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Name: ________________________ Class: ___________________ Date: __________
ID: A
Final Exam Practice
Multiple Choice
Identify the choice that best completes the statement or answers the question.
____
1. Compared to the graph of the base function f(x) = | x|, the graph of the function g(x) + 5 = |x| is translated
A 5 units to the right
C 5 units down
B 5 units up
D 5 units to the left
____
2. Compared to the graph of the base function f(x) = |x| , the graph of the function g(x) = |x + 9| is translated
A 9 units to the right
C 9 units down
B 9 units up
D 9 units to the left
____
3. What is the equation of the transformed function, g(x), after the transformations are applied to the graph of
the base function f (x) = x 2 , shown in blue, to obtain the graph of g(x), shown in red?
A
g (x) + 3 = (x − 5)
B
g (x) = (x + 3) − 5
2
2
C
g (x) − 5 = (x + 3)
D
g (x) = (x − 5) + 3
1
2
2
Name: ________________________
____
4. The two functions in the graph shown are reflections of each other. Select the type of reflection(s).
A
B
____
ID: A
a reflection in the line y = x
a reflection in the x-axis and the y-axis
C
D
a reflection in the y-axis
a reflection in the x-axis
5. When a function is reflected in the x-axis, the coordinates of point (x, y) become
A (x, –y)
C (–x, –y)
B (–x, y)
D (x, y)
2
Name: ________________________
____
ID: A
6. Which of the graphs shown below represents the base function f(x) = x 2 and the stretched function g(x) =
1
− x 2?
5
A
C
B
D
3
Name: ________________________
____
7. Which is the graph of the function f(x) = (x − 6) + 3?
A
C
2
B
____
ID: A
D
8. In the graph shown, which transformations must be applied to the blue curve to obtain the red curve?
A
B
C
D
a reflection in the x-axis and a translation of 5 units down
a reflection in the y-axis and a translation of 5 units up
a reflection in the x-axis and a translation of 5 units up
a reflection in the y-axis and a translation of 5 units down
4
Name: ________________________
____
ID: A
9. Which of the following graphs represents the graph of the function f (x) = |x| transformed to
f (x) = 2 |−2x + 4| + 2?
A
C
B
D
5
Name: ________________________
ID: A
____ 10. When the function f (x) = |x| is transformed to f (x) = −4 |x + 3| + 2, the graph looks like
A
C
B
D
____ 11. Which of the following functions is the correct inverse for the function f(x) = 3x + 5?
1 5
1 5
A f −1 (x) = x−
C f −1 (x)= − x−
3 3
3 3
1 5
1 5
B f −1 (x) = − x+
D f −1 (x) = x+
3 3
3 3
9
____ 12. Which of the following functions is the correct inverse for the function f(x) = − x + 6?
2
2 4
2 4
A f −1 (x) = − x+
C f −1 (x) = − x−
9 3
9 3
9
4
9
4
B f −1 (x)= x+
D f −1 (x) = x−
2 3
2 3
____ 13. Which of the following functions is the correct inverse for the function f(x) = x 2 + 7,
{x | x ≥ 0, x ∈ R}?
2
A f −1 (x) = (x − 7)
C f −1 (x) = x − 7
B
f −1 (x) =
x +7
D
6
f −1 (x) =
x+7
Name: ________________________
ID: A
____ 14. Which graph represents the inverse of the graph shown?
A
C
B
D
7
Name: ________________________
ID: A
____ 15. Which graph represents the inverse of the function shown?
A
C
B
D
____ 16. Compared to the graph of the base function f(x) = x , the graph of the function g(x) =
A 2 units up
C 2 units to the left
B 2 units to the right
D 2 units down
x − 2 is translated
____ 17. Compared to the graph of the base function f(x) = x , the graph of the function g(x) + 8 =
A 8 units to the left
C 8 units to the right
B 8 units up
D 8 units down
8
x is translated
Name: ________________________
ID: A
____ 18. Compared to the graph of the base function f(x) = x , the graph of the function g(x) =
A 5 units down
C 5 units right
B 5 units left
D 5 units up
____ 19. When b < 0, the function g(x) = bx has what relationship to the base function f(x) =
A f(x) is stretched horizontally by a factor of 1/|b|
B f(x) is stretched horizontally by a factor of 1/|b| and reflected in the y-axis
C f(x) is stretched vertically by a factor of |b|
D f(x) is stretched vertically by a factor of |b| and reflected in the x-axis
x − 5 is translated
x?
____ 20. In the following graph, what transformations must be applied to the blue curve to obtain the red curve?
A
B
C
D
a reflection in the x-axis, a vertical translation 5 units up, and a horizontal translation 3
units to the right
a reflection in the x-axis, a vertical translation 5 units down, and a horizontal translation
3 units to the right
a reflection in the x-axis, a vertical translation 3 units up, and a horizontal translation 5
units to the left
a reflection in the x-axis, a vertical translation 3 units up, and a horizontal translation 5
units to the right
9
Name: ________________________
ID: A
____ 21. The two functions in the graph shown are reflections of each other. Select the type of reflection(s).
A
B
a reflection in the y-axis
a reflection in the x-axis and the y-axis
C
D
a reflection in the line y = x
a reflection in the x-axis
____ 22. Which is the graph of the square root of the function f(x) = (x − 5) 2 − 2?
A
C
B
D
10
Name: ________________________
ID: A
____ 23. Which of the following functions is the correct inverse for the function f(x) =
A
f (x) =
x+2
C
f (x) = x + 2
B
f −1 (x) =
x +2
D
f −1 (x) = (x − 2)
−1
−1
____ 24. Which graph represents the square root of the graph shown?
A
C
B
D
11
2
2
x − 2 , {x | x ≥ 0, x ∈ R}?
Name: ________________________
ID: A
____ 25. Which graph shows the graphical solution to the radical equation 0 = 2 (x − 5) − 2?
C
A
B
D
____ 26. Which radical equation can be solved using the graph shown below?
A
B
− 4−x = x +2
4−x = x +2
C
D
12
x +2 = − 4+x
4+x = x +2
Name: ________________________
ID: A
____ 27. What is the solution to the radical equation 0 = x + 9 − 3?
A 18
C –18
B 36
D 0
____ 28. What is the solution to the radical equation 0 = 2 2(x + 4) − 8?
C 4
A –4
B 12
D 128
____ 29. Which of the following is a polynomial function?
A y = −4x 4 + 4x 3 − 7x 2 + 9x
C
B
f (x) = −4 x − 7
D
g (x) = x + 4
−4x + 9
y=
x2
____ 30. Which graph represents an odd-degree polynomial function with two x-intercepts?
C
A
B
D
____ 31. If −9x 3 + 9x 2 + 4 is divided by 6x + 5, then the restriction on x is
5
6
A x≠
C x≠−
5
6
5
6
B x≠
D x≠−
6
5
13
Name: ________________________
ID: A
____ 32. If −2x 3 − 6x 2 + 5x − 7 is divided by x − 7 to give a quotient of −2x 2 − 20x − 135 and
a remainder of –952, then which of the following is true?
A (x − 7) (−2x 2 − 20x − 135) = –952
B −2x 3 − 6x 2 + 5x − 7 = (x − 7) (−2x 2 − 20x − 135) – 952
C
D
(x − 7) (−2x 2 − 20x − 135) = 952
−2x 3 − 6x 2 + 5x − 7= (x − 7) (−2x 2 − 20x − 135) + 952
____ 33. When P (x) = 5x 3 − 2x + 2 is divided by 5x − 2, the remainder is
12
A x2 + x +
C P(5 / 2) = 601 / 8
5
B P(–2) = –34
D P(2 / 5) = 38 / 25
____ 34. For a polynomial P(x), if P (6) = 0, then which of the following must be a factor of P(x)?
C x2 + 6
A x2 − 6
B x+6
D x−6
____ 35. Which of the following binomials is a factor of x 3 + 12x 2 + 29x + 18 ?
A x−2
C x−1
B x−9
D x+2
____ 36. Determine the value of k so that x + 2 is a factor of x 3 + 10x 2 + 23x + k .
A k = –1
C k = 14
B k = –14
D k=1
____ 37. Which of the following is the fully factored form of x 3 + 2x 2 − 23x − 60?
A (x + 3) (x − 4) (x + 5)
C (x − 3) (x + 4) (x + 5)
B (x + 3) (x + 4) (x − 5)
D (x − 3) (x − 4) (x − 5)
____ 38. Which of the following is the fully factored form of x 3 + 9x 2 − 4x − 36?
2
A (x − 2) (x + 9)
C (x + 2) (x − 2) (x − 9)
B
(x − 2) (x − 9)
2
D
(x + 2) (x − 2) (x + 9)
____ 39. One root of the equation x 3 + 7x 2 − 33x − 135 = 0 is
A –3
C 9
B 3
D –5
14
Name: ________________________
ID: A
____ 40. Which of the following graphs of polynomial functions corresponds to a cubic polynomial equation with
roots 4, 1, and 3?
A
C
B
D
15
Name: ________________________
ID: A
____ 41. Which of the following graphs of polynomial functions corresponds to a polynomial equation with zeros –6
(multiplicity of 2) and –1 (multiplicity of 2)?
C
A
B
D
____ 42. Determine the equation of a circle with centre at (3, –3) and radius 10.
A (x − 3) 2 + (y + 3) 2 = 100
C (x − 3) 2 + (y + 3) 2 = 10
B
(x − 3) 2 + (y + 3) 2 = 20
D
(x − 3) 2 + (y + 3) 2 =
10
____ 43. If the angle θ is –5000° in standard position, it can be described as having made
8
7
A 13 rotations
C 27 rotations
9
9
8
7
B −13 rotations
D −27 rotations
9
9
____ 44. If the angle θ is 1600° in standard position, in which quadrant does it terminate?
A quadrant III
C quadrant II
B quadrant IV
D quadrant I
16
Name: ________________________
ID: A
____ 45. A ball is riding the waves at a beach. The ball’s up and down motion with the waves can be described using
ÊÁ πt ˆ˜
the formula h = 2.3sinÁÁÁÁ ˜˜˜˜ , where h is the height, in metres, above the flat surface of the water and t is the
Ë 3¯
time, in seconds. What is the height of the ball, to the nearest hundredth of a metre, after t = 17 s?
C –1.99 m
A –0.87 m
D 1.99 m
B –2.66 m
____ 46. A tricycle has a front wheel that is 30 cm in diameter and two rear wheels that are each 12 cm in diameter. If
the front wheel rotates through a angle of 32°, through how many degrees does each rear wheel rotate, to the
nearest tenth of a degree?
A 32.0°
C 80.0ℑ
B 40.0ℑ
D 160.0ℑ
____ 47. The point P(0.391, 0.921) is the point of intersection of a unit circle and the terminal arm of an angle θ in
standard position. What is the equation of the line passing through the centre of the circle and the point P?
Round the slope to two decimal places.
C y = 2.36x + 0.92
A y = 2.36x
B y = 0.42x
D y = 2.36x + 0.39
____ 48. Which function, where x is in radians, is represented by the graph shown below?
A
B
y = −cos x
y = sin x
C
D
y = cos x
y = −sin x
____ 49. The period (in degrees) of the graph of y = cos 4x is
A 270°
C 90°
B 180°
D 45°
17
Name: ________________________
ID: A
____ 50. Which function is represented by the graph shown below, where θ is in radians?
A
B
5
y = − sin(−2x)
4
5
y =−2 sin(− x)
4
C
D
5
y =−2 cos(− x)
4
5
y = − cos(−2x)
4
____ 51. The graph of y = sin x can be obtained by translating the graph of y = cos x
π
π
A
units to the right
C
units to the right
4
3
π
B
units to the right
D π units to the right
2
ÁÊÁ ÊÁ
π ˆ˜ ˜ˆ˜
____ 52. What is the period of the sinusoidal function y = −cos ÁÁÁ 8 ÁÁÁ x − ˜˜˜˜ ˜˜˜ − 2?
Á ÁË
2 ¯˜
Ë
¯
1
1
A
π
C
π
8
4
1
B 4π
D
π
2
____ 53. Which of the following is not an asymptote of the function f (θ) = tanθ?
7
5
A x=− π
C x=− π
2
2
9
B x=− π
D x = −π
2
____ 54. Which function has zeros only at θ = nπ,n ∈ I?
2
A y = tan(θ + π)
C
3
B
y = tan (θ − π )
D
18
7
π)
6
5
y = tan(θ + π)
4
y = tan(θ −
Name: ________________________
ID: A
____ 55. Given the trigonometric function y = tanx, which is the x-coordinate at which the function is undefined?
9
1
π
C − π
A
2
3
7
3
B − π
D
π
6
4
____ 56. Given the trigonometric function y = tan x, find the value of the y-coordinate of the point with x-coordinate
–1200°.
3
C 1
A
B −1
D undefined
1
____ 57. What are the solutions for sin 2 x − = 0 in the interval 0° ≤ x ≤ 360°?
2
A x = 45° and 225° and 315° and 135°
C x = 90° and 270° and 225°
B x = 30° and 210° and 135°
D x = 60° and 240° and 45°
Use the following information to answer the questions.
The height, h, in metres, above the ground of a car as a Ferris wheel rotates can be modelled by the function
ÊÁ πt ˆ˜
h (t) = 18 cos ÁÁÁÁ ˜˜˜˜ + 19, where t is the time, in seconds.
Ë 80 ¯
____ 58. What is the radius of the Ferris wheel?
A 9m
B 18 m
C
D
19 m
36 m
____ 59. How long does it take for the wheel to revolve once?
π
A
s
C 160 s
80
80
B 80 s
D
s
π
____ 60. What is the minimum height of a car?
A 19 m
B 9m
C
D
160 m
80 m
____ 61. What is the maximum height of a car?
A 19 m
B 80 m
C
D
160 m
31 m
19
Name: ________________________
ID: A
Use the following information to answer the questions.
The height, h, in centimetres, of a piston moving up and down in an engine cylinder can be modelled by the
function h (t) = 14 sin (80πt) + 14, where t is the time, in seconds.
____ 62. What is the period?
7
A
s
40
B
8s
____ 63. Which expression is equivalent to
A
B
C
cos θ (1 + sinθ)
1 + sin 2 θ
cos θ
1 − sinθ
D
1
s
40
1
s
14
cos θ
?
1 + sinθ
C
D
1 − sinθ
cos θ
1 + sinθ
cos θ
____ 64. Which expression is equivalent to tanθ + cot θ?
A
1
C
B
cos θ sinθ
D
____ 65. Which expression is equivalent to
A
B
tan(A + B)
cot(A + B)
1
cos θ sinθ
ÊÁ sinθ ˆ˜
˜˜
2 ÁÁÁÁ
˜˜
cos
θ
Ë
¯
tanA − tanB
?
1 + tanA tanB
C tan(A − B)
D cot(A − B)
____ 66. Simplify sin168° cos 143° − cos 168° sin143°. Round your answer to the nearest hundredth.
A 0.47
C –0.75
B –1.15
D 0.42
____ 67. What is the general solution, in degress, to the equation 2cos x cos 2x − 2 sinx sin2x = −1?
A 40° + 180n° and 80° + 180n°. where n ∈ I C 40° + 120n° and 80° + 120n°, where n ∈ I
B 40° + 120n°, where n ∈ I
D 220° + 120n° and 330° + 120n°, where
n∈I
1
____ 68. What is the general solution, in radians, to the equation (4cos 2 2θ + 1) sin θ = 0?
3
A 2πn where n ∈ I
C 3πn where n ∈ I
π
n where n ∈ I
B no solution
D
3
____ 69. Which set of properties does the function y = 2 x have?
A no x-intercept, no y-intercept
C no x-intercept, y-intercept is 1
B x-intercept is 1, no y-intercept
D x-intercept is 0, y-intercept is 0
20
Name: ________________________
ID: A
____ 70. Which choice best describes the function y = 6 x ?
A both increasing and decreasing
C
B decreasing
D
increasing
neither increasing nor decreasing
x
ÁÊ 1 ˜ˆ
____ 71. Which set of properties is correct for the function y = ÁÁÁÁ ˜˜˜˜ ?
Ë 9¯
A domain {x| x ∈ R}, range
C domain {x| x ∈ R}, range
{y| y > 0, y ∈ R}
{y| y ≤ 0, y ∈ R}
B domain {x| x ∈ R}, range
D domain {x| x ∈ R}, range
{y| y ≥ 0, y ∈ R}
{y| y < 0, y ∈ R}
____ 72. Which exponential equation matches the graph shown?
A
B
ÊÁ 1 ˆ˜ x
y = ÁÁÁÁ ˜˜˜˜
Ë 8¯
C
y = 8x
D
ÊÁ 1 ˆ˜ x
y = − ÁÁÁÁ ˜˜˜˜
Ë 8¯
y = −8x
____ 73. A bacteria colony initially has 1500 cells and doubles every week. Which function can be used to model the
population, p, of the colony after t days?
t
A
p (t) = 1500 (3)
t
C
p (t) = 1500 (2) 7
B
p (t) = 1500 (2)
t
D
p (t) = 1500 (3) 7
t
21
Name: ________________________
ID: A
____ 74. To the nearest year, how long would an investment need to be left in the bank at 5%, compounded annually,
for the investment to triple?
A 15 years
C 28 years
B 26 years
D 23 years
____ 75. Which function results when the graph of y = 6 x is translated 2 units down?
A
B
y = 6x − 2
y = 6x + 2
C
D
y = 6x − 2
y = 6x + 2
____ 76. What is the exponential equation for the function that results from the transformations listed being applied to
the base function y = 9 x ?
• a reflection in the y-axis
• a vertical stretch by a factor of 6
• a horizontal stretch by a factor of 7
x
A
y = −7(9) 6
B
y = 6 (9)
−x
7
x
C
y = 7 (9) 6
D
y = −6(9) 7
x
22
Name: ________________________
ID: A
ÊÁ 7 ˆ˜ x
____ 77. Which graph represents the function y = 2 ÁÁÁÁ ˜˜˜˜ ?
Ë 9¯
A
C
B
D
____ 78. Which equation can be used to model the given information, where the population has been rounded to the
nearest whole number?
