Name: ________________________ Class: ___________________ Date: __________ ID: A

1

Final Exam Practice

Multiple ChoiceIdentify 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 translatedA 5 units to the right C 5 units downB 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 translatedA 9 units to the right C 9 units downB 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( ) = x2, shown in blue, to obtain the graph of g(x), shown in red?

A g x( ) + 3 = x − 5( )2 C g x( ) − 5 = x + 3( )

2

B g x( ) = x + 3( )2 − 5 D g x( ) = x − 5( )

2 + 3

Name: ________________________ ID: A

2

____ 4. The two functions in the graph shown are reflections of each other. Select the type of reflection(s).

A a reflection in the line y = x C a reflection in the y-axisB a reflection in the x-axis and the y-axis D a reflection in the x-axis

____ 5. When a function is reflected in the x-axis, the coordinates of point (x, y) becomeA (x, –y) C (–x, –y)B (–x, y) D (x, y)

Name: ________________________ ID: A

3

____ 6. Which of the graphs shown below represents the base function f(x) = x2 and the stretched function g(x) =

−15

x2?

A C

B D

Name: ________________________ ID: A

4

____ 7. Which is the graph of the function f(x) = x − 6( )2 + 3?

A C

B D

____ 8. In the graph shown, which transformations must be applied to the blue curve to obtain the red curve?

A a reflection in the x-axis and a translation of 5 units downB a reflection in the y-axis and a translation of 5 units upC a reflection in the x-axis and a translation of 5 units upD a reflection in the y-axis and a translation of 5 units down

Name: ________________________ ID: A

5

____ 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

Name: ________________________ ID: A

6

____ 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?

A f −1(x) = 13

x− 53

C f −1(x)= −13

x− 53

B f −1(x) = −13

x+ 53

D f −1(x) = 13

x+ 53

____ 12. Which of the following functions is the correct inverse for the function f(x) = −92

x + 6?

A f −1(x) = −29

x+ 43

C f −1(x) = −29

x− 43

B f −1(x)= 92

x+ 43

D f −1(x) = 92

x− 43

____ 13. Which of the following functions is the correct inverse for the function f(x) = x2 + 7, { x | x ≥ 0, x ∈ R}?

A f −1(x) = x − 7( )2 C f −1(x) = x − 7

B f −1(x) = x + 7 D f −1(x) = x + 7

Name: ________________________ ID: A

7

____ 14. Which graph represents the inverse of the graph shown?

A C

B D

Name: ________________________ ID: A

8

____ 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) = x − 2 is translatedA 2 units up C 2 units to the leftB 2 units to the right D 2 units down

____ 17. Compared to the graph of the base function f(x) = x , the graph of the function g(x) + 8 = x is translatedA 8 units to the left C 8 units to the rightB 8 units up D 8 units down

Name: ________________________ ID: A

9

____ 18. Compared to the graph of the base function f(x) = x , the graph of the function g(x) = x − 5 is translatedA 5 units down C 5 units rightB 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) = 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-axisC 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

____ 20. In the following graph, what transformations must be applied to the blue curve to obtain the red curve?

A a reflection in the x-axis, a vertical translation 5 units up, and a horizontal translation 3 units to the right

B a reflection in the x-axis, a vertical translation 5 units down, and a horizontal translation 3 units to the right

C a reflection in the x-axis, a vertical translation 3 units up, and a horizontal translation 5 units to the left

D a reflection in the x-axis, a vertical translation 3 units up, and a horizontal translation 5 units to the right

Name: ________________________ ID: A

10

____ 21. The two functions in the graph shown are reflections of each other. Select the type of reflection(s).

A a reflection in the y-axis C a reflection in the line y = xB a reflection in the x-axis and the y-axis D 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

Name: ________________________ ID: A

11

____ 23. Which of the following functions is the correct inverse for the function f(x) = x − 2, {x | x ≥ 0, x ∈ R}?

A f −1(x) = x + 2 C f −1(x) = x2 + 2

B f −1(x) = x + 2 D f −1(x) = x − 2( )2

____ 24. Which graph represents the square root of the graph shown?

A C

B D

Name: ________________________ ID: A

12

____ 25. Which graph shows the graphical solution to the radical equation 0 = 2 (x − 5) − 2?

A C

B D

____ 26. Which radical equation can be solved using the graph shown below?

A − 4− x = x + 2 C x + 2 = − 4+ x

B 4− x = x + 2 D 4+ x = x + 2

Name: ________________________ ID: A

13

____ 27. What is the solution to the radical equation 0 = x + 9 − 3?A 18 C –18B 36 D 0

____ 28. What is the solution to the radical equation 0 = 2 2(x + 4) − 8?

A –4 C 4B 12 D 128

____ 29. Which of the following is a polynomial function?

A y = −4x4 + 4x3 − 7x2 + 9x C g x( ) = x + 4

B f x( ) = −4x − 7 D y = −4x + 9

x2

____ 30. Which graph represents an odd-degree polynomial function with two x-intercepts?A C

B D

____ 31. If −9x3 + 9x2 + 4 is divided by 6x + 5, then the restriction on x is

A x ≠ 65

C x ≠ −56

B x ≠ 56

D x ≠ −65

Name: ________________________ ID: A

14

____ 32. If −2x3 − 6x2 + 5x − 7 is divided by x − 7 to give a quotient of −2x2 − 20x − 135 anda remainder of –952, then which of the following is true?

A x − 7( )(−2x2 − 20x − 135) = –952

B −2x3 − 6x2 + 5x − 7 = x − 7( )(−2x2 − 20x − 135) – 952

C x − 7( )(−2x2 − 20x − 135) = 952

D −2x3 − 6x2 + 5x − 7= x − 7( )(−2x2 − 20x − 135) + 952

____ 33. When P x( ) = 5x3 − 2x + 2 is divided by 5x − 2, the remainder is

A x2 + x + 125

C P(5 / 2) = 601/ 8

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)?

A x2 − 6 C x2 + 6B x + 6 D x − 6

____ 35. Which of the following binomials is a factor of x3 + 12x2 + 29x + 18 ?A x − 2 C x − 1B x − 9 D x + 2

____ 36. Determine the value of k so that x + 2 is a factor of x3 + 10x2 + 23x + k.A k = –1 C k = 14B k = –14 D k = 1

____ 37. Which of the following is the fully factored form of x3 + 2x2 − 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 x3 + 9x2 − 4x − 36?

A x − 2( ) 2 x + 9( ) C x + 2( ) x − 2( ) x − 9( )

B x − 2( ) 2 x − 9( ) D x + 2( ) x − 2( ) x + 9( )

____ 39. One root of the equation x3 + 7x2 − 33x − 135= 0 isA –3 C 9B 3 D –5

Name: ________________________ ID: A

15

____ 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

Name: ________________________ ID: A

16

____ 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)?A C

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

A 1389

rotations C 2779

rotations

B −1389

rotations D −2779

rotations

____ 44. If the angle θ is 1600° in standard position, in which quadrant does it terminate?A quadrant III C quadrant IIB quadrant IV D quadrant I

Name: ________________________ ID: A

17

____ 45. A ball is riding the waves at a beach. The ball’s up and down motion with the waves can be described using

the formula h = 2.3sinπt3

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃ , where h is the height, in metres, above the flat surface of the water and t is the

time, in seconds. What is the height of the ball, to the nearest hundredth of a metre, after t = 17 s?A –0.87 m C –1.99 mB –2.66 m D 1.99 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 80.0ℑB 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.A y = 2.36x C y = 2.36x + 0.92B 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 y = −cos x C y = cos xB y = sin x D y = −sin x

____ 49. The period (in degrees) of the graph of y = cos 4x isA 270° C 90°B 180° D 45°

Name: ________________________ ID: A

18

____ 50. Which function is represented by the graph shown below, where θ is in radians?

A y = −54

sin(−2x) C y =−2 cos(−54

x)

B y =−2 sin(−54

x) D y = −54

cos(−2x)

____ 51. The graph of y = sinx can be obtained by translating the graph of y = cosx

Aπ4

units to the right Cπ3

units to the right

Bπ2

units to the right D π units to the right

____ 52. What is the period of the sinusoidal function y = −cos 8 x − π2

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

Ê

Ë

ÁÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃̃ − 2?

A18

π C14

π

B 4π D12

π

____ 53. Which of the following is not an asymptote of the function f θ( ) = tanθ?

A x = −72

π C x = −52

π

B x = −92

π D x = −π

____ 54. Which function has zeros only at θ = nπ,n ∈ I?

A y = tan(θ + 23

π) C y = tan(θ − 76

π)

B y = tan θ − π( ) D y = tan(θ + 54 π)

Name: ________________________ ID: A

19

____ 55. Given the trigonometric function y = tanx, which is the x-coordinate at which the function is undefined?

A92

π C −13

π

B −76

π D34

π

____ 56. Given the trigonometric function y = tanx, find the value of the y-coordinate of the point with x-coordinate –1200°.

A 3 C 1B −1 D undefined

____ 57. What are the solutions for sin2x −12

= 0 in the interval 0° ≤ x ≤ 360°?

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

h t( ) = 18cosπt80

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃ + 19, where t is the time, in seconds.

____ 58. What is the radius of the Ferris wheel?A 9 m C 19 mB 18 m D 36 m

____ 59. How long does it take for the wheel to revolve once?

Aπ80

s C 160 s

B 80 s D80π s

____ 60. What is the minimum height of a car?A 19 m C 160 mB 9 m D 80 m

____ 61. What is the maximum height of a car?A 19 m C 160 mB 80 m D 31 m

Name: ________________________ ID: A

20

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( ) = 14sin 80πt( ) + 14, where t is the time, in seconds.

____ 62. What is the period?

A740

s C140

s

B 8 s D114

s

____ 63. Which expression is equivalent to cosθ

1+ sinθ ?

Acosθ 1+ sinθ( )

1+ sin2θ C1− sinθcosθ

Bcosθ

1− sinθ D1+ sinθcosθ

____ 64. Which expression is equivalent to tanθ + cotθ?

A 1 C1

cosθ sinθ

B cosθ sinθ D 2sinθcosθ

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

____ 65. Which expression is equivalent to tanA− tanB1+ tanA tanB

?

A tan(A + B) C tan(A − B)B cot(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 –0.75B –1.15 D 0.42

____ 67. What is the general solution, in degress, to the equation 2cosxcos 2x − 2sinxsin2x = −1?A 40° + 180n° and80° + 180n°. where n ∈ I C 40° + 120n° and80° + 120n°, where n ∈ IB 40° + 120n°, where n ∈ I D 220° + 120n° and 330° + 120n°, where

n ∈ I

____ 68. What is the general solution, in radians, to the equation (4cos22θ + 1)sin13

θ = 0?

A 2πn where n ∈ I C 3πn where n ∈ I

B no solution Dπ3

n where n ∈ I

____ 69. Which set of properties does the function y = 2x have?A no x-intercept, no y-intercept C no x-intercept, y-intercept is 1B x-intercept is 1, no y-intercept D x-intercept is 0, y-intercept is 0

Name: ________________________ ID: A

21

____ 70. Which choice best describes the function y = 6x?A both increasing and decreasing C increasingB decreasing D neither increasing nor decreasing

____ 71. Which set of properties is correct for the function y = 19

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

x

?

A domain {x|x ∈ R}, range {y|y > 0,y ∈ R}

C domain {x|x ∈ R}, range {y|y ≤ 0,y ∈ R}

B domain {x|x ∈ R}, range {y|y ≥ 0,y ∈ R}

D domain {x|x ∈ R}, range {y|y < 0,y ∈ R}

____ 72. Which exponential equation matches the graph shown?

A y = 18

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

x

C y = − 18

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

x

B y = 8x D 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?

A p t( ) = 1500 3( )t C p t( ) = 1500 2( )

t

7

B p t( ) = 1500 2( )t D p t( ) = 1500 3( )

t

7

Name: ________________________ ID: A

22

____ 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 yearsB 26 years D 23 years

____ 75. Which function results when the graph of y = 6x is translated 2 units down?

A y = 6x − 2 C y = 6x − 2

B y = 6x + 2 D 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 = 9x?• a reflection in the y-axis• a vertical stretch by a factor of 6• a horizontal stretch by a factor of 7

A y = −7 9( )x

6 C y = 7 9( )x

6

B y = 6 9( )−x

7 D y = −6 9( )x

7

Name: ________________________ ID: A

23

____ 77. Which graph represents the function y = 279

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

x

?

