Pure Maths Unit 1 Paper 1 2013
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Pure Math 2013 Paper 1 Past Paper...
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l:EST CODE
FORM TP 201323
02134010
MAY/JUNE 2013
CARIBBEAN
EXAMINATIONS
COUNCIL
CARIBBEAN ADVANCED PROFICIENCY EXAMINATION® PURE MATHEMATICS ALGEBRA, GEOMETRY AND CALCULUS Unit 1- Paper 01
1 hour 30 minutes ( 12 JUNE 2013 (p.m.)) READ THE FOLLOWING INSTRUCTIONS CAREFULLY. 1.
This test consists of 45 items. You will have 1 hour and 30 minutes to answer them.
2.
In addition to this test booklet, you should have an answer sheet.
3.
Do not be concerned that the answer sheet provides spaces for more answers than there are items in this test.
4.
Each item in this test has four suggested answers lettered (A), (B), (C), (D). Read each item you are about to answer and decide which choice is best.
5.
On your answer sheet, find the number which corresponds to your item and shade the space having the same letter as the answer you have chosen. Look at the sample item below. Sample Item The expression (1 + (A) (B)
4 10
(C)
1+3.J3 4 + 2.J3
(D)
.J3 ) is equivalent to 2
The best answer to this item is "4 + 2
---
i iiiiii
---==
Sample Answer
.J3 ",so answer space (D) has been shaded.
6.
If you want to change your answer, be sure to erase it completely before you fill in your new choice.
7.
When you are told to begin, turn the page and work as quickly and as carefully as you can. If you cannot answer an item, omit it and go on to the next one. You can return later to the item omitted. Your score will be the total number of correct answers.
8.
You may do any rough work in this booklet.
9.
The use of silent, non-programmable scientific calculators is allowed.
iiiii
i iiiiii
Examination Materials: A list of mathematical formulae and tables. (Revised 2012)
! !!!!
DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO. Copyright © 2010 Caribbean Examinations Council All rights reserved. 0213401 0/CAPE 2013
-2-
1.
.J8 + ..J32- .Jl62 can be simplified as (A)
4.
(C) (D) 5.
I. II. III.
-p>-q pz > pq p - I< q- I
(A) (B) (C) (D)
I only II only I and III only II and III only
I-2J2
(B)
3-2J2
(C)
I+Ji
(D)
I+2J2
If a remainder of 7 is obtained when
(A) (B) (C) (D)
6.
4 x 4 +8x3 -2x 2 -6x-4? I.
Two roots of the cubic equation 2x3 + 3x2 - 5x- 6 are -I and -2. The THIRD root is
(C)
(D)
-3 2
I
-II -IO IO II
Which of the following are factors of
IV.
X+ I X- I X+ 2 X- 2
(A) (B) (C) (D)
I and II only II and III only I and III only I and IV only
(A) (B) (C) (D)
(a- b)(ct- a 3 b + a 2 b 2 - ab 3 + b4 ) (a- b)(a4 + a 3 b + a 2 b 2 + ab 3 + b 4) (a+ b)(a4 - a 3 b + a 2 b2 - ab 3 + b4 ) (a+ b)(a4 + a 3 b + a 2 b 2 + ab 3 + b4 )
II.
(B)
gives
x3 - 3x + k is divided by x- 3, then k equals
If p and q are positive integers such that p < q, then which of the following statements is/are correct?
(A)
v2 +I
(A)
III. 3.
J2 -I
~
-2J2
(B)
2.
. I. . Rat10na Ismg
7.
2
3 2
3
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-3-
8.
Which of the following mapping diagrams does NOT represent a function? y
9.
If g(x) is the inverse of.f(x) then the correct diagram is (A)
(A)
y
t__
(B) (B) X
y
(C) (C) X
~
y
r: r
L
(D) (D) X
~
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-410.
