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

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