Year (x) Population (y)
0
100
1
104
2
108
3
112
4
117
5
122
A
y = 100 (1.04)
B
y = 100 (1.4)
x
x
C
y = 100 (1.04)
D
y = 100 (1.4)
23
x−1
x−1
Name: ________________________
ID: A
____ 79. Solve for x, to one decimal place.
7333 = 5 x
A 1466.6
B 11.1
C
D
36 667.0
5.5
C
D
6
3.0
____ 80. Solve for x.
(36) = 216
A 0.3
B 7
3x
(x + 7 )
t
ÊÁ 1 ˆ˜ 45
____ 81. The half-life of a radioactive element can be modelled by M = M 0 ÁÁÁÁ ˜˜˜˜ , where M 0 is the initial mass of
Ë 32 ¯
the element; t is the elapsed time, in hours; and M is the mass that remains after time t. The half-life of the
element is
A
B
11 h
10 h
C
D
18 h
9h
____ 82. Another way of writing 5 5 = 3125 is
A
log 5 5 = 3125
C
log 3125 5 = 5
B
log 5 3125 = 5
D
log 5 5 = 3125
____ 83. Another way of writing 7 −3 =
A
B
ÁÊ 1 ˜ˆ
log 7 ÁÁÁÁ ˜˜˜˜ = −343
Ë 3¯
ÁÊ 1 ˜ˆ
log 3 ÁÁÁÁ − ˜˜˜˜ = 343
Ë 7¯
1
is
343
C
D
ÁÊ 1 ˜ˆ˜
˜˜ = −3
log 7 ÁÁÁÁ
˜
Ë 343 ¯
ÁÊ 1 ˜ˆ˜
˜˜ = −3
log 7 ÁÁÁÁ −
˜
Ë 343 ¯
____ 84. Compared to the graph of the base function y = log 10 x, the graph of the function y = log 10 x + 4 is translated
A 4 units to the left
C 4 units up
B 4 units down
D 4 units to the right
24
Name: ________________________
ID: A
____ 85. Which graph represents the function y = −3log 3 [(x − 2)] − 3?
C
A
B
D
____ 86. Which if the following is equivalent to the expression log 4 sw 10 y?
A log 4 s + 10log 4 w + log 4 y
C log 4 s + log 4 w + 10log 4 y
B 10log 4 s − 10log 4 w + log 4 y
D 10log 4 s + log 4 w + log 4 y
____ 87. Solve 8 x = 486. Round your answer to two decimal places.
A 3.59
C 1.78
B 2.97
D 0.34
1
5
as x → − + (right to left)?
4x + 5
4
f(x) → 0
f(x) is undefined
____ 88. What is true about the behaviour of the function f(x) =
A
B
f(x) → −∞
f(x) → +∞
C
D
25
Name: ________________________
____ 89. What is the x-intercept of f(x) =
A
B
ID: A
1
?
2x + 4
There is no x-intercept.
1
−
2
C
−2
D
0
A
−4
− 5?
x−9
C
B
D
____ 90. Which graph represents the function f(x) =
26
Name: ________________________
ID: A
____ 91. Which function represents the graph shown below?
A
B
9
+3
x−8
−9
f(x) =
+3
x−8
f(x) =
C
D
−9
−8
x+3
9
f(x) =
−8
x+3
f(x) =
____ 92. Which of the following functions has a slant asymptote when graphed?
5x 3 − 10x 2 − 15x
5x 3 − 10x 2 − 15x
f(x)
=
C
A f(x) =
x 2 − 3x − 4
x 2 − 3x
B
f(x) =
5x 3 − 10x 2 − 15x
x 2 − 2x − 3
D
all of the above
6
____ 93. Which function has vertical asymptotes with equations x = −9 and x = − ?
7
7x + 6
1
A f(x) = 2
C f(x) = 2
x + 15x + 54
7x − 69x + 54
1
−9
B f(x) = 2
D f(x) = 2
7x + 69x + 54
x + 15x + 54
____ 94. Which function has a point of discontinuity at x = 3?
x−3
x−3
A f(x) = 2
C f(x) = 2
2x − 2x − 12
x − 6x − 12
x+3
x+3
B f(x) = 2
D f(x) = 2
x − 6x − 12
x − 6x + 9
27
Name: ________________________
____ 95. Which graph represents f(x) =
ID: A
x−3
?
5x − 23x + 24
2
A
C
B
D
____ 96. Which function has a y-intercept of −
A
B
−8
x − 12x − 27
8
f(x) =
(−8x + 3)(x + 9)
f(x) =
2
8
?
27
−8
x + 12x + 27
C
f(x) =
D
all of the above
2
2
____ 97. Which function has a horizontal asymptote with equation y = ?
7
−2x − 3
7x − 3
A f(x) =
C f(x) =
7x + 8
2x + 8
7x + 8
2x − 3
B f(x) =
D f(x) =
2x − 3
7x + 8
28
Name: ________________________
1
____ 98. Which function has an x-intercept of ?
3
−6x − 2
A f(x) =
5x − 3
5x − 3
B f(x) =
6x − 2
ID: A
C
D
6x − 2
5x − 3
5x − 2
f(x) =
6x − 3
f(x) =
____ 99. What is the equation for the vertical asymptote of the graph of the function shown?
A
B
x=2
x=3
C
D
y=7
y=6
____ 100. Which function has a graph in the shape of a parabola?
(x − 3) 2 (x − 7)
x−3
A f(x) =
C f(x) =
(x − 3)(x − 7)
(x − 3) 3 (x − 7)
B
f(x) =
(x − 3) 2 (x − 7)
x−3
____ 101. What are the x-intercepts of the graph of f(x) =
A
B
–7, –5
2, –9
D
none of the above
x 2 + 7x − 18
?
x 2 + 12x + 35
C 7, 5
D –2, 9
29
Name: ________________________
____ 102. Solve the equation 0 =
A
B
no solution
x = −1
ID: A
6x 3 + 6
graphically.
−x 3 − 8x 2
C
D
0
x = −8
____ 103. Given the functions f (x) = x 2 − 3 and g (x) = −9 − x , determine the equation for the combined function
y = f (x) + g (x) .
A y = x 2 − 27x − 12
C y = x 2 + 27x + 6
B
y = x2 − x + 6
D
y = x 2 − x − 12
____ 104. Given the functions f (x) = x 2 − 8 and g (x) = −2 − x 2 , determine the equation for the combined function
y = f (x) + g (x) .
A
B
−6 − 2x 2
−10
C
D
30
16 − x 4
4−x
Name: ________________________
ID: A
For the following question(s), assume that x is in radians, if applicable.
____ 105. Given the functions f (x) = cosx and g (x) = −x, a graph of the combined function h (x) = f (x) + g (x) most
likely resembles
A
C
B
D
____ 106. Given the functions f (x) = 9 x and g (x) = 9sin x, what is the range of the composite function h (x) = f (x)g (x) ?
A {y|− 9 ≤ y ≤ 9, y ∈ R}
C {y| y ∈ R}
B cannot be determined
D {y| y > 9, y ∈ R}
31
Name: ________________________
ID: A
____ 107. Shown is the graph of h (x) = f(g(x)), where f (x) = sinx and g (x) is a function of the form g (x) = a(x + b).
What equation represents g (x) ?
1
A g (x) = (x + 9)
2
B
g (x) = 2(x − 9)
C
g (x) = 2(x + 9)
D
1
g (x) = (x − 9)
2
32
Name: ________________________
ID: A
____ 108. Given the functions f (x) = 0.6 x and g (x) = cosx, the graph of the combined function h (x) = f (x)g (x) most
likely resembles
A
C
B
D
33
Name: ________________________
ID: A
____ 109. Given the functions f (x) = x 2 − 4 and g (x) = x − 4, a graph of the combined function h (x) =
resembles
A
C
B
D
34
f (x )
most likely
g (x)
Name: ________________________
ID: A
____ 110. An equation for the graph shown is most likely
A
B
4 x + cosx
4 x cosx
C
D
35
4 x + sinx
4 sinx
Name: ________________________
ID: A
____ 111. An equation for the graph shown is most likely
A
f (x) =
B
2 sinx
cosx
x
____ 112. Given the functions f (x) = x + 3 and g (x) =
A
B
−8
,x ≠ 0
x
3x − 8
(f û g)(x) =
, x ≠ −3
x+3
(f û g)(x) =
C
sinx
cosx
D
f (x) =
sinx
x
1
, what is the simplified form of (f û g)(x)?
x−3
3x − 8
,x ≠ 3
C (f û g)(x) =
x−3
1
D (f û g)(x) = , x ≠ 0
x
____ 113. Given f (x) = 9x 2 + 7x and g (x) = 2 − x, determine 5g (x) − f (x) .
A 45x 2 + 36x − 2
C 45x 2 + 6x − 5
B 9x 2 + 6x + 2
D −9x 2 − 12x + 10
36
Name: ________________________
____ 114. Given the functions f (x) =
y = f(g(x))?
A
B
ID: A
1
x and g (x) = − (x − 5) , which of the following is most likely the graph of
3
C
D
37
Name: ________________________
ID: A
____ 115. Given the functions f (x) = log x and g (x) = x + 4, which of the following is most likely the graph of
h (x) = f(g(x))?
A
C
B
D
____ 116. Solve for the variable:
5 P r = 20
A 5
B 2
C
D
60
7
____ 117. An orchestra has 2 violinists, 3 cellists, and 4 harpists. Assume that the players of each instrument have to sit
together, but they can sit in any position in their own group. In how many ways can the conductor seat the
members of the orchestra in a line?
A 144
C 24
B 72
D 1728
____ 118. For a mock United Nations, 6 boys and 7 girls are to be chosen. If there are 12 boys and 9 girls to choose
from, how many groups are possible?
A 846 720
C 960
B 33 264
D 120 708 403 200
____ 119. For which of the following terms is a = 55 in the expansion of (x + y)11?
A ax 3 y 8
C ax 2 y 9
B
ax 8 y 3
D
38
ax 11
Name: ________________________
ID: A
____ 120. The leadership committee at a high school has 4 grade 10 students, 2 grade 11 students, and 6 grade 12
students. This year, 12 grade 10, 8 grade 11, and 10 grade 12 students applied for the committee. How many
ways are there to select the committee?
A 2 910 600
C 733
B 100 590 336 000
D 163 136
Short Answer
1. Create a graph of g(x) = f (x − 1) + 2 for each base function given, using transformations.
a) f(x) = x 2
b) f(x) = |x|
2. Determine the equation, in standard form, of each parabola after being transformed from f(x) = x 2 by the
given translations.
a) 4 units to the right and 3 units up
b) 2 units to the left and 1 unit up
c) 2 units down and 7 units to the left
39
Name: ________________________
ID: A
3. Given the graph of a function, sketch the resulting graph after the specified transformation.
a) reflection in the x-axis
b) reflection in the y-axis
c) reflection in the x-axis and the y-axis
4. Determine the equation of the function g(x) after the indicated reflection.
2
a) f(x) = (x − 1) + 2, in the x-axis
b) f(x) = |x| + 1, in the y-axis
5. a) Sketch the graph of g(x) = 2f (2x) for each base function.
i) f(x) = x
ii) f(x) = x 2
iii) f(x) = |x|
b) Write the equation for g(x) to represent a single stretch that results in the same graph as in each function in
part a).
c) Describe how each stretch affects the domain and range for each function.
40
Name: ________________________
ID: A
6. For each g(x), describe, in the appropriate order, the combination of transformations that must be applied to
the base function f(x) = x .
a) g(x) = −
2(x + 1) − 2
b) g(x) = 2 x − 3 − 4
1
5−x +1
c) g(x) = −
2
7. For each of the following, describe the combination of transformations that must be applied to the graph of
f(x) = x 2 (shown in blue) to obtain the graph of g(x) (shown in red).
a)
b)
c)
41
Name: ________________________
ID: A
8. For each function f(x),
i) determine f −1 (x)
ii) graph f(x) and its inverse
5
a) f(x) = x − 3
2
b) f(x) = 3 (x − 2) − 3
2
9. Determine the equation of each radical function, which has been transformed from f(x) = x by the given
translations.
a) vertical stretch by a factor of 5, then a horizontal translation of 6 units right
1
b) horizontal stretch by a factor of , then a vertical translation of 4 units down
6
c) horizontal reflection in the y-axis, then a vertical translation of 9 units up and horizontal translation of 2
units right
1
2
d) horizontal stretch by a factor of , vertical reflection in the x-axis, and vertical stretch by a factor of
3
3
10. Sketch the graph of f(x) =
11. Solve the equation
−2x + 3
and use it to sketch the graph of y =
6
f(x) .
3x − 6 = 12 graphically.
12. Jim states that the equations
Ê
x 2 = 25 and ÁÁÁ
Ë
ˆ2
x ˜˜˜ = 25 have the same solution. Is he correct? Justify your
¯
reasoning.
13. A student designs a special container as part of an egg drop experiment. She believes that the container can
withstand a fall as long as the speed of the container does not exceed 80 ft/s. She uses the equation
v=
(v 0 ) 2 + 2ad to model the velocity, v, in feet per second, as a function of constant acceleration, a, in
feet per second squared and the drop distance, d, in feet. Assuming the student’s specifications are correct,
will the egg break if the student drops the egg from shoulder height (5 ft) off a building 80 ft high? What is
the maximum height the egg can be dropped from? (Note: The acceleration due to gravity is 32 ft/s2.)
14. Solve the equation
x 3 − 4 = 2 graphically.
15. Factor fully.
a) x3 + 6x2 + 11x + 6
b) 4x3 – 11x2 – 3x
c) x4 – 81
16. Factor fully.
a) x2(x – 2)(x + 2) + 3x + 6
b) 16x4 – (x + 1)2
42
Name: ________________________
ID: A
17. Factor 2x3 + 5x2 – 14x – 8 fully
18. Solve.
a) 3x3 + 2x2 – 8x + 3 = 0
b) 2x3 + x2 – 10x – 5 = 0
c) 5x4 = 7x2 – 2
19. Solve by factoring.
a) x4 + 3x2 – 28 = 0
b) 2x4 – 54x = 0
20. Solve by graphing using technology. Round answers to one decimal place.
a) x3 – 7 > 0
b) (x + 14) 3 ≤ 1
21. A child swings on a playground swing set. If the length of the swing’s chain is 3 m and the child swings
π
through an angle of , what is the exact arc length through which the child travels?
9
22. A 3-m ladder is leaning against a vertical wall such that the angle between the ground and the ladder is
π
.
3
What is the exact height that the ladder reaches up the wall?
ÊÁ π ˆ˜
23. Given that sinx = cos ÁÁÁ ˜˜˜˜ and that x lies in the first quadrant, determine the exact measure of angle x.
ÁË 5 ¯
24. Without using a calculator, determine two angles between 0° and 360° that have a cosecant of −
2
.
3
Include an explanation of how you determined the two angles.
25. Given a circle of diameter 21 cm, determine the arc length subtended by a central angle of 1.2 radians.
26. Angles A and B are located in the first quadrant. If sin A =
2
3
and cos B =
, determine the exact value
2
2
of sec A + sec B.
27. Determine the exact measures for all angles where tanθ = −
3 in the domain −180° ≤ θ ≤ 180°.
28. A grandfather clock shows a time of 7 o’clock. What is the exact radian measure of the angle between the
hour hand and the minute hand?
29. Explain how you could graph the function y = cos x given a table of values containing ordered pairs for the
function y = sinx.
43
Name: ________________________
ID: A
30. Describe the transformations that, when applied to the graph of y = cos x, result in the graph of
ÈÍ Ê
˘
π ˆ˜˜˜ ˙˙˙˙
ÍÍÍ 1 ÁÁÁ
y = −2cos ÍÍ ÁÁ x − ˜˜ ˙˙ + 1.
ÍÍ 8 Ë
3 ¯ ˙˙
Î
˚
31. A pebble is embedded in the tread of a rotating bicycle wheel of diameter 60 cm. If the wheel rotates at 4
revolutions per second, determine a relationship between the height, h, in centimetres, of the pebble above
the ground as a function of time, t, in seconds.
32. A population, p, of bears varies according to p (t) = 250 + 30cos t, where t is the time, in years, and angles are
measured in radians.
a) What are the maximum and minimum populations?
b) What is the first interval, in years and months, over which the population is increasing?
33. A girl jumps rope such that the height, h, in metres, of the middle of the rope can be approximated by the
equation h = 0.7sin (72t + 9) + 0.75, where t is the time, in seconds.
a) What is the amplitude of this function?
b) How many revolutions of the rope does the girl make in 1 min?
34. Use a counterexample to show that cos(x + y) = cos x + cos y is not an identity.
35. What is the solution for 2cos x −
3 = 0 for 0 ≤ x ≤ 2π ?
36. Solve cot 2 θ + cot θ = 0. State the solution in general form.
37. Solve sec 2 θ − 2tanθ − 3 = 0. State the general solution to the nearest degree.
44
Name: ________________________
ID: A
38. a) Determine the type of function shown in each graph.
i)
ii)
iii)
b) Describe what you would expect to see in the first differences column of a table of values for each graph
in part a).
39. Sketch the graph of an exponential function with all of the following characteristics:
• domain {x| x ∈ R}
• range {y| y > 0, y ∈ R}
• y-intercept of 3
• no x-intercept
• the function is always decreasing
45
Name: ________________________
ID: A
1
x−2
(3) ,
2
a) describe the transformations of the function when compared to the function y = 3 x
b) sketch the graph of the given function and y = 3 x on the same set of axes
c) state the domain, the range, and the equation of the asymptote
40. For the function y =
41. Write the equation for the function that results from each transformation or set of transformations applied to
the base function y = 5 x .
a) reflect in the y-axis
b) shift 3 units to the right
c) shift 1 unit down and 4 units to the left
d) reflect in the x-axis and shift 2 units down
42. Match each exponential scatter plot with the corresponding equation of its curve of best fit.
a)
b)
c)
i) y = 2 (1.6)
x
ii) y = 40 (0.6)
x
iii) y = 10 (1.8)
x
4n − 1
ÁÊ 1 ˜ˆ
43. Solve for n: 9 n − 1 = ÁÁÁÁ ˜˜˜˜
Ë 3¯
44. Graph the function f(x) = −log(x + 2) − 1. Identify the domain, the range, and the equation of the vertical
asymptote.
46
Name: ________________________
ID: A
45. Given log 2 7 ≈ 2.8074, find the value of log 2 14.
46. Solve the equation 6 3x + 1 = 22x − 3 . Leave your answer in exact form.
47. a) Determine an equation in the form f(x) =
1
for a function with a vertical asymptote at x = 2 and a
kx − c
1
y-intercept of − .
8
b) Sketch the graph of the function.
47
Name: ________________________
ID: A
3
.
4x − 5
a) Determine the key features of the function:
i) domain and range
ii) intercepts
iii) equations of any asymptotes
b) Sketch the graph of the function.
48. Consider the function f(x) =
48
Name: ________________________
ID: A
3x + 8
.
x−2
a) Determine the key features of the function:
i) domain and range
ii) intercepts
iii) equations of any asymptotes
b) Sketch the graph of the function.
49. Consider the function f(x) =
x+3
.
x − x − 12
a) Determine the key features of the function:
i) domain and range
ii) intercepts
iii) equations of any asymptotes
b) Sketch the graph of the function.
50. Consider the function f(x) =
2
51. Given the functions f (x) = x + 1 and g (x) = x 2 + 3x + 1, determine a simplified equation for
h (x) = f (x) + g (x) .
52. Given the functions f (x) = x 2 − 4 and g (x) = x 2 − 3x + 2, determine a simplified equation for h (x) =
49
f (x )
.
g (x)
Name: ________________________
ID: A
ÁÊÁÁ 2
˜ˆ
ÁÁÁ x + 1 ˜˜˜˜˜
˜¯
ËÁ
.
53. a) Graph the functions f (x) = sin (x) and g (x) = 2
b) Use the graphs to graph the function h (x) = f (x)g (x) .
54. Determine the equation(s) of the vertical asymptote(s) of the function y =
0.9 x
.
2x 2 − 3
55. Given the functions f (x) = x 2 − 7 and g (x) = 2 − x 3 , what is the value of f(g(2))?
50
Name: ________________________
ID: A
56. Joe wants to travel from his home to school. The school is 6 blocks east and 6 blocks north. How many routes
can Joe take from his house to school if he only moves east and north.
ÊÁ a b ˆ˜ 4
57. Use the binomial theorem to expand ÁÁÁÁ − ˜˜˜˜ .
Ë2 3¯
58. Simplify the expression
(2n + 2)!