Which of the following is true if a., fi and y are roots of the cubic equation 3x3- 4x2 -7x- 10 = 0?
a+ fi + r
(B)
a+ fi+r=-, afi+ fir+ra = -
(C)
a + fi + r
=- '
a + f3 + r
4 = -'
=-,
3
afi + fir + ra = -
-3
-7
4
3
3 4 3
7 3
afi + fir + ra = af3 + f3r + ra
(D)
5 16 log2 16 log 2 30
1
4
2
25
(A)
I 36 -log-
(B)
log-
(C)
0
(D)
1
2
(A)
a x a 3
(B)
-a x < - o r x >a 3
(C)
x > - a andx I x I, a> 0, are
3
The annual growth, g(x), (in thousands) of the population over x years is represented by g(x) = 2x. Over how many years will an annual growth of 32 thousand be achieved? (A) (B) (C)
12.
-7
(A)
(D)
11.
4
13.
converse tautology contradiction contra positive
The statement -(p v (- p 1\ q)) is logically equivalent to (A) (B) (C) (D)
pA-q p :::::> -q
-pA-q -p:::::>-q
25
25 4
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-516.
A
vector equation ts gtven as
s[ -~)+ tG) =[ -n.
20.
If~ is an acute angle and cos ~ = 2._ , then 13
sec~=
The values of sand
tare, respectively
(A)
5 13
(A) (B) (C) (D)
-2 -2 2 2
and -1 and 1 and 1 and -1
(B)
(C) (D)
17.
sin (30°- A) is equal to -1 cosA - -sm A
(B)
-1 cosA + -sm A
(D)
18.
2 sin (A) (B) (C) (D)
19.
J3 .
(A)
(C)
12 13 13 12 13 5 ,
2
2
21.
J3 .
2
J3
-
2
J3
-
2
x2 + y- 1Ox + 2y + 1 = 0.
2
The coordinates of the other end of the diameter are
1 . A cosA + -sm
2
(A) (B) (C) (D)
1 sm . A cosA - -
2
e cos ~ is equivalent to sin (8 + ~) + sin (8- ~) sin(8+~)-sin(8-~) cos(8+~)+cos(8-~) cos(8+~)-cos(8-~)
The point (2, 3) is at one end of a diameter of the circle whose equation is
22.
(-12, -5) (-12, -1 ) (8, -5) (8, -1)
The value of sin[;+ (A) (B) (C) (D)
p)
is
- sinp - cosp sinp cosp
The equation of the circle whose centre has coordinates (4, I) and whose radius is 7 units is (A) (B)
(C) (D)
x2 + y + 8x + y- 49 = 0 x 2 + y- 8x- 2y- 32 = 0 x 2 + y - 8x- y + 49 = 0 x 2 + y + 8x + 2y + 66 = 0
GO ON TO THE NEXT PAGE
0213401 0/CAPE 2013
-623.
What value of e, 0 :S e :S n, satisfies the equation 2 cos 2 e + 3 cos e - 2 = 0?
27.
(A) (B) (C)
7(
(A)
(B)
(c) (D)
Ifp = 2i+ j andq = /.. i+6j are perpendicular vectors, then the value of/.. is
6 7(
(D)
4
-3 • -1 0 2
7(
3
28.
7(
JC. The general solution for sin 29 = sm-ts 6
2 (A)
2nJC +. ff6 B= 5JC (2n+1)16
(B)
B=
'{
24.
With respect to an ongm 0, A has coordinates (3, -2). The position vector of3 OA is (A)
(3 , -6)
(B)
(9, - 2)
(C)
(-~J
(D)
25.
(B) (C)
(D) 26.
5JC nJC+ 12
B= {
(D)
nJC+ ff B= 6 5JC (n+1)
(_:)
sin lOA -2 cos 2A 2 cos SA sin A 2 sin 5AcosA
1 + cos 4A - sin 4A = (A) (B) (C)
(D)
1 +cos 4A 2cos2 A cos2A 2 cos 2 A sin 2 A
nJC +12 ff 5JC (2n7r) 12
(C)
The expression sin 6A + sin 4A may be written as (A)
{M+~12
{
6
29.
