  y   6 
  

 y   6 
  
1  y
n is th e lin e y  6
Or use of a minimum of two points
M1
A1
[2]
Allow both marks for y   x www
Use of a general point or two different points leading to
M1
 6 


 6 
B1
[2]
y=6; if seen alone M1B1
4755
Mark Scheme
Question
9 (v)
Answer
d e t M  0  M is singular (or ‘no inverse’).
The transformation is many to one.
9
 0
R  QM  
 1
(vi)
0

0
10

00
1  0
  
1  0
1

1
Marks
B1
E1
[2]
M1
Attempt to multiply in correct order
Or argue by rotation of the line y   x
1 x   y 
    
1 y   y 
q is th e lin e y  x
Guidance
www
Accept area collapses to 0, or other equivalent statements
A1
[2]
y  x SC B1 following M0
June 2013