.
(2n − 2)!0!
59. A neon sign with the words “Espresso Coffee” on it has 5 letters burnt out. In how many ways can you select
3 good letters and 2 burnt-out letters?
60. A math teacher is preparing a quiz for all of the students in grade 12. She wants to give each student the same
questions, but have each student’s questions appear in a different order. If there are 128 students in the grade
12 class, what is the least number of questions the quiz must contain so everyone gets a test with the
questions in a different order.
Problem
1. An object falls to the ground from a height of 25 m. The height, h, in metres, of the object above the ground
1
can be modelled by the function h(t) = − at 2 + 25, where a is the acceleration due to gravity, in metres per
2
second squared, and t is the time, in seconds.
a) Write an equation for the height of the object on Earth given a = 9.8 m/s2.
b) Write an equation for the height of the object on Mars given a = 3.7 m/s2.
c) Graph both functions on the same set of axes.
d) What scale factor can be applied to the Earth function to transform it to the Mars function?
51
Name: ________________________
ID: A
2. The base function f(x) = x is reflected in the x-axis, stretched horizontally by a factor of 2, compressed
1
vertically by a factor of , and translated 3 units to the left and 5 units down.
3
a) Write the equation of the transformed function g(x).
b) Graph the original function and the transformed function on the same set of axes.
c) Which transformations must be done first but in any order?
d) Which transformations must be done last but in any order?
3. The cost of renting a car for a day is a flat fee of $50 plus $0.12 for each kilometre driven. Let C represent
the total cost of renting a car for a day if it is driven a distance, x, in kilometres.
a) Write the total cost function for the car rental.
b) Determine the inverse of this function.
c) What does this inverse function represent?
d) Give an example of how this function can be used.
4. For f(x) = 5 −x and g(x) = −2 6(x + 2) − 3, do the following.
a) Graph f(x) and g(x) on the same set of axes.
b) Determine the domain and range of each function.
c) Explain which transformations would need to be applied to the graph of f(x) to obtain the graph of g(x).
5. Two groups of students are conducting a lab to determine the relationship between the period, p, in seconds,
of a pendulum and the length, l, in metres, of the string. The curves of best fit from the experiment are shown
on the graph.
a) When asked the type of function that could be used to model their findings, both groups argue that a
radical function can be used. Do you agree with each group?
b) How do these graphs differ from the graph of f(x) = x ?
c) Write a function to approximate the graph for each group.
d) What may have caused the differences in the data between the two groups? Justify your answer in terms of
transformations.
52
Name: ________________________
ID: A
1 2
mv , where m
2
represents the mass of the object, in kilograms, and v represents its speed, in metres per second.
a) Determine the general equation for the velocity of a mass as a function of its kinetic energy.
b) Find the speed of an object of mass 12 kg moving with a kinetic energy of
i) 200 J
ii) 420 J
c) Graph the function if the mass is 12 kg.
d) John conducts an experiment and graphs the data, resulting in the graph below. What is the mass of the
object?
6. The kinetic energy (energy of motion), E, in joules, of an object is given by the equation E =
7. Factor 2x4 – 7x3 – 41x2 – 53x – 21 fully.
8. Show that x + a is a factor of the polynomial P(x) = (x + a)4 + (x + c)4 – (a – c)4.
9. The height of a square-based box is 4 cm more than the side length of its square base. If the volume of the
box is 225 cm3, what are its dimensions?
10. Solve −x 3 + 5x 2 − 8x + 4 ≥ 0 algebraically and graphically.
11. Determine an equation in expanded form for the polynomial function represented by the graph.
53
Name: ________________________
ID: A
12. Two billiard balls collide and then separate from one another at the same, constant speed. Assume the billiard
table is frictionless. The angle between the balls is 1.25 radians. After 2 s, the distance between the balls is 1
m. How fast are the balls moving, to the nearest hundredth of a metre per second?
13. To support a new 2.5-m wall in the construction of a home, the carpenters nail a piece of wood from the top
of the wall to the floor, with the piece of wood forming the hypotenuse of the right triangle it makes with the
wall and floor. The piece of wood is nailed to the ground such that it makes a 30° angle with the floor.
a) Represent this situation with a diagram.
b) Which trigonometric ratio can be used to determine the length of the piece of wood?
c) Determine the length of the piece of wood.
14. a) Without using a calculator, determine two angles between 0° and 360° that have a sine ratio of −
1
.
2
b) Use a calculator and a diagram to verify your answers to part a).
15. When a pendulum that is 0.5 m long swings back and forth, its angular displacement, θ, in radians, from rest
1 ÊÁ π ˆ˜
position is given by θ = sin ÁÁÁÁ t ˜˜˜˜ , where t is the time, in seconds. At what time(s) during the first 4 s is the
4 Ë2 ¯
pendulum displaced 1 cm vertically above its rest position? (Assume the pendulum is at its rest position at t =
0.)
16. The table shows the hours of daylight measured on the first day of each month, over a 1-year period in a
northern Ontario city.
Month
Hours of Daylight
(h:min)
1
8:25
2
9:55
3
11:35
4
13:30
5
15:48
6
16:15
7
15:25
8
14:26
9
12:35
10
10:39
11
9:01
12
8:00
a) Graph the table data.
b) Use the graph and the table to develop a sinusoidal model to represent the information.
c) Graph the model on the same set of axes as the data. Comment on the fit.
d) Use your model to estimate the number of hours of daylight, to the nearest tenth of an hour, on January 15,
and verify the solution using the graph.
54
Name: ________________________
ID: A
17. Wilson places a measuring tape on a pillar of a dock to record the water level in his local coastal community.
He finds that a high tide of 1.77 m occurs at 5:17 a.m., and a low tide of 0.21 m occurs at 11:38 a.m.
a) Estimate the period of the fluctuation of the water level.
b) Estimate the amplitude of the pattern.
c) Predict when the next two high tides will occur.
d) Predict when the next two low tides will occur.
18. The graph of y = cos x is transformed so that the amplitude becomes 2 and the x-intercepts coincide with the
maximum values.
a) What is the equation of the transformed function?
b) What phase shift of the transformed function will produce a y-intercept of –1?
c) What is the equation of the function after the transformation in part b)?
d) Verify your solution to part c) by graphing.
19. Prove the identity 1 + cos θ =
sin 2 θ
.
1 − cos θ
ÁÊ π
˜ˆ ÁÊ π
˜ˆ
20. Prove the identity sin ÁÁÁÁ − x ˜˜˜˜ cot ÁÁÁÁ + x ˜˜˜˜ = −sinx .
Ë2
¯ Ë2
¯
21. Prove the identity
1 − cos 2θ + sin2θ
= tanθ .
1 + cos 2θ + sin2θ
22. Prove the identity
cos 2 θ − sin 2 θ
= 1 − tanθ .
cos 2 θ + sinθ cos θ
23. An angle satisfies the relation (sec θ) (cot θ) = 1.
a) Use the definition of the reciprocal trigonometric ratios to express the left side of the relation in terms of
the sine and/or cosine ratios.
b) Determine the value(s) for the angle. Do not use a calculator.
c) Verify your answer to part b) using a calculator.
d) Show your answer to part b) using a unit circle.
24. Prove the identity
sin3θ + sinθ
= tan2θ.
cos 3θ + cos θ
25. Solve sin3x + sinx = cos x for 0 ≤ x ≤ 2π .
26. What is the general solution to tanx (csc x + 2) = 0?
27. A radioactive sample with an initial mass of 72 mg has a half-life of 10 days.
a) Write a function to relate the amount remaining, A, in milligrams, to the time, t, in days.
b) What amount of the radioactive sample will remain after 20 days?
c) What amount of the radioactive sample was there 30 days ago?
d) How long, to the nearest day, will it take for there to be 0.07 mg of the initial sample remaining?
55
Name: ________________________
ID: A
28. a) Rewrite the function y = 2 −2x + 4 + 6 in the form y = a(2) b(x − h) + k .
b) Describe the transformations that must be applied to the graph of y = 2 x to obtain the graph of the given
function.
c) Graph the function.
d) Determine the equation of the function that results after the graph in part c) is reflected in the x-axis.
e) Graph the function from part d).
29. Solve the equation
3
256 2 × 16x = 64 x − 3 .
30. Solve the equation 2 3x = 4.
31. A $21 500 investment earns 5.25% interest, compounded quarterly.
a) Determine the value of the investment in 5 years.
b) How long will it take the original investment to double in value?
32. A chemist has a 20-mg sample of polonium-218. He needs approximately 81.5% of it for an experiment.
Given that the half-life of polonium-218 is approximately 3.1 min, how many seconds will it take for the
sample to decay to the desired mass?
33. An investment offers a bonus of 2% of the principal after being invested for 5 years. If $50 000 is invested at
4.75%, compounded annually, for 10 years, describe how the graph of the investment with the bonus differs
from the graph of the investment without the bonus.
34. Given log7 ≈ 0.8451 and log2 ≈ 0.3010, find the value of log28.
ÊÁ b ˆ˜
Á 1˜
35. The stellar magnitude scale compares the brightness of stars using the equation m2 − m1 = log ÁÁÁÁ ˜˜˜˜ , where
Á b2 ˜
Ë ¯
m1 and m2 are the apparent magnitudes (how bright the stars appear in the sky) of the two stars being
compared, and b 1 and b 2 are their brightness (how much light they emit).
a) The brightest appearing star in our sky, Sirius, has an apparent magnitude of −1.5. How much brighter
does Sirius appear than Betelgeuse, whose apparent magnitude is 0.12? Round your answer to the nearest
whole number.
b) The Sun appears about 1.3×10 10 times as bright in the sky as does Sirius. What is the apparent magnitude
of the Sun, to the nearest tenth?
36. Prove that log a + log a 2 + log a 3 − log a 6 = log1.
37. Show that
1
log a b
= log b a .
38. Prove that log q 5 p 5 = log q p .
56
Name: ________________________
ID: A
39. a) Graph the function y = log 3 x.
b) Graph the following functions on the same graph:
y = log 3 3x
y = log 3 9x
y = log 3 27x
c) Explain the effect of the constant k in the function y = log 3 kx.
40. For his dream car, Bruce invested $18 000 at 7.8% interest, compounded monthly, for 5 years. After the 5
years, he still did not have enough money. How much longer will he have to invest the money at 5% interest,
compounded daily, to have a total of $35 000? Round to the nearest tenth of a year.
ÈÍ
˘˙
Í1
˙
41. Sketch the graph of the function f(x) = −2 log ÍÍÍÍ (x + 1) ˙˙˙˙ − 1. Determine the x-intercept algebraically, to the
ÍÎ 2
˙˚
nearest hundredth.
42. The time, t, in hours, that it takes Alistair to jog 5 km is inversely proportional to his average speed, v, in
kilometres per hour.
a) Write a function to represent the time as a function of the speed.
b) Sketch the graph of this function.
c) If Alistair jogs at 4.5 km/h, how long does it take him to complete a 5-km run, to the nearest minute?
43. The pressure exerted on the floor by the heel of someone’s shoe is inversely proportional to the square of the
width of the heel of the shoe. When Megumi wears 2-cm-wide heels, she exerts a pressure of 400 kPa.
a) Determine a function to represent the pressure, p, exerted by Megumi if she wears heels of width w.
b) Sketch the graph of this function.
c) If she wears spike heels with a width of 0.5 cm, what pressure does she exert?
44. A photographer uses a light meter to measure the intensity of light from a flash bulb. The intensity, I, in lux,
of the flash bulb is a function of the distance, d, in metres, from the light and can be represented by
10
I(d) = 2 , d > 0.
d
a) Determine the following, to two decimal places:
i) the intensity of light 3 m from the flash bulb
ii) the average rate of change in the intensity of light for the interval 1 < d < 3
b) What does the sign of your answer to part a)ii) indicate about the light intensity?
57
Name: ________________________
ID: A
45. Write an equation for the graph of the rational function shown. Explain your reasoning.
46. Write an equation for a rational function whose graph has all of the following features:
• vertical asymptote with equation x = 3
• horizontal asymptote with equation y = 2
• hole at x = 1
• no x-intercepts
47. a) Use the asymptotes and intercepts to make a quick sketch of the function f(x) =
x+1
and its reciprocal
x−5
x−5
on the same set of axes.
x+1
b) Describe the symmetry in the graphs in part a).
c) Determine the equation of the mirror line in your graph from part a).
d) Determine intervals where f is positive and where f is negative. Determine intervals where g is positive
and where g is negative. How do the two sets of intervals compare to each other?
x+b
x+d
e) Does the pattern from part d) occur for all pairs of functions f(x) =
and g(x) =
, b ≠ d ? Explain
x+d
x+b
why or why not.
g(x) =
48. An airplane makes a 990-mi flight with a tail wind and returns, flying into the wind. The total flying time is 3
h 20 min, and the plane’s airspeed is 600 mph. What is the wind speed?
58
Name: ________________________
ID: A
49. A ski club charters a bus for a ski trip at a cost of $480. In an attempt to lower the bus fare per skier, the club
invites non-members to go along. After five non-members join the trip, the fare per skier decreases by $4.80.
How many club members are going on the trip?
50. Given the functions f (x) =
1
and g (x) = sin x, determine the equation for h (x) = f(g(x)).
1 − x2
51. Given the functions f (x) = x 2 − 3x − 10 and g (x) = x 2 − 5x, graph the function h (x) =
intercepts and identify any asymptotes and/or points of discontinuity.
52. What are the domain and range of the function y = 2sin x , where x is in radians?
59
f (x )
. Label all
g (x)
Name: ________________________
ID: A
53. Use the graph of the combined function f (x) = 2 x − x 2 to determine an approximate solution to the inequality
2x > x 2 .
54. Jenny and Jimmy are a married couple who work at the same store. Jimmy’s total weekly salary, in dollars, if
he sells x items is given by S (x) = 10.0 + 5x , and Jenny’s total weekly salary, in dollars, if she sells x items is
given by S (x) = 8.0 + 6x .
a) Assuming that they sell the same number of items in a week, what is the minimum number of items they
have to sell so that Jenny’s weekly salary is at least $100 more than Jimmy’s?
b) Assuming that they sell the same number of items in a week, what is the minimum number of items they
each need to sell to make their combined weekly salary greater than $1000?
60
Name: ________________________
ID: A
55. The heights, h, of two balls, in metres, for a horizontal distance of x metres are shown in the graph.
What was the difference in height of the two balls when the horizontal distance was 0 m?
56. The dimensions of a window are shown.
a) What function in simplest form represents the area of the entire window?
b) If the width, x, of the window is 1.2 m, what is the total area of the window, to the nearest tenth of a
square metre?
7!
.
3!2!
a) What is the smallest number that can be created that meets these conditions? Explain your reasoning.
b) What is the difference between the largest number and the smallest number? Explain your reasoning.
57. The number of different permutations using all of the 1-digit numbers of a set is given by
61
Name: ________________________
ID: A
58. To win the grand prize in lottery A, a player must select all six of the winning numbers drawn from the
numbers 1 to 49. To win in lottery B, a player must select all seven of the winning numbers drawn from 1 to
49. Bernadette argues that the chances of randomly selecting the winning number for lottery A are seven
times as good as winning for lottery B. Create an argument to agree or disagree with this statement.
59. Tanya goes to a fast food stand at the beach. There are 4 types of burgers, 3 sizes of French fries, and either
orange pop or root beer to drink.
a) Create a tree diagram to show the possible choices of lunch if one of each item can be selected.
b) In how many ways can Tanya buy 2 burgers, 2 fries, and 2 drinks for her and her friend?
c) If Tanya does not like root beer and her friend does not like orange pop, how many possible choices are
there?
62
Name: ________________________
ID: A
60. On a Saturday, Charlie has to go to the library to study for a few hours, and then to the school to play a
volleyball game.
a) How many routes are there for Charlie to go from home to the library if she only moves south and east?
b) How many routes are there for her to go from home to school moving only south and east?
c) Assuming Charlie moves south and east going from home to school and north and west going from school
to home, how many routes are there for her to complete the round trip?
d) If Charlie could walk each route from home to school in 40 min, how long would it take her and her 22
classmates to walk all of the routes? Consider only the route from home to school, not the round trip.
61. Prove the identity
1 + tanθ 1 − tanθ
=
.
1 + cot θ cot θ − 1
62. Solve the equation log
3
2
x 2 + 48x = .
