Download exam revision s1 2012 solns

Practice Exam Semester 1
2012
2
The next three questions refer to the
diagram below.
B
Solutions
Name: ___________________________
1.
Vectors
shown:
a
~
and
b
~
C
are represented as
An expression for
b
~ is:
ab
~
~
A
 a b
~
~
B
ab
~
~
C
ab
~
~
D
a
~
E
Their sum is best represented by:
A
B
3
C
D
4
E
The expression
represented by:
A (10, 0)
B (5, 1)
C (1, 3)
D (1, 3)
E (1, 3)
c
in terms of
~
ab
~
~
a
~
may be
c
A vector parallel to ~ but half its
magnitude may be represented by:
A (1, 2)
B (0.5, 2)
C (2, .05)
D (1, 4)
E (1, 2)
and
A
D
5




C
Simplify OA AC  BC  OB
A 0
9

B
OB
C
2OB


AB
D

CA
E
6
A vector represented by a displacement
of (5, 12) has a magnitude of:
119
A
B 13
C 13
D  119
B
10
C
10
D
4 2
E
2
a  5 i  12 j
b  5 i  12 j
~
~
~ and ~
~ , then
If ~
which of the following is true?
ab
~
~
A
C
D
7
A vector has a displacement of (5, 2). A
The direction it makes with the positive
x-axis is:
A 21.8
B 68.2
C 68.2
D 111.8
E 158.2
a 2i  4 j
8
~
If
3a  b
~
A
B
C
D
E
~
~
is:
5i  9 j
b i  3 j
and
~
~
~
~
~
7 i  15 j
~
~
5i  j
~
~
7i9 j
~
~
5 i  15 j
~
~
B
a
~
a
~
~
//
>
b
~
b
~
a
E
11
~
b
>
~
The length x is:
B
then
~
~
B
ab
B
2 17
E
c  3i  j
2c
~
~
~ then the magnitude of
If ~
is:
A 4
B 2 10
A
B
C
D
E
5.14
6.13
9.53
10.44
12.44
B
12
2The angle  to the nearest degree, is:
A
B
C
D
E
13
14
D
16
32°
38°
51°
58°
60°
cos 39 is equal to:
A sin 39
B cos 39
C tan 39
D tan 51
E sin 51
A
B
C
D
E
17
3From a vertical fire tower 70 m high, an D
observer records the angle of depression
to a small fire as 5°.
The distance of the fire from the foot of
the tower to the nearest metre is:
A 6m
B 70 m
C 350 m
D 800 m
E 803 m
4A bearing of 290°T is the same as:
A N20°W
B S20°W
C S70°W
D N70°W
E N20°E
31°
44°
57°
63°
78°
E
6The length of the side marked x is:
A
B
C
D
E
18
15
5The smallest angle to the nearest degree B
is:
6.8
11.5
12.5
21.3
22.1
7The area of the triangle is:
D
A
B
C
D
E
D
5.9 square units
8.8 square units
11.5 square units
13.2 square units
17.6 square units
B
19
8The length of the arc which subtends an C
angle of 135° with radius 8 cm is:
A
B
C
D
E
20
5.2 cm
16.8 cm
18.8 cm
33.5 cm
75.4 cm
9The sector has an area of 150 cm2 and
subtends an angle of 70°. What is the
radius of the circle correct to 1 decimal
place?
A
B
C
D
E
21
6.1 cm2
11.1 cm2
15.7 cm2
19.1 cm2
20.2 cm2
1Which is the area of the shaded
0segment, to the nearest whole number?
A
B
C
D
E
C
A
9 cm2
18 cm2
35 cm2
51.5 cm2
61.4 cm2
22
11.56 radians is closest to:
1A 45°
B 89°
C 140°
D 179°
E 280°
B
23
1A clock has an hour hand 40 cm long.
2The area swept after 30 minutes is
nearest to:
A 5 cm2
B 10 cm2
C 2513 cm2
D 5026 cm2
E 1838 cm2
C
24
Which of the following is NOT a
polynomial?
A
3x  4
3
 2x  6
5
B
C
8x  7
D
E
1

 3x  
x

2
3x  2 x  1
4
D
25
3
2
If 3x  x  5 x  6
  2a  b  x3   c  a  x2   c  a  x  6,
then:
A a  1, b  2, c  3
B a  3, b  2, c  1
C
C
D
E
26
27
28
3x 2  5 x  4
x 2  2 x  3 in partial fractions is:
3x
4

x  3 x 1
A
4
3x

x  3 x 1
B
4
3x

x  3 x 1
C
4x
3

x  3 x 1
D
4x
3

x  3 x 1
E
B
30
2 x 2  3x  4
x3  2 x 2  x =
4
2
3


x x  1 ( x  1) 2
A
E
a  2, b  1, c  3
a   2, b  1, c   3
a  2, b  1, c  3
3x  5
If x  4 is expressed in the form
b
A
x  4 then:
A A  3 and b  15
B A  3 and b  12
E

