OXFORD CAMBRIDGE AND RSA EXAMINATIONS
Advanced Subsidiary General Certificate of EducationAdvanced General Certificate of Education
MEI STRUCTURED MATHEMATICS 4755Further Concepts For Advanced Mathematics (FP1)
Wednesday 18 JANUARY 2006 Afternoon 1 hour 30 minutes
Additional materials:8 page answer bookletGraph paperMEI Examination Formulae and Tables (MF2)
TIME 1 hour 30 minutes
INSTRUCTIONS TO CANDIDATES
• Write your name, centre number and candidate number in the spaces provided on the answerbooklet.
• Answer all the questions.
• You are permitted to use a graphical calculator in this paper.
• Final answers should be given to a degree of accuracy appropriate to the context.
INFORMATION FOR CANDIDATES
• The number of marks is given in brackets [ ] at the end of each question or part question.
• You are advised that an answer may receive no marks unless you show sufficient detail of theworking to indicate that a correct method is being used.
• The total number of marks for this paper is 72.
HN/3© OCR 2006 [D/102/2663] Registered Charity 1066969 [Turn over
This question paper consists of 4 printed pages.
Section A (36 marks)
1 You are given that
(i) Calculate, where possible, 2B, CA and [5]
(ii) Show that matrix multiplication is not commutative. [2]
2 (i) Given that express and in terms of a and b. [2]
(ii) Prove that [3]
3 Find expressing your answer in a fully factorised form. [6]
4 The matrix equation represents two simultaneous linear equations in x and y.
(i) Write down the two equations. [2]
(ii) Evaluate the determinant of
What does this value tell you about the solution of the equations in part (i)? [3]
5 The cubic equation has roots
(i) Write down the values of and [2]
(ii) Find the cubic equation with roots simplifying your answer as far as possible.[4]
6 Prove by induction that [7]Sn
r�1
1r(r � 1) �
n
n � 1.
2a, 2b and 2g ,
abg .ab � bg � gaa � b � g ,
a, b and g .x3 � 3x 2 � 7x � 1 � 0
6 23 1
--
ÊË
ˆ¯ .
6 23 1
--
ÊË
ˆ¯
ÊË
ˆ¯ = Ê
ˈ¯
xy
ab
Sn
r�1 (r � 1) (r � 1),
zz* � � z �2 � 0.
z*� z �z � a � bj,
A � B .A � C,
C =-Ê
ËÁÁ
ˆ
¯˜˜
1 10 20 1
.B =-Ê
ˈ¯
2 31 4
,A = ÊË
ˆ¯
4 31 2
,
2
4755 January 2006
Section B (36 marks)
7 A curve has equation .
(i) Show that y can never be zero. [1]
(ii) Write down the equations of the two vertical asymptotes and the one horizontal asymptote.[3]
(iii) Describe the behaviour of the curve for large positive and large negative values of x, justifyingyour description. [2]
(iv) Sketch the curve. [3]
(v) Solve the inequality [4]
8 You are given that the complex number satisfies the equation ,where p and q are real constants.
(i) Find and in the form Hence show that and [6]
(ii) Find the other two roots of the equation. [3]
(iii) Represent the three roots on an Argand diagram. [2]
q � 10.p � �8a � bj .a 3a 2
z3 � 3z2 � pz � q � 0a � 1 � j
3 � x2
4 � x2 � –2.
y �3 � x2
4 � x2
3
4755 January 2006 [Turn over
9 A transformation T acts on all points in the plane. The image of a general point P is denoted by
always lies on the line and has the same y-coordinate as P. This is illustrated in Fig. 9.
Fig. 9
(i) Write down the image of the point under transformation T. [1]
(ii) P has coordinates State the coordinates of [2]
(iii) All points on a particular line l are mapped onto the point Write down the equation ofthe line l. [1]
(iv) In part (iii), the whole of the line l was mapped by T onto a single point. There are an infinitenumber of lines which have this property under T. Describe these lines. [1]
(v) For a different set of lines, the transformation T has the same effect as translation parallel tothe x-axis. Describe this set of lines. [1]
(vi) Find the matrix which represents the transformation. [3]
(vii) Show that this matrix is singular. Relate this result to the transformation. [3]
2 � 2
(3, 6).
P�.(x, y).
(10, 50)
x
y
¢P1
P2
P1
¢P2
y � 2xP�
P�.
4
4755 January 2006
Section A
1(i)
1(ii)
4 62
2 8−⎛ ⎞
= ⎜ ⎟⎝ ⎠
B , A + C is impossible,
3 12 41 2
⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠
CA , A - B = 2 60 2⎛ ⎞⎜ ⎟−⎝ ⎠
4 3 2 3 11 01 2 1 4 4 5
2 3 4 3 5 01 4 1 2 8 11
−⎛ ⎞⎛ ⎞ ⎛ ⎞= =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠
−⎛ ⎞⎛ ⎞ ⎛ ⎞= =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠
≠
AB
BA
AB BA
B1 B1
M1, A1 B1
[5]
M1
E1
[2]
CA 3 2× matrix M1 Or AC impossible, or student’s own correct example. Allow M1 even if slip in multiplication Meaning of commutative
2(i)
2(ii)
( )2 2z a b= + , * jz a b= −
( )( )
( )
* 2 2
2* 2 2 2 2 0
zz a bj a bj a b
zz z a b a b
= + − = +
⇒ − = + − + =
B1 B1
[2]
M1
M1 A1
[3]
Serious attempt to find *zz , consistent with their *z ft their z in subtraction All correct
3 ( )( ) ( )
( )( )
( )( )[ ]
( )
( )( )
1
2
1
2
1 1
11 2 1
61
1 2 1 66
1
12 3 5
61
2 5 16
n
r
n
r
r r
n n n n
n n n
r
n n n
n n n
= =
+ −
= + + −
= + + −
= −
= + −
= + −
∑ ∑
M1
M1,
A1, A1
M1
A1 [6]
Condone missing brackets Attempt to use standard results Each part correct Attempt to collect terms with common denominator c.a.o.