3
63
ID: A
Final Exam Practice
Answer Section
MULTIPLE CHOICE
1. ANS:
NAT:
KEY:
2. ANS:
NAT:
KEY:
3. ANS:
NAT:
KEY:
4. ANS:
NAT:
5. ANS:
NAT:
6. ANS:
NAT:
7. ANS:
NAT:
KEY:
8. ANS:
NAT:
KEY:
9. ANS:
NAT:
KEY:
10. ANS:
NAT:
KEY:
11. ANS:
NAT:
KEY:
12. ANS:
NAT:
KEY:
13. ANS:
NAT:
KEY:
14. ANS:
NAT:
15. ANS:
NAT:
C
PTS: 1
DIF: Average
OBJ: Section 1.1
RF2
TOP: Horizontal and Vertical Translations
vertical translation
D
PTS: 1
DIF: Easy
OBJ: Section 1.1
RF2
TOP: Horizontal and Vertical Translations
horizontal translation
B
PTS: 1
DIF: Average
OBJ: Section 1.1
RF2
TOP: Horizontal and Vertical Translations
horizontal translation | vertical translation
D
PTS: 1
DIF: Average
OBJ: Section 1.2
RF5
TOP: Reflections and Stretches
KEY: reflection
A
PTS: 1
DIF: Easy
OBJ: Section 1.1
RF5
TOP: Reflections and Stretches
KEY: reflection
A
PTS: 1
DIF: Average
OBJ: Section 1.2
RF3 | RF5
TOP: Reflections and Stretches
KEY: graph | vertical stretch | reflection
C
PTS: 1
DIF: Easy
OBJ: Section 1.3
RF4
TOP: Combining Transformations
graph | horizontal translation | vertical translation
C
PTS: 1
DIF: Average
OBJ: Section 1.3
RF4
TOP: Combining Transformations
graph | vertical translation | reflection
D
PTS: 1
DIF: Difficult
OBJ: Section 1.3
RF4 | RF5
TOP: Combining Transformations
graph | vertical translation | horizontal translation | stretch | reflection
C
PTS: 1
DIF: Average
OBJ: Section 1.3
RF4 | RF5
TOP: Combining Transformations
graph | vertical translation | horizontal translation | stretch | reflection
A
PTS: 1
DIF: Easy
OBJ: Section 1.4
RF6
TOP: Inverse of a Relation
inverse of a function | function notation
A
PTS: 1
DIF: Average
OBJ: Section 1.4
RF6
TOP: Inverse of a Relation
inverse of a function | function notation
C
PTS: 1
DIF: Average
OBJ: Section 1.4
RF6
TOP: Inverse of a Relation
inverse of a function | function notation
D
PTS: 1
DIF: Average
OBJ: Section 1.4
RF6
TOP: Inverse of a Relation
KEY: graph | inverse of a function
D
PTS: 1
DIF: Easy
OBJ: Section 1.4
RF6
TOP: Inverse of a Relation
KEY: graph | inverse of a function
1
ID: A
16. ANS:
NAT:
KEY:
17. ANS:
NAT:
KEY:
18. ANS:
NAT:
KEY:
19. ANS:
NAT:
KEY:
20. ANS:
NAT:
KEY:
21. ANS:
NAT:
KEY:
22. ANS:
NAT:
23. ANS:
NAT:
KEY:
24. ANS:
NAT:
25. ANS:
NAT:
KEY:
26. ANS:
NAT:
KEY:
27. ANS:
NAT:
KEY:
28. ANS:
NAT:
KEY:
29. ANS:
NAT:
KEY:
30. ANS:
NAT:
KEY:
31. ANS:
NAT:
32. ANS:
NAT:
D
PTS: 1
DIF: Easy
OBJ: Section 2.1
RF13
TOP: Radical Functions and Transformations
vertical translation
D
PTS: 1
DIF: Average
OBJ: Section 2.1
RF13
TOP: Radical Functions and Transformations
vertical translation
C
PTS: 1
DIF: Easy
OBJ: Section 2.1
RF13
TOP: Radical Functions and Transformations
horizontal translation
B
PTS: 1
DIF: Average
OBJ: Section 2.1
RF13
TOP: Radical Functions and Transformations
horizontal stretch | reflection
D
PTS: 1
DIF: Average
OBJ: Section 2.1
RF13
TOP: Radical Functions and Transformations
graph | horizontal translation | vertical translation | reflection
A
PTS: 1
DIF: Average
OBJ: Section 2.1
RF13
TOP: Radical Functions and Transformations
reflection
D
PTS: 1
DIF: Average
OBJ: Section 2.2
RF13
TOP: Square Root of a Function
KEY: graph
C
PTS: 1
DIF: Easy
OBJ: Section 2.1 | Section 2.2
RF13
TOP: Radical Functions and Transformations | Square Root of a Function
inverse of a radical function
D
PTS: 1
DIF: Average
OBJ: Section 2.2
RF13
TOP: Square Root of a Function
KEY: graph | square root | of a function
A
PTS: 1
DIF: Average
OBJ: Section 2.3
RF13
TOP: Solving Radical Equations Graphically
graphical solution
B
PTS: 1
DIF: Easy
OBJ: Section 2.3
RF13
TOP: Solving Radical Equations Graphically
graphical solution
D
PTS: 1
DIF: Average
OBJ: Section 2.3
RF13
TOP: Solving Radical Equations Graphically
algebraic solution
C
PTS: 1
DIF: Difficult
OBJ: Section 2.3
RF13
TOP: Solving Radical Equations Graphically
algebraic solution
A
PTS: 1
DIF: Easy
OBJ: Section 3.1
RF12
TOP: Characteristics of Polynomial Functions
polynomial function
B
PTS: 1
DIF: Average
OBJ: Section 3.1
RF12
TOP: Characteristics of Polynomial Functions
odd-degree | x-intercepts
C
PTS: 1
DIF: Average
OBJ: Section 3.2
RF12
TOP: The Remainder Theorem
KEY: restriction
B
PTS: 1
DIF: Average
OBJ: Section 3.2
RF11
TOP: The Remainder Theorem
KEY: quotient | remainder
2
ID: A
33. ANS:
NAT:
34. ANS:
NAT:
35. ANS:
NAT:
KEY:
36. ANS:
NAT:
37. ANS:
NAT:
KEY:
38. ANS:
NAT:
KEY:
39. ANS:
NAT:
KEY:
40. ANS:
NAT:
KEY:
41. ANS:
NAT:
KEY:
42. ANS:
NAT:
43. ANS:
NAT:
NOT:
44. ANS:
NAT:
45. ANS:
NAT:
KEY:
46. ANS:
NAT:
47. ANS:
NAT:
48. ANS:
NAT:
KEY:
49. ANS:
NAT:
KEY:
50. ANS:
NAT:
KEY:
D
PTS: 1
DIF: Difficult +
OBJ: Section 3.2
RF11
TOP: The Remainder Theorem
KEY: remainder theorem | remainder
D
PTS: 1
DIF: Easy
OBJ: Section 3.3
RF11
TOP: The Factor Theorem
KEY: factor theorem | factor
D
PTS: 1
DIF: Average
OBJ: Section 3.3
RF11
TOP: The Factor Theorem
factor theorem | integral zero theorem | factor
C
PTS: 1
DIF: Average
OBJ: Section 3.3
RF11
TOP: The Factor Theorem
KEY: factor theorem | factor
B
PTS: 1
DIF: Easy
OBJ: Section 3.3
RF11
TOP: The Factor Theorem
factored form | factor theorem | factor
D
PTS: 1
DIF: Average
OBJ: Section 3.3
RF11
TOP: The Factor Theorem
factored form | factor theorem | factor
A
PTS: 1
DIF: Average
OBJ: Section 3.4
RF12
TOP: Equations and Graphs of Polynomial Functions
polynomial equation | roots
B
PTS: 1
DIF: Average
OBJ: Section 3.4
RF12
TOP: Equations and Graphs of Polynomial Functions
polynomial equation | roots | graph
C
PTS: 1
DIF: Average
OBJ: Section 3.4
RF12
TOP: Equations and Graphs of Polynomial Functions
polynomial equation | zeros | graph | multiplicity
A
PTS: 1
DIF: Difficult +
OBJ: Section 4.2
T2
TOP: Unit Circle
KEY: unit circle | unit circle equation
B
PTS: 1
DIF: Average
OBJ: Section 4.1
T1
TOP: Angles and Angle Measure
KEY: rotations | standard position
Mixed numbers
C
PTS: 1
DIF: Average
OBJ: Section 4.1
T1
TOP: Angles and Angle Measure
KEY: rotations | standard position
C
PTS: 1
DIF: Easy
OBJ: Section 4.4
T4
TOP: Introduction to Trigonometric Equations
trigonometric ratios
C
PTS: 1
DIF: Difficult +
OBJ: Section 4.1
T1
TOP: Angles and Angle Measure
KEY: arc length | degrees
A
PTS: 1
DIF: Difficult
OBJ: Section 4.3
T3
TOP: Trigonometric Ratios
KEY: unit circle | trigonometric ratios
A
PTS: 1
DIF: Easy
OBJ: Section 5.1
T4
TOP: Graphing Sine and Cosine Functions
graph | sinusoidal function
C
PTS: 1
DIF: Easy
OBJ: Section 5.1
T4
TOP: Graphing Sine and Cosine Functions
period | sinusoidal function
D
PTS: 1
DIF: Average
OBJ: Section 5.1
T4
TOP: Graphing Sine and Cosine Functions
function | amplitude | period | sinusoidal function
3
ID: A
51. ANS:
NAT:
KEY:
52. ANS:
NAT:
KEY:
53. ANS:
NAT:
54. ANS:
NAT:
55. ANS:
NAT:
56. ANS:
NAT:
57. ANS:
NAT:
KEY:
58. ANS:
NAT:
KEY:
59. ANS:
NAT:
KEY:
60. ANS:
NAT:
KEY:
61. ANS:
NAT:
KEY:
62. ANS:
NAT:
KEY:
63. ANS:
NAT:
KEY:
64. ANS:
NAT:
KEY:
65. ANS:
NAT:
KEY:
66. ANS:
NAT:
KEY:
67. ANS:
NAT:
KEY:
B
PTS: 1
DIF: Easy
OBJ: Section 5.2
T4
TOP: Transformations of Sinusoidal Functions
translation | primary trigonometric function
C
PTS: 1
DIF: Average
OBJ: Section 5.2
T4
TOP: Transformations of Sinusoidal Functions
period | sinusoidal function
D
PTS: 1
DIF: Easy
OBJ: Section 5.3
T4
TOP: The Tangent Function
KEY: asymptote | tangent function
B
PTS: 1
DIF: Difficult +
OBJ: Section 5.3
T4
TOP: The Tangent Function
KEY: zeros | transformation
A
PTS: 1
DIF: Average
OBJ: Section 5.3
T4
TOP: The Tangent Function
KEY: undefined | tangent function
A
PTS: 1
DIF: Average
OBJ: Section 5.3
T4
TOP: The Tangent Function
KEY: coordinate | tangent function
A
PTS: 1
DIF: Difficult +
OBJ: Section 5.4
T4
TOP: Equations and Graphs of Trigonometric Functions
quadratic trigonometric equation
B
PTS: 1
DIF: Easy
OBJ: Section 5.4
T4
TOP: Equations and Graphs of Trigonometric Functions
amplitude | sinusoidal function
C
PTS: 1
DIF: Average
OBJ: Section 5.4
T4
TOP: Equations and Graphs of Trigonometric Functions
period | sinusoidal function
B
PTS: 1
DIF: Average
OBJ: Section 5.4
T4
TOP: Equations and Graphs of Trigonometric Functions
minimum | sinusoidal function
D
PTS: 1
DIF: Average
OBJ: Section 5.4
T4
TOP: Equations and Graphs of Trigonometric Functions
maximum | sinusoidal function
C
PTS: 1
DIF: Easy
OBJ: Section 5.4
T4
TOP: Equations and Graphs of Trigonometric Functions
period | sinusoidal function
C
PTS: 1
DIF: Average
OBJ: Section 6.1
T6
TOP: Reciprocal, Quotient, and Pythagorean Identities
trigonometric identity
C
PTS: 1
DIF: Average
OBJ: Section 6.1
T6
TOP: Reciprocal, Quotient, and Pythagorean Identities
trigonometric identity
C
PTS: 1
DIF: Average
OBJ: Section 6.2
T6
TOP: Sum, Difference, and Double-Angle Identities
tangent | sum identities | difference identities
D
PTS: 1
DIF: Average
OBJ: Section 6.2
T6
TOP: Sum, Difference, and Double-Angle Identities
sum identities | difference identities | evaluate
C
PTS: 1
DIF: Difficult
OBJ: Section 6.4
T6
TOP: Solving Trigonometric Equations Using Identities
double-angle identities | general solutions
4
ID: A
68. ANS:
NAT:
KEY:
69. ANS:
NAT:
KEY:
70. ANS:
NAT:
KEY:
71. ANS:
NAT:
KEY:
72. ANS:
NAT:
KEY:
73. ANS:
NAT:
KEY:
74. ANS:
NAT:
75. ANS:
NAT:
KEY:
76. ANS:
NAT:
KEY:
77. ANS:
NAT:
KEY:
78. ANS:
NAT:
KEY:
79. ANS:
NAT:
KEY:
80. ANS:
NAT:
KEY:
81. ANS:
NAT:
82. ANS:
NAT:
NOT:
83. ANS:
NAT:
NOT:
C
PTS: 1
DIF: Difficult
OBJ: Section 6.4
T6
TOP: Solving Trigonometric Equations Using Identities
double-angle identities | general solutions
C
PTS: 1
DIF: Easy
OBJ: Section 7.1
RF9
TOP: Characteristics of Exponential Functions
intercepts | exponential function
C
PTS: 1
DIF: Easy
OBJ: Section 7.1
RF9
TOP: Characteristics of Exponential Functions
increasing | decreasing
A
PTS: 1
DIF: Average
OBJ: Section 7.1
RF9
TOP: Characteristics of Exponential Functions
domain | range
A
PTS: 1
DIF: Average
OBJ: Section 7.1
RF9
TOP: Characteristics of Exponential Functions
equation | graph | exponential function
C
PTS: 1
DIF: Average
OBJ: Section 7.2
RF9
TOP: Transformations of Exponential Functions
modelling | exponential growth
D
PTS: 1
DIF: Easy
OBJ: Section 7.3
RF10
TOP: Solving Exponential Equations
KEY: compound interest
C
PTS: 1
DIF: Easy
OBJ: Section 7.2
RF9
TOP: Transformations of Exponential Functions
transformations of exponential functions
B
PTS: 1
DIF: Easy
OBJ: Section 7.2
RF9
TOP: Transformations of Exponential Functions
transformations of exponential functions
A
PTS: 1
DIF: Average
OBJ: Section 7.2
RF9
TOP: Transformations of Exponential Functions
graph | transformations of exponential functions
A
PTS: 1
DIF: Difficult
OBJ: Section 7.1
RF9
TOP: Characteristics of Exponential Functions
modelling | exponential function
D
PTS: 1
DIF: Average
OBJ: Section 7.3
RF10
TOP: Solving Exponential Equations
exponential equation | systematic trial
B
PTS: 1
DIF: Average
OBJ: Section 7.3
RF10
TOP: Solving Exponential Equations
exponential equation | equate exponents
D
PTS: 1
DIF: Difficult
OBJ: Section 7.3
RF10
TOP: Solving Exponential Equations
KEY: half-life | exponential decay
B
PTS: 1
DIF: Easy
OBJ: Section 8.1
RF7
TOP: Understanding Logarithms
KEY: logarithm | exponential function
Draft
C
PTS: 1
DIF: Easy
OBJ: Section 8.1
RF7
TOP: Understanding Logarithms
KEY: logarithm | exponential function
Draft
5
ID: A
84. ANS:
NAT:
KEY:
85. ANS:
NAT:
KEY:
86. ANS:
NAT:
87. ANS:
NAT:
KEY:
88. ANS:
NAT:
KEY:
89. ANS:
NAT:
KEY:
90. ANS:
NAT:
KEY:
91. ANS:
NAT:
KEY:
C
PTS: 1
DIF: Easy
OBJ: Section 8.2
RF8
TOP: Transformations of Logarithmic Functions
vertical translation | transformation
D
PTS: 1
DIF: Average
OBJ: Section 8.2
RF8
TOP: Transformations of Logarithmic Functions
horizontal translation | vertical translation | vertical stretch | horizontal stretch
A
PTS: 1
DIF: Easy
OBJ: Section 8.3
RF9
TOP: Laws of Logarithms
KEY: product law | laws of logarithms
B
PTS: 1
DIF: Easy
OBJ: Section 8.4
RF10
TOP: Logarithmic and Exponential Equations
exponential equation
B
PTS: 1
DIF: Average
OBJ: Section 9.1
RF14
TOP: Exploring Rational Functions Using Transformations
reciprocal of linear function | behaviour at non-permissible values
A
PTS: 1
DIF: Easy
OBJ: Section 9.1
RF14
TOP: Exploring Rational Functions Using Transformations
reciprocal of linear function | x-intercept
B
PTS: 1
DIF: Average
OBJ: Section 9.1
RF14
TOP: Exploring Rational Functions Using Transformations
reciprocal of linear function | graph from function
A
PTS: 1
DIF: Average
OBJ: Section 9.1
RF14
TOP: Exploring Rational Functions Using Transformations
reciprocal of linear function | function from graph
6
ID: A
92. ANS: A
93.
94.
95.
96.
97.
PTS:
TOP:
ANS:
NAT:
KEY:
ANS:
NAT:
KEY:
ANS:
NAT:
KEY:
ANS:
NAT:
KEY:
ANS:
NAT:
KEY:
1
DIF: Difficult +
OBJ: Section 9.1 NAT: RF14
Exploring Rational Functions Using Transformations
KEY: slant asymptote | hole | factor
B
PTS: 1
DIF: Average
OBJ: Section 9.2
RF14
TOP: Analysing Rational Functions
reciprocal of quadratic function | vertical asymptote
A
PTS: 1
DIF: Average
OBJ: Section 9.2
RF14
TOP: Analysing Rational Functions
reciprocal of quadratic function | vertical asymptote
C
PTS: 1
DIF: Difficult
OBJ: Section 9.2
RF14
TOP: Analysing Rational Functions
rational function | discontinuity | hole
C
PTS: 1
DIF: Average
OBJ: Section 9.2
RF14
TOP: Analysing Rational Functions
reciprocal of quadratic function | y-intercept
D
PTS: 1
DIF: Average
OBJ: Section 9.2
RF14
TOP: Analysing Rational Functions
linear expressions in numerator and denominator | horizontal asymptote
7
ID: A
98. ANS:
NAT:
KEY:
99. ANS:
NAT:
KEY:
100. ANS:
NAT:
101. ANS:
NAT:
KEY:
102. ANS:
C
PTS: 1
DIF: Average
OBJ:
RF14
TOP: Analysing Rational Functions
linear expressions in numerator and denominator | x-intercept
B
PTS: 1
DIF: Easy
OBJ:
RF14
TOP: Analysing Rational Functions
quadratic denominator | vertical asymptote
B
PTS: 1
DIF: Average
OBJ:
RF14
TOP: Analysing Rational Functions
KEY:
B
PTS: 1
DIF: Average
OBJ:
RF14
TOP: Connecting Graphs and Rational Equations
rational function | x-intercept
B
Section 9.2
PTS:
TOP:
ANS:
NAT:
KEY:
ANS:
NAT:
KEY:
ANS:
NAT:
KEY:
ANS:
NAT:
KEY:
ANS:
NAT:
KEY:
1
DIF: Difficult +
OBJ: Section 9.3 NAT:
Connecting Graphs and Rational Equations
KEY:
D
PTS: 1
DIF: Easy
OBJ:
RF1
TOP: Sums and Differences of Functions
add functions | subtract functions
B
PTS: 1
DIF: Easy
OBJ:
RF1
TOP: Sums and Differences of Functions
subtract functions | add functions
A
PTS: 1
DIF: Difficult
OBJ:
RF1
TOP: Sums and Differences of Functions
add functions | graph | subtract functions
C
PTS: 1
DIF: Easy
OBJ:
RF1
TOP: Products and Quotients of Functions
multiply functions | range
A
PTS: 1
DIF: Difficult
OBJ:
RF1
TOP: Composite Functions
composite functions | transformations | graph
RF14
rational equation | graph
Section 10.1
103.
104.
105.
106.
107.
8
Section 9.2
Section 9.2
hole
Section 9.3
Section 10.1
Section 10.1
Section 10.2
Section 10.3
ID: A
108. ANS:
NAT:
KEY:
109. ANS:
NAT:
KEY:
110. ANS:
NAT:
KEY:
111. ANS:
NAT:
KEY:
112. ANS:
NAT:
113. ANS:
NAT:
KEY:
114. ANS:
NAT:
KEY:
115. ANS:
NAT:
KEY:
116. ANS:
NAT:
117. ANS:
NAT:
118. ANS:
NAT:
119. ANS:
NAT:
KEY:
120. ANS:
NAT:
KEY:
B
PTS: 1
DIF: Difficult
OBJ:
RF1
TOP: Products and Quotients of Functions
multiply functions | graph
C
PTS: 1
DIF: Average
OBJ:
RF1
TOP: Products and Quotients of Functions
divide functions | graph
A
PTS: 1
DIF: Average
OBJ:
RF1
TOP: Sums and Differences of Functions
add functions | graph
D
PTS: 1
DIF: Difficult
OBJ:
RF1
TOP: Products and Quotients of Functions
divide functions | graph
C
PTS: 1
DIF: Average
OBJ:
RF1
TOP: Composite Functions
KEY:
D
PTS: 1
DIF: Average
OBJ:
RF1
TOP: Sums and Differences of Functions
add functions
D
PTS: 1
DIF: Average
OBJ:
RF1
TOP: Composite Functions
composite functions | transformations | graph
C
PTS: 1
DIF: Average
OBJ:
RF1
TOP: Composite Functions
composite functions | transformations | graph
B
PTS: 1
DIF: Easy
OBJ:
PC2
TOP: Permutations
KEY:
D
PTS: 1
DIF: Average
OBJ:
PC2
TOP: Permutations
KEY:
B
PTS: 1
DIF: Difficult
OBJ:
PC3
TOP: Combinations
KEY:
C
PTS: 1
DIF: Average
OBJ:
PC4
TOP: The Binomial Theorem
binomial expansion | binomial theorem
A
PTS: 1
DIF: Difficult
OBJ:
PC3
TOP: Combinations
combinations | fundamental counting principle
9
Section 10.2
Section 10.2
Section 10.1
Section 10.2
Section 10.2
composite functions | notation
Section 10.1
Section 10.3
Section 10.3
Section 11.1
permutations
Section 11.1
fundamental counting principle
Section 11.2
combinations
Section 11.3
Section 11.2
ID: A
SHORT ANSWER
1. ANS:
2
a) The graph of f(x) = x 2 is shown in blue and the graph of g(x) = (x − 1) + 2 is shown in red.
b) The graph of f(x) = |x| is shown in blue and the graph of g(x) = |x − 1| + 2 is shown in red.
