J12/5/MATME/SP1/ENG/TZ0/XX
22145011
MATHEMATICS
STANDARD LEVEL
PAPER 1
NAME: ________________________
CLASS: ________________________
Friday 12th June 2015 (morning)
1 hour
INSTRUCTIONS TO CANDIDATES


Do not open this examination paper until instructed to do so.
You are not permitted access to any calculator for this paper.


Answer all questions in the space provided.
Unless otherwise stated in the question, all numerical answer should be given exactly or
correct to three significant figures.
The maximum mark for this examination paper is [60 marks].
There are two sections in this paper which consists of 13 pages.


J12/5/MATME/SP1/ENG/TZ0/XX
Full marks are not necessarily awarded for a correct answer with no working. Answers musbe
supported by working and/or explanations. Where an answer is incorrect, some marks may be given
for a correct method, provided this is shown by written working. You are therefore advised to show
all working.
SECTION A
1.
[Maximum mark: 5]
The sets P, Q and U are defined as
U = {Real Numbers} , P = {Positive Numbers} and Q = {Rational Numbers}.
Put the following numbers in the correct region on the Venn diagram.
6
, 3 × 10–4 , cos (30) , - 16 , –
11
U
P
Q
[5 marks]
P.2
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2.
[Maximum mark: 6]
Three points L(4, 3), M(14, 3) and N(14, y) are given, where y  0. If LM = 2NM, find
(a)
the value of y, and
[2 marks]
(b)
the equation of the perpendicular bisector of LN in the form of ax + by = c, where a, b
and c are constants.
[4 marks]
P.3
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3.
[Maximum mark: 6]
There are 3 black carp and 7 white carp in a pond. The carp are caught one by one at
random without replacement from the pond. Find the probabilities of the following events
happening.
(a)
Two black carp are caught.
[2 marks]
(b)
Two different colours carp are caught.
[2 marks]
(c)
Two white carp are caught given that at least one of the carp is white.
[2 marks]
P.4
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4.
[Maximum mark: 6]
It is given that the equation x 2 - 2kx + 3 = 2k has 1 repeated real root, find the possible
values of k.
[6 marks]
P.5
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5.
[Maximum mark: 7]
Part of the graph of f (x) = -x 2 + bx + c is shown below.
y
x
(a)
Write down the value of c.
(b)
The graph of y = f (x) has an x-intercept = -2 , find the value of b.
(c)
Find the vertex of the graph of y = f (x).
[1 mark]
[2 marks]
[4 marks]
P.6
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P.7
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SECTION B
6.
[Maximum mark: 15]
Let f(t) = a cos b(t – c) + d, t  0. Part of the graph of y = f(t) is given below.
When t = 3, there is a maximum value of 29, at M.
When t = 9, there is a minimum value of 15.
(a)
(i)
Find the value of a.
(ii)
Show that b =

6
.
(iii) Find the value of d.
(iv) Write down a value for c.
The transformation P is given by a horizontal stretch of a scale of
[5 marks]
1
, followed by a
2
 3 

transformation of 
 −10 
(b) Let M’ be the image of M under P. Find the coordinates of M’.
[2 marks]
The graph of g is the image of the graph of f under P.
(c) Find g(t) in the form g(t) = 7 cos B(t – C) + D.
(d) Solve g(t) =
7 3 + 24
on 0  t  10.
2
[4 marks]
[4 marks]
P.8
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P.9
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P.10
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7.
[Maximum mark: 15]
The sides of a square are 8 cm in length. The midpoints of the sides of this square are joined to
form a new square and four triangles (diagram 1), the process is repeated twice, as shown in
diagrams 2 and 3.
8 cm
x2
x1
A1
Diagram 1
Diagram 2
Diagram 3
Let xn denote the length of the equal sides of each new triangle.
Let An denote the area of each new triangle.
(a) The following table gives the values of xn and An for 1 £ n £ 4 , copy and complete the
table. (construct another table on the space provided. )
n
1
xn
An
4 cm
2
3
4
2 cm
2 cm2
1 cm2
[4 marks]
(b) The process described above is repeated, find log2 A1 + log 2 A2 + ×××+ log 2 A19 .
[5 marks]
(c) Consider an initial square of side length k cm. The process described above is repeated
indefinitely. If the total area of all shaded regions is k cm2, find the value of k.
[6 marks]
P.11
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P.12
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P.13
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Solutions:
1.
U
Q
P
6
11
cos (30)
3 × 10–4
- 16
–
A1 A1 A1 A1 A1
2.
(a)
LM = 2NM
(14 – 4) = 2(y – 3)
10 = 2(y – 3)
5=y–3
y=8
(b)
M1
A1
æ 4 + 14 3 + 8 ö
Coordinates of the mid-point of LN = ç
,
÷
è 2
2 ø
= (9, 5.5)
y - 5.5 8 - 3
´
= -1
x - 9 14 - 4
y - 5.5 1
´ = -1
x-9 2
y – 5.5 = –2(x – 9)
2x + y = 23.5
\The required equation is 2x + y = 23.5
3.
(a)
The required probability =
=
A1
M1A1
A1
3 2