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 1001 1042 1083 1124 1175 122

A y = 100 1.04( ) x C y = 100 1.04( ) x − 1

B y = 100 1.4( ) x D y = 100 1.4( ) x − 1

Name: ________________________ ID: A

24

____ 79. Solve for x, to one decimal place.

7333= 5x

A 1466.6 C 36 667.0B 11.1 D 5.5

____ 80. Solve for x.

36( ) 3x = 216 x + 7( )

A 0.3 C 6B 7 D 3.0

____ 81. The half-life of a radioactive element can be modelled by M = M0

132

Ê

Ë

ÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜

t

45, where M 0 is the initial mass of

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 11 h C 18 hB 10 h D 9 h

____ 82. Another way of writing 55 = 3125 is

A log55 = 3125 C log31255 = 5

B log53125= 5 D log55 = 3125

____ 83. Another way of writing 7−3 = 1343

is

A log7

13

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃ = −343 C log7

1343

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃ = −3

B log3 −17

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃ = 343 D log7 − 1

343

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃ = −3

____ 84. Compared to the graph of the base function y = log10x, the graph of the function y = log10x + 4 is translated

A 4 units to the left C 4 units upB 4 units down D 4 units to the right

Name: ________________________ ID: A

25

____ 85. Which graph represents the function y = −3log3[(x − 2)] − 3?

A C

B D

____ 86. Which if the following is equivalent to the expression log4sw10y?

A log4s+ 10log4w + log4y C log4s+ log4w + 10log4y

B 10log4s− 10log4w + log4y D 10log4s+ log4w + log4y

____ 87. Solve 8x = 486. Round your answer to two decimal places.A 3.59 C 1.78B 2.97 D 0.34

____ 88. What is true about the behaviour of the function f(x) = 14x + 5

as x → −54

+ (right to left)?

A f(x) → −∞ C f(x) → 0B f(x) → +∞ D f(x) is undefined

Name: ________________________ ID: A

26

____ 89. What is the x-intercept of f(x) = 12x + 4

?

A There is no x-intercept. C −2

B −12

D 0

____ 90. Which graph represents the function f(x) = −4x − 9

− 5?

A C

B D

Name: ________________________ ID: A

27

____ 91. Which function represents the graph shown below?

A f(x) = 9x − 8

+ 3 C f(x) = −9x + 3

− 8

B f(x) = −9x − 8

+ 3 D f(x) = 9x + 3

− 8

____ 92. Which of the following functions has a slant asymptote when graphed?

A f(x) = 5x3 − 10x2 − 15x

x2 − 3x − 4C f(x) = 5x3 − 10x2 − 15x

x2 − 3x

B f(x) = 5x3 − 10x2 − 15x

x2 − 2x − 3D all of the above

____ 93. Which function has vertical asymptotes with equations x = −9 and x = −67

?

A f(x) = 7x + 6

x2 + 15x + 54C f(x) = 1

7x2 − 69x + 54

B f(x) = 1

7x2 + 69x + 54D f(x) = −9

x2 + 15x + 54

____ 94. Which function has a point of discontinuity at x = 3?

A f(x) = x − 3

2x2 − 2x − 12C f(x) = x − 3

x2 − 6x − 12

B f(x) = x + 3

x2 − 6x − 12D f(x) = x + 3

x2 − 6x + 9

Name: ________________________ ID: A

28

____ 95. Which graph represents f(x) = x − 3

5x2 − 23x + 24?

A C

B D

____ 96. Which function has a y-intercept of − 827

?

A f(x) = −8

x2 − 12x − 27C f(x) = −8

x2 + 12x + 27

B f(x) = 8(−8x + 3)(x + 9)

D all of the above

____ 97. Which function has a horizontal asymptote with equation y = 27

?

A f(x) = −2x − 37x + 8

C f(x) = 7x − 32x + 8

B f(x) = 7x + 82x − 3

D f(x) = 2x − 37x + 8

Name: ________________________ ID: A

29

____ 98. Which function has an x-intercept of 13

?

A f(x) = −6x − 25x − 3

C f(x) = 6x − 25x − 3

B f(x) = 5x − 36x − 2

D f(x) = 5x − 26x − 3

____ 99. What is the equation for the vertical asymptote of the graph of the function shown?

A x = 2 C y = 7B x = 3 D y = 6

____ 100. Which function has a graph in the shape of a parabola?

A f(x) =(x − 3)2(x − 7)(x − 3)(x − 7)

C f(x) = x − 3

(x − 3)3(x − 7)

B f(x) =(x − 3)2(x − 7)

x − 3D none of the above

____ 101. What are the x-intercepts of the graph of f(x) = x2 + 7x − 18

x2 + 12x + 35?

A –7, –5 C 7, 5B 2, –9 D –2, 9

Name: ________________________ ID: A

30

____ 102. Solve the equation 0 = 6x3 + 6

−x3 − 8x2 graphically.

A no solution C 0B x = −1 D x = −8

____ 103. Given the functions f x( ) = x2 − 3 and g x( ) = −9− x, determine the equation for the combined function y = f x( ) + g x( ).

A y = x2 − 27x − 12 C y = x2 + 27x + 6

B y = x2 − x + 6 D y = x2 − x − 12

____ 104. Given the functions f x( ) = x2 − 8 and g x( ) = −2− x2, determine the equation for the combined function y = f x( ) + g x( ).

A −6− 2x2 C 16− x4

B −10 D 4 − x

Name: ________________________ ID: A

31

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 resemblesA C

B D

____ 106. Given the functions f x( ) = 9x and g x( ) = 9sinx, 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}

Name: ________________________ ID: A

32

____ 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( )?

A g x( ) = 12

x + 9( ) C g x( ) = 2 x + 9( )

B g x( ) = 2(x − 9) D g x( ) = 12

(x − 9)

Name: ________________________ ID: A

33

____ 108. Given the functions f x( ) = 0.6x and g x( ) = cosx, the graph of the combined function h x( ) = f x( )g x( ) most likely resemblesA C

B D

Name: ________________________ ID: A

34

____ 109. Given the functions f x( ) = x2 − 4 and g x( ) = x − 4, a graph of the combined function h x( ) =f x( )g x( )

most likely

resemblesA C

B D

Name: ________________________ ID: A

35

____ 110. An equation for the graph shown is most likely

A 4x + cosx C 4x + sinx

B 4x cosx D 4sinx

Name: ________________________ ID: A

36

____ 111. An equation for the graph shown is most likely

A f x( ) = cosxx

Csinxcosx

B 2sinx D f x( ) = sinxx

____ 112. Given the functions f x( ) = x + 3 and g x( ) = 1x − 3

, what is the simplified form of (f û g)(x)?

A (f û g)(x) = − 8x

, x ≠ 0 C (f û g)(x) = 3x − 8x − 3

, x ≠ 3

B (f û g)(x) = 3x − 8x + 3

, x ≠ −3 D (f û g)(x) = 1x

, x ≠ 0

____ 113. Given f x( ) = 9x2 + 7x and g x( ) = 2− x, determine 5g x( ) − f x( ).

A 45x2 + 36x − 2 C 45x2 + 6x − 5B 9x2 + 6x + 2 D −9x2 − 12x + 10

Name: ________________________ ID: A

37

____ 114. Given the functions f x( ) = x and g x( ) = −13

x − 5( ), which of the following is most likely the graph of

y = f(g(x))?A C

B D

Name: ________________________ ID: A

38

____ 115. Given the functions f x( ) = logx 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:

5Pr = 20

A 5 C 60B 2 D 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 24B 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 960B 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 ax3y8 C ax2y9

B ax8y3 D ax11

Name: ________________________ ID: A

39

____ 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 733B 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) = x2

b) f(x) = x| |

2. Determine the equation, in standard form, of each parabola after being transformed from f(x) = x2 by the given translations.a) 4 units to the right and 3 units upb) 2 units to the left and 1 unit upc) 2 units down and 7 units to the left

Name: ________________________ ID: A

40

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.

a) f(x) = x − 1( )2 + 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) = xii) f(x) = x2

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.

Name: ________________________ ID: A

41

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

c) g(x) = −12

5− x + 1

7. For each of the following, describe the combination of transformations that must be applied to the graph of f(x) = x2 (shown in blue) to obtain the graph of g(x) (shown in red).a)

b)

c)

Name: ________________________ ID: A

42

8. For each function f(x), i) determine f −1(x)ii) graph f(x) and its inverse

a) f(x) = 52

x − 3

b) f(x) = 3 x − 2( )2 − 3

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

b) horizontal stretch by a factor of 16

, then a vertical translation of 4 units down

c) horizontal reflection in the y-axis, then a vertical translation of 9 units up and horizontal translation of 2 units right

d) horizontal stretch by a factor of 13

, vertical reflection in the x-axis, and vertical stretch by a factor of 23

10. Sketch the graph of f(x) = −2x + 36

and use it to sketch the graph of y = f(x) .

11. Solve the equation 3x − 6 = 12 graphically.

12. Jim states that the equations x2 = 25 and xÊ

ËÁÁÁ

ˆ

¯˜̃˜

2

= 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 = (v0)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 x3 − 4 = 2 graphically.

15. Factor fully.a) x3 + 6x2 + 11x + 6b) 4x3 – 11x2 – 3xc) x4 – 81

16. Factor fully. a) x2(x – 2)(x + 2) + 3x + 6b) 16x4 – (x + 1)2

Name: ________________________ ID: A

43

17. Factor 2x3 + 5x2 – 14x – 8 fully

18. Solve. a) 3x3 + 2x2 – 8x + 3 = 0b) 2x3 + x2 – 10x – 5 = 0c) 5x4 = 7x2 – 2

19. Solve by factoring.a) x4 + 3x2 – 28 = 0b) 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

through an angle of π9

, what is the exact arc length through which the child travels?

22. A 3-m ladder is leaning against a vertical wall such that the angle between the ground and the ladder is π3

.

What is the exact height that the ladder reaches up the wall?

23. Given that sinx = cosπ5

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃ and that x lies in the first quadrant, determine the exact measure of angle x.

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

2 and cos B=

32

, determine the exact value

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 = cosx given a table of values containing ordered pairs for the function y = sinx.

Name: ________________________ ID: A

44

30. Describe the transformations that, when applied to the graph of y = cos x, result in the graph of

y = −2cos18

x − π3

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

È

Î

ÍÍÍÍÍÍÍÍÍ

˘

˚

˙̇˙̇˙̇˙̇̇

+ 1.

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+ 30cost, 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) = cosx + cosy is not an identity.

35. What is the solution for 2cosx − 3 = 0 for 0 ≤ x ≤ 2π?

36. Solve cot2θ + cotθ = 0. State the solution in general form.

37. Solve sec2θ − 2tanθ − 3 = 0. State the general solution to the nearest degree.

Name: ________________________ ID: A

45

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

Name: ________________________ ID: A

46

40. For the function y = 12

3( )x − 2,

a) describe the transformations of the function when compared to the function y = 3x

b) sketch the graph of the given function and y = 3x on the same set of axesc) state the domain, the range, and the equation of the asymptote

41. Write the equation for the function that results from each transformation or set of transformations applied to

the base function y = 5x.a) reflect in the y-axisb) shift 3 units to the rightc) shift 1 unit down and 4 units to the leftd) 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

43. Solve for n: 9n − 1 = 13

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

4n − 1

44. Graph the function f(x) = −log(x + 2) − 1. Identify the domain, the range, and the equation of the vertical asymptote.

Name: ________________________ ID: A

47

45. Given log27 ≈ 2.8074, find the value of log214.

46. Solve the equation 63x + 1 = 22x − 3. Leave your answer in exact form.

47. a) Determine an equation in the form f(x) = 1kx− c

for a function with a vertical asymptote at x = 2 and a

y-intercept of −18

.

b) Sketch the graph of the function.

Name: ________________________ ID: A

48

48. Consider the function f(x) = 34x − 5

.

a) Determine the key features of the function: i) domain and range ii) intercepts iii) equations of any asymptotesb) Sketch the graph of the function.

Name: ________________________ ID: A

49

49. Consider the function f(x) = 3x + 8x − 2

.

a) Determine the key features of the function: i) domain and range ii) intercepts iii) equations of any asymptotesb) Sketch the graph of the function.

50. Consider the function f(x) = x + 3

x2 − x − 12.

a) Determine the key features of the function: i) domain and range ii) intercepts iii) equations of any asymptotesb) Sketch the graph of the function.

51. Given the functions f x( ) = x + 1 and g x( ) = x2 + 3x + 1, determine a simplified equation for h x( ) = f x( ) + g x( ) .

52. Given the functions f x( ) = x2 − 4 and g x( ) = x2 − 3x + 2, determine a simplified equation for h x( ) =f x( )g x( )

.

Name: ________________________ ID: A

50

53. a) Graph the functions f x( ) = sin x( ) and g x( ) = 2x

2+ 1

Ê

Ë

ÁÁÁÁÁÁÁ

ˆ

¯

˜̃̃˜̃̃˜

.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 = 0.9x

2x2 − 3.

55. Given the functions f x( ) = x2 − 7 and g x( ) = 2− x3, what is the value of f(g(2))?

Name: ________________________ ID: A

51

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.

57. Use the binomial theorem to expand a2

− b3

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

4

.

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

can be modelled by the function h(t) = − 12

at2 + 25, where a is the acceleration due to gravity, in metres per

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?

Name: ________________________ ID: A

52

2. The base function f(x) = x is reflected in the x-axis, stretched horizontally by a factor of 2, compressed

vertically by a factor of 13

, and translated 3 units to the left and 5 units down.

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 −xand 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.

Name: ________________________ ID: A

53

6. The kinetic energy (energy of motion), E, in joules, of an object is given by the equation E = 12

mv2, where m

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 ofi) 200 Jii) 420 Jc) 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?

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 −x3 + 5x2 − 8x + 4 ≥ 0 algebraically and graphically.

11. Determine an equation in expanded form for the polynomial function represented by the graph.

Name: ________________________ ID: A

54

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 − 12

.

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

position is given by θ = 14

sinπ2

tÊ

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃ , where t is the time, in seconds. At what time(s) during the first 4 s is the

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 8:252 9:553 11:354 13:305 15:486 16:157 15:258 14:269 12:35

10 10:3911 9:0112 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.