4(i)
4(ii)
6 23x y a
x y b− =
− + =
Determinant = 0 The equations have no solutions or infinitely many solutions.
B1 B1
[2]
B1
E1 E1
[3]
No solution or infinitely many solutions Give E2 for ‘no unique solution’ s.c. 1: Determinant = 12, allow ‘unique solution’ B0 E1 E0
s.c. 2: Determinant = 10
give
maximum of B0 E1 5(i)
5(ii)
3α β γ+ + = − , 7αβ βγ γα+ + = − , 1αβγ = − Coefficients A, B and C
2 2 2 2 3 6
2 2 2 2 2 2 4 7 28
2 2 2 8 1 8
BA
CA
DA
α β γ
α β β γ γ α
α β γ
−+ + = × − = − =
× + × + × = × − = − =
−× × = × − = − =
3 26 28 8 0x x x⇒ + − + = OR
3 2
3 2
3 2
22
3 7 1 02 2 2
3 7 1 08 4 2
6 28 8 0
x x ωω
ω ω ω
ω ω ω
ω ω ω
= ⇒ =
⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ − + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⇒ + − + =
⇒ + − + =
B2 [2]
M1
A3 [4]
M1 A1
A1
A1 [4]
Minus 1 each error to minimum of 0 Attempt to use sums and products of roots ft their coefficients, minus one each error (including ‘= 0’ missing), to minimum of 0 Attempt at substitution Correct substitution Substitute into cubic (ft) c.a.o.
6
( )1
11 1
n
r
nr r n=
=+ +
∑
n = 1, LHS = RHS = 12
Assume true for n = k
Next term is ( )( )
11 2k k+ +
Add to both sides
( )( )( )
( )( )
( )( )( )
( )( )
2
2
1RHS1 1 2
2 11 2
2 11 2
11 2
12
kk k k
k kk k
k kk k
kk k
kk
= ++ + +
+ +=
+ +
+ +=
+ +
+=
+ +
+=
+
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
B1
E1
B1
M1
A1
E1
E1
[7]
Assuming true for k (must be explicit) (k + 1)th term seen c.a.o.
Add to 1
kk +
(ft)
c.a.o. with correct working True for k, therefore true for k + 1
(dependent on 12
kk++
seen)
Complee argument
Section A Total: 36
7(i)
7(ii)
7(iii)
7(iv)
7(v)
Section B 23 0x+ ≠ for any real x.
y = -1, 2, 2x x= = − Large positive x, 1y −→ − (e.g. consider 100x = ) Large negative x, 1y −→ − (e.g. consider 100x = − ) Curve 3 branches correct Asymptotes labelled Intercept labelled
2
2 22
2
3 2 3 8 24
11
( ) 11
x x xx
x
x
+= − ⇒ + = − +
−⇒ =
⇒ = ±
From graph, 11 2 or 2 11x x− ≤ < − < ≤
E1
[1]
B1, B1
B1 [3]
M1 B1
[2]
B1 B1
B1 [3]
M1
A1
B1 A1
[4]
Evidence of method required From below on each side c.a.o. Consistent with (i) and their (ii), (iii) Consistent with (i) and their (ii), (iii) Labels may be on axes Lose 1 mark if graph not symmetrical May be written in script Reasonable attempt to solve Accept 11
2x < − and 2 x< seen c.a.o.
8(i)
8(ii)
8(iii)
( )( )
22
3
1 j 2 j
1 j 2 j 2 2 j
α
α
= + =
= + = − +
( )( )
3 23 02 j 2 3 2 j 1 j 08 2 0
8 and 2 0 10
z z pz qp q
p j p qp p q q
+ + + =⇒ − + × + + + =
⇒ + + + − =
= − + − = ⇒ =
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
A1
[6]
B1 M1 A1
[3]
B2
[2]
Substitute their 2α and 3α into cubic Equate real and imaginary parts to 0 Results obtained correctly Any valid method c.a.o. Argand diagram with all three roots clearly shown; minus 1 for each error to minimum of 0 ft their real root
Section B (continued)
9(i)
9(ii) 9(iii)
9(iv)
9(v)
9(vi)
9(vii)
( )25,50
1 ,2
y y⎛ ⎞⎜ ⎟⎝ ⎠
6y = All such lines are parallel to the x-axis. All such lines are parallel to 2y x= .
102
0 1
⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠
10 1det 0 1 0 02
20 1
⎛ ⎞⎜ ⎟ = × − × =⎜ ⎟⎜ ⎟⎝ ⎠
Transformation many to one.
B1 [1]
B1, B1
[2]
B1
[1]
B1 [1]
B1
[1]
B3 [3]
M1
E2 [3]
Or equivalent 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 May be awarded without previous M1 Allow E1 for ‘transformation has no inverse’ or other partial explanation
Section B Total: 36 Total: 72