The cosine of the angle between the vectors -6 j and i + j is (A)
(B)
(C)
(D)
-1
J2 1
J2 -5
J2 6
J2
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-- ---
r
- -----------------------:r-------------------------;~
-7Item 30 refers to the diagram below. y
33.
l=x
~(x 3 sin x) dx
(A) (B) (C) (D)
34.
x2 x2 x2 x2
may be expressed as
(cos x + 3 sin x) (x cos x- 3 sin x) (3 cos x + sin x) (x cos x + 3 sin x)
The function g is defined as 3x + 5 for x < 3 g(x)= { px+2 for x~3 For the function to be continuous at x the value of 'p' should be
30.
In the diagram above showing NOT defined for (A) (B) (C) (D)
31.
lim x~3
(A) (B) (C) (D)
y = x, y
(A) (B) (C) (D)
is
0 X~ 0 x> 0 X< 0 X =
35.
2
(A)
-21 (3 -4x) 2
(B)
21 (3 -4x) 2
x- 3
(C)
Given that lim sin x = 1 , where x is measx-+O X • 3 . . Jim Sin X • ured In radians, then x---+0 ~ IS
27-8x (3-4xf
00
(D)
32.
-3 -1 4 12
If y = x - 6 then dy is 3-4x dx
X -9. --IS
0 6
= 3,
36.
If y
-27 -8x (3 -4x) 2
= -J2x + 1 then
2
d Y is dx 3
1 (A) (B)
(C) (D)
. 3 sm 2 sin3x 2x 2 3 3 2
(A)
(2x+
1)( -J2x+ 1) -1
(B)
(2x +
1)(-J2x + 1)
(C) (D)
(2x + 1)
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- 837.
38.
If y =tan 6x then dy is dx 2 (A) 6 tan 6x (B) sec 2 6x (C) 6 sec 2 6x (D) sec 6x tan 6x
(A)
(B)
(B)
y = sin x + k y =cos X+ k
(C) (D)
y = - COS X + k y =-sin x + k
(C)
(D)
f"(x) = 6x, then given that f'(O) = 0, and cis a constant,j(x) =
If
(A) (B) (C) (D)
Given that
3x2 + x + c x3 + x + c 3x2 + c x3 + c
42.
3 4
9, 4
27 4
The gradient of the normal to the curve = 3x 2 - 2x + 1 at x = 1 is
y
The path ofan object is given parametrically as x = sin t + 2, y = cos t + I . The slope of the tangent at t (A)
-I
(B) (C) (D)
0
is
4
1
(A) 40.
J: 4f(x)dx =9 , the value of
J: 3f(x)d;c
If dy =cos x then dx (A)
39.
41.
=-1t
4
4 (B)
2
is (C)
. -I
(D)
4
undefined
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-943.
Water is leaking from a tank. The rate of change in volume of the water in the tank with respect to time, t, is inversely proportional to the volume, V, of water in the tank. If k is a positive constant of proportionality, then the equation that models this situation
44.
Given dy = 2x, then possible sketches of dx the graph of y are
•
I.
y
II.
y
III.
y
IV.
y
lS
(A)
-k V =-
(B)
---
(C)
dV =-k.JV dt V =-kt
(D)
.Ji
dV dt
-k
v
-----+--~--+---~~X
-1
(A) (B) (C) (D)
0
1
I and II only III and IV only I, III and IV only II, III and IV only
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- 1045.
The radius of a circle is increasing at a rate of O.lcm s- 1_ At the instant when the radius is 3 em, the rate of increase of the area in cm2 s- 1 is 2
(A)
-Jr
(B)
-Jr
5 3
5
(C)
2n
(D)
47t
END OF TEST
IF YOU FINISH BEFORE TIME IS CALLED, CHECK YOUR WORK ON THIS TEST.
0213401 0/CAPE 2013
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