Name: ________________________ ID: A

55

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 = cosx 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+ cosθ = sin2θ1− cosθ .

20. Prove the identity sinπ2

− xÊ

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃ cot

π2

+ xÊ

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃ = −sinx.

21. Prove the identity 1− cos 2θ + sin2θ1+ cos 2θ + sin2θ = tanθ.

22. Prove the identity cos2θ − sin2θ

cos2θ + sinθ cosθ= 1− tanθ.

23. An angle satisfies the relation secθ( ) cotθ( ) = 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 + sinθcos 3θ + cosθ = tan2θ.

25. Solve sin3x + sinx = cosx for 0 ≤ x ≤ 2π.

26. What is the general solution to tanx cscx + 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?

Name: ________________________ ID: A

56

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 = 2xto 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 25623 × 16x = 64x − 3.

30. Solve the equation 23x = 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.

35. The stellar magnitude scale compares the brightness of stars using the equation m2 − m1 = logb1

b2

Ê

Ë

ÁÁÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃˜̃, where

m1 and m2 are the apparent magnitudes (how bright the stars appear in the sky) of the two stars being

compared, and b1 and b2 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 1.3×1010 times as bright in the sky as does Sirius. What is the apparent magnitude of the Sun, to the nearest tenth?

36. Prove that loga + loga2 + loga3 − loga6 = log1.

37. Show that 1

logab

= logba.

38. Prove that logq5 p5 = logqp.

Name: ________________________ ID: A

57

39. a) Graph the function y = log3x.

b) Graph the following functions on the same graph: y = log33x

y = log39x

y = log327x

c) Explain the effect of the constant k in the function y = log3kx.

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.

41. Sketch the graph of the function f(x) = −2log12

x + 1( )

È

Î

ÍÍÍÍÍÍÍÍ

˘

˚

˙̇˙̇˙̇˙̇

− 1. Determine the x-intercept algebraically, to the

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

I(d) = 10d2

, d > 0.

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 < 3b) What does the sign of your answer to part a)ii) indicate about the light intensity?

Name: ________________________ ID: A

58

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 + 1x − 5

and its reciprocal

g(x) = x − 5x + 1

on the same set of axes.

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?

e) Does the pattern from part d) occur for all pairs of functions f(x) = x + bx + d

and g(x) = x + dx + b

, b ≠ d? Explain

why or why not.

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?

Name: ________________________ ID: A

59

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

1− x2 and g x( ) = sinx, determine the equation for h x( ) = f(g(x)).

51. Given the functions f x( ) = x2 − 3x − 10 and g x( ) = x2 − 5x, graph the function h x( ) =f x( )g x( )

. Label all

intercepts and identify any asymptotes and/or points of discontinuity.

52. What are the domain and range of the function y = 2sinx, where x is in radians?

Name: ________________________ ID: A

60

53. Use the graph of the combined function f x( ) = 2x − x2 to determine an approximate solution to the inequality

2x > x2.

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?

Name: ________________________ ID: A

61

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?

57. The number of different permutations using all of the 1-digit numbers of a set is given by 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.

Name: ________________________ ID: A

62

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?

Name: ________________________ ID: A

63

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+ tanθ1+ cotθ = 1− tanθ

cotθ − 1.

62. Solve the equation log x2 + 48x3 = 23

.

ID: A

1

Final Exam PracticeAnswer Section

MULTIPLE CHOICE

1. ANS: C PTS: 1 DIF: Average OBJ: Section 1.1NAT: RF2 TOP: Horizontal and Vertical Translations KEY: vertical translation

2. ANS: D PTS: 1 DIF: Easy OBJ: Section 1.1NAT: RF2 TOP: Horizontal and Vertical Translations KEY: horizontal translation

3. ANS: B PTS: 1 DIF: Average OBJ: Section 1.1NAT: RF2 TOP: Horizontal and Vertical Translations KEY: horizontal translation | vertical translation

4. ANS: D PTS: 1 DIF: Average OBJ: Section 1.2NAT: RF5 TOP: Reflections and Stretches KEY: reflection

5. ANS: A PTS: 1 DIF: Easy OBJ: Section 1.1NAT: RF5 TOP: Reflections and Stretches KEY: reflection

6. ANS: A PTS: 1 DIF: Average OBJ: Section 1.2NAT: RF3 | RF5 TOP: Reflections and Stretches KEY: graph | vertical stretch | reflection

7. ANS: C PTS: 1 DIF: Easy OBJ: Section 1.3NAT: RF4 TOP: Combining Transformations KEY: graph | horizontal translation | vertical translation

8. ANS: C PTS: 1 DIF: Average OBJ: Section 1.3NAT: RF4 TOP: Combining Transformations KEY: graph | vertical translation | reflection

9. ANS: D PTS: 1 DIF: Difficult OBJ: Section 1.3NAT: RF4 | RF5 TOP: Combining Transformations KEY: graph | vertical translation | horizontal translation | stretch | reflection

10. ANS: C PTS: 1 DIF: Average OBJ: Section 1.3NAT: RF4 | RF5 TOP: Combining Transformations KEY: graph | vertical translation | horizontal translation | stretch | reflection

11. ANS: A PTS: 1 DIF: Easy OBJ: Section 1.4NAT: RF6 TOP: Inverse of a Relation KEY: inverse of a function | function notation

12. ANS: A PTS: 1 DIF: Average OBJ: Section 1.4NAT: RF6 TOP: Inverse of a Relation KEY: inverse of a function | function notation

13. ANS: C PTS: 1 DIF: Average OBJ: Section 1.4NAT: RF6 TOP: Inverse of a Relation KEY: inverse of a function | function notation

14. ANS: D PTS: 1 DIF: Average OBJ: Section 1.4NAT: RF6 TOP: Inverse of a Relation KEY: graph | inverse of a function

15. ANS: D PTS: 1 DIF: Easy OBJ: Section 1.4NAT: RF6 TOP: Inverse of a Relation KEY: graph | inverse of a function

ID: A

2

16. ANS: D PTS: 1 DIF: Easy OBJ: Section 2.1NAT: RF13 TOP: Radical Functions and Transformations KEY: vertical translation

17. ANS: D PTS: 1 DIF: Average OBJ: Section 2.1NAT: RF13 TOP: Radical Functions and Transformations KEY: vertical translation

18. ANS: C PTS: 1 DIF: Easy OBJ: Section 2.1NAT: RF13 TOP: Radical Functions and Transformations KEY: horizontal translation

19. ANS: B PTS: 1 DIF: Average OBJ: Section 2.1NAT: RF13 TOP: Radical Functions and Transformations KEY: horizontal stretch | reflection

20. ANS: D PTS: 1 DIF: Average OBJ: Section 2.1NAT: RF13 TOP: Radical Functions and Transformations KEY: graph | horizontal translation | vertical translation | reflection

21. ANS: A PTS: 1 DIF: Average OBJ: Section 2.1NAT: RF13 TOP: Radical Functions and Transformations KEY: reflection

22. ANS: D PTS: 1 DIF: Average OBJ: Section 2.2NAT: RF13 TOP: Square Root of a Function KEY: graph

23. ANS: C PTS: 1 DIF: Easy OBJ: Section 2.1 | Section 2.2NAT: RF13 TOP: Radical Functions and Transformations | Square Root of a FunctionKEY: inverse of a radical function

24. ANS: D PTS: 1 DIF: Average OBJ: Section 2.2NAT: RF13 TOP: Square Root of a Function KEY: graph | square root | of a function

25. ANS: A PTS: 1 DIF: Average OBJ: Section 2.3NAT: RF13 TOP: Solving Radical Equations Graphically KEY: graphical solution

26. ANS: B PTS: 1 DIF: Easy OBJ: Section 2.3NAT: RF13 TOP: Solving Radical Equations Graphically KEY: graphical solution

27. ANS: D PTS: 1 DIF: Average OBJ: Section 2.3NAT: RF13 TOP: Solving Radical Equations Graphically KEY: algebraic solution

28. ANS: C PTS: 1 DIF: Difficult OBJ: Section 2.3NAT: RF13 TOP: Solving Radical Equations Graphically KEY: algebraic solution

29. ANS: A PTS: 1 DIF: Easy OBJ: Section 3.1NAT: RF12 TOP: Characteristics of Polynomial Functions KEY: polynomial function

30. ANS: B PTS: 1 DIF: Average OBJ: Section 3.1NAT: RF12 TOP: Characteristics of Polynomial Functions KEY: odd-degree | x-intercepts

31. ANS: C PTS: 1 DIF: Average OBJ: Section 3.2NAT: RF12 TOP: The Remainder Theorem KEY: restriction

32. ANS: B PTS: 1 DIF: Average OBJ: Section 3.2NAT: RF11 TOP: The Remainder Theorem KEY: quotient | remainder

ID: A

3

33. ANS: D PTS: 1 DIF: Difficult + OBJ: Section 3.2NAT: RF11 TOP: The Remainder Theorem KEY: remainder theorem | remainder

34. ANS: D PTS: 1 DIF: Easy OBJ: Section 3.3NAT: RF11 TOP: The Factor Theorem KEY: factor theorem | factor

35. ANS: D PTS: 1 DIF: Average OBJ: Section 3.3NAT: RF11 TOP: The Factor Theorem KEY: factor theorem | integral zero theorem | factor

36. ANS: C PTS: 1 DIF: Average OBJ: Section 3.3NAT: RF11 TOP: The Factor Theorem KEY: factor theorem | factor

37. ANS: B PTS: 1 DIF: Easy OBJ: Section 3.3NAT: RF11 TOP: The Factor Theorem KEY: factored form | factor theorem | factor

38. ANS: D PTS: 1 DIF: Average OBJ: Section 3.3NAT: RF11 TOP: The Factor Theorem KEY: factored form | factor theorem | factor

39. ANS: A PTS: 1 DIF: Average OBJ: Section 3.4NAT: RF12 TOP: Equations and Graphs of Polynomial FunctionsKEY: polynomial equation | roots

40. ANS: B PTS: 1 DIF: Average OBJ: Section 3.4NAT: RF12 TOP: Equations and Graphs of Polynomial FunctionsKEY: polynomial equation | roots | graph

41. ANS: C PTS: 1 DIF: Average OBJ: Section 3.4NAT: RF12 TOP: Equations and Graphs of Polynomial FunctionsKEY: polynomial equation | zeros | graph | multiplicity

42. ANS: A PTS: 1 DIF: Difficult + OBJ: Section 4.2NAT: T2 TOP: Unit Circle KEY: unit circle | unit circle equation

43. ANS: B PTS: 1 DIF: Average OBJ: Section 4.1NAT: T1 TOP: Angles and Angle Measure KEY: rotations | standard positionNOT: Mixed numbers

44. ANS: C PTS: 1 DIF: Average OBJ: Section 4.1NAT: T1 TOP: Angles and Angle Measure KEY: rotations | standard position

45. ANS: C PTS: 1 DIF: Easy OBJ: Section 4.4NAT: T4 TOP: Introduction to Trigonometric Equations KEY: trigonometric ratios

46. ANS: C PTS: 1 DIF: Difficult + OBJ: Section 4.1NAT: T1 TOP: Angles and Angle Measure KEY: arc length | degrees

47. ANS: A PTS: 1 DIF: Difficult OBJ: Section 4.3NAT: T3 TOP: Trigonometric Ratios KEY: unit circle | trigonometric ratios

48. ANS: A PTS: 1 DIF: Easy OBJ: Section 5.1NAT: T4 TOP: Graphing Sine and Cosine Functions KEY: graph | sinusoidal function

49. ANS: C PTS: 1 DIF: Easy OBJ: Section 5.1NAT: T4 TOP: Graphing Sine and Cosine Functions KEY: period | sinusoidal function

50. ANS: D PTS: 1 DIF: Average OBJ: Section 5.1NAT: T4 TOP: Graphing Sine and Cosine Functions KEY: function | amplitude | period | sinusoidal function

ID: A

4

51. ANS: B PTS: 1 DIF: Easy OBJ: Section 5.2NAT: T4 TOP: Transformations of Sinusoidal Functions KEY: translation | primary trigonometric function

52. ANS: C PTS: 1 DIF: Average OBJ: Section 5.2NAT: T4 TOP: Transformations of Sinusoidal Functions KEY: period | sinusoidal function

53. ANS: D PTS: 1 DIF: Easy OBJ: Section 5.3NAT: T4 TOP: The Tangent Function KEY: asymptote | tangent function

54. ANS: B PTS: 1 DIF: Difficult + OBJ: Section 5.3NAT: T4 TOP: The Tangent Function KEY: zeros | transformation

55. ANS: A PTS: 1 DIF: Average OBJ: Section 5.3NAT: T4 TOP: The Tangent Function KEY: undefined | tangent function

56. ANS: A PTS: 1 DIF: Average OBJ: Section 5.3NAT: T4 TOP: The Tangent Function KEY: coordinate | tangent function

57. ANS: A PTS: 1 DIF: Difficult + OBJ: Section 5.4NAT: T4 TOP: Equations and Graphs of Trigonometric FunctionsKEY: quadratic trigonometric equation