C A  3 and b  12

D A  3 and b  15

E A  3 and b  17

3
2
If 3x  12 x  36 x  ax( x  b)( x  c),
then a, b and c respectively are:
A −3, 12, 36
B −3, 12, 3
C −3, −12, 36
D 3, 12, 36
E −3, -−2, −36
C
If
A
2
3
2
( x  4)(ax  bx  c)  2 x  7 x  x  20
then a, b and c respectively are:
A 2, −1, 5
B 2, 1, −5
C −2, −1, −5
D 1, 2, 5
E −1, 2, 5

29
31
B
4
2
3


x x  1 ( x  1) 2
C
2
3
4


x x  1 ( x  1) 2
D
3
4
2


x x  1 ( x  1) 2
E
4
2
3


x x  1 ( x  1) 2
The point(s) of intersection between the A
2
parabola y  x  2 x  5 and the
straight line y  3x  12 is/are:
A (3.65, 1.05), (4.65, 25.95)
B (3.65, 1.05)
C (4.65, 25.96)
D (2.32, 5.03)
E (2.32, 5.03), (7.32, 33.97)
32
The distance between (5, 1) and
(4, 3) is found by calculating:
A
B
C
E
33
36
(4  5) 2  (  3  1) 2
A
(4  5) 2  (  3  1) 2

( 4  5)  ( 3  1)
2
The gradient of the line perpendicular to B
2 x  5 y  1  0 is:

(4  5) 2  (3  1) 2

D
E
B
2
C
(4  5) 2  (  3  1) 2
The midpoint of the line segment
between the point (2, 8) and (4, 6)
is:
A (2, 2)
B (1, 1)
C (3, 7)
D (1, 7)
E (1, 1)
The equation of the line that contains
the points (2, 5) and (5, 10) is:
5x  3 y  5  0
A
B
C
D
E
35
5
2
2
5

D
E
D
37
2
5
3
The equation of the line parallel to
A
5 x  2 y  9  0 and passing through the
point (1, 2) is:
A 5x  2 y  1  0
B
C
34
5
2
E
D
E
2x  5 y  9  0
5x  2 y  9  0
5x  2 y  9  0
2x  5 y  8  0
5x  3 y  5  0
3x  5 y  5  0
38
Josh walks in a line from A(4, 1) to
A
B(1,7), then directly from B to
C(2, 7). The total distance travelled is:
A 9.54
B 8.54
C 8.94
D 9.94
E 15.94
39
The points (1, 3), (2, 5) and (5, y) are
collinear. The value of y is:
A 11
B 25
C 20
D 29
E 35
3x  5 y  5  0
5x  3 y  5  0
M divides AB externally in the ratio
3:2. Given A(5, 2) and B(5, 8), M is
the point:
A (25, 28)
B (0, 3)
C (1, 2)
D (12, 18)
E (25, 28)
E
D
40
Joining the points A(3, 4), B(0, 1)
and C(8, 7) forms:
A a straight line
B an isosceles triangle
C an equilateral triangle
D a right-angled triangle at B
E a right-angled triangle at C
A
41
PQRS is a parallelogram with P(2, 3),
Q(1, 7) and R(6, 3). The coordinates of
S are:
A (3, 1)
B (5, 7)
C (9, 2)
D (1, 3)
E (2, 1)
A
42
The equation of the line perpendicular
to 3 y  4 x  13  0 and has and x-
A
intercept of -3.
A 4 y  3x  9  0
B
C
D
E
43
4 y  3x  9  0
3 y  4 x  12  0
3 y  3x  6  0
4 y  3x  3  0
The equation of the perpendicular
bisector of the line joining the points
(−10, −2) and (2, 0).
A y  6x  3  0
B
C
D
E
y  6x  3  0
y  6 x  23  0
6y  x  2  0
6 y  6 x  23  0
C
Name: ___________________________
Section B Short/Extended answer
1
The angle of depression from a lighthouse
200 m high to a cargo ship in the sea is 3°.
What is the distance of the ship from the foot of
the lighthouse?
2
opp
adj
200
tan 3 
x
200
x
tan 3
x = 3816.2 m (correct to 1 decimal place)
tan  
2
The angle of elevation from a Melbourne
Airport runway to a plane at an altitude of
300 m is 7°. How far is the plane from the
runway, correct to 1 decimal place?
3
opp
adj
300
tan 7 
d
300
d
tan 7
d = 2443.3 m (correct to 1 decimal place)
tan  
3
A cross-country runner starts at checkpoint A
and runs for 8 km on a bearing of 50°T to reach
checkpoint B, then heads directly east for
10 km to checkpoint C. How far is checkpoint
C from the starting point A?
3
b2 = a2  c2  (2ac cos B)
d2 = 82 + 102 – (2  8  10 cos 140°)
d2 = 64 + 100 – (–122.6)
d2 = 286.6
d = 286.6
d = 16.93 km (correct to 2 decimal places)
4
A boat sails 14 km west, then 12 km south.
Find the bearing from its starting point.
2
14
12
 14 
  tan 1  
 12 
 = 49.4°
Bearing is (180° + 49.4°)T = 229.4°T
or
S49.4 W
tan  
5
A helicopter flies 25 km in the direction of
N52°W. How far west of the starting point is
it, correct to 1 decimal place?
2
adj
hyp
d
cos 38° =
25
d = 25 × cos 38°
d = 19.7 km
cos  
6
In triangle ABC, a = 12, c = 8 and C = 35°.
Find two possible values of A.
Case 1
a
c