PTS: 1
DIF: Average
TOP: Combining Transformations
2. ANS:
2
a) g(x) = (x − 4) + 3
OBJ: Section 1.3
KEY: translation
NAT: RF4
OBJ: Section 1.3
KEY: translation
NAT: RF4
= x 2 − 8x + 16 + 3
= x 2 − 8x + 19
b) g(x) = (x + 2) + 1
2
= x 2 + 4x + 4 + 1
= x 2 + 4x + 5
c) g(x) = (x + 7) − 2
2
= x 2 + 14x + 49 − 2
= x 2 + 14x + 47
PTS: 1
DIF: Average
TOP: Combining Transformations
10
ID: A
3. ANS:
a)
b)
c)
PTS: 1
DIF: Difficult +
TOP: Reflections and Stretches
4. ANS:
a) g (x) = −f(x)
OBJ: Section 1.2 NAT: RF3 | RF5
KEY: graph | reflection
= − (x − 1) − 2
b) g(x) = f (−x)
2
= |−x| + 1
= |x | + 1
PTS: 1
DIF: Average
TOP: Combining Transformations
OBJ: Section 1.3 NAT: RF4 | RF5
KEY: reflection | translation
11
ID: A
5. ANS:
a) i) The graph of f(x) = x is shown in blue and the graph of g(x) = 2(2x) is shown in red.
ii) The graph of f(x) = x 2 is shown in blue and the graph of g(x) = 2(2x)2 is shown in red.
iii) The graph of f(x) = |x| is shown in blue and the graph of g(x) = 2 |2x| is shown in red.
b) i) g(x) = 4x (vertical stretch by a factor of 4)
ii) g(x) = 8x 2 (vertical stretch by a factor of 8)
iii) g(x) = 4 |x| (vertical stretch by a factor of 4)
c) The stretches do not affect the domain or range of any of the functions.
PTS: 1
DIF: Average
TOP: Reflections and Stretches
OBJ: Section 1.2 NAT: RF3
KEY: stretch | graph
12
ID: A
6. ANS:
a) a reflection in the x-axis, a horizontal compression by a factor of
1
, and then a translation of 1 unit to the
2
left and 2 units down
b) a vertical stretch by a factor of 2, and then a translation of 3 units to the right and 4 units down
1
c) reflections in the x-axis and the y-axis, a vertical compression by a factor of , and then a translation of 5
2
units to the right and 1 unit up
PTS: 1
DIF: Difficult +
OBJ: Section 1.3 NAT: RF4 | RF5
TOP: Combining Transformations
KEY: translation | stretch | reflection
7. ANS:
a) a reflection in the x-axis, and then a translation of 2 units to the right and 3 units up
b) a vertical stretch by a factor of 2, and then a translation of 1 unit to the left and 1 unit down
1
c) a vertical compression by a factor of , a reflection in the x-axis, and then a translation of 2 units to the
2
right and 2 units up
PTS: 1
DIF: Average
TOP: Combining Transformations
OBJ: Section 1.3 NAT: RF4
KEY: graph | transformation
13
ID: A
8. ANS:
a) i)
y=
5
x−3
2
x=
5
y−3
2
x+3=
5
y
2
2x + 6 = 5y
y=
2
6
x+
5
5
2
6
x+
5
5
ii) The graph of f(x) is shown in blue and the graph of f −1 (x) is shown in red.
f −1 (x) =
y = 3(x − 2) 2 − 3
b) i)
x = 3(y − 2) 2 − 3
x + 3 = 3(y − 2) 2
x+3
= (y − 2) 2
3
±
x+3
= y−2
3
y = 2±
x+3
3
x+3
3
ii) The graph of f(x) is shown in blue and the graph of f −1 (x) is shown in red.
f −1 (x) = 2 ±
14
ID: A
PTS: 1
DIF: Difficult
OBJ: Section 1.4 NAT: RF6
TOP: Inverse of a Relation
KEY: inverse of a function | graph | function notation
9. ANS:
Substitute values into the general equation g(x) = a b(x − h) + k .
a) g(x) = 5 x − 6
b) g(x) =
6x − 4
c) g(x) =
−(x − 2) + 9
or
g(x) =
d) g(x) = −
−x + 2 + 9
2
3
1
x
1
3
or
g(x) = −
2
3
3x
PTS: 1
DIF: Average
OBJ: Section 2.1
TOP: Radical Functions and Transformations
15
NAT: RF13
KEY: transformations
ID: A
10. ANS:
The graph of y = f(x) is shown in black, and the graph of y =
PTS: 1
DIF: Difficult
TOP: Square Root of a Function
11. ANS:
OBJ: Section 2.2 NAT: RF13
KEY: graph | square root of a function
PTS: 1
DIF: Average
OBJ: Section 2.3
TOP: Solving Radical Equations Graphically
12. ANS:
No, Jim is not correct.
f(x) is shown in blue.
NAT: RF13
KEY: graphical solution
Ê
x 2 = 25 has two possible solutions of ±25, and ÁÁÁ
Ë
ˆ2
x ˜˜˜ = 25 has only one solution,
¯
+25.
PTS: 1
DIF: Difficult
OBJ: Section 2.3
TOP: Solving Radical Equations Graphically
16
NAT: RF13
KEY: algebraic solution
ID: A
13. ANS:
The velocity can be calculated by using the height from which the egg is dropped. The height is 80 ft plus the
height of the student, or 85 ft.
v=
(v 0 ) 2 + 2ad
=
0 + 2(32)(85)
=
5440
v = 73.76
Since the speed is less than 80 ft/s, the egg will not crack.
The maximum height is limited by a velocity of 80 ft/s, so
v=
(v 0 ) 2 + 2ad
80 =
0 + 2(32)d
80 =
64d
6400 = 64d
100 = d
The maximum height is 100 ft.
PTS: 1
DIF: Average
OBJ: Section 2.3
TOP: Solving Radical Equations Graphically
14. ANS:
NAT: RF13
KEY: algebraic solution
PTS: 1
DIF: Difficult
OBJ: Section 2.3
TOP: Solving Radical Equations Graphically
NAT: RF13
KEY: graphical solution
17
ID: A
15. ANS:
a) The possible factors are (x ± 1), (x ± 2), and (x ± 3). Try x = −1 using the factor theorem.
P(x) = x 3 + 6x 2 + 11x + 6
P(−1) = (−1) 3 + 6(−1) 2 + 11(−1) + 6
= −1 + 6 − 11 + 6
P(−1) = 0
Thus, (x + 1) is a factor of P(x).
Use synthetic division to find the quadratic factor.
1
1
−
×
1
6
11
6
1
5
6
5
6
0
Thus,
x 3 + 6x 2 + 11x + 6 = (x + 1)(x 2 + 5x + 6)
= (x + 1)(x + 2)(x + 3)
b) 4x − 11x − 3x = x(4x 2 − 11x − 3)
3
2
= x(4x 2 − 12x + x − 3)
= x[4x(x − 3) + (x − 3)]
= x(x − 3)(4x + 1)
c) x − 81 = (x − 9)(x 2 + 9)
4
2
= (x − 3)(x + 3)(x 2 + 9)
PTS: 1
DIF: Average
OBJ: Section 3.3 NAT: RF11
TOP: The Factor Theorem
KEY: factor theorem | factor
NOT: A variety of factoring techniques is required.
18
ID: A
16. ANS:
a) x 2 (x − 2)(x + 2) + 3x + 6 = x 2 (x − 2)(x + 2) + 3(x + 2)
= (x + 2)[x 2 (x − 2) + 3]
= (x + 2)(x 3 − 2x 2 + 3)
Use the factor theorem on the second factor. Try x = −1.
P(x) = x 3 − 2x 2 + 3
P(−1) = (−1) 3 − 2(−1) 2 + 3
= −1 − 2 + 3
P(−1) = 0
Divide.
1
1
−
×
1
−2
0
3
1
−3
3
−3
3
0
The quotient is not factorable. Thus,
x 2 (x − 2)(x + 2) + 3x + 6 = (x + 2)(x + 1)(x 2 − 3x + 3)
b) 16x 4 − (x + 1) 2 = (4x 2 ) 2 − (x + 1) 2
= [4x 2 − (x + 1)][4x 2 + (x + 1)]
= (4x 2 − x − 1)(4x 2 + x + 1)
PTS:
TOP:
KEY:
NOT:
1
DIF: Average
OBJ: Section 3.3 NAT: RF11
The Factor Theorem
factor theorem | factor | integral zero theorem | grouping | rational zero theorem
A variety of factoring techniques is required.
19
ID: A
17. ANS:
1
Possible values of x in the factor theorem are ±1, ± , ±2, ±4, and ±8.
2
Try x = 2.
P(x) = 2x 3 + 5x 2 − 14x − 8
P(2) = 2(2) 3 + 5(2) 2 − 14(2) − 8
= 16 + 20 − 28 − 8
P(2) = 0
Thus, x − 2 is a factor of P(x).
Divide.
−2
2
−
×
2
5
−14
−8
−4
−18
−8
9
4
0
Thus,
2x 3 + 5x 2 − 14x − 8 = (x − 2)(2x 2 + 9x + 4)
= (x − 2)(2x 2 + x + 8x + 4)
= (x − 2)[x(2x + 1) + 4(2x + 1)]
= (x − 2)(2x + 1)(x + 4)
PTS:
TOP:
KEY:
NOT:
1
DIF: Difficult +
OBJ: Section 3.3 NAT: RF11
The Factor Theorem
factor theorem | factor | integral zero theorem | grouping | rational zero theorem
A variety of factoring techniques is required.
20
ID: A
18. ANS:
a) Try x = 1 in the factor theorem.
P(x) = 3x 3 + 2x 2 − 8x + 3
P(1) = 3(1) 3 + 2(1) 2 − 8(1) + 3
= 3+2−8+3
P(1) = 0
Thus, x − 1 is a factor.
Use synthetic division to find another factor.
−1
3
−
×
3
2
−8
3
−3
−5
3
5
−3
0
Another factor is 3x2 + 5x − 3. Thus,
3x 3 + 2x 2 − 8x + 3 = 0
(x − 1)(3x 2 + 5x − 3) = 0
x = 1 or 3x 2 + 5x − 3 = 0
Use the quadratic formula to find the other solutions.
3x 2 + 5x − 3 = 0
x=
=
−5 ±
5 2 − 4(3)(−3)
2(3)
−5 ± 61
6
−5 − 61
−5 + 61
, 1,
.
6
6
2x 3 + x 2 − 10x − 5 = 0
The solutions are x =
b)
x 2 (2x + 1) − 5(2x + 1) = 0
(x 2 − 5)(2x + 1) = 0
x=
c)
5, −
5, −
1
2
5x 4 = 7x 2 − 2
5x 4 − 7x 2 + 2 = 0
5x 4 − 5x 2 − 2x 2 + 2 = 0
5x 2 (x 2 − 1) − 2(x 2 − 1) = 0
(x 2 − 1)(5x 2 − 2) = 0
x = −1, 1, −
2
,
5
2
5
21
ID: A
PTS:
TOP:
KEY:
NOT:
19. ANS:
a)
1
DIF: Average
OBJ: Section 3.3 NAT: RF11
The Factor Theorem
polynomial equation | factor theorem | factor | roots | integral zero theorem
A variety of factoring techniques is required.
x 4 + 3x 2 − 28 = 0
(x 2 + 7)(x 2 − 4) = 0
(x 2 + 7)(x − 2)(x + 2) = 0
x = 2, − 2
b) 2x − 54x = 0
4
2x(x 3 − 27) = 0
2x(x − 3)(x 2 + 3x + 9) = 0
x = 0, 3
PTS:
NAT:
KEY:
NOT:
1
DIF: Average
OBJ: Section 3.3 | Section 3.4
RF11
TOP: The Factor Theorem | Equations and Graphs of Polynomial Functions
polynomial equation | factor theorem | factor | integral zero theorem | rational zero theorem | roots
A variety of factoring techniques is required.
22
ID: A
20. ANS:
a)
x > 1.9
b)
x ≤ −13
PTS: 1
DIF: Difficult +
OBJ: Section 3.4
TOP: Equations and Graphs of Polynomial Functions
KEY: technology | inequality | graph | roots
21. ANS:
a = θr
=
a=
NAT: RF12
π
(3)
9
π
3
The child travels through an arc length of
PTS: 1
DIF: Easy
TOP: Angles and Angle Measure
π
m.
3
OBJ: Section 4.1
KEY: arc length
23
NAT: T1
ID: A
22. ANS:
Use the trigonometry of right triangles. The hypotenuse is the length of the ladder, or 3 m. The angle between
π
the ladder and the ground is . The opposite side to the angle is the height the ladder reaches up the wall. Let
3
this height be h.
ÊÁ π ˆ˜
h
= sin ÁÁÁÁ ˜˜˜˜
3
Ë3¯
ÊÁ π ˆ˜
h = 3sin ÁÁÁÁ ˜˜˜˜
Ë3¯
h=
3 3
2
The height the ladder reaches up the wall is
PTS: 1
DIF: Average
TOP: Trigonometric Ratios
23. ANS:
ÊÁ π ˆ˜
sinx = cos ÁÁÁÁ ˜˜˜˜
Ë5¯
3 3
m.
2
OBJ: Section 4.3 NAT: T3
KEY: special angles | trigonometric ratios
ÊÁ π 3π ˆ˜
˜˜
sinx = cos ÁÁÁÁ −
˜˜
2
10
Ë
¯
ÊÁ 3π ˆ˜
sinx = sin ÁÁÁÁ ˜˜˜˜
Ë 10 ¯
x=
3π
10
PTS: 1
DIF: Average
OBJ: Section 4.4
TOP: Introduction to Trigonometric Equations
KEY: equivalent trigonometric expression | exact value
24. ANS:
NAT: T4
3
3
2
, sinθ = −
. Since sin60° =
, the reference angle is 60°. The ratio is negative in
2
2
3
quadrants III and IV.This means that the angle can be found by looking for reflections of 60° that lie in these
quadrants.
quadrant III: 180° + 60° = 240°
quadrant IV: 360° – 60° = 300°
Since csc θ = −
PTS: 1
DIF: Average
OBJ: Section 4.3 NAT: T2
TOP: Trigonometric Ratios
KEY: reference angle | reciprocal trigonometric ratios | unit circle
24
ID: A
25. ANS:
d
r=
2
=
21
2
= 10.5
a = rθ
a = (10.5)(1.2)
= 12.6
The arc length is 12.6 cm.
PTS: 1
DIF: Average
TOP: Angles and Angle Measure
26. ANS:
sin A =
2
2
cos B =
3
2
π
6
ÊÁ π ˆ˜
ÊÁ π ˆ˜
sec A + sec B = sec ÁÁÁÁ ˜˜˜˜ + sec ÁÁÁÁ ˜˜˜˜
Ë4¯
Ë6¯
∠A =
π
4
OBJ: Section 4.1 NAT: T1
KEY: arc length | central angle
∠B =
=
=
2
2
+
2
3
2 3 +2 2
6
=
2 18 + 2 12
6
=
2×3 2 +2×2 3
6
=
3 2 +2 3
3
PTS: 1
DIF: Average
OBJ: Section 4.3 NAT: T3
TOP: Trigonometric Ratios
KEY: exact value | reciprocal trigonometric ratios
27. ANS:
The tangent ratio is negative in quadrants II and IV. In quadrant II for the domain 0° ≤ θ ≤ 180°, θ = 120°. In
quadrant IV for the domain −180° ≤ θ ≤ 0°, θ = −60°.
PTS: 1
DIF: Difficult
TOP: Trigonometric Ratios
OBJ: Section 4.3 NAT: T3
KEY: primary trigonometric ratios | exact value
25
ID: A
28. ANS:
5π
6
PTS: 1
DIF: Easy
OBJ: Section 4.1 NAT: T1
TOP: Angles and Angle Measure
KEY: radian
29. ANS:
π
You would need to subtract or 90° from each x-value for y = sin x and plot the points using the
2
corresponding y-values. The zeros of the sine function would become the maximum or minimum values of
the cosine function.
PTS: 1
DIF: Easy
OBJ: Section 5.2 NAT: T4
TOP: Graphing Sine and Cosine Functions
KEY: phase shift
30. ANS:
a reflection in the x-axis, a vertical stretch by a factor of 2, a horizontal stretch by a factor of 8, a phase shift
π
of to the right, and a vertical translation of 1 unit up
3
PTS: 1
DIF: Average
OBJ: Section 5.2
TOP: Transformations of Sinusoidal Functions
KEY: transformations | sinusoidal function
31. ANS:
Solutions may vary. Sample solution:
NAT: T4
The amplitude is half the diameter, or 30 cm. The maximum height of the pebble is 60 cm, so the vertical
1
displacement must be 30 cm. The wheel rotates at 4 revolutions per second, so the period is s. Thus, the
4
2π
value of b is
or 8π .
1
4
Thus, the relationship between the height of the pebble above the ground and time is
h = 30 sin (8πt) + 30
PTS: 1
DIF: Easy
OBJ: Section 5.4
TOP: Equations and Graphs of Trigonometric Functions
26
NAT: T4
KEY: sinusoidal function | modelling
ID: A
32. ANS:
a) From the function, the maximum and minimum populations are 250 + 30 or 280 bears and 250 − 30 or 220
bears.
b) Graph the function p (t) = 250 + 30cos t.
The graph is first increasing over the interval [3.14,6.28] or [π, 2π], or from approximately 3 years 1
months to 6 years 3
2
3
1
months.
3
PTS: 1
DIF: Average
OBJ: Section 5.4 NAT: T4
TOP: Equations and Graphs of Trigonometric Functions
KEY: sinusoidal function | maximum | minimum
33. ANS:
a) 0.7 m
2π
2π
π
b) Since b = 72 and period =
, then period =
or
s. The number of revolutions of the rope is the
b
72
36
36
reciprocal of the period, or , or 11.46 rev/s. Multiply by 60 to get 688 revolutions/min.
π
PTS: 1
DIF: Difficult
OBJ: Section 5.4
TOP: Equations and Graphs of Trigonometric Functions
KEY: period | amplitude | sinusoidal function
27
NAT: T4
ID: A
34. ANS:
Answers may vary. Sample answer:
π
Use x = 0 and y = :
2
L.S. = cos(x + y)
R.S. = cos x + cos y
ÊÁ
π ˆ˜
= cos ÁÁÁ 0 + ˜˜˜˜
ÁË
2¯
= cos
= cos 0 + cos
π
2
= 1+0
π
2
=1
=0
L.S. ≠ R.S.