58. ANS: B PTS: 1 DIF: Easy OBJ: Section 5.4NAT: T4 TOP: Equations and Graphs of Trigonometric FunctionsKEY: amplitude | sinusoidal function

59. ANS: C PTS: 1 DIF: Average OBJ: Section 5.4NAT: T4 TOP: Equations and Graphs of Trigonometric FunctionsKEY: period | sinusoidal function

60. ANS: B PTS: 1 DIF: Average OBJ: Section 5.4NAT: T4 TOP: Equations and Graphs of Trigonometric FunctionsKEY: minimum | sinusoidal function

61. ANS: D PTS: 1 DIF: Average OBJ: Section 5.4NAT: T4 TOP: Equations and Graphs of Trigonometric FunctionsKEY: maximum | sinusoidal function

62. ANS: C PTS: 1 DIF: Easy OBJ: Section 5.4NAT: T4 TOP: Equations and Graphs of Trigonometric FunctionsKEY: period | sinusoidal function

63. ANS: C PTS: 1 DIF: Average OBJ: Section 6.1NAT: T6 TOP: Reciprocal, Quotient, and Pythagorean IdentitiesKEY: trigonometric identity

64. ANS: C PTS: 1 DIF: Average OBJ: Section 6.1NAT: T6 TOP: Reciprocal, Quotient, and Pythagorean IdentitiesKEY: trigonometric identity

65. ANS: C PTS: 1 DIF: Average OBJ: Section 6.2NAT: T6 TOP: Sum, Difference, and Double-Angle IdentitiesKEY: tangent | sum identities | difference identities

66. ANS: D PTS: 1 DIF: Average OBJ: Section 6.2NAT: T6 TOP: Sum, Difference, and Double-Angle IdentitiesKEY: sum identities | difference identities | evaluate

67. ANS: C PTS: 1 DIF: Difficult OBJ: Section 6.4NAT: T6 TOP: Solving Trigonometric Equations Using IdentitiesKEY: double-angle identities | general solutions

ID: A

5

68. ANS: C PTS: 1 DIF: Difficult OBJ: Section 6.4NAT: T6 TOP: Solving Trigonometric Equations Using IdentitiesKEY: double-angle identities | general solutions

69. ANS: C PTS: 1 DIF: Easy OBJ: Section 7.1NAT: RF9 TOP: Characteristics of Exponential Functions KEY: intercepts | exponential function

70. ANS: C PTS: 1 DIF: Easy OBJ: Section 7.1NAT: RF9 TOP: Characteristics of Exponential Functions KEY: increasing | decreasing

71. ANS: A PTS: 1 DIF: Average OBJ: Section 7.1NAT: RF9 TOP: Characteristics of Exponential Functions KEY: domain | range

72. ANS: A PTS: 1 DIF: Average OBJ: Section 7.1NAT: RF9 TOP: Characteristics of Exponential Functions KEY: equation | graph | exponential function

73. ANS: C PTS: 1 DIF: Average OBJ: Section 7.2NAT: RF9 TOP: Transformations of Exponential Functions KEY: modelling | exponential growth

74. ANS: D PTS: 1 DIF: Easy OBJ: Section 7.3NAT: RF10 TOP: Solving Exponential Equations KEY: compound interest

75. ANS: C PTS: 1 DIF: Easy OBJ: Section 7.2NAT: RF9 TOP: Transformations of Exponential Functions KEY: transformations of exponential functions

76. ANS: B PTS: 1 DIF: Easy OBJ: Section 7.2NAT: RF9 TOP: Transformations of Exponential Functions KEY: transformations of exponential functions

77. ANS: A PTS: 1 DIF: Average OBJ: Section 7.2NAT: RF9 TOP: Transformations of Exponential Functions KEY: graph | transformations of exponential functions

78. ANS: A PTS: 1 DIF: Difficult OBJ: Section 7.1NAT: RF9 TOP: Characteristics of Exponential Functions KEY: modelling | exponential function

79. ANS: D PTS: 1 DIF: Average OBJ: Section 7.3NAT: RF10 TOP: Solving Exponential Equations KEY: exponential equation | systematic trial

80. ANS: B PTS: 1 DIF: Average OBJ: Section 7.3NAT: RF10 TOP: Solving Exponential Equations KEY: exponential equation | equate exponents

81. ANS: D PTS: 1 DIF: Difficult OBJ: Section 7.3NAT: RF10 TOP: Solving Exponential Equations KEY: half-life | exponential decay

82. ANS: B PTS: 1 DIF: Easy OBJ: Section 8.1NAT: RF7 TOP: Understanding Logarithms KEY: logarithm | exponential functionNOT: Draft

83. ANS: C PTS: 1 DIF: Easy OBJ: Section 8.1NAT: RF7 TOP: Understanding Logarithms KEY: logarithm | exponential functionNOT: Draft

ID: A

6

84. ANS: C PTS: 1 DIF: Easy OBJ: Section 8.2NAT: RF8 TOP: Transformations of Logarithmic Functions KEY: vertical translation | transformation

85. ANS: D PTS: 1 DIF: Average OBJ: Section 8.2NAT: RF8 TOP: Transformations of Logarithmic Functions KEY: horizontal translation | vertical translation | vertical stretch | horizontal stretch

86. ANS: A PTS: 1 DIF: Easy OBJ: Section 8.3NAT: RF9 TOP: Laws of Logarithms KEY: product law | laws of logarithms

87. ANS: B PTS: 1 DIF: Easy OBJ: Section 8.4NAT: RF10 TOP: Logarithmic and Exponential Equations KEY: exponential equation

88. ANS: B PTS: 1 DIF: Average OBJ: Section 9.1NAT: RF14 TOP: Exploring Rational Functions Using TransformationsKEY: reciprocal of linear function | behaviour at non-permissible values

89. ANS: A PTS: 1 DIF: Easy OBJ: Section 9.1NAT: RF14 TOP: Exploring Rational Functions Using TransformationsKEY: reciprocal of linear function | x-intercept

90. ANS: B PTS: 1 DIF: Average OBJ: Section 9.1NAT: RF14 TOP: Exploring Rational Functions Using TransformationsKEY: reciprocal of linear function | graph from function

91. ANS: A PTS: 1 DIF: Average OBJ: Section 9.1NAT: RF14 TOP: Exploring Rational Functions Using TransformationsKEY: reciprocal of linear function | function from graph

ID: A

7

92. ANS: A

PTS: 1 DIF: Difficult + OBJ: Section 9.1 NAT: RF14TOP: Exploring Rational Functions Using Transformations KEY: slant asymptote | hole | factor

93. ANS: B PTS: 1 DIF: Average OBJ: Section 9.2NAT: RF14 TOP: Analysing Rational Functions KEY: reciprocal of quadratic function | vertical asymptote

94. ANS: A PTS: 1 DIF: Average OBJ: Section 9.2NAT: RF14 TOP: Analysing Rational Functions KEY: reciprocal of quadratic function | vertical asymptote

95. ANS: C PTS: 1 DIF: Difficult OBJ: Section 9.2NAT: RF14 TOP: Analysing Rational Functions KEY: rational function | discontinuity | hole

96. ANS: C PTS: 1 DIF: Average OBJ: Section 9.2NAT: RF14 TOP: Analysing Rational Functions KEY: reciprocal of quadratic function | y-intercept

97. ANS: D PTS: 1 DIF: Average OBJ: Section 9.2NAT: RF14 TOP: Analysing Rational Functions KEY: linear expressions in numerator and denominator | horizontal asymptote

ID: A

8

98. ANS: C PTS: 1 DIF: Average OBJ: Section 9.2NAT: RF14 TOP: Analysing Rational Functions KEY: linear expressions in numerator and denominator | x-intercept

99. ANS: B PTS: 1 DIF: Easy OBJ: Section 9.2NAT: RF14 TOP: Analysing Rational Functions KEY: quadratic denominator | vertical asymptote

100. ANS: B PTS: 1 DIF: Average OBJ: Section 9.2NAT: RF14 TOP: Analysing Rational Functions KEY: hole

101. ANS: B PTS: 1 DIF: Average OBJ: Section 9.3NAT: RF14 TOP: Connecting Graphs and Rational Equations KEY: rational function | x-intercept

102. ANS: B

PTS: 1 DIF: Difficult + OBJ: Section 9.3 NAT: RF14TOP: Connecting Graphs and Rational Equations KEY: rational equation | graph

103. ANS: D PTS: 1 DIF: Easy OBJ: Section 10.1NAT: RF1 TOP: Sums and Differences of Functions KEY: add functions | subtract functions

104. ANS: B PTS: 1 DIF: Easy OBJ: Section 10.1NAT: RF1 TOP: Sums and Differences of Functions KEY: subtract functions | add functions

105. ANS: A PTS: 1 DIF: Difficult OBJ: Section 10.1NAT: RF1 TOP: Sums and Differences of Functions KEY: add functions | graph | subtract functions

106. ANS: C PTS: 1 DIF: Easy OBJ: Section 10.2NAT: RF1 TOP: Products and Quotients of Functions KEY: multiply functions | range

107. ANS: A PTS: 1 DIF: Difficult OBJ: Section 10.3NAT: RF1 TOP: Composite Functions KEY: composite functions | transformations | graph

ID: A

9

108. ANS: B PTS: 1 DIF: Difficult OBJ: Section 10.2NAT: RF1 TOP: Products and Quotients of Functions KEY: multiply functions | graph

109. ANS: C PTS: 1 DIF: Average OBJ: Section 10.2NAT: RF1 TOP: Products and Quotients of Functions KEY: divide functions | graph

110. ANS: A PTS: 1 DIF: Average OBJ: Section 10.1NAT: RF1 TOP: Sums and Differences of Functions KEY: add functions | graph

111. ANS: D PTS: 1 DIF: Difficult OBJ: Section 10.2NAT: RF1 TOP: Products and Quotients of Functions KEY: divide functions | graph

112. ANS: C PTS: 1 DIF: Average OBJ: Section 10.2NAT: RF1 TOP: Composite Functions KEY: composite functions | notation

113. ANS: D PTS: 1 DIF: Average OBJ: Section 10.1NAT: RF1 TOP: Sums and Differences of Functions KEY: add functions

114. ANS: D PTS: 1 DIF: Average OBJ: Section 10.3NAT: RF1 TOP: Composite Functions KEY: composite functions | transformations | graph

115. ANS: C PTS: 1 DIF: Average OBJ: Section 10.3NAT: RF1 TOP: Composite Functions KEY: composite functions | transformations | graph

116. ANS: B PTS: 1 DIF: Easy OBJ: Section 11.1NAT: PC2 TOP: Permutations KEY: permutations

117. ANS: D PTS: 1 DIF: Average OBJ: Section 11.1NAT: PC2 TOP: Permutations KEY: fundamental counting principle

118. ANS: B PTS: 1 DIF: Difficult OBJ: Section 11.2NAT: PC3 TOP: Combinations KEY: combinations

119. ANS: C PTS: 1 DIF: Average OBJ: Section 11.3NAT: PC4 TOP: The Binomial Theorem KEY: binomial expansion | binomial theorem

120. ANS: A PTS: 1 DIF: Difficult OBJ: Section 11.2NAT: PC3 TOP: Combinations KEY: combinations | fundamental counting principle

ID: A

10

SHORT ANSWER

1. ANS:

a) The graph of f(x) = x2 is shown in blue and the graph of g(x) = x − 1( )2 + 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 OBJ: Section 1.3 NAT: RF4TOP: Combining Transformations KEY: translation

2. ANS:

a) g(x) = x − 4( )2 + 3

= x2 − 8x + 16+ 3

= x2 − 8x + 19

b) g(x) = x + 2( )2 + 1

= x2 + 4x + 4+ 1

= x2 + 4x + 5

c) g(x) = x + 7( )2 − 2

= x2 + 14x + 49− 2

= x2 + 14x + 47

PTS: 1 DIF: Average OBJ: Section 1.3 NAT: RF4TOP: Combining Transformations KEY: translation

ID: A

11

3. ANS: a)

b)

c)

PTS: 1 DIF: Difficult + OBJ: Section 1.2 NAT: RF3 | RF5TOP: Reflections and Stretches KEY: graph | reflection

4. ANS: a) g x( ) = −f(x)

= − x − 1( )2 − 2

b) g(x) = f −x( )

= −x| | + 1

= x| | + 1

PTS: 1 DIF: Average OBJ: Section 1.3 NAT: RF4 | RF5TOP: Combining Transformations KEY: reflection | translation

ID: A

12

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) = x2 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) = 8x2 (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 OBJ: Section 1.2 NAT: RF3TOP: Reflections and Stretches KEY: stretch | graph

ID: A

13

6. ANS:

a) a reflection in the x-axis, a horizontal compression by a factor of 12

, and then a translation of 1 unit to the

left and 2 units downb) a vertical stretch by a factor of 2, and then a translation of 3 units to the right and 4 units down

c) reflections in the x-axis and the y-axis, a vertical compression by a factor of 12

, and then a translation of 5

units to the right and 1 unit up

PTS: 1 DIF: Difficult + OBJ: Section 1.3 NAT: RF4 | RF5TOP: 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 upb) a vertical stretch by a factor of 2, and then a translation of 1 unit to the left and 1 unit down

c) a vertical compression by a factor of 12

, a reflection in the x-axis, and then a translation of 2 units to the

right and 2 units up

PTS: 1 DIF: Average OBJ: Section 1.3 NAT: RF4TOP: Combining Transformations KEY: graph | transformation

ID: A

14

8. ANS:

a) i) y = 52

x − 3

x = 52

y − 3

x + 3 = 52

y

2x + 6 = 5y

y = 25

x + 65

f −1(x) = 25

x + 65

ii) The graph of f(x) is shown in blue and the graph of f −1(x) is shown in red.

b) i) y = 3(x − 2)2 − 3

x = 3(y − 2)2 − 3

x + 3 = 3(y − 2)2

x + 33

= (y − 2)2

± x + 33

= y − 2

y = 2± x + 33

f −1(x) = 2± x + 33

ii) The graph of f(x) is shown in blue and the graph of f −1(x) is shown in red.