sin A sin C
sin A sin C

a
c
a sin C
sin A 
c
12  sin 35
sin A 
8
sin A = 0.8604
A = 59.4° or (180° – 59.4°)T = 120.6°
Case 2
3
7
Find the largest angle in the triangle with sides
4 cm, 5 cm and 6 cm.
2
a2  b2  c2
2ab
2
4  52  6 2
cos C =
2 45
5
cos C =
40
cos C = 0.125
C = cos1 0.125
C = 82.8°
cos C =
8
Convert the following angles to radian
measure.
(a) 135
(b) 280
(c) 328 35
(a) 135 = 135 
135
180
3
135 =
4
3

180
135 =
(b) 280 = 280 
280
=
180
14
=
9

180
(c) 328 35 = 328 35 
= 5.7349
c

180
9
Convert the following radian measures to
degrees and minutes (where appropriate).
(a)
10
8
3
3
(a)
180
8
8
=


3
3
1440
=
3
= 480
(b) 2.573c
(b) 2.573c = 2.573 
180
(c) 0.625 c
(c) 0.625c = 0.625 
180

= 147.4220407
= 14725

= 35.8098622
= 3549
A goat farm is in the shape of a sector with a
radius 90 m and a subtended angle of 78°. Find
the length of fencing needed to enclose the
farm.
4
78 c
= 1.36 radians
78 
180
l = r
l = 90  1.36
l = 122.4 m
Length of fence = (90  122.4  90)
= 302.4 m
11
A playing field is designed in the shape of a
sector with an area of 8000 m2, and subtends an
angle of 80°.
Find the radius of the sector.
5
80 c
= 1.40 radians
80 
180
1
A  r 2
2
1
8000  r 2  1.40
2
8000
2
r2 
1.40
2
r  11 428.57
r  11 428.57
r  106.9 m
(correct to 1 decimal place)
12
From a point A on the level ground, the angle
of elevation of the top of a tree is 30°. From a
point B on the ground 20 m away from point A
and in line with A, and the foot of the tree, the
angle of elevation to the top of the tree is 60°.
How tall is the tree, to the nearest metre?
6
tan 60° =
h
x
[1]
tan 30° =
h
20  x
[2]
h
tan 60
Substituting into [2] gives:
h
tan 30° =
h
20 
tan 60
h 

tan 30 20 
h
tan 60 

h tan 30
20 tan 30 
h
tan 60
20 tan 30° tan 60°  h tan 30° = h tan 60°
h tan 60°  h tan 30° = 20 tan 30° tan 60°
h (tan 60°  tan 30°) = 20 tan 30° tan 60°
20 tan 30  tan 60
h
tan 60  tan 30
h = 17 m (to the nearest metre)
From [1], x =
13
In a cross-country run, the competitors run
200 m along a track from the starting point A,
directly north to a checkpoint B. From
checkpoint B they make their way across to
checkpoint C, then back to starting point A.
The distance between C and A is 250 m. The
bearing of C from starting point A has been
recorded as 068°T.
8
(a) A = 68°
(b) a2 = b2  c2  (2bc cos A)
a2 = 2002  2502  (2  200  250 
cos 68°)
= 40000  62500 – 37460.7
= 65039.3
a = 65 039.3
= 255 m (to the nearest metre)
Distance BC = 212 m
(a) Write the value of angle A.
(b) Find the distance between the checkpoints
B and C.
(c)
Bearing required is 180  B.
(c) Determine the bearing of C from B to the
nearest degree.
b
a