Thus, cos(x + y) = cos x + cos y is not an identity.
PTS: 1
DIF:
TOP: Proving Identities
35. ANS:
2cos x − 3 = 0
2cos x =
3
cos x =
3
2
Average
OBJ: Section 6.3 NAT: T6
KEY: counterexample
ÊÁ
ˆ
ÁÁ 3 ˜˜˜
Á
˜˜
x = cos ÁÁ
ÁÁ 2 ˜˜˜
Ë
¯
−1
=
π
6
Since cosine is also positive in quadrant IV, another solution is 2π −
PTS: 1
DIF: Easy
OBJ: Section 6.4
TOP: Solving Trigonometric Equations Using Identities
36. ANS:
cot 2 θ + cot θ = 0
π 11π
=
.
6
6
NAT: T6
KEY: exact value
cot θ (cot θ + 1) = 0
cot θ = 0 or cot θ = −1
θ=
π
3π
or θ =
2
4
Since the period for cot θ is π, the solution in general form is θ =
π
3π
+ nπ and θ =
+ nπ , where n ∈ I.
2
4
PTS: 1
DIF: Difficult
OBJ: Section 6.4
TOP: Solving Trigonometric Equations Using Identities
NAT: T6
KEY: exact value | general solutions
28
ID: A
37. ANS:
sec 2 θ − 2tan θ − 3 = 0
(1 + tan 2 θ) − 2tan θ − 3 = 0
tan 2 θ − 2tan θ − 2 = 0
Use the quadratic formula.
tan θ =
2±
= 1±
12
2
3
≈ 2.732 or − 0.732
θ = tan −1 (2.732) or tan −1 (−0.732)
≈ 70° or − 36°
Since the period for tanθ is 180°, a positive solution corresponding to −36° is −36° + 180° or 144°. The
general solution is 70° + 180°n and 144° + 180°n , where n ∈ I.
PTS: 1
DIF: Difficult
OBJ: Section 6.4 NAT: T6
TOP: Solving Trigonometric Equations Using Identities
KEY: exact value | general solutions | quadratic formula
38. ANS:
a) i) quadratic
ii) exponential
iii) linear
b) i) successive values would be increasing by a constant amount
ii) successive values would be increasing by a constant factor
iii) all values would be constant
PTS: 1
DIF: Average
OBJ: Section 7.1
TOP: Characteristics of Exponential Functions
KEY: linear | quadratic | exponential function
29
NAT: RF9
ID: A
39. ANS:
Answers may vary.
ÊÁ 1 ˆ˜ x
Sample answer: The graph of y = 3 ÁÁÁÁ ˜˜˜˜ :
Ë 3¯
PTS:
NAT:
TOP:
KEY:
40. ANS:
1
DIF: Average
OBJ: Section 7.1 | Section 7.2
RF9
Characteristics of Exponential Functions | Transformations of Exponential Functions
domain | range | intercepts | exponential function
1
and a translation of 2 units to the right
2
1
x−2
b) The graph of y = 3 x is shown in blue and the graph of y = (3)
is shown in red.
2
a) a vertical compression by a factor of
c) domain {x| x ∈ R}, range {y| y > 0, y ∈ R}, y = 0
PTS: 1
DIF: Average
OBJ: Section 7.2
TOP: Transformations of Exponential Functions
KEY: graph | transformations of exponential functions
30
NAT: RF9
ID: A
41. ANS:
a) y = 5 −x
b) y = 5 x − 3
c) y = 5 x + 4 − 1
d) y = −5 x − 2
PTS: 1
DIF: Average
OBJ: Section 7.2
TOP: Transformations of Exponential Functions
KEY: equation | transformations of exponential functions
42. ANS:
a) ii)
b) i)
c) iii)
NAT: RF9
PTS: 1
DIF: Average
OBJ: Section 7.1
TOP: Characteristics of Exponential Functions
KEY: graph | modelling | exponential function
43. ANS:
4n − 1
ÁÊ 1 ˜ˆ
9n − 1 = ÁÁÁÁ ˜˜˜˜
Ë3¯
NAT: RF9
n−1
Ê ˆ 4n − 1
ÁÊÁ 3 2 ˜ˆ˜
= ÁÁ 3 −1 ˜˜
Ë ¯
Ë ¯
3 2n − 2 = 3 1 − 4n
Equate the exponents:
2n − 2 = 1 − 4n
6n = 3
n=
1
2
PTS: 1
DIF: Average
TOP: Solving Exponential Equations
OBJ: Section 7.3 NAT: RF10
KEY: change of base
31
ID: A
44. ANS:
domain: {x| x > −2,x ∈ R}
range: {y| y ∈ R}
equation of vertical asymptote: x = –2
PTS: 1
DIF: Difficult
OBJ: Section 8.2
TOP: Transformations of Logarithmic Functions
KEY: transformation | vertical translation | asymptote | graph
45. ANS:
log 2 14 = log2 (2 × 7)
NAT: RF8
= log2 2 + log 2 7
≈ 1 + 2.8074
= 3.8074
PTS: 1
DIF: Difficult
TOP: Laws of Logarithms
46. ANS:
6 3x + 1 = 2 2x − 3
OBJ: Section 8.3 NAT: RF9
KEY: power law | laws of logarithms
log(6 3x + 1 ) = log(2 2x − 3 )
(3x + 1) log 6 = (2x − 3) log2
3x log6 + log 6 = 2x log 2 − 3log2
x(3log 6 − 2log2) = −3log2 − log6
x=
−(3log2 + log6)
3log6 − 2log2
PTS: 1
DIF: Average
OBJ: Section 8.3 | Section 8.4
NAT: RF9
TOP: Laws of Logarithms | Logarithmic and Exponential Equations
KEY: exponential equation | laws of logarithms
32
ID: A
47. ANS:
a) f(x) =
1
4x − 8
b)
PTS: 1
DIF: Average
OBJ: Section 9.1 NAT: RF14
TOP: Exploring Rational Functions Using Transformations
KEY: reciprocal of linear function | vertical asymptote | y-intercept | graph
33
ID: A
48. ANS:
ÏÔÔ
¸ÔÔ
Ô
5
Ô
a) i) ÔÌ x| x ≠ , x ∈ R Ô˝ , {y| y ≠ 0, y ∈ R}
ÔÔÓ
ÔÔ˛
4
3
ii) x-intercept: none, y-intercept: −
5
5
iii) x = , y = 0
4
b)
PTS: 1
DIF: Average
OBJ: Section 9.2 | Section 9.3
NAT: RF14
TOP: Analysing Rational Functions | Connecting Graphs and Rational Equations
KEY: reciprocal of linear function | key features | graph
34
ID: A
49. ANS:
a) i) {x| x ≠ 2, x ∈ R}, {y| y ≠ 3, y ∈ R}
8
ii) x-intercept: − , y-intercept: −4
3
iii) x = 2, y = 3
b)
PTS: 1
DIF: Average
OBJ: Section 9.2 | Section 9.3
NAT: RF14
TOP: Analysing Rational Functions | Connecting Graphs and Rational Equations
KEY: linear expressions in numerator and denominator | key features | graph
35
ID: A
50. ANS:
a) i) {x| x ≠ −3,x ≠ 4, x ∈ R}, {y| y ≠ 0, y ∈ R}
1
ii) x-intercept: none, y-intercept: −
4
iii) x = 4, y = 0
ÊÁ
1 ˆ˜
b) Note the hole at the point ÁÁÁÁ −3, − ˜˜˜˜ .
7¯
Ë
PTS: 1
DIF: Difficult
OBJ: Section 9.2 | Section 9.3
NAT: RF14
TOP: Analysing Rational Functions | Connecting Graphs and Rational Equations
KEY: rational function | key features | graph
51. ANS:
h (x) = f (x) + g (x)
= x + 1 + x 2 + 3x + 1
= x 2 + 4x + 2
PTS: 1
DIF: Easy
OBJ: Section 10.1 NAT: RF1
TOP: Sums and Differences of Functions
KEY: add functions
36
ID: A
52. ANS:
h (x) =
f (x)
g (x)
=
x2 − 4
x 2 − 3x + 2
=
(x − 2) (x + 2)
(x − 2) (x − 1)
=
(x + 2)
, x ≠ 2, x ≠ 1
(x − 1)
PTS: 1
DIF: Average
OBJ: Section 10.2 NAT: RF1
TOP: Products and Quotients of Functions
KEY: divide functions | restrictions
53. ANS:
f(x) is in blue, g(x) is in red, and h(x) is in black.
PTS: 1
DIF: Difficult
OBJ: Section 10.2 NAT: RF1
TOP: Products and Quotients of Functions
KEY: multiply functions | graph
37
ID: A
54. ANS:
2x 2 − 3 = 0
2x 2 = 3
x2 =
3
2
x=±
3
2
x=±
6
2
PTS: 1
DIF: Difficult +
OBJ: Section 10.2 NAT: RF1
TOP: Products and Quotients of Functions
KEY: divide functions | vertical asymptotes
55. ANS:
3
g (2) = 2 − (2)
= −6
f (−6) = (−6) − 7
2
= 29
f(g(2)) = 29
PTS: 1
DIF: Average
OBJ: Section 10.3 NAT: RF1
TOP: Composite Functions
KEY: composite functions | evaluate
56. ANS:
The number of different routes can be calculated using the following permutation:
12!
= 924
6!6!
Joe can take one of 924 different routes to travel from home to school.
PTS: 1
DIF: Average
OBJ: Section 11.1 NAT: PC1
TOP: Permutations
KEY: permutations
57. ANS:
ÊÁ a b ˆ˜ 4
Ê ˆ4Ê ˆ0
Ê ˆ3Ê ˆ1
Ê ˆ2Ê ˆ2
Ê ˆ1Ê ˆ3
Ê ˆ0Ê ˆ4
ÁÁ − ˜˜ = C ÁÁÁ a ˜˜˜ ÁÁÁ − b ˜˜˜ + C ÁÁÁ a ˜˜˜ ÁÁÁ − b ˜˜˜ + C ÁÁÁ a ˜˜˜ ÁÁÁ − b ˜˜˜ + C ÁÁÁ a ˜˜˜ ÁÁÁ − b ˜˜˜ + C ÁÁÁ a ˜˜˜ ÁÁÁ − b ˜˜˜
ÁÁ 2 3 ˜˜ 4 0 ÁÁ 2 ˜˜ ÁÁ 3 ˜˜ 4 1 ÁÁ 2 ˜˜ ÁÁ 3 ˜˜ 4 2 ÁÁ 2 ˜˜ ÁÁ 3 ˜˜ 4 3 ÁÁ 2 ˜˜ ÁÁ 3 ˜˜ 4 4 ÁÁ 2 ˜˜ ÁÁ 3 ˜˜
Ë
¯
Ë ¯ Ë ¯
Ë ¯ Ë ¯
Ë ¯ Ë ¯
Ë ¯ Ë ¯
Ë ¯ Ë ¯
ÊÁ a ˆ˜ 4 ÊÁ b ˆ˜ 0
ÊÁ a ˆ˜ 3 ÊÁ b ˆ˜ 1
ÊÁ a ˆ˜ 2 ÊÁ b ˆ˜ 2
ÊÁ a ˆ˜ 1 ÊÁ b ˆ˜ 3
ÊÁ a ˆ˜ 0 ÊÁ b ˆ˜ 4
Á
˜
Á
˜
Á
˜
Á
˜
Á
˜
Á
˜
Á
˜
Á
˜
= 1 ÁÁÁ ˜˜˜ ÁÁÁ − ˜˜˜ + 4 ÁÁÁ ˜˜˜ ÁÁÁ − ˜˜˜ + 6ÁÁÁ ˜˜˜ ÁÁÁ − ˜˜˜ + 4 ÁÁÁ ˜˜˜ ÁÁÁ − ˜˜˜ + 1 ÁÁÁÁ ˜˜˜˜ ÁÁÁÁ − ˜˜˜˜
Ë2¯ Ë 3¯
Ë2¯ Ë 3¯
Ë2¯ Ë 3¯
Ë2¯ Ë 3¯
Ë2¯ Ë 3¯
=
a 4 a 3 b a 2 b 2 2ab 3 b 4
−
+
−
+
16
6
6
27
81
PTS: 1
DIF: Average
TOP: The Binomial Theorem
OBJ: Section 11.3 NAT: PC4
KEY: binomial expansion | binomial theorem
38
ID: A
58. ANS:
(2n + 2)!
1 ⋅ 2 ⋅ 3 ⋅ … ⋅ (2n − 2)(2n − 1)(2n)(2n + 1)(2n + 2)
=
(2n − 1)!0!
1 ⋅ 2 ⋅ 3 ⋅ … ⋅ (2n − 2)(2n − 1) ⋅ 1
= (2n)(2n + 1)(2n + 2)
= 2n(4n 2 + 6n + 2)
= 8n 3 + 12n 2 + 4n
PTS: 1
DIF: Average
OBJ: Section 11.1 NAT: PC1
TOP: Permutations
KEY: factorial
59. ANS:
There are 14 letters in total. 5 of these are burnt out.
( 9 C 3 )( 5 C 2 ) = 84 × 10
= 840
There are 840 ways to select 3 good letters and 2 burnt-out letters.
PTS: 1
DIF: Average
OBJ: Section 11.1 NAT: PC3
TOP: Combinations
KEY: combinations
60. ANS:
Assume there are two questions on the test. They could appear in 2! possible ways. This can be carried on for
additional questions.
For 3 questions, there are 3! = 6 possible orders.
For 4 questions, there are 4! = 24 possible orders.
For 5 questions, there are 5! = 120 possible orders.
For 6 questions, there are 6! = 720 possible orders.
There must be 6 questions for everyone to get a test with the questions in a different order.
PTS: 1
DIF:
TOP: Permutations
Average
OBJ: Section 11.1 NAT: PC1
KEY: fundamental counting principle
39
ID: A
PROBLEM
1. ANS:
a) h(t) = −4.9t 2 + 25
b) h(t) = −1.85t 2 + 25
c) The Earth function is shown in blue and the Mars function is shown in red.
d) The scale factor that can be applied to the Earth function to transform it to the Mars function is
4.9 ×
37
.
98
37
= 1.85
98
PTS: 1
DIF: Difficult
TOP: Reflections and Stretches
2. ANS:
a) g(x) = −
1
3
OBJ: Section 1.2 NAT: RF3
KEY: graph | stretch | function notation
1
(x + 3) − 5
2
b)
c) The reflection, horizontal stretch, and vertical compression must be done first, but can be done in any
order.
d) The translations to the left and down must be done last, but can be done in any order.
PTS: 1
DIF: Difficult +
TOP: Combining Transformations
OBJ: Section 1.3 NAT: RF4
KEY: graph | transformation | function notation
40
ID: A
3. ANS:
a) C(x) = 50 + 0.12x
C = 50 + 0.12x
b)
C − 50 = 0.12x
x=
C − 50
0.12
C − 50
0.12
c) This function represents the distance the car can be driven for a given rental cost.
d) Answers may vary. Sample answer:
If you have $65 to spend on the car rental, for how many kilometres can you drive the car?
65 − 50
f −1 (65) =
0.12
f −1 (C) =
= 125
You can drive a total of 125 km for the $65 rental fee.
PTS: 1
DIF: Average
TOP: Inverse of a Relation
4. ANS:
a)
OBJ: Section 1.4 NAT: RF6
KEY: inverse of a function | function notation
b) f(x): Domain: {x| x ≤ 0, x ∈ R}; Range: {y| y ≥ 0, y ∈ R}
g(x): Domain: {x| x ≥ −2, x ∈ R}; Range: {y| y ≤ −3, y ∈ R}
2
c) The vertical stretch changes from 5 to 2, which means a vertical compression by a factor of . There is a
5
1
horizontal compression by a factor of . The graph is reflected in both the x-axis and the y-axis. The graph is
6
translated 2 units left and 3 units down.
PTS: 1
DIF: Difficult
OBJ: Section 2.1
TOP: Radical Functions and Transformations
KEY: graph | transformations | domain | range
41
NAT: RF13
ID: A
5. ANS:
a) In both cases, the functions are in the shape of a transformed radical function.
b) In both cases, the function is stretched horizontally. The graph created by group 1 has a larger horizontal
stretch than the graph created by group 2. Vertical and horizontal translations are applied to both graphs.
c) Answers may vary. Sample answer:
Group 1: f(x) = 2(x − 1) + 1.6
Group 2: g(x) = 3(x − 1) + 1.6
d) Answers may vary. Sample answer:
Group 1 may have pushed the pendulum or Group 2 may have started the timer early. Students should note
that the only difference between the two groups can be accounted for by a horizontal stretch in the function.
PTS: 1
DIF: Difficult
OBJ: Section 2.1
TOP: Radical Functions and Transformations
42
NAT: RF13
KEY: graph | transformations
ID: A
6. ANS:
E=
a)
1 2
mv
2
2E = mv 2
2E
= v2
m
2E
m
b) i) Substitute m = 12 and E = 200 in the equation.
2E
v=
m
v=
=
2(200)
12
v ≈ 5.8
The speed of the object is 5.8 m/s.
ii) Substitute m = 12 and E = 420 in the equation.
2E
v=
m
=
2(420)
12
v ≈ 8.4
The speed of the object is 8.4 m/s.
c) When m = 12,
2E
v=
m
=
v=
2E
12
E
6
43
ID: A
d) The point (20, 2) is on the graph. Substitute this into the equation v =
v=
2E
m
2=
2(20)
m
4=
2E
and solve for m.
m
40
m
m = 10
The mass of the object is 10 kg.
PTS: 1
DIF: Average
OBJ: Section 2.1
TOP: Radical Functions and Transformations
44
NAT: RF13
KEY: graph | horizontal stretch
ID: A
7. ANS:
Let P(x) = 2x4 – 7x3 – 41x2 – 53x – 21.
Test x = –1 in the factor theorem.
P(x) = 2x 4 − 7x 3 − 41x 2 − 53x − 21
P(−1) = 2(−1) 4 − 7(−1) 3 − 41(−1) 2 − 53(−1) − 21
= 2 + 7 − 41 + 53 − 21
P(−1) = 0
Thus, x + 1 is a factor. Divide to determine another factor.
1
2
−
×
2
−7
−41
−53
−21
2
−9
−32
−21
−9
−32
−21
0
Thus,
P(x) = (x + 1)(2x 3 − 9x 2 − 32x − 21)
Now factor the cubic. Test x = –1 in the factor theorem. Let Q(x) = 2x 3 − 9x 2 − 32x − 21.
Q(x) = 2x 3 − 9x 2 − 32x − 21
Q(−1) = 2(−1) 3 − 9(−1) 2 − 32(−1) − 21
= −2 − 9 + 32 − 21
Q(−1) = 0
Thus, x + 1 is a factor. Divide to determine another factor.
1
2
−
×
2
−9
−32
−21
2
−11
−21
−11
−21
0
Thus,
P(x) = (x + 1)(x + 1)(2x 2 − 11x − 21)
= (x + 1) 2 (2x 2 − 14x + 3x − 21)
= (x + 1) 2 [2x(x − 7) + 3(x − 7)]
= (x + 1) 2 (x − 7)(2x + 3)
PTS: 1
DIF: Average
TOP: The Factor Theorem
OBJ: Section 3.3 NAT: RF11
KEY: factor theorem | integral zero theorem | factor
45
ID: A
8. ANS:
By the factor theorem, x + a is a factor of P(x) if P(–a) = 0.
P(−a) = (−a + a) 4 + (−a + c) 4 − (a − c) 4
= 0 + [−(a − c)] 4 − (a − c) 4
= (a − c) 4 − (a − c) 4
P(−a) = 0
PTS: 1
DIF: Average
OBJ: Section 3.3 NAT: RF11
TOP: The Factor Theorem
KEY: factor theorem | factor
9. ANS:
Let x represent the side length of the base. Then, V(x) = x2(x + 4).
Solve x2(x + 4) = 225.
Since 25 is a factor of 225, and 25 is a square, try x = 5 as a solution.
L.S. = x 2 (x + 4)
= 5 2 (5 + 4)
= 25(9)
= 225
= R.S.
Thus, x = 5 is a solution. The dimensions of the box could be 5 cm by 5 cm by 9 cm.
Rewrite the equation in the form P(x) = 0.
x 2 (x + 4) = 225
x 3 + 4x 2 − 225 = 0
Since x = 5 is a solution, x − 5 is a factor of P(x). Divide to find another factor.
−5
1
−
×
1
4
0
−225
−5
−45
−225
9
45
0
x 3 + 4x 2 − 225 = 0
(x − 5)(x 2 + 9x + 45) = 0
x2 + 9x + 45 = 0 has no real solutions. So the only solution is x = 5. The dimensions of the box are 5 cm by 5
cm by 9 cm.
PTS: 1
DIF: Average
OBJ: Section 3.3 | Section 3.4
NAT: RF11
TOP: The Factor Theorem | Equations and Graphs of Polynomial Functions
KEY: factor theorem | integral zero theorem | polynomial equation | root
46
ID: A
10. ANS:
Multiply the equation by −1 to clear the negative leading coefficient.
−x 3 + 5x 2 − 8x + 4 ≥ 0
−1(−x 3 + 5x 2 − 8x + 4) ≤ 0
x 3 − 5x 2 + 8x − 4 ≤ 0
Factor x 3 − 5x 2 + 8x − 4 using the factor theorem. Let P(x) = x 3 − 5x 2 + 8x − 4. Try x = 1 in the factor
theorem.
P(x) = x 3 − 5x 2 + 8x − 4
P(1) = (1) 3 − 5(1) 2 + 8(1) − 4
= 1−5+8−4
P(1) = 0
Thus, x − 1 is a factor of P(x). Divide to find another factor.
−1
1
−
×
1
−5
8
−4
−1
4
−4
−4
4
0
Thus,
x 3 − 5x 2 + 8x − 4 ≤ 0
(x − 1)(x 2 − 4x + 4) ≤ 0
(x − 1)(x − 2) 2 ≤ 0
Construct a table.
x>2
x<1
1<x<2
+
+
x−1
−
+
−
−
x−2
+
−
−
x−2
(x − 1)(x − 2)(x − 2)
+
+
−
The expression is equal to zero at x = 1 and x = 2. Thus, the solution is x ≤ 1 and x = 2.
Use a graphing calculator to graph the corresponding polynomial function, y = −x 3 + 5x 2 − 8x + 4. Then, use
the Zero operation.
47
ID: A
The zeros are 1 and 2. From the graph, −x 3 + 5x 2 − 8x + 4 ≥ 0 when x ≤ 1 or x = 2.
PTS: 1
DIF: Difficult +
OBJ: Section 3.3 | Section 3.4
NAT: RF11 | RF12 TOP: The Factor Theorem | Equations and Graphs of Polynomial Functions
KEY: polynomial inequality | factor theorem | factor | integral zero theorem | root
11. ANS:
The graph has a single zero at x = 0 and a double zero at x = 2. The graph also passes through the point (3, 6).
Thus, the graph is of the form y = ax(x − 2)2. Substitute the point (3, 6) to find a.
y = ax(x − 2) 2
6 = a(3)(3 − 2) 2
6 = 3a
a=2
Thus,
y = 2x(x − 2) 2
= 2x(x 2 − 4x + 4)
y = 2x 3 − 8x 2 + 8x
PTS: 1
DIF: Average
OBJ: Section 3.4
TOP: Equations and Graphs of Polynomial Functions
48
NAT: RF12
KEY: polynomial equation | graph | zeros
ID: A
12. ANS:
Use the cosine law.
1 2 = d 2 + d 2 − 2d (d ) cos 1.25
1 = 2d 2 − 2d 2 cos 1.25
= d 2 (2 − 2cos 1.25)
d=
1
2 − 2cos 1.25
Recall that s =
s=
d
. The balls have travelled a distance, d, in metres, in a time, t, of 2 s.
t
1
2 − 2cos 1.25
2
≈ 0.43
Therefore, after 2 s, the billiard balls are moving at approximately 0.43 m/s.
PTS: 1
DIF: Difficult +
OBJ: Section 4.3 | Section 4.4
NAT: T3 | T5
TOP: Trigonometric Ratios | Introduction to Trigonometric Equations
KEY: trigonometric ratios | radians | cosine law
49
ID: A
13. ANS:
a)
b) Use the sine ratio, since the side opposite the given angle is known and the hypotenuse is needed.
2.5
c) sin 30° =
x
x=
=
2.5
sin 30°
2.5
1
2
=5
The length of the piece of wood is 5 m.
PTS: 1
DIF: Easy
OBJ: Section 4.3 | Section 4.4
NAT: T3 | T5
TOP: Trigonometric Ratios | Introduction to Trigonometric Equations
KEY: trigonometric ratios | special angles | trigonometric equations
50
ID: A
14. ANS:
1
, the reference angle is 30°. The sine ratio is negative in quadrants III and IV. Look for
2
reflections of the 30° angle in these quadrants.
quadrant III: 180° + 30° = 210°
quadrant IV: 360° – 30° = 330°
1
1
b) Using a calculator, sin 210° = − and sin 330° = − .
2
2
a) Since sin 30° =
PTS: 1
DIF: Average
OBJ: Section 4.2 | Section 4.3
NAT: T2 | T3
TOP: Unit Circle | Trigonometric Ratios
KEY: sine ratio | reference angle | unit circle
51
ID: A
15. ANS:
<fix tech art so 49 cm arrow goes only to bottom of right-angle triangle, not to black dot>
ÊÁ 49 ˆ˜
θ = cos −1 ÁÁÁ ˜˜˜˜
ÁË 50 ¯
≈ 0.20033
Graph the function θ =
1 ÊÁÁÁ π
sin
4 ÁÁË 2
ˆ˜
t ˜˜˜ and determine the points at which θ = ±0.20033.
˜¯
Therefore, during the first 4 s the pendulum is displaced 1 cm vertically at approximately 0.6 s and 1.4 s to
one side and at approximately 2.6 s and 3.4 s to the other side.
PTS: 1
DIF: Difficult
OBJ: Section 5.4
TOP: Equations and Graphs of Trigonometric Functions
52
NAT: T4
KEY: linear trigonometric equation
ID: A
16. ANS:
Answers may vary. Sample answer:
a) Convert h:min to hours.
Month
Hours of Daylight
1
8.42
2
9.92
3
11.58
4
13.5
5
15.8
6
16.25
7
15.42
8
14.43
9
12.58
10
10.65
11
9.02
12
8
16.25 − 8
, or a = 4.125. The vertical shift is 8 + 4.125, or c = 12.125.
2
The sine wave starts at 12.125, at approximately 3.29 months. So, d = 3.29.
360
The period is
, or k = 30.
12
An equation to represent the data is t = 4.125 sin [30 (m − 3.29)] + 12.125.
c)
b) The amplitude is
There seems to be a strong correlation between the graph and the data.
d) Substitute m = 1.5 (halfway between January 1 and February 1) into the equation:
53
ID: A
t = 4.125 sin [30 (m − 3.29)] + 12.125
= 4.125 sin [30 (1.5 − 3.29)] + 12.125
≈ 8.8
There are approximately 8.8 h, or 8 h 48 min, of daylight on January 15. The graph verifies this solution.
PTS: 1
DIF: Difficult +
OBJ: Section 5.4 NAT: T4
TOP: Equations and Graphs of Trigonometric Functions
KEY: model | sinusoidal function | graph
17. ANS:
a) 11:38 – 5:17 = 6:21
So, 6 h 21 min represents half of a full cycle. Therefore, a full cycle would have a period of 12 h 42 min, or
12.7 h.
1.77 − 0.21
b) The amplitude is
, or 0.78 m.
2
c) The next high tide will occur 12 h 42 min after 5:17 a.m., or at 5:59 p.m. The next after this will be 12 h 42
min after 5:59 p.m., or at 6:41 a.m.
d) The next low tide will occur 12 h 42 min after 11:38 a.m., or at 12:20 a.m. The next after this will be 12 h
42 min after 12:20 a.m., or at 1:02 p.m.
PTS: 1
DIF: Average
OBJ: Section 5.1 | Section 5.2
NAT: T4
TOP: Graphing Sine and Cosine Functions | Transformations of Sinusoidal Functions
KEY: model | periodic function | predict
54
ID: A
18. ANS:
a) y = 2 cos x − 2
b) Let c represent the phase shift.
y = 2 cos(x − c) − 2
−1 = 2 cos(0 − c) − 2
1 = 2 cos(−c)
1
= cos(−c)
2
−c = 60°
c = −60°
The phase shift is 60° to the left.
c) y = 2 cos (x + 60°) − 2
d)
PTS: 1
DIF: Average
OBJ: Section 5.2
TOP: Transformations of Sinusoidal Functions
KEY: graph | transformations | cosine function | equation
55
NAT: T4
ID: A
19. ANS:
L.S. = 1 + cos θ
R.S. =
sin 2 θ
1 − cos θ
=
1 − cos 2 θ
1 − cos θ
=
(1 − cos θ) (1 + cos θ)
1 − cos θ
= 1 + cos θ
L.S. = R.S.
sin 2 θ
Therefore, 1 + cos θ =
.
1 − cos θ
PTS: 1
DIF: Average
OBJ: Section 6.1 | Section 6.3
NAT: T6
TOP: Reciprocal, Quotient, and Pythagorean Identities | Proving Identities
KEY: Pythagorean identities | proof
20. ANS:
ÊÁ π
ˆ˜ ÊÁ π
ˆ˜
L.S. = sin ÁÁÁÁ − x ˜˜˜˜ cot ÁÁÁÁ + x ˜˜˜˜ R.S. = −sin x
Ë2
¯ Ë2
¯
= cos x(−tanx)
ÊÁ sin x ˆ˜
˜˜
= cos x ÁÁÁ −
ÁË cos x ˜˜¯
= −sinx
L.S. = R.S.
ÁÊÁ π
˜ˆ ÁÊ π
˜ˆ
Therefore, sin ÁÁÁ − x ˜˜˜˜ cot ÁÁÁÁ + x ˜˜˜˜ = −sinx .
Ë2
¯ Ë2
¯
PTS: 1
DIF: Average
OBJ: Section 6.1 | Section 6.2 | Section 6.3
NAT: T6
TOP: Reciprocal, Quotient, and Pythagorean Identities | Sum, Difference, and Double-Angle Identities |
Proving Identities
KEY: difference identities | sum identities | quotient identities | proof
56
ID: A
21. ANS:
L.S. =
=
1 − cos 2θ + sin2θ
1 + cos 2θ + sin2θ
L.S. = tanθ
1 − (1 − 2sin2 θ) + 2 sinθ cos θ
1 + (2cos 2 θ − 1) + 2sinθ cos θ
=
2sin 2 θ + 2sinθ cos θ
2 cos 2 θ + 2 sinθ cos θ
=
2sinθ (sinθ + cos θ)
2 cos θ (sin θ + cos θ)
=
sinθ
cos θ
= tanθ
L.S. = R.S
1 − cos 2θ + sin2θ
= tanθ .
Therefore,
1 + cos 2θ + sin2θ
PTS: 1
DIF: Average
OBJ: Section 6.2 | Section 6.3
NAT: T6
TOP: Sum, Difference, and Double-Angle Identities | Proving Identities
KEY: double-angle identities | proof
22. ANS:
cos 2 θ − sin 2 θ
L.S. =
R.S. = 1 − tanθ
cos 2 θ + sinθ cos θ
sinθ
= 1−
(cos θ − sinθ) (cos θ + sinθ)
cos
θ
=
cos θ (cos θ + sinθ)
cos θ − sinθ
=
cos θ − sinθ
cos θ
=
cos θ
L.S. = R.S.
PTS: 1
DIF:
TOP: Proving Identities
Average
OBJ: Section 6.3 NAT: T6
KEY: trigonometric identity | proof
57
ID: A
23. ANS:
a) sec θ cot θ =
=
b)
ÊÁ 1 ˆ˜ ÊÁ cos θ ˆ˜
ÁÁ
˜Á
˜
ÁÁ cos θ ˜˜˜ ÁÁÁ sinθ ˜˜˜
Ë
¯Ë
¯
1
sinθ
1
=1
sinθ
1 = sinθ
θ = 90°
c) This is verified using a calculator.
d)
PTS: 1
DIF: Average
OBJ: Section 6.4
TOP: Solving Trigonometric Equations Using Identities
KEY: reciprocal trigonometric ratios | unit circle
58
NAT: T6
ID: A
24. ANS:
L.S. =
=
=
sin 3θ + sinθ
cos 3θ + cos θ
R.S. = tan2θ
sin2θ cos θ + cos 2θ sinθ + sinθ
cos 2θ cos θ − sin 2θ sinθ + cos θ
2sinθ cos 2 θ + (2cos 2 θ − 1) sinθ + sinθ
(2cos 2 θ − 1) cos θ − 2sin 2 θ cos θ + cos θ
ÊÁ
ˆ
ÁÁ 2cos 2 θ + 2cos 2 θ − 1 + 1 ˜˜˜
ÁÁ
˜˜
ÁÁ 2 cos 2 θ − 1 − 2sin 2 θ + 1 ˜˜
Ë
¯
Ê
ˆ
sinθ ÁÁÁÁ 4cos 2 θ ˜˜˜˜
=
cos θ ÁÁÁ 2cos 2θ ˜˜˜
Ë
¯
sinθ
=
cos θ
=
4sin θ cos 2 θ
2cos 2θ cos θ
=
2sinθ cos θ
cos 2θ
=
sin2θ
cos 2θ
= tan 2θ
sin3θ + sinθ
= tan2θ.
This proves that
cos 3θ + cos θ
L.S. = R.S.
PTS: 1
DIF: Difficult
OBJ: Section 6.2 | Section 6.3
NAT: T6
TOP: Sum, Difference, and Double-Angle Identities | Proving Identities
KEY: sum identities | proof
59
ID: A
25. ANS:
sin3x + sinx = cos x
sin2x cos x + cos 2x sinx + sinx = cos x
2sinx cos 2 x + (2cos 2 x − 1) sinx + sinx = cos x
2sinx cos 2 x + 2cos 2 x sinx − sinx + sinx = cos x
4sinx cos 2 x = cos x
4 sinx cos 2 x − cos x = 0
cos x(4sinx cos x − 1) = 0
cos x(2 sin2x − 1) = 0
cos x = 0 or sin2x =
x=
1
2
π 3π
π 5π 13π 17π
,
or 2x = ,
,
,
2 2
6 6
6
6
π 3π π 5π 13π 17π
,
, ,
,
,
2 2 12 12 12 12
π 5π 13π 17π π 3π
,
,
,
, , .
The solution is x =
12 12 12 12 2 2
x=
PTS: 1
DIF: Difficult +
OBJ: Section 6.4
TOP: Solving Trigonometric Equations Using Identities
26. ANS:
tanx (csc x + 2) = 0
tanx = 0
csc x + 2 = 0
x = 0 + nπ
= nπ
NAT: T6
KEY: general solutions | equation
csc x = −2
sinx = −
x=−
1
2
π
11π
or x =
6
6
sin x is also negative in quadrant III, so find a reflection of x =
11π
in this quadrant. Another value for
6
1
7π
sinx = − is x =
.
2
6
Since the period of tan x is π and the period of sin x is 2π , the general solution is is
7π
11π
x = nπ or x =
+ n π or x =
+ n π , where n ∈ I.
6
6
PTS: 1
DIF: Average
OBJ: Section 6.4
TOP: Solving Trigonometric Equations Using Identities
60
NAT: T6
KEY: general solutions | equation
ID: A
27. ANS:
t
ÊÁ 1 ˆ˜ 10
a) A = 72 ÁÁÁÁ ˜˜˜˜
Ë 2¯
t
ÊÁ 1 ˆ˜ 10
b) A = 72 ÁÁÁÁ ˜˜˜˜
Ë 2¯
20
ÊÁ 1 ˆ˜ 10
= 72 ÁÁÁÁ ˜˜˜˜
Ë 2¯
ÊÁ 1 ˆ˜ 2
= 72 ÁÁÁÁ ˜˜˜˜
Ë 2¯
= 18
There will be 18 mg remaining after 20 days.
t
ÁÊ 1 ˜ˆ 10
c) A = 72 ÁÁÁÁ ˜˜˜˜
Ë 2¯
ÊÁ 1 ˆ˜
= 72 ÁÁÁÁ ˜˜˜˜
Ë 2¯
−30
10
ÊÁ 1 ˆ˜ −3
= 72 ÁÁÁÁ ˜˜˜˜
Ë 2¯
= 576
There was 576 mg 30 days ago.
t
d)
ÊÁ 1 ˆ˜ 10
A = 72 ÁÁÁÁ ˜˜˜˜
Ë 2¯
t
ÊÁ 1 ˆ˜ 10
0.07 = 72 ÁÁÁÁ ˜˜˜˜
Ë 2¯
t
0.07 ÊÁÁÁ 1 ˆ˜˜˜ 10
= Á ˜˜
ÁË 2 ¯
72
Use systematic trial.
0.07
= 0.000 972
72 Ö
ÊÁ 1 ˆ˜ 10
For t = 100, ÁÁÁÁ ˜˜˜˜ =
Ö 0.000 977.
Ë2¯
It will take approximately 100 days for there to be 0.07 mg remaining.
PTS: 1
NAT: RF9 | RF10
DIF: Average
OBJ: Section 7.2 | Section 7.3
TOP: Transformations of Exponential Functions | Solving Exponential Equations
61
ID: A
KEY: modelling | evaluate exponential functions
28. ANS:
a) y = 2
−2 (x − 2 )
+6
1
b) Reflect in the y-axis, compress horizontally by a factor of , and translate 2 units to the right and 6 units
2
up.
c)
d) y = −2 −2(x − 2) − 6
e)
PTS: 1
DIF: Average
TOP: Exponential Functions
29. ANS:
3
OBJ: Section 7.2 NAT: RF9
KEY: graph | transformations of exponential functions
256 2 × 16 x = 64 x − 3
2
(2 8 ) 3 × 24x = 26x − 18
4x +
2
4x +
16
3
= 26x − 18
16
= 6x − 18
3
−2x = −
x=
70
3
35
3
PTS: 1
DIF: Average
TOP: Solving Exponential Equations
OBJ: Section 7.3 NAT: RF10
KEY: exponential equation | change of base
62
ID: A
30. ANS:
2 3x = 2 2
3x = 2
x=
2
3
PTS: 1
DIF: Easy
OBJ: Section 7.3 NAT: RF10
TOP: Solving Exponential Equations
KEY: exponential equation | change of base
31. ANS:
5.25
a) The quarterly interest rate is
or 1.3125, and the number of compounding periods is 4 × 5 or 20.
4
A = P (1 + i)
n
= 21 500 (1 + 0.013 125)
20
≈ 27 906.10
The value of the investment after 5 years is $27 906.10.
n
b) 43 000 = 21 500 (1 + 0.013125)
2 = (1.013125)
Using systematic trial, n = 53.2 quarters or approximately 13.3 years.
n
PTS: 1
DIF: Average
OBJ: Section 7.3 NAT: RF10
TOP: Solving Exponential Equations
KEY: exponential equation | compound interest formula
32. ANS:
81.5% of 20 mg is 16.3 mg, and 3.1 min is 186 s.
t
ÊÁ 1 ˆ˜ 186
16.3 = 20 ÁÁÁÁ ˜˜˜˜
Ë 2¯
0.815 = 0.5
t
186
Using systematic trial,
t
= 0.3. Therefore, the time is approximately 56 s.
186
PTS: 1
DIF: Average
TOP: Solving Exponential Equations
33. ANS:
OBJ: Section 7.3 NAT: RF10
KEY: exponential equation | half-life
same shape, vertical translation up by $1000
PTS: 1
DIF: Easy
OBJ: Section 7.2 NAT: RF10
TOP: Characteristics of Exponential Functions | Transformations of Exponential Functions
KEY: exponential equation | compound interest formula | transformations
63
ID: A
34. ANS:
log 28 = log(7 × 2 2 )
= log7 + 2log2
≈ 0.8451 + 2(0.3010)
= 1.4471
PTS: 1
DIF: Difficult
OBJ: Section 8.3 NAT: RF9
TOP: Laws of Logarithms
KEY: power law of logarithms
35. ANS:
Let m1 represent the apparent magnitude of Sirius, b1 represent the brightness of Sirius, m2 represent the
apparent magnitude of the Sun, and b2 represent the brightness of the Sun.
ÊÁ b ˆ˜
Á 1˜
a) m2 − m1 = log ÁÁÁÁ ˜˜˜˜
Á b2 ˜
Ë ¯
ÊÁ b ˆ˜
Á 1˜
0.12 + 1.5 = log ÁÁÁÁ ˜˜˜˜
Á b2 ˜
Ë ¯
ÊÁ b ˆ˜
Á 1˜
1.62 = log ÁÁÁÁ ˜˜˜˜
Á b2 ˜
Ë ¯
10 1.62 =
b1
b2
b1
≈ 41.69
b2
Sirius is approximately 42 times as bright as the Sun.
ÊÁ b ˆ˜
Á 2˜
b) m1 − m2 = log ÁÁÁÁ ˜˜˜˜
Á b1 ˜
Ë ¯
−1.5 − m2 = log(1.3×10 10 )
m2 ≈ −1.5 − 10.11
m2 ≈ −11.61
The apparent magnitude of the Sun is –11.6.
PTS:
NAT:
TOP:
KEY:
1
DIF: Difficult
OBJ: Section 8.1 | Section 8.3 | Section 8.4
RF7 | RF9
Understanding Logarithms | Laws of Logarithms | Logarithmic and Exponential Equations
logarithmic equation | exponential function
64
ID: A
36. ANS:
L. S. = log a + loga 2 + loga 3 − loga 6
R. S. = log1
= log a + 2log a + 3log a − 6log a
=0
= (1 + 2 + 3 − 6) loga
= 0loga
=0
L.S. = R.S.
Thus, log 5 + log5 + log5 − log5 = log1.
2
3
6
PTS: 1
DIF: Average
TOP: Laws of Logarithms
37. ANS:
1
L. S. =
R. S. = log b a
log a b
OBJ: Section 8.3 NAT: RF9
KEY: power law | laws of logarithms
= 1 ÷ log a b
= 1÷
logb
loga
= 1×
loga
logb
= log b a
L.S. = R.S.
Thus,
1
log a b
= log b a .
PTS: 1
DIF: Difficult +
TOP: Laws of Logarithms
OBJ: Section 8.3 NAT: RF9
KEY: change of base formula | laws of logarithms
65
ID: A
38. ANS:
L. S. = log q 5 p 5
=
=
R. S. = log q p
log q p 5
log q q 5
5log q p
5
= log q p
L.S. = R.S.
Thus, log q p = log q p .
5
5
PTS: 1
DIF: Difficult +
TOP: Laws of Logarithms
39. ANS:
a), b)
OBJ: Section 8.3 NAT: RF10
KEY: laws of logarithms
c) The constant k will result in a vertical translation up by log 3 k . Since k is of the form 3n, where n is a whole
number, the translation is up n units.
PTS: 1
DIF: Average
OBJ: Section 8.2
TOP: Transformations of Logarithmic Functions
66
NAT: RF8
KEY: vertical translation
ID: A
40. ANS:
For the first 5 years, the investment is compounded monthly for a total of 5 × 12, or 60, periods.
A = 18 000(1 + 0.0065) 60
= 26 552.12
Solve for the remaining time—compounded daily for n years is 365n periods.
ÊÁ
ˆ˜ 365n
0.05
Á
˜˜
35 000 = 26 552.12 ÁÁÁ 1 +
˜˜
365
Ë
¯
ÊÁ 365.05 ˆ˜ 365n
35 000
˜˜
= ÁÁ
26 552.12 ÁÁË 365 ˜˜¯
ÊÁ 35 000 ˆ˜
Ê
ˆ
˜˜ = 365n log ÁÁÁ 365.05 ˜˜˜
log ÁÁÁÁ
˜
Á
˜
Á 365 ˜˜
Ë 26 552.12 ¯
Ë
¯
n ≈ 5.53
Bruce will need invest for approximately 5.5 more years.
PTS: 1
DIF: Difficult
OBJ: Section 8.4
TOP: Exponential and Logarithmic Equations
KEY: exponential function | logarithmic equation
67
NAT: RF10
ID: A
41. ANS:
Find the x-intercept by substituting y = 0 and solving for x.
ÍÈÍ 1
˙˘˙
0 = −2log ÍÍÍÍ (x + 1) ˙˙˙˙ − 1
ÍÎ 2
˙˚
ÈÍ
˘˙
Í1
˙
1
− = log ÍÍÍÍ (x + 1) ˙˙˙˙
2
ÍÎ 2
˙˚
−
10
−
2(10
1
2
1
2
=
1
(x + 1)
2
)−1= x
x ≈ −0.37
The x-intercept is −0.37.
PTS:
NAT:
TOP:
KEY:
1
DIF: Average
OBJ: Section 8.2 | Section 8.4
RF8 | RF10
Transformations of Logarithmic Functions | Logarithmic and Exponential Equations
transformation | logarithmic equation
68
ID: A
42. ANS:
a) t(v) =
5
,v > 0
v
b)
c) 1 h and 7 min
PTS:
NAT:
TOP:
KEY:
1
DIF: Easy
OBJ: Section 9.1 | Section 9.3
RF14
Exploring Rational Functions Using Transformations | Connecting Graphs and Rational Equations
word problem | reciprocal of linear function | graph
69
ID: A
43. ANS:
a) Since the pressure, p, exerted by a person’s shoe is inversely proportional to the width, w, of the shoe, you
k
k
can write p(w) = 2 for some k. Since p(2) = 400, substituting these values into p(w) = 2 gives k = 1600.
w
w
1600
p(w) = 2 , w > 0
w
b)
c) p(w) =
1600
w2
p(0.5) =
1600
0.5 2
= 6400
Megumi exerts 6400 kPa of pressure.
PTS: 1
DIF: Average
OBJ: Section 9.3
TOP: Connecting Graphs and Rational Equations
KEY: reciprocal of quadratic function | graph
70
NAT: RF14
ID: A
44. ANS:
a) i) I(d) =
10
d2
I(3) =
10
32
≈ 1.11
The intensity is 1.11 lux.
10
I(d) = 2
ii)
d
10 10
−
I(3) − I(1) 3 2 1 2
=
3−1
2
= −4.44
The rate of change of intensity is –4.44 lux/m.
b) As the distance from the light increases, the intensity decreases.
PTS: 1
DIF: Average
OBJ: Section 9.1 | Section 9.3
NAT: RF14
TOP: Exploring Rational Functions Using Transformations | Connecting Graphs and Rational Equations
KEY: reciprocal of quadratic function | average rate of change
45. ANS:
Answers may vary. Sample answer:
1
f(x) = 2
x −4
k
, k > 0, is a reasonable candidate since it is difficult to tell from the
Any function of the form f(x) = 2
x −4
graph how stretched the function is.
PTS: 1
DIF: Average
TOP: Analysing Rational Functions
46. ANS:
Answers may vary. Sample answer:
2(x − 1) 2
f(x) =
(x − 1)(x − 3)
OBJ: Section 9.2 NAT: RF14
KEY: reciprocal of quadratic function
PTS: 1
DIF: Difficult
OBJ: Section 9.2 | Section 9.3
NAT: RF14
TOP: Analysing Rational Functions | Connecting Graphs and Rational Equations
KEY: rational function | hole
71
ID: A
47. ANS:
a) For f(x) =
For g(x) =
ÊÁ
x+1
1 ˆ˜
: asymptotes x = 5, y = 1; intercepts (−1,0), ÁÁÁÁ 0,− ˜˜˜˜
x−5
5¯
Ë
x−5
: asymptotes x = −1, y = 1; intercepts (5,0), (0,−5)
x+1
b) There is reflective symmetry in this graph, about a vertical mirror line that runs exactly halfway between
the two vertical asymptotes.
c) x = 2
d) f(x) > 0 for x < −1 or x > 5 and f(x) < 0 for −1 < x < 5.
g(x) > 0 for x < −1 or x > 5 and g(x) < 0 for −1 < x < 5.
The intervals where f and g are positive and negative are identical.
x+b
x+d
e) Yes, this pattern is true for all pairs of functions f(x) =
and g(x) =
because when you take the
x+d
x+b
reciprocal of any y-value, it does not change sign (i.e., the positive y-values of f remain positive when their
reciprocals are taken).
PTS: 1
DIF: Difficult
OBJ: Section 9.2 | Section 9.3
NAT: RF14
TOP: Analysing Rational Functions | Connecting Graphs and Rational Equations
KEY: linear expressions in numerator and denominator | graph | reciprocal
72
ID: A
48. ANS:
Let w represent the wind speed.
Then, 600 + w is the ground speed with the wind
600 − w is the ground speed against the wind
10
3 h 20 min is equal to
h.
3
Therefore,
990
990
10
+
=
600 + w 600 − w
3
(3)990 (600 − w) + (3)990 (600 + w) = 10 (600 + w) (600 − w)
(3)99(600 − w) + (3)99 (600 + w) = (600 + w) (600 − w)
178 200 − 297w + 178 200 + 297w = 360 000 − w 2
w 2 = 3600
w = 60
The wind speed is 60 mph.
PTS: 1
DIF: Average
OBJ: Section 9.3
TOP: Connecting Graphs and Rational Equations
49. ANS:
Let x represent the number of members going on the ski trip.
480 480
−
= 4.8
x
x+5
ÊÁ 480 480 ˆ˜
˜˜ = 4.8 (x) (x + 5)
(x) (x + 5) ÁÁÁÁ
−
˜˜
x
x
+
5
Ë
¯
NAT: RF14
KEY: rational equation
480 (x + 5) − 480x = 4.8 (x) (x + 5)
100(x + 5) − 100x = x(x + 5)
100x + 500 − 100x = x 2 + 5x
x 2 + 5x − 500 = 0
(x + 25) (x − 20) = 0
x = −25, x = 20
The negative answer cannot be used in the context of this problem, and it is excluded.
20 members of the ski club are going on the trip.
PTS: 1
DIF: Average
OBJ: Section 9.3 NAT: RF14
TOP: Connecting Graphs and Rational Equations
KEY: linear expressions in numerator and denominator | x-intercept | excluded solution
73
ID: A
50. ANS:
h (x) =
=
1
1 − sin 2 x
1
cos 2 x
= sec 2 x
PTS: 1
DIF: Average
TOP: Composite Functions
51. ANS:
OBJ: Section 10.3 NAT: RF1
KEY: composite functions
PTS: 1
DIF: Difficult
OBJ: Section 10.2 NAT: RF1
TOP: Products and Quotients of Functions
KEY: divide functions | vertical asymptotes | hole
52. ANS:
Domain: {x| x ∈ R}
ÏÔÔ
¸ÔÔ
1
Ô
Range: ÔÌ y | ≤ y ≤ 2, y ∈ R ÔÔ˝
ÔÓÔ 2
Ô˛Ô
PTS: 1
DIF: Average
TOP: Composite Functions
OBJ: Section 10.3 NAT: RF1
KEY: composite functions | domain | range
74
ID: A
53. ANS:
From the graph, the x-values that result in the combined function having a value greater than zero (positive),
which is equivalent to 2 x > x 2 , are −0.8 < x < 2 or x > 4.
PTS: 1
DIF: Difficult +
OBJ: Section 10.1 NAT: RF1
TOP: Sums and Differences of Functions
KEY: combined function | inequality | graph
54. ANS:
a) D (x) = (8.0 + 6x) − (10.0 + 5x)
= −2.0 + x
This difference must be at least 100.
−2 + x ≥ 100
x ≥ 102
The minimum number sold is 102 items.
b) T (x) = (8.0 + 6x) + (10.0 + 5x)
= 18.0 + 11x
This total must be at least 1000.
18 + 11x ≥ 1000
11x ≥ 982
x ≥ 89.272…
The minimum number sold is 90 items.
PTS: 1
DIF: Difficult
OBJ: Section 10.1 NAT: RF1
TOP: Sums and Differences of Functions
KEY: add functions | subtract functions
55. ANS:
The difference in the y-intercepts is 18 − 12 = 6.
One ball was 6 m above the other.
PTS: 1
DIF: Easy
OBJ: Section 10.1 NAT: RF1
TOP: Sums and Differences of Functions
KEY: subtract functions | graph
75
ID: A
56. ANS:
a) Area of rectangle = 2x 2
2
ÁÊ x ˜ˆ
π ÁÁÁÁ ˜˜˜˜
Ë 2¯
Area of semi-circle =
2
=
Total area = 2x 2 +
=
πx 2
8
πx 2
8
16x 2 + πx 2
8
x 2 (16 + π )
8
2
x (16 + π )
b) Area =
8
=
=
1.2 2 (16 + π )
8
≈ 3.4
The area of the window is approximately 3.4 m2.
PTS: 1
DIF: Average
OBJ: Section 10.1 NAT: RF1
TOP: Sums and Differences of Functions
KEY: add functions
57. ANS:
a) The single digits range from 0 to 9. One digit is used 3 times, one digit is used twice, and the rest are used
only once. Since there are seven digits in total, the smallest digits make the smallest number. Therefore, the
number is 1 000 123.
b) The largest number can be found in a similar way using the largest possible digits. The largest number is 9
998 876. The difference between the numbers is 9 998 876 − 1 000 123 or 8 998 753.
PTS: 1
DIF:
TOP: Permutations
Difficult
OBJ: Section 11.1 NAT: PC2
KEY: permutations
76
ID: A
58. ANS:
The total number of possible tickets for lottery A is 49 C 6 or 13 983 816.
Therefore, the probability of randomly selecting the winning numbers is
1
P(A) =
13 983 816
The total number of possible tickets for lottery B is 49 C 7 or 85 900 584.
Therefore, the probability of randomly selecting the winning numbers is
1
P(B) =
85 900 584
Compare the probabilities:
1
P(A) 13 983 816
=
P(B)
1
85 900 584
=
85 900 584
13 983 816
≈ 6.14
Bernadette is not correct. The probability of winning lottery A is only 6.14 times as great.
PTS: 1
DIF:
TOP: Combinations
Difficult
OBJ: Section 11.2 NAT: PC3
KEY: combinations
77
ID: A
59. ANS:
a)
There are 24 possible choices for lunch.
b) The total numbers of ways is (24)(24) = 576.
c) Tanya and her friend will both have half of their choices eliminated. The total number of choices
is (12)(12) or 144.
PTS: 1
DIF:
TOP: Permutations
Easy
OBJ: Section 11.1 NAT: PC1
KEY: tree diagram
78
ID: A
60. ANS:
7!
5040
a)
=
3!4!
144
= 35
There are 35 possible routes from home to the library.
ÊÁ 7! ˆ˜ ÊÁ 15! ˆ˜
˜˜ ÁÁ
˜
b) ÁÁÁÁ
˜˜ ÁÁ 4!11! ˜˜˜ = (35)(1365)
3!4!
Ë
¯Ë
¯
= 47 775
There are 47 775 possible routes from home to school.
c) There are 47 775 possible routes in each direction, so the total number of routes from home to school and
back is 47 7752 or 2 282 450 625.
d) The total time to walk all of the routes for 1 person is (40)(47 775) or 1 911 000 min.
Charlie and her classmates (23 people in all) can walk the routes in 1 911 000 ÷ 23 or about
83 087.0 min.
Convert to days:
83 087.0
≈ 57.7
(60)(24)
It would take about 58 days for the 23 students to walk all of the routes from home to school.
PTS: 1
DIF:
TOP: Permutations
61. ANS:
1 + tanθ
L.S. =
1 + cot θ
Difficult
OBJ: Section 11.1 NAT: PC1 | PC2
KEY: permutations | fundamental counting principle
R.S. =
1 − tanθ
cot θ − 1
sin θ
cos θ
=
cos θ
1+
sin θ
sinθ
cos θ
=
cos θ
−1
sinθ
cos θ + sin θ
cos θ
=
sinθ + cos θ
sinθ
cos θ − sinθ
cos θ
=
cos θ − sinθ
sinθ
ÁÊ cos θ + sinθ ˜ˆ˜ ÁÊÁ
˜ˆ˜
sinθ
˜˜ ÁÁ
˜˜
= ÁÁÁÁ
˜
Á
˜
cos θ
Ë
¯ Ë cos θ + sin θ ¯
ÁÊ cos θ − sinθ ˜ˆ˜ ÁÊÁ
˜ˆ˜
sinθ
˜˜ ÁÁ
˜˜
= ÁÁÁÁ
˜
Á
˜
cos θ
Ë
¯ Ë cos θ − sinθ ¯
1+
=
1−
sinθ
cos θ
=
sinθ
cos θ
L.S. = R.S.
PTS: 1
DIF: Average
OBJ: Section 6.1 | Section 6.3
NAT: T6
TOP: Reciprocal, Quotient, and Pythagorean Identities | Proving Identities
KEY: quotient identities | proof
79
ID: A
62. ANS:
log
x 2 + 48x =
3
log(x + 48x)
2
1
3
=
2
3
2
3
1
2
log(x 2 + 48x) =
3
3
log(x 2 + 48x) = 2
log(x 2 + 48x) = log100
x 2 + 48x = 100
x 2 + 48x − 100 = 0
(x + 50)(x − 2) = 0
x = −50 or x = 2
Check the values for extraneous roots.
In this case, both values are possible and solve the equation, so they are both valid.
PTS: 1
DIF: Difficult
OBJ: Section 8.4
TOP: Logarithmic and Exponential Equations
KEY: logarithmic equation | extraneous roots
80
NAT: RF10