10 9
M1
1
15
A1
P.14
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(b)
The required probability =
=
(c)
5.
M1
7
15
A1
7 6
´
10
9
The required probability =
1
115
=
4.
3 7 7 3
 + 
10 9 10 9
M1
1
2
A1
x 2 - 2kx + 3- 2k = 0 has 1 repeated real root.
D=0
2
(-2k) - 4(3- 2k) = 0
4k 2 + 8k -12 = 0
4(k -1)(k + 3) = 0
k = 1 or – 3
M1
M1 A1
M1
A1 A1
(a)
c=8
A1
(b)
Sub (-2,0) into the graph y = -x 2 + bx + 8,
0 = -(-2)2 + b(-2) + 8
M1
b=2
A1
(c)
The equation of axis of symmetry is: x =
-2
2(-1)
x =1
M1
A1
When x = 1,
y = -(1)2 + 2(1) + 8
M1
The vertex = (1,9)
A1
y=9
6.
(a)
(i)
a=
29 − 15
2
=7
A1
P.15
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Period = (9 – 3) × 2
(ii)
M1
= 12
2
12
b=
=

6
d=
(iii)
A1
29 + 15
2
= 22
(b)
A1
(iv) c = 3 (accept c = 3 ± 12k , k  Z )
A1
1
The x-coordinate of M’ = 3  + 3
2
M1
= 4.5
The y-coordinate of M’ = 29 – 10
= 19
\The coordinates of M’ are (4.5, 19).
(c)
B=
=

6

1
2

3
A1
C = 4.5 (accept c = 4.5 ± 6k , k  Z )
D = 22 – 10
= 12
3
g(t) =
7 3 + 24
2
(t − 4.5) + 12 =
7 3
+ 12
2
(d)

3
A1
A1

\g(t) = 7 cos (t − 4.5) + 12
7 cos
A1
7 cos

3
(t − 4.5) =
A1
M1
7 3
2
P.16
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cos
\

3

3
(t − 4.5) =
(t − 4.5) =
t − 4 .5 =

3
2

or
6
3

1
2
3
(t − 4.5) =
M1
6
11
6
11
2
A1
t = 10
A1
t − 4.5 =
t=5
7.

(t − 4.5) = 2 −
(a)
N
1
2
3
xn
An
4 cm
2 2 cm
4 cm2
2 cm
8 cm2
2 cm2
4
2 cm
1 cm2
A1 A1 A1 A1
(b)
log2 A1 + log 2 A2 + ×××+ log2 A19 = 3+ 2 +1+ 0 + ...+ (-15)
=
1
[ 2(3) + (19 -1)(-1)] ´19
2
= -114
(c)
1 æ kö
1 term of the geometric sequence = = ´ ç ÷
2 è 2ø
1
Common ratio =
2
st
1 æ kö
´ç ÷
2 è 2ø
Sum of all shaded region =
1
12
1 æ kö
´ç ÷
2 è 2ø
k=
1
12
k=4
M1 A1
M1 A1
A1
2
A1
A1
2
M1 A1
2
M1
A1
P.17