sin B sin A
sin B sin A

b
a
b sin A
sin B 
a
250  sin 68
sin B 
255
sin B = 0.9090
B = sin1 (0.9090)
B = 65°
The bearing of C from B = (180°  65°)
= 115°T
or
S 65 E
The next three questions refer to
b
~  (3, 2).
14
15
a
~
 (4, 2) and
On the same set of axes, draw the vectors
a
b
~ and ~ .
Calculate
16
3a  2b
~
.
~
a b
Calculate
~
~
2
3(4, 2)  2(3, 2)
 (12, 6)  (6, 4)
 (6, 10)
2
ab
2
~
.
a b
~
17


 (4, 2)  (3, 2)  (7, 0)
~


Show that AB  BC  AD  DC .
~
 7 0
 49
7
2




2

LHS  AB  BC  AD
 AC  AD




 AC  DA
 DA  AC

 DC
 RHS.
3
18
Find the horizontal and vertical components of Horizontal component:
a vector of magnitude 6, which makes an angle  6 cos (120)
of 120 with the positive x-axis.
1
6    3
2

Vertical component:
 6 sin (120)
3
6
3 3
2

19
If
d  2 i  7 j e  2 i  j
f  i 2 j
~
~
~
~
~ , ~
~ and ~
~ ,
calculate
d  e  f  5 i  10 j
~
~
d  e f
~
~
~
~
.
~
~
d  e f
~
3
~
3
~
(5) 2  10 2

 125
5 5
3i  3 j
20
~
What angle does the vector
the positive x-axis?

21
~
~
make with
3
 1
3
, 4th quadrant
  45
tan  

OA  7 i  3 j
If
~
OB   i  5 j
and
~
~


, find AB .

AB 

OB  OA  8 i  8 j
~
~
3
2
22
7
A hiker is at a position (5, 15) where the
coordinates represent the distances in
kilometres east and north of O, respectively. If
the campsite is at a position given by (17, 25),
find the distance and the direction the hiker
must take to reach the campsite.

Let OH  (5, 15)

OC (17, 25)

HC  (17, 25)  (5, 15)  (22, 10)

HC  222  102  24.17
10
22
  24.44  2427
24.17 km to the campsite, on a course
E24 27 N (or N 6533 E)
tan  
23
Analysis
A triathlon course is in the shape of a triangle.
O is the start and finish point of the event. The
swimming leg goes from O to A. The cycling
leg is from A to B. The final leg, running from
B to O. The coordinates of A and B are (3, 1)
and (1, 14) respectively. The coordinates
represent the distance in kilometres east and
j
i
north of O. Take ~ and ~ as the unit vectors

9

along OX and OY .


(a) Express the vectors OA and OB in terms (a)
j
i
of ~ and ~ .

~
~

(c)

3i  j
~
~

OB  i  14 j

~
~


AB  OB  OA  2 i  15 j
AB  2 i  15 j
(b) Hence show that

OA 
(b)
.

Calculate the magnitudes of OA , OB and (c)
~
~

OA  3 2  12  10  3.16

OB  12  14 2  197  14.04
AB .

AB  2 2  15 2  229  15.13
Total distance  3.16  14.04  15.13
 32.33 km
1

tan 1  18.4
(e) Find the angle that OA makes with the x- (e)
3
axis.
15
tan 1
 82.41

2
(f)
(3rd quadrant)
AB
(f) Calculate the angle that
makes with
(that
is,
262.41)
the x-axis and hence show that the bearing of B
Bearing from A  (90  82.41)
from A is S7.59W.
(d) Find the distance of the course.
(d)
S7.59W
24

2
In the regular hexagon shown, AB  a and
~

FA  b .
~
Express the following in terms of a and b .
~

(a)
OB
~

(a)

(b)
OC
OB  b
~

(b)
OC  a
~

(c)
FC
(d)
OA
(e)
AD



(c)
FD
~

(d)

(f)
FC  2 a
OA  b  a
~
~

(e)
AD  2 a  2 b
~
~

(f)
25.
Find the values of a, b and c if
( x  2)(ax 2  bx  c)   2 x3  5x 2  6 x  8
FD  2 a  b
~
~
LHS  ax 3  bx 2  cx  2ax 2  2bx  2c
 ax 3   b  2a  x 2   c  2b  x  2c
hence a   2, c   4
b  2a  5 so b  1
2
26.
Find the values of a, b and c if
ax  x  b  x  c   7  x3  5x2  6x  7