ID: A

15

PTS: 1 DIF: Difficult OBJ: Section 1.4 NAT: RF6TOP: 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) = −x + 2 + 9

d) g(x) = −23

113

x

or

g(x) = −23

3x

PTS: 1 DIF: Average OBJ: Section 2.1 NAT: RF13TOP: Radical Functions and Transformations KEY: transformations

ID: A

16

10. ANS:

The graph of y = f(x) is shown in black, and the graph of y = f(x) is shown in blue.

PTS: 1 DIF: Difficult OBJ: Section 2.2 NAT: RF13TOP: Square Root of a Function KEY: graph | square root of a function

11. ANS:

PTS: 1 DIF: Average OBJ: Section 2.3 NAT: RF13TOP: Solving Radical Equations Graphically KEY: graphical solution

12. ANS:

No, Jim is not correct. x2 = 25 has two possible solutions of ±25, and xÊ

ËÁÁÁ

ˆ

¯˜̃˜

2

= 25 has only one solution,

+25.

PTS: 1 DIF: Difficult OBJ: Section 2.3 NAT: RF13TOP: Solving Radical Equations Graphically KEY: algebraic solution

ID: A

17

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 = (v0)2 + 2ad

= 0+ 2(32)(85)

= 5440

v = 73.76Since 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 = (v0)2 + 2ad

80= 0+ 2(32)d

80= 64d

6400= 64d

100= dThe maximum height is 100 ft.

PTS: 1 DIF: Average OBJ: Section 2.3 NAT: RF13TOP: Solving Radical Equations Graphically KEY: algebraic solution

14. ANS:

PTS: 1 DIF: Difficult OBJ: Section 2.3 NAT: RF13TOP: Solving Radical Equations Graphically KEY: graphical solution

ID: A

18

15. ANS: a) The possible factors are (x ± 1), (x ± 2), and (x ± 3). Try x = −1 using the factor theorem.

P(x) = x3 + 6x2 + 11x + 6

P(−1) = (−1)3 + 6(−1)2 + 11(−1) + 6

= −1+ 6− 11+ 6

P(−1) = 0Thus, (x + 1) is a factor of P(x).Use synthetic division to find the quadratic factor.

1 1 6 11 6

− 1 5 6

× 1 5 6 0

Thus,

x3 + 6x2 + 11x + 6 = (x + 1)(x2 + 5x + 6)

= (x + 1)(x + 2)(x + 3)

b) 4x3 − 11x2 − 3x = x(4x2 − 11x − 3)

= x(4x2 − 12x + x − 3)

= x[4x(x − 3) + (x − 3)]

= x(x − 3)(4x + 1)

c) x4 − 81= (x2 − 9)(x2 + 9)

= (x − 3)(x + 3)(x2 + 9)

PTS: 1 DIF: Average OBJ: Section 3.3 NAT: RF11TOP: The Factor Theorem KEY: factor theorem | factorNOT: A variety of factoring techniques is required.

ID: A

19

16. ANS:

a) x2(x − 2)(x + 2) + 3x + 6 = x2(x − 2)(x + 2) + 3(x + 2)

= (x + 2)[x2(x − 2) + 3]

= (x + 2)(x3 − 2x2 + 3)Use the factor theorem on the second factor. Try x = −1.

P(x) = x3 − 2x2 + 3

P(−1) = (−1)3 − 2(−1)2 + 3

= −1− 2+ 3

P(−1) = 0Divide.

1 1 −2 0 3

− 1 −3 3

× 1 −3 3 0

The quotient is not factorable. Thus,

x2(x − 2)(x + 2) + 3x + 6 = (x + 2)(x + 1)(x2 − 3x + 3)

b) 16x4 − (x + 1)2 = (4x2)2 − (x + 1)2

= [4x2 − (x + 1)][4x2 + (x + 1)]

= (4x2 − x − 1)(4x2 + x + 1)

PTS: 1 DIF: Average OBJ: Section 3.3 NAT: RF11TOP: The Factor Theorem KEY: factor theorem | factor | integral zero theorem | grouping | rational zero theoremNOT: A variety of factoring techniques is required.

ID: A

20

17. ANS:

Possible values of x in the factor theorem are ±1, ±12

, ±2, ±4, and ±8.

Try x = 2.

P(x) = 2x3 + 5x2 − 14x − 8

P(2) = 2(2)3 + 5(2)2 − 14(2)− 8

= 16+ 20− 28− 8

P(2) = 0Thus, x − 2 is a factor of P(x).Divide.

−2 2 5 −14 −8

− −4 −18 −8

× 2 9 4 0

Thus,

2x3 + 5x2 − 14x − 8 = (x − 2)(2x2 + 9x + 4)

= (x − 2)(2x2 + x + 8x + 4)

= (x − 2)[x(2x + 1) + 4(2x + 1)]

= (x − 2)(2x + 1)(x + 4)

PTS: 1 DIF: Difficult + OBJ: Section 3.3 NAT: RF11TOP: The Factor Theorem KEY: factor theorem | factor | integral zero theorem | grouping | rational zero theoremNOT: A variety of factoring techniques is required.

ID: A

21

18. ANS: a) Try x = 1 in the factor theorem.

P(x) = 3x3 + 2x2 − 8x + 3

P(1) = 3(1)3 + 2(1)2 − 8(1)+ 3

= 3+ 2− 8+ 3

P(1) = 0Thus, x − 1 is a factor.Use synthetic division to find another factor.

−1 3 2 −8 3

− −3 −5 3

× 3 5 −3 0

Another factor is 3x2 + 5x − 3. Thus,

3x3 + 2x2 − 8x + 3 = 0

(x − 1)(3x2 + 5x − 3) = 0

x = 1 or 3x2 + 5x − 3 = 0Use the quadratic formula to find the other solutions.

3x2 + 5x − 3 = 0

x =−5± 52 − 4(3)(−3)

2(3)

=−5± 61

6

The solutions are x = −5− 61

6, 1,

−5+ 616

.

b) 2x3 + x2 − 10x − 5 = 0

x2(2x + 1) − 5(2x + 1) = 0

(x2 − 5)(2x + 1) = 0

x = 5, − 5, − 12

c) 5x4 = 7x2 − 2

5x4 − 7x2 + 2 = 0

5x4 − 5x2 − 2x2 + 2 = 0

5x2(x2 − 1) − 2(x2 − 1) = 0

(x2 − 1)(5x2 − 2) = 0

x = −1, 1,− 25

,25

ID: A

22

PTS: 1 DIF: Average OBJ: Section 3.3 NAT: RF11TOP: The Factor Theorem KEY: polynomial equation | factor theorem | factor | roots | integral zero theoremNOT: A variety of factoring techniques is required.

19. ANS:

a) x4 + 3x2 − 28= 0

(x2 + 7)(x2 − 4) = 0

(x2 + 7)(x − 2)(x + 2) = 0

x = 2, − 2

b) 2x4 − 54x = 0

2x(x3 − 27) = 0

2x(x − 3)(x2 + 3x + 9) = 0

x = 0, 3

PTS: 1 DIF: Average OBJ: Section 3.3 | Section 3.4NAT: RF11 TOP: The Factor Theorem | Equations and Graphs of Polynomial FunctionsKEY: polynomial equation | factor theorem | factor | integral zero theorem | rational zero theorem | rootsNOT: A variety of factoring techniques is required.

ID: A

23

20. ANS: a)

x > 1.9b)

x ≤ −13

PTS: 1 DIF: Difficult + OBJ: Section 3.4 NAT: RF12TOP: Equations and Graphs of Polynomial Functions KEY: technology | inequality | graph | roots

21. ANS: a = θr

= π9

(3)

a = π3

The child travels through an arc length of π3

m.

PTS: 1 DIF: Easy OBJ: Section 4.1 NAT: T1TOP: Angles and Angle Measure KEY: arc length

ID: A

24

22. ANS: Use the trigonometry of right triangles. The hypotenuse is the length of the ladder, or 3 m. The angle between

the ladder and the ground is π3

. The opposite side to the angle is the height the ladder reaches up the wall. Let

this height be h.h3

= sinπ3

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

h = 3sinπ3

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

h =3 3

2

The height the ladder reaches up the wall is 3 3

2 m.

PTS: 1 DIF: Average OBJ: Section 4.3 NAT: T3TOP: Trigonometric Ratios KEY: special angles | trigonometric ratios

23. ANS:

sinx = cosπ5

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

sinx = cosπ2

− 3π10

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

sinx = sin3π10

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

x = 3π10

PTS: 1 DIF: Average OBJ: Section 4.4 NAT: T4TOP: Introduction to Trigonometric Equations KEY: equivalent trigonometric expression | exact value

24. ANS:

Since cscθ = − 2

3, sinθ = −

32

. Since sin60° =3

2, the reference angle is 60°. The ratio is negative in

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°

PTS: 1 DIF: Average OBJ: Section 4.3 NAT: T2TOP: Trigonometric Ratios KEY: reference angle | reciprocal trigonometric ratios | unit circle

ID: A

25

25. ANS:

r = d2

= 212

= 10.5

a = rθ

a = (10.5)(1.2)

= 12.6The arc length is 12.6 cm.

PTS: 1 DIF: Average OBJ: Section 4.1 NAT: T1TOP: Angles and Angle Measure KEY: arc length | central angle

26. ANS:

sin A =2

2

∠A = π4

cos B=3

2

∠B = π6

sec A+ sec B= secπ4

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃ + sec

π6

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

= 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: T3TOP: 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 OBJ: Section 4.3 NAT: T3TOP: Trigonometric Ratios KEY: primary trigonometric ratios | exact value

ID: A

26

28. ANS: 5π6

PTS: 1 DIF: Easy OBJ: Section 4.1 NAT: T1TOP: Angles and Angle Measure KEY: radian

29. ANS:

You would need to subtract π2

or 90° from each x-value for y = sin x and plot the points using the

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: T4TOP: 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

of π3

to the right, and a vertical translation of 1 unit up

PTS: 1 DIF: Average OBJ: Section 5.2 NAT: T4TOP: Transformations of Sinusoidal Functions KEY: transformations | sinusoidal function

31. ANS: Solutions may vary. Sample solution:

The amplitude is half the diameter, or 30 cm. The maximum height of the pebble is 60 cm, so the vertical

displacement must be 30 cm. The wheel rotates at 4 revolutions per second, so the period is 14

s. Thus, the

value of b is 2π14

or 8π.

Thus, the relationship between the height of the pebble above the ground and time ish = 30sin 8πt( ) + 30

PTS: 1 DIF: Easy OBJ: Section 5.4 NAT: T4TOP: Equations and Graphs of Trigonometric Functions KEY: sinusoidal function | modelling

ID: A

27

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+ 30cost.

The graph is first increasing over the interval [3.14,6.28] or [π, 2π], or from approximately 3 years 123

months to 6 years 313

months.

PTS: 1 DIF: Average OBJ: Section 5.4 NAT: T4TOP: Equations and Graphs of Trigonometric Functions KEY: sinusoidal function | maximum | minimum

33. ANS: a) 0.7 m

b) Since b = 72 and period = 2πb

, then period = 2π72

or π36

s. The number of revolutions of the rope is the

reciprocal of the period, or 36π , or 11.46 rev/s. Multiply by 60 to get 688 revolutions/min.

PTS: 1 DIF: Difficult OBJ: Section 5.4 NAT: T4TOP: Equations and Graphs of Trigonometric Functions KEY: period | amplitude | sinusoidal function

ID: A

28

34. ANS: Answers may vary. Sample answer:

Use x = 0 and y = π2

:

L.S. = cos(x + y)

= cos 0+ π2

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

= cosπ2

= 0

R.S.= cosx + cosy

= cos 0+ cosπ2

= 1+ 0

= 1

L.S. ≠ R.S.Thus, cos(x + y) = cosx + cosy is not an identity.

PTS: 1 DIF: Average OBJ: Section 6.3 NAT: T6TOP: Proving Identities KEY: counterexample

35. ANS:

2cosx − 3 = 0

2cosx = 3

cosx =3

2

x = cos−1 32

Ê

Ë

ÁÁÁÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃˜̃̃

= π6

Since cosine is also positive in quadrant IV, another solution is 2π − π6

= 11π6

.

PTS: 1 DIF: Easy OBJ: Section 6.4 NAT: T6TOP: Solving Trigonometric Equations Using Identities KEY: exact value

36. ANS:

cot2θ + cotθ = 0

cotθ cotθ + 1( ) = 0

cotθ = 0 or cotθ = −1

θ = π2

or θ = 3π4

Since the period for cotθ is π, the solution in general form is θ = π2

+ nπ and θ = 3π4

+ nπ, where n ∈ I.