LHS  ax x 2  cx  bx  bc  7
3
 ax 3  acx 2  abx 2  abcx  7
 ax 3   ac  ab  x 2  abcx  7
hence a  1
ac  ab   5
b  c  5
b  5  c
abc  6
bc  6




5c  c 2  6
5c c  6
c 2  5c  6  0
(c  2)(c  3)  0
c   2, b   3 or
c   3, b   2
27
7 x  14
Express x  3 x  10 as partial fractions.
2
7 x  14
7 x  14

x  3 x  10 ( x  5)( x  2)
A
B


x5 x 2
7 x  14  A( x  2)  B ( x  5)
3
2
7 x  14  ( A  B ) x  2 A  5 B
A  B  7 [1]
2 A  5 B  14 [2]
2  [1]  [2]
7 B  28
B4
A3
28
Solve the following equations simultaneously:
y  4x  5
2 y  4x  2
2(4 x  5)  4 x  2
2
8 x  10  4 x  2
4 x  12
x3
y7
29
Find the point(s) of intersection between the
2 x  x 2  9  10  0
parabola
x2  2 x  1  0
2
y  x  9 and the straight line with equation 2 x  y  10  0
( x  1) 2  0
x  1
point of intersection is ( 1,  8)
2
30
(2, 5) is the midpoint of the segment AB. If A
 x1  x2 y1  y2 
,


has coordinates (w, v) and B has coordinates
2 
Midpoint  2
(3, 6) find w and v.
w3
2
2
w=1
v  6

5
2
v = 4
w = 1 and v = 4
3
31
If the distance between A(3, 4) and B(3, p) is
13 units, find the value of p.
5
(3  3)2  (  4  p)2  169
(  4  p)2  169

4  p  13
p   4 13
p  17 or 9
32
Find the equation of the line passing through
P(4, 1) and parallel to the line joining A(2,
5) and B(4, 9).
33
Find the equation of the perpendicular bisector
of the segment AB, where A has coordinates
(2, 3) and B has coordinates (4, 7)
34
Show that triangle ABC is isosceles given
A(1, 5), B(2, 1) and C(4, 1).
9   5 14
mAB 
 7
42
2

y  1  7( x  4)
y  1  7 x  28
7 x  y  29  0
4
 2 4 37 

,

  3, 2
2
2




5
Midpoint

7  3 10 
mAB 

 5
42
2
1
m 
5
1
y  2  ( x  3)
5
5 y  10  x  3
x  5 y  13  0
AB  (1   2) 2  (5  1) 2  25  5
AC  (4  1) 2  (1  5) 2  25  5
BC  (4   2) 2  (1  1) 2  36  6
Since AB  AC, then triangle is isosceles.
3
35
Find the equation of the line perpendicular to
3x  2 y  8  0 and having the same
y-intercept.
36
Find the coordinates of the point C that divides
the line joining A(7, 4) and B(1, 2) internally
in the ratio 2:5.
y-intercept: (0, 4)
2 y  3x  8
3
y  x4
2
3
m
2

2
m 
3

2
y4
x
3

2
y
x4
3
or 2x + 3y  12 = 0
5
2 1  5  7 37
2

5
25
7
7


2 2  5 4
16  2
y

 2
25
7
7
4
x
 2  2
5 , 2 
7.
Coordinates of point C are  7
37
If the points A(7, 2), B(5, 4) and C(4, z) are
collinear, find the value of z.
mAB  mAC

4

4 2 z  2


57 47

1 z2
 
2
3
2z + 4 = 3
2z = 1
1
z=2
38
A(7, 2), B(2, 14), C(10, 9) and D(5, 3)
form the quadrilateral ABCD. Find:
(a)
the gradient of AB
(a)
(b) the gradient of DC
(c)
(b)
(e) the length of AB
(f)

9  14
10   2

5

12
mBC 
the length of BC.
Describe the quadrilateral ABCD.

39
5  10
12

5
mDC 
the gradient of BC
(d) the gradient of AD
14  2
2  7
12

5
mAB 
(c)
(d)
(e)

3 2
5  7

5

12
mAD 
d AB  (  2   7) 2  (14  2) 2  13
d BC  (10   2) 2  (9  14) 2  13
(f)
mAB  mDC  AB//DC
mBC  mAD  BC//AD
12  5 
mAB  mBC  
 1  AB  BC
5 12
d AB  d BC  13  ABCD
is a square.
8