PTS: 1 DIF: Difficult OBJ: Section 6.4 NAT: T6TOP: Solving Trigonometric Equations Using Identities KEY: exact value | general solutions

ID: A

29

37. ANS:

sec2θ − 2tanθ − 3 = 0

(1+ tan2θ) − 2tanθ − 3 = 0

tan2θ − 2tanθ − 2 = 0

Use the quadratic formula.

tanθ =2± 12

2

= 1± 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: T6TOP: Solving Trigonometric Equations Using Identities KEY: exact value | general solutions | quadratic formula

38. ANS: a) i) quadraticii) exponentialiii) linearb) i) successive values would be increasing by a constant amountii) successive values would be increasing by a constant factoriii) all values would be constant

PTS: 1 DIF: Average OBJ: Section 7.1 NAT: RF9TOP: Characteristics of Exponential Functions KEY: linear | quadratic | exponential function

ID: A

30

39. ANS: Answers may vary.

Sample answer: The graph of y = 313

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

x

:

PTS: 1 DIF: Average OBJ: Section 7.1 | Section 7.2NAT: RF9 TOP: Characteristics of Exponential Functions | Transformations of Exponential FunctionsKEY: domain | range | intercepts | exponential function

40. ANS:

a) a vertical compression by a factor of 12

and a translation of 2 units to the right

b) The graph of y = 3x is shown in blue and the graph of y = 12

3( )x − 2 is shown in red.

c) domain {x|x ∈ R}, range {y|y > 0,y ∈ R} , y = 0

PTS: 1 DIF: Average OBJ: Section 7.2 NAT: RF9TOP: Transformations of Exponential Functions KEY: graph | transformations of exponential functions

ID: A

31

41. ANS:

a) y = 5−x

b) y = 5x − 3

c) y = 5x + 4 − 1

d) y = −5x − 2

PTS: 1 DIF: Average OBJ: Section 7.2 NAT: RF9TOP: Transformations of Exponential Functions KEY: equation | transformations of exponential functions

42. ANS: a) ii) b) i) c) iii)

PTS: 1 DIF: Average OBJ: Section 7.1 NAT: RF9TOP: Characteristics of Exponential Functions KEY: graph | modelling | exponential function

43. ANS:

9n − 1 = 13

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

4n − 1

32ÊËÁÁÁ

ˆ¯˜̃̃

n − 1

= 3−1ÊËÁÁÁ

ˆ¯˜̃̃

4n − 1

32n − 2 = 31 − 4n

Equate the exponents:2n − 2 = 1− 4n

6n = 3

n = 12

PTS: 1 DIF: Average OBJ: Section 7.3 NAT: RF10TOP: Solving Exponential Equations KEY: change of base

ID: A

32

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 NAT: RF8TOP: Transformations of Logarithmic Functions KEY: transformation | vertical translation | asymptote | graph

45. ANS: log214= log2(2× 7)

= log22+ log27

≈ 1+ 2.8074

= 3.8074

PTS: 1 DIF: Difficult OBJ: Section 8.3 NAT: RF9TOP: Laws of Logarithms KEY: power law | laws of logarithms

46. ANS: 63x + 1 = 22x − 3

log(63x + 1) = log(22x − 3)

(3x + 1) log6= (2x − 3) log2

3x log6+ log6= 2x log2− 3log2

x(3log6− 2log2)= −3log2− log6

x =−(3log2+ log6)3log6− 2log2

PTS: 1 DIF: Average OBJ: Section 8.3 | Section 8.4NAT: RF9 TOP: Laws of Logarithms | Logarithmic and Exponential EquationsKEY: exponential equation | laws of logarithms

ID: A

33

47. ANS:

a) f(x) = 14x − 8

b)

PTS: 1 DIF: Average OBJ: Section 9.1 NAT: RF14TOP: Exploring Rational Functions Using Transformations KEY: reciprocal of linear function | vertical asymptote | y-intercept | graph

ID: A

34

48. ANS:

a) i) x|x ≠ 54

, x ∈ RÏ

ÌÓ

ÔÔÔÔÔÔÔÔ

¸

˝˛

ÔÔÔÔÔÔÔÔ, {y|y ≠ 0,y ∈ R}

ii) x-intercept: none, y-intercept: −35

iii) x = 54

, y = 0

b)

PTS: 1 DIF: Average OBJ: Section 9.2 | Section 9.3NAT: RF14 TOP: Analysing Rational Functions | Connecting Graphs and Rational EquationsKEY: reciprocal of linear function | key features | graph

ID: A

35

49. ANS: a) i) {x|x ≠ 2,x ∈ R}, { y|y ≠ 3,y ∈ R}

ii) x-intercept: −83

, y-intercept: −4

iii) x = 2,y = 3 b)

PTS: 1 DIF: Average OBJ: Section 9.2 | Section 9.3NAT: RF14 TOP: Analysing Rational Functions | Connecting Graphs and Rational EquationsKEY: linear expressions in numerator and denominator | key features | graph

ID: A

36

50. ANS: a) i) {x|x ≠ −3,x ≠ 4,x ∈ R}, { y|y ≠ 0,y ∈ R}

ii) x-intercept: none, y-intercept: −14

iii) x = 4,y = 0

b) Note the hole at the point −3,− 17

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃.

PTS: 1 DIF: Difficult OBJ: Section 9.2 | Section 9.3NAT: RF14 TOP: Analysing Rational Functions | Connecting Graphs and Rational EquationsKEY: rational function | key features | graph

51. ANS: h x( ) = f x( ) + g x( )

= x + 1+ x2 + 3x + 1

= x2 + 4x + 2

PTS: 1 DIF: Easy OBJ: Section 10.1 NAT: RF1TOP: Sums and Differences of Functions KEY: add functions

ID: A

37

52. ANS:

h x( ) =f x( )g x( )

= x2 − 4x2 − 3x + 2

=x − 2( ) x + 2( )x − 2( ) x − 1( )

=x + 2( )x − 1( )

, x ≠ 2,x ≠ 1

PTS: 1 DIF: Average OBJ: Section 10.2 NAT: RF1TOP: 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: RF1TOP: Products and Quotients of Functions KEY: multiply functions | graph

ID: A

38

54. ANS:

2x2 − 3 = 0

2x2 = 3

x2 = 32

x = ± 32

x = ±6

2

PTS: 1 DIF: Difficult + OBJ: Section 10.2 NAT: RF1TOP: Products and Quotients of Functions KEY: divide functions | vertical asymptotes

55. ANS:

g 2( ) = 2− 2( ) 3

= −6

f −6( ) = −6( ) 2 − 7

= 29f(g(2)) = 29

PTS: 1 DIF: Average OBJ: Section 10.3 NAT: RF1TOP: Composite Functions KEY: composite functions | evaluate

56. ANS: The number of different routes can be calculated using the following permutation:12!6!6!

= 924

Joe can take one of 924 different routes to travel from home to school.

PTS: 1 DIF: Average OBJ: Section 11.1 NAT: PC1TOP: Permutations KEY: permutations

57. ANS:

a2

− b3

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

4

=4C0

a2

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

4

−b3

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

0

+4C1

a2

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

3

−b3

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

1

+4C2

a2

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

2

−b3

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

2

+4C3

a2

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

1

−b3

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

3

+4C4

a2

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

0

−b3

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

4

= 1a2

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

4

−b3

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

0

+ 4a2

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

3

−b3

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

1

+ 6a2

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

2

−b3

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

2

+ 4a2

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

1

−b3

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

3

+ 1a2

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

0

−b3

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

4

= a4

16− a3b

6+ a2b2

6− 2ab3

27+ b4

81

PTS: 1 DIF: Average OBJ: Section 11.3 NAT: PC4TOP: The Binomial Theorem KEY: binomial expansion | binomial theorem

ID: A

39

58. ANS: (2n + 2)!

(2n − 1)!0!=

1 ⋅ 2 ⋅ 3 ⋅ … ⋅ (2n − 2)(2n − 1)(2n)(2n + 1)(2n + 2)1 ⋅ 2 ⋅ 3 ⋅ … ⋅ (2n − 2)(2n − 1) ⋅ 1

= (2n)(2n + 1)(2n + 2)

= 2n(4n2 + 6n + 2)

= 8n3 + 12n2 + 4n

PTS: 1 DIF: Average OBJ: Section 11.1 NAT: PC1TOP: Permutations KEY: factorial

59. ANS: There are 14 letters in total. 5 of these are burnt out.(9C3)(5C2) = 84× 10

= 840There are 840 ways to select 3 good letters and 2 burnt-out letters.

PTS: 1 DIF: Average OBJ: Section 11.1 NAT: PC3TOP: 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: Average OBJ: Section 11.1 NAT: PC1TOP: Permutations KEY: fundamental counting principle

ID: A

40

PROBLEM

1. ANS: a) h(t) = −4.9t2 + 25b) h(t) = −1.85t2 + 25c) 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 3798

.

4.9× 3798

= 1.85

PTS: 1 DIF: Difficult OBJ: Section 1.2 NAT: RF3TOP: Reflections and Stretches KEY: graph | stretch | function notation

2. ANS:

a) g(x) = −13

12

x + 3( ) − 5

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 + OBJ: Section 1.3 NAT: RF4TOP: Combining Transformations KEY: graph | transformation | function notation

ID: A

41

3. ANS: a) C(x) = 50+ 0.12xb) C = 50+ 0.12x

C − 50= 0.12x

x = C − 500.12

f −1(C) = C − 500.12

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?

f −1(65) = 65− 500.12

= 125You can drive a total of 125 km for the $65 rental fee.

PTS: 1 DIF: Average OBJ: Section 1.4 NAT: RF6TOP: Inverse of a Relation KEY: inverse of a function | function notation

4. ANS: a)

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}

c) The vertical stretch changes from 5 to 2, which means a vertical compression by a factor of 25

. There is a

horizontal compression by a factor of 16

. The graph is reflected in both the x-axis and the y-axis. The graph is

translated 2 units left and 3 units down.

PTS: 1 DIF: Difficult OBJ: Section 2.1 NAT: RF13TOP: Radical Functions and Transformations KEY: graph | transformations | domain | range

ID: A

42

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 NAT: RF13TOP: Radical Functions and Transformations KEY: graph | transformations

ID: A

43

6. ANS:

a) E = 12

mv2

2E = mv2

2Em

= v2

v = 2Em

b) i) Substitute m = 12 and E = 200 in the equation.

v = 2Em

=2(200)

12

v ≈ 5.8The speed of the object is 5.8 m/s.ii) Substitute m = 12 and E = 420 in the equation.

v = 2Em

=2(420)

12

v ≈ 8.4The speed of the object is 8.4 m/s.c) When m = 12,

v = 2Em

= 2E12

v = E6

ID: A

44

d) The point (20, 2) is on the graph. Substitute this into the equation v = 2Em

and solve for m.

v = 2Em

2 =2(20)

m

4 = 40m

m= 10The mass of the object is 10 kg.

PTS: 1 DIF: Average OBJ: Section 2.1 NAT: RF13TOP: Radical Functions and Transformations KEY: graph | horizontal stretch

ID: A

45

7. ANS: Let P(x) = 2x4 – 7x3 – 41x2 – 53x – 21.Test x = –1 in the factor theorem.

P(x) = 2x4 − 7x3 − 41x2 − 53x − 21

P(−1) = 2(−1)4 − 7(−1)3 − 41(−1)2 − 53(−1) − 21

= 2+ 7− 41+ 53− 21

P(−1) = 0Thus, x + 1 is a factor. Divide to determine another factor.

1 2 −7 −41 −53 −21

− 2 −9 −32 −21

× 2 −9 −32 −21 0

Thus,

P(x) = (x + 1)(2x3 − 9x2 − 32x − 21)

Now factor the cubic. Test x = –1 in the factor theorem. Let Q(x) = 2x3 − 9x2 − 32x − 21.

Q(x) = 2x3 − 9x2 − 32x − 21

Q(−1) = 2(−1)3 − 9(−1)2 − 32(−1) − 21

= −2− 9+ 32− 21

Q(−1) = 0Thus, x + 1 is a factor. Divide to determine another factor.

1 2 −9 −32 −21

− 2 −11 −21

× 2 −11 −21 0

Thus,

P(x) = (x + 1)(x + 1)(2x2 − 11x − 21)

= (x + 1)2(2x2 − 14x + 3x − 21)

= (x + 1)2[2x(x − 7) + 3(x − 7)]

= (x + 1)2(x − 7)(2x + 3)

PTS: 1 DIF: Average OBJ: Section 3.3 NAT: RF11TOP: The Factor Theorem KEY: factor theorem | integral zero theorem | factor

ID: A

46

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: RF11TOP: 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.= x2(x + 4)

= 52(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.

x2(x + 4) = 225

x3 + 4x2 − 225= 0Since x = 5 is a solution, x − 5 is a factor of P(x). Divide to find another factor.

−5 1 4 0 −225

− −5 −45 −225

× 1 9 45 0

x3 + 4x2 − 225= 0

(x − 5)(x2 + 9x + 45) = 0x2 + 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 3.4NAT: RF11 TOP: The Factor Theorem | Equations and Graphs of Polynomial FunctionsKEY: factor theorem | integral zero theorem | polynomial equation | root

ID: A

47

10. ANS: Multiply the equation by −1 to clear the negative leading coefficient.

−x3 + 5x2 − 8x + 4 ≥ 0

−1(−x3 + 5x2 − 8x + 4) ≤ 0

x3 − 5x2 + 8x − 4 ≤ 0

Factor x3 − 5x2 + 8x − 4 using the factor theorem. Let P(x) = x3 − 5x2 + 8x − 4. Try x = 1 in the factor theorem.

P(x) = x3 − 5x2 + 8x − 4

P(1) = (1)3 − 5(1)2 + 8(1)− 4

= 1− 5+ 8− 4

P(1) = 0Thus, x − 1 is a factor of P(x). Divide to find another factor.

−1 1 −5 8 −4

− −1 4 −4

× 1 −4 4 0

Thus,

x3 − 5x2 + 8x − 4 ≤ 0

(x − 1)(x2 − 4x + 4) ≤ 0

(x − 1)(x − 2)2 ≤ 0Construct a table.

x < 1 1 < x < 2 x > 2x − 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 = −x3 + 5x2 − 8x + 4. Then, use the Zero operation.

ID: A

48

The zeros are 1 and 2. From the graph, −x3 + 5x2 − 8x + 4 ≥ 0 when x ≤ 1 or x = 2.

PTS: 1 DIF: Difficult + OBJ: Section 3.3 | Section 3.4NAT: RF11 | RF12 TOP: The Factor Theorem | Equations and Graphs of Polynomial FunctionsKEY: 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 = 2Thus,

y = 2x(x − 2)2

= 2x(x2 − 4x + 4)

y = 2x3 − 8x2 + 8x

PTS: 1 DIF: Average OBJ: Section 3.4 NAT: RF12TOP: Equations and Graphs of Polynomial Functions KEY: polynomial equation | graph | zeros

ID: A

49

12. ANS:

Use the cosine law.

12 = d2 + d2 − 2d d( ) cos 1.25

1 = 2d2 − 2d2 cos 1.25

= d2 2− 2cos 1.25( )

d = 12− 2cos 1.25

Recall that s = dt. The balls have travelled a distance, d, in metres, in a time, t, of 2 s.

s =

12− 2cos 1.25

2

≈ 0.43Therefore, after 2 s, the billiard balls are moving at approximately 0.43 m/s.

PTS: 1 DIF: Difficult + OBJ: Section 4.3 | Section 4.4NAT: T3 | T5 TOP: Trigonometric Ratios | Introduction to Trigonometric EquationsKEY: trigonometric ratios | radians | cosine law

ID: A

50

13. ANS: a)

b) Use the sine ratio, since the side opposite the given angle is known and the hypotenuse is needed.

c) sin 30° = 2.5x

x = 2.5sin 30°

= 2.512

= 5The length of the piece of wood is 5 m.

PTS: 1 DIF: Easy OBJ: Section 4.3 | Section 4.4NAT: T3 | T5 TOP: Trigonometric Ratios | Introduction to Trigonometric EquationsKEY: trigonometric ratios | special angles | trigonometric equations

ID: A

51

14. ANS:

a) Since sin 30° = 12

, the reference angle is 30°. The sine ratio is negative in quadrants III and IV. Look for

reflections of the 30° angle in these quadrants.quadrant III: 180° + 30° = 210°quadrant IV: 360° – 30° = 330°

b) Using a calculator, sin 210° = − 12

and sin 330° = − 12

.

PTS: 1 DIF: Average OBJ: Section 4.2 | Section 4.3NAT: T2 | T3 TOP: Unit Circle | Trigonometric Ratios KEY: sine ratio | reference angle | unit circle

ID: A

52

15. ANS: <fix tech art so 49 cm arrow goes only to bottom of right-angle triangle, not to black dot>

θ = cos−1 4950

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

≈ 0.20033

Graph the function θ = 14

sinπ2

tÊ

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃ 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 NAT: T4TOP: Equations and Graphs of Trigonometric Functions KEY: linear trigonometric equation

ID: A

53

16. ANS: Answers may vary. Sample answer:a) Convert h:min to hours.Month Hours of Daylight

1 8.422 9.923 11.584 13.55 15.86 16.257 15.428 14.439 12.58

10 10.6511 9.0212 8

b) The amplitude is 16.25− 8

2, or a = 4.125. The vertical shift is 8 + 4.125, or c = 12.125.

The sine wave starts at 12.125, at approximately 3.29 months. So, d = 3.29.

The period is 36012

, or k = 30.

An equation to represent the data is t = 4.125 sin [30m− 3.29( )] + 12.125.c)

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:

ID: A

54

t = 4.125 sin [30m− 3.29( )] + 12.125

= 4.125 sin [30 1.5− 3.29( )] + 12.125

≈ 8.8There 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: T4TOP: Equations and Graphs of Trigonometric Functions KEY: model | sinusoidal function | graph

17. ANS: a) 11:38 – 5:17 = 6:21So, 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.

b) The amplitude is 1.77− 0.21

2, or 0.78 m.

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 5.2NAT: T4 TOP: Graphing Sine and Cosine Functions | Transformations of Sinusoidal FunctionsKEY: model | periodic function | predict

ID: A

55

18. ANS: a) y = 2 cosx − 2b) Let c represent the phase shift.y = 2cos(x − c) − 2

−1 = 2cos(0− c) − 2

1 = 2cos(−c)

12

= cos(−c)

−c = 60°

c = −60°The phase shift is 60° to the left.c) y = 2 cos x + 60°( ) − 2d)

PTS: 1 DIF: Average OBJ: Section 5.2 NAT: T4TOP: Transformations of Sinusoidal Functions KEY: graph | transformations | cosine function | equation

ID: A

56

19. ANS: L.S. = 1+ cosθ

R.S.= sin2θ1− cosθ

= 1− cos2θ1− cosθ

=1− cosθ( ) 1+ cosθ( )

1− cosθ

= 1+ cosθL.S. = R.S.

Therefore, 1+ cosθ = sin2θ1− cosθ .

PTS: 1 DIF: Average OBJ: Section 6.1 | Section 6.3NAT: T6 TOP: Reciprocal, Quotient, and Pythagorean Identities | Proving IdentitiesKEY: Pythagorean identities | proof

20. ANS:

L.S. = sinπ2

− xÊ

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃ cot

π2

+ xÊ

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

= cosx(−tanx)

= cosx − sinxcosx

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

= −sinx

R.S.= −sinx

L.S. = R.S.

Therefore, sinπ2

− xÊ

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃ cot

π2

+ xÊ

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃ = −sinx.

PTS: 1 DIF: Average OBJ: Section 6.1 | Section 6.2 | Section 6.3NAT: T6 TOP: Reciprocal, Quotient, and Pythagorean Identities | Sum, Difference, and Double-Angle Identities | Proving Identities KEY: difference identities | sum identities | quotient identities | proof

ID: A

57

21. ANS:

L.S. = 1− cos 2θ + sin2θ1+ cos 2θ + sin2θ

=1− (1− 2sin2θ) + 2sinθ cosθ1+ (2cos2θ − 1) + 2sinθ cosθ

= 2sin2θ + 2sinθ cosθ2cos2θ + 2sinθ cosθ

=2sinθ sinθ + cosθ( )

2cosθ sinθ + cosθ( )

= sinθcosθ

= tanθ

L.S. = tanθ

L.S. = R.S

Therefore, 1− cos 2θ + sin2θ1+ cos 2θ + sin2θ = tanθ.

PTS: 1 DIF: Average OBJ: Section 6.2 | Section 6.3NAT: T6 TOP: Sum, Difference, and Double-Angle Identities | Proving IdentitiesKEY: double-angle identities | proof

22. ANS:

L.S. = cos2θ − sin2θcos2θ + sinθ cosθ

=cosθ − sinθ( ) cosθ + sinθ( )

cosθ cosθ + sinθ( )

= cosθ − sinθcosθ

R.S.= 1− tanθ

= 1− sinθcosθ

= cosθ − sinθcosθ

L.S. = R.S.

PTS: 1 DIF: Average OBJ: Section 6.3 NAT: T6TOP: Proving Identities KEY: trigonometric identity | proof

ID: A

58

23. ANS:

a) secθ cotθ = 1cosθ

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

cosθsinθ

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

= 1sinθ

b) 1

sinθ = 1

1 = sinθ

θ = 90°c) This is verified using a calculator.d)

PTS: 1 DIF: Average OBJ: Section 6.4 NAT: T6TOP: Solving Trigonometric Equations Using Identities KEY: reciprocal trigonometric ratios | unit circle

ID: A

59

24. ANS:

L.S. = sin3θ + sinθcos 3θ + cosθ

= sin2θ cosθ + cos 2θ sinθ + sinθcos 2θ cosθ − sin2θ sinθ + cosθ

=2sinθ cos2θ + (2cos2θ − 1)sinθ + sinθ(2cos2θ − 1)cosθ − 2sin2θ cosθ + cosθ

= sinθcosθ

2cos2θ + 2cos2θ − 1+ 1

2cos2θ − 1− 2sin2θ + 1

Ê

Ë

ÁÁÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃˜̃

= sinθcosθ

4cos2θ2cos 2θ

Ê

Ë

ÁÁÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃˜̃

= 4sinθ cos2θ2cos 2θ cosθ

= 2sinθ cosθcos 2θ

= sin2θcos 2θ

= tan2θ

R.S.= tan2θ

L.S. = R.S.

This proves that sin3θ + sinθcos 3θ + cosθ = tan2θ.

PTS: 1 DIF: Difficult OBJ: Section 6.2 | Section 6.3NAT: T6 TOP: Sum, Difference, and Double-Angle Identities | Proving IdentitiesKEY: sum identities | proof

ID: A

60

25. ANS: sin3x + sinx = cosx

sin2xcosx + cos 2xsinx + sinx = cosx

2sinxcos2x + (2cos2x − 1)sinx + sinx = cosx

2sinxcos2x + 2cos2xsinx − sinx + sinx = cosx

4sinxcos2x = cosx

4sinxcos2x − cosx = 0

cosx(4sinxcosx − 1) = 0

cosx(2sin2x − 1) = 0

cosx = 0 or sin2x = 12

x = π2

,3π2

or 2x = π6

,5π6

,13π6

,17π6

x = π2

,3π2

,π12

,5π12

,13π12

,17π12

The solution is x = π12

,5π12

,13π12

,17π12

,π2

,3π2

.

PTS: 1 DIF: Difficult + OBJ: Section 6.4 NAT: T6TOP: Solving Trigonometric Equations Using Identities KEY: general solutions | equation

26. ANS: tanx cscx + 2( ) = 0tanx = 0

x = 0+ nπ

= nπ

cscx + 2 = 0

cscx = −2

sinx = −12

x = −π6

orx = 11π6

sin x is also negative in quadrant III, so find a reflection of x = 11π6

in this quadrant. Another value for

sinx = −12

is x = 7π6

.

Since the period of tan x is π and the period of sin x is 2π, the general solution is is

x = nπ or x = 7π6

+ nπ or x = 11π6

+ nπ, where n ∈ I.

PTS: 1 DIF: Average OBJ: Section 6.4 NAT: T6TOP: Solving Trigonometric Equations Using Identities KEY: general solutions | equation

ID: A

61

27. ANS:

a) A = 7212

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

t

10

b) A = 7212

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

t

10

= 7212

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

20

10

= 7212

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

2

= 18There will be 18 mg remaining after 20 days.

c) A = 7212

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

t

10

= 7212

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

−30

10

= 7212

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

−3

= 576There was 576 mg 30 days ago.

d) A = 7212

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

t

10

0.07= 7212

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

t

10

0.0772

= 12

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

t

10

Use systematic trial.0.0772

=Ö 0.000 972

For t = 100, 12

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

10

=Ö 0.000 977.

It will take approximately 100 days for there to be 0.07 mg remaining.

PTS: 1 DIF: Average OBJ: Section 7.2 | Section 7.3NAT: RF9 | RF10 TOP: Transformations of Exponential Functions | Solving Exponential Equations

ID: A

62

KEY: modelling | evaluate exponential functions 28. ANS:

a) y = 2−2 x − 2( ) + 6

b) Reflect in the y-axis, compress horizontally by a factor of 12

, and translate 2 units to the right and 6 units

up.c)

d) y = −2−2(x − 2) − 6e)

PTS: 1 DIF: Average OBJ: Section 7.2 NAT: RF9TOP: Exponential Functions KEY: graph | transformations of exponential functions

29. ANS:

25623 × 16x = 64x − 3

28( )2

3 × 24x = 26x − 18

24x +

16

3 = 26x − 18

4x + 163

= 6x − 18

−2x = −703

x = 353

PTS: 1 DIF: Average OBJ: Section 7.3 NAT: RF10TOP: Solving Exponential Equations KEY: exponential equation | change of base

ID: A

63

30. ANS:

23x = 22

3x = 2

x = 23

PTS: 1 DIF: Easy OBJ: Section 7.3 NAT: RF10TOP: Solving Exponential Equations KEY: exponential equation | change of base

31. ANS:

a) The quarterly interest rate is 5.25

4 or 1.3125, and the number of compounding periods is 4× 5 or 20.

A = P 1+ i( ) n

= 21 500 1+ 0.013 125( ) 20

≈ 27 906.10The value of the investment after 5 years is $27 906.10.

b) 43 000= 21 500 1+ 0.013125( ) n

2 = 1.013125( ) n

Using systematic trial, n = 53.2 quarters or approximately 13.3 years.

PTS: 1 DIF: Average OBJ: Section 7.3 NAT: RF10TOP: 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.

16.3= 2012

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

t

186

0.815= 0.5

t

186

Using systematic trial, t

186= 0.3. Therefore, the time is approximately 56 s.

PTS: 1 DIF: Average OBJ: Section 7.3 NAT: RF10TOP: Solving Exponential Equations KEY: exponential equation | half-life

33. ANS:

same shape, vertical translation up by $1000

PTS: 1 DIF: Easy OBJ: Section 7.2 NAT: RF10TOP: Characteristics of Exponential Functions | Transformations of Exponential FunctionsKEY: exponential equation | compound interest formula | transformations

ID: A

64

34. ANS: log28= log(7× 22)

= log7+ 2log2

≈ 0.8451+ 2(0.3010)

= 1.4471

PTS: 1 DIF: Difficult OBJ: Section 8.3 NAT: RF9TOP: 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.

a) m2 − m1 = logb1

b2

Ê

Ë

ÁÁÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃˜̃

0.12+ 1.5= logb1

b2

Ê

Ë

ÁÁÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃˜̃

1.62= logb1

b2

Ê

Ë

ÁÁÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃˜̃

101.62 =b1

b2

b1

b2

≈ 41.69

Sirius is approximately 42 times as bright as the Sun.

b) m1 − m2 = logb2

b1

Ê

Ë

ÁÁÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃˜̃

−1.5− m2 = log(1.3×1010)

m2 ≈ −1.5− 10.11

m2 ≈ −11.61

The apparent magnitude of the Sun is –11.6.

PTS: 1 DIF: Difficult OBJ: Section 8.1 | Section 8.3 | Section 8.4NAT: RF7 | RF9 TOP: Understanding Logarithms | Laws of Logarithms | Logarithmic and Exponential EquationsKEY: logarithmic equation | exponential function

ID: A

65

36. ANS: L .S.= loga + loga2 + loga3 − loga6

= loga + 2loga + 3loga − 6loga

= (1+ 2+ 3− 6) loga

= 0loga

= 0

R.S.= log1

= 0

L.S. = R.S.Thus, log5+ log52 + log53 − log56 = log1.

PTS: 1 DIF: Average OBJ: Section 8.3 NAT: RF9TOP: Laws of Logarithms KEY: power law | laws of logarithms

37. ANS:

L .S.= 1

logab

= 1÷ logab

= 1÷logbloga

= 1×logalogb

= logba

R.S.= logba

L.S. = R.S.

Thus, 1

logab

= logba.

PTS: 1 DIF: Difficult + OBJ: Section 8.3 NAT: RF9TOP: Laws of Logarithms KEY: change of base formula | laws of logarithms

ID: A

66

38. ANS:

L .S.= logq5 p5

=logqp5

logqq5

=5logqp

5

= logqp

R.S.= logqp

L.S. = R.S.Thus, logq

5 p5 = logqp.

PTS: 1 DIF: Difficult + OBJ: Section 8.3 NAT: RF10TOP: Laws of Logarithms KEY: laws of logarithms

39. ANS: a), b)

c) The constant k will result in a vertical translation up by log3k. 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 NAT: RF8TOP: Transformations of Logarithmic Functions KEY: vertical translation

ID: A

67

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

= 26 552.12Solve for the remaining time—compounded daily for n years is 365n periods.

35 000= 26 552.12 1+ 0.05365

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

365n

35 00026 552.12

= 365.05365

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

365n

log35 000

26 552.12

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃ = 365n log

365.05365

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

n ≈ 5.53Bruce will need invest for approximately 5.5 more years.

PTS: 1 DIF: Difficult OBJ: Section 8.4 NAT: RF10TOP: Exponential and Logarithmic Equations KEY: exponential function | logarithmic equation

ID: A

68

41. ANS:

Find the x-intercept by substituting y = 0 and solving for x.

0 = −2log12

x + 1( )

È

Î

ÍÍÍÍÍÍÍÍ

˘

˚

˙̇˙̇˙̇˙̇

− 1

−12

= log12

x + 1( )

È

Î

ÍÍÍÍÍÍÍÍ

˘

˚

˙̇˙̇˙̇˙̇

10−

1

2 = 12

(x + 1)

2(10−

1

2 ) − 1 = x

x ≈ −0.37The x-intercept is −0.37.

PTS: 1 DIF: Average OBJ: Section 8.2 | Section 8.4NAT: RF8 | RF10 TOP: Transformations of Logarithmic Functions | Logarithmic and Exponential EquationsKEY: transformation | logarithmic equation

ID: A

69

42. ANS:

a) t(v) = 5v

,v > 0

b)

c) 1 h and 7 min

PTS: 1 DIF: Easy OBJ: Section 9.1 | Section 9.3NAT: RF14 TOP: Exploring Rational Functions Using Transformations | Connecting Graphs and Rational EquationsKEY: word problem | reciprocal of linear function | graph

ID: A

70

43. ANS: a) Since the pressure, p, exerted by a person’s shoe is inversely proportional to the width, w, of the shoe, you

can write p(w) = k

w2 for some k. Since p(2) = 400, substituting these values into p(w) = k

w2 gives k = 1600.

p(w) = 1600w2

, w > 0

b)

c) p(w) = 1600w2

p(0.5)= 16000.52

= 6400Megumi exerts 6400 kPa of pressure.

PTS: 1 DIF: Average OBJ: Section 9.3 NAT: RF14TOP: Connecting Graphs and Rational Equations KEY: reciprocal of quadratic function | graph

ID: A

71

44. ANS:

a) i) I(d) = 10d2

I(3) = 1032

≈ 1.11The intensity is 1.11 lux.

ii) I(d) = 10d2

I(3) − I(1)3− 1

=

1032 − 10

12

2

= −4.44The 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 9.3NAT: RF14 TOP: Exploring Rational Functions Using Transformations | Connecting Graphs and Rational EquationsKEY: reciprocal of quadratic function | average rate of change

45. ANS: Answers may vary. Sample answer:

f(x) = 1

x2 − 4

Any function of the form f(x) = kx2 − 4

, k > 0, is a reasonable candidate since it is difficult to tell from the

graph how stretched the function is.

PTS: 1 DIF: Average OBJ: Section 9.2 NAT: RF14TOP: Analysing Rational Functions KEY: reciprocal of quadratic function

46. ANS: Answers may vary. Sample answer:

f(x) =2(x − 1)2

(x − 1)(x − 3)

PTS: 1 DIF: Difficult OBJ: Section 9.2 | Section 9.3NAT: RF14 TOP: Analysing Rational Functions | Connecting Graphs and Rational EquationsKEY: rational function | hole

ID: A

72

47. ANS:

a) For f(x) = x + 1x − 5

: asymptotes x = 5,y = 1; intercepts (−1,0), 0,−15

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

For g(x) = x − 5x + 1

: asymptotes x = −1,y = 1; intercepts (5,0), (0,−5)

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 orx > 5 and f(x) < 0 for −1 < x < 5.g(x) > 0 for x < −1 orx > 5 and g(x) < 0 for −1 < x < 5.The intervals where f and g are positive and negative are identical.

e) Yes, this pattern is true for all pairs of functions f(x) = x + bx + d

and g(x) = x + dx + b

because when you take the

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 9.3NAT: RF14 TOP: Analysing Rational Functions | Connecting Graphs and Rational EquationsKEY: linear expressions in numerator and denominator | graph | reciprocal

ID: A

73

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

3 h 20 min is equal to 103

h.

Therefore,990

600+ w+ 990

600− w= 10

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− w2

w2 = 3600

w = 60The wind speed is 60 mph.

PTS: 1 DIF: Average OBJ: Section 9.3 NAT: RF14TOP: Connecting Graphs and Rational Equations KEY: rational equation

49. ANS: Let x represent the number of members going on the ski trip.

480x

− 480x + 5

= 4.8

x( ) x + 5( )480x

− 480x + 5

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃ = 4.8 x( ) x + 5( )

480 x + 5( ) − 480x = 4.8 x( ) x + 5( )

100(x + 5) − 100x = x(x + 5)

100x + 500− 100x = x2 + 5x

x2 + 5x − 500= 0

x + 25( ) x − 20( ) = 0

x = −25, x = 20The 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: RF14TOP: Connecting Graphs and Rational Equations KEY: linear expressions in numerator and denominator | x-intercept | excluded solution

ID: A

74

50. ANS:

h x( ) = 1

1− sin2x

= 1

cos2x

= sec2x

PTS: 1 DIF: Average OBJ: Section 10.3 NAT: RF1TOP: Composite Functions KEY: composite functions

51. ANS:

PTS: 1 DIF: Difficult OBJ: Section 10.2 NAT: RF1TOP: Products and Quotients of Functions KEY: divide functions | vertical asymptotes | hole

52. ANS: Domain: {x|x ∈ R}

Range: y |12

≤ y ≤ 2,y ∈ RÏ

ÌÓ

ÔÔÔÔÔÔÔÔ

¸

˝˛

ÔÔÔÔÔÔÔÔ

PTS: 1 DIF: Average OBJ: Section 10.3 NAT: RF1TOP: Composite Functions KEY: composite functions | domain | range

ID: A

75

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 2x > x2, are −0.8< x < 2 or x > 4.

PTS: 1 DIF: Difficult + OBJ: Section 10.1 NAT: RF1TOP: Sums and Differences of Functions KEY: combined function | inequality | graph

54. ANS: a) D x( ) = 8.0+ 6x( ) − 10.0+ 5x( )

= −2.0+ xThis difference must be at least 100.−2+ x ≥ 100

x ≥ 102The minimum number sold is 102 items.b) T x( ) = 8.0+ 6x( ) + 10.0+ 5x( )

= 18.0+ 11xThis 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: RF1TOP: 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: RF1TOP: Sums and Differences of Functions KEY: subtract functions | graph

ID: A

76

56. ANS:

a) Area of rectangle = 2x2

Area of semi-circle =π x

2

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

2

2

= πx2

8

Total area = 2x2 + πx2

8

= 16x2 + πx2

8

=x2 16+ π( )

8

b) Area=x2 16+ π( )

8

=1.22 16+ π( )

8

≈ 3.4The area of the window is approximately 3.4 m2.

PTS: 1 DIF: Average OBJ: Section 10.1 NAT: RF1TOP: 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: Difficult OBJ: Section 11.1 NAT: PC2TOP: Permutations KEY: permutations

ID: A

77

58. ANS: The total number of possible tickets for lottery A is 49C6 or 13 983 816.

Therefore, the probability of randomly selecting the winning numbers is

P(A) = 113 983 816

The total number of possible tickets for lottery B is 49C7 or 85 900 584.

Therefore, the probability of randomly selecting the winning numbers is

P(B) = 185 900 584

Compare the probabilities:

P(A)P(B)

=

113 983 816

185 900 584

= 85 900 58413 983 816

≈ 6.14Bernadette is not correct. The probability of winning lottery A is only 6.14 times as great.

PTS: 1 DIF: Difficult OBJ: Section 11.2 NAT: PC3TOP: Combinations KEY: combinations

ID: A

78

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: Easy OBJ: Section 11.1 NAT: PC1TOP: Permutations KEY: tree diagram

ID: A

79

60. ANS:

a) 7!

3!4!= 5040

144

= 35There are 35 possible routes from home to the library.

b) 7!

3!4!

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

15!4!11!

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃ = (35)(1365)

= 47 775There 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 about83 087.0 min.Convert to days:83 087.0(60)(24)

≈ 57.7

It would take about 58 days for the 23 students to walk all of the routes from home to school.

PTS: 1 DIF: Difficult OBJ: Section 11.1 NAT: PC1 | PC2TOP: Permutations KEY: permutations | fundamental counting principle

61. ANS:

L.S. = 1+ tanθ1+ cotθ

=1+ sinθ

cosθ

1+ cosθsinθ

=

cosθ + sinθcosθ

sinθ + cosθsinθ

= cosθ + sinθcosθ

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

sinθcosθ + sinθ

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

= sinθcosθ

R.S.= 1− tanθcotθ − 1

=1− sinθ

cosθcosθsinθ − 1

=

cosθ − sinθcosθ

cosθ − sinθsinθ

= cosθ − sinθcosθ

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

sinθcosθ − sinθ

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃

= sinθcosθ

L.S. = R.S.

PTS: 1 DIF: Average OBJ: Section 6.1 | Section 6.3NAT: T6 TOP: Reciprocal, Quotient, and Pythagorean Identities | Proving IdentitiesKEY: quotient identities | proof

ID: A

80

62. ANS:

log x2 + 48x3 = 23

log(x2 + 48x)

1

3 = 23

13

log(x2 + 48x) = 23

log(x2 + 48x) = 2

log(x2 + 48x) = log100

x2 + 48x = 100

x2 + 48x − 100= 0

(x + 50)(x − 2) = 0

x = −50 orx = 2Check 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 NAT: RF10TOP: Logarithmic and Exponential Equations KEY: logarithmic equation